An introduction to nonlinear finite element analysis (J N Reddy)

An Introduction to Nonlinear FiniteElement Analysis

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J. N. REDDYDistinguished ProfessorDepartment of Mechanical EngineeringTexas A&M University, College StationTX 77843–3123, USA

An Introduction to NonlinearFinite Element Analysis

1

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ISBN 0-19-852529-X 978-0-19-852529-5

10 9 8 7 6 5 4 3 2

To

To my beloved teacher

Professor John Tinsley Oden


About the Author

J. N. Reddy is Distinguished Professor and the Holder of Oscar S. WyattEndowed Chair in the Department of Mechanical Engineering at Texas A&MUniversity, College Station, Texas. Prior to the current position, he was the Clifton C.Garvin Professor in the Department of Engineering Science and Mechanics at VirginiaPolytechnic Institute and State University (VPI&SU), Blacksburg, Virginia.

Dr. Reddy is internationally known for his contributions to theoretical andapplied mechanics and computational mechanics. He is the author of over 300 journalpapers and dozen other books, including Introduction to the Finite Element Method(Second Edition), McGraw-Hill, l993; Theory and Analysis of Elastic Plates, Taylor& Francis, 1999; Energy Principles and Variational Methods in Applied Mechanics(Second Edition), John Wiley, 2002; Mechanics of Laminated Plates and Shells:Theory and Analysis, (second Edition) CRC Press, 2004; An Introduction to theMathematical Theory of Finite Elements (coauthored with J. T. Oden), John Wiley,l976; Variational Methods in Theoretical Mechanics (coauthored with J. T. Oden),SpringerVerlag, 1976; The Finite Element Method in Heat Transfer and FluidDynamics (Second Edition; coauthored with D. K. Gartling), CRC Press, 2001; and

Professor Reddy is the recipient of the Walter L. Huber Civil EngineeringResearch Prize of the American Society of Civil Engineers (ASCE), the WorcesterReed Warner Medal and the Charles Russ Richards Memorial Award of the AmericanSociety of Mechanical Engineers (ASME), the 1997 Archie Higdon DistinguishedEducator Award from the American Society of Engineering Education (ASEE), the1998 Nathan M. Newmark Medal from the American Society of Civil Engineers, the2000 Excellence in the Field of Composites from the American Society of Composites,the 2000 Faculty Distinguished Achievement Award for Research and the 2003 BushExcellence Award for Faculty in International Research from Texas A&M University,and the 2003 Computational Solid Mechanics Award from the U.S. Association ofComputational Mechanics (USACM).

Dr. Reddy is a Fellow of ASCE, ASME, the American Academy of Mechanics(AAM), the American Society of Composites, the U.S. Association of ComputationalMechanics, the International Association of Computational Mechanics (IACM),and the Aeronautical Society of India (ASI). Professor Reddy is the Editor-in-Chief of Mechanics of Advanced Materials and Structures, International Journal ofComputational Engineering Science and International Journal of Structural Stabilityand Dynamics, and he serves on the editorial boards of over two dozen other journals,including International Journal for Numerical Methods in Engineering, ComputerMethods in Applied Mechanics and Engineering, and International Journal of Non-Linear Mechanics.


ix

Contents

Preface xvii

1 Introduction 1

1.1 Mathematical Models 1

1.2 Numerical Simulations 3

1.3 The Finite Element Method 5

1.4 Nonlinear Analysis 71.4.1 Introduction 71.4.2 ClassiÞcation of Nonlinearities 7

1.5 The Big Picture 11

References 12

2 The Finite Element Method: A Review 13

2.1 Introduction 13

2.2 One-Dimensional Problems 132.2.1 Governing Differential Equation 132.2.2 Finite Element Approximation 142.2.3 Derivation of the Weak Form 162.2.4 Interpolation Functions 182.2.5 Finite Element Model 22

2.3 Two-Dimensional Problems 242.3.1 Governing Differential Equation 242.3.2 Finite Element Approximation 242.3.3 Weak Formulation 262.3.4 Finite Element Model 272.3.5 Interpolation Functions 282.3.6 Assembly of Elements 33

2.4 Library of Two-Dimensional Finite Elements 362.4.1 Introduction 362.4.2 Triangular Elements 372.4.3 Rectangular Elements 38

2.5 Numerical Integration 402.5.1 Preliminary Comments 402.5.2 Coordinate Transformations 412.5.3 Integration Over a Master Rectangular Element 442.5.4 Integration Over a Master Triangular Element 45

x

2.6 Computer Implementation 462.6.1 General Comments 462.6.2 One-Dimensional Problems 482.6.3 Two-Dimensional Problems 52

2.7 Closure 53

Problems 57

References 59

3 Heat Transfer and Other Field Problems in One Dimension 61

3.1 Model Differential Equation 61

3.2 Weak Formulation 62

3.3 Finite Element Model 62

3.4 Solution Procedures 643.4.1 General Comments 643.4.2 Direct Iteration Procedure 653.4.3 Newtons Iteration Procedure 68

3.5 Computer Implementation 733.5.1 Introduction 733.5.2 Preprocessor Unit 733.5.3 Processor Unit 76

3.6 Closing Remarks 82

Problems 82

References 85

4 Nonlinear Bending of Straight Beams 87

4.1 Introduction 87

4.2 EulerBernoulli Beams 884.2.1 Basic Assumptions 884.2.2 Displacement Field and Strains 884.2.3 Weak Forms 894.2.4 Finite Element Model 954.2.5 Iterative Solutions of Nonlinear Equations 974.2.6 Load Increments 1004.2.7 Membrane Locking 1014.2.8 Computer Implementation 102

4.3 Timoshenko Beams 1104.3.1 Displacement Field and Strains 1104.3.2 Weak Forms 1114.3.3 General Finite Element Model 1134.3.4 Shear and Membrane Locking 1154.3.5 Tangent Stiffness Matrix 117

xi

Problems 124

References 126

5 Heat Transfer and Other Fields Problems in Two Dimensions 127

5.1 Model Equation 127

5.2 Weak Form 128

5.3 Finite Element Model 129

5.4 Solution Procedures 1315.4.1 Direct Iteration 1315.4.2 NewtonRaphson Iteration 131

5.5 Computer Implementation 1325.5.1 Introduction 1325.5.2 Numerical Integration 1335.5.3 Element Calculations 135

Problems 139

References 140

6 Nonlinear Bending of Elastic Plates 141

6.1 Introduction 141

6.2 Classical Plate Theory 1416.2.1 Assumptions of the Kinematics 1416.2.2 Displacement Field and Strains 142

6.3 Variational Formulation of CPT 1446.3.1 Virtual Work 1446.3.2 Weak Forms 1476.3.3 Equilibrium Equations 1486.3.4 Boundary Conditions 1496.3.5 Stress ResultantDeßection Relations 152

6.4 Finite Element Models of CPT 1536.4.1 General Formulation 1536.4.2 Tangent Stiffness Coefficients 1576.4.3 Some Plate Finite Elements 161

6.5 Computer Implementation Aspects and Numerical Resultsof CPT Elements 1646.5.1 Computer Implementation 1646.5.2 Results of Linear Analysis 1666.5.3 Results of Nonlinear Analysis 168

6.6 FirstOrder Shear Deformation Plate Theory 1736.6.1 Introduction 1736.6.2 Displacement Field 1736.6.3 Weak Formulation 174

xii

6.7 Finite Element Models of FSDT 1776.7.1 Virtual Work Statements 1776.7.2 Finite Element Model 1786.7.3 Tangent Stiffness Coefficients 1826.7.4 Shear and Membrane Locking 184

6.8 Computer Implementation and Numerical Resultsof FSDT Elements 1846.8.1 Computer Implementation 1846.8.2 Results of Linear Analysis 1856.8.3 Results of Nonlinear Analysis 189

6.9 Theory of Doubly-Curved Shells 1966.9.1 Introduction 1966.9.2 Geometric Description 1976.9.3 StrainDisplacement Relations 2016.9.4 Stress Resultants 2026.9.5 Equations of Motion 205

6.10 Finite Element Analysis of Shells 2066.10.1 Weak Forms 2066.10.2 Finite Element Model 2076.10.3 Linear Results 2096.10.4 Nonlinear Results 217

Problems 222

References 225

7 Flows of Viscous Incompressible Fluids 229

7.1 Introduction 2297.2 Governing Equations 230

7.2.1 Introduction 2307.2.2 Conservation of Mass 2317.2.3 Conservation of Momenta 2327.2.4 Conservation of Energy 2327.2.5 Constitutive Equations 2337.2.6 Boundary Conditions 234

7.3 Governing Equations in Terms of Primitive Variables 2357.3.1 Vector Form 2357.3.2 Cartesian Component Form 236

7.4 VelocityPressure Finite Element Model 2377.4.1 Weak Form 2377.4.2 Finite Element Model 239

7.5 Penalty Finite Element Models 2417.5.1 Introduction 2417.5.2 Penalty Function Method 242

xiii

7.5.3 Reduced Integration Penalty Model 2447.5.4 Consistent Penalty Model 245

7.6 Computational Aspects 2467.6.1 Properties of the Matrix Equations 2467.6.2 Choice of Elements 2477.6.3 Evaluation of Element Matrices in Penalty Models 2497.6.4 Post-Computation of Stresses 250

7.7 Computer Implementation 2517.7.1 Mixed Model 2517.7.2 Penalty Model 254

7.8 Numerical Examples 2547.8.1 Preliminary Comments 2547.8.2 Fluid Squeezed Between Parallel Plates 2557.8.3 Flow of a Viscous Lubricant in a Slider Bearing 2587.8.4 Wall-Driven Cavity Flow 2617.8.5 Backward-Facing Step 267

7.9 Least-Squares Finite Element Models 2697.9.1 Introduction 2697.9.2 Finite Element Model 2727.9.3 Computational Aspects 2747.9.4 Numerical Examples 275

Problems 283

References 283

8 Nonlinear Analysis of TimeDependent Problems 287


8.2 Time Approximations 2888.2.1 Introduction 2888.2.2 Parabolic Equations 2898.2.3 Hyperbolic Equations 292

8.3 Stability and Accuracy 2958.3.1 Preliminary Comments 2958.3.2 Stability Criteria 296

8.4 Transient Analysis of Nonlinear Problems 2978.4.1 Introduction 2978.4.2 Heat Transfer 2978.4.3 Flows of Viscous Incompressible Fluids 2988.4.4 Plate Bending (FSDT) 300

8.5 Computer Implementation 304

xiv

8.6 Numerical Examples 3078.6.1 Linear Problems 3078.6.2 Nonlinear Problems 314

Problems 323

References 325

9 Finite Element Formulations of Solid Continua 327

9.1 Introduction 3279.1.1 Background 3279.1.2 Descriptions of Motion 328

9.2 Strain and Stress Measures 3299.2.1 Deformation Gradient Tensor 3299.2.2 Green and Almansi Strain Tensors 3319.2.3 Polar Decomposition 3349.2.4 Stress Tensors 3349.2.5 EnergeticallyConjugate Stresses and Strains 335

9.3 Strain and Stress Measures Between ConÞgurations 3379.3.1 Notation 3379.3.2 Conservation of Mass 3399.3.3 Green Strain Tensors for Various ConÞgurations 3409.3.4 Euler Strain Tensor 3429.3.5 Relationships Between Various Stress Tensors 343

9.4 Constitutive Equations 346

9.5 Total Lagrangian and Updated Lagrangian Formulationsof Continua 3479.5.1 Principle of Virtual Displacements 3479.5.2 Total Lagrangian Formulation 3489.5.3 Updated Lagrangian Formulation 350

9.6 Finite Element Models of Two-Dimensional Continua 3539.6.1 Introduction 3539.6.2 Total Lagrangian Formulation 3539.6.3 Updated Lagrangian Formulation 3609.6.4 Computer Implementation 3619.6.5 Numerical Results 361

9.7 Shell Finite Elements 3699.7.1 Introduction 3699.7.2 Incremental Equations of Motion 3699.7.3 Finite Element Models of a Continuum 3709.7.4 Shell Finite Element 3729.7.5 Numerical Examples 378

Problems 381

References 387

xv

10 Material Nonlinearities and Coupled Problems 389


10.2 Nonlinear Elastic Problems 390

10.3 Small Deformation Theory of Plasticity 39110.3.1 Introduction 39110.3.2 Ideal Plasticity 39210.3.3 Strain Hardening Plasticity 39310.3.4 ElasticPlastic Analysis of a Bar 395

10.4 Non-Newtonian Fluids 40410.4.1 Introduction 40410.4.2 Governing Equations in Cylindrical Coordinates 40510.4.3 Power-Law Fluids 40710.4.4 WhiteMetzner Fluids 40910.4.5 Numerical Examples 412

10.5 Coupled Fluid Flow and Heat Transfer 41710.5.1 Finite Element Models 41710.5.2 Numerical Examples 419

References 423

Appendix A1: Solution Procedures for Linear AlgebraicEquations 425

A1.1 Introduction 425

A1.2 Direct Methods 427A1.2.1 Preliminary Comments 427A1.2.2 Symmetric Solver 428A1.2.3 Unsymmetric Solver 430

A1.3 Iterative Methods 430A1.3.1 General Comments 430A1.3.2 Solution Algorithms 432

References 433

Appendix A2: Solution Procedures for Nonlinear AlgebraicEquations 439

A2.1 Introduction 439

A2.2 Picard Iteration Method 440

A2.3 NewtonRaphson Iteration Method 444

A2.4 Riks and ModiÞed Riks Schemes 448

References 457

Subject Index 459


Preface

The objective of this book is to present the theory and computerimplementation of the Þnite element method as applied to simple nonlinearproblems of heat transfer and similar Þeld problems, ßuid mechanics, and solidmechanics. Both geometric as well as material nonlinearities are considered,and static and transient (i.e. time-dependent) responses are studied. Theguiding principle in writing the book was to make the presentation suitablefor (a) adoption as a text book for a Þrst course on nonlinear Þnite elementanalysis (or for a second course following an introductory course on the Þniteelement method), and (b) for use by engineers and scientists from variousdisciplines for self study and practice.

There exist a number of books on nonlinear Þnite elements. Most of thesebooks contain a good coverage of the topics of structural mechanics, and fewaddress topics of ßuid dynamics and heat transfer. While these books serveas good references to engineers or scientists who are already familiar with thesubject but wish to learn advanced topics or latest developments, they arenot suitable as textbooks for a Þrst course or for self study on nonlinear Þniteelement analysis.

The motivation and encouragement that led to the writing of the presentbook have come from the users of the authors book, An Introduction tothe Finite Element Method (McGrawHill, 1984; Second Edition, 1993; thirdedition scheduled for 2004), who have found the approach presented there tobe most suitable for any one irrespective of their scientiÞc background interested in learning the method, and also from the fact that there does notexist a book that is suitable as a textbook for a Þrst course on nonlinear Þniteelement analysis. The author has taught a course on nonlinear Þnite elementanalysis many times during the last twenty years, and the present book isan outcome of the lecture notes developed during this period. The sameapproach as that used in the aforementioned book, namely, the differentialequation approach, is adopted in the present book to introduce the theory,formulation, and computer implementation of the Þnite element method asapplied to nonlinear problems of science and engineering.

Beginning with a model (i.e. typical) second-order, nonlinear differentialequation in one dimension, the book takes the reader through increasinglycomplex problems of nonlinear beam bending, nonlinear Þeld problems in twodimensions, nonlinear plate bending, nonlinear formulations of solid continua,ßows of viscous incompressible ßuids in two dimensions (i.e. NavierStokesequations), time-approximation schemes, continuum formulations of shells,and material nonlinear problems of solid mechanics.

As stated earlier, the book is suitable as a textbook for a Þrst courseon nonlinear Þnite elements in civil, aerospace, mechanical, and mechanicsdepartments as well as in applied sciences. It can be used as a referenceby engineers and scientists working in industry, government laboratories andacademia. Introductory courses on the Þnite element method, continuummechanics, and numerical analysis should prove to be helpful.

The author has beneÞted in writing the book by the encouragement andsupport of many colleagues around the world who have used his book, AnIntroduction to the Finite Element Method, and students who have challengedhim to explain and implement complicated concepts and formulations in simpleways. While it is not possible to name all of them, the author expresses hissincere appreciation. In particular, it is a pleasure to acknowledge the helpof the authors students Juan P. Pontaza with the least-squares Þnite elementanalysis of ßuid ßow problems in Chapters 7 and 8, and Goy Teck Lim withthe plasticity example in Chapter 10. The author expresses his deep sense ofgratitude to his teacher, Professor J. T. Oden (University of Texas at Austin),to whom this book is dedicated and without whose advice, mentorship andsupport it would not have been possible for the author to modestly contributeto the Þeld of applied mechanics in general and theory and application of theÞnite element method in particular, through authors teaching, research, andtechnical writings.

J. N. ReddyCollege Station, Texas

1

Introduction

1.1 Mathematical Models

One of the most important thing engineers and scientists do is to model naturalphenomena. They develop conceptual and mathematical models to simulatephysical events, whether they are aerospace, biological, chemical, geological,or mechanical. The mathematical models are developed using laws of physicsand they are often described in terms of algebraic, differential, and/or integralequations relating various quantities of interest.A mathematical model can be broadly deÞned as a set of relationships

among variables that express the essential features of a physical system orprocess in analytical terms. The relationships that govern the system take theform of algebraic, differential, and integral equations. Mathematical models ofphysical phenomena are often based on fundamental scientiÞc laws of physicssuch as the principle of conservation of mass, the principle of conservation oflinear momentum, and the principle of conservation of energy. Mathematicalmodels of biological and other phenomena may be based on observations andaccepted theories. Keeping the scope of the present study in mind, we limitour discussions to engineering systems that are governed by laws of continuummechanics.Mathematical models of engineering systems are often characterized by very

complex equations posed on geometrically complicated regions. Consequently,many of the mathematical models, until the advent of electronic computation,were drastically simpliÞed in the interest of analytically solving them. Overthe last three decades, however, the computer has made it possible, withthe help of mathematical models and numerical methods, to solve manypractical problems of science and engineering. There now exists a new andgrowing body of knowledge connected with the use of numerical methods andcomputers to analyze mathematical models of physical systems, and this bodyis known as computational mechanics. Major established industries such asthe automobile, aerospace, chemical, pharmaceutical, petroleum, electronicsand communications, as well as emerging industries such as biotechnology, relyon computational mechanicsbased capabilities to simulate complex systems

2 NONLINEAR FINITE ELEMENT ANALYSIS

Here, we consider the familiar example of a simple pendulum to illustratehow a mathematical model of the motion of the pendulum can be constructedusing the principle of conservation of linear momentum, that is, Newtonssecond law of motion. Numerical analysis of the problem will be consideredin the sequel.

Example 1.1.1

As an example of the mathematical model development, consider the problem of a simplependulum. The system consists of a bob of mass m attached to one end of a rod of length land the other end is pivoted to a Þxed point O, as shown in Figure 1.1.1. The primary goalof the mathematical model to be derived here is to have a means to determine the motion(i.e. angular displacement as a function of time) of the bob. Keeping the goal of the analysisin mind, we make several assumptions. (1) The bob as well as the rod connecting the bobto the Þxed point O are rigid; (2) the mass of the rod is negligible relative to the mass of thebob; and (3) there is no friction at the pivot. These assumptions may be removed to obtaina mathematical model that describes the system more accurately.

The equation governing the motion can be derived using the principle of conservationof linear momentum, also known as Newtons second law of motion, which states that thevector sum of externally applied forces is equal to the time rate of change of the linearmomentum of the body:

F =ma (1.1.1)

where F is the vector sum of all forces acting on a system, m is the mass of the system, anda is the acceleration of the system.

Applying Newtons second law in the x direction (see Figure 1.1.1), we have

Fx = −mg sin θ, v = ldθ

dt(1.1.2)

where θ is the angular displacement, v is the velocity along x, and t denotes time. Then,the equation for motion about the pivot becomes

−mg sin θ =mld2θ

dt2or

d2θ

dt2+g

lsin θ = 0 (1.1.3)

Figure 1.1.1 Simple pendulum.

m

θ l

x

y

R

mg

ma=mθ l¨

O

INTRODUCTION 3

Thus, the problem at hand involves solving the nonlinear differential equation

d2θ

dt2+g

lsin θ = 0, 0 < t ≤ T (1.1.4)

subjected to the initial (i.e. at time t = 0) conditions

θ(0) = θ0,dθ

dt(0) = v0 (1.1.5)

where θ0 and v0 are the initial values of angular displacement and velocity, respectively.Mathematically, the problem is called an initial-value problem. If the amplitude θ is notsmall, the restoring moment is proportional to sinθ, and Eq. (1.1.3) represents a nonlinearequation. For small θ, sin θ is approximately equal to the angle θ, and the motion is describedby the linear equation

d2θ

dt2+ λ2θ = 0, λ2 =

g

l(1.1.6)

which represents a simple harmonic motion.

The general analytical solution to the linear equation (1.1.6) is given by

θ(t) = A sinλt+B cosλt, λ =qg

l(1.1.7)

where A and B are constants to be determined using the initial conditions in (1.1.5). Weobtain

A =v0λ, B = θ0 (1.1.8)

and the solution to the linear problem is

θ(t) =v0λsinλt+ θ0 cosλt (1.1.9)

For zero initial velocity and non-zero initial position θ0, we have

θ(t) = θ0 cosλt (1.1.10)

1.2 Numerical Simulations

By a numerical simulation of a process, we mean the solution of thegoverning equations (or mathematical model) of the process using a numericalmethod and a computer. While the derivation of the governing equationsfor most problems is not unduly difficult, their solution by exact methodsof analysis is a formidable task. In such cases, numerical methods ofanalysis provide an alternative means of Þnding solutions. Numerical methodstypically transform differential equations to algebraic equations that are to besolved using computers. For example, the mathematical formulation of thesimple pendulum resulted in the nonlinear differential equation (1.1.4), whoseanalytical solution cannot be obtained. Therefore, one must consider usinga numerical method to solve it. Even linear problems may not admit exactsolutions due to geometric and material complexities, but it is relatively easy


to obtain approximate solutions using numerical methods. These ideas areillustrated below using the simple pendulum problem of Example 1.1.1. TheÞnite difference method is used as the numerical method.

Example 1.2.1

Here we consider the numerical solution of Eq. (1.1.4) governing a simple pendulum using theÞnite difference method. In the Þnite difference method, the derivatives are approximated bydifference quotients (or the function is expanded in a Taylor series) that involve the unknownvalue of the solution at time ti+1 and the known value of the solution at ti. For example,consider the Þrst-order equation

du

dt= f(t, u) (1.2.1)

We approximate the derivative at ti by

³du

dt

´ ¯¯t=ti

≈ u(ti+1)− u(ti)ti+1 − ti

= f(ui, ti) (1.2.2)

orui+1 = ui +∆t f(ui, ti) (1.2.3a)

whereui = u(ti), ∆t = ti+1 − ti (1.2.3b)

Equation (1.2.3a) can be solved, starting from the known value u0 of u(t) at t = 0, foru1 = u(t1) = u(∆t). This process can be repeated to determine the values of u at timest = ∆t,2∆t, . . . , n∆t. This is known as Eulers explicit method, also known as the forwarddifference scheme. Note that we are able to convert the ordinary differential equation (1.2.1)to an algebraic equation (1.2.3a) that needs to be evaluated at different times to constructthe time history of u(t).

Eulers explicit method can be applied to the nonlinear second-order equation (1.1.4).First we rewrite Eq. (1.1.4) as a pair of Þrst-order equations

dθ

dt= v,

dv

dt= −λ2 sin θ (1.2.4)

which are coupled (i.e. one cannot be solved without the other). Applying the scheme ofEq. (1.2.4) to the equations at hand, we obtain

θi+1 = θi +∆t vi; vi+1 = vi −∆t λ2 sin θi (1.2.5)

The expressions for θi+1 and vi+1 in Eq. (1.2.5) are repeatedly computed using the knownsolution (θi, vi) from the previous time step. At time t = 0, we use the known initial values(θ0, v0). Thus, one needs a computer and a computer language like Fortran (77 or 90) towrite a computer program to compute numbers.

The numerical solutions of equation (1.2.5) for two different time steps, ∆t = 0.05 and∆t = 0.025, along with the exact linear solution (1.1.10) (with θ0 = π/4) are presentedin Figure 1.2.1. The numerical solutions of the nonlinear problem are dependent on thetime step, and smaller the time step more accurate the solution is. This is because theapproximation of the derivative in Eq. (1.2.2) tends to the exact derivative with ∆t→ 0.

0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5Time, t

-3.0

-2.0

-1.0

0.0

1.0

2.0

Ang

ular

dis

plac

emen

t

Linear (exact)

Nonlinear (0.025)Nonlinear (0.05)

(∆t = 0.05)(∆t = 0.025)

INTRODUCTION 5

Figure 1.2.1 Analytical and numerical solutions of the simple pendulum.

1.3 The Finite Element Method

As illustrated in the previous section, numerical methods are extremelypowerful tools for engineering analysis. With the advent of computers, therehas been a tremendous explosion in the development and use of numericalmethods. Of these, the Þnite difference methods and the Þnite element methodand their variants are the most commonly used methods in the analysis ofpractical engineering problems. In Þnite difference methods, derivatives ofvarious order are approximated using Taylors series. The traditional Þnitedifference methods suffer from two major drawbacks: (1) applying boundaryconditions of the gradient type requires additional approximation; (2) Þnitedifference formulas are traditionally developed for rectangular grids, making itdifficult to use them for irregular domains. Advances have been made in recentyears to overcome these drawbacks but the remedies are problem dependent.The Þnite element method is based on the idea that every system is physicallycomposed of different parts and hence its solution may be represented in parts.In addition, the solution over each part is represented as a linear combinationof undetermined parameters and known functions of position and possiblytime. The parts can differ from each other in shape, material properties, andphysical behavior. Even when the system is of one geometric shape and madeof one material, it is simpler to represent its solution in a piecewise manner.


In recent years, generalizations of the Þnite element method have emerged(e.g. the generalized Þnite element method and element-free methods ormeshless methods); however, in this study we limit our discussion to thetraditional Þnite element method [9].The traditional Þnite element method is endowed with three basic features.

First, a domain of the system is represented as a collection of geometricallysimple subdomains, called Þnite elements. Second, over each Þnite element,the unknown variables are approximated by a linear combination of algebraicpolynomials and undetermined parameters, and algebraic relations amongthe parameters are obtained by satisfying the governing equations, often ina weighted-integral sense, over each element. The undetermined parametersrepresent the values of the unknown variables at a Þnite number of preselectedpoints, called nodes, in the element. Third, the algebraic relations from allelements are assembled using continuity and equilibrium considerations.There are several reasons why an engineer or scientist should study the

Þnite element method; these are listed below.

1. The Þnite element method is the most powerful numerical method everdevised for the analysis of engineering problems. It is capable of handlinggeometrically complicated domains, a variety of boundary conditions,nonlinearities, and coupled phenomena that are common in practicalproblems. The knowledge of how the method works greatly enhances theanalysis skill and provides a greater understanding of the problem beingsolved.

2. Commercial software packages or canned computer programs based onthe Þnite element method are often used in industrial, research, andacademic institutions for the solution of a variety of engineering andscientiÞc problems. The intelligent use of these programs and a correctinterpretation of the output is often predicated on knowledge of the basictheory underlying the method.

3. It is not uncommon to Þnd mathematical models derived in personalresearch and development that cannot be evaluated using canned programs.In such cases, an understanding of the Þnite element method and knowledgeof computer programming can help design programs to evaluate themathematical models.

The basic ideas underlying the Þnite element method are reviewed inChapter 2 using linear differential equations involving a single variable in oneand two dimensions. The main objective there is to introduce the terminologyand steps involved, e.g. weak formulation of differential equations over aelement, derivation of the interpolation functions, and numerical evaluationof coefficients and so on. Readers who are familiar with the Þnite elementmethod as applied to linear differential equations may skip Chapter 2 and gostraight to Chapter 3.

INTRODUCTION 7

1.4 Nonlinear Analysis

1.4.1 Introduction

Recall from the simple pendulum problem of Examples 1.1.1 and 1.2.1, thatnonlinearity naturally arises in a true, rigorous mathematical formulation ofphysical problems. Based on assumptions of smallness of certain quantitiesof the formulation, the problem may be reduced to a linear problem. Linearsolutions may be obtained with considerable ease and less computational costwhen compared to nonlinear solutions. Further, linear solutions due to variousboundary conditions and load cases may be scaled and superimposed. Inmany instances, assumptions of linearity lead to reasonable idealization of thebehavior of the system. However, in some cases assumption of linearity mayresult in an unrealistic approximation of the response. The type of analysis,linear or nonlinear, depends on the goal of the analysis and errors in thesystems response that may be tolerated. In some cases, nonlinear analysis isthe only option left for the analyst as well as the designer (e.g. high-speedßows of inviscid ßuids around solid bodies).Nonlinear analysis is a necessity, for example, in (a) designing high

performance and efficient components of certain industries (e.g. aerospace,defense and nuclear), (b) assessing functionality (e.g., residual strength andstiffness of structural elements) of existing systems that exhibit some types ofdamage and failure, (c) establishing causes of system failure, (d) simulatingtrue material behavior of processes, and (e) research to gain a realisticunderstanding of physical phenomena.The following features of nonlinear analysis should be noted (see [17]):

The principle of superposition does not hold. Analysis can be carried out for one load case at a time. The history (or sequence) of loading inßuences the response. The initial state of the system (e.g. prestress) may be important.

1.4.2 ClassiÞcation of Nonlinearities

There are two common sources of nonlinearity: (1) geometric and (2) material.The geometric nonlinearity arises purely from geometric consideration (e.g.nonlinear straindisplacement relations), and the material nonlinearity is dueto nonlinear constitutive behavior of the material of the system. A third typeof nonlinearity may arise due to changing initial or boundary conditions. Wewill discuss various types of nonlinearities through simple examples [5].The simple pendulum problem of Example 1.1.1 is an example of geometric

nonlinearity. A common example of geometric nonlinearity is provided by (seeHinton [5]) a rigid link supported by a linear elastic torsional spring at oneend and subjected to a vertical point load at the other end, as shown in Figure1.4.1. Moment equilibrium (i.e. summing the moments about the hinge) and

θ

l

F

M = kT θ

F

θ

l

F

kT

°

, θ

, F l/

KT

0.0 0.5 1.0 1.5

Rotation

0

4

8

12

16

20

Load

par

amet

er

Nonlinear

Linear


the linear momentrotation relationship give

M − Fl cos θ = 0, M = kT θ or F =kT θ

l cos θ(1.4.1)

where kT is the torsional spring constant, l length of the link, and M is themoment experienced by the torsional spring due to the angular displacementθ. If θ is small, cos θ → 1 and the governing equation is reduced to a linearequation

F =kT θ

l, θ <

π

2(1.4.2)

The nonlinear and linear responses are shown in Figure 1.4.2. Clearly, thenonlinearity in the present case is due to the change of geometry, and thenonlinear deßection is less than the linear deßection as the load is increased.Such a nonlinearity is known as the hardening type.

Figure 1.4.1 Rigid linklinear torsional spring cantilever.

Figure 1.4.2 Geometric nonlinear response of a rigid linklinear torsionalspring cantilever.

0.0 0.5 1.0 1.5

Rotation

0.0

0.2

0.4

0.6

0.8

1.0

1.2

1.4

1.6

Load

par

amet

er

Linear

Nonlinear

, θ

, Fl /

KT

INTRODUCTION 9

The material nonlinearity may be introduced into the problem if themomentrotation relationship is nonlinear

M = kT (θ)θ, say kT = k0 − k1θ (1.4.3)

where k0 and k1 are material parameters that are determined through tests. Ifwe use the relationship (1.4.3) in (1.4.1), we obtain a nonlinear equation thatcontains both geometric and material nonlinearities; if Eq. (1.4.3) is used in(1.4.2), the resulting nonlinearity is only due to the material behavior.Figure 1.4.3 contains plots of load F versus rotation θ for the materially

nonlinear case. Note that the material nonlinearity in the present case is dueto the change (reduction) in the torsional spring stiffness, and the nonlineardeßection is greater than the linear deßection as the load is increased. Such anonlinearity is known as the softening type. In the present case, the geometricnonlinearity dominates if both nonlinearities are included.An example of another type of nonlinearity is provided by (see Hinton [5])

the axial deformation of an isotropic, homogeneous, linear elastic rod withconstrained end displacement, as shown in Figure 1.4.4. The rod is of length2L, uniform cross-sectional area of A, and loaded with an axial force P at itsmidpoint. The lower end of the rod is constrained so that it can at most havean axial displacement of u0 (which is assumed to be very small compared tothe length L).

Figure 1.4.3 Material nonlinear response of a rigid linklinear torsionalspring cantilever.

u0

A

B

C

L

L

P

uB

P

u0


Figure 1.4.4 Axial deformation of a rod with constrained end displacement.

The governing equation and boundary condition at A are given by

EAd2u

dx2= 0, u(0) = 0 (1.4.4)

The displacement at point B can be determined using the boundary conditionat point C, which is dependent on whether the displacement of point C equalsu0. Thus, we haveµ

EAdu

dx

¶B− P = 0, if uB = uC < u0 (1.4.5)

uC = 0, if uB ≥ u0 (1.4.6)

In the former case, the displacement is given by

u(x) =Px

EA, uB =

PL

EA, if uC < u0 (1.4.7)

In the latter case (by solving the governing equation in the two intervals andusing the continuity and boundary condition with L+u0 ≈ L) the displacementis given by

u(x) =

⎧⎨⎩Px2EA , 0 ≤ x ≤ L,

P (2L−x)2EA , L ≤ x ≤ 2L

(1.4.8a)

and

uB =PL

2EA(1.4.8b)

Thus the forcedisplacement relationship of the rod is bilinear. Such problemsare called contact or nonlinear boundary problems [5].

INTRODUCTION 11

1.5 The Big Picture

Engineers design and manufacture various types of systems. Engineeringdesign is the process of altering dimensions, shapes, and materials to Þndthe best (optimum) conÞguration of the system to carry out certain speciÞcfunction. Analysis is an aid to design and it involves (1) mathematical modeldevelopment, (2) data acquisition by measurements, (3) numerical simulation,and (4) evaluation of the results in light of known information and correctionsto the mathematical model. The mathematical model is developed using lawsof physics and assumptions concerning the process behavior. The data includesthe actual system parameters (geometry, loading, and boundary conditions)and constitutive properties. The constitutive properties such as the modulus,conductivity, and so on are determined in laboratory experiments. Themathematical model, in most practical cases, does not admit analytical(or exact) solution due to the geometric complexity and/or nonlinearities.Nonlinearities in a mathematical model arise from changing geometry ormaterial behavior. It is necessary to employ numerical methods to compute anapproximate solution to the mathematical model. The Þnite element method isa powerful numerical method that can be used to analyze engineering problems[115].A typical Þnite element analysis exercise begins with the actual physical

system or part thereof to be analyzed [5]. Then we form a set of objectives forthe analysis. If the analysis objective is to help develop a preliminary design ofthe system, the analysis can be very simple. On the other hand, if the analysisobjective is to verify and certify the Þnal design of a system, the analysis mustbe the most sophisticated that can be carried out. Thus, the objectives willdictate the type of idealization of the system to be adopted; for example,should we model as a two-dimensional or three-dimensional problem?, analyzeas a linear or nonlinear problem and what type of nonlinearities to beconsidered?, what type constitutive model to be used?, how are the loadsand boundary conditions of the actual system are idealized?, what couplingeffects, if any, to be considered?, and so on.Once the system idealization is complete (i.e. mathematical model is in

place), one must decide on type of numerical approximation (and software tobe used). This involves (1) choice of unknowns, which in turn dictates the typeof Þnite element model, (2) type of elements, (3) type of mesh, and if nonlinearanalysis is to be carried out, select (4) magnitude of load increments, (5)type of iterative method of solution, (6) error criterion, (7) error tolerance, and(8) maximum allowable number of iterations for termination of the program.The Þnal step in creating a computational model is veriÞcation of the

code and validation of the mathematical model. VeriÞcation is the processof determining if the computational model is an accurate discrete analog ofthe mathematical model [15]. Thus, if the round-off errors introduced due toÞnite arithmetic in a computer are negligible, the computational model should


give the exact solution of the mathematical model. Thus veriÞcation involvescomparing the numerical results with known exact and/or experimental resultsof benchmark problems. On the other hand, validation is the process ofdetermining the degree to which the mathematical model (hence the computercode that is veriÞed) represents the physical reality of the system from theperspective of the intended uses of the model. Obviously, the validationexercise can be deÞned only in relation to the intended uses of the model.For example, a mathematical model based on linear elasticity is adequate fordetermining linear elastic solutions of a solid but inadequate for determiningits nonlinear response. The validation exercise allows one to modify themathematical model to include the missing elements that make the computedresponse come closer to the physical response. In fact, a mathematical modelcan never be validated; it can only be invalidated. It is always a goodidea, when developing new software, to undertake the veriÞcation exercise.Validation is a must when studying new and mulit-physics problems.

References

1. Bathe, K. J., Finite Element Procedures, PrenticeHall, Englewood Cliffs, NJ (1996).

2. Belytschko, T., Liu, W. K., and Moran, B., Nonlinear Finite Elements for Continua andStructures, John Wiley, Chichester, UK (2000).

3. CrisÞeld, M. A., Non-Linear Finite Element Analysis of Solids and Structures, Vol. 1:Essentials, John Wiley, Chichester, UK (1991).

4. CrisÞeld, M. A., Non-Linear Finite Element Analysis of Solids and Structures, Vol. 2:Advanced Topics, John Wiley, Chichester, UK (1997).

5. Hinton, E. (ed.) NAFEMS Introduction to Nonlinear Finite Element Analysis,NAFEMS, Glasgow, UK (1992).

6. Oden, J. T., Finite Elements of Nonlinear Continua, McGrawHill, New York (1972).

7. Owen, D. R. J. and Hinton, E., Finite Elements in Plasticity: Theory and Practice,Pineridge Press, Swansea, UK (1991).

8. Reddy, J. N., Energy and Variational Methods in Applied Mechanics, John Wiley, NewYork (1984).

9. Reddy, J. N., An Introduction to the Finite Element Method, 2nd edn, McGrawHill,New York (1993).

10. Reddy, J. N., Energy Principles and Variational Methods in Applied Mechanics, 2nd edn,John Wiley, New York (2002).

11. Reddy, J. N. and Gartling, D. K., The Finite Element Method in Heat Transfer and FluidDynamics, 2nd edn, CRC Press, Boca Raton, FL (1999).

12. Yang, Y.-B. and Kuo, S.-R., Theory & Analysis of Nonlinear Framed Structures,PrenticeHall, Singapore (1994).

13. Zienkiewicz, O. C. and Taylor, R. L., The Finite Element Method, Fourth Edition, Vol.1: Basic Formulation and Linear Problems, McGrawHill, London, UK (1989).

14. Zienkiewicz, O. C. and Taylor, R. L., The Finite Element Method, 4th edn, Vol. 2: Solidand Fluid Mechanics, Dynamics and Non-linearity, McGrawHill, London, UK (1989).

15. Sargent, R. G., An overview of veriÞcation and validation of simulation models,Proceedings of the 1987 Winter Simulation Conference, Society of Computer Simulation(1987).

2

The Finite Element Method:A Review

2.1 Introduction

The main ideas of the Þnite element method were presented in Chapter 1. Tosummarize, the Þnite element method has the following three basic features:

1. Divide the whole (i.e. domain) into parts, called Þnite elements.

2. Over each representative element, develop the relations among thesecondary and primary variables (e.g. forces and displacements,heats and temperatures, and so on).

3. Assemble the elements (i.e. combine the relations of all elements) to obtainthe relations between the secondary and primary variables of the wholesystem.

In the present chapter, we review the basic steps of the Þnite element modeldevelopment as applied to one- and two-dimensional problems described bytypical second-order differential equations. The main objective is to familiarizethe reader with the speciÞc steps involved in the Þnite element formulation andits applications. Readers who are already familiar with the authors approachto Þnite element modeling may skip this chapter.

2.2 One-Dimensional Problems

2.2.1 Governing Differential Equation

Consider the differential equation

− d

dx

µadu

dx

¶+ cu = f for 0 < x < L (2.2.1)

where a = a(x), c = c(x), and f = f(x) are the data (i.e. known quantities) ofthe problem, and u(x) is the solution to be determined. The data depends onthe material properties, geometry, and loads or source. The equation arisesin a number of Þelds, and it must be solved subject to appropriate boundaryconditions. Table 2.2.1 contains a list of Þelds, by no means exhaustive, inwhich Eq. (2.2.1) arises.


2.2.2 Finite Element Approximation

The domain (0, L) of the problem consists of all points between x = 0 andx = L. The points x = 0 and x = L are the boundary points of the totaldomain. In the Þnite element method, the domain (0, L) is divided into a set ofintervals. A typical interval, called a Þnite element, is of length he and locatedbetween points x = xa and x = xb, where xa and xb denote the coordinates ofthe end points of the Þnite element with respect to the coordinate x.

Table 2.2.1 List of Þelds in which the model equation (2.2.1) arises, withmeaning of various parameters and variables (see the bottomof the table for the meaning of some parameters*).

Field of Primary Problem data Secondarystudy variable variable

u a c f Q

Heat Tempe- Thermal Surface Heat Heattransfer rature conductance convection generation

T − T∞ kA Apβ f Q

Flow Fluid- Permea- InÞl- Pointthrough head bility tration sourceporousmedium φ µ 0 f Q

Flow Pressure Pipe Pointthrough resistance sourcepipes P 1/R 0 0 Q

Flow of Velocity Viscosity Pressure Shearviscous gradient stressßuids vz µ 0 −dP/dx σxz

Elastic Displa- Tension Transverse Pointcables cement force force

u T 0 f P

Elastic Displa- Axial Axial Pointbars cement stiffness force force

u EA 0 f P

Torsion Angle of Shear Torqueof twist stiffnessbars θ GJ 0 0 T

Electro- Electrical Dielectric Charge Electricstatics potential constant density ßux

φ ε 0 ρ E

*k = thermal conductance; β = convective Þlm conductance; p = perimeter; P = pressure orforce; T∞ = ambient temperature of the surrounding ßuid medium; R = 128µh/(πd4) withµ being the viscosity; h, the length and d the diameter of the pipe; E = Youngs modulus;A = area of cross-section; J = polar moment of inertia.

f(x)

u(xa) = u1 u(xb) = u2

1 2aQ

xxdxdua

a

==

⎟⎠⎞

⎜⎝⎛− bQ

xxdxdua

b

==

⎟⎠⎞

⎜⎝⎛

THE FINITE ELEMENT METHOD: A REVIEW 15

In the Þnite element method, we seek an approximate solution to Eq.(2.2.1) over each Þnite element; a typical Þnite element is shown in Figure2.2.1. The Þnite element approximation ueh(x) is sought in the form

u(x) ≈ ueh(x) = ce1ϕe1(x) + ce2ϕe2(x) + . . .+ cenϕen(x)

=nXj=1

cejϕej(x) (2.2.2)

where ϕej(x) are functions to be selected and cej are constants to be determined

such that Eq. (2.2.2) satisÞes the differential equation (2.2.1) and appropriateend conditions over the element. Since there are n unknown parameters, weneed n relations to determine them. Substituting the approximate solution(2.2.2) into the left-hand side of Eq. (2.2.1), we obtain an expression that, ingeneral, will not be equal to the right-hand side of the equation, f(x). Thedifference between the two sides of the equation is called the residual

− d

dx

µaduehdx

¶+ cueh(x)− f(x) ≡ Re(x, ce1, ce2, . . . , cen) 6= 0 (2.2.3)

We wish to determine cej (j = 1, 2, · · · , n) such that the residual is zero, insome sense, over the element.One way of making the residual zero is in weighted-integral senseZ xb

xawei (x)R

e(x, c1, c2, . . .) dx = 0, i = 1, 2, . . . , n (2.2.4)

where wei (i = 1, 2, . . . , n) are weight functions. Equation (2.2.4) provides aset of n algebraic relations among the parameters cej (j = 1, 2, . . . , n). The setwe1(x), we2(x), . . . , wen(x) must be linearly independent so that the algebraicequations (2.2.4) are also linearly independent and invertible.There are other choices of wei that may be used. In the present study

we take wei (x) to be the same as the approximation functions ϕei (x). This

particular choice is known as the Galerkin method. Different choice of theweight functions will result in a different set of algebraic equations or differentÞnite element models of the same differential equation.

Figure 2.2.1 A typical Þnite element in one dimension.


2.2.3 Derivation of the Weak Form

To weaken the continuity required of ueh(x), we must trade some of thedifferentiation in (2.2.4) from ueh to wei such that both ueh and wei aredifferentiated equally, once each in the present case. The resulting integralform is termed the weak form of Eq. (2.2.1) because it allows approximationfunctions with less (or weaker) continuity (or differentiability).A three-step procedure of constructing the weak form of Eq. (2.2.1) is

presented next.

Step 1. The Þrst step is to write the weighted-residual statement as in Eq.(2.2.4)

0 =

Z xb

xawei

∙− d

dx

µaduehdx

¶+ cueh − f

¸dx (2.2.5)

Step 2. The second step is to trade differentiation from ueh to wei , using

integration by parts. We obtain

0 =

Z xb

xa

µadweidx

duehdx

+ cweiueh − wei f

¶dx−

∙wei · a

duehdx

¸xbxa

(2.2.6)

Step 3. Examining the boundary term appearing in the weak form (2.2.6),namely, the expression ∙

wei · adu

dx

¸xbxa

The coefficient of the weight function wei in the boundary expression, a(du/dx),is called the secondary variable, and its speciÞcation constitutes the naturalor Neumann boundary condition. The primary variable is the dependentunknown of the differential equation, u, in the same form as the weightfunction in the boundary expression (i.e. replace wei with u). The speciÞcationof a primary variable on the boundary constitutes the essential or Dirichletboundary condition. For the model equation at hand, the primary andsecondary variables are

Primary variable: u Secondary variable: adu

dx≡ Q (2.2.7)

In writing the Þnal weak form, we denote the secondary variables at theends of the element as

Qea = −µadu

dx

¶xa

, Qeb =

µadu

dx

¶xb

(2.2.8)

The primary and secondary variables at the nodes are shown on the typicalelement in Figure 2.2.1. Students of engineering who have taken a course in


mechanics of deformable bodies (or strength of materials) recognize that thisÞgure shows the free-body diagram of a typical but arbitrary portion of a bar,with Qea and Q

eb denoting the axial forces; Q

ea is a compressive force while Q

eb

is a tensile force (algebraically, both are in the positive x direction, as shownin Figure 2.2.1). For heat conduction problems, Qea denotes the heat input atthe left end and Qeb the heat output from the right end of the element. Withthe notation in (2.2.8), the Þnal expression for the weak form is given by

0 =

Z xb

xa

µadweidx

duehdx

+ cweiueh − wei f

¶dx− wei (xa)Qea − wei (xb)Qeb (2.2.9)

This completes the three-step procedure of constructing the weak form.

Remarks

1. The weak form in (2.2.9) contains two types of expressions: thosecontaining both wei and u

eh, and those containing only w

ei . The expression

containing both wei and ueh is called the bilinear form (i.e. linear in w

ei and

linear in ueh):

B(wei , ueh) ≡

Z xb

xa

µadweidx

duehdx

+ cweiueh

¶dx (2.2.10)

Similarly, the expression containing only wei (but not ueh) is called the linear

form:

`(wei ) =

Z xb

xawei f dx+ w

ei (xa)Q

ea +w

ei (xb)Q

eb (2.2.11)

More formally, we say that B(u, v) is linear in both u and v, and `(v) islinear in v, if and only if the following conditions hold:

B(c1u1 + c2u2, v) = c1B(u1, v) + c2B(u2, v)

B(u, c1v1 + c2v2) = c1B(u, v1) + c2B(u, v2)(2.2.12a)

`(c1v1 + c2v2) = c1`(v1) + c2`(v2) (2.2.12b)

where c1 and c2 are constants and u, v, u1, u2, v1, and v2 are dependentvariables.

2. In view of the above deÞnitions, the weak form (2.2.9) of Eq. (2.2.1) cannow be expressed as

B(wei , ueh) = `(w

ei ) (2.2.13)

which is called the variational problem associated with Eq. (2.2.1). As willbe seen later, the bilinear form B(wei , u

eh) results directly in the element

coefficient matrix, and the linear form `(wei ) leads to the right-hand sidecolumn vector of the Þnite element equations.


3. Those who have a background in applied mathematics or solid andstructural mechanics will appreciate the fact that the weak form (2.2.9)or the variational problem (2.2.13) is nothing but the statement of theprinciple of the minimum total potential energy Π(ueh) (applied to the barelement):

0 = δΠ = B(δueh, ueh)− `(δueh)

where δ is the variational symbol and Π(ueh) is the quadratic functionaldeÞned by

Π(ueh) =1

2B(ueh, u

eh)− `(ueh) (2.2.14)

=

Z xb

xa

"a

2

µdueidx

¶2+c

2(ueh)

2 − uehf#dx− ueh(xa)Qea − ueh(xb)Qeb

(2.2.15)

Equation (2.2.14) holds only when the bilinear form B(w, u) is symmetricin u and w,

B(w, u) = B(u,w) (2.2.16)

and `(u) is linear in u. The expression 12B(u

eh, u

eh) represents the elastic

strain energy stored in the bar Þnite element and `(ueh) represents the workdone by applied distributed force f(x) and point loads Qea and Q

eb.

2.2.4 Interpolation Functions

Recall that the weak form over an element is equivalent to the differentialequation, and it contains the end conditions on the forces Qei (see Figure2.2.1). Therefore, the approximate solution ueh(x) should be selected such thatthe differentiability (or continuity) conditions implied by the weak form aremet and the end conditions on the primary variables u(xi) = u

ei are satisÞed.

Since the weak form contains the Þrst-order derivative of ueh, any functionwith a non-zero Þrst derivative would be a candidate for ueh. Thus, the Þniteelement approximation ueh of u(x) can be an interpolant, that is, must be equalto uea at xa and u

eb at xb. Thus, a linear polynomial (see Figure 2.2.2)

ueh(x) = ce1 + c

e2x (2.2.17)

is admissible if we can select ce1 and ce2 such that

ueh(xa) = ce1 + c

e2xa = u

ea, ueh(xb) = c

e1 + c

e2xb = u

eb

or ∙1 xa1 xb

¸½ce1ce2

¾=

½ueaueb

¾→ ce1 =

uebxb − ueaxaxb − xa , ce2 =

ueb − ueaxb − xa (2.2.18)

he

1x

2

u(x) True solution

x

ea xxx 1==

eb xxx 2==

)(11 xLu ee

)()()(

2211

21

xLuxLuxccxu

eeee +=+≈

)(22 xLu ee

eu1

eu2


Substitution of (2.2.18) for cei into (2.2.17) yields

ueh(x) = Le1(x)u

e1 + L

e2(x)u

e2 =

2Xj=1

Lej(x)uej (2.2.19)

where

Le1(x) =xb − xxb − xa , Le2(x) =

x− xaxb − xa (2.2.20)

are the linear Lagrange interpolation functions, and

ue1 = uea, ue2 = u

eb (2.2.21)

are the nodal values of ueh(x) at x = xa and x = xb, respectively. Note thatLei (x) satisfy the interpolation property

Lei (xej) =

½1, if i = j0, if i = j

(2.2.22)

where xe1 = xa and xe2 = xb (see Figure 2.2.2). In addition, the Lagrange

interpolation functions satisfy the property, known as the partition of unity:

nXj=1

Lej(x) = 1 (2.2.23)

Figure 2.2.2 Linear Þnite element approximation over an element.


If we wish to approximate u(x) with a quadratic polynomial, we write

ueh(x) = ce1 + c

e2x+ c

e3x2 (2.2.24)

Since there are three parameters ce1, ce2, and c

e3, we must identify one more

nodal point in the element to express all three c,s in terms of the values of uehat three nodes. Of course, we can also carry the nodal values uea and u

eb (so

that they can be used to join adjacent elements) and the parameter ce3 as theunknowns of the approximation. Identifying the third node at the center ofthe element [see Figure 2.2.3(a)], we can write

ueh(xe1) ≡ ue1 = ce1 + ce2xe1 + ce3(xe1)2

ueh(xe2) ≡ ue2 = ce1 + ce2xe2 + ce3(xe2)2 (2.2.25)

ueh(xe3) ≡ ue3 = ce1 + ce2xe3 + ce3(xe3)2

where

xe1 = xa, xe2 = xa +he2, xe3 = xa + he = xb (2.2.26)

Solving Eqs. (2.2.33) for cei in terms of uei , we obtain

ueh(x) = Le1(x)u

e1 + L

e2(x)u

e2 + L

e3(x)u

e3 =

3Xj=1

Lej(x)uej (2.2.27)

where Lei (x) are the quadratic Lagrange interpolation functions [see Figure2.2.3(b)]

Le1(x) =

µx− xe2xe1 − xe2

¶µx− xe3xe1 − xe3

¶Le2(x) =



¶(2.2.28)

Le3(x) =



¶Higher-order Lagrange interpolation of u(x) can be developed along the

similar lines. Thus, an (n− 1)st degree Lagrange interpolation of u(x) can bewritten as

ueh(x) = Le1(x)u

e1 + L

e2(x)u

e2 + . . .+ L

en(x)u

en =

nXj=1

Lej(x)uej (2.2.29)

where the Lagrange interpolation functions of degree n− 1 are given by

Lej(x) =nY

i=1,i6=j

Ãx− xeixej − xei

!(2.2.30)

1.0 1.01.0

x

1 2 3x he

(a)

(b)

1 2 3

)(2 xLe)(1 xLe

)(3 xLe

)(xLei

ex3ex2

ex1

x


Figure 2.2.3 (a) Quadratic Þnite element. (b) Quadratic Lagrangeinterpolation functions.

The Þnite element solution ueh(x) must fulÞll certain requirements in orderthat it be convergent to the actual solution u(x) as the number of elements(h reÞnement) or the degree of the polynomials (p reÞnement) is increased.These are:

1. The approximate solution should be continuous and differentiable asrequired by the weak form.

2. It should be a complete polynomial, that is, include all lower-order termsup to the highest order term used.

3. It should be an interpolant of the primary variables at the nodes of theÞnite element (at least interpolate the solution at the end points).

The reason for the Þrst requirement is obvious; it ensures that every termof the governing equation has a non-zero contribution to the coefficient matrix.The second requirement is necessary in order to capture all possible states, thatis, constant, linear and so on, of the actual solution. For example, if a linearpolynomial without the constant term is used to represent the temperaturedistribution in a one-dimensional system, the approximate solution can neverbe able to represent a uniform state of temperature Þeld in the element. Thethird requirement is necessary in order to enforce continuity of the primaryvariables at the end points where the element is connected to other elements.


2.2.5 Finite Element Model

Substitution of (2.2.29) into (2.2.9) will give the necessary algebraic equationsamong the nodal values uei and Q

ei of the element. In order to formulate

the Þnite element model based on the weak form (2.2.9), it is not necessaryto decide a priori the degree of approximation of ueh(x). The model can bedeveloped using an arbitrary degree of interpolation. For n > 2, the weakform in Eq. (2.2.9) must be modiÞed to include non-zero secondary variables,if any, at interior nodes:

0 =

Z xb

xa

µadw

dx

duehdx

+ cwueh

¶dx−

Z xb

xawf dx−

nXi=1

w(xei )Qei (2.2.31)

where xei is the global coordinate of the ith node of element Ωe. If nodes 1 and

n denote the end points of the element, then Qe1 and Qen represent the unknown

point sources, and all other Qei are the point sources at nodes 2, 3, . . . , n, whichare always known.Substituting Eq. (2.2.29) for ueh and w

e1 = Le1, w

e2 = Le2, . . . , w

ei =

Lei , . . . , wen = L

en into the weak form (2.2.31), we obtain n algebraic equations.

The ith algebraic equation can be written as

0 =

Z xb

xa

⎡⎣adLeidx

⎛⎝ nXj=1

uejdLejdx

⎞⎠+ cLei⎛⎝ nXj=1

uejLej

⎞⎠− Leif⎤⎦ dx

−nXj=1

Lei (xej)Q

ej

=nXj=1

"Z xb

xa

ÃadLeidx

dLejdx

+ cLeiLej

!dx

#uej −

Z xb

xaLeif dx−Qei

0 =nXj=1

Keiju

ej − fei −Qei (2.2.32)

for i = 1, 2, . . . , n, where

Keij =

Z xb

xa

ÃadLeidx

dLejdx

+ cLeiLej

!dx = B(Lei , L

ej), fei =

Z xb

xafLei dx

(2.2.33)Note that the interpolation property (2.2.22) is used to write

nXj=1

Lej(xei )Q

ej = Q

ei (2.2.34)

In matrix notation, these algebraic equations can be written as

[Ke]ue = fe+ Qe (2.2.35)


The matrix [Ke] is called the coefficient matrix, or stiffness matrix in structuralmechanics applications. The column vector fe is the source vector, or forcevector in structural mechanics problems.Equation (2.2.35) has 2n unknowns (ue1, u

e2, . . . , u

en) and (Q

e1, Q

e2, . . . , Q

en),

called primary and secondary element nodal degrees of freedom; hence, itcannot be solved without having an additional n conditions. Some of theseequations are provided by the boundary conditions of the problem and theremaining by the balance of the secondary variables Qei at nodes commonto elements. The balance of equations can be implemented by putting theelements together (i.e. assembling the element equations) to form the totalsystem. Upon assembly and imposition of boundary conditions, we shallobtain exactly the same number of algebraic equations as the total number ofunknown primary (uei ) and secondary (Q

ei ) degrees of freedom.

The coefficient matrix [Ke], which is symmetric, and source vector fecan be evaluated for a given interpolation and data (a, c, and f). When a, c,and f are functions of x, it may be necessary to evaluate [Ke] and fe usingnumerical integration. We will discuss the numerical integration concepts inthe sequel. Here we give the exact values of [Ke] and fe for linear as wellas quadratic interpolations for element-wise constant values of a, c, and f .Suppose that ae, ce, and fe denote the element-wise constant values of a(x),c(x), and f(x). Then the following matrices can be derived by evaluating theintegrals exactly.

Linear element

L1(x) = 1− x

he, L2(x) =

x

heµaehe

∙1 −1

−1 1

¸+cehe6

∙2 11 2

¸¶½ue1ue2

¾=fehe2

½11

¾+

½Qe1Qe2

¾(2.2.36)

Quadratic element

L1(x) =

µ1− 2x

he

¶µ1− x

he

¶, L2(x) =

4x

he

µ1− x

he

¶, L3(x) = − x

he

µ1− 2x

he

¶

Ãae3he

⎡⎣ 7 −8 1−8 16 −81 −8 7

⎤⎦+ cehe30

⎡⎣ 4 2 −12 16 2

−1 2 4

⎤⎦!⎧⎨⎩ue1ue2ue3

⎫⎬⎭=fehe6

⎧⎨⎩141

⎫⎬⎭+⎧⎨⎩Qe1Qe2Qe3

⎫⎬⎭ (2.2.37)

Note that the contribution of uniform source to the nodes in a quadraticelement is non-uniform, that is, fei 6= fehe/3.


2.3 Two-Dimensional Problems

2.3.1 Governing Differential Equation

Consider the problem of Þnding u(x, y) such that the following partialdifferential equation is satisÞed

−∙∂

∂x

µaxx

∂u

∂x

¶+∂

∂y

µayy∂u

∂y

¶¸= f(x, y) in Ω (2.3.1)

where Ω is a two-dimensional domain with boundary Γ. Here axx and ayy arematerial coefficients in the x and y directions, respectively, and f(x, y) is theknown source. For example, in a heat transfer problem, u denotes temperatureT , axx and ayy denote the conductivities, kxx and kyy, and f is the internalheat generation. For an isotropic medium, we set kxx = kyy = k. Similarly, fora ground water ßow problem u denotes the water head (i.e. velocity potential),axx and ayy are the permeabilities in the x and y directions, respectively, andf(x, y) is distributed water source. Equation (2.3.1) also arises in other Þeldsof science and engineering, and some of them are listed in Table 2.3.1.Equation (2.3.1) must be solved in conjunction with speciÞed boundary

conditions of the problem. The following two types of boundary conditionsare assumed:

u = u(s) on Γu (2.3.2)

qn =

µaxx

∂u

∂xnx + ayy

∂u

∂yny

¶+ qc = q(s) on Γq (2.3.3)

where Γu and Γq are disjoint portions of the boundary Γ such that Γ = Γu∪Γq,qc refers to the convective component of ßux (e.g. in heat transfer problems)

qc = hc(u− uc) (2.3.4)

and (nx, ny) denote the direction cosines of the unit normal vector on theboundary. In Eq. (2.3.4), hc denotes the convective heat transfer coefficient.The radiative heat transfer boundary condition (which is a nonlinear functionof u) is not considered here. However, radiation boundary condition will beconsidered in the nonlinear analysis.

2.3.2 Finite Element Approximation

In the Þnite element method, the domain Ω = Ω ∪ Γ is divided into a setof subdomains Ωe = Ωe ∪ Γe, called Þnite elements (see Figure 2.3.1). Anygeometric shape for which the approximation functions can be derived uniquelyqualiÞes as an element. We shall discuss simple geometric shapes and ordersof approximation shortly. To keep the formulative steps very general (i.e. notconÞne the formulation to a speciÞc geometric shape or order of the element),we have denoted the domain of a typical element by Ωe and its boundary by Γe.


The element Ωe can be a triangle or quadrilateral in shape, and the degree ofinterpolation over it can be linear, quadratic, and so on. The non-overlappingsum of all elements Ωe is denoted by Ωh, and it is called the Þnite elementmesh of the domain Ω. In general, Ωh may not equal Ω when the boundary Γis curved. Of course, for polygonal domains, the Þnite element mesh exactlyrepresents the actual domain.

Table 2.3.1 List of Þelds in which the model equation (2.3.1) arises, withmeaning of various parameters and variables (see the bottom ofthe table for the meaning of some parameters*).

Field Variable Data Data Data Variableu axx ayy f qn

Heat Tempe- Thermal Thermal Heat Heattransfer rature conductance conductance generation ßux

T kxx kyy f qn

Flow Fluid- Permea- Permea- InÞl- Fluxthrough head bility bility trationporousmedium φ µxx µyy f qn

Torsion of Warping 1 1 0 ∂φ∂n

cylindrical functionmembers φ

Torsion of Stress 1 1 2Gθ ∂φ∂n

cylindrical functionmembers ψ

Deßection Displa- Tension Tension Transverse qnof cement forcemembranes u T T f ∂u

∂n

Flows of Velocity 1 1 0 qninviscid potentialßows φ ∂φ

∂n

Flows of Stream 1 1 0 qninviscid functionßows ψ ∂ψ

∂n

Electro- Electrical Dielectric Dielectric Charge Electricstatics potential constant constant density ßux

φ ε ε ρ ∂ψ∂n

* k = thermal conductance; β = convective Þlm conductance; T∞ = ambient temperatureof the surrounding ßuid medium; G = shear modulus; θ = angle of twist.

x

y

Γ

Γ

Ω e

e


Figure 2.3.1 Finite element discretization of a domain.

Suppose that the dependent unknown u is approximated over a typicalÞnite element Ωe by the expression

u(x, y) ≈ ueh(x, y) =nXj=1

uejLej(x, y) (2.3.5)

where ueh(x, y) represents an approximation of u(x, y) over the element Ωe,

parameters uej denote the values of the function ueh(x, y) at a selected number

of points, called element nodes, in the element Ωe, and Lej are the Lagrangeinterpolation functions associated with the element. As we shall see shortly,the interpolation functions depend not only on the number of nodes in theelement, but also on the shape of the element. The shape of the element mustbe such that its geometry is uniquely deÞned by a set of nodes. A triangle(n = 3) is the simplest two-dimensional geometric shape in two dimensions.

2.3.3 Weak Formulation

The n parameters (or nodal values) uej in Eq. (2.3.5) must be determinedsuch that the approximate solution ue(x, y) satisÞes the governing Eq. (2.3.1)and boundary conditions of the problem. As in the case of a variationaland weighted-residual method, we seek to satisfy the governing differentialequation in a weighted-integral sense, as described in Section 2.2. The type ofÞnite element model depends on the weighted-integral form used to generatethe algebraic equations. Thus, if one uses the weak form, the resulting modelwill be different from those obtained with a weighted-residual statement inwhich the weight function can be any one of several choices. In the remainderof this chapter, we shall be primarily concerned with the weak form Þniteelement models.


The weak form of a differential equation is a weighted-integral statementthat is equivalent to both the governing differential equation as well as theassociated natural boundary conditions. We shall develop the weak formof Eqs. (2.3.1) and (2.3.3) over the typical element Ωe using the three-stepprocedure. The Þrst step is to take all non-zero expressions in Eq. (2.3.1) toone side of the equality, multiply the resulting equation with a weight functionw, and integrate the equation over the element domain Ωe:

0 =

ZΩew

∙− ∂

∂x

µaxx

∂ueh∂x

¶− ∂

∂y

µayy∂ueh∂y

¶− f(x, y)

¸dx dy (2.3.6)

The expression in the square brackets of the above equation represents aresidual of the approximation of the differential equation (2.3.1), becauseueh(x, y) is only an approximation of u(x, y). For n independent choices ofw, we obtain a set of n linearly independent algebraic equations.In the second step, we distribute the differentiation among u and w equally,

so that both u and w are required to be differentiable only once with respectto x and y. To achieve this we use the component form of the gradient (ordivergence) theorem,Z

Ωe

∂

∂x(wF1) dx dy =

IΓe(wF1)nx ds (2.3.7a)

ZΩe

∂

∂y(wF2) dx dy =

IΓe(wF2)ny ds (2.3.7b)

where nx and ny are the components (i.e. the direction cosines) of the unitnormal vector

n = nxex + nyey = cosα ex + sinα ey (2.3.8)

on the boundary Γe, and ds is the arc length of an inÞnitesimal line elementalong the boundary. We obtain

0 =

ZΩe

∙axx

∂w

∂x

∂u

∂x+ ayy

∂w

∂y

∂u

∂y− wf

¸dx dy−

IΓew

∙axx

∂u

∂xnx + ayy

∂u

∂yny

¸ds

(2.3.9)From an inspection of the boundary term in Eq. (2.3.9), we note that u is

the primary variable, and speciÞcation of u constitutes the essential boundarycondition. The coefficient of the weight function in the boundary expression,namely

qn = axx∂u

∂xnx + ayy

∂u

∂yny (2.3.10)

is the secondary variable. Its speciÞcation constitutes the natural boundarycondition. By deÞnition qn is positive outward from the surface as we movecounterclockwise along the boundary Γe. The secondary variable qn denotesthe ßux normal to the boundary of the element.


The third and last step of the formulation is to use the deÞnition (2.3.10)in Eq. (2.3.9) and write it as

0 =

ZΩe

µaxx

∂w

∂x

∂u

∂x+ ayy

∂w

∂y

∂u

∂y− wf

¶dx dy −

IΓew qn ds (2.3.11)

or

0 = B(w, u)− `(w) (2.3.12)

where the bilinear form B(·, ·) and linear form `(·) are deÞned by

B(w, u) =

ZΩe

µaxx

∂w

∂x

∂u

∂x+ ayy

∂w

∂y

∂u

∂y

¶dx dy (2.3.13a)

`(w) =

ZΩew f dx dy +

IΓewqn ds (2.3.13b)

Note that the bilinear form is symmetric in its arguments (w, u)

B(w, u) = B(u,w)

and `(w) is linear in w. Therefore, it is possible to construct the associatedquadratic functional from the formula

I(u) =1

2B(u, u)− `(u) (2.3.14)


The weak form in Eq. (2.3.11) requires that the approximation chosen for ushould be at least linear in both x and y so that there are no terms in (2.3.11)that become identically zero. Suppose that u is approximated over a typicalÞnite element Ωe by the expression of the form (2.3.5). Substituting the Þniteelement approximation (2.3.5) for u into the weak form (2.3.11), we obtain

0 =nXj=1

(ZΩe

"∂w

∂x

Ãaxx

∂Lej∂x

!+∂w

∂y

Ãayy

∂Lej∂y

!− w f

#dx dy

)uej−

IΓew qn ds

(2.3.15)This equation must hold for any weight function w. Since we need nindependent algebraic equations to solve for the n unknowns, ue1, u

e2, ..., u

en,

we choose n independent functions for w: w = Le1, Le2, ..., L

en. This particular

choice of weight functions is a natural one when the weight function is viewedas a virtual variation of the dependent unknown (i.e. w = δue =

Pni=1 δu

ei L

ei ).


For each choice of w we obtain an algebraic relation among (ue1, ue2, ..., u

en).

The ith algebraic equation is obtained by substituting w = Lei into Eq.(2.3.15):

nXj=1

Keiju

ej = Q

ei + q

ei (2.3.16)

where the coefficients Keij , Q

ei , and q

ei are deÞned by

Keij =

ZΩe

Ãaxx

∂Lei∂x

∂Lej∂x

+ ayy∂Lei∂y

∂Lej∂y

!dx dy (2.3.17a)

Qei =

ZΩef Lei dx dy, qei =

IΓeqn L

ei ds (2.3.17b)

We note that Keij = Ke

ji (i.e. [Ke] is symmetric). The symmetry of thecoefficient matrix is due to the symmetry of the bilinear form, which in turnis due to the weak form development. In matrix notation, Eq. (2.3.16) takesthe form

[Ke]ue = Qe+ qe ≡ F e (2.3.18)

This completes the Þnite element model development. Before we discussassembly of elements, it is informative to determine the interpolation functionsLei for certain basic two-dimensional Þnite elements.

2.3.5 Interpolation Functions

The Þnite element approximation ue(x, y) of u(x, y) over an element Ωe

must satisfy the following conditions in order for the approximate solutionto converge to the true solution:

1. ue(x, y) must be continuous as required in the weak form of the problem(i.e. all terms in the weak form are represented as non-zero values).

2. The polynomials used to represent ue(x, y) must be complete (i.e. allterms, beginning with a constant term up to the highest order used inthe polynomial should be included in the expression of ue(x, y)).

3. All terms in the polynomial should be linearly independent.

The number of linearly independent terms in the representation of ue dictatesthe shape and number of degrees of freedom of the element. Here we reviewthe interpolation functions of linear triangular and rectangular elements.An examination of the variational form (2.3.11) and the Þnite element

matrices in Eq. (2.3.17a) shows that the Lei should be at least linear functionsof x and y. The polynomial

ue(x, y) = ce1 + ce2x+ c

e3y (2.3.19)


is the lowest-order polynomial that meets the requirements. It contains threelinearly independent terms, and it is linear in both x and y. The polynomial iscomplete because the lower-order term, namely, the constant term, is included.To write the three constants (ce1, c

e2, c

e3) in terms of the nodal values of u

e, wemust identify three points or nodes in the element Ωe. The three nodes mustbe such that they uniquely deÞne the geometry of the element and allow theimposition of inter-element continuity of the variable ue(x, y). Obviously, thegeometric shape deÞned by three points in a two-dimensional domain is atriangle. Thus the polynomial in Eq. (2.3.19) is associated with a triangularelement and the three nodes are identiÞed as the vertices of the triangle.On the other hand, the polynomial

ue(x, y) = ce1 + ce2x+ c

e3y + c

e4xy (2.3.20)

contains four linearly independent terms, and is linear in x and y, with abilinear term in x and y. This polynomial requires an element with fournodes. It is a rectangle with nodes at the four corners of the rectangle.The interpolation functions for linear triangular and rectangular elements

are given below. Higher-order two-dimensional elements (i.e. element withhigher-order interpolation polynomials) will be discussed in Section 2.4.

Linear triangular element

The linear interpolation functions for the three-node triangle [see Figure2.3.2(a)] are (see Reddy [1, pp. 304307])

Lei (x, y) =1

2Ae(αei + β

ei x+ γ

ei y), (i = 1, 2, 3) (2.3.21)

where Ae is the area of the triangle, and αei , β

ei , and γ

ei are geometric constants

known in terms of the nodal coordinates (xi, yi):

αei = xjyk − xkyj ; βei = yj − yk; γei = −(xj − xk) (2.3.22)

for i 6= j 6= k, and i, j, and k permute in a natural order. Note that (x, y)are the global coordinates used in the governing equation (2.3.1) over thedomain Ω. The interpolation functions Lei (i = 1, 2, . . . , n) satisfy the followinginterpolation properties:

(i) Lei (xj , yj) = δij , (i, j = 1, 2, 3); (ii)3Xi=1

Lei (x, y) = 1 (2.3.23)

and they are called the Lagrange interpolation functions. Note that use of thelinear interpolation functions Lei of a triangle will result in the approximationof the curved surface u(x, y) by a planar function ueh(x, y) =

P3i=1 u

eiLei (x, y).

y

x

1

2

3

ΩΓe

e

y

x

b

1

4 3

2a

x_

y_

Ω Γe e


Figure 2.3.2 The linear (a) triangular and (b) rectangular Þnite elements.

The integrals in the deÞnition of Keij and Q

ei can be evaluated for given

data: axx, ayy, and f . For example, for element-wise constant values of thedata, that is, axx = a

exx, ayy = a

eyy, and f = f

e, we have (see Reddy [1, pp.311313]) the following results:

Keij =

1

4Ae(aexxβ

ei βej + a

eyyγ

ei γej ); Qei =

feAe3

(2.3.24)

where Ae is the area of the triangular element, and βei and γ

ei are known in

terms of the global nodal coordinates of the element nodes, as given in Eq.(2.3.22). For a right-angled triangular element with base a and height b, andnode 1 at the right angle (nodes are numbered counterclockwise), [Ke] takesthe form (see Reddy [1, p. 387])

[Ke] =aexx2

⎡⎣ α −α 0−α α 00 0 0

⎤⎦+ aeyy2

⎡⎣ β 0 −β0 0 0

−β 0 β

⎤⎦ (2.3.25)

where α = b/a and β = a/b. Of course, for cases in which the conductivitiesare functions of (x, y), numerical integration can be used to evaluate thecoefficients (see Section 2.5.3).The evaluation of boundary integrals of the type

qei =

IΓeqenL

ei (s) ds (2.3.26)

where qen is a known function of the distance s along the boundary Γe, involves

evaluation of line integrals. It is necessary to compute such integrals only when


Γe, or a portion of it, coincides with the boundary Γq of the total domain Ωon which the ßux is speciÞed. On portions of Γe that are in the interior ofthe domain Ω, qen on side (i, j) of element Ω

e cancels with qfn on side (p, q) ofelement Ωf when sides (i, j) of element Ωe and (p, q) of element Ωf are thesame (i.e. at the interface of elements Ωe and Ωf ). This can be viewed as thebalance of the internal ßux. When Γe falls on the boundary Γu of the domainΩ, qen is not known there and can be determined in the post-computation.Note that the primary variable u is speciÞed on Γu. For additional details, seeReddy [1, pp. 313318].

Linear rectangular element

For a linear rectangular element, we have

u(x, y) =4Xi=1

ueiLei (x, y) (2.3.27)

where Li are the Lagrange interpolation functions expressed in terms of theelement coordinates (x, y) (see Reddy [1, pp. 308311])

Le1 = (1−x

a)(1− y

b), Le2 =

x

a(1− y

b)

Le3 =x

a

y

b, Le4 = (1−

x

a)y

b(2.3.28)

and (x, y) denote the local coordinates with origin located at node 1 ofthe element, and (a, b) denote the horizontal and vertical dimensions of therectangle [see Figure 2.3.2(b)].The integrals in the deÞnition of Ke

ij and Qei can be easily evaluated over a

rectangular element of sides a and b. For example, for element-wise constantvalues of the data, that is, axx = aexx, ayy = aeyy, and f = fe, we have (seeReddy [1, p. 313; p. 387]) the following results:

[Ke] = aexx[S11] + aeyy[S

22], Qei =feab

4(2.3.29)

where

[S11] =1

6

⎡⎢⎢⎣2α −2α −α α

−2α 2α α −α−α α 2α −2αα −α −2α 2α

⎤⎥⎥⎦ (2.3.30a)

[S22] =1

6

⎡⎢⎢⎣2β β −β −2ββ 2β −2β −β

−β −2β 2β β−2β −β β 2β

⎤⎥⎥⎦ (2.3.30b)

1

2

3

14

5

2

1

2

32

1

3

4

Global node numbers

Element (local) node numbers


and α = b/a and β = a/b. Again, for cases in which the conductivities arefunctions of (x, y), numerical integration is used to evaluate the coefficients,as discussed in Section 2.5.4. When the element is non-rectangular, that is,a quadrilateral, we use coordinate transformations to represent the integralsover a square geometry and then use numerical integration to evaluate them.

2.3.6 Assembly of Elements

The assembly of Þnite elements to obtain the equations of the entire domainis based on the following two rules:

1. Continuity of the primary variable (i.e. temperature)

2. Balance of secondary variables (i.e. heat ßux)

We illustrate the assembly procedure by considering a Þnite element meshconsisting of a triangular element and a quadrilateral element (see Figure2.3.3).Let K1

ij (i, j = 1, 2, 3) denote the coefficient matrix corresponding to the

triangular element, and let K2ij (i, j = 1, 2, 3, 4) denote the coefficient matrix

corresponding to the quadrilateral element. The nodes of the Þnite elementmesh are called global nodes. From the mesh shown in Figure 2.3.3, it isclear that the following correspondence between global and element nodesexists: nodes 1, 2, and 3 of element 1 correspond to global nodes 1, 2, and3, respectively. Nodes 1, 2, 3, and 4 of element 2 correspond to global nodes2, 4, 5, and 3, respectively. Hence, the correspondence between the local andglobal nodal values of temperature is

u11 = u1, u12 = u

21 = u2, u

13 = u

24 = u3, u

22 = u4, u

23 = u5 (2.3.31)

which amounts to imposing the continuity of the primary variables at thenodes common to elements 1 and 2. Note that the continuity of the primaryvariables at the inter-element nodes guarantees the continuity of the primaryvariable along the entire inter-element boundary.

Figure 2.3.3 Globallocal correspondence of nodes for assembly of elements.


Next, we consider the balance of secondary variables at the interelementboundaries. At the interface between the two elements, the ßux from thetwo elements should be equal in magnitude and opposite in sign. For thetwo elements shown in Figure 2.3.3, the interface is along the side connectingglobal nodes 2 and 3. Hence, the internal ßux q1n on side 23 of element 1should balance the ßux q2n on side 41 of element 2 (recall the sign conventionon qen):

(q1n)2−3 = (q2n)4−1 or (q1n)2−3 = (−q2n)1−4 (2.3.32)

In the Þnite element method, the above relation is imposed in a weighted-integral sense:Z

h123

q1nL12 ds = −

Zh214

q2nL21 ds,

Zh123

q1nL13 ds = −

Zh214

q2nL24 ds (2.3.33)

where hepq denotes length of the side connecting node p to node q of elementΩe.Now we are ready to assemble the element equations for the two-element

mesh. The element equations of the two elements are written separately Þrst.For the triangular element, the element equations are of the form

K111u

11 +K

112u

12 +K

113u

13 = Q

11 + q

11

K121u

11 +K

122u

12 +K

123u

13 = Q

12 + q

12 (2.3.34a)

K131u

11 +K

132u

12 +K

133u

13 = Q

13 + q

13

For the rectangular element the element equations are given by

K211u

21 +K

212u

22 +K

213u

23 +K

214u

24 = Q

21 + q

21

K221u

21 +K

222u

22 +K

223u

23 +K

224u

24 = Q

22 + q

22

K231u

21 +K

232u

22 +K

233u

23 +K

234u

24 = Q

23 + q

23 (2.3.34b)

K241u

21 +K

242u

22 +K

243u

23 +K

244u

24 = Q

24 + q

24

In order to impose the balance condition in (2.3.32), it is necessary to add thesecond equation of element 1 to the Þrst equation of element 2, and also addthe third equation of element 1 to the fourth equation of element 2:

(K121u

11 +K

122u

12 +K

123u

13) + (K

211u

21 +K

212u

22 +K

213u

23 +K

214u

24)

= (Q12 + q12) + (Q

21 + q

21)

(K131u

11 +K

132u

12 +K

133u

13) + (K

241u

21 +K

242u

22 +K

243u

23 +K

244u

24)

= (Q13 + q13) + (Q

24 + q

24)


Using the local-global nodal variable correspondence in Eq. (2.3.1), we canrewrite the above equations as

K121u1 + (K

122 +K

211)u2 + (K

123 +K

214)u3 +K

212u4 +K

213u5

= Q12 +Q21 + (q

12 + q

21)

K131u1 + (K

132 +K

241)u2 + (K

133 +K

244)u3 +K

242u4 +K

243u5

= Q13 +Q24 + (q

13 + q

24)

Now we can impose the conditions in Eq. (2.3.33) by setting appropriateportions of the expressions in parenthesis on the right-hand side of theabove equations to zero (or a speciÞed non-zero value). In general, whenseveral elements are connected, the assembly of the elements is carried out byputting element coefficients Ke

ij , Qei , and q

ei into proper locations of the global

coefficient matrix and right-hand column vectors. This is done by means ofthe connectivity relations, that is, correspondence of the local node numberto the global node number.The assembly procedure described above can be used to assemble elements

of any shape and type. The procedure can be implemented in a computer withthe help of the local-global nodal correspondence.For heat conduction problems that involve convection heat transfer at

the boundary, that is, when heat is transferred from one medium to thesurrounding medium (often, a ßuid) by convection, the Þnite element modeldeveloped earlier requires some modiÞcation. For a convection boundary, thenatural boundary condition is a balance of energy transfer across the boundarydue to conduction and/or convection (i.e. Newtons law of cooling):

(axx∂u

∂xnx + ayy

∂u

∂yny) + hc (u− uc) = qn (2.3.35)

where hc is the convective conductance (or the convective heat transfercoefficient), uc is the (ambient) temperature of the surrounding ßuid medium,and qn is the speciÞed heat ßux. The Þrst term accounts for heat transfer byconduction, the second by convection, and the third accounts for the speciÞedheat ßux, if any. It is the presence of the term hc(u− uc) that requires somemodiÞcation of the weak form in Eq. (2.3.11). To include the convectiveboundary condition (2.3.35), the boundary integral in Eq. (2.3.9) should bemodiÞed. Instead of replacing the coefficient of w in the boundary integralwith qn, we use Eq. (2.3.35):

0 =

ZΩe

∙axx

∂w

∂x

∂u

∂x+ ayy

∂w

∂y

∂u

∂y−wf

¸dx dy −

IΓew

∙axx

∂u

∂xnx + ayy

∂u

∂yny

¸ds

=

ZΩe

µaxx

∂w

∂x

∂u

∂x+ ayy

∂w

∂y

∂u

∂y− wf

¶dx dy −

IΓew[qn − hc(u− uc)] ds

(2.3.36)


or0 = B(w, u)− `(w) (2.3.37)

where w is the weight function, and B(·, ·) and `(·) are the bilinear and linearforms

B(w, u) =

ZΩe

µaxx

∂w

∂x

∂u

∂x+ ayy

∂w

∂y

∂u

∂y

¶dx dy +

IΓehcwu ds (2.3.38a)

`(w) =

ZΩefw dxdy +

IΓehcucw ds+

IΓeqnw ds (2.3.38b)

Note that the unknown surface temperature in the convective boundarycondition has been made part of the bilinear form B(·, ·) while all the knownquantities remain part of the linear form `(·). The Þnite element model of Eq.(2.3.36) is (see Reddy [1], pp. 341346) given by

[Ke]ue = F e (2.3.39)

where

Keij =

ZΩe

µaxx

∂Li∂x

∂Lj∂x

+ ayy∂Li∂y

∂Lj∂y

¶dx dy +

IΓehcLiLj ds (2.3.40a)

Fi =

ZΩefLi dx dy +

IΓehcucLi ds+

IΓeqnLi ds (2.3.40b)

The Þnite element model (2.3.39) is valid for both conductive andconvective heat transfer boundary conditions. For problems with no convectiveboundary conditions, the convective contributions to the element coefficientsare omitted. Indeed, these contributions have to be included only for thoseelements whose sides fall on the boundary with speciÞed convection heattransfer. The contribution due to convective boundaries to the elementcoefficient matrix and source vector can be computed by evaluating lineintegrals, as discussed in Reddy [1], pp. 342345.

2.4 Library of Two-Dimensional Finite Elements

2.4.1 Introduction

The objective of this section is to present a library of two-dimensionaltriangular and rectangular elements of the Lagrange family, that is, elementsover which only the function not its derivatives are interpolated. Oncewe have elements of different shapes and order at our disposal, we canchoose appropriate elements and associated interpolation functions for a givenproblem. The interpolation functions are developed here for regularly shapedelements, called master elements. These elements can be used for numerical

3

5

1

P

4

2

6

L 1 = 0

L 1 = 0.5

L 1 = 1

L 12

3

1

sP

b

hA 1

A 2

A 3


evaluation of integrals deÞned on irregularly shaped elements. This requires atransformation of the geometry from the actual element shape to its associatedmaster element. We will discuss the numerical evaluation of integrals inSection 2.6.

2.4.2 Triangular Elements

The three-node triangular element was developed in Section 2.3.5. Higher-order triangular elements (i.e. triangular elements with interpolation functionsof higher degree) can be systematically developed with the help of the so-calledarea coordinates. For triangular elements, it is possible to construct three non-dimensional coordinates Li (i = 1, 2, 3), which vary in a direction normal tothe sides directly opposite each node (see Figure 2.4.1). The coordinates aredeÞned such that

Li =AiA, A =

3Xi=1

Ai (2.4.1)

where Ai is the area of the triangle formed by nodes j and k and an arbitrarypoint P in the element, and A is the total area of the element. For example, A1is the area of the shaded triangle which is formed by nodes 2 and 3 and pointP . The point P is at a perpendicular distance of s from the side connectingnodes 2 and 3. We have A1 = bs/2 and A = bh/2. Hence

L1 =A1A=s

h(2.4.2)

Clearly, L1 is zero on side 23 (hence, zero at nodes 2 and 3) and has a value

of unity at node 1. Thus, L1 is the interpolation function associated with node1. Similarly, L2 and L3 are the interpolation functions associated with nodes2 and 3, respectively. In summary, we have

Li = Li (2.4.3)

Figure 2.4.1 DeÞnition of area coordinates Li for triangular elements.


The area coordinates Li can be used to construct interpolation functionsfor higher-order triangular elements. For example, a higher-order element withk nodes per side (equally spaced on each side) has a total of n nodes

n =k−1Xi=0

(k − i) = k + (k − 1) + · · ·+ 1 = k

2(k + 1) (2.4.4)

and its degree is equal to k−1. The explicit forms of the interpolation functionsfor the linear and quadratic elements are recorded below:

Le =⎧⎨⎩L1L2L3

⎫⎬⎭ ; Le =

⎧⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎩

L1(2L1 − 1)L2(2L2 − 1)L3(2L3 − 1)4L1 L24L2 L34L3 L1

⎫⎪⎪⎪⎪⎪⎪⎪⎬⎪⎪⎪⎪⎪⎪⎪⎭(2.4.5)

Note that the order of the interpolation functions in the above arrayscorresponds to the node numbers shown in Figure 2.4.2(a). Thus, the Þrstthree rows of the vectors in Eq. (2.4.5) correspond to the Þrst three nodes ofthe linear and quadratic elements, which correspond to the three vertices ofthe triangular element. The last three rows of the second vector in Eq. (2.4.5)associated with the quadratic element correspond to the mid-side nodes of thetriangular element. A similar node-numbering scheme is used for rectangularelements, which are discussed next.

2.4.3 Rectangular Elements

The Lagrange interpolation functions associated with rectangular elementscan be obtained from the tensor product of corresponding one-dimensionalLagrange interpolation functions. We take a local coordinate system (ξ, η)such that −1 ≤ (ξ, η) ≤ 1. This choice of local coordinate system is dictatedby the Gauss quadrature rule used in the numerical evaluation of integralsover the element (see Section 2.5).The linear and quadratic interpolation functions are given in Eqs. (2.4.6)

and (2.4.7), respectively (see Figure 2.4.2(b) for the node numbers).

Le = 1

4

⎧⎪⎪⎨⎪⎪⎩(1− ξ)(1− η)(1 + ξ)(1− η)(1 + ξ)(1 + η)(1− ξ)(1 + η)

⎫⎪⎪⎬⎪⎪⎭ (2.4.6)

1

2

3

14

3

2

56

(a)

(b)

ξ ξ

η η

1 2

34

1 2

34

5

6

7

8 9


Figure 2.4.2 Linear and quadratic (a) triangular and (b) rectangularelements.

Le = 1

4

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

(1− ξ)(1− η)(−ξ − η − 1) + (1− ξ2)(1− η2)(1 + ξ)(1− η)(ξ − η − 1) + (1− ξ2)(1− η2)(1 + ξ)(1 + η)(ξ + η − 1) + (1− ξ2)(1− η2)(1− ξ)(1 + η)(−ξ + η − 1) + (1− ξ2)(1− η2)

2(1− ξ2)(1− η)− (1− ξ2)(1− η2)2(1 + ξ)(1− η2)− (1− ξ2)(1− η2)2(1− ξ2)(1 + η)− (1− ξ2)(1− η2)2(1− ξ)(1− η2)− (1− ξ2)(1− η2)

4(1− ξ2)(1− η2)

⎫⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎬⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎭(2.4.7)

The serendipity elements are those rectangular elements that have nointerior nodes. These elements have fewer nodes compared with the higher-order Lagrange elements. The interpolation functions of the serendipityelements are not complete, and they cannot be obtained using tensor productsof one-dimensional Lagrange interpolation functions. Instead, an alternativeprocedure must be employed, as discussed in Reddy [1]. The interpolationfunctions for the two-dimensional quadratic serendipity element are given by


(see Figure 2.4.3)

Le = 1

4

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

(1− ξ)(1− η)(−ξ − η − 1)(1 + ξ)(1− η)(ξ − η − 1)(1 + ξ)(1 + η)(ξ + η − 1)(1− ξ)(1 + η)(−ξ + η − 1)

2(1− ξ2)(1− η)2(1 + ξ)(1− η2)2(1− ξ2)(1 + η)2(1− ξ)(1− η2)

⎫⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎬⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎭(2.4.8)

Figure 2.4.3 Quadratic rectangular serendipity element.

2.5 Numerical Integration

2.5.1 Preliminary Comments

An accurate representation of irregular domains (i.e. domains with curvedboundaries) can be accomplished by the use of reÞned meshes and/orirregularly shaped curvilinear elements. For example, a non-rectangularregion cannot be represented using rectangular elements; however, it canbe represented by quadrilateral elements. Since the interpolation functionsare easily derivable for a rectangular element and it is easier to evaluateintegrals over rectangular geometries, we transform the Þnite element integralstatements deÞned over quadrilaterals to a rectangle. The transformationresults in complicated expressions for the integrands in terms of thecoordinates used for the rectangular element. Therefore, numerical integrationis used to evaluate such complicated integrals. The numerical integrationschemes, such as the GaussLegendre numerical integration scheme, requirethe integral to be evaluated on a speciÞc domain or with respect to a speciÞccoordinate system.

ξ

η

1 2

34

5

6

7

8

ξ

η

ξ=1

η=1

x

y

eΩ

1Ω2Ω

3ΩΩ


2.5.2 Coordinate Transformations

Gauss quadrature requires the integral to be expressed over a square regionΩ of dimension 2× 2 with respect to the coordinate system, (ξ, η) to be suchthat −1 ≤ (ξ, η) ≤ 1. The transformation of the geometry and the variablecoefficients of the differential equation from the problem coordinates (x, y) tothe local coordinates (ξ, η) results in algebraically complex expressions, andthey preclude analytical (i.e. exact) evaluation of the integrals. Thus, thetransformation of a given integral expression, deÞned over element Ωe, to oneon the domain Ω facilitates the numerical integration. Each element of theÞnite element mesh is transformed to Ω, only for the purpose of numericallyevaluating the integrals (see Figure 2.5.1). The element Ω is called a masterelement. For example, every quadrilateral element can be transformed to asquare element with a side of length 2 and −1 ≤ (ξ, η) ≤ 1 that facilitatesthe use of GaussLegendre quadrature to evaluate integrals deÞned over thequadrilateral element.The transformation between a typical element Ωe in the mesh and the

master element Ω [or equivalently, between (x, y) and (ξ, η)] is accomplishedby a coordinate transformation of the form

x =mXj=1

xejφej(ξ, η) , y =

mXj=1

yejφej(ξ, η) (2.5.1)

where φj denote the Þnite element interpolation functions of the master

element Ω. The coordinates in the master element are chosen to be thenatural coordinates (ξ, η) such that −1 ≤ (ξ, η) ≤ 1. This choice is dictatedby the limits of integration in the Gauss quadrature rule used to evaluatethe integrals. For this case, the φej denote the interpolation functions of thefour-node rectangular element shown in Figure 2.4.2(b) (i.e. m = 4).

Figure 2.5.1 Transformation of quadrilateral elements to the masterrectangular element for numerical evaluation of integrals.


The transformation (2.5.1) maps, for example, the line ξ = 1 in Ω to the linedeÞned parametrically by x = x(1, η) and y = y(1, η) in the xy-plane. In other

words, the master element Ω is transformed, under the linear transformation,into a quadrilateral element (i.e. a four-sided element whose sides are notparallel) in the xy-plane. Conversely, every quadrilateral element of a mesh

can be transformed to the same four-noded square (master) element Ω in the(ξ, η)-plane.In general, the dependent variable(s) of the problem are approximated by

expressions of the form

u(x, y) =nXj=1

uejLej(x, y) (2.5.2)

The interpolation functions Lej used for the approximation of the dependentvariable, in general, are different from φej used in the approximation of thegeometry. Depending on the relative degree of approximations used for thegeometry [see Eq. (2.5.1)] and the dependent variable(s) [see Eq. (2.5.2)], theÞnite element formulations are classiÞed into three categories:

1. Superparametric (m > n). The approximation used for the geometry ishigher order than that used for the dependent variable.

2. Isoparametric (m = n). Equal degree of approximation is used for bothgeometry and dependent variables.

3. Subparametric (m < n). Higher-order approximation of the dependentvariable is used.

It should be noted that the transformation of a quadrilateral element ofa mesh to the master element Ω is solely for the purpose of numericallyevaluating the integrals (see Figure 2.5.1). No transformation of the physicaldomain or elements is involved in the Þnite element analysis. The resultingalgebraic equations of the Þnite element formulation are always in terms of thenodal values of the physical domain. Different elements of the Þnite elementmesh can be generated from the same master element by assigning appropriateglobal coordinates to each of the elements. Master elements of a differentorder deÞne different transformations and hence different collections of Þniteelements within the mesh. For example, a quadratic rectangular masterelement can be used to generate a mesh of quadratic curvilinear quadrilateralelements. The transformations of a master element should be such that nospurious gaps exist between elements, and no element overlaps occur. Forexample, consider the element coefficients

Keij =

ZΩe

"axx(x, y)

∂Lei∂x

∂Lej∂x

+ ayy(x, y)∂Lei∂y

∂Lej∂y

#dx dy (2.5.3)


The integrand (i.e. the expression in the square brackets under the integral) isa function of the global coordinates x and y. We must rewrite it in terms of ξand η using the transformation (2.5.1). Note that the integrand contains notonly functions but also derivatives with respect to the global coordinates (x, y).

Therefore, we must relate (∂Lei∂x ,

∂Lei∂y ) to (

∂Lei∂ξ ,

∂Lei∂η ) using the transformation

(2.5.1).The functions Lei (x, y) can be expressed in terms of the local coordinates

(ξ, η) by means of the transformation (2.5.1). Hence, by the chain rule ofpartial differentiation, we have

∂Lei∂ξ

=∂Lei∂x

∂x

∂ξ+∂Lei∂y

∂y

∂ξ;

∂Lei∂η

=∂Lei∂x

∂x

∂η+∂Lei∂y

∂y

∂η

or, in matrix notation, ( ∂Lei∂ξ∂Lei∂η

)=

"∂x∂ξ

∂y∂ξ

∂x∂η

∂y∂η

#( ∂Lei∂x∂Lei∂y

)(2.5.4)

which gives the relation between the derivatives of Lei with respect to theglobal and local coordinates. The matrix in Eq. (2.5.4) is called the Jacobianmatrix of the transformation (2.5.1):

[Je] =

"∂x∂ξ

∂y∂ξ

∂x∂η

∂y∂η

#(2.5.5)

Note from the expression given for Keij in Eq. (2.5.3) that we must relate

(∂Lei∂x ,

∂Lei∂y ) to (

∂Lei∂ξ ,

∂Lei∂η ), whereas Eq. (2.5.4) provides the inverse relations.

Therefore, Eq. (2.5.4) must be inverted. We have( ∂Lei∂x∂Lei∂y

)= [Je]−1

( ∂Lei∂ξ∂Lei∂η

)(2.5.6)

This requires that the Jacobian matrix [Je] be non-singular.Using the transformation (2.5.1), we can write

∂x

∂ξ=

mXj=1

xj∂φej∂ξ

,∂y

∂ξ=

mXj=1

yj∂φej∂ξ

(2.5.7a)

∂x

∂η=

mXj=1

xj∂φej∂η

,∂y

∂η=

mXj=1

yj∂φej∂η

(2.5.7b)

and by means of Eq. (2.5.5) one can compute the Jacobian matrix and thenits inverse. Thus, given the global coordinates (xj , yj) of element nodes and


the interpolation functions φej used for geometry, the Jacobian matrix can be

evaluated using Eq. (2.5.5). A necessary and sufficient condition for [J ]−1to exist is that the determinant J , called the Jacobian, be non-zero at everypoint (ξ, η) in Ω :

Je ≡ det[Je] = ∂x

∂ξ

∂y

∂η− ∂x∂η

∂y

∂ξ6= 0. (2.5.8)

From Eq. (2.5.8) it is clear that the functions ξ(x, y) and η(x, y) must becontinuous, differentiable, and invertible. Moreover, the transformation shouldbe algebraically simple so that the Jacobian matrix can be easily evaluated.Transformations of the form in Eq. (2.5.1) satisfy these requirements and therequirement that no spurious gaps between elements or overlapping of elementsoccur.Returning to numerical evaluation of integrals, we have from Eq. (2.5.6),( ∂Lei

∂x∂Lei∂y

)= [J ]−1


)≡ [J∗]


)(2.5.9)

where J∗ij is the element in position (i, j) of the inverse of the Jacobian matrix[Je]. The element area dA = dx dy in element Ωe is transformed to

dA = |[Je]| dξ dη (2.5.10)

in the master element Ω.Equations (2.5.7)-(2.5.10) provide the necessary relations to transform

integral expressions on any element Ωe to an associated master element Ω.For instance, consider the integral expression in Eq. (2.5.3), where axx andayy are functions of x and y. Suppose that the Þnite element Ωe can be

generated by the master element Ωe. Under the transformation (2.5.1) we canwrite

Keij =

ZΩe

"axx(x, y)

∂Lei∂x

∂Lej∂x

+ ayy(x, y)∂Lei∂y

∂Lej∂y

#dx dy

=

ZΩeFij(ξ, η) dξ dη (2.5.11)

The discussion presented above is valid for master elements of bothrectangular and triangular geometry.

2.5.3 Integration Over a Master Rectangular Element

Integrals deÞned over a rectangular master element ΩR can be numericallyevaluated using the GaussLegendre quadrature formulasZ

ΩR

F (ξ, η) dξ dη =

Z 1

−1

Z 1

−1F (ξ, η) dξ dη ≈

MXI=1

NXJ=1

F (ξI , ηJ) WIWJ (2.5.12)


whereM andN denote the number of Gauss quadrature points, (ξI , ηJ) denotethe Gauss point coordinates, and WI andWJ denote the corresponding Gaussweights as shown in Table 2.5.1 (from Table 7.2 in Reddy [1]).

Table 2.5.1 Gauss quadrature points and weights for rectangular elements.R 1−1 F (ξ) dξ =

PN

I=1 F (ξI)WI

N Points, ξI Weights, WI

1 0.0000000000 2.0000000000

2 ± 0.5773502692 1.0000000000

3 0.0000000000 0.8888888889± 0.7745966692 0.5555555555

4 ± 0.3399810435 0.6521451548± 0.8611363116 0.3478548451

5 0.0000000000 0.5688888889± 0.5384693101 0.4786286705± 0.9061798459 0.2369268850

6 ± 0.2386191861 0.4679139346± 0.6612093865 0.3607615730± 0.9324695142 0.1713244924

The selection of the number of Gauss points is based on the formulaN = int[(p+1)/2]+1, where p is the polynomial degree to which the integrandis approximated. In most cases, the interpolation functions are of the samedegree in both ξ and η, and therefore one has M = N . When the integrand isof a different degree in ξ and η, we use max(M,N). The minimum allowablequadrature rule is one that yields the area or volume of the element exactly.The maximum degree of the polynomial refers to the degree of the highestpolynomial in ξ or η that is present in the integrands of the element matricesof the type in Eq. (2.5.3). Note that the polynomial degree of coefficients aswell as Jeij should be accounted for in determining the total polynomial degreeof the integrand. Of course, the coefficients axx, ayy, and J

eij , in general,

may not be polynomials. In those cases, their functional variations must beapproximated by a suitable polynomial (e.g. by a binomial series) in order todetermine the polynomial degree of the integrand.

2.5.4 Integration Over a Master Triangular Element

In the preceding section, we discussed numerical integration on quadrilateralelements which can be used to represent very general geometries as well asÞeld variables in a variety of problems. Here we discuss numerical integration


on triangular elements. Since quadrilateral elements can be geometricallydistorted, it is possible to distort a quadrilateral element to obtain a requiredtriangular element by moving the position of the corner nodes, and the fourthcorner in the quadrilateral is merged with one of the neighboring nodes. Inactual computation, this is achieved by assigning the same global node numberto two corner nodes of the quadrilateral element. Thus, master triangularelements can be obtained in a natural way from associated master rectangularelements. Here we discuss the transformations from a master triangularelement to an arbitrary triangular element.We choose the unit right isosceles triangle (see Table 2.5.2) as the master

element. An arbitrary triangular element Ωe can be generated from the mastertriangular element ΩT by transformation of the form (2.6.1). The derivativesof Lei with respect to the global coordinates can be computed from Eq. (2.5.6),which take the form⎧⎪⎨⎪⎩

∂Lei∂x

∂Lei∂y

⎫⎪⎬⎪⎭ = [Je]−1⎧⎪⎨⎪⎩∂Lei∂ L1

∂Lei∂ L2

⎫⎪⎬⎪⎭ , [Je] =

⎡⎢⎣∂x∂ L1

∂y

∂ L1

∂x∂ L2

∂y

∂ L2

⎤⎥⎦ (2.5.13)

Note that only L1 and L2 are treated as linearly independent coordinatesbecause L3 = 1− L1 − L2.After transformation, integrals on ΩT have the formZ

ΩeG(ξ, η) dξ dη =

ZΩeG(L1, L2, L3) dL1 dL2 (2.5.14)

which can be approximated by the quadrature formula

ZΩeG(L1, L2, L3) dL1 dL2 ≈

NXI=1

G(SI)WI (2.5.15)

where WI and SI denote the weights and integration points of the quadraturerule. Table 2.5.2 contains the location of integration points and weights forone-, three-, and four-point quadrature rules over triangular elements.

2.6 Computer Implementation

2.6.1 General Comments

In this section, computer implementation of Þnite element calculations ispresented to illustrate the ease with which theoretical ideas can be transformedinto practice. The material presented here is based on a more detailed accountpresented by Reddy [1] and Reddy and Gartling [4]. We begin with somegeneral comments on a typical Þnite element program.

a

ba

c

a

b

c

d


Table 2.5.2 Quadrature weights and points for triangular elements.

Number Degree ofof polynomial Location of integration pointsintegration Order of thepoints residual L1 L2 L3 W Geometric locations

11 O(h2) 1/3 1/3 1/3 1 a

2 1/2 0 1/2 a3 O(h3) 1/2 1/2 0 1/3 b

0 1/2 1/2 c

3 1/3 1/3 1/3 −27/48 a0.6 0.2 0.2 25/48 b

4 O(h4) 0.2 0.6 0.2 25/48 c0.2 0.2 0.6 25/48 d

A typical computer program consists of three basic parts:

1. Preprocessor

2. Processor

3. Postprocessor

In the preprocessor part of a program, the input data of the problemare read in and/or generated. This includes the geometry of the domain,analysis option (e.g. static, eigenvalue, or transient analysis), the data ofthe problem (e.g. deÞnition of the coefficients appearing in the differentialequation), boundary conditions, Þnite element analysis information (e.g.element type, number of elements, geometric information required to generatethe Þnite element mesh and element connectivity), and indicators for variouspostprocessing options (e.g. print, no print, types of quantities to becalculated, and so on). In the postprocessor part of the program, the solution iscomputed by interpolation at points other than the nodes, secondary variablesthat are derivable from the solution are also computed, and the output data are


processed in a desired format for printout and/or plotting. The preprocessorand postprocessor computer modules may contain a few Fortran statements toread and print pertinent information, simple subroutines (e.g. subroutines togenerate mesh and compute the gradient of the solution), or complex programslinked to other units via disk and tape Þles.The processor module, where typically large amounts of computing time

are spent, may consist of several subroutines, each having a special purpose.The main modules include:

1. Generation of the element matrices using numerical integration.

2. Assembly of element equations.

3. Imposition of the boundary conditions.

4. Solution of the algebraic equations for the nodal values of the primaryvariables (see Appendix 1).

The degree of sophistication and the complexity of a Þnite element programdepend on the general class of problems being programmed, the generality ofthe data in the equation, and the intended user of the program. Figure 2.6.1contains the ßow chart of a typical Þnite element computer program. A typicalbut simple Þnite element programs are included in the book by Reddy [1] andinterested reader may wish to study Chapters 7 and 13 of [1], where detailsof the implementation of one- and two-dimensional problems of heat transfer,ßuid mechanics, and solid mechanics are discussed.

2.6.2 One-Dimensional Problems

Here we discuss the main ideas behind the calculation of element coefficientmatrices, [Ke], [Me], and fe for the model problem discussed in Sections2.2 and 2.3 (see Chapter 8 for transient problems). Recall that the coefficientmatrices are of the form

Keij =

Z xb

xa

"a(x)

dLeidx

dLejdx

+ c(x)LeiLej

#dx

Meij =

Z xb

xact(x)L

eiLej dx, fei =

Z xb

xaf(x)Lei (x) dx (2.6.1)

We use the coordinate transformation of the form [see Eq. (2.5.1)]

x =MXj=1

xejφej(ξ) (2.6.2)

to express the integrals posed over a typical element Ωe = (xa, xb) as thoseover the interval, −1 ≤ ξ ≤ 1, so that we can use the Gauss quadrature to

PreprocessorInitialize global

[K], [M], F

DO e = 1 to n

Impose boundary conditions

Transfer global information(material properties, geometry and solution)

to element

Solve the equationsPostprocessor

Processor

CALL ELKMF to calculate Kij(e)

Mij(e), and fi

(e), and assemble to form global [K], [M], and F


Figure 2.6.1 The ßow chart of a typical Þnite element program.

evaluate them numerically. In Eq. (2.6.2), xej denote the global coordinatesof node j of element ωe = (xa, xb), and φ

ej are the approximation functions

used to approximate the geometry. For example, if we use linear interpolationfunctions φej(ξ) = Lej(ξ) to represent the geometry of the element, we havexe1 = xa, x

e2 = xb and Eq. (2.6.2) becomes

x = xaLe1(ξ) + xbL

e2(ξ) =

xa + xb2

+xb − xa2

ξ = xa +he2ξ (2.6.3)

The Jacobian of transformation is given by

Je ≡ dx

dξ=he2

(2.6.4)

The above transformation is exact for all (straight) line elements.The derivatives of Lei (x) with respect to the global coordinate x is given by

dLeidx

=dLeidξ

dξ

dx=dLeidξJ−1e


Then the integrals in Eq. (2.6.1) become

Keij =

Z xb

xa

"a(x)

dLeidx

dLejdx

+ c(x)LeiLej

#dx

=

Z 1

−1

"a(ξ)

µdLeidξJe−1¶Ã

dLejdξJ−1e

!+ c(ξ)LeiL

ej

#Je dξ

Meij =

Z xb

xact(x)L

eiLej dx =

Z 1

−1

³ct(ξ)L

eiLej

´Je dξ

fei =

Z xb

xaf(x)Lei (x) dx =

Z 1

−1f(ξ)LeiL

ej Je dξ (2.6.5)

where a(ξ) = a(x(ξ)) and so on. Each of the integral expressions above canbe evaluated using the Gauss quadratureZ 1

−1F (ξ)Je dξ =

NGPXNI=1

F (ξNI)JeWNI (2.6.6)

where NGP is the number of Gauss points, ξNI is the NIth Gauss point, andWNI is the NIth Gauss weight.To implement the above development into a computer subroutine for

arbitrary degree of Lei , we must Þrst create a subroutine of all interpolationsfunctions and their derivatives with respect to ξ that we intend to use in ouranalysis. For the present discussion, we limit them to the linear and quadraticLagrange family of functions. These functions and their derivatives are givenbelow.

Linear

L1(ξ) =1

2(1− ξ) , dL1

dξ= −0.5; L2(ξ) =

1

2(1 + ξ) ,

dL2dξ

= 0.5 (2.6.7)

Quadratic

L1(ξ) = −12ξ (1− ξ) , dL1

dξ= 0.5 (2ξ − 1)

L2(ξ) =³1− ξ2

´,

dL2dξ

= −2ξ

L3(ξ) =1

2ξ (1 + ξ) ,

dL3dξ

= 0.5 (1 + 2ξ) (2.6.8)

The following variables names are used in the subroutine (see INTRPL1Din Box 2.6.1):

SFL(i) = Li, DSFL(i) =dLidξ, GDSFL(i) =

dLidx

SUBROUTINE INTRPL1D(ELX,GJ,IEL,NPE,XI)C __________________________________________________________________ C C The subroutine computes shape functions and their derivatives for C Hermite cubic and Lagrange linear, quadratic, and cubic elements C C X......... Global (i.e. problem) coordinate C XI ...... Local (i.e. element) coordinate C H........ Element length C SFL.... Interpolation (or shape) functions C DSFL..... First derivative of SF with respect to XI C GDSFL.. First derivative of SF with respect to X C GJ...... Jacobian of the transformation C __________________________________________________________________ C

IMPLICIT REAL*8(A-H,O-Z) COMMON /SHP/SFL(4),GDSFL(4) DIMENSION DSFL(4),ELX(3)

C C SHAPE FUNCTIONS AND THEIR DERIVATIVES C C Linear functions C

IF(IEL.EQ.1)THENSFL(1)=0.5*(1.0-XI) SFL(2)=0.5*(1+XI) DSFL(1)=-0.5 DSFL(2)=0.5

ENDIF C C Quadratic functionsC

IF(IEL.EQ.2)THENSFL(1)=0.5*(XI-1.0)*XI SFL(2)=1.0-XI*XI SFL(3)=0.5*(XI+1.0)*XI DSFL(1)=XI-0.5 DSFL(2)=-2.0*XI DSFL(3)=XI+0.5

ENDIFGJ=0.0 DO 10 I=1,NPE

10 GJ=GJ+DSFL(I)*ELX(I) DO 30 I=1,NPE

GDSFL(I)=DSFL(I)/GJ 30 CONTINUEC

RETURN END


Box 2.6.1 Listing of subroutine INTRPL1D.

The notation should be transparent to the reader: SFL = Shape Functionsof the Lagrange family; DSFL = Derivative of SFL with respect to ξ; andGDSFL = Global Derivative (i.e. derivative with respect to x) of SFL. Allof them are n × 1 arrays, where n is the Nodes Per Element, NPE (i.e. n =NPE). In addition, GJ is used for the Jacobian J and XI for ξ.


Next, we implement the steps to evaluate the matrix coefficients Kij ,Mij ,and fi (i = 1 to NPE) into a subroutine called ELEKMF1D. To this end,we must assume some form of the data, a(x), c(x), ct(x) and f(x). Althoughwe can use any integrable functions, we restrict in the present discussion ourchoice of these coefficients to linear polynomials for the entire domain of theproblem:

a(x) = a0 + a1x, c(x) = c0 + c1x, ct(x) = ct0 + ct1x, f(x) = f0 + f1x(2.6.9)

Obviously, the coefficients of the polynomials must be read in the preprocessorand transferred to the subroutine ELEKMF1D through a common block (orthe argument list).Since the evaluation of the matrix coefficients involves summation on the

number of Gauss points, the arrays used for [K], [M ], and f must beinitialized outside the do-loop on NGP, the Number of Gauss Points. Thefollowing notation is used:

ELK(i,j) = Keij , ELM(i,j) =Me

ij , ELF(i) = fei , i = 1, 2, . . . , NPE(2.6.11)

The Gauss points are arranged in a matrix form so that the Jth columncorresponds to the Jth order Guass rule. The same notation is used for theGauss weights: GAUSPT (NI,NJ) = ξNI , NIth Gauss point of the NJ-point Gauss rule; GAUSWT (NI,NJ) = WNI , NIth Gauss weight of theNJ weight Gauss rule. Inside the do-loop on NI = 1 to NGP , we mustcall Subroutine INTRPL1D to compute SFL(i) and GDSFL(i) at the NIthGauss point and then compute all necessary quantities. Box 2.6.2 contains alisting of Subroutine ELEKMF1D.

2.6.3 Two-Dimensional Problems

The ideas presented in Section 2.6.2 for one-dimensional problems extend in astraightforward way to two-dimensional problems. The main differences are:(a) numerical integration over two-dimensional elements; (b) the Jacobian Jeis the determinant of the Jacobian matrix [Je] deÞned in Eq. (2.5.5); and(c) global derivatives of the interpolation functions are determined using Eq.(2.5.9).The Fortran statements of generating element coefficient matrices [Ke],

[Me], and Qe are provided in the form of subroutine ELEKMF2D in Box2.6.3, along with the subroutine INTRPL2D for interpolation functionsof rectangular elements and subroutine TEMPORAL for generating thecoefficient matrices [ Ke] and F e deÞned in Eq. (8.2.10) for transientproblems.

SUBROUTINE ELEKMF1D(MODEL,NDF,NGP,NPE)C __________________________________________________________________ CC IEL...... Element TYPE (1, Linear; 2, Quadratic) C H...... Element length C X...... Global (i.e. problem) coordinate C XI .... Local (i.e. element) coordinate C [GAUSPT].. 4×4 matrix of Gauss points: Nth column corresponds C to the N-point Guass rule C [GAUSWT]. 4×4 matrix of Gauss weights (see the comment above) C [ELK].. Element coefficient matrix [K] C ELF.... Element source vector f C ELX.... Vector of the global coordinates of element nodes C

IMPLICIT REAL*8(A-H,O-Z) COMMON/STF1/ELK(9,9),ELF(9),ELX(4) COMMON/STF2/AX0,AX1,CX0,CX1,FX0,FX1 COMMON/SHP/SF(4),GDSF(4) DIMENSION GAUSPT(5,5),GAUSWT(5,5)

C DATA GAUSPT/5*0.0D0,−0.57735027D0,0.57735027D0,3*0.0D0,−0.77459667D0,

* 0.0D0,0.77459667D0,2*0.0D0,−0.86113631D0,−0.33998104D0,0.33998104D0, * 0.86113631D0,0.0D0,−0.906180D0,−0.538469D0,0.0D0,0.538469D0,0.906180D0/

DATA GAUSWT/2.0D0,4*0.0D0,2*1.0D0,3*0.0D0,0.55555555D0,0.88888888D0, * 0.55555555D0,2*0.0D0,0.34785485D0,2*0.65214515D0,0.34785485D0,0.0D0, * 0.236927D0,0.478629D0,0.568889D0,0.478629D0,0.236927D0/

C H = ELX(NPE) - ELX(1)IEL = NPE-1

C Initialize the arrays DO 10 J=1,NPE ELF(J) = 0.0 DO 10 I=1,NPE

10 ELK(I,J)=0.0 C DO-LOOP on number of Gauss points begins here

DO 40 NI=1,NGP XI=GAUSS(NI,NGP) CALL INTRPL1D(ELX,GJ,IEL,NPE,XI) CNST=GJ*WT(NI,NGP) X=ELX(1)+0.5*(1.0+XI)*HAX=AX0+AX1*XCX=CX0+CX1*XFX=FX0+FX1*X

C Calculate element coefficients DO 20 I=1,NPE

ELF(I)=ELF(I)+FX*SFL(I)*CNST DO 20 J=1,NPE

S00=SFL(I)*SFL(J)*CNSTS11=GDSFL(I)*GDSFL(J)*CNST ELK(I,J)=ELK(I,J)+AX*S11+CX*S00

20 CONTINUE 40 CONTINUE

RETURN END


Box 2.6.2 Listing of subroutine ELEKMF1D.

2.7 Closure

The present chapter was devoted to a study of (1) the Þnite elementmodels of one- and two-dimensional problems involving Poissons equation,(2) a derivation of interpolation functions for basic one- and two-dimensional elements, (3) numerical evaluation of integrals, and (4) computerimplementation ideas. An understanding of the topics presented in thischapter is a prerequisite for the subsequent chapters of this book.


Box 2.6.3 Listings of subroutines ELEKMF2D, INTRPL2D, andTEMPORAL.

SUBROUTINE ELEKMF2D(NPE,NN,ITEM) C ________________________________________________________________ C C Element calculations based on linear and quadratic rectangular elements C and isoparametric formulation are carried out for the heat transfer and C penalty model of viscous incompressible fluid flow. C ________________________________________________________________ C

IMPLICIT REAL*8(A-H,O-Z) COMMON/STF/ELF(18),ELK(18,18),ELM(18,18),ELXY(9,2),ELU(18),A1,A2 COMMON/PST/A10,A1X,A1Y,A20,A2X,A2Y,C0,CX,CY,F0,FX,FY COMMON/SHP/SF(9),GDSF(2,9),SFH(16) COMMON/PNT/IPDF,IPDR,NIPF,NIPR DIMENSION GAUSPT(5,5),GAUSWT(5,5) COMMON/IO/IN,ITT

C DATA GAUSPT/5*0.0D0, −0.57735027D0, 0.57735027D0, 3*0.0D0, 2 −0.77459667D0, 0.0D0, 0.77459667D0, 2*0.0D0, −0.86113631D0, 3 −0.33998104D0, 0.33998104D0, 0.86113631D0, 0.0D0, −0.90617984D0, 4 −0.53846931D0,0.0D0,0.53846931D0,0.90617984D0/

C DATA GAUSWT/2.0D0, 4*0.0D0, 2*1.0D0, 3*0.0D0, 0.55555555D0, 2 0.88888888D0, 0.55555555D0, 2*0.0D0, 0.34785485D0, 3 2*0.65214515D0, 0.34785485D0, 0.0D0, 0.23692688D0, 4 0.47862867D0, 0.56888888D0, 0.47862867D0, 0.23692688D0/

C NDF = NN/NPE NET=NPE

C C Initialize the arrays C

DO 120 I = 1,NN ELF(I) = 0.0 DO 120 J = 1,NN IF(ITEM.NE.0) THEN

ELM(I,J)= 0.0 ENDIF

120 ELK(I,J)= 0.0 C C Do-loops on numerical (Gauss) integration begin here. Subroutine C INTRPL2D is called here C

DO 200 NI = 1,IPDF DO 200 NJ = 1,IPDF XI = GAUSPT(NI,IPDF) ETA = GAUSPT(NJ,IPDF) CALL INTRPL2D (NPE,XI,ETA,DET,ELXY,NDF) CNST = DET*GAUSWT(NI,IPDF)*GAUSWT(NJ,IPDF) X=0.0 Y=0.0 DO 140 I=1,NPE X=X+ELXY(I,1)*SF(I)

140 Y=Y+ELXY(I,2)*SF(I) C

SOURCE=F0+FX*X+FY*Y IF(ITEM.NE.0) THEN

CT=C0+CX*X+CY*Y ENDIF A11=A10+A1X*X+A1Y*Y A22=A20+A2X*X+A2Y*Y

C II=1 DO 180 I=1,NET JJ=1 DO 160 J=1,NET S00=SF(I)*SF(J)*CNST S11=GDSF(1,I)*GDSF(1,J)*CNST S22=GDSF(2,I)*GDSF(2,J)*CNST

C C Heat transfer and like problems (i.e. single DOF problems):_______ C

ELK(I,J) = ELK(I,J) + A11*S11 + A22*S22 + A00*S00 IF(ITEM.NE.0) THEN

ELM(I,J) = ELM(I,J) + CT*S00 ENDIF

160 JJ = NDF*J+1 C Source of the form fx = F0 + FX*X + FY*Y is assumed

L=(I-1)*NDF+1 ELF(L) = ELF(L)+CNST*SF(I)*SOURCE

180 II = NDF*I+1 200 CONTINUE


IF(ITEM.NE.0) THEN C C Compute the coefficient matrices of the final algebraic equations C (i.e. after time approximation) in the transient analysis:_______ C

CALL TEMPORAL(NN) ENDIF RETURN END

SUBROUTINE INTRPL2D(NPE,XI,ETA,DET,ELXY,NDF) C ________________________________________________________________ C C The subroutine evaluates the interpolation functions (SF(I)) and C their derivatives with respect to global coordinates (GDSF(I,J)) C for Lagrange linear & quadratic rectangular elements, using the C isoparametric formulation. The subroutine also evaluates HermiteC interpolation functions and their global derivatives using the C subparametric formulation. C C SF(I).......Interpolation function for node I of the element C DSF(J,I)..Derivative of SF(I) with respect to XI if J=1 and C and ETA if J=2 C GDSF(J,I)...Derivative of SF(I) with respect to X if J=1 and C and Y if J=2 C XNODE(I,J)...J-TH (J=1,2) Coordinate of node I of the element C NP(I)......Array of element nodes (used to define SF and DSF) C GJ(I,J)....Jacobian matrix C GJINV(I,J).Inverse of the jacobian matrix C ________________________________________________________________ C

IMPLICIT REAL*8 (A-H,O-Z) DIMENSION ELXY(9,2),XNODE(9,2),NP(9),DSF(2,9),GJ(2,2),GJINV(2,2) COMMON/SHP/SF(9),GDSF(2,9) COMMON/IO/IN,ITT DATA XNODE/-1.0D0, 2*1.0D0, -1.0D0, 0.0D0, 1.0D0, 0.0D0, -1.0D0,

* 0.0D0, 2*-1.0D0, 2*1.0D0, -1.0D0, 0.0D0, 1.0D0, 2*0.0D0/ DATA NP/1,2,3,4,5,7,6,8,9/

C FNC(A,B) = A*B IF(NPE.EQ.4) THEN

C C LINEAR Lagrange interpolation functions for FOUR-NODE element C

DO 10 I = 1, NPE XP = XNODE(I,1) YP = XNODE(I,2) XI0 = 1.0+XI*XP ETA0=1.0+ETA*YP SF(I) = 0.25*FNC(XI0,ETA0) DSF(1,I)= 0.25*FNC(XP,ETA0)

10 DSF(2,I)= 0.25*FNC(YP,XI0) ELSE

IF(NPE.EQ.8) THEN C C QUADRATIC Lagrange interpolation functions for EIGHT-NODE element C

DO 20 I = 1, NPE NI = NP(I) XP = XNODE(NI,1) YP = XNODE(NI,2) XI0 = 1.0+XI*XP ETA0 = 1.0+ETA*YP XI1 = 1.0-XI*XI ETA1 = 1.0-ETA*ETA IF(I.LE.4) THEN

SF(NI) = 0.25*FNC(XI0,ETA0)*(XI*XP+ETA*YP-1.0) DSF(1,NI) = 0.25*FNC(ETA0,XP)*(2.0*XI*XP+ETA*YP) DSF(2,NI) = 0.25*FNC(XI0,YP)*(2.0*ETA*YP+XI*XP)

ELSE IF(I.LE.6) THEN

SF(NI) = 0.5*FNC(XI1,ETA0) DSF(1,NI) = -FNC(XI,ETA0) DSF(2,NI) = 0.5*FNC(YP,XI1)

ELSE SF(NI) = 0.5*FNC(ETA1,XI0) DSF(1,NI) = 0.5*FNC(XP,ETA1) DSF(2,NI) = -FNC(ETA,XI0)

ENDIF ENDIF

20 CONTINUE ELSE

C C QUADRATIC Lagrange interpolation functions for NINE-NODE element C

DO 30 I=1,NPE NI = NP(I) XP = XNODE(NI,1) YP = XNODE(NI,2) XI0 = 1.0+XI*XP ETA0 = 1.0+ETA*YP XI1 = 1.0-XI*XI ETA1 = 1.0-ETA*ETA XI2 = XP*XI ETA2 = YP*ETA IF(I .LE. 4) THEN

SF(NI) = 0.25*FNC(XI0,ETA0)*XI2*ETA2 DSF(1,NI)= 0.25*XP*FNC(ETA2,ETA0)*(1.0+2.0*XI2) DSF(2,NI)= 0.25*YP*FNC(XI2,XI0)*(1.0+2.0*ETA2)

ELSE IF(I .LE. 6) THEN

SF(NI) = 0.5*FNC(XI1,ETA0)*ETA2 DSF(1,NI) = -XI*FNC(ETA2,ETA0) DSF(2,NI) = 0.5*FNC(XI1,YP)*(1.0+2.0*ETA2)

ELSE IF(I .LE. 8) THEN

SF(NI) = 0.5*FNC(ETA1,XI0)*XI2 DSF(2,NI) = -ETA*FNC(XI2,XI0) DSF(1,NI) = 0.5*FNC(ETA1,XP)*(1.0+2.0*XI2)

ELSE SF(NI) = FNC(XI1,ETA1) DSF(1,NI) = -2.0*XI*ETA1 DSF(2,NI) = -2.0*ETA*XI1

ENDIF ENDIF

ENDIF 30 CONTINUE

ENDIF ENDIF

C C Compute the Jacobian matrix [GJ] and its inverse [GJINV], and [GDSF] C

DO 40 I = 1,2 DO 40 J = 1,2 GJ(I,J) = 0.0 DO 40 K = 1,NPE

40 GJ(I,J) = GJ(I,J) + DSF(I,K)*ELXY(K,J) C

DET = GJ(1,1)*GJ(2,2)-GJ(1,2)*GJ(2,1) GJINV(1,1) = GJ(2,2)/DET GJINV(2,2) = GJ(1,1)/DET GJINV(1,2) = -GJ(1,2)/DET GJINV(2,1) = -GJ(2,1)/DET DO 50 I = 1,2 DO 50 J = 1,NPE GDSF(I,J) = 0.0 DO 50 K = 1, 2

50 GDSF(I,J) = GDSF(I,J) + GJINV(I,K)*DSF(K,J) RETURN END

SUBROUTINE TEMPORAL(NN) C ____________________________________________________________________________________C C The subroutine computes the algebraic equations associated with the parabolic differential equationsC by using the α-family of approximations. A constant source is assumed. C ____________________________________________________________________________________ C

IMPLICIT REAL*8(A-H,O-Z) COMMON/STF/ELF(18),ELK(18,18),ELM(18,18),ELXY(9,2),ELU(18),A1,A2

CC The α-family of time approximation for parabolic equations C

DO 20 I=1,NN SUM=0.0 DO 10 J=1,NN SUM=SUM+(ELM(I,J)-A2*ELK(I,J))*ELU(J)

10 ELK(I,J)=ELM(I,J)+A1*ELK(I,J) 20 ELF(I)=(A1+A2)*ELF(I)+SUM

RETURN END



Problems

2.1 (Least-Squares Method). In the least-squares method, the residual Re [see Eq. (2.2.3)]is minimized in the following sense:

δ

Z xb

xa

[Re]2dx = 0 or∂

∂ci

Z xb

xa

[Re(x, c1, c2, · · · , cn)]2dx = 0, i = 1, 2, · · · , n

(a) Identify the weight function wei if the least-squares method is to be deduced fromEq. (2.2.4), (b) develop the least-squares Þnite element model, and (c) discuss thetype of Þnite element approximation of u that may be used.

2.2 Consider the differential equation

− d

dx

³adu

dx

´+d2

dx2

µbd2u

dx2

¶+ cu = f

where a, b, c, and f are known functions of position x. (a) Develop the weak formover a typical element Ωe = (xa, xb) such that the bilinear form is symmetric, (b)identify the bilinear and linear forms and construct the quadratic functional, (c)develop the Þnite element model of the equation, and (d) discuss the type of Þniteelement approximation of u that may be used.

2.3 Derive the Lagrange cubic interpolation functions for a four-node (one-dimensional)element (with equally spaced nodes) using the alternative procedure based oninterpolation properties (2.2.22). Use the local coordinate x for simplicity.

2.4 Derive the Þnite element model of the differential equation

− d

dx

³a(x)

du

dx

´= f(x) for 0 < x < L

for the boundary conditions

u(0) = u0,h³a(x)

du

dx

´+ ku

ix=L

= P

2.5 The following differential equation arises in connection with heat transfer in aninsulated rod:

− d

dx

³kdT

dx

´= q for 0 < x < L

T (0) = T0,hkdT

dx+ β(T − T∞) + q

i ¯x=L

= 0

where T is the temperature, k the thermal conductivity, and q the heat generation.Take the following values for the data: q = 0, q = 0, L = 0.1 m, k = 0.01 W m−1 C−1,β = 25 W m−2C−1. T0 = 50C, and T∞ = 5C. Solve the problem using two linearÞnite elements for temperature values at x = 1

2L and L. Answer: U2 = 27.59C,

U3 = 5.179C, Q(1)1 = 4.482 W m−2 = −Q(2)2 .

2.6 An insulating wall is constructed of three homogeneous layers with conductivitiesk1, k2, and k3 in intimate contact (see Figure P2.6). Under steady-state conditions,the temperatures at the boundaries of the layers are characterized by the externalsurface temperatures T1 and T4 and the interface temperatures T2 and T3. Formulatethe problem to determine the temperatures Ti (i = 1, . . . ,4) when the ambienttemperatures T0 and T5 and the (surface) Þlm coefficients β0 and β5 are known.

x

L

1 2 321 3 4

k1 = 90 W/(m ºC) k2 = 75 W/(m ºC)k3 = 50 W/(m ºC)h1 = 0.03 mh2 = 0.04 mh3 = 0.05 mβ = 500 W/(m2 ºC)T∞ = 20ºC

h1 h2 h3


Assume that there is no internal heat generation and that the heat ßow is one-dimensional (∂T/∂y = 0). Answer: U1 = 84.489C, U2 = 68.977C, U3 = 50.881C,U4 = 45.341C, (Q11)def = 217.16 W-m

−2, (Q32)def = −155.11 W-m−2.

Figure P2.6

2.7 Consider the steady laminar ßow of a viscous ßuid through a long circular cylindricaltube. The governing equation is

−1r

d

dr

³rµdw

dr

´=P0 − PLL

≡ f0, 0 < r < R0

where w is the axial (i.e. z) component of velocity, µ is the viscosity, and f0 isthe gradient of pressure (which includes the combined effect of static pressure andgravitational force). The boundary conditions are³

rdw

dr

´ ¯r=0

= 0, w(R0) = 0

Using the symmetry and (a) two linear elements, (b) one quadratic element, determinethe velocity Þeld and compare with the exact solution at the nodes:

we(r) =f0R204µ

∙1−³r

R0

´2¸2.8. In the problem of the ßow of a viscous ßuid through a circular cylinder (see Problem

2.7), assume that the ßuid slips at the cylinder wall; that is, instead of assuming thatw = 0 at r = R0, use the boundary condition that

kw = −µdwdr

at r = R0

in which k is the coefficient of sliding friction. Solve the problem with two linearelements.

P = 50 k

20 in.

SteelAluminum

P = 50 k

16 in.

A BC


2.9 The two members in Figure P2.9 are fastened together and to rigid walls. Ifthe members are stress free before they are loaded, what will be the stresses anddeformations in each after the two 50,000 lbs. loads are applied? Use Es = 30× 106psi and Ea = 107 psi; the aluminum rod is 2 in. in diameter and the steel rod is 1.5 in.

in diameter. Answer: U2 = 0.0134 in., P(1)2 = −21, 052.6 lb, and σ(1) = 6701.25 psi.

Figure P2.9

2.10 Evaluate the coefficients Keij and F

ei of Eqs. (2.3.40a,b) for a linear triangular element.

2.11 Repeat Problem 2.10 for a linear rectangular element.2.12 Consider the partial differential equation governing heat transfer in an axisymmetric

geometry

−1r

∂

∂r

³rkrr

∂T

∂r

´− ∂

∂z

³kzz

∂T

∂z

´= f(r, z) (1)

where (krr, kzz) and f are the conductivities and internal heat generation per unitvolume, respectively. In developing the weak form, we integrate over the elementalvolume of the axisymmetric geometry: r dr dθ dz. Develop the weak form andassociated Þnite element model over an element.

References1. Reddy, J. N., An Introduction to the Finite Element Method, 2nd edn, McGrawHill,New York (1993).

2. Reddy, J. N. and Gartling, D. K., The Finite Element Method in Heat Transfer andFluid Dynamics, 2nd edn, CRC Press, Boca Raton, FL (2001).


3

Heat Transfer andOther Field Problems

in One Dimension

3.1 Model Differential Equation

We shall consider a model nonlinear differential equation governing one-dimensional problems involving a single unknown to illustrate the Þniteelement model development and discuss solution methods to solve the resultingnonlinear algebraic equations. As we shall see, the weak form and Þniteelement model developed in Section 2.2 for a linear model equation are alsovalid for the nonlinear case. The main difference is that the Þnite elementequations are nonlinear, and therefore iterative methods must be used to solvethe nonlinear Þnite element equations.Consider the differential equation (see Table 2.2.1 for Þelds of study)

− d

dx

∙a(x, u)

du

dx

¸+ b(x, u)

du

dx+ c(x, u)u = f(x), 0 < x < L (3.1.1)

subjected to boundary conditions of the form

nx adu

dx+ β(x, u) (u− u∞) = Q, or u = u (3.1.2)

at a boundary point. Here u(x) denotes the dependent variable to bedetermined, a, b, and c are known functions of x and u (and possibly derivativesof u), f is a known function of x, u∞ and u are known parameters, Q thesecondary variable, and nx is the cosine of the angle between the positive x-axis and the outward normal to the edge at the node (note that nx = −1 atx = xa and nx = 1 at x = xb). The second boundary condition is a specialcase of the Þrst (with u = u∞) in the limit β → ∞. We wish to solve thisnonlinear differential equation using the Þnite element method.Clearly, the source of nonlinearity is in a, b and/or c, which in most

engineering systems include geometric and material parameters that may befunctions of the dependent variable u. For example, for heat conduction in arod with convective heat transfer through the surface, one has a = kA, b = 0,


and c = Pβ, where k denotes conductivity, β the convective heat transfercoefficient, A the cross-sectional area, and P the perimeter of the rod. Thenthe nonlinearity arises from the conductivity and heat transfer coefficientsbeing functions of temperature u.

3.2 Weak Formulation

Suppose that the domain Ω = (0, L) is divided into N elements. A typicalelement is denoted as Ωe = (xa, xb), where xa and xb denote the globalcoordinates of the end nodes of the element. The weak form of Eq. (3.1.1)over the element is given by

0 =

Z xb

xa

∙adw

dx

du

dx+ bw

du

dx+ cwu− wf

¸dx−

∙w

µadu

dx

¶¸xbxa

=

Z xb

xa

∙a(x, u)

dw

dx

du

dx+ b(x, u)w

du

dx+ c(x, u)wu− wf(x)

¸dx

− [Qe1 − βa (u(xa)− ua∞)]w(xa)−hQe2 − βb

³u(xb)− ub∞

íw(xb) (3.2.1)

where w(x) is a weight function, and [see Eq. (3.1.2)]

−∙adu

dx

¸x=xa

= Qe1 − βa [u(xa)− ua∞]∙adu

dx

¸x=xb

= Qe2 − βbhu(xb)− ub∞

i (3.2.2)

Here (ua∞, ub∞) denote the reference values and (βa,βb) denote the Þlmcoefficients at the left and right ends of the element, respectively. The weakform (3.2.1) suggests that u is the primary variable and Q is the secondaryvariable of the formulation. Recall that specifying a primary variable is calledthe essential (or geometric) boundary condition and specifying a secondaryvariable is termed the natural (or force) boundary condition. The Þrstboundary condition in Eq. (3.1.2) is of the natural type and it is nonlinear;the second boundary condition in Eq. (3.1.2) one is of the essential type.

3.3 Finite Element Model

Suppose that the dependent unknown u(x) is approximated over element Ωe

by the Þnite element approximation of the form

u(x) ≈ ueh(x) =nXj=1

uejLej(x) (3.3.1)

Substituting the approximation (3.3.1) for u and w = Lei into the weak form(3.2.1), we obtain the Þnite element model

[Ke]ue = fe+ Qe (3.3.2)

NONLINEAR EQUATIONS IN SINGLE VARIABLE IN ONE DIMENSION 63

where

Keij =

Z xb

xa

"a(x, uh)

dLeidx

dLejdx

+ b(x, uh)Lei

dLejdx

+ c(x, uh)LiLj

#dx

+ βaLei (xa)L

ej(xa) + βbL

ei (xb)L

ej(xb) (3.3.3a)

fei =

Z xb

xaf(x)Lei dx+ βau

a∞L

ei (xa) + βbu

b∞L

ei (xb) (3.3.3b)

Note that the coefficient matrix is a nonlinear function of the unknown nodalvalues uej , and it is an unsymmetric matrix when b 6= 0; when b = 0, Ke

ij is asymmetric matrix. Also, the coefficients involving β are only present at theboundary nodes where convection type boundary condition is speciÞed.

Example 3.3.1

To gain more insight into the make up of the coefficient matrix, suppose that a(x) = a0u(x),b = c = 0 and β = 0, where a0 may be a function of x only. Then we have

Keij =

Z xb

xa

ae0

ÃnX

k=1

uekLek

!dLeidx

dLejdx

dx

=

nXk=1

uek

Z xb

xa

a0Lek

dLeidx

dLejdx

dx (i)

Now consider, as an example, linear approximation (n = 2) of u(x), and a0 is a constantwithin an element, say ae0. Then we have

Keij =

nXk=1

ae0uek

Z xb

xa

LekdLeidx

dLejdx

dx

=

nXk=1

ae0uek(−1)i+j

1

h2e

Z xb

xa

Lek dx

= (−1)i+j ae0

2he

ÃnX

k=1

uek

!= (−1)i+j a

e0

2he(ue1 + u

e2) (ii)

or

[Ke] =ae0(u

e1 + u

e2)

2he

∙1 −1

−1 1

¸(iii)

The assembly of element equations follows the same procedure as in linearÞnite element analysis. If we denote the global nodal vector by U, theassembled system of equations can be written as

[K(U)]U = F (3.3.4)

where [K] and F denote the global coefficient matrix and the right-handside vector, respectively. In the example above, [K] is a linear function of


the nodal values Ui and it is a symmetric matrix. Consequently, the resultingÞnite element equations are nonlinear; in the present case, they are quadraticin Ui. Example 3.3.2 illustrates these ideas.

Example 3.3.2

The assembled equations (3.3.4) associated with a mesh of two linear elements of equallength, and with the data used in Example 3.3.1, are

1

2h

"(U1 +U2) −(U1 +U2) 0

−(U1 +U2) (U1 + 2U2 +U3) −(U2 +U3)0 −(U2 +U3) (U2 +U3)

#(U1U2U3

)

=

⎧⎨⎩f(1)1

f (1)2 + f (2)1

f(2)2

⎫⎬⎭+

⎧⎨⎩Q(1)1

Q(1)2 +Q(2)1Q(2)2

⎫⎬⎭ (i)

Now suppose that the assembled system of equations is subjected to the boundaryconditions

u(0) = u0,hadu

dx+ β (u− u∞)

ix=L

= Q (ii)

withβ = β0 + β1u (iii)

These conditions for the present mesh of two linear elements imply

U1 = u0, Q(2)2 = Q− (β0 + β1U3) (U3 − u∞) (iv)

Hence, after imposing the boundary conditions and balance of the secondary variables

Q(1)2 +Q(2)1 = 0, the assembled equations in (i) become

1

2h

"(U1 +U2) −(U1 +U2) 0

−(U1 +U2) (U1 +2U2 +U3) −(U2 +U3)0 −(U2 +U3) (U2 +U3) + 2h(β0 + β1U3)

#(u0U2U3

)

=

⎧⎨⎩f (1)1

f (1)2 + f (2)1

f (2)2

⎫⎬⎭+

⎧⎨⎩Q(1)1

Q(1)2 +Q(2)1Q+ (β0 + β1U3)u∞

⎫⎬⎭ (v)

3.4 Solution Procedures

3.4.1 General Comments

The numerical procedures used to solve nonlinear algebraic equations (3.3.4)are iterative in nature. Two iterative procedures are outlined here for theproblem at hand. Some general features of iterative methods used for nonlinearequations are discussed before getting into the details of each method (seeAppendix 2 for additional discussion).Suppose that we wish to solve the nonlinear matrix equation

[A(U)]U = F


We begin by assuming that the solution U(r−1) at the (r − 1)st iteration isknown, and we wish to seek the solution U(r) at the rth iteration. At thebeginning of the iteration, that is, when r = 1, the solution U(0) is assumedor guessed consistent with the problem data. Using the solution from the(r−1)st iteration, we compute the coefficient matrix [A(U(r−1))]. Since [A] isevaluated using estimated vector U, in general, [A(U(r−1))]U(r) 6= F.Hence, we are left with a residual

R ≡ [A(U(r−1))]U(r) − F (3.4.1)

The objective of the iteration process is to reduce this residual to a very small,negligible value, ²: vuut NX

I=1

R2I ≤ ² (3.4.2)

Alternatively, the iteration may proceed until the difference between solutionsfrom two consecutive iterations, measured with the Euclidean norm, is lessthan the tolerance ²: vuutPN

I=1 |U (r)I − U (r−1)I |2PNI=1 |U (r)I |2

≤ ² (3.4.3)

3.4.2 Direct Iteration Procedure

The direct iteration technique, also known as the Picard iteration method ofsuccessive substitution, is the simplest of the two methods discussed here. Inthe direct iteration procedure, the solution at the rth iteration is determinedfrom the equation

[K(U(r−1))]U(r) = F (3.4.4)

where the coefficient matrix [K] is evaluated using the known solution fromthe (r−1)st iteration. It is assumed that the coefficient matrix [K] is invertibleafter the imposition of boundary conditions. Thus, the initial guess vectorU0 should be such that (a) it satisÞes the speciÞed essential boundaryconditions and (b) [K] is invertible. This means, in the present case, thatUI 6= 0 for at least one value of I.Figure 3.4.1 depicts the general idea of the direct iteration procedure for a

single degree-of-freedom system. Here K denotes the slope of the line joiningthe origin to the point K(U) on the curve F = K(U)U ≡ f(U). Note thatK(U) is not the slope of the tangent to the curve at U . The direct iterationconverges if the nonlinearity is mild and it diverges if the nonlinearity is severe.Divergence is more likely for hardening type nonlinearity. Acceleration ofconvergence for some types of nonlinearities may be achieved by using a

K(U)U ≡ f(U)

F

U

FC

UCU0

K(U0)

U1

K(U1)

U2

K(U2)

U3

°

UC − Converged solution

U0 − Initial guess solution


Figure 3.4.1 Convergence of the direct iteration procedure (calculation ofU for a speciÞed source value F ).

weighted-average of solutions from the last two iterations rather than thesolution from the last iteration:

U ≡ ρU(r−2) + (1− ρ)U(r−1), 0 ≤ ρ ≤ 1 (3.4.5)

where ρ is called the acceleration parameter and r is the current iterationnumber (i.e. when we seek solution U(r)). The value of ρ depends on thenature of nonlinearity.

Example 3.4.1

We wish to solve the nonlinear differential equation

− d

dx

³udu

dx

´= f0, 0 < x < 1 (i)

subjected to the boundary conditionshudu

dx

ix=0

= Q, u(1) = u (ii)

For a mesh of two linear elements with h1 = h2 = h, the assembled equations are given by

Eq. (i) of Example 3.3.2. After imposing the boundary conditions Q(1)1 = Q, Q(1)2 +Q(2)1 = 0


and U3 = u, Eq. (i) of Example 3.3.2 becomes

1

2h

"(U1 + U2) −(U1 + U2) 0

−(U1 + U2) (U1 + 2U2 + U3) −(U2 + U3)0 −(U2 + U3) (U2 + U3)

#(U1U2u

)(r)

=

⎧⎨⎩f(1)1

f (1)2 + f (2)1

f(2)2

⎫⎬⎭+

( Q0Q(2)2

)(iii)

or, omitting the last equation, we obtain the following condensed set of equations:

1

2h

∙(U1 + U2) −(U1 + U2)

−(U1 + U2) (U1 + 2U2 + U3)

¸½U1U2

¾(r)=

½f(1)1

f (1)2 + f (2)1

¾+

½Q

(U2+U3)2h u

¾(iv)

where UI = U(r−1)I (I = 1,2,3) denote the nodal values from the previous iteration.

As a speciÞc problem, we use the following data

Q = 0, u =√2 = 1.4142, f0 = −1, h = L/2 = 0.5 (v)

and select the initial guess vector, which must be consistent with speciÞed essential boundaryconditions (i.e. U3 = u =

√2), to be

U(0)1 = 1.0, U(0)2 = 1.0, U(0)3 =√2 = 1.4142 (vi)

Then Eq. (iii) becomes (fei = −h/2)"2 −2 0

−2 4.4142 −2.41420 −2.4142 2.4142

#(U1U2√2

)(1)= −

(0.250.50.25

)+

(00Q(2)2

)(vii)

and eliminating the third equation, we obtain the following global set of equations at thebeginning of the Þrst iteration [see Eq. (iv)]:∙

2 −2−2 4.4142

¸½U1U2

¾(1)= −

½0.250.5

¾+

½0

2.4142√2

¾=

½−0.252.9142

¾(viii)

The solution at the end of Þrst iteration is

U(1)1 = 0.9785, U(1)2 = 1.1035, U(1)3 = 1.4142 (ix)

For the second iteration, the equations solved are (ρ = 0)∙2.0821 −2.0821

−2.0821 4.5998

¸½U1U2

¾(2)= −

½0.250.5

¾+

½0

3.5607

¾(x)

and the solution becomes

U(2)1 = 0.9962, U(2)2 = 1.1163, U(2)3 = 1.4142 (xi)


The root-mean-square error in the solution (with respect to solution obtained in the Þrstiteration) [see Eq. (3.4.1)] is 0.0106.

The exact solution to this problem, with the data given in Eq. (ii), is (one can verify bydirect substitution)

u(x) =p1 + x2 (xii)

The convergence tolerance is taken to be ² = 10−3, and the maximum number of iterations isprescribed to be 10. The Þnite element solutions obtained with the linear (L) and quadratic

(Q) elements are compared with the exact solution in Table 3.4.1. The Þnite element solution

converges to the exact solution in four iterations.

Table 3.4.1 Comparison of Þnite element solutions with the exact solutionof the problem in Example 3.4.1 (direct iteration).

x Iteration 2L 4L 1Q 2Q Exact

0.00 1 0.9785 0.9517 0.9699 0.94152 0.9962 0.9903 0.9954 0.98813 0.9995 0.9988 0.9995 0.99864 0.9999 0.9999 0.9999 0.9999 1.0000

0.25 1 − 0.9830 − 0.97272 − 1.0226 − 1.02083 − 1.0299 − 1.02984 − 1.0307 − 1.0308 1.0308

0.50 1 1.1036 1.0767 1.1013 1.06652 1.1163 1.1136 1.1168 1.11273 1.1178 1.1176 1.1118 1.11764 1.1180 1.1180 1.1187 1.1180 1.1180

0.75 1 − 1.2330 − 1.22852 − 1.2489 − 1.24883 − 1.2499 − 1.25004 − 1.2500 − 1.2500 1.2500

1.00 1.4142 1.4142 1.4142 1.4142 1.4142

3.4.3 Newtons Iteration Procedure

In Newtons method, also referred to as the NewtonRaphson method, Eq.(3.4.4) is written as

R ≡ [K]U− F = 0 (3.4.6)

where R is the residual vector. We expand R in Taylors series about the(known) solution at the ith iteration:

R(U) = R(U(r−1))+µ∂R∂U

¶(r−1)· δU+ . . . (3.4.7)


Omitting the terms of order 2 and higher, we obtain

µ∂R∂U

¶(r−1)· δU = −R(U(r−1)) (3.4.8a)

[T (U(r−1))]δU = −R(U(r−1)) (3.4.8b)

where [T ] is called the tangent matrix

[T (U(r−1))] ≡µ∂R∂U

¶(r−1)(3.4.9)

The component deÞnition of the tangent matrix at the element level is

T eij ≡∂Rei∂uej

=∂

∂uej

ÃnX

m=1

Keimu

em − F ei

!

=nX

m=1

Ã∂Ke

im

∂uejuem +K

eim

∂uem∂uej

!=

nXm=1

∂Keim

∂uejuem +K

eij (3.4.10)

The residual vector after the (r − 1)st iteration is given by

−R(U(r−1)) = F− [K(U(r−1))]U(r−1) (3.4.11)

The solution at the rth iteration is then given by

Ur = U(r−1) + δU (3.4.12)

The symmetry or unsymmetry of the direct matrix or the tangent matrixdepends on the original differential equation as well as the weak form usedto develop the Þnite element equations; the symmetry or unsymmetry ofthe tangent matrix depends on the direct matrix as well as on the natureof nonlinearity included in a(x, u) and c(x, u).

Figure 3.4.2 shows the convergence of the NewtonRaphson iterationprocedure for a single degree-of-freedom system. Here T (U) denotes theslope of the tangent to the curve F = K(U)U at U . The NewtonRaphsoniteration converges for hardening as well as softening type nonlinearities. Forhardening type, convergence may be accelerated using the underrelaxation inEq. (3.4.5). The method may diverge for a saddle point behavior. Severalcomments regarding the NewtonRaphson procedure are in order.

At the end of each iteration, the procedure gives an increment of thesolution as opposed to the total solution in the direct iteration procedure.

K(U)U − F ≡ R(U)

F

U

FC

U0

T(U0)

T(U1)

T(U2)

UC = U3

°

UC − Converged solution

U0 − Initial guess solutionδU1 δU2

U1 = δU1 + U0 U2 = δU2 + U0


Figure 3.4.2 Convergence of the Newton iteration procedure (calculation ofδU for a speciÞed residual value R).

Since the actual speciÞed essential boundary conditions are included in theinitial guess vector, the incremental vector δU is subjected to only thehomogeneous form of the speciÞed essential boundary conditions.

The tangent matrix does not have to be exact; an approximate [T ] canalso provide the solution but it may take more iterations. In any case, theresidual vector will be computed using the deÞnition R = F− [K]U.

When the tangent matrix is updated only once in a certain numberof iterations (to save computational time), the procedure is known asthe modiÞed NewtonRaphson method. Generally, the modiÞed NewtonRaphson iteration method takes more iterations to converge than the fullNewtons iteration, and it may even diverge if the nonlinearity is severe.

Example 3.4.2

Consider the element coefficient matrix in Eq. (3.3.3a). Assume that a(x, u) = ae(x), b = 0,c(x, u) = c0(x) + c1(x)u+ c2(x)u

2, βa = 0, βb = β0 + β1uen. Then we have

Keij =

Z xb

xa

½ae(x)

dLeidx

dLejdx

+£c0(x) + c1(x)uh(x) + c2(x)u

2h(x)

¤LeiL

ej

¾dx

+ [β0 + β1uh(xb)]Lei (xb)L

ej(xb) (i)

Note that in this case [Ke] is symmetric. We wish to compute the element tangent coefficient


matrix. We have

T eij =

nXm=1

∂Keim

∂uejuem +K

eij

=

nXm=1

∂

∂uej

(Z xb

xa

£c0(x) + c1(x)uh + c2(x)u

2h

¤LeiL

emdx

+ [β0 + β1uh(xb)]Lei (xb)L

em(xb)

)uem +K

eij

=

nXm=1

uem

"Z xb

xa

¡c1L

ej + 2c2uhL

ej

¢LeiL

emdx+ β1L

ej(xb)L

ei (xb)L

em(xb)

#+Ke

ij

=

Z xb

xa

¡c1uh +2c2u

2h

¢LeiL

ejdx+ β1uh(xb)L

ei (xb)L

ej(xb) +K

eij (ii)

where the identities

∂uh∂uej

= Lej ,

nXm=1

uemLem = uh(x)

are used in arriving at the last line. Note that the tangent matrix [T e] is also symmetric.

Example 3.4.3

Here we wish to solve the nonlinear Þnite element equations of Example 3.4.1 using theNewtonRaphson iterative technique. First, we compute the tangent matrix at the elementlevel and then assemble the equations. For an element, we have

T eij =

nXm=1

∂Keim

∂uejuem +K

eij

=

nXm=1

∂

∂uej

µZ xb

xa

uhdLeidx

dLemdx

dx

¶uem +K

eij

=

Z xb

xa

∂uh∂uej

dLeidx

ÃnX

m=1

uemdLemdx

!dx+Ke

ij

=

Z xb

xa

duhdx

dLeidx

Lej dx+Keij ≡ Ke

ij +Keij (i)

where the identitynX

m=1

uemdLemdx

=duhdx

is used in arriving at the last line. We have,

Keij =

Z xb

xa

duhdx

dLeidx

Lej dx =ue2 − ue12

Z xb

xa

dLeidx

Lej dx


or

[ Ke] =ue2 − ue12he

∙−1 −11 1

¸Thus, the tangent matrix becomes

[T e] = [Ke] + [ Ke] =(ue1 + u

e2)

2he

∙1 −1

−1 1

¸+(ue2 − ue1)2he

∙−1 −11 1

¸where uei denote the nodal values known from the previous iteration. Alternatively,

[T e] = [Ke] +

⎡⎣ ∂Ke11

∂ue1ue1 +

∂Ke12

∂ue1ue2

∂Ke11

∂ue2ue1 +

∂Ke12

∂ue2ue2

∂Ke21

∂ue1ue1 +

∂Ke22

∂ue1ue2

∂Ke21

∂ue2ue1 +

∂Ke22

∂ue2ue2

⎤⎦uei=ue

i

=(ue1 + u

e2)

2he

∙1 −1

−1 1

¸+

1

2he

∙ue1 − ue2 ue1 − ue2

−ue1 + ue2 −ue1 + ue2

¸=(ue1 + u

e2)

2he

∙1 −1

−1 1

¸+(ue2 − ue1)2he

∙−1 −11 1

¸(ii)

Note that the tangent coefficient matrix is not symmetric.With the data in Eqs. (ii) and (iii) of Example 3.4.1, at the beginning of the Þrst iteration

we have (after assembly and imposition of the boundary conditions)∙2 −2

−2 4

¸½δU1δU2

¾= −

½0.250.50

¾−½

2− 2−2 + 4.4142− 3.4142

¾=

½−0.250.50

¾(iii)

The solution to these equations is

δU1 = 0.0, δU2 = 0.125, δU3 = 0.0 (iv)

and the complete solution becomes

U(1)1 = 1.0000, U(1)2 = 1.1250, U(1)3 = 1.4142 (v)

The solution at the second iteration is obtained using∙2 −2.25

−2 4.50

¸½δU1δU2

¾=

½0.0156

−0.0312

¾(vi)

The solution to these equations is

δU1 = 0.0, δU2 = −0.0007, δU3 = 0.0 (vii)

and the complete solution becomes

U(2)1 = 1.0000, U(2)2 = 1.1180, U(2)3 = 1.4142 (viii)

The root-mean-square error in the solution (with respect to the solution obtained in the Þrstiteration) [see Eq. (3.4.2)] is 0.0034.


Table 3.4.2 contains the numerical results obtained with the NewtonRaphson iterationmethod. The NewtonRaphson method gives converged solution in only three iterations asopposed to four iterations taken by the direct iteration method. For the mesh of 1 quadraticelement, the error criterion is met while the solution did not actually coincide with the exactsolution.

Table 3.4.2 Comparison of Þnite element solutions with the exact solutionof the problem in Example 3.4.3 (NewtonRaphson iteration).

x Iteration 2L 4L 1Q 2Q Exact

0.00 1 1.0000 1.0000 1.0000 1.00002 1.0000 1.0000 1.0000 1.00003 1.0000 1.0000 1.0000 1.0000 1.0000

0.25 1 1.0312 1.03122 1.0308 1.03083 1.0308 1.0308 1.0308

0.50 1 1.1250 1.1250 1.1241 1.12502 1.1180 1.1180 1.1187 1.11803 1.1180 1.1180 1.1187 1.1180 1.1180

0.75 1 1.2812 1.27662 1.2504 1.25023 1.2500 1.2500 1.2500


3.5.1 Introduction

The nonlinear formulations and solution procedures described above can beimplemented on a computer using Fortran language. Fortran is chosen heremainly because of its transparent nature (i.e. one can see how the statementscorrespond to the theory) compared to C++ language. Of course, once thelogic of implementation is understood, it is easier to carry out the actualcomputations in any suitable computational languages. The logical statementsfor each of the iterative methods are described here. It is necessary for thereader to familiarize with programs FEM1DV2 and FEM2DV2 described inAppendix 1 of the Þnite element book by Reddy [1] to fully understand thelogic and variable names used here. The listing of FEM1DV2 contains Fortranstatements for assembly, imposition of boundary conditions, and solution ofequations (banded symmetric and unsymmetric equation solvers are listed inAppendix 1 of this book). A general ßow chart of the nonlinear analysisprogram is given in Figure 3.5.1.

3.5.2 Preprocessor Unit

The following information is read in the preprocessor.

Initialize global Kij, fi

Iter = Iter + 1

DO 1 to n

Iter = 0

Impose boundary conditionsand solve the equations

CALL ELKF to calculate Kij(n)

and fi(n), and assemble to form

global Kij and Fi

Transfer global information(material properties, geometry, and solution)

to element

yes Iter < Itmax no

Error < ε yesno

STOP

Print solution

Write a message

STOP


1. Read or deÞne the model equation data

(a) DeÞne the coefficients a(x, u), b(x, u), and c(x, u) and source f(x) in themodel differential equation (3.1.1) (they are assumed to be polynomialsof certain degree); for example, assume

a(x, u) = ax0 + ax1x+ au1u+ au2u2 + aux1

du

dx+ aux2

µdu

dx

¶2b(x, u) = bx0 + bx1x+ bu1u+ bu2u

2 + bux1du

dx+ bux2

µdu

dx

¶2

Figure 3.5.1 A ßow chart of the nonlinear analysis program for the solutionof the model equation (3.1.1).


c(x, u) = cx0 + cx1x+ cu1u+ cu2u2 + cux1

du

dx+ cux2

µdu

dx

¶2f(x) = fx0 + fx1x+ fx2x

2 (3.5.1)

so that a large class of nonlinear one-dimensional problems may be solvedwith the program. The choice in Eq. (3.5.1) requires us to read the followingparameters (assuming continuous data for the whole domain)

AX0, AX1, AU1, AU2, AUX1, AUX2; BX0, BX1, BU1, etc.

(b) Read geometric and analysis data

X0 = global coordinate of the Þrst node of the domain (assumed to bea straight line),AL = domain size (length),IEL = element type (= 1, linear; = 2, quadratic),NEM = number of elements in the mesh,NONLIN = ßag for type of analysis (= 0,linear; > 0, nonlinear)EPS = allowable error tolerance ² for the convergence test,ITMAX = maximum allowable number of iterations for convergence.

(c) Read speciÞed boundary and initial data: NSPV , NSSV ,NSMB, ISPV (I, J), V SPV (I), ISSV (I, J), V SSV (I), ISMB(I, J),BETA0(I), BETAU(I), UREF (I), and GU0(I), whereNSPV = number of speciÞed primary degrees of freedom (U),NSSV = number of speciÞed secondary degrees of freedom (Q ≡nxa

dudx),

NSMB = number of speciÞed mixed boundary conditions[Q+ (β0 + βuU)(U − Uref)],

ISPV (I, J) = array of speciÞed primary degrees of freedom (I =1, 2, . . . , NSPV and J = 1, 2); ISPV (I, 1)= global node number atwhich the primary variable is speciÞed, and ISPV (I, 2) is the degree offreedom number at that node which is speciÞed (in the class of problemsdiscussed in this chapter, NDF = 1 and therefore ISPV (I, 2) = 1 forall cases).V SPV (I) = array of speciÞed values of primary variable DOF inISPV (I, J) [V SPV (I) should be in the same sequence as the speciÞeddegrees of freedom in ISPV (I)],ISSV (I, J) = similar to ISPV (I, J), except for secondary variables,V SSV (I) = similar to V SPV (I), except for secondary variables,ISMB(I, J) = array of speciÞed mixed boundary conditions (I =1, 2, . . . , NSSV ) and has meaning similar to arrays ISPV (I, J) andISSV (I, J),BETA0(I) = array of speciÞed values of β0,


BETA1(I) = array of speciÞed values of βu,UREF (I) = array of speciÞed values of Uref ,GU0(I) = initial (guess) solution vector (I = 1, 2, . . ., NEQ);NEQ=number of total primary degrees of freedom in the mesh.

2. Write out the input data and necessary warning messages.

3.5.3 Processor Unit

Here we discuss mainly the calculation of element coefficients, imposition ofvarious types of boundary conditions, and iterative algorithm. The number ofGauss points needed to evaluate the coefficients is determined by the highestpolynomial degree, p [NGP = (p+ 1)/2]. In the present case, the expressionthat dictates the degree of the polynomial is the integral involving c(x, u).If we assume c to be a linear function of u, and if u is approximated usinglinear polynomials, the integrand of the above integral expression is cubic(p = 3). Hence, a two-point integration is needed to evaluate it; for quadraticinterpolation of u, the integrand is a Þfth degree polynomial (p = 5), andhence it requires three-point Gauss quadrature. Of course, the polynomialdegrees goes up if a is a higher degree polynomial of x.The element coefficients are evaluated using the Gauss rule. Based on the

discussion presented in Section 2.7, the element coefficients in (3.3.3a), forβ = 0, can be expressed as

ELK(I, J) = ELK(I, J) + (AX ∗GDSFL(I) ∗GDSFL(J)+BX ∗ SFL(I) ∗GDSFL(J) + CX ∗ SFL(I) ∗ SFL(J))∗GJ ∗GAUSWT (NI,NGP )

which is evaluated inside a loop on NI = 1, 2, . . . , NGP and inner loops onI, J = 1, 2, . . . , NPE. Boxes 3.5.1 and 3.5.2 show the Fortran subroutines forthe calculation of [Ke] and fe.The imposition of boundary conditions on a banded system of non-

symmetric equations is discussed next. The assembled coefficient matrix[GLK] is stored a full band width (NBW ) form, and it is of the orderNEQ×NBW , the last column GLK(I,NBW ) being reserved for the sourcevector. Note that GLK(I,NHBW ) is the main diagonal of the assembledmatrix.

Essential boundary conditions

Now suppose that UI is speciÞed to be UI . Then the Ith equation of thesystem is replaced with the equation UI = UI . This is done as follows:

GLK(I, J) = 0.0 for all J = 1, 2, . . . , NBW and J 6= IGLK(I,NHBW ) = 1.0, GLK(I,NBW ) = UI (3.5.2)

This is repeated for all NSPV conditions.

SUBROUTINE COEFF (IEL,NPE,F0) IMPLICIT REAL*8(A-H,O-Z) DIMENSION GAUSS(5,5),WT(5,5) COMMON /SHP/ SFL(4),GDSFL(4) COMMON /STF/ STIF(3,3),ELF(3),ELX(3),ELU(3),AX0,AX1,AU

C DATA GAUSS/5*0.0D0,-0.57735027D0,0.57735027D0,3*0.0D0,

1 -0.77459667D0,0.0D0,0.77459667D0,2*0.0D0,-0.86113631D0,2 -0.33998104D0,0.33998104D0,0.86113631D0,0.0D0,3 -0.906180D0,-0.538469D0,0.0D0,0.538469D0,0.906180D0/

DATA WT /2.0D0,4*0.0D0,2*1.0D0,3*0.0D0,0.55555555D0,1 0.88888888D0,0.55555555D0,2*0.0D0,0.34785485D0,2 2*0.65214515D0,0.34785485D0,0.0D0,0.2369227D0,3 0.478629D0,0.568889D0,0.478629D0,0.236927D0/

C NGP=IEL+1EL=ELX(IEL+1)-ELX(1) DO 10 I=1,NPE

ELF(I)=0.0 DO 10 J=1,NPE

10 ELK(I,J)=0.0 C

DO 80 NI=1,NGP XI=GAUSS(NI,NGP) CALL SHAPE (EL,ELX,GJ,IEL,NPE,XI) CNST=GJ*WT(NI,NGP) X=0.0U=0.0 DO 20 I=1,NPE

X=X+SFL(I)*ELX(I) U=U+SFL(I)*ELU(I)

20 CONTINUEC

AX=AX0+AX1*X+AU*UDO 40 I=1,NPE ELF(I)=ELF(I)+F0*SFL(I)*CNST

DO 40 J=1,NPE S00=GDSFL(I)*GDSFL(J)*CNST ELK(I,J)=ELK(I,J)+AX*S00

40 CONTINUE 80 CONTINUE

RETURN END


Box 3.5.1 Subroutine for the generation of the coefficient matrices in thedirect iteration method.

SUBROUTINE COEFF (IEL,NPE,F0) IMPLICIT REAL*8(A-H,O-Z) DIMENSION GAUSS(5,5),WT(5,5),TANG(3,3) COMMON /SHP/ SFL(4),GDSFL(4) COMMON /STF/ STIF(3,3),ELF(3),ELX(3),ELU(3),AX0,AX1,AU

C DATA GAUSS/5*0.0D0,−0.57735027D0,0.57735027D0,3*0.0D0,

1 −0.77459667D0,0.0D0,0.77459667D0,2*0.0D0,−0.86113631D0,2 −0.33998104D0,0.33998104D0,0.86113631D0,0.0D0,3 −0.906180D0,−0.538469D0,0.0D0,0.538469D0,0.906180D0/

DATA WT /2.0D0,4*0.0D0,2*1.0D0,3*0.0D0,0.55555555D0,1 0.88888888D0,0.55555555D0,2*0.0D0,0.34785485D0,2 2*0.65214515D0,0.34785485D0,0.0D0,0.2369227D0,3 0.478629D0,0.568889D0,0.478629D0,0.236927D0/

C NGP=IEL+1EL=ELX(IEL+1)-ELX(1) DO 10 I=1,NPE

ELF(I)=0.0 DO 10 J=1,NPE

TANG(I,J)= 0.010 ELK(I,J) = 0.0

DO 80 NI=1,NGP XI=GAUSS(NI,NGP) CALL SHAPE (EL,ELX,GJ,IEL,NPE,XI) CNST=GJ*WT(NI,NGP) X=0.0U=0.0 DU=0.0 DO 20 I=1,NPE

X=X+SFL(I)*ELX(I) U=U+SFL(I)*ELU(I)DU=DU+GDSFL(I)*ELU(I)

20 CONTINUEAX=AX0+AX1*X+AU*UDO 40 I=1,NPE ELF(I)=ELF(I)+F0*SFL(I)*CNST

DO 40 J=1,NPE S00=GDSFL(I)*GDSFL(J)*CNST S10=GDSFL(I)*SFL(J)*CNSTELK(I,J)=ELK(I,J)+AX*S00 TANG(I,J)=TANG(I,J)+AU*DU*S10

40 CONTINUE 80 CONTINUEC C Compute the residual vector and tangent matrix C

DO 100 I=1,NPE DO 100 J=1,NPE

100 ELF(I)=ELF(I)-ELK(I,J)*ELU(J) DO 120 I=1,NPE

DO 120 J=1,NPE 120 ELK(I,J)=ELK(I,J)+TANG(I,J) C

RETURN END


Box 3.5.2 Subroutine for the generation of the coefficient matrices in theNewtonRaphson iteration method.


Natural boundary conditions

Next, suppose that QI is speciÞed to be QI . Then the force in the Ith equationis augmented with QI :

GLK(I,NBW ) = GLK(I,NBW ) + QI (3.5.3)

This is repeated for all NSSV conditions.

Mixed boundary conditions

Finally, we consider mixed boundary conditions of the form

QI + βI (UI − Uref) = 0 (3.5.4)

where βI is a function of u, say βI = β0I +β

1IUI . The boundary condition may

be implemented as follows:

GLK(I,NHBW ) = GLK(I,NHBW ) +³β0I + β

1IUI

´GLK(I,NBW ) = GLK(I,NBW ) +

³β0I + β

1IUI

Úref (3.5.5)

The above statement modiÞes the assembled direct stiffness matrix and sourcevector in direct iteration procedure. In the case of NewtonRaphson iteration,the assembled tangent matrix is modiÞed as follows:

GLK(I,NHBW ) = GLK(I,NHBW ) +³β0I + 2β

1IUI

´− β1IUref (3.5.6)

GLK(I,NBW ) = GLK(I,NBW )−³β0I + β

1IUI

ÚI +

³β0I + β

1IUI

Úref

Box 3.5.3 contains a listing of the subroutine that imposes various types ofboundary conditions on an banded, unsymmetric system of equations.The major steps of the two iterative algorithms are summarized below.

1. Read in the convergence tolerance TOLR, maximum allowable number ofiterations ITMAX, and the initial guess vector GUC(I).

2. Initialize the previous solution vector GUP = 0.3. Set the iteration counter, ITER = 0.

4. Begin the iteration counter on ITER = ITER+ 1.

5. Initialize the global coefficient matrix [GLK] (and source vector GLF ifused); when unsymmetric banded solver is used, the last column of [GLK]takes the place of GLF.

6. Calculate the element coefficient matrices [ELK] using the last iterationsolution GUC (or a weighted-average of GUC and GUP whenunderrelaxation is used) and element source vectors ELF and assembleto obtain [GLK] (and GLF).

SUBROUTINE BNDRYUNSYM(ITYPE,MXNEQ,MXFBW,MXEBC,MXNBC,NDF,NHBW,1 GLK, GLU,NSPV,NSSV,NSMB,ISPV,ISSV,ISMB,VSPV,VSSV,BETA0,BETAU,UREF)

C __________________________________________________________________ C C The subroutine is used to implement specified boundary conditions C on BANDED UNSYMMETRIC system of finite element equations C __________________________________________________________________ C

IMPLICIT REAL*8 (A-H,O-Z) DIMENSION ISPV(MXEBC,2),ISSV(MXNBC,2),ISMB(MXNBC,2),VSPV(MXEBC),

1 VSSV(MXEBC),UREF(MXNBC),BETA0(MXNBC),BETAU(MXNBC),2 GLU(MXNEQ),GLK(MXNEQ,MXFBW)

CNBW=2*NHBW

C C Include specified PRIMARY degrees of freedom C

IF(NSPV.NE.0)THEN DO 120 NP=1,NSPV

NB=(ISPV(NP,1)-1)*NDF+ISPV(NP,2) DO 110 J=1,NBW

110 GLK(NB,J)=0.0D0 GLK(NB,NHBW)=1.0D0

120 GLK(NB,NBW)=VSPV(NP)ENDIF

C C Modify the source vector to include specified non-zero SECONDARY VARIABLES C

IF(NSSV.NE.0)THEN DO 130 NS=1,NSSV

NB=(ISSV(NS,1)-1)*NDF+ISSV(NS,2) 130 GLK(NB,NBW)=GLK(NB,NBW)+VSSV(NS)

ENDIF C C Implement the specified MIXED BOUNDARY CONDITIONS C

IF(NSMB.NE.0)THEN DO 150 MB=1,NSMB NB=(ISMB(MB,1)-1)*NDF+ISMB(MB,2)IF(ITYPE.LE.1)THEN

GLK(NB,NHBW)=GLK(NB,NHBW)+BETA0(MB)+BETAU(MB)*GLU(NB)GLK(NB,NBW)=GLK(NB,NBW)+UREF(MB)*(BETA0(MB)+BETAU(MB)*GLU(NB))

ELSEGLK(NB,NHBW)=GLK(NB,NHBW)+BETA0(MB)+2.0*BETAU(MB)*GLU(NB)

* -UREF(MB)*BETAU(MB)GLK(NB,NBW)=GLK(NB,NBW)+UREF(MB)*(BETA0(MB)+BETAU(MB)*GLU(NB))

* -(BETA0(MB)+BETAU(MB)*GLU(NB))*GLU(NB)ENDIF

150 CONTINUEENDIFRETURNEND


Box 3.5.3 Subroutine for the implementation of various types of boundaryconditions.


7. Impose the boundary conditions (the essential boundary conditions mustbe homogeneous).

8. Solve the equations (solution is returned in the last column of [GLK]).

9a. In direct iteration, update the current and previous solution vectorsGUP = GUC and GUC = GLF (GLF = [GLK]last).

9b. In NewtonRaphson iteration, update the current and previous solutionvectors GUP = GUC and GUC = GUC+ GLF.

10. Calculate the residual or solution error, ERROR [see Eqs. (3.4.2)(3.4.3)].

11. If ERROR ≤ TOLR print ITER,ERROR, and GUC and stop. IfERROR > TOLR, continue.

12. If ITER > ITMAX, stop; otherwise, repeat steps 4 through 11.

Example 3.5.1

Consider heat transfer in an isotropic bar of length L = 0.18 m. The surface of the baris insulated so that there is no convection from the surface. The governing equation of theproblem is the same as Eq. (3.1.1) with b = c = 0, u = T , the temperature. The conductivitya(x,T ) = k is assumed to vary according to the relation

k = k0 (1 + βT ) (3.5.7)

where k0 is the constant thermal conductivity [k0 = 0.2 W/(mK)] and β the temperature

coefficient of thermal conductivity [β = 2× 10−3 (C−1)]. Suppose that there is no internalheat generation (i.e. f = 0) and the boundary conditions are

T (0) = 500K, T (L) = 300K (3.5.8)

Table 3.5.1 shows the linear and nonlinear solutions T (x). The results obtained with thedirect iteration method and NewtonRaphson method with ² = 0.001 are tabulated in Table3.5.1. In both methods, the convergent solution was obtained for three iterations. Bothmethods and both meshes give the same solution.

Table 3.5.1 Finite element solutions of a nonlinear heat conduction equation.

DI/NR Direct Iteration NewtonRaphson

x Linear 8L 4Q 8L 4Q

0.0000 500.00 500.00 500.00 500.00 500.000.0225 475.00 477.24 477.24 477.24 477.240.0450 450.00 453.94 453.94 453.94 453.940.0675 425.00 430.06 430.06 430.05 430.050.0900 400.00 405.54 405.54 405.54 405.540.1125 375.00 380.35 380.35 380.34 380.340.1350 350.00 354.40 354.40 354.40 354.400.1575 325.00 327.65 327.65 327.65 327.650.1800 300.00 300.00 300.00 300.00 300.00


3.6 Closing Remarks

In this chapter, Þnite element formulations of nonlinear boundary-valueproblems in one dimension are discussed, iterative methods of solving thenonlinear algebraic equations are studied, and computer implementation ofthe nonlinear Þnite element analysis is outlined. The simple class of problemsdiscussed herein should give the reader a clear understanding of the workingsof nonlinear Þnite element analysis steps. In the next chapter we will considermulti-variable, one-dimensional, nonlinear equations governing bending andstretching of straight beams.

Problems

3.1 Consider the second-order differential equation

− d

dx

³µdu

dx

´= f(x), µ = µ0

³du

dx

ń−1(a)

where u(x) is the dependent unknown, f(x) is a known function of position x, and µis a function of the dependent variable, as given in Eq. (a). Write the Þnite elementmodel and derive the tangent stiffness matrix coefficients.

3.2 Compute the tangent matrix for the case a(x,u) = ae0(x)dudx, c(x, u) = 0, and β = 0

(see Example 3.4.2).

3.3 Consider the nonlinear differential equation

− d

dx

h(u+

√2)du

dx

i= 1, 0 < x < 1 (a)

du

dx(0) = 0, u(1) = 0 (b)

Analyze the nonlinear problem using the Þnite element method with direct iterationprocedure. Tabulate the nodal values of u(x) for 4 and 8 linear elements and 2 and 4quadratic elements.

3.4 Formulate Problem 3.2 with NewtonRaphson iteration procedure, and compute thetangent coefficient matrix. Tabulate the nodal values of u(x) for 4 and 8 linear elementsand 2 and 4 quadratic elements.

3.5 Formulate the nonlinear differential equation

−d2u

dx2+ 2u3 = 0, 1 < x < 2 (a)

u(1) = 1,hdu

dx+ u2

ix=2

= 0 (b)

using the Þnite element method, and solve the problem using direct iterationprocedure. Tabulate the nodal values of u(x) for 4 and 8 linear elements and 2 and 4quadratic elements. The exact solution is given by u(x) = 1/x.

3.6 Compute the tangent stiffness matrix associated with Problem 3.5, and solve it withNewtonRaphson iteration procedure. Tabulate the nodal values of u(x) for 4 and 8linear elements and 2 and 4 quadratic elements.



−d2u

dx2− 2udu

dx= 0, 0 < x < 1 (a)

u(0) = 1, u(1) = 0.5 (b)

using the Þnite element method, and solve the problem using direct iterationprocedure. Tabulate the nodal values of u(x) for 4 and 8 linear elements and 2 and 4quadratic elements. The exact solution is given by u(x) = 1/(1 + x).


3.9 Formulate the nonlinear differential equation in Problem 3.7 subject to the boundaryconditions

u(0) = 1,hdu

dx+ u2

ix=1

= 0 (a)



3.11 Formulate the nonlinear differential equation in Problem 3.7 subject to the boundaryconditions h

du

dx+ 2u

ix=0

= 1,hdu

dx+ u2

ix=1

= 0 (a)




−d2u

dx2−³du

dx

´3= 0, 0 < x < 1 (a)

hdu

dx+ uix=0

=3√2,

hdu

dx

ix=1

= 0.5 (b)

using the Þnite element method, and solve the problem using direct iterationprocedure. Tabulate the nodal values of u(x) for 4 and 8 linear elements and 2 and 4

quadratic elements. The exact solution is given by u(x) =p2(1 + x).

3.14 Consider simultaneous steady-state conduction and radiation in a plate. Themathematical formulation of the problem in non-dimensional form is given by

− ddξ

³k(θ)

dθ

dξ

´= 0, 0 < ξ < ξ0 (a)

θ(0) = θ0, θ(ξ0) = 1.0 (b)


where ξ is the non-dimensional thickness coordinate, θ is the non-dimensionaltemperature, and k is the conductivity

k(θ) = k0

³1 +

4

3Nθ3´

(c)

and N is called the conduction-to-radiation parameter. Develop the Þnite elementmodel of the problem and compute the element tangent matrix.

3.15 Analyze the nonlinear problem in Problem 3.14 using the Þnite element method withNewtonRaphson iteration procedure. Take k0 = 1, N = 0.01, ξ0 = 1, and θ0 = 0.5.Tabulate and plot the nodal values of θ(ξ) for 8 linear elements and 4 quadraticelements.

3.16 The explosion of a solid explosive material in the form of an inÞnite cylinder may bedescribed by [3]

−1r

d

dr

³rdu

dr

´= 2eu, 0 < r < 1

subject to the boundary conditions

At r = 0 :du

dr= 0; At r = 1 : u = 0

The exact solution of the nonlinear equation is

u(r) = ln4

(1 + r2)2

Analyze the nonlinear problem using the Þnite element method with (a) NewtonRaphson and (b) direct iteration procedure. Tabulate and plot the nodal values of u(r)for 8 linear elements and 4 quadratic elements and compare with the exact solution.

3.17 Redo Problem 3.16 when the right-hand side is replaced by eu. This problem has twosolutions

ui(r) = ln8λi

(1 + λir2)2, i = 1,2

where λi are the roots of the equation

8λ

(1 + λ)2= 1

3.18 Heat and mass transfer within a porous catalyst particle is described by [3]

d2u

dr2+a

r

du

dr= b2u exp

∙c(1− u)

1 + d(1− u)

¸, 0 < r < 1


At r = 0 :du

dr= 0; At r = 1 : u = 1

Here a, b, c, and d are the problem parameters. Analyze the nonlinear problem usingthe Þnite element method with the direct iteration procedure. Take a = 0, b = 1, c = 2and d = 0.1, and tabulate and plot the solutions u(r) for meshes of 8 linear elementsand 4 quadratic elements.

3.19 Repeat Problem 3.18 with the data a = 2, b = 2, c = 4 and d = 0.2.


3.20 Repeat Problem 3.18 with the data a = 0, b = 0.16, c = 14 and d = 0.7 The problemhas three possible numerical solutions.

3.21 Axial mixing in an isothermal tubular reactor where a second-order reaction occurs isdescribed by [3]

− 1

Pe

d2u

dx2+du

dx+Da u2 = 0, 0 < x < 1


At x = 0 : u = 1+1

Pe

du

dx; At x = 1 :

du

dx= 0

where Pe and Da are the problem parameters. Analyze the nonlinear problem usingthe Þnite element method with the direct iteration procedure. Take Pe=5 and Da=1,and tabulate and plot the solution u(x) for 8 linear elements.

3.22 Consider the pair of nonlinear differential equations

d3u

dx3+1

6

µd2u

dx2

¶2vdv

dx= 0,

d2v

dx2+1

2

du

dx

³dv

dx

´3= 0


u(0) = v(0) = 1, u(1) = 16,dv

dx(0) = −1, v(1) = 0.5

Formulate the Þnite element model of the equations and compute the tangentcoefficient matrix.

3.23 Consider the pair of nonlinear differential equations

− 1

Pe

d2u

dx2+du

dx+ β (u− uc) = αDa (1− v) exp

µu

1 + u/γ

¶,

− 1

Pe

d2v

dx2+dv

dx= Da (1− v) exp

µu

1 + u/γ

¶in (0,1) and subject to the boundary conditions

At x = 0 : Pe u =du

dx, Pe v =

dv

dx

At x = 1 :du

dx= 0,

dv

dx= 0

Formulate the Þnite element model of the equations and compute the tangentcoefficient matrix.


2. Sachdev, P. L., A Compendium on Nonlinear Ordinary Differential Equations, JohnWiley, New York (1997).

3. Kubicek, M. and Hlavacek, V., Numerical Solution of Nonlinear Boundary ValueProblems with Applications, PrenticeHall, Englewood Cliffs, NJ (1983).


4

Nonlinear Bendingof Straight Beams

4.1 Introduction

In this chapter, we consider a slightly more complicated one-dimensionalnonlinear problem than that was discussed in Chapter 1. A beam is astructural member whose length to cross-sectional dimensions is very large andit undergoes not only stretching along its length but also bending about anaxis transverse to the length. When the applied loads on the beam are large,the linear loaddeßection relationship ceases to be valid because the beamdevelops internal forces that resist deformation, and the magnitude of internalforces increases with the loading as well as the deformation. This nonlinearloaddeßection response of straight beams is the topic of this section.

In developing a general nonlinear formulation of beams, straight orcurved, one must deÞne the measures of stress and strain consistent withthe deformations accounted for in the formulation. Such a formulation,called a continuum formulation, will be discussed in Chapter 9. The presentnonlinear formulation of straight beams is based on assumptions of largetransverse displacements, small strains and small to moderate rotations. Theseassumptions allow us to use the stress measure of force per unit undeformedarea and strain measure of change in length to the original length (and inthe case of shear strain, change in the angle from π/2). The changes in thegeometry are small so that no distinction between the PiolaKirchoff andCauchy stresses (to be discussed in detail later) will be made. The nonlinearityin the formulation comes solely from the inclusion of the inplane forces thatare proportional to the square of the rotation of the transverse normal to thebeam axis.

Two different theories to model the kinematic behavior of beams areconsidered here: (1) the EulerBernoulli beam theory (EBT) that neglectsthe transverse shear strain, and (2) the Timoshenko beam theory (TBT),which accounts for the transverse shear strain in the simplest way. In each


case, we begin with an assumed displacement Þeld and compute strains thatare consistent with the kinematic assumptions of the theory. Then we developthe weak forms using the principle of virtual displacements and the associateddisplacement Þnite element model. We also discuss certain computationalaspects (e.g. membrane and shear locking) and iterative methods for theproblems at hand. Computer implementation issues are also presented.Discussion of other linear Þnite element models of the Timoshenko beamtheory are also presented for completeness.

4.2 EulerBernoulli Beams

4.2.1 Basic Assumptions

For the sake of completeness, the governing equations of the nonlinear bendingof beams are developed from basic considerations. The classical beam theoryis based on the EulerBernoulli hypothesis that plane sections perpendicularto the axis of the beam before deformation remain (a) plane, (b) rigid (notdeform), and (c) rotate such that they remain perpendicular to the (deformed)axis after deformation. The assumptions amount to neglecting the Poissoneffect and transverse strains. A reÞned theory is that due to Timoshenko, andit will be discussed in the sequel. The principle of virtual displacements willbe used to formulate the variational problem and associated Þnite elementmodel.

4.2.2 Displacement Field and Strains

The bending of beams with moderately large rotations but with small strainscan be derived using the displacement Þeld

u1 = u0(x)− zdw0dx, u2 = 0, u3 = w0(x) (4.2.1)

where (u1, u2, u3) are the total displacements along the coordinate directions(x, y, z), and u0 and w0 denote the axial and transverse displacements of apoint on the neutral axis.

Using the nonlinear strain-displacement relations (sum on repeatedsubscripts is implied; see Chapter 9)

εij =1

2

Ã∂ui∂xj

+∂uj∂xi

!+1

2

Ã∂um∂xi

∂um∂xj

!(4.2.2)

and omitting the large strain terms but retaining only the square of du3/dx(which represents the rotation of a transverse normal line in the beam), weobtain

NONLINEAR BENDING OF STRAIGHT BEAMS 89

ε11 = εxx =du0dx− z d

2w0dx2

+1

2

µdw0dx

¶2=

"du0dx

+1

2

µdw0dx

¶2#− z

Ãd2w0dx2

!≡ ε0xx + zε

1xx (4.2.3a)

ε0xx =du0dx

+1

2

µdw0dx

¶2, ε1xx = −

d2w0dx2

(4.2.3b)

and all other strains are zero. Note that the notation x1 = x, x2 = y, andx3 = z is used. These strains are known as the von Karman strains.

4.2.3 Weak Forms

The weak form of structural problems can be directly derived (i.e. withoutknowing the governing differential equations) using the principle of virtualdisplacements. The principle states that if a body is in equilibrium, thetotal virtual work done by actual internal as well as external forces inmoving through their respective virtual displacements is zero. The virtualdisplacements are arbitrary except that they are zero where displacementsare prescribed. The analytical form of the principle over a typical elementΩe = (xa, xb) (see Figure 4.2.1) is given by (see Reddy [2])

δW e ≡ δW eI − δW e

E = 0 (4.2.4)

where δW eI is the virtual strain energy stored in the element due to actual

stresses σij in moving through the virtual strains δεij , and δW eE is the work

done by externally applied loads in moving through their respective virtualdisplacements. Here σij and εij denote the Cartesian components of the stressand the Green strain tensors, respectively. Due to the assumption of smallstrains, no distinction will be made here between the Cauchy and secondPiola—Kirchhoff stress tensors (see Chapter 9).

For the beam element, we have

δW eI =

ZV e

δεij σij dV

δW eE =

Z xb

xaq δw0 dx+

Z xb

xaf δu0 dx+

6Xi=1

Qei δ∆ei

(4.2.5)

where V e denotes the element volume, q(x) is the distributed transverse load(measured per unit length), f(x) is the distributed axial load (measured perunit length), Qei are the generalized nodal forces, and δ∆ei are the virtual

2he

∆1 ∆4

∆3

∆2 ∆5

∆6

1 2

11Q

2Q

3Q

4Q

5Q

6Q

(a) (b)

he


generalized nodal displacements of the element (see Figure 4.2.1) deÞned by

∆e1 = u0(xa), ∆e2 = w0(xa), ∆

e3 =

∙−dw0dx

¸xa

≡ θ(xa)

∆e4 = u0(xb), ∆e5 = w0(xb), ∆

e6 =

∙−dw0dx

¸xb

≡ θ(xb) (4.2.6a)

Qe1 = −Nxx(xa), Qe4 = Nxx(xb)

Qe2 = −∙dw0dxNxx +

dMxx

dx

¸xa

, Qe5 =

∙dw0dxNxx +

dMxx

dx

¸xb

Qe3 = −Mxx(xa), Qe6 =Mxx(xb) (4.2.6b)

In view of the explicit nature of the assumed displacement Þeld (4.2.5) inthe thickness coordinate z and its independence of coordinate y, the volumeintegral can be expressed as a product of integrals over the length and area ofthe element: Z

V e(·) dV =

Z xb

xa

ZAe(·) dA dx

Therefore, the expression for the virtual strain energy can be simpliÞed asfollows (only non-zero components of strain and stress are ε11 ≡ εxx andσ11 ≡ σxx)

δW eI =

Z xb

xa

ZAeδεxx σxx dA dx =

Z xb

xa

ZAe

³δε0xx + zδε

1xx

´σxx dA dx

=

Z xb

xa

ZAe

"µdδu0dx

+dw0dx

dδw0dx

¶− z d

2δw0dx2

#σxx dA dx

=

Z xb

xa

"µdδu0dx

+dw0dx

dδw0dx

¶Nxx − d

2δw0dx2

Mxx

#dx (4.2.7)

Figure 4.2.1 The EulerBernoulli beam Þnite element with generalizeddisplacement and force degrees of freedom. (a) Nodaldisplacements. (b) Nodal forces.


whereNxx is the axial force (measured per unit length) andMxx is the moment(measured per unit length)

Nxx =

ZAeσxx dA , Mxx =

ZAeσxxz dA (4.2.8)

The virtual work statement in Eq. (4.2.7) becomes

0 =

Z xb

xa

"µdδu0dx

+dw0dx

dδw0dx

¶Nxx − d

2δw0dx2

Mxx

#dx

−Z xb

xaq(x) δw0(x) dx−

Z xb

xaf(x) δu0(x) dx−

6Xi=1

Qei δ∆ei (4.2.9)

The above weak form is equivalent to the following two statements, whichare obtained by collecting terms involving δu0 and δw0 separately [see thedeÞnitions in Eqs. (4.2.6a,b)]:

0 =

Z xb

xa

µdδu0dx

Nxx − δu0f(x)¶dx−Qe1 δ∆e1 −Qe4 δ∆e4 (4.2.10a)

0 =

Z xb

xa

"dδw0dx

µdw0dxNxx

¶− d

2δw0dx2

Mxx − δw0 q(x)#dx

−Qe2 δ∆e2 −Qe3 δ∆e3 −Qe5 δ∆e5 −Qe6 δ∆e6 (4.2.10b)

The differential equations governing nonlinear bending of straight beamscan be obtained, although not needed for Þnite element model development,from the virtual work statement in (4.2.9), equivalently, the weak forms(4.2.10a,b), or from a vector approach in which forces and moments aresummed over a typical beam element.Integration by parts of the expressions in (4.2.9) to relieve δu0 and δw0 of

any differentiation results in

0 =

Z xb

xa

(µ−dNxxdx

− f¶δu0 −

"d

dx

µdw0dxNxx

¶+d2Mxx

dx2+ q

#δw0

)dx

+

∙Nxxδu0 +

µdw0dxNxx +

dMxx

dx

¶δw0 −Mxx

dδw0dx

¸xbxa

−6Xi=1

Qei δ∆ei

Since δu0 and δw0 are arbitrary and independent of each other in xa < x < xbas well as at x = xa and x = xb (independently), it follows that the governingequations of equilibrium, known as the Euler equations, are

δu0 : − dNxxdx

= f(x) (4.2.11a)

δw0 : − d

dx

µdw0dxNxx

¶− d

2Mxx

dx2= q(x) (4.2.11b)

q(x)

z

x

f(x)

Mxx + ∆Mxx

q(x)

Mxx

V + ∆VV

dwdx

x

z

Nxx

Nxx+∆Nxx

∆x


In view of the deÞnitions (4.2.6a), deÞnitions (4.2.6b) are obtained as thenatural (or force) boundary conditions

Qe1 +Nxx(xa) = 0, Qe4 −Nxx(xb) = 0Qe2 +

∙dw0dxNxx +

dMxx

dx

¸xa

= 0, Qe5 −∙dw0dxNxx +

dMxx

dx

¸xb

= 0

Qe3 +Mxx(xa) = 0, Qe6 −Mxx(xb) = 0 (4.2.12)

The vector approach involves identifying a typical beam element of length∆x with all its forces and moments, summing them, and taking the limit∆x → 0. Consider the beam element shown in Figure 4.2.2, where Nxx isthe internal axial force, V (x) is the internal vertical shear force, Mxx is theinternal bending moment, f(x) is the external axial force, and q(x) is externaldistributed transverse load. Summing the forces in the x and z coordinatedirections, and moments about the y axis, we obtain

XFx = 0 : −Nxx + (Nxx +∆Nxx) + f(x)∆x = 0XFz = 0 : − V + (V +∆V ) + q(x)∆x = 0XMy = 0 : −Mxx + (Mxx +∆Mxx)− V∆x+Nxx∆xdw0

dx+ q(x)∆x(c∆x) = 0

Figure 4.2.2 A typical beam element with forces and moments to deriveequations of equilibrium using the vector approach.


Taking the limit ∆x → 0, we obtain the following three equations:

dNxxdx

+ f(x) = 0

dV

dx+ q(x) = 0

dMxx

dx− V +Nxx

dw0dx

= 0

which are equivalent to the two equations in (4.2.11a,b). Note that V is theshear force on a section perpendicular to the x-axis, and it is not equal to theshear force Q(x) acting on the section perpendicular to the deformed beam.In fact, one can show that V = Q+Nxx(dw0/dx).If one starts with the governing equations (4.2.11a,b), their weak forms can

be developed using the usual three-step procedure:

0 =

Z xb

xav1

µ−dNxxdx− f

¶dx

=

Z xb

xa

µdv1dxNxx − v1f

¶dx− [v1Nxx]xbxa

=

Z xb

xa

µdv1dxNxx − v1f

¶dx− v1(xa)[−Nxx(xa)]− v1(xb)Nxx(xb) (4.2.13a)

0 =

Z xb

xav2

"− ddx

µdw0dxNxx

¶− d

2Mxx

dx2− q

#dx

=

Z xb

xa

"dv2dx

µdw0dxNxx

¶− d

2v2dx2

Mxx − v2q#dx

−∙v2

µdw0dxNxx +

dMxx

dx

¶¸xbxa

−∙µ−dv2dx

¶Mxx

¸xbxa

=

Z xb

xa

"dv2dx

µdw0dxNxx

¶− d

2v2dx2

Mxx − v2q#dx

− v2(xa)∙−µdw0dxNxx −

dMxx

dx

¶¸xa

− v2(xb)∙dw0dxNxx +

dMxx

dx

¸xb

−∙−dv2dx

¸xa

[−Mxx(xa)]−∙−dv2dx

¸xb

Mxx(xb) (4.2.13b)

where v1 and v2 are the weight functions, whose meaning is obvious if theexpressions fv1dx and qv2dx are to represent the work done by external forces.We see that v1 ∼ δu0 and v2 ∼ δw0. Clearly, Eqs. (4.2.13a,b) are the same,with the definitions (4.2.6b) or (4.2.12), as those in Eqs. (4.2.10a,b).The resultant force Nxx and momentMxx can be expressed in terms of the

displacements once the constitutive behavior is assumed. Suppose that the


beam is made of a linear elastic material. Then the total stress is related tothe total strain by Hooke’s law

σxx = Eeεxx (4.2.14)

Then we have

Nxx =

ZAe

σxx dA =

ZAeEeεxx dA

=

ZAeEe"du0dx

+1

2

µdw0dx

¶2− z d

2w0dx2

#dA

= Aexx

"du0dx

+1

2

µdw0dx

¶2#−Bexx

d2w0dx2

(4.2.15a)

Mxx =

ZAe

σxx z dA =

ZAeEeεxx z dA

=

ZAeEe"du0dx

+1

2

µdw0dx

¶2− z d

2w0dx2

#z dA

= Bexx

"du0dx

+1

2

µdw0dx

¶2#−Dexx

d2w0dx2

(4.2.15b)

where Aexx, Bexx, and Dexx are the extensional, extensional-bending, and

bending stiffnesses of the beam element

(Aexx, Bexx, D

exx) =

ZAeEe³1, z, z2

´dA (4.2.16)

For beams made of an isotropic material, the extensional—bending stiffnessBexx is zero when the x-axis is taken along the geometric centroidal axis. Wehave Bexx = 0, A

exx = E

eAe, and Dexx = EeIe, where Ae and Ie are the cross-

sectional area and second moment of inertia (about the y-axis) of the beamelement. For simplicity, we shall omit the element label e on the variables.In general, Axx, Bxx, and Dxx are functions of x whenever the modulus Eand/or the cross-sectional area is a function of x.The virtual work statements (4.2.10a,b) can be expressed in terms of the

generalized displacements (u0, w0) by using Eqs. (4.2.15a,b). We have

0 =

Z xb

xaAxx

dδu0dx

"du

dx+1

2

µdw0dx

¶2#dx−

Z xb

xaf(x)δu0 dx

−Q1 δu0(xa)−Q4 δu0(xb) (4.2.17a)

0 =

Z xb

xa

(Axx

dδw0dx

dw0dx

"du0dx

+1

2

µdw0dx

¶2#+Dxx

dδ2w0dx2

d2w0dx2

)dx

−Z xb

xaqδw0 dx−Q2 δw0(xa)−Q3 δθ(xa)−Q5 δw0(xb)−Q6 δθ(xb)

(4.2.17b)


where it is assumed that the coupling coefficient Bxx is zero because of thechoice of the coordinate system; that is, the x-axis is assumed to coincide withthe geometric centroidal axis

RA z dA = 0.


Let the axial displacement u0(x) and transverse deflection w0(x) areinterpolated as [θ = −(dw0/dx)]

u0(x) =2Xj=1

ujψj(x) , w0(x) =4Xj=1

∆jφj(x) (4.2.18)

∆1 ≡ w0(xa), ∆2 ≡ θ(xa), ∆3 ≡ w0(xb), ∆4 ≡ θ(xb) (4.2.19)

and ψj are the linear Lagrange interpolation functions, and φj are the Hermitecubic interpolation functions. For a linear problem, this element gives exactnodal displacements ui and ∆i for any f(x) and q(x) when Axx and Dxxare element-wise constants. Then the element is said to be a superconvergentelement.Substituting Eq. (4.2.18) for u0(x), (4.2.19) for w0(x), and δu0(x) = ψi(x)

and δw0(x) = φi(x) (to obtain the ith algebraic equation of the model) intothe weak forms (4.2.17a,b), we obtain

0 =2Xj=1

K11ij uj +

4XJ=1

K12iJ ∆J − F 1i (i = 1, 2) (4.2.20a)

0 =2Xj=1

K21Ij uj +

4XJ=1

K22IJ∆J − F 2I (I = 1, 2, 3, 4) (4.2.20b)

where

K11ij =

Z xb

xaAxx

dψidx

dψjdx

dx , K12iJ =

1

2

Z xb

xa

µAxx

dw0dx

¶dψidx

dφJdx

dx

K21Ij =

Z xb

xaAxx

dw0dx

dφIdx

dψjdx

dx , K21Ij = 2K

12jI

K22IJ =

Z xb

xaDxx

d2φIdx2

d2φJdx2

dx+1

2

Z xb

xa

"Axx

µdw0dx

¶2# dφIdx

dφJdxdx

F 1i =

Z xb

xaf ψi dx+ Qi , F 2I =

Z xb

xaq φI dx+ QI (4.2.21)

for (i, j = 1, 2) and (I, J = 1, 2, 3, 4), where Q1 = Q1, Q2 = Q4, Q1 = Q2,Q2 = Q3, Q3 = Q5, and Q4 = Q6. See Eq. (4.2.6b) for the definitions of Qi.Note that the coefficient matrices [K12], [K21] and [K22] are functions of the


unknown w0(x). Also, note that [K12]T 6= [K21]; hence, the element direct

stiffness matrix is unsymmetric.

The above deÞnition of coefficients Kαβij is based on a particular

linearization of Eqs. (4.2.17a,b) , which is probably the most natural. Otherforms of linearization are possible. For example, if consider Eq. (4.2.17a).the coefficient of dδu0/dx contains a linear term and a nonlinear term. Topreserve the linear bar stiffness, the linear term should be kept as a part ofthe stiffness matrix. The nonlinear term can be either included in the stiffnesscoefficient, as is done in the deÞnition given in Eq. (4.2.21), or the wholenonlinear term may be assumed to be known from the previous iteration.In the latter case, the term ends up in the load vector F 1. This choiceof linearization is known to slow down the convergence. In the case of Eq.(4.2.17b), we know that the term dw0/dx outside the square brackets is due tothe nonlinear strain. Hence, it was linearized (i.e. calculated using the solutionfrom the previous iteration) in deÞningK21

ij of Eq. (4.2.21). One may linearize

Eq. (4.2.17b) such that du0/dx+0.5(dw0/dx)2 is calculated using the solution

from the previous iteration. In that case K21ij = 0 and K

22ij will have additional

contribution. Thus, it is possible to computationally decouple the equationsfor u and ∆ and solve the two equations iteratively, feeding the solutionfrom one equation to the other. However, such a strategy often results innonconvergence.Equations (4.2.20a,b) can be written compactly as

2Xγ=1

Xp=1

Kαγip ∆

γp = F

αi , or

2Xp=1

Kα1ip up +

4XP=1

Kα2iP ∆P = F

αi (4.2.22)

In matrix form, we have∙[K11] [K12][K21] [K22]

¸½ ∆1∆2

¾=

½ F 1F 2

¾(4.2.23)

where∆1i = ui, i = 1, 2; ∆2i = ∆i, i = 1, 2, 3, 4 (4.2.24)

Note that the direct stiffness matrix is unsymmetric only due to the factthat [K12] contains the factor 1/2 whereas [K21] does not. One way to make[K21]T = [K12] is to split the linear strain du0/dx in Eq. (4.2.17) into twoequal parts and take one of the two parts as known from a previous iteration:Z xb

xa

(Aexx

dδw0dx

dw0dx

"du0dx

+1

2

µdw0dx

¶2#)dx

=1

2

Z xb

xaAexx

(dw0dx

dδw0dx

du0dx

+

"du0dx

+

µdw0dx

¶2# dδw0dx

dw0dx

)dx

(4.2.25)


The Þrst term of the above equation constitutes [K21] and the second oneconstitutes a part of [K22]. The symmetrized equations are∙

[K11] [K12][K21] [K22]

¸½ u∆

¾=

½ F 1F 2

¾(4.2.26)

where

K11ij = K

11ij =

Z xb

xaAexx

dψidx

dψjdx

dx

K12iJ = K

12iJ =

1

2

Z xb

xa

µAexx

dw0dx

¶dψidx

dφJdx

dx

K21Ij =

1

2

Z xb

xa

µAexx

dw0dx

¶dφIdx

dψjdx

dx , K21Ij = K

12jI

K22IJ =

Z xb

xaDexx

d2φIdx2

d2φJdx2

dx

+1

2

Z xb

xaAexx

"du0dx

+

µdw0dx

¶2# dφIdx

dφJdx

dx (4.2.27)

Note that in the symmetrized case, we must assume that u0(x) is also knownfrom a previous iteration.

4.2.5 Iterative Solutions of Nonlinear Equations

The direct iteration and NewtonRaphson methods introduced in Chapter 3are revisited here in connection with the nonlinear Þnite element equations ofthe EBT. Consider the nonlinear equations (4.2.23), which can be written as

[Ke(∆e)]∆e = F e (4.2.28)

where

∆e1 = u1, ∆e2 = ∆

e1, ∆

e3 = ∆

e2, ∆

e4 = u2, ∆

e5 = ∆

e3 , ∆

e6 = ∆

e4 (4.2.29a)

F e1 = F11 , F

e2 = F

21 , F

e3 = F

22 , F

e4 = F

12 , F

e5 = F

23 , F

e6 = F

24 (4.2.29b)

The system (4.2.28) of nonlinear algebraic equations can be linearized using thedirect iteration and NewtonRaphson iterative methods of Section 3.4. Theseare presented next. Note that the linearized equations may be symmetricor unsymmetric, depending on the formulation, and therefore an appropriateequation solver must be used. On the other hand, an unsymmetric bandedequations solver may be used in all cases.


Direct iteration procedure

In the direct iteration procedure, the solution at the rth iteration is determinedfrom the assembled set of equations

[K(∆(r−1))]∆r = F or [K(∆(r−1))]∆r = F (4.2.30)

where the direct stiffness matrix [Ke] is evaluated at the element level usingthe known solution ∆e(r−1) at the (r − 1)st iteration.NewtonRaphson iteration procedure

In the NewtonRaphson procedure, the linearized element equation is of theform

[T (∆(r−1)]∆r = −R(∆(r−1)) = F− ([Ke]∆e)(r−1) (4.2.31)

where the tangent stiffness matrix [T e] associated with the EulerBernoullibeam element is calculated using the deÞnition

[T ] ≡µ∂R∂∆

¶(r−1), or T eij ≡

Ã∂Rei∂∆ej

!(r−1)(4.2.32)

The solution at the rth iteration is then given by

∆r = ∆(r−1) + δ∆ (4.2.33)

Although the direct stiffness matrix [Ke] is unsymmetric, it can be shownthat the tangent stiffness matrix [T e] is symmetric. Further, it can be shownthat the tangent stiffness matrix is the same whether one uses [Ke] or [Ke][see Eqs. (4.2.23) and (4.2.26)].The coefficients of the element tangent stiffness matrix [T e] can be

computed using the deÞnition in (4.2.32). In terms of the components deÞnedin Eq. (4.2.22), we can write

Tαβij =

⎛⎝∂Rαi∂∆βj

⎞⎠(r−1) (4.2.34)

for α,β = 1, 2. The components of the residual vector can be expressed as

Rαi =2Xγ=1

Xp=1

Kαγip ∆

γp − Fαi

=2Xp=1

Kα1ip ∆

1p +

4XP=1

Kα2iP∆

2P − Fαi

=2Xp=1

Kα1ip up +

4XP=1

Kα2iP ∆P − Fαi (4.2.35)


Note that the range of p is dictated by the size of the matrix [Kαβ]. We have

Tαβij =

⎛⎝∂Rαi∂∆βj

⎞⎠ = ∂

∂∆βj

⎛⎝ 2Xγ=1

Xp=1

Kαγip ∆

γp − Fαi

⎞⎠=

2Xγ=1

Xp=1

⎛⎝Kαγip

∂∆γp

∂∆βj+∂Kαγ

ip

∂∆βj∆γp

⎞⎠= Kαβ

ij +2Xp=1

∂

∂∆βj

³Kα1ip

úp +

4XP=1

∂

∂∆βj

³Kα2iP

´∆P (4.2.36)

We compute the tangent stiffness matrix coefficients Tαβij explicitly as shownbelow:

T 11ij = K11ij +

2Xp=1

∂K11ip

∂ujup +

4XP=1

∂K12iP

∂uj∆P

= K11ij +

2Xp=1

0 · up +4X

P=1

0 · ∆P (4.2.37)

Since∂Kαβ

ij

∂uk= 0 for all α,β, i, j and k (4.2.38)

the coefficients [T 11] and [T 21] of the tangent stiffness matrix are the same asthose of the direct stiffness matrix:

[T 11] = [K11] , [T 21] = [K21] (4.2.39)

Next consider

T 12iJ = K12iJ +

2Xp=1

Ã∂K11

ip

∂∆J

!up +

4XP=1

Ã∂K12

iP

∂∆J

!∆P

= K12iJ + 0 +

4XP=1

"Z xb

xa

1

2Axx

∂

∂∆J

µdw0dx

¶dψidx

dφPdx

dx

#∆P

= K12iJ +

4XP=1

"Z xb

xa

1

2Axx

∂

∂∆J

Ã4XK

∆KdφKdx

!dψidx

dφPdx

dx

#∆P

= K12iJ +

4XP=1

"Z xb

xa

1

2Axx

dφJdx

dψidx

dφPdx

dx

#∆P

= K12iJ +

Z xb

xa

1

2Axx

dψidx

dφJdx

Ã4X

P=1

dφPdx

∆P

!dx


= K12iJ +

Z xb

xa

µ1

2Axx

dw0dx

¶dψidx

dφJdx

dx

= K12iJ +K

12iJ = 2K

12iJ = K

21Ji (4.2.40)

T 22IJ = K22IJ +

2Xp=1

Ã∂K21

Ip

∂∆J

!up +

4XP=1

Ã∂K22

IP

∂∆J

!∆P

= K22IJ +

2Xp=1

"Z xb

xaAxx

∂

∂∆J

Ã4XK

∆KdφKdx

!dφIdx

dψpdx

dx

#up

+4X

P=1

"Z xb

xa

1

2Axx

∂

∂∆J

µdw0dx

¶2 dφIdx

dφPdx

dx

#∆P

= K22IJ +

Z xb

xaAxx

dφIdx

dφJdx

⎛⎝ 2Xp=1

dψpdxup

⎞⎠ dx

+

Z xb

xaAxx

µdw0dx

¶dφIdx

dφJdx

Ã4X

P=1

∆PdφPdx

!dx

= K22IJ +

Z xb

xaAxx

µdu0dx

+dw0dx

dw0dx

¶dφIdx

dφJdx

dx (4.2.41)

4.2.6 Load Increments

Examining the expression (4.2.15a) for the internal axial force Nxx, it is clearthat the rotation of a transverse normal contributes to tensile componentof Nxx irrespective of the sign of the load. As a result, the beam becomesincreasingly stiff with an increase in load. Hence, for large loads thenonlinearity may be too large for the numerical scheme to yield convergentsolution. Therefore, it is necessary to divide the total load F into severalsmaller load increments δF1, δF2, . . . , δFN such that

F =NXi=1

δFi (4.2.42)

For the Þrst load step, the iterative procedure outlined earlier can be used todetermine the deßection. If it does not converge within a reasonable number ofiterations, it may be necessary to further reduce the load increment F1 = δF1.Once the solution for the Þrst load increment is obtained, it is used as theinitial guess vector for the next load F2 = δF1+δF2. This is continued untilthe total load is reached.Another way to accelerate the convergence is to use a weighted average of

the solutions from the last two iterations in evaluating the stiffness matrix at


the rth iteration:

∆∗r−1 = γ∆r−2 + (1− γ)∆r−1, 0 ≤ γ ≤ 1 (4.2.43)

where γ is called the acceleration parameter. A value of γ = 0.5 is suggestedwhen the iterative scheme experiences convergence difficulty. Otherwise, oneshould use γ = 0.

4.2.7 Membrane Locking

For the linear case, the axial displacement u0 is uncoupled from the bendingdeßection w0, and they can be determined independently from the Þniteelement models [see Eqs. (4.2.20a,b)]

[K11]u = F 1, K11ij =

Z xb

xaAexx

dψidx

dψjdx

dx (4.2.44)

[K22(L)]∆ = F 2, K22(L)IJ =

Z xb

xaDexx

d2φIdx2

d2φJdx2

dx (4.2.45)

respectively. Here the superscript L signiÞes the linear stiffness coefficients.Under the assumptions of linearity, if a beam is subjected to only bendingforces and no axial loads, then u0(x) = 0 when u0 is speciÞed to be zero at (atleast) one point. In other words, a hingedhinged beam and a pinnedpinnedbeam (see Figures 4.2.3(a) and (b), respectively) will have the same deßectionw0(x) under the same loads and u0(x) = 0 for all x. However, this is not thecase when the beam undergoes nonlinear bending. The coupling between u0and w0 will cause the beam to undergo axial displacement even when thereare no axial forces, and the solution (u0, w0) will be different for the two casesshown in Figure 4.2.3.First, we note that the hingedhinged beam does not have any end

constraints on u0. If the geometry, boundary conditions, and loading aresymmetric about the center, then u0 = 0 there. Consequently, the beam doesnot experience any axial strain, that is, ε0xx = 0 (because the beam is freeto slide on the rollers to accommodate transverse deßection). On the otherhand, the pinnedpinned beam is constrained from axial movement at x = 0and x = L. As a result, it will develop axial strain to accommodate thetransverse deßection. The former beam will have larger transverse deßectionthan the latter, as the latter offers axial stiffness to stretching, and the axialstiffness increases with the load.Thus, for a hingedhinged beam, the element should experience no

stretching:

ε0xx ≡du0dx

+1

2

µdw0dx

¶2= 0 (membrane strain) (4.2.46)

(a) (b)00

=≠

wu

00

=≠

wu

00

==

wu

00

==

wu


Figure 4.2.3 Nonlinear bending of (a) hingedhinged and (b) pinnedpinned beams.

In order to satisfy the constraint in (4.2.46), we must have

du0dx

∼µdw0dx

¶2(4.2.47)

The similarity is in the sense of having the same degree of polynomial variationof du0/dx and (dw0/dx)

2. For example, when u0 is interpolated using linearfunctions and w0 with cubic, the constraint in Eq. (4.2.47) is clearly notmet and the resulting element stiffness matrix is excessively stiff (hence,results in zero displacement Þeld), and the element is said to lock. Thisphenomenon is known as the membrane locking. In fact, unless a very higher-order interpolation of u0 is used, the element will not satisfy the constraint.A practical way to satisfy the constraint in Eq. (4.2.47) is to use the

minimum interpolation of u0 and w0 (i.e. linear interpolation of u0 andHermite cubic interpolation of w0) but treat ε

0xx as a constant. Since du0/dx

is constant, it is necessary to treat (dw0/dx)2 as a constant in numerically

evaluating the element stiffness coefficients. Thus, if Axx is a constant,all nonlinear stiffness coefficients should be evaluated using one-point Gaussquadrature, that is, use the reduced integration. These coefficients includeK12ij , K

21ij , T

12ij , T

21ij , and the nonlinear parts of K

22 and T 22ij . All other termsmay be evaluated exactly using two-point quadrature for constant values ofAxx and Dxx.

4.2.8 Computer Implementation

The ßow chart for nonlinear bending of beams is shown in Figure 4.2.4. Notethat there is an outer loop on load increments (NLS=number of load steps).Except for the deÞnition of the stiffness coefficients, much of the logic remainsthe same as that shown in Box 3.5.1.The element stiffness matrix in Eq. (4.2.22) is deÞned by submatrices

[K11], [K12], and [K22], and the solution vector ∆ is partitioned into theaxial displacement vector u and vector ∆ of transverse displacements.

Initialize global Kij, fi

Iter = Iter + 1

DO 1 to n

Iter = 0


CALL ELKF to calculate Kij(n)

and fi(n), and assemble to form global Kij and Fi

Transfer global information(material properties, geometry, and solution)

to element

yes Iter < Itmax no

Error < ε yesno

STOP

Print solution

Write a message

Load loopDO L=1, NLS

F=F+DF


Figure 4.2.4 A computer ßow chart for the nonlinear Þnite element analysisof beams.


In practice, it is desirable to rearrange the solution vector as

∆ = u1, w1 = ∆1, θ1 = ∆2, u2, w2 = ∆3, θ2 = ∆4T (4.2.48)

This in turn requires rearrangement of the stiffness coefficients such that theoriginal symmetry, if any, is preserved. For linear interpolation of u0(x) andHermite cubic interpolation of w0(x), the submatrix [K

11] is of the order 2×2,[K12] is 2×4, and [K22] is 4×4. Therefore, the total size of the stiffness matrixis 6× 6. Thus Eq. (4.2.23) has the speciÞc matrix form⎡⎢⎢⎢⎢⎢⎢⎣

K1111 K11

12 K1211 K12

12 K1213 K12

14

K1121 K11

22 K1221 K12

22 K1223 K12

24

K2111 K21

12 K2211 K22

12 K2213 K22

14

K2121 K21

22 K2221 K22

22 K2223 K22

24

K2131 K21

32 K2231 K22

32 K2233 K22

34

K2141 K21

42 K2241 K22

42 K2243 K22

44

⎤⎥⎥⎥⎥⎥⎥⎦

⎧⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎩

ue1ue2∆e1∆e2∆e3∆e4

⎫⎪⎪⎪⎪⎪⎪⎬⎪⎪⎪⎪⎪⎪⎭=

⎧⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎩

F 11F 12F 21F 22F 23F 24

⎫⎪⎪⎪⎪⎪⎪⎬⎪⎪⎪⎪⎪⎪⎭(4.2.49)

Rearranging the equations according to the displacement vector in Eq.(4.2.48), we obtain⎡⎢⎢⎢⎢⎢⎢⎣

K1111 K12

11 K1212 K11

12 K1213 K12

14

K2111 K22

11 K2212 K21

12 K2213 K22

14

K2121 K22

21 K2222 K21

22 K2223 K22

24

K1121 K12

21 K1222 K11

22 K1223 K12

24

K2131 K22

31 K2232 K21

32 K2233 K22

34

K2141 K22

41 K2242 K21

42 K2243 K22

44

⎤⎥⎥⎥⎥⎥⎥⎦

⎧⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎩

ue1∆e1∆e2ue2∆e3∆e4

⎫⎪⎪⎪⎪⎪⎪⎬⎪⎪⎪⎪⎪⎪⎭=

⎧⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎩

F 11F 21F 22F 12F 23F 24

⎫⎪⎪⎪⎪⎪⎪⎬⎪⎪⎪⎪⎪⎪⎭(4.2.50)

The computer implementation of such rearrangement of element coefficientsis presented in Box 4.2.1, where NDF denotes the degrees of freedom per node(=3) and NPE the nodes per element (=2). This rearrangement is carriedout after the element coefficients [K11], [K12], [K22], F 1 = 0, and F 2are computed inside loops on Gauss quadrature. There are two loops, one forfull integration and another for reduced integration.The number of full Gauss points (NGP) is determined by the highest

polynomial degree p of all integrands of the linear stiffness coefficients (recallthat reduced integration is to be used for the nonlinear terms): NGP=(p+1)/2.For example, if linear interpolation of u0 and Hermite cubic interpolation ofw0 is used, the integrands of the stiffness coefficients deÞned in Eq. (4.2.21)have the following polynomial degrees (assuming that the nonlinear terms aretreated as if they are constant):

K11ij = degree of Axx, K12

iJ = degree of Axx,

K22(1)IJ = degree of Dexx + 2, K

22(2)IJ = degree of Aexx + 0

F 1i = degree of f(x) + 1, F 2I = degree of q(x) + 3

CC Rearranging of the element matrix coefficients for C the EULER−BERNOULLI beam element (EBE)C

II=1DO 200 I=1,NPE

I0=2*I-1ELF(II+1)=ELF2(I0)ELF(II+2)=ELF2(I0+1)JJ=1DO 100 J=1,NPE

J0=2*J-1ELK(II,JJ) = ELK11(I,J)ELK(II,JJ+1) = ELK12(I,J0)ELK(II,JJ+2) = ELK12(I,J0+1)ELK(II+1,JJ) = ELK21(I0,J)ELK(II+2,JJ) = ELK21(I0+1,J)ELK(II+1,JJ+1) = ELK22(I0,J0)ELK(II+1,JJ+2) = ELK22(I0,J0+1)ELK(II+2,JJ+1) = ELK22(I0+1,J0)ELK(II+2,JJ+2) = ELK22(I0+1,J0+1)

100 JJ=NDF*J+1200 II=NDF*I+1


Box 4.2.1 Fortran statements to rearrange stiffness coefficients.

In particular, for constant values of AXX = Aexx, DXX = Dexx, FX = f , andQX = q, we have NGP = (3 + 1)/2 = 2 (dictated by F 2I ) and the number ofreduced integration points is LGP = 1. The full Gauss quadrature is used toevaluate [K11], [K22(1)], F 1, and F 2, whereas the reduced integration isused to evaluate [K12] and [K22(2)].The computation of the direct stiffness coefficients and force vectors deÞned

in Eq. (4.2.21) is straightforward. For example, we have

ELF1(i) = ELF1(i) + FX ∗ SFL(i) ∗ CNSTELF2(I) = ELF2(I) +QX ∗ SFH(I) ∗ CNST

ELK11(i, j) = ELK11(i, j)

+AXX ∗GDSFL(i) ∗GDSFL(j) ∗ CNSTELK22(I, J) = ELK22(I, J)

+DXX ∗GDDSFH(I) ∗GDDSFH(J) ∗ CNSTin the full integration loop, and

ELK12(i, J) = ELK12(i, J) + 0.5 ∗AXX ∗DW∗GDSFL(i) ∗GDSFH(J) ∗ CNST


ELK21(I, j) = ELK21(I, j) +AXX ∗DW∗GDSFH(I) ∗GDSFL(j) ∗ CNST

ELK22(I, J) = ELK22(I, J) + 0.5 ∗AXX ∗DW ∗DW∗GDSFH(I) ∗GDSFH(J) ∗ CNST

in the reduced integration loop. Here, SFL(i) = ψi, SFH(I) = φI ,

GDSFH(I) = dφIdx , GDDSFH(I) =

d2φIdx2

, GDSFL(i) = dψidx , and DW =

(dw0/dx) for i, j = 1, 2 and I, J = 1, 2, 3, 4. Similarly, the extra terms thatneed to be added to the direct stiffnesses can be computed in the reducedintegration loop as [see Eq. (4.2.41)]

TANG12(i, J) = TANG12(i, J) + 0.5 ∗AXX ∗DW∗GDSFL(i) ∗GDSFH(J) ∗ CNST

TANG22(I, J) = TANG22(I, J) +AXX ∗ (DU +DW ∗DW )∗GDSFH(I) ∗GDSFH(J) ∗ CNST

where DU = (du0/dx).

Example 4.2.1

Consider a beam of length L = 100 in., 1 in. × 1 in. cross-sectional dimensions, hingedat both ends, made of steel (E = 30 msi), and subjected to uniformly distributed load ofintensity q0 lb/in. Using the symmetry about x = L/2, one-half of the domain is used as thecomputational domain. The geometric boundary conditions for the computational domainare

w0(0) = u0(L/2) =³dw0dx

´x=L

2

= 0 (4.2.51)

The load is divided into load increments of equal size ∆q0 = 1 lb/in. A tolerance of² = 10−3 and maximum allowable iterations of 30 (per load step) are used in the analysis.The initial solution vector is chosen to be the zero vector, so that the Þrst iteration solutioncorresponds to the linear solution

u0(x) = 0, w0(x) =q0L

4

24Dxx

µx

L− 2 x

3

L3+x4

L4

¶(4.2.52)

In particular, the center deßection is (for q0 = 1)

w0(L

2) =

5q0L4

384Dxx= 0.5208 in. (4.2.53)

For the four element mesh, the linear stiffness matrix, force vector, and the global linearsolution vector are given by (with the speciÞed boundary conditions ∆2 = 0, ∆13 = 0 and∆15 = 0)

[Ke] = 105

⎡⎢⎢⎢⎢⎣24 0.0000 0.00 −24 0.0000 0.000 0.1536 −0.96 0 −0.1536 −0.960 −0.9600 8.00 0 0.9600 4.00

−24 0.0000 0.00 24 0.0000 0.000 −0.1536 0.96 0 0.1536 0.960 −0.9600 4.00 0 0.9600 8.00

⎤⎥⎥⎥⎥⎦


F e =

⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩0.0006.250

−13.0210.0006.25013.021

⎫⎪⎪⎪⎪⎬⎪⎪⎪⎪⎭,

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎩

∆3∆5∆6∆8∆9∆11∆12∆14

⎫⎪⎪⎪⎪⎪⎪⎪⎪⎬⎪⎪⎪⎪⎪⎪⎪⎪⎭=

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎩

−0.016660.20223

−0.015230.37109

−0.011460.48218

−0.006120.52083

⎫⎪⎪⎪⎪⎪⎪⎪⎪⎬⎪⎪⎪⎪⎪⎪⎪⎪⎭Table 4.2.1 contains the results of the nonlinear analysis obtained with the direct iteration

procedure as well as the NewtonRaphson iteration (acceleration parameter, γ = 0). The

Gauss ruleM×N has the meaning thatM Gauss points are used for the evaluation of linear

stiffness coefficients as well as the force components, and N Gauss points are used to evaluate

the nonlinear stiffness coefficients. As discussed earlier, the problem should not exhibit any

nonlinearity. The correct solution (4.2.53) is predicted by the use of 2×1 Gauss rule (see thelast column of Table 4.2.1). Both 4 and 8 element meshes and direct and NewtonRaphson

methods predicted the same result. The 2 × 2 Gauss rule not only yields incorrect results,but it takes more iterations to converge.

Table 4.2.1 Finite element results for the deßections of a hingedhingedbeam under uniformly distributed load.

Load q0 Direct iteration (DI) NewtonRaphson (NR) DINR(2× 2) (2× 2) 2× 1

4 elem. 8 elem. 4 elem. 8 elem. 4 and 8

1.0 0.5108 (3)∗ 0.5182 (3) 0.5108 (4) 0.5182 (4) 0.5208 (3)2.0 0.9739 (5) 1.0213 (3) 0.9739 (4) 1.0213 (4) 1.0417 (3)3.0 1.3763 (6) 1.4986 (4) 1.3764 (4) 1.4986 (4) 1.5625 (3)4.0 1.7269 (7) 1.9451 (4) 1.7265 (4) 1.9453 (4) 2.0833 (3)5.0 2.0356 (9) 2.3609 (5) 2.0351 (4) 2.3607 (4) 2.6042 (3)6.0 2.3122 (11) 2.7471 (5) 2.3116 (3) 2.7467 (3) 3.1250 (3)7.0 2.5617 (14) 3.1054 (6) 2.5630 (2) 3.1074 (2) 3.6458 (3)8.0 2.7936 (17) 3.4418 (7) 2.7930 (2) 3.4422 (2) 4.1667 (3)9.0 3.0049 (22) 3.7570 (7) 3.0060 (3) 3.7564 (2) 4.6875 (3)10.0 3.2063 (29) 4.5013 (8) 3.2051 (3) 4.0523 (2) 5.2083 (3)

∗ Number of iterations taken to converge.

Example 4.2.2

Next, we consider the straight beam of Example 4.2.1 with (a) pinned ends, and (b) clampedends, and under uniformly distributed transverse load. Noting the symmetry of the solutionabout x = L/2, one-half of the domain is used as the computational domain. The geometricboundary conditions for the computational domain of the two problems are

pinned: u0(0) = w0(0) = u0(L

2) =

dw0dx

|x=L

2= 0 (4.2.54)

clamped: u0(0) = w0(0) =dw0dx

|x=0 = u0(L

2) =

dw0dx

|x=L

2= 0 (4.2.55)


The load increments of∆q0 = 1.0 lb/in., a tolerance of ² = 10−3, and maximum allowable

iterations of 25 (per load step) are used in the analysis. The initial solution vector is chosento be the zero vector. Solutions to the linear problems are

pinned: u0(x) = 0, w0(x) =q0L

4

24Dxx

x2

L2

³1− x

L

´2(4.2.56)

clamped: u0(x) = 0, w0(x) =q0L

4

24Dxx

µx

L− 2 x

3

L3+x4

L4

¶(4.2.57)

and the maximum deßections occurs at L/2. For q0 = 1 lb/in., L = 100 in., and E = 30×106psi, they are given by (Dxx = EH3/12, H = 1)

pinned: w0(L

2) =

5q0L4

384Dxx= 0.5208 in. (4.2.58)

clamped: w0(L

2) =

q0L4

384Dxx= 0.1042 in. (4.2.59)

The linear nodal displacements obtained using four elements in half beam are⎧⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎩

∆3∆5∆6∆8∆9∆11∆12∆14

⎫⎪⎪⎪⎪⎪⎪⎪⎪⎬⎪⎪⎪⎪⎪⎪⎪⎪⎭pinned

=

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎩

−0.016660.20223

−0.015230.37109

−0.114580.48218

−0.006120.52083

⎫⎪⎪⎪⎪⎪⎪⎪⎪⎬⎪⎪⎪⎪⎪⎪⎪⎪⎭,

⎧⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎩

∆5∆6∆8∆9∆11∆12∆14

⎫⎪⎪⎪⎪⎪⎪⎬⎪⎪⎪⎪⎪⎪⎭clamped

=

⎧⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎩

0.01994−0.002730.05859

−0.003130.09155

−0.001950.10417

⎫⎪⎪⎪⎪⎪⎪⎬⎪⎪⎪⎪⎪⎪⎭Tables 4.2.2 and 4.2.3 contain the results of the nonlinear analysis of pinnedpinned and

clampedclamped beams, respectively; the results were obtained with the NewtonRaphsoniteration method. The direct iteration method did not converge even for 100 iterations perload step when ∆q = 1.0. It is possible to Þnd a value of ∆q and ITMAX for which onecan obtain converged solutions.

Table 4.2.2 Finite element results for the deßections of a pinnedpinnedbeam under uniform load (NR).

2× 2 2× 1Load q0 4 elements 8 elements 4 elements 8 elements

1.0 0.3669 (5)∗ 0.3680 (5) 0.3687 (5) 0.3685 (5)2.0 0.5424 (4) 0.5446 (4) 0.5466 (4) 0.5457 (4)3.0 0.6601 (3) 0.6629 (3) 0.6663 (4) 0.6645 (4)4.0 0.7510 (3) 0.7543 (3) 0.7591 (4) 0.7564 (4)5.0 0.8263 (3) 0.8299 (3) 0.8361 (4) 0.8324 (4)6.0 0.8912 (3) 0.8950 (3) 0.9027 (4) 0.8979 (4)7.0 0.9485 (3) 0.9525 (3) 0.9617 (4) 0.9558 (4)8.0 1.0002 (3) 1.0043 (3) 1.0150 (4) 1.0080 (4)9.0 1.0473 (3) 1.0516 (3) 1.0638 (4) 1.0557 (4)10.0 1.0908 (3) 1.0952 (3) 1.1089 (4) 1.0997 (4)


0.0 1.0 2.0 3.0 4.0 5.0 6.0 7.0 8.0 9.0 10.0Load,

0.00

0.10

0.20

0.30

0.40

0.50

0.60

0.70

0.80

0.90

1.00

1.10

1.20

Def

lect

ion

,

Clamped-clamped

Pinned-pinned

q0

w0


There is no signiÞcant difference between the solutions obtained with the two integrationrules for this problem. Figure 4.2.5 shows the loaddeßection curves for the two beams. Ifthe axial displacement degrees of freedom are suppressed (i.e. equivalent to setting u0 = 0at every point of the beam) in the nonlinear analysis of beams, the beam will behave verystiff, and the deßections experienced will be less than those shown in Tables 4.2.2 and 4.2.3and Figure 4.2.5.

Table 4.2.3 Finite element results for the deßections of a clampedclampedbeam under uniform load (NR and 2× 1 Gauss rule).

Direct iteration NewtonRaphson iteration

Load q0 4 elements 8 elements 4 elements 8 elements

1.0 0.1033 (3)∗ 0.1034 (3) 0.1034 (3) 0.1034 (3)2.0 0.2022 (4) 0.2023 (4) 0.2022 (3) 0.2023 (3)3.0 0.2938 (4) 0.2939 (4) 0.2939 (3) 0.2939 (3)4.0 0.3773 (5) 0.3774 (5) 0.3773 (3) 0.3774 (3)5.0 0.4529 (5) 0.4531 (5) 0.4528 (3) 0.4530 (3)6.0 0.5213 (6) 0.5215 (6) 0.5214 (3) 0.5216 (3)7.0 0.5840 (7) 0.5842 (7) 0.5839 (3) 0.5841 (3)8.0 0.6412 (8) 0.6412 (8) 0.6413 (3) 0.6414 (3)9.0 0.6945 (9) 0.6944 (9) 0.6943 (3) 0.6943 (3)10.0 0.7433 (10) 0.7431 (10) 0.7435 (3) 0.7433 (3)


Figure 4.2.5 Load versus deßection curves.

z

x

u

w0

γxz φx

dwdx

− 0

0

dwdx

− 0


4.3 Timoshenko Beams


The EBT is based on the assumption that a straight line transverse to theaxis of the beam before deformation remains (i) straight, (ii) inextensible, and(iii) normal to the mid-plane after deformation. In the TBT, the Þrst twoassumptions are kept but the normality condition is relaxed by assuming thatthe rotation is independent of the slope (w0,x) of the beam.The displacement Þeld of the beam in the TBT can be expressed as

u1 = u0(x) + zφx(x), u2 = 0, u3 = w0(x) (4.3.1)

where (u1, u2, u3) are the displacements of a point along the (x, y, z)coordinates, (u0, w0) are the displacements of a point on the mid-plane ofan undeformed beam, and φx is the rotation (about the y-axis) of a transversestraight line (see Figure 4.3.1).The only non-zero strains are

εxx =∂u1∂x

+1

2

µdw0dx

¶2=du0dx

+1

2

µdw0dx

¶2+ z

dφxdx

≡ ε0xx + zε1xx (4.3.2)

γxz =∂u1∂z

+∂u3∂x

= φx +dw0dx

≡ γ0xz (4.3.3)

The virtual strains are

δε0xx =dδu0dx

+dw0dx

dδw0dx

, δε1xx =dδφxdx

, δγ0xz = δφx +dδw0dx

(4.3.4)

Figure 4.3.1 Kinematics of a beam in the TBT.


4.3.2 Weak Forms

Next, we use the principle of virtual displacements to develop the necessaryweak statements of the TBT. We have

0 = δW e ≡ δW eI + δW e

E (4.3.5)

δW eI =

Z xb

xa

ZAe(σxxδεxx + σxzδγxz) dAdx

=

Z xb

xa

ZAe

hσxx

³δε0xx + zδε

1xx

´+ σxzδγ

0xz

idAdx

=

Z xb

xa

³Nxxδε

0xx +Mxxδε

1xx +Qxδγ

0xz

´dx (4.3.6a)

δW eE = −

"Z xb

xaqδw0 dx+

Z xb

xaf δu0 dx+

6Xi=1

Qei δ∆ei

#(4.3.6b)

where q is the distributed transverse load, Qei are the element generalizedforces, ∆ei element generalized displacements, and

Nxx =

ZAσxx dA, Mxx =

ZAσxxz dA, Qx = Ks

ZAσxz dA (4.3.7)

and Ks is the shear correction coefficient introduced to account for thedifference between the shear energy calculated using equilibrium stresses andthat predicted by the Timoshenko beam theory on account of constant stateof shear stress through the beam height.For example, consider a homogeneous beam with rectangular cross-section,

with width b and height h. The actual shear stress distribution through thethickness of the beam is given by

σcxz =3Q02bh

"1−

µ2z

h

¶2#, −h

2≤ z ≤ h

2

where Q0 is the transverse load. The transverse shear stress in the first-ordertheory is a constant, σfxz = Q0/bh. The strain energies due to transverse shearstresses in the two theories are

U cs =1

2G13

ZA(σcxz)

2 dA =3Q205G13bh

, Ufs =1

2G13

ZA

³σfxz

´2dA =

Q202G13bh

The shear correction factor is the ratio of Ufs to Ucs , which gives Ks = 5/6.

The shear correction factor, in general, depends on the geometry and materialproperties.


The EulerLagrange equations are given by

δu0 : − dNxxdx

= f(x) (4.3.8)

δφ : − dMxx

dx+Qx = 0 (4.3.9)

δw0 : − dQxdx

− d

dx

µNxx

dw0dx

¶= q(x) (4.3.10)

It is clear that (u0, w0,φx) are the primary variables and

Nxx, Qx, and Mxx (4.3.11)

are the secondary variables. Thus the pairing of the primary and secondaryvariables is as follows:

(u0, Nxx), (w0, Qx), (φx,Mxx) (4.3.12)

Only one member of each pair may be speciÞed at a point in the beam.If we assume linear elastic behavior

σxx = Eεxx, σxz = Gγxz (4.3.13a)

the axial force Nxx, the shear force Qx, and bending moment Mxx can beexpressed in terms of the generalized displacements (u0, w0,φx) using thedeÞnitions (4.3.7). We obtain

Nxx = Axx

"du0dx

+1

2

µdw0dx

¶2#+Bxx

dφxdx

Mxx = Bxx

"du0dx

+1

2

µdw0dx

¶2#+Dxx

dφxdx

(4.3.13b)

Qx = Sxx

µdw0dx

+ φx

¶where Axx, Bxx, and Dxx are deÞned in Eq. (4.2.16), and Sxx is the shearstiffness

Sxx = Ks

ZAG dA = KsGA (4.3.13c)

G being the shear modulus, Ks the shear correction coefficient, and A thearea of cross-section. As discussed earlier, the stiffnesses Axx, Bxx, and Dxxare functions of x whenever the modulus E and/or cross-sectional area is afunction of x.


The equations of equilibrium of the TBT for the isotropic case can beexpressed in terms of the generalized displacements as

− d

dx

(Axx

"du0dx

+1

2

µdw0dx

¶2#)= f (4.3.14)

− d

dx

∙Sxx

µdw0dx

+ φx

¶¸− d

dx

(Axx

dw0dx

"du0dx

+1

2

µdw0dx

¶2#)= q (4.3.15)

− d

dx

µDxx

dφxdx

¶+ Sxx

µdw0dx

+ φx

¶= 0 (4.3.16)

4.3.3 General Finite Element Model

The Þnite element model of the Timoshenko beam equations can beconstructed using the virtual work statement in Eqs. (4.3.5), where the axialforce Nxx, the shear force Qx, and bending moment Mxx are known in termsof the generalized displacements (u0, w0,φx) by Eq. (4.3.13b). The virtualwork statement (4.3.5) is equivalent to the following three statements

0 =

Z xb

xa

(Axx

dδu0dx

"du0dx

+1

2

µdw0dx

¶2#+ fδu0

)dx

−Qe1δu0(xa)−Qe4δu0(xb) (4.3.17)

0 =

Z xb

xa

dδw0dx

(Sexx

µdw0dx

+ φx

¶+Aexx

dw0dx

"du0dx

+1

2

µdw0dx

¶2#

−δw0q)dx−Qe2δw0(xa)−Qe5δw0(xb) (4.3.18)

0 =

Z xb

xa

∙Dexx

dδφxdx

dφxdx

+ Sexx δφx

µdw0dx

+ φx

¶¸dx

−Qe3 δφx(xa)−Qe6 δφx(xb) (4.3.19)

where δu0, δw0, and δφx are the virtual displacements. The Qei have thesame physical meaning as in the EulerBernoulli beam element, and theirrelationship to the horizontal displacement u0, transverse deßection w0, androtation φx, is

Qe1 = −Nxx(xa), Qe4 = Nxx(xb)

Qe2 = −∙Qx +Nxx

∂w0∂x

¸x=xa

, Qe5 =

∙Qx +Nxx

∂w0∂x

¸x=xb

Qe3 = −Mxx(xa), Qe6 =Mxx(xb) (4.3.20)


An examination of the virtual work statements (4.3.17)(4.3.19) suggeststhat u0(x), w0(x), and φx(x) are the primary variables and therefore mustbe carried as nodal degrees of freedom. In general, u0, w0, and φxneed not be approximated by polynomials of the same degree. However,the approximations should be such that possible deformation modes (i.e.kinematics) are represented correctly. We will return to this point shortly.It is also possible to develop the weak forms (4.3.17)(4.3.19) using the

governing equations (4.3.8)(4.3.10) and introducing the secondary variables(4.3.20). It is left as an exercise to the reader.Suppose that the displacements are approximated as

u0(x) =mXj=1

uejψ(1)j , w0(x) =

nXj=1

wejψ(2)j , φx(x) =

pXj=1

sejψ(3)j (4.3.21)

where ψ(α)j (x) (α = 1, 2, 3) are Lagrange interpolation functions of degree

(m− 1), (n− 1), and (p− 1), respectively. At the moment, the values of m, n,and p are arbitrary, that is, arbitrary degree of polynomial approximations ofu0, w0, and φx may be used. Substitution of (4.3.21) for u0, w0, and φx, and

δu0 = ψ(1)i , δw0 = ψ

(2)i , and δφx = ψ

(3)i into Eqs. (4.3.17)(4.3.19) yields the

Þnite element model

0 =mXj=1

K11ij u

ej +

nXj=1

K12ij w

ej +

pXj=1

K13ij s

ej − F 1i (4.3.22)

0 =mXj=1

K21ij u

ej +

nXj=1

K22ij w

ej +

pXj=1

K23ij s

ej − F 2i (4.3.23)

0 =mXj=1

K31ij u

ej +

nXj=1

K32ij w

ej +

pXj=1

K33ij s

ej − F 3i (4.3.24)

K11ij =

Z xb

xaAxx

dψ(1)i

dx

dψ(1)j

dxdx, K12

ij =1

2

Z xb

xaAxx

dw0dx

dψ(1)i

dx

dψ(2)j

dxdx

K21ij =

Z xb

xaAxx

dw0dx

dψ(2)i

dx

dψ(1)j

dxdx, K13

ij = 0, K31ij = 0

K22ij =

Z xb

xaSxx

dψ(2)i

dx

dψ(2)j

dxdx+

1

2

Z xb

xaAxx

µdw0dx

¶2 dψ(2)idx

dψ(2)j

dxdx

K23ij =

Z xb

xaSxx

dψ(2)i

dxψ(3)j dx = K32

ji

K33ij =

Z xb

xa

⎛⎝Dxxdψ(3)idx

dψ(3)j

dx+ Sxxψ

(3)i ψ

(3)j

⎞⎠ dx (4.3.25a)


F 1i =

Z xb

xaψ(1)i f dx+Q

e1ψ

(1)i (xa) +Q

e4ψ

(1)i (xb)

F 2i =

Z xb

xaψ(2)i q dx+Q

e2ψ

(2)i (xa) +Q

e5ψ

(2)i (xb)

F 3i = Qe3ψ

(3)i (xa) +Q

e6ψ

(3)i (xb) (4.3.25b)

The element equations (4.3.22)(4.3.24) can be expressed in matrix form as⎡⎣ [K11] [K12] [K13][K21] [K22] [K23][K31] [K32] [K33]

⎤⎦⎧⎨⎩uws

⎫⎬⎭ =⎧⎨⎩F 1F 2F 3

⎫⎬⎭ (4.3.26)

The choice of the approximation functions ψ(α)i dictates different Þnite element

models. The choice of linear polynomials ψ(1)i = ψ

(2)i is known to yield a

stiffness matrix that is nearly singular. This will be discussed further in the

next section. When ψ(1)i are quadratic and ψ

(2)i are linear, the stiffness matrix

is 5× 5. It is possible to eliminate the interior degree of freedom for w0 and

obtain 4× 4 stiffness matrix. This element behaves well. When ψ(1)i are cubic

and ψ(2)i are quadratic, the stiffness matrix is 7 × 7. If the interior nodal

degrees of freedom are eliminated, one obtains 4 × 4 stiffness matrix that isknown to yield the exact solution at the nodes in the linear case when theshear stiffness and bending stiffnesses are element-wise constant. More detailsof various Timoshenko beam elements will be given in the sequel.

4.3.4 Shear and Membrane Locking

A number of Timoshenko beam Þnite elements for the linear case (i.e. withoutvon Karman nonlinearity) have appeared in the literature. They differ fromeach other in the choice of approximation functions used for the transversedeßection w0 and rotation φx, or in the variational form used to develop theÞnite element model. Some are based on equal interpolation and others onunequal interpolation of w0 and φx.The Timoshenko beam Þnite element with linear interpolation of both w0

and φx is the simplest element. Linear interpolation of w0 means that theslope dw0/dx is constant (see Figure 4.3.2). In thin beam limit, that is, as thelength-to-thickness ratio becomes large (say, 100), the slope should be equal to−φx, which is also represented as linear as opposed to being a constant. On theother hand, a constant representation of φx results in zero bending energy whilethe transverse shear is non-zero. This inconsistency in the representation ofthe kinematics through linear approximation of both w0 and φx results in zerodisplacements and rotations, which trivially satisfy the Kirchhoff constraintφx = −dw0/dx, and the element is said be very stiff in the thin beam limit.Such behavior is known as shear locking.

he

1he

1w1

2 2

w

ss2 2

1


Figure 4.3.2 Kinematics of deformation of the Timoshenko beam elementwhen both w0 and φx are interpolated linearly.

To overcome the locking, one may use equal interpolation for bothw0 and φx but treat φx as a constant in the evaluation the shear strain,γxz = (dw0/dx) + φx. This is often realized by using selective integration,in which one-point (reduced) Gauss quadrature is used to evaluate thestiffness coefficients associated with the transverse shear strain, and all othercoefficients of the stiffness matrix are evaluated using exact (full) integration.A more detailed discussion on the alleviation of shear locking and membranelocking in Timoshenko beam element is presented below.

Since applied distributed loads are represented as point loads in the Þniteelement method, Eq. (4.3.15) with q = 0 and constant Sexx implies that

(shear strain) γ0xz ≡ φx +dw0dx

= constant (4.3.27)

Similarly, for a problem that involves only bending deformation, the elementshould experience no stretching [see Eq. (4.3.14) with f = 0]:

(membrane strain) ε0xx ≡du0dx

+1

2

µdw0dx

¶2= 0 (4.3.28)

In order to satisfy the above constraints, we must have

φx ∼ dw0dx

(4.3.29)

du0dx

∼µdw0dx

¶2(4.3.30)

The similarity is in the sense of having the same degree of polynomial variation.For example, when φx is linear and w0 is quadratic, the constraint in Eq.(4.3.27) is clearly met. Similarly, when both u0 and w0 are linear, theconstraint in Eq. (4.3.28) is automatically met; however, when quadratic


interpolation is used for both u0 and w0, du0/dx is linear and (dw0/dx)2

is quadratic and the resulting element experiences locking, known as themembrane locking. If u0 is interpolated using a cubic polynomial, w0 witha quadratic polynomial, and φx with a linear polynomial, we have

φx (linear) ∼ dw0dx

(linear) (4.3.31)

du0dx

(quadratic) ∼µdw0dx

¶2(quadratic) (4.3.32)

In summary, the constraints would be satisÞed for the following two cases:

u0, φx, and w0 all linear, with constant representation of γxz; the lattercan be accomplished by using reduced integration to evaluate the shearstiffness coefficients.

u0 is cubic, w0 is quadratic, and φx is linear. However, this will result in a9× 9 stiffness matrix with different degrees of freedom at different nodes,making it difficult (but not impossible) to implement into a computerprogram.

The membrane locking can also be avoided, in addition to usingappropriate interpolation of the variables, by using selective Gauss quadrature,as discussed in Section 4.2.7. In the present study, we shall use equal linearor quadratic interpolation of the variables (u0, w0,φ) with reduced integrationof all nonlinear terms (to avoid membrane locking) and shear terms, thatis, stiffness coefficients involving Sxx (to avoid shear locking) of the stiffnessmatrix.

4.3.5 Tangent Stiffness Matrix

Returning to the nonlinear Þnite element model (4.3.26) of Section 4.3.3, wecompute the tangent stiffness matrix of the Timoshenko beam element. Muchof the computer implementation discussion presented in Section 4.2.8 is alsovalid for the Timoshenko beam element.

The tangent matrix coefficients are deÞned by

Tαβij = Kαβij +

3Xγ=1

nXk=1

∂

∂∆βj

¡Kαγik

¢∆γk (4.3.33)

In particular, we have

T 11ij = K11ij + 0

T 12ij = K12ij +

1

2

Z xb

xaAxx

dw0dx

dψ(1)i

dx

dψ(2)j

dxdx = 2K12

ij

he

1

he

1

w1

2 2

w s s2 21

he

1

he

1

w1

3 3

w w s3 s312

2 2

s2


T 13ij = K13ij = 0

T 21ij = K21ij + 0 = K

21ij

T 22ij = K22ij +

Z xb

xaAxx

"du0dx

+

µdw0dx

¶2# dψ(2)idx

dψ(2)j

dxdx

T 23ij = K23ij + 0 = K

23ij

T 31ij = K31ij + 0 = K

31ij

T 32ij = K32ij + 0 = K

32ij

T 33ij = K33ij + 0 = K

33ij (4.3.34)

where the direct stiffness coefficients Kαβij are deÞned by Eq. (4.3.25a).

As discussed earlier, we must use reduced integration on nonlinear stiffnesscoefficients and shear stiffness coefficients, while the remaining stiffnesscoefficients may be evaluated using full integration. For example, considerthe following integral expression:

Z xb

xaAxx

⎡⎣du0dx

dψ(2)i

dx

dψ(2)j

dx+3

2

µdw0dx

¶2 dψ(2)idx

dψ(2)j

dx

⎤⎦ dx (4.3.35)

If quadratic interpolation of both u0 and w0 is used (see Figure 4.3.3) andAxx is constant, then the Þrst term in the integrand is a cubic polynomialwhile the second term is a fourth-order polynomial. Thus, exact evaluationof the Þrst term requires two-point Gauss quadrature while the second termrequires three-point Gauss quadrature. If we use two-point Gauss quadratureto evaluate T 22ij (and K

22ij ), the Þrst term in the coefficient of Axx is integrated

exactly while the second term is integrated approximately. This amounts toapproximating (dw0/dx)

2 as a linear polynomial. Consequently, the constraintε0xx = 0 [see Eq. (4.3.28)] is satisÞed.

The rearrangement of the elements of nodal displacement vector, as shown

in Eq. (4.2.50), requires rearranging of Kαβij of Eq. (4.3.26). Box 4.3.1 shows

the logic for this rearrangement.

Figure 4.3.3 Linear and quadratic Timoshenko beam Þnite elements.

C Rearranging of the element stiffness coefficientsC of the TIMOSHENKO beam element (TBT)C ELK11, ELF1, etc are defined by Eqs... (4.3.25a,b)C

II=1DO 160 I=1,NPE

JJ=1ELF(II) = ELF1(I)ELF(II+1) = ELF2(I)DO 150 J=1,NPE

ELK(II,JJ) = ELK11(I,J)ELK(II,JJ+1) = ELK12(I,J)ELK(II+1,JJ) = ELK21(I,J)ELK(II+1,JJ+1) = ELK22(I,J)ELK(II+1,JJ+2) = ELK23(I,J)ELK(II+2,JJ+1) = ELK32(I,J)ELK(II+2,JJ+2) = ELK33(I,J)

150 JJ=J*NDF+1160 II=I*NDF+1


Box 4.3.1 Fortran statements to rearrange stiffness coefficients.

Example 4.3.1

Consider the hingedhinged beam of Example 4.2.1. Using the symmetry about x = L/2,one-half of the domain is used as the computational domain. The geometric boundaryconditions for the computational domain are

w0(0) = u0(L

2) = φx(

L

2) = 0 (4.3.36)

The load is divided into load increments of equal size ∆q0 = 1 lb/in. A tolerance of ² = 10−3

is used in the analysis. The initial solution vector is chosen to be the zero vector, so thatthe Þrst iteration solution corresponds to the linear problem. The exact linear solution is(u0(x) = 0)

w0(x) =q0L

4

24Dxx

µx

L− 2 x

3

L3+x4

L4

¶+q0L

2

2Sxx

µx

L− x2

L2

¶, φx(x) =

q0L3

24Dxx

µ1− 6 x

2

L2+ 4

x3

L3

¶(4.3.37)

The center deßection of the linear problem is (q0 = 1 lb/in., E = 30× 106 psi, ν = 0.25,and Ks = 5/6)

w0(L

2) =

5q0L4

384Dxx+q0L

2

8Sxx= 0.5208 + 0.0125× 10−4 = 0.5208 in. (4.3.38)

Thus, the effect of shear deformation is negligible. It should be noted that the reduced-integration (Timoshenko) element (RIE), in general, does not give exact nodal values evenfor the linear problem (i.e. not a superconvergent element). For a reÞned mesh and/orhigher-order elements, one may expect to get the exact linear solution.


For the mesh of four reduce integration elements, the linear stiffness matrix, force vector,and the global linear solution vector are (with ν = 0.25 and speciÞed boundary conditions∆2 = 0, ∆13 = 0 and ∆15 = 0)

[Ke] = 105

⎡⎢⎢⎢⎢⎣24 0.0 0.0 −24 0.0 0.00 8.0 −50.0 0 −8.0 −50.00 −50.0 314.5 0 50.0 310.5

−24 0.0 0.0 24 0.0 0.00 −8.0 50.0 0 8.0 50.00 −50.0 310.5 0 50.0 314.5

⎤⎥⎥⎥⎥⎦

F e =

⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩0.006.250.000.006.250.00

⎫⎪⎪⎪⎪⎬⎪⎪⎪⎪⎭,

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎩

∆3∆5∆6∆8∆9∆11∆12∆14

⎫⎪⎪⎪⎪⎪⎪⎪⎪⎬⎪⎪⎪⎪⎪⎪⎪⎪⎭=

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎩

−0.016410.19659

−0.015040.36142

−0.011330.47009

−0.006050.5079

⎫⎪⎪⎪⎪⎪⎪⎪⎪⎬⎪⎪⎪⎪⎪⎪⎪⎪⎭Table 4.3.1 contains the results obtained with the NewtonRaphson iteration; FI denotes

full integration (NGP = LGP = 2 for linear elements NGP = LGP = 3 for quadratic

elements) and RI denotes reduced integration (NGP = 2 and LGP = 1 for linear elements

and NGP = 3 and LGP = 2 for quadratic elements). The convergence is reached in 3

iterations. It is clear that the quadratic elements are not as sensitive as the linear elements

to locking; also, the effect of locking on the solution becomes less with reÞned meshes. The

convergence of the solution with mesh reÞnement is also clear.

Table 4.3.1 Finite element results for the deßections of a hingedhingedbeam under uniform load.

Load q0 Integration 4L 2Q 8L 4Q 16L 8Qrule

1.0 RI 0.5079 0.5210 0.5177 0.5210 0.5201 0.5210FI 0.0101 0.4943 0.0370 0.5149 0.1223 0.5198

2.0 RI 1.0159 1.0419 1.0354 1.0419 1.0403 1.0419FI 0.0201 0.9817 0.0741 1.0294 0.2447 1.0395

3.0 RI 1.5238 1.5629 1.5531 1.5629 1.5604 1.5629FI 0.0302 1.4560 0.1111 1.5428 0.3670 1.5592

4.0 RI 2.0318 2.0838 2.0708 2.0838 2.0806 2.0838FI 0.0403 1.9130 0.1482 2.0548 0.4893 2.0788

5.0 RI 2.5397 2.6048 2.5885 2.6048 2.6007 2.6048FI 0.0504 2.3502 0.1852 2.5654 0.6117 2.5983

10.0 RI 5.0794 5.2096 5.1770 5.2096 5.2014 5.2096FI 0.1007 4.2312 0.3704 5.0728 1.2233 5.1927


Example 4.3.2

Here we consider the pinnedpinned beam of Example 3.2.2. The geometric boundaryconditions for the computational domain of the problem are

u0(0) = w0(0) = u0(L

2) = φx(

L

2) = 0 (4.3.39)

The load increments of ∆q0 = 1.0 lb/in., a tolerance of ² = 10−3 is used in the analysis. The

initial solution vector is chosen to be the zero vector.

Table 4.3.2 contains the results for pinnedpinned beam; the results were obtained withthe NewtonRaphson iteration (γ = 0) method and reduced integration. The convergenceof the solution with mesh reÞnement and higher-order elements is apparent from the results.Also, higher-order elements and reÞned meshes are less sensitive to shear locking.

Table 4.3.2 Finite element results for the deßections of a pinnedpinnedbeam under uniformly distributed load.

Load q0 4L 2Q 8L 4Q

1.0 0.3654 (5)∗ 0.3687 (5) 0.3677 (5) 0.3685 (5)2.0 0.5439 (4) 0.5458 (4) 0.5451 (4) 0.5454 (4)3.0 0.6637 (3) 0.6644 (3) 0.6639 (3) 0.6640 (3)4.0 0.7562 (3) 0.7560 (3) 0.7557 (3) 0.7555 (3)5.0 0.8327 (3) 0.8318 (3) 0.8316 (3) 0.8312 (3)6.0 0.8985 (3) 0.8970 (3) 0.8969 (3) 0.8964 (3)7.0 0.9567 (3) 0.9546 (3) 0.9546 (3) 0.9540 (3)8.0 1.0090 (3) 1.0065 (3) 1.0066 (3) 1.0058 (3)9.0 1.0568 (3) 1.0538 (3) 1.0540 (3) 1.0531 (3)10.0 1.1009 (3) 1.0975 (3) 1.0977 (3) 1.0967 (3)


Table 4.3.3 contains nonlinear deßections, w = wmax102(Dxx/L4), of pinnedpinnedbeams for various length-to-thickness ratios L/H. The effect of shear deformation is clearfrom the results; thicker the beams, larger the shear strains and deßections w. Of course, themaximum deßection wmax will be smaller with smaller L/H ratio (or thicker beams), becausethicker beams have larger stiffness. Thus they also exhibit less geometric nonlinearity, ascan be seen from Figure 4.3.4, where the deßection at the center, w0(L/2), is plotted as afunction of the load q0 for various values of L/H ratio. The deßections predicted by theTBT are the same as those predicted by the EBT when L/H ≥ 100.

Lastly, we consider the clampedclamped beam of Example 4.2.3. The geometricboundary conditions for the computational domain of the problem are

u0(0) = w0(0) = φx(0) = u0(L

2) = φx(

L

2) = 0 (4.3.40)

All other parameters are taken to be the same as in the case of the pinnedpinned beam.Table 4.3.4 contains the nonlinear analysis results for clampedclamped beam; the resultswere obtained with the direct iteration as well as the NewtonRaphson iteration methodusing (a) mesh of 8 linear Timoshenko beam elements, and (b) 4 quadratic Timoshenkobeam elements. The results obtained with both methods and meshes are virtually the same,although the direct iteration scheme takes more iterations to converge.

L = Length, H = Height of the beam

q

w

= 100LH

= 50LH

= 10LH

(in

.)

(psi.)


Table 4.3.3 The effect of length-to-thickness ratio on the deßections w =wmax10

2Dxx/L4 of a pinnedpinned beam under uniformly

distributed load (4Q element mesh).

Load q0 Length-to-thickness ratio, L/H

10 20 25 100

1.0 1.335 (2)∗ 1.310 (2) 1.307 (2) 0.921 (5)2.0 2.669 (2) 2.620 (2) 2.614 (2) 1.364 (4)3.0 4.004 (2) 3.931 (2) 3.921 (2) 1.660 (3)4.0 5.338 (2) 5.241 (2) 5.228 (2) 1.889 (3)5.0 6.673 (2) 6.551 (2) 6.534 (2) 2.078 (3)6.0 8.008 (2) 7.861 (2) 7.840 (2) 2.241 (3)7.0 9.342 (2) 9.171 (2) 9.145 (2) 2.385 (3)8.0 10.677 (2) 10.480 (2) 10.450 (2) 2.515 (3)9.0 12.011 (2) 11.790 (2) 11.753 (2) 2.633 (3)10.0 13.346 (2) 13.099 (2) 13.056 (2) 2.742 (3)


Figure 4.3.4 Load versus deßection curves for pinnedpinned beam.

0 2 4 6 8 10Load, q (psi)

0.0

0.1

0.2

0.3

L/H=100

L/H=10

EBT = Euler−Bernoulli beam theoryTBT = Timoshenko beam theory

EBTTBT

TBT

w =

wEH

3 /qL4

Non

dim

ensi

onal

defl

ecti

on


Table 4.3.4 Finite element results for the deßections of a clampedclampedbeam under uniformly distributed load (ν = 0.25).

Direct iteration NewtonRaphson Iteration

Load q0 8L 4Q 8L 4Q

1.0 0.1019 (3)∗ 0.1035 (3) 0.1019 (3) 0.1035 (3)2.0 0.1997 (4) 0.2025 (4) 0.1997 (3) 0.2025 (3)3.0 0.2906 (4) 0.2943 (4) 0.2906 (3) 0.2943 (3)4.0 0.3738 (5) 0.3778 (5) 0.3737 (3) 0.3778 (3)5.0 0.4493 (5) 0.4535 (5) 0.4492 (3) 0.4534 (3)6.0 0.5178 (6) 0.5219 (6) 0.5179 (3) 0.5220 (3)7.0 0.5806 (7) 0.5846 (7) 0.5805 (3) 0.5845 (3)8.0 0.6379 (8) 0.6416 (8) 0.6380 (3) 0.6418 (3)9.0 0.6908 (8) 0.6947 (9) 0.6910 (3) 0.6946 (3)10.0 0.7406 (9) 0.7434 (10) 0.7403 (3) 0.7436 (3)


Figure 4.3.5 Loaddeßection response predicted by the EBT and TBT forclampedclamped, thin (L/H = 100) and thick (L/H = 10)beams.

We close this chapter with a note that the geometric nonlinearity consideredin the analysis of beams in this chapter is one where the strains are assumed tobe small while the rotations are moderately large. In Chapter 9 on continuumformulations, we will revisit this topic in the context of plane elasticity.


Problems4.1 Consider the nonlinear differential equations

−dNxx

dx= f(x) (a)

d2

dx2

µDxx

d2w0dx2

¶− d

dx

³Nxx

dw0dx

´= q(x) (b)

where q(x) is the distributed transverse force (positive upward), and

Nxx = Axx

∙du0dx

+1

2

³dw0dx

´2¸(c)

Rewrite the equations (by introducing the bending moment M(x) as a dependentvariable) as a set of second-order equations

− d

dx

½Axx

∙du0dx

+1

2

³dw0dx

´2¸¾− f(x) = 0 (d)

−d2w0dx2

− Mxx

Dxx= 0 (e)

−d2Mxx

dx2− d

dx

½dw0dx

Axx

∙du0dx

+1

2

³dw0dx

´2¸¾= q(x) (f)

Develop: (a) the weak form, and (b) the Þnite element model using interpolation ofthe form

u0 =

mXj=1

uejψ(1)j (x), w0 =

nXj=1

wejψ(2)j (x), Mxx =

rXj=1

Mej ψ

(3)j (x) (g)

4.2 Consider the problem of (linear) bending of beams according to the EulerBernoullibeam theory. The principle of minimum total potential energy states that if thebeam is in equilibrium then the total potential energy associated with the equilibriumconÞguration is the minimum; that is, the equilibrium displacements are those whichmake the total potential energy a minimum. Thus, solving the equations governing theequilibrium of the EulerBernoulli beam is equivalent to minimizing the total potentialenergy

Π(u0,w0) =

Z xb

xa

½Axx2

∙du0dx

+1

2

³dw0dx

´2¸2+Dxx

2

µd2w0dx2

¶2¾dx

−Z xb

xa

(fu0 + qw0) dx (a)

where Axx = ExA and Dxx = ExIyy are the extensional and bending stiffnesses.The necessary condition for the minimum of a functional is that its Þrst variation bezero: δΠ = 0, which yields the governing equations of equilibrium. As you know, thestatement δΠ = 0 is the same as the weak forms of the governing equations of theEulerBernoulli beam theory. The weak form requires Hermite cubic interpolationof the transverse deßection w0. Now suppose that we wish to relax the continuityrequired of the interpolation used for w0(x) by introducing the relation

dw0dx

= ϕ(x) (b)


Then the total potential energy functional takes the form

Π(u0,w0,ϕ) =

Z xb

xa

½Axx2

∙du0dx

+1

2

³dw0dx

´2¸2+Dxx

2

³dϕ

dx

´2¾dx

−Z xb

xa

(fu0 + qw0) dx−6Xi=1

∆eiQ

ei (c)

Since the functional now contains only the Þrst derivative of u0 and ϕ, linear Lagrange(minimum) interpolation can be used. Thus the original problem is replaced with thefollowing equivalent problem:Minimize Π(u0,w0,ϕ) in Eq. (c) subjected to the constraint

dw0dx

−ϕ(x) = 0 (d)

Develop the penalty function formulation of the constrained problem by deriving (a)the weak form, and (b) the Þnite element model.

4.3 Develop the weak forms of the governing equations (4.3.8)(4.3.10); make use of thedeÞnitions in Eq. (4.3.20).

4.4 Analyze a clamped (at both ends) beam under uniformly distributed load (of intensity,q0) using the NewtonRaphson iteration method and (a) the EulerBernoulli beamelement, and (b) the Timoshenko beam element. Plot the non-dimensional maximumdeßection, w = wmax/L versus the load parameter, P = q0L

3/EI, with at least twelvepoints on the graph. Use 8 elements in the beam, with

EA = 30× 106h1 + 0.5

x

L

i, EI =

30× 10612

h1 + 0.5

x

L

i3(a)

and L = 100. Investigate the effect of numerical integration rule on the accuracy ofthe results.

4.5 The principle of minimum total potential energy for axisymmetric bending of polarorthotropic plates according to the Þrst-order shear deformation theory requiresδΠ(w0,φ) = 0, where

δΠm = 2

Z a

b

"³D11

dφ

dr+D12

φ

r

´dδφ

dr+1

r

³D12

dφ

dr+D22

φ

r

´δφ

+A55

³φ+

dw0dr

´³δφ+

dδw0dr

´− 2qδw0

#rdr (a)

where b is the inner radius and a the outer radius. Derive the displacement Þniteelement model of the equations. In particular, show that the Þnite element model isof the form (i.e. deÞne the matrix coefficients of the following equation)∙

[K11] [K12][K12]T [K22]

¸½wφ

¾=

½F0

¾(b)

4.6 Implement the displacement Þnite element model into a computer program and verifywith analytical solutions of simply supported and clamped circular plates (see Reddy[3]).


4.7 Include the geometric nonlinearity (by accounting for the von Karman nonlinearstrains) in the total potential energy functional of Problem 4.5, and develop the Þniteelement model.

4.8 Implement the nonlinear Þnite element model of Problem 4.7 into a computer programand validate it with other published results (see Chapter 6).

References


2. Reddy, J. N., Energy Principles and Variational Methods in Applied Mechanics, 2nd edn,John Wiley, New York (2002).

3. Reddy, J. N., Theory and Analysis of Elastic Plates, Taylor and Francis, Philadelphia,PA (1999).

5

Heat Transfer andOther Field Problemsin Two Dimensions

5.1 Model Equation

The Þnite element analysis of nonlinear two-dimensional problems involves thesame basic steps as those described for one-dimensional problems in Chapter3. Finite element formulation of a model second-order equation was presentedin Section 2.3. Here, we extend that development to problems in which thecoefficients of the differential equation are possibly functions of the dependentvariable and its derivatives.Consider the problem of Þnding the solution u of the following second-

order partial differential equation (a slightly more general equation than theone considered in Section 2.3)

− ∂

∂x

µaxx

∂u

∂x

¶− ∂

∂y

µayy∂u

∂y

¶+ a00u = f(x, y) in Ω (5.1.1)

where axx, ayy, and a00 are known functions of position (x, y) and thedependent unknown u and its derivatives, and f is a known function of positionin a two-dimensional domain Ω with boundary Γ [see Figure 5.1.1(a)]. Forexample, axx may be assumed to be of the form

axx = axx(x, y, u,∂u

∂x,∂u

∂y) (5.1.2)

The equation is subject to certain boundary conditions, whose form will beapparent from the weak formulation.In the Þnite element method, the domain Ω is discretized into a collection

of elements Ωe [see Figure 5.1.1(b)]

Ω ≈ Ωh =N[e=1

Ωe, Ω = Ω ∪ Γ, Ωe = Ωe ∪ Γe (5.1.3)

Triangular element Quadrilateral elementDomain Boundary

(a) (b)

Ω Γ


Figure 5.1.1 Finite element discretization of a two-dimensional domain Ωand a typical Þnite element Ωe.

5.2 Weak Form

We use a representative element domain Ωe to derive the weak form of themodel equation, with the assumption that each element has a unique geometricshape and associated interpolation functions.Following the steps of Section 2.3, the weak form of Eq. (5.2.1) over a

typical Þnite element Ωe, whether triangular or quadrilateral shape, can bedeveloped. The Þrst step is to multiply Eq. (5.2.1) with a weight function w,which is assumed to be differentiable once with respect to x and y, and thenintegrate the equation over the element domain Ωe :

0 =

ZΩew

∙− ∂

∂x

µaxx

∂u

∂x

¶− ∂

∂y

µayy

∂u

∂y

¶+ a00u− f

¸dx dy (5.2.1)

In the second step we distribute the differentiation among u and w equally.To achieve this, we integrate the Þrst two terms in Eq. (5.2.1) by parts usingthe component forms of the GreenGauss theorem (gradient or divergencetheorem).

0 =

ZΩe

µaxx

∂w

∂x

∂u

∂x+ ayy

∂w

∂y

∂u

∂y+ a00wu− wf

¶dx dy

−IΓew

µaxx

∂u

∂xnx + ayy

∂u

∂yny

¶ds (5.2.2)

where nx and ny are the components (i.e. the direction cosines) of the unitnormal vector

n = nxi+ ny j = cosα i+ sinα j (5.2.3)

Boundary, Γ

Domain , Ω

nn

nx

y

y

x

dsdx

dy

^

TWO-DIMENSIONAL PROBLEMS IN SINGLE VARIABLE 129

on the boundary Γe, and ds is the arc length of an inÞnitesimal line elementalong the boundary (see Figure 5.2.1). The circle on the boundary integraldenotes integration over the closed boundary Γe. From an inspection of theboundary term in Eq. (5.2.2), we note that u is the primary variable. Thecoefficient of the weight function w in the boundary expression is

axx∂u

∂xnx + ayy

∂u

∂yny ≡ qn (5.2.4)

and it constitutes the secondary variable. Thus, the weak form of Eq. (5.2.1)is

0 =

ZΩe

µaxx

∂w

∂x

∂u

∂x+ ayy

∂w

∂y

∂u

∂y+ a00wu− w f

¶dx dy−

IΓew qn ds (5.2.5)

The function qn = qn(s) denotes the outward ßux from the boundary aswe move counter-clockwise along the boundary Γe. The secondary variable qnis of physical interest in most problems. For example, in the case of the heattransfer through an anisotropic medium, aij denotes the conductivities of themedium, and qn denotes the heat ßux normal to the boundary of the element.The weak form (or weighted-integral statement) in (5.2.5) forms the basis ofthe Þnite element model of Eq. (5.2.1).

5.3 Finite Element Model

The weak form in Eq. (5.2.5) requires that the approximation chosen for ushould be at least linear in both x and y so that every term in Eq. (5.2.5)has a non-zero contribution to the integral. Since the primary variable is justu, which must be made continuous between elements, the Lagrange family ofinterpolation functions is admissible. Hence, u is approximated over a typical

Figure 5.2.1 A typical two-dimensional domain with a curved boundary.


Þnite element Ωe by the expression

u(x, y) ≈ ueh(x, y) =nXj=1

uejψej (x, y) (5.3.1)

where uej is the value of ueh at the jth node of the element, and ψ

ej are the

Lagrange interpolation functions, which have the property

ψei (xj , yj) = δij (5.3.2)

where (xj , yj) are the global coordinates of the jth node of the element Ωe.

Substituting the Þnite element approximation (5.3.1) for u into the weakform (5.2.5), we obtain

0 =

ZΩe

"∂w

∂x

⎛⎝axx nXj=1

uj∂ψej∂x

⎞⎠+ ∂w∂y

⎛⎝ayy nXj=1

uj∂ψej∂y

⎞⎠+ a00w nXj=1

ujψej

− wf#dx dy −

IΓew qn ds

=nXj=1

Keiju

ej − fei −Qei (5.3.3)

where

Keij =

ZΩe

Ãaxx

∂ψei∂x

∂ψej∂x

+ ayy∂ψei∂y

∂ψej∂y

+ a00ψeiψ

ej

!dx dy

fei =

ZΩeψei f dx dy (5.3.4)

Qe =

IΓeψei qn ds

Note that Keij = K

eji (i.e. [K] is symmetric), even though it may be a function

of the unknown nodal values uej . The set of n nonlinear algebraic equationscan be written in matrix form as

[Ke]ue = fe+ Qe (5.3.5)

Equation (5.3.5) represents the Þnite element model of Eq. (5.1.1). Thiscompletes the Þnite element model development. The usual tasks of assemblyof element equations, imposition of boundary conditions, and solution of linearalgebraic equations (after an iterative method is applied) are standard andtherefore not discussed here (see Reddy [1]).


5.4 Solution Procedures

5.4.1 Direct Iteration

The assembled form of the nonlinear equation (5.3.5) is

[K(U)]U = F (5.4.1)

where [K] is the assembled coefficient matrix, U the vector of global nodalvalues, and F the assembled source vector. In the direct iteration procedure,the solution at the rth iteration is determined from the equation

[K(U(r−1))]U(r) = F (5.4.2)

where the direct matrix [K] is evaluated using the solution (known) at the(r−1)st iteration. The direct iteration procedure can be applied to the elementequation (5.3.5), as was done in Chapters 3 and 4. We obtain

[Ke(ue(r−1))]ue(r) = F e (5.4.3)

5.4.2 NewtonRaphson Iteration

In the NewtonRaphson procedure, we solve the equation

[T (U(r−1))]δU = −R(U(r−1)) (5.4.4)

where R is the residual vector

−R(U(r−1)) = F− [K(U(r−1))]U(r−1) (5.4.5)

and [T ] is the tangent matrix

[T (U(r−1))] ≡µ∂R∂U

¶(r−1)(5.4.6)

The solution at the end of the rth iteration is then given by

Ur = U(r−1) + δU (5.4.7)

At the element level, Eq. (5.4.4) takes the form

[T e(ue(r−1))]δue = F e(ue(r−1))− [Ke(ue(r−1)]ue(r−1) (5.4.8)

The coefficients of the tangent matrix can be computed using the deÞnition

T eij ≡∂Rei∂uej

=nX

m=1

∂Keim

∂ujuem +K

eij (5.4.9)


Example 5.4.1

Suppose that aexx and aeyy have the form

aexx = aex0(x, y) + a

exu · u+ aexux ·

∂u

∂x+ aexuy ·

∂u

∂y

aeyy = aey0(x, y) + a

eyu · u+ aeyux ·

∂u

∂x+ aeyuy ·

∂u

∂y(5.4.10)

where ax0, axu, and so on are functions of only x and y. In addition, we assume that a00 isonly a function of x and y. Then we have

T eij =Keij +

nXm=1

∂Keim

∂ujuem (5.4.11)

where

nXm=1

∂Keim

∂ujuem =

nXm=1

∙ZΩe

µ∂aexx∂uj

∂ψei∂x

∂ψem∂x

+∂aeyy∂uj

∂ψei∂y

∂ψem∂y

+∂ae00∂uj

ψeiψem

¶dxdy

¸uem

=

nXm=1

½ZΩe

∙µaexuψ

ej + a

exux

∂ψej∂x

+ aexuy∂ψej∂y

¶∂ψei∂x

∂ψem∂x

+

µaeyuψ

ej + a

eyux

∂ψej∂x

+ aeyuy∂ψej∂y

¶∂ψei∂y

∂ψem∂y

#dxdy

)uem

=

ZΩe

∙∂u

∂x

∂ψei∂x

µaexuψ

ej + a

exux

∂ψej∂x

+ aexuy∂ψej∂y

¶+∂u

∂y

∂ψei∂y

µaeyuψ

ej + a

eyux

∂ψej∂x

+ aeyuy∂ψej∂y

¶#dxdy (5.4.12)

Note that, although [Ke] is symmetric, the symmetry of [T e] depends on the nature of thenonlinearity.


5.5.1 Introduction

An accurate representation of non-rectangular domains and domains withcurved boundaries can be accomplished by the use of reÞned meshes and/orirregularly shaped elements. For example, a non-rectangular region can berepresented more accurately by triangular and quadrilateral elements thanrectangular elements. However, it is easy to derive interpolation functions fora rectangular element, and it is easier to evaluate integrals over rectangulargeometries than over irregular geometries. Therefore, it is practical to usequadrilateral elements with straight or curved sides but have a means toevaluate the integrals involved in the deÞnitions of the coefficient matricesKeij and T

eij over the quadrilateral elements. A coordinate transformation


between the coordinates (x, y) used in the formulation of the problem,called global coordinates, and another coordinate system (ξ, η), called localcoordinate system, that is convenient in deriving the interpolation functionsand evaluating the integrals is needed. The choice of the local coordinatesystem is dictated by the choice of numerical integration method. Here weshall use, as was done in one-dimensional problems, the Gauss quadraturemethod to evaluate integrals deÞned over two-dimensional elements. The mainsteps in the numerical evaluation of Þnite element matrices are reviewed here(see Section 2.5 for more details).

5.5.2 Numerical Integration

The transformation between Ωe and Ω is accomplished by a coordinatetransformation of the form [cf. Eqs. (2.5.1) and (2.5.2)]

x =mXj=1

xejψej (ξ, η) , y =

mXj=1

yejψej (ξ, η) (5.5.1)

while a typical dependent variable u(x, y) is approximated by

u(x, y) =nXj=1

uejψej (x, y) =

nXj=1

uejψej (x(ξ, η), y(ξ, η)) (5.5.2)

where ψej denotes the interpolation functions of the master elementΩ and

ψej are interpolation functions of a typical element Ωe over which u isapproximated. The transformation (5.5.1) maps a point (x, y) in a typical

element Ωe of the mesh to a point (ξ, η) in the master element Ω and viceversa, if the Jacobian of the transformation is positive-deÞnite. The positive-deÞnite requirement of the Jacobian dictates admissible geometries of elementsin a mesh.Recall that a Þnite element model is nothing but a system of algebraic

equations among the nodal values of the primary variables and secondaryvariables. The coefficients of these algebraic equations contain integrals of thephysical parameters (e.g. material properties) and approximation functions.The integral expressions are, in general, complicated algebraically due tothe spatial variation of the parameters and their dependence on the solutionand possibly its derivatives as well as due to the coordinate transformations.Therefore, the integrals are evaluated numerically, which requires evaluationof the integrand at a selective number of points in the domain, multiplytheir values with suitable weights and summing. Here we discuss the Gaussquadrature to evaluate integrals over quadrilateral elements.


We illustrate the essential elements of the Gauss quadrature by consideringthe integral expression [see Eq. (5.3.4)]

Keij =

ZΩe

∙axx(x, y)

∂ψei∂x

∂ψej∂x

+ ayy(x, y)∂ψei∂y

∂ψej∂y

+ a00(x, y)ψeiψ

ej

¸dx dy

(5.5.3)

We wish to transform the integral from Ωe to the master element Ω = (ξ, η) :−1 ≤ ξ ≤ 1, −1 ≤ η ≤ 1 so that the Gauss quadrature can be used. FromEq. (2.5.6), we have ( ∂ψei

∂x∂ψei∂y

)= [J ]−1

( ∂ψei∂ξ∂ψei∂η

)(5.5.4)

which gives the derivatives of ψei with respect to the global coordinates (x, y)in terms of the derivatives of ψei with respect to the local coordinates (ξ, η).The matrix [Je] is the Jacobian matrix of the transformation (5.5.1)

[J ]e =

"∂x∂ξ

∂y∂ξ

∂x∂η

∂y∂η

#e(5.5.5)

and its determinant |J | is the Jacobian, which must be greater than zero inorder to invert Eq. (5.5.4). Negative non-zero values of |J | imply that aright-hand coordinate system is transformed to a left-hand coordinate system,which should be avoided. The Jacobian matrix can be computed using thetransformation (5.5.1) in Eq. (5.5.5). We have

[J ]e =

"∂x∂ξ

∂y∂ξ

∂x∂η

∂y∂η

#e=

⎡⎣Pmi=1 xi

∂ ψi∂ξ

Pmi=1 yi

∂ ψi∂ξPm

i=1 xi∂ ψi∂η

Pmi=1 yi

∂ ψi∂η

⎤⎦e

=

⎡⎣ ∂ ψ1∂ξ

∂ ψ2∂ξ · · · ∂ ψm

∂ξ

∂ ψ1∂η

∂ ψ2∂η · · · ∂ ψm

∂η

⎤⎦e⎡⎢⎢⎢⎣x1 y1x2 y2...

...xm ym

⎤⎥⎥⎥⎦e

(5.5.6)

Thus, given the global coordinates (xj , yj) of element nodes and the

interpolation functions ψej used for geometry, the Jacobian matrix can be

evaluated using Eq. (5.5.6). Note that ψej are different, in general, from ψejused in the approximation of the dependent variables. The Jacobian is givenby

|J | = J11J22 − J12J21 (5.5.7)

We have from Eq. (5.5.4)⎧⎪⎨⎪⎩∂ψei∂x

∂ψei∂y

⎫⎪⎬⎪⎭ = [J ]−1⎧⎪⎨⎪⎩∂ψei∂ξ

∂ψei∂η

⎫⎪⎬⎪⎭ ≡ [J∗]⎧⎪⎨⎪⎩∂ψei∂ξ

∂ψei∂η

⎫⎪⎬⎪⎭ (5.5.8)


where

J∗11 =J22|J | , J

∗12 = −

J12|J | , J

∗22 =

J11|J | , J

∗21 = −

J21|J | (5.5.9)

Returning to the coefficients Keij in Eq. (5.5.3), we can write it now in

terms of the natural coordinates (ξ, η) as

Keij =

ZΩ

(axx(ξ, η)

ÃJ∗11∂ψei∂ξ

+ J∗12∂ψei∂η

!ÃJ∗11∂ψej∂ξ

+ J∗12∂ψej∂η

!

+ ayy(ξ, η)

ÃJ∗21∂ψej∂ξ


!ÃJ∗21∂ψej∂ξ


!

+ a00(ξ, η)ψeiψ

ej

)|J | dξ dη

≡ZΩFij(ξ, η) dξ dη (5.5.10)

where the element area dA = dxdy in element Ωe is transformed to |J | dξ dηin the master element Ω.Using the Gauss quadrature formulas for integrals deÞned over a

rectangular master element Ω, which are the same as those for the one-dimensional quadrature, we obtain

ZΩFij(ξ, η) dξ dη =

Z 1

−1

"Z 1

−1Fij(ξ, η) dη

#dξ ≈

Z 1

−1

"NXJ=1

Fij(ξ, ηJ)WJ

#dξ

≈MXI=1

NXJ=1

Fij(ξI , ηJ)WIWJ (5.5.11)

where M and N denote the number of Gauss quadrature points in the ξ andη directions, (ξI , ηJ) denote the Gauss points, and WI and WJ denote thecorresponding Gauss weights (see Table 2.5.1).As already discussed, the number of Gauss points required to evaluate

an integral accurately is based on the following rule: if the integrand is apolynomial of degree p, it is integrated exactly by employing NGP ≡ N =int[12(p+1)]; that is, the smallest integer greater than

12(p+1). In most cases,

the interpolation functions are of the same degree in both ξ and η, and wetake NGP =MGP ≡M .

5.5.3 Element Calculations

Calculation of element coefficients require evaluation of interpolation functionsand their derivatives. The Fortran statements for the calculation of the


Jacobian matrix and determinant are given in the subroutine SHPRCT (seeAppendix 2 of Reddy [1]). The notation used is as follows:

SF (i) = ψei , DSF (1, i) =∂ψei∂ξ

, DSF (2, i) =∂ψei∂η

[ELXY ] = array of the global coordinates of element nodes

ELXY (i, 1) = xi, ELXY (i, 2) = yi

[GJ ] = the Jacobian matrix, [J ]; [GJ ] = [DSF ][ELXY ]

[GJINV ] = inverse of the Jacobian matrix, [GJ ]−1

DET = the determinant of the Jacobian matrix, |J |GDSF (1, i) =

∂ψei∂x

, GDSF (2, i) =∂ψei∂y

[GDSF ] = [GJINV ][DSF ]

The above calculations are carried out in the subroutine SHPRECT.The subroutine ELECOFNT that calculates [Ke], [T e], and fe for the

direct iteration and NewtonRaphson iteration is listed in Box 5.5.1. Thefollowing notation is used (n = NPE):

ELF (i) = fei , ELK(i, j) = Keij , TANG(i, j) =

nXm=1

∂Keim

∂ujuem

GAUSPT (I, J) = Ith Gauss point in the Jth Gauss-point rule (I ≤ J)GAUSWT (I, J) = Ith Gauss weight in the Jth Gauss-point rule (I ≤ J)

ELU(I) = uei from the previous iteration

Other variables have the same meaning as indicated earlier. It is assumed thatthe coefficients ax0 and ay0 are linear functions of x and y

ax0 = ax00 + ax0x x+ ax0y y, ay0 = ay00 + ay0x x+ ay0y y (5.5.12)

Of course, other variations of the coefficients may be assumed and implementedwithout difficulty.

Example 5.5.1

Consider heat transfer in an isotropic medium [axx = ayy in Eq. (5.2.1)] of rectangular shapea × b = 0.18 × 0.1 m. The conductivity axx = ayy = k is assumed to vary according to therelation (a00 = 0)

k = k0 (1 + βT ) (5.5.13)

where k0 is the constant thermal conductivity, β the temperature coefficient of thermalconductivity, and T the temperature. Suppose that there is no internal heat generation (i.e.f = 0) and the boundary conditions are

T (0, y) = 500K, T (a, y) = 300K (5.5.14a)

∂T

∂y= 0 at y = 0, b for any x (5.5.14b)

SUBROUTINE ELECOEF(NPE,NGP,ITYPE,NONLN)C ________________________________________________________________CC Element calculations based on linear and quadratic rectangularC elements and isoparametric formulation are carried out for theC model equation in (4.2.1). C C NPE − Nodes per element (4: linear; 8: serendipity quadratic,C 9: complete quadratic) C NGP − Number of Gauss points.C ITYPE − Type of iterative method used:C ITYPE=1, Direct iteration, ITYPE>1, Newton−Raphson iterationC ________________________________________________________________C

IMPLICIT REAL*8(A-H,O-Z)COMMON/STF/ELF(7),ELK(9,9),ELXY(9,2),ELU(9)COMMON/PST/A10,A1X,A1Y,A20,A2X,A2Y,A00,F0,FX,FY,

* A1U,A1UX,A1UY,A2U,A2UX,A2UYCOMMON/SHP/SF(9),GDSF(2,9)DIMENSION GAUSPT(5,5),GAUSWT(5,5),TANG(9,9)

CDATA GAUSPT/5*0.0D0, -0.57735027D0, 0.57735027D0, 3*0.0D0,

2 -0.77459667D0, 0.0D0, 0.77459667D0, 2*0.0D0, -0.86113631D0,3 -0.33998104D0, 0.33998104D0, 0.86113631D0, 0.0D0, -0.90617984D0,4 -0.53846931D0,0.0D0,0.53846931D0,0.90617984D0/

CDATA GAUSWT/2.0D0, 4*0.0D0, 2*1.0D0, 3*0.0D0, 0.55555555D0,

2 0.88888888D0, 0.55555555D0, 2*0.0D0, 0.34785485D0,3 2*0.65214515D0, 0.34785485D0, 0.0D0, 0.23692688D0,4 0.47862867D0, 0.56888888D0, 0.47862867D0, 0.23692688D0/

CC Initialize the arraysC

DO 100 I = 1,NPEELF(I) = 0.0

DO 100 J = 1,NPEIF(ITYPE.GT.1)THEN

TANG(I,J)=0.0ENDIF

100 ELK(I,J)= 0.0CC Do-loops on numerical (Gauss) integration begin here. SubroutineC SHPRCT (SHaPe functions for ReCTangular elements) is called hereC

DO 200 NI = 1,NGP DO 200 NJ = 1,NGP

XI = GAUSPT(NI,NGP)ETA = GAUSPT(NJ,NGP)CALL SHPRCT (NPE,XI,ETA,DET,ELXY)CNST = DET*GAUSWT(NI,NGP)*GAUSWT(NJ,NGP)X=0.0Y=0.0U = 0.0UX= 0.0UY= 0.0


Box 5.5.1 Subroutine (in Fortran) ELECOFNT for the calculation of element

matrices [Ke], [T e], and fe of the model problem in Eq. (5.2.1).


This is essentially a one-dimensional problem (in the x coordinate), and can be solved assuch (see Table 3.5.1).

Table 5.5.1 shows the linear and nonlinear solutions T (x, y0) for any y0. Direct iterationis used to solve the problem. It took two iterations to converge (² = 0.01). Also, the solutionis independent of the mesh in the y-direction. The present results were found to be identicalto those obtained with the one-dimensional model (for the same tolerance of ² = 0.01).

DO 140 I=1,NPEU=U+ELU(I)*SF(I)UX=UX+ELU(I)*GDSF(1,I)UY=UY+ELU(I)*GDSF(2,I)X=X+ELXY(I,1)*SF(I)

140 Y=Y+ELXY(I,2)*SF(I)CC Define the coefficients of the differential equationC

FXY=F0+FX*X+FY*YAXX=AX0+AXX*X+AXY*YAYY=AY0+AYX*X+AYY*Y IF(NONLN.GT.0)THEN

AXX=AXX+AXU*U+AXUX*UX+AXUY*UYAYY=AYY+AYU*U+AYUX*UX+AYUY*UY

ENDIFCC Define the element source vector and coefficient matrixC

DO 180 I=1,NPEELF(I)=ELF(I)+FXY*SF(I)*CNSTDO 160 J=1,NPE

S00=SF(I)*SF(J)*CNSTS11=GDSF(1,I)*GDSF(1,J)*CNSTS22=GDSF(2,I)*GDSF(2,J)*CNSTELK(I,J)=ELK(I,J)+A11*S11+A22*S22+A00*S00

CC Define the part needed to add to [K] to define [T]C

IF(ITYPE.GT.1)THENS10=GDSF(1,I)*SF(J)*CNSTS20=GDSF(2,I)*SF(J)*CNSTS12=GDSF(1,I)*GDSF(2,J)*CNSTS21=GDSF(2,I)*GDSF(1,J)*CNSTTANG(I,J)=TANG(I,J)

* +UX*(A1U*S10+AXUX*SXX+AXUY*S12)* +UY*(A2U*S20+AYUX*SYX+AYUY*S22)

ENDIF160 CONTINUE180 CONTINUE200 CONTINUE

CC Compute the residual vector and tangent matrix C

IF(ITYPE.GT.1)THENDO 220 I=1,NPEDO 220 J=1,NPE

220 ELF(I)=ELF(I)-ELK(I,J)*ELU(J)DO 240 I=1,NPEDO 240 J=1,NPE

240 ELK(I,J)=ELK(I,J)+TANG(I,J)ENDIFRETURNEND

Box 5.5.1 (continued)


Table 5.5.1 Finite element solutions of a nonlinear heat conduction equation

[k0 = 0.2 W/(mK) and β = 2× 10−3 (K−1)].

x Linear 4× 4L* 2× 2Q9 8× 8L4 4× 4Q9

0.0000 500.00 500.00 500.00 500.00 500.000.0225 475.00 − − 477.31 477.310.0450 450.00 454.02 454.03 454.03 454.030.0675 425.00 − − 430.12 430.120.0900 400.00 405.56 405.57 405.57 405.570.1125 375.00 − − 380.32 380.320.1350 350.00 354.33 354.34 354.34 354.340.1575 325.00 − − 327.58 327.580.1800 300.00 300.00 300.00 300.00 300.00

* m× nL4, for example, denotes a mesh of the four-node linear (L) elements.

Problems

5.1 Consider the nonlinear problem of Example 5.5.1 (heat transfer in two dimensions). Usethe uniform 4 × 4 nine-node quadratic element mesh to analyze the problem using thefollowing data and boundary conditions:

a = 0.18 m, b = 0.1 m, f0 = 0 W/m3 (a)

k = k0 (1 + βT ) , k0 = 25 W/(mC) (b)

T (0, y) = 100 C , T (a, y) = 50 C

k∂T

∂n+ hc(T − T∞) = 0 at y = 0, b (c)

Use β = 0.2, T∞ = 10C, and hc = 50 W/(m2 C).5.2 The energy equation for simultaneous conduction and radiation in a participating

medium can be expressed by−∇ · [ke(T )∇T ] = g

where

ke(T ) = k+16σn2T 3

3β

Here T is the temperature, g is the internal heat generation, n denotes the refractiveindex of the medium, σ is the StefanBoltzman constant, and β is the Roseland meanextinction coefficient (see Ozisik [4]). Develop the Þnite element model of the equationand determine the tangent coefficient matrix for a planar (two-dimensional) domain.

5.3 Repeat Problem 5.2 for a radially axisymmetric domain.

5.4 Suppose that the boundary of a typical Þnite element is subject to both convective andenclosed radiation heat transfer [cf. Eq. (2.3.36)]:

(axx∂u

∂xnx + ayy

∂u

∂yny) + hc(u− uc) + σ²(u4 − u4c ) = qn

where σ is the StefanBoltzman constant and ² is the emissivity of the surface.Reformulate the Þnite element equations in (2.3.39) to account for the radiation term.Hint: σ²(u4 − u4c ) = σ²(u2 + u2c )(u+ uc)(u− uc) ≡ hr(u)(u− uc).


5.5 Compute the tangent coefficient matrix associated with the nonlinear radiation boundarycondition of Problem 5.4.


2. Reddy, J. N. and Gartling, D. K., The Finite Element Method in Heat Transfer andFluid Dynamics, 2nd edn, CRC Press, Boca Raton, FL (2001).

3. Reddy, J. N., Energy Principles and Variational Methods in Applied Mechanics, 2ndedn, John Wiley, New York (2002).

4. Ozisik, M. N., Finite Difference Methods in Heat Transfer, CRC Press, Boca Raton,FL (1994).

5. Holman, J. P., Heat Transfer, 7th edn, McGrawHill, New York (1990).

6

Nonlinear Bending ofElastic Plates and Shells

6.1 Introduction

A plate is a ßat structural element with planform dimensions that are largecompared to its thickness and is subjected to loads that cause bendingdeformation in addition to stretching. In most cases, the thickness is nogreater than one-tenth of the smallest in-plane dimension. A shell is muchlike a plate except that it is not ßat but curved. Because of the smallness ofthickness dimension, it is often not necessary to model them using three-dimensional elasticity equations. Simple two-dimensional plate and shelltheories can be developed to study the deformation and stresses in platestructures undergoing small strains, small to moderate rotations, and largedisplacements (i.e. w0/h ≥ 1).Here we derive governing equations of the classical and Þrst-order theories

of plates and shells with the von Karman strains, and develop theirdisplacement Þnite element models. The principle of virtual displacementsis used to derive the weak forms, and the displacement Þnite element modelsare developed using the weak forms.

6.2 Classical Plate Theory

6.2.1 Assumptions of the Kinematics

The classical plate theory (CPT) is one in which the displacement Þeld isselected so as to satisfy the Kirchhoff hypothesis. The Kirchhoff hypothesishas the following three assumptions (see Figure 6.2.1):

(1) Straight lines perpendicular to the mid-surface (i.e. transverse normals)before deformation, remain straight after deformation.

(2) The transverse normals do not experience elongation (i.e. they are in-extensible).

(3) The transverse normals rotate such that they remain perpendicular to themid-surface after deformation.

z

y

x

z

xw0

u0

∂w0

∂x−

∂w0

∂x−


Figure 6.2.1 Undeformed and deformed geometries of an edge of a plateunder the Kirchhoff assumptions.


Let us denote the undeformed mid-plane of the plate with the symbol Ω0. Thetotal domain of the plate is the tensor product Ω0×(−h/2, h/2). The boundaryof the total domain consists of surfaces St(z = h/2) and Sb(z = −h/2), andthe edge Γ ≡ Γ× (−h/2, h/2). In general, Γ is a curved surface, with outwardnormal n = nxex+nyey, where nx and ny are the direction cosines of the unitnormal.The Kirchhoff hypothesis implies the following form of the displacement

Þeld (see Reddy [13] and Figure 6.2.1)

u(x, y, z) = u0(x, y)− z∂w0∂x

v(x, y, z) = v0(x, y)− z∂w0∂y

(6.2.1)

w(x, y, z) = w0(x, y)

where (u0, v0, w0) denote the displacements of a material point at (x, y, 0) in(x, y, z) coordinate directions. The nonlinear strains are given by

NONLINEAR BENDING OF ELASTIC PLATES AND SHELLS 143

E11 =∂u

∂x+1

2

"µ∂u

∂x

¶2+

µ∂v

∂x

¶2+

µ∂w

∂x

¶2#

E12 =1

2

µ∂u

∂y+∂v

∂x+∂u

∂x

∂u

∂y+∂v

∂x

∂v

∂y+∂w

∂x

∂w

∂y

¶E13 =

1

2

µ∂u

∂z+∂w

∂x+∂u

∂x

∂u

∂z+∂v

∂x

∂v

∂z+∂w

∂x

∂w

∂z

¶E22 =

∂v

∂y+1

2

"µ∂u

∂y

¶2+

µ∂v

∂y

¶2+

µ∂w

∂y

¶2#

E23 =1

2

µ∂v

∂z+∂w

∂y+∂u

∂y

∂u

∂z+∂v

∂y

∂v

∂z+∂w

∂y

∂w

∂z

¶E33 =

∂w

∂z+1

2

"µ∂u

∂z

¶2+

µ∂v

∂z

¶2+

µ∂w

∂z

¶2#(6.2.2)

If the components of the displacement gradients are of the order ², i.e.

∂u

∂x,∂u

∂y,∂v

∂x,∂v

∂y,∂w

∂z= O(²) (6.2.3)

then the small strain assumption implies that terms of the order ²2 are omittedin the strains. If the rotations of transverse normals are moderate (say10 − 15), then the following terms are small but not negligible comparedto ² µ

∂w

∂x

¶2,

µ∂w

∂y

¶2,∂w

∂x

∂w

∂y(6.2.4)

Thus for small strains and moderate rotations, the straindisplacementrelations (6.2.2) take the form

εxx =∂u

∂x+1

2

µ∂w

∂x

¶2, εxy =

1

2

µ∂u

∂y+∂v

∂x+∂w

∂x

∂w

∂y

¶εxz =

1

2

µ∂u

∂z+∂w

∂x

¶, εyy =

∂v

∂y+1

2

µ∂w

∂y

¶2εyz =

1

2

µ∂v

∂z+∂w

∂y

¶, εzz =

∂w

∂z(6.2.5)

where, for this special case of geometric nonlinearity (i.e. small strains butmoderate rotations), the notation εij is used in place of Eij . The correspondingSecondPiola Kirchhoff stresses will be denoted σij .For the displacement Þeld in Eq. (6.2.1), ∂w/∂z = 0. In view of the

assumptions in Eqs. (6.2.3) and (6.2.4), the strains (6.2.5) for the displacement


Þeld (6.2.1) reduce to

εxx =∂u0∂x

+1

2

µ∂w0∂x

¶2− z∂

2w0∂x2

εxy =1

2

Ã∂u0∂y

+∂v0∂x

+∂w0∂x

∂w0∂y

− 2z ∂2w0∂x∂y

!(6.2.6)

εyy =∂v0∂y

+1

2

µ∂w0∂y

¶2− z∂

2w0∂y2

εxz =1

2

µ−∂w0∂x

+∂w0∂x

¶= 0

εyz =1

2

µ−∂w0∂y

+∂w0∂y

¶= 0 (6.2.7)

εzz = 0

The strains in Eqs. (6.2.6) and (6.2.7) are called the von Karman strains, andthe associated plate theory is termed the classical plate theory with the vonKarman strains. Note that the transverse strains (εxz, εyz, εzz) are identicallyzero in the classical plate theory.In matrix notation, Eq. (6.2.6) can be written as⎧⎨⎩

εxxεyyγxy

⎫⎬⎭ =⎧⎨⎩ε0xxε0yyγ0xy

⎫⎬⎭+ z⎧⎨⎩ε1xxε1yyγ1xy

⎫⎬⎭ (6.2.8a)

⎧⎨⎩ε0xxε0yyγ0xy

⎫⎬⎭ =⎧⎪⎪⎪⎨⎪⎪⎪⎩

∂u0∂x +

12

³∂w0∂x

´2∂v0∂y +

12

³∂w0∂y

´2∂u0∂y +

∂v0∂x +

∂w0∂x

∂w0∂y

⎫⎪⎪⎪⎬⎪⎪⎪⎭ ,

⎧⎨⎩ε1xxε1yyγ1xy

⎫⎬⎭ =⎧⎪⎪⎨⎪⎪⎩−∂2w0

∂x2

−∂2w0∂y2

−2∂2w0∂x∂y

⎫⎪⎪⎬⎪⎪⎭ (6.2.8b)

6.3 Variational Formulation of CPT

6.3.1 Virtual Work

Here, the weak form based on the principle of virtual displacements appliedto the classical plate theory is derived for a typical Þnite element Ωe. Inthe derivation, we account for thermal (and hence, moisture) effects onlywith the understanding that the material properties are independent of thetemperature and that the temperature T is a known function of position(hence, δT = 0). Thus, the temperature enters the formulation only throughconstitutive equations.As noted earlier, the transverse strains (γxz, γyz, εzz) are identically zero in

the classical plate theory. Consequently, the transverse stresses (σxz,σyz,σzz)

z

x

y

q(x,y)

ns

Boundary


do not enter the formulation because the virtual strain energy of these stressesis zero (due to the fact that kinematically consistent virtual strains must bezero):

δεxz = 0, δεyz = 0, δεzz = 0 (6.3.1)

Whether the transverse stresses are accounted for or not in a theory, they arepresent in reality to keep the plate in equilibrium. In addition, these stresscomponents may be speciÞed on the boundary. Thus, the transverse stressesdo not enter the virtual strain energy expression but must be accounted for inthe boundary conditions and equilibrium of forces.The principle of virtual displacements is

0 = δW e ≡ (δUe + δV e) (6.3.2)

where Ue is the strain energy stored and V e is the work done by appliedforces in an element. Suppose that q is the distributed force at z = −h/2, and(σnn,σns,σnz) are the stress components on the boundary Γ

e = Γe × (−h2 ,h2 )

of the plate element (see Figure 6.3.1).The virtual strain energy in element Ωe is given by

δUe =

ZΩe

Z h2

−h2

(σxxδεxx + σyyδεyy + 2σxyδεxy) dz dx dy

=

ZΩe

µNxxδε

0xx +Mxxδε

1xx +Nyyδε

0yy +Myyδε

1yy

+Nxyδγ0xy +Mxyδγ

1xy

¶dx dy (6.3.3)

where (Nxx, Nyy, Nxy) are the forces per unit length and (Mxx,Myy,Mxy) arethe moments per unit length (see Figure 6.3.2):⎧⎨⎩

NxxNyyNxy

⎫⎬⎭ =Z h

2

−h2

⎧⎨⎩σxxσyyσxy

⎫⎬⎭ dz ,⎧⎨⎩Mxx

Myy

Mxy

⎫⎬⎭ =Z h

2

−h2

⎧⎨⎩σxxσyyσxy

⎫⎬⎭ z dz (6.3.4)

Figure 6.3.1 Geometry of a plate element with curved boundary.

x

z

y

Myy

Qx

Nxx

Nyy

Mxx

Mxy

Qy

QnNnn

NnsMns

Mxy

Nxy

Nxy

Mnn

MyyQx

Nxx

Nyy

Mxx

Qy

Qn

Nns

Mns

Mxy

Nnn

Mnn

Nxy

Nxy

Mxy


The virtual work done by the distributed transverse load q(x, y) applied atz = −h/2, the transverse reaction force of an elastic foundation at z = h/2,in-plane normal stress σnn, in-plane tangential stress σns, and transverse shearstress σnz is

δV e = −½Z

Ωeq(x, y)δw(x, y,

−h2)dx dy +

ZΩeFs(x, y)δw(x, y,

h

2) dxdy

+

IΓe

Z h2

−h2

[σnnδun + σnsδus + σnzδw] dzds

¾(6.3.5)

= −½Z

Ωe(q − kw0) δw0 dx dy +

IΓe

µNnnδu0n −Mnn

∂δw0∂n

+Nnsδu0s −Mns∂δw0∂s

+Qnδw0

¶ds

¾(6.3.6)

where Fs = −kw0, k is the foundation modulus, u0n, u0s, and w0 arethe displacements along the normal, tangential, and transverse directions,respectively, and (see Figure 6.3.2)½

NnnNns

¾=

Z h2

−h2

½σnnσns

¾dz,

½Mnn

Mns

¾=

Z h2

−h2

½σnnσns

¾z dz (6.3.7a)

Figure 6.3.2 Forces and moments per unit length on a plate element.


Qn =

Z h2

−h2

σnz dz (6.3.7b)

The stresses (σnn,σns) on the boundary Γe are related to (σxx,σyy,σxy) in the

interior of Ωe by the transformation½σnnσns

¾=

∙n2x n2y 2nxny

−nxny nxny n2x − n2y

¸⎧⎨⎩σxxσyyσxy

⎫⎬⎭ (6.3.8)

6.3.2 Weak Forms

Substituting for δUe and δV e from Eqs. (6.3.3) and (6.3.6) into the virtualwork statement in Eq. (6.3.2), we obtain the weak form

0 =

ZΩe

µNxxδε

0xx +Mxxδε

1xx +Nyyδε

0yy +Myyδε

1yy +Nxyδγ

0xy

+Mxyδγ1xy + kw0δw0 − qδw0

¶dx dy

−IΓe

µNnnδu0n +Nnsδu0s −Mnn

∂δw0∂n

−Mns∂δw0∂s

+Qnδw0

¶ds

=

ZΩe

∙µ∂δu0∂x

+∂w0∂x

∂δw0∂x

¶Nxx +

µ∂δv0∂y

+∂w0∂y

∂δw0∂y

¶Nyy

+

µ∂δu0∂y

+∂δv0∂x

+∂w0∂x

∂δw0∂y

+∂w0∂y

∂δw0∂x

¶Nxy

− ∂2δw0∂x2

Mxx − ∂2δw0∂y2

Myy − 2∂2δw0∂x∂y

Mxy + kδw0w0 − δw0q¸dx dy

−IΓe


∂δw0∂n

−Mns∂δw0∂s

+Qnδw0

¶ds

(6.3.9)

The statement (6.3.9) is equivalent to the following three weak forms:

0 =

ZΩe

µ∂δu0∂x

Nxx +∂δu0∂y

Nxy

¶dx dy −

IΓeNnnδu0n ds (6.3.10)

0 =

ZΩe

µ∂δv0∂x

Nxy +∂δv0∂y

Nyy

¶dx dy −

IΓeNnsδu0s ds (6.3.11)

0 =

ZΩe

∙∂δw0∂x

µ∂w0∂x

Nxx +∂w0∂y

Nxy

¶+∂δw0∂y

µ∂w0∂x

Nxy +∂w0∂y

Nyy

¶− ∂

2δw0∂x2

Mxx − ∂2δw0∂y2

Myy − 2∂2δw0∂x∂y

Mxy + kδw0w0 − δw0q¸dx dy

−IΓe

µ−Mnn

∂δw0∂n

−Mns∂δw0∂s

+Qnδw0

¶ds (6.3.12)


6.3.3 Equilibrium Equations

To obtain the governing equations of the CPT, integrate expressions in Eqs.(6.3.10)(6.3.12) by parts to relieve the virtual displacements (δu0, δv0, δw0)in Ωe of any differentiation (so that we can use the fundamental lemma ofvariational calculus); we obtain

0 =

ZΩe

"− (Nxx,x +Nxy,y) δu0 − (Nxy,x +Nyy,y) δv0

−ÃMxx,xx + 2Mxy,xy +Myy,yy +N − kw0 + q

!δw0

#dx dy

+

IΓe

"(Nxxnx +Nxyny) δu0 + (Nxynx +Nyyny) δv0

+

ÃMxx,xnx +Mxy,ynx +Myy,yny +Mxy,xny + P

!δw0

− (Mxxnx +Mxyny)∂δw0∂x

− (Mxynx +Myyny)∂δw0∂y

¸ds

−IΓe


∂δw0∂n

−Mns∂δw0∂s

+Qnδw0

¶ds

(6.3.13)

where a comma followed by subscripts denotes differentiation with respect tothe subscripts: Nxx,x = ∂Nxx/∂x, and so on, and

N (u0, v0, w0) = ∂

∂x

µNxx

∂w0∂x

+Nxy∂w0∂y

¶+∂

∂y

µNxy

∂w0∂x

+Nyy∂w0∂y

¶

P(u0, v0, w0) =µNxx

∂w0∂x

+Nxy∂w0∂y

¶nx +

µNxy

∂w0∂x

+Nyy∂w0∂y

¶ny

(6.3.14)The equations of equilibrium are obtained by setting the coefficients of δu0,δv0, and δw0 in Ω

e to zero:

δu0 :∂Nxx∂x

+∂Nxy∂y

= 0 (6.3.15)

δv0 :∂Nxy∂x

+∂Nyy∂y

= 0 (6.3.16)

δw0 :∂2Mxx

∂x2+ 2

∂2Mxy

∂y∂x+∂2Myy

∂y2

+N (u0, v0, w0)− kw0 + q = 0 (6.3.17)


6.3.4 Boundary Conditions

To cast the boundary conditions on an arbitrary edge whose normal is n,we express all generalized displacements (u0, v0, w0,

∂w0∂x ,

∂w0∂x )from the (x, y, z)

system in terms of the corresponding generalized displacements in the normal,tangential, and transverse directions. We have

u0 = u0nnx − u0sny, v0 = u0nny + u0snx (6.3.18a)

∂w0∂x

=∂w0∂n

nx − ∂w0∂sny,

∂w0∂y

=∂w0∂n

ny +∂w0∂snx (6.3.18b)

The boundary expression of Eq. (6.3.13) takes the form

IΓe

"(Nxxnx +Nxyny) (δu0nnx − δu0sny)

+ (Nxynx +Nyyny) (δu0nny + δu0snx)

+

ÃMxx,xnx +Mxy,ynx +Myy,yny +Mxy,xny + P

!δw0

− (Mxxnx +Mxyny)

µ∂δw0∂n

nx − ∂δw0∂s

ny

¶− (Mxynx +Myyny)

µ∂δw0∂n

ny +∂δw0∂s

nx

¶#ds

−IΓe


∂δw0∂n

−Mns∂δw0∂s

+Qnδw0

¶ds

=

IΓe

(³Nxxn

2x + 2Nxynxny +Nyyn

2y −Nnn

´δu0n

+h(Nyy −Nxx)nxny +Nxy(n2x − n2y)−Nns

iδu0s

+ (Mxx,xnx +Mxy,ynx +Myy,yny +Mxy,xny + P −Qn) δw0−³Mxxn

2x + 2Mxynxny +Myyn

2y −Mnn

´ ∂δw0∂n

−h(Myy −Mxx)nxny +Mxy(n

2x − n2y)−Mns

i ∂δw0∂s

)ds (6.3.19)

The natural boundary conditions are obtained by setting the coefficients ofδu0, δv0, δw0,

∂δw0∂n and ∂δw0

∂s on Γe to zero:

δu0n : Nnn = Nxxn2x + 2Nxynxny +Nyyn

2y

δu0s : Nns = (Nyy −Nxx)nxny +Nxy(n2x − n2y)δw0 : Qn =Mxx,xnx +Mxy,ynx +Myy,yny +Mxy,xny + P

(6.3.20a)


∂δw0∂n

: Mnn =Mxxn2x + 2Mxynxny +Myyn

2y

∂δw0∂s

: Mns = (Myy −Mxx)nxny +Mxy(n2x − n2y)

(6.3.20b)

From Eq. (6.3.20), it is clear that the primary variables (i.e. generalizeddisplacements) and secondary variables (i.e. generalized forces) of the theoryare:

Primary variables: u0n, u0s, w0,∂w0∂n

,∂w0∂s

Secondary variables: Nnn, Nns, Qn, Mnn, Mns

(6.3.21)

We note that the equations of equilibrium in Eqs. (6.3.15)(6.3.17) havethe total spatial differential order of eight. In other words, if the governingequations are expressed in terms of the displacements (u0, v0, w0), they wouldcontain second-order spatial derivatives of u0 and v0 and fourth-order spatialderivatives of w0. This implies that there should be only eight (four essentialand four natural) boundary conditions, whereas Eq. (6.3.21) shows Þveessential and Þve natural boundary conditions, giving a total of ten boundaryconditions. To eliminate this discrepancy, one may integrate the tangentialderivative term by parts to obtain the boundary term

−IΓMns

∂δw0∂s

ds =

IΓ

∂Mns

∂sδw0 ds− [Mnsδw0]Γ (6.3.22)

The term [Mnsδw0]Γ is zero when the end points of a closed curve coincideor when Mns = 0. If Mns = 0 is not speciÞed at corners of the boundary Γof a polygonal plate, concentrated forces of magnitude Fc = −2Mns will beproduced at the corners. The factor of 2 appears because Mns from two sidesof the corner are added there.The remaining boundary term in Eq. (6.3.22) is added to the shear force

Qn (because it is a coefficient of δw0 on Γe) to obtain the effective shear force

Vn ≡ Qn + ∂Mns

∂s(6.3.23)

The speciÞcation of this effective shear force Vn is known as the Kirchhoff free-edge condition. Finally, the correct boundary conditions of the CPT involvespecifying the following quantities:

Generalized displacements: u0n, u0s, w0,∂w0∂n

Generalized forces: Nnn, Nns, Vn, Mnn

(6.3.24)

Thus, at every boundary point one must know u0n or Nnn, u0s or Nns, w0 orVn, and ∂w0/∂n or Mnn. On an edge parallel to the x-axis (i.e. s = x and


n = y), for example, the above boundary conditions become

u0n = v0, u0s = u0, w0,∂w0∂n

=∂w0∂y

(6.3.25a)

Nnn = Nyy, Nns = Nyx, Vn = Vy, Mnn =Myy (6.3.25b)

Next we discuss some common types of boundary conditions for the linearbending of a rectangular plate with edges parallel to the x and y coordinates.Here we use the edge at y = 0 (nx = 0 and ny = −1) to discuss the boundaryconditions (see Figure 6.3.2). It should be noted that only one element ofeach of the four pairs may (and should) be speciÞed on an edge of a plate.The force boundary conditions may be expressed in terms of the generalizeddisplacements using the plate constitutive equations discussed in the nextsection.

Free edge, y = 0: A free edge is one that is geometrically not restrained inany way. Hence, we have

u0 6= 0, v0 6= 0, w0 6= 0, ∂w0∂y

6= 0 (6.3.26a)

However, the edge may have applied forces and/or moments

Nxy = Nxy, Nyy = Nyy, Vn ≡ −Qy − ∂Mxy

∂x= Vn, Myy = Myy (6.3.26b)

where quantities with a hat are speciÞed forces/moments. For free rectangularplates, Mxy = 0, hence no corner forces are developed.

Fixed (or clamped) edge, y = 0: A Þxed edge is one that is geometricallyfully restrained

u0 = 0, v0 = 0, w0 = 0,∂w0∂y

= 0 (6.3.27)

Therefore, the forces and moments on a Þxed edge are not known a priori (i.e.they are reactions to be determined as a part of the analysis). For clampedrectangular plates, Mxy = 0, hence no corner forces are developed.

Simply supported edge y = 0: The phrase simply supported does notuniquely deÞne the boundary conditions, and one must indicate what it means,especially when both in-plane and bending deßections are involved. Here wedeÞne two types of simply supported boundary conditions:

SS1: u0 = 0, w0 = 0; Nyy = Nyy, Myy = Myy (6.3.28)

SS2: v0 = 0, w0 = 0; Nxy = Nxy, Myy = Myy (6.3.29)


6.3.5 Stress ResultantDeßections Relations

To express the forces and moments (N, M) per unit length in termsof the generalized displacements (u0, v0, w0), we must invoke appropriatestressstrain relations. In the CPT, all three transverse strain components(εzz, εxz, εyz) are zero by deÞnition. Since εzz = 0, the transverse normalstress σzz, though not zero identically, does not appear in the virtual workstatement and hence in the equations of motion. Consequently, it amountsto neglecting the transverse normal stress. Thus we have, in theory, a case ofboth plane strain and plane stress. However, from practical considerations, athin to moderately thick plate is in a state of plane stress because the thicknessis small compared to the in-plane dimensions. Hence, the plane stress-reducedconstitutive relations may be used.For an orthotropic material with principal materials axes (x1, x2, x3)

coinciding with the plate coordinates (x, y, z), the plane stress-reducedthermoelastic constitutive equations can be expressed as (see Reddy [3,40])⎧⎨⎩

σxxσyyσxy

⎫⎬⎭ =⎡⎣Q11 Q12 0Q12 Q22 00 0 Q66

⎤⎦⎧⎨⎩εxx − α1 ∆Tεyy − α2 ∆T

γxy

⎫⎬⎭ (6.3.30)

where Qij are the plane stress-reduced stiffnesses

Q11 =E1

1− ν12ν21 , Q12 =ν12E2

1− ν12ν21 =ν21E1

1− ν12ν21 (6.3.31a)

Q22 =E2

1− ν12ν21 , Q66 = G12 (6.3.31b)

and (σi, εi) are the stress and strain components, respectively, α1 and α2 arethe coefficients of thermal expansion, and ∆T is the temperature incrementfrom a reference state, ∆T = T − T0. The moisture strains are similarto thermal strains (i.e. for moisture strains replace ∆T and αi with themoisture concentration increment and coefficients of hygroscopic expansion,respectively).The plate constitutive equations relate the forces and moments per unit

length in Eq. (6.3.4) to the strains (6.2.8b) of the plate theory. For a platemade of a single or multiple orthotropic layers, the plate constitutive relationsare obtained using the deÞnitions in Eq. (6.3.4). For plates laminated ofmultiple orthotropic layers whose material axes are arbitrarily oriented withrespect to the plate axes, the plate constitutive relations couple the in-planedisplacements to the out-of-plane displacements even for linear problems (seeReddy [40] for details). For a single orthotropic layer, the plate constitutiverelations are greatly simpliÞed. They are⎧⎨⎩

NxxNyyNxy

⎫⎬⎭ =Z h

2

−h2

⎡⎣Q11 Q12 0Q12 Q22 00 0 Q66

⎤⎦⎧⎨⎩ε0xx + zε

1xx − α1∆T

ε0yy + zε1yy − α2∆T

γ0xy + zγ1xy

⎫⎬⎭ dz


=

⎡⎣A11 A12 0A12 A22 00 0 A66

⎤⎦⎧⎨⎩ε0xxε0yyγ0xy

⎫⎬⎭−⎧⎨⎩NTxx

NTyy

0

⎫⎬⎭ (6.3.32)

⎧⎨⎩Mxx

Myy

Mxy

⎫⎬⎭ =Z h

2

−h2

⎡⎣Q11 Q12 0Q12 Q22 00 0 Q66

⎤⎦⎧⎨⎩ε0xx + zε

1xx − α1∆T

ε0yy + zε1yy − α2∆T

γ0xy + zγ1xy

⎫⎬⎭ z dz=

⎡⎣D11 D12 0D12 D22 00 0 D66


⎫⎬⎭−⎧⎨⎩MTxx

MTyy

0

⎫⎬⎭ (6.3.33)

where Aij are extensional stiffnesses and Dij are bending stiffnesses, which aredeÞned in terms of the elastic stiffnesses Qij as

(Aij ,Dij) =

Z h2

−h2

Qij(1, z2)dz or Aij = hQij , Dij =

h3

12Qij (6.3.34)

and NT and MT are thermal stress resultants½NTxx

NTyy

¾=

½Q11α1 +Q12α2Q12α1 +Q22α2

¾Z h2

−h2

∆T (x, y, z) dz (6.3.35a)

½MTxx

MTyy

¾=

½Q11α1 +Q12α2Q12α1 +Q22α2

¾Z h2

−h2

∆T (x, y, z) z dz (6.3.35b)

where α1 and α2 are the thermal coefficients of expansion, and ∆T is thetemperature change (above a stress-free temperature), which is a knownfunction of position. For isotropic plates, Eqs. (6.3.35a,b) simplify to

NTxx = N

Tyy =

NT(1− ν) , NT = Eα

Z h2

−h2

∆T dz (6.3.36a)

MTxx =M

Tyy =

MT

(1− ν) , MT = Eα

Z h2

−h2

∆T z dz (6.3.36b)

6.4 Finite Element Models of CPT

6.4.1 General Formulation

In this section, the displacement Þnite element model of Eqs. (6.3.15)(6.3.17) governing plates according to the CPT is developed. The virtualwork statements of the CPT over a typical orthotropic plate Þnite element Ωe


are given by [from Eqs. (6.3.10)(6.3.12)]

0 =

ZΩe

Ã∂δu0∂x

(A11

"∂u0∂x

+1

2

µ∂w0∂x

¶2#+A12

"∂v0∂y

+1

2

µ∂w0∂y

¶2#)

+A66∂δu0∂y

∙∂u0∂y

+∂v0∂x

+∂w0∂x

∂w0∂y

¸!dx dy

−ZΩe

∂δu0∂x

NTxxdx dy −

IΓeNnδu0n ds (6.4.1a)

0 =

ZΩe

Ã∂δv0∂y

(A12

"∂u0∂x

+1

2

µ∂w0∂x

¶2#+A22

"∂v0∂y

+1

2

µ∂w0∂y

¶2#)

+A66∂δv0∂x

∙∂u0∂y

+∂v0∂x

+∂w0∂x

∂w0∂y

¸!dx dy

−ZΩe

∂δv0∂y

NTyydx dy −

IΓeNsδu0s ds (6.4.1b)

0 =

ZΩe

(∂δw0∂x

"∂w0∂x

(A11

"∂u0∂x

+1

2

µ∂w0∂x

¶2#+A12

"∂v0∂y

+1

2

µ∂w0∂y

¶2#)

+A66∂w0∂y

µ∂u0∂y

+∂v0∂x

+∂w0∂x

∂w0∂y

¶#

+∂δw0∂y

"∂w0∂y

(A12

"∂u0∂x

+1

2

µ∂w0∂x

¶2#+A22

"∂v0∂y

+1

2

µ∂w0∂y

¶2#)

+A66∂w0∂x

µ∂u0∂y

+∂v0∂x

+∂w0∂x

∂w0∂y

¶#

+∂2δw0∂x2

ÃD11

∂2w0∂x2

+D12∂2w0∂x2

!+∂2δw0∂y2

ÃD12

∂2w0∂x2

+D22∂2w0∂x2

!

+ 4D66∂2δw0∂x∂y

∂2w0∂x∂y

+ kδw0w0 − δw0q)dx dy

+

ZΩe

Ã∂2δw0∂x2

MTxx +

∂2δw0∂y2

MTyy

!dx dy −

IΓe

µVnδw0 −Mn

∂δw0∂n

¶ds

(6.4.1c)

where

Nn = Nxxnx +Nxyny, Ns = Nxynx +Nyyny

Vn = Qn +∂Mns

∂s(6.4.2)

Mn =Mxxnx +Mxyny, Ms =Mxynx +Myyny


and (nx, ny) denote the direction cosines of the unit normal on the elementboundary Γe. We note from the boundary terms in Eqs. (6.4.1a-c) thatu0, v0, w0, and ∂w0/∂n are used as the primary variables (or generalized

displacements), and Nn, Ns, Vn, and Mn as the secondary degrees offreedom (or generalized forces). Thus, Þnite elements based on the CPTrequire continuity of the transverse deßection w0 and its derivatives acrosselement boundaries (i.e. C1-continuity of w0). Also, to satisfy the constantdisplacement (rigid body mode) and constant strain requirements, thepolynomial expansion for w0 should be a complete quadratic. The in-planedisplacements u0 and v0 need only be C

0 continuous.Assume Þnite element approximation of the form

u0(x, y) =mXj=1

uejψej (x, y)

v0(x, y) =mXj=1

vejψej (x, y) (6.4.3)

w0(x, y) =nXj=1

∆ejϕej(x, y)

where ψej are the Lagrange interpolation functions, ∆ej are the values of w0

and its derivatives at the nodes, and ϕej are the interpolation functions, thespeciÞc form of which will depend on the geometry of the element and thenodal degrees of freedom interpolated. Substituting approximations (6.4.3)for (u0, v0, w0) and (ψ

ei ,ψ

ei ,ϕ

ei ) for the virtual displacements (δu0, δv0, δw0)

into Eqs. [6.4.1(a)(c)], we obtain⎡⎣ [K11] [K12] [K13][K21] [K22] [K23][K31] [K32] [K33]

⎤⎦⎧⎨⎩uv∆

⎫⎬⎭ =⎧⎨⎩F 1F 2F 3

⎫⎬⎭+⎧⎨⎩F 1TF 2TF 3T

⎫⎬⎭ (6.4.4)

The stiffness coefficients Kαβij (not symmetric) and force vectors Fαi and F

αTi

(α,β = 1, 2, 3) are deÞned as follows:

K11ij =

ZΩe

ÃA11

∂ψei∂x

∂ψej∂x

+A66∂ψei∂y

∂ψej∂y

!dx dy

K12ij =

ZΩe

ÃA12

∂ψei∂x

∂ψej∂y

+A66∂ψei∂y

∂ψej∂x

!dx dy = K21

ji

K13ij =

1

2

ZΩe

"∂ψei∂x

ÃA11

∂w0∂x

∂ϕej∂x

+A12∂w0∂y

∂ϕej∂y

!

+A66∂ψei∂y

Ã∂w0∂x

∂ϕej∂y

+∂w0∂y

∂ϕej∂x

!#dx dy


K22ij =

ZΩe

ÃA66

∂ψei∂x

∂ψej∂x

+A22∂ψei∂y

∂ψej∂y

!dx dy

K23ij =

1

2

ZΩe

"∂ψei∂y

ÃA12

∂w0∂x

∂ϕej∂x

+A22∂w0∂y

∂ϕej∂y

!

+A66∂ψei∂x

Ã∂w0∂x

∂ϕej∂y

+∂w0∂y

∂ϕej∂x

!#dx dy

K31ij =

ZΩe

"∂ϕei∂x

ÃA11

∂w0∂x

∂ψej∂x

+A66∂w0∂y

∂ψej∂y

!

+∂ϕei∂y

ÃA66

∂w0∂x

∂ψej∂y

+A12∂w0∂y

∂ψej∂x

!#dx dy

K32ij =

ZΩe

"∂ϕei∂x

ÃA12

∂w0∂x

∂ψej∂y

+A66∂w0∂y

∂ψej∂x

!

+∂ϕei∂y

ÃA66

∂w0∂x

∂ψej∂x

+A22∂w0∂y

∂ψej∂y

!#dx dy

K33ij =

ZΩe

"D11

∂2ϕei∂x2

∂2ϕej∂x2

+D22∂2ϕei∂y2

∂2ϕej∂y2

+D12

Ã∂2ϕei∂x2

∂2ϕej∂y2

+∂2ϕei∂y2

∂2ϕej∂x2

!

+ 4D66∂2ϕei∂x∂y

∂2ϕej∂x∂y

+ kϕeiϕej

#dx dy

+1

2

ZΩe

("A11

µ∂w0∂x

¶2+A66

µ∂w0∂y

¶2# ∂ϕei∂x

∂ϕej∂x

+

"A66

µ∂w0∂x

¶2+A22

µ∂w0∂y

¶2# ∂ϕei∂y

∂ϕej∂y

+ (A12 +A66)∂w0∂x

∂w0∂y

Ã∂ϕei∂x

∂ϕej∂y

+∂ϕei∂y

∂ϕej∂x

!)dx dy (6.4.5)

F 1i =

IΓeNnψ

ei ds, F

2i =

IΓeNsψ

ei ds

F 3i =

ZΩeqϕei dx dy +

IΓe

µVnϕ

ei −Mn

∂ϕei∂n

¶ds

F 1Ti =

ZΩeNTxx

∂ψei∂x

dxdy, F 2Ti =

ZΩeNTyy

∂ψei∂y

dx dy

F 3Ti = −ZΩe

Ã∂2ϕei∂x2

MTxx +

∂2ϕei∂y2

MTyy

!dx dy (6.4.6)


where NTxx,M

Txx, etc. are the thermal forces and moments deÞned in Eqs.

(6.3.35a,b).This completes the general Þnite element model development of the classical

plate theory. The Þnite element model in Eq. (6.4.4) is called a displacementÞnite element model because it is based on equations of motion expressed interms of the displacements, and the generalized displacements are the primarynodal degrees of freedom.

6.4.2 Tangent Stiffness Coefficients

The solution of nonlinear algebraic equations arising in the analysis ofstructural problems, one often uses the NewtonRaphson method or itsimprovements. To this end it useful to derive the tangent stiffness coefficientsassociated with the CPT.The NewtonRaphson iterative method involves solving equations of the

form ⎡⎣ [T 11] [T 12] [T 13][T 21] [T 22] [T 23][T 31] [T 32] [T 33]

⎤⎦⎧⎨⎩δ∆1δ∆2δ∆3

⎫⎬⎭ = −⎧⎨⎩R1R2R3

⎫⎬⎭ (6.4.7)

where

∆1i = ui, ∆2i = vi, ∆

3i = ∆

3i (6.4.8)

the coefficients of the submatrices [Tαβ] are deÞned by

Tαβij =∂Rαi

∂∆βj(6.4.9)

the components of the residual vector Rα are

Rαi =3Xγ=1

n∗Xk=1

Kαγik ∆

γk − Fαi (6.4.10)

and n∗ denotes n or m, depending on the nodal degree of freedom. Thus, wehave

Tαβij =∂

∂∆βj

⎛⎝ 3Xγ=1

n∗Xk=1

Kαγik ∆

γk − Fαi

⎞⎠ = 3Xγ=1

n∗Xk=1

∂Kαγik

∂∆βj∆γk +K

αβij (6.4.11)

The only coefficients that depend on the displacements are K13ij , K

23ij , K

31ij ,

K32ij , and K

33ij , and they are functions of only the transverse displacement

∆3i = ∆3i . Hence, derivatives of all submatrices with respect to ∆

1j = uj and


∆2j = vj are zero. We have

T 11ij =3Xγ=1

n∗Xk=1

∂K1γik

∂uj∆γk +K

11ij = K

11ij , T 12ij =

3Xγ=1

n∗Xk=1

∂K1γik

∂vj∆γk +K

12ij = K

12ij

T 21ij =3Xγ=1

n∗Xk=1

∂K2γik

∂uj∆γk +K

21ij = K

21ij , T 22ij =

3Xγ=1

n∗Xk=1

∂K2γik

∂vj∆γk +K

22ij = K

22ij

T 31ij =3Xγ=1

n∗Xk=1

∂K3γik

∂uj∆γk +K

31ij = K

31ij , T 32ij =

3Xγ=1

n∗Xk=1

∂K3γik

∂vj∆γk +K

32ij = K

32ij

T 13ij =3Xγ=1

n∗Xk=1

∂K1γik

∂∆3j∆γk +K

13ij =

nXk=1

∂K13ik

∂∆3j∆3k +K

13ij

=1

2

nXk=1

∆3k∂

∂∆3j

½ZΩe

∙∂ψei∂x

µA11

∂w0∂x

∂ϕek∂x

+A12∂w0∂y

∂ϕek∂y

¶

+A66∂ψei∂y

µ∂w0∂x

∂ϕek∂y

+∂w0∂y

∂ϕek∂x

¶#dx dy

¾+K13

ij

=1

2

nXk=1

∆3k

½ZΩe

"∂ψei∂x

ÃA11

∂ϕej∂x

∂ϕek∂x

+A12∂ϕej∂y

∂ϕek∂y

!

+A66∂ψei∂y

Ã∂ϕej∂x

∂ϕek∂y

+∂ϕej∂y

∂ϕek∂x

!#dx dy

¾+K13

ij

=1

2

ZΩe

"∂ψei∂x

ÃA11

∂w0∂x

∂ϕej∂x

+A12∂w0∂y

∂ϕej∂y

!

+A66∂ψei∂y

Ã∂w0∂x

∂ϕej∂y

+∂w0∂y

∂ϕej∂x

!#dx dy +K13

ij

= K13ij +K

13ij = 2K

13ij (= T

31ji )

T 23ij =3Xγ=1

n∗Xk=1

∂K2γik

∂∆3j∆γk +K

23ij =

nXk=1

∂K23ik

∂∆3j∆3k +K

23ij

=1

2

nXk=1

∆3k∂

∂∆3j

½ZΩe

∙∂ψei∂y

µA12

∂w0∂x

∂ϕek∂x

+A22∂w0∂y

∂ϕek∂y

¶

+A66∂ψei∂x

µ∂w0∂x

∂ϕek∂y

+∂w0∂y

∂ϕek∂x

¶#dx dy

¾+K23

ij

=1

2

nXk=1

∆3k

½ZΩe

"∂ψei∂y

ÃA12

∂ϕej∂x

∂ϕek∂x

+A22∂ϕej∂y

∂ϕek∂y

!

+A66∂ψei∂x

Ã∂ϕej∂x

∂ϕek∂y

+∂ϕej∂y

∂ϕek∂x

!#dx dy

¾+K13

ij


=1

2

ZΩe

"∂ψei∂y

ÃA12

∂w0∂x

∂ϕej∂x

+A22∂w0∂y

∂ϕej∂y

!

+A66∂ψei∂x

Ã∂w0∂y

∂ϕej∂x

+∂w0∂x

∂ϕej∂y

!#dx dy +K23

ij

= K23ij +K

23ij = 2K

23ij (= T

32ji ) (6.4.12)

The computation of T 33ij requires the calculation of three parts:

T 33ij =3Xγ=1

n∗Xk=1

∂K3γik

∂∆3j∆γk +K

33ij

=n∗Xk=1

Ã∂K31

ik

∂∆3juk +

∂K32ik

∂∆3jvk +

∂K33ik

∂∆3j∆3k

!+K33

ij (6.4.13)

We shall compute these terms Þrst. We have

n∗Xk=1

uk∂K31

ik

∂∆3j

=mXk=1

uk∂

∂∆3j

½ZΩe

∙∂ϕei∂x

µA11

∂w0∂x

∂ψek∂x

+A66∂w0∂y

∂ψek∂y

¶

+∂ϕei∂y

µA66

∂w0∂x

∂ψek∂y

+A12∂w0∂y

∂ψek∂x

¶#dx dy

¾

=mXk=1

uk

½ZΩe

"∂ϕei∂x

ÃA11

∂ϕej∂x

∂ψek∂x

+A66∂ϕej∂y

∂ψek∂y

!

+∂ϕei∂y

ÃA66

∂ϕej∂x

∂ψek∂y

+A12∂ϕej∂y

∂ψek∂x

!#dx dy

¾

=

ZΩe

"∂ϕei∂x

ÃA11

∂u0∂x

∂ϕej∂x

+A66∂u0∂y

∂ϕej∂y

!

+∂ϕei∂y

ÃA66

∂u0∂y

∂ϕej∂x

+A12∂u0∂x

∂ϕej∂y

!#dx dy (6.4.14)

n∗Xk=1

vk∂K32

ik

∂∆3j

=mXk=1

vk∂

∂∆3j

½ZΩe

∙∂ϕei∂x

µA12

∂w0∂x

∂ψek∂y

+A66∂w0∂y

∂ψek∂x

¶

+∂ϕei∂y

µA66

∂w0∂x

∂ψek∂x

+A22∂w0∂y

∂ψek∂y

¶#dx dy

¾


=mXk=1

vk

½ZΩe

"∂ϕei∂x

ÃA12

∂ϕej∂x

∂ψek∂y

+A66∂ϕej∂y

∂ψek∂x

!

+∂ϕei∂y

ÃA66

∂ϕej∂x

∂ψek∂x

+A22∂ϕej∂y

∂ψek∂y

!#dx dy

¾

=

ZΩe

"∂ϕei∂x

ÃA12

∂v0∂y

∂ϕej∂x

+A66∂v0∂x

∂ϕej∂y

!

+∂ϕei∂y

ÃA66

∂v0∂x

∂ϕej∂x

+A22∂v0∂y

∂ϕej∂y

!#dx dy (6.4.15)

n∗Xk=1

∆3k∂K33

ik

∂∆3j

=1

2

nXk=1

∆3k∂

∂∆3j

½ZΩe

("A11

µ∂w0∂x

¶2+A66

µ∂w0∂y

¶2# ∂ϕei∂x

∂ϕek∂x

+

"A66

µ∂w0∂x

¶2+A22

µ∂w0∂y

¶2# ∂ϕei∂y

∂ϕek∂y

+ (A12 +A66)∂w0∂x

∂w0∂y

µ∂ϕei∂x

∂ϕek∂y

+∂ϕei∂y

∂ϕek∂x

¶)dx dy

¾

=1

2

nXk=1

∆3k

½ZΩe

"Ã2A11

∂w0∂x

∂ϕej∂x

+ 2A66∂w0∂y

∂ϕej∂y

!∂ϕei∂x

∂ϕek∂x

+

Ã2A66

∂w0∂x

∂ϕej∂x

+ 2A22∂w0∂y

∂ϕej∂y

!∂ϕei∂y

∂ϕek∂y

+ (A12 +A66)

×Ã∂w0∂x

∂ϕej∂y

+∂w0∂y

∂ϕej∂x

!µ∂ϕei∂x

∂ϕek∂y

+∂ϕei∂y

∂ϕek∂x

¶#dx dy

¾

=

ZΩe

"A11

µ∂w0∂x

¶2 ∂ϕei∂x

∂ϕej∂x

+A66∂w0∂x

∂w0∂y

∂ϕei∂x

∂ϕej∂y

+A66∂w0∂x

∂w0∂y

∂ϕei∂y

∂ϕej∂x

+A22

µ∂w0∂y

¶2 ∂ϕei∂y

∂ϕej∂y

+

µA12 +A66

2

¶Ã∂w0∂x

∂ϕej∂y

+∂w0∂y

∂ϕej∂x

!µ∂ϕei∂x

∂w0∂y

+∂ϕei∂y

∂w0∂x

¶#dx dy

=

ZΩe

"A11

µ∂w0∂x

¶2 ∂ϕei∂x

∂ϕej∂x

+A22

µ∂w0∂y

¶2 ∂ϕei∂y

∂ϕej∂y

+A66∂w0∂x

∂w0∂y

Ã∂ϕei∂x

∂ϕej∂y

+∂ϕei∂y

∂ϕej∂x

!+

µA12 +A66

2

¶


×µ∂ϕei∂x

∂w0∂y

+∂ϕei∂y

∂w0∂x

¶Ã∂w0∂x

∂ϕej∂y

+∂w0∂y

∂ϕej∂x

!#dx dy (6.4.16)

Therefore, T 33ij is given by combining the expressions in Eqs. (6.4.14)(6.4.16)

and K33ij . We obtain

T 33ij =

ZΩe

"D11

∂2ϕei∂x2

∂2ϕej∂x2

+D22∂2ϕei∂y2

∂2ϕej∂y2

+D12

Ã∂2ϕei∂x2

∂2ϕej∂y2

+∂2ϕei∂y2

∂2ϕej∂x2

!

+ 4D66∂2ϕei∂x∂y

∂2ϕej∂x∂y

+ kϕeiϕej

#dx dy

+

ZΩe

½(Nxx +N

Txx)∂ϕei∂x

∂ϕej∂x

+ (Nyy +NTyy)∂ϕei∂y

∂ϕej∂y

+Nxy

Ã∂ϕei∂x

∂ϕej∂y

+∂ϕei∂y

∂ϕej∂x

!

+ (A12 +A66)∂w0∂x

∂w0∂y

Ã∂ϕei∂x

∂ϕej∂y

+∂ϕei∂y

∂ϕej∂x

!

+

"A11

µ∂w0∂x

¶2+A66

µ∂w0∂y

¶2# ∂ϕei∂x

∂ϕej∂x

+

"A66

µ∂w0∂x

¶2+A22

µ∂w0∂y

¶2# ∂ϕei∂y

∂ϕej∂y

¾dx dy (6.4.17)

Clearly, the tangent stiffness matrix of an element is symmetric (while theoriginal element stiffness is not symmetric).

6.4.3 Some Plate Finite Elements

There exists a large body of literature on triangular and rectangular platebending Þnite elements of isotropic or orthotropic plates based on the CPT(e.g. see [1320]). There are two kinds of plate bending elements of theCPT. A conforming element is one in which the inter-element continuity ofw0, θx ≡ ∂w0/∂x, and θy ≡ ∂w0/∂y (or ∂w0/∂n) are satisÞed, and a non-conforming element is one in which the continuity of the normal slope, ∂w0/∂n,is not satisÞed.An effective non-conforming triangular element (the BCIZ triangle) was

developed by Bazeley et al. [14], and it consists of three degrees of freedom(w0, θx, θy) at the three vertex nodes (see Figure 6.4.1). The element performsvery well in bending as well as vibration problems (with a consistent massmatrix).

z,

∂x,

∂y∂w0 ∂w 0w0

y, η yb=

1

3

2

x, ξ xa=

z

x, ξ xa=

,∂x

,∂y

∂w0 ∂w0w0

y, η yb

=

1

3

2

a

c

b

4

∂n∂w0


Figure 6.4.1 A non-conforming triangular element with three degrees offreedom (w0, ∂w0/∂x,∂w0/∂y) per node.

A conforming triangular element due to Clough and Tocher [15] is anassemblage of three triangles as shown in Figure 6.4.2. The normal slopecontinuity is enforced at the mid-side nodes between the sub-triangles. Ineach sub-triangle, the transverse deßection is represented by the polynomial(i = 1, 2, 3),

wi0(x, y) = ai + bi ξ + ci η + di ξη + ei ξ2 + fi η

2 + gi ξ3 + hi ξ

2η + ki ξη2 + ì η

3

(6.4.18)where (ξ, η) are the local coordinates, as shown in Figure 6.4.2. The thirtycoefficients are reduced to nine, three (w0, ∂w0/∂x, ∂w0/∂y) at each vertex ofthe triangle, by equating the variables from the vertices of each sub-triangleat the common points and normal slope between the mid-side points of sub-triangles.

Figure 6.4.2 A conforming triangular element.

a

b

z

x

4 3

21

,∂x

,∂y

∂w0 ∂w0w0u0, v0,

non-conforming element

y

a

b

ξ = x/a, η = y/b


A non-conforming rectangular element has w0, θx, and θy as the nodalvariables (see Figure 6.4.3). The element was developed by Melosh [16] andZienkiewicz and Cheung [17]. The normal slope variation is cubic along anedge, whereas there are only two values of ∂w0/∂n available on the edge.Therefore, the cubic polynomial for the normal derivative of w0 is not thesame on the edge common to two elements. The interpolation functions forthis element can be expressed compactly as

ϕei = gi1 (i = 1, 4, 7, 10); ϕei = gi2 (i = 2, 5, 8, 11)

ϕei = gi3 (i = 3, 6, 9, 12)

gi1 =1

8(1 + ξ0)(1 + η0)(2 + ξ0 + η0 − ξ2 − η2) (6.4.19)

gi2 =1

8ξi(ξ0 − 1)(1 + η0)(1 + ξ0)2, gi3 =

1

8ηi(η0 − 1)(1 + ξ0)(1 + η0)2

ξ = (x− xc)/a, η = (y − yc)/b, ξ0 = ξξi, η0 = ηηiwhere 2a and 2b are the sides of the rectangle, (xc, yc) are the globalcoordinates of the center of the rectangle, and (ξi, ηi) are the coordinates of thenodes in (ξ, η) coordinate system [e.g. (ξ1, η1) = (−1,−1), (ξ2, η2) = (1,−1),etc.].A conforming rectangular element with w0, ∂w0/∂x, ∂w0/∂y, and

∂2w0/∂x∂y as the nodal variables (see Figure 6.4.4) was developed by Bogneret al. [18]. The interpolation functions for this element are

ϕei = gi1 (i = 1, 5, 9, 13); ϕei = gi2 (i = 2, 6, 10, 14)

ϕei = gi3 (i = 3, 7, 11, 15); ϕei = gi4 (i = 4, 8, 12, 16)

gi1 =1

16(ξ + ξi)

2(ξ0 − 2)(η + ηi)2(η0 − 2)

Figure 6.4.3 A non-conforming rectangular element.

a

b

z

x

4 3

21

conforming element

y

a

b

,∂x

,∂w0

∂y∂w0w0u0, v0, ,

∂y∂ w0

∂x

2

ξ = x/a, η = y/b


Figure 6.4.4 A conforming rectangular element.

gi2 =1

16ξi(ξ + ξi)

2(1− ξ0)(η + ηi)2(η0 − 2)

gi3 =1

16ηi(ξ + ξi)

2(ξ0 − 2)(η + ηi)2(1− η0)

gi4 =1

16ξiηi(ξ + ξi)

2(1− ξ0)(η + ηi)2(1− η0) (6.4.20)

The conforming element has four degrees of freedom per node, whereasthe non-conforming element has three degrees of freedom per node. For theconforming rectangular element the total number of in-plane and bendingnodal degrees of freedom per element is 8+16 = 24, and for the non-conformingelement, the total number is 8 + 12 = 20.

6.5 Computer Implementation Aspects andNumerical Results of CPT Elements


The conforming and non-conforming rectangular finite elements developedin this chapter are implemented into a computer program using bilinearinterpolation of (u0, v0) and Hermite cubic interpolation of w0. The elementgeometry is represented using bilinear interpolation functions. In view ofthe different interpolation of the in-plane displacements and the transversedeflection, one must compute both types of interpolation functions and theirfirst and second derivatives (see Problem 6.24) in each call of the shapefunctions subroutine. In addition, one must rearrange the stiffness coefficientssuch that the finite element nodal displacement vector is of the form (tominimize the bandwidth of the stiffness matrix)

∆ = u01, v01, w01, θx1, θy1, θxy1, u02, v02, w02, θx2, θy2, θxy2, . . .


where (u0i, v0i, w0i, θxi, θyi, θxyi) are the displacements at node i. The samelogic as in the case of EulerBernoulli beams may be used to rearrange thestiffness coefficients of the CPT elements (see Box 4.2.1) as discussed next.

First, one must compute all element forces Fαi , element stiffnesses Kαβij ,

and extra stiffness terms to form the total tangent matrix. Two separatedo-loops on Guass quadrature are required to compute all the force andstiffness coefficients. The full integration loop is used to evaluate all forcecomponents and all linear stiffnesses. The reduced integration loop is usedto compute the nonlinear stiffness terms. One may use separate arrays (say,EXT13(I, J), EXT33(I, J), and so on) to store the extra terms (to those of

Kαβij ) of the tangent stiffness coefficients. Next, the stiffness coefficients K

31ij

and K32ij , for example, are rearranged as follows:

II = 1

DO 300 I = 1, NPE

DO 200 K = 1, NWD

K0 = (I − 1) ∗NWD +KKK = II +K + 1

JJ = 1

DO 100 J = 1, NPE

ELK(KK,JJ) = ELK31(I, J)

ELK(KK,JJ + 1) = ELK32(I, J)

100 JJ = J ∗NDF + 1200 CONTINUE

300 II = I ∗NDF + 1

whereNWD is equal to 3 for non-conforming element and 4 for the conformingelement; it denotes the number of degrees of freedom per node associatedwith the transverse deßection. Similar logic may be used to rearrange thecoefficients K13

ij , K23ij , and K

33ij . The logic to rearrange K

11ij , K

12ij , K

21ij , and

K22ij is the same as shown in Box 4.3.1.Note that the addition of the extra terms to the direct stiffness matrix [K]

in order to obtain the tangent stiffness matrix, [K]tan, must be carried outonly after the imbalance (or residual) vector is computed:

−R = [K]∆− F→ ELF (I) = ELF (I)−ELK(I, J) ∗ELU(J)[K]tan = [K] + [EXT ]→ ELK(I, J) = ELK(I, J) +EXT (I, J)

where [EXT ] in the present discussion is used only for the extra terms that areadded to [K] to obtain the tangent stiffness matrix. The above two operationsmust be carried out sequentially in separate do-loops.

x

y

a2

b 2

b 2

a2

= 0∂w0

∂xat 0x= ;= u0 = 0∂w0

∂yat 0y== v0

= w0 = 0∂w0

∂x= u0

= w0 = 0∂w0

∂y= v0

= w0 = 0∂w0

∂x= u0= w0 = 0∂w0

∂y= v0

Symm. B.C.:


Solution symmetries available in a problem should be taken advantage of toidentify the computational domain because they reduce computational effort.For example, a 2 × 2 mesh in a quadrant of the plate is the same as 4 × 4mesh in the total plate, and the results obtained with the two meshes wouldbe identical, within the round-off errors of the computation, if the solutionexhibits biaxial symmetry. A solution is symmetric about a line only if (a)the geometry, including boundary conditions, (b) the material properties, and(c) the loading are symmetric about the line. The boundary conditions alonga line of symmetry should be correctly identiÞed and imposed in the Þniteelement model. The boundary conditions along the edges and the symmetrylines of a simply supported rectangular plate are shown in Figure 6.5.1. Inthe case of the conforming element, we may also set θxy ≡ ∂2w0/∂x∂y = 0 atthe center of the plate. When one is not sure of the solution symmetry, it isadvised that the whole plate be modeled.

6.5.2 Results of Linear Analysis

We consider the bending of rectangular plates with various edge conditions toevaluate the elements developed herein. The foundation modulus k is set tozero in all examples. The linear stiffness coefficients are evaluated using 4× 4Gauss rule, while the stresses were computed at the center of the elements(i.e. one-point quadrature is used). The effect of the integration rule on theaccuracy of solutions will be examined in the sequel.

Figure 6.5.1 Boundary conditions for rectangular plates with biaxialsymmetry.


Example 6.5.1

Consider a simply supported (SS-1) rectangular plate under uniformly distributed load. Thegeometric boundary conditions of the computational domain (see the shaded quadrant inFigure 6.5.1) are

u0 =∂w0∂x

= 0 at x = 0; v0 =∂w0∂y

= 0 at y = 0 (6.5.1)

v0 = w0 =∂w0∂y

= 0 at x =a

2; u0 = w0 =

∂w0∂x

= 0 at y =b

2

∂2w0∂x∂y

= 0 at x = y = 0 (for conforming element only) (6.5.2)

Table 6.5.1 shows a comparison of non-dimensionalized Þnite element solutions with

the analytical solutions (see Reddy [3]) of isotropic and orthotropic square plates under

uniformly distributed transverse load q0. The stresses were evaluated at the center of the

element. Hence, the locations of the maximum normal stresses are (a/8, b/8), (a/16, b/16),

and (a/32, b/32) for uniform meshes 2× 2, 4× 4, and 8× 8, respectively, while those of σxyare (3a/8,3b/8), (7a/16, 7b/16), and (15a/32,15b/32) for the three meshes. The analytical

solutions were evaluated using m,n = 1,3, . . . ,19. The exact maximum deßection occurs

at x = y = 0, maximum stresses σxx and σyy occur at (0,0, h/2), and the maximum shear

stress σxy occurs at (a/2, b/2,−h/2).

Table 6.5.1 Maximum transverse deßections and stresses* of simplysupported square plates under a uniformly distributed load q0(linear analysis).

Non-conforming ConformingAnalytical

Variable 2× 2 4× 4 8× 8 2× 2 4× 4 8× 8 solution

Isotropic plate (ν = 0.25)

w × 102 4.8571 4.6425 4.5883 4.7619 4.5952 4.5739 4.5698σxx 0.2405 0.2673 0.2740 0.2239 0.2637 0.2731 0.2762σxy 0.1713 0.1964 0.2050 0.1688 0.1935 0.2040 0.2085

Orthotropic plate (E1/E2 = 25, G12 = G13 = 0.5E2, ν12 = 0.25)

w × 102 0.7082 0.6635 0.6531 0.7710 0.6651 0.6522 0.6497σxx 0.7148 0.7709 0.7828 0.5560 0.7388 0.7743 0.7866σyy 0.0296 0.0253 0.0246 0.0278 0.0249 0.0245 0.0244σxy 0.0337 0.0421 0.0444 0.0375 0.0416 0.0448 0.0463

*w = w0E2h3/(q0a

4), σ = σh2/(q0a2).

Example 6.5.2

Here we consider a clamped square plate under uniformly distributed load. The boundaryconditions are taken to be

u0 =∂w0∂x

= 0 at x = 0; v0 =∂w0∂y

= 0 at y = 0 (6.5.3a)


u0 = v0 = w0 =∂w0∂x

=∂w0∂y

= 0 at x =a

2; u0 = v0 = w0 =

∂w0∂x

=∂w0∂y

= 0 at y =b

2(6.5.3b)

∂2w0∂x∂y

= 0 on clamped edges (for conforming element only) (6.5.4)

Table 6.5.2 contains the non-dimensionalized deßections and stresses. The locations ofthe normal stresses reported for the three meshes are:

2× 2 : (a

8,b

8); 4× 4 : (

a

16,b

16); 8× 8 : (

a

32,b

32)

and shear stresses reported for the three meshes are

2× 2 : (3a

8,3b

8); 4× 4 : (

7a

16,7b

16); 8× 8 : (

15a

32,15b

32)

These stresses are not necessarily the maximum ones in the plate. For example,

for an 8 × 8 mesh, the maximum normal stress in the isotropic plate is found to

be 0.2300 at (0.46875a,0.03125b,−h/2) and the maximum shear stress is 0.0226 at

(0.28125a, 0.09375b,−h/2) for the non-conforming element. The conforming element yieldsslightly better solutions than the non-conforming element for deßections but not for the

stresses, and both elements show good convergence.

Table 6.5.2 Maximum transverse deßections and stresses* of clamped(CCCC), isotropic and orthotropic, square plates (a = b) undera uniformly distributed load q0 (linear analysis).

Variable Non-conforming Conforming

2× 2 4× 4 8× 8 2× 2 4× 4 8× 8Isotropic plate (ν = 0.25)

w × 102 1.5731 1.4653 1.4342 1.4778 1.4370 1.4249σxx 0.0987 0.1238 0.1301 0.0861 0.1197 0.1288σxy 0.0497 0.0222 0.0067 0.0489 0.0224 0.0068

Orthotropic plate (E1/E2 = 25, G12 = G13 = 0.5E2, ν12 = 0.25)

w × 102 0.1434 0.1332 0.1314 0.1402 0.1330 0.1311σxx 0.1962 0.2491 0.2598 0.1559 0.2358 0.2576σyy 0.0085 0.0046 0.0042 0.0066 0.0047 0.0043σxy 0.0076 0.0046 0.0019 0.0083 0.0048 0.0020

*w = w0E2h3/(q0a

4), σ = σh2/(q0a2).

6.5.3 Results of Nonlinear Analysis

Here we investigate geometrically nonlinear response of plates using theconforming and non-conforming plate Þnite elements. The nonlinear termsare evaluated using reduced integration. Full integration (F) means 4 × 4Gauss rule and reduced integration (R) means 1× 1 Gauss rule.


Example 6.5.3

First, we consider the nonlinear bending of an isotropic (ν = 0.3) square plate underuniformly distributed transverse load, q0 (see Levy [32], Wang [33], and Kawai and Yoshimura[34]). The following simply supported (SS3) geometric boundary conditions are used:

u0 = v0 = w0 = 0 on all four edges (6.5.5)

Since θx = (∂w0/∂x) and θy = (∂w0/∂y) are not speciÞed in SS3, it follows that thefollowing boundary conditions are

on y = 0, b : Mxy =Myy = 0; on x = 0, a : Mxx =Mxy = 0 (6.5.6)

satisÞed in the integral sense [see Eq. (6.4.7)].

Using the biaxial symmetry, only a quadrant is modeled with a uniform 4×4 or 8×8 meshof rectangular elements. The boundary conditions along the lines of symmetry are shownin Figure 6.5.1. The following geometric and material parameters are used, although thenon-dimensional transverse deßection and stresses presented here are independent of them(but may depend on ν):

a = b = 10 in., h = 0.1 in., E = 30× 106 psi, ν = 0.3 (6.5.7)

The notation FF means full integration (4×4 Gauss rule) is used for the numerical evaluationof all coefficients, while FR means full integration is used for all but nonlinear terms andreduced integration (1× 1 Gauss rule) is used for the nonlinear terms. The number in frontof FF or FR stands for the mesh (4 refers to 4×4 mesh and so on). Stresses are evaluatedat the center of the element (i.e. 1× 1 Gauss point). A load increment of ∆q0 = 7.5, whichis equal to the increment of load parameter, ∆P ≡∆q0a4/Eh4 = 25, is used along with theconvergence tolerance of ² = 10−2. Except for the Þrst load step, which took 5 iterations,the convergence was achieved for 2 or 3 iterations.

The center deßection, w = w0/h, and total stresses (i.e. membrane and ßexuralcontributions included), σxx = σxx(A,A,h/2)(a2/Eh2) and σxy = σxy(B,B,−h/2)(a2/Eh2)as functions of the load parameter, P = q0a

4/Eh4 are presented in Table 6.5.3. The location(A,A,h/2) refers to the Gauss point nearest to the center (x = y = 0) but at the top of theplate, while (B,B,−h/2) refers to the Gauss point nearest to the corner x = y = a/2, atthe bottom of the plate (see Figure 6.5.1). There is very little difference between the resultsobtained with reduced and full integration of the nonlinear stiffness coefficients.

Plots of the load parameter P versus the deßection w and P versus various stresses

are presented in Figure 6.5.2. Although the linear and nonlinear values of σxx for the load

parameter P = 25 are maximum at the center of the plate, the location of the maximum

normal stress σxx in the nonlinear analysis changes as the load value is increased. For

example, the maximum value of σxx at P = 250 in the Þnite element analysis occurs at

(x, y)=(2.8125,0.3125), and its value is found to be 21.177. It is clear that the membrane

stresses are a signiÞcant part of the total stresses.

Example 6.5.4

This example is concerned with the bending of a simply supported (SS1) orthotropic squareplate under a uniformly distributed transverse load q0. The geometry and material propertiesused are given below.

a = b = 12 in., h = 0.138 in., E1 = 3× 106 psi, E2 = 1.28× 106 psiG12 = G23 = G13 = 0.37× 106 psi., ν12 = ν23 = ν13 = 0.32 (6.5.8)


A load increment of ∆q0 = 0.2 psi and a uniform mesh of 4× 4 in a quarter plate was used.Figure 6.5.3 shows plots of the center deßection versus the intensity of the distributed load.The Þnite element results are in close agreement with the experimental results of Zaghlouland Kennedy [35].

Table 6.5.3 Maximum transverse deßections and stresses* of simplysupported (SS3), isotropic (ν = 0.3) square plates (a = b) undera uniformly distributed load q0 (nonlinear analysis).

Non-conforming ConformingLoadparameter 4FF 4FR 8FR 4FF 4FR 8FR

Linear w → 1.127 1.127 1.113 1.116 1.116 1.110

25 w 0.670 0.673 0.670 0.669 0.670 0.669σxx 5.279 5.321 5.423 5.252 5.324 5.426σxy 3.014 3.028 3.179 2.994 3.019 3.172

50 w 0.944 0.951 0.946 0.943 0.949 0.945σxx 8.035 8.142 8.227 8.018 8.219 8.247σxy 4.510 4.552 4.818 4.484 4.566 4.810

75 w 1.124 1.136 1.128 1.123 1.135 1.127σxx 10.057 10.231 10.271 10.048 10.390 10.309σxy 5.600 5.674 6.036 5.569 5.719 6.029

100 w 1.262 1.280 1.268 1.262 1.279 1.267σxx 11.731 11.974 11.961 11.729 12.217 12.017σxy 6.493 6.599 7.049 6.457 6.679 7.042

125 w 1.377 1.400 1.383 1.377 1.400 1.383σxx 13.197 13.509 13.440 13.201 13.838 13.513σxy 7.265 7.403 7.933 7.224 7.521 7.927

150 w 1.475 1.504 1.483 1.476 1.505 1.483σxx 14.519 14.902 14.777 14.529 15.317 14.867σxy 7.951 8.122 8.726 7.905 8.280 8.722

175 w 1.563 1.597 1.571 1.563 1.598 1.571σxx 15.737 16.191 16.009 15.752 16.693 16.117σxy 8.573 8.778 9.451 8.524 8.979 9.449

200 w 1.641 1.681 1.651 1.642 1.683 1.651σxx 16.874 17.401 17.162 16.894 17.989 17.287σxy 9.147 9.385 10.123 9.094 9.631 10.123

225 w 1.713 1.758 1.724 1.713 1.761 1.724σxx 17.946 18.546 18.251 17.970 19.222 18.393σxy 9.681 9.952 10.751 9.625 10.244 10.754

250 w 1.779 1.830 1.791 1.779 1.834 1.791σxx 18.965 19.638 19.287 18.993 20.401 19.446σxy 10.182 10.486 11.344 10.122 10.827 11.349

*w = w0/h, σ = σ(a2/Eh2).


Figure 6.5.2 (a) Load versus deßection and (b) load versus stress plotsfor simply supported (SS3) isotropic (ν = 0.3) square platesunder uniformly distributed transverse load q0.

0 50 100 150 200 250 300

Load parameter

0.0

0.2

0.4

0.6

0.8

1.0

1.2

1.4

1.6

1.8

2.0D

efle

ctio

n,

(a)

4×4 mesh8×8 mesh

w

, P

a = b = 10 in., h = 0.1 in.E = 30×106, ν = 0.3

4

400 ,

Ehaq

Ph

ww ==

(b)

0 50 100 150 200 250 300

Load parameter,

0

5

10

15

20

25

Str

esse

s,

4×4 mesh

8×8 mesh

xxσ

xyσ

(membrane)σxx

σ xx

P

⎟⎟⎠

⎞⎜⎜⎝

⎛=2

2

Eha

xxxx σσ

0.0 0.4 0.8 1.2 1.6 2.0 2.4Load parameter,

0.00

0.05

0.10

0.15

0.20

0.25

0.30D

efle

ctio

n,

SS-1

Clamped

(4×4 C)

(8×8 NC)

w0(

0,0)

P


Figure 6.5.3 Loaddeßection curves for simply supported and clampedorthotropic square plates under uniform load.

Example 6.5.5

The last example of this section is concerned with the bending of a clamped square plateunder uniformly distributed load q0. The geometric and material parameters used are thesame as those in Eq. (6.5.8). The boundary conditions of a clamped edge are taken to be

u0 = v0 = w0 =∂w0∂x

=∂w0∂y

= 0 (6.5.9)

Of course, for conforming element, one may also impose (∂2w0/∂x∂y) = 0.

A uniform mesh of 8× 8 non-conforming elements in a quarter plate is used, and a loadincrements of ∆q0 = 0.05, 0.05, 0.1, 0.2, 0.2, . . . ,0.2 psi was used. The linear solution atq0 = 0.05 is found to be w0(0,0) = 0.00302 in. A plot of the center deßection versus theintensity of the distributed load for the clamped orthotropic plate is included in Figure 6.5.3(see [3638]).

We close this section with a note that the plate bending elements of theCPT discussed here are adequate for most engineering applications, whichinvolve thin, isotropic plate structures and shear deformation is negligible. Inthe coming sections, we discuss plate elements based on the Þrst-order sheardeformation plate theory (FSDT), which can be used to analyze both thin andthick plates.


6.6 First-Order Shear Deformation Plate Theory

6.6.1 Introduction

The preceding sections of the book were devoted to the study of bendingof plates using the CPT, in which transverse normal and shear stresses areneglected. The FSDT extends the kinematics of the CPT by relaxing thenormality restriction (see Section 6.2) and allowing for arbitrary but constantrotation of transverse normals.In this chapter, we develop displacement Þnite element models of the FSDT.

As we shall see in the sequel, the formulation requires only C0 interpolationof all generalized displacements. Consequently, the element is much simplerto implement on a computer. Of course, the element can be used to analyzethick as well as thin plates. We begin with the theoretical formulation of thetheory (see Reddy [3] for additional details).

6.6.2 Displacement Field

Under the same assumptions and restrictions as in the classical laminate theorybut relaxing the normality condition, the displacement Þeld of the FSDT canbe expressed in the form

u(x, y, z) = u0(x, y) + zφx(x, y)

v(x, y, z) = v0(x, y) + zφy(x, y) (6.6.1)

w(x, y, z) = w0(x, y)

where (u0, v0, w0,φx,φy) are unknown functions to be determined. As before,(u0, v0, w0) denote the displacements of a point on the plane z = 0 and φxand φy are the rotations of a transverse normal about the y- and x-axes,respectively (see Figure 6.6.1). The quantities (u0, v0, w0,φx,φy) are calledthe generalized displacements.The notation that φx denotes the rotation of a transverse normal about

the y-axis and φy denotes the rotation about the x-axis may be confusing tosome because they do not follow the right-hand rule. However, the notationhas been used extensively in the literature, and we will not depart from it. If(βx,βy) denote the rotations about the x- and y-axes, respectively, that followthe right-hand rule, then

βx = −φy , βy = φx (6.6.2)

For thin plates, that is, when the plate in-plane characteristic dimension tothickness ratio is on the order of 50 or greater, the rotation functions φx andφy should approach the respective slopes of the transverse deßection [21]:

φx = −∂w0∂x

, φy = −∂w0∂y

(6.6.3)

z

y

x

z

x

u0

w0

γxz φx

∂w0

∂x−

∂w0

∂x−


Figure 6.6.1 Undeformed and deformed geometries of an edge of a plateunder the assumptions of the FSDT.

The von Karman nonlinear strains associated with the displacement Þeld(6.6.1) are (εzz = 0)

⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩εxxεyyγyzγ0xzγxy

⎫⎪⎪⎪⎪⎬⎪⎪⎪⎪⎭ =⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩ε0xxε0yyγ0yzγxzγ0xy

⎫⎪⎪⎪⎪⎬⎪⎪⎪⎪⎭+ z⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩ε1xxε1yy00γ1xy

⎫⎪⎪⎪⎪⎬⎪⎪⎪⎪⎭ =⎧⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎩

∂u0∂x +

12(∂w0∂x )

2

∂v0∂y +

12(∂w0∂y )

2

∂w0∂y + φy∂w0∂x + φx

∂u0∂y +

∂v0∂x +

∂w0∂x

∂w0∂y

⎫⎪⎪⎪⎪⎪⎪⎬⎪⎪⎪⎪⎪⎪⎭+ z

⎧⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎩

∂φx∂x∂φy∂y00

∂φx∂y +

∂φy∂x

⎫⎪⎪⎪⎪⎪⎬⎪⎪⎪⎪⎪⎭(6.6.4)

Note that the strains (εxx, εyy, γxy) are linear through the plate thickness,while the transverse shear strains (γxz, γyz) are constant.

6.6.3 Weak Formulation

The weak form of the FSDT can be derived using the principle of virtualdisplacements

0 = δW e ≡ δUe + δV e (6.6.5)

where the virtual strain energy δUe and the virtual work done by applied


forces δV e in an element Ωe are given by

δUe =

ZΩe

(Z h2

−h2

hσxx

³δε0xx + zδε

1xx

´+ σyy

³δε0yy + zδε

1yy

´+ σxy

³δγ0xy + zδγ

1xy

´+ σxzδγ

0xz + σyzδγ

0yz

idz

)dx dy (6.6.6)

δV e = −(Z

Γe

Z h2

−h2

[σnn (δu0n + zδφn) + σns (δu0s + zδφs) + σnzδw0] dz ds

+

ZΩe(q − kw0) δw0 dx dy

)(6.6.7)

where Ωe denotes the undeformed mid-plane of a typical plate element, hthe total thickness, ρ0 the density of the plate, k the modulus of the elasticfoundation (if any), and (σnn,σns,σnz) are the edge stresses along the (n, s, z)coordinates.Substituting for δUe and δV e from Eqs. (6.6.6) and (6.6.7) into the virtual

work statement in Eq. (6.6.5) and integrating through the thickness, we obtain

0 =

ZΩe

hNxxδε

0xx +Mxxδε

1xx +Nyyδε

0yy +Myyδε

1yy +Nxyδγ

0xy

+Mxyδγ1xy +Qxδγ

0xz +Qyδγ

0yz + kw0δw0 − qδw0 ] dx dy

−ZΓe(Nnnδu0n +Nnsδu0s +Mnnδφn +Mnsδφs +Qnδw0) ds (6.6.8)

where the stress resultants (Nxx, Nyy, Nxy,Mxx,Myy,Mxy) were deÞned inEq. (6.3.4), (Nnn, Nns,Mnn,Mns, Qn) in Eqs. (6.3.6a,b)(6.3.7), and thetransverse forces per unit length (Qx,Qy) are deÞned by½

QxQy

¾=

Z h2

−h2

½σxzσyz

¾dz (6.6.9)

and φn and φs are the rotations of a transverse normal about s and −ncoordinates, respectively.Since the transverse shear strains are represented as constant through the

laminate thickness, it follows that the transverse shear stresses will also beconstant. It is well known from elementary theory of homogeneous beams thatthe transverse shear stress variation is parabolic through the beam thickness.This discrepancy between the actual stress state and the constant stress statepredicted by the FSDT is often corrected in computing the transverse shearforces (Qx, Qy) by multiplying the integrals in Eq. (6.6.9) with a parameterKs, called shear correction coefficient:½

QxQy

¾= Ks

Z h2

−h2

½σxzσyz

¾dz (6.6.10)


This amounts to modifying the plate transverse shear stiffnesses. The factorKs is computed such that the strain energy due to transverse shear stressesin Eq. (6.6.10) equals the strain energy due to the true transverse stressespredicted by the three-dimensional elasticity theory.To obtain the governing equations of equilibrium, Þrst we substitute for

the virtual strains in terms of the virtual displacements into Eq. (6.6.8), andthen integrate-by-parts the expressions to relieve the virtual displacements(δu0, δv0, δw0, δφx, δφy) in Ω

e of any differentiation. We obtain

0 =

ZΩe

"− (Nxx,x +Nxy,y) δu0 − (Nxy,x +Nyy,y) δv0− (Mxx,x +Mxy,y −Qx) δφx − (Mxy,x +Myy,y −Qy) δφy− (Qx,x +Qy,y − kw0 +N + q) δw0

#dx dy

+

IΓe

"(Nxxnx +Nxyny) δu0 + (Nxynx +Nyyny) δv0

+ (Mxxnx +Mxyny) δφx − (Mxynx +Myyny) δφy

+

ÃQxnx +Qyny + P

!δw0

#ds

−IΓe(Nnnδu0n +Nnsδu0s +Mnnδφn +Mnsδφs +Qnδw0) ds (6.6.11)

where N and P are deÞned by Eqs. (6.3.14a,b). The boundary terms can beexpressed in terms of the normal and tangential components u0n, u0s, φn, andφs using Eqs. (6.3.18a,b) and

φx = nxφn − nyφs , φy = nyδφn + nxδφs (6.6.12)

This will yield the natural boundary conditions given in Eq. (6.3.20), whichrelate the forces and moments on an arbitrary edge to those on edges parallelto the coordinates (x, y, z).The EulerLagrange equations are

δu0 :∂Nxx∂x

+∂Nxy∂y

= 0 (6.6.13)

δv0 :∂Nxy∂x

+∂Nyy∂y

= 0 (6.6.14)

δw0 :∂Qx∂x

+∂Qy∂y

− kw0 +N + q = 0 (6.6.15)

δφx :∂Mxx

∂x+∂Mxy

∂y−Qx = 0 (6.6.16)

δφy :∂Mxy

∂x+∂Myy

∂y−Qy = 0 (6.6.17)


The primary and secondary variables of the theory are

primary variables: u0n, u0s, w0, φn, φs

secondary variables: Nnn, Nns, Qn, Mnn, Mns (6.6.18)

The plate constitutive equations in Eqs. (6.3.32) and (6.3.33) are valid alsofor the Þrst-order plate theory. In addition, we have the following constitutiveequations for transverse shear forces of an orthotropic plate::½

QyQx

¾= Ks

Z h2

−h2

½σyzσxz

¾dz = Ks

∙A44 00 A55

¸½γyzγxz

¾(6.6.19)

where the extensional stiffnesses A44 and A55 are deÞned by

(A44, A55) =

Z h2

−h2

(Q44, Q55) dz, Q44 = G23, Q55 = G13 (6.6.20)

The stress resultants in an orthotropic plate are related to the generalizeddisplacements (u0, v0, w0,φx,φy) by [see Eqs. (6.3.32) and (6.3.33)]⎧⎨⎩

NxxNyyNxy

⎫⎬⎭ =⎡⎣A11 A12 0A12 A22 00 0 A66


⎫⎬⎭−⎧⎨⎩NTxx

NTyy

0

⎫⎬⎭ (6.6.21)

⎧⎨⎩Mxx

Myy

Mxy

⎫⎬⎭ =⎡⎣D11 D12 0D12 D22 00 0 D66


⎫⎬⎭−⎧⎨⎩MTxx

MTyy

0

⎫⎬⎭ (6.6.22)

½QyQx

¾= Ks

∙A44 00 A55

¸½γ0yzγ0xz

¾(6.6.23)

6.7 Finite Element Models of FSDT

6.7.1 Virtual Work Statements

Using the weak form (6.6.9) developed in Section 6.6, we can constructthe Þnite element models of the equations governing the FSDT. The stressresultants in Eq. (6.6.9) are understood to be known in terms of thegeneralized displacements (u0, v0, w0,φx,φy) via Eqs. (6.6.22)(6.6.24). Thevirtual work statement (6.6.9) is equivalent to (collecting the terms involvingδu0, δv0, δw0, δφx, and δφy separately) the following Þve weak forms

0 =

ZΩe

µ∂δu0∂x

Nxx +∂δu0∂y

Nxy

¶dx dy −

IΓe(Nxxnx +Nxyny) δu0 ds (6.7.1)


0 =

ZΩe

µ∂δv0∂x

Nxy +∂δv0∂y

Nyy

¶dx dy −

IΓe(Nxynx +Nyyny) δv0 ds (6.7.2)

0 =

ZΩe

∙∂δw0∂x

Qx +∂δw0∂y

Qy +∂δw0∂x

µNxx

∂w0∂x

+Nxy∂w0∂y

¶+∂δw0∂y

µNxy

∂w0∂x

+Nyy∂w0∂y

¶− δw0q + kw0δw0

#dx dy

−IΓe

"µQx +Nxx

∂w0∂x

+Nxy∂w0∂y

¶nx

+

µQy +Nxy

∂w0∂x

+Nyy∂w0∂y

¶ny

#δw0 ds (6.7.3)

0 =

ZΩe

µ∂δφx∂x

Mxx +∂δφx∂y

Mxy + δφxQx

¶dx dy

−IΓe(Mxxnx +Mxyny) δφx ds (6.7.4)

0 =

ZΩe

µ∂δφy∂x

Mxy +∂δφy∂y

Myy + δφyQy

¶dx dy

−IΓe(Mxynx +Myyny) δφy ds (6.7.5)

We note from the boundary terms in Eqs. (6.7.1)(6.7.5) that u0, v0, w0,φx, φy are used as the primary variables (or generalized displacements) asopposed to u0n, u0s, w0, φn, φs. Unlike in the CPT, the rotations (φx,φy) areindependent of w0. Since no derivatives of (u0, v0, w0, φx, φy) appear in thelist of the primary variables, all generalized displacements may be interpolatedusing the Lagrange interpolation functions. Hence, the element is called C0

element with respect to all dependent unknowns.We identify the secondary variables of the formulation as

Nn ≡Nxxnx +Nxyny, Ns ≡ Nxynx +Nyyny (6.7.6a)

Mn ≡Mxxnx +Mxyny, Ms ≡Mxynx +Myyny (6.7.6b)

Qn ≡µQx +Nxx

∂w0∂x

+Nxy∂w0∂y

¶nx

+

µQy +Nxy

∂w0∂x

+Nyy∂w0∂y

¶ny (6.7.7)


The virtual work statements in Eqs. (6.7.1)(6.7.5) contain at the most onlythe Þrst derivatives of the dependent variables (u0, v0, w0,φx,φy). Therefore,


they can all be approximated using the Lagrange interpolation functions. Inprinciple, (u0, v0), w0, and (φx,φy) can be approximated with differing degreesof Lagrange interpolation functions. Let

u0(x, y) =mXj=1

ujψ(1)j (x, y), v0(x, y) =

mXj=1

vjψ(1)j (x, y) (6.7.8)

w0(x, y) =nXj=1

wjψ(2)j (x, y) (6.7.9)

φx(x, y) =pXj=1

S1jψ(3)j (x, y), φy(x, y) =

pXj=1

S2jψ(3)j (x, y) (6.7.10)

where ψ(α)j (α = 1, 2, 3) are Lagrange interpolation functions. One can use

linear, quadratic, or higher-order interpolations of these variables. Althoughthe development is general, in the implementation of this element, we shalluse equal interpolation of all variables.Substituting Eqs. (6.7.8)(6.7.10) for (u0, v0, w0,φx,φy) into Eqs. (6.7.1)

(6.7.5), we obtain the following Þnite element model:⎡⎢⎢⎢⎢⎣[K11] [K12] [K13] [K14 [K15][K21] [K22] [K23] [K24] [K25][K31] [K32] [K33] [K34] [K35][K41] [K42] [K43] [K44] [K45][K51] [K52] [K53] [K54] [K55]

⎤⎥⎥⎥⎥⎦⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩ueveweS1S2

⎫⎪⎪⎪⎪⎬⎪⎪⎪⎪⎭ =⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩F 1F 2F 3F 4F 5

⎫⎪⎪⎪⎪⎬⎪⎪⎪⎪⎭+⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩F 1TF 2T0F 4TF 5T

⎫⎪⎪⎪⎪⎬⎪⎪⎪⎪⎭(6.7.11)

or, in generic matrix form

[Ke]∆e = F e (6.7.12)

where the coefficients of the submatrices [Kαβ] and vectors Fα and FαTare deÞned for (α,β = 1, 2, 3, 4, 5) by the expressions

K11ij =

ZΩe

⎛⎝A11∂ψ(1)i∂x

∂ψ(1)j

∂x+A66

∂ψ(1)i

∂y

∂ψ(1)j

∂y

⎞⎠ dx dy

K12ij =

ZΩe

⎛⎝A12∂ψ(1)i∂x

∂ψ(1)j

∂y+A66

∂ψ(1)i

∂y

∂ψ(1)j

∂x

⎞⎠ dx dy

K13ij =

1

2

ZΩe

⎡⎣∂ψ(1)i∂x

⎛⎝A11∂w0∂x

∂ψ(2)j

∂x+A12

∂w0∂y

∂ψ(2)j

∂y

⎞⎠+A66

∂ψ(1)i

∂y

⎛⎝∂w0∂x

∂ψ(2)j

∂y+∂w0∂y

∂ψ(2)j

∂x

⎞⎠#dx dy


K22ij =

ZΩe

⎛⎝A66∂ψ(1)i∂x

∂ψ(1)j

∂x+A22

∂ψ(1)i

∂y

∂ψ(1)j

∂y

⎞⎠ dx dyK23ij =

1

2

ZΩe

⎡⎣∂ψ(1)i∂y

⎛⎝A12∂w0∂x

∂ψ(2)j

∂x+A22

∂w0∂y

∂ψ(2)j

∂y

⎞⎠+A66

∂ψ(1)i

∂x

⎛⎝∂w0∂x

∂ψ(2)j

∂y+∂w0∂y

∂ψ(2)j

∂x

⎞⎠#dx dyK31ij =

ZΩe

⎡⎣∂ψ(2)i∂x

⎛⎝A11∂w0∂x

∂ψ(1)j

∂x+A66

∂w0∂y

∂ψ(1)j

∂y

⎞⎠+∂ψ

(2)i

∂y

⎛⎝A66∂w0∂x

∂ψ(1)j

∂y+A12

∂w0∂y

∂ψ(1)j

∂x

⎞⎠ #dx dyK32ij =

ZΩe

⎡⎣∂ψ(2)i∂x

⎛⎝A12∂w0∂x

∂ψ(1)j

∂y+A66

∂w0∂y

∂ψ(1)j

∂x

⎞⎠+∂ψ

(2)i

∂y

⎛⎝A66∂w0∂x

∂ψ(1)j

∂x+A22

∂w0∂y

∂ψ(1)j

∂y

⎞⎠ #dx dyK33ij =

ZΩe

⎛⎝KsA55∂ψ(e)i∂x

∂ψ(2)j

∂x+KsA44

∂ψ(2)i

∂y

∂ψ(2)j

∂y+ kψ

(e)i ψ

(2)j

⎞⎠ dx dy

+1

2

ZΩe

⎧⎨⎩"A11

µ∂w0∂x

¶2+A66

µ∂w0∂y

¶2# ∂ψ(e)i∂x

∂ψ(2)j

∂x

+

"A66

µ∂w0∂x

¶2+A22

µ∂w0∂y

¶2# ∂ψ(2)i∂y

∂ψ(2)j

∂y

+ (A12 +A66)∂w0∂x

∂w0∂y

⎛⎝∂ψ(2)i∂x

∂ψ(2)j

∂y+∂ψ

(2)i

∂y

∂ψ(2)j

∂x

⎞⎠)dx dyK34ij =

ZΩeKsA55

∂ψ(2)i

∂xψ(3)j dx dy, K35

ij =

ZΩeKsA44

∂ψ(2)i

∂yψ(3)j dx dy

K44ij =

ZΩe

⎛⎝D11∂ψ(3)i∂x

∂ψ(3)j

∂x+D66

∂ψ(3)i

∂y

∂ψ(3)j

∂y+KsA55ψ

(3)i ψ

(3)j

⎞⎠ dx dy

K45ij =

ZΩe

⎛⎝D12∂ψ(3)i∂x

∂ψ(3)j

∂y+D66

∂ψ(3)i

∂y

∂ψ(3)j

∂x

⎞⎠ dx dy


K55ij =

ZΩe

⎛⎝D66∂ψ(3)i∂x

∂ψ(3)j

∂x+D22

∂ψ(3)i

∂y

∂ψ(3)j

∂y+KsA44ψ

(3)i ψ

(3)j

⎞⎠ dx dy

F 1i =

IΓeNnψ

(1)i ds, F 2i =

IΓeNsψ

(1)i ds

F 3i =

ZΩeqψ

(2)i dxdy +

IΓeQnψ

(2)i ds

F 4i =

IΓeMnψ

(3)i ds, F 5i =

IΓeMsψ

(3)i ds

F 1Ti =

IΩe

∂ψ(1)i

∂xNTxx ds, F

2Ti =

IΩe

∂ψ(1)i

∂yNTyy ds

F 4Ti =

IΩe

∂ψ(3)i

∂xMTxx ds, F

5Ti =

IΩe

∂ψ(3)i

∂yMTyy ds

K21ij = K

12ji ; K

43ij = K

34ji ; K

54ij = K

45ji (6.7.13)

and all other stiffness coefficients are zero. Here NTxx and N

Tyy denote thermal

forces and MTxx and M

Tyy the thermal moments. It is noted that the element

stiffness matrix is not symmetric.The displacement-based C0 plate bending element of Eq. (6.7.12) is often

referred to in the Þnite element literature as the Mindlin plate element, whichis labeled in this book as the Þrst-order shear deformation theory (FSDT)element. When the bilinear rectangular element is used for all generalizeddisplacements (u0, v0, w0,φx,φy), the element stiffness matrices are of theorder 20 × 20; and for the nine-node quadratic element they are 45 × 45 (seeFigure 6.7.1).

Figure 6.7.1 Linear and nine-node quadratic rectangular elements for theFSDT.

a

b

z

x

y

4 3

2

1

u0, v0 , w0, φx , φy

a

b

z

x

y

4 3

21

7

6

5

8 9

u0, v0 , w0, φx , φy


6.7.3 Tangent Stiffness Coefficients

The tangent stiffness matrix coefficients needed for the NewtonRaphsonmethod of solving the nonlinear equations in Eq. (6.7.12) can be computedusing the deÞnition. Since the source of nonlinearity in the CPT and FSDTis the same, the nonlinear parts of the tangent stiffness coefficients derived forthe CPT are also applicable to the FSDT.Suppose that the tangent stiffness matrix is of the same form as the direct

stiffness matrix in Eq. (6.7.1). Then the coefficients of the submatrices [Tαβ]are deÞned by

Tαβij =∂Rαi

∂∆βj(6.7.14)

where the components of the residual vector Rα are given by

Rαi =5Xγ=1

n∗Xk=1

Kαγik ∆

γk − Fαi (6.7.15)

∆1i = ui, ∆2i = vi, ∆

3i = wi, ∆

4i = S

1i , ∆

5i = S

2i (6.7.16)

and n∗ denotes n, m, or p, depending on the nodal degree of freedom. Thus,we have

Tαβij =∂

∂∆βj

⎛⎝ 5Xγ=1

n∗Xk=1

Kαγik ∆

γk − Fαi

⎞⎠ = 5Xγ=1

n∗Xk=1

∂Kαγik

∂∆βj∆γk +K

αβij (6.7.17)

It should be noted that only coefficients that depend on the solution are K13ij ,

K23ij , K

31ij , K

32ij , and K

33ij . Further, they are functions of only w0 (or functions

of wj). Hence, derivatives of all submatrices with respect to uj , vj , S1j , and

S2j are zero. Thus, we have

T 11ij =5Xγ=1

n∗Xk=1

∂K1γik

∂uj∆γk +K

11ij = K

11ij , T 12ij =

5Xγ=1

n∗Xk=1

∂K1γik

∂vj∆γk +K

12ij = K

12ij

T 13ij =5Xγ=1

n∗Xk=1

∂K1γik

∂wj∆γk +K

13ij =

nXk=1

∂K13ik

∂wjwk +K

13ij

=1

2

ZΩe

⎡⎣∂ψ(1)i∂x

⎛⎝A11∂w0∂x

∂ψ(2)j

∂x+A12

∂w0∂y

∂ψ(2)j

∂y

⎞⎠+A66

∂ψ(1)i

∂y

⎛⎝∂w0∂x

∂ψ(2)j

∂y+∂w0∂y

∂ψ(2)j

∂x

⎞⎠#dx dy +K13ij


= K13ij +K

13ij = 2K

13ij (= T

31ji = K

31ji )

T 14ij =5Xγ=1

n∗Xk=1

∂K1γik

∂S1j∆γk +K

14ij = K

14ij , T 15ij =

5Xγ=1

n∗Xk=1

∂K1γik

∂S2j∆γk +K

15ij = K

15ij

T 21ij =5Xγ=1

n∗Xk=1

∂K2γik

∂uj∆γk +K

21ij = K

21ij , T 22ij =

5Xγ=1

n∗Xk=1

∂K2γik

∂vj∆γk +K

22ij = K

22ij

T 23ij =5Xγ=1

n∗Xk=1

∂K2γik

∂wj∆γk +K

23ij =

nXk=1

∂K23ik

∂wjwk +K

23ij

=1

2

ZΩe

⎡⎣∂ψ(1)i∂y

⎛⎝A12∂w0∂x

∂ψ(2)j

∂x+A22

∂w0∂y

∂ψ(2)j

∂y

⎞⎠+A66

∂ψ(1)i

∂x

⎛⎝∂w0∂y

∂ψ(2)j

∂x+∂w0∂x

∂ψ(2)j

∂y

⎞⎠#dx dy +K23ij

= K23ij +K

23ij = 2K

23ij (= T

32ji = K

32ji )

T 24ij =5Xγ=1

n∗Xk=1

∂K2γik

∂S1j∆γk +K

24ij = K

24ij , T 25ij =

5Xγ=1

n∗Xk=1

∂K2γik

∂S2j∆γk +K

25ij = K

25ij

T 31ij =5Xγ=1

n∗Xk=1

∂K3γik

∂uj∆γk +K

31ij = K

31ij , T 32ij =

5Xγ=1

n∗Xk=1

∂K3γik

∂vj∆γk +K

32ij = K

32ij

T 33ij =5Xγ=1

n∗Xk=1

∂K3γik

∂wj∆γk +K

33ij =

n∗Xk=1

Ã∂K31

ik

∂wjuk +

∂K32ik

∂wjvk +

∂K33ik

∂wjwk

!+K33

ij

=

ZΩe

⎛⎝KsA55∂ψ(2)i∂x

∂ψ(2)j

∂x+KsA44

∂ψ(2)i

∂y

∂ψ(2)j

∂y+ kψ

(2)i ψ

(2)j

⎞⎠ dx dy+

ZΩe

½(Nxx +N

Txx)∂ψ

(2)i

∂x

∂ψ(2)j

∂x+ (Nyy +N

Tyy)∂ψ

(2)i

∂y

∂ψ(2)j

∂y

+Nxy

⎛⎝∂ψ(2)i∂x

∂ψ(2)j

∂y+∂ψ

(2)i

∂y

∂ψ(2)j

∂x

⎞⎠+ (A12 +A66)

∂w0∂x

∂w0∂y

⎛⎝∂ψ(2)i∂x

∂ψ(2)j

∂y+∂ψ

(2)i

∂y

∂ψ(2)j

∂x

⎞⎠+

"A11

µ∂w0∂x

¶2+A66

µ∂w0∂y

¶2# ∂ψ(2)i∂x

∂ψ(2)j

∂x

+

"A66

µ∂w0∂x

¶2+A22

µ∂w0∂y

¶2# ∂ψ(2)i∂y

∂ψ(2)j

∂y

¾dx dy


T 34ij =5Xγ=1

n∗Xk=1

∂K3γik

∂S1j∆γk +K

34ij = K

34ij , T 35ij =

5Xγ=1

n∗Xk=1

∂K3γik

∂S2j∆γk +K

35ij = K

35ij

T 44ij =5Xγ=1

n∗Xk=1

∂K4γik

∂S1j∆γk +K

44ij = K

44ij , T 45ij =

5Xγ=1

n∗Xk=1

∂K4γik

∂S2j∆γk +K

45ij = K

45ij

T 55ij =5Xγ=1

n∗Xk=1

∂K5γik

∂S2j∆γk +K

55ij = K

55ij (6.7.18)

Note that all nonlinear coefficients are the same as those derived for the CPT.Once again, we note that the tangent stiffness matrix of the FSDT element issymmetric.

6.7.4 Shear and Membrane Locking

The C0−plate bending elements based on the FSDT are among the simplestavailable in the literature. Unfortunately, when lower order (quadratic orless) equal interpolation of the generalized displacements is used, the elementsbecome excessively stiff in the thin plate limit, yielding displacements that aretoo small compared to the true solution. As discussed earlier for beams, thistype of behavior is known as shear locking. There are a number of papers onthe subject of shear locking and elements developed to alleviate the problem(see [2328]). A commonly used technique is to use selective integration[29,30]: use full integration to evaluate all linear stiffness coefficients anduse reduced integration to evaluate the transverse shear stiffnesses (i.e. all

coefficients inKαβij that containA44 andA55) and nonlinear stiffnesses. Higher-

order elements or reÞned meshes of lower-order elements experience relativelyless locking, but sometimes at the expense of the rate of convergence. Withthe suggested Gauss rule, highly distorted elements tend to have slower ratesof convergence but they give sufficiently accurate results.

6.8 Computer Implementation andNumerical Results of FSDT Elements


The FSDT element is quite simple to implement, and the implementationfollows the same ideas as discussed in Chapter 5 for single-variables problemsin two dimensions. The main difference is that the number of degrees offreedom (NDF) is 5, and one must rearrange the coefficients using the Fortranstatements included in Box 4.3.1. Thus, the element displacement vector is ofthe form

∆ = u01, v01, w01,φx1,φy1, u02, v02, w02,φx2,φy2, . . .


Two separate do-loops on Guass quadrature are required to compute allthe force and stiffness coefficients. The full integration loop is used to evaluateall force components and all linear, except for the transverse shear stiffnesses.The reduced integration loop has two parts: one for the transverse shearterms and the other for nonlinear terms. One may use separate arrays (say,EXT13(I, J), EXT33(I, J), and so on to store the extra terms (to those

of Kαβij ) of the tangent stiffness coefficients. The stiffness coefficients are

rearranged as in Box 4.3.1.

6.8.2 Results of Linear Analysis

The effect of the integration rule and the convergence characteristics ofthe FSDT plate element based on equal interpolation (bilinear as well asbiquadratic elements) is illustrated through several examples.

Example 6.8.1

Consider a simply supported (SS1) isotropic (ν = 0.25 and Ks = 5/6) square plateunder uniformly distributed transverse load q0. The geometric boundary conditions of thecomputational domain (see the shaded quadrant in Figure 6.8.1) are

u0 = φx = 0 at x = 0; v0 = φy = 0 at y = 0 (symm. lines) (6.8.1)

v0 = w0 = φy = 0 at x =a

2; u0 = w0 = φx = 0 at y =

b

2(6.8.2)

The following non-dimensionalizations of the quantities are used:

w = w0(0,0)E2h

3

a4q0, σxx = σxx(0,0,

h

2)h2

b2q0, σyy = σyy(0, 0,

h

4)h2

b2q0

σxy = σxy(a

2,b

2,−h2)h2

b2q0, σxz = σxz(

a

2,0,−h

2)h

bq0, σyz = σyz(0,

b

2,h

2)h

bq0(6.8.3)

where the origin of the coordinate system is taken at the center of the plate, 0 ≤ x ≤ a/2,0 ≤y ≤ b/2, and −h/2 ≤ z ≤ h/2. The stresses in the Þnite element analysis are computed at thereduced Gauss points, irrespective of the Gauss rule used for the evaluation of the elementstiffness coefficients. The Gauss point coordinates A and B are shown in Table 6.8.1. TheÞnite element solutions are compared with the analytical solutions from [2,3] in Table 6.8.2for two side-to-thickness ratios a/h = 10 and 100. The notation nL stands for n×n uniformmesh of linear rectangular elements and nQ9 for n×n uniform mesh of nine-node quadraticelements in a quarter plate.

The stresses are evaluated at the Gauss points as indicated below:

σxx(A,A,h

2), σxy(B,B,−h

2), σxz(B,A,−h

2) (6.8.4)

Thus, as mesh is reÞned or higher-order elements are used, the Gauss point locations getcloser to the points at which the analytical solutions for stresses are evaluated.

x

y

a2

b 2

b 2

a2 = w0= u0 φx = 0= w0= v0 φy = 0

= w0= u0 φx = 0

= w0= v0 φy = 0

at 0x= ; φy = 0 at 0y=φx = 0Symm. B.C.: = u0 = v0


Figure 6.8.1 Geometric boundary conditions for SS1 type simplysupported rectangular plates.

Table 6.8.1 The Gauss point locations at which the stresses are computedin the Þnite element analysis of simply supported plates.

Point 2L 4L 8L 1Q9 2Q9 4Q9

A 0.125a 0.0625a 0.03125a 0.10566a 0.05283a 0.02642aB 0.375a 0.4375a 0.46875a 0.39434a 0.44717a 0.47358a

The following notation is used in Table 6.8.2: F = full integration; R = reducedintegration; S = selective integration: full integration of all except the transverse shearcoefficients, which are evaluated using reduced integration rule. The CPT solution isindependent of side-to-thickness ratio, a/h.

The results of Table 6.8.2 indicate that the FSDT Þnite element with equal interpolationof all generalized displacements does not experience shear locking for thick plates even whenfull integration rule is used for the evaluation of all stiffness coefficients. Shear locking isevident when the element is used to model thin plates (a/h ≥ 100) with full integration rule(F). Also, higher-order elements are less sensitive to locking but exhibit slower convergence.The element behaves uniformly well for thin and thick plates when the reduced (R) orselectively reduced integration (S) rule is used. The selective integration rule gives the mostaccurate solutions.

Example 6.8.2

Next consider a clamped isotropic (ν = 0.25 and Ks = 5/6) square plate under uniformlydistributed load. The non-dimensionalizations used are the same as in Eq. (6.8.3), exceptfor the location of the stresses. The analytical stresses were non-dimensionalized as follows:

σxx = σxx(a

2, 0,−h

2)h2

b2q0, σxz = σxz(

a

2,0,−h

2)h

bq0(6.8.5)


Table 6.8.2 Effect of integration rule on the linear deßections w and stressesσ of simply supported, isotropic (ν = 0.25 and Ks = 5/6),square plates under a uniformly distributed load q0.

a/h Mesh w × 102 σxx σxy σxz

Finite element solutions (FSDT)

10 2LF 2.4742 0.1185 0.0727 0.26272LS 4.7120 0.2350 0.1446 0.27502LR 4.8887 0.2441 0.1504 0.27501QF 4.5304 0.2294 0.1610 0.28131QS 4.9426 0.2630 0.1639 0.28471QR 4.9711 0.2645 0.1652 0.28864LF 3.8835 0.2160 0.1483 0.33664LS 4.7728 0.2661 0.1850 0.33564LR 4.8137 0.2684 0.1869 0.33562QF 4.7707 0.2699 0.1930 0.34372QS 4.7989 0.2715 0.1939 0.34242QR 4.8005 0.2716 0.1943 0.34258LF 4.5268 0.2590 0.1891 0.37008LS 4.7966 0.2743 0.2743 0.20148LR 4.7866 0.2737 0.2737 0.20084QF 4.7897 0.2749 0.2044 0.37374QS 4.7916 0.2750 0.2043 0.37354QR 4.7917 0.2750 0.2044 0.3735

Anal. solns [3] 4.7914 0.2762 0.2085 0.3927

Finite element solutions

100 2LF 0.0469 0.0024 0.0014 0.26352LS 4.4645 0.2350 0.1446 0.27502LR 4.6412 0.2441 0.1504 0.27501QF 4.0028 0.2040 0.1591 0.27331QS 4.7196 0.2629 0.1643 0.28371QR 4.7483 0.2645 0.1652 0.28864LF 0.1819 0.0108 0.0071 0.34624LS 4.5481 0.2661 0.1850 0.33564LR 4.5890 0.2684 0.1869 0.33562QF 4.4822 0.2644 0.1893 0.34852QS 4.5799 0.2715 0.1941 0.34142QR 4.5815 0.2716 0.1943 0.34258LF 0.6497 0.0401 0.0275 0.38478LS 4.5664 0.2737 0.2008 0.36918LR 4.5764 0.2743 0.2014 0.36914QF 4.5530 0.2741 0.2020 0.37494QS 4.5728 0.2750 0.2044 0.37344QR 4.5729 0.2750 0.2044 0.3735

Finite element solutions (CPT)

4× 4C 4.5952 0.2637 0.1935 −8× 8C 4.5734 0.2732 0.2040 −

Anal. solns [3] 4.5698 0.2762 0.2085 0.3927


while the Þnite element solutions were non-dimensionalized as

σxx = σxx(A,B,−h2)h2

b2q0, σxz = σxz(A,B,−h

2)h

bq0(6.8.6)

where the values of A and B for different meshes are given in Table 6.8.3. Table 6.8.4

contains the non-dimensionalized displacements and stresses. The FSDT element with

selective integration or reduced integration is accurate in predicting the bending response.

Table 6.8.3 The Gauss point locations at which the stresses are computedin the Þnite element analysis of clamped plates (Example 6.8.2).

Point 4L 8L 2Q9 4Q9

A 0.4375a 0.46875a 0.44717a 0.47358a

B 0.0625a 0.03125a 0.05283a 0.02642a

Table 6.8.4 Effect of integration rule on the linear deßections w and stressesσ of clamped, isotropic (ν = 0.25 and Ks = 5/6), square platesunder a uniform load q0.

a/h Integ. Variable 4× 4L 2× 2Q9 8× 8L 4× 4Q9

10 F w × 102 1.2593 1.5983 1.5447 1.6685σxx 0.1190 0.1568 0.2054 0.2301σxz 0.3890 0.4193 0.4463 0.4578

10 S w × 102 1.6632 1.6880 1.6721 1.6758σxx 0.1689 0.1813 0.2275 0.2357σxz 0.4056 0.4118 0.4511 0.4566

10 R w × 102 1.6854 1.6903 1.6776 1.6760σxx 0.1718 0.1817 0.2204 0.2358σxz 0.4045 0.4120 0.4509 0.4566

100 F w × 102 0.0386 1.1222 1.3982 1.3546σxx 0.0041 0.0947 0.0204 0.1825σxz 0.3734 0.4732 0.4229 0.4862

100 S w × 102 1.4093 1.4382 1.4219 1.4268σxx 0.1731 0.1846 0.2337 0.2417σxz 0.4236 0.4255 0.4729 0.4776

100 R w × 102 1.4334 1.4417 1.4279 1.4271σxx 0.1762 0.1853 0.2346 0.2418σxz 0.4234 0.4284 0.4727 0.4779

100 CPT(C) 4× 4 w × 102 = 1.4370; σxx = 0.1649

x

y

a2

b 2

b 2

a2


w0= u0 = v0 = 0

w0= u0 = v0 = 0

w0= u0 = v0 = 0

w0= u0 = v0 = 0


6.8.3 Results of Nonlinear Analysis

Here we consider several examples of the nonlinear bending of rectangularplates using the nonlinear FSDT element. The effect of the integration rule toevaluate the nonlinear and transverse shear stiffness coefficients is investigatedin the Þrst example. Unless stated otherwise, a uniform mesh of 4 × 4 nine-node quadratic elements is used in a quarter plate for the FSDT. For thischoice of mesh, full integration (F) is to use 3 × 3 Gauss rule and reducedintegration (R) is to use 2×2 Gauss rule. Stresses are calculated at the centerof the element. The shear correction coefficient is taken to be Ks = 5/6.

Example 6.8.3

Consider an isotropic, square plate with

a = b = 10 in., h = 1 in., E = 7.8× 106 psi, ν = 0.3 (6.8.7)

Two types of simply supported boundary conditions are studied. The displacement boundaryconditions used for SS1 and SS3 are (see Figures 6.8.1 and 6.8.2, respectively)

SS1: At x = a/2 : v0 = w0 = φy = 0; At y = b/2 : u0 = w0 = φx = 0(6.8.8)

SS3: u0 = v0 = w0 = 0 on simply supported edges (6.8.9)

Uniformly distributed load of intensity q0 is used. The boundary conditions along thesymmetry lines for both cases are given by Eq. (6.8.1). It is clear that SS3 providesmore edge restraint than SS1 and therefore should produce lower transverse deßections.

Using the load parameter introduced earlier, P ≡ q0a4/E2h4, the incremental load vectoris chosen to be

∆P = 6.25, 6.25, 12.5, 25.0,25.0, . . . , 25.0A tolerance of ² = 10−2 is used for convergence in the NewtonRaphson iteration scheme tocheck for convergence of the nodal displacements.

Figure 6.8.2 Geometric boundary conditions used for SS3 type simplysupported rectangular plates.


Table 6.8.5 contains the deßections w0(0,0) and normal stresses σxx = σxx(a2/Eh2)obtained with a uniform mesh of 4× 4Q9 FSDT elements for various integration rules (alsosee Figure 6.8.3). The number of iterations taken for convergence are listed in parenthesis.The linear FSDT plate solution for load q0 = 4875psi (or P = 6.25) is w0 = 0.2917in.for SS1 and w0 = 0.3151in. for SS3 (when reduced integration is used to evaluate thetransverse shear stiffnesses). As discussed earlier, the 4 × 4Q9 meshes are not sensitive toshear or membrane locking, and therefore the results obtained with various integration rulesare essentially the same.

Table 6.8.5 Center deßection w and stresses σxx of a simply supported (SS1 and SS3)plates under uniformly distributed load (Example 6.8.3).

SS3 SS1

P RR∗ FR FF RR FR FF

Deßections, w0(0,0)

6.25 0.2790 (3) 0.2790 (4) 0.2780 (3) 0.2813 (3) 0.2813 (3) 0.2812 (3)12.5 0.4630 (3) 0.4630 (3) 0.4619 (3) 0.5186 (3) 0.5186 (3) 0.5185 (3)25.0 0.6911 (3) 0.6911 (3) 0.6902 (3) 0.8673 (4) 0.8673 (4) 0.8672 (4)50.0 0.9575 (3) 0.9575 (3) 0.9570 (3) 1.3149 (4) 1.3149 (4) 1.3147 (4)75.0 1.1333 (3) 1.1333 (3) 1.1330 (3) 1.6241 (3) 1.6239 (3) 1.6237 (3)100.0 1.2688 (3) 1.2688 (3) 1.2686 (3) 1.8687 (3) 1.8683 (3) 1.8679 (3)125.0 1.3809 (2) 1.3809 (2) 1.3808 (2) 2.0758 (2) 2.0751 (2) 2.0746 (2)150.0 1.4774 (2) 1.4774 (2) 1.4774 (2) 2.2567 (2) 2.2556 (2) 2.2549 (2)175.0 1.5628 (2) 1.5629 (2) 1.5629 (2) 2.4194 (2) 2.4177 (2) 2.4168 (2)200.0 1.6398 (2) 1.6399 (2) 1.6399 (2) 2.5681 (2) 2.5657 (2) 2.5645 (2)225.0 1.7102 (2) 1.7103 (2) 1.7103 (2) 2.7056 (2) 2.7023 (2) 2.7009 (2)250.0 1.7752 (2) 1.7753 (2) 1.7754 (2) 2.8338 (2) 2.8296 (2) 2.8279 (2)

Normal stressess, σxx(0.625,0.625, h/2)

6.25 1.861 1.861 1.856 1.779 1.779 1.78012.5 3.305 3.305 3.300 3.396 3.396 3.39825.0 5.319 5.320 5.317 5.882 5.882 5.88550.0 8.001 8.002 8.001 9.159 9.162 9.16575.0 9.983 9.984 9.983 11.458 11.462 11.465100.0 11.633 11.634 11.634 13.299 13.307 13.308125.0 13.084 13.085 13.085 14.878 14.890 14.889150.0 14.396 14.398 14.398 16.278 16.293 16.290175.0 15.608 15.610 15.610 17.553 17.572 17.567200.0 16.741 16.743 16.743 18.733 18.755 18.748225.0 17.811 17.813 17.812 19.837 19.863 19.854250.0 18.828 18.831 18.829 20.880 20.909 20.898

* The Þrst letter refers to the integration rule used for the nonlinear terms while the secondletter refers to the integration rule used for the shear terms.

0 50 100 150 200 250Load parameter,

0.0

0.5

1.0

1.5

2.0

2.5

3.0D

efle

ctio

n, SS-3 (CPT)

SS-1 (CPT)

SS-1 (FSDT)

SS-3 (FSDT)

(a)

w

P

4

400 ,

Ehaq

Ph

ww ==

0 50 100 150 200 250Load parameter,

0

4

8

12

16

20

24

Str

esse

s, SS-3 (CPT)

SS-1 (CPT)

SS-1 (FSDT)

SS-3 (FSDT)

(b)

σ xx

P

Membrane stresses

⎟⎟⎠

⎞⎜⎜⎝

⎛=2

2

Eha

xxxx σσ


Figure 6.8.3 Plots of (a) center deßection w0 versus load P and (b) centernormal stress σxx versus load P for isotropic (ν = 0.3),simply supported square plates under uniformly distributedload (4× 4Q9 for FSDT and 8× 8C for CPT).


Example 6.8.4

Orthotropic plates subjected to uniformly distributed transverse load (i.e. q = q0=constant)are analyzed. The geometric and material parameters used are

a = b = 12 in., h = 0.138 in., E1 = 3× 106 psi, E2 = 1.28× 106 psiG12 = G13 = G23 = 0.37× 106psi, ν12 = 0.32 (6.8.10)

A uniform mesh of 4 × 4Q9 elements with reduced integration is used in a quadrant. Theincremental load vector is chosen to be

∆P = 0.05,0.05, 0.1,0.2, 0.2, . . . ,0.2

Twelve load steps are used, and a tolerance of ² = 0.01 is used for convergence.

Plots of load q0 (psi) versus center deßection w0 (in.) and q0 versus normal stress (totalas well as membrane) σxx = σxx(a2/E2h

2) are shown in Figure 6.8.4 for SS1 and SS3 plates.The Þgures also show the results obtained using 8 × 8 mesh of conforming CPT elements.Table 6.8.6 contains the center deßection and total normal stress as a function of the load forthe two boundary conditions. The linear FSDT solution for load q0 = 0.05 is w0 = 0.01132for SS1 and w0 = 0.01140 for SS3.

Table 6.8.6 Center deßection w0 and normal stress σxx forsimply supported orthotropic square plates under uniformlydistributed load (4× 4Q9; Example 6.8.4).

SS1 SS3

q0 CPT FSDT FSDT CPT FSDT FSDTw0 w0 σxx w0 w0 σxx

0.05 0.0113 (2) 0.0113 1.034 0.0112 0.0113 1.0560.10 0.0224 (2) 0.0224 2.070 0.0217 0.0218 2.1160.20 0.0438 (3) 0.0439 4.092 0.0395 0.0397 4.0580.40 0.0812 (3) 0.0815 7.716 0.0648 0.0650 7.1030.60 0.1116 (3) 0.1122 10.702 0.0823 0.0824 9.4060.80 0.1367 (3) 0.1377 13.169 0.0957 0.0959 11.2841.00 0.1581 (2) 0.1594 15.255 0.1068 0.1069 12.8941.20 0.1767 (2) 0.1783 17.050 0.1162 0.1162 14.3161.40 0.1932 (2) 0.1951 18.631 0.1245 0.1244 15.6021.60 0.2081 (2) 0.2103 20.044 0.1318 0.1318 16.7831.80 0.2217 (2) 0.2241 21.324 0.1385 0.1384 17.8802.00 0.2343 (2) 0.2370 22.495 0.1447 0.1445 18.909

Example 6.8.5

Here, we analyze an orthotropic plate with clamped edges, that is, all generalizeddisplacements are zero on the boundary (see Figure 6.8.5) The boundary conditions of aclamped edge are taken to be

u0 = v0 = w0 = φx = φy = 0 (6.8.11)

The geometric and material parameters used are the same as those listed in Eq. (6.8.10). Auniformly distributed load of intensity q0 is used.


Figure 6.8.4 Center deßection w0(0, 0) and stresses σxx as functions of theload q0 for simply supported, orthotropic, square plates underuniformly distributed load (Example 6.8.4).

0.0 0.4 0.8 1.2 1.6 2.0 2.4

Load,

0

4

8

12

16

20

24

28

32

Str

esse

s,

SS-3 (CPT)

SS-1 (CPT)

SS-1 (FSDT)

SS-3 (FSDT)

(b)

σ xx

P

Membrane stresses

⎟⎟⎠

⎞⎜⎜⎝

⎛=2

2

Eha

xxxx σσ

0.0 0.4 0.8 1.2 1.6 2.0 2.4

Load,

0.00

0.05

0.10

0.15

0.20

0.25

0.30D

efle

ctio

n,

SS-1 (CPT)

SS-1 (FSDT)

SS-3 (CPT & FSDT)

(a)

w0

q0


x

y

a2

b 2

b 2

a2

= w0= u0 = v0 = 0φyφx =

= w0= u0 = v0 = 0φyφx=

= w0= u0 = v0 = 0φyφx = = w0= u0 = v0 = 0φyφx=

q0

(8×8 NC)

(4×4 Q9)

(4×4 Q9)

w0

x

y

0 4 8 12 16 20 24

Load,

0.00

0.05

0.10

0.15

0.20

0.25

0.30

0.35

0.40

Def

lect

ion

,

CPT (orthotropic)

FSDT (orthotropic)

FSDT (isotropic)


Figure 6.8.5 Boundary conditions for a rectangular plate with clampededges.

The linear solution for load q0 = 0.5 is w0 = 0.0301. Figure 6.8.6 contains a plot of

load versus center deßection, and Table 6.8.7 contains center deßections and stresses for

the problem (see [3638]). Figure 6.8.6 also contains plots of the CPT deßections obtained

using 8× 8 mesh of the non-conforming elements and FSDT deßections of an isotropic plate(h = 0.138 in., E = 1.28× 106 psi, and ν = 0.3) obtained with 4× 4Q9 mesh.

Figure 6.8.6 Nonlinear center deßection w0 versus load parameter q0 forclamped, orthotropic, square plates under uniform load.

x

y

a = 100 in.

E = 106 psi, ν = 0.3

h = 10 in.

0at0,0

===

yyv φ

0at0,0

===

xxu φ

⎪⎩

⎪⎨⎧

=====

edge clamped on the00

yxwuv

φφ

1 23

4

5 10 2015

25

26

27

2829

9

24

19

14


Table 6.8.7 Center deßection w0 and normal stress σxx for clampedorthotropic square plates under uniformly distributed load(4× 4Q9).

q0 w0 σxx q0 w0 σxx

0.5 0.0294 (3) 4.317 12.0 0.2450 (2) 46.0011.0 0.0552 (3) 8.467 14.0 0.2610 (2) 49.8512.0 0.0948 (3) 15.309 16.0 0.2754 (2) 53.4314.0 0.1456 (3) 24.811 18.0 0.2886 (2) 56.8006.0 0.1795 (3) 31.599 20.0 0.3006 (2) 59.9988.0 0.2054 (3) 37.078 22.0 0.3119 (2) 63.05310.0 0.2268 (2) 41.793 24.0 0.3224 (2) 65.986

Example 6.8.6

The last example is concerned with the nonlinear bending of a clamped, isotropic, circularplate under uniformly distributed load. Recall that axisymmetric bending of circular platesmay be studied using one-dimensional bending elements (see Problems 4.54.8).

Figure 6.8.7 shows the Þnite element mesh and boundary conditions of a quadrant of acircular plate. The exact center deßection for the linear case is given by (see Reddy [2,3])

w0(0) =q0a

4

64D

µ1 +

8

3(1− ν)Ks

h2

a2

¶(6.8.12)

where a denotes the radius of the plate. For ν = 0.3, Ks = 5/6, and a/h = 10, the centerdeßection is given by w0(0) = 1.0457(q0a

4/64D). The linear FSDT solution obtained withthe mesh of Þve nine-node quadratic elements is w0(0) = 1.0659(q0a

4/64D) (¡ 2% error).Figure 6.8.8 shows the loaddeßection curve for the problem (a = 100 in., h = 10 in., E = 106

psi, and ν = 0.3).

Figure 6.8.7 Mesh and boundary conditions used for a clamped circularplate.

w0/h

(q0a4/Eh4)

E = 106 psi, ν = 0.3a = 100 in., h = 10 in.

0 20 40 60 80 100 120

Load parameter,

0.0

1.0

2.0

3.0

4.0D

efle

ctio

n,

Mesh of 5-Q9 elements


Figure 6.8.8 Loaddeßection curve for clamped isotropic circular plateunder uniform load.

6.9 Theory of Doubly-Curved Shells

6.9.1 Introduction

In this section, we review the governing equations of shells. A number of shelltheories exist in the literature, and many of these theories were developedoriginally for thin shells and are based on the KirchhoffLove kinematichypothesis that straight lines normal to the undeformed midsurface remainstraight and normal to the middle surface after deformation. A detailed studyof thin isotropic shells can be found in the monographs by Ambartsumyan[4143], Flugge [44], Kraus [45], Timoshenko and Woinowsky-Krieger [46] andDym [47]. The Þrst-order shear deformation theory of shells, also known asthe Sanders shell theory [48,49], can be found in Kraus [45].

Following this introduction, the equilibrium equations and straindisplacement relations are presented. For additional details, one may consult[40, 4452]. The Þnite element models of doubly-curved shells is presented inSection 6.10.


6.9.2 Geometric Description

Figure 6.9.1(a) shows a uniform thickness shell, where (ξ1, ξ2, ζ) denote theorthogonal curvilinear coordinates such that ξ1 and ξ2 curves are the lines ofcurvature on the middle surface (ζ = 0). The position vector of a typical point(ξ1, ξ2, 0) on the middle surface is denoted by r, and the position of an arbitrarypoint (ξ1, ξ2, ζ) in the shell is denoted by R, as shown in Figure 6.9.2(b). Thesquare of the distance ds between points (ξ1, ξ2, 0) and (ξ1 + dξ1, ξ2 + dξ2, 0)is determined by (see Reddy [40, 50])

(ds)2 = dr · dr = a21(dξ1)2 + a22(dξ2)2 (6.9.1a)

dr = g1 dξ1 + g2 dξ2 , gα =∂r

∂ξα, gαβ = gα · gβ (6.9.1b)

where the vectors g1 and g2 are tangent to the ξ1 and ξ2 coordinate lines, gαβ(α,β = 1, 2) is called the surface metric tensor and aα (α = 1, 2) are

aα =√gαα, (no sum on α) (6.9.2)

Note that g1 · g2 = 0 when the lines of principal curvature coincide with thecoordinate lines.

The unit vector normal to the middle surface can be determined from

n =g1 × g2a1a2

(6.9.3)

Further, we have the WeingartenGauss relations

∂n

∂ξα=gαRα, (no sum on α) (theorem of Rodrigues) (6.9.4)

∂

∂ξ2

µa1R1

¶=

1

R2

∂a1∂ξ2

,∂

∂ξ1

µa2R2

¶=

1

R1

∂a2∂ξ1

(Codazzi conditions) (6.9.5)

The values of the principal radii of curvature of the middle surface are denotedby R1 and R2 [see Figure 6.9.1(c)]. In general, n, R1 and R2 are functions ofξ1 and ξ2.

The position vector R of a point at a distance ζ from the middle surfacecan be expressed in terms of r and n by [see Figure 6.9.1(b)]

R = r+ ζn (6.9.6)

By differentiation we have

∂R

∂ξα= gα + ζ

∂n

∂ξα≡Gα (6.9.7)


Figure 6.9.1 Geometry of a doubly-curved shell [50]. (a) Shell geometry.(b) Position vectors of points on the midsurface and abovethe midsurface. (c) A differential element of the shell (dS1and dS2 denote the arc lengths).

hξ1

ξ2

ζ

(a)

R2 R1

ξ1ξ2

ζdξ2

dξ1

1111

11 1 ξξ dAdRz

adS =⎟⎟⎠

⎞⎜⎜⎝

⎛+= 222

222 1 ξξ dAd

Rz

adS =⎟⎟⎠

⎞⎜⎜⎝

⎛+=

dS1dS2

(c)

ξ2

R2

n

ζ

rR

h

(b)

x1

x2

x3

e1

e3∧

e2

∧∧


and using Eq. (6.9.4) we obtain

Gα =∂R

∂ξα=

µ1 +

ζ

Rα

¶gα, (no sum on α) (6.9.8a)

andGαβ ≡Gα ·Gβ (6.9.8b)

Hence, the square of the distance dS between points (ξ1, ξ2, ζ) and (ξ1 +dξ1, ξ2 + dξ2, ζ + dζ) is given by

(dS)2 = dR · dR = A21(dξ1)2 +A22(dξ2)

2 +A23(dζ)2 (6.9.9a)

in whichdR =G1 dξ1 +G2 dξ2 + n dζ, (6.9.9b)

and A1, A2, and A3 are the Lame coefficients [see Fig. 6.9.1(c)]

A1 = a1

µ1 +

ζ

R1

¶=pG11, A2 = a2

µ1 +

ζ

R2

¶=pG22, A3 = 1

(6.9.10)Note that vector Gα is parallel to the vector gα. In view of the Codazziconditions (6.9.5) and Eq. (6.9.10), it can be shown that the following relationsbetween the derivatives of aα and Aα hold:

1

A2

∂A1∂ξ2

=1

a2

∂a1∂ξ2

,1

A1

∂A2∂ξ1

=1

a1

∂a2∂ξ1

(6.9.11)

From Figure 6.9.1(c) the elements of area of the cross sections are

dS1dζ = A1 dξ1dζ = a1

µ1 +

ζ

R1

¶dξ1 dζ,

dS2dζ = A2 dξ2dζ = a2

µ1 +

ζ

R2

¶dξ2 dζ (6.9.12)

An elemental area of the middle surface (ζ = 0) is determined by [see Figure6.9.2(a)]

dA0 = dr1 × dr2 · n =µ∂r

∂ξ1× ∂r

∂ξ2· n¶dξ1 dξ2 = a1a2 dξ1 dξ2 (6.9.13)

and an elemental area of the surface at ζ is given by [see Figure 6.9.2(b)]

dAζ = dR1 × dR2 · n =µ∂R

∂ξ1× ∂R∂ξ2

· n¶dξ1 dξ2 = A1A2 dξ1 dξ2 (6.9.14)

(b)

ξ1 ξ2

ζ

Surface at +ζ

dR1 dR2n

dAζ

x1

x2

x3

e1∧

e3∧

e2∧

R2

R1

RG2

G1

dAζ = A1 A2 dξ1 dξ2

ζ

ξ1 ξ2

ζ

Middle surface

dr1 dr2n

dA0

x1

x2

x3

e1

e3∧

e2∧

r2r1

rg2

g1

dA0 = a1 a2 dξ1 dξ2

(a)

∧


The volume of a differential element above the midsurface is given by

dV = dR1 × dR2 · ndζ = dAζdζ = A1A2 dξ1 dξ2 dζ (6.9.15)

Figure 6.9.2 Surface area elements of a doubly-curved shell [50]. (a) Areaelement on the midsurface. (b) Area element on a surface at+ζ.


6.9.3 StrainDisplacement Relations

The engineering components of the GreenLagrange strain tensor in anorthogonal curvilinear coordinate system are given by (no sum on repeatedindices; see [51])

εi =∂

∂ξi

µuiAi

¶+1

Ai

3Xk=1

ukAk

∂Ai∂ξk

+1

2

"∂

∂ξi

µuiAi

¶+1

Ai

3Xk=1

ukAk

∂Ai∂ξk

#2

+1

2A2i

3Xk=1,k 6=i

µ∂uk∂ξi

− uiAk

∂Ai∂ξk

¶2(6.9.16a)

γij =AiAj

∂

∂ξj

µuiAi

¶+AjAi

∂

∂ξi

ÃujAj

!

+3X

k=1,k 6=i

1

AiAj

µ∂uk∂ξi

− uiAk

∂Ai∂ξk

¶Ã∂uk∂ξj

− ujAk

∂Aj∂ξk

!

+1

Aj

Ã∂ui∂ξj

− ujAi

∂Aj∂ξi

!"∂

∂ξi

µuiAi

¶+1

Ai

3Xk=1

ukAk

∂Ai∂ξk

#

+1

Ai

Ã∂uj∂ξi

− uiAj

∂Ai∂ξj

!"∂

∂ξj

ÃujAj

!+1

Aj

3Xk=1

ukAk

∂Aj∂ξk

#(6.9.16b)

where i 6= j in Eq. (6.9.16b), and

ξ3 = ζ, A1 = a1

µ1 +

ζ

R1

¶, A2 = a2

µ1 +

ζ

R2

¶, A3 = a3 = 1 (6.9.17)

Substituting equation (6.9.17) into (6.9.16a,b) and making use of conditions(6.9.10) and (6.9.11), one obtains

ε1 =1

A1

µ∂u1∂ξ1

+1

a2

∂a1∂ξ2

u2 +a1R1u3

¶+

1

2A21

"µ∂u1∂ξ1

+u2a2

∂a1∂ξ2

+a1R1u3

¶2+

µ∂u2∂ξ1

− u1a2

∂a1∂ξ2

¶2+

µ∂u3∂ξ1

− a1R1u1

¶2#

ε2 =1

A2

µ∂u2∂ξ2

+1

a1

∂a2∂ξ1

u1 +a2R2u3

¶+

1

2A22

"µ∂u2∂ξ2

+u1a1

∂a2∂ξ1

+a2R2u3

¶2+

µ∂u1∂ξ2

− u2a1

∂a2∂ξ1

¶2+

µ∂u3∂ξ2

− a2R2u2

¶2#

ε3 =∂u3∂ζ

+1

2

"µ∂u1∂ζ

¶2+

µ∂u2∂ζ

¶2+

µ∂u3∂ζ

¶2#


γ23 =1

A2

∂u3∂ξ2

+A2∂

∂ζ

µu2A2

¶+1

A2

"∂u2∂ζ

µ∂u2∂ξ2

+u1a1

∂a2∂ξ1

+a2R2u3

¶

+∂u1∂ζ

µ∂u1∂ξ2

− u2a1

∂a2∂ξ1

¶+∂u3∂ζ

µ∂u3∂ξ2

− a2R2u2

¶#

γ13 =1

A1

∂u3∂ξ1

+A1∂

∂ζ

µu1A1

¶+1

A1

"∂u1∂ζ

µ∂u1∂ξ1

+u2a2

∂a1∂ξ2

+a1R1u3

¶

+∂u2∂ζ

µ∂u2∂ξ1

− u1a2

∂a1∂ξ2

¶+∂u3∂ζ

µ∂u3∂ξ1

− a1R1u1

¶#

γ12 =A2A1

∂

∂ξ1

µu2A2

¶+A1A2

∂

∂ξ2

µu1A1

¶+

1

A1A2

"µ∂u1∂ξ2

− u2a1

∂a2∂ξ1

¶µ∂u1∂ξ1

+1

a2

∂a1∂ξ2

u2 +a1R1u3

¶+

µ∂u2∂ξ1

− u1a2

∂a1∂ξ2

¶µ∂u2∂ξ2

+1

a1

∂a2∂ξ1

u1 +a2R2u3

¶+

µ∂u3∂ξ1

− a1R1u1

¶µ∂u3∂ξ2

− a2R2u2

¶#(6.9.18)

6.9.4 Stress Resultants

Next, we introduce the stress resultants acting on a shell element. The tensileforce measured per unit length along a ξ2 coordinate line on a cross sectionperpendicular to a ξ1 coordinate line [see Figure 6.9.1(c)] is σ11 dS2. The totaltensile force on the differential element in the ξ1 direction can be computedby integrating over the entire thickness of the shell:

h/2Z−h/2

σ11 dS2 dζ = a2

⎡⎢⎣ h/2Z−h/2

σ11

µ1 +

ζ

R2

¶dζ

⎤⎥⎦ dξ2 ≡ N11 a2 dξ2 (6.9.19)

where h the total thickness of the shell, ζ = −h/2 and ζ = h/2 denote thebottom and top surfaces of the shell, and N11 is the membrane force per unitlength in ξ1 direction, acting on a surface perpendicular to the ξ1-coordinate(see Figure 6.9.3):

N11 =

h/2Z−h/2

σ11

µ1 +

ζ

R2

¶dζ (6.9.20a)

Analogously, the moment of the force σ11 dS2 about the ξ2-axis is

M11 =

h/2Z−h/2

ζσ11

µ1 +

ζ

R2

¶dζ (6.9.20b)

N22 = N2

ξ1 ξ2

ζ

Q2

N21 = N6N11 = N1 N12 = N6

Q1

M12 = M6

M11 = M1

M21 = M6

M22 = M2

All resultants on these sides are equal but with opposite signs to those on the parallel edges


Thus, the stress resultants per unit length can be deÞned as follows (N1 = N11,N6 = N12 = N21, N2 = N22, M1 = M11, M6 = M12 = M21, M2 = M22,σ1 = σ11, σ12 = σ6, etc.; see Figure 6.9.3):

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

N11N22N12N21M11

M22

M12

M21

⎫⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎬⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎭=

h/2Z−h/2

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

σ1³1 + ζ

R2

´σ2³1 + ζ

R1

´σ6³1 + ζ

R2

´σ6³1 + ζ

R1

´ζσ1

³1 + ζ

R2

´ζσ2

³1 + ζ

R1

´ζσ6

³1 + ζ

R2

´ζσ6

³1 + ζ

R1

´

⎫⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎬⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎭

dζ (6.9.21)

The shear forces Qi are deÞned as

½Q1Q2

¾= Ks

h/2Z−h/2

⎧⎨⎩σ5³1 + ζ

R2

´σ4³1 + ζ

R1

´⎫⎬⎭ dζ (6.9.22)

where Ks is the shear correction factor Ks.

In developing a moderately thick shell theory we make certain assumptions(as we did in the case of plates). They are outlined next.

Figure 6.9.3 Stress resultants on a shell element.


1. the transverse normal is inextensible (i.e. ε33 ≈ 0),2. normals to the reference surface of the shell before deformation remainstraight but not necessarily normal after deformation (a relaxed KirchhoffLove hypothesis),

3. the shell deßections are small and strains are inÞnitesimal, and

4. as in the case of plates, we assume that σ3 is negligible and use the planestress-reduced constitutive relations.

Consistent with the assumptions of a moderately thick shell theory, weassume the following form of the displacement Þeld:

u1(ξ1, ξ2, ζ, t) = u0(ξ1, ξ2, t) + ζφ1(ξ1, ξ2, t)

u2(ξ1, ξ2, ζ, t) = v0(ξ1, ξ2, t) + ζφ2(ξ1, ξ2, t)

u3(ξ1, ξ2, ζ, t) = w0(ξ1, ξ2, t) (6.9.23)

in which (u0, v0, w0) are the displacements of a point (ξ1, ξ2, 0) on themidsurface of the shell, and (φ1,φ2) are the rotations of a normal to thereference surface.

The Sanders [48] nonlinear straindisplacement relations associated withthe displacement Þeld in Eq. (6.9.19) are

ε = ε0+ ζε1 (6.9.24a)

where

⎧⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎩

ε0xxε0yyγ0xyγ0xzγ0yz

⎫⎪⎪⎪⎪⎪⎬⎪⎪⎪⎪⎪⎭=

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎩

∂u0∂x +

w0R1+ 1

2

³∂w0∂x − u0

R1

´2∂v0∂y +

w0R2+ 1

2(∂w0∂y − v0

R2)2

∂u0∂y +

∂v0∂x +

³∂w0∂x − u0

R1

´³∂w0∂y − v0

R2

´∂w0∂x − u0

R1+ φx

∂w0∂y − v0

R2+ φy

⎫⎪⎪⎪⎪⎪⎪⎪⎪⎬⎪⎪⎪⎪⎪⎪⎪⎪⎭,

⎧⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎩

ε1xxε1yyγ1xyγ1xzγ1yz

⎫⎪⎪⎪⎪⎪⎬⎪⎪⎪⎪⎪⎭=

⎧⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎩

∂φx∂x∂φy∂y

∂φx∂y +

∂φy∂x

00

⎫⎪⎪⎪⎪⎪⎪⎬⎪⎪⎪⎪⎪⎪⎭(6.9.24b)

and dx = a1 dξ1, dy = a2 dξ2, and dz = dζ.

In the remainder of this development, we omit the term z/R in thedeÞnition of the stress resultants and assume that aα,β = 0 (α,β = 1, 2)(i.e. constant radii of curvatures). Thus, for thin shallow shells, we have

µ1 +

ζ

R1

¶≈ 1,

µ1 +

ζ

R2

¶≈ 1 (6.9.25)

and we have N12 = N21 and M12 =M21.


The stress resultants are related to the strains, for an orthotropic shell inthe absence of thermal and other inßuences, as follows:

⎧⎨⎩N1N2N6

⎫⎬⎭ =⎡⎣A11 A12 0A12 A22 00 0 A66

⎤⎦⎧⎪⎪⎨⎪⎪⎩

∂u0∂x +

w0R1+ 1

2(∂w0∂x − u0

R1)2

∂v0∂y +

w0R2+ 1

2(∂w0∂y − v0

R1)2

∂u0∂y +

∂v0∂x +

³∂w0∂x − u0

R1

´³∂w0∂y − v0

R2

´⎫⎪⎪⎬⎪⎪⎭

(6.9.26a)⎧⎨⎩M1

M2

M6

⎫⎬⎭ =⎡⎣D11 D12 0D12 D22 00 0 D66

⎤⎦⎧⎪⎪⎨⎪⎪⎩

∂φx∂x∂φy∂y

∂φx∂y +

∂φy∂x

⎫⎪⎪⎬⎪⎪⎭ (6.9.26b)

½Q2Q1

¾= Ks

∙A44 00 A55

¸( ∂w0∂y − v0

R2+ φy

∂w0∂x − u0

R1+ φx

)(6.9.26c)

6.9.5 Equations of Motion

The equations of motion of the Sanders shell theory are

−µ∂N1∂x

+∂N6∂y

¶− Q1 +N1

R1+ I0

∂2u0∂t2

+ I1∂2φx∂t2

= 0 (6.9.27)

−µ∂N6∂x

+∂N2∂y

¶− Q2 +N2

R2+ I0

∂2v0∂t2

+ I1∂2φy∂t2

= 0 (6.9.28)

−µ∂Q1∂x

+∂Q2∂y

¶+N1R1

+N2R2

−N3(u0, v0, w0)− q + I0∂2w0∂t2

= 0 (6.9.29)

−µ∂M1

∂x+∂M6

∂y

¶+Q1 + I2

∂2φx∂t2

+ I1∂2u0∂t2

= 0 (6.9.30)

−µ∂M6

∂x+∂M2

∂y

¶+Q2 + I2

∂2φy∂t2

+ I1∂2v0∂t2

= 0 (6.9.31)

where

Ii =NXk=1

Z zk+1

zk

ρ(k)ζi dζ (6.9.32)

N1(u0, v0, w0) = N1µ∂w0∂x

− u0R1

¶+N6

µ∂w0∂y

− v0R2

¶N2(u0, v0, w0) = N6

µ∂w0∂x

− u0R1

¶+N2

µ∂w0∂y

− v0R2

¶(6.9.33)

N3(u0, v0, w0) = ∂N1∂x

+∂N2∂y

ρ being the mass density.


6.10 Finite Element Analysis of Shells

6.10.1 Weak Forms

The displacement Þnite element model of the equations governing doubly-curved shells, Eqs. (6.9.27)(6.9.31), can be derived in a manner similar tothat of plates. In fact, the Þnite element model of doubly-curved shells isidentical to that of FSDT with additional terms in the stiffness coefficients.For the sake of completeness, the main equations are presented here.

We begin with the weak forms of Eqs. (6.9.27)(6.9.31):

0 =

ZΩe

"∂δu0∂x

N1 +∂δu0∂y

N6 − δu0Q1 +N1R1

+ I0δu0∂2u0∂t2

+ I1δu0∂2φ1∂t2

#dx dy

−IΓeP1δu0 ds (6.10.1a)

0 =

ZΩe

"∂δv0∂x

N6 +∂δv0∂y

N2 − δv0Q2 +N2R2

+ I0δv0∂2v0∂t2

+ I1δv0∂2φ2∂t2

#dx dy

−IΓeP2δv0 ds (6.10.1b)

0 =

ZΩe

"∂δw0∂x

Q1 +∂δw0∂y

Q2 + δw0

µN1R1

+N2R2

¶− δw0q + I0δw0∂

2w0∂t2

+∂δw0∂x

N1 + ∂δw0∂y

N2#dx dy −

IΓeVnδw0 ds (6.10.1c)

0 =

ZΩe

Ã∂δφ1∂x

M1 +∂δφ1∂y

M6 + δφ1Q1 + I2δφ1∂2φ1∂t2

+ I1δφ1∂2u0∂t2

!dx dy

−IΓeT1δφ1 ds (6.10.1d)

0 =

ZΩe

Ã∂δφ2∂x

M6 +∂δφ2∂y

M2 + δφ2Q2 + I2δφ2∂2φ2∂t2

+ I1δφ2∂2v0∂t2

!dxdy

−IΓeT2δφ2 ds (6.10.1e)

where the stress resultants Ni, Mi and Qi are deÞned by Eqs. (6.9.21) and(6.9.22). We note from the boundary terms in Eq. (6.10.1ae) that u0, v0,w0, φ1, and φ2 are the primary variables. Therefore, we can use the C0

interpolation of the displacements. The secondary variables are

P1 ≡N1n1 +N6n2, P2 ≡ N6n1 +N2n2T2 ≡M1n1 +M6n2, T2 ≡M6n1 +M2n2

Vn ≡ (Q1 +N1)n1 + (Q2 +N2)n2 (6.10.2)


where (n1, n2) are the direction cosines of the unit normal to the surface. Notethat in the case of shells, surface displacements are coupled to the transversedisplacement even for linear analysis of isotropic shells.


Using interpolation of the form

u0(x, y, t) =mXj=1

uj(t)ψej (x,y), v0(x, y, t) =

mXj=1

vj(t)ψej (x, y) (6.10.3)

w0(x, y, t) =nXj=1

wj(t)ψej (x, y) (6.10.4)

φ1(x, y, t) =pXj=1

S1j (t)ψej (x,y), φ2(x, y, t) =

pXj=1

S2j (t)ψej (x, y) (6.10.5)

where ψej are Lagrange interpolation functions. In the present study, equalinterpolation (m = n = p) of Þve displacements, with p = 1, 2, . . . is used. Notethat the Þnite element model developed here for doubly-curved shells containsthe FSDT plate element as a special case (set 1/R1 = 0 and 1/R2 = 0).Substituting Eqs. (6.10.3)(6.10.5) for (u0, v0, w0,φ1,φ2) into the weak

forms in Eqs. (6.10.1ae), we obtain the semidiscrete Þnite element model ofthe Þrst-order shear deformation shell theory:⎛⎜⎜⎝⎡⎢⎢⎣[K11] [K12] [K13] [K14] [K15][K12]T [K22] [K23] [K24] [K25][K13]T [K23]T [K33] [K34] [K35][K14]T [K24]T [K34]T [K44] [K45][K15]T [K25]T [K35]T [K45]T [K55]

⎤⎥⎥⎦+⎡⎢⎢⎣[0] [0] [0] [0] [0][0] [0] [0] [0] [0][0] [0] [G] [0] [0][0] [0] [0] [0] [0][0] [0] [0] [0] [0]

⎤⎥⎥⎦⎞⎟⎟⎠⎧⎪⎪⎨⎪⎪⎩ueveweS1S2

⎫⎪⎪⎬⎪⎪⎭

+

⎡⎢⎢⎣I0[M ] [0] [0] I1[M ] [0][0] I0[M ] [0] [0] I1[M ][0] [0] I0[M ] [0] [0]

I1[M ] [0] [0] I2[M ] [0][0] I1[M ] [0] [0] I2[M ]

⎤⎥⎥⎦⎧⎪⎪⎨⎪⎪⎩ueveweS1S2

⎫⎪⎪⎬⎪⎪⎭ =

⎧⎪⎪⎨⎪⎪⎩F1F2F3F4F5

⎫⎪⎪⎬⎪⎪⎭ (6.10.6)

where the coefficients of the submatrices [Kαβ] are deÞned as follows:

K1αij =

ZΩe

Ã∂ψei∂x

Nα1j +

∂ψei∂yNα6j − ψei

Qα1j +Nα1j

R1

!dx dy

K2αij =

ZΩe

Ã∂ψei∂x

Nα6j +

∂ψei∂yNα2j − ψei

Qα2j +Nα2j

R2

!dx dy

K3αij =

ZΩe

"∂ψei∂x

Qα1j +∂ψei∂yQα2j + ψ

ei

ÃNα1j

R1+Nα2j

R2

!#dx dy


K4αij =

ZΩe

µ∂ψei∂x

Mα1j +

∂ψei∂yMα6j + ψ

eiQ

α1j

¶dx dy

K5αij =

ZΩe

µ∂ψei∂x

Mα6j +

∂ψei∂yMα2j + ψ

eiQ

α2j

¶dx dy (6.10.7)

for α = 1, 2, . . . , 5. The coefficients NαIj , M

αIj , and Q

αIj for α = 1, 2, . . . , 5 and

I = 1, 2, 6 are given by

N11j = A11

"∂ψej∂x

+

µu02R21

− 1

R1

∂w0∂x

¶ψej

#

N21j = A12

"∂ψej∂y

+

µv02R22

− 1

R2

∂w0∂x

¶ψej

#

N31j =

µA11R1

+A12R2

¶ψej +

1

2

ÃA11

∂w0∂x

∂ψej∂x

+A12∂w0∂y

∂ψej∂y

!

N16j = A66

"∂ψej∂y

− 1

R1

µ∂w0∂y

− 12

v0R2

¶ψej

#

N26j = A66

"∂ψej∂x

− 1

R2

µ∂w0∂x

− 12

u0R1

¶ψej

#

N36j =

A662

Ã∂w0∂x

∂ψej∂y

+∂w0∂y

∂ψej∂x

!

N12j = A12

"∂ψej∂x

+

µu02R21

− 1

R1

∂w0∂x

¶ψej

#

N22j = A22

"∂ψej∂y

+

µv02R22

− 1

R2

∂w0∂y

¶ψej

#

N32j =

µA12R1

+A22R2

¶ψej +

1

2

ÃA12

∂w0∂x

∂ψej∂x

+A22∂w0∂y

∂ψej∂y

!N46j = 0, N

56j = 0, N

41j = 0, N

51j = 0, N

42j = 0, N

52j = 0

Q11j = −KsA55R1

ψej , Q31j = KsA55

∂ψej∂x

, Q41j = KsA55ψej

Q22j = −KsA44R2

ψej , Q32j = KsA44

∂ψej∂y, Q42j = KsA55ψ

ej

Q21j = 0, Q51j = 0, Q

12j = 0, Q

52j = 0

N 11j = −

N1R1ψej , N 2

1j = −N6R2ψej , N 3

1j = N1∂ψej∂x

+N6∂ψej∂y

N 12j = −

N6R1ψej , N 2

2j = −N2R2ψej , N 3

2j = N6∂ψej∂x

+N2∂ψej∂y

N 41j = 0, N 5

1j = 0, N 42j = 0, N 5

2j = 0


M11j =M

21j =M

31j = 0, M4

1j = D11∂ψej∂x

, M51j = D12

∂ψej∂y

M12j =M

22j =M

32j = 0, M4

2j = D12∂ψej∂x

, M52j = D22

∂ψej∂y

M16j =M

26j =M

36j = 0, M4

6j = D66∂ψej∂y, M5

6j = D66∂ψej∂x

(6.10.8b)

Mij =

ZΩeψeiψ

ej dx dy (6.10.8c)

F 1i =

IΓeP1ψ

ei dx dy, F 2i =

IΓeP2ψ

ei dx dy

F 3i =

ZΩeqψei dx dy +

IΓeQnψ

ei ds

F 4i =

IΓeT1ψ

ei dx dy, F 5i =

IΓeT2ψ

ei dxdy (6.10.8d)

The tangent stiffness coefficients can be computed as in the case of plates.

6.10.3 Linear Results

Here, numerical results are presented for linear analysis of some benchmarkproblems from the literature [50]. Quadrilateral elements are used withselective integration rule to evaluate the stiffness coefficients (full integrationfor bending terms and reduced integration for bending-membrane couplingterms and transverse shear terms). Results are compared with those availablein the literature.

Example 6.10.1 (Clamped cylindrical shell)

Consider the deformation of a cylindrical shell with internal pressure [40]. The shell isclamped at its ends (see Figure 6.10.1). The geometric and material parameters used are:

R1 = 1030³1

R1≈ 0´, R2 = R = 20 in., a = 20 in., h = 1 in. (6.10.9a)

E1 = 7.5× 106 psi, E2 = 2× 106 psi, G12 = 1.25× 106 psiG13 = G23 = 0.625× 106 psi, ν12 = 0.25 (6.10.9b)

The pressure is taken to be p0 = (6.41/π) ksi. The numerical results obtained using 4 × 4mesh of four-node (linear) quadrilateral elements (4 × 4Q4) and 2 × 2 mesh of nine-node(quadratic) quadrilateral elements (2×2Q9) in an octant (u0 = φ1 = 0 at x1 = 0; v0 = φ2 = 0at x2 = 0,πR/2; and u0 = v0 = w0 = φ1 = φ2 = 0 at x1 = a/2) of the shell are presented inTable 6.10.1. The reference solutions by Rao [53] and Timoshenko and Woinowsky-Krieger[46] did not account for the transverse shear strains.

x1

ζ = x3

Rp0

a

ζ = x3

x2

p0


Figure 6.10.1 Clamped cylindrical shell with internal pressure.

Table 6.10.1 Maximum radial deßection (w0 in.) of a clamped cylindricalshell with internal pressure.

Present Solutions

Laminate 4× 4Q4 2× 2Q9 Ref. [53] Ref. [46]

0 0.3754 0.3727 0.3666 0.367

Example 6.10.2 (Doubly-curved shell panel)

Next, we consider a spherical shell panel (R1 = R2 = R) under central point load [40]. Theshell panel is simply supported at edges (see Figure 6.10.2). The geometric and materialparameters used are:

R1 = R2 = R = 96 in., a = b = 32 in., h = 0.1 in. (6.10.10a)

E1 = 25E2, E2 = 106 psi, G12 = G13 = 0.5E2, G23 = 0.2E2, ν12 = 0.25 (6.10.10b)

The point load is taken to be F0 = 100 lbs. The numerical results obtained using variousmeshes of linear and quadratic elements in a quadrant of the shell are presented in Table6.10.2. The Þnite element solution converges with reÞnement of the mesh to the seriessolution of Vlasov [52], who did not consider transverse shear strains in his analysis.

The remaining example problems of this chapter are analyzed using variousp levels [see Eq. (6.10.3)(6.10.5)]. With Þve degrees of freedom at each node,the number of degrees of freedom per element for different p values is given onthe next page.

hξ1

ξ2

ζ

x1

x2

x3

R2 R1

b a

F0


Figure 6.10.2 Simply supported spherical shell panel under central pointload.

Table 6.10.2 Maximum radial deßection (−w0 × 10 in.) of a simplysupported spherical shell panel under central point load.

Present Solutions

Material 4× 4Q4 2× 2Q9 4× 4Q9 4× 4Q9 Ref. [53] Ref. [52]Uniform Uniform Uniform Nonuniform

Isotropic 0.3506 0.3726 0.3904 0.3935 0.3866 0.3956

Orthotropic 0.9373 1.0349 − 1.2644 − −

Table: Number of degrees of freedom per element for different p values.

Element type p level DoF per element

Q4 1 20Q9 2 45Q25 4 125Q49 6 245Q81 8 405


The numerical integration rule (Gauss quadrature) used is I×J×K, whereK denotes the number of Gauss points (i.e. K×K Gauss rule) used to evaluatethe transverse shear terms (i.e. those containing A44 and A55), J denotes thenumber of Gauss points to evaluate the bending-membrane coupling terms(which are zero for the linear analysis of plates), and I denotes the numberof Gauss points used to evaluate all remaining terms in the stiffness matrix.One may use full integration for all terms, reduced integration for all terms,or selective integration where reduced integration for transverse shear andcoupling terms and full integration for all other terms in the stiffness matrix.The values of I, J and K used in the present study for different p levels andintegration rules are listed below.

Table: The Gauss quadrature rule used for various terms.

p level Full Selective Reducedintegration integration integration

1 2×2×2 2×1×1 1×1×12 3×3×3 3×2×2 2×2×24 5×5×5 5×4×4 4×4×46 7×7×7 7×6×6 6×6×68 9×9×9 9×8×8 8×8×8

Example 6.10.3 (Clamped cylindrical shell panel)

Here we consider an isotropic cylindrical shell panel with the following geometric and materialparameters and subjected to uniformly distributed transverse (normal to the surface) loadq (see Figure 6.10.3):

α =0.1 rad., R = 100 in., a = 20 in., h = 0.125 in. (6.10.11a)

E = 0.45× 106 psi, ν = 0.3, q = 0.04psi (6.10.11b)

Two sets of uniform meshes, one with 81 nodes (405 DoF) and the other with 289 nodes(1,445 DoF), are used in a quadrant of the shell with different p levels. For example, forp = 1 the mesh is 8 × 8Q4, for p = 2 the mesh is 4 × 4Q9, and for p = 8 the mesh is1× 1Q81 all meshes have a total of 81 nodes. Doubling the above meshes will have 289nodes. The vertical displacement at the center of the shell obtained with various meshesand integration rules are presented in Table 6.10.3. The results obtained with selective andreduced integrations are in close agreement with those of Palazotto and Dennis [51] andBrebbia and Connor [54].

The next example deals with the well-known benchmark problem ofScordelisLo roof [55]. A solution to this problem was Þrst discussed by Cantinand Clough [56] (who used ν = 0.3).

y, v

x, u

a

b

z, w

q

R

h

αb = 2αR

AB


Figure 6.10.3 Clamped cylindrical shell panel under uniform transverseload.

Table 6.10.3 Vertical deßection (−wA × 102 in.) at the center of theclamped cylindrical panel under uniform transverse load.

Mesh of 81 nodes Mesh of 289 nodes

p Full Selective Reduced Full Selective Reducedlevel integ. integ. integ. integ. integ. integ.

1 0.3378 1.1562 1.1577 0.7456 1.1401 1.14042 1.1721 1.1351 1.1352 1.1427 1.1349 1.13494 1.1347 1.1349 1.1349 1.1349 1.1349 1.13498 1.1349 1.1348 1.1348 1.1348 1.1349 1.1349

Palazotto and Dennis [51] reported −1.144×10−2 in. while Brebbia and Connor[54] reported a value of −1.1× 10−2 in.

Example 6.10.4 (Barrel vault)

The problem consists of a cylindrical roof with rigid supports at edges x = ±a/2 while edgesat y = ±b/2 are free. The shell is assumed to deform under its own weight (i.e. q actsvertically down, not perpendicular to the surface of the shell). The geometric and materialdata of the problem is (see Figure 6.10.4)

α = 40 (0.698 rad.), R = 300 in., a = 600 in., h = 3 in. (6.10.12)

E = 3× 106 psi, ν = 0.0, qy = q siny

R, qz = −q cos y

R, q = 0.625psi

The boundary conditions on the computational domain are

At x = 0 : u0 = φ1 = 0, At x = a/2 : v0 = w0 = φ2 = 0

At y = 0 : v0 = φ2 = 0, At y = b/2 : Free (6.10.13)

y, v

x, u

a

b

z,w

R

h

q

α

A

B

wB

C

D


Figure 6.10.4 A cylindrical shell roof under its own weight.

Two sets of uniform meshes, one with 289 nodes (1,445 DoF) and the other with1,089 nodes (5,445 DoF), are used in a quadrant of the shell with different p levels. Thedisplacement at y = ±b/2 (middle of the free edge) of the shell, obtained with various meshesand integration rules, are presented in Table 6.10.4. To avoid shear and membrane lockingone must use at least a mesh of 4 × 4Q25 (p = 4). The results obtained with selective andreduced integrations are in close agreement with those reported by Simo, Fox and Rifai [57].

Table 6.10.4 Vertical deßection (−wB in.) at the center of the free edgeof a cylindrical roof panel under its own weight.

Mesh of 289 nodes Mesh of 1,089 nodes

p Full Selective Reduced Full Selective Reducedlevel integ. integ. integ. integ. integ. integ.

1 0.9002 3.2681 3.6434 1.8387 3.5415 3.64312 3.6170 3.6393 3.6430 3.6367 3.6425 3.64284 3.6374 3.6430 3.6430 3.6399 3.6428 3.64288 3.6392 3.6429 3.6429 3.6419 3.6429 3.6429

Simo, Fox and Rifai [57] reported wref = −3.6288 in. for deep shells.

Figure 6.10.5 shows the variation of the vertical deßections w0(0, y) and u0(300, y) as afunction of y, while Figure 6.10.6 shows the convergence of the vertical displacement wB forp = 1, 2, 4, 8. Figure 6.10.5 also contains the results of Zienkiewicz [58].

φ = (y/R)(180/π)

(a)

φ

w0

(in

.)u

0 (i

n.)

(b)

φ

-4.0

-3.5

-3.0

-2.5

-2.0

-1.5

-1.0

-0.5

0.0

0.5

1.0

Ver

tica

l de

flec

tion

,

4x4Q25 - Present

4x4Q81 - Present

Zienkiewicz [58]

0 10 20 30 40Angular distance,

-0.06

-0.03

0.00

0.03

0.06

0.09

0.12

0.15

0.18

Dis

plac

emen

t,

4x4Q25 - Present

4x4Q81 - Present

Zienkiewicz [58]

0 10 20 30 40Angular distance,


Figure 6.10.5 (a) Deßection w0(0, y). (b) Displacement u0(300, y).

Example 6.10.5 (Pinched cylinder)

This is another well-known benchmark problem [44, 59, 60]. The circular cylinder with rigidend diaphragms is subjected to a point load at the center on opposite sides of the cylinder,as shown in Figure 6.10.7. The geometric and material data of the problem is

α =π

2rad., R = 300 in., a = 600 in., h = 3 in.

E = 3× 106 psi, ν12 = 0.3 (6.10.15)

wB

wre

f

0.900.910.920.930.940.950.960.970.980.991.001.01

Rel

ativ

e de

flec

tion

,

0 5 10 15 20 25 30 35Number of nodes per edge

Q4 - SI

Q9 - FI

Q25 - FI

Q81 - FI

P

a/2

h

P

x

Rigid Diaphragm

z

R

A y

RigidDiaphragm


Figure 6.10.6 Convergence of the relative vertical deßection, wB/wref .

Figure 6.10.7 Geometry of the pinched circular cylinder problem.

The boundary conditions used are:

At x = 0 : u0 = φ1 = 0, At x = a/2 : v0 = w0 = φ2 = 0

At y = 0, b/2 : v0 = φ2 = 0 (6.10.16)

Three different meshes with 81 nodes, 289 nodes and 1,089 nodes (with different p values)

are used in the octant of the cylinder. Table 6.10.6 contains radial displacement at the point

of load application. The solution of Flugge [44] is based on classical shell theory. It is clear

that the problem requires a high p level to overcome shear and membrane locking. Figure

6.10.8 shows the convergence characteristics of the problem.

0.60

0.65

0.70

0.75

0.80

0.85

0.90

0.95

1.00

1.05

Rel

ativ

e de

flec

tion

,

0 5 10 15 20 25 30 35Number of nodes per edge

Q4 - SI

Q9 - SI

Q25 - FI

Q81 - FI

wA

wre

f


Table 6.10.6 Radial displacement (−wA × 105) at node 1 of the pinchedcircular cylinder problem.

Mesh of 81 nodes Mesh of 289 nodes Mesh of 1,089 nodes

p level Full Selec. Reduc. Full Selec. Reduc. Full Selec. Reduc.

1 0.1282 1.5784 1.8453 0.2785 1.7724 1.8600 0.6017 1.8432 1.86902 0.4184 1.7247 1.8451 1.2238 1.8395 1.8596 1.6844 1.8636 1.86774 1.1814 1.8108 1.8438 1.7574 1.8510 1.8586 1.8335 1.8648 1.86678 1.7562 1.8309 1.8415 1.8325 1.8548 1.8579 1.8471 1.8653 1.8661

The analytical solution of Flugge [44] is −1.8248 × 10−5 in.; The value given by Cho andRoh [59] is wref = −1.8541× 10−5 in.

Figure 6.10.8 Convergence of the relative radial deßection, wA/wref .

6.10.4 Nonlinear Results

Here present some few examples of nonlinear bending of shells. Thethin shell approximation is used and the results presented are based on thenonlinear strains in Eqs. (6.9.26ac).

Example 6.10.6 (Clamped shallow cylindrical panel)

Here, we consider nonlinear bending of a shallow shell panel clamped on all its four sides, asshown in Figure 6.10.9. The geometric and material parameters used are:

E = 0.45× 106 psi, ν = 0.3, a = 20 in., R = 100 in., h = 0.125 in., α = 0.1 rad.Uniformly distributed load, with a load step of ∆q0 = −0.02 psi (for a total of 20 load steps)is used. The boundary conditions of the computational domain are

At ξ1 = 0 : u1 = φ1 = 0; At ξ2 = 0 : u2 = φ2 = 0 (symmetry)

ξ1

a

bR

α

A

BC

D

h

q0

Rξ2

ζ

Clamped

Clamped

Clamped

Clamped


At ξ1 =a

2and ξ2 = α : u1 = u2 = u3 = φ1 = φ2 = 0 (6.10.17)

Figure 6.10.10 shows the center deßection versus applied load for various meshes. Full

integration is used on all terms. The results are in close agreement with those of Palazotto

and Dennis [51].

Figure 6.10.9 Geometry and computational domain of the cylindricalshell panel.

Figure 6.10.10 Center deßection versus load for the clamped cylindricalshell panel.

0.0

0.1

0.2

0.3

0.4

Pres

sure

load

(psi

)

0.0 0.1 0.2 0.3 0.4Deflection at the center of the panel (in)

1x1Q25

2x2Q25

4x4Q25

8x8Q [51]

ξ1

a

bR α

A

B

C

DhRξ2

Hinged

Hinged

PFree

Free

0

200

400

600

800

1000

1200

Poin

t loa

d (lb

)

1x1Q25

2x2Q25

4x4Q25

Sabir and Lock [61]

0.0 0.2 0.4 0.6 0.8 1.0 1.2Deflection at the center of the panel (in)


Example 6.10.7 (Hinged shallow cylindrical panel)

Consider a shallow shell panel hinged on straight edges and free on curved edges, as shownin Figure 6.10.11. The geometric and material parameters used are the same as those in Eq.(6.10.17) except h = 1.0 in. Point load at the center of the panel is used with a load stepof ∆P = −100 lbs (for a total of 12 load steps). Full integration is used on all terms. Theboundary conditions of the computational domain are

At ξ1 = 0 : u1 = φ1 = 0; At ξ2 = 0 : u2 = φ2 = 0 (symmetry)

At ξ2 = α : u1 = u2 = u3 = φ1 = 0; At ξ1 = a/2 : Free(6.10.18)

Figure 6.10.12 shows the center deßection versus applied load for various meshes. The resultsare in close agreement with those of Sabir and Lock [61].

Figure 6.10.11 Geometry and computational domain of the shell panel.

Figure 6.10.12 Deßection versus load for the clamped shell panel.

0.0

1.0

2.0

3.0

4.0

5.0

6.0

7.0

8.0

9.0

10.0

Dea

d w

eigh

t (ps

i)

Linear

1x1Q25

2x2Q25

4x4Q25

0.0 4.0 8.0 12.0 16.0 20.0 24.0Vertical deflection (in)


Example 6.10.8 (The Barrel Vault problem)

Here, we consider the Barrel Vault problem of Example 6.10.4 for nonlinear analysis (seeFigure 6.10.4). The geometric and material parameters and the boundary conditions usedare the same as those in Eqs. (6.10.12) and (6.10.13). Sixteen load steps of ∆q0 = −0.625 psiare used. Figure 6.10.13 shows the center deßection versus applied load for various meshes.For the total load (10 psi) considered, the shell experiences no snap through behavior.

Example 6.10.9 (Clamped orthotropic cylinder)

The last example of this chapter is concerned with the nonlinear bending of an orthotropiccylinder clamped at its ends (see Figure 6.10.14). The following geometric and materialparameters (those of glassepoxy Þber-reinforced composite material) are used:

E1 = 7.5×106 psi, E2 = 2.0×106 psi, G12 = 1.25× 106 psi, G13 = G23 = 0.625× 106 psi

ν12 = 0.25, a = 10 in., R = 20 in., h = 1.0 in., α =π

8rad. (6.10.19)

The boundary conditions of the computational domain are

At ξ1 = 0 : u1 = φ1 = 0; At ξ2 = 0,α : u2 = φ2 = 0 (symmetry)

At ξ1 =a

2: u1 = u2 = u3 = φ1 = φ2 = 0 (6.10.20)

Twenty load steps of ∆q0 = 500 psi are used. Figure 6.10.15 shows the center deßection

versus applied load for various meshes. The results are in close agreement with those of

Kreja, Schmidt, and Reddy [60].

Figure 6.10.13 Center deßection versus load for a cylindrical shell panelunder its own weight.

a/2

hR A

B

C

Dξ1

ζ

Rξ2

Clamped

q

0.0

1.0

2.0

3.0

4.0

5.0

6.0

7.0

8.0

9.0

10.0

Inte

rnal

pre

ssur

e (k

si)

2x2Q25

4x4Q25

1x1Q9 Kreja et al. [60]

4x1Q9 Kreja et al. [60]

0.0 0.2 0.4 0.6 0.8 1.0 1.2 1.4Central radial deflection (in)


Figure 6.10.14 Geometry and computational domain of a clamped circularcylinder.

Figure 6.10.15 Center deßection versus load for the clamped orthotropiccylinder.


Problems

6.1 Develop the weak forms of the following linear equations governing the classical platetheory:

−µ∂2Mxx

∂x2+ 2

∂2Mxy

∂x∂y+∂2Myy

∂y2

¶= q (a)

−∂2w0∂x2

−¡D22Mxx − D12Myy

¢= 0 (b)

−∂2w0∂y2


¢= 0 (c)

−2∂2w0∂x∂y

− (D66)−1Mxy = 0 (d)

where

Dij =Dij

D0, D0 =D11D22 −D212 (e)

Dij being the bending stiffnesses of an orthotropic plate

Dij = Qijh3

12, (i = 1,2,6), (f)

and Qij are the elastic stiffnesses deÞned in terms of the principal moduli (E1,E2),shear modulus G12, and Poissons ratio ν12 as

Q11 =E1

1− ν12ν21, Q12 =

ν12E21− ν12ν21

, Q22 =E2

1− ν12ν21,

Q66 =1

S66= G12, ν21 = ν12

E2E1

(g)

6.2 Develop the (mixed) Þnite element model associated with the equations in Problem6.1. Assume approximation of the form

w0 =

rXi=1

wiψ(1)i , Mxx =

sXi=1

Mxiψ(2)i

Myy =

pXi=1

Myiψ(3)i , Mxy =

qXi=1

Mxyiψ(4)i

where ψ(α)i , (α = 1,2,3,4) are appropriate interpolation functions. Discuss theminimum requirements of the interpolation functions.

6.3 Assume

ψ1i = ψ2i = ψ

3i = ψ

4i = bilinear functions of a rectangular element (a)

and develop the Þnite element program of the mixed model of Problems 1 and 2.The resulting stiffness matrix is of the order 16 by 16. Analytical computation of thecoefficient matrices for a rectangular element is simple and straightforward.


6.4 Develop the weak forms of the following nonlinear equations governing the classicalplate theory:

−³∂Nxx

∂x+∂Nxy

∂y

´= 0 (a)

−³∂Nxy∂x

+∂Nyy

∂y

´= 0 (b)

−µ∂2Mxx

∂x2+ 2

∂2Mxy

∂x∂y+∂2Myy

∂y2

¶−N (u0, v0,w0) = q (c)

−∂2w0∂x2


¢= 0 (d)

−∂2w0∂y2


¢= 0 (e)

−2∂2w0∂x∂y

− (D66)−1Mxy = 0 (f)

where (Nxx,Nyy,Nxy,N ) are known in terms of (u0, v0, w0) and their derivatives byEqs. (6.3.32) and (6.3.14a). Note that the dependent unknowns are u0, v0, w0, Mxx,Myy, and Mxy.

6.5 Develop the (mixed) nonlinear Þnite element model of the equations in Problem 6.4.Assume interpolation of the form

u0 =

rXi=1

uiψ(1)i , v0 =

rXi=1

viψ(1)i , w0 =

rXi=1

wiψ(2)i

Mxx =

sXi=1

Mxiψ(3)i , Myy =

pXi=1

Myiψ(4)i , Mxy =

qXi=1

Mxyiψ(5)i

6.6 Implement the mixed Þnite element model of Problem 6.5 and validate your programwith the example problems presented in Section 6.5.

6.7 Develop the weak forms of the following nonlinear equations governing the classicalplate theory:

−³∂Nxx

∂x+∂Nxy

∂y

´= 0, −

³∂Nxy

∂x+∂Nyy

∂y

´= 0

−∂2Mxx

∂x2+ 4D66

∂4w0∂x2∂y2

− ∂2Myy

∂y2− ∂

∂x

³Nxx

∂w0∂x

+Nxy∂w0∂y

´− ∂

∂y

³Nxy

∂w0∂x

+Nyy∂w0∂y

´= q

−∂2w0∂x2

−¡D22Mxx + D12Myy

¢= 0, −∂

2w0∂y2

−¡D12Mxx + D11Myy

¢= 0

where (Nxx,Nyy,Nxy) are known in terms of (u0, v0,w0) and their derivatives by Eqs.(6.3.32). Note that the dependent unknowns are u0, v0, w0, Mxx, and Myy.

6.8 Develop the (mixed) nonlinear Þnite element model of the equations in Problem 6.7by assuming approximations of the form

u0 =

rXi=1

uiψ(1)i , v0 =

rXi=1

viψ(1)i , w0 =

rXi=1

wiψ(2)i


Mxx =

sXi=1

Mxiψ(3)i , Myy =

pXi=1

Myiψ(4)i

6.9 Implement the nonlinear Þnite element model of Problem 6.8 into a computer programand validate it with the known results of this chapter.

6.10 Develop the weak forms of the following nonlinear equations governing the Þrst-ordershear deformation plate theory:

−³∂Nxx

∂x+∂Nxy

∂y

´= 0, −

³∂Nxy

∂x+∂Nyy

∂y

´= 0

−A55∂

∂x

³∂w0∂x

+ φx

´−A44

∂

∂y

³∂w0∂y

+ φy

´− ∂

∂x

³Nxx

∂w0∂x

+Nxy∂w0∂y

´− ∂

∂y

³Nxy

∂w0∂x

+Nyy∂w0∂y

´= q

−∂Mxx

∂x−D66

∂

∂y

³∂φx∂y

+∂φy∂x

´+KsA55

³∂w0∂x

+ φx

´= 0

−D66∂

∂x

³∂φx∂y

+∂φy∂x

´− ∂Myy

∂y+KsA44

³∂w0∂y

+ φy

´= 0

where (Nxx,Nyy,Nxy) are known in terms of (u0, v0,w0) and their derivatives by Eqs.(6.3.32). Note that the dependent unknowns are (u0, v0,w0,φx,φy,Mxx,Myy).

6.11 Develop the (mixed) nonlinear Þnite element model of the equations in Problem 6.10by assuming approximations of the form

u0 =

rXi=1

uiψ(1)i , v0 =

rXi=1

viψ(1)i , w0 =

rXi=1

wiψ(2)i

φx =

rXi=1

S1i ψ(3)i , φy =

rXi=1

S2i ψ(3)i

Mxx =

sXi=1

Mxiψ(4)i , Myy =

pXi=1

Myiψ(5)i

6.12 Determine the loaddeßection behavior of a rectangular plate with two opposite edgessimply supported (SS1) and the other two edges free, and subjected to uniformlydistributed transverse load. Use a = 10 in., h = 1 in., E1 = 7.8×106 psi, E2 = 2.6×106psi, ν12=0.25, G12 = G13 = G23 = 1.3 × 106 psi, and 8 × 8 uniform mesh ofnonconforming CPT elements in a quadrant of the plate. Use load steps such thatthe load parameter P ≡ q0a4/E2h4 is equal to 5, and use 12 load steps. Plot (a) loadversus center deßection (w0/h versus P ), and (b) load versus maximum stress (σxxand σxy). Use a convergence tolerance of ε = 10−2 and a reasonable value of ITMAX.

6.13 Repeat Problem 6.12 for a rectangular plate with two opposite edges simply supported(SS1) and the other two edges clamped.

6.14 Repeat Problem 6.12 for a rectangular plate with two opposite edges clamped and theother two edges free.

6.15 Repeat Problem 6.12 using 4× 4Q9 mesh of the FSDT elements.6.16 Repeat Problem 6.13 using 4× 4Q9 mesh of the FSDT elements.6.17 Repeat Problem 6.14 using 4× 4Q9 mesh of the FSDT elements.6.18 Analyze the circular plate problem of Example 6.8.6 when the edge is simply supported

(SS3). Use all other parameters as in Example 6.8.6.


6.19 Verify the strain—displacement relations in Eq. (6.9.18).

6.20 Verify the strain—displacement relations associated with the displacement field in Eq.(6.9.19) are given by (6.9.20a,b).

6.21 Derive the equations of motion in Eqs. (6.9.27)—(6.9.31).

6.22 Derive the tangent stiffness coefficients associated with the finite element model inEqs. (6.10.6)—(6.10.8).

6.23 Analyze the circular cylinder problem of Example 6.10.12 by assuming the fiberdirection is ξ2 (as opposed to ξ1) and compare the results with those in Figure 6.10.15.

6.24 Show that the second derivatives of a function f(ξ,η) with respect to the globalcoordinates (x, y) are related to its derivatives with respect to the local coordinates(ξ, η) by [see Eqs. (5.5.4)—(5.5.8)]

⎧⎪⎨⎪⎩∂2f∂x2

∂2f∂y2

∂2f∂x∂y

⎫⎪⎬⎪⎭ =

⎡⎢⎣¡∂x∂ξ

¢2 ¡∂y∂ξ

¢22∂x∂ξ

∂y∂ξ¡

∂x∂η

¢2 ¡∂y∂η

¢22∂x∂η

∂y∂η

∂x∂ξ

∂x∂η

∂y∂ξ

∂y∂η

∂x∂η

∂y∂ξ+ ∂y

∂η∂x∂ξ

⎤⎥⎦−1⎛⎜⎝

⎧⎪⎨⎪⎩∂2f∂ξ2

∂2f∂η2

∂2f∂ξ∂η

⎫⎪⎬⎪⎭−

⎡⎢⎣∂2x∂ξ2

∂2y∂ξ2

∂2x∂η2

∂2y∂η2

∂2x∂ξ ∂η

∂2y∂ξ∂η

⎤⎥⎦½ ∂f∂x∂f∂y

¾⎞⎟⎠

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7

Flows of ViscousIncompressible Fluids

7.1 Introduction

Fluid mechanics is one of the oldest branches of physics, and is concerned withthe motion of gases and liquids and their interaction with the surroundings.For example, the ßight of birds in the air and the motion of Þsh in the watercan be understood by the principles of ßuid mechanics. Such understandinghelps us design airplanes and ships. The formation of tornadoes, hurricanes,and thunderstorms can also be explained with the help of the equations ofßuid mechanics.A ßuid state of matter is characterized by the relative mobility of its

molecules. Very strong intermolecular attractive forces exist in solids whichare responsible for the property of relative rigidity (or stiffness) in solids. Theintermolecular forces are weaker in liquids and extremely small in gases. Thestress in a solid body is proportional to the strain (i.e. deformation per unitlength), while the stress in a ßuid is proportional to the time rate of strain (i.e.rate of deformation per unit length). The proportionality parameter in thecase of ßuids is known as the viscosity. It is a measure of the intermolecularforces exerted as layers of ßuid attempt to slide past one another. The viscosityof a ßuid, in general, is a function of the thermodynamic state of the ßuid andin some cases the strain rate.Fluid mechanics is a very broad area and is traditionally divided into

smaller topical areas based on characteristics of the ßuid properties or thebasic nature of the ßow. An inviscid ßuid is one where the viscosity isassumed to be zero. An incompressible ßuid is one with constant densityor an incompressible ßow is one in which density variations (compared to areference density) are negligible. An inviscid and incompressible ßuid is termedan ideal or a perfect ßuid. A real ßuid is one with Þnite viscosity, and it mayor may not be incompressible. When the viscosity of a ßuid depends only onthermodynamic properties, and the stress is linearly related to the strain rate,the ßuid is said to be Newtonian. A non-Newtonian ßuid is one which does notobey the Newtonian (i.e. linear) stress-strain rate relation. A non-Newtonianconstitutive relation can be algebraic, differential, or integral type.


The ßow of viscous ßuids can also be classiÞed into two major types: asmooth, orderly motion is called laminar ßow, and a random, ßuctuatingmotion is called turbulent ßow. Viscous ßows can be characterized by a non-dimensional parameter known as the Reynolds number, Re = ρUL/µ, whichis deÞned as the ratio of inertial forces ρU2 to viscous forces µU/L. Here ρdenotes the density of the ßuid, µ the ßuid viscosity, U the characteristic ßowvelocity, and L is a characteristic dimension of the ßow region. High viscosityßuids and/or small velocities produce relatively small Reynolds numbers anda laminar ßow. The case of Re << 1 corresponds to the ßow (called Stokesßow) in which the inertial effects are small compared to the viscous effects andtherefore neglected. The ßow of less viscous ßuids and/or higher velocities leadto higher Reynolds numbers and a turbulent ßow. High Reynolds number ßowscontain regions of both laminar and turbulent ßows.The motion of a ßuid is governed by the global laws of conservation of

mass, momenta, and energy. These equations consist of a set of nonlinearpartial differential equations in terms of the velocity components, temperature,and pressure. The equations of motion resulting from the application of theconservation of linear momentum principle are known as the NavierStokesequations. When temperature effects are not important, the energy equation isuncoupled from the NavierStokes equations. Thus for isothermal ßows, onlythe NavierStokes equations and continuity equation are solved. One may non-dimensionalize the variables in the NavierStokes equations to express themin terms of the Reynolds number. We shall work with physical quantities inthis study.In the present chapter we review the governing equations of ßows of

incompressible ßuids, develop Þnite element models based on alternativeformulations, and discuss computer implementation of the Þnite elementmodels developed herein. Simple examples of applications of the Þnite elementmodels are also included.

7.2 Governing Equations

7.2.1 Introduction

There are two alternative descriptions used to express the conservation lawsin analytical form. In the Þrst, one considers the motion of all matterpassing through a Þxed spatial location. Here one is interested in variousproperties (e.g. velocity, pressure, temperature, density, and so on) of thematter that instantly occupies the Þxed spatial location. This descriptionis called the Eulerian description or spatial description. In the second, onefocuses attention on a set of Þxed material particles, irrespective of their spatiallocations. The relative displacements of these particles and the stress causedby external forces and temperature are of interest in this case. This descriptionis known as the Lagrangian description or material description. The Eulerian

FLOWS OF VISCOUS INCOMPRESSIBLE FLUIDS 231

description is most commonly used to study ßuid ßows and coupled ßuidßow and heat transfer, while the Lagrangian description is generally used tostudy stress and deformation of solid bodies. Here we present the governingequations of a continuous medium based on the Eulerian description. For aderivation of the equations, the reader may consult the books on continuummechanics (e.g. Bird, Stewart, Lightfoot [1], Malvern [2], and Reddy andRasmussen [3]), heat transfer (e.g. see Bejan [4], Holman [5], and Ozisik[6,7]), and ßuid mechanics (e.g. Batchelor [8] and Schlichting [9]).

7.2.2 Conservation of Mass

The principle of conservation of mass can be stated as the time rate of changeof mass in a Þxed volume is equal to the rate of inßow of mass through thesurface. Application of the principle to an element of the region (called controlvolume) results in the following equation, known as the continuity equation:

∂ρ

∂t+∇ · (ρv) = 0 (7.2.1)

where ρ is the density (kg/m3) of the medium, v the velocity vector (m/s),and ∇ is the vector differential operator. Introducing the material derivativeor Eulerian derivative operator D/Dt

D

Dt=∂

∂t+ v ·∇ (7.2.2)

Eq. (7.2.1) can be expressed in the alternative form

Dρ

Dt+ ρ∇ · v = 0 (7.2.3)

For steady-state conditions, the continuity equation becomes

∇ · (ρv) = 0 (7.2.4)

When the density changes following a ßuid particle are negligible, thecontinuum is termed incompressible and we have Dρ/Dt = 0. The continuityequation (7.2.3) then becomes

∇ · v = 0 (7.2.5)

which is often referred to as the incompressibility condition.The incompressibility condition (7.2.5) expresses the fact that the volume

change is zero for an incompressible ßuid during its deformation. Since thevelocity Þeld u is required to satisfy the equations of motion derived in the nextsection, Eq. (7.2.5) is known as a constraint among the velocity components.


7.2.3 Conservation of Momenta

The principle of conservation of linear momentum (or Newtons Second Lawof motion) states that the time rate of change of linear momentum of a givenset of particles is equal to the vector sum of all external forces acting onthe particles of the set, provided Newtons Third Law of action and reactiongoverns the internal forces. Newtons Second Law can be written as

ρDv

Dt= ∇ · σ + ρf (7.2.6)

where σ is the Cauchy stress tensor (N/m2) and f is the body force vector,measured per unit mass. Equation (7.2.6) is known as the NavierStokesequations.The principle of conservation of angular momentum can be stated as the

time rate of change of the total moment of momentum of a given set of particlesis equal to the vector sum of the moments of the external forces acting onthe system. In the absence of distributed couples, the principle leads to thesymmetry of the stress tensor:

σ = (σ)T (7.2.7)

where the superscript T denotes the transpose of the enclosed quantity.

7.2.4 Conservation of Energy

The principle of conservation of energy (or the First Law of Thermodynamics)states that the time rate of change of the total energy is equal to the sum ofthe rate of work done by applied forces and the change of heat content perunit time. For an incompressible ßuid, the First Law of Thermodynamics canbe expressed as

ρcvDT

Dt= −∇ · q+Q+Φ (7.2.8)

where T denotes the temperature ( C), q the heat ßux vector (W/m2), Q isthe internal heat generation (W/m3), Φ is the viscous dissipation function

Φ = τ : D (7.2.9)

and cv is the speciÞc heat [J/(kg ·C)] at constant volume. In Eq. (7.2.9), τdenotes the viscous part of the stress tensor σ and D the strain rate tensor,as discussed below. Other types of internal heat generation may arise fromother physical processes such as chemical reactions and Joule heating.


7.2.5 Constitutive Equations

For most of this study we assume the ßuid to be Newtonian (i.e. theconstitutive relations are linear). Non-Newtonian ßuids will be consideredin Chapter 10. Further, the ßuids are assumed to be incompressible, and theßow is laminar. For ßows involving buoyancy forces, an extended form of theBoussinesq approximation (see Bejan [4]) may be invoked, which allows thedensity ρ to vary with temperature T according to the relation

ρ = ρ0[1− β(T − T0)] (7.2.10)

where β is the coefficient of thermal expansion (m/m/C) and the subscriptzero indicates a reference condition. The variation of density as given in Eq.(7.2.10) is permitted only in the description of the body force; the density inall other situations is assumed to be that of the reference state, ρ0.For viscous incompressible ßuids the total stress σ can be decomposed into

hydrostatic and viscous parts:

σ = τ + (−P )I (7.2.11)

where P is the hydrostatic pressure, I the unit tensor, and τ is the viscousstress tensor. For Newtonian ßuids, the viscous stress tensor is related to thestrain rate tensor D by

τ = C : D (7.2.12)

where C is the fourth-order tensor of ßuid properties. The strain rate tensorD is deÞned by

D =1

2

h(∇v) + (∇v)T

i(7.2.13)

For an isotropic ßuid (i.e. whose material properties are independentof direction), the fourth-order tensor C can be expressed in terms of twoconstants, λ and µ, called the Lame constants, and Eq. (7.2.12) takes theform

τ = λ(tr[D])I+ 2µD (7.2.14)

where (tr[D]) denotes the trace (or sum of the diagonal elements) of the matrix[D], which consists of the elements of the second-order tensor D. For anincompressible ßuid, we have tr[D] = 0 and Eq. (7.2.14) becomes

τ = 2µD (7.2.15)

The Fourier heat conduction law states that the heat ßux is proportional tothe gradient of temperature

q = −k ·∇T (7.2.16)


where k denotes the conductivity tensor of order two. The negative sign inEq. (7.2.16) indicates that heat ßows from high-temperature regions to low-temperature regions. For an isotropic medium, k is of the form

k = kI (7.2.17)

where k denotes the thermal conductivity [W/(m·C)] of the medium.The material coefficients, µ, cv,β, and k are generally functions of position

and temperature. Conductivity tensor is a symmetric, second-order tensor(i.e. kT = k). The volumetric heat source for the ßuid and/or solid maybe a function of temperature, time, and spatial location. In developing theÞnite element models, the dependence of the material properties on the spatiallocation is assumed. The dependence of the viscosity and conductivity ontemperature and strain rates is discussed in [10].

7.2.6 Boundary Conditions

The boundary conditions for the ßow problem (i.e. momentum equations) aregiven by [10]

v = v on Γv (7.2.18a)

t ≡ n · σ = t on Γσ (7.2.18b)

where n is the unit normal to the boundary and Γv and Γσ are the boundaryportions on which the velocity and tractions are speciÞed, respectively (seeFigure 7.2.1).The boundary conditions for the heat transfer problem (i.e. energy

equation) are

T = T on ΓT (7.2.19a)

qn ≡ n · q = qn on Γq (7.2.19b)

where ΓT and Γq are the boundary portions on which the temperature andheat ßux are speciÞed, respectively. A more general boundary condition forthe heat transfer problem is given by

qcond + qconv + qrad = qn on Γq (7.2.20)

where qcond, qconv, and qrad are the conductive, convective, and radiative partsof heat ßux, respectively,

qcond = −k ·∇T, qconv = hc(T − Tc), qrad = hr(T − Tr) (7.2.21)

Here hc and hr denote the heat transfer coefficients associated with convectiveand radiative heat transfer, respectively. In general, they are functions ofposition, time, and temperature.

Ω

ΓT

Γq

(b)

Ω

Γv

Γσ

(a)


Figure 7.2.1 A schematic of various boundary portions. (a) For ßuid ßow.(b) For heat transfer.

The various portions of the boundary Γ must satisfy the requirements

Γv ∪ Γσ = Γ, Γv ∩ Γσ = 0 (7.2.22a)

ΓT ∪ Γq = Γ, ΓT ∩ Γq = 0 (7.2.22b)

where 0 denotes the empty set. Equations (7.2.22a,b) imply that the totalboundary is composed of two disjoint portions, Γv and Γσ, or ΓT and Γq, whoseunion is the total boundary. Of course, in the analysis of practical problemsone may encounter situations where, for example, both velocity and tractionare known at a point. This is a mathematical singularity, and one must pickone of the two conditions but not both. Often the primary variables, i.e. vand T are picked over the secondary variables t and qn.

7.3 Governing Equations in Terms ofPrimitive Variables

7.3.1 Vector Form

Equations (7.2.3), (7.2.6), and (7.2.8) can be expressed in terms of theprimitive variables (v, P, T ) by means of equations (7.2.12)(7.2.16). Theresults are summarized below for isotropic, Newtonian, viscous, incompressibleßuids in the presence of buoyancy forces:

∇ · v = 0 (7.3.1)

ρ0

µ∂v

∂t+ v ·∇v

¶= −∇P + µ∇ ·

h(∇v) + (∇v)T

i+ ρ0f + ρ0gβ(T − T0) (7.3.2)

ρ0cv

µ∂T

∂t+ v ·∇T

¶= ∇ · (k∇T ) +Q+ Φ (7.3.3)


where v represents the velocity vector, ρ0 the density, g the gravity forcevector per unit mass, T the temperature, cv the speciÞc heat of the ßuid atconstant volume, and Q the rate of heat generation.The above equations are valid for the ßuid region Ωf . In the solid region

Ωs, the ßuid velocity is zero, v = 0, and the only relevant equation is (7.3.3).The energy equation (7.3.3) for the solid region is given by

ρsCs∂T

∂t= ∇ · (ks∇T ) +Qs (7.3.4)

In writing Eq. (7.3.4) it is assumed that the solid is stationary with respectto the coordinate frame such that the convective transport of energy [i.e. thenonlinear part of Eq. (7.3.3)] need not be considered. The Þnite elementmodel of Eq. (7.3.4) was discussed in Chapter 4 when the conductivity ks isa function, in general, of x, y, and T .

7.3.2 Cartesian Component Form

The vector form of the equations in (7.3.1)(7.3.3) allows us to expressthem in any coordinate system by expressing the vector operator ∇ and allother vectors in that coordinate system. In the Cartesian coordinate system(x1, x2, x3), the kinematic and constitutive relations take the form

Dij =1

2

Ã∂vi∂xj

+∂vj∂xi

!(7.3.5)

σij = −P δij + τij ; τij = 2µDij (7.3.6)

The conservation equations can be expressed as

∂vi∂xi

= 0 (7.3.7)

ρ0

Ã∂vi∂t+ vj

∂vi∂xj

!=

∂

∂xj

"−P δij + µ

Ã∂vi∂xj

+∂vj∂xi

!#+ ρ0fi − ρ0giβ(T − T0) (7.3.8)

ρ0C

Ã∂T

∂t+ vj

∂T

∂xj

!=

∂

∂xi

µk∂T

∂xi

¶+Q+ 2µDijDij (7.3.9)

for the ßuid region Ωf and

ρsCs∂T

∂t=

∂

∂xi

µks∂T

∂xi

¶+Qs (7.3.10)

for the solid region Ωs. Equations (7.3.5)(7.3.10) are written for a Cartesiangeometry in an Eulerian reference frame, with the indices i, j = 1, 2, 3 (or


i, j = 1, 2 for two-dimensional problems); the Einstein summation conventionon repeated indices is used (see Reddy and Rasmussen [3], pp. 1820).It is possible to express the equations in terms of the stream function and

vorticity functions. For example, in the two-dimensional case, we could writeequations (7.3.7) and (7.3.8) in terms of the stream function ψ and vorticityω which are deÞned in a two-dimensional case by the relations:

v2 ≡ − ∂ψ∂x1

, v1 ≡ ∂ψ

∂x2, ω ≡ ∂v1

∂x2− ∂v2∂x1

= ∇2ψ (7.3.11)

Here we chose to use the standard notation for the stream function andvorticity. The symbol ψ should not be confused with ψi used for interpolationfunctions. Of course, it would be clear in the context of the discussion.In the present study, we shall consider two different Þnite element models

of equations (7.3.7) and (7.3.8). The Þrst one is a natural formulation inwhich the weak forms of equations (7.3.7) and (7.3.8) are used to constructthe Þnite element model. The resulting Þnite element model is termed thevelocitypressure model or mixed model. The phrase mixed is used becausevelocity variables are mixed with the force-like variable, pressure, and bothtypes of variables are retained in a single formulation. The second model isbased on the interpretation that the continuity equation (7.3.7) is an additionalrelation among the velocity components (i.e. a constraint among the vi), andthe constraint is satisÞed in a least-squares (i.e. approximate) sense. Thisparticular method of including the constraint in the formulation is knownas the penalty function method, and the model is termed the penaltyÞniteelement model. In this case, the pressure variable is effectively eliminatedfrom the formulation. It is informative to note that the velocitypressure(or mixed) formulation is the same as the Lagrange multiplier formulation,wherein the constraint is included by means of the Lagrange multiplier method.The Lagrange multiplier turns out to be the negative of the pressure.There exist in computational ßuid dynamics literature hundreds of papers

on Þnite element models of incompressible ßows and their applications.Interested readers may consult the books by Reddy and Gartling [10] andGresho and Sani [11] for many of these references. Some of these works willbe cited at appropriate places of this book.

7.4 VelocityPressure Finite Element Model

7.4.1 Weak Form

The starting point for the development of the Þnite element models of Eqs.(7.3.7) and (7.3.8) is their weak forms. Here we consider the two-dimensionalcase, and the three-dimensional case follows in a straightforward manner.First, we write Eqs. (7.3.7) and (7.3.8) for the two-dimensional case using


the notation v1 = vx, v2 = vy, x1 = x, and x2 = y. We have

∂vx∂x

+∂vy∂y

= 0 (7.4.1)

ρ0

µvx∂vx∂x

+ vy∂vx∂y

¶− ∂

∂x

µ2µ∂vx∂x

¶− ∂

∂y

∙µ

µ∂vx∂y

+∂vy∂x

¶¸+∂P

∂x= ρ0fx

(7.4.2)

ρ0

µvx∂vy∂x

+ vy∂vy∂y

¶− ∂

∂y

µ2µ∂vy∂y

¶− ∂

∂x

∙µ

µ∂vx∂y

+∂vy∂x

¶¸+∂P

∂y= ρ0fy

(7.4.3)

where (fx, fy) are the components of the body force vector.The weighted-integral statements of these equations over a typical element

Ωe are given by

ZΩeQ

µ∂vx∂x

+∂vy∂y

¶dx dy = 0 (7.4.4)Z

Ωewx

½ρ0

µvx∂vx∂x

+ vy∂vx∂y

¶− ∂

∂x

µ2µ∂vx∂x

¶− ∂

∂y

∙µ

µ∂vx∂y

+∂vy∂x

¶¸+∂P

∂x− ρ0fx

¾dx dy = 0 (7.4.5)Z

Ωewy

½ρ0

µvx∂vy∂x

+ vy∂vy∂y

¶− ∂

∂y

µ2µ∂vy∂y

¶− ∂

∂x

∙µ

µ∂vx∂y

+∂vy∂x

¶¸+∂P

∂y− ρ0fy

¾dx dy = 0 (7.4.6)

where (Q,wx, wy) are weight functions, which will be equated, in the RitzGalerkin Þnite element models, to the interpolation functions used for(P, vx, vy), respectively (see Reddy and Gartling [10] for details).Equation (7.4.4) will remain unaffected, as integration-by-parts does not

help reduce differentiability on (vx, vy). In addition, the boundary termsobtained will be in conßict with the physical boundary conditions. The samecomment applies to the Þrst expression in Eqs. (7.4.5) and (7.4.6). Note thattrading of differentiability between the weight functions and problem variablesis subjected to the restriction that the resulting boundary expressions arephysically meaningful. Otherwise, the secondary variables of the formulationmay not be the quantities the physical problem admits as the boundaryconditions. An examination of the boundary stress components tα in equation(7.2.18b) shows that the pressure term occurs as a part of the expression [alsosee Eq. (7.3.6)]. Therefore, the pressure term must also be integrated-by-partsto keep the boundary stresses in tact. The integration-by-parts also allows thepressure variable to have lower-order approximation.


The Þnal expressions for the weak forms are given byZΩeQ

µ∂vx∂x

+∂vy∂y

¶dx dy = 0 (7.4.7)Z

Ωe

∙wxρ0

µvx∂vx∂x

+ vy∂vx∂y

¶+ 2µ

∂wx∂x

∂vx∂x

+∂wx∂x

P

+ µ∂wx∂y

µ∂vx∂y

+∂vy∂x

¶− ρ0wxfx

¸dx dy −

IΓewxtx ds = 0 (7.4.8)Z

Ωe

∙wyρ0

µvx∂vy∂x

+ vy∂vy∂y

¶+ 2µ

∂wy∂y

∂vy∂y

+∂wy∂y

P

+ µ∂wy∂x

µ∂vx∂y

+∂vy∂x

¶− ρ0wyfy

¸dx dy −

IΓewyty ds = 0 (7.4.9)

where (tx, ty) are the boundary stress (or traction) components

tx = (2µ∂vx∂x

− P )nx + µ(∂vx∂y

+∂vy∂x)ny

ty = µ(∂vx∂y

+∂vy∂x)nx + (2µ

∂vy∂y

− P )ny(7.4.10)

and (nx, ny) are the direction cosines of the unit normal vector n on theboundary Γe.


Since we are developing the RitzGalerkin Þnite element models, the choiceof the weight functions is restricted to the spaces of approximation functionsused for the pressure and velocity Þelds. Suppose that the dependent variables(vx, vy, P ) are approximated by expansions of the form

vx(x, y, t) =MXm=1

ψm(x, y)vmx (t), vy(x, y, t) =

MXm=1

ψm(x, y)vmy (t) (7.4.11a)

P (x, y, t) =NXn=1

φn(x, y)Pn(t) (7.4.11b)

where ψ and φ are interpolation (or shape) functions, and (vmx , vmy , Pn) are

nodal values of (vx, vy, P ). The weight functions (wx, wy,Q) have the followingcorrespondence (see Reddy [12,13] for further details):

Q ≈ P, wx ≈ vx, wy ≈ vy (7.4.12)

Substitution of Eqs. (7.4.11a,b) into Eqs. (7.4.7)(7.4.9) results in thefollowing Þnite element equations [10]:⎡⎣ [M ] [0] [0]

[0] [M ] [0][0] [0] [0]

⎤⎦⎧⎨⎩ úvx úvy úP

⎫⎬⎭+⎡⎣ [C(v)] [0] [0]

[0] [C(v)] [0][0] [0] [0]

⎤⎦⎧⎨⎩vxvyP

⎫⎬⎭


+

⎡⎣ 2[Sxx] + [Syy] [Syx] −[Sx0][Sxy] [Sxx] + 2[Syy] −[Sy0]−[Sx0]T −[Sy0]T [0]

⎤⎦⎧⎨⎩vxvyP

⎫⎬⎭ =⎧⎨⎩F 1F 20

⎫⎬⎭ (7.4.13)

The coefficient matrices shown in Eqs. (7.4.13) are defined by

Mij =

ZΩe

ρ0ψeiψ

ej dx dy

Ceij(v) =

ZΩe

ρ0ψei

Ãvx∂ψej∂x

+ vy∂ψej∂y

!dx dy

Sξηij =

ZΩeµ∂ψei∂ξ

∂ψej∂η

dx dy ; ξ, η = x, y

Sξ0ij =

ZΩe

∂ψei∂ξ

φej dx dy ; ξ = x, y

F 1 =

ZΩe

ρ0ψei fx dx dy +

IΓeψei tx ds

F 2 =

ZΩe

ρ0ψei fy dx dy +

IΓeψei ty ds

(7.4.14)

The two sets of interpolation functions used in Eqs. (7.4.11a,b) should beof the Lagrange type, that is, derived by interpolating only the values of thefunctions — and not their derivatives. There are two different finite elementsassociated with the two sets of field variables (vx, vy) and P , and hence thereare two different finite element meshes corresponding to the two variables overthe same domain, Ω. If one of the meshes contains the other mesh as a subset,then we choose to display the first mesh and indicate the nodal degrees offreedom associated with the nodes of a typical element of the mesh.The interpolation used for the pressure variable should be different from

that used for the velocities, because the weak forms in Eqs. (7.4.7)—(7.4.9)contain only the first derivatives of the velocities vx and vy and no derivativesof the pressure P . In addition, the essential boundary conditions of theformulation do not include specification of the pressure; it enters the boundaryconditions as a part of the natural boundary conditions. This implies that thepressure variable need not be carried as a variable that is continuous acrossinterelement boundaries. These observations lead to the conclusion that thepressure variable should be interpolated with functions that are one orderless than those used for the velocity field and that the approximation may bediscontinuous (i.e. not continuous from one element to other). Thus, quadraticinterpolation of vi and discontinuous linear interpolation of P are admissible.Models that use equal interpolation of the velocities and pressure with thisformulation are known to give inaccurate results (see [14—26]).


7.5 Penalty Finite Element Models

7.5.1 Introduction

The penalty function method, like the Lagrange multiplier method (see [3,10, 13]), allows us to reformulate a problem with constraints as one withoutconstraints. In order to use the penalty function method for the ßow of aviscous incompressible ßuid, Þrst it is reformulated as a variational problemsubjected to a constraint. For the purpose of describing the penalty functionmethod, we consider the steady Stokes ßow problem (i.e. without time-dependent and nonlinear terms) in two dimensions. Then the penalty methodis applied to the variational problem with a constraint. The development willthen be extended to unsteady NavierStokes equations.Consider the weak forms in Eqs. (7.4.7)(7.4.9), and omit the time

derivative and nonlinear terms. These can be expressed in the form

B((wx, wy,Q), (vx, vy, P )) = `(wx, wy, Q) (7.5.1)

where (wx, wy, Q) are the weight functions used for the momentum andcontinuity equations, respectively, B(·, ·) is a bilinear form [i.e. an expressionthat is linear in (wx, wy, Q) as well as (vx, vy, P )] and `(·) is a linear form,deÞned by

B((wx, wy, Q), (vx, vy, P ))

=

ZΩeµ

∙2

µ∂wx∂x

∂vx∂x

+∂wy∂y

∂vy∂y

¶+

µ∂wx∂y

+∂wy∂x

¶µ∂vx∂y

+∂vy∂x

¶¸dx dy

−ZΩe

∙µ∂wx∂x

+∂wy∂y

¶P +

µ∂vx∂x

+∂vy∂y

¶Q

¸dx dy (7.5.2a)

`(wx, wy, Q) =

ZΩeρ0 (wxfx + wyfy) dx dy +

IΓe(wxtx + wyty) ds (7.5.2b)

and (tx, ty) are the boundary stress components deÞned in Eq. (7.4.10). Thestatement in Eq. (7.5.1) is known as the variational problem associated withsteady-state Stokes problem.The Þnite element model based on the variational problem (7.5.1) is a

special case of the mixed Þnite element model in Eq. (7.4.13). Equation(7.4.13) is more general than the problem at hand in that Eq. (7.4.13)is valid for time-dependent NavierStokes equations. To make it simpleto understand the penalty function method, only the steady-state Stokesproblem is considered. The inertial (i.e. time-dependent) and convective (i.e.nonlinear) terms may be added to the equations of the penalty formulation asthey are not connected to the divergence-free (i.e. incompressibility) condition,which is central to the penalty formulation.


7.5.2 Penalty Function Method

Suppose that the velocity Þeld (vx, vy) is such that the continuity equation(7.4.1) is satisÞed identically. Then the weight functions (wx, wy), being virtualvariations of the velocity components, also satisfy the continuity equation

∂wx∂x

+∂wy∂y

= 0 (7.5.3)

As a result, the second integral expression in the bilinear form (7.5.2a) dropsout, and the pressure, and hence the weight function Q, does not appearexplicitly in the variational problem (7.5.1). The resulting variational problemnow can be stated as follows: among all vectors v = vxex+ vyey that satisÞesthe continuity equation (7.4.1), Þnd the one that satisÞes the variationalproblem

B0((wx, wy), (vx, vy)) = `0(wx, wy) (7.5.4)

for all admissible weight functions w = wxex + wyey [i.e. that which satisÞesthe condition ∇ · w [see Eq. (7.5.3)]. The bilinear and linear forms in Eq.(7.5.4) are deÞned by

B0(w,v) =ZΩeµ

∙2

µ∂wx∂x

∂vx∂x

+∂wy∂y

∂vy∂y

¶+

µ∂wx∂y

+∂wy∂x

¶µ∂vx∂y

+∂vy∂x

¶¸dx dy

(7.5.5a)

`0(w) =

ZΩeρ0 (wxfx + wyfy) dx dy +

IΓe(wxtx + wyty) ds (7.5.5b)

The variational problem in Eq. (7.5.4) is a constrained variational problem,because the solution vector v is constrained to satisfy the continuity equation.We note that B0(·, ·) is symmetric

B0(w,v) = B0(v,w) (7.5.6)

and linear in w as well as v, that is, the following relations hold for any vectorsw1,w2,v1, and v2 that satisfy the incompressibility condition [see Eq. (7.5.3)]and arbitrary constants α and β:

B0(αw1 + βw2,v) = αB0(w1,v) + βB0(w2,v) (7.5.7a)

B0(w,αv1 + βv2) = αB0(w,v1) + βB0(w,v2) (7.5.7b)

Thus, B0(·, ·) is called bilinear if and if only it satisÞes conditions in Eq.(7.5.7a,b). Similarly `0(·) is called linear in w if and only if it satisÞes thecondition

`0(αw1 + βw2) = α`0(w1) + β`0(w2) (7.5.8)


Whenever the bilinear form of a variational problem is symmetric in itsarguments, it is possible to construct a quadratic functional such that theminimum of the quadratic functional is equivalent to the variational problem(see Reddy [12,13]). The quadratic functional is given by the expression

I0(v) =1

2B0(v,v)− `0(v) (7.5.9)

Now we can state that Eqs. (7.4.1)(7.4.3) governing the steady ßow of viscousincompressible ßuids are equivalent to minimizing the quadratic functionalI0(v) [v = (vx, vy)] subjected to the constraint

G(v) ≡ ∂vx∂x

+∂vy∂y

= 0 (7.5.10)

At this point it should be remembered that the discussion presented inthis section thus far is to reformulate the problem as one of a constrainedproblem so that the penalty function method can be used. The advantage ofthe constrained problem is that the pressure variable P does not appear inthe formulation.In the penalty function method, the constrained problem is reformulated

as an unconstrained problem as follows: minimize the modiÞed functional

IP (v) ≡ I0(v) + γe2

ZΩe[G(v)]2dx (7.5.11)

where γe is called the penalty parameter. Note that the constraint is included ina least-squares sense into the functional. Seeking the minimum of the modiÞedfunctional IP (v) is equivalent to seeking the minimum of both I0(v) and G(v),the latter with respect to the weight γe. The larger the value of γe, the moreexactly the constraint is satisÞed. The necessary condition for the minimumof IP is

δIP = 0 (7.5.12)

We have

0 =

ZΩe

∙2µ∂δvx∂x

∂vx∂x

+ µ∂δvx∂y

µ∂vx∂y

+∂vy∂x

¶¸dx dy −

ZΩeρ0δvxfx dx dy

−IΓeδvxtx ds+

ZΩeγe∂δvx∂x

µ∂vx∂x

+∂vy∂y

¶dx dy (7.5.13a)

0 =

ZΩe

∙2µ∂δvy∂y

∂vy∂y

+ µ∂δvy∂x

µ∂vx∂y

+∂vy∂x

¶¸dx dy −

ZΩeρ0fyδvy dx dy

−IΓeδvyty ds+

ZΩeγe∂δvy∂y

µ∂vx∂x

+∂vy∂y

¶dx dy (7.5.13b)

These two statements provide the weak forms for the penalty Þnite elementmodel with δvx = wx and δvy = wy. We note that the pressure does not


appear explicitly in the weak forms (7.5.13a,b), although it is a part of theboundary stresses. An approximation for the pressure can be post-computedfrom the relation

P = −γe (∂vx∂x

+∂vy∂y) (7.5.14)

where v = v(γe) is the solution of Eqs. (7.5.13a,b). The time derivative termsand nonlinear terms can be added to equations (7.5.13a,b) without affectingthe above discussion. We obtain

0 =

ZΩe

∙δvxρ0

µ∂vx∂t

+ vx∂vx∂x

+ vy∂vx∂y

¶+ 2µ

∂δvx∂x

∂vx∂x

+ µ∂δvx∂y

µ∂vx∂y

+∂vy∂x

¶+ γe

∂δvx∂x

µ∂vx∂x

+∂vy∂y

¶¸dx dy

−ZΩeρ0δvxfx dx dy −

IΓeδvxtx ds (7.5.15a)

0 =

ZΩe

∙δvyρ0

µ∂vy∂t

+ vx∂vy∂x

+ vy∂vy∂y

¶+ 2µ

∂δvy∂y

∂vy∂y

+ µ∂δvy∂x

µ∂vx∂y

+∂vy∂x

¶+ γe

∂δvy∂y

µ∂vx∂x

+∂vy∂y

¶¸dx dy

−ZΩeρ0fyδvy dx dy −

IΓeδvyty ds (7.5.15b)

7.5.3 Reduced Integration Penalty Model

The penalty Þnite element model is obtained from Eqs. (7.5.15a,b) bysubstituting Þnite element interpolation (7.4.11a) for the velocity Þeld, andδvx = ψ

ei and δvy = ψ

ei :∙

[C(v)] [0][0] [C(v)]

¸½ vxvy

¾+

∙2[Sxx] + [Syy] [Syx]

[Sxy] [Sxx] + 2[Syy]

¸½ vxvy

¾+

∙[Sxx] [Sxy][Syx] [Syy]

¸½ vxvy

¾+

∙[M ] [0][0] [M ]

¸½ úvx úvy

¾=

½ F 1F 2

¾(7.5.16)

where [M ], [C(v)], [Sξη], F 1, and F 2 are the same as those deÞned in Eq.(7.4.14), and

Sξη =

ZΩeγe∂ψei∂ξ

∂ψej∂η

dx dy (7.5.17)

Equation (7.5.16) can be expressed symbolically as (v = vx,vyT)£C(ρ,v) +K(µ) + K(γ)

¤v +M úv = F (7.5.18)

The numerical construction of the penalty terms presented in Eq. (7.5.18)requires special consideration, the details of which are given in Section 7.5.5.


7.5.4 Consistent Penalty Model

An alternative formulation (see [10, 15, 17]) of the penalty Þnite element modelis based directly on the use of Eq. (7.5.14). In this formulation, the pressurein the momentum equations (7.4.2) and (7.4.3) is replaced using Eq. (7.5.14)and Eq. (7.4.1) is replaced with Eq. (7.5.14). The weak forms of Eq. (7.5.14)is

0 =

ZΩeδP

∙P + γe

µ∂vx∂x

+∂vy∂y

¶¸dx dy (7.5.19)

The Þnite element model of Eq. (7.5.19), with P interpolated as in Eq.(7.4.11b), is given by

[Mp]P+ γe[[S0x] [S0y]]½ vxvy

¾= 0 (7.5.20a)

where

Mpij =

ZΩeφiφj dx dy, S

0xij =

ZΩeφi∂ψj∂x

dx dy, S0yij =

ZΩeφi∂ψj∂y

dx dy

(7.5.20b)Since [Mp] is invertible, we can write

P = −γe[Mp]−1[S0x] [S0y]½ vxvy

¾(7.5.21)

Next, we write Eq. (7.4.13) in the alternative form∙[M ] [0][0] [M ]

¸½ úvx úvy

¾+

∙[C(v)] [0][0] [C(v)]

¸½ vxvy

¾−½[S0x]T

[S0y]T

¾P

+



¸½ vxvy

¾=

½ F 1F 2

¾(7.5.22)

−[S0x]T [S0y]T½ vxvy

¾= 0 (7.5.23)

When Eq. (7.4.21) is substituted for the pressure into Eq. (7.5.22), we obtain∙[M ] [0][0] [M ]

¸½ úvx úvy

¾+

∙[C(v)] [0][0] [C(v)]

¸½ vxvy

¾

+ γe∙[S0x]T[Mp]−1[S0x] [S0x]T[Mp]−1[S0y][S0y]T[Mp]−1[S0x] [S0y]T[Mp]−1[S0y]

¸½ vxvy

¾

+



¸½ vxvy

¾=

½ F 1F 2

¾(7.5.24)


which can be written in compact form as

M úv + [C(v) +Kµ +Kγ ]v = F (7.5.25)

where [see Eqs. (7.4.14) for the deÞnition of Mij and Sξηij ]

[M] =

∙[M ] [0][0] [M ]

¸, [C] =

∙[C(v)] [0][0] [C(v)]

¸

[Kγ] = γe∙[S0x]T[Mp]−1[S0x] [S0x]T[Mp]−1[S0y][S0y]T[Mp]−1[S0x] [S0y]T[Mp]−1[S0y]

¸(7.5.26)

[K]µ =



¸Equation (7.5.25) has the same general form as Eq. (7.5.18). The overall

size of the penalty Þnite element model in Eq. (7.5.18) or (7.5.25) is reducedin comparison to the mixed Þnite element model in Eq. (7.4.13). To recoverthe pressure, the inverted form of (7.5.21) is used with the velocity Þeld thatis obtained from Eq. (7.5.25). The penalty Þnite element model describedhere is commonly termed a consistent penalty model because it is derivedfrom the discretized form of equation (7.5.19). This is in contrast to thereduced penalty model described earlier, which falls into the category knownas the reduced integration penalty (RIP) methods (see Oden [19]). Althoughthe two Þnite element models are mathematically identical, there are subtledifferences that affect the numerical implementation. The successful numericalimplementation of the consistent penalty method relies on the ability toefficiently construct the Kp matrix, that is, invert M

p at the element level.This restricts the choice of the basis functions used to represent the pressure.

7.6 Computational Aspects

7.6.1 Properties of the Matrix Equations

Some of the properties of the matrix equations in (7.4.13), (7.5.18), and(7.5.25) are listed below because they greatly inßuence the choice of a solutionprocedure for the various types of problems [10].

1. The matrix equations (7.4.13), (7.5.18), and (7.5.25) represent discreteanalogs of the basic conservation equations with each term representing aparticular physical process. For example,M represents the mass matrix, Crepresents the velocity dependent convective transport term,Kµ representsthe viscous terms, and Kγ represents the divergence-free condition. Theright-hand side F contains body forces and surface forces.

2. An inspection of the structure of the individual matrices in (7.4.14) showsthat M and Kµ are symmetric, while C is unsymmetric. This makes the


coefficient matrices of the vector v in Eqs. (7.4.13), (7.5.18), and (7.5.25)unsymmetric, and the solution procedure must deal with an unsymmetricsystem. When material properties are constant and ßow velocities aresufficiently small, the convective terms are negligible and the equations arelinear and symmetric.

3. An additional difficulty of the mixed Þnite element model is the presenceof zeroes on the matrix diagonals corresponding to the pressure variables[see Eq. (7.4.13)]. Direct equation solving methods must use some type ofpivoting strategy, while the use of iterative solvers is severely handicappedby poor convergence behavior attributable mainly to the form of theconstraint equation.

4. Equations (7.4.13), (7.5.18), and (7.5.25) represent a set of ordinarydifferential equations in time. The fact that the pressure does not appearexplicitly in the continuity equation [see Eq. (7.4.13)] makes the systemtime-singular in the pressure and precludes the use of purely explicit time-integration methods.

5. The choice of the penalty parameter is largely dictated by the ratio ofthe magnitude of penalty terms to the viscous and convective terms (orcompared to the Reynolds number, Re), the mesh, and the precision of thecomputer. The following range of γ is used in computations

γ = 104Re to γ = 1012Re (7.6.1)

7.6.2 Choice of Elements

As is clear from the weak statements, the Þnite element models of conductiveheat transfer as well as viscous incompressible ßows require only the C0-continuous functions to approximate the Þeld variables (i.e. temperature,velocities, and pressure). Thus, any of the Lagrange and serendipity familyof interpolation functions are admissible for the interpolation of the velocityÞeld in mixed and penalty Þnite element models.The choice of interpolation functions used for the pressure variable

in the mixed Þnite element model is further constrained by the specialrole the pressure plays in incompressible ßows. Recall that the pressurecan be interpreted as a Lagrange multiplier that serves to enforce theincompressibility constraint on the velocity Þeld. From Eq. (7.4.11b) it isseen that the approximation functions φn used for pressure is the weightingfunction for the continuity equation. In order to prevent an overconstrainedsystem of discrete equations, the interpolation used for pressure must be atleast one order lower than that used for the velocity Þeld (i.e. unequal orderinterpolation). Further, pressure need not be made continuous across elementsbecause the pressure variable does not constitute a primary variable of the


weak form presented in Eqs. (7.4.7)(7.4.9). Note that the unequal orderinterpolation criteria can be relaxed for certain stabilized formulations suchas the methods proposed by Hughes et al. [26], which will not be discussedhere.Convergent Þnite element approximations of problems with constraints

are governed by the ellipticity requirement and the LadyzhenskayaBabuskaBrezzi (LBB) condition (see Reddy [13, pp. 454461], Oden [19], Oden andCarey [20], and others [2123]). It is by no means a simple task to rigorouslyprove whether every new element developed for the viscous incompressibleßows satisÞes the LBB condition. The discussion of the LBB condition isbeyond the scope of the present study and will not be discussed here.Commonly used elements for two-dimensional ßows of viscous

incompressible ßuids are shown in Figure 7.6.1. In the case of linear elements,pressure is treated as discontinuous between elements; otherwise, the wholedomain will have the same pressure. Two different pressure approximationshave been used when the velocities are approximated by quadratic Lagrangefunctions. The Þrst is a continuous bilinear approximation, in which thepressure is deÞned at the corner nodes of the element and is made continuousacross element boundaries. The second pressure approximation involves adiscontinuous (between elements) linear variation deÞned on the element by

Φ = φ =⎧⎨⎩1xy

⎫⎬⎭ (7.6.2)

Figure 7.6.1 The triangular and quadrilateral elements used for the mixedand penalty Þnite element models.

Nodes with u and vNodes with u, v, and P Nodes with P only


Here the unknowns are not nodal point values of the pressure but correspondto the coefficients in P = a · 1 + b · x+ c · y. In Eq. (7.6.2) the interpolationfunctions are written in terms of the global coordinates (x, y) for the problem.When the eight-node quadratic element is used to represent the velocity

Þeld, a continuous-bilinear pressure approximation may be selected. When adiscontinuous pressure variation is utilized with this element, the constantpressure representation over each element must be used. The quadraticquadrilateral elements shown in Figure 7.6.1 are known to satisfy the LBBcondition and thus give reliable solutions for velocity and pressure Þelds. Otherelements may yield acceptable solutions for the velocity Þeld but the pressureÞeld is often in error.

7.6.3 Evaluation of Element Matrices in Penalty Models

The numerical evaluation of the coefficient matrices appearing in equation(7.5.18) requires special consideration [10]. This aspect is discussed here for thesteady-state case. For the steady-state ßows with constant material properties,Eq. (7.5.18) is of the form

(ρ C(v) + µ K+ γK)v = F (7.6.3)

where C is the contribution due to the convective terms, K is the contributionfrom the viscous terms, and K is from the penalty terms, which come fromthe incompressibility constraint. In theory, as we increase the value of γ, theconservation of mass is satisÞed more exactly. However, in practice, for somelarge value of γ, the contribution from the viscous terms would be negligiblysmall compared to the penalty terms in a computer. Thus, if K is a non-singular (i.e. invertible) matrix, the solution of Eq. (7.6.3) for a large value ofγ is trivial, v = 0. While the solution satisÞes the continuity equation, itdoes not satisfy the momentum equations. In this case the discrete problem(7.6.3) is said to be overconstrained or locked. If K is singular, then the

sum (ρ C+µ K+ γK) is non-singular (because ρ C+µ K is non-singular), anda non-trivial solution to the problem is obtained.The numerical problem described above is eliminated by proper evaluation

of the integrals in C, K, and K. It is found that if the coefficients of K(i.e. penalty matrix coefficients) are evaluated using a numerical integrationrule of an order less than that required to integrate them exactly, the Þniteelement equations (7.6.3) give acceptable solutions for the velocity Þeld. Thistechnique of under-integrating the penalty terms is known in the literature asreduced integration. For example, if a linear quadrilateral element is used toapproximate the velocity Þeld, the matrix coefficients C and K (as well asMfor unsteady problems) are evaluated using the 2× 2 Gauss quadrature, andK is evaluated using the one-point (1× 1) Gauss quadrature. The one-pointquadrature yields a singular K. Therefore, Eq. (7.6.3) can be solved because


(ρ C + µ K + γK) is non-singular and can be inverted (after assembly andimposition of boundary conditions) to obtain a good Þnite element solutionof the original problem. When a quadratic quadrilateral element is used, the3×3 Gauss quadrature is used to evaluate C, K, andM, and the 2×2 Gaussquadrature is used to evaluate K. Of course, as the degree of interpolationgoes up, or very reÞned meshes are used, the resulting equations become lesssensitive to locking.Concerning the post-computation of pressure in the penalty model, the

pressure should be computed by evaluating Eq. (7.5.14) at integration pointscorresponding to the reduced Gauss rule. This is equivalent to using aninterpolation for pressure that is one order less than the one used for thevelocity Þeld. The pressure computed using equation (7.5.14) at the reducedintegration points is not always reliable and accurate. The pressures predictedusing the linear elements, especially for coarse meshes, are seldom acceptable.Quadratic elements are known to yield more reliable results. In general,triangular elements do not yield stable solutions for pressures. Varioustechniques have been proposed in the literature to obtain accurate pressureÞelds (see [2729]). A procedure for the post-computation of pressure isdiscussed in [10,29].

7.6.4 Post-Computation of Stresses

The analysis of a ßow problem generally includes calculation of not only thevelocity Þeld and pressure but also the computation of the stress Þeld. A briefdiscussion of the stress calculation is presented next [10].For a plane two-dimensional ßow, the stress components (σxx,σyy,σxy) are

given by

σxx = 2µ∂vx∂x

− P, σyy = 2µ∂vy∂y

− P, σxy = µµ∂vx∂y

+∂vy∂x

¶(7.6.4)

where µ is the viscosity of the ßuid. Substitution of the Þnite elementapproximations (7.4.11a,b) for the velocity Þeld and pressure into Eqs. (7.6.4)yields

σxx = 2µMXj=1

∂ψj∂xvjx − P, σyy = 2µ

MXj=1

∂ψj∂yvjy − P

σxy = µMXj=1

µ∂ψj∂yvjx +

∂ψj∂x

vjy

¶ (7.6.5)

where P is calculated from [see Eq. (7.5.14)]

P (x, y) =NXj=1

φj(x, y)Pj (7.6.6a)


for the mixed model, and from

Pγ(x, y) = −γMXj=1

µ∂ψj∂xvjx +

∂ψj∂yvjy

¶(7.6.6b)

for the penalty model.The spatial derivatives of the interpolation functions in Eqs. (7.6.5a-c)

and (7.6.6b) must be evaluated using the reduced Gauss point rule. Thus, thestresses (as well as the pressure) are computed using the one-point Gauss rulefor linear elements and with the 2× 2 Gauss rule for the quadratic elements.The stresses computed at interior integration points can extrapolated tothe nodes by a simple linear extrapolation procedure, and they may beappropriately averaged between adjacent elements to produce a continuousstress Þeld.


7.7.1 Mixed Model

The computer implementation of the mixed model is some what complicatedby the fact that the element contains variable degrees of freedom and thecoefficient matrix is not positive-deÞnite due to the appearance of zeros onthe diagonal. Here, we discuss computer implementation of mixed modelwith quadratic approximation of the velocity Þeld and bilinear continuousapproximation of the pressure. A eight- or nine-node element depicting allnodal values of the formulation will have three degrees of freedom (vx, vy, P )at the corner nodes and two degrees of freedom (vx, vy) at the mid-side andinterior nodes. This complicates the calculation of element matrices as wellthe assembly of element equations to form the global system of equations.Fortran statements of the element calculations for the mixed Þnite element

model are given in Box 7.7.1. Here ELV denotes the element velocityvector, AMU the viscosity µ, and RHO the density ρ. The meaningof other variables remain the same as deÞned in earlier discussions (e.g.NGP = number of Gauss points; GAUSSPT (I, J)= array of Gauss points;GAUSSWT (I, J)= array of Gauss weights; DET = determinant of theJacobian matrix; NPE = number of nodes per element, 8 or 9; SF (I) =ψi, Lagrange interpolation functions of the quadratic element; SFL(I) =Lagrange interpolation functions of the bilinear element; and GDSF (α, i) =∂ψi/∂xα, etc.).To facilitate the assembly, we create a companion array NFD(I, J) to the

connectivity array NOD(I, J). This array is similar to the array NOD but itconnects the degrees of freedom rather than the global nodes associated withthe element nodes:

NFD(I, J) = The last global degree of freedom number associated with theJth node of the Ith element.


The word last refers to the third degree of freedom at the corner nodes andthe second degree of freedom at the mid-side and interior nodes. To see themeaning of array [NFD], consider the mesh of six nine-node elements shownin Figure 7.7.1. First note that the connectivity array for the mesh is givenby

1 2 3 4 5 6 7 8 9

[NOD] =

⎡⎢⎢⎢⎢⎢⎢⎣

1 3 17 15 2 10 16 8 93 5 19 17 4 12 18 10 115 7 21 19 6 14 20 12 1315 17 31 29 16 24 30 22 2317 19 33 31 18 26 32 24 3519 21 35 33 20 28 34 26 27

⎤⎥⎥⎥⎥⎥⎥⎦

123456

The NFD array is

1 2 3 4 5 6 7 8 9

[NFD] =

⎡⎢⎢⎢⎢⎢⎢⎣

3 8 40 35 5 24 37 20 228 13 45 40 10 28 42 24 2613 18 50 45 15 32 47 28 3035 40 72 67 37 56 69 52 5440 45 77 72 42 60 74 56 5845 50 82 77 47 64 79 60 62

⎤⎥⎥⎥⎥⎥⎥⎦

123456

In addition, we deÞne an array of the total degrees of freedom of nodes 1through 9 of the element

NFR= 3 3 3 3 2 2 2 2 2

Now with the help of these arrays, we can assemble the element matrices.Fortran statements of the assembly are given in Box 7.7.2.

Figure 7.7.1 Element and global degrees of freedom of a mesh of nine-nodeelements used for the mixed formulation.

(a)

(b)

1 2 3 4 5 6 7

8 14

21

28

3529

22

15

30 31 32 33 34

1-3 4,5 6-8 11-139,10 16-1814,15

19,20 31,32

48-50

80-82

63,64

65-67 68,69 70-7273,74

75-7778,79

51,52

33-3521,22 23,24

36,37 38-40

1

4

2 3

5 6

1 2

34

56

7

8 9

Global node numbers

Global D.O.F. numbersElement numbers

Element node numbers

(vx, vy, P); (vx, vy)


Box 7.7.1 Fortran statements of the mixed element calculations.

DO 100 NI=1,NGP DO 100 NJ=1,NGP XI=GAUSS(NI,NGP) ETA=GAUSS(NJ,NGP) CALL SHP2DPV(NPE,XI,ETA,ELXY,DET) CONST=DET*WT(NI,NGP)*WT(NJ,NGP)

C VX=0.0VY=0.0DO I=1,NPE

L=2*I-1VX=VX+SF(I)*ELV(L)VY=VY+SF(I)*ELV(L+1)

ENDDO C

II=1 KI = 3 DO 80 I = 1,NPE

JJ=1 KJ = 3 DO 70 J = 1,NPE CONV = SF(I)*(VX*GDSF(1,J)+VY*GDSF(2,J))*CONST SX = GDSF(1,I)*GDSF(1,J)*CONST SY = GDSF(2,I)*GDSF(2,J)*CONST SXY= GDSF(1,I)*GDSF(2,J)*CONST SYX= GDSF(2,I)*GDSF(1,J)*CONST IF (J.LE.4)THEN

GX = GDSF(1,I)*SFL(J)*CONST GY = GDSF(2,I)*SFL(J)*CONST

ENDIF IF (I.LE.4)THEN

GXT= GDSF(1,J)*SFL(I)*CONST GYT= GDSF(2,J)*SFL(I)*CONST

ENDIF C

ELK(II,JJ) = ELK(II,JJ) +AMU*(2.0*SX+SY)+RHO*CONV ELK(II,JJ+1) = ELK(II,JJ+1) +AMU*SYX ELK(II+1,JJ+1) = ELK(II+1,JJ+1)+AMU*(SX+2.0*SY)+RHO*CONV ELK(II+1,JJ) = ELK(II+1,JJ) +AMU*SXY IF(J .LE. 4)THEN

ELK(II,JJ+2) = ELK(II,JJ+2) - GX ELK(II+1,JJ+2) = ELK(II+1,JJ+2) - GY

ENDIF IF(I .LE. 4)THEN

ELK(II+2,JJ) = ELK(II+2,JJ) - GXT ELK(II+2,JJ+1) = ELK(II+2,JJ+1) - GYT

ENDIF IF(J .GT. 4)KJ=2

70 JJ = JJ + KJ IF(I .GT. 4)KI=2

80 II = II + KI 100 CONTINUE


Box 7.7.2 Fortran statements of the assembly procedure used for themixed model.

7.7.2 Penalty Model

Computer implementation of the penalty Þnite element model is quitestraightforward and is the same as any multi-degree of freedom systems (seeChapters 2 and 5, and Box 7.7.1), and hence not discussed further. Onlynote should be made of the fact that the element calculations involve twoGauss loops: a full integration loop for the evaluations of all terms exceptfor the penalty terms; and the other one is a reduced integration loop for theevaluation of the penalty terms of the coefficient matrix [K] [see Eq. (7.6.3)].

7.8 Numerical Examples


In this section, a small number of ßow problems solved using the Þnite elementmodels developed herein are presented. Additional examples may be found inthe book by Reddy and Gartling [10]. The examples presented herein weresolved using the RIP Þnite element model and mixed Þnite element model. The

C Assembly of element coefficient matrix and right-hand column vectorC for a typical (Nth) element.C

DO 90 I=1,NPE NDFI=NFR(I) NR=NFD(N,I)-NDFI DO 90 II=1,NDFI

NR=NR+1 IF(I.LE.5)THEN

L=(I-1)*3+II ELSE

L=12+(I-5)*2+IIENDIF GLF(NR) = GLF(NR)+ELF(L) DO 90 J=1,NPE

NDFJ=NFR(J) NCL=NFD(N,J)-NDFJ DO 90 JJ=1,NDFJ

NCL=NCL+1 IF(J.LE.5)THEN

M=(J-1)*3+JJ ELSE M=12+(J-5)*2+JJ ENDIF

GLK(NR,NCL)=GLK(NR,NCL)+ELK(L,M) 90 CONTINUE


objective of the first several examples is to evaluate the accuracy of the penaltyand mixed finite element models by comparing with the available analytical ornumerical results and to illustrate the effect of the penalty parameter on theaccuracy of the solutions. The remaining examples are for Reynolds numbersgreater than unity (i.e. convective terms are included), and the results wereobtained using the RIP finite element model.

7.8.2 Fluid Squeezed Between Parallel Plates

Consider the slow flow of a viscous incompressible material squeezed betweentwo long parallel plates [see Figure 7.8.1(a)]. When the length of the plates isvery large compared to both the width of and the distance between the plates,we have a case of plane flow. Although this is a moving boundary problem, wewish to determine the velocity and pressure fields for a fixed distance betweenthe plates, assuming that a state of plane flow exists.Let V0 be the velocity with which the two plates are moving toward each

other (i.e. squeezing out the fluid), and let 2b and 2a denote, respectively, thedistance between and the length of the plates [see Figure 7.8.1(a)]. Due to thebiaxial symmetry present in the problem, it suffices to model only a quadrantof the domain. As a first mesh, we use a 5× 3 nonuniform mesh of nine-nodequadratic elements in the mixed model, and a 10 × 6 mesh of the four-nodelinear elements and 5×3 mesh of nine-node quadratic elements in the penaltymodel [see Figure 7.8.1(b)]. The non-uniform mesh, with smaller elementsnear the free surface (i.e. at x = a), is used to approximate accurately thesingularity in the shear stress at the point (a, b) = (6, 2). The mesh used forthe penalty model has exactly the same number of nodes as the mesh used forthe nine-node mesh of the mixed model. The velocity boundary conditionsare shown in Figure 7.8.1(b). The velocity field at x = 6 (outflow boundary)

Figure 7.8.1 (a) Geometry, computational domain, and (b) the finiteelement mesh used for the analysis of slow flow of viscousincompressible fluid between parallel plates.

x

(a)v0

Computational domain

y

2b2a

v0y

(b)

•

Linear elementQuadratic element

vx = 0 ty = 0

vx = 0, vy = −v0

tx = 0 ty = 0

vy = 0, tx = 0

•••

x


is not known; if we do not impose any boundary conditions there, it amountsto requiring tx = ty = 0 in the integral sense. In the mixed Þnite elementmodel, it is necessary to specify the pressure at least at one node. In thepresent case, the node at (x, y) = (a, 0) is speciÞed to have zero pressure. Anapproximate analytical solution to this two-dimensional problem is providedby Nadai [30] (also see [12,13]), and it is given by

vx(x, y) =3V0x

2b

Ã1− y

2

b2

!, vy(x, y) = −3V0y

2b

Ã3− y

2

b2

!

P (x, y) =3µV02b3

(a2 + y2 − x2)(7.8.1)

The velocities vx(x, 0) obtained with the two Þnite element models comparewell with the analytical solution (see Reddy [11]), as shown in Table 7.8.1.The nine-node element gives very good results for both the penalty and mixedmodels. The inßuence of the penalty parameter on the accuracy of the solutionis clear from the results. Whether the element is linear or quadratic, it isnecessary to use a large value of the penalty parameter.

Table 7.8.1 Comparison of Þnite element solution vx(x, 0) with theanalytical solution for ßuid squeezed between plates.

γ = 1.0 γ = 100 γ = 108 Mixedmodel Series

x Four Nine* Four Nine Four Nine Nine solutionx -node -node -node -node -node -node -node

1.00 0.0303 0.0310 0.6563 0.6513 0.7576 0.7505 0.7497 0.75002.00 0.0677 0.0691 1.3165 1.3062 1.5135 1.4992 1.5031 1.50003.00 0.1213 0.1233 1.9911 1.9769 2.2756 2.2557 2.2561 2.25004.00 0.2040 0.2061 2.6960 2.6730 3.0541 3.0238 3.0203 3.00004.50 0.2611 0.2631 3.0718 3.0463 3.4648 3.4307 3.4292 3.37505.00 0.3297 0.3310 3.4347 3.3956 3.8517 3.8029 3.8165 3.75005.25 0.3674 0.3684 3.6120 3.5732 4.0441 3.9944 3.9893 3.93755.50 0.4060 0.4064 3.7388 3.6874 4.1712 4.1085 4.1204 4.12505.75 0.4438 0.4443 3.8316 3.7924 4.2654 4.2160 4.2058 4.31256.00 0.4793 0.4797 3.8362 3.7862 4.2549 4.1937 4.2364 4.5000

*The 3× 3 Gauss rule for non-penalty terms and the 2× 2 Gauss rule for penalty terms areused for quadratic elements.

Figure 7.8.2 contains plots of the velocity vx(x, y) for x = 4 and x = 6,and Figure 7.8.3 contains plots of pressure P (x, y), for y = y0, where y0 isthe y−coordinate of the Gauss point nearest to the centerline or top plate.These results were obtained with two different meshes: 5× 3 and 10× 8. Thepressure in the penalty model was computed using Eq. (7.5.19) with the 2×2Gauss rule for the quadratic rectangular element and the one-point formula

, vx

vx at x = 4

vx at x = 6

Analytical solution

0 1 2 3 4

Horizontal velocity

0.0

0.5

1.0

1.5

2.0

Dis

tan

ce, y

Analytical solution (y = 2)

FEM solution

(y = 0.1875)

Analytical solution (y = 0)

FEM solution

(y = 0.0625)

0 1 2 3 4 5 6Distance, x

0

1

2

3

4

5

6

7

8

9

10

Pre

ssu

re, P

0 1 2 3 4 5 6Distance, x

-1

0

1

2

3

4

5

6

7

8

Pre

ssu

re, P


for the linear element, whereas in the mixed model (as well as the analyticalsolution) it is computed at the nodes. If the pressure in the penalty modelwere computed using the full quadrature rule for rectangular elements, wewould have obtained erroneous values. In general, the same integration ruleas that used for the evaluation of the penalty terms in the coefficient matrixmust be used to compute the pressure. The oscillations in pressure computednearest to the top plate are due to the singularity in the boundary conditionsat (x, y) = (6, 2).

Figure 7.8.2 Velocity Þelds for ßuid squeezed between parallel plates.

Figure 7.8.3 Pressures for ßuid squeezed between parallel plates.

h1

h2

h(x)

x

y

x

P = P0 = 0 P = P0 = 0

V0 = 1

vx = vy = 0

vx = V0 , vy = 0

(a)

h1

h2

x

y

x

P = P0 = 0 P = P0 = 0

V0 = 1

vx = vy = 0

vx = V0 , vy = 0

(b)


7.8.3 Flow of a Viscous Lubricant in a Slider Bearing

The slider (or slipper) bearing consists of a short sliding pad moving at avelocity u = V0 relative to a stationary pad inclined at a small angle withrespect to the stationary pad, and the small gap between the two pads isÞlled with a lubricant [see Figure 7.8.4(a)]. Since the ends of the bearingare generally open, the pressure there is atmospheric, P0. If the upper padis parallel to the base plate, the pressure everywhere in the gap must beatmospheric (because dP/dx is a constant for ßow between parallel plates),and the bearing cannot support any transverse load. If the upper pad isinclined to the base pad, a pressure distribution, in general, a function of xand y, is set up in the gap. For large values of V0, the pressure generated canbe of sufficient magnitude to support heavy loads normal to the base pad.When the width of the gap and the angle of inclination are small, one

may assume that vy = 0 and the pressure is not a function of y. Assuming atwo-dimensional state of ßow and a small angle of inclination, and neglectingthe normal stress gradient (in comparison with the shear stress gradient), theequations governing the ßow of the lubricant between the pads can be reducedto (see Schlichting [9])

Figure 7.8.4 Schematic and the Þnite element mesh for slider bearing.


µ∂2vx∂y2

=dP

dx, 0 < x < L (7.8.2a)

where the pressure gradient is given by

dP

dx=6µV0h2

µ1− H

h

¶, H =

2h1h2h1 + h2

(7.8.2b)

The solution of Eqs. (7.8.2a,b), subject to the boundary conditions

vx(0, 0) = V0, vx(x, h) = 0 (7.8.2c)

is

vx(x, y) =

ÃV0 − h2

2µ

dP

dx

y

h

!µ1− y

h

¶(7.8.3a)

P (x) =6µV0L(h1 − h)(h− h2)

h2(h21 − h22)(7.8.3b)

σxy(x, y) = µ∂vx∂y

=dP

dx

µy − h

2

¶− µV0

h(7.8.3c)

where

h(x) = h1 +h2 − h1L

x (7.8.4)

In the Þnite element analysis we do not make any assumptions concerningvy and the pressure gradient, and solve the Stokes equations [i.e. neglect theconvective terms in Eqs. (7.4.2) and (7.4.3)], with the following choice ofparameters:

h1 = 2h2 = 8× 10−4 ft, L = 0.36 ft, µ = 8× 10−4 lb/ft2, V0 = 30 ft(7.8.5)

We use a mesh (Mesh 1) of 18× 8 linear quadrilateral elements to analyze theproblem. The mesh and boundary conditions are shown in Figure 7.8.4(b).The penalty parameter is chosen to be γ = µ × 108. Table 7.8.2 containsa comparison of the Þnite element solutions and analytical solutions forthe velocity, pressure, and shear stress. Figure 7.8.5 contains plots of thehorizontal velocity vx at x = 0 ft, x = 0.18 ft, and x = 0.36 ft. Figure 7.8.6contains plots of pressure and shear stress as a function of x at y = 0. TheÞnite element solutions for the pressure and shear stress were computed atthe center of the Þrst row of elements along the moving block. The resultsare in good agreement with the analytical solutions (7.8.3ac), validating theassumptions made in the development of the analytical solution.


Table 7.8.2 Comparison of Þnite element solutions velocities with theanalytical solutions for viscous ßuid in a slider bearing.

vx(0, y) vx(0.18, y) vx(0.36, y)

y FEM Analy. y FEM Analy. y FEM Analy.

0.0 30.000 30.000 0.00 30.000 30.000 0.00 30.000 30.0001.0 22.923 22.969 0.75 25.139 25.156 0.50 29.564 29.5312.0 16.799 16.875 1.50 20.596 20.625 1.00 28.182 28.1253.0 11.626 11.719 2.25 16.372 16.406 1.50 25.853 25.7814.0 7.403 7.500 3.00 12.465 12.500 2.00 22.577 22.5005.0 4.130 4.219 3.75 8.874 8.906 2.50 18.354 18.2816.0 1.805 1.875 4.50 5.600 5.625 3.00 13.184 13.1257.0 0.429 0.469 5.25 2.642 2.656 3.50 7.066 7.0318.0 0.000 0.000 6.00 0.000 0.000 4.00 0.000 0.000

Analytical solution FEM Solution

x P (x, 0) −σxy(x, 0) x y P −σxy

0.01 7.50 59.99 0.1125 0.4922 8.46 56.610.03 22.46 59.89 0.3375 0.4766 25.46 56.600.05 37.29 59.67 0.5625 0.4609 42.31 56.470.07 51.89 59.30 0.7875 0.4453 58.76 56.170.09 66.12 58.77 1.0125 0.4297 74.69 55.690.27 129.60 38.40 2.5875 0.3203 134.40 41.770.29 118.57 32.71 2.8125 0.3047 125.60 36.930.31 99.58 25.70 3.0375 0.2891 107.60 30.760.33 70.30 17.04 3.2625 0.2734 77.39 22.890.35 27.61 6.31 3.4875 0.2578 30.80 12.82

x = 10x, y = y × 104, P = P × 10−2.

Figure 7.8.5 Velocity distributions for the slider bearing problem.

vx(x0,y)

x0 = 0.0x0 = 0.18x0 = 0.36

h(x

)

FEM

Analytical

0 10 20 30Velocity,

0.0E+0

1.0E-4

2.0E-4

3.0E-4

4.0E-4

5.0E-4

6.0E-4

7.0E-4

8.0E-4

Hei

ght,


Figure 7.8.6 Pressure and shear stress distributions for the slider bearingproblem.

7.8.4 Wall-Driven Cavity Flow

Consider the laminar ßow of a viscous, incompressible ßuid in a square cavitybounded by three motionless walls and a lid moving at a constant velocity inits own plane (see Figure 7.8.7). Singularities exist at each corner where themoving lid meets a Þxed wall. This example is one that has been extensivelystudied by analytical, numerical, and experimental methods (see [3133],among others), and it is often used as a benchmark problem to test a newnumerical method or formulation. In solving this problem, the mesh usedshould be such that the boundary layer thickness is resolved. The boundary

layer thickness is of the order of Re−12 , where Re = ρv0a/µ is the Reynolds

number and v0 is the lid velocity and a is the dimension of the cavity.Assuming a unit square and that the velocity of the top wall is unity, we

can discretize the ßow region using a uniform, 8×8 mesh of linear elements or4× 4 of nine-node quadratic elements. At the singular points, namely at thetop corners of the lid, we assume that vx(x, 1) = v0 = 1.0. The linear solutionfor the horizontal velocity along the vertical centerline obtained with the twomeshes is shown in Figure 7.8.8, and the variation of pressure along the topwall (computed at the reduced Gauss points) is shown in Figure 7.8.9. Thenumerical values of the velocity Þeld are tabulated in Table 7.8.3. It is clearthat the value of the penalty parameter between γ = 102 and γ = 108 hassmall effect on the solution.

x

Shear stressPressure × 10-2

Analytical

−σ x

y

P ×

10-2

0.00 0.04 0.08 0.12 0.16 0.20 0.24 0.28 0.32 0.36

Distance,

0

25

50

75

100

125

150P

ress

ure

, S

hea

r st

ress

,


Figure 7.8.7 Wall-driven cavity problem.

Figure 7.8.8 Plots of horizontal velocity vx(0.5, y) along the verticalcenterline of the cavity.

Table 7.8.3 Velocity vx(0.5, y) obtained with various values of the penaltyparameter γ (linear solution).

y Mesh: 8× 8L4 Mesh: 4× 4Q9γ = 102 γ = 108 γ = 102 γ = 108

0.125 −0.0557 −0.0579 −0.0589 −0.06150.250 −0.0938 −0.0988 −0.0984 −0.10390.375 −0.1250 −0.1317 −0.1320 −0.13940.500 −0.1354 −0.1471 −0.1442 −0.15630.625 −0.0818 −0.0950 −0.0983 −0.11180.750 0.0958 0.0805 0.0641 0.04810.875 0.4601 0.4501 0.4295 0.4186

x

vx = 1, vy = 0

vx = 0, vy = 0

vx = 0 vy = 0vx = 0 vy = 0

y

a

a

0.5 a

8×8L4 mesh (γ = 102)4×4Q9 mesh (γ = 102)

4×4Q9 mesh (γ = 108)

8×8L4 mesh (γ = 108)

y

vx(0.5,y)-0.2 -0.1 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0

Horizontal velocity,

0.0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1.0

Dis

tan

ce,

x

P8×8L4 mesh (γ = 102)

4×4Q9 mesh (γ = 102)

4×4Q9 mesh (γ = 108)

8×8L4 mesh (γ = 108)

0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0Horizontal distance,

-15.0

-10.0

-5.0

0.0

5.0

10.0

15.0

Pre

ssu

re,


Figure 7.8.9. Plots of pressure P (x, y0) along the top wall of the cavity.

Next, we consider the wall-driven cavity problem for nonlinear analysis (i.e.solve NavierStokes equations). The role of the load parameter is played bythe Reynolds number (Re = ρv0a/µ). For the problem at hand, the Reynoldsnumber can be varied by varying the density while keeping the viscosityconstant. Thus, we take (in addition to the the choice of the characteristicvelocity v0 = 1 and characteristic length a = 1) µ = 1 so that Re = ρ.The problem is solved using uniform 8 × 8 mesh of linear elements as wellas 4× 4 mesh of nine-node quadratic elements, and the results are presentedin Table 7.8.4 for Re = 100, 500, and 700 (γ = 108 and ² = 10−2). Theconverged nonlinear solution of the preceding Reynolds number is used as theinitial guess in the Þrst iteration of the next Reynolds number. In general,for very high Reynolds numbers underrelaxtion must be used to accelerate theconvergence by using the weighted average of velocities from two consecutiveiterations

vr = βv(r) + (1− β)v(r−1) (7.8.6)

to compute the coefficient matrix. Here β is the acceleration parameter.

Figure 7.8.10 shows plots of the horizontal velocity along the cavitycenterline obtained with 8 × 8 and 16 × 20 mesh of bilinear elements forRe = 0 and 500 (β = 0). The pressure obtained with the two meshes isshown in Figure 7.8.11. Clearly, the pressure exhibits oscillations. Figure7.8.12 contains plots of the horizontal velocity obtained with 16 × 20 meshfor Re = 500, 103, 5 × 103, 104 (the increment of Re is taken to be 500 andβ = 0.5).

vx(0.5,y)

y

Re = 0 (8×8 mesh)

Re = 0 (16×20 mesh)

Re = 500 (8×8 mesh)

Re = 500 (16×20 mesh)

-0.4 -0.2 0.0 0.2 0.4 0.6 0.8 1.0Horizontal velocity,

0.0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1.0

Dis

tan

ce,


Table 7.8.4 Velocity vx(0.5, y) obtained with linear and quadratic elementsand for various values of the Reynolds number.

y 8× 8L 4× 4Q9Re → 100(5) 500(8) 700(9) 100(5) 500(8) 700(10)∗

0.125 −0.0498 −0.0242 −0.0140 −0.0554 −0.0141 −0.01060.250 −0.0870 −0.0503 −0.0345 −0.0968 −0.0540 −0.00890.375 −0.1164 −0.0733 −0.0564 −0.1313 −0.1143 −0.06720.500 −0.1231 −0.0700 −0.0586 −0.1414 −0.1252 −0.11810.625 −0.0635 0.0027 0.0039 −0.0814 −0.0455 −0.08310.750 0.0649 0.0389 0.0354 0.0486 0.1045 0.08080.875 0.3750 0.1761 0.1241 0.3629 0.2113 0.1628

* Number in parentheses denotes the number of iterations taken for convergence.

Figure 7.8.10 Velocity vx(0.5, y) vversus y for various Reynolds numbers(8× 8 and 16× 20 meshes of bilinear elements).

Next, a comparison of the present results with the Þnite difference solutionsof Ghia et al. [34] are presented. The mesh of bilinear elements used in thepresent study is shown in Figure 7.8.13; a penalty parameter of γ = 108Reand convergence tolerance of ε ≤ 10−2 were used. Convergence was achievedwith 3 iterations at each Reynolds number step (Re = 100, 400, 103). Figures7.8.14 and 7.8.15 show the results for Re = 400 and Re = 103.

0.0 0.2 0.4 0.6 0.8 1.0Horizontal distance,

-50

0

50

100

150

200P

ress

ure

, P(x

,y0)

x

y0 = center of the last row of elements

Re = 0 (16×20 mesh)

Re = 500 (16×20 mesh)

Re = 0 (8×8 mesh)

Re = 500 (8×8 mesh)

-0.3-0.2-0.1 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0


0.0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1.0

Dis

tan

ce,

vx(0.5,y)

y

Re = 500

Re = 1,000

Re = 5,000

Re = 10,000

16×20 mesh of bilinear elements(∆Re = 500, β = 0.5, ε = 10−2)


Figure 7.8.11 Plots of pressure P (x, y0) along the top wall of the cavity(8× 8 and 16× 20 meshes).

Figure 7.8.12 Velocity vx(0.5, y) versus y for various Reynolds numbers(16× 20 mesh of bilinear elements).

u = 0v = 0

u = 0v = 0

u = 0, v = 0

u = 1, v = 0

StreamlinesPressure

x

v-

velo

city

0.00 0.25 0.50 0.75 1.00-0.5

-0.4

-0.3

-0.2

-0.1

0.0

0.1

0.2

0.3

0.4

0.5

u - velocity

y

-0.50 -0.25 0.00 0.25 0.50 0.75 1.000.0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1.0


Figure 7.8.13 Mesh used for the wall-driven cavity problems.

Figure 7.8.14 Normalized horizontal and vertical velocity proÞles alongthe vertical and horizontal mid-sections of the cavity (computed, Ghia et al. [34]) and pressure contours andstreamlines (Re = 400).

StreamlinesPressure

x

v-

velo

city

0.00 0.25 0.50 0.75 1.00-0.6

-0.4

-0.2

0.0

0.2

0.4

0.6

u - velocity

y

-0.50 -0.25 0.00 0.25 0.50 0.75 1.000.0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1.0


Figure 7.8.15 Horizontal and vertical velocity proÞles along the verticaland horizontal mid-sections of the cavity for Re = 103

( computed, Ghia et al. [34]). Pressure contours andstreamlines.

7.8.5 BackwardFacing Step

Here we analyze the well-known backwardfacing problem (see Gartling [35]and Reddy and Gartling [10]) using the penalty Þnite element model. Thegeometry and boundary cnditions of the computatinal domain are shown inFigure 7.8.16. A penalty parameter of γ = 108Re and convergence toleranceof ε ≤ 10−2 are used. Convergence is achieved with 5 iterations.Figure 7.8.17 shows contour plots of the stremlines and pressure, while

Figure 7.8.18 contains pressure proÞles along the upper and lower walls forReynolds number Re = 800. The results compare well with those of Gartling[35]. This problem will be revisited in the next section in the context of theleast-squares Þnite element model.

vx = vx (y)

vy = 0

vx = 0, vy = 0

vx = 0, vy = 0

Outflow

( 0.0, 0.5 )

( 0.0, -0.5 )

( 30.0, 0.5 )

( 30.0, -0.5 )

0 1 2 3 4 5 6 7 8 9 10

Pressure

0 1 2 3 4 5 6 7 8 9 10

Streamlines


Figure 7.8.16 Geometry and boundary conditions for ßow over abackwardfacing step.

Figure 7.8.17 Streamlines and pressure contours for ßow over abackwardfacing step (Re = 800).

x

Pres

sure

0 5 10 15 20 25 30-0.15

-0.10

-0.05

0.00

0.05

0.10

Lower wall

Upper wall


Figure 7.8.18 Pressure profiles along the upper and lower walls of thechannel for flow over a backward—facing step (Re = 800).

7.9 Least-Squares Finite Element Models

7.9.1 Introduction

In this section, we present least-squares finite element model of the Navier—Stokes equations [36—39]. The least-squares method has the property ofminimizing the residuals in the differential equations. For the Navier—Stokesequations, which have no underlying minimum principles, the least-squaresmethod provides a variational framework [37,38]. The main ideas of the least-squares method are described using the steady Stokes flow problem.Consider the following vector form of equations governing the steady, slow

flow of a viscous incompressible fluid:

−µ∇ ·h(∇v) + (∇v)T

i+∇P − ρf = 0 (7.9.1)

∇ · v = 0 (7.9.2)

where v is the velocity vector, P is the pressure and f is the body forcevector. Choosing a suitable finite element approximation of (v,P ) and theirsubstitution into the governing equations results in residuals

R1 = −µ∇ ·h(∇v) + (∇v)T

i+∇P − ρf (7.9.3a)

R2 = ∇ · v (7.9.3b)


In the least-squares method, the sum of the squares of the residuals in thegoverning equations is minimized: minimize I, where

I(v, P ) ≡ 12

ZΩe

³R21 +R22

´=1

2

Ãk− µ∇ ·

h(∇v) + (∇v)T

i+∇P − ρf k20 + k∇ · v k20

!(7.9.4)

and k · k0 denotes the L2(Ωe)-norm of the enclosed quantity

kuk20 =ZΩe|u|2 dΩ (7.9.5)

The minimization problem δI = 0 is equivalent to the variational problem:find (v, P ) in a suitable vector space such that for all (w, Q) in the same vectorspace the following equation holds:

B((v, P ), (w, Q)) = `((w, Q)) (7.9.6)

where B(·, ·) is a symmetric bilinear form and `((w, Q)) is the linear form

B((v, P ), (w, Q))

=

ZΩe

n−µ∇ ·

h(∇v) + (∇v)T

i+∇P

o·n−µ∇ ·

h(∇w) + (∇w)T

i+∇Q

odΩ

+

ZΩe(∇ · v)(∇ ·w)dΩ (7.9.7a)

`((w, Q)) =

ZΩe

ρf ·n−µ∇ ·

h(∇w) + (∇w)T

i+∇Q

odΩ (7.9.7b)

The variational problem in (7.9.6) has some nice mathematical as wellcomputational properties. For example, the finite element model associatedwith (7.9.6) has a positive-definite coefficient matrix.Since the variational problem (7.9.6) is based on the differential equation

rather than the weak form, it involves the same order derivatives as thoseappearing in the governing equations, and the variational form does not includethe natural boundary terms. Hence, the approximation functions selectedmust be such that both natural and essential boundary conditions can beimposed. This requires the use of at least C1-continuous functions for thevelocity field v.In contrast to the weak form finite element models developed earlier, the

least-squares finite element model based on the variational problem (7.9.6) hashigher differentiability requirements which could be perceived as a practicaldisadvantage. A way to avoid the use of C1-continuous functions is to re-writethe governing equations as an equivalent set of first-order equations. The most


common transformation to an equivalent Þrst-order system is to introduce thevorticity vector. In two-dimensional problems, the total number of variables isincreased by one, but one has the beneÞt of directly solving for the vorticity.Yet another approach is to introduce the stresses as independent variables.In two-dimensional problems the total number of variables is increased bythree. A third option is to introduce all partial derivatives of the velocityvector Þeld as independent variables. In two-dimensional problems the totalnumber of variables is increased by four and one has the added beneÞt of easilycomputing physical quantities of interest (in the post-processing stage) thatare linear combinations of the partial derivatives of the velocity vector Þeld;that is, vorticity and stresses. For additional details, see Pontaza and Reddy[37,38].Here we present a formulation based on the Þrst-order system of equations

involving velocities, pressure, and vorticity. To write the second-orderequations in (7.9.1) in an equivalent set of Þrst-order equations, we introducethe vorticity vector

ω = ∇× v (7.9.8)

Making use of the vector identities

∇×∇×v = −∇2v+∇ (∇ · v) , ∇·h(∇v) + (∇v)T

i= ∇2v+∇ (∇ · v) (7.9.9)

and the incompressibility condition ∇ · v, Eqs. (7.9.1) and (7.9.2) can bereplaced by the following Þrst-order system:

µ∇× ω +∇P − ρf = 0 (7.9.10a)

∇ · v = 0 (7.9.10b)

ω −∇× v = 0 (7.9.10c)

The least-squares functional associated with the Þrst-order system(7.9.10ac) is given by

I(v, P,ω) =1

2

Ãkµ∇× ω +∇P − ρf k20 + k∇ · vk20 + kω −∇× vk20

!(7.9.11)

Like before, we can deÞne the discrete problem by minimizing the functionalin (7.9.11) with respect to the chosen approximating spaces. The minimumrequirement on approximation functions is that they all be Lagrange family offunctions (C0-continuity). Because the formulation is based on a variationalframework there are no compatibility restrictions between the velocity andpressure approximation spaces, so the same Lagrange basis can be used for allprimary variables (v, P,ω).



We will now develop the least-squares Þnite element model of the steady, two-dimensional ßows of viscous incompressible ßuids. The model is developed forthe vorticity based equivalent Þrst-order system, which in dimensionless formis written as

∂vx∂x

+∂vy∂y

= 0 (7.9.12)

vx∂vx∂x

+ vy∂vx∂y

+∂P

∂x+1

Re

∂ωz∂y

= fx (7.9.13)

vx∂vy∂x

+ vy∂vy∂y

+∂P

∂y− 1

Re

∂ωz∂x

= fy (7.9.14)

ωz +∂vx∂y

− ∂vy∂x

= 0 (7.9.15)

where Re = ρUL/µ is the Reynolds number. Note that for the two-dimensional case the other two component of the vorticity vector are identicallyzero, ω = (0, 0,ωz).To develop the least-squares Þnite element model we deÞne the least-

squares functional of the residuals over a typical element Ωe:

Ie =1

2

ZΩe

³R21 +R22 +R23 +R24

´dxdy (7.9.16)

where

R1 = ∂vx∂x

+∂vy∂y

R2 = vx∂vx∂x

+ vy∂vx∂y

+∂P

∂x+1

Re

∂ωz∂y

− fx

R3 = vx∂vy∂x

+ vy∂vy∂y

+∂P

∂y− 1

Re

∂ωz∂x

− fy

R4 = ωz + ∂vx∂y

− ∂vy∂x

(7.9.17)

and the primary variables (vx, vy, P,ωz) are approximated by expansions ofthe form

vx(x, y) =NXi=1

ψi(x, y)vix, vy(x, y) =

NXi=1

ψi(x, y)viy

P (x, y) =NXi=1

ψi(x, y)Pi, ωz(x, y) =NXi=1

ψi(x, y)ωiz

(7.9.18)

where ψi are the Lagrange family of interpolation functions, and (vix, v

iy, Pi,ω

iz)

are nodal values of (vx, vy, P,ωz).


In developing the finite element model, we assume that the convectiveterm in the residuals associated with the two momentum equations has beenlinearized; that is, vx and vy in the convective terms are evaluated from knownvalues of vix and v

iy (from the preceding iteration). In essence this amounts to

linearizing the least-squares functional prior to minimization.Minimizing the least-squares functional in Eq. (7.9.16) with respect to the

nodal values of velocities, pressure, and vorticity, we obtain

δIe =∂Ie

∂vxδvx +

∂Ie

∂vyδvy +

∂Ie

∂PδP +

∂Ie

∂ωzδωz = 0 (7.9.19)

which yields four sets of N equations each over a typical element:

∂Ie

∂vix= 0,

∂Ie

∂viy= 0,

∂Ie

∂Pi= 0,

∂Ie

∂ωiz= 0 (7.9.20)

for i = 1, 2, . . . , N . The resulting finite element equations are given by⎡⎢⎣ [S11 +S22] [S12 −S21] [0] [S20][S21 − S12] [S11 + S22] [0] − [S10]

[0] [0] [S11 + S22] 1Re [S

12 − S21][S02] − [S01] 1

Re [S21 −S12] 1

Re2 [S11 + S22] + [S00]

⎤⎥⎦⎧⎪⎨⎪⎩vxvyP ωz

⎫⎪⎬⎪⎭+

⎡⎢⎣[C00(v)] [0] [C01(v)] 1

Re [C02(v)]

[0] [C00(v)] [C02(v)] − 1Re [C

01(v)][C10(v)] [C20(v)] [0] [0]1Re [C

20(v)] − 1Re [C

10(v)] [0] [0]

⎤⎥⎦⎧⎪⎨⎪⎩vxvyP ωz

⎫⎪⎬⎪⎭=⎧⎪⎨⎪⎩F 1F 2F 3F 4

⎫⎪⎬⎪⎭(7.9.21)

where the coefficient matrices are defined by

C00ij (v) =

ZΩeCiCj dx dy, Ci = vx

∂ψi∂x

+ vy∂ψi∂y

C01ij (v) =

ZΩeCi∂ψj∂x

dx dy, C02ij (v) =

ZΩeCi∂ψj∂y

dx dy

C10ij (v) =

ZΩe

∂ψi∂xCj dx dy, C20ij (v) =

ZΩe

∂ψi∂yCj dx dy

S00ij =

ZΩe

ψiψj dx dy

S01ij =

ZΩe

ψi∂ψj∂x

dxdy, S02ij =

ZΩe

ψi∂ψj∂y

dxdy

S10ij =

ZΩe

∂ψi∂x

ψj dxdy, S20ij =

ZΩe

∂ψi∂y

ψj dxdy

S11ij =

ZΩe

∂ψi∂x

∂ψj∂x

dx dy, S22ij =

ZΩe

∂ψi∂y

∂ψj∂y

dx dy


S11ij =

ZΩe

∂ψi∂x

∂ψj∂x

dxdy, S22ij =

ZΩe

∂ψi∂y

∂ψj∂y

dx dy

S12ij =

ZΩe

∂ψi∂x

∂ψj∂y

dx dy, S21ij =

ZΩe

∂ψi∂y

∂ψj∂x

dxdy

F 1i =

ZΩeCi fx dxdy, F 2i =

ZΩeCi fy dx dy

F 3i =

ZΩe

µ∂ψi∂xfx +

∂ψi∂yfy

¶dx dy

F 4i =

ZΩe

1

Re

µ∂ψi∂yfx − ∂ψi

∂xfy

¶dx dy (7.9.22)

Note that Eq. (7.9.21) cannot be solved until they are assembled and boundaryconditions are imposed.

7.9.3 Computational Aspects

1. Inspection of the structure of the Þnite element equations in (7.9.21) revealsthat the system is symmetric and positive-deÞnite, even in the convectiondominated limit 1/Re → 0. In contrast with the mixed Þnite elementmodel, where the resulting system of discrete equations is unsymmetricand indeÞnite (zeroes along the diagonal), the least-squares Þnite elementmodel offers great advantages from a computational point of view.

2. The symmetric positive-deÞniteness property allows the use of robustiterative methods for the solution of the discrete system of equations.Iterative solution techniques such as preconditioned conjugate gradientmethods can be implemented without the need of global assembly. Largescale problems can be solved using element-by-element solution proceduresin a fully parallel environment.

In the context of least-squares Þnite element models for the incompressibleNavierStokes equations, predominantly low-order expansions, i.e. linearor quadratic Lagrange functions, have been used. Although not commonlyemphasized, low-order approximations tend to lock, and reduced integrationtechniques must be used to obtain acceptable numerical results. When enoughredundant degrees of freedom are constrained, the least-squares Þnite elementmodel using reduced integration yields a collocation Þnite element model.However, the collocation Þnite element model may not always be reliableand the least-squares functional cannot be used to measure the quality ofthe solution. Moreover, the collocation solution may not be smooth at thenodes and post-processing is needed to recover nodal values from the reducedintegration points.


Appropriate minimization of the least-squares functional is done usingfull integration and p-reÞnement (see Pontaza and Reddy [37,38] andWinterscheidt and Surana [39]). The quality of the numerical solution may bejudged by the value of the least-squares functional, which decays exponentiallyfast as the expansion order of the basis is increased (see Pontaza and Reddy[38]). Commonly used elements suited for p-reÞnement are either of the nodalor modal type. A nodal expansion is of the Lagrange type; when the nodespacing is chosen such that the nodes coincide with the location of the rootsof a Jacobi polynomial the basis is known as a spectral basis. Modal basisare nodeless expansions whose coefficients are associated with modes of ahierarchical basis. Multi-dimensional nodal and modal basis can be easilyconstructed by taking tensor product of the one-dimensional basis. Details onthe construction of both the nodal and modal expansions can be found in thework of Warburton et al. [40].

7.9.4 Numerical Examples

Kovasznay ßow

We consider two-dimensional, steady ßow in Ω = [−0.5, 1.5]× [−0.5, 1.5]. Weuse Kovasznays exact solution to the stationary incompressible NavierStokesequations to verify exponentially fast decay of the L2 least-squares functionaland L2 error norms. The solution is given by

vx = 1− eλx cos(2πy), vy =λ

2πeλx sin(2πy), P =

1

2

³1− e2λx

´(7.9.23)

where λ = Re/2−(Re2/4+4π2)1/2. Figure 7.9.1(a) shows vx-velocity contoursof the exact solution for Re = 40 and Figure 7.9.1(b) shows the discretizationof the domain using a non-uniform mesh of 8 quadrilateral Þnite elements.

The exact solution is used to compute the velocity boundary conditions onΓ and pressure is speciÞed at a point. No boundary conditions on vorticityare necessary. The resulting discrete system is solved using Newtons methodwith Cholesky factorization at each Newton step. Convergence is declaredwhen the normalized norm of the residual in velocities, k∆vk/kvk, was lessthan 10−4, which typically required 5 Newton iterations. A plot of the L2least-squares functional and L2 error of the velocity and vorticity Þelds as afunction of the expansion order in a logarithmic-linear scale is shown in Figure7.9.2. Exponentially fast decay of the L2 least-squares functional and L2 erroris observed.

-0.5 0.0 0.5 1.0 1.5

-0.5

0.0

0.5

1.0

1.5

x

y

( b )-0.5 0.0 0.5 1.0 1.5

-0.5

0.0

0.5

1.0

1.5

x

y

( a )

Expansion order, P

L2

norm

2 3 4 5 6 7 8 9 1010-18

10-16

10-14

10-12

10-10

10-8

10-6

10-4

10-2

100

LS functional

|| vx - v px ||

|| vy - v py ||

|| ωz - ω pz ||


Figure 7.9.1 Kovasznay ßow: (a) vx-velocity component contours of theexact solution for Re = 40; (b) computational domain using8 quadrilateral elements.

Figure 7.9.2 Decay of the least-squares functional and convergence of thevelocity and vorticity Þelds to the exact Kovasznay solution.


Flow over a backwardfacing step

We consider two-dimensional, steady ßow over a backwardfacing step atRe = 800. The geometry and boundary conditions are taken from thebenchmark solution of Gartling [35] and are shown in Figure 7.8.16. As shownin Figure 7.8.16 the standard step geometry was simpliÞed by excluding thechannel portion upstream of the step. The boundary conditions for the stepgeometry include the no-slip condition at all solid surfaces and a parabolicinlet velocity proÞle given by vx(y) = 24y(0.5 − y) for 0 ≤ y ≤ 0.5. TheReynolds number is based on the mean inlet velocity.

Instead of imposing an outßow boundary condition in a strong sense weimpose it in a weak sense through the least-squares functional. For example,if we use the vorticity based Þrst-order system the L2 least-squares functionalis given by

I(v, P,ω) =1

2

µk (v ·∇)v+∇P + 1

Re∇× ω − f k20 + kω −∇× vk20

+ k∇ · vk20 + kn · σ − f sk20,Γoutßow¶

(7.9.24)

where σ is a pseudo-stress (see Gresho [35]), σ = −P I + (1/Re)∇v, and f sare the prescribed pseudo-tractions, typically taken to be zero at an outßowboundary.

The domain, Ω = [0, 30]× [−0.5, 0.5], is discretized using 20 Þnite elements:two elements along the height of the channel and 10 uniform elements alongthe length of the channel. The numerical simulation is performed usingthe two-dimensional incompressible NavierStokes equations in the velocitygradient based Þrst-order form (see Pontaza and Reddy [38]). A 11th ordermodal expansion is used in each element and the resulting discrete systemis solved using Newtons method. At each Newton step, the linear systemof algebraic equations is solved using the conjugate gradient method witha symmetric GaussSeidel preconditioner. Convergence of the conjugategradient method was declared when the norm of the residual was less than10−6. Nonlinear convergence was declared when the normalized norm of theresidual in velocities, k∆vk/kvk, was less than 10−4, which typically requiredfour Newton iterations. The analysis starts with Re = 100 and steps toRe = 800 using a solution continuation technique with increments of Re = 100.Away from the corner of the step at (x, y) = (0, 0), the L2 least-squaresfunctional remained below 10−5 through the Reynolds number stepping.Figure 7.9.3 shows the streamlines, the vector velocity Þeld, and pressure

contours for 0 ≤ x ≤ 10, where most of the interesting ßow structures occur.The ßow separates at the step corner and forms a large recirculation regionwith a reattachment point on the lower wall of the channel at approximatelyx = 6. A second recirculation region forms on the upper wall of the channelbeginning near x = 5 with a reattachment point at approximately x = 10.5.

0.0 1.0 2.0 3.0 4.0 5.0 6.0 7.0 8.0 9.0 10.0-1.0

-0.5

0.0

0.5

1.0(a)

(b)

(c)


Figure 7.9.3 Flow over a backwardfacing step at Re = 800: (a)streamlines, (b) vector velocity Þeld, and (c) pressure Þeld.

Figure 7.9.4 shows vx-velocity proÞles along the channel height at x = 7and x = 15. We compare with tabulated values from the benchmark solutionof Gartling [35] and Þnd excellent agreement. Gartlings benchmark solutionis based on a mixed Galerkin formulation using grid systems ranging from6× 120 to 40× 800 biquadratic elements. Figure 7.9.5 shows pressure proÞlesalong the length of the channel walls. The slopes of the pressure proÞlesbecome constant near the exit plane, meaning that the ßow has recovered tofully developed conditions at the exit.

Flow past a circular cylinder at low Reynolds number

The last example deals with the two-dimensional ßow of an incompressibleßuid past a circular cylinder [37]. The cylinder is of unit diameter and isplaced in the Þnite region Ω = [−15.5, 30.5] × [−20.5, 20.5]. The center ofthe cylinder lies at (x, y) = (0, 0), so that the inßow boundary is locatedat 10.5 cylinder diameters left (or in front) of the center of the cylinderand the outßow boundary is at 30.5 cylinder diameters downstream of thecenter of the cylinder. The top and bottom boundaries are located each at20.5 cylinder diameters above and below the center of the cylinder. Havingconsidered a large computational domain allows us to impose free-streamboundary conditions at the top and bottom of the domain without noticeablyaffecting the solution.

x

Pres

sure

0 5 10 15 20 25 30-0.20

-0.15

-0.10

-0.05

0.00

0.05

0.10

Lower wall

Upper wall

vx

y

-0.2 0.0 0.2 0.4 0.6 0.8 1.0 1.2-0.50

-0.25

0.00

0.25

0.50

x = 7

x = 15

Gartling


Figure 7.9.4 Flow over a backwardfacing step at Re = 800: Horizontalvelocity proÞles along the height of the channel at x = 7and x = 15. Comparison with the benchmark solution ofGartling [35].

Figure 7.9.5 Flow over a backwardfacing step at Re = 800: PressureproÞles along lower and upper walls of the channel.

x

y

-10 0 10 20 30

-20

-10

0

10

20

(a) (b)


The boundary conditions include a speciÞed value of 1.0 for the x-component of velocity at the inßow, top, and bottom boundaries; that is,the free-stream velocity u∞ is speciÞed to be unity. At these boundaries they-component of velocity is set to zero. The outßow boundary conditions areimposed in a weak sense through the least-squares functional. The Reynoldsnumbers considered here are 20 and 40, for which a steady-state solutionexists. The Reynolds number is based on the free-stream velocity and cylinderdiameter.

The vorticity based Þnite element model in Eq. (7.9.21) is used with sixth-order nodal expansions in each element. The Þnite element mesh consists of501 Þnite elements (see Figure 7.9.6), where a close-up view of the geometricdiscretization around the circular cylinder is also shown. To accuratelyrepresent the circular arc, the same approximation for the geometry and thesolution (i.e. isoparametric formulation) is used. The total number of degreesof freedom for the problem is 73,344. The storage of the assembled systemof equations in banded or in compressed sparse row/column format for suchlarge size problems is prohibitively expensive in terms of computer memory.Therefore matrix-free techniques, also known as element-by-element solutionalgorithms, are implemented in a matrix-free version of the conjugate gradientmethod with a Jacobi preconditioner.

Figure 7.9.6 Computational domain and mesh for ßow past a circularcylinder. (a) Computational mesh. (b) Close-up view of thegeometric discretization around the circular cylinder.

θ (degrees)

Cp

0 30 60 90 120 150 180-1.5

-1.0

-0.5

0.0

0.5

1.0

1.5

Re = 20

Re = 40

Grove et al. (Re = 40)

LE TE


At each Newton step the linear system of equations is solved using thematrix-free conjugate gradient algorithm with a Jacobi preconditioner andconvergence tolerance for the norm of the residual to be 10−6. Nonlinearconvergence is declared when the relative norm of the residual in velocitiesbetween two consecutive iterations was less than 10−4, which required lessthan six Newton iterations.

Figure 7.9.7 shows the computed surface pressure coefficient distributionsalong the cylinder surface for Re = 20 and 40, together with experimentalmeasurements of Grove et al. [41] for Re = 40. The Þnite element resultsare in good agreement with the experimental measurements. The computeddrag coefficients for Re = 20 and Re = 40 are CD = 2.0862 and CD = 1.5537,respectively. Good agreement is found between the computed drag coefficientsand the experimental mean curve of Tritton [42], where the correspondingvalues are CD = 2.05 and CD = 1.56.

Figure 7.9.8 shows computed pressure contours and streamlines in the wakeregion for Re = 20 and Re = 40. The predicted wake extends 1.86 and 4.55cylinder radii measured from the back of the cylinder. The values for thewake lengths are in good agreement with the numerical solution of Dennis andChang [43], whose computed wake lengths for Re = 20 and 40 were reportedas 1.88 and 4.69 cylinder radii, respectively. Better agreement for the caseRe = 40 is found with the numerical solution of Kawaguti and Jain [44], whoreported a computed wake length of 4.50 cylinder radii.

Figure 7.9.7 Flow past a circular cylinder: comparison of the computedpressure coefficient distributions along the cylinder surfacewith experimental results of Grove et al. [41] for Re = 40.

x

y

-2 -1 0 1 2 3

-2

-1

0

1

2

p0.600

0.525

0.450

0.375

0.300

0.225

0.150

0.075

-0.000

-0.075

-0.150

-0.225

-0.300

-0.375

-0.450

(b)

x

y

-2 -1 0 1 2 3

-2

-1

0

1

2

p0.675

0.591

0.507

0.423

0.339

0.255

0.171

0.087

0.004

-0.080

-0.164

-0.248

-0.332

-0.416

-0.500

(a)


Figure 7.9.8 Flow past a circular cylinder at (a) Re = 20 and (b) Re = 40:pressure contours and streamlines in the wake region.


Problems7.1 Consider the vector equations (7.3.1) and (7.3.2). Develop the weak statements of the

equations in vector form.7.2 Consider equations (7.4.1)(7.4.3) in cylindrical coordinates (r, θ, z). For axisymmetric

viscous incompressible ßows (i.e. the ßow Þeld is independent of the θ coordinate),and when the convective (nonlinear) terms are neglected, we have

ρ∂vx∂t

=1

r

∂

∂r(rσr)− σθ

r+∂σrz∂z

+ fr (i)

ρ∂vz∂t

=1

r

∂

∂r(rσrz) +

∂σzz∂z

+ fz (ii)

1

r

∂

∂r(rvx) +

∂vz∂z

= 0 (iii)

where

σr = −P + 2µ∂vx∂r

, σθ = −P + 2µvxr

σz = −P + 2µ∂vz∂z

, σrz = µ(∂vx∂z

+∂vz∂r)

Develop the semidiscrete mixed Þnite element model of the equations.7.3 Develop the semidiscrete penalty Þnite element model of the equations in Problem 7.2.7.4 The equations governing unsteady slow ßow (i.e. Stokes ßow) of viscous incompressible

ßuids in the x-y plane can be expressed in terms of vorticity ω and stream functionψ :

ρ∂ω

∂t− µ∇2ω = 0

−ω −∇2ψ = 0Develop the semidiscrete Þnite element model of the equations. Discuss the meaningof the secondary variables. Use the α-family of approximation to reduce the ordinarydifferential equations to algebraic equations.

7.5 Compute the tangent coefficient matrix for the penalty Þnite element model in equation(7.5.16).

7.6 Verify Eqs. (7.9.10)(7.9.12).7.7 Develop the least-squares Þnite element model with velocity Þeld and pressure as

variables of the Navier-Stokes equations governing axisymmetric ßows (see Problem7.2).

7.8 Develop the least-squares Þnite element model with velocity Þeld, pressure, andvorticity as variables of the Stokes equations governing axisymmetric ßows.

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42. Tritton, D. J., “Experiments on the Flow Past a Circular Cylinder at Low ReynoldsNumbers,” Journal of Fluid Mechanics, 6, 547—567 (1959).

43. Dennis, S. C. R. and Chang, G. Z., “Numerical Solutions for Steady Flow Past a CircularCylinder at Reynolds Numbers up to 100,” Journal of Fluid Mechanics, 42, 471—489(1970).

44. Kawaguti, M. and Jain, P., “Numerical Study of a Viscous Fluid Past a CircularCylinder,” Journal of the Physical Society of Japan, 21, 2055—2063 (1966).


8

Nonlinear Analysis ofTime-Dependent Problems

8.1 Introduction

In this chapter, we develop the Þnite element models of time-dependentproblems with nonlinearities and describe some standard time approximationschemes. All classes of problems discussed in the previous chapters will berevisited in the context of transient analysis. We begin with the generaldiscussion of the Þnite element modeling of time-dependent problems.The Þnite element formulation of time-dependent problems involves

following two stages:

1. Spatial approximation, where the solution u(x, t) of the equation underconsideration is approximated by expressions of the form

u(x, t) ≈ Ue(x, t) =nXj=1

uej(t)ψej (x) (8.1.1)

and the spatial Þnite element model of the equation is developed usingthe procedures of static or steady-state problems, while carrying all time-dependent terms in the formulation. This step results in a set of ordinarydifferential equations (i.e. a semidiscrete system of differential equations)in time for the nodal variables uej(t) of the element. Equation (8.1.1)represents the spatial approximation of u for any time t. When the solutionis separable into functions of time only and space only, u(x, t) = T (t)X(x),the approximation (8.1.1) is justiÞed for the overall transient response of astructure, in contrast to wave propagation type solutions.

2. Temporal approximation, where the system of ordinary differentialequations in time are further approximated in time, often using Þnitedifference formulae for the time derivatives. This step allows conversion ofthe system of ordinary differential equations into a set of algebraic equationsamong uej at time ts+1 = (s+ 1)∆t, where ∆t is the time increment and sis an integer.


At the end of the Þrst step (i.e. after spatial approximation using the Þniteelement method), we obtain, in general, a matrix differential equation of theform

[Ce] úue+ [Me]ue+ [Ke]ue = F e (8.1.2)

at the element level, which represents a system of ordinary differentialequations in time. Here u represents a m × 1 vector of nodal values, and[Ce], [Me], and [Ke] are m×m matrices and F e is m×1 vector, m being thenumber of nodal degrees of freedom per Þnite element. The matrices appearingin Eq. (8.1.2) may be functions of the unknown u(x, t), making (8.1.2) a setof nonlinear differential equations. Next, using a time approximation scheme,Eq. (8.1.2) is reduced to a set of nonlinear algebraic equations, as will beshown in the sequel, of the form

[ Ke]s+1ues+1 = F es,s+1 (8.1.3)

where [ Ke] and F e are known in terms of [Ce], [Me], [Ke], F e, ues,and úues. The subscript s + 1 refers to the time, ts+1, at which thesolution is sought. Equation (8.1.3) is then assembled and solved using knownboundary conditions and initial conditions. Thus, at the end of the two-stageapproximation, one has a continuous spatial solution at discrete intervals oftime:

ue(x, ts) =nXj=1

uej(ts)ψej (x) (s = 0, 1, . . .) (8.1.4)

In the next section, we study time approximation schemes using Eq. (8.1.2).We discuss stability and accuracy of time approximation schemes in Section8.3. The two-step procedure is illustrated for all major problems of theprevious sections, namely, nonlinear heat conduction (or Þeld problems ofthat type), bending of plates, and viscous incompressible ßows. Of course, theprocedure described here can be applied to other problems.

8.2 Time Approximations

8.2.1 Introduction

All time approximation schemes are broadly classiÞed as implicit and explicit.In explicit schemes, we Þnd uj at time ts+1 using the known value of uj attime ts. The implicit schemes are based on Þnding uj at time ts+1 using notonly the known value of uj at time ts but also values at ts+1. Explicit schemesare conditionally stable, that is, the time step size is limited approximately tothe time taken for an elastic wave to cross the smallest element dimension inthe mesh. Implicit schemes have no such restriction and the time steps usedin these schemes can be one or two orders of magnitude larger than the timesteps used in simple explicit schemes. However, the accuracy of the implicit

TIME-DEPENDENT PROBLEMS 289

schemes deteriorates as the time step size increases relative to the period ofresponse of the system. The economy of the two schemes depends on (1) thestability limit of the explicit scheme, (2) the cost of the implicit scheme, (3)the relative size of time increments that can yield acceptable accuracy withthe implicit scheme compared to the stability limit of the explicit scheme, and(4) the size of the computational model.In this section, we consider time approximation schemes to reduce Eq.

(8.1.2) to Eq. (8.1.3). The case of [M ] = [0] arises in heat transferand ßuid dynamics, and the equation is known as the parabolic equation.Equation (8.1.2), in its general form is known as an hyperbolic equation, andit arises in structural dynamics with damping ([C] 6= [0]) and without damping([C] = [0]); [C] denotes the damping matrix and [M ] the mass matrix. Thetime approximation of (8.1.2) for parabolic and hyperbolic equations will bederived separately. Equation of the form (8.1.2) is also obtained by otherspatial approximations methods like the Þnite difference method and boundaryelement method, among others. Therefore, the discussion of converting matrixequations of the type (8.1.2) is equally valid for methods used to approximatethe spatial variation of the solution.

8.2.2 Parabolic Equations

Consider the parabolic equation [i.e. set [M ] equal to zero in Eq. (8.1.2)]

[Ce] úue+ [Ke]ue = F e (8.2.1a)

which arises in heat transfer and ßuid dynamics problems. The global solutionvector is subject to the initial condition

u(0) = u0 (8.2.1b)

where u0 denotes the value of the enclosed quantity u at time t = 0.The eigenvalue problem associated with Eq. (8.2.1a) is obtained by

assuming the solution u(t) to decay with time

u = u0e−λt, F = F 0e−λt (8.2.2a)

where u0 is the vector of amplitudes (independent of time) and λ is theeigenvalue. Substitution of Eq. (8.2.2a) into Eq. (8.2.1a) gives

[−λ[C] + [K]) u0 = F 0 (8.2.2b)

The most commonly used method for solving (8.2.1a,b) is the α-familyof approximation, in which a weighted average of the time derivative of adependent variable is approximated at two consecutive time steps by linearinterpolation of the values of the variable at the two steps:


(1− α) úus + α úus+1 ≈ us+1 − usts+1 − ts for 0 ≤ α ≤ 1 (8.2.3)

where s, for example, refers to the value of the enclosed quantity at timet = ts. In the interest of brevity, the element label e on various quantitiesis omitted (i.e. time approximation scheme is used at the element level).Equation (8.2.3) can be expressed alternatively as

us+1 = us +∆t úus+α úus+α = (1− α) úus + α úus+1 for 0 ≤ α ≤ 1 (8.2.4)

where ∆t = ts+1 − ts.Note that when α = 0, Eq. (8.2.3) reduces, for any nodal value ui(t), toµ

duidt

¶t=ts

≈ ui(ts+1)− ui(ts)ts+1 − ts (8.2.5)

Clearly, it amounts to replacing the time derivative of ui(t) at t = ts with theÞnite difference of its values at t = ts+1 (i.e. value from a time step ahead)and t = ts. Equation (8.2.5) is nothing but an approximation of the derivativeof a function, since ∆t = ts+1 − ts is Þnite, i.e. not make it approach zero.The approximation in Eq. (8.2.5) is known as the forward difference becauseit uses the function value ahead of the current position in computing the slope(see Figure 8.2.1). One may also use the value of the function from a timestep behind µ

duidt

¶t=ts

≈ ui(ts)− ui(ts−1)ts − ts−1 (8.2.6a)

or µduidt

¶t=ts+1

≈ ui(ts+1)− ui(ts)ts+1 − ts (8.2.6b)

which is the same as that in Eq. (8.2.3) when α = 1. Equation (8.2.6b) isknown as the backward difference. If we use the values ahead and behind incomputing the slope µ

duidt

¶t=ts

≈ ui(ts+1)− ui(ts−1)ts+1 − ts−1 (8.2.7)

it is called the centered difference, which is not a special case of Eq. (8.2.3).

Actual tangent line at t = ts

ts−1 ts ts+1t0 tNt

u(t)

us−1 us us+1

Actual slope,dudt

u(ts−1) = us−1

u(ts) = us

u(ts+1) = us+1


Figure 8.2.1 Approximation of the Þrst derivative of a function.

For different values of α, we obtain the following well-known numericalintegration schemes from Eq. (8.2.3):

α =

⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩0, the forward difference scheme (conditionally stable);

Order of accuracy = O(∆t)12 , the CrankNicolson scheme (stable); O((∆t)2)23 , the Galerkin method (stable); O((∆t)2)1, the backward difference scheme (stable); O(∆t)

(8.2.8)

The phrases stability and conditional stability will be discussed in Section8.3.Equation (8.2.4) can be used to reduce ordinary differential equations

(8.2.1a,b) to algebraic equations among the uj at time ts+1. Assuming that[C] is independent of time t, we obtain (see Problem 8.1)

[ K]s+1us+1 = Fs,s+1 (8.2.9)

where

[ K]s+1 = [C] + a1[K]s+1, [K]s = [C]− a2[K]s (8.2.10a)

Fs,s+1 = [K]sus + a1Fs+1 + a2Fs (8.2.10b)

a1 = α∆t, a2 = (1− α)∆t (8.2.10c)

Equation (8.2.9) provides a means to compute for us+1 whenever us isknown. Of course, [C], [K], and F are known for all times (parts of F


may not be known at the element level, but after assembly they are knownwhenever the corresponding u are unknown).Equations (8.2.9) is valid for a typical element. The assembly, imposition

of boundary conditions, and solution of the assembled equations are the sameas that used for static (or steady-state) problems. Calculation of [ K] and Fat time t = 0 requires knowledge of the initial conditions u0 and the timedependency of [C], [K], and F.Note that for α = 0 (the forward difference scheme), Eq. (8.2.10a) gives

[ Ke] = [Ce]. When the matrix [Ce] is diagonal, Eq. (8.2.9) becomes explicit

in the sense that one can solve for us+1 directly without inverting [ K].However, in spatial approximation by the Þnite element method, [Ce] is derivedusing a weak form, and it is never a diagonal matrix. The matrix [Ce]derived using a weak form is called the consistent matrix. Thus, the Þniteelement equations with consistent (mass) matrix [Ce] never are explicit (inthe sense that no inversion of the coefficient matrix is required). In a Þnitedifference method, the matrix [Ce] is diagonal, and therefore an explicit timeintegration scheme (such as the forward difference method) results in explicitset of equations, which are quite inexpensive to solve at each time step. Tohave the advantage of less computational time in dynamic/transient analysesby the Þnite element method, it is desirable to have [Ce] diagonalized (see

Problem 8.8). Thus, explicit (in the sense that [ K] is diagonal) Þnite elementequations can be obtained only when (a) the time approximation scheme isexplicit, and (b) the mass matrix [C] is diagonal.

8.2.3 Hyperbolic Equations

Consider the second-order equation

[Me]ue+ [Ce] úue+ [Ke]ue = F e (8.2.11)

which arises in structural dynamics; [Ce] denotes the damping matrix, [Me]the mass matrix, and [Ke] the stiffness matrix. The global displacement vectoru is subject to the initial conditions that the displacement and velocity areknown at time t = 0

u(0) = u0, úu(0) = v0 (8.2.12)

There are several numerical integration methods available to integrate second-order (i.e. hyperbolic) equations [2,4,5]. Among these, the Newmark familyof time integration schemes [4] is widely used in structural dynamics. Othermethods, such as the Wilson method and the Houbolt method [5], can be usedto develop the algebraic equations from the second-order differential equations(8.2.11).


In the Newmark method, the function and its time derivatives areapproximated according to

us+1 = us +∆t úus + 12(∆t)2us+γ (8.2.13a)

úus+1 = úus + us+α∆t (8.2.13b)

us+α = (1− α)us + αus+1 (8.2.13c)

and α and γ are parameters that determine the stability and accuracy of thescheme. For α = 0.5, the following values of γ deÞne various well-knownschemes:

γ =

⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩

12 , the constantaverage acceleration method (stable)13 , the linear acceleration method (conditionally stable)0, the central difference method (conditionally stable)85 , the Galerkin method (stable)2, the backward difference method (stable)

(8.2.14)

The set of ordinary differential equations in (8.2.11) can be reduced, withthe help of Eqs. (8.2.13ac), to a set of algebraic equations relating us+1 tous (see Problem 8.2). We have

[ K]s+1us+1 = Fs,s+1 (8.2.15)

where

[ K]s+1 = [K]s+1 + a3[M ]s+1 + a6[C]s+1

Fs,s+1 = Fs+1 + [M ]s+1As + [C]s+1Bs (8.2.16)

As = a3us + a4 úus + a5us, Bs = a6us + a7 úus + a8us

and ai, i = 1, 2, . . . , 8, are deÞned as

a1 = α∆t, a2 = (1− α)∆t (8.2.17a)

a3 =1

β(∆t)2, a4 = a3∆t, a5 =

1

γ− 1, γ = 2β (8.2.17b)

a6 =α

β∆t, a7 =

α

β− 1, a8 = ∆t

µα

γ− 1

¶(8.2.17c)

The following remarks concerning the Newmark scheme are in order:

1. The calculation of [ K] and F in Newmarks scheme requires knowledgeof the initial conditions u0, úu0, and u0. In practice, one does notknow u0. As an approximation, it can be calculated from the assembled


system of equations associated with (8.2.15) using initial conditions on u, úu, and F (often F is assumed to be zero at t = 0):

u0 = [M ]−1 (F0 − [K]u0 − [C] úu) (8.2.18)

2. At the end of each time step, the new velocity vector úus+1 andacceleration vector us+1 are computed using the equations

us+1 = a3(us+1 − us)− a4 úus − a5us úus+1 = úus + a2us + a1us+1 (8.2.19)

where a1 and a2 are deÞned in Eq. (8.2.17a).

3. Equation (8.2.15) is not valid for the centered difference scheme (γ = 2β =0), as some of the parameters ai are not deÞned for this scheme. Analternative algebraic manipulation of the equations is required. It can beshown that (see Problem 8.7)

[H]s+1us+1 = Fs+1 − [K]s+1As − [C]s+1Bs (8.2.20)

[H]s+1 = β(∆t)2[K]s+1 + α∆t[C]s+1 + [M ]s+1

As = us +∆t úus + 1− γ2(∆t)2us (8.2.21)

Bs = úus + (1− α)∆tusand the displacements and velocities are updated using the relations

us+1 = us +∆t úus + (∆t)2

2[(1− γ)us + γus+1] (8.2.22)

úus+1 = úus +∆t [(1− α)us + αus+1] (8.2.23)

4. The centered difference scheme (γ = 2β = 0) with α = 0 yields [see Eqs.(8.2.20) and (8.2.21)]

[M ]s+1us+1 = Fs+1 − [K]s+1µus +∆t úus + 1

2(∆t)2us

¶− [C]s+1

µ úus +∆tus

¶(8.2.24)

Thus, if the mass matrix is diagonalized, the system in Eq. (8.2.24) becomesexplicit (no inversion of the coefficient matrix is required).

5. For natural vibration, the forces and the solution is assumed to be periodic

u = u0eiωt, F = F 0eiωt, i =√−1 (8.2.25)


where u0 is the vector of amplitudes (independent of time) and ω is thefrequency of natural vibration of the system. Substitution of Eq. (8.2.25)into Eq. (8.2.11) yieldsh

−ω2[M ] + iω[C] + [K]ú0 = F 0 (8.2.26)

Equation (8.2.26) is called an eigenvalue problem, which may have complexeigenvalues when damping [C] is included.

8.3 Stability and Accuracy


In general, the application of a time approximation scheme to an initial-valueproblem results in equation of the type

[ K]us+1 = [K]us or us+1 = [A]us (8.3.1)

where [A] = [ K]−1[K] is called the ampliÞcation matrix, and [ K] and [K]are matrix operators that depend on the problem parameters, for example,geometric and material properties and Þnite element mesh parameter, andus+1 is the solution vector at time ts+1.Since Eq. (8.2.3), for example, represents an approximation that is used

to derive equation of the type (8.3.1), error is introduced into the solutionus+1 at each time step. Since the solution us+1 at time ts+1 dependson the solution us at time ts, the error can grow with time. The timeapproximation scheme is said to be stable if the error introduced in us doesnot grow unbounded as Eq. (8.3.1) is solved repeatedly for s = 0, 1, . . .. Inorder that the error to remain bounded, it is necessary and sufficient that thelargest eigenvalue of the ampliÞcation matrix [A] is less than or equal to unity:

|λmax| ≤ 1 (8.3.2)

where λmax is the largest value that satisÞes the equation³[A]− λA[I]

ú = 0 (8.3.3)

Equation (8.3.3) represents an eigenvalue problem. If condition (8.3.2) issatisÞed for any value of ∆t, the scheme is said to be unconditionally stable orsimply stable. If Eq. (8.3.2) places a restriction on the size of the time step∆t, the scheme is said to be conditionally stable.Accuracy of a numerical scheme is a measure of the closeness between the

approximate solution and the exact solution, whereas stability of a solutionis a measure of the boundedness of the approximate solution with time. As


one might expect, the size of the time step can inßuence both accuracy andstability. When we construct an approximate solution, we like it to converge tothe true solution when the number of elements or the degree of approximationis increased and the time step∆t is decreased. A time approximation scheme issaid to be convergent if, for Þxed ts and∆t, the numerical value us convergesto its true value u(ts) as ∆t→ 0. Accuracy is measured in terms of the rateat which the approximate solution converges. If a numerical scheme is stableand consistent, it is also convergent.

8.3.2 Stability Criteria

The α-family of approximations is stable for all numerical schemes in whichα < 1

2 , only if the time step satisÞes the following (stability) condition:

∆t < ∆tcr ≡ 2

(1− 2α)λmax (8.3.4)

where λmax is the largest eigenvalue of the Þnite element equations (8.2.2b):

([Ke]− λ[Ce]) u0e = F 0e (8.3.5a)

Note that the same mesh as that used for the transient analysis must be usedto calculate the eigenvalues of the assembled system (8.3.5a); after assemblyand imposition of boundary conditions, the eigenvalue problem in Eq. (8.3.5a)becomes homogeneous

([K]− λ[C]) U0 = 0 (8.3.5b)

The stability criterion in Eq. (8.3.4) is arrived using Eq. (8.3.2). TheampliÞcation matrix for the α-family of approximations is given by

[A] = [ K]−1[K] = ([C] + a1[K])−1([C]− a2[K]) (8.3.6)

Let λmax be the maximum eigenvalue of Eq. (8.3.5b). Then it can be shownthat (using spectral decomposition of [A]) the maximum eigenvalue of [A] isequal to

λAmax =

¯1− (1− α)∆tλmax1 + α∆tλmax

¯≤ 1 (8.3.7)

from which it follows that the α-family of approximations is unconditionallystable if α ≥ 1

2 . In the case α <12 , the method is stable only if condition

(8.3.4) is satisÞed.Similarly, all Newmark schemes in which γ < α and α ≥ 1

2 , the stabilityrequirement is

∆t ≤ ∆tcr =∙1

2ω2max(α− γ)

¸−1/2(8.3.6)


where ωmax is the maximum natural frequency of the undamped system(8.2.26) ³

[K]− ω2[M ]´∆ = 0 (8.3.7)

8.4 Transient Analysis of Nonlinear Problems

8.4.1 Introduction

Here we discuss time approximations of the problems considered in heattransfer, ßuid mechanics, and solid mechanics with nonlinearities. Thediscussion mainly focused on developing the fully discretized Þnite elementequations for transient response. It should be noted that the stability criteriadiscussed in the previous sections for conditionally stable schemes are validonly for linear problems, and no such estimates are available for nonlinearproblems.

8.4.2 Heat Transfer

Consider the model equation (4.2.1). For time-dependent problems, it takesthe form

c0∂u

∂t− ∂

∂x

µa11∂u

∂x

¶− ∂

∂y

µa22∂u

∂y

¶+ a00u = f(x, y, t) (8.4.1)

where aij are, in general, functions of position and time. In addition, aij(i, j = 1, 2) are functions of u, ∂u/∂x, and ∂u/∂y. However, we assume thatc0 is only a function of position but not time. The semidiscretization followsthe same steps as in the steady-state case (see Chapter 4). The weak form ofEq. (8.4.1) is given by

0 =

ZΩe

µc0w

∂u

∂t+ a11

∂w

∂x

∂u

∂x+ a22

∂w

∂y

∂u

∂y+ a00wu−wf

¶dx dy −

IΓewqn ds

(8.4.2)The Þnite element approximation is assumed to be of the form

u(x, y, t) ≈ ueh(x, y, t) =nXj=1

uej(t)ψej (x, y) (8.4.3)

where the nodal values uej are now assumed to be functions of time.Substitution of Eq. (8.4.3) into Eq. (8.4.2) gives rise to the Þnite elementequations

[Ce] úue+ [Ke]ue = fe+ Qe (8.4.4)


where

Ceij =

ZΩec0ψ

eiψ

ej dx dy

Keij =

ZΩe

Ãa11∂ψei∂x

∂ψej∂x

+ a22∂ψei∂y

∂ψej∂y

+ a00ψeiψ

ej

!dx dy

fei =

ZΩeψei f dx dy, Qe =

IΓeψei qn ds (8.4.5)

The fully discretized set of equations associated with (8.4.4) is given by

[ K(us+1)]us+1 = [K(us)]us + Fs,s+1 ≡ F (8.4.6)

where

[ K(us+1)] = [C] + a1[K(us+1)], [K(us)] = [C]− a2[K(us)] Fs,s+1 = a1Fs+1 + a2Fs, a1 = α∆t, a2 = (1− α)∆t (8.4.7)

Note that only [K] is a function of the nodal unknowns u.When the direct iteration is used to solve the nonlinear equations, at the

(r + 1)st iteration we solve the equation

[ K(urs+1)]ur+1s+1 = F (8.4.8a)

with

F = ([C]− a2[K(us)]) us + a1Fs+1 + a2Fs (8.4.8b)

Note that F remains unchanged during the nonlinear iteration for a giventime step, whereas [ K] changes during the iteration due to the latest knownsolution ur+1s+1.

8.4.3 Flows of Viscous Incompressible Fluids

Weak form Þnite element model

The weak form (RitzGalerkin) Þnite element models developed in Chapter7 already include time-dependent terms [see Eqs. (7.4.13), and (7.5.16) or(7.5.18)]. The fully discretized equations of (7.5.18) are discussed here. Wehave

[ K(vs+1)]vs+1 = [ K(vs)]vs + Fs,s+1 (8.4.9a)

where

[ K(vs+1)] = [M] + a1¡[C(vs+1)] + [K(vs+1)] + [K]

¢[ K(vs)]s = [M]− a2

¡[C(vs)] + [K(vs)] + [K]

¢(8.4.9b)

Fs,s+1 = a1Fs+1 + a2Fs


where [M], [C], [K], and [K] are deÞned in Eqs. (7.4.14) and (7.5.17).Dependence of [C] on the nodal values v is due to the convective terms,

and dependence of [K] on v is due to viscosity being a function of the strainrates (e.g. power-law ßuids). If viscosity is constant, that is, Newtonian ßuids,then [K] is independent of the nodal values v of the velocity Þeld. For directiteration solution, Eq. (8.4.9a) takes the form

[ K(vrs+1)]vr+1s+1 = [K(vs)]vs + Fs,s+1 (8.4.10)

Least-squares Þnite element models

The least-squares Þnite element models of time-dependent NavierStokesequations (see Section 7.9) are discussed next. We consider two alternativeleast-squares formulations [6,7]: (1) velocitypressurevorticity model, and (2)velocitypressurevelocity gradient model.The velocitypressurevorticity formulation is based on the minimization

of the least-squares functional

I1(v, P,ω) =1

2

" °°°°∂v∂t + (v ·∇)v +∇P − 1

Re∇× ω − f

°°°°20

+ kω −∇× vk20 + k∇ · vk20 + k∇ · ωk20#

(8.4.11)

where k · k0 denotes the L2(Ω)-norm

kuk0 =ZΩe|u(x, t)|2 dΩ, t ∈ (0, τ) (8.4.12)

The Þnite element model based on minimization of I1 requires that we chooseat least piecewise bilinear (in two dimensions) or trilinear (in three dimensions)polynomials for v, P , and ω. The temporal terms are approximated using thebackward (Euler) difference scheme.The

velocitypressurevelocity gradient formulation is based on minimization ofthe functional

I2(v, P,U) =1

2

" °°°°∂v∂t + (v ·U)T +∇P − 1

Re(∇ ·U)T − f

°°°°20

+ kU− (∇v)Tk20 + k∇ · vk20 + k∇(trU)k20#

(8.4.13)

where U is the velocity gradient tensor. The Þnite element model based onminimization of I2 also requires, at the minimum, piecewise bilinear (in twodimensions) or trilinear (in three dimensions) polynomials for v, P , and U.


8.4.4 Plate Bending (FSDT)

Here we consider the FSDT with von Karman nonlinearity (see Chapter 6).We begin with the following equations of motion of the theory (see Reddy[57]):

I0∂2u0∂t2

− ∂Nxx∂x

− ∂Nxy∂y

= 0, I0∂2v0∂t2

− ∂Nxy∂x

− ∂Nyy∂y

= 0

I0∂2w0∂t2

− ∂

∂x

µNxx

∂w0∂x

+Nxy∂w0∂y

¶− ∂

∂y

µNxy

∂w0∂x

+Nyy∂w0∂y

¶− ∂Qx∂x

− ∂Qy∂y

− q = 0

I2∂2φx∂t2

− ∂Mxx

∂x− ∂Mxy

∂y+Qx = 0, I2

∂2φy∂t2

− ∂Mxy

∂x− ∂Myy

∂y+Qy = 0

(8.4.14a)

where I0 and I2 are the principal and rotatory inertias

I0 = ρh, I2 =ρh3

12(8.4.14b)

and the inplane forces (Nxx, Nxy, Nyy), transverse shear forces (Qx, Qy), andmoments (Mxx,Mxy,Myy) were deÞned in Eqs. (6.3.4) and (6.6.11).The weak forms of Eqs. (8.4.14a) are given by

0 =

ZΩe

Ã∂ψi∂xNxx +

∂ψi∂yNxy + I0ψi

∂2u0∂t2

!dx dy −

IΓeNnnψi ds (8.4.15a)

0 =

ZΩe

Ã∂ψi∂xNxy +

∂ψi∂yNyy + I0ψi

∂2v0∂t2

!dx dy −

IΓeNnsψi ds (8.4.15b)

0 =

ZΩe

∙∂ψi∂x

µQx +Nxx

∂w0∂x

+Nxy∂w0∂y

¶+∂ψi∂y

µQy +Nxy

∂w0∂x

+Nyy∂w0∂y

¶+ I0ψi

∂2w0∂t2

− ψiq¸dx dy −

IΓeQnψi ds (8.4.15c)

0 =

ZΩe

Ã∂ψi∂xMxx +

∂ψi∂yMxy + ψiQx + I2ψi

∂2φx∂t2

!dx dy −

IΓeMnnψi ds

(8.4.15d)

0 =

ZΩe

Ã∂ψi∂xMxy +

∂ψi∂yMyy + ψiQy + I2ψi

∂2φy∂t2

!dx dy −

IΓeMnsψi ds

(8.4.15e)


where (Nnn, Nns,Mnn,Mns,Qn) are the stress resultants on an edge with unitnormal n, and they are deÞned in Eqs. (6.3.7a,b), (6.3.9), and (6.4.2). Notethat the nonlinearity in Eqs. (8.4.15ae) is solely due to w0, and the nonlinearterms are present only in (Nxx, Nxy, Nyy).For convenience of writing the Þnite element equations, the linear and

nonlinear parts of the inplane forces (Nxx, Nxy, Nyy) are denoted as

N = N0+ N1 (8.4.16)

where N0 is the linear part and N1 is the nonlinear part. For anorthotropic plate (with the principal material axes coinciding with the plateaxes), the forces and moments are related to the generalized displacements(u0, v0, w0,φx,φy) as follows:

N0xx = A11

∂u0∂x

+A12∂v0∂y, N0

yy = A12∂u0∂x

+A22∂v0∂y

N0xy = A66

µ∂u0∂y

+∂v0∂x

¶, Mxx = D11

∂φx∂x

+D12∂φy∂y

Myy = D12∂φx∂x

+D22∂φy∂y, Mxy = D66

µ∂φx∂y

+∂φy∂x

¶Qx = A55

µφx +

∂w0∂x

¶, Qy = A44

µφy +

∂w0∂y

¶N1xx =

1

2

"A11

µ∂w0∂x

¶2+A12

µ∂w0∂y

¶2#

N1yy =

1

2

"A12

µ∂w0∂x

¶2+A22

µ∂w0∂y

¶2#

N1xy = A66

∂w0∂x

∂w0∂y

(8.4.17a)

and Aij and Dij are the plate stiffnesses [see Eq. (6.3.34)]

Aij = hQij , Dij =h3

12Qij (8.4.17b)

and Qij are deÞned in Eqs. (6.3.31a,b).The Þnite element model based on equal interpolation of all generalized

displacements

u0(x, y, t) =nXj=1

∆1j (t)ψj(x, y), v0(x, y, t) =nXj=1

∆2j (t)ψj(x, y)

w0(x, y, t) =nXj=1

∆3j (t)ψj(x, y) (8.4.18)

φx(x, y, t) =nXj=1

∆4j (t)ψj(x, y), φy(x, y, t) =nXj=1

∆5j (t)ψj(x, y)


can be expressed as

0 =5Xβ=1

nXj=1

Mαβij ∆

βj +

5Xβ=1

nXj=1

Kαβij ∆

βj − Fαi ≡ Rαi (8.4.19)

for α = 1, 2, . . . , 5, where ∆αi denotes the value of the αth variable, in theorder (u0, v0, w0,φx,φy), at the ith node (i = 1, 2, . . . , n) of the element. Thenon-zero coefficients of Eq. (8.4.19) are deÞned by

M11ij =

ZΩeI0ψiψj dx dy, M

22ij =M

33ij =M

11ij

M44ij =M

55ij =

ZΩeI2ψiψj dx dy

F 1i =

IΓeNnnψi ds, F

2i =

IΓeNnsψi ds

F 3i =

ZΩeqψi dx dy +

IΓeQnψi ds

F 4i =

IΓeMnnψi ds, F

5i =

IΓeMnsψi ds

K1αij =

ZΩe

µ∂ψi∂xNα1j +

∂ψi∂yNα6j

¶dx dy

K2αij =

ZΩe

µ∂ψi∂xNα6j +

∂ψi∂yNα2j

¶dx dy

K3αij =

ZΩe

∙∂ψi∂x

µQα1j +

∂w0∂x

Nα1j +

∂w0∂y

Nα6j

¶+∂ψi∂y

µQα2j +

∂w0∂x

Nα6j +

∂w0∂y

Nα2j

¶¸dx dy

K4αij =

ZΩe

µQα1jψi +

∂ψi∂xMα1j +

∂ψi∂yMα6j

¶dx dy

K5αij =

ZΩe

µQα2jψi +

∂ψi∂xMα6j +

∂ψi∂yMα2j

¶dx dy (8.4.20)

and Kαβij denotes the coefficient matrix of the αth variable in the βth equation

(α,β = 1, 2, 3, 4, 5). Thus, N Iij denotes the contribution from NI(= N

0I +N

1I )

to the αth variable, where j denotes the node number. The non-zero terms

appearing in the deÞnition of Kαβij are given below:

N11j = A11

∂ψj∂x

, N16j = A66

∂ψj∂y

N21j = A12

∂ψj∂y, N2

6j = A66∂ψj∂x

N31j =

1

2

µA11

∂w

∂x

∂ψj∂x

+A12∂w

∂y

∂ψj∂y

¶


N36j =

A662

µ∂w

∂x

∂ψj∂y

+∂w

∂y

∂ψj∂x

¶N12j = A12

∂ψj∂x

, N22j = A22

∂ψj∂y

N32j =

1

2

µA12

∂w

∂x

∂ψj∂x

+A22∂w

∂y

∂ψj∂y

¶Q31j = A55

∂ψj∂x

, Q32j = A44∂ψj∂y

Q41j = A55ψj , Q52j = A44ψj

M41j = D11

∂ψj∂x

, M46j = D66

∂ψj∂y

M42j = D12

∂ψj∂x

, M51j = D12

∂ψj∂y

M56j = D66

∂ψj∂x

, M52j = D22

∂ψj∂y

(8.4.21)

Equation (8.4.19), when generalized to include structural damping, has theform

[M ]∆+ [C] ú∆+ [K]∆ = F (8.4.22)

The fully discretized equations are

[ K(∆s+1)]∆s+1 = Fs,s+1 (8.4.23)

where

[ K(∆s+1)] = [K(∆s+1)] + a3[M ]s+1 + a6[C]s+1 Fs,s+1 = Fs+1 + [M ]s+1As + [C]s+1Bs

As = a3∆s + a4 ú∆s + a5∆sBs = a6∆s + a7 ú∆s + a8∆s

(8.4.24a)

and ai are deÞned as (γ = 2β)

a1 = α∆t, a2 = (1− α)∆t, a3 =1

β(∆t)2, a4 = a3∆t,

a5 =1

γ− 1, a6 =

α

β∆t, a7 =

α

β− 1, a8 = ∆t

µα

γ− 1

¶ (8.4.24b)

At the end of each time step, the new velocity vector ú∆s+1 andacceleration vector ∆s+1 are computed using the equations

∆s+1 = a3(∆s+1 − ∆s)− a4 ú∆− a5∆s (8.4.25a)

ú∆s+1 = ú∆s + a2∆s + a1∆s+1 (8.4.25b)


Solution of Eq. (8.4.23) by the NewtonRaphson iteration method resultsin the following linearized equations for the incremental solution at the (r+1)stiteration

δ∆ = −[ KT (∆rs+1)]−1Rrs+1 (8.4.26)

[ KT (∆rs+1)] ≡∙∂R∂∆

¸rs+1

, Rrs+1 = [ K(∆rs+1)]∆rs+1 − Fs,s+1(8.4.27)

The total solution is obtained from

∆r+1s+1 = ∆rs+1 + δ∆ (8.4.28)

Note that the tangent stiffness matrix is evaluated using the latest knownsolution, while the residual vector contains contributions from the latest knownsolution in computing [ K(∆rs+1)]∆rs+1 and previous time step solution incomputing Fs,s+1. The velocity and acceleration vectors are updated usingEqs. (8.4.25a,b) only after convergence is reached for a given time step.


Computer implementation of nonlinear time-dependent problems iscomplicated by the fact that one must keep track of the solution vectorsat different loads, times, and iterations. Thus, there are three levels ofcalculations. Often, for a Þxed value of load (or Reynolds number in thecase of ßuid ßows), one wishes to obtain the transient solution. Therefore, theouter loop is on the number of load steps, followed by a loop on the number oftime steps, and the inner most loop being on nonlinear iterations. Flow chartshown in Figure 8.5.1 illustrates the general idea.As an example, Fortran statements showing the transfer of global solution

vectors from previous time step and current iteration of the current timestep to subroutine to calculate element matrices and residual vector are givenbelow. Here variable IDY N is a ßag: IDY N = 0 is for static analysis andIDY N > 0 is for transient analysis. Also we have

W0(I) = element displacement vector at time ts

W1(I) = element velocity vector at time ts

W2(I) = element acceleration vector at time ts

W (I) = element displacement vector at ts+1 in the latest iteration.

Fortran statements showing the transfer of global solution vectors to elementsolution vectors are given in Box 8.5.1. Statements showing the residual vectorcalculation inside the subroutine are given in Box 8.5.2.


Figure 8.5.1 Flow chart of the nonlinear transient analysis of a problem.

DO e = 1 to n

error < ε

STOPnoyes

no

yes


Transfer global information(material properties, geometry and solution)

to element

Initialize global Kij, Mij, and fi

iter = 0

CALL ELKMF to calculate Kij(e),

Mij(e), and fi(e), and assemble to form

global Kij, Mij, and Fi

Update velocities,accelerations, and

print solution

Write a message

iter = iter + 1

Time loopDO nt =1, ntime

Load loopDO nl =1, nload

F=F+∆F

iter < itmax

CC Initialize the global coefficient matrices and vectorsC

DO 180 I=1,NEQGLF(I)=0.0DO 180 J=1,NHBW

180 GLK(I,J)=0.0CC Do-loop on the number of ELEMENTS to compute element matricesC and their assembly begins here (IDYN > 0, flag for dynamic anal.)C

DO 200 N=1,NEM L=0 DO 190 I=1,NPE NI=NOD(N,I) ELXY(I,1)=X(NI) ELXY(I,2)=Y(NI) LI=(NI-1)*NDF DO 190 J=1,NDF LI=LI+1 L=L+1 IF(IDYN.GT.0)THEN

ELP(L)=GLP(LI) !GLP - Global solution vector at time ts ELV(L)=GLV(LI) !GLV - Global velocity vector at time tsELA(L)=GLA(LI) !GLA - Global acceleration vector at time ts

ENDIFELU(L)=GLU(LI) !GLU - Global solution vector at ts+1 & current iteration.

190 CONTINUE CC Call subroutine to compute [ELK], [ELM], [ELC] and ELFC

. . . . . . . .

. . . . . . . .200 CONTINUE

CC ELM(I,J) mass matrix; ELK(I,J) current coefficient matrix (updated in eachC iteration for nonlinear problems); ELKP(I,J) coefficient matrix based onC previous step solution (ELK=ELKP for linear problems); ELP(J) solution vector C from previous time step; ELU(J) solution vector at current time and iteration. C

IF(IDYN.EQ.1)THEN !Parabolic equationsDO 120 I=1,NN SUM=0.0 DO 100 J=1,NN SUM=SUM + (ELM(I,J)-A2*ELKP(I,J))*ELP(J)

100 ELK(I,J)=ELM(I,J)+A1*ELK(I,J) 120 ELF(I)=(A1+A2)*ELF(I) + SUM !Assumed time-independent sourceC

ELSE !Hyperbolic equationsC C ELC(I,J) - Damping matrix; ELP, ELV, ELA are displacement, velocity, andC acceleration vectors from previous time step (do not change during iteration). C

DO 150 I=1,NN SUM=0.0 DO 140 J=1,NN SUM=SUM+ELM(I,J)*(A3*ELP(J)+A4*ELV(J)+A5*ELA(J))

* +ELC(I,J)*(A6*ELP(J)+A7*ELV(J)+A8*ELA(J)) 140 ELK(I,J)=ELK(I,J)+A3*ELM(I,J)+A6*ELC(I,J) 150 ELF(I)=ELF(I)+SUM

ENDIF RETURN END


Box 8.5.1 Fortran statements showing the transfer of global solutionvectors to element solution vectors.

Box 8.5.2 Fortran statements for the calculation of [ K] and F.


8.6 Numerical Examples

8.6.1 Linear Problems

Here we consider several representative examples of time-dependent linearproblems. We begin with a heat conduction problem.

Example 8.6.1

Consider the transient heat conduction equation

∂θ

∂t−µ∂2θ

∂x2+∂2θ

∂y2

¶= 1 in Ω (8.6.1)

where θ is the non-dimensional temperature and Ω is the square domain of side 2. Theboundary condition is that θ = 0 on the boundary for t > 0 (see Figure 8.6.1). The initialcondition is that θ(x, y,0) = 0. We wish to Þnd the temperature Þeld inside the domain fort > 0.

In view of the biaxial symmetry, it is sufficient to model one quadrant of the domain.The boundary conditions along the lines of symmetry require that the heat ßux be zerothere. Thus, the boundary conditions of the computational domain are

∂θ

∂x(0, y, t) = 0,

∂θ

∂y(x,0, t) = 0, θ(1, y, t) = 0, θ(x,1, t) = 0 (8.6.2)

We choose a uniform mesh 8 × 8 of linear rectangular elements to model the domain,and investigate the stability and accuracy of various schemes. Since the CrankNicolson(α = 0.5) and backward difference (α = 1.0) methods are unconditionally stable schemes,one can choose any value of ∆t. However, if ∆t is too large, the solution may not be accurateeven when it is stable. In order to estimate the time step, one must calculate the maximumeigenvalue for the mesh used in the transient analysis. The solution of the eigenvalue problemassociated with the 8×8 mesh of linear elements yields 64 eigenvalues, of which the maximumeigenvalue is λmax = 1492.56. Therefore, the critical time step for the forward differencescheme (α = 0.0) is given by ∆tcr = (2/1492.56) = 0.00134.

Figure 8.6.2 shows plots of the temperature θ(0,0, t) versus time t for a∆t = 0.002, which

is greater than the critical time step. For very small times, θ(0,0, t) ≈ t, and both backwarddifference and CrankNicolson schemes show stable behavior while the forward difference

is unstable. For ∆t = 0.001, the forward difference too gives the same result as the stable

schemes using ∆t = 0.05. Table 8.6.1 shows the numerical values of θ(0,0, t) predicted by

various schemes and two different meshes. One must note that the critical times step for the

4×4Q9 mesh is different as the maximum eigenvalue is different (∆tcr is likely to be smaller

for the mesh of quadratic elements). Figure 8.6.3 contains plots of the evolution of θ(0,0, t)

with time, reaching a steady-state at around t = 1.2. Finally, Figure 8.6.4 shows plots of

θ(x,0, t) versus x for t = 0.2, 0.5, and 1.0. The difference between the solutions obtained by

the two meshes and different time approximation schemes cannot be seen in the plots.

2.0

2.0 x

y

0=θ

0=θ

0=θ

0=θ

x

y

1.0

1.0

0=∂∂

xθ

0=∂∂

yθ

0=θ

0=θ


Figure 8.6.1 Actual and computational domains of the transient heattransfer problem.

Figure 8.6.2 Transient response predicted by various schemes.

0.00 0.01 0.02 0.03 0.04 0.05Time,

0.00

0.01

0.02

0.03

0.04

0.05

0.06

0.07

0.08

0.09

0.10

Tem

pera

ture

t

, θ Forward difference∆t = 0.002

Crank−Nicolson andBackward difference

∆t = 0.002


Table 8.6.1 Evolution of θ(0, 0, t) obtained with various timeapproximation schemes (∆t = 0.05 for CrankNicolson andbackward difference schemes, and ∆t = 0.001 for the forwarddifference scheme).

8× 8L 4× 4Q9

Crank- Backward Forward Crank- Backward ForwardTime Nicolson difference difference Nicolson difference difference

0.05 0.0497 0.0480 0.0500 0.0496 0.0479 0.05000.10 0.0975 0.0916 0.0983 0.0971 0.0913 0.09790.15 0.1398 0.1294 0.1400 0.1390 0.1288 0.13930.20 0.1740 0.1612 0.1737 0.1730 0.1604 0.17280.25 0.2006 0.1873 0.2004 0.1996 0.1864 0.19940.30 0.2215 0.2085 0.2213 0.2205 0.2075 0.22020.35 0.2379 0.2257 0.2376 0.2368 0.2247 0.23650.40 0.2506 0.2395 0.2503 0.2495 0.2385 0.24930.45 0.2605 0.2506 0.2603 0.2594 0.2496 0.25920.50 0.2682 0.2595 0.2680 0.2672 0.2585 0.26700.55 0.2743 0.2667 0.2741 0.2732 0.2656 0.27310.60 0.2790 0.2724 0.2788 0.2779 0.2714 0.27780.65 0.2826 0.2770 0.2825 0.2816 0.2760 0.28150.70 0.2855 0.2807 0.2854 0.2845 0.2797 0.28440.75 0.2877 0.2837 0.2876 0.2867 0.2827 0.28660.80 0.2895 0.2860 0.2894 0.2885 0.2850 0.28840.85 0.2908 0.2879 0.2908 0.2898 0.2870 0.28980.90 0.2919 0.2895 0.2918 0.2909 0.2885 0.29090.95 0.2927 0.2907 0.2926 0.2917 0.2897 0.29171.00 0.2933 0.2916 0.2933 0.2924 0.2907 0.2924

Example 8.6.2

Here we consider the transient response u(x, y, t) of a square membrane Þxed (u = 0) on itsboundary. The governing equation is

∂2u

∂t2−µ∂2u

∂x2+∂2u

∂y2

¶= 1 in Ω (8.6.3)

See Figure 8.6.1 for the domain and boundary conditions with θ replaced by u. We use 8×8mesh of linear elements in the quarter of the domain to determine the response. The criticaltime step for the linear acceleration method (α = 0.5 and γ = 1/3) is (λmax = 1492.56)∆tcr = 0.0897.

Figure 8.6.5 shows plots of the transverse deßection u(0,0, t) versus time t for a ∆t = 0.1,which is greater than the critical time step. The linear acceleration method (α = 0.5and γ = 1/3) gives almost the same response as the constant-average acceleration method(α = 0.5 and γ = 0.5) when ∆t = 0.05 (see Table 8.6.2).

, θ

(∆t=0.05)

(∆t=0.001)

(∆t=0.05)

0.0 0.2 0.4 0.6 0.8 1.0 1.2 1.4Time, t

0.00

0.05

0.10

0.15

0.20

0.25

0.30

0.35T

empe

ratu

re,

Crank-Nicolson

Forward difference

Backward difference

t = 1.0

t = 0.5

t = 0.2

α = 0.5 and 1.0∆ t = 0.05 (8×8L and 4×4Q9)

, θ

0.0 0.2 0.4 0.6 0.8 1.0Distance, x

0.00

0.05

0.10

0.15

0.20

0.25

0.30

0.35

Tem

pera

ture


Figure 8.6.3 Evolution of the temperature θ(0, 0, t).

Figure 8.6.4 Transient response of the heat conduction problem ofExample 8.6.1.

0.0 1.0 2.0 3.0 4.0 5.0 6.0Time, t

-0.40

-0.20

0.00

0.20

0.40

0.60

0.80

1.00D

ispl

acem

ent,

u(x

,0,t

)

• Constant−average accelerationmethod (∆t = 0.1)

♦ linear acceleration method (∆t = 0.1)

! linear acceleration method (∆t = 0.05)


Figure 8.6.5 Evolution of the deßection u(x, 0, t) of the membrane.

Table 8.6.2 Deßection u(x, 0, t) versus time t for a square membraneÞxed on its edges and subjected to uniform load (8× 8L4).

t u u u t u u uCAM* LAM* LAM CAM* LAM* LAM

0.1 0.0025 0.0017 0.0029 1.7 0.5624 0.5578 0.54240.2 0.0125 0.0117 0.0154 1.8 0.5025 0.5023 0.48580.3 0.0325 0.0317 0.0379 1.9 0.4419 0.4397 0.42600.4 0.0625 0.0617 0.0704 2.0 0.3833 0.3872 0.36610.5 0.1025 0.1017 0.1129 2.1 0.3243 0.3122 0.30690.6 0.1525 0.1517 0.1657 2.2 0.2655 0.2841 0.25110.7 0.2125 0.2117 0.2269 2.3 0.2105 0.1740 0.19590.8 0.2825 0.2812 0.2989 2.4 0.1601 0.2250 0.14270.9 0.3624 0.3626 0.3896 2.5 0.1131 −0.0078 0.10141.0 0.4500 0.4565 0.4876 2.6 0.0706 0.2833 0.06931.1 0.5378 0.5482 0.5701 2.7 0.0343 −0.3422 0.02891.2 0.6110 0.6161 0.6301 2.8 0.0038 0.6653 −0.01961.3 0.6550 0.6546 0.6619 2.9 −0.0204 −1.1816 −0.05531.4 0.6656 0.6658 0.6608 3.0 −0.0348 1.9455 −0.06251.5 0.6492 0.6482 0.6359 3.1 −0.0339 −3.5061 −0.03581.6 0.6133 0.6079 0.5939 3.2 −0.0102 5.9403 0.0207

*CAM = constantaverage acceleration method (∆t = 0.1); LAM = linear accelerationmethod (∆t = 0.1).

LAM with ∆t = 0.05.

, t (µs)

w

h = 1cmh = 2.5cm ∆ t = 20 µs

0 500 1000 1500 2000Time

0

20

40

60

80

100

Def

lect

ion

,


Example 8.6.3

Next, we consider the transient response w0(x, y, t) of a simply supported (SS1), isotropicsquare plate under uniform load of intensity q(x, y, t) = q0H(t), where H(t) denotes theHeaviside step function. We use 4 × 4Q9 mesh in the quarter of the domain to determinethe response. The constantaverage acceleration method (α = 0.5, γ = 0.5) is used with∆t = 20 µs = 20× 10−6 s. The geometric and material parameters used are

a = b = 25 cm, h = 1 or 2.5 cm, E1 = E2 = 2.1× 106 N/cm2

ν12 = 0.25, ρ = 8× 10−6 N-s2/cm4(8.6.4)

Figure 8.6.6 shows plots of the non-dimensionalized transverse deßection w =103w0(0, 0, t)E2h

3/q0a4 versus time t (in µs) for thin (h = 1 cm) and thick (h = 2.5 cm),

simply supported, isotropic (ν = 0.25) plates (4×4Q9 mesh; ∆t = 20µs). Note that the effectof shear deformation is to increase the amplitude and reduce the period of the transversedeßection (see Table 8.6.3).

Example 8.6.4

Lastly, we study the motion (for Re = 0) of a viscous ßuid inside a wall-driven cavity (seeSection 7.8.4). We use 16 × 20Q4 non-uniform mesh (of four-node rectangular elements) inthe domain. The mesh size in each coordinate direction are given by

DX = 0.0625, 0.0625, . . . , 0.0625, DY = 0.0625, . . . , 0.0625, 0.03125, . . . , 0.03125

The Crank-Nicolson method (α = 0.5) with two different time steps ∆t = 0.01 and∆t = 0.001 are used. Table 8.6.4 contains the velocity Þeld vx(0.5, y, t) × 10 for timest = 0.01, 0.05, and 0.1. The solution reaches the steady state (² = 10−2) at time t = 0.1when ∆t = 0.01 is used. Figure 8.6.7 shows the evolution of vx(0.5, y, t) (∆t = 0.01).

Figure 8.6.6 Evolution of the transverse deßection w of a simply supported,isotropic plate under uniform load.


Table 8.6.3 Non-dimensionalized center deßection w versus time t forsimply supported, isotropic, square plates.

t w t w

(µs) h = 1 h = 2.5 (µs) h = 1 h = 2.5

20 0.067 0.426 360 50.661 76.73940 0.334 2.193 380 54.768 68.26360 0.860 6.074 400 58.922 58.35280 1.626 12.813 420 63.078 46.337100 2.674 22.770 440 67.238 33.574120 4.189 34.696 460 71.368 22.403140 6.405 46.408 480 75.488 13.751160 9.348 56.971 500 79.533 7.423180 12.871 66.677 520 83.218 3.213200 16.819 75.708 540 86.276 0.605220 21.044 84.185 560 88.635 −0.511240 25.459 91.721 580 90.348 0.728260 30.002 96.806 600 91.472 5.332280 34.532 98.062 620 92.084 13.441300 38.861 95.669 640 92.323 23.727320 42.878 90.826 660 92.251 34.641340 46.718 84.384 680 91.849 45.351

Table 8.6.4 The horizontal velocity Þeld vx(0.5, y, t)× 10 versus time t forthe wall-driven cavity problem (16× 20Q4 mesh).

y t = 0.01 t = 0.01 t = 0.05 t = 0.05 t = 0.10 Steady∆t = 0.01 ∆t = 0.001 ∆t = 0.01 ∆t = 0.001 ∆t = 0.01 state

0.0625 −0.1342 −0.1953 −0.3103 −0.3247 −0.3655 −0.36880.1250 −0.1936 −0.3140 −0.5624 −0.5841 −0.6558 −0.66310.1875 −0.2314 −0.3940 −0.7888 −0.8163 −0.9108 −0.91980.2500 −0.2691 −0.4651 −1.0122 −1.0435 −1.1499 −1.15930.3125 −0.3157 −0.5475 −1.2346 −1.2746 −1.3802 −1.38860.3750 −0.3759 −0.6536 −1.4790 −1.5053 −1.5967 −1.60280.4375 −0.4516 −0.7902 −1.6964 −1.7151 −1.7793 −1.78200.5000 −0.5435 −0.9605 −1.8536 −1.8643 −1.8906 −1.88950.5625 −0.6465 −1.1577 −1.8846 −1.8878 −1.8700 −1.86520.6250 −0.7474 −1.3479 −1.7011 −1.6946 −1.6336 −1.62500.6875 −0.8097 −1.4428 −1.1889 −1.1653 −1.0700 −1.05720.7500 −0.7536 −1.1523 −0.2093 −0.1693 −0.0520 −0.03820.7813 −0.6325 −0.7744 0.5100 0.5471 0.6713 0.68200.8125 −0.4077 −0.1695 1.4014 1.4197 1.5467 1.55260.8438 −0.0054 0.7336 2.4885 2.4716 2.5918 2.59650.8750 0.6329 1.9318 3.7259 3.6716 3.7646 3.78240.9063 1.7000 3.5232 5.1185 5.0707 5.1198 5.16160.9375 3.3334 5.3837 6.5139 6.5756 6.6082 6.64100.9688 5.9470 7.5970 7.9975 8.2488 8.3805 8.2838

vx(0.5,y,t)

yStokes flow

16×20Q4 mesh∆t = 0.001

-0.2 0.0 0.2 0.4 0.6 0.8 1.0


0.0

0.2

0.4

0.6

0.8

1.0

Dis

tan

ce,

t = 0.01t = 0.02

Steady-state

t = 0.03

t = 0.10t = 0.05


Figure 8.6.7 Evolution of the horizontal velocity vx(0.5, y, t) inside a wall-driven cavity.

8.6.2 Nonlinear Problems

The nonlinear results of time-dependent problems are presented for heattransfer, plate bending, and ßuid ßow problems.

Example 8.6.5Consider the equation

∂T

∂t− ∂

∂x

³k∂T

∂x

´− ∂

∂y

³k∂T

∂y

´= 0 in Ω (8.6.5)

where T is the temperature and k is the conductivity. The domain Ω is a rectangle ofdimensions a = 0.18 m and b = 0.1 m along the x and y coordinates, respectively; theconductivity k is of the form

k = k0 (1 + βT ) (8.6.6)

where k0 is the constant thermal conductivity, and β is the temperature coefficient of thermalconductivity. We take k0 = 0.2 W/(m C) and β = 2 × 10−3 (C−1), and the boundaryconditions to be

T (0, y, t) = 500C , T (a, y, t) = 300C ,∂T

∂y= 0 at y = 0, b (8.6.7)

The initial condition is assumed to be

T (x, y, 0) = 0C (8.6.8)

This is essentially a one-dimensional problem.


Table 8.6.5 contains numerical results of the Þnite element analysis using 4× 2Q9 mesh;Figures 8.6.8(a) and (b) show the evolution of the temperature at different locations of the

domain.

Table 8.6.5 The temperature Þeld T (x, y, t) (for any Þxed y) of the heattransfer problem (4× 2Q9 mesh, ∆t = 0.005).

x t = 0.01 t = 0.02 t = 0.03 t = 0.05 t = 0.10 S-State

0.0225 426.96 456.51 471.49 477.91 477.55 477.310.0450 289.94 386.79 427.71 448.57 453.08 454.030.0675 178.69 343.34 401.45 427.63 430.45 430.120.0900 119.17 311.90 375.40 402.23 404.97 405.570.1125 118.00 289.10 350.39 377.39 380.54 380.320.1350 167.10 278.42 326.87 350.13 353.99 354.340.1575 250.20 293.20 316.46 327.12 327.80 327.58

Example 8.6.6

Here we consider the nonlinear transient response of the plate considered in Example 8.6.3.

The problem data used is the same as given in Eq. (8.6.4), except h = 2.5 cm and∆t = 10 µs.

Figure 8.6.9 shows plots of the linear and nonlinear center deßections versus time for q0 = 103,

q0 = 5× 103, and q0 = 104 (also see Table 8.6.6). The deßection is non-dimensionalized asw = 103w0(0,0, t)E2h

3/q0a4.

Example 8.6.7

The next example is concerned with the nonlinear transient analysis of the wall-driven cavity

problem of Example 8.6.4. A mesh of 16× 20 four-node quadrilateral elements is used with∆t = 0.001. The transient response is calculated for Reynolds numbers Re = 1,000 and

Re = 2,500 separately. Figures 8.6.10(a) and (b) show plots of the nonlinear steady-state

and transient center horizontal velocity vx(0.5, y, t) versus y for various times.

Example 8.6.8 (Least-squares model)

The last example of this chapter deals with the transient ßow past a circular cylinder[6,7]. The interest of this problem is in the periodic ßow pattern that develops when thefree stream Reynolds number is greater than Rec ≈ 46 and the modeling of the outßowboundary conditions. The outßow boundary condition is a particularly challenging one sincethe computational domain must be truncated in a region where the vortex street is fullydeveloped.


Figure 8.6.8 Evolution of the temperature T (x, y, t) for the nonlinear heattransfer problem. (a) Temperature, T (x, y0, t) for any value ofy0. (b) Evolution of temperatures for different values of x.

We consider a circular cylinder of unit diameter placed in the Þnite region Ω : −8 ≤x ≤ 25,−8 ≤ y ≤ 8. The boundary conditions include a speciÞed value of 1.0 for the x-component of velocity at the inßow, top, and bottom boundaries. At these boundaries they-component of velocity is set to zero, see Figure 8.6.11(a). At these boundaries no boundaryconditions for vorticity or velocity gradients are speciÞed. We consider ßow at a free streamReynolds number of 100. The Þnite element mesh used for the computations is shown inFigure 8.6.11(b) and consists of 6, 052 bilinear elements and 6,226 nodes. A collocation

0.00 0.04 0.08 0.12 0.16 0.20

Distance, x

100

150

200

250

300

350

400

450

500T

empe

ratu

re,

(a)

Τ(x,

y 0,t

)

t = 0.1, 0.2, 0.3

For any y4×2Q9 mesh

Crank−Nicolson(∆t = 0.005)

t = 0.01

t = 0.02

t = 0.03

0.00 0.10 0.20 0.30Time,

0

50

100

150

200

250

300

350

400

450

500

Tem

pera

ture

,

(b)

Τ(x,

y 0,t

)

t

x = 0.0225

x = 0.0675x = 0.1125

x = 0.1575

0 250 500 750 1000 1250Time

0

20

40

60

80

100

120

Def

lect

ion

,

, t

w

q0 = 1 (Linear)

q0 = 5×103 N/cm2

q0 = 104 N/cm2

4×4Q9∆t = 10 µs


solution will be sought for this problem. One collocation point per element is used, withthe collocation point located at the center of each element. For the bilinear elements, thecollocation point coincides with the reduced integration point.

Figure 8.6.9 Plots of the linear and nonlinear center deßections w versustime t (µs) for simply supported square plate under uniformload (4× 4Q9 mesh, ∆t = 10µs).

Table 8.6.6 Center deßections w versus time t (µs) for simply supportedsquare plate under uniform load.

t Linear Nonlinear

(µs) q0 = 1 q0 = 103 q0 = 5× 103 q0 = 10

4

10 0.105 0.105 0.105 0.10520 0.525 0.525 0.525 0.52540 2.637 2.637 2.637 2.63780 15.119 15.119 15.116 15.108100 26.282 26.277 26.147 25.746200 78.605 77.565 61.334 42.258300 93.457 86.558 30.263 4.352400 53.509 41.715 −2.343 7.872500 6.760 2.214 25.179 42.854600 9.727 17.908 62.625 18.889700 59.089 69.535 33.608 0.115800 95.215 90.271 0.151 36.143900 73.660 51.497 23.921 37.009

1,000 20.968 5.232 60.393 1.060


In the weak form (Galerkin) Þnite element model, the outßow boundary conditions [seeFigure 8.6.11(a)]

−P + 1

Re

∂vx∂x

= 0,∂vy∂x

= 0 (8.6.9)

are imposed in a weak sense rather than in a pointwise manner. They are imposed in aleast-squares sense in the least-squares Þnite element model.

The velocitypressurevorticity formulation

In the velocitypressurevorticity formulation we are left to model the outßow boundarycondition by specifying both velocity components or a velocity component and either pressureor vorticity. We can set a reference pressure of P = 0 at the outßow boundary; however, wecan say nothing about either of the velocity components or about the vorticity. Just settingP = 0 at the outßow boundary is not sufficient and would lead to a mathematically ill-posedproblem.

Taking advantage of the ßexibility of the least-squares method, we can modify thevelocity-pressure-vorticity L2 least-squares functional I1 of Eq. (8.4.11) to enforce theoutßow boundary conditions (8.6.9) in a weak sense:

I1(v, P,ω) =1

2

"°°°∂v∂t+ (v ·∇)v+∇P + 1

Re∇× ω − f

°°°20+ kω −∇× vk20 + k∇ · vk20

+ k∇ · ωk20 +°°°−P + 1

Re

∂vx∂x

°°°20,Γoutflow

+°°°∂vy∂x

°°°20,Γoutflow

#(8.6.10)

We consider a space-time decoupled formulation, where the temporal terms arediscretized using the trapezoidal rule (see [7] for details on space-time coupled and decoupledleast-squares formulations). The collocation solution is most accurate at the collocationpoints, i.e. at the reduced integration points. In a post-processing stage the nodal values forall degrees of freedom are recovered by taking an average of abutting elements to a node.

We choose the point (x, y) = (2,0) to trace the change of the velocity component vywith time. We use a Þxed time increment of ∆t = 0.1 for the simulation. Even though theleast-squares method is stable for large time increments, a small time increment is desirablefor accuracy reasons. When a small time increment is used, the solution at the previoustime step serves as a very good initial guess for the solution at the current time step, thusthe conjugate gradient method takes only a few iterations to converge.

Figure 8.6.12 shows the time history of the velocity component vy at the point (x, y) =(2, 0). From the Þgure we see that shedding starts around t = 50. The shedding periodmeasured from Figure 8.6.12 is found to be T = 6.10, which gives a dimensionless sheddingfrequency of St = 0.164. Our result is in good agreement with the experimental resultSt = 0.166 of Hammache and Gharib [16].

Figure 8.6.13 shows contour plots of the vx- and vy-velocity components. From visualinspection of the contour plots we see that the outßow boundary condition is modelled ina satisfactory manner. By deÞnition, an outßow boundary condition should permit theßow to exit the domain gracefully and passively and not have any effect on the behaviorof the solution in the domain near the open boundary and especially far from it (see Saniand Gresho [14]). Clearly, the outßow boundary condition did not have any effect on thebehavior of the solution immediately behind the cylinder or near the open boundary. Theimposition of the outßow boundary condition through the least-squares functional is notonly efficient but very elegant.

-0.3-0.2-0.1 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0


0.0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1.0D

ista

nce

,

t = 0.001

t = 0.01

t = 0.02

t = 0.05

Steady state

Re = 1,000 (16x20 Q4 mesh)

(b)

y

vx(0.5,y,t)

∆t = 0.001

-0.3-0.2-0.1 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0


0.0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1.0

Dis

tan

ce,

t = 0.001

t = 0.01

t = 0.02

t = 0.1 and Steady state

Re = 2,500 (16x20 Q4 mesh)

(a)

y

vx(0.5,y,t)

∆t = 0.001

-

-


Figure 8.6.10 Evolution of the horizontal velocity Þeld vx(0.5, y, t) for thewall-driven cavity problem (nonlinear analysis).

(a)

(b)

vx = vy =

vx =vy =

vx = vy =

vx =vy =

x

yOutflowboundary

(- 8 , -8)

(-8 , 8)

(25 , -8)

(25 , 8)

10

00

1, 0

1, 0


Figure 8.6.11 (a) Geometry and boundary conditions for ßow past acircular cylinder. (b) Finite element mesh for ßow past acircular cylinder.

Figure 8.6.12 Time history of vy-velocity component at (x, y) = (2, 0).

0 50 100 150 200-0.8

-0.6

-0.4

-0.2

0.0

0.2

0.4

0.6

0.8

time

Vel

ocity

,v y


Figure 8.6.13 Numerical results for cylinder in cross ßow using the velocity-pressure-vorticity formulation: (a) vx-velocity contours and(b) vy-velocity contours.

Yet another approach is to carry out the simulation using a space-timecoupled formulation [7]. In such an approach the least-squares functional

I1(v, P,ω) =1

2

Ãk∂v∂t+ (v ·∇)v+∇P + 1

Re∇× ω − fk20,Ω×(0,τ ]

+ kω −∇× vk20,Ω×(0,τ ] + k∇ · vk20,Ω×(0,τ ] + k∇ · ω k20,Ω×(0,τ ]!

is minimized in space-time, where k · k0,Ω×(0,τ ] denotes the L2 norm of theenclosed quantity in space-time, i.e.,

kuk20,Ω×(0,τ ] =τZ0

ZΩ

|u(x, t)|2 dΩ dt

Details and applications of such a formulation to this and other problems canbe found in Pontaza and Reddy [7].

(a)

(b)

0 50 100 150 200-0.8

-0.6

-0.4

-0.2

0.0

0.2

0.4

0.6

0.8

time

Vel

ocity

,v y

(a)

(b)


The velocitypressurevelocity gradient formulation

In the velocitypressurevelocity gradient formulation we have the added freedom ofspecifying any of the velocity gradients at the outßow boundary. Here we set P = 0 andU12 = ∂vy/∂x = 0 at the outßow boundary to model the outßow boundary condition. Again,we choose the point (x, y) = (2,0) to trace the change of the vy-velocity component withtime and use a Þxed time increment of ∆t = 0.1 for the simulation.

Figure 8.6.14 shows the time history of the vy-velocity component at the point (x, y) =(2,0). Like in the velocitypressurevorticity formulation, the outßow boundary conditionis satisfactory (see Figure 8.6.15) and the shedding period is found to be T = 6.10, whichgives a dimensionless shedding frequency of St = 0.164.

Figure 8.6.14 Time history of vy-velocity component at (x, y) = (2, 0).

Figure 8.6.15 Numerical results for cylinder in cross ßow using the velocitypressurevelocity gradient formulation: (a) vx-velocity contoursand (b) vy-velocity contours.


Problems8.1 Derive Eq. (8.2.9) for parabolic equations. What if [C] is a function of time?

8.2 Derive Eq. (8.2.15) for hyperbolic equations.

8.3 The α−family of approximation and a general class of time approximation schemescan be derived using the Þnite element method in time domain. Consider a Þrst-orderdifferential equation of the form

Cdu

dt+Ku(t) = F (t) (i)

The weighted-integral statement of Eq. (i) over a typical time interval (ts, ts+1) isgiven by Z ts+1

ts

w(t) (C úu+Ku− F )dt (ii)

where w is a weight function, which is arbitrary at the moment. Suppose that u isapproximated as

u(t) ≈nXj=1

Ujψj(t) (iii)

where Uj denotes the value of u(t) at time node t = tj and ψj(t) is the associatedinterpolation function. In general, to obtain n independent relations w(t) can be chosento be n independent functions φi 6= ψj (j = 1,2, . . . , n). Such a scheme is known asthe Petrov-Galerkin method [10]. When φj = ψj , it is known as the Galerkin method.Obtain (a) the Petrov-Galerkin Þnite element model of the weighted-integral statement(ii), and (b) for the choice of n = 2 in Eq. (iii) [i.e. linear interpolation of u(t)] and

φ1(t) = −1 + 3t+ 3α(1− 2t), φ2(t) = 2− 3t− 3α(1− 2t) (iv)

show that the Þnite element equations are given byµC

2

∙−1 1−1 1

¸+K∆t

2

∙α 1− α

1− α α

¸¶½U1U2

¾=∆t

2

½αF1 + (1− α)F2(1− α)F1 +αF2

¾(v)

where F (t) is also interpolated as

F (t) = F1ψ1(t) +F2ψ2(t) (vi)

Solve the second equation of (v) for U2 = us+1 in terms of U1 = us, and obtain

[C + α∆tK]us+1 = [C − (1− α)∆tK]us +∆t [(1− α)Fs + αFs+1] (vii)

8.4 Use quadratic approximation of u(t) with φi = ψi (i.e. Galerkins method) in the weakform (ii) of Problem 8.1 and arrive at the Þnite element equationsÃC

6

"−3 4 −1−4 0 41 −4 3

#+K∆t

30

"4 2 −12 16 2

−1 2 4

#!(U1U2U3

)=∆t

30

"4 2 −12 16 2

−1 2 4

#(F1F2F3

)(i)

whereU1 = us, U2 = us+ 1

2, U3 = us+1 (ii)


and similar deÞnition holds for Fi. Determine the values of U2 and U3 in terms of U1.

8.5 Consider the second-order equation

Md2u

dt2+C

du

dt+Ku = F (i)

The weighted-integral statement over an element is given byZ ts+1

ts

w(t) (Mu+C úu+Ku− F )dt (ii)

The weak form is given byZ ts+1

ts

(−M úw úu+Cw úu+Kwu−wF )dt =Mw(ts)v(ts)−Mw(ts+1)v(ts+1) (iii)

where v(t) = úu(t). The weighted-integral form in Eq. (ii) requires quadratic or higher-order interpolation while that in Eq. (iii) admits linear or higher-order approximations.Use the weighted-integral statement in Eq. (ii) with w(t) = φi = ψi and n = 3 (i.e.quadratic approximation) and obtainÃ

M

3∆t

"2 −4 28 −16 8

−2 4 −2

#+C

6

"−3 4 −1−4 0 41 −4 3

#

+K∆t

30

"4 2 −12 16 2

−1 2 4

#!(U1U2U3

)=∆t

30

"4 2 −12 16 2

−1 2 4

#(F1F2F3

)(iv)

whereU1 = us, U2 = us+ 1

2, U3 = us+1 (v)

Equation (iv) can be solved for us+ 12and U3 = us+1 in terms of us.

8.6 Use the weak form (iii) of Problem 8.5 with w(t) = φi = ψi and n = 2 (i.e. linearapproximation), and obtainÃ

−M∆t

∙1 −1

−1 1

¸+C

2

∙−1 1−1 1

¸+K∆t

6

∙2 11 2

¸!½U1U2

¾=∆t

6

∙2 11 2

¸½F1F2

¾+M

½úU1

− úU2

¾(i)

and then solve for U2 = us+1 in terms of quantities known at time ts. Hint: ReplaceF1 = Fs with Fs =Mus +C úus +Kus, and solve for us+1µ6M

(∆t)2+3C

∆t+K

¶us+1 =M

µ6

(∆t)2us +

6

∆túus + 2us

¶+C

³3

∆tus + 2 úus

´+Fs+1

(ii)

which is the same as the Newmark scheme with α = 1/2 and γ = 1/3 (and a7 = 0).

8.7 Use the Newmark scheme in Eqs. (8.2.22) and (8.2.23) to reduce Eq. (8.2.11) to thatin Eq. (8.2.20). Hint: Use Eqs. (8.2.22), (8.2.23), and (8.2.11) to eliminate us+1and úus+1 to arrive at the desired equation.


8.8 The mass matrices of the Þnite element formulations are always non-diagonal. Theymay be replaced with rationally derived diagonal matrices, also called lumped massmatrices, for use in explicit formulations. The row-sum lumping and proportionallumping techniques are two ways to compute diagonal matrices. In row-sum lumpingthe sum of the coefficients of each row of the consistent mass matrix is used as thediagonal element and the off-diagonal elements are set to zero:

Meii =

nXj=1

ZΩeρψeiψ

ej dx =

ZΩeρψei dx, Mij = 0 for i 6= j (i)

where the propertyPn

j=1 ψej = 1 of the interpolation functions is used. In proportional

lumping the diagonal elements are computed to be proportional to the diagonalelements of the consistent mass matrix while conserving the total mass of the element:

Meii = α

ZΩeρψeiψ

ei dx, α =

RΩeρ dxPn

i=1

RΩeρψeiψ

ei dx

(ii)

Compute the lumped mass matrices for linear and quadratic one-dimensional elementsand show that the critical time step computed using lumped mass matrices is greaterthan that computed using the consistent mass matrix.

8.9 The equation governing bending of beams according to the EulerBernoulliassumptions and in the presence of viscous (velocity-dependent) damping is givenby

∂2

∂x2

µEI∂2w0∂x2

+ csI∂3w0∂x2∂t

¶+ I0

∂2w0∂t2

+ c∂w0∂t

= q(x, t) (i)

where I denotes the moment of inertia, I0 the principal mass inertia, cs the viscousresistance to strain velocity, and c the viscous resistance to transverse displacement.Develop the semidiscrete Þnite element model and fully discretized Þnite element modelusing the Newmark scheme.

8.10 Establish the stability condition in Eq. (8.3.4).

References1. Argyris, J. H. and Scharpf, O. W., Finite Elements in Time and Space, AeronauticalJournal of the Royal Society, 73, 10411044 (1969).

2. Bathe, K. J. , Finite Element Procedures, PrenticeHall, Englewood Cliffs, NJ (1996).

3. Engelman, M. S. and Jamnia, M. A., Transient Flow Past a Circular Cylinder: aBenchmark Solution, International Journal for Numerical Methods in Fluids, 11, 9851000 (1990).

4. Newmark, N. M., A Method for Computation of Structural Dynamics, Journal ofEngineering Mechanics, 85, 6794 (1959).

5. Nickell, R. E., On the Stability of Approximation Operators in Problems of StructuralDynamics, International Journal of Solids and Structures, 7, 301319 (1971).

6. Pontaza, J. P. and Reddy, J. N., Spectral/hp Least-Squares Finite ElementFormulation for the Navier-Stokes Equations, Journal of Computional Physics, 190(2),523549 (2003).

7. Pontaza, J. P. and Reddy, J. N., Space-Time Coupled Spectral/hp Least-SquaresFinite Element Formulation for the Incompressible NavierStokes Equations, Journalof Computional Physics (to appear).


8. Reddy, J.N. and Gartling, D. K., The Finite Element Method in Heat Transfer andFluid Dynamics, 2nd edn, CRC Press, Boca Raton, FL (2000).


10. Reddy, J. N., Applied Functional Analysis and Variational Methods in Engineering,McGrawHill, New York (1986); reprinted by Krieger Publishing, Melbourne, FL(1991).

11. Reddy, J. N., Energy Principles and Variational Methods in Applied Mechanics, SecondEdition, John Wiley, New York (2002).


13. Reddy, J. N., Mechanics of Laminated Composite Plates and Shells. Theory andAnalysis, 2nd edn, CRC Press, Boca Raton, FL (2004).

14. Sani, R. L. and Gresho, P. M., Resume and Remarks on the Open Boundary ConditionMinisymposium, International Journal for Numerical Methods in Fluids, 18, 9831008(1994).

15. Wood, W. L., Control of CrankNicolson Noise in the Numerical Solution of the HeatConduction Equation, International Journal for Numerical Methods in Engineering,11, 10591065 (1977).

16. Hammache, M. and Gharib, M., An Experimental Study of the Parallel and ObliqueVortex Shedding from Circular Cylinders, Journal of Fluid Mechanics, 232, 567590(1991).

9

Finite Element Formulationsof Solid Continua

9.1 Introduction

9.1.1 Background

In the linear description of the motion of solid bodies one assumes that thedisplacements and strains are very small and that the material is linearlyelastic. In addition, the equilibrium equations were derived using theundeformed conÞguration of the body. In geometrically nonlinear analysisof beams and plates that was considered in earlier chapters the assumptionof small strains allowed us to ignore the changes in the geometry of thebody and the distinction between various measures of stress and strain,as we proceeded to determine the deformation for the next load. In thischapter, we shall study geometrically nonlinear behavior in which changesin geometry, however large or small, have a signiÞcant effect on the loaddisplacement characteristics of solid bodies. When the geometric changes aresigniÞcant, that is, displacements and strains are large, the geometry of thebody must be updated to determine the new position x of the material pointX.Consequently, it becomes necessary to distinguish between various measuresof stress and strain, and descriptions of motion. Intuition tells us that if strainenergy is based on the product of stress and strain, it is not expected to changesimply because of our choice of stress or strain measure. Thus, energeticallyconjugate pairs of stress and strain that produce the same strain energy mustbe used [16].Before we develop geometrically nonlinear Þnite element formulations of

solid continua, it is useful to familiarize ourselves with the associated nonlinearcontinuum mechanics. The knowledge of nonlinear continuum mechanicsenables us to understand better the restrictions placed on the model, betterinterpret and apply the results of the analysis, and prevent the use of acomputer program beyond the range of its applicability.

X

x

u

C0

(Undeformed configuration)

x3 , X3

C = C1

(Deformed configuration)

x2 , X2

x1 , X1

Particle X (occupying position X)

Particle X (occupying position x)


9.1.2 Descriptions of Motion

Consider a deformable body of known geometry, constitution, and loading.For a given geometry and loading, the body will undergo deformation (i.e.macroscopic geometric changes within the body). If the applied loads aretime-dependent, the deformation of the body will be a function of time, thatis, the geometry of the body will change continuously with time. If the loadsare applied slowly so that the deformation is only dependent on the loads, thebody will take a deÞnitive shape at the end of each load application. Whetherthe deformation is time-dependent or not, the forces in the deformed bodywill be in equilibrium at all times.Suppose that the body initially occupies a conÞguration C0, in which a

particle X of the body occupies the position X, referred to a rectangularCartesian system (X1,X2,X3). Note that X is the name of the particle thatoccupies the location X in conÞguration C0, and therefore (X1,X2,X3) arecalled the material coordinates. After the application of the loads, the bodydeforms and assumes a new conÞguration C. The particle X now occupies theposition x in the deformed conÞguration C (see Figure 9.1.1).An analytical description of the deformation of a continuous body follows

one of the two approaches [13,5,6]. In the Þrst approach, called the materialor Lagrangian description, the motion of the body is referred to a referenceconÞguration CR, which is often chosen to be the undeformed conÞguration,CR = C0. Thus, in the Lagrangian description, the current coordinates(x1, x2, x3) are expressed in terms of the reference coordinates (X1,X2,X3):

x = x(X, t) (9.1.1)

Figure 9.1.1 Reference and deformed conÞgurations of a body.

FINITE ELEMENT FORMULATIONS OF SOLID CONTINUA 329

and the variation of a typical variable φ over the body is described with respectto the material coordinates (X1,X2,X3):

φ = φ(X, t) (9.1.2)

In the spatial or Eulerian description, the motion is referred to the currentconÞguration C occupied by the body, and φ is described with respect to theposition (x1, x2, x3) in space, currently occupied by material particle X:

φ = φ(x, t), X = X(x, t) (9.1.3)

The coordinates (x1, x2, x3) are termed the spatial coordinates.Equations (9.1.2) and (9.1.3) each convey a different information [13]. In

Eq. (9.1.2), a change in time t implies that the same material particle X,occupying position X in C0, has a different value φ. Thus the attention isfocused on the material particle X. In Eq. (9.1.3), a change in time t impliesthat a different value φ is observed at the same spatial location x, now probablyoccupied by a different material particle X. Hence, attention is focused on aspatial position x.In the study of solid bodies, the Eulerian description is less useful since

the conÞguration C is unknown. On the other hand, it is the preferreddescription for the study of motion of ßuids because the conÞguration is knownand remains unchanged, and we wish to determine the changes in the ßuidvelocities, pressure, density and so on. Thus, in the Eulerian description,attention is focused on a given region of space instead of a given body ofmatter. The development of continuum equations in the Eulerian descriptionof ßuid ßows was presented in Chapter 7. Here we focus our attention onthe Lagrangian description of the motion of solid bodies undergoing geometricchanges.

9.2 Strain and Stress Measures

9.2.1 Deformation Gradient Tensor

Consider two material particles P and Q in the neighborhood of each otherin the reference conÞguration C0 (see Figure 9.2.1). The positions of P and Qare denoted by XP and XQ, respectively. The position of Q relative to P isgiven by the elemental vector dX in C0:

dX = XQ −XPAfter deformation the material particles P and Q occupy spatial positions xPand xQ, respectively in C; they are now labeled as P and Q. The position ofQ relative to P is denoted by dx and it is given by

dx = xQ − xP

XQ

Q

P Q_

XP

xP P_

xQ

uQ

uP

0C3x X3

dx

dX C

2x X2

1x X1

(time t = 0)

(time t)


Figure 9.2.1 Deformation of a line segment PQ in a continuous medium.

The displacements of the material particles P and Q are given by

uP = xP −XP , uQ = xQ −XQ (9.2.1)

One of the key quantities in Þnite deformation analysis is the deformationgradient tensor F, which gives the relationship of a material line dX beforedeformation to the line dx after deformation. It is deÞned as [13]

dx = F · dX = dX ·FT where F =

µ∂x

∂X

¶T≡ (∇ox)T (9.2.2)

and ∇o is the gradient operator with respect to X. We also have

dX = F−1 · dx = dx · F−T, where F−T =∂X

∂x≡ ∇X (9.2.3)

and ∇ is the gradient operator with respect to x. In indicial notation, Eqs.(9.2.2) and (9.2.3) can be written as

F = FiIei EI , FiI =∂xi∂XI

(9.2.4a)

F−1 = F−1Ii EIei, F−1Ii =∂XI∂xi

(9.2.4b)

Here the lower case indices refer to the current (spatial) Cartesian coordinates,whereas upper case indices refer to the reference (material) Cartesian


coordinates. The determinant of F is called the Jacobian of the motion and itis denoted by J . The deformation tensor F can be expressed in terms of thedisplacement vector as

F = (∇ox)T = (∇ou+ I)T , FT = ∇ox = ∇ou+ I (9.2.5)

The deformation tensor conveys no information about the translation of thebody. Further, if F=I everywhere in the body, then the body is not rotatedand is undeformed. If F has the same value at every material point in a body,then the mapping x = x(X, t) is said to be a homogeneous motion of the body.

9.2.2 Green and Almansi Strain Tensors

Next we discuss a general measure of deformation, independent of bothtranslation and rotation. Consider two material particles P and Q in theneighborhood of each other (separated by dX) in the reference conÞguration(see Figure 9.2.1). In the deformed conÞguration the material points P and Qare denoted by P and Q, and they are separated by dx. We wish to determinethe change in the distance dX between the material points P and Q as thebody deforms and the material points move to the new locations P and Q.The distances between points P and Q and points P and Q are given,

respectively, by

(dS)2 = dX · dX (9.2.6a)

(ds)2 = dx · dx = (F · dX) · (F · dX) = dX · (FT · F) · dX≡ dX ·C · dX (9.2.6b)

where C is the right CauchyGreen deformation tensor

C = FT · F (9.2.7)

The change in the squared lengths that occurs as the body deforms fromthe initial to the current conÞguration can be expressed relative to the originallength as

(ds)2 − (dS)2 = 2 dX ·E · dX (9.2.8)

where E is called the GreenLagrange strain tensor or simply the Green straintensor, which can be expressed as

E =1

2

³FT · F− I

´=

1

2(C− I) (9.2.9)

=1

2

h(∇ou)T +∇ou+ (∇ou)T · (∇ou)

i(9.2.10)

Clearly, the Green strain tensor is symmetric. Also, the change in the squaredlengths is zero if and only if E = 0.


Alternatively, the change in the squared lengths that occurs as the bodydeforms from the initial to the current conÞguration can be expressed relativeto the current length as

(ds)2 − (dS)2 = 2 dx · e · dx (9.2.11)

where e is called the AlmansiHamel (Eulerian) strain tensor or simply theEuler strain tensor, which can be expressed as

e =1

2

³I−F−T · F−1

´=

1

2

³I−B−1

´(9.2.12)

=1

2

h(∇u)T +∇u− (∇u)T · (∇u)

i(9.2.13)

where B = F ·FT is the Cauchy strain tensor, and its inverse is called the leftCauchyGreen or Finger tensor.In the Cartesian component form, we can write

E = EIJ EI EJ (9.2.14)

e = eijeiej (9.2.15)

with components

EIJ =1

2

µ∂xm∂XI

∂xm∂XJ

− δIJ¶

(9.2.16a)

=1

2

µ∂uI∂XJ

+∂uJ∂XI

+∂uK∂XI

∂uK∂XJ

¶(9.2.16b)

eij =1

2

Ãδij − ∂XK

∂xi

∂XK∂xj

!(9.2.17a)

=1

2

Ã∂ui∂xj

+∂uj∂xi

− ∂uk∂xi

∂uk∂xj

!(9.2.17b)

In expanded notation, the Green strain components, for example, are givenby

E11 =∂u1∂X1

+1

2

"µ∂u1∂X1

¶2+

µ∂u2∂X1

¶2+

µ∂u3∂X1

¶2#

E22 =∂u2∂X2

+1

2

"µ∂u1∂X2

¶2+

µ∂u2∂X2

¶2+

µ∂u3∂X2

¶2#

E33 =∂u3∂X3

+1

2

"µ∂u1∂X3

¶2+

µ∂u2∂X3

¶2+

µ∂u3∂X3

¶2#

E12 =1

2

µ∂u1∂X2

+∂u2∂X1

+∂u1∂X1

∂u1∂X2

+∂u2∂X1

∂u2∂X2

+∂u3∂X1

∂u3∂X2

¶E13 =

1

2

µ∂u1∂X3

+∂u3∂X1

+∂u1∂X1

∂u1∂X3

+∂u2∂X1

∂u2∂X3

+∂u3∂X1

∂u3∂X3

¶E23 =

1

2

µ∂u2∂X3

+∂u3∂X2

+∂u1∂X2

∂u1∂X3

+∂u2∂X2

∂u2∂X3

+∂u3∂X2

∂u3∂X3

¶(9.2.18)

e0

e0

a

b

AA

B

C

B

D

DC e0

e0

x1

x2 X2

X1


Example 9.2.1

Consider a rectangular block of dimensions a × b × h, where h is very small compared toa and b [5]. Suppose that the block is deformed into the diamond shape shown in Figure9.2.2. By inspection, the geometry of the deformed body can be described as follows: let(X1,X2,X3) denote the coordinates of a material point in the undeformed conÞguration.Thus the coordinate mapping and its inverse are given by

x1 =X1 +e0bX2, X1 =

ab

ab− e20x1 −

ae0ab− e20

x2

x2 =X2 +e0aX1, X2 = − be0

ab− e20x1 +

ab

ab− e20x2

x3 =X3, X3 = x3.

Thus, the displacement components of a material point in the Lagrangian description are

u1 = x1 −X1 =e0bX2, u2 = x2 −X2 =

e0aX1, u3 = x3 −X3 = 0.

The only non-zero Green strain tensor components are given by

E11 =1

2

³e0a

´2, E12 =

e02b+e02a, E22 =

1

2

³e0b

´2.

The Almansi strain tensor components are

e11 = − e20ab− e20

− 12

∙e20(e

20 + b

2)

(ab− e20)2

¸,

e12 =e0(a+ b)

(ab− e20)+e30(a+ b)

(ab− e20)2,

e22 = − e20ab− e20

− 12

∙e20(e

20 + a

2)

(ab− e20)2

¸.

Figure 9.2.2 Undeformed and deformed rectangular block.


9.2.3 Polar Decomposition

Recall that the deformation gradient tensor F transforms a material vectordX into the corresponding spatial vector dx, and it forms an essential part ofthe deÞnition of any strain measure. Another role of F in connection with thestrain measures is discussed with the help of the polar decomposition theoremof Cauchy, which enables F to be written as [13]

F = R ·U = V ·R (9.2.19)

where R is an orthogonal rotation tensor and U and V are symmetric stretchtensors. An orthogonal rotation tensor R is one that satisÞes RT ·R = I.

To evaluate the tensors R and U, we recall the deÞnition of C:

C = FT · F = UT ·RT ·R ·U = UT ·U (9.2.20)

To compute U using Eq. (9.2.20), it is necessary to write C in terms of itseigenvalues and eigenvectors:

C =3Xα=1

λ2αNαNα (9.2.21)

where λ2α are the eigenvalues and Nα are the eigenvectors of C. Then

U =3Xα=1

λα NαNα (9.2.22)

and the rotation tensor R can be obtained from Eq. (9.2.19) as R = F ·U−1.

9.2.4 Stress Tensors

The equations of motion or equilibrium must be derived for the deformedconÞguration of the structure at time t. However, since the geometry of thedeformed conÞguration is not known, the equations must be written in terms ofthe known reference conÞguration. In doing so we introduce various measuresof stress. They emerge in a natural way as we transform volumes and areasfrom the deformed conÞguration to undeformed conÞguration [13].

First we introduce the true stress, that is, stress in the deformedconÞguration. Consider a deformed body at its current position. If we denoteby df(n) the force on a small area nda located at the position x, the stressvector can be deÞned as t(n) = df

da . The Cauchy stress tensor σ is deÞned tobe the current force per unit deformed area:

df = t da = da · σ, where da = da n (9.2.23a)


where Cauchys formula, t = σ · n, is used (see Problem 9.5).

To express df in terms of a stress times the initial undeformed area dArequires a new stress tensor P,

df = dA ·P, where dA = dA N (9.2.23b)

where N is the unit normal to the undeformed area dA. The stress tensorP is called the Þrst PiolaKirchhoff stress tensor, and it gives the currentforce per unit undeformed area. The Þrst PiolaKirchhoff stress tensor is notsymmetric.

The second PiolaKirchhoff stress tensor S, which is used in the totalLagrangian formulation of geometrically nonlinear analysis, is introduced asfollows. Recall from Eq. (9.2.3) that dX = F−1 · dx. Analogously, we cantransform the force df on the deformed elemental area da to the force dF onthe undeformed elemental area dA (not to be confused between the force dFand deformation tensor F)

dF = F−1 · df = F−1 · (dA ·P) = dA ·P · F−T ≡ dA · S (9.2.24)

Thus, the second PiolaKirchhoff stress tensor gives the transformed currentforce per unit undeformed area. The second PiolaKirchhoff stress tensor issymmetric whenever the Cauchy stress tensor is symmetric.

In summary, we have the following relationships between various stressmeasures (J denotes the determinant of F):

P = J F−1 · σ = S · FT (9.2.25a)

σ =1

JF ·P = 1

JF · S · FT (9.2.25b)

S = J F−1 · σ ·F−T = P · F−T (9.2.25c)

9.2.5 Energetically-Conjugate Stresses and Strains

The rate of internal work done (power) in a continuous medium in the currentconÞguration can be expressed as (see [3])

W =1

2

Zvσ : d dv (9.2.26)

where σ is the Cauchy stress tensor and d is the symmetric part of the velocitygradient tensor

d =1

2

h(∇v)T +∇v

i, v =

dx

dt(9.2.27)


The pair (σ,d) is said to be energetically-conjugate since it produces the(strain) energy stored in the deformable medium. We can show that thesecond PiolaKirchhoff stress tensor S is energetically-conjugate to the rate ofGreenLagrange strain tensor úE:

W =1

2

ZVS : úE dV (9.2.28)

The proof of the above statement requires the use of a number of identities[3], which are presented next.

The Þrst one is a relation between the rate of deformation gradient tensorúF and ∇0v. We have

úF ≡ dF

dt=d

dt

µ∂x

∂X

¶T=

∙∂

∂X

µdx

dt

¶¸T=

µ∂v

∂X

¶T= (∇0v)T , dx

dt= v

(9.2.29)Next, we note that

úF =

µ∂v

∂X

¶T=

µ∂v

∂x

¶T·µ∂x

∂X

¶T= L ·F, L ≡

µ∂v

∂x

¶T= úF ·F−1 (9.2.30)

where L is the velocity gradient tensor.

The time derivative of the Lagrangian strain tensor, úE, is known as thematerial strain rate tensor. We have

úE ≡ dE

dt=1

2

d

dt

³FT · F− I

´=1

2

³úFT · F+FT · úF

´(9.2.31)

The symmetric part of the velocity gradient tensor can be related to the strainrate tensor

d =1

2

³LT + L

´=1

2

∙³úF · F−1

´T+ úF · F−1

¸=1

2

³F−T · úFT + úF · F−1

´=1

2F−T ·

³úFT +FT · úF · F−1

´=1

2F−T ·

³úFT ·F+FT · úF

´· F−1

= F−T · úE · F−1 (9.2.32)

Now returning to Eq. (9.2.26) for the rate of work done, we have

1

2

Zvσ :d dv =

1

2

Zvσ : L dv (by symmetry of σ and d)


=1

2

ZVJ σ :

³úF · F−1

´dV [by Eq. (9.2.30)]

=1

2

ZVJ (σijeiej) :

h³úFpIep EI

´·³(F−1)Jq EJeq

ídV

=1

2

ZVJ σij úFjI(F

−1)Ii dV =1

2

ZVJ σji úFjI(F

−1)Ii dV (since σij = σji)

=1

2

ZVJ σij úFiI(F

−1)Ij dV (by renaming i to be j and j to be i)

=1

2

ZVJh³(F−1)Iq EIeq

´· (σijeiej)

i:³úFpJep EJ

´dV

=1

2

ZV

³J F−1 · σ

´: úF dV =

1

2

ZVP : úF dV (9.2.33)

Thus, the Þrst PiolaKirchhoff stress tensor P is work conjugate to the rateof the deformation gradient tensor úF. Similarly,

1

2

Zvσ : d dv

=1

2

ZVJ σ :

³F−T · úE · F−1

´dV [using Eq. (9.2.32)]

=1

2

ZVJ (σijeiej) :

³(F−1)Ipep úEIJ (F−1)Jq eq

´dV

=1

2

ZVJ³(F−1)Ii σij (F−1)Jj

´úEIJ dV (EIJ = EJI)

=1

2

ZVJ³(F−1)Ii σij (F−1)Jj

´úEJI dV

=1

2

ZVJ

∙³(F−1)Ip EIep

´· (σijeiej) ·

³(F−1)Jq EJeq

´T¸:³úEPQ EP EQ

´dV

=1

2

ZVJ³F−1 · σ · F−T

´: úE dV =

1

2

ZVS : úE dV (9.2.34)

Thus, the second PiolaKirchhoff stress tensor S is work conjugate to the rateof the GreenLagrangian strain tensor úE.

9.3 Strain and Stress MeasuresBetween ConÞgurations

9.3.1 Notation

The determination of the Þnal conÞguration of a solid body undergoing largedeformation is not an easy task. A practical way of determining the ÞnalconÞguration C from a known initial conÞguration C0 is to assume that the totalload is applied in increments so that the body occupies several intermediateconÞgurations, Ci (i = 1, 2, . . .), prior to occupying the Þnal conÞguration.

xx x10 233 3

xx x10 222 2

xx x10 211 1

Q

P

ds

1C

Q

P

ds

0C Q

P

2C

dsu u0

u02

1

1

2

0


The magnitude of load increments should be such that the computationalmethod used is capable of predicting the deformed conÞguration at each loadstep. In the determination of an intermediate conÞguration Ci, the Lagrangiandescription of motion can use any of the previously known conÞgurations C0,C1, . . . , Ci−1 as the reference conÞguration CR. If the initial conÞguration isused as the reference conÞguration with respect to which all quantities aremeasured, it is called the total Lagrangian description. If the latest knownconÞguration Ci−1 is used as the reference conÞguration, it is called the updatedLagrangian description.

Here we introduce the notation used for positions, displacements, strains,stresses, etc. in the rest of this chapter. We consider three equilibriumconÞgurations of the body, namely, C0, C1, and C2, which correspond to threedifferent loads. As shown in Figure 9.3.1, the three conÞgurations of thebody can be thought of as the initial undeformed conÞguration C0, the lastknown deformed conÞguration C1, and the current deformed conÞguration C2to be determined. It is assumed that all variables, such as the displacements,strains, and stresses are known up to the C1 conÞguration. We wish to developa formulation for determining the displacement Þeld of the body in the currentdeformed conÞguration C2. It is assumed that the deformation of the bodyfrom C1 to C2 due to an increment in the load is small, and the accumulateddeformation of the body from C0 to C1 can be arbitrarily large but continuous(i.e. neighborhoods move into neighborhoods).

Figure 9.3.1 Initial and two consecutive conÞgurations of a body.


The following notation is adopted in the present study (see Bathe [4]). Aleft superscript on a quantity denotes the conÞguration in which the quantityoccurs, and a left subscript denotes the conÞguration with respect to whichthe quantity is measured. Thus i

jQ indicates that the quantity Q occursin conÞguration Ci but measured in conÞguration Cj . When the quantityunder consideration is measured in the same conÞguration in which it occurs,the left subscript may not be used. The left superscript will be omitted onincremental quantities that occur between conÞgurations C1 and C2. Of course,right subscript(s) refer to components in rectangular Cartesian coordinatesystems, as will be clear in the context of the discussion. Although somewhatcumbersome, the notation conveys the meaning more directly.

Since we are dealing with three different conÞgurations, it is necessary tointroduce the following symbols in the three conÞgurations [4,25]:

conÞguration: C0 C1 C2coordinates of a point: 0x 1x 2x

volumes: 0V 1V 2V

areas: 0A 1A, 2A (9.3.1)

density: 0ρ 1ρ 2ρ

total displacements of a point: 10u

20u

We use a rectangular Cartesian coordinate system to formulate theequations. We rewrite the equations presented earlier in a form suitablefor the current study. When the body deforms under the action ofexternal forces, a particle X occupying the position 0x = (0x1,

0x2,0x3) =

(X1,X2,X3) in conÞguration C0 moves to a new position 1x = (1x1,1x2,

1x3)in conÞguration C1 and position 2x = (2x1,

2x2,2x3) in conÞguration C2. The

total displacements of the particle X in the two conÞgurations C1 and C2 canbe written as

10u =

1x− 0x or 10ui =

1xi − 0xi (i = 1, 2, 3) (9.3.2)20u =

2x− 0x or 20ui =

2xi − 0xi (i = 1, 2, 3) (9.3.3)

and the displacement increment of the point from C1 to C2 is

u = 20u− 1

0u or ui =20ui − 1

0ui (i = 1, 2, 3) (9.3.4)

9.3.2 Conservation of Mass

The relations among the mass densities 0ρ, 1ρ, and 2ρ of materials inconÞgurations C0, C1, and C2, respectively, can be established using the


principle of conservation of mass. The principle requires that the mass ofa material body be conserved in moving through different conÞgurationsZ

2V

2ρ d 2V =

Z1V

1ρ d 1V =

Z0V

0ρ d 0V (9.3.5)

where iV is the volume of the body when it occupies the conÞguration Ci. Bya change of coordinates of integration from 2xi to

0xi, we haveZ2V

2ρ d 2V =

Z0V

2ρ 20J d0V (9.3.6)

where 20J is the determinant of the deformation gradient tensor

20F (or the

Jacobian of the transformation)

20J = det

³20F´=

¯¯∂ 2xi∂ 0xj

¯¯ =

¯¯ ∂

2x1∂ 0x1

∂ 2x1∂ 0x2

∂ 2x1∂ 0x3

∂ 2x2∂ 0x1

∂ 2x2∂ 0x2

∂ 2x2∂ 0x3

∂ 2x3∂ 0x1

∂ 2x3∂ 0x2

∂ 2x3∂ 0x3

¯¯ (9.3.7)

Equation (9.3.5) implies, in view of Eq. (9.3.6), that

0ρ = 2ρ 20J (9.3.8)

Similarly, we have1ρ = 2ρ 21J (9.3.9)

9.3.3 Green Strain Tensors for Various ConÞgurations

The Cartesian components of the GreenLagrange strain tensors, 10Eij and20Eij , in the two conÞgurations C1 and C2 are deÞned by

10Eij =

1

2

Ã∂ 1xk∂ 0xi

∂ 1xk∂ 0xj

− δij!

(9.3.10)

20Eij =

1

2

Ã∂ 2xk∂ 0xi

∂ 2xk∂ 0xj

− δij!

(9.3.11)

The strain components in Eqs. (9.3.10) and (9.3.11) can be expressed interms of the displacement components 10ui and

20ui. First, using Eqs. (9.3.2)

and (9.3.3), we can write

∂ 1xk∂ 0xi

=∂ 10uk∂ 0xi

+ δki,∂ 2xk∂ 0xi

=∂ 20uk∂ 0xi

+ δki (9.3.12)


Note that 0xi = Xi. Substituting into Eqs. (9.3.10) and (9.3.11), we obtain

10Eij =

1

2

Ã∂ 10ui∂ 0xj

+∂ 10uj∂ 0xi

+∂ 10uk∂ 0xi

∂ 10uk∂ 0xj

!(9.3.13)

20Eij =

1

2

Ã∂ 20ui∂ 0xj

+∂ 20uj∂ 0xi

+∂ 20uk∂ 0xi

∂ 20uk∂ 0xj

!(9.3.14)

GreenLagrange Incremental Strain Tensor

It is useful in the sequel to deÞne the incremental strain components 0εij , thatis, strains induced in moving from conÞguration C1 to conÞguration C2. TheGreenLagrange strain increment tensor is deÞned as

2 0εij d0xi d

0xj = (2ds)2 − (1ds)2

= [(2ds)2 − (0ds)2]− [(1ds)2 − (0ds)2]= 2

³20Eij − 1

0Eij´d 0xi d

0xj (9.3.15)

≡ 2 (0eij + 0ηij) d0xi d

0xj (9.3.16)

where 0eij are linear components of strain increment tensor

0eij =1

2

Ã∂ui∂ 0xj

+∂uj∂ 0xi

+∂ 10uk∂ 0xi

∂uk∂ 0xj

+∂uk∂ 0xi

∂ 10uk∂ 0xj

!(9.3.17)

and 0ηij are the nonlinear components

0ηij =1

2

∂uk∂ 0xi

∂uk∂ 0xj

(9.3.18)

The linearity of 0eij and nonlinearity of 0ηij is understood to be in terms ofthe incremental displacement components ui.

For geometrically linear analysis, that is, when strains are inÞnitesimal,only two conÞgurations C1 = C0 and C2 are involved, 1ui = 0, 2ui = ui,and ui,k are small enough to neglect their products. Consequently, the linearcomponents of strain increment tensor 0eij become the same as the componentsof the GreenLagrange strain tensor 20Eij , and both reduce to the inÞnitesimalstrain components

0eij =1

2

Ã∂ui∂ 0xj

+∂uj∂ 0xi

!(9.3.19)


Updated GreenLagrange Strain Tensor

The GreenLagrange strain tensor 20Eij introduced earlier is useful in the totalLagrangian formulation. With the updated Lagrangian formulation in mind,we deÞne the components of GreenLagrange strain tensor with respect toconÞguration C1. Such a strain tensor, denoted as 21εij , is called the updatedGreenLagrange strain tensor. It is deÞned by

2(21εij)d1xi d

1xj =³2ds

´2 − ³1ds´2 (9.3.20)

Using (note that all are Þrst-order derivatives)

d 2xi =∂ 2xi∂ 1xj

d 1xj , (2ds)2 =∂ 2xk∂ 1xi

∂ 2xk∂ 1xj

d 1xi d1xj (9.3.21)

ui =2xi − 1xi,

∂ 2xk∂ 1xi

=∂ uk∂ 1xi

+ δki (9.3.22)

we can write

21εij =

1

2

Ã∂ 2xk∂ 1xi

∂ 2xk∂ 1xj

− δij!

=1

2

Ã∂ui∂ 1xj

+∂uj∂ 1xi

+∂uk∂ 1xi

∂uk∂ 1xj

!(9.3.23)

≡ 1eij + 1ηij (9.3.24)

where

1eij =1

2

Ã∂ui∂ 1xj

+∂uj∂ 1xi

!, 1ηij =

1

2

Ã∂uk∂ 1xi

∂uk∂ 1xj

!(9.3.25)

Note that the deÞnition of 21εij involves the components of the displacementincrement vector u. Therefore, 21εij are also called the updated GreenLagrangestrain increment components.

9.3.4 Euler Strain Tensor

Suppose that a body occupying the undeformed conÞguration C0 takesa number of intermediate conÞgurations before occupying the deformedconÞguration C1, and we wish to determine the next conÞguration C2 in asingle incremental step. Although the accumulated displacements 1ui of thebody from conÞguration C0 to conÞguration C1 can be large, the incrementaldisplacements ui within the incremental step from C1 to C2 are assumedto be small. Hence, we may refer the strains to conÞguration C2. The


strains occurring in the body at conÞguration C2 and measured in the sameconÞguration are deÞned by

2 11εij d1xi d

1xj =³1ds

´2 − ³0ds´2 (9.3.26)

2 22εij d2xi d

2xj =³2ds

´2 − ³1ds´2 (9.3.27a)

=

Ãδij − ∂

1xk∂ 2xi

∂ 1xk∂ 2xj

!d 2xi d

2xj (9.3.27b)

These strains are called the Euler strain tensor components. Using

1xk =2xk − uk, ∂ 1xk

∂ 2xj= δkj − ∂uk

∂ 2xj(9.3.28)

we obtain (22εij ≡ 2εij)

2εij =1

2

Ã∂ui∂ 2xj

+∂uj∂ 2xi

− ∂uk∂ 2xi

∂uk∂ 2xj

!(9.3.29)

The linear part of the Euler strain tensor 2εij is called the inÞnitesimal straintensor and denoted 2eij

2eij =1

2

Ã∂ui∂ 2xj

+∂uj∂ 2xi

!(9.3.30)

which is identical in form to the inÞnitesimal strain tensor 0eij given in Eq.(9.3.19), except that the reference conÞguration is changed to C2. These straincomponents are energetically-conjugate to the Cauchy stress components.

9.3.5 Relationships Between Various Stress Tensors

The Cauchy stress components σij in conÞgurations C1 and C2 are denoted by1σij =

11σij ,

2σij =22σij (9.3.31)

The Cauchy stress components 1σij , for example, can be depicted in theconÞguration C1 by selecting a rectangular parallelepiped with faces parallelto the 1x1− 1x2,

1x2− 1x3,1x3− 1x1 planes much the same way as in small

strain formulations.

Recall that the second PiolaKirchhoff stress tensor characterizes thecurrent force in C2 but transformed to C0 and measured per unit area in C0:

(0n · S)d 0A = d 0f (9.3.32)


where 0n denotes the unit normal to the area element d 0A in C0. The forced 0f is related to the force d 2f by

d 0f = 20F

−1 · d 2f , 20F =

∂ 2x

∂ 0x(9.3.33)

where 20F is the deformation gradient tensor between conÞgurations C0 and C2.It is useful in the updated Lagrangian formulation to deÞne another kind

of stress tensor called the updated Kirchhoff stress tensor. Consider theinÞnitesimal rectangular parallelepiped containing the point P in conÞgurationC1 enclosed by the six surfaces

1xi = const.,1xi + d

1xi = const. (i = 1, 2, 3) (9.3.34)

The Cauchy stress components on this rectangular parallelepiped are 1σij . Asthe body deforms from conÞguration C1 to C2, this rectangular parallelepipedwill deform into, in general, a non-rectangular parallelepiped. Using theconÞguration C1 as the reference, we can deÞne the updated Kirchhoffstress components 21Sij as the internal forces per unit area acting along thenormal and two tangential directions of each of the side surfaces of theparallelepiped in the C2 conÞguration. The updated Kirchhoff stresses 21Sijcan be decomposed as

21Sij =

1σij + 1Sij (9.3.35)

where 1σij =11Sij are the Cauchy stress components in C1 conÞguration and

1Sij are the updated Kirchhoff stress increment tensor components.

The second PiolaKirchhoff stress tensor components in the C1 and C2conÞgurations are denoted by 10Sij and

20Sij , respectively. The components in

the two conÞgurations are related by

20Sij =

10Sij + 0Sij (9.3.36)

where 0Sij are the components of the Kirchhoff stress increment tensor.

Recall from Eqs. (9.2.25a,b) that the stress tensors 2σ and 20S are related

by

2σ =³20J´−1

F · 20S · FT, 2σij =³20J´−1Ã ∂ 2xi

∂ 0xm

!20Smn

Ã∂ 2xj∂ 0xn

!(9.3.37)

20S =

20JF

−1 · 2σ · F−T, 20Sij =

20J

Ã∂ 0xi∂ 2xm

!2σmn

Ã∂ 0xj∂ 2xn

!(9.3.38)

where 20J is the determinant of20F.


The relations between the Cauchy stress tensor components 2σij and thesecond PiolaKirchhoff stress tensor components 20Sij can also be written as[0ρ = 2ρ 20J ; see Eq. (9.3.8)]

2σij =2ρ0ρ

∂ 2xi∂ 0xp

∂ 2xj∂ 0xq

20Spq (9.3.39)

20Sij =

0ρ2ρ

∂ 0xi∂ 2xp

∂ 0xj∂ 2xq

2σpq (9.3.40)

where 0ρ and 2ρ represent the mass densities of the material in conÞgurationsC0 and C2, respectively.The Cauchy stress tensor components 2σij in C2 can be related to the

updated Kirchhoff stress tensor components 21Sij by the following formulae:

21Sij =

1ρ2ρ

∂ 1xi∂ 2xp

∂ 1xj∂ 2xq

2σpq (9.3.41)

2σij =2ρ1ρ

∂ 2xi∂ 1xp

∂ 2xj∂ 1xq

21Spq (9.3.42)

The transformation between the second PiolaKirchhoff stress tensorcomponents in different reference conÞgurations is provided by the followingrelations:

20Sij =

0ρ1ρ

∂ 0xi∂ 1xp

∂ 0xj∂ 1xq

21Spq (9.3.43)

10Sij =

0ρ1ρ

∂ 0xi∂ 1xp

∂ 0xj∂ 1xq

1σpq (9.3.44)

Finally, from Eqs. (9.3.43) and (9.3.44), we obtain the relations between theincremental stresses 0Sij and 1Spq:

0Sij =0ρ1ρ

∂ 0xi∂ 1xp

∂ 0xj∂ 1xq

1Spq (9.3.45)

1Sij =1ρ0ρ

∂ 1xi∂ 0xp

∂ 1xj∂ 0xq

0Spq (9.3.46)

These relations are useful in the calculation of the material coefficients forlarge strain problems in the updated Lagrangian formulation, as will be seenshortly.


9.4 Constitutive Equations

Materials for which the constitutive behavior is only a function of the currentstate of deformation are known as elastic. In the special case in which thework done by the stresses during a deformation is dependent only on theinitial state and the current conÞguration, the material is called hyperelastic.For hyperelastic materials we assume that there exists a stored strain energyfunction, U0 per unit undeformed volume, such that the components of thematerial elasticity tensor C are given by

Cijk` =∂Sij∂Ek`

(9.4.1)

In the derivation of Þnite element models of incremental nonlinear analysisof solid continua, it is necessary to specify the stressstrain relations inincremental form. In the total Lagrangian formulation, for example, theconstitutive relations can be expressed in terms of the Kirchhoff stressincrement tensor components 0Sij and GreenLagrange strain incrementtensor components 0εij as

0Sij = 0Cijk` 0εk` (9.4.2)

In the updated Lagrangian formulation, it can be expressed in terms ofthe updated Kirchhoff stress increment tensor components 1Sij and updatedGreenLagrange strain increment tensor components 1εij :

1Sij = 1Cijk` 1εk` (9.4.3)

where 0Cijkl and 1Cijkl denote the incremental constitutive tensors withrespect to the conÞgurations C0 and C1, respectively. It can be shown that thefollowing transformation rules hold between the components of the elasticitytensor C in different conÞgurations:

0Cijk` =0ρ1ρ

∂ 0xi∂ 1xp

∂ 0xj∂ 1xq

∂ 0xk∂ 1xr

∂ 0x`∂ 1xs

1Cpqrs (9.4.4)

1Cijkl =1ρ0ρ

∂ 1xi∂ 0xp

∂ 1xj∂ 0xq

∂ 1xk∂ 0xr

∂ 1x`∂ 0xs

0Cpqrs (9.4.5)

With these two equations, only one set of coefficients, namely, 0Cijkl or 1Cijkl,need to be known or given. The other set of coefficients can be obtained simplyby transformation. We can therefore ensure that the material propertiesimplied by the total and updated Lagrangian formulations are identical toeach other, provided that the same step sizes are used in both formulations.

Often, the incremental material laws (9.4.2) and (9.4.3) with identicalcoefficients, that is, 0Cijk` = 1Cijk`, are employed in the total and updated


Lagrangian Þnite element formulations. Of course, the error introduced bythe assumption can be negligible if the strains are relatively small but thedifference can be signiÞcant in large deformation problems, for example, inhigh velocity impact problems.

9.5 Total Lagrangian and Updated LagrangianFormulations of Continua

9.5.1 Principle of Virtual Displacements

The equations of the Lagrangian incremental description of motion can bederived from the principles of virtual work (i.e. virtual displacements, virtualforces, or mixed virtual displacements and forces). Since our ultimate objectiveis to develop the Þnite element model of the equations governing a body, wewill not actually derive the differential equations of motion but utilize thevirtual work statements to develop the Þnite element models.

The displacement Þnite element model is based on the principle of virtualdisplacements. The principle requires that the sum of the external virtualwork done on a body and the internal virtual work stored in the body shouldbe equal to zero:

δW ≡Z2V

2σ : δ(2e) d2V − δ 2R = 0 (9.5.1a)

=

Z2V

2σij δ(2eij) d2V − δ 2R = 0 (9.5.1b)

where δ 2R denotes the virtual work done by applied forces

δ 2R =

Z2V

2f · δu d 2V +Z2S

2t · δu d 2S (9.5.2a)

=

Z2V

2fiδui d2V +

Z2S

2tiδui d2S (9.5.2b)

and d 2S denotes surface element and 2f is the body force vector (measured perunit volume) and 2t is the boundary stress (or traction) vector (measured perunit surface area) in conÞguration C2. The variational symbol δ is understoodto operate on unknown displacement variables (2ui and ui).

Equation (9.5.1) cannot be solved directly since the conÞguration C2 isunknown. This is an important difference compared with the linear analysisin which we assume that the displacements are inÞnitesimally small so thatthe conÞguration of the body does not change. In a large deformation analysisspecial attention must be given to the fact that the conÞguration of the bodyis changing continuously. This change in conÞguration can be dealt withby deÞning appropriate stress and strain measures. The objective of their


introduction into the analysis is to express the internal work in Eq. (9.5.1) interms of an integral over a conÞguration that is known. The stress and strainmeasures that we shall use are the 2nd PiolaKirchhoff stress tensor and theGreenLagrange strain tensor, which are energetically congugate to eachother [see Eq. (9.2.34)].

9.5.2 Total Lagrangian Formulation

In the total Lagrangian formulation, all quantities are measured with respectto the initial conÞguration C0. Hence, the virtual work statement in Eq. (9.5.1)must be expressed in terms of quantities referred to the reference conÞguration.We use the following identities [4]Z

2V

2σij δ(2eij) d2V =

Z0V

20Sij δ(

20Eij) d

0V (9.5.3)Z2V

2fi δui d2V =

Z0V

20fi δui d

0V (9.5.4)Z2S

2ti δui d2S =

Z0S

20ti δui d

0S (9.5.5)

where 20fi and

20ti are the body force and boundary traction components

referred to the conÞguration C0. Using Eqs. (9.5.3)(9.5.5) in Eq. (9.5.1b) wearrive at Z

0V

20Sij δ(

20Eij) d

0V − δ(20R) = 0 (9.5.6)

where

δ(20R) =

Z0V

20fi δui d

0V +

Z0S

20ti δui d

0S (9.5.7)

Next we simplify the virtual work statement (9.5.6). First, we note that[see Eqs. (9.3.15) and (9.3.16)]

δ(20Eij) = δ(10Eij) + δ(0εij) = δ(0εij)

= δ(0eij) + δ(0ηij) (9.5.8)

where δ(10Eij) = 0 because it is not a function of the unknown displacements.The virtual strains are given by

δ(0eij) =1

2

Ã∂δui∂ 0xj

+∂δuj∂ 0xi

+∂δuk∂ 0xi

∂ 10uk

∂ 0xj+∂ 10uk

∂ 0xi

∂δuk∂ 0xj

!(9.5.9)

δ(0ηij) =1

2

Ã∂δuk∂ 0xi

∂uk∂ 0xj

+∂uk∂ 0xi

∂δuk∂ 0xj

!(9.5.10)


Substituting Eqs. (9.5.8) for δ(20Eij) and Eq. (9.3.36) for20Sij into Eq. (9.5.6),

we arrive at the expression

0 =

Z0V

20Sij δ(

20Eij) d

0V − δ(20R)

=

Z0V

³10Sij + 0Sij

´δ(0εij) d

0V − δ(20R)

=

Z0V

n0Sij δ(0εij) +

10Sij [δ(0eij) + δ(0ηij)]

od 0V − δ(20R)

=

Z0V

0Sij δ(0εij) d0V +

Z0V

10Sij δ(0ηij) d

0V + δ(10R)− δ(20R)(9.5.11)

where δ(10R) is the virtual internal energy (in moving the actual internal forcesthrough virtual displacements) stored in the body at conÞguration C1

δ(10R) =

Z0V

10Sij δ(0eij) d

0V (9.5.12)

Since the body is in equilibrium at conÞguration C1, by the principle of virtualwork applied to conÞguration C1 we have

0 =

Z0V

10Sij δ(0eij) d

0V −Z0V

10fi δui d

0V −Z0S

10tiδui d

0S (9.5.13)

and therefore

δ(10R) =

Z0V

10fi δui d

0V +

Z0S

10ti δui d

0S (9.5.14)

Equation (9.5.11) forms the basis for the Þnite element model. We onlyneed to replace 0Sij in terms of the strains and ultimately the displacementincrements using an appropriate constitutive relation. The Þrst term of Eq.(9.5.11) represents the change in the virtual strain energy due to the virtualincremental displacements ui between conÞgurations C1 and C2. The secondterm represents the virtual work done by forces due to initial stresses 10Sij . Thelast two terms together denote the change in the virtual work done by appliedbody forces and surface tractions in moving from C1 to C2. This is primarilydue to the geometric changes that take place between the two conÞgurations.Equation (9.5.11) represents the statement of virtual work for the incrementaldeformation between the conÞgurations C1 and C2, and no approximations aremade in arriving at it.

Towards constructing the displacement Þnite element model of Eq. (9.5.11),we invoke the constitutive relation (9.4.2) to express 0Sij in terms of theincremental strain components 0εij . Equation (9.5.11) takes the formZ

0V0Cijk` 0εk` δ(0εij) d

0V +

Z0V

10Sij δ(0ηij) d

0V = δ(20R)−δ(10R) (9.5.15)


Equation (9.5.15) is nonlinear in the displacement increments ui. To makeit computationally tractable, we assume that the displacements ui are small(which is indeed the case provided the load step is small in moving from C1 toC2) so that the following approximations hold:

0Sij ≈ 0Cijk` 0ek` and δ(0εij) ≈ δ(0eij) (9.5.16)

Then Eq. (9.5.15) can be simpliÞed toZ0V

0Cijk` 0ek` δ(0eij) d0V +

Z0V

10Sij δ(0ηij) d

0V = δ(20R)−δ(10R) (9.5.17)

Equation (9.5.17) is the weak form for the development of the Þnite elementmodel based on the total Lagrangian formulation. The total stress components10Sij are evaluated using the constitutive relation

10Sij = 0Cijk`

10Ek` (9.5.18)

where 10Ek` are the GreenLagrange strain tensor components deÞned in Eq.(9.3.13).

A summary of all the pertinent equations of the total Lagrangianformulation is presented in Table 9.5.1.

9.5.3 Updated Lagrangian Formulation

In the updated Lagrangian formulation, all quantities are referred to the latestknown conÞguration, namely C1. Hence, the virtual work statement in Eq.(9.5.1b) must be recast in terms of quantities referred to C1. We use theidentities Z

2V

2σij δ(2eij) d2V =

Z1V

21Sij δ(

21εij) d

1V (9.5.19a)Z2V

2fi δui d2V =

Z1V

21fi δui d

1V (9.5.19b)Z2S

2ti δui d2S =

Z1S

21ti δui d

1S (9.5.19c)

where 21fi and

21ti are the body force and boundary traction components

referred to the conÞguration C1, and 21εij are the components of the updated

GreenLagrange strain tensor components deÞned in Eq. (9.3.23). Using Eqs.(9.5.19ac), Eq. (9.5.1b) can be expressed asZ

1V

21Sij δ(

21εij) d

1V − δ(21R) = 0 (9.5.20)


where

δ(21R) =

Z1V

21fi δui d

1V +

Z1S

21ti δui d

1S (9.5.21)

Table 9.5.1 Summary of equations of the total Lagrangian formulation.

1. Weak form

0 =

Z0V

20Sij δ(

20Eij) d

0V − δ(20R) (9.5.6)

20Sij =

20J

µ∂ 0xi∂ 2xm

¶2σmn

µ∂ 0xj∂2xn

¶(9.3.40)

20Eij =

1

2

µ∂ 20ui

∂ 0xj+∂ 20uj

∂ 0xi+∂ 20uk

∂ 0xi

∂ 20uk

∂ 0xj

¶(9.3.14)

δ(20R) =

Z0V

20fiδui d

0V +

Z0S

20tiδui d

0S (9.5.7)

2. Incremental decompositions

20Sij =

10Sij + 0Sij (9.3.36)

20Eij =

10Eij + 0εij , 0εij =0 eij + 0ηij (9.3.15)&(9.3.16)

10Sij = 0Cijk`

10Ek`, 0Sij = 0Cijk` 0εk` (9.4.2)

10Eij =

1

2

µ∂ 10ui

∂ 0xj+∂ 10uj

∂ 0xi+∂ 10uk

∂ 0xi

∂ 10uk

∂ 0xj

¶(9.3.13)

0eij =1

2

µ∂ui∂ 0xj

+∂uj∂ 0xi

+∂ 10uk

∂ 0xi

∂uk∂ 0xj

+∂uk∂ 0xi

∂ 10uk

∂ 0xj

¶(9.3.17)

0ηij =1

2

∂uk∂ 0xi

∂uk∂ 0xj

(9.3.18)

3. Weak form with incremental decompositionsZ0V

0Sij δ(0εij) d0V +

Z0V

10Sij δ(0ηij) d

0V = δ(20R)−Z0V

10Sij δ(0eij) d

0V

(9.5.11)&(9.5.12)

4. Linearized weak form with incremental decompositions Use the approximations

0Sij = 0Cijk` 0εk` ≈ 0Cijk` 0ek`, δ(0εij) ≈ δ(0eij) (9.5.16)

to rewrite the weak form asZ0V


Z0V

10Sij δ(0ηij) d

0V

= δ(20R)−Z0V

10Sij δ(0eij) d

0V (9.5.17)


The virtual strains are given by δ(21εij) = δ(1eij) + δ(1ηij) [see Eq. (9.3.24)]where

δ(1eij) =1

2

Ã∂δui∂ 1xj

+∂δuj∂ 1xi

!(9.5.22)

δ(1ηij) =1

2

Ã∂δuk∂ 1xi

∂uk∂ 1xj

+∂uk∂ 1xi

∂δuk∂ 1xj

!(9.5.23)

Now using Eqs. (9.3.24) and (9.3.35), we can write Eq. (9.5.20) as

0 =

Z1V

21Sij δ(

21εij) d

1V − δ(21R)

=

Z1V

³1σij + 1Sij

´δ(21εij) d

1V − δ(21R)

=

Z1V

n1Sij δ(

21εij) +

1σij [δ(1eij) + δ(1ηij)]od 1V − δ(21R)

=

Z1V

1Sij δ(21εij) d

1V +

Z1V

1σij δ(1ηij) d1V + δ(11R)− δ(21R)(9.5.24)

where δ(11R) is the virtual internal energy stored in the body at conÞgurationC1

δ(11R) =

Z1V

1σij δ(1eij) d1V (9.5.25)

Since the body is in equilibrium at conÞguration C1, the principle of virtualwork applied to conÞguration C1 yields

δ(11R) =

Z1V

11fi δui d

1V +

Z1S

11ti δui d

1S (9.5.26)

Next, we invoke the constitutive relation (9.4.3) to express 1Sij in terms ofthe incremental strain components 21εij . Equation (9.5.24) takes the formZ

1V1Cijk`

21εk` δ(

21εij) d

1V +

Z1V

1σij δ(1ηij) d1V

= δ(21R)−Z1V

1σij δ(1eij) d1V (9.5.27)

As before, we assume that the displacements ui are small so that the followingapproximations hold:

1Sij ≈ 1Cijk` 1ek` and δ(21εij) ≈ δ(1eij) (9.5.28)

Then Eq. (9.5.27) takes the formZ1V


Z1V

1σij δ(1ηij) d1V

= δ(21R)−Z1V

1σij δ(1eij) d1V (9.5.29)


Equation (9.5.29) is the weak form for the development of the Þnite elementmodel based on the updated Lagrangian formulation. The total Cauchy stresscomponents 1σij are evaluated using the constitutive relation

1σij = 1Cijk`11εk` (9.5.30)

where 11εk` are the Almansi strain tensor components deÞned by [see Eq.

(9.3.26)]

2 11εij d1xi d

1xj =³1ds

´2 − ³0ds´2 (9.5.31)

which gives

11εij =

1

2

Ã∂ 10ui

∂ 1xj+∂ 10uj

∂ 1xi− ∂

10uk

∂ 1xi

∂ 10uk

∂ 1xj

!(9.5.32)

A summary of all the pertinent equations of the updated Lagrangianformulation is presented in Table 9.5.2.

9.6 Finite Element Models of Two-DimensionalContinua

9.6.1 Introduction

Here we discuss Þnite element models based on the total and updatedLagrangian formulations presented in the last section. Attention is focusedhere on two-dimensional problems under the assumption of linear elasticbehavior. Further, we assume orthotropic behavior.

9.6.2 Total Lagrangian Formulation

We introduce the notation

0x1 = x,0x2 = y,

1u1 = u,1u2 = v, u1 = u, u2 = v (9.6.1)

We can write the Þrst expression of Eq. (9.5.17) in the alternate formZ0V

0Cijk` 0ek` δ(0eij) d0V =

Z0Vδ0eT[0C]0e d 0V

=

Z0VδuT ([D] + [Du])T [0C] ([D] + [Du]) u d 0V (9.6.2)

where

0e =⎧⎨⎩ 0exx

0eyy2 0exy

⎫⎬⎭ =⎧⎪⎨⎪⎩

∂u∂x∂v∂y

∂u∂y +

∂v∂x

⎫⎪⎬⎪⎭+⎧⎪⎨⎪⎩

∂u∂x

∂u∂x +

∂v∂x

∂v∂x

∂u∂y∂u∂y +

∂v∂y∂v∂y

∂u∂x

∂u∂y +

∂v∂x

∂v∂y +

∂u∂x

∂u∂y +

∂v∂x

∂v∂y

⎫⎪⎬⎪⎭


=

⎛⎜⎝⎡⎢⎣∂∂x 0

0 ∂∂y

∂∂y

∂∂x

⎤⎥⎦+⎡⎢⎣

∂u∂x

∂∂x

∂v∂x

∂∂x

∂u∂y

∂∂y

∂v∂y

∂∂y

∂u∂y

∂∂x +

∂u∂x

∂∂y

∂v∂y

∂∂x +

∂v∂x

∂∂y

⎤⎥⎦⎞⎟⎠½ u

v

¾

≡ ([D] + [Du]) u (9.6.3)

Table 9.5.2 Summary of equations of the updated Lagrangian formulation.

1. Weak form

0 =

Z1V

21Sij δ(

21εij) d

1V − δ(21R) (9.5.20)

21Sij =

1ρ2ρ

∂1xi∂2xp

∂1xj∂2xq

2σpq (9.3.41)

21εij =

1

2

µ∂ui∂ 1xj

+∂uj∂ 1xi

+∂uk∂ 1xi

∂uk∂ 1xj

¶(9.3.23)

δ(21R) =

Z1V

21fi δui d

1V +

Z1S

21ti δui d

1S (9.5.21)

2. Incremental decompositions

21Sij =

1σij + 1Sij ,11Sij =

1σij (9.3.35)21εij = 1eij + 1ηij (9.3.24)1σij = 1Cijk`

11εk`, 1Sij = 1Cijk`

21εk` (9.5.30)

11εij =

1

2

µ∂ 10ui

∂ 1xj+∂ 10uj

∂ 1xi− ∂

10uk

∂ 1xi

∂ 10uk

∂ 1xj

¶(9.5.32)

1eij =1

2

µ∂ui∂ 1xj

+∂uj∂ 1xi

¶, 1ηij =

1

2

µ∂uk∂ 1xi

∂uk∂ 1xj

¶(9.3.25)

3. Weak form with incremental decompositionsZ1V

1Sij δ(21εij) d

1V +

Z1V

1σij δ(1ηij) d1V = δ(21R)−

Z1V

1σij δ(1eij) d1V

(9.5.24)&(9.5.25)

4. Linearized weak form with incremental decompositions Use the approximations

1Sij = 1Cijk`21εk` ≈ 1Cijk` 1ek`, δ(1εij) ≈ δ(1eij) (9.5.28)

to rewrite the weak form asZ1V


Z1V

1σij δ(1ηij) d1V

= δ(21R)−Z1V

1σij δ(1eij) d1V (9.5.29)


δ 0e =⎧⎨⎩δ 0exxδ 0eyy2δ 0exy

⎫⎬⎭ = ([D] + [Du]) δu (9.6.4)

[0C] =

⎡⎣ 0C11 0C12 0

0C12 0C22 00 0 0C66

⎤⎦ (9.6.5)

The second expression of Eq. (9.5.17) can be written asZ0V

10Sij δ(0ηij) d

0V =

Z0Vδ 0ηT10S d 0V (9.6.6)

0η =⎧⎨⎩ 0ηxx

0ηyy2 0ηxy

⎫⎬⎭ = 1

2

⎧⎪⎨⎪⎩∂u∂x

∂u∂x +

∂v∂x

∂v∂x

∂u∂y∂u∂y +

∂v∂y∂v∂y

2³∂u∂x

∂u∂y +

∂v∂x

∂v∂y

´⎫⎪⎬⎪⎭

=1

2

⎡⎢⎣∂u∂x

∂∂x

∂v∂x

∂∂x

∂u∂y

∂∂y

∂v∂y

∂∂y

∂u∂y

∂∂x +

∂u∂x

∂∂y

∂v∂y

∂∂x +

∂v∂x

∂∂y

⎤⎥⎦⎧⎨⎩u

v

⎫⎬⎭=1

2[Du]u (9.6.7)

δ 0η =⎧⎨⎩δ 0ηxxδ 0ηyy2δ 0ηxy

⎫⎬⎭ =⎧⎪⎨⎪⎩

∂δu∂x

∂u∂x +

∂δv∂x

∂v∂x

∂δu∂y

∂u∂y +

∂δv∂y

∂v∂y

∂δu∂x

∂u∂y +

∂u∂x

∂δu∂y +

∂δv∂x

∂v∂y +

∂v∂x∂δv∂y

⎫⎪⎬⎪⎭= [Du]δu = [Dδu]u (9.6.8)

10S =⎧⎨⎩10Sxx10Syy10Sxy

⎫⎬⎭ =⎡⎣ 0C11 0C12 0

0C12 0C22 00 0 0C66

⎤⎦⎧⎨⎩10Exx10Eyy2 10Exy

⎫⎬⎭ (9.6.9a)

10E =⎧⎨⎩

10Exx10Eyy2 10Exy

⎫⎬⎭ =⎧⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎩

∂u∂x +

12

∙³∂u∂x

´2+³∂v∂x

´2¸∂v∂y +

12

∙³∂u∂y

´2+³∂v∂y

´2¸∂u∂y +

∂v∂x +

³∂u∂x

∂u∂y +

∂v∂x

∂v∂y

´

⎫⎪⎪⎪⎪⎪⎬⎪⎪⎪⎪⎪⎭(9.6.9b)

However, the expression (9.6.6) is not convenient for the Þnite elementdevelopment since the vector δ 0η is a nonlinear function of the vector


of displacement increments u. It must be written in an alternate way tofacilitate the construction of Þnite element matrices. We haveZ

0Vδ 0ηT10S d 0V

=

Z0V

∙10Sxx

µ∂δu

∂x

∂u

∂x+∂δv

∂x

∂v

∂x

¶+ 1

0Syy

µ∂δu

∂y

∂u

∂y+∂δv

∂y

∂v

∂y

¶+ 1

0Sxy

µ∂δu

∂x

∂u

∂y+∂δv

∂x

∂v

∂y+∂u

∂x

∂δu

∂y+∂v

∂x

∂δv

∂y

¶¸d 0V

=

Z0V

⎧⎪⎪⎪⎨⎪⎪⎪⎩∂δu∂x∂δu∂y∂δv∂x∂δv∂y

⎫⎪⎪⎪⎬⎪⎪⎪⎭T ⎡⎢⎢⎣

10Sxx

10Sxy 0 0

10Sxy

10Syy 0 0

0 0 10Sxx

10Sxy

0 0 10Sxy

10Syy

⎤⎥⎥⎦⎧⎪⎪⎪⎨⎪⎪⎪⎩∂u∂x∂u∂y∂v∂x∂v∂y

⎫⎪⎪⎪⎬⎪⎪⎪⎭ d0V

=

Z0VδuT[D]T[ 10S][D]u d 0V (9.6.10)

where

[10S] =

⎡⎢⎢⎣10Sxx

10Sxy 0 0

10Sxy

10Syy 0 0

0 0 10Sxx

10Sxy

0 0 10Sxy

10Syy

⎤⎥⎥⎦ (9.6.11a)

10S =⎧⎨⎩10Sxx10Syy10Sxy

⎫⎬⎭ , [D] =

⎡⎢⎢⎢⎣∂∂x 0∂∂y 0

0 ∂∂x

0 ∂∂y

⎤⎥⎥⎥⎦½uv

¾(9.6.11b)

This completes the development of alternate form of the expressions in thevirtual work statement (9.5.17).

Suppose that the total and incremental displacement Þelds are interpolatedas

u =½uv

¾=

½Pnj=1 ujψj(x)Pnj=1 vjψj(x)

¾= [Ψ]∆ (9.6.12)

u =½uv

¾=

½Pnj=1 ujψj(x)Pnj=1 vjψj(x)

¾= [Ψ]∆ (9.6.13)

where

[Ψ] =

∙ψ1 0 ψ2 0 · · · ψn 00 ψ1 0 ψ2 · · · 0 ψn

¸(9.6.14a)

∆T = u1, v1, u2, v2, · · · , un, vn (9.6.14b)

∆T = u1, v1, u2, v2, · · · , un, vn (9.6.14c)


We haveZ0V

0δ0eT[0C]0e d 0V

=

Z0VδuT ([D] + [Du])T [0C] ([D] + [Du]) u d 0V

=

Z0Vδ∆T ([D] + [Du]) [Ψ]T [0C] ([D] + [Du]) [Ψ]∆ d 0V

=

Z0Vδ∆T[BL]T[0C][BL]∆ d 0V (9.6.15)

Z0Vδ 0ηT10S d 0V

=

Z0VδuT[D]T[ 10S][D]u d 0V

=

Z0Vδ∆T[BNL]T[ 10S][BNL]∆ d 0V (9.6.16)

δ(10R) =

Z0V

10Sij δ(0eij) d

0V

=

Z0Vδ 0eT10S d 0V

=

Z0Vδ∆T[BL]T10S d 0V (9.6.17)

δ(20R) =

Z0V

20fiδui d

0V +

Z0S

20tiδui d

0S

=

Z0Vδ∆T[Ψ]T20f d 0V +

Z0Sδ∆T[Ψ]T20t d 0S (9.6.18)

where [BL] is the 3× 2n matrix and [BNL] is the 4× 2n matrix deÞned by[BL] = ([D] + [Du]) [Ψ], [BNL] = [D][Ψ] (9.6.19)

Substitution of Eqs. (9.6.15)(9.6.18) into Eq. (9.5.17), and use of thefundamental Lemma of calculus of variations (i.e. δ∆ arbitrary variations),yields the following Þnite element model associated with the total Lagrangianformulation of two-dimensional nonlinear continua:

([KL] + [KNL]) ∆ = 20F− 10F (9.6.20)

where

[KL] =

Z0V[BL]

T[0C][BL] d0V (9.6.21)

[KNL] =

Z0V[BNL]

T[ 10S][BNL] d0V (9.6.22)


10F =Z0V[BL]

T10S d 0V (9.6.23)

20F =Z0V[Ψ]T20f d 0V +

Z0S[Ψ]T20t d 0S (9.6.24)

20f =½ 20fx20fy

¾, 20t =

½ 20tx20ty

¾(9.6.25)

Note that the stiffness matrix [K] = [KL] + [KNL] is symmetric since[ 10S] and [0C] are symmetric. Also, the total and updated Lagrangianformulations are incremental formulations, that is, determining δ∆ ≡ ∆,the stiffness matrix in Eq. (9.6.20) is the tangent stiffness matrix. The directstiffness matrix is implicit in the vector 10F. For a linear analysis, we have∆ = ∆, 10F = 0, and [KNL] = [0]. The formulation presented aboveis easily extendable to three-dimensional problems.

For two-dimensional problems, the matrices [BL] and [BNL] are given by

[BL] = [B0L] + [B

uL] + [B

vL] (9.6.26)

[B0L] =

⎡⎢⎣∂ψ1∂x 0 ∂ψ2

∂x 0 · · · ∂ψn∂x 0

0 ∂ψ1∂y 0 ∂ψ2

∂y · · · 0 ∂ψn∂y

∂ψ1∂y

∂ψ1∂x

∂ψ2∂y

∂ψ2∂x · · · ∂ψn

∂y∂ψn∂x

⎤⎥⎦ (9.6.27a)

[BuL] =

⎡⎢⎣∂u∂x

∂ψ1∂x 0 ∂u

∂x∂ψ2∂x · · · ∂u

∂x∂ψn∂x 0

∂u∂y∂ψ1∂y 0 ∂u

∂y∂ψ2∂y · · · ∂u

∂y∂ψn∂y 0

∂u∂x

∂ψ1∂y +

∂u∂y∂ψ1∂x 0 ∂u

∂x∂ψ2∂y +

∂u∂y∂ψ2∂x · · · ∂u

∂x∂ψn∂y +

∂u∂y∂ψn∂x 0

⎤⎥⎦(9.6.27b)

[BvL] =

⎡⎢⎣ 0∂v∂x

∂ψ1∂x 0 ∂v

∂x∂ψ2∂x · · · ∂v

∂x∂ψn∂x

0 ∂v∂y∂ψ1∂y 0 ∂v

∂y∂ψ2∂y · · · ∂v

∂y∂ψn∂y

0 ∂v∂x

∂ψ1∂y +

∂v∂y∂ψ1∂x 0 ∂v

∂x∂ψ2∂y +

∂v∂y∂ψ2∂x · · · ∂v

∂x∂ψn∂y +

∂v∂y∂ψn∂x

⎤⎥⎦(9.6.27c)

[BNL] =

⎡⎢⎢⎢⎢⎣∂ψ1∂x 0 ∂ψ2

∂x 0 · · · ∂ψn∂x 0

∂ψ1∂y 0 ∂ψ2

∂y 0 · · · ∂ψn∂y 0

0 ∂ψ1∂x 0 ∂ψ2

∂x · · · 0 ∂ψn∂x

0 ∂ψ1∂y 0 ∂ψ2

∂y · · · 0 ∂ψn∂y

⎤⎥⎥⎥⎥⎦ (9.6.28)

The Þnite element equations (9.6.20) can be written in explicit form as∙[K11L] + [K11N ] [K12L]

[K21L] [K22L] + [K22N ]

¸½ uv

¾=

½ 20F 1− 10F 120F 2− 10F 2

¾(9.6.29)


where

K11Lij = he

ZΩe

½0C11

µ1 +

∂u

∂x

¶2 ∂ψi∂x

∂ψj∂x

+ 0C22

µ∂u

∂y

¶2 ∂ψi∂y

∂ψj∂y

+ 0C12

µ1 +

∂u

∂x

¶∂u

∂y

µ∂ψi∂x

∂ψj∂y

+∂ψi∂y

∂ψj∂x

¶+ 0C66

∙µ1 +

∂u

∂x

¶∂ψi∂y

+∂u

∂y

∂ψi∂x

¸×∙µ1 +

∂u

∂x

¶∂ψj∂y

+∂u

∂y

∂ψj∂x

¸¾dx dy

K12Lij = he

ZΩe

½1C11

µ1 +

∂u

∂x

¶∂v

∂x

∂ψi∂x

∂ψj∂x

+ 0C22

µ1 +

∂v

∂y

¶∂u

∂y

∂ψi∂y

∂ψj∂y

+ 0C12

∙µ1 +

∂u

∂x

¶µ1 +

∂v

∂y

¶∂ψi∂x

∂ψj∂y

+∂u

∂y

∂v

∂x

∂ψi∂y

∂ψj∂x

¸+ 0C66

∙µ1 +

∂u

∂x

¶∂ψi∂y

+∂u

∂y

∂ψi∂x

¸×∙µ1 +

∂v

∂y

¶∂ψj∂x

+∂v

∂x

∂ψj∂y

¸¾dx dy = K21L

ji

K22Lij = he

ZΩe

½0C11

µ∂v

∂x

¶2 ∂ψi∂x

∂ψj∂x

+ 0C22

µ1 +

∂v

∂y

¶2 ∂ψi∂y

∂ψj∂y

+ 0C12

µ1 +

∂v

∂y

¶∂v

∂x

µ∂ψi∂x

∂ψj∂y

+∂ψi∂y

∂ψj∂x

¶+ 0C66

∙µ1 +

∂v

∂y

¶∂ψi∂x

+∂v

∂x

∂ψi∂y

¸×∙µ1 +

∂v

∂y

¶∂ψj∂x

+∂v

∂x

∂ψj∂y

¸¾dx dy

K11Nij = he

ZΩe

∙10Sxx

∂ψi∂x

∂ψj∂x

+ 10Sxy

µ∂ψi∂y

∂ψj∂x

+∂ψi∂x

∂ψj∂y

¶+ 1

0Syy∂ψi∂y

∂ψj∂y

¸dx dy = K22N

ij (9.6.30a)

20F

1i = he

ZΩe

20fxψi dx dy + he

IΓe

20txψi ds

20F

2i = he

ZΩe

20fyψi dx dy + he

IΓe

20tyψi ds

10F

1i = he

ZΩe

½µ1 +

∂u

∂x

¶∂ψi∂x

10Sxx +

∂u

∂y

∂ψi∂y

10Syy

+

∙µ1 +

∂u

∂x

¶∂ψi∂y

+∂u

∂y

∂ψi∂x

¸10Sxy

¾dx dy


10F

2i = he

ZΩe

½∂v

∂x

∂ψi∂x

10Sxx +

µ1 +

∂v

∂y

¶∂ψi∂y

10Syy

+

∙µ1 +

∂v

∂y

¶∂ψi∂x

+∂v

∂x

∂ψi∂y

¸10Sxy

¾dx dy (9.6.30b)

where he is the thickness of the element (for the plane elastic problem).

9.6.3 Updated Lagrangian Formulation

In view of the detailed discussion of the Þnite element model development forthe total Lagrangian formulation and the similarity between Eqs. (9.5.17)and (9.5.29), the Þnite element model based on the updated Lagrangianformulation can be simply written as

([KL] + [KNL]) ∆ = 21F− 11F (9.6.31)

where

[KL] =

Z0V[B0L]

T[1C][B0L] d

1V (9.6.32)

[KNL] =

Z1V[BNL]

T[ 1σ][BNL] d1V (9.6.33)

11F =Z1V[B0L]

T1σ d 1V (9.6.34)

21F =Z1V[Ψ]T21f d 1V +

Z1S[Ψ]T21t d 1S (9.6.35)

21f =½ 21fx21fy

¾, 21t =

½ 21tx21ty

¾(9.6.36)

where [B0L] and [BNL] are deÞned by Eqs. (9.6.27a) and (9.6.28), respectively,and

[1σ] =

⎡⎢⎢⎣1σxx

1σxy 0 01σxy

1σyy 0 00 0 1σxx

1σxy0 0 1σxy

1σyy

⎤⎥⎥⎦ (9.6.37a)

1σ =⎧⎨⎩1σxx1σyy1σxy

⎫⎬⎭ =⎡⎣ 1C11 1C12 0

1C12 1C22 00 0 1C66

⎤⎦⎧⎨⎩11εxx11εyy2 11εxy

⎫⎬⎭ (9.6.37b)

11ε =⎧⎨⎩

11εxx11εyy2 11εxy

⎫⎬⎭ =⎧⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎩

∂u∂x − 1

2

∙³∂u∂x

´2+³∂v∂x

´2¸∂v∂y − 1

2

∙³∂u∂y

´2+³∂v∂y

´2¸∂u∂y +

∂v∂x −

³∂u∂x

∂u∂y +

∂v∂x

∂v∂y

´

⎫⎪⎪⎪⎪⎪⎬⎪⎪⎪⎪⎪⎭(9.6.37c)


The Þnite element equations (9.6.31) can be written in explicit form as∙[K11L] + [K11N ] [K12L]

[K21L] [K22L] + [K22N ]

¸½ uv

¾=

½ 21F 1− 11F 121F 2− 11F 2

¾(9.6.38)

where

K11Lij = he

ZΩe

µ1C11

∂ψi∂x

∂ψj∂x

+ 1C66∂ψi∂y

∂ψj∂y

¶dx dy

K12Lij = he

ZΩe

µ1C12

∂ψi∂x

∂ψj∂y

+ 1C66∂ψi∂y

∂ψj∂x

¶dx dy = K21L

ji

K22Lij = he

ZΩe

µ1C66

∂ψi∂x

∂ψj∂x

+ 1C22∂ψi∂y

∂ψj∂y

¶dx dy

K11Nij = he

ZΩe

∙1σxx

∂ψi∂x

∂ψj∂x

+ 1σxy

µ∂ψi∂y

∂ψj∂x

+∂ψi∂x

∂ψj∂y

¶+ 1σyy

∂ψi∂y

∂ψj∂y

¸dx dy = K22N

ij (9.6.39a)

21F

1i = he

ZΩe

21fxψi dx dy + he

IΓe

21txψi ds

21F

2i = he

ZΩe

21fyψi dx dy + he

IΓe

21tyψi ds

11F

1i = he

ZΩe

µ∂ψi∂x

1σxx +∂ψi∂y

1σxy

¶dx dy

11F

2i = he

ZΩe

µ∂ψi∂x

1σxy +∂ψi∂y

1σyy

¶dx dy (9.6.39b)

where he is the thickness of the element1x1 = x and

1x2 = y.


The computer implementation of the two nonlinear formulations discussed inthis chapter follows along the same lines as discussed for the NewtonRaphsonprocedure used for plate bending. Box 9.6.1 contains the portion of the mainprogram where element information is passed on to the element subroutine anderror check, while Box 9.6.2 contains the main parts of the element subroutine.

9.6.5 Numerical Results

Here we present numerical results obtained for a cantilever beam underuniformly distributed load using the total and updated Lagrangianformulations. Suppose that the beam is of length a = 10 in., height b = 1in., and thickness h = 1 in. A mesh of Þve eight-node quadratic elements areused, and load increment of ∆q0 = −0.5 psi (acting downward) is used. The


Box 9.6.1 Fortran statements showing the transfer of element informationand error check.

P0=0.0 NFLAG=0 DO 500 LOAD=1,NLS P0=P0+DP(LOAD) ITER=0

220 ITER=ITER+1 ...

C C Compute and assemble element matrices C

DO 340 N=1,NEM DO 260 I=1,NPE

NI=NOD(N,I) ELXY(I,1)=X(NI) ELXY(I,2)=Y(NI) L=NI*NDF-1 K=I*NDF-1 ELU(K) =GLU(L) ELU(K+1)=GLU(L+1)

260 CONTINUE C270 CALL STIFF (IEL,NGP,NN,NPE,P,NAXIS,ITER,NEWTON,LFORM)

.

.

.340 CONTINUE C C Impose the specified displacement and force dof and solve the equations (GF is theC incremental displacement vector). Update the total solution vector GT and coordinates. C

DO 420 I=1,NNM L=(I-1)*NDF+1 GLU(L) =GLU(L) +GF(L) GLU(L+1)=GLU(L+1)+GF(L+1) IF(LFORM.GT.1)THEN

X(I)=X(I)+GF(L) Y(I)=Y(I)+GF(L+1)

ENDIF420 CONTINUEC

SNORM=0.0 ENORM=0.0 DO 430 I=1,NEQ

SNORM=SNORM+GLU(I)*GLU(I) 430 ENORM=ENORM+GF(I)*GF(I)

TOL=DSQRT(ENORM/SNORM) IF (TOL.GT.EPS) THEN

IF (ITER.GT.ITMAX) THENwrite a message STOP

ELSE GOTO 220

ENDIFENDIF. ..

500 CONTINUE


Box 9.6.2 Fortran statements showing the element matrix calculations forthe total and updated Lagrangian formulations.

SUBROUTINE STIFF(IEL,IDYN,NGP,NN,NPE,P,NAXIS,ITER,NEWTON,LFORM) C _________________________________________________________CC STIFFNESS MATRIX FOR ISOPARAMETRIC QUADRILATERAL ELEMENTS C (Total and Updated Lagrangian Formulations)CC LFORM.....Indicator for the type of formulation: C LFORM=1, TOTAL Lagrangian formulationC LFORM>1, UPDATED Lagrangian formulationC _________________________________________________________

.

.

.C C Initialize element force vector and stiffness matrix C

DO 10 I=1,NN F(I)=0.0 DO 10 J=1,NN ELSTIF(I,J)=0.0

10 CONTINUE C C Numerical integration to evaluate the element matricesC

DO 60 NI=1,NGP DO 60 NJ=1,NGP XI =GAUSSPT(NI,NGP) ETA=GAUSSPT(NJ,NGP) CALL SHAPE (NPE,XI,ETA,SF,GDSF,DET,ELXY) CONST=DET*GAUSSWT(NI,NGP)*GAUSSWT(NJ,NGP)

C C Define the gradients of displacements and strains C

X=0.0 U=0.0 UX=0.0 UY=0.0 VX=0.0 VY=0.0 DO 20 I=1,NPE L=(I-1)*NDF+1 X=X+SF(I)*ELXY(I,1) U=U+SF(I)*ELU(L) UX=UX+GDSF(1,I)*ELU(L) UY=UY+GDSF(2,I)*ELU(L) VX=VX+GDSF(1,I)*ELU(L+1)

20 VY=VY+GDSF(2,I)*ELU(L+1) UX2=UX*UXUY2=UY*UYVX2=VX*VXVY2=VY*VY


CIF(LFORM.EQ.1)THEN

UXP1=1.0+UXUXP2=UXP1*UXP1VYP1=1.0+VYVYP2=VYP1*VYP1

CEX=UX+0.5*(UX2+VX2) EY=VY+0.5*(UY2+VY2) EXY=UY+VX+UX*UY+VX*VY S11=C(1,1)*EX+C(1,2)*EY S22=C(1,2)*EX+C(2,2)*EY S12=C(3,3)*EXY

ELSEC C Compute the Almansi strains and Cauchy stresses C

EX=UX-0.5*(UX2+VX2) EY=VY-0.5*(UY2+VY2) EXY=UY+VX-UX*UY-VX*VY S11=C(1,1)*EX+C(1,2)*EY S22=C(1,2)*EX+C(2,2)*EY S12=C(3,3)*EXY

ENDIFC C Compute element force vector and stiffness matrixC

II=1 DO 50 I=1,NPE

CC Imbalance force coefficients for the TOTAL Lagrangian formulationC

IF(LFORM.EQ.1)THENF(II) =F(II) +(P*SF(I)-UXP1*GDSF(1,I)*S11-UY*GDSF(2,I)*S22

* -(UXP1*GDSF(2,I)+UY*GDSF(1,I))*S12)*CONST F(II+1)=F(II+1)+(P*SF(I)-VX*GDSF(1,I)*S11-VYP1*GDSF(2,I)*S22

* -(VYP1*GDSF(1,I)+VX*GDSF(2,I))*S12)*CONST CC Imbalance force coefficients for the UPDATED Lagrangian formulationC

ELSEF(II)=F(II)+(P*SF(I)-GDSF(1,I)*S11-GDSF(2,I)*S12)*CONST F(II+1)=F(II+1)+(P*SF(I)-GDSF(1,I)*S12-GDSF(2,I)*S22)*CONST

ENDIFJJ=1 DO 40 J=1,NPE SIG=S11*GDSF(1,I)*GDSF(1,J)+S22*GDSF(2,I)*GDSF(2,J)

* +S12*(GDSF(1,I)*GDSF(2,J)+GDSF(2,I)*GDSF(1,J)) C

IF(LFORM.EQ.1)THEN

(Box 9.6.2 continued)

CC Stiffness coefficients for the TOTAL Lagrangian formulationC

ELSTIF(II,JJ)=ELSTIF(II,JJ)+(C(1,1)*UXP2*GDSF(1,I)*GDSF(1,J)* +C(2,2)*UY2*GDSF(2,I)*GDSF(2,J) * +C(1,2)*UXP1*UY*(GDSF(1,I)*GDSF(2,J)+GDSF(2,I)*GDSF(1,J))* +C(3,3)*(UXP1*GDSF(2,I)+UY*GDSF(1,I))** (UXP1*GDSF(2,J)+UY*GDSF(1,J))+SIG)*CONST

ELSTIF(II+1,JJ+1)=ELSTIF(II+1,JJ+1)+(C(1,1)*VX2*GDSF(1,I)* *GDSF(1,J)+C(2,2)*VYP2*GDSF(2,I)*GDSF(2,J)* +C(1,2)*VYP1*VX*(GDSF(1,I)*GDSF(2,J)+GDSF(2,I)*GDSF(1,J))* +C(3,3)*(VYP1*GDSF(1,I)+VX*GDSF(2,I))** (VYP1*GDSF(1,J)+VX*GDSF(2,J))+SIG)*CONST

ELSTIF(II,JJ+1)=ELSTIF(II,JJ+1)+(C(1,1)*UXP1*VX*GDSF(1,I)* *GDSF(1,J)+C(2,2)*VYP1*UY*GDSF(2,I)*GDSF(2,J)* +C(1,2)*(UXP1*VYP1*GDSF(1,I)*GDSF(2,J)* +UY*VX*GDSF(2,I)*GDSF(1,J))* +C(3,3)*(UXP1*GDSF(2,I)+UY*GDSF(1,I))** (VYP1*GDSF(1,J)+VX*GDSF(2,J)))*CONST

ELSTIF(II+1,JJ)=ELSTIF(II+1,JJ)+(C(1,1)*UXP1*VX*GDSF(1,J)* *GDSF(1,I)+C(2,2)*VYP1*UY*GDSF(2,J)*GDSF(2,I)* +C(1,2)*(UXP1*VYP1*GDSF(1,J)*GDSF(2,I)* +UY*VX*GDSF(2,J)*GDSF(1,I))* +C(3,3)*(UXP1*GDSF(2,J)+UY*GDSF(1,J))** (VYP1*GDSF(1,I)+VX*GDSF(2,I)))*CONST

CC Stiffness coefficients for the UPDATED Lagrangian formulationC

ELSEELSTIF(II,JJ)=ELSTIF(II,JJ)+(C(1,1)*GDSF(1,I)*GDSF(1,J)

* +C(3,3)*GDSF(2,I)*GDSF(2,J)+SIG)*CONSTELSTIF(II+1,JJ+1)=ELSTIF(II+1,JJ+1)+(C(3,3)*GDSF(1,I)

* *GDSF(1,J)+C(2,2)*GDSF(2,I)*GDSF(2,J)+SIG)*CONSTELSTIF(II,JJ+1)=ELSTIF(II,JJ+1)+(C(1,2)*GDSF(1,I)*GDSF(2,J)

* +C(3,3)*GDSF(2,I)*GDSF(1,J))*CONSTELSTIF(II+1,JJ)=ELSTIF(II+1,JJ)+(C(1,2)*GDSF(2,I)*GDSF(1,J)

* +C(3,3)*GDSF(1,I)*GDSF(2,J))*CONSTENDIF

40 JJ=NDF*J+1 50 II=NDF*I+1 60 CONTINUE

RETURN END

(Box 9.6.2 continued)


material properties used are

E = 1.2× 104 psi, ν = 0.2 (9.6.40)

The beam assumed to be in the plane state of stress so that (no distinction ismade between 1Cij and 0Cij for the two formulations, and they are assumedto remain constant during the deformation).


Table 9.6.1 contains the numerical results obtained with the updatedLagrangian formulation for the center deßection at the free end (also seeFigure 9.6.1), while Table 9.6.2 contains (also see Figure 9.6.2) the Cauchyand second PiolaKirchhoff stresses at the Þxed end of the beam (at the Gausspoint located nearest to the top left end). The load was distributed equally atthe top and bottom of the beam. Surprisingly, the solution without iterationis very good, except for initial oscillations. It took only 2 or 3 iterations toconverge for each load step.

Table 9.6.1 Transverse deßections of a cantilevered beam under uniformload (applied equally at the top and bottom), obtained withthe updated Lagrangian formulation (5× 1Q8 mesh).

−q0 x y u(10,0.5) v(10, 0.5)

1.0 9.914 −0.730 −0.086 −1.230*9.968 −0.287 −0.032 −0.787

2.0 9.674 −1.884 −0.326 −2.3849.822 −1.291 −0.178 −1.791

3.0 9.327 −2.893 −0.673 −3.3939.530 −2.373 −0.470 −2.873

4.0 8.920 −3.750 −1.080 −4.2509.145 −3.327 −0.855 −3.827

5.0 8.493 −4.460 −1.507 −4.9608.699 −4.153 −1.301 −4.653

6.0 8.070 −5.045 −1.930 −5.5458.244 −4.831 −1.756 −5.331

8.0 7.286 −5.926 −2.714 −6.4267.391 −5.831 −2.609 −6.331

10.0 6.609 −6.540 −3.391 −7.0406.656 −6.506 −3.344 −7.006

12.0 6.035 −6.981 −3.965 −7.4816.052 −6.971 −3.948 −7.471

14.0 5.550 −7.309 −4.450 −7.8095.557 −7.305 −4.443 −7.805

16.0 5.137 −7.562 −4.863 −8.0625.142 −7.559 −4.858 −8.059

*The Þrst row corresponds to iteration and the second row to the no iteration.

0.0 2.0 4.0 6.0 8.0

Transverse deflection,

0.0

4.0

8.0

12.0

16.0

Load

,

Without iteration

With iteration

v(10,0.5)

q 0


Figure 9.6.1 Transverse deßection v(10, 0.5) versus load q0 for a cantileverbeam under uniform load of intensity q0.

Table 9.6.2 A comparison of the stresses 1σxx and10Sxx obtained with the

updated Lagrangian formulation.

−q0 1σxx 10Sxx

1σyy 10Syy

1.0 153.55 161.76 23.15 28.882.0 300.14 332.20 43.65 65.413.0 434.44 503.43 61.50 107.054.0 555.38 671.46 77.11 151.825.0 662.83 833.34 90.87 198.026.0 758.08 988.19 103.17 244.668.0 918.14 1277.32 124.61 337.1610.0 1046.77 1542.59 143.15 427.1912.0 1152.26 1788.70 159.78 514.3314.0 1240.32 2019.60 175.11 598.7216.0 1314.87 2238.26 189.50 680.62

Table 9.6.3 contains the numerical results for the center deßection at thefree end and the second PiolaKirchhoff stresses at the Þxed end of the beam(at the Gauss point located nearest to the top left end). The results wereobtained with the total Lagrangian description and with nonlinear iteration.Results are included only for selective loads. Only 2 to 3 iterations were takento converge at each load step.

xxσ

q 0

0 400 800 1200 1600 2000 2400

Axial stress,

0

4

8

12

16

Loa

d,

Second Piola-Kirchhoff stress (five Q8 elements)

Cauchy stress (five Q8 elements)

(updated Lagrangian formulation)


Table 9.6.3 Transverse deßections of a cantilevered beam obtained withthe total Lagrangian formulation.

−q0* u(10, 0.5) v(10, 0.5) 1σxx 10Sxx

1.0 −0.087 −1.230 143.84 151.972.0 −0.331 −2.382 263.28 294.794.0 −1.091 −4.237 426.04 538.126.0 −1.942 −5.516 509.29 728.248.0 −2.721 −6.380 539.85 877.7510.0 −3.391 −6.979 536.14 998.5112.0 −3.957 −7.408 509.23 1098.7814.0 −4.434 −7.726 465.82 1184.0416.0 −4.839 −7.971 410.10 1257.97

*The linear solution at q0 = −0.5 is v(10, 0.5) = −0.6227.

Figure 9.6.2 Normal stresses versus load for a cantilever beam underuniform load of intensity q0.

This completes the development of Þnite element models of nonlinearcontinua. The procedures developed herein can be readily extended toaxisymmetric problems, three-dimensional problems, problems with assumeddisplacements and/or strains [79], and composite plates and shells [10]. Theextension of the formulations to transient problems is straightforward.


9.7 Shell Finite Elements

9.7.1 Introduction

In previous chapters, the beam and plate elements were developed usingbeam and plate theories that were derived from an assumed displacementÞeld. Such theories are limited to geometrically linear analyses and nonlinearanalysis with small strains and moderate rotations. The Þnite element modelsto be developed in this section are based on three-dimensional elasticityequations, and the geometry and the displacement Þelds of the structureare directly discretized by imposing some geometric and static constraintsto satisfy the assumptions of a beam or shell theory. Both shell and beamelements can have a variable number of nodes, and the shell element canbe modiÞed as transition elements to model shell intersections or solid-to-shelltransition regions. Such formulations would appear to be especially applicableto material and geometrical nonlinear analysis of shell-type structures in whichlarge displacements and rotations are experienced (see [2,1027]).

9.7.2 Incremental Equations of Motion

Consider the motion of a body in a Þxed Cartesian coordinate system.Assuming that the body may experience large displacements and rotations,we wish to determine deformed conÞgurations of the body for differenttimes/loads. We assume that all conÞgurations from time t = 0 to the currenttime t, both inclusive, have been determined, and the conÞguration C2 for timet = t+∆t is sought next. The total Lagrangian description with the principleof virtual displacements (9.5.17) is used to express the dynamic equilibriumof the body in the conÞguration C2.For dynamic analysis, the principle of virtual displacements (9.5.17) must

be modiÞed to include inertial terms. In this case we haveZ2V

2ρ 2ui δ2ui d

2V =

Z0V

0ρ 2ui δ2ui d

0V (9.7.1)

and hence the mass matrix can be evaluated using the initial conÞguration ofthe body. Using Hamiltons principle we obtain the equations of motion of themoving body at time t+∆t in the variational form asZ

0V

0ρ 2ui δ(2ui) d

0V +

Z0V

0Cijrs 0ers δ 0eij d0V +

Z0V

10Sij δ 0ηij d

0V

= 2R−Z0V

10Sij δ 0eij d

0V (9.7.2)

where δ(2ui) = δui.


9.7.3 Finite Element Model of a Continuum

Equation (9.7.2) can be used to develop the nonlinear displacement Þniteelement model for any continuum. The basic step in deriving the Þniteelement equations for a shell element is the selection of proper interpolationfunctions for the displacement Þeld and geometry. In the case of beam andshell elements, the approximation for the geometry is chosen such that thebeam or shell kinematic hypotheses are realized. First, we derive the Þniteelement model of a continuum and then specialize it to shells [2426].

It is important that the coordinates and displacements are interpolatedusing the same interpolation functions (isoparametric formulation) so thatthe displacement compatibility across element boundaries can be preserved inall conÞgurations. Let

0xi =nXk=1

ψk0xki ,

1xi =nXk=1

ψk1xki ,

2xi =nXk=1

ψk2xki (9.7.3)

1ui =nXk=1

ψk1uki , ui =

nXk=1

ψk uki (i = 1, 2, 3) (9.7.4)

where the right superscript k indicates the quantity at nodal point k, ψk is theinterpolation function corresponding to nodal point k, and n is the number ofelement nodal points.

Substitution of Eqs. (9.7.3) and (9.7.4) in Eq. (9.7.2) yields the Þniteelement model

10[M ]∆e+ (10[KL] + 1

0[KNL])∆e = 2R− 10F (9.7.5)

where ∆e is the vector of nodal incremental displacements from time t totime t+∆t in an element, and 10[M ]∆e, 10[KL]∆e, 10[KNL]∆e, and 10Fare obtained by evaluating the integrals, respectively:Z

0V

0ρ 2ui δ2ui d

0V,

Z0V

0Cijrs 0ers δ 0eij d0VZ

0V

10Sij δ 0ηij d

0V,

Z0V

10Sij δ 0eij d

0V

Various matrices are deÞned by

10[KL] =

Z0A

10[BL]

T0[C]

10[BL] d

0V (9.7.6a)

10[KNL] =

Z0V

10[BNL]

T0[S]

10[BNL] d

0V (9.7.6b)

10[M ] =

Z0V

0ρ 1[H]T 1[H] d 0V (9.7.6c)

10F =

Z0V

10[BL]

T 10 S d 0V (9.7.6d)


In the above equations, 10[BL] and10[BNL] are the linear and nonlinear

straindisplacement transformation matrices, 0[C] is the incremental stress-strain material property matrix, 10[S] is a matrix of 2nd PiolaKirchhoff stress

components, 10 S is a vector of these stresses, and 1[H] is the incrementaldisplacement interpolation matrix. All matrix elements correspond to theconÞguration at time t and are deÞned with respect to the conÞguration attime t = 0. It is important to note that Eq. (9.7.5) is only an approximationto the actual solution to be determined in each time step [see Eq. (9.5.15)].Therefore, it may be necessary to iterate in each time step until Eq. (9.5.15),with inertia terms, is satisÞed to a required tolerance.

The Þnite element equations (9.7.5) are second-order differential equationsin time. In order to obtain numerical solutions at each time step, Eq. (9.7.5)needs to be converted to algebraic equations using a time approximationscheme, as explained in Chapter 8. In the present study the Newmark scheme(see Section 8.2.3 for details) is used.

In the Newmark time integration scheme, the displacements and velocitiesare approximated by [see Eqs. (8.2.22) and (8.2.23)]

t+∆t∆ = ∆t t∆+ t ú∆+ (∆t)2[(12− β) t∆+ β t+∆t∆]

t+∆t ú∆ = t ú∆+∆t[(1− α) t∆+ α t+∆t∆] (9.7.7)

where α = 12 and β =

14 for the constantaverage acceleration method, and ∆t

is the time step. Rearranging Eqs. (9.7.5) and (9.7.7), we obtain [10[M ] ≈ 20[M ];

see Eq. (8.2.15)]10[ K]∆ = 2 R (9.7.8)

where ∆ is the vector of nodal incremental displacements at time t, ∆ =t+∆t∆− t∆, and

10[ K] = a3

10[M ] +

10[KL] +

10[KNL] (9.7.9a)

2 R = 2R− 10F+ 1

0[M ]³a3

t∆+ a4 t ú∆+ a5 t∆´(9.7.9b)

a3 =1

β(∆t)2, a4 = a3∆t, a5 =

1

2β− 1 (9.7.10)

Once Eq. (9.7.8) is solved for ∆ at time t+∆t, the acceleration and velocityvectors are obtained using [a1 = α∆t and a2 = (1− α)∆t; see Eqs. (8.2.19)]

t+∆t∆ = a3∆− a4 t ú∆− a5 t∆t+∆t ú∆ = t ú∆+ a1 t+∆t∆+ a2 t∆ (9.7.11)


The Þnite element equations (9.7.8) are solved, after assembly andimposition of boundary conditions, iteratively at each time step until Eq.(9.5.15) is satisÞed within a required tolerance. The NewtonRaphson methodwith RiksWempner algorithm (see Appendix A1) is used in the present study.

9.7.4 Shell Finite Element

The shell element is deduced from the three-dimensional continuum elementby imposing two kinematic constraints: (1) straight line normal to the mid-surface of the shell before deformation remains straight but not normal afterdeformation; (2) the transverse normal components of strain and stress areignored in the development. However, the shell element admits arbitrarilylarge displacements and rotations but small strains since the shell thicknessis assumed not to change and the normal is not allowed to distort [22, 2426,28].

Consider the solid three-dimensional element shown in Figure 9.7.1. Let(ξ, η) be the curvilinear coordinates in the middle surface of the shell and ζbe the coordinate in the thickness direction. The coordinates (ξ, η, ζ) arenormalized such that they vary between −1 and +1. The coordinates of atypical point in the element can be written as

xi =nXk=1

ψk(ξ, η)

∙1 + ζ

2(xki )top +

1− ζ2(xki )bottom

¸(9.7.12)

where n is the number of nodes in the element, and ψk(ζ, η) is the Þniteelement interpolation function associated with node k. If ψk(ξ, η) are derivedas interpolation functions of a parent element, square or triangular in-plane,then compatibility is achieved at the interfaces of curved space shell elements.DeÞne

V k3i = (xki )top − (xki )bottom, ek3 = V

k3/|Vk

3 | (9.7.13)

where Vk3 is the vector connecting the upper and lower points of the normal

at node k. Equation (9.7.12) can be rewritten as

xi =nXk=1

ψk(ξ, η)

∙(xki )mid +

ζ

2V k3i

¸=

nXk=1

ψk(ξ, η)

∙(xki )mid +

ζ

2hke

k3i

¸(9.7.14)

where hk = |Vk3 | is the thickness of the shell element at node k. Hence,

the coordinates of any point in the element at time t are interpolated by theexpression

1xi =nXk=1

ψk(ξ, η)

∙1xki +

ζ

2hk1ek3i

¸(9.7.15)

k3e

k1e

k2e

3E

1E

2E Element nodes

x3

x2

x1


Figure 9.7.1 Geometry and coordinate system of a curved shell element.

and the displacements and incremental displacements by

1ui =1xi − 0x1 =

nXk=1

ψk(ξ, η)

∙1uki +

ζ

2hk³1ek3i − 0ek3i

´¸(9.7.16)

ui =2ui − 1ui =

nXk=1

ψk(ξ, η)

∙uki +

ζ

2hk³2ek3i − 1ek3i

´¸(9.7.17)

Here 1uki and uki denote, respectively, the displacement and incrementaldisplacement components in the xi-direction at the kth node and time t. Forsmall rotation dΩ at each node, we have

dΩ = θk21ek1 + θ

k11ek2 + θ

k31ek3 (9.7.18)

the increment of vector 1ek3 can be written as

∆1ek3 =2ek3 − 1ek3 = dΩ× 1ek3 = θ

k11ek1 − θk2 1ek2 (9.7.19)

Then Eq. (9.7.17) becomes

ui =nXk=1

ψk(ξ, η)

∙uki +

ζ

2hk³θk1

1ek1i − θk2 1ek2i´¸

(i = 1, 2, 3) (9.7.20)

The unit vectors 1ek1 and1ek2 at node k can be obtained from the relations

1ek1 =E2 × 1ek3|E2 × 1ek3|

, 1ek2 =1ek3 × 1ek1 (9.7.21)


where Ei are the unit vectors of the stationary global coordinate system(0x1,

0x2,0x3). Equation (9.7.35) can be written in matrix form as

u = u1 u2 u3T = 1[H]3×5n∆e5n×1 (9.7.22)

where ∆e = uki θk1 θk2T, (i = 1, 2, 3, k = 1, 2, . . . , n, and n is the number ofnodes) is the vector of nodal incremental displacements (Þve per node), and1[H] is the incremental displacement interpolation matrix

1[H]3×5n =

⎡⎣ . . . ψk 0 0 12ψkζhk

1ek11 −12ψkζhk

1ek21 . . .

. . . 0 ψk 0 12ψkζhk

1ek12 −12ψkζhk

1ek22 . . .

. . . 0 0 ψk12ψkζhk

1ek13 −12ψkζhk

1ek23 . . .

⎤⎦(9.7.23)

For each time step or iteration step one can Þnd three unit vectors at eachnode from Eqs. (9.7.19) and (9.7.21).

From Eq. (9.3.17) the linear strain increments 0e = 0e11 0e22 0e33 20e1220e13 20e23T can be expressed as

0e = 1[A]0u (9.7.24a)

where 0u is the vector of derivatives of increment displacements,

0u = 0u1,1 0u1,2 0u1,3 0u2,1 0u2,2 0u2,3 0u3,1 0u3,2 0u3,3T

1[A]6×9 =

⎡⎢⎢⎢⎢⎣1 + 1

0u1,1 0 0 . . . 10u3,1 0 0

0 10u1,2 0 . . . 0 1

0u3,2 00 0 1

0u1,3 . . . 0 0 1 + 10u3,3

10u1,2 1 + 1

0u1,1 0 . . . 10u3,2

10u3,1 0

10u1,3 0 1 + 1

0u1,1 . . . 1 + 10u3,3 0 1

0u3,10 1

0u1,310u1,2 . . . 0 1 + 1

0u3,310u3,2

⎤⎥⎥⎥⎥⎦(9.7.24b)

The dots correspond to the entries (not displayed due to the page width)

. . .

. . .

. . .

. . .

. . .

. . .

=

10u2,1 0 00 1 + 1

0u2,2 00 0 1

0u2,31 + 1

0u2,210u3,1 0

10u2,3 0 1

0u2,10 1

0u2,3 1 + 10u2,2

and 0ui,j = ∂ui/∂0xj . The vectors 0u and 0e are related to the

displacement increments at nodes by

0u = [N ] u = [N ] 1[H]∆e0e = 1[A]0u = 1[A][N ] 1[H]∆e ≡ 1

0[BL]∆e (9.7.25a)10[BL] =

1[A][N ] 1[H]


where [N ]T is the operator of differentials

[N ]T =

⎡⎣ ∂∂ 0x1

∂∂ 0x2

∂∂ 0x3

0 0 0 0 0 0

0 0 0 ∂∂ 0x1

∂∂ 0x2

∂∂ 0x3

0 0 0

0 0 0 0 0 0 ∂∂ 0x1

∂∂ 0x2

∂∂ 0x3

⎤⎦ (9.7.25b)

The components of 1[A] include 10ui,j . From Eq. (9.7.16) the global

displacements are related to the natural curvilinear coordinates (ξ, η) and thelinear coordinate ζ. Hence the derivatives of these displacements 10ui,j withrespect to the global coordinates 0x1,

0x2, and0x3 are obtained through the

relation

[10ui,j ] =

⎡⎢⎢⎣∂ 1u1∂ 0x1

∂ 1u2∂ 0x1

∂ 1u3∂ 0x1

∂ 1u1∂ 0x2

∂ 1u2∂ 0x2

∂ 1u3∂ 0x2

∂ 1u1∂ 0x3

∂ 1u2∂ 0x3

∂ 1u3∂ 0x3

⎤⎥⎥⎦ = 0[J ]−1

⎡⎢⎢⎣∂ 1u1∂ξ

∂ 1u2∂ξ

∂ 1u3∂ξ

∂ 1u1∂η

∂ 1u2∂η

∂ 1u3∂η

∂ 1u1∂ζ

∂ 1u2∂ζ

∂ 1u3∂ζ

⎤⎥⎥⎦ (9.7.26)

The Jacobian matrix 0[J ] is deÞned as

0[J ] =

⎡⎢⎢⎣∂ 0x1∂ξ

∂ 0x2∂ξ

∂ 0x3∂ξ

∂ 0x1∂η

∂ 0x2∂η

∂ 0x3∂η

∂ 0x1∂ζ

∂ 0x2∂ζ

∂ 0x3∂η

⎤⎥⎥⎦ (9.7.27)

and is computed from the coordinate deÞnition of Eq. (9.7.15). The derivativesof displacements 1ui with respect to the coordinates ξ, η, and ζ can becomputed from Eq. (9.7.16). In the evaluations of element matrices in Eqs

(9.7.6a-d), the integrands of 10[BL], 0[C],10BNL],

10[S],

1[H], and 10 S should beexpressed in the same coordinate system, namely the global coordinate system(0x1,

0x2,0x3) or the local curvilinear system (x01, x02, x03).

The number of stress and strain components are reduced to Þve since weneglect the transverse normal components of stress and strain. Hence, theglobal derivatives of displacements, [10ui,j ] which are obtained in Eq. (9.7.26),are transformed to the local derivatives of the local displacements along theorthogonal coordinates by the following relation

⎡⎢⎢⎢⎣∂ 1u01∂x01

∂ 1u02∂x01

∂ 1u03∂x01

∂ 1u01∂x02

∂ 1u02∂x02

∂ 1u03∂x02

∂ 1u01∂x03

∂ 1u02∂x03

∂ 1u03∂x03

⎤⎥⎥⎥⎦ = [θ]T3×3[10ui,j ] [θ]3×3 (9.7.28)

where [θ]T is the transformation matrix between the local coordinate system(x01, x02, x03) and the global coordinate system (0x1,

0x2,0x3) at the integration


point. The transformation matrix [θ] is obtained by interpolating the threeorthogonal unit vectors (1e1,

1e2,1e3) at each node:

[θ] =

⎡⎣Pnk=1 ψk

1ek11Pnk=1 ψk

1ek21Pnk=1 ψk

1ek31Pnk=1 ψk

1ek12Pnk=1 ψk

1ek22Pnk=1 ψk

1ek32Pnk=1 ψk

1ek13Pnk=1 ψk

1ek23Pnk=1 ψk

1ek33

⎤⎦ (9.7.29)

Since the element matrices are evaluated using numerical integration, thetransformation must be performed at each integration point during thenumerical integration.

In order to obtain 10[BL], the vector of derivatives of incremental

displacements u0 needs to be evaluated. Equations (9.7.26) and (9.7.28)can be used again except that 1ui are replaced by ui and the interpolationequation for ui, Eq. (9.7.17), is applied.

The development of the matrix of material stiffness, 0[C0], is discussed next.

Here we wish to present it for shell element composed of orthotropic materiallayers with the principal material coordinates (x1, x2, x3) oriented arbitrarilywith respect to the local coordinate system (x01, x02, x03) (with x3 = x03). For akth lamina of a laminated composite shell the matrix of material stiffnesses isgiven by

0[C0](k) =

⎡⎢⎢⎢⎢⎣C 011 C 012 C016 0 0C 012 C 022 C026 0 0C 016 C 026 C066 0 00 0 0 C 044 C 0450 0 0 C 045 C 055

⎤⎥⎥⎥⎥⎦ (9.7.30)

where

C 011 = m4Q11 + 2m

2n2(Q12 + 2Q66) + n4Q22

C 012 = m2n2(Q11 +Q22 − 4Q66) + (m4 + n4)Q12

C 016 = mn[m2Q11 − n2Q22 − (m2 − n2)(Q12 + 2Q66)]

C 022 = n4Q11 + 2m

2n2(Q12 + 2Q66) +m4Q22

C 026 = mn[n2Q11 −m2Q22 + (m

2 − n2)(Q12 + 2Q66)]C 066 = m

2n2(Q11 +Q22 − 2Q12) + (m2 − n2)2Q66C 044 = m

2Q44 + n2Q55, C

045 = mn(Q55 −Q44)

C 055 = m2Q55 + n

2Q44

m = cos θ(k), n = sin θ(k) (9.7.31)

where Qij are the plane stress-reduced stiffnesses of the kth orthotropic laminain the material coordinate system. The Qij can be expressed in terms ofengineering constants of a lamina

Q11 =E1

1− ν12ν21 , Q12 =ν12E2

1− ν12ν21 , Q22 =E2

1− ν12ν21


Q44 = G23, Q55 = G13, Q66 = G12 (9.7.32)

where Ei is the modulus in the xi-direction, Gij (i 6= j) are the shear moduliin the xi-xj plane, and νij are the associated Poissons ratios (see Reddy [10]).

To evaluate element matrices in Eqs. (9.7.6ad), we employ the Gaussquadrature. Since we are dealing with laminated composite structures,integration through the thickness involves individual lamina. One way is touse Gauss quadrature through the thickness direction. Since the constitutiverelation 0[C] is different from layer to layer and is not a continuous functionin the thickness direction, the integration should be performed separately foreach layer. This increases the computational time as the number of layers isincreased. An alternative way is to perform explicit integration through thethickness and reduce the problem to a two-dimensional one. The Jacobianmatrix, in general, is a function of (ξ, η, ζ). The terms in ζ may be neglectedprovided the thickness to curvature ratios are small. Thus the Jacobian matrix0[J ] becomes independent of ζ and explicit integration can be employed. If ζterms are retained in 0[J ], Gauss points through the thickness should be added.In the present study we assume that the Jacobian matrix is independent of ζin the evaluation of element matrices and the internal nodal force vector.

Since the explicit integration is performed through the thickness, theexpression for "

∂1u0i∂x0j

#, 1

0[A0], 0u0, 1[H], 1

0[B0], 10ε0ij

are now expressed in an explicit form in terms of ζ. Hence, we can use exactintegration through the thickness and use the Gauss quadrature to performnumerical integration on the mid-surface of the shell element.

For thin shell structures, in order to avoid locking we use reducedintegration scheme to evaluate the stiffness coefficients associated with thetransverse shear deformation. Hence we split the constitutive matrix 0[C

0]into two parts, one without transverse shear moduli 0[C

0]B, and the otherwith only transverse shear moduli 0[C

0]S . Full integration is used to evaluatethe stiffness coefficients containing 0[C

0]B, and reduced integration is used forthose containing 0[C

0]S.If a shell element is subjected to a distributed load (such as the weight or

pressure), the corresponding load vector 2R from Eq. (9.7.6) is given by

2R5n×1 =Z0A

1[H]T

⎧⎨⎩2P12P22P3

⎫⎬⎭ d 0A (9.7.33)

where 2Pi is the component of distributed load in the0xi-direction at time

t + ∆t, 0A is the area of upper or middle or bottom surface of the shell


element depending on the position of the loading and the loading is assumeddeformation-independent.

Substituting 1[H] into Eq. (9.7.33) yields

2R5n×1 =Z

0A

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

. . . . . . . . .ψk 0 00 ψk 00 0 ψk

12ζψkhk

1ek1112ζψkhk

1ek12 −12ζψkhk

1ek13−12ζψkhk

1ek2112ζψkhk

1ek22 −12ζψkhk

1ek23. . . . . . . . .

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦⎧⎨⎩2P12P22P3

⎫⎬⎭ d 0A

=NGPXr=1

NGPXs=1

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

. . .ψk

2P1ψk

2P2ψk

2P312ζψkhk

P3i=1

2Pi1ek1i

−12ζψkhk

P3i=1

2Pi1ek2i

. . .

⎫⎪⎪⎪⎪⎪⎪⎪⎪⎪⎬⎪⎪⎪⎪⎪⎪⎪⎪⎪⎭WξrWηs | 0J |(ξr,ηs)

2

h(9.7.34)

where h =PNPEk=1 ψk(ξ, η)hk is the shell thickness at each Gauss point, and W

is the weight at each Gauss point, and |0J | is the determinant of the Jacobianmatrix in Eq. (9.7.27) at each Gauss point. Here the ζ terms are retainedin Jacobian matrix and let ζ equal to 1, −1, or 0, respectively, when thedistributed loading is at the top, bottom, or middle surface.


A number of numerical examples of isotropic and orthotropic plates andshells are presented. Only static bending problems of plates and shells areincluded. The RiksWempner method is employed for tracing the nonlinearloaddeßection path (see Appendix 1). For most of the problems thereduced/selective integration scheme is used to evaluate the element stiffnesscoefficients. For additional examples, the reader may refer to [22,2426].

Simply-supported (BC1) orthotropic plate under uniform load

Figure 9.7.2(a) shows the plate and material properties used. A quarter ofthe plate with the boundary and symmetry conditions shown in the Þgure ismodeled with four nine-node shell elements. The results are shown in Figure9.7.2(b) along with the experimental results of Zaghloul and Kennedy [29]. Forthis simply-supported plate, the Þnite element results are in good agreementwith the experimental results of Zaghloul and Kennedy [29].


Figure 9.7.2 Geometrically nonlinear response of an orthotropic plate.(a) Geometry and mesh. (b) Load-deßection curves.

Four-layer (0/90/90/0) clamped plate under uniform load

Figure 9.7.3(a) shows a clamped, symmetrically laminated, square plate underuniform load. The material properties of a typical layer and Þnite elementmesh are also shown in the Þgure. A quarter of the plate is modeled usingfour nine-node elements. The present results along with the experimental

h = 0.138 in.

32.01037.0

1028.11036

231312

62

61

=×===

×=×=

12psi,

psi,psi,

νGGG

EE

• • •

•••

• • •

•

•

•

•

•

•

• • • • •

• ••• •x

y

b = 6 in.

a = 6 in.

(a)

(b)

2.0

1.5

1.2

0.8

0.4

0.0

Linear solution

Experiment [29]Present

0.0 0.1 0.2 0.3 0.4 0.5

Loa

d in

ten

sity

(ps

i)

Center deflection, u3 (in)

!

h = 0.096 in.

2395.0103125.0

108315.1108282.16

231312

62

61

=×===

×=×=

12psi,

psi,psi,

νGGG

EE

• • •

•••

• • •

•

•

•

•

•

•

• • • • •

• ••• •x

y

b = 6 in.

a = 6 in.

(a)

(b)

2.0

1.5

1.2

0.8

0.4

0.0

Linear solution

Experiment [29]Present (close to [30])

0.0 0.1 0.2 0.3 0.4 0.5

Loa

d in

ten

sity

(ps

i)

Center deflection, u3 (in)

!


results of Zaghloul and Kennedy [29] are shown in Figure 9.7.3(b). The tworesults are not in good agreement in this case. The difference between thetheoretical and experimental results is attributed to possible difference in thesupport conditions used in the Þnite element analysis and those used in theexperiment (i.e. the exact nature of clamped boundary conditions used inthe test may not be the ones used in the Þnite element analysis), because thepresent results are veriÞed against independent Þnite element study by Putchaand Reddy [30].

Figure 9.7.3 Nonlinear response of a clamped cross-ply laminated plate.(a) Geometry and mesh. (b) Loaddeßection curves.


Cylindrical shell roof under self-weight

Here we consider the linear analysis of a cylindrical panel under its own weight[see Figure 9.7.4(a)]. This is often used as a benchmark problem [27] in theliterature to validate numerical methods (i.e. assessment of the performanceof shell Þnite elements). It was shown in [28] that with the uniform (i.e.for all stiffness coefficients) reduction of integration order the quadratic andcubic shell elements show a more rapid convergence and better accuracy thanwith the reduced integration order applied to transverse shear terms only.The linear results obtained in the present study with the nine-node quadraticelement and uniform reduced integration order are shown in Figures 9.7.4(b)and 9.7.4(c). Even with one element the results show good agreement withthe exact solution [27], and for further mesh reÞnement the results are closeto the exact one.

Simply-supported spherical shell under point load

A simply supported isotropic spherical shell panel under central point load[see Figure 9.7.5(a)] is analyzed for its large displacement response using ameshes of sixteen four-node elements and four nine-node elements in a quarterof the shell. Figure 9.7.5(b) shows the responses calculated, including thepostbuckling range, with the modiÞed RiksWempner method. The Þgurealso includes the results of Bathe and Ho [17].

Shallow cylindrical shell under point load

An isotropic shallow cylindrical shell hinged along the longitudinal edges andfree at the curved boundaries and subjected to a point load is analyzed [seeFigure 9.7.6(a)]. One-quarter of the shell is modeled with four nine-nodeshell elements. The structure exhibits snap-through as well as snap-backphenomena, as shown in 9.7.6(b). The solution obtained by CrisÞeld [18]is also shown in Figure 9.7.6(b) to be compared with the present results.

Problems9.1 Consider the uniform deformation of a square of side 2 units initially centered at

X = (0,0). The deformation is given by the mapping [3]

x1 =1

4(18 + 4X1 + 6X2), x2 =

1

4(14 + 6X2)

(a) Sketch the deformed conÞguration of the body.

(b) Compute the components of the deformation gradient tensor F and its inverse(display them in matrix form).

(c) Compute the components of the right and left CauchyGreen deformationtensors (display them in matrix form).

(d) Compute the Greens and Almansis strain tensor components (display them inmatrix form).

-

-

-

25 ft.

25 ft.

R = 25 ft.

x1, u1

x2, u2

x3, u3

40°

h = 3 in.

Free edge

Free edge

Simply supportedrigid diaphramu1 = u3 = φ1 = 0

Symmetry lineu1 = φ1 = 0

u2 = φ2 = 0(a)

E1 = 3×106 psi., ν = 0.0ρgh = 0.625 lb/in2.

(c) ♦°°

°

♦♦ ♦

♦

40°

30°30°20°10°

− 0.005

− 0.01

u2

θ

(b)

MeshIntegrationrule 2 × 2

1 × 1

2 × 2

3 × 3

°

°♦♦

♦

°♦

♦

− 0.1

− 0.2

− 0.3

20°10° 30°40° θ

u3

♦°


Figure 9.7.4 A cylindrical shell roof under self-weight. (a) Geometry andthe Þnite element mesh. (b) Vertical displacement on mid-section. (c) Axial displacement at support.


Figure 9.7.5 A simply supported spherical shell panel. (a) Geometry andÞnite element mesh. (b) Loaddeßection curves.

- -2a

x2--

x1

x3

R

h = 3.9154 in.2a

Simply supported:u1 = u2 = u3 = φ1 = 0

Symmetry line: u1 = φ1 = 0

P


Symmetry line:u2 = φ2 = 0

E = 104 psi, ν = 0.3R = 100 in., a = 30.9017 in.

(a)

- -- - - -

0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5

-

20.0

17.5

15.0

12.5

10.0

7.5

5.0

2.5

0.0

-

-

-

-

Bathe and Ho [17]9-node elements4-node elements Present

Loa

d, P

(kip

s)

Transverse deflection, u3/h

(b)

E = 3103 Ν/mm2, ν = 0.3R = 2540 mm, L = 254 mm.

-

-

-

L

R

x1, u1

x2, u2

x3, u3

h = 6.35 mm

Free edge


Symmetry line: u1 = φ1 = 0


0.1

L

Symmetry line:u2 = φ2 = 0

P

(a)

0.7

0.6

0.5

0.4

0.3

0.2

0.1

0.0

−0.1

−0.2

−0.3

−0.4

−0.50 5 10 15 20 25 30

Center deflection, u3 (mm)

Loa

d, P

(kN

)

Crisfield [18]

! Present

(b)


Figure 9.7.6 Geometrically nonlinear response of a shallow cylindricalshell. (a) Geometry and Þnite element mesh. (b) Loaddeßection curves.

9.2 Consider the deformation given by the mapping [3]

x1 =1

4[4X1 + (9− 3X1 − 5X2 −X1X2) t] , x2 =

1

4[4X2 + (16+ 8X1) t]

(a) For X = (0,0) and t = 1 determine the deformation gradient tensor F and rightCauchyGreen tensor C.

x1

x3

x2

-t3

-t2 ntn

-t1


(b) Find the eigenvalues (stretches) λ1 and λ2 and the associated eigenvectors N1 andN2. Partial answer: λ1 = 2.2714, N1 = 0.8385 0.5449T.

(c) Use the polar decomposition to determine the symmetric stretch tensor U androtation tensor R.

9.3 Determine the displacements and strains in the (x1, x2) system for the bodies shown inFigure P9.3.

9.4 Determine the displacements and strains in the (x1, x2) system for the bodies shown inFigure P9.4.

Figure P9.3 Figure P9.4

9.5 Consider the inÞnitesimal tetrahedron in Cartesian coordinates shown in Figure P9.5. If−t1,−t2,−t3, and t denote the stress vectors in the outward directions on the faces ofthe inÞnitesimal tetrahedron whose areas are ∆s1, ∆s2, ∆s3, and ∆s, respectively, wehave by Newtons second law for the mass inside the tetrahedron,

t∆s− t1∆s1 − t2∆s2 − t3∆s3 + ρ∆vf = ρ∆va (a)

where ∆v is the volume of the tetrahedron, ρ the density, f the body force per unit mass,and a the acceleration. Establish the Cauchy formula by writing

ti = σi1e1 + σi2e2 + σi3e3 = σijej (b)

∆s1 = (n · e1)∆s, ∆s2 = (n · e2)∆s, ∆s3 = (n · e3)∆s (c)

∆v =∆h

3∆s (d)

where ∆h is the perpendicular distance from the origin to the slant face.

Figure P9.5

e0

a

b

A B

D DC e0

x1, X1

x2X2

C

A B

e0

a

b

A B

DC e0

D

x1, X1

x2

X2

C

A B

x1=k x22


9.6 Establish the following relationships between area elements and volume elements ofreference conÞguration and current conÞguration:

(a) Nansons formula: nda = F−T · NJ dA.

(b) dv = J dV .

where dA and dV are area and volume elements, respectively, in the referenceconÞguration and da and dv are area and volume elements, respectively, in the deformedconÞguration.

9.7 Let an arbitrary region in a continuous medium be denoted by v and the bounding closedsurface of this region be continuous and denoted by s. Let each point on the boundingsurface move with the velocity vs. It can be shown that the time derivative of the volumeintegral over some continuous function Q(x, t) is given by

D

Dt

Zv

Q(x, t)dv ≡Zv

∂Q

∂tdv +

Is

Qvs · n ds (a)

This expression for the differentiation of a volume integral with variable limits issometimes known as the three-dimensional Leibniz rule. The material derivative operatorD/Dt corresponds to changes with respect to a Þxed mass, that is, ρdv is constantwith respect to this operator. Show formally by means of Leibnizs rule, the divergencetheorem, and conservation of mass that

D

Dt

Zv

ρφ dv ≡Zv

ρDφ

Dtdv (b)

9.8 Newtons second law of motion applied to a continuum states that the rate of change ofmomentum following a material region of Þxed mass is equal to the sum of all the forceson the region. When the forces are divided into surface forces and body forces, Newtonssecond law reads:

D

Dt

Zv

ρv dv =

Is

n ·↔σ ds+Zv

ρb dv (a)

where↔σ is the surface stress tensor, b is the body force per unit mass, ρ is the mass

density, and v is the material velocity. Since the material particle mass ρdv is constantwith respect to the material time derivative D/Dt, make use of the divergence theoremand obtain the differential form of Newtons second law of motion for a continuum (seeChapter 7):

ρDv

Dt= div

↔σ + ρb (b)

9.9 Let e denote the thermodynamic internal energy per unit mass of a material. Then theequation of change for total energy of a material region can be written (see Chapter 7):

D

Dt

Zv

ρ

µe+

v2

2

¶dv =

Is

n ·↔σ · v ds+Zv

ρb · v dv −Is

q · n ds (a)

The Þrst two terms on the right-hand side describe the rate of work done on the materialregion by the surface stresses and the body forces. The third integral describes the netoutßow of heat from the region, causing a decrease of energy inside the region. Theheat-ßux vector q describes the magnitude and direction of the ßow of heat energy perunit time and per unit area.

e0

a

b

A B

D DC e0

x1, X1

x2X2

C


By suitable operations obtain the differential form of the energy equation:

ρD

Dt

µe+

v2

2

¶= div

³↔σ · v

´+ ρb · v − div q (b)

Subtract the contribution from kinetic energy and obtain

ρDe

Dt= div

³↔σ · v

´− v · div ↔

σ − div q (c)

This is called the thermodynamic form of the energy equation for a continuum.

9.10 Determine the two-dimensional displacement Þeld and the inÞnitesimal strain Þeld forthe simple shear of a rectangular block shown in the Þgure below. Determine the normaland shear strain in the diagonal line element of the rectangular block shown in FigureP9.10.

Figure P9.10

References1. Malvern, L. E., Introduction to the Mechanics of a Continuous Medium, PrenticeHall,Englewood Cliffs, NJ (1969).

2. Hinton, E. (ed), NAFEMS Introduction to Nonlinear Finite Element Analysis, NAFEMS,Glasgow, UK (1992).

3. Bonet, J. and Wood, R. D., Nonlinear Continuum Mechanics for Finite ElementAnalysis, Cambridge University Press, New York (1997).

4. Bathe, K. J. , Finite Element Procedures, PrenticeHall, Englewood Cliffs, NJ (1996).

5. Reddy, J. N., Energy Principles and Variational Methods in Applied Mechanics, SecondEdition, John Wiley, New York (2002).

6. Reddy, J. N. and Rasmussen, M. L., Advanced Engineering Analysis, John Wiley, NewYork, 1982; reprinted by Krieger, Melbourne, FL (1990).

7. Heyliger, P. R. and Reddy, J. N., A Mixed Updated Lagrangian Formulation for PlaneElastic Bodies, Journal of Composites Technology & Research, 9(4), 131140 (1987).

8. Heyliger, P. R. and Reddy, J. N., On a Mixed Finite Element Model for LargeDeformation Analysis of Elastic Solids, International Journal of Non-Linear Mechanics,23(2), 131145 (1988).

9. Roy, S. and Reddy, J. N., Non-Linear Analysis of Adhesively Bonded Joints,International Journal of Non-Linear Mechanics, 23(2), 97112 (1988).

10. Reddy, J. N.,Mechanics of Laminated Composite Plates and Shells: Theory and Analysis,2nd edn, CRC Press, Boca Raton, FL (2004).


11. Horrigmoe, G. and Bergan, P. G., Incremental Variational Principle and FiniteElement Models for Nonlinear Problems, Computer Methods in Applied Mechanics andEngineering, 7, 201217 (1976).

12. Wunderlich, W., Incremental Formulations for Geometrically Nonlinear Problems, inFormulations and Algorithm in Finite Element Analysis, K.J. Bathe, J.T. Oden, and W.Wunderlich (eds), 193239, MIT Press, Boston, MA (1977).

13. Ramm, E., A Plate/Shell Element for Large Deßections and Rotations, in Formulationsand Computational Algorithms in Finite Element Analysis, K.J. Bathe, J. T. Oden, andW. Wunderlich (eds), MIT Press, Boston, MA (1977).

14. Bergan, P. G. and Soreide, T. H., Solution of Large Displacement and InstabilityProblems Using the Current Stiffness Parameter, Finite Elements in NonlinearMechanics, 647649, Tapir Press (1978).

15. Rao, K. P., A Rectangular Laminated Anisotropic Shallow Thin Shell Finite Element,Computer Methods in Applied Mechanics and Engineering, 15, 1333 (1978).

16. Bathe, K. J. and Bolourchi, S., A Geometric and Material Nonlinear Plate and ShellElement, Computers and Structures, 11, 2348 (1980).

17. Bathe, K. J. and Ho, L. W., A Simple and Effective Element for Analysis of GeneralShell Structures, Computers and Structures, 13, 673681 (1981).

18. CrisÞeld, M. A., A Fast Incremental/Iterative Solution Procedure That Handles Snap-Through, Computers and Structures, 13, 5562 (1981).

19. Chang, T. Y. and Sawamiphakdi, K., Large Deformation Analysis of Laminated ShellsBy Finite Element Method, Computers and Structures, 13, 331340 (1981).

20. Kanok-Nukulchai, W., Taylor, R. L., and Hughes, T. J. R., A Large DeformationFormulation for Shell Analysis by the Finite Element Method, Computers andStructures, 13, 1927 (1981).

21. Surana, K. S.,Geometrically Nonlinear Formulation for the Three Dimensional Solid-Shell Transition Finite Elements, Computers and Structures, 15, 549566 (1982).

22. Chao, W. C. and Reddy, J. N., Analysis of Laminated Composite Shells Using aDegenerated 3-D Element, International Journal for Numerical Methods in Engineering,20, 19912007 (1984).

23. Stanley, G. M. and Felippa, C. A., Computational Procedures for Postbuckling forComposite Shells, in Finite Element Methods for Nonlinear Problems, P. G. Bergan, K.J. Bathe, and W. Wunderlich (eds.), 359385, SpringerVerlag, Berlin (1986).

24. Liao, C. L., Reddy, J. N., and Engelstad, S. P., A Solid-Shell Transition Element forGeometrically Nonlinear Analysis of Laminated Composite Structures, InternationalJournal for Numerical Methods in Engineering, 26, 18431854 (1988).

25. Liao, C. L. and Reddy, J. N., A Continuum-Based Stiffened Composite Shell Elementfor Geometrically Nonlinear Analysis, AIAA Journal, 27(1), 95101 (1989).

26. Liao, C. L. and Reddy, J. N., Analysis of Anisotropic, Stiffened Composite LaminatesUsing a Continuum Shell Element, Computers and Structures, 34(6), 805815 (1990).

27. Scordelis, A. C. and Lo, K. S., Computer Analysis of Cylindrical Shells, ACI Journal,61, 539561 (1964).

28. Zienkiewicz, O. C., Taylor, R. L., and Too, J. M., Reduced Integration Techniques inGeneral Analysis of Plates and Shells, International Journal for Numerical Methods inEngineering, 3, 275290 (1971).

29. Zaghloul, S. A. and Kennedy, J. B., Nonlinear Behavior of Symmetrically LaminatedPlates, Journal of Applied Mechanics, 42, 234236 (1975).

30. Putcha, N. S. and Reddy, J. N., A ReÞned Mixed Shear Flexible Finite Element forthe Nonlinear Analysis of Laminated Plates, Computers and Structures, 22, 529538(1986).

10

Material Nonlinearitiesand Coupled Problems

10.1 Introduction

Recall that nonlinearities arise from two independent sources. (1) Nonlinearitydue to changes in the geometry or position of the material particles of acontinuum, which is called the geometric nonlinearity. (2) Nonlinearity due tothe nonlinear material behavior, which is called material nonlinearity. In solidmechanics, the geometric nonlinearity arises from large strains and/or largerotations, and these enter the formulation through the straindisplacementrelations as well as the equations of motion. In ßuid mechanics and coupledßuid ßow and heat transfer, the geometric nonlinearity arises as a result ofthe spatial (or Eulerian) description of motion, and they enter the equationsof motion through material time derivative term. Material nonlinearity inall disciplines of engineering arise from nonlinear relationship between thekinetic and kinematic variables, for example, stressstrain relations, heat ßuxtemperature gradient relations, and so on. In general, material nonlinearitiesarise due to the material parameters (e.g. moduli, viscosity, conductivity,etc.) being functions of strains (or their rates), temperature, and other basicvariables.The Þnite element formulations presented in the previous chapters were

largely based on geometric nonlinearity. However, the nonlinearity in the one-and two-dimensional Þeld problems discussed in Chapters 3 and 5 could havecome from either sources. In this chapter, material nonlinear formulationsare given attention. This Þeld is very broad and special books are devotedto various types of nonlinearities (e.g. plasticity, viscoelasticity, and non-Newtonian materials). The objective here is to brießy discuss nonlinear elasticand elasticplastic material models for solids and power-law model for viscousincompressible ßuids.


10.2 Nonlinear Elastic Problems

Materials for which the constitutive behavior is only a function of the currentstate of deformation are known as elastic. In the special case in which the workdone by the stresses during a deformation is dependent only on the initial stateand the current conÞguration, the material is called hyperelastic, that is, thereexists a strain energy density function U0(Eij) such that

Sij =∂U0∂Eij

(10.2.1)

where Sij ard Eij are the components of the second PiolaKirchhoff stresstensor and GreenLagrange strain tensor, respectively. When U0 is a nonlinearfunction of the strains, the body is said to be nonlinearly elastic. A nonlinearlyelastic material has the following features: (a) U0 is a nonlinear function ofstrains, (b) all of the deformation is recoverable on removal of loads causingthe deformation, and (c) there is no loss of energy (i.e. loading and unloadingis along the same stress-strain path; see Figure 10.2.1).Here we consider a one-dimensional problem to discuss the Þnite element

formulation of a nonlinear elastic material for the case of kinematicallyinÞnitesimal strains. Consider the nonlinear uniaxial stressstrain relation

σxx = E F(εxx) (10.2.2)

where εxx is the inÞnitesimal strain, E is a material constant, and F is anonlinear function of the strain.The virtual work expression for the axial deformation of a bar made of a

nonlinear elastic material is

0 =

ZA

Z xb

xaσxxδεxx dxdA−

Z xb

xafδu dx− P e1 δu(xa)− P e2 δu(xb)

=

Z xb

xa[EA F(εxx)δεxx − fδu] dx− P e1 δu(xa)− P e2 δu(xb) (10.2.3)

The residual vector for the Þnite element model is

Rei =

Z xb

xa

∙EA F(εxx)dψ

ei

dx− fψei

¸dx− P ei (10.2.4)

and the tangent stiffness matrix is

Keij =

∂Rei∂uej

= EA

Z xb

xa

∂F∂εxx

∂εxx∂uej

dψeidx

dx = EA

Z xb

xa

µ∂F∂εxx

¶dψeidx

dψejdx

dx

(10.2.5)where small strain assumption is used in arriving at the last step.

Stress

Strain

Linear elastic

Nonlinear elastic

Loading and unloading

Hardening type

Softeningtype

MATERIAL NONLINEARITIES AND COUPLED PROBLEMS 391

Figure 10.2.1 A nonlinear elastic stressstrain curve.

An example of the nonlinear elastic response is provided by RombergOsgood model

F(!xx) = (!xx)n , "F"!xx

= n (!xx)n−1 (10.2.6)

where n > 0 is a material parameter. The value of n = 1 yields the linearelastic case. This discussion can be extended to multi-axial case, whereF = F(!ij).

10.3 Small Deformation Theory of Plasticity

10.3.1 Introduction

Plasticity refers to non-recoverable deformation and non-unique stress pathsin contrast to nonlinear elasticity, where the entire loaddeßection path isunique and the strains are recovered on load removal. The mathematicaltheory of plasticity is of a phenomenological nature on the macroscopic scale,and the objective of the theory is to provide a theoretical description ofthe relationship between stress and strain for a material that exhibits anelasto-plastic response. The plastic behavior is characterized by irreversibilityof stress paths and the development of permanent (i.e. non-recoverable)deformation (or strain), known as yielding (or plastic ßow).If uniaxial behavior of a material is considered, a nonlinear stress-strain

relationship on loading alone does not determine if nonlinear elastic or plasticbehavior is exhibited. Unloading part of the curve determines if it is elastic orplastic [see Figure 10.3.1(a) and (b)]; the elastic material follows the same pathin loading and unloading, while the plastic material shows a history-dependentpath unloading.

Stress

Strain

Plastic strain

Loading

Unloading

(b)

Stress

Strain

Loading

Unloading

(a)


Figure 10.3.1 Stressstrain behavior of (a) ideal plasticity, and (b) strainhardening plasticity.

The theory of plasticity deals with an analytical description of the stressstrain relations of a deformed body after a part or all of the body has yielded.The stressstrain relations must contain:

1. The elastic stress-strain relations.

2. The stress condition (or yield criterion) which indicates onset of yielding.

3. The stressstrain or stressstrain increment relations after the onset ofplastic ßow.

10.3.2 Ideal Plasticity

Many materials exhibit an ideal plastic (or elasticperfectlyplastic) behavior,as shown in Figure 10.3.1(a). In this case, there exists a limiting stress, calledyield stress, denoted by #Y , at which the strains are indeterminate. For all


stresses below the yield stress, a linear (or nonlinear) stressstrain relation isassumed:

σij < σY linear elastic behavior

σij ≥ σY plastic deformation (not recoverable) (10.3.1)

10.3.3 Strain Hardening Plasticity

A hardening plastic material model provides a reÞnement of the ideal plasticmaterial model. In this model, it is assumed that the yield stress depends onsome parameter κ (e.g. plastic strain εp), called the hardening parameter. Thegeneral yield criterion is expressed in the form

F (σij ,κ) = 0 (10.3.2)

This yield criterion can be viewed as a surface in the stress space, with theposition of the surface dependent on the instantaneous value of the hardeningparameter κ. Since any yield criterion should be independent of the orientationof the coordinate system used, F should be a function of the stress invariantsonly. Experimental observations indicate that plastic deformation in metals isindependent of hydrostatic pressure. Therefore, F must be a function of thestress invariants of the deviatoric stress tensor σ0:

F (J 02, J03,κ) = 0, J

02 =

1

2σ0ijσ

0ij , J

03 =

1

3σ0ijσ

0jkσ

0ki (10.3.3)

Two of the most commonly used yield criteria are given next.

The Tresca yield criterion

F = 2σ cos θ − Y (κ) = 0, σ =qJ 02 (10.3.4)

The Hubervon Mises yield criterion

F =q3J 02 − Y (κ) = 0 (10.3.5)

where Y is the yield stress from uniaxial tests, θ is the angle between the line

of pure shear and the principal stress σ1, and σ =qJ 02 is called the effective

stress.After initial yielding, the stress level at which further plastic deformation

occurs may be dependent on the current degree of plastic straining, knownas strain hardening. Thus, the yield surface will vary (i.e. expand) at eachstage of plastic deformation. When the yield surface is independent of thedegree of plasticity, the material is said to be ideally (or perfectly) plastic.

Stress

Strain

dσp

E

σY

dε

dεp

dεe

Slope ET - Elastic−plastic tangent modulus


If the subsequent yield surfaces are a uniform expansion of the original yieldsurface, the hardening model is said to be isotropic. On the other hand, if thesubsequent yield surfaces preserve their shape and orientation but translate inthe stress space, kinematic hardening is said to take place.Consider the uniaxial stressstrain curve shown in Figure 10.3.2. The

behavior is initially linear elastic with slope E (Youngs modulus) until onsetof yielding at the uniaxial yield stress #Y . Thereafter, the material response iselastoplastic with the local tangent to the curve, ET , called the elasto-plastictangent modulus, continually changing.At some stress level # in the plastic range, if the load is increased to induce

a stress of d#, it results in a corresponding strain d!. This increment of straincontains two parts: elastic d!e (recoverable) and plastic d!p (non-recoverable):

d! = d!e + d!p, d!e =d#

E,

d#

d!= ET (10.3.6)

The strain-hardening parameter, H, is deÞned by

H =d#

d!p=

dσdε

1− dεe

dε

=ET

1− ETE

(10.3.7)

The element stiffness for the linear elastic portion is, say [Ke]:

[Ke] =

Z xb

xa[B]T [De][B] dx (10.3.8)

where [De] is the linear elasticity matrix (De = E for the uniaxial case). When

the element deforms plastically, [De] reßects the decreased stiffness. This iscomputed, for uniaxial material behavior, by the following procedure: The

Figure 10.3.2 A strain hardening plastic behavior for the uniaxial case.


increment in load dF causes an incremental displacement du

du = hedεxx = he (dεe + dεp) , dF = Adσ = AeHdε

p (10.3.9)

where he is the length and Ae the area of cross-section of the element. Theeffective stiffness is

Eep =dF

du=

AeHdεp

he (dεe + dεp)=EAehe

∙1− E

(E +H)

¸(10.3.10)

The element stiffness for plastic range becomes,

[Kep] =

Z xb

xa[B]T [Dep][B]dx (10.3.11)

where [Dep] is the material stiffness in the plastic range. For uniaxial caseDep = Eep.Equation (10.3.8) is valid when σ < σY and Eq. (10.3.11) is valid for

σ > σY . Note that dσ = σ − σY when σ > σY .

10.3.4 ElasticPlastic Analysis of a Bar

Here we present a detailed computational procedure for the analysis of anelastoplastic problem. The procedure is described via a one-dimensionalelastoplastic bar problem. We shall consider a linear strain-hardeningmaterial subjected to an increasing uniaxial load.

Update of stresses

At a load-step number r where the deformation is elastic, the stress in a typicalelement with the strain increment ∆εr can be readily updated as

σri = σ(r−1)i +Ei∆ε

r (10.3.12)

where Ei is the elastic modulus of element i. This linear elastic behaviorwill continue up till a point where the resulting strain increment will initiateplastic yielding of the material. Now the updating of the stress in the elementis not as straighforward as given in Eq. (10.3.12), and it can get complicatedwhen the deformation is partly elastic and partly elasto-plastic, as shown fromPoints A to B in the stressstrain curve of Figure 10.3.3.To update the stress state from Points A to B, one can Þrst assume that the

deformation is elastic and compute the corresponding elastic stress, commonlyreferred to as the elastic stress predictor. Using Eq. (10.3.12), the elastic stresspredictor σe can be calculated as

σei = σ(r−1)i +Ei∆ε

ri (10.3.13)

1

RA

B

CD

eiσσ

riσ

1−riσYσ

riε∆

ε

A′C′


Computing the elastic stress predictor brings the stress state from Point A toA0. A correction is made to transfer the stress state back to the elasto-plasticstate at Point B. We introduce a correction factor R (see Figure 10.3.3)

R =#ei − #y#ei − #(r−1)i

(10.3.14)

so that the stress at point B can be written as

#ri = #(r−1)i + [(1−R)Ei +RET ]∆!ri (10.3.15)

Here ET denotes the elasticplastic tangent modulus, which is related to theelastic modulus E and strain-hardening parameter H by Eq. (10.3.7). In thecase where the element has already yielded in previous load steps, as illustratedby point C in Figure 10.3.3, the approach of determining the elastic stresspredictor and making correction to the stress state at point D still applieswith R = 1 in Eq. (10.3.15):

#ri = #(r−1)i +ET∆!

ri (10.3.16)

Update of plastic strain

The extent of plastic ßow in a deformed material can be readily characterizedby the measure of plastic strain. To determine the plastic strain in an elementat point B of Figure 10.3.3, it will be useful to rewrite Eq. (10.3.15) as

#ri = #Y +ET (R∆!ri ) ≡ #Y +∆#ri (10.3.17)

Figure 10.3.3 Transition of elastic to elastoplastic behavior.

H1

Yσ

pε

ryiσ


Equation (10.3.17) can be interpreted as that adjusts the stress state atpoint A to the yield stress before predicting the elastic stress and its correction.This will allow one to isolate the stress component ∆#ri and strain R!

ri that

are involved in the plastic ßow. With Eq. (10.3.6), the plastic strain incrementis

∆!rpi = R∆!ri −

∆#riEi

=

µ1− ET

Ei

¶R∆!ri (10.3.18)

Equation (10.3.18) can also be used for elements that have already yielded inprevious load steps by setting R = 1.

Update of yield stress limit

Besides assessing the extent of plastic deformation, the measure of the plasticstrain will become especially crucial for strain-hardening materials where theyield limit is a function of the plastic strain. A plot of yield limit against theplastic strain for a typical linear strain-hardening material is shown in Figure10.3.4. Once the plastic strain occurs, the yield limit will be modiÞed andupdated as

#ryi = #Y +H∆!rpi (10.3.19)

IdentiÞcation of deformation modes

The updated yield limit will come in handy when one is to check the type ofdeformation an element is undergoing. Once the correct type of deformationis identiÞed, the stress and strain values can then be updated according toEqs. (10.3.l5) and (10.3.18). There are four types of deformation:

Figure 10.3.4 Stressstrain behavior of a strainhardening material.

i i+1

N

FNFi+1Fi

∆F


(a) Elastic Loading: (an element that has not yielded previously continues todeform elastically)

|#r−1i | < |#r−1yi | and |#ei| < |#r−1yi | (10.3.20a)

(b) Elastic-Plastic Loading: (an element that has not yielded previously willdeform elasto-plastically)

|#r−1i | < |#r−1yi | and |#ei| > |#r−1yi | (10.3.20b)

(c) Plastic Loading: (an element that previously yielded will continue todeform plastically)

|#r−1i | > |#r−1yi | and |#ei| > |#r−1i | (10.3.20c)

(d) Elastic Unloading: (an element previously yielded is now unloadingelastically)

|#r−1i | > |#r−1yi | and |#ei| < |#r−1i | (10.3.20d)

Force equilibrium

Since the displacement Þnite element model is based on the principle of virtualdisplacements, the solution satisÞes the equilibrium equations, provided thedeformation is linearly elastic. However, in the Þnite element analysis ofelastoplastic problems equilibrium equations may not be satisÞed during theperiod when stresses are adjusted to account for plastic strains. Adjustmentsmust be made to achieve equilibrium at each step by redistributing the forcesneighboring elements.For example, consider a node N at the interface of element i that has

yielded and the adjacent element i+ 1 that is still elastic (see Figure 10.3.5).At this node, the force equilibrium will be violated during the analysis becausethe force in element i is limited such that the stress in the element does notexceed the yield stress. The difference between the force calculated using theelastic analysis and the plastic force must now be taken up by all other elasticelements in the mesh. Thus to restore equilibrium of forces, a force correctionmust be made at the node N :

∆F = Fi − Fi+1 − FN (10.3.21)

Figure 10.3.5 Force equilibrium at node N .


However, to preserve the Þnite element equations of the original problem(to retain the same forces in other unaffected elements), the force correctioncannot be imposed as a nodal force. Instead, the force correction may beapplied as a nodal displacement

∆ucN =∆F LiETAi

(10.3.22)

where Li and Ai are the length and cross-sectional area of element i. Thiscorrection procedure will continue until force equilibrium at all nodes isrestored, within an acceptable error of tolerance.Figure 10.3.6 contains the ßow chart of various steps in the elastoplastic

analysis of a typical problem. A subroutine that updates stresses and plasticstrains is also given in Box 10.3.1.

A numerical example

Consider a bar of length 5 m that is Þxed at one end and is subjected to auniform body force f . The material properties of the bar are taken as

E = 104 N/m2, A = 1 m2, σy = 5 N/m2, H = 103 N/m2

The bar is discretized using a mesh of Þve linear elements. The elastoplasticiterative scheme discussed in this section was implemented and the resultsare presented in Tables 10.3.1 and 10.3.2. At the start of the analysis whenelements are still elastic, a nominal body force of 0.005 N/m was imposed toÞnd the maximum stress induced in the elements. The critical load Fcr for theÞrst element to yield was computed from this maximum element stress and isimposed in the next load-step:

fcr =σY f

max|σi| (10.3.23)

Here i is the element number and N is the total number of unyielded elements.In the new load-step where f is 1.1111 N/m, the computed results reveal thatthe Þrst element (element 1 in this example) had just yielded; up to this point,the analysis is still elastic. Then another nominal body force of 0.005 N/m isadded to calculate the critical load for the next element to yield. The stiffnessof the yielded Element 1 is reduced in this load-step and the results in Table10.3.2 indicate a violation of force equilibrium at the nodes connecting theyielded element, except for the node that is Þxed. Corrections to the nodaldisplacements were made until equilibrium was satisÞed at all nodes. Onlythen, the critical load for the next element to yield could be calculated andthe same procedure is repeated until all elements yield.Reaction forces at the Þxed end of the bar against the free-end

displacements are plotted in Figure 10.3.7, together with the results from

♦ Load increment

♦ Solve

START

Compute element strain increment

Compute element elastic stress predictor and yield stress limit

(assume elasticdeformation)

(Element has not yielded in the previous load-step)

(Element yielded in the previous load-step)

(Element is unloading elastically)

(Element is deforming elastically)

Update element stress and plastic strain

(Element has just yielded)

(Element continues to yield)

Is equilibrium satisfied at each

node?

Modify nodal displacements

(Go to the next load-step)

riε∆

⎟⎟⎠

⎞⎜⎜⎝

⎛

+=∆+=

−−

−

11

1

rpiY

ryi

iriei

H

E

εσσεσσ

YESNO11 −− > r

yiri σσ

[ ] FuK r =

fff rr ∆+= −1

1−> ryiei σσ 1−> r

iei σσ0=RNO

YES

( )[ ]ri

i

Trpi

rpi

riTi

ri

ri

REE

ERER

εεε

εσσ

∆⋅⎟⎟⎠

⎞⎜⎜⎝

⎛−+=

∆⋅+−+=

−

−

1

1

1

1

NO

YES

1=R

NO YES1+= rr

1−−=∆

∆+=ri

rii

Ci

ri

ri

uuu

uuu

1−−−=

riei

YeiRσσσσ


Figure 10.3.6 Flow chart for elastoplastic analysis of a bar.


Box 10.3.1 Fortran statements for the subroutine to calculate the stressand plastic strain of a bar element.

SUBROUTINE STRESS(N,MXELM,NDF,NPE,IELEM,ELSIG,PSTRAIN)

C The subroutine evaluates the stress at the center of the element and then updateC the stress and plastic strain against the yield criteria.C SIGMA Updated elastic stress YOUNG Elastic modulusC TYIELD Current yield limit TYOUNG Elasto−plastic tangent modulusC ELSIG(N) Elastic stress of element N YSTRESS Yield stressC PSTRAIN(N) Plastic strain of element N HP Strain hardening parameter

.

. Definition of global and local variables and common blocks

.C Compute the total strain increment at the center of element

H = ELX(NPE) − ELX(1)CALL SHP1D (H,IELEM,NPE,0.D0)UX=0.D0DO K=1,NPE

UX=UX+EPV((K−1)*NDF+1)*GDSF(K)ENDDO

C Update the elastic stress and current yield limitSIGMA = ELSIG(N)+YOUNG*UXIF(ELSIG(N).GT.0.0)TYIELD=YSTRESS+HP*PSTRAIN(N)IF(ELSIG(N).LE.0.0)TYIELD=−YSTRESS+HP*PSTRAIN(N)

C Check the updated elastic stress for yieldingIF (ABS(ELSIG(N)) .GT. ABS(TYIELD)) THEN

IF (ABS(SIGMA) .GT. ABS(ELSIG(N))) THENR = 1.D0

ELSER = 0.D0

ENDIFELSE

IF (ABS(SIGMA) .GT. ABS(TYIELD)) THENR = (ABS(SIGMA)-ABS(TYIELD))/(ABS(SIGMA)-ABS(ELSIG(N)))

ELSER = 0.D0

ENDIFENDIF

C Correct the elastic stress and calculate the plastic strainELSIG(N)= ELSIG(N)+((1.D0-R)*YOUNG+R*TYOUNG)*UXPSTRAIN(N)= PSTRAIN(N)+(R/(1.D0+HP/YOUNG))*UXRETURNEND

C _____________________________________________________________________________

C _____________________________________________________________________________

1

2

3

4

5

R

f

0

10

20

30

40

50

60

70

80

90

100

0.00 0.05 0.10 0.15 0.20 0.25

Nodal displacement at the bar end

Rea

ctio

n f

orce

,R

Present results (5 elements)

ABAQUS (5 elements)

R

f

-

----

--

----

••

•••

••

•

•••

••

••

••

••

••

••

- - - -

E = 10,000, A = 1, L = 5,H = 1,000, σy = 5


Table 10.3.1 Nodal displacements for various load steps.

Body force f Node no. Nodal displacement*

0.0050 2 2.2500×10−63 4.0000×10−64 5.2500×10−65 6.0000×10−66 6.2500×10−6

1.1111 2 5.0000×10−43 8.8889×10−44 1.1667×10−35 1.3333×10−36 1.3889×10−3

1.1161 2 5.2475×10−4 (5.2475×10−4)3 9.1539×10−4 (9.1539×10−4)4 1.1944×10−3 (1.1944×10−3)5 1.3618×10−3 (1.3618×10−3)6 1.4176×10−3 (1.4176×10−3)

10.0050 2 4.9525×10−2 (4.4525×10−2)3 8.8044×10−2 (7.8044×10−2)4 1.1556×10−1 (1.0056×10−1)5 1.3207×10−1 (1.1207×10−1)6 1.3257×10−1 (1.1257×10−1)

* Values in parenthesis are corrected to satisfy force equilibrium at each node.

Figure 10.3.7 Reaction forces versus nodal displacements at bar end.


Table 10.3.2 Element stresses and strains for various load steps.

Body force Element Total strain Plastic strain Bar forcef number ε εp |F | = σA

0.0050 1 2.2500×10−6 0 0.02252 1.7500×10−6 0 0.01753 1.2500×10−6 0 0.01254 0.7500×10−6 0 0.00755 0.2500×10−6 0 0.0025

1.1111 1* 5.0000×10−4 0 5.00002 3.8889×10−4 0 3.88893 2.7778×10−4 0 2.77784 1.6667×10−4 0 1.66675 0.5556×10−4 0 0.5556

1.1161 1 5.5248×10−4 4.5680×10−3 9.56802 3.9064×10−4 0 3.90643 2.7903×10−4 0 2.79034 1.6742×10−4 0 1.67425 0.5581×10−4 0 0.5581

Corrected 1 5.2475×10−4 0.2250×10−4 5.0225to satisfy 2 3.9064×10−4 0 3.9064force 3 2.7903×10−4 0 2.7903equilibrium 4 1.6742×10−4 0 1.6742

5 0.5581×10−4 0 0.5581. . . . . . . . . . . . . . .

10.0050 1 4.9525×10−2 4.4568×10−2 49.56802 3.8519×10−2 3.4563×10−2 39.56303 2.7514×10−2 2.4558×10−2 29.55804 1.6508×10−2 1.4553×10−2 19.55305 0.5503×10−2 0.4548×10−2 9.5480

Corrected 1 4.4525×10−2 4.0023×10−2 45.0225to satisfy 2 3.3519×10−2 3.0018×10−2 35.0175force 3 2.2514×10−2 2.0013×10−2 25.0125equilibrium 4 1.1508×10−2 1.0008×10−2 15.0075

5 5.0275×10−4 0.0025×10−4 5.0025

* Element just yielded at that load step.

the commercial Þnite element software ABAQUS. There is a very goodagreement with the solutions generated by the iterative scheme discussedand those obtained with ABAQUS. A 20-element mesh also produced resultsidentical to those in Figure 10.3.7; the results are also veriÞed using ABAQUS.Figure 10.3.8 contains the true stressstrain diagram of Element 1, where onemay note that the elastoplastic material curve is recovered.

909.08

0

5

10

15

20

25

30

35

40

45

50

0.0 0.005 0.010 0.015 0.020 0.025 0.030 0.035 0.040 0.045 0.050

Total strain, ε

Str

ess,

σ Element 1

E = 10,000, A = 1, L = 5,H = 1,000, σy = 5

1

2

3

4

5

R

1


Figure 10.3.8 True stressstrain curve of Element 1.

10.4 Non-Newtonian Fluids

10.4.1 Introduction

Some of the most challenging and important areas currently underinvestigation in computational ßuid mechanics concern ßows of non-Newtonianßuids. A non-Newtonian ßuid is deÞned to be the one whose constitutivebehavior is nonlinear (i.e., stresses are nonlinear functions of strain rates).Such ßuids may or may not have memory of past deformation (i.e. viscoelasticor not). Practical examples of such ßuids are multigrade oils, liquid detergents,paints, and printing inks. Polymer solutions and polymer melts also fall withinthis category. All such ßows are extremely important in forming processes ofvarious kinds applied to metals, plastics, or glass.

Numerical approaches used for analyzing ßows of non-Newtonian ßuidsdiffer vary little from the ones used for Newtonian ßuids. When the shearviscosity is a function of the rate of deformation tensor (i.e., for so-calledpower-law ßuids), the equation of motion can still be written explicitly interms of velocity components. For such ßuids, the same formulations as thoseof the Newtonian ßuids can be used. For example, the pressurevelocity andpenalty Þnite element models discussed in Chapter 7 can be used for power-law ßuids. The constitutive equations of a viscoelastic ßuid can be describedin terms of the extra stress components and are given in terms of eitherdifferential equations or as integral equations. In a differential constitutivemodel, the extra stress components and their derivatives are related to the


velocity components and their derivatives [69]. The computational modelsof viscoelastic ßuids obeying a differential model require us to treat stresscomponents as independent variables along with the velocity components andpressure.

Most existing works in the Þnite-element analysis of viscoelastic ßuidshave considered different types of Maxwell/Oldroyd models, in which bothviscosity and relaxation time are constants. The analyses were based on mixedformulations in which the velocities, pressure, and stresses are treated as nodalvariables. An important class of ßuids, which are characterized by shearviscosity and elasticity require additional study from numerical simulationpoint of view. The viscoelastic effects along with memory effects of such ßuidscan be studied using the WhiteMetzner model [7].

The objective of this section is to study penalty Þnite element modelsof power-law and viscoelastic WhiteMetzner type ßuids. As a part of thisdiscussion, we present Þnite element models of equations governing ßows ofviscous incompressible ßuids through axisymmetric geometries. Therefore,Þrst we rewrite the conservation of mass and momentum in the cylindricalcoordinate system.

10.4.2 Governing Equations in Cylindrical Coordinates

Equations (7.3.1)(7.3.3) can be expressed in a cylindrical coordinate system,(r, θ, z), by writing all vectors and tensors, including the del operator, interms of components in a cylindrical coordinate system. For example, thedel operator and the material time derivative operators in the cylindricalcoordinate system are given by

∇ = er ∂∂r+ eθ

1

r

∂

∂θ+ ez

∂

∂z(10.4.1)

D

Dt=∂

∂t+ vr

∂

∂r+vθr

∂

∂θ+ vz

∂

∂z(10.4.2)

where (er,eθ,ez) are the unit basis vectors and (vr, vθ, vz) are the velocitycomponents in the r, θ, and z directions, respectively. Note that the basisvector ez is constant while the vectors er and eθ depend on the angularcoordinate θ. Thus the derivatives of the basis vectors er and eθ with respectto the coordinates r and z are zero, and the derivatives with respect to θ aregiven by

∂er∂θ

= eθ;∂eθ∂θ

= −er (10.4.3)

The components of the strain rate tensor D in Eq. (7.2.13) and constitutiverelations (7.2.11) and (7.2.15) take the form

Drr =∂vr∂r; Dθθ =

1

r

∂vθ∂θ

+vrr


Dzz =∂vz∂z

; 2Drθ =∂vθ∂r

− vθr+1

r

∂vr∂θ

(10.4.4)

2Dθz =1

r

∂vz∂θ

+∂vθ∂z

; 2Dzr =∂vr∂z

+∂vz∂r

σrr = −P + 2µDrr ; σθθ = −P + 2µDθθ ; σzz = −P + 2µDzzσrθ = 2µDrθ ; σθz = 2µDθz ; σzr = 2µDzr (10.4.5)

The governing equations (7.3.1)(7.3.3) expressed in the cylindricalcoordinate system are

1

r

∂

∂r(rvr) +

1

r

∂vθ∂θ

+∂vz∂z

= 0 (10.4.6)

ρ

ÃDvrDt

− v2θ

r

!= ρfr +

1

r

∙∂(rσrr)

∂r+∂σrθ∂θ

+∂(rσzr)

∂z

¸− σθθ

r(10.4.7a)

ρ

µDvθDt

+vrvθr

¶= ρfθ +

1

r

∙∂(rσrθ)

∂r+∂σθθ∂θ

+∂(rσθz)

∂z

¸+σrθr

(10.4.7b)

ρ

µDvzDt

¶= ρfz +

1

r

∙∂(rσzr)

∂r+∂σzθ∂θ

+∂(rσzz)

∂z

¸(10.4.7c)

ρCsDT

Dt=1

r

∂

∂r

µrkrr

∂T

∂r

¶+1

r2∂

∂θ

µkθθ∂T

∂θ

¶+∂

∂z

µkzz∂T

∂z

¶+Φ+Q (10.4.8)

where the stress components are known in terms of the velocity componentsvia equations (10.4.4) and (10.4.5). The viscous dissipation Φ is given by

Φ = 2µ³D2rr +D

2θθ +D

2zz

´+ 4µ

³D2rθ +D

2rz +D

2zθ

´(10.4.9)

and the material time derivative becomes

D

Dt=∂

∂t+ vr

∂

∂r+vθr

∂

∂θ+ vz

∂

∂z(10.4.10)

Note that the time derivative of a vector (or tensor) in a rotating referenceframe is given by (see Reddy and Rasmussen [5], pp. 6974)∙

D(·)Dt

¸nonrot

=

∙D(·)Dt

¸rot+ ω × (·) (10.4.11)

where ω = dθdtez =

vθr ez is the angular velocity vector of the rotating frame of

reference. Therefore, we haveµDv

Dt

¶nonrot

=

µDv

Dt

¶rot+ (vθrez)× v

= er

µDvrDt

− vθ vθr

¶+ eθ

µDvθDt

+ vrvθr

¶+ ez

DvzDt

(10.4.12)


Equation (10.4.11) could be used in conjunction with (10.4.7ac) to provide adescription of ßuid motion in a rotating cylindrical coordinate system.

For axisymmetric conditions (i.e. all variables are independent of θ, andfθ = vθ = 0), the governing equations are simplied to

1

r

∂

∂r(rvr) +

∂vz∂z

= 0 (10.4.13)

ρ

µ∂vr∂t

+ vr∂vr∂r

+ vz∂vr∂z

¶= ρfr +

1

r

∂(rσrr)

∂r+∂σzr∂z

− σθθr(10.4.14a)

ρ

µ∂vz∂t

+ vr∂vz∂r

+ vz∂vz∂z

¶= ρfz +

1

r

∂(rσzr)

∂r+∂σzz∂z

(10.4.14b)

ρCs

µ∂T

∂t+ vr

∂T

∂r+ vz

∂T

∂z

¶=1

r

∂

∂r

µrkrr

∂T

∂r

¶+∂

∂z

µkzz∂T

∂z

¶+Φ+Q (10.4.15)

10.4.3 Power-Law Fluids

Many ßuids used in industrial applications are characterized by so-calledpower-law constitutive behavior. Power-law ßuids exhibit nonlinear materialbehavior according to the relation

σ = τ − P I, τ = 2µ(I2)D (10.4.16)

µ = µ0 (I2)(n−1)2 (10.4.17a)

I2 =1

2DijDij (sum on i and j) (10.4.17b)

where τ is the viscous part of the stress tensor σ, I is the unit tensor, I2 isthe second invariant of D, and parameters µ0 and n characterize the ßuid(determined experimentally). For many non-Newtonian ßuids, the viscosity µdecreases with increasing shear rate. These are called shear thinning ßuids andhave the power law index n < 1. Fluids with power law index n > 1 are calledshear-thickening ßuids. For such ßuids, µ increases with increasing shear rate.For n = 1, Eq. (10.4.17a) gives the Newtonian relation (µ = µ0, constant).The power-law constitutive relation in (10.4.16) makes the problem nonlinearfor n 6= 1, even for the case where the convective term v ·∇v is negligible.To illustrate how the power-law constitutive equation affects the Þnite

element equations, we consider steady, isothermal, axisymmetric ßows ofviscous incompressible ßuids. Assuming negligible viscous dissipation andbody forces, The governing equations (10.4.13)(10.4.15) can be expressed


in terms of the velocity components (vr, vz) as

1

r

∂

∂r(rvr) +

∂vz∂z

= 0 (10.4.18)

ρ

µvr∂vr∂r

+ vz∂vr∂z

¶=1

r

∂

∂r

µ2µr

∂vr∂r

¶+∂

∂z

∙µ

µ∂vr∂z

+∂vz∂r

¶¸− 2µvr

r2− ∂P∂r

(10.4.19)

ρ

µvr∂vz∂r

+ vz∂vz∂z

¶=1

r

∂

∂r

∙µr

µ∂vr∂z

+∂vz∂r

¶¸+∂

∂z

µ2µ∂vz∂z

¶− ∂P∂z

(10.4.20)

The penaltyÞnite element model of Eqs. (10.4.18)(10.4.20) isstraightforward. After replacing P in Eqs. (10.4.19) and (10.4.20) with

P = −γpµ∂vr∂r

+vrr+∂vz∂z

¶(10.4.21)

one may construct their week forms and go on to develop the Þnite elementmodel [note that Eq. (10.4.18) is no longer used]. The value of the penaltyparameter γp that is most suitable for this class problems must be determinedby conducting numerical experiments with some benchmark problems. TheÞnite element model has the form

[C(v) +K(µ) +Kp]v = F or [K(v)]v = F (10.4.22)

where [C], [K], and [Kp] are the convective, diffusive, and penaltycontributions to the coefficient matrix. Note that [C] and [K] both depend onthe unknown velocity Þeld.

The direct iteration scheme for this case is given by

[ K(v(r−1))]vr = F (10.4.23)

where [ K] is evaluated using the viscosity µ = µ(Dij) computed according toEq. (10.4.17a) at the rth iteration. It would be more appropriate to compute

µ at the Gauss points of the element and use it in the evaluation of Kij ,than to assume µ is element-wise constant. For example, consider the elementcoefficients K11

ij in the diffusion/viscous portion of the coefficient matrix:

K11ij =

ZΩe

µ2µ∂ψi∂x

∂ψj∂x

+ µ∂ψi∂y

∂ψj∂y

¶dxdy (10.4.24)


For numerical evaluation of K11ij we use the Gauss quadrature rule

K11ij =

NGPXI,J=1

AMU(I, J) ∗ [2.0 ∗GDSF (1, i) ∗GDSF (1, j)

+GDSF (2, i) ∗GDSF (2, j) ] ∗ CONST (10.4.25)

where AMU(I, J) is the value of µ at the (I, J)th Gauss point. Obviously,AMU(I, J) must be evaluated using Eq. (10.4.17a) prior to using it in Eq.(10.4.25).

10.4.4 WhiteMetzner Fluids

The general constitutive equation for ßuids dominated by shear viscosity isgiven by (see [79])

τ + λ∂τ∂t+ v ·∇τ − [(∇v)T · τ + τ · (∇v)] = 2ηD (10.4.26)

where λ is the relaxation time, η is the shear viscosity of the ßuid, and D isthe rate of deformation tensor.

To characterize the WhiteMetzner ßuid, it is necessary to know theviscosity curve as a function of shear rate and the Þrst normal stress difference.The extra stress tensor τ is separated into purely viscous part τ2 andviscoelastic part τ1:

τ = τ1 + τ2 (10.4.27)

τ1 + λ

(∂τ1

∂t+ v ·∇τ1 −

h(∇v)T · τ1 + τ1 · (∇v)

i)= 2η1D (10.4.28)

τ2 = 2η2D (10.4.29)

and η1 and η2 can be deÞned as the fractions of the shear viscosity η. Theßuid which obeys Eq. (10.4.27) is characterized by η1, η2, and λ, which arethe functions of rate of deformation tensor, D.

Here, the relaxation time λ is assumed to depend on the shear rate γaccording to

λ(γ) = a+ b(log γ) + c(log γ)2, γ =p4I2 (10.4.30)

All constants in Eq. (10.4.30) are evaluated by curve Þtting data of polymericmelts (see [16]). Also, η is assumed to be of the particular form

η = η(I2), I2 =1

2DijDij (10.4.31)


Although Eq. (10.4.31) gives the general functional form for the viscosityfunction, experimental observation and a theoretical basis must be used toprovide a speciÞc model for non-Newtonian viscosities. A variety of modelscan be used to calculate viscosity. Here, the power law model in Eq. (10.,4.17a)is used (replace µ with η).

In view of Eq. (10.4.27), the equation of motion (in the absence of bodyforces) for the WhiteMetzner viscoelastic ßuids can be written as

∇ · (τ1 + 2η2D)−∇P = ρµ∂v

∂t+ v ·∇v

¶(10.4.32)

For simplicity, we drop the superscript 1 from τ1 in Eq. (10.4.32) as well as inEq. (10.4.28). Equations (10.4.32) and (10.4.28), together with the continuityequation ∇ ·v = 0, represent the system of governing equations for the White-Metzner ßuids. The viscosity η1 is calculated using Eq. (10.4.28) and λ fromEq. (10.4.30). Viscosity η2 is often taken as a function of η1. Note that thepower-law constitutive equation can be obtained as a special case from Eq.(10.4.28).

For axisymmetric ßows of WhiteMetzner ßuids, the governing equationsare given by

∂u

∂r+u

r+∂w

∂z= 0 (10.4.33)

∂

∂r

µτrr + 2η2

∂u

∂r

¶+1

r

µτrr + 2η2

∂u

∂r− τθθ − 2η2u

r

¶+∂

∂r

∙τrz + η2

µ∂u

∂z+∂w

∂r

¶¸− ∂P∂r

= ρ

µ∂u

∂t+ u

∂u

∂r+ w

∂u

∂z

¶(10.4.34)

∂

∂r

∙τrz + η2

µ∂u

∂z+∂w

∂r

¶¸+1

r

∙τrs + η2

µ∂u

∂z+∂w

∂r

¶¸+∂

∂z

µτzz + 2η2

∂w

∂z

¶− ∂P∂z

= ρ

µ∂w

∂t+ u

∂w

∂r+ w

∂w

∂z

¶(10.4.35)

τrr + λ

∙∂τrr∂t

+ u∂τrr∂r

+ w∂τrr∂z

− 2µ∂u

∂rτrr +

∂u

∂zτrz

¶¸= 2η1

∂u

∂r(10.4.36a)

τzz + λ

∙∂τzz∂t

+ u∂τzz∂r

+ w∂τzz∂z

− 2µ∂w

∂zτzz +

∂w

∂rτrz

¶¸= 2η1

∂w

∂z(10.4.36b)

τrz + λ

∙∂τrz∂t

+ u∂τrz∂r

+ w∂τrz∂z

− 2µ∂u

∂rτrz +

∂w

∂zτrz +

∂u

∂zτzz +

∂w

∂rτrr

¶¸= η1

µ∂w

∂r+∂u

∂z

¶(10.4.36c)

τθθ + λ

∙∂τθθ∂t

+ u∂τθθ∂r

+ w∂τθθ∂z

− 2urτθθ

¸= 2η1

u

r(10.4.36d)

where vr = u and vz = w denote the velocity components in the radial andaxial directions, respectively.


The (mixed) penaltyÞnite element model of the above equations canbe developed by constructing the weak forms of Eqs. (10.4.34)(10.4.36)[after replacing the pressure with Eq. (10.4.21)] and interpolating velocities(u,w) and extra stress components (τrr, τθθ, τzz, τrz) as follows (with equalinterpolation of all variables):

u(r, z, t) =NXj=1

uj(t)ψj(r, z), w(r, z, t) =NXj=1

wj(t)ψj(r, z)

τrr(r, z, t) =LXj=1

τ jrr(t)ψj(r, z), τzz(r, z, t) =LXj=1

τ jzz(t)ψj(r, z) (10.4.37)

τrz(r, z, t) =LXj=1

τ jrz(t)ψj(r, z), τθθ(r, z, t) =LXj=1

τ jθθ(t)ψj(r, z)

where, (uj , wj) are the nodal velocities, (τjrr, τ

jzz, τ

jrz, τ

jθθ) are the nodal extra

stress components, and ψj are the Lagrange interpolation functions. The Þniteelement model has the general form

[Me] ú∆e+ [Ke]∆e = F e (10.4.38)

where

∆e =

⎧⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎩

uwτrrτzzτrzτθθ

⎫⎪⎪⎪⎪⎪⎪⎬⎪⎪⎪⎪⎪⎪⎭(10.4.39)

The nonzero coefficient matrices are listed below.

K11ij =

ZΩe

∙ρψi

µu∂ψj∂r

+w∂ψj∂z

¶+ η2

µ2∂ψi∂r

∂ψj∂r

+∂ψi∂z

∂ψj∂z

+ψiψjr2

¶¸r dr dz

+ γp

ZΩe

µ∂ψi∂r

+1

rψi

¶µ∂ψj∂r

+1

rψj

¶r dr dz

K12ij =

ZΩe

µη2∂ψi∂z

∂ψj∂r

¶r dr dz + γp

ZΩe

µ∂ψi∂r

+1

rψi

¶∂ψj∂z

r dr dz

K13ij =

ZΩe

∂ψi∂rψj r dr dz, K31

ij = −2ZΩeη2ψi

∂ψj∂r

r dr dz

K21ij =

ZΩe

µη2∂ψi∂r

∂ψj∂z

¶r dr dz + γp

ZΩe

µ∂ψj∂r

+1

rψj

¶∂ψi∂z

r dr dz

K22ij =

ZΩe

∙ρψi

µu∂ψj∂r

+w∂ψj∂z

¶+ η2

µ∂ψi∂r

∂ψj∂r

+ 2∂ψi∂z

∂ψj∂z

¶¸r dr dz

+ γp

ZΩe

∂ψi∂z

∂ψj∂z

r dr dz


K24ij =

ZΩe

∂ψi∂zψj r dr dz, K25

ij =

ZΩe

∂ψi∂rψj r dr dz

K33ij =

ZΩe

(ψiψj + λ

∙ψi

µu∂ψj∂r

+ w∂ψj∂z

¶− 2ψi

µ∂u

∂rψj

¶¸)r dr dz

K35ij = 2

ZΩeλ∂u

∂zψiψj r dr dz, K42

ij = −2ZΩeη2ψi

∂ψj∂z

r dr dz

K44ij =

ZΩe

(ψiψj + λ

∙ψi

µu∂ψj∂r

+ w∂ψj∂z

¶− 2∂w

∂zψiψj

¸)r dr dz

K45ij = −2

ZΩeλ∂w

∂rψiψj r dr dz, K51

ij = −2ZΩeη1ψi

∂ψi∂z

r dr dz

K52ij = −2

ZΩeη1ψi

∂ψj∂r

r dr dz, K53ij = −2

ZΩeλ∂w

∂rψiψj r dr dz

K54ij = −2

ZΩeλ∂u

∂zψiψj r dr dz, K61

ij = −2ZΩeη1ψi

ψjrr dr dz

K55ij =

ZΩe2

(ψiψj + λ

µuψj∂r+ w

∂ψj∂z

¶−µ∂u

∂r+∂w

∂z

¶ψiψj

)r dr dz

K66ij =

ZΩe

(ψiψj + λ

∙ψi

µu∂ψj∂r

+ w∂ψj∂z

¶− 2u

rψiψj

¸)r dr dz

M11ij =

ZΩeρψiψj r dr dz, M22

ij =

ZΩeρψiψj r dr dz

M33ij =

ZΩeλψiψj r dr dz, M44

ij =

ZΩeλψiψj r dr dz

M55ij =

ZΩeλψiψj r dr dz, M66

ij =

ZΩeλψiψj r dr dz (10.4.40)

Bar over u and w indicate that they are to be calculated using values fromthe previous iteration.

For viscoelastic ßuids, speciÞcation of velocities is insufficient on accountof ßuid memory. If the boundary of the domain contains an entry region, thenfully developed ßow conditions may be assumed. All extra stress componentsmust be speciÞed as essential boundary conditions along the entry region.Failing to do so may lead to the propagation of errors throughout the ßowdomain when relaxation time λ becomes large.


Consider the steady ßow of a power-law ßuid in a uniform pipe (R = 1, L = 6,w0 = 10, n = 0.2, µ0 = 1.0, and γp = 108). The geometry and boundaryconditions of the computational domain are shown in Figure 10.4.1. Threedifferent meshes shown in Figure 10.4.2 are used. The Þnite element resultsobtained using the velocity Þnite element model in Eq. (10.4.22) along with

Mesh 2: 6×10 (uniform)

Mesh 1: 3×5 (uniform)

Mesh 3: 12×15 (non-uniform)

z

u = w = 0

u = 0

u = 0w = w0

u = 0

r

L

Rz

r


the analytical results for the axial velocities as a function of the radial distanceare presented in Table 10.4.1.

Figure 10.4.1 Geometry and boundary conditions for ßow through a pipe.

Figure 10.4.2 Three different meshes used for the pipe ßow.


Table 10.4.1 A comparison of the velocity Þeld w(r, 6) for the ßow of apower-law ßuid in a pipe (R = 1, L = 6, w0 = 10, n = 0.2,µ0 = 1.0, and γp = 10

8).

Velocity, w(r,6)

r Mesh 1 Mesh2 Mesh 3 Analytical*

0.00 11.894 12.833 13.402 13.3330.05 − − − 13.3330.10 − 12.832 13.401 13.3330.15 − − − 13.3330.20 11.865 12.826 13.395 13.3320.25 − − − 13.3300.30 − 12.795 13.360 13.3240.35 − − − 13.3090.40 11.721 12.686 13.243 13.2790.45 − − − 13.2230.50 − 12.401 12.940 13.1250.55 − − 12.679 12.9640.60 10.852 11.783 12.305 12.7110.65 − − 11.787 12.3280.70 − 10.610 11.086 11.7650.75 − − 10.156 10.9600.80 7.916 8.563 8.946 9.8380.85 − − 7.396 8.3050.90 − 5.133 5.451 6.2480.95 − − 2.993 3.5321.00 0.000 0.000 0.000 0.000

* w(r,6) = w0¡3n+1n+1

¢h1−¡rR

¢1+ 1n

i.

Next, the same problem (with R = 1 and L = 5) is studied with themixed Þnite element model in Eq. (10.4.38). Both power-law (n = 0.25,η0 = 10

4, η2 = 0) and WhiteMetzner (n = 0.25, η0 = 104, η2 = 0, a = 0.435,

b = −0.453, c = 0.1388) ßuids are analyzed. A uniform mesh of 10×6 bilinearelements is used. In addition to the velocity boundary conditions shown inFigure 10.4.1, the stresses are speciÞed to be zero at the entrance. The penaltyparameter is taken to be γp = 10

8η0.

Figures 10.4.3 and 10.4.4 contain plots of the axial velocity proÞles atz = 2.0 and z = 5.0, respectively. We note that the velocity proÞles obtainedwith the velocity model (VM) in Eq. (10.4.22) and mixed model (MM) in Eq.(10.4.38) are essentially the same when stress boundary conditions are notimposed. The stress boundary conditions (which can be imposed point-wiseonly in the mixed model) do have an effect: speciÞcation of the stresses at theinlet increases the centerline velocity for both power-law ßuids as well as


Figure 10.4.3 Axial velocity proÞles for the ßow of power-law ßuid througha pipe.

Figure 10.4.4 Axial velocity proÞles for the ßow of WhiteMetzner ßuidthrough a pipe.

0.0

0.2

0.4

0.6

0.8

1.0

r

0.0 0.5 1.0

w(r, 0)

w(r, 2) w(r, 5)

0

0.2

0.4

0.6

0.8

1.0

0 0.5 1.0 1.5 2.0

r r

0.0 0.5 1.0 1.5 2.0

Axial velocity, w(r,z)

0.0

0.2

0.4

0.6

0.8

1.0

Nonlinear (MM) without SBCLinear (MM) with SBCNonlinear (MM) with SBC

°SBC = Stress boundary conditions

Linear (VM, MM) without SBCNonlinear (VM)

w(r, 2) w(r, 5)

r r

1.0

0.8

0.6

0.4

0.2

0.00.0 0.5 1.0 1.5 2.0 2.5

1.0

0.8

0.6

0.4

0.2

0.00.0 0.5 1.0 1.5 2.0 2.5

Axial velocity, w(r,z) Axial velocity, w(r,z)

Nonlinear with SBC°SBC = Stress Boundary ConditionsLinear without SBCLinear with SBC

0.0

0.2

0.4

0.6

0.8

1.0r

0 5 10 15 20 25Axial velocity, w(r,6)

γp= 100

γp= 101 γp= 102

γp= 102 γp= 1012

γp= 103

γp= 104

γp= 108 − 1012

NewtonianNon-Newtonian


WhiteMetzner ßuids. If the stress boundary conditions are not imposed, themixed model for the WhiteMetzner ßuid does not yield converged solution.Figure 10.4.5 shows the effect of the penalty parameter on the velocity Þeld.The results are self-explanatory.

Table 10.4.2 Effect of the convective terms on the velocity Þeld for theßow of a power-law ßuid in a pipe (R = 1, L = 5, w0 = 10,n = 0.2, and 10× 6Q4 mesh).

Velocity, w(r,5)

Non-Newtonian

r Newtonian Re = 0 Re = 102 Re = 103

0.0 1.8358 1.3357 1.1312 1.02670.1 1.8001 1.3355 1.1302 1.02650.2 1.7398 1.3338 1.1289 1.02660.3 1.6461 1.3271 1.1270 1.02680.4 1.5186 1.3081 1.1241 1.02710.5 1.3578 1.2657 1.1197 1.02750.6 1.1651 1.1849 1.1097 1.02740.7 0.9397 1.0468 1.0799 1.02650.8 0.6712 0.8273 0.9809 1.02090.9 0.3079 0.4795 0.6709 0.89671.0 0.0000 0.0000 0.0000 0.0000

* Re = ρw0Rη0

.

Figure 10.4.5 The effect of the penalty parameter on the ßow of a power-law ßuid.

x

1.0

0.8

0.6

0.4

0.2

0.00 2 4 6 8 10 12 14 16 18 20

Axial velocity, v(x,6)

°

Newtonian fluid (Re = 0)Non-Newtonian (Re = 0)Non-Newtonian (Re = 12.5)Non-Newtonian (Re = 25.0)


The effect of the convective terms on the velocity Þeld is also investigatedfor the power-law ßuid (n = 0.25, η0 = 10

4, η2 = 0, R = 1, L = 5, and w0 = 1)using the velocity model in Eq. (10.4.22). The results are presented in Table10.4.2. The effect is to ßatten the velocity proÞle from a parabolic one. Figure10.4.6 shows the axial velocity proÞle for ßow through a plane channel (a = 6,b = 1, n = 0.2, η0 = η1 = 1). The results are in good agreement with those in[9].

Figure 10.4.6 Axial velocity proÞles for ßow of power-law ßuid through aplane channel.

10.5 Coupled Fluid Flow and Heat Transfer

10.5.1 Finite Element Models

There exist many engineering systems where the ßuid ßow is affected by theheat transfer to or from the system and vice versa. Convective cooling of aheated body (like in an internal combustion engine) is a generic example of thecoupling. In such cases, we solve the governing equations of ßuid ßow (i.e. theNavierStokes equations and continuity equation) as well as the heat transfer(i.e. the energy equation), as discussed in Section 7.3. The energy equation isgiven by [see Eq. (7.3.3)]

ρC

∙µu∂T

∂x+ v

∂T

∂y

¶+∂T

∂t

¸=∂

∂x

µk11∂T

∂x

¶+∂

∂y

µk22∂T

∂y

¶+Φ+ q (10.5.1)

where C is speciÞc heat, Φ is the dissipation energy

Φ = σijεij (10.5.2)

and q is internal heat generation.


The Þnite element model of the energy equation is given by

[Ce] úT e+ [He]T e = Qe (10.5.3)

where

Heij = ρC

ZΩeψ(2)i

⎛⎝u∂ψ(2)j∂x

+ v∂ψ

(2)j

∂y

⎞⎠ dxdy+

ZΩe

⎛⎝k11∂ψ(2)i∂x

∂ψ(2)j

∂x+ k22

∂ψ(2)i

∂y

∂ψ(2)j

∂y

⎞⎠ dxdyCeij = ρC

ZΩeψ(2)i ψ

(2)j dxdy

Qei =

ZΩeψ(2)i (q +Φ)dxdy +

IΓeqnψ

(2)i ds (10.5.4)

u and v being the velocity components (that couple to the ßuid ßow

problem), and ψ(2)i the Lagrange interpolation functions used to interpolate

the temperature Þeld.

Equation (10.5.3) must be solved along with the ßow equations

[Me] ú∆e+ [Ke]∆e = F e (10.5.5)

developed earlier (see Chapters 7 and 8). Both equations, Eqs. (10.5.3)and (10.5.5), must be solved iteratively, using the latest velocity Þeld andtemperature Þeld to compute the coefficient matrices. It should be noted thatßuid ßow equations are coupled to the heat transfer equation through thebody force terms fx and fy in the NavierStokes equations. For buoyancy-driven ßows, the body force in the direction of the gravity is a function of thetemperature, that is, fy = −ρgβ(T − T0) if the y-axis is taken vertically up.Thus, the momentum equations are fully coupled with the energy equation.

The following strategy is found suitable for the convective heat transferproblems. Solve the energy equation (10.5.3) with an assumed velocity Þeld(say, zero). Then use the assumed velocity and temperature Þelds in (10.5.5)and solve for the new velocity Þeld. The initial guess of the velocity Þeld canbe either the linear (Stokes) solution (or the solution of a problem at lowerRayleigh number when solving for high Rayleigh number ßows). The Rayleighnumber Ra and Prandtl number Pr (charactristic numbers used in convectiveheat transfer) are deÞned by

Ra =β g L3∆T

κ ν, ν =

µ

ρ, κ =

k

ρcp, Pr =

ν

κ

T = TH

T = TC

L

L Insulated

u = v = 0u = v = 0

u = v = 0

u = v = 0

(a) (b)

x

y


where g is acceleration due to gravity, β is thermal expansion coefficient, Lthe characteristic dimension of the ßow region, ∆T the temperature differencebetween hot and cold walls, ρ the density, µ the viscosity, k the conductivity,and cp is the speciÞc heat at constant pressure.


Here we include couple of sample problems, taken from Reddy and Gartling[15], to illustrate the ideas presented above.

Heated cavity

Consider a closed square cavity Þlled with a viscous incompressible ßuid. Thetop and bottom faces of the cavity are assumed to be insulated while thevertical faces subjected to different temperatures, as shown in Figure 10.5.1(a).A typical 16 × 16 mesh of eight-node quadratic elements is shown in Figure10.5.1(b).

This problem was solved using the earliest versions of NACHOS code [24],which is based on velocitypressure (mixed) formulation and used the directiteration (Picard) scheme. More recent solutions have been obtained usingNewtons method for the combined equation set equivalent to (10.5.3) and(10.5.5). The streamline and isotherm plots are shown in Figures 10.5.2 and10.5.3 for Rayleigh numbers of Ra = 104 and 106, respectively (Pr = ν/κ =0.71). For the lower Rayleigh number, the ßow is relatively weak and thethermal Þeld is only slightly perturbed from a conduction solution. At thehigher Rayleigh number, the ßow Þeld develops a considerable structure whilethe thermal Þeld becomes vertically stratiÞed in the core of the cavity withhigh heat ßux regions along the vertical boundaries [24].

Figure 10.5.1 (a) Geometry of a heated square cavity. (b) A typical Þniteelement mesh.

Streamlines Isotherms



Solar receiver

Figure 10.5.4 shows a cross section of an annular solar receiver tube surroundedby an eccentrically located glass envelope. The inner tube carries a heattransfer ßuid that is heated by a ßux that varies with position around thetube. The incident ßux is due to solar energy being concentrated on the tubeby a parabolic trough collector. The glass envelope provides a shield to reducethe forced convection (wind) heat loss from the collector tube [15, 24].

Figure 10.5.2 Streamlines and isotherms for natural convection in a squarecavity Þlled with viscous ßuid (Ra = 104, Pr = 0.71).

Figure 10.5.3 Streamlines and isotherms for natural convection in a squarecavity Þlled with viscous ßuid (Ra = 106, Pr = 0.71).

+

T = constantu = v = 0

Glass envelop

T = constantu = v = 0

Receiver tube

IsothermsStreamlines(×10−4cm2/sec)

(a)

48101

153 470417365

312259206

528

583º K

556º K472500º K

444

417

389

361

333

Streamlines(×10−4cm2/sec)

Isotherms

(b)

593º K

−402−357−312−266−221−176−132−86−41

636º K679º K

549506

463

420

377


Figures 10.5.510.5.7 contain streamline and isotherm plots for an air-Þlled annulus for various temperature and geometric conÞgurations. The ßowpattern and heat ßux distribution are quite sensitive to variations in theseparameters even though the Rayleigh number is the same for all cases [15, 24].

Figure 10.5.4 Mesh and boundary conditions for the annular solar receiver.

Figure 10.5.5 Plots of streamlines and isotherms for the solar receiver. (a)uniform wall temperature, Ra = 1.2 × 104. (b) asymmetricwall temperature, hot on top (Ra = 1.2× 104).


−2.33−1.25

−0.38−5.49×10−3

4.52 3.542.56

1.580.60

677.8

755.6833.2

911.1908.9

1066.7

1144.41222.2

Streamlines(×10−3 in2/sec)

Isotherms

1010.5936.3

862.1

787.9

713.7 639.5565.4

491.2417

11.9110.589.448.216.98

5.75

0.82

4.522.05

3.29


Figure 10.5.6 Plots of streamlines and isotherms for the solar receiver;uniform wall temperature, hot on bottom (Ra = 1.2× 104).

Figure 10.5.7 Plots of streamlines and isotherms for the solar receiver;uniform wall temperature, eccentric geometry (Ra = 1.2 ×104).

Additional details on the formulation as well as applications of coupledheat transfer and ßuid ßow can be found in [1724] (in particular, see [23,24]and Chapter 5 of the book by Reddy and Gartling [15] and reference therein).


References

1. Hill, R., The Mathematical Theory of Plasticity, Oxford University Press, Oxford, U.K.,1950.

2. Johnson, W. and Mellor, P. B., Engineering Plasticity, John Wiley, New York (1983).

3. Owen, D. R. J. and Hinton, E., Finite Elements in Plasticity: Theory and Practice,Pineridge Press, UK (1980) (see pp. 26-29 for one-dimensional bar problems; pp.129-148 for one-dimensional Timoshenko beams; pp. 215-281 for two-dimensionalproblems).

4. Zienkiewicz, O. C. and Taylor, R. L., The Finite Element Method, 4th edn, Vol. 2:Solid and Fluid Mechanics, Dynamics and Non-Linearity, McGrawHill, London, UK(1989) (see Chapter 7).

5. Reddy, J. N. and Rasmussen, M. L., Advanced Engineering Analysis, John Wiley, NewYork (1982); reprinted by Krieger Publishing, Melbourne, FL (1991).

6. Phan Thien, N. and Tanner, R. I., A New Constitutive Equation Derived FromNetwork Theory, Journal of Non-Newtonian Fluid Mechanics, 2, 353365 (1977).

7. White, J. L. and Metzner, A. B., Development of Constitutive Equations for PolymericMelts and Solutions, Journal of Applied Polymer Science, 7, 18671889 (1963).

8. Crochet, M. J. and Walters, K., Numerical Methods in Non-Newtonian FluidMechanics, Annual Review of Fluid Mechanics, 15, 241260 (1983).

9. Crochet, M. J., Davies, A. R., andWalters, K., Numerical Simulation of Non-NewtonianFlow, Elsevier, Amsterdam, The Netherlands (1984).

10. Gartling, D. K. and Phan Thien, N., A Numerical Simulation of a Plastic Fluid in aParallel-Plate Plastometer, Journal of Non-Newtonian Fluid Mechanics, 14, 347360(1984).

11. Reddy, J. N. and Padhye, V. A., A Penalty Finite Element Model for AxisymmetricFlows of Non-Newtonian Fluids, Numerical Methods for Partial Differential Equations,4, 3356 (1988).

12. Iga, M. and Reddy, J. N., Penalty Finite Element Analysis of Free Surface Flows ofPower-Law Fluids, International Journal of Non-Linear Mechanics, 24(5), 383399(1989).

13. Reddy, M. P. and Reddy, J. N., Finite Element Analysis of Flows of Non-Newtonian Fluids in Three-Dimensional Enclosures, International Journal of Non-Linear Mechanics, 27(1), 926 (1992).

14. Reddy, M. P. and Reddy, J. N., Numerical Simulation of Forming Processes Using aCoupled Fluid Flow and Heat Transfer Model, International Journal for NumericalMethods in Engineering, 35, 807833 (1992).

15. Reddy, J. N. and Gartling, D. K., The Finite Element Method in Heat Transfer andFluid Dynamics, Second Edition, CRC Press, Boca Raton, FL (2001).

16. Coleman, C. J., A Finite Element Routine for Analyzing Non-Newtonian Fluids, Part1, Journal of Non-Newtonian Fluid Mechanics, 7, 289301 (1980).

17. Gartling, D. K., Convective Heat Transfer Analysis by the Finite Element Method,Computer Methods in Applied Mechanics and Engineering, 12, 365382 (1977).

18. Marshall, R. S., Heinrich, J. C., and Zienkiewicz, O. C., Natural Convection in aSquare Enclosure by a Finite Element, Penalty Function Method, Using Primitive FluidVariables, Numerical Heat Transfer, 1, 315330 (1978).

19. Reddy, J. N. and Satake, A., A Comparison of Various Finite Element Models ofNatural Convection in Enclosures, Journal of Heat Transfer, 102, 659666 (1980).


20. Reddy, J. N., Penalty-Finite-Element Analysis of 3-D Navier-Stokes Equations,Computer Methods in Applied Mechanics and Engineering, 35, 87106 (1982).

21. de Vahl Davis, G. and Jones, I. P., Natural Convection in a Square Cavity: AComparison Exercise, International Journal for Numerical Methods in Fluids, 3, 227248 (1983).

22. Dhaubhadel, M., Telionis, D., and Reddy, J. N., Finite Element Analysis of FluidFlow and Heat Transfer for Staggered Bundle of Cylinders in Cross Flow, InternationalJournal for Numerical Methods in Fluids, 7, 13251342 (1987).

23. Pelletier, D. H., Reddy, J. N., and Schetz, J. A., Some Recent Developments andTrends in Finite Element Computational Natural Convection, in Annual Review ofNumerical Fluid Mechanics and Heat Transfer, Vol. 2, C. L. Tien and T. C. Chawla(eds.), Hemisphere, New York, 3985 (1989).

24. Gartling, D. K., Convective Heat Transfer Analysis by the Finite Element Method,Computer Methods in Applied Mechanics and Engineering, 12, 365382 (1977).

AppendixA1

Solution Procedures forLinear Algebraic Equations

A1.1 Introduction

All Þnite element equations, after assembly and imposition of boundary (andinitial) conditions, can be expressed as a set of linear algebraic equations thatmust be solved. In general, the Þnite element equations are of the form

[A]X = B (A1.1.1)

where [A] is the coeffienct matrix resulting from the assembly of elementmatrices, X column of unknowns (typically nodal values), and B isthe known column vector. In the Þnite element method, [A] is a bandedsparse matrix that may be either symmetric or unsymmetric depending on thecharacteristics of the governing differential equation(s) describing the physicalproblem, and possibly the nonlinear solution method used (see Appendix A2).A banded matrix is one in which all elements beyond a diagonal parallel

to the main diagonal are zero. The maximum number of non-zero diagonalsabove or below the main diagonal plus 1 (to account for the main diagonal) iscalled half bandwidth (NHBW) of the matrix [see Figure A1.1.1(a)]. When thematrix is symmetric, it is sufficient to store elements in upper or lower halfbandwidth of the matrix [A] [see Figure A1.1.1(b)] and write solvers to takenote of the fact aij = aji for all rows i and columns j of the matrix. Whenthe matrix is not symmetric, one must store all elements in the full bandwidth(2*NHBW-1) of the matrix. Here we include the Fortran subroutines of asymmetric banded equation solver (SYMSOLVR) and unsymmetric bandedequation solver (UNSYMSLV).A linear set of algebraic equations can be solved by either a direct or

iterative method. Direct methods, like the Gauss elimination method, providethe solution after a Þxed number of steps and are less sensitive to the

a11 a12 . . . a1k 0 0 0 . . . 0 a21 a22 . . . a2k a2k+1 0 0 . . . 0 a31 a32 . . . a3k a3k+1 a3k+2 0 0 . . .. . . . . . . . .

0 . . . 0 ank . . . ank . . . ann

[A] =

NHB1

NHB2

NHBW= max NHB1, NHB2

A parallel diagonal

Main diagonal

All zeros

All zeros

(a)

NEQ = n

a11 a12 . . . a1k 0 0 0 . . . 0 a21 a22 . . . a2k a2k+1 0 0 . . . 0 a31 a32 . . . a3k a3k+1 a3k+2 0 0 . . .. . . . . . . . .

0 . . . 0 ank . . . ank . . . ann

[A] =(symm)

(b)

[A] =(half banded

form)

a11 a12 . . . a1k

a22 a23 . . . a2k+1

a33 a34 . . . a3k+2

. . . . . . .

NHBW

ann 0 . . . 0

an-1, n-1 an-1, n...0

NHBW

NEQ = n

Main diagonal

Last diagonal beyond which all coefficients are zero


conditioning of the matrix (the condition number of a matrix is the ratio ofthe largest to smallest eigenvalue associated with the matrix). Apart from theround-off errors introduced in the computer, direct methods yield the exactsolution to the equations. However, the main deÞciency of the direct solversis that they require the coefficient matrix be stored in an ordered format tocarry out the elementary operations (i.e. multiplication, division, etc.) onthe coefficients of the augmented matrix [A|B] and obtain the solution. Inrecent years, the direct solvers have been reÞned to reduce this deÞciencythrough innovative data management techniques (e.g. frontal solvers, skylinesolvers, and others; see Carey and Oden [1]). These improvements enableusers to solve moderately large systems of equations efficiently. However, theyhave been found to be unsuitable for solving very large systems of equations(especially in three-dimensional problems) because they demand out-of-corestorage of the equations and hence require large data transfers. In addition,direct methods are difficult to organize for efficient use on multiprocessor,parallel computers and are therefore seeing reduced utilization.

Figure A1.1.1 Banded symmetric and unsymmetric matrices.

SOLUTION PROCEDURES FOR LINEAR ALGEBRAIC EQUATIONS 427

The limitations on CPU time and storage requirements preclude the useof direct solvers for complex problems with more than a quarter millionequations, and iterative methods are found to be more efficient in that theyrequire less storage and CPU time. Iterative methods are approximate andonly converged solutions will be close to the true solution. In iterativemethods, the global matrix formation may be avoided. The major operationin iterative methods is the matrix-vector multiplication as compared to thematrix reduction (by elementary operations) in direct methods. A signiÞcantadvantage of iterative methods is that a given set of equations can be dividedinto as many subsets of equations as there are processors and calculations canbe performed in parallel on the array of processors. However, convergencecharacteristics of iterative methods depend on the condition number of thesystem of linear equations, and a suitable preconditioner is a must to achieveconvergence.Following this introduction, a brief review of some methods for the solution

of linear algebraic equations is presented. For additional details, the readermay consult [24].

A1.2 Direct Methods

A1.2.1 Preliminary Comments

Direct methods are those in which simultaneous linear algebraic equationsare solved exactly (within the computational round-off error) by successiveelimination of variables and back substitution. The Gauss elimination methodis a Þxed-step technique [28], and frontal [5] and skyline [6] solution methodsare examples of direct solution methods that use the Gauss eliminationtechnique efficiently. The direct methods are the most commonly usedtechniques when the number of equations involved is not too large. Thenumber of elementary operations for Gauss elimination is of the ordern3/3 + O(n2), where n denotes the number of unknowns.The frontal solution procedure [5] is faster than most direct solvers; it

requires less core space as long as active variables can be kept in the core, andit allows for partial pivoting. An additional advantage is that no speciÞc nodenumbering scheme is needed, though a judicious element numbering helps tominimize the front width.Due to the fact that the approximation functions are deÞned only within an

element, the coefficient matrix in the Þnite element method is banded; that is,aij = 0 for j > i+ nb, where nb is the half bandwidth of the matrix [A]. Thisgreatly reduces the number of operations in solving the equations, if we makenote of the fact that elements outside the bandwidth are zero. Of course, thebandwidth size depends on the global node numbering. The skyline techniqueis one in which bandedness of the Þnite element equations is exploited bystoring the row number mj of the Þrst nonzero element in column j. The


variables mi, i = 1, 2, · · · , n, deÞne the skyline of the matrix. For additionaldetails see Bathe [6].Here, we include the Fortran subroutines of a symmetric banded equation

solver (SYMSOLVR) and unsymmetric banded equation solver (UNSYMSLV)that are used in most of the calculations presented in this book [7].

A1.2.2 Symmetric Solver

In the case of symmetric solver the global coefficient matrix [A] ≡ [GLK]is stored in symmetric banded form (NEQ × NHBW ), where NEQ is thenumber of global equations and NHBW is the half bandwidth. Fortranstatements for the assembly of element coefficient matrices [ELK] and elementsource vector ELF for the symmetric case are included in Box A1.2.1.Note that the global source vector is stored in array GLF. After Gausselimination is completed in SYMSOLVR (see Box A1.2.2), the solution isreturned in array GF.

Box A1.2.1 Fortran statements for the calculation of the half bandwidthand assembly of element coefficient matrices.

DO 140 N=1,NEM DO 110 I=1,NPE NI=NOD(N,I) ELU(I)=GAMA*GP2(NI)+(1.0-GAMA)*GP1(NI)

110 ELX(I)=GLX(NI) CALL ELEKMF(IEL,NPE,NONLIN,F0) DO 130 I=1,NPE NR=(NOD(N,I)-1)*NDF DO 130 II=1,NDF

NR=NR+1 L=(I-1)*NDF+II GLF(NR)=GLF(NR)+ELF(L) DO 120 J=1,NPE

NCL=(NOD(N,J)-1)*NDF DO 120 JJ=1,NDF M=(J-1)*NDF+JJ NC=NCL-NR+JJ+1 IF(NC.GT.0)THEN

GLK(NR,NC)=GLK(NR,NC)+ELK(L,M) ENDIF

120 CONTINUE 130 CONTINUE 140 CONTINUE

GAMA = Acceleration parameter, ρ [see Eq. (3.4.5)]

GP2 = Solution from the(r − 2) iterationGP1 = Solution from (r − 1) iteration


Box A1.2.2 Subroutine for solution of banded symmetric equations.

SUBROUTINE SYMSOLVR(NRM,NCM,NEQNS,NBW,BAND,RHS,IRES) C C The subroutine solves a banded, symmetric, system of algebraic equations [BAND]U = RHSC using the Gauss elimination method: The coefficient matrix is input as BAND(NEQNS,NBW) andC the column vector is input as RHS(NEQNS), where NEQNS is the actual number of equations andC NBW is the half band width. The true dimensions of the matrix [BAND] in the calling program, areC NRM by NCM. When IRES is greater than zero, the right hand elimination is skipped. C

IMPLICIT REAL*8(A-H,O-Z) DIMENSION BAND(NRM,NCM),RHS(NRM) MEQNS=NEQNS-1 IF(IRES.LE.0) THEN

DO 30 NPIV=1,MEQNS NPIVOT=NPIV+1 LSTSUB=NPIV+NBW-1 IF(LSTSUB.GT.NEQNS) THEN

LSTSUB=NEQNS ENDIF DO 20 NROW=NPIVOT,LSTSUB NCOL=NROW-NPIV+1 FACTOR=BAND(NPIV,NCOL)/BAND(NPIV,1) DO 10 NCOL=NROW,LSTSUB ICOL=NCOL-NROW+1 JCOL=NCOL-NPIV+1

10 BAND(NROW,ICOL)=BAND(NROW,ICOL)-FACTOR*BAND(NPIV,JCOL) 20 RHS(NROW)=RHS(NROW)-FACTOR*RHS(NPIV) 30 CONTINUE

ELSE 40 DO 60 NPIV=1,MEQNS

NPIVOT=NPIV+1 LSTSUB=NPIV+NBW-1 IF(LSTSUB.GT.NEQNS) THEN

LSTSUB=NEQNS ENDIF DO 50 NROW=NPIVOT,LSTSUB NCOL=NROW-NPIV+1 FACTOR=BAND(NPIV,NCOL)/BAND(NPIV,1)

50 RHS(NROW)=RHS(NROW)-FACTOR*RHS(NPIV) 60 CONTINUE

ENDIF CC Back substitutionC

DO 90 IJK=2,NEQNS NPIV=NEQNS-IJK+2 RHS(NPIV)=RHS(NPIV)/BAND(NPIV,1) LSTSUB=NPIV-NBW+1 IF(LSTSUB.LT.1) THEN

LSTSUB=1 ENDIF

NPIVOT=NPIV-1 DO 80 JKI=LSTSUB,NPIVOT NROW=NPIVOT-JKI+LSTSUB NCOL=NPIV-NROW+1 FACTOR=BAND(NROW,NCOL)

80 RHS(NROW)=RHS(NROW)-FACTOR*RHS(NPIV) 90 CONTINUE

RHS(1)=RHS(1)/BAND(1,1) RETURN END

Continued in the inset

DO 140 N=1,NEM DO 110 I=1,NPE NI=NOD(N,I) ELU(I)=GAMA*GP2(NI)+(1.0-GAMA)*GP1(NI)

110 ELX(I)=GLX(NI) CALL ELEKMF(IEL,NPE,NONLIN,F0) DO 130 I=1,NPE NR=(NOD(N,I)-1)*NDF DO 130 II=1,NDF

NR=NR+1 L=(I-1)*NDF+II GLK(NR,NBW)=GLK(NR,NBW)+ELF(L) DO 120 J=1,NPE NCL=(NOD(N,J)-1)*NDF DO 120 JJ=1,NDF

M=(J-1)*NDF+JJ NC=NCL-NR+JJ+NHBW IF(NC.GT.0)THEN

GLK(NR,NC)=GLK(NR,NC)+ELK(L,M)ENDIF

120 CONTINUE 130 CONTINUE 140 CONTINUE


A1.2.3 Unsymmetric Solver

In the case of unsymmetric solver the matrix [GLK] is stored in full bandwidthform (NEQ × NBW ), where NEQ is the number of global equations andNBW = 2∗NHBW is twice the half bandwidth. Since there is an additionalcolumn in NBW , we use it to store the global source vector; that is,GLF (I) → GLK(I,NBW ). Fortran statements for the assembly of elementmatrices for the unsymmetric case are included in Box A1.2.3. After Gausselimination is completed in UNSYMSLV (see Box A1.2.4), the solution isreturned in the last column of the coefficient matrix GLK(I,NBW ).

Box A1.2.3 Fortran statements for assembly of the element matrices andsource vector into a banded unsymmetric global matrix.

A1.3 Iterative Methods

A1.3.1 General Comments

Among the various iterative methods that are available in the literature, theConjugate Gradient (CG) method [9] is most widely used because it is a Þnitestep method (i.e., apart from round-off errors, the solution is achieved ina Þxed number of iterations) and it can be used to determine the inverse.However, the number of iterations required depends on the condition numberof the coefficient matrix. The convergence of the conjugate gradient method,and iterative methods in general, can be improved by preconditioning and/orscaling the equations [1012].


Box A1.2.4 Subroutine for solution of banded unsymmetric equations.

SUBROUTINE UNSYMSLV(A,NRMAX,NCMAX,N,ITERM) C ________________________________________________________________C C Solver for BANDED UNSYMMETRIC system of algebraic equations C ________________________________________________________________ C

IMPLICIT REAL*8 (A-H,O-Z) DIMENSION A(NRMAX,NCMAX) CERO=1.0D-15 PARE=CERO**2 NBND=2*ITERM NBM=NBND-1

C C Begin elimination of the lower leftC

DO 80 I=1,N IF (DABS(A(I,ITERM)).LT.CERO) GO TO 10 GO TO 20

10 IF (DABS(A(I,ITERM)).LT.PARE) GO TO 110 20 JLAST=MIN0(I+ITERM-1,N)

L=ITERM+1 DO 40 J=I,JLAST L=L-1 IF (DABS(A(J,L)).LT.PARE) GO TO 40 B=A(J,L) DO 30 K=L,NBND

30 A(J,K)=A(J,K)/B IF (I.EQ.N) GO TO 90

40 CONTINUE L=0 JFIRST=I+1 IF (JLAST.LE.I) GO TO 80 DO 70 J=JFIRST,JLAST L=L+1 IF (DABS(A(J,ITERM-L)).LT.PARE) GO TO 70 DO 50 K=ITERM,NBM

50 A(J,K-L)=A(J-L,K)-A(J,K-L) A(J,NBND)=A(J-L,NBND)-A(J,NBND) IF (I.GE.N-ITERM+1) GO TO 70 DO 60 K=1,L

60 A(J,NBND-K)=-A(J,NBND-K) 70 CONTINUE 80 CONTINUE 90 L=ITERM-1

DO 100 I=2,N DO 100 J=1,L IF (N+1-I+J.GT.N) GO TO 100 A(N+1-I,NBND)=A(N+1-I,NBND)-A(N+1-I+J,NBND)*A(N+1-I,ITERM+J)

100 CONTINUE RETURN

110 WRITE (6,140) I,A(I,ITERM) STOP

140 FORMAT (/,2X,'Computation stopped in UNSYMSLV because zero appears * on the main diagonal *** Eqn no. and value:',I5,E12.4) END


The limitations on storage can be overcome by solving the equations atthe element level, that is, use the GaussSeidel iteration idea for the setof variables associated with the element. This approach avoids assembly ofelement matrices to form the global coefficient matrix. This idea of using theelement-by-element data structure of the coefficient matrix was Þrst pointedout by Fox and Stanton [13] and Fried [1416]. The phrase element-by-element refers to a particular data structure for Þnite element techniqueswherein information is stored and maintained at the element level rather thanassembled into a global data structure. In this method the matrixvectormultiplications are carried out at the element level and the assembly is carriedout on the resultant vector. This idea proves to be very attractive whensolving large problems, because the matrixvector multiplication can be donein parallel on a series of processors. Another advantage of this method is thatthe resultant savings in storage, compared to direct solvers, allows solution oflarge problems on small computers.For iterative solution methods, the advantages of the element-by-element

data structure over assembling the global coefficient matrix are

1. the need for formation and storage of a global matrix is eliminated, andtherefore the total storage and computational costs are low,

2. the amount of storage is independent of the node numbering and meshtopology and depends on the number and type of elements in the mesh,and

3. the element-by-element solution algorithms can be vectorized for efficientuse on supercomputers.

The major disadvantage of the element-by-element data structure is thelimited number of preconditioners that can be formulated from theunassembled matrices. This becomes of critical importance when the linearsystem is not well-conditioned as in the mixed method, incompressible ßowmodel. A review of the literature on element-by-element algorithms canbe found in [17], and the methods have been investigated by numerousinvestigators [1832].

A1.3.2 Solution Algorithms

In this section, we review three iterative solvers from [17] that are applicableto nonsymmetric, positive deÞnite equation systems that are typical ofisothermal ßow algorithms. The three iterative solution schemes used hereare the Biorthogonal Conjugate Gradient method (BCG) [11], the LanczosORTHORES [11], and the GMRES [31].The conjugate gradient method for solving a system of equations can

be interpreted as the search for the minimum of the energy E of thesystem. The energy of the system is a minimum when the residual vector


r = F − KU∗ vanishes. The algorithm for the biorthogonal conjugategradient method (also known as two-term form of the steepest descent method,Lanczos/ORTHOMIN) for unsymmetric systems of equations [11,17] is givenin Table A1.3.1, and the steps involved in the Lanczos ORTHORES solutionalgorithm [11,17] are given in Table A1.3.2.The third iterative solver uses the GMRES solution algorithm. For an

approximate solution of the form U0 + z, where U0 is the initial guessvector and z is a member of the Krylov space K of dimension k, theGMRES algorithm determines the vector z such that k F − K(U0 + z) kis minimized, where k · k denotes the L2−norm. The Krylov space is givenby K = span

nU0, KU0, K

2U0, · · · Kk−1U0

o. Therefore, when solving large

systems of equations, as the value of k increases, the amount of storage requiredalso increases. This drawback can be overcome by employing the GMRESalgorithm iteratively by using a smaller value for k and restarting the algorithmafter every k steps. The restart version of the GMRES algorithm [31,17] isexplained in Table A1.3.3.The presence of a penalty matrix in the global coefficient matrix of the

penalty model spoils the condition number. This results in slow convergencewhen using iterative solvers. However, the convergence of the iterativesolvers can be improved by preconditioning the system. In [17], the systemof equations is transformed using diagonal scaling matrix (Jacobi/diagonalpreconditioning). Accordingly,

KU∗ = F becomes KU = F (A1.3.1)

K =W−1/2 KW−1/2; U =W1/2U∗; F =W−1/2F (A1.3.2)

where Wii = K−1ii is a diagonal matrix. During the matrix multiplication,

the element-by-element data structure is exploited and the multiplicationsare carried out at the element level and the residuals are then assembled toform the global vector. The diagonal terms are always positive because of theviscous and penalty terms.

References

1. Carey, G. F. and Oden, J. T., Finite Elements: Computational Aspects, Prentice Hall,Englewood Cliffs, NJ (1984).

2. Atkinson, K. E., An Introduction to Numerical Analysis, John Wiley, New York (1978).

3. Carnahan, B., Luther, H. A., and Wilkes, J. O., Applied Numerical Methods, JohnWiley, New York (1969).

4. Fadeev, D. K. and Fadeeva, V. N., Computational Methods of Linear Algebra, Freeman,San Francisco, CA (1963).

5. Hood, P., Frontal Solution Program for Unsymmetric Matrices, International Journal

for Numerical Methods in Engineering, 10, 379399 (1976); also see 10, 1055 (1976) for

a correction.


Table A1.3.1 Steps involved in using biorthogonal conjugate gradientmethod (Lanczos/ORTHOMIN solver).

Repeat the following steps for each nonlinear iteration:

I. Initial Calculations

(1) Form the element stiffness matrix Ke and force vectorFe.

(2) Apply essential and/or natural boundary conditions, and modify Ke and Fe.

(3) Store the element matrices in A (whose dimensions are nem, neleq, neleq)*.

(4) Store the inverse of the diagonal terms of the global system in W (Wii =Pneme=1

K−1ii ).

(5) Assemble the global force vector.

II. Preconditioning

Form the preconditioned system of equations

KU = F; K =W−1/2 KW−1/2 , U =W1/2U∗ , F =W−1/2F

III. Lanczos ORTHOMIN Algorithm

(1) For known initial solution vector U0, compute:

r0 = F − KU0 , P0 = r0 , r0 = P0 = r0 ,

α0 = 0 , λ0 =

¡r0,r0

¢¡KP0 ,r0

¢ , U1 = U0 + λ0P0

(2) For each ORTHOMIN iteration m = 1,2,3, . . ., compute [(a, b) =Paibi]:

λm =(rm,rm)¡KPm,rm

¢ , αm =(rm,rm)

(rm−1,rm−1),

Pm = rm + αmPm−1, Pm = r + αmPm−1 ,

rm+1 = rm − λmKPm , rm+1 = rm − λmKT Pm−1 ,

Um+1 = Um + λmPm

(3) Convergence criterion kUm+1k/kr0k ≤ 10−6

(4) If convergence criterion is satisÞed U∗ =W−1/2Um+1

* nem = number of elements in the Þnite element mesh, neleq = number of element equations.


Table A1.3.2 Steps involved in using Lanczos/ORTHORES solver.



(1) Form the element stiffness matrix Ke and force vector Fe.

(2) Apply the boundary conditions, and modify Ke and Fe.

(3) Store the element matrices in A (whose dimensions are nem, neleq, neleq).

(4) Store inverse of the diagonal terms of the global system inW

Wii =

nemXe=1

K−1ii


II. Preconditioning:

Form the preconditioned system of equations

KU = F , K =W−1/2 KW−1/2 , U =W1/2 U , F =W−1/2F

III. Lanczos ORTHORES Algorithm

(1) For known initial solution vector U0, compute:

r0 = F − KU0

(2) For each ORTHORES iteration m = 0,1,2, . . ., compute (r = r0,λ0 = 0):

λm+1 =(rm,rm)¡Krm,rm

¢βm+1 =

(1 ; if m = 0h

1− λm+1

λm(rm,rm)

(rm−1,rm−1)1βm

i−1; if m ≥ 1

rm+1 = βm+1¡rm − λm+1Krm

¢+¡1− βm+1

¢rm−1

rm+1 = βm+1¡rm − λm+1Krm

¢+¡1− βm+1

¢rm−1

Um+1 = βm+1¡Um + λm+1rm

¢+¡1− βm+1

¢Um−1

(3) Check convergencekUm+1k/kr0k ≤ 10−6

(4) If convergence criterion is satisÞed, set

U∗ =W−1/2Um+1


Table A1.3.3 Steps involved in using the GMRES solver.



(1) Form the element stiffness matrix Ke and force vector Fe.

(2) Apply the boundary conditions, and modify Ke and Fe.

(3) Store the element matrices in A (whose dimensions are nem, neleq, neleq).

(4) Store inverse of the diagonal terms of the global system inW: Wii =Pnem

e=1K−1ii


II. Preconditioning: Form the preconditioned system of equations

KU = F , K =W−1/2 KW−1/2 , U =W1/2 U , F =W−1/2F

III. GMRES Algorithm

(1) Start Choose U0 and compute

r0 = F− KU0, andv1 = U0/k U0 k

(2) Iterate: For j = 1,2..., k do:

hi,j = (Kvj , vi), i = 1, 2, ..., j,

vj+1 = Kvj −Pj

i=1 hi,jvi,

hj+1,j =k vj+1 k, andvj+1 = vj+1/hj+1,j

(3) Form the approximate solution

Uk = U0 +Vy, where y minimizes

k e− Hy k,y ∈Rk.

(4) Restart

Compute rk = F− KUk;

check convergence; if satisÞed stop; otherwise, compute

U0 := Uk, v1 := Uk/|Uk|, and go to step 2.

(5) Convergence criterion kUm+1k/kr0k ≤ 10−6

(6) If convergence criterion is satisÞed U∗ =W−1/2U

where V is a N × k matrix whose columns are l2 orthonormal basis vectorsv1,v2, . . . ,vk, e = k U0 k, 0, . . . ,0, and H is the upper k × k Hessenberg matrixwhose entries are the scalars hi,j . When using the restart version of the GMRESalgorithm, the total number of iteration m can be computed from the number ofrestarts and the dimension of k.


6. Bathe, K. J., Finite Element Procedures, Prentice Hall, Englewood Cliffs, New Jersey(1998).


8. Reddy, J. N. and Gartling, D. K., The Finite Element Method in Heat Transfer andFluid Dynamics, CRC Press, Boca Raton, FL (1994); 2nd edn (2000).

9. Hestenes, M. R. and Stiefel, E. L., Methods of Conjugate Gradients for Solving LinearSystems, National Bureau of Standards Journal of Research, 49, 409436 (1952).

10. Golub, G. H. and Van Loan, C. F., Matrix Computations, 2nd edn, The Johns HopkinsUniversity Press, Baltimore, MD (1989).

11. Jea, K. C. and Young, D. M., On the SimpliÞcation of GeneralizedConjugateGradient Methods for Nonsymmetrizable Linear Systems, Linear AlgebraApplications, 52, 399417 (1983).

12. Evans, D. J., Use of Preconditioning in Iterative Methods for Solving LinearEquations with Symmetric Positive DeÞnite Matrices, Computer Journal, 4, 7378(1961).

13. Fox, R. L. and Stanton, E. L., Developments in Structural Analysis by Direct EnergyMinimization, AIAA Journal, 6, 10361042 (1968).

14. Fried, I., Gradient Methods for Finite Element Eigenproblems, AIAA Journal, 7,739741 (1969).

15. Fried, I., More on Generalized Iterative Methods in Finite Element Analysis, AIAAJournal, 7, 565567 (1969).

16. Fried, I., A Gradient Computational Procedure for the Solution of Large ProblemsArising from the Finite Element Discretization Method, International Journal forNumerical Methods in Engineering, 2, 477494 (1970).

17. Reddy, M. P., Reddy, J. N., and Akay, H. U., Penalty Finite Element Analysisof Incompressible Flows Using Element by Element Solution Algorithms, ComputerMethods in Applied Mechanics and Engineering, 100, 169205 (1992).

18. Wathen, A. J., An Analysis of Some Element-by-Element Techniques, ComputerMethods in Applied Mechanics and Engineering, 74, 271287 (1989).

19. Hughes, T. J. R., Levit, I., and Winget, J., An Element-by-Element ImplicitAlgorithm for Heat Conduction, Journal of the Engineering Mechanics Division,ASCE, 109(2), 576585 (1983).

20. Winget, J. M. and Hughes, T. J. R., Solution Algorithms for Nonlinear Transient HeatConduction Analysis Employing Element-by-Element Iterative Strategies, ComputerMethods in Applied Mechanics and Engineering, 52, 711815 (1985).

21. Carey, G. F. and Jiang, B., Element-by-Element Linear and Nonlinear SolutionSchemes, Communications in Applied Numerical Methods, 2, 145153 (1986).

22. Hayes, J. L. and Devloo, P., An Element-by-Element Block Iterative Method forLarge Non-Linear Problems, in Innovative Methods for Nonlinear Behavior, W. K.Liu et al. (eds.), Pineridge Press, Swansea, UK, 5162 (1985).

23. Buratynski, E. K.,An Element-by-Element Method for Heat Conduction CAEIncluding Composite Problems, International Journal for Numerical Methods inEngineering, 26, 199215 (1988).

24. Levit, I., Element By Element Solvers of Order N, Computers & Structures, 27(3),357360 (1987).

25. de La Bourdinnaye, A., The Element-by-Element Method as a Preconditioner forLinear Systems Coming from Finite Element Models, International Journal ofSupercomputer Applications, 3, 6068 (1989).


26. Prakhya, K. V. G., Some Conjugate Gradient Methods for Symmetric andNonsymmetric Systems, Communications in Applied Numerical Methods, 4, 531539(1988).

27. Carey, G. F., Barragy, E., R. Mclay, and Sharma, M., Element-by-Element Vectorand Parallel Computations, Communications in Applied Numerical Methods, 4, 299307 (1988).

28. Hayes, L. J. and Devloo, P., A Vectorized Version of a Sparse Matrix-VectorMultiplication, International Journal for Numerical Methods in Engineering, 23,10431056 (1986).

29. Shakib, F., Hughes, T. J. R., and Johan, Z., A Multi-Element Group PreconditionedGMRES Algorithm for Nonsymmetric Systems Arising in Finite Element Analysis,Computer Methods in Applied Mechanics and Engineering, 75, 415456 (1989).

30. Tezduyar, T. E. and Liou, J., Grouped Element-By-Element Iteration Schemes forIncompressible Flow Computations, Computational Physics Communications, 53,441453 (1989).

31. Saad, Y. and Schultz, M. H., GMRES: A Generalized Minimal Residual Algorithmfor Solving Nonsymmetric Linear Systems, SIAM Journal of ScientiÞc and StatisticalComputations, 7(3), 856869 (1986).

32. Mitchell, J. A. and Reddy, J. N., A Multilevel Hierarchical Preconditioner for ThinElastic Solids, International Journal for Numerical Methods in Engineering, 43,13831400 (1998).

AppendixA2

Solution Procedures forNonlinear Algebraic Equations

A2.1 Introduction

Finite element formulations of nonlinear differential equations lead tononlinear algebraic equations for each element of the Þnite element mesh.The element equation is of the form,

[Ke(ue)]ue = F e (A2.1.1)

where

[Ke] − element coefficient matrix (or stiffness matrix), whichdepends on the solution vectorue,

ue − column vector of element nodal values, and (A2.1.2)

F e − column vector of element nodal forces.

When [Ke] is independent of ue, the matrix coefficients can be evaluated forall elements, and the assembled equations can be solved for the global nodalvalues u after imposing boundary conditions. When [Ke] depends on theunknown solution vector ue, the matrix coefficients cannot be evaluated.If we can Þnd an approximation to ue, say ue1, then [Ke(ue1)] can beevaluated and assembled. This amounts to linearizing the nonlinear equations,Eq. (A2.1.1). Then a next approximation to the solution can be obtained bysolving the assembled equations,

u2 = [K(u1)]−1F (A2.1.3)

This procedure can be repeated until the approximate solution comes close tothe actual solution in some measure. Such a procedure is called an iterativeprocedure.

Displacement, u

Linear, K(0)

Equilibrium pathR(u,F) = 0

Tangent to the path

secant

F1

Loa

d, F

°

u1


Here we discuss the following commonly used iterative procedures:

1. The Picard Iteration (or Direct Iteration) Method

2. The NewtonRaphson Iteration Method

3. The Riks Method

The details of these methods are discussed next, with the aid of a singlenonlinear equation. For a more complete presentation of these methods, thereader may consult the references at the end of this appendix.Consider the nonlinear equation,

K(u) · u = F or R(u) = 0 (A2.1.4)

where u is the solution to be determined, K(u) is a known function of u, F isthe known force, and R is the residual

R(u) = K(u) · u− F (A2.1.5)

A plot of the equilibrium path, R(u, F ) = 0, is shown in Figure A2.1.1. Forany value u1, K(u1) denotes the secant of the curve at u = u1, and (

∂R∂u )|u1

denotes the tangent of the curve at u = u1.

A2.2 Picard Iteration Method

In the Picard iteration, also known as the direct iteration method, we beginwith an initial guess for u, say u(0), (u(0) = 0 in Figure A2.1.1) and determinea Þrst approximation to u by solving the equation

u(1) =³K(u(0))

´−1F (A2.2.1)

Figure A2.1.1 Typical forcedisplacement curve.

SOLUTION PROCEDURES FOR NONLINEAR ALGEBRAIC EQUATIONS 441

u(1) 6= u, and a second approximation for u is sought by using the lastapproximation to evaluate K

u(2) =³K(u(1))

´−1F (A2.2.2)

This procedure is continued until the difference between two consecutiveapproximations of u differ by a preselected value. Thus, the algorithm andcriterion for convergence may be written as

Algorithm u(r) =³K(u(r−1))

´−1F (A2.2.3)

Convergence Criterion

s(u(r) − u(r−1))2

(u(r))2< ² (A2.2.4)

where ² denotes the convergence tolerance and r denotes the iteration number.A geometric interpretation of the procedure described above is illustrated in

Figure A2.2.1(a) for an initial guess of u(0) = 0. At the beginning of iterationr, the secant of the curve R(u) = 0 is found at the point u = u(r−1) andthe solution u(r) is computed using Eq. (A2.2.3). Figure A2.2.1(a) shows theconvergence to the true solution uc, whereas Figure A2.2.1(b) shows a possibledivergence of the algorithm. Thus, the success of the algorithm depends onthe nature of the nonlinear curve R(u) = 0, the initial guess, and the loadincrement.In the direct iteration method discussed above, the secant is evaluated at

each iteration and inverted to obtain the next approximate solution. This canbe computationally very expensive when the number of algebraic equations tobe solved is large, that is, when K is a matrix and [K]−1 is its inverse. WhenK has a linear portion, and in most problems of interest to us it does, analternative direct iteration algorithm can be formulated. Let

K(u) = KL +KN (u) (A2.2.5)

where KL and KN(u) are the linear and nonlinear parts of K. Note that KLis the slope at u = 0 of the curve R(u) = 0. Then we can write

u(r) = (KL)−1[F −KN(u(r−1)) · u(r−1)] (A2.2.6)

This scheme involves evaluating the nonlinear partKN at each iteration, whichis computationally less expensive when compared to evaluating [K(ur−1))] andinverting it. The inversion of KL is required only once, and it should be savedfor subsequent use. The criterion in Eq. (A2.2.4) can be used to check forconvergence.Geometrically, the alternative direct iteration algorithm uses the initial

slope, that is, slope at the origin of the curve for all iterations, while updatingthe effective force

F ≡ F −KN(u(r−1)) · u(r−1)

Displacement, u

u(1) = linear solution = F/K0

F

Loa

d, F

u(1)

uc = Converged solutionu(i) = Solution at the end

of the ith iteration

uc

K(0)=K0

°

°

K(u(1))=K1

°

(a)

u(2) u(3)

(b)

Displacement, u

F

Loa

d, F Possible divergence

°

°

°

°

u(4) u(2) u(1)u(5)u(3)


at each iteration (see Figure A2.2.2). The rate of convergence of this algorithm,if at all it converges, is slower than that in Eq. (A2.2.3).

Figure A2.2.1 Direct iteration scheme. (a) Case of convergence. (b) Caseof divergence.

♦

F− F1^

Displacement, u

F

Loa

d, F

u1 u2 u3

ui = solution at the end of the ith iteration

uc

°

◊

♦

◊

♦

◊

♦

u4u5

◊


Figure A2.2.2 ModiÞed direct iteration scheme.

As applied to the assembly of Þnite element equations (A2.1.1), the twoalgorithms take the following forms:

Algorithm 1 ur = [K(u(r−1))]−1F (A2.2.7)

Algorithm 2 ur = [KL]−1³F− [KNL(u(r−1))]u(r−1)

´(A2.2.8)

Convergence Criterion

vuutPNI=1(u

(r)I − u(r−1)I )2PNI=1(u

(r)I )

2< ² (A2.2.9)

The global matrices [K], [KL] and [KNL] are assembled from the correspondingelement matrices:

[K] is assembled using [Ke(ue(r−1))][KL] is assembled using [K

eL] (A2.2.10)

[KNL] is assembled using [KeNL(ue(r−1))]

where[Ke] = [Ke

L] + [KeNL] (A2.2.11)

The rate of convergence of the iterative procedure can be accelerated,in certain type of nonlinear behavior, by a relaxation procedure in which aweighted average of the last two solutions is used to evaluate [Ke] (or [Ke

NL]):


Ke(ue), where ue = γue(r−2) + (1 − γ)ue(r−1), and 0 ≤ γ ≤ 1 iscalled the relaxation or acceleration parameter. In this case Eqs. (A2.2.7) and(A2.2.8) take the form,

Algorithm 1 ur = [K(u)]−1F (A2.2.12)

Algorithm 2 ur = [KL]−1(F− [KNL(u)]u) (A2.2.13)

u = γu(r−2) + (1− γ)u(r−1) (A2.2.14)

The actual value of γ varies from problem to problem.The computational algorithm of the direct iteration is summarized below.

At each load level follow the steps:

1. Compute element matrices [Ke] and F e (for transient problems, [Ke]

and F e are to be replaced by [ Ke] and F e) using the solution u(r−1)from the previous iteration (of current load and/or time). For the Þrstiteration of the subsequent load steps, use uc, the converged solution ofthe last load step. For the Þrst load step use uc = 0, provided [Ke]−1exists, to compute the linear solution.

2. Assemble the element matrices [Ke] and F e (or [ Ke] and F e).3. Apply the boundary conditions on the assembled set of equations.

4. Solve the assembled equations.

5. Check for convergence using Eq. (A2.2.9).

6a. If the convergence criterion is satisÞed, increase the load to next level,initialize the counter on iterations, and repeat Steps 15.

6b. If the convergence criterion is not satisÞed, check if the maximum numberof allowable iterations is exceeded. If yes, terminate the computationprinting a message to that effect. If the number of iterations did notexceed the maximum allowed, update u(r−1) and u(r) and repeatSteps 15.

A2.3 NewtonRaphson Iteration Method

Suppose that we know solution of Eq. (A2.1.1) at (r − 1)st iteration andinterested in seeking solution at the rth iteration. We expand R(u) about theknown solution u(r−1) in Taylors series,

R(u) = R(u(r−1)) +µ∂R

∂u

¶ ¯u(r−1)

· δu+ 12

Ã∂2R

∂u2

! ¯u(r−1)

· (δu)2 + · · · = 0(A2.3.1)

where δu is the increment,

δu(r) = u(r) − u(r−1) (A2.3.2)


Assuming that the second- and higher-order terms in δu are negligible, we canwrite Eq. (A2.3.1) as

δu(r) = −³KT (u

(r−1))´−1 ·R(u(r−1))

=³KT (u

(r−1))´−1 ³

F −K(u(r−1)) · u(r−1)´

(A2.3.2)

where KT is the slope (tangent) of the curve R(u) at u(r−1):

KT =∂R

∂u

¯u(r−1)

(A2.3.3)

The residual or imbalance force, R(u(r−1)) is gradually reduced to zero if theprocedure converges. Equation (A2.3.2) gives the increment of u at the rthiteration so that the total solution is

u(r) = u(r−1) + δu(r) (A2.3.4)

The iteration is continued until a convergence criterion, say Eq. (A2.2.4), issatisÞed. Other convergence criteria include checking the magnitude of theimbalance force.A geometrical interpretation of the NewtonRaphson procedure is shown

in Figure A2.3.1(a). For most problems, the method has faster convergencecharacteristics. Figure A2.3.1(b) illustrates possible divergence of the iterativeprocedure for certain problems.The NewtonRaphson method requires that the tangent KT be computed

at each iteration. This can be very expensive when many degrees of freedomare involved. A modiÞed NewtonRaphson technique involves, for a Þxed loadstep, either keeping KT Þxed while updating the imbalance force at eachiteration (see Figure A2.3.2) or updating KT only at each preselected numberof iterations while updating the imbalance force at each iteration. There areseveral other modiÞcations of the procedure.The NewtonRaphson and modiÞed NewtonRaphson procedures take the

following forms when applied to the assembly of element equations (A2.3.2):

NewtonRaphson Procedure

δU = −[KT ]−1R (A2.3.5)

ModiÞed NewtonRaphson Procedure

δU = −[KT ]−1R (A2.3.6)

where [KT ] is the tangent matrix

[KT ] =∂R∂U

¯Ur−1

or (KT )ij =∂Ri∂Uj

= Kij +nXk=1

∂Kik∂Uj

Uk (A2.3.7a)

(b)

Displacement, u

F

Loa

d, F

u1

Possible divergence

°

Displacement, u

δu(1) = u(1) = linear solution = F/K0

F

Loa

d, F

uc = Converged solutionδu(i) = Solution increment at the

end of the ith iterationu(i) = Total solution at the end

of the ith iteration (for load F )

uc

KT(0) = K(0) = K0

°

°

KT(u(1))

°

(a)

u(1) u(2)

u(3)

δu(1) δu(2) δu(3)


and

R = [K(U(r−1))]U(r−1) − F (A2.3.7b)

[KT ] = [KT (U)], [K] = [K(U)] (A2.3.8)

and U is the solution at the beginning of the current load step, andU(r) = U(r−1) + δU(r) (A2.3.9)

The relaxation procedure described in Eq. (A2.3.9) can be used to accelerateconvergence for certain nonlinear problems.

Figure A2.3.1 The NewtonRaphson scheme. (a) A case of convergence.(b) A case of divergence.

Displacement, u

u(1) = Linear solution = F/K0

F

Loa

d, F

uc = Converged solutionu(i) = Solution at the end

of the ith iteration

uc

KT(0) = K(0) = K0

°

°

KT(u(1))

°

°

u(1) u(3)u(2)

u(4)


Figure A2.3.2 ModiÞed NewtonRaphson scheme.

Since only the increment of the solution is computed in each iteration of theNewtonRaphson iteration, the incremental equations in (A2.3.5) and (A2.3.6)are subject to homogeneous form of speciÞed essential boundary conditions ofthe problem. Thus after the Þrst iteration of the Þrst load step, any speciÞednon-zero values of U should be set to zero so that in the subsequent iterationsand loads, δu is subjected to homogeneous boundary conditions.For each load step, the following computations are required for the Newton

Raphson or modiÞed NewtonRaphson procedure:

1. Evaluate element matrices [Ke] and F e (or [ Ke] and F e for thetransient problems), and compute [Ke

T ] and Re using Eqs. (A2.3.7a,b)for an element.

2. Assemble element matrices [KeT ] and Re; for the modiÞed Newton

Raphson iteration procedure, save either assembled [KT ] or its inversefor use in subsequent iterations.

3. Apply the boundary conditions on the assembled set of equations.Note: Set the speciÞed boundary conditions on U to zero after Step 3in the Þrst iteration of the Þrst load step.

4. Solve the assembled equations.

5. Update the solution vector using Eq. (A2.3.9).


6. Check for convergence.

7a. If the convergence criterion is satisfied, increase the load, initialize theiteration counter, and repeat Steps 1—6. For the modified Newton—Raphson iteration, compute F e − [K]Ue(r−1) in Step 1 and go toStep 2.

7b. If the convergence criterion is not satisfied, check if the maximum numberof iterations allowed is exceeded. If it is, terminate the computation byprinting a message. If the maximum allowable number of iterations isnot exceeded, go to Step 1.

In order to reduce the number of operations per iteration, in the modifiedNewton—Raphson method the same system matrices are used for severaliterations. These matrices are updated only at the beginning of each loadstep or only when the convergence rate becomes poor. The modified Newton—Raphson method may require more iterations to reach a new equilibrium point.

A2.4 Riks and Modified Riks Schemes

The Newton—Raphson method and its modifications are often used to tracenonlinear solution paths. However, the Newton—Raphson method fails to tracethe nonlinear equilibrium path through the limit point (see Figure A2.4.1),because in the vicinity of a limit point the tangent matrix [KT ] becomessingular, and the iteration procedure diverges. Riks [1] and Wempner [2]suggested a procedure to predict the nonlinear equilibrium path throughlimit points. The method provides the Newton—Raphson method and itsmodifications with a technique to control progress along the equilibrium path.The theoretical development of this method and its modification can be foundin [3—6]. In the modified Riks method (see [3—6]) the load increment for eachload step is considered to be an unknown (see [3]) and solved as a part of thesolution.The basic idea of the Riks technique can be described for a single nonlinear

equation as follows (see Figure A2.4.2): The length ∆s of the tangent tothe current equilibrium point is prescribed, and the new point is found as theintersection of the normal to the tangent with the equilibrium path [see FigureA2.4.2(a)]. Then iteration is performed along the normal toward the newequilibrium point, as illustrated in Figure A2.4.2(a). Crisfield [4] suggestedusing a circular arc in place of the normal [see Figure A2.4.2(b)]. The centerof the circle is at the current equilibrium point and ∆s is its radius. Formultidimensional problems the normal and circular arcs become a plane andsphere, respectively, Crisfield [4] updated the tangent stiffness matrix only atthe beginning of each load increment (i.e. modified Newton—Raphson method).In the present study we describe the modified Riks method due to Crisfield[4].

Displacement, u

Loa

d, F

Limit points (KT = 0)


Figure A2.4.1 Loaddeßection curve with limit points.

We wish to solve Eq. (A2.1.4) for u as a function of the source term F . IfF is independent of the geometry, we can write it as

F = λF (A2.4.1)

where λ is a scalar, called load parameter, which is considered as an unknownparameter. Equation (A2.1.5) becomes

R = K(u) · u− λF (A2.4.2)

Now suppose that the solution (u(r−1)n ,λ

(r−1)n ) at (r− 1)st iteration of the nth

load step is known and we wish to determine the solution (u(r)n ,λ

(r)n ) at the

rth iteration. Expanding R, which is now a function of λ and u, in Taylorsseries about the known solution, we have,

R(u(r)n ,λ(r)n ) = R(u

(r−1)n ,λ(r−1)n ) +

µ∂R

∂λ

¶(r−1)δλ(r)n +

µ∂R

∂u

¶(r−1)δu(r)n + · · ·

= 0

Omitting the higher-order terms involving the increments δλ(r)n and δu

(r)n , we

obtain0 = R(r−1)n − F · δλ(r)n + (KT )

(r−1) · δu(r)n (A2.4.3)

Displacement, uu1 u2

°

λ

λ1

∆s1

u11

∆s2

λ2 °

∆si = Distance to the ith plane

u21

u12

Normal plane

(a)

Displacement, uu1 u2

°

λ

λ1

∆s1

u11

∆s2

λ2 °

∆si = Radius of the ith arc

u21

u12

Circular arc (b)


Figure A2.4.2 The Riks scheme. (a) Normal plane method. (b) Circuararc method.

where KT = ∂R/∂u is the tangent matrix [see Eq. (A2.3.3)]. The incrementalsolution at the current iteration of the nth load step is given by

δu(r)n = −K−1T (R(r−1)n − F · δλ(r)n≡ δu(r)n + δλ(r)n · δun (A2.4.4a)

where δu(r)n is the usual increment in displacement due to known out-of-balance

force vector R(r−1)n with known λ

(r−1)n and KT is the tangent at the beginning

of the current load increment (i.e. Modified Newton—Raphson method is used)

δu(r)n = −K−1T R(r−1)n (A2.4.4b)

uu0

°λ0

u1

δλ11

°

u11

Arc length

λ1

1

∆u11

u12

u13

for N−R

∆u11 for N−R

∆u12

δu11

λ1λ10

23

° °

δu11

δu11∧ . δλ1

λ

1


and δun is the tangential solution (see Figure A2.4.3)

δun = K−1T F (A2.4.4c)

Note that KT is evaluated using the converged solution un−1 of the last loadstep,

KT =

µ∂R

∂u

¶ ¯u=un−1

= K(un−1) +µ∂K

∂u

¶ ¯u=un−1

· un−1 (A2.4.5)

and δun is computed at the beginning of each load step.The solution at the rth iteration of the current load step is given by

un = un−1 +∆u(r)n (A2.4.6a)

∆u(r)n = ∆u(r−1)n + δu(r)n , λ(r)n = λ(r−1)n + δλ(r)n (A2.4.6b)

For the very Þrst iteration of the Þrst load step, we assume u = u0, and avalue for the incremental load parameter δλ01, and solve the equation

δu1 = (KT )−1F (A2.4.7)

and compute δu(0)1 = δλ01 · δu1.

A2.4.3 ModiÞed Riks scheme.


Select the arc length ∆s to be the length of the vector

(∆s)2 = ∆u(r)n ·∆u(r)n (A2.4.8)

Substituting for ∆u(r)n from Eqs. (A2.4.6b) [and δu

(r)n from Eq. (A2.4.4a)]

we obtain the following quadratic equation for the increment in the load

parameter, δλ(r)n :

a1[δλ(r)n ]

2 + 2a2[δλ(r)n ] + a3 = 0 (A2.4.9a)

where

a1 = δun · δuna2 = (∆u

(r−1)n + δu(r)n ) · δun

a3 = (∆u(r−1)n + δu(r)n ) · (∆u(r−1)n + δu(r)n )− (∆s)2 (A2.4.9b)

Let us denote the roots of this quadratic equation as δλ(r)n1 and δλ

(r)n2 ). To avoid

tracing back the equilibrium path (i.e. going back on the known equilibriumpath), we require the angle between the incremental solution vectors at two

consecutive iterations, ∆u(r−1)n and ∆u

(r)n , be positive. Corresponding to the

two roots, δλ(r)n1 and δλ

(r)n2 , there correspond to two values of ∆u

(r)n , denoted

∆u(r)n1 and ∆u

(r)n2 . The root that gives the positive angle is the one we select

from (δλ(r)n1 , δλ

(r)n2 ). The angle is deÞned to be product of the vector ∆u

(r−1)n

and ∆u(r)n . Then we check to see which one of the following two products is

positive:

∆u(r−1)n ·∆u(r)n1 and ∆u(r−1)n ·∆u(r)n2 (A2.4.10a)

If both roots are positive, then we select the ones closest to the linear solution,

δλ(r)n = − a32a2

(A2.4.10b)

The Þrst arc length ∆s is computed using Eqs. (A2.4.7) and (A2.4.8):

∆s = δλ01pδu1 · δu1 (A2.4.11)

To control the number of iterations taken to converge in the subsequent loadincrements, δs can be scaled,

∆sn = ∆sn−1 · IdI0

(A2.4.12)

where ∆sn−1 is the arc length used in the last iteration of the (n − 1)stload step, Id is the number of desired iterations (usually < 5) and I0 is thenumber of iterations required for convergence in the previous step. Equation


(A2.4.12) will automatically give small arc lengths in the areas of the mostsevere nonlinearity and longer lengths when the response is linear or nearlylinear. To avoid convergence of the solution at a higher equilibrium path,maximum arc lengths should be speciÞed.

For all load steps after the Þrst, the initial incremental load parameter δλ(0)n

is calculated fromδλ(0)n = ±∆sn · [δun · δun]− 1

2 (A2.4.13)

The plus sign is for continuing the load increment in the same direction as theprevious load step and the negative sign is to reverse the load step. The signfollows that of the previous increment unless the value of determinant of thetangent matrix has changed in sign.The modiÞed Riks procedure described above for a single equation can be

extended to the Þnite element equations in (A2.1.1). We introduce a loadparameter λ as an additional dependent variable,

F = λF (A2.4.14)

In writing Eq. (A2.4.14), loads are assumed to be independent of thedeformation. The assembled equations associated with Eq. (A2.1.1) become

R(U,λ) = [K]U− λF = 0 (A2.4.15)

The residual vector R is now considered as a function of both U and λ.Now suppose that the solution (U(r−1)n ,λ

(r−1)n ) at the (r − 1)st iteration

of the nth load step is known. Expanding R in Taylors series about(U(r−1)n ,λ

(r−1)n ), we have

Ri = Ri(U(r−1)n ,λ(r−1)n )+

µ∂Ri∂λ

¶Ur−1j ,λr−1n

·δλ(r)−Ã∂Ri∂Uj

!Ur−1j ,λr−1n

δU(r)j +· · ·

where the subscript n is omitted for brevity. Omitting second- and higher-

order terms involving δλ(r) and δU(r)j , we can write

0 = R(r−1) − δλ(r)n · F+ [KT ]δU(r)

or

δU(r)n = −[KT ]−1R(r−1) + δλ(r)n [KT ]−1F (A2.4.16a)

≡ δU(r)n + δλ(r)n δ Un (A2.4.16b)

where F is the load vector, R(r−1)n the unbalanced force vector at iteration(r − 1),

δU(r)n = −[KT ]−1R(r−1)n , (A2.4.17a)


and δλ(r)n the load increment [given by Eq. (A2.4.10)], and

δ Un = [KT ]−1F (A2.4.17b)

For the Þrst iteration of any load step, δλ(0)n is computed from [see Eq.

(A2.4.13)]:

δλ(0)n = ±∆sn(δ UTnδ Un)−1/2 (A2.4.18)

where [see Eq. (A2.4.12)]

(∆s)n = (∆s)n · Id/I0 (A2.4.19a)

and (∆s)n is the arc length computed using the relation

(∆s)n =q∆UTn−1∆Un−1 (A2.4.19b)

∆Un−1 being the converged solution increment of the previous load step.For the Þrst iteration of the Þrst load step, we use:

∆s = δλ01

qδ UT1 · δ U1 (A2.4.20a)

δ U1 = [KT ]−1F, [KT ] = [KT (U0)] (A2.4.20b)

where δλ01 is an assumed load increment and U0 is an assumed solutionvector (often we assume δλ01 = 1 and U0 = 0).The solution increment is updated using

∆Urn = ∆U(r−1)n + δU(r)n (A2.4.21)

and the total solution at the current load step is given by

Un = Un−1 + ∆U(r)n (A2.4.22)

The constants in Eq. (A2.4.9b) are computed using

a1 = δ UTnδ Una2 = (∆U(r−1)n + δU(r)n )Tδ Un (A2.4.23)

a3 = (∆U(r−1)n + δU(r)n )T (∆U(r−1)n + δU(r)n )− (∆s)2nThe computational algorithm of the modiÞed Riks method is summarized

next.


First iteration of Þrst load step

(i) Choose a load increment δλ01 (say, δλ01 = 1) and solution vector U0 (say, U0 = 0).

(ii) Form element matrices [KeT ] and

Re = [Ke]Ue0 − F e

(iii) Assemble element matrices.

(iv) Apply the boundary conditions.

(v) Solve for δ U1 and δU(1)1 using Eqs. (A2.4.17a,b).

(vi) Compute the solution increment [see Eqs. (A2.4.16) and (A2.4.21)] and update thesolution

δU(1)1 = δU(1)1 + δλ01δ U1; ∆U(1)1 = δU(1)1U1 = U0 + δU(1)1

(vii) Update the load increment λ(1)1 = δλ01.

(viii) Compute the arc length [see Eq. (A2.4.20a)]

∆s = δλ0

qδ UT1 δ U1

(ix) Go to Step 9.

First iteration of any load step except the Þrst

1. Calculate the system matrices [Ke], [KeT ] and F e.

2. Assemble the element matrices.

3. Apply the boundary conditions.

4. Compute the tangential solution

δ Un−1 = [KT ]−1F

5. Compute the initial incremental load parameter δλ(0)n by Eq. (A2.4.18):

δλ(0)n = ±(∆s)n[δ UTn · δ Un]−1/2 (A2.4.24)

6. Compute the incremental solution using Eq. (A2.4.17a):

U(r)n = −[KT ]−1R(0)n

7. Update the total solution vector and load parameter:

δU(1)n = δU(1)n + δλ(0)n δ U(1)n

Un = Un−1 + δU(1)n

λ(1)n = λ(0)n + δλ(0)n , ∆U(1)n = δU(1)n (A2.4.25)


8. Check for convergence [see Eq. (A2.4.28)]. If convergence is achieved, go to step 15below. If not, continue with the 2nd iteration by going to step 9.

The rth iteration of any load step (r = 2,3, . . .)

9. Update the external load vector

F(r−1) = λ(r−1)n F (A2.4.26)

10. Update the system matrices (skip forming of [KT ] for modiÞed Newton-Raphsoniteration).

11. Solve for δU(r)n and δ Un from the two sets of equations in (A2.4.17a,b); for themodiÞed NewtonRaphson method Eq. (A2.4.17b) need not be resolved.

12. Compute the incremental load parameter δλ[= δλ(r)n ] from the following quadraticequation:

a1(δλ)2 + 2a2 δλ+ a3 = 0

a1 = δ UTn · δ Una2 = (δU(r)n + ∆U(r−1)n )T · δ Una3 = (δU(r)n + ∆U(r−1)n )T · (δU(r) + ∆U(r−1)n )− (∆s)2n

and ∆s is the arc length of the current load step. Two solutions δλ1 and δλ2 of

this quadratic equation are used to compute two corresponding vectors ∆U(r)n1 and

∆U(r)n2 . The δλ that gives positive value to the product ∆U

(r−1)n ·∆U(r)n is selected.

If both δλ1 and δλ2 give positive values of the product, we use the one giving the smallestvalue of (−a3/a2).

13. Compute the correction to the solution vector

δU(r)n = δU(r)n + δλ · δ Un,

and update the incremental solution vector, the total solution vector and the loadparameter:

∆U(r)n = ∆U(r−1)n + δU(r)nUn = Un−1 + ∆U(r)n (A2.4.27)

λ(r)n = λ(r−1)n + δλ(r)n

14. Repeat steps 913 until the following convergence criterion is satisÞed:"(U(r)n − U(r−1)n )T · (U(r)n − U(r−1)n )

(U(r)n )T U(r)n

# 12

< ² (A2.4.28)

15. Adjust the arc length for the subsequent load steps by [see Eq. (A2.4.9a)] ∆s =∆s(Id/I0).

16. Start a new load step by returning to Step 1.


References

1. Riks, E., The Application of Newtons Method to the Problem of Elastic Stability,Journal of Applied Mechanics, 39, 10601066 (1972).

2. Wempner, G. A., Discrete Approximations Related to Nonlinear Theories of Solids,International Journal of Solids and Structures, 7, 15811599 (1971).

3. Batoz, J. L. and Dhatt, G.,Incremental Displacement Algorithms for Non-linearProblems, International Journal for Numerical Methods in Engineering, 14, 12621267 (1979).

4. CrisÞeld, M. A., A Fast Incremental/Iterative Solution Procedure that Handles Snap-Through, Computers and Structures, 13, 5562 (1981).

5. Bergan, P. G., Horrigmoe, G., Krakeland, B., and Soreide, T. H., Solution Techniquesfor Non-linear Finite Element Problems, International Journal for Numerical Methodsin Engineering, 12, 16771696 (1978)

6. Padovan, J. and Tovichakchaikul, S., Self-Adaptive PredictorCorrector Algorithmsfor Static Nonlinear Structural Analysis, Computers & Structures, 15, 365377 (1982).

7. Newton, I., De analysis per aequationes inÞnitas (1690), in The Mathematical Papersof Isaac Newton, Vol. 11 (1667-1670), D. T. Whiteside (ed.), Cambridge UniversityPress, Cambridge, UK, 207247 (1968).

8. Raphson, J., Analysis Aequationum universalis seu ad aequationes algebraicasresolvendas methodus generalis et expedita, ex nove inÞnitarum serierum doctrinadeducta ac demonstrata, London (1690) (original in British Library, London).

9. Cajori, F., Historical Note on the NewtonRaphson Method of Approximation,American Mathematical Monthly, 18, 2932 (1911).

10. Bicanic, N. and Johnson, K. H., Who was -Raphson?, International Journal forNumerical Methods in Engineering, 148152 (1979).


Subject Index

SUBJECT INDEX -159

Acceleration of convergence, (i ijAcceleration parameter. (i(i. 101, 107.

444( t - f a n n l y of approximation, 289, 296Alsori l imi . -111 . .143, 444Almansi -Namel strain Iriiwor, 332.

:!33, :t5;iA l l e r n a t i v e direct i t e r a t ion algorithm,

-•141Amplification mat r ix , 295Area coordinates, 37Assembly of element, 33, 251Axial force. !H. 112

backward difference: method, 290. 291293, 3f)7

Ranis vectors. 405BCI7 n-ian^le. I U 1Beam:

hoiwlins of. 8SEuler Bernoulli tlieory of. 88Timoshi'iiko theory of, 1 JO

Liil i iu-f i i - form, 17, 28, 2 4 J , '270li iort l iojronal Conjugate Gradient-

method. -132Boundary eondilions:

Diriehlel . I f it 'MM<!]]t.i;ll. 1 C . 7f)

forco. 92mixed, 70mil nra]. 1C, 79, 92Neumann. Hiof f lu ids , 23-1

rioitssinesfi approximation. 2;i:i

t'1 • -f.'onl i n u i l y . 155("anchv-'f Jreeti < ie f ( j r ina t io i ] I.ensor.

3;u. :«2Cfiudiy strain tensor. 332Oauehy stress. 334Ceut.ni] different-e method. 290. 292Clamped. 107, 121 . !51. 1.67. 192, 195

209. 212. 217, 220Classical Ix ' i i in t.h<sjry, 88Classical plate I heory (CPT):

assumptions of. 1-11boundary conditions, 149displacement, field, 142oqnntions of motion, 1-18finite element model of. l">.'istrains. 142

Cndtv/M conditions. 197Cooflifit'iUs:

of ( l ierma! expansion, 152Collocation method. 274Compulational nioclianios, 1Conditionally stable, 2i)">C'onfigiiration. :i28. 337. 338. 339Conforming element. IG1CoTiju^f i le gradient method, 274. 277.

280. -130Consistent m;iss m a t r i x , 292Consistent, penally model. 2-15. 24(>Const ant -averapje-aceelerat ion

method. 29:!. 309, 312, 371Constitutive relations, 40!)Continuity equation. 231. 242Cont inuum formulation. 87Convergence cri ter ion, 441 , 4-13

NONLINEAR FINITE ELEMENT ANALYSIS

Convergence tolerance, 441CPT, see: Classical plate theoryCrank-Nicolson scheme, 291, 307, 31Cylindrical coordinate system, 405Cylindrical shell, 209, 212

Deformation gradient tensor, 330Direct iteration, 65, 77, 98, 107, 12

131, 408, 440Direct methods, 427Displacement finite element model,

157Doubly-curved shell, 196, 21

displacement field, 204equations of motion, 205finite element model of, 206, 207strains, 204stress resultants, 203

Eigenvalue problem, 289Eigenvalues, 334Elasticity tensor, 346Elastic-perfectly-plastic, 392Elasto-plastic tangent modulus, 394Element-by-element, 432Energetically-conjugate, 336Engineering constants, 376Equations of equilibrium,

Euler-Bernoulli beam theory, 91Timoshenko beam theory, 113

Essential boundary condition, 76see Boundary conditions

Euler strain tensor, 332, 343Euler-Bernoulli beam theory, 88

displacement field of, 88Euler-Bernoulli hypotheses, 88Euler-Lagrange equations, 112, 176Eulerian description, 230Euler's explicit method, 4Explicit scheme, 288Extensional stiffnesses, 94, 153

Finger tensor, 332Finite difference method, 5Finite element, 5, 13Finite element mesh, 25Finite element method, 5, 290

Finite element model of:beams (EBT), 95beams (TBT), 113heat transfer, 297, 418isotropic, Newtonian, viscous,

incompressible fluids, 235plates (CPT), 153plates (FSDT), 177power law fluids, 407, 411

First-order shear deformation theory:displacement field, 173equations of motion, 177finite element model of, 177, 17generalized displacement, 173strains, 174

First law of thermodynamics, 232First Piola-Kirchhoff stress, 335Fixed edge, 151Flow over a backward-facing step, 277Flow past a circular cylinder, 278Fluid mechanics, 229Force boundary condition, 92Forward difference scheme, 4, 290, 291Fourier's heat conduction law, 233Free-body diagram, 17Free edge, 151Frontal solution, 427FSDT, see: First-order shear

deformation theoryFull integration, 104, 116, 118, 120, 1

Galerkin's method, 15, 291, 293Gauss elimination method, 427Gauss points, 45, 135, 185, 251Gauss quadrature, 45, 102, 116, 118, 135Gauss weights, 45, 135, 251Gauss-Legendre quadrature, 44Generalized nodal displacements, 90Generalized nodal forces, 89Geometric nonlinearity, 7, 389Global coordinates, 30, 130, 133GMRES, 432Governing Equations:

of Newtonian, viscous,incompressible fluids, 235

of non-Newtonian fluids, 406Green-Lagrange incremental strain, 341

460

SUBJECT INDEX

Green-Lagrange strain tensor,201, 331, 333, 340, 390

/i-refinement, 218Half bandwidth, 425Hamilton's principle, 369Hardening, 8Heated cavity, 419Heat flux, 233Hermite cubic interpolation

functions, 95Hinged, 119, 219Homogeneous motion, 331Hydrostatic pressure, 233Hyperelastic material, 346, 390

Ideal plastic, 392Imbalance force, 445Implicit scheme, 288Incompressibility condition, 231Incompressible flow, 229Infinitesimal strain tensor, 341, 343Initial conditions, 288, 292Initial-value problem, 3Interpolation property, 19, 30Inviscid fluid, 229Isoparametric formulation, 42Iterative method, 274, 430

Jacobian, 44, 134, 331, 340Jacobian matrix, 43, 134, 251

Kinetic hardening, 394Kirchhoff assumptions, 141Kirchhoff free-edge condition, 150Kirchhoff stress increment tensor, 344Kovasznay flow, 275

Ladyzhenskaya-Babuska-Brezzi (LBB)condition, 248

Lagrange interpolation, 30, 32, 38,114, 130, 155, 179, 207, 411, 418

Higher-order, 20Linear, 19, 95Quadratic, 20

Lagrange multiplier method, 241Lagrangian description, 230, 328

Lame constants, 199, 233Laminar flow, 230Lanczos orthores, 433Least squares finite element model,

267, 269, 272, 299, 315Least squares functional, 271, 272,

275, 277, 299, 318Linear acceleration method, 293Linear element, 23Linear form, 17, 28, 241, 270Linear rectangular element, 32Linear triangular element, 30Linearly independent, 15Local coordinates, 133Locking, 274

membrane, 102, 117shear, 115, 184

Load increments, 100, 106, 108, 119, 172,361

Master element, 36, 41Material coordinates, 328Material description, 230, 328Material nonlinearity, 7, 389Material stiffnesses, 376Material strain rate tensor, 336Mathematical model, 1Matrix,

coefficient, 23stiffness, 23, 358tangent stiffness, 358

Membrane locking, see LockingMetric, 197Mindlin plate theory, 181Mixed finite element model, 237, 246,

254, 274Moment, 91

Natural boundary condition, 16, 79,92

Navier-Strokes equation, 230, 232Newmark's integration scheme, 293,

296, 371Newton-Raphson iteration scheme,

68, 78, 98, 107, 121, 131, 157,189, 439, 444

modified, 70, 445

461

NONLINEAR FINITE ELEMENT ANALYSIS

Newtonian fluid, 229, 233Newton's iteration procedure, 68Nodal degrees of freedom, 23Nodes, 6, 26

global, 33Nonconforming element, 161Nonlinear analysis of:

Euler-Bernoulli beams, 88Timoshenko beams, 110

Non-Newtonian fluid, 229, 404Numerical integration, 40Numerical simulation, 3p-refinement, 21Parabolic equations, 289Penalty finite element model, 237,

241, 246, 405Penalty parameters, 243, 247, 256,

261Picard iteration, 65, 440Pinned, 107, 121Plane stress, 365Plane stress-reduced stiffnesses, 152,

376Plastic flow, 391Plasticity, 391Power-law fluid, 404, 407Primary variables, 16, 27, 62, 112,

150, 177, 206, 235Primitive variables, 235Principal radii of curvature, 197Principle of:

conservation of angularmomentum, 232

conservation of energy, 232conservation of linear momentum,

232conservation of mass, 231, 340minimum total potential energy, 18virtual displacements, 89, 111, 145,

174, 347, 369virtual work, 349, 352

Quadratic element, 23

Rate of deformation gradient tensor,336

Rectangular elements, 38

Reduced integration, 102, 104, 118, 120,168, 212, 214, 249, 254, 274, 377

Reduced integration penalty model,244

Relaxation, 443Residual, 15Riks, 440, 448Riks-Wempner method, 378

modified, 381Romberg-Osgood model, 391Rotation tensor, 334

Sanders' shell theory, 196, 205Second Piola-KirchhofT stress, 335,

390Secondary variables, 16, 27, 62, 112,

150, 177, 206, 235Serendipity elements, 39Shallow cylindrical panel, 217, 219Shear correction coefficient, 111, 175Shear correction factors, 203Shear force, 112, 203Shear locking, see LockingShear thickening fluids, 407Shear thinning fluids, 407Simply supported, 151, 167, 169, 185

189, 210, 312, 378, 381Skyline technique, 427Softening, 9Solar receiver, 420Space-time coupled formulation, 321Spatial approximation, 287Spatial description, 230, 329Specific heat, 236Stability, 295, 296Stiffnesses:

bending, 94, 153extensional, 94, 153extensional-bending, 94

Strokes flow, 230, 241, 269Strain-displacement relations, 143

nonlinear, 88Strain energy, 89, 174Strain energy density, 390Strain hardening, 393, 394Strain rate tensor, 232, 405Stream function, 237

462

SUBJECT INDEX

Stress,Cauchy, 334deviatoric, 393invariants, 393tensor. 232vector. 334

Stretch tensor, 334Subparametric formulation, 42Superconvergent element, 95Superparametric formulation, 42

Tangent matrix, 69, 131, 445Tangent stiffness matrix, 98, 117, 157

182, 390Theorem of Rodrigues, 197Thermal coefficients of expansion, 153Thermal stress resultant, 153Timoshenko beam theory, 110

displacement field of, 110finite element model of, 113

Tolerance, 65, 108, 119, 169, 189, 1Total Lagrangian description, 338Total Lagrangian formultaion, 348,

353Transverse force, 2, 175Tresca yield criterion, 393Triangular elements, 37Turbulent flow, 230

Unconditionally stable, 295Updated Green-Lagrange strain

tensor, 342Updated Kirchhoff stress increment

tensor, 344Updated Kirchhoff stress tensor, 344Updated Lagrangian description, 338Updated Lagrangian formulation,

350, 360

Validation, 12Variational problem, 17, 241, 242, 270Vectors, 23Velocity gradient tensor, 335Velocity-Pressure model, 237Velocity--Pressure-Velocity model,

299Velocity-Pressure-Vorticity model,

299Verification, 11Viscoelastic fluid, 404Viscosity, 229Viscous incompressible fluid, 407Viscous stress tensor, 233von Karman nonlinearity, 300von Karman strains, 89, 144, 174von Mises yield criterion, 393Vorticity, 237, 271

Weak form, 16, 26, 62, 128, 239Weak forms for:

Doubly-curved shell, 206Euler-Bernoulli beam, 89heat transfer, 297plates (CPT), 147plates (FSDT), 174, 300steady Stokes flow, 241Timoshenko beam, 111

Weight functions, 15Weighted-integral, 15Weingarten-Gauss relations, 197White-Metzner model, 409

Yield criterion, 392Yield stress, 392Yielding, 391

463