Finite Element Method (Jorge Chavez)

M. Gosz

Finite Element MethodApplications in Solids, Structures, and Heat Transfer

Contents

1 Introduction 1

2 Mathematical Preliminaries 5

2.1 Matrix algebra . . . . . . . . . . . . . . . . . . . . . . . . . . 52.1.1 Multiplication of a matrix by a scalar . . . . . . . . . 62.1.2 Transpose of a matrix . . . . . . . . . . . . . . . . . . 62.1.3 Matrix multiplication . . . . . . . . . . . . . . . . . . 62.1.4 The identity matrix . . . . . . . . . . . . . . . . . . . 72.1.5 Determinant of a matrix . . . . . . . . . . . . . . . . . 82.1.6 Inverse of a matrix . . . . . . . . . . . . . . . . . . . . 92.1.7 Linear algebraic equations . . . . . . . . . . . . . . . . 102.1.8 Integration of matrices . . . . . . . . . . . . . . . . . . 11

2.2 Vectors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 122.2.1 Vector dot product . . . . . . . . . . . . . . . . . . . . 132.2.2 Vector cross product . . . . . . . . . . . . . . . . . . . 142.2.3 Vector transformation . . . . . . . . . . . . . . . . . . 15

2.3 Second-order tensors . . . . . . . . . . . . . . . . . . . . . . . 172.3.1 Tensor transformation . . . . . . . . . . . . . . . . . . 202.3.2 Eigenvalues and eigenvectors of second-order tensors . 222.3.3 Common operations with vectors and tensors . . . . . 24

2.4 Calculus . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 282.4.1 Fundamental theorem of calculus . . . . . . . . . . . . 282.4.2 Integration by parts . . . . . . . . . . . . . . . . . . . 282.4.3 Taylor series expansion . . . . . . . . . . . . . . . . . . 302.4.4 Gradient and divergence . . . . . . . . . . . . . . . . . 322.4.5 The level surface . . . . . . . . . . . . . . . . . . . . . 332.4.6 Divergence theorem . . . . . . . . . . . . . . . . . . . 34

2.5 Newton’s method . . . . . . . . . . . . . . . . . . . . . . . . . 382.5.1 Nonlinear equations of several variables . . . . . . . . 40

2.6 Kinematics of motion . . . . . . . . . . . . . . . . . . . . . . . 452.7 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47

3 One-Dimensional Problems 51

3.1 The weak form . . . . . . . . . . . . . . . . . . . . . . . . . . 533.2 Finite element approximations . . . . . . . . . . . . . . . . . 54

3.2.1 The linear-u element . . . . . . . . . . . . . . . . . . . 55

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3.2.2 The quadratic-u element . . . . . . . . . . . . . . . . . 573.3 Plugging in the trial and test functions . . . . . . . . . . . . . 593.4 Algorithm for matrix assembly . . . . . . . . . . . . . . . . . 693.5 One-dimensional elasticity . . . . . . . . . . . . . . . . . . . . 73

3.5.1 Strain and stress calculation . . . . . . . . . . . . . . . 753.5.2 Finite element results . . . . . . . . . . . . . . . . . . 76

3.6 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 79

4 Linearized Theory of Elasticity 83

4.1 Cauchy’s law . . . . . . . . . . . . . . . . . . . . . . . . . . . 834.2 Principal stresses . . . . . . . . . . . . . . . . . . . . . . . . . 894.3 Equilibrium equation . . . . . . . . . . . . . . . . . . . . . . . 934.4 Small-strain tensor . . . . . . . . . . . . . . . . . . . . . . . . 96

4.4.1 Relative stretch . . . . . . . . . . . . . . . . . . . . . . 964.4.2 Angle change . . . . . . . . . . . . . . . . . . . . . . . 974.4.3 Small-displacement gradients . . . . . . . . . . . . . . 99

4.5 Hooke’s law . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1024.5.1 Engineering constants . . . . . . . . . . . . . . . . . . 1034.5.2 Hooke’s law for plane stress and plane strain . . . . . 1064.5.3 Hooke’s law for thermal stress problems . . . . . . . . 108

4.6 Axisymmetric problems . . . . . . . . . . . . . . . . . . . . . 1094.7 Weak form of the equilibrium equation . . . . . . . . . . . . . 1154.8 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 118

5 Steady-State Heat Conduction 121

5.1 Derivation of the steady-state heat equation . . . . . . . . . . 1225.2 Fourier’s law . . . . . . . . . . . . . . . . . . . . . . . . . . . 1245.3 Boundary conditions . . . . . . . . . . . . . . . . . . . . . . . 1255.4 Weak form of the steady-state heat equation . . . . . . . . . . 1285.5 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 129

6 Continuum Finite Elements 131

6.1 Three-node triangle . . . . . . . . . . . . . . . . . . . . . . . . 1326.1.1 The B-matrix . . . . . . . . . . . . . . . . . . . . . . . 135

6.2 Development of an arbitrary quadrilateral . . . . . . . . . . . 1376.2.1 Compatibility issues . . . . . . . . . . . . . . . . . . . 1386.2.2 The bi-unit square . . . . . . . . . . . . . . . . . . . . 1416.2.3 The parent domain . . . . . . . . . . . . . . . . . . . . 1436.2.4 The B-matrix . . . . . . . . . . . . . . . . . . . . . . . 1456.2.5 Derivatives of the shape functions . . . . . . . . . . . . 1476.2.6 Area change . . . . . . . . . . . . . . . . . . . . . . . . 147

6.3 Four-node tetrahedron . . . . . . . . . . . . . . . . . . . . . . 1506.3.1 The B-matrix . . . . . . . . . . . . . . . . . . . . . . . 153

6.4 Eight-node brick . . . . . . . . . . . . . . . . . . . . . . . . . 1546.4.1 The B-matrix . . . . . . . . . . . . . . . . . . . . . . . 156

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6.4.2 Volume change . . . . . . . . . . . . . . . . . . . . . . 1576.5 Element matrices and vectors . . . . . . . . . . . . . . . . . . 1596.6 Gauss quadrature . . . . . . . . . . . . . . . . . . . . . . . . . 1636.7 Bending of a cantilever beam . . . . . . . . . . . . . . . . . . 1766.8 Analysis of a plate with hole . . . . . . . . . . . . . . . . . . . 1816.9 Thermal stress analysis of a composite cylinder . . . . . . . . 186

6.9.1 Heat conduction analysis . . . . . . . . . . . . . . . . . 1866.9.2 Thermal stress analysis . . . . . . . . . . . . . . . . . 1916.9.3 Results . . . . . . . . . . . . . . . . . . . . . . . . . . 195

6.10 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 196

7 Structural Finite Elements 203

7.1 Space truss . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2037.1.1 Strain-displacement relationship . . . . . . . . . . . . 2057.1.2 Element matrix . . . . . . . . . . . . . . . . . . . . . . 2107.1.3 Element vector . . . . . . . . . . . . . . . . . . . . . . 212

7.2 Euler-Bernoulli beams . . . . . . . . . . . . . . . . . . . . . . 2217.2.1 Kinematic assumptions . . . . . . . . . . . . . . . . . . 2227.2.2 Finite element approximations . . . . . . . . . . . . . 2247.2.3 Element matrix . . . . . . . . . . . . . . . . . . . . . . 2277.2.4 Element vector . . . . . . . . . . . . . . . . . . . . . . 229

7.3 Mindlin-Reissner plate theory . . . . . . . . . . . . . . . . . . 2377.3.1 Assumptions . . . . . . . . . . . . . . . . . . . . . . . 2387.3.2 Strain-displacement relations and Hooke’s law . . . . . 2397.3.3 Finite element approximations . . . . . . . . . . . . . 2417.3.4 Element matrix . . . . . . . . . . . . . . . . . . . . . . 2437.3.5 Element vector . . . . . . . . . . . . . . . . . . . . . . 246

7.4 Deflection of a clamped plate . . . . . . . . . . . . . . . . . . 2497.5 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 252

8 Linear Transient Analysis 257

8.1 Derivation of the equation of motion . . . . . . . . . . . . . . 2588.1.1 Weak form . . . . . . . . . . . . . . . . . . . . . . . . 259

8.2 Semi-discrete equations of motion . . . . . . . . . . . . . . . . 2598.2.1 Properties of the mass matrix . . . . . . . . . . . . . . 2628.2.2 Natural frequencies and normal modes . . . . . . . . . 263

8.3 Central difference method . . . . . . . . . . . . . . . . . . . . 2668.3.1 Dynamic response of a simple pendulum . . . . . . . . 2698.3.2 Stability of central difference method . . . . . . . . . . 2718.3.3 Elastic wave propagation in a one-dimensional bar . . 275

8.4 Trapezoidal rule . . . . . . . . . . . . . . . . . . . . . . . . . . 2828.4.1 Dynamic response of a cantilever beam . . . . . . . . . 288

8.5 Unsteady heat conduction . . . . . . . . . . . . . . . . . . . . 2918.5.1 Weak form of unsteady heat equation . . . . . . . . . 2938.5.2 Finite element approximations . . . . . . . . . . . . . 294

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8.5.3 Backward difference method . . . . . . . . . . . . . . . 2968.5.4 Stability of the backward difference method . . . . . . 2968.5.5 Unsteady heat conduction in a composite cylinder . . 298

8.6 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 300

9 Small-Strain Plasticity 303

9.1 Basic concepts . . . . . . . . . . . . . . . . . . . . . . . . . . 3039.2 Yield condition . . . . . . . . . . . . . . . . . . . . . . . . . . 306

9.2.1 Dependence on hydrostatic pressure . . . . . . . . . . 3089.2.2 The von Mises yield condition . . . . . . . . . . . . . . 3109.2.3 Geometry of the von Mises yield surface . . . . . . . . 3119.2.4 Simple experiments . . . . . . . . . . . . . . . . . . . . 312

9.3 Flow and hardening rules . . . . . . . . . . . . . . . . . . . . 3159.4 Derivation of the elastoplastic tangent . . . . . . . . . . . . . 3179.5 Finite element implementation . . . . . . . . . . . . . . . . . 327

9.5.1 Convergence criteria . . . . . . . . . . . . . . . . . . . 3299.5.2 Radial return stress update scheme . . . . . . . . . . . 331

9.6 One-dimensional elastoplastic deformation of a bar . . . . . . 3389.7 Elastoplastic analysis of a thick-walled cylinder . . . . . . . . 3419.8 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 346

10 Treatment of Geometric Nonlinearities 349

10.1 Large-deformation kinematics . . . . . . . . . . . . . . . . . . 35010.1.1 The Green-Lagrange strain tensor . . . . . . . . . . . 352

10.2 Weak form in the original configuration . . . . . . . . . . . . 35810.2.1 Internal virtual work . . . . . . . . . . . . . . . . . . . 35810.2.2 External virtual work . . . . . . . . . . . . . . . . . . 362

10.3 Linearization of the weak form . . . . . . . . . . . . . . . . . 36310.3.1 Internal virtual work . . . . . . . . . . . . . . . . . . . 36610.3.2 External virtual work . . . . . . . . . . . . . . . . . . 368

10.4 Snap-through buckling of a truss structure . . . . . . . . . . . 37110.4.1 Element tangent matrix . . . . . . . . . . . . . . . . . 37310.4.2 Element internal force vector . . . . . . . . . . . . . . 37810.4.3 Numerical results . . . . . . . . . . . . . . . . . . . . . 379

10.5 Uniaxial tensile test of a rubber dog-bone specimen . . . . . . 38210.5.1 Element tangent matrix . . . . . . . . . . . . . . . . . 38410.5.2 Element internal force vector . . . . . . . . . . . . . . 38610.5.3 Numerical results . . . . . . . . . . . . . . . . . . . . . 387

10.6 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 390

Bibliography 392

Index 397

To my wonderful children Madeline and David,

and to my lovely wife Mary Lee, who makes each day

an exciting adventure.

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Preface

Tell me and I’ll forget. Show me, and I may not remember. Involve me, and

I’ll understand.—Native American Saying

Learning the finite element method can be a real mathematical challengefor both students and practicing engineers who have been in the workplacefor longer than two years. The author believes, however, that anyone whohas successfully completed the first two years of an undergraduate programin engineering or the physical sciences has the potential to understand andapply the concepts of the finite element method to real-world problems.

This book covers topics in linear, linear dynamic, and nonlinear finite ele-ment procedures. It provides a thorough treatment written in simple languageof the mathematics required to fully understand all of these finite elementtopics. This book is designed to be a learning aid for practicing engineersand students who wish to understand the fundamentals of the finite elementmethod and apply it to practical problems in industry. It is not intended tobe a comprehensive volume. One of the helpful features of the book is that itcontains a collection of case studies. The case studies define a problem, discussappropriate solution strategies, and warn against common pitfalls. They alsoemphasize how the results change when important parameters in the problemare altered, such as material properties, geometry, and mesh refinement.

There is no question that the finite element method is a powerful analysistool. It is so prevalent throughout the world that most medium- and large-sized manufacturing companies own or lease commercial finite element soft-ware. Often practicing engineers are forced to use these codes (even thoughthey do not have the background and/or experience) in order to get resultsquickly. They lack confidence about the assumptions that are made, about theboundary conditions that are chosen, and whether or not the proper failurecriterion is being used. This book is intended to help the practicing engineergain confidence in properly using the finite element method. The book is alsointended as a textbook for advanced undergraduates and first-year graduatestudents in engineering and the physical sciences.

One of the most common concerns of engineers and students interested inthe finite element method is whether or not they have the requisite mathbackground to learn the material. Chapter 2 identifies and reviews the basicmathematical building blocks that are absolutely necessary for understandingthe finite element method. The topics include matrix algebra, fundamental

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concepts from calculus such as the Taylor series expansion and the divergencetheorem, and an introduction to vectors and tensors and how they transformfrom one coordinate system to another. The chapter ends with a discussionof the mechanics of a continuous medium, which is especially important forunderstanding the concepts presented later in the book on nonlinear finiteelement procedures.

In Chapter 3, the finite element method is introduced by considering second-order ordinary differential equations with constant coefficients. In this chap-ter, the very important concepts of trial functions, test functions, and theweak form of the differential equation are introduced. The degree of con-tinuity of trial and test functions is also discussed. Chapter 3 also coversthe finite element assembly process, and it is shown how this process arisesnaturally from the weak form. The various types of boundary conditionsthat are encountered in the finite element method are introduced, and someone-dimensional applications are presented. The accuracy of the approximatesolutions is assessed for different element types (linear and quadratic) as thefinite element mesh is refined.

The fundamental concepts that are needed to understand how the finite ele-ment method is applied to problems in solid mechanics are presented in Chap-ter 4. Here an introduction to the linearized theory of elasticity is presented.The chapter begins with a derivation of Cauchy’s law, and then proceedswith a discussion on principal stresses and a derivation of the equilibriumequations. The chapter then provides a thorough treatment of the small-strain tensor and derives Hooke’s law for three-dimensional, plane stress, planestrain, thermoelastic, and axisymmetric formulations. The chapter concludeswith a derivation of the weak form of the equilibrium equation.

In Chapter 5 another important field equation in engineering is derived —the steady-state heat equation. The chapter then goes on to discuss Fourier’slaw and the boundary conditions relevant to heat conduction. The chapterends with an example of heat conduction in a composite sphere followed by aderivation of the weak form of the steady-state heat equation. Both Chapters4 and 5 emphasize that partial differential equations that arise in engineeringand the physical sciences come from balance laws. Stated simply, a forcebalance gives the equilibrium equation, and an energy balance give the heatequation.

A widely used class of finite elements called continuum finite elements isintroduced in Chapter 6. The chapter begins with a discussion on what ismeant by finite element shape functions and ends with an explanation ofthe isoparametric finite elements. Element types include the three-node tri-angle, the arbitrary four-node quadrilateral, the four-node tetrahedron, andthe eight-node brick. The computation of the finite element stiffness matrixfor continuum elements involves evaluating integrals over the domain of theelement. The method for evaluating such integrals in finite element codesis called Gauss quadrature. Several examples are given near the end of thechapter that clearly illustrate how Gauss quadrature works and its accuracy.

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The chapter concludes with three case studies. The first case study considersthe problem of a cantilever beam subjected to a concentrated end load. Thecase study examines the use of the four-node quadrilateral element for planeelasticity applications and discusses the numerical problem of locking that canoccur under certain conditions and how to avoid it. Next, the problem of aplate with a centrally located hole subjected to a uniform distributed load isconsidered. Here symmetry boundary conditions are explained in detail, anda mesh convergence study is carried out to see how the approximate solutionto the problem changes as the mesh is further refined. The last case studyconsiders a thermal stress problem of a copper wire surrounded by a ceramiccoating. In the study, the copper wire is heated from some stress- free refer-ence temperature. The temperature distribution in the composite as well asthe resulting thermal stress distribution is obtained.

Various types of structural finite elements are considered in Chapter 7.These include the space truss, the Euler-Bernoulli beam element, and theMindlin-Reissner plate element. During the formulation of these elements,the engineering assumptions that place certain restrictions on the displace-ment field, called kinematic constraints, are highlighted. The chapter con-tains several examples that illustrate the use of truss and beam elements andconcludes with a case study of a square plate with clamped edges subjectedto a uniform pressure distribution. During the case study, the performanceof the Mindlin-Reissner plate element is examined. In addition, the numeri-cal problem of shear locking, which can be encountered as the plate becomesvery thin compared to the in-plane dimensions, is demonstrated. It is shownthat the problem of shear locking can be avoided by using a technique calledselective reduced integration.

An introduction to finite element procedures for linear transient analysis isprovided in Chapter 8. A transient analysis is nothing more than an attemptto solve a partial differential equation that depends on time. Examples of suchequations are the equations of motion and the unsteady heat equation. Sub-stitution of the finite element shape functions into the weak form discretizesthe problem in space, yielding a system of ordinary differential equations thatvary continuously in time. To obtain the complete time history for the prob-lem of interest, a suitable time integration scheme must be adopted. In thischapter, two of the most popular time integration schemes are covered. Theseare the central difference method and the trapezoidal rule. A section of thechapter is also devoted to the topic of assessing the stability of a particulartime integrator. The chapter ends with some general guidelines for time stepselection. The chapter also includes two case studies. In the first case studythe problem of a one-dimensional bar subjected to an impact loading is con-sidered, and the nature of stress wave propagation in the bar is studied as afunction of mesh refinement and time step size. In the second case study thedynamic response of a cantilever beam is obtained using the trapezoidal rule.

Often in solid mechanics we encounter problems in which the relationshipbetween strain and displacement is linear, but the stress-strain relationship

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is nonlinear. This type of problem is often referred to as a materially-onlynonlinear problem. In chapter 9, due to its practical importance in solid me-chanics and engineering design, the theory of small-strain, rate-independentplasticity is presented. The chapter begins with a basic treatment of one-dimensional idealized stress-strain behavior and then proceeds to discuss themain ingredients that are needed in any plasticity theory: yield condition,flow rule, and hardening rule. The chapter ends with the derivation of theelastoplastic tangent, which is needed for the finite element implementationof small- strain plasticity based on Newton’s method. The chapter ends byproviding a detailed description of the general procedure for solving materiallynonlinear problems. The main ideas are applied by presenting a simple ex-ample of a one-dimensional elastic-plastic bar. Finally,the chapter ends witha case study of a thick-walled, elastic-plastic cylinder subjected to an internalpressure loading.

Problems in solid mechanics that involve arbitrarily large translations, ro-tations, and/or deformations are classified as geometrically nonlinear. In ge-ometrically nonlinear problems, there exists a nonlinear relationship betweenstrain and displacement. A problem can be geometrically nonlinear even ifthe deformation in the solid remains small enough so that the material re-mains in the linearly elastic region. An introduction to the rich subject oftreatment of geometrically nonlinear problems is presented in Chapter 10.The chapter begins with a general discussion on large-strain kinematics andlarge-strain measures. The finite element formulation for geometrically non-linear problems is then presented. The formulation begins by writing downthe principle of virtual work over some fixed reference configuration. Duringthe process, which is carried out through a simple change of variables, thealternative stress measures (first Piola-Kirchhoff and second Piola-Kirchhoffstress tensors) naturally emerge. The concept of work conjugacy is also dis-cussed. The chapter ends with a description of the total-Lagrangian procedurefor solving geometrically nonlinear problems and two case studies. In the firstcase study, the snap-through buckling problem of a simple truss structure isconsidered, and the second study investigates the large deformation responseof a neo-Hookean test specimen.

This book can be used as a textbook or for self study. The material pre-sented in Chapters 1–7 forms a semester-long course for advanced undergrad-uates and first-year graduate students. The course could be entitled, Intro-

duction to the Finite Element Method. The second part of the book, Chapters8–10, form the subject matter of a semester-long sequel to the introductorycourse. The course could be called, Dynamic and Nonlinear Finite Element

Procedures. The material presented in the book can also be used to teach shortcourses in the finite element method. Most of the material that is presentedin this book is taken from my course notes and experiences teaching two four-day short courses intended for professional engineers. Each short course offersa series of lectures, breaks, and hands-on problem-solving sessions. Duringthe last two hours of each day, the students engage in a computer laboratory

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experience. The case studies presented throughout the book can be used forthis purpose. The first four-day course covers most of the material presentedin Chapters 1–7. The second short course covers the material presented inChapters 8–10.

A unique feature of this book is that every figure throughout the book wasgenerated by writing and running an individual MATLAB program. Doingthis allows the three-dimensional figures to be viewed in an interactive vir-tual reality environment. This is accomplished simply by opening the HTMLfile on the publisher’s web site. The file contains a gallery of figures fromthe book. Clicking on the caption of an individual figure opens up a virtualreality modeling language (VRML) file allowing students to see and explorein a three-dimensional, virtual world. The use of the VRML files requires aVRML web browser plug-in. The plug-in can be downloaded, for example,at http://www.parallelgraphics.com/products/cortona/. An instructor’s re-source file containing detailed solutions to the problems presented at the endof each chapter can also be obtained through the publisher’s web site.

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Acknowledgments

I thank my past and present students and my colleagues at the Illinois Instituteof Technology, especially Sudhakar Nair, Kevin Cassel, and John Way, fortheir enthusiasm and insightful ideas that helped form my way of thinking overthe years. I also thank Ted Belytschko, Brian Moran, and Jan Achenbach atNorthwestern University for inspiring me to pursue a career in computationalmechanics. Finally, I would like to thank my wife Mary Lee, and childrenMadeline and David, for putting up with me during the past year.

About the author

Mike Gosz was raised in the small Wisconsin town of Kaukauna. He grad-uated from Marquette Univesity with a B.S. degree (summa cum laude) inmechanical engineering. During college, he spent two years working as amanufacturing engineer at a General Motors assembly plant in Janesville,Wisconsin. He received his M.S. degree in 1989 and Ph.D. in 1993, both intheoretical and applied mechanics, from Northwestern University.

He is currently an associate professor in the Mechanical, Materials, andAerospace Engineering Department at the Illinois Institute of Technology,where he has been on the faculty since 1996. His first faculty position (1993–1996) was as an assistant professor in the Mechanical Engineering Departmentat the University of New Hampshire.

Professor Gosz teaches a wide variety of undergraduate and graduate coursesin the areas of solid mechanics and computational mechanics. He also regu-larly teaches short courses in the finite element method to practicing engineersfrom companies such as Lockheed Martin, Caterpillar, BP, Motorola, Molex,just to name a few.

Professor Gosz has over 30 technical publications in the areas of finite el-ement methods, fracture mechanics, composite materials, and fluid structureinteraction. He received the IBM Faculty Development Award in 1994 fromthe T.J. Watson Research center in Yorktown Heights NY, and the RalphBarnett teaching award in 1998. He is a member of ASME and the AmericanAcademy for Mechanics.

1

Introduction

The finite element method made its debut after a series of papers was pub-lished by Turner in 1959 [1]. Before 1965, the title, “Finite Element Method,”did not even exist; the method was referred to as the direct stiffness method(DSM). For the most part, the method was confined to the structural me-chanics community and the aerospace industry. An excellent article on thehistory of the finite element method has been published by Felippa [2]. Sincethe genesis of the method during the early 1960s, thousands of articles havebeen published on the subject. Today, the method has reached such a stateof maturity that it is now thought of as a method for solving general fieldproblems in all areas of engineering and the physical sciences.

A modern definition of the finite element method might state that it issimply a numerical technique for obtaining approximate solutions to partialdifferential equations. To illustrate this idea, suppose that an engineer wantsto obtain the temperature distribution in a thin plate subjected to some sortof external heat source. To accomplish this task, the engineer must makea variety of assumptions, and develop an appropriate mathematical modelthat is both physically realistic (retains the essential physics in the problem)and is yet amenable to solution. As an example, the engineer may assumesteady-state conditions (the temperature field in the plate does not changewith time), the top and bottom of the plate are perfectly insulated, the heatconduction is isotropic (the material’s resistance to heat flow is the same inall directions), etc. Under such conditions, the temperature field in the platesatisfies the two-dimensional Laplace’s equation. That is,

∇2T = 0

or∂2T

∂X2+

∂2T

∂Y 2= 0 (1.1)

For simple geometries, a closed-form (analytical) solution to Laplace’s equa-tion can be obtained; however, for complicated geometries it is usually nec-essary to resort to numerical techniques. The main idea in the finite elementmethod is to discretize or “break up” the domain of interest into a collectionof points and subdomains called nodes and elements. This idea is depictedin Figure 1.1. Once the domain of interest is discretized, the field variableof interest, temperature in this case, is interpolated throughout the area or

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2 Finite Element Method

FIGURE 1.1

Finite element mesh.

volume of the element using simple polynomial interpolation functions calledshape functions. Loosely speaking, for linear field problems the process ofdiscretizing the domain into nodes and elements leads to a system of linearalgebraic equations that can be readily solved on a digital computer. For non-linear problems the technique is very much the same, except the equations aresolved in an iterative manner, with each iteration solving a system of linearalgebraic equations.

The finite element procedure

The finite element procedure involves three general phases: a pre-processingphase, an analysis phase, and a post-processing phase. The three phases areillustrated in Figure 1.2. Perhaps the most time consuming and difficult ofthe three phases is the pre-processing phase. Given a physical problem ofinterest, the engineer must use sound judgement to transform the real-worldproblem into a mathematical model that captures the essential physics andis amenable to numerical solution. After making reasonable assumptions andcoming up with a suitable mathematical model, usually involving a partial dif-ferential equation along with appropriate boundary conditions, he must thendecide what type of analysis to perform: linear, nonlinear, transient, steady-state, etc. He must also decide upon suitable material properties, choose theappropriate type of finite element, and then create a finite element mesh thatis sufficiently refined in regions where large gradients in the field variable areexpected. Mesh generation by itself is a time-consuming process. Fortunately,with the advent of modern commercial pre-processing software, the amountof time spent in the pre-processing phase has been cut considerably and madeeasier than ever before.

For the most part, the analysis phase of the finite element procedure islike feeding the data created in the pre-processing phase into a black box.During the analysis phase, a computer program reads the data and solves alinear system of algebraic equations, either once for linear problems, or manytimes (often iteratively) for nonlinear and/or dynamic problems. When theanalysis phase is finished, the results of interest such as nodal displacements

Introduction 3

Pre-processing phase Post-processing phase

Analysis phase

Kd = F

is σ ≤ σy?

FIGURE 1.2

Flow chart of the finite element procedure.

and temperatures are output to a data file or files. Element quantities suchas stress, strain, and heat flux can also be output at the user’s request.

In the final phase of the finite element procedure, the engineer faces the diffi-cult task of interpreting the results of the analysis phase. The post-processingphase has been made significantly easier throughout the years through thedevelopment of user-friendly software packages with excellent graphics capa-bilities. The post-processing phase still, however, requires a good workingknowledge of the physics of the problem of interest. For example, in a stressanalysis, the engineer may want to know if a solid body under a given set of ex-ternal loads will fail. To answer this question, the engineer must decide uponan appropriate failure criterion and then carefully investigate the stress fieldsin the solid. The engineer typically does this by observing color contour plots.At this time the engineer must assess whether or not the nice-looking contourplots make sense. In particular, he must be able to answer questions such as:Are the boundary conditions satisfied? Is the finite element mesh sufficientlyrefined? Was the assumption of linear material behavior appropriate? Is amore complicated analysis required?

Where do we go from here?

The finite element method is a rich and exciting subject. Learning the finiteelement method can be difficult, but ultimately very satisfying if we pay at-tention to the details of the method along the way. Let us now begin our


journey in the next chapter by studying the mathematical background thatis needed to obtain an excellent understanding of the finite element method.Starting the journey in this fashion may seem slow at first, but stay patient.You will be rewarded.

2

Mathematical Preliminaries

To climb steep hills requires a slow pace at first.—Shakespeare

2.1 Matrix algebra

A matrix is a tool for arranging numerical values and/or functions in anorganized manner. A matrix contains rows and columns. An element of amatrix is the number (or function) that occupies a specific row and column.As an example, consider the following matrix:

A =

1 34 25 6

(2.1)

In this example, and throughout the remainder of the book, we will use bold-face type to denote that the quantity A is a matrix. The elements of thematrix in this example are numbers ranging from one to six. The matrix hasthree rows and two columns. Hence, we say that the order of the matrix is3 × 2. Often it is required to carefully keep track of the order of matrices.Occasionally throughout the book, the order will be labeled underneath theboldface symbol, e.g.,

A3 × 2

Keeping track of the order of matrices is especially important when performingmatrix multiplication, which will be discussed below.

Often a matrix has the special property of being square and/or symmetric.A square matrix is a matrix with the same number of rows as columns. Asquare matrix is said to be symmetric if

Aij = Aji (2.2)

where the subscripts i and j represent the row and column number, respec-tively. The elements A11, A22, A33, etc. are said to lie on the diagonal ofthe matrix. The elements that lie above and below the diagonal are said tobelong to the upper and lower triangles, respectively.

5


2.1.1 Multiplication of a matrix by a scalar

One of the simplest operations that can be performed on a matrix is multipli-cation by a scalar. A scalar is a quantity that can be characterized by a singlevalue at every point in space. A scalar can be a function of position, but itdoes not have direction associated with it. Temperature is an example of ascalar quantity. Throughout the book, scalar quantities will be denoted byeither lowercase alphabetic or lowercase Greek letters, unless noted otherwise.Multiplication of the matrix, A, by a scalar, α, can be defined as follows. If

C = αA (2.3)

then the elements of C areCij = αAij (2.4)

The elements of the matrix C are obtained by multiplying each element ofthe matrix A by the scalar α.

2.1.2 Transpose of a matrix

The transpose of a matrix is obtained by interchanging its rows and columns.For example, if a matrix A is given as

A =

1 35 60 1

(2.5)

then

AT =

[

1 5 03 6 1

]

(2.6)

where the superscript T is used to denote the transpose. The order of thematrix A in the above example is 3 × 2, whereas the order of AT is 2 × 3.

2.1.3 Matrix multiplication

Suppose that the matrix C is defined as the product of two matrices A andB, i.e.,

C = AB (2.7)

Matrix multiplication is said to be conformable if the number of columns inthe matrix A equals the number of rows in the matrix B. Let’s suppose thatthe order of A is 2 × 3. In order for the multiplication to be conformable,the matrix B must be of order 3 × 2. An easy way to check whether or nota matrix multiplication is conformable is to write down the order of the twomatrices below each matrix. So in the example above, one could write

C3 × 3

=A

3 × 2B

2 × 3(2.8)

Mathematical Preliminaries 7

Focusing on the right-hand side of equation (2.8), if the inner dimensions areequal, then the multiplication is conformable. The outer dimensions indicatethe order of the resulting matrix, C, which is 3 × 3 in this case.

Provided that the matrix multiplication is conformable, the elements of thematrix C are obtained as follows:

Cij =

K∑

k=1

AikBkj (2.9)

where K is the number of columns in A (or number of rows in B).In order to demonstrate a hand calculation, consider the following matrices

A and B:

A =

1 34 25 6

; B =

[

1 3 24 2 1

]

. (2.10)

Using equation (2.9) to calculate the product of A and B yields

C11 = A11B11 + A12B21

= (1)(1) + (3)(4)

= 13

C12 = A11B12 + A12B22

= (1)(3) + (3)(2)

= 9

etc. (2.11)

The result is

C =

13 9 512 16 1029 27 16

(2.12)

To conclude this section, we emphasize that matrix multiplication, in gen-eral, is not commutative. That is, AB 6= BA. A useful identity, however, thatinvolves taking the transpose of the product of two matrices can be stated asfollows:

(AB)T = BT AT (2.13)

This identity will be useful later on when it is time to substitute finite elementshape functions into the weak form, discussed in Chapter 3.

2.1.4 The identity matrix

Throughout the book, the bold symbol I will be used to denote the identitymatrix. The identity matrix is a square matrix whose elements are one in thediagonal positions and zero everywhere else. The components of the identity


matrix can be written succinctly using the Kronecker delta symbol. TheKronecker delta symbol is defined as

δij =

1 : if i = j

0 : if i 6= j

So, for example, the 3 × 3 identity matrix is written as

I3 × 3

=

1 0 00 1 00 0 1

2.1.5 Determinant of a matrix

Throughout the book, the determinant of a square matrix, A, will be denotedas detA. Suppose that A is a 2× 2 matrix. The determinant of A is definedas

detA

2 × 2=

∣

∣

∣

∣

A11 A12

A21 A22

∣

∣

∣

∣

= A11A22 − A12A21 (2.14)

Once we have established the definition (2.14), the determinant of higher ordermatrices can be defined as

detA

n × n=

n∑

j=1

AijCij (2.15)

Here in equation (2.15), the subscript i identifies the row of the matrix. Itcan take on any value between 1 and n. The matrix Cij is called the cofactormatrix. The elements of the cofactor matrix are defined as

Cij = −1(i+j)Dij (2.16)

where Dij is the determinant of the sub-matrix of C that is obtained by cross-ing out row i and column j. As an example, let us compute the determinantof the following 3 × 3 matrix:

A =

1 2 32 4 −23 −2 1

(2.17)

Letting i = 1 in equation (2.15), we obtain

detA = A11C11 + A12C12 + A13C13

= 1

∣

∣

∣

∣

4 −2−2 1

∣

∣

∣

∣

− 2

∣

∣

∣

∣

2 −23 1

∣

∣

∣

∣

+ 3

∣

∣

∣

∣

2 43 −2

∣

∣

∣

∣

= −64 (2.18)


Undergraduate engineering students usually learn the operation carried outin (2.18) when evaluating the cross product of two vectors. In that case,the first row of A contains three unit vectors i, j, and k. Computing thedeterminant of any matrix larger than 3× 3 by hand is very time consuming,because the operation requires a large number of multiplications. Fortunately,the determinant of a matrix is rarely explicitly computed in finite elementprocedures. On the other hand, having the ability to perform a quick handcalculation to find the determinant of small (2× 2 or 3× 3) matrices is usefulfor finding the inverse of these matrices required in some of the problemspresented throughout the book. The inverse of a square matrix is defined inthe next section. If the determinant of a square matrix is found to be zero,that matrix is said to be singular.

2.1.6 Inverse of a matrix

When a matrix A is multiplied by its inverse, A−1, the result is the identitymatrix, i.e.,

AA−1 = A−1A = I

We emphasize that only square matrices have an inverse, and the inverse ofa singular matrix does not exist. The inverse of a square matrix, A, can befound from the following formula:

A−1 =1

detACT (2.19)

where CT is the transpose of the cofactor matrix. As an example, we canquickly compute the inverse of a 2 × 2 matrix as follows. Suppose

A =

[

2 −1−1 1

]

It can be easily seen that detA = 1. The cofactor matrix is

C =

[

1 11 2

]

Notice that in this example, the cofactor matrix happens to be symmetric,i.e., C = CT , and hence

A−1 =1

1

[

1 11 2

]

=

[

1 11 2

]

We can check our work by making sure that A−1A = I, i.e.,[

2 −1−1 1

] [

1 11 2

]

=

[

1 00 1

]


So for 2 × 2 symmetric matrices, the inverse is obtained by flip-flopping thediagonal terms, switching the sign of each off-diagonal term, and dividing eachelement by the determinant of the original 2 × 2 matrix.

2.1.7 Linear algebraic equations

Consider the following system of three equations:

x1 − x2 = 10

−x1 + 2x2 − x3 = 0

−x2 + x3 = 1 (2.20)

In each equation above, the unknowns, x1, x2, and x3 appear linearly. Thismeans, for example, that the unknowns are not raised to a power, and theydo not appear as arguments in nonlinear functions such as sine and cosine.Solving systems of linear algebraic equations is the major computational taskin every finite element analysis. Even nonlinear finite element procedures boildown to solving linear algebraic equations, albeit over and over.

A method for solving systems of linear algebraic equations that is amenableto numerical computation employs matrix methods. To illustrate how theprocedure works, let us write the system of equations (2.20) in the form Ax =b as follows:

1 −1 0−1 2 −1

0 −1 1

x1

x2

x3

=

001

(2.21)

If we multiply both sides of (2.21) by A−1 we obtain

A−1Ax = A−1b =⇒

Ix = A−1b =⇒

x = A−1b

The system of equations can be solved in the above fashion, as long as thedeterminant of the matrix A is not zero. If detA = 0, then A is singular andthere would be infinitely many vectors x that satisfy equation (2.20).

Let us now proceed and attempt to solve the matrix problem. The deter-minant of the matrix A is computed as

detA = 1

∣

∣

∣

∣

2 −1−1 1

∣

∣

∣

∣

+ 1

∣

∣

∣

∣

−1 −10 1

∣

∣

∣

∣

+ 0

∣

∣

∣

∣

−1 20 −1

∣

∣

∣

∣

= 0

Because detA = 0, we cannot solve for the vector x. Let us suppose, however,that the value of the variable x1 is given. This is referred to as a constraintor boundary condition. For the moment, let us take x1 = 1. Because x1 isknown, the matrix problem (2.20) can be reduced to two equations with only


two unknowns. To obtain the reduced problem, we expand row two and rowthree as follows:

−1(1) + 2x2 − x3 = 0

−x2 + x3 = 1 (2.22)

We now write (2.22) back into matrix form, obtaining[

2 −1−1 1

]

x2

x3

=

11

(2.23)

The resulting 2 × 2 system of equations can now be solved, because the de-terminant of the matrix on the left-hand side is no longer zero. Becausethe constraint was imposed on x1, the reduced system of equations could alsohave been obtained by simply crossing out the first row and first column of theoriginal matrix problem, and modifying the right-hand side (due to the factthat the prescribed value of x1 was nonzero). If the problem were changed byinsisting instead that x1 = 0, the reduced system could be obtained by cross-ing out the first row and column, but leaving the right-hand side unaltered.Finally, solving the reduced system, we get

x2

x3

=

[

1 11 2

]

11

=

23

(2.24)

2.1.8 Integration of matrices

Sometimes it is convenient to store functions of one or several variables ina matrix. The definite integral of a matrix is obtained by integrating eachelement (function) of the matrix. The result is another matrix of the sameorder. For example, suppose that the matrix A is defined as

A =

x2 0 x3

x x4 x + 20 1 x5

and we wish to evaluate the following integral:

∫

1

0

Adx (2.25)

The result is obtained by integrating each of the nine elements of A withrespect to x from zero to one. The result is

∫

1

0

Adx =

1/3 0 1/41/2 1/5 5/2

0 1 1/6


Integration of matrices is a common operation in the derivation of finite el-ement matrices, discussed in Chapter 3. Because the operation typically re-quires the evaluation of many individual integrals, a symbolic math softwarepackage is a very useful tool for integrating matrices.

2.2 Vectors

A vector is a physical quantity that has both magnitude and direction. Forceis an example of a vector quantity. Both magnitude and direction are neededto completely describe it. A vector u is illustrated in Figure 2.1. The X1,X2, X3 axes labeled in the figure form what is called a Cartesian coordinatesystem. There are three other vectors shown in the figure. These are labelede1, e2, and e3. They represent unit vectors (otherwise known as base vectors)pointing in the X1, X2, and X3 directions, respectively. It turns out that anyvector u can be expressed in terms of the base vectors as follows:

u = u1 e1 + u2 e2 + u3 e3 (2.26)

X1

X3

X2

u

u3

u1

u2

e1

e2

e3

FIGURE 2.1

Cartesian components of a vector.


The quantities u1, u2, and u3 in equation (2.26) are called components of u.Each component represents the the projection of u onto one of the coordinateaxes. For example, the component u1 is the projection of u onto the X1 axis.Note that the components of a vector are scalar quantities; they do not havedirection associated with them.

2.2.1 Vector dot product

The dot product between two vectors u and v is defined as

u · v = |u||v| cos θ (2.27)

where |u| and |v| are the magnitudes of the two vectors, and cos θ is the cosineof the angle formed between them. Alternatively, the vector dot product canbe written as

u · v =

3∑

i=1

uivi = u1v1 + u2v2 + u3v3 (2.28)

The limit on top of the summation symbol on the right-hand side of (2.28)can be either two or three, depending on whether we are working in two-dimensional or three-dimensional space.

Throughout the book, we will adopt a convention called indicial notation.The use of indicial notation will allow us to write complex expressions verycompactly. One of the fundamental rules of indicial notation is that whenevertwo subscripts are repeated in a single expression, such as in equation (2.28)above, the summation symbol is omitted, but it is still implied. Hence, us-ing indicial notation, the dot product operation between two vectors can bewritten succinctly as

u · v = uivi (2.29)

Having introduced indicial notation, the vector u can now be expressed as

u = uiei (2.30)

The components of u are obtained by taking the dot product of u with eachof the base vectors, i.e.,

ui = u · ei (2.31)

Just to be completely clear, let us now carry out the operation spelled out inequation (2.31) above. Expanding the right-hand side, we obtain

u · ei = (uj ej) · ei

= uj(ej · ei)

= ujδji

= u1δ1i + u2δ2i + u3δ3i

= ui (2.32)


There are several subtle points that should be mentioned concerning the oper-ations carried out in (2.32) above. First of all, in the first step, we were carefulnot to let a subscript (or index) appear more than twice in each expression. Ifan index appears more than twice in a single expression, then that expressionbecomes ambiguous. In the second step, notice that the dot product betweenthe two base vectors ei and ej gives the Kronecker delta. This is becausethe dot product between distinct base vectors is zero, and the dot productbetween any base vector and itself is one. The final result in (2.32) stemsfrom the fact that the subscript i is a free index. A free index can take on anyvalue ranging from one to three (in three-dimensional space). So notice thatthe equality stated in the last line holds for any value of i, due to the specialproperty of the Kronecker delta. From now on, if we see an expression likeuiδij , we will simply replace the repeated index, i, with the free index, j, andget rid of the Kronecker delta symbol, i.e., uiδij = uj .

2.2.2 Vector cross product

Another very important operation with vectors is the vector cross product.If we consider any two vectors u and v, the vector cross product u × v givesanother vector that points in the direction (according to the right-hand rule)that is perpendicular to the common plane in which both vectors lie. Theresult is another vector, c say, whose magnitude is

|c| = |u||v| sin φ (2.33)

where φ is the angle formed between u and v. The vector c can be obtainedby evaluating the following determinant

c =

∣

∣

∣

∣

∣

∣

e1 e2 e3

u1 u2 u3

v1 v2 v3

∣

∣

∣

∣

∣

∣

(2.34)

where e1, e2, and e3 are unit base vectors. Expanding the determinant (2.34)we get

c = (u2v3 − u3v2) e1 − (u1v3 − u3v1)e2 + (u1v2 − u2v1)e3 (2.35)

Another convenient way to express the components of c is through the use ofthe permutation symbol. In order to define the components of the permutationsymbol, it is helpful to look at Figure 2.2. If we travel counterclockwise aroundthe circle, we would pass the numbers (depending on the starting location)(2, 1, 3), (3, 2, 1), and (1, 3, 2). These numbers inside the parentheses are calledodd permutations of the indices (i, j, k). On the other hand, if we travelclockwise, we would pass (1, 2, 3), (2, 3, 1), and (3, 1, 2). These are called even

permutations of (i, j, k). Having defined odd and even permutations of the


3

2

1

3

2

1

FIGURE 2.2

Odd and even permutations of (1, 2, 3).

indices, the permutation symbol can now be defined as

ǫijk =

−1 : if i, j, k are odd permutations of 1, 2, 3

0 : if any i, j, k are equal

+1 : if i, j, k are even permutations of 1, 2, 3

So, for example, ǫ123 = 1, ǫ213 = −1, ǫ113 = 0. It turns out that the compo-nents of c can be expressed in terms of the permutation symbol as follows:

ci = ǫijkujvk (2.36)

Expanding equation (2.36) above gives

c1 = ǫ123u2v3 + ǫ132u3v2

= u2v3 − u3v2

c2 = ǫ213u1v3 + ǫ231u3v1

= −u1v3 + u3v1

c3 = ǫ312u1v2 + ǫ321u2v1

= u1v2 − u2v1

Note that the components c1, c2, and c3 above are precisely the same compo-nents as those given in (2.35).

2.2.3 Vector transformation

Often it is necessary to obtain the components of a vector in a new coordinatesystem that differs from the fixed Cartesian coordinate system by a rotationof the coordinate axes. The idea of forming a new set of axes by rotating thefixed Cartesian axes is shown in Figure 2.3. We will refer to the new set ofaxes as the primed set. The primed set is identified by the dashed lines in thefigure. The base vectors that are associated with the primed set of axes are e′

1,

e′2, and e′

3, or simply as e′

i. Now let us focus on the vector u. The projection

of u onto each of the primed axes X ′1, X ′

2, and X ′

3yields the components of


X1

X2

e1

e2

e′1

e′2

u

u′1

u′2

X ′1

X ′2

FIGURE 2.3

Components of a vector in a rotated (primed) coordinate system.

the vector in the primed coordinate system. These primed components canbe obtained by taking the dot product of the vector u with each of the primedbase vectors, i.e.,

u′i= u · e′

i(2.37)

We emphasize here that the vector u is the same vector, no matter whatcoordinate system its components are described in. We can express the vectorin terms of the primed base vectors,

u = u′ie′

i(2.38)

or, alternatively, we can express it in terms of the unprimed base vectors,

u = uiei (2.39)

If we now substitute the expression (2.39) into equation (2.37), we obtain anexpression for the primed components of the vector u in terms of the unprimedcomponents,

u′i= (ujej) · e

′i

= uj(ej · e′i)

= (e′i· ej)uj (2.40)

The quantity e′i· ej gives the projection of the primed base vectors onto the

unprimed axes. If we leta

j

i= e′

i· ej (2.41)

then equation (2.40) can be written as

u′i= a

j

iuj (2.42)


The inverse of the relation (2.42) gives the unprimed components ui in termsof the primed components, i.e.,

ui = (u′ke′

k) · ei

= ai

ku′k

= ai

ju′j

(2.43)

2.3 Second-order tensors

While the physical properties of a vector (magnitude and direction) are easyto think about, the physical meaning of a second-order tensor is a little moredifficult to grasp. Mathematically speaking, a second-order tensor is a func-tion that operates on a vector to produce another vector. This idea can bedepicted by the following equation:

v −→ T −→ u (2.44)

In order to clearly define the conceptual operation above, we need to be ableto express the components of the tensor T in terms of the base vectors e1,e2, and e3. It turns out that any second-order tensor, T, can be expressed interms of its components and base vectors as follows:

T = Tijeiej (2.45)

The quantity eiej involving the two base vectors lying side by side is called adyadic product. Expanding (2.45) above we get

T = T11e1e1 + T12e1e2 + T13e1e3

+T21e2e1 + T22e2e2 + T23e2e3

+T31e3e1 + T32e3e2 + T33e3e3 . (2.46)

So it is now clear that a second-order tensor has nine components. The ninecomponents can be conveniently arranged in matrix form as follows:

[T] =

T11 T12 T13

T21 T22 T23

T31 T32 T33

(2.47)

The tensor operation (2.44) is very much like the vector dot product. Theresult, however, is a vector instead of a scalar. The operation can be moreformally written as

T · v = u (2.48)


By substituting the expression (2.45) into equation (2.48) we obtain

u = (Tijeiej) · (vkek) (2.49)

The operation defined above is an example of what is called a tensor product.In order to perform the operation, we treat the components Tij and vk asscalars. These can be moved around as follows:

u = (Tijvkei)(ej · ek) (2.50)

Next, in equation (2.50) above, the usual vector dot product operation iscarried out for the two base vectors ej and ek. Doing this we get

u = Tijvkδjkei = Tikvkei (2.51)

Hence, the components of the resulting vector u are

ui = Tikvk = Tijvj (2.52)

The components Tij of any second-order tensor T can be obtained as fol-lows:

Tij = ei · T · ej

= ei · (Tklekel) · ej

= (ei · ek)Tkl(el · ej)

= δikTklδlj = Tij (2.53)

From equation (2.53) we can see that the elements in the first row of thematrix (2.47) are actually components of a the vector e1 ·T. The second andthird rows are the components of e2 · T, and e3 · T respectively.

One of the fundamental second-order tensors in the study of mechanics isthe stress tensor σ. When we take the tensor product e1 · σ of the stresstensor and the base vector e1, the result is a vector having units of force perunit area. This vector acts on a plane whose unit outward normal∗points inthe direction of the X1 axis. The vector has components σ11, σ12, and σ13.Physically, these components represent the normal and shear stresses actingon the plane perpendicular to the X1 axis.

Example 2.1

In continuum mechanics, the components of second-order tensors are conve-niently stored in 3×3 matrices. The determinants of such matrices often have

∗A unit outward normal is a unit vector passing through a point on a surface that isperpendicular to the surface and is directed out of the body


a

b

ca× b

FIGURE 2.4

Volume of a parallelopiped, V = (a × b) · c.

physical significance. As an example, suppose that a 3 × 3 matrix A is givenas

A =

a1 a2 a3

b1 b2 b3

c1 c2 c3

(2.54)

Now let a, b, and c be vectors whose components are defined by the first,second, and third rows of A respectively, e.g.,

a = a1 e1 + a2 e2 + a3 e3 = ai ei

b = b1 e1 + b2 e2 + b3 e3 = bi ei

c = c1 e1 + c2 e2 + c3 e3 = ci ei

Having defined these vectors, it can be easily shown that the determinant ofthe matrix A can be expressed as

detA = (a × b) · c

= (c × a) · b

= (b × c) · a (2.55)

The operation carried out in each line of (2.55) is called a scalar triple prod-uct. Referring to Figure 2.4, we can see that each triple product (and hencedetA) defines the volume of a parallelopiped whose edges are defined by thevectors a, b, and c.† Another important observation that we can make hereis that if the vectors a, b, and c all point in the same direction, the resultingscalar triple product, and hence detA, is zero. This will have important im-plications when we study eigenvalues and eigenvectors of matrices in Section2.3.2.

†a×b gives a vector whose magnitude is the area of the base of the parallelopiped. Taking

the dot product of this vector with c yields the base times the height.


The scalar triple product can be expressed in indicial notation as

(b × c) · a = (ǫijkbjck ei) · (al el)

= ǫijkbjckalδil

= ǫijkbjckai

= ǫijkaibjck. (2.56)

As a final remark, if Aij are the components of A, we can let ai = A1i,bj = A2j , and ck = A3k which yields an alternative expression for detA interms of the permutation symbol,

detA = ǫijkA1iA2jA3k (2.57)

2.3.1 Tensor transformation

One of the definitions of a second-order tensor states that if a physical quantitytransforms under a rotation of the coordinate axes according to the tensortransformation law, then that physical quantity is a second-order tensor. Thisdefinition is akin to saying if something walks like a duck, then it is a duck.

We emphasize here that a tensor is a physical quantity independent of thecoordinate system to which its components are referred. Because of this wecan write the tensor T in terms of base vectors in the primed and unprimedcoordinate systems as follows:

Tij = ei · T · ej

T′ij

= e′i· T · e′

j(2.58)

Using equation (2.45) along with (2.58) above, we can write the primed com-ponents of the tensor T in terms of the unprimed components. Doing this weget

T′ij

= e′i· (Tkl ekel) · e

′j

= ak

ia

l

jTkl

= ap

ia

q

jTpq, (2.59)

and vice versa,

Tij = ei · (T′kl

e′ke′

l) · ej

= ai

ka

j

lT

′kl

(2.60)

Equation (2.59) (or equivalently (2.60)) is referred to as the tensor transfor-mation law. If the components of a physical quantity transform according toone of these relations, then that quantity is a second-order tensor.


Example 2.2

As an example, suppose that the components of a second-order tensor in theunprimed coordinate system are given as

T =

10 −2 0−2 20 0

0 0 1

(2.61)

Now suppose we want to find the components of this tensor in a primedcoordinate system that is obtained by rotating the unprimed axes about theX3 axis by an amount θ = π/4. The components, a

j

i, of the transformation

tensor can now be calculated as follows:

a1

1= e′

1· e1 =

√2/2

a2

1= e′

1· e2 =

√2/2

a3

1= e′

1· e3 = 0

a1

2= e′

2· e1 = −

√2/2

a2

2= e′

2· e2 =

√2/2

a3

2= e′

2· e3 = 0

a1

3= e′

3· e1 = 0

a2

3= e′

3· e2 = 0

a3

3= e′

3· e3 = 1 (2.62)

It is convenient to store the components aj

iin a 3 × 3 matrix Q, i.e.,

Q =

√2/2

√2/2 0

−√

2/2√

2/2 00 0 1

(2.63)

The tensor transformation operation (2.59) can now be performed by carryingout the following matrix multiplications:

T′ = QTQT

=

√2/2

√2/2 0

−√

2/2√

2/2 00 0 1

10 −2 0−2 20 0

0 0 1

√2/2 −

√2/2 0

√2/2

√2/2 0

0 0 1

=

13 5 05 17 00 0 1

(2.64)

Keep in mind that while the components of a second-order tensor are differentdepending on what coordinate system they are referred to, the second-ordertensor itself is still a physical quantity that is independent of the coordinatesystem to which it is referred, just like a vector.


2.3.2 Eigenvalues and eigenvectors of second-order tensors

Recall from the definition of a second-order tensor, T, that if we take thetensor product of T and some vector u, the result is another vector, say v.In general, both the magnitude and direction of v are not the same as thoseof u.

Now suppose that we try to find a unit vector n, such that when we takethe tensor product of T and n, we get another vector that points in thesame direction as n (but with a different magnitude). This statement can beexpressed mathematically as

Tn = λn (2.65)

or in component formTijnj = λni (2.66)

In other words, we are looking for a unit vector n such that T · n gives avector that is proportional to n. The constant of proportionality, λ, is calledan eigenvalue of T.

By performing some algebraic manipulations, equation (2.66) can be writtenas

Tijnj = λnjδij ⇒

(Tij − λδij)nj = 0 (2.67)

To see things more clearly, let us now write equation (2.67) in matrix notation.Doing this gives

T11 − λ T12 T13

T21 T22 − λ T23

T31 T32 T33 − λ

n1

n2

n3

=

000

. (2.68)

Recalling our discussion in Section 2.3, it is now enlightening to view the el-ements in each row of the matrix on the left-hand side of (2.68) as componentsof three vectors a, b, and c, e.g.,

a = (T11 − λ) e1 + T12 e2 + T13 e3 = aiei

b = T21 e1 + (T22 − λ) e2 + T23 e3 = bi ei

c = T31 e1 + T32 e2 + (T33 − λ) e3 = ci ei (2.69)

Using the definitions for a, b, and c above, equation (2.68) can now be recastas

a · n = 0

b · n = 0

c · n = 0 (2.70)

The only way that all three of the dot products in (2.70) can be zero (forarbitrary nonzero vectors n) is if all three vectors a, b, and c all point in the


same direction. If this is the case, the scalar triple product (a×b) · c is zero,and hence the following determinant must also be equal to zero:

∣

∣

∣

∣

∣

∣

T11 − λ T12 T13

T21 T22 − λ T23

T31 T32 T33 − λ

∣

∣

∣

∣

∣

∣

= 0. (2.71)

Equation (2.71) can also be written in component form as

|Tij − λδij | = 0 (2.72)

So it turns out that the eigenvalues of a second-order tensor T are the rootsof the cubic equation obtained by expanding the determinant equation (2.71)(or equivalently (2.72)). Because the equation is a cubic polynomial, thereare three eigenvalues. Once the eigenvalues are found, we can go back toequation (2.68) and solve for the eigenvectors, as demonstrated in the followingexample.

Example 2.3

It is desired to find the eigenvalues and eigenvectors of a second-order tensor,T, whose components are given as

T =

7 −1 2−1 6 4

2 4 8

(2.73)

To get the eigenvalues, we must solve the following equation:∣

∣

∣

∣

∣

∣

7 − λ −1 2−1 6 − λ 4

2 4 8 − λ

∣

∣

∣

∣

∣

∣

= 0 (2.74)

By expanding the determinant in equation (2.74) we get the following char-acteristic equation:

−λ3 + 21λ2

− 124λ + 176 = 0 (2.75)

Solving (2.75) for the three roots λ1, λ2, and λ3 yields

λ1 ≈ 11.36108849

λ2 ≈ 7.60076487

λ3 ≈ 2.03814664 (2.76)

To compute the eigenvectors, it is necessary to plug in the eigenvalues reportedin equation (2.76) above into the following matrix equation:

7 − λ −1 2−1 6 − λ 4

2 4 8 − λ

n1

n2

n3

=

000

(2.77)


and solving for n1, n2, and n3. Doing this for λ = 11.36108849 yields

−4.36108849n1 − n2 + 2n3 = 0

−n1 − 5.36108849n2 + 4n3 = 0

2n1 + 4n2 − 3.36108849 = 0 (2.78)

Next, we solve the second equation in (2.77) above for n1 in terms of n2 andn3, and plug the result back into the first equation. Doing this we get

22.38018131n2 − 15.44435396n3 = 0 ⇒

n2 = 0.6900906541n3

n1 = 0.300362937n3 (2.79)

Finally, we make the eigenvector a unit vector by insisting that nini = 1.Normalizing in this way we get

n(1)

1

n(1)

2

n(1)

3

=

0.239987730.551377250.79899249

, (2.80)

where the superscript refers to the fact that we have just solved for the com-ponents of the eigenvector associated with the largest eigenvalue. Followingthe same procedure for the other two eigenvalues we obtain

n(2)

1

n(2)

2

n(2)

3

=

0.89189235−0.45021797

0.04279981

;

n(3)

1

n(3)

2

n(3)

3

=

0.383319620.70234386

−0.59981594

(2.81)

2.3.3 Common operations with vectors and tensors

For the sake of completeness, in this section we will go over some common op-erations using tensor notation that arise in solid mechanics and heat transfer.The operations begin with the tensor product and then progress to the tensorscalar products. The concepts of symmetric and skew second-order tensorswill also be covered. This subsection ends with a table summarizing thesecommon operations.

Tensor product

The tensor product of any two second-order tensors, A and B, is defined as

A · B = (Aijeiej) · (Bklekel)

= AijBklδjkeiel

= AikBkleiel

= AikBkjeiej (2.82)


By carefully following each step in equation (2.82) above, one will noticethat the result of the tensor product operation is another tensor, C, whosecomponents are defined as

Cij = AikBkj (2.83)

The operation involves writing down, side by side, both tensors A and B interms of the Cartesian base vectors, and then taking the dot product of theinner two base vectors.

Tensor scalar product

As the title suggests, the tensor scalar product between two second-ordertensors produces a scalar. The operation is very common in the derivation offinite element equations. There are actually two tensor scalar products thatwe can define. The first scalar product is written as

α = A : B (2.84)

and the second is written asβ = A · ·B (2.85)

In each case, the result is a scalar, but not necessarily the same. To describethe operation, the scalar product (2.84) is obtained by first writing down, sideby side, the two tensors in terms of the Cartesian base vectors, making surethat a subscript does not appear more than twice in the expression. In thesecond step we take the dot product of the left base vector associated with A

and the left base vector associated with B. Multiplying everything togethergives the desired scalar α, i.e.,

α = A : B

= (Aijeiej) : (Bklekel)

= AijBklδikδjl

= AilBil

= AikBik (2.86)

The second scalar product (2.85) is very similar to the first, except we takethe dot product of the left base vector associated with A with the right basevector associated with B. The result is

β = A · ·B

= (Aijeiej) : (Bklekel)

= AijBklδilδjk

= AikBki (2.87)

As an example of the physical significance of this operation, the tensor scalarproduct between the Cauchy stress tensor and the small-strain tensor givestwice the strain energy density (recoverable elastic energy per unit volume)at a point in a linearly elastic solid.


Symmetric and skew second-order tensors

It turns out that any second-order tensor, T, can be decomposed into symmet-ric and skew (antisymmetric) parts. The symmetric part, Tsym, is obtainedby calculating the average of the tensor T and its transpose as follows:

Tsym =1

2

(

T + TT)

(2.88)

The skew part, Tskew, is obtained by subtracting TT from T as follows:

Tskew =1

2

(

T − TT)

(2.89)

Note that adding the symmetric and skew parts of T yields the tensor T, i.e.,

Tsym + Tskew =1

2

(

T + TT)

+1

2

(

T − TT)

=1

2

(

T + TT + T − TT)

= T (2.90)

An important operation that comes up frequently in the derivation of finite ele-ment equations is the tensor scalar product between a symmetric second-ordertensor, D, and another second-order tensor T, not necessarily symmetric. Wecan easily show that

D : T = D : Tsym

as follows:

D : Tsym =1

2D :

(

T + TT)

=1

2(DikTik + DikTki)

=1

2(DikTik + DkiTki)

=1

2(DikTik + DikTik)

= DikTik

= D : T (2.91)

Note that we used the that fact that Dki = Dik in the third line of (2.91)above since D is symmetric. Note that in the fourth line we just replaced thedummy index k with i and vice versa.

We end this section by listing some common tensor and vector operations inTable 2.1 below. The last row in Table 2.1 requires some further explanation.The identity can be proved as follows:

(A · B) : C = (AikBkj eiej) : (Cmn emen)

= AikBkjCmnδimδjn

= AikBkjCij (2.92)


On the other hand,

A : (C · BT ) = (Aij eiej) : (CklBml ekem)

= AijCklBmlδikδjm

= AkmCklBml

= AikCijBkj

= AikBkjCij (2.93)

and

B : (AT· C) = (Bij eiej) : (AlkClm ekem)

= BijAlkClmδikδjm

= AliBimClm

= AliBijClj

= AlkBkjClj

= AikBkjCij (2.94)

The fact that the last lines in equations (2.92), (2.93), and (2.94) are all equalproves the identity.

TABLE 2.1

Common operations with vectors and tensors.

Description Symbol Result

Vector dot product a · b

= (aiei) · (bkek)= aibkδik

= aibi

Magnitude of a vector |v| =√

vivi

Scalar triple product a × b · c = ǫijkaibjck

Trace of a tensor trT= T : I= Tikδik

= Tkk

Tensor transpose TT = Tijejei = Tjieiej

Determinant detT = ǫijkT1iT2jT3k

A useful identity A · B : C = A :(

C · BT)

= B :(

AT · C)


2.4 Calculus

The calculus is the fundamental mathematical tool that has allowed numericaltechniques such as the finite element method to come into being. This sectioncovers the essential ingredients that are needed for the development of bothlinear and nonlinear finite element procedures.

2.4.1 Fundamental theorem of calculus

One of the most important theorems that is presented in a first course incalculus is the fundamental theorem of calculus. The theorem tells us thatthe definite integral of a function f(x) from a to b is obtained by finding theantiderivative, F (x), of the function and evaluating the difference, F (b)−F (a),i.e.,

∫

b

a

f(x) dx = F (b) − F (a) (2.95)

In order for the theorem (2.95) to be applicable, the function F (x) must becontinuous in the interval a ≤ x ≤ b.

An informal proof of the theorem can be illustrated as follows. Considerthe plot of some function F (x) as shown in Figure 2.5. The function f(x) rep-resents the slope of the function F (x) at each point x. The quantity f(x)∆x

represents the change in F (x) due to a movement ∆x in the x direction. Thesummation of all the changes (in the limit as ∆x → 0) is by definition thedefinite integral of f(x) from a to b. It gives the total change in F from point

a to point b. Hence∫

b

af(x) dx = F (b) − F (a).

2.4.2 Integration by parts

When deriving finite element equations, it is necessary to evaluate integralsthat have the following form:

∫

b

a

f(x)g(x) dx (2.96)

A technique that can be used to evaluate such integrals is called integrationby parts.

The first step in the procedure involves taking the derivative of the productof F (the antiderivative of f) and the function g as follows:

d

dx[F (x)g(x)] = [F (x)g(x)]

′

= f(x)g(x) + F (x)g′(x) (2.97)


x

F (x)

F ′(x)∆x = f(x)∆x

a b

∆x

FIGURE 2.5

Fundamental theorem of calculus.

and hence,

f(x)g(x) = [F (x)g(x)]′− F (x)g′(x) (2.98)

where we have used the product rule to get the last line in (2.97).The second step is to replace f(x)g(x) in (2.96) with the right-hand side of

(2.98). Doing this we get

∫

b

a

f(x)g(x) dx =

∫

b

a

[F (x)g(x)]′dx −

∫

b

a

F (x)g′(x) dx (2.99)

Next, we apply the fundamental theorem of calculus introduced in the previoussection to the first integral on the right-hand side of (2.99) yielding

∫

b

a

[F (x)g(x)]′dx = F (x)g(x)|

b

a

= F (b)g(b) − F (a)g(a) (2.100)

Notice that the antiderivative of [F (x)g(x)]′

is simply F (x)g(x). Finally,combining (2.99) and (2.100) gives

∫

b

a

f(x)g(x) dx = −

∫

b

a

F (x)g′(x) dx + F (b)g(b) − F (a)g(a) (2.101)

Integration by parts is used in the development of the finite element method.Specifically, it comes into play when obtaining the weak form of the differentialequation of interest. This will be discussed at great length in the next chapter.


2.4.3 Taylor series expansion

A key mathematical concept behind modern nonlinear finite element proce-dures is the Taylor series expansion. Engineering students are usually intro-duced to this concept in a first course in calculus. Although powerful, theidea is quite simple, as will be explained below.

We begin by assuming that some function, f(x), can be expressed in theform of a power series as follows:

f(x) = a0 + a1(x − x0) + a2(x − x0)2 + a3(x − x0)

3 + ... (2.102)

where a0, a1, a2, etc. are unknown constants, and x0 is some point alongthe x axis. Here we tacitly assume that f(x0) exists. Once the unknownconstants are found, the resulting expression is the Taylor series expansion ofthe function f(x) about the point x0.

The unknown constants can be found by repeatedly differentiating bothsides of equation (2.102). To begin, notice that if we let x = x0, all of theterms on the right-hand side vanish, except for a0. Hence we conclude thata0 = f(x0). Differentiating both sides of equation (2.102) with respect to x,and again setting x = x0, we get

f′(x0) = a1 (2.103)

Next, taking the second derivative of both sides of (2.102), we get

f′′(x0) = 2a2 (2.104)

Repeating this process by taking the third derivative etc. gives the entire ex-pansion

f(x) = f(x0)+ f′|x0

(x−x0)+f ′′|x0

2!(x−x0)

2 +f ′′′|x0

3!(x−x0)

3 + ... (2.105)

where all derivatives in equation (2.105) are evaluated at the point x0. For agiven value of x, if the right-hand side of equation (2.105) tends toward thecorrect value of f(x) as the number of terms in the series is increased, theseries is said to converge.

If we retain only the first two terms of the Taylor series expansion (2.105),the approximate representation for f(x) about x0 becomes

f(x) = f(x0) + f′|x0

∆x (2.106)

The right-hand side of (2.106) is the equation of a straight line that is tangentto the curve f(x) and passes through the point f(x0). The second term on theright-hand side is the approximate change in the function f(x) with respect toa movement to the left or right of the point x0. This “change” will be calledDf . We note that the approximate change Df becomes exact as we get closer


and closer to the point x0. The process of expanding a function in a two-term Taylor series is called linearization. The operator D is the differentialoperator and follows the same rules as standard differentiation. Of practicalimportance is the fact that it obeys the product rule, i.e.,

D(fg) = (Df)(g) + (f)(Dg) (2.107)

This property will aid in the linearization of functions that come from non-linear field problems discussed in Chapter 10.

2.4.3.1 Nonlinear functions of several variables

Now suppose that we have a function of two variables f(x, y). As in theprevious section, we will express this function as a power series as follows:

f(x, y) = a0 + a1(x − x0) + a2(y − y0)

+a3(x − x0)2 + a4(y − y0)

2

+a5(x − x0)3 + a6(y − y0)

3 + . . . (2.108)

If we evaluate both sides of equation (2.108) at the point (x0, y0), it is clearthat a0 = f(x0, y0). If we take the partial derivative of both sides with respectto x and evaluate the resulting expressions at the point (x0, y0) we get

a1 =∂f

∂x

∣

∣

∣

∣

(x0,y0)

(2.109)

Similarly, taking the partial derivative of both sides of (2.108) with respect toy we obtain

a2 =∂f

∂y

∣

∣

∣

∣

(x0,y0)

(2.110)

By continuing this process of taking partial derivatives of both sides of equa-tion (2.108) with respect to x and y we finally obtain

f(x, y) = f(x0, y0) +∂f

∂x(x − x0) +

∂f

∂y(y − y0)

+

(

∂2f

∂x2

)

2!(x − x0)

2 +

(

∂2f

∂y2

)

2!(y − y0)

2 + . . . (2.111)

In equation (2.111) above, it is understood that all partial derivatives areevaluated at the point (x0, y0).

In the solution of nonlinear finite element equations and nonlinear equationsof several variables, it is common practice to approximate the function ofseveral variables by the first three terms of the Taylor series (2.111), i.e.,

f(x, y) ≈ f(x0, y0) +∂f

∂x

∣

∣

∣

∣

(x0,y0)

∆x +∂f

∂y

∣

∣

∣

∣

(x0,y0)

∆y (2.112)


The Taylor series approximation (2.112) above for a function of two variablesis an equation for a plane that passes through point (x0, y0) and is tangent tothe surface z = f(x, y) at point (x0, y0). Again we can write

f(x, y) ≈ f(x0, y0) + Df (2.113)

where Df =∂f

∂x∆x+

∂f

∂y∆y is the approximate change in the function f(x, y)

as a result of moving away from the point (x0, y0). Again, the operator, D,obeys the usual rules of differentiation.

2.4.4 Gradient and divergence

In this section we define the gradient operator and explain what is meant bygradient of a scalar and gradient of a vector. We then address the concept ofdivergence.

To begin, let us consider a scalar function f(x, y, z) of three variables x andy and z. If we use a linear Taylor series to approximate the function f in thevicinity of some point P having coordinates (x0, y0, z0), we get

f(x, y, z) ≈ f(x0, y0, z0) +∂f

∂x∆x +

∂f

∂y∆y +

∂f

∂z∆z (2.114)

where it is understood that the partial derivatives in equation (2.114) areevaluated at (x0, y0, z0).

The partial derivatives of f with respect to x, y, and z can be viewed ascomponents of the vector ∇f , i.e.,

∇f =

∂f

∂x

∂f

∂y

∂f

∂z

(2.115)

Similarly, the increments ∆x = x−x0, ∆y = y−y0, and ∆z = z−z0 can alsobe interpreted as components of the vector ∆x. The Taylor series (2.114) cannow be written as

f(x, y, z) ≈ f(x0, y0, z0) + ∇f · ∆x (2.116)

From our work in the previous section, we recognize that the second term onthe right-hand side of (2.116) gives an approximation for the change in f dueto a movement in the direction of ∆x.

Now suppose that n is a unit vector that points in the direction of ∆x, andlet ǫ be an infinitesimal parameter. Then ǫn = dx is an infinitesimal vector


x

y

z

Ptangent plane

level surface

FIGURE 2.6

Level surface and tangent plane.

that points in the same direction as ∆x. Taking the dot product of the twovectors ∇f and dx we get

∇f · dx = ǫ∇f · n

= ǫDnf (2.117)

Here in equation (2.117), the quantity Dnf is called the directional derivativeof the function f in the direction of n. The directional derivative is the rateof change of f in the direction of n. Hence, ǫDnf gives the change in f dueto a small movement ǫ in the n direction. So we now see that the gradient ofa scalar function is a very useful quantity. From ∇f we can get the rate ofchange of f in any direction simply by computing the directional derivativeDnf defined in (2.117).

2.4.5 The level surface

Let us imagine for the moment that there is a physical scalar quantity, for ex-ample temperature, associated with each point that lies on the curved surfaceshown in Figure 2.6. If the temperature has a constant value, T0 say, at eachpoint on the surface, then the surface can also be represented mathematicallyas

T (x, y, z) = T0 (2.118)

Such a surface, on which the physical quantity does not change from point topoint, is called a level surface.

Now let n be any unit vector that lies in the tangent plane to the surfaceat point P . By definition of a level surface, the rate of change of the functionT (x, y, z) in the direction of n is zero, i.e.,

DnT = ∇T · n = 0 (2.119)


The statement (2.119) tells us that the gradient ∇T , evaluated at point P ,must be perpendicular to the level surface at point P .

Recognizing that the level surface T (x, y, z) = T0 comes from a fully three-dimensional temperature field, T (x, y, z), we also conclude from (2.119) thatthe gradient acts in the direction in which that rate of change of temperatureis maximum. To see this more clearly, if we march tangent to the level surface,the rate of change of temperature is zero. Because the rate of change is thetemperature gradient ∇T dotted with a unit vector n, that dot product ismaximum (and hence the rate of change of temperature) when n happens tobe parallel to ∇T .

2.4.6 Divergence theorem

Having defined the gradient of a scalar function, from now on, we can treatthe gradient operator as just another vector. Letting x = X1, y = X2, andz = X3, the gradient vector can be expressed as

∇ =∂

∂Xi

ei (2.120)

Taking the dot product of ∇ and some other vector, say u, gives what iscalled the divergence of u, i.e.,

∇ · u = (∂

∂Xi

ei) · (uj ej)

=∂uj

∂Xi

ei · ej

=∂uj

∂Xi

δij

=∂ui

∂Xi

= ui,i (2.121)

The divergence theorem is an extremely useful tool for deriving partial dif-ferential equations that arise from balance laws. It also is often used whenderiving finite element equations that arise from partial differential equations.The purpose of this section is to present an informal proof of the theorem.Hopefully, this will make the interested reader feel comfortable using the the-orem later on beginning in Chapter 4.

To begin, consider a square domain as shown in Figure 2.7. Let us assumethat the domain is of width ∆x, height ∆y, and thickness t (into the page).The vector q represents the flux of something (e.g., heat or mass) going intoand out of the square element as indicated by the arrows in the figure. We notethat the flux of a quantity is the amount of that quantity flowing through aunit area per unit time. The components of q in this two-dimensional situation


qy(y)

qy(y + ∆y)

qx(x) qx(x + ∆x)x

y

FIGURE 2.7

Flux going in and out of a square element.

are qx and qy. For now, we will assume that no flux can flow into or out ofthe page. The quantity qx(x + ∆x, y) is the flux passing through the rightface, and qx(x, y) is the flux passing through the left face. The quantitiesqy(x, y) and qy(x, y + ∆y) are the flux components in the y direction flowinginto the bottom face and out the top face, respectively. For the remainder ofthis section, we will think of q as being heat flux, the energy flowing througha unit area per unit time.

Having defined heat flux, let us now calculate the total energy per unittime, E, flowing out of the square element shown in Figure 2.7. To do this weneed to multiply the flux on each face by the cross-sectional area, and thenadd up the contributions from each face. Doing this yields

E = [qx(x + ∆x, y) − qx(x, y)] t∆y

+ [qy(x, y + ∆y) − qy(x, y)] t∆x (2.122)

We now let W be the total energy per unit volume flowing out of the square.The quantity W is obtained by dividing both sides of equation (2.122) by thevolume of the square element, t∆x∆y. The result, in the limit as ∆x and ∆y

go to zero, can be expressed as

W =∂qx

∂x+

∂qy

∂y(2.123)

The quantity W is called the energy density. From our work in the previoussection, the energy density can be expressed in terms of the gradient operatoras follows:

W = ∇ · q, (2.124)

which is the divergence of the vector q. We now see that the divergence of thevector q gives the total energy flowing out of an infinitesimally small region,per unit time, per unit volume.


e e + 1

e + M

FIGURE 2.8

Flux going in and out of square domain.

If the energy density is known at each point inside an arbitrary domain, Ω,bounded by the surface, Γ, the total energy flowing out of Ω can be writtenas

E =

∫

Ω

WdΩ

=

∫

Ω

∇ · qdΩ (2.125)

Let us now calculate the total energy per unit time leaving an arbitrarydomain Ω in a different way. For the sake of simplicity, the domain of interestwill be composed of an assemblage of square elements as shown in Figure 2.8.The square elements are arranged in M rows and N columns. Focusing onsome element e labeled in the figure, let xe

Iand xe

I+1represent the x positions

of the left and right faces. Similarly let ye

Jand ye

J+1denote the y positions

of the bottom and top faces. As long as no energy is created or destroyed,the energy flowing out of element e through edge xe

I+1must be equal to the

energy flowing into element e + 1 through edge xe+1

I. In other words, the

energy flowing out of an element through its right edge, must be equal to theenergy flowing into the adjacent element through its left edge. One can alsoargue that the energy flowing out of element e through the top edge ye

J+1

must be equal to the energy flowing into element e + M through the bottomedge y

e+M

J. Here, M is the number of elements in a given row. Following this

argument, if we perform an energy balance on the entire assemblage, on anelement-by-element basis, the only contributions of energy flowing into and


out of the domain are indicated by the solid arrows in Figure 2.8. All of thedashed arrows cancel out of the total energy balance!

Let us now add up the total energy flowing out of the domain, as indicatedby the solid arrows. For the bottom, right, top, and left edges we get:

E = −

M∑

e=1

qy(x, ye

1)t∆x +

N∑

e=1

qx(xe

M+1, y)t∆y

+

M∑

e=1

qy(x, ye

N+1)t∆x −

N∑

e=1

qx(xe

1, y)t∆y (2.126)

In the limit as the element size is shrunk to zero (and the number of elementsapproaches infinity) the energy balance statement (2.126) is equivalent to thefollowing contour integral:

E =

∫

Γ1+Γ2+Γ3+Γ4

q · ndΓ

+

∫

Γ

q · ndΓ (2.127)

where Γ1 to Γ4 in (2.127) represent the bottom, right, top, and left edges ofthe square domain, and Γ is the union of these four edges. In the derivation ofthe contour integral (2.127), we have used the following relationships betweenthe flux vector q and the unit vector n that acts perpendicular to each edge:

bottom edge: q · n = −qy

right edge: q · n = +qx

top edge: q · n = +qy

left edge: q · n = −qx

(2.128)

Finally, comparing equation (2.125) with equation (2.127) we arrive at thefollowing result known as the divergence theorem:

∫

V

∇ · qdV =

∫

Γ

n · qdΓ (2.129)

Equation (2.129) above can be written in a more general form as

∫

V

∇ · AdV =

∫

Γ

n · AdΓ (2.130)

where the quantity A can be a scalar, a vector, or a tensor. As long as thequantity A is “sufficiently differentiable,”‡ then the volume integral of the

‡see Malvern [4] for a more detailed discussion on the continuity requirements of A for thedivergence theorem to hold.


divergence of A is equivalent to the surface integral of n · A. The divergencetheorem is needed in the derivation of partial differential equations in engi-neering and physics, as well as in deriving the weak forms of such equationsnecessary for a finite element implementation. For additional reading on vec-tor calculus and the divergence theorem, see, e.g., Schey [5].

2.5 Newton’s method

Suppose we want to find the value of x that satisfies the following equation:

f(x) = 0 (2.131)

If f(x) is a nonlinear function of the variable x, finding the solution to equation(2.131) is not necessarily an easy task. A very powerful method for obtainingsolutions to nonlinear equations is Newton’s method. The first step is toapproximate the function f(x) by a linear Taylor series expansion. Equation(2.131) then becomes

f(x0) + f′|x=x0

(x − x0) = 0 (2.132)

where x0 is called an initial guess. Letting ∆x = (x − x0) and then solving(2.132) for ∆x we get

∆x = −f(x0)

f ′|x=x0

(2.133)

Once ∆x is obtained, a (hopefully) better estimate for the solution to (2.131)is x1 = x0 + ∆x. The quantity x1 can then be used as a next guess, andthe entire process can be repeated. The iterative procedure can be writtensuccinctly in two lines as follows:

∆xi = −f(xi)

f ′|x=xi

xi+1 = xi + ∆xi (2.134)

where i = 0, 1, 2, 3 . . . .Newton’s method for nonlinear functions of one variable can be described

graphically as shown in Figure 2.9. As shown in the figure, after makingthe initial guess, x0, one can check how much the function at this location,f(x0), deviates from zero. This difference represents the error associated withthe initial guess. The error is referred to as the residual. If the residual islarger than some user-defined tolerance, then the next guess, x1, is computedusing (2.134). The new residual, f(x1), is computed and checked against thetolerance. If it is still unacceptable, the process is repeated again. The process


x0 x1x2

f(x0)

f(x1)

f(x)

x

f(x2)

FIGURE 2.9

Newton’s method.

continues until the residual is small compared with the user-defined tolerance.Newton’s method for functions of one variable is illustrated in the followingexample.

Example 2.4

In this example, we consider the problem of a block of mass m attached toa nonlinear spring as shown in Figure 2.10. The coordinate z shown in thefigure is measured positive downward. The position z = 0 corresponds to theposition of the block when the spring is unstretched (i.e., when there is notension or compression in the spring).

z

FIGURE 2.10

Spring-mass example.


Now suppose that the tension in the spring, t, is related to its elongation,z, by the following equation:

t = 2z5 + 1000z0.6 (2.135)

We desire to find the equilibrium position of the spring when the block issubjected to the influence of gravity. To solve the problem we will take m =1.0 kg and g = 9.81m/s2. To be consistent with our units, it is understoodthat the elongation of the spring will be in meters, and the tension in thespring will be in Newtons.

Starting from some configuration in which the spring is stretched, equilib-rium of the block in the z direction requires that

−t + mg = 0

and hence2z

5 + 1000z0.6− 9.81 = 0 (2.136)

Letting f(z) = 2z5 + 1000z0.6 − 9.81 and f ′(z) = 10z4 + 600z−0.4 we cannow begin the Newton’s method iterations. A logical starting point is to pickz0 = 0 (the location of the mass when the spring is unstretched) as the initialguess. In this problem, however, the slope of the tension versus elongationcurve tends to infinity as z goes to zero. This guess would lead to numericaldifficulties when the algorithm attempts to divide by zero in the first iteration.To alleviate this problem, we will choose a small positive value z0 = 0.0001 asthe initial guess and proceed. The numerical results for the first five iterationsare tabulated in Table 2.2 below.

2.5.1 Nonlinear equations of several variables

Now consider two nonlinear functions f(x, y) and g(x, y). Suppose that wewish to find the values of x and y such that

f(x, y) = 0

g(x, y) = 0 (2.137)

Here we will illustrate Newton’s method for getting approximate solutions tothe system of equations (2.137) above. The first step involves expressing the

TABLE 2.2

Numerical results for spring-mass example.

i zi × 10−4 (m) f(zi) (N) f ′(zi) (N/m) ∆zi (m)

0 1.000000000 −5.828928294 23886.43023 2.440267648 × 10−4

1 3.440267648 −1.454829074 14571.83878 9.983840033 × 10−5

2 4.438651652 −0.074670893 13159.84655 5.674146179 × 10−6

3 4.495393114 −1.897806887 × 10−4 13093.15110 1.449465352 × 10−8

4 4.495538061 −1.223813939 × 10−9 13092.98224 9.347098441 × 10−14


functions f(x, y) and g(x, y) as linear Taylor series expansions. This resultsin the following two linear equations:

f(x0, y0) +∂f

∂x∆x +

∂f

∂y∆y = 0

g(x0, y0) +∂g

∂x∆x +

∂g

∂y∆y = 0 (2.138)

Here in equation (2.138), the initial guess is the ordered pair (x0, y0). It isunderstood that the partial derivatives in (2.138) are to be evaluated at theinitial guess. The quantities ∆x and ∆y are defined as

∆x = x − x0

∆y = y − y0

The two linearized equations (2.138) for f(x, y) and g(x, y) both representplanar surfaces. One planar surface is tangent to the surface z = f(x, y) atpoint (x0, y0), and the other is tangent to the surface z = g(x, y) at point(x0, y0). The intersection of each planar surface with the plane z = 0 formsa pair of intersecting straight lines. The point where the two straight linesintersect is an approximate solution to the system of equations (2.137). InNewton’s method, this approximate solution is subsequently used as the initialguess, and the process is repeated until the solution “converges” to the exactsolution.

To proceed further, let us now write the linearized system of equations(2.138) in matrix notation as follows:

∂f

∂x

∂f

∂y

∂g

∂x

∂g

∂y

∆x

∆y

=

−f(x0, y0)−g(x0, y0)

(2.139)

The 2× 2 matrix on the left-hand side of equation (2.140) above is called theJacobian matrix J. The iterative incremental equations can be written as

∂f

∂x

∂f

∂y

∂g

∂x

∂g

∂y

i

∆x

∆y

i

=

−f(xi, yi)−g(xi, yi)

i

(2.140)

where i = 0, 1, 2, 3, . . . is the iteration counter. The entire procedure is illus-trated in the following example.


x

y

z

z = f(x, y)

z = g(x, y)

z = 0

FIGURE 2.11

Newton’s method for functions of several variables.

Example 2.5

Consider the following two equations:

x2 + y

2− 1 = 0 (2.141)

x2− y = 0 (2.142)

Both equations (2.141) and (2.142) represent a surface in three-dimensionalspace. These surfaces are shown in Figure 2.11. Notice that the intersectionof the function z = x2 + y2 − 1 with the plane z = 0 is a circle of radius one.The intersection of the surface z = x2 − y with the plane z = 0 is a parabola.

Let us now look for (x, y) pairs in the first quadrant that satisfy bothequations (2.141) and (2.142) simultaneously. We will start with the point(1, 1) as our initial guess. The Jacobian matrix for the system of equations(2.141) and (2.142) is obtained by taking partial derivatives of the functionsf(x, y) and g(x, y) with respect to x and y. The linearized system of equationsis then written as

[

2x 2y

2x −1

]

∆x

∆y

=

−(x2 + y2 − 1)−(x2 − y)

(2.143)

The linearized representations of the functions f and g are planar surfaces,as shown in Figure 2.12. In this figure, the dark shaded surface representsthe plane z = 0. The medium-shaded plane is the linearized representationof f , and the light shaded plane is the linearized representation of g. Noticethat the intersection of each plane with the plane z = 0 is a line, and the


x

yz

z = 0

z = f(x, y)

z = g(x, y)

FIGURE 2.12

Intersecting planes.

intersection of the two lines is the solution to the linear system of equations(2.143).

In Newton’s method, equation (2.143) is used in an iterative manner toobtain successively better estimates for the actual solution, starting with aninitial guess (x0, y0). During the first iteration, the Jacobian matrix is setup and evaluated at the initial guess. The right-hand side is then set up andevaluated at the initial guess. The right-hand side is often called the residualvector. As the guesses for x and y become better and better, the right-handside of (2.143) gets closer and closer to zero. After the right-hand side isevaluated at the initial guess, the linear system of equations is solved for ∆x

and ∆y. Finally, the new guess is obtained by adding these increments to theprevious guess.

In the present example, the first three iterations of Newton’s method arecarried out in Table 2.3 below. The iterations stop when the residual is smallcompared with a user-defined tolerance. A very attractive feature of Newton’smethod from a computational point of view is that if the initial guess is nottoo far off from a solution to the system of nonlinear equations, and if the

TABLE 2.3

Newton’s method example.

i J R ∆x x

0

[

2.0000 2.00002.0000 −1.0000

]

−1.00000.0000

−0.1667−0.3333

0.83330.6667

1

[

1.6667 1.33331.6667 −1.0000

]

−0.1389−0.0278

−0.0452−0.0476

0.78810.6190

2

[

1.5762 1.23811.5762 −1.0000

]

−0.0043−0.0020

−0.0019−0.0010

0.78620.6180


i

|R|

1 2 3 4 50

0.2

0.4

0.6

0.8

1.0

FIGURE 2.13

Quadratic convergence.

Jacobian matrix is formulated correctly, then the method exhibits quadraticconvergence.

To give a brief explanation of the what is meant by quadratic convergence,let us examine the residual vector R. The magnitude of this vector is definedas

|R| =√

RaRa (2.144)

where Ra is an element of R, and the summation convention is implied. In thepresent example, the magnitude of the residual vector |R| is plotted versusthe iteration number as shown in Figure 2.13. Notice the rapid rate at withthe magnitude of the residual vector decreases. The relative residual can bedefined as

Ri =

|R|i+1

|R|i(2.145)

Quadratic convergence occurs when the relative residual decays quadratically,i.e.,

Ri+1

≈ (Ri)2 (2.146)

or in other words, when the new relative residual computed within iterationi + 1 is approximately equal to the square of the previous relative residualcomputed in iteration i.


P

p

dX

dx

x

xp

XXP

FIGURE 2.14

The undeformed and deformed configurations.

2.6 Kinematics of motion

Fundamental to the study of continuum mechanics is the notion that a con-tinuous body is made up of infinitely many material points. When the bodyis deformed by the action of external forces, for example, the motion of thematerial points can be described by a mathematical function that depends onthe initial location of the material points and time. This concept is illustratedin Figure 2.14 below. In the figure, the position of material point P in the un-deformed configuration is defined by the position vector XP . At any instantof time, the position of this same material point p in the deformed configu-ration is defined by the position vector xp. Given the general position, X, ofa material point in the undeformed configuration, the location of a materialpoint, x, in the deformed configuration is given by the following function ormapping:

x = x(X). (2.147)

Equation (2.147) can be thought of as three equations (in general nonlinear)of the three independent variables X1, X2, and X3, i.e.,

x1 = x1(X1,X2,X3)

x2 = x2(X1,X2,X3)

x3 = x3(X1,X2,X3) (2.148)

In other words, each component of the position vector x is a function of X1,X2, and X3. Unless stated otherwise, we will assume that the mapping of


material points from the undeformed configuration to the deformed config-uration is one to one. That is, each material point is mapped to a uniquelocation in the deformed configuration (one material point cannot occupy twodifferent locations in the deformed configuration).

Let us now look at other material points in a small region surrounding pointP . If this region is infinitesimal, the motion of these points can be describedby a two-term Taylor series expansion of equation (2.148) as follows:

x1 = x1(XP

1,X

P

2,X

P

3)

+∂x1

∂X1

(X1 − XP

1) +

∂x1

∂X2

(X2 − XP

2) +

∂x1

∂X3

(X3 − XP

3)

x2 = x2(XP

1,X

P

2,X

P

3)

+∂x2

∂X1

(X1 − XP

1) +

∂x2

∂X2

(X2 − XP

2) +

∂x2

∂X3

(X3 − XP

3)

x3 = x3(XP

1,X

P

2,X

P

3)

+∂x3

∂X1

(X1 − XP

1) +

∂x3

∂X2

(X2 − XP

2) +

∂x3

∂X3

(X3 − XP

3) (2.149)

where all partial derivatives are to be evaluated at point P . The three equa-tions in (2.149) can be written in matrix notation as

x1

x2

x3

=

xp

1

xp

2

xp

3

+

∂x1

∂X1

∂x1

∂X2

∂x1

∂X3

∂x2

∂X1

∂x2

∂X2

∂x2

∂X3

∂x3

∂X1

∂x3

∂X2

∂x3

∂X3

X1 − XP

1

X2 − XP

2

X3 − XP

3

(2.150)

or

x − xp = J · (X − XP ) ⇒

dx = J · dX (2.151)

where J is the 3×3 Jacobian matrix associated with the mapping between theundeformed and deformed configurations, dX = X − Xp is an infinitesimalvector in the undeformed configuration pointing from XP to some point X inthe neighborhood of P , and dx = x − xp is an infinitesimal vector pointingfrom xp to x in the deformed configuration.

In the study of kinematics of deformable bodies, the Jacobian matrix associ-ated with the mapping between the undeformed and deformed configurationsis known as the deformation gradient tensor F. The deformation gradienttensor contains information about the motion of material points within aninfinitesimal region surrounding point XP . Hence equation (2.151) can equiv-alently be written as

dx = F · dX (2.152)


In terms of its Cartesian components, the deformation gradient tensor can bewritten as

Fij =∂xi

∂Xj

(2.153)

The deformation gradient tensor is a very useful quantity, as we will see later inour study of finite element methods for large strain and deformation problemsin Chapter 10. As an illustration, let ds be the length of a line segment dx inthe deformed configuration. The length ds can be expressed in terms of F asfollows:

ds2 = dx · dx

= (F · dX) · (F · dX)

= dX · (FT· F) · dX

= dX · C · dX (2.154)

where C is the right Cauchy-Green deformation tensor. So given any vectordX in the undeformed configuration, the length of this vector in the deformedconfiguration can be computed from equation (2.154).

2.7 Problems

Problem 2.1

Letting

K =

1 −1 0−1 2 −1

0 −1 1

; B =

[

2 3 40 1 5

]

; d =

2−2

1

perform the following matrix multiplications:

(a) Kd (b) Bd

(c) BT B (d) KB

Problem 2.2

Let

A =

1 2 02 4 21 4 −1

; B =

[

1 −1−1 3

]

Determine the following:

(a) detA (b) detB

(c) A−1 (d) B−1


Problem 2.3

Consider the following system of linear algebraic equations:

d1 − d2 = R1

−d1 + 2d2 − d3 = 0

−d2 + d3 = 1

where R1 is an unknown constant. Solve the system of equations for d2 andd3 subject to the constraint d1 = 0. After finding d2 and d3, plug d2 backinto the first equation and solve for the constant R1.

Problem 2.4

Let

A =

x 0 3x3

3 x2 + 1 x4

2x 6x3 − x 1

Evaluate∫

1

−1Adx.

Problem 2.5

The Cartesian components of a second-order tensor, T, are provided in thematrix below,

[Tij ] =

7 −1 −6−1 6 −3−6 −5 11

With the help of Table 2.1, evaluate the following:

(a) detT (b) T : T (c) T · T

(d) Tsym (e) trT (f) Tskew

Problem 2.6

The Cartesian components of a symmetric second-order tensor, T, are givenas

[Tij ] =

2 −1 0−1 1 −1

0 −1 1

Determine the components of T in a primed coordinate system that is createdby rotating the fixed Cartesian axes through an angle of π/6 about the X3

axis.

Problem 2.7

Compute the eigenvalues and eigenvectors of the tensor defined in Problem2.6.


Problem 2.8

Consider the following nonlinear function:

f(x) = x4− 3x + 2

Perform two iterations of Newton’s method to obtain an estimate for the rootof f(x) in the vicinity of x = 1.

Problem 2.9

Consider the following system of nonlinear equations:

x2

4+ 4y2 = 1

x2 + y

2 = 1

Perform two iterations of Newton’s method to obtain an estimate for thelocation where the two curves intersect in the first quadrant.

Problem 2.10

Using the divergence theorem, show that the volume, V , enclosed by thesurface Γ can be obtained by evaluating the following surface integral:

V =1

3

∫

Γ

n · xdΓ

where x is a position vector from the origin of a Cartesian coordinate systemto a point on the surface, and n is the unit outward normal vector to thesurface at that point.

Problem 2.11

Consider the following two-dimensional mapping between the undeformed anddeformed configurations:

x1 = X2

1+ X2

x2 = X1 + 2X2

where x1 and x2 represent the coordinates of a point in the deformed config-uration that originally occupied point (X1,X2) in the original configuration.Evaluate the components of the deformation gradient tensor F at the location(X1 = 0,X2 = 0).

Problem 2.12

Write out the Taylor series expansions for the following functions:


(a) cos θ

(b) sin θ

(c) eiθ

From the results obtained in parts (a), (b), and (c) show that

eiθ = cos θ + i sin θ

where i is the imaginary number, i.e., i2 = −1. The identity in the aboveequation is known as Euler’s equation.

3

One-Dimensional Problems

Everything should be made as simple as possible, but not simpler.—Albert Einstein

In order to demonstrate the mathematical concepts behind the finite elementmethod, we begin by applying the method to solve second-order, ordinarydifferential equations with constant coefficients. Employing such a powerfulnumerical method to solve such simple equations is akin to using a hammerto kill a fly. Nevertheless, with an understanding of the concepts presentedin this chapter, the serious reader will be able to get through the rest of thebook with relative ease.

To begin, consider the following differential equation:

au′′ + bu + c = 0; 0 ≤ x ≤ L (3.1)

along with the boundary conditions

u(0) = 0; u′(L) = 1

When students are first introduced to differential equations, they are given theexercise of “plugging in” a function into a differential equation and checkingwhether or not the function is a solution to the equation. In the finite elementmethod, such a candidate function is called a trial function. Throughout thischapter, the trial function will be denoted as u(x).

One of the main ideas in the finite element method is to break up the intervalof interest, i.e., 0 ≤ x ≤ L in the example above, into subintervals calledelements. For the sake of simplicity, it is desirable to use trial functions thatare continuous piecewise. A continuous piecewise trial function is a functionthat does not have any “jumps,” but it can have “kinks” (locations where theslope of the trial function is not continuous).

An illustration of what is meant by continuous piecewise trial functions isshown in Figure 3.1. Here, we have broken up the interval 0 ≤ x ≤ L intothree elements, all of equal length. The solid line in the figure is an exampleof a piecewise linear trial function. The dashed line represents a piecewisequadratic trial function. The derivative of each type of trial function, u′, isplotted in the lower plot in Figure 3.1. Notice the jumps in the curves atthe junctions between each element. These junctions are referred to as nodallocations, or simply nodes.

51


u

u′

1 2 3 4

linear

quadratic

1 2 3 4

FIGURE 3.1

Continuous piecewise trial functions.

When one tries to play the game of plugging in continuous piecewise trialfunctions into the differential equation (3.1), he encounters a problem. Whatis it? One can differentiate the trial function once, as shown in the lower plotof Figure 3.1, but the second derivative does not exist! The trial functionsshown in Figure 3.1 are classified as C0 continuous. This means that thefunctions are continuous, but their first derivatives are piecewise continuous.

There are different classes of trial functions that are commonly used in thefinite element method having varying degrees of continuity. For example, aC1 trial function has a continuous first derivative, but its second derivative ispiecewise continuous. Trial functions that are C1 continuous are commonlyused in the formulation of beam elements discussed in Chapter 5.

In order to remedy the situation of not being able to plug in the trialfunctions directly into the differential equation, it is necessary to derive whatis called the weak form of the differential equation. The differential equationalong with the boundary conditions is referred to as the strong form. Theweak form is a mathematically equivalent statement to the strong form. Onemay like to think of it as the strong form in disguise. It turns out that thetrial functions can be substituted into the weak form without any difficulty.After that, the game will be to find trial functions that satisfy the weak form.

One-Dimensional Problems 53

3.1 The weak form

In the differential equation (3.1) above, because it is second order, there mustbe two boundary conditions attached to it. The first boundary conditionimposes a direct restriction on the function u at the left end of the interval,i.e., u(x = 0) = 0. This type of boundary condition is called a Dirichletboundary condition, named after mathematician Lejeune Dirichlet. In thefinite element literature, it is called an essential boundary condition. Theboundary condition at x = L is a condition on the slope (or first derivative)of the function u. A boundary condition that places a restriction on the firstderivative of the field variable is called a Neumann boundary condition aftermathematician John von Neumann. A Neumann condition is also called anatural boundary condition.

Now that we have discussed the various types of boundary conditions thatcan be associated with the differential equation, we now introduce what iscalled a test function. A test function is a mathematical concoction with thefollowing properties:

1. w(x) is C0 continuous in 0 ≤ x ≤ L.

2. w(x) = 0 at all essential boundaries

In other words, a test function comes from a family of functions that arecontinuous on the interval of interest and are zero when evaluated at theessential boundary points.

Having defined the test function, the next step in the derivation of the weakform is to multiply each term in the differential equation (3.1) by w(x), i.e.,

au′′w + buw + cw = 0 (3.2)

and then to integrate both sides from 0 to L. Doing this we obtain

a

∫

L

0

u′′w dx + b

∫

L

0

uw dx + c

∫

L

0

w dx = 0 (3.3)

Equation (3.3) is still not in a suitable form, because it contains the u′′ terminside the first integral on the left-hand side. Hence, we still cannot plug in C0

trial functions into this equation. To remedy the situation, we can integratethe first integral in equation (3.3) by parts. The goal of integrating by partsis to remove one of the derivatives on u(x) and put it on the function w(x),without changing the value of the integral. Using the techniques presented inChapter 2, we now express the first integral on the left-hand side of (3.3) asfollows:

a

∫

L

0

u′′w dx = a

∫

L

0

(u′w)′ dx − a

∫

L

0

u′w

′ dx (3.4)


where we have used the relation

(u′w)′ = u

′′w + u

′w

′⇒

u′′w = (u′

w)′ − u′w

′ (3.5)

From the fundamental theorem of calculus discussed in Section 2.4.1 in Chap-ter 2, we see that the first integral on the right-hand side of (3.4) has theform

∫

L

0

f′(x) dx = f(L) − f(0),

and hence,∫

L

0

(u′w)′ dx = u

′(L)w(L) − u′(0)w(0) (3.6)

Because the test function w(x) is defined to be zero at the essential boundarypoints, we see that for the present differential equation defined by equation(3.1) we must have w(0) = 0. The value of the test function at all naturalboundary points (at x = L in the present case) is not known and thereforethe value w(x = L) is arbitrary.

Equation (3.3) can now be written as

a

∫

L

0

u′w

′ dx = b

∫

L

0

uw dx + c

∫

L

0

w dx + au′(L)w(L) (3.7)

Finally, from the natural boundary condition in (3.1) (u′(L) = 1) we obtainthe weak form of the differential equation as

a

∫

L

0

u′w

′ dx = b

∫

L

0

uw dx + c

∫

L

0

w dx + aw(L) (3.8)

While the weak form is mathematically equivalent to the differential equation(3.1), it has the advantage that only one derivative appears on the function u,and therefore continuous piecewise trial functions can now be substituted intothe weak form without any difficulty. Notice that we can integrate the firstderivative of a continuous piecewise trial function even though it has jumps(see Figure 3.1).

3.2 Finite element approximations

In this section we will now formally define some common C0 trial functionsthat can be substituted into equation (3.8).


x

ue

1 2

u1

u2

l

FIGURE 3.2

Linear-u element.

3.2.1 The linear-u element

We will begin by defining piecewise linear trial functions. Let us consider thesubinterval, or element, as shown in Figure 3.2. As shown in the figure, node 1is located at the left end of the element, and node 2 is located at the right end.Over the length l of the element, the trial function varies linearly as shown inthe figure. This portion of the trial function (defined over a single element)will be denoted as ue. Here we use the subscript e to emphasize the fact thatthe trial function is defined over a single element. The superposed tilde isused to emphasize the fact that the trial function, in general, is approximate.The trial function has the value u1 at the left node and u2 at the right node.To facilitate the rest of our discussion, it is convenient to introduce the localcoordinate x measured from the left end of the element as shown in Figure 3.2.We are now in position to express the trial function ue as a function of x, andthe values of the trial function u1 and u2 at the nodal points. In order toaccomplish this, since ue is a linear function of x we can write

ue(x) = c1 + c2x (3.9)

where c1 and c2 are constants. To determine the constants, we use the fol-lowing end conditions:

ue(0) = u1

ue(l) = u2 (3.10)

Plugging in the first condition in (3.10) into equation (3.9) we obtain u1 = c1.Substituting the second condition into (3.9) we get u2 = u1 + c2l, or c2 =(u2 − u1)/l. Plugging the expressions for c1 and c2 into equation (3.9) andsimplifying we obtain

ue(x) = (1 −x

l)u1 +

x

lu2 (3.11)


1N1 N2

1 2

x

FIGURE 3.3

Shape functions for linear-u element.

The two functions of x that appear in equation (3.11) are interpolation func-tions called shape functions. Notice that for the linear-u element, both shapefunctions are linear in x.

It is standard procedure in the finite element method to store the shapefunctions associated with a single element in a matrix called Ne. For thelinear-u element we have

Ne =[

1 − x/l x/l]

=[

N1(x) N2(x)]

(3.12)

where N1(x) and N2(x) are the shape functions. These shape functions areshown in Figure 3.3. Notice that the first shape function is equal to one whenevaluated at the left end of the element (node 1) and is equal to zero whenevaluated at the right end (node 2). The second shape function is equal tozero at node 1 and one at node 2. All standard finite element shape functionshave this property. The property can be stated succinctly as

NI(xJ) = δIJ (3.13)

Equation (3.13) states that the shape function NI evaluated at node J, whichoccupies position xJ , is equal to δIJ , the Kronecker delta defined in Chapter 2.Here the subscripts I and J take on the values 1 to nen, where nen is thenumber of nodes owned by the element (two nodes in this case). The one-dimensional shape functions can also be derived by choosing a local coordinateto be zero at the center of a linear-u element of length le. The domain ofthe element then becomes −le/2 ≤ x ≤ le/2. Letting ξ = 2x/le, the onedimensional shape functions become

N1 =1

2(1 − ξ)

N2 =1

2(1 + ξ)


Having introduced the shape function matrix Ne, it is convenient to rewriteequation (3.11) in matrix notation as follows:

ue(x) =[

N1(x) N2(x)]

(

u1

u2

)

=Ne

1 × 2ue

2 × 1(3.14)

Later on, when it comes time to plug the trial function into the weak form,it will be necessary to take the derivative of the trial function ue(x) withrespect to x. Since x = x − x1, where x1 is the location of node 1, we have

u′e

=due

dx

=due

dx

dx

dx

= −1

lu1 +

1

lu2 (3.15)

Next, we write equation (3.15) in matrix form as

u′e

=[

−1/l 1/l]

u1

u2

=Be

1 × 2ue

2 × 1(3.16)

The matrix Be in equation (3.16) is commonly referred to as the B-matrix.The B-matrix operates on the nodal unknowns (u1 and u2 in this case), yield-ing the derivative of the trial function at each point along the length of theelement.

3.2.2 The quadratic-u element

Another important one-dimensional finite element is the quadratic-u element.In this element we assume that the trial function varies quadratically overthe length of the element. We therefore need three nodes to uniquely specifya quadratic function over the element. The trial function is expressed as aquadratic function of x as follows

u(x) = c1 + c2x + c3x2 (3.17)

The quadratic-u element is shown in Figure 3.4. The displacement values ateach of the three nodes are labeled u1, u2, and u3, respectively. Like thelinear-u element, it is desirable to express the trial function in terms of shapefunctions and the nodal displacements. To derive the shape functions for the


x

1 2 3

u1

u2

u3

FIGURE 3.4

Quadratic-u element.

quadratic-u element, we impose the following three conditions at the nodallocations:

ue(x = −le) = u1

ue(x = 0) = u2

ue(x = +le) = u3 (3.18)

After substituting the conditions (3.18) into equation (3.17) we obtain

u1 = c1 − c2le + c3l2

e

u2 = c1

u3 = c1 + c2le + c3l2

e. (3.19)

Solving equation (3.17) for c1, c2, and c3 yields

ue(ξ) = −1

2ξ(1 − ξ)u1 + (1 − ξ

2)u2 +1

2ξ(1 + ξ) (3.20)

where we have used the change of variables

ξ = 2x/le

We can write equation (3.20) in matrix form as

ue =

[

−1

2ξ(1 − ξ) 1 − ξ2

1

2ξ(1 + ξ)

]

u1

u2

u3

=Ne

1 × 3ue

3 × 1(3.21)


ξ = 0 ξ = +1ξ = −1

1

ξ

N1

N2

N3

FIGURE 3.5

Shape functions for quadratic-u element.

Notice that the variable ξ, also known as a natural or dimensionless coordi-nate, ranges from −1 ≤ ξ ≤ +1. The functions that depend on the naturalcoordinate are the shape functions. Notice also that for the quadratic-u el-ement, there are three shape functions, one associated with each node. Thethree shape functions are plotted in Figure 3.5. Notice again that the shapefunctions obey the Kronecker delta property (3.13). If we now differentiatethe trial function with respect to x we obtain

u′e

=due

dx

=due

dξ

dξ

dx

=2

le

[

−1

2+ ξ −2ξ

1

2+ ξ

]

u1

u2

u3

=Be

1 × 3ue

3 × 1(3.22)

3.3 Plugging in the trial and test functions

For a given type of finite element, whether it be linear, quadratic, or whatever,we are now in a position to substitute the desired trial function into the weakform (3.8). To do this, it is convenient to break up the interval from 0 ≤ x ≤ L


into many smaller subintervals called elements. The integrals from 0 to L

that appear in (3.8) can be evaluated by summing up the integrals over theindividual elements. After substituting the trial function and the test functioninto the weak form we obtain

a

nel∑

e=1

∫

Ωe

u′ew

′edx = b

nel∑

e=1

∫

Ωe

uewedx

+c

nel∑

e=1

∫

Ωe

wedx + awn, (3.23)

where we = we(x) is the test function, defined over an individual element,and Ωe is the domain of an individual element (either 0 ≤ x ≤ le or −le/2 ≤

x ≤ le/2). Here in equation (3.23), n is the number of nodal points alongthe interval 0 ≤ x ≤ L, and the quantity wn is the value of the test functionevaluated at the node located at x = L. The quantity nel refers to the totalnumber of elements along the interval 0 ≤ x ≤ L.

Because the test function w(x) is arbitrary (just as long as it meets the tworequirements for a test function) it is natural to assume that it varies alongeach element in the same manner as the trial function. That is, for givenvalues w1, w2, . . ., wn of the test function at the nodal locations, the sameshape functions that were used to interpolate the trial function are used tointerpolate the test function nodal point data over the length of the element.This procedure of using the same interpolation functions for both the trialand test functions is consistent with a Galerkin method [1]. Hence we take

we(x) = Newe

w′e(x) = Bewe (3.24)

where Ne is the matrix containing the element shape functions, and Be is theelement B-matrix. Again, the quantity we is a vector containing the valuesof the test function at the nodal points. Incidentally, both Ne and Be are1× 2 matrices for the linear-u element and 1× 3 matrices for the quadratic-uelement.

Before we proceed further, it is useful to write down the following identities:

u′ew

′e

= w′eu′e

= (Bewe)(Beue)

= (Bewe)T (Beue)

= wT

eBT

eBeue (3.25)

Similarly, we can write

uewe = weue

= (Newe)(Neue)

= (Newe)T (Neue)

= wT

eNT

eNeue (3.26)


1 32 4

1 2e

FIGURE 3.6

Local and global node numbers.

and finally

we(x) = Newe

= (Newe)T

= wT

eNT

e(3.27)

When performing the manipulations (3.25) through (3.27), one must recognizethat both the trial function and the test function are scalar quantities. Hence,the product of the trial function and the test function obeys the commutativelaw of multiplication. The identity (2.13) in Chapter 2 on page 7 was used(i.e., the transpose of the product of two matrices is equal to the product ofthe transpose of each matrix in reverse order).

Since wT

eand ue contain nodal point data, they do not depend on x. Thus,

these quantities can be taken outside of the integrals. We must, however,be careful to take wT

eout to the left and ue out to the right, so as to keep

the matrix multiplications conformable. Keeping this in mind, the identities(3.25) to (3.27) can be substituted into equation (3.23) to obtain

a

nel∑

e=1

wT

e

∫

Ωe

BT

eBe dxue = b

nel∑

e=1

wT

e

∫

Ωe

NT

eNe dxue

+c

nel∑

e=1

wT

e

∫

Ωe

NT

edx + awn (3.28)

The final step in this section is to define a relationship between the vectorsue and we, which contain the values of the trial and test functions at the localnodes (nodal points owned by an individual element), and the global vectorsu and w containing the values of the trial and test functions at all of thenodes within the interval 0 ≤ x ≤ L. This relationship can be accomplishedby defining a Boolean matrix Le (a matrix containing ones and zeros) for eachelement. The matrix Le contains information on how the local node numbersfor a given element are related to the global node numbers. As an example,suppose that the interval 0 ≤ x ≤ L is broken up into three elements as shownin Figure 3.6. The global nodes can be numbered in any order and will beemphasized by bold text. Now suppose that each interval is a linear-u element(discussed in the previous section). The local node numbers are chosen suchthat local node 1 is always located at the left end of the element and local node


2 is always located at the right end. Focusing on the middle element in thefigure, local node 1 corresponds to global node 2 and local node 2 correspondsto global node 4. In this example, the vector ue can be expressed in terms ofthe vector u as follows:

u1

u2

=

[

0 1 0 00 0 0 1

]

u1

u2

u3

u4

, (3.29)

or in general

ue = Leu (3.30)

Similarly, for the test function we can write

w1

w2

=

[

0 1 0 00 0 0 1

]

w1

w2

w3

w4

(3.31)

Notice that the matrix Le is of order nen × n, where nen is the number ofnodes owned by the element, and n is the total number of nodes in the interval0 ≤ x ≤ L.

Recognizing that wT

e= wT LT

eand ue = Leu, we can now substitute these

expressions into equation (3.28) to obtain

wT

(

nel∑

e=1

LT

e

[

a

∫

Ωe

BT

eBe dx − b

∫

Ωe

NT

eNe dx

]

Le

)

u

= wT

(

nel∑

e=1

LT

e

[

c

∫

Ωe

NT

edx

]

+ LT

ΓNT

Γ

)

(3.32)

The last term in equation (3.32) deserves some further explanation. Here LΓ

is an nen × n Boolean matrix associated with the element that touches thenatural boundary point. Recall that the natural boundary is the locationwhere the derivative of the field variable (u in this case) is specified. Thematrix NΓ is the matrix containing the shape functions for the element thattouches the natural boundary point. This shape function matrix is evaluatedat the right end of the element (at the x value associated with local node 2).

The interested reader will observe that equation (3.32) has the form

wT (Ku − F) = 0 (3.33)

where

K =

(

nel∑

e=1

LT

e

[

a

∫

Ωe

BT

eBe dx − b

∫

Ωe

NT

eNe dx

]

Le

)

(3.34)


u

u2

u3 u4

u5

1 2 3 4 5 6

FIGURE 3.7

Family of trial functions for a given finite element mesh.

is the global system matrix and

F =

(

nel∑

e=1

LT

e

[

c

∫

Ωe

NT

edx

]

+ LT

ΓNT

Γ

)

(3.35)

is the global force vector. Equations (3.34) and (3.35) define the finite elementassembly process.

Since the test function is arbitrary, the vector w will not in general be zero.Hence the only way that equation (3.33) can be satisfied is for the quantityinside the parentheses to vanish, i.e.,

Ku − F = 0 (3.36)

orKu = F (3.37)

Equation (3.36) is a consequence of the function scalar product theorem. Theargument goes as follows. Consider the dot product between two vectors a

and b. Suppose that we seek a vector b that satisfies the equation

a · b = 0 (3.38)

for all possible vectors a. This is a strong statement. We are seeking avector b that satisfies equation (3.38) for every conceivable vector a havingany magnitude and orientation. One possibility is that equation (3.38) canbe satisfied if the vector b is chosen to be perpendicular to a. However, theonly way equation (3.38) can be satisfied for all arbitrary vectors a is for the


vector b to be identically zero. Note that wT in equation (3.33) is analogousto b and Ku − F is analogous to a.

We have now come to the point where we can make a very powerful state-ment. When the interval 0 ≤ x ≤ L is discretized into a number of nodes andelements, the resulting finite element mesh spawns a family of trial functions.Each trial function within the family is similar in shape and satisfies the es-sential boundary conditions but has different values at the nodes. This idea isillustrated in Figure 3.7, where we have specified that u = 0 at global nodes1 and 6. Which is the best trial function for the given finite element mesh?The best trial function within the family is the one that satisfies the equationKu = f . Note that the trial function that satisfies this equation is not the ex-act solution to the differential equation, but rather an approximate solution.In general, the approximate solution gets better and better by refining thefinite element mesh, i.e., dividing the domain of interest into more and moreelements. In order to solve actual problems using the theory developed inthis chapter, it is necessary to evaluate the integrals that appear in equation(3.32). To simplify matters, we first write equation (3.37) as follows:

(

nel∑

e=1

LT

e[Ke + Me]Le

)

u =

(

nel∑

e=1

LT

efe

)

+ LT

ΓfΓ (3.39)

where

K =

(

nel∑

e=1

LT

e[Ke + Me]Le

)

(3.40)

and

F =

(

nel∑

e=1

LT

efe

)

+ LT

ΓfΓ (3.41)

Here in equation 3.39, Ke and Me are called element matrices. Each elementhas associated with it element matrices that contribute to the global systemmatrix K. The quantities fe and fΓ are called element vectors. The assemblyof all the element matrices and vectors yields the global system matrix andglobal vector on the right-hand side. Equations (3.40) and (3.41) representmatrix assembly and vector assembly, respectively. For the purpose of per-forming the examples and problems at the end of this chapter, it is convenientto tabulate the element matrices and vectors for both the linear-u element andthe quadratic-u element in Table 3.1 below. The resulting integrals can ei-ther be obtained by lengthy hand calculations, or through the use of symbolicmath packages such as Maple [6] or Mathematica [7].

We end this section with a short discussion on the element vector fΓ. Recallthat this vector arises when one or both ends of a one-dimensional domain is anatural boundary point. In that event only the elements that touch a naturalboundary point are affected. If an element touches a natural boundary point,then the element vector fe for that element must be augmented by the vector


fΓ. The total element vector for an element that touches a natural boundarypoint is fe + fΓ. The element vector fΓ is given in Table 3.2 below for boththe linear-u and quadratic-u elements.

Example 3.1

In order to illustrate the machinery developed in the previous sections, let usconsider the following differential equation

u′′ + 1 = 0

u(0) = 0

u′(3) = 1

0 ≤ x ≤ 3 (3.42)

We see that the differential equation above is just a special case of equation(3.1) with a = 1, b = 0, c = 1,and L = 3. Therefore, the weak form of (3.42)can be written as

∫

3

0

u′w

′ dx =

∫

3

0

w dx + w(x = 3) (3.43)

TABLE 3.1

Element matrices and vectors.

Matrix Ke Me fe

a

∫

Ωe

BT

eBe dx b

∫

Ωe

NT

eNe dx c

∫

Ωe

NT

edx

Linear-ua

le

[

1 −1−1 1

]

−b le

6

[

2 11 2

]

c le

2

11

Quadratic-ua

3 le

[

7 −8 1−8 16 −8

1 −8 7

]

−b le

30

[

4 2 −12 16 2

−1 2 4

]

c le

6

141

TABLE 3.2

Additional element vector.

fΓ aNT

Γ(local node 1) aNT

Γ(local node 2)

Linear-u a

10

a

01

Quadratic-u a

100

a

001


1 2 3 4

1 2 1 2 1 2

¤

£

¡

¢1

¤

£

¡

¢2

¤

£

¡

¢3

FIGURE 3.8

Interval discretized with three linear-u elements. The global node numbersare emphasized in bold.

Let us now break up the interval 0 ≤ x ≤ 3 into three linear-u elements, eachof unit length. Because b = 0, and x = 3 is a natural boundary point, therelevant element matrices are Ke, fe, and fΓ. Recall that these are given inTables 3.1 and 3.2.

Now let us suppose that the four global nodes along the interval are labeled(in bold text) sequentially from left to right as shown in Figure 3.8. The localnodes 1 and 2 are labeled at the top (left and right edges) of each element.The Boolean matrix Le for each element is given as

L1 =

[

1 0 0 00 1 0 0

]

L2 =

[

0 1 0 00 0 1 0

]

L3 =

[

0 0 1 00 0 0 1

]

Next, plugging in the Boolean matrices into equation (3.39) we get(

LT

1K1L1 + LT

2K2L2 + LT

3K3L3

)

u

= LT

1f1 + LT

2f2 + LT

3

[

f3 + NT

3(x = 1)

]

(3.44)

Because the length of each element in this example is the same, and becauseall of the elements are linear-u elements, the element matrices and vectors areall the same, i.e.,

K1 = K2 = K3 =

[

1 −1−1 1

]

(3.45)

and

f1 = f2 = f3 =

1/21/2

(3.46)

Notice that element 3 touches the natural boundary point at x = 3 (or equiv-alently x = 1). So we have included the vector fΓ in the global right-hand-sidevector. The vector fΓ in the present example is given as

fΓ =

01

(3.47)


If we now perform the required matrix multiplications we obtain

1 −1 0 0−1 1 0 0

0 0 0 00 0 0 0

+

0 0 0 00 1 −1 00 −1 1 00 0 0 0

+

0 0 0 00 0 0 00 0 1 −10 0 −1 1

u1

u2

u3

u4

=

1/21/200

+

01/21/20

+

00

1/21/2

+

0001

(3.48)

and after assembly (matrix addition), the global system of equations is finallyobtained,

1 −1 0 0−1 2 −1 0

0 −1 2 −10 0 −1 1

u1

u2

u3

u4

=

1/211

3/2

(3.49)

Notice that the resulting 4×4 left-hand side system matrix is symmetric, andit is also singular, i.e., detK = 0. When we impose the boundary conditionat node 1, u1 = 0, we obtain a reduced system matrix that is nonsingular.The reduced system of equations can be written as

2 −1 0−1 2 0

0 −1 1

u2

u3

u4

=

11

3/2

(3.50)

and upon solution we get

u1

u2

u3

u4

=

07/26

15/2

(3.51)

The finite element solution is plotted versus the exact (analytical) solutionin Figure 3.9. In the upper graph, the field variable u is plotted versus positionx. The finite element approximation is represented by the dashed line, andthe exact solution is conveyed by the solid line. It turns out that the finiteelement solution is in perfect agreement with exact solution at the nodalpoints, as evidenced by the fact that the circles lie right on top of the solidcurve. This is typical of one dimensional-problems, but not for all problemsin general; see Hughes [8] for a more detailed treatment of this topic. Onthe other hand, because we have used linear-u elements, the finite elementsolution varies linearly between the nodal points, and hence, the finite elementsolution exhibits error in the regions between the nodes. In the lower plot,the derivative u′ is plotted versus x for both the finite element (dashed line)


PSfrag

u

u′

x

x

0 0.5 1 1.5 2 2.5 3

0 0.5 1 1.5 2 2.5 3

0

1

2

3

4

0

2

4

6

8

FIGURE 3.9

Finite element versus exact solution.

and exact solution (solid line). Notice that the finite element approximationfor u′ is discontinuous at the nodal points. Because u varies linearly alongeach element, the finite element solution predicts that u′ is constant at everypoint within the element.

Finally, it is worth mentioning that the essential boundary condition, u(0) =0, is satisfied exactly by the finite element solution. The natural boundarycondition is not, however. Notice the error between the finite element predic-tion and the exact solution in the lower plot at x = 3. This error gets smallerand smaller as the mesh is refined.

For the sake of completeness, the exact solution to the differential equa-tion (3.42) can be obtained by repeatedly taking the antiderivative of bothsides of the equation (in other words, integrating both sides of the equation).Integrating once we obtain

u′ = −x + C1

where C1 is a constant of integration. The constant C1 can be obtained by


enforcing the natural boundary condition at x = 3. Thus we have

u′(3) = 1 ⇒

1 = −3 + C1, and therefore,

C1 = 4

Having found the constant C1, we can integrate again to obtain

u = −x2

2+ 4x + C2

where C2 is another constant. Finally, after enforcing the boundary condition,u(0) = 0, we see that C2 = 0, and the exact solution is

u(x) = −x2

2+ 4x (3.52)

3.4 Algorithm for matrix assembly

In the example in the previous section, the finite element assembly process(the process of getting the global system matrix and right-hand-side vectorfrom the element matrices and vectors) involved several steps. To get theglobal system matrix, for example, it was necessary to set up each elementmatrix, and then to define the Boolean matrix Le associated with each el-ement. The augmented matrix for each element was obtained by pre- andpost-multiplying the element matrix by the Boolean matrix. The system ma-trix was finally obtained by adding up all of the augmented matrices. Thisprocedure was presented for educational purposes only; it is never actuallyused in commercial finite element codes because of the large number of multi-plications that are involved, which (for large matrices) is very computationallyexpensive. In this section we present the approach that is used in practice.This computationally efficient approach can be taught like a recipe, memo-rized, and then easily followed.

To begin, we introduce a connectivity list. A connectivity list is a matrixcontaining the same number of rows as there are elements in the finite ele-ment mesh. The element number is typically stored in the first column. Theremaining columns contain the global node numbers that are associated witheach local node number attached to the element. In the example in the previ-ous section, the finite element mesh contained three elements. Each elementwas a two-node linear-u element. Since the nodes were numbered sequentiallyfrom left to right, the connectivity list would look as follows: In Table 3.3, wenotice that for element 1, local node 1 corresponds to global node 1, and localnode 2 corresponds to global node 2. Similarly, for element 2, local node 1


TABLE 3.3

Connectivity list.

Element 1 21 1 22 2 33 3 4

1 -1

-1 1

1

2

1 2

element 1

1 -1

-1 1

2

3

2 3

element 2

1 -1

-1 1

3

4

3 4

element 3

1 2 3 4

1

2

3

4

1 -1 0 0

-1 1+1 -1 0

0 -1 1+1 -1

0 0 -1 1

FIGURE 3.10

Matrix assembly.

corresponds to global node 2 and local node 2 corresponds to global node 3,etc. In this section, the connectivity list will be stored in an array (matrix)labeled ix(nen, nel), where nen is the number of nodes owned by the element,and nel is the number of elements in the mesh.

For the case where there is only one unknown per node (one degree of free-dom per node), the finite element matrix assembly process can be illustratedgraphically in Figure 3.10. In the diagram, each element matrix is writtendown side by side and labeled. Focusing our attention on element 1, therecipe goes as follows. The first and second row, and the first and secondcolumns are labeled 1 and 2 respectively. From the connectivity list, we seethat these are the global node numbers that correspond to local nodes 1 and


1

1

2

1

element 1

1

1

3

2

element 2

1

1

4

3

element 3

1

1+1

1+1

1

1

2

3

4

FIGURE 3.11

Vector assembly.

2 in element 1. Global node 1 is stored in ix(1, 1) and global node 2 is storedin ix(2, 1). Note that the global node number that is attached to local node i

in element j is stored in ix(i, j). Because there are a total of four degrees offreedom in the finite element mesh, the global system matrix is of order 4×4.The rows and columns of the global system matrix are labeled consecutivelyfrom 1 to 4 as shown in the bottom portion of the figure. To describe thematrix assembly process, the number that occupies row I and column J inthe 2×2 element matrix gets added to the global matrix in row I column J asindicated by the arrows in the figure. Starting with element 1, this process isrepeated for each element until the global system matrix is completely assem-bled. Note that all of the numbers in the global stiffness matrix are initiallyzero, and then numerical values are added to it (element by element) untilit is fully assembled. The procedure for vector assembly is the same as thatfor matrix assembly as depicted in Figure 3.11. The algorithm for the matrixand vector assembly process is given in Table 3.4 below.

The algorithm for element matrix and vector assembly begins by definingthe parameter nst = nen ∗ ndf which is the number of nodes owned by theelement times the number of degrees of freedom (unknowns) per node. So,for the one-dimensional linear-u element with one degree of freedom per node,nst = 2, corresponding to the number of rows in the element matrix. Afterdefining nst, each element of the global stiffness matrix is zeroed, and webegin looping over the number of elements in the model.


TABLE 3.4

Algorithm for element matrix and vector assembly.

Algorithm : Matrix and vector assembly(K,F)

nst = nen ∗ ndf

K ← 0 zero the global systemF ← 0 zero the global right-hand-side vector

for iel ← 1 to nel

k ← Ke set up the element matrixf ← fe set up the element vectorfor j ← 1 to nen loop over local node numbers

ll ← 0 initialize counter ll to zerofor i ← 1 to ndf loop over degrees of freedom per node

ll = ll + 1ld(ll) = ndf ∗ (ix(j, iel) − 1) + i

continuecontinue

for ii ← 1 to nst

F (ld(ii)) = f(ii)for jj ← 1 to nst

K(ld(ii), ld(jj)) = K(ld(ii), ld(jj)) + k(ii, jj)continuecontinue

continue

Once inside the outer element loop, the first step is to set up the elementmatrix. This is typically done in practice by calling a function or subroutinefor the required element type. The next step is to set up an array, which wecall ld. This array stores the global equation numbers associated with eachlocal node owned by the element that we are currently working on. So in thepresent example, when iel = 1, the values stored in the ld array are ld(1) = 1and ld(2) = 2. When iel = 2, ld(1) = 2 and ld(2) = 3, etc.

After the element matrix and ld arrays are set up, the algorithm proceeds tothe second nested loop. This nested loop takes the numerical values stored inthe element matrix and element vector and adds them to the proper locationsin the global system matrix and right-hand-side vector. Once the elementmatrix and vector are assembled, the algorithm proceeds to the next element.The process continues until the global matrix and right-hand-side vector arefully assembled.


b = ρg

L/2

FIGURE 3.12

Composite bar subjected to gravity.

3.5 One-dimensional elasticity

We end this chapter by considering a more practical example. Consider a barwith a solid circular cross section as shown in Figure 3.12. Let us suppose thatthe bar is made of two different materials, material 1 and material 2, labeledin the figure. The total length of the bar is 4m, and each material section is oflength 2m. The two different materials are assumed to be perfectly bonded atthe interface located at x = 2m. The bar is subjected to the force of gravityacting in the axial direction. The material properties, Young’s modulus, E,and mass density, ρ, for each section are taken to be those given in Table 3.5.

Next, let us assume that the normal stress distribution on any cross sectionalong the length of the bar is uniform. This means that at any given crosssection, the normal stress has the same magnitude at every point on that crosssection. Recall that stress has units of force per unit area. Let us also assumethat all points within the bar can displace in the axial direction only. Theseassumptions make the problem one-dimensional.

Thus far, we have been applying the finite element method to obtain approx-imate solutions to second-order, ordinary differential equations. The questionnow is: what is the governing differential equation to the problem at hand,and how do we derive it?

To derive the differential equation that governs the response of the one-

TABLE 3.5

Mechanical properties forcomposite bar.

Material E (GPa) ρ (kg/m3)

1 2 2000

2 200 7850


σ(x) σ(x + ∆x)

x

∆x

FIGURE 3.13

Section of a one-dimensional bar.

dimensional bar described above, let us consider a small section of the bar oflength ∆x as shown in Figure 3.13. The left end of the section is located atposition x, and the right end of the section is located at x + ∆x. The normalstress acting at the right end is σ(x + ∆x), and the normal stress acting atthe left end is σ(x). The force of gravity generates a net force acting on thesmall segment of magnitude ρg∆x, where ρ is the mass density of the sectionmaterial (either ρ1 or ρ2), g is the gravitational constant, and A is the cross-sectional area. The governing differential equation that we are seeking comesfrom a simple force balance. Summing forces in the x direction gives

A σ(x + ∆x) − σ(x) + Aρg∆x = 0 (3.53)

If we now divide both sides of equation (3.53) by A∆x we obtain

σ(x + ∆x) − σ(x)

∆x+ ρg = 0 (3.54)

Equation (3.54) must hold for arbitrarily small ∆x. Hence, in the limit as ∆x

goes to zero, we get

lim∆x→0

σ(x + ∆x) − σ(x)

∆x+ ρg = 0 (3.55)

Notice that the first term on the left-hand side of (3.55) is the definition ofthe derivative, dσ/dx. Thus we can write

dσ

dx+ ρg = 0 (3.56)

For linearly elastic material behavior, the normal stress, σ, is related to theextensional strain, ǫ, through the one-dimensional form of Hooke’s law, i.e.,

σ = Eǫ (3.57)


where E is Young’s modulus. The extensional strain at point x is defined asthe change in length divided by the original length of a small section of thebar of length ∆x in the limit as ∆x goes to zero, i.e.,

ǫ = lim∆x→0

∆u

∆x=

du

dx(3.58)

where u = u(x) is the axial displacement. Equation (3.58) is called a strain-displacement relationship. Plugging the strain-displacement relationship (3.58)along with the constitutive equation (3.57) into (3.56) yields

Eu′′ + ρg = 0 (3.59)

Comparing equation (3.59) above with equation (3.1) at the very beginningof this chapter, we see that equation (3.59) is just a special case of (3.1)with a = E, and c = ρg. Because both ends of the bar are clamped, nodisplacement can occur at either end. Thus the boundary conditions for theproblem are u(0) = 0 and u(L) = 0.

The finite element matrices and vectors for the present one-dimensionalelasticity problem are obtained by writing down the weak form of equation(3.59) over a single element and then substituting the finite element approx-imations into this expression. If we model the problem, for example, withtwo linear-u elements, the element matrices and vectors for elements insidematerial 1 and material 2 are given in Table 3.6. We can ensure continuity ofdisplacement at the interface between the two different materials by placinga single node at the interface where x = 2m. Given the element matrices andvectors, the global system of equations is obtained using the finite element as-sembly process described in the previous section, and enforcing the prescribeddisplacement boundary conditions at both ends of the bar.

3.5.1 Strain and stress calculation

After the solution to the global system of equations is obtained, the resultingnodal displacements are used to obtain the strain and stress in each element

TABLE 3.6

Element matrices and vectors for linear-uelements inside material 1 and material 2.

Region Ke fe

material 1E1

le

[

1 −1−1 1

]

ρ1gle

2

11

material 2E2

le

[

1 −1−1 1

]

ρ2gle

2

11


as a post-processing step. To illustrate the procedure, suppose we model theproblem with linear-u elements. The nodal displacements associated with thelocal left and right nodes are then u1 and u2, respectively. The strain in theelement is obtained by multiplying the 1 × 2 element B-matrix by the 2 × 1vector containing the local nodal displacements. Doing this we get

ǫ = Beue

=[

−1/le 1/le

]

u1

u2

=u2 − u1

le(3.60)

where le is the length of the element. We see here that for the case of the linear-u element, the strain turns out to be constant along the length of the element.For this reason, the element is often referred to as a constant-strain element.Once the strain in the element is obtained, the stress is easily computed usingHooke’s law. We must make sure, however, to plug in the correct Young’smodulus, depending on whether the element lies in material 1 or material 2.

If we model the problem instead with quadratic-u elements, for example,the strain calculation is the same, except we would use the element B-matrixdefined in (3.22). Letting u1, u2, and u3 be the displacements at local nodes1, 2, and 3 of an individual quadratic-u element, the strain in the element iscomputed as

ǫ =Be

1 × 3ue

3 × 1

=2

le

[

−1

2+ ξ −2ξ

1

2+ ξ

]

u1

u2

u3

where ξ = 2x/le is the natural (dimensionless) coordinate defined in Fig-ure 3.4, and x is the local coordinate measured positive to the right startingfrom local node 2 located at the middle of the element. Notice that the strainalong the length of a quadratic-u element varies linearly with ξ and hencewith x. When the constitutive behavior is governed by the one-dimensionalHooke’s law, σ = Eǫ, the stress also varies linearly along the length of aquadratic-u element.

3.5.2 Finite element results

We end this section by presenting some finite element results for the casewhere the composite bar is modeled with linear-u elements. The numericalresults for the case where the mesh is composed of four linear-u elements ineach material (for a total of eight elements) are presented in Figure 3.14.

In the upper plot in Figure 3.14, the displacement, u, normalized withrespect to the quantity E1/(ρ1g) is plotted versus the dimensionless length,

One-Dimensional Problems 77u/(

E1

ρ1g)

x/L

ǫ×

10−

6σ/(ρ

1gL

)

0 0.2 0.4 0.6 0.8 1

0 0.2 0.4 0.6 0.8 1

0 0.2 0.4 0.6 0.8 1

-400

-300

-200

-100

0

100

-100

0

100

200

0

2

4

6

FIGURE 3.14

Finite element solution (eight linear-u elements).

x/L, where L is the total length of the composite bar. As shown in theplot, the displacement varies linearly along the length of each element, andthe displacement is continuous at x/L = 0.5, the interface between the twodifferent materials. Notice that the essential boundary conditions at x = 0and x = L are satisfied exactly by the finite element solution.

The strain, ǫ, along the length of the bar is plotted versus x/L in themiddle plot in Figure 3.14. Because we are using linear-u elements, the plotreveals that the strain field is discontinuous, and the strain is constant alongthe length of each individual element. The normal stress, σ, normalized withrespect to ρ1gL is plotted versus x/L in the lower plot in Figure 3.14. Like


PSfrag

u/(

E1

ρ1g)

x/L

ǫ×

10−

6σ/(ρ

1gL

)

0 0.2 0.4 0.6 0.8 1

0 0.2 0.4 0.6 0.8 1

0 0.2 0.4 0.6 0.8 1

-400

-300

-200

-100

0

100

-100

0

100

200

-2

0

2

4

6

FIGURE 3.15

Finite element solution (96 linear-u elements).

the strain field, the stress field is also discontinuous, and it is constant alongeach individual element. The magnitude of the jump (discontinuity) in stress(and/or strain) can be used as a measure of the accuracy of the numericalresults. In general, as the mesh is refined, the magnitude of the jump de-creases and accuracy increases. The finite element results for a much finermesh composed of 96 linear-u elements are shown in Figure 3.15. In the finemesh, the discontinuity in strain and stress between each element is almostindistinguishable, revealing that the finite element solution will not changeappreciably with further mesh refinement. When this happens, the solutionis said to have converged. Notice that with the fine mesh, the strain and stress


distributions are linear in each material section, but there is a discontinuityin the strain field at x = L/2. This is due to the fact that the bar is made oftwo different materials, which produces a kink in the displacement field at thislocation. Notice that the stress field is continuous, consistent with Newton’sthird law.

3.6 Problems

Problem 3.1

Consider the following differential equation:

u′′ + 2u = 0

0 ≤ x ≤ 1

u′(0) = 3

u(1) = 0

(a) Derive the weak form.

(b) Derive the element matrix and vector for a linear-u elementof length l = 1.

(c) Derive the element matrix and vector for a quadratic-u elementof length l = 1.

(d) Mesh the problem with two linear-u elements, and assemble theglobal system of equations. Enforce the boundary condition,u3 = 0, at the right end to obtain the reduced system. Solve forthe unknowns u1 and u2.

(e) Repeat part (d) using a mesh of one quadratic-u element.

Problem 3.2

Consider the problem of a rod with a circular solid cross section that is con-nected to a rotating shaft as shown in Figure 3.16. The shaft is spinning witha constant angular velocity ω = 50 rad/s. Assume that the rod is made ofsteel with Young’s modulus E = 200GPa, and mass density ρ = 7850 kg/m3.The length of the rod is L = 1m. Model the rod with two linear-u elementsand determine:

(a) The displacement in the X direction at the end ofthe rod.

(b) The stress distribution σ(X) along the length of therod.


Y

ω

X

FIGURE 3.16

Problem 3.2. Rod connected to a rotating shaft.

Hint: For a rod rotating about a fixed axis, the acceleration in the X directionat any point along the length of the rod is ω2X. Treat the problem as astatic problem with a linearly varying body force, b = ρω2X, acting in the X

direction. Assume that the rod can displace only in the X direction and thatthe rod is fixed at the left end.

Problem 3.3

X

E(X) = C1eC2X

FIGURE 3.17

Problem 3.3. Functionally graded bar subjected to a constant body force.

Consider the problem of an inhomogeneous bar with a constant circularcross section that is fixed at both ends as shown in Figure 3.17. The bar oflength L is subjected to a constant body force b(x) = C acting in the positiveX direction. The Young’s modulus, E(X) of the bar material varies with


TABLE 3.7

Problem 3.3. Numerical values.

E1 (GPa) E2 (GPa) C (N/m3) L (m)

66.0 200.0 78.0 1.0

position along the length of the bar and is given as

E(X) = C1eC2X

C1 = E1

C2 = ln

(

E2

E1

)

/L

where E1 is the value of Young’s modulus at the left end of the bar, and E2

is the value at the right end. The numerical values for E1, E2, L, and C aregiven in Table 3.7. The diameter of the bar is d = 0.01m.

(a) Determine the element matrix and vector for a linear-u elementinside the domain 0 ≤ X ≤ 1m.

(b) Write a one-dimensional finite element program to obtain an approx-imate solution to the problem.

(c) Obtain plots of the displacement, u, versus X for severaldifferent meshes. Refine your mesh until your solutionno longer changes appreciably with the number of elements.

(d) From your most refined mesh used in part (c), obtain a plot of thenormal stress in the column σ(X), as a function of X.

(e) If the ultimate compressive strength of the bar material is 10MPa,do you expect the column to fail?

Problem 3.4

A bar with a tapered cross section is shown in Figure 3.18. The cross section

X

D1

D2

FIGURE 3.18

Problem 3.4. Bar with a tapered cross section.


is circular with diameter D1 at the left end and D2 at the right end, andhence the cross-sectional area is a function of X, i.e., A = A(X). The barhas length L and Young’s modulus E. Discretize the bar with one linear-uelement and calculate the following:

(a) The element matrix in terms of D1, D2, E, and L.

(b) The element vector for the case when a constant body forceb(X) = C acts in the X direction.

4

Linearized Theory of Elasticity

This chapter covers the fundamentals of the linearized theory of elasticity.The material is essential for understanding the application of the finite ele-ment method to problems in the area of solid mechanics. The chapter beginswith the derivation of Cauchy’s law, which is a very useful tool for assess-ing the accuracy of finite element solutions. We then go on to deriving theequilibrium equation, the governing partial differential equation for problemsin solid mechanics in the absence of inertia. The chapter also covers the im-portant topics of the small strain tensor, Hooke’s law, plane stress and planestrain, and axisymmetric problems. The chapter ends with a derivation ofthe weak form of the equilibrium equation, which is the starting point for thefinite element implementation presented in Chapter 6.

4.1 Cauchy’s law

Consider the solid body shown in Figure 4.1. The surface of the body islabeled Γ, and Ω is the interior volume (or area, depending on whether weare working in two or three dimensions). As shown in the figure, the body isloaded by a distribution of surface tractions over a portion of its boundary.A surface traction is a vector quantity that has units of force per unit area.

Ω

Γu

ti

Γt

b

FIGURE 4.1

Solid body subjected to surface tractions and body force field.

83


If we multiply a traction vector by a differential surface area, dΓ, we get adifferential surface force acting in the same direction as the traction vector.The body can also be subjected to a body force distribution. A body force isa vector quantity that acts at each point throughout the volume of a body.A body force has units of force per unit volume. When a body force vectoris multiplied by a differential volume element, dΩ, the result is a differentialforce vector acting in the same direction as the body force.

We can decompose the surface, Γ, into two parts, Γu and Γt. The quantityΓu represents a portion of the boundary where the displacement components,ui, are specified. Likewise, Γt is the portion of the boundary where the com-ponents, ti, of the traction vector are prescribed. We mention here that onecannot specify both the traction vector and the displacement vector at thesame point. As an example, one cannot specify both t1 and u1 at the samepoint on Γ. If t1 is specified at a point, nothing can be said about u1 at thatpoint. Additional information is needed about the material properties be-fore the displacement field can be fully determined for a given set of specifiedboundary tractions.

Having defined the traction vector, we now focus on the general state ofstress at a point and seek a relationship between the state of stress at a pointand the traction vector at that same point. This relationship is known asCauchy’s law. To begin, consider an infinitesimal tetrahedron that is cut outof a material in such a way that three of its sides have unit outward normals−e1, −e2, and −e3, respectively, as shown in Figure 4.2. One of the faces(the inclined face) has area, A, and unit outward normal n. The tractionvector acting on the inclined face is labeled t(n). Now let t(e1), t(e2), and t(e3)

be traction vectors acting on the planes whose unit outward normals are e1,e2, and e3. As a consequence of Newton’s third law (for every action thereis an equal and opposite reaction), the three traction vectors acting on theremaining (negative) faces of Cauchy’s tetrahedron are −t(e1), −t(e2), and−t(e3). The three traction vectors, t(ei) each have three components. We willcall these components σi1, σi2, and σi3. Hence, we can write

t(ei) = σij ej (4.1)

The sum of all external force vectors acting on the faces of the tetrahedronmust be equal to zero. Letting b represent the body force acting throughoutthe volume of the tetrahedron, the equilibrium statement can be written as

−t(e1)A1 − t(e2)A2 − t(e3)A3 +1

3bhA + t(n)

A = 0 (4.2)

where here we have used the fact that the volume, V , of the tetrahedron isequal to one-third its base, A, times its height, h, i.e.,

V =1

3Ah

Linearized Theory of Elasticity 85

n

X1

X2

X3

−e1

−e2

−e3

h

A

FIGURE 4.2

Cauchy’s tetrahedron.

We note that the height of the tetrahedron is the shortest distance from theorigin to a point on the inclined face. The quantities A1, A2, and A3 inequation (4.2) are the areas of the faces perpendicular to e1, e2, and e3,respectively.

In order to proceed further, it is necessary to derive a relationship betweenthe surface area of the inclined face and the areas, Ai, of the other three faces.To begin, let a be the length of the edge of the tetrahedron that lies on theX1 axis. Now consider the vector a e1. The height, h, of the tetrahedron isequal to the projection of a e1 onto the n axis, i.e.,

h = a (e1 · n) (4.3)

Hence,

a =h

n1

The volume, V = 1/3Ah, can now be equivalently written as

V =1

3A1a

=1

3A1

h

n1

(4.4)

Comparing the last line in (4.4) with the volume expression that we startedwith, we see that

A1 = An1


where n1 is the component of n in the e1 direction. Using this same argumentwith the remaining faces, one can readily see that

Ai = Ani (4.5)

Let us now return to the balance statement (4.2). By substituting (4.5)into (4.2) we get

−t(e1)An1 − t(e2)An2 − t(e3)An3 +1

3bhA + t(n)

A = 0 (4.6)

The last step in the derivation of Cauchy’s law is to divide both sides ofequation (4.6) by A and take the limit as h → 0. Doing this we obtain,

t(n) = t(e1)n1 + t(e2)n2 + t(e3)n3

= t(ei)ni

= niσijej (4.7)

Equation (4.7) is known as Cauchy’s law. It can be written in tensor notationas

t(n) = n · σ (4.8)

or in component form as

t(n)

j= niσij (4.9)

The component form of equation (4.8) comes from

t(n) = n · σ

= (nk ek) · (σij eiej)

= nkδkiσij ej

= niσij ej (4.10)

Throughout the remainder of the book, when we write Cauchy’s law (eitherin tensor or component form) we will drop the superscript (n) and it willbe understood implicitly that the traction vector acts on a plane whose unitoutward normal is n. We will simply write

t = σ · n or tj = niσij (4.11)

Cauchy’s law (4.11) reveals several important facts. First of all, recallingthe definition of a second-order tensor given in Chapter 2 on page 17, it re-veals that the nine components, σij , of the three traction vectors, t(ei), forma second-order tensor, σ. This second-order tensor is called the Cauchy stresstensor. Cauchy’s law also tells us how the components of a traction vectoracting at a point on any inclined plane are related to the stress components


X1

X2

X3

σ11

σ12

σ22

σ33

σ13

σ23

σ31

σ32

σ21

FIGURE 4.3

Components of the Cauchy stress tensor.

at that point to ensure equilibrium of an infinitesimal volume element sur-rounding that point.

The nine components of the Cauchy stress tensor are shown in Figure 4.3.As shown in the figure, the stress component σ11 is the normal stress actingon the face whose unit outward normal points in the X1 direction. This face isoften referred to as the positive X1 face. Similarly, the stress components σ22

and σ33 are the normal stress components acting on the positive X2 and X3

faces, respectively. The shear stress components are also shown in the figure.The stress component σ12 is the shear stress acting on the positive X1 facepointing in the X2 direction. For any particular stress component, the firstsubscript refers to the face on which the stress component acts. The secondsubscript refers to the direction in which it acts.

We end this section by making the important observation that the Cauchystress tensor is symmetric. Recall from our study of Chapter 2 that if σij =σji, then the tensor, σ, is symmetric. One can see that the Cauchy stresstensor is symmetric by referring back to Figure 4.3 and performing individualmoment balances about the X1, X2, and X3 axes, respectively. For example,a moment balance about the X1 axis gives

σ32∆X1∆X2∆X3 − σ23∆X1∆X3∆X2 = 0 ⇒

σ32 = σ23 (4.12)

where we have used ∆X1, ∆X2, and ∆X3 to denote the lengths of the edgesof the cube. Similarly, by summing moments about the X2 and X3 axes weget

σ21 = σ12

σ31 = σ13

88 Finite Element Methodt2 = 1.0 psi

X1

X2

A

FIGURE 4.4

Application of Cauchy’s law.

Because the Cauchy stress tensor is symmetric, Cauchy’s law (4.11) can alsobe expressed as

tj = niσij

= niσji

= σjini

Changing the free index from j to i in the previous equation and using j forthe repeated index, we get

ti = σijnj (4.13)

Example 4.1

To illustrate the use of Cauchy’s law and its utility in the interpretation offinite element results, consider the problem of a 10 in. × 10 in. plate with acentrally located hole as shown in Figure 4.4. The plate is loaded by a uniformtraction distribution t2 = 1.0 psi along the top and bottom edges. A contourplot of the stress component σ22 is superposed on top of the plate geometry.For the given dimensions of the plate and the applied traction distribution,the maximum value of σ22 in the plate is slightly above 3.0 psi. The ratioof applied stress versus the maximum stress found in the plate is called thestress concentration factor. Here the stress concentration factor is

σmax

22

t2≈ 3.0


The location of the stress concentration is labeled A in the figure. Let us nowinvestigate the values of the stress components σ11 and σ12 at this location.At point A, the unit outward normal to the surface is parallel to the X1 axis.Hence, n1 = 1 and n2 = 0. Because the inside of the hole is a traction-freesurface, ti = 0 at all points on the inside of the circular hole. ApplyingCauchy’s law (4.13) we get

t1 = 0 ⇒

σ11n1 + σ12n2 = 0 ⇒

σ11(1) + σ12(0) = 0 ⇒

σ11 = 0 (4.14)

Similarly, because t2 also equals zero at this point we get

t2 = 0 ⇒

σ21n1 + σ22n2 = 0 ⇒

σ21(1) + σ22(0) = 0 ⇒

σ21 = σ12 = 0 (4.15)

We emphasize that the stress component σ22 at point A cannot be obtainedfrom Cauchy’s law. This stress value must be obtained from the solution ofthe elasticity problem.

4.2 Principal stresses

An understanding of principal stresses and principal directions is very impor-tant in the analysis of solids and structures. The procedure for finding princi-pal stresses and principal directions is identical to that for finding eigenvaluesand eigenvectors of second-order tensors described in Chapter 2 in Section2.3.2.

Suppose that the Cauchy stress tensor is known at some point P in thesolid shown in Figure 4.5. Now imagine making a variety of “cuts” throughpoint P . Here we use the word “cut” to represent a plane, passing throughP , whose unit outward normal is n as shown in the figure. Cauchy’s lawtells us that if we take the dot product of the stress tensor, σ, and the unitoutward normal, n, the result is a traction vector pointing in some direction.We now seek a very special direction, such that the dot product, σ ·n, gives atraction vector that is parallel to n. Stated mathematically, we seek to solvethe following equation:

σ · n = σn (4.16)


Pn

FIGURE 4.5

Inclined plane passing through solid.

Assuming for the moment that equation (4.16) can be solved, the scalar quan-tity σ represents the magnitude of the normal stress acting on a plane havingunit outward normal n. Note that no shear stress acts on this plane. Thequantity σ is called a principal stress, and the unit vector n is the correspond-ing principal direction. The principal stresses and directions are eigenvaluesand eigenvectors of the stress tensor.

Next, it is convenient to express equation (4.16) as follows:

[σ − σI] · n = 0 (4.17)

where I is the second-order identity tensor. As discussed in Chapter 2, equa-tion (4.17) can be satisfied (for nonzero vectors n) only if the following deter-minate is equal to zero,

|σ − σI| = 0 (4.18)

Expanding equation (4.18), we get the following cubic equation:

−σ3 + I1σ

2 + I2σ − I3 = 0 (4.19)

where

I1 = tr(σ)

I2 =1

2

[

σ : σ − tr(σ)2]

I3 = detσ (4.20)

are scalar coefficients called stress invariants. They are called invariants be-cause no matter what coordinate system the components of the stress tensorare referred to, one will always get the same coefficients in equation (4.19).


Once the three roots of equation (4.19) are found, the corresponding princi-pal directions are obtained by successively plugging in each principal stressback into equation (4.17) and solving for the vector n. By convention, theprincipal stresses are arranged in order of algebraically largest to smallest val-ues. We emphasize here that the state of stress at a point can be completelycharacterized by the three principal stresses along with the three principaldirections.

Example 4.2

To illustrate the procedure for calculating principal stresses and principaldirections, consider the following state of stress at a point:

σ11 σ12 σ13

σ12 σ22 σ23

σ13 σ23 σ33

=

1 −10 0−10 1 0

0 0 0

MPa (4.21)

Dropping the units for the intermediate calculations, the determinant equationto be solved is then

∣

∣

∣

∣

∣

∣

1 − σ −10 0−10 1 − σ 0

0 0 0 − σ

∣

∣

∣

∣

∣

∣

= 0

Expanding the determinant above we obtain

−σ3 + 2σ2 + 99σ = 0 (4.22)

The three roots of the characteristic equation (4.22) are found to be

σ1 = 11MPa

σ2 = 0MPa

σ3 = −9MPa (4.23)

Notice that the three principal stresses are labeled in the order of algebraicallylargest to smallest values.

Next, in order to calculate the principal directions, we begin by substitutingthe first principal stress into equation (4.17) and solving for the componentsof the vector n. Doing this we get

−10 −10 0−10 −10 0

0 0 −11

n1

n2

n3

=

000

(4.24)

Expanding the first row of (4.24) we see that

−10n1 − 10n2 = 0 ⇒

n2 = −n1, and

n3 = 0 (4.25)


Notice that we would have come to the same conclusion as (4.25) if we hadexpanded the second row of (4.24). If we insist that the vector n be a unitvector, it follows that its magnitude must be 1, i.e.,

n2

1+ n

2

2+ n

2

3= 1 (4.26)

Finally, by substituting (4.25) into (4.26) we get

n1 =

√2

2

n2 = −

√2

2⇒

n(1) =

√2

2e1 −

√2

2e2 (4.27)

where the superscript (1) emphasizes that the principal direction that we justcalculated is associated with the first principal stress, i.e, n(1) is the principaldirection associated with σ1. Following the same procedure, the other twoprincipal directions are found to be

n(2) = e3

n(3) =

√2

2e1 +

√2

2e2 (4.28)

If we carefully study the three principal directions obtained in this section,we will notice that the three unit vectors n(1), n(2), and n(3) are mutuallyperpendicular, i.e., n(i) · n(j) = δij . This happens to be a property of sym-metric, second-order tensors whose components are real. Such tensors havereal eigenvalues and mutually perpendicular eigenvectors.

In order to better illustrate the results that we have just obtained, imaginecutting out an infinitesimal stress element surrounding point P in the followingway. First, make two cuts that are parallel to the X1 axis. Second, make twocuts parallel to the X2 axis. Finally, make two cuts parallel to the X3 axis.Doing this we can extract a stress element that is lined up with the X1,X2, and X3 axes. The state of stress (4.21) can be represented graphically bydrawing the normal and shear stress components acting on each of the six facesas shown in Figure 4.6 (a). Note that for the state of stress considered in thepresent example, the normal and shear stresses are zero on faces perpendicularto the X3 axis. Hence we can represent the state of stress by drawing a two-dimensional stress element.

By making cuts perpendicular to the n(1), n(2), and n(3) principal direc-tions, we can also extract a stress element surrounding point P as shown inFigure 4.6 (b). Note that we can also represent this element in two dimen-sions, because the principal direction n(2) happens to be parallel to the X3

axis. If we now let the axes X ′1, X ′

2, and X ′

3line up with the principal direc-

tions n(1), n(3), and n(2), respectively, then the stress components referred to


this primed set of axes can be expressed in matrix notation as

σ =

11 0 00 −9 00 0 0

MPa (4.29)

If we compute the three invariants of the stress tensor from the stress compo-nents in the principal orientation (4.29) we get

I1 = tr(σ) = 11 − 9 + 0 = 2

I2 =1

2

[

σ : σ − tr(σ)2]

=1

2

[

(11)2 + (−9)2 − (2)2]

= 99

I3 = detσ = (11)(−9)(0) = 0

which is the same as the result that we obtained using the stress components(4.21); see equation (4.22).

11 MPa

−9MPa

π

4

1 MPa

1 MPa

−10 MPa

X1

(a) (b)

FIGURE 4.6

Stress element in the aligned orientation (a) and the principal orientation (b).

4.3 Equilibrium equation

Equilibrium of a solid body requires that the net external force vector actingon the body is equal to zero. The net external force acting on the body comesfrom the surface traction distribution and the body force field. The net forcefrom the traction distribution can be obtained by “summing up” all of thedifferential force vectors, ti dΓ, acting on the surface. Likewise, the net forcevector due to the body force distribution comes from the summation of thedifferential force vectors, bi dΩ. So the equilibrium statement for the body


can be written as∫

Γ

ti dΓ +

∫

Ω

bi dΩ = 0 (4.30)

We saw in Section 4.1 that the traction vector, t, acting at a point P on Γ isrelated to the Cauchy stress tensor at that point through the relation

t = n · σ ⇒

ti = σijnj

where n = nj ej is the unit outward normal to the boundary at point P .The key to deriving a partial differential equation from a balance law is

to convert all integrals in the balance statement into volume integrals. Tothis end, we now recognize that the first integral in the equilibrium statement(4.30) can be recast as a volume integral by using the divergence theoremcovered in Chapter 2, Section 2.4.6. The divergence theorem states that thesurface integral,

∫

Γ

n · σ dΓ

can be replaced by the following volume integral:∫

Ω

∇ · σ dΩ

The vector quantity ∇ · σ is called the divergence of the stress tensor. Thecomponents of this vector can be obtained using the techniques that wereintroduced in Chapter 2, Section 2.3.3, i.e.,

∇ · σ =

(

∂

∂Xi

ei

)

· (σjk ej ek)

=∂σjk

∂Xi

δij ek

= σjk,j ek

= σji,j ei

= σij,j ei

where the comma denotes partial differentiation with respect to the spatialvariables, and we have used the fact that the Cauchy stress tensor is symmetric(σij = σji).

The equilibrium statement, in terms of volume integrals, can now be writtenas

∫

Ω

(σij, j + bi) dΩ = 0 (4.31)

Because the statement (4.31) must be true for all subdomains inside Ω, onecan argue that the integrand in (4.31) must be identically zero at every pointinside Ω. This fact is actually a subtle point that requires further explanation.


Γ3

Γ

Γ1Γ2

∫

Ω2f(x, y) dΓ 6= 0!

Ω

FIGURE 4.7

Illustration showing that the equilibrium statement must be true for all sub-domains inside Ω.

The argument can be better understood if we take a look at Figure 4.7. Herein the figure, we are considering a square (two-dimensional) domain withboundary Γ. Let us imagine for the moment (for the sake of simplicity)that the integrand in (4.31) is a scalar function represented by the surfaceplot shown in the figure. Now suppose that the function is zero everywhereexcept inside the subdomains Ω2 and Ω3, which are bounded by Γ2 and Γ3

labeled in the figure. In addition, suppose that the function inside Ω2 is equalin magnitude but opposite in sign to the function inside Ω3. Obviously, theintegral of such a function over the entire domain, Ω, would be zero, seeminglysatisfying equation (4.31). However, the equilibrium statement must be truefor all subdomains. If we cut out an arbitrary portion of the body, we requirethat portion to be in equilibrium. Notice for example that the integral ofthe function over Ω2 is not zero, violating equation (4.31). The fact thatthe integrand in (4.31) is actually a vector field does not alter this argument.One can imagine a vector field being zero everywhere inside Ω except for thesubdomains Ω2 and Ω3, in which we could define nonzero vector fields thatare equal in magnitude but opposite in direction.

So the governing partial differential equation (the equilibrium equation)now emerges from the balance statement (4.31) and is written as

σij,j + bi = 0 in Ω

ti = t∗i

on Γt

ui = u∗i

on Γu (4.32)

The partial differential equation (4.32) and boundary conditions that go alongwith it are referred to as the strong form. Here the superposed star indicatesthat the quantity (either traction or displacement component) is specified ata point on the boundary.


4.4 Small-strain tensor

In this section, we present the mathematical and physical arguments requiredto gain an understanding of the meaning of the small-strain tensor and howits components are related to the displacement components in a continuousbody.

4.4.1 Relative stretch

We begin by defining the relative stretch, ε(n), at a point in the body. Therelative stretch is defined as the change in length divided by the originallength of an infinitesimal line segment that initially lies in a direction parallelto the unit vector n. Let us consider an infinitesimal line segment dX(n), withmagnitude dS(n) in the original configuration as shown in Figure 4.8. Afterdeformation, this vector maps into the vector dx(n) whose magnitude is ds(n).The relative stretch can then be written as

ε(n) =

ds(n) − dS(n)

dS(n)

= λ(n)

− 1 (4.33)

where λ(n) = ds(n)/dS(n) is called the stretch of the line segment. The quan-tity λ(n) is also commonly referred to as the stretch ratio.

Now suppose that we choose a line segment dX(1) that is initially parallelto the X1 axis. The relative stretch of this line segment is

ε(1) =

ds(1) − dS(1)

dS(1)(4.34)

It turns out that the relative stretch ε(1) can be expressed in terms of thedeformation gradient tensor, F, introduced at the end of Chapter 2. The

X1X1

X2X2

dX(n) dx(n)

FIGURE 4.8

Relative stretch of infinitesimal line segment.


argument utilizes that fact that an infinitesimal line segment in the originalconfiguration, dX, maps into an infinitesimal line segment dx in the deformedconfiguration through the relation

dx = F · dX

= dX · FT (4.35)

Hence, the magnitude of the infinitesimal line segment in the deformed con-figuration, ds(1), can be expressed as

ds(1) =

√

dx(1) · dx(1)

=√

dX(1) · (FT · F) · dX(1)

=√

dX(1) · C · dX(1)

whereC = FT

· F (4.36)

is called the Green deformation tensor. The stretch, ε(1), can now be writtenas

ε(1) =

√dX(1) · C · dX(1) − dS(1)

dS(1)(4.37)

The trick to getting equation (4.37) into a manageable form is to divide boththe numerator and denominator by dS(1) and to recognize that the vectordX(1)/dS(1) is simply the unit base vector e1. Doing this yields

ε(1) =

√

e1 · C · e1 − 1

=√

C11 − 1 (4.38)

4.4.2 Angle change

Let us now consider two infinitesimal line segments dX(1) and dX(2) thatare initially parallel to the X1 and X2 axes in the undeformed configurationas shown in Figure 4.9. Since these vectors are perpendicular, the anglebetween them is π/2. These vectors become dx(1) and dx(2) in the deformedconfiguration. The quantity θ12 is the angle between these two vectors in thedeformed configuration. The engineering shear strain, γ12, is defined as thedifference between the original angle, π/2, and the angle θ12, i.e.,

γ12 =π

2− θ12 (4.39)

It is now instructive to look at the relationship between θ12 and the in-finitesimal vectors dx(1) and dx(2). To begin, the dot product between thesetwo vectors gives

dx(1)· dx(2) = ds

(1)ds(2) cos θ12


X1

X2

X1

X2

dX(1)

dX(2)

π/2

dx(1)

dx(2)

θ12

FIGURE 4.9

Relative stretch and angle change of two infinitesimal line segments.

Next, let us divide both sides of the previous equation by dS(1) and dS(2),the magnitudes of the infinitesimal line segments in the original configuration.Doing this we get

dx(1)

dS(1)

dx(2)

dS(2)= λ1λ2 cos θ12 (4.40)

where λ1 and λ2 are the stretch ratios in the X1 and X2 directions, respec-tively. Equation (4.40) can now be written in terms of the deformation gra-dient tensor, F, as follows:

dX(1)

dS(1)·

FT· F

·dX(2)

dS(2)= λ1λ2 cos θ12 (4.41)

Recognizing that dX(1)/dS(1) = e1, dX(2)/dS(2) = e2, and FT · F = C,equation (4.41) becomes

e1 · C · e2 = λ1λ2 cos θ12 (4.42)

and hence,

cos θ12 =C12

λ1λ2

(4.43)

The connection between the engineering shear strain, γ12, and the quantity,cos θ12, can be established by taking the sine of both sides of equation (4.39).Doing this we obtain

sin γ12 = sin[

π

2− θ12

]

= sinπ

2cos θ12 + cos

π

2sin θ12

= cos θ12 (4.44)

where the second line in (4.44) follows from the trigonometric double-angleformula. From equation (4.44), we can express γ12 in terms of C12 and the


stretch ratios as follows:

sin γ12 =C12

λ1λ2

(4.45)

4.4.3 Small-displacement gradients

Recall from Chapter 2 that the deformation gradient tensor, F, contains in-formation about how individual points are mapped from the original config-uration to the deformed configuration. The components of F can be writtenas

Fij =∂xi

∂Xj

=∂

∂Xj

(ui + Xi)

=∂ui

∂Xj

+ δij

= ui,j + δij (4.46)

where, again, the comma denotes partial differentiation with respect to thespatial X1 − X2 − X3 coordinates.

When the displacement gradients (partial derivatives of the displacementcomponents) are small, we can neglect the products of these gradients thatwould otherwise be nonzero components of the Green deformation tensor. Ifwe neglect these products, the components of the Green deformation tensorbecome

Cij = FkiFkj

≈

2u1,1 + 1 u1,2 + u2,1 u1,3 + u3,1

u1,2 + u2,1 2u2,2 + 1 u2,3 + u3,2

u1,3 + u3,1 u2,3 + u3,2 2u3,3 + 1

(4.47)

Having assumed small-displacement gradients, let us now compute the relativestretch ε(1). From equation (4.38) we can write

ε(1) =

√

C11 − 1

≈√

2u1,1 + 1 − 1 (4.48)

A Taylor series expansion of the first term about u1,1 = 0 yields

√

2u1,1 + 1 ≈ 1 +1

2(1)−

12 (2)u1,1

= 1 + u1,1 (4.49)

and hence,

ε(1)

≈ u1,1


The stretches ε(2) and ε(3) can be found using the same procedure. The resultis

ε(2)

≈ u2,2

ε(3)

≈ u3,3

The small-strain tensor is defined so that the stretch in the direction of anarbitrary unit vector n is obtained by pre- and post-multiplying the straintensor by the unit vector n. This operation can be written as

ε(n) = n · ǫ · n (4.50)

or in matrix notation,ε(n) = nT

ǫn (4.51)

Having made this definition, the components of the small-strain tensor canbe written as

[ǫij ] =

u1,11

2(u1,2 + u2,1)

1

2(u1,3 + u3,1)

1

2(u1,2 + u2,1) u2,2

1

2(u2,3 + u3,2)

1

2(u1,3 + u3,1)

1

2(u1,2 + u2,1) u3,3

(4.52)

Note that the diagonal components of the small-strain tensor defined in (4.52)contain the relative stretches in the X1, X2, and X3 directions, respectively.For small-displacement gradients, the off-diagonal terms are related to theengineering shear strains γ12, γ13, and γ23. To see this, let us recall therelationship (4.45) between the engineering shear strain component, γ12, andthe quantities C12, λ1, and λ2, i.e.,

sin γ12 =C12

λ1λ2

=C12

√C11C22

(4.53)

Expressing equation (4.53) in terms of the displacement gradients we get

sin γ12 =u1,2 + u2,1

√

(2u1,1 + 1)(2u2,2 + 1)(4.54)

If we view the right-hand side of (4.54) as a function of two independentvariables, u1,1 and u2,2, and expand this function as a two-term Taylor seriesabout u1,1 = 0 and u2,2 = 0 we obtain

sin γ12 = (u1,2 + u2,1) + (u1,2 + u2,1)(u1,1 + u2,2) (4.55)

Hence, when the displacement gradients are small, we can neglect the secondterm on the right-hand side of (4.55). In addition, when the displacement


gradients are small, the quantity, γ12, is much less than unity. Therefore,sin γ12 ≈ γ12, and the expression for the engineering shear strain reduces to

γ12 ≈ u1,2 + u2,1 (4.56)

Following the same argument, one can conclude that under the condition ofsmall-displacement gradients, the engineering shear strain in the X2−X3 andX1 − X3 planes are related to the displacement gradients as follows:

γ23 ≈ u2,3 + u3,2

γ13 ≈ u1,3 + u3,1

Notice that the small-strain components ǫ12, ǫ13, and ǫ23 in (4.52) above areequal to one-half the engineering shear strains γ12, γ13, and γ23. The 1/2factor is needed so that ǫ transforms according to the tensor transformationlaw given in Chapter 2, in Section 2.3.1.

The components of the small-strain tensor can be written down on one lineusing indicial notation as follows:

ǫij =1

2(ui,j + uj,i) (4.57)

Notice that ǫ is symmetric, and therefore we need to only keep track of six ofits components. These components are typically stored in a 6 × 1 vector as

ǫ =

ǫ11

ǫ22

ǫ33

γ23

γ13

γ12

(4.58)

Example 4.3

A two-dimensional deformation from the original configuration to the de-formed configuration is defined as

x1 = X1 + αX2

x2 = X2 (4.59)

The deformation described in (4.59) above is called simple shear. Let us nowcompute the relative stretch ε(n), where

n =

√

(2)/2√

(2)/2

From equation (4.59) we can compute the components of the deformationgradient tensor,

F =

[

1 α

0 1

]


and hence,

C =

[

1 α

α α2 + 1

]

The stretch in the direction of n can now be obtained directly from C. Weget

ε(n) =

√n · C · n − 1

= 1/2√

4 + 4α + 2α2 − 1

≈α

2(4.60)

The last line in (4.60) follows if one assumes that α is small, and expands thesecond line as a Taylor series about α = 0.

Let us now compute the same quantity ε(n) using the small-strain tensor.Hopefully we will get the same answer. Noting that ui = xi − Xi, the small-strain components turn out to be

ǫ =

[

u1,1 1/2(u1,2 + u2,1)1/2(u1,2 + u2,1) u2,2

]

=

[

0 α/2α/2 0

]

Computing the stretch in the n direction from (4.51) we get

ε(n) =

[√2/2

√2/2

]

[

0 α/2α/2 0

] [√2/2

√2/2

]

=α

2(4.61)

which, of course, is the correct result. Note that we would not get this result ifwe forgot the 1/2 factor in the definition of the small shear strain components.

4.5 Hooke’s law

Much of the linearized theory of elasticity hinges on the assumptions thatthe material is linearly elastic and isotropic. Loosely speaking, an isotropicmaterial is one whose mechanical properties are independent of the orientationin which the material is tested. In other words, an isotropic material does nothave any preferred directions.

Hooke’s law for linearly elastic materials states that there is a linear relation-ship between the Cauchy stress components and the small-strain components.This relationship can be written as

σij = Cijklǫkl (4.62)


where Cijkl represents the components of a fourth-order tensor called theelasticity tensor.

If a material is linearly elastic and isotropic, then the components of theelasticity tensor are invariant (remain unchanged) under a rotation of thecoordinate axes. An isotropic fourth-order tensor has the form

Cijkl = λδijδkl + 2µδikδjl (4.63)

where λ and µ are Lame’s constants. After substituting equation (4.63) into(4.62) we obtain

σij = λδijδkl + 2µδikδjl ǫkl

= λǫkkδij + 2µǫij (4.64)

Hence, from now on, we will write the component form of Hooke’s law forisotropic materials as

σij = λǫkkδij + 2µǫij (4.65)

The inverse of equation (4.65) can be obtained by writing the relationshipin tensor notation

σ = λ tr(ǫ)I + 2µǫ (4.66)

and taking the tensor scalar product of both sides with the second-orderidentity tensor, i.e.,

σ : I = λ tr(ǫ)I : I + 2µǫ : I ⇒

σkk = λǫkkδijδij + 2µǫkk

= 3λǫkk + 2µǫkk ⇒

ǫkk =σkk

3λ + 2µ(4.67)

Finally, by substituting equation (4.67) into (4.65) and simplifying we obtain

ǫij =1

2µσij −

λ

2µ (3λ + 2µ)σkkδij (4.68)

4.5.1 Engineering constants

In engineering practice, it common to use what are called engineering con-stants instead of the Lame’s constants that appear in equation (4.65). In thissection we define the engineering constants (Young’s modulus, Poisson’s ratio,and shear modulus) and present Hooke’s law in matrix notation in terms ofthese constants. At the end of this section, we provide relationships betweenthe engineering constants and Lame’s constants.

To begin, consider a state of stress at a point in a linearly elastic andisotropic material in which the only nonzero stress component is σ11. The


resulting extensional strain in the X1 direction is then ǫ11. Under these con-ditions, the normal stress σ11 and the extensional strain ǫ11 are related by

σ11 = Eǫ11 (4.69)

where E is Young’s modulus. Similarly, if the only nonzero stress componentis σ22 or σ33, then we would have σ22 = Eǫ22, or σ33 = Eǫ33. Due to thePoisson effect, the application of the stress σ11 causes the material to strainlaterally in both the X2 and X3 directions. For an isotropic material, themagnitude of the lateral contraction in both the X2 and X3 directions is thesame. Poisson’s ratio, ν, is defined as the ratio of the lateral strain to thelongitudinal strain, i.e.,

ν = −ǫ22

ǫ11=

ǫ33

ǫ11(4.70)

Note that the strains in the X2 and X3 directions as a result of the extensionalstrain ǫ11 are negative, and hence, for typical engineering materials, Poisson’sratio is a positive constant ranging from 0 ≤ ν ≤ 0.5.

The strain components ǫ22 and ǫ33 can now be written in terms of theapplied stress, Young’s modulus, and Poisson’s ratio as follows:

ǫ22 = −ν

Eσ11

ǫ33 = −ν

Eσ11

(4.71)

Using the definitions of Poisson’s ratio and Young’s modulus, the generalizedHooke’s law can now be written as

ǫ11 =σ11

E−

ν

Eσ22 −

ν

Eσ33

ǫ22 = −ν

Eσ11 +

σ22

E−

ν

Eσ33

ǫ33 = −ν

Eσ11 −

ν

Eσ22 +

σ33

E

γ23 =1

Gσ23

γ13 =1

Gσ13

γ12 =1

Gσ12

These six equations can be written in matrix notation as:

ǫ11

ǫ22

ǫ33

γ23

γ13

γ12

=

1/E −ν/E −ν/E 0 0 0−ν/E 1/E −ν/E 0 0 0−ν/E −ν/E 1/E 0 0 0

0 0 0 1/G 0 00 0 0 0 1/G 00 0 0 0 0 1/G

σ11

σ22

σ33

σ23

σ13

σ12

(4.72)


The 6 × 6 matrix in equation (4.72) is called the compliance matrix. For thesake of completeness, we can invert the matrix equation (4.72) to obtain

σ11

σ22

σ33

σ23

σ13

σ12

=

E (1 − ν) Eν Eν 0 0 0

Eν E (1 − ν) Eν 0 0 0

Eν Eν E (1 − ν) 0 0 00 0 0 G 0 00 0 0 0 G 00 0 0 0 0 G

ǫ11

ǫ22

ǫ33

γ23

γ13

γ12

(4.73)

where

E =E

(1 + ν) (1 − 2ν)

A comparison of equations (4.73) and (4.65) reveals the following relationshipsbetween Lame’s constants and the engineering constants

λ =Eν

(1 + ν)(1 − 2ν)

µ = G (4.74)

Relationship between G, E, and ν

As discussed in the previous section, only two elastic constants are required tocompletely characterize the elastic behavior of a linearly elastic and isotropicsolid. One could use Lame’s constants, or the engineering constants. In thissection, we derive the relationship between the shear modulus and Young’smodulus and Poisson’s ratio to emphasize that only two of these constantsare independent.

To begin, consider a stress element on the surface of a body that is linedup with the X1 and X2 axes. Suppose that this stress element is subjectedto a state of pure shear, where the faces of the element are subjected to theshear stress τ . From the generalized Hooke’s law, we can write

τ = Gγ (4.75)

where γ = 2ǫ12 is the engineering shear strain.We know from our study of principal stresses and directions that the stress

element in the principal orientation is obtained by rotating the stress elementcounterclockwise through an angle θ = π/4. If we now set up a primedcoordinate system to be lined up with the principle directions X ′

1, X ′

2.

We emphasize that the action of the shear stress τ causes no extensionalstrain, i.e., ǫ11 = ǫ22 = 0.

The extensional strain ǫ′11

in the primed coordinate system is related toYoung’s modulus and Poisson’s ratio by

ǫ′11

=τ

E−

ν

E−τ =

τ

E(1 + ν) (4.76)


Because the small-strain tensor is a second-order tensor, its componentstransform according to equation (2.59), the tensor transformation law. Hencethe components of the strain tensor in the primed coordinate system arerelated to the components in the unprimed system through the relation

ǫ′ij

= ap

ia

q

jǫpq (4.77)

From equation (4.77) we can write

ǫ′11

= ap

1a

q

1ǫpq

= a1

1a1

1ǫ11 + a

1

1a2

1ǫ12 + a

2

1a1

1ǫ22

= a1

1a2

1(2ǫ12)

=γ

2(4.78)

Equating equations (4.78) and (4.76) we get

γ

2=

τ

E(1 + ν) (4.79)

and hence,

G =E

2 (1 + ν)(4.80)

4.5.2 Hooke’s law for plane stress and plane strain

On occasion, a three-dimensional problem can be reduced to a two-dimensionalproblem by making the assumption of either plane stress or plane strain. Inthis section, we will explain these two assumptions and express Hooke’s lawfor each case.

To describe the assumption of plane stress, let us consider the example ofa thin plate as shown in Figure 4.10. The plate lies in the X1 −X2 plane andhas a constant thickness, t. The upper and lower surfaces of the plate aretraction free, but tractions can act on the edges of the plate. The smallest in-plane dimension of the plate is labeled W . Now consider a point, P , locatedon the top surface of the plate. At this location, since the top surface istraction free, the stress components σ33, σ13, and σ23 are all identically zero.The same is true for points that lie on the bottom surface of the plate. Ifthe thickness of the plate is small compared to W , then it is reasonable toassume that the out-of-plane stress components do not have a chance to buildup appreciably through the thickness of the plate, i.e., they remain smallcompared to σ11, σ22, and σ12. Hence, under the assumption of plane stresswe assume σ33 = σ13 = σ23 ≈ 0. As a general rule of thumb, the assumptionof plane stress is justified when t/W ≤ 0.1.


W

X1

X2

X3

t

FIGURE 4.10

The assumption of plane stress.

Hooke’s law for the case of plane stress can be written as

ǫ11

ǫ22

γ12

=

1/E −ν/E 0−ν/E 1/E 0

0 0 1/G

σ11

σ22

σ12

ǫ33 = −ν

E(σ11 + σ22) (4.81)

It is important to notice that under the assumption of plane stress, the out-of-plane strain component, ǫ33, is not in general zero. It can be calculatedfrom the in-plane stress components σ11 and σ22 and the elastic constants.Inverting the relation (4.83) we obtain

σ11

σ22

σ12

=

E νE 0νE E 00 0 G

ǫ11

ǫ22

γ12

(4.82)

where E = E/(

1 − ν2)

.In engineering practice, the assumption of plane strain is commonly made

when the cross-sectional geometry of the structure does not vary in one di-rection, and the loading also does not vary in that same direction. Undersuch conditions, the analysis of a single cross section gives the behavior of theentire structure, except in the near vicinity of the ends.

To explain further, let us consider the structure depicted in Figure 4.11. Thegeometry of the structure’s cross section does not vary in the X3 direction,and the characteristic dimension, W , labeled in the figure is large comparedto the X1−X2 in-plane dimensions of the cross section. The loading does notvary in the X3 direction. The structure shown in the figure could be viewedas a dam designed to hold back water. The hydrostatic pressure exerted by


X1

X2

X3

W

FIGURE 4.11

The assumption of plane strain.

the water on the dam does not vary appreciably in the X3 direction, althoughit varies linearly in the X2 direction.

Now let us consider a section of the structure that is sufficiently far fromthe ends outlined in the figure. Ordinarily, if the front and back surfaces ofthis section were free surfaces, the section would have the tendency to expandin the X3 direction due to the compressive distributed load. The constraintof the surrounding material, however, prevents this from happening, and theextensional strain, ǫ33, is approximately zero. The surrounding material alsoprevents transverse shear. Hence, for the case of plane strain, we assumeǫ33 = ǫ13 = ǫ23 ≈ 0. We note that under these conditions, the normal stress,σ33, is in general nonzero. The normal stress is needed to prevent the sectionfrom elongating or shortening in the X3 direction.

Hooke’s law for the case of plane strain can be written as

σ11

σ22

σ12

=

E (1 − ν) Eν 0

Eν E (1 − ν) 00 0 G

ǫ11

ǫ22

γ12

σ33 = Eν (ǫ11 + ǫ22) (4.83)

where again

E =E

(1 + ν) (1 − 2ν)

4.5.3 Hooke’s law for thermal stress problems

When a solid body made of typical engineering materials is heated, the bodyexpands. Similarly, when the body is cooled, it contracts. For an uncon-


strained body, this expansion or contraction does not cause any stress todevelop inside the body. On the other hand, when the body is constrained,there are forces present trying to prevent the expansion. Under such condi-tions, thermal stresses develop due to the presence of elastic strain. As anexample, suppose that a stress-free bar that is clamped at both ends is sub-sequently heated uniformly. Because the bar is clamped at both ends, thetotal strain in the bar remains zero. The thermal strain however is positive.Hence, the elastic strain is negative, and the resulting normal stress in thebar is compressive.

The total strain, ǫij , at a point in a body can be decomposed into elasticand thermal parts. The relation can be written as

ǫij = ǫe

ij+ ǫ

T

ij(4.84)

where ǫe

ijis the elastic part, and ǫT

ijis the thermal part. The thermal strain

at a point in the body can be written as

ǫT

ij= αij∆T (4.85)

where ∆T is the temperature change from some stress-free reference temper-ature, and αij are coefficients of thermal expansion. For isotropic materials,the amount of thermal expansion is the same in all directions, and hence forthat case we have

ǫT

ij= α∆Tδij (4.86)

Here in equation (4.86) α is simply called the coefficient of thermal expansionhaving units of 1/C.

We emphasize that stress develops in a body due to the presence of elasticstrains. Hence, it should be remembered that the strain components to beused in Hooke’s law are the elastic strain components, i.e., σij = Dijklǫ

e

kl.

The substitution of the strain decomposition (4.84) along with the definitionof the thermal strain components (4.86) into Hooke’s law gives the followingmodified version of Hooke’s law for thermoelastic problems:

σij = Dijkl (ǫkl − αδkl∆T ) (4.87)

4.6 Axisymmetric problems

Axisymmetric problems are another important class of problems that are two-dimensional simplifications of fully three-dimensional problems. In order for aproblem to be considered axisymmsetric, both the geometry and the loadingmust possess the same axial symmetry about a fixed axis. The geometry ofan axisymmetric problem is obtained by revolving a two-dimensional domain


r

z

FIGURE 4.12

Axisymmetric analysis of fiber pull-out problem.

about a fixed axis, creating a solid of revolution. The loading is also obtainedby revolving some type of load condition such as a point load or distributedload about the same fixed axis. For example, a uniform surface traction dis-tribution acting on the top of a circular plate could be generated by revolvinga line load about an axis passing through the center of the plate. If a problempossesses such axial symmetry (both geometry and loading), then it can besolved by performing a two-dimensional axisymmetric analysis.

A nice example that illustrates the ideas above is the fiber pull-out prob-lem illustrated in Figure 4.12. An inner core (fiber) is embedded inside anouter cylinder (matrix). The three-dimensional geometry is shown on the left.The top of the fiber is subjected to a uniform traction distribution in the z

direction, and the bottom of the cylinder is constrained from moving in thez direction. The traction distribution applied to the fiber has the tendencyto ”pull out” the fiber from the surrounding matrix. Because this problempossesses the requisite axial symmetry (both in geometry and loading), theproblem can be reduced to a two-dimensional problem shown in the centerof the figure. For the sake of interest, the deformed shape of the compositecylinder is shown to the right, and a contour plot of the stress component σzz

is superposed on top of the geometry. The goal of this section is to derive therelationship between strain and displacement for axisymmetric problems. Tobegin, it is necessary to define a cylindrical coordinate system. A cylindri-cal coordinate system is formed by three mutually perpendicular coordinatecurves. The coordinate curves are the intersections of coordinate surfaces, aswe will discuss in the following paragraphs.


X1

X2

X3

ξ1

ξ2

ξ3

(a) (b)

P

rz

θ

FIGURE 4.13

Coordinate curves.

The intersection of a cylindrical surface of radius r and two perpendicularplanar surfaces is shown in Figure 4.13. The cylinder is aligned with the X3

axis. The two planes shown in the figure can be defined as

X3 = z; θ = θc (4.88)

where z is the vertical distance from some point that lies on the X3 axis topoint P labeled in the figure. From now on we will refer to the plane θ = θc asthe θ-plane. The coordinate curve ξ1 is the intersection of the plane X3 = z

and the θ-plane. The intersection of the cylinder and the plane X3 = z formsthe coordinate curve ξ2. Finally, the intersection of the cylinder and the θ-plane forms the coordinate curve ξ3. The point P in the figure is located atthe intersection of all three coordinate curves. The location of this point canalso be defined by the cylindrical coordinates r, θ, and z labeled in the figure.

The gradient operator in general curvilinear coordinates is defined as

∇ =∂

∂ξi

gi (4.89)

where gi are the natural unit base vectors in the curvilinear coordinate system.The natural base vectors are defined as

gi =∂X

∂ξi

(4.90)

In other words, each base vector is formed by calculating the vector, dX,which represents an infinitesimal change in the position vector X with respect


to a differential movement along the ξi curve, and dividing that vector by itsmagnitude. Incidentally, the base vectors gi are unit vectors, because themagnitude of dX is dξi, the arc length traveled by the tip of the vector X

during the infinitesimal movement along the coordinate curve. Because thebase vectors gi are mutually perpendicular, the curvilinear coordinate systemis called an orthogonal curvilinear coordinate system.

In cylindrical coordinates, the partial derivatives with respect to the curvi-linear coordinates are related to partial derivatives with respect to the param-eters r, θ, and z as follows:

∂

∂ξ1

=∂

∂r

∂

∂ξ2

=1

r

∂

∂θ

∂

∂ξ3

=∂

∂z(4.91)

Note that the result in the second line in equation (4.91) is a result of the factthat the arc length of the differential movement dξ2 is equal to r dθ. Finally,letting er = g1, eθ = g2, and ez = g3, we obtain the familiar form

∇ =∂

∂rer +

1

r

∂

∂θeθ +

∂

∂zez (4.92)

In general, the displacement vector, u, in cylindrical coordinates has threecomponents, i.e.,

u = ur er + uθ eθ + uz ez (4.93)

In axisymmetric problems, however, because both the geometry and the load-ing possess axial symmetry, the displacement field does not vary in the θ

direction, and hence we can write

u = ur er + uz ez (4.94)

The main task that is left in this section is to evaluate the displacementgradient tensor using the gradient operator (4.92) and the axisymmetric form(4.94) of the displacement vector. Before doing this, however, it is convenientto investigate the derivatives of the base vectors er, eθ, and ez with respectto r, θ, and z. We will need these derivatives later in the derivation of ∇u.

To begin, let us write down the base vectors er, eθ, and ez in terms of thefixed Cartesian base vectors e1, e2, and e3 as follows:

er = cos θ e1 + sin θ e2

eθ = − sin θ e1 + cos θ e2

ez = e3 (4.95)


Differentiating the first line of (4.95) with respect to θ gives

er,θ = − sin θ e1 + cos θ e2

= eθ (4.96)

Similarly, differentiating the second line with respect to θ yields

eθ,θ = − cos θ e1 − sin θ e2

= − er (4.97)

The remaining derivatives of the base vectors are all zero. The results aresummarized in Table 4.1 below.

Getting the components of ∇u in cylindrical coordinates is just a lengthyexercise in the use of the chain rule once the derivatives of the base vectorsare known. The operation to be carried out can be written as

∇u =

(

∂

∂rer +

1

r

∂

∂θeθ +

∂

∂zez

)

⊗ (ur er + uθ eθ + uz ez) (4.98)

To perform the tensor operation (4.98), each term on the left side of the ⊗

symbol must operate on each term on the right side of the symbol. Eachoperation requires the use of the chain rule. For example, if the first term onthe left operates on the first term on the right we get

er

∂

∂r(ur er) = er ur,r er + er ur

∂er

∂r

= ur,r erer (4.99)

Note that the last term in (4.99) drops out because the direction of er doesnot change with respect to a change in r. With the help of Table 4.1, ∇u cannow be written as

∇u = ur,r erer + uθ,r ereθ + uz,r erez

+

1

r[ur,θ eθer + ur eθeθ] +

1

r[uθ,θ eθeθ − uθ eθer]

+1

r[uz,θ eθez]

+ ur,z ezer + uθ,zezeθ + uz,z ezez (4.100)

TABLE 4.1

Derivatives of basevectors.

∂ er eθ ez

∂/∂r 0 0 0

∂/∂θ eθ − er 0

∂/∂z 0 0 0


For axisymmetric problems, the displacement component uθ is equal to zero,and all derivatives of the displacement components with respect to θ are alsozero. Hence, for axisymmetric problems, the components of ∇u can be writtenin matrix form as

∇u =

ur,r 0 uz,r

01

rur 0

ur,z 0 uz,z

(4.101)

Armed with the displacement gradient matrix (4.101) for axisymmetricproblems, we can now compute the components of the relevant deformationtensors. For example, the components of the deformation tensor F are com-puted as follows:

x = X + u ⇒

F = ∇u + I (4.102)

where I is the second-order identity tensor. Using the relationship (4.102)between ∇u and F, the components of F can be written as

F =

ur,r + 1 0 uz,r

01

rur + 1 0

ur,z 0 uz,z + 1

(4.103)

If we now assume that the displacement gradients are small and neglectproducts of displacement gradients such as u2

r,r, then the components of Green

deformation tensor, C = FT F, are approximately equal to the following:

C ≈

2ur,r + 1 0 uz,r + ur,z

02ur

r+ 1 0

uz,r + ur,z 0 2uz,z + 1

(4.104)

Using the relationship between the components of C and the small straintensor ǫ discussed in Section 4.4.3, the components of the small strain tensorcan finally be written as

ǫ =

ǫrr

ǫθθ

ǫzz

γrz

=

ur,r

1

rur

uz,z

uz,r + ur,z

(4.105)

To end this section, we note that there are four nonzero strain componentsfor the case of axisymmetric problems. The strain components ǫrr, ǫθθ, and ǫzz

are the relative stretches in the radial, circumferential, and axial directions,respectively. The strain component ǫθθ, which gives the relative stretch in the


FIGURE 4.14

Continuous piecewise surface.

circumferential direction, is also known as the hoop strain. The engineeringshear strain γrz gives the decrease in angle experienced by two fibers that areoriginally perpendicular to each other and both lying in the r − z plane.

For linearly elastic and axisymmetric problems, Hooke’s law becomes

σrr

σθθ

σzz

σrz

=

2µ + λ λ λ 0λ 2µ + λ λ 0λ λ 2µ + λ 00 0 0 µ

ǫrr

ǫθθ

ǫzz

ǫrz

(4.106)

where Lame’s constants in (4.106) above are defined in equation (4.74) interms of Young’s modulus and Poisson’s ratio.

4.7 Weak form of the equilibrium equation

Recall from our work in Section 4.3 that the equilibrium equation can bewritten as

σij,j + bi = 0 (4.107)

In a displacement-based finite element method, we seek to use continuouspiecewise trial functions as candidate solutions to the partial differential equa-tion (4.107). The trial function is a vector-valued function having two or threecomponents, depending on whether we are working in two or three dimensions.At the moment, one might wonder what is meant by a continuous piecewisedisplacement field in the context of two- and three-dimensional solid mechan-ics problems. In order to illustrate, suppose that a series of measurements are


made for the displacement component, uX , at a large number of points locatedon the surface of a thin plate. If this surface plot turned out to be continuouspiecewise, it would have the characteristics shown in Figure 4.14. Notice thatthe surface is continuous, i.e., there are no gaps or overlaps, but there existcreases (lines along which the slope of the surface in a given direction is dis-continuous). Mathematically speaking, at a particular point lying on one ofthe creases, the partial derivative of the function representing the surface withrespect to one or more of the spatial coordinates will be discontinuous. Sec-ond partial derivatives would not exist at such points. Because the divergenceof the stress (the first term in the equilibrium equation) indirectly involvessecond partial derivatives of the displacement field, we cannot directly plugin continuous piecewise trial functions into the equilibrium equation. In orderto remedy this situation, we proceed like we did in Chapter 3 and derive theweak form of the equilibrium equation.

In order to derive the weak form, the first step is to introduce a vector-valuedtest function whose components are wi having the following properties:

1. wi = C0 continuous in Ω

2. wi = 0 on Γu (4.108)

In other words, we insist that the components of the test function be at leastcontinuous piecewise over the domain of interest, and require wi to be zero atpoints on the boundary wherever ui is specified. The next step is to multiplythe equilibrium equation (4.107) by the test function and then integrate theresulting expression over the domain of interest, Ω. Doing this gives

∫

Ω

σij,jwi dΩ +

∫

Ω

biwi dΩ = 0 (4.109)

Next, in order to get rid of the term involving the divergence of the stress onthe left-hand side of equation (4.109) above, it is necessary to integrate thefirst integral by parts. The goal is to replace the integral with an equivalentintegral that does not involve first partial derivatives of the stress field. Theintegration by parts can be readily carried out with the help of the followingidentity:

(σijwi),j= σij,jwi + σijwi,j (4.110)

where the right-hand side of equation (4.110) above is obtained using theproduct rule. Notice that the first term on the right-hand side of (4.110) isthe integrand of the first integral in (4.109). Hence using equation (4.110) wecan rewrite equation (4.109) as follows:

∫

Ω

(σijwi),jdΩ −

∫

Ω

σijwi,j dΩ +

∫

Ω

biwi dΩ = 0 (4.111)

We now recognize that the integral on the left-hand side of equation (4.111)is a volume integral whose integrand is the scalar quantity, ∇ · (w · σ), the


divergence of the vector w · σ. To see this more clearly, it is helpful to gothrough the exercise of translating expressions from indicial notation intoequivalent expressions in tensor notation and vice versa. For example,

w · σ = (wiei) · (σklekel)

= wiσklδikel

= wkσklel (4.112)

and hence,

∇ · (w · σ) =

(

∂

∂Xi

ei

)

· (wkσklel)

=∂wk

∂Xi

σklδil

=∂wk

∂Xl

σkl

= σijwi,j (4.113)

We can now employ the divergence theorem (2.129) and recast the first integralon the left-hand side of (4.111) into the following surface integral:

∫

Ω

∇ · (w · σ) dΩ =

∫

Γ

n · (w · σ) dΓ (4.114)

Finally, noting that

n · (w · σ) = (niei) · (wkσklel)

= nlwkσkl

= σijnjwi

= tiwi (4.115)

and rearranging terms, equation (4.111) can be written as∫

Ω

σijwi,j dΩ =

∫

Ω

biwi dΩ +

∫

Γ

tiwi dΓ (4.116)

where ti are the components of the traction vector at a point on Γt.Equation (4.116) is the weak form of the equilibrium equation, or the prin-

ciple of virtual work. The left-hand side of this equation is commonly referredto as the internal virtual work. The right-hand side is the external virtualwork. If we view the test function as having units of length, then the left-handside of (4.116) has units of stress times volume, and the right-hand side hasunits of force times length, both equivalent to work. If we want to, at thisstage we can plug the stress-strain law into the left-hand side of the weakform, but we will defer this exercise until after the finite element approxima-tions are substituted into the weak form, which will be covered in the nextchapter.


4.8 Problems

Problem 4.1

Show that for an isotropic and linearly elastic material Hooke’s law can bewritten as

σij = 2G

[

ǫij +ν

1 − 2νǫkkδij

]

where G is the shear modulus and ν is Poisson’s ratio

Problem 4.2

Using the form of Hooke’s law obtained in Problem 4.1, show that the equilib-rium equation (4.32) can be written in terms of the displacement componentsui as follows:

G

[

∂ui

∂Xj∂Xj

+1

1 − 2ν

∂2uj

∂Xj∂Xi

]

+ bi = 0

Problem 4.3

A solid body is subjected to all-around hydrostatic pressure of magnitude P .

(a) State the boundary condition on the surface of the body.

(b) If the body is homogeneous, determine a stress field inside the bodythat satisfies the equilibrium equation and the boundary conditions.

Problem 4.4

Consider the following state of stress at a point:

σ =

10 −5 0−5 5 0

0 0 0

MPa

(a) Determine the three stress invariants I1, I2, and I3.

(b) Determine the three principal stresses.

(c) Determine the three principal directions.

(d) Indicate on a sketch a stress element in the principal orientation.


Problem 4.5

A two-dimensional deformation from the original configuration to the de-formed configuration is defined as

x1 = 3X2

1+ X2

x2 = X1 + X3

2

(a) Determine the components of the deformation gradient tensor F.

(b) Determine the components of the Green deformation tensor C.

(c) Determine the relative stretch in the e1 and e2 directions at the point(X1,X2) = (1, 1).

5

Steady-State Heat Conduction

I have learned through bitter experience the one supreme lesson to conserve

my anger, and as heat conserved is transmitted into energy, even so our anger

controlled can be transmitted into a power that can move the world.—Mahatma Gandhi

The temperature distribution in solids is needed, for example, to predict ther-mal stresses that can develop in constrained bodies. In this chapter, we con-sider the phenomenon of steady-state heat conduction. We will restrict ourattention to isotropic heat conduction, where it is assumed that a material’sresistance to heat flow is the same in all directions.

An energy balance is performed by calling upon the principle of conservationof energy. The principle states that the energy that flows into a body per unittime minus the energy that flows out per unit time is equal to the change ofinternal energy inside the body per unit time. When the internal energy inthe body stops changing with time, the body is said to have reached steadystate. Under steady-state conditions, the temperature field no longer dependson time, but it can vary spatially throughout the body.

We will begin this chapter by deriving the partial differential equation thatgoverns steady-state heat conduction. During the process, we will define therelevant physical quantities: heat flux, internal heat source, temperature gra-dient, and thermal conductivity. The units associated with each of thesequantities will also be discussed. After deriving the partial differential equa-tion from an energy balance, we will discuss the relevant boundary conditions,such as forced convection. Finally, we will go ahead and derive the weak formwhich turns out to be the starting point for a temperature-based finite ele-ment formulation. The chapter also includes a discussion of symmetry, whichunder certain circumstances can be exploited, reducing computational costs.

Under conditions of small strain and deformation, it is reasonable to as-sume that the heat produced by mechanical work inside a body is negligiblecompared to other sources. Under these conditions, we can perform an un-coupled analysis, whereby the steady-state heat equation is first solved forthe temperature field, and then the temperature field is used as input to aseparate stress analysis. The temperature field dictates the thermal straindistribution in the solid, which in turn gives rise to thermal stresses when thesolid is constrained. In other words, the thermal strain distribution loads thesolid, and the resulting displacement, elastic strains, and stresses are obtained

121


S(Xi)

Ω

Γ

q

Γq

ΓT

FIGURE 5.1

Solid body subjected to internal heat source and surface flux distribution.

by performing a separate finite element analysis.

5.1 Derivation of the steady-state heat equation

In order to derive the steady-state heat equation, it is helpful to observeFigure 5.1. The figure illustrates a solid body defined by the interior volumeΩ and outer surface Γ. The body is subjected to a prescribed temperaturedistribution over a portion of the surface labeled ΓT . In addition, a heat fluxdistribution is prescribed over the portion of the boundary labeled Γq. A heatflux is a vector quantity that describes the amount of energy flowing through aunit area per unit time. Hence the flux vector, q, at a point on the surface hasthe SI units of Joules per meter squared per second, i.e., J/(m2 · s). In the signconvention adopted here, positive flux points out of the body. Heat can alsobe generated (or dissipated) internally by means of a heat source distributionS(X1,X2,X3) inside Ω. The heat source has units of energy per unit volumeper unit time, i.e., J/(m3 · s). The relevant quantities and associated units forsteady-state heat conduction are summarized in Table 5.1. Note that in thetable we have used the unit Watt instead of Joules per second, i.e., W = J/s.

Conservation of energy (assuming steady-state conditions have been reached)states that the energy flowing into the body Ω per unit time must be equalto the energy flowing out of the body through Γ per unit time. The energyflowing into (or out of) the body due to the heat source can be accountedfor by integrating the heat source distribution over the volume Ω. The fluxvector at a point on the surface can be decomposed into a component that isperpendicular to the surface and one that is tangential to the surface (normaland tangential components). The tangential component of the flux does not

Steady-State Heat Conduction 123

contribute to the energy balance because this energy is neither flowing intoor out of the body. Hence, the net flow of energy out of the surface of thebody is obtained by evaluating the surface integral of the normal component,qn = q · n, of the heat flux vector, where n is the unit outward normal to thesurface. The energy balance statement can now be written as

∫

Ω

S dΩ −

∫

Γ

n · qdΓ = 0 (5.1)

Just like in our derivation of the equilibrium equation in Section 4.3, thekey to massaging the balance statement (5.1) above into a partial differentialequation is to convert the surface integral involving the normal component ofthe flux vector into an equivalent volume integral. This can be accomplishedby employing the divergence theorem, i.e.,

∫

Γ

n · qdΓ =

∫

Ω

∇ · qdΩ (5.2)

Hence the energy balance now becomes∫

Ω

S − ∇ · q dΩ = 0 (5.3)

Because the balance statement (5.3) must hold for all subdomains inside Ω,the integrand must be identically zero at all points within Ω. Hence,

S − ∇ · q = 0 (5.4)

The partial differential equation (5.4) is often referred to as the heat equa-tion. The second term on the left-hand side of equation (5.4) is called thedivergence of the flux. This term can be written in indicial notation as fol-lows:

∇ · q =

(

∂

∂Xi

ei

)

· (qjej)

=∂qj

∂Xi

δij

= qi,i (5.5)

TABLE 5.1

Relevant quantities and units needed forsteady-state heat conduction.

Symbol Name SI units

k Thermal conductivity[

W

m C

]

q Heat flux vector[

W

m2

]

S Heat source[

W

m3

]


and hence, the governing partial differential equation can also be written incomponent form as

S − qi,i = 0 (5.6)

5.2 Fourier’s law

Thus far, the partial differential equation that we derived in the previoussection involves partial derivatives of the flux vector. In this section we in-troduce the constitutive relation between heat flux and temperature that isknown as Fourier’s law. Armed with this relation, we will be able to recast thedifferential equation in terms of temperature as the unknown field variable.

Fourier’s law is an experimentally determined relationship that states thatthe heat flux at a point in a body is proportional to the temperature gradientat that point. Mathematically, Fourier’s law can be written as

q = −K∇T (5.7)

where K is a second-order tensor whose Cartesian components are the direc-tionally dependent thermal conductivities of the material. The quantity K

can be written in matrix notation as follows:

K =

k11 k12 k13

k21 k22 k23

k31 k32 k33

(5.8)

The components of the thermal conductivity tensor have units of J/(m ·C · s).For the sake of simplicity, throughout the remainder of this book, we willrestrict our attention to materials whose resistance to heat flow is the samein all directions. Such materials are called isotropic. For the case of isotropicheat conduction, the components of the thermal conductivity tenor can bewritten as

Kij = kδij (5.9)

where k is referred to as the thermal conductivity. Fourier’s law can then bewritten as

q = −k∇T (5.10)

or in component formqi = −kT,i (5.11)

By substituting Fourier’s law (5.11) into the heat equation (5.6) we get

kT,ii + S = 0 or

∇2T + S = 0 (5.12)


a b

c d

T (X1, X2) = T (−X1, X2)

T (X1, X2) = T (−X1, X2)

X1

X2

FIGURE 5.2

Exploiting symmetry.

It is worth mentioning here that when the heat source is zero, the heat equa-tion reduces to simply ∇2T = 0, which is called Laplace’s equation. Whenthe heat source is present, the equation is known as the Poisson equation.

5.3 Boundary conditions

In the present chapter, we will consider three different types of boundary con-ditions that are encountered in steady-state heat conduction analyses. Firstof all, the temperature, T , can be prescribed over a portion of the body.Mathematically, this condition can be stated as

T = T∗ on ΓT (5.13)

where ΓT is the portion of the boundary over which the temperature is a pre-scribed function of the spatial variables. The prescribed temperature bound-ary condition is the essential boundary condition. It is a direct conditionplaced on the field variable. Secondly, the normal component of the heat flux,q∗n, can be specified on Γq. This condition can be written as

n · q = q∗n

on Γq (5.14)

Because q = −k∇T from Fourier’s law, the flux condition (5.14) poses acondition on the gradient (partial derivatives) of the field variable. The fluxboundary condition is the natural boundary condition.


FIGURE 5.3

Heat transfer problem of an inner core embedded in a sphere.

Closely related to the prescribed flux condition is forced convection. Inforced convection, heat transfer takes place through the surface of the bodyinto a surrounding fluid flowing at a certain velocity across the body. Theforced convection condition can be stated as

q∗n

= h (T − T∞) (5.15)

where h is the film coefficient. It turns out that the finite element implemen-tation of steady-state heat conduction with the forced convection conditioncauses no difficulties and still leads to a linear system of equations.

Symmetry can often be exploited in steady-state heat conduction problems,reducing the computational cost. To illustrate the idea, consider the two-dimensional solid shown in Figure 5.2. A two-dimensional solid in the contextof heat conduction means that no heat can flow in the X3 direction (into or outof the page). As shown in the figure, the geometry of the body is symmetricabout both the X1 and X2 axes. Now suppose that temperature distributionsare prescribed along edges a-b and c-d labeled in the figure in such a mannerthat both are symmetric about the X2 axis. The temperature distribution inthe body would then be an even function of X1, and hence, the heat flux inthe X1 direction would be zero along the line X1 = 0. It would then onlybe necessary to model one-half of the plate (say the right half), enforcing thezero-flux condition along the left edge, similar to a traction-free edge in solidmechanics.

Radiation is another way in which heat can be transferred from one bodyto another. The net heat flux flowing through the surfaces of radiating bodiesis proportional to the temperature at the surface of the body raised to thefourth power. Thus, radiation gives rise to a nonlinear heat transfer problemand will not be considered here. A nice introduction to this topic can be foundin Lienhard [8].

Example 5.1

To apply some of the important concepts described in this chapter, we nowconsider the problem of a solid spherical core embedded inside a much larger


(1225 km/C)1

r− 225 C

FIGURE 5.4

Inner core heated to a uniform temperature of 1000 C, while the outer surfaceis held at 20 C.

spherical body as shown in Figure 5.3. The diameter of the inner core is1000m, and the diameter of the outer sphere is 5000m. Initially, the tem-perature distribution in both the inner core and outer sphere is held at auniform temperature of 20 C. If the surface of the outer sphere is maintainedat a constant temperature of 20 C and the inner core is subsequently heatedto a uniform temperature of 1000 C, it is desired to obtain the steady-statetemperature distribution in the outer sphere. Due to the spherical symmetryof the problem (both the geometry and loading), the temperature field in theouter sphere will only depend on the radial coordinate r, measured positive ra-dially outward from the center of the inner core. Hence T (X1,X2,X3) = T (r).

Recall from the previous section that the steady-state heat equation can bewritten as

k∇2T + S = 0 (5.16)

Because there is no heat source present, the steady-state heat equation reducesto the Laplace equation, i.e., ∇2T = 0. To proceed with the solution, it isconvenient to work in a (r, θ, and φ) spherical coordinate system. The gradientoperator in spherical coordinates is given as

∇ =∂

∂rer +

1

r

∂

∂φeφ +

1

r sinφ

∂

∂θeθ (5.17)

where er, eφ, and eθ are unit base vectors. Because the temperature distribu-tion only depends on r, the heat equation reduces to the following second-orderdifferential equation:

1

r2

d

dr

(

r2dT

dr

)

= 0 (5.18)

The boundary conditions can be written as

T (r = ri) = 1000 C

T (r = ro) = 20 C (5.19)

where ri and ro are the inner and outer surfaces, respectively, of the outercore. The exact solution to equation (5.19) above can be obtained by direct


integration. Integrating once yields

r2dT

dr= C1 (5.20)

where C1 is an unknown constant. Integrating a second time gives

−C1

r+ C2 (5.21)

where C2 is another constant of integration. The constants C1 and C2 arefound by enforcing the two boundary conditions specified in equation (5.19).Doing this yields the following temperature distribution:

T (r) = C3/r + C4 (5.22)

where C3 = 1225 km/C and C4 = 225 C. A contour plot of the temperaturedistribution is shown in Figure 5.4.

5.4 Weak form of the steady-state heat equation

Because the heat equation involves second partial derivatives of the temper-ature field, the desire to use C0 trial functions makes it necessary to derivethe weak form of the equation as our starting point for a finite element imple-mentation. In order to derive the weak form, we follow the same proceduredescribed in Chapter 4, Section 4.7.

The weak form of the heat equation is actually easier to derive than theweak form of the equilibrium equation because of the fact that temperatureis a scalar quantity. To begin, we introduce a scalar-valued test function,w(X1,X2,X3). The test function is assumed to be sufficiently smooth and tobe zero at any points on the boundary where the temperature is specified, i.e.,the test function is defined to be zero on all essential boundaries. Next, wemultiply the heat equation by the test function, and integrate the resultingexpression over the domain of interest Ω. Doing this gives

k

∫

Ω

T,iiw dΩ +

∫

Ω

Sw dΩ = 0 (5.23)

In order to get the expression (5.23) into the desired form, the first integralon the left-hand side needs to be integrated by parts. Noting that

(T,iw),i

= T,iiw + T,iw,i (5.24)

we can rewrite equation (5.23) as

k

∫

Ω

(T,iw),i

dΩ − k

∫

Ω

T,iw,i dΩ +

∫

Ω

Sw dΩ = 0 (5.25)


It is now instructive to write the above equation in tensor notation. Doingthis yields

k

∫

Ω

∇ · (w∇T ) dΩ − k

∫

Ω

∇T · ∇w dΩ +

∫

Ω

Sw dΩ = 0 (5.26)

Note that the integrand of the first integral on the left-hand side is the diver-gence of the vector w∇T . The second integral involves the dot product of thetemperature gradient vector with the gradient of the test function. Finally,the third integral involves the product of the two scalar functions S and w,i.e., the heat source and the test function. Employing the divergence theoremto recast the first volume integral on the left-hand side of equation (5.26) asan equivalent surface integral yields

k

∫

Γ

n · (w∇T ) dΓ − k

∫

Ω

∇T · ∇w dΩ +

∫

Ω

Sw dΩ = 0 (5.27)

where n is the unit outward normal to the boundary.The final step in the derivation of the weak form involves substituting

Fourier’s law, q = −k∇T , into the expression (5.27) and recognizing thatn ·q = q∗

nis the normal component of the heat flux on the boundary Γq. The

final result is

k

∫

Ω

∇w · ∇T dΩ =

∫

Ω

wS dΩ −

∫

Γq

wq∗n

dΓ (5.28)

Note that the surface integral in (5.28) above only needs to be carried outover Γq, because the test function w is zero on ΓT .

The weak form of the heat equation contains only first partial deriva-tives of the temperature field. Thus, continuous piecewise trial functionscan be substituted into the weak form without any difficulty, leading to atemperature-based finite element method for steady-state heat conduction, aswill be demonstrated in the next chapter.

5.5 Problems

Problem 5.1

Consider the steady-state heat conduction problem of a fin that is insulatedon its outer surface so that heat can only flow in one dimension. The lengthof the fin is L = 0.1m, and its thermal conductivity is k = 1.0 J/(m ·C · s).Suppose that the bar is subjected to the following boundary conditions:

T (X = 0) = 200 C

q∗n(X = L) = h(T − T∞)


where h is the film coefficient and T∞ is the remote temperature of the sur-rounding fluid. Letting h = 10 J/(m2 · s) and T∞ = 20 C, perform the fol-lowing:

(a) Derive the governing differential equation.

(b) Derive the weak form.

(c) Model the problem with two linear-temperature elementsto obtain an estimate for the temperature of the fin at X = L.

6

Continuum Finite Elements

Although this may seem a paradox, all exact science is dominated by the idea

of approximation. When a man tells you that he knows the exact truth about

anything, you are safe in inferring that he is an inexact man.

—Bertrand Russell

In this chapter we will explore a few of the most popular finite elementsfor use in continuum problems. For two-dimensional problems, we will con-sider the three-node triangle and the arbitrary four-node quadrilateral. Thearbitrary four-node quadrilateral falls within a class of elements called isopara-metric finite elements. In the chapter we will discuss the isoparametric con-cept and emphasize why it is important. We will also consider two importantthree-dimensional elements: the four-node tetrahedron and the eight-nodebrick.

The development of the various element types will be performed in thecontext of solid mechanics. The development of all elements for heat transferis essentially the same. The only difference is that the field variable in solidmechanics is displacement, whereas in heat transfer it is temperature. Intwo- and three-dimensional solid mechanics problems, displacement is a vectorfield. The displacement at each point has both magnitude and direction. Onthe other hand, temperature is a scalar field. So actually, the formulation ofthe element types for heat transfer is somewhat easier than for solid mechanics.

There is some notation that will be used in this chapter that is best de-fined here. We will be dealing with finite elements in two configurations: theundeformed configuration, and the deformed configuration. The undeformedconfiguration of an element is its initial geometry, before any loads are applied.The deformed configuration is the geometry of the element after it undergoesdeformation. A position vector acting from the origin of a Cartesian coor-dinate system to a point inside the undeformed element will be defined asX. After the element is deformed, the same point that originally occupiedposition X gets transported to position x inside the deformed element. Thedisplacement u of point X is defined as the deformed position minus the orig-inal position, i.e., u = x − X. This relationship between displacement andposition can be written in component form as ui = xi − Xi.

131


1

2

3

d1X , d1Y

A1

A2

A3d2X , d2Y

d3X , d3Y

(X,Y )

FIGURE 6.1

The three-node triangle.

6.1 Three-node triangle

The use of the three-node triangular element for plane stress elasticity prob-lems was first reported in 1956 in a paper by Turner et al. [9]. At that time, itwas necessary to do everything possible to reduce three-dimensional problemsinto two-dimensional problems due to lack of computational resources. Thethree-node triangle is still popular today, primarily because of the ease withwhich complicated geometries can be meshed with triangles. A schematic of athree-node triangle is shown in Figure 6.1. The numbers labeled in the figurerepresent the local node numbers. The numbering scheme is defined such thatlocal node one is chosen to be any one of the three vertices. The remainingnodes are then numbered sequentially in a counterclockwise fashion. It isconvenient to arrange the displacement components at the nodal points in a6 × 1 vector as follows:

de =

d1X

d1Y

d2X

d2Y

d3X

d3Y

(6.1)

where daX and daY are the displacement components associated with node a

in the X and Y directions, respectively. One of the main ideas in the finiteelement method is to interpolate the field variable inside the element usingpolynomial functions. For the case of the three-node triangle, we assume thatthe displacement components vary linearly in X and Y . Mathematically, thefinite element approximations for the displacement components can be written

Continuum Finite Elements 133

as

uX

uY

e

=

α1 + α2X + α3Y

β1 + β2X + β3Y

(6.2)

where α1, . . . , β3 are constants. It is common practice to express equation(6.2) in terms of shape functions and the displacement components at thenodal points. To do this, we impose the following three conditions:

ui(X1, Y1) = d1i

ui(X2, Y2) = d2i

ui(X3, Y3) = d3i (6.3)

where i can be either X or Y it doesn’t make any difference which one youchoose. Letting i = X and substituting the constraints (6.3) into the first lineof equation (6.2) yields

d1X

d2X

d3X

=

1 X1 Y1

1 X2 Y2

1 X3 Y3

α1

α2

α3

(6.4)

The algebra is rather lengthy, but equation (6.4) can be solved for α1, α2, andα3 without too much trouble. The resulting expressions can be put back intoequation (6.2), and after collecting terms, the final expression can be writtenas

uX = N1(X,Y )d1X + N2(X,Y )d2X + N3(X,Y )d3X (6.5)

where the functions N1, N2, and N3 are the three shape functions. These canbe written in terms of the spatial variables and nodal coordinates as follows:

N1(X,Y ) =1

2A(X32Y − Y32X + X2Y3 − Y2X3)

N2(X,Y ) =1

2A(−X31Y + Y31X − X1Y3 + Y1X3)

N3(X,Y ) =1

2A(X21Y − Y21X + X1Y2 − Y1X2) (6.6)

Here in equation (6.6), X1, Y1 . . . X3, Y3 are the X and Y coordinates of thenodes. The quantities X21, X32, etc., represent differences in the nodal co-ordinates, i.e., X21 = X2 − X1. The quantity A is the area of the triangularelement. The area can be easily computed using the cross-product operation.Let X21 and Y21 be the components of a vector X(1) that points from localnode one to local node two. Similarly, let X31 and Y31 be the componentsof a vector X(2) that points from local node one to local node three. Themagnitude of the cross product of these two vectors yields twice the area ofthe triangle, i.e.,

2A =∣

∣

∣X(1)

× X(2)

∣

∣

∣

= X21Y31 − Y21X31


N1 N2 N3

111 222

333

FIGURE 6.2

Shape functions for the three-node triangle.

We point out that the three shape functions are all linear in X and Y , andthey obey the Kronecker delta property that was introduced in Chapter 3,i.e.,

NI(XJ ) = δIJ (6.7)

The three shape functions are plotted in Figure 6.2. Note that each surfaceis planar.

An alternative way of getting the shape functions for the three-node triangleis through the use of area coordinates. The area coordinates are labeled A1,A2, and A3 in Figure 6.1. As the point (X,Y ) moves around inside thetriangle, each of these three areas changes continuously. It is left as an exerciseat the end of this chapter to compute the area of the three area coordinatesthrough the use of the cross-product operation to show that

N1(X,Y ) = A1(X,Y )/A

N2(X,Y ) = A2(X,Y )/A

N3(X,Y ) = A3(X,Y )/A (6.8)

It is easy to see that the shape functions (6.8) satisfy the Kronecker deltaproperty and are linear in X and Y . As point (X,Y ) approaches local nodea, the area coordinate Aa becomes the total area A, and hence

Aa(Xa, Ya)/A = 1

The vector-valued trial function u can now be expressed as the product ofa 2 × 6 matrix containing shape functions and a 6 × 1 vector containing thenodal displacements as follows:

uX

uY

e

=

[

N1 0 N2 0 N3 00 N1 0 N2 0 N3

]

d1X

d1Y

d2X

d2Y

d3X

d3Y

(6.9)


orue

2 × 1=

Ne

2 × 6de

6 × 1(6.10)

6.1.1 The B-matrix

Proceeding along, it is now necessary to come up with a relationship betweenthe in-plane strain components, ǫXX , ǫY Y , and γXY , defined in Chapter 4,and the nodal displacement vector de, defined above. Recall that the in-planestrain components are related to the displacement components as follows:

ǫXX = uX,X

ǫY Y = uY,Y

γXY = uX,Y + uY,X (6.11)

where, again, the comma denotes partial differentiation with respect to thespatial variables. For the sake of convenience, we will store the three in-planestrain components in a 3 × 1 vector, ǫ, as follows:

ǫ =

ǫXX

ǫY Y

γXY

(6.12)

Having defined the strain vector, we now seek a relationship of the form

ǫe

3 × 1=

uX,X

uY,Y

uX,Y + uY,X

e

=Be

3 × 6de

6 × 1(6.13)

where Be is the B-matrix for the element. The B-matrix operates on the nodaldisplacement vector, giving an approximation for the strain components at apoint within the element. The strain calculation is carried out as a post-processing step in the finite element method after the nodal displacementsare found.

The components of the B-matrix can be obtained by differentiating thedisplacement components uX and uY as follows:

uX,X =

3∑

a=1

Na,XdaX

uY,Y =

3∑

a=1

Na,Y daY

uX,X + uY,X =

3∑

a=1

Na,Y daX + Na,XdaY (6.14)


and then writing the resulting expression in matrix notation as

ǫe =

N1,X 0 N2,X 0 N3,X 00 N1,Y 0 N2,Y 0 N3,Y

N1,Y N1,X N2,Y N2,X N3,Y N3,X

d1X

d1Y

d2X

d2Y

d3X

d3Y

(6.15)

To obtain a final expression for the B-matrix associated with the three-nodetriangle, we must differentiate the shape functions. Doing this we get

N1,X = −Y32

2A

N1,Y =X32

2A

N2,X =Y31

2A

N2,Y = −X31

2A

N3,X = −Y21

2A

N3,Y =X21

2A(6.16)

The final result is

Be

3 × 6=

1

2A

−Y32 0 Y31 0 −Y21 00 X32 0 −X31 0 X21

X32 −Y32 −X31 Y31 X21 −Y21

(6.17)

It turns out that the B-matrix for the three-node triangle depends only onthe nodal coordinates. In other words, the elements of the B-matrix containsimply constants. As a consequence, the finite element approximation forthe strain field is constant at each point within the element. The three-nodetriangle is often referred to as a constant-strain triangle.

An important feature of the B-matrix is that each of its rows sum to zero.For example, adding up the first row gives

Y2 − Y3 + Y3 − Y1 + Y1 − Y2 = 0

The fact that the rows of the B-matrix sum to zero makes sense from thefollowing physical argument. Suppose that all of the nodal displacements daX

and daY were chosen to have a value of 1. This type of displacement is calleda rigid-body translation. If an element moves as a rigid body, no strain shoulddevelop in the element. This condition is satisfied if each row of Be sums tozero.


X

Y

1 2

3

4

FIGURE 6.3

Arbitrary four-node quadrilateral.

6.2 Development of a four-node arbitrary quadrilateral

Let us now consider a slightly more complicated two-dimensional element,the four-node quadrilateral. A four-node quadrilateral is shown in Figure 6.3.Like the three-node triangle, the local node numbering is chosen such thatlocal node 1 is located at any one of the corners. The remaining nodes arethen numbered sequentially in a counterclockwise fashion. There are two dis-placement components associated with each node, so the nodal displacementscan be stored in an 8 × 1 vector as follows:

de =

d1X

d1Y

d2X

d2Y

d3X

d3Y

d4X

d4Y

(6.18)

Let us begin by interpolating the trial function with a four-term polynomialas follows:

uX

uY

e

=

α1 + α2X + α3Y + α4XY

β1 + β2X + β3Y + β4XY

(6.19)

To derive the shape functions for the element, we proceed like we did withthe three-node triangle and enforce the following conditions at the nodes:

ui(X1, Y1) = d1i

ui(X2, Y2) = d2i

ui(X3, Y3) = d3i

ui(X4, Y4) = d4i (6.20)


where i can denote either X or Y . Letting i = X and enforcing the fourconditions stated in (6.20) yields

d1X

d2X

d3X

d4X

=

1 X1 Y1 X1Y1

1 X2 Y2 X2Y2

1 X3 Y3 X3Y3

1 X4 Y4 X4Y4

α1

α2

α3

α4

(6.21)

Except for simple element geometries like squares and rectangles, the solutionof equation (6.21) is very messy. For the moment, let us assume that itcan be solved in principle to obtain the following relationship between thecomponents of the trial function and the nodal displacements:

ue =

uX

uY

e

=

[

N1 0 N2 0 N3 0 N4 00 N1 0 N2 0 N3 0 N4

]

d1X

d1Y

d2X

d2Y

d3X

d3Y

d4X

d4Y

=Ne

2 × 8de

8 × 1(6.22)

where the shape functions, N1 to N4, are functions of X and Y , i.e., Na =Na(X,Y ). In the next section we will illustrate the shortcomings of the finiteelement interpolation defined in equation (6.19).

6.2.1 Compatibility issues

A mesh composed of two four-node quadrilateral elements is shown in Fig-ure 6.4. The original shape of each element is illustrated in Figure 6.4(a), andthe initial nodal coordinates are given in Table 6.1. The nodes are displacedaccording to the values reported in the table. Each element is identified bya number inside a rounded box. Local node 1 for each element is chosen tobe the node occupying the lower left corner. The remaining local nodes arenumbered counterclockwise. The global node numbers are indicated in bold.

Following the procedure outlined in the previous section, let us now obtainthe shape functions for each element. Focusing on element 1, equation (6.21)becomes

d1X

d2X

d3X

d4X

=

1 0 0 01 1 0 01 2 1 21 1 1 1

α1

α2

α3

α4

(6.23)


X

Y

X

Y

¤

£

¡

¢1

¤

£

¡

¢2

1 2 3

4 5 6

1 1

(a) (b)

FIGURE 6.4

Compatibility issues with the four-node quadrilateral.

where the subscripts 1 to 4 refer to local node numbers. Note that in equation(6.23) above, we are inserting the local nodal coordinates; for example, X1

refers to the X coordinate of local node one1. Solving (6.23) for α1 to α4

yields

α1 = d1X

α2 = −d1X + d2X

α3 = −d1X − 3d3X + 2d4X

α4 = d1X − d2X + d3X − d4X

After plugging in these values into the first line of equation (6.19) and collecting

TABLE 6.1

Nodal coordinates anddisplacements.

Node X Y daX daY

1 0 0 0.0 0.02 1 0 -0.1 0.13 2 0 0.2 0.24 1 1 -0.3 -0.15 2 1 -0.2 -0.16 3 1 -0.2 0.1


terms, we obtain the following four shape functions for element 1:

N1(X,Y ) = 1 − X − Y + XY

N2(X,Y ) = X − XY

N3(X,Y ) = −Y + XY

N4(X,Y ) = 2Y − XY (6.24)

Following the same procedure, the shape functions for element 2 are found tobe

N1(X,Y ) = 2 − X − 2Y + XY

N2(X,Y ) = −1 + X + Y − XY

N3(X,Y ) = −2Y + XY

N4(X,Y ) = 3Y − XY (6.25)

Armed with the shape functions for both elements, we can now interpolatethe displacement field within each element as follows:

u1 = N1d1

u2 = N2d2

where the subscripts 1 and 2 refer to elements 1 and 2, respectively.We are now able to investigate the motion of individual points that lie inside

an undeformed element. Let x1 give the position of a point in the deformedconfiguration that originally occupied position X1 inside the undeformed el-ement 1. Similarly, let x2 identify the deformed position of point X2 insideelement 2. The deformed positions are related to the original positions andthe nodal displacements as follows:

x1 = X1 + N1d1

x2 = X2 + N2d2 (6.26)

The deformed positions of points that were originally located inside theundeformed elements are shown in Figure 6.4(b). This figure deserves somefurther discussion. The lines defined by global nodes 4–5, 5–6, 1–2, and 2–3are originally horizontal and remain straight lines after deformation. On theother hand, lines 1–4, 2–5, and 3–6, which are originally straight but slanted,are no longer straight lines in the deformed configuration. This causes a severeproblem along the edge 2–5. Notice that even though the deformed mesh isconnected at all the nodes, there is a gap that opens up between the twoelements along edge 2–5. This is called an incompatibility. This incompatibledeformation would give rise to a response that is too compliant, i.e., not stiffenough due to the fact that the material points between nodes 2 and 5 becomedisconnected.


X

Y

(-1,-1) (1,-1)

(1,1)(-1,1)

1 2

34

FIGURE 6.5

Bi-unit square mapped into arbitrary quadrilateral.

It turns out that if all of the element edges in the original configuration areeither horizontal or vertical (no slanted edges), the deformation (no matterwhat the nodal displacements are) would end up being compatible. This istrue for four-node quadrilaterals that are either square or rectangular. Un-fortunately, complicated geometries are difficult to mesh with squares and/orrectangles. It is therefore desirable to come up with a four-node element that,when connected to other four-node elements in a mesh, results in a compat-ible displacement field. Fortunately, this problem has been overcome by thedevelopment of a class of elements called isoparametric finite elements. Thistopic will be discussed in the next section.

6.2.2 The bi-unit square

A square element is shown in Figure 6.5. The square is centered at the originand has sides of length two units. As in the last example, we will interpolatethe displacement components over the element using a four-term polynomialaccording to equation (6.19). With the nodal coordinates defined in Fig-ure 6.5, the equation for the constants α1 to α4 is

d1X

d2X

d3X

d4X

=

1 −1 −1 11 1 −1 −11 1 1 11 −1 1 −1

α1

α2

α3

α4

(6.27)


Solving equation (6.27) yields

α1 =1

4(d1X + d2X + d3X + d4X)

α2 =1

4(−d1X + d2X + d3X − d4X)

α3 =1

4(−d1X + −d2X + d3X + d4X)

α4 =1

4(d1X − d2X + d3X − d4X)

Finally, plugging α1, . . . , α4 back into the first line of equation (6.19) andcollecting terms gives

N1 =1

4(1 − X)(1 − Y )

N2 =1

4(1 + X)(1 − Y )

N3 =1

4(1 + X)(1 + Y )

N4 =1

4(1 − X)(1 + Y ) (6.28)

The shape functions defined in equation (6.28) above happen to be productsof one-dimensional shape functions derived in Chapter 3. For this reason,they are often referred to as bi-linear shape functions.

Let us now investigate the performance of the bi-linear shape functions.To begin, let pe be an 8 × 1 vector that stores the nodal coordinates of thedeformed element. Let Pe be an 8×1 vector that stores the nodal coordinatesof the undeformed bi-unit square. Now let Xe be a position vector definingthe location of a point inside the undeformed bi-unit square, and let xe definethe location of a point inside the deformed element. The vectors xe and Xe

are then related through the following equations:

xe = Xe + ue

= Xe + Nede

= NePe + Nede

= Ne(Pe + de)

= Nepe

Suppose that we prescribe the nodal displacements given in Table 6.2. Thebi-unit square in this deformed configuration is shown in Figure 6.5. The dotsshow the locations of points that were originally located inside the undeformedbi-unit square. Notice that the interpolation gives rise to a deformed elementwhose edges remain straight. This reveals that the shape functions for thebi-unit square can be used as interpolation functions to find the interior points


TABLE 6.2

Nodal displacementsfor the bi-unitsquare.

Node daX daY

1 2.0 2.62 1.9 3.13 2.3 2.94 1.5 2.5

of an arbitrary quadrilateral, given the quadrilateral’s nodal point locations.The interpolation yields a deformed element whose edges are all straight lines.Another way of saying this is that the bi-linear shape functions can be usedto “map” the bi-unit square into an arbitrary quadrilateral. This mapping iscommonly written as

x =

4∑

a=1

Naxa

y =

4∑

a=1

Naya ⇒

xe = Nepe

6.2.3 The parent domain

In order to complete the development of the arbitrary four-node quadrilateral,it is necessary to define the bi-unit square in a ξ−η coordinate system as shownin Figure 6.6. The bi-unit square in the ξ−η coordinate system is called theparent domain. The shape functions for the bi-unit square can be written interms of the dimensionless variables, ξ and η, as follows:

N1(ξ, η) =1

4(1 − ξ)(1 − η)

N2(ξ, η) =1

4(1 + ξ)(1 − η)

N3(ξ, η) =1

4(1 + ξ)(1 + η)

N4(ξ, η) =1

4(1 − ξ)(1 + η) (6.29)

A surface plot of the bi-linear shape function Na(ξ, η) is shown in Figure 6.7.Note that the surface, which in general is not planar, is linear along the fourelement edges. It takes on the value of 1 at local node a, and it is zero at theremaining nodes.


ξ

η

X

Y

x

X

FIGURE 6.6

Bi-unit square in ξ−η coordinate system.

From our previous discussion of the bi-unit square, given the nodal coor-dinates, we can now construct arbitrary quadrilaterals with straight edges inthe X−Y coordinate system. The deformed element in the X−Y coordinatesystem is called the physical domain. The mapping of the bi-unit square fromthe parent domain into the physical domain can be written as

Xe = Ne(ξ, η)Pe

where Ne is a 2×8 matrix containing the shape functions (6.29), and Pe is an8×1 vector that contains the nodal coordinates of the element in the physicaldomain. Let us now assume that the undeformed element in the physicaldomain undergoes a deformation defined by the nodal displacements. A pointxe, inside the deformed element that once occupied position Xe in the physicaldomain, can be obtained from the following mapping:

xe = Ne(ξ, η)pe

where pe is an 8×1 vector that contains the nodal coordinates of the deformedelement. The use of the bi-linear shape functions ensures that the edges ofthe deformed element remain straight lines.

The displacement ue is defined as the deformed position xe minus the un-deformed position Xe, i.e.,

ue = xe − Xe

= Ne(pe − Pe)

= Nede (6.30)


ξ

η

Na(ξ, η)

FIGURE 6.7

Surface plot of bi-linear shape function evaluated at node a.

The last line in (6.30) reveals the remarkable fact that the same bi-linear shapefunctions that are used to map the bi-unit square in the parent domain into anarbitrary quadrilateral in the physical domain can also be used to interpolatethe displacement field throughout an arbitrary quadrilateral. Because thebi-linear shape functions map straight lines into straight lines, the resultingdeformation of an assemblage of four-node quadrilaterals will be compatible.

To emphasize this concept, let us revisit the two-element mesh shown inFigure 6.4. If the bi-linear shape functions (6.29) were used instead of theshape functions (6.24) and (6.25), the resulting deformation would look likethat depicted in Figure 6.8. Notice that all of the edges of the deformedelements are straight, and the resulting displacement field is compatible, i.e.,there is no longer a gap along edge 2–5.

We end this section by stating that the arbitrary four-node quadrilateralbelongs to a class of elements called isoparametric finite elements. An elementis said to be isoparametric if the shape functions used to interpolate thedisplacement field are the same (iso) as those used for mapping (parametric)the element from the parent domain to the physical domain.

6.2.4 The B-matrix

The goal of this section is to find a relationship between the in-plane straincomponents stored in the vector ǫe and the vector de containing the nodal dis-placements of the four-node quadrilateral. Just as we did for the three-nodetriangle, the strain components are obtained by differentiating the displace-


X

Y

X

Y

¤

£

¡

¢1

¤

£

¡

¢2

1 2 3

4 5 6

1 2

(a) (b)

FIGURE 6.8

Compatible deformation of arbitrary quadrilaterals.

ment components as follows:

ǫXX

ǫY Y

γXY

e

=

uX,X

uY,Y

uX,Y + uY,X

e

=

∑

4

a=1Na,XdaX

∑

4

a=1Na,Y daY

∑

4

a=1(Na,Y daX + Na,XdaY )

(6.31)

We can write equation (6.31) in matrix notation as

ǫe = Bede

where Be is the 3 × 8 B-matrix. Note that the B-matrix contains derivativesof the shape functions with respect to X and Y , i.e.,

Be =

N1,X 0 N2,X 0 N3,X 0 N4,X 00 N1,Y 0 N2,Y 0 N3,Y 0 N4,Y

N1,Y N1,X N2,Y N2,X N3,Y N3,X N4,Y N4,X

(6.32)

Because the shape functions N1 to N4 are functions of the natural coordinatesξ and η, care must be exercised in taking the derivatives of these functionswith respect to X and Y , as discussed the next section.


6.2.5 Derivatives of the shape functions

When asked to differentiate the bi-linear shape functions (6.29) with respectto X and Y , the natural thing to do is to recognize that X and Y are functionsof ξ and η and employ the chain rule as follows:

Na,X = Na,ξξ,X + Na,ηη,X

Na,Y = Na,ξξ,Y + Na,ηη,Y

While taking the derivatives of the shape functions with respect to ξ andη poses no difficulty, evaluating derivatives ξ,X , ξ,Y , η,X , and η,Y can beproblematic. The reason is that, even though the forward mapping fromthe parent domain to the physical domain is explicitly defined by equation(6.30), the reverse mapping from the physical domain to the parent domain,in general, is not known in closed form. To alleviate this problem, one caninstead use implicit differentiation as follows:

Na,ξ = Na,XX,ξ + Na,Y Y,ξ

Na,η = Na,XX,η + Na,Y Y,η (6.33)

In equation (6.33) above, the derivatives of the spatial variables X and Y withrespect to the natural coordinates can be readily evaluated given the mapping(6.30). Treating Na,X and Na,Y as unknowns and writing (6.33) in matrixform yields

Na,ξ

Na,η

=

[

X,ξ Y,ξ

X,η Y,η

]

Na,X

Na,Y

(6.34)

The matrix on the right-hand side of equation (6.34) is called the Jacobianmatrix and is commonly referred to as J. The physical meaning of the com-ponents of J will be discussed in the next section when we discuss how aninfinitesimal area in the physical domain is related to the corresponding in-finitesimal area in the parent domain.

By inverting equation (6.34) we obtain

Na,X

Na,Y

= J−1

Na,ξ

Na,η

(6.35)

So in other words, the derivatives of the shape functions with respect to X andY at some point inside the element in the physical domain are obtained by firstcomputing the Jacobian matrix, J, at that point, then computing its inverse,J−1, and finally multipling J−1 by the vector containing the derivatives of theshape functions with respect to ξ and η.

6.2.6 Area change

In this section, we investigate the relationship between an infinitesimal areain the physical domain and the corresponding area in the parent domain. The


ξ

η

ξ(1)

ξ(2)

X

YX

(1)X(2)

FIGURE 6.9

Area change between infinitesimal elements in parent and physical domains.

relationship will be needed later in the formulation of the element matricesfor isoparametric elements.

To begin, consider an infinitesimal area in the parent domain depicted inFigure 6.9. Points inside this infinitesimal area in the parent domain map topoints inside an infinitesimal area in the physical domain through the mapping

X = X(ξ, η)

Y = Y (ξ, η) (6.36)

Recalling our study of the Taylor series expansion in Chapter 2, we nowexpand each of the two functions in equation (6.36) as a linear Taylor seriesabout the point (ξ0, η0), which lies inside the infinitesimal area in the parentdomain as follows:

X ≈ X(ξ0, η0) +∂X

∂ξ(ξ − ξ0) +

∂X

∂η(η − η0)

Y ≈ Y (ξ0, η0) +∂Y

∂ξ(ξ − ξ0) +

∂Y

∂η(η − η0) (6.37)

where the partial derivatives are understood to be evaluated at point (ξ0, η0).If we restrict our attention to points X and Y that are infinitesimally close tothe points X(ξ0, η0) and Y (ξ0, η0), then the linear Taylor series representationsof the functions X and Y are exact. Letting X − X(ξ0, η0) = dX, Y −

Y (ξ0, η0) = dY , ξ − ξ0 = dξ, and η − η0 = dη, we can write equation (6.37)in matrix notation as follows:

dX

dY

=

[

X,ξ X,η

Y,ξ Y,η

]

dξ

dη

(6.38)


It is now instructive to view dX and dY as components of an infinitesi-mal vector dX in the physical domain, and dξ and dη as components of aninfinitesimal vector dξ in the parent domain. Equation (6.38) can then bewritten as

dX = JT dξ (6.39)

where J is the Jacobian matrix, i.e.,

J =

[

X,ξ Y,ξ

X,η Y,η

]

(6.40)

One can think of equation (6.39) as the matrix equation that maps an in-finitesimal line segment dξ in the parent domain into the infinitesimal linesegment dX in the physical domain.

Now let the bottom edge of the area in the parent domain be defined by thevector dξ

(1), and let the left edge of the area be defined by the vector dξ(2).

The components of these two vectors in the ξ−η coordinate system are givenas

dξ(1) =

dξ

0

; dξ(2) =

0dη

(6.41)

The magnitude of the cross product between the two vectors dξ(1) and dξ

(2)

yields the area, dA, of the infinitesimal element in the parent domain, i.e.,

dA =∣

∣

∣dξ

(1)× dξ

(2)

∣

∣

∣= dξdη

The vectors dξ(1) and dξ

(2) map into the vectors dX(1) and dX(2) respectively.Using the relation (6.39), we can write the components of these two vectorsin the physical domain as

dX(1) =

X,ξ dξ

Y,ξ dξ

; dX(2) =

X,η dη

Y,η dη

(6.42)

Finally, the infinitesimal area, da, in the physical domain is equal to themagnitude of the cross product dX(1) × dX(2); hence,

da =∣

∣

∣dX(1)

× dX(2)

∣

∣

∣=

∣

∣

∣

∣

∣

∣

∣

e1 e2 e3

X,ξdξ Y,ξdξ 0

X,ηdη Y,ηdη 0

∣

∣

∣

∣

∣

∣

∣

= (X,ξ dξ)(Y,η dη) − (Y,ξ dξ)(X,η dη)

= detJdξ dη

= j dξ dη (6.43)

Here in equation (6.43) above, j is the determinant of the Jacobian matrix,and e1, e2, and e3 are Cartesian base vectors. We now see that

da = j dA (6.44)


1

2

3

4

FIGURE 6.10

Four-node tetrahedron.

As a final remark, the interested reader should make the connection be-tween the Jacobian matrix and the deformation gradient tensor, F, defined inChapter 2. If one views the parent domain as the original configuration andthe physical domain as the deformed configuration, then JT defines the de-formation of infinitesimal line segments originally lying in the parent domain.

6.3 Four-node tetrahedron

A close cousin to the three-node triangle is the four-node tetrahedron. Theelement is the simplest among the three-dimensional continuum elements. Asshown in Figure 6.10, it has four nodes and four faces. We interpolate thefield variable throughout the volume of the element using interpolants thatare linear in X, Y , and Z. Hence, for solid mechanics problems, we assume

uX

uY

uZ

e

=

α1 + α2X + α3Y + α4Z

β1 + β2X + β3Y + β4Z

γ1 + γ2X + γ3Y + γ4Z

(6.45)

The 12 × 1 vector containing the nodal displacements can be written as


de

12 × 1=

d1X

d1Y

d1Z

d2X

d2Y

d2Z

d3X

d3Y

d3Z

d4X

d4Y

d4Z

(6.46)

The shape functions for the element are derived by insisting that the trialfunction be equal to the nodal displacements when evaluated at the nodalpoints, i.e.,

uX(X1, Y1, Z1) = d1X

uX(X2, Y2, Z2) = d2X

uX(X3, Y3, Z3) = d3X

uX(X4, Y4, Z4) = d4X (6.47)

Inserting the constraints (6.47) into the first line of (6.45) gives the followingfour equations in four unknowns:

1 X1 Y1 Z1

1 X2 Y2 Z2

1 X3 Y3 Z3

1 X4 Y4 Z4

α1

α2

α3

α4

=

d1X

d2X

d3X

d4X

(6.48)

It turns out that the algebra required to solve the system of equations (6.48) israther lengthy. It is much easier to derive the shape functions for this elementusing volume coordinates.

Consider a point, P , that lies at some location, (X,Y,Z), inside the volumeof the tetrahedron. Given this location, we can define four volumes. The firstvolume, V1, is the volume of the tetrahedron having one of its vertices at pointP , and the other three lying on the face defined by the local nodes 2, 3, and4. The volume coordinate, V2, is defined by the point P and local nodes 1,3, and 4, and so on. The vertices of each volume coordinate are provided inTable 6.3.

We now let a be a vector pointing from local node 1 to local node 2. Sim-ilarly, let b be another vector pointing from node 1 to 3, and c be a vectorpointing from node 1 to 4. These three vectors can be written in terms of


TABLE 6.3

Volume coordinates and vertices.

Volume Vertex 1 Vertex 2 Vertex 3 Vertex 4

V1 P 2 3 4V2 P 1 3 4V3 P 1 2 4V4 P 1 2 3

Cartesian base vectors as follows:

a = (X2 − X1)eX + (Y2 − Y1)eY + (Z2 − Z1)eZ

b = (X3 − X1)eX + (Y3 − Y1)eY + (Z3 − Z1)eZ

c = (X4 − X1)eX + (Y4 − Y1)eY + (Z4 − Z1)eZ (6.49)

Having defined the vectors a, b, and c, the volume, V , of the tetrahedron canbe calculated by employing the scalar triple-product operation described inChapter 2, Section 2.3. Doing this gives

V =1

6[a × b] · c

=1

6(aXbZ − aZbY ) cX − (aXbZ − aZbX) cY

+ (aXbY − aY bX) cZ (6.50)

The volume coordinates V1 to V4 can also be expressed in terms of the nodalcoordinates and the spatial variables using the scalar triple-product operationcarried out in (6.50) above. The expressions can be obtained by defining thecomponents of a, b, and c according to the definitions listed in Table 6.4, andthen carrying out the operation (6.50). Note that in the table we have usedthe notation X21 = X2 − X1, Y21 = Y2 − Y1, etc.

The shape functions for the four-node tetrahedron can be written compactlyin terms of the volume coordinates as follows:

Na =Va

V(6.51)

TABLE 6.4

Definition of the components of a, b, and c for the volume coordinatecalculations.

Volume aX aY aZ bX bY bZ cX cY cZ

V1 X32 Y32 Z32 X42 Y42 Z42 X − X2 Y − Y2 Z − Z2

V2 X31 Y31 Z31 X41 Y41 Z41 X − X1 Y − Y1 Z − Z1

V3 X21 Y21 Z21 X41 Y41 Z41 X − X1 Y − Y1 Z − Z1

V4 X21 Y21 Z21 X31 Y31 Z31 X − X1 Y − Y1 Z − Z1

V X21 Y21 Z21 X31 Y31 Z31 X41 Y41 Z41


where a is the node number, which can take on any value between 1 and 4.Note that the volume coordinates, and hence shape functions, are linear inX, Y , and Z. The shape functions also possess the Kronecker delta property,i.e., NI(XJ) = δIJ . For example, as the point P moves inside the volume,each of the four volumes changes continuously. When the point P approachesnode 1, the volume coordinate V4 approaches zero. When the point P movesto nodes 2, 3, or 4, the volume V4 also goes to zero. So in this case wehave V4(X1, Y1, Z1)/V = 0, V4(X2, Y2, Z2)/V = 0, V4(X3, Y3, Z3)/V = 0, andV4(X4, Y4, Z4)/V = 1

6.3.1 The B-matrix

The B-matrix for the four-node tetrahedron can be derived by recalling thedefinition of the small-strain components in terms of the displacement com-ponents. The 6× 1 strain vector at a point within the element can be writtenas

ǫe =

ǫXX

ǫY Y

ǫZZ

γY Z

γXZ

γXY

e

=

uX,X

uY,Y

uZ,Z

uY,Z + uZ,Y

uX,Z + uZ,X

uX,Y + uY,X

e

=

Na,XdaX

Na,Y daY

Na,ZdaZ

Na,ZdaY + Na,Y daZ

Na,ZdaX + Na,XdaZ

Na,Y daX + Na,XdaY

(6.52)

Writing equation (6.52) in the form σe = Bede yields the following 6 × 12matrix:

Be =

N1,X 0 0 N2,X 0 0 N3,X 0 0 N4,X 0 0

0 N1,Y 0 0 N2,Y 0 0 N3,Y 0 0 N4,Y 0

0 0 N1,Z 0 0 N2,Z 0 0 N3,Z 0 0 N4,Z

0 N1,Z N1,Y 0 N2,Z N2,Y 0 N3,Z N3,Y 0 N4,Z N4,Y

N1,Z 0 N1,X N2,Z 0 N2,X N3,Z 0 N3,X N4,Z 0 N4,X

N1,Y N1,X 0 N2,Y N2,X 0 N3,Y N3,X 0 N4,Y N4,X 0

The derivatives of the shape functions with respect to X, Y , and Z are givenin Table 6.5. Notice that the derivatives of the shape functions are constant,and therefore the four-node tetrahedron is a constant-strain element. While inprinciple, complicated geometries are easy to mesh with four-node tetrahedralelements, in general, the volume must be meshed with a large number ofelements to obtain good accuracy.


TABLE 6.5

Derivatives of the shape functions for the four-node tetrahedron.

Na ∂/∂X ∂/∂Y ∂/∂Z

N1

1

6V(X32Z42 − Z32Y42)

1

6V(Z32X42 − Z42X32)

1

6V(X32Y42 − Y32X42)

N2

1

6V(X32Z42 − Z32Y42)

1

6V(X32Z42 − Z32Y42)

1

6V(X32Z42 − Z32Y42)

N3

1

6V(X32Z42 − Z32Y42)

1

6V(X32Z42 − Z32Y42)

1

6V(X32Z42 − Z32Y42)

6.4 Eight-node brick

The arbitrary isoparametric hexahedron, otherwise known as simply the eight-node brick, is a modern-day workhorse for both linear and nonlinear contin-uum problems. Even though the element is considered low order, it gives goodaccuracy and can be used to mesh arbitrary volumes with relative ease.

To begin the isoparametric formulation, we consider a tri-unit cube in aξ − η − ζ parent domain as shown in Figure 6.11 The mapping between theparent domain and the physical domain is chosen to have the following form:

X = α1 + α2ξ + α3η + α4ζ + α5ξη + α6ξζ + α7ηζ + α8ξηζ

Y = β1 + β2ξ + β3η + β4ζ + β5ξη + β6ξζ + β7ηζ + β8ξηζ

Z = γ1 + γ2ξ + γ3η + γ4ζ + γ5ξη + γ6ξζ + γ7ηζ + γ8ξηζ (6.53)

The shape functions for the eight-node brick can be derived by insisting thatthe values of X, Y , and Z, when evaluated at the nodes, be equal to Xa, Ya,and Za, respectively, where a denotes the local node number and can takeon any value between 1 and 8. Local node 1 is chosen to occupy position(ξ = −1, η = −1, ζ = −1), and the remaining nodes are numbered accordingto the convection depicted in Figure 6.11. Focusing on the X componentsonly, the constants α1 to α8 can be obtained by solving the following systemof equations:

X1

X2

X3

X4

X5

X6

X7

X8

=

1 −1 −1 −1 1 1 1 −11 1 −1 −1 −1 −1 1 11 1 1 −1 1 −1 −1 −11 −1 1 −1 −1 1 −1 11 −1 −1 1 1 −1 −1 11 1 −1 1 −1 1 −1 −11 1 1 1 1 1 1 11 −1 1 1 −1 −1 1 −1

α1

α2

α3

α4

α5

α6

α7

α8

(6.54)

If we now solve equation (6.54) for α1 to α8 and insert the resulting constants


ξ

η

ζ

1

2

3

4

5

6

7

8

X

Y

Z

FIGURE 6.11

Eight-node brick in parent and physical domains.

into the first line of equation (6.53) and collect terms, we get the followingeight shape functions:

N1(ξ, η, ζ) = 1/8(1 − ξ)(1 − η)(1 − ζ)

N2(ξ, η, ζ) = 1/8(1 + ξ)(1 − η)(1 − ζ)

N3(ξ, η, ζ) = 1/8(1 + ξ)(1 + η)(1 − ζ)

N4(ξ, η, ζ) = 1/8(1 − ξ)(1 + η)(1 − ζ)

N5(ξ, η, ζ) = 1/8(1 − ξ)(1 − η)(1 + ζ)

N6(ξ, η, ζ) = 1/8(1 + ξ)(1 − η)(1 + ζ)

N7(ξ, η, ζ) = 1/8(1 + ξ)(1 + η)(1 + ζ)

N8(ξ, η, ζ) = 1/8(1 − ξ)(1 + η)(1 + ζ) (6.55)

The shape functions defined in (6.55) above are often expressed succinctly inone line as

Na =1

8(1 ± ξ)(1 ± η)(1 ± ζ) (6.56)

where the ± sign convention follows (6.55). The shape functions (6.55) areoften called the tri-linear shape functions, because they are products of theone-dimensional shape functions derived in Chapter 3.

Having derived the shape functions, the mapping between the parent and


physical domains can be written as

X = Na(ξ, η, ζ)Xa

Y = Na(ξ, η, ζ)Ya

Z = Na(ξ, η, ζ)Za (6.57)

It turns out that equation (6.57), involving the tri-linear shape functions,maps the tri-unit cube in the parent domain into an arbitrary hexahedronin the physical domain. The resulting arbitrary hexahedron in the physicaldomain has straight edges. The resulting element faces in the physical domainare in general not planar surfaces.

Let us now use the same tri-linear shape functions defined in (6.55) aboveto interpolate the displacement field within an element. The displacementinterpolation can be written as

uX

uY

uZ

e

=

Na(ξ, η, ζ)daX

Na(ξ, η, ζ)daY

Na(ξ, η, ζ)daZ

(6.58)

It turns out that the displacement interpolation (6.58) gives rise to compatibledeformation between adjacent elements.

6.4.1 The B-matrix

Recall that for three-dimensional applications, the small-strain componentscan be stored in a 6 × 1 vector as follows:

ǫXX

ǫY Y

ǫZZ

γY Z

γXZ

γXY

=

uX,X

uY,Y

uZ,Z

uY,Z + uZ,Y

uX,Z + uZ,X

uX,Y + uY,X

(6.59)

To derive the B-matrix for the eight-node brick element, we seek a matrixequation of the form

ǫe

6 × 1=

Be

6 × 24de

24 × 1(6.60)

where Be is the element B-matrix, and de is the vector containing the nodaldisplacements, i.e.,

de =

d1X

d1Y

d1Z

...d8Z

(6.61)


The B-matrix provides the connection between the approximate strain fieldwithin the element and the nodal displacements. Using the strain-displacementrelationship (6.59), the 6 × 24 B-matrix can be expressed as

Be =

[

Be1

6 × 3Be2

6 × 3. . .

Be8

6 × 3

]

(6.62)

where

Bea =

Na,X 0 00 Na,Y 00 0 Na,Z

0 Na,Z Na,Y

Na,Z 0 Na,X

Na,Y Na,X 0

(6.63)

is the 6 × 3 submatrix associated with node a.The derivatives of the shape functions with respect to the global X, Y , and

Z coordinates are obtained by using the technique of implicit differentiationexplained in Section 6.2.5. Doing this yields

Na,ξ = Na,XX,ξ + Na,Y Y,ξ + Na,ZZ,ξ

Na,η = Na,XX,η + Na,Y Y,η + Na,ZZ,η

Na,ζ = Na,XX,ζ + Na,Y Y,ζ + Na,ZZ,ζ (6.64)

or in matrix notation

Na,ξ

Na,η

Na,ζ

=

X,ξ Y,ξ Z,ξ

X,η Y,η Z,η

X,ζ Y,ζ Z,ζ

Na,X

Na,Y

Na,Z

(6.65)

The 3× 3 matrix defined in (6.65) above is the Jacobian matrix. The deriva-tives of the shape functions with respect to the global coordinates are obtainedby obtaining the inverse relationship as follows:

Na,X

Na,Y

Na,Z

= J−1

Na,ξ

Na,η

Na,ζ

(6.66)

6.4.2 Volume change

To end this section, we now illustrate how an infinitesimal volume elementin the physical domain is related to the corresponding infinitesimal volumeelement in the parent domain. This relationship will be needed later for theevaluation of the element matrix, which involves evaluating a volume integralover the parent domain.

We begin by writing down the mapping between the parent domain andthe physical domain as follows:

X = f (ξ, η, ζ) (6.67)


X

Y

Z

dξ

dX(1)

ξ

ηζdv

dV

FIGURE 6.12

Infinitesimal volume element in parent and physical domains.

Equation (6.67) above simply states that the position vector, X, of a pointin the physical domain, which once occupied position (ξ, η, ζ) in the parentdomain, is a vector-valued function of the natural coordinates. From our workin Chapter 2, Section 2.6, we recall that given the components dξ, dη, anddζ of an infinitesimal vector in the parent domain, the corresponding vector,dX, in the physical domain can be obtained as

dX =∂X

∂ξdξ +

∂X

∂ηdη +

∂X

∂ζdζ (6.68)

Let us now define dξ, dη, and dζ to be infinitesimal vectors in the par-ent domain that are mutually perpendicular and are parallel to the ξ, η, andζ axes, respectively. To avoid clutter, only the vector dξ is labeled in Fig-ure 6.12.The figure shows an infinitesimal volume element, dV , in the parentdomain whose edges are defined by the vectors dξ, dη, and dζ. The cor-responding infinitesimal volume in the physical domain, dv, is also shownin the figure. Its edges are defined by the three vectors dX(1), dX(2), anddX(3). Only dX(1) is labeled in the figure. The three infinitesimal vectorsin the physical domain can be written in terms of their components and theCartesian base vectors eX , eY , and eZ as

dX(1) =∂X

∂ξdξ

=∂X

∂ξdξ eX +

∂Y

∂ξdξ eY +

∂Z

∂ξdξ eZ

dX(2) =∂X

∂ηdη

=∂X

∂ηdη e1 +

∂Y

∂ηdη e2 +

∂Z

∂ηdη e3,


dX(3) =∂X

∂ζdζ

=∂X

∂ζdζ e1 +

∂Y

∂ζdζ e2 +

∂Z

∂ζdζ e3 (6.69)

From our study of Chapter 2, we recall that the volume, dv, defined by thethree edges, dX(1), dX(2), and dX(3), can be evaluated by carrying out thefollowing scalar triple product:

dv =

dX(1)× dX(2)

· dX(3) (6.70)

The volume, dV , of the element in the parent domain can be evaluated in thesame manner, i.e.,

dV = dξ × dη · dζ

= dξdηdζ (6.71)

Finally, the scalar triple product (6.70) above (and hence dv) is equal to thefollowing determinant:

dv =

∣

∣

∣

∣

∣

∣

∣

X,ξdξ Y,ξdξ Z,ξdξ

X,ηdη Y,ηdη Z,ηdη

X,ζdζ Y,ζdζ Z,ζdζ

∣

∣

∣

∣

∣

∣

∣

= j dξdηdζ (6.72)

After expanding the determinant (6.72) and observing the result (6.71), wesee that the two volumes dv and dV are related through the determinant, j,of the Jacobian matrix defined in the previous section. In other words,

dv = j dV (6.73)

6.5 Element matrices and vectors

In the previous sections, we focused on deriving the shape functions and B-matrices for a variety of two- and three-dimensional continuum elements. Theprocess of going through these derivations is needed to gain a sound under-standing of continuum element behavior. In this section we go one step furtherand illustrate the process of obtaining the element matrix and element vectorfor continuum elements. For illustration purposes, we will focus on the arbi-trary four-node quadrilateral and assume the condition of either plane strainor plane stress.

From our discussion in Chapter 4, the weak form of the equilibrium equationcan be written over a single element as follows:

∫

Ωe

σijwi,j dΩ =

∫

Ωe

biwi dΩ +

∫

Γe

tiwi dΓ (6.74)


where wi are the components of the test function, σij are the Cauchy stresscomponents, bi are the components of the body force vector, and ti are thetraction components. Here in equation (6.74), Ωe refers to the interior of theelement, and Γe is the surface on which the traction vector is specified.

Recall that the components of the test function are chosen to be sufficientlysmooth inside the domain of interest, and zero at points on the boundarywhere the displacement vector is specified. For example, if the displacementcomponent u1 is specified at some point located on the boundary, then thecorresponding component of the test function, w1, would also be zero at thatpoint. For two-dimensional problems, it is sufficient to consider a family oftest functions that varies only with the spatial coordinates X1 and X2. Inthe following derivations, we will assume that wi = wi(X1,X2), and hence,all derivatives of the test function with respect to X3 are zero.

Having made the above restrictions on the test function, let us now expandthe integrand on the left-hand side of equation (6.74) above. Doing this yields

σijwi,j = σ11w1,1 + σ12w1,2

+σ21w2,1 + σ22w2,2 (6.75)

Next, noting that the Cauchy stress tensor is symmetric, we can combineterms on the right-hand side of equation (6.75) as follows:

σijwi,j = σ11w1,1 + σ12 (w1,2 + w2,1) + σ22w2,2

= δǫTσ (6.76)

where, for the sake of convenience, we have arranged the in-plane stress com-ponents and the derivatives of the test function in the following 3× 1 vectors:

σ =

σ11

σ22

σ12

; δǫ =

w1,1

w2,2

w1,2 + w2,1

(6.77)

Having made these definitions, we can now write the principle of virtual workin matrix notation as

∫

Ωe

δǫT

1 × 3σ

3 × 1dΩ =

∫

Ωe

wT

1 × 2b

2 × 1dΩ +

∫

Γe

wT

1 × 2t∗

2 × 1dΓ (6.78)

where w is a 2×1 vector containing the components of the test function, andb and t∗ are 2×1 vectors containing the components of the body force vectorand specified traction vector, respectively.

Employing the Galerkin procedure introduced in Chapter 3, we now inter-polate the test function, w , over the domain of the element using the samefinite element shape functions that are used to interpolate the displacementfield. The finite element approximation for the test function, we, then be-comes

we = Newe (6.79)


where, for the case of the four-node arbitrary quadrilateral, Ne is the 2 × 8matrix containing the shape functions defined in equation (6.29), and we isan 8 × 1 vector containing the values of the test function at the nodes.

The finite element approximation for, δǫe, the vector containing the deriva-tives of the test function, can be written as

δǫe =

Na,1wa1

Na,2wa2

Na,2wa1 + Na,1wa2

(6.80)

where the summation convention is implied (a = 1 to 4 for the case of the four-node quadrilateral). Here wai refers to the ith component of the test functionevaluated at local node a. Writing equation (6.80) in matrix notation gives

δǫe =

N1,X 0 N2,X 0 N3,X 0 N4,X 00 N1,Y 0 N2,Y 0 N3,Y 0 N4,Y

N1,Y N1,X N2,Y N2,X N3,Y N3,X N4,Y N4,X

w1X

w1Y

w2X

w2Y

w3X

w3Y

w4X

w4Y

= Bewe (6.81)

where the subscript X refers to the first component (i = 1) and Y refers tothe second component (i = 2). Here in the last line of equation (6.81), Be isthe 3×8 B-matrix defined in equation (6.32) containing the derivatives of theshape functions with respect to X and Y .

Let us now substitute the finite element approximations (6.79) and (6.81)into the weak form (6.78). Using the identities wT = wT

eNT

eand δǫ

T

e=

wT

eBT

egives

nel∑

e=1

wT

e

∫

Ωe

BT

eσe dΩ

=

nel∑

e=1

wT

e

∫

Ωe

NT

ebdΩ +

∫

Γe

NT

et∗ dΓ

(6.82)

Expressing the vector we in terms of the Boolean localization matrix, Le,yields

wT

nel∑

e=1

LT

e

(∫

Ωe

BT

eσe dΩ

)

= wT

nel∑

e=1

LT

e

(∫

Ωe

NT

ebdΩ +

∫

Γe

NT

et∗ dΓ

)

(6.83)


After invoking the arbitrariness of the global test function vector, w, we finallyobtain

nel∑

e=1

LT

e

(∫

Ωe

BT

eσe dΩ

)

=nel∑

e=1

LT

e

(∫

Ωe

NT

ebdΩ +

∫

Γe

NT

et∗ dΓ

)

(6.84)

We now recognize that the integral on the left-hand side of equation (6.84)is an element vector, f int

e, that contains the internal nodal forces. Note that

the internal force vector depends upon the stress field σe within the element,and we have not assumed anything as of yet for the constitutive (stress-strain)law. The sum of the integrals on the right-hand side is the element vector,fext

e, containing the external nodal forces. The fully assembled discretized

equations can then be written as

f int = fext (6.85)

When the material behavior is linearly elastic and governed by Hooke’s law,the vector σe containing the stress components within the element can beexpressed as

σe = Ceǫe (6.86)

Hence, the finite element approximation for the stress field within the elementcan be written as

σe = CeBede

= CeBeLed (6.87)

where Ce is the 3 × 3 elasticity matrix for the case of either plane strainor plane stress defined in Chapter 4, and d is the global vector containingthe nodal displacements. Substituting the last line in equation (6.86) intoequation (6.84) gives

nel∑

e=1

LT

e

(∫

Ωe

BT

eCeBe dΩLe

)

d

=nel∑

e=1

LT

e

(∫

Ωe

NT

ebdΩ +

∫

Γe

NT

et∗ dΓ

)

(6.88)

The element stiffness matrix, Ke, comes from the integral on the left-handside of equation (6.88). The element external force vector comes from theintegral on the right-hand side. To summarize we write

Ke =

∫

Ωe

BT

eCeBe dΩ (6.89)


fext

e=

∫

Ωe

NT

ebdΩ +

∫

Γe

NT

et∗ dΓ (6.90)

In practice, the integrals defined in equations (6.89) and (6.90) are carriedout over the area (and, if specified tractions are present, edges) of the bi-unitsquare in the parent domain. For example, in order to get an equivalent areaintegral in the parent domain, we must employ the area-change relationshipderived in Section 6.2.6. A differential volume in the physical domain can beexpressed as a differential volume in the parent domain as

dΩe = t jdξdη (6.91)

where in the present formulation for plane problems, t is the out-of-planethickness. Hence, the element stiffness matrix can be expressed as the follow-ing double integral in the parent domain:

Ke = t

∫

1

−1

∫

1

−1

BT

eCeBe jdξdη (6.92)

In finite element codes, the integrals over the parent domain are evaluatedusing a technique called Gauss quadrature. The technique gives the exactvalue of an integral whose integrand is a polynomial function, as explained indetail in the next section followed by some examples that demonstrate howto evaluate element matrices and vectors using Gauss quadrature.

6.6 Gauss quadrature

Gauss quadrature is often misunderstood to be a numerical integration tech-nique for obtaining approximate values to integrals. While Gauss quadraturecan be used to obtain approximate values of integrals, the main idea is tofind the exact values of integrals whose integrands are polynomial functions.This is accomplished using tabulated point and weight values. In this sec-tion we will describe the method of Gauss quadrature for evaluating bothone-dimensional and multidimensional integrals.

To begin, consider the following integral whose integrand is a cubic poly-nomial

∫

1

−1

(

α1 + α2ξ + α3ξ2 + α4ξ

3)

dξ (6.93)

where α1 to α4 are arbitrary constants. The exact value, I, of the integral is

I = 2α1 +2

3α3 (6.94)


We now seek to find integration points, ξI , and weights, WI , such that

nint∑

I=1

f(ξI)WI = I (6.95)

where nint is the total number of integration points that are needed for thesum to give the exact value of the integral. Here in equation (6.95), theintegration points, ξI , (also referred to as quadrature points) are locationswithin the interval −1 ≤ ξ ≤ 1. The quantities, WI , are numerical weightvalues associated with each integration point.

Let us now try to satisfy equation (6.95) with two integration points ξ1 andξ2. Equation (6.95) then becomes

I = f(ξ1)W1 + f(ξ2)W2

= (α1 + α2ξ1 + α3ξ2

1+ α4ξ

3

1)W1

+(α1 + α2ξ2 + α3ξ2

2+ α4ξ

3

2)W2

= (W1 + W2)α1 + (ξ1W1 + ξ2W2)α2

+(ξ2

1W1 + ξ

2

2W2)α3 + (ξ3

1W1 + ξ

3

2W2)α4 (6.96)

We now have the following two results for the same integral I:

I = (W1 + W2)α1 + (ξ1W1 + ξ2W2)α2 + (ξ2

1W1 + ξ

2

2W2)α3

+(ξ3

1W1 + ξ

3

2W2)α4

I = 2α1 +2

3α3 (6.97)

By comparing coefficients in equation (6.97) above we conclude that

W1 + W2 = 2

ξ1W1 + ξ2W2 = 0

ξ2

1W1 + ξ

2

2W2 = 2/3

ξ3

1W1 + ξ

3

2W2 = 0 (6.98)

We immediately see that the first equation in (6.98) can be satisfied if welet W1 = W2 = 1. The second and fourth equations are then satisfied auto-matically if we choose the quadrature point locations to be symmetric aboutξ = 0, i.e., ξ1 = −ξ2. The third line in equation (6.98) then becomes

2ξ2

1= 2/3 ⇒

ξ1 = ±1√

3(6.99)

Hence, the exact value of an integral of the form (6.93) whose integrand is acubic polynomial is

I = f(−1/√

3)(1) + f(1/√

3)(1) (6.100)


We can also obtain Gauss points and weights for higher order polynomialsusing the procedure described above. For example, suppose that the integrandis the following fifth-order polynomial:

f(ξ) = α1 + α2ξ + α3ξ2 + α4ξ

3 + α5ξ4 + α6ξ

5 (6.101)

The resulting equations to be solved are then

W1 + W2 + W3 = 2

ξ1W1 + ξ2W2 + ξ3W3 = 0

ξ2

1W1 + ξ

2

2W2 + ξ

2

3W3 = 2/3

ξ3

1W1 + ξ

3

2W2 + ξ

3

3W3 = 0

ξ4

1W1 + ξ

4

2W2 + ξ

4

3W3 = 2/5

ξ5

1W1 + ξ

5

2W2 + ξ

5

3W3 = 0 (6.102)

Let us for the moment assume that equations (6.102) can be solved by as-suming ξ2 = 0, and that the remaining integration points are symmetricallylocated about ξ = 0, i.e., ξ1 = −ξ3. After making this assumption, we seethat the second equation in (6.102) is satisfied if W1 = W3. We are now leftwith the following three equations in three unknowns:

2W1 + W2 = 2

2ξ2

1W1 = 2/3

2ξ4

1W1 = 2/5 (6.103)

Solving the second line in (6.103) for W1 and then plugging the result intothe third line gives

ξ1 = ±√

3/5 (6.104)

Next, inserting the known value of ξ1 back into the second line yields W1 =5/9, and hence, W2 = 8/9. The exact value of the integral is

I = f(−√

3/5)(5/9) + f(0)(8/9) + f(√

3/5)(5/9) (6.105)

The procedure that we have been following for generating integration pointsand weights can be carried out in principle for polynomials up to any order.The integration points and weights for polynomials up to fifth order are tab-ulated in Table 6.6 below. For the sake of interest, the locations of thesepoints are depicted in Figure 6.13.

It turns out that n quadrature points are required to obtain the exact valueof an integral whose integrand is a polynomial of order 2n−1. So, for example,for a fourth-order polynomial we have 2n − 1 = 4 or n = 5/2. Since n mustbe an integer, three quadrature points are required.


ξ1 = 0

ξ1 = −1/√

3 ξ1 = 1/√

3

ξ1 = −√

3/5 ξ2 = 0 ξ3 =√

3/5

FIGURE 6.13

Quadrature point locations and weights.

TABLE 6.6

Quadrature points and weights.

n 2n − 1 ξ1 ξ2 ξ3 w1 w2 w3

1 1 0 – – 2 – –

2 3 −1/√

3 1/√

3 – 1 1 –

3 5 −√

3/5 0√

3/5 5/9 8/9 5/9

Multidimensional integrals

Once the quadrature points and weights are obtained for one-dimensional in-tegrals, it is straightforward to apply these rules to evaluate multidimensionalintegrals such as area and volume integrals. To illustrate, let us consider thefollowing area integral:

I =

∫

1

−1

∫

1

−1

f(ξ, η) dξ dη (6.106)

As we recall from calculus, double integration is carried out by first evaluat-ing the inner integral with respect to ξ (keeping η fixed) and then computingthe outer integral with respect to η (keeping ξ fixed). In other words, dou-ble integration simply requires evaluating two one-dimensional integrals. Toillustrate, let us carry out the double integration (6.106) using a two-pointGauss quadrature rule in both the ξ and η directions as follows:

I =

∫

1

−1

[Wξ1f(ξ1, η) + Wξ2

f(ξ2, η)] dη

= Wη1[Wξ1

f(ξ1, η1) + Wξ2f(ξ2, η1)]

+Wη2[Wξ1

f(ξ1, η2) + Wξ2f(ξ2, η2)]


TABLE 6.7

Points and weights for 2 × 2quadrature.

I ξI ηI WI

1 −1/√

3 −1/√

3 1

2 1/√

3 −1/√

3 1

3 1/√

3 1/√

3 1

4 −1/√

3 1/√

3 1

and hence

I = Wη1Wξ1

f(ξ1, η1) + Wη1Wξ2

f(ξ2, η1)

+Wη2Wξ1

f(ξ1, η2) + Wη2Wξ2

f(ξ2, η2) (6.107)

Employing a two-point rule in both the ξ and η directions is referred to as2 × 2 quadrature. The operation can be written succinctly as

I =4

∑

I=1

WIf(ξI , ηI)

where the quadrature point locations and weights are given in Table 6.7.For the sake of interest, the mapping of the quadrature points (for 2 ×

2 quadrature) from the parent domain to the physical domain is shown inFigure 6.14. The X and Y locations of the quadrature points are be obtainedfrom the isoparametric mapping (6.30), i.e.,

XI =

4∑

a=1

Na(ξI , ηI)Xa

YI =

4∑

a=1

Na(ξI , ηI)Ya

where Xa and Ya are the coordinates of local node a in the physical domain.

Example 6.1

To illustrate the process of evaluating one-dimensional integrals using Gaussquadrature, let us consider the following integral:

∫

X2

X1

f(X) dX (6.108)

Notice that the upper and lower limits are not from minus one to one, sothe quadrature rule formulas derived above do not directly apply here. It is


1

1

2

2

3

3

4

4

I = 1

I = 2

I = 3

I = 4

ξ

η

X

Y

FIGURE 6.14

Quadrature point locations in parent and physical domains.

possible, however, through a change of variables, to recast the integral intothe proper form. To do this, we rely on the one-dimensional shape functionsto map the interval from −1 to 1 into a generic interval from X1 to X2 asfollows:

X = N1X1 + N2X2

=1

2(1 − ξ)(X1) + 1/2(1 + ξ)(X2) (6.109)

Notice that when the value of ξ is equal to −1, X is equal to X1. When ξ isequal to 1, the value of X is X2. Now suppose we are interested in evaluatingthe following integral using Gauss quadrature:

∫

5

4

(1 + X2) dX (6.110)

We will begin by using equation (6.109) to get the integral into the properform. The relationship between X and ξ is

X =1

2(1 − ξ)(4) +

1

2(1 + ξ)(5)

=9

2+

1

2ξ (6.111)


Given this relationship, we can now evaluate the relationship between dX anddξ, i.e.,

dX

dξ=

1

2⇒

dX =1

2dξ (6.112)

The integral (6.110) can now be written as

∫

1

−1

f(ξ) dξ

where

f(ξ) =1

2

[

1 +

9

2+

1

2ξ

2]

Because the integrand f(ξ) is quadratic in ξ, we have

2n − 1 = 2 ⇒

n = 3/2 (6.113)

or in other words, two quadrature points are required to obtain the exactvalue of the integral. Carrying out the required two-point quadrature ruleyields

I = W1f(ξ1) + W2f(ξ2)

= (1)f

(

−1√

3

)

+ (1)f

(

1√

3

)

= (1)

1

2

[

1 +

(

9

2−

1

2

1√

3

)2]

+ (1)

1

2

[

1 +

(

9

2+

1

2

1√

3

)2]

=64

3

If we perform a quick hand calculation as a check on our work, we notice thatthe result above agrees with the exact value of the integral, i.e.,

∫

5

4

(

1 + X2)

dX =

(

X +X3

3

)∣

∣

∣

∣

5

4

= (5 + 125/3) − (4 + 64/3)

=64

3


1 2

3

4

t∗ =

tX

tY

X

Y

FIGURE 6.15

Uniform traction distribution acting over an element edge.

Example 6.2

Recall that the element vector that arises from a prescribed traction (natural)boundary condition applied along an element edge is given as

fe =

∫

Γe

NT

et∗ dΓ (6.114)

where, for two-dimensional plane problems, Γe refers to the element edge (oredges) along which the traction is prescribed, Ne is the 2 × 8 shape functionmatrix, and t∗ is the 2 × 1 traction vector acting at a point on Γe. Noticethat the resulting element vector is 8 × 1.

To illustrate how to calculate the element vector, let us consider the situa-tion illustrated in Figure 6.15. Here, a constant traction distribution,

t∗ =

tX

tY

is applied along element edge 3–4 of an arbitrary four-node quadrilateral inthe undeformed configuration. The nodal coordinates for the element aregiven in Table 6.8.

Because the shape functions for the arbitrary four-node quadrilateral aregiven in terms of the ξ and η coordinates, it is necessary to recast the inte-gral (6.114) into an equivalent integral in the parent domain. The tractiondistribution acts along edge 3–4, so we only need to study how edge 3–4 mapsfrom the parent domain to the physical domain. Notice that edge 3–4 is thetop edge of the bi-unit square where η = 1.

In order to evaluate the integral in the parent domain, we need to know therelationship between dΓ and an infinitesimal movement, dξ, along the line 3-4in the parent domain.


TABLE 6.8

Nodal coordinates offour-node quadrilateral.

Local node Xa Ya

1 0.0 0.02 1.0 0.03 1.5 1.54 0.0 1.0

We begin by recalling Section 6.2.6, where we derived a relationship betweenan infinitesimal vector d ξ in the parent domain and an infinitesimal vectordX in the physical domain. This relationship is given as

dX = JT dξ

where J is the Jacobian matrix, i.e.,

J =

[

X,ξ Y,ξ

X,η Y,η

]

Along edge 3–4 (η = 1) in the parent domain, the infinitesimal movement d ξ

is just a movement in the ξ direction only. Hence,

d ξ =

d ξ

0

and, as a result, the infinitesimal vector in the physical domain reduces to

dX =

X,ξ

Y,ξ

dξ

The magnitude of the infinitesimal vector, dΓ, in the physical domain can beexpressed as

dΓ = t |dX|

= t

√

X2

,ξ+ Y 2

,ξdξ (6.115)

where in the present two-dimensional formulation, t is the thickness of theelement (assumed to be constant).

We can now write the element vector as an integral in the parent domainas

fe = t

∫

1

−1

NT

e(ξ, η = 1)t∗

√

X2

,ξ+ Y 2

,ξdξ (6.116)

The last step in this example is to go through the algebraically lengthy taskof evaluating the integral (6.116). The process is simplified if one notices that


the first two shape functions N1 and N2 are identically zero along edge 3–4where η = 1. We also note that the relationship between dΓ and dξ happensto be very simple in this example. Evaluating the derivatives X,ξ and Y,ξ weobtain

X,ξ =1

8(5 + η); Y,ξ =

1

8(1 + η) (6.117)

and hence,

dΓ = t

√

0.752 + 0.52 dξ

= tl34

2dξ (6.118)

where l34 is the length of edge 3–4 in the physical domain, i.e.,

l34 =√

(X4 − X3)2 + (Y4 − Y3)2 (6.119)

The final result is

fe = t

∫

1

−1

0000

1

2(1 + ξ)tX

1

2(1 + ξ)tY

1

2(1 − ξ)tX

1

2(1 − ξ)tY

l34

2dξ

= tl34

2

0000tX

tY

tX

tY

(6.120)

A few closing remarks are in order here. The quantities in the element vectorare called equivalent nodal loads. Equivalent nodal loads are point loads thatare applied at the nodes as a result of the traction distribution acting along anedge of the element. The point loads are computed by evaluating integrals thatcome directly from the weak form of the governing differential equation. It isalso worth mentioning that the shape functions N3 and N4 in this example,when evaluated along the line η = 1, reduce to the one-dimensional shapefunctions that depend only on ξ. Because N1 and N2 are zero along this line,the equivalent nodal loads could alternatively be obtained by simply carrying


X

Y

(0,0) (1,0)

(2,1)(0,1)

22 26

3031

FIGURE 6.16

Element matrix computation for arbitrary quadrilateral.

out the following integrals:

f3i = t

∫

1

−1

1

2(1 − ξ)til34 dξ

f4i = t

∫

1

−1

1

2(1 + ξ)til34 dξ

where i can be either X or Y .

Example 6.3

In this example we demonstrate the process of computing the 8 × 8 elementmatrix for the arbitrary four-node quadrilateral. For the sake of simplicity, wewill assume the condition of plane stress. The element in the physical domainis shown in Figure 6.16. The nodal coordinates and global node numbers arelabeled in the figure. The connectivity list for the element is given in Table 6.9below. We emphasize that a correct connectivity list begins with any one ofthe global node numbers and then continues in a counterclockwise fashion.

The first step in the process of computing the element matrix is to writedown the mapping between the parent domain and the physical domain. The

TABLE 6.9

Connectivity list.

Element 1 2 3 4

1 26 30 31 22


isoparametric mapping is written as

X(ξ, η) =4

∑

a=1

Na(ξ, η)Xa

Y (ξ, η) =4

∑

a=1

Na(ξ, η)Ya (6.121)

where Xa and Ya are the X and Y coordinates of local node a, respectively.Plugging the nodal coordinates into equation (6.121) gives

X = N1(1) + N2(2) + N3(0) + N4(0) = N1 + 2N2

Y = N1(0) + N2(1) + N3(1) + N4(0) = N2 + N3 (6.122)

Recall that the shape functions for the four-node quadrilateral are defined as

N1 =1

4(1 − ξ) (1 − η)

N2 =1

4(1 + ξ) (1 − η)

N3 =1

4(1 + ξ) (1 + η)

N4 =1

4(1 − ξ) (1 + η) (6.123)

After simplifying the algebra, the mapping (6.122) can be written succinctlyas

X =1

4(3 + ξ)(1 − η)

Y =1

2(1 + ξ) (6.124)

The next step in the procedure is to first set up the Jacobian matrix asso-ciated with the mapping between the parent and physical domains, and thento compute the determinant of the Jacobian matrix. Doing this we obtain

J =

[

X,ξ Y,ξ

X,η Y,η

]

=

[

1/4(1 − η) 1/2−1/4(3 + ξ) 0

]

⇒

j =1

8(3 + ξ) (6.125)

Notice that the determinant of Jacobian matrix is not constant within theelement but varies linearly with the natural coordinate ξ.


Armed with the Jacobian matrix, the matrix, Be, containing the derivativesof the shape functions with respect to the global X and Y coordinates, turnsout to be

Be =

1 − ξ

3 + ξ

01 + ξ

3 + ξ

0−1 − ξ

3 + ξ

0−1 + ξ

3 + ξ

0

02(−1 + η)

3 + ξ

01 − η

3 + ξ

02 + ξ + η

3 + ξ

0−1 − ξ − 2η

3 + ξ

2(−1 + η)

3 + ξ

1 − ξ

3 + ξ

1 − η

3 + ξ

1 + ξ

3 + ξ

2 + ξ + η

3 + ξ

−1 − ξ

3 + ξ

−1 − ξ − 2η

3 + ξ

−1 + ξ

3 + ξ

Assuming unit thickness, the element matrix can be expressed as a doubleintegral in the parent domain as

Ke =

∫

1

−1

∫

1

−1

BT

eCeBej dξdη (6.126)

It turns out that the exact value of the integral on the right-hand side ofequation (6.126) can be obtained using a 2×2 quadrature rule. Starting withthe first quadrature point, the technique involves evaluating the integrandat the quadrature point and then multiplying the result by a weight factor.The sum of the contributions from each quadrature point is the value of theintegral — an 8 × 8 matrix in this case. Just to keep things as simple aspossible, let’s take Young’s modulus to be E = 1 and ν = 0. The plane stresselasticity matrix is then

Ce =

E 0 00 E 00 0 E/2

(6.127)

Evaluating the matrix product BT

eCeBe j at the quadrature points (ξa ±

1/√

3, ηa = ±1/√

3) yields the following contributions to the element matrix:

Ke1 =

0.385121 −0.128374 −0.093976 −0.034398 −0.103193 0.034398 −0.187952 0.128374

−0.128374 0.577681 0.064187 −0.239548 0.034398 −0.154789 0.029789 −0.183344

−0.093976 0.064187 0.073404 0.017199 0.025181 −0.017199 −0.004608 −0.064187

−0.034398 −0.239548 0.017199 0.132982 0.009217 0.064187 0.007982 0.042380

−0.103193 0.034398 0.025181 0.009217 0.027650 −0.009217 0.050362 −0.034398

0.034398 −0.154789 −0.017199 0.064187 −0.009217 0.041476 −0.007982 0.049127

−0.187952 0.029789 −0.004608 0.007982 0.050362 −0.007982 0.142199 −0.029789

0.128374 −0.183344 −0.064187 0.042380 −0.034398 0.049127 −0.029789 0.091837

Ke2 =

0.180116 −0.023295 −0.063642 −0.086937 −0.133526 0.086937 0.017053 0.023295

−0.023295 0.350869 0.011647 −0.162227 0.014768 −0.232111 −0.003121 0.043469

−0.063642 0.011647 0.130406 0.043469 −0.031821 −0.043469 −0.034942 −0.011647

−0.086937 −0.162227 0.043469 0.130406 0.055116 0.066763 −0.011647 −0.034942

−0.133526 0.014768 −0.031821 0.055116 0.156821 −0.055116 0.008526 −0.014768

0.086937 −0.232111 −0.043469 0.066763 −0.055116 0.183237 0.011647 −0.017889

0.017053 −0.003121 −0.034942 −0.011647 0.008526 0.011647 0.009363 0.003121

0.023295 0.043469 −0.011647 −0.034942 −0.014768 −0.017889 0.003121 0.009363

Ke3 =

0.018725 −0.006242 0.017053 −0.023295 −0.069884 0.023295 0.034106 0.006242

−0.006242 0.028088 0.003121 −0.000836 0.023295 −0.104826 −0.020174 0.077574

0.017053 0.003121 0.090058 0.011647 −0.063642 −0.011647 −0.043469 −0.003121

−0.023295 −0.000836 0.011647 0.049710 0.086937 0.003121 −0.075290 −0.051995

−0.069884 0.023295 −0.063642 0.086937 0.260811 −0.086937 −0.127285 −0.023295

0.023295 −0.104826 −0.011647 0.003121 −0.086937 0.391217 0.075290 −0.289511

0.034106 −0.020174 −0.043469 −0.075290 −0.127285 0.075290 0.136647 0.020174

0.006242 0.077574 −0.003121 −0.051995 −0.023295 −0.289511 0.020174 0.263932

Ke4 =

0.146807 −0.034398 0.025181 −0.009217 −0.078012 0.009217 −0.093976 0.034398

−0.034398 0.101054 0.017199 −0.001235 0.081386 −0.104428 −0.064187 0.004608

0.025181 0.017199 0.013825 0.004608 0.012590 −0.004608 −0.051596 −0.017199

−0.009217 −0.001235 0.004608 0.013825 0.021807 0.039006 −0.017199 −0.051596

−0.078012 0.081386 0.012590 0.021807 0.112410 −0.021807 −0.046988 −0.081386

0.009217 −0.104428 −0.004608 0.039006 −0.021807 0.210994 0.017199 −0.145572

−0.093976 −0.064187 −0.051596 −0.017199 −0.046988 0.017199 0.192560 0.064187

0.034398 0.004608 −0.017199 −0.051596 −0.081386 −0.145572 0.064187 0.192560


TABLE 6.10

Flow chart for computing Ke.

1. Zero the element matrix Ke

2. Set up the elasticity matrix Ce

3. Set up the Gauss point locations and weights4. Loop over the quadrature points, a = 1 to nint

5. Compute Kea = BT

e(ξa, ηa)CeBe(ξa, ηa) j(ξa, ηa)Wa

6. Ke = Ke + Kea

7. Continue looping over quadrature points

Finally, adding things together yields the total contribution to the elementmatrix, i.e.,

Ke = Ke1 + Ke2 + Ke3 + Ke4.

The result is

Ke =

0.730769 −0.192308 −0.115385 −0.153846 −0.384615 0.153846 −0.230769 0.192308

−0.192308 1.057692 0.096154 −0.403846 0.153846 −0.596154 −0.057692 −0.057692

−0.115385 0.096154 0.307692 0.076923 −0.057692 −0.076923 −0.134615 −0.096154

−0.153846 −0.403846 0.076923 0.326923 0.173077 0.173077 −0.096154 −0.096154

−0.384615 0.153846 −0.057692 0.173077 0.557692 −0.173077 −0.115385 −0.153846

0.153846 −0.596154 −0.076923 0.173077 −0.173077 0.826923 0.096154 −0.403846

−0.230769 −0.057692 −0.134615 −0.096154 −0.115385 0.096154 0.480769 0.057692

0.192308 −0.057692 −0.096154 −0.096154 −0.153846 −0.403846 0.057692 0.557692

The flow chart outlining the procedure for the element matrix computationusing Gauss quadrature is given in Table 6.10.

6.7 Bending of a cantilever beam

To apply some of the concepts introduced in this chapter, let us now considerthe problem of a cantilever beam loaded at the right end by a concentratedforce, P = 10 kN, shown in Figure 6.17. The length of the beam is 3m,and it has a rectangular, 20mm × 200mm, cross section. We will assumethat the beam material is linearly elastic and isotropic with Young’s modulusE = 200GPa. For the moment, let’s take Poisson’s ratio to be ν = 0.3.

To begin, we will assume plane stress, and construct a finite element meshusing rectangular four-node elements. From our previous discussion in Sec-tion 6.2, the displacement interpolation within an element then has the form

uX

uY

e

=

α1 + α2X + α3Y + α4XY

β1 + β2X + β3Y + β4XY

(6.128)

where α1, . . . , β4 are constants. The strain components within the element areobtained by taking the partial derivatives of the displacement components.The extensional (bending) strain in the element has the form

ǫXX = α2 + α4Y (6.129)


P = 10 kN

3m

FIGURE 6.17

Cantilever beam.

From elementary mechanics of materials theory, the deflection curve for acantilever beam loaded at the right end by a concentrated force, P , is givenas

uY =PL3

6EI

[

3

(

X

L

)2

−

(

X

L

)3]

(6.130)

where L is the length of the beam, E is Young’s modulus, and I is the momentof inertia of the beam’s cross section about the neutral axis. Notice that thedeflection curve from mechanics of materials theory is a cubic polynomial inX. The second line in (6.128) indicates that the finite element approximationfor the vertical displacement within an element varies linearly with X. Thissuggests that a relatively large number of elements will be required in theX direction in order to achieve an accurate solution for the deflection of thebeam.

The normal stress distribution, σXX(Y ), acting on some cross section lo-cated a distance X from the left end is also available from mechanics of ma-terials theory. The elementary expression can be written as

σXX =−M(X)Y

I(6.131)

where M(X) is the net bending moment acting on the cross section, and Y isthe distance from the beam’s neutral surface (surface on which the extensionalstrain is zero). Because the normal stress is proportional to Y and the materialis linearly elastic, the extensional strain must also vary linearly with Y . Thefact that the finite element approximation for the extensional strain varieslinearly with Y suggests that only a few elements are needed in the Y direction.

In the present analysis, to gain experience with how the solution changesas the mesh is refined, we will consider the four different meshes shown inFigure 6.18. The finite element solution for the end deflection obtained foreach mesh is reported in Table 6.11. The elementary mechanics of materialsprediction for the end deflection can be obtained by substituting X = L into


(a) (b)

(c) (d)

FIGURE 6.18

2 × 5 mesh (a), 2 × 20 mesh (b), 2 × 40 mesh (c), and 4 × 40 mesh (d).

equation (6.130) above. Doing this gives

uY =PL3

3EI

=(10 kN)(3m)3

3(200 × 109 N/m2)

[

(0.02m)(0.2m)3

12

]

= 33.75mm (6.132)

Comparing the elementary solution with the deflection values reported in thetable, the percent difference between the finite element solution for case (c)and the elementary solution is approximately 6%. Hence, as we expected, afine mesh is required to obtain good accuracy.

The elementary mechanics of materials solution (6.131) for the bendingstress in the beam is valid at cross sections that are sufficiently far fromclamped ends and concentrated loads. Hence, for comparison purposes, thefinite element solution for the bending stress is compared to the elementarysolution at a cross section located 1.5m from the left end. The result for the2 × 40 mesh is plotted in Figure 6.19. Here in the figure, the bending stress,σXX , is plotted versus distance, Y , from the neutral surface. The solid curverepresents the solution from elementary theory, and the circular data pointsrepresent the finite element solution. The stress was output at five equally-spaced points located along the right edge of the elements contiguous to theline X = 1.5m. Excellent agreement is obtained between the finite element

TABLE 6.11

End deflection of beam with increasingmesh density.

Mesh density uY (mm) Percent error

2 × 5 7.5259 77.72 × 20 27.2109 19.42 × 40 31.3006 7.34 × 40 31.8934 5.5


Y (m)

σX

X(M

Pa)

-0.1 -0.05 0 0.05 0.1-150

-100

-50

0

50

100

150

FIGURE 6.19

Bending stress σXX along the line X = 1.5m for the 2 × 40 mesh.

solution and elementary theory, even with only two elements through thethickness of the beam. For the sake of interest, a contour plot of the bendingstress, superposed on top of the beam’s (magnified) deformed geometry, isshown in Figure 6.20.

Employing the 4 × 40 mesh, let us now run the problem again for twodifferent values of Poisson’s ratio, ν = 0.3 and ν = 0.499. In the linearizedtheory of elasticity discussed in Chapter 4, the Lame’s constant, λ, approachesinfinity as ν approaches 1/2. In the three-dimensional version of Hooke’s law,λ is the coefficient in front of ǫkk, the trace of the small-strain tensor. Thetrace of the small-strain tensor physically represents the volumetric straindefined as the change in volume of an infinitesimal element divided by theoriginal volume. As Poisson’s ratio approaches 1/2, the material’s resistanceto volumetric deformation goes to infinity. Under such conditions, the materialis said to be incompressible. An incompressible material can deform in a shear

P = 10 kN

FIGURE 6.20

Contour plot of the bending stress σXX for the 4 × 40 refined mesh.


TABLE 6.12

End deflection of beam for case of plane strain withν = 0.3 and ν = 0.499.

Poisson’s ratio(quadrature rule)

ν = 0.3(2 × 2)

ν = 0.499(2 × 2)

ν = 0.499(1 × 1)

End deflection (mm) 28.83 2.76 22.92

mode, but the volume remains constant during deformation.

The plane-strain results for the two values of Poisson’s ratio (ν = 0.3 andν = 0.499) are reported in Table 6.12. Also reported in the table is the enddeflection for the case when ν = 0.499, but only a 1 × 1 quadrature rulewas used to integrate the stiffness matrix. For the fully integrated (2 × 2quadrature) element, when ν = 0.499, the end deflection is extremely smallcompared to the case when ν = 0.3. This behavior is referred to as locking.For the case of plane strain when Poisson’s ratio is nearly 1/2, locking willoccur no matter how much the mesh is refined! The same locking behaviorcan occur when using the three-node triangle under conditions of plane strain,and also for the eight-node brick under nearly incompressible conditions. Wenote that locking does not occur in the present problem when ν = 0.499 underthe condition of plane stress.

When the element stiffness matrix is integrated using a 1 × 1 quadraturerule, the locking behavior is no longer present. The technique of underinte-grating the stiffness matrix is called reduced integration. It is widely used tohelp eliminate the problem of locking in the analysis of nearly incompressiblematerials. The numerical problem of locking has been addressed by manyresearchers and can be remedied in a variety of ways. The locking of the arbi-trary four-node quadrilateral can be taken care of under many circumstancesby simply employing reduced integration. Unfortunately, under certain con-ditions, underintegrated elements (the four-node quad being one of them)possess what are called spurious zero-energy modes. Spurious zero-energymodes are deformed configurations of the element, resembling the shape of anhourglass, in which the total energy in the element is zero. For static prob-lems, this can result in a singular global stiffness matrix. Elements can bemade more robust by using the technique of reduced integration along withwhat is called hourglass control or hourglass stabilization. An element withhourglass control has additional stiffness terms included to prevent it fromgoing into such zero-energy modes. For further discussion on this topic seeFlanagan and Belytschko [10] and Belytschko et al. [11].


6.8 Analysis of a plate with hole

Let us now consider the problem of a 2m× 2m plate with a centrally locatedhole as shown in Figure 6.21. The top and bottom edges of the plate aresubjected to a uniform traction distribution of magnitude σ0 acting in theY direction. The radius of the hole is r = 0.1m. In the present study, wewill begin by assuming that the thickness of the plate is small compared tothe width of the plate so that the condition of plane stress prevails. Unlessotherwise stated, we will assume E = 200GPa and ν = 0.3. Before weproceed, it is necessary to discuss the boundary conditions associated with theproblem. Because both the geometry and the loading are symmetric aboutthe X and Y axes, it is only necessary to model one-quarter of the plate (saythe upper-right quadrant). We still however need to be careful to apply thecorrect prescribed displacement boundary conditions on the left and bottomedges of the quarter model.

To give a mathematical explanation of the boundary conditions, it is nec-essary to discuss the concept of odd and even functions. A function f(X) issaid to be even if f(X) = f(−X). On the other hand, the function f(X) isdefined to be odd if f(X) = −f(−X). Having defined odd and even func-tions, let us now consider a point in the upper-right quadrant of the plate. Ifthis point displaces in the positive X direction, then, due to symmetry, themirror image of that point in the lower-right quadrant must also move thesame amount in the X direction. Hence, we conclude that the displacementcomponent uX is an even function of Y , i.e., uX(X,Y ) = uX(X,−Y ). If the

σ0

2m

2mX

Y

FIGURE 6.21

Plate with a hole.


(a) (b)

FIGURE 6.22

Coarse mesh (a) and refined mesh (b).

same point in the upper-right quadrant moves in the positive Y direction,then the mirror image point in the lower-right quadrant must displace thesame distance in the −Y direction. Hence uY is an odd function in Y , i.e.,uY (X,Y ) = −uY (X,−Y ). Because uY is odd in Y , we conclude that thisdisplacement component must be zero along the line Y = 0. Following thesame argument, we conclude that the displacement component uX is an oddfunction of X, i.e., uX(X,Y ) = −uX(−X,Y ). Therefore uX must be zeroalong the line X = 0. These symmetry boundary conditions are enforced inpractice by setting the Y degree of freedom equal to zero at nodes that lie onthe X axis and setting the X degree of freedom equal to zero at nodes thatlie on the Y axis.

Having discussed the symmetry boundary conditions to be applied along theleft and bottom edges of the plate, let us now begin our analysis by employingthe two meshes composed of arbitrary four-node (plane stress) quadrilateralsshown in Figure 6.22. The coarse mesh (a) is composed of 70 elements, andthe refined mesh (b) has 200 elements. To enforce the boundary conditions, wesimply constrain the nodes lying on the X axis from moving in the Y directionand the nodes on the Y axis from moving in the X direction. The uniformtraction distribution is applied along the top edge by identifying the elementedges that lie on this edge. By identifying an element edge and specifyingthe magnitude of the traction acting on that edge, the element vector can becalculated. Recall that the element vector is defined as

fe = t

∫

Γte

NT

et∗ dΓ (6.133)

where Γteis the edge along which the traction distribution is applied. In

commercial finite element codes, the value of this integral is stored in a li-brary, and the element vector is set up inside the element subroutine. The


(a) (b)FIGURE 6.23

Sparse structure of the global stiffness matrix (a) and banded strucutre (b).

element vector is then assembled into the global force vector following thefinite element assembly process.

After the boundary conditions are defined, all that is left to do is to definethe material properties for each element and run the analysis. Before runningthe analysis however, it is a good idea for computational efficiency to check thenode numbering of the mesh. When using a direct solver, based on Gaussianelimination, the manner in which the mesh is numbered can make a hugeimpact on computer run time. To illustrate this idea, let us focus our attentionon Figure 6.23. The figure shows the structure of the global stiffness matrix forthe coarse mesh before renumbering (a), and the same mesh after renumberingusing the Cuthill-McKee algorithm [12]. The structure of the global stiffnessmatrix is revealed by the positions in the matrix occupied by nonzero numbers.The matrix (a) to the left is said to be sparse, because it has many nonzerovalues occupying positions far from the diagonal. On the other hand, thematrix (b) on the right is called banded, because most of the nonzero valuesare clustered about the diagonal. The banded matrix requires significantlyless computer storage and, hence, a shorter run time.

The number of columns away from the diagonal (either to the right or tothe left) occupied by nonzero numbers is called the half bandwidth b. Thehalf bandwidth can be calculated by the following formula:

b = ndf(m + 1) (6.134)

where ndf is the number of degrees of freedom per node, and the parameterm is the maximum difference between node numbers within an element. Toget m, one needs to look at the connectivity list for each element and recordthe maximum difference between node numbers for each element. The global


σY

Y/σ

0

σY

Y/σ

0

X (m) X (m)0 0.5 10 0.5 1

0

0.5

1

1.5

2

2.5

3

0

0.5

1

1.5

2

2.5

3

FIGURE 6.24

Stress distribution σY Y /σ0 plotted versus distance X along the line Y = 0.

maximum can then be extracted by comparing the element maximum values.Using a direct solver, the computer run time in seconds can be estimatedusing the following formula:

time ≈nb2

2α(6.135)

where n is the total number of equations in the model, and α is a parameterthat defines the speed of the computer, e.g., in million floating-point opera-tions per second (MFLOPS). The MFLOP rating gives an indication of thenumber of multiplications that can be performed per second. Modern-dayPC’s are capable of achieving between 2000 and 6000 MFLOPS.

As an example, suppose that a two-dimensional finite element mesh is com-posed of a rectangular 500×500 node grid. The largest (worst) possible valueof m would be 500 × 500 − 1 = 249999. Assuming a MFLOP rating of 2000,the computer run time for the worst-case scenario would be approximately 90days! On the other hand, numbering sequentially, row by row, would give ahalf bandwidth of 502 − 1 = 501. The run time for that case would be onlyabout 30 seconds.

A contour plot of the stress distribution σY Y /σ0 is superposed on top ofthe plate geometry for the case of the course mesh back in Figure 6.21. Thestress concentration factor, σmax

Y Y/σ0, near the hole obtained with the course

mesh is approximately 2.4. For the fine mesh, the stress concentration factorturns out to be nearly 3.1. The analytical solution for the stress concentrationfactor has been reported by Howland [13] to be 3.14. Hence the fine mesh


(a) (b)

FIGURE 6.25

Contour plot of σY Y /σ0. Plane strain locked mesh with ν = 0.499 (a) andplane stress mesh with ν = 0.3.

gives excellent results.

Exploring further, Figure 6.25 shows an X−Y plot of σY Y /σ0 versus X

along the line Y = 0. Five output points per element were chosen to makethe curves. The results for the coarse mesh are shown to the left (a) andthe fine-mesh results are shown to the right (b). For the coarse mesh thereare six elements having an edge lying on the line Y = 0. Notice that thestress distribution along Y = 0 is piecewise continuous, but there are jumpsat the nodal points. The presence of locations where the jump in stress islarge suggests that mesh refinement is needed in these regions. The curveobtained for the fine mesh is significantly improved, but a jump in the stressis still present at the location near the hole. At this location, a single nodelies on the bottom edge, just adjacent to the element contiguous to the holeboundary.

To conclude this study, just out of curiosity, let’s switch from plane stressto plane strain and run the problem again with the refined mesh setting Pois-son’s ratio equal to ν = 0.499. Recall that as ν approaches 1/2, a linearlyelastic and isotropic material becomes incompressible. It can deform in ashear mode, but not volumetrically. A contour plot of σY Y /σ0 is shown inFigure 6.25(a). The plot exhibits a nonsensical “checkerboard” pattern andexhibits unreasonable values for the stress components. These symptoms in-dicate that locking has occurred. As discussed in the previous analysis of acantilever beam, locking can be avoided for the plane-strain, nearly incom-pressible case by underintegrating the element, i.e., using 1 × 1 quadrature,and employing hourglass stabilization. The contour plot in Figure 6.25(b) is


z

FIGURE 6.26

Representative section of a copper core surrounded by a ceramic coating.

for the plane stress case with ν = 0.3. This plot shows the correct nature ofthe stress distribution.

6.9 Thermal stress analysis of a composite cylinder

To end this chapter, due to the practical importance of thermal stress prob-lems, we perform a thermal stress analysis of a long composite cylinder com-posed of an inner copper core and a ceramic coating. A representative sectionof the composite cylinder is shown in Figure 6.26. The diameter of the innercore is 0.1m, and the thickness of the outer ceramic coating is also 0.1m. Inthe analysis we will assume that the inner copper core is heated uniformly toa temperature of 200 C, and air at 20 C flows over the outer surface of thecomposite, causing heat transfer by forced convection. In the analysis, we willtake the film coefficient to be h = 10 J/(m2 ·C · s), and we will assume thatthe inner copper core and the outer ceramic cylinder are perfectly bonded atthe interface.

After steady-state conditions are reached, we desire to determine the ther-mal stress distribution in the composite. We will assume that the ultimatetensile strength of the ceramic coating is σu = 150MPa. During the post-processing phase of the analysis, we will address the question of whetheror not we expect the coating to fail under the present circumstances. Themechanical properties for the copper and ceramic used in the analysis arereported in Table 6.13 below.

6.9.1 Heat conduction analysis

The first step in the analysis is to determine the steady-state temperaturedistribution in the composite. Because both the geometry of the compositeand the prescribed temperature distribution possess axial symmetry aboutthe z axis, it is only necessary to perform a two-dimensional, axisymmetric


TABLE 6.13

Mechanical properties for copper and ceramic.

Material E (GPa) ν α ( 10−6

C) k ( J

m·C·s)

Copper 110 0.33 17 400

Ceramic 400 0.22 5 1

heat conduction analysis. In addition, because the prescribed temperaturedistribution does not vary in the z direction and because we have assumedthat the composite cylinder is “long,” the solution for the temperature fieldshould not depend on z, except in the near vicinity of the ends. Therefore, itis only necessary to consider the quarter model of the representative sectionas shown in Figure 6.27.

To explain the boundary conditions applied to the quarter model, a tem-perature of T = 200 C is specified at all points inside the copper cylinder. Toprevent heat from flowing in the z direction, we specify a zero-flux condition,i.e., qz = 0, on the top and bottom of the ceramic cylinder as shown in thefigure. The forced convection condition, T = h(T − T∞), is applied along theright edge.

Finite element implementation

In the absence of an internal heat source, the weak form of the steady-stateheat equation can be written over a single element as follows:

k

∫

Ωe

∇w · ∇T dΩ = −

∫

Γe

wq∗n

dΓ (6.136)

r

z

ceramic

qz = 0

qz = 0

T=

200C

T = h (T − T∞)

FIGURE 6.27

Quarter model and associated boundary conditions for the heat conductionanalysis.


where ∇w and ∇T represent the gradient of the test function and temperaturefield, respectively, and q∗

nis the specified normal component of the heat flux

acting on Γe. The gradient operator in cylindrical coordinates can be writtenas

∇ =∂

∂rer +

1

r

∂

∂θeθ +

∂

∂zez (6.137)

where er, eθ, and ez are unit base vectors. In axisymmetric heat conductionproblems, it is assumed that the temperature does not vary in the circumfer-ential direction. The partial derivative of the temperature field with respectto θ is then zero, and the components of the temperature gradient can bewritten as

∇T =

∂T

∂r

∂T

∂z

(6.138)

Let us now proceed and employ a four-node arbitrary quadrilateral ele-ment for heat conduction analysis. The temperature field over the four-nodequadrilateral is interpolated using the bi-linear shape functions defined inSection 6.2.3, equation (6.29). The interpolation can be written as

Te = Te(r, z) = NeTe (6.139)

where Ne is a 1 × 4 matrix containing the bi-linear shape functions, andTe is a 4 × 1 vector containing the nodal temperatures. The finite elementapproximation for the temperature gradient is then

∇T =

∂T

∂r

∂T

∂z

= BeTe (6.140)

where Be is the 2 × 4 B-matrix containing partial derivatives of the shapefunctions with respect to the global r and z directions, i.e.,

Be

2 × 4=

[

N1,r N2,r N3,r N4,r

N1,z N2,z N3,z N4,z

]

(6.141)

Using the Galerkin procedure, we now interpolate the test function over theelement using the same shape functions that were employed for the tempera-ture field. Doing this yields

we = Newe

∇we = Bewe (6.142)

where we is a 4 × 1 vector containing the values of the test function at thenodes. By substituting the finite element approximations (6.140) and (6.142)


into the weak form (6.136) we get

wT

e

k

∫

Ωe

BT

eBe dΩ

Te = −wT

e

∫

Γe

NT

eq∗ndΓ

(6.143)

Equation (6.143) above can be written succinctly as

wT

eKeTe = −wT

efe (6.144)

where sandwiched between wT

eand Te on the left-hand side is the 4 × 4

element matrix,Ke

4 × 4= k

∫

Ωe

BT

eBe dΩ (6.145)

The element matrix for heat conduction applications is referred to as theconductance matrix. The element vector appears on the right-hand side ofequation (6.144) and is given as

fe = −

∫

Γe

NT

eq∗n

dΓ (6.146)

In the present problem, the normal component of the heat flux vector, q∗n,

was specified to be zero on the top and bottom edges of the ceramic region,and hence, for elements having an edge lying on one of these boundaries, theelement vector is zero.

For elements with an edge lying on the line r = 0.15, the right boundary,the forced-convection condition needs to be enforced. On a forced-convectionboundary, the normal component of the flux is given as

q∗n

= h (T − T∞) (6.147)

where h is the film coefficient, T is the unknown temperature at a pointon the boundary, and T∞ is the temperature of the surrounding fluid faraway from the boundary. Recall that in the present problem we have chosenthe remote temperature to be T∞ = 20 C. Substituting equations (6.145)through (6.147) into equation (6.144) and invoking the arbitrariness of thetest function yields

k

∫

Ωe

BT

eBe dΩ + h

∫

Γe

NT

eNedΓ

Te = hT∞

∫

Ωe

NT

edΓ (6.148)

After assembling the element matrices and vectors following the usual finiteelement assembly procedure, the global system of equations to be solved canbe written simply as

KT = F (6.149)

We see from equation (6.148) that the total contribution to the elementmatrix is composed of the conductance matrix, plus a contribution from the


elements that are contiguous to the forced-convection boundary. Only theelements that touch the outer surface of the ceramic get both contributionsto the element matrix.

In the present axisymmetric formulation, the element matrix and vectorare evaluated by carrying out integrals over the ξ−η parent domain. This isaccomplished by recognizing that a differential volume, dΩe, in the physicaldomain can be related to a differential volume in the parent domain as follows:

dΩe = 2πr drdz

= 2π(Nara) j dξdη (6.150)

Note that the second line in equation (6.150) follows from the mapping be-tween the parent domain and the physical domain, i.e.,

r = Na(ξ, η)ra

z = Na(ξ, η)za (6.151)

where the summation convention is implied, and ra and za are the r and z

coordinates of local node a. The 4×4 element matrix can be evaluated exactlyusing a 2 × 2 quadrature rule.

The line integrals that appear in equation (6.148) can be evaluated in theparent domain by replacing the differential surface area, dΓ, by the corre-sponding area in the parent domain. For example, if local nodes 2 and 3 of anelement lie on the forced-convection boundary, then dΓ would be expressedas

dΓ = 2πr

√

r2,η

+ z2,η

dη (6.152)

where r,η and z,η are components of the Jacobian matrix, i.e.,

J =

[

r,ξ z,ξ

r,η z,η

]

(6.153)

The shape function matrix that appears inside the surface integrals wouldthen have to be evaluated along the line ξ = 1, using a sufficient quadraturerule.

The finite element mesh employed in the heat conduction analysis is com-posed of 130 four-node axisymmetric quadrilaterals. It is shown in Fig-ure 6.28(b). The mesh was designed with the subsequent thermal stressanalysis in mind. In the thermal stress analysis, we expect a large discon-tinuity in the stress components σzz and σθθ across the interface due to themismatch in mechanical properties between the coper and ceramic. Hence,the mesh employed in the thermal stress analysis needs to be focused in thisregion. Actually, we could get away with using a much coarser mesh forthe heat conduction analysis, but it is convenient to use the same mesh forboth analyses. The nodal temperatures can then be directly transferred fromthe heat-transfer mesh to the structural mesh. The temperature distributionis needed to calculate the thermal strain distribution and resulting thermalstress distribution, as described in the next section.


ur

=0

uz = 0

t∗ = 0

(a) (b)

uz = C

zz

rr

q∗n = h (T − T∞)

q∗n = 0

q∗n = 0

FIGURE 6.28

Finite element mesh for thermal stress analysis (a), and heat conduction anal-ysis (b).

6.9.2 Thermal stress analysis

Now that the temperature distribution in the composite is known, let usproceed with the thermal stress analysis. To explain the finite element imple-mentation for linear thermoelastic problems, we begin by writing down thediscrete version of the weak form. Due to the absence of body forces andsurface tractions in the present example, the statement can be written simplyas

∫

Ωe

BT

eσ dΩ = 0 (6.154)

The next step is to substitute Hooke’s law for thermoelasticity problems givenby equation (4.87) in Chapter 4 into equation (6.154).

Hooke’s law for thermoelastic axisymmetric problems can be written as

σrr

σθθ

σzz

σrz

= E

1 − ν ν ν 0ν 1 − ν ν 0ν ν 1 − ν 0

0 0 01 − 2ν

2

ǫrr − α∆T

ǫθθ − α∆T

ǫzz − α∆T

γrz

(6.155)

where

E =E

(1 + ν) (1 − 2ν)

and α is the coefficient of thermal expansion. The quantity ∆T is the temper-ature change from a stress-free reference state. We emphasize that the strain


components on the right-hand side are the total strain components. Equation(6.155) above can be written compactly as

σ

4 × 1=

C4 × 4

[

ǫ

4 × 1−

α∆T4 × 1

]

(6.156)

As discussed in Chapter 4, Section 4.6, the strain-displacement relationshipfor axisymmetric problems can be written as

ǫrr = ur,r

ǫθθ =1

rur

ǫzz = uz,z

γrz = uz,r + ur,z (6.157)

Armed with this relationship, we can now express the strain at a point withinthe element in terms of the element B-matrix and the nodal displacementvector. Doing this yields

ǫe

4 × 1=

Be

4 × 8de

8 × 1(6.158)

The 4× 8 B-matrix can be written in terms of the derivatives of the bi-linearshape functions with respect to r and z as follows:

Be =

N1,r 0 N2,r 0 N3,r 0 N4,r 01

rN1,r 0

1

rN2,r 0

1

rN3,r 0

1

rN4,r 0

0 N1,z 0 N2,z 0 N3,z 0 N4,z

N1,z N1,r N2,z N2,r N3,z N3,r N4,z N4,r

(6.159)

Hence, the finite element approximation for the stress field within an elementcan be written as

σe = Ce [Bede − α∆Te] (6.160)

Substituting equation (6.160) into equation (6.154) and rearranging termsgives

∫

Ωe

BT

eCeBe dΩ

de = α

∫

Ωe

BT

eCe∆Te dΩ (6.161)

Upon assembly of the element matrices and vectors, the system of equationsto be solved has the usual form Kd = F.

From the right-hand side of equation (6.161) we see that the temperaturefield within the element is required in order to evaluate the element vector.Given the temperature values at the nodal points, the temperature at a pointinside the element is computed as follows:

Te(ξ, η) = Na(ξ, η)Ta (6.162)


where Ta is the temperature at node a. Both integrals in equation (6.161) canbe carried out in the parent domain as follows:

Ke = 2π

∫

1

−1

BT

eCeBe (Nara)jdξdη

fe = 2πα

∫

1

−1

BT

eCe∆Te (Nara)jdξdη (6.163)

where

∆T

NaTa − Tref

NaTa − Tref

NaTa − Tref

0

(6.164)

is the vector containing the difference between the temperature at the quadra-ture point and the stress-free reference temperature Tref . Note that here inthe present axisymmetric analysis we have employed the relation

dΩe = 2πrj dξdη

= 2π(Nara)j dξdη (6.165)

Mechanical boundary conditions

Before presenting the results, it is instructive to discuss the mechanical bound-ary conditions to be applied to the representative section, as shown in Fig-ure 6.28(a). As shown in the figure, the displacement component, uz, isconstrained to be zero along the line z = 0. The plane z = 0 can be consid-ered a plane of symmetry in the present problem, and hence the displacementin the z direction must be zero at all nodes lying on this plane. Due to thefact that the problem is axisymmetric, the nodes lying on the line r = 0 areconstrained from moving in the r direction.

Because we have assumed that the cylinder is “long,” the deformation ineach representative composite section making up the cylinder must deform inan identical manner. The resulting deformation must also be compatible. Forboth of these conditions to be met, all points lying on the top and bottomsurfaces of each section must displace the same amount in the z direction.

The boundary condition, uz = C, along the top edge where C is an un-known constant, is called a periodic boundary condition. If this conditionwere not enforced, the deformation between adjacent sections would end upbeing incompatible, such as that depicted in Figure 6.29.

To explain how periodic boundary conditions can be implemented in a finiteelement code, suppose that the three elements shown in Figure 6.30 all havean edge lying on the top surface of the representative section. The top leftnode (5) is chosen to be the master node. Equations 9 and 10 correspond tothe r and z degrees of freedom attached to this node. Now let the remainingnodes (6,7, and 8) lying on the top edge be slave nodes. The equation numbers


FIGURE 6.29

Incompatible deformation between adjacent cells.

associated with the r and z degrees of freedom of the slave nodes are (11,12),(13,14), and (15,16), as labeled in the figure.

The next step is simply to change equation numbers 12, 14, and 16 toequation number 10. This, in effect, ties the z degree of freedom associatedwith the slave nodes to the z degree of freedom of the master node. Once theequations are renumbered, the element matrices and vectors are assembled inthe usual manner. Before the equations can be solved, however, it is necessaryto zero the right-hand side of equations 12, 14, and 16 and set the pivotlocations that are equal to zero on the left-hand side equal to one. The pivotpositions K12 12, K14 14, and K16 16 will all be zero after assembly due to thefact that we removed equations 12, 14, and 16. Next, upon solution of theequation Kd = F, the z degrees of freedom associated with the slave nodes willall be zero. These degrees of freedom must subsequently be set equal to thez degree of freedom of the master node. After this displacement update, thestrain in each element can calculated in the usual way, i.e., ǫe = Bede. Finally,the stress components are obtained from the modified version of Hooke’s law,σe = Ce(ǫe − α∆Te).

r

z

1 2 3 4

5 6 7 8

¤

£

¡

¢1

¤

£

¡

¢2

¤

£

¡

¢3

(9,10) (11,12) (13,14) (15,16)

FIGURE 6.30

Implementation of periodic boundary conditions.


r (m)

(a) (b)

σθθ(M

Pa)

r (m)T

(C

)0 0.05 0.1 0.150 0.05 0.1 0.15

80

100

120

140

160

180

200

-600

-400

-200

0

200

400

FIGURE 6.31

Hoop stress distribution (a), and temperature distribution (b).

6.9.3 Results

The results for the thermal stress analysis are shown in Figure 6.31(a) andFigure 6.32. The hoop stress, σθθ, is plotted versus the radial distance, r,in Figure 6.31(a). As shown in the figure, the hoop stress is constant andcompressive within the inner copper region. There exists a sharp discontinuityin the hoop stress across the interface, where it suddenly attains a high tensilevalue of approximately 325MPa and then gradually decreases as r approachesthe outer boundary. Because the hoop stress exceeds the ultimate tensile stressof 150MPa, we would expect the ceramic coating to fail. The failure wouldlikely initiate in the form of radial cracks starting at the interface and thenpropagating outward until catastrophic failure.

(a) (b)

FIGURE 6.32

Contour plot of the hoop stress in the ceramic coating (a), and deformedgeometry when the periodic boundary condition along the top edge is notenforced (b).


The deformed geometry of the composite (magnified 100×) is shown inFigure 6.32(a). The dashed lines represent the geometry of the undeformedcomposite. Just for the sake of interest, the deformed mesh without enforcingthe periodic boundary condition is shown in Fig 6.32(b).

6.10 Problems

Problem 6.1

The nodal coordinates for a three-node triangle for steady-state heat conduc-tion are given in Table 6.14 below. A uniform heat source, S = 30 J/(m3 · s),acts over the area of the element, and a uniform heat flux distribution,qn = 10 J/(m2 · s), acts over edge 2–3. Assume that the thermal conductivityis k = 10 J/(m ·C · s), and consider a unit thickness.

(a) Derive the 3 × 3 element conductance matrix.

(b) Derive the element vector.

TABLE 6.14

Problem 6.1.Nodal coordinates.

Node X Y

1 0.0 0.0

2 1.0 0.0

3 0.5 0.5

Problem 6.2

Consider a four-node isoparametric quadrilateral for plane-stress elasticity ap-plications. The nodal coordinates, connectivity list, and nodal displacementsare reported in Tables 6.15–6.17 below. Determine the stress componentsσXX , σY Y , and σXY at the location (ξ = 0, η = 0). Take E = 30 × 106 psiand ν = 0.3. Assume unit thickness.


TABLE 6.15

Problem 6.2.Nodal coordinatesin inches.

Node X Y

20 1.5 1.5

22 0.0 1.0

26 0.0 0.027 1.0 0.0

TABLE 6.16

Problem 6.2. Connectivitylist.

Element 1 2 3 4

1 26 27 20 22

TABLE 6.17

Problem 6.2. Nodaldisplacements ininches.

node daX daY

20 0.00 0.00

22 −0.01 0.01

26 0.00 0.0027 0.00 0.00

Problem 6.3

The nodal coordinates for a constant-strain triangle are given in Table 6.18below. From a steady-state heat conduction analysis, the nodal temperaturesare found to be T1 = 50 C, T2 = 70 C, and T3 = 45 C. If the stress-freereference temperature of the element is 20 C, determine the 6× 1 equivalentnodal load vector. Assume plane strain, and take E = 200GPa, ν = 0.3, andα = 17 × 10−6/C.


TABLE 6.18

Problem 6.3.Nodal coordinatesin meters.

Node X Y

1 0.0 0.0

2 1.0 0.0

3 0.5 0.5

1 2

3 4

56

78

(0,0) (1,0)

(0,1) (1,1)

(0.2,0.3)(0.7,0.25)

(0.8,0.7)(0.25,0.8)

FIGURE 6.33

Problem 6.4. Patch test.

Problem 6.4

Consider a mesh composed of five arbitrary quadrilateral elements for planeelasticity applications as shown in Figure 6.33. Suppose that the displacementcomponents uX and uY are prescribed at nodes 1 through 4 according to thefollowing equation:

uX = α1 + α2X + α3Y

uY = β1 + β2X + β3Y (6.166)

where α1, . . . , β3 are coefficients defined in Table 6.19.

TABLE 6.19

Problem 6.4. Coefficients.

α1 α2 α3 β1 β2 β3

0.0 0.002 0.005 0.0 0.003 0.001

Using a finite element program, run the problem for the case of plane straintaking E and ν to be any values that you like. Observation of the outputshould reveal that the displacement components at nodes 5 through 6 should


obey the linear displacement field defined in equation (6.166), i.e.,

d5X = (0.0) + (.002)(0.2) + (0.005)(0.3) = 0.0019

d5Y = (0.0) + (.003)(0.2) + (0.001)(0.3) = 0.0009

etc.

In addition, the strain components at each quadrature point within every

element should be those obtained differentiating the displacement componentsin equation (6.166) with respect to X and Y , i.e.,

ǫXX = uX,X = α2 = 0.002

ǫY Y = uY,Y = β3 = 0.001

γ12 = uX,Y + uY,X = α3 + β2 = 0.008

The finite element analysis carried out above is called a patch test. Becausethe arbitrary quadrilateral is capable of capturing a linear field, passing ofthe test requires that the displacements at nodes 5 through 8, as well as theresulting strains at the quadrature points, be correct up to machine precision.

Problem 6.5

The shape functions for the three-node triangle can be expressed in terms ofthe areas A1, A2, and A3 depicted in Figure 6.1. Compute the area coordinateA3 as follows:

A3 =1

2

∣

∣

∣X(1)

× X

∣

∣

∣

whereX(1) = (X2 − X1)e1 + (Y2 − Y1)e2,

andX = (X − X1)e1 + (Y − Y1)e2

Show thatN3 = A3/A

where N3 is the shape function defined in equation (6.6).

Problem 6.6

A uniform pressure of magnitude P is applied to one of the faces of an eight-node brick element. The coordinates of the local nodes are given in Table 6.20.The element faces are defined such that the first face lies on the plane ξ = −1in the parent domain. The face definitions are given as

Face 1: ξ = −1Face 2: ξ = +1Face 3: η = −1Face 4: η = +1Face 5: ζ = −1Face 6: ζ = +1


TABLE 6.20

Problem 6.6. Local nodalcoordinates for eight-nodebrick.

Local node X Y Z

1 0.0 0.0 0.0

2 1.0 0.0 0.0

3 1.5 1.0 0.0

4 0.0 1.0 0.0

5 0.0 0.0 0.5

6 1.0 0.0 0.5

7 1.2 0.0 0.5

8 0.0 1.0 0.5

Determine the 24×1 element vector containing the equivalent nodal loads forthe case when the pressure acts over the second face, i.e., where ξ = 1 in theparent domain.

Problem 6.7

A three-node triangle is shown in Figure 6.34. The local node numberingand nodal coordinates (in inches) are labeled in the figure. A concentratedforce of magnitude P = 1000 lb is applied at local node 2. Determine thedisplacement components d2X and d2Y if d1X = d1Y = d3X = d3Y = 0.Assume plane strain and take E = 30 × 106 psi and ν = 0.3.

1 2

3

(0,0) (1,0)

(0,1)

X

Y

P

FIGURE 6.34

Problem 6.7. Three-node triangle subjected to concentrated force.


Problem 6.8

Compute the exact values of the following integrals using Gauss quadrature:

(a)∫

3

0(x2 + x) dx

(b)∫

1

−1(x3 + x5) dx

Problem 6.9

A one-dimensional, isoparametric quadratic-u element is shown in the parentdomain in Figure 6.35. The shape functions in terms of the natural coordinate,ξ, are given as

N1 =1

2ξ (ξ − 1)

N2 = 1 − ξ2

N3 =1

2ξ (ξ + 1)

The nodal coordinates of the element in the physical domain are labeled inthe figure. If the nodal displacements are given as d1 = 0, d2 = 1, and d3 = 2,plot the strain distribution ǫ(X) along the length of the element, where X

is measured from the left end of the element. Does the result make physicalsense?

Xξ11 22 33

ξ = −1 ξ = 0 ξ = 1 X = 0X = 0.25

X = 1

FIGURE 6.35

Problem 6.9. Quadratic-u element in parent and physical domains.

7

Structural Finite Elements

The ability to simplify means to eliminate the unnecessary so that the neces-

sary may speak.—Hans Hofmann

Most books on finite element analysis present structural finite elements be-fore the continuum elements covered in the previous chapter. In the author’sopinion, it is beneficial to introduce structural finite elements after contin-uum elements. By doing so, the formulation of structural elements can beginby first writing down the three-dimensional strain-displacement relationships,and then simplifying these relationships by making kinematic assumptions.The present approach requires knowledge of isoparametric finite elements dis-cussed in the previous chapter.

Kinematic assumptions are restrictions placed on the deformation of a solidbased on sound engineering judgement. For example, in the study of axiallyloaded bars, it is assumed that the bar can only resist axial forces. Thisassumption, in turn, prompts us to assume that the member can only undergoaxial extension or compression and not, for example, bending. All derivationsof structural elements begin with such kinematic assumptions. Structuralelements are a very powerful tool. The simplifying assumptions often enableengineers to analyze very complicated structures that otherwise could not bemodeled using continuum elements due to the high computational cost.

In the present chapter, we will begin our study of structural elements byconsidering the simplest case, the space truss. We will then go on to Euler-Bernoulli beams, and then to Mindlin-Reissner plates. The chapter is pre-sented within the framework of small-strain kinematics.

7.1 Space truss

The main assumption in the development of the space truss is that the elementcan only extend or shorten in the axial direction. It is also assumed that theonly nonzero stress in the element is the normal stress acting perpendicular toany given cross section. The normal stress distribution acting on each crosssection is assumed to be constant. These assumptions prohibit the element

203


X

Y

rX

1

1

2

2

FIGURE 7.1

Space truss in parent and physical domains.

from resisting shear and bending deformation. We begin the formulationby considering a two-node space truss as shown in Figure 7.1. The parentdomain of the element is chosen to be parallel to the X axis. The length, L,and cross-sectional area, A, of the element in the parent domain correspondto the length and area of the undeformed element.

To study the kinematics of the space truss, it is necessary to study themotion of points that lie on the X axis in the parent domain. These points canbe mapped onto an arbitrary line in the physical domain using the followingone-dimensional shape functions:

N1(X) =1

2

(

1 −2X

L

)

N2(X) =1

2

(

1 +2X

L

)

(7.1)

Notice that while the parent domain is a one-dimensional domain definedin the interval (−L/2 ≤ X ≤ L/2), the physical domain is a line in three-dimensional space. Introducing the natural (dimensionless) coordinate, ξ =2X/L, the shape functions can be expressed as

N1(ξ) =1

2(1 − ξ)

N2(ξ) =1

2(1 + ξ) (7.2)

The mapping of the parent domain into the physical domain can now be

Structural Finite Elements 205

written as

X = N1X1 + N2X2

Y = N1Y1 + N2Y2

Z = N1Z1 + N2Z2 (7.3)

By writing equation (7.3) in matrix notation we get

X

Y

Z

=

N1 0 0 N2 0 00 N1 0 0 N2 00 0 N1 0 0 N2

X1

Y1

Z1

X2

Y2

Z2

or

Xe = NePe (7.4)

Here in equations (7.3) and (7.4), Xa, Ya, and Za represent the coordinatesof node a in the physical domain. These coordinates (six of them) are conve-niently stored in a 6 × 1 vector, Pe. For the sake of convenience, we definethe local coordinate, r, to be the distance along the centerline of the elementfrom local node 1 to some point Xe along the length of the element, as shownin Figure 7.1.

7.1.1 Strain-displacement relationship

Having defined the mapping between the parent and physical domains, it isnow necessary to find a relationship between the extensional strain, ǫrr, ateach point along the length of the element and the nodal displacements. Thedisplacement vector, u, at some location, r, along the length of the spacetruss is shown in Figure 7.2. The displacement vector at this location is thedifference between the position vector x in the deformed configuration minusthe position vector X of this same point in the undeformed configuration. Thedisplacement vector can be written in terms of Cartesian base vectors as

u = uX eX + uY eY + uZ eZ (7.5)

where uX , uY , and uZ are the components of u labeled in Figure 7.2. Theprojection of u onto the local r axis is obtained by taking the dot product ofu and the unit base vector er, i.e.,

ur = er · u

= muX + nuY + luZ (7.6)


XY

Z

er

uX

u uY

uZ

X1

2

FIGURE 7.2

Projection of the displacement vector u onto the local r axis.

where the quantities m, n, and l are the direction cosines between the local r

axis and the X, Y , and Z axes. In other words,

m = er · eX

n = er · eY

l = er · eZ (7.7)

We can write equation (7.6) in matrix notation as

ur = AT u (7.8)

where

AT =[

m n l]

(7.9)

is a 1 × 3 matrix containing the direction cosines, and u is a 3 × 1 vectorcontaining the displacement components uX , uY , and uZ .

As discussed in Chapter 4, the extensional strain, ǫrr, at a point along thelength of the space truss is the rate of change of ur with respect to a movementin the r direction, i.e.,

ǫrr =dur

dr(7.10)

In the present analysis of the two-node space truss, the strain-displacementrelation (7.10) gives good accuracy when the axial extension of the truss issmall enough to be within the linear region of the uniaxial stress-strain curveof the material, and when the rotation of the truss is “small.” A discussion ofthe error associated with the strain-displacement relation (7.10) as a functionof rotation angle is presented in an example at the end of this section.


To proceed further, let us now introduce a 6 × 1 matrix, pe, that containsthe nodal coordinates of the element in some deformed configuration, i.e.,

pe =

x1

y1

z1

x2

y2

z2

(7.11)

If we subtract Pe from pe, we get, de, the vector containing the nodal dis-placements. The finite element approximation for the displacement vector,ue, at some point r along the element is related to de through the shapefunction matrix as follows:

ue = Nede (7.12)

where Ne is the 3×6 shape function matrix defined in (7.1). The relationship(7.10) between the extensional strain and the displacement component ur

can now be written in terms of the shape function matrix and the nodaldisplacements. Doing this yields

ǫrr =dur

dr⇒

ǫrre=

d

dr

(

AT Nede

)

= ATdNe

drde

= Bede (7.13)

where Be is the 1 × 6 B-matrix yet to be determined.To complete the derivation of the strain-displacement relationship for the

two-node space truss element, we need to differentiate the shape functionmatrix with respect to r. The technique for doing this makes use of the chainrule. The technique can be explained clearly by introducing a position vectorXe(ξ) as shown in Figure 7.3. This vector is directed from the origin to apoint in the physical domain that occupies position ξ in the parent domain.Let us now introduce another position vector, Xe(ξ + ∆ξ), pointing from theorigin to the point in the physical domain that occupies location ξ+∆ξ in theparent domain. If we now look at the difference between these two vectorsand divide by ∆ξ, in the limit as ∆ξ goes to zero, we get

lim∆ξ→0

Xe(ξ + ∆ξ) − Xe(ξ)

∆ξ=

dXe

dξ(7.14)

Note that dXe/dξ is a vector that points in the local r direction. The mag-nitude of this vector, |dXe/dξ|, is the rate of change of r with respect to a


Xe(ξ)

Xe(ξ + ∆ξ)

ξξ + ∆ξ

1 2

FIGURE 7.3

The position vectors Xe(ξ) and Xe(ξ + ∆ξ).

movement in the ξ direction, i.e.,

∣

∣

∣

∣

dXe

dξ

∣

∣

∣

∣

=dr

dξ(7.15)

Next, invoking the chain rule, we write

dNe

dξ=

dNe

dr

dr

dξ⇒

dNe

dr=

dNe/dξ

dr/dξ(7.16)

Referring to the last line in (7.16), the numerator on the right-hand side is easyto evaluate, since the expressions for the shape functions are given explicitlyin terms of ξ in equation (7.2). The denominator, however, requires somefurther thought. Since dr/dξ = |dXe/dξ|, we can write

dXe

dξ=

dNe

dξPe

=

dN1/dξ 0 0 dN2/dξ 0 00 dN1/dξ 0 0 dN2/dξ 00 0 dN1/dξ 0 0 dN2/dξ

X1

Y1

Z1

X2

Y2

Z2


Hence,

dXe

dξ=

−1/2 0 0 1/2 0 00 −1/2 0 0 1/2 00 0 −1/2 0 0 1/2

X1

Y1

Z1

X2

Y2

Z2

=1

2

X2 − X1

Y2 − Y1

Z2 − Z1

(7.17)

The magnitude of dXe/dξ can be computed as follows:∣

∣

∣

∣

dXe

dξ

∣

∣

∣

∣

=dr

dξ

=1

2

√

(X2 − X1)2 + (Y2 − Y1)2 + (Z2 − Z1)2

=L

2(7.18)

where, again, L is the length of the undeformed element in the parent domain.Therefore, the derivative of the shape function matrix with respect to r canbe written simply as

dNe

dr=

2

L

dNe

dξ(7.19)

Referring back to equation (7.13), the extensional strain in the r direction cannow be written as

ǫrre=

2

LAT

dNe

dξde

= Bede (7.20)

where Be is the 1× 6 B-matrix for the two-node space truss. After differenti-ating the shape function matrix with respect to ξ and performing the requiredmatrix multiplications, the B-matrix can finally be expressed as

Be =2

L

[

m n l]

−1/2 0 0 1/2 0 00 −1/2 0 0 1/2 00 0 −1/2 0 0 1/2

=1

L

[

−m −n −l m n l]

(7.21)

where

m = (X2 − X1)/L

n = (Y2 − Y1)/L

l = (Z2 − Z1)/L (7.22)


7.1.2 Element matrix

Having obtained the strain-displacement relationship for the two-node spacetruss in the previous section, we can now derive the finite element matricesand vectors for the element by starting with the weak form of the equilibriumequation discussed in Chapter 4. In this section, we will follow the practice ofexpressing the weak form of the equilibrium equation over a single element.Doing this, we will see that the element matrix emerges from the internalvirtual work (the left-hand side of the weak form), and the element vectorcomes from the external virtual work (right-hand side). After deriving the el-ement matrix and vector, we will be in a position to solve problems involvingassemblages of space truss elements. The process of solving such problemsinvolves following the finite element assembly procedure covered in Chapter3, and solving systems of linear algebraic equations. Once the linear alge-braic equations are solved, the stress and strain in each truss element can becalculated as a post-processing step, once the nodal displacements are known.

To begin, the general expression for the principle of virtual work over asingle element is written as

∫

Ωe

σijwi,j dΩ =

∫

Ωe

biwi dΩ +

∫

Γe

tiwi dΓ (7.23)

Here in equation (7.23), σij are the Cauchy stress components, wi are thecomponents of a vector-valued test function, and bi are the components ofthe body force vector acting at points inside the element volume, Ωe. Thequantity ti represents the traction components acting on the outer surface,Γe. It is worth mentioning here that the quantities σij , wi, bi, and ti arefunctions of the position vector Xe pointing from the origin to a point alongthe length of the truss in the physical domain. In the following discussion, forthe sake of simplicity, we will ignore the body force.

Let us now specialize the principle of virtual work statement for the case ofthe space truss element. Due to the cylindrical geometry of the space truss,the volume integrals in equation (7.23) can be replaced by line integrals. Thiscan be done by recognizing that

dΩe = Ae dr

where Ae is the cross-sectional area of the element in the physical domain,and dr is a differential axial segment of the truss. In addition, because wehave assumed that the only nonzero stress component is the normal stresscomponent, σrr, the internal virtual work, δWint

e, can be written as

δWint

e= Ae

∫

Le/2

−Le/2

σrrwr,r dr (7.24)

where Le is the length of the element in the physical domain.


It is now convenient to express the integral in equation (7.24) as an equiva-lent integral over the parent domain through a change of variables. Using thefollowing relationships:

dr =Le

2dξ

σrr = Eeǫrr (7.25)

the integral can be written as

δWint

e= AeEe

Le

2

∫

1

−1

ǫrrwr,r dξ (7.26)

where Ee is the Young’s modulus of the truss (assumed to be constant alongthe length of the element).

We are now in a position to substitute the finite element approximations forthe extensional strain and the test function into the internal work expression(7.26). Recalling equation (7.20), the finite element approximation for theextensional strain is

ǫrre= Bede (7.27)

Similarly, if we interpolate the test function, w, along the length of the ele-ment using the same linear shape functions as were used to interpolate thedisplacement field, we can write

wr = AT w ⇒

wre= AT Newe (7.28)

and hence,

wr,re= AT

dNe

drwe

= Bewe (7.29)

Substituting the finite element approximations (7.27) and (7.29) into the in-ternal work statement (7.26) gives

δWint

e= wT

e

AeEe

Le

2

∫

1

−1

BT

eBe dξ

de (7.30)

We now recognize that the element matrix is the 4×4 matrix inside the curlybrackets in (7.30) above, i.e.,

Ke =AeEeLe

2

∫

1

−1

BT

eBe dξ (7.31)


The integral in equation (7.31) can be carried out as follows:

Ke =AeEeLe

2

1

L2e

∫

1

−1

−m

−n

−l

m

n

l

[

−m −n −l m n l]

dξ (7.32)

and hence,

Ke

4 × 4=

AeEe

Le

m2 mn ml −m2 −mn −ml

mn n2 nl −mn −n2 −nl

ml nl l2 −ml −nl −l2

−m2 −mn −ml m2 mn ml

−mn −n2 −nl mn n2 nl

−ml −nl −l2 ml nl l2

(7.33)

7.1.3 Element vector

As mentioned in the previous section, the element vector is derived from theexternal virtual work expression. In the absence of body forces, the externalvirtual work for a single element can be written as

δWext

e=

∫

Γe

tiwi dΓ (7.34)

where ti are the components of traction vector, wi are the components ofthe test function, and Γe represents the outer surface of the truss. For thecase of the truss element, we now replace the traction distribution aroundthe perimeter of the truss by the net force per unit length, q, acting at somelocation, r, along the length of the truss. If we restrict our attention to thecase where the distributed load is uniform along the length of the truss, thenthe external virtual work expression becomes

δWext

e=

∫

Le

0

qws dr (7.35)

where q is the magnitude of the distributed load, and ws is the component ofthe test function in the local s direction. As shown in Figure 7.4, the local s

axis lies in the same plane and acts in the same direction as the distributedload, and it is perpendicular to the axial r direction. The component of thetest function ws can be obtained by taking the dot product of the unit basevector es and the test function vector w, i.e.,

ws = es · w

= mmwx + nnwy + ll wz

= TT w (7.36)


s

X

Y

Z

q

FIGURE 7.4

Truss element subjected to uniform distributed load q.

Here in equation (7.36), the quantities mm, nn, and ll are the direction cosinesbetween the fixed Cartesion base vectors eX , eY , and eZ and the base vectores. In other words,

mm = es · eX

nn = es · eY

ll = es · eZ

The quantity TT in equation (7.36) is the transpose of a transformation matrixthat contains the direction cosines, i.e.,

T =

mm

nn

ll

(7.37)

Assuming that the direction and magnitude of the distributed load are bothknown, then the unit base vector es and hence the direction cosines can bereadily calculated.

The last step in the derivation of the element vector for the two-node spacetruss acted upon by a uniform distributed load is to substitute equation (7.36)into (7.35) and interpolate the test function w along the length of the element


using the linear shape functions defined in (7.2). Doing this gives

δWext

e=

∫

Le

0

qTT Newe dr

= wT

e

∫

r

0

qNT

eTdr

= wT

e

∫

1

−1

qNT

eT

Le

2dξ

(7.38)

where we is the vector containing the nodal values of the test function. Notethat in the last line of equation (7.38) above, we have performed a change ofvariables, dr = (Le/2) dξ, where Le is the length of the element. We nowidentify the element vector as the 6 × 1 vector inside the curly brackets in(7.38), i.e.,

fe6 × 1

=qLe

2

∫

1

−1

NT

e

6 × 3T

3 × 1dξ (7.39)

After performing the required matrix multiplication in the integrand of (7.39)and carrying out the integral we obtain

fe =qLe

2

∫

1

−1

1/2(1 − ξ) 0 00 1/2(1 − ξ) 00 0 1/2(1 − ξ)

1/2(1 + ξ) 0 00 1/2(1 + ξ) 00 0 1/2(1 + ξ)

mm

nn

ll

dξ

=qLe

2

mm

nn

ll

mm

nn

ll

(7.40)

As a closing remark, if a uniform body force acts over the element, the sameprocedure as that employed for the case of a distributed load is used to get thebody force contribution to the element vector. The body force contributionis identical to the distributed load contribution, except that the quantity q isreplaced by Ab, where A is the cross-sectional area of the truss, and b is themagnitude of the body force. Note that the quantity Ab has units of forceper unit length. For the case of the body force, the quantities mm, nn, andll would then refer to the direction cosines between the body force vector andthe Cartesian base vectors. The total element vector, when both a tractiondistribution and a body force distribution act over the element, is simply thesum of the individual element vectors, i.e., fe = fq

e+ f b

e.


1 2

3

X

Y

P = 1000 lb

¤

£

¡

¢1

¤

£

¡

¢2

FIGURE 7.5

Assemblage of space trusses.

Example 7.1

To demonstrate the use of space truss elements for structural applications,let us consider an assemblage of axially loaded members connected by pinnedjoints as shown in Figure 7.5. The joints, labeled 1 and 2 in the figure, areprevented from translating in all three directions. A point load (P = 1000 lb)is applied at joint 3, which is free to move in any direction. Because bothmembers lie in the same plane, and because the load acts in this plane, theproblem can be treated as two-dimensional. The goal of this exercise is toperform a hand calculation to obtain the displacement of joint 3. In addition,we seek to find the stress and strain in each axially loaded member.

To begin, we will assume that each member is linearly elastic with Young’smodulus E = 30 × 106 psi, and we will take the cross-sectional area for eachmember to be A = 2 in.2. The coordinates of each joint are given in Table 7.1below.

The procedure for solving the problem involves several steps. The firststep is to write down the element matrix for each truss element and assem-ble the global system matrix. Because we are treating the problem as two-dimensional, the order of the element matrices will be 4× 4, and the order ofthe global matrix will be 6 × 6. Because joints 1 and 2 are constrained frommoving in the X and Y directions, the reduced global matrix, after enforcingthe boundary conditions, will be 2 × 2.

TABLE 7.1

Nodal coordinates for thetruss assemblage.

Joint X (in.) Y (in.)

1 0.0 0.02 30.0 -1.03 14.0 14.0


We will model the assemblage shown in Figure 7.5 with two linear spacetruss elements. Each space truss has cross-sectional area A and Young’s mod-ulus E. We will denote the length of each element as L1 and L2, respectively.The element matrix for the two-dimensional truss element can be obtainedfrom the three-dimensional matrix (7.33) by crossing out the rows and columnsthat correspond to the fixed degrees of freedom, i.e., row three, column three,row six, and column six. The result is

Ke =AeEe

Le

m2 mn −m2 −mn

mn n2 −mn −n2

−m2 −mn m2 mn

−mn −n2 mn n2

(7.41)

Using the global node numbering shown in Figure 7.5, the connectivity in-formation for each element is summarized in Table 7.2. The local-to-globalnode-numbering scheme is such that for element 1, local node 1 correspondsto global node 1, and local node 2 corresponds to global node 3. For element2, local node 1 refers to global node 3, and local node 2 is attached to globalnode 2. Notice that, for each element in the model, the unit vector er pointsin the direction from local node 1 to local node 2. The direction cosines, m

and n, for each element are the projections of this unit vector onto the X andY axes, i.e.,

m1 = e1

r· eX = X21/L1

n1 = e1

r· eY = Y21/L1

m2 = e2

r· eX = X21/L2

n2 = e2

r· eY = Y21/L2 (7.42)

where e1

rand e2

rare the unit vectors pointing in the axial direction for elements

1 and 2, respectively, and m1, n1, m2, and n2 are the direction cosines. Herein equation (7.42) the notation X21, and Y21 represents the differences X2−X1

and Y2 − Y1, respectively, where Xa and Ya are the X and Y coordinates oflocal node a attached to the element of interest.

TABLE 7.2

Connectivity list.

Element Local node 1 Local node 2¤

£

¡

¢1 1 3

¤

£

¡

¢2 3 2


The element matrices can now be written as

K1 =AE

(L1)3

196 196 −196 −196 1196 196 −196 −196 2

−196 −196 196 196 5−196 −196 196 196 6

1 2 5 6

K2 =AE

(L2)3

256 −240 −256 240 5−240 225 240 −225 6−256 240 256 −240 3

240 −225 −240 225 45 6 3 4

In the element matrices above, the global equation numbers that are asso-ciated with each row and column of the element matrices are labeled in themargins. The global equation numbers are obtained from the connectivityinformation reported in Table 7.2. For example, because there are two un-knowns (degrees of freedom) at each node, there are two global equations(2n − 1 and 2n) associated with each global node number n. In other words,focusing on element one, local node one corresponds to global equations oneand two, while local node two corresponds to global equations five and six.Having labeled the rows and columns of the element matrices, we can nowgo ahead and assemble the global matrix. Just as a reminder, the elementsof the global matrix are obtained by summing the element contributions asfollows:

KIJ =

nel∑

e=1

Ke

ij(7.43)

where I and J are the global equation numbers associated with local equationsi and j that belong to element e. For example, the quantity that occupiesrow five and column five of the global matrix is

K55 = K1

33+ K

2

11

=196AE

(L1)3+

256AE

(L2)3(7.44)

When the assembly process is finished, upon substitution of the numericalvalues for A, E, L1, and L2, we get the following for the global system matrix


(to six digits beyond the decimal place):

K = K1 + K2

=

.151523 .151523 .000000 .000000 −.151523 −.151523

.151523 .151523 .000000 .000000 −.151523 −.151523

.000000 .000000 .000000 .000000 .000000 .000000

.000000 .000000 .000000 .000000 .000000 .000000

−.151523 −.151523 .000000 .000000 .151523 .151523

−.151523 −.151523 .000000 .000000 .151523 .151523

+

.000000 .000000 .000000 .000000 .000000 .000000

.000000 .000000 .000000 .000000 .000000 .000000

.000000 .000000 .145604 −.136504 −.145604 .136504

.000000 .000000 −.136504 .127972 .136504 −.127972

.000000 .000000 −.145604 .136504 .145604 −.136504

.000000 .000000 .136504 −.127972 −.136504 .127972

=

.151523 .151523 .000000 .000000 −.151523 −.151523

.151523 .151523 .000000 .000000 −.151523 −.151523

.000000 .000000 .145604 −.136504 −.145604 .136504

.000000 .000000 −.136504 .127972 .136504 −.127972

−.151523 −.151523 −.145604 .136504 .297127 .015019

−.151523 −.151523 .136504 −.127972 .015019 .279495

× 107

(

lb

in.

)

The second step in the solution procedure is to form the right-hand-sidevector, enforce the essential boundary conditions, and solve for the nodal dis-placements. In the present example, since there are no body forces or externaltractions present, the right-hand-side vector just contains the external pointload applied at node 3. The vector also contains four unknown reactions as-sociated with the fixed degrees of freedom. Hence, the external force vectorin the present problem is

f =

R1X

R1Y

R2X

R2Y

0−P

(7.45)

The fixed conditions at nodes 1 and 2 are enforced by crossing out the cor-responding rows and columns of the unconstrained system. Doing this givesthe following reduced system of equations to be solved:

[

2971270 150190150190 2794950

]

d3X

d3Y

=

0−1000

(7.46)

Finally, solving for the displacements at node 3 yields

d3X = 1.81 × 10−5 in.; d3Y = −3.59 × 10−4 in.


TABLE 7.3

Stress and strain results forthe truss assemblage.

Element ǫ × 10−7 σ (psi)¤

£

¡

¢1 −122 −365

¤

£

¡

¢2 −118 −354

Now that the nodal displacements are known, we can go ahead and computethe strain and stress in each element as a post-processing step. Recalling therelationship (7.20) between the extensional strain in the element and the localnodal displacement vector, we obtain the following for the strain in element1:

ǫ1 = B1d1

=1

L1

[

−m1 −n1 m1 n1

]

d1X

d1Y

d3X

d3Y

=1

14√

2

[

−√

2/2 −√

2/2√

2/2√

2/2]

00

1.81 × 10−5

3.59 × 10−4

≈ −12.2 × 10−6 (7.47)

The normal stress in element 1 is now obtained by simply multiplying theextensional strain by Young’s modulus, i.e., σ1 = E1ǫ1 ≈ −365 psi. Note thatthe minus sign in front of the strain and stress indicates that element 1 is ina state of compression. The strain and stress in element 2 is found using thesame procedure. The results for both elements are summarized in Table 7.3.

We emphasize that the strain (and hence the stress) along the entire lengthof the two-node space truss element is constant. For this reason, commercialfinite element packages typically report the values of stress and strain at justthe centroid of the each individual space truss element.

Example 7.2

In the derivation of the strain-displacement relationship for the two-node spacetruss, we assumed that the extensional strain was equal to the derivative of theaxial displacement, ur, with respect to r — the axial direction of the truss inits original configuration. In this example, we investigate the range of validityof that assumption by examining the strain in the truss as it undergoes ageneral rigid-body rotation.


To begin, let us consider the problem of a two-node truss element as shownin Figure 7.6. The element having original length, L, lies in the X−Y plane andis parallel with the X axis in its undeformed state. The element is subjectedto a stretch

λ = l/L = 1.01 (7.48)

where l is the final length of the element. Keeping the stretch fixed, theelement is then rotated about the Z axis. We wish to determine the angle,θ, that the bar can be rotated at which the error in the strain computationusing equation (7.20) reaches 10%.

After the initial stretch of the element, the extensional strain is ǫ0 = λ−1 =0.01. The nodal displacement vector in the deformed configuration is givenas

de =

000

l cos θ − L

l sin θ

0

(7.49)

and the B-matrix associated with the undeformed configuration is

Be =

[

−1

L0 0

1

L0 0

]

(7.50)

The strain in the element as computed from equation (7.20) is then

ǫ = Bede

= λ cos θ − 1 (7.51)

θ

X

Y

FIGURE 7.6

Truss element subjected to a stretch and rigid body rotation.


λ = 1.01

λ = 1.02

λ = 1.03

θ

ǫ/ǫ0

0 0.01 0.02 0.03 0.04 0.050.86

0.88

0.9

0.92

0.94

0.96

0.98

1

FIGURE 7.7

Plot of error in strain versus angle of rotation for several values of initialstretch.

A plot of the extensional strain, ǫ, in the truss obtained from equation (7.20)(normalized with respect to ǫ0 = λ− 1) is plotted in Figure 7.7. For the caseλ = 1.01, the error in the strain calculation reaches 10% at an angle of onlyθ = 0.045 radians ≈ 2.6 degrees. As shown in the figure, the error decreasesas the stretch increases.

7.2 Euler-Bernoulli beams

In this section, we present the formulation for a two-node Euler-Bernoullibeam element. The element is extremely useful for the analysis of beams andframed structures. In order to simplify the formulation, we will restrict ourattention to two dimensions, and assume that deformation can only occurin the X−Y plane. In addition, we will assume that all loads act in theX−Y plane, and the beam’s cross section is symmetric about the Y axis soas to prevent out-of-plane deflection. After deriving the element matrices andvectors for the Euler-Bernoulli beam element, we will present some examplesillustrating the proper use of the element at the end of this section.


7.2.1 Kinematic assumptions

When a beam is subjected to bending, as shown in Figure 7.8, a portion ofthe beam experiences positive extensional strain (tension) and a portion ex-periences negative extensional strain (compression). Because the extensionalstrain makes a transition from tension to compression, there must exist a sur-face somewhere throughout the height of the beam on which the extensionalstrain is zero. That surface is called the neutral surface. The intersectionof the neutral surface and the cross section of the beam is called the neutralaxis. In the present analysis, the coordinate axes are chosen such that the X

direction points in the axial direction of the beam (to the right), and the Y

direction is chosen positive downward. The Z axis then goes into the page toform a right-handed coordinate system.

Before any loads are applied, it is assumed that any given cross section alongthe length of the beam is planar and perpendicular to the neutral surface.The main kinematic assumption in Euler-Bernoulli beam theory is that crosssections must remain planar and perpendicular to the neutral surface. Thiskinematic assumption is illustrated in Figure 7.9.

Having made the kinematic assumptions, we can now write down the strain-displacement relationship. To this end, let us focus on point A labeled inFigure 7.9. Before deformation, point A is located some distance, Y , from theneutral surface (here the distance is negative because point A lies above theneutral surface). Because we have assumed that the beam is only subjectedto transverse loads (loads parallel to the Y direction) and that cross sectionsremain planar and perpendicular to the neutral surface, the cross sectionsdepicted in Figure 7.9 move downward and rotate clockwise about the Z axis.The displacement, uX , of point A in the X direction is then

uX = −Y sin θ (7.52)

where θ is the angle measured positive clockwise from the vertical undeformed

X

Y

Z

neutral surface

neutral axis

FIGURE 7.8

Neutral surface and neutral axis of a beam subjected to bending.


Y

Y

A

A

uX

X

θ

FIGURE 7.9

Kinematic assumptions.

cross section to the rotated cross section. If we restrict our attention to smallrotations, then sin θ ≈ θ, and we get

uX = −Y θ (7.53)

In the present two-dimensional formulation, the vertical deflection will bedenoted as uY = v. The theory assumes that the vertical deflection is afunction of X only, i.e., v = v(X).

The extensional strain in the X direction can now be evaluated by differ-entiating the displacement component uX with respect to X, i.e.,

ǫXX = uX,X = −Y θ,X (7.54)

Similarly, the transverse shear strain γXY can be computed as

γXY = uX,Y + uY,X

= −θ + v,X

= −θ + θ = 0 (7.55)

The last line in equation (7.55) follows from the fact that the slope of thedeflection curve, v,X , is equal to the tangent of the angle θ, i.e.,

v,X = tan θ (7.56)

When the angle θ is much smaller than unity, then tan θ ≈ θ and hence, therotation, θ, depicted in Figure 7.9 is approximately equal to the slope of thedeflection curve.


θ1 θ2

v1 v2

X

L

FIGURE 7.10

Two-node Euler-Bernoulli beam element.

We conclude that Euler-Bernoulli beam theory does not allow for the pos-sibility of transverse shear strains. In addition, we note that the extensionalstrain, ǫXX , varies linearly with the distance, Y , from the neutral surface,and the extensional strain in the Y direction is zero because we have assumedthat the vertical deflection is a function of X only.

In addition to the kinematic assumptions, Euler-Bernoulli beam theory as-sumes that the material is linearly elastic and that the only non-zero stresscomponent is σXX , the normal stress in the X direction. If σXX is the onlynonzero stress component, it then follows that

σXX = EǫXX

= −EY θ,X

(7.57)

where E is Young’s modulus.

It is worth mentioning here that by assuming σXX is the only nonzerostress, we violate the kinematic assumptions described above. Because wehave assumed that the vertical deflection, v, is a function of X only, thenthe extensional strain in the Y direction is zero. The normal stress in theY direction cannot also be zero from our knowledge of Hooke’s law for thecase of plane strain discussed in Chapter 4. Realizing this contradiction, weproceed anyway, as such contradictions are common in structural theories.

7.2.2 Finite element approximations

A deflection curve represents the vertical deflection of a beam as a function ofdistance X from the left end. In the present section we will let v(X) be thedeflection curve. As discussed in the previous subsection, when the rotationof the beam’s cross section is small, the angle, θ, is approximately equal to theslope v,X of the deflection curve. A two-node Euler-Bernoulli beam elementof length L is shown in Figure 7.10. As shown in the figure, each local nodehas two degrees of freedom, a vertical deflection in the Y direction, and arotation. Hence the displacement vector, de, for this element can be written


as

de =

v1

θ1

v2

θ2

(7.58)

where va and θa are the vertical deflection and rotation at local node a. Inthe derivation of the finite element shape functions for the two-node Euler-Bernoulli beam element, we insist that both the vertical deflection and therotation be continuous across element boundaries. It is necessary that therotation be continuous to avoid violating the assumption that planes remainperpendicular to the neutral surface. Under the present assumption of smallrotations, this is equivalent to insisting that both v and v,X be continuousacross element boundaries.

It turns out that the lowest order polynomial that assures continuity ofboth v and v,X is cubic, i.e.,

v(X) = α1 + α2X + α3X2 + α4X

3 (7.59)

where α1, . . . , α4 are constants. In order to derive the shape functions, it isnecessary to solve for these constants in terms of the nodal quantities (7.58).To get the four equations needed, we insist that

v(X = 0) = v1

v,X(X = 0) = θ1

v(X = L) = v2

v,X(x = L) = θ2

By using equation (7.59) and invoking the above boundary conditions we get

1 0 0 00 1 0 01 L L2 L3

0 1 2L 3L2

α1

α2

α3

α4

=

v1

θ1

v2

θ2

(7.60)

Finally, after solving the system of equations (7.60) and substituting the resultback into (7.59) we obtain

v(X) = N1(X)v1 + N2(X)θ1 + N3(X)v2 + N4(X)θ2 (7.61)

where N1, . . . , N4 are the finite element shape functions. Introducing thedimensionless variable, ξ = X/L, the four shape functions can be written as

N1 = 1 − 3ξ2 + 2ξ3

N2 = Lξ(1 − 2ξ + ξ2)

N3 = ξ2(3 − 2ξ)

N4 = Lξ2(ξ − 1) (7.62)


N1

N2

N3

N4

ξ

1

1

0

0

FIGURE 7.11

Shape functions for two-node Euler-Bernoulli beam element.

The four shape functions defined in equation (7.62) are plotted in Figure 7.11.They give rise to trial functions that have continuous first derivatives andpiecewise continuous second derivatives. Such shape functions are classifiedas C1 continuous. The shape functions given in (7.62) are known as Hermitepolynomials.

To end this subsection, we now derive the B-matrix for the two-node Euler-Bernoulli beam element. Recall that the element B-matrix provides the linkbetween the state of strain in the element and the element nodal unknowns.For the Euler-Bernoulli beam element, the only relevant strain component isthe extensional strain ǫXX . The finite element approximation for the exten-sional strain is obtained as follows:

ǫXX = −Y θ,X

= −Y v,XX ⇒

ǫXXe

= −Y Ne,XXde

= −Y Bede (7.63)

The matrix, Be, in the last line of equation (7.63) contains the second deriva-tives of the shape functions with respect to X. These derivatives can be easilyevaluated using the definition of the shape functions (7.62) and by employingthe chain rule, e.g.,

Ne,X =dNe

dξ

dξ

dX

=Le

2Ne,ξ (7.64)

After evaluating all the derivatives, we get

Be

1 × 4=

1

L2e

[

12ξ − 6 Le(6ξ − 4) −12ξ + 6 Le(6ξ − 2)]

(7.65)


As a check on the B-matrix reported in equation (7.65), in the event thatthe beam element undergoes a rigid body translation in the Y direction, noextensional strain should develop in the element. If the element undergoes arigid-body translation of magnitude 1, the resulting nodal displacement vectorwould be

de =

1010

(7.66)

The resulting extensional strain is then

ǫXXe

= −Y1

L2e

[

12ξ − 6 Le(6ξ − 4) −12ξ + 6 Le(6ξ − 2)]

1010

= −Y

L2e

[(12ξ − 6)(1) + (−12ξ + 6)(1)] = 0 (7.67)

which gives the desired result. One would come to the same conclusion (thatno extensional strain develops) if the element undergoes a rigid-body rotation.This can be accomplished, for example, by letting

de =

0101

(7.68)


Like the two-node truss element, we derive the element matrix and vectorfor the Euler-Bernoulli beam element by starting with the principle of virtualwork. We begin by writing the internal virtual work over a single beamelement as follows:

δWint

e=

∫

Ωe

σijwi,j dΓ (7.69)

where σij is the Cauchy stress, wi are the components of a vector-valued testfunction, and Ωe is the volume of the beam element.

In the present two-dimensional formulation, the test function has two com-ponents, wX and wY . Let us now restrict our attention to test functions thatare C1 continuous and obey the same kinematic assumptions as the displace-ment components. Doing this we write

wX = −Y δθ(X)

wY = δv(X) (7.70)


where the test function δθ can be thought of as a virtual rotation, and δv(X)can be viewed as a virtual deflection in the Y direction. Assuming that thevirtual rotations are small, the two test functions, δv and δθ, are related asfollows:

δθ = δv,X

Because the only nonzero stress component in the problem is σXX , theinternal virtual work statement simplifies to the following:

δWint

e=

∫

Ωe

σXXwX,X dΩ

= −

∫

Ae

Y

∫

Le

0

σXXδθ,X dX

dA (7.71)

Here in equation (7.71), we have assumed that the beam has a constant cross-sectional area along the length, X, of the beam. Hence, the differential volumebecomes dΩe = Ae dX.

The next step is to recall the relationship (7.57) that provides the connectionbetween the normal stress and the slope of the deflection curve, i.e., σXX =−EY θ,X . By substituting this relationship into (7.71) we obtain

δWint

e= Ee

∫

Ae

Y2 dA

∫

Le

0

θ,Xδθ,X dX (7.72)

The first integral on the right-hand side of (7.72) is, by definition, the areamoment of inertia of the beam’s cross section about the neutral axis, i.e.,

∫

Ae

Y2 dA = IZ (7.73)

As an example, for a beam with a rectangular cross section of width, b, andheight, h, the area moment of inertia about the neutral axis is

IZ =bh3

12(7.74)

The internal work statement can now be written as

δWint

e= EeIZ

∫

Le

0

θ,Xδθ,X dX (7.75)

Substituting the following finite element approximations

θ,X = Bede

δθ,X = Bewe (7.76)

into equation (7.75) yields

δWint

e= wT

e

EeIZLe

∫

1

0

BT

eBe dξ

de (7.77)


X

Y

Z

Γq

FIGURE 7.12

Traction distribution acting on Γq.

where we have performed the change of variables, X = Leξ, and hence,dX = Le dξ. The element matrix is the 4× 4 matrix inside the curly brackets(sandwiched in between the nodal displacement vector and the vector contain-ing the nodal values of the test functions) on the right-hand side of (7.77).Upon performing the required matrix multiplications and integrating over thelength of the element, the element matrix is finally obtained as

Ke =EeIZ

L3e

12 6Le −12 6Le

6Le 4L2

e−6Le 2L2

e

−12 −6Le 12 −6Le

6Le 2L2

e−6Le 4L2

e

(7.78)


Let us now focus on the derivation of the element vector for the two-nodeEuler-Bernoulli beam element. Recall that the element vector arises from theright-hand side of the principle of virtual work statement (the external virtualwork). In the absence of body forces, the external virtual work can be writtenas

δWext =

∫

Γ

tiwi dΓ (7.79)

where ti are the components of the traction vector acting on the outer surface,Γ, of the beam. In the present formulation, we have assumed that the beamis subjected to loads that act only in the X−Y plane. Hence, the surfaces onwhich tractions can act are the top and bottom surfaces of the beam, as wellas the right and left ends. Note that tractions can be applied on the rightand left ends as long as the resultant force does not have a component in theaxial, X, direction.

Let us for the moment focus on the top surface of the beam, which is labeledΓq, in Figure 7.12. We assume that the traction distribution acting on thissurface can vary in the X direction, but remains constant in the Z direction.


ΓM

XY

Z

FIGURE 7.13

Traction distribution acting on ΓM .

If we let b be the thickness of the beam in the Z direction, then the force perunit length, q(X), acting on the beam is just the vertical traction tY timesthe thickness, b, i.e.,

q(X) = btY (X) (7.80)

Notice that the force per unit length, q(X), is only a function of X. It iscommonly referred to as a distributed load. The contribution to the externalwork from the distributed load acting on the top surface can be written as

∫

Γq

tiwi dΓ = b

∫

l

0

tY wY dX

=

∫

L

0

q(X)δv(X) dX (7.81)

where L is the length of the beam.

Next, let us focus on the traction distribution acting on the right face of thebeam as shown in Figure 7.13. Here we assume that the traction distributionis such that it produces no net force in the axial direction of the beam. Thisrequirement can be stated mathematically as follows:

∫

ΓM

tX dΓ = 0 ⇒

b

∫

h/2

−h/2

σXX dY = 0 (7.82)

where we have assumed that the shape of the beam’s cross section is rectan-gular, and h is the total height of the beam in the Y direction.

Let us now investigate the external virtual work of the traction distributionacting on the right surface of the beam, ΓM . The external virtual work can


be written as∫

ΓM

tXwX dΓ =

∫

ΓM

(σXX)(−Y δθ) dΓ

= −b

∫

h/2

−h/2

σXXY δθ dY

= M∗δθ (7.83)

where M∗ is the net bending moment that is prescribed on ΓM .If shear tractions are prescribed on one of the ends of the beam, we will

call this end ΓV . The contribution to the external virtual work due to sheartractions acting on ΓV can be written as

∫

ΓV

tY wY dΓ = b

∫

h/2

−h/2

tY w,dY

= V∗w (7.84)

where V ∗ is the net shear force acting on ΓV . Finally, the external virtualwork due to all of the tractions acting on the surface of the beam combinedcan be written as

δWext =

∫

L

0

qδv dX + M∗δθ + V

∗δv (7.85)

We end this subsection by deriving the element vector for the case wherethe element is subjected to a uniform distributed load of magnitude q. Fornow we will assume that the element does not touch either of the boundariesΓM or ΓV . Under these circumstances, the external virtual work reduces to

δWext

e=

∫

Le

0

qδv dX (7.86)

Next, we interpolate the test function δv over the length, Le, of the elementusing the same shape functions that we used to interpolate the trial function.Doing this gives

δv

e

= Newe (7.87)

where Ne is the 1×4 shape function matrix that containes the shape functionsdefined in equation (7.62), and we is a 4 × 1 vector containing the values ofthe test functions at the nodal points, i.e.,

we =

δv1

δθ1

δv2

δθ2

(7.88)


After plugging the finite element approximation (7.87) into the external workexpression (7.86) we get

δWext

e= wT

e

qLe

∫

1

0

NT

edξ

(7.89)

where, again, we have used the natural coordinate ξ = X/Le. We now recog-nize that the 4 × 1 vector inside the curly brackets on the right-hand side ofequation (7.89) is the element vector, i.e,

fe = qLe

∫

1

0

NT

edξ (7.90)

Evaluating the integral on the right-hand side of equation (7.90) above yields

fe = qLe

1/2Le/121/2

−Le/12

(7.91)

We observe from the element vector (7.91) that in order to correctly modelthe situation of a uniform distributed load acting over the span of an element,we must apply both nodal forces in the Y direction as well as equal andopposite concentrated couples at the nodal locations. As a closing remark, ifan element does in fact touch one of the boundaries (either ΓM or ΓV ) thenthe element vector for that element is simply augmented by the concentratedcouple or concentrated force at the local node that touches the boundary. Theentire procedure will be demonstrated in the following example.

Example 7.3

In order to demonstrate the use of the Euler-Bernoulli beam element, let usconsider an example of a cantilever beam of length L = 10m loaded by auniform distributed load, q = 1000N/m. A concentrated force, P = 1000N,acts at the right end as shown in Figure 7.14. The concentrated force can beviewed as the resultant force of a shear traction distribution applied at theright end of the beam. The cross section of the beam is rectangular with baseb = 100mm, and height h = 400mm. We will assume that the beam is madeof steel with Young’s modulus E = 200GPa.

To get an idea of how many elements to use along the length of the beam, itis instructive to look at the bending moment diagram. The bending momentdiagram for the present example is shown in Figure 7.15. As shown in thefigure, the maximum bending moment, Mmax = −60, 000N · m, occurs at theleft end of the beam, and the bending moment goes to zero at the right endas expected. From elementary mechanics of materials theory, the bending


X

Y

P = 1000N

q = 1000N/m

FIGURE 7.14

Cantilever beam subjected to uniform distributed load and concentrated endload.

moment at position X along the length of the beam is related to the verticaldeflection at that location through the relation

EIZv,XX = −M(X) (7.92)

where E is Young’s modulus, and IZ is the area moment of inertia of the crosssection about the neutral axis. Employing the finite element interpolationfunction (7.61), and differentiating the trial function twice with respect to X

X (m)

M (N·m)

0 2 4 6 8 10-60,000

-40,000

-20,000

0

FIGURE 7.15

Bending moment diagram.


gives

v,XXe

= Bede

=1

L2e

[(12ξ − 6)v1 + Le(6ξ − 4)θ1

+ (−12ξ + 6)v2 + Le(6ξ − 2)θ2] (7.93)

where ξ = X/Le. We see from equation (7.93) above that the finite elementapproximation for v,XX is capable of capturing a linearly varying bendingmoment. For this reason, we can get away with using only one beam elementin regions where the bending moment varies linearly with X. In the presentexample, the bending moment varies quadratically along the length of thebeam, and hence, more than one element will be necessary to obtain goodaccuracy.

Let us now begin the process of obtaining an approximate finite elementsolution by discretizing the beam along its length with two Euler-Bernoullibeam elements (each of equal length) as shown in Figure 7.16. The elementsare identified by numbers inside rounded boxes. The global node numbers arelabeled in bold, and the global equation numbers associated with each node arelisted below the global node numbers. Following the finite element assemblyprocess described in Chapter 3, we begin by writing down the element matricesand indicating the local-to-global equation numbering to the right of thesematrices. Doing this we obtain

K1 =EIZ

L3

12 6L −12 6L 16L 4L2 −6L 2L2 2−12 −6L 12 −6L 36L 2L2 −6L 4L2 4

1 2 3 4

K2 =EIZ

L3

12 6L −12 6L 36L 4L2 −6L 2L2 4−12 −6L 12 −6L 56L 2L2 −6L 4L2 6

3 4 5 6

Upon assembly of the element matrices, the 6× 6 global system matrix turnsout to be

K =EIZ

L3

12 6L −12 6L 0 06L 4L2 −6L 2L2 0 0−12 −6L 24 0 −12 6L

6L 2L2 0 8L2 −6L 2L2

0 0 −12 −6L 12 −6L

0 0 6L 2L2 −6L 4L2

(7.94)

¤

£

¡

¢1

¤

£

¡

¢2

1,2 2,3 3,41 2 3

FIGURE 7.16

Two-element model of cantilever beam.


Next, from equation (7.90), we can write down the element vectors as follows:

f1 = qL

1/2 1L/12 21/2 3

−L/12 4

f2 = qL

1/2 3L/12 41/2 5

−L/12 6

+

0 30 4P 50 6

Note that we have augmented the element vector f2 by the concentrated load,P , acting at local node 2 of the second element. Assembling the elementvectors yields

f =

qL/2qL2/12

qL

0qL/2 + P

−qL2/12

(7.95)

The global system of equations can now be written as

EIZ

L3

12 6L −12 6L 0 06L 4L2 −6L 2L2 0 0−12 −6L 24 0 −12 6L

6L 2L2 0 8L2 −6L 2L2

0 0 −12 −6L 12 −6L

0 0 6L 2L2 −6L 4L2

v1

θ1

v2

θ2

v3

θ3

=

qL/2qL2/12

qL

0qL/2 + P

−qL2/12

(7.96)

Before solving the global system of equations (7.96), it is necessary to en-force the essential boundary conditions at global node 1. Because node 1 islocated at a clamped end, both the deflection, v1, and the rotation, θ1, arezero at this location. These boundary conditions can be enforced by crossingout the first and second rows and columns of the global system of equations.Doing this, we obtain the following reduced system to be solved:

EIZ

L3

24 0 −12 6L

0 8L2 −6L 2L2

−12 −6L 12 −6L

6L 2L2 −6L 4L2

v2

θ2

v3

θ3

=

qL

0qL/2 + P

−qL2/12

(7.97)

Solving the reduced system (7.97) yields (to eight digits beyond the decimalpoint)

v1

θ1

v2

θ2

v3

θ3

=

.00000000

.00000000

.00512695

.00171875

.01484375

.00203125

(7.98)


Having obtained the nodal displacements and rotations, let us now calculatethe bending moment along the length of each element from the finite elementsolution. Utilizing equation (7.92), the bending moment at location X alongthe length of the beam can be written as

M(X) = −EIZv,XX

= −EIZθ,X ⇒

Me = −EeIZBede (7.99)

where in the last line we have plugged in the finite element approximationθ,X = Bede. The bending moment along the length of element 1 then becomes

M1(ξ) = −EIZB1d1

=−EIZ

L2

[

12ξ − 6 L(6ξ − 4) −12ξ + 6 L(6ξ − 2)]

v1

θ1

v2

θ2

Next, noting that the length of element 1 is 5m, and the local displacementvector is

d1 =

0.000000000.000000000.005126950.00171875

(7.100)

we get the following equation for the bending moment along the length ofelement 1:

M1(ξ) = 42499.8400ξ − 57916.5866 (7.101)

When evaluating the bending moment along the length of the element, wemust remember that the natural coordinate, ξ = X/l, is measured from theleft end of the element. Following the same procedure, we get the followingfor the bending moment along the length of the second element:

M2(ξ) = 17500.1600ξ − 15416.7467 (7.102)

The bending moment obtained from the finite element method is plottedversus distance, X, along the length of the beam in Figure 7.17. The finiteelement approximation is represented by the dashed line in the figure, andthe exact solution is depicted by the solid line. Notice that the finite elementapproximation predicts a linearly varying bending moment along the lengthof each element. Also notice that the natural boundary condition, M∗ = 0,at the right end of the beam is only approximately satisfied by the presentfinite element solution.

If instead, we wanted to obtain the normal stress distribution in each beamelement, we would follow essentially the same procedure that we used tocompute the bending moment, i.e., σe = −EeY Bede. The maximum tensile


X (m)

M (N · m)

0 2 4 6 8 10-60,000

-40,000

-20,000

0

FIGURE 7.17

Bending moment versus distance, X, along the length of the beam. Thedashed line represents the finite element approximation.

and compressive stress occurs at the top and bottom of the beam at the leftend, where the bending moment is maximum.

As a closing remark, it is worth mentioning that very accurate finite solu-tions to beam and frame problems can be obtained using just a few Euler-Bernoulli beam elements. In the present two- element example, the finiteelement solution for the end deflection agrees with the exact solution to eightdigits!

7.3 Mindlin-Reissner plate theory

Due to the importance of plates and shells in engineering analysis and design,in this section we provide an introduction to the rich subject of plate and shellelement technology by presenting Mindlin-Reissner plate theory and its finiteelement implementation. One of the big advantages of the Mindlin-Reissnertheory is that it does not neglect transverse shear strains (as was the case inEuler-Bernoulli beam theory). Hence, the theory allows for the analysis ofboth thin and thick plates using low-order elements such as four-node quadri-laterals. After presenting the theory and finite element implementation, weend the section with a case study in which we analyze a simply supportedplate with a uniform distributed load acting over the top surface. In the casestudy we explore the performance of the Mindlin-Reissner plate element andpoint out some numerical difficulties that can arise when using the elementand how to avoid them in practice. For additional information on this subject,


X

Y

Z

FIGURE 7.18

Mindlin-Reissner plate in undeformed and deformed configurations.

the reader is encouraged to visit Hughes and Tezduyar [14].

7.3.1 Assumptions

To help understand the assumptions that are made in Mindlin-Reissner platetheory, let us focus our attention on Figure 7.18. The plate with thickness t

originally lies in the X−Y plane. The midsurface of the plate is the shadedsurface in the undeformed (upper) configuration. The plate can be thoughtof as being composed of infinitely many inextensible fibers, or line segments,that are originally perpendicular to the midsurface. Each fiber goes from theoriginal configuration to the deformed configuration (below) by translating inthe Z direction and rotating about its center. The midsurface of the plate isalso shaded in the deformed configuration. Notice that the midsurface in thedeformed configuration is not in general planar, and the fibers do not have toremain perpendicular to it.

The kinematics of an individual fiber is illustrated in Figure 7.19. Aftertranslating in the Z direction, the fiber gets into the deformed configurationby first undergoing a positive (right-hand rule) rotation about the Y axis,followed by a negative rotation about the X axis. The resulting displacementcomponents in the X and Y directions respectively are labeled uX and uY

in the figure. Let us now focus our attention on a point that lies somewherealong the length of the fiber. If the rotations θX and θY are assumed to besmall, then the displacement components uX and uY at this point can beexpressed as

uX = ZθX

uY = ZθY (7.103)

where Z is the distance along the fiber from the fiber center to the point ofinterest. The coordinate Z is positive at points above the midsurface and


X

Y

Z

uX uY

FIGURE 7.19

Kinematics of a single fiber.

negative at points below. The displacement component uZ can be written as

uZ = v(X,Y ) (7.104)

Equation (7.104) states that the vertical displacement, v, of any point alongan individual fiber, is a function of X and Y only.

Another important assumption in Mindlin-Reissner plate theory is that thenormal stress in the Z direction is zero, i.e., σZZ = 0. This assumption doesnot follow from the kinematic assumptions stated above. We will see later,when we introduce Hooke’s law into the formulation, that assuming σZZ = 0gives rise to a nonzero extensional strain component in the Z direction. Thisis in contradiction to equation (7.104), which implies ǫZZ = ∂v/∂Z = 0!Again, such contradictions are common in the development of structural ele-ments. They are justifiable when they simplify the theory and good accuracyis still achieved between the theory and experiment. Note that the assump-tion σZZ = 0 alone is not identical to the plane-stress assumption discussedin Chapter 4. The plane-stress assumption assumes that the transverse shearstresses, σXZ and σY Z , are also zero. In Mindlin-Reissner plate theory, equa-tion (7.103) allows for the possibility of transverse shear strains, and theassociated transverse shear stresses are not ignored.

7.3.2 Strain-displacement relations and Hooke’s law

In the linearized theory of elasticity discussed in Chapter 4, the small-straincomponents, ǫij , are related to the displacement components, ui, through the


relation

ǫij =1

2(ui,j + uj,i) (7.105)

Assuming the small-strain kinematics (7.105), the relevant strain componentsin Mindlin-Reissner plate theory are obtained by differentiating the displace-ment components given in equations (7.103) and (7.104) with respect to X

and Y , respectively. Doing this gives

ǫXX = uX,X = ZθX,X

ǫY Y = uY,Y = ZθY,Y

γY Z = uY,Z + uZ,Y = θY + v,Y

γXZ = uX,Z + uZ,X = θX + v,X

γXY = uX,Y + uY,X = Z (θX,Y + θY,X) (7.106)

Next, assuming linearly elastic and isotropic material behavior, the nonzerostress components can be expressed in terms of the strain components definedin (7.106) above using the generalized Hooke’s law. To do this, we store thefive nonzero stress components in two vectors as follows:

σb =

σXX

σY Y

σXY

; σs =

σXZ

σY Z

(7.107)

The 3 × 1 stress vector σb contains the in-plane components of the stress,

and σs contains the transverse shear stress components. The in-plane stress

components can be written in terms of the in-plane strain components as

σXX

σY Y

σXY

=

E νE 0

νE E 0

0 0 G

ǫXX

ǫY Y

γXY

= Z

E νE 0

νE E 0

0 0 G

θX,X

θY,Y

θX,Y + θY,X

= ZCbκ (7.108)

where E = E/(

1 − ν2)

, Cb is the 3×3 plane stress elasticity matrix, and κ isa 3 × 1 vector containing the partial derivatives of the rotations. The vectorκ is called the curvature vector. The transverse shear stress components can


ξ

η

Z

1

2

3

4

FIGURE 7.20

Mindlin-Reissner plate element in parent domain.

be written in terms of the transverse shear strain components as

σXZ

σY Z

=

[

G 00 G

]

γXZ

γY Z

=

[

G 00 G

]

θX + v,X

θY + v,Y

= Gγ (7.109)

where G is the shear modulus, and γ is a 2×1 vector containing the transverseshear strains.

Having defined both the strain-displacement and stress-strain relationships,in the following sections we derive the finite element matrices and vectorsassociated with the four-node Mindlin-Reissner plate element.


A Mindlin-Reissner plate element in the parent domain is shown in Fig-ure 7.20. The neutral surface in the parent domain is a planar bi-unit squarethat lies in the plane Z = 0. The four nodes are located at the corners of thebi-unit square, and the local node numbering is chosen such that the nodes arenumbered counterclockwise from 1 to 4 as shown in the figure. The elementhas thickness t, and all of the fibers in the parent domain are perpendicularto the neutral surface. There are three unknowns associated with each node:vertical deflection, va, and two rotations, θXa and θY a. The two rotationsare often referred to as nodal rotations. They really refer to the rotation ofthe fibers passing through the nodes about the coordinate axes. Having saidthis, we can now define a 3 × 1 vector, dea, containing the nodal unknowns


(or degrees of freedom) at node a as follows:

dea

3 × 1=

va

θXa

θY a

(7.110)

The 12 × 1 vector containing the nodal unknowns associated with all fournodes can then be written as

de

12 × 1=

de1

de2

de3

de4

(7.111)

Next, we employ the bi-linear shape functions to interpolate the vertical de-flection and nodal rotations over the element. Doing this yields the followingfinite element approximations:

ve(X,Y ) = Na(ξ, η)va

θXe(X,Y ) = NaθXa

θY e(X,Y ) = NaθY a, (7.112)

where Na(ξ, η) are the bilinear shape functions, i.e.,

Na(ξ, η) =1

4(1 ± ξ)(1 ± η) (7.113)

The finite element approximation for the curvature vector defined in equation(7.106) can now be written as

κe =

Na,XθXa

Na,Y θY a

Na,Y θXa + Na,XθY a

(7.114)

or

κe = Bb

ede (7.115)

where Bb

eis a 3× 12 matrix widely referred to as the bending B-matrix. The

bending B-matrix is obtained by writing equation (7.114) in matrix notation.The result is

Bb

e

3 × 12=

[

Be1 Be2 Be3 Be4

]

(7.116)

where

Bb

ea=

0 Na,X 00 0 Na,Y

0 Na,Y Na,X

(7.117)


We end this section by writing down the finite element approximation forthe vector γ containing the transverse shear strains. From equation (7.109)we write

γ =

θX + v,X

θY + v,Y

⇒

γe

=

NaθXa + Na,Xva

NaθY a + Na,Y va

(7.118)

The finite element approximation, γ, can be written in matrix notation asfollows:

γ = Bs

ede (7.119)

where Bs

eis called the shear B-matrix. The shear B-matrix is defined as

Bs

e

2 × 12=

[

Bs

e1Bs

e2Bs

e3Bs

e4

]

(7.120)

where

Bs

ea=

[

Na,X Na 0Na,Y 0 Na

]

(7.121)

is the contribution to the shear B-matrix from local node a.


From our study of Chapter 6, we recall that the element matrix (stiffnessmatrix) comes from the internal virtual work statement written over a sin-gle element defined by the domain Ωe. For a plate element with constantthickness, t, the internal virtual work statement can be written as

δWint

e=

∫

t/2

−t/2

∫

Ae

σijwi,j dA (7.122)

where Ae is the area of the element, and wi are the components of a vector-valued test function that will be defined below. By expanding the integrandin (7.122) we get

σijwi,j = σXXwX,X + σY Y wY,Y + σXY (wX,Y + wY,X)

+σXZ(wX,Z + wZ,X) + σY Z(wY,Z + wZ,Y ) (7.123)

where we have used the that fact that the stress tensor is symmetric. Noticethat only five out of the six stress components appear in (7.123), because wehave assumed that σZZ=0.

In defining the test function, as usual, we insist that the test function becontinuous and identically zero at points on the boundary of the domain wherethe displacement components are prescribed. In our present theory for plates,


we also insist that the test function obey the same kinematic assumptions asthe displacement field. Doing this we can write

wX = ZδθX(X,Y )

wY = ZδθY (X,Y )

wZ = δv(X,Y ) (7.124)

Here in equation (7.124), Z is the vertical distance from the midsurface ofthe plate to the point of interest. The quantities δθX , δθY , and δv(X,Y )are scalar test functions. They obey the following conditions at the essentialboundary points:

δθX = 0 : at boundary points where θX is specified

δθY = 0 : at boundary points where θY is specified

δv = 0 : at boundary points where v is specified

Differentiating the test functions (7.124) with respect to X and Y gives

wX,X = ZδθX,X

wY,Y = ZδθY,Y

wX,Y + wY,X = Z (δθX,Y + δθY,X)

wX,Z + wZ,X = δθX + δv,X

wY,Z + wZ,Y = δθY + δv,Y (7.125)

It is convenient to store the derivatives of the test function defined in (7.125)in two vectors as follows:

wX,X

wY,Y

wX,Y + wY,X

= Z

δθX,X

δθY,Y

δθX,Y + δθY,X

= Zδκ

and

wX,Z + wZ,X

wY,Z + wZ,Y

=

δθX + δv,X

δθY + δv,Y

= δγ

where the 3 × 1 vector, δκ, can be viewed as a virtual curvature vector, andthe 2×1 vector, δγ, can be viewed as a virtual transverse shear strain vector.

The virtual internal work expression can now be written in matrix notationin terms of the stress vectors σ

b and σs as follows:

δWint

e=

∫

t/2

−t/2

∫

Ae

Z2

δκT

1 × 3σ

b

3 × 1

dA + t

∫

Ae

δγT

1 × 2σ

s

2 × 1

dA (7.126)


where the first integral on the right-hand side is the bending contribution,and the second integral is the shear contribution.

In order to complete the derivation of the element matrices for the Mindlin-Reissner plate element, it is necessary to substitute the finite element approx-imations into the internal virtual work (7.126). The finite element approx-imations for the virtual curvature and shear strain vectors are obtained bydefining a vector, we, containing the nodal values of the components of thetest function. This can be done as follows:

we

12 × 1=

we1

we2

we3

we4

(7.127)

where

wea =

δva

δθXa

δθY a

(7.128)

is the 3 × 1 contribution from each node a. If we now interpolate the com-ponents of the test function over the element using the same bilinear shapefunctions that were used to interpolate the displacement field we get

δv

δθX

δθY

e

=

Naδva

NaδθXa

NaδθXa

(7.129)

The finite element approximations for the virtual curvature vector and virtualshear strain vector are then written as

δκ = Bb

ewe (7.130)

andδγ = Bs

ewe (7.131)

where Bb

eand Bs

eare the bending B-matrix and the shear B-matrix defined

in equations (7.116) and (7.120). Next, plugging in the finite element approx-imations into the internal virtual work statement (7.126) yields

δWint

e= wT

e

∫

t/2

−t/2

∫

Ae

Z2BbT

eCb

eBb

edA + tG

∫

Ae

BsT

eBs

edA

de (7.132)

We recognize that the element stiffness matrix, Ke, is the 12 × 12 matrixinside the curly brackets in equation (7.132). Carrying out the integrationwith respect to Z in closed-form, the stiffness matrix can be written as

Ke =t3

12

∫

Ae

BbT

eCb

eBb

edA + t

∫

Ae

BsT

eCs

eBs

edA (7.133)


The first integral on the right-hand side of equation (7.133) above is referredto as the bending stiffness matrix. The second integral is the shear stiffnessmatrix. Both integrals can be fully integrated by employing a 2×2 quadraturerule on the integrals expressed in the parent domain. It turns out that fullyintegrating the shear stiffness matrix leads to numerical difficulties when thethickness of the plate is much less than the characteristic in-plane width. Thisissue is explored at the end of this chapter, where we consider the problem ofa square plate with clamped edges subjected to uniform pressure loading onthe top surface.


In the absence of body forces, the external virtual work statement can bewritten as

δWext =

∫

Γt

t∗iwi dΓ (7.134)

The element vector for a Mindlin-Reissner plate element is obtained by writingthe external virtual work expression over a single element and then substitut-ing the finite element approximations into the resulting expression. In doingthis, we recognize that, in general, the traction distribution can act over thetop and bottom surfaces of the plate element. In addition, if the elementhas an edge that is contiguous to the boundary of the plate as shown in Fig-ure 7.21, then that edge may also have surface tractions acting upon it. Inthe present theory we assume that the edge tractions are distributed in such away that the resultant force per unit length in the direction perpendicular tothe boundary is zero. This means that there can be no in-plane extension orcompression of the neutral surface. In the present derivation we assume thatthe tractions acting on the top and bottom surfaces of the plate act normalto the surface, i.e., no shear tractions act on the top and bottom.

Letting q be the traction component in the Z direction, the external virtualwork statement can now be written over a single element as

δWext

e=

∫

Ae

qδv dA +

∫

Γe

M∗δθn dΓ +

∫

Γe

V∗δv dΓ (7.135)

Here in equation (7.135), the first integral on the right-hand side gives thevirtual work contribution from the normal traction distribution acting overthe top and bottom surfaces of the element. Recall that the quantity δv isthe component of the test function in the Z direction. Hence, the productqδv gives force per unit length. The second integral on the right-hand side isthe external work contribution for the external normal tractions acting on theelement edge Γe. The quantity M∗ is the moment per unit length obtainedby integrating the normal traction distribution through the thickness of theplate. The moment per unit length is obtained by multiplying the normaltraction tn by the wn = Zδθn and integrating through the thickness of the


Ae Γe

n

s

Z

V ∗M∗

FIGURE 7.21

Traction distribution acting on plate.

plate, i.e.,

∫

Γe

∫

t/2

−t/2

t∗nwn dZdΓ =

∫

Γe

∫

t/2

−t/2

Zt∗nδθn dZdΓ

=

∫

Γe

M∗δθn dΓ (7.136)

Here in equation (7.136) above, the quantity δθn can be viewed as a virtualright-hand-rule rotation about the n axis labeled in Figure 7.21. The lastintegral on the right-hand side of (7.135) comes from multiplying the verticalshear traction acting on the edge by the test function, δv, and integratingover the edge of the plate element. Doing this we get

∫

Γe

∫

t/2

−t/2

t∗Zδv dZdΓ =

∫

Γe

V∗δv dΓ (7.137)

where V ∗ is the vertical shear force per unit length acting on the elementedge.

The element vector can now be obtained by substituting the finite elementapproximations into the external virtual work expression (7.135). To do this,we need to write the finite element approximations for the test function δv andδθn in terms of shape function matrices and the 12×1 vector, we, which con-tains the nodal values of the test function. The finite element approximationfor δv can be written as

δv = Nv

ewe (7.138)

where Nv

eis a 1 × 12 shape function matrix which is given as

Nv

e=

[

N1 0 0 N2 0 0 N3 0 0 N4 0 0]

(7.139)


The finite element approximation for the component of the test function,wn, can be obtained as follows. The normal component of the test functionis given as

wn = en · w, (7.140)

where en is a unit base vector pointing in the n direction labeled in Figure 7.21.Because the n axis lies in the X−Y plane, the dot product operation (7.140)above is simply

wn = mmwX + nnwY

=[

mm nn]

wX

wY

(7.141)

where mm and nn are the direction cosines between the local n axis and theX and Y axes, i.e., mm = en · eX and nn = en · eY . Next, substitutingwX = ZδθX and wY = ZδθY into equation (7.141) yields

wn = Z[

mm nn]

δθX

δθY

(7.142)

Finally, because wn = Zδθn, where δθn corresponds to a positive (right-handrule) rotation about the s axis shown in Figure 7.21, we conclude that

δθn =[

mm nn]

δθX

δθY

(7.143)

Equation (7.143) can be written in matrix notation as follows:

δθn = ANθ

ewe (7.144)

where A =[

mm nn]

is a 1 × 2 matrix containing the direction cosines,

Nθ

e=

[

0 N1 0 0 N2 0 0 N3 0 0 N4 00 0 N1 0 0 N2 0 0 N3 0 0 N4

]

(7.145)

is a matrix containing the shape functions, and

we =

we1

we2

we3

we4

where

wea =

δva

δθXa

δθY a

(7.146)


The last step in the derivation of the element vector for the four-nodeMindlin-Reissner plate element is to substitute the following finite elementapproximations:

δve = wT

eNvT

e

δθne = wT

eNθT

eAT (7.147)

into the external virtual work statement. Doing this yields

δWext

e= wT

e

∫

Ae

NvT

eq dA +

∫

Γe

NθT

eAT

M∗ dΓ +

∫

Γe

NvT

eV

∗ dΓ

The element vector is the quantity inside the curly brackets in the equationabove. Hence,

fe =

∫

Ae

NvT

eq dA +

∫

Γe

NθT

eAT

M∗ dΓ +

∫

Γe

NvT

eV

∗ dΓ (7.148)

To end this section, we note that the area integral and the line integrals thatappear in equation (7.148) can be readily calculated as equivalent area andline integrals in the parent domain using the techniques covered in Chapter 6(see Section 6.2.6 and Example 6.2).

7.4 Deflection of a clamped plate

To illustrate the practical use of the Mindlin-Reissner plate element, let usconsider the problem of a 2m×2m square plate with clamped edges as shownin Figure 7.22. The plate is loaded by a uniform traction distribution actingin the negative Z direction and having magnitude q = 100 kPa. We will beginthe analysis by taking the thickness of the plate to be t = 0.1m, and we willassume that the plate is linearly elastic and isotropic with E = 200GPa andν = 0.3.

It is helpful to refer to analytical solutions, if available, to study the per-formance and convergence of finite element types. An analytical solution tothe problem of interest, based on Poisson-Kirchhoff (thin plate) theory, canbe obtained by Hencky’s method.[15] The result for the deflection, vc, at thecenter of the plate as reported by Taylor and Govindjee [16] can be writtenas

vc = α

qa4

D(7.149)

where, to ten digits,

α = 1.265319087 × 10−3 (7.150)


XY

Z

q = 100 kPa

FIGURE 7.22

Deflection of a square plate with clamped edges.

and

D =E

1 − ν2

t3

12(7.151)

is the stiffness of the plate.

Because the plate has clamped edges, the boundary conditions along thelines Y = 0, Y = 2m, X = 0, and X = 2m are simply

v = 0; θX = 0; and θY = 0 (7.152)

The finite element mesh employed in the analysis is a 20× 20 uniform grid offour-node quadrilaterals as shown in Figure 7.23.

X

Yv = 0, θX = 0, θY = 0

FIGURE 7.23

Mesh employed for clamped plate: 20 × 20 uniform grid.


0 0.5 1 1.5 20 0.5 1 1.5 20

0.2

0.4

0.6

0.8

1

0

0.2

0.4

0.6

0.8

1

X (m) X (m)v/v

exact

v/v

exact

(b)(a)FIGURE 7.24

Deflection along the line Y = 1m, t = 0.1m (a) and t = 0.01m (b).

Numerical Results

The numerical results for the vertical deflection along the center of the clampedplate are illustrated in Figure 7.24. Numerical results for the case when theplate thickness is t = 0.1m are shown in Figure 7.24(a). The dashed line inthe figure shows the vertical deflection normalized with respect to the ana-lytical solution (7.149) along the line Y = 1m. Here, full integration (2 × 2)quadrature was employed for both the bending and shear stiffness matricesgiven in equation (7.133). The solid line in Figure 7.24(a) shows the verticaldeflection along the line Y = 1m for the case when the bending stiffness ma-trix is fully integrated, and one-point quadrature was employed to integratethe shear stiffness matrix. The practice of using different quadrature rules onvarious parts of the stiffness matrix is known as selective reduced integration(SRI).

Numerical results for the vertical deflection of the plate along the line Y =1m are shown for the case when the plate thickness is reduced to t = 0.01m inFigure 7.24 (b). Again, the vertical deflection is normalized with respect to theanalytical solution (7.149). Here the solid line represents the vertical deflectionusing the technique of selective reduced integration, i.e., 2× 2 quadrature forthe bending stiffness and 1× 1 quadrature for the shear stiffness matrix. Thedashed line in the figure represents the case when both stiffness matrices arefully integrated. We observe from the figure that the response of the fullyintegrated element is extremely stiff in bending. This behavior is referredto as shear locking. On the other hand, the element that employs selectivereduced integration performs very well.

It turns out that under most circumstances, the Mindlin-Reissner plate ele-ment with selective reduced integration gives excellent results. Unfortunately,the underintegrated element can still, under certain circumstances, undergozero-energy modes as discussed in Chapter 6, Section 6.7. Hourglass controlcan be added to the element to make it more robust (see, e.g., Flanagan andBelytscko [10], Hughes [7], and Belytschko et al. [11] for more details). For


FIGURE 7.25

Contour plot of vertical deflection.

the sake of interest, a contour plot of the vertical deflection superposed ontop of the deformed geometry (magnified 50×) is shown in Figure 7.25.

7.5 Problems

Problem 7.1

element 1 element 2

1 2X

Y 3

P

FIGURE 7.26

Problem 7.1. Truss assemblage.

Consider the truss assemblage shown in Figure 7.26. The cross section ofeach element is rectangular with dimensions 2 in.×1 in.. The nodal coordinatesare give in Table 7.4, and the material properties (Young’s modulus and initialyield stress) are provided in Table 7.5. Model the problem with two trusselements and perform the following:


TABLE 7.4

Problem 7.1: Nodalcoordinates.

Node X(in.) Y (in.)

1 0 0

2 48 0

3 24 36

TABLE 7.5

Problem 7.1. Elementproperties.

Element E(psi) σY (psi)

1 30 × 106 30,000

2 10 × 106 10,000

(a) If the load P = 1000 lb, determine the displacementcomponents d3X and d3Y .

(b) Determine the normal stress in each element.

(c) Determine the critical load Pc that can be applied tothe structure before the onset of yield in any one of the elements.

Problem 7.2

A two-dimensional truss assembly loaded by a force of magnitude P = 1000 lbis supported by a cable as shown in Figure 7.27. The force vector is given as

P = 1000

(√2

2e1 +

√2

2e2

)

.

The cable has a diameter of D = 0.25 in. and is made of steel with Young’smodulus E = 30 × 106 psi. Each truss is made of aluminum with Young’smodulus 10 × 106 psi and cross-sectional area A = 1.0 in.2. The coordinatesof nodes 1–4 labeled in the figure are given in Table 7.6. Model the problemwith three truss elements and determine the following:

(a) The displacement components d3X and d3Y

(b) The tension in the cable


1

2

X

Y

3

4

P

FIGURE 7.27

Problem 7.2. Truss assemblage supported by a cable.

TABLE 7.6

Problem 7.2. Nodalcoordinates.

Node X(in.) Y (in.)

1 0 0

2 20 0

3 3 10

Problem 7.3

Consider the problem of a simply supported beam as shown in Figure 7.28.The beam is loaded by a uniform distributed load of magnitude q = 1000 kN/m.The beam is 8m long and has a square 100mm×100mm cross section. Modelthe problem with two Euler-Bernoulli beam elements, each of equal length, toestimate the center deflection of the beam. Also obtain an estimate for thebending moment at the center of the beam.


q = 1000 kN/m

FIGURE 7.28

Problem 7.3. Simply supported beam.

8

Linear Transient Analysis

The first pull on the cord always sends the drapes in the wrong direction.—Boyle’s Other Law

An important decision that must be made during the analysis of solids andstructures is whether to perform a static analysis, a quasi-static analysis, ora dynamic analysis. In a static analysis, all variables in the problem, such asapplied load and material properties, are assumed to be independent of time.Inertia is also neglected. In a quasi-static analysis, the loads and materialproperties can depend on time, but again inertia is ignored. Viscoelasticbehavior and creep are examples of time-dependent material models. In adynamic analysis all variables in the problem can depend on time, and inertiaeffects are included.

In general, it is appropriate to perform a static analysis when the naturalfrequency of the external loading is much lower than the lowest natural fre-quency of the structure. A rule of thumb is that a static analysis is sufficientwhen the frequency of the loading is less than one-tenth of the lowest naturalfrequency of the structure. When the frequency of the loading is above thisthreshold, inertia effects become important and should not be ignored.

Transient problems in solids and structures can be separated into two cat-egories: wave propagation and structural dynamics. In wave-propagationproblems, the frequency of the external loading is very high. Blast and im-pact loading are common examples. In wave-propagation analyses, the timeduration of the event is, in general, very small — on the order of millisecondsto microseconds. Hence, it may be important to capture the high-frequencyinformation such as stress wave propagation that occurs within these smalltime scales. In structural dynamics, the main concern is the overall motionof the structure without regard to the high-frequency (small time scale) in-formation. Here the events are on the order of seconds to minutes, or longer.

At the end of this chapter, we also include a discussion on transient heatconduction. Given an initial temperature distribution, it is often necessaryto determine how long it takes for the temperature distribution in a solidbody to reach steady state. In addition, heat transfer phenomena may becoupled with structural analysis, such as thermal stress problems where heattransients are important and stress information is desired during the transientheat-conduction event.

257


Ω

Γu

ti

Γt

b

FIGURE 8.1

Newton’s second law applied to a continuum.

8.1 Derivation of the equation of motion

The derivation of the equation of motion is very similar to the derivationof the equilibrium equation derived in Chapter 4. To begin, consider thesolid body shown in Figure 8.1. The body is loaded by surface tractions anda body force distribution. The equation of motion is obtained by applyingNewton’s second law to the continuum. Assuming that the mass of the bodydoes not change with time, the balance statement says that the net externalforce vector acting on the body is equal to its total mass times acceleration.Mathematically, the statement can be written as

∫

Γ

ti dΓ +

∫

Ω

bi dΩ =

∫

Ω

ρui dΩ (8.1)

where ρ is the mass density (mass per unit volume) of a point in the contin-uum, and ui are the components of acceleration at that point. Throughoutthis chapter, superposed dots indicate differentiation with respect to time.

The equation of motion can be obtained by converting all integrals in thebalance statement (8.1) above into volume integrals, and then invoking thefact that the domain of integration, Ω, is arbitrary as discussed in Section 4.3.Recall that the surface integral on the left-hand side can be converted into avolume integral by utilizing Cauchy’s law and then employing the divergencetheorem. Doing this gives

∫

Ω

(σijnj + bi − ρui) dΩ = 0 (8.2)

and hence the equation of motion emerges as

σij,j + bi = ρui (8.3)

Ing. Civil

Nota

La aplicación de los teoremas del cálculo vectorial a este término exige mucha mas claridad en la exposición. Es necesario dar una explicación "para brutos".

Ing. Civil

Óvalo

Linear Transient Analysis 259

It is worth mentioning here that the partial differential equation (8.3) is alsoknown as the wave equation. We seek a displacement field, ui = ui(Xi, t), thatis a function of both space and time and satisfies both the partial differentialequation and the boundary conditions. To describe the boundary conditions,the displacement components can be prescribed functions of time on Γu. Thisis the essential boundary condition, i.e., ui(Xi, t) = u∗

i(Xi, t) on Γu. The

traction distribution can also be specified functions of time on Γt. This is thenatural boundary condition. The essential and natural boundary conditionscan be summarized as follows:

ui(Xi, t) = u∗i(Xi, t) on Γu

ti(Xi, t) = t∗i(Xi, t) on Γt (8.4)

The partial differential equation (8.3) along with the boundary conditions(8.4) is known as the strong form of the equation of motion.

8.1.1 Weak form

The weak form of the equation of motion is obtained by multiplying thepartial differential equation (8.3) by a vector-valued test function, wi, andthen integrating over the domain of the body. The process of deriving theweak form can be written down as follows:

∫

Ω

σij,jwi dΩ +

∫

Ω

biwidΩ =

∫

Ω

ρui dΩ ⇒

∫

Ω

(σijwi),jdΩ −

∫

Ω

σijwi,j dΩ +

∫

Ω

biwi dΩ =

∫

Ω

ρui dΩ (8.5)

The last step is to apply the divergence theorem to the first integral on thesecond line of (8.5) above and then to recognize that σijnj = t∗

ion Γt from

Cauchy’s law. Finally, the weak form can be written as∫

Ω

σijwi,jdΩ +

∫

Ω

ρuiwi dΩ =

∫

Ω

biwi dΩ +

∫

Γt

t∗iwi dΓ (8.6)

Note that it is only necessary to carry out the integration over Γt, becausethe test function is defined to be zero on Γu.

8.2 Semi-discrete equations of motion

Let us now proceed and substitute the usual finite element approximationsinto the weak form. It is convenient to begin the process by writing downthe weak form over a single element and then substituting the approximate

Ing. Civil

Nota

No se da mayor explicación acerca de esta función y su interpretación física. Del principio del trabajo virtual, se puede identificar a esta función como el "desplazamiento virtual" del sistema.

Ing. Civil

Óvalo


fields into that statement. This procedure provides the element matricesand element vectors that are needed to form the global discretized systemof equations.

The first step in obtaining the discretized equations is to approximate thevector-valued test function, w, as follows:

w = Newe = wT

eNT

e(8.7)

where, again, we contains the values of the test function at the nodal points.We emphasize that the test function does not depend on time; it only dependson the spatial variables. Next, the acceleration field, u, within the element isapproximated as

¨u = Nede (8.8)

where de is a vector containing the nodal accelerations for the element. Thatis,

dea =

daX

daY

daZ

(8.9)

where a ranges from 1 to the number of nodes within the element. Substitutingequations (8.7) and (8.8) into the weak form written over a single elementyields

wT

e

∫

Ωe

BT

eσ dΩ +

∫

Ωe

ρNT

eNe dΩ de

= wT

e

∫

Ωe

NT

ebdΩ +

∫

Γt

NT

et∗ dΓ

(8.10)

For general three-dimensional solid mechanics problems, the quantity σ in(8.10) above is a 6 × 1 vector containing the stress components σ11, σ22, σ33,σ23, σ13, and σ12. The quantities Be and Ne are 6 × 3a and 3 × 3a matricesrespectively. For example, for the case of the eight-node brick element, theorder of the B-matrix would be 6 × 24.

Invoking the arbitrariness of we in equation (8.10) above yields

Mede + f int

e= fext

e(8.11)

where

Me =

∫

Ωe

ρNT

eNe dΩ (8.12)

is called the consistent mass matrix. In addition, the vector

f int

e=

∫

Ωe

BT

eσ dΩ (8.13)


is an element vector that contains the internal nodal forces, and

fext

e=

∫

Ωe

NT

eb,dΩ +

∫

Γt

NT

et∗ dΓ (8.14)

is the element vector that contains the equivalent external nodal forces. Wenote that the element vector containing the internal nodal forces is not com-pletely discretized, because we have not yet made any assumptions as to howthe stress field varies within the element. If we assume linearly elastic behav-ior, i.e.,

σ = Cǫ (8.15)

where C is the elasticity matrix, then the approximate stress field, σ, wouldbe

σ = CeBede (8.16)

Hence, for linearly elastic materials, the internal force vector is equal to theelement stiffness matrix times the vector containing the nodal displacements.In other words,

f int

e=

(∫

Ωe

BT

eCeBe dΩ

)

de = Kede (8.17)

The element matrices and vectors that are needed in dynamic analysis ofsolids and structures are summarized in Table 8.1.

After matrix and vector assembly, the global system of equations can bewritten as

Md + f int = fext (8.18)

For linear dynamic problems the equation is written as

Md + Kd = fext (8.19)

The system of equations (8.18), or equivalently (8.19) for linear problems,is referred to as the semi-discrete form. This means that the equations havebeen discretized in space using the finite element shape functions, but they arestill continuous in time. The semi-discrete form is a system of second-order,ordinary differential equations that must be integrated in time to obtain thedesired time history.

TABLE 8.1

Element matrices and vectors for dynamic analysis.

Me f int

efext

e

∫

Ωe

ρNT

eNe dΩ

∫

Ωe

BT

eσ dΩ or

Kede if linear

∫

Ωe

NT

ebdΩ +

∫

Γt

NT

et∗ dΓ


8.2.1 Properties of the mass matrix

In the study of linear algebra, a quadratic form is defined as

α = xT Qx (8.20)

where x is an n × 1 vector, Q is an n × n matrix, and α is the scalar thatresults after performing the required matrix multiplication. Having definedthe quadratic form, the matrix Q is defined as positive definite if the quantityα is greater than zero for all vectors, x, that are not identically zero. In otherwords, Q is defined to be positive definite if

xT Qx > 0 for all x 6= 0 (8.21)

The kinetic energy of a body in motion is defined as

KE =1

2

∫

Ω

ρu2 dΩ (8.22)

where u is the velocity field. The finite element approximation for the velocityvector over a single element can be written as

˙u

e

= Nede (8.23)

where de are the nodal velocities. Substituting the finite element approxima-tion for the velocity field into the kinetic energy expression (8.22) yields thefollowing expression for the kinetic energy of an element:

KE

e

=1

2

∫

Ωe

ρ

(

Nede

) (

Nede

)

dΩ

=1

2de

T

∫

Ωe

ρNT

eNe dΩ

de

=1

2de

T

Me de (8.24)

where we recognize that the last line in equation (8.24) above is a quadraticform involving the nodal velocity vector and the element mass matrix. Be-cause the kinetic energy of an element is always greater than zero (except forthe special case when the nodal velocities are all zero), we conclude that theelement mass matrix is positive definite.

An important consequence of the fact that the element mass matrix ispositive definite is that it is nonsingular. This means that the global massmatrix that results after assembly is also nonsingular. Thus, the dynamicanalysis of an unconstrained body does not cause numerical difficulties —unlike static analysis, which requires degrees of freedom to be constrainedafter assembly to prevent the system matrix from being singular.


8.2.2 Natural frequencies and normal modes

In the absence of external forces acting on a body, the right-hand side of thesemi-discrete equations of motion is zero, and the system of equations is calledhomogeneous. The homogeneous system of equations is also often referred toas a free-vibration problem. When the system of equations is linear, i.e.,f int = Kd, then the global homogeneous system can be written as

Md + Kd = 0 (8.25)

where we emphasize that the vector, d = d(t), is a continuous function oftime.

Let us now assume that the solution, d(t), to the system of equations (8.25)is oscillatory in time and has the form

d(t) = Aeiωt (8.26)

where i is the imaginary number, and ω is called the frequency of oscillation(having units of rad/s). The quantity A in equation (8.26) above is a m × 1vector containing unknown real constants, where m is the number of equa-tions in the linear system. The vector A can be viewed as the amplitude ofoscillation. It turns out that the time-dependent function, eiωt, can be writtenin terms of the sine and cosine functions as follows:

eiωt = cos ωt + i sin ωt (8.27)

Equation (8.27) is called Euler’s equation. It can be derived by expanding thefunctions cosωt, sin ωt, and eiωt as Taylor series and then judiciously combin-ing the resulting expressions (see Chapter 2, Problem 2.12). Differentiatingequation (8.26) first once, and then twice, with respect to time yields

d = Aiωeiωt

d = −Aω2eiωt (8.28)

After inserting the derivatives (8.28) back into equation (8.25) we obtain[

K − ω2M

]

Aeiωt = 0 (8.29)

We now observe that equation (8.29) above is beginning to look like aneigenvalue problem introduced in Chapter 2. In order to get this equationinto standard form, we can pre-multiply both sides by the inverse of the massmatrix, M−1. Doing this yields

[

M−1K − ω2I

]

Aeiωt = 0 (8.30)

where I is an m × m identity matrix. We now seek values of ω that satisfyequation (8.30). The only way this equation can be satisfied for nontrivial vec-tors Aeiωt is when the determinant of the resulting matrix inside the bracketsis zero, i.e.,

∣

∣M−1K − ω2I

∣

∣ = 0 (8.31)


Hence, finding the natural frequencies of the linear system (8.25) boils downto finding the eigenvalues of the resulting m × m matrix

[

M−1K − ω2I]

. Inpractice, the eigenvalues can be obtained by an eigenvalue extraction routinesuch as the Lanczos algorithm [18]. Once the eigenvalues are obtained, theycan be substituted back into equation (8.28), one by one, to obtain the indi-vidual eigenvectors, A(i), that correspond to each eigenvalue ωi. The resultingvectors, A(i), are widely referred to as a normal modes or mode shapes.

Example 8.1

To illustrate the process of calculating natural frequencies and normal modesby the finite element method, let us consider an unconstrained, two-nodelinear-u element. Recall from our work in Chapter 3 that the element stiffnessmatrix for the linear-u element is

Ke =AeEe

Le

[

1 −1−1 1

]

(8.32)

where Ae, Le, and Ee are the cross-sectional area, length, and Young’s mod-ulus, respectively. The element mass matrix is obtained by evaluating thefollowing integral over the volume of the element:

Me = Aeρe

∫

Le

0

NT

eNe dX (8.33)

where ρe is the mass density of the element (assumed to be constant along itslength), and Ne is the 1 × 2 matrix containing the shape functions, i.e.,

Ne

1 × 2=

[

1−X

Le

X

Le

]

If we now perform the requisite matrix multiplications and integrate the re-sulting matrix we obtain

Me =ρeAeLe

6

[

2 11 2

]

(8.34)

It is interesting that the sum of all of the elements inside the mass matrix(8.33) above equals the total mass of the element. Note that the total massof the element equals the mass density of the element times the volume of theelement. Hence, the total mass is equal to ρeAeLe.

Armed with the element stiffness and mass matrices, we can now go aheadand compute the eigenvalues of

[

M−1

eKe − ω2I

]

. Multiplying the elementstiffness matrix by the inverse of the element mass matrix and then subtractingω2I yields

[

M−1

eKe − ω

2I]

=

6Ee

ρeL2e

− ω2 −6Ee

ρeL2e

−6Ee

ρeL2e

6Ee

ρeL2e

− ω2

(8.35)


Evaluating the determinant of the matrix (8.35) above and setting the resultequal to zero gives the following characteristic equation:

ω2(

−12Ee + ω2ρeL

2

e

)

= 0 (8.36)

Solving equation (8.36) for ω gives

ω1 = 0

ω2 =2√

3 c

Le

(8.37)

The first natural frequency, ω1 = 0, corresponds to a rigid-body translation ofthe element. The second natural frequency, ω2, corresponds to an oscillatoryextension/compression mode of vibration. The quantity, c =

√

Ee/ρe, thatappears on the second line of equation (8.37) is the longitudinal wave speed.The longitudinal wave speed refers to the velocity at which a longitudinalstress wave propagates in a linearly elastic medium with Young’s modulus Ee

and mass density ρe. The wave speed in structural materials is very high. Forexample, for steel with E = 200GPa and ρ = 7850 kg/m3, the wave speed isapproximately 5048m/s!

The normal modes of vibration for the unconstrained two-node linear-uelement can now be obtained by substituting the natural frequencies, ω1 andω2, back into equation (8.30) and solving for the normal modes, A(1) andA(2). For the first natural frequency, ω1 = 0, we obtain

6Ee

ρeL2e

−6Ee

ρeL2e

−6Ee

ρeL2e

6Ee

ρeL2e

A(1)

1

A(1)

2

=

00

(8.38)

Expanding the first row in (8.38) above yields

6Ee

ρeL2e

A(1)

1−

6Ee

ρeL2e

A(2)

1= 0 ⇒

A(1)

1= A

(2)

1(8.39)

We come to the same conclusion after expanding the second row. We seefrom equation (8.39) that the components of the eigenvector A(1) are equal,but the magnitude is arbitrary. It is customary to normalize each eigenvector(normal mode) so that the magnitude is unity. Doing this for A(1) gives

A(1) =

√2

2√

2

2

(8.40)

Ing. Civil

Nota

por qué concluye que "c" es la velocidad de propagación de la onda???

Ing. Civil

Óvalo


Following the same procedure, the normalized components of the second nor-mal mode are found to be

A(2) =

√2

2

−

√2

2

(8.41)

8.3 Central difference method

The central difference method is a widely used method for integrating thesemi-discrete equations of motion derived in Section 8.2. The main idea is to“break up” or discretize the equations in time leading to a system of equationsthat must be solved at each time step. Marching forward in time, from timestep to time step, then provides an approximation of the time history of thefield variable in question.

In order to describe the central difference method, let us focus our attentionon Figure 8.2. A generic plot of velocity, v, at a given node in the model versustime, t, is shown in Figure 8.2 (a). The discrete instants of time n − 1, n,and n + 1 are labeled in the figure. Intermediate instants of time, n − 1/2and n + 1/2, are also labeled in the figure. The acceleration of a nodal pointis the derivative of the velocity of that node with respect to time. Keepingthis in mind, the first step in the central difference method is to approximatethe acceleration of the node at time n by the slope of the straight line drawnbetween the velocity at times n− 1/2 and n + 1/2 as illustrated in the figure.

v d

∆t

n − 1n − 1/2

nn + 1/2

n + 1n − 1n − 1/2

nn + 1/2

n + 1

(a) (b)FIGURE 8.2

Central difference approximations for velocity (a) and displacement (b) versustime.


Mathematically, this approximation can be written as

a = d =dv

dt≈

vn+1/2 − vn−1/2

∆t(8.42)

where ∆t is the time step (the width of the time interval between time n−1/2and n + 1/2).

Next, we need to use the fact that the nodal velocity is the derivativeof the displacement with respect to time. This will allow us to obtain anapproximation for the acceleration at time n in terms of the displacement atdiscrete instants of time. To do this, we now focus on Figure 8.2 (b). Here,a generic plot of the displacement versus time of the same node is illustrated.The velocity at time n + 1/2 is approximated as the slope of the straight linefrom dn to dn+1. Hence we can write

vn+1/2 ≈dn+1 − dn

∆t(8.43)

Similarly we make the approximation that the velocity at time n− 1/2 is theslope of the straight line from dn−1 to dn, i.e.,

vn−1/2 ≈dn − dn−1

∆t(8.44)

Finally, substituting equations (8.43) and (8.44) into equation (8.42) yields

an = dn =≈

dn+1 − dn

∆t

−

dn − dn−1

∆t

∆t

or

an =dn+1 − 2dn + dn−1

∆t2(8.45)

The semi-discrete form at some time n is written as

Mdn = Fn − fn (8.46)

where Fn is the external force vector at time n, and fn is the internal forcevector. Note that for linear dynamic problems, the internal force vector isequal to Kdn. Substituting the central difference approximation (8.45) forthe nodal accelerations at time n yields

1

∆t2Mdn+1 = Fn − fn +

1

∆t2M(2dn − dn−1) (8.47)

Next, multiplying both sides of equation (8.47) by ∆t2M−1 gives

dn+1 = ∆t2M−1 (Fn − fn) + 2dn − dn−1 (8.48)

Notice that in equation (8.48) above, if the mass matrix is diagonal, thecalculation of the nodal displacement vector at the future instant of time,


TABLE 8.2

Central difference algorithm.

1. Set up initial conditions, dn−1 = dn, dn−1/2 = dn

2. Set up the time independent mass matrix,

M =∑

e

∫

Ωe

ρeNT

eNe dΩ

3. Loop over time steps

3.1 Compute the internal force vector, fn =∑

e

∫

Ωe

BT

eσ dΩ

3.2 Compute the external force vector,

Fn =∑

e

∫

Ωe

NT

ebdΩ +

∫

Γe

NT t∗ dΓ

3.3 Solve for dn+1

3.4 Compute nodal velocities, dn+1/2 = (dn+1 − dn) /∆t

3.5 Compute nodal accelerations, dn =(

dn+1/2 − dn−1/2

)

/∆t

3.6 Update displacement and velocity vectors,

dn = dn+1, dn−1/2 = dn+1/2

4. Continue time loop

n + 1, does not involve solving a matrix problem at each time step. Thecalculation simply requires knowledge of the external force vector, the internalforce vector, and the nodal displacement vector at the historical instants oftime n and n − 1. For this reason, the central difference time-integrationscheme is referred to as an explicit method.

A flow chart of the computational procedure for implementation of thecentral difference method is given in Table 8.2. The central difference time-integration scheme turns out to be a very robust and efficient method forintegrating the semi-discrete equations in time. The method can even handlenonlinear material behavior without any difficulty, because the method onlyrequires that the internal force vector be calculated at the known equilibriumconfiguration of the body at time n. On the other hand, one of the majordrawbacks of the method is that it is conditionally stable. This means thatwe must pay particular attention to time-step size selection. If the time stepis chosen to be above a critical value, the solution becomes unstable, i.e., theapproximate finite element solution for the nodal displacements tends towardinfinity as time marches on. This, of course, is physically unrealistic.

In the following subsections we provide some examples to illustrate the useof the central difference method and to discuss accuracy of the numericalresults. We also provide a discussion on the stability of the method andprovide guidelines for proper time-step selection.


θ

eθ

l

mg

FIGURE 8.3

Simple pendulum.

8.3.1 Dynamic response of a simple pendulum

To demonstrate the use of the central difference method to obtain the dynamicresponse of mechanical systems, we consider the example of a simple pendulumas shown in Figure 8.3. A point mass, m, is attached to the end of a massless,rigid link of length l. Given the initial angle, θ0, between the rigid link andthe vertical dashed line shown in the figure, it is desired to obtain the responseof the mass subjected only to the force of gravity.

The equation of motion for the point mass can be obtained by summingforces in the direction of the unit vector eθ labeled in the figure (tangentialdirection) and setting the resultant force equal to the mass times the acceler-ation in the tangential direction. Doing this yields

−mg sin θ = ms (8.49)

where g is the gravitational constant, and s represents the distance traveled bythe mass from some initial starting location along the circular path depictedby the dashed arc. Given the relationship between arc length and the angleθ, i.e., s = lθ, equation (8.49) can be recast as

mlθ + mg sin θ = 0 (8.50)

When the angle θ is much less than one radian, then sin θ ≈ θ, and equation(8.50) reduces to

θ +g

lθ = 0 (8.51)

Equation (8.51) above is a second-order ordinary differential equation intime. We now seek the time history, θ(t), subjected to a set of initial condi-


tions. In the present example we will take the initial starting angle at timet = 0 to be θ0 = π/30, and we will assume that the pendulum is initially atrest. We will take the gravitational constant and length of the pendulum armto be g = 9.81m/s2, and l = 1m, respectively.

To obtain an approximate solution to equation (8.51) using the centraldifference method, we begin by writing down the central difference approx-imation for θ at some discrete instant of time n, and then substituting theapproximation back into equation (8.51). The equation of motion at time n

is written asθn +

g

lθn = 0 (8.52)

Substituting the central difference approximation,

θn ≈1

∆t2(θn+1 − 2θn + θn−1) (8.53)

into equation (??) above and simplifying gives

θn+1 +

(

g∆t2

l− 2

)

θn + θn−1 = 0 (8.54)

By selecting a time step size and taking n = 0 to represent the orientationof the pendulum at time zero, we can now solve equation (8.54) above for thenew angle θn+1. The central difference equation to be solved can be writtenas

θn+1 = −

(

g∆t2

l− 2

)

θn − θn−1 (8.55)

Notice that the new angle, θn+1, depends only on historical information on theright-hand side of equation (8.55). Once the new angle is calculated, the newangle then becomes a historical value, and the entire process can be repeated,yielding θn+2, θn+3, etc. as the solution marches forward in time.

The numerical solution is plotted versus the exact solution of equation (8.51)for four different time steps in Figure 8.4. As shown in the figure, the solutionis obtained up to t = 5 s for ∆t = 0.05 s, ∆t = 0.1 s, ∆t = 0.4 s, and ∆t =0.6386 s. In each plot, the numerical result is represented by the circular datapoints, and the exact solution is depicted by the solid line.

Notice that for the smallest time step considered, the agreement betweenthe numerical results and the exact solution is excellent. As the time step issubsequently increased, however, notice that the period (time between suc-cessive peaks) becomes shorter. This is known as period error. Notice alsothat for the largest time step selected, ∆t = 0.6386, the angle, θ, becomesunbounded as time marches forward. It turns out that the critical time stepin the present problem (to eight digits beyond the decimal point) is

∆tcrit =2

√

g/l= 0.63855086 (8.56)

Linear Transient Analysis 271θ(r

ad)

θ(r

ad)

θ(r

ad)

θ(r

ad)

t (s) t (s)

t (s) t (s)

∆t = 0.05 ∆t = 0.1

∆t = 0.4 ∆t = 0.6386

0 1 2 3 4 5 0 1 2 3 4 5

0 1 2 3 4 50 1 2 3 4 5

-0.2

-0.1

0

0.1

0.2

-2

-1

0

1

2

-0.2

-0.1

0

0.1

0.2

-0.2

-0.1

0

0.1

0.2

FIGURE 8.4

Angle θ plotted versus time for four time steps: ∆t = 0.05 s, ∆t = 0.1 s,∆t = 0.4 s, and ∆t = 0.6386 s.

If we choose a time step that is above the critical time step (even just slightlyabove) the solution is unstable.

A discussion on the stability of the central difference method and the cal-culation of the critical time step is given in the following subsection.

8.3.2 Stability of central difference method

In order to assess the stability of the central difference method, it is onlynecessary to analyze the following single differential equation:

d + ω2d = 0 (8.57)

The exact solution to the differential equation (8.57) above can be written as

d(t) = Aeiωt

= A cos ωt + B sinωt (8.58)

where ω is the natural frequency, and A and B are constants that, in general,can be real or complex.


Let us now consider the finite difference equation corresponding to equation(8.57) at time n as follows:

dn + ω2dn = 0 (8.59)

Substituting the central difference approximation (8.45) into (8.59) yields

dn+1 +(

ω2∆t

2− 2

)

dn + dn−1 = 0 (8.60)

Next, letting Ω2 = ω2∆t2 and simplifying, equation (8.60) above can bewritten as

dn+1 +(

Ω2− 2

)

dn + dn−1 = 0 (8.61)

where the quantity Ω = ω∆t is called the sampling frequency.The stability of the central difference method can be examined by assuming

that the solution to the difference equation (8.61) has the form

dn = eαn∆t = φn (8.62)

where ∆t is the time step size, and n is the discrete time counter. By com-paring equations (8.26) and (8.62), we observe that in order for the differencesolution (8.62) to have the same oscillatory nature as the exact solution, theparameter α needs to be a complex constant.

Substituting the assumed form (8.62) into the difference equation (8.61)gives

φn+1 +

(

Ω2− 2

)

φn + φ

n−1 = 0 (8.63)

Next, dividing both sides of (8.63) by φn−1 yields the following characteristicequation:

φ2 +

(

Ω2− 2

)

φ + 1 = 0 (8.64)

The characteristic equation (8.64) is simply a quadratic equation. The solu-tion for φ is then

φ = −(

Ω2− 2

)

±

√

(Ω2 − 2)2− 4

2(8.65)

In order for the solution to the difference equation to have the desiredoscillatory behavior, the quantity φ must have both real and imaginary parts.The first term on the right-hand side of equation (8.65) is real, because Ω2 isa real positive constant. Hence, in order for φ to have an imaginary part, thequantity inside the radical must be less than zero. Thus,

(

Ω2− 2

)2

< 4 ⇒

Ω2− 2 < 2 or

Ω < 2 (8.66)


Equation (8.66) above reveals that the largest time step that still gives riseto an oscillatory solution is

∆tcrit =2

ω(8.67)

where ∆tcrit is called the critical time step.Following the procedure outlined above for finding the critical time step

for the single equation (8.57), one can see by induction that for a system ofordinary differential equations, i.e.,

d + ω2

1d = 0

d + ω2

2d = 0

d + ω2

3d = 0

...

d + ω2

md = 0

(8.68)

where m is the number of equations in the system, the critical time stepcomes from the equation with the largest natural frequency. Hence for thecentral difference method, the critical time step for the semi-discrete systemof equations (8.18) is

∆tcrit =2

ωmax(8.69)

where ωmax is the largest natural frequency of the system.In the pendulum problem discussed in the previous subsection, the natural

frequency of the single-degree-of-freedom system was

ω =√

g/l

where g is the gravitational constant and l is the length of the rigid link. Fromequation (8.69), the critical time step for that problem is

∆tcrit =2

√

g/l

Plugging in the numerical values g = 9.81m/s2 and l = 1m yields ∆tcrit =0.63855086 to eight digits. Recall that when the time step was chosen to belarger than this critical value (even just slightly larger) the numerical solutioneventually became unbounded.

In general, the calculation of the critical time step for a system of equationsinvolves the use of an eigenvalue extraction routine to determine the largesteigenvalue of the matrix [M−1 − ω2I]. This can be very time consuming forlarge systems. Fortunately, it turns out that the critical time step can beestimated with good accuracy using the element eigenvalue inequality [19].The element eigenvalue inequality states that the largest eigenvalue of thesystem is bounded by the largest eigenvalue of the element in the mesh with


the smallest characteristic dimensions. Mathematically, the inequality can bewritten as

0 ≤ ωmax

≤ ωmax

e(8.70)

Hence, a slightly conservative estimate for the critical time step is

∆tcrit ≈2

ωmaxe

(8.71)

Example 8.2

As an example of the use of the element eigenvalue inequality for estimatingthe critical time step, consider a mesh composed of three linear-u elementsof length L, 2L, and 3L. From our work in Example 8.1, the largest naturalfrequency for the a unconstrained linear-u element is

ωmax

e=

2√

3c

Le

where c =√

Ee/ρe is the wave speed, and Le is the length of the element. Inthe present three-element model, the largest element natural frequency comesfrom the element with the shortest length, L in this case. Hence, the criticaltime is

∆tcrit ≈2

ωmaxe

=2

2√

3c

L

=L

√3c

(8.72)

We note that the estimate for the critical time step given in equation (8.72)above employs a consistent mass matrix for the computation of the naturalfrequencies. A lumped mass matrix can be obtained using the technique ofrow summing, i.e., each diagonal term in the mass matrix is assigned to bethe sum of that row, and the off-diagonal terms are all assigned to be zero.For the two-node linear-u element, the lumped mass matrix is given as

Mlumped

e=

ρeAeLe

2

[

2 11 2

]

(8.73)

The nonzero natural frequency for the linear-u element with a lumped massmatrix turns out to be ωmax

e= 2c/Le. The critical time step in the present

example for the lumped mass matrix case is then

∆tcrit =2

ωmaxe

= L/c (8.74)


The quantity L/c physically represents the time that it takes for a longitudinalwave to traverse an element of length L. The ratio,

c∆t

L

is referred to as the Courant-Friedrichs-Lewy (CFL) number. Note that whenthe time step is equal to the critical time step, the CFL number is equal to1. In transient finite element analyses, it is typical to begin an analysis bychoosing the CFL number to be one-half.

A way of estimating the critical time step for two- and three-dimensionalproblems is to take the critical time step to be the time it takes for a wave topropagate across the smallest element in the mesh, i.e.,

∆tcrit ≈l

c(8.75)

where, again, c is the wave speed and l is the characteristic length of thesmallest element.

8.3.3 Elastic wave propagation in a one-dimensional bar

To demonstrate the use of the central difference method for obtaining approx-imate solutions to wave propagation problems in solid mechanics, we considera long bar with a constant circular cross section subjected to an impact load-ing at the left end as shown in Figure 8.5. We will take the bar to be 1000mlong with cross-sectional area A = 1m2, and we will take Young’s modulusand mass density to be E = 200GPa and ρ = 7850 kg/m3, respectively. Thedisplacement at the right end of the bar is fixed. In this example we will treatthe problem as one-dimensional so that points lying on any given cross sectioncan displace only in the X direction. The time-dependent pressure loading,P (t), acting at the left end is described by the following equation:

P (t) = σ0H(t) (8.76)

where H(t) is the Heaviside unit step function. In the present example, wewill take σ0 = 50MPa.

X

P (t) = σ0H(t)

FIGURE 8.5

Elastic bar subjected to impact loading.

Ing. Civil

Lápiz

Ing. Civil

Lápiz

Ing. Civil

Lápiz

Ing. Civil

Lápiz

Ing. Civil

Lápiz


u (m)

v (m/s)

a (m/s2)

t (s)0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4

0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4

0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4

-500

0

500

-1

0

1

2

0

0.2

0.4

FIGURE 8.6

Displacement, velocity, and acceleration versus time at the location X =500m (20 equally spaced elements).

Numerical results

In order to study how the spatial discretization of the bar affects the numericalresults, we consider two different meshes composed of 20 and 100 equallyspaced linear-u elements. We begin the numerical analysis by employing alumped mass matrix and a time step size of ∆t = 0.5L/c, where L is theelement length and c =

√

E/ρ is the wave speed. Recall from our work in theprevious section that the critical time step for a linear-u element of length L

with a lumped mass matrix is ∆tcrit = L/c.Numerical results for the two different meshes considered are shown in Fig-

ure 8.6 and Figure 8.7. Here we fix our attention at a point in the middle ofthe bar, X = 500m, and monitor the displacement, velocity, and accelerationat this point as functions of time.

From the upper plot in Figure 8.6, we observe that it takes approximately0.1 s for the wave to hit the point of observation located 500m from the leftend. The time that it takes for the wave to reach this point can be calculatedby taking the length, 500m, and dividing by the wave speed, i.e.,

t =500m

√

(200 × 109 N/m2) / (7850 kg/m3)≈ 0.1 s

Note that the displacement is zero at the point of observation until the wavehits. Subsequently, the displacement increases linearly with time until ap-proximately t = 0.3 s (the time that it takes for the wave to traverse theentire length of the bar, reflect off the fixed end, and then propagate back tothe observation point). Note that it takes approximately 0.2 s for the waveto travel the entire length of the bar. It turns out that the present numerical


u (m)

v (m/s)

a (m/s2)

t (s)0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4

0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4

0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4

-500

0

500

-1

0

1

2

0

0.2

0.4

FIGURE 8.7

Displacement, velocity, and acceleration versus time at the location X =500m (100 equally spaced elements). Lumped-mass results are depicted bythe solid curves. Consistent-mass results are indicated by the dashed curves.

solution for the displacement versus time is in excellent agreement with theanalytical solution to the one-dimensional wave equation with the boundaryconditions described above.

The velocity and acceleration at the point X = 500m are plotted versustime in the lower two plots of Figure 8.6. From the displacement-versus-timecurve, we expect the velocity to be zero until the wave hits, and then changeabruptly to a constant velocity of approximately 1.25m/s. The velocity shouldbecome zero again after the wave reflects off the fixed end and propagatesback to the observation point. The oscillations that appear in the figure aredue to numerical error. These are widely referred to as spurious oscillations.The spurious oscillations are more pronounced in the acceleration-versus-timecurve shown in the lower plot.

The corresponding results for the refined mesh (100 elements) is shownin Figure 8.7. In this plot we have provided numerical results using both alumped and consistent mass matrix. The lumped mass matrix results aredepicted by the solid curves, and the consistent mass matrix results are in-dicated by the dashed curves. It can be observed from the figures that meshrefinement does not remove the spurious oscillations, but the period of the os-cillations does become shorter. We note that decreasing the time step does notremove the oscillations either. Notice also that the consistent versus lumpedmass matrix results for the displacement versus time are virtually indistin-guishable. The use of the consistent mass matrix, however, gives rise to agreater amount of noise, as shown in the velocity and acceleration plots. As a


u (m)

v (m/s)

a (m/s2)

t (s)

×104

0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4

0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4

0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4

-5

0

5

-50

0

50

0

0.2

0.4

FIGURE 8.8

Displacement, velocity, and acceleration versus time at the location X =500m (100 equally spaced elements, ∆t = 1.00001L/c).

general rule of thumb, when employing the central difference method to inte-grate the equations of motion in time, the use of a lumped mass matrix givesbetter accuracy. The use of the lumped mass matrix also improves computa-tional efficiency due to the fact that it is not necessary to solve simultaneousequations at each time step.

For the sake of interest numerical results for the refined mesh using a timestep of ∆t = 1.00001L/c (slightly above the critical time step for the lumpedmass matrix) are shown in Figure 8.8. Notice that the solution is very quicklybecoming unbounded as time increases.

For the coarse mesh composed of 20 linear-u elements, the normal stress(normalized with respect to σ0) versus position along the length of the baris plotted at three instants of time (t = 0.5 s, t = 0.25 s, and t = 0.4 s) inFigure 8.9. Focusing on the upper plot for the case when t = 0.05 s, weobserve that the stress wave has just reached the X = 250m location, andthe magnitude of the normal stress is equal to σ0 for X < 250m. Next,referring to the middle plot when t = 0.25 s, we observe that the stress wavehas reflected off the fixed end and it has begun to propagate back towardthe left end. Note that the amplitude of the stress wave doubles as the wavereflects off the fixed end. In the lower plot when t = 0.4 s, the stress wave hasnearly propagated back to the left end.


σ/σ0

t = .05 s

σ/σ0

t = .25 s

σ/σ0

t = .4 s

X (m)

0 200 400 600 800 1000

0 200 400 600 800 1000

0 200 400 600 800 1000

-3

-2

-1

0

-3

-2

-1

0

-3

-2

-1

0

FIGURE 8.9

Stress wave propagation in a bar.

Numerical results with internal damping

Thus far, in our study of wave propagation in a one-dimensional bar subjectedto impact loading, we have ignored internal damping. In actual structures,internal damping is always present due to, for example, friction generated bythe relative motion of adjacent atoms. The presence of friction causes energyto be dissipated in the form of heat.

A common way of adding damping to the present one-dimensional wavepropagation problem is to augment the constitutive law by a viscous dampingterm. Doing this, the stress-strain law can be written as

σ = E (ǫ + αǫ) (8.77)

where E is Young’s modulus, α is a viscous damping coefficient having unitsof seconds, and ǫ is the time rate of change of extensional strain.

The internal force vector for a single linear-u element is written as

f int

e= Ae

∫

Le

0

BT

eσ dX, (8.78)

where Be is the element B-matrix, i.e.,

Be =1

Le

[

−1 1]

(8.79)

σ is the normal stress, and Ae and Le are the length and cross-sectional area ofthe element, respectively. Using the stress-strain law (8.77), the finite element


approximation for the normal stress in the element can be written as

σe = EeBede + EαeBede (8.80)

where de is the vector containing the nodal displacements, and de is a vec-tor containing the nodal velocities. Next, by substituting the finite elementapproximation (8.80) back into equation (8.78) we get

f int

e=

(

AeEe

∫

Le

0

BT

eBe dX

)

de

+

(

AeEeαe

∫

Le

0

BT

eBe dX

)

de (8.81)

From equation (8.81), we now recognize that the element internal forcevector can be written in terms of the element stiffness matrix as follows:

f int

e= Kede + αeKede (8.82)

Hence, for a one-dimensional wave propagation problem with internal damp-ing of the form given by equation (8.77), the fully assembled semi-discreteequations can be written as

Md + αKd + Kd = F (8.83)

where F is the external force vector, and we have assumed that the dampingcoefficient, α, is the same for each element in the model. The type of dampingleading to the semi-discrete form (8.83) is known as stiffness proportionaldamping.

Substituting the central difference approximation for the nodal accelerationvector at time n into the semi-discrete form (8.83) yields

1

∆t2M (dn+1 − 2dn + dn−1) + αKdn + Kdn = Fn (8.84)

Next, it is necessary to come up with an approximation for the nodal velocityvector that appears on the left-hand side of equation (8.84) above. In orderto keep the method explicit, we employ a backward difference approximationfor the first derivative as follows:

dn ≈dn − dn−1

∆t(8.85)

Finally, plugging in the backward difference approximation (8.85) into equa-tion (8.83) and rearranging terms yields

dn+1 = M−1

∆t2Fn −

(

α∆t + ∆t2)

fn + α∆tfn−1

+ 2dn − dn−1 (8.86)


u (m)

v (m/s)

a (m/s2)

t (s)0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4

0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4

0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4

-50

0

50

100

-1

0

1

2

0

0.2

0.4

FIGURE 8.10

Displacement, velocity, and acceleration versus time at the location X =500m (100 equally spaced elements, α = 0.001 s, and ∆t = 0.5L/c).

where we have let fn = Kdn and fn−1 = Kdn−1. Note that the time in-tegration method that employs a central difference approximation for theacceleration and a backward difference approximation for the velocity leadsto a fully explicit method as long as we employ a lumped mass matrix.

Numerical results with internal damping are shown in Figure 8.10. Herethe displacement, velocity, and acceleration at X = 500m are plotted versustime for the case when α = 0.001 and ∆t = 0.5L/c. Notice that the presenceof a sufficient amount of internal damping smooths the spurious oscillationsthat were observed for the undamped case.

The present algorithm, which employs a central difference approximationfor the second time derivative and a backward difference approximation forthe first derivative, results in a decrease in the critical time step size. Thecritical time step for the damped linear-u element with a lumped mass matrixhas been reported by Belytschko [20]. The result is

∆tcrit =(

√

1 + µ2 − µ

)

L

c(8.87)

where µ = αc/L, and L is the length of the shortest element in the mesh.In the present analysis with L = 10m, α = 0.001, E = 200GPa, and

ρ = 7850 kg/m3, the critical time step (to four digits beyond the decimalplace) is

∆tcrit = 0.6154L

c(8.88)

The numerical results when the time step is chosen to be slightly larger thanthe critical time step are shown in Figure 8.11.


u (m)

v (m/s)

a (m/s2)

t (s)0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4

0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4

0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4

-200

0

200

-1

0

1

2

0

0.2

0.4

FIGURE 8.11

Displacement, velocity, and acceleration versus time at the location X =500m (100 equally spaced elements, α = 0.001 s, and ∆t = 0.617L/c).

8.4 Trapezoidal rule

A popular method for integrating the semi-discrete equations for structuraldynamics applications is the trapezoidal rule. To describe the method, con-sider a generic plot of acceleration versus time as shown in Figure 8.12. Let usrestrict our attention to a small time interval between time n and time n + 1

a an

an+1

∆t

n n + 1

FIGURE 8.12

Trapezoidal rule approximation.


labeled in the figure. The area under the acceleration curve during this timeinterval can be obtained from the fundamental theorem of calculus introducedin Chapter 2, Section 2.4.1, as follows:

∫

n+1

n

a(t) dt =

∫

n+1

n

v dt = vn+1 − vn (8.89)

where vn+1 and vn are the values of the velocity at times n and n+1, respec-tively. Similarly, the area under the velocity curve between times n and n+1is

∫

n+1

n

v(t) dt =

∫

n+1

n

d dt = dn+1 − dn (8.90)

Let us now approximate each integral on the left-hand side of equations(8.89) and (8.90) by the area of a trapezoid (see Figure 8.12). The area of thetrapezoid that approximates the area under the acceleration curve is

∫

n+1

n

a(t) dt ≈∆t

2(an+1 + an) (8.91)

Similarly, the area under the velocity curve can be approximated as∫

n+1

n

v(t) dt ≈∆t

2(vn+1 + vn) (8.92)

Substituting the approximations (8.91) and (8.92) into the respective equa-tions (8.89) and (8.90) yields

vn+1 − vn =∆t

2(an+1 + an)

dn+1 − dn =∆t

2(vn+1 + vn) (8.93)

The final step in the derivation of the trapezoidal rule time integrationscheme is to manipulate equation (8.93) until we get a relationship betweenthe acceleration at time n and the nodal displacement. To do this, we firstconsider the second line in equation (8.93) and solve for vn+1 in terms of theother variables. Doing this gives

vn+1 =2

∆t(dn+1 − dn) − vn (8.94)

Next, substituting the expression (8.94) for vn+1 into the first line in equation(8.93) gives

an+1 =4

∆t2(dn+1 − dn) −

4

∆tvn − an

=4

∆t2

(

dn+1 − dn − ∆tvn −∆t2

4an

)

=4

∆t2

(

dn+1 − dn

)

(8.95)


TABLE 8.3

Trapezoidal rule algorithm.

1. Prescribe initial conditions, i.e., set up dn and an

2. Set up the global mass and stiffness matrices,

M =∑

e

∫

Ωe

ρeNT

eNe dΩ

, K =∑

e

∫

Ωe

BT

eCeBe dΩ

3. Solve for acceleration at time n, an = M−1 (Fn − fn)


4.1 Compute dn = dn + ∆tvn + ∆t2

4an

4.2 Compute the external force vector,

Fn+1 =∑

e

∫

Ωe

NT

ebdΩ +

∫

Γe

NT t∗ dΓ

4.3 Solve for nodal displacements at time n + 1,

dn+1 = K−1

eff

(

Fn+1 + 4

∆t2Mdn

)

, where Keff = 4

∆t2M + K.

4.4 Compute the new velocity, vn+1 = 2

∆t(dn+1 − dn) − vn.

4.5 Compute the new acceleration,

an+1 = 4

∆t2(dn+1 − dn)

4.6 Output results

4.7 Update displacement, velocity and acceleration,dn = dn+1, vn = vn+1, and an = an+1


where

dn = dn + ∆tvn +∆t2

4an (8.96)

For an undamped linear system, the semi-discrete form of the equations ofmotion at time n + 1 can be written as

Mdn+1 + Kdn+1 = Fn+1 (8.97)

where M is the global mass matrix, K is the global stiffness matrix, and Fn+1

is the global external force vector at time n + 1. Plugging in the trapezoidalrule approximation (8.95) into the semi-discrete form (8.97) yields

[

4

∆t2M + K

]

dn+1 = Fn+1 +4

∆t2Mdn (8.98)

A flow chart detailing the computational procedure for the trapezoidal rule isgiven in Table 8.3.

The resulting algorithm for the trapezoidal rule is classified as an implicitmethod, because at each time step we must solve a system of linear algebraic


equations. This is due to the fact that the global stiffness matrix (not diago-nal) must be added to the mass matrix. Solution of linear algebraic equationsduring each time step can be very time consuming. It turns out, however, thatthe trapezoidal rule is unconditionally stable. This means that the numeri-cal solution will remain oscillatory no matter how large the time step. Thefact that the trapezoidal rule is unconditionally stable somewhat offsets thehigh computational cost of solving linear equations, because often sufficientaccuracy in structural dynamics problems can be obtained using relativelylarge time steps. This is fortunate, because structural dynamics calculationsare typically carried out on the order of minutes, unlike wave propagationproblems, which are on the order of seconds or less. It is worth mention-ing here that the trapezoidal rule time-integration scheme is a special case ofNewmark’s beta method [21].

We end this section with two examples in which we demonstrate the use ofthe trapezoidal rule. We will consider strategies for time step selection anddiscuss the accuracy of the numerical results.

Example 8.3

To illustrate the trapezoidal rule time-integration scheme for structural dy-namics problems, we first consider the single-degree-of-freedom system com-posed of a mass, m, attached to two linear (massless) springs with stiffnessk1 and k2 as shown in Figure 8.13.

Let us begin by considering a time step of ∆t = 0.001 s, and we will gothrough a hand calculation to determine the position, velocity, and accelera-tion of the mass at time t = 0.002 s. For the calculation, we will choose the

X

FIGURE 8.13

Spring mass system.


system parameters and initial conditions to be as follows:

m = 1.0 kg

k1 = 1.0N/m

k2 = 2.0N/m

x(t = 0) = 1.0m

x(t = 0) = 1.0m/s

In order to obtain the semi-discrete equation of motion, we will modelthe springs with two massless linear-u elements having stiffness k1 and k2,respectively. The element stiffness matrices are then

K1 =

[

k1 −k1

−k1 k1

]

; K2 =

[

k2 −k2

−k2 k2

]

.

If the connectivity list is that given in Table 8.4, then, upon assembly of theelement stiffness matrices given above, the global stiffness matrix turns outto be

K =

k1 −k1 0−k1 k1 + k2 −k2

0 −k2 k2

(8.99)

Because we are taking each spring to be massless, the mass, m, of the blockis lumped at global node 2. The global mass matrix is then simply

M =

0 0 00 m 00 0 0

(8.100)

The fixed end conditions in the problem are enforced by eliminating degreesof freedom 1 and 3. After doing this, we get the following equation of motion:

mX + (k1 + k2)X = 0 (8.101)

where X is the equilibrium position of the block measured from the fixed endas shown in Figure 8.13.

The next step in the solution procedure is to substitute the trapezoidal ruleapproximations into the equation of motion (8.101). The trapezoidal ruleapproximations for the present spring mass example can be written as

Xn+1 =4

∆t2

(

Xn+1 − Xn

)

(8.102)

TABLE 8.4

Connectivity list.

Element 1 2

1 1 22 2 3


TABLE 8.5

Numerical results up to t = 0.002 s using ∆t = 0.001 s.

t (s) X (m) X (m/s) X (m/s2) X (m)

0.000 1.00000000000 0.00000000000 -3.00000000000 1.001000750000.001 1.00099999925 0.99999850000 -3.00299999800 1.001999247000.002 1.00199849550 0.99699400226 -3.00599548630 1.00299473800

where

Xn = Xn + ∆tXn +∆t2

4Xn (8.103)

After inserting (8.102) and (8.103) into equation (8.101), we get the followingincremental equation:

(k1 + k2 +4

∆t2m)Xn+1 = −

4

∆t2Xn (8.104)

After imposing the initial conditions,

X(t = 0) = 1.0m

X(t = 0) = 1.0m/s (8.105)

and solving the incremental equation (8.104), we report the numerical resultsup to t = 0.002 s in Table 8.5.

The natural frequency of the spring mass system can be obtained by solvingthe following equation:

(k1 + k2) − ω2m = 0. (8.106)

Solving for ω then yields

ω =

√

k1 + k2

m(8.107)

Substituting the numerical values (k1 = 1.0N/m, k2 = 2.0N/m, and m =1.0 kg) into equation (8.107) above yields ω = 1.732 rad/s. The period, T ,of the spring mass system represents the time that it takes the mass to gothrough one complete cycle. In the present example, the period is then

T =2π

ω≈ 3.63 s (8.108)

The numerical results for position, velocity, and acceleration of the blockup to t = 4.0 s are shown in Figure 8.14. The solid line in the plots is for thecase when ∆t = T/30. The dashed line represents a time step selection of∆t = T/5. We note that the numerical results remain virtually unchanged inthe present example when the time step is chosen to be smaller than T/30.For larger time steps, however, we notice that we are still capturing the correctamplitude of oscillation, but the period is getting longer. This type of error


d(t)

v(t)

a(t)

t (s)

0 0.5 1 1.5 2 2.5 3 3.5 4

0 0.5 1 1.5 2 2.5 3 3.5 4

0 0.5 1 1.5 2 2.5 3 3.5 4

-5

0

5

-2

0

2

-2

0

2

FIGURE 8.14

Numerical solution ∆t = T/30 (solid line) and ∆t = T/5 (dashed line).

is known as period elongation error. As a general rule of thumb, if the lowestnatural frequency is known, a time step of ∆t = T/30 generally gives goodresults in structural dynamics problems that are dominated by the first fewlowest natural frequencies.

8.4.1 Dynamic response of a cantilever beam

In this section we revisit the cantilever beam problem that was studied inChapter 6, Section 6.7. Here we will investigate the dynamic response of thebeam by employing the trapezoidal rule time-integration scheme covered inthe previous sections.

Employing the same geometry and material properties that were chosen inSection 6.7, we will take the length of the beam to be L = 3m and assumethat it has a 20mm× 200mm rectangular cross section. We will assume thatthe beam is linearly elastic and isotropic, with Young’s modulus E = 200GPaand Poisson’s ratio ν = 0.3. In the present dynamic analysis, the beam willbe loaded by a concentrated force at the right end, as shown in Figure 8.15.The entire magnitude of the load, P = 10 kN, is applied suddenly at timet = 0 s and then remains throughout the dynamic event.

Several factors must be addressed before we can begin the analysis. Inparticular, we must select a finite element mesh that is sufficiently refined,and we must choose a time step size that is small enough to achieve goodaccuracy, yet not too small so as to avoid excessive computational cost. Aftermaking these decisions, we will explore the end deflection versus time usingseveral different time step sizes.


P = 10 kN

FIGURE 8.15

Dynamic response of a cantilever beam.

Mesh selection

In structural dynamics problems, the dynamic response is dominated by thefirst few lowest natural frequencies. Hence, it is necessary to make sure thatthe finite element mesh for the cantilever beam is sufficiently refined to ac-curately capture the low frequencies. Recall that the natural frequencies of astructure are calculated by solving the following equation:

∣

∣M−1K − ω2I

∣

∣ = 0 (8.109)

where M and K are the global mass and stiffness matrices, respectively, andI is an identity matrix. We note that K and M are the reduced matrices thataccount for the nodal constraints.

Finite element results for the lowest natural frequency of the beam forseveral different meshes are depicted in Figure 8.16. The meshes consideredwere the 2×5, 2×20, 2×40, and 4×40 rectangular grids shown in Figure 6.18.In the analysis, four-node (plane stress) isoparametric quadrilaterals wereemployed with a consistent mass matrix.

Employing Euler-Bernoulli beam theory, the analytical result for the lowestnatural frequency can be expressed as

ωanalytical

1=

1.8752

2πL2

√

EIZ

ρA(8.110)

where L is the length of the beam, A is the cross-sectional area, E is Young’smodulus, and IZ is the moment of inertia of the beam’s cross section aboutthe neutral axis (see, e.g., [22] for details). Plugging in the numerical valuesfor the parameters in the present problem into equation (8.110) above gives

ωanalytical

1≈ 18.12Hz. The finite element solution using the 4×40 grid yields

ω1 ≈ 18.6Hz, about 2.6 % above the analytical result.


ω (Hz)

mesh density2 × 5 2 × 20 2 × 40 4 × 40

20

25

30

35

40

FIGURE 8.16

Lowest natural frequency, ω1, versus mesh refinement.

Time step selection

As a general rule of thumb, reasonable accuracy for the dynamic responsecan be obtained by making sure that the period of oscillation associated withthe lowest natural frequency is approximately 30 times the time step size. Inthe present example, the period, T , of oscillation is the inverse of the lowestnatural frequency reported in Hz. Hence,

T =1

18.6s ≈ 0.054 s (8.111)

According to the general rule stated above, a time step size of

∆t =0.054

30≈ 0.0018 s (8.112)

should then be adequate.The finite element approximation for the end deflection, v, normalized with

respect to the static deflection, vstatic, is plotted versus time in Figure 8.17.Here, the numerical results are presented for three different time steps (∆t =0.002 s, ∆t = 0.001 s, and ∆t = 0.01 s). The dynamic response for ∆t = 0.002is represented by the solid curve with circular data markers. The responsefor ∆t = 0.001 is depicted by the dashed curve with triangular data markers,and the largest time step considered is shown by the dashed-dotted line withsquare data markers.

If we estimate the period from the solid curve (time between successivepeaks) we get slightly greater than 0.05 s, which is in good agreement withour initial prediction using 18.6Hz for the lowest natural frequency. As the


t (s)

v/v

static

0 0.02 0.04 0.06 0.08 0.1 0.12 0.140

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2

FIGURE 8.17

Dynamic end deflection of cantilever beam for ∆t = 0.002 s, ∆t = 0.001 s, and∆t = 0.01 s.

time step size is increased, the amplitude of oscillation remains the same,but the period gets longer. This type of numerical error is known as periodelongation error.

We mention as a closing remark that the maximum end deflection obtainedduring the dynamic response is approximately twice the static deflection. Thisphenomenon is known as dynamic amplification.

8.5 Unsteady heat conduction

In this section we derive the partial differential equation that governs tran-sient, i.e., time-dependent, heat conduction. The equation is widely referredto as the the unsteady heat equation. Like all the previous partial differentialequations derived throughout the book, the unsteady heat equation comesfrom a balance law — an energy balance.

Referring to a fixed control volume, conservation of energy requires thatthe amount of energy flowing into the control volume per unit time, minusthe amount of energy flowing out of the volume per unit time, must equal thechange of internal energy inside the volume per unit time.

To begin the derivation, consider the control volume shown in Figure 8.18.As shown in the figure, the heat flux vector is prescribed over a portion of thesurface, Γq. A continuous heat source distribution, S(X1,X2,X3) = S(X),


S(X)

U = ρcpT

Ω

Γ

q

Γq

ΓT

FIGURE 8.18

Solid body subjected to internal heat source and surface flux distribution.

acts throughout the volume. The change in internal energy per unit time,U(X), at some point, X, inside the volume is given as

U = ρcpT (8.113)

where ρ is the mass density at point X, cp is the specific heat, and T is therate of change of temperature with respect to time. Note that the specificheat of a substance is the amount of energy per unit mass required to raisethe temperature at a point by 1 C.

The energy balance statement can now be written as

∫

Ω

S(X) dΩ −

∫

Γq

n · qdΓ =

∫

Ω

ρcpT dΩ (8.114)

Throughout the rest of this section, we will assume that the temperaturerange of interest is small enough to justify the assumption that the massdensity and specific heat are constants. In other words, we will assume thatthese parameters do not depend on temperature or time.

The next step in the derivation of the unsteady heat equation is to expressall of the integrals in the balance statement (8.114) above as volume integrals.To do this, we need to employ the divergence theorem on the integral involvingthe surface flux distribution. Doing this yields

∫

Γq

n · qdΓ =

∫

Ω

∇ · qdΩ (8.115)

and hence,∫

Ω

S − ∇ · q − ρcpTdΩ = 0 (8.116)

Conservation of energy requires that the statement (8.116) above be truefor all control subvolumes within the control volume. Hence, the integrand,


as discussed in Chapter 4, must vanish at all points, X, within the controlvolume. The unsteady heat equation then emerges as

−∇ · q + S = ρcpT (8.117)

Alternatively, the partial differential equation (8.117) can be written in com-ponent form as

−qi,i + S = ρcpT (8.118)

where summation is implied, and the dummy subscript i ranges, in general,from 1 to 3.

As discussed in Chapter 5, the heat flux vector, q, is related to the tem-perature gradient, ∇T , through Fourier’s law. Recall that for isotropic heatconduction, this constitutive relationship can be written as

q = −k∇T (8.119)

where k is the thermal conductivity. Substituting Fourier’s law into the dif-ferential equation (8.117) yields

k∇ · ∇T + S = ρcpT or

k∇2T + S = ρcpT (8.120)

The last line in equation (8.120) above can be written in component form as

kT,ii + S = ρcpT (8.121)

8.5.1 Weak form of unsteady heat equation

In this section we will give a brief derivation of the weak form for the caseof isotropic, unsteady heat conduction. The derivation follows our work inChapter 5. To begin, we simply multiply the differential equation (8.121) bya scalar test function w that depends only on the spatial variables. The testfunction is defined to be sufficiently smooth inside the domain of interest andis zero at any points on the boundary where the temperature is specified.Mathematically speaking, the test function is defined as

w(X) =

at least C0 : in Ω

0 : on ΓT

Multiplying the differential equation (8.121) by the test function, w, andintegrating over the volume Ω yields

k

∫

Ω

T,iiw dΩ +

∫

Ω

Sw dΩ = ρcp

∫

Ω

Tw dΩ (8.122)

Because the first integral on the left-hand side of equation (8.122) involvessecond partial derivatives with respect to the spatial variables, and because


we desire to use continuous piecewise trial functions for the temperature dis-tribution, it is necessary to integrate the first integral on the left-hand side ofequation (8.122) by parts. Noting that

(T,iw),i

= T,iiw + T,iw,i (8.123)

we obtain

−k

∫

Ω

T,iw,i dΩ + k

∫

Γq

niT,iw dΓ +

∫

Ω

Sw dΩ = ρcp

∫

Ω

Tw dΩ, (8.124)

where the second integral on the left-hand side of equation (8.124) is a con-sequence of the divergence theorem. Here, ni are the components of the unitoutward normal to the boundary Γ. This integral only needs to be evaluatedover the portion of the boundary where the heat flux is specified (naturalboundary), because the test function is defined to be zero over the portion ofthe boundary where the temperature is specified (essential boundary). Theweak form (8.124) above can equivalently be written as

k

∫

Ω

T,iw,i dΩ + ρcp

∫

Ω

Tw dΩ = −

∫

Γq

q∗nw dΓ +

∫

Ω

Sw dΩ, (8.125)

where we have used the fact that the quantity, −kniT,i = q∗n, is the normal

component of the heat flux vector specified on Γq.


Starting with the weak form (8.125), let us now express this statement overa single element inside Ω, and then plug in the usual finite element approx-imations. The finite element approximation for the temperature field withinan element is written as

Te = NeTe (8.126)

where Ne is the 1× nen matrix containing the shape functions and Te is thenen × 1 vector containing the nodal values of the temperature field. Here,nen refers to the number of nodes per element. Similarly, the finite elementapproximation for the time rate of change of the temperature field, T , overan element is written as

˙Te = NeTe (8.127)

The finite element approximation for the temperature gradient is obtained bysimply differentiating equation (8.126) with respect to the spatial variables.Doing this yields

∇Te = BeTe (8.128)

where, assuming that we are working in three dimensions, Be is the 3 ×

nen B-matrix. Next, employing the same finite element shape functions to


interpolate the test function over an element we obtain

we = Newe

∇we = Bewe (8.129)

where, again, we is a nen × 1 vector containing the nodal values of the testfunction.

Substituting the finite element approximations above into the weak formover a single element yields

wT

e

(

ke

∫

Ωe

BT

eBe dΩ

)

Te +

(

ρecpe

∫

Ωe

NT

eNe dΩ

)

Te

= wT

e

−

∫

Γq

NT

eq∗n

dΓ +

∫

Ω

NT

eS dΩ

(8.130)

Finally, letting

Ke = ke

∫

Ωe

BT

eBe dΩ

Ce = ρecpe

∫

Ωe

NT

eNe dΩ

fe = −

∫

Γq

NT

eq∗n

dΓ +

∫

Ω

NT

eS dΩ (8.131)

and invoking the arbitrariness of the vector, we, we obtain

KeTe + CeTe = fe (8.132)

Here in equation (8.132) above, the matrix, Ke, is the conductance matrix,Ce is referred to as the capacitance matrix, and fe is a vector containing theamount of energy per unit time entering the element at each node. We willsimply refer to the vector fe as the equivalent nodal load vector. Note thesimilarity between the conductance and capacitance matrices and the stiffnessand mass matrices for solid mechanics applications.

After assembling the finite element equations using the standard assemblyprocess we get

KT + CT = F (8.133)

Notice that equation (8.133) above is a system of first-order differential equa-tions. The equations are semi-discrete, because they have been discretizedin space, but they are still continuous in time. The final step to getting ap-proximate transient solutions to the unsteady heat equation is to employ anappropriate approximation for the time derivative in equation (8.133), yield-ing a system of linear equations that can be solved incrementally at eachinstant of time. Treatment of this topic is given in the next section.


8.5.3 Backward difference method

The backward difference method is a widely used method for integrating theunsteady heat equation in time. To describe the method, let us begin bywriting down the semi-discrete equations at some instant of time, n, at whichthe nodal temperatures are known as follows:

KTn + CTn = Fn (8.134)

The backward difference approximation for the time rate of change of thenodal temperatures is given as

Tn ≈Tn − Tn−1

∆t(8.135)

In other words, equation (8.135) above states that the slope of the temperature-versus-time curve is approximately equal to the slope of the straight line drawnbetween the temperature at time n− 1 and the temperature at time n. Afterplugging in the backward difference approximation into the semi-discrete formwe get

[

K +1

∆tC

]

Tn = Fn +1

∆tCTn−1 (8.136)

We observe from equation (8.136) that the use of the backward differenceapproximation leads to an implicit method that involves the solution of si-multaneous equations at each time step. This is due to the fact that thematrix on the left-hand side is, unfortunately, not diagonal. On the otherhand, the method turns out to be unconditionally stable, which means thatthe solution will remain bounded no matter how large a time step is selected.The stability of the backward difference method is discussed briefly in thenext section.

8.5.4 Stability of the backward difference method

To assess the stability of the backward difference method described in theprevious section, it is only necessary to study the behavior of the followingsingle homogeneous equation:

T + βT = 0 (8.137)

where β is a real positive constant. Due to the physical nature of heat con-duction, we seek a solution to equation (8.137) that decays with time, i.e., weseek a transient solution that eventually reaches steady state once the timedependence of the external heat sources is removed. To this end, let us assumean exponentially decaying solution of the form

T (t) = e−αt (8.138)


TABLE 8.6

Backward difference method algorithm.

1. Prescribe initial conditions, i.e., set up Tn−1

2. Set up the global conductance and capacitance matrices,

C =∑

e

ρecpe

∫

Ωe

NT

eNe dΩ

, K =∑

e

ke

∫

Ωe

BT

eBe dΩ


3.1 Compute the external load vector,

Fn =∑

e

−∫

Γe

NT

eq∗n

dΩ +∫

Ωe

NT S dΩ

3.2 Solve for nodal temperatures at time n,Tn = K−1

eff

(

Fn + 1

∆tCTn−1

)

, where Keff = K + 1

∆tC.

3.3 Output results

3.4 Update nodal temperatures,Tn−1 = Tn


where α is a real positive constant.The first step in the stability analysis is to write down equation (8.137) at

time n, and then substitute the backward difference approximation into thatresulting expression. Doing this gives

Tn + β

(

Tn − Tn−1

∆t

)

= 0 (8.139)

or(

1 +β

∆t

)

Tn −β

∆tTn−1 = 0 (8.140)

Let us now assume that the solution to the difference equation (8.140) abovehas the form

Tn = e−αn∆t = φn (8.141)

Before moving forward, we mention that in the present analysis, stabilityrequires that the function φ continually decrease with increasing n. Next,substituting (8.141) into (8.140) yields

(

1 +β

∆t

)

φn−

β

∆tφ

n−1 = 0 (8.142)

If we now divide both sides of equation (8.142) by φn−1 we get(

1 +β

∆t

)

φ −β

∆t= 0 (8.143)

orφ =

α

∆t + α(8.144)


From equation (8.144) above, we conclude that for the method to be stable,we must have

α

∆t + α< 1 (8.145)

or∆t > 0 (8.146)

Hence, the backward difference approach yields a time integration scheme thatis unconditionally stable. A flowchart illustrating the finite element procedurefor solving the unsteady heat equation using the backward difference timeintegration scheme is provided in Table 8.6.

8.5.5 Unsteady heat conduction in a composite cylinder

Let us now revisit the problem discussed in Chapter 6, Section 6.9, of along composite cylinder composed of a copper core surrounded by a ceramiccoating. Recall that in Chapter 6, we obtained the steady-state temperaturedistribution and resulting thermal stress distribution in the composite. Inthe present section, we will investigate the transient temperature distributionthroughout the cylinder. We will pay particular attention to the time thatit takes for the temperature distribution to reach steady state. We will alsostudy the effect of time step selection on the transient response.

The cross section of the composite cylinder is shown in Figure 8.19. Theradius, r, of the inner copper core and the thickness, t, of the ceramic coatingwere chosen such that r = t = 0.1m. The thermal properties employed in thepresent analysis are given in Table 8.7. The composite cylinder is assumedto be long compared to the outer radius, and the ends are assumed to be

Copper

Ceramic

q∗n

= h(T − T∞)

X

Y

FIGURE 8.19

Copper/ceramic composite cylinder. The finite element mesh of the quartermodel is composed of 200 four-node quadrilateral elements.


TABLE 8.7

Thermal properties for copper and ceramic.

Material cp ( J

kg·C) ρ

(

kg

m3

)

k ( J

m·C·s)

Copper 385 8950 400

Ceramic 800 2000 1

insulated so that no heat can flow along the axial direction of the cylinder.We recognize that the solution to the present problem could be obtained bysolving a one-dimensional unsteady heat equation. However, for illustrativepurposes, we will go ahead and treat the problem as two-dimensional andemploy the mesh shown in Figure 8.19 composed of 200 planar, four-nodequadrilateral elements. Due to symmetry, only one-quarter of the mesh ismodeled. This is accomplished by enforcing a zero normal flux conditionalong the edges X = 0 and Y = 0. At time t = 0 the entire composite is heldat a constant temperature of 20 C. Subsequently, the inner core is suddenlyheated to a uniform temperature of 200 C. Throughout the transient heatconduction event, air at 20 C flows over the outer surface of the cylinder,causing heat transfer by forced convection. As in our study in Chapter 6, thefilm coefficient in the present analysis was taken to be h = 10 J/(m2 · C · s).Note that it is not necessary to model the inner core in the present problem.The effect of the inner core is included in the analysis by prescribing thetemperature at the nodes located at r = 0.1m to be 200 C at the beginning

T C

time (hours)0 1 2 3 4 5

20

30

40

50

60

70

80

90

100

FIGURE 8.20

Temperature-versus-time curve at the location r = 0.2m for ∆t = 100 s,∆t = 1000 s, and ∆t = 2000 s.


of the first time step. Numerical results are shown in Figure 8.20. Here, thetemperature at r = 0.2m (at the outer surface of the ceramic) is plotted versustime in hours. The circular data markers represent the numerical result forthe case when ∆t = 100 s. The square data markers represent the case when∆t = 1000 s, and the diamond data markers are for the case when ∆t = 2000 s.We observe from the figure, that it takes approximately five hours for thetemperature at the outer surface of the cylinder to reach steady state. Thesteady-state temperature is slightly above 90 C, which agrees with the resultthat we obtained in Chapter 6, Section 6.9, see Figure 6.31(b). We note thatthe numerical results do not change appreciably for time steps smaller than100 s. For the larger time steps considered, as expected, all results eventuallyconverge to the correct steady-state temperature, but some detail about thetransient event, especially near the beginning of the time interval considered,is lost.

8.6 Problems

Problem 8.1

Determine the consistent mass matrix for a two-node linear-u element hav-ing cross-sectional area, A, Young’s modulus E, and mass density, ρ. Also,determine the natural frequencies and normal modes.

Problem 8.2

Consider a one-dimensional wave propagation problem discretized with threetwo-node linear-u elements of length l, 2l, and 3l. Use a lumped mass matrixand estimate the critical time step size ∆tcrit if the central difference methodis to be employed.

Problem 8.3

Prove that the trapezoidal rule time-integration scheme is unconditionallystable.

Problem 8.4

Consider the problem of a one-dimensional elastic bar of length 2L, constantcross-sectional area A, Young’s modulus E, and mass density ρ.

(a) Model the problem with one quadratic-u element. Using a lumped

mass matrix, obtain a finite element approximation for the lowest


1 2

34

(0,0) (0.1,0)

(0.1,0.1)(0,0.1)

FIGURE 8.21

Problem 8.6: One-element mesh for plane unsteady heat conduction.

three natural frequencies of the bar.

(b) Now model the problem with two linear-u elements. Again, using alumped mass matrix, obtain a finite element approximation for thethree lowest natural frequencies. Compare these frequencies withthose obtained in part (a).

Problem 8.5

Consider a mesh composed of one four-node quadrilateral for plane, unsteadyheat conduction as shown in Figure 8.21. The nodal coordinates in metersare labeled in the figure. The initial temperature distribution in the elementis uniform (T = 20 C). Suddenly, at time t = 0, the temperature at nodes 1and 4 is prescribed to be 200 C and is held fixed at these nodes thereafter.A forced convection condition is applied along edge 2–3.

(a) Derive the 4 × 4 conductance matrix.

(b) Derive the 4 × 4 capacitance matrix.

(c) Obtain the 2 × 2 semi-discrete system of equations.

(d) Employ a backward difference approximation to obtain the nodaltemperatures at nodes 2 and 3 at time t = 60 s. Take∆t = 30 s, ρ = 2000 kg/m3, and cp = 800 J/(kg ·C).

9

Small-Strain Plasticity

Creative power, is that receptive attitude of expectancy which makes a mold

into which the plastic and as yet undifferentiated substance can flow and take

the desired form.—Thomas Troward

Knowledge of the classical theory of plasticity is essential for the design andanalysis of solids and structures. In the present chapter, we provide an intro-duction to the topic of rate-independent, small-strain plasticity. The subjectalso serves as an excellent avenue through which to present a standard finiteelement procedure for solving nonlinear problems in solid mechanics.

We begin the chapter in the next section by defining some fundamentalconcepts that are needed in the study of small-strain plasticity theory. Wethen move ahead and talk about the basic ingredients in the theory: yieldcondition, hardening rule, and flow rule.

The basic finite element procedure for solving small-strain plasticity prob-lems is based on Newton’s method, in which the resulting linearized system ofalgebraic equations is solved in an iterative manner. Given the state of stressand deformation in a solid body at some time, n, after the body is subjectedto a change in external loading, the goal is to find the new state of stress andstrain in the body at time n + 1, such that equilibrium, yield condition, flowrule, and hardening rule are all satisfied.

After the finite element procedure is described, we present some examplesillustrating how it is implemented in practice. We conclude the chapter bystudying the elastic-plastic response of a thick-walled cylinder subjected toan internal pressure.

9.1 Basic concepts

Idealized stress-versus-strain response

As the external loads acting on a ductile solid are gradually increased, eventu-ally permanent plastic deformation takes place. In other words, upon removalof the external loads, the body does not return to its original shape. Under

303


ǫpǫe

A

B

E

Et

σ

ǫ

FIGURE 9.1

Idealized stress-versus-strain response.

such conditions, the stress-versus-strain response of the material is nonlinear,even in the small-strain regime.

An idealized uniaxial stress-versus-strain curve for a ductile solid is shownin Figure 9.1. As shown in the figure, the initial stress-versus-strain responseis linear with slope, E, Young’s modulus. As the applied load is subsequentlyincreased, eventually point A labeled in the figure is reached, defining theonset of plastic (permanent) deformation. This point is referred to as theyield point. As the load is increased further to point B, plastic deformationcontinues to takes place. The slope of the stress-versus-strain curve afterinitial yield is widely referred to as the tangent modulus Et.

Let us now investigate what happens when the load is gradually removedfrom state B. From experimental observations, during unloading, the stress-versus-strain response proceeds along the path indicated by the dashed linein the figure. Notice that the dashed line is parallel to the initial elasticresponse. When the specimen is finally unloaded back to a state of zero stress,permanent plastic deformation remains. The strain at point B is referred to asthe total strain, ǫ. The permanent plastic strain that remains after unloadingis denoted as ǫp. The elastic strain that is recovered during unloading iscalled ǫe. Under small-strain conditions, we can decompose the total straininto elastic and plastic parts as follows:

ǫ = ǫe + ǫ

p (9.1)

Returning to point B on the stress-strain curve, now suppose that the stressis increased by some small amount dσ. The resulting strain increment, dǫ, iscomposed of both elastic and plastic parts, dǫe and dǫp. From Hooke’s law,

Small-Strain Plasticity 305

the stress increment is related to the elastic strain increment by

dσ = Edǫe

= E (dǫ − dǫp) (9.2)

Alternatively, the stress increment can be related to the tangent modulus asfollows:

dσ = Etdǫ

(9.3)

By combining equations (9.2) and (9.3) we get a relationship between thestress increment, dσ, and the plastic strain increment dǫp. Doing this yields

dǫ =dσ

Et⇒

dσ = E

(

dσ

Et− dǫ

p

)

⇒

dσ =1

1

Et−

1

E

dǫp (9.4)

The last line in equation (9.4) above is commonly written as

dσ = H dǫp (9.5)

The parameter H in equation (9.5) above is called the hardening parameter.It represents the slope of the uniaxial stress-versus-plastic-strain curve. Whenthe hardening parameter is zero at all strain levels beyond initial yield, thematerial response is said to be perfectly plastic. Equation (9.5) above is auniaxial representation of a hardening rule. The hardening rule describes howthe plastic strain changes due to a change in the stress.

Multiaxial state of stress and strain

Having discussed the idealized stress-versus-strain response of ductile mate-rials, let us now focus on the general three-dimensional state of stress andstrain at a point in a solid. Because plastic deformation occurs due to rela-tive slip between crystallographic planes, in order to adequately describe thisphysical phenomenon, it is necessary to decompose the Cauchy stress tensorinto deviatoric (shear) and volumetric (mean) parts.

The mean stress, σm, is defined as one-third of the trace of the Cauchystress tensor, i.e., σm = 1/3 tr(σ). As an example, if a body is subjectedto all-around hydrostatic pressure of magnitude p, then the state of stress atthat point would be

σ =

p 0 00 p 00 0 p

(9.6)


For this case the mean stress is σm = 1/3(3p) = p. The mean stress can beexpressed in many different ways as follows:

σm =1

3σkk

=1

3tr(σ)

=1

3σ : I

=1

3(σ1 + σ2 + σ3) (9.7)

where I is the second-order identity tensor, and σ1, σ2, and σ3 are the threeprincipal stresses. The deviatoric stress components, σ′

ij, are obtained by

subtracting the mean stress from the Cauchy stress components as follows:

σ′ij

= σij −1

3σkkδij or

σ′ = σ − σm (9.8)

One of the major objectives in any plasticity theory is to come up witha multiaxial relationship between the stress increments and the total strainincrements during plastic loading. Such a relationship can be written as

dσij = Dep

ijkldǫkl (9.9)

where Dep

ijklare the components of a fourth-order tensor called the elastoplas-

tic tangent.

9.2 Yield condition

Arguably the main ingredient in any mathematical theory of plasticity isthe yield condition. The yield condition is also commonly referred to as ayield function or yield surface. In general, a yield condition depends on thesix stress components, the yield stress, and other variables that for now wewill refer to as history variables. History variables depend on the way inwhich the material was loaded, accumulated permanent strain, etc. Givena set of history variables and yield stress, the yield condition describes thecombinations of the Cauchy stress components at a point that give rise toinitial yield (or subsequent yielding) at that point.

A general form of the yield condition can be written as

f(σij , σy, α1, . . . , αn) = 0 (9.10)


(a) (b) (c)

τ

FIGURE 9.2

Dependence of yield on principal stresses.

where σy is the yield stress (which in general is not constant) and α1, . . . , αn

are the history variables.

For a given state of material deformation described by the yield stress andset of history variables, the yield condition can be thought of as a level surfacein six-dimensional stress space. The level surface in stress space would corre-spond to the locus of points, σij , that satisfy equation (9.10). Unfortunatelywe are incapable of illustrating such a six-dimensional space.

An important yield condition for ductile metals is the von Mises yield con-dition developed by Richard von Mises in 1913. One of the main assumptionsin the derivation of the von Mises yield condition is that the material behavioris isotropic. This means, roughly speaking, that both the initial elastic be-havior and the initial yield behavior do not depend on direction. Under thisassumption, the yield condition only depends on the principal stresses (as wellas the yield stress and history variables). To illustrate this concept, considera two-dimensional stress element subjected to a state of pure shear as shownin Figure 9.2(a). The magnitude of the shear stress is σ12 = τ . The principalorientation of this stress element is obtained by rotating the element coun-terclockwise through an angle of 45 degrees as shown in Figure 9.2(b). Theprincipal stresses are σ1 = τ and σ2 = −τ . We emphasize that the two statesof stress, (a) and (b), are equivalent. Now consider another stress elementthat is subjected to only normal stresses, σ11 = −τ and σ22 = τ , as shown inFigure 9.2(c). Clearly, the state of stress (c) is not equivalent to the state ofstress (a), because even though the principal stresses are equal, the principalorientation is different. Now suppose that the state of stress (a) causes initialyield. The state of stress (c) may or may not cause the material to yield, de-pending on whether or not the material possesses a preferred orientation. Asan example, the distance between the centers of closest neighboring circles inthe figure could represent crystal lattice spacing. Looking carefully, one willobserve that the lattice spacing in the direction of σ1 in (b) is slightly largerthan the lattice spacing in the direction of σ1 in state (c). Because of this,just because state (a), or equivalently (b), is sufficient to cause initial yield,does not necessarily mean that state (c) will also cause initial yield. On theother hand, if the material is isotropic, with no preferred orientation, then the


material cannot distinguish the difference between states (a) and (c) as far asinitial yield is concerned, and the two states can be considered equivalent.

Having assumed isotropy, the yield condition can now be written as

f(σ1, σ2, σ3, σy, α1, . . . , αn) = 0 (9.11)

Hence, for a given yield stress and set of history variables, the yield conditioncan be represented as a level surface in three-dimensional principal stressspace. This space, known as the Haigh-Westergaard stress space, will bediscussed in detail in the next section.

9.2.1 Dependence on hydrostatic pressure

Recall from our study in Chapter 4, that given a state of stress, σij , theprincipal stresses, σ1, σ2, and σ3, are obtained by expanding the followingdeterminant equation

|σij − λδij | = 0

or∣

∣

∣

∣

∣

∣

σ11 − λ σ12 σ13

σ12 σ22 − λ σ23

σ13 σ23 σ33 − λ

∣

∣

∣

∣

∣

∣

= 0

Doing this yields

−λ3 + (σ11 + σ22 + σ33)λ

2

+ (σ2

12− σ11σ22 − σ11σ33 + σ

2

13+ σ

2

23)λ

+ +σ11σ22σ33 − σ2

13σ22 − σ

2

12σ33 − σ11σ

2

23

+ 2σ12σ13σ23 = 0 (9.12)

Recalling the definitions of the stress invariants introduced in Chapter 4, andafter some algebraic manipulations, we can express equation (9.12) as

−λ3 + I1λ

2 + I2λ + I3 (9.13)

where

I1 = tr(σ)

I2 =1

2

(

σ : σ − tr(σ)2)

I3 = detσ (9.14)

are the stress invariants. It is worth mentioning here that if the principalstresses are known, the stress invariants can be written in terms of the prin-cipal stresses as

I1 = σ1 + σ2 + σ3

I2 = − (σ1σ2 + σ1σ3 + σ2σ3)

I3 = σ1σ2σ3 (9.15)


Referring back to the yield condition, if the yield condition depends onlyon the principal stresses, it can also be expressed as a function of the threeinvariants, i.e.,

f(I1, I2, I3, σy, α1, . . . , αn) = 0 (9.16)

If the yield behavior does not depend on the mean stress, then the yieldcondition is independent of I1. So for a fixed set of history variables σy,α1, . . . , αn, the yield condition then reduces to a level surface in three-dimensionalprincipal stress space that only depends on I2 and I3.

If the yield condition does not depend on the mean stress, σm = 1/3I1, wecan equivalently rewrite the yield condition in terms of the deviatoric stresscomponents. To do this, recall that the deviatoric stress components areobtained by taking the Cauchy stress components and subtracting the meanstress, i.e.,

σ′ij

= σij −1

3σkkδij (9.17)

Next, if we consider a general state of stress in the principal orientation andsubtract off the mean stress we get

σ1 − σm 0 00 σ2 − σm 00 0 σ3 − σm

=1

3

2σ1 − σ2 − σ3 0 00 −σ1 + 2σ2 − σ3 00 0 −σ1 − σ2 + 2σ3

=

σ′1

0 00 σ′

20

0 0 σ′3

(9.18)

Equation (9.18) above reveals that the deviatoric stress tensor has the sameprincipal directions as the stress tensor.

The three invariants of the state of stress on the right-hand side of equation(9.18) are

I1 = 0

I2 =1

2

(

σ′21

+ σ′22

+ σ′23

)

I3 = σ′1σ′2σ′3

(9.19)

To distinguish between the invariants of the deviatoric stress tensor and thestress tensor, the invariants of the deviatoric stress tensor are usually labeledJ1, J2, and J3, i.e.,

J1 = tr(σ′) = 0

J2 =1

2σ′ij

σ′ij

J3 = detσ′ = σ

′1σ′2σ′3

(9.20)


Because the first invariant of the deviatoric stress tensor is zero, the yieldcondition can be expressed as a function of only J2 and J3, along with thehistory variables and yield stress.

9.2.2 The von Mises yield condition

A widely used yield condition employed to model the behavior of ductilemetals is the von Mises yield condition. The von Mises criterion ignoresthe third invariant of the deviatoric stress tensor and assumes that the yieldfunction only depends on J2 and the history variables. The von Mises yieldcondition can be written as

J2 − k2 = 0 (9.21)

where k is a scalar quantity that can, in general, depend on the history vari-ables.

When the yield behavior is assumed to be isotropic, a general state of stressat a point can be written in terms of the principal stresses as

σ1 0 00 σ2 00 0 σ3

(9.22)

without regard to the principal directions. The deviatoric part of the stress(9.22) is obtained by subtracting off the mean stress as follows:

σ′ = σ −

1

3σkkδij

=

2σ1 − σ2 − σ3 0 00 2σ2 − σ1 − σ3 00 0 2σ3 − σ1 − σ2

(9.23)

The quantity J2 then turns out to be

J2 =1

3

(

σ2

1+ σ

2

2+ σ

2

3− σ1σ2 − σ1σ3 − σ2σ3

)

(9.24)

and the von Mises yield condition can be written in terms of the three principalstresses as

(

σ2

1+ σ

2

2+ σ

2

3− σ1σ2 − σ1σ3 − σ2σ3

)

+ 3k2 = 0 (9.25)

The von Mises effective stress, σ, is defined as

σ =√

3J2 (9.26)

Using the definition (9.26), the von Mises yield condition is often written as

σ −√

3k = 0 (9.27)

The physical significance of the parameter σ will be discussed in the nextsection.


σ1

σ3

σ2

octahedral plane

σ3 = 0

σ1 = σ2 = σ3

FIGURE 9.3

Von Mises yield surface.

9.2.3 Geometry of the von Mises yield surface

The geometry of the von Mises yield surface in principal stress space is shownin Figure 9.3. As shown in the figure, the yield surface is a circular cylinderwhose axis of revolution is the line σ1 = σ2 = σ3. Planes perpendicularto this axis of revolution are referred to as octahedral planes. Note thatthe hydrostatic stress or mean stress is the same at all points lying on anoctahedral plane. The plane σ3 = 0 represents states of biaxial stress. Theintersection of the plane σ3 = 0 and the yield surface forms an ellipse. Hencefor states of biaxial stress, the von Mises yield condition can be described byan ellipse lying in the σ1−σ2 plane.

Another view of the von Mises yield surface is provided in Figure 9.4. Thisview illustrates more clearly the intersection of the yield surface and the oc-tahedral plane. It turns out that the intersection of the yield surface andthe octahedral plane is a circle of radius r =

√2k. To see this, consider the

octahedral plane that passes through the point σ1 = σ2 = σ3 = 0. This planeis called the π-plane. The state of stress at points that lie on the π plane ispurely deviatoric. Focusing only on points that lie in the π-plane, the dis-tance, r, between the origin and a point that lies on the yield surface can becomputed using the distance formula, i.e.,

r =√

σ′21

+ σ′22

+ σ′23

=√

2J2 (9.28)


σ1

σ2σ3

r

σ3 = 0

σ1 = σ2 = σ3 = 0

FIGURE 9.4

Von Mises yield surface.

From the definition of J2, we can now see that

r2

2= J2 = k

2 (9.29)

and hence,r =

√2k (9.30)

9.2.4 Simple experiments

Consider the von Mises yield condition

J2 − k2 = 0 (9.31)

where k is a constant that depends on the material properties and, in general,the material history variables. The parameter, k, can be found by perform-ing mechanical experiments as discussed below. Let us first consider a solidsubjected to uniaxial extension as shown in Figure 9.5. The test specimenshown in the figure is commonly referred to as a dog-bone specimen. The testis designed so that in the section of the specimen with the smaller diameter,the state of stress and deformation is homogeneous, i.e., the state of stressand strain in the narrow section of the specimen does not vary with position.Under such conditions, the stress matrix at a point in the test section can be


σ

FIGURE 9.5

Dog-bone test specimen subjected to uniaxial tension.

written as

σ =

σ 0 00 0 00 0 0

(9.32)

where σ is the normal stress in the direction of the applied load. The deviatoricstress is obtained by subtracting the mean stress as follows:

σ′ = σ − σmI

=

2

3σ 0 00 − 1

3σ 0

0 0 − 1

3σ

(9.33)

The value of J2 at a point in the test section then turns out to be

J2 =1

2σ′ij

σ′ij

=1

2(4/9 + 1/9 + 1/9)σ2

=σ2

3(9.34)

In the present uniaxial tension test, yield initiates when the normal stressreaches the yield stress, i.e., when σ = σy. Plugging in this condition into theexpression (9.34) for J2 yields

1

3σ

2

y− k

2 = 0 ⇒ k =1√

3σy (9.35)

The von Mises yield condition can now be written as

J2 −1

3σ

2

y= 0 (9.36)

It is convenient to recast equation (9.36) above by multiplying both sides by3, moving σ2

yover to the right-hand side, and then taking the square root of

both sides. Doing this yields√

3J2 − σy = 0 (9.37)


Next, letting√

3J2 = σ, the yield conditions can be written as

σ − σy = 0 (9.38)

The quantity, σ, in the von Mises yield condition (9.38) is called the vonMises effective stress. The yield condition in this form is convenient to usein practice, because, for example, one can assess whether a component canbe expected to yield under a given set of external loading conditions by per-forming a linear finite element analysis and then observing a contour plot ofthe von Mises effective stress. If σ ≥ σy at a particular location, then thematerial can be expected to yield at that location.

Now suppose we perform an experiment with a solid circular bar undertorsion. The state of stress at a point on the surface of the bar is one of pureshear. The stress matrix at this point can be written as

0 τ 0τ 0 00 0 0

(9.39)

where τ is the magnitude of the shear stress. Note that the state of stressabove is purely deviatoric. Hence,

J2 = 2τ2 (9.40)

The bar under torsion begins to yield when the shear stress, τ , reaches theinitial yield stress in shear, τy. The yield condition then becomes

τ2

y− k

2 = 0 (9.41)

Solving for the parameter k yields

k = τy (9.42)

In order for the parameter k to be a material parameter, we should get thesame value of k, independent of the test performed. Comparing equations(9.35) and (9.42), we conclude that in order for k to be a material parameter,we must have

τy =σy√

3(9.43)

In other words, the initial yield stress in shear must be related to the initialyield stress in tension through equation (9.43). Fortunately, the relation-ship (9.43) above has been borne out for ductile metals through experimen-tation. As an example, the initial yield stress in tension for ASTM-A36 steelis approximately 36, 000 psi. The initial yield stress in shear is approximately21, 000 psi, which is in very good agreement with the relation (9.43) (see, e.g.,[22]).


9.3 Flow and hardening rules

Now that we have discussed the notion of a yield condition, in this sectionwe discuss the widely used flow rules and hardening rules in small-strain in-cremental plasticity theory. Recall that a flow rule provides a relationshipbetween the plastic strain increments and the stress increments. A hardeningrule describes how the yield stress of a material evolves with plastic straining.In this section we will only discuss the most commonly used flow rules andhardening rules. In particular, we will discuss the associated and nonassoci-ated flow rules and the isotropic kinematic hardening rules.

Flow rules

In the development of the classical small-strain plasticity theory, von Misesproposed the existence of a plastic potential function, Q(σij). He proposedthat the plastic strain increments are related to the plastic potential functionas follows:

dǫp

ij= dλ

∂Q

∂σij

(9.44)

where dλ is a scalar constant of proportionality. In general, the function Q

must be viewed as a function of the six independent stress components in six-dimensional stress space. The local of points, σij , in this space that renderthe function Q constant define a level surface.

The partial derivatives of the function Q with respect to the stress com-ponents can be viewed as the components of the gradient of the functionQ. Recall, from our work in Chapter 2, that the gradient of a level surfaceyields a vector that is perpendicular to that surface. Hence, the quantity,ddQσij , is a vector in six-dimensional space that is normal to the level surfaceQ(σij) = constant. Note that equation (9.44) implies that the plastic strainincrements are normal to the level surface Q = constant.

A flow rule is said to be associated if the plastic potential function, Q(σij),is equal to the yield function F (σij). Under these circumstances, the flow rulecan be written as

dǫp

ij= dλ

∂F

∂σij

(9.45)

Hence, for the case of an associated flow rule, the plastic strain increments arenormal to the yield surface. If a plastic potential function is not equal to theyield function, the flow rule is said to be nonassociated. From experimentalobservations, the plastic deformation of metals can be characterized quitewell by an associated flow rule. For porous materials such as rocks, concrete,and soil, a nonassoiciated flow rule provides a better representation of plasticdeformation.


The physical basis for the associated flow rule comes from the postulate of

maximum plastic dissipation. This postulate was proposed independently byvon Mises [23], Taylor [24], and Hill [25]. For hardening and perfectly plasticmaterials, the consequences of the postulate of maximum plastic resistanceare that:

1. The plastic strain increments are normal to the yield surface.

2. The yield surface is convex.

Roughly speaking, a hardening material is one whose tangent modulus in-creases with plastic deformation. A nonhardening material (also known asperfectly plastic) is one whose tangent modulus is zero during plastic defor-mation. Hence, for hardening and perfectly plastic materials, an associatedflow rule of the form given by equation (9.45) is an appropriate choice.

Hardening rules

When a ductile material is subjected to tensile loading above the initial yieldpoint, if the specimen is subsequently unloaded and again reloaded in tension,it is well known that the subsequent yield point is higher than the initial yieldpoint. This behavior is known as work hardening. The final ingredient neededin a small-strain plasticity theory is the hardening rule. In this section we willbriefly discuss the isotropic and kinematic hardening rules.

In the isotropic hardening model, it is assumed that the material is isotropicin some stress-free reference state and that the degree of hardening depends onthe complete history of plastic deformation from this stress-free state. It is alsoassumed that the yield surface uniformly expands during plastic deformationwithout changing its shape.

The von Mises yield condition with an isotropic hardening rule can bewritten as

σ − σy(α) = 0 (9.46)

where σ is the von Mises effective stress, and α is a variable that is a measureof the degree of hardening at a given state of stress. The parameter α isknown as a hardening parameter. Two commonly used hardening parametersfor metals are the total plastic work per unit volume, and the total equivalentplastic strain. The total plastic work per unit volume, Wp, is defined as

Wp =

∫

ǫp

ij

0

σijdǫp

ij(9.47)

Similarly, the total equivalent plastic strain, ǫp, is defined as

ǫp =

∫

ǫp

ij

0

dǫp, (9.48)


where

dǫp =

√

2

3dǫ

p

ijdǫ

p

ij(9.49)

The total equivalent plastic strain is defined such that the parameter, ǫp, re-duces to ǫp for the case of uniaxial tensile loading. To show this it is necessaryto use the fact that plastic deformation is volume preserving, and hence,

tr(dǫp) = 0 ⇒ dǫ

p

11+ dǫ

p

2+ dǫ

p

3= 0 (9.50)

During uniaxial tensile loading of an isotropic material we have

dǫp

22= dǫ

p

33= −

1

2dǫ

p

11(9.51)

We will leave it to the reader to work out the remainder of the details.In the kinematic hardening model, it is assumed that during plastic loading

the yield surface translates in stress space while maintaining its shape. Thegeneral form of the yield surface with kinematic hardening takes the form

F (σ − α) − k = 0 (9.52)

Here in equation (9.52), σ is the Cauchy stress tensor, and α is a second-ordertensor called the back stress. The evolution equation for α can be obtained,for example, using Ziegler’s model [26]. The parameter k in equation (9.52) isa material parameter that governs the size of the yield surface. It is typicallytaken to be a function of either the total plastic work per unit volume, or thetotal accumulated plastic strain.

When a ductile material is loaded in tension above the initial yield point,some degree of work hardening occurs. When the work-hardened material isthen loaded in compression until the compressive yield point, one finds thatthe compressive yield stress is lower than the compressive yield stress beforethe work hardening took place. This behavior is known as the Bauchinger ef-fect. The kinematic work-hardening model takes into account the Bauchingereffect, and hence, it more accurately models the hardening behavior of ductilemetals. On the other hand, the isotropic hardening model is still widely usedbecause of its simplicity.

9.4 Derivation of the elastoplastic tangent

In this section, we derive the elastoplastic tangent matrix for a material whoseplastic behavior can be described with a von Mises yield surface, an associatedflow rule, and an isotropic hardening rule as described in the previous section.The elastoplastic tangent is one of the key ingredients needed in the finite


element implementation of any plasticity theory. In the derivation, we willbegin by writing down the following form of the von Mises yield condition:

F (σ, ǫp) = σ

2− σ

2

y(ǫp) = 0 (9.53)

where σ is the von Mises effective stress and ǫp is the total equivalent plasticstrain.

An associated flow rule requires that the direction of the plastic strainincrement be normal to the yield surface. This statement can be expressedmathematically as

dǫp

ij= dλ

∂F

∂σij

(9.54)

where dλ is a constant of proportionality. The partial derivative of the yieldfunction with respect to the stress components can be obtained in severalsteps. The first step is to employ the chain rule as follows:

∂F

∂σkl

=∂σ2

∂σkl

= 2σ∂σ

∂σkl

(9.55)

The second step is to recall the definition of the von Mises effective stress andtake the partial derivative of σ2 with respect to σkl. Doing this yields

∂σ2

∂σkl

=∂

∂σkl

(

3

2σ′ij

σ′ij

)

=3

2

∂σ′ij

∂σkl

σ′ij

+ σ′ij

∂σ′ij

∂σkl

= 3∂σ′

ij

∂σkl

σ′ij

(9.56)

Next, equating the last line in equation (9.55) with the last line in (9.56) weobtain the intermediate results

2σ∂σ

∂σkl

= 3∂σ′

ij

∂σkl

σ′ij

(9.57)

and hence,∂σ

∂σkl

=3

2σ

∂σ′ij

∂σkl

σ′ij

(9.58)

The last step is to evaluate the partial derivatives of the deviatoric stresscomponents with respect to the Cauchy stress components. This can easilybe accomplished by first expressing σ′

ijin terms of σkl as follows:

σ′ij

= σij −1

3σkkδij

= σklδikδlj −1

3σklδlkδij (9.59)


where δij is the Kronecker delta, and then simply differentiating the last linewith respect to σkl. Doing this gives

∂σ′ij

∂σkl

= δikδlj −1

3δlkδij (9.60)

Finally, by substituting the expression (9.60) into (9.59) we get

∂σ

∂σkl

=3

2σσ′ij

(

δikδlj −1

3δlkδij

)

=3

2σ

(

σ′kl−

1

3δlkσ

′jj

)

=3

2σσ′kl

(9.61)

where last line in equation (9.61) above follows from the fact that the trace ofthe deviatoric stress tensor is zero, i.e. σ′

jj= σ′

kk= 0. Substituting equation

(9.61) into (9.55) yields∂F

∂σkl

= 3σ′kl

(9.62)

Having obtained the partial derivatives of the yield function with respectto the stress components, we now go ahead and substitute the result (9.62)into the expression (9.54) for the plastic strain increments. Doing this we get

dǫp

ij= dλ

′σ′ij

(9.63)

where we have used the notation dλ′ = 3dλ.The effective plastic strain increment, dǫp, is defined as

dǫp =

√

2

3dǫ

p

ijdǫ

p

ij(9.64)

From this definition, we can express the constant, dλ′, in terms of the effectiveplastic strain increment by performing a contraction operation as follows:

dǫp

ijdǫ

p

ij= dλ

′2σ′ij

σ′ij⇒

(dǫp)

2=

2

3dλ

′2σ′ij

σ′ij

(9.65)

From the definition of the von Mises effective stress, i.e., σ2 = 3/2σ′ij

σ′ij

, theresult is

dλ′ =

3

2σdǫ

p (9.66)

In order to proceed further, we must now invoke what is called the consis-tency condition. The consistency condition states that during plastic loading,the change in the yield function, dF , with respect to its arguments must be


zero. The yield function is a function of the von Mises effective stress, which,in turn, depends on the deviatoric stress components. The yield function alsodepends on the yield stress, which is a function of the effective plastic strain,ǫp. Hence, the change in the yield function is written as

dF =∂F

∂σ

∂σ

∂σ′ij

dσ′ij

+∂F

∂σy

dσy

dǫpdǫ

p = 0 (9.67)

where it is understood that the derivatives in equation (9.67) above are to beevaluated at some known state of stress that lies on the yield surface. Next,writing the yield function as

F = σ2− σ

2

y(ǫp),

the derivatives on the right-hand side of equation (9.67) turn out to be

∂F

∂σ= 2σ;

∂σ

∂σ′ij

=3

2σσ′ij

;∂F

∂σy

= −2σy;∂σy

∂ǫp= H (9.68)

where the quantity, H, in the last line of (9.68) above is the hardening param-eter, i.e., the slope of the uniaxial stress versus plastic strain curve. Pluggingin the derivatives defined in (9.68) into the consistency condition (9.67) yields

3σ′ij

dσ′ij− 2σy(ǫp)Hdǫ

p = 0 (9.69)

Solving equation (9.69) above for dǫp gives

dǫp =

3

2Hσy(ǫp)σ′ij

dσ′ij

(9.70)

Finally, an expression for the plastic strain increments in terms of the devi-atoric stress increments can be obtained by substituting equation (9.66) intoequation (9.63) and then using the relation (9.70). Doing this gives

dǫp

ij=

9

4σ2yH

σ′kl

dσ′kl

σ′ij

(9.71)

where we have used the fact that σ = σy for states of stress that lie on theyield surface.

Recall that the goal of this section is to derive the components of the elasto-plastic tangent. In other words, we seek a relationship between the stressincrements and the total strain increments of the form

dσij = Dep

ijkldǫkl (9.72)

where Dep

ijklare the components of the fourth-order elastoplastic tangent. De-

composing the total strain increments into elastic and plastic parts, the stress


increments can be expressed in terms of the fourth-order elasticity tensorintroduced in Chapter 4 and the plastic strain increments as follows:

dσij = Cijkldǫe

kl

= Cijkl dǫkl − dǫp

kl

= Cijkl

dǫkl −1

HPmndσ

′mn

Pkl

(9.73)

where we have employed equation (9.71) along with the definition

Pkl =3

2σy

σ′kl

(9.74)

The components Pkl are often referred to as the components of the plasticflow direction tensor. Next, we use the identity

Pmndσ′mn

= Pmndσmn (9.75)

to obtain

dσij = Cijkl

dǫkl −1

HPmndσmnPkl

⇒

[

dσij +1

HCijklPmnPkldσmn

]

= Cijkldǫkl ⇒

dσmn

[

δimδjn +1

HCijklPmnPkl

]

= Cijkldǫkl (9.76)

The identity (9.75) above follows from equation (9.60), i.e.,

Pmndσ′mn

= Pmn

∂σ′mn

∂σkl

dσkl

= Pmn

(

δmkδln −1

3δlkδmn

)

dσkl

= Pmn

(

dσmn −1

3dσkkδmn

)

= Pmndσmn −1

3Pmmdσkk

= Pmndσmn (9.77)

The last line in equation (9.77) is due to the fact that the trace of the plasticflow direction tensor is zero, i.e., Pmm = 0. Next, letting

rij =1

HCijklPkl (9.78)

equation (9.76) above can be recast as

dσmn δimδjn + rijPmn = Cijkldǫkl (9.79)


Proceeding further, we notice that equation (9.79) can be written in tensornotation as

A : dσ = C : dǫ (9.80)

where A is a fourth-order tensor with components

Aijmn = δimδjn + rijPmn (9.81)

Let us now perform a contraction operation to both sides of equation (9.80)above as follows:

B : (A : dσ) = B : (C : dǫ) ⇒

(B : A) : dσ = B : (C : dǫ) (9.82)

The contraction operation between two fourth-order tensors yields a second-order tensor. The components of the tensor B : A are obtained as follows:

(Bijkleiejekel) : (Arsmneresemen) = BijklArsmnδksδlreiejemen (9.83)

Hence,B : A

ijmn= BijklAklmn (9.84)

The second line in equation (9.82) can now be written in component form as

BijklAklmndσmn = BijmnCmnkldǫkl (9.85)

From equation (9.85) we now recognize that the elastoplastic tangent can bereadily obtained if we can find a fourth-order tensor, B, such that

BijklAklmn = δimδjn (9.86)

Once the components of B are obtained, the components of the elastoplastictangent are then

Dep

ijkl= BijmnDmnkl (9.87)

Next, let us assume that the components of B are a linear combination ofthe components of A. Thus,

Bijkl = δikδjl + αrijPkl (9.88)

where α is an unknown constant, and rij is defined in equation (9.78). Ex-panding the product BijklAklmn gives

BijklAklmn = (δikδjl + αrijPkl) (δkmδln + rklPmn)

= δimδjn + rijPmn + αrijrklPklPmn (9.89)

For the resulting expression (9.89) above to have the desired form, we musthave

(1 + α + αrpqPpq) = 0 (9.90)


Solving equation (9.94) above for α yields

α =−1

1 + rpqPpq

(9.91)

and hence,

Bijkl = δikδjl −1

1 + rpqPpq

rijPkl (9.92)

Finally, by substituting equation (9.92) into equation (9.87) we get

BijmnCmnkl = (δimδjn + αrijPmn) Cmnkl

=

Cijkl −1

1 + rpqPpq

CmnklrijPmn

(9.93)

Hence, the elastoplastic tangent for von Mises plasticity with isotropic hard-ening is

Dep

ijkl=

Cijkl −1

1 + rpqPpq

Cmnkl rijPmn

(9.94)

The evaluation of the elastoplastic tangent matrix at a point in a solid insome known equilibrium configuration at time n requires knowledge of Young’smodulus, Poisson’s ratio, the hardening parameter H, and the current yieldstress, which is a function of the total accumulated plastic strain ǫp. Theprocess of evaluating the elastoplastic tangent is described in the followingexample.

Example 9.1

The implementation of equation (9.94) above looks intimidating. Actually,given a state of stress and the total accumulated plastic strain, it is not thatdifficult to compute the components of the elastoplastic tangent. To demon-strate the process, suppose that the uniaxial stress-strain curve for the mate-rial of interest can be represented by the Ramberg-Osgood model [27]. TheRamberg-Osgood equation can be written as

ǫ =σ

E+

σ0α

E

(

σ

σ0

)m

(9.95)

where E is Young’s modulus (the initial slope of the stress-strain curve), andσ0, α, and m, are material parameters. In the present example, the materialparameters are taken to be those that are typical for an aluminum alloy.They are given in Table 9.1. The resulting stress-versus-strain curve is shownin Figure 9.6. Recall that the hardening parameter, H, is the slope of theuniaxial stress-versus-plastic-strain curve. From our work in Section 9.1, itcan be written in terms of Young’s modulus and the tangent modulus asfollows:

H =1

1

Et+

1

E

(9.96)


0 0.01 0.02 0.030

10,000

20,000

30,000

40,000

50,000

σ (psi)

ǫ

FIGURE 9.6

Ramberg-Osgood stress-strain curve.

where the tangent modulus, Et, is the slope of the uniaxial stress-strain curve.The quantity Et can be obtained by differentiating both sides of equation(9.95) with respect to the total strain ǫ. Doing this yields

dǫ

dǫ= 1 =

1

E

dσ

dǫ+

αm

E

(

σ

σ0

)m−1dσ

dǫ⇒

Et =

dσ

dǫ=

1

1

E+

αm

E

(

σ

σ0

)m−1(9.97)

Now that we have an expression for the hardening parameter, let us nowgo ahead and assume that the following state of plane stress exists at somepoint in a solid body:

σ =

30, 000 −10, 000 0−10, 000 40, 000 0

0 0 0

psi. (9.98)

We will assume that the state of stress above was achieved by subjectingthe solid to monotonically increasing proportional loading. This means that,

TABLE 9.1

Ramberg-Osgood materialparameters.

σ0 (psi) α m E (psi) ν

38, 000 3

710 10 × 106 0.22


starting from some zero state of stress, the load is increased in such a way thatthe ratios of the stress components remain constant during loading. Havingsaid this, the accumulated effective plastic strain associated with this levelof applied load is obtained by first computing the von Mises effective stress,and then substituting that quantity into the stress-strain law to get the totalstrain. The plastic strain can then be found once the total strain is known.

We now recall from the previous section that the incremental stress-strainrelation for the case of a von Mises yield surface and isotropic hardening rulecan be written as

dσij =

Cijkl −1

1 + rmnPmn

CmnklrijPmn

dǫkl (9.99)

where

Pij =3

2σσ′ij

are the components of the plastic flow direction tensor, and

rij =1

HCijklPkl

Next, substituting the expression

Cijkl = λδijδkl + 2µδikδjl

into equation (9.99) and simplifying, we obtain

dσij = λδij dǫkk + 2µdǫij −2µ

1 + rpqPpq

rijPkldǫkl (9.100)

where we have used the fact that the trace of the plastic flow direction tensoris zero, i.e., Pmm = 0.

It is now convenient to express equation (9.100) in matrix notation. Forplane problems, it is necessary to define the following 4 × 1 vectors:

dσ =

dσ11

dσ22

dσ33

dσ12

; dǫ =

dǫ11

dǫ22

dǫ33

2dǫ12

P =

P11

P22

P33

2P12

; P =

P11

P22

P33

P12

; r =

r11

r22

r33

r12

(9.101)

Three-dimensional problems pose no added difficulty. The vectors σ, ǫ, P,P, and r would be 6 × 1 in that case. Having defined the vectors in (9.101)


TABLE 9.2

Effective strain quantities.

ǫ ǫe ǫp

0.004 0.00672 0.00272

above, we leave it to the reader to verify that equation (9.100) can be writtenin matrix notation as follows:

dσ =

[

C −2µ

1 + PT rrPT

]

dǫ (9.102)

where C is a 4 × 4 elasticity matrix given as

C =

λ + 2µ λ λ 0λ λ + 2µ λ 0λ λ λ + 2µ 00 0 0 µ

(9.103)

and the quantities λ and µ are Lame’s constants defined in Chapter 4, i.e.,

λ =Eν

(1 + ν)(1 − 2ν); µ =

E

2(1 + ν)

To get numerical values for the 4×4 elastoplastic tangent matrix defined inequation (9.102), let us now proceed as follows. The deviatoric stress can beobtained for the state of stress defined in (9.98) by subtracting off the meanstress. Doing this yields

σ′ =

6, 667 −10, 000 0−10, 000 16, 667 0

0 0 23, 333

psi (9.104)

Next, recalling that σ =√

3/2σ′ij

σ′ij

, the von Mises effective stress turns out

to be σ = 40, 000 psi. To get the total accumulated plastic strain, we first needto determine the elastic strain. The elastic strain is simply the effective stressdivided by Young’s modulus, i.e., ǫe = σ/E. The total strain is then foundby plugging in the effective stress into the Ramberg-Osgood equation. Doingthis gives the total accumulated plastic strain values (to five digits beyondthe decimal point) reported in Table 9.2. Finally, the hardening parameter isobtained by substituting the effective stress into equation (9.96). Doing thisyields Et = 1.28 × 106 psi and H = 1.14 × 106 psi. Carrying out the requiredmatrix multiplications in equation (9.102) gives the following elastoplastictangent matrix:

Dep

4 × 4=

11104227. 2438553. 4314363. 468953.

2438553. 9462892. 5955698. 1172382.

4314363. 5955698. 7587082. −1641334.

468953. 1172382. −1641334. 3394932.

psi (9.105)


9.5 Finite element implementation

Once the elastoplastic tangent matrix for the plasticity model of interest isknown, the finite element implementation is relatively straightforward. Theprocess begins by writing down the discretized finite element equations thatequate the internal nodal force vector with the external nodal force vector atsome particular instant of time n. Because the stress-strain relationship isnonlinear, the resulting equations are nonlinear. The equations can be lin-earized, however, by expanding the nonlinear equations in a two-term Taylorseries using the procedure described in Chapter 2. The resulting linearizedsystem of equations can then be solved in an iterative manner using New-ton’s method. One of the tricky steps in the process involves what is calleda stress update scheme. After each iteration of Newton’s method, once thenodal displacement increments are found, it is necessary to compute the stresscomponents within an element. In this chapter we will describe a popular andeffective stress update scheme called the radial return method [28].

We begin by writing down the discretized system of equations as follows:

∑

e

∫

Ωe

BT

eσe dΩ

n

=∑

e

∫

Γte

NT

et∗ dΓ

n

(9.106)

where, for the sake of simplicity, we have neglected body forces. The quantity,t∗, is the specified traction vector at points of the boundary, Γte

, of an element,and n represents some discrete instant of time. Alternatively, equation (9.106)above can be simply written as

f int

n= fext

n(9.107)

It is worth mentioning here that equation (9.106) above is not completelydiscretized, because the stress components σe are still continuous functionsof the spatial variables within an element.

Let us now assume that the equilibrium configuration of the solid body isknown at time n, and we wish to find a new equilibrium configuration due toa change in the external nodal load vector. In other words, we wish to findthe displacement, strain, and stress fields within the body when the externalforce vector is changed from fext

nto some known value fext

n+1. The system of

equations to be solved can then be written as

∑

e

∫

Ωe

BT

eσe dΩ

n+1

=∑

e

∫

Γte

NT

et∗ dΓ

n+1

(9.108)

Equation (9.108) above is nonlinear, because the unknown stress field at timen+1 is a nonlinear function of the strain field. Note that the stress field is the


only nonlinear function in the equation, because the B-matrix for an elementjust depends on the derivatives of the shape functions with respect to thenodal coordinates of the body in some known (fixed) reference configuration.

We can linearize the system of equations by approximating the stress fieldas a two-term Taylor series expansion as follows:

σen+1≈ σen

+∂σe

∂ǫe

∣

∣

∣

∣

n

∆ǫen(9.109)

where it is understood that all the quantities on the right-hand side of equation(9.109) are evaluated at time n. The partial derivative of the element stressvector with respect to the element strain vector evaluated at time n gives thenst×nst elastoplastic tangent matrix, Dep

n. The order of Dep

ndepends on the

number of stress and strain components that are relevant to the problem ofinterest. For example, Dep

nwould be 3×3 for the case of plane stress, and 6×6

for fully three-dimensional problems. The quantity ∆ǫen= ǫen+1

−ǫen

is the vector containing the strain increments. The strain increment vectorcan be written in terms of nodal displacement increments by invoking theusual relationship between strain and displacement as follows:

∆ǫen= Be ∆den

(9.110)

and hence,σen+1

≈ σen+ Dep

eBe∆den

(9.111)

The next step is to substitute the linearized expression (9.111) for the stressvector back into equation (9.108). Doing this gives

∑

e

∫

Ω

BT

eDep

eBe dΩ

n

∆den=

∑

e

∫

Γte

NT

et∗ dΓ

n+1

−∑

e

∫

Ω

BT

eσe dΩ

n

(9.112)

The fully assembled global system of equations can now be written as

Ktan

n∆dn = fext

n+1− f int

n(9.113)

where Ktan

nis the tangent stiffness matrix (Jacobian matrix), fext

n+1is the global

external load vector at time n + 1, and f int

nis the global vector containing

the internal nodal forces evaluated at time n. These matrices and vectors aresummarized in Table 9.3.

As discussed in Chapter 2, Section 2.5, Newton’s method involves iteratingupon the linearized system of equations (9.113) until convergence is achieved.The equation to be iterated upon can be written as

Ki

n+1∆di

n= Fn+1 − f i

n+1(9.114)


TABLE 9.3

Finite element matrices and vectors.

Ktan

nfext

n+1f int

n

∑

e

∫

Ωe

BT

eDep

eBe dΩ

n

∑

e

∫

Γte

NT

et∗ dΓ

n+1

∑

e

∫

Ωe

BT

eσe dΩ

n

where we have used K to denote the tangent stiffness matrix, F to denotethe external force vector, and f to identify the internal force vector. Here inequation (9.114), i is the iteration counter, i.e., i = 0, 1, 2, . . . , niter, whereniter is the total number of iterations to be performed. It is understood thatwhen the counter is equal to zero, the “zeroth” iteration, the matrices andvectors are to be evaluated at time n, i.e., K0

n+1= Kn etc.

9.5.1 Convergence criteria

In practice, while iterating upon equation (9.114) above, there are two con-vergence criteria that are commonly employed. The first is an out-of-balanceforce criterion, and the second is a displacement-based criterion.

To describe the first criterion, we begin by defining the residual, or out-of-balance, force vector R. During a typical iteration, i, the residual vector isdefined as the difference between the external force vector at time n + 1 andthe internal force vector evaluated at the beginning of the ith iteration, i.e.,

Ri

n+1= Fn+1 − f i

n+1(9.115)

Iteration continues until the norm (magnitude) of the residual vector is smallcompared to a user-defined tolerance. The norm of the residual vector isdefined as

‖R‖ =√

RkRk (9.116)

where summation is implied, and k goes from 1 to the number of equationsin the global system (9.114). In practice, it is advantageous to use a relativeresidual criterion whereby iteration continues until the residual evaluated atthe beginning of the ith iteration is small compared to the residual computedat the beginning of some previous iteration. For example, the relative residualcriterion could be stated as

‖Ri

n+1‖

‖Ri−1

n+1‖≤ ǫ (9.117)

where ǫ is a user-defined tolerance, typically taken to be a small value like1×10−6, and we have normalized the residual at the beginning of the ith iter-ation with respect to the residual computed at the beginning of the previousiteration.

The displacement-based criterion is very similar to the out-of-balance forcecriterion, except that the iteration continues until the norm of the global


displacement increment vector is small compared to a user-defined tolerance.Hence, in a displacement-based criterion, iteration proceeds until, e.g.,

‖di

n+1‖

‖di−1

n+1‖≤ ǫ (9.118)

In the displacement-based criterion, the norm of ∆de is computed at the endof a typical iteration i, after the solution to the linear system of equations isobtained. The norm, for example, can then be compared to the same normcomputed at the end of the previous iteration.

Whether we are using the out-of-balance force criterion or the displacement-based criterion, when convergence is finally achieved, either the process isstopped, or the load can be changed (say increased), and the iterations canthen begin again for the next load step. When a small-strain plasticity anal-ysis goes well, convergence is typically achieved in two to four iterations. Onthe other hand, when the Newton’s method strategy fails to converge duringa particular load step, it is often necessary to reduce the size of the load stepand take a larger number of load steps to get to the the final desired load. Thetechnique of reducing the size of the load step in order to achieve convergenceis akin to making a new initial guess in Newton’s method that is hopefullybetter than the previous guess. A flowchart of the incremental iterative pro-cedure based on Newton’s method for solving small-strain plasticity problemsis presented in Table 9.4.

TABLE 9.4

Algorithm for small-strain plasticity

1. Compute the internal force vector f i

n+1

2. Compute the residual vector Ri

n+1

3. Check for convergence. Is, e.g.,‖Ri

n+1‖

‖Ri−1

n+1‖≤ ǫ?

4. If convergence is not achieved, then go to step 5

5. Compute the tangent stiffness matrix Ki

n+1

6. Solve the linear system of equations Ki

n+1∆di

n= Ri

n+1

7. Loop over the elements7.1 Compute the strain field in each element7.2 Compute the stress field in each element

8. If desired, increment the load, and repeat the process starting fromstep 1 above


9.5.2 Radial return stress update scheme

In the nonlinear elastic-plastic finite element procedure outlined in the pre-vious section, one of the crucial steps is the stress update scheme after thenodal displacement increments have been obtained at the end of a particularNewton’s method iteration. In this section we describe an effective and widelyused procedure called the radial return method [28].

To begin, suppose that the state of stress and strain in the body are knownat some time n. We will assume that the state of stress at time n lies on theyield surface. Given the total strain field in the body at time n + 1, the goalof the radial return method is to find the state of stress at time n + 1 suchthat the yield condition, flow rule, and hardening rule are all satisfied.

The first step in the formulation is to decompose the unknown stress σn+1

into deviatoric and hydrostatic parts as follows:

σn+1 = σ′n+1

+1

3tr (σn+1) I

= σ′n+1

+ κ tr(

ǫe

n+1

)

I

= σ′n+1

+ κ tr (ǫn+1) I (9.119)

Here in equation (9.119), ǫe

n+1refers to the elastic part of the total strain

tensor at time n + 1, I is the second-order identity tensor, and κ is the bulkmodulus. The bulk modulus is defined as the constant of proportionalitybetween the mean stress and the volumetric strain. This relationship canbe obtained by writing down Hooke’s law for isotropic and linearly elasticmaterial behavior as follows:

σij = 2µǫe

ij+ λǫ

e

kkδij ⇒

σkk = (2µ + 3λ) ǫe

kk⇒

1

3σkk =

2µ + 3λ

3ǫe

kk

= κǫe

kk(9.120)

where the superscript e emphasizes that the stress components are related tothe elastic strain components. The relationship between the bulk modulusand the engineering constants (Young’s modulus and Poisson’s ratio) is givenas

κ =E

3 (1 − 2ν)(9.121)

The last line in equation (9.119) follows from the fact that the trace of theplastic strain tensor is zero, i.e., tr

(

ǫp

n+1

)

= 0.

The next step is to express the deviatoric stress tensor, σ′n+1

, in terms ofthe plastic strain increment tensor ∆ǫ

p

n. To do this we need to write down

the Hooke’s law relationship between the deviatoric stress components and


the elastic part of the deviatoric strain components as follows:

σ′ij

= σij −1

3σkkδij

= 2µǫe

ij+ λǫ

e

kkδij − κǫ

e

kkδij

= 2µ

(

ǫ′eij

+1

3ǫe

kkδij

)

+ (λ − κ) ǫe

kkδij

= 2µǫ′eij

(9.122)

Next, from the last line in equation (9.122) and noting that the elastic partof the strain is equal to the total strain minus the plastic part, we can write

σ′n+1

= 2µ[

ǫ′n+1

− ǫ′pn+1

]

(9.123)

Because the volumetric part of the plastic strain is zero, the deviatoric stresscan be written in terms of both the deviatoric strain and the plastic strain as

σ′n+1

= 2µ[

ǫ′n+1

− ǫp

n+1

]

(9.124)

Let us now decompose the plastic strain tensor at time n + 1 as follows:

ǫp

n+1= ǫ

p

n+ ∆ǫ

p

n(9.125)

where ǫp

nis the known plastic part of the strain at time n, and ∆ǫ

p

nis the

unknown plastic strain increment tensor. Substituting equation (9.125) intoequation (9.124) yields

σ′n+1

= 2µ

ǫ′n+1

− ǫp

n− ∆ǫ

p

n

(9.126)

The elastic trial stress, σ′∗n+1

, is defined as

σ′∗n+1

= 2µ

ǫ′n+1

− ǫp

n

(9.127)

Note that if the loading situation is such that the plastic strain incrementsare zero, then the actual stress state at time n + 1 would be equal to the trialstress. An equivalent expression for the elastic trial stress that perhaps issomewhat easier to implement can be written in terms of the deviatoric strainincrement and the deviatoric strain at time n as follows:

σ′∗n+1

= 2µǫ′n

+ 2µ∆ǫ′n

= σ′n

+ 2µ∆ǫ′n

(9.128)

During general plastic loading conditions, the deviatoric stress, σ′n+1

, can bewritten in terms of the elastic trial stress and the plastic strain increments as

σ′n+1

= σ′∗n+1

− 2µ∆ǫp

n(9.129)


We now seek to find ∆ǫp

nsuch that the yield condition, flow rule, and hard-

ening rule are satisfied.In the radial return method, the flow rule is satisfied in an approximate

manner by assuming that the plastic strain increment tensor, ∆ǫp

n, is normal

to the yield surface at time n + 1. This statement can be expressed as

∆ǫp

n= ∆λ

∂Fn+1

∂σn+1

(9.130)

where ∆λ is a constant. If we assume a von Mises yield condition with anisotropic hardening rule, the yield function at time n + 1 can be written as

Fn+1 = σ2

n+1− σy

(

ǫp

n+1

)

= 0 (9.131)

From our work in Section 9.4, we recall that the partial derivative of theyield function (9.130) above with respect to the stress tensor at time n + 1 isproportional to the deviatoric stress at time n + 1, i.e.,

∂Fn+1

∂σn+1

=3

2

∂

σ′n+1

: σ′n+1

∂σn+1

= 3σ′n+1

(9.132)

Plugging the result (9.132) into (9.130) and then substituting the resultingexpression into equation (9.129) yields

σ′∗n+1

= [1 + 6µ∆λ] σ′n+1

(9.133)

If we view the deviatoric stress as a vector in three-dimensional principalstress space, equation (9.133) reveals that the unknown stress σ

′n+1

acts inthe same direction as the trial stress. The stress σ

′n+1

is obtained by marchingalong a radial line from the trial stress to the point σ

′n+1

, which lies on theupdated yield surface Fn+1 = 0.

The next step is to substitute equation (9.132) into equation (9.130). Doingthis we get

∆ǫp

n= 3∆λ σ

′n+1

(9.134)

In order to solve for ∆λ, loosely speaking, we square both sides of equation(9.134) as follows:

∆ǫp

n: ∆ǫ

p

n= 9∆λ

2σ

′n+1

: σ′n+1

⇒

∆λ =∆ǫp

n

2σn+1

(9.135)

where σ is the von Mises effective stress at time n + 1, i.e.,

σn+1 =

√

3

2σ

′n+1

: σ′n+1


σ′∗n+1

σ1 σ2

σ3

σ′n

σ′n+1

FIGURE 9.7

Graphical interpretation of the radial return method.

and

ǫp

n=

√

2

3∆ǫ

p : ∆ǫp

is the effective plastic strain increment. Substituting equation (9.135) intoequation (9.133) yields

σ′∗n+1

=

1 +3µ∆ǫp

n

σn+1

σ′n+1

(9.136)

Finally, recalling the definitions of the von Mises effective stress and the ef-fective plastic strain increment, we square both sides of equation (9.136) andrearrange to obtain

σn+1 + 3µ∆ǫp

n− σ

∗n+1

= 0 (9.137)

The resulting equation (9.137) is in general a nonlinear function of the ef-fective plastic strain increment ∆ǫp. The equation can be solved, for example,by carrying out separate Newton’s method iterations within the stress updateportion of the flow chart given in Table 9.4. Once the quantity ∆ǫp

nis ob-

tained, the effective stress, σn+1, can be obtained from the yield condition(9.131). Note that the effective plastic strain at time n + 1 is obtained byadding the effective plastic strain increment to the effective plastic strain attime n. The updated yield stress is obtained from the uniaxial stress-straincurve. Once σn+1 is found, we can get σ

′n+1

from equation (9.130). It thenfollows that

σn+1 = σ′n+1

+ κ tr ǫn+1I (9.138)

A graphical interpretation of the radial return method is shown in Fig-ure 9.7. The deviatoric stress, σ

′n, lies on the yield surface at time n. After

computation of the elastic trial stress, σ′∗n+1

, the deviatoric stress at time n+1is obtained by returning along a radial line to the updated yield surface.


(0,0) (1,0)

(1,1)(0,1)

ǫ

σ (psi)

dσ

dǫ

0 0.002 0.004 0.006

10,000

30,000

50,000

FIGURE 9.8

Radial return example.

Example 9.2

In order to illustrate how the radial return stress-update scheme is imple-mented in a finite element code, in this section we consider a simple one-element example in which the element is initially stress free and also free ofresidual plastic strain. Nodal displacement increments are then prescribed,and it is desired to compute the new state of stress in the element.

To begin, consider a four-node quadrilateral element as shown in Figure 9.8.The nodal coordinates of the undeformed element are labeled in the figure.The element is then subjected to a simple shear deformation, such that thenodal displacement vector at time n + 1 is given as

dn+1 =

0.0000.0000.0000.0000.0050.0000.0050.000

(9.139)

We will take the uniaxial stress-strain response of the material to be thebilinear hardening idealization as shown in Figure 9.8. Young’s modulus,E = 30 × 106 psi, is the slope of the initial linear region, and the slope of thestress-versus-strain curve during plastic loading (after the onset of yielding)is Et = E/4. In the present example we will take Poisson’s ratio to beν = 0.3, and the initial yield stress to be σ0

y= 30, 000 psi. These properties are

summarized in Table 9.5. We emphasize that, during plastic loading, the slope


of the stress-versus-total-strain curve, Et, is not the same as the hardeningparameter H. Recall that H is the slope of the stress-versus-plastic-straincurve defined as

H =1

1/Et − 1/E(9.140)

The first step in the radial return method is to obtain the components of thetotal strain tensor, ǫe, within the element. Recall that the superposed tildehas been used throughout the book to emphasize that the quantity is a finiteelement approximation. The subscript e indicates that the quantity is definedwithin an element. Throughout the rest of this example, both the superposedtilde and subscript e will be dropped in order to simplify the notation. Thetotal strain calculation is accomplished by following the procedure describedin Chapter 6. It requires setting up the element B-matrix and then multiplyingthat matrix by the nodal displacement vector. As an intermediate step, theJacobian matrix turns out to be

J =

[

1/2 00 1/2

]

(9.141)

The resulting B-matrix, containing the derivatives of the shape functions withrespect to X1 and X2, is then

B =1

2

−1 + η 0 1 − η 0 1 + η 0 −1 − η 00 −1 + ξ 0 −1 − ξ 0 1 + ξ 0 1 − ξ

−1 + ξ −1 + η −1 − ξ 1 − η 1 + ξ 1 + η 1 − ξ −1 − η

Multiplying the B-matrix by the displacement vector yields

ǫn+1 =

ǫ11

ǫ22

ǫ33

γ12

=

000

0.005

(9.142)

where only the engineering shear strain, γ12, is nonzero as expected. Thecomponents of the total strain tensor can now be stored in matrix form asfollows:

ǫn+1 =

0 0.0025 00.0025 0 0

0 0 0

(9.143)

TABLE 9.5

Example 9.3 mechanical properties.

E (psi) Etan (psi) ν σ0

y(psi)

30 × 106 7.5 × 106 0.3 30, 000


TABLE 9.6

Example 9.2 results.

∆ǫp σn+1 (psi) σ′n+1

(psi) σn+1 (psi)

0.001567 45, 673

000

26, 369

000

26, 369

In the present example, the volumetric part of the total strain tensor is zero;hence, the deviatoric strain happens to be equal to the total strain, i.e., ǫ

′n+1

=ǫn+1.

Because the element is initially free from residual plastic strain, the totalplastic strain in the element ǫ

p

nis zero. The elastic trial stress is then simply

σ′∗n+1

= 2µǫ′n+1

= 2µǫn+1

=

0 57, 692.3 057, 692.3 0 0

0 0 0

psi (9.144)

The next step is to compute the effective trial stress σ∗n+1

. Recalling that

σ∗n+1

=

√

3

2σ

′∗n+1

: σ′∗n+1

,

we obtain

σ∗n+1

=

√

3

2(57692.32 + 57692.32)

= 99, 926.0 psi (9.145)

In the present example, the von Mises effective stress at time n + 1, whichis equal to the updated yield stress, can be written in terms of the initial yieldstress, the hardening parameter, and the effective plastic strain increment asfollows:

σn+1 = σy + H∆ǫp (9.146)

Substituting the results (9.145) and (9.146) for the effective trial stress andupdated yield stress into equation (9.137) yields the following value for theeffective plastic strain increment:

∆ǫp = 0.001567 (9.147)

Once the plastic strain increment is known, equation (9.136) gives the cor-rected deviatoric stress components. Finally, by substituting the deviatoricstress components into equation (9.138) we obtain the corrected values for thestress components. The results are summarized in Table 9.6.


9.6 One-dimensional elastoplastic deformation of a bar

In order to demonstrate the salient features of the finite element procedurefor material-only nonlinear problems, in this section we consider the problemof a bar loaded by two concentrated forces as shown in Figure 9.9. The barhas a constant circular cross section, A = 1.0 in.2, and length L = 48 in.. Wewill assume that the bar is made out of a linearly elastic, linear hardeningmaterial with the properties given in Table 9.7. In Table 9.7, σ0

yis the initial

yield stress, ǫ0y

is the total extensional strain at which yield initiates, E isYoung’s modulus, and Etan is the tangent modulus, the slope of the stress-versus-total-strain curve after initial yield.

In the present example, we will model the bar with two linear-u elements.The concentrated forces, P1 = 15, 000 lb and P2 = 25, 000 lb, will be appliedin one load step, and we will go through two iterations of Newton’s methodto obtain an estimate for the equilibrium positions of nodes 2 and 3 labeledin Figure 9.9.

The incremental iterative equation to be solved can be written as

Ktani

n+1∆di

n= Fn+1 − f i

n+1; i = 0, 1, 2 . . . (9.148)

where Ktani

n+1is the tangent stiffness matrix to be set up at the beginning

of the ith iteration, ∆di

nare the unknown nodal displacement increments,

Fn+1 is the global vector containing the external nodal loads, and f i

n+1is the

vector containing the internal nodal forces.

We note that when the iteration counter i is equal to zero, the tangentstiffness matrix and global vectors are to be evaluated at the known state, n,

1 2 3

P1 = 15, 000 lb P2 = 25, 000 lb

FIGURE 9.9

Elastic-plastic deformation of a bar loaded by concentrated forces.

TABLE 9.7

Mechanical properties of elastic-plasticbar.

σ0

y(psi) ǫ0

yE (psi) Etan (psi)

20,000 0.001 20 × 106 5 × 106


which, in the present example, is the unloaded, stress-free state, i.e.,

Ktan0

n+1=

Ktan

n

f0

n+1= fn

Fn+1 =

15, 00025, 000

lb (9.149)

The next step in the solution procedure is to set up the global tangentstiffness matrix. To do this, we write down the tangent stiffness matrix foreach element, and then assemble the element matrices following the usualassembly procedure. One of the things to keep in mind when evaluatingthe element tangent stiffness matrix is that, in the present one-dimensionalproblem, we have assumed that the only nonzero stress in the bar is the normalstress in the axial direction. Hence, during elastic loading, the slope of thestress-strain curve is, E, Young’s modulus. During plastic loading, the slopeis, Etan, the tangent modulus. In other words,

dσ

dǫ=

E : during elastic loading

Et : during elastic-plastic loading(9.150)

The element tangent stiffness matrix for a linear-u element can now be writtenas

Kei

n+1= AeE

′e

∫

le

0

BT

e

2 × 1Be

1 × 2dx

=AeE

′e

le

[

1 −1−1 1

]

(9.151)

where Ae and le are the cross-sectional area and length of the element respec-tively, and

E′e

=dσ

dǫ

∣

∣

∣

∣

e

is the slope of the stress-strain curve defined in equation (9.150) associatedwith the total strain in element e.

Having obtained the element tangent stiffness matrix, the element vectorcontaining the internal nodal forces can be written as

fei

n+1= Ae

∫

le

0

BT

e(σe)

i

n+1dx

= Ae(σe)i

n+1

−11

(9.152)


TABLE 9.8

Numerical results for the elastic-plastic bar.

i σ1

(psi)

σ2

(psi)

Etan

1

(×106 psi)

Etan

2

(×106 psi)

Ktan

(×106 lb/in)

f

(lb)

R

(lb)

d

(in.)

0 0 0 20 20

[

1.6667 −.8333

−.8333 .8333

]

0

0

15, 000

25, 000

.048

.078

1 25,000 21,250 5 5

[

.4167 −.2083

−.2083 .2083

]

3, 750

21, 250

11, 250

3, 750

.120

.168

2 40,000 25,000 5 5

[

.4167 −.2083

−.2083 .2083

]

15, 000

25, 000

0

0

.120

.168

After assembling the element matrices and vectors defined in equations (9.151)and (9.152) above, we obtain the following global matrices and vectors:

Ktani

n+1=

2A

L

E′1

−E′1

0

−E′1

E′1

+ E′2

−E′2

0 −E′2

E′2

i

n+1

Fn+1 =

R1

15, 00025, 000

f i

n+1= A

−σ1

σ1 − σ2

σ2

i

n+1

(9.153)

where we have used le = L/2, and Ae = A given in the problem statement.The quantities E′

1, E′

2, σ1, and σ2 refer to the slope of the stress-strain curve

and stress in element 1 and 2, respectively. Note that because the left end ofthe bar is fixed, there is an unknown reaction, R1, at global node 1.

The last step in the process is to enforce the nodal constraints to obtain thereduced system of equations to be solved iteratively using Newton’s method.Because d1 is constrained to be zero, the 2 × 2 reduced system is obtainedby crossing out the first row and first column of the global system (9.153).Hence, the system of equations to be iterated upon is

2A

L

[

E′1

+ E′2

−E′2

−E′2

E′2

]i

n+1

∆d2

∆d3

i

n+1

=

15, 00025, 000

− A

σ1 − σ2

σ2

i

n+1

(9.154)

The numerical results are tabulated for each iteration in Table 9.8. As can beobserved from the table, the residual, or out-of-balance force vector, R, goesto zero (exactly) after the second iteration.


P

FIGURE 9.10

Thick-walled cylinder subjected to internal pressure.

9.7 Elastoplastic analysis of a thick-walled cylinder

As an application of the theory and finite element methods presented in thischapter, let us now consider the problem of a thick-walled cylinder subjectedto internal pressure as shown in Figure 9.10. The inner radius of the cylinderis a = 1.0m and the outer radius is b = 2.0m. We will assume that thematerial behavior is bilinear hardening with the properties given in Table 9.9.

Linear elastic solution

The analytical solution to the linearly elastic problem of a thick-walled cylin-der subjected to internal pressure is called Lame’s solution. The solution canbe found in any textbook on the theory of elasticity (see, for example, Timo-shenko and Goodier [29]). For the present case, when the cylinder is subjectedto an internal pressure, P , the analytical expressions for the radial and hoopstress components as a function of r, the radial distance from the longitudinalaxis, can be written as

σrr = −Pa2

b2 − a2

[

(b/r)2− 1

]

σθθ =Pa2

b2 − a2

[

(b/r)2

+ 1]

(9.155)

TABLE 9.9

Material parameters for thethick-walled cylinder.

E (GPa) ν Et (GPa) σ0

y(MPa)

200 0.3 67 100


The out-of-plane normal stress can be obtained from our knowledge of Hooke’slaw for the case of plane strain. From equation (4.83) in Section 4.5.2 weobtain

σzz =ν(1 − ν)

(1 + ν)(1 − 2ν)(σrr + σθθ) (9.156)

After plugging in the numerical values (a = 1m, b = 2m, ν = 0.3, andP = 1Pa) into equations (9.155) and (9.156), we get the following state ofstress at a point located on the inner surface of the cylinder:

σrr 0 00 σθθ 00 0 σzz

=

−1 0 00 1.6667 00 0 0.2692

Pa (9.157)

Armed with the analytical solution for the state of stress at points on theinner surface, let us now compute the von Mises effective stress at the innersurface. Recall that the von Mises effective stress is defined as

σ =

√

3

2σ′

ijσ′

ij

We will leave it to the interested reader to show that the effective stress turnsout to be σ = 2.31Pa (to two digits beyond the decimal point).

A contour plot of the von Mises effective stress obtained by the finite ele-ment method is shown in Figure 9.11. A plane strain analysis was performedusing four-node isoparametric quadrilateral elements on the quarter model.The mesh is composed of 10 elements in the radial direction and 15 elementsin the circumferential direction for a total of 150 elements. Symmetry bound-ary conditions were enforced so that nodes lying on the line Y = 0 wereconstrained in the Y direction, and nodes lying on the line X = 0 were con-strained from moving in the X direction. The internal pressure was incorpo-rated into the analysis by assembling the element vectors associated with theelements contiguous to the inner surface. A uniform distributed load of mag-nitude 1Pa was assumed to act over these element edges. The finite elementresult for the von Mises effective stress at points lying on the inner surfaceis σ = 2.21Pa. The fact that the finite element solution for σ is about 4 %below the analytical result suggests that the mesh could use some refinementin the region near the inner surface due to the presence of high stress gradi-ents in the radial direction. From the linear finite element solution, becausethe solution scales linearly with the applied internal pressure, we expect thecylinder to yield when the internal pressure reaches approximately 45MPa.Note that this estimate is obtained by dividing the initial yield stress of thematerial (100MPa) by the maximum von Mises effective stress in the cylinder(σ = 2.21Pa) obtained for a unit internal pressure.

In the next section, we will carry out the elastic-plastic analysis. For thesake of interest, we will increase the internal pressure up to 100MPa and thencompletely unload the cylinder. It turns out that after removal of the load, a


1

1.2

1.4

1.6

1.8

2

2.2

FIGURE 9.11

Contour plot of the von Mises effective stress (Pa) for P = 1.0Pa.

state of residual stress exists in the cylinder due to the presence of permanentplastic strains that remain after the load is removed.

Elastic-plastic solution

For the elastic-plastic analysis, the internal pressure was increased from 0MPato 100MPa in 100 equal load steps. The internal pressure was then decreasedfrom 100MPa back down to 0MPa, again, in 100 equal load steps.

To present the elastic-plastic numerical results, we begin by focusing ourattention on a quadrature point that is nearest to the inner radius at r =a. The von Mises effective stress, σ, at this location is plotted versus totalaccumulated plastic strain, ǫp, as shown in Figure 9.12. During the initialportion of the loading cycle, the response at the point of interest is linearlyelastic. This is evident due to the fact that ǫp is zero until the initial yieldstress of 100MPa is reached. Subsequently, as the load is further increased,the response is elastic-plastic. Because we have taken the material behavior tobe bilinear hardening, the effective stress versus accumulated effective plastic


0.0 0.0002 0.0004 0.0006 0.0008

20

40

60

80

100

120

140

160

180

σ(M

Pa)

ǫp

FIGURE 9.12

Effective stress, σ, versus total accumulated plastic strain, ǫp, at a point nearthe inner radius of the cylinder.

strain response is linear with slope H = 100MPa.

After the internal pressure reaches 100MPa, the pressure is then completelyremoved in 100 equal load steps. During the removal of the load, the responseis again elastic, with no further increase in the total accumulated plastic strain.The effective stress decreases to approximately 10MPa and then increases to avalue of approximately 40MPa by the time the internal pressure is completelyremoved. The effective stress when the internal pressure reaches 0MPa isemphasized by the data marker with the larger radius in Figure 9.12.

For the sake of interest, the hoop stress, σθθ, at the quadrature point near-est the inner radius is plotted versus load step, n, in Figure 9.13. Noticethat the initial response is linear until the initial yield point is reached af-ter approximately 45 load steps. The hoop stress versus load step response isthen nonlinear during elastic-plastic loading until the internal pressure reaches100MPa. During unloading, the response is linear until the maximum com-pressive hoop stress, ≈ −25MPa, is reached. As the internal pressure contin-ues to be decreased, the stress distribution in the cylinder redistributes, andthe final value of the hoop stress (still compressive) is nearly −10MPa.

To study the numerical performance of the present nonlinear Newton’smethod procedure, the norm of the out-of-balance force vector, ‖R‖, is plottedversus load step in Figure 9.14. At a given load step, the norm after the firstiteration is indicated by the circular data markers. The norm after the seconditeration is depicted by the diamond data markers, and the norm after thethird iteration is shown by the square data markers. In the present analysisa relative residual convergence criterion was chosen. The solution was taken


0 50 100 150 200-40

-20

0

20

40

60

80

100

σθθ(M

Pa)

Load step n

FIGURE 9.13

Hoop stress, σθθ, near the inner radius of the cylinder versus load step n.

‖R‖

n20 40 60 80 100 120 140 160 1800

0.005

0.01

0.015

0.02

0.025

0.03

0.035

0.04

0.045

0.05

FIGURE 9.14

Norm of the out-of-balance force vector, ‖R‖, at each iteration versus loadstep. For a given load step, the norm after the first iteration is indicated bythe circular data markers. The norm after the second iteration is depicted bythe diamond data markers, and the norm after the third iteration is indicatedby the square data markers.

to have converged when‖Ri‖

‖R1‖< 1 × 10−2

During elastic loading, convergence was achieved in one iteration, as expected.During elastic-plastic loading, convergence was achieved in three to four iter-ations.

A contour plot of the von Mises effective stress in the cylinder is shown in


90

100

110

120

130

140

150

160

170

(MPa)

(a)

(MPa)

10

15

20

25

30

35

40

45

(b)

FIGURE 9.15

Contour plot of the von Mises effective stress when the internal pressurereaches 100MPa (a), and contour plot of the effective stress after removalof the load (b).

Figure 9.15. The contour plot when the internal pressure reaches 100MPa isshown in Figure 9.15 (a), and the plot after the load is completely removedis shown in Figure 9.15 (b). In both plots, the deformation is magnified 100times.

9.8 Problems

Problem 9.1

Plot the yield surface in σ1–σ2 space for the case of biaxial stress. Assumeplane stress and a von Mises yield condition. Hint: σXX and σY Y are theonly nonzero stresses acting on the stress element.

Problem 9.2

The following yield condition is proposed that includes the effect of hydrostaticpressure:

√

1

2σ′

ijσ′

ij= τy − µσkk

where µ is a friction coefficient (typically having a value ranging from zero toone) and τy is the yield stress shear.

(a) Create a surface plot of the yield surface in principal stress space.

(b) Display the intersection of the yield surface and the plane σ3 = 0.


Problem 9.3

X

Y

FIGURE 9.16

Problem 9.3. Elastic-plastic compression of a block.

A block is constrained by rigid, frictionless supports as shown in Figure 9.16.Assume that the block material obeys an associated flow rule and the followingvon Mises yield condition:

σ − σy (ǫp) = 0 (9.158)

where σy is the yield stress taken to be a function of the accumulated effectiveplastic strain ǫp. Assume that the uniaxial stress-strain curve is bilinear withE = 30 × 106 psi and ν = 0.3. The initial yield stress is σ0

y= 30, 000 psi

and the hardening parameter is H = E/3. Write a computer program thatemploys the radial return method to calculate the normal stress σXX as afunction of the total strain ǫXX in the range 0 ≤ ǫXX ≤ 0.002. Display yourresult graphically on an X-Y plot. Assume plane strain.

10

Treatment of Geometric Nonlinearities

God would not be so unkind, said a prominent engineer, as to make the equa-

tions of nature nonlinear.∗

So far we have restricted our attention to bodies that undergo small defor-mations and rotations. Under these conditions, it is appropriate to use thesmall-strain tensor introduced in Chapter 4. In the present chapter we willconsider the general case of bodies that are allowed to undergo arbitrarilylarge deformations as well as rigid-body translations and rotations. When adeformable body undergoes general motion, the relationship between strainand displacement is no longer linear, even when the deformation in the bodyis small and the stress-strain response is linear. This type of nonlinearityis called a geometric nonlinearity. When the strain and deformation is smallbut the material response is nonlinear, as is the case for small-strain plasticitytheory discussed in the previous chapter, we have only a material nonlinearity.

In the development of a finite element procedure for solving problems withboth material and geometric nonlinearities, we need to clearly make the dis-tinction between the deformed and original configurations of the body. Theprocess begins by writing down the principle of virtual work over the de-formed configuration. Because the deformed configuration keeps changing, itis convenient to express the principle of virtual work over a fixed referenceconfiguration. For geometrically nonlinear problems, the principle of virtualwork can be viewed as a nonlinear function of the deformed configuration ofthe body. In this chapter we will employ an approach that is similar to thattaken by Bonet [30] and Holzapfel [31], among others, to obtain a Newton’smethod procedure for solving the resulting nonlinear equations. In the ap-proach, the principle of virtual work over the original configuration is firstlinearized, and then the finite element approximations are substituted intothe resulting expression. Invoking the arbitrariness of the test function thenyields a system of linear algebraic equations that can be solved in an iterativemanner. The approach of formulating the geometrically nonlinear problem ina fixed reference configuration is called a total-Lagrangian formulation. Often,when constitutive equations are expressed in incremental form involving theCauchy stress tensor, it is convenient to work in the deformed configuration.

∗Stewart, Ian. Does God Play Dice? The Mathematics of Chaos. Basil Blackwell, Cam-bridge, MA 1989.

349


In this case, the linearized version of the principle of virtual work in the orig-inal configuration can be recast into an equivalent expression in the deformedconfiguration through a change of variables. This approach is known as anupdated-Lagrangian formulation.

To introduce the subject of geometrically nonlinear problems, in the presentchapter we will restrict our attention to the total-Lagrangian formulation.We will begin with a discussion on large-deformation kinematics and definethe relevant deformation tensors in the next section. We will then proceedand illustrate the process of recasting the principle of virtual work from thedeformed configuration into an equivalent expression in the original configu-ration. During this process, the alternative stress measures (first and secondPiola-Kirchhoff stress tensors) as well as the important concept of work con-jugacy naturally emerge.

Armed with the principle of virtual work in the original configuration, wethen go through the linearization process. Substituting the finite elementapproximations into the linearized version of the principle of virtual workyields a system of linear algebraic equations that can be solved in an iterativemanner by Newton’s method.

The chapter concludes with two case studies. In the first case study, weconsider a simple truss structure and observe the phenomenon of snap-throughbuckling after the applied load reaches a critical value. In the second casestudy, the reader is taken through a large-deformation analysis of a nonlinearlyelastic (rubber) test specimen.

10.1 Large-deformation kinematics

Recall, from our study of Chapter 2, that the fundamental idea behind thestudy of the motion of continuous bodies is that at any given instant of time, t,there exists a sufficiently smooth function that maps individual material pointsfrom an undeformed reference configuration to a deformed configuration. Themapping can be written as

x = x(X) (10.1)

where x is the position vector that defines the location of the point in the de-formed configuration, and X is the position vector that defines the location ofthat same point in the undeformed configuration. In Chapter 2 we discoveredthat an infinitesimal line segment, dx, having magnitude ds in the deformedconfiguration is related to the corresponding infinitesimal line segment in theundeformed configuration through the relation

dx = F · dX (10.2)

Treatment of Geometric Nonlinearities 351

X1

X2

dx

dX

FIGURE 10.1

Infinitesimal line segment in reference and deformed configurations.

where F is the deformation gradient tensor. It is instructive to think of equa-tion (10.2) as the link between the locus of points that make up an infinitesi-mal line segment in the original configuration and that same set of points thatmake up an infinitesimal line segment in the deformed configuration. Recallthat the components, Fij , of the deformation gradient tensor are defined as

Fij =∂xi

∂Xj

(10.3)

The idea of mapping an infinitesimal line segment from the reference config-uration to the deformed configuration is depicted graphically in Figure 10.1.

The displacement of a point in a solid is defined as the difference betweenthe location x of the point in the deformed configuration and the location X

of this same point in the reference configuration, i.e.,

u = x − X or ui = xi − Xi (10.4)

From the definition (10.3), we can express the components of the deformationgradient tensor in terms of the displacement components as follows:

Fij =∂

∂Xj

ui + Xi

=∂ui

∂Xj

+ δij (10.5)

where the last line in equation (10.5) follows from the fact that ∂X1/∂X1 = 1,∂X1/∂X2 = 0, etc.


10.1.1 The Green-Lagrange strain tensor

Having defined the components of the deformation gradient tensor above, letus now focus on an infinitesimal line segment, dx, in the deformed configu-ration. We will use the quantity ds to represent the magnitude of this linesegment. Evaluating the square of ds gives

ds2 = dx · dx

= (F · dX) · (F · dX)

= dX ·(

FT· F

)

· dX

= dX · C · dX (10.6)

where C is the Green deformation tensor. Evaluating the difference betweenthe squared magnitude of the infinitesimal line segment in the deformed andundeformed configurations yields

ds2− dS

2 = dX · C · dX − dX · dX

= dX · [C − I] · dX (10.7)

where I is the second-order identity tensor.The Green-Lagrange strain tensor, E, is defined through the following re-

lationship:ds

2− dS

2 = 2dX · E · dX (10.8)

Hence, from equation (10.7) we can write

E =1

2[C − I] (10.9)

Dividing both sides of equation (10.8) by the quantity dS2 yields

ds2 − dS2

dS2= λ

(n)− 1

= 2n · E · n (10.10)

From equation (10.10) above we see that the operation of pre- and post-multiplying the Green strain by a unit vector n (defining some direction inthe undeformed configuration) gives the stretch, λ(n), minus 1.

It is both useful and insightful to express the components of the Green-Lagrange strain tensor in terms of the components of the deformation gradienttensor and/or the components of the displacement vector. Expressing thecomponents of E in terms of F gives

E =1

2[C − I]

=1

2

[

FT· F − I

]

⇒

Eij =1

2[FkiFkj − δij ] (10.11)


X1

X2

P

p

θ

φ

FIGURE 10.2

Two-dimensional body undergoing a rigid-body rotation.

Next, by substituting equation (10.5) into (10.11) we obtain

Eij =1

2

[(

∂uk

∂Xi

+ δki

) (

∂uk

∂Xj

+ δkj

)

− δij

]

=1

2

[

∂uk

∂Xi

∂uk

∂Xj

+∂uj

∂Xi

+∂ui

∂Xj

+ δij − δij

]

=1

2

[

∂ui

∂Xj

+∂uj

∂Xi

+∂uk

∂Xi

∂uk

∂Xj

]

(10.12)

Recalling the definition of the small-strain tensor introduced in Chapter 4,i.e.,

ǫij =1

2

(

∂ui

∂Xj

+∂uj

∂Xi

)

we notice that the components of the Green-Lagrange strain tensor tend to-ward the components of the small strain tensor when the displacement gradi-ents are small.

Example 10.1

Suppose that the two-dimensional body shown in Figure 10.2 undergoes arigid-body rotation about the X3 axis. Because the body only undergoesrigid-body motion, it is not deformed during the process, and the state ofstrain at each point in the body should be identically zero.

Let us now focus on point P labeled in the figure. The location of thispoint in the undeformed configuration is defined by the position vector X,


and its location in the deformed configuration is represented by the vector x.The orientation of each vector is defined by the angles θ and φ labeled in thefigure. One can see that the components of the position vectors x and X canbe expressed in terms of θ and φ as follows:

X1

X2

=

cos θ

sin θ

;

x1

x2

=

cos (θ + φ)sin (θ + φ)

(10.13)

Next, using the trigonometric addition formulas, the components of the posi-tion vector x can equivalently be expressed as

x1

x2

=

cos θ cos φ − sin θ sin φ

sin θ cos φ + cos θ sin φ

(10.14)

Comparing equations (10.13) and (10.14), we can now write the componentsof x in terms of the components of X as follows:

x1

x2

=

[

cos φ − sin φ

sinφ cos φ

]

X1

X2

or

x = RX (10.15)

The 2 × 2 matrix R in equation (10.15) above is called the rotation matrix.Equation (10.15) can be written in component form as

xi = RijXj (10.16)

The components of the deformation gradient tensor describing the rigid-bodyrotation are obtained by differentiating equation (10.16) with respect to Xj .Doing this gives

Fij =∂xi

∂Xj

= Rij (10.17)

Using equation (10.11), the components of the Green strain tensor can nowbe evaluated as follows:

Eij =1

2[RkiRkj − δij ] (10.18)

and hence,

E11 =1

2[Rk1Rk1 − 1]

=1

2[R11R11 + R21R21 − 1]

=1

2

[

cos 2φ + sin 2

φ − 1]

= 0 (10.19)

We leave it to the reader to verify that the remaining components of the Greenstrain tensor also vanish.


The present example illustrates the important fact that the componentsof the Green-Lagrange strain tensor vanish under a rigid-body motion. Thisindicates that the Green strain tensor is a true measure of strain, independentof any rigid-body motion. On the other hand, if we evaluate the componentsof the small-strain tensor under the present rigid-body rotation, we get thefollowing:

ui = xi − Xi ⇒

∂ui

∂Xj

= Fij − δij

∂uj

∂Xi

= Fji − δji ⇒

ǫij =1

2(Fij + Fji − 2δij)

=1

2(Rij + Rji − 2δij) (10.20)

By evaluating the components of the small-strain tensor we get

[ǫij ] =

[

cos φ − 1 00 cos φ − 1

]

(10.21)

From equation (10.21) above, we observe that the components of the small-strain tensor vanish only when the rigid-body rotation defined by the angle φ

is much smaller than unity. For the general case of large rigid-body rotations,the small-strain tensor would incorrectly predict nonzero extensional strainsin the X1 and X2 directions.

Example 10.2

A four-node isoparametric quadrilateral is shown in Figure 10.3. The mappingbetween the parent domain and the original configuration is given as

Xi =

4∑

a=1

Na(ξ, η)Xa

i(10.22)

where Na(ξ, η) are the shape functions, i.e.,

N1 =1

4(1 − ξ)(1 − η)

N2 =1

4(1 + ξ)(1 − η)

N3 =1

4(1 − ξ)(1 + η)

N4 =1

4(1 + ξ)(1 + η)


X1

X2

1

2

3

4

FIGURE 10.3

Four-node quadrilateral in undeformed and deformed configurations.

and Xa

iare the coordinates of local node a in the original configuration. The

local nodal coordinates in the original and deformed configurations are givenin Table 10.1.

The goal of the present example is to determine the components of thedeformation gradient tensor, F, at the point (ξ, η) = (0, 0) in the parentdomain. We will also illustrate the process of computing the components ofthe Green strain tensor, E, at this same location. We will treat the presentproblem as two-dimensional, so the F and E matrices will be 2 × 2.

The first step in the solution is to recall how the components of the de-formation gradient tensor are related to the displacement components. Therelationship can be written as

Fij = ui,j + δij (10.23)

The finite element approximation for the displacement field within the elementis written as

uie= Nadai (10.24)

where summation is implied (a=1 to 4) and dai refers to the ith displacementcomponent at local node a.

The nodal displacements in the present example are summarized in Ta-ble 10.2. Because local node 2 is the only node that is displaced, the displace-

TABLE 10.1

Nodal coordinates in undeformed and deformedconfigurations (inches)

Configuration 1 2 3 4

Original (1.0,1.0) (2.0,2.0) (1.5,3.0) (0.5,2.0)

Deformed (1.0,1.0) (3.0,2.0) (1.5,3.0) (0.5,2.0)


TABLE 10.2

Nodaldisplacements ininches.

a daX daY

1 0.0 0.02 1.0 0.03 0.0 0.04 0.0 0.0

ment interpolation turns out to be quite simple. From Equation (10.24) weget

uXe

= N2d2X = 1/4(1 + ξ)(1 − η)(1.0)

uY e

= 0

Differentiating the displacement components above yields

uX,Xe

= N2,Xd2X

uX,Y e

= N2,Y d2X

uY,Xe

= 0

uY,Y e

= 0

With d2X = 1 and d2Y = 0, the finite element approximation for the com-ponents of the deformation gradient tensor within the element turns out tobe

Fe =

[

N2,X + 1 N2,Y

0 1

]

(10.25)

Next, we need to evaluate the derivatives of the shape function N2 with respectto the X and Y coordinates in the original configuration. After plugging inthe nodal coordinates in the original configuration, the mapping between theparent domain and the original configuration can be written as

X = N1X1 + N2X2 + N3X3 + N4X4

= N1(1.0) + N2(2.0) + N3(1.5) + N4(0.5)

=5

4−

1

4η +

1

2ξ

Y = N1Y1 + N2Y2 + N3Y3 + N4Y4

= N1(1.0) + N2(2.0) + N3(3.0) + N4(2.0)

= 2 +1

2η +

1

2ξ


The Jacobian matrix can then be calculated as follows:[

X,ξ Y,ξ

X,η Y,η

]

=

[

1/2 1/2−1/4 1/2

]

Hence,

N2,X

N2,Y

=

[

1/2 1/2−1/4 1/2

]−1

N2,ξ

N2,η

=

[

4/3 −4/3−2/3 4/3

]

1

4(1 − η)

− 1

4(1 + ξ)

=

1

3(2 + ξ − η)

1

6(−1 − η − 2ξ)

(10.26)

The final step is to evaluate Fe at the point (ξ,η)=(0,0) and compute the finiteelement approximation for the Green strain using the relation E = 1

2(C− I).

The result is

Fe =

[

5/3 −1/60 1

]

Ee =

[

8/9 −5/36−5/36 1/72

]

(10.27)

10.2 Weak form in the original configuration

From our study of linearized elasticity in Chapter 4, we recall that the weakform of the equilibrium equation can be written as

∫

v

σijwi ,j dv =

∫

v

biwi dv +

∫

s

tiwi ds (10.28)

where the integrals are all volume integrals over the deformed configurationv. The surface of v is denoted as s. We emphasize that the comma denotespartial differentiation with respect to the spatial variables xi. The goal ofthis section is to recast the integrals in (10.28) as equivalent integrals over theoriginal configuration. This can be accomplished simply through a change ofvariables.

10.2.1 Internal virtual work

Let us begin with the internal virtual work expression on the left-hand sideof (10.28). In order to express this integral as an equivalent integral over theoriginal configuration, it is necessary to express each term in the integrandas a function of the spatial coordinates, Xi, in the original configuration.


From our work in Chapter 2, we recall that a differential volume element inthe deformed configuration is related to a differential volume in the originalconfiguration through the relation

dv = j dV (10.29)

where j is the determinant of the deformation gradient tensor, i.e., j = detF.Let us now look at the next term, the gradient of the test function wi,j . Wecan use the chain rule to express this quantity in terms of derivatives withrespect to the spatial coordinates as follows:

wi,j =∂wi

∂xj

=∂wi

∂Xk

∂Xk

∂xj

(10.30)

Equivalently, equation (10.30) can be expressed in tensor notation as

∇xw = ∇w · F−1 (10.31)

We point out here that the quantity F−1 is the inverse of the deformationgradient tensor. If there exists a one-to-one mapping from the original config-uration to the deformed configuration, then there must also exist a one-to-oneinverse mapping. In other words, a differential line segment in the deformedconfiguration can be mapped (or pulled back) to the original configurationusing the relation

dX = F−1· dx (10.32)

To check the tensorial expression on the right-hand side of equation (10.31),we can write the expression back in component form as follows:

∇w · F−1 =

(

∂wi

∂Xk

ei ek

)

·

(

∂Xl

∂xj

el ej

)

=∂wi

∂Xk

∂Xl

∂xj

δkl ei ej

=

(

∂wi

∂Xk

∂Xk

∂xj

)

ei ej (10.33)

Next, substituting equations (10.29) and (10.30) back into the left-hand sideof equation (10.28) yields

δWint =

∫

v

σijwi,j dv

=

∫

V

σij

∂wi

∂Xk

∂Xk

∂xj

j dV

=

∫

V

(

jσij

∂Xk

∂xj

)

∂wi

∂Xk

dV (10.34)


Contained inside the parentheses in equation (10.34) above are the compo-nents, Pij , of the first Piola-Kirchhoff stress tensor. In tensor notation it isdefined as

P = jσ · F−T (10.35)

where j = detF, σ is the Cauchy stress tensor, and F−T is the transpose ofthe inverse of the deformation gradient tensor. From a dimensional analysiswe can readily see that the first Piola-Kirchhoff stress components have unitsof force per unit undeformed area. Letting l be a characteristic length inthe deformed configuration and L be a characteristic length in the originalconfiguration, the quantity

Pij = jσij

∂Xk

∂xj

has units of

[Pij ] =

[(

l3

L3

) (

F

l2

) (

L

l

)]

=

[

F

L2

]

(10.36)

From further observation of (10.35), we can also see that the first Piola-Kirchhoff stress tensor is not symmetric. To show this, it is necessary to showthat P and PT are not equal as follows:

P = jσ · F−T⇒

PT = jF−1· σ

T = jF−1· σ 6= P (10.37)

A symmetric tensor can be constructed by pre-multiplying the first Piola-Kirchhoff stress tensor by the inverse of the deformation gradient tensor. Do-ing this yields

S = jF−1· σ · F−T (10.38)

The symmetric second-order tensor, S, defined in equation (10.38) above iscalled the second Piola-Kirchhoff stress tensor.

Noting that P = F · S, we can obtain an expression for the internal virtualwork over the original configuration in terms of the second Piola-Kirchhoffstress as follows:

δWint =

∫

V

P : ∇w dV

=

∫

V

(F · S) : ∇w dV (10.39)


Next, employing the identity introduced in Chapter 2, Table 2.1, the internalwork expression (10.39) can be written as

δWint =

∫

V

S :(

FT· ∇w

)

dV

=

∫

V

Sij

∂xk

∂Xi

∂wk

∂Xj

dV (10.40)

To end this subsection, it is important to recognize that the tensor scalarproduct of the first Piola-Kirchhoff stress and ∇w yields a scalar quantityhaving units of work per unit original volume. This can be seen by performingthe following dimensional analysis:

[P : ∇w] =

[

F

L2

] [

l

L

]

=

[

F · l

L3

]

, (10.41)

where, F refers to force, l is a characteristic length in the deformed config-uration, and L is a characteristic length in the original configuration. Thequantities P and ∇w are hence called work conjugate pairs. The scalar prod-uct of P and ∇w, integrated over the original volume, gives work (F × l).Likewise, the quantities S and FT · ∇w are also work conjugate pairs.

The relationships between the stress measures introduced in this sectionand the various work conjugate pairs are summarized in Table 10.3. In theliterature on nonlinear finite element procedures, the test function, w, is oftenthought of as a fictitious infinitesimal change in the deformed configuration,i.e., w = δx. The gradient of the test function would then correspond to thefirst variation of the deformation gradient tensor, δF. Similarly, the quantityFT · ∇w = FT · δF is related to the variation of the Green-Lagrange strain.Note that from our work in Chapter 2, the tensor scalar product of a sym-metric second-order tensor with another second-order tensor is equal to thescalar product of the symmetric tensor and the symmetric part of the othertensor. For example,

S : δE = S :

[

1

2

(

δFT· F + FT

· δF)

]

= S :(

F · δFT)sym

= S :(

F · δFT)

(10.42)

Likewise, the variation in the small-strain tensor, δǫ, is the symmetric part of∇xw. This can be seen as follows:

δǫ =1

2

(

∇xwT + ∇xw)

= ∇xwsym (10.43)

Hence, σ and δǫ are work conjugate pairs.


TABLE 10.3

Alternative stress measures and work conjugate pairs.

Stress measure σ P S

Conjugate pair ∇xw or δǫ ∇w or δF FT·∇w or δE or F

T δF

Relationship with σ σ jσ · F−T jF−1· σ · F

−T

10.2.2 External virtual work

Let us now focus on the external virtual work. The expression in the deformedconfiguration is written as

δWext =

∫

v

biwi dv +

∫

s

tiwi ds (10.44)

where bi are the components of the body force vector with units of force perunit deformed volume, ti are the components of the surface traction vectorhaving units of force per unit deformed area, and wi are the components ofthe test function. Because the external virtual work expression is composed ofa body force contribution and a contribution from external surface tractions,it is convenient to write the expression as

δWext =

δWext

b

+

δWext

t

(10.45)

The body force contribution can be written as an integral over the originalconfiguration as follows:

δWext

b

=

∫

V

biwi jdV (10.46)

where, again, j is the determinant of the deformation gradient tensor. Toevaluate the body force vector at a point in the original configuration, it isnecessary to first determine the location of that point in the deformed config-uration using the mapping between the deformed and original configurations.The body force vector at the point in the deformed configuration is thenapplied at the corresponding point in the original configuration.

To express the traction contribution as an area integral in a fixed referenceconfiguration, it is convenient to choose a two-dimensional ξ−η coordinatesystem as a parent domain as shown in Figure 10.4. The relationship betweena differential surface area, ds, in the deformed configuration and a differentialsurface area, dA = dξdη, in the parent domain is given as

ds =

∣

∣

∣

∣

∂x

∂ξ×

∂x

∂η

∣

∣

∣

∣

dξdη (10.47)

where the vectors ∂x/∂ξ and ∂x/∂η are differential changes in the positionvector x, labeled in the figure, with respect to differential movements in the


X1

X2

X3

ξ

η

dξ

dηdx(1)

dx(2)

FIGURE 10.4

Differential areas in deformed configuration and parent domain.

ξ and η directions in the parent domain. From our work in Chapter 6, werecall that the position vector x can be viewed as a function of the naturalcoordinates ξ and η. An infinitesimal change in the position vector x due toan infinitesimal movement in the ξ direction is then

dx(1) =∂x

∂ξdξ (10.48)

Similarly, an infinitesimal change in the position vector x due to an infinites-imal movement in the η direction is

dx(2) =∂x

∂ηdη (10.49)

The differential area, ds, in the deformed configuration is then the magnitudeof the cross product of the two vectors dx(1) and dx(1) as seen in Figure 10.4.

The surface traction contribution to the external virtual work can finallybe expressed as an integral in the parent domain as

δWext

t

=

∫

A

tiwe

∣

∣

∣

∣

∂x

∂ξ×

∂x

∂η

∣

∣

∣

∣

dξdη (10.50)

10.3 Linearization of the weak form

The principle of virtual work statement (whether it be written in the deformedconfiguration or the original configuration) can be viewed simply as a scalar-valued function of the test function w and x, the locus of points that define


the volume, v, of the deformed configuration. The function, w, is assumed tobe a kinematically admissible test function that is imposed on the deformedconfiguration at some known instant of time n. It is understood that the testfunction does not in any way depend on the actual deformation of the body.

In problems that involve large strains and displacements, both the internalvirtual work and, in general, the external virtual work are highly nonlinearfunctions of x. For this reason, in an attempt to find a deformed configurationof the body that satisfies the principle of virtual work for all kinematicallyadmissible test functions, we need to resort to a suitable nonlinear solutionstrategy. In this chapter we will demonstrate the nonlinear solution procedureusing Newton’s method described in Chapter 2.

To begin, consider the following nonlinear equation

δWint

n+1− δW

ext

n+1= 0 (10.51)

where the subscript n + 1 refers to a discrete instant of time. Just to becompletely clear, the deformed configuration of the body is defined by xn atsome instant of time n. At some future instant of time, n + 1, the body isassumed to occupy xn+1. Equation (10.51) can be linearized by performingan operation that is very similar to a two-term Taylor series expansion ofthe functions δWint and δWext about the deformed configuration at time n.Letting

δWint

n+1≈ δW

int

n+ DδW

int

n

δWext

n+1≈ δW

ext

n+ DδW

ext

n(10.52)

the linearized version of equation (10.51) can be written as

D(δWint

n− δW

ext

n) = δW

ext

n− δW

int

n(10.53)

In other words, equation (10.53) says that the “change” in the internal virtualwork, minus the “change” in the external work, from some known configura-tion at time n, is equal to the external virtual work at time n, minus theinternal work at time n. Throughout the rest of the book, DA will denote thechange in the quantity A with respect to one of its arguments. The quantityA can be a scalar, a vector, or a tensor.

Let us now be more precise as to what this “change” actually means. Thevirtual work (internal or external) is a scalar function of the vector-valuedfunctions x and w. Suppose that the position of each point, x, in the deformedconfiguration of the body at time n is changed by an amount ∆x. The quantity∆x is a vector-valued function that gives the change in the position of eachpoint x. The new configuration of the body is then x + ∆x. The quantityDδW represents the change in the virtual work due to a change in x, whilekeeping the test function, w, fixed. This change can be computed by findingthe rate of change of δW in the direction of ∆x, and then multiplying thisrate by the magnitude of ∆x.


ǫ∆x

δW(x)

δW(x + ǫ∆x)

δW

FIGURE 10.5

Surface illustrating the change in virtual work due to a change δx in thedeformed configuration.

Visual aids are very helpful in developing an understanding of this concept.Focusing our attention on Figure 10.5, let us imagine that each point lying onthe plane shown in the figure represents a kinematically admissible deformedconfiguration, x, that can be substituted into δWint or δWext. The renderedsurface in the figure represents the scalar value of the virtual work whenevaluated at x, and some test function — it doesn’t matter which one. Nowsuppose we move from configuration x in the direction of ∆x as shown in thefigure. The rate of change, dδW, of the virtual work in the direction ∆x canbe written as

dδW = limǫ→ 0

δW(x + ǫ∆x) − δW(x)

ǫ |∆x|

(10.54)

where ǫ is a parameter, not to be confused with the small-strain tensor intro-duced in Chapter 4. This rate can be interpreted as the slope of the line thatis tangent to the surface at x labeled in the figure. Finally, the change in thevirtual work, DδW, due to a change, ∆x, of the deformed configuration of thebody can be written as

DδW = dδW |∆x|

= limǫ→ 0

δW(x + ǫ∆x) − δW(x)

ǫ

(10.55)

As an example of the use of equation (10.55), suppose that A(x) = x · x.


The change, DA, is then computed as follows:

DA = limǫ→ 0

(x + ǫ∆x) · (x + ǫ∆x) − x · x

ǫ

= limǫ→ 0

x · x + 2ǫx · ∆x + ǫ2∆x · ∆x − x · x

ǫ

= 2x · ∆x (10.56)

Notice that the change in the nonlinear quantity A = x·x results in a quantitythat is linear in ∆x.

It turns out that the change operator, D, obeys the usual properties of dif-ferentiation. Thus, for example, we can use the product rule of differentiationin order to compute the change of products of functions, i.e.,

D(AB) = (DA)(B) + (A)(DB)

Using the procedure described above, the change of important quantities innonlinear solid mechanics such as F, S, E, etc. can all be readily calculated.The changes in these quantities are needed in the linearization of the principleof virtual work. Let us now proceed and linearize the internal virtual workstatement in the next subsection.

10.3.1 Internal virtual work

To begin the process of linearizing the internal virtual work statement, webegin by writing down the internal virtual work as an integral over the originalconfiguration as follows:

δWint =

∫

V

S :(

FT· ∇w

)

dV (10.57)

Because the integral in equation (10.57) is over the original configuration,which does not change with respect to a change in the deformed configuration,it is only necessary to evaluate the change of the integrand with respect to achange, ∆x, in the deformed configuration. Doing so involves the use of theproduct rule as follows:

D

S :(

FT· ∇w

)

= DS :(

FT· ∇w

)

+ S : D(

FT· ∇w

)

(10.58)

Another application of the product rule to the second term on the right-handside of equation (10.58) above yields

D(

FT· ∇w

)

= DFT· ∇w + FT

· D∇w

= FT· ∇w (10.59)

The last line in equation (10.59) above follows from the fact that the testfunction is independent of the deformed configuration.


The final step in the linearization of the internal virtual work involves eval-uating the changes DF and DS. The change in the deformation gradient canbe readily obtained using the definition (10.55) as follows:

DF = limǫ→0

[

∂

∂X(x + ǫ∆x) −

∂x

∂X

]

ǫ

=∂∆x

∂X(10.60)

Hence the components of DF are ∂∆xi/∂Xj . Physically, the tensor DF rep-resents the gradient of the change in the deformed configuration.

The change in the second Piola-Kirchhoff stress with respect to a changein the deformed configuration can be obtained if there exists a relationship ofthe form

DS = D : DE (10.61)

For elastic materials (not necessarily linearly elastic), the quantity D is afourth-order tensor known as the Lagrangian or material elasticity tensor.The components of the material elasticity tensor can be determined if a con-stitutive relationship between S and E is known. Note that S and E arework conjugate pairs, so it is appropriate to form a constitutive relationshipbetween these two quantities. We leave it to the reader to show that S andE are work conjugate pairs in an exercise at the end of this chapter.

Having obtained the changes in the relevant quantities, the change in theinternal virtual work can now be written as

DδWint =

∫

V

DS :(

FT· ∇w

)

dV +

∫

V

S :(

δFT· ∇w

)

dV (10.62)

where we have used the notation

δF =∂∆x

∂X(10.63)

It is worth mentioning here that in nonlinear finite element procedures, thefirst integral on the right-hand side of equation (10.62) gives rise to what iswidely referred to as the material tangent stiffness matrix. The geometrictangent stiffness matrix, also known as the initial stress tangent stiffness ma-trix, emerges from the second integral on the right-hand side. The linearizedinternal virtual work statement in the original configuration can be expressedin component form as

DδWint =

∫

V

DSij

∂xm

∂Xk

∂wm

∂Xl

dV +

∫

V

Sij

∂∆xm

∂Xi

∂wm

∂Xj

dV (10.64)

Substituting the relation DS = D : DE into equation (10.64) gives analternative expression for the linearized internal virtual work. The change in


the Green strain tensor is defined as

DE =1

2D

[

FT· F − I

]

(10.65)

where I is the second-order identity tensor. Carrying out the operation definedin (10.65) above yields

DE =1

2

[

DFT· F + FT

· DF]

=

DFT· F

sym

⇒

D : DE = D :(

DFT· F

)

(10.66)

The last line in equation (10.66) is a result of the fact that the materialelasticity tensor is symmetric. Finally, substituting the component form ofequation (10.66) into (10.64) yields

DδWint =

∫

V

Dijkl

∂∆xm

∂Xk

∂xm

∂Xl

∂xn

∂Xi

∂wn

∂Xj

dV

+

∫

V

Sij

∂∆xk

∂Xi

∂wk

∂Xj

dV. (10.67)

10.3.2 External virtual work

In this section we will go through the process of linearizing the external vir-tual work statement. For the sake of simplicity, we will assume a constantbody force — one that does not change with respect to a change in the de-formed configuration of the body. For this case, the change in the body forcecontribution to the external virtual work is zero, i.e.,

D

δWext

b

= 0 (10.68)

Let us now focus on the contribution to the external virtual work due tospecified tractions acting on the surface of the body. Here we will assume thatthe traction vector at each point on the surface has a constant magnitude p

and remains normal to the surface as the body deforms. Note that whilethe magnitude of the traction vector remains constant, its direction changescontinuously with the deformation of the body. This type of loading is widelyreferred to as a follower force. The follower force, in general, is a highlynonlinear function of the deformed configuration, and hence, the tractioncontribution to the external virtual work requires linearization.

Having made the assumptions on the surface traction distribution statedabove, the traction components at a point on the deformed surface can bewritten in terms of the components, ni, of the unit outward normal to thesurface at that point as

ti = −pni (10.69)


x1

x2

x3ξ

η

1

2

3

4

t = −pn

FIGURE 10.6

Distributed follower force.

where p is the magnitude of the traction vector (pressure). The tractioncontribution to the external virtual work can now be written as

δWext

t

= −p

∫

s

wini ds

= −p

∫

A

w ·

(

∂x

∂ξ×

∂x

∂η

)

dξdη (10.70)

Note that in the second line of equation (10.70) above, we have expressedthe virtual work contribution as an area integral over the ξ−η parent domainshown in Figure 10.6. The result follows from the fact that at each point onthe surface

nds =

(

∂x

∂ξ×

∂x

∂η

)

dξdη (10.71)

The result (10.71) comes from our discussion in Section 10.2.2, where weshowed that the cross product of two infinitesimal vectors, dx(1) and dx(2),shown in Figure 10.4, yields a vector having magnitude ds that points in thedirection normal to the surface at point x.

Equation (10.71) can be written in component form using the permutationtensor introduced in Chapter 2. The result is

δWext

t

= −p

∫

A

wiǫijk

∂xj

∂ξ

∂xk

∂ηdξdη (10.72)

The traction contribution to the external virtual work can be readily linearizedstarting with equation (10.72). The operation requires the use of the product


rule as follows:

D

δWext

t

= −p

∫

A

wiǫijkD

(

∂xj

∂ξ

∂xk

∂η

)

dξdη

= −p

∫

A

wiǫijk

[

D

(

∂xj

∂ξ

)

∂xk

∂η+

∂xj

∂ξD

(

∂xk

∂η

)]

dξdη

(10.73)

Finally, the linearized expression can be written as

D

δWext

t

= −p

∫

A

wiǫijk

(

∂∆xj

∂ξ

∂xk

∂η+

∂xj

∂ξ

∂∆xk

∂η

)

dξdη (10.74)

For the sake of interest, let us now go ahead and substitute the finite elementapproximations into equation (10.74). The finite element approximations canbe written as

∂xi

∂ξ

e

= Na,ξxai ;

∂∆xi

∂ξ

e

= Na,ξ∆dai

∂xi

∂η

e

= Na,ηxai ;

∂∆xi

∂η

e

= Na,η∆dai

wi = Nawai (10.75)

where Na is the shape function associated with local node a, ∆dai are thedisplacement increment components, and xai are the components of the nodalcoordinates in the deformed configuration.

Substituting the finite element approximations (10.75) into equation (10.74)yields an expression of the form

D

δWext

t

=wT

e

1 × 3a

−p

∫

A

NT

e

3a × 3he

3 × 1dA

(10.76)

Here in equation (10.76), we is a 3a × 1 vector containing the nodal valuesof the test function. Note that the finite element approximation for the testfunction is

we = Newe (10.77)

where Ne is the matrix containing the shape functions. After some lengthyalgebra, the three components of the vector he appearing in equation (10.76)turn out to be

h1 = (Na,ξ∆da2Nb,ηxb3 + Nb,ξxb2Na,η∆da3)

− (Na,ξ∆da3Nb,ηxb2 + Nb,ξxb3Na,η∆da2)

h2 = − (Na,ξ∆da1Nb,ηxb3 + Nb,ξxb1Na,η∆da3)

+ (Na,ξ∆da3Nb,ηxb1 + Nb,ξxb3Na,η∆da1)

h3 = (Na,ξ∆da1Nb,ηxb2 + Nb,ξxb1Na,η∆da2)

− (Na,ξ∆da2Nb,ηxb1 + Nb,ξxb2Na,η∆da1) (10.78)


where the summation convention is implied, and the indices a and b rangefrom 1 to the number of nodes per element. Finally, from equations (10.76)and (10.78), the discretized version of equation (10.74) can be written as

D

δWext

t

= wT

e

−p

∫

A

NT

eHe dA

∆de (10.79)

whereHe =

[

H1 H2 H3 . . . Hnen

]

(10.80)

Ha =

0 xb3Bab −xb2Bab

−xb3Bab 0 xb1Bab

xb2Bab −xb1Bab 0

(10.81)

whereBab = (Na,ξNb,η − Nb,ξNa,η) (10.82)

and nen is the number of nodes per element.To end this section, we note that the 3a× 3a matrix inside the curly brack-

ets in equation (10.79) above is an element tangent matrix that arises dueto the presence of surface tractions that change with the deformation of thebody. This matrix is commonly referred to as a load correction tangent ma-trix. In the nonlinear solution strategy using Newton’s method, this matrixis subtracted from the sum of the material and geometric tangent matrices.The element load correction tangent matrix can be written as

Kload

e= −

∫

A

NT

eHe dξdη (10.83)

Note that the load correction tangent matrix is not symmetric, and hence theresulting global system matrix is not symmetric. Under such conditions, wemust resort to a solver that is designed to handle nonsymmetric matrices. Thisdoes not pose any difficulty, but a nonsymmetric solver, in general, requiresmore in-core storage due to the fact that both the upper and lower trianglesof the system matrix need to be stored in memory.

The finite element implementation of the total Lagrangian approach is pre-sented in detail through two examples in the following sections.

10.4 Snap-through buckling of a truss structure

To illustrate the total Lagrangian procedure for solving geometrically non-linear problems, we consider an assemblage of space trusses subjected to avertical load as shown in Figure 10.7. This problem is widely known as asnap-through buckling problem. Referring to Figure 10.7, we will assume


1 2

3

X1 = X

P

X2 = Y

FIGURE 10.7

Truss assemblage loaded by a concentrated force.

that both degrees of freedom at node 1 are fixed. Node 2 will be allowedto translate in the X1 direction but not in the X2 direction. Similarly, node3 is constrained in the X1 direction, but allowed to move freely in the X2

direction. The length of each truss is 20 in., and we will take both trusses tohave the same material properties, with Young’s modulus E = 10 × 106 psiand cross-sectional area 0.0625 in.2. The load, P , shown in the figure will beapplied incrementally. Starting from zero, the load will be gradually increasedby equal load steps until buckling behavior is observed.

To begin the formulation, let us write down the change in the internalvirtual work given by equation (10.64) as follows:

DδWint =

∫

V

DSij

∂xm

∂Xi

∂wm

∂Xj

dV +

∫

V

Sij

∂∆xm

∂Xi

∂wm

∂Xj

dV (10.84)

For the case of a space truss element, we assume that the only nonzero stresscomponent is the normal stress acting perpendicular to any cross section ofthe truss. This implies that the only nonzero component of the first Piola-Kirchhoff stress tensor is P11. Because P = F · S, we can see that the onlynonzero component of the second Piola-Kirchhoff stress tensor is S11. If S11

is the only nonzero component of S, the change in the internal virtual work(10.84) then takes the following simple form:

DδWint =

∫

V

DS11

∂xm

∂X1

∂wm

∂X1

dV +

∫

V

S11

∂∆xm

∂X1

∂wm

∂X1

dV (10.85)

Next, let us take the original (unstressed) configuration of each truss el-ement to be the domain −L/2 ≤ X1 ≤ L/2, where L is the undeformed


length of the truss. In other words, we assume that each truss originally lieson the X1 axis, centered at the origin. A differential volume in the originalconfiguration is then

dV = AdX1 (10.86)

where A is the undeformed cross-sectional area of the truss. The change inthe internal virtual work written over the original configuration then becomes

DδWint =

∑

e

Ae

∫

Le/2

−Le/2

DS11

∂xm

∂X1

∂wm

∂X1

dX1

+ Ae

∫

Le/2

−Le/2

S11

∂∆xm

∂X1

∂wm

∂X1

dX1

(10.87)

In the present truss problem, because the structure is loaded by a con-centrated force whose magnitude and direction does not change as the trussassemblage deforms, the change in the external virtual work due to a changein the configuration of the truss assemblage is zero. Hence, the nonlinearequation to be solved can be written as

δWint

n+1= δW

ext

n(10.88)

where time n represents some known equilibrium configuration of the assem-blage, and n+1 represents a future instant of time at which we seek to find theequilibrium configuration of the assemblage. Finally, the linearized version tobe iterated upon is

δWint

n+ DδW

int

n= δW

ext

n(10.89)

orDδW

int

n= δW

ext

n− δW

int

n(10.90)

10.4.1 Element tangent matrix

In order to obtain a system of linear algebraic equations to be solved itera-tively using Newton’s method, it is necessary to introduce the finite elementapproximations and substitute these approximations into the linearized ver-sion of the weak form.

The mapping of an element between the original configuration and thedeformed configuration at time n is given as

xe

3 × 1=

Ne

3 × 6pe

6 × 1(10.91)

Here in equation (10.91), xe represents a position vector that points fromthe origin to some point along the length of a truss element in the deformedconfiguration that occupied the location X1 in the original configuration. The3 × 6 matrix Ne contains the linear shape functions, and pe is a 6 × 1 vector


that contains the nodal coordinates in the deformed configuration. The shapefunction matrix is given as

Ne =

N1 0 0 N2 0 00 N1 0 0 N2 00 0 N1 0 0 N2

(10.92)

where

N1(ξ) =1

2(1 − ξ)

N2(ξ) =1

2(1 + ξ) (10.93)

and we have let ξ = 2X1/Le

Let us now start the process of plugging the finite element approximationsdefined above into the linearized version of the principle of virtual work. Tobegin, we will focus our attention on the second integral on the right-handside of equation (10.87). By carefully observing the integrand, we recognizethat the quantity

∂∆xm

∂X1

∂wm

∂X1

is simply the dot product between two vectors, yielding a scalar. Consistentwith a standard Galerkin procedure, the vector-valued test function, w, isapproximated using the linear shape functions. The finite element approxi-mation for the test function along the length of the element in the deformedconfiguration at time n then becomes

we = Newe (10.94)

where we is a 6 × 1 vector containing the nodal values of the test function.The finite element approximation for the derivative of the test function withrespect to X1 is obtained by differentiating equation (10.94). Doing this yields

∂we

∂X1

= Bewe (10.95)

where

Be =

∂N1

∂X1

0 0∂N2

∂X1

0 0

0∂N1

∂X1

0 0∂N2

∂X1

0

0 0∂N1

∂X1

0 0∂N2

∂X1

(10.96)

contains the derivatives of the shape functions with respect to X1. Notingthat

∂Na

∂X1

=∂Na

∂ξ

∂ξ

∂X1


the B-matrix defined in equation (10.96) turns out to be

Be =1

Le

−1 0 0 1 0 00 −1 0 0 1 00 0 −1 0 0 1

(10.97)

Next, we recognize that the vector ∆x represents a change in displacementfrom the deformed configuration at time n. The finite element approximationfor this vector can be written as

∆xe = Ne∆de (10.98)

where ∆de is a 6 × 1 vector containing the nodal displacement increments.The derivative of ∆xe with respect to X1 is then

∂∆xe

∂X1

= Be∆de (10.99)

Finally, substituting the finite element approximations into the second integralon the right-hand side of equation (10.87) yields

wT

e

Ae

∫

Le/2

−Le/2

S11BT

eBe dX1

∆de (10.100)

The 6×6 matrix sandwiched in between the curly brackets above is the elementgeometric tangent stiffness matrix. It can be written as

KG

e= Ae

∫

Le/2

−Le/2

S11BT

eBe dX1 (10.101)

Because the matrix Be is constant in this case, it can be taken outside of theintegral, and the geometric stiffness matrix can be evaluated by performing asimple hand calculation. The result is

KG

e=

AeS11

Le

1 0 0 −1 0 00 1 0 0 −1 00 0 1 0 0 −1

−1 0 0 1 0 00 −1 0 0 1 00 0 −1 0 0 1

(10.102)

Let us now turn to the first integral on the right-hand side of equation(10.87). For the present problem involving an assemblage of space trusses,we will assume a linear relationship between the second Piola-Kirchhoff stressand the Green-Lagrange strain of the form

S = λtr(E) + 2µE (10.103)


where λ and µ are material constants. The constitutive relation (10.103)above is known as the St. Venant-Kirchhoff model. The model is useful forrepresenting linearly elastic material behavior when the relative stretch ateach point in the body remains small. Note that it is possible for the relativestretch at a point in a body to remain small while the body undergoes generallarge rigid body rotations and translations.

Using the St. Venant-Kirchhoff model, the component S11 of the secondPiola-Kirchhoff stress can be written as

S11 = (λ + 2µ) E11 + λ (E22 + E33) (10.104)

The change, DS11, can then be computed as follows:

DS11 = (λ + 2µ) DE11 + λ (DE22 + DE33) (10.105)

Using the definition of the Green-Lagrange strain tensor in terms of the de-formation gradient tensor, F, we can write

E =1

2

(

FT· F − I

)

(10.106)

The first component, E11, of the Green-Lagrange strain tensor can be writtenin component form as

E11 =1

2

(

∂xk

∂X1

∂xk

∂X1

− 1

)

(10.107)

Using the definition (10.55), the change in E11 turns out to be

DE11 = limǫ→0

1

ǫ

[

1

2

(

∂xk

∂X1

+ ǫ∂∆xk

∂X1

) (

∂xk

∂X1

+ ǫ∂∆xk

∂X1

)

− 1

−1

2

∂xk

∂X1

∂xk

∂X1

− 1

]

(10.108)

Taking the limit as the parameter ǫ goes to zero and simplifying gives

DE11 =∂∆xk

∂X1

∂xk

∂X1

(10.109)

We leave it to the reader to verify that the changes DE22 and and DE33 thatappear in equation (10.105) are zero. It is due to the fact that in the presenttruss formulation, x and ∆x only depend on X1. Hence,

DS11 = (λ + 2µ)∂∆xk

∂X1

∂xk

∂X1

= E∂∆xk

∂X1

∂xk

∂X1

(10.110)


where we have let E = λ + 2µ. Finally, substituting the expression (10.110)into the first integral on the right-hand side of equation (10.85) yields

Ae

∫

Le/2

−Le/2

(

DS11

∂xk

∂X1

∂wk

∂X1

)

dX1

= AeEe

∫

Le/2

−Le/2

(

∂∆xk

∂X1

∂xk

∂X1

∂xk

∂X1

∂wk

∂X1

)

dX1 (10.111)

Let us now go ahead and substitute the finite element approximations intothe integral (10.111) above. The finite element approximation for the positionvector x is given by the mapping (10.91). Hence, the derivative of the positionvector with respect to X1 is

∂xe

∂X1

= Bepe (10.112)

where Be is the B-matrix defined in equations (10.96) and (10.97), and pe isa 6×1 vector containing the nodal coordinates in the deformed configuration.Finite element approximations for the remaining quantities in the integrandof (10.111) have already been defined in equations (10.95) and (10.99). Sub-stituting these approximations into (10.111) yields

AeEe

∫

Le/2

−Le/2

(

∂∆xk

∂X1

∂xk

∂X1

∂xk

∂X1

∂wk

∂X1

)

dX1

= wT

e

AeEe

∫

Le/2

−Le/2

(

BT

eBepep

T

eBT

eBe

)

dX1

∆de (10.113)

The integral in-between the curly brackets in equation (10.113) above is widelyreferred to as the material tangent stiffness matrix, KM

e. In the present truss

element formulation, the material tangent stiffness matrix can readily be eval-uated in closed form. The result is,

KM

e=

AeEe

Le

(

x21

Le

)2x21y21

L2e

x21z21

L2e

−

(

x21

Le

)2

−x21y21

L2e

−x21z21

L2e

x21y21

L2e

(

y21

Le

)2y21z21

L2e

−x21y21

L2e

−

(

y21

Le

)2

−y21z21

L2e

x21z21

L2e

y21z21

L2e

(

z21

Le

)2

−x21z21

L2e

−y21z21

L2e

−

(

z21

Le

)2

−

(

x21

Le

)2

−x21y21

L2e

−x21z21

L2e

(

x21

Le

)2x21y21

L2e

x21z21

L2e

−x21y21

L2e

−

(

y21

Le

)2

−y21z21

L2e

x21y21

L2e

(

y21

Le

)2y21z21

L2e

−x21z21

L2e

−y21z21

L2e

−

(

z21

Le

)2x21z21

L2e

y21z21

L2e

(

z21

Le

)2


where we have let xa, ya, and za be the local coordinates of node a in thedeformed configuration, and Ee = (λ + 2µ)e. The remaining quantities in thematerial tangent stiffness matrix are defined as follows:

x21 = x2 − x1

y21 = y2 − y1

z21 = z2 − z1

We note that the material tangent stiffness matrix for the two-node spacetruss element defined in the previous equation reduces to the element matrixfor the space truss element derived in Chapter 7, Section 7.1.2, for small-strain/rotation applications, as seen in equation (7.33).

10.4.2 Element internal force vector

The element internal force vector at some instant of time n emerges from theinternal virtual work statement. The expression at time n can be written overa single element as

δWint

n

e= Ae

∫

Le/2

−Le/2

(

S11

∂xk

∂X1

∂wk

∂X1

)

dX1 (10.114)

Substituting the finite element approximations introduced in the previoussection into equation (10.114) above gives

δWint

n

e= wT

e

AeS11

∫

Le/2

−Le/2

BT

eBepe dX1

(10.115)

The element internal force vector is then the 6 × 1 vector inside the curlybrackets in equation (10.115) above, i.e.,

f int

e

6 × 1= AeS11

∫

Le/2

−Le/2

BT

eBepe dX1 (10.116)

After carrying out the matrix multiplication and integrating we get

f int

e= AS11

x21

L

y21

L

z21

L

x21

L

y21

L

z21

L

(10.117)


where, again, x21 = x2 − x1, etc.

It is worth mentioning that because we have employed the linear shapefunctions to map the truss element from the original to the deformed config-uration, the Green-Lagrange strain component, E11, turns out to be constantalong the length of the element. Since we are assuming that S11 = (λ+2µ)E11,the stress component, S11, is also constant along the length of the elementand, hence, can be taken outside the integral (10.117).

10.4.3 Numerical results

In this section we investigate the response of the truss assemblage as thevertical load, P , is increased incrementally until buckling is observed. Fromour work in the previous section, the system of equations to be iterated uponusing Newton’s method can be written as

[

KM + KG]i

n+1∆di

n= Fn+1 − f i

n+1(10.118)

where KM and KG are the global material and geometric tangent stiffnessmatrices, F is the global external force vector, and f is the global internalforce vector. After the nodal constraints are enforced, i.e.,

d1X = 0

d1Y = 0

d1Z = 0

d2Y = 0

d2Z = 0

d3X = 0

d3Z = 0

the order of the reduced system of equations is only 2 × 2. After matrix andvector assembly, the reduced system to be iterated upon turns out to be

[

K44 K48

K48 K88

]i

n+1

∆d2X

∆d3Y

i

n

=

0P

n+1

−

f2X

f3Y

i

n+1

(10.119)


where

K44 =A1E1

L1

(

x(1)

21

L1

)2

+A2E2

L2

(

x(2)

21

L2

)2

K48 = −A2E2

L2

x(2)

21y(2)

21

L2

2

K88 = −A2E2

L2

(

y(2)

21

L2

)2

f2X = A1S(1)

11

x(1)

21

L1

f3Y = A2S(2)

11

y(2)

21

L2

and P is the magnitude of the applied load. Note that in the definitionsabove we are employing the superscripts (1) and (2) to distinguish betweenthe nodal coordinates of each element.

The numerical results for the present snap-through buckling problem arepresented in Table 10.4 and Figure 10.8. In the numerical analysis, the totalload of 40, 000 lb was applied in 40 equal load steps.

The numerical results from the first load step are summarized in Table 10.4.In the table, the reduced tangent stiffness matrix, internal force vector, andresidual are reported at the beginning of each iteration. The residual, R, isdefined as

R =‖Ri

n+1‖

‖R1

n+1‖

(10.120)

whereRi

n+1= Fn+1 − f i

n+1(10.121)

is the out-of-balance force vector computed at the beginning of the ith itera-tion, and

‖Ri

n+1‖ =

√

Ri

n+1· Ri

n+1(10.122)

is the norm of the out-of-balance force vector. In the present analysis, conver-gence was achieved when the norm of the out-of-balance force at the beginningof the ith iteration, divided by the norm of the out-of-balance force at the be-ginning of the first iteration, is less than a user-specified tolerance, i.e.,

‖Ri

n+1‖

‖R1

n+1‖

< tol (10.123)

Here the tolerance was chosen to be 1 × 10−5. From the table, we observethat during the first load increment, convergence was achieved in only threeiterations.


δ (in.)

F (lb)

0 5 10 15 20 25

0.5

1

1.5

2

2.5

3

3.5

4×104

FIGURE 10.8

Force versus downward displacement of upper joint (40 equal load steps ∆F =1000 lb).

The numerical results for the vertical deflection, δ, of the upper joint (node3) is plotted versus applied load as shown in Figure 10.8. From the figure weobserve that the truss structure becomes unstable at a load of approximately15, 000 lb. This is widely referred to as the bifurcation point. At this load step,Newton’s method fails to converge in ten iterations. After the bifurcationpoint, the analysis gets back on track, and convergence is achieved withinfour to five iterations. Note that during the post-buckling behavior, it takesincreasing load to produce additional deflection. For the sake of interest, the

TABLE 10.4

Numerical results during first load step (results reported at theend of the ith iteration).

iteration K × 10−4 (lb/in.) f (lb) R

1

[

3.1250 0.00000.0000 0.7812

]

0.00000.0000

1.0000

2

[

3.1151 0.00000.0000 0.7614

]

0.0000−980.8819

0.0191

3

[

3.1148 0.00000.0000 0.7610

]

0−999.7432

2.5679 × 10−4

4 —

0−999.9966

3.4194 × 10−6


S11(p

si)

P (lb)0 1 2 3 4

×104

-8

-6

-4

-2

0

2

4

6

8

10×105

FIGURE 10.9

Normal stress in element two versus applied load.

normal stress in element 2 (the inclined truss element) is plotted versus appliedload in Figure 10.9. As shown in the figure, the normal stress is compressiveuntil after the bifurcation point, when it suddenly becomes tensile. A typicalyield stress for aluminum is somewhere between 20, 000 and 80, 000 psi (see,e.g., [22]). From the figure, we then conclude that yielding can be expectedto initiate well before the onset of buckling.

10.5 Uniaxial tensile test of a rubber dog-bone specimen

In order to demonstrate the total Lagrangian finite element procedure for fullythree-dimensional problems, in this section we study the behavior of a rubber-like test specimen clamped at one end and subjected to uniform tension at theother, as shown in Figure 10.10. The geometry of the specimen was chosen sothat the length and width of the gage section are 1.0 in. and 0.25 in., respec-tively. The total length of the specimen is 3.5 in., and the largest characteristicwidth is 1.25 in.. The out-of-plane thickness of the specimen was chosen to be0.1 in. The goal of this section is to demonstrate the stress-versus-elongationresponse of the specimen employing a neo-Hookean material model describedbelow.


σ3.5

0.251.25

1.0

FIGURE 10.10

Dog-bone specimen (compressible neo-Hookean material) clamped at one endand loaded by uniform tension at the other (dimensions in inches).

Finite element formulation

To begin the finite element formulation, let us write down the linearized ver-sion of the principle of virtual work as follows:

DδWint

n+ DδW

ext

n= δW

ext

n− δW

int

n(10.124)

where, again, n refers to some known equilibrium configuration of the body.In the present example, in the absence of body forces, and when both themagnitude and direction of the traction vector remain constant as the load isincreased, the change in the external work, DWext, is equal to zero. Hence,we only need to be concerned with evaluating the change in the internalvirtual work and then plugging in the finite element approximations into thatresulting expression in order to get the linearized system of equations to beiterated upon.

Recall from Section 10.2 that the internal virtual work expression can bewritten as

δWint =

∫

V

S : (FT· ∇w) dV

=

∫

V

Sij

∂xk

∂Xi

∂wk

∂Xj

dV (10.125)

The change, DδWint, in the internal virtual work is then

DδWint =

∫

V

Dijkl

∂∆xm

∂Xi

∂xm

∂Xj

∂xn

∂Xk

∂wn

∂Xl

dV

+

∫

V

Sij

∂∆xk

∂Xi

∂wk

∂Xj

dV, (10.126)


where we have assumed that

DSij = DijklDEkl (10.127)

Here in equation (10.127), Dijkl is the fourth-order material tangent elasticitytensor and

DEkl =∂∆xn

∂Xk

∂xn

∂Xl

(10.128)

are the components of the change in the Green-Lagrange strain tensor withrespect to a change in the deformed configuration of the body.

A widely used constitutive law for modeling rubber like materials is theneo-Hookean material model. For a compressible neo-Hookean material, thesecond Piola-Kirchhoff stress tensor is taken to be a function of the Greendeformation tensor. It is assumed that the state of stress at a point in thebody depends only on the final state of deformation of the body, independentof the way in which the body was loaded to that final state. Such a materialfalls in a class of materials called hyperelastic. The compressible neo-Hookeanconstitutive law can be written as

S = µ(

I − C−1)

+ λ (ln j)C−1 (10.129)

where I is the second-order identity tensor, C−1 is the inverse of the Greendeformation tensor, and j is the determinant of the deformation gradienttensor, i.e., j = detF . The quantities µ and λ are material parameters. Forthe neo-Hookean material model defined in equation (10.129), the componentsof the material elasticity tensor are given as

Dijkl = λC−1

ijC

−1

kl+ 2 (µ − λ ln j) C

−1

ikC

−1

jl(10.130)

See, for example, Holzapfel [31], Chapter 6, for a detailed treatment of hyper-elastic material models.

10.5.1 Element tangent matrix

In this section we derive the material and geometric tangent stiffness matricesthat are needed for the geometrically nonlinear analysis of the neo-Hookeantest specimen described above. To accomplish this, it is necessary to sub-stitute the finite element approximations into the linearized version of theinternal virtual work written over a single element. In this section we willspecialize the formulation for the case of the isoparametric eight-node brickelement discussed in Chapter 6, Section 6.4.

To begin, the mapping between the element in the parent ξ−η−ζ domainand the element in the deformed configuration at time n can be written as

xe = Nepe (10.131)

where xe is a position vector pointing from the origin of a Cartesian coordinatesystem to a point in the deformed configuration, Ne is the 3 × 24 matrix


containing the tri-linear shape functions defined in equation (6.55), and pe is a24× 1 vector containing the nodal coordinates in the deformed configuration.Similarly, the mapping between the element in the parent domain and theelement in the original configuration can be written as

Xe = NePe (10.132)

where Pe is the 24 × 1 vector containing the nodal coordinates in the origi-nal configuration. In addition, the finite element approximation for the testfunction can be written as

we = Newe (10.133)

where we is a 24 × 1 vector containing the nodal values of the test functionin the deformed configuration.

Having defined the mapping between the parent domain and the originaland deformed configurations and having introduced the finite element approx-imation for the test function, we can now go ahead and begin substituting thefinite element approximations into the linearized internal virtual work expres-sion. To facilitate the process, we point out that the quantity

∂xk

∂Xi

∂wk

∂Xj

defines the components of a second-order tensor. The finite element approxi-mation for the components of this tensor can be expressed in matrix notationas follows:

∂xk

∂Xi

∂wk

∂Xj

e

=wT

e

1 × 24

BT

ej

24 × 3

Bei

3 × 24pe

24 × 1(10.134)

where the subscripts i and j take on the values ranging from 1 to 3. Here inequation (10.134) above, Bei is a 3 × 24 matrix defined as follows:

Bei

3 × 24=

[

Be1

3 × 3Be2

3 × 3Be3

3 × 3. . .

Be8

3 × 3

]

(10.135)

where

[Bei]a3 × 3

=

∂Na

∂Xi

0 0

0∂Na

∂Xi

0

0 0∂Na

∂Xi

(10.136)

contains the derivatives of the shape functions with respect to the spatialcoordinates, Xi.

From equation (10.134) we can now see that the finite element approxima-tion for the components

∂∆xm

∂Xe

∂wm

∂Xj


can be written in matrix notation as

∂∆xm

∂Xi

∂wm

∂Xj

e

= wT

eBT

ejBei∆de (10.137)

Substituting the finite element approximations (10.134) and (10.137) into theintegrals on the right-hand side of equation (10.126) yields

DδWint

e= wT

e

KM

e+ KG

e

∆de (10.138)

where

KM

e=

∫

V

Dijkl

(

BT

ejBeipe

) (

pT

eBT

elBek

)

dV

KG

e=

∫

V

SijBT

ejBei dV (10.139)

are the material and geometric tangent stiffness matrices. Here in equation(10.139), the summation convention is implied and, again, the indices i and j

range from 1 to 3.For the sake of convenience, the material tangent stiffness matrix can be

written entirely in matrix notation. After some lengthy algebra we obtain

KMe

24 × 24=

∫

V

BMT

e

24 × 6De

6 × 6BM

e

6 × 24dV, (10.140)

where

BMe

6 × 24=

pT

eBT

e1Be1

pT

eBT

e2Be2

pT

eBT

e3Be3

pT

eBT

e2Be3 + pT

eBT

e3Be2

pT

eBT

e1Be3 + pT

eBT

e3Be1

pT

eBT

e1Be2 + pT

eBT

e2Be1

(10.141)

is the material B-matrix and

De =

D1111 D1122 D1133 D1123 D1113 D1112

D2211 D2222 D2233 D2223 D2213 D2212

D3311 D3322 D3333 D3323 D3313 D3312

D2311 D2322 D2333 D2323 D2313 D2312

D1311 D1322 D1333 D1323 D1313 D1312

D1211 D1222 D1233 D1223 D1213 D1212

(10.142)

is the 6 × 6 material tangent elasticity matrix.

10.5.2 Element internal force vector

To complete the formulation, it is necessary to come up with the finite elementapproximation for the internal virtual work evaluated at time n. Recall that


the internal virtual work at time n is given as

δWint

n=

∫

V

Sij

∂xk

∂Xi

∂wk

∂Xj

dV

n

(10.143)

From our work in the previous section, the finite element approximation for theintegrand on the right-hand side of equation (10.143) above can be expressedover a single element as

Sij

∂xk

∂Xi

∂wk

∂Xj

e

= wT

e

(

SijBT

ejBeipe

)

(10.144)

where the matrix, Bei, is defined in equation (10.136). The expression (10.144)above can be expanded and then written in matrix notation as

Sij

∂xk

∂Xi

∂wk

∂Xj

e

= wT

eBMT

eSe (10.145)

where Se is a 6 × 1 vector containing the components of the second Piola-Kirchhoff stress tensor, and BM

eis the 6 × 24 material B-matrix defined in

equation (10.141). The vector Se is given as

Se =

S11

S22

S33

S23

S13

S12

(10.146)

where it is understood that it is to be evaluated at time n.After substituting the finite element approximation (10.145) into the inter-

nal virtual work expression at time n, we obtain

δWint

e

n= wT

e

∫

V

BMT

eSe dV

(10.147)

Finally, we recognize that the 24 × 1 vector inside the curly brackets on theright-hand side of equation (10.147) above is the element internal force vector,i.e.,

f int

e=

∫

V

BMT

eSe dV (10.148)

10.5.3 Numerical results

Having obtained the element matrices and vectors, the final equation to beiterated upon using Newton’s method can now be written as

[

KM + KG]i

n+1∆di

n= Fn+1 − f i

n+1(10.149)


FIGURE 10.11

Dog-bone specimen in undeformed and deformed configurations.

The finite element mesh employed for the present dog-bone specimen is com-posed of 184 eight-node brick elements (one element through the thickness).The top view of the mesh in both the undeformed and deformed configurationsis shown in Figure 10.11. A uniform tensile traction distribution was appliedat the right end, and all three degrees of freedom were fixed for nodes lyingon the left edge. The magnitude of the applied traction was increased from0psi to 150 psi in 30 equal steps. Note that a traction of 150 psi correspondsto a total load of 18.75 lb based on the undeformed geometry of the surfaceon which the tractions are applied, i.e., (150 psi)(1.25 in)(0.1 in) = 18.75 lb.The material parameters, λ and µ, were chosen such that λ = 278 psi andµ = 417 psi. These material parameters would correspond to a Young’s mod-ulus of 1000 psi and a Poisson’s ratio of 0.2 for a linearly elastic and isotropicmaterial. A 2 × 2 × 2 quadrature rule was employed to fully integrate thetangent stiffness matrices and the internal force vector.

The convergence criterion was chosen to be a relative residual criterion.Recall that the out-of-balance force at the beginning of the ith iteration isdefined as the global external force vector minus the internal force vector, i.e.,

Ri = Fn+1 − f i

n+1(10.150)

In the present analysis, the solution was assumed to have converged when

‖Ri‖

‖R1‖≤ 0.1 (10.151)

where ‖Ri‖ is the norm of the out-of-balance force vector at the beginning ofthe ith iteration, and ‖R1‖ is the norm of that vector at the beginning of thefirst iteration. In the present numerical simulation, the convergence criterion(10.151) above was achieved within one to five iterations.

It turns out that when the external traction reaches 150 psi, the final lengthof the specimen is approximately twice the initial length, yielding a stretch of


λ

stre

ssm

easu

re(p

si)

σ11

P11

S11

1 1.2 1.4 1.6 1.8 2 2.20

200

400

600

800

1000

1200

FIGURE 10.12

Stress measures.

approximately 2. For the sake of interest, the various stress measures, evalu-ated near the middle of the narrow portion of the specimen, are plotted versusstretch in Figure 10.12. The solid line represents, σ11, the first component ofthe Cauchy stress. The dashed line represents, P11, the first component of thefirst Piola-Kirchhoff stress tensor, and the dashed-dotted line represents S11,the first component of the second Piola-Kirchhoff stress tensor. As shown inthe figure, there is a linear relationship between the Cauchy stress and thestretch. This is characteristic of neo-Hookean material behavior. When theexternal traction reaches 150 psi, we can also see that the first Piola-Kirchhoffstress reaches approximately 750 psi. We leave it to the reader to verify thatthis stress value is reasonable based on the fact that the first component ofthe first Piola-Kirchhoff stress physically represents the total load divided bythe original cross-sectional area. In addition, by observing the Cauchy stresswith stretch, one can infer that a 27% reduction in cross-sectional area of thenarrow portion of the specimen has occurred.

The present example not only illustrates the total Lagrangian finite ele-ment procedure for geometrically nonlinear problems, but it also emphasizesthe importance of understanding the alternative stress measures that can bereported during post-processing. In the present study, if we were to mistak-enly report the first or second Piola-Kirchhoff stress instead of the Cauchystress, we would significantly underestimate the physically meaningful stress.


10.6 Problems

Problem 10.1

(3,3)

X1

X2

ξ

(1,1)

(2,3)

FIGURE 10.13

Problem 10.1. Mapping of a quadratic-u element from parent to physicaldomains.

A three-node quadratic-u element is mapped from an original (unstretched)configuration to a deformed configuration as shown in Figure 10.13. Theoriginal length of the element is L. The mapping from the original to thedeformed configurations can be written as

X1 =

3∑

a=1

Naxa

1

X2 =

3∑

a=1

Naxa

2

where a is the local node number, xa

1and xa

2are the nodal coordinates in

the deformed configuration, and Na are the three shape functions. The shapefunctions can be written in terms of the natural coordinate, ξ = 2X1/L, asfollows:

N1(ξ) =1

2ξ(ξ − 1)

N2(ξ) = 1 − ξ2

N3(ξ) =1

2ξ(ξ + 1)


(a) Determine the coordinates of the point located at ξ = 0 in theoriginal configuration.

(b) Determine the stretch, λ, at the point ξ = 0 in the originalconfiguration.

(c) Assume that the material is incompressible. Derive a relationshipbetween the cross-sectional area, a, of the element in the deformedconfiguration and the cross-sectional area, A, in the originalconfiguration.

(d) Derive the geometric tangent stiffness matrix for the elementin the deformed configuration. Assume that S11 is the only nonzerocomponent of the second Piola-Kirchhoff stress tensor, and take therelationship between S11 and E11, the first component of theGreen-Lagrange strain tensor, to be linear.

Problem 10.2

Show that S and E are work conjugate pairs.

Problem 10.3

Consider a finite element mesh composed of one three-node triangle for planestrain applications. The local nodal coordinates in both the original anddeformed configurations are provided in Table 10.5.

(a) Determine the components of the deformation gradient tensor at apoint within the element.

(b) Determine the components of the Green-Lagrange strain at a pointwithin the element.

(c) Derive the 6 × 6 geometric tangent stiffness matrix for the element.

(d) Derive the 6 × 1 internal force vector. Assume a St.Venant-Kirchhoff lawand take E = 30 × 106 psi and ν = 0.3.

TABLE 10.5

Problem 10.3. Nodal coordinates(in inches) in original anddeformed configurations.

Local node X Y x y

1 0.0 0.0 0.0 0.02 0.5 0.2 0.6 0.33 0.3 0.5 0.3 0.5

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Index

Accumulated effective plastic strain,317

Alternative stress measures, 350, 389Arbitrary four-node quadrilateral, 131Area coordinates, 134Associated flow rule, 315Axisymmetric problems, 83, 110, 187

B-matrix, 60, 135, 136Back stress, 317Backward difference approximation,

281, 296Balance law, 94Banded matrix, 183Bauchinger effect, 317Bending B-matrix, 242Bending stiffness matrix, 246Biaxial stress, 311Bifurcation, 381Bi-unit cube, 154Body force, 84Boris Grigorievich Galerkin, 60Bulk modulus, 331

Candidate function, see trial func-tion

Capacitance matrix, 295Cartesian coordinate system, 12Cauchy-Green deformation tensor, 47Cauchy’s law, 84, 86Cauchy stress tensor, 86Cauchy’s tetrahedron, 84Central difference method, 266CFL number, 275Coefficient of thermal expansion, 109,

191Commutative, 7

Commutative law, 61Compliance matrix, 105Compliant, 140Conditionally stable, 268Conductance matrix, 189, 295Conformable, see matrix multiplica-

tionConnectivity list, 69Conservation of energy, 121, 291Consistency condition, 319Consistent mass matrix, 260Constant-strain element, 76Constant-strain triangle, 136Constitutive law, 162Control volume, 291Convergence criteria, 329Coordinate curves, 110Courant-Friedrichs-Lewy number, see

CFL numberCritical time step, 273Curvature vector, 240Cuthill-McKee algorithm, 183Cylindrical coordinates, 110

Deformation gradient tensor, 46, 351Deformed configuration, 131Degrees of freedom, 217Deviatoric stress, 306, 309Differential equation, 51Directional derivative, 33Dirichlet boundary condition, 53Distributed load, 230Divergence of a vector, 34Divergence theorem, 37, 94Ductile solid, 304Dyadic product, 17Dynamic amplification, 291

397

Eight-node brick, 131, 154Elasticity tensor, 103Elastoplastic tangent, 306, 317Element eigenvalue inequality, 273Energy density, 35Equilibrium equation, 83, 94Equivalent nodal loads, 172Essential boundary condition, 53, 125Euler’s equation, 50, 263Explicit time integration, 268Extensional strain, 75

Fiber pull-out problem, 110Film coefficient, 126, 189Finite element locking, 180First Piola-Kirchhoff stress, 360Follower force, 368Forced convection, 126, 189Fourier’s law, 124, 293Four-node square, 141Four-node tetrahedron, 131Free vibration, 263

Galerkin method, 60Galerkin procedure, 188Gaussian elimination, 183Gauss quadrature, 163

cubic polynomial, 163fifth-order polynomial, 165integration points, 164multi-dimensional integrals, 166quadrature point locations in phys-

ical domain, 167quadrature points, 164required number of quadrature

points, 165weight values, 164

Geometric nonlinearity, 349Geometric tangent stiffness matrix,

367, 375Global equation number, 217Green deformation tensor, 97, 352Green-Lagrange strain tensor, 352

Haigh-Westergaard stress space, 308

Hardening parameter, 305, 316Heat conduction

isotropic, 121steady-state, 121

Heat equation, 123Heat flux, 122Heat source, 122Heaviside unit step function, 275Hencky’s method, 249Hermite polynomials, 226Homogeneous deformation, 312Homogeneous equation, 263Hooke’s law, 83Hoop strain, 115Hoop stress, 195Hourglass control, 180Hyperelastic, 384

Idealized stress-strain curve, 304Implicit method, 296Implicit time integration, 284Incompatible deformation, 140Incompressible material, 179Indicial notation, 13Initial stress tangent stiffness matrix,

367In-plane strain components, 135Integration by parts, 28Interpolation functions, 2Isoparametric, 145Isoparametric finite elements, 131, 141Isoparametric hexahedron, 154Isotropic, 103Isotropic hardening, 317Isotropic heat conduction, 124, 293Isotropic:isotropic tensor, 103

Jacobian matrix, 41, 147John von Neumann, 53

Kinematic hardening, 317Kinematically admissible, 365Kronecker delta, 8, 134

Lame’s constants, 103, 179, 326

Lame’s solution, 341Laplace’s equation, 1, 125Lejeune Dirichlet, 53Level surface, 33Linear algebraic equations, 10

constraint, 10Linearization, 31Linearized theory of elasticity, 83Linear-u element, 60, 69Load correction tangent matrix, 371Local node numbering, 137Locking, 180

Master node, 193Material elasticity tensor, 367Material nonlinearity, 349Material points, 140Material tangent elasticity tensor, 384Material tangent stiffness matrix, 367,

377Matrix

cofactor matrix, 8conformable matrix multiplica-

tion, 6determinant of a square matrix,

8diagonal, 5element of a matrix, 5identity matrix, 7inverse, 9lower triangle, 5, 371multiplication, 6multiplication by a scalar, 6nonsymmetric, 371order of a matrix, 5singular, 9square, 5symmetric, 5transpose, 6upper triangle, 5, 371

Mean stress, 305Mindlin-Reissner plate theory, 238

inextensible fiber, 238kinematic assumptions, 238midsurface, 238

shear locking, 251

Natural base vector, 111Natural boundary condition, 53, 125Natural frequency, 264Neo-Hookean material, 384, 389Neutral axis, 222Neutral surface, 222Newton’s method, 38, 340, 364

quadratic convergence, 44Nodes, 51Nonassociated flow rule, 315Normal mode, 264Normal stress, 74

Octahedral plane, 311Odd and even functions, 181Orthogonal curvilinear coordinates,

112Out-of-balance force, 329

Parent domain, 143Partial differential equation, 94Patch test, 199Perfectly plastic, 305Period elongation error, 288, 291Period error, 270Periodic boundary condition, 193Permutation symbol, 14

even permutation, 15odd permutation, 15

Permutation tensor, 369Physical domain, 144Piola-Kirchhoff stress

first, 350second, 350

Plane strain, 106Plane stress, 106Plastic deformation, 304Poisson equation, 125Poisson-Kirchhoff theory, 249Positive definite, 262Postulate of maximum plastic dissi-

pation, 316Power series, 30

Principle of virtual work, 117Product rule, 29Proportional loading, 324Pullback, 359

Quadratic form, 262Quadratic-u element, 60Quasi-static, 257

Radial return method, 327, 331Ramberg-Osgood model, 323Reduced integration, 180Relative stretch, 96Residual, 38, 329Residual vector, 43

norm, 329Rigid-body translation, 136Rotation matrix, 354

Sampling frequency, 272Scalar triple product, 19, 152, 159Second-order tensor

skew, 26symmetric, 26

Second Piola-Kirchhoff stress, 360Selective reduced integration, 251Semi-discrete form, 261, 296Shape functions, 2, 133, 138Shear B-matrix, 243Shear stiffness matrix, 246Simple pendulum, 269Simple shear, 101Slave node, 193Small-strain tensor, 83, 96Snap-through buckling, 371Solid of revolution, 110Sparse matrix, 183Spherical coordinates, 127

gradient operator, 127Spurious oscillations, 277Stiffness proportional damping, 280Strain-displacement relationship, 75Stress concentration factor, 89Stress invariants, 308Stress update scheme, 327

Stretch, 96Stretch ratio, see stretchStrong form, 52, 95St.Venant-Kirchhoff material, 376Surface traction, 83

Tangent modulus, 304, 339Taylor series, 30Tensor transformation law, 20Test function, 53Thermal conductivity, 124, 293Thermal strain, 121Three-node triangle, 131, 136Total-Lagrangian formulation, 349Total strain, 304Transverse shear stress and strain,

239Trapezoidal rule, 282Trial function, 51, 137

C0 continuous, 52, 293C1 continuous, 52, 226continuous piecewise, 51piecewise linear, 51piecewise quadratic, 51vector valued, 134

Tri-linear shape functions, 155Tri-unit cube, 154Two-node space truss, 204

Unconditionally stable, 285, 298Undeformed configuration, 131, 140Unsteady heat equation, 291Updated-Lagrangian formulation, 350

Volume coordinates, 151Volumetric strain, 179, 331Von Mises effective stress, 314, 342Von Mises, Richard, 307

Wave equation, 259Weak form, 52Work conjugate, 350

Yield condition, 306Yield point, 304Young’s modulus, 75, 104, 304, 323

Index

Accumulated effective plastic strain,317

Alternative stress measures, 350, 389Arbitrary four-node quadrilateral, 131Area coordinates, 134Associated flow rule, 315Axisymmetric problems, 83, 110, 187

B-matrix, 60, 135, 136Back stress, 317Backward difference approximation,

281, 296Balance law, 94Banded matrix, 183Bauchinger effect, 317Bending B-matrix, 242Bending stiffness matrix, 246Biaxial stress, 311Bifurcation, 381Bi-unit cube, 154Body force, 84Boris Grigorievich Galerkin, 60Bulk modulus, 331

Candidate function, see trial func-tion

Capacitance matrix, 295Cartesian coordinate system, 12Cauchy-Green deformation tensor, 47Cauchy’s law, 84, 86Cauchy stress tensor, 86Cauchy’s tetrahedron, 84Central difference method, 266CFL number, 275Coefficient of thermal expansion, 109,

191Commutative, 7

Commutative law, 61Compliance matrix, 105Compliant, 140Conditionally stable, 268Conductance matrix, 189, 295Conformable, see matrix multiplica-

tionConnectivity list, 69Conservation of energy, 121, 291Consistency condition, 319Consistent mass matrix, 260Constant-strain element, 76Constant-strain triangle, 136Constitutive law, 162Control volume, 291Convergence criteria, 329Coordinate curves, 110Courant-Friedrichs-Lewy number, see

CFL numberCritical time step, 273Curvature vector, 240Cuthill-McKee algorithm, 183Cylindrical coordinates, 110

Deformation gradient tensor, 46, 351Deformed configuration, 131Degrees of freedom, 217Deviatoric stress, 306, 309Differential equation, 51Directional derivative, 33Dirichlet boundary condition, 53Distributed load, 230Divergence of a vector, 34Divergence theorem, 37, 94Ductile solid, 304Dyadic product, 17Dynamic amplification, 291

397

Eight-node brick, 131, 154Elasticity tensor, 103Elastoplastic tangent, 306, 317Element eigenvalue inequality, 273Energy density, 35Equilibrium equation, 83, 94Equivalent nodal loads, 172Essential boundary condition, 53, 125Euler’s equation, 50, 263Explicit time integration, 268Extensional strain, 75

Fiber pull-out problem, 110Film coefficient, 126, 189Finite element locking, 180First Piola-Kirchhoff stress, 360Follower force, 368Forced convection, 126, 189Fourier’s law, 124, 293Four-node square, 141Four-node tetrahedron, 131Free vibration, 263

Galerkin method, 60Galerkin procedure, 188Gaussian elimination, 183Gauss quadrature, 163

cubic polynomial, 163fifth-order polynomial, 165integration points, 164multi-dimensional integrals, 166quadrature point locations in phys-

ical domain, 167quadrature points, 164required number of quadrature

points, 165weight values, 164

Geometric nonlinearity, 349Geometric tangent stiffness matrix,

367, 375Global equation number, 217Green deformation tensor, 97, 352Green-Lagrange strain tensor, 352

Haigh-Westergaard stress space, 308

Hardening parameter, 305, 316Heat conduction

isotropic, 121steady-state, 121

Heat equation, 123Heat flux, 122Heat source, 122Heaviside unit step function, 275Hencky’s method, 249Hermite polynomials, 226Homogeneous deformation, 312Homogeneous equation, 263Hooke’s law, 83Hoop strain, 115Hoop stress, 195Hourglass control, 180Hyperelastic, 384

Idealized stress-strain curve, 304Implicit method, 296Implicit time integration, 284Incompatible deformation, 140Incompressible material, 179Indicial notation, 13Initial stress tangent stiffness matrix,

367In-plane strain components, 135Integration by parts, 28Interpolation functions, 2Isoparametric, 145Isoparametric finite elements, 131, 141Isoparametric hexahedron, 154Isotropic, 103Isotropic hardening, 317Isotropic heat conduction, 124, 293Isotropic:isotropic tensor, 103

Jacobian matrix, 41, 147John von Neumann, 53

Kinematic hardening, 317Kinematically admissible, 365Kronecker delta, 8, 134

Lame’s constants, 103, 179, 326

Lame’s solution, 341Laplace’s equation, 1, 125Lejeune Dirichlet, 53Level surface, 33Linear algebraic equations, 10

constraint, 10Linearization, 31Linearized theory of elasticity, 83Linear-u element, 60, 69Load correction tangent matrix, 371Local node numbering, 137Locking, 180

Master node, 193Material elasticity tensor, 367Material nonlinearity, 349Material points, 140Material tangent elasticity tensor, 384Material tangent stiffness matrix, 367,

377Matrix

cofactor matrix, 8conformable matrix multiplica-

tion, 6determinant of a square matrix,

8diagonal, 5element of a matrix, 5identity matrix, 7inverse, 9lower triangle, 5, 371multiplication, 6multiplication by a scalar, 6nonsymmetric, 371order of a matrix, 5singular, 9square, 5symmetric, 5transpose, 6upper triangle, 5, 371

Mean stress, 305Mindlin-Reissner plate theory, 238

inextensible fiber, 238kinematic assumptions, 238midsurface, 238

shear locking, 251

Natural base vector, 111Natural boundary condition, 53, 125Natural frequency, 264Neo-Hookean material, 384, 389Neutral axis, 222Neutral surface, 222Newton’s method, 38, 340, 364

quadratic convergence, 44Nodes, 51Nonassociated flow rule, 315Normal mode, 264Normal stress, 74

Octahedral plane, 311Odd and even functions, 181Orthogonal curvilinear coordinates,

112Out-of-balance force, 329

Parent domain, 143Partial differential equation, 94Patch test, 199Perfectly plastic, 305Period elongation error, 288, 291Period error, 270Periodic boundary condition, 193Permutation symbol, 14

even permutation, 15odd permutation, 15

Permutation tensor, 369Physical domain, 144Piola-Kirchhoff stress

first, 350second, 350

Plane strain, 106Plane stress, 106Plastic deformation, 304Poisson equation, 125Poisson-Kirchhoff theory, 249Positive definite, 262Postulate of maximum plastic dissi-

pation, 316Power series, 30

Principle of virtual work, 117Product rule, 29Proportional loading, 324Pullback, 359

Quadratic form, 262Quadratic-u element, 60Quasi-static, 257

Radial return method, 327, 331Ramberg-Osgood model, 323Reduced integration, 180Relative stretch, 96Residual, 38, 329Residual vector, 43

norm, 329Rigid-body translation, 136Rotation matrix, 354

Sampling frequency, 272Scalar triple product, 19, 152, 159Second-order tensor

skew, 26symmetric, 26

Second Piola-Kirchhoff stress, 360Selective reduced integration, 251Semi-discrete form, 261, 296Shape functions, 2, 133, 138Shear B-matrix, 243Shear stiffness matrix, 246Simple pendulum, 269Simple shear, 101Slave node, 193Small-strain tensor, 83, 96Snap-through buckling, 371Solid of revolution, 110Sparse matrix, 183Spherical coordinates, 127

gradient operator, 127Spurious oscillations, 277Stiffness proportional damping, 280Strain-displacement relationship, 75Stress concentration factor, 89Stress invariants, 308Stress update scheme, 327

Stretch, 96Stretch ratio, see stretchStrong form, 52, 95St.Venant-Kirchhoff material, 376Surface traction, 83

Tangent modulus, 304, 339Taylor series, 30Tensor transformation law, 20Test function, 53Thermal conductivity, 124, 293Thermal strain, 121Three-node triangle, 131, 136Total-Lagrangian formulation, 349Total strain, 304Transverse shear stress and strain,

239Trapezoidal rule, 282Trial function, 51, 137

C0 continuous, 52, 293C1 continuous, 52, 226continuous piecewise, 51piecewise linear, 51piecewise quadratic, 51vector valued, 134

Tri-linear shape functions, 155Tri-unit cube, 154Two-node space truss, 204

Unconditionally stable, 285, 298Undeformed configuration, 131, 140Unsteady heat equation, 291Updated-Lagrangian formulation, 350

Volume coordinates, 151Volumetric strain, 179, 331Von Mises effective stress, 314, 342Von Mises, Richard, 307

Wave equation, 259Weak form, 52Work conjugate, 350

Yield condition, 306Yield point, 304Young’s modulus, 75, 104, 304, 323

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Finite Element Method (Jorge Chavez)