Basic Theory of Finite Element Method 9782759829392

Table of contents :
Preface
About the Author
Contents
CHAPTER 1 Introduction of Finite Element Method
CHAPTER 2 Fundamentals of Elasticity Mechanics
CHAPTER 3 Weak Form of Equivalent Integration
CHAPTER 4 Elements and Shape Functions
CHAPTER 5 Isoparametric Element and Numerical Integration
CHAPTER 6 Finite Element Computation Scheme of Elasticity Problems
CHAPTER 7 Solutions of Linear Algebraic Equations
CHAPTER 8 Error Estimation and Adaptive Analysis
CHAPTER 9 Programs of Finite Element Method
Appendix A. Keyword Index
Appendix B. Matrix Calculation
Appendix C. Summary of Elements and Shape Functions
Appendix D. Gaussian Integration Points and Weights
Appendix E. Exercise Solutions
References

Citation preview

Textbooks for Tomorrow’s Scientists

Yongliang WANG

Basic Theory of Finite Element Method

Printed in France

EDP Sciences – ISBN(print): 978-2-7598-2938-5 – ISBN(ebook): 978-2-7598-2939-2 DOI: 10.1051/978-2-7598-2938-5 All rights relative to translation, adaptation and reproduction by any means whatsoever are reserved, worldwide. In accordance with the terms of paragraphs 2 and 3 of Article 41 of the French Act dated March 11, 1957, “copies or reproductions reserved strictly for private use and not intended for collective use” and, on the other hand, analyses and short quotations for example or illustrative purposes, are allowed. Otherwise, “any representation or reproduction – whether in full or in part – without the consent of the author or of his successors or assigns, is unlawful” (Article 40, paragraph 1). Any representation or reproduction, by any means whatsoever, will therefore be deemed an infringement of copyright punishable under Articles 425 and following of the French Penal Code. The printed edition is not for sale in Chinese mainland. Customers in Chinese mainland please order the print book from Science Press. ISBN of the China edition: Science Press 978-7-03-075291-8 Ó Science Press, EDP Sciences, 2023

To My Students

Preface

This book covers the finite element method and is intended to be used in courses taught in English or in bilingual courses. It is suitable for the knowledge system of undergraduate and postgraduate students majoring in engineering mechanics and similar majors. Most of the material in the book is based on the lecture notes of a bilingual course in computational mechanics given by the author. The lecture notes were used in undergraduate engineering mechanics majors from 2017 to 2022 and the related graduate civil engineering, energy power, and mining engineering majors from 2019 to 2022 at China University of Mining and Technology (Beijing). The lecture notes have been revised and optimised several times to form this book. The book introduces the basic theory of the finite element method, formulates a finite element computation scheme for elasticity problems, and provides typical computer programs. To enable readers to quickly master the knowledge system and solution process of the finite element method, the book focuses on the basic concepts, algorithm theory, and key techniques of this method. From the perspective of reforming teaching methods and enriching the curriculum system, this book meets the following four requirements: (1) English instruction: The book is written in English. Simple and common English words are used to facilitate the interpretation of keywords, thus reducing the “double challenges” in the English barrier of non-native readers and the knowledge comprehension of new curriculum. (2) Systematic knowledge: The book only requires basic knowledge of elasticity. Starting with the simple weak form of the one-dimensional elasticity problem and the finite element solution process, it introduces the basic concepts and key techniques, namely, element, isoparametric transformation, numerical integration, and computation scheme of two-dimensional and three-dimensional

DOI: 10.1051/978-2-7598-2938-5.c901 Ó Science Press, EDP Sciences, 2023

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Preface

elasticity problems, to enable readers to quickly grasp the knowledge system of the finite element method. (3) Practice exercises: Carefully selected homework exercises at the end of each chapter and programming projects are well designed to form a set of exercise libraries matched with the knowledge topics of the book for testing the student’s understanding of the material and the corresponding exercise solutions are provided. Mastering the basic theory of the finite element method and having certain programming skills are the basic requirements of a computational mechanics course. In addition to the theoretical knowledge, this book provides typical application programs so that readers can further practice by implementing the finite element method in specific problems. The electronic courseware related to this book can be available by contacting the author through the email: [email protected]. (4) Advanced applications: The book briefly introduces error estimation and the adaptive finite element method. This can serve as a reference for more advanced readers interested in high-performance computation and analysis techniques. This work was supported by the National Natural Science Foundation of China (Grant Nos. 41877275 and 51608301), Beijing Natural Science Foundation (Grant No. L212016), the Teaching Reform and Research Projects of Undergraduate Education of China University of Mining and Technology, Beijing (Grant Nos. J210613, J200709, and J190701), and the Yue Qi Young Scholar Project Foundation of China University of Mining and Technology, Beijing (Grant No. 2019QN14). Some of the material in this book refers to classical textbooks on the finite element method, and the pioneer authors and researchers are sincerely respected and appreciated. Because this book is restricted by the limited knowledge of its author, a few errors are unavoidable. The author hopes that experts, scholars, and other readers of this book will provide helpful suggestions for the book’s improvement. Dr. Yongliang Wang Department of Engineering Mechanics School of Mechanics and Civil Engineering State Key Laboratory of Coal Resources and Safe Mining China University of Mining and Technology (Beijing) July 2022

About the Author

Dr. Yongliang Wang is currently a researcher in the Department of Engineering Mechanics, School of Mechanics and Civil Engineering, State Key Laboratory of Coal Resources and Safe Mining, at China University of Mining and Technology (Beijing), and the head of computational mechanics group. He obtained his Ph.D. degree from the Department of Civil Engineering at Tsinghua University in 2014. In 2015, 2016, 2017, and 2019, he successively visited the Zienkiewicz Centre for Computational Engineering at Swansea University, UK, the Applied and Computational Mechanics Center at Cardiff University, UK, and the Rockfield Software Ltd, UK, to carry out cooperative research. In 2022 and 2023, he visited University of California, Berkeley and San Diego, USA, as visiting scholar. His research interests include high-performance adaptive finite element method, computation and analysis of rock damage and fracture, and structural vibration and stability. He has also taught computational mechanics at the undergraduate level and the basic theory of the finite element method, computational solid mechanics, and rock fracture mechanics, and frontier and progress in mechanics at the graduate level. He is a board member of the Soft Rock Branch of the Chinese Society for Rock Mechanics and Engineering and a member of the Chinese Society of Theoretical and Applied Mechanics, China Civil Engineering Society, and China Coal Society; moreover, he serves as a project expert of the National Natural Science Foundation of China, and project expert of the Degree Center of Ministry of Education of China. As the person in charge, he presided over eighteen research projects, including the National Natural Science Foundation of China, Beijing Natural Science Foundation, Teaching Reform and Research Projects of Undergraduate Education, the China Postdoctoral Science Foundation, Fundamental Research Funds for the Central Universities, Ministry of Education of China, Key Laboratory Open Project Foundation of Soft Soil Characteristics and Engineering Environment, and Yue Qi Young Scholar Project Foundation, CUMTB. He participated in twelve research projects, including the Major Scientific Research Instrument Development of the

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About the Author

National Natural Science Foundation and the National Key Research and Development Program of China. He published four books in English and more than 100 academic papers; furthermore, he obtained more than 50 software copyrights. He has received the Rock Mechanics Education Award, Chinese Society for Rock Mechanics and Engineering (2021), the Excellent Supervisor Award, CUMTB (2019, 2020), the Science and Technology Award, China Coal Industry Association (2019), the Yue Qi Young Scholar Award, CUMTB (2019), the Emerald Literate Highly Commended Paper Award (2018), and the Frontrunner 5000 (F5000) Top Articles in Outstanding S&T Journals of China (2016).

Contents Preface . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . V About the Author . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . VII CHAPTER 1 Introduction of Finite Element Method . . . . . . . . . . . . . . 1.1 Development Process of Finite Element Method . . 1.2 Computation Procedure of Finite Element Method 1.3 Main Contents of the Book . . . . . . . . . . . . . . . . . . 1.4 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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1 1 3 5 6

Fundamentals of Elasticity Mechanics . . . . . . . . . . . . . . . . . . . . . . 2.1 Displacements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.2 Strains . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.3 Stresses . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.4 Geometric Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.5 Constitutive Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.6 Equilibrium Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.6.1 Three-Dimensional Problems . . . . . . . . . . . . . . . . . . 2.6.2 Two-Dimensional Plane Stress and Strain Problems . 2.6.3 Two-Dimensional Axisymmetric Problems . . . . . . . . 2.7 Boundary Conditions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.8 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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Form of Equivalent Integration . . . . . . . . . . . . . . . . . . . . . . . . . Weak Form of Equivalent Integration for Differential Equations Weak Form of One-Dimensional Elasticity Problems . . . . . . . . . Finite Element Computation Based on Weak Form . . . . . . . . . . 3.3.1 Galerkin Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.3.2 Finite Element Computation . . . . . . . . . . . . . . . . . . . . . Global Assembly from One-Dimensional Elements . . . . . . . . . . . Treatments on Boundary Conditions . . . . . . . . . . . . . . . . . . . . . Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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CHAPTER 2

CHAPTER 3 Weak 3.1 3.2 3.3

3.4 3.5 3.6

Contents

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CHAPTER 4 . . . . . . . . . . . . . . . . . . . . . . . .

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Isoparametric Element and Numerical Integration . . . . . . . . . . . . . . . . 5.1 Isoparametric Element . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.1.1 One-Dimensional Isoparametric Lagrange Element . . . . . 5.1.2 Two-Dimensional Isoparametric Triangle Element . . . . . 5.1.3 Two-Dimensional Isoparametric Rectangle Element . . . . 5.1.4 Three-Dimensional Isoparametric Tetrahedron Element . 5.1.5 Three-Dimensional Isoparametric Hexahedron Element . 5.1.6 Requirements of Isoparametric Element . . . . . . . . . . . . . 5.2 Numerical Integration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.2.1 One-Dimensional Integration for Lagrange Element . . . . 5.2.2 Two-Dimensional Integration for Triangle Element . . . . 5.2.3 Two-Dimensional Integration for Rectangle Element . . . 5.2.4 Three-Dimensional Integration for Tetrahedron Element 5.2.5 Three-Dimensional Integration for Hexahedron Element . 5.2.6 Required Order of Numerical Integration . . . . . . . . . . . . 5.3 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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59 59 59 60 63 64 67 69 70 70 73 74 75 76 78 78

Elements and Shape Functions . . . . . . . . . . . . . . . . . . . . . . 4.1 One-Dimensional Lagrange Element . . . . . . . . . . . . . 4.1.1 Linear Element with Two Nodes . . . . . . . . . . 4.1.2 Higher-Order Lagrange Element . . . . . . . . . . 4.1.3 Quadratic Lagrange Element . . . . . . . . . . . . . 4.2 Two-Dimensional Triangle Element . . . . . . . . . . . . . . 4.2.1 Triangle with Three Nodes . . . . . . . . . . . . . . . 4.2.2 Higher-Order Triangle Element . . . . . . . . . . . 4.2.3 Quadratic Triangle Element . . . . . . . . . . . . . . 4.2.4 Cubic Triangle Element . . . . . . . . . . . . . . . . . 4.3 Two-Dimensional Rectangle Element . . . . . . . . . . . . 4.3.1 Linear Rectangle Element with Four Nodes . . 4.3.2 Higher-Order Rectangle Element . . . . . . . . . . 4.3.3 Quadratic Rectangle Element . . . . . . . . . . . . 4.4 Three-Dimensional Tetrahedron Element . . . . . . . . . . 4.4.1 Linear Tetrahedron Element with Four Nodes 4.4.2 Higher-Order Tetrahedron Element . . . . . . . . 4.4.3 Quadratic Tetrahedron Element . . . . . . . . . . . 4.4.4 Cubic Tetrahedron Element . . . . . . . . . . . . . . 4.5 Three-Dimensional Hexahedron Element . . . . . . . . . . 4.5.1 Hexahedron with Eight Nodes . . . . . . . . . . . . 4.5.2 Higher-Order Hexahedron Element . . . . . . . . 4.5.3 Quadratic Hexahedron Element . . . . . . . . . . . 4.6 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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CHAPTER 5

Contents

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CHAPTER 6 Finite Element Computation Scheme of Elasticity Problems . . 6.1 Weak Form for General Elasticity Problems . . . . . . . . . 6.2 Finite Element Method for Solving Elasticity Problems . 6.3 Global Assembly from High-Dimensional Elements . . . . 6.4 Treatments on Boundary Conditions . . . . . . . . . . . . . . . 6.5 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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CHAPTER 7 Solutions of Linear Algebraic Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101 7.1 LU Decomposition Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101 7.2 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 106 CHAPTER 8 Error Estimation and Adaptive Analysis . . . . . . . . . . . . . . . . 8.1 Error Estimation of Finite Element Solutions . . . . . . . 8.1.1 Error of Finite Element Solutions . . . . . . . . . . 8.1.2 Superconvergent Patch Recovery Method . . . . . 8.2 Adaptive Finite Element Method . . . . . . . . . . . . . . . . 8.2.1 Categories of Adaptive Finite Element Method 8.2.2 h-Version Adaptive Finite Element Method . . . 8.2.3 hp-Version Adaptive Finite Element Method . . 8.3 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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CHAPTER 9 Programs of Finite Element Method . . . . . . . . . . . . . . . . . 9.1 One-Dimensional Program of Beam Deformation . . 9.1.1 Main Program . . . . . . . . . . . . . . . . . . . . . . . 9.1.2 Numerical Example . . . . . . . . . . . . . . . . . . . 9.1.3 Interactive Interface . . . . . . . . . . . . . . . . . . . 9.2 Two-Dimensional Program of Plane Strain Problem 9.2.1 Main Program . . . . . . . . . . . . . . . . . . . . . . . 9.2.2 Numerical Example . . . . . . . . . . . . . . . . . . . 9.2.3 Interactive Interface . . . . . . . . . . . . . . . . . . . 9.3 Three-Dimensional Program of Solid Compression . . 9.3.1 Main Program . . . . . . . . . . . . . . . . . . . . . . . 9.3.2 Numerical Example . . . . . . . . . . . . . . . . . . . 9.3.3 Interactive Interface . . . . . . . . . . . . . . . . . . . 9.4 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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Contents

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Appendix A. Keyword Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 165 Appendix B. Matrix Calculation . . . . . B.1 Definition . . . . . . . . . . . . . . . . . B.2 Matrix Addition or Subtraction B.3 Transpose . . . . . . . . . . . . . . . . . B.4 Transpose of a Product . . . . . . B.5 Inverse . . . . . . . . . . . . . . . . . . . B.6 Symmetric Matrices . . . . . . . . . B.7 Partitioning . . . . . . . . . . . . . . .

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169 169 170 171 172 172 172 172

Appendix C. Summary of Elements and Shape Functions C.1 One-Dimensional Lagrange Element . . . . . . . . . . . C.2 Two-Dimensional Triangle Element . . . . . . . . . . . C.3 Two-Dimensional Rectangle Element . . . . . . . . . . C.4 Three-Dimensional Tetrahedron Element . . . . . . . C.5 Three-Dimensional Hexahedron Element . . . . . . .

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Appendix D. Gaussian Integration Points and Weights . . . . . . . . . . . . . . . . . 179 Appendix E. Exercise Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 189 References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 207

Chapter 1 Introduction of Finite Element Method 1.1

Development Process of Finite Element Method

Understanding the behaviour and mechanisms of nature is fundamental in scientific research. Owing to the limited capacity of the human mind, we fail to grasp the complexity of the world in its entirety. Thus, engineers, scientists, or even economists subdivide a system under investigation into individual components, or ‘elements,’ the behaviour of which is easy to understand and reconstruct the original system from these components so that its behaviour can be naturally studied. In many cases, we can use a limited number of well-defined components to get an effective model, and such problems are called ‘discrete’. In other cases, the subdivision continues indefinitely, and the problem can only be defined by the fictitious concept of infinitesimals (which are not rigorously defined in standard mathematical analysis). This leads to differential equations or equivalent statements, which contain an infinite number of elements. Such systems are correspondingly called ‘continuous’. With the development of digital computers, discrete problems can usually be solved easily, even with a large number of elements. Since the capacity of any computer is limited, continuous problems can only be solved accurately by mathematical manipulations. However, the mathematical techniques available for exact solutions are usually limited to oversimplified situations. In order to solve realistic continuous problems, various discretisation methods have been put forward. All these methods involve an approximation that, ideally, as the number of discrete variables increases, it approaches the limit of the true continuum solution. Because most classical mathematical approximation methods and various direct approximation methods used in engineering belong to this category, it is difficult to determine the origin of the finite element method and the exact time of its invention. These approximate numerical methods definitely lay a crucial foundation for the generation of the finite element method, and promote the development of relevant numerical algorithms and numerical models in computational mechanics. The historical development of approximate numerical methods involved in the finite element method can also be obtained in some review papers. Figure 1.1 shows the evolution process, which leads to the concept of finite element analysis: DOI: 10.1051/978-2-7598-2938-5.c001 © Science Press, EDP Sciences, 2023

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Basic Theory of Finite Element Method

FIG. 1.1 – Historical development of approximate numerical methods involved in finite element method.

(1) Mathematicians and engineers have different views on the discretization of continuous problems. Since 1870, some mathematicians have developed general techniques that can be directly applied to differential equations, such as the trial functions (Ritz, 1908; Rayleigh, 1870), variational methods (Ritz, 1908; Rayleigh, 1870), weighted residuals (Biezeno and Koch, 1923; Galerkin, 1915; Gauss, 1795). Furthermore, using the trial function in each partition domain, the piecewise continuous trial functions were developed (Zienkiewicz and Cheung, 1964; Argyris and Kelsey, 1960; Prager and Synge, 1947; Courant, 1943). (2) The differential equations can be approximately expressed by difference equations. In 1910, the classical difference method was established (Southwell, 1946; Liebman, 1918; Richardson, 1910). Then, based on the variational principle, the variational difference method was developed (Feng, 1965; Wilkins, 1964; Varga, 1962). It should be emphasised that, in China, K. Feng of the Chinese Academy of Science also proposed a discretisation numerical method based on the variational principle for solving elliptic partial differential equations in 1965. He is regarded as one of the pioneers in the basic theory of the finite element method: ‘Independently of parallel developments in the West, he (Feng) created a theory of the finite element method. He was instrumental in both the implementation of the method and the creation of its theoretical foundation using estimates in Sobolev spaces’ (Lax, 1993). (3) Some engineers and technicians directly used the method of analogy structure analysis to discretize the continuous domain, and the structural analog substitution was developed (Newmark, 1949; McHenry, 1943; Hrenikoff, 1941).

Introduction of Finite Element Method

3

Subsequently, the method of direct continuum elements was developed, and the representative research works are from Turner et al. (1956) and Clough (1960). Turner et al. proved that a more direct but equally intuitive replacement of properties could be made significantly and more effectively by considering that small portions or ‘elements’ in a continuum behave in a simplified way. Clough used the term ‘finite element,’ which was born based on the view of ‘direct analogy’ in engineering, implying the direct use of a standard methodology applicable to discrete systems. This concept is extremely important, as it allows for improved understanding from the perspective of conceptual; meanwhile, it provides a unified method for various problems and the development of standard computing programs. With the development of the above methods, various schemes for analysing problems of discrete nature have developed into a standard finite element method. When studying a structure, civil engineers first determine the force–displacement relations for each element of the structure and then assemble the structure by following a well-defined procedure to establish a local equilibrium at each ‘node’ or connecting point of the structure. The equations obtained can solve the unknown displacements. All of these analyses follow a standard pattern that is generally applicable to discrete systems. Thus, it is possible to define a standard discrete system, and this section focuses on establishing the procedures applicable to such systems. Using a unified treatment of standard discrete problems, the finite element process can be defined as a method of approximation to continuum problems as follows: (a) The continuum is divided into a finite number of elements whose behaviour is specified by a finite number of parameters. (b) As an assembly of its elements, the solution for the complete system precisely follows the same rules as those applied to standard discrete problems. The purpose of this book is to use the conventional and standard finite element method as a general discretisation procedure for constructing standard discrete problems from continuum problems. Furthermore, some high-performance finite element algorithms (such as the adaptive optimization algorithm) have been developed; readers can further comprehend based on this basis of the standard method.

1.2

Computation Procedure of Finite Element Method

The reader should be aware that the finite element solution of the problem always follows the standard method, in which the process of solving any type of problem is always carried out by the following steps: (1) Define the problem to be solved by the governing differential equations. Based on the differential equations in the analysis of continuous systems, the equivalent integration form of the problem is constructed as virtual work, variational or weak form.

4

Basic Theory of Finite Element Method

(2) Select the type and order of the finite element. These elements and corresponding shape functions will be used in the analysis of discrete systems, which will be substituted into the equivalent integral form. (3) Define a set of mesh for the problem. This involves the description of the distribution of nodes and elements and the description of the boundary conditions. (4) Compute and assemble the matrixes from elements. The relationships in all elements are defined and assembled by the equivalent integral form of the problem; for example, the global stiffness matrix and load vector are set from each element using the element location vector. In this way, the continuous system is meshed and discretized to form an approximate discrete system. In this way, the differential equations representing continuous systems are transformed into algebraic equations of discrete systems. (5) Solve the system of algebraic equations. For widespread static problems, the algebraic equations are especially linear, which could be solved effectively by some well-developed mathematical techniques or specialized solvers. The solutions of the system of algebraic equations are solutions on all nodes. (6) Derive the solutions of variables in the global domain. Using the solutions on the nodes and the solutions of displacement, stress, and strain (derivatives of displacement) in each element domain can be obtained by further applying the interpolation of shape functions. Based on the above analysis steps, this book introduces the solution strategy and basic process of the finite element method. This book introduces the widely-used displacement-type finite element method. This method takes displacement u as the basic unknown variable. As shown in figure 1.2, in the continuous system, the governing equations of mechanical behaviors are the differential equations Lu = 0 (where L is the differential operator) and the equivalent integral equation G (w, u) = 0 (where G is the integral operator and w is an arbitrary function used) is obtained through equivalent transformation (such as weak form transformation). Then through discretization (using element and shape function), the algebraic equations Ku = f (where K and f are global stiffness matrix and load vector, respectively) on the approximate discrete system are obtained. Solving the algebraic equations can

FIG. 1.2 – Computation procedure from differential equations to integral equations and then to algebraic equations.

Introduction of Finite Element Method

5

derive the approximate solutions of the basic unknown displacement u. This book provides, more specifically, the mathematical basis derived from these classic ideas.

1.3

Main Contents of the Book

This book introduces the basic theory of the finite element method for solving elastic mechanics problems and presents computer programs for finite element analysis. It is divided into nine chapters covering the above three aspects: elasticity mechanics theory, finite element methodology, and finite element program. In particular, the finite element methodology is the focus of this book, including basic techniques and computation schemes. Figure 1.3 presents a structural chart of the main contents and chapters: (1) In chapter 1, the historical development, as well as the computation procedure of the finite element method are briefly discussed, and a contents list is provided. (2) In chapter 2, the fundamentals of elasticity mechanics are introduced. (3) In chapter 3, the weak form of equivalent integration is presented. (4) In chapter 4, the elements and shape functions for one-, two-, and threedimensional problems are discussed.

FIG. 1.3 – Structural chart of main contents and chapters in this book.

Basic Theory of Finite Element Method

6

(5) In chapter 5, the isoparametric element and numerical integration are introduced. (6) In chapter 6, the finite element computation scheme of elasticity problems is described. (7) In chapter 7, an effective numerical method for solving linear algebraic equations is introduced. (8) In chapter 8, the error estimation and adaptive analysis for high-performance algorithms are presented. (9) In chapter 9, some typical computer programs for finite element analysis are provided, and their use is illustrated.

1.4

Exercises

1.4.1 Describe the treatment of ‘standard discrete problems’. 1.4.2 Summarise the computation procedure of the finite element method.

Chapter 2 Fundamentals of Elasticity Mechanics This chapter covers equations for the elasticity theory and field theory, which are pretty common in diverse applications in engineering. First, the general three-dimensional elastic cases are studied and simplified to special two-dimensional problems. In the elastic equations given here, it is assumed that the strain is less than 1. It is assumed that the reader has got linear elasticity theory. Yet, for the sake of comprehensiveness, the basic equations for the different categories to be considered are summarized. The reader can consult standard references and technical books about the theme for a more general discussion (Love, 2011; Ciarlet, 1988; Timoshenko and Goodier, 1970; Lekhnitskii, 1963; Sokolnikoff, 1956; Muskhelishvili, 1953) and refer to the literature on large strain effects (Zienkiewicz et al., 2013a; Belytschko et al., 2000). In some situations, the indicial form of equations can be prior. Yet, in this part, the basic equations are expressed in the form of a matrix and later applied to establish finite element development. The basic equation of elasticity theory is explained by stress, strain, displacement, initial conditions, boundary conditions, and the constitutive relationship between stress and strain. First, each equation group of a general three-dimensional problem in a Cartesian coordinate system is designated, as displayed in figure 2.1. In addition, two-dimensional forms are considered. The issues of the two dimensions are as follows: (a) Plane stress case. There is only nonzero stress in the problem plane here and no stress in the direction orthogonal to the thin plate, as exhibited in figure 2.2a. (b) Plane strain case. It is assumed that the strain perpendicular to the plane under consideration is zero. This may occur in a prism, as shown in figure 2.2b, where the load perpendicular to the plane remains unchanged. (c) Axisymmetric case. In the cylindrical coordinate system r − z − θ, the angle θ in the plane considered is constant, as displayed in figure 2.2c. It is assumed that all components of stress, strain, and displacement are only related to r and z.

DOI: 10.1051/978-2-7598-2938-5.c002 © Science Press, EDP Sciences, 2023

Basic Theory of Finite Element Method

8

FIG. 2.1 – Three-dimensional elasticity problems.

FIG. 2.2 – Two-dimensional elasticity problems.

2.1

Displacements

In terms of the three-dimensional issue, the field of displacement is provided by: 8 9 < u ðx; y; z Þ = uðxÞ ¼ v ðx; y; z Þ ; ð2:1Þ : ; w ðx; y; z Þ where the position is given in Cartesian coordinates x, y, z, which may also be expressed in the form of a matrix as: 8 9

> x > > > 6 > > > @ 7 e > > 78 9 6 y > > 0 = 6 0 < 7< u = @z 7 ez 6 ð2:12Þ ¼6 e¼ 7 v : cxy > > 7: ; 6 @ @ > > w > > 0 7 > 6 > > 7 > 6 > cyz > ; : 7 6 @y @x czx 7 6 @ @ 7 6 7 6 0 6 @z @y 7 7 6 4 @ @ 5 0 @z @x In order to give thought to the three kinds of two-dimensional issues in a coincident way, four strain components are included in e, hence, 3 2 @ 0 7 8 9 2 3 6 7 6 @x ex 0> 7  > 6 @ > = < > 7 6 ey 7 6 0 u 0 7 6 7¼6 @y e¼6 þ ¼ S p u þ ez ð2:13Þ 7 4 ez 5 6 e > > 7 v > ; : z> 7 6 0 0 cxy 0 7 6 4 @ @ 5 @y

@x

in terms of the problems of the plane (where ez denotes that plane strain is zero, note that it is not suitable for plane stress), and 3 2 @ 0 7 6 8 9 6 @r 7 7 6 @ e > > r > > 7  6 0 < = 6 7 u ez @z 7 6 ¼6 1 e¼ ð2:14Þ 7 v ¼ Sa u e > > h > > 6 0 7 : ; 6 7 crz 7 6 r 5 4 @ @ @z @r

Basic Theory of Finite Element Method

12

in terms of the issues of axisymmetry, as displayed in figure 2.4. The difference between the types of two-dimensional problems only exists in plane stress problems εz and the existence of component εθ in the case of axisymmetry.

2.5

Constitutive Equations

All the above equations have nothing to do with the material properties under consideration. In order to implement this theory, the relationship between stress and strain should be established, which may reflect the properties of different materials. The relations are called stress–strain equations or constitutive equations, which can arise in complex forms, yet in this procedure, the simplest linear elastic case is taken into consideration. Under this assumption, the equations of stress–strain for a linear elastic material are given as: r ¼ De

ð2:15aÞ

e ¼ D1 r;

ð2:15bÞ

or

where the matrix D is called the elasticity matrix, and its inverse, D−1, is the elasticity matrix of compliances (Timoshenko and Goodier, 1970). In terms of materials of elasticity, the stress can also be inferred from the energy storage function W displayed for the strains, therefore r¼

@W : @e

ð2:16Þ

It is applicable to both linear and nonlinear materials, hence, it can be utilized to generalise the above forms. In terms of the linear elastic equations in equation (2.15a), the stored energy function is as: 1 W ðeÞ ¼ eT De: 2

ð2:17Þ

The complementary energy U(σ) can also be defined from which the strain can be derived as: e¼

@U ; @r

ð2:18Þ

where in the linear elastic case, 1 U ðrÞ ¼ rT D1 r: 2

ð2:19Þ

The application of a complementary or stored energy form also guarantees that D and D−1 are symmetric.

Fundamentals of Elasticity Mechanics

13

In isotropic materials, there is no change in the matrix of elasticity when a coordinate alternation is used. The general expression of isotropic materials is written with six stress and strain terms. In terms of isotropic materials, any two independent elastic constants can be used. For instance, under Cartesian coordinates, the form is offered by: 9 9 8 38 2 rx > ex > 1 v v 0 0 0 > > > > > > > > > > > 7> 6 v 1 v 0 0 0 ey > ry > > > > > > > > 7> = 16 < < 7 rz = 6 v v 1 0 0 0 ez 7 6 ¼ ð2:20Þ 7> sxy >: 0 0 2ð 1 þ v Þ 0 0 > E6 > cxy > > > 7> 6 0 > > > > > 5> 4 0 0 0 0 2ð1 þ v Þ 0 c > s > > > > > > > ; ; : yz > : yz > 0 0 0 0 0 2ð 1 þ v Þ czx szx The elasticity matrix of moduli is obtained by inverting: 3 2 ð1  v Þ v v 0 0 0 7 6 v ð1  v Þ v 0 0 0 7 6 7 E6 v v ð 1  v Þ 0 0 0 7; ð2:21Þ 6 D¼ 6 7 0 0 0 ð 1  2v Þ=2 0 0 d6 7 5 4 0 0 0 0 ð1  2v Þ=2 0 0 0 0 0 0 ð1  2v Þ=2 where d = (1 + v)(1 − 2v). In order to keep all the parameters of materials positive, the form of d limits the assigned material parameter values of v. Hence, 1 1\v\ : 2

ð2:22Þ

The value range of the parameter is − 1 < v < 1/2. In terms of isotropic materials, the form of a two-dimensional problem is constituted using four stress and strain terms via ignoring the last two columns and rows in equation (2.20). Via Cartesian coordinates, the form turns. 9 9 8 2 38 ex > rx > 1 v v 0 > > > > > > rz >: ez > 0 > E v v 1 > > > > ; ; : : cxy sxy 0 0 0 2ð 1 þ v Þ In the example of plane stress, rz must be set to zero to calculate the matrix D. Thus  v rx þ ry : ez ¼  ð2:24Þ E The inverse of equation (2.23) can be obtained. 9 9 8 2 38 ex > rx > 1 v 0 0 > > > > > > = < < 6v 1 0 7 ey = E ry 0 6 7 ¼ 5 > ez > : rz > 0 > ð1  v 2 Þ 4 0 0 0 > > > > ; ; : : cxy sxy 0 0 0 ð1  v Þ=2

ð2:25Þ

Basic Theory of Finite Element Method

14

The third row and column of array D are filled to obtain the stresses. In the plane strain and axisymmetric problems, the inverse is directly obtained via equation (2.23), as all stress components are nonzero. The following expression can be derived in Cartesian coordinates: 9 9 8 2 38 ex > rx > ð1  v Þ v v 0 > > > > > > = E6 < < 7 ey = ry v ð1  v Þ v 0 7 ¼ 6 ð2:26Þ 5 > ez > ; rz > v ð1  v Þ 0 > d4 v > > > > ; ; : : cxy sxy 0 0 0 ð1  2v Þ=2 which is consistent with the problem of three dimensions if the last two rows and columns of each array in equation (2.22) are ignored. It is noticed that the case where ez is zero is handled by only substituting this value when the stresses are computed, yet rz is always nonzero unless the Poisson’s ratio v is zero.

2.6 2.6.1

Equilibrium Equations Three-Dimensional Problems

The linear momentum or equilibrium equations for the solid behaviour of three dimensions can be described in Cartesian coordinates, given by @rx @syx @szx þ þ þ bx ¼ 0 @x @y @z @sxy @ry @szy þ þ þ by ¼ 0 @x @y @z @sxz @syz @rz þ þ þ bz ¼ 0 @x @y @z

ð2:27Þ

The stress is symmetric for angular momentum (moment equilibrium). Using only the independent components of the symmetric stress offered in equation (2.8), the equilibrium equations, which are in Cartesian coordinates, can be expressed by: ST r þ b ¼ 0;

ð2:28Þ

where, in the system of Cartesian coordinates, S denotes the differential operator, which is identical to that for strain in equation (2.11), and b represents the vector of body forces, shown by: b ¼ ½ bx

2.6.2

by

bz T :

ð2:29Þ

Two-Dimensional Plane Stress and Strain Problems

In the two-dimensional plane problems, τyz, τzx, and bz are ignored. It is noted that σz does not relate to the z coordinate. Hence, the stress matrix is expressed by equation (2.9), and the two equilibrium equations left can be provided as:

Fundamentals of Elasticity Mechanics @rx @syx þ þ bx ¼ 0 @x @y @sxy @ry þ þ by ¼ 0 @x @y

15

ð2:30Þ

By noting that τxy = τyx, the equilibrium equations can be displayed by: ST p r þ b ¼ 0;

ð2:31Þ

where Sp is identical to that in equation (2.13), and b and u have only two components.

2.6.3

Two-Dimensional Axisymmetric Problems

In axisymmetric problems, the stress is displaced in equation (2.10), and body force is displayed by b = [br bz]T. Via the stresses on the volume element expressed in figure 2.3 and summing the forces in the directions of r and z, the two equilibrium equations are shown as: @rr rr  rh @szr þ þ þ br ¼ 0 @r r @z @srz @rz þ þ bz ¼ 0 @r @z

ð2:32Þ

By moment equilibrium (angular momentum), τrz = τzr is obtained again. Note that in the axisymmetric problem, the equilibrium differential operator is shown as: 2 3 @ 1 1 @ þ 0  6 @r r r @z 7 T 7; ð2:33Þ Sa ¼ 6 4 5 @ @ 0 0 @z @r and is unequal to ST a . This distinction is owing to the system of curvilinear coordinates. This is a disadvantage if the differential equation form is used, yet for approaches applied in the finite element analysis, it is exhibited that the distinction vanishes, and expressions in the curvilinear coordinate system allow significant simplifications.

2.7

Boundary Conditions

The equilibrium and strain–displacement equations are valid for all body domain points, denoted by Ω. Boundary conditions (denoted as Γ) are appointed for each point on the solid surface. There are two kinds of boundary conditions to be thought of: (a) displacement boundary conditions and (b) traction boundary conditions. The boundaries can be sorted into two classifications, containing the displacement

Basic Theory of Finite Element Method

16

conditions on the domain Γu and the traction conditions on the domain Γt. Hence, the entire boundary is the collection of the displacement domain and traction domain. The condition of the displacement boundary is appointed at each point of the boundary Γu as: u ¼ uðxÞ;

x 2 Cu ;

ð2:34Þ

where u is the given value and x is the boundary point. The traction boundary conditions are designated for all boundary points on Γt and provided for the following stress equation: t ¼ GT r ¼ tðxÞ;

x 2 Ct ;

where, as to three-dimensional problems, GT denotes 2 nx 0 0 ny 0 GT ¼ 4 0 ny 0 nx nz 0 0 nz 0 ny

ð2:35Þ the matrix provided by: 3 nz 0 5; ð2:36Þ nx

with nx, ny, and nz being the direction cosines for the outward-pointing normal n to the boundary Γt. Note that matrices G and S are with identical nonzero structures. In two-dimensional plane problems, G is reduced to:

 nx 0 0 n y T Gp ¼ ; ð2:37Þ 0 ny 0 nx where nx and ny are the components of the unit outward normal vectors on the boundary. In the case of axial symmetry, nx = nr and ny = nz are obtained, meanwhile, Ga is also nonzero-like Gp . Both types of boundary conditions are vectors, constraining all of these points on the boundaries. It is promising to designate some components by displacement conditions and other components by traction conditions.

2.8

Exercises

2.8.1 Summarise the basic variables and equations of three- and two-dimensional elasticity. 2.8.2 Using the formulation presented in this chapter, provide the basic variables and equations of axial deformation for the one-dimensional elastic cantilever beam shown in figure 2.5. The coordinates of the two ends are x = a and x = b, and the axial uniform load is bx.

FIG. 2.5 – Axial deformation of one-dimensional elastic cantilever beam.

Chapter 3 Weak Form of Equivalent Integration 3.1

Weak Form of Equivalent Integration for Differential Equations

The previous chapter lists governing partial differential equations of elastic mechanics and the boundary conditions. The general solutions obtained by this method are rarely suitable and even challenging to solve in science and engineering. Therefore, we need to find other solutions; for example, we can use other forms to characterize the identical problem in order to get a more accurate approximate solution with this method. In this chapter, we will introduce the weak form of differential equations. Through the following steps, we can obtain a weak form for a set of differential equations: (1) (2) (3) (4)

Multiply each differential equation by an appropriate arbitrary function. Integrate the product over the space domain of the problem. Reduce the order of the involved derivatives using the integration by parts. Introduce the boundary conditions reasonably.

In this chapter, we only integrate into the space domain, therefore it is unnecessary to express the time dependency and initial conditions. Its integral time domain can also be considered (Katona and Zienkiewicz, 1985; Wood, 1984; Zienkiewicz et al., 1984), but it is not necessary to study here. It can construct a weak form when decomposing any differential equation by using the above steps. An arbitrary function is one that can take any conceivable value. The arbitrary function can use a polynomial, a trigonometric function, a Heaviside function, a Dirac delta function, or any other function to take any conceivable value. The characteristic of arbitrary function is virtual displacement, which may be the reader already knows. Here, let’s consider a weak form for the one-dimensional equilibrium equation of elasticity first, which can help us understand the above process.

3.2

Weak Form of One-Dimensional Elasticity Problems

First, a one-dimensional elasticity problem with a single component x in a Cartesian coordinate system is considered a representative example. The displacement DOI: 10.1051/978-2-7598-2938-5.c003 © Science Press, EDP Sciences, 2023

Basic Theory of Finite Element Method

18

u(x) can be expressed as a function of the coordinate x. It can be assumed that the problem is defined in a domain X, and also have some interval that a < x < b. So we write the displacement to u(x) and found that all strain components to be zero as defined in the previous chapter except ex ¼

@u : @x

ð3:1Þ

Reduce the system of multiple equilibrium equations into a single equation: @rx þ bx ¼ 0; @x

ð3:2Þ

and derive the single constitutive equation as follows: rx ¼ Eex ;

ð3:3Þ

where E is Young’s modulus of elasticity. The following boundary conditions will be introduced: u¼u

or tx ¼ t x ¼ nx rx ;

on x ¼ a; b;

ð3:4Þ

where nx is the unit outward normal, which is +1 at b and −1 at a. We can combine the above equations by substituting the constitutive and strain– displacement equations into the equilibrium equations. Then we can get the equilibrium equation expressed by displacement as:
 @ @u E ð3:5Þ þ bx ¼ 0; @x @x where E can be a function of x. There is only a single dependent variable u(x), it is an irreducible form of the differential equation. The set of equations (3.1)–(3.4) is referred to as the strong form of the one-dimensional elasticity problem. We can use these equations to formulate a set of discrete equations obtained by the finite difference method (Hildebrand, 1987; Mitchell and Griffiths, 1980; Forsythe and Wasow, 1960). However, a better way is to use the steps described above to introduce an integral weak form of the equations. Then the discrete equations are derived based on this weak form. First, in domain X described by an interval a < x < b, we introduce an arbitrary function w(x). By multiplying the equilibrium equation (3.2) by this function, we get that
 @rx g ðx; u; rx Þ ¼ w ðx Þ bx þ ¼ 0: ð3:6Þ @x It is more convenient to have a form equal to zero because if w(x) is zero everywhere except where x = xc, then for equation (3.6) to be zero, at x = xc, it has to satisfy the differential equation. By repeating all points, it is a conclusion that at all points in the domain differential, the equation is satisfied, provided that w(x) is arbitrariness. Then, in the next step, we obtain an integral form that is also zero by integrating over the domain:

Weak Form of Equivalent Integration
 @rx G ðw; u; rx Þ ¼ wðxÞ bx þ dx ¼ 0: @x X

19

Z

In the third step, we can integrate the stress term into parts as follows: Z Z x¼b @rx @w  rx dx dx ¼ ½wðxÞrx  x¼a  wðxÞ @x X X @x Z @w rx dx ¼ ½wðxÞrx jx¼b  ½wðxÞrx jx¼a  X @x Z @w rx dx ¼ ½wðxÞrx nx jx¼b þ ½wðxÞrx nx jx¼a  X @x Z @w rx dx ¼ ½wðxÞrx nx jC  X @x

ð3:7Þ

ð3:8Þ

where Г is the boundary of X, and we have nx(a) = − 1 and nx(b) = 1. The boundary term can be expressed as follows by tractive force: ½wðxÞrx nx jC ¼ ½wðxÞtx jC :

ð3:9Þ

We also denote Гu by the part of the boundary where u ¼ u and Гt by the other part of the boundary where tx ¼ t x ; and the entire boundaries are C ¼ Cu [ Ct . With this symbol, we can express the weak form of the equilibrium: Z Z @w rx dx þ wtx jCu þ wtx jCt wðxÞbx dx  G ðw; u; rx Þ ¼ @x ZX ZX @w ð3:10Þ rx dx þ wtx jCu þ wt x jCt ¼ wðxÞbx dx  @x X X ¼0 For simplicity, by the special properties of the arbitrary function, specify the place of displacement, we set wjCu ¼ 0 to eliminate the need of tx. Z Z @w rx dx þ wt x jCt ¼ 0: G ðw; u; rx Þ ¼ wðxÞbx dx  ð3:11Þ X X @x To obtain a pure displacement or ‘irreducible’ form of the equations, we can introduce the constitutive equation and strain displacement equation. Thus, we got that Z Z @w @u E dx þ wt x jCt ¼ 0: G ðw; u Þ ¼ wðxÞbx dx  ð3:12Þ @x X X @x This form is known as a displacement model or an irreducible form. By comparing equation (3.12) with equation (3.5), we found that the strong form requires the second derivatives of u and w with respect to the coordinate x, but the weak form only needs the specification of the first derivatives. Construction of approximate solution further needs to be based on weak form.

Basic Theory of Finite Element Method

20

3.3 3.3.1

Finite Element Computation Based on Weak Form Galerkin Method

We multiply a specified set of functions by unknown parameters to express the displacement uðx; tÞ, thus the approximate solution is constructed. Similarly, we use the same amount of specified functions and arbitrary parameters to write an arbitrary function. These can be expressed as follows: u ðx Þ  u^ ðx Þ ¼

N X n¼1

^ ðx Þ ¼ w ðx Þ  w

un ðx Þan þ ub ðx Þ

N X m¼1

ð3:13Þ wm ðx Þwm

We assume that in the above form, both un ðx Þ and wm ðx Þ are zero at the specified displacement boundaries. Then it specifies the function ub ðx Þ as any function satisfying the displacement boundary condition. For instance, if the displacement boundary condition is u ðLÞ ¼ d on the domain 0 < x < L, the function can be defined as: x ub ðx Þ ¼ d ¼ ub ðx Þd: L The Galerkin method is considerably older than the finite element method. The latter mainly uses locally based (element) functions in the expansion of equation (3.12), but the general procedures are identical. Since this process produces equations with integral form, these equations can be obtained by weighted summation of the subdomains; all weighted residual approximations are collectively called generalised finite element method. Occasionally, we find that it may be useful to use ‘local’ and ‘global’ trial functions. The approximate Galerkin solution to the elasticity problem can be obtained by inserting equation (3.13) into equation (3.12): " # Z Z N N N X X X du dw du n m b ^ ðw; ^ u^ Þ ¼ G E an þ wm wm bx dx  wm d dx dx dx X X dx m¼1 m¼1 n¼1 ð3:14Þ N X þ wm wm ðx Þt x jCt ¼ 0 m¼1

Because the functions, un , wm , and ub are known, the integrals can be evaluated as:

Z

dwm dun E dx dx X dx Z Z dwm dub E fm ¼ wm bx dx  ddx þ wm t x jCt dx X X dx

Kmn ¼

ð3:15Þ

Weak Form of Equivalent Integration

21

The array Kmn is the stiffness matrix, and fm is the specified load matrix. Because the parameters wm are arbitrary, the expression multiplied by each wm must be zero. This leads to the following set of equations: N X

Kmn an ¼ fm ;

m ¼ 1; 2; . . .; N :

ð3:16Þ

n¼1

The problem is thus expressed in the matrix form of the stiffness equation: Ka ¼ f ; where

2

K11 6 K21 6 K ¼ 6 .. 4 .

KN 1

K12 K22 KN 2





K1N K1N .. .

KNN

ð3:17Þ 8 f1 > > > < 7 f2 7 7 and f ¼ .. > 5 . > > : fN 3

9 > > > = > > > ;

:

The formal solution to the problem is now given by: a ¼ K 1 f :

ð3:18Þ

However, the inverse of K is never computed in practical situations. Instead, the solution can be obtained either by directly solving equation (3.17) or using an iterative approach. Once the parameters an are known, we can use equation (3.13) to compute the approximation for the displacement, and that for stresses from: ! N X dub ðx Þ @^ u ðxÞ dun ¼E an þ d : rx ðxÞ ¼ E dx @x dx n¼1

ð3:19Þ

Example 3.1. Solutions of Galerkin method for a one-dimensional elastic cantilever beam. For instance, we take a static problem with a length of 10 units and E = 1000, as shown in figure 3.1.

FIG. 3.1 – One-dimensional elastic cantilever beam in example 3.1.

Basic Theory of Finite Element Method

22 The loading is given by:

( bx ¼

10 0

for 0\x\5 ; for 5\x\10

and the boundary conditions are u ð 0Þ ¼ 0

and

t x ð10Þ ¼ 25:

The weak form of this problem is Z 10 Z 5 @w @u 1000 dx  w ðx Þ10dx  w ð10Þ25 ¼ 0: @x @x 0 0

ð3:20Þ

We consider an approximate solution given by: N  n X x

u^ ðx Þ ¼

n¼1

10

an

and

^ ðx Þ ¼ w

N  m X x m¼1

10

wm :

Note that ub ðx Þ ¼ 0 for this problem. The solution to the weak form is given by: Z 5  m N Z 10  x m1  x n1  X x 10mn dx an ¼ 10dx þ 25; ð3:21Þ 10 10 10 0 0 n¼1 where m = 1, 2,..., N. The stiffness matrix and the load vector are Z 10  x m þ n2 100mn ; 10mn dx ¼ Kmn ¼ 10 m þn  1 0 Z fm ¼

5

 x m 10

0


 100 1 m þ 1 10dx þ 25 ¼ þ 25: mþ1 2

For example, if M = N = 2, the stiffness matrix and the load matrix are    

k11 k12 100 100 f1 75=2 K¼ ; f ¼ : ð3:22Þ ¼ ¼ k21 k22 f2 100 400=3 175=6 Using equation (3.18), the unknown parameters can be obtained as follows:

0:62500 a1 ¼ : ð3:23Þ a¼ a2 0:25000 The unknown parameters obtained from the above solutions are substituted into equation (3.13), and the displacement function can be obtained as follows: u ðx Þ  u^ ðx Þ ¼

2  n X x n¼1

10

an ¼

 x 2 x  0:625 þ ð0:25Þ 10 10

¼ 0:0025x þ 0:0625x 2

ð3:24Þ

Weak Form of Equivalent Integration

23

Using equation (3.18), the stress function is obtained as follows: ! 2 X @^ u ðxÞ dun ^x ðxÞ ¼ E ¼E an ¼ 5x þ 62:5: r @x dx n¼1

ð3:25Þ

Similarly, the solutions using one to five terms are presented in table 3.1, in which the symbol N represents the number of terms. TAB. 3.1 – Parameters for one-dimensional elastic cantilever beam in example 3.1. N 1 2 3 4 5

a1 0.37500 0.62500 0.78125 0.78125 0.73437

a2

a3

a4

a5

−0.25000 −0.71875 −0.71875 −0.25000

0.31250 0.31250 −1.09275

0.00000 1.64063

0.65625

We can find that there are some critical aspects in constructing approximate solutions from the results of the above problem. First, we note that the solutions using three and four terms are the same. This means that in the weak form, terms that contribute to the solution can be discarded. Another conclusion is when the solution is constructed in a global manner, it does not result in convergent behaviour for individual parameters. Thus, it is necessary to use more stable approximation procedures, such as the finite element method (this will be discussed later). The solutions using one to three terms for displacement and stress are obtained and shown in figure 3.2. It can be observed that the displacement at the free end is the same, no matter how many terms are used. This is only more common and valid in

FIG. 3.2 – Displacement and stress solutions in example 3.1 based on Galerkin method using N terms: (a) displacement and (b) stress. Notes: The symbol N represents the number of terms, and the symbol Exact represents the exact analytical solutions.

Basic Theory of Finite Element Method

24

one-dimensional static problems and less applicable in higher-dimensional problems. The displacement solution converges faster than the stress solution, but we found that some points are more accurate than others. These points are called superconvergent points, which play a significant role in future research on error estimation and adaptive mesh refinement.

3.3.2

Finite Element Computation

It is a more convenient way to construct the approximating functions wm and un by dividing the domain to be analyzed into small regular-shaped regions. For example, between a and b, we can divide a one-dimensional region into ‘M ’ finite small segments by defining a set of N points xi such that x1 ¼ a;

xi \xi þ 1

and

xN ¼ b:

For a one-dimensional problem, it is also possible to have each increment define a finite element domain (or, more simply, an element), and groups of points define nodes. The element division and node distribution are the basic components of the finite element method for describing what is referred to as the finite element mesh or simply the mesh for the problem. By using the subdivision above, we can define a simple group of continuous polynomial approximating functions: 8 0; x\xi1 > > > > x  xi1 > > > < x  x ; xi1  x  xi i i1 ð3:26Þ ui ¼ x  x > i þ 1 > > ; x \x  x i iþ1 > > xi þ 1  xi > > : 0; x [ xi þ 1 The plot of these functions and their first derivative is shown in figure 3.3. Only when the function is continuous in x, but the first derivative is piecewise continuous, and the discontinuities are located on the nodes we call it C0 function. In subsequent chapters, we considered selecting appropriate C0 functions for higher spatial dimensions. Here, we just study the description of general forms in one dimension. It can be seen from the figure the end functions can be used as the spatial form of the ub functions if necessary. We assume that wi ¼ ui , thus, over the domain, all integrals defined in the weak form functional can be obtained. Indeed, by noting that, we can usually evaluate the integrals on each element individually Z M Z xi þ 1 X XZ ðÞdx ¼ ðÞdx  ðÞdx: ð3:27Þ X

i¼1

xi

e

Xe

Considering any interval [xi, xi+1] shown in figure 3.4, we note that it is defined by the same two local functions N1 and N2 , which are called the shape functions for

Weak Form of Equivalent Integration

25

FIG. 3.3 – One-dimensional finite element approximation for ui : (a) functions and (b) derivatives.

FIG. 3.4 – One-dimensional finite element shape functions: (a) functions and (b) derivatives. the element. We also define the local node coordinates in each element as x1e and x2e to simplify the notation. With the above notation, the displacements and the arbitrary weight function also can be expressed as: u^ e ¼ N1 ðx 0 Þ u^1e þ N2 ðx 0 Þ u^2e ^ 1e þ N2 ðx 0 Þ w ^ 2e ^ e ¼ N1 ð x 0 Þ w w

ð3:28Þ

In each element, we define the local coordinate system as x 0 ¼ x  x1e , and let he ¼ x2e  x1e be the element length; the shape functions can be expressed as: N1 ð x 0 Þ ¼ 1 

x0 he

and

N2 ð x 0 Þ ¼

x0 ; he

ð3:29Þ

and they are the same for all elements. The derivatives of the shape functions are given by the following formula: dN1 dN1 1 ¼ ¼ he dx dx 0

and

dN2 dN2 1 ¼ ¼ : he dx dx 0

ð3:30Þ

Basic Theory of Finite Element Method

26

Thus, we can write the approximation of the weak form as: ^ ðw; ^ r ðw; ^ f ðw; ^ u^ Þ ¼ G ^ u^ Þ  G ^ u^ Þ  w ^ ðx Þt x jCt ; G

ð3:31Þ

^ r , and the contribution of body loading where the contribution to internal stress is G ^ to force is Gf . Using equation (3.28), each term is defined as follows: 9 8 dN1 > > >  >  e

Z M  X he < dx 0 = u^1 dN1 dN2 e e ^ ^ u^ Þ ¼ ^1 w ^2 Ee Gr ðw; w dx 0 ; 0 0 u^2e > > dx dx 0 > e¼1 ; : dN2 > dx 0 ^ f ðw; ^ u^ Þ ¼ G

M X e¼1

^ 1e ½w

^ 2e  w

Z

he 0



N1 b dx 0 : N2 x

Each element can be evaluated as: 9 8 dN1 > > >  >   e Z he > = < dx 0 > K11 dN1 dN2 e 0 E K ¼ dx ¼ e 0 0 > > K21 dx dx 0 > > > dN2 > ; : dx 0 e

Z he

f1 N1 e 0 f ¼ bx dx ¼ : f2e N 0 2

ð3:32Þ

e K12 e K22

 ; ð3:33Þ

For the shape functions given in equation (3.29), if we make Ee and bx in each element constant, the matrices are  

Ee 1 1 1 1 e e and f ¼ bx he : ð3:34Þ K ¼ 1 1 1 2 he

3.4

Global Assembly from One-Dimensional Elements

In the one-dimensional beam model shown in figure 3.5, N − 1 elements (‘M’ small finite segments) and N nodes are used.

FIG. 3.5 – Element and node numbers of one-dimensional problem.

Weak Form of Equivalent Integration

27

Then, we define the relation between the local and global node numbers for each element, as indicated in table 3.2. TAB. 3.2 – Local-to-global node numbering for two-end elements for one-dimensional elastic beam. Local node number

Element number 1 2

1 1 2

2 2 3

  

3 3 4

N−1 N−1 N

According to the node numbers of each element, the corresponding element location vector can be provided to determine the relation between local and global node locations as follows: k1 ¼ f 1

2 g T;

k2 ¼ f 2

3 g T ; . . .; kN 1 ¼ f N  1

N g T:

ð3:35Þ

To globally assemble one-dimensional elements, the node numbers are shown on the left and upper sides of the matrix, and the expanded form of the stiffness matrix is given through the element location vector: Node number

1 1 K11 1 6 K21 6 6 0 6 6 6 6 N 1 4 N 1 2 3 .. .

2

9 8 f1 > > > > > > > f2 > > > > > > = < f3 > ¼ .. > . > > > > > > > > > f > > N 1 > > ; : fN

2 1 K12 1 2 K22 þ K11 2 K21

3  0 2

2 K12 3  K22 þ K11 N 1 K21

N 1

N 38 9 u^1 > > > > > > 7> 0 u^2 > > > > > 7 > 3 = 7< u^3 > K12 7 7 .. .. > 7> > > . > 7>

N 1 . N  > > > N 5> ^ u > K12 K22 þ K11 N 1 > > > ; : N N u^N K21 K22

ð3:36Þ

Similarly, the node numbers are shown on the left of the load vector, and the expanded load is given by: 9 8 9 8 f11 f1 > 1 > > > Node > > > > > > > > > f2 > > f21 þ f12 > > > > > 2 number > > > > > = > = < f3 > < f2 þf3 > 3 2 1 ¼ : ð3:37Þ .. .. . .. > > > > . . > > > > > > > > > > > N 2 > > N 1 > > > > > > f2 f þ f1N 1 > > > ; > ; : N 1 > : N fN f2N 1

Basic Theory of Finite Element Method

28

The summation indicated in equation (3.36) and subsequent expressions result in a standard linear problem, with the final stiffness equations being Ku ¼ f where

2

K11 6 K21 6 K ¼ 6 .. 4 .

K12 K22

KN 1

KN 2





K1N K1N .. .

KNN

ð3:38Þ 8 f1 > > > < f2 7 7 7 and f ¼ .. > 5 . > > : fN 3

9 > > > = > > > ;

:

Now, the formal solution to the problem is: u ¼ K 1 f :

ð3:39Þ

But it is difficult for us to compute the inverse of K in practical situations. Instead, the solution is obtained either by directly solving equation (3.38) or by an iterative method. Once the displacements at the nodes u are known, the approximation of the displacement for each element may be computed as: u^ e ðx 0 Þ ¼ N1 ðx 0 Þ u^1e þ N2 ðx 0 Þ u^2e ;

ð3:40Þ

and that for stresses from as: ^ex ðx 0 Þ ¼ E r

3.5


 @^ u e ðx 0 Þ dN1 ðx 0 Þ e dN2 ðx 0 Þ e ^ ^ ¼E u1 þ u2 : @x 0 dx 0 dx 0

ð3:41Þ

Treatments on Boundary Conditions

This section is concerned with the boundary conditions of the one-dimensional beam model. Especially, we discuss the displacement and traction boundary conditions for representative one-dimensional elastic problems. Other issues can be treated similarly at the corresponding nodes. (1) Displacement boundary conditions Using the above finite element form, because the parameters are now all physical values, it may be easy to implement the displacement boundary conditions. That is, they satisfy u^ ðxa Þ  u a :

ð3:42Þ

Thus, to impose the displacement condition u ð0Þ  u, we set u^1  u and rewrite equation (3.36) as:

Weak Form of Equivalent Integration 2

1 60 6 60 6 6 6 6 4

0 1 2 K22 þ K11 2 K21

0 2 K

2 12 3  K22 þ K11 N 1 K21

29

9 8 9 38 u1 u^1 > > > > > > > > > > > 1 > > 7> > > 0 f  K u ^ u > > > > 2 1 2 21 > > > > 7 > > > > 3 = = < < 7 K12 f ^ u 3 3 7 ¼ 7 . .. .. .. > > > 7> . > > > > > > > 7>

N 1 . N  > > > > N 5> > > ^ u > > fN 1 > K12 > K22 þ K11 N 1 > > > > > ; ; : : N N u^N fN K21 K22 ð3:43Þ

This is equivalent to setting w1  0. To simplify, the row and column of known displacement at the corresponding nodes can be deleted, thus reducing the size of the stiffness matrix. (2) Traction boundary conditions Similarly at x ¼ L, impose the traction condition only requires the modification: f N ! f N þ t x ðLÞ:

ð3:44Þ

These physical characteristics highlight the obvious advantages of using the finite element method to approximate variables. Example 3.2. Solutions of finite element method for one-dimensional elastic cantilever beam. The example shown in figure 3.1 is considered again here. The domain was divided into four equal elements, as shown in figure 3.6.

FIG. 3.6 – Four-element mesh for one-dimensional elastic cantilever beam in example 3.2. For example, using four elements, the stiffness matrix, and the load matrix are       Ee 1 1 1000 1 1 400 400 ¼ ; K1 ¼ K2 ¼ K3 ¼ K4 ¼ ¼ 400 400 2:5 1 1 he 1 1



1 1 1 12:5 1 f ¼ f ¼ b x he ¼ ; ¼  10  2:5 1 12:5 1 2 2 1

2

Basic Theory of Finite Element Method

30





1 1 1 0 1 ¼ : ¼  0  2:5 f ¼ f ¼ b x he 1 0 1 2 2 3

ð3:45Þ

4

Using the element location vector, the global stiffness matrix may be derived as follows: Node number

1 2 3 4 5

2

1

1 K11 6 K1 6 21

6 6 4

2

1 K12

3  2

1 K22 þ K11 2 K21

2

400 400 6 400 400 þ 400 6 ¼6 400 6 4

400 400 þ 400 400

400 800 400

5

2 K12  3 3 þ K11

3 K12 4  3 K21 K22 þ K11 4 K21

2 K22

2

400 400 6 400 800 6 ¼6 400 6 4

4

400 800 400

400 400 þ 400 400 3

3

7 7 7 7 4 5 K12 4 K22 3

7 7 7 7 400 5 400

ð3:46Þ

7 7 7 7 400 5 400

Similarly, the load matrix may be derived as follows: Node 1 number 2 3 4 5

9 9 8 9 8 8 9 8 12:5 > 12:5 f1 > f11 > > > > > > > > > > > > > > > > > > > > > > > = < 25 > = > < 12:5 þ 12:5 > = > = > < f2 > < f21 þ f12 > ¼ 12:5 12:5 þ 0 f3 ¼ f22 þ f13 ¼ > > > > > > > 0þ0 > > > > > > >f3 þf4 > 0 > > > > > 1 > > > > > > > > > > f4 > > 2 ; : ; : ; ; : : 0 0 f5 f24

The matrix form of the stiffness equation can be expressed as follows: 9 38 9 8 2 400 400 12:5 > u^1 > > > > > > > > > > > > > > 7> 6 400 800 400 7< u^2 = < 25 = 6 7 6 400 800 400 7> u^3 > ¼ > 12:5 > 6 4 400 800 400 5> u^ > > > 0 > > > > > > ; : ; > : 4> 400 400 0 u^5

ð3:47Þ

ð3:48Þ

Then, the displacement boundary conditions u^1  u 1  0 can be introduced into the stiffness equation, and the global stiffness equation can be rewritten as follows: 9 2 38 9 8 0 > 1 0 u^1 > > > > > > > > > > > > > > 6 0 800 400 7> 6 7< u^2 = < 25 = 6 7 ð3:49Þ 400 800 400 6 7> u^3 > ¼ > 12:5 > > > > 4 ^ 0 400 800 400 5> u > > > > 4 > > ; : ; > : > 0 400 400 u^5

Weak Form of Equivalent Integration

31

To simplify, the row and column of known displacement at the corresponding node 1 can be deleted, thus reducing the dimension of the stiffness matrix: 9 2 38 9 8 25 > 800 400 u^2 > > > > > > > = < = < 6 400 800 400 7 u^3 12:5 6 7 ð3:50Þ ¼ 4 0 > 400 800 400 5> u^ > > > > ; : ; > : 4> 0 400 400 u^5 The traction boundary conditions f5 ! f5 þ t x ð10Þ ¼ 25 can be introduced into the load equation, and the global stiffness equation is then rewritten as follows: 9 2 38 9 8 25 > 800 400 u^2 > > > > > > > = < = < 6 400 800 400 7 u^3 12:5 6 7 ð3:51Þ ¼ 4 400 800 400 5> > > 0 > > > > u^4 > ; : ; : 25 400 400 u^5 Using equation (3.39), the unknown displacements at the nodes are obtained as follows: 9 8 9 8 u^2 > > 0:15625 > > > > = < = > < > 0:25 u^3 ð3:52Þ ¼ 0:3125 > u^ > > > > > ; : ; > : 4> 0:375 u^5 The unknown displacements on the nodes obtained from the above solutions are then substituted into equation (3.40), and the displacement function for each element can be obtained as follows:
 x0 x0 Element 1 : u^ 1 ðx 0 Þ ¼ 1  ð3:53aÞ  0 þ  0:15625 ¼ 0:0625x 0 he he Element 2 : u^ 2 ðx 0 Þ ¼


 x0 x0 1  0:15625 þ  0:25 ¼ 0:0375x 0 þ 0:15625 ð3:53bÞ he he

0

Element 3 :

u^ ðx Þ ¼

Element 4 :

u^ 4 ðx 0 Þ ¼

3




x0 1 he

  0:25 þ

x0  0:3125 ¼ 0:025x 0 þ 0:25 he

ð3:53cÞ


 x0 x0 1  0:3125 þ  0:375 ¼ 0:025x 0 þ 0:3125 ð3:53dÞ he he

Using equation (3.41), the stress function for each element can be obtained as follows:
 @^ u 1 ðx 0 Þ 1 1 1 0 ^x ðx Þ ¼ E Element 1 : r ¼ E   0 þ  0:15625 ¼ 62:5 ð3:54aÞ @x 0 he he

Basic Theory of Finite Element Method

32

Element 2 :

^2x ðx 0 Þ r


 @^ u 2 ðx 0 Þ 1 1 ¼E ¼ E   0:15625 þ  0:25 ¼ 37:5 ð3:54bÞ @x 0 he he

Element 3 :

^3x ðx 0 Þ r


 @^ u 3 ðx 0 Þ 1 1 ¼E ¼ E   0:25 þ  0:3125 ¼ 25 @x 0 he he

Element 4 :

^4x ðx 0 Þ ¼ E r


 @^ u 4 ðx 0 Þ 1 1 ¼ E   0:3125 þ  0:375 ¼ 25 @x 0 he he

ð3:54cÞ

ð3:54dÞ

The solution above is shown in figure 3.7, in which the symbol n represents the number of finite elements. Similarly, it is also shown the solution uses eight elements. Comparing the two solutions, we note that the displacements converge more quickly than the stresses once again, but they have some superconvergent points.

FIG. 3.7 – Displacement and stress solutions in example 3.2 based on finite element method using n elements: (a) displacement and (b) stress. Notes: The symbol n represents the number of finite elements, and the symbol Exact represents the exact analytical solutions.

3.6

Exercises

3.6.1 Describe the implementation steps for the weak form of equivalent integration. 3.6.2 Using the Galerkin method, for example 3.1 with E(x) = Ex, bx(x) = bxx, compute the global stiffness matrix K and load matrix f. 3.6.3 Using the finite element method, for example 3.2 with E(x) = Ex, bx(x) = bxx, compute the global stiffness matrix K and load matrix f. 3.6.4 Summarise the functions of the element location vector.

Chapter 4 Elements and Shape Functions To compute the element matrices, it is necessary to select appropriate shape functions. Previously, we introduced the basic strategy of finite element computation based on the weak form using a one-dimensional element with two nodes, the simplest low-order linear element. In this chapter, we introduce a high-order Lagrange element for one-dimensional problems. Further, for two-dimensional problems, the triangle and rectangle elements are provided; for three-dimensional problems, the tetrahedron and hexahedron elements are also presented.

4.1 4.1.1

One-Dimensional Lagrange Element Linear Element with Two Nodes

This linear element has only two end nodes corresponding to the endpoints of a line segment, as shown in figure 4.1, and the displacements at the nodes are u^1e and u^2e . Such elements are called one-dimensional elements with two nodes.

FIG. 4.1 – One-dimensional element with two nodes.

For this one-dimensional problem, we write the displacement as: u e
u^ e ¼

2 X a¼1

Na ðx 0 Þ u^ae ¼ N1 ðx 0 Þ u^1e þ N2 ðx 0 Þ u^2e :

ð4:1Þ

DOI: 10.1051/978-2-7598-2938-5.c004 © Science Press, EDP Sciences, 2023

Basic Theory of Finite Element Method

34

Previously, a local coordinate system in each element was defined as x 0 ¼ x  x1e , and the element length is he ¼ x2e  x1e ; the shape functions are given by: N1 ð x 0 Þ ¼ 1 

x0 he

and

N2 ð x 0 Þ ¼

x0 he

ð4:2Þ

and are the same for every element. The derivatives of the shape functions are dN1 dN1 1 ¼ ¼ he dx dx 0

and

dN2 dN2 1 ¼ ¼ : he dx dx 0

ð4:3Þ

In equation (4.1), the displacement function is linear in x; therefore, the element is called a one-dimensional linear element.

4.1.2

Higher-Order Lagrange Element

It is difficult to describe complex displacement changes using the aforementioned linear elements. Accordingly, higher-order shape functions and elements with more nodes are introduced. A higher-order element with more nodes is shown in figure 4.2.

FIG. 4.2 – One-dimensional higher-order element.

The displacement is written as: u e
u^ e ¼

n X a¼1

Na ðnÞ u^ae ¼ N1 ðnÞ u^1e þ N2 ðnÞ u^2e þ    þ Nn ðnÞ u^ne ; 1  n  1; ð4:4Þ

where n is the total number of given functions used in the element, u^ae are unknown parameters to be determined, and n is a new local coordinate. The local coordinate selected here is for the convenience of numerical integration in the local coordinate system in the next computation procedure. The local coordinate n is related to the global coordinate x as follows: x¼

n X a¼1

Na ðnÞ xae ¼ N1 ðnÞ x1e þ N2 ðnÞ x2e þ    þ Nn ðnÞ xne ; 1  n  1:

ð4:5Þ

The global and the local coordinate in the above transformation are in one-to-one correspondence, which is not necessarily linear. Only when the shape functions are linear, as in the case of the two-node elements obtained previously, is the transformation linear.

Elements and Shape Functions

35

To ensure that the unknown parameters to be determined are the displacements at the nodes, the following equation should hold true:  1; na ¼ nb Na ðnb Þ ¼ dab ¼ ; ð4:6Þ 0; na 6¼ nb where nb is the local coordinate with position xb. Then, through the values of equation (4.4) on the node coordinates, we can get that the unknown parameters are exactly the displacements on the nodes. In order that equation (4.6) can be satisfied, the following simple construction for higher-order shape functions may be used, based on the Lagrange interpolation formula: lap ðnÞ ¼

n Y

ðn  nb Þ ðna  nb Þ

b¼1 b 6¼ a ðn  n1 Þðn  n2 Þð  Þðn  na1 Þðn  na þ 1 Þð  Þðn  nn Þ ; ¼ ðna  n1 Þðna  n2 Þð  Þðna  na1 Þðna  na þ 1 Þð  Þðna  nn Þ

ð4:7Þ

where the order of the polynomial lap ðnÞ is p = n−1. The coordinates of internal points of na are uniformly located between the endpoints. For one-dimensional elements, the shape functions are defined as follows: Na ðnÞ ¼ lap ðnÞ

ð4:8Þ

Further, the completeness condition requires that u^ e ðnÞ should contain any constant c (displacement of rigid body), yielding. u^ e ðnÞ ¼

n X

Na ð nÞ c ¼ c

or

a¼1

n X

Na ðnÞ ¼ 1:

ð4:9Þ

a¼1

In particular, in the linear example above, we set n1 ¼ 1 and n2 ¼ 1 with n = 2 (p = 1); then, we obtain the two shape functions using equation (4.7): N1 ðnÞ ¼ l11 ðnÞ ¼

n1 1 ¼ ð1  nÞ and 1  1 2

N2 ðnÞ ¼ l21 ðnÞ ¼

nþ1 1 ¼ ð1 þ nÞ: 1þ1 2 ð4:10Þ

The global coordinate x can be expressed using the local coordinate n as: x¼

2 X

1 1 Na ðnÞ xae ¼ N1 ðnÞ x1e þ N2 ðnÞ x2e ¼ ð1  nÞ x1e þ ð1 þ nÞ x2e : 2 2 a¼1

ð4:11Þ

Using equation (4.11), the derivative of the coordinate function is given by: je ¼

@x he ¼ : @n 2

ð4:12Þ

Basic Theory of Finite Element Method

36

The above derivative between the global and the local coordinates is Jacobian, which is represented as je . The derivatives of the shape functions are given by:

4.1.3

@Na 1 @Na ¼ ; a ¼ 1; 2; je ðnÞ @n @x

ð4:13aÞ

@N1 1 @N1 1 @N2 1 @N2 1 ¼ ¼ ; ¼ ¼ : je @n he @x je @n he @x

ð4:13bÞ

Quadratic Lagrange Element

For one-dimensional quadratic shape functions, the local coordinates of nodes are located at: n1 ¼ 1;

n2 ¼ 1;

and

n3 ¼ 0;

ð4:14Þ

in which the order of the polynomial is p = 3 − 1 = 2 (n = 3). The quadratic Lagrange element with three nodes is shown in figure 4.3.

FIG. 4.3 – One-dimensional quadratic Lagrange element with three nodes.

We obtain the three shape functions using the Lagrange interpolation formula (4.7) as: ð n  1Þ ð n  0Þ 1 ¼ nð n  1Þ ð1  1Þð1  0Þ 2 ðn þ 1Þðn  0Þ 1 N2 ðnÞ ¼ l22 ðnÞ ¼ ¼ nð n þ 1Þ ð1 þ 1Þð1  0Þ 2 ðn þ 1Þðn  1Þ N3 ðnÞ ¼ l32 ðnÞ ¼ ¼ 1n2 ð0 þ 1Þð0  1Þ N1 ðnÞ ¼ l12 ðnÞ ¼

ð4:15Þ

If we let the global coordinates for the element be given by x1e, x2e , and x3e (x1e and are the boundary end nodes), the global coordinate x can be expressed using the local coordinate n as: x2e

x ¼ N1 ðnÞx1e þ N2 ðnÞx2e þ N3 ðnÞx3e :

ð4:16Þ

Elements and Shape Functions

The Jacobian is now given by:       @x 1 e 1 e 1 ¼ n  x1 þ n þ x2  2nx3e ¼ he þ n x1e þ x2e  2x3e je ðnÞ ¼ @n 2 2 2

37

ð4:17Þ

The value of this Jacobian function is constant only when the coordinate x3e for node 3 is at the middle point of the element. The derivatives of the shape functions are given by: @Na 1 @Na ¼ ; a ¼ 1; 2; . . .; n: je ðnÞ @n @x

4.2 4.2.1

ð4:18Þ

Two-Dimensional Triangle Element Triangle with Three Nodes

The finite element domain is defined by dividing the two-dimensional continuum domain into a set of mesh composed of some triangular elements, as shown in figure 4.4a. The linear functions over the triangle with three nodes, as shown in figure 4.4b, are constructed by linear polynomials. It is worth noting that this two-dimensional triangle element was proposed and used by Courant (1943) and Turner et al. (1956) to analyze the plane elasticity problems in the initial development stage of finite element method.

FIG. 4.4 – Two-dimensional domain discretization into triangle elements: (a) triangle mesh, (b) triangle element.

Basic Theory of Finite Element Method

38

The approximate linear function on the triangular element with three nodes in the Cartesian coordinate system is expressed as: 8 9 < a1 = ^ e ¼ ½ 1 x y  a2 ¼ a1 þ a2 x þ a3 y; ð4:19Þ ue
u : ; a3 ^ e is the displacement in the element, which can be the displacement u in the where u x-direction or the displacement v in the y-direction; the unknown parameters a1 , a2 , and a3 are computed by the displacements of the three vertices of the triangle. According to equation (4.19), the displacement on the three vertices can be expressed using the corresponding coordinates as follows: 8 e9 2 38 9 ^1 = 1 x1 y1 < a1 =

1 = < a1 > a2 ¼ 4 1 > ; : > 1 a3

x1 x2 x3

31 8 e 9 2 ^1 = a y1

> e > ^a > > > aa u > > > > 38 e 9 > > a¼1 > > ^1 = a3 < u = < 3 P 1 e e ^2 ¼ b3 5 u ^a : ba u > : e ; 2D > a¼1 > > ^3 u c3 > > > > 3 > > P > > e > ^a > ca u ; : a¼1

ð4:22Þ The geometric interpretation of the parameters ba and ca is shown in figure 4.4b. By substituting the expression equation (4.22) of parameters αa into equation (4.19), the displacement function in the element expressed by nodal displacement can be obtained:

Elements and Shape Functions

39

^ e ¼ a 1 þ a2 x þ a3 y u ¼

3 X 1 ^ a þ ba u ^ a x þ ca u ^ a yÞ ðaa u 2D a¼1

¼

3 X 1 ^ ea ðaa þ b a x þ c a y Þ u 2D a¼1

ð4:23Þ

According to the above formula, the shape functions can be obtained: Na ðx; y Þ ¼

1 ðaa þ ba x þ ca y Þ; 2D

a ¼ 1; 2; 3:

ð4:24Þ

The shape function for a = 3 is shown in figure 4.5a, where it can be seen that its value at node 3 is 1 and 0 at nodes 1 or 2. More generally, these shape functions satisfy.  1; a ¼ b Na ðxb ; yb Þ ¼ dab ¼ ; ð4:25Þ 0; a 6¼ b where ðxb ; yb Þ are the local coordinates. Moreover, the completeness condition then ^ e ðxb ; yb Þ contain any constant c (displacement of rigid body), yielding requires that u ^ e ðx; y Þ ¼ u

3 X

Na ðx; y Þ c ¼ c

a¼1

or

3 X

Na ðx; y Þ ¼ 1:

ð4:26Þ

a¼1

Using these shape functions, the approximate function of the element is expressed as: ^e ¼ u

3 X a¼1

^ ea : Na ðx; y Þ u

FIG. 4.5 – Shape function N3 for two-dimensional triangle with three nodes.

ð4:27Þ

Basic Theory of Finite Element Method

40

For example, as shown in figure 4.6a, the two-dimensional plane problem is illustrated in which a set of simple linear functions for a parameter u can be constructed using linear polynomials on a three-node triangle. From a geometric point of view, the numerator of the shape function is twice the area of the triangle determined by the nodal coordinates xa, ya of the two nodes, and the point (x, y). So, in figure 4.6a, we can define 2Da ðx; y Þ ¼ ðaa þ ba x þ ca yÞ

ð4:28Þ

and then, the shape functions can be written as follows: Na ¼

Da ðx; y Þ ; D

a ¼ 1; 2; 3

ð4:29Þ

This ratio also defines a simple and convenient triangular coordinate system, which is called area coordinates (figure 4.6b). We denote these as: La ¼

Da ; D

a ¼ 1; 2; 3:

ð4:30Þ

In addition, it is important to note that these three coordinates are limited by the total area such that L1 þ L 2 þ L 3 ¼ 1

ð4:31Þ

The shape function of the three-node triangle can be obtained by using area coordinates, as shown in the following formula: Na ¼ L a ;

a ¼ 1; 2; 3

ð4:32Þ

FIG. 4.6 – Triangle element with area coordinates: (a) subareas of triangle, (b) area coordinates.

4.2.2

Higher-Order Triangle Element

For triangular elements with more than three nodes, the above analysis process for three-node triangles also applies. Figure 4.7 shows the first three members of the triangular element family. In addition, for the complete set of two-dimensional

Elements and Shape Functions

41

polynomials, we can use Pascal triangle, as shown in figure 4.8, to represent them. Since it is known from the Pascal triangle that the complete order of any polynomial can be exactly matched with the number of nodes of the triangle of the same order, the shape function of the six-node triangle can be obtained by using the quadratic polynomial as follows: 8 9 a1 > > > > > > > > > a2 > > > > > > > = <   a 3 e e 2 2 ^ ¼ 1 x y x xy y u
u > a4 > ð4:33Þ > > > > > > > > > > a > 5> > ; : > a6 ¼ a1 þ xa2 þ ya3 þ x 2 a4 þ xya5 þ y 2 a6 This method requires inversing a 6 × 6 matrix to obtain the expression for the ^ ea in terms of the parameters αa. The area coordinates used to nodal parameter u define shape functions can eliminate the necessity for inversing the matrix.

FIG. 4.7 – Triangle element family: (a) linear, (b) quadratic, (c) cubic.

FIG. 4.8 – Pascal triangle.

Basic Theory of Finite Element Method

42

It is simple to directly represent an arbitrary triangle of order M to derive shape functions for higher-order triangle elements (Irons et al., 1968). As shown in figure 4.9, denoting a typical node a by three numbers I, J, and K corresponding to the position of coordinates L1a, L2a, and L3a, then the shape function in terms of three Lagrangian interpolations can be written as: Na ¼ lII ðL1 Þ lJJ ðL2 Þ lKK ðL3 Þ

with I þ J þ K ¼ M ;

ð4:34Þ

where lII , lJJ , and lKK are given by equation (4.7), with L1, L2, and L3 taking the place of ξ. It is easy to verify the results of the above expressions: Na ¼ 1

at L1 ¼ L1I ;

L2 ¼ L2J ;

L3 ¼ L3K ;

ð4:35Þ

and 0 at all other nodes. The highest term occurring in the expansion is LI1 LJ2 LK 3 , and, as I + J + K ≡ M for any point, the polynomial is also of order M. Equation (4.34) is valid for quite an arbitrary node distribution in the graph shown in figure 4.9, and if the nodal lines are equally spaced (i.e., 1/M), the formula can be simplified (Taylor, 1972; Silvester, 1969; Argyris et al., 1968b). The reader can easily verify the second-and third-order shape functions as given below and deduce the shape functions for higher-order elements.

FIG. 4.9 – A general triangular element.

Elements and Shape Functions

4.2.3

43

Quadratic Triangle Element

Corner nodes: Na ¼ ð2La  1ÞLa ;

a ¼ 1; 2; 3:

ð4:36Þ

Mid-side nodes: N4 ¼ 4L1 L2 ;

N5 ¼ 4L2 L3 ;

N6 ¼ 4L3 L1 :

ð4:37Þ

The concept of the quadratic triangle element was first proposed by de Veubeke (1965) and later used in the plane stress analysis problems by Argyris (1965).

4.2.4

Cubic Triangle Element

Corner nodes: 1 Na ¼ ð3La  1Þð3La  2ÞLa ; 2

a ¼ 1; 2; 3:

ð4:38Þ

Mid-side nodes: 9 N4 ¼ L1 L2 ð3L1  1Þ; 2 9 N7 ¼ L2 L3 ð3L3  1Þ; 2

9 N5 ¼ L1 L2 ð3L2  1Þ; 2 9 N8 ¼ L3 L1 ð3L3  1Þ; 2

9 N6 ¼ L2 L3 ð3L2  1Þ 2 9 N9 ¼ L3 L1 ð3L1  1Þ 2

ð4:39Þ

Internal node: N10 ¼ 27L1 L2 L3 :

ð4:40Þ

The ‘bubble’ function arises in the last shape function for an internal node with no contribution along boundaries.

4.3 4.3.1

Two-Dimensional Rectangle Element Linear Rectangle Element with Four Nodes

Now we consider the second example of two-dimensional shape functions, the rectangle shown in figure 4.10. The rectangular element considered has side lengths of 2a in the x-direction and 2b in the y-direction. To make it easier to derive the shape functions, we can use a local Cartesian system x 0 ; y 0 defined by: x 0 ¼ x  x0

and

y 0 ¼ y  y0 ;

ð4:41Þ

Basic Theory of Finite Element Method

44

FIG. 4.10 – Rectangle element geometry and node numbers.

where x0 ¼

4 1X xa 4 a¼1

and

y0 ¼

4 1X ya 4 a¼1

ð4:42Þ

in which x0, y0 are located at the centre o0 of the rectangle, and xa and ya (a = 1, 2, 3, and 4) are the coordinates of the nodes. For each displacement component, four functions are now required to uniquely define the shape function. Its expression is as follows: 8 9 a1 > > > = < > a2 e e 0 0 0 0 ^ ¼ ½1 x y x y  ¼ a1 þ x 0 a2 þ y 0 a3 þ x 0 y 0 a4 ; u
u ð4:43Þ a3 > > > > : ; a4 ^ e is displacement in the element, which can be the displacement u in the xwhere u direction or the displacement v in the y-direction; the unknown parameters a1 to a4 can be derived using the displacements at each of the four vertices of the rectangle. The coefficients αa may be obtained by using equation (4.43) at each vertex node: 8 e9 2 38 9 ^1 > u 1 a b ab > > > > > > a1 > < e= 6 7 < a2 = ^2 u 1 a b ab 6 7 ¼ : ð4:44Þ ^e > 4 1 a u b ab 5> a > > > > ; ; : 3e > : 3> ^4 u 1 a b ab a4

Elements and Shape Functions

45

By inverting the coefficient matrix and using substitutions, as in equations (4.21) and (4.22), we can again solve for αa in terms of the nodal displacements to obtain       1 x0 y0 e 1 x0 y0 e e ^ ¼ ^1 þ ^2 1 1þ u 1 u 1þ u 4 4 a b a b         ð4:45Þ 1 x0 y0 e 1 x0 y0 e ^3 þ ^4 1þ 1 þ 1þ 1 u u 4 4 a b a b which yields the four shape functions:    1 x0 y0 1 N1 ¼ 1 ; 4 a b     1 x0 y0 1þ N3 ¼ 1þ ; 4 a b

   1 x0 y0 1þ N2 ¼ 1 4 a b     0 1 x y0 1 N4 ¼ 1þ 4 a b

ð4:46Þ

To use numerical integration to evaluate the integrals in weak form and obtain the rectangular element geometry and the local node numbers shown in figure 4.11, we let. n¼

x0 a

and

g¼

y0 ; b

ð4:47Þ

which are products of one-dimensional Lagrange interpolations using the coordinates −1 ≤ ξ, η ≤ 1. Then, the shape functions may be written as: 1 N1 ¼ ð1  nÞð1  gÞ; 4 1 N3 ¼ ð1 þ nÞð1 þ gÞ; 4

1 N2 ¼ ð1 þ nÞð1  gÞ 4 1 N4 ¼ ð 1  nÞ ð 1 þ gÞ 4

ð4:48Þ

FIG. 4.11 – Rectangular element geometry and local node numbers: (a) global coordinates, (b) local coordinates.

Basic Theory of Finite Element Method

46

The shape function for a = 3 is shown in figure 4.12, where it can be seen that its value is 1 at node 3 and 0 at nodes 1, 2, and 4. More generally, these shape functions satisfy.  1; a ¼ b ; ð4:49Þ Na ðnb ; gb Þ ¼ dab ¼ 0; a 6¼ b where ðnb ; gb Þ are the local coordinates. Moreover, the completeness condition then ^ e ðn; gÞ contain any constant c (displacement of rigid body), yielding. requires that u ^ e ðn; gÞ ¼ u

4 X a¼1

Na ðn; gÞ c ¼ c

or

4 X

Na ðn; gÞ ¼ 1:

ð4:50Þ

a¼1

FIG. 4.12 – Shape function N3 for two-dimensional rectangle with four nodes.

4.3.2

Higher-Order Rectangle Element

According to the above illustrations, a systematic and simple way to generate shape functions of any order can be obtained by the product of Lagrange polynomials, including the two local coordinates (Zienkiewicz et al., 1969; Argyris et al., 1968a; Ergatoudis et al., 1968). Specifically, if a node is marked in two dimensions with its column number I and row number J, then the following formula can be obtained by using the definition of Lagrange interpolation in equation (4.7): Na  NIJ ¼ lIn ðnÞ lJm ðgÞ;

ð4:51Þ

in which n and m refer to the number of subdivisions in each direction. For the rectangle with four nodes, as shown in figure 4.13a, the mappings of I and J to node number a are shown in table 4.1.

Elements and Shape Functions

47

TAB. 4.1 – Numbering for rectangle with four nodes. Label

Node number

a I J

1 1 1

2 2 1

3 2 2

4 1 2

Figure 4.13 shows a rectangle element with m = n. When m = n = 1, we can easily obtain the following result: 1 Na ¼ ð1 þ na nÞ ð1 þ ga gÞ; 4

ð4:52Þ

in which ξa, ηa are the normalised coordinates at node a.

FIG. 4.13 – Rectangle element family: (a) linear, (b) quadratic.

4.3.3

Quadratic Rectangle Element

Table 4.2 shows the mappings of I and J to the node numbers a for quadratic rectangle elements, as shown in figure 4.13b. TAB. 4.2 – Numbering for quadratic rectangle element. Label a I J

Node number 1 1 1

2 3 1

3 3 3

4 1 3

5 2 1

6 3 2

7 2 3

8 1 2

9 2 2

The shape functions of quadratic rectangle elements yielding by the products of one-dimensional quadratic Lagrange interpolations are as follows: Corner nodes: 1 Na ¼ ng ðn þ na Þ ðg þ ga Þ; 4

a ¼ 1; 2; 3; 4

ð4:53Þ

Basic Theory of Finite Element Method

48

Mid-side nodes:  1  na ¼ 0; Na ¼ g 1  n2 ðg þ ga Þ; 2   1 ga ¼ 0; Na ¼ n ðn þ na Þ 1  g2 ; 2 Centre node:

4.4 4.4.1

a ¼ 5; 7 a ¼ 6; 8

   N9 ¼ 1  n2 1  g 2

ð4:54Þ

ð4:55Þ

Three-Dimensional Tetrahedron Element Linear Tetrahedron Element with Four Nodes

As shown in figure 4.14, a simple element for three-dimensional problems is a tetrahedron with four nodes.

FIG. 4.14 – Tetrahedron element geometry and node numbers.

Similarly, a compatible displacement field derived from a complete linear polynomial expansion is as follows: 8 9 a1 > > > = < > a2 e e ^ ¼ ½1 x y z  ¼ a1 þ xa2 þ ya3 þ za4 ; u
u ð4:56Þ a > > > ; : 3> a4

Elements and Shape Functions

49

^ e is the displacement in the element, which can be the displacement u in where u the x-direction, the displacement v in the y-direction, or the displacement w in the z-direction; the unknown parameters a1 to a4 may be evaluated in terms of the displacements at each vertex of the tetrahedron. The coefficients αa may be obtained by using equation (4.56) at each vertex node: 8 e9 2 38 9 ^ > u 1 x1 y1 z1 > > > > > a1 > = 6 < 1e > 7 < a2 = ^2 u 1 x y z 2 2 2 6 7 ¼ : ð4:57Þ ^ e > 4 1 x3 y3 z3 5 > u a > > > > ; ; : 3e > : 3> ^4 u 1 x4 y4 z4 a4 The inverse is 2

1 61 6 41 1

x1 x2 x3 x4

y1 y2 y3 y4

where

31 z1 z2 7 7 ¼ 1 z3 5 6V z4 2

1 61 6V ¼ det6 41 1

x1 x2 x3 x4

2

a1 6 b1 6 4 c1 d1

y1 y2 y3 y4

a2 b2 c2 d2

a3 b3 c3 d3

3 z1 z2 7 7; z3 5 z4

with V being the volume of the tetrahedron, and 2 3 2 x2 y2 z2 1 6 7 6 a1 ¼ det4 x3 y3 z3 5; b1 ¼  det4 1 x4 y4 z4 2 3 1 x2 z2 6 7 c1 ¼ det4 1 x3 z3 5; 1 x4 z4

3 a4 b4 7 7; c4 5 d4

ð4:59Þ

y2 y3

3 z2 7 z3 5

1

y4

z4

1 6 d1 ¼  det4 1

x2 x3

3 y2 7 y3 5

1

x4

y4

2

ð4:58Þ

ð4:60Þ

with the other constants defined by cyclic interchange of the subscripts in the order 1, 2, 3, 4. The above solution for the parameters αa allows for rewriting the element interpolations as nodal parameters: ^ e ¼ a1 þ a2 x þ a3 y þ a4 z u ¼ ¼

4 X 1 ^ a þ ba u ^ a x þ ca u ^ a y þ da u ^azÞ ðaa u 6V a¼1 4 X

1 ^ ea ðaa þ b a x þ c a y þ da z Þ u 6V a¼1

ð4:61Þ

Basic Theory of Finite Element Method

50

This yields the shape functions: Na ðx; y; z Þ ¼

1 ðaa þ ba x þ ca y þ da z Þ; 6V

a ¼ 1; 2; 3; 4:

ð4:62Þ

As discussed previously, the tetrahedral element family shown in figure 4.15 is similar in properties to those of the triangular family. First, for each stage, we can also get complete polynomials in three coordinates. Besides, since the faces are divided in the same way as the previous triangles, the same order is obtained for the polynomials in two coordinates in the plane of the face and thus ensuring the compatibility of elements. In addition, the polynomials contain no redundant terms.

FIG. 4.15 – Tetrahedron element family: (a) linear, (b) quadratic, (c) cubic. The above form allows the area coordinates to be generalized to the volume coordinates represented by L1, L2, L3, and L4, as shown in figure 4.16, where La ¼

Va ; V

a ¼ 1; 2; 3; 4;

ð4:63Þ

in which V1, V2, V3, and V4 are the sub-volumes of the tetrahedra 234P, 413P, 421P, and 123P, and V is the volume of the tetrahedron with vertices 1, 2, 3, and 4, respectively. These sub-volumes are obtained according to the following formula:

Elements and Shape Functions

51

6Va ¼ aa þ ba x þ ca y þ da z:

ð4:64Þ

The shaded region in figure 4.16b is the sub-volume V1. The volume coordinates must satisfy the constraint: 4 X

Li ¼ L1 þ L2 þ L3 þ L4 ¼ 1:

ð4:65Þ

i¼1

The shape functions of the volumetric coordinate for a four-node tetrahedron are as follows: Na ¼ L a ;

a ¼ 1; 2; 3; 4:

ð4:66Þ

FIG. 4.16 – Tetrahedron element with volume coordinates: (a) four-node tetrahedron, (b) V1 for 234P.

4.4.2

Higher-Order Tetrahedron Element

By establishing appropriate Lagrange interpolations similar to equation (4.7), it can be seen that the formulae for the shape functions of the higher-order tetrahedron are exactly the same as that for triangles. The result is Na ¼ lII ðL1 Þ lJJ ðL2 Þ lKK ðL3 Þ lLL ðL4 Þ with I þ J þ K þ L ¼ M ;

ð4:67Þ

in which M is the order of the tetrahedron. For the following shape functions for the quadratic and cubic cases, the readers can verify them directly.

4.4.3

Quadratic Tetrahedron Element

Corner nodes: Na ¼ La ð2La  1Þ;

a ¼ 1; 2; 3; 4

ð4:68Þ

Basic Theory of Finite Element Method

52

Mid-side nodes: N5 ¼ 4L2 L1 ; N8 ¼ 4L2 L3 ;

4.4.4

N6 ¼ 4L3 L1 ; N9 ¼ 4L3 L4 ;

N7 ¼ 4L4 L1 N10 ¼ 4L4 L2

ð4:69Þ

a ¼ 1; 2; 3; 4

ð4:70Þ

Cubic Tetrahedron Element

Corner nodes: 1 Na ¼ La ð3La  1Þ ð3La  2Þ; 2 Mid-side nodes: 9 N5 ¼ L1 L2 ð3L1  1Þ; 2 9 N8 ¼ L1 L2 ð3L4  1Þ; 2 9 N11 ¼ L2 L3 ð3L3  1Þ; 2 9 N14 ¼ L3 L4 ð3L2  1Þ; 2

9 9 N6 ¼ L1 L2 ð3L2  1Þ; N7 ¼ L1 L2 ð3L3  1Þ 2 2 9 9 N9 ¼ L2 L3 ð3L1  1Þ; N10 ¼ L2 L3 ð3L2  1Þ 2 2 ð4:71Þ 9 9 N12 ¼ L2 L3 ð3L4  1Þ; N13 ¼ L3 L4 ð3L1  1Þ 2 2 9 9 N15 ¼ L3 L4 ð3L3  1Þ; N16 ¼ L3 L4 ð3L4  1Þ 2 2

Mid-face nodes:

4.5 4.5.1

N17 ¼ 27L1 L2 L3 ;

N18 ¼ 27L4 L1 L2

N19 ¼ 27L3 L4 L1 ;

N20 ¼ 27L2 L3 L4

ð4:72Þ

Three-Dimensional Hexahedron Element Hexahedron with Eight Nodes

For a three-dimensional problem, we take the hexahedron shown in figure 4.17 as an example for analysis. The derivation of the shape functions for the hexahedron is the same as the procedure used for the four-node rectangle. In the present case, it is simple and convenient to use a local Cartesian system x 0 ; y 0 ; z 0 defined by: x 0 ¼ x  x0 ;

y 0 ¼ y  y0 ;

z 0 ¼ z  z0 ;

ð4:73Þ

where x0 ¼

8 1X xa ; 8 a¼1

y0 ¼

8 1X ya ; 8 a¼1

z0 ¼

8 1X za ; 8 a¼1

ð4:74Þ

in which x0, y0, z0 are located at the centre of the hexahedron, and xa, ya, za is the coordinates of the nodes.

Elements and Shape Functions

53

FIG. 4.17 – Hexahedron element geometry and node numbers.

We may write a polynomial expression for ue as: 8 9 a1 > > > > > > > > > a2 > > > > > > > > > > a3 > > > > > > =

4 e e 0 0 0 0 0 0 0 0 0 0 0 0 ^ ¼ ½1 x y z x y y z z x x y z  u
u > a5 > > > > > > > > > > > a > > 6 > > > > > > > > a > 7 > > > ; : a8

ð4:75Þ

¼ a1 þ x 0 a2 þ y 0 a3 þ z 0 a4 þ x 0 y 0 a5 þ y 0 z 0 a6 þ z 0 x 0 a7 þ x 0 y 0 z 0 a8 ^ e is displacement in the element, which can be the displacement u in the where u x-direction, the displacement v in the y-direction, or the displacement w in the z-direction; the unknown parameters a1 to a8 may be evaluated in terms of the displacements at each of the eight vertices of the hexahedron. To use numerical integration for the integrals in weak form and obtain the geometry and local node numbers of the hexahedron element, as shown in figure 4.18, we suppose n¼

x0 y0 z0 ; g ¼ ; and f ¼ ; a b c

ð4:76Þ

which are the local coordinates, with the range of − 1 ≤ ξ, η, ζ ≤ 1. The coefficients αa may be obtained analogously as the derivation process of the case of the two-dimensional rectangle element. By inverting the coefficient matrix and using substitutions, we obtain the eight shape functions:

Basic Theory of Finite Element Method

54

FIG. 4.18 – Hexahedron element geometry and local node numbers: (a) global coordinates, (b) local coordinates.

1 1 N1 ¼ ð1  nÞ ð1  gÞ ð1  fÞ; N5 ¼ ð1  nÞ ð1  gÞ ð1 þ fÞ 8 8 1 1 N2 ¼ ð1 þ nÞ ð1  gÞ ð1  fÞ; N6 ¼ ð1 þ nÞ ð1  gÞ ð1 þ fÞ 8 8 1 1 N3 ¼ ð1 þ nÞ ð1 þ gÞ ð1  fÞ; N7 ¼ ð1 þ nÞ ð1 þ gÞ ð1 þ fÞ 8 8 1 1 N4 ¼ ð1  nÞ ð1 þ gÞ ð1  fÞ; N8 ¼ ð1  nÞ ð1 þ gÞ ð1 þ fÞ 8 8

4.5.2

ð4:77Þ

Higher-Order Hexahedron Element

We can describe the equivalent Lagrange-family elements of the three-dimensional hexahedron type in a precisely analogous way to that given. Furthermore, we also can use a direct product of three Lagrange polynomials to generate the shape functions for such elements. By extending the notation of equation (4.51), we now have the interpolations: Na  NIJK ¼ lIn ðnÞ lJm ðgÞ lKp ðfÞ

ð4:78Þ

for n, m, and p subdivisions along each side, and with 1  n; g; f  1:

ð4:79Þ

Table 4.3 shows the mappings of I, J, and K to the node numbers a for a hexahedron with eight nodes.

TAB. 4.3 – Numbering for hexahedron with eight nodes. Label a I J K

Node number 1 1 1 1

2 2 1 1

3 2 2 1

4 1 2 1

5 1 1 2

6 2 1 2

7 2 2 2

8 1 2 2

Elements and Shape Functions

55

Figure 4.19 shows a hexahedron element with m = n = p. For m = n = p = 1, we obtain the simple result: 1 Na ðn; g; fÞ ¼ ð1 þ na nÞ ð1 þ ga gÞ ð1 þ fa fÞ; 8

ð4:80Þ

in which ξa, ηa, ζa are the normalised coordinates at node a. The higher-order hexahedral elements were proposed by Zienkiewicz et al. (1969) and then elaborated on, particularly by Argyris et al. (1968a). All the comments about internal nodes apply here. Figure 4.19 shows the first two elements in the three-dimensional Lagrange family.

FIG. 4.19 – Hexahedron element family: (a) linear, (b) quadratic.

4.5.3

Quadratic Hexahedron Element

The shape functions of quadratic hexahedron elements yielding by the products of one-dimensional quadratic Lagrange interpolations are as follows: Corner nodes: 1 Na ¼ ngf ðn þ na Þ ðg þ ga Þ ðf þ fa Þ; 8

a ¼ 1; 2; . . .; 8:

ð4:81Þ

Mid-side nodes:  1  na ¼ 0; Na ¼ gf 1  n2 ðg þ ga Þ ðf þ fa Þ; 4   1 ga ¼ 0; Na ¼ nf ðn þ na Þ 1  g2 ðf þ fa Þ; 4   1 fa ¼ 0; Na ¼ ng ðn þ na Þ ðg þ ga Þ 1  f2 ; 4

a ¼ 9; 10; 11; 12 a ¼ 13; 14; 15; 16 a ¼ 17; 18; 19; 20

ð4:82Þ

Basic Theory of Finite Element Method

56 Mid-face nodes:

  1  na ¼ ga ¼ 0; Na ¼ f 1  n2 1  g2 ðf þ fa Þ; a ¼ 21; 22 2    1 ga ¼ fa ¼ 0; Na ¼ n ðn þ na Þ 1  g2 1  f2 ; a ¼ 23; 24 2    1  fa ¼ na ¼ 0; Na ¼ g 1  n2 ðg þ ga Þ 1  f2 ; a ¼ 25; 26 2 Centre node:

4.6

    Na ¼ 1  n2 1  g2 1  f 2 ;

a ¼ 27

ð4:83Þ

ð4:84Þ

Exercises

4.6.1 One-dimensional element of line with two nodes. (1) Using the equations of the shape functions, verify the following:  1; na ¼ nb Na ðnb Þ ¼ dab ¼ ; 0; na 6¼ nb 2 X

Na ðnÞ ¼ 1:

ð4:85Þ

ð4:86Þ

a¼1

(2) Using the equations of the shape functions, compute the following derivatives: @Na @ 2 Na ; ; a ¼ 1; 2: @n @n2

ð4:87Þ

4.6.2 Two-dimensional element of triangle with three nodes. (1) Using the equations of the shape functions, verify the following:  1; a ¼ b ; Na ðxb ; yb Þ ¼ dab ¼ 0; a 6¼ b 3 X

Na ðx; y Þ ¼ 1:

ð4:88Þ

ð4:89Þ

a¼1

(2) Using the equations of the shape functions, compute the following derivatives: @Na @Na @ 2 Na @ 2 Na @ 2 Na ; ; ; ; ; a ¼ 1; 2; 3: @x @y @xy @x 2 @y 2

ð4:90Þ

Elements and Shape Functions

57

4.6.3 Two-dimensional element of rectangle with four nodes (1) Using the equations of the shape functions, verify the following:  1; a ¼ b ; Na ðnb ; gb Þ ¼ dab ¼ 0; a 6¼ b 4 X

Na ðn; gÞ ¼ 1:

ð4:91Þ

ð4:92Þ

a¼1

(2) Using the equations of the shape functions, compute the following derivatives: @Na @Na @ 2 Na @ 2 Na @ 2 Na ; ; ; ; ; @n @g @ng @g2 @n2

a ¼ 1; 2; 3; 4:

ð4:93Þ

4.6.4 Three-dimensional element of tetrahedron with four nodes. (1) Using the equations of the shape functions, verify the following:  1; a ¼ b ; Na ðxb ; yb ; zb Þ ¼ dab ¼ 0; a 6¼ b 4 X

Na ðx; y; z Þ ¼ 1:

ð4:94Þ

ð4:95Þ

a¼1

(2) Using the equations of the shape functions, compute the following derivatives: @Na @Na @Na @ 2 Na @ 2 Na ; ; ; ; @x @y @z @xy @yz @ 2 N a @ 2 Na @ 2 Na @ 2 Na @ 3 N a ; ; ; ; ; @zx @x 2 @y 2 @z 2 @xyz

ð4:96Þ a ¼ 1; 2; 3; 4

4.6.5 Three-dimensional element of hexahedron with eight nodes. (1) Using the equations of the shape functions, verify the following:  1; a ¼ b Na ðnb ; gb ; fb Þ ¼ dab ¼ ; 0; a 6¼ b 8 X a¼1

Na ðn; g; fÞ ¼ 1:

ð4:97Þ

ð4:98Þ

58

Basic Theory of Finite Element Method

(2) Using the equations of the shape functions, compute the following derivatives: @Na @Na @Na @ 2 Na @ 2 Na ; ; ; ; @n @g @f @ng @gf @ 2 Na @ 2 N a @ 2 Na @ 2 Na @ 3 Na ; ; a ¼ 1; 2; . . .; 8 ; ; ; @fn @g2 @n2 @f2 @ngf

ð4:99Þ

4.6.6 Summarise the functions of Lagrange interpolation for constructing one-, two-, and three-dimensional elements.

Chapter 5 Isoparametric Element and Numerical Integration 5.1 5.1.1

Isoparametric Element One-Dimensional Isoparametric Lagrange Element

When the finite element method is applied to one-dimensional problems, elements with different shapes and sizes are often used. To unify elements with arbitrary length in a one-dimensional problem, a method for mapping elements in the global coordinate system to standard elements in the unified local parent coordinate system is developed. Then, all computations in an element are converted to the standard element for processing. A representative isoparametric mapping of a one-dimensional Lagrange element with two nodes is shown in figure 5.1.

FIG. 5.1 – Isoparametric mapping of one-dimensional Lagrange element with two nodes: (a) global coordinates, (b) local parent coordinates.

The relation between the global and local parent coordinates for a one-dimensional Lagrange element is given by: xe ¼

n X a¼1

Na ðnÞ xae :

ð5:1Þ

DOI: 10.1051/978-2-7598-2938-5.c005 © Science Press, EDP Sciences, 2023

Basic Theory of Finite Element Method

60

The use of isoparametric mappings was introduced by Taig (1962). The types of parametric mapping that can be considered in the analysis are defined by the relation between the coordinate interpolation and the dependent-variable interpolation. We denote the mapping by: xe ¼

n X

Na ðnÞ xae ¼ N1 ðnÞ x1e þ N2 ðnÞ x2e þ    þ Nn ðnÞ xne ; 1  n  1

ð5:2aÞ

e Na ðnÞ u^ae ¼ N1 ðnÞ u^1e þ N2 ðnÞ u^2e þ    þ Nm ðnÞ u^m ; 1  n  1

ð5:2bÞ

a¼1

u^ e ¼

m X a¼1

Once the shape functions are available in terms of the parent coordinates, we may immediately use the concept of parametric mapping. We have the following three cases: (1) Sub-parametric interpolation: The order n of the interpolation for x is lower than that order m for u. (2) Isoparametric interpolation: The order n of the interpolation for x is the same as that order m for u. (3) Super-parametric interpolation: The order n of the interpolation for x is higher than that order m for u. Indeed, we need not always use the same interpolation for coordinates and dependent variables; however, an isoparametric mapping is the most convenient because nodal coordinates and nodal dependent variables are placed at the same physical locations. Owing to the transformation of the above coordinate system, the derivatives of several functions are involved in finite element computation. This requires switching between different coordinate systems. The derivatives of functions on element e with respect to the two coordinate systems are related as follows: n @f @f @x @f X @Na ðnÞ e @f ¼ ¼ xa ¼ j; @n @x @n @x a¼1 @n @x

ð5:3Þ

P a ðnÞ where j ¼ na¼1 @N@n xae is known as the Jacobian for the transformation and by the relation defining the local parent coordinates, it can be computed explicitly in terms of the local coordinates.

5.1.2

Two-Dimensional Isoparametric Triangle Element

The concepts of isoparametric mapping and isoparametric element can be used in two-dimensional triangle element elements to obtain a two-dimensional isoparametric triangle element. A representative isoparametric mapping of a two-dimensional triangle element with three nodes is shown in figure 5.2, in which x and y are the global coordinates and L1 ; L2 ; L3 are the local parent coordinates.

Isoparametric Element and Numerical Integration

61

FIG. 5.2 – Isoparametric mapping of two-dimensional triangle element with three nodes: (a) global coordinates, (b) local area coordinates.

The relation between the global and local parent coordinates for a two-dimensional triangle element is given by xe ¼

n X a¼1

Na ðL1 ; L2 ; L3 Þ xae :

ð5:4Þ

The isoparametric mappings for the coordinate interpolation and the dependent-variable interpolation are xe ¼

n X a¼1

^e ¼ u

n X a¼1

Na ðL1 ; L2 ; L3 Þ xae ;

0  L1 ; L2 ; L3  1

^ ea ; Na ð L1 ; L2 ; L3 Þ u

0  L 1 ; L2 ; L3  1

and

3 X

Li ¼ 1

ð5:5aÞ

Li ¼ 1

ð5:5bÞ

i¼1

and

3 X i¼1

where L1 ; L2 ; L3 are the local area coordinates defined above; x e is the global ^ e is the coordinate in the element, which can be the coordinate x or y, and u displacement in the element, which can be the displacement u in the x-direction or the displacement v in the y-direction. By the usual chain rule of partial differentiation, the x and y derivatives of any function f may be written in matrix form as: 9 8 9 8 3 P @f @Li > @f > > > > > > > > > > = < @x = < i¼1 @Li @x > ¼ : 3 @f @L > > > @f > P > > > i> > > > > ; : ; : @y i¼1 @Li @y

ð5:6Þ

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62

The derivatives of the area coordinates may be obtained by differentiating the three equations: 9 8 3 P > > > > L > > i 8 9 > > > > i¼1 > =

: e; > a¼1 > > y > > n > > P > > > ; : Na ðL1 ; L2 ; L3 Þ yae > a¼1

Using the chain rule, the derivatives of f with respect to x and y may be written as:

2



@f @x

@f @y



2

0 ¼ 41 0

3 2 0 1 0 5 ¼ 4 X1 Y1 1

@L1 6 @x 36 1 6 6 @L2 X 3 56 6 @x Y3 6 6 4 @L3 @x

1 X2 Y2

3 @L1 @y 7 7 7 @L2 7 7; @y 7 7 7 @L3 5 @y

ð5:8Þ

where we denote the derivatives by: Xi ¼

n X @Na a¼1

@Li

Solving equation (5.8) yields 3 2 @L1 @L1 6 @x @y 7 7 2 6 7 6 1 1 6 @L2 @L2 7 7 ¼ 4 X1 X2 6 6 @x @y 7 7 6 Y1 Y2 7 6 4 @L3 @L3 5 @x

xae

and

Yi ¼

n X @Na a¼1

31 2 1 0 X3 5 4 1 0 Y3

@Li

yae :

3 2 B 0 1 4 1 05 ¼ B2 2D 1 B3

ð5:9Þ

3 C1 C2 5; C3

ð5:10Þ

@y

where B 1 ¼ Y2  Y3 ;

C 1 ¼ X3  X 2

B 2 ¼ Y3  Y1 ; B 3 ¼ Y1  Y2 ;

C 2 ¼ X1  X 3 C 3 ¼ X2  X 1

ð5:11Þ

and D ¼ ðX1 B1 þ X2 B2 þ X3 B3 Þ=2. Through the above analysis, the derivatives of the area coordinates derived from equation (5.10) are substituted into equation (5.6), and the x and y derivatives of any function f can be obtained.

Isoparametric Element and Numerical Integration

5.1.3

63

Two-Dimensional Isoparametric Rectangle Element

The concepts of isoparametric mapping and isoparametric element can be used in two-dimensional rectangle elements to obtain a two-dimensional isoparametric rectangle element. A representative isoparametric mapping of a two-dimensional rectangular element with four nodes is shown in figure 5.3, in which x and y are the global coordinates, and ξ and η are the local parent coordinates.

FIG. 5.3 – Isoparametric mapping of two-dimensional rectangle element with four nodes: (a) global coordinates, (b) local parent coordinates.

The relation between the global and local parent coordinates for a two-dimensional rectangular element is given by: xe ¼

n X a¼1

Na ðn; gÞ xae :

ð5:12Þ

The isoparametric mappings for the coordinate interpolation and the dependent-variable interpolation are given by: xe ¼

n X a¼1

^e ¼ u

n X a¼1

Na ðn; gÞ xae ;

1  n; g  1

ð5:13aÞ

^ ea ; Na ðn; gÞ u

1  n; g  1

ð5:13bÞ

where ξ and η are the local parent coordinates defined above, x e is the global ^ e is the coordinate in the element, which can be the coordinate x or y, and u displacement in the element, which can be the displacement u in x-direction or the displacement v in the y-direction. By the usual chain rule of partial differentiation, the ξ and η derivatives of any function f (ξ, η) may be written in matrix form as:

Basic Theory of Finite Element Method

64 8 9 2 @f > @x > > > > > < @n = 6 6 @n ¼6 > > 4 @x > @f > > ; : > @g @g

8 9 38 9 @y > @f > @f > > > > > > > > = = < @x > < @x > @n 7 7 ¼J : 7 > @y 5> > > > @f > > @f > > > > > : ; : ; @g @y @y

ð5:14Þ

The array J is known as the Jacobian matrix of the transformation, and by the relation defining the parent coordinates, it can be computed explicitly in terms of the local coordinates. In terms of the mapping defining the coordinate transformation, we have 2 n 3 n @N P @Na P a x y a a 6 a¼1 @n 7 a¼1 @n 6 7 J¼6 n ð5:15Þ 7: n @N 4 P @Na 5 P a xa ya a¼1 @g a¼1 @g On the other hand, the x and y derivatives of any function f (ξ, η) can be expressed as: 8 9 8 9 @f > @f > > > > > > > > > = > = < < @n > @x 1 ¼J : ð5:16Þ > > @f > @f > > > > > > > > > : ; : ; @y @g The inverse of the Jacobian matrix is easily obtained from equation (5.4) and is given by: 2 3 @y @y  @n 7 16 6 @g 7 J 1 ¼ 6 ð5:17Þ 7; j 4 @x @x 5  @g @n where j ¼ det J ¼

@x @y @x @y  @n @g @g @n

ð5:18Þ

is the determinant of the Jacobian matrix.

5.1.4

Three-Dimensional Isoparametric Tetrahedron Element

For a three-dimensional tetrahedron element with four nodes, the isoparametric mapping is shown in figure 5.4, in which x, y, and z are the global coordinates and L1 ; L2 ; L3 ; L4 are the local parent coordinates.

Isoparametric Element and Numerical Integration

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FIG. 5.4 – Isoparametric mapping of three-dimensional tetrahedron element with four nodes: (a) global coordinates, (b) local volume coordinates.

The relation between the global and local parent coordinates for a three-dimensional tetrahedral element is given by: xe ¼

n X a¼1

Na ðL1 ; L2 ; L3 ; L4 Þ xae :

ð5:19Þ

The isoparametric mappings for the coordinate interpolation and the dependent variable interpolation are given by: xe ¼

n X a¼1

^e ¼ u

n X a¼1

Na ðL1 ; L2 ; L3 ; L4 Þ xae ;

0  L1 ; L2 ; L3 ; L4  1

^ ea ; Na ðL1 ; L2 ; L3 ; L4 Þ u

0  L1 ; L2 ; L3 ; L4  1 and

and

4 X

Li ¼ 1; ð5:20aÞ

i¼1 4 X

Li ¼ 1; ð5:20bÞ

i¼1

where L1 ; L2 ; L3 ; L4 are the local volume coordinates defined above, x e is the global ^ e is the coordinate in the element, which can be the coordinate x, y, or z, and u displacement in the element, which can be the displacement u in the x-direction, the displacement v in the y-direction, or the displacement w in the z-direction. By the usual chain rule of partial differentiation, the x, y, and z derivatives of any function f may be written in matrix form as: 9 8 8 9 >P 4 @f @L > i> @f > > > > > > > > > > > > > > > i¼1 @Li @x > @x > > > > > > > > > > > > > < =

> > @L @y > > > > > > > @y > > i¼1 i > > > > > > > > > 4 @f @L > > > > > @f > >P i> ; : > > > ; : @z @L @z i i¼1

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The derivatives of the volume coordinates may now be obtained by differentiating the following four equations: 9 8 4 P > > > > Li > > > > > i¼1 8 9 > > > > > n > >P 1 > > > e> > > > Na ðL1 ; L2 ; L3 ; L4 Þ xa > = < e= < x : ð5:22Þ ¼ a¼1 f ¼ e n P y > > > e> > > > > ; > : e> N ð L ; L ; L ; L Þ y a 1 2 3 4 a> > z > > > > > > a¼1 > > n P > > > e ; : N a ð L1 ; L2 ; L3 ; L4 Þ z a > a¼1

Using the chain rule, the derivatives of f with respect to x, written as: 2 @L1 6 @x 36 2 3 2 0 0 0 1 1 1 1 6 6 @L2   6 1 0 0 7 6 X1 X2 X3 X4 76 @x @f @f @f 76 7 6 ¼6 4 0 1 0 5 ¼ 4 Y1 Y2 Y3 Y4 56 @L3 @x @y @z 6 Z1 Z2 Z3 Z4 6 0 0 1 6 @x 4 @L4 @x

y, and z may now be 3 @L1 @z 7 7 @L2 7 7 @z 7 7; ð5:23Þ @L3 7 7 @z 7 7 @L4 5

@L1 @y @L2 @y @L3 @y @L4 @y

@z

where we denote the parent derivatives by: Xi ¼

n X @Na a¼1

@Li

xae ;

Yi ¼

n X @Na a¼1

@Li

Solving equation (5.23) yields 3 @L1 @L1 @L1 6 @x @y @z 7 7 6 7 6 6 @L2 @L2 @L2 7 2 1 1 1 7 6 7 6 6 @x @y @z X X X 7 6 1 6 2 3 7¼ 6 6 @L3 @L3 @L3 7 4 Y1 Y2 Y3 7 6 Z1 Z2 Z3 6 @x @y @z 7 7 6 7 6 4 @L4 @L4 @L4 5

yae ;

and

Zi ¼

n X @Na a¼1

@Li

zae :

ð5:24Þ

2

@x

@y

31 2 1 0 61 X4 7 7 6 Y4 5 4 0 0 Z4

0 0 1 0

3 0 07 7¼ 1 0 5 6V 1

2

B1 6 B2 6 4 B3 B4

C1 C2 C3 C4

3 D1 D2 7 7; D3 5 D4

@z ð5:25Þ

where

2

1 Y2 B1 ¼  det4 1 Y3 1 Y4

3 Z2 Z3 5; Z4

2

1 C1 ¼ det4 1 1

X2 X3 X4

3 Z2 Z3 5; Z4

2

1 X2 D1 ¼  det4 1 X3 1 X4

3 Y2 Y3 5; Y4 ð5:26Þ

Isoparametric Element and Numerical Integration

67

with the other constants defined by a cyclic interchange of the subscripts in the order 1, 2, 3, 4, and 2 3 1 X1 Y1 Z 1 6 1 X2 Y2 Z 2 7 1 7 V ¼ det6 ð5:27Þ 4 1 X3 Y3 Z 3 5 : 6 1 X4 Y4 Z 4 Through the above analysis, the derivatives of the area coordinates derived from equation (5.25) are substituted into equation (5.21), and the x, y, and z derivatives of any function f can be obtained.

5.1.5

Three-Dimensional Isoparametric Hexahedron Element

For a three-dimensional isoparametric hexahedron element with eight nodes, the isoparametric mapping is shown in figure 5.5, in which x, y, and z are the global coordinates, and ξ, η, and ζ is the local parent coordinates.

FIG. 5.5 – Isoparametric mapping of three-dimensional hexahedron element with eight nodes: (a) global coordinates, (b) local parent coordinates. The relation between the global and local parent coordinates for a three-dimensional hexahedral element is given by: xe ¼

n X a¼1

Na ðn; g; fÞ xae :

ð5:28Þ

The isoparametric mappings for the coordinate interpolation and the dependent-variable interpolation are given by: xe ¼

n X a¼1

Na ðn; g; fÞ xae ;

1  n; g; f  ; 1

ð5:29aÞ

Basic Theory of Finite Element Method

68

^e ¼ u

n X a¼1

^ ea ; Na ðn; g; fÞ u

1  n; g; f  1

ð5:29bÞ

where ξ, η, and ζ are the local parent coordinates defined above, x e is the global ^ e is the coordinate in the element, which can be the coordinate x, y, or z, and u displacement in the element, which can be the displacement u in the x-direction, the displacement v in the y-direction, or the displacement w in the z-direction. By the usual chain rule of partial differentiation, the ξ, η, and ζ derivatives of any function f (ξ, η, ζ) may be written in matrix form as: 8 9 2 @f > @x > > > > > 6 @n > > > > @n > > > 6 > > = 6 < @f > 6 @x ¼6 > > @g > 6 > 6 @g > > > > 6 > > > > @f > 4 @x > > ; : > @f @f

@y @n @y @g @z @f

3 9 8 9 @z 8 @f > @f > > > > > > > 7 > > > > > > > @n 7> @x @x > > > > > > > > > > > 7> = = < < @z 7 @f @f 7 ¼ J : > @g 7 @y > @y > > > > > 7> > > > > > > 7> > > > > > > > > @z 5> > > ; ; : @f > : @f > @f @z @z

ð5:30Þ

The array J is known as the Jacobian matrix of the transformation, and by the relation defining the parent coordinates, it can be computed explicitly in terms of the local coordinates. In terms of the mapping defining the coordinate transformation, we have 2

n @N P a 6 a¼1 @n xa 6 6 n 6 P @Na J¼6 6 a¼1 @g xa 6 6 n 4 P @Na xa a¼1 @f

n @N P a ya @n a¼1 n @N P a ya a¼1 @g n @N P a ya @f a¼1

3 n @N P a za 7 a¼1 @n 7 7 n @N 7 P a za 7 7: a¼1 @g 7 7 n P @Na 5 za a¼1 @f

ð5:31Þ

On the other hand, the x, y, and z derivatives of any function f (ξ, η, ζ) can be expressed as: 8 9 8 9 @f > @f > > > > > > > > > > > @x > > > > > > @n > > > > > > > > > > = = < > < @f > @f 1 : ð5:32Þ ¼J > > @y > > @g > > > > > > > > > > > > > > > > > > > @f > > > > ; : @f > > ; : > @f @z The inverse of the Jacobian matrix can be easily obtained from equation (5.31), and is given by:

Isoparametric Element and Numerical Integration

J 1 where

2 a 14 1 ¼ b1 j c1

2

3 n @N P a za 7 a¼1 @g 7 7; n P @Na 5 za a¼1 @f

2

3 n @N P a ya 7 a¼1 @g 7 7 n @N 5 P a ya a¼1 @f

n @N P a 6 a¼1 @g ya 6 a1 ¼ det6 n 4 P @Na ya a¼1 @f n @N P a 6 a¼1 @g xa 6 c1 ¼ det6 n 4 P @Na xa a¼1 @f

a2 b2 c2

69

3 a3 b3 5 c3 2

n @N P a 6 a¼1 @g xa 6 b1 ¼  det6 n 4 P @Na xa a¼1 @f

ð5:33Þ

3 n @N P a za 7 a¼1 @g 7 7 n P @Na 5 za a¼1 @f

ð5:34Þ

with the other constants defined by the cyclic interchange of the subscripts in the order 1, 2, 3; j ¼ det J

ð5:35Þ

is the determinant of the Jacobian matrix.

5.1.6

Requirements of Isoparametric Element

As noted above, the mapping from parent to global coordinates in an element must always be single-valued. That is, each point in the parent coordinates should have unique global coordinates. The parametric mapping is controlled by the placement of nodal coordinates for each element. The condition that ensures a one-to-one mapping is that the Jacobian determinant j (computed by the chain rule) is positive: j [0

ð5:36Þ

For lower-order elements, one can establish criteria to ensure that equation (5.36) is satisfied everywhere in the element. However, as the order increases, this becomes more difficult. To ensure that the condition is satisfied, the following must be checked: (1) (2) (3) (4)

Placement of edge nodes to make j positive. Placement of face nodes to make j positive. Placement of internal nodes to make j positive. All vertex angles are less than 180°.

Checking that the value of j is positive at each node of an element is sufficient to ensure that the above conditions are satisfied, and thus the specified parametric mapping produces a valid element.

Basic Theory of Finite Element Method

70

5.2

Numerical Integration

As the shape functions and derivatives are known, the next step requires the evaluation of integrals in the weak form to obtain the element matrices.

5.2.1

One-Dimensional Integration for Lagrange Element

Because the shape functions are expressed in terms of the local parent coordinate, it is also convenient to express the integrals in terms of the local parent coordinate. The differential is given in terms of the Jacobian as follows: dx ¼

@x dn ¼ j ðnÞdn: @n

ð5:37Þ

Thus, the integrals may be expressed as: Z xe Z 1 2 f ðx Þdx ¼ f ðnÞj ðnÞdn; x1e

1

ð5:38Þ

where f ðnÞ ¼ f ðxðnÞÞ. If the integrand is complex, it is difficult to compute the above definite integral. In another manner, numerical integration is used to approximate the integrals. An efficient formula for polynomial forms is the Gauss quadrature or Gauss–Legendre quadrature, which is defined on the interval −1 ≤ ξ ≤ 1. The integral is replaced by the following form: Z 1 n X   f ðnÞ dn  f nj w j ; ð5:39Þ 1

j¼1

with an appropriate choice of integration points nj and weights wj . The Gauss– Legendre quadrature formula is exact in the case of polynomials of degree at most 2n−1. Table 5.1 provides typical locations of the integration points and weights. TAB. 5.1 – Gaussian integration points and weights for one-dimensional Lagrange element. n nj wj 1 2 3

4

0 pffiffiffi 1= 3 pffiffiffiffiffiffiffi  0:6 0.0 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffi  3 þ 4:8 7 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffi  3  4:8 7

2.0 1.0 5/9 8/9 1 1  pffiffiffiffiffiffiffi 2 3 4:8 1 1 þ pffiffiffiffiffiffiffi 2 3 4:8

Isoparametric Element and Numerical Integration

71

Example 5.1. Stiffness for a one-dimensional element with two nodes by numerical integration. The shape functions and Jacobian for a one-dimensional element with two nodes are 1 N1 ¼ ð1  nÞ; 2

1 N2 ¼ ð1 þ nÞ; 2

j¼

@x he ¼ : @n 2

ð5:40Þ

The stiffness for a two-node element can be computed as follows: 9 9 8 8 @N1 @n >  @N   Z he > Z 1> =  = < 1> < @N1 @n @N2 @n @N1 @N2 @n @x e 0 @x Ee Ee dx ¼ K ¼ jðnÞdn @n @x @n @x @x @x 1 > 0 > ; ; : @N2 > : @N2 @n > @n @x 9 8 @x @N 1 1 > > >  Z 1> =  @N 1 < @N2 1 1 @n jðnÞ Ee jðnÞdn ¼ @N2 1 > @n jðnÞ @n jðnÞ 1 > > > ; : @n jðnÞ 9 8 @N 1 2 >   Z 1 Z 1> =  < @n he E @N1 2 @N2 2 1 h dn ¼ Ee 1 1 dn ¼ e e 2he 1 1 @n he @n he 2 1 > 1 ; : @N2 2 > @n he which is identical to that obtained previously. Numerical integration is performed using the one-point quadrature formula because the shape-function derivatives are constant. The numerical integration is as follows: Z 1 1 X   dn ¼ f nj wj ¼ f ð0Þw1 ¼ 1  2 ¼ 2; ð5:41Þ 1

j¼1

which is obviously the correct answer. For this problem, quadrature was not required; however, for higher-order elements, quadrature greatly simplifies the calculations. Example 5.2. Stiffness for a one-dimensional element with three nodes by numerical integration. The shape functions and Jacobian for a one-dimensional element with three nodes are 1 N1 ¼ nðn  1Þ; 2 @x ¼ jðnÞ ¼ @n

1 N2 ¼ nðn þ 1Þ; 2

N3 ¼ 1n2 ;

      1 e 1 e 1 n  x1 þ n þ x  2nx3e ¼ he þ n x1e þ x2e  2x3e : 2 2 2 2

ð5:42Þ

ð5:43Þ

Basic Theory of Finite Element Method

72

For the case where the coordinate for node 3 is centred between nodes 1 and 2, the derivatives are linear in ξ, and the Jacobian is constant and equal to 12 he . The stiffness for a three-node element involves the integral: 9 8 @N1 > > > > > > > >  Z he > =  < @x > @N @N1 @N2 @N3 2 e Ee K ¼ dx 0 > @x @x @x @x 0 > > > > > > > > ; : @N3 > @x 9 8 @N1 @n > > > > > @n @x > >  >  Z 1> = < @N @n > @N1 @n @N2 @n @N3 @n 2 Ee jðnÞdn ¼ > > @n @x > @n @x @n @x @n @x 1 > > > > > > ; : @N3 @n > @n @x 9 8 @N1 1 > > > > > > > > j @n ðnÞ > > e >  >  Z 1< @N2 1 = @N1 1 @N2 1 @N3 1 Ee ¼ jðnÞdn @n je ðnÞ > @n je ðnÞ @n je ðnÞ @n je ðnÞ 1 > > > > > > > @N3 1 > > > > ; : @n je ðnÞ 9 8 @N1 2 > > > > > > @n he > > > Z 1> =  @N 2 @N 2 @N 2  1 < @N 2 2 1 2 3 Ee he dn ¼ > > h h h he 2 @n @n @n @n e e e 1 > > > > > > > ; : @N3 2 > @n he 3 2        1 2 1 1 1 n n nþ n  ð2nÞ 7 6 2 2 2 2 7 Z 16     2   7 6 2Ee 7 6 1 1 1 1 ¼ 7 dn 6 n n þ n þ n þ ð2nÞ 7 6 he 1 2 2 7 6  2   2 5 4 1 1 nþ ð2nÞ ð2nÞ2 n  ð2nÞ 2 2 In this case, the exact stiffness is recovered using a two-point quadrature formula. For example, K11 is  Z  2 X   2Ee 1 1 2 K 11 ¼ n dn  f nj w j 2 he 1 j¼1 !  2   2Ee 1 1 1 1 2 ð5:44Þ ¼  pffiffiffi  1 þ pffiffiffi  1 he 3 2 3 2 ¼

7Ee 3he

Isoparametric Element and Numerical Integration

73

It can be seen that this result is exactly the same as the result of direct integration, as shown below:   Z  Z  2Ee 1 1 2 2Ee 1 1 2 K 11 ¼ n dn ¼ n nþ dn 2 4 he 1 he 1   2Ee n3 n2 n 1 ð5:45Þ  þ ¼ 4 h 3 2 e

1

7Ee ¼ 3he The reader can verify that the final result is 2 3 7 1 8 E e 4 Ke ¼ 1 7 8 5: 3he 8 8 16

ð5:46Þ

If node 3 is not centred between nodes 1 and 2, the Jacobian is not equal to 12 he and should be regarded as a part of the integrand.

5.2.2

Two-Dimensional Integration for Triangle Element

The numerical integration formula in terms of the area coordinates is given by (Anderson et al., 1968; Hammer et al., 1956; Radau, 1880): Z 1 Z 1L1 n X   I ¼ f ðL1 ; L2 ; L3 Þ dL2 dL1  f L1j ; L2jj ; L3j wj ; 0 0 ð5:47Þ j¼1 L3 ¼ 1  L1  L 2 : A series of necessary integration points and weights are listed in table 5.2 (Lyness and Jespersen, 1975; Cowper, 1973; Abramowitz and Stegun, 1965). TAB. 5.2 – Gaussian integration points and weights for a two-dimensional triangle element. Order Linear (n = 1)

Quadratic (n = 3)

Figure

Integration points

Area coordinates

Weights

a

1 1 1 ; ; 3 3 3

1

a

1 1 ; ; 0 2 2 1 1 ; 0; 2 2 1 1 0; ; 2 2 2 1 1 ; ; 3 6 6 1 2 1 ; ; 6 3 6 1 1 2 ; ; 6 6 3

1 3 1 3 1 3 1 3 1 3 1 3

b c a

Quadratic (n = 3)

b c

Basic Theory of Finite Element Method

74

TAB. 5.2 – (continued). Order

Cubic (n = 4)

Quintic (n = 7)

5.2.3

Figure

Integration points

Area coordinates

a

1 1 1 ; ; 3 3 3

b

0:6; 0:2; 0:2

c

0:2; 0:6; 0:2

d

0:2; 0:2; 0:6

a

1 1 1 ; ; 3 3 3 a1 ; b1 ; b1 b1 ; a1 ; b1 b1 ; b1 ; a1 a2 ; b2 ; b2 b2 ; a2 ; b2 b2 ; b2 ; a2 with a1 ¼ 0:0597158717 b1 ¼ 0:4701420641 a2 ¼ 0:07974269853 b2 ¼ 0:1012865073

b c d e f g

Weights 27 48 25 48 25 48 25 48



0.2250000000 0.1323941527 0.1323941527 0.1323941527 0.1259391805 0.1259391805 0.1259391805

Two-Dimensional Integration for Rectangle Element

Even though integrals for rectangular four-node elements may be calculated quite easily, the integration becomes considerably more difficult for axisymmetric and mapped elements using the parent coordinates. Thus, in general, it is best to use quadrature (numerical integration) as a general computational tool. To use isoparametric mappings, it is first necessary to change the domain over which the elements are integrated. This is a standard procedure that is taught in integral calculus. Thus, for a two-dimensional problem involving rectangular elements, the integral over the element domain is converted to an integral over the parent domain −1 ≤ ξ, η ≤ 1. The infinitesimal area elements in the global and local coordinate systems are related as follows: dxdy ¼ dn  dg     @x @y @x @y dn i þ dn j  dg i þ dg j ¼ @n @n @g @g     @x @y @x @y ¼ iþ j dn  iþ j dg @n @n @g @g @x @y @n @n ¼ dndg ¼ jdndg @x @y @g @g

ð5:48Þ

Isoparametric Element and Numerical Integration

Using the isoparametric form, the integral is transformed as: Z ye Z x e Z 1Z 1 2 2 f ðx; y Þdxdy ¼ f ðn; gÞj ðn; gÞdndg; y1e

x1e

1

1

75

ð5:49Þ

where j ðn; gÞ is the determinant of the Jacobian matrix described above. With the above choice for the domain of the parent element, it is now convenient to evaluate integrals using Gaussian quadrature. Accordingly, we can integrate into two directions using Z 1Z 1 N X M X f ðn; gÞje ðn; gÞdndg  f ðnm ; gn Þje ðnm ; gn Þwm wn ; ð5:50Þ 1

1

n¼1 m¼1

where m denotes the quadrature points in the ξ direction, and n the quadrature points in the η direction. The location and weight of the quadrature points in each direction are given in table 3.1 for one-dimensional integrations. Example 5.3. For a two-dimensional rectangle element with four nodes, compute  R1 R1  numerically the integral 1 1 n2 þ g2 dndg. In this case, the integral is recovered using a two-point quadrature formula in two directions: Z 1Z 1 2 X 2 X  2  n þ g2 dndg  g ðnm ; gn Þwm wn 1

1

n¼1 m¼1

¼ g ðn1 ; g1 Þw1 w1 þ g ðn1 ; g2 Þw1 w2 þ g ðn2 ; g1 Þw2 w1 þ g ðn2 ; g2 Þw2 w2     1 1 1 1 ð5:51Þ ¼ g  pffiffiffi ;  pffiffiffi  1  1 þ g  pffiffiffi ; pffiffiffi  1  1 3 3 3 3     1 1 1 1 þ g pffiffiffi ;  pffiffiffi  1  1 þ g pffiffiffi ; pffiffiffi  1  1 3 3 3 3 8 ¼ 3 It can be seen that this result is exactly the same as that of direct integration: Z 1Z 1 Z 1Z 1 Z 1Z 1  2  4 4 8 2 2 n þ g dndg ¼ n dndg þ g2 dndg ¼ þ ¼ : ð5:52Þ 3 3 3 1 1 1 1 1 1

5.2.4

Three-Dimensional Integration for Tetrahedron Element

The numerical integration formula in terms of the volume coordinates is given by:

Basic Theory of Finite Element Method

76 Z

1

I ¼ 0

Z 0

1L1

Z

1L1 L2

f ðL1 ; L2 ; L3 ; L4 Þ dL3 dL2 dL1 

0

n X   f L1j ; L2jj ; L3j ; L4j wj ; j¼1

L4 ¼ 1  L1  L 2  L 3 ð5:53Þ Furthermore, table 5.3 presents some integration points and weights.

TAB. 5.3 – Gaussian integration points and weights for three-dimensional tetrahedron element. Order Figure Integration points Volume coordinates Weights Linear (n = 1)

Quadratic (n = 4)

a

1 1 1 1 ; ; ; 4 4 4 4

a

a; b; b; b

b

b; a; b; b

c

b; b; a; b

d

a b Cubic (n = 5)

c d e

5.2.5

b; b; b; a a ¼ 0:58541020 b ¼ 0:13819660 1 1 1 1 ; ; ; 4 4 4 4 1 1 1 1 ; ; ; 2 6 6 6 1 1 1 1 ; ; ; 6 2 6 6 1 1 1 1 ; ; ; 6 6 2 6 1 1 1 1 ; ; ; 6 6 6 2

1 1 4 1 4 1 4 1 4

4 5 9 20 9 20 9 20 9 20



Three-Dimensional Integration for Hexahedron Element

Integration for hexahedron elements is performed as in the case of quadrilateral shapes. For a three-dimensional problem, the quadrature over the element domain is converted to an integral over the parent domain −1 ≤ ξ, η, ζ ≤ 1.

Isoparametric Element and Numerical Integration

77

The infinitesimal volume elements in the global and local coordinate systems are related as follows: dxdydz ¼ dn  ðdg  dfÞ   @x @y @z dn i þ dn j þ dn k  ¼ @n @n @n     @x @y @z @x @y @z dg i þ dg j þ dg k  df i þ df j þ df k @g @g @g @f @f @f   @x @y @z ¼ iþ jþ k dn @n @n @n      ð5:54Þ @x @y @z @x @y @z iþ jþ k dg  iþ jþ k df @g @g @g @f @f @f @x @y @z @n @n @n @x @y @z dndgdf ¼ jdndgdf ¼ @g @g @g @x @y @z @f @f @f Using the parametric form, the integral is transformed as follows: Z z e Z ye Z x e Z 1Z 1Z 1 2 2 2 f ðx; y; z Þdxdydz ¼ f ðn; g; fÞj ðn; g; fÞdndgdf z1e

y1e

x1e

1

1

1

ð5:55Þ

where je ðn; g; fÞ is the determinant of the Jacobian matrix for the three-dimensional hexahedron element. The numerical integration formula is Z 1Z 1Z 1 f ðn; g; fÞj ðn; g; fÞdndgdf 1

1

1



L X N X M X l¼1 n¼1 m¼1

ð5:56Þ f ðnm ; gn ; fl Þj ðnm ; gn ; fl Þwm wn wl

where m, n, and l denote quadrature points in the ξ, η, and ζ directions, respectively. The location and weight of the quadrature points in each direction are given in table 3.1 for one-dimensional integrations. Example 5.4. For a three-dimensional hexahedron element with eight nodes, com R1 R1 R1  pute numerically the integral 1 1 1 n2 þ g2 þ f2 dndgdf. In this case, the integral is recovered using a two-point quadrature formula in three directions:

Basic Theory of Finite Element Method

78 Z

1

1

Z

1

1

Z

1

1



2 X 2 X 2 X  n2 þ g2 þ f2 dndgdf  g ðnm ; gn ; fl Þwm wn wl l¼1 n¼1 m¼1

¼ g ðn1 ; g1 ; f1 Þw1 w1 w1 þ g ðn1 ; g1 ; f2 Þw1 w1 w2 þ g ðn1 ; g2 ; f1 Þw1 w2 w1 þ g ðn1 ; g2 ; f2 Þw1 w2 w2 þ g ðn2 ; g1 ; f1 Þw2 w1 w1 þ g ðn2 ; g1 ; f2 Þw2 w1 w2 þ g ðn2 ; g2 ; f1 Þw2 w2 w1 þ g ðn2 ; g2 ; f2 Þw2 w2 w2     1 1 1 1 1 1 ¼ g  pffiffiffi ;  pffiffiffi ;  pffiffiffi  1  1  1 þ g  pffiffiffi ;  pffiffiffi ; pffiffiffi  1  1  1 3 3 3 3 3 3     1 1 1 1 1 1 þ g  pffiffiffi ; pffiffiffi ;  pffiffiffi  1  1  1 þ g  pffiffiffi ; pffiffiffi ; pffiffiffi  1  1  1 3 3 3 3 3 3     1 1 1 1 1 1 þ g pffiffiffi ;  pffiffiffi ;  pffiffiffi  1  1  1 þ g pffiffiffi ;  pffiffiffi ; pffiffiffi  1  1  1 3 3 3 3 3 3     1 1 1 1 1 1 þ g pffiffiffi ; pffiffiffi ;  pffiffiffi  1  1  1 þ g pffiffiffi ; pffiffiffi ; pffiffiffi  1  1  1 3 3 3 3 3 3 ¼8

ð5:57Þ

It can be seen that this result is exactly the same as that of direct integration: Z 1Z 1Z 1  2  n þ g2 þ f2 dndgdf 1 1 1 Z 1Z 1Z 1 Z 1Z 1Z 1 Z 1Z 1Z 1 n2 dndgdf þ g2 dndgdf þ f2 dndgdf ð5:58Þ ¼ 1

1

1

1

1

1

1

1

1

8 8 8 ¼ þ þ ¼8 3 3 3

5.2.6

Required Order of Numerical Integration

For one-dimensional problems, with the appropriate choice of points ξj and weights wj, the numerical integration formula is exact for polynomials of degree at most 2n−1. For two-dimensional and three-dimensional problems, the number of integration points in each dimension need not be the same, but for convenience, the same number is often selected.

5.3

Exercises

5.3.1 Based on the parametric mapping defined by the relation between the coordinate interpolation and the dependent-variable interpolation for a one-dimensional Lagrange element, describe sub-parametric, isoparametric, and super-parametric interpolation. 5.3.2 Consider the two-dimensional rectangle isoparametric element as shown in figure 5.6.

Isoparametric Element and Numerical Integration

79

(a) Write the expression for an isoparametric coordinate mapping in this element. (b) Determine the location of the local coordinates ξ and η that define the centroid of this element. (c) Compute the expression for the Jacobian matrix J for this element and evaluate the Jacobian matrix at the centroid. (d) Compute the derivatives of the shape function N3 at the centroid.

FIG. 5.6 – Two-dimensional rectangle element.

5.3.3 Compute the stiffness K13 for a one-dimensional element with three nodes as follows (a) By numerical integration using one, two, and three integration points. (b) By direct integration. 5.3.4 For a two-dimensional rectangle element with four nodes, compute numerically and by direct integration the integral: Z 1Z 1 ðn þ gÞdndg: ð5:59Þ 1

1

Chapter 6 Finite Element Computation Scheme of Elasticity Problems In this chapter, we consider two- and three-dimensional elasticity problems (Timoshenko and Goodier, 1970) and describe all the necessary steps for the finite element analysis of general solids, including the weak form, finite element method for solving elasticity problems, global assembly using the element location vector, and treatments on boundary conditions.

6.1

Weak Form for General Elasticity Problems

We begin by considering the weak form of three-dimensional problems in elasticity. The weak form of the equilibrium equations for linear elasticity may be written as: Z   G ðw; u; rÞ ¼ w T b þ ST r dX ¼ 0: ð6:1Þ X

Expanding the equations for the three-dimensional problem in Cartesian coordinates yields 91 8 0 @rx @syx @szx > > > > þ þ >C > 8 9 B8 9 > @x > @y @z > > > Z < w u = T B< b x = > =C < C B @s @r @s xy y zy CdX ¼ 0: B by þ ð6:2Þ G ðw; u; rÞ ¼ þ þ wv C B > @x @y @z > X : w ; B: b ; > > C > > w z > A @ @sxz @syz @rz > > > > > þ þ ; : @x @y @z For the three-dimensional problem in Cartesian coordinates, the infinitesimal volume element is given by dX ¼ dxdydz. Integrating the terms involving the derivatives of stresses and using Green’s theorem yields

DOI: 10.1051/978-2-7598-2938-5.c006 © Science Press, EDP Sciences, 2023

Basic Theory of Finite Element Method

82

9 8 @rx @syz @szx > > > > þ þ > > 9T > 8 > > @x @y @z > > > w > Z > = = > < @s < u> @r @s xy y zy dX þ þ wv > @x @y @z > X> > ; > > : > > ww > > > @sxz @syz @rz > > > > > þ þ ; : @x @y @z 9T 2 8 > wu > Z < = nx 6 ¼ wv 4 0 > > C: ; 0 ww

0

0

ny

0

ny 0

0 nz

nx 0

nz ny

@ B6 @x B6 B6 B6 0 B6 B6 B6 6 Z B B6 0 B6  B6 X B6 @ B6 B6 @y B6 B6 B6 0 B6 B6 @4 @ @z

1T

3

02

0 @ @y 0 @ @x @ @z 0

9 8 rx > > > > > > > > 3> ry > > > > nz > > > = < r z 7 0 5 dC > sxy > > > > > nx > > > > > syz > > > > > ; : szx

ð6:3Þ

0

C 7 C 7 C 7 C 0 7 C 7 C 7 9 8 7 @ 7 > w >C C 7 < u =C @z 7 C 7 wv C > C 7> 0 7: ww ;C C 7 C 7 C @ 7 C 7 C 7 @y 7 C A 5 @ @x

9 8 rx > > > > > > > > > ry > > > > > > > < rz = dX > > > sxy > > > > > > > > syz > > > > > ; : szx

where dC ¼ dA is the infinitesimal area on the boundary surface. To illustrate the transformation process in the above formula, the derivation and transformation process for the first term is presented in detail: Z Z zb Z yb Z xb @rx @rx wu wu dX ¼ dxdydz @x @x X za ya xa Z zb Z y b Z zb Z yb Z xb @wu ½wu rx jx¼b  ½wu rx jx¼a dydz ¼ rx dxdydz þ za ya xa @x za ya Z zb Z y b Z zb Z yb Z xb @wu ½wu rx nx jx¼b þ ½wu rx nx jx¼a dC rx dxdydz þ ¼ z ya xa @x za ya Z zb Z yb Za @wu wu rx nx dC rx dX þ ¼ X @x z ya Za Z @wu wu rx nx dC ð6:4Þ rx dX þ ¼ X @x C Z

Α key transformation in the above formula is Z zb Z y b zb Z y b ½wu rx jx¼b  ½wu rx jx¼a dydz ¼ ½wu rx nx jx¼b þ ½wu rx nx jx¼a dC: ð6:5Þ

za

ya

za

ya

Finite Element Computation Scheme of Elasticity Problems

83

This equation can be proved using the projection of the boundary surface of the three-dimensional solid on the boundaries x = b and x = a in the coordinate plane y–o–z, as shown in figure 6.1. The following geometric relation holds true for each infinitesimal boundary surface: dydz ¼ nx dC; dydz ¼ nx dC;

on the boundary x ¼ b on the boundary x ¼ a

ð6:6aÞ ð6:6bÞ

FIG. 6.1 – Geometric relationship for the integration on the boundary surface of the three-dimensional solid.

Subsequently, we note that 2 @ 6 @x 0 6 @ 6 6 0 6 @y 6 6 6 0 0 6 6 @ @ 6 6 6 @y @x 6 @ 6 0 6 @z 6 4 @ 0 @z

3 0 7 7 7 0 7 7 7 9 @ 78 7< wu = @z 7 7 wv ¼ Sw ¼ ew : : ; 0 7 7 ww 7 @ 7 7 7 @y 7 @ 5 @x

ð6:7Þ

Basic Theory of Finite Element Method

84

Thus, this is merely the strain variation expressed in terms of displacement derivatives. We also note that the boundary integral term may be written as: 9 8 rx > > > > 8 9 2 > > ry > 3> > > > Z < wu =T nx 0 0 ny 0 nz > = < r z 4 5 0 ny 0 nx nz 0 dC wv > sxy > C: w ; > ð6:8Þ 0 0 nz 0 ny nx > > > w > > syz > > > > ; : szx Z R ¼ w T GT rdC ¼ C w T tdC C

where t ¼ G r defines the boundary traction vector. If we split the boundary into two parts in which traction and displacement boundary conditions are defined, we may insert the specified traction boundary conditions as: Z Z Z Z Z w T tdC ¼ w T tdC þ w T tdC ¼ w T tdC þ w T tdC: ð6:9Þ T

C

Cu

Ct

Cu

Ct

Generally, we impose the displacement boundary condition u ¼ u directly and set w ¼ 0 on Cu . Thus, the weak form for the linear elastic equilibrium equations may be written compactly in matrix form as: Z Z  T  G ðw; u; rÞ ¼ w b  eT r dX þ w T t dC ¼ 0: ð6:10Þ w X

Ct

The weak form of linear elasticity may be written completely in terms of displacements to compute strains using geometric equations: Z Z Z G ðw; uÞ ¼ w T bdX  eT DedX þ w T tdC ¼ 0; ð6:11Þ w X

X

Ct

where it is also understood that the strains are expressed in terms of displacement derivatives. This is called the displacement method or irreducible form of the problem, as it is only necessary to define the distribution of the displacement field to solve a problem.

6.2

Finite Element Method for Solving Elasticity Problems

This basic structure is fundamental to all finite element methods. As the strains involve the first derivatives of displacement components, the approximation to these components needs only be continuous in X. A similar result holds for the relation between the strain variation and the displacement variation. Thus, in all multidimensional elasticity problems, we only require C0 continuous functions of the type

Finite Element Computation Scheme of Elasticity Problems

85

presented in the previous chapter to approximate the displacement and virtual displacement. Accordingly, we assume that the displacements are given in each element by: ^¼ ww

n X a¼1

^¼ uu

n X b¼1

^ ea Na ð xÞ w

ð6:12aÞ

^ eb Nb ð xÞ u

ð6:12bÞ

^ ea and u ^ eb are the displacements where Na ðxÞ and Nb ðxÞ are the shape functions, and w at the nodes. Inserting the approximation for displacements into the weak form yields X ^ e ðw; ^ ðw; ^ u ^ Þ ¼ 0: ^ u ^Þ ¼ ð6:13Þ G G ðw; uÞ  G e

The expression for each element domain Xe is given by: Z Z Z T e T ^ ^ u ^Þ ¼ ^ bdX ^ T tdC; G ðw; ew DedX þ w w Xe

Xe

ð6:14Þ

Cet

in which the strains are expressed as: ^ew ¼ ^e ¼

n X a¼1 n X b¼1

^ ea ¼ SðNa Þw SðNb Þ^ u eb

¼

n X a¼1 n X b¼1

^ ea Ba w ð6:15Þ

^ eb Bb u

where Ba is the three-dimensional strain–displacement matrix: 3 2 @Na 0 0 7 6 @x 7 6 @Na 7 6 6 0 0 7 7 6 @y 7 6 @Na 7 6 7 6 0 0 6 @z 7 Ba ¼ 6 @N @N 7: a 6 a 0 7 7 6 @x 7 6 @y 6 @Na @Na 7 7 6 0 7 6 @z @y 7 6 4 @N @Na 5 a 0 @z @x

ð6:16Þ

Evaluating the weak form for an individual element yields ^e

G ¼

n X n X a¼1 b¼1

e ^ eb w eT a K ab u

þ

n X a¼1

e w eT a fa

¼

n X a¼1

w eT a

n X b¼1

! ^ eb Þ þ f ea ðK eab u

ð6:17Þ

Basic Theory of Finite Element Method

86

where the element stiffness and load matrices are computed as: K eab

Z ¼

Xe

BT a DBb dX;

f ea

Z ¼

Z

Xe

Na bdX þ

Cet

Na tdC:

ð6:18Þ

In this form, we again obtain a standard discrete system in which the total problem is assembled from the contribution of each element. After the assembly of all elements, the weak form for the approximation is ^ ¼ G

n X a¼1

wT a

n X b¼1

! ^b  f a K ab u

¼0

ð6:19Þ

with K ab ¼

X e

K eab ;

fa ¼

X e

f ea ;

ð6:20Þ

and as w a is arbitrary, the governing equation of the problem becomes n X b¼1

^ b ¼ f b; K ab u

for all a

ð6:21Þ

We can derive the stiffness equations: ^ ¼ f; Ku

ð6:22Þ

which, after the boundary conditions are formally inserted, yields the solution for nodal displacements: ^ ¼ K 1 f ; u

ð6:23Þ

in which K 1 is the matrix inverse of K. Once the displacements are known, the approximation for strain and stresses in the element may be computed as: ^e ¼

n X b¼1

6.3

SðNb Þ^ u eb ¼

n X b¼1

^ eb Bb u

^¼ r

n X b¼1

DSðNb Þ^ u eb ¼

n X b¼1

^ eb DBb u

ð6:24Þ

Global Assembly from High-Dimensional Elements

The global assembly of two- or three-dimensional elements is exactly the same as that in the previous one-dimensional problem according to the element location vector. For expository purposes, a typical two-dimensional example is provided to illustrate the global assembly process and computation procedure. Some steps in the computation are presented in detail; experienced advanced readers may skip them.

Finite Element Computation Scheme of Elasticity Problems

87

Example 6.1. Solutions of finite element method for two-dimensional plane strain problems. A two-dimensional plane strain model is shown in figure 6.2. The dimensions of the model are a = 30 in the horizontal x-direction and b = 10 in the vertical y-direction. The material parameters are Young’s modulus E of 1 × 106 and Poisson’s ratio v of 0.3. The boundary conditions are that the upper left and upper right vertices, as well as the lower parts, are fixed. The vertical concentrated load F acting on the middle domain is −1 × 106.

FIG. 6.2 – Element and node numbers of two-dimensional problem. Then, for each element in the local parent coordinate system, a standard two-dimensional rectangle element with four nodes is used, as shown in figure 6.3. The relation of the local nodes to the global node number is indicated in table 6.1. According to the node numbers of each element, the corresponding element location vector can be obtained to determine the relation between local and global node locations as follows: k1 ¼ f 2

1 3

4 gT ; k2 ¼ f 4 3

5

6 gT :

ð6:25Þ

To obtain the global stiffness matrix, the stiffness matrix of each element should be computed according to the formula introduced in the previous section: K eab

Z ¼

Xe

BT a DBb dX:

ð6:26Þ

By substituting the corresponding parameters of this model, the stiffness matrices of the two elements shown below can be obtained. Here, the final results are given directly:

88

2

K e11

6 Ke 6 K e ¼ 6 21 4 K e31 K e41 2

K e22 K e32

K e23 K e33

K e24 7 7 7 K e34 5

K e42

K e43

K e44

240384:60

42735:03

758546:95

Symmetric

48076:92

245726:48

240384:60

202991:45

48076:92

630341:82 240384:60

379273:48

48076:92

491480:77

240384:60 758546:95

202991:45 48076:92

48076:92 251068:35

245726:48 240384:60

491480:77

240384:60 758546:95

42735:03 48076:92 491480:77

48076:92

3

251068:35 7 7 7 240384:60 7 7 379273:48 7 7 7 48076:92 7 7 630341:82 7 7 7 240384:60 5 758546:95

ð6:27Þ

Basic Theory of Finite Element Method

6 6 6 6 6 6 6 ¼6 6 6 6 6 6 4

K e13

491480:77

K e14

3

K e12

Finite Element Computation Scheme of Elasticity Problems

89

FIG. 6.3 – Two-dimensional rectangle element with four nodes in local parent coordinate system. TAB. 6.1 – Local-to-global node numbering for two-dimensional problem. Element number 1 2 3 4

Local node number

1 2 1 3 4

2 4 3 5 6

To explain the derivation of the above element stiffness matrix, the computation of the first coefficient 491480.77 in K 133 for element 1 is presented in detail. First, the matrix forms of the element stiffness matrix are expanded: Z Z K eab ¼ BT DB dX ¼ ST ðNa ÞDST ðNb ÞdX b a Xe

Xe

3 82 @Na @Na > > 0 0 > > 6 @x @y 7 > 6 7 > > 5 >4 @N @N > a a > > 0 0 > > @y @x > > > Z > < 2 ¼ ð1  v Þ v v Xe > > > >E 6 v > ð1  v Þ v > 6 > > 6 > > 4 d v v ð 1  vÞ > > > > > 0 0 0 > > : 8 2 > > > 6 > > 6 > > > 6 > > 6 Z >

6 d Xe > > 6 > > 6 > > 6 > > 4 > > :

@Na @Nb ð1  v Þ þ @x @x @Na @Nb ð1  2v Þ=2 @y @y @Na @Nb vþ @y @x @Na @Nb ð1  2v Þ=2 @x @y

9 > > > > > > > > > > > > > > > 3> 2 > = @Nb 0 36 @x dX 7 0 7> 6 > > 7 6 @N b7> > 76 0 0 > 76 > > @y 7 76 > 7 > 56 0 7 > ð6:28Þ 0 7> > 6 0 > ð1  2v Þ=2 4 @Nb @Nb 5 > > > ; @y @x 39 @Na @Nb > > vþ 7> > @x @y 7> > > > 7 > @Na @Nb 7 > ð1  2v Þ=2 7> = @y @x 7 7 dX 7> @Na @Nb > 7> ð1  v Þ þ > 7> @y @y > 7> > > 5 > @Na @Nb > ð1  2v Þ=2 ; @x @x

Basic Theory of Finite Element Method

90

where d = (1 + v)(1 − 2v). It can be seen that the above formula involves the derivatives of the shape functions with respect to the global coordinates x and y, which can be converted to the local coordinate system n and g as follows: 8 8 9 9 9 2 38 @y @y > @Na > @Na > @Na > > > > > > > > >  < < < = 1 6 @g = = @n 7 @n @x 7 @n ¼ 6 ¼ J 1 4 5 @Na > @N > j @N > > > @x @x > > > > > : a> : a> : ; ; ;  @y @g @g @g @n 1 ¼   X  X  X ! Pn @Na n @Na n @Na n @Na xa y  x y a¼1 a¼1 @g a a¼1 @g a a¼1 @n a @n 8 9 2 P Pn @Na 3> @Na > @Na n > > y y  a a < a¼1 a¼1 6 7 @n = @g @n 6 7 4 Pn @Na 5> @Na > Pn @Na > : ; xa xa >  a¼1 a¼1 @g @g @n 1 ¼   X  X  X ! Pn @Na n @Na n @Na n @Na xa y  x y a¼1 a¼1 @g a a¼1 @g a a¼1 @n a @n 3 2    X Pn @Na @Na n @Na @Na ya  y 6 a¼1 a¼1 @n a @g @n @g 7 7 6   X ð6:29Þ 7 6  Pn @Na 4 @Na n @Na @Na 5 x þ x  a a¼1 a¼1 @n a @g @n @g where @x @y @x @y  ¼ j ¼ det J ¼ @n @g @g @n X  @Na n y : a¼1 @n a

X

@Na x a¼1 @n a n

 X

 X  @Na n @Na y  x a¼1 @g a a¼1 @g a n

Here, the following shape functions are used: 1 N1 ¼ ð1  nÞ  ð1  gÞ; 4 1 N3 ¼ ð1 þ nÞ  ð1 þ gÞ; 4

1 N2 ¼ ð 1  nÞ  ð 1 þ gÞ 4 1 N4 ¼ ð 1 þ nÞ  ð 1  gÞ 4

ð6:30Þ

The derivatives of the shape functions are @N1 1 @N1 1 ¼  ð1gÞ; ¼  ð1nÞ; 4 4 @n @g @N3 1 @N3 1 ¼ ð1 þ gÞ; ¼ ð1 þ nÞ; 4 4 @n @g

@N2 1 @N2 1 ¼  ð1 þ gÞ; ¼ ð1nÞ 4 4 @n @g @N4 1 @N4 1 ¼ ð1gÞ; ¼  ð 1 þ nÞ 4 4 @n @g

ð6:31Þ

Finite Element Computation Scheme of Elasticity Problems

91

Moreover, the global coordinates of the nodes are ðx1 ; y1 Þ ¼ ð0; 10Þ; ðx2 ; y2 Þ ¼ ð0; 0Þ; ðx3 ; y3 Þ ¼ ð15; 0Þ; ðx4 ; y4 Þ ¼ ð15; 10Þ: ð6:32Þ Using the derivatives of the shape functions and these coordinates, we obtain n X @Na a¼1

1 1 1 1 15 xa ¼  ð1gÞ  0 ð1 þ gÞ  0 þ ð1 þ gÞ  15 þ ð1gÞ  15 ¼ 4 4 4 4 2 @n

a¼1

@g

a¼1

@g

a¼1

@n

n X @Na n X @Na n X @Na

1 1 1 1 ya ¼  ð1nÞ  ð10Þ þ ð1nÞ  0 þ ð1 þ nÞ  0 ð1 þ nÞ  ð10Þ ¼ 5 4 4 4 4 1 1 1 1 xa ¼  ð1nÞ  0 þ ð1nÞ  0 þ ð1 þ nÞ  15 ð1 þ nÞ  15 ¼ 0 4 4 4 4 1 1 1 1 ya ¼  ð1gÞ  ð10Þ ð1 þ gÞ  0 þ ð1 þ gÞ  0 þ ð1gÞ  ð10Þ ¼ 0 4 4 4 4 ð6:33Þ

Using the above formula to compute the derivative of the shape function N3 and substituting the above results, we obtain 9 8 @N3 > > > > = < 1 @x ¼   X  X  X ! @N > > 3 P > > @N n @N n @N n @N a a a a n ; : xa y  x y a¼1 @y a¼1 @g a a¼1 @g a a¼1 @n a @n 3 2    X Pn @Na @N3 n @Na @N3 y  y a 6 a¼1 a¼1 @n a @g @n @g 7 7 6   X 7 6  Pn @Na 4 @N3 n @Na @N3 5 x þ x  a a a¼1 a¼1 @n @g @n @g 3 2 1 1 ð 1 þ g Þ  0  ð 1 þ n Þ 5  1 7 4 4 6 ¼ 5 4 1 15 1 15 0  ð 1 þ g Þ þ ð 1 þ n Þ 500 4 2 4 2 9 2 5 3 8 1 > > 2 6 4 ð1 þ gÞ 7 < 30 ð1 þ gÞ = ¼ 4 ð6:34Þ 5¼ > 75 15 ; : 1 ð1 þ nÞ > ð1 þ nÞ 8 20

Moreover, it is seen that the determinant of the Jacobian matrix is j ¼ 75 2.

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By substituting the derivatives of shape function N3 and the material parameters into equation (6.18), the expression for the first coefficient in K 133 of element 1 is  Z E @N3 @N3 @N3 @N3 ð1  v Þ þ ð1  2v Þ=2 dX @x @x @y @y Xe d " #  2  2 Z E 1 1 2 2 ¼ ð 1 þ gÞ ð 1  v Þ þ ð1 þ nÞ ð1  2v Þ=2 jdndg 30 20 Xe ð1 þ v Þð1  2v Þ "  #  2 Z 106 1 2 1 75 2 2 dndg ð1 þ gÞ 0:7 þ ð1 þ nÞ 0:2 ¼ 30 20 2 Xe ð1 þ 0:3Þð1  0:6Þ "  #  2 Z 106 1 2 1 75 2 2 dndg ¼ ð1 þ gÞ 0:7 þ ð1 þ nÞ 0:2 30 20 2 ð 1 þ 0:3 Þ ð 1  0:6 Þ Xe Z h i ¼ 1923076:92 0:02917  ð1 þ gÞ2 þ 0:01875  ð1 þ nÞ2 dndg Xe

ð6:35Þ The integrand in the above formula is a quadratic polynomial of n and g. Thus, at least two integral points should be used in each direction, and there are four integral points in the element. Numerical integration is used to solve the problem:  Z E @N3 @N3 @N3 @N3 ð1  v Þ þ ð1  2v Þ=2 dX @x @x @y @y Xe d Z h i ¼ 1923076:92  0:02917  ð1 þ gÞ2 þ 0:01875  ð1 þ nÞ2 dndg X " e     # 1 2 1 2 11 ¼ 1923076:92  0:02917  1  pffiffiffi þ 0:01875  1  pffiffiffi 3 3 "     # 1 2 1 2 ð6:36Þ 11  0:02917  1 þ pffiffiffi þ 0:01875  1  pffiffiffi 3 3 "     # 1 2 1 2 11 þ 0:02917  1  pffiffiffi þ 0:01875  1 þ pffiffiffi 3 3 "    # 1 2 1 2 11 þ 0:02917  1 þ pffiffiffi þ 0:01875  1 þ pffiffiffi 3 3 ¼ 491480:77 In this way, the stiffness coefficient can be obtained by numerical integration, further, other stiffness coefficients can be obtained similarly by the same computation process. Through the above steps, the values of stiffness matrices of each element can be computed. For the global assembly of two-dimensional elements, the node numbers are marked on the left and upper sides of the matrix, and the expanded form of the stiffness matrix is given using the element location vector:

Finite Element Computation Scheme of Elasticity Problems

Node number 1 2 3 4 5 6

1 K 122 6 K 112 6 1 6 K 32 6 1 6K 6 42 4 2

2 K 121 K 111 K 131 K 141

3 K 123 1  1 K 13 2  þ K 22  K 33 K 143 þ K 212 K 232 K 242

4 K 124 1  1 K 14 2  þ K 21  K 34 K 144 þ K 211 K 231 K 241

5 K 223 K 213 K 233 K 243

93

6

38 9 8 9 ^1 > > f1 > u > > > > > > >^ > > > > 7> f2 > u > > > > 2 > > > > 7 = = < 2 7< K 24 7 u ^3 f3 ; ¼ 2 7 ^ > > f4 > u K 14 7> > > > > > > 4> > > > ^ f5 > u > > > > K 234 5> 5 > > > > ; ; : : 2 ^6 f6 u K 44 ð6:37Þ

where u and f are the global displacement and load matrices, respectively. The loads in the element can be obtained by integrating the product of the body forces and the shape functions: Z e fa ¼ Na bdX: ð6:38Þ Xe

Similarly, the node numbers are marked on the left of the load vector, and the expanded load is given using the element location vector: Node 1 number 2 3 4 5 6

6.4

9 8 9 8 f1 > f 12 > > > > > > > > > > > > > f2 > f 11 > > > > > > > > > = < = < 1 f3 f 3 þ f 22 ¼ : > > > f4 > f 14 þ f 21 > > > > > > > > > f > > > > > > f 23 > > > ; > ; : 5> : 2 f6 f4

ð6:39Þ

Treatments on Boundary Conditions

Here, we focus on the displacement and traction boundary conditions for representative two-dimensional elastic problems. Other issues can be treated similarly at the corresponding nodes. (1) Displacement boundary conditions Using the above finite element form, it is very simple to impose displacement boundary conditions, as the parameters are now physical values, that is, they satisfy ^1 ¼ u ^2 ¼ u ^4 ¼ u ^5 ¼ u ^ 6  u  0: u

ð6:40Þ

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94

Thus, the stiffness equation (6.37) can be expressed as: Node number 1 2 3 4 5 6

3 21 2 I 0 0 60 I 0 6   6 0 0 K1 þ K2 33 22 6 60 0 0 6 4 0 0

4 5 0 0 0 0 I 0 0 I 0 0

6 38 9 8 9 ^1 > 0> >u > > > > > > > > > > > 7> ^ 0> u > > > > 2> > > 7> = = < < ^ f 07 u 3 ; 3 7 ¼ ^4 > 07 0> u > > > > > 7> > > > > > >u >0> ^ > 0 5> > > > ; > ; : 5> : > ^6 I 0 u

ð6:41Þ

where I is the identity matrix. To simplify, rows and columns in the stiffness matrix of known displacement at the corresponding nodes can be deleted, thus reducing the size of the stiffness matrix:  1  ^ 3 ¼ f 3: K 33 þ K 222 u ð6:42Þ The matrices and vectors involved are expanded, leading to    f3x u3 982961:54 0 ¼ : f3y v3 0 1517093:9

ð6:43Þ

(2) Traction boundary conditions To solve the above matrix equation, we should know the load at node 3. As there is no body force at present, the load matrix is   f3x 0 ¼ : ð6:44Þ f3y 0 Furthermore, the force acting on the boundary is considered. Imposing a traction condition at node 3 only requires the modification f3y ! f3y þ t ¼ 1  106 : The stiffness equation (6.42) to be solved becomes the following form:    0 982961:54 0 u3 ¼ ; v3 1000000 0 1517093:9 which is solved directly, and the following results can be obtained   0 u3 ¼ : v3 0:6592

ð6:45Þ

ð6:46Þ

ð6:47Þ

The displacement functions for each element can be obtained by combining the shape functions and node displacements:

Finite Element Computation Scheme of Elasticity Problems

u^ 1 ðn; gÞ ¼

4 X a¼1

95

Na ðn; gÞ u^a1

¼ N1  0 þ N2  0 þ N3  0 þ N 4  0

ð6:48aÞ

¼0 Element 1: ^ v 1 ðn; gÞ ¼

4 X a¼1

Na ðn; gÞ ^ va1

¼ N1  0 þ N2  0 þ N3  ð0:6592Þ þ N4  0

ð6:48bÞ

¼ 0:1648ð1 þ nÞð1 þ gÞ

u^ 2 ðn; gÞ ¼

4 X a¼1

Na ðn; gÞ u^a2

¼ N1  0 þ N2  0 þ N3  0 þ N 4  0

ð6:49aÞ

¼0 Element 2: ^ v 2 ðn; gÞ ¼

4 X a¼1

Na ðn; gÞ ^ va2

¼ N1  0 þ N2  ð0:6592Þ þ N3  0 þ N4  0

ð6:49bÞ

¼ 0:1648ð1  nÞð1 þ gÞ Once the displacements are known, the element stresses can be approximated as: ^¼ r

4 X b¼1

^ eb ¼ DSðNb Þ u

4 X b¼1

^ eb : DBb u

ð6:50Þ

In the computation process, the derivatives of each shape function are required. They can be computed as in the case of N3 above, as shown in equation (6.33). The derivatives of the other shape functions are 9 8 8 9 8 @N 9 8 1 9 @N1 > 1 > > 2> > > > = > = <  ð 1  gÞ > <  ð 1 þ gÞ > = > = > < < @x @x 30 30 ; ; ¼ ¼ @N1 > > > @N2 > > > 1 > 1 > > > > ; :  ð 1  nÞ ; > ; : : : ð1  n Þ ; @y @y 20 20 9 8 8 ð6:51Þ 9 @N 1 > > > 4> > ð1  gÞ > = = < < @x 30 ¼ @N4 > 1 > > > : > > ; :  ð 1 þ nÞ ; @y 20

Basic Theory of Finite Element Method

96

The stress functions for each element can be obtained by combining the derivatives of the shape functions and node displacements: Element 1: ^1 ¼ r

r1x

r1y

r1z

s1xy

T

¼

4 X b¼1

^ 1b ¼ DSðNb Þ u

4 X b¼1

^ 1b DBb u

    ^ 11 þ B2 u ^ 12 þ B3 u ^ 13 þ B4 u ^ 14 ¼ D B1 0 þ B2 0 þ B3 u ^ 13 þ B4 0 ¼ DB3 u ^ 13 ¼ D B1 u 3 2 @N3 0 2 36 @x 7 ð1  v Þ v v 0 7( ) 6 7 6 @N 3 6 7 7 u^31 6 v ð 1  v Þ v 0 E6 76 0 ¼ 6 @y 7 76 7 ^v 1 56 d4 v v ð1  v Þ 0 3 0 7 7 6 0 0 0 0 ð1  2v Þ=2 4 @N3 @N3 5 @y @x 3 2 1 ð 1 þ gÞ 0 2 3 7 6 30 ð1  v Þ v v 0 7 6 1 7 6 6 7 0 ð1  v Þ v 0 E6 v 7 76 0 ð 1 þ n Þ ¼ 6 7 76 20 7 0:6592 56 d4 v v ð1  v Þ 0 7 6 0 0 5 4 0 0 0 ð1  2v Þ=2 1 1 ð 1 þ nÞ ð 1 þ gÞ 20 30 3 2 1 1 ð1 þ nÞv 7 6 30 ð1 þ gÞð1  v Þ 20 7 6 7 6 1 1 6  ð1 þ gÞv ð 1 þ nÞ ð 1  v Þ 7 7 0 E6 30 20 7 6 ¼ 6 7 0:6592 1 1 d6 7 ð1 þ gÞv ð1 þ nÞv 7 6 30 20 7 6 5 4 1 1 ð1 þ nÞð1  2v Þ=2 ð1 þ gÞð1  2v Þ=2 20 30 9 8 1 > > > > ð 1 þ n Þv > > > > 20 > > > > > > > > 1 > > > > ð 1 þ n Þ ð 1  v Þ = < E 20 ¼ 0:6592 1 > d> > > > > ð1 þ nÞv > > > > 20 > > > > > > > > 1 > ; : ð1 þ gÞð1  2v Þ=2 > 30 9 8 8 9 0:15ð1 þ nÞ > > 19015:38ð1 þ nÞ > > > > > > > = > < 0:35ð1 þ nÞ > = < 44369:23ð1 þ nÞ > ¼ ¼ 126769:231 > > 0:15ð1 þ nÞ > > > > > 19015:38ð1 þ nÞ > > > > ; > : ; : 0:067ð1 þ gÞ 8451:28ð1 þ gÞ ð6:52Þ

Finite Element Computation Scheme of Elasticity Problems

97

Element 2: ^2 ¼ r

r2x

r2y

r2z

s2xy

T

¼

4 X b¼1

^ 2b ¼ DSðNb Þ u

4 X b¼1

^ 2b DBb u

    ^ 21 þ B2 u ^ 22 þ B3 u ^ 23 þ B4 u ^ 24 ¼ D B1 0 þ B2 u ^ 22 þ B3 0 þ B4 0 ¼ DB2 u ^ 22 ¼ D B1 u 3 2 @N2 0 2 36 @x 7 ð1  v Þ v v 0 7( ) 6 6 @N2 7 76 0 7 u^22 v ð 1  v Þ v 0 E6 6 76 ¼ 6 @y 7 76 7 ^v 2 56 d4 v v ð1  v Þ 0 0 7 7 2 6 0 0 0 0 ð1  2v Þ=2 4 @N2 @N2 5 @y @x 3 2 1 0 2 3  ð 1 þ gÞ 7 6 30 ð1  v Þ v v 0 7 6 1 7 6 6 7 0 ð1  v Þ v 0 E6 v 7 76 0 ð 1n Þ ¼ 6 7 76 20 7 0:6592 56 d4 v v ð1  v Þ 0 7 6 0 0 5 4 0 0 0 ð1  2v Þ=2 1 1 ð1nÞ  ð 1 þ gÞ 20 30 3 2 1 1 ð1nÞv 7 6  30 ð1 þ gÞð1  v Þ 20 7 6 7 6 1 1 7 6  ð 1 þ g Þv ð 1n Þ ð 1  v Þ 7 6 0 E 30 20 7 ¼ 6 7 0:6592 1 1 d6 7 6  ð1 þ gÞv ð1nÞv 7 6 30 20 7 6 5 4 1 1 ð1nÞð1  2v Þ=2  ð1 þ gÞð1  2v Þ=2 20 8 30 9 1 > > > > ð 1n Þv > > > > 20 9 8 > > > > > > 0:15ð1nÞ > > > > 1 > > > > > > > ð1nÞð1  v Þ = < 0:35ð1nÞ > = E< 20 ¼ 126769:231 ¼ 0:6592 1 > > d> 0:15ð1nÞ > > > > > > > > > ð1nÞv ; : > > > > 20 > > 0:067 ð 1 þ g Þ > > > > 1 > > > ; :  ð1 þ gÞð1  2v Þ=2 > 30 9 8 19015:38ð1nÞ > > > > > = < 44369:23ð1nÞ > ¼ > > > > > 19015:38ð1nÞ > ; : 8451:28ð1 þ gÞ ð6:53Þ Through the solutions of the above stress functions, the stress values at the integration points can be obtained by substituting the coordinates of the integration

Basic Theory of Finite Element Method

98

points of the specified element. For example, the stress at the integral point ðn; gÞ ¼    p1ffiffi3 ; p1ffiffi3 of element 1 is  9 8 1 > > > > p ffiffi ffi 19015:38 1  > > > > > 3 > > 9 > 8 > >   > > > 19015:38ð1 þ nÞ > > 1 > > > > > > > > > > > p ffiffi ffi 44369:23 1  = < 44369:23ð1 þ nÞ =
> 1 > 19015:38ð1 þ nÞ > > > > > > > 19015:38 1  pffiffiffi > > > ; > : > > > 3 > 8451:28ð1 þ gÞ > > > >   > > > > 1 > > > ; : 8451:28 1 þ pffiffiffi > 3 9 8 8036:84 > > > > > = < 18752:63 > ¼ > > > > > 8036:84 > ; : 13330:63 ð6:54Þ Furthermore, by substituting the node coordinates of the specified element into the solution of the above stress function, the stress values at the nodes can be obtained. For example, the stresses at node 2 of ðn; gÞ ¼ ð1; 1Þ of element 1 are as follows:

^1 ¼ r

¼

r1x

8 > > > < > > > :

r1y 0 0

9 9 8 8 19015:38ð1 þ nÞ > > 19015:38ð1  1Þ > > > > > > > > > > T < 44369:23ð1 þ nÞ = < 44369:23ð1  1Þ = 1 1 rz sxy ¼ ¼ > > > 19015:38ð1  1Þ > > > > 19015:38ð1 þ nÞ > > > > ; : ; > : 8451:28ð1 þ 1Þ 8451:28ð1 þ gÞ 9 > > > =

> 0 > > ; 16902:56 ð6:55Þ

6.5

Exercises

6.5.1 Derive the weak form for general elasticity problems by following the steps below:

Finite Element Computation Scheme of Elasticity Problems (a) (b) (c) (d)

99

Multiply each equation by an appropriate arbitrary function. Integrate this product over the space domain of the problem. Use integration by parts to reduce the order of the derivatives to a minimum. Introduce boundary conditions if possible.

6.5.2 In example 6.1, compute the value of the last coefficient in the stiffness matrix K 133 of element 1: Z   E @N3 @N3 @N3 @N3 ð1  v Þ þ ð1  2v Þ=2 dX: ð6:56Þ d @y @y @x @x Xe

Chapter 7 Solutions of Linear Algebraic Equations By using the finite element method, a standard discrete system is obtained. The basic stiffness equations of this system are linear. In general, a large number of such equations is difficult to solve by directly inverting the stiffness matrix. Herein, we introduce the commonly used LU decomposition method or triangular decomposition method, also known as the Gaussian elimination method, including the forward elimination and back substitution procedures.

7.1

LU Decomposition Method

Consider a set of linear algebraic equations representing the global stiffness equations in the finite element method, given by: K^ u ¼ f;

ð7:1Þ

where K is an n × n square matrix with known values, representing the global ^ is a vector of unknown parameters, representing the global stiffness matrix, u displacement vector, and f is a vector with known values, representing the global load vector. The matrix K is symmetric and positive definite; thus, it can be written as the product of a lower triangular matrix L with unit diagonals and an upper triangular matrix U as follows: K ¼ LU; where

2

1 6 L21 6 L ¼ 6 .. 4 .

Ln1

0 1 Ln2

ð7:2Þ   .. .



3 0 07 7 .. 7 .5

ð7:3Þ

1

DOI: 10.1051/978-2-7598-2938-5.c007 © Science Press, EDP Sciences, 2023

Basic Theory of Finite Element Method

102

and

2

U11 6 0 6 U ¼ 6 .. 4 .

U12 U22

0

0

  .. .

3 U1n U2n 7 7 .. 7: . 5

ð7:4Þ

   Unn

This form is called a LU decomposition or triangular decomposition of K. The solution to the equations can now be obtained by sequentially solving the pair of equations (Demmel, 1997; Taylor, 1985; Strang, 1976; Wilkinson and Reinsch, 1971; Ralston, 1965): Ly ¼ f

ð7:5Þ

U^ u ¼ y;

ð7:6Þ

and

where y is introduced to facilitate the separation. In terms of the individual equations, the solution is given by: yi ¼ fi ; yi ¼ fi 

i¼1 i1 X

Lij yj ; i ¼ 2; 3; . . .; n

ð7:7Þ

j¼1

and u^i ¼

yi ; Uii

1 u^i ¼ Uii

i¼n yi 

n X

! Uij u^j ;

i ¼ n  1; n  2; . . .; 1

ð7:8Þ

j¼i þ 1

where equation (7.7) is commonly called forward elimination, and equation (7.8) is called back substitution. As indicated in figure 7.1, it is convenient to subdivide the coefficient array in the stiffness matrix K into three parts: (1) Reduced zone: the region that is fully reduced. (2) Active zone: the region currently being reduced, where the jth column above the diagonal and the jth row to the left of the diagonal constitute the active zone. (3) Unreduced zone: the region that contains the original unreduced coefficients.

Solutions of Linear Algebraic Equations

103

FIG. 7.1 – Reduced, active, and unreduced zones in LU decomposition of stiffness matrix K.

The coefficients in the decomposed matrices L and U can be stored in the active zone, respectively, as shown in figure 7.2. Through the above processes, the stiffness matrix is decomposed into two matrix parts.

FIG. 7.2 – Matrices L and U stored in active zone.

The algorithm for the LU decomposition of an n × n square matrix can be described using figure 7.3 as follows:

Basic Theory of Finite Element Method

104

FIG. 7.3 – LU decomposition of stiffness matrix K. The first row and column are activated as: U11 ¼ K11 ; L11 ¼ 1:

ð7:9Þ

For each active zone j from 2 to n, Lj1 ¼

Kj1 ; U11

U1j ¼ K1j

and 1 Lji ¼ Uii

Kji 

Uij ¼ Kij 

! Ljm Umi

m¼1 i1 X m¼1

and finally, with Ljj = 1,

i1 X

ð7:10Þ

Lim Umj

ð7:11Þ i ¼ 2; 3; . . .; j  1

Solutions of Linear Algebraic Equations

105 j1 X

Ujj ¼ Kjj 

Ljm Umj :

ð7:12Þ

m¼1

Example 7.1 Using the triangular decomposition of the stiffness matrix K, solve the displacements in the stiffness equations: 8 9 2 3 4 2 1 1) READ(10, *) etype READ(10, *) x_coords, y_coords, z_coords nf=1 READ(10, *) nr, (k, nf(:, k), i=1, nr) CALL formnf(nf) neq=MAXVAL(nf) ALLOCATE(loads(0:neq), kdiag(neq)) kdiag=0 ! – – – – – – – – – – –loop the elements to find global arrays sizes – – – elements_1: DO iel=1, nels CALL hexahedron_xz(iel, x_coords, y_coords, z_coords, coord, num) CALL num_to_g(num, nf, g) g_num(:, iel)=num g_coord(:, num)=TRANSPOSE(coord) g_g(:, iel)=g CALL fkdiag(kdiag, g) END DO elements_1 DO i=2, neq kdiag(i)=kdiag(i)+kdiag(i-1) END DO

Programs of Finite Element Method (continued) ALLOCATE(kv(kdiag(neq))) WRITE(11, ‘(2(A, I5))’) & " There are", neq, " equations and the skyline storage is", kdiag(neq) ! – – – – – – – – – – –element stiffness integration and assembly – – – CALL sample(element, points, weights) kv=zero elements_2: DO iel=1, nels CALL deemat(dee, prop(1, etype(iel)), prop(2, etype(iel))) num=g_num(:, iel) g=g_g(:, iel) coord=TRANSPOSE(g_coord(:, num)) km=zero gauss_pts_1: DO i=1, nip CALL shape_der(der, points, i) jac=MATMUL(der, coord) det=determinant(jac) CALL invert(jac) deriv=MATMUL(jac, der) CALL beemat(bee, deriv) km=km+MATMUL(MATMUL(TRANSPOSE(bee), dee), bee)*det*weights(i) END DO gauss_pts_1 CALL fsparv(kv, km, g, kdiag) END DO elements_2 loads=zero READ(10, *)loaded_nodes IF(loaded_nodes/=0) READ(10, *) (k, loads(nf(:, k)), i=1, loaded_nodes) READ(10, *) fixed_freedoms IF(fixed_freedoms/=0) THEN ALLOCATE(node(fixed_freedoms), sense(fixed_freedoms), & value(fixed_freedoms), no(fixed_freedoms)) READ(10, *) (node(i), sense(i), value(i), i=1, fixed_freedoms) DO i=1, fixed_freedoms no(i)=nf(sense(i), node(i)) END DO kv(kdiag(no))=kv(kdiag(no))+penalty loads(no)=kv(kdiag(no))*value END IF ! – – – – – – – – – – equation solution – – – – – – – – – – – – – – – – CALL sparin(kv, kdiag) CALL spabac(kv, loads, kdiag) loads(0)=zero WRITE(11, ‘(/A)’)" Node x-disp y-disp z-disp" DO k=1, nn WRITE(11, ‘(I5, 3E12.4)’) k, loads(nf(:, k)) END DO ! – – – – – – – – – – –recover stresses at nip integrating points – – – WRITE(11, ‘(/A, I2, A)’)" The integration point (nip=", nip, ") stresses are:" WRITE(11, ‘(A, /, A)’)" Element x-coord y-coord z-coord", & "sig_x sig_y sig_z tau_xy tau_yz tau_zx" elements_3: DO iel=1, nels CALL deemat(dee, prop(1, etype(iel)), prop(2, etype(iel))) num=g_num(:, iel) coord=TRANSPOSE(g_coord(:, num)) g=g_g(:, iel)

147

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148

(continued) eld=loads(g) gauss_pts_2: DO i=1, nip CALL shape_der(der, points, i) CALL shape_fun(fun, points, i) gc=MATMUL(fun, coord) jac=MATMUL(der, coord) CALL invert(jac) deriv=MATMUL(jac, der) CALL beemat(bee, deriv) sigma=MATMUL(dee, MATMUL(bee, eld)) WRITE(11, ‘(I8, 4X, 3E12.4)’) iel, gc WRITE(11, ‘(6E12.4)’) sigma END DO gauss_pts_2 END DO elements_3 ! – – – – – – – – – – –recover stresses at node points – – – points(1,1)=1, points(2,1)=1, points(3,1)=1, points(4,1)=1 points(5,1)=-1, points(6,1)=-1, points(7,1)=-1, points(8,1)=-1 points(1,2)=1, points(2,2)=1, points(3,2)=-1, points(4,2)=-1 points(5,2)=1, points(6,2)=-1, points(7,2)=1, points(8,2)=-1 points(1,3)=1, points(2,3)=-1, points(3,3)=1, points(4,3)=-1 points(5,3)=1, points(6,3)=1, points(7,3)=-1, points(8,3)=-1 WRITE(11,'(/A,I2,A)')" The Node point stresses are:" WRITE(11,'(A,/,A)')" Element x-coord y-coord z-coord", & " sig_x sig_y sig_z tau_xy tau_yz tau_zx" elements_4: DO iel=1,nels CALL deemat(dee,prop(1,etype(iel)),prop(2,etype(iel))) num=g_num(:,iel) coord=TRANSPOSE(g_coord(:,num)) g=g_g(:,iel) eld=loads(g) gauss_pts_3: DO i=1,nip CALL shape_der(der,points,i) CALL shape_fun(fun,points,i) gc=MATMUL(fun,coord) jac=MATMUL(der,coord) CALL invert(jac) deriv=MATMUL(jac,der) CALL beemat(bee,deriv) sigma=MATMUL(dee,MATMUL(bee,eld)) WRITE(11,'(I8,4X,3E12.4)')iel,gc WRITE(11,'(6E12.4)')sigma END DO gauss_pts_3 END DO elements_4 STOP END PROGRAM p3DFEM

9.3.2

Numerical Example

The input parameters for the above geometric model and basic conditions are contained in file 3DFEM.dat; two elements with uniform size are used; the input data and their interpretation are listed in table 9.3.

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TAB. 9.3 – Input data and interpretation of the three-dimensional program of solid compression. Input data 8 1 2 1 8 1 1.0e6 0.3 0.0 0.5 0.0 1.5 3.0 0.0 -2.0 10 1000 2000 3000 4000 7000 8 0 0 0 9 0 0 0 10 0 0 0 11 0 0 0 12 0 0 0 2 5 0.0 0.0 -1.0e6 6 0.0 0.0 -1.0e6 0

Interpretation element-node element in x, element in y, element in z, integration points, material E v x_coords y_coords z_coords nr displacement BC k, nf(:,k), i=1, nr nr traction BC k, nf(:,k), i=1, nr fixed_freedoms

Based on running the three-dimensional program of solid compression, the output solutions for the above geometric model and basic conditions are contained in file 3DFEM.res, and are shown below: Node x-disp y-disp z-disp 1 0.0000E+00 0.0000E+00 0.0000E+00 2 0.0000E+00 0.0000E+00 0.0000E+00 3 0.0000E+00 0.0000E+00 0.0000E+00 4 0.0000E+00 0.0000E+00 0.0000E+00 5 -0.3438E+00 0.1201E-15 -0.4332E+01 6 0.3438E+00 0.1728E-15 -0.4332E+01 7 0.0000E+00 0.0000E+00 0.0000E+00 8 0.0000E+00 0.0000E+00 0.0000E+00 9 0.0000E+00 0.0000E+00 0.0000E+00 10 0.0000E+00 0.0000E+00 0.0000E+00 11 0.0000E+00 0.0000E+00 0.0000E+00 12 0.0000E+00 0.0000E+00 0.0000E+00 The integration point (nip= 8) stresses are: Element x-coord y-coord z-coord sig_x sig_y sig_z tau_xy tau_yz tau_zx 1 0.3943E+00 0.1183E+01 -0.4226E+00 0.1661E+02 -0.4920E+02 -0.1806E+03 0.4014E+01 -0.8760E+02 0.3011E+01 1 0.3943E+00 0.1183E+01 -0.1577E+01 -0.6770E+02 -0.8533E+02 -0.2167E+03 0.1076E+01 -0.2347E+02 0.3011E+01 1 0.3943E+00 0.3170E+00 -0.4226E+00 0.4450E+01 -0.1318E+02 -0.4839E+02 0.4014E+01 -0.8760E+02 0.8068E+00 1 0.3943E+00 0.3170E+00 -0.1577E+01 -0.1814E+02 -0.2286E+02 -0.5807E+02 0.1076E+01 -0.2347E+02 0.8068E+00 1 0.1057E+00 0.1183E+01 -0.4226E+00

150

Basic Theory of Finite Element Method 0.1661E+02 -0.4920E+02 -0.1806E+03 -0.4014E+01 -0.8760E+02 -0.3011E+01 1 0.1057E+00 0.3170E+00 -0.4226E+00 0.4450E+01 -0.1318E+02 -0.4839E+02 -0.4014E+01 -0.8760E+02 -0.8068E+00 1 0.1057E+00 0.1183E+01 -0.1577E+01 -0.6770E+02 -0.8533E+02 -0.2167E+03 -0.1076E+01 -0.2347E+02 -0.3011E+01 1 0.1057E+00 0.3170E+00 -0.1577E+01 -0.1814E+02 -0.2286E+02 -0.5807E+02 -0.1076E+01 -0.2347E+02 -0.8068E+00 2 0.3943E+00 0.2683E+01 -0.4226E+00 0.4450E+01 -0.1318E+02 -0.4839E+02 -0.4014E+01 0.8760E+02 0.8068E+00 2 0.3943E+00 0.2683E+01 -0.1577E+01 -0.1814E+02 -0.2286E+02 -0.5807E+02 -0.1076E+01 0.2347E+02 0.8068E+00 2 0.3943E+00 0.1817E+01 -0.4226E+00 0.1661E+02 -0.4920E+02 -0.1806E+03 -0.4014E+01 0.8760E+02 0.3011E+01 2 0.3943E+00 0.1817E+01 -0.1577E+01 -0.6770E+02 -0.8533E+02 -0.2167E+03 -0.1076E+01 0.2347E+02 0.3011E+01 2 0.1057E+00 0.2683E+01 -0.4226E+00 0.4450E+01 -0.1318E+02 -0.4839E+02 0.4014E+01 0.8760E+02 -0.8068E+00 2 0.1057E+00 0.1817E+01 -0.4226E+00 0.1661E+02 -0.4920E+02 -0.1806E+03 0.4014E+01 0.8760E+02 -0.3011E+01 2 0.1057E+00 0.2683E+01 -0.1577E+01 -0.1814E+02 -0.2286E+02 -0.5807E+02 0.1076E+01 0.2347E+02 -0.8068E+00 2 0.1057E+00 0.1817E+01 -0.1577E+01 -0.6770E+02 -0.8533E+02 -0.2167E+03 0.1076E+01 0.2347E+02 -0.3011E+01 The node point stresses are: Element x-coord y-coord z-coord sig_x sig_y sig_z tau_xy tau_yz tau_zx 1 0.5000E+00 0.1500E+01 0.0000E+00 0.6018E+02 -0.4561E+02 -0.2122E+03 0.8816E+01 -0.1111E+03 0.6612E+01 1 0.5000E+00 0.1500E+01 -0.2000E+01 -0.1250E+03 -0.1250E+03 -0.2916E+03 0.0000E+00 0.3323E-14 0.6612E+01 1 0.5000E+00 0.0000E+00 0.0000E+00 0.6645E-14 0.1551E-13 0.6645E-14 0.8816E+01 -0.1111E+03 0.0000E+00 1 0.5000E+00 0.0000E+00 -0.2000E+01 0.0000E+00 0.0000E+00 0.0000E+00 0.0000E+00 0.0000E+00 0.0000E+00 1 0.0000E+00 0.1500E+01 0.0000E+00 0.6018E+02 -0.4561E+02 -0.2122E+03 -0.8816E+01 -0.1111E+03 -0.6612E+01 1 0.0000E+00 0.0000E+00 0.0000E+00 0.4618E-14 0.1078E-13 0.4618E-14 -0.8816E+01 -0.1111E+03 0.0000E+00 1 0.0000E+00 0.1500E+01 -0.2000E+01 -0.1250E+03 -0.1250E+03 -0.2916E+03 0.0000E+00 0.2309E-14 -0.6612E+01 1 0.0000E+00 0.0000E+00 -0.2000E+01 0.0000E+00 0.0000E+00 0.0000E+00 0.0000E+00 0.0000E+00 0.0000E+00 2 0.5000E+00 0.3000E+01 0.0000E+00 -0.6645E-14 -0.1551E-13 -0.6645E-14 -0.8816E+01 0.1111E+03 0.0000E+00 2 0.5000E+00 0.3000E+01 -0.2000E+01 0.0000E+00 0.0000E+00 0.0000E+00 0.0000E+00 0.0000E+00 0.0000E+00

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2 0.5000E+00 0.1500E+01 0.0000E+00 0.6018E+02 -0.4561E+02 -0.2122E+03 -0.8816E+01 0.1111E+03 0.6612E+01 2 0.5000E+00 0.1500E+01 -0.2000E+01 -0.1250E+03 -0.1250E+03 -0.2916E+03 0.0000E+00 0.3323E-14 0.6612E+01 2 0.0000E+00 0.3000E+01 0.0000E+00 -0.4618E-14 -0.1078E-13 -0.4618E-14 0.8816E+01 0.1111E+03 0.0000E+00 2 0.0000E+00 0.1500E+01 0.0000E+00 0.6018E+02 -0.4561E+02 -0.2122E+03 0.8816E+01 0.1111E+03 -0.6612E+01 2 0.0000E+00 0.3000E+01 -0.2000E+01 0.0000E+00 0.0000E+00 0.0000E+00 0.0000E+00 0.0000E+00 0.0000E+00 2 0.0000E+00 0.1500E+01 -0.2000E+01 -0.1250E+03 -0.1250E+03 -0.2916E+03 0.0000E+00 0.2309E-14 -0.6612E+01

9.3.3

Interactive Interface

Figure 9.35 shows that this interface includes the parameter input window on the upper left, the model diagram on the upper right, the summarised input data on the lower left, and the output data on the lower right.

FIG. 9.35 – Interface of the three-dimensional program of solid compression. (1) Step 1 Figure 9.36 shows one parameter in the first line of the parameter input window: Element node: node number of the element. The value entered in this figure was 8.

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FIG. 9.36 – Element node in the three-dimensional program of solid compression.

(2) Step 2 Figure 9.37 shows five parameters in the second line of the parameter input window: Element number in x: element number in x-direction. The value entered in this figure was 1. Element number in y: element number in y-direction. The value entered in this figure was 2. Element number in z: element number in z-direction. The value entered in this figure was 1. Integration points: number of integration points. The value entered in this figure was 8. Material: number of material types. The value entered in this figure was 1. (3) Step 3 Figure 9.38 shows two parameters in the third line of the parameter input window: Young’s modulus E: Young’s modulus. The value entered in this figure was 1.0e6. Poisson’s ratio v: Poisson’s ratio. The value entered in this figure was 0.3.

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FIG. 9.37 – Element number in x, y, z, integration points, and material in the threedimensional program of solid compression.

FIG. 9.38 – Young’s modulus and Poisson’s ratio in the three-dimensional program of solid compression.

154

Basic Theory of Finite Element Method

(4) Step 4 Figure 9.39 shows one parameter in the fourth line of the parameter input window: x-coordinate of node: the x-coordinate of each node. The x-coordinate value of each node can be input sequentially, and the option ‘Add’ is selected. Then, the window ‘Add successfully’ appears, and the option ‘Agree’ is selected to confirm the addition. The values entered in this figure were 0 and 0.5.

FIG. 9.39 – The x-coordinate of node in the three-dimensional program of solid compression. (5) Step 5 Figure 9.40 shows one parameter in the fifth line of the parameter input window: y-coordinate of node: The y-coordinate of each node. The y-coordinate value of each node can be input sequentially, and the option ‘Add’ is selected. Then, the window ‘Add successfully’ appears, and the option ‘Agree’ is selected to confirm the addition. The values entered in this figure were 0, 1.5, and 3. (6) Step 6 Figure 9.41 shows one parameter in the sixth line of the parameter input window: z-coordinate of node: The z-coordinate of each node. The z-coordinate value of each node can be input sequentially, and the option ‘Add’ is selected. Then, the window ‘Add successfully’ appears, and the option ‘Agree’ is selected to confirm the addition. The values entered in this figure were 0 and −2.

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FIG. 9.40 – The y-coordinate of node in the three-dimensional program of solid compression.

FIG. 9.41 – The z-coordinate of node in the three-dimensional program of solid compression.

156

Basic Theory of Finite Element Method

(7) Step 7 Figure 9.42 shows one parameter in the seventh line of the parameter input window: Dis. BC no.: the number of displacement boundary conditions. The value entered in this figure was 10.

FIG. 9.42 – Number of displacement boundary conditions in the three-dimensional program of solid compression.

(8) Step 8 Figure 9.43 shows two parameters in the eighth line of the parameter input window: Node number of Dis. BC: node number of displacement boundary, specifying the node at which a specific displacement occurs. The values entered in this figure were 1, 2, 3, 4, 7, 8, 9, 10, 11, and 12. Value of Dis. BC: values of displacement boundary, specifying the values used for displacement boundary in x and y directions, respectively. The value of the displacement boundary of each specifying node can be input sequentially, and the option ‘Add’ is selected. Then, the window ‘Add successfully’ appears, and the option ‘Agree’ is selected to confirm the addition. The values entered in this figure were 0, 0, and 0.

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FIG. 9.43 – Node number and value of displacement boundary conditions in the three-dimensional program of solid compression.

(9) Step 9 Figure 9.44 shows one parameter in the ninth line of the parameter input window: Tra. BC no.: the number of traction boundary conditions. The value entered in this figure was 2. (10) Step 10 Figure 9.45 shows two parameters in the tenth line of the parameter input window: Node number of Tra. BC: node number of traction boundary, specifying the node at which specific traction occurs. The values entered in this figure were 5 and 6. Value of Tra. BC: values of traction boundary, specifying the values used for traction boundary in x-, y-, and z-directions, respectively. The value of the traction boundary of each specifying node can be input sequentially, and the option ‘Add’ is selected. Then, the window ‘Add successfully’ appears, and the option ‘Agree’ is selected to confirm the addition. The value entered in this figure were 0, 0, and −1.0e6.

158

Basic Theory of Finite Element Method

FIG. 9.44 – Number of traction boundary condition in the three-dimensional program of solid compression.

FIG. 9.45 – Node number and value of traction boundary condition in the three-dimensional program of solid compression.

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(11) Step 11 Figure 9.46 shows one parameter in the eleventh line of the parameter input window: Fixed freedom: the number of nodes with the specified displacement. The value entered in this figure was 0 by default.

FIG. 9.46 – Fixed freedom in the three-dimensional program of solid compression.

(12) Step 12 As shown in figure 9.47, the option ‘Display input data’ is selected, and the aforementioned input dataset is summarised on the lower left of the interface. (13) Step 13 As shown in figure 9.48, the option ‘Generate file 3DFEM.dat’ is selected, and the input dataset is recorded in the file 3DFEM.dat, which is saved in the path of the current file. Subsequently, the window ‘Generate successfully’ appears, and the option ‘Agree’ is selected for confirmation.

160

Basic Theory of Finite Element Method

FIG. 9.47 – Input data display in the three-dimensional program of solid compression.

FIG. 9.48 – Input data generation in the three-dimensional program of solid compression.

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161

(14) Step 14 As shown in figure 9.49, the option ‘Compute’ is selected, and the computation program starts to conduct operations using the aforementioned input data. Once the computation is completed, the window ‘Compute successfully’ appears, and the option ‘Agree’ is selected for confirmation. The output dataset above is recorded in the file 3DFEM.res, which was saved in the path of the current file.

FIG. 9.49 – Computation control in the three-dimensional program of solid compression.

(15) Step 15 As shown in figure 9.50, the option ‘Display output data’ is selected, and the output data of computed results are displayed on the lower right of the interface. The option ‘Return’ in the interactive interface, will return to the interface for selecting the computation program shown in figure 9.2. The option ‘Exit’ will exit the current interactive interface.

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FIG. 9.50 – Output data display in the three-dimensional program of solid compression.

9.4

Exercises

9.4.1 Using the one-, two-, and three-dimensional programs provided above, compute the displacements and stresses for the following one-, two-, and three-dimensional models with various values of the parameters a, b, and c, as shown in figure 9.51. 9.4.2 Using the analytical results of the theoretical models, verify the correctness of the programs in this chapter, and discuss the influence of different mesh densities on the accuracy of the solutions. 9.4.3 Based on the one-dimensional program of beam deformation, use shape functions, and numerical integration to compute the results. 9.4.4 Based on the two-dimensional program of plane strain problem, compute the solutions for the plane stress model.

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FIG. 9.51 – One-, two-, and three-dimensional models with variable geometric sizes (a, b, and c).

Appendix A. Keyword Index

The following is a list of frequently used keywords in this book, which can be quickly indexed to the specified chapters. It should be noted that some keywords appear and are used in multiple chapters, which will be given simultaneously.

Chapter 1 Computational mechanics

Numerical method

Numerical algorithm

Numerical model

Continuous system

Discrete system

Finite element method

Finite element analysis

Element

Mesh

Chapter 2 Three-dimensional problem

Two-dimensional problem

Plane stress problem

Plane strain problem

Axisymmetric problem

Displacement

Strain

Stress

Geometric equations

Constitutive equations

Equilibrium equations

Boundary conditions

Chapter 3 Strong form Differential

Weak form Integration

Arbitrary function

Integration by parts

Derivative

Galerkin method

Shape function

Stiffness

Load

Element location vector

Appendix A. Keyword Index

166

Chapter 4 Lagrange element

Hexagon

Triangle

Hexagon element

Triangle element

Area coordinate

Rectangle

Volume coordinate

Rectangle element

Linear element

Tetrahedron

Quadratic element

Tetrahedron element

Cubic element

Chapter 5 Isoparametric mapping

Jacobian matrix

Isoparametric element

Numerical integration

Global coordinate

Gaussian quadrature

Local coordinate

Integration point

Derivative

Weight

Chapter 6 Elasticity problem

Three-dimensional problem

Two-dimensional problem

Displacement

Strain

Stress

Geometric equations

Constitutive equations

Equilibrium equations

Boundary conditions

Weak form

Stiffness equations

Chapter 7 Algebraic equations Stiffness matrix

Upper triangular matrix Lower triangular matrix

Decomposition

Gaussian elimination

Row

Forward elimination

Column

Back substitution

Appendix A. Keyword Index

167

Chapter 8 Error estimation

Superconvergent solutions

Adaptive finite element method

Superconvergent patch recovery

Error tolerance

Solution refinement

Chapter 9 Program

Modules

Geometric model

Input data

Arrays

Output solutions

Appendix B. Matrix Calculation

A feature of the finite element method is that systems of equations are compactly expressed in matrix form, and thus matrix operations can be used to simplify the calculations. The basic matrix concepts and operations used in this book include matrix definitions, addition or subtraction, transpose of a matrix, transpose of a product, inverse, symmetric matrices, and partitioning of matrices.

B.1 Definition The following linear relations between a set of variables xi and bj (i = 1, 2, 3; j = 1, 2) a11 x1 þ a12 x2 þ a13 x3 ¼ b1 a21 x1 þ a22 x2 þ a23 x3 ¼ b2

ðB:1Þ

½A]fxg =fbg

ðB:2Þ

Ax¼b

ðB:3Þ

can be written as:

or

where
a A  ½A ¼ 11 a21

a12 a22

8 9    < x1 = b1 a13 ; x  fxg ¼ x2 ; b  fbg ¼ a23 b2 : ; x3

ðB:4Þ

The above notation involves the concept of a matrix and the operation of matrix multiplication. Matrices are defined as ‘arrays of number’ of the type in equation (B.4). Matrices containing a single column of numbers are often referred to as vector or column matrices, whereas a matrix with multiple columns and rows is called a rectangular matrix. The multiplication of a matrix by a column vector, as in

Appendix B. Matrix Calculation

170

equation (B.2), is defined by the equivalence of the left and right sides of equations (B.1) and (B.2). Boldface characters are used to denote both vectors and matrices. If another relationship to equation (B.1), using the same a, but a different set of x and b, we have the following linear relations: a11 x10 þ a12 x20 þ a13 x30 ¼ b01

ðB:5Þ

a21 x10 þ a22 x20 þ a23 x30 ¼ b02 then equations (B.1) and (B.5) can be simultaneously expressed as: ½A½X ¼ ½B

or AX ¼ B

where
A  ½A ¼

a11 a21



a12 a22

B  ½B ¼ or equivalently




ðB:6Þ 2

x1

a13 6 ; X  ½X ¼4 x2 a23 x3
0  b1 b1 b2

x10

3

7 x20 5; x30

ðB:7Þ

b02


a11 x1 þ    a11 x10 þ    b ¼ B  ½B ¼ 1 a21 x1 þ    a21 x10 þ    b2

b01 b02

 ðB:8Þ

Two matrices are equal if and only if their corresponding entries are equal. The multiplication of full matrices is defined in equation (B.6), and evidently, it is meaningful only if the number of columns in A is equal to the number of rows in X. One property that distinguishes matrix multiplication is that, in general, AX 6¼ XA That is, unlike scalar multiplication, matrix multiplication is not commutative.

B.2 Matrix Addition or Subtraction If relations of the form from equations (B.1) and (B.5) are added, then we have: a11 ðx1 þ x10 Þ þ a12 ðx2 þ x20 Þ þ a13 ðx3 þ x30 Þ ¼ b1 þ b01

a21 ðx1 þ x10 Þ þ a22 ðx2 þ x20 Þ þ a23 ðx3 þ x30 Þ ¼ b2 þ b02

ðB:9Þ

which can also be written as: Ax þ Ax0 ¼ b þ b0 if we define the addition of two matrices as the matrix, each entry of which is the sum of the corresponding entries of the two matrices. Clearly, this can be done only if the matrices have the same dimension (same number of rows and columns). For example,

Appendix B. Matrix Calculation 2

a11 4 a21 a31

3 2 a12 b11 a22 5 þ 4 b21 a32 b31

171

3 2 b12 a11 þ b11 b22 5 ¼ 4 a21 þ b21 b32 a31 þ b31

3 2 a12 þ b12 c11 a22 þ b22 5 ¼ 4 c21 a32 þ b32 c31

3 c12 c22 5 c32

or AþB ¼ C

ðB:10Þ

implies that every entry of C is equal to the sum of the corresponding entries of A and B. Matrix subtraction is defined similarly.

B.3 Transpose The transpose of a matrix is obtained by interchanging the rows and columns of the given matrix and is denoted by the superscript T. For example, 2 3
T a11 a21 a11 a12 a13 ðB:11Þ ¼ 4 a12 a22 5 a21 a22 a23 a13 a23 In mechanics problems, we often encounter a number of quantities, such as force, which can be expressed in vector form: 8 9 f1 > > > > > = < f2 > ðB:12Þ f ¼ .. > . > > > > > : ; fn This, in turn, is often associated with the same number of displacements given by another vector: 8 9 u1 > > > = < u2 > u¼ ðB:13Þ .. > > > ; : . > un It is known that work is represented as the sum of the products of force and displacement: S¼

n X

fk uk

k¼1

Clearly, the transpose is useful here, as, by the rule of matrix multiplication, we have 8 9 u1 > > > = < u2 > ¼ f T u ¼ uT f ðB:14Þ S ¼ ff1 f2    fn gT .. > > . > ; : > un

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172

B.4 Transpose of a Product An operation that sometimes occurs is taking the transpose of a matrix product. The following equation can be proved by examining the corresponding entries of the matrices on the two sides: ðABÞT ¼ BT AT

ðB:15Þ

B.5 Inverse If in equation (B.3), the matrix A is ‘square’, that is, it represents the coefficients of simultaneous equation (B.1), the number of which is equal to the number of unknowns x, then, under certain assumptions, it is possible to solve for unknowns in terms of the known coefficients b. This solution can be written as: x ¼ A1 b

ðB:16Þ

where matrix A−1 is known as the ‘inverse’ of the square matrix B. Clearly, A−1 is also square and of the same dimension as B. We can obtain equation (B.12) by multiplying both sides of equation (B.3) by A−1: A1 A ¼ I ¼ AA1

ðB:17Þ

where I is the ‘identity’ matrix, that is, it has 0 at all off-diagonal positions and 1 on the diagonal.

B.6 Symmetric Matrices Symmetric matrices are often encountered in structural problems. If the general entry of a matrix A is denoted by aij, then A is called symmetric if aij ¼ aji

or A ¼ AT :

ðB:18Þ

A symmetric matrix must be square. It can be shown that the inverse of a symmetric matrix is also symmetric: A1 ¼ ðA1 ÞT  AT :

ðB:19Þ

B.7 Partitioning The product of two matrices A and B can also be expressed in terms of their blocks. For instance, if

Appendix B. Matrix Calculation

2

a11 A ¼ 4 a21 a31

a12 a22 a32

173

a13 a23 a33

a14 a24 a34

2

b11 6 b21 a15 6 a25 5; B ¼ 6 6 b31 4 b41 a35 b51 3

3 b12 b22 7 7 b32 7 7 b42 5 b52

ðB:20Þ

then A and B can be multiplied by partitioning each matrix into submatrices, indicated by the lines, and applying the rules of matrix multiplication first to each of these submatrices, as if it were a scalar, and then carrying out further multiplication in the usual manner. Specifically, A and B can be written as:

 A11 A12 B1 A¼ ;B¼ ðB:21Þ A21 A22 B2 and then, the product AB can be expressed as:
 A11 B1 þ A12 B2 AB ¼ A21 B1 þ A22 B2

ðB:22Þ

This product can be verified as representing the complete product by further multiplication. The essential feature of partitioning is that the size of subdivisions has to be such as to make the products of the type A11 B1 meaningful, i.e., the number of columns in A11 must be equal to the number of rows in B1 , etc. If the above definition holds, then all further operations can be conducted on partitioned matrices, treating each partition as if it were a scalar. It should be noted that any matrix can be multiplied by a scalar (number). Here, obviously, the requirements of equality of appropriate rows and columns no longer apply: 2 3 ka11 ka12 ka13 ka14 ka15 Ak ¼ kA ¼ 4 ka21 ka22 ka23 ka24 ka25 5 ðB:23Þ ka31 ka32 ka33 ka34 ka35 If a symmetric matrix is divided into an equal number of submatrices Aij in rows and columns, then we have Aij ¼ AT ji

ðB:24Þ

Appendix C. Summary of Elements and Shape Functions

The following are the elements and shape functions for the one-, two-, and three-dimensional cases introduced in this book. For the convenience of the reader, they are summarised to be indexed and compared.

C.1 One-Dimensional Lagrange Element Linear element with two nodes N1 ðx 0 Þ ¼ 1 

x0 he

and

N2 ðx 0 Þ ¼

x0 ; he

with

x 0 ¼ x  x1e

and

he ¼ x2e  x1e ðC:1Þ

Quadratic Lagrange element ð n  1Þ ð n  0Þ 1 ¼ nð n  1Þ ð1  1Þð1  0Þ 2 ðn þ 1Þðn  0Þ 1 N2 ðnÞ ¼ l22 ðnÞ ¼ ¼ nð n þ 1Þ ð1 þ 1Þð1  0Þ 2 ðn þ 1Þðn  1Þ N3 ðnÞ ¼ l32 ðnÞ ¼ ¼ 1n2 ð0 þ 1Þð0  1Þ N1 ðnÞ ¼ l12 ðnÞ ¼

C.2 Two-Dimensional Triangle Element Triangle with three nodes Na ðx; y Þ ¼

1 ðaa þ ba x þ ca y Þ; 2D

a ¼ 1; 2; 3

ðC:2Þ

Appendix C. Summary of Elements and Shape Functions

176

a1 ¼ x2 y3  x3 y2 ; b1 ¼ y2  y3 ; c1 ¼ x3  x2 with a2 ¼ x3 y1  x1 y3 ; b2 ¼ y3  y1 ; c2 ¼ x1  x3 a3 ¼ x1 y2  x2 y1 ; b3 ¼ y1  y2 ; c3 ¼ x2  x1 and D ¼ ðx1 b1 þ x2 b2 þ x3 b3 Þ=2

ðC:3Þ

Quadratic triangle element Corner nodes: Na ¼ ð2La  1ÞLa ;

a ¼ 1; 2; 3

ðC:4Þ

Mid-side nodes: N4 ¼ 4L1 L2 ;

N5 ¼ 4L2 L3 ;

N6 ¼ 4L3 L1

ðC:5Þ

Cubic triangle element Corner nodes: 1 Na ¼ ð3La  1Þð3La  2ÞLa ; 2

a ¼ 1; 2; 3

ðC:6Þ

Mid-side nodes: 9 N4 ¼ L1 L2 ð3L1  1Þ; 2 9 N7 ¼ L2 L3 ð3L3  1Þ; 2

9 N5 ¼ L1 L2 ð3L2  1Þ; 2 9 N8 ¼ L3 L1 ð3L3  1Þ; 2

9 N6 ¼ L2 L3 ð3L2  1Þ 2 9 N9 ¼ L3 L1 ð3L1  1Þ 2

ðC:7Þ

Internal node: N10 ¼ 27L1 L2 L3

ðC:8Þ

C.3 Two-Dimensional Rectangle Element Linear rectangle element with four nodes 1 N1 ¼ ð1  nÞð1  gÞ; 4 1 N3 ¼ ð1 þ nÞð1 þ gÞ; 4

1 N2 ¼ ð1 þ nÞð1  gÞ 4 1 N4 ¼ ð 1  nÞ ð 1 þ gÞ 4

ðC:9Þ

Quadratic rectangle element Corner nodes: 1 Na ¼ ng ðn þ na Þ ðg þ ga Þ; 4

a ¼ 1; 2; 3; 4

ðC:10Þ

Appendix C. Summary of Elements and Shape Functions

177

Mid-side nodes:  1  na ¼ 0; Na ¼ g 1  n2 ðg þ ga Þ; 2   1 ga ¼ 0; Na ¼ n ðn þ na Þ 1  g2 ; 2 Center node:

a ¼ 5; 7 a ¼ 6; 8

   N9 ¼ 1  n2 1  g 2

ðC:11Þ

ðC:12Þ

C.4 Three-Dimensional Tetrahedron Element Linear tetrahedron element with four nodes Na ðx; y; z Þ ¼

1 ðaa þ ba x þ ca y þ da z Þ; 6V

a ¼ 1; 2; 3; 4

ðC:13Þ

Quadratic tetrahedron element Corner nodes: Na ¼ La ð2La  1Þ;

a ¼ 1; 2; 3; 4

ðC:14Þ

Mid-side nodes: N5 ¼ 4L2 L1 ; N8 ¼ 4L2 L3 ;

N6 ¼ 4L3 L1 ; N9 ¼ 4L3 L4 ;

N7 ¼ 4L4 L1 N10 ¼ 4L4 L2

ðC:15Þ

a ¼ 1; 2; 3; 4

ðC:16Þ

Cubic tetrahedron element Corner nodes: 1 Na ¼ La ð3La  1Þ ð3La  2Þ; 2 Mid-side nodes: 9 N5 ¼ L1 L2 ð3L1  1Þ; 2 9 N8 ¼ L1 L2 ð3L4  1Þ; 2 9 N11 ¼ L2 L3 ð3L3  1Þ; 2 9 N14 ¼ L3 L4 ð3L2  1Þ; 2

9 9 N6 ¼ L1 L2 ð3L2  1Þ; N7 ¼ L1 L2 ð3L3  1Þ 2 2 9 9 N9 ¼ L2 L3 ð3L1  1Þ; N10 ¼ L2 L3 ð3L2  1Þ 2 2 ðC:17Þ 9 9 N12 ¼ L2 L3 ð3L4  1Þ; N13 ¼ L3 L4 ð3L1  1Þ 2 2 9 9 N15 ¼ L3 L4 ð3L3  1Þ; N16 ¼ L3 L4 ð3L4  1Þ 2 2

Appendix C. Summary of Elements and Shape Functions

178

Mid-face nodes: N17 ¼ 27L1 L2 L3 ; N19 ¼ 27L3 L4 L1 ;

N18 ¼ 27L4 L1 L2 N20 ¼ 27L2 L3 L4

ðC:18Þ

C.5 Three-Dimensional Hexahedron Element Hexahedron with eight nodes 1 1 N1 ¼ ð1  nÞ ð1  gÞ ð1  fÞ; N5 ¼ ð1  nÞ ð1  gÞ ð1 þ fÞ 8 8 1 1 N2 ¼ ð1 þ nÞ ð1  gÞ ð1  fÞ; N6 ¼ ð1 þ nÞ ð1  gÞ ð1 þ fÞ 8 8 1 1 N3 ¼ ð1 þ nÞ ð1 þ gÞ ð1  fÞ; N7 ¼ ð1 þ nÞ ð1 þ gÞ ð1 þ fÞ 8 8 1 1 N4 ¼ ð1  nÞ ð1 þ gÞ ð1  fÞ; N8 ¼ ð1  nÞ ð1 þ gÞ ð1 þ fÞ 8 8

ðC:19Þ

Quadratic hexahedron element Corner nodes: 1 Na ¼ ngf ðn þ na Þ ðg þ ga Þ ðf þ fa Þ; 8

a ¼ 1; 2; . . .; 8

ðC:20Þ

Mid-side nodes:  1  na ¼ 0; Na ¼ gf 1  n2 ðg þ ga Þ ðf þ fa Þ; 4   1 ga ¼ 0; Na ¼ nf ðn þ na Þ 1  g2 ðf þ fa Þ; 4   1 fa ¼ 0; Na ¼ ng ðn þ na Þ ðg þ ga Þ 1  f2 ; 4

a ¼ 9; 10; 11; 12 a ¼ 13; 14; 15; 16

ðC:21Þ

a ¼ 17; 18; 19; 20

Mid-face nodes:   1  na ¼ ga ¼ 0; Na ¼ f 1  n2 1  g2 ðf þ fa Þ; a ¼ 21; 22 2    1 ga ¼ fa ¼ 0; Na ¼ n ðn þ na Þ 1  g2 1  f2 ; a ¼ 23; 24 2    1  fa ¼ na ¼ 0; Na ¼ g 1  n2 ðg þ ga Þ 1  f2 ; a ¼ 25; 26 2 Center node:

    Na ¼ 1  n 2 1  g 2 1  f 2 ;

a ¼ 27

ðC:22Þ

ðC:23Þ

Appendix D. Gaussian Integration Points and Weights

For one-dimensional problems, Gaussian integration is given by: Z 1 n X   f ðnÞ dn  f nj wj 1

ðD:1Þ

j¼1

with an appropriate choice of the integration points nj and weights wj . The Gaussian quadrature formula in equation (D.1) using n integration points is exact in the case of polynomials of degree at most 2n − 1. For high-dimensional problems, the number of integration points in each dimension abides by the above rules. Accordingly, for two-dimensional problems, we can integrate into two directions using Z 1Z 1 N X M X f ðn; gÞje ðn; gÞdndg  f ðnm ; gn Þje ðnm ; gn Þwm wn ðD:2Þ 1

1

n¼1 m¼1

where m denotes quadrature points in the ξ direction and n quadrature points in the η direction. For three-dimensional problems, we can integrate into three directions using Z 1Z 1Z 1 f ðn; g; fÞj ðn; g; fÞdndgdf 1

1

1



L X N X M X l¼1 n¼1 m¼1

ðD:3Þ f ðnm ; gn ; fl Þj ðnm ; gn ; fl Þwm wn wl

where m denotes quadrature points in the ξ direction, n quadrature points in the η direction, and l quadrature points in the ζ direction. For complex integrands expressed by higher-order polynomials, more integration points are required to obtain exact formulas. The following decimal forms of these integration points and weights are provided.

180

Appendix D. Gaussian Integration Points and Weights

n=1 Integration points

Weights

0.0000000000000000

2.0000000000000000

n=2 Integration points

Weights

–0.5773502691896250

1.0000000000000000

0.5773502691896250

1.0000000000000000

n=3 Integration points

Weights

–0.7745966692414830

0.5555555555555550

0.0000000000000000

0.8888888888888880

0.7745966692414830

0.5555555555555550

n=4 Integration points

Weights

–0.8611363115940520

0.3478548451374530

–0.3399810435848560

0.6521451548625460

0.3399810435848560

0.6521451548625460

0.8611363115940520

0.3478548451374530

n=5 Integration points

Weights

–0.9061798459386640

0.2369268850561890

–0.5384693101056830

0.4786286704993660

0.0000000000000000 0.5384693101056830

0.5688888888888880 0.4786286704993660

0.9061798459386640

0.2369268850561890

Appendix D. Gaussian Integration Points and Weights

n=6 Integration points

Weights

–0.9324695142031520

0.1713244923791700

–0.6612093864662640

0.3607615730481380

–0.2386191860831960

0.4679139345726910

0.2386191860831960

0.4679139345726910

0.6612093864662640

0.3607615730481380

0.9324695142031520

0.1713244923791700

n=7 Integration points

Weights

–0.9491079123427580

0.1294849661688690

–0.7415311855993940

0.2797053914892760

–0.4058451513773970

0.3818300505051180

0.0000000000000000

0.4179591836734690

0.4058451513773970

0.3818300505051180

0.7415311855993940

0.2797053914892760

0.9491079123427580

0.1294849661688690

n=8 Integration points

Weights

–0.9602898564975360

0.1012285362903760

–0.7966664774136260

0.2223810344533740

–0.5255324099163290

0.3137066458778870

–0.1834346424956490

0.3626837833783620

0.1834346424956490

0.3626837833783620

0.5255324099163290

0.3137066458778870

0.7966664774136260

0.2223810344533740

0.9602898564975360

0.1012285362903760

181

182

Appendix D. Gaussian Integration Points and Weights

n=9 Integration points

Weights

–0.9681602395076260

0.0812743883615744

–0.8360311073266350

0.1806481606948570

–0.6133714327005900

0.2606106964029350

–0.3242534234038080

0.3123470770400020

0.0000000000000000

0.3302393550012590

0.3242534234038080

0.3123470770400020

0.6133714327005900

0.2606106964029350

0.8360311073266350

0.1806481606948570

0.9681602395076260

0.0812743883615744

n = 10 Integration points

Weights

–0.9739065285171710

0.0666713443086881

–0.8650633666889840

0.1494513491505800

–0.6794095682990240

0.2190863625159820

–0.4333953941292470

0.2692667193099960

–0.1488743389816310

0.2955242247147520

0.1488743389816310

0.2955242247147520

0.4333953941292470

0.2692667193099960

0.6794095682990240

0.2190863625159820

0.8650633666889840

0.1494513491505800

0.9739065285171710

0.0666713443086881

n = 11 Integration points

Weights

–0.9782286581460570

0.0556685671161736

–0.8870625997680950

0.1255803694649040

–0.7301520055740490

0.1862902109277340

–0.5190961292068110

0.2331937645919900

–0.2695431559523440

0.2628045445102460

0.0000000000000000 0.2695431559523440

0.2729250867779000 0.2628045445102460

0.5190961292068110

0.2331937645919900

0.7301520055740490

0.1862902109277340

0.8870625997680950

0.1255803694649040

0.9782286581460570

0.0556685671161736

Appendix D. Gaussian Integration Points and Weights

n = 12 Integration points

Weights

–0.9815606342467190

0.0471753363865118

–0.9041172563704740

0.1069393259953180

–0.7699026741943040

0.1600783285433460

–0.5873179542866170

0.2031674267230650

–0.3678314989981800

0.2334925365383540

–0.1252334085114680

0.2491470458134020

0.1252334085114680

0.2491470458134020

0.3678314989981800

0.2334925365383540

0.5873179542866170

0.2031674267230650

0.7699026741943040

0.1600783285433460

0.9041172563704740

0.1069393259953180

0.9815606342467190

0.0471753363865118

n = 13 Integration points

Weights

–0.9841830547185880

0.0404840047653158

–0.9175983992229770

0.0921214998377284

–0.8015780907333090

0.1388735102197870

–0.6423493394403400

0.1781459807619450

–0.4484927510364460

0.2078160475368880

–0.2304583159551340

0.2262831802628970

0.0000000000000000

0.2325515532308730

0.2304583159551340

0.2262831802628970

0.4484927510364460

0.2078160475368880

0.6423493394403400

0.1781459807619450

0.8015780907333090

0.1388735102197870

0.9175983992229770

0.0921214998377284

0.9841830547185880

0.0404840047653158

183

184

Appendix D. Gaussian Integration Points and Weights

n = 14 Integration points

Weights

–0.9862838086968120

0.0351194603317518

–0.9284348836635730

0.0801580871597602

–0.8272013150697650

0.1215185706879030

–0.6872929048116850

0.1572031671581930

–0.5152486363581540

0.1855383974779370

–0.3191123689278890

0.2051984637212950

–0.1080549487073430

0.2152638534631570

0.1080549487073430

0.2152638534631570

0.3191123689278890

0.2051984637212950

0.5152486363581540

0.1855383974779370

0.6872929048116850

0.1572031671581930

0.8272013150697650

0.1215185706879030

0.9284348836635730

0.0801580871597602

0.9862838086968120

0.0351194603317518

n = 15 Integration points

Weights

–0.9879925180204850

0.0307532419961172

–0.9372733924007060

0.0703660474881081

–0.8482065834104270

0.1071592204671710

–0.7244177313601700

0.1395706779261540

–0.5709721726085380

0.1662692058169930

–0.3941513470775630

0.1861610000155620

–0.2011940939974340

0.1984314853271110

0.0000000000000000

0.2025782419255610

0.2011940939974340

0.1984314853271110

0.3941513470775630

0.1861610000155620

0.5709721726085380

0.1662692058169930

0.7244177313601700

0.1395706779261540

0.8482065834104270 0.9372733924007060

0.1071592204671710 0.0703660474881081

0.9879925180204850

0.0307532419961172

Appendix D. Gaussian Integration Points and Weights

n = 16 Integration points

Weights

–0.9894009349916490

0.0271524594117540

–0.9445750230732320

0.0622535239386478

–0.8656312023878310

0.0951585116824927

–0.7554044083550030

0.1246289712555330

–0.6178762444026430

0.1495959888165760

–0.4580167776572270

0.1691565193950020

–0.2816035507792580

0.1826034150449230

–0.0950125098376374

0.1894506104550680

0.0950125098376374

0.1894506104550680

0.2816035507792580

0.1826034150449230

0.4580167776572270

0.1691565193950020

0.6178762444026430

0.1495959888165760

0.7554044083550030

0.1246289712555330

0.8656312023878310

0.0951585116824927

0.9445750230732320

0.0622535239386478

0.9894009349916490

0.0271524594117540

n = 17 Integration points

Weights

–0.9905754753144170

0.0241483028685479

–0.9506755217687670

0.0554595293739872

–0.8802391537269850

0.0850361483171791

–0.7815140038968010

0.1118838471934030

–0.6576711592166900

0.1351363684685250

–0.5126905370864760

0.1540457610768100

–0.3512317634538760

0.1680041021564500

–0.1784841814958470

0.1765627053669920

0.0000000000000000

0.1794464703562060

0.1784841814958470

0.1765627053669920

0.3512317634538760

0.1680041021564500

0.5126905370864760 0.6576711592166900

0.1540457610768100 0.1351363684685250

0.7815140038968010

0.1118838471934030

0.8802391537269850

0.0850361483171791

0.9506755217687670

0.0554595293739872

0.9905754753144170

0.0241483028685479

185

186

Appendix D. Gaussian Integration Points and Weights

n = 18 Integration points

Weights

–0.9915651684209300

0.0216160135264833

–0.9558239495713970

0.0497145488949698

–0.8926024664975550

0.0764257302548890

–0.8037049589725230

0.1009420441062870

–0.6916870430603530

0.1225552067114780

–0.5597708310739470

0.1406429146706500

–0.4117511614628420

0.1546846751262650

–0.2518862256915050

0.1642764837458320

–0.0847750130417353

0.1691423829631430

0.0847750130417353

0.1691423829631430

0.2518862256915050

0.1642764837458320

0.4117511614628420

0.1546846751262650

0.5597708310739470

0.1406429146706500

0.6916870430603530

0.1225552067114780

0.8037049589725230

0.1009420441062870

0.8926024664975550

0.0764257302548890

0.9558239495713970

0.0497145488949698

0.9915651684209300

0.0216160135264833

n = 19 Integration points

Weights

–0.9924068438435840

0.0194617882297264

–0.9602081521348300

0.0448142267656996

–0.9031559036148170

0.0690445427376412

–0.8227146565371420

0.0914900216224500

–0.7209661773352290

0.1115666455473330

–0.6005453046616810

0.1287539625393360

–0.4645707413759600

0.1426067021736060

–0.3165640999636290

0.1527660420658590

–0.1603586456402250

0.1589688433939540

0.0000000000000000

0.1610544498487830

0.1603586456402250

0.1589688433939540

0.3165640999636290 0.4645707413759600

0.1527660420658590 0.1426067021736060

0.6005453046616810

0.1287539625393360

0.7209661773352290

0.1115666455473330

0.8227146565371420

0.0914900216224500

0.9031559036148170

0.0690445427376412

0.9602081521348300

0.0448142267656996

0.9924068438435840

0.0194617882297264

Appendix D. Gaussian Integration Points and Weights

n = 20 Integration points

Weights

–0.9931285991850940

0.0176140071391521

–0.9639719272779130

0.0406014298003869

–0.9122344282513260

0.0626720483341090

–0.8391169718222180

0.0832767415767047

–0.7463319064601500

0.1019301198172400

–0.6360536807265150

0.1181945319615180

–0.5108670019508270

0.1316886384491760

–0.3737060887154190

0.1420961093183820

–0.2277858511416450

0.1491729864726030

–0.0765265211334973

0.1527533871307250

0.0765265211334973

0.1527533871307250

0.2277858511416450

0.1491729864726030

0.3737060887154190

0.1420961093183820

0.5108670019508270

0.1316886384491760

0.6360536807265150

0.1181945319615180

0.7463319064601500

0.1019301198172400

0.8391169718222180

0.0832767415767047

0.9122344282513260

0.0626720483341090

0.9639719272779130

0.0406014298003869

0.9931285991850940

0.0176140071391521

187

Appendix E. Exercise Solutions

Chapter 1 1.4.1 Describe the treatment of ‘standard discrete problems’. Solutions: The existence of a unified treatment of ‘standard discrete problems’ leads us to the first definition of the finite element process as a method of approximation to continuum problems such that: (a) the continuum is divided into a finite number of parts (elements), the behaviour of which is specified by a finite number of parameters, and (b) the solution for the complete system as an assembly of its elements precisely follows the same rules as those applicable to standard discrete problems. 1.4.2 Summarise the computation procedure of the finite element method. Solutions: It should be clear to the reader that the finite element solution of a problem always follows a standard methodology. The solution process for any problem type is always performed by the following steps: (1) Define the problem to be solved in terms of differential equations. Construct the integral form for the problem as virtual work, variational, or weak formulation. (2) Select the type and order of the finite elements to be used in the analysis. (3) Define the mesh for the problem. This involves the description of the node and element layout, as well as the specification of boundary conditions and parameters for the formulation used. (4) Compute and assemble the element arrays. The particular virtual work, variational, or weak form provides the basis for computing the specific relationships of each element. (5) Solve the resulting set of linear algebraic equations for the unknown parameters. For static problems, this requires the solution of linear algebraic equations. (6) Output the results for the nodal and element variables. Graphical outputs are also useful in this step.

Appendix E. Exercise Solutions

190

Chapter 2 2.8.1 Summarise the basic variables and equations of three- and two-dimensional elasticity. Solutions: 1. Three-dimensional elasticity Basic variables: (1) Displacement components: u, v, w (2) Strain components: ex , ey , ez , cxy , cyz , czx (3) Stress components: rx , ry , rz , sxy , syz , szx Basic equations: (1) Geometric equations:

2

@ 6 @x 6 9 6 8 6 0 ex > 6 > > > > 6 > > > e > > 6 y > > = 6 0 < ez 6 ¼6 @ e¼ c > > 6 xy > > > > 6 > > c > > 6 @y yz > > ; 6 : czx 6 0 6 6 4 @

0 @ @y 0 @ @x @ @z

@z

0

3 0 7 7 7 0 7 7 7 @ 78 9 7< u = @z 7 7 v : ; 0 7 7 w 7 @ 7 7 7 @y 7 @ 5 @x

(2) Constitutive equations: r ¼ De 2

ð1  v Þ 6 v 6 E6 v with D ¼ 6 d6 6 0 4 0 0

v ð1  v Þ v 0 0 0

v v ð1  v Þ 0 0 0

0 0 0 ð1  2v Þ=2 0 0

3 0 0 7 0 0 7 7 0 0 7 7 0 0 7 5 ð1  2v Þ=2 0 0 ð1  2v Þ=2

(3) Equilibrium equations: @rx @syx @szx þ þ þ bx ¼ 0 @x @y @z @sxy @ry @szy þ þ þ by ¼ 0 @x @y @z @sxz @syz @rz þ þ þ bz ¼ 0 @x @y @z

Appendix E. Exercise Solutions

191

2. Two-dimensional elasticity (Plane stress and stain case) Basic variables: (1) Displacement components: u, v (2) Strain components: ex , ey , ez , cxy (3) Stress components: rx , ry , rz , sxy Basic equations: (1) Geometric equations: 3

2

@ 6 @x 2 3 6 ex 6 6 ey 7 6 6 0 6 7 ¼ e¼4 ez 5 6 6 6 0 cxy 6 4 @ @y

0

7 8 9 7 0> @ 7  > > = < > 7 u 0 @y 7 þ ¼ S p u þ ez 7 e > > 7 v > ; : z> 0 7 0 7 @ 5 @x

(2) Constitutive 8 equations: 9 2 rx > 1 > > > = < 6v E ry 6 ¼ Plane stress: rz > > ð1  v 2 Þ 4 0 > > ; : s 0 8 xy 9 2 r ð 1  v Þ > x > > > = E6 < ry v ¼ 6 Plane strain: rz > > d4 v > > ; : sxy 0 (3) Equilibrium equations:

9 38 ex > 0 0 > > > 7< ey = 0 0 7 5> ez > 0 0 > > ; : cxy 0 ð1  v Þ=2 9 38 ex > v v 0 > > > 7< ey = ð1  v Þ v 0 7 5> ez > v ð1  v Þ 0 > > ; : cxy 0 0 ð1  2v Þ=2

v 1 0 0

@rx @syx þ þ bx ¼ 0 @x @y @sxy @ry þ þ by ¼ 0 @x @y

2.8.2 Using the formulation presented in this chapter, provide the basic variables and equations of axial deformation for the one-dimensional elastic cantilever beam shown in figure 2.5. The coordinates of the two ends are x = a and x = b, and the axial uniform load is bx. Solutions: 1. Basic variables: (1) Displacement component:ux (2) Strain component: ex (3) Stress component: rx

Appendix E. Exercise Solutions

192

Basic equations: @u @x (2) Constitutive equation: rx ¼ Eex

(1) Geometric equation: ex ¼

(3) Equilibrium differential equation:

@rx þ bx ¼ 0 @x

Chapter 3 3.6.1 Describe the implementation steps for the weak form of equivalent integration Solutions: A weak form for a set of differential equations is obtained through the following steps: (1) (2) (3) (4)

Multiply each equation by an appropriate arbitrary function. Integrate this product over the space domain of the problem. Use integration by parts to reduce the order of the derivatives to a minimum. Introduce boundary conditions if possible.

3.6.2 Using the Galerkin method, for example 3.1 with EðxÞ ¼ E  x; bxðxÞ ¼ bx  x, compute the global stiffness matrix K and load matrix F. Solutions: The conditions of these problems are  10x for 0\x\5 Loading: bx ¼ 0 for 5\x\10 Boundary conditions: uð0Þ ¼ 0 and t x ð10Þ ¼ 25 Z 10 Z 10 @w @u Ex  w ðx Þbx xdx  w ð10Þ25 ¼ 0 Weak form: @x @x 0 0 Consider an approximate solution: uðxÞ  u^ ðxÞ ¼

N X n¼1

u^ ðxÞ ¼

^ un ðxÞan þ ub ðxÞ; wðxÞ  wðxÞ ¼ N  n X x n¼1

10

^ an ; wðxÞ ¼

N  m X x m¼1

10

N X m¼1

wm ðxÞwm

wm

The weak form is Z 10  m N Z 10  x n1  x m1 X E x mn an xdx ¼ bx xdx þ 25 100 10 10 10 0 n¼1 0 The coefficients of stiffness matrix and load matrix are Z 10  x m þ n2 E Emn mn xdx ¼ Kmn ¼ 100 10 m þn 0

Appendix E. Exercise Solutions

Z

5

fm ¼

 x m 10

0

193

10xdx þ 25 ¼

 1000 1 m þ 2 þ 25 mþ2 2

For example, m = n = 2, E = 1000, bx ¼ 10 the stiffness matrix and load matrix are given by: 9 8 200 >

   > = < k k12 500 667 f1 3 K ¼ 11 ¼ ;f ¼ ¼ 325 k21 k22 f2 667 1000 > > ; : 8 3.6.3 Using the finite element method, for example 3.2 with EðxÞ ¼ E  x; bxðxÞ ¼ bx  x, compute the global stiffness matrix K and load matrix f. Solutions: We divided the domain into four equal finite elements, so the length of the element is he ¼ x2e  x1e ¼ 2:5;

x 0 ¼ x  xie

x0 he The derivatives of shape functions are

i ¼ 1; 2; 3; 4

N2 ð x 0 Þ ¼

The shape functions are N1 ðx 0 Þ ¼ 1 

dN1 dN2 1 1 ¼ 0:4; ¼ ¼ ¼ 0 he 2:5 dx dx

x0 . he

dN2 dN2 1 1 ¼ 0:4 ¼ ¼ ¼ 0 he 2:5 dx dx

The element stiffness matrix and load matrix are 9 8 dN1   Z he > Z 2:5  =
< 0 > 0:4 dN1 dN2 dx 0 Ex ½k e ¼ ¼ dx Ex f0:4 0:4 dx 0 dx 0 0 > 0 ; : dN2 > dx 0  Z 2:5
0:16 0:16 ¼ Ex dx 0 0:16 0:16 0

fe ¼

Z

he

0





N1 b xdx 0 ¼ N2 x

Z

2:5 0

0:4 gdx 0

9 8 x0 > > =

> ; : x 2:5

(1) Element stiffness matrix Element 1: Z K ¼ 1

0

2:5

Ex

0






 E 1 1 500 0:16 0:16 0 ¼ dx ¼ 500 0:16 0:16 2 1 1

500 500

Element 2:


 Z 2:5 3E 1 1 1500 0:16 0:16 K2 ¼ ¼ dx 0 ¼ E ðx 0 þ 2:5Þ 1500 0:16 0:16 2 1 1 0



1500 1500



Appendix E. Exercise Solutions

194

Element 3: Z 3 K ¼

2:5

0


 5E 1 0:16 0 dx ¼ 0:16 2 1




0:16 E ðx þ 5Þ 0:16 0


 1 2500 2500 ¼ 1 2500 2500

Element 4:


 Z 2:5 7E 1 1 3500 0:16 0:16 4 0 0 K ¼ ¼ E ðx þ 7:5Þ dx ¼ 3500 0:16 0:16 2 1 1 0

3500 3500



(2) Load matrix Element 1: Z

2:5

f1 ¼ 0

Element 2: Z

2:5

f2 ¼ 0

Element 3:

9 8 x0 > >   =

> ; : x 2:5

9 8 x0 > >   =

> ; : x 2:5

9 8 x0 > >   = < 2:5 1 2:5 bx ðx 0 þ 5Þdx 0 ¼ 0 f3 ¼ 0 0 > > 0 ; : x 2:5 Z

Element 4: Z f4 ¼ 0

2:5

9 8 x0 > >   =

> ; : x 2:5

(3) Global stiffness matrix and load matrix Element location vector: k1 ¼ f 1 2 gT ; k2 ¼ f 2

3 g T ; k3 ¼ f 3

4 gT ; k4 ¼ f 4

5 gT

The element stiffness matrix and load matrix can be integrated into the global stiffness matrix and load matrix:

Appendix E. Exercise Solutions

1

2

2 1 500 26 6 500 K ¼ 36 6 0 44 0 5 0 2

500 6 500 6 ¼6 6 0 4 0 0

195

500 500 þ 1500 1500 0 0

3

4

0 0 1500 0 1500 þ 2500 2500 2500 2500 þ 3500 0 3500

500 0 2000 1500 1500 4000 0 2500 0 0

0 0 2500 6000 3500

5 3 0 0 7 7 0 7 7 3500 5 3500

3 0 0 7 7 0 7 7 3500 5 3500

9 8 9 8 10:42 > 1> 10:42 > > > > > > > > > > 2> = < 62:5 > < 20:83 þ 41:67 > = > ¼ 52:08 f ¼ 3 52:08 þ 0 > > > > > > 4> 0þ0 > > 0 > > > > > > ; : ; > : 0 5 0 3.6.4 Summarise the functions of the element location vector Solutions: Omission.

Chapter 4 4.6.1 One-dimensional element of line with two nodes (1) Solutions: Omission. (2) Using the equations of the shape functions, compute the following derivatives: Solutions: The shape functions: N1 ðnÞ ¼ l11 ðnÞ ¼

n  n2 1n n  n1 1þn ; N2 ðnÞ ¼ l21 ðnÞ ¼ ¼ ¼ 2 2 n1  n2 n 2  n1

The derivatives: @N1 1 @N2 1 @ 2 N1 @ 2 N2 ¼ ; ¼ ; ¼ 0; ¼0 2 @n 2 @n2 @n @n2 4.6.2 Two-dimensional element of triangle with three nodes. (1) Solutions: Omission. (2) Using the equations of the shape functions, compute the following derivatives:

Appendix E. Exercise Solutions

196

Solutions: The shape functions: Na ðx; y Þ ¼

1 ðaa þ ba x þ ca y Þ; 2D

a ¼ 1; 2; 3

The derivatives: @Na ba @Na ca @ 2 N a @ 2 Na @ 2 Na ¼ ; ¼ ; ¼ 0; ¼ 0; ¼0 @x 2D @y 2D @xy @x 2 @y 2 4.6.3 Two-dimensional element of rectangle with four nodes (1) Solutions: Omission. (2) Using the equations of the shape functions, compute the following derivatives: Solutions: The shape functions: 1 N1 ¼ ð1  nÞð1  gÞ; 4 1 N3 ¼ ð1 þ nÞð1 þ gÞ; 4

1 N2 ¼ ð1 þ nÞð1  gÞ 4 1 N4 ¼ ð 1  nÞ ð 1 þ gÞ 4

The derivatives: @N1 1 @N2 1 @N3 1 @N1 1 ¼ ðg  1Þ; ¼ ð1  gÞ; ¼ ðg þ 1Þ; ¼  ð g þ 1Þ 4 4 4 4 @n @n @n @n @N1 1 @N2 1 @N3 1 @N4 1 ¼ ðn  1Þ; ¼  ðn þ 1Þ; ¼ ðn þ 1Þ; ¼ ð 1  nÞ 4 4 4 4 @g @g @g @g @ 2 N1 1 @ 2 N2 1 @ 2 N 3 1 @ 2 N4 1 ¼ ; ¼ ; ¼ ; ¼ 4 @ng 4 @ng 4 @ng 4 @ng @ 2 N1 @ 2 N2 @ 2 N3 @ 2 N4 ¼ 0; ¼ 0; ¼ 0; ¼0 @n2 @n2 @n2 @n2 @ 2 N1 @ 2 N2 @ 2 N3 @ 2 N4 ¼ 0; ¼ 0; ¼ 0; ¼0 2 2 2 @g @g @g @g2 4.6.4 Three-dimensional element of tetrahedron with four nodes. (1) Solutions: Omission. (2) Using the equations of the shape functions, compute the following derivatives: Solutions: The shape function: Na ðx; y; z Þ ¼

1 ðaa þ ba x þ ca y þ da z Þ; 6V

a ¼ 1; 2; 3; 4

Appendix E. Exercise Solutions

197

The derivatives: @Na ba @Na ca @Na d a @ 2 Na @ 2 Na @ 2 Na ¼ ; ¼ ; ¼ ; ¼ 0; ¼ 0; ¼ 0; @x 6V @y 6V @z 6V @xy @yz @zx @ 3 Na @ 2 Na @ 2 Na @ 2 Na ¼ 0; ¼ 0; ¼ 0; ¼0 @xyz @x 2 @y 2 @z 2 4.6.5 Three-dimensional element of hexahedron with eight nodes. (1) Solutions: Omission. (2) Using the equations of the shape functions, compute the following derivatives: Solutions: The shape function: 1 Na ðn; g; fÞ ¼ ð1 þ na nÞ ð1 þ ga gÞ ð1 þ fa fÞ 8 The derivatives: @Na @n @Na @f 2 @ Na @gf @ 2 Na @n2

1 @Na 1 ¼ na ð1 þ ga gÞð1 þ fa fÞ; ¼ ga ð1 þ na nÞð1 þ fa fÞ; 8 8 @g 1 @ 2 Na 1 ¼ fa ð1 þ na nÞð1 þ ga gÞ; ¼ na ga ð1 þ fa fÞ; 8 8 @ng 2 1 @ Na 1 ¼ ga fa ð1 þ na nÞ; ¼ na fa ð1 þ ga gÞ; 8 8 @nf 2 2 @ Na @ Na @ 3 Na 1 ¼ n g f ¼ 0; ¼ 0; ¼ 0; @g2 @ngf 8 a a a @f2

4.6.6 Summarise the functions of Lagrange interpolation for constructing one-, two-, and three-dimensional elements. Solutions: Omission.

Chapter 5 5.3.1 Based on the parametric mapping defined by the relation between the coordinate interpolation and the dependent-variable interpolation for a one-dimensional Lagrange element, describe sub-parametric, isoparametric, and super-parametric interpolation. Solutions: The types of parametric mapping that can be considered in an analysis are defined by the relation between the coordinate interpolation and the dependent-variable interpolation. We denote the mapping by: xe ¼

n X a¼1

Na ðnÞ xae ¼ N1 ðnÞ x1e þ N2 ðnÞ x2e þ    þ Nn ðnÞ xne ; 1  n  1

ð5:2aÞ

Appendix E. Exercise Solutions

198

u^ e ¼

m X a¼1

e Na ðnÞ u^ae ¼ N1 ðnÞ u^1e þ N2 ðnÞ u^2e þ    þ Nm ðnÞ u^m ; 1  n  1

ð5:2bÞ

Once the shape functions are available in terms of the parent coordinates, we may immediately use the concept of parametric mapping. We have the following three cases: (1) Sub-parametric interpolation: The order n of the interpolation for x is lower than that order m for u. (2) Isoparametric interpolation: The order n of the interpolation for x is the same as that order m for u. (3) Super-parametric interpolation: The order n of the interpolation for x is higher than that order m for u. 5.3.2 Consider the two-dimensional rectangle isoparametric element as shown in figure 5.6. (a) Write the expression for an isoparametric mapping of coordinates in this element. (b) Determine the location of the local coordinates ξ and η, which define the centroid of this element. (c) Compute the expression for the Jacobian matrix J of this element, and evaluate the Jacobian at the centroid. (d) Compute the derivatives of the shape function N3 at the centroid. Solutions: (a) Write the expression for an isoparametric mapping of coordinates in this element. The shape functions are 1 N1 ¼ ð1  nÞð1  gÞ; 4 1 N3 ¼ ð1 þ nÞð1 þ gÞ; 4

1 N2 ¼ ð1 þ nÞð1  gÞ 4 1 N4 ¼ ð 1  nÞ ð 1 þ gÞ 4

The mapping between the global and the local coordinate system is expressed as: x ðn; gÞ ¼

X a

Na ðn; gÞxa ¼

X ð 1 þ n a nÞ ð 1 þ ga gÞ a

4

xa

ð1  nÞð1  gÞ ð1 þ nÞð1  gÞ ð1 þ nÞð1 þ gÞ  0þ  6þ 3 4 4 4 ð1  nÞð1 þ gÞ þ 0 4 3ð3  gÞð1 þ nÞ ¼ 4 ¼

Appendix E. Exercise Solutions

y ðn; gÞ ¼

X a

199

Na ðn; gÞya ¼

X ð1 þ na nÞð1 þ g gÞ a ya 4 a

ð1  nÞð1  gÞ ð1 þ nÞð1  gÞ ð 1 þ nÞ ð 1 þ gÞ  0þ  0þ 3 4 4 4 ð1  nÞð1 þ gÞ þ 3 4 3ð 1 þ gÞ ¼ 2 ¼

(b) Determine the location of the local coordinates ξ and η, which define the centroid of this element. The centre of the irregular quadrilateral is located at: x¼

0þ6þ3þ0 9 ¼ 4 4

y¼

0þ0þ3þ3 3 ¼ 4 2

Using the mapping relationship, the local coordinates of the centre are. 8  9 3 > ( > < ¼ 1 þ n ð 3  gÞ n¼0 4 4 and > 3 3 g¼0 > : ¼ ð 1 þ gÞ 2 2 (c) Compute the expression for the Jacobian matrix J of this element and evaluate the Jacobian matrix at the centroid. For this two-dimensional problem, the Jacobian is.  2 3 2

3 @x @y 9 1 1  g 0 6 @n@n 7 6 4 7 3 7 6 7 J ¼6 4 @x @y 5 ¼ 4 3 35  ð 1 þ nÞ @g@g 4 2 The Jacobian at the centroid is  2

9 1 1  g 64 3 J ¼6 4 3   1þn 4

3 2 9 07 6 7¼4 4 3 35  4 2

3 0 3 2

7 5

Appendix E. Exercise Solutions

200

(d) Compute the derivatives of the shape function N3 at the centroid. The derivatives of the shape function N3 at the centroid are 9 8 9 2 8 38 1 þ g 9 2 38 9 8 1 9 @N3 > 4 4 @N3 > 1> > > > > > > > > > > > > > > > > 0 0 = > = < 4 = 69

< @n = 6 9

< @x = 7 7 1 6 6 7 7 ¼ ¼ ¼4 ¼J 5 > 4 2 2 5> > > > > > @N > @N3 > 2 2 > > > > > > ; ; > ; ; > :1þn> :1> : 3> :2> > ; : @g @y 9 9 3 9 3 4 4 using the inverse of the Jacobian matrix 2 3 2 4 3 0 6 7 6 8 2 7 ¼ 69 J 1 ¼ 6 27 4 3 9 5 4 2 9 4 4

3 0 2 3

7 7 5

5.3.3 Compute stiffness K 13 for one-dimensional element with three nodes using the following manners (a) Numerical integration by one, two, and three integration points, respectively. (b) Direct integration. Solutions: (a) Numerical integration The shape functions and Jacobian of one-dimensional element with three nodes are 1 N1 ¼ nðn  1Þ; 2 jðnÞ ¼

@x ¼ @n

n

1 N2 ¼ nðn þ 1Þ; 2

N3 ¼ 1n2



   1 e 1 e 1 x1 þ n þ x2  2nx3e ¼ he þ n x1e þ x2e  2x3e 2 2 2

For the case where the coordinate for node 3 is centered between nodes 1 and 2, the derivatives are linear in ξ, and the Jacobian is constant 12 he . The stiffness for the three-node element involves the integral of 9 8 @N1 > > > > > > > > > @x > > > > >  Z he < =
@N2 @N1 @N2 @N3 e Ee dx 0 K ¼ > @x @x @x @x 0 > > > > > > > > > > > @N 3 > > ; : @x 3 2



 



 1 2 1 1 1 n n nþ n ð2nÞ 7 6 2 2 2 2 7 6 7 6 Z 



2 



7 2Ee 1 6 1 1 1 1 7 dn 6 ¼ n þ n þ ð2nÞ n þ n  7 6 he 1 6 2 2 2 2 7 7 6



 5 4 1 1 ð2nÞ nþ ð2nÞ ð2nÞ2 n 2 2

Appendix E. Exercise Solutions

201

Therefore, the stiffness parameter to be obtained is K13

2Ee ¼ he

 1 n  ð2nÞdn 2 1

Z

1

Using a one-point quadrature formula, we can obtain that 2Ee ¼ he

K13



 1 2Ee 1 n  ð2nÞdn ¼ 0  ð2  0Þ  2 ¼ 0 2 2 he 1

Z

1

Using a two-point quadrature formula, we can obtain that K13 ¼

2Ee he

2Ee ¼ he ¼

 1 n  ð2nÞdn 2 1

Z

1




1 1 pffiffiffi  3 2





   2 1 1 2 pffiffiffi  1  pffiffiffi  1 þ  pffiffiffi  3 3 2 3

8Ee 3he

Using a three-point quadrature formula, we can obtain that K13

2Ee ¼ he 2Ee ¼ he ¼

Z

1 1






 1 n  ð2nÞdn 2





  pffiffiffiffiffiffiffi 1  pffiffiffiffiffiffiffi 5 pffiffiffiffiffiffiffi 1  pffiffiffiffiffiffiffi 5 1 8 þ  0:6  þ 0  ð2  0Þ 0:6  2 0:6 2 0:6 2 9 2 9 2 9

8Ee 3he

(b) Direct integration K13

2Ee ¼ he

Z

1



1

 1 8Ee n  ð2nÞdn ¼  2 3he

5.3.4 For two-dimensional rectangle element with four nodes by numerical integration and direct integration, respectively, compute the following integration Z

1 1

Z

1

1

ðn þ gÞdndg

Appendix E. Exercise Solutions

202

Solutions: (1) Numerical integration Z 1Z 1 2 X 2 X ðn þ gÞdndg ¼ ðnm þ gn Þwm wn 1

1

m¼1 n¼1

¼ ðn1 þ g1 Þw1 w1 þ ðn1 þ g2 Þw1 w2 þ ðn2 þ g1 Þw2 w1 þ ðn2 þ g2 Þw2 w2





 1 1 1 1 1 1 ¼  pffiffiffi  pffiffiffi  1  1 þ  pffiffiffi þ pffiffiffi  1  1 þ pffiffiffi  pffiffiffi 3 3 3 3 3 3

 1 1  1  1 þ pffiffiffi þ pffiffiffi  1  1 3 3 ¼0

(2) Direct integration Z 1Z 1 1


2 1 Z 1 n þ ng dg ¼ ðn þ gÞdndg ¼ 2gdg ¼ 0 1 1 2 1 1 Z

1

Chapter 6 6.5.1 Derive the weak form for general elasticity problems by following the steps below: (a) (b) (c) (d)

Multiply each equation by an appropriate arbitrary function. Integrate this product over the space domain of the problem. Use integration by parts to reduce the order of the derivatives to a minimum. Introduce boundary conditions if possible.

Solutions: Omission. 6.5.2 In example 6.1, compute the value of the last coefficient in the stiffness matrix K 133 of element 1:  Z 
E @N3 @N3 @N3 @N3 ð1  v Þ þ ð1  2v Þ=2 dX d @y @y @x @x Xe Solutions: The derivatives of shape functions (as shown in example 6.1) are 9 8 8 9 @N3 > 1 > > > > > > > > > ð 1 þ g Þ = < @x = < 30 ¼ > > > > > ; : 1 ð 1 þ nÞ > > > > @N3 > ; : @y 20 and the determinant of the Jacobian matrix is j ¼ formulas into the integral equation, we can get

75 . By substituting these 2

Appendix E. Exercise Solutions

203


 E @N3 @N3 @N3 @N3 ð 1  mÞ þ ð1  2mÞ=2 dX d @y @y @x @x Xe "  #

2 Z E 1 2 1 2 2 ¼ ð 1 þ nÞ ð 1  m Þ þ ð1 þ gÞ ð1  2mÞ=2 jdndg 20 30 Xe ð1 þ mÞð1  2mÞ "  #

2 Z 106 1 2 1 75 2 2 dndg ð1 þ nÞ 0:7 þ ð1 þ gÞ 0:2 ¼ 20 30 2 Xe ð1 þ 0:3Þð1  2  0:3Þ Z h i ¼ 1923076:92 0:065625  ð1 þ nÞ2 þ 0:008333  ð1 þ gÞ2 dndg Xe 9 8"



 # > > 1 2 1 2 > > > 0:065625  1  pffiffiffi þ 0:008333  1  pffiffiffi  11þ > > > > > > > 3 3 > > > > > > " # > >

2

2 > > > > 1 1 > > > > > > p ffiffi ffi p ffiffi ffi þ 0:008333  1 þ  1  1 þ 0:065625  1  > > = < 3 3 # ¼ 1923076:92  "



 > > > > 1 2 1 2 > > > p ffiffi ffi p ffiffi ffi þ 0:008333  1   1  1þ > 0:065625  1 þ > > > > > > 3 3 > > > >" > > # > >

 > > 2  > > 2 1 > > 1 > > > > 11 ; : 0:065625  1 þ pffiffiffi þ 0:008333  1 þ pffiffi3 3

Z

¼ 758543:6

Chapter 7 7.2.1 Using the LU decomposition of the stiffness matrix K, compute the displacements: 8 9 2 3 6 2 1