Basic Finite Element
Method as Applied to
Injury Biomechanics
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Basic Finite Element
Method as Applied to
Injury Biomechanics
Edited by
King-Hay Yang
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Contents
List of Contributors............................................................................... xvii
Foreword.............................................................................................. xix
Preface ................................................................................................ xxi
PART 1 BASIC FINITE ELEMENT METHOD AND ANALYSIS
AS APPLIED TO INJURY BIOMECHANICS
CHAPTER 1 Introduction ............................................................ 3
King H. Yang
1.1 Finite Element Method and Analysis ..................................... 3
1.2 Calculation of Strain and Stress From the FE Model ................ 7
1.2.1 Average Strain and Point Strain ....................................7
1.2.2 Normal and Shear Strain............................................ 12
1.2.3 Calculation of Stress................................................. 16
1.3 Sample Matrix Structural Analysis ...................................... 21
1.3.1 Element Stiffness Matrix of a Linear Spring ................. 21
1.3.2 Element Stiffness Matrix of a Linear Spring Not
in Line With the X-axis............................................. 23
1.3.3 Element Stiffness Matrix of a Homogeneous Linear
Elastic Bar .............................................................. 27
1.3.4 Global Stiffness Matrix of Multiple Inline Linear
Springs or Bars ........................................................ 29
1.3.5 Global Stiffness Matrix of a Simple Biomechanics
Problem.................................................................. 32
1.3.6 Global Stiffness Matrix of a Simple Truss Bridge .......... 36
1.3.7 Gaussian or Gauss Elimination ................................... 39
1.4 From MSA to a Finite Element Model................................. 44
Exercises............................................................................... 46
References............................................................................. 49
CHAPTER 2 Meshing, Element Types, and Element Shape
Functions...............................................................51
King H. Yang
2.1 Structure Idealization and Discretization .............................. 51
2.2 Node.............................................................................. 55
2.3 Element .......................................................................... 56
2.3.1 Simplest Element Types ............................................ 58
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Contents
2.3.2 1D Element Type ..................................................... 59
2.3.3 2D Element Type ..................................................... 61
2.3.4 3D Element Type ..................................................... 67
2.4 Formation of Finite Element Mesh ...................................... 67
2.5 Element Shape Functions and [B] Matrix ............................. 69
2.5.1 1D, 2-Node Element Shape Functions.......................... 69
2.5.2 2D, 3-Node Linear Triangular Element ........................ 79
2.5.3 4-Node Rectangular Bilinear Plane Element With
Edges Parallel to the Coordinate Axes ......................... 87
2.5.4 2D, 4-Node Plate Element Shape Functions With
Edges Parallel to the Coordinate Axes ......................... 96
2.5.5 3D, 4-Node Shell Element ....................................... 104
2.5.6 3D, 8-Node Trilinear Element Shape Functions ........... 105
Exercises..............................................................................108
References............................................................................109
CHAPTER 3 Isoparametric Formulation and Mesh Quality ......... 111
King H. Yang
3.1 Introduction....................................................................111
3.2 Natural Coordinate System ...............................................111
3.3 Isoparametric Formulation of 1D Elements..........................112
3.3.1 1D Linear Bar Element Isoparametric Shape
Functions.............................................................. 112
3.3.2 1D Beam Element Isoparametric Shape Functions ....... 115
3.4 Isoparametric Formulation of 2D Element ...........................121
3.4.1 Isoparametric Formulation of 2D Triangular
Element................................................................ 121
3.4.2 Isoparametric Formulation of 2D Bilinear
Element................................................................ 123
3.4.3 Determine the [B] Matrix Based on Isoparametric
Formulation........................................................... 125
3.5 Isoparametric Formulation of 3D Element ...........................128
3.5.1 Constant-Strain Tetrahedral Element.......................... 128
3.5.2 Trilinear Hexahedral Element................................... 128
3.6 Transfer Mapping Function for 2D Element.........................131
3.7 Jacobian Matrix and Determinant of Jacobian Matrix ............133
3.8 Element Quality (Jacobian, Warpage, Aspect Ratio, etc.) .......137
3.8.1 Jacobian and Normalized Jacobian ............................ 138
3.8.2 Internal and Skew Angles ........................................ 141
3.8.3 Warpage ............................................................... 142
Contents
3.8.4 Aspect Ratio.......................................................... 142
3.8.5 Distortion.............................................................. 142
3.8.6 Stretch.................................................................. 143
3.8.7 Generation of High-Quality Mesh ............................. 143
3.9 Saint-Venant Principle and Patch Test .................................145
Exercises..............................................................................147
References............................................................................149
CHAPTER 4 Element Stiffness Matrix ...................................... 151
King H. Yang
4.1 Introduction....................................................................151
4.2 Direct Method ................................................................152
4.2.1 Direct Formation of Structure Stiffness Matrix ............ 152
4.2.2 Direct Method for a 2-Node Beam Element ................ 157
4.3 Strong Formulation..........................................................163
4.4 Weak Formulation ...........................................................165
4.4.1 Variational Method ................................................. 165
4.4.2 Weighted Residuals Method..................................... 179
4.4.3 Section Summary ................................................... 183
4.5 Derive Element Stiffness Matrix From Shape Functions.........183
4.5.1 Gauss Quadrature................................................... 184
4.5.2 1D Element Stiffness Matrix Using Gauss
Quadrature ............................................................ 188
4.5.3 Gauss Integration Points for 2D and 3D Elements........ 190
4.5.4 2D and 3D Element Stiffness Matrices Using Gauss
Quadrature ............................................................ 192
4.5.5 Full and Reduced Integration.................................... 197
4.5.6 Zero-Energy Mode ................................................. 198
4.6 Method of Superposition ..................................................201
4.6.1 Stiffness Matrix of a 2D Frame Element .................... 201
4.6.2 Stiffness Matrix of a 2-Node, Pseudo-3D Frame
Element................................................................ 202
4.7 Coordinate Transformation................................................203
4.7.1 2D Transformation of Vector.................................... 204
4.7.2 2D Transformation of Stiffness Matrix....................... 206
4.7.3 2D Transformation of Inclined Boundary
Condition.............................................................. 210
4.7.4 3D Rotation........................................................... 215
4.8 Chapter Summary Using a Numerical Example ....................216
Exercises..............................................................................227
References............................................................................230
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Contents
CHAPTER 5 Material Laws and Properties ............................... 231
King H. Yang
5.1 Material Laws.................................................................231
5.1.1 Linear Elastic Material............................................ 232
5.1.2 ElasticePlastic Material .......................................... 235
5.1.3 Hyperelastic Material.............................................. 236
5.1.4 Viscoelastic Material............................................... 237
5.1.5 Orthotropic Material ............................................... 239
5.1.6 Foam Material ....................................................... 240
5.1.7 Material Defined by Equation of State ....................... 241
5.2 Material Test Strategy and Associated Property ....................241
5.2.1 Experimental Types for Biological Tissue Testing........ 242
5.2.2 Reverse Engineering Methodology ............................ 247
5.2.3 List of Common Material Properties of Biological
Tissues ................................................................. 249
5.3 Building Laboratory-Specific Material Property Library.........251
Exercises..............................................................................254
References............................................................................255
CHAPTER 6 Prescribing Boundary and Loading Conditions
to Corresponding Nodes ....................................... 257
King H. Yang
6.1 Essential and Natural Boundary Conditions .........................257
6.2 Nodal Constraint and Prescribed Displacement.....................258
6.2.1 Nodal Constraint .................................................... 258
6.2.2 Prescribed Displacement.......................................... 259
6.2.3 Penalty Method...................................................... 261
6.2.4 Symmetric Finite Element Modeling Through Nodal
Constraint ............................................................. 263
6.3 Natural Boundary/Loading Conditions ................................264
6.3.1 Concentrated Loads ................................................ 265
6.3.2 Distributed Load .................................................... 271
6.3.3 Initial Velocity and Acceleration ............................... 277
Exercises..............................................................................277
CHAPTER 7 Stepping Through Finite Element Analysis ............. 281
King H. Yang
7.1 Introduction....................................................................281
7.2 Iterative Procedures Versus Gaussian Elimination .................281
7.2.1 Jacobi or Simultaneous Displacement Method............. 283
7.2.2 GausseSeidel or Successive Displacement Method ...... 284
Contents
7.3 Verification and Validation................................................288
7.3.1 Historical Aspects .................................................. 288
7.3.2 Verification............................................................ 290
7.3.3 Validation ............................................................. 292
7.3.4 Quantifying the Extent of Validation.......................... 295
7.3.5 Uncertainty Qualification......................................... 301
7.4 Response Variables..........................................................302
7.4.1 Principal Stress ...................................................... 302
7.4.2 Maximum Shear Stress............................................ 305
7.4.3 Von Mises Stress.................................................... 305
Exercises..............................................................................306
References............................................................................307
CHAPTER 8 Modal and Transient Dynamic Analysis ................. 309
King H. Yang
8.1 Introduction....................................................................309
8.2 Element Mass Matrix.......................................................310
8.2.1 Consistent Mass Matrix........................................... 311
8.2.2 Lumped Mass Matrix.............................................. 318
8.3 Modal Analysis...............................................................320
8.3.1 Free Vibration of a MasseSpring System ................... 321
8.3.2 Forced Vibration .................................................... 328
8.3.3 Numerical Methods for Finding Eigenvalues and
Eigenvectors.......................................................... 330
8.3.4 Section Remarks .................................................... 354
8.4 Damping........................................................................355
8.4.1 Coulomb Damping ................................................. 355
8.4.2 Viscous Damping ................................................... 357
8.5 Direct Integration Methods ...............................................362
8.5.1 Central Difference Method....................................... 364
8.5.2 The Newmark Method ............................................ 370
8.6 Implicit and Explicit Solvers .............................................376
8.6.1 Implicit Solver....................................................... 377
8.6.2 Explicit Solver....................................................... 378
8.6.3 Using FE Solvers ................................................... 378
Exercises..............................................................................380
References............................................................................381
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Contents
PART 2 MODELING HUMAN BODY FOR INJURY
BIOMECHANICS ANALYSIS
Introduction .............................................................. 385
CHAPTER 9 Developing FE Human Models From
Medical Images ................................................... 389
Anil Kalra
9.1 Introduction....................................................................389
9.2 Biomedical Images for Finite Element Mesh Development.....390
9.2.1 X-ray Imaging ....................................................... 390
9.2.2 Computed Tomography Imaging ............................... 391
9.2.3 Magnetic Resonance Imaging................................... 392
9.2.4 Positron Emission Tomography................................. 392
9.2.5 Ultrasound Imaging ................................................ 392
9.3 Physics Behind 3D Segmentation of Medical Images.............394
9.4 Meshing Human Body .....................................................395
9.4.1 Pre-mesh Segment.................................................. 396
9.4.2 Finite Element Mesh Development............................ 402
9.4.3 Post-mesh Segment................................................. 407
9.5 Exemplary Whole Body FE Mesh Development ...................412
Acknowledgments..................................................................413
References............................................................................413
CHAPTER 10 Parametric Human Modeling .............................. 417
Jingwen Hu
10.1 Introduction..................................................................417
10.1.1 What Is a Parametric Human Model?......................417
10.1.2 Why Are Parametric Human Models Needed?..........417
10.1.3 Need for Parametric Human Model.........................421
10.2 Current State-of-The-Art FE, Whole-Body, Human
Models ........................................................................421
10.3 How to Build a Parametric Human Model .........................423
10.3.1 Method Overview ................................................423
10.3.2 Statistical Models of Human Geometry ...................425
10.3.3 Mesh Morphing...................................................430
10.3.4 Example of Parametric, Whole-Body, Human
Model................................................................434
10.3.5 Tissue Material Properties for a Parametric,
Human Model.....................................................436
Contents
10.4 How to Validate a Parametric Human Model......................436
10.5 Chapter Conclusion........................................................438
Acknowledgments..................................................................438
References............................................................................438
CHAPTER 11
Modeling Passive and Active Muscles ................ 447
Masami Iwamoto
11.1 Introduction..................................................................447
11.2 Methods for Modeling Passive Muscle ..............................450
11.3 Methods for Modeling Muscular Activation .......................453
11.3.1 Estimation of Muscle Activation Based
on EMG Data .....................................................453
11.3.2 Estimation of Muscle Activation Using PID
Controller...........................................................456
11.3.3 Estimation of Muscle Activation Using
Reinforcement Learning .......................................460
11.3.4 Discussion for Better Estimation of Muscle
Activation ..........................................................463
11.4 Application of Muscle Models.........................................463
11.5 Chapter Conclusion........................................................465
References............................................................................466
CHAPTER 12
Modeling the Head for Impact Scenarios ............ 469
Haojie Mao
12.1 Why Is Numerical Modeling of the Human Head
Essential?.....................................................................469
12.2 Introduction of Corresponding Anatomy............................470
12.2.1 Understanding human head Anatomy While
Keeping Head Biomechanics and Injury in Mind ......470
12.2.2 Anatomy of the Human Head ................................470
12.3 Injury Mechanism..........................................................480
12.3.1 Focusing on Brain Injury ......................................480
12.3.2 So, How Do These Linear and Rotational
Accelerations of the Head Affect the Brain?.............481
12.3.3 Anything Else Besides Linear and Rotational
Accelerations?.....................................................482
12.3.4 Instead of Accelerations, Describing Brain
Tissue-Level Response Is Key in Understanding
Injury Mechanisms ..............................................483
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Contents
12.4 Material Models............................................................484
12.4.1 Brain Material Models..........................................484
12.4.2 Skull, Flesh, and Scalp Material Models..................487
12.5 Material Properties ........................................................487
12.6 Test Data Available for Model Validation ..........................488
12.6.1 Brain Pressure.....................................................488
12.6.2 Brain Motion ......................................................491
12.6.3 Skull Response....................................................492
12.6.4 Facial Response...................................................493
12.7 Brief Overview of Human Head Models............................495
12.8 Discussion....................................................................497
12.9 Concluding Remarks......................................................499
References............................................................................499
CHAPTER 13 Modeling the Neck for Impact Scenarios ............ 503
Duane S. Cronin, Dilaver Singh, Donata Gierczycka,
Jeffery Barker, David Shen
13.1 Introduction..................................................................503
13.2 Anatomy of the Neck .....................................................504
13.3 Neck Anthropometrics....................................................508
13.3.1 Curvature of the Cervical Spine .............................509
13.3.2 Vertebral Geometry..............................................510
13.3.3 Intervertebral Disc and Facet Joints Dimensions .......510
13.3.4 External Measurements.........................................512
13.4 Neck Injury ..................................................................513
13.5 Material Models and Properties for Tissues........................515
13.5.1 Cortical and Cancellous Bone................................516
13.5.2 Ligaments ..........................................................516
13.5.3 Intervertebral Disc ...............................................517
13.5.4 Cartilage ............................................................518
13.5.5 Musculature........................................................519
13.6 Test Data for Computational Model Verification and
Validation.....................................................................520
13.7 Computational Neck Models ...........................................522
13.8 Closure........................................................................531
References............................................................................532
CHAPTER 14 Modeling the Thorax for Impact Scenarios .......... 539
King H. Yang, Barbara R. Presley
14.1 Introduction and Corresponding Anatomy..........................539
Contents
14.2 Injury Types and Mechanisms..........................................541
14.2.1 Bony Injuries ......................................................541
14.2.2 Heart and Lung Injuries........................................542
14.2.3 Great Vessel Injuries ............................................544
14.2.4 Soft Tissue Injuries ..............................................545
14.3 Factors Affecting the Thorax Modeling .............................545
14.3.1 Geometric Variations............................................545
14.3.2 Rib Angles of Males Versus Females ......................547
14.3.3 Bone Material Properties.......................................550
14.3.4 Aortic Geometric Variations ..................................557
14.4 FE Thorax Models.........................................................559
14.4.1 FE Thorax Model Before 2005 ..............................559
14.4.2 FE Thorax Model After 2005 ................................563
14.4.3 Selection of Rib Material Properties .......................574
14.5 Concluding Remarks......................................................574
Acknowledgments..................................................................575
References............................................................................576
CHAPTER 15
Modeling the Lower Torso for Impact
Scenarios .......................................................... 585
King H. Yang
15.1 Introduction and Corresponding Anatomy..........................585
15.2 Injury Severity and Experimentally Derived Material
Properties.....................................................................587
15.3 Computational Abdomen Models .....................................595
15.4 Test Data Available for Model Validation ..........................598
15.4.1 Frontal Rigid Bar Test ..........................................598
15.4.2 Frontal Rigid Bar, Seat Belt, and Distributed
Load Test ...........................................................599
15.4.3 Frontal Impacts for the Abdomen With a Simulated
Steering Wheel....................................................600
15.4.4 Lateral/Oblique Pendulum Impact ..........................601
15.4.5 Lateral Drop Tests ...............................................601
15.4.6 Lateral Sled Tests ................................................601
15.4.7 Pretensioner Tests................................................603
15.5 Concluding Remarks......................................................604
Acknowledgments..................................................................604
References............................................................................605
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Contents
CHAPTER 16 Modeling the Spine and Upper and Lower
Extremities for Impact Scenarios ........................ 609
King H. Yang
16.1 Introduction and Corresponding Anatomy..........................609
16.1.1 Anatomy of the Spine...........................................611
16.1.2 Anatomy of the Upper Extremity ...........................613
16.1.3 Anatomy of the Pelvis and Lower Extremity ............614
16.2 Injury Types .................................................................616
16.2.1 Spine and Pelvis Injuries.......................................616
16.2.2 Injuries Near Joints in the Upper and Lower
Extremities .........................................................619
16.3 Factors May Affect Spine and Extremity Modeling .............625
16.3.1 Cortical and Trabecular Bones in Vertebrae..............625
16.3.2 Spinal Angle and Facet-Joint Orientation.................626
16.3.3 Strain-Rate Effects on Long Bones .........................626
16.4 Finite Element Spine and Extremity Models.......................628
16.4.1 Spine Models......................................................628
16.4.2 Upper Extremity Models.......................................629
16.4.3 Lower Extremity Models ......................................633
16.5 Concluding Remarks......................................................644
Acknowledgments..................................................................645
References............................................................................645
CHAPTER 17 Modeling of Vulnerable Subjects ........................ 655
Xin Jin
17.1 Introduction and Background...........................................655
17.2 Modeling of Pediatric Subjects ........................................656
17.2.1 Introduction........................................................656
17.2.2 Geometry and Composition Characteristics ..............657
17.2.3 Mesh Generation .................................................662
17.2.4 Determination of Material Properties ......................663
17.2.5 Model Validation.................................................663
17.2.6 Summary ...........................................................663
17.3 Modeling of Elderly Female Subjects................................668
17.3.1 Introduction........................................................668
17.3.2 Geometry and Composition Characteristics ..............670
17.3.3 Mesh Generation and Material Property
Determination .....................................................672
17.3.4 Model Validation.................................................677
17.3.5 Summary ...........................................................681
17.4 Chapter Conclusion........................................................682
References............................................................................682
Contents
CHAPTER 18
Modeling of Blast Wave and Its Effect on
the Human/Animal Body ..................................... 689
Feng Zhu
18.1 Basic Blast Physics........................................................689
18.2 Blast Wave Modeling Strategies in the Numerical
Simulations ..................................................................690
18.2.1 Defining the PulseeTime Curve Directly.................690
18.2.2 Defining Blast Loads Using Blast Pressure
Functions ...........................................................690
18.2.3 Modeling the Explosive as a Material .....................691
18.3 Simulations of Blast Wave Effect on the
BodydCase Studies ......................................................693
18.3.1 Case Study 1dSimulation of Blast Effect on the
Lower Extremity .................................................694
18.3.2 Case Study 2dSimulation of Blast Effect
on the Brain........................................................696
References............................................................................701
Concluding Remarks ..............................................................................703
Index...................................................................................................707
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List of Contributors
Jeffery Barker
The University of Waterloo, Waterloo, Ontario, Canada
Duane S. Cronin
The University of Waterloo, Waterloo, Ontario, Canada
Donata Gierczycka
The University of Waterloo, Waterloo, Ontario, Canada
Jingwen Hu
University of Michigan, Ann Arbor, Michigan, United States
Masami Iwamoto
Toyota Central Research & Development Laboratories, Inc., Nagakute-city, Japan
Xin Jin
Wayne State University, Detroit, Michigan, United States
Anil Kalra
Ford Motor Company, Dearborn, Michigan, United States
Haojie Mao
Western University, London, Ontario, Canada
Barbara R. Presley
Wayne State University, Detroit, Michigan, United States
David Shen
The University of Waterloo, Waterloo, Ontario, Canada
Dilaver Singh
The University of Waterloo, Waterloo, Ontario, Canada
King H. Yang
Wayne State University, Detroit, Michigan, United States
Feng Zhu
Embry-Riddle Aeronautical University, Daytona Beach, Florida, United States
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Foreword
It is both an honor and a pleasure to write this foreword to Dr. King Yang’s book on
finite element analysis applied to injury biomechanics. I have worked closely with
Dr. Yang for over 30 years on a large number of injury biomechanics projects
involving experimentation or modeling or both. Dr. Yang is a pioneer in the
modeling of impact biomechanics, and together with his students and postdoctoral
fellows, he has formulated models of the entire human body, including head and
brain, neck, shoulder, thorax, abdomen, pelvis, and lower extremities. He has also
been teaching courses in finite element methods for many years. This book is a
culmination of his research and teaching experience, containing a vast amount of information related to the theory of the finite element method and its application to
impact injury to the human.
The first part of the book is for students who have not been introduced to the
finite element method of structural analysis, which is usually a semester-length
course by itself. The material is presented succinctly but with rigor. It also serves
as an excellent review resource for those who have taken a first course in finite
element methods. For the second part, Dr. Yang collaborated with his colleagues
and former students to describe the state-of-the-art research on modeling of human
and animal responses to impact, including details of the formulated models. To the
best of my knowledge, the material presented in Part II of the book is unique and up
to date.
Although he may not have specifically stated this in his book, it is a fact that
every model he has ever published was validated in some fashion against available
experimental data. It has been his intent right from the start of his career that he
will publish only validated models. This is now a policy in Dr. Yang’s Advanced
Human Modeling Lab. However, not all journals require model validation, and it
becomes very difficult for the reader to ascertain the validity and reliability of
an unvalidated model.
The Stapp Car Crash Journal will only publish models that have been validated. This not only raises the quality of the journal, but also gives the reader
and user the confidence that results from simulations using these models will be,
at the very least, reasonable and fairly accurate. As the complexity of the models
increase along with the cost of doing impact testing, it is the responsibility of all
modelers to validate their models and to include validation as part of their
publication.
Finally, let me echo the same sentiments expressed by Dr. Yang and acknowledge
with gratitude those who have donated their bodies to science, and specifically for
impact biomechanics research. Their generosity has made it possible for the
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Foreword
development of crash-test dummies and for the validation of impact models. Looking to the future, we can envision the day when the human computer model becomes
more humanlike than the crash dummies. It is my hope that the biomechanics community will design cars that are safe for people and not dummies, and opt to use
modeling as a replacement for all crash testing, including the crash dummies.
Albert I. King
Member, US National Academy of Engineering
Preface
The purpose of this book is to provide basic principles behind the finite element
method for static and dynamic analyses, and to augment this material with practical
applications in the modeling of biological tissues, organs, and the whole body. The
aim is to aid senior undergraduate students and beginning graduate students,
especially those in civil engineering, mechanical engineering, bioengineering, or
biomedical engineering programs, who use the finite element method to analyze
their designs for setting up research projects. In order to encourage self-study, a brief
refresher in fundamental principles taught in earlier engineering curriculum is
included; therefore students in this targeted audience do not need review their
previous textbooks.
In recent years, there has been a change in what knowledge is necessary to
perform finite element modeling. Published research results and our own
experiences indicate that, as a result of increasingly fast computational speeds, it
is more advantageous to use a large number of the simplest types of elements
than a smaller number of elements based on higher-order interpolation functions.
Additionally, the need to teach students how to write finite element analysis code
is no longer as critically needed as it was in the 1970s, because advanced finite
element software packages are readily available at little or no cost to university
students. As such, only those theories deemed necessary for understanding finite
element formulations and interpretations of analysis results are presented.
Part I of this book covers basic engineering principles pertinent to the finite
element method. Basic components forming a finite element model and the concepts
of strain and stress relevant to finite element analysis are reviewed in Chapter 1. The
rest of Part I is organized to follow the sequential order of developing a finite
element model: (1) idealizing the geometry to develop a finite element mesh and
establishing various element types and element shape functions (Chapter 2), (2)
formulating element shape functions based on isoparametric formulation and
ensuring high-quality mesh (Chapter 3), (3) setting up element and global stiffness
matrices (Chapter 4), (4) implementing material laws and properties (Chapter 5), (5)
setting up proper boundary and loading conditions (Chapter 6), (6) stepping through
static solutions using the finite element method (Chapter 7), and (7) describing
issues related to dynamic solutions (Chapter 8).
In Part II, examples of component and subsystem models of biological tissues are
presented. This part is specially designed for readers who need to develop finite
element models to solve impact biomechanics-related problems. Chapter 9 covers
the general procedures for converting medical images to finite element meshes. In
Chapter 10, methods related to parametric modeling of human body models are
addressed. The methods for modeling passive and active muscles are introduced
in Chapter 11. From Chapters 12 to 16, injury mechanisms, material laws, and
material properties relevant for developing finite element models of various human
body regions are discussed. The need for modeling the most vulnerable populations
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Preface
is covered in Chapter 17. Due to the recent prevalence of blast-induced injuries,
fundamentals of blast modeling are presented lastly in Chapter 18.
The first part of this book is partially based on lecture notes developed by the
author over the past 30 years. Additional contributions have come from several
outstanding colleagues with whom I have had the privilege of collaborating over
the years. We are very grateful for the support of Altair, Engineering Technology
Associates, ESI Group, ESTECO, LSTC, and Materialise for providing education
and research software packages, and to graduate students at Wayne State University
who have provided feedback through the development of this textbook. In particular,
I would like to thank Mr. Dominic Isopi for making exercise problems, and Dr.
Clifford C. Chou for his valuable suggestions. Last but not least, I am greatly
indebted to Mrs. Jane Yang, my wife and best friend, for her excellent illustration
work, and Ms. Barbara R. Presley for her editorial assistance and critique.
King H. Yang
Detroit, Michigan
PART
Basic Finite Element
Method and Analysis
as Applied to Injury
Biomechanics
1
King H. Yang
This page intentionally left blank
CHAPTER
Introduction
1
King H. Yang
Wayne State University, Detroit, Michigan, United States
1.1 FINITE ELEMENT METHOD AND ANALYSIS
The finite element (FE) method comprises a set of numerical procedures for obtaining solutions to many continuum mechanics problems, with an accuracy acceptable
to engineers. In classical continuum mechanics, problems are described with partial
differential equations. As long as the geometry can be described with a simple equation, the chance of finding the exact solution through the use of the classical method
is reasonably high. Unfortunately, real-world problems tend to involve complex
geometry and loading conditions. As such, most real-world problems cannot be
solved analytically. In contrast, the FE method provides answers to almost all structural mechanics problems. However, the accuracy of the results depends on how the
FE model is set up to represent the problem. In general, an FE model that consists of
a large number of interconnected subregions (to be described in Chapter 2 as
elements) will yield better accuracy, but a model with more elements will require
more computing resources. Thus, engineers need to balance the acceptable extent
of accuracy with computational costs, which should include the cost of downtime
while waiting for the results.
The word “continuum” is defined in the Cambridge dictionary as “something that
changes in character gradually or in very slight stages without any clear dividing
points.” In contrast, “discrete” is defined as “having an independent existence or
form apart from other similar things; separate.” The field of continuum mechanics
deals with the analysis of mechanical behaviors of a material, which can be represented as a continuous mass, as opposed to discrete particles. As such, materials
that cannot be represented as a continuous mass cannot be analyzed readily with
continuum mechanics methods.
The continuum requirement does not mean that only one material can be analyzed at
one instance. We will use trabecular bone, which consists of trabeculae and marrow, to
illustrate different situations for applying the concept of continuum. An FE model made
of 3 mm or larger typical size elements representing the trabecular bone will not allow
the modeler to distinguish trabeculae and marrow within one element. In this case, it
would be appropriate for analyzing the trabecular bone with this model. This is
because the element size is about 20 times that of a typical trabecula (around
Basic Finite Element Method as Applied to Injury Biomechanics. http://dx.doi.org/10.1016/B978-0-12-809831-8.00001-5
Copyright © 2018 Elsevier Inc. All rights reserved.
3
4
CHAPTER 1 Introduction
150 mm). At this relatively large ratio, each element consists of approximately the
same volume of trabeculae and marrow, and hence can be considered as a continuum.
If a small section of the same trabecular bone is observed with a very powerful
magnifying glass (such as a microcomputed tomography (CT) scanner with a
resolution of 10 mm), all features of bone marrow, trabeculae, and void spaces will
be prominently visible. To represent this small section, each trabecula and bone
marrow will need to be explicitly represented, because material properties for these
two components are significantly different. In this case, a 20 mm or smaller typical
element size FE model will be needed to properly analyze the structure at this scale
without violating the continuum assumption. In other words, we can model a structure
consisting of two (or more) materials, as long as each material satisfies its own
continuum assumption. Obviously, a model with such detailed information would
be computationally very costly and would provide little benefit over modeling with
continuum mechanics methods for understanding the overall behaviors of trabecular
bones.
There are two predecessors to the FE method, and the first was the slopedeflection method. George A. Maney (1888e1947, according to the American
Society of Civil Engineers) derived the slope-deflection method to analyze beam
and bending frame responses (Maney, 1914), and this method was considered to
be the predecessor of the matrix structural analysis (MSA). The MSA method was
initiated in the 1930s and became mature by the 1970s. At that time, this type of analysis was mostly conducted manually, with assistance of a slide rule and a simple
digital calculator. According to Felippa (2001), major contributors in this field
include Duncan and Collar (1934) who formulated discrete aeroelasticity in matrix
form, Argyris who analyzed structural responses using energy theorems (Argyris
and Kelsey, 1960), and Turner (1959) who proposed the direct stiffness method.
The FE method, derived from the MSA method, has become more popular than
MSA in recent years, because MSA is limited to solving one-dimensional (1D) truss,
beam, and frame (or a combination of truss and beam) problems, whereas the FE
method can be used for two-dimensional (2D) area and three-dimensional (3D) volume elements in addition to 1D elements. Initially, the FE method was mainly used
in the aerospace industry, where there were sufficient financial resources to afford
large-scale mainframe computers for engineering analysis. In the past five decades,
numerous research articles have been published addressing various fundamental
formulations associated with the FE method. Based on these new theoretical
developments, numerous FE software packages have been developed through either
public institutions or commercial entities. More importantly, these FE solution packages are available for college students at little or no cost.
Finite element analysis (FEA) is now a standard practice that is used routinely in
numerous fields of engineering. Compared to a couple of decades ago, general
purpose FE software packages (solvers) are now readily available for use in both
academic institutions and engineering industries. These packages are frequently
updated to incorporate the newest advancements in the FE method. Considering
the availability of these powerful FE solvers, along with the ever-increasing
computational power at very low costs, this book is written with the assumption
1.1 Finite Element Method and Analysis
that students will be using an FE software package for problem solving while
concurrently learning the theoretical background of the FE method. Thus, in Part I
(Chapters 1e8) of this book, we will use a step-by-step approach to describe theories
behind the FE method. This will be done in accordance with the steps required to
create an FE model, including identifying material properties, applying boundary
and loading conditions, and finding solutions for FEA using the FE method.
In Chapters 2e6, the fundamental knowledge needed to develop a static FE
model is presented. To enhance true understanding of the knowledge, conceptual
and real-world practical examples are illustrated to show how the theories are
applied. For the remaining chapters in Part I, emphasis is placed on solving static
and dynamic problems, validating models, and analyzing results. We hope that
the study of this book will lead not only to the capability of developing highquality FE models, but also to greater awareness of basic theories and drawbacks
behind the FE method. If you are an advanced student who wishes to develop
new fundamental theories in order to augment the capabilities of existing FE
solution packages, we recommend that you read other publications dedicated to
in-depth theoretical aspects of the FE method.
A real-world example that emulates an FE mesh is a truss bridge, as shown in
Fig. 1.1. The MSA method is frequently used to solve problems related to trussbridge designs. A member of an FE mesh possesses connecting points, or “nodes,”
and is considered an “element.” A truss is a structure in which only axial forces are
relevant. A 1D truss element contains two nodes. Connections of truss members
(elements) to form a bridge can be numerically idealized as an FE mesh. This
mesh consists of a number of 1D straightline elements interconnected at nodes.
Because the number of elements must be limited in order for the problem to be
solved by a digital computer, Professor Ray Clough at UC Berkeley proposed the
terminology “finite element method” in 1957, and this terminology is still used
(Clough and Wilson, 1999).
In an idealized truss bridge, all joints are pinned together (i.e., free to rotate) and
the weight of each truss member is neglected, because external forces are typically
much larger than body forces induced by gravity. For MSA or FEA, all external
loads are applied at the nodes, because any nonnodal loads cannot be calculated
by either of the methods. Upon loading, only axial forces are developed within
each truss member. In the real world, truss members are either bolted or welded
together, and hence each truss member may be subjected to bending moments and
FIGURE 1.1
Ravenswood bridge, a truss bridge over Ohio River on highway US 33.
5
6
CHAPTER 1 Introduction
shear forces. Because the bending moments and shear forces are much smaller than
the axial forces that are developed, these bending moments and shear forces can be
neglected during analysis.
The FE method has capabilities of finding solutions to 2D and 3D element types
that are not available in the MSA method. In addition, the FE method is excellent for
solving problems that involve complex geometries, multiple material compositions,
and complex boundary and loading conditions; none of which can be easily calculated using analytical methods. Additional advantages include capabilities for
displaying displacement, strain, and stress contours, reducing the time of design
cycles, and eliminating or reducing developmental prototypes by allowing new
designs to be evaluated on a computer prior to creating a physical prototype. It
should be noted that only in very limited cases (such as a simple plane truss problem)
can the FE method be used to find exact solutions. However, if the model is correctly
formulated and a sufficient number of elements are used to develop the FE model,
the solutions will be very close to exact solutions.
As mentioned previously, FEA is now routinely used for analyzing problems
related to a number of engineering fields, such as structural mechanics, biomechanics, heat transfer, fluid dynamics, and fluidesolid interface problems. For
simplicity, illustrations in this book will be limited to those relevant to structural
mechanics and injury biomechanics, unless otherwise specified. For these two fields,
FEA comprises a set of discrete numerical procedures that are used to solve for strain
and stress distributions, and other response variables, such as displacement, velocity,
acceleration, rotation, von Mises stress, and strain rate. Some applications of such
analyses include examining risks of injury in human bodies subjected to impacts,
designing new parts or systems of parts, modifying existing products, reducing
weight while increasing the load-carrying capacity, and determining if a structure
is safe prior to manufacturing. Because a high quantity of numerical calculations
is frequently involved, a multicore, high-performance computer is needed for
real-world FEA. Also, several software packages are needed to develop meshes,
calculate corresponding stresses/strains, and analyze the results.
The toll of traffic accidents on human suffering and associated costs are substantial, and the aforementioned characteristics of the FE method make it a perfect tool
for studying risks and prevention in this important area of unintentional injury and
death. Traffic accidents are among the top 10 causes of death in the United States. In
2010 alone, 32,999 people died and 3.9 million people were injured in such accidents in the United States (Blincoe et al., 2015). Further increasing the significance
of this type of loss is that victims of fatal automotive crashes are more likely to be in
their productive years than are people who die from other causes. Further, the total
social and economic costs, which include lost productivity, medical costs, legal and
court costs, emergency service costs, insurance administration costs, traffic congestion costs, property damage, and workplace losses, from these crashes were
estimated to be $242 billion, or about 1.6% of the annual gross domestic product.
In the fields of injury biomechanics and impact biomechanics, principles of mechanics are applied to study the impact responses and injury tolerances of biological
1.2 Calculation of Strain and Stress From the FE Model
materials or systems under extreme loading conditions. A good understanding of this
discipline is required if one wishes to design protective devices or countermeasures
to prevent or reduce the severity of injuries due to automotive crashes or other types
of impact. Using experimental methods, researchers in the discipline of injury
biomechanics study overall responses, such as globally or locally measurable impact
forces or accelerations at regions of interest (RoIs) within a body. These data are
used to establish injury thresholds, or the safety (or tolerance) limits, for those
body regions. However, the risk of injury is governed by internal responses of the
tissue or body part in response to the impact, not the impact force or acceleration.
Currently, FEA is the most suitable way to study internal responses, such as stress
and strain within a body part. The same set of procedures that are used for studying
the risk of injury can be applied for studying the integrity of orthopedic implant designs, determining the efficacy of protective equipment for sports activities,
designing an age- and gender-specific personal protection system for minimizing injuries resulting from local impacts, automotive crashes, among other applications.
1.2 CALCULATION OF STRAIN AND STRESS FROM
THE FE MODEL
When a material is loaded with a force, a stress is produced. At the same time, the
stress will deform the material and induce a strain. Finding stress/strain distributions
is the ultimate goal for most structural mechanics problems, because failure of a material depends on internal responses (e.g., reaching the failure stress at a specific
point in a specified direction), not the overall external input (force, acceleration,
etc.). In the FE method, we calculate the nodal displacement first, and then the strain
is calculated using the strainedisplacement relationship. Finally, we compute the
stress by using the corresponding constitutive equation(s) specific to that material.
In the following sections, we define different components of strain and stress.
1.2.1 AVERAGE STRAIN AND POINT STRAIN
Axial loading is defined as applying a force on a structure directly along an axis of
the structure. As an example, we start with a one-dimensional (1D) truss member
formed by points P1 and P2, with an initial length of L (Fig. 1.2) and a deformed
length of L0 , after axial loading is applied. The average axial engineering strain
is defined as the total amount of deformation (L0 L) divided by the initial length
(L) of the truss member (Eq. 1.1). Typically, the Greek letter ε (Epsilon) is used
to denote strain. Based on this definition, the average axial strain (also known as
the normal strain or extensional strain) has a unit of measure of inch/inch or
m/m, or is simply dimensionless. The word “normal” may cause confusion, because
we use the term “normal vector” to define a vector that is perpendicular to the
surface at a specific point in mathematics. In mechanics, the term “normal stress/
strain” indicates that the stress/strain component is along the axial direction in a
7
8
CHAPTER 1 Introduction
FIGURE 1.2
The original (undeformed) and deformed configurations of a truss member inline with the
x-axis. The truss member is loaded at point P2 toward the right side, while point P1 is
restrained from any movement. Upon axial loading, the deformed length of the truss
member becomes L0 .
one-dimensional (1D) problem, or the stress/strain components are along x-, y-, or
z-direction of the coordinate system in a 3D problem.
When the axial loading is in tension, the deformed length L0 is greater than the
original length L. Hence, the tensile (extensional) strain is defined to have a positive
value. Conversely, a compressive loading results in a negative strain.
L0 L
(1.1)
L
The concept of an average strain over the entire truss member given in Eq. (1.1)
is different from that of a point strain, which is defined as the strain measured at a
particular point inside the truss member. Fig. 1.3 illustrates a truss member with two
internal points (P and Q) that are separated by a small distance dx in the undeformed
configuration. These two points are deformed to P0 and Q0 after loading.
We will begin introducing the concept of the FE method by gradually adding
some FE terminology. In Fig. 1.3, we will consider that the entire truss member is
the whole structure to be analyzed, P and Q represent two nodes within the structure,
and element PeQ is one of the constituent elements that form the structure. We will
now analyze the strain for element PeQ from the displacements measured at nodes
P and Q. In other words, instead of measuring the overall deformed and original
lengths (L0 and L) of the truss member shown in Fig. 1.2 to determine the average
strain, we will quantify the displacements (u) at points P and Q given in Fig. 1.3
to define the point strain.
From the bottom part of Fig. 1.3, point P is axially displaced by uP to P0 and point
Q is displaced by uQ to Q0 . We define Du as the difference in the axial displacements
uQ and uP, that is, Du ¼ uQ uP. We will now prove that Du (the difference in point
ε¼
1.2 Calculation of Strain and Stress From the FE Model
FIGURE 1.3
The point strain at point P can be calculated from the difference in point displacements
(uP and uQ) of the deformed configuration and the original length (Dx, length between
points P and Q).
displacement) is the same as LP0 eQ0 LPeQ (i.e., the difference in lengths formed by
P0 eQ0 and PeQ). From the figure, we depict that LPeQ ¼ Dx and LP0 eQ0 ¼
Dx þ uQ uP ¼ Dx þ Du. Thus, LP0 eQ0 LPeQ ¼ Du. Based on the definition
for the average strain, we find that ε ¼ Du
Dx .
When the distance Dx between points P and Q approaches zero, the point strain
εP at point P is defined by taking the limit of the average strain over the segment
PeQ, as shown in the following equation.
εP ¼ lim
Dx/0
Du du
¼
Dx dx
(1.2)
This equation is called the 1D strainedisplacement equation. As its name
implies, this equation describes the relationship between the difference in nodal
displacements and the point strain for a 1D truss member. For FE terminology,
the term “point” becomes “node,” the expression “truss member” or “segment” becomes “element,” the phrase “point displacements” becomes “nodal displacements,”
and the corresponding nodal strains can be computed from nodal displacements by
using the strainedisplacement equation.
In contrast to an engineering strain in which the magnitude of deformation is
divided by the original length, a true strain in the axial direction is defined as the
magnitude of deformation divided by the current (original plus deformation) length.
Because large strains cannot be reached instantaneously, the overall true strain is
approximated by summing the true strain at each step, which is calculated by
dividing the change in length by the current length. If the loading is continuous
9
10
CHAPTER 1 Introduction
(i.e., step size is infinitesimally small), integration rather than summation should be
used to determine the true strain.
P
Z Lcurrent
dL ln Lcurrent
DL
¼
z
;
εT ¼
L
L
L
0
current
L0
where ln designates the natural log.
Using different step sizes, we demonstrate the differences between calculating
true strain based on summation and integral methods, respectively. Here the word
“step” could mean the time-step needed in a dynamic problem for advancing the
time, or the load-step related to solving a nonlinear problem in a step-by-step
manner. For a step size of x in engineering strain using the summation method,
the true strain can be approximated as
x
;
ðεT Þi ¼ ðεT Þi1 þ 1 þ ðεT Þi1
where i is the step number and it is assumed that (εT)0 ¼ 0 (i.e., no initial strain
exists). Using the summation method, a step increment of 5% in compressive engineering strain (x ¼ 0.05) results in a true strain of 5% (0.05/1 ¼ 0.05) in the
first step, and 10.26% (0.05 þ (0.05)/(1 0.05) ¼ 0.1026) in the second
step. Note that directional changes (i.e., from compression to tension) result in
different absolute magnitudes when the summation method is used for determining
engineering strain. For example, a constant increment of 5% in tensile strain
(x ¼ 0.05) results in a true strain of 5% in the first step and 9.76% in the second
step. In contrast, the same increment of 5% in compressive deformation creates a
true strain of 5% and 10.26%, respectively, for the first and second steps.
When loading is continuous, the integration method provides an accurate calculation of the true strain. The difference in results between the integration and summation methods becomes smaller as the step size becomes smaller. For example, a
5% shortening in the original length results in εT ¼ ln(0.95/1) ¼ 5.13% using the
integration method, which represents a difference of 0.13% compared to that calculated from the first step of the summation method. Calculating compressive engineering strain using five steps in which each step is 1%, rather than one step at
5% in the previous example, results in an approximate true strain of 5.10% (see
Table 1.1). This represents a difference of only 0.03% compared to that calculated
from the integration method.
Because strains are calculated from nodal displacements at each step in the FE
method, such strains must be true strains in nature. Fig. 1.4 shows strain contours
computed by an FE model representing a rectangular block subjected to uniform
compressive displacement loading on the left side while the right side is restrained
from any movement. The prescribed 4-step loading conditions are that the block is
compressed by 5%, 10%, 15%, and 20%, respectively. By the definition of engineering strain, the strain at each load step is the ratio of the change in length
(i.e., 5%, 10%, 15%, and 20%) and the original length (100%), with
compression being negative. As such, the engineering strains should
be 5%, 10%, 15%, and 20%, respectively, for the four steps.
1.2 Calculation of Strain and Stress From the FE Model
Table 1.1 Comparison of Engineering Strain Versus True Strains Calculated
by Summation and Integration Methods for 1%e5% Engineering Strain With
1% Increments and 5%e25% Engineering Strain With 5% Increments in
Compression
Engineering
Strain (%)
Strain by
Summation
(%)
Strain by
Integration
(%)
Engineering
Strain (%)
Strain by
Summation
(%)
Strain by
Integration
(%)
1
1.00
1.01
5
5.00
5.13
2
3
2.01
3.03
2.02
3.05
10
15
10.26
15.83
10.54
16.25
4
4.06
4.08
20
21.77
22.31
5
5.10
5.13
25
28.16
28.77
Contour Plot
Strain(vonMises, Max)
Global System
Advanced Average
7.000E-02
6.500E-02
6.000E-02
5.500E-02
5.000E-02
No result
Contour Plot
Strain(vonMises, Max)
Global System
Advanced Average
1.700E-01
1.650E-01
1.600E-01
1.550E-01
1.500E-01
No result
Contour Plot
Strain(vonMises, Max)
Global System
Advanced Average
1.200E-01
1.150E-01
1.100E-01
1.050E-01
1.000E-01
No result
Contour Plot
Strain(vonMises, Max)
Global System
Advanced Average
2.200E-01
2.150E-01
2.100E-01
2.050E-01
2.000E-01
No result
FIGURE 1.4
FE model computations of strain contours subjected to uniformly distributed
displacement loading on the left edge with up to 20% engineering strain in four steps. The
contours clearly show that the strains calculated with the FE model are
approximately 5%, 10.5%, 16%, and 21.5%, respectively, for steps 1e4. This
exercise demonstrates that strains calculated with an FE model are true strains in nature,
not engineering strain.
In the first step (the upper-left block), the calculated strain for the entire FE
model is 5% (observed from the contour legend). For the second step (the block
on the upper right), it is seen that the strain is 10.5%. For the third (the block
on the lower left) and fourth (the block on the lower right) steps, the respective
strains are 16% and 21.5%. These values indicate that strains computed using
an FE software package are not engineering strain. Rather, they are closer to true
11
12
CHAPTER 1 Introduction
strains computed from the summation method as only four steps are prescribed.
Compared to Table 1.1, the contour values shown in Fig. 1.4 are not accurate,
because of the limited number of contour brackets and large overall range (2%)
selected when drawing the contour plots. Note that when a homogeneous block is
subjected to infinitesimal strain, the differences among the engineering strain, point
strain, and true strain are negligible.
1.2.2 NORMAL AND SHEAR STRAIN
Normal strain (also known as extensional strain) is different from shear strain in that
normal strain is the ratio of the change in length to the original length along the x-, y, or z-direction, whereas shear strain is a measure of a change in angle in an x-y, y-z,
or z-x plane. Normal strain is dimensionless, whereas the unit of measure for shear
strain is radians. For axially loaded 1D elements, only normal strains exist. For 2D or
3D problems, strains need to be decomposed into normal and shear components.
We shall consider two sidelines of a 2D infinitesimal rectangular element that are
formed by nodal points P0 and P2 in the x-direction and P0 and P1 in the y-direction.
The sizes of this element are dx along the x-direction and dy along the y-direction
(Fig. 1.5). The angles a and b are intentionally magnified in order to better illustrate
the angle changes. After deformation, the horizontal sideline displaces to P00 eP02 and
+
'
+
α
′
β
′
+
+
FIGURE 1.5
Two sidelines P0eP2 and P0eP1 of an infinitesimal element with dimensions of dx and dy
are deformed to P00 eP20 and P00 eP10 . The displacement along the x-axis is denoted by u
while the displacement along the y-axis is denoted by v. The angles a and b are greatly
exaggerated to better highlight the angle changes.
1.2 Calculation of Strain and Stress From the FE Model
the vertical sideline moves to P00 eP01 . In this figure, u represents the displacement
along the x-axis and v represents the displacement along the y-axis (i.e., the displacements needed to move from point P0 to point P00 ). Although not shown in the figure,
the next letter in the alphabet, w, is chosen to represent the displacement along the
z-axis when dealing with a 3D problem. Based on Pythagoras’ theorem, the
deformed length P00 eP02 can be calculated using the following equation:
sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
ffi
2
2
0
vu
vv
(1.3)
length P0 P02 ¼
dx þ u þ dx u þ v þ dx v :
vx
vx
In the y-direction, after cancelling out identical terms and eliminating the square
of the derivative
vv dx
vx
2
, because the angle b is very small, the length P00 P02 can
be approximated by dx þ vu
vx dx. Because the normal strain along the x-direction (εxx)
is defined as the change in length along the x-axis divided by the original length, εxx
can be calculated as
vu
0
dx þ dx dx
0
length P0 eP2 lengthðP0 eP2 Þ
vu
vx
¼ :
(1.4)
εxx ¼
¼
dx
vx
length ðP0 eP2 Þ
It is fairly common to see εxx written as εx in other engineering books. Because
shear strain requires two subindices (e.g., εxy) to fully describe its association with
the axes involved, a double x is chosen for consistency. Using the same approach as
given in Eq. (1.4), we can determine that εyy ¼ vv
vy, where v is the displacement along
the y-direction. Extending the 2D illustration to 3D, we can see that εzz ¼ vw
vz , where
w is the displacement along the z-direction. These three expressions
vv
vw
εxx ¼ vu
vx; εyy ¼ vy ; and εzz ¼ vz are the normal strainedisplacement equations
for a 3D element.
A simple shear can be thought of as a change that would happen to a rectangular
element if the top surface was pushed lightly toward the right, as demonstrated in
Fig. 1.6. We can see from this figure that the volume of the element remains constant
(i.e., isochoric). In this configuration, the change in angle (q) is defined as the average
shear strain, gavg, that is, gavg ¼ q. A positive average shear strain means that there is
a clockwise rotation of the vertical sideline. This simple shear configuration is a special case of the general shear configuration, which involves two angle changes (one
with respect to the x-axis and one with respect to the y-axis), as shown in Fig. 1.5.
The total shear strain (also known as the engineering shear strain) is generally
denoted by the Greek letter g (Gamma). The xy component of the total shear strain,
gxy, is the sum of the changes in angles with respect to both x- and y-directions. In
other words, gxy ¼ a þ b, as shown in Fig. 1.5.
From trigonometry, the small angle approximation states that for a very small
angle, the angle and the tangent of the angle are approximately the same. Thus,
from Fig. 1.5, b can be approximated as tan b. If vu
vx dx (i.e., normal strain is
13
14
CHAPTER 1 Introduction
Δ
θ
FIGURE 1.6
A rectangular element (solid lines) is gently pushed on the top surface to the right side
(dashed lines). The magnitude of the average shear strain (q, expressed in radians in this
configuration) is positive, because the angle changes from p2 (vertical line) to a smaller
angle (clockwise rotation).
much smaller than the axial length), we can derive b from Fig. 1.5 as given in Eq.
(1.5).
vv
vv
v þ dx v
dx
vv
vx
¼
z vx
.
(1.5)
bztan b ¼
vu
dx
vx
dx þ u þ dx u
vx
Similarly, a can be derived as
uþ
a z tan a ¼
vu
dy u
vu
vy
¼
;
dy
vy
(1.6)
and gxy is
gxy ¼ a þ b ¼
vu vv
þ ¼ 2εxy .
vy vx
where the shear strain εxy is the average of two strains, that is, εxy ¼ 12
(1.7)
vv
vx
þ vu
vy ,
which is equivalent to one half (1/2) of the engineering shear strain gxy. The shear
strain εxy discussed here is also known as Cauchy’s shear strain and is applicable
to problems involving only small deformations. Students may wonder why we
need to have two different definitions (εxy, gxy) for “shear strain.” This is because
engineering strain (gxy) does not possess the quality of tensor while Cauchy’s strain
is a tensor, which is needed in FE formulation. For materials that may exhibit large
1.2 Calculation of Strain and Stress From the FE Model
deformations, such as soft biological tissues, the constitutive law (see Section 1.2.3)
based on Cauchy’s strain will not be suitable. As such, you are encouraged to study
Green strain and Almansi strain, which are suitable for both small and large (finite)
deformations, but are not in the scope of this book.
For a 3D element, the shear strainedisplacement equations due to small deformations are illustrated in the equation below:
gxy ¼
vv vu
vv vw
vw vu
þ
¼ 2εxy ; gyz ¼ þ
¼ 2εyz ; gzx ¼
þ
¼ 2εzx :
vx vy
vz vy
vx vz
(1.8)
Based on the principle of symmetry, we can show that gxy ¼ gyx, gyz ¼ gzy, and
gzx ¼ gxz. If we combine all normal and shear components, the strainedisplacement
equation can be written in matrix form, as shown in Eq. (1.9) for a 1D element,
Eq. (1.10) for a 2D element, and Eq. (1.11) for a 3D element, where u, v and w
are displacements along the x-, y-, and z-direction, respectively.
εxx ¼
2
du
dx
v
6
8
9 6 vx
6
>
< εxx >
= 6
εyy ¼ 6
60
>
>
:
; 6
6
gxy
6v
4
vy
2
8
εxx
>
>
>
>
>
>
εyy
>
>
>
>
< ε
zz
>
g
> xy
>
>
>
>
>
gyz
>
>
>
:
gzx
8
εxx
>
>
>
>
>
>
εyy
>
>
>
>
< ε
zz
¼
>
>
2ε
> xy
> >
>
>
>
>
>
>
>
>
>
>
>
> 2εyz
>
>
:
; >
2εzx
9
>
>
>
>
>
>
>
>
>
>
=
6
6
6
6
6
6
9 6
6
6
>
>
6
>
>
6
>
>
6
>
>
>
6
>
= 6
6
¼6
>
6
>
>
6
>
>
6
>
>
6
>
>
6
>
; 6
6
6
6
6
6
6
6
4
(1.9)
3
07
7
7
v7
7 u
7
vy 7 v
7
v7
5
vx
(1.10)
3
v
vx
0 7
7
7
7
7
7
v
0
0 7
7
vy
7
7
7
8 9
v 7
7> u >
0 0
7
vz 7< =
7 v
7>
v v
;
: >
0 7
7 w
vy vx
7
7
7
v v 7
7
0
vz vy 7
7
7
7
v
v 7
5
0
vz
vx
0
(1.11)
15
16
CHAPTER 1 Introduction
1.2.3 CALCULATION OF STRESS
As described in entry level mechanics courses, the intensity of a force distributed
over a given cross-sectional area is defined as the stress on that section. Stress is
generally denoted by the Greek letter s (sigma). Typically, a positive sign is used
to indicate a tensile stress (member in tension), and a negative sign to indicate a
compressive stress (member in compression). As shown in the previous section,
stresses can also have normal and shear components just as strains can have normal
and shear components. In the FE method, stresses are calculated from strains based
on the constitutive equation(s) rather than from calculations in which the force
components are divided by corresponding cross-sectional areas.
In accordance with System International (SI) metric units, stress is expressed in
units of Pascals (Pa).
1 Pascal (Pa) ¼ 1 N/m2
1 kPa ¼ 103 Pa ¼ 103 N/m2
1 MPa ¼ 106 Pa ¼ 106 N/m2 ¼ 1 N/mm2
1 GPa ¼ 109 Pa ¼ 109 N/m2
In accordance with U.S. customary imperial units, stress is expressed in pounds
per square in (psi).
1 psi ¼ 1 lb/in2 ¼ 6.895 kPa, and
1 ksi ¼ 1 kilopounds/in2 ¼ 6.895 MPa
A constitutive equation describes relationships among two or more physical
measures (such as stress vs. strain, stress vs. strain with strain rate effect, etc.)
that are specific to a material. In the field of mechanics of materials, the relationship
between the stress and the strain of a specific type of material is described as the
constitutive relationship of that particular material. While it is commonly implied
that constitutive equations can be derived mathematically, the reality is that we
must rely on obtaining experimental data and using curve fitting procedures for
finding the material constants associated with these equations.
Assuming that all stress and strain components are linearly related for a material,
the constitutive equation, or the stressestrain relationship of this material, has the
general form
sij ¼ Eijkl εkl
(1.12)
where Eijkl is a fourth-order tensor of the material (also known as the material stiffness matrix) and i, j, k, and l range from 1 to 3 for 3D problems and from 1 to 2 for 2D
problems. We can determine that Eijkl possesses 81 (3 3 3 3) separate components for a 3D problem. Obviously, finding the 81 constants needed to fully describe
such a material is a daunting task. As a result, this material type is seldom used. Note
that the terms “material stiffness matrix” expressed here and “element stiffness
matrix,” to be introduced in Section 1.3 and Chapter 4, are different but related. The
material stiffness matrix describes the stressestrain relationship, while the element
stiffness matrix is associated with the size and material properties of the element.
1.2 Calculation of Strain and Stress From the FE Model
An important distinction between standard FE procedures and classical
mechanics is that in FE methods stresses are calculated from strains as shown in
Eq. (1.12), whereas in classical mechanics strains are calculated from stresses. A
similar expression in terms of the strain-stress (as opposed to stressestrain) relationship is shown in Eq. (1.13), where C is a symmetric tensor of a material compliance
matrix. By multiplying [E]1 to both sides of Eq. (1.12), the relationship between the
material stiffness and compliance matrices can be easily depicted. Note that in standard FE procedures, strains are calculated from nodal displacements before stresses
are computed from strains, and in classical mechanics it is taught that strain is
induced by stress, and stress is the result of force. Therefore, Eq. (1.13) is seldom
used in FE methods.
εkl ¼ Cijkl sij
1
Cijkl ¼ Eijkl
(1.13)
We can show that both shear stresses and shear strains are symmetric in nature.
In other words, sxy ¼ syx, syz ¼ szy, and szx ¼ sxz, and likewise εxy ¼ εyx, εyz ¼ εzy,
and εzx ¼ εxz. Due to the symmetry of the tensor, the indices for the stress (i and j)
can be swapped, and so can the indices for the strain (k and l):
Cijkl ¼ Cjikl ¼ Cijlk ¼ Cjilk. This symmetrical condition reduces the number of independent elastic components from 81 to only 36 for full description of the stiffness
and compliance tensors. The 36 independent components can be expressed by
assigning sxx ¼ s1, syy ¼ s2, szz ¼ s3, syz ¼ s4, sxz ¼ s5, sxy ¼ s6 and εxx ¼ ε1,
εyy ¼ ε2, εzz ¼ ε3, gyz ¼ 2εyz ¼ ε4, gxz ¼ 2εxz ¼ ε5, gxy ¼ 2εxy ¼ ε6, and then
Eq. (1.11) can be written as
si ¼ Eij εj ;
where i and j each ranges from 1
9 2
8
E11 E12
s1 >
>
>
>
>
>
6
>
>
>
>
6 E21 E22
>
s2 >
>
>
>
>
6
>
>
>
>
<s = 6
6 E31 E32
3
¼6
6
>
> 6 E41 E42
> s4 >
>
>
>
>
6
>
> 6E
>s >
>
>
5
>
>
4 51 E52
>
>
>
>
;
:
s6
E61 E62
(1.14)
to 6. We write Eq. (1.14) in component form as
9
38
E13 E14 E15 E16 > ε1 >
>
>
>
>
7>
>
>
>
>
E23 E24 E25 E26 7
ε
2 >
>
>
>
7>
>
>
>
7>
<
7
E33 E34 E35 E36 7 ε3 =
:
(1.15)
E43 E44 E45 E46 7
> ε4 >
>
7>
>
>
>
7>
>
>ε >
>
>
E53 E54 E55 E56 7
5
>
>
5>
>
>
>
>
:
;
E63 E64 E65 E66
ε6
The advantage of using the tensor format is that one single tensor equation can
be used to represent a system of equations. Additionally, equations written in
tensor notation can be directly incorporated into computer code with little effort.
Considering the limited scope of this book, and that the targeted readers are
unlikely to know much about tensors, descriptions of the tensor notation and
how to manipulate tensors are not covered. Unless a tensor is specifically needed,
we use matrix notation to describe the stressestrain relationship.
Using the complicated constitutive equations, shown in Eq. (1.15), is not the only
means for describing the stressestrain relationship. Commonly available engineering materials can be expressed with much simpler forms of constitutive equations.
17
18
CHAPTER 1 Introduction
For example, the most widely used isotropic linear elastic material was derived from
Hooke’s law. Robert Hooke (Jul. 1635eMar. 1703) found that the extension of a
spring was proportional to the force applied to it. Hooke’s birth and death dates
are based on the MacTutor History of Mathematics Archive, a wonderfully organized database compiled by O’Connor and Robertson (2017). Looking at the birth
and death dates in this database gives you an appreciation that the FE method is
not at all a new branch of engineering.
Because extension is related to strain and stress is related to force, Hooke’s law
became the constitutive equation for a 1D, linear elastic material. To expand from
the 1D Hooke’s law to a perfectly isotropic elastic 3D material, two nonzero components are required to fully describe the relationship between stresses and strains as
shown in Eq. (1.16), where l and m are the Lamé’s constants, named after a French
mathematician Gabriel Lamé (Jul. 1795eMay 1870).
8
9
9
38
sxx > 2
>
>
>
l þ 2m
l
l
0 0 0 > εxx >
>
>
>
>
>
>
>
>
>
6
>
>
>
syy >
>
>
>
>
l þ 2m
l
0 0 07
εyy >
>
>
>
>
7
6
>
>
>
>
>
>
7>
<
<s >
= 6
=
7
6
l
þ
2m
0
0
0
ε
zz
zz
7
¼6
(1.16)
> 2εxy >
> sxy >
> 6
>
m 0 07
>
7>
6
>
>
>
>
>
>
>
7>
6
>
>
>
>
>
>
4
Symm
m 0 5>
2εyz >
>
>
>
>
s
yz
>
>
>
>
:
>
>
;
>
>
:
;
m
2εzx
szx
The dimensions (units) for m and l are the same as pressure (e.g., Pascal). In the
theory of elasticity, the second Lamé constant (m) has the same definition as that of
the shear modulus (G). However, the first Lamé constant (l) has no specific physical
meaning, although m and l are related to the speed of elastic wave. For this reason,
the Lamé constants are not directly measurable. As such, Young’s (elastic) modulus
(E) and Poisson’s ratio (n), which can be directly measured from experiments, are
more common than the Lamé constants when describing the stressestrain relationship for an isotropic, linear elastic material. The Lamé constants are related to
Young’s modulus and Poisson’s ratio as:
m¼G¼
E
2ð1 þ nÞ
(1.17)
and
l¼
En
:
ð1 þ nÞð1 2nÞ
(1.18)
From Eq. (1.16), we find that the summation of three normal stresses, which is a
quantity related to the overall volume change, is
sxx ¼ ðl þ 2mÞεxx þ lðεyy þ εzz Þ
syy ¼ ðl þ 2mÞεyy þ lðεxx þ εzz Þ
szz ¼ ðl þ 2mÞεzz þ lðεyy þ εxx Þ
sxx þ syy þ szz ¼ ð3l þ 2mÞðεxx þ εyy þ εzz Þ:
(1.19)
1.2 Calculation of Strain and Stress From the FE Model
Fluid-like materials, such as water, brain tissues, or the nucleus of an intervertebral disc, provide no resistance to shear loading. This material type has a Poisson’s
ratio of 0.5, which is known to be an incompressible material to express the fact that
it exhibits zero volume change when loaded. Because normal stresses are equal
(sxx ¼ syy ¼ szz) for fluid-like materials in static equilibrium, pressure induced by
external forces is used to determine the response.
Recall that hydrostatic pressure in liquid is the pressure due to the force of gravity. Thus, hydrostatic pressure increases in proportion to the depth of the fluid. In
most real-world problems, the magnitude of the hydrostatic pressure is much smaller
than the force-induced pressure. Hence, pressure within a fluid-like material is the
key response to study.
Based on Eq. (1.18), the corresponding l value for an incompressible fluid-like
material (n ¼ 0.5) would be infinitely large. Thus, it is not physically possible to
solve such a problem using the FE method. To avoid this difficulty, many FE modelers have used a Poisson’s ratio of 0.4999, which represents a nearly incompressible
fluid, when modeling materials such as water, brain, or nucleus. As a result of setting
Poisson’s ratio very close but not equal to 0.5, the corresponding normal stress components are not equal, that is (sxx s syy s szz). In this case, the pressure is derived
from Eq. (1.19) as
1
ð3l þ 2mÞ
ðεxx þ εyy þ εzz Þ:
P ¼ ðsxx þ syy þ szz Þ ¼
(1.20)
3
3
Through Eq. (1.20), we find that DP ¼ 3lþ2m
3 ðεxx þ εyy þ εzz Þ when compared
with the zero loading state. We will introduce a term bulk modulus (K) to refer to
a property that is defined as the ratio of pressure change (DP) to the magnitude of
volume change (DV) divided by the original volume (V0), as given in Eq. (1.21).
Here DV
V0 zεxx þ εyy þ εzz , if all second order and higher terms are neglected.
Thus, the bulk modulus can be expressed as a function of the Lamé’s constants.
Eq. (1.21) also describes K in terms of the more commonly known elastic modulus
and Poisson’s ratio, which will be discussed next.
K¼
DP
3l þ 2m
E
¼
¼
DV=V0
3
3ð1 2nÞ
(1.21)
The bulk modulus is measurable and frequently listed in engineering handbooks
for fluid-like materials, such as oil, honey, and gasoline. As such, it is a commonly
prescribed material property for fluid-like materials. Eqs. (1.17) and (1.18) express
the two Lamé constants as functions of the elastic modulus (E) and Poisson’s ratio
(n) for isotropic, linear elastic materials. These two properties are discussed in
almost all fundamental textbooks and corresponding values for many materials
are reported in numerous engineering handbooks and online sources. Hence, these
two constants are used frequently in setting up FE models for structural analysis.
For 2D analysis, there are two cases, plane stress and plane strain, that deserve
special mention. For a plane stress problem, no stresses are perpendicular to the
cross section. An example is a flat plate that lies in the x-y plane with loading forces
19
20
CHAPTER 1 Introduction
only along the direction of the plane. In this loading condition, the assumption is that
szz ¼ syz ¼ szx ¼ 0. Notice that the shear stress syz is used interchangeably as syz displayed in Eq. (1.16), because the Greek letter s emphasizes the origin of the stress is
due to shear, while s underscores the fact that it is a component of the stress tensor.
Eq. (1.22) shows the stressestrain equation for this plane stress problem.
2
3
8
9
9
8
1 n
0
εxx >
>
6
7>
< sxx >
=
=
<
E 6
0 7
6n 1
7 εyy
syy ¼
(1.22)
7
26
>
>
>
:
; 1n 4
;
:
1 n 5>
g
sxy
xy
0 0
2
For a plane strain problem, no strains normal to the cross section are allowed. An
example problem that is commonly solved using this method involves a long dam
with its long axis lying along the z-direction. This dam has the same cross section
throughout the entire length, and is loaded by water along the surface. In this loading
configuration, all strain components involving the z-axis must vanish, that is,
εzz ¼ gyz ¼ gzx ¼ 0. Eq. (1.23) shows the stressestrain relationships of a plane strain
problem.
2
3
9
9
8
8
1n
n
0
εxx >
>
6
7>
=
=
< sxx >
<
6 n
E
1n
0 7
6
7 εyy
syy ¼
(1.23)
6
7
>
>
>
; ð1 þ nÞð1 2nÞ 4
;
:
:
1 2n 5>
g
sxy
xy
0
0
2
For a general 3D problem, the constitutive equation for an isotropic linear elastic
material in terms of E and n has the form of
2
3
1
n
n
n
0
0
0
6
7
6
78
8
9
6
7 ε 9
s
n
1
n
n
0
0
0
6
7>
>
>
xx
xx >
>
>
>
6
7>
>
>
>
>
>
>
>
>
6
7
>
>
>
>
>
>
>
>
s
ε
6
7
>
>
>
>
yy
yy
n
n
1
n
0
0
0
>
>
>
>
>
>
>
>
6
7>
>
>
>
>
>
>
>
6
7
<
< szz =
6
7 εzz =
E
1
2n
6
7
¼
:
0
0
0
0
0 7
>
ð1 þ nÞð1 2nÞ 6
2
sxy >
2εxy >
>
>
>
>
6
7>
>
>
>
>
>
>
>
6
7>
>
>
>
>
>
>
>
6
7>
1 2n
>
>
>
>
>
syz >
2εyz >
6 0
7>
>
>
>
>
0
0
0
0
>
>
>
>
6
7
>
>
>
>
2
:
:
;
;
6
7
szx
6
7 2εzx
6
7
1 2n 5
4 0
0
0
0
0
2
(1.24)
Eq. (1.24) describes an idealized isotropic linear material, which does not exist in
the real world, since there are always microscopic differences. Nevertheless, this
1.3 Sample Matrix Structural Analysis
material remains the most commonly used because of its simplicity. Other materials
are more complex than the isotropic linear elastic material model can describe. The
corresponding constitutive equations of those special materials will be listed when
they become relevant in later chapters.
1.3 SAMPLE MATRIX STRUCTURAL ANALYSIS
For this section, we shall work through several examples to appreciate the similarities and differences between the MSA and FEA methods. The MSA method, also
known as the direct stiffness method, is in essence a tool for computing forces
and displacements within each truss or frame element of the structure. This direct
stiffness method is the most commonly used to describe the fundamental theory
behind the FE method. By knowing the stiffness-force relationship of each of the
elements that make up an entire structure, a matrix form of the global stiffness
equations can be assembled from the stiffness matrices of the individual elements.
The equilibrium relationship between the global stiffness matrix and externally
applied forces can then be solved to determine any unknown displacements and
reaction forces of the structure.
1.3.1 ELEMENT STIFFNESS MATRIX OF A LINEAR SPRING
Consider a weightless and linear spring element, aligned along the x-direction, having a spring constant k (Fig. 1.7). This spring element is formed by two nodes (P1
and P2), and each node allows only one axial displacement for a total of two degrees
of freedom (DOFs). It is further assumed that a positive value indicates that both the
force and displacement are pointing toward the right side. Based on Hook’s law, the
static equilibrium equation for a linear spring can be written as
F ¼ k Dx;
(1.25)
where F is the applied force and Dx is the axial displacement of the spring.
FIGURE 1.7
The free-body diagram of a linear spring element. P1 and P2 represent the two nodes, and
F1 and F2 are nodal forces. The corresponding nodal displacements are u1 and u2.
21
22
CHAPTER 1 Introduction
Assuming that the spring is fixed at the left-hand side, that is, u1 ¼ 0, we can
write the following equation based on Hooke’s law:
F 2 ¼ k u2 :
(1.26)
We
P then calculate the reaction force from the static equilibrium equation
F ¼ 0 as:
F1 ¼ F2 ¼ k u2 :
(1.27)
Rewriting these two equations in matrix form yields:
k
F1
0 k
u1 ¼ 0
u2 ¼
or
¼
k
0 k
u2
F2
F1
.
F2
(1.28)
Similarly, if the spring is fixed at the right side, the three corresponding equations
are:
F1 ¼ k u1 ;
(1.29)
F2 ¼ F1 ¼ k u1 ; and
F1
k 0
u1
k
or
¼
u1 ¼
k 0
F2
u2 ¼ 0
k
F1
(1.30)
F2
:
(1.31)
Because both u1 and u2 need not be zero, Eqs. (1.28) and (1.31) can be assembled to
become two simultaneous equations in matrix form, as shown in Eq. (1.32). For
clarity
be represented
" on how# we arrive at Eq. (1.32), let the matrix in Eq. (1.28)
"
#
as
1
k11
1
k12
and the matrix in Eq. (1.31) be represented as
2
k11
2
k12
, where
1
1
2
2
k21
k22
k21
k22
the numbers in the exponent position do not represent an exponent, but is “u1 ¼ 0”
for Eq. (1.28) and “u2 ¼ 0” for Eq. (1.31). The assembly
of these two matrices
"
#
is the same as adding the matrices, which results in
1
2
þ k11
k11
1
2
k12
þ k12
. How1
2
1
2
k21
þ k21
k22
þ k22
ever, this assembly process is not totally a summation process. If it were, the results
on the right-hand side of Eq. (1.32) would be 2F1 and 2F2. The reason we use the
term “assemble” is that only the stiffness portions of the equations are summed at
each respective matrix location, while terms related to vectors [forces F1 and F2
from Eqs. 1.28 and 1.31] are not summed to form Eq. (1.32).
Another way to look at this assembly process is that F1 and F2 in each of Eqs.
(1.28) and (1.31) show only partial contributions of the total forces of F1 and F2.
Each contribution is relative to the specific configurations, where u1 and u2 are alternatively assumed to be zero at the two different instances, respectively. Eq. (1.32)
explicitly highlights the step-by-step nature of the assembly process for easy
understanding.
0 þ k k þ 0
u1
k k
u1
F1
¼
¼
(1.32)
0k kþ0
k k
u2
u2
F2
1.3 Sample Matrix Structural Analysis
In Eq. (1.32), the symmetric matrix
k
k
describes the forceedisplacement
k k
relationship and is known as the element stiffness matrix [k]. This matrix is different
from a global stiffness matrix, denoted as [K], which will be introduced later.
In addition to being symmetric, each column of [k] has a specific physical meaning. When a specific nodal DOF (e.g., u1 or u2) is assumed to have a value of unity
(i.e., one) while all other DOFs are assumed to be zero, the column of [k] matching
the specific DOF with the value of unity describes the nodal load needed to generate
that particular deformation state. In other words, if the displacement field is
u1 ¼ 1
, we place u1 ¼ 1 and u2 ¼ 0 in the left-hand side of Eq. (1.32), which
u2 ¼ 0
k k
k
1
becomes
¼
. As such, the entries in column two
k k 22 0 21
k
k11
are immaterial, since u2 ¼ 0. Thus, the first column
of Eq. (1.32) describe
k21
1
the forces needed to generate the displacement field
. Similarly, if the displace0
0
k21
ment field is
, the second column
of Eq. (1.32) describes the forces
1
k22
needed to generate this displacement field. These characteristics, along with the
static equilibrium condition, will be applied in several examples shown later to
derive the element stiffness matrix.
We must note that Eq. (1.32) is a singular matrix, and therefore no solution can
be found. The two equations that form this matrix are essentially the same, due to the
need to satisfy the equilibrium condition F1 ¼ F2. A proper boundary condition
needs to be set up before the displacement can be determined.
1.3.2 ELEMENT STIFFNESS MATRIX OF A LINEAR SPRING NOT
IN LINE WITH THE X-AXIS
In a real-world problem, a spring element may not be in line with any particular axis in
the global coordinate system. Fig. 1.8 shows a linear spring element in a typical position.
The rotation angle q is defined as the angle between the horizontal line drawn through
P1 and the line formed by P1 and P2, representing the spring. A counterclockwise
rotation indicates a positive rotation angle. This sign convention will be used throughout
the book for consistent formulation of the 2-node linear element stiffness matrix. Any
deviation from this definition will result in a different expression of the element stiffness
matrix. For example, the rotation angle is q for element P1eP2, because this angle is
formed by a horizontal line (drawn from P1 towards the right hand side) and P1eP2.
If the element were marked as P2eP1 (the first horizontal line would be drawn through
P2 towards right), and the angle of rotation would be p þ q.
Although a spring element allows only one DOF (axial displacement) per node,
as mentioned previously, a linear spring element not in line with the x-axis or y-axis
has two pseudo-DOFs per node. This designation is aimed for easier assembly with
other elements that are described in the x-y coordinate system. Because there is a
total of four DOFs (u1, v1, u2, and v2), the order of the corresponding element
23
24
CHAPTER 1 Introduction
θ
FIGURE 1.8
A linear spring is rotated about P1 from its original position along the x-axis shown in
Fig. 1.7 in a counterclockwise direction by an angle q. Although a spring element
possesses only one DOF per node, two pseudo-DOFs are used for each node for easier
assembly with other elements that are described in the global coordinate system.
stiffness matrix [k] is 44. We will calculate the forces needed to generate a
displacement field, in which the nth DOF has unity displacement, while displacements for all other DOFs are zero. The resulting forces will be the same as the
nth column of the element stiffness matrix. By repeating this process four times,
the complete 44 element stiffness matrix can be determined.
Step 1:
As stated earlier, the first column of the element stiffness matrix ([k]44) can be
identified by assuming u1 ¼ 1, v1 ¼ 0, u2 ¼ 0, and v2 ¼ 0, and then finding the
corresponding forces needed to form this configuration (Fig. 1.9).
′
θ
′
FIGURE 1.9
A free-body diagram showing a spring element with a unity horizontal displacement u1,
while all other DOFs (v1, u2, and v2) are assumed to be zero. Because the displacement is
assumed to be infinitesimal, the angle q formed by the axis of the undeformed spring and
the x-axis is assumed to be the same as the angle formed by the deformed spring and the
x-axis. Also, the shortening of the spring in this loading configuration is approximated as
“d,” which is defined as the distance between P1 and the perpendicular intersection of P10
with respect to the axis of the original spring (P1eP2).
1.3 Sample Matrix Structural Analysis
Keeping in mind the definitions of opposite, adjacent, and hypotenuse with respect
adj
to the angle of interest from basic trigonometry, recall that cos q ¼ hyp
. Therefore, the
shortening of the spring, d, can be calculated as u1 cos q. As such, the axial force
needed to create this deformed configuration is F ¼ k u1 cos q. This force can be
decomposed into a horizontal component Q1 and a vertical component R1. Because
the horizontal displacement u1 is assumed to be very small, the difference in angle
between the original and deformed configurations is considered to be negligible.
From the free-body diagram shown in Fig. 1.9, we find that
Q1 ¼ Q2 ¼ F cos q ¼ k u1 cos2 q and
(1.33)
R1 ¼ R2 ¼ F sin q ¼ k u1 cos q sin q:
(1.34)
If we let cos q ¼ C and sin q ¼ S, the static equilibrium conditions shown in
Eqs. (1.33) and (1.34) can be written in matrix form as
8
9
2 2 3
Q1 >
C
>
>
>
>
>
>
6
7
<R >
=
6 CS 7
1
6
7
:
(1.35)
k6
u
¼
1
2 7
>
Q2 >
>
>
4 C 5
>
>
>
>
:
;
R2
CS
Step 2:
Next, we find the second column of [k] by assuming that u1 ¼ 0, v1 ¼ 1, u2 ¼ 0,
and v2 ¼ 0. Fig. 1.10 shows the configurations before and after deformation.
Using a similar procedure to the one used for the preceding segment, in which the
only nodal displacement was u1 ¼ 1, we calculate the shortening of the spring, d,
from v1 and angle b based on Fig. 1.10 as d ¼ v1 sin b. As such, the force needed
to create this deformed configuration is F ¼ k v1 sin b. Again, this force can be
decomposed into a horizontal component Q1 and a vertical component R1, as previously shown. From basic trigonometry, we notice from Fig. 1.10 that there are two
right triangles. The first contains b and is formed by the x-axis, y-axis, and the line
drawn from P01 perpendicular to the line P1eP2. The second contains q and is formed
by the x-axis, the line drawn from P01 perpendicular to the line P1eP2, and line
P1eP2. Thus, these two right triangles share one common angle, the angle opposite
b from the first right triangle and the angle opposite q from the second right triangle.
Because there is a total of 180 degrees in a triangle, angles denoted by q and b must
be of the same magnitude. As such, F ¼ k v1 sin b ¼ k v1 sin q.
Note that the angle q, formed by the deformed spring and x-axis, is analogous to
the angle q shown previously in Fig. 1.9. By choosing the angle q, in contrast to
angle b, we will be using the same geometric configuration as that used in Step 1.
For the remaining steps, this same configuration of angle q will be used for consistency. We now decompose the force F to four components based on the static equilibrium conditions (Q1 ¼ Q2 and R1 ¼ R2), as shown in the following equations:
Q1 ¼ Q2 ¼ F cos q ¼ k v1 sin q cos q and
(1.36)
25
26
CHAPTER 1 Introduction
FIGURE 1.10
A free-body diagram showing a spring element with a unit vertical displacement of v1 ¼ 1,
while all other DOFs are assumed zero. Again, the angle q is formed by the axis of the
undeformed spring and the x-axis. We can easily determine from the corresponding rightangled (90 degrees) triangle (also known as rectangled triangle) that the angle b has the
same magnitude as the angle q.
R1 ¼ R2 ¼ F sin q ¼ k v1 sin2 q:
(1.37)
Again, we let cos q ¼ C and sin q ¼ S. Combining Eqs. (1.36) and (1.37), we write
the result in matrix form as
8
9
3
2
Q1 >
CS
>
>
>
>
>
>
6 2 7
<R >
=
6 S 7
1
7
6
:
(1.38)
v
k6
¼
1
7
>
Q2 >
>
>
4 CS 5
>
>
>
>
:
;
S2
R2
Step 3:
Similarly, when assuming u1 ¼ 0, v1 ¼ 0, u2 ¼ 1, and v2 ¼ 0, we write the corresponding forceedisplacement equations as
8
9
2
3
Q1 >
C 2
>
>
>
>
>
>
6
7
<R >
=
6 CS 7
1
7 u2 ¼
:
(1.39)
k6
6 2 7
>
Q >
>
4 C 5
> 2>
>
>
>
:
;
R2
CS
1.3 Sample Matrix Structural Analysis
Step 4:
If u1 ¼ 0, v1 ¼ 0, u2 ¼ 0, and v2 ¼ 1, the corresponding forceedisplacement
equations are represented by
8
9
3
2
Q1 >
CS
>
>
>
>
>
>
7
6
<R >
=
6 S2 7
1
7
(1.40)
k6
6 CS 7v2 ¼ > Q >:
>
5
4
> 2>
>
>
>
:
;
S2
R2
After this exercise, all four columns of the element stiffness matrix become
available. By assembling all four columns, we write the element stiffness matrix
[k] as
9 8
9
2
38
Q1 >
C2 CS
u1 >
C2 CS >
>
>
>
>
>
>
>
>
>
>
> >
>
6
2 7
6 CS S2
7< v1 = < R1 =
CS
S
7
k6
¼
(1.41)
6
7 u2 > > Q2 >:
2
C2 CS 5>
>
>
>
>
4 C CS
>
>
>
>
>
>
>
:
:
; >
;
v2
R2
CS S2
CS S2
The element stiffness matrix shown in Eq. (1.41) is a singular matrix, just like the
one given in Eq. (1.32). Unless we implement proper boundary conditions, no solution can be found for this equation. As a quick check for a special case, where q ¼ 0
(i.e., the spring is along the x-axis), and therefore cos q ¼ 1 and sin q ¼ 0, we can
write Eq. (1.41) as Eq. (1.42). We can easily see that Eq. (1.42) is the same as
Eq. (1.32), which is for the single spring aligned along the x-axis.
9 8
9
38
2
u1 >
Q1 >
1 0
1 0 >
>
>
>
>
>
>
>
>
>
> >
>
7>
6
7< v1 = < R1 =
6 0 0
0
0
7
6
k6
¼
(1.42)
1 0 7
u > >
Q >
>
5>
4 1 0
> 2>
> 2>
> >
>
>
>
>
>
:
; :
;
0 0
0 0
v2
R2
1.3.3 ELEMENT STIFFNESS MATRIX OF A HOMOGENEOUS LINEAR
ELASTIC BAR
Now, we consider a very similar problem, where a weightless straight bar, of length
L, elastic modulus E, and cross-sectional area A is aligned along the x-axis, as shown
in Fig. 1.11. This bar can be represented by a 2-node element for which each node
(P1 and P2) allows only one DOF. That is, there is an axial displacement at each node
with a total of two DOFs per element. We further assume that a positive value indicates that both the force and displacement are pointing to the right. In this figure, the
top part shows the bar in its undeformed (original) configuration, the middle portion
shows that the point P1 is displaced by u1, and the bottom shows that the point P2 is
displaced by u2.
27
28
CHAPTER 1 Introduction
FIGURE 1.11
A linear isotropic bar element of length L is represented by nodes P1 and P2 (top). Point P1
is deformed by a magnitude of u1 to P10 , while point P2 is restrained from any movement
(middle). Point P1 is restrained from any movement, while point P2 is deformed to P20 by a
displacement of u2 (bottom).
We know from beginner mechanics courses that for any homogeneous linear bar
of constant cross-sectional area, the cross-sectional area A, force F, length L, strain ε,
stress s, and Young’s modulus E have the following relationships:
F
DL
; and ε ¼
(1.43)
A
L
where DL is the magnitude of deformation along the axis of the bar. From these relationships, we determine that
s ¼ E ε; s ¼
AE
DL:
(1.44)
L
From Eq. (1.44), we can easily see the similarity between the linear isotropic bar
element and the spring element. If we replace the spring constant k by AE
L , the two
equations are identical. Again, assuming the nodal displacements u1 ¼ 1 and
u2 ¼ 0, the first column of the element stiffness matrix can be determined. If
u1 ¼ 0 and u2 ¼ 1, the second column of the element stiffness matrix can be identified. As such, the forceedisplacement equations of a linear isotropic bar along
the x-axis can be assembled as Eq. (1.45).
3
2
AE
AE
6 L
u
1 1
F1
L 7
7 1 ¼ AE
6
(1.45)
4 AE AE 5 u
L 1 1
F2
2
L
L
F¼
1.3 Sample Matrix Structural Analysis
Using the same definition for the rotation angle q as previously presented for the
spring element shown in Section 1.3.2, the element forceedisplacement equations
with four pseudo-DOFs of u1, v1, u2, and v2 can be written as Eq. (1.46),
9 8
9
2
38
Q1 >
u1 >
C2 CS
C 2 CS >
>
>
>
>
>
>
>
> >
>
>
>
>
6
2
2 7
7< v1 = < R1 =
CS
S
CS
S
AE 6
6
7
(1.46)
7 u2 > ¼ > Q 2 >
2
L 6
C2
CS 5>
>
>
>
>
4 C CS
>
>
>
>
>
>
> :
>
:
;
;
v2
R2
CS S2
CS
S2
where cos q ¼ C and sin q ¼ S.
1.3.4 GLOBAL STIFFNESS MATRIX OF MULTIPLE INLINE LINEAR
SPRINGS OR BARS
Top portion of Fig. 1.12 shows a structure consisting of two inline springs, with
spring constants k1 and k2. Similarly, the middle portion of the figure shows a structure consisting of two inline bars, with a constant Young’s moduli (E1 and E2) and
constant cross-sectional areas (A1 and A2). Both structures can be represented by the
same FE model consisting of three nodes (P1, P2, and P3) and two elements, as
shown in the bottom portion of the figure. For the loading condition, the net force
can be applied by any combination of the three nodal forces F1, F2, and F3. Finally,
the nodal displacements due to the applied forces are represented by u1, u2, and u3.
1
2
FIGURE 1.12
Top: A two-spring structure with corresponding spring constants of k1 and k2. Middle: A 2bar structure with constant cross-sectional areas A1 and A2 and constant elastic moduli
E1 and E2, respectively. Bottom: An FE model, made up of three nodes (P1, P2, and P3)
and two elements, representing the above two structures. Numbers in circles indicate the
element numbers.
29
30
CHAPTER 1 Introduction
The force equilibrium condition requires that F1 þ F2 þ F3 ¼ 0, and we can
determine the following from the free-body diagram representing this system:
F1 ¼ k1 ðu1 u2 Þ; F3 ¼ k2 ðu3 u2 Þ; and F2 ¼ k1 ðu1 u2 Þ k2 ðu3 u2 Þ:
(1.47)
We write these three equations in matrix format
8 9 2
k1
k1
>
=
< F1 >
6
F2 ¼ 4 k1 k1 þ k2
>
;
: >
F3
0
k2
as
38 9
>
=
< u1 >
7
k2 5 u2 .
>
;
: >
k2
u3
0
(1.48)
Eq. (1.48) is called the global forceedisplacement equation, and the matrix is
called the global stiffness matrix, denoted as [K]. The method mentioned above is
quite straight forward. However, as the number of springs increases, writing these
individual equations from the free-body diagram becomes complex. As an alternative, we can write the stiffness matrix of each individual element according to the
single spring example shown in Section 1.3.1. This alternative is shown as follows.
Step 1: Form element stiffness matrices
From Eq. (1.32), the stiffness matrices for elements 1 and 2 are written as:
(
½kelement 1
½kelement 2
u1
)
¼
u2
u2
u3
"
¼
k1
k1
k1
k1
k2
k2
k2
k2
#(
38 9 8 9
0 >
> u1 >
> F1 >
>
> >
u1
7< = < =
6
7
6
04 k1 k1 0 5 u2 ¼ F2 and
>
> >
>
u2
>
: >
;
; >
: >
0
0
0
u3
0
(1.49)
38 9 8 9
2
0
0
0 >
= >
< 0 >
=
< u1 >
u2
7
6
04 0 k2 k2 5 u2 ¼ F2 :
>
u3
; >
: >
;
: >
0 k2 k2
u3
F3
)
2
k1
k1
(1.50)
Step 2: Assemble element stiffness matrices into the global stiffness matrix
The right-hand sides of Eqs. (1.49) and (1.50) contain one additional row and one
additional column (those added “0” terms) not seen on the left-hand sides of the
same equations. By adding an extra row and column in each equation, it is much
easier to assemble these element stiffness matrices into a global stiffness matrix
[K]. We simply place the corresponding items (or entries) into proper positions
within the matrix. Eq. (1.51) shows the global forceedisplacement relationship representing the entire structure obtained through the assembly of the two element stiffness matrices. As we expected, Eqs. (1.48) and (1.51) are identical, even though we
used two different approaches to derive the global stiffness matrix.
8 9 2
3 8 9 8 9
k1
k1
0
>
>
< u1 >
=
= >
< F1 >
=
< u1 >
6
7
½K u2 ¼ 4 k1 k1 þ k2 k2 5
(1.51)
u 2 ¼ F2
>
>
: >
;
; >
: >
;
: >
u3
0
k2
k2
u3
F3
1.3 Sample Matrix Structural Analysis
As stated earlier regarding the singular nature of this matrix, no solution can be
found for Eqs. (1.48) or (1.51), because boundary and loading conditions are not
provided. Assuming a boundary condition where the structure is fixed on the left
side at P1 (i.e., u1 ¼ 0), a loading condition with a force of 100 dimensionless units
toward the right side is applied at P3 (i.e., F3 ¼ 100), and the two dimensionless
spring constants are k1 ¼ 10 and k2 ¼ 15, we write the forceedisplacement
equations as Eq. (1.52). Note that F2 is written as zero, because no force is applied
at that node.
9
2
3 8 9 8
10
10
0
>
= >
< F1 >
<0>
=
6
7
u2 ¼
0
(1.52)
4 10 10 þ 15 15 5
>
>
; >
:
: >
;
0
15
15
100
u3
Under the fixed boundary condition at P1 (u1 ¼ 0), Eq. (1.52) can be decomposed
into three equations:
10u2 ¼ F1
25u2 15u3 ¼ 0
15u2 þ 15u3 ¼100:
We can clearly see that the first equation cannot be solved, because both sides of the
equation contain unknown variables. However, the second and third linear equations
with two unknowns (u2 and u3) can easily be solved simultaneously. Since we know
u1 ¼ 0, there is no need to solve for it. We can obtain all the displacement information we need (u2 and u3) by solving the second and third equations, and the first
equation is not needed. We deduce from this example that the nth row and nth column can be eliminated from further consideration when a fixed (zero) boundary condition is prescribed at the nth DOF. We eliminate the first row and first column of Eq.
(1.52) for further calculations. With the remaining two equations, we find u2 ¼ 10
and u3 ¼ 16.67. We can
P find the reaction force F1 ¼ 100 either by using the force
equilibrium equation F ¼ 0, or by using Eq. (1.52) with the calculated values for
u2 and u3 directly plugged into the first equation in the global forceedisplacement
equation.
For the structure with two bars shown in Fig. 1.12, we can represent the corresponding forceedisplacement equations with Eq. (1.53), which is essentially Eq.
(1.51) with each spring constant set to the stiffness AE
L of the corresponding bar.
2
3
A1 E1
A1 E1
0 7
8 9 6
L1
6 L1
7 8 9 8 9
u
>
>
6
7 >
1
< u1 >
< = 6 A E A E
= >
< F1 >
=
A2 E2
A2 E2 7
1 1
1 1
6
7
½K u2 ¼ 6 ¼
(1.53)
u
F
þ
2
2
>
L1
L1
L2
L2 7
: >
: >
; 6
; >
;
: >
7 >
u3
u3
F3
6
7
4
A2 E2
A2 E2 5
0
L2
L2
31
32
CHAPTER 1 Introduction
1.3.5 GLOBAL STIFFNESS MATRIX OF A SIMPLE BIOMECHANICS
PROBLEM
Example 1.1
Fig. 1.13 shows a simplified arm with humerus, radius (and ulna), and elbow
joint. A handbag weighs 50 N is hung at P2. Neglecting the weight of the arm,
calculate the biceps force needed to maintain equilibrium using (1) methods
learned in statics, and (2) an FE model formulated with bar elements.
FIGURE 1.13
The biceps brachii (element 2) is activated to resist vertical load of 50 N applied at
insertion point P2. Elements 1 and 3 represent a section of radius and humerus,
respectively. The center of rotation for the elbow joint is located at P1. P3 is the origin of
the biceps brachii. The distance between P1 and P2 is 3 cm and between P1 and P3 is
20 cm.
Solution
Problems related to calculating
the muscle
force are typically solved using the
P
P
equilibrium equations F ¼ 0 and M ¼ 0 learned in statics. Here, the FE
method is used to find the bicep force using a 3-node, 3-element FE model to
calculate the muscle force.
A muscle produces force along the axial direction. For the biceps to rotate the
forearm about the elbow joint, a moment is needed. Here moment is defined as the
magnitude of the tendency of force to rotate a body about a specific point or axis,
and it is equal to the product of the linear force and moment arm. A moment arm
is the perpendicular distance between the line of action of the force and the center
1.3 Sample Matrix Structural Analysis
of rotation. For this example, the center of rotation is located at P1, and the
moment arm is the perpendicular distance between line P2eP3 and point P1.
To maintain this static equilibrium configuration, the clockwise moment (usually denoted as negative moment) caused by the handbag needs to be resisted by
the counterclockwise moment (designated as positive moment) generated by the
biceps. Using the free-body diagram shown on theright-hand side of Fig. 1.13,
the angle P1eP3eP2 is calculated as q ¼ tan1
3
20
¼ 8:53 . The moment
arm of the biceps can be calculatedP
as 3 sin(90 8.53 ) ¼ 2.967 cm. Based
on the static equilibrium equation M ¼ 0, we calculate from Eq. (1.54) that
the biceps’ contraction force is 50.556 N:
X
M ¼ Fbiceps 2:967 50 3 ¼ 0 and Fbiceps ¼ 50:556 N.
(1.54)
We shall now solve the same problem using the FE method. As mentioned
previously, the global stiffness matrix [K] can be obtained by assembling the stiffness matrices of the individual elements. We let the axial stiffness of the radius be
k1, the axial stiffness of the biceps muscle be k2, and that for the humorous be k3.
Using these assumptions, Eqs. (1.55)e(1.57) show the three corresponding
element stiffness matrices with the associated DOFs.
u1
1 1
u1
½kelement 1
¼ k1
(1.55)
1 1
u2
u2
When considering values to plug into Eq. (1.46), care must be taken to ensure
that a proper angle is entered. For the bar element formed by points P2eP3, a
good way to ensure correctness is to first draw a line from P2 along the positive
horizontal axis, and then measure the angle from this horizontal axis to line
P2eP3. Since element 2 is 8.53 degrees beyond the vertical axis, the angle of
rotation for element 2 is q ¼ 98.53 degrees. As such, cos q ¼ 0.148 and sin
q ¼ 0.989. Plugging these values into Eq. (1.46) results in
8
9
9
38
2
u2 >
u2 >
0:022 0:146 0:022 0:146 >
>
>
>
>
>
>
>
>
>
>
>
>
7>
6
<
7 < v2 =
6 0:146 0:978
0:146
0:978
v2 =
7
¼ k2 6
.
½kelement 2
6 0:022 0:146
>
0:022 0:146 7
u3 >
u3 >
>
>
>
>
5>
4
>
>
>
>
>
>
>
>
:
:
;
;
0:146 0:978 0:146 0:978
v3
v3
(1.56)
For element 3, the angle of rotation q is 90 degrees, cos q ¼ 0, and sin q ¼ 1.
1 1
v1
½kelement 3 ¼ k3
(1.57)
1 1
v3
33
34
CHAPTER 1 Introduction
Now, we assemble the three element stiffness matrices shown in Eqs. (1.55)
e(1.57), as described in Step 1 of Section 1.3.4, to form the global stiffness matrix. Eq. (1.58) shows the resulting global stiffness matrix and its corresponding
nodal DOFs.
9
2
38
u1 >
k1
0
k1
0
0
0
>
>
>
>
>
6
7>
>
>
6 0
7>
>
>
0
0
k
k
0
v
3
3
1
>
>
>
6
7>
>
>
>
>
6
7<
=
6 k1 0 k1 þ 0:022k2
7
0:146k
0:022k
0:146k
u
2
2
2
2
6
7
6 0
> v2 >
>
0
0:146k2
0:978k2
0:146k2
0:978k2 7
6
7>
>
>
>
6
7>
>
>
>
>
6 0
7
>
0
0:022k
0:146k
0:022k
0:146k
u
2
2
2
2 5>
3 >
>
4
>
>
>
>
:
;
0 k3 0:146k2
0:978k2 0:146k2 0:978k2 þ k3
v3
(1.58)
Because the purpose of this example is to determine the biceps force needed to
balance the moment generated by the handbag, it is assumed that the elbow joint
has a fixed support while the humerus has a roller support. We can apply the zero
displacement boundary conditions at u1, v1, and u3 to represent the pinned and
roller supports as discussed in Step 2 of Section 1.3.4 by eliminating the first,
second, and fifth columns and rows. The forceedisplacement equation becomes
8 9 2
38 9
k1 þ 0:022k2 0:146k2
u2 >
0:146k2
>
>
>
>
> u2 >
>
< = 6
7< =
6
7
½K v2 ¼ 4 0:146k2
0:978k2
0:978k2 5 v2
>
>
>
>
>
>
;
;
: >
: >
v3
0:146k2
0:978k2 0:978k2 þ k3
v3
9
8
9 8
(1.59)
0 >
F2H >
>
>
>
>
>
>
=
<
= <
¼ F2V ¼ 50
>
>
>
>
>
>
>
;
:
:
; >
0
F3V
where the subscript H indicates that the force is in a horizontal direction, and the
subscript V indicates a vertical force. There are a number of ways to solve Eq.
(1.59), such as using Cramer’s rule, the matrix inversion method, or Gaussian
elimination.
For simplicity, let us assume that the axial stiffness of the biceps (k2) is k and
the axial stiffness for each of the two bones (k1 and k3) is 1000k. With these assumptions, Eq. (1.59) becomes
9
9 8
2
38 9 8
1000:022 0:146
0:146
>
=
= >
< F2H >
= >
< 0 >
< u2 >
6
7
k4 0:146
0:978
0:978 5 v2 ¼ F2V ¼ 50 : (1.60)
>
>
>
;
; >
:
; >
:
: >
0
0:146
0:978 1000:978
v3
F3V
1.3 Sample Matrix Structural Analysis
For this example, we first multiply both sides of the equation by [K]1 (the
inverse of the [K] matrix) and then we solve for u2, v2, and v3. Because [K]1 is
8 2
391
2
3
0:146
0:001 0:00015
0
>
>
< 1000:022 0:146
=
6
7
6
7
k4 0:146
0:978
0:978 5
¼ 1k 4 0:00015 1:0235 0:001 5,
>
>
:
;
0:146
0:978 1000:978
0
0:001 0:001
1
multiplying [K] to both sides of Eq. (1.60) produces
9
9
8
8 9
2
38
0 >
0 >
u2 >
0:001
0:00015
0
>
>
>
>
>
>
>
>
>
=
= 16
<
< =
7<
7 50
0:00015
1:0235
0:001
v2 ¼ ½K1 50 ¼ 6
4
5
>
>
>
>
>
>
k
>
>
>
>
>
;
;
:
;
:
: >
0
0
0
0:001
0:001
v3
9
8
0:0075 >
>
>
>
=
<
1
¼
51:176 .
>
k>
>
>
;
:
0:05
(1.61)
To determine the force generated by the biceps, we need to first find the length
change of this muscle. Assuming the biceps has a unit stiffness (i.e., k ¼ 1 N/cm),
the nodal displacements determined from Eq. (1.61) are u2 ¼ 0.0075,
v2 ¼ 51.176, and v3 ¼ 0.05, where all units are in cm. From these nodal displacements, we calculate that P2, which was originally located at (3, 0), has
moved a distance of u2 horizontally and v2 vertically to P02 , located at (2.9925,
51.176); and P3 is moved from (0, 20) a distance of v3 vertically to P03 at
(0, 19.95). Next, we calculate the length change by finding the original length
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
of the biceps P2 P3 ¼ 32 þ 202 ¼ 20:224 and the deformed length of
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
0
0
the same muscle
P2 P3 ¼ 2:99252 þ ð51:176 19:95Þ2 ¼ 71:19 .
Using Hooke’s law, we compute the muscle force as Fmuscle ¼ k (71.19
20.224) ¼ 50.966 N. This solution, in which the stiffness k is assumed to
have a unit value, differs by 0.41 N (less than 1% error) from the solution found
using statics. With larger values of k, the nodal displacements are much smaller
(and more realistic), and the results approach those obtained using statics. Eq.
(1.62) shows the calculated nodal displacements with the value of k assumed to
be 1000. As expected, the corresponding muscle force is 50.3 N, which is
much closer to the value found using statics (less than 0.5% error).
fu2 v2 v3 gT ¼ 103 f 0:0075 51:176 0:05gT
(1.62)
Part of this difference came from roundoff errors introduced when manual calculations were performed. Also, perfect equilibrium (i.e., all nodal displacements
are zero) is assumed when using the static solution, whereas the FE method is
based on deformable mechanics and must allow some displacements or deformations of the elements to occur.
35
36
CHAPTER 1 Introduction
1.3.6 GLOBAL STIFFNESS MATRIX OF A SIMPLE TRUSS BRIDGE
Example 1.2
A simple truss bridge has a total span of 8 m, and a height of 3 m, and is subjected
to a vertical downward force of 50 kN and a horizontal wind force of 20 kN toward the right side at P4. At point P1, the bridge is pinned to its foundation while
it is on a roller support at point P3. Each truss member has a cross-sectional
dimension of 0.1 0.1 m, a Young’s modulus of 200 GPa, and an ultimate
strength of 400 MPa. As such, the stiffness
AE
L
values for the five elements
are: 5 108, 5 108, 4 108, 4 108, and 6.67 108 N/m, respectively, for elements 1, 2, 3, 4, and 5. Fig. 1.14 shows the FE model created to represent this
bridge. Establish the global forceedisplacement equation.
FIGURE 1.14
A truss-bridge FE model with four nodes (P1, P2, P3, and P4) and five elements. The
numbers in circles indicate the element numbers. Each node possesses two
translational DOFs, for a total of eight DOFs. The boundary conditions eliminate three
DOFs (u1, v1, and v3). The angles formed by elements 1 and 4 and by elements 2 and
3 are 36.9 degrees, based on the geometric relationships.
Solution
Based on Section 1.3.3, we can easily write the element stiffness matrices and
their corresponding DOFs for elements 1 and 2 formed by P1eP2 and P2eP3,
respectively, because both elements are parallel to the x-axis. For element 3
formed by P3eP4, the angle is 126.9 degrees (i.e., cos q ¼ 0.6 and sin
q ¼ 0.8). Similarly, the angle is 36.9 degrees for element 4 formed by P1eP4
1.3 Sample Matrix Structural Analysis
(i.e., cos q ¼ 0.8 and sin q ¼ 0.6), and the angle is 90 degrees for element 5
formed by P2eP4 (i.e., cos q ¼ 0 and sin q ¼ 1). From Eqs. (1.45) and (1.46),
we write the element-based equations as follows:
8
9
8
u1 >
>
>
>
>
>
>
>
>
>
>
>
>
>
< v1 >
=
<
¼ ½kelement P1 P2
½kelement 1
> u2 >
>
>
>
>
>
>
>
>
>
>
>
>
>
:
;
:
v2
9
u1 >
>
>
>
>
v1 =
8
9
8
u2 >
>
>
>
>
>
>
>
>
>
>
>
>
>
>
< v2 =
<
¼ ½kelement P2 P3
½kelement 2
> u3 >
>
>
>
>
>
>
>
>
>
>
>
>
>
:
;
:
v3
9
u2 >
>
>
>
>
v2 =
2
0
6
6 0 0
¼ 5 10 6
6 1 0
u2 >
>
>
4
>
>
;
0 0
v2
86
2
1
6
6 0
6
¼ 5 108 6
6 1
u3 >
>
>
4
>
>
;
0
v3
8
9
8
9
u3 >
u3 >
>
>
>
>
>
>
>
>
>
>
>
>
<v >
=
<v >
=
3
3
¼ ½kelement P3 P4
½kelement 3
> u4 >
>
>
u4 >
>
>
>
>
>
>
>
>
>
>
>
:
;
:
;
v4
v4
2
0:36 0:48
6
6 0:48 0:64
¼ 4 108 6
6 0:36 0:48
4
0:48
8
9
8
u1 >
>
>
>
>
>
>
>
>
>
>
<v >
=
<
1
½kelement 4
¼ ½kelement P1 P4
>
>
u4 >
>
>
>
>
>
>
>
>
>
:
;
:
v4
2
0:64
6
6 0:48
¼ 4 108 6
6 0:64
4
0:48
and
1
0:64
0
0
0
0
0:36
0:48
0:36
0:48
9
u1 >
>
>
>
v =
9
38
u1 >
>
>
>
>
>
>
7>
>
< v1 >
=
0 0 7
7
;
7
> u2 >
1 0 7
>
>
>
5>
>
>
>
>
:
;
0 0
v2
(1.63)
9
38
1 0 > u2 >
>
>
>
>
>
7>
>
< v2 >
=
0 0 7
7
;
7
> u3 >
1 0 7
>
>
>
5>
>
>
>
>
:
;
0 0
v3
(1.64)
1
0
9
38
u3 >
>
>
>
>
>
7>
<v >
=
0:64 7
3
7
;
0:48 7
u4 >
>
>
5>
>
>
>
>
:
;
0:64
v4
(1.65)
0:48
1
u4 >
>
>
>
;
v4
0:48
0:64
0:36
0:48
0:48
0:64
0:36
0:48
38
0:48 >
>
>
7>
<
0:36 7
7
0:48 7
>
5>
>
>
:
0:36
9
u1 >
>
>
>
v1 =
;
u4 >
>
>
>
;
v4
(1.66)
37
38
CHAPTER 1 Introduction
8
9
8
u2 >
>
>
>
>
>
>
>
>
>
>
>
>
>
< v2 >
=
<
¼ ½kelement P2 P4
½kelement 5
>
>
u4 >
>
>
>
>
>
>
>
>
>
>
>
>
:
;
:
v4
9
u2 >
>
>
>
>
v2 =
2
0
0
6
6 0
1
¼ 6:67 10 6
>
6
u4 >
0
>
40
>
>
;
v4
0 1
86
9
38
u2 >
>
>
>
>
>
>
7>
>
=
< v2 >
0 1 7
7
:
7
> u4 >
0
0 7
>
>
>
5>
>
>
>
>
;
:
v4
0
1
(1.67)
0
0
The first step toward finding the solution is to write the forceedisplacement
equilibrium equations by assembling all five element stiffness matrices into a
global stiffness matrix [K]. Eq. (1.68) lists this matrix. Students should make it
a habit to ensure that it is a symmetrical matrix as the first step to partially ensure
that no mistakes were made.
3
2
7:56
1:92 5:00 0:00
0:00
0:00 2:56 1:92
6 1:92
1:44
0:00
0:00
0:00
0:00 1:92 1:44 7
7
6
7
6
6 5:00 0:00
10:00
0:00 :5:00 0:00
0:00
0:00 7
7
6
6
0:00
0:00
6:67
0:00
0:00
0:00 6:67 7
7
8 6 0:00
½K ¼ 10 6
7
7
6 0:00
1:92
1:44
1:92
0:00
5:00
0:00
6:44
7
6
6 0:00
0:00
0:00
0:00
1:92
2:56
1:92 2:56 7
7
6
7
6
4 2:56 1:92 0:00
0:00
1:44
1:92
4:00
0:00 5
1:92
1:44
0:00
6:67
1:92
2:56
0:00
10:67
(1.68)
The second step is to look at the boundary conditions. With pinned and roller
boundary conditions, we can eliminate the first, second, and sixth columns and
rows. We write the remaining 55 matrix as
9
8
38 u 9
2
u2 >
2 >
>
>
10
0
5
0
0
>
>
>
>
>
>
>
>
>
>
>
>
>
7
>
>
>
6
>
>
>
>
v
v
0
6:67
0
0
6:67
2
2
>
>
>
>
7
6
<
=
=
<
7
6
86
¼ 10 6 5
0
6:44 1:44 1:92 7
½K u3
u
3
7
>
>
>
>
>
>
>
7>
6
>
0
1:44
4
0 5>
>
>
>
>
4 0
u4 >
u4 >
>
>
>
>
>
>
>
>
>
>
>
>
:
;
;
:
0
6:67
1:92
0
10:67
v4
v4
9
8
9 8
F2H > >
0
>
>
>
>
>
>
>
>
>
> >
>
>
>
>
>
>
>
>
>
>
>
0
F
>
>
=
< 2V >
<
= >
:
(1.69)
¼
¼
0
>
> F3H >
>
> >
>
>
>
>
>
>
>
>
> F4H >
> 20000 >
> >
>
>
>
>
>
>
>
;
:
:
; >
50000
F4V
1.3 Sample Matrix Structural Analysis
Solving the remaining five simultaneous equations for the five unknowns reveals all of the nodal displacements, u2, v2, u3, u4, and v4, in Eq. (1.69). However,
as the number of DOFs becomes larger, it becomes increasingly difficult to solve
these kinds of problems manually. A computer program could be written to
resolve this issue. In the previous example, the inverse stiffness matrix [K]1 is
multiplied to both sides of the equation before the nodal displacements are found.
We can use the same set of procedures for this example. We could also use the
Gaussian elimination method taught in linear algebra classes. In the next section,
we briefly review this method.
1.3.7 GAUSSIAN OR GAUSS ELIMINATION
Gaussian elimination (also known as Gauss elimination) is a commonly used
method for solving systems of linear equations with the form of [K] {u} ¼ {F}.
In matrix operations, there are three common types of manipulation that serve to
produce a new matrix that possesses the same characteristics as the original:
1. Interchange any two rows.
2. Multiply each entry in any row by a nonzero constant value.
3. Add the values from each entry of one row to each entry of another row.
The goal of using Gaussian elimination is to produce a new matrix with the same
properties as the original [K], but in a format in which only the upper triangle has
nonzero entries. Using the previous 55 matrix as an example, the upper triangle
consists of entries in the upper right triangle of the matrix and includes the entries
in the right diagonal line, with the form
2
3
m11 m12 m13 m14 m15
6
7
m22 m23 m24 m25 7
6 0
6
7
6 0
0
m33 m34 m35 7
6
7.
6
7
0
0
m44 m45 5
4 0
0
0
0
0
m55
We achieve the goal of Gaussian elimination by properly applying one of the
aforementioned three operations at a time. After the upper triangular matrix is
formed, we use the backward substitution method to solve for the last variable first.
The reason to call this method “backward substitution” is that the last row of the
upper triangular matrix needs to be solved first. Because there is only one nonzero
entry in the last row of the upper triangular matrix, we can find the unknown variable
by simple arithmetic division, that is, from
39
40
CHAPTER 1 Introduction
9
8
u2 > 2 m
>
>
>
11
>
>
>
>
>
>
6
>
>
v
0
>
= 6
< 2 >
6
¼6
½K u3
6 0
>
>
>
>
6
>
>
>
4 0
u4 >
>
>
>
>
>
>
;
:
0
v
4
m12
m22
m13
m23
m14
m24
0
0
m33
0
m34
m44
0
0
0
38
m15 >
>
>
>
7>
m25 7>
>
7<
m35 7
7>
7>
m45 5>
>
>
>
>
m55 :
9 8
u2 > >
>
>
>
>
> >
>
>
>
v2 >
>
<
= >
¼
u3 > >
> >
>
>
>
u4 >
>
>
>
>
>
:
; >
v4
9
F2H >
>
>
>
>
F2V >
>
=
F4V
F3H >/v4 ¼ m :
55
>
>
F4H >
>
>
>
;
F4V
Having the value of v4, we solve for the second to the last variable. Since
45 v4
m44u4 þ m45v4 ¼ F4H, we can solve for u4 as u4 ¼ F4H m
. We repeatedly apply
m44
the same set of procedures until the values of all variables have been found.
We will use a typical 64-bit computer to illustrate a critical issue when using
Gaussian elimination. It is well known that such a computer stores a real (decimal)
number in a floating-point format using 64 bits: 1 bit to represent the sign (plus or
minus), 52 bits to represent the number of precision digits (mantissa), and 11 bits
to represent the exponent. When dividing a number by another very small number,
the available digits in the mantissa may not be sufficient to maintain the needed precision, that is, round-off error may occur. In Gaussian elimination, a pivot point or
pivot position is the position in a row that coincides with the right diagonal line. The
values at pivot points are used as denominators in forming the upper triangular matrix. To eliminate round-off errors introduced by division with a very small number,
the first type of manipulation is used to move the row with a very small number at the
pivot point to another row. This is done by simply interchanging rows such that large
numbers are located at the pivot positions. We use the second and third operations to
obtain zeros in the lower left part of the matrix, which is necessary to obtain the
upper triangular matrix.
A modified version of the Gauss elimination method is the GausseJordan elimination method. The goal of GausseJordan elimination is to end up with a matrix
that has a right diagonal line of all ones (1’s) with all other positions of the
matrix containing zeros. This is accomplished by using the same three types
of matrix manipulations used in the Gauss elimination method. Because the square
matrix consists of only unit values in the diagonal entries, the solutions for all the
unknowns become readily available. One disadvantage of the GausseJordan method
lies in the fact that it is computationally more expensive than the Gauss elimination
method. As such, it is only useful for solving problems by manual calculation when
there are a small number of simultaneous equations. By using the Gaussian elimination method rather than the GausseJordan method, we avoid many additional steps.
Because the FE method usually involves a large system, the Gauss elimination
method is more commonly used.
In the following section, we demonstrate, in a step-by-step manner, the processes
in the Gauss elimination method. Of course, a computer program should be written
and used instead of manual calculations. Using the previous example as the starting
point, Eq. (1.68) is repeated below.
1.3 Sample Matrix Structural Analysis
9
8
2
u2 >
>
10
0
5
0
>
>
>
>
>
>
6
>
>
>
>
>
6 0
v2 >
6:67
0
0
>
>
>
>
6
=
<
6
¼ 108 6
0
6:44 1:44
½K u3
6 5
>
>
>
>
6
>
>
>
>u >
6 0
0
1:44
4
>
4 >
>
4
>
>
>
>
>
>
;
:
0 6:67 1:92
0
v4
9
8
9 8
F2H > >
0
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
> >
>
>
>
>
>
>
>
>
0
F
2V
>
>
>
>
>
>
>
=
<
<
= >
¼
¼
0
F3H > >
>
>
>
>
>
>
>
>
>
>
>
>
>
>
> F4H >
> 20000 >
> >
>
>
>
>
>
>
>
>
>
>
>
>
> >
;
:
:
;
50000
F4V
38
>
>
>
>
7>
>
>
6:67 7
>
7>
7<
7
1:92 7
>
7>
>
>
0 7
>
5>
>
>
>
10:67 :
0
9
u2 >
>
>
>
>
>
v2 >
>
>
=
u3 >
>
>
>
u4 >
>
>
>
>
;
v4
The first step is to make all but the first entry in column one equal to 0. We notice
that row three in this column contains the only nonzero value. In order to manipulate
row three to make the leading number 0, we must multiply the existing number (5)
by a value such that adding the result to the first entry in row one (10) produces 0.
Using rule two, we multiply each entry in row three by 2:
2
10
6
6 0
6
86
10 6 10
6
4 0
0
0
5
0
6:67
0
0
0
0
12:88
1:44
2:88
4
6:67
1:92
0
38
>
>
>
>
7>
6:67 7>
>
7<
7
3:84 7
>
7>
0 5>
>
>
>
>
10:67 :
0
9
9 8
u2 > >
0
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
0
v2 >
>
>
=
= <
:
0
u3 > ¼ >
>
>
>
>
>
>
>
>
>
>
u4 > > 20000 >
>
>
>
>
>
;
:
; >
50000
v4
Next, we add row 1 to row 3, but we do not affect row 1:
2
10
0
6
6:67
6 0
6
86
10 6 0
0
6
0
4 0
0 6:67
5
0
0
0
7:88
2:88
1:44
1:92
4
0
38
>
0
>
>
>
7>
6:67 7>
>
7<
7
3:84 7
>
7>
>
0 5>
>
>
>
10:67 :
9
9 8
u2 > >
0
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
0
v2 >
>
>
=
= <
:
0
u3 > ¼ >
>
>
>
>
>
>
>
>
>
>
u4 > > 20000 >
>
>
>
>
>
;
:
; >
50000
v4
Now that all but the first value in column one are equal to 0, and we apply a
similar process to column two. We want all but the second value in column two
to equal 0, and this means we must address 6.67 in the final row. This can be
changed to 0 simply by adding the values from row 2.
41
42
CHAPTER 1 Introduction
2
10
6
6 0
6
108 6
6 0
6
4 0
5
0
0
6:67
0
0
0
0
0
7:88 2:88
1:44
4
0
1:92
0
38
>
0
>
>
>
7>
6:67 7>
>
7<
3:84 7
7>
7>
0 5>
>
>
>
>
:
4
9
9 8
0
u2 > >
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
0
v2 >
>
>
=
<
= >
¼
0
u3 > >
>
>
> >
>
>
>
>
>
20000
u4 >
>
>
>
>
>
>
>
>
>
;
; :
50000
v4
Two operations, rules two and three, are needed to transform the entry in row
four, column three to 0. We first multiply row four by 7:88
1:44 (notice this operation
also applies to the force vector):
2
5
0
10
0
6
6 0 6:67
6
0
108 6
6 0
6
0
4 0
7:88
7:88
0
1:92
0
38
>
0
>
>
>
7>
6:67 7>
>
7<
7
2:88 3:84 7
>
7>
21:89
0 5>
>
>
>
>
:
0
4
0
0
9 8
0
u2 > >
>
>
>
>
>
>
>
> >
0
v2 >
>
<
= >
0
u3 > ¼ >
>
>
>
>
> 1:094 105
> >
u4 >
>
>
>
:
; >
v4
50000
9
>
>
>
>
>
>
>
=
;
>
>
>
>
>
>
>
;
and then add the values from row three to these results to form the new row four:
2
10
6
6 0
6
86
10 6 0
6
4 0
0
0
6:67
0
5
0
0
0
7:88 2:88
0
0
19:01
0
1:92
0
38
>
>
>
>
7>
6:67 7>
>
7<
7
3:84 7
>
7>
3:84 5>
>
>
>
>
:
4
0
9 8
0
u2 > >
>
>
>
>
>
>
>
>
>
0
v2 >
>
<
= >
0
u3 > ¼ >
>
>
>
>
>
>
u4 > > 1:094 105
>
>
>
:
; >
v4
50000
9
>
>
>
>
>
>
>
=
>
>
>
>
>
>
>
;
:
Likewise, we multiply row five by 7:88
1:92, and then add the values from row three
to form the new row five:
2
10
0
6
6 0 6:67
6
0
108 6
6 0
6
0
4 0
0
0
5
0
0
0
7:88 2:88
0
0
19:01
2:88
38
>
0
>
>
>
7>
>
6:67 7>
7<
7
3:84 7
>
7>
3:84 5>
>
>
>
>
12:58 :
9 8
0
u2 > >
>
>
>
>
>
>
>
> >
0
v2 >
>
<
= >
0
u3 > ¼ >
>
>
>
>
5
u4 >
> 1:094 10
> >
>
>
>
>
; :
v4
2:052 106
9
>
>
>
>
>
>
>
=
>
>
>
>
>
>
>
;
:
The pattern should be apparent by now that multiplying by a value in another
row and then dividing by the value in the current row gives a result that can be
subtracted from that other row to produce 0. To see this process once again, we
1.3 Sample Matrix Structural Analysis
multiply row five by
row five:
2
10
6
6 0
6
108 6
6 0
6
4 0
0
0
19:01,
2:88
5
then add the values from row four to produce the new
0
6:67
0
0
0
0
7:88
0
2:88
19:01
0
0
0
38
>
>
>
>
7>
6:67 7>
>
7<
3:84 7
7>
7>
3:84 5>
>
>
>
>
79:20 :
0
9 8
0
u2 > >
>
>
>
>
>
>
>
>
>
0
v2 >
>
<
= >
0
u3 > ¼ >
> >
>
>
>
u4 >
1:094
105
>
> >
>
>
>
; :
v4
1:464 106
9
>
>
>
>
>
>
>
=
>
>
>
>
>
>
>
;
:
The matrix now has the form of an upper triangular matrix, meaning all values
below and to the left of the right diagonal line are zeros. At this point, we apply the
backward substitution method to determine nodal displacements.
We start with the last row, which contains 108 ½ 0 0 0 0 79:20 , and we
multiply sequential values in this row by sequential values in the nodal-displacement
vector:
108 ðð0Þðu2 Þ þ ð0Þðv2 Þ þ ð0Þðu3 Þ þ ð0Þðu4 Þ þ ð79:20Þðv4 ÞÞ ¼ 1:464 106
We can do this more simply by recognizing that only the last value in the row is
nonzero, and therefore v4 is simply the final value in the force vector divided by the
last entry in the upper triangular matrix [K]:
1:464 106
¼ 1:849 104
v4 ¼
ð79:2 108 Þ
We can use v4 to find u4, and so on. Following are the computations for the values
of the nodal displacements in m:
u4 ¼
1:094 105 108 3:84 v4 1:804 105
¼
¼ 0:949 104 ;
19:01 108
19:01 108
u3 ¼
108 ð2:88Þ u4 108 3:84 v4 9:837 104
¼
¼ 1:248 104 ;
108 7:88
7:88
v2 ¼
108 6:67 v4
¼ 1:849 104 ; and
108 6:67
108 5 u3
¼ 0:624 104 .
108 10
The nodal displacements calculated using the MSA or direct stiffness method
are exactly the same as the exact solutions for truss-related problems. For element
types other than truss or spring, the nodal solutions are not likely to have the same
values as the exact solutions. A simple rule of thumb is that the more the elements
are used to represent the structure of interest, the more closely the results
will approach the exact solutions. More descriptions for other element types are
provided in Chapter 2.
u2 ¼
43
44
CHAPTER 1 Introduction
1.4 FROM MSA TO A FINITE ELEMENT MODEL
As demonstrated in the previous section, the MSA method is convenient for
analyzing framed structures. Although that section is mainly used to demonstrate
the MSA method, the concepts of nodes and elements in the FE method are intentionally embedded to highlight the similarities between these two methods. However, the MSA method is limited to only solving truss or framed structures. Real-world
problems tend to consist of surface (2D) or solid (3D) components in addition to
framed structures. For these structures, the FE method is needed. Also, demonstrated
in Section 1.3, both the MSA and FE methods can be formed directly by using the
forceedisplacement relationship of each truss member to form the element stiffness
matrix before the corresponding structure stiffness matrix is assembled. As such,
both the MSA and FE methods yield the exact solution at the nodes. Because we
may not be able to directly obtain such displacement functions for other types of
structures, other principles of mechanics, such as the workeenergy principle, are
needed to formulate the element stiffness matrix. These other relationships will be
explored in Chapter 4.
The formulations for establishing the element stiffness matrices for all elements
and then assembling these matrices into the structure (global) stiffness matrix are
embedded in software packages and are transparent to software users. For this
reason, some users may not pay attention to the element formulation details, and
this may lead to misuse of the software or misinterpretation of the results. Now
that we have reviewed formulations of a truss structure, we will go through the
four key steps in developing an FE truss-structure model. These same four steps
can be used to develop any FE model.
An FE model is basically a numerical exemplification of a real-world structure
subjected to loading. There are four key steps in the development of any FE model:
1.
2.
3.
4.
Idealize and discretize the structure into meshes.
Select the governing material laws.
Establish boundary conditions.
Input loading conditions.
Each of these four steps will be separately described in the chapters that follow.
The following sections briefly discuss the basic concepts of forming an FE mesh for
structural analysis and solving injury biomechanics problems.
In the 1960s, a number of open-source FE software programs were freely available for solving practical problems in the fields of aerospace, mechanical, and civil
engineering. These programs were created to allow the transfer of research findings
to other engineers and included the National Aeronautics and Space Administration
(NASA) sponsored FE solver NASA STRucture ANalysis (NASTRAN) developed at
the Computer Sciences Corporation, and the Structural Analysis Program (SAP)
developed at the University of California in Berkeley (Wilson, 1970). Because of
the widespread usage of these programs, many of the terms initiated during development of these computer programs are still in use today. One of these terms is
1.4 From MSA to a Finite Element Model
“mesh,” which is basically a list of nodes, elements, material types, and the loading
and boundary conditions that define the computational domain of the problem.
For each node, the required pieces of information are the nodal coordinates in
space and any descriptions of the associated nodal constraints in proper degrees
of freedom (DOFs). Any loading conditions (such as nodal forces or moments)
are also applied through nodes, but such applications are typically noted in another
section of the input data deck (e.g., the loading condition section).
For each element, the required pieces of information are the connecting sequence
of the nodes and the material law associated with the element. The connecting
sequence is dependent upon the particular set of shape functions (to be described
in Chapter 2) chosen to formulate the element. Also, the element stiffness matrix
(to be described in Chapter 4) can be formed from the material law and associated
properties, before all element stiffness matrices are assembled into the structure
stiffness matrix. After these steps, the FE solution can be initiated to calculate nodal
displacements.
When using FEA software, we are responsible for defining our problems based
on a consistent system of units, because no software assigns a set of default units.
The three base units used in the FE method are for the descriptions of mass, length,
and time. Other units are derived from these base units. For example,
•
•
•
•
•
1
1
1
1
1
acceleration unit ¼ 1 length unit/(1 time unit)2.
velocity unit ¼ 1 length unit/(1 time unit).
force unit ¼ 1 mass unit 1 acceleration unit.
density unit ¼ 1 mass unit/(1 length unit)3.
stress unit ¼ 1 mass unit gravity constant/(1 length unit)2.
As a result, selecting a set of base units in kilograms (kg), meters (m), and seconds (s) would yield an acceleration unit of m/s2, a velocity unit of m/s, a force unit
of N, a density unit of kg/m3, and a stress unit of Pa. If the set of base units are in kg,
mm, and ms, then the acceleration unit is mm/ms2 (equivalent to 1000 m/s2), the velocity unit is mm/ms (or m/s), the force unit is kN, the density unit becomes kg/mm3,
and the stress unit is kN/mm2 or GPa.
For a typical mild steel with a density 7.88 g/cm3, we need to input 7880 into the
designated space for describing the density when the kg, m, and s base unit set is
selected, because 7.88 g/cm3 is equivalent to 7880 kg/m3. Similarly, a value of
7.88 106 needs to be used when the kg, mm, and ms base unit set is chosen.
In Section 1.3.6, the five-member weightless, truss-bridge example prescribed
the basic unit in length as m, and the derived units for stress and modulus as Pa
(1 GPa ¼ 109 Pa) and for force as N (1 kN ¼ 103 N). Notice that in this static problem, units representing mass and time are not considered anywhere in the calculations. If we are to choose mm as the input unit for length, then the unit for stress/
modulus should be in GPa and the unit for force should be in kN.
A better practice is to preselect a consistent unit set when constructing an FE
model. The exercise below is designed for us to become familiar with choosing a
consistent unit set when creating an FE model. An input data deck, based on a set
45
46
CHAPTER 1 Introduction
Table 1.2 A Sample Input Data Deck for the 5-Member Truss-Bridge Example
Node
No.
xCoordinate
(m)
yCoordinate
(m)
zCoordinate
(m)
1
0.0
0.0
u1 ¼ 0 and
v1 ¼ 0
2
3
4
4.0
8.0
4.0
0.0
0.0
3.0
v3 ¼ 0
Boundary
Condition
Loading
Condition (N)
F4x ¼ 20000
F4y ¼ 50000
Element No.
1st Node
2nd Node
Young’s Modulus (Pa)
Area (m2)
1
2
3
4
5
1
2
3
1
2
2
3
4
4
4
200 109
200 109
200 109
200 109
200 109
0.01
0.01
0.01
0.01
0.01
of kg, m, and s base units, is used to describe the same five-member, truss-bridge
problem (Table 1.2). Because this is a 2D problem (i.e., only x- and y-coordinates
in m are needed), it does not include any dynamic effect (i.e., only static loading)
and the mass of all truss members are negligible (i.e., no gravitational forces).
Thus, the input force unit chosen is Newtons, the unit for Young’s modulus is Pascals, and the cross-sectional area needs to be in meters squared. Similarly, the final
nodal displacements calculated from this data deck will be in units of meters. From
Table 1.2, we can see that the length and stiffness of each truss member needed in the
FEA can easily be calculated by any software program.
EXERCISES
1. Express the Lamé constants l in terms of E and y, K and G, y and G, and K and
y, and m in terms of E and y, K and l, y and l, and K and y.
2. A consistent system of units is required for preparing the input data deck of a
finite element model. Complete the table below based on the three known
base units for mass, length, and time.
1.4 From MSA to a Finite Element Model
Mass
Length
Time
Velocity
Acceleration
Force
Stress
kg
kg
g
ton
slug
m
mm
mm
mm
ft
s
ms
ms
s
s
m/s
m/s2
N
Pa
3. An element has endpoints at Points 1 and 2, which lie on the x-axis. Point 1 is
at x ¼ 6 and Point 2 is at x ¼ 2. After a force is applied, Point 1 is at x ¼ 4
and Point 2 is at x ¼ 9. What is the average strain of the element?
4. An element is connected by P0 at (0, 0) and P2 at (5, 0). These points move
after a set of forces is applied to the element. The new coordinates of the
element are P0 at (2, 2) and P2 at (4, 4). Use Eq. (1.3) to calculate the
length of the element after the forces are applied.
5. Using the same situation as described in Exercise 1.4, calculate the normal
strain of the element after forces are applied. Do this problem in two
different ways: (1) Assume the element is an infinitesimal element, and (2)
do not assume an infinitesimal element.
6. A rectangular 2D element has the nodal points located at (0, 0), (2, 0), (2, 4),
and (0, 4). (1) After a set of forces is applied, the point locations are (0, 0),
(2, 0), (3, 4), and (1, 4). What is the shear strain? (2) If the points are (0, 0),
(2, 0.1), (2.1, 4.2), and (0.1, 4.1), what is the average shear strain and the
total shear strain?
7. Using the isotropic elastic material and Eq. (1.16), build the constitutive
matrix of steel. Assume Young’s modulus is 200 GPa and Poisson’s Ratio is
0.3.
8. A spring element is connected by points P1 and P2 and both are located on the
x-axis. Initially, P1 is at 2 m and P2 is at 5 m. P2 remains stationary, while a
120 N force is applied to P1. The spring has a stiffness of 56 N/m. Use the
direct stiffness method to solve for the displacement of P1.
9. A steel bar has two points located at P1 and P2. P1 is at (1, 1) and constrained
in the x- and y-directions. Point 2 is at (6, 0.5). A 7 kN force is applied at a
45 degrees angle above the horizontal at P2. What are the x- and
y-displacements of the bar? Here, elastic modulus of steel is 200 GPa and the
area of the bar is 0.1 m2.
10. Using
2
82
6 91
6
6
4 13
Gauss elimination, solve the matrix equation below.
38 9 8 9
64 96 96 > u1 > > 43 >
>
>
> >
>
< >
=
< 92 >
= >
10 97 49 7
7 v1
¼
7
28 16 81 5>
u2 >
>
>
> 80 >
> >
>
: >
;
: >
; >
96
92 55 98 15
v2
11. Four aluminum bars (elastic modulus ¼ 70 GPa, Poisson’s ratio ¼ 0.35,
area ¼ 100 mm2) are arranged as shown in the figure below. P1 is at (0, 0),
47
48
CHAPTER 1 Introduction
P2 is at (3, 3), P3 is at (0, 3), and P4 is at (3, 6). P1 and P3 are constrained in
the x- and y-directions. A force of 3000 N is applied in the x-direction. Find
the displacements of the nodes.
3
4
2
1
12. The figure below shows a simplified arm with the origin of the biceps brachii
located on the humerus at a vertical distance of 25 cm from the elbow joint,
and the insertion point located on the radius at a distance of 4 cm distal to the
elbow joint. Instead of 50 N assumed in Example 1.1, a grocery bag
weighing 135 N is located at the insertion point. Calculate the biceps force.
13. Prove why Eq. (1.46) cannot be used for a single element with only two
constraints.
References
REFERENCES
Argyris, J.H., Kelsey, S., 1960. Energy Theorems and Structural Analysis. Butterworth
Scientific, London. ISBN:978-1-4899-5852-55.
Blincoe, L.J., Miller, T.R., Zaloshnja, E., Lawrence, B.A., 2015. The Economic and Societal
Impact of Motor Vehicle Crashes, 2010 (Revised) Report No. DOT HS 812013. National
Highway Traffic Safety Administration, Washington, DC.
Clough, R.W., Wilson, E.L., 1999. Early finite element research at Berkeley. In: Proceedings
of the Fifth U.S. National Conference on Computational Mechanics, Boulder, Colorado
August 4e6, 1999.
Duncan, W.J., Collar, A.R., 1934. A method for the solution of oscillations problems by
matrices. Philosophical Magazine Series 7 (17), 865.
Felippa, C.A., 2001. A historical outline of matrix structural analysis:a play in three acts.
Computers and Structures 79, 1313e1324.
Maney, G.A., 1914. An Investigation of the Stresses in Cantilever Flat Slabs. MS thesis in
Theoretical and Applied Mechanics. University of Illinois.
O’Connor, J.J., Robertson, E.F., 2017. MacTutor History of Mathematics Archive. URL:
http://www-history.mcs.st-andrews.ac.uk/index.html.
Turner, M.J., 1959. The direct stiffness method of structural analysis. In: Structural and
Materials Panel Paper, AGARD Meeting, Aachen, Germany, 1959.
Wilson, E.L., 1970. SAP-a General Structural Analysis Program. UCB/SESM Report No.
70/21. University of California, Berkeley.
49
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CHAPTER
Meshing, Element Types,
and Element Shape
Functions
2
King H. Yang
Wayne State University, Detroit, Michigan, United States
2.1 STRUCTURE IDEALIZATION AND DISCRETIZATION
Before analyzing the response of a structure subjected to loading, an FE model must
be created. As demonstrated in Table 1.2, the computer software only has access to
the nodal coordinates, element connectivity, boundary conditions, and loading conditions from the input data set. Based on these data, the software forms the element
stiffness matrices [k], assembles them into the structure stiffness matrix [K], applies
the boundary and loading conditions, and then uses the Gauss elimination or an
equivalent method to find the nodal displacements. In other words, it is the user’s
responsibility to discretize the structure, numerically idealize each member, prescribe boundary conditions, and apply loading conditions. In Example 1.2, the whole
bridge structure is discretized into five truss members. Each truss member is idealized into a 2-node, 1D bar element. For each element, the forming nodes need to be
arranged in a specific order, based on the way the element stiffness matrix is formulated. In FE terminology, the word mesh is defined as the collection of nodes (which
contain information related to geometric locations) and elements (which prescribe
the order of connectivity between nodes). After the mesh is created, boundary and
loading conditions are applied at corresponding nodes. Finally, the Gauss elimination method is used to calculate nodal displacements.
An example of the numerical discretization and idealization is shown in Fig. 2.1,
where the cross-sectional view A-A of a long 3D dam is idealized as a 2D plane
strain problem before the cross section is discretized into a finite number of triangular elements. Clearly, developing a 3D FE model of the entire dam is an option,
if there are no concerns regarding costs associated with model development or
computational resources. Even then, it would take a long time to simulate the
response of a large model, which may not be desirable. Since the strength of a
dam depends on the weakest section of the entire dam, it is reasonable and computationally more efficient to analyze only a representative 2D section of the entire 3D
dam.
Assuming that the z-axis is set to be along the axial direction of the dam, we can
see from a transverse slice of the dam that force is applied to the xey plane. Because
Basic Finite Element Method as Applied to Injury Biomechanics. http://dx.doi.org/10.1016/B978-0-12-809831-8.00002-7
Copyright © 2018 Elsevier Inc. All rights reserved.
51
52
CHAPTER 2 Meshing, Element Types, and Element Shape Functions
FIGURE 2.1
Left: A sketch of a long 3D dam used for water storage. Inset: The cross-sectional
view A-A, representing the weakest cross section of the dam, can be used to analyze the
structure integrity of this dam under load. Right: This cross section is discretized into a
finite number of triangular plane strain elements. Further discussion regarding the
advantages and disadvantages of using triangular element type is presented in
Section 2.3.
a dam is much thicker along the z-direction than along the cross-sectional directions,
it is understandable that strain components involving z-axis are much smaller than
those involving x- or y-axes. In other words, it is safe to assume that
εzz ¼ gyz ¼ gzx ¼ 0 for a plane strain element. As a side note, a plane stress element
(which represents a plate with a much smaller dimension in the z-direction as
compared to those in the x- and y-directions) subjected to biaxial loading along
the x- or y-axes would have nearly zero stress involving the z-axis. In other words,
szz ¼ syz ¼ szx ¼ 0 are assumed for a plane stress element. The equations needed
to represent plane stress and plane strain elements are provided in Section 1.2.3,
Eqs. (1.22) and (1.23), respectively.
Unlike the example problem shown above, all real-world engineering problems
are 3D. Before modern, high-performance computers became readily available,
numerous theories were developed and reported to reduce the total number of
DOFs so problems could be solved with the relatively low-speed and low corememory computers of that time period. Examples aimed to reduce the total DOFs
include the use of the axisymmetric solution principle, which is defined as solving
a 2D symmetric problem when the structure can be formed by rotating a 1D line
about a single point or solving a 3D problem by rotating a 2D edge about an axis
of rotation. For example, a 2D circular disk can be simplified by rotating a straight
line 360 degrees about the origin, and a 3D cylindrical structure with varying diameters over the vertical length can be simplified by rotating a 2D plate 360 degrees
about the vertical axis (Fig. 2.2).
2.1 Structure Idealization and Discretization
FIGURE 2.2
An axisymmetric approach can be used to model a circular disk by rotating a line element
360 degrees about the center or to study a 3D vase by rotating a slice made of 2D
elements 360 degrees about its axis.
True axisymmetric problems require all the geometry, material, and loading conditions to be symmetric about the same point or axis. As an example, the axisymmetric approach can be used to solve an aircraft turbine blade loading under
normal operations. However, when attempting to solve a problem related to a couple
of pieces of debris hitting one of the turbine blades, axisymmetric 2D method will
not work because the load is asymmetric. Therefore, despite the savings in computational resources gained by reducing real-world 3D problems into axisymmetric 2D
problems, finely meshed 3D models are preferred by modern FE model developers.
Another commonly used method to reduce computational time is to capitalize on
symmetric conditions. For example, if a symmetric plate is loaded under symmetric
plane stress loading conditions, only a quarter of a plate is needed to solve the problem. Again, in order to use this type of simplification, there needs to be a perfectly
symmetric condition. Considering that in most real-world problems loading is asymmetric, and modern computers are equipped with large amounts of core memory and
large clusters of high-speed processors, saving computational resources becomes a
less critical issue, unless an extremely large model is presented. Another advantage
of using a full model instead of simplification based on symmetric conditions is the
ability to detect artifacts. If a problem was truly symmetric, then the results would be
symmetric. If asymmetric results were found, it would be obvious that there were
numerical artifacts associated with poor meshing.
Another consideration related to idealizing a structure includes omissions of
some details (such as small holes or fillets) that are insignificant for the particular
53
54
CHAPTER 2 Meshing, Element Types, and Element Shape Functions
FIGURE 2.3
A finer mesh size is used near the region of interest (ROI), located at the lower left corner,
to calculate the stress increase due to concentrated loading (not shown) near the ROI. For
the areas away from the ROI, the mesh size is much larger, because the stress variation
would be much lower. This approach is advantageous when the computational resources
are limited. The disadvantage of this approach is that a new model needs to be developed
each time the ROI is changed.
problem. By skipping these details, an FE model can be created more quickly. Additionally, in the early days FE models were developed for specific types of engineering analyses. For example, at one time it was common practice to use small-sized
elements near the areas of interest while coarse meshes were implemented for all
other elements (e.g., Fig. 2.3). Again, problems that called for special approaches
were mainly due to lack of computational resources. A model created with one specific analysis in mind could only be used for finding a solution for that particular purpose. A new model would need to be created if the finely meshed area were no
longer of concern for a new analysis type.
With the aid of meshing software, modern, general-purpose FE models tend to
be created using a uniformly distributed mesh with a very fine mesh size, so one
model is sufficient to study different boundary and loading conditions. Fig. 2.4
shows a general purpose, detailed FE car mesh developed at the National Crash
Analysis Center for vehicle crashworthiness evaluations. The model can be used
to simulate frontal, side, and rear-end impacts, which is quite an advantage over
creating three separate FE models for the three different impact directions. The
next two sections describe the fundamental nature of nodes and elements in
more detail.
2.2 Node
FIGURE 2.4
A cut-away view of a Ford Taurus whole car-FE model developed at the National Crash
Analysis Center (NCAC) of the George Washington University.
2.2 NODE
The required information for a node, also known as a grid point, is the spatial location in a global coordinate system. In a global coordinate system, the location of a
point is defined in a rectangular or spherical space. Commonly used coordinate systems include the Cartesian coordinate system, polar coordinate system, and spherical coordinate system.
At each node, any constrained DOFs can be prescribed, and the responding
DOFs due to applied forces and moments can be calculated. In theory, all nodes
possess six DOFs, that is, three translations along the x-, y-, and z-axes and three rotations about the x-, y-, and z-axes. Typically, when these six DOFs are constrained,
they are denoted as 1, 2, 3, 4, 5, and 6, respectively. For example, a 1, 2, 3, 5
constraint means there are no translations along any of the three axes (denoted as
1, 2, 3) and no rotation is allowed about the y-axis (denoted as 5).
For some element types, certain DOFs are prohibited from happening, and these
prohibitions need to be carefully considered. For example, 1D, 2-node truss, spring,
and cable elements all allow three translational DOFs, but no rotational DOFs are
permitted in 3D analysis. On the other hand, a 2-node beam element, which is
geometrically idealized the same way as a 1D truss or spring element, admits all
six DOFs at each node. Similarly, a 4-node membrane element contains no rotational
DOFs (i.e., only three translational DOFs are allowed), but a 4-node plate element
allows all three translational DOFs and two rotational DOFs about the two axes
in the plane of the element (i.e., rotation about the axis perpendicular to the element
surface is not calculated). Similar to a 4-node plate element, a 4-node shell element
in certain FEA software packages does not allow in-plane rotation, which is also
known as the drilling DOF. The reason for neglecting the drilling DOF is that there
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CHAPTER 2 Meshing, Element Types, and Element Shape Functions
can be incompatibilities with adjacent elements. However, the in-plane rotational
DOF is permitted in some FEA software packages. Lastly, an 8-node hexahedral
brick element allows only three translational DOFs.
All forces that load the FE model need to be applied through the corresponding
nodal translational DOFs, and moments need to be applied at the corresponding nodal
rotational DOFs. The 8-node, hexahedral brick elements do not permit any rotational
DOFs, and therefore application of moments to this element type has no effect. Similarly, any translational displacement constraint (e.g., zero nodal displacements in the
x-, y- and/or z-axes) and any rotational constraint can only be applied at the node of the
corresponding nodal DOF. If one particular element type allows no rotational DOFs,
then applying any constraints to those DOFs will have no effect on the FEA results.
For example, prohibiting zero translational and rotational movements of an 8-node
hexahedral brick element will yield the same results as prohibiting zero translational
DOFs alone. In some cases, the loading condition is described as a fixed amount of
motion (such as compressing a tube for a fixed distance) or a constant velocity
(such as a car-FE model hitting a rigid wall at a speed of 50 km/h), these prescribed
motions need to be applied to the node(s), as well.
For ease of use, some software packages allow users to apply distributed forces to
a surface area. Internally, these packages still need to redistribute the surface loads
into nodal loads. For FE solvers without such an option, users need to distribute the
surface loads according to the corresponding element shape functions, to be
described in Chapter 6.
2.3 ELEMENT
Elements are the basic building blocks used to idealize structures. Because a building block can be made of any material types (e.g., metal, wood, bone), information
related to its material properties must be contained within the element. Geometrically speaking, elements can be in the form of lines (1D trusses or beams), areas
(2D membrane and plate), or solids (tetrahedral or brick). They are formed by connecting nodes that are arranged in a specific order, based on the mathematical relationships (shape functions) among these nodes. A wrong sequence provided to the
input data deck will result in erroneous results. Unless stated otherwise, all nodes
forming a 1D element are arranged from left to right, and all nodes in a 2D element
are arranged in a counterclockwise order. For a 3D brick element representing an
8-node, 2-layer solid element, the nodes that form the lower layer of the element
(P1 to P4) are arranged first, in a counterclockwise order, and the upper layer (P5
to P8) is subsequently arranged in the same counterclockwise manner. Note that
the starting points for the bottom (P1) and top (P5) layers initiate at the same corner
but are located at different heights along the z-axis.
The set of element shape functions in the FE method is a collection of multipurpose interpolation functions needed for determining physical values of non-nodal
points based on known nodal coordinates within an element. These functions can
also be used to redistribute uniformly or nonuniformly distributed surface loads to
2.3 Element
nodal loads, and to find stress/strain contours. The order of a shape function depends
on how many nodes there are in an element. For example, a linear (first order) interpolation is sufficient for a 2-node, 1D bar element, while a quadratic (second order)
interpolation is needed for a 3-node, 1D bar element.
Limited by the low computational power available for FEA in the early days,
there was a tendency to use higher order interpolation element types in order to
lower the number of elements, which in turn, reduced the total number of DOFs
for obtaining solutions on these limited-power computers. These complicated or
special higher order elements are more difficult to formulate than lower order elements, despite the fact that they are capable of satisfying a higher order of continuity.
A simple continuity of the displacement field is called a C0 continuity. In most
structural mechanics and injury biomechanics problems, strain is the most sought
after response variable. Because strain can be calculated from the first derivatives
of the displacement fields, a C0 continuity is sufficient for most structural mechanics
and biomechanics problems. There are other problems which require continuity of
the first derivatives of the displacement fields across the element edges. This type
of continuity is known as C1 continuity. For example, some plate and shell elements
based on KirchhoffeLove theories require C1 continuity in order to be valid. Fig. 2.5
graphically explains the essence of C0 and C1 continuity.
FIGURE 2.5
The column on the left depicts an element with C0 continuity, that is, the displacement or
deflection field is continuous, but the velocity or slope field du
dx is not. For an element
with C1 continuity, both the displacement/deflection and velocity/slope fields are
continuous, but the acceleration/curvature field is not.
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CHAPTER 2 Meshing, Element Types, and Element Shape Functions
2.3.1 SIMPLEST ELEMENT TYPES
With ever-increasing computational power, modern FE models are built with
extremely large numbers of elements to ensure close resemblance of the structures
of interest. As the size of the elements becomes smaller and smaller, the difference in
solutions between lower and higher order elements diminishes to a point of insignificance, in the majority of cases. As such, present-day FE modelers prefer the use of
the simplest types of element with very fine mesh resolutions. Because higher order
elements are not commonly used in contemporary FEA, theories behind the element
types used for possessing a higher order of continuity will not be discussed in this
book; only the simplest types of element will be discussed.
At an early design stage, a coarse mesh model may be useful for capturing trends
in the magnitudes of response variables that occur due to parameter changes. Simulations can be run quickly when using such a coarse mesh, and hence many parametric studies can be completed in a short period of time. A finer mesh model, which
guarantees numerical convergence, is eventually needed to obtain a more accurate
solution than that obtained from a coarse mesh model.
In the FE method, the word “convergence” is related to how much discrepancy
exists between the solutions calculated by the FE model compared to those obtained
analytically or experimentally when the mesh is continuously refined. Without
convergence, we cannot be confident that the FEA represents the real-world scenario. The source of discrepancy may come from lack of iteration convergence
and mesh convergence.
Regarding iteration convergence, some FE solutions are obtained through iterative procedures (e.g., the Jacobi method, to be described in Section 7.2). The iterative procedures will not stop until the error between two successive iterations is
smaller than a preset value. Using such a method, the numerical and analytical/
experimental solutions become closer with a higher number of numerical iterations.
If the numerical analysis is allowed to run long enough, the two solutions will be
nearly identical. Hence, a lack of iteration convergence can be easily resolved by
reducing the allowable error between two successive iterations.
In terms of mesh convergence, all FE models involve some degree of simplification when creating a mesh to represent the structure of interest. Obviously, a finer
mesh will result in a closer representation of the real-world structure to be modeled.
In this case, it is understandable that the more elements are used to develop a model,
the more accurate the results will be. After several mesh refinements, the mesh is
considered “converged” if the differences in solutions between the current refinement and its predecessor are smaller than a preset value. In some occasions,
continuing increase of the mesh density actually moves the solution away from
convergence. This phenomenon usually indicates that the model is not properly
defined. One possible reason may come from oversimplification of the model. As
mentioned previously, omitting some fine details (such as not modeling small holes)
is a common practice to reduce the total DOFs. In this case, the high stress concentration that occurs near the small hole cannot be exhibited unless the hole is
2.3 Element
–.15
0
.15
.3
.45
.6
.75
.9
1.05
FIGURE 2.6
The stress distribution for a linear elastic plate subjected to a concentrated load is twice as
high for the model meshed with rectangular elements as compared to models meshed with
triangular elements. Note that only eight divisions are set up to highlight the differences in
stress distributions when different element types are used. As such, the peak element
stresses discussed in the previous paragraph cannot be displayed in this figure. Those
values are directly output from the FE solver. Even then readers can see that the “high
stress” regions are different when different element types are selected to model the plate.
explicitly represented in the mesh. Other possibilities include the selection of
different (and sometimes wrong) element types. For example, using a beam element
type is more efficient than other element types when modeling a beam (See Section
2.5.3, Fig. 2.21, and Table 2.4).
In addition to the requirement for convergence, FE models created using
different element types or arranged in different manners could result in different solutions. Fig. 2.6 shows a rectangular plate loaded at the center of the left edge by a
point (concentrated) load and fixed at the right edge of the plate. Even though the
stress contours for all three models look highly similar, the magnitudes of the
peak stresses are quite different. For the model developed with quadrilateral elements, the peak stress (a dimensionless magnitude of 3.9) is found to be more
than twice that of the model built with right-angled triangle elements (a peak magnitude of 1.8) and equilateral triangle elements (a peak magnitude of 1.7). Here, no
units (such as MPa) are provided for the stress, because the purpose of this exercise
is to find relative values when the same material properties are used for all three
models, without referencing any particular material.
2.3.2 1D ELEMENT TYPE
The 1D element type includes, but is not limited to, a 2-node linear or a 3-node
quadratic element. jThe 2-node elements, considered to be the simplest 1D element
type (or pseudo 2D or 3D line elements when they are not in line with global axes),
include the structure types of truss, spring, cable, beam, and frame.
The only load a 2-node truss member can support is axial (i.e., no bending load is
allowed). As such, no resistance can be generated from a truss member when a transverse force (vertical load) is applied. However, when a number of truss members are
connected together by pin joints, the structure as a whole can resist applied forces
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CHAPTER 2 Meshing, Element Types, and Element Shape Functions
FIGURE 2.7
A taper truss structure (left) can be idealized as several truss elements (right), each with a
constant cross-sectional area.
that are not aligned with the axial direction of the truss member. Because truss members are pinned together, these connecting members are free to rotate about the axes
of the pins, and hence there is no resistance to moments. Additionally, this structure
type needs to have a constant cross-sectional area, a constant elastic modulus, and
can be oriented in any direction. For idealizing a truss member with varying
cross-sectional areas, as shown in Fig. 2.7 (left), a series of elements, each with a
different constant cross-sectional area, need to be used (Fig. 2.7 right). The number
of elements used for idealizing a truss is based on the desired degree of accuracy.
A spring can only support an axial load (tension or compression) and has only
one DOF for each node along the axis of the spring. Sometimes when the axis of
the spring is not aligned with any of the axes of the global coordinate system, two
or three pseudo-DOFs per node are assumed for 2D or 3D problems, respectively.
In such a case, the two or three DOFs are mathematically related and can be calculated from the angles formed by the axis of the spring and any of the axes of the
global coordinate system. Unlike a truss element, in which a constant crosssectional area, a length, and an elastic modulus are required in order for all the entries of the element stiffness matrix to be calculated, only a single spring constant is
needed to form the element stiffness matrix of a spring.
A cable element can also support an axial load, but it may be further decomposed
such that it becomes either a tension-only or compression-only cable element. A
tension-only cable element can provide load resistance when it is elongated, but becomes slack and provides no load resistance when it is being shortened. Computationally, this is done by detaching the stiffness when the tension-only cable element
goes into compression. A compression-only cable element has an effect opposite to a
tension-only cable element. Because a muscle can only generate force when it is in
contraction, a compression-only cable element needs to be used to model an active
muscle.
2.3 Element
In classical mechanics, a beam is defined as having two of the three dimensions
significantly smaller than the third one. The main purpose of a beam is to transmit
transverse force through bending. Hence, a beam element has two DOFs per node,
a vertical deflection and a rotation, to support both shear force and moment.
For this reason, a C1 continuity is required to formulate a classical plane beam
element.
A beam element is significantly different from a truss element, which supports
only axial loading. Another significant difference between a beam and a truss
element is that a truss element can either support compression or tension, but not
both at the same time. On the other hand, there exists a neutral axis (a plane where
all stresses are zero) for a beam element, with one side of the neutral axis subjected
to tension while the opposite side is subjected to compression.
Although only two nodes are needed to fully describe the geometric orientation
of a beam element, some software packages require that a third reference node be
added in order to designate the directions for the width and height of the beam.
This is useful for proper cross-sectional moments of inertia to be calculated. With
this information, the stress within the beam can be calculated at integration points
along the direction of the height on the cross section. Other software packages
only need the user to directly input the elastic modulus, cross-sectional area, and
moment of inertia, and hence no such reference node is needed. In such a case,
no stress within the beam would be calculated, due to the lack of cross-sectional
shape.
A frame element, which can be formulated with the principle of superposition, is
in essence the combination of a truss element and a classical plane beam element.
This structure is more realistic than classical truss and beam elements, because
real-world structures are rarely found to carry purely axial or transverse loading.
Many software packages describe a frame element as a general beam element.
Modern FEA software packages are versatile and powerful. However, these
packages are sometimes so sophisticated, it is difficult to use them properly. In order
to create proper models, we need to read the User Manuals as well as the Theoretical
Manuals, so the various subtle differences among the structure types can be understood. Table 2.1 summarizes some critical aspects of the aforementioned 1D structure types. Note that these summaries may not be applicable for all FE software
packages.
2.3.3 2D ELEMENT TYPE
The 2D element type is a 3D structure with one of the dimensions much smaller than
the other two (e.g., a plate subjected to in-plane loading). This element type includes, but is not limited to, 3-node linear triangular, 6-node quadratic triangular,
4-node bilinear quadrilateral, 9-node biquadratic quadrilateral, and 8-node
serendipity quadrilateral elements (Fig. 2.8). Clearly, the two simplest element types
are the 3-node linear triangular and 4-node bilinear quadrilateral elements.
The 3-node element type is the easiest one to use for the purpose of meshing.
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CHAPTER 2 Meshing, Element Types, and Element Shape Functions
Table 2.1 One-Dimensional Structure Types That Can Be Modeled With the
Simplest 2-Node Element
Structure
Type
DOFs per
Nodea
External Loads Resisted
DOFs
Bar, Truss,
Spring
Cable
1
Tension and compression
Axial displacement
1
Axial displacement
Beam
2
Frame
5
Either tension or
compression
Transverse force and
bending moment
Tension/compression/
bending moment
a
Transverse displacement
and rotation
Three displacements and
two rotations
Pseudo-DOFs used for orientation are not included.
(A)
(D)
(B)
(C)
(E)
FIGURE 2.8
Example 2D element types include: (A) 3-node linear, (B) 6-node quadratic, (C) 4-node
bilinear quadrilateral, (D) 9-node biquadratic quadrilateral, and (E) 8-node serendipity
quadrilateral elements.
However, many more 3-node elements need to be employed in order to achieve the
same accuracy that a 4-node element type can provide.
Typically, a constant value is prescribed in the input data deck to describe the
thickness (depth) of the entire element. For cases where nonuniform thicknesses
are required, some software packages allow different thicknesses to be assigned at
2.3 Element
each node within the element. Importantly, the out-of-plane thickness will not be
graphically displayed in any pre- and postprocessing software packages, regardless
of the thickness. Allowable limits for the ratio of thickness to in-plane lateral dimensions (width and length) are problem dependent and cannot be definitively prescribed. As a rule of thumb for qualifying the use of this element type, the
thickness along the z-axis should be less than 10% of the dimensions in the x- or
y-direction.
All of these 2D surface element types can be used to represent membrane, plane
stress, plane strain, plate, and shell structures. The distinctions among these five
structure types are that membrane, plane stress, and plane strain elements possess
strength only along the surface of the plane, a plate element has only out-of-plane
stiffness (i.e., support load through bending), and a shell element can provide both
in-plane and out-of-plane stiffness. All 2D element types can be used independently
or in conjunction with 3D element types for modeling 3D structures. For example, a
vertebra can be represented by a layer of 2D shell elements to represent the thin outer
cortical bone and 3D solid elements to represent the trabecular bone within.
The thickness of a membrane element is considered too thin to resist any
compressive loading. Despite the thinness, both membrane and plane stress elements
can take tensile load. For plane stress elements, any stress components involving an
out-of-plane axis must be zero. That is, szz, syz, and szx are all zero for an element
that lies on the xey plane. Similarly, a plane strain element in the xey plane postulates that εzz, εyz, and εzx are all zero.
In classical mechanics, a shell must have a curved surface (e.g., the outer surface
of an oil storage tank or the haul of a submarine), while a plate has a flat surface.
However, because of the large number of elements routinely employed in modern
FE models, the distinction between curved and flat surfaces has become blurry; a
curved surface can be well approximated by joining a large number of flat surfaces.
Also, in some software packages, the plate element is allowed to provide stiffness to
resist both in-plane and out-of-plane loading. Since the distinction between a shell
and a plate is sometimes not as clear in the modern FE method as it was when analytical methods were used, we need to be well aware of the ways plate and shell elements are formulated in our chosen software package, in order to correctly select
the proper 2D element type to develop an FE model.
Because membrane, plane stress, and plane strain elements provide only in-plane
stiffness, all rotational DOFs are constrained, while translational DOFs are allowed.
As such, these elements possess two DOFs (two in-plane displacements) per node.
Typical applications of membrane elements are related to modeling thin fabrics,
such as a seat belt or an airbag. Similarly, the mesothelium, which forms the lining
of the pleura (thoracic cavity), pericardium (the membrane enclosing the heart), and
peritoneum (abdominal cavity), needs to be modeled using membrane elements. On
the other hand, plane stress elements are commonly used for solving problems
involving a thin plate subjected to only in-plane loading. For example, a seat-belt
buckle, as shown in Fig. 2.9, is commonly modeled using plane stress elements,
while a seat belt is commonly modeled with membrane elements.
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CHAPTER 2 Meshing, Element Types, and Element Shape Functions
FIGURE 2.9
A seat-belt buckle modeled using plane stress elements, while a seat belt is modeled with
membrane elements.
As explained above, a 2D plate element supports loading that is orthogonal to the
plane surface through combined bending and shear. Hence, a 2D plate element has
three DOFs per node (a vertical displacement and two in-plane rotations). An
example that would be well modeled with plate elements is loading on a cover of
a utility access manhole by passing cars. This is because external forces applied
to this road structure are, in general, perpendicular to the surface of the cover, which
in turn produces bending load. Plate elements can be further categorized into thin
and thick plates. Although the distinction between a thin versus a thick plate is
not well defined, a thickness to width or length ratio of lower than 10% is generally
considered a thin plate, while a ratio greater than 10% is regarded as a thick plate.
The Kirchhoff plate theory or KirchhoffeLove plate theory that was derived to
calculate the deformations and stresses within a plate is the foundation generally
adopted for the formulation of a thin plate. The Love theory (Love, 1888) was developed by Augustus Edward Hough Love (Apr. 1863eJun. 1940), based on a series
of lectures given by Gustav R. Kirchhoff (Mar. 1824eOct. 1887) at the University
of Heidelberg. Here the birth and death dates are based on the MacTutor History of
Mathematics Archive wonderfully compiled by O’Connor and Robertson (2017).
Unless otherwise stated, all birth and death dates listed are according to this online
source. The assumptions based on Kirchhoff’s proposal require that a straight line
that is perpendicular to the midsurface of the plate remains straight and orthogonal
to the deformed midsurface (Fig. 2.10). As such, the transverse shear deformation is
neglected.
For a thick plate, the formulation generally follows the ReissnereMindlin plate
theory, which accounts for shear behavior not considered in the Kirchhoff plate theory. The basic differences between Mindlin and Reissner’s plate theories are:
•
•
Eric Reissner (Jan. 1913eNov. 1996) assumed that the displacement across the
plate (i.e., out-of-plane) may not be linear, and the thickness of the plate may
change with loading (Reissner, 1945).
Raymond D. Mindlin (Sep. 1906eNov. 1987) assumed a linear variation in the
displacement across the plate thickness, and the thickness of the plate remained
unchanged with loading. Also, any normal stress through the thickness is
neglected (Mindlin, 1951).
2.3 Element
θ
θ
θ=
FIGURE 2.10
A cross-sectional view of a thin plate before and after bending deformation. The Kirchhoff
plate theory assumes that a straight line that is perpendicular to the midsurface of the
undeformed plate remains straight and perpendicular to the deformed midsurface. The
slope or rotation angle is equal to the first derivative of the deflection (i.e., q ¼ dw
dx ).
Despite the slight variation between Mindlin and Reissner’s theories, both
considered nonzero out-of-plane shear deformations. As such, it is frequently
referred to as ReissnereMindlin thick plate theory. This thick plate theory is in
contrast to KirchhoffeLove’s thin plate theory, in which zero out-of-plane deformation is assumed. In general, the thick plate formulation is recommended, because this
method results in a more accurate solution. However, if there are a large number of
elements with high aspect ratios (the ratio of length to width), the thick plate formulation should not be considered.
Because the ReissnereMindlin theory is more versatile than other theories, in
almost all commercial software packages, such as ABAQUS, ANSYS, LS-DYNA,
and PAM-CRASH, the element libraries are based on the ReissnereMindlin theory.
In addition to different assumptions used in the derivations of different plate theories, ways of implementing these theories into the element library have also varied.
For example, the first type of plate element implemented in LS-DYNA was the
Hughes-Liu shell element, while the most computationally efficient type was the
Belytschko-Tsay shell element (Hallquist, 2006). As software users, we need to understand which implemented element type best approximates the solution we wish to
obtain.
A shell element, in essence, is a combination of a plane stress (two DOFs) and a
plate element (three DOFs) into one element with five DOFs. Hence, formulating a
shell element can be done through superposition of a plane stress element and a plate
element. As noted above, in some software packages, deformation of a plate element
may include in-plane stretch or shortening and out-of-plane deflection due to
bending. To the best of the author’s knowledge, no FE software packages allow users
to output deformations due to individual in-plane or out-of-plane loadings. Instead,
the in-plane and out-of-plane deformations are summed and reported as the total
deformation. Table 2.2 summarizes some critical aspects of the aforementioned
2D structure types.
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CHAPTER 2 Meshing, Element Types, and Element Shape Functions
Table 2.2 Structure Types That Can Be Modeled With the Simplest 4-Node,
2D Surface Element
4-Node
Surface
Element
Stiffness
Matrix
DOF per Node
Available DOF
Membrane,
plane stress,
and plane
strain
2
In-plane
translations
88
Plate
3
One out-of-plane
translation and two
in-plane rotations
1212
Shell
5
Three translations
and two in-plane
rotations
2020
In the explicit FE solution package LS-DYNA (LSCT, Livermore, CA), the shell
element is used for modeling shells, plates, membranes, plane stress, and plane
strain, all using the keyword *ELEMENT_SHELL augmented with a large number
of element formulation options (ELFORM), such as Belytschko-Tsay membrane
(ELFORM 5), plane stress (ELFORM 12), plane strain (ELFORM 13),
Belytschko-Wong-Chiang shell (ELFORM 10), etc. As a user, this fact further emphasizes the need to become familiar with the methods of element formulation in
software packages before the proper element type is chosen for development of
FE models. Due to the limited scope of the current book, additional discussion on
this rather complex subject of implementations of various shell element types can
2.4 Formation of Finite Element Mesh
(A)
(B)
(C)
(D)
FIGURE 2.11
Example 3D element types include: (A) 4-node tetrahedral, (B) 8-node trilinear, (C) 6node prismatic, and (D) 5-node pyramidal elements.
not be provided. If you are interested in details on this topic, we recommend reading
the LS-DYNA theory manual (Hallquist, 2006).
2.3.4 3D ELEMENT TYPE
The three-dimensional (3D) solid element type includes, but is not limited to, 4-node
tetrahedral linear, 8-node trilinear, 6-node prismatic or wedge-like, and 5-node pyramidal elements (Fig. 2.11). Higher order elements such as the 10-node quadratic
tetrahedral element and the 20-node hexahedron tri-quadratic (serendipity) elements
are more complex and not shown in this figure. Each node of the 3D element type
possesses only three translational DOFs. Accordingly, moment cannot be applied
to this element type, and no rotation will ever occur. Again, the simplest 4-node
tetrahedral and 8-node trilinear elements are recommended for the development
of 3D solid elements in FE models.
The most adaptable element type for any 3D geometric shape is the tetrahedral
elements, and these elements are very easy to form using automatic meshing algorithms. However, unless a very large number of elements is used, this element type
frequently stiffens the response. On the other hand, high-quality hexahedral elements are more difficult to form, especially for very complex geometry, but usually
provide a more accurate solution without a great number of elements.
2.4 FORMATION OF FINITE ELEMENT MESH
The two previous sections describe the node and element needed to form the mesh of an
FE model. A mesh is a collection of elements interconnected at nodes used to idealize
a structure of interest. When a mesh is augmented with proper boundary and loading
conditions, it becomes an FE model. For simple problems, such as those highlighted in
Fig. 1.14, the mesh can be easily generated by hand, as shown in Table 1.1. A more
complex geometry requires the application of an FE preprocessor to develop the mesh.
Most modern FE model preprocessors allow users to interchange different coordinate systems (such as the Cartesian, polar, and spherical) for easy development of the
model. Nevertheless, the end product is always displayed in Cartesian coordinates.
Only a concise definition of the Cartesian coordinate system is listed here. We
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CHAPTER 2 Meshing, Element Types, and Element Shape Functions
recommend that you become familiar with other coordinate systems through personal
studies. The other coordinate systems could be useful for identifying the nodal coordinates of objects with special shapes, such as setting nodes around a circle.
The Cartesian coordinate system, also known as the rectangular coordinate
system, was first used by René Descartes (Mar. 1596eFeb. 1650). A 2D Cartesian
plane consists of two perpendicular axes crossing each other at the origin. In
general, the horizontal axis is called the x-axis, and the vertical axis is called
the y-axis. For a 3D Cartesian space, a third axis (z-axis) is defined by adding a
line through the origin and perpendicular to the xey plane. Using Cartesian coordinates, the x-, y-, and z-coordinates indicate how far away the point is located
from the origin.
It must be emphasized that the orientation of the element coordinate system is the
one that is the most convenient for that element. In other words, the element coordinate system may not need to coincide with the global coordinate system that is
used to describe nodal locations. This element-based coordinate system is defined
specifically to closely relate to the material behaviors. For example, a femur is a
transverse isotropic material that has a higher elastic modulus along its axial direction than on the transverse plane. In this case, a local element coordinate system is
needed to define which axis is along the axis of the femur and which other axes
define the transverse plane, so that proper material properties can be implemented.
Many software packages are available to generate FE meshes. A list of public
domain, downloadable, and university developed Automatic Mesh Generation
Methods and Software, maintained by a German engineer Robert Schneiders, can
be found on the Internet (Schneiders, 2017). Although it is convenient to use these
packages to generate FE meshes, we must understand some fundamental background
related to the selection of proper element types to correctly develop an FE model.
When meshing an FE model, it is essential that any adjacent elements are only
connected to neighboring elements through their nodes. In other words, “floating”
or “isolated” nodes are not allowed. Fig. 2.12 below demonstrates two neighboring
Floating Node
FIGURE 2.12
Example of a floating node created when two incompatible elements are connected side
by side. In this case, continuity for the displacement and strain become questionable.
2.5 Element Shape Functions and [B] Matrix
elements with element 1 on the left side, formed by a 4-node bilinear element, and
element 2 on the right side, made of an 8-node serendipity element. We can easily
observe that the common edge (the junction between the two elements) consists of
two nodes on element 1 and three nodes on element 2. Clearly, there is a floating
node. Because element 1 uses a linear interpolation while element 2 uses a quadratic
interpolation, the strains may become incompatible along this edge. While some software packages provide warning messages whenever there is a floating node present,
other packages may allow users to run the FE model without questioning this potential
mistake. Due to the incorrectly formulated input data deck in the latter case, incorrect
outputs will be generated.
2.5 ELEMENT SHAPE FUNCTIONS AND [B] MATRIX
As briefly mentioned in Section 2.3, the element shape functions are used to interpolate the nodal coordinates to identify the coordinate values of points located
anywhere within the element. The same shape functions can also use the modelcalculated nodal results to interpolate physical values located anywhere within the
element. Eq. (2.1) explains the concept of the shape functions in mathematical
terms,
4x;y;z ¼
n
X
Ni 4i
(2.1)
1
where 4 is a physical quantity that could be the nodal coordinate values, nodal
temperature, or nodal displacement, 4x,y,z is the 4 value at the coordinate point
(x, y, z) within the element, Ni are the shape functions, and 4i are the 4 values at
all distinct nodes. The value n is the number of DOFs constituted the element.
For example, a 2-node beam element has two DOFs per node for a total of
four DOFs, while a 2-node truss element has one DOF per node for a total of
two DOFs.
2.5.1 1D, 2-NODE ELEMENT SHAPE FUNCTIONS
The simplest 1D element to represent the truss, spring, and cable is the 2-node linear
element. A similar 2-node configuration can also be used to represent a beam. However, shape functions for these two categories of elements are very different.
2.5.1.1 2-Node Linear Bar Element
Fig. 2.13 shows a 1D element formed by two nodal points, P1 and P2, with displacement at P1 as u1 and displacement at P2 as u2. For a 2-node element, Eq. (2.1)
becomes
4x ¼ N1 41 þ N2 42 .
(2.2)
69
70
CHAPTER 2 Meshing, Element Types, and Element Shape Functions
1
FIGURE 2.13
A 2-node linear bar element with nodal displacements u1 at P1 and u2 at P2.
Example 2.1
Assume that the x coordinates for P1 is 4 and P2 is 6. We can intuitively identify
the coordinates at one-quarter distance from the left to be 4.5, midpoint to be 5,
and three-quarters distance to be 5.5. Obviously, it would be much easier to use a
computer program to compute the needed values. The algorithm needed for such
a computer program is embedded in the ideas of shape functions. Show the algorithm for finding the shape functions.
Solution
Assuming a linear interpolation exists within this element, 4 can be expressed as
4x ¼ a0 þ a1 x.
(2.3)
41 ¼ a0 þ a1 4;
(2.4)
42 ¼ a0 þ a1 6:
(2.5)
At P1 (x ¼ 4),
and at P2 (x ¼ 6),
To find a1, we subtract Eq. (2.4) from Eq. (2.5). In doing so, we have
4 41
.
(2.6)
a1 ¼ 2
2
We then insert the results from Eq. (2.6) into either Eq. (2.4) or (2.5). We choose
Eq. (2.5):
4 41
4 1 ¼ a0 þ 2
40a0 ¼ 341 242 .
(2.7)
2
Finally, we insert a0 and a1 back into Eq. (2.3):
4 41
x.
(2.8)
4x ¼ 341 242 þ 2
2
The physical meaning of this equation can be explained as the value of any
physical quantity 4 (such as the temperature, nodal coordinate, or nodal displacement) at any location x within the element (between P1 and P2), and it can be
calculated using Eq. (2.8). As a trivial example, assuming that the temperatures
2.5 Element Shape Functions and [B] Matrix
are 4 at P1 and 6 at P2, what is the temperature at x ¼ 5? By inserting x ¼ 5,
41 ¼ 4 , and 42 ¼ 6 into Eq. (2.8), we can easily find the result that
4x¼5 ¼ 341 242 þ
42 41
ð6 Þ ð4 Þ
5 ¼ 5 .
x ¼ ð3Þð4 Þ ð2Þð6 Þ þ
2
2
(2.9)
For a nontrivial example, if the displacements u1 ¼ 41 ¼ 35 mm and
u2 ¼ 42 ¼ 47 mm, we can find the displacement at x ¼ 5 from Eq. (2.8) as
4x¼5 ¼ 341 242 þ
42 41
ð47Þ ð35Þ
5 ¼ 41 mm.
x ¼ ð3Þð35Þ ð2Þð47Þ þ
2
2
(2.10)
The next step seems simple, but it is an essential step is necessary for understanding subsequent material. To find the shape functions, we rearrange Eq. (2.8)
by collecting all 41 terms together and all 42 terms together.
4x ¼ ð3 0:5xÞ41 þ ð0:5x 2Þ42
(2.11)
By comparing terms in Eq. (2.11) with those in Eq. (2.2), we can determine that the
two shape functions ½ N1 N2 are of the values N1 ¼ 3 0.5x and N2 ¼ 0.5x 2.
If we plug the same two 4 values (35 and 47 mm) and x ¼ 5 into Eq. (2.11)
4x¼5 ¼ ð3 0:5 5Þ35 þ ð0:5 5 2Þ47 ¼ 41 mm.
We find that the displacement value at the same coordinate is identical to that
calculated from Eq. (2.8) and illustrated in Eq. (2.10). The same outcome is easily
understandable, because Eq. (2.11) is simply a rearrangement of Eq. (2.8). By
writing it in the form of Eqs. (2.2) and (2.11), the element shape functions are
easier to visualize.
The shape functions shown in Eq. (2.11) are only valid for the 2-node element
with P1(x ¼ 4) and P2(x ¼ 6). For different P1 and P2 values, a new set of shape
functions are needed. Because deriving shape functions for each element with
different coordinate values is time-consuming, the computational cost would
be prohibitively high if the FE model were to consist of a great number of
elements. A more general approach is to derive the shape functions based on
the length of the element. From Fig. 2.13, we define the length of the bar to be
L, that is, x(P2) x(P1) ¼ L.
Example 2.2
Assuming a local coordinate system with x(P1) ¼ 0 and x(P2) ¼ L, use the linear
interpolation algorithm for finding the shape functions.
Solution
Let 4x ¼ a0 þ a1x, as previously shown for the linear interpolation in Eq. (2.3).
At P1,
(2.12)
41 ¼ a0 þ a1 0; and
71
72
CHAPTER 2 Meshing, Element Types, and Element Shape Functions
at P2,
42 ¼ a0 þ a1 L.
(2.13)
Solving these two simultaneous equations, we have
4 41
.
(2.14)
a0 ¼ 41 and a1 ¼ 2
L
Inserting the values of a0 and a1 into Eq. (2.3), we find that
4 41
x
x
4x ¼ 41 þ 2
x ¼ 1 41 þ 42 .
(2.15)
L
L
L
And the element shape functions ½ N1 N2 are the coefficients listed in
Eq. (2.15):
N1 ¼
Lx
x
and N2 ¼
L
L
(2.16)
Example 2.3
Similar to the previous two examples, we can assume a different local coordinate
L
system where xðP1 Þ ¼ L
2 and xðP2 Þ ¼ 2, that is, the new origin of the coordinate
system is located at the center of the bar. Determine the algorithm for finding the
shape functions.
Solution
At P1,
41 ¼ a0 þ a1 L
.
2
(2.17)
At P2,
L
42 ¼ a0 þ a 1 .
2
Solving these two simultaneous equations, we find that
a0 ¼
41 þ 4 2
4 41
and a1 ¼ 2
; and
2
L
41 þ 42 4 2 4 1
L 2x
L þ 2x
41 þ
42 .
þ
x¼
2L
2L
2
L
Based on Eq. (2.2), the two element shape functions are
4x ¼
N1 ¼
L 2x
L þ 2x
and N2 ¼
.
2L
2L
(2.18)
(2.19)
(2.20)
(2.21)
2.5 Element Shape Functions and [B] Matrix
These three examples show that using different element coordinate systems and
different ways to describe the length of the element changes the expressions of the
element shape functions. Eq. (2.11) demonstrates that for a bar element with P1 ¼ 4
and P2 ¼ 6, the corresponding shape functions are N1 ¼ 3 0.5x and
x
N2 ¼ 0.5x 2. From Eq. (2.16), we see that N1 ¼ Lx
L and N2 ¼ L for an element
with P1 ¼ 0 and P2 ¼ L. Lastly, Eq. (2.21) shows that for a bar element with
P1 ¼ L2 and P2 ¼ L2, the corresponding shape functions are N1 ¼ L2x
and
2L
L
þ
2x
N2 ¼ 2L . As we can see, element shape functions are different when different coordinate systems are chosen to describe the nodal positions.
Now, we observe that the sum of N1 and N2 yields a unit value (1) for all three
choices of the corresponding coordinate system. If the point of evaluation is located
at P1, we can see that N1 ¼ 1 and N2 ¼ 0 for all three example cases. Similarly, if P2
is selected as the evaluation point, we can see that N1 ¼ 0 and N2 ¼ 1 for all cases.
From these exercises, we can deduce the following properties for the element shape
functions related to a 2-node bar element:
•
•
Summation of all the element shape functions is equal to 1.
For the two nodal points P1 and P2, N1 ¼ 1 and N2 ¼ 0 when evaluation is at
point P1. When the evaluation point is P2, N1 ¼ 0 and N2 ¼ 1.
Later, we can demonstrate that these characteristics also hold true for other
element types. An important note is that the sequence of nodal arrangements can
greatly affect the outcomes of the calculations for shape functions. In all three examples, the first node is located on the left-hand side of the element, while the second node is located on the right-hand side. Eq. (2.21) is a more general form as
compared to Eqs. (2.11) and (2.16), because the origin of its local coordinate system
is defined at the center of the element, the same as needed for Gauss quadrature,
which is a numerical integration procedure commonly used in the FE method to
identify element-related parameters (see Section 4.5.1). Using this coordinate system, the shape functions for different elements of different lengths would still be
different. In Section 3.2, the concept of isoparametric shape functions is introduced,
which can be used to represent the same type of elements for easy numerical
manipulations.
Now, we change 4 (a physical quantity) to u (nodal displacement) in Eq. (2.20)
and by using the shape functions described in Eq. (2.21), we can write
L 2x
L þ 2x
u1 þ
u2 ¼ ½ N1 N2 f u1 u2 gT . (2.22)
2L
2L
This equation allows the calculation of the displacement at any point (i.e., u(x)) from
nodal displacements u1 and u2, as long as N1 and N2 are known. By definition, the
element strain for an axial element can be calculated by differentiating Eq. (2.22)
with respect to x:
T
du df ½ N1 N2 f u1 u2 g
¼
εxx ¼
dx
dx
u1
u1
d½ N1 N2 u1
1 1
¼
¼
¼ ½B
.
(2.23)
dx
u2
u2
u2
L L
uðxÞ ¼ N1 u1 þ N2 u2 ¼
73
74
CHAPTER 2 Meshing, Element Types, and Element Shape Functions
The difference between nodal displacements f u1 u2 gT and displacements anywhere {u(x)} must be kept in mind to avoid confusion. From Eq. (2.23), we can find
the element strain based on the displacements, or more precisely, nodal displacements. Thus, we call this equation the strainedisplacement equation. Along with
this equation, we introduce a new [B] matrix (or the strainedisplacement matrix)
commonly used in the FE method. This matrix describes the relationship between
the strain and nodal displacements and can be directly derived from the shape functions of a bar element by the following equation:
d
d½N
½ N1 N2 ¼
.
(2.24)
dx
dx
In theory, the strainedisplacement [B] matrix should be named strainenodal
displacement matrix, while εxx ¼ du
dx is better suited for the name straine
displacement matrix. The obvious difference arises from the fact that u(x) represents
the displacement anywhere within the element, while f u1 u2 gT are nodal displacements. However, the terminology for the [B] matrix has been used universally,
and hence we will continue using the term strainedisplacement matrix to represent
the [B] matrix. As mentioned, the equation εxx ¼ du
dx is, by definition, the straine
displacement equation. Because finding the solution to εxx ¼ du
dx requires analytical
differentiation, it may be difficult to incorporate all needed algorithms into a computer program to do the calculation. Eq. (2.23) implies that the axial strain εxx can be
calculated by simple multiplication and summation from the [B] matrix and nodal
displacements f u1 u2 gT . Thus, εxx ¼ ½Bf u1 u2 gT is commonly known as the
strainedisplacement equation.
Therefore, in addition to interpolating nodal values, the element shape functions
play another very important role in the FE method, that is, they can be used to calculate the element strain from the [B] matrix and nodal displacements. A final note
regarding the constant [B] matrix for a bar element: we can easily deduce from
Eq. (2.23) that the axial strain within the 2-node bar element is constant with a
1
magnitude of εxx ¼ u2 u
L .
½B ¼
2.5.1.2 2-Node Beam Element
Although a beam element uses the same 2-node configuration as the bar, truss,
and cable elements, the interpolation functions are quite different. This is
because each node of the beam element possesses two DOFs instead of one in
any of these elements. The two DOFs are vertical deflection along the z-axis
and rotation about the y-axis. Aside from the two nodal points P1 and P2, a third
reference node may be needed to prescribe the direction of the vertical z-axis.
As mentioned in Section 2.2, this element type requires C1 continuity, and hence
shape functions based on Hermite interpolation, which is named after Charles
Hermite (Dec. 1822eJan. 1901), are needed. Fig. 2.14 shows a beam element of
length L along the x-axis.
2.5 Element Shape Functions and [B] Matrix
θ
θ
FIGURE 2.14
A 2-node beam element of length L along the x-axis has two DOFs (vertical deflection
along the z-axis and rotation about the y-axis) per node. The sign convention for a positive
moment or rotation is in the counterclockwise direction about the y-axis. Also, a positive
force or deflection is along the positive z-axis.
The Hermite interpolation has the form
4ðxÞ ¼ a1 þ a2 x þ a3 x2 þ a4 x3 .
(2.25)
If we let the physical quantity 4(x) represent the vertical deflection w, then
wðxÞ ¼ a1 þ a2 x þ a3 x2 þ a4 x3 ¼ N1 w1 þ N2 q1 þ N3 w2 þ N4 q2 .
(2.26)
1
Because of the C continuity, both the vertical deflection (w) and its slope (q,
small rotation about the y-axis), which is the first derivative of the deflection,
must be continuous. For a very small angle of rotation, qztan q ¼ dw
dx , which can
be thought of as the slope of w with respect to x. Thus, to find q, we differentiate
Eq. (2.26):
qðxÞ ¼ a2 þ 2a3 x þ 3a4 x2 .
(2.27)
Applying the boundary conditions yields:
wðx ¼ 0Þ ¼ w1 ¼ a1 ;
(2.28)
qðx ¼ 0Þ ¼ q1 ¼ a2 ;
(2.29)
wðx ¼ LÞ ¼ w2 ¼ a1 þ a2 L þ a3 L2 þ a4 L3 ¼ w1 þ q1 L þ a3 L2 þ a4 L3 ; and
(2.30)
qðx ¼ LÞ ¼ q2 ¼ a2 þ 2a3 L þ 3a4 L2 ¼ q1 þ 2a3 L þ 3a4 L2 .
(2.31)
We write Eqs. (2.30) and (2.31) in matrix form in terms of two unknowns a3 and a4
as
"
#( ) (
)
a3
w2 w1 q1 L
L2 L3
¼
.
2L 3L2
q2 q1
a4
75
76
CHAPTER 2 Meshing, Element Types, and Element Shape Functions
Using Cramer’s rule, named after Gabriel Cramer (Jul. 1704eJan 1752), the two unknowns a3 and a4 can be easily calculated from Eqs. (2.30) and (2.31) written in matrix form,
a3 ¼
w2 w1 q1 L
L3
q2 q1
3L2
2
a4 ¼
¼
3
3w2 3w1 2Lq1 Lq2
and
L2
L
2L
L
3L2
L2
w2 w1 q1 L
q2 q1
2L
2
L
2L
¼
3
L
3L2
2w1 2w2 þ Lq1 þ Lq2
.
L3
(2.32)
(2.33)
For students who are not familiar with Cramer’s rule, consider that we have two
simultaneous linear equations:
ax þ by ¼ c
dx þ ey ¼ f :
We can rewrite these two equations in matrix form as
a b
x
c
¼
.
d e
y
f
y gT are solved by finding the
Based on Cramer’s rule, the two unknowns f x
ratios of determinants as shown below:
x¼
c
b
a
f
a
d
e
d
; y¼
b
a
e
d
c
f
;
b
e
where j j represents the determinant of the matrix. More specifically, the denominaa b
tor for both unknowns are the determinant of the matrix
. To find the first
d e
a
unknown x, we make the numerator by replacing the first column
of the ded
a b
c
nominator matrix
with
and then calculate the determinant of the
d e
f
new matrix. Similarly, the numerator for the second unknown y is calculated by
c
replacing the second column of the matrix with
and then computing the
f
determinant.
2.5 Element Shape Functions and [B] Matrix
Placing the four constants (a1, a2, a3, and a4) back to Eq. (2.26) give us
3w2 3w1 2Lq1 Lq2 2 2w1 2w2 þ Lq1 þ Lq2 3
x þ
x .
L2
L3
Rearranging the terms in the form of N1w1 þ N2q1 þ N3w2 þ N4q2, as seen in Eq.
(2.26), the beam element shape functions are expressed as
wðxÞ ¼ w1 þ q1 x þ
wðxÞ ¼
4
X
Ni wi ¼
i¼1
þ
L3 3Lx2 þ 2x3
L3 x 2L2 x2 þ Lx3
3Lx2 2x3
w1 þ
q1 þ
w2
3
3
L
L
L3
L2 x2 þ Lx3
q2 .
L3
(2.34)
where w(x) is known as the generalized displacement, which includes both the vertical deflection and the rotation. From Eq. (2.34), the four beam element shape functions Ni can be directly visualized. To find the slope q, we need to take the derivative
vw ¼ vN1 w þ vN2 q þ vN3 w þ vN4 q . For this reason, we list both N and N for easy
i
i,x
vx 1
vx 1
vx 2
vx 2
vx
references.
L3 3Lx2 þ 2x3
6Lx þ 6x2
and
N
¼
;
1;x
L3
L3
(2.35)
L3 x 2L2 x2 þ Lx3
L3 4L2 x þ 3Lx2
and N2;x ¼
;
3
L
L3
(2.36)
3Lx2 2x3
6Lx 6x2
and
N
¼
; and
3;x
L3
L3
(2.37)
N1 ¼
N2 ¼
N3 ¼
L2 x2 þ Lx3
2L2 x þ 3Lx2
and N4;x ¼
.
(2.38)
3
L
L3
For x ¼ 0 to L, N1 to N4 can be calculated from Eqs. (2.35)e(2.38) and plotted as
shown in Fig. 2.15. We can see that all four curves are continuous. We can also envision that the slopes of these four curves are continuous, as well. Hence, this set of
shape functions exhibits C1 continuity (i.e., both the deflection and slope of the
deflection are continuous). We can see from Fig. 2.15 that N1 ¼ 1, N2 ¼ 0,
N3 ¼ 0, and N4 ¼ 0 at x ¼ 0 (i.e., at P1). Similarly, N1 ¼ 0, N2 ¼ 0, N3 ¼ 1, and
N4 ¼ 0 can be observed at x ¼ L (i.e., at P2). We can also observe from the plots
for N1, N3, and N4 that the slopes (tangential lines) of these three shape-function
curves at P1 are zero. Additionally, the slope at P2 is 1 (a positive slope indicates
that the rotation is in the counterclockwise direction), as seen from the plot for
N2. Similarly, the slope observed from the plot for N4 at P2 is 1, while the slopes
seen in the plots at P2 for N1, N2, and N3 are zero. Thus, this set of functions fits
the basic characteristics of shape functions.
From Eqs. (2.35)e(2.38), we can express the vertical deflection and rotation at
any point within the element using the following equations:
N4 ¼
wðxÞ ¼ ½ N1
qðxÞ ¼ ½ N1;x
N2
N2;x
N3
N4 f w1
N3;x
q1
N4;x f w1
w2
q1
q2 gT and
(2.39)
T
(2.40)
w2
q2 g .
77
78
CHAPTER 2 Meshing, Element Types, and Element Shape Functions
N1
1
0.8
0.8
0.6
0.6
0.4
0.4
0.2
0.2
0
0
0
0.2
0.4
0.6
0.8
1
N3
1
0
0.2
0.4
0.6
0.8
1
0.6
0.8
1
-0.2
N4
1
0.8
0.8
0.6
0.6
0.4
0.4
0.2
0.2
0
0
N2
1
0
0.2
0.4
0.6
0.8
1
0
0.2
0.4
-0.2
FIGURE 2.15
Shape functions for a 2-node beam element plotted as the ratio of the x coordinates from
0 to L.
Now we shall look into the longitudinal (axial) stress generated during the
bending process. Consider an elastic, isotropic, and straight beam subjected to a vertical load at the right end, as shown in Fig. 2.16. From classical mechanics, we know
that there exists a neutral axis at which the longitudinal stress vanishes. If the cross
section of the beam is symmetric, then the neutral axis is located at the center of the
beam. In the configuration shown in Fig. 2.16, any points above the neutral axis
(concave side) are in compression while those points below the neutral axis (convex
side) are in tension.
If we consider the fact that only two nodes are used to represent a beam element,
we can intuitively understand that this line element must coincide with the neutral
axis of the beam. As such, no longitudinal stress would be observed within this
element. Because compressive and tensile stresses described earlier do occur under
bending, it would be incorrect not to consider their existence.
From Fig. 2.16, the neutral axis is located at half the height of the beam
throughout the entire beam. If the bending angle q is small, this angle can be approximated by the slope of the beam, that is, q ¼ dw
dx . Now, the axial elongation u at a distance z from the neutral axis in the tensile side can be calculated by multiplying the
distance z and bending angle as
uðxÞ ¼ z tan qzz q ¼ z
dw
.
dx
(2.41)
2.5 Element Shape Functions and [B] Matrix
θ
x
FIGURE 2.16
A beam element with a height of h is bent with a vertical deflection w. It is assumed that a
straight line perpendicular to the neutral axis in the undeformed beam remains straight
and perpendicular to the deformed neutral axis. The slope of the deformed neutral axis is
q ¼ dw
dx . Consider a point at a distance z below the neutral axis, the axial elongation (u) at
this height is u ¼ z q, and the maximum axial elongation is located at z ¼ h2 and
umax ¼ 12 h q.
From Eq. (2.41), the longitudinal strainedisplacement equation at a distance z from
the neutral axis can now be determined as
εxx ¼
du
d2 w
d2
¼ z 2 ¼ z 2 ½ N1
dx
dx
dx
¼ z½Bf w1
q1
w2
N2
N3
N4 f w1
q1
w2
q2 gT
(2.42)
T
q2 g .
The above equation shows the axial strainedisplacement relationship. In other
words, the axial strain at a distance z from the neutral axis is the product of z and
d 2 w, which is also known as the curvature of the curve. A small curvature indicates
dx2
very little change in the bending angle (slope), while a large curvature shows a large,
sharp turn. The strainedisplacement (or curvatureedisplacement) matrix [B] for a
2-node beam element can be calculated as shown in the equation below.
d2
½ N1 N2 N3 N4 dx2
6 12x
4 6x
6 12x
2 6x
2þ 3 þ 2
3 þ 2
L L
L L
L
L
L2
L
½B14 ¼
¼
(2.43)
2.5.2 2D, 3-NODE LINEAR TRIANGULAR ELEMENT
The simplest 2D elements are the 3-node triangular element and 4-node rectangular
plane stress element. These elements provide stiffness due to in-plane loading.
2.5.2.1 3-Node Linear Triangular Element
Fig. 2.17 shows a triangular element formed by three nodal points, P1, P2, and P3,
with two DOFs (translations along the x- and y-axes, i.e., u and v) at each of the
nodal points, for a total of six DOFs. Again, the counterclockwise arrangement of
the nodal points is followed, as previously prescribed.
79
80
CHAPTER 2 Meshing, Element Types, and Element Shape Functions
,
,
)
)
,
)
FIGURE 2.17
A 3-node triangular element. The coordinates (x,y) and displacements u(x,y) and v(x,y) of
any point P within the element can be determined from the respective nodal values.
For the horizontal displacement (i.e., displacement along the x-direction, u(x,y))
at any point (x,y) within the triangle, the following interpolation function can be
assumed:
uðx; yÞ ¼ a1 þ a2 x þ a3 y ¼ N1 u1 þ N2 u2 þ N3 u3 .
(2.44)
The same relationship can also be assumed for the vertical displacement anywhere within the element:
vðx; yÞ ¼ b1 þ b2 x þ b3 y ¼ N1 v1 þ N2 v2 þ N3 v3 .
(2.45)
Based on Eq. (1.10), which is repeated here for convenience
2
3
v
6
07
8
9 6 vx
7
6
7( )
>
< εxx >
= 6
v7
6
7 u
εyy ¼ 6 0
;
7
vy 7 v
>
>
:
; 6
6
7
gxy
6v v7
4
5
vy vx
vv
the axial and shear strains can be calculated as εxx ¼ vu
vx ¼ a2 ; εyy ¼ vy ¼ b3 ;
vu
vv
and gxy ¼ vy þ vx ¼ a3 þ b2 . As we can see from these calculations, all strain components within the element are constant values. For this reason, the linear 3-node
triangular element is also known as the constant strain triangle (CST). Due to this
nature, an FE model formed by linear triangular elements typically behaves more
stiffly than the physical structure that is being modeled. The only way to avoid this
extra stiffness is to provide needed strain variations across the structure by using many
elements. For this reason, this type of element is not recommended for complex
loading conditions, except for the case where a very fine mesh is used. There is no
2.5 Element Shape Functions and [B] Matrix
universal rule for determining how many triangular elements are needed to achieve an
acceptable solution. But it can be seen later from Fig. 2.21 and Table 2.4 in the end of
Section 2.5.3 that a simulated beam model made of 1288 CST elements did not
perform as well as one made of 40 quadrilateral elements.
From Eq. (2.44), three equations for nodal displacements u1, u2, and u3 at the
three nodal points P1, P2, and P3, respectively, can be derived as
u1 ¼ a1 þ a2 x 1 þ a3 y 1 ;
(2.46)
u2 ¼ a1 þ a2 x2 þ a3 y2 ; and
(2.47)
u3 ¼ a1 þ a2 x3 þ a3 y3 .
(2.48)
Because nodal values u1, u2, and u3 are calculated from FE solutions, the magnitudes of u1, u2, and u3 are considered to be known constant values, not variables.
Additionally, x1, x2, and x3 and values y1, y2, and y3 are nodal coordinates, and hence
all are known constant values. As such, we have three equations to solve for three
unknowns: a1, a2, and a3.
There are a number of ways to find these unknowns. Here, the method involving
only the matrix operation is discussed. The above three equations can be written in
matrix form as
8 9
8 9 2
38a 9
a1 >
>
>
>
1
u
1
x
y
>
>
>
>
>
1
1 <
< >
=
=
< 1=
6
7
(2.49)
u2 ¼ 4 1 x2 y2 5 a2 ¼ ½F a2 .
>
>
>
>
>
>
>
>
: >
;
>
:
;
;
:
u3
1 x3 y3
a3
a3
To find a1, a2, and a3, we can simply invert the [F] matrix and then multiply it to
both sides of the equations:
8 9
8 9
>
>
=
< u1 >
=
< a1 >
(2.50)
a2 ¼ ½F1 u2 .
>
>
;
: >
;
: >
a3
u3
As we know from linear algebraerelated courses on matrix operations, the inverse of any matrix [A] can be expressed in terms of its determinant (det) and adjoint
matrix (adj) as shown in Eq. (2.51). Note here that a classical adjoint of a square
matrix is also known as an adjugate or adjunct of the matrix.
½A1 ¼
1
adjðAÞ
detðAÞ
(2.51)
where
adjðAÞ ¼ ðcofactor matrix of AÞT .
(2.52)
To find the cofactor matrix of a 33 matrix [A], we assume it has the following
form:
2
3
a11 a12 a13
6
7
½A ¼ 4 a21 a22 a23 5.
(2.53)
a31 a32 a33
81
82
CHAPTER 2 Meshing, Element Types, and Element Shape Functions
Then, the cofactor matrix of [A] can be expressed as
2
3
A11 A12 A13
6
7
cofactor matrix ðAÞ ¼ 4 A21 A22 A23 5
A31 A32 A33
2
a22 a23
a21 a23
a21
6þ
þ
6 a
a31 a33
a31
32 a33
6
6
6 a12 a13
a11 a13
a11
6
þ
¼ 6
6 a32 a33
a31 a33
a31
6
6
6 a12 a13
a11 a13
a11
4þ
þ
a22 a23
a21 a23
a21
a22
3
7
a32 7
7
7
a12 7
7
7.
a32 7
7
7
a12 7
5
a22
(2.54)
Note that every other entry has the opposite sign. The process for determining
this cofactor matrix is to delete the row and column that is at the position of interest
in Eq. (2.54). For example, to find A11, eliminate row 1 and column 1, and then find
a22 a23
with proper sign conventhe determinant of the remaining 22 matrix
a32 a33
tion. Similarly, we calculate A12 by eliminating row 1 and column 2, finding the
determinant of the remaining matrix, and then multiplying by negative 1. Thus,
a21 a23
. Example 2.4 provides an exercise to go through these
A12 ¼ a31 a33
processes.
2
1
Example 2.4
6
Find the cofactor matrix of A ¼ 4 6
7
Solutions
A11 ¼
5
4
8
9
A21 ¼ A31 ¼
3
8
9
3
5
4
5
8
¼ 13; A12 ¼ 2
2
2
¼ 6; A22 ¼
2
13
6
Thus, cofactor matrixðAÞ ¼ 4 6
7
7
4 5.
9
6 4
¼ 26; A13 ¼
7 9
1 3
7 9
¼ 7; A32 ¼ 3
3
¼ 12; A23 ¼ 1
3
6
4
26
13
¼ 14; A33 ¼
3
7
12 6 5.
14 7
6
5
7
8
1 2
7 8
1
2
6
5
¼ 13
¼6
¼ 7
2.5 Element Shape Functions and [B] Matrix
Next, we find the adjoint by transposing the cofactor matrix based on Eq. (2.52).
3
2
A11 A21 A31
7
6
7
adjðAÞ ¼ ðcofactor ðAÞÞT ¼ 6
4 A12 A22 A31 5
A13 A23 A31
2
3 2
a22 a23
a22 a23
a12 a13
a13 a12
a12 a13
6
7 6
6
6 a
a32 a33
a22 a23 7
a33 a32
6
7 6 a32 a33
32 a33
6
7 6
6
6 a
7
a11 a13
a11 a13
a11 a13 7 6 a23 a21
21 a23
6
7¼6
¼6
6
7 6
6
6 a31 a33
a31 a33
a21 a23 7
a31 a33
6
7 6 a33 a31
6
7 6
6 a21 a22
a11 a12
a11 a12 7 6 a21 a22
a12 a11
4
5 4
a31
a32
a31
a32
a21
a22
a31
a32
a32
a31
(2.55)
The last part of Eq. (2.55) is aimed at eliminating the minus signs by swapping columns. However, there is no need to do so, except for the cosmetic reasons. To
explain this procedure, consider that the determinant for the matrix
a b
¼ ad bc. By swapping the first and second columns,
c d
b a
¼ ðad bcÞ. Thus, swapping columns results in negative determinants,
d c
which in turn allows the removal of the negative signs.
From Eq. (2.50), the three constants a1, a2, and a3 can be found by using the
following equation:
8 9
8 9 2
31 8 9
1 x1 y1
>
>
>
< a1 >
=
< u1 >
=
< u1 >
=
6
7
(2.56)
a2 ¼ ½F1 u2 ¼ 4 1 x2 y2 5
u2 .
>
>
>
: >
;
: >
;
: >
;
a3
u3
1 x3 y3
u3
From Eqs. (2.51) and (2.55), we find
2
½F1
6
6
6
6
1
1 6
6
adjðFÞ ¼
¼
6
detðFÞ
detðFÞ 6
6
6
6
4
F22
F23
F13
F12
F12
F32
F33
F33
F32
F22
F23
F21
F11
F13
F13
F33
F31
F31
F33
F23
F21
F22
F12
F11
F11
F31
F32
F32
F31
F21
F13
3
7
F23 7
7
7
F11 7
7
7.
F21 7
7
7
F12 7
5
F22
(2.57)
83
a12
a22
a13
a23
a11
a21
a13
3
7
a23 7
7
7
a11 7
7
7
7
a21 7
7
7
a12 7
5
a22
84
CHAPTER 2 Meshing, Element Types, and Element Shape Functions
Replacing the known F11 to F33 values in Eq. (2.57) results in
2
x2 y2
y1 x1
x1 y1
6
6 x y
y3 x3
x2 y2
6 3 3
6
6
y1 1
1 y1
1 6 y2 1
½F1 ¼
6
detðFÞ 6 y3 1
y2 1
1 y3
6
6
6 1 x2
x1 1
1 x1
4
x3 1
1 x3
1 x2
3
7
7
7
7
7
7
7.
7
7
7
7
5
(2.58)
Finally, we simplify the matrix by solving each of the determinants as follows:
2
3
x2 y3 x3 y2 x3 y1 x1 y3 x1 y2 x2 y1
1
6
7
½F1 ¼
(2.59)
y3 y1
y1 y2 5
4 y2 y3
detðFÞ
x3 x2
x1 x3
x2 x1
We know that the determinant of the [F] matrix shown in Eq. (2.59) is related to
the area of the triangle. The processes involved in deriving the formula are relatively lengthy and will not be covered in this book. However, the results are as
follows:
2
3
1 x1 y1
1
1 6
7
Area of a triangle ¼ A ¼ det½F ¼ det4 1 x2 y2 5.
(2.60)
2
2
1 x3 y3
From Eqs. (2.56), (2.59) and (2.60), we have
8 9
8 9
u1 >
a1 >
>
>
>
>
>
>
>
>
>
>
>
>
=
< >
=
< >
1
a2 ¼ ½F
u2
>
>
>
>
>
>
>
> >
>
> >
>
>
;
: >
;
: >
a3
u3
2
x2 y3 x3 y2 x3 y1 x1 y3
6
6
1
6 y2 y3
¼
y3 y1
detðFÞ 6
4
x3 x2
x1 x3
2
¼
1 6
4
2A
x2 y3 x3 y2
x3 y1 x1 y3
y2 y3
y3 y1
x3 x2
x1 x3
ðEq. 2:56Þ
38 9
x1 y2 x2 y1 >
> u1 >
>
> >
>
7>
7< =
7
ðEq. 2:59Þ
y1 y2 7 u2
>
>
>
5>
>
>
>
: >
;
x2 x1
u3
38 9
x1 y2 x2 y1 >
=
< u1 >
7
y 1 y 2 5 u2
>
;
: >
x2 x1
u3
ðEq. 2:60Þ
2.5 Element Shape Functions and [B] Matrix
Inserting a1, a2, and a3 into Eq. (2.44) and then rearranging terms in the form of
u(x) ¼ N1u1 þ N2u2 þ N3u3, we have
uðxÞ ¼ a1 þ a2 x þ a3 y ¼
¼
1
½ðx2 y3 x3 y2 Þu1 þ ðx3 y1 x1 y3 Þu2 þ ðx1 y2 x2 y1 Þu3 2A
þ
1
½ðy2 y3 Þu1 þ ðy3 y1 Þu2 þ ðy1 y2 Þu3 x
2A
þ
1
½ðx3 x2 Þu1 þ ðx1 x3 Þu2 þ ðx2 x1 Þu3 y
2A
1
½ðx2 y3 x3 y2 Þ þ ðy2 y3 Þx þ ðx3 x2 Þyu1
2A
þ
1
½ðx3 y1 x1 y3 Þ þ ðy3 y1 Þx þ ðx1 x3 Þyu2
2A
þ
1
½ðx1 y2 x2 y1 Þ þ ðy1 y2 Þx þ ðx2 x1 Þyu3
2A
Compared with Eq. (2.44), the three shape functions of a linear triangular
element can be expressed as
N1 ¼
1
f ðx2 y3 x3 y2 Þ þ ðy2 y3 Þx þ ðx3 x2 Þyg;
2A
(2.61)
1
(2.62)
f ðx3 y1 x1 y3 Þ þ ðy3 y1 Þx þ ðx1 x3 Þyg; and
2A
1
(2.63)
N3 ¼
f ðx1 y2 x2 y1 Þ þ ðy1 y2 Þx þ ðx2 x1 Þyg.
2A
As previously mentioned, this same set of shape functions can be used to find the
coordinates, horizontal displacement u(x,y), and vertical displacement v(x,y) of
any point P(x,y) within the element. The three shape functions have geometric meanings. Fig. 2.17 shows a point P(x,y) that divides the whole triangle into three subtriangles with areas A1, A2, and A3 opposite to points P1, P2, and P3, respectively.
From Eq. (2.60),
3
2
1 x y
7
6
2A1 ¼ det4 1 x2 y2 5 ¼ ðx2 y3 x3 y2 Þ þ ðy2 y3 Þx þ ðx3 x2 Þy.
(2.64)
N2 ¼
1
x3
y3
Note that Eqs. (2.64) and (2.61) have the same attributes. In other words,
A1
.
A
Similarly, N2 and N3 can be expressed in terms of A2 and A3 as
N1 ¼
N2 ¼
A2
A3
and N3 ¼ .
A
A
(2.65)
(2.66)
85
86
CHAPTER 2 Meshing, Element Types, and Element Shape Functions
FIGURE 2.18
Three area coordinates N1, N2, and N3 of a triangular element. We can see from this figure
that N1 þ N2 þ N3 ¼ 1 holds true at any point within the element.
These three equations demonstrate that the shape functions for a 3-node linear constant strain triangle are related to the ratios of the three subareas to the entire area of
the triangle. Hence, these shape functions are referred to as the area coordinates.
Note that there are only three shape functions, and together they comprise the entire
area of the triangle. Fig. 2.18 graphically shows the magnitudes of these three area
coordinates at different locations within the triangle.
The strainedisplacement equations {ε} ¼ [B]{u} for this triangular element can
be derived from u and v using Eq. (1.10) restated for a triangular element as listed
below.
εxx ¼
uðx; yÞ ¼ N1 u1 þ N2 u2 þ N3 u3
(2.67)
vðx; yÞ ¼ N1 v1 þ N2 v2 þ N3 v3
(2.68)
vu vN1
vN2
vN3
¼
u1 þ
u2 þ
u3
vx
vx
vx
vx
1
½ ðy2 y3 Þu1 þ ðy3 y1 Þu2 þ ðy1 y2 Þu3 ¼
2A
εyy ¼
vv vN1
vN2
vN3
¼
v1 þ
v2 þ
v3
vy
vy
vy
vy
1
½ ðx3 x2 Þv1 þ ðx1 x3 Þv2 þ ðx2 x1 Þv3 ¼
2A
gxy ¼
¼
(2.69)
(2.70)
vu vv vN1
vN2
vN3
vN1
vN2
vN3
þ ¼
u1 þ
u2 þ
u3 þ
v1 þ
v2 þ
v3
vy vx
vy
vy
vy
vx
vx
vx
1
½ ðx3 x2 Þu1 þ ðy2 y3 Þv1 þ ðx1 x3 Þu2 þ ðy3 y1 Þv2
2A
þ ðx2 x1 Þu3 þ ðy1 y2 Þv3 (2.71)
2.5 Element Shape Functions and [B] Matrix
In matrix form, the strainedisplacement equations and the strainedisplacement
matrix [B] are expressed as
8
9
2
3
ðy2 y3 Þ
0
ðy3 y1 Þ
0
ðy1 y2 Þ
0
>
< εxx >
=
1 6
7
εyy ¼
0
ðx3 x2 Þ
0
ðx1 x3 Þ
0
ðx2 x1 Þ 5
4
>
>
2A
:
;
gxy
ðx3 x2 Þ ðy2 y3 Þ ðx1 x3 Þ ðy3 y1 Þ ðx2 x1 Þ ðy1 y2 Þ
8
9
u1 >
>
>
>
>
>
>
>
>
>
>
>
v
1
>
>
>
>
>
>
>
<u >
=
2
.
>
v2 >
>
>
>
>
>
>
>
>
>
> u3 >
>
>
>
>
>
>
>
:
;
v3
(2.72)
Recalling that the strainedisplacement matrix [B] describes the relationship
between the8 strain
tensor and nodal displacements, we write
9
u
>
>
1
>
>
>
>
>
>
>
> v1 >
>
8
9
>
>
>
>
>
>
>
>
< εxx >
< u2 >
=
=
εyy ¼ ½B
, and therefore
v2 >
>
>
>
:
>
>
;
>
>
gxy
>
> u3 >
>
>
>
>
>
>
>
>
>
>
>
:
;
v3
2
3
ðy2 y3 Þ
0
ðy3 y1 Þ
0
ðy1 y2 Þ
0
1 6
7
½B ¼
0
ðx3 x2 Þ
0
ðx1 x3 Þ
0
ðx2 x1 Þ 5.
4
2A
ðx3 x2 Þ ðy2 y3 Þ ðx1 x3 Þ ðy3 y1 Þ ðx2 x1 Þ ðy1 y2 Þ
(2.73)
2.5.3 4-NODE RECTANGULAR BILINEAR PLANE ELEMENT WITH
EDGES PARALLEL TO THE COORDINATE AXES
Fig. 2.19 shows a 4-node rectangular plane element formed by points P1, P2, P3, and
P4, with the origin of the coordinate system located at the geometric center of the
element. Again, these four nodes are arranged in a counterclockwise manner, as
described earlier. Each node has two DOFs, translations along the x-axis (u) and
along the y-axis (v). Further, it is assumed that the same shape functions are used
to interpolate nodal coordinates, horizontal displacement (u), and vertical displacement (v), at any point within the element from corresponding nodal values. We assume that the following polynomials are used to interpolate u and v:
uðx; yÞ ¼ a1 þ a2 x þ a3 y þ a4 xy ¼ N1 u1 þ N2 u2 þ N3 u3 þ N4 u4 and
(2.74)
vðx; yÞ ¼ b1 þ b2 x þ b3 y þ b4 xy ¼ N1 v1 þ N2 v2 þ N3 v3 þ N4 v4 .
(2.75)
87
88
CHAPTER 2 Meshing, Element Types, and Element Shape Functions
FIGURE 2.19
A 4-node rectangular bilinear element with dimensions 2a 2b with the origin of the
coordinate system located at the center of the element.
where a1, a2, a3, and a4 and b1, b2, b3, and b4 are constants, and N1, N2, N3, and N4 are
the element shape functions. Note that the interpolation equations involving constants a’s and b’s have the same form regardless of which origin of the Cartesian coordinate system chosen. However, selecting a different coordinate system would
result in a different set of shape functions, as noted in the descriptions for bar elements (see Section 2.5.1). This element type is called the bilinear element because
the u and v displacements are linear along the x-axis when a constant y-coordinate is
chosen and linear along the y-axis when a constant value of the x-coordinate is
selected. For nodal displacements u1 through u4, the following equations can be written from Eq. (2.74) by plugging in the respective nodal coordinates:
u1 ¼ uða; bÞ ¼ a1 þ a2 ðaÞ þ a3 ðbÞ þ a4 ab
(2.76)
u2 ¼ uða; bÞ ¼ a1 þ a2 ðaÞ þ a3 ðbÞ þ a4 ðabÞ
(2.77)
u3 ¼ uða; bÞ ¼ a1 þ a2 ðaÞ þ a3 ðbÞ þ a4 ab
(2.78)
u4 ¼ uða; bÞ ¼ a1 þ a2 ðaÞ þ a3 ðbÞ þ a4 ðabÞ
(2.79)
The four unknowns (a1, a2, a3, and a4) can be found by solving these four
3 þu4
2 þu3 u4
simultaneous equations. These values, a1 ¼ u1 þu2 þu
, a2 ¼ u1 þu4a
,
4
u1 u2 þ u3 þ u4
u1 u2 þ u3 u4
, and a4 ¼
, are then inserted back into Eq. (2.74).
a3 ¼
4b
4ab
Then we rearrange all terms in the order of u1, u2, u3, and u4, so that horizontal
displacement at any point, u(x,y), within the element can be found using the
following equation:
uðx; yÞ ¼ N1 u1 þ N2 u2 þ N3 u3 þ N4 u4
¼
ða xÞðb yÞ
ða þ xÞðb yÞ
u1 þ
u2
4ab
4ab
þ
ða þ xÞðb þ yÞ
ða xÞðb þ yÞ
u3 þ
u4 .
4ab
4ab
(2.80)
2.5 Element Shape Functions and [B] Matrix
That is, the shape functions for a 4-node bilinear element are displayed as
ða xÞðb yÞ
4ab
ða þ xÞðb yÞ
N2 ¼
4ab
(2.81)
ða þ xÞðb þ yÞ
N3 ¼
4ab
ða xÞðb þ yÞ
.
N4 ¼
4ab
Because Eqs. (2.74) and (2.75) are of the same form, we can easily deduce
that the same set of shape functions can be used to find the vertical displacement
v(x,y) at any of the points within the element. Again, it can be checked that this
set of shape functions fits the criteria that: (1) the sum of all shape functions
is equal to 1 and (2) at point P1(a,b), N1 ¼ 1, while N2 N3 ¼ N4 ¼ 0, etc.
We can now derive the strainedisplacement [B] matrix from Eqs. (2.74), (2.75),
and (2.81) as
N1 ¼
3
2
3
8P
9
4
v
v
6
0 7>
>
>
Ni ui >
07
>
>
6
7
6
vx
>
8
9 6 vx
>
>
7
7>
6
i¼1
<
=
ε
8
9
>
6
7
7
xx >
6
>
>
>
>
7<u= 6
7
<
= 6
6
6
v7
v 7>
>
>
4
7
7>X
6
εyy ¼ 6
>
>
>
6 0 vy 7 : ; ¼ 6 0 vy 7 >
>
>
>
>
N
v
:
>
>
6
7 v
7
6
i i;
>
>
:
; 6
7
7 i¼1
6
gxy
6
7
7
6
4v v5
6v v7
5
4
vy vx
vy vx
2
2
vN
6 1
6 vx
6
6
6
¼6 0
6
6
6 vN1
4
vy
2
1 6
¼
4
4ab
0
vN2
vx
0
vN3
vx
0
vN4
vx
vN1
vy
0
vN2
vy
0
vN3
vy
0
vN1
vx
vN2
vy
vN2
vx
vN3
vy
vN3
vx
vN4
vy
ðb yÞ
0
ðb yÞ
0
ða xÞ
0
ða xÞ ðb yÞ ða þ xÞ
8 9
u1 >
>
>
>
>
> >
>
>
>
>
>
v
1
>
>
3>
> >
>
>
>
>
>
>
>
>
> u2 >
>
0 7>
>
>
7>
>
>
7 < v2 =
vN4 7
7> >
7 > u3 >
vy 7 >
> >
>
>
7>
>
>
>
>
>
v3 >
>
>
vN4 7
5>
>
>
>
>
>
vx >
>
>
u4 >
>
>
>
>
>
: >
;
v4
8 9
u1 >
>
>
>
>
>
>
>
>
>
>
v1 >
>
>
>
>
>
>
>
>
>
>
>
>
u
>
>
2
>
>
>
>
>
>
3>
>
<
v2 =
0
ðb þ yÞ
0
ðb þ yÞ
0
7> >
ða þ xÞ
0
ða þ xÞ
0
ða xÞ 5 >
u >
>
> 3>
>
>
>
>
ðb yÞ ða þ xÞ ðb þ yÞ ða xÞ ðb þ yÞ >
>
>
>
> v3 >
>
>
>
>
>
>
>
>
> u4 >
>
>
>
>
>
>
: >
;
v4
89
90
CHAPTER 2 Meshing, Element Types, and Element Shape Functions
2
1 6
½B ¼
4
4ab
ðb yÞ
0
ðb yÞ
0
ðb þ yÞ
0
ðb þ yÞ
0
ða xÞ
0
ða þ xÞ
0
ða þ xÞ
0
ða xÞ ðb yÞ ða þ xÞ
ðb yÞ
ða þ xÞ ðb þ yÞ
ða xÞ
0
3
7
ða xÞ 5.
ðb þ yÞ
(2.82)
Another commonly used method to obtain the element shape functions is the
application of the Lagrange interpolation, published by Joseph-Louis Lagrange
(Jan. 1736eApr. 1813). According to Wolfram Math World (2017), the Lagrange
interpolation was first published by Waring in 1779, rediscovered by Euler in 1783,
and then published by Lagrange in 1795 (Jeffreys and Jeffreys, 1988). Using this
interpolation method, a polynomial curve of up to (n 1)th degrees can be found
to pass through a set of n data points on the xey plane. For example, a set of three
data points will need a second degree polynomial to pass through all three points.
Assume a set of n data points are located at (x1, y1), (x2, y2), ., (xn, yn) on the
xey plane, described as yi ¼ f(xi). Then, the Lagrange interpolation of this function
has the form of
f ðxÞ ¼ Q1 y1 þ Q2 y2 þ . þ Qn yn ;
where
ðx x2 Þðx x3 Þ.ðx xn Þ
;
ðx1 x2 Þðx1 x3 Þ.ðx1 xn Þ
(2.84)
ðx x1 Þðx x3 Þ.ðx xn Þ
; and
ðx2 x1 Þðx2 x3 Þ.ðx2 xn Þ
(2.85)
ðx x1 Þðx x2 Þ.ðx xn1 Þ
.
ðxn x1 Þðxn x2 Þ.ðxn xn1 Þ
(2.86)
Q1 ¼
Q2 ¼
(2.83)
Qn ¼
Note that if x ¼ x1, the numerator and denominator are the same for Eq. (2.84), and
thus Q1 ¼ 1. All other equations have x x1 in the numerator, which makes Q2 through
Qn equal to 0. The same findings also apply to x ¼ x2, x ¼ x3, ., x ¼ xn. Finally, the
sum of Q1 through Qn equals 1. Thus, functions Q1 through Qn fit the characteristics
of the shape functions. For this reason, the Lagrange interpolation method is frequently
used to define the element shape functions, especially for those high-order elements.
Example 2.5
Find a polynomial equation f(x) that passes through three points on the xey plane
with coordinates of (1, 1), (2, 4), and (3, 9).
Solution
Because only three points are involved, Eqs. (2.84)e(2.86) become
ðx x2 Þðx x3 Þ
Q1 ¼
ðx1 x2 Þðx1 x3 Þ
Q2 ¼
ðx x1 Þðx x3 Þ
ðx2 x1 Þðx2 x3 Þ
Q3 ¼
ðx x1 Þðx x2 Þ
ðx3 x1 Þðx3 x2 Þ
2.5 Element Shape Functions and [B] Matrix
We plug the three points into Eqs. (2.84)e(2.86):
Q1 ¼
ðx 2Þðx 3Þ x2 5x þ 6
¼
;
ð1 2Þð1 3Þ
2
Q2 ¼
ðx 1Þðx 3Þ x2 4x þ 3
¼
; and
ð2 1Þð2 3Þ
1
Q3 ¼
ðx 1Þðx 2Þ x2 3x þ 2
¼
.
ð3 1Þð3 2Þ
2
We now have our three Q values needed for Eq. (2.83), and the y values are
y1 ¼ 1, y2 ¼ 4, and y3 ¼ 9. By plugging these three values into Eq. (2.83), we find
that
x2 5x þ 6
x2 4x þ 3
x2 3x þ 2
1þ
4þ
9
2
1
2
1
9
5
27
¼
4 þ x2 þ þ 16 x þ ð3 12 þ 9Þ ¼ x2 .
2
2
2
2
f ðxÞ ¼
For each x value on a polynomial equation, there is only one corresponding y
value. As such, x1 s x2 s x3.sxn as long as there are no repeated points on the
polynomial equation. Hence, it is not possible for the denominator to become
zero, and there is no need to worry about the problem associated with division
to zero.
To find the shape functions for a 2D, 4-node plane element, we need to employ the
Lagrange interpolation twice, first along the x-direction and then along the y-direction,
or vice versa. Finally, we integrate the shape functions obtained from each axis into
one set of shape functions for this 2D element. The following section shows a
step-by-step approach to determine the 2D, 4-node plane element shape functions.
Step 1:
We first determine the two shape functions to be used for interpolating any
physical value 4 from left to right. Considering x1 ¼ a and x2 ¼ a, the two shape
functions needed for interpolation along the x-direction are
N1x ¼
x x2
xa
xa
x x1
xþa xþa
¼
and N2x ¼
¼
;
¼
¼
2a
2a
x1 x2 a a
x2 x1 a þ a
(2.87)
where the “1” in N1x represents node 1 and the “2” in N2x represents node 2.
Step 2:
Next, we determine the two shape functions to be used for interpolating any
physical value 4 from bottom to top. Considering y1 ¼ b and y4 ¼ b, the two shape
functions needed for interpolation along the y-direction are
N1y ¼
y y2
yb
yb
y y1
yþb yþb
¼
and N4y ¼
¼
;
¼
¼
2b
2b
y1 y2 b b
y4 y1 b þ b
(2.88)
where the “1” in N1y represents node 1, and the “4” in N4x represents node 4.
91
92
CHAPTER 2 Meshing, Element Types, and Element Shape Functions
Step 3:
For a line parallel to the x-axis at y ¼ b (i.e., line P1eP2 in Fig. 2.19), a physical value anywhere on this line (4a) can be written using the shape functions N1x and
N2x and the physical values at node 1 (41) and node 2 (42) as
xa
xþa
41 þ
4 .
(2.89)
2a
2a 2
We can use the same two shape functions N1x and N2x to interpolate the physical
values for any points on the line P4eP3 (4b) using the physical values at node 4 (44)
and node 3 (43) as
4a ¼ 4at any point on P1 P2 ¼ N1x 41 þ N2x 42 ¼
4b ¼ 4at any point on P4 P3 ¼ N1x 44 þ N2x 43 ¼
xa
xþa
44 þ
4 .
2a
2a 3
(2.90)
Step 4:
Next, we interpolate from 4a (corresponding to the bottom line of the rectangle,
i.e., line P1eP2) to 4b (corresponding to the top line of the rectangle, i.e., line
P4eP3) along the y-axis so that the physical value for any point within the element
can be obtained:
yb xa
xþa
yþb xa
xþa
41 þ
42 þ
44 þ
43
4 ¼ N1y 4a þ N4y 4b ¼
2b 2a
2a
2b
2a
2a
¼
ðy bÞðx aÞ
ðy bÞðx þ aÞ
ðy þ bÞðx aÞ
ðy þ bÞðx þ aÞ
41 42 44 þ
43 .
4ab
4ab
4ab
4ab
(2.91)
Step 5:
Finally, we organize the terms in Eq. (2.91), so the x is the second component
in the first set of parentheses, and y is the second component in the second set.
Additionally, we rectify the negative signs associated with 42 and 44. After these
procedures, the resulting four shape functions are
N1 ¼
ða xÞðb yÞ
;
4ab
(2.92)
N2 ¼
ða þ xÞðb yÞ
;
4ab
(2.93)
ða þ xÞðb þ yÞ
; and
4ab
(2.94)
N3 ¼
ða xÞðb þ yÞ
.
(2.95)
4ab
This set of four shape functions is identical to the one shown in Eq. (2.81), which
was found using a different approach. In addition to the aforementioned applications, shape functions are useful for calculating distributions of any physical
quantities, such as the temperature contours on a weather map.
N4 ¼
2.5 Element Shape Functions and [B] Matrix
Example 2.6
The weather report shows that the temperatures in the cities of Novi, Royal Oak,
Troy, and Commerce Township are 60, 64, 68, and 62 F, respectively. Assume
that the coordinates, in miles, of these four cities are (10, 5), (10, 5),
(10, 5), and (10, 5), respectively. Draft the temperature contours based on the
locally measured temperatures and geometric locations based on bilinear
interpolation.
Solution
Based on the coordinates of the four cities, it is clear that these four cities are
located at the four corners of a rectangle, and the origin of the coordinate system
is located exactly at the center of the rectangle. Thus, the rectangle formed by the
four cities is analogous to the rectangle shown in Fig. 2.19, with a ¼ 10 and
b ¼ 5. From Eqs. (2.92)e(2.95), we find that
ð10 xÞð5 yÞ
ð10 þ xÞð5 yÞ
ð10 þ xÞð5 þ yÞ
; N2 ¼
; N3 ¼
; N4
200
200
200
ð10 xÞð5 þ yÞ
.
¼
200
N1 ¼
The physical values (temperatures) at the four corners are 41 ¼ 60 F,
42 ¼ 64 F, 43 ¼ 68 F, and 44 ¼ 62 F and the interpolation equation is
T(x,y) ¼ N141 þ N242 þ N343 þ N444. Combining the values with the equation
gives us
ð10 xÞð5 yÞ
ð10 þ xÞð5 yÞ
ð10 þ xÞð5 þ yÞ
60 þ
64 þ
68
Tðx; yÞ ¼
200
200
200
þ
ð10 xÞð5 þ yÞ
62
200
ð50 5x 10y þ xyÞ 60 þ ð50 þ 5x 10y xyÞ 64
þð50 þ 5x þ 10y þ xyÞ 68 þ ð50 5x þ 10y xyÞ 62
¼
200
¼
12; 700 þ 50x þ 60y þ 2xy
200
To draft the contours, let us first identify locations where the temperature is the
same based on the above equation. We start with 62 F.
12; 700 þ 50x þ 60y þ 2xy
¼ 62 050x þ 60y þ 2xy ¼ 300
200
Because we have only one equation to solve for two unknowns, multiple combinations of (x, y) could satisfy the equation listed above. Rewrite this equation in
the form of
ð60 þ 2xÞy ¼ 300 50x0y ¼
300 50x
.
60 þ 2x
93
94
CHAPTER 2 Meshing, Element Types, and Element Shape Functions
Using this equation, we can assume an x value within the ranges of
10 x 10, and then calculate the corresponding y value. If the calculated y
value is outside the boundary of 5 y 5, then this point is not valid.
FIGURE 2.20
The left figure shows several northwestern suburbs of Detroit based on a Google map.
Assume that the four cities are at the four corners of a rectangle of 20 10 miles in
size. The 4-node plane shape functions can be used to find the isotherm lines shown
on the right, based on locally measured temperatures at the corner cities.
Table 2.3 Geometric Points for 62 F Isotherm
x
10.0
9.0
8.0
7.0
6.0
5.0
4.0
3.0
2.0
1.0
0.0
y
5.0
3.57
2.27
1.09
0.0
1.0
1.92
2.78
3.57
4.31
5.0
For illustration purposes, we assume that x ¼ 10, 9, ., 0, 1, ., 10 and
then calculate the corresponding y values. For an isotherm of 62 F, locations
listed in Table 2.3 satisfy the equation above. Also, for any locations with
x > 0, the corresponding y values are out of the range and cannot be used. We
repeat the procedure to find the geometric points associated with the 64 and
66 F isotherm lines. Connecting these points for the isotherms yields the contour
map shown on the right of Fig. 2.20.
2.5.3.1 Comparison of CST and Bilinear Quadrilateral Element
Example 2.7
Assume a cantilever beam of length 1 m is loaded at the free end by a downward
vertical force of 1000 N (Fig. 2.21 top). The beam has a Young’s modulus of
100 GPa, a cross-sectional width of 0.05 m, and height of 0.1 m, which result
2.5 Element Shape Functions and [B] Matrix
FIGURE 2.21
A cantilever beam (top) is fixed on the left side and vertically loaded on the right side.
Two example FE models consisting only of triangular and quadrilateral plane
elements are presented to represent the beam. The model made of triangular
elements has a total of 82 elements (middle), while the model made of quadrilateral
elements has a total of 640 elements (bottom).
5 105 m4 . In theory, a beam should be
in a cross-sectional moment of inertia of 12
modeled with beam elements to secure the best result. However, in certain circumstances plane elements integrate better with other elements within the
same structure. Calculate the free-end deflection (maximum deflection) using
(1) an analytical approach and (2) FE models made of 2-node beam elements,
3-node constant strain triangular elements, and 4-node bilinear quadrilateral elements of varying element sizes. Calculate the maximum deflection of each model
using an FE software package and tabulate the results.
Solution
For a cantilever beam of length L, elastic modulus E, moment of inertia I, and
load P at the free end (Fig. 2.21 top), the maximum deflection based on analytical
solution is PL3/3EI. Thus, the exact solution results in the maximum deflection of
0.8 mm.
Two example cantilever beam FE models consisting of a total of 82 2D, triangular surface elements and 640 2D, quadrilateral surface elements are shown in
the middle and bottom of Fig. 2.21, respectively. Additional models based on
2-node beam elements, 3-node CST elements, and 4-node quadrilateral elements
have also been created, as listed in Table 2.4. An FE software package ANSYS
14.5 (Canonsburg, PA) was used to calculate the maximum deflections for all
95
96
CHAPTER 2 Meshing, Element Types, and Element Shape Functions
Table 2.4 Comparisons of Maximum Deflections Calculated by an FE
Model Against Analytical Analysis
Element Type
2-Node beam element
3-Node CST plane
element
4-Node bilinear plane
element
No. of
Row/Element
1/2
1/4
1/8
2/82
4/324
8/1288
2/40
4/160
8/640
Max. Deflection
(mm)
0.765823
0.796087
0.803653
0.537724
0.716967
0.782914
0.805085
0.806810
0.807759
Difference (%)
4.27
0.49
0.46
32.78
10.34
2.14
0.64
0.85
0.97
Models consisting of beam elements are very efficient, as demonstrated by the fact that only four
elements are needed to achieve an accuracy within 0.5%, while models made of triangular elements
require many more elements to match accuracy obtained when other element types are selected.
models, and results are compared to the analytical solutions (Table 2.4). From
these results, it is clear that the beam element is the most efficient one to use.
In theory, a structure consists of only beam elements and modeling with only
beam elements should provide the same result as that calculated using analytical
method. However, roundoff error and the number of integration points selected
contribute to the small error observed. Nevertheless, the beam element is the
best element type to choose for modeling beam structures. Also, it can be seen
that the model based on beam elements converges much quicker than the model
that consists of CST or bilinear elements.
For models meshed entirely with CST elements, the error is greater than 2%,
even with a fine mesh of 1288 elements. This element type converges slowly, and
hence a model of this type requires a large number of elements to predict acceptable solutions. On the other hand, quadrilateral elements converge much quicker,
and fewer elements are needed to attain the same results as what can be achieved
using the CST element type. This simple example demonstrates that the CST
element should be used sparsely.
2.5.4 2D, 4-NODE PLATE ELEMENT SHAPE FUNCTIONS WITH EDGES
PARALLEL TO THE COORDINATE AXES
2.5.4.1 Use Pascal’s Triangle to Select Polynomial Terms
A plate is actually a 3D structure component that is typically simplified using the 2D
element type. As mentioned in Section 2.3, a plate element possesses three DOFs, a
2.5 Element Shape Functions and [B] Matrix
vertical deflection (w) and two in-plane rotations (qx and qy), per node for a total of
12 DOFs. Instead of having a separate plate element type, some software packages
simply use a “generalized shell element” with several options for representing the
membrane element, plate element, and shell element. Considering the lumping of
these three element types together into one generalized shell element, there can
be a great deal of confusion for software users. In addition, significant differences
exist in the shell element formulations among different software packages. It is
therefore necessary for users of software to read and comprehend the contents discussed in the theory and user manuals published by the software vendors. In other
words, it is in the user’s best interest to understand the fundamental theory behind
the formulation of each element type, before the software is used to find solutions
for the problem at hand. With so many research papers published on different
ways of formulating plate and shell elements, it would be impossible to include
all variations in this book; only the most fundamental ones related to plate element
are listed here.
Earlier in this chapter, a second order, 4-term polynomial equation is used to
interpolate the in-plane displacements u and v for a plane stress element. The selection of the interpolation equation (a1 þ a2x þ a3y þ a4xy) makes this element type a
bilinear element, as explained previously. In addition to the xy term, there are other
second order terms: x2 and y2 shown in Fig. 2.22 (right). The triangular shaped
figure, shown in Fig. 2.22 (left), is commonly known as Pascal’s triangle, which
is credited to Blaise Pascal (Jun. 1623eAug. 1662), who published the concept. It
is worth noting that in a number of old civilizations, such as China, India, and Persia,
the same concept was reported long before Pascal’s publication. In addition to the
triangle, Blaise Pascal was noted for his discoveries that the sum of the three internal
angles of a triangle is equal to 180 degrees and the atmospheric pressure decreases as
the height increases, because a vacuum exists in outer space.
Pascal’s triangle is used to determine the number of terms in a complete nth order, two variables polynomial and the coefficients of all components with the form
of (x þ y)n. First, the Pascal’s triangle on the right-hand side of Fig. 2.22 can be used
to determine the number of complete, two-variable polynomial terms when given the
order of the equation. For instance, a complete polynomial with the exponent n ¼ 0
has only one constant term (a0), which corresponds to the first row in the triangle.
For n ¼ 1, there are three terms (a0 þ a1x þ a2y), corresponding to the first and
second rows in the triangle. For a second order exponent (n ¼ 2), the complete
Exponent
0
1
Pascal's Triangle
1
1
1
2
1
3
1
4
5
1
1
2
3
4
5
X
1
3
6
10
(X+Y)
1
X
1
4
10
X
1
5
X
1
X
Y
XY
XY
XY
XY
FIGURE 2.22
Pascal’s triangle and corresponding polynomial terms for (x þ y)n.
Y
XY
XY
XY
Y
XY
XY
Y
XY
Y
97
98
CHAPTER 2 Meshing, Element Types, and Element Shape Functions
polynomial has the form of a0 þ a1x þ a2y þ a3x2þa4xy þ a5y2, as seen in the first
three rows. This same procedure can be used to determine the complete polynomial
terms for any two-variable equations with higher order exponents.
This feature is used to select the appropriate terms to form an interpolation function needed to derive the element shape functions. For instance, an 8-node serendipity plane element requires eight polynomial terms and the eight associated
coefficients needed to calculate the eight shape functions. However, the complete
polynomial for a second order exponent (n ¼ 2) has a total of six terms, which is
two terms short for a serendipity element. On the other hand, a third order exponent
(n ¼ 3) has a total of ten terms, which is two terms more than that needed. If all six
terms in the second order exponent are selected, code developers can decide which
two terms selected from (x3, x2y, xy2, y3) are used to form the interpolation function.
Alternatively, the code developer may want to choose combinational forms (e.g.,
x3 þ x2y and xy2 þ y3) for use in the calculation of the element shape functions.
As such, shape functions for a serendipity element may differ a great deal due to
different choices of polynomial terms.
The second application of Pascal’s triangle is to determine the coefficients (a0,
a1, a2, ., an) for a partial polynomial. A partial polynomial is related to only the
terms within the same order. For example, a second order two-variable polynomial
(x þ y)2 is related to only the x2, xy and y2 terms (i.e., the third row of the right Pascal’s triangle). The coefficients for the exponent n ¼ 2 are found in the third row of
the left Pascal’s triangle, where the values are 1, 2, and 1. Hence, the resulting second order partial polynomial is (x þ y)2 ¼ x2þ2xy þ y2. Eq. (2.96) illustrates this
concept up to the third order. If the component polynomials include coefficients,
such as (2x þ y)2, the corresponding coefficients can be easily obtained by replacing
x with 2x in Eq. (2.96). Based on this concept, we find (2x þ y)2 ¼ (2x)2 þ 2(2x)
y þ y2 ¼ 4x2 þ 4xy þ y2.
ðx þ yÞ0 ¼ 1; ðx þ yÞ1 ¼ x þ y; ðx þ yÞ2 ¼ x2 þ 2xy þ y2 ;
ðx þ yÞ3 ¼ x3 þ 3x2 y þ 3xy2 þ y3 ; etc.
(2.96)
2.5.4.2 Select Polynomial Functions to Interpolate
a Four-Node Plate Element
For a 4-node plate element, there are a total of 12 DOFs, because each node has three
DOFs. As we can see from the previous example using Pascal’s triangle, there are
only 10 xy polynomial terms for a complete third order exponent. To obtain the
12 polynomial terms needed for the 4-node plate element, two additional terms
from the fourth order polynomial must be involved. Because the complete fourth
order polynomial has 15 xy terms, educated selections must be made based on
individual or combined xy polynomial terms to obtain the necessary 12 terms. As
mentioned earlier, different choices of polynomial terms will yield different interpolation equations and hence different shape functions. In this section, only the selection listed below is discussed. If you are solving problems in which plate and shell
2.5 Element Shape Functions and [B] Matrix
elements are of great importance, you should consider reading more research papers
regarding the pros and cons of different polynomial terms selected to make up the
interpolation function. For now, assume that a 12-term polynomial used to interpolate the vertical deflection w(x,y) of a 4-node plate element has the form
wðx; yÞ ¼ a1 þ a2 x þ a3 y þ a4 x2 þ a5 xy þ a6 y2 þ a7 x3 þ a8 x2 y þ a9 xy2 þ a10 y3
þ a11 x3 y þ a12 xy3 .
(2.97)
Then, equations for the two slopes (in-plane rotations) are
vwðx; yÞ
¼ a3 þ a5 x þ 2a6 y þ a8 x2 þ 2a9 xy þ 3a10 y2 þ a11 x3 þ 3a12 xy2 ; and
vy
(2.98)
vwðx; yÞ
¼ a2 þ 2a4 x þ a5 y þ 3a7 x2 þ 2a8 xy þ a9 y2 þ 3a11 x2 y þ a12 y3 . (2.99)
vx
Written in matrix form, these three equations are
fwðx; yÞg31 ¼ ½F312 fag121 0
8
9
>
>
wðx;
yÞ
>
>
>
>
>
>
2
>
>
>
>
>
>
1 x y x2 xy
>
>
>
< vwðx; yÞ >
= 6
6
vy > ¼ 6 0 0 1 0 x
>
4
>
>
>
>
>
>
>
>
>
>
0 1 0 2x y
>
vwðx; yÞ >
>
>
>
: vx >
;
y2
x3
x2 y
xy2
y3
2y
0
x2
2xy
3y2
0
3x2
2xy
y2
0
9
a1 >
>
>
>
>
>
>
>
a2 >
>
>
>
>
>
>
>
a3 >
>
>
>
>
>
>
>
>
>
a4 >
>
>
>
>
>
a5 >
>
>
>
>
>
>
>
>
a6 >
=
;
>
a7 >
>
>
>
>
>
>
>
>
a8 >
>
>
>
>
>
>
>
a9 >
>
>
>
>
>
>
>
>
a10 >
>
>
>
>
>
>
>
a11 >
>
>
>
>
>
>
;
a12
(2.100)
8
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
3
3
3 >
>
x y
xy
>
>
<
7>
7
x3
3xy2 7
5>
>
>
>
2
3
>
>
3x y y
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
:
where {w(x,y)}31 represents the generalized displacements (a vertical deflection
and two in-plane rotations) at any point (x,y) within the element.
99
100
CHAPTER 2 Meshing, Element Types, and Element Shape Functions
2.5.4.3 Identify 12 Constants for the Interpolation Polynomial
We begin to find the 12 constants a1 through a12 by selecting the origin of the coordinate system to coincide with P1 and letting P2 be located at a distance on the positive x-axis, as shown in Fig. 2.23. To determine these 12 constants, let us first select
the edge P1eP2. As we can see, all y-coordinates on this edge are zero. Hence,
Eq. (2.97) becomes
wðx; yÞ ¼ a1 þ a2 x þ a4 x2 þ a7 x3 .
(2.101)
We now consider a 4-node plate element as an analogue to a 2-node beam
element. For the 2-node, 1D beam element described in Section 2.5.1, each node
has two DOFs: vertical deflection w and rotation qy. For a 4-node, 2D plate element
located on the xey plane, each node allows three DOFs: vertical deflection along the
z-axis and two in-plane rotations, qy about the y-axis and qx about the x-axis.
Eq. (2.101) describes the vertical deflection on the P1eP2 edge of a 4-node plate;
hence, it is analogous to a beam element. Next, we take the derivative of the vertical
deflection w, shown in Eq. (2.101), with respect to x at the same edge P1eP2 (where
y ¼ 0).
vwðx; yÞ
¼ qy ¼ a2 þ 2a4 x þ 3a7 x2
(2.102)
vx
Eqs. (2.101) and (2.102) describe w and qy of a 4-node plate element on the edge of
y ¼ 0. These two equations are identical to the Hermite interpolation, shown in Eq.
(2.25), used to formulate a 2-node beam element. Thus, the four needed constants,
a1, a2, a4, and a7, can be solved as previously described, by applying the known
nodal deflections (w1 and w2) and the slopes for rotation about the y-axis (q1y and
q2y) to the interpolation function.
As a quick summary, we have used two vertical deflections (w1 and w2) and
two rotations about the y-axis (q1y and q2y) to find the four constants a1, a2, a4,
FIGURE 2.23
A plate of thickness h is represented by a 4-node surface element situated on the xey
plane. Each node allows three DOFs: the vertical deflection wi, rotation qix about the xaxis, and rotation qiy about the y-axis.
2.5 Element Shape Functions and [B] Matrix
and a7. Let us now consider the slope vw
vy , as shown in Eq. (2.98), at the same edge
(P1eP2), where y ¼ 0. This results in
vwðx; yÞ
¼ qx ¼ a3 þ a5 x þ a8 x2 þ a11 x3 .
vy
(2.103)
We have four unknowns in Eq. (2.103), but only two slopes for the rotation about
the x-axis (q1x and q2x) are left to be used. Because there are more unknowns
than available equations, the slopes about the x-axis rotation have a potential for
discontinuity. Therefore, we can conclude that the selection of the polynomial
terms described in Eq. (2.97) allows a deflection continuity along the interelement
boundaries, but not necessarily a continuous slope across the adjacent elements.
As such, this formulation method may not provide the most accurate solutions.
Despite this deficiency, it has been shown that plate elements formed from
this set of polynomial terms provide acceptable solutions in a great number of
cases.
From Fig. 2.23 we can see that a positive vertical deflection (w) will generate a
rotation about the “negative” y-axis. In other words, a positive w results in a counterclockwise rotation about the y-axis, which is opposite to the clockwise rotation
shown based on the right-hand rule to represent “positive” rotation. We do not
need to add any sign to the rotation about the x-axis because a positive w generates
a rotation in the same way as that designated by the right-hand rule. As such, a negavw ðx;yÞ
tive sign is usually added to the matrix of the vx portion (i.e., the third row) of
Eq. (2.100) to have matching sign conventions.
Note that Eq. (2.100) is a 312 matrix that can be used to find w, qx, and qy at any
point (x, y) within the plate element. Because we have four nodal points (x1, y1), (x2,
y2), (x3, y3), and (x4, y4), inserting these four sets of (xi, yi) values into Eq. (2.100)
yields a 1212 matrix. We then have 12 equations (1212 matrix) to solve for
12 unknowns, a1 through a12. We cannot identify a special set of nodal points that
allows us to express the 12 constants in an elegant way, such as that shown for
the shape functions of a 4-node rectangular element. Hence, sections below describe
a numerical method to calculate the 12 constants.
An alternative method for finding the 12 constants is to first solve the Hermite
interpolation to identify a1, a2, a4, and a7, and then reduce the matrix size to 88
before finding the remaining eight constants from eight equations. Once all constants are found, they can be inserted back into Eq. (2.97), where the terms are
arranged by nodal values {w1, q1x, q1y, w2, q2x,., q4y}, before the element shape
functions are determined.
Even with the number of equations being reduced to eight for finding the eight
unknowns, manual calculations would be quite lengthy for obtaining this set of
shape functions for a plate element. Use of a computer program is advised for
finding the solutions. Procedures needed to write this computer program are outlined below. After finding the 12 constants, the element shape functions can be
obtained.
101
102
CHAPTER 2 Meshing, Element Types, and Element Shape Functions
2.5.4.4 Find Shape Functions for a 4-Node Plate Element
Step 1:
vw
From Eq. (2.100), we can identify w; vw
vy ; and vx (a vertical deflection and two
rotations) as functions of 12 constants for each of the 4 nodes, because the nodal coordinates are known physical values. At node 1, the nodal coordinate is assumed to be
(x1, y1), and the corresponding generalized displacements are (w1, q1x, q1y). Copying
from Eq. (2.100), the top three rows of Eq. (2.104) can be written. As described earlier,
we added a negative sign to the third row of this matrix to have a consistent way to
enable the same direction of rotation due to the same sign of deflection. By repeating
the same procedures for node 2 (x2, y2), we have the next three equations for (w2, q2x,
q2y). Similarly, we obtain the six additional equations for (w3, q3x, q3y) and (w4, q4x,
q4y). Eq. (2.104) shows the combination of these 12 equations in matrix form. In other
words, the 312 matrix shown in Eq. (2.100) will become a 1212 matrix, and this
new matrix is assigned as the [G] matrix.
nodalfwg121
2
1
6
6
60
6
6
60
6
6
6
61
6
¼6
60
6
6
6
6:
6
6
6
6:
4
0
8
9
w1 >
>
>
>
>
>
>
>
>
>
>
>
>
>
q
>
1x >
>
>
>
>
>
>
>
>
>
>
>
>
>
>
q
1y
>
>
>
>
>
>
>
>
>
>
>
< w2 >
=
¼ ½G1212 fag121 0
>
>
>
>
>
> q2x >
>
>
>
>
>
>
>
>
>
>
>
:
>
>
>
>
>
>
>
>
>
>
>
>
>
>
:
>
>
>
>
>
>
>
>
>
>
:
;
q4y
x1
y1
x1 2
x1 y1
y1 2
x1 3
x1 2 y1
x1 y1 2
y1 3
x1 3 y1
0
1
0
x1
2y1
0
x1 2
2x1 y1
3y1 2
x1 3
1
0
2x1
y1
0
3x1 2
2x1 y1
y1 2
0
3x1 2 y1
x2
y2
x2 2
x2 y2
y2 2
x2 3
x2 2 y2
x2 y2 2
y2 3
x2 3 y2
0
1
0
x2
2y2
0
x2 2
2x2 y2
3y2 2
x2 3
:
:
:
:
:
:
:
:
:
:
:
:
:
:
:
:
:
:
:
:
1
0
2x4
y4
0
3x4 2
2x4 y4
y4 2
0
3x4 2 y4
38
9
a1 >
>
>
>
>
>
7>
>
>
>
>
>
27
>
a2 >
>
>
3x1 y1 7
>
>
7>
>
>
7>
>
>
>
>
3 7>
>
>
a
y1 7>
3 >
>
>
>
7>
>
>
>
>
>
>
3 7
=
< a4 >
x2 y2 7
7
7
>a >
>
3x2 y2 2 7
>
7>
>
> 5 >
>
7>
>
>
>
7>
>
>
>
>
7
:
: 7>
>
>
>
>
>
>
7>
>
>
>
7>
>
:
>
>
: 7>
>
>
>
>
5>
>
>
>
;
:
3
a12
y4
(2.104)
x1 y1 3
2.5 Element Shape Functions and [B] Matrix
Step 2:
We can easily find [G]1 numerically from Eq. (2.104) because (x1, y1) through
(x4, y4) are known. By multiplying [G]1 to both sides of Eq. (2.104), we find the 12
constants, a1 through a12, as shown in Eq. (2.105).
fag ¼ ½G1 f w1
q1x
q1y
.
q4y gT
(2.105)
Step 3:
We now write the generalized displacements w at any point (x, y) within the
element as a function of nodal displacements by inserting Eq. (2.105) into Eq.
(2.100), as shown in Eq. (2.106).
8
9
>
>
wðx;
yÞ
>
>
>
>
>
>
>
>
>
>
>
< vwðx; yÞ >
=
¼ ½F312 fag121
fwðx; yÞg ¼
> vy >
>
>
>
>
>
> vwðx; yÞ >
>
>
>
>
: vx >
;
¼ ½F312 ½G1 1212 f w1
q1x
q1y
.
q4y gT 121
(2.106)
By definition, the element shape functions are used to derive a physical value at
any point within the element; that is, fwðx; yÞg ¼ ½Nf w1 q1x q1y . q4y gT .
Thus, the shape functions for a 4-node plate element, with the interpolation function
selected as described above, is expressed as [N] ¼ [F][G]1.
2.5.4.5 Determine StraineDisplacement Matrix
As described for a beam element, the strainedisplacement relationship for a plate
element is equivalent to the curvatureedisplacement relationship. By recalling
how the curvature was derived for a beam element, we can deduce that the generalized curvatures for a plate element have the form
v2 w
¼ 2a4 þ 6a7 x þ 2a8 y þ 6a11 xy;
vx2
(2.107)
v2 w
¼ 2a6 þ 2a9 x þ 6a10 y þ 6a12 xy; and
vy2
(2.108)
2v2 w
¼ 2a5 þ 4a8x þ 4a9 y þ 6a11 x2 þ 6a12 y2 .
vxvy
(2.109)
kx ¼
ky ¼
kxy ¼
Note that in some textbooks, a negative sign is assigned to the curvatures, but for
consistency with the definition used in the formulation of the [B] matrix for the beam
element, the negative sign convention is not adopted here.
103
104
CHAPTER 2 Meshing, Element Types, and Element Shape Functions
Step 4:
Rewrite the generalized curvatures shown in Eqs. (2.107)e(2.109) in matrix
form as a function of the 12 constants, a1 through a12. This new matrix, [H], consists
of all the coefficients, as shown in Eq. (2.110).
8
9
>
< kx >
=
fkg ¼ ½Hfag0 ky
>
>
:
;
kxy
2
0
6
6
¼6 0
4
0
2
0
0
2
0
0
6x 2y
0
0
6xy
0
0
0
0
2
0
0
2x
6y
0
0
0
0
2
0
0
4x
4y
0
6x2
3
a1
6a 7
6 2 7
7
36
6 a3 7
0
6
7
76 a 7
76 4 7
6xy 76
7
56 : 7
6
7
7
6y2 6
6 : 7
6
7
4 : 5
a12
(2.110)
Step 5:
Insert {a} from Eq. (2.105) into Eq. (2.110), to obtain Eq. (2.111).
fkg ¼ ½Hfag ¼ ½H½G1 f w1
¼ ½Bf w1
q1x
q1y
q1x
q1y
. q4y gT
. q4y gT
(2.111)
Hence, the strainedisplacement or curvatureedisplacement matrix [B] for a 4node plate element can be written as [H][G]1.
2.5.5 3D, 4-NODE SHELL ELEMENT
A 4-node shell element is usually formed by the superposition of a 4-node plane stress
element and a 4-node plate element. Because a plane stress element has two DOFs per
node and a plate element has three DOFs per node, a shell element has a total of five
DOFs per node. The only DOF not accounted for in a shell element is the rotation
about the z-axis (i.e., in-plane rotation), also known as the drilling DOF.
This five-DOFs shell element is problematic, because a real-world structure usually involves connecting a shell to a beam or another shell not in the same reference
surface. Hence, it is more convenient to artificially make a shell element having a
total of six DOFs (three translational and three rotational DOFs). This treatment
equips a 4-node shell element with a total of 24 DOFs per element. So, the order
of the stiffness matrix for this augmented shell element is 24 24. Eq. (2.112)
2.5 Element Shape Functions and [B] Matrix
shows the results of superposition of a plane stress element, a plate element, and zero
stiffness for the four drilling DOFs.
3
2
0
0
½kplane 88
7
6
(2.112)
½k2424 ¼ 4
0
½kplate 1212
0
5
0
0
044
The shell element formulated as shown above contains no stiffness for in-plane
rotation, hence Eq. (2.112) is a singular matrix. To resolve this problem, a small
amount of artificial torsional stiffness is supplemented in some software packages.
The obvious question is how much artificial stiffness is small enough to ensure accuracy from the model.
Because there may be different polynomial terms selected for a plate element
during formulation, there are several plate elements that exist. Similarly, there are
numerous shell elements available. One such method is presented by Rengarajan
et al. (1995). However, discussion about different ways a shell element can be
formulated is beyond the scope of this book. Advanced readers may want to look
into scientific literature or study the theoretical manual provided by the software
vendor in order to better understand the pros and cons regarding how to select the
proper element type.
2.5.6 3D, 8-NODE TRILINEAR ELEMENT SHAPE FUNCTIONS
Fig. 2.24 shows an 8-node solid brick element with dimensions of 2a by 2b by 2c.
The origin of this brick element is located at its geometric center. Again, the
numbering sequence for this element is important. In this case, P1 is located at
the bottom layer, at coordinates (a, b, c), while P2, P3, and P4 are formed
through a counterclockwise rotation. For the top layer, P5 is located at coordinates
(a, b, c), and P6 through P8 are formed using the same counterclockwise
rotation.
Assume the interpolation function for this solid element has the form
4ðx; y; zÞ ¼ a1 þ a2 x þ a3 y þ a4 z þ a5 xy þ a6 yz þ a7 zx þ a8 xyz ¼
8
X
Ni 4i .
i¼1
(2.113)
This is a trilinear interpolation equation, because it varies linearly along the
x-axis if both the y and z values are held constant. Similarly, it behaves linearly along
the y-axis if the x and z values are held constant, and linearly along the z-axis if the x
and y values are held constant.
By now, we are familiar with a couple of methods used to identify the
element shape functions. Although it is quite tedious, we have enough information thus far to be able to derive this set of shape functions. As such, detailed
105
106
CHAPTER 2 Meshing, Element Types, and Element Shape Functions
FIGURE 2.24
An 8-node trilinear element of size 2a 2b 2c. The origin of the local coordinate
system is located at the geometric center. The nodes are arranged such that P1 through
P4 are at the bottom layer while P5 through P8 are at the top layer. In both layers, these
nodes are arranged in a counterclockwise manner.
derivations are not provided. Instead, only the resulting element shape functions
are listed below.
ða xÞðb yÞðc zÞ
ða þ xÞðb yÞðc zÞ
; N2 ¼
8abc
8abc
ða þ xÞðb þ yÞðc zÞ
ða xÞðb þ yÞðc zÞ
N3 ¼
; N4 ¼
8abc
8abc
(2.114)
ða xÞðb yÞðc þ zÞ
ða þ xÞðb yÞðc þ zÞ
N5 ¼
; N6 ¼
8abc
8abc
ða þ xÞðb þ yÞðc þ zÞ
ða xÞðb þ yÞðc þ zÞ
N7 ¼
; N8 ¼
8abc
8abc
We can use the same shape functions for all three translational DOFs and rewrite
Eq. (2.113) in matrix format as
8 9
u1 >
>
>
>
>
>
>
>
>
> v1 >
>
>
>
2
3
9
8
>
>
>
N1 0
0
N2 0
0
: : 0
> w1 >
>
uðx;
y;
zÞ
>
>
>
>
<
=
<
= 6
7
6
7
vðx; y; zÞ ¼ 4 0 N1 0
0 N2 0
: : 0 5
u2
.
>
>
>
>
>
>
:
;
>
>
wðx; y; zÞ
> : >
>
0
0 N1
0
0 N2 : : N8 324>
>
>
>
> : >
>
>
>
>
>
>
: >
;
w8 241
(2.115)
N1 ¼
2.5 Element Shape Functions and [B] Matrix
107
By inserting this equation into Eq. (1.11), the strainedisplacement equation, the
following 3D strainedisplacement equations can be established:
8
9 8
9
εxx > > εxx >
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
ε
ε
yy
yy
>
>
>
>
>
>
> >
>
>
>
>
>
>
>
< ε = < ε >
=
zz
zz
¼
>
>
gxy >
2εxy >
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
> 2εyz >
> gyz >
> >
>
>
>
>
>
>
>
>
>
>
:
:
; >
;
gzx
2εzx
2
3
v
0
0
6 vx
7
6
7
8 9
6
7
u1 >
6 0 v 0 7
>
>
vy
6
7
> >
>
>
6
7
>
>
v1 >
>
>
6
7 2
>
>
3
>
>
>
>
6
7
v
N
0
0
N
0
0
:
:
0
>
>
1
2
w
>
6 0 0 vz 7
1>
<
=
6
7 6
7
6
7
6
7
¼ 6 v v 0 7 4 0 N1 0
0 N2 0
: : 0 5
u2
.
> >
>
>
6 vy vx
7
>
>
: >
6
7
0
0 N1
0
0 N2 : : N8 324>
>
>
>
>
6
7
>
>
>
>
6 0 v v 7
>
>
:
>
>
6
7
vz vy
>
: >
;
6
7
w8 241
6
7
6
7
4 v
v 5
0 vx
vz
63
(2.116)
Finally, the strainedisplacement matrix [B] can be written as
3
2
v
0
0
7
6 vx
7
6
7
6
6 0 v 0 7
vy
7
6
7
6
7 2
6
7
6
v
N1 0
0
N2 0
0
: :
6 0 0 vz 7
7 6
6
7 6 0 N1 0
½B ¼ 6
0 N2 0
: :
v
v
7 4
6 vy
vx 0 7
6
7
6
0
0 N1
0
0 N2 : :
7
6
6 0 v v 7
6
vz vy 7
7
6
7
6
7
6
5
4 v
v
0 vx
vz
63
0
3
7
0 7
.
5
N8 324
(2.117)
108
CHAPTER 2 Meshing, Element Types, and Element Shape Functions
EXERCISES
1. Consider a 4-node surface element with P1 (0, 0), P2 (2, 0), P3 (2, 3.1), and P4
(0, 3). Find the element shape functions in this global coordinate system.
2. Consider a 4-node bilinear element having nodal coordinates P1 (0, 0), P2
(6, 0), P3 (6, 4), and P4 (0, 4). (i) Calculate the average shear strain after
deformation which causes the coordinates of P3 and P4 to become (6.1, 4.0)
and (0.1, 4.0), (ii) Assuming the origin of the coordinate system is located at
P1, show step-by-step derivations of the four element shape functions [N],
(iii) Determine
matrix
[B] based on the equation
2 the strainedisplacement
2
3
3
8 9
u1 >
v
v
>
>
>
9
6
6
>
>
07
0 78 P
4
>
8
9 6 vx
>
6 vx
>
7
7>
v1 >
>
>
>
>
>
>
>
6
6
7
7
N
u
i i >
>
>
>
>
6
<
<
< εxx >
=
= 6
7
7
v7 u
v7
1
u2 =
6
6
0
εyy ¼ 6 0
, and
¼ ½B
¼6
7
7 X
4
vy 7 v
vy 7>
>
>
>
>
6
: >
>
>
:
>
; 6
>
>
>
>
6
6
7
7>
gxy
N
v
>
:
>
;
i i
>
6v v7
6v v7
>
>
: >
>
>
4
4
5
5
1
>
: >
;
vy vx
vy vx
v4
(iv). Calculate the element strain from the [B] matrix when the coordinates
of P3 and P4 become (6.1, 4.0) and (0.1, 4.0).
3. Create a table of all the 1D and 2D elements, and some 3D elements with the
number of degrees of freedom, type of degrees of freedom, what kinds of
loads are resisted, and the stiffness matrix dimensions.
4. Find the shape functions for a 1D bar element with the two points P1 and P2
located at coordinates x ¼ 3 and x ¼ 5.
5. Find the strainedisplacement [B] matrix of the element in problem 4.
6. Find the shape functions of a beam element that has endpoints of x ¼ 3.5
and x ¼ 7.2.
7. Create graphs of the four shape functions of a beam element that has
endpoints of x ¼ 3.5 and x ¼ 7.2.
8. Find the [B] matrix of the beam element used in problems 6 and 7.
9. If a constant strain triangle element has points P1 (0, 0), P2 (4, 0), and P3
(2, 5), what are the shape functions of this element?
10. Find a polynomial equation (x) that passes through three points on the xey
plane with coordinates of (0.6, 12), (6.2, 3.9), and (10.1, 9).
References
11. A room 15 40 m in size has points P1 through P4 arranged in a
counterclockwise manner in the corners of the room. The air pressures at
these points are 100, 105, 108, and 98, respectively. Find and plot the
contour lines of 100, 103, and 106 kPa over the room area.
12. Use Lagrange interpolation to find the shape functions for a 3D, 8-node
trilinear element with dimensions 2a 2b 2c.
REFERENCES
Hallquist, J.O., 2006. LS-DYNA Theoretical Manual. ISBN:0-9778540-0-0, Available
through LSTC website: http://lstc.com/download/manuals.
Jeffreys, H., Jeffreys, B.S., 1988. Lagrange’s interpolation formula. x9.011. In: Methods of
Mathematical Physics, third ed. Cambridge University Press, Cambridge, England, p. 260.
Love, A.E.H., 1888. The small free vibrations and deformation of a thin elastic shell. Philosophical Transactions of the Royal Society of London A 179, 491e546.
Mindlin, R.D., 1951. Influence of rotatory inertia and shear on flexural motions of isotropic,
elastic plates. ASME Journal of Applied Mechanics 18, 31e38.
O’Connor, J.J., Robertson, E.F., 2017. MacTutor History of Mathematics Archive. URL:
http://www-history.mcs.st-andrews.ac.uk/index.html.
Reissner, E., 1945. The effect of transverse shear deformation on the bending of elastic plates.
ASME Journal of Applied Mechanics 12, A68eA77.
Rengarajan, G., Aminpour, M.A., Knight Jr., N.F., 1995. Improved assumed-stress hybrid
shell element with drilling degrees of freedom for linear stress, buckling and free vibration
analyses. International Journal for Numerical Methods in Engineering 38, 1917e1943.
Schneiders, R., 2017. URL: http://www.robertschneiders.de/meshgeneration/software.html.
Wolfram Math World, 2017. URL: http://mathworld.wolfram.com.
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CHAPTER
Isoparametric Formulation
and Mesh Quality
3
King H. Yang
Wayne State University, Detroit, Michigan, United States
3.1 INTRODUCTION
For an FE model to predict acceptable results, a high-quality mesh is necessary.
However, developing a high-quality mesh requires considerable resources, especially for structures with irregular shapes, such as the human body. On most occasions, the job of meshing is so time-consuming and tedious that very few
engineers or students prefer to do such assignments. Frequently, the leastexperienced engineers are assigned to do most of the mesh development. Because
an ill-formulated model will not provide acceptable results, the paradigm of using
the least-experienced people to develop FE models needs to be altered. For many
research projects, meshless approaches, such as the smoothed-particle hydrodynamics (SPH) and element-free Galerkin (EFG) methods, are used with an aim to
reduce the burden of developing the mesh. However, the technology is not yet
mature. Until then, high-quality mesh will be continuously needed for achieving
better and more accurate model predictions.
3.2 NATURAL COORDINATE SYSTEM
In 1968, Bruce Irons (1924eDec. 1983) and Olgierd (Olek) Zienkiewicz (May
1921eJan. 2009) presented the concept of isoparametric formulation at the Royal
Aeronautical Society meeting in London (Irons and Zienkiewicz, 1968). This idea
was accepted immediately and has been widely used in formulating element shape
functions [N], strainedisplacement matrices [B], and the element stiffness matrices
[k], among other things. The key to isoparametric formulation is the utilization of a
natural coordinate system or intrinsic coordinate system instead of the global coordinate system, as described previously in Section 2.5. Natural coordinates are
dimensionless and are defined with reference to the element rather than the global
coordinate system.
The term “isoparametric” indicates that a template geometry is used for all
elements of the same type to describe the nodal coordinates of the element, regardless
of the size of the element or where in space the element is located. Also, the same set
Basic Finite Element Method as Applied to Injury Biomechanics. http://dx.doi.org/10.1016/B978-0-12-809831-8.00003-9
Copyright © 2018 Elsevier Inc. All rights reserved.
111
112
CHAPTER 3 Isoparametric Formulation and Mesh Quality
of shape functions is used to define the element geometric shape as well as displacements or other DOFs. More importantly, this natural coordinate system accommodates elements that do not have a perfect geometry, such as those shown previously
in Section 2.5. For example, a 2D, 4-node surface element previously used to derive
the element shape functions has a perfect rectangular shape with all internal angles set
at 90 degrees. Any deviations from this geometry would have a different shape function from those presented. By using isoparametric formulations, it is much easier for
computer programmers to formulate the element stiffness matrices of varying element
geometries, because each and every element has identical natural coordinates. Using
the isoparametric approach, the need to formulate each element separately becomes
obsolete. An additional advantage in using the isoparametric formulation is that
curved edges can be used in the formulation. Since this is only applicable for highorder element types, details of this advantage are not covered in this book.
A natural coordinate system is a locally defined system such that any points within
an element are expressed by a set of dimensionless numbers that are within 1. For
the natural coordinate system that is analogous to the Cartesian coordinate system
based on x-, y-, and z-coordinates, symbols used in the natural coordinate system
are x (xi), h (eta), and z (zeta). There are also other natural coordinate systems,
such as the area coordinate system used for triangular elements and the volume
coordinate system used for tetrahedral elements.
As previously stated, regardless of the coordinate system on which the element is
based, the final input deck is always described in a global coordinate system. To use
isoparametric formulation effectively, a transformation function (also known as a
transfer mapping function) to map from a natural coordinate system to a global
coordinate system needs to be determined.
3.3 ISOPARAMETRIC FORMULATION OF 1D ELEMENTS
For the following sections, the geometries of each element type are presented in
x-, h-, and z-coordinates, unless otherwise specified. Each of these coordinate values
ranges from 1 to 1. By maintaining this value range, each element type is given an
identical shape, and this is why the process is called isoparametric formulation.
3.3.1 1D LINEAR BAR ELEMENT ISOPARAMETRIC SHAPE
FUNCTIONS
Fig. 3.1 shows a 2-node bar element defined in a global coordinate system and a
x-based natural coordinate system. The origin of this natural coordinate system is
positioned at the center of the element, such that point P1 has a coordinate value
of x ¼ 1 and P2 has a coordinate value of x ¼ 1. When mapping from the global
coordinate system to the natural coordinate system, the magnitudes corresponding
to the positive and negative sides of the element may have different lengths in the
natural coordinate system as compared to the global, as shown in this figure. The
advantage of having this arrangement may not be obvious, especially for a
3.3 Isoparametric Formulation of 1D Elements
FIGURE 3.1
A 2-node linear, 1D element expressed in a global coordinate system (top) and a natural
coordinate system (bottom).
2-node, 1D element. However, when one side of the element requires more details
than the other, this unequal layout between the left and right sides of the origin
may present certain advantages. For example, if the left side of a 3-node quartic
element requires a shorter length resolution than the right side, the center node
(i.e., the origin of the natural coordinate system) of the element needs to be set at
a point near the left-side node to afford a higher resolution.
As described in Section 2.5.1, a linear interpolation is used to define the 2-node
bar element. Using the same concept, the interpolation equation defined in a natural
coordinate system of a physical quantity 4 is written as follows:
4ðxÞ ¼ a0 þ a1 x ¼ N1 41 þ N2 42 .
(3.1)
Let 4 represent the displacement u(x). At x ¼ 1 and x ¼ 1, the corresponding nodal
displacements can be expressed as
u1 ¼ a0 a1
(3.2)
u2 ¼ a0 þ a1 .
(3.3)
and
Solving these two simultaneous equations yields
u1 þ u2
u2 u1
and a1 ¼
.
(3.4)
2
2
Inserting these two constants back into Eq. (3.1) and rearranging the terms yields
a0 ¼
u1 þ u2 u2 u1
1x
1þx
u1 þ
u2 .
þ
x¼
(3.5)
2
2
2
2
Comparing Eq. (3.1) to Eq. (3.5), the shape functions for this 2-node, 1D bar element
are
uðxÞ ¼
1x
1þx
and N2 ¼
.
(3.6)
2
2
We can graphically illustrate N1 and N2 by calculating the values for N1 and N2 from
x ¼ 1 (P1) to x ¼ 1 (P2), as shown in Fig. 3.2. From this figure, we can clearly see
that N1 and N2 fit the characteristics of a set of shape functions.
N1 ¼
113
114
CHAPTER 3 Isoparametric Formulation and Mesh Quality
FIGURE 3.2
The magnitude of N1 and N2 graphically displayed in x coordinates for an isoparametric
2-node element.
3.3.1.1 1D Transfer Mapping Functions and Interpolations
As mentioned earlier, there are several applications of the element shape functions.
One application is the mapping from the natural (e.g., x) to the global (e.g., x)
coordinate system, and vice versa. They can also be used for interpolating the
coordinates and displacements or identifying the strainedisplacement [B] matrix,
which is then used to determine displacements anywhere within the element from
nodal displacements.
We first illustrate how to use the element shape functions through interpolation to
find the coordinate or displacement anywhere within the element. Based on the characteristics of the 2-node element shape functions shown in Eq. (3.6), we can write
the following two interpolation equations:
x1
x1
1x
1þx
1x 1þx
¼
½
N
x1 þ
x2 (3.7)
N
¼
¼
fxg
1
2
2
2
x2
x2
2
2
and
fug ¼ ½ N1
N2 u1
u2
1x 1þx
¼
2
2
u1
u2
¼
1x
1þx
u1 þ
u2 .
2
2
(3.8)
Second, we show how to apply the element shape functions to identify the transfer mapping function. Eqs. (3.7) and (3.8) show that we can find x or u anywhere
within the element through interpolation of the nodal values, as long as the x location
is known. That is, Eq. (3.7) is the transfer mapping function to map from x to x-coordinate system. By rearranging Eq. (3.7), we have x ¼ 12 ½ðx1 þ x2 Þ þ ðx2 x1 Þx.
2xðx þx Þ
1
2
Hence, the transfer mapping function from x to x coordinate system is x ¼ x2 x
.
1
Third, we show how to use the natural coordinate systemebased element shape
functions to find the strain displacement matrix [B]. From Eq. (2.23), εxx is the axial
strain, or the change in displacement with respect to x εxx ¼ du
dx . In Section 2.5.1,
1 1
the strainedisplacement matrix ½B ¼
is obtained from shape functions
L L
derived from the global coordinate system, and we can use that information to
u1
du
.
find the strain εxx ¼ dx ¼ ½B
u2
3.3 Isoparametric Formulation of 1D Elements
du dx
In the current section, we apply the chain rule du
dx ¼ dx dx to find the straine
displacement matrix [B] that is based on the natural coordinate system. To apply
the chain rule, we use information obtained from Eqs. (3.7) and (3.8). Rearranging
2
1
Eq. (3.8) gives us fug ¼ u1 þu
þ u2 u
2
2 x, and taking the derivative results in
du ¼ u2 u1 . Rearranging Eq. (3.7) gives us fxg ¼ x1 þx2 þ x2 x1 x, and taking the
2
2
2
dx
derivative results in
dx
dx
1
¼ x2 x
2 . Thus,
du du dx u2 u1
2
u2 u1
¼
¼
.
(3.9)
¼
dx dx dx
x2 x1
2
L
u1
1 1
1
By expressing Eq. (3.9) in matrix form εxx ¼ 1
, the
L u1 þ L u2 ¼
u2
L L
strainedisplacement matrix [B] can be obtained as
εxx ¼
fεxx g ¼
1 1
L L
u1
u2
¼ ½B
u1
u2
.
(3.10)
It is of no surprise that Eq. (3.10) has the same form as Eq. (2.23). To obtain the
stress, we must apply the constitutive equations that are partially described in
Chapter 1 (see Section 1.2.3).
3.3.2 1D BEAM ELEMENT ISOPARAMETRIC SHAPE FUNCTIONS
As described in Section 2.5.1, a 2-node, 1D beam element has a total of four DOFs
(two vertical deflections w1 and w2, and two rotations, also known as slopes, q1 and
q2). This element type needs four shape functions and C1 continuity to fully describe
the behaviors (Fig. 3.3). To determine the shape functions, the previously described
Hermite interpolation can be used.
FIGURE 3.3
Top: A 2-node beam element described in a Cartesian coordinate system. This element
has a total of four DOFs, two deflections, and two rotations. Bottom: The nodal coordinates
of an isoparametric beam element described in the natural coordinate system always
range from 1 to 1.
115
116
CHAPTER 3 Isoparametric Formulation and Mesh Quality
One way to obtain the element shape functions in natural coordinates is to map
the shape functions determined in the global coordinate system to the natural coordinate system. With reference to Fig. 3.3, we need to map x1 ¼ 0 to x1 ¼ 1 and
x2 ¼ L to x2 ¼ 1. Despite that a beam element requires a C1 continuity, and hence
the use of Hermite interpolation, variations between the two coordinates (x1 and
x2) are still linearly related. We can insert x1 ¼ 0 and x2 ¼ L in Eq. (3.7) and find
the transfer mapping function as
L
(3.11)
x ¼ ð1 þ xÞ.
2
dx ¼ L along with the inverse dx ¼ 2 will be used
Eq. (3.11) and the derivative dx
dx
2
L
throughout the remainder of the book to convert between natural and global systems
for 2-node bar and beam elements. Now, by substituting this transfer mapping func
3
2
þ2x3 , we have
tion for x (i.e., Eq. 3.11) in Eq. (2.35) N1 ¼ L 3Lx
L3
2
3
L
L
3 ð1 þ xÞ
2 ð1 þ xÞ
L3 3Lx2 þ 2x3
2
2
¼
1
þ
N1 ¼
3
2
L
L
L3
3
3
1
2 3x þ x
.
(3.12)
¼ 1 ð1 þ xÞ2 þ ð1 þ xÞ3 ¼
4
4
4
This same method can be applied to find the rest of the shape functions described
in the natural coordinate system. This exercise shows that it does not matter which
coordinate system is used to derive the element shape functions, as long as we have a
transfer mapping function available for interchanging the shape functions between
the coordinate systems.
In this section, we use a different approach to promote greater appreciation of the
characteristics of the shape functions. As shown before, one of the characteristics of
shape functions is that when the shape functions are evaluated at the nth node, the
nth shape function yields a unity value (1), and all other shape functions result in
zero value. Using a 2-node beam element with a total of four DOFs as an example,
we repeat this procedure until each of the four DOFs is set to unity once. The
following four configurations, (a) through (d) demonstrate the four steps laid out
for the new method. Fig. 3.4 graphically displays these configurations.
(a)
(b)
(c)
(d)
N1
N2
N3
N4
can
can
can
can
be obtained
be obtained
be obtained
be obtained
by
by
by
by
setting
setting
setting
setting
w1 ¼ 1 and q1 ¼ w2 ¼ q2 ¼ 0.
q1 ¼ 1 and w1 ¼ w2 ¼ q2 ¼ 0.
w2 ¼ 1 and w1 ¼ q1 ¼ q2 ¼ 0.
q2 ¼ 1 and w1 ¼ w2 ¼ q1 ¼ 0.
We will step through an example of determining the shape functions for a 2-node
beam element. Since the coefficients are unique at each step, the letters that represent them progress from (a) to (d), just as the steps are labeled (a) to (d).
3.3 Isoparametric Formulation of 1D Elements
FIGURE 3.4
The shape functions for a 2-node beam element expressed in the natural coordinate
system. One of the key characteristics for a set of shape functions is that one of the four
DOFs is set to unity, and all other DOFs are zero.
Step (a):
Assume N1 has the form of N1 ¼ a1 þ a2x þ a3x2 þ a4x3. This equation must
satisfy the following two conditions:
1. at x ¼ 1, N1 ¼ w1 ¼ 1 and N1,x ¼ q1 ¼ 0, and
2. at x ¼ 1, N1 ¼ w2 ¼ 0 and N1,x ¼ q2 ¼ 0.
dx ¼ L and dx ¼ 2 . By substituting
From Eq. (3.11) we can easily determine that dx
dx
2
L
this derivative and the two conditions above into the assumed interpolation equation
for N1, we have
N1 ðx ¼ 1Þ ¼ 1 ¼ a1 a2 þ a3 a4
N1;x ðx ¼ 1Þ ¼ 0 ¼
dN1 dx
2
¼ ða2 2a3 þ 3a4 Þ
L
dx dx
N1 ðx ¼ 1Þ ¼ 0 ¼ a1 þ a2 þ a3 þ a4
N1;x ðx ¼ 1Þ ¼ 0 ¼ ða2 þ 2a3 þ 3a4 Þ
2
L
Solving these four simultaneous equations results in a1 ¼ 0.5, a2 ¼ 0.75, a3 ¼ 0,
and a4 ¼ 0.25. The L term drops out. Thus,
N1 ¼ 0:5 0:75x þ 0:25x3 ¼
2 3x þ x3 ð1 xÞ2 ð2 þ xÞ
.
¼
4
4
(3.13)
117
118
CHAPTER 3 Isoparametric Formulation and Mesh Quality
As expected, Eq. (3.13) is the same as Eq. (3.12), which was derived from the
transfer mapping.
Step (b):
As in Step (a), assume N2 has the form N2 ¼ b1 þ b2x þ b3x2 þ b4x3. This equation must satisfy two conditions:
1. at x ¼ 1, N2 ¼ w1 ¼ 0 and N2,x ¼ q1 ¼ 1, and
2. at x ¼ 1, N2 ¼ w2 ¼ 0 and N2,x ¼ q2 ¼ 0.
By applying the above two conditions and the derivative into the assumed equation for N2, we have
N2 ðx ¼ 1Þ ¼ 0 ¼ b1 b2 þ b3 b4
N2;x ðx ¼ 1Þ ¼ 1 ¼
dN2 dx
2
¼ ðb2 2b3 þ 3b4 Þ
L
dx dx
N2 ðx ¼ 1Þ ¼ 0 ¼ b1 þ b2 þ b3 þ b4
N2;x ðx ¼ 1Þ ¼ 0 ¼
dN2 dx
2
¼ ðb2 þ 2b3 þ 3b4 Þ
L
dx dx
Solving these four simultaneous equations results in b1 ¼ L8; b2 ¼ L8; b3 ¼ L8;
and b4 ¼ L8. Thus,
N2 ¼
Lð1 xÞ2 ð1 þ xÞ
L
.
1 x x2 þ x3 ¼
8
8
(3.14)
Step (c):
Similarly, assume N3 has the form N3 ¼ c1 þ c2x þ c3x2 þ c4x3. This equation
must satisfy two conditions:
1. at x ¼ 1, N3 ¼ w1 ¼ 0 and N3,x ¼ q1 ¼ 0, and
2. at x ¼ 1, N3 ¼ w2 ¼ 1 and N3,x ¼ q2 ¼ 0.
Once again, by applying the conditions into the assumed equation for N3, we
have
N3 ðx ¼ 1Þ ¼ 0 ¼ c1 c2 þ c3 c4
N3;x ðx ¼ 1Þ ¼ 0 ¼
dN3 dx
2
¼ ðc2 2c3 þ 3c4 Þ
L
dx dx
N3 ðx ¼ 1Þ ¼ 1 ¼ c1 þ c2 þ c3 þ c4
N3;x ðx ¼ 1Þ ¼ 0 ¼
dN3 dx
2
¼ ðc2 þ 2c3 þ 3c4 Þ
L
dx dx
Solving these four simultaneous equations results in c1 ¼ 0.5, c2 ¼ 0.75, c3 ¼ 0, and
c4 ¼ 0.25. Thus,
N3 ¼ 0:5 þ 0:75x 0:25x3 ¼
2 þ 3x x3 ð1 þ xÞ2 ð2 xÞ
.
¼
4
4
(3.15)
3.3 Isoparametric Formulation of 1D Elements
Step (d):
Similar to Step (b), assume N4 has the form N4 ¼ d1 þ d2x þ d3x2 þ d4x3. This
equation must satisfy two conditions:
1. at x ¼ 1, N4 ¼ w1 ¼ 0 and N4,x ¼ q1 ¼ 0, and
2. at x ¼ 1, N4 ¼ w2 ¼ 0 and N4,x ¼ q2 ¼ 1.
Finally, applying the conditions into the assumed equation for N4 yields
N4 ðx ¼ 1Þ ¼ 0 ¼ d1 d2 þ d3 d4
N4;x ðx ¼ 1Þ ¼ 0 ¼
dN4 dx
2
¼ ðd2 2d3 þ 3d4 Þ
L
dx dx
N4 ðx ¼ 1Þ ¼ 0 ¼ d1 þ d2 þ d3 þ d4
N4;x ðx ¼ 1Þ ¼ 1 ¼
dN4 dx
2
¼ ðd2 þ 2d3 þ 3d4 Þ
L
dx dx
Solving
these
four
simultaneous
equations
results
in:
L
L
L
L
d1 ¼ 8 ; d2 ¼ 8 ; d3 ¼ 8 ; and d4 ¼ 8. Thus,
L 1 x þ x2 þ x3
Lð1 þ xÞ2 ðx 1Þ
(3.16)
¼
N4 ¼
8
8
Eqs. (3.13)e(3.16) are the shape functions for the 2-node beam element derived
using the isoparametric formulation. They are collectively listed here for future
reference.
2 3x þ x3 ð1 xÞ2 ð2 þ xÞ
¼
4
4
2
3
L 1xx þx
Lð1 xÞ2 ð1 þ xÞ
N2 ¼
¼
8
8
N1 ¼
(3.17)
2 þ 3x x3 ð1 þ xÞ2 ð2 xÞ
N3 ¼
¼
4
4
2
3
L 1 x þ x þ x
Lð1 þ xÞ2 ðx 1Þ
N4 ¼
¼
8
8
The curvatureedisplacement matrix [B]14 in a beam element is equivalent to
d 2 ½N
the strainedisplacement in a bar element. Hence, we can derive it from dx2 i .
Only the derivation related to N1 is provided here. Again, when x1 ¼ 0, and
x2 ¼ L, we can easily identify the transfer mapping function from discussion of
dx
1 þx2 Þ
2
¼ 2x
Eq. (3.7) as x ¼ 2xðx
x2 x1
L 1. As before, the length ratio, dx ¼ L from Eq.
(3.11), remains unchanged regardless of which coordinate values are used.
119
120
CHAPTER 3 Isoparametric Formulation and Mesh Quality
2
To solve B11 ¼ ddxN21 , we start with N1 ¼ 14 2 3x þ x3 from Eq. (3.17). The
2
1
1
derivative is dN
dx ¼ 4 3 þ 3x . The next step is to take the derivative of N1 with
respect to x, which we execute by making use of the chain rule:
2 3 þ 3x2
dN1 dN1 dx 1 ¼ 3 þ 3x2 ¼
¼
.
L
dx
dx dx 4
2L
2
2
dx
d 3 þ 3x
d 3 þ 3x
Using the chain rule one more time, dx
¼ dx
dx, which results in
2L
2L
d
dx
3þ3x2
2L
B11
2
L
¼ 6x
L2 . To summarize,
0 1
3
d
2 3x þ x
d2 N1
d B
B 4
¼
¼
B
dx @
dx
dx2
¼
d 3 þ 3x2
dx
2L
1
2C
d 3 þ 3x2
C
C¼
LA dx
2L
dx 6x
¼ .
dx L2
(3.18)
As mentioned, the curvatureedisplacement matrix in a beam element is equivalent to the strainedisplacement in a bar element, and both matrices are collectively
known as the [B] matrix. Recall from Eq. (3.11), x ¼ L2 ð1 þ xÞ, we find the transfer
mapping function from x to x as x ¼ 2x
L 1. We can easily see that by making substitutions for x into Eq. (3.18) as follows, B11 of the curvatureedisplacement matrix
has the same form as previously shown in Eq. (2.43).
6x
6 2x
1
¼
L2 L2 L
6 12x
¼ 2þ 3
L
L
You are encouraged to work on your own to derive the rest of the [B] matrix in the
natural coordinate system from the global coordinate system using the transfer mapping function x ¼ L2 ð1 þ xÞ. In case you need more assistance before you have
enough confidence, please refer to the derivation in Exercise 1 of this chapter.
You can also practice your arithmetic skills by transferring from the [B] matrix in
the natural coordinate system, as seen in the top portion of Eq. (3.19), to the global
coordinate system by using x ¼ 2x
L 1. Here the [B] matrices based on both the natural and global coordinate systems are listed for you to check your work.
6x 3x 1 6x 3x þ 1
½BNatural
¼
14
L
L
L2
L2
(3.19)
6 12x
4 6x
6 12x
2 6x
Global
½B14 ¼ 2 þ 3 þ 2
3 þ 2
L L
L L
L
L
L2
L
3.4 Isoparametric Formulation of 2D Element
FIGURE 3.5
The global coordinates of the three vertices of the triangular plane element for Example
3.1.
3.4 ISOPARAMETRIC FORMULATION OF 2D ELEMENT
The simplest types of 2D elements are the 3-node triangular element and 4-node
plane element. The isoparametric formulation of these two element types is shown
in the following sections.
3.4.1 ISOPARAMETRIC FORMULATION OF 2D TRIANGULAR ELEMENT
As seen in Fig. 3.5, connecting a point P(x,y) within the triangle to the three vertices
results in three subareas A1, A2 and A3, where A1 is the area of the triangle directly
opposite to P1, A2 is the area of the triangle directly opposite to P2, and A3 is the area
of the triangle directly opposite to P3. The area coordinates are then defined as the
ratios of A1, A2 and A3 to the area of the entire triangle (Eq. 3.20). In other words,
P(x, y) in the Cartesian coordinate system is equivalent to P(x1, x2, x3) in the area
coordinate system. Coincidently, this set of area coordinates x1, x2 and x3 can also
be used as the natural coordinates of a triangle as well as the element shape
functions.
A1
A2
A3
; x2 ¼ N2 ¼ ; x3 ¼ N3 ¼
(3.20)
A
A
A
Note that the summation of these three natural coordinates is equal to unity (1),
that is, these functions satisfy the first characteristic of shape functions. If point P
coincides with P1, Eq. (3.20) yields x1 ¼ 1, x2 ¼ 0, x3 ¼ 0. Similarly, if point P
coincides with P2, we can see that x1 ¼ 0, x2 ¼ 1, x3 ¼ 0, and if point P coincides
with P3, then x1 ¼ 0, x2 ¼ 0, x3 ¼ 1. These calculations show that the three natural
coordinates also satisfy the second characteristic for shape functions.
x1 ¼ N 1 ¼
121
122
CHAPTER 3 Isoparametric Formulation and Mesh Quality
Example 3.1
Fig. 3.5 shows a triangle formed by P1(0, 0), P2(8, 2), and P3(4, 6) with a point
P(3, 2) located inside the triangle. Finding the area coordinates at point P within
the triangle numerically proves that the three area coordinates can be used as the
element shape functions.
Solution
The total area of a triangle can be calculated using the general equation
2
1 x1
6
2A ¼ det4 1 x2
1 x3
y1
3
2
1 x1
1 6
7
y2 50A ¼ det4 1 x2
2
y3
1 x3
y1
3
7
y2 5
y3
A, the area of the larger, outer triangle, can be calculated from the three sets of
nodal coordinates for P1, P2, and P3. Similarly, the three respective subareas A1,
A2, and A3 can be calculated by replacing the coordinates for P as the first node
along with the coordinates for the second and third nodes, using the following
equations:
3
2
3
1 x1 y1
1 0 0
1 6
7
7 1 6
A ¼ det4 1 x2 y2 5 ¼ det4 1 8 2 5 ¼ 20;
2
2
1 4 6
1 x3 y3
3
2
2
3
1 x y
1 3 2
1 6
7
7 1 6
A1 ¼ det4 1 x2 y2 5 ¼ det4 1 8 2 5 ¼ 10;
2
2
1 4 6
1 x3 y3
3
3
2
2
1 x y
1 3 2
1 6
1
7
7
6
A2 ¼ det4 1 x3 y3 5 ¼ det4 1 4 6 5 ¼ 5; and
2
2
1 x1 y1
1 0 0
3
2
2
3
1 x y
1 3 2
1 6
7
7 1 6
A3 ¼ det4 1 x1 y1 5 ¼ det4 1 0 0 5 ¼ 5:
2
2
1 8 2
1 x2 y2
2
(3.21)
(3.22)
(3.23)
(3.24)
Notice that the sum of the areas of the three interior triangles is equal to the area
of the large triangle: 10 þ 5 þ 5 ¼ 20.
Now each x value is the proportion of the outer triangle that is taken by the
represented inner triangle:
x1 ¼
10
5
5
¼ 0:5 x2 ¼
¼ 0:25 x3 ¼
¼ 0:25:
20
20
20
(3.25)
3.4 Isoparametric Formulation of 2D Element
We can deduce that the nodal coordinates P1, P2, and P3 can be used to calculate the coordinates at any point within the triangle. From Fig. 3.5, the x- and
y-coordinates of point P can be calculated from x1, x2, and x3 shown in Eq.
(3.25), resulting in
x ¼ x1 x1 þ x2 x2 þ x3 x3 ¼ 3 and y ¼ x1 y1 þ x2 y2 þ x3 y3 ¼ 2
(3.26)
The results of these calculations show that indeed the area coordinates can be
used to find the coordinate values of any location within the triangle; this is an
important application of the shape functions. You are encouraged to select other
locations within the triangle and show that all of them fit the same description. By
using the characteristics of shape functions, Eqs. (3.27) and (3.28) demonstrate
that nodal coordinates and displacements can both be directly determined from
the area coordinates and nodal values as
"
x1
x
¼
y
0
0
x2
0
x3
x1
0
x2
0
9
8
x1 >
>
>
>
>
>
>
>
>
>
>
y
1 >
>
>
>
>
>
#>
>
>
0 < x2 =
y2 >
x3 >
>
>
>
>
>
>
>
>
>x >
>
3 >
>
>
>
>
>
>
;
:
y3
(3.27)
8
9
u1 >
>
>
>
>
>
>
>
>
>
>
>
v
1
>
>
>
>
>
#>
>
>
<
0
u2 =
x3 >
> v2 >
>
>
>
>
>
>
>
>
>
>
u
3 >
>
>
>
>
>
>
:
;
v3
(3.28)
and
"
x1
u
¼
v
0
0
x2
0
x3
x1
0
x2
0
Note that a 3-node triangular element is a constant-strain triangle, and hence
there is no need to derive the strainedisplacement matrix [B].
3.4.2 ISOPARAMETRIC FORMULATION OF 2D BILINEAR ELEMENT
Using isoparametric formulation eliminates difficulties that would arise if the elements were defined in the global coordinate system with a shape that is not a rectangle. As described before, a 4-node plane stress element allows two DOFs per
node, a total of eight DOFs. Fig. 3.6 shows the mapping of a typical nonrectangular
quadrilateral surface element with nodes described in a global coordinate system
mapped into the natural coordinate system. In this particular instance, none of the
123
124
CHAPTER 3 Isoparametric Formulation and Mesh Quality
FIGURE 3.6
A typical quadrilateral element mapped into a square element on the x-h plane.
internal angles are right angles (90 degrees) for the element that is described in the
global coordinate system. We can imagine that the shape functions for this quadrilateral element would differ from the shape functions for a rectangular element,
which is described in Section 2.5.2 as having dimensions of 2a by 2b. Compared
to rectangular elements, quadrilateral elements require much greater computational
resources to calculate and store all the shape functions for each and every element in
the model. This problem does not exist in the isoparametric formulation using the
natural coordinates, because the transfer mapping effectively equips every element
with the same idealized geometry where x and h both range from 1 to 1 along the
horizontal and vertical axes, respectively.
In Fig. 3.6, we can see that regardless of the actual values of x and y in the global
coordinate system, the point in the lower left corner is mapped to P1 ¼ (1, 1).
Moving counterclockwise, the point in the lower right corner is mapped to
P2 ¼ (1, 1). The upper right point is mapped to P3 ¼ (1, 1) and the upper left to
P4 ¼ (1, 1).
By now, several methods have been used to derive the element shape functions
Ni. Intuitively we can see the associations between the shape functions and the
points; N1 is associated with P1, such that both x and h are subtracted, corresponding
to the negative values of each coordinate. However, you are encouraged to work on
your own to prove that the shape functions for a 4-node bilinear element in the
natural coordinate system have the forms
1
N1 ¼ ð1 xÞð1 hÞ;
4
1
N2 ¼ ð1 þ xÞð1 hÞ;
4
1
N3 ¼ ð1 þ xÞð1 þ hÞ; and
4
1
N4 ¼ ð1 xÞð1 þ hÞ.
4
(3.29)
(3.30)
(3.31)
(3.32)
3.4 Isoparametric Formulation of 2D Element
3.4.3 DETERMINE THE [B] MATRIX BASED ON ISOPARAMETRIC
FORMULATION
Based on the characteristics of the element shape functions, displacements at any
points within the element are the product of element shape functions (described in
natural coordinate system) and nodal displacements, as shown below.
8 9
u1 >
>
>
>
>
>v >
>
>
>
1
>
>
>
#>
>
>
" N
<
0
N2 0
N3 0
N4 0
1
uðx; hÞ
u2 =
(3.33)
¼
vðx; hÞ
: >
>
>
0 N1
0 N2
0 N3
0 N4 >
>
>
>
>
>
>
>
: >
>
>
>
: >
;
2
3
v4
v
6
0 7
8
9 6 vx
7
6
7 ε
> xx =
> 6
<
7 u
v
6
7
From Eq. (1.10), εyy ¼ 6 0
, we know that there is a need to
7
vy
>
>
6
7 v
:
; 6
7
gxy
6 v v 7
4
5
vy vx
vv vu
vv
first evaluate vu
vx ; vy; vy ; and vx before we determine the strainedisplacement matrix [B] in the natural coordinate system. Because both u and v are expressed in terms
vu
vv
vv
of x and h in Eq. (3.33), we need to first identify vu
vx; vh; vx ; and vh. For now, we
vu
vu
will concentrate on finding vx and vh through the application of the chain rule.
vu vu vx vu vy
vu vu vx vu vy
¼
þ
and
¼
þ
vx vx vx vy vx
vh vx vh vy vh
(3.34)
The two simultaneous equations listed in Eq. (3.34) can be rewritten, in matrix form,
as
8 9 2
38 9
vu >
vx vy >
vu >
>
>
>
>
> =
> >
> 6
>
<
=
7<
vx
vx
vx
6
7 vx
¼6
.
(3.35)
7
>
4 vx vy 5>
vu >
vu >
>
>
>
>
>
>
>
>
: ;
: ;
vy
vh
vh vh
2
3
vx vy
6 vx vx 7
6
7
Here the matrix on the right-hand side of Eq. (3.35), i.e., 6
7, is known as
4 vx vy 5
vh vh
the 2D Jacobian matrix [J]. By inspecting the Jacobian matrix, we can see that all
entries are related to the scaling factors between the x- and y-coordinates with
vu
respect to the x- and h- coordinates. The two unknowns vu
vx and vy in Eq. (3.35)
125
126
CHAPTER 3 Isoparametric Formulation and Mesh Quality
can be solved by using one of the many methods. We will use Cramer’s rule, as
described in Section 2.5.1, for solving these two unknowns as follows:
vu
vx
vu
vh
vu
¼
vx
vx
vx
vx
vh
vy
vx
vx
vx
vu
vx
vy
vh
vu
and
¼
vy
vy
vx
vx
vh
vx
vx
vu
vh
.
vy
vx
vy
vh
vx
vh
vy
vh
Using the same approach, we now replace u by v in Eq. (3.36) to find
vv
vx
vv
vh
vv
¼
vx
vx
vx
vx
vh
vy
vx
vx
vx
vv
vx
vy
vh
vv
and
¼
vy
vy
vx
vx
vh
vx
vx
vv
vh
.
vy
vx
vy
vh
vx
vh
vy
vh
(3.36)
vv
vx
and
vv
vy
as
(3.37)
The above listed four equations (Eqs. 3.36 and 3.37) involve a common denominator
vx vy
vx vx
, which is known as the determinant of the Jacobian matrix j½Jj. Knowing
vx vy
vh vh
a b
¼ ad bc, we rewrite the four equations shown
the determinant of a matrix
c d
in Eqs. (3.36) and (3.37) as
vu
1 vu vy vu vy
¼
;
vx j½Jj vx vh vh vx
(3.38)
vu
1 vu vx vu vx
¼
;
vy j½Jj vh vx vx vh
(3.39)
vv
1 vv vy vv vy
¼
; and
vx j½Jj vx vh vh vx
(3.40)
vv
1 vv vx vv vx
¼
.
vy j½Jj vh vx vx vh
(3.41)
3.4 Isoparametric Formulation of 2D Element
127
As a quick reminder, u and v are the horizontal and vertical displacements anywhere within the element, ui and vi are nodal displacements along the horizontal and
4
4
P
P
Ni ui and v ¼
Ni vi for a 4-node bilinear element.
vertical directions, and u ¼
i¼1
i¼1
With the four derivatives described in Eqs. (3.38)e(3.41), we write the strain vector
εxx εyy gxy T from Eq. (1.10) in a step-by-step fashion with appropriate notes
written at the bottom of the corresponding matrix as
2
3
2
3
vy v vy v
v
0
6
78
07
9
8
9 6
4
6 vh vx vx vh
7
P
6 vx
7
6
7>
ε
>
>
>
6
7( )
xx >
N i ui >
>
>
>
>
6
7
>
>
>
7
< i¼1
=
<
= 6
6
7>
6
v7 u
1
vx
v
vx
v
6
7
7
0
εyy ¼ 6
¼
0
6
7
6
vy 7
>
>
>
4
vx vh vh vx 7 >
j½Jj 6
X
>
>
>
6
7 v
>
>
>
>
6
7>
>
>
:
; 6
7
N
v
:
;
6
7
i
i
gxy
6
7
6 vx v
7
4v v5
vx v vy v vy v 5
i¼1
4
P
u¼
N i ui
vy vx
vx vh vh vx vh vx vx vh
P
Eq. 1:10
2
vy v vy v
6
6 vh vx vx vh
6
6
1 6
6
0
¼
6
j½Jj 6
6
6
6 vx v
vx v
4
vx vh vh vx
v¼
Eqs. 3:38e3:41
3
7
7
7
7
vx v
vx v 7
7
7
vx vh vh vx 7
7
7
vy v vy v 7
5
vh vx vx vh
0
"
N1
0
N2
: :
0
N1
0
: : N4
0
N i vi
8 9
u1 >
>
>
> >
>
>
>
>
>
>
>
>
v1 >
>
>
>
>
>
>
>
>
>
>
>
<u >
=
#
2
>
> : >
>
>
>
>
>
>
>
>
>
>
>
>
>
:
>
>
>
>
>
>
>
: >
;
v4 81
28>
>
32
u¼N1 u1 þN2 u2 þN3 u3 þN4 u4
v¼N1 v1 þN2 v2 þN3 v3 þN4 v4
(3.42)
Thus, the strainedisplacement matrix [B] can be expressed as
2
3
vy v vy v
0
6
7
6 vh vx vx vh
7
6
7 6
N1 0 N2
1 6
vx v
vx v 7
7
0
½B38 ¼
6
7
vx vh vh vx 7
j½Jj 6
0 N1 0
6
7
6 vx v
7
vx v vy v vy v 5
4
vx vh vh vx vh vx vx vh
: :
0
: :
N4
.
28
32
(3.43)
Explicitly writing the [B] matrix is quite tedious, and ideally a computer could be
used for such jobs. As such, Eq. (3.43) will not be expanded here. Rather, it will be
used as a homework assignment with the answers provided for you to check your
work.
.
128
CHAPTER 3 Isoparametric Formulation and Mesh Quality
3.5 ISOPARAMETRIC FORMULATION OF 3D ELEMENT
3.5.1 CONSTANT-STRAIN TETRAHEDRAL ELEMENT
The two simplest element types for 3D elements are the tetrahedral elements and
trilinear hexahedral solid elements. The natural coordinate system selected for the
3D tetrahedral element is based on the volume ratio, which is analogous to the area
ratio selected for the 2D triangular element. The tetrahedron is a useful element
type for filling in regions with very complex geometry without sacrificing the element
quality. Also, this element type can be automatically generated in most software packages. However, like the constant-strain triangle, poor performance can be associated
with tetrahedral elements. Thus, this element type is not frequently used for accurate
FEA.
Each node of a tetrahedral element has three translational DOFs (u, v and w), for
a total of 12 DOFs. The numbering scheme for the tetrahedron follows a counterclockwise direction at the triangular base (P1, P2, and P3). Additions of vertical
layers (which consist of only one point in this case) are added from bottom to
top, and the top node (P4) is at the vertex (Fig. 3.7).
To establish the natural coordinates, we drop a line from P4 to somewhere within
the tetrahedron, and call this point P. We can visualize from Fig. 3.7(left) that
PeP1eP2eP4 forms a subtetrahedron. Note that in this case, PeP1eP2 forms the
base, and as such, the points are considered to be on the same vertical layer, with
only P4 forming a new layer at the vertex. Similarly, we can identify three additional
tetrahedrons PeP2eP3eP4, PeP3eP1eP4, and PeP1eP2eP3. If we assume that
the total volume of the tetrahedron is V and the volume for each subtetrahedron is
Vi, where i ¼ 1 to 4, then volume ratios VVi ¼ xi are denoted as the natural coordinates
4
P
of the tetrahedron. Obviously,
xi ¼ 1.
i¼1
3.5.2 TRILINEAR HEXAHEDRAL ELEMENT
Fig. 3.7 (right) shows the mapping of a 3D hexahedral solid element in a global
coordinate system to a cubical element in the natural coordinate system. Because
FIGURE 3.7
Left: A tetrahedron which is divided into four subtetrahedrons, where the volume ratio is
used to define its natural coordinate system. Right: An 8-node, 3D hexahedral solid
element in a global coordinate system mapped onto a cubical element in the natural
coordinate system, and vice versa.
3.5 Isoparametric Formulation of 3D Element
129
it takes eight nodes to form a hexahedral element, we must choose an eight-term
polynomial equation to interpolate any physical quantity 4 within this element
from nodal values. Each node of such a solid element has three translational
DOFs. So each 3D hexahedral element has a total of 24 DOFs.
Eq. (3.44) shows the polynomial equation chosen to interpolate this element type.
There are multiple ways to derive the element shape functions, as demonstrated previously. You are encouraged to derive these shape functions using your preferred
method. Eq. (3.45) shows the shape functions for the hexahedral 8-node brick
element. The choice of using a positive or negative sign in the equation is dependent
upon whether there is a positive or negative value where the nodal point is located. For
example, P1 has natural coordinates of x ¼ 1, h ¼ 1, and z ¼ 1, and hence the
shape function is N1 ¼ 18 ð1 xÞð1 hÞð1 zÞ. Similarly, P6 has natural coordinates
of x ¼ 1, h ¼ 1, and z ¼ 1, and the shape function is N6 ¼ 18 ð1 þ xÞð1 hÞð1 þ zÞ.
4ðx; h; zÞ ¼ a1 þ a2 x þ a3 h þ a4 z þ a5 xh þ a6 hz þ a7 zx þ a8 xhz ¼
8
X
Ni 4i
i¼1
(3.44)
1
N1 ¼ ð1 xÞð1 hÞð1 zÞ
8
1
N2 ¼ ð1 þ xÞð1 hÞð1 zÞ
8
1
N3 ¼ ð1 þ xÞð1 þ hÞð1 zÞ
8
1
N4 ¼ ð1 xÞð1 þ hÞð1 zÞ
8
(3.45)
1
N5 ¼ ð1 xÞð1 hÞð1 þ zÞ
8
1
N6 ¼ ð1 þ xÞð1 hÞð1 þ zÞ
8
1
N7 ¼ ð1 þ xÞð1 þ hÞð1 þ zÞ
8
1
N8 ¼ ð1 xÞð1 þ hÞð1 þ zÞ
8
The displacement at any point within the element has the following form:
2
N1
9
8
>
6
=
< uðx; h; zÞ >
6
vðx; h; zÞ
¼6
6 0
>
>
;
:
4
wðx; h; zÞ 31
0
0
0
N2
0
0
:
:
N8
0
N1
0
0
N2
0
:
:
0
N8
0
N1
0
0
N2
:
:
0
0
8 9
u1 >
>
>
>
>
>
>
>
>
v1 >
>
>
>
>
3
>
>
>
>
>
>
w
1
>
>
0 7
>
> >
>
<
7
u2 =
0 7
.
7
>
: >
>
>
5
>
>
>
>
N8
>
> : >
>
>
>
>
324>
>
>
>
>
>
v
> 8>
>
>
: >
;
w8 241
(3.46)
130
CHAPTER 3 Isoparametric Formulation and Mesh Quality
The Jacobian
2
vx
6
6 vx
6
6 vx
6
½J ¼ 6
6 vh
6
6 vx
4
vz
matrix has the following form:
3 2 P
P
vð Ni xi Þ vð Ni yi Þ
vy vz
7 6
vx
vx
vx vx 7 6
7 6 P
P
6
7
vy vz 7 6 vð Ni xi Þ vð Ni yi Þ
7¼6
vh vh 7 6
vh
vh
7 6 P
P
6
7
vy vz 5 4 vð Ni xi Þ vð Ni yi Þ
vz vz
vz
vz
3
Ni zi Þ
7
7
vx
7
P
vð Ni zi Þ 7
7
7.
7
vh
7
P
7
vð Ni zi Þ 5
vz
vð
P
(3.47)
By combining Eqs. (1.11) and (3.46), the strainedisplacement equation can be
written as
2
8
εxx
>
>
>
>
>
>
>
>
εyy
>
>
>
>
>
>
>
< εzz
9
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
=
>
>
gxy
>
>
>
>
>
>
>
>
gyz
>
>
>
>
>
:
gzx
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
;
¼
8
εxx
>
>
>
>
>
>
>
>
εyy
>
>
>
>
>
>
>
< εzz
>
>
2εxy
>
>
>
>
>
>
>
>
2εyz
>
>
>
>
>
:
2εzx
6
6
6
6
6
6
9 6
6
>
6
>
>
6
>
>
6
>
>
6
>
>
6
>
>
>
6
>
>
6
>
= 6
6
¼6
>
6
>
>
6
>
>
6
>
> 6
>
>
6
>
>
6
>
>
6
>
>
; 6
6
6
6
6
6
6
6
4
3
v
vx
0
0
0
v
vy
0
0
0
v
vz
v v
vy vx
0
0
v
vz
v
vy
v
vz
0
v
vx
2
7
6
7
6
7
6
7
6
7
6
7
6
7
6
7
6
7
6
7
6
7
6
7
6
78 9 6
7> u > 6
7>
6
> >
>
= 6
7<
7
6
7 v ¼6
>
7>
6
> >
> 6
7>
7: w ; 6
7
6
7
6
7
6
7
6
7
6
7
6
7
6
7
6
7
6
7
6
7
6
7
6
5
4
3
v
vx
0 7
7
7
7
7
7
v
7
0 7
78 8
vy
9
7 P
7>
N i ui >
>
>
7>
>
>
>
> i¼1
>
>
v 7
>
7>
>
>
>
0
>
7>
>
8
<X
=
vz 7>
7
N
v
i i
0
7
>
7>
>
i¼1
>
v
>
>
>
>
0 7
>
7>
>
vx
>X
>
7>
>
> 8
>
7>
>
:
7>
Ni wi ;
7
i¼1
v v 7
7
7
vz vy 7
7
7
7
v 7
5
0
vx
0
0
0
v
vy
0
v
vz
Eq. 111
3
2
8
εxx
>
>
>
>
>
>
>
>
εyy
>
>
>
>
>
>
>
< εzz
>
>
gxy
>
>
>
>
>
>
>
>
gyz
>
>
>
>
>
:
gzx
6
6
6
6
6
6
9 6
6
>
6
>
>
6
>
>
6
>
>
6
>
>
6
>
>
>
6
>
>
6
>
= 6
6
¼6
>
6
>
>
6
>
>
6
>
>
6
>
>
6
>
>
6
>
>
6
>
>
; 6
6
6
6
6
6
6
6
4
v
vx
0
0
v
vy
0
0
v
vy
v
vx
0
v
vz
v
vz
0
7
7
7
0 7
7
7
7
7
0 7
7
7
7
7
v 7
vz 7
7
7
7
0 7
7
7
7
7
v 7
vy 7
7
7
7
7
v 7
7
vx 7
7
5
2
N1
6
6
6 0
6
4
0
0
0
N2
:
:
N8
0
N1
0
0
:
:
0
N8
0
N1
0
:
:
0
0
0
3
8 9
u1 >
>
>
> >
>
>
>
>
>
>
>
>
> v1 >
>
>
>
>
>
>
>
>
>
>
>
>
>
>
w
1>
>
>
>
>
>
> >
>
>
>
>
< u2 >
=
7
7
0 7
.
7
>
>
>
5
: >
>
>
>
>
> >
>
>
>
N8 324>
>
>
>
: >
>
>
>
>
>
>
>
>
>
>
>
>
>
>
v
>
>
8
>
>
>
>
>
>
>
: >
;
w8 241
63
(3.48)
3.6 Transfer Mapping Function for 2D Element
The strainedisplacement matrix [B]624 is the product of the first two terms of
the right-hand side of Eq. (3.48). Because the values Ni are expressed in x-h-z, the
chain rule needs to be applied to complete the partial differentiation.
3.6 TRANSFER MAPPING FUNCTION FOR 2D ELEMENT
The purpose of a transfer mapping function is to allow us to map an element from a
global coordinate system to a natural coordinate system, and vice versa. For
example, Fig. 3.6 in Section 3.4 shows the two-way transfer mapping of a 4-node
plane quadrilateral element. This function can be derived from the element shape
functions.
Eq. (3.7) allows us to calculate the corresponding natural coordinates from the
x-coordinates of P1 and P2. We can invert this equation to map from the natural coordinate (x) to the corresponding coordinate (x) in the global coordinate system.
Thus, the two equations shown in Eq. (3.49) are the transfer mapping functions
for the 2-node, 1D element.
x¼
2x ðx1 þ x2 Þ
1
5x ¼ ½ðx1 þ x2 Þ þ ðx2 x1 Þx
x2 x1
2
(3.49)
In Section 3.3.1.1, we discuss a special case where x1 ¼ 0 and x2 ¼ L. In this special case, the transfer mapping functions are x ¼ L2 ð1 þ xÞ and x ¼ 2x
L 1. To obtain
the transfer mapping functions for a 2D, 4-node plane element, recall that the
element shape functions can be used to find coordinates as well as horizontal and
vertical displacements for any location within the element. Thus, the coordinate
value for a point P(x, y) can be determined from nodal coordinates as shown below.
8 9
x1 >
>
>
>
>
>
>
>
>
y1 >
>
>
>
>
>
>x >
>
>
>
2
>
>
>
#>
>
>
" N
<
0
N2 0
N3 0
N4 0
1
x
y2 =
0
¼
y
x3 >
>
>
0 N1
0 N2
0 N3
0 N4 >
>
>
>
>
>
>
>
y3 >
>
>
>
> >
>
>
>
>
>
>
> x4 >
>
:
;
y4
1
x ¼ ½ð1 xÞð1 hÞx1 þ ð1 þ xÞð1 hÞx2 þ ð1 þ xÞð1 þ hÞx3
4
þ ð1 xÞð1 þ hÞx4 (3.50)
1
y ¼ ½ð1 xÞð1 hÞy1 þ ð1 þ xÞð1 hÞy2 þ ð1 þ xÞð1 þ hÞy3
4
þ ð1 xÞð1 þ hÞy4 (3.51)
131
132
CHAPTER 3 Isoparametric Formulation and Mesh Quality
Example 3.2
The input data deck lists a 4-node quadrilateral plane element with its nodal
values in the global coordinate system as P1(0, 0), P2(5, 0), P3(5.5, 3), and
P4(0, 3). (1) Determine the global coordinates for a point P(x, y) that corresponds
to x ¼ 0.6 and h ¼ 0.4 in the natural coordinate system. (2) Determine the natural
coordinates for a point that corresponds to x ¼ 4.28 and y ¼ 2.1 in the global
coordinate system.
Solution
Part (1): From Eqs. (3.50) and (3.51), we can find that:
1
x ¼ ½ð1 0:6Þð1 0:4Þðx1 ¼ 0Þ þ ð1 þ 0:6Þð1 0:4Þðx2 ¼ 5Þ
4
þ ð1 þ 0:6Þð1 þ 0:4Þðx3 ¼ 5Þ þ ð1 0:6Þð1 þ 0:4Þðx4 ¼ 0Þ ¼ 4:28 and
1
y ¼ ½ð1 0:6Þð1 0:4Þðy1 ¼ 0Þ þ ð1 þ 0:6Þð1 0:4Þðy2 ¼ 0Þ
4
þ ð1 þ 0:6Þð1 þ 0:4Þðy3 ¼ 3Þ þ ð1 0:6Þð1 þ 0:4Þðy4 ¼ 3Þ ¼ 2:1:
Alternatively, Eq. (3.50) can be expanded as
1
x ¼ ½x1 ð1 x h þ xhÞ þ x2 ð1 þ x h xhÞ þ x3 ð1 þ x þ h þ xhÞ
4
þ x4 ð1 x þ h xhÞ0
1
x ¼ ½ðx1 þ x2 þ x3 þ x4 Þ þ ð x1 þ x2 þ x3 x4 Þx þ ð x1 x2 þ x3 þ x4 Þh
4
þ ðx1 x2 þ x3 x4 Þxh.
(3.52)
Similarly, Eq. (3.51) can be expanded as
1
y ¼ ½ðy1 þ y2 þ y3 þ y4 Þ þ ð y1 þ y2 þ y3 y4 Þx þ ð y1 y2 þ y3 þ y4 Þh
4
þ ðy1 y2 þ y3 y4 Þxh.
(3.53)
Eqs. (3.52) and (3.53) with x ¼ 0.6 and h ¼ 0.4 also yield the same global coordinates (4.28, 2.1) as previously shown.
Part (2): To find the natural coordinates (x, h) from the global coordinates
(4.28, 2.1), we apply the known coordinates x1, x2, x3 and x4 into Eqs. (3.52)
and (3.53). This results in:
1
1
x ¼ 4:28 ¼ ½10:5 þ 10:5x þ 0:5h þ 0:5xh and y ¼ 2:1 ¼ ½6 þ 6h.
4
4
Solving these two simultaneous equations yields x ¼ 0.6 and h ¼ 0.4.
So, the transfer mapping functions for a 4-node, 2D plane element are
expressed in Eqs. (3.50) and (3.51) and in Eqs. (3.52) and (3.53). You are encouraged to derive the transfer mapping functions for the 8-node, 3D solid element
from the element shape functions on your own.
3.7 Jacobian Matrix and Determinant of Jacobian Matrix
3.7 JACOBIAN MATRIX AND DETERMINANT OF JACOBIAN
MATRIX
The Jacobian matrix [J] is named after the 19th century German mathematician Carl
Jacobi (Dec. 1804eFeb. 1851). One of the many applications for the Jacobian matrix is to transfer mapping from one coordinate system to another, such as the transformation from a Cartesian to natural coordinate system, spherical to Cartesian
coordinate system, polar to Cartesian coordinate system, and vice versa for each.
In the FE method, the most common forms of the Jacobian matrix used in 1D,
2D, and 3D transformation are listed below.
2
3
vx vy vz
6
7
3
2
6 vx vx vx 7
vx vy
6
7
6 vx vy vz 7
6 vx vx 7
vx
6
7
7
6
½J1D ¼
; ½J2D ¼ 6
7; ½J3D ¼ 6
7
6 vh vh vh 7
4 vx vy 5
vx
6
7
6 vx vy vz 7
vh vh
4
5
vz vz vz
Physically, the meaning of the 1D Jacobian matrix
dx
dx
can be described as
the ratio of the x-coordinate to x-coordinate. For the previously described 1D
bar element with x1 ¼ 0 and x2 ¼ L and Eq. (3.49) where x ¼ 12 ½ðx1 þ x2 Þþ
ðx2 x1 Þx, the Jacobian matrix is calculated as:
dx x1 x2 x2 x1 L
¼
þ ¼
¼
dx
2
2
2
2
This exercise demonstrates that the scaling factor for each unit of x value corresponds to the x value. Note that this scale factor of L2 remains constant as long as the
length of the element remains L. In other words, while (x1, x2) may be (3L, 4L) or
(L, 0), the Jacobian is always the same.
The determinant of Jacobian matrix is known as the Jacobian determinant j½Jj,
which is frequently referred to as “the Jacobian.” The diagonal entries of the Jacobian matrix are related to the scale factors between the two coordinates involved (x
vs. x and y vs. h), while the off-diagonal terms are related to how much skewness the
element shape represents. Additionally, the Jacobian represents the length, area, and
volume ratios for 1D, 2D, and 3D elements, respectively. The following examples
explain how we calculate the Jacobian based on a 2D element.
Example 3.3
A 2D, 4-node plane element has the global nodal coordinates of P1 (0, 0), P2
(6, 0), P3 (6, 4), and P4 (0, 4). Using isoparametric formulation, map this
element to a natural coordinate system, then find the Jacobian matrix [J] and
its determinant j½Jj.
133
134
CHAPTER 3 Isoparametric Formulation and Mesh Quality
Solution
The same shape functions listed in Eqs. (3.29)e(3.32) can be used to identify the
coordinates at any location within the element. By entering the four sets of nodal
coordinates [(0, 0), (6, 0), (6, 4), (0, 4)] into the calculations, the four entries of the
2D Jacobian matrix can be explicitly written as
vx
v
¼ ðN1 x1 þ N2 x2 þ N3 x3 þ N4 x4 Þ
vx vx
1
¼ ½ ð1 hÞx1 þ ð1 hÞx2 þ ð1 þ hÞx3 ð1 þ hÞx4 ¼ 3;
4
vy
v
¼ ðN1 y1 þ N2 y2 þ N3 y3 þ N4 y4 Þ
vx vx
1
¼ ½ ð1 hÞy1 þ ð1 hÞy2 þ ð1 þ hÞy3 ð1 þ hÞy4 ¼ 0;
4
vx
v
¼ ðN1 x1 þ N2 x2 þ N3 x3 þ N4 x4 Þ
vh vh
1
¼ ½ ð1 xÞx1 ð1 þ xÞx2 þ ð1 þ xÞx3 þ ð1 xÞx4 ¼ 0; and
4
vy
v
¼ ðN1 y1 þ N2 y2 þ N3 y3 þ N4 y4 Þ
vh vh
1
¼ ½ ð1 xÞy1 ð1 þ xÞy2 þ ð1 þ xÞy3 þ ð1 xÞy4 ¼ 2:
4
3 0
Thus the Jacobian matrix is ½J ¼
, and the determinant of the Jaco0 2
bian matrix is j½Jj ¼ 6.
Note that the entries of [J] shown here have physical meaning. J11 is the scale
factor from the coordinate x to x, J22 is the scale factor from the coordinate h to y,
and j½Jj is the ratio of the area measured in the x-y coordinate system to the area
measured in the x-h coordinate system. Because there are no off-diagonal terms,
this is a simple mapping from a rectangle in the global coordinate system to a
square in the natural coordinate system. Note here that the Jacobian matrix consists of no x or h terms when a rectangular-shaped element is mapped to the natural coordinate system. If the element in the global coordinate system is not a
rectangle, as shown in the next example, some x or h terms will not vanish.
Example 3.4
Repeat Example 3.3 with the coordinates of P3 changed to (6, 4.1).
Solution
Using the same procedures as in the previous example, we have:
vx
¼ 3;
vx
vy
vx
¼ 0:025ð1 þ hÞ;
¼ 0;
vx
vh
vy
¼ 2:025 þ 0:025x.
vh
3.7 Jacobian Matrix and Determinant of Jacobian Matrix
Hence, ½J ¼
3
0:025ð1 þ hÞ
0 2:025 þ 0:025x
and j½Jj ¼ 6:075 þ 0:075x.
We can see from the nonvanishing, off-diagonal term J12 that this mapping is
not rectangular to square mapping. The area of the trapezoid can be calculated as
24.3. As such, the area ratio is 6.075, which can be compared to the determinant
of the Jacobian matrix 6.075 þ 0.075x. Any extra terms involving x and h will be
considered during the selection of the integration points, which will be covered in
Section 4.5.
Example 3.5
Take the same four geometric points as in Example 3.3, but arrange them differently as P1(0, 0), P2(0, 4), P3(6, 4), and P4(6, 0) (see Fig. 3.8 left). In other words,
the four nodes in this example are arranged in a clockwise rotation, as compared
to the counterclockwise arrangements P1 (0, 0), P2 (6, 0), P3 (6, 4), and P4 (0, 4)
used in Example 3.3. Find the Jacobian matrix and Jacobian.
Solution
As before, by inserting the nodal coordinates into appropriate equations, we have
vx
v
¼ ðN1 x1 þ N2 x2 þ N3 x3 þ N4 x4 Þ
vx vx
1
¼ ½ ð1 hÞx1 þ ð1 hÞx2 þ ð1 þ hÞx3 ð1 þ hÞx4 ¼ 0;
4
vy
v
¼ ðN1 y1 þ N2 y2 þ N3 y3 þ N4 y4 Þ
vx vx
1
¼ ½ ð1 hÞy1 þ ð1 hÞy2 þ ð1 þ hÞy3 ð1 þ hÞy4 ¼ 2;
4
(A)
(0,4)
(6,4)
(0,0)
(6,0)
(B)
FIGURE 3.8
Left A 4-node element with its connectivity arranged in the clockwise direction. Right:
an ill-formed 4-node quadrilateral element.
135
136
CHAPTER 3 Isoparametric Formulation and Mesh Quality
vx
v
¼ ðN1 x1 þ N2 x2 þ N3 x3 þ N4 x4 Þ
vh vh
1
¼ ½ ð1 xÞx1 ð1 þ xÞx2 þ ð1 þ xÞx3 þ ð1 xÞx4 ¼ 3; and
4
vy
v
¼ ðN1 y1 þ N2 y2 þ N3 y3 þ N4 y4 Þ
vh vh
1
¼ ½ ð1 xÞy1 ð1 þ xÞy2 þ ð1 þ xÞy3 þ ð1 xÞy4 ¼ 0:
4
0 2
3 0
So the Jacobian matrix becomes
example illustrates that a
and the Jacobian is 6. This
negative Jacobian indicates that nodes are not arranged properly in accordance
with the way by which the element is formulated.
Example 3.6
Fig. 3.8 (right) shows a 4-node element with nodal coordinates as indicated. The
locations of P3 and P4 are reversed to represent an element with its nodal coordinates incorrectly entered into the input data deck. Although it is obvious
from this figure that this is not a standard quadrilateral element, a computer would
not be able to detect this visual effect, and the calculations would be made without
a flag being raised. Find the Jacobian matrix and Jacobian of this element.
Solution
As shown in the previous three examples, the four entries of the Jacobian matrix
are calculated first. Because a computer program does not have visual capability
to ascertain that the element has no similarity to a quadrilateral element, it will
treat the four nodal coordinates as provided. In other words, the four sets of nodal
coordinates as realized by the computer would be [(0, 0), (6, 0), (0, 4), and (6, 4)].
Hence,
vx
v
¼ ðN1 x1 þ N2 x2 þ N3 x3 þ N4 x4 Þ
vx vx
1
¼ ½ ð1 hÞð0Þ þ ð1 hÞð6Þ þ ð1 þ hÞð0Þ ð1 þ hÞð6Þ ¼ 3h
4
Similarly, we can find
matrix is
vy
vx
¼ 0;
vx
vh
vy
¼ 3x; and vh
¼ 2. Thus, the Jacobian
3h 0
½J ¼
3x 2
and j½Jj ¼ 6h.
This result shows that a wrongly ordered element yields a negative Jacobian.
Again, a negative Jacobian indicates that the numbering of nodes does not follow
the proper way the element is formulated.
3.8 Element Quality (Jacobian, Warpage, Aspect Ratio, etc.)
3.8 ELEMENT QUALITY (JACOBIAN, WARPAGE, ASPECT
RATIO, ETC.)
From the listing in Table 1.2, Section 1.4, we understand that an FE model, in
essence, is nothing but a list of nodes (consisting of node numbers and nodal coordinates), and element types (e.g., 1D, 2-node bar, or 2D, 4-node bilinear). The sequences of nodes are connected within an element (element connectivity, as
discussed in Examples 3.3e3.6), and there are material properties (Young’s
modulus, Poisson’s ratio, etc.) for the elements. From these pieces of information,
an FE solver (FE computer program) forms each element stiffness matrix individually and then assembles all element stiffness matrices into a global (also known as
structure) stiffness matrix. When the appropriate boundary and loading conditions
are added, the numerical FE solver calculates nodal displacements, which in turn
are used to determine the strains and stresses for all elements.
In Chapter 2, we described some element types, such as 1D line, 2D triangular, 2D
rectangular, 3D tetrahedral, and 3D brick. We also noted that each of these element
types has a specific geometric shape associated with it. For any elements with geometries not conforming to what is prescribed for these element types, it is too difficult
to identify the shape functions needed to interpolate physical values within the
element. With isoparametric formulation, we have the capabilities to formulate
element shape functions by mapping an element of any geometric shape to a standardized isoparametric element. For example, the quadrilateral element shown in
Example 3.4 can be mapped to an isoparametric, squared element. We notice from
this example that the Jacobian of a nonrectangular element would have x and/or h
terms, not seen in the Jacobians calculated for rectangular-shaped elements, to designate that the mapping is not a rectangular to square mapping. This minor detail points
to a fact that the similarity between an element in the global coordinate system and an
idealized element can be quantified by some indexes, such as the Jacobian.
In isoparametric formulation, element shape functions are formed based on
idealized shapes (e.g., an equilateral triangle for a triangular-shaped element and
a square for a 4-sided polygonal element). As such, an element that does not
conform to these idealized shapes is considered a less-than-perfect (could range
from acceptable to poor quality) element. Because real-world structures need not
be in the form of idealized geometry, the isoparametric formulation is adopted in
almost all modern FE solvers, mostly because of its flexibility in constructing
elements with nonidealized shapes. An additional advantage of using isoparametric
elements is that the mathematical equations derived from the shape functions,
element stiffness [k] matrix (to be discussed in Chapter 4), etc. can be directly
imported into FE code with minimal effort. On the other hand, the disadvantage
of using isoparametric elements is the high probability that some elements with
extremely poor quality will create inaccuracies or even errors, which is not desirable
in engineering analysis. For this reason, the overall quality of the mesh becomes an
important topic in the development of FE meshes. Quality must be checked and
reported along with the analysis of model predictions.
137
138
CHAPTER 3 Isoparametric Formulation and Mesh Quality
There are no universally accepted, minimally required quality standards for an
FE mesh. This is partly because there are multiple ways to judge the quality of an
element. Also, quality is a relative term, and there may be reasons to accept different
levels of quality for different problems. Lastly, quality alone cannot guarantee the
accuracy of an FEA, as we have emphasized several times that an FE model can
only provide acceptable, not exact, solutions.
In modeling structures with highly irregular shapes, such as the human body, it
may be adequate to do analysis with an FE model containing some elements with
lower quality. For example, if the ill-quality elements are located away from the
area of interest or the number of ill-quality elements is below a small percentage
of the total elements in the same component, then this approach may be acceptable.
This is an important concept, because trying to make sure every element is within a
preset limit is very time-consuming. Since we are not sacrificing too much on the
accuracy of the model predictions, an approach that allows some ill-quality elements
is acceptable. In the following sections, we describe several quantifiable measures
commonly used to judge the quality of an element.
3.8.1 JACOBIAN AND NORMALIZED JACOBIAN
In Section 3.7, some examples are used to demonstrate that a distorted element will
have off-diagonal terms in the Jacobian matrix, and hence a lower magnitude of the
Jacobian (determinant of the Jacobian matrix) as compared to that for an ideally
shaped element. This means that any distortion of the element shape, which may
introduce additional errors in the FE solution, will lower the Jacobian value. For
this reason, the Jacobian value could easily be used as an indicator for the quality
of the mesh.
In Section 3.6, we indicate that the Jacobian is the length ratio of a 1D element,
area ratio of a 2D element, and volume ratio of a 3D element. It follows from this
description that a larger element will have a larger Jacobian. For example, the Jacobians for a 2D, 24-by-24 squared element (J ¼ 144) and a 2D, 6-by-6 element (J ¼ 9)
are quite different. Yet, both elements conform perfectly to the idealized 2-by-2 isoparametric element. Because of this, use of the Jacobian value as a parameter for
checking element quality is inadequate. Instead, a normalized Jacobian (also known
as the Jacobian ratio) is provided in most FE preprocessing software packages. Here,
the normalized Jacobian is defined as the ratio between the smallest and largest values
of the determinant of the Jacobian matrix, evaluated either at the nodal locations or
integration points (to be introduced in Chapter 4). Based on this definition, the
normalized Jacobian ranges from 0 to 1, with 1 representing a perfect confirmation.
As long as the nodes forming the element are in correct order and the element is not
heavily distorted, the minimum Jacobian ratio should be greater than 0. Most FE
solvers will not run when there is a negative Jacobian element (e.g., Example 3.5
or 3.6). Other solvers, such as Abaqus (Simulia, Providence, RI), take it a step further
and will not run if the minimum Jacobian is below 0.2. A perfectly mapped element
3.8 Element Quality (Jacobian, Warpage, Aspect Ratio, etc.)
from the global to the natural coordinate system will have a Jacobian ratio of 1. Thus,
a Jacobian ratio that is closer to 1 indicates a better element quality. In general, it is
recommended that the minimal Jacobian ratio should be 0.6 or higher. Note that
different software packages may use different methods for calculating the Jacobian.
Despite only minute differences between different packages, it is still a good practice
to find out which method is adopted by the chosen software.
Example 3.7
The table below shows the Jacobian values calculated using HyperMesh 12.0
(Altair, Troy, MI) for twelve 4-node plane elements with similar geometry. Black
rectangles have been added to make the differences in the red elements easier to
see. Determine the method used by HyperMesh for the calculation of the
Jacobian.
Case No.
Nodal Coordinates
Jacobian
1
(0,0), (6,0), (6,4), (0,4)
1
2
(0,0), (6,0), (6,4.2), (0,4)
0.9722
3
(0,0), (6,0), (6,4.4), (0,4)
0.9465
4
(0,0), (6,0), (6,4.6), (0,4)
0.9226
5
(0,0), (6,0), (6,4.8), (0,4)
0.9003
6
(0,0), (6,0), (6,5), (0,4)
0.8794
7
(0,0), (5.8,0), (6,4), (0,4)
0.9806
8
9
(0,0), (6.2,0), (6,4), (0,4)
(0,0), (6.4,0), (6,4), (0,4)
0.9812
0.9634
10
(0,0), (6.6,0), (6,4), (0,4)
0.9465
11
(0,0), (6.8,0), (6,4), (0,4)
0.9303
12
(0,0), (7,0), (6,4), (0,4)
0.9150
Element Shape
Solution
From Examples 3.4 and 3.6, we observe that x and h terms could exist in the equation that represents the Jacobian of an element. In these cases, we need appropriate
values representing x and h to calculate the normalized Jacobian. As the concept
of Gauss integration has not yet been introduced, we will simply state that
x and h should be evaluated at four Gauss integration points located at
x; h ¼ p1ffiffi ¼ 0:5774 for a 2D, 4-node element. That is, the coordinates of the
3
139
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CHAPTER 3 Isoparametric Formulation and Mesh Quality
four (22) integration points are (x1 ¼ 0.5774, h1 ¼ 0.5774), (x2 ¼ 0.5774,
h2 ¼ 0.5774), (x3 ¼ 0.5774, h3 ¼ 0.5774), and (x4 ¼ 0.5774, h4 ¼ 0.5774).
Note that it is customary that Gauss points are organized in a clockwise manner.
Because Case 1 has the same nodal coordinates as shown in Example 3.3, the
3 0
four entries of the Jacobian matrix are ½J ¼
and the determinant of the
0 2
Jacobian matrix j½Jj ¼ 6. Because there are no x or h terms in the Jacobian matrix, it is clear that the Jacobian evaluated at any of the four Gauss integration
points is 6. Hence, the ratio of the maximum Jacobian (¼6) and minimum Jacobian (¼6) is 1. Note that the area of a 6-by-4 rectangle (¼24) is 6 times of an
idealized 2-by-2 square (¼4). If the Jacobian reported by HyperMesh is the
area ratio, then the Jacobian should be 6. Since HyperMesh states a Jacobian
of 1, we understand that it is the Jacobian ratio that is reported.
In Case 2, we will calculate the four entries of the Jacobian matrix before the
equation for the Jacobian is determined.
J11 ¼
vx 1
¼ ½ ð1 hÞx1 þ ð1 hÞx2 þ ð1 þ hÞx3 ð1 þ hÞx4 ¼ 3
vx 4
J12 ¼
vy 1
¼ ½ ð1 hÞy1 þ ð1 hÞy2 þ ð1 þ hÞy3 ð1 þ hÞy4 ¼ 0:05ð1 þ hÞ
vx 4
J21 ¼
vx 1
¼ ½ ð1 xÞx1 ð1 þ xÞx2 þ ð1 þ xÞx3 þ ð1 xÞx4 ¼ 0
vh 4
J22 ¼
vy 1
¼ ½ ð1 xÞy1 ð1 þ xÞy2 þ ð1 þ xÞy3 þ ð1 xÞy4 ¼ 2:05 þ 0:05x
vh 4
j½Jj ¼ J11 J22 J12 J21 ¼ 3 ð2:05 þ 0:05xÞ
It is clear that this equation involves only x terms. We know that Gauss points
3 and 4 have the same x value of 0.5774, while Gauss points 1 and 2 have the
same x value of 0.5774. For x ¼ 0:5774; j½Jj ¼ 6:23661 and for x ¼ 0:5774;
j½Jj ¼ 6:06339. Thus, the Jacobian ratio ¼ 6:06339
6:23661 ¼ 0:9722.
In Case 6, J11 ¼ 3, J12 ¼ 0.25(1þh), J21 ¼ 0, and J22 ¼ 2.25 þ 0.25x; and
j½Jj ¼ J11 J22 J12 J21 ¼ 3 ð2:25 þ 0:25xÞ.
Again, this equation involves only the x term. For x ¼ 0:5774; j½Jj ¼ 7:18305
and x ¼ 0:5774; j½Jj ¼ 6:31695. Thus, the Jacobian ratio ¼ 6:31695
7:18305 ¼ 0:8794.
In Case 12, J11 ¼ 3.25 0.25h, J12 ¼ 0, J21 ¼ 0.25(1 þ x), and J22 ¼ 2; and
j½Jj ¼ J11 J22 J12 J21 ¼ ð3:25 0:25hÞ 2:
For h ¼ 0:5774; j½Jj ¼ 6:2113 and h ¼ 0:5774; j½Jj ¼ 6:7887. Thus, the
Jacobian ratio ¼ 6:2113
6:7887 ¼ 0:9150.
Based on the 2 2 Gauss integration points, the four calculated Jacobian ratios (Cases 1, 2, 6, and 12) match those obtained using the HyperMesh software.
Hence, it is safe to assume that this software package uses Gauss integration
3.8 Element Quality (Jacobian, Warpage, Aspect Ratio, etc.)
points for calculating the Jacobian (ratio). You are encouraged to run through the
rest of the cases to test this hypothesis.
As triangular and tetrahedral elements are constant strain in nature, these
element types should be used sparsely. By definition, the Jacobian ratio for these
elements is always 1, and hence there is no need for checking it. Other parameters
for quality checks must be performed for these two element types.
3.8.2 INTERNAL AND SKEW ANGLES
The maximum and minimum internal angles are two parameters used to determine
how different the element is from the idealized element. Here, the internal angles
represent the internally measured angles formed by any two edges of the element.
Generally accepted internal angles range from 30 to 120 degrees for triangular
and tetrahedral elements. For quadrilateral and hexahedral elements, the internal angles are recommended to be within 45e135 degrees.
In a triangular element, the skew angle is related to the four angles formed by the
intersection of two lines, one by linking one vertex of the triangle to the midpoint of
its opposite side, and the other by linking the midpoints of the other two sides of the
triangle. This process is repeated for all three vertices, for a total of 12 angles. The
skew angle is defined as 90 degrees minus the minimum of the 12 angles (Fig. 3.9A).
For a quadrilateral element, we first find the minimum of the four angles formed by
the two lines joining the opposite midpoints of the element (Fig. 3.9B). Ninety degrees minus the minimum of the four angles defines the skew angle for the 4-node
element. For a 3D brick element, the skew angle is defined in the same fashion as a
4-node element, but using all six faces of the element. The generally accepted skew
angle is lower than 45 degrees.
(A)
(B)
FIGURE 3.9
(A) The four angles formed by connecting the line linked by the vertex P1 to the midpoint
of line P2eP3 and the line linked by the two midpoints on lines P1eP2 and P1eP3 of a
triangular element. This process is repeated for vertices P2 and P3, until all 12 angles are
found. (B) The four angles formed by connecting the two midpoints of opposite edges of a
4-node quadrilateral element.
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CHAPTER 3 Isoparametric Formulation and Mesh Quality
3.8.3 WARPAGE
Warpage is a measure of the maximum deviation angle of any of the faces of an
element relative to a single plane. From geometry classes, we know that any three
distinct points in 3D space determine a plane, as long as they do not lie in a straight
line. For a 4-node surface element lying on one plane, the warpage is zero. In the
case that the fourth node does not lie in the same plane as the other three nodes,
the element is said to have warpage. In this case, the first warpage angle measures
the angle formed by two adjacent triangles cut out of a 4-node element via the
dash line P2eP4 (Fig. 3.10). The angle a, formed by P3eP4eP2 plane with respect
to the P1eP2eP4 plane, is one of the warpage angles of the element. The same
method is used to find the second warpage angle with the exception that the dashed
line is drawn from P1 to P3. The angle measured between the P4eP1eP3 plane and
P1eP2eP3 plane is the second warpage angle. The maximum of the two warpage
angles is designated as the warpage of the 4-node surface element. The same method
is used to determine the warpage of a hexahedral element for each of the six surfaces. The generally accepted warpage angle should be less than 15 degrees.
3.8.4 ASPECT RATIO
The aspect ratio of an element is calculated by dividing the longest edge by the
shortest edge of the element. The generally accepted aspect ratio is less than 3 for
95% or more of all the elements in the same structural component, and less than
5 for all elements.
3.8.5 DISTORTION
Distortion is defined as the product of the minimum Jacobian and the area measured
in the natural coordinate system (equal to 4 for a 4-node surface element) divided by
the area measured in the global coordinate system, as shown in Eq. (3.54). As shown
in Examples 3.4 and 3.6, an element that is not a rectangular shape has some x and/or
h components. Therefore, the calculated Jacobian ratio will depend on which Gauss
FIGURE 3.10
The warpage angle for a 4-node plane element is defined as the angle between the planes
formed by two adjacent triangles (P1eP2eP4 and P2eP3eP4) of the quadrilateral
element. The maximum of the two angles is defined as the warpage of the element.
3.8 Element Quality (Jacobian, Warpage, Aspect Ratio, etc.)
integration point is chosen. Assuming the area of a 4-node plane element measured
in the global coordinate system is 25 and the minimum Jacobian is 5.6, the distortion
is calculated to be 0.896. The ideal value of distortion is 1 (i.e., no distortion), but
any value over 0.6 is considered acceptable.
Distortion ¼
minimum determinant of the Jacobian matrix 4
area measured in the global coordinate system
(3.54)
3.8.6 STRETCH
Stretch is in some ways related to the aspect ratio, but deals with internal length measurements. Examples of internal length include the radius of a circle, which is
tangentially fitted into a triangular element, and the maximum diagonal length of
a quadrilateral element (Fig. 3.11). For a triangular element, the stretch is defined
pffiffi
pffiffiffiffi
2
as RLmax12, while for a quadrilateral element it is defined as Lmin
Dmax . The ideal stretch
value is 1, but anything larger than 0.3 is considered acceptable.
3.8.7 GENERATION OF HIGH-QUALITY MESH
As mentioned in Chapter 1, many software packages are available to generate FE
meshes. These packages allow users to automatically generate constant-strain triangular or tetrahedral elements. However, these element types are less efficient in
achieving convergence, and hence they are not the preferred choice when developing
FE models. High-quality quadrilateral and hexahedral elements, on the other hand,
are more difficult to generate, but these element types tend to produce better results.
No matter what types of elements are chosen, one common problem is that
changing the typical mesh size to afford a convergence analysis is not a trivial
job. In light of this difficulty, many laboratories and industrial users prefer the use
of a multiblock approach. Instead of developing the mesh directly from the computer
aided design (CAD) files, engineering drawings, or segmented human anatomy, an
intermediate step called “blocking” is added before the mesh is constructed. Briefly,
(A)
(B)
FIGURE 3.11
(A) The radius of a circle that tangentially fits into a triangular element. (B) The maximum
diagonal length of a quadrilateral element.
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CHAPTER 3 Isoparametric Formulation and Mesh Quality
FIGURE 3.12
A midsagittal view of the blocks used in a multiblock approach to generate high-quality
hexahedral elements for a human head model.
Reused from Mao, H., Gao, H., Cao, L., Genthikattia, V.V., Yang, K.H., 2011. Development of high-quality
hexahedral human brain meshes using feature-based multiblock approach. Computer Methods in Biomechanics
and Biomedical Engineering, 1e9 with permission.
a general set of blocks is used to roughly cover the entire structure of interest. As an
example, Fig. 3.12 shows the blocks used by Mao et al. (2011) when constructing
their human head model. These blocks explicitly represent several major brain components, such as the cerebrum, cerebellum, and brainstem. Lastly, the vertices of the
block structures are translated and rotated to match the brain geometry segmented
from medical images.
Further adjustment of the blocks is required so that all blocks are of high quality
and confirm to the anatomical features. After that a split function in multiblock software can easily be applied to mesh the entire structure with a good quality mesh of
any selected element sizes. Upon completion of the mesh generation, a quality check
usually identifies some elements of lesser quality that require further attention.
There are some preprocessing mesh generation packages that provide optimization
tools to automatically improve the quality of the mesh. While such tools reduce the
manpower needed for the quality improvement, none are totally satisfactory when
dealing with complex geometry, such as the human body. In such cases, manual
adjustment is frequently needed. This kind of task requires years of experience,
and sometimes it is more of an art than science. Unfortunately, there are not enough
experienced people who want to develop mesh for a career. As such, this task is
generally given to those who are new to the field. This paradigm needs to be altered
so that only high-quality mesh will be used for the FEA.
The multiblock approach could significantly increase the initial efforts needed to
develop the first mesh. However, when the mesh density needs to be increased, for
3.9 Saint-Venant Principle and Patch Test
FIGURE 3.13
The CAD file (top left) divided into blocks (top right) before medium and high density FE
meshes are developed to simulate the femoral component of a total knee replacement.
Picture courtesy of Professor Nicole M. Grosland, Center for Computer Aided Design, The University of Iowa.
reasons such as accuracy improvement, constructing a higher density mesh becomes
almost effortless. In Fig. 3.13, the bottom two images show two different density
meshes created from multiple blocks of the femoral component of a total knee
replacement.
The multiblock method is also extremely useful for constructing subject-specific
FE human models that account for geometric variations among different subjects.
Although detailed descriptions of multiblock meshing are beyond the scope of
this chapter, you are encouraged to look into descriptions of block meshing provided
by software vendors. Three such vendors are ICEM CFD (ANSYS, Canonsburg,
PA), HyperMesh (Altair, Troy, MI), and TrueGrid (XYZ Scientific Applications,
Inc., Houston, TX). All have been successfully adopted in our laboratory for a number of years. Fig. 3.14 shows a completed FE model that was constructed using the
multiblock approach. This model represents a 10-year-old child in a seated position.
This particular model can be obtained free of charge for use at any academic institution from: https://automotivesafety.wayne.edu/models.
3.9 SAINT-VENANT PRINCIPLE AND PATCH TEST
The Saint-Venant principle states that the effects of loading with the same magnitude
but different distributions dissipate quickly as distance increases. In other words,
145
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CHAPTER 3 Isoparametric Formulation and Mesh Quality
FIGURE 3.14
A seated 10-year-old version of the Collaborative Human Advanced Research Models
(CHARM) generated using a multiblock approach.
as the distance from the point of loading becomes greater, the local effects are
reduced such that they can be considered not to be present. This principle is named
after the French engineer Adhémar Jean Claude Barré de Saint-Venant (Aug.
1797eJan. 1886). As an example of this principle, a point load and a distributed
load of the same total force magnitude will produce different displacements if these
displacements are evaluated at the region near the point of loading, but there will be
no difference between the effects for the two loading conditions if the displacements
are evaluated at a faraway region. Using this principle, the region at a distance from
the point load, which is applied at the left-central location as demonstrated in
Fig. 2.6, exhibits a uniformly distributed response, regardless of which element
type is selected.
The Patch test is a method that makes use of the Saint-Venant principle in the FE
method. Regardless of the element type or the quality of the element, responses
generated by an FE model should display a uniformly distributed pattern where
the responses are evaluated at a distance from the applied load. According to Professor Olgierd (Olek) Cecil Zienkiewicz, the “Patch Test” is one of the many terms,
such as “Isoparametric” and “Serendipity” introduced by Professor Bruce Irons
(Zienkiewicz, 1984). To illustrate a patch test, let us mesh a homogeneous rectangular plate with perfect-quality 4-node plane stress elements (Fig. 3.15). We then
3.9 Saint-Venant Principle and Patch Test
FIGURE 3.15
A patch consisting of triangular and 4-sided polygon elements (right) is used to replace
part of the FE model made of perfectly rectangular elements (left) for conducting a patch
test.
replace some of the elements in the mid-region of the plane with a patch, which is a
combination of several different element types, such as significantly distorted
4-sided polygonal elements and triangular elements with the same material properties. If no differences in the model-predicted responses are observed between the
meshes with or without the patch, then the patch test passes. Otherwise, we need
to replace the poor-quality elements with better ones, and then redo the test. In doing
so, we can also establish the minimum quality of the mesh. If improving the mesh
quality still produces error, we need to check the fidelity of the FE code, if it was not
done prior to using the software.
It is important to note that with modern technologies made available by highpower computers, the patch test is not commonly needed. As the typical element
size becomes smaller, more FE models are created with many of the simplest types
of elements. Increasingly, tools are available for model developers to check the quality of the FE mesh, and more often the patch test is used only for cases where no
more than a handful of elements are used to represent the structure of interest.
EXERCISES
1. Prove that the shape functions for a 2-node beam element derived from the
natural coordinates are the same as those derived from the global coordinate
system.
Solutions: The transfer mapping function is x ¼ 2x
L 1, and the [B] matrix in
the natural
6x 3x 1 6x 3x þ 1
Natural
coordinate system is ½B14 ¼
. Thus,
L
L
L2
L2
B11
B12
6x
¼ 2¼
L
6
3x 1
¼
¼
L
3
2x
1
L
2
L
6 12x
¼ 2þ 3
L
L
2x
1 1
4 6x
L
¼ þ 2
L
L L
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CHAPTER 3 Isoparametric Formulation and Mesh Quality
B13
B14
6x
¼ 2 ¼
L
3x þ 1
¼
¼
L
6
3
2x
1
L
2
L
¼
6 12x
3
L2
L
2x
1 þ1
2 6x
L
¼ þ 2
L
L L
2. Prove that the curvatureedisplacement [B] matrix for a 2-node beam element
derived from the natural coordinates are the same as those derived from the
global coordinate system.
3. Show step-by-step derivation of the element shape functions for a 4-node
bilinear element based on the natural coordinate system.
4. Identify some 4-node plane elements that possess a Jacobian ratio of 0.6 or
less, and graphically show their element shapes.
5. Compute the Jacobian matrix for a 4-node quadrilateral element in the natural
coordinate system with nodal coordinates P1(0, 0), P2(2, 0), P3(3, 2), P4(0,
1.5).
6. (1) Write transfer functions for both the natural and global coordinate systems
for a 2D quad element with corners at (2, 1), (5, 0), (4, 3), and (1, 4).
(2) Determine the global coordinates for a point P that corresponds to
x ¼ 0.6 and h ¼ 0.4 in the natural coordinate system. (3) Determine the
natural coordinates for a point Q that corresponds to x ¼ 2.67 and y ¼ 1.05
in the global coordinate system.
7. Expand the [B] matrix shown in Eq. (3.43) into one equation for a
quadrilateral element.
8. Show that all off-diagonal terms are zero in a Jacobian matrix when mapping
a rectangle of length 2a (parallel to the x-axis) and width 2b (parallel to the
y-axis) from a global coordinate system to a natural coordinate system.
9. Write transfer functions for both the natural and global coordinate systems for
a bar element with endpoints at 2 and 7.
References
10. Find the Jacobian matrix of the element in Problem 6 and the determinant of
that matrix.
11. Use Microsoft Excel or other software to calculate the Jacobians for all of the
cases in Example 3.7. Add the following elements to the list.
Case No.
P1
P2
P3
P4
13
14
(0, 0)
(0, 0)
(6, 0)
(10, 0)
(6, 6)
(6, 4)
(0, 4)
(0, 4)
12. Use isoparametric shape functions to solve the bar element in Problem 9.
Point 1 is constrained in the x direction and a force of þ100 N force is
applied to Point 2. The elastic modulus is 70 GPa and the cross-sectional
area of the bar is 0.05 m2.
REFERENCES
Irons, B.M., Zienkiewicz, O.C., 1968. The isoparametric finite element system e a new
concept in finite element analysis. In: Proc. Conf. Recent Advances in Stress Analysis,
Royal Aeronautical Society, London.
Mao, H., Gao, H., Cao, L., Genthikattia, V.V., Yang, K.H., 2011. Development of high-quality
hexahedral human brain meshes using feature-based multi-block approach. Computer
Methods in Biomechanics and Biomedical Engineering 1e9.
Zienkiewicz, Olgierd C., 1984. Obituary: Professor Bruce Irons. International Journal for
Numerical Methods in Engineering 20, 1167e1168. http://dx.doi.org/10.1002/
nme.1620200615.
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CHAPTER
Element Stiffness Matrix
4
King H. Yang
Wayne State University, Detroit, Michigan, United States
4.1 INTRODUCTION
Only three FE analysis types are covered in this book: static, modal, and transient
dynamic. The first step for all three analyses involves the creation of an element stiffness matrix [k] for every element. To formulate a [k] matrix, we introduce strong and
weak formulations. Once created, these element stiffness matrices are assembled
into a global (or so-called structure) stiffness matrix [K ] in static analysis. Then
the loading conditions, expressed in terms of nodal load vectors {f}, are calculated
from element shape functions. Finally, the forceedisplacement equations, arranged
in matrix form as shown in Eq. (4.1) for a static analysis problem, are numerically
solved to obtain nodal displacements by using a computer and software with
methods such as Gauss elimination.
½Kfug ¼ ff g;
(4.1)
where {u} is the set of nodal displacement vectors f u v w qx qy qz gT ,
with u, v, and w corresponding to the three translations (nodal displacements) along
the x-, y-, and z-axes, respectively, and qx, qy, and qz corresponding to the three nodal
rotation angles about the x-, y-, and z-axes.
The term “element stiffness matrix” describes the relationships of actions (such
as applied force) and responses in terms of the degrees-of-freedom (DOFs) for the
set of nodes that form the element. By the late 1950s, the key concepts of individual
element stiffness matrices and assembly of these matrices to form a global stiffness
matrix already existed (Turner, 1959), and these same concepts are still in use today.
In Chapter 1, we describe the element stiffness matrix in the context of the matrix
structural analysis (MSA) method. This same concept also applies to the FE method.
There are three common ways to formulate the element stiffness matrix:
•
•
Direct methoddbased on strong formulation (see Section 4.3), such as the force
equilibrium equations. That is, the stiffness matrix is derived directly from force
and moment equilibrium conditions. This method applies only to some very
simple element types, such as a bar or a beam.
Variational methoddbased on weak formulation (see Section 4.4). This method
originated from the principle of minimum potential energy. Several different
element stiffness matrices have been derived using this method. However, the
Basic Finite Element Method as Applied to Injury Biomechanics. http://dx.doi.org/10.1016/B978-0-12-809831-8.00004-0
Copyright © 2018 Elsevier Inc. All rights reserved.
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CHAPTER 4 Element Stiffness Matrix
•
method is limited to those structures in which the potential energy equation is
available.
Weighted residuals methoddalso based on a weak formulation. By assuming a
set of approximate solutions in conjunction with different selections of
weighting functions, approximate solutions that are reasonably close to the
exact solutions can be found. The most commonly used weighted residuals
method is the Galerkin’s.
In the following sections, explanations are given on how the direct method is used
to form the structure stiffness matrix of a simple truss structure as well as element
stiffness matrices for a 2-node bar element and a 2-node beam element. To highlight
the essence of the variational and Galerkin weak form methods, the stiffness matrix of
a 2-node bar element is created using both methods. With the exception of the direct
method, the isoparametric formulation and strainedisplacement matrix [B]
described in Chapter 3 are used to formulate the element stiffness matrices.
4.2 DIRECT METHOD
The direct method can be used to directly construct the structure stiffness matrix [K ],
if only a handful of elements are needed to characterize the behavior of the structure.
As an example, we generate the [K ] matrix for the entire 3-node, 2-spring system
from its FBD as shown in Section 1.3.4. But the same method would be more difficult to adopt when generating [K ] for the 5-element system in Example 1.2, Section
1.3.6. When the number of elements becomes much larger than that shown in
Example 1.2, application of this method becomes very difficult, if not impossible.
Instead, we use the direct method to derive the element stiffness matrix [k] before
all individual element stiffness matrices [k]i are assembled into the global stiffness
matrix [K ]. Even with this alternative application of the direct method, it can only be
used for a few of the simple element types, such as a bar, truss, cable, or beam
element. Previously in Section 1.3, the direct method was used to derive the element
stiffness matrices for spring, bar, and truss members. In this section, direct methods
used to construct the [K ] matrix of a truss structure with only three elements and to
construct the [k] matrix of a 2-node beam element are presented.
4.2.1 DIRECT FORMATION OF STRUCTURE STIFFNESS MATRIX
Consider a 3-node, 3-element truss structure as shown in Fig. 4.1. Each truss member has the same Young’s modulus E and constant cross-sectional area A. The
respective axial stiffness for each truss member (ki) is marked on the figure. Note
that all truss members are pinned at the connecting nodes, and hence cannot sustain
rotational loads. For this reason, each node has two translational DOFs, and the
entire truss structure has a total of six DOFs. Thus, the structure stiffness matrix
[K ] must be a 66 matrix.
4.2 Direct Method
FIGURE 4.1
A 3-node, 3-element bar structure with a total of six DOFs.
As described in Chapter 1, the first column of the stiffness matrix can be identified by finding the force needed to have a unity displacement in the first DOF, while
displacements in all other DOFs are assigned to be zero. Note that a unity displacement in a DOF is small compared to the size of the truss member. Hence, the angles
do not appreciably change. This same principle is applied to the remaining DOFs
sequentially, to identify all other columns needed to create the structure stiffness matrix [K ]. Let the forceedisplacement equation representing this system be
fFg61 ¼ ½K66 fdg61 , where {d} represents three horizontal and three vertical
displacements, {F} is the force vector, and [K ] is the structure stiffness matrix.
(a) Column 1 of [K]: set u1 ¼ 1, v1 ¼ u2 ¼ v2 ¼ u3 ¼ v3 ¼ 0.
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CHAPTER 4 Element Stiffness Matrix
To accommodate this displacement configuration, P1 needs to move in
the þx-direction (i.e., to the right) by one unit to P10 , while P2 and P3 remain at
their original positions. As such, both elements 1 and 3 need to be elongated. As
stated earlier, the unit displacement is greatly exaggerated in the figure to
illustrate the change. Thus, all internal angles remain approximately the same
upon deformation. As element 3 is stretched to the right by 1 unit (u1 ¼ 1), the
accompanied elongation for element 1 must be equal to
dzu1 cosð45 Þ ¼ 0:707, based on the geometric relationship shown. Additionally, we know that element 1 is fixed at P2 in this configuration. Thus, the
axial force needed to stretch element 1 for a displacement of 0.707 units from
P1 to P10 is FP1 ¼ 0.707k1, according to Hooke’s law. The reaction force acting
at P2 for element 1 is of equal magnitude but opposite direction, that is
FP2 ¼ 0.707k1, based on the equilibrium requirement. Similarly, the equal
and opposite axial forces needed to elongate element 3 by 1 unit are FP1 ¼ k3 at
P1 and FP3 ¼ k3 at P3. Because the axial forces for elements 1 and 3 are of
different directions, and vectors in different directions cannot be directly
combined, we now decompose these axial forces into horizontal and vertical
components for both elements 1 and 3 as follows:
Element 1:
F1H ¼ 0:707k1 cosð45 Þ ¼ 0:5k1 and F1V ¼ 0:707k1 sinð45 Þ ¼ 0:5k1 ; and
F2H ¼ 0:707k1 cosð45 Þ ¼ 0:5k1 and F2V ¼ 0:707k1 sinð45 Þ ¼ 0:5k1 ;
where the index 1 or 2 is used to distinguish node P1 from P2, and H and V represent
the horizontal and vertical components, respectively.
Element 3:
F1H ¼ k3 and F3H ¼ k3
Combining F1H, F1V, F2H, F2V, and F3H force components yields the first column
of the global stiffness matrix corresponding to
f F1H F1V F2H F2V F3H F3V gT as
½Kfirst column ¼ f 0:5k1 þ k3
0:5k1
0:5k1
0:5k1
(b) Column 2 of [K ]: set v1 ¼ 1,u1 ¼ u2 ¼ v2 ¼ u3 ¼ v3 ¼ 0.
k3
0 gT .
4.2 Direct Method
In this displacement field, the unit vertical displacement at P1 (v1 ¼ 1) would
affect both elements 1 and 3 as shown. Again, the unit displacement is much
smaller than the length of element 3, but is exaggerated for illustration purpose.
As such, the angle q formed by the deformed and undeformed element 3 would
be so small that sin q z 0 and cos q z 1. Hence, the change in length for
element 3 is negligible, that is, no force is needed to achieve this deformed
configuration for element 3. For element 1, the length of the element is
shortened by dzv1 cosð45 Þ ¼ 0:707, based on the geometric relationship as
shown. Therefore, the axial force needed at P1 to shorten this element is
FP1 ¼ 0.707k1, and the reaction force at P2 is FP2 ¼ 0.707k1. We now
decompose the axial forces at nodes P1 and P2 for element 1 into horizontal and
vertical components as
F1H ¼ 0:707k1 cosð45 Þ ¼ 0:5k1 ;
F1V ¼ 0:707k1 sinð45 Þ ¼ 0:5k1 .
Similarly, F2H ¼ 0.5k1 and F2V ¼ 0.5k1; and
½Ksecond column ¼ f 0:5k1
0:5k1
0:5k1
0:5k1
0
0 gT :
(c) Column 3 of [K]: set u2 ¼ 1, u1 ¼ v1 ¼ v2 ¼ u3 ¼ v3 ¼ 0.
½Kthird column ¼ f 0:5k1
0:5k1
0:5k1
0:5k1
0 gT .
0
(d) Column 4 of [K]: set v2 ¼ 1, u1 ¼ v1 ¼ u2 ¼ u3 ¼ v3 ¼ 0.
½Kfourth column ¼ f 0:5k1
0:5k1
0:5k1
0:5k1 þ k2
0
k2 gT .
(e) Column 5 of [K]: set u3 ¼ 1, u1 ¼ v1 ¼ u2 ¼ v2 ¼ v3 ¼ 0.
½Kfifth column ¼ f k3
0 0
0
k3
0 gT .
(f) Column 6 of [K]: set v3 ¼ 1, u1 ¼ v1 ¼ u2 ¼ v2 ¼ u3 ¼ 0.
½Ksixth column ¼ f 0
0
0
k2
0
k 2 gT .
The complete structure stiffness matrix [K] then becomes
3
2
0:5k1 þ k3 0:5k1 0:5k1
0:5k1 k3 0
7
6
6 0:5k1
0:5k1
0:5k1
0:5k1
0
0 7
7
6
7
6
7
6 0:5k1
0:5k
0:5k
0:5k
0
0
1
1
1
7.
½K ¼ 6
6 0:5k1 0:5k1 0:5k1
0:5k1 þ k2 0 k2 7
7
6
7
6
6
k3
0
0
0
k3
0 7
5
4
0
0
0
k2
0
k2
155
156
CHAPTER 4 Element Stiffness Matrix
Now we can check this result against the result obtained from assembling
the three element stiffness matrices as that shown in Chapter 1. For
element 1, the element is rotated by an angle of 135 degrees from horizontal,
and we have:
ðcos 135 Þ2 ¼ 0:5;
8
u1
>
>
>
>
<v
1
½kelement 1
>
u2
>
>
>
:
v2
ðcos 135 Þðsin 135 Þ ¼ 0:5; and ðsin 135 Þ2 ¼ 0:5:
9
9
38
2
u1 >
0:5 0:5 0:5 0:5 >
>
>
>
>
>
>
>
>
>
7>
6
=
7< v1 =
6 0:5 0:5
0:5
0:5
7
6
¼ k1 6
>
0:5 0:5 7
u2 >
>
>
>
5>
4 0:5 0:5
>
>
>
>
>
>
:
;
;
0:5 0:5 0:5 0:5
v2
For the vertically orientated element 2 and horizontally orientated element 3, we
have
8
9
9
38
2
u2 >
u2 >
0 0
0 0 >
>
>
>
>
>
>
>
>
>
>
7>
6
<v >
<v >
=
=
7
6
0
1
0
1
2
2
7
6
½kelement 2
¼ k2 6
and
>
0 0 7
u3 >
u3 >
>
>
>
>
5>
40 0
>
>
>
>
>
>
>
>
:
:
;
;
0 1
0 1
v3
v3
8
9
9
38
2
u1 >
u1 >
1 0
1 0 >
>
>
>
>
>
>
>
>
>
>
7>
6
<v >
<v >
=
=
7
6
0
0
0
0
1
1
7
6
½kelement 3
¼ k3 6
:
>
1 0 7
u3 >
u3 >
>
>
>
>
5>
4 1 0
>
>
>
>
>
>
>
>
:
:
;
;
0 0
0 0
v3
v3
Assembling all three element stiffness matrices yields the global stiffness matrix
2
3
0:5k1 þ k3 0:5k1 0:5k1
0:5k1 k3 0
6
7
6 0:5k1
0:5k1
0:5k1
0:5k1
0
0 7
6
7
6
7
6 0:5k1
7
0:5k
0:5k
0:5k
0
0
1
1
1
6
7.
½K ¼ 6
7
0:5k
0:5k
þ
k
0
k
0:5k
0:5k
1
1
1
1
2
2 7
6
6
7
6
k3
0
0
0
k3
0 7
4
5
0
0
0
k2
0
k2
This exercise shows that the structure stiffness matrix can be directly derived,
but this approach may be too cumbersome to do as the numbers of nodes
and elements increase. However, it is quite simple to create a structure stiffness
matrix by assembling individual stiffness matrices already developed for each
element, especially if we use a computer program to achieve this task.
Hence, the latter is the preferred method for developing the structure stiffness
matrix.
4.2 Direct Method
4.2.2 DIRECT METHOD FOR A 2-NODE BEAM ELEMENT
Consider a beam element that has a constant Young’s modulus E, a cross-sectional
moment of inertia I, and a length L. The element stiffness matrix of this beam
element can be derived directly from these physical properties. For background, a
classical bar, spring, truss, or rod can take axial (tensile or compressive) forces
but no transverse loads. On the contrary, a classical beam is subjected to transverse
(bending) loads but cannot take axial forces. For this reason, a bar element requires
the cross-sectional area (A) while a beam element needs the bending (crosssectional) moment of inertia (I) to define the respective mechanical behaviors. As
mentioned in Section 2.5.1, a 2-node beam element has two DOFs (a vertical deflection and a rotation) at each node, for a total of four DOFs (Fig. 4.2).
4.2.2.1 Brief Review
Recall from courses related to the strength of material or mechanics of material, the
static equilibrium equation for a 2-node beam element subjected to no axial or
torsional load can be expressed as
8
9
>
>
d3 wðxÞ
>
>
>
>
> EI dx3 >
>
>
>
>
x¼0
>
>
>
>
>
>
>
>
8
9 8
9 >
>
>
>
2
>
>
V
w
>
>
> 1 >
> 1>
> >
> >
d wðxÞ
>
>
>
>
>
>
>
>
EI
>
>
>
>
>
>
2
=
< q = <M = <
dx
x¼0
1
1
;
(4.2)
¼
¼
½k
3
>
>
w > >
V > >
>
>
> EI d wðxÞ
>
> 2 >
> 2>
> >
> >
>
>
>
>
>
>
>
:
; :
; >
>
dx3 x¼L >
>
>
q2
M2
>
>
>
>
>
>
>
>
>
>
>
>
>
>
2 wðxÞ
>
>
d
>
>
>
>
: EI
;
2
dx x¼L
θ
θ
FIGURE 4.2
A 2-node beam element with two DOFs (a vertical deflection w and a rotation q) per node,
for a total of four DOFs.
157
158
CHAPTER 4 Element Stiffness Matrix
where [k] is the element stiffness matrix, w(x) represents the vertical deflection as a
function of the x-coordinate, q(x) is the rotation, V represents the nodal shear force,
M stands for the nodal bending moment, 1 and 2 represent the node numbers, E is
Young’s modulus, I is the moment of inertia, and L is the length of the element.
The shape functions for this element are derived using the Hermite interpolation,
as introduced in Section 2.5.1. The calculated element shape functions are recapped
here, and Eqs. (2.35)e(2.38) are relisted as Eqs. (4.3)e(4.6).
N1 ¼
L3 3Lx2 þ 2x3
6Lx þ 6x2
and N1;x ¼
3
L
L3
(4.3)
N2 ¼
L3 x 2L2 x2 þ Lx3
L3 4L2 x þ 3Lx2
and
N
¼
2;x
L3
L3
(4.4)
N3 ¼
3Lx2 2x3
6Lx 6x2
and
N
¼
3;x
L3
L3
(4.5)
L2 x2 þ Lx3
2L2 x þ 3Lx2
and N4;x ¼
(4.6)
3
L
L3
By assigning the generalized displacement function w(x) to represent both
the deflection and rotation anywhere within the element, the magnitude can
be computed from the element shape functions and wi (nodal generalized
displacement) as
N4 ¼
wðxÞ ¼ N1 w1 þ N2 q1 þ N3 w2 þ N4 q2
9
8
w1 >
>
>
>
>
>
>
>
>
>
>
>
>
=
< q1 >
.
¼ ½ N1 N2 N3 N4 >
>
>
w2 >
>
>
>
>
>
>
>
>
>
>
;
:
q2
(4.7)
By combining Eqs. (4.2) and (4.7), we can write
2 3
3
d
6 dx3 ðN1 w1 þ N2 q1 þ N3 w2 þ N4 q2 Þ
7
x¼0 7
6
8
9
6
7
6 d2
7
>
> V1 >
>
6
>
>
ðN1 w1 þ N2 q1 þ N3 w2 þ N4 q2 Þ 7
>
6
7
<M >
=
6 dx2
x¼0 7
1
6
.
¼ EI 6
7
3
>
V2 >
7
>
>
6
7
d
>
>
>
>
6
7
:
;
6 dx3 ðN1 w1 þ N2 q1 þ N3 w2 þ N4 q2 Þx¼L 7
M2
6
7
6
7
4 d2
5
ðN1 w1 þ N2 q1 þ N3 w2 þ N4 q2 Þ
2
dx
x¼L
(4.8)
4.2 Direct Method
2
d ðN ; N ; N ; N Þ and
To simplify Eq. (4.8), we need to first identify dx
2
1
2
3
4
d 3 ðN ; N ; N ; N Þ from Eqs. (4.3)e(4.6). After some simple derivations, we
3
1
2
3
4
dx
can write the following equations:
d2
ð12x 6LÞ d 2
ð6x 4LÞ d 2
ð12x þ 6LÞ
N1 ¼
;
N2 ¼
;
N3 ¼
;
2
3
2
L
L2
L3
dx
dx
dx2
2
d
ð6x 2LÞ
N4 ¼
and
2
L2
dx
(4.9)
d3
12 d 3
6
d3
12 d3
6
N
¼
;
N
¼
;
N
¼
;
N4 ¼ 2 .
(4.10)
1
2
3
L3 dx3
L2 dx3
L3
L
dx3
dx3
Now, we insert the results from Eqs. (4.9) and (4.10) into Eq. (4.8) and write them in
matrix form as Eq. (4.11). Notice Eq. (4.8) has EI pulled out of the matrix for simplification. To avoid L in denominators within the matrix in Eq. (4.11), L13 is also
pulled out.
9
9
8
2
38
w1 >
V1 >
12 6L
12 6L >
>
>
>
>
>
>
>
>
>
>
>
>
= EI 6
<M >
2 7
6 6L 4L2
7< q1 =
6L
2L
1
6
7
¼ 36
(4.11)
>
L 4 12 6L
V2 >
12 6L 7
w2 >
>
>
>
>
5>
>
>
>
>
>
>
>
>
:
;
;
:
6L
2L2
6L 4L2
M2
q2
Hence, the element stiffness matrix for a 2-node beam element aligned along the
x-axis is
2
3
12 6L
12 6L
6
7
6L 4L2
6L 2L2 7
EI 6
6
7.
½k ¼ 3 6
(4.12)
L 4 12 6L
12 6L 7
5
6L
2L2
6L
4L2
As stated earlier, this element stiffness matrix is a singular matrix, and no solutions can be obtained without proper boundary and loading conditions. Note that the
beam element type can only take vertical loading, which in turn generates vertical
deflection and rotation. As such, it provides no resistance to axial and torsional
loading, and we shall not expect to see any axial or torsional responses, even if a
beam element is subjected to axial or torsional loads. In some software packages,
generalized beam elements may allow axial and torsional loading in addition to
bending. Thus, it is important to understand which element type is in order to avoid
unintentional errors. Finally, if a beam element is not oriented along one of the axes
of the global coordinate system, a transformation from this local coordinate systeme
based element stiffness matrix (to be presented in Section 4.7) needs to be performed
before it can be incorporated with the stiffness matrices created for other elements.
159
160
CHAPTER 4 Element Stiffness Matrix
Example 4.1
Consider a cantilever beam with a length of 100 in., cross-sectional moment of
inertia of 100 in.4, and Young’s modulus of 10 106 psi. Note that the selection
of imperial units in this example is intended to demonstrate that as long as a set of
consistent units is used, the calculated results are consistent. (1) Find the deflections and rotation angles at nodes P2 and P3 (Fig. 4.3A) when the cantilever beam
is subjected to a vertical downward force of 100 lb at the midpoint. (2) The same
beam is modeled as one element (Fig. 4.3B) with loading to the four DOFs by
T
P
PL
P
PL
, where P ¼ 100 lb, and L ¼ 100 in. A counterclock2 8 2 8
wise moment is designated as a positive moment.
Solution
(1) Because loading can only be applied at a node that exists, we idealize this
structure with two beam elements, where element 1 is formed by P1eP2 and
element 2 is formed by P2eP3 (see Fig. 4.3A). By dividing the structure into
two elements, the length becomes 50 in. for each element, and the concentrated
load can be directly applied at P2. Using the beam element stiffness matrix shown
in Eq. (4.12), the stiffness matrices for both elements are expressed as
2
½kelement 1 ¼
12
300
300
5000
6
10 106 100 6
6 300 10000
6 12 300
503
4
12
300
3
7
300 5000 7
7 and
12
300 7
5
300 10000
(A)
(B)
FIGURE 4.3
(A) A cantilever beam vertically loaded at the center modeled as a 3-node, 2element FE model. (B) The same beam modeled as a 2-node, 1-element FE model
and loaded.
4.2 Direct Method
2
½kelement 2 ¼
12
300
300
5000
6
10 106 100 6
6 300 10000
6 12 300
503
4
12
3
300
7
300 5000 7
7.
12
300 7
5
300 10000
If we assemble these two element stiffness matrices by carefully aligning the
corresponding DOFs, we have the structure stiffness matrix of this beam problem
as follows:
2
12
300
12
6
6 300 10000 300
6
6
6
24
10 10 100 6
6 12 300
½K ¼
6 300 5000
503
0
6
6
6 0
0
12
4
0
0
300
300
0
5000
0
0
12
20000 300
300
12
5000
300
3
0
7
7
7
7
300 7
7;
5000 7
7
7
300 7
5
0
10000
and the forceedisplacement equation becomes:
8
9
2
F1 >
>
>
>
>
>
6
>
>
>
6
>
>
M1 >
>
>
>
>
6
>
>
>
6
< F >
=
6
2
¼ 80006
6
>
>
M
2
>
>
6
>
>
>
>
6
>
>
>
6
> F3 >
>
>
>
4
>
>
>
>
:
;
M3
12
300
12
300
10000
300
12 300
300 5000
24
0
0
0
12
0
0
300
38
>
>
>
7>
7>
>
5000
0
0
7>
>
>
7>
<
0
12
300 7
7
7
20000 300 5000 7>
>
>
7>
>
>
300
12
300 7
>
5>
>
>
:
5000 300 10000
300
0
0
9
w1 >
>
>
>
>
q1 >
>
>
>
>
w =
2
q2 >
>
>
>
>
>
w3 >
>
>
>
;
q3
.
The boundary conditions at the fixed end require that w1 ¼ q1 ¼ 0, hence the first
two rows and first two columns can be eliminated from the calculations. With
F2 ¼ 100 and M2 ¼ F3 ¼ M3 ¼ 0, solving the remaining four equations with
four unknowns is trivial. We can use the Gaussian elimination, as explained in
Section 1.3.7, or any other method to find the following final solutions:
8
9
2
F2 >
24
0
12
>
>
>
>
>
<M =
6 0
20000 300
2
6
¼ 80006
>
4 12 300
F3 >
12
>
>
>
>
:
;
M3
300 5000 300
8
9
0:00417
>
>
>
>
>
< 0:000125 >
=
¼
;
>
0:01042 >
>
>
>
>
:
;
0:000125
38 9 8 9
300 > w2 > > w2 >
> >
>
>
>
= >
<q >
=
< >
5000 7
2
7 q2
0
7
w3 >
300 5>
> >
> w3 >
>
>
>
; >
: >
;
: >
q3
q3
10000
where the deflections are in inches and rotation angles are in radians.
(2) Because this beam is represented by only one element (see Fig. 4.3B), the
length of the element is now 100 in. Therefore the forceedisplacement equation
becomes
161
162
CHAPTER 4 Element Stiffness Matrix
9
8
>
P >
>
>
>
>
>
>
>
>
>
>
2
>
>
>
>
>
>
2
>
>
>
>
>
>
>
PL >
>
>
>
= 10 106 100 6
< >
6
8
6
¼
6
3
>
>
100
P
>
>
4
>
> >
>
>
>
>
> 2 >
>
>
>
>
>
>
>
>
>
> PL >
>
>
>
>
>
;
: 8 >
12
6L
6L 4L2
12 6L
6L
2L2
9
38
w1 >
>
>
>
>
>
7>
=
< q >
6L 2L2 7
1
7
.
12 6L 7
w2 >
>
>
5>
>
>
>
>
;
:
6L 4L2
q2
12
6L
Applying the two boundary conditions w1 ¼ q1 ¼ 0 reduces the above equation to
9
8
P>
>
>
=
< >
w2
50
12
12 6L
2
0
¼
1000
¼ 1000
2
>
>
1250
600
q2
6L 4L
>
;
: PL >
8
600
40000
w2
.
q2
Solving these two equations results in an end deflection w2 ¼ 0.01042 in. and
rotation q2 ¼ 0.000125 rad. These results are identical to the w3 and q3 calculated in part (1), where two beam elements are used to represent the cantilever
structure. In part (2), we use only one 2-node beam element to represent the
cantilever beam and apply vertical forces and moments to P1 and P2, as shown in
Fig. 4.3B. Surprisingly, the calculated nodal deflections and rotations at the end
nodes (P3 in part (1) and P2 in part (2)) are identical. The fact that a concentrated
load applied at the center has an equivalent effect as the two forces and two
moments applied at the two end nodes will be discussed in Section 6.3.1. For now,
we can accept the information that a midpoint vertical downward force of P is
comparable to applying four nodal load vectors with magnitudes of
T
P; PL; P; PL
to the four corresponding DOFs, where L is the length of
2
8
2 8
the element.
From this simple exercise, we also notice that only a handful of elements are
sufficient to provide the exact answers at all nodal locations. Because the element
stiffness matrix is derived in the same way as the analytical method, beam elements always provide the same nodal solutions as those obtained from analytical
methods, regardless of the number of elements used. Obviously, we cannot calculate the deflection and rotation at the midpoint when using only one element.
Thus, more elements must be used to show the curved nature of the beam deflections and rotations.
4.3 Strong Formulation
4.3 STRONG FORMULATION
Formulating the element stiffness matrix using the direct method or based on direct
formulation from the governing differential equations is regarded as strong formulation. These governing differential equations are generally derived from basic physical principles, such as the conservation of mass, energy, and momentum. In Chapter
1, the stiffness matrix [k] for a bar element with a constant Young’s modulus E and
cross-sectional area A is directly derived from force equilibrium based on the MSA
method. Thus, formulation of element stiffness matrices in this manner is based on
strong formulation, and the calculated nodal responses are identical to those derived
analytically.
The stiffness matrix [k] for a bar element is formulated as having a constant
cross-sectional area, and a bar with varying cross-sectional areas needs to be approximated by several bar elements, each with a different constant cross-sectional area,
as shown in Fig. 2.7 in Section 2.3.2. We shall now look back into continuum mechanics for finding solutions for a bar with varying cross sections. Consider that a
slender bar of length L is fixed on the left edge and loaded by a constant distributed
load per unit length q, as shown in Fig. 4.4. Now, we choose a thin slice, with a
thickness of Dx, off from the bar at a distance x from the fixed end. Next, we
draw a free-body diagram of this slice. If we let the net force on the left-hand
side be Q(x) (pointing to the left) and on the right-hand side be Q(x þ Dx) (pointing
to the right), then the force equilibrium condition requires that
Qðx þ DxÞ QðxÞ
QðxÞ ¼ Qðx þ DxÞ þ qDx0
þ q ¼ 0:
(4.13)
Dx
Now, by gradually shrinking the slice thickness until Dx approaches zero, we can
write
lim
Dx/0
Qðx þ DxÞ QðxÞ dQ
¼
.
Dx
dx
(4.14)
Δ
Δ
FIGURE 4.4
A bar of varying cross-sectional area loaded by a constant distributed load per unit length
q. The free-body diagram shows a thin slice taken at a distance x from the fixed end. Note
that a positive force is pointing to the right.
163
164
CHAPTER 4 Element Stiffness Matrix
Then, for a constant force per unit length q, the equilibrium condition shown in Eq.
(4.13) can be rewritten as
dQ
þ q ¼ 0:
(4.15)
dx
For an isotropic linear elastic bar with a constant elastic modulus E and a
point displacement u, the stressestrain relationship as shown in Chapter 1 is
repeated here as
du
.
(4.16)
dx
As explained in Section 1.4, we need to select a set of base units before the problem is formulated. For example, selecting a base set of kilogram, meter, and second
results in a force unit of N and a stress unit of Pascal (Pa). After replacing the point
force Q in Eq. (4.15) with the product of the constant cross-sectional area A of the
slice and the stress s (i.e., Q ¼ As), we have
s ¼ Eε ¼ E
dQ
dAs
þq¼
þ q.
dx
dx
Further, we can replace s with E du
dx from Eq. (4.16) to obtain
dAs
d
du
þq¼
AE
þ q.
dx
dx
dx
Finally, we simplify and summarize the combination of Eqs. (4.15) and (4.16) as
dQ
d2 u
þ q ¼ EA 2 þ q ¼ 0:
(4.17)
dx
dx
Eq. (4.17) is the governing differential equation for this bar problem. To solve it, we
need to apply the boundary conditions. As shown in Fig. 4.4, the left edge of this bar
with varying cross sections is fixed; hence, the displacement is zero at this edge.
Also, the right edge of this bar is free of restraint; hence, the slope at the free end
is zero. Thus, the boundary conditions can be summarized as
du
ðx ¼ LÞ ¼ 0:
(4.18)
dx
The combination of the governing differential equation shown in Eq. (4.17) and
boundary conditions displayed in Eq. (4.18) is called the strong formulation of this
bar problem. This set of equations can be easily solved by those of you who have
learned partial differential equations. This strong-form approach provides the exact
solution everywhere but with one significant limitation. That is, most real-world
structures are too intricate for proper mathematical descriptions of geometries.
Additionally, real-world problems usually involve high complexities related to
multiple material compositions and complex loading and boundary conditions.
Hence, analytical or strong-form solutions for real-world problems are not practical
uðx ¼ 0Þ ¼ 0 and
4.4 Weak Formulation
in most cases. To ease the limitation that exists when using the strong-form solution,
a less rigid solution can be used. In the next section, we explain the method capable
of providing acceptable engineering solutions based on the weak formulation.
4.4 WEAK FORMULATION
Many location- or time-dependent numerical methods (e.g., the method used to identify
a second-degree polynomial, y ¼ a þ bx þ cx2, from experimental data through a
curve fitting procedure or the method used to find acceleration at certain time points
from time-dependent displacements, velocities, and applied forces of a dynamic system) offer approximate solutions as a function of geometric locations and/or time. Like
these numerical methods, the FE method provides nodal solutions with acceptable accuracy, depending on the way the problem is formulated by strong or weak form. In
strong form described in Section 4.3, compulsory solutions are automatically satisfied
at certain geometric locations (e.g., nodal deflections or rotations of a beam problem).
In this section, a more popular weak form is described to solve the problem on an
“averaged” sense. As the name implies, the weak form of a problem is basically a problem declaration that results in a weaker solution than that from a strong-form declaration of the same problem, because the weak form lacks the nodewise enforcement.
Solving the governing differential equations in strong form can be quite rigorous,
and weak formulation is used to simplify these rigors, making the problems much
easier to solve. Some problems that cannot be solved through strong formulation
become solvable with weak formulation, due to the relaxation in the way an element
is formulated. Hence, weak formulation is far more popular to use in the FE method
than strong formulation.
There are two different approaches commonly used in weak formulation, the
variational method and weighted residuals method. Both methods implicitly contain
the governing differential equations, and hence can provide acceptable solutions. In
the variational method, the main objective is to find the displacements that minimize
a preselected functional. In most cases, this functional is the total potential energy
based on the virtual work principle. For the weighted residuals method, a test function is multiplied to the equations of residuals deduced from the strong-form differential equations. The displacements can be found by using integration by parts in
such a manner that there are minimal or zero residuals. These two methods are
essentially the same when a proven variational principle (such as the minimum potential energy) exists. Because the weighted residuals method applies the element
shape functions as part of the way the problem is formulated, it has become much
more popular than the variational method.
4.4.1 VARIATIONAL METHOD
For a linear elastic bar subjected to axial loading, the area underneath the forcee
deformation curve of the loading phase is the strain energy stored in the body, which
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CHAPTER 4 Element Stiffness Matrix
is equivalent to the work potential produced by external forces. Similarly, the area
underneath the stressestrain curve for a given strain is called the strain energy density (also known as the internal energy density), which is defined as the strain energy
per unit volume. We can do a quick check on the validity of this definition by finding
the ratio of the strain energy (e.g., N$m based on SI units) and volume (m3). The
resulting unit of energy density measure is N/m2, which is equivalent to that of
the stress (N/m2) times the strain (dimensionless).
For an isotropic, linear elastic bar with a constant elastic modulus E and crosssectional area A, the stressestrain relationship is s ¼ Eε or s ¼ E du
dx (Eq. 4.16), if
the energy related to shear deformation is neglected. Because lower case “u” is
used in this book to represent displacement, contrary to other mechanics books
which designate it for strain energy density, “s” is chosen to indicate the strain energy density to avoid confusion. When a material is continuously loaded, the strain
energy density is defined as the area underneath the stressestrain curve from 0 to a
preselected strain magnitude. For an isotropic, linear elastic material, the area of the
triangle, formed by the linear stressestrain curve, the x-axis, and the vertical line
drawn from a given strain magnitude ε (i.e., x ¼ ε), can be written as
1
1
s ¼ sε ¼ Eε2 .
(4.19)
2
2
A reversed way of looking at Eq. (4.19) is that if the strain energy density is known,
we can use it to find the constitutive equation. That is, stress can be derived by differds ¼ Eε. With ε ¼ du, the total
entiating the strain energy with respect to strain as s ¼ dε
dx
internal energy U for a 1D problem can be found by integrating the strain/internal energy density s (shown in Eq. 4.19) over the total volume V of the bar:
Z
Z
Z
1
1 L du
1 L du
du
EA dx.
(4.20)
U¼
sεdV ¼
E εAdx ¼
2 V
2 0
dx
2 0 dx
dx
For a 1D problem, each stress and strain vector contains only one component, sxx
and εxx, respectively. As such, the dot product fsg$fεg ¼ sε. In 2D or 3D problems,
both stress and strain are tensors with vector forms fsg ¼ f sxx syy sxy gT and
T
fεg ¼ εxx εyy gxy . Obviously, we cannot “multiply” the 3 1 stress “vector” with another 3 1 strain “vector,” because this product cannot be defined. In
this case, the product of the stress and strain vectors fsg$fεg is written as
fsgT fεg. Hence, the equation for strain energy needs to be written as
R
U ¼ 12 V sT εdV for 2D or 3D problems. As in the example used to demonstrate
strong formulation (Fig. 4.4), we assume that there is a constant distributed load
per unit length, and this force is represented by q. We know from physics that
work is the product of the force and distance. The force, in this case, is the distributed load. Due to this load, the work potential W can be expressed as the product of
the constant distributed load and the displacement integrated with respect to x:
Z L
W¼
quðxÞdx.
(4.21)
0
4.4 Weak Formulation
Thus, the total potential energy functional for the bar is the difference between
the total internal energy U from Eq. (4.20) and the work potential W from Eq. (4.21):
Z
Z L
1 L du
du
EA dx quðxÞdx.
(4.22)
PP ¼ U W ¼
2 0 dx
dx
0
In Eq. (4.22), no contributions from concentrated or end-surface stresses are added,
because the initial problem in the strong formulation is set up only for distributed
loading. If a concentrated force or end stress is added, Eqs. (4.21) and (4.22) would
need to be updated to include such effects. For example, if a concentrated force P is
applied at x ¼ L on the bar shown in Fig. 4.4 (a positive force indicates a force pointing towards the positive x-direction), then the total potential energy functional
becomes
PP ¼ U Wdistributed Wconcentrated
Z L
Z
1 L du
du
EA dx ¼
quðxÞdx Puðx ¼ LÞ.
2 0 dx
dx
0
(4.22a)
Variational forms of Eq. (4.22) that result from varying loading conditions are too
numerous to illustrate individually. You will need to address additional energy terms
in the total potential energy functional as necessary. Note that all terms in Eq. (4.22)
are functions of u(x). Therefore we can rewrite Eq. (4.22) more explicitly to highlight the fact that u(x) is the primary variable for the total internal energy (Eq.
4.20), work potential (Eq. 4.21), and total potential energy functional (Eq. 4.22):
PP ½uðxÞ ¼ U½uðxÞ W½uðxÞ.
(4.23)
It must be emphasized that u(x) shown in Eq. (4.23) must be admissible, that is,
u(x) must be continuous (because it is a continuous bar) and must also satisfy all
essential boundary conditions. There are two different types of boundary conditions,
essential and natural. More descriptions on these two types of boundary conditions
are provided in Chapter 6. Briefly, the essential boundary conditions describe conditions that are directly related to nodes, such as a fixed nodal displacement or rotation. In this example, the only essential boundary condition is that the left-hand side
of the bar is fixed. Hence, the final solutions for u(x) must satisfy the condition
u(x ¼ 0) ¼ 0. As stated for the principle of minimum potential energy, a structure
will be in a stable equilibrium if the total potential energy functional PP is minimized for any admissible displacements u(x). Based on this principle, a rolling
golf ball will slowly roll to the bottom of a valley and stop there, because the bottom
has the least potential energy. As the geometry becomes more complex, loading and
boundary conditions become more complicated, and this principle becomes the
method of choice for finding the approximate solutions. In the example shown
below, we form the structure stiffness matrix of a system of springs to demonstrate
the use of this principle.
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CHAPTER 4 Element Stiffness Matrix
Example 4.2
For the system of springs shown in Fig. 4.5, use the principle of minimum potential energy to determine the forceedisplacement equations.
Solution
Recall from basic physics, the potential energy stored in a spring with a spring
constant k is PE ¼ 12 kx2 . Since we know the boundary condition u1 ¼ 0, we
can write equations for the internal energy U, work potential W, and total potential
energy PP for the system. The total internal energy U is the sum of the internal
energy of each spring. Spring 1 is elongated at P2 by a displacement of u2.
Therefore U1 ¼ 12k1 u2 2 . Spring 2 is elongated by the difference in the displacements of P2 and P3: u3 u2, and therefore U2 ¼ 12 k2 ðu3 u2 Þ2 . Likewise,
U3 ¼ 12 k3 ðu4 u2 Þ2 . The total internal energy of the system is
1
1
1
U ¼ k1 u2 2 þ k2 ðu3 u2 Þ2 þ k3 ðu4 u2 Þ2 .
2
2
2
Work potential is the product of force and displacement, which is elongation.
There are two forces, F3 and F4, with the corresponding elongations u3 and u4,
respectively. Therefore
W ¼ F3 u3 þ F4 u4 :
The total potential energy PP is the difference between the internal energy U
and the work potential W:
1
1
1
PP ¼ U W ¼ k1 u2 2 þ k2 ðu3 u2 Þ2 þ k3 ðu4 u2 Þ2 F3 u3 F4 u4 .
2
2
2
2
1
3
FIGURE 4.5
A 4-node, 3-element system of springs fixed on the left-hand side and axially loaded
with F3 at P3 and F4 at P4.
4.4 Weak Formulation
To minimize PP, we must make all of the partial derivatives with respect to ui
(where i ¼ 2 to 4) equal to zero. That is,
vPP
¼ k1 u2 k2 ðu3 u2 Þ k3 ðu4 u2 Þ ¼ 0;
vu2
vPP
¼ k2 ðu3 u2 Þ F3 ¼ 0; and
vu3
vPP
¼ k3 ðu4 u2 Þ F4 ¼ 0:
vu4
By rewriting these three equations in matrix form, we have
2
k1 þ k2 þ k3
6
k2
4
k3
k2
k2
0
38 9 8 9
k3 >
< u2 >
= >
=
< 0 >
7
0 5 u3 ¼ F3 .
>
: >
; >
;
: >
k3
u4
F4
This example highlights the direct use of the minimum potential energy principle to identify the structural forceedisplacement equation. One prerequisite for
using this method is that the equations needed to calculate the internal energy and
work potential must be available. Instead of finding the structure stiffness matrix
[K ] as shown in this example, the same principle can be used to derive the
element stiffness matrix [k] for application in the FE method, as demonstrated
in the following sections.
Based on the principle of virtual work, all virtual movements in static equilibrium produce zero virtual work. Using this virtual work principle, a small change
in the total potential energy, dPP, can be obtained by perturbing an admissible but
infinitesimal axial displacement du(x) to the axial displacement u(x) in the total potential energy functional. We let du(x) ¼ ej(x), where e is a very small number (note
that the meaning for e here is different from the symbol used to represent strain, “ε”)
and j(x) is an admissible displacement that also satisfies the essential boundary
conditions.
In Example 4.2, the total potential energy PP is a only function of the displacement u, that is, PP ¼ PP(u). Thus, a small perturbation in displacement (du) would
result in a small change in the total potential energy (dPP). In other words, we can
write that PP(u þ du) ¼ PP(u) þ PP(du) ¼ PP(u) þ dPP. Rearranging terms in
this equation, the small change in total potential energy is written as
dPP ¼ PP ðu þ duÞ PP ðuÞ.
(4.24)
The small change in displacement, du(x), and small change in total potential energy, dPP, are the variations of u(x) and PP, respectively. Because variations are key
to solving this class of problems, the term variational method is utilized. As
observed from Eq. (4.24), we need to find PP(u þ du) and PP(u) before deriving
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CHAPTER 4 Element Stiffness Matrix
an equation for the small potential energy variation dPP. Because PP(u), in terms of
a constant Young’s modulus E and cross-sectional area A, has been previously
derived in Eq. (4.22), we write PP(u þ du) from Eq. (4.22) as
Z L
Z
1 L dðu þ duÞ
dðu þ duÞ
EA
dx qðu þ duÞdx.
PP ðu þ duÞ ¼
2 0
dx
dx
0
Applying PP(u þ du) derived above and PP(u) derived in Eq. (4.22), we find dPP
from Eq. (4.24) as
Z
Z L
Z
1 L dðu þ duÞ
dðu þ duÞ
1 L du
du
EA
dx EA dx
qðu þ duÞdx dPP ¼
2 0
dx
dx
2 0 dx
dx
0
Z L
þ
qudx.
0
Rearranging terms and factoring EA
2 out of the integral result in
2
3
Z Z L 2
EA 4 L dðu þ εjÞ 2
du
dx dx5
dPP ¼
2
dx
dx
0
0
2
4
Z
L
0
Z
qðu þ εjÞdx 3
L
qudx5.
0
Next, we combine terms within common integrals:
2 #
Z "
Z L
EA L du
dj 2
du
þε
dPP ¼
qεjdx.
dx 2 0
dx
dx
dx
0
Expanding terms results in
2 #
Z " EA L du 2
du dj
dj 2
du
þ ε
þ 2ε
dPP ¼
dx
2 0
dx
dx dx
dx
dx
Z L
½qu þ qεj qudx;
0
And simplifying gives
EA
dPP ¼
2
Z L"
0
2 #
Z L
du dj
2 dj
þε
qjdx.
2ε
dx ε
dx dx
dx
0
Finally, expanding again and then simplifying results in
Z Z L
Z L
du dj
ε2 EA L dj 2
dx þ
dPP ¼ εEA
dx ε
qjdx.
2
dx
0 dx dx
0
0
(4.25)
Based on the principle of minimum potential energy, we know that dPP must be
greater than zero (dPP > 0), because PP is at its minimal stage. Hence, any dPP
will make the current PP larger. Note that the second item in the right-hand side
4.4 Weak Formulation
2
of Eq. (4.25) involves two squared terms e2 and
dj
dx
. Thus, this item can only be
positive. For any positive or negative values of e, the only way to satisfy a stable
equilibrium with dPP > 0 is to guarantee that the remaining two items in the
right-hand side of Eq. (4.25) vanish. In other words,
Z L
Z L
du dj
dx ε
qjdx ¼ 0:
εEA
0 dx dx
0
Multiplying through by 1ε results in
Z L
Z L
du dj
dx qjdx ¼ 0:
EA
0 dx dx
0
(4.26)
Eq. (4.26) is, in essence, the principle of virtual work. According to this principle,
any admissible, infinitesimal movements of the system from a static equilibrium produce zero virtual work. In other words, variations of the internal energy and work
potential must be equal. Now, by expanding Eq. (4.26) based on integration by parts,
we have
L
Z L
Z L
Z L 2 Z L
du dj
du
d u
EA
dx jðxÞ EA
qjdx ¼ EA
qjdx
jdx
dx
dx2
0 dx dx
0
0
0
0
¼ 0:
(4.27)
We know that j(x) must be admissible for the same reason u(x) must be admissible.
Hence, the fixed boundary condition on the left-hand side of the bar implies that
j(0) ¼ 0. If we let j(L) ¼ jL, then the above equation becomes
Z L 2 du
d u
EA jL EA
þ q jdx ¼ 0:
(4.28)
dx
dx2
0
Because j(x) is an arbitrary function and jL is an arbitrary number, the only condition that satisfies Eq. (4.28) must fulfill the following two equations:
du
¼ 0 and
dx
2 d u
EA
þ q ¼ 0:
dx2
(4.29)
(4.30)
Note that Eq. (4.30), derived from the principle of minimum potential energy, is
identical to Eq. (4.17), the strong formulation of the same example problem. Similarly, Eq. (4.29) is the same as Eq. (4.18), even though these two equations are
derived through two different approaches. The derivation shown in this section for
the bar/spring problem demonstrates that the variational method yields the same
set of equations as the differential equations described for the strong formulation.
In other words, nodal displacements calculated from the FE method formulated using strong or weak form would yield the same values. For element types other than
171
172
CHAPTER 4 Element Stiffness Matrix
bar or spring, it is unusual to come up with the same set of equations by using two
different formulations. From Eqs. (4.24)e(4.30), we can easily deduce that if u(x) is
the exact solution, then dPP ¼ 0. On the other hand, if u(x) is not the exact solution
then dPP > 0.
So far, the entire bar is treated as a continuum. The next step is to divide this continuum into a finite number of elements for use in the FE method. Fig. 4.6 shows the bar
as a continuum divided into three 2-node, 1D bar elements. Then, the internal energy,
work potential, and total potential energy can be written as
U ¼ Uelement 1 þ Uelement 2 þ Uelement 3 ;
(4.31)
W ¼ Welement 1 þ Welement 2 þ Welement 3 ; and
(4.32)
PP ¼ PP element 1 þ PP element 2 þ PP element 3 .
(4.33)
Because all elements are created equal (i.e., of the same shape despite different
lengths), we can show that with Eqs. (4.31)e(4.33), listed above, there is an implication that the total potential energy within any element must be zero, based on the
principle of virtual work that U¼ W for any element. In this instance, we choose the
second element to demonstrate how to apply the variational method:
PP element 2 ¼ Uelement 2 Welement 2 ¼ 0
(4.34)
FIGURE 4.6
Top: A continuum bar discretized into an FE mesh with three 2-node elements, for a total
of four nodes. Each node has one DOF. Element 2 in the middle, which is not affected by
any boundary conditions, is used to demonstrate how the variational method is applied to
formulate the element stiffness matrix of a 2-node bar element. Bottom: A hypothetical
axial displacement profile shows that u must be a continuous function. Note that the axial
displacement u should be along the x-axis.
4.4 Weak Formulation
Eq. (4.34) is the basis for developing the element stiffness matrix and consistent
nodal force based on nodal DOFs in the FE method. We shall now assign the length
of element 2 as L0 , so that there is no confusion with the total length of the continuum
bar L. The two sets of equations shown below are a recap of Eqs. (3.6) and (3.10) for
a 2-node linear bar element. These equations are (1) the element shape functions
based on the isoparametric formulation with natural coordinate of x ranging
from 1 to 1, and (2) the strainedisplacement matrix [B]:
1x
1þx
d½N dx
1 1
and N2 ¼
; ½ B ¼
¼
N1 ¼
.
2
2
dx dx
L0 L0
Eq. (4.20) for the total internal energy within element 2, with L0 instead of L, can
be written as
Z 0
1 L du
du
EA dx
Uelement 2 ¼
2 0 dx
dx
Note that this equation is based on the global coordinate system. We now transfer
it from the global coordinate system to the local isoparametric coordinate system.
dx, that is, dx ¼ Jdx.
From Chapter 3, the Jacobian (length ratio) is defined as J ¼ dx
Also, the strainedisplacement equation (Eq. 2.23) has a form of
u1
¼
½B
. Finally, the integration limits for x ¼ 0 to L0 needs to be
εxx ¼ du
dx
u2
changed to x ¼ 1 to 1. Inserting all these values into the above equation, we have
Z T
u1
u1
1 1
½B
ðJdxÞ.
Uelement 2 ¼
EA½B
2 1
u2
u2
As previously described, the inner product of two vectors {q} and {r} is the product of {q} transpose ({q}T) and {r}. This same concept can be extended to matrix
T
u1
u1
products. As such, the first ½B
.
term needs to be changed to ½B
u2
u2
Additionally, the transpose of the product of two matrices equals the product of their
T T
u1
u1
T
T T
transposes in reverse order, that is, (QR) ¼ R Q , or ½B
¼
½BT
u2
u2
in this case. Finally, u1 and u2 are nodal displacements (constants) and can be moved
out of the integration. Based on these facts, we further rearrange the above equation
as
T Z 1
u1
1 u1
Uelement 2 ¼
½½BT ðEAÞ½BJdx
.
(4.35)
2 u2
u
1
2
173
174
CHAPTER 4 Element Stiffness Matrix
For a spring, the stored energy has the form of 12 kx2 , where k is the spring constant and x is the deformation. Eq. (4.35) is intentionally made to be analogous to the
stored energy equation of a spring with a form of 12fugT ½kfug. That is, we equate
R1
T
length of
1 ½½B ðEAÞ½BJdx to be the element stiffness matrix [k]. Because the
0
1 1
element 2 is L0 , the Jacobian J ¼ L2 . Inserting the Jacobian and ½B ¼
L0 L0
R1
into 1 ½½BT ðEAÞ½BJdx produces
2
3
1
0 Z 1
6 L0 7
L
1 1
6
7
dx .
½k ¼
4 1 5EA L0 L0
2
1
L0
Because all constant terms do not need to be integrated, we can write the above
equation as
2
3
1
6 L 0 7
Z 1
7 1 1
EAL0 6
6
7
½k ¼
dx
7 0
0
2 6
4 1 5 L L 1
L0
2
!
!3
(4.36)
1
1
7
6
2
3
2
6 ðL0 Þ2
ðL0 Þ 7
1 1
7
EAL0 6
EA 4
7
6
5.
ð2Þ
¼
¼
7
6
!
!7
2 6
L0 1 1
7
6 1
1
5
4
2
0 2
ðL Þ
ðL0 Þ
Recall that for this example we chose to use L0 as opposed to L to avoid
confusion between calculations for a portion of the continuum bar with those for
the total bar. Except for this minor difference that we artificially imposed, Eq.
(4.36) is identical to Eq. (1.43), in which the direct method is used to derive the
element stiffness matrix for a 2-node bar. From this example, we know that the
element stiffness matrix [k] can be obtained using the variational method by integration of the [B] matrix. Hence, we discover yet another application of the element
shape functions, in addition to using them to find coordinates (position vectors),
to calculate generalized displacements anywhere within the element, and to generate
contours of any physical quantity. In a more general form, the element stiffness matrix for a 2-node bar element with a length L, constant elastic modulus E, and crosssectional area A can be written as
Z
EAL 1 T
½k ¼
½B ½Bdx.
(4.37)
2
1
4.4 Weak Formulation
From Eq. (4.21), we shall now consider the work potential for element 2. By definition, Welement n ¼ fugT ff gelement n , where {f} is the element load vector. This step
of obtaining the element load vector is necessary for cases where the load is applied
in between nodes. For better understanding of the element load vector, recall that
loading in the FE method can only be applied through the nodal DOFs. To avoid
any confusion, recall that u(x) represents the displacement anywhere within the
u1
element. That is, u(x) ¼ N1u1 þ N2u2 or ½ N1 N2 for a 2-node bar element.
u2
8 9
2 3
Z
N
< u1 = L0
qL0 1 4 1 5
dx ¼ f u1 u2 g
¼
quðxÞdx ¼
q½ N1 N2 dx
: ;2
2 1
1
0
u2
N2
8 09
8 09
3
2
1x
qL >
qL >
>
>
>
>
>
>
>
>
>
>
>
>
>
7
<
<
=
=
0 Z 16
2
2
2 >
7
6
qL
7dx ¼ f u1 u2 g
6
0ff gelement 2 ¼
¼ f u1 u2 g
7
6
>
>
>
2 1 4
>
>
>
0>
>
> qL0 >
>
>
1 þ x5
>
>
>
: qL >
:
;
;
2
2
2
(4.38)
Z
Welement 2
L0
Z
1
Validation of the above calculations can be done through a quick check of the force
equilibrium. Because the total applied load is qL0 , one half of the total load applied at
each node would satisfy the force equilibrium equation. In a more general form, the
consistent load vector for a 2-node bar element with a length of L and a uniformly
distributed force per unit length q, the element load vector {fe} corresponding to the
nodal DOFs can be written as
8 9
qL >
>
>
Z 1 =
< >
N1
2
.
(4.39)
ffe g ¼
q
j½Jjdx ¼
>
>
N2
1
>
;
: qL >
2
Eq. (4.37) shows that the element stiffness matrix can be directly derived from
the element strainedisplacement matrix [B], which in turn is derived from the
element shape functions [N]. Also, Eq. (4.39) demonstrates that another application
of element shape functions is to determine the consistent element load vectors. More
examples related to distributing load vectors that are not directly applied at the nodal
locations are provided in Chapter 6. As a quick summary of the purposes of element
shape functions described so far, they allow users to
•
•
•
•
Interpolate nodal coordinates and displacements
Draw contours of physical quantity
Derive element stiffness matrices
Derive nodal force vectors
We now use an example to expand the variational method from 1D to 2D plane
stress element.
175
176
CHAPTER 4 Element Stiffness Matrix
Example 4.3
The 2a by 2b rectangular plane stress element previously shown in Fig. 2.19 is
reused for this example. Using the principle of minimum potential energy, determine the element stiffness matrix.
Solution
For a 2D, 4-node plane stress element, each node has 2 DOFs corresponding to
the horizontal and vertical displacements. That is, the nodal displacement
u3 v3 u4 v4 gT . Let the strain and stress
fdg ¼ f u1 v1 u2 v2
T
vectors have the form fεg ¼ εxx εyy gxy
and fsg ¼ f sxx syy sxy gT ,
respectively.
The total potential energy functional (PP) within this element is the strain energy (U) minus the work potential (W), which may arise due to any combination
of concentrated force, body force, and surface traction. The body force is
frequently assumed zero in the FE method, because it is already considered as
prestress in the initial equilibrium. Additionally, external forces are typically
much larger than the body force, and hence the effect due to the body force is
negligible.
From Eq. (1.10), we know that the strainedisplacement equation is
2
v
6
8
9 6 vx
6
>
< εxx >
= 6
εyy ¼ 6
60
>
>
:
; 6
6
gxy
6v
4
vy
3
07
7
7 v7
7 u
;
7
vy 7 v
7
v7
5
vx
where {u} and {v} are horizontal and vertical displacements, respectively, anywhere within the element. Because both {u} and {v} are unknown, our task is to
4.4 Weak Formulation
find the approximated values based on the minimum potential energy principle.
We will approach the solution in five steps:
Step 1: Identify the generalized displacement equation {u} as a function of
element shape functions [N]
In this step, we approximate the horizontal and vertical displacement anywhere within the element in terms of shape functions. These interpolation functions were described previously in Sections 2.5 and 3.4. Therefore we know that
u ¼ N1 u1 þ N2 u2 þ N3 u3 þ N4 u4
v ¼ N1 v1 þ N2 v2 þ N3 v3 þ N4 v4 .
In matrix form, the generalized displacement {u} can be written as a function
of nodal displacements {d} as
fug ¼
u
N1
¼ ½Nfdg ¼
0
v
0
N1
N2
0
0
N2
N3
0
0
N3
N4
0
8 9
u1 >
>
>
>
>
> v1 >
>
>
>
>
>
>
>
>
>
>
>
>
u2 >
>
>
>
>
> >
0 < v2 =
.
>
N4 >
>
> u3 >
>
>
>
>
> v3 >
>
>
>
>
>
>
>
>
> u4 >
>
>
>
>
: >
;
v4
Step 2: Identify the strainedisplacement matrix [B]
By inserting the generalized displacement equation found in Step 1 into the strain
edisplacement equation, the strain vector in this element can be approximated as
2
v
8
9 6
vx
6
εxx >
>
6
>
>
>
>
<
= 6
6
εyy ¼ 6
60
>
>
>
>
6
>
>
:
; 6
gxy
6
4v
vy
2
vN1
9 6
8
6 vx
εxx >
6
>
>
>
>
>
= 6
<
6
εyy ¼ 6
6 0
>
>
>
>
6
>
>
; 6
:
gxy
6
4 vN1
vy
3
8
>
>
>
>
>
>
>
>
>
>
>
<
vN1
vN2
vN3
vN4
u1 þ
u2 þ
u3 þ
u4
vx
vx
vx
vx
9
>
>
>
>
>
>
>
>
>
>
>
=
07
7
7( )
7
v7 u
vN1
vN2
vN3
vN4
7
v1 þ
v2 þ
v3 þ
v4
¼
; or
7
vy
vy
vy
vy
vy 7 v
>
>
>
>
>
>
>
>
7
>
>
>
>
7
>
>
>
>
v5
vN
vN
vN
vN
1
1
2
4
>
>
>
>
u
v
u
v
þ
þ
þ
.
þ
>
>
1
1
2
4
:
vy
vx
vy
vx ;
vx
8 9
u1 >
>
>
>
> >
>
>
>
>
>
>
>
>
v
>
1
3>
>
> >
>
>
vN2
vN3
vN4
>
> >
>
>
0
0
0
0 7>
>
> u2 >
>
vx
vx
vx
7>
>
>
> >
7>
>
>
>
>
>
7
vN1
vN2
vN3
vN4 7< v2 =
7
0
0
0
¼ B fdg;
vy
vy
vy
vy 7
>
>
>
> u3 >
7>
>
7>
>
>
>
7>
>
>
>
vN1 vN2 vN2 vN3 vN3 vN4 vN4 5>
>
>
v
>
3
>
> >
>
>
>
vx
vy
vx
vy
vx
vy
vx >
>
>
>
>
>
u
>
> 4>
>
>
>
>
>
>
>
;
: >
v4
177
178
CHAPTER 4 Element Stiffness Matrix
where [B] is the strain-(nodal) displacement matrix in the form of:
3
2
½B38
v
6
6 vx
6
6
6
¼6 0
6
6
6v
4
vy
07
7
7
v7
7
7
vy 7
7
v7
5
vx
0
N1
N1
0
N2
0
0
N2
N3
0
0
N3
N4
0
0
N4
:
28
32
Step 3: Calculate the strain energy U
By inserting {ε} ¼ [B]{d} found in Step 2 into Eq. (1.22), the stressestrain
relationship for a plane stress element (repeated below as the equation with the
first equal sign), the stress vector within the element is
3
2
8
8
9
9
9
8
1 n
0
>
>
7>
6
< εxx >
< sxx >
=
=
=
< εxx >
7
E 6
0 7 εyy ¼ ½D εyy ¼ ½D½B fdg ;
6n 1
syy ¼
81
38
7
26
>
>
>
:g >
:
; 1n 4
;
;
:g >
1 n 5>
sxy
xy
xy
0 0
2
2
1
n
E 6n
1n2 6
1
3
0
7
7
0
7.
where ½D ¼
7
4
1 n5
0 0
2
With this information, we determine U as follows:
6
6
U¼
¼
1
2
1
2
Z
fsgT fεgdV
Z
V
ð½D½BfdgÞT
f½BfdggdV
i.e.; fsgT
i.e.; fεg
V
1
¼ fdgT
2
Z
ð½BÞT ½DT f½BgdVfdg
V
1
¼ fdgT ½kfdg;
2
R
where ½k88 ¼ V ½BT83 ½D33 ½B38 dV, since [D]T ¼ [D].
Step 4: Calculate the work potential W
Loads applied to the element could be any combination of concentrated load,
body force, and surface traction. In the FE method, loads can only be applied to
nodes at the corresponding nodal DOFs. Because no specifics are provided
regarding these loads, we shall collectively call these force vectors {f}. Thus,
the work potential is
W ¼ fdgT ff g
4.4 Weak Formulation
Step 5: Determine the total potential energy functional and minimize it
The total potential energy functional, PP, is the strain energy U minus the
work potential W. Inserting the results from Steps 3 and 4, we have
1
PP ¼ U W ¼ fdgT ½kfdg fdgT ff g.
2
In Step 2, we found that [B] can be directly calculated from the element shape
functions [N]. Entries for [D] are directly obtained from the constitutive equation
for the linear elastic, plane stress element. As such, the element stiffness matrix
[k] can be calculated through integration. However, integrating these terms
analytically is not only prone to errors, but also not easily programmed into an
FE analysis software package. We shall defer finding the entries for [k] until
we introduce the Gauss quadrature scheme in Section 4.5.1.
To minimize PP, we find the point where the derivative with respect to {d} is
equal to 0, which results in
vPP
¼ ½kfdg ff g ¼ f0g0½kfdg ¼ ff g.
vfdg
Based on the minimum total potential energy principle, we find that the result
has the same form as the static equilibrium equation. This result again proves that
the principle of minimum total potential energy can be used to derive the element
stiffness matrix.
4.4.2 WEIGHTED RESIDUALS METHOD
The direct method we discuss in Section 4.2 applies to only a handful of very simple
1D element types. Similarly, the strong formulation discussed in Section 4.3 is not
easily applied to real-world problems with complex geometries. The variational
method, described in Section 4.4.1, is a common weak-formulation method based
on minimizing the total potential energy, and it only applies to those problems in
which the potential energy equations can be described. Because the FE method
has been widely utilized to all kinds of engineering problems, we can imagine
that many other more versatile methods must exist for the formulation of different
types of elements. One of these methods is the weighted residuals method.
In brief, for the weighted residuals method, we attempt to find approximate nodal
displacements in structural mechanics problems through a set of trial functions.
Although not exact, these approximate solutions are very close to the exact solutions
and satisfy all the prescribed boundary conditions. As a first step, a set of trial solutions is proposed and inserted into the governing differential equations. Because this
set of trial solutions does not contain the exact answers, some nonzero leftovers (i.e.,
residuals) must be present. We then integrate the product of weighting functions and
the residuals. If any choice of weighting functions results in a zero integrated value,
179
180
CHAPTER 4 Element Stiffness Matrix
then the residuals must be approaching zero, and the trial solutions will be very close
to the exact solutions.
There are a handful of associated submethods, which are distinguished by the
ways in which the weighting functions are chosen, based on the principle of
weighted residuals. These submethods include, but are not limited to, the Galerkin,
subdomain, least squares, and collocation. Among these, the most commonly used is
the Galerkin weighted residuals method. This method has been credited to Russian
mathematician Boris Galerkin (Mar. 1871eJun. 1945), according to the MacTutor
History of Mathematics archive maintained by Drs. John J O’Connor and Edmund
F Robertson.
2
Take the governing differential equation (strong formulation) EA ddxu2 þ q ¼ 0
(shown in Eq. 4.17) for a bar problem with an essential boundary condition
¼ 0. It is easy to understand
u(x ¼ 0) ¼ 0 and a natural boundary condition du
dx
x¼L
that if ueðxÞ, which is a set of approximate solutions that differs from the set of exact
solutions, is input into the governing differential equation, nonzero residuals must
exist. In other words,
d2 ue
þ qðxÞ ¼ RðxÞ;
(4.40)
dx2
where ueðxÞ is the approximate solutions set, q(x) is the uniform distributed load, and
R(x) is the “equation of residuals.” Before minimizing the residuals R(x) to determine the solution for the bar problem, we first integrate the product of the residuals
R(x) and a weighting function W(x) over the entire element length. When presenting
the differences between strong and weak formulations in the beginning of Section
4.4, we did not explicitly show the procedures used to make a formulation “weaker”
than its original form. With integration, the strong formulation becomes weak
because integration tends to smooth the response. Hence, the accuracy can only
be characterized on an average sense as accuracy at each point is not reflected.
In theory, if the differential equation(s) formed from strong formulation can be
solved, the calculated results must be the exact solution. However, imperfections
are always present in real-world problems (e.g., simplifications are made when characterizing material properties), the original formulation is necessarily flawed. Thus,
the exact solutions to the flawed formulations do not exactly reflect real-world
results, and average results calculated from weak equations may actually be closer
to the underlying physics than what the strong formulation
can provide. Eq. (4.41)
RL
shows the condition where the integrated result 0 RðxÞWðxÞdx becomes zero for
any of the weighting functions.
Z L
d 2 ue
EA 2 þ q WðxÞdx ¼ 0
dx
0
(4.41)
Z L 2 Z L
d ue
0
EA
ðqÞWðxÞdx
WðxÞdx ¼ dx2
0
0
EA
4.4 Weak Formulation
If the approximate solution ueðxÞ satisfies Eq. (4.41), then ueðxÞ is very close to the
exact solution. While the weighting function W(x) can be any function, it still needs
to satisfy all of the boundary conditions. There are several ways for selecting the
weighting functions, as previously mentioned. For the Galerkin method, the preference is to use a linear combination of a set of n trial weighting functions Wi ¼ ji(x)
as the approximate solution set ueðxÞ, as shown in the following equation:
ueðxÞ ¼
n
X
c i ji ;
(4.42)
i¼1
where ci are constants. This trial solution looks identical to that shown in Eq. (2.1)
P
4x;y;z ¼ n1 Ni 4i , which is the interpolation shape functions that can be used to
interpolate any physical quantity 4. Comparing Eqs. (2.1) and (4.42), ci reflect the
shape functions, and ji mirror the nodal physical quantities. Indeed, this is the
same concept used in Galerkin’s method, which is the base for solving FE problems,
to be explained later in this section. Using the integration by parts method with
constants E, A, and q, as well as the two boundary conditions listed above, the
left-hand and right-hand sides of Eq. (4.41) become
Z L 2 Z L
d ue
de
u de
u dji
LHS: EA
dx
ðxÞdx
¼
EA
j
ðxÞ
EA
j
i
i
2
dx x from 0 to L
dx
0
0 dx dx
Z L
Z L
de
u de
u de
u dji
de
u dji
¼ EA ji ðLÞ
dx
¼
EA
dx
EA
j
ð0Þ
EA
i
dx x¼L
dx x¼ 0
0 dx dx
0 dx dx
(4.43)
and
Z
RHS : In Eq. (4.43), the term EA
0
L
Z
qji ðxÞdx ¼ q
u
ji ðLÞ ddxe
L
ji ðxÞdx.
(4.44)
0
vanishes because of the natural boundary
de
u condition du
ðx
¼
LÞ
¼
0,
while
the
term
EA
j
ð0Þ
disappears because of
i
dx
dx x¼L
x¼ 0
the essential boundary condition u(x ¼ 0) ¼ 0. By combining Eqs. (4.43) and
(4.44), we have
Z L
Z L
de
u dji
dx ¼ q
ji ðxÞdx.
(4.45)
EA
0 dx dx
0
181
182
CHAPTER 4 Element Stiffness Matrix
PN
dji
u¼
de
u
From Eq. (4.42), we can easily see that ddxe
i¼1 ci dx .Now by inserting dx into Eq.
(4.45), we have
! Z L X
Z L
N
djj
dji
EA
ci
jj ðxÞdx
dx ¼ q
dx
dx
0
0
i¼1
(4.46)
Z L Z L
N
X
djj
dji
0
ci EA
jj ðxÞdx.
dx ¼ q
dx
dx
0
0
i¼1
dj
dj
j
i
Note that the term dj
dx in the LHS of Eq. (4.45) is changed to dx for the second dx
entry in Eq. (4.46). In index notation, the same index is not allowed to appear
more than twice for any single entry. Because the index i has been used twice in
dj
i
the term ci dj
dx to signify the need for summation, the second dx entry has become
djj
dx . For those of you who are still confused about the interchangeable nature of index
notation, please refer to any textbooks with descriptions of continuum mechanics.
According to the index notation method, any repeated index (i in this case) represents a summation. Using this convention, there is (in essence) no need to keep the
summation sign in the equation. Still, this summation sign is intentionally made
redundant for those who have not yet studied the index notation commonly used
in continuum mechanics.
RL
djj
i
The term EA 0 dj
dx
dx dx in the bottom part of Eq. (4.46) can be interpreted
as the structure stiffness matrix of a bar element. The same idea can be applied to
only one element at a time within the whole structure that is discretized into a
number of elements. That is, Eq. (4.46) can also be used to determine the element
stiffness matrix [k] in the same manner as explained for the variational method.
Now, recall from Eq. (3.6) for the 2-node linear element, that the shape functions
based on the isoparametric formulation with natural coordinates of x ranging
from 1 to 1 are represented as N1 ¼ 1x
and N2 ¼ 1þx
2
2 , and the 1D Jacobian is
dx
L
expressed as dx ¼ 2.
Assume the approximate solution set ue within the element is
ueðxÞ ¼ N1 u1 þ N2 u2 .
(4.47)
In other words, the trial weighting functions ji(x) are the same as the element
shape functions Ni(x), and the nodal displacements u1 and u2 are of constant values.
To find [k], we need to first take the derivatives of Ni(x):
dN1 dN1 dx
1
¼ ;
¼
L
dx
dx dx
(4.48)
dN2 dN2 dx 1
¼ ; and
¼
dx
dx dx L
(4.49)
4.5 Derive Element Stiffness Matrix From Shape Functions
Z L Z L
dNj
dNi
½k ¼ EA
BT Bdx
dx ¼ EA
dx
dx
0
0
2
3
1
Z L 6 L 7
6
7 1
6
7
¼ EA
6
7
0 4 1 5 L
L
2
x
L2
6
6
1
dx ¼ EA6
6 x
L
4
L2
3L
x
L2 7
x
L2
2
1
7
EA
4
7 ¼
7
L
5
1
1
3
(4.50)
5.
1
0
Also, the element load vector is
8 9
3
1x
qL >
>
>
>
<
=
L
16
7
2
6 2 7 Ldx ¼
.
jj ðxÞdx ¼ q
ffe g ¼ q
41 þ x52
>
0
1
> qL >
>
:
;
2
2
Z
Z
2
(4.51)
As expected, Eqs. (4.50) and (4.51) are the same as Eqs. (4.36) and (4.39), even
though different methods are used to derive the element stiffness matrix and nodal
load vector.
4.4.3 SECTION SUMMARY
Based on these exercises, a quick summary of the findings can be made as follows:
•
•
•
•
•
Strong formulation provides the exact answers at nodal locations but is rarely
applicable, mostly due to the complex geometry.
Weak formulation is based on an integral of the strong formulation.
There are two popular weak-formulation methods used in the FE method, the
variational method and weighted residuals method.
The element shape functions can be used to derive the element stiffness matrix
and nodal load vector.
When both are applicable, the variational and weighted residuals methods result
in the same element stiffness matrix and element nodal load vectors.
4.5 DERIVE ELEMENT STIFFNESS MATRIX FROM SHAPE
FUNCTIONS
We have shown that the element stiffness matrix can be directly derived from the
element shape functions for a 1D, 2-node bar element. The same concept is applicable for all other element types. To prove this statement would require extensive
work and is beyond the scope of “basic finite element method” intended for this
book. A second aspect of this section is the usage of numerical integration to complete the formation of the element stiffness matrix and load vectors. In Section 4.4,
integration is performed based on fundamental calculus. However, there are all kinds
183
184
CHAPTER 4 Element Stiffness Matrix
of integration rules that might be encountered when using this approach. Thus, it
would be difficult to program all the rules into a computer language. Because the
FE method comprises a set of numerical procedures, it only makes sense to apply
numerical integration techniques instead of analytical methods.
4.5.1 GAUSS QUADRATURE
For a 2-node bar element, we can conclude from both Eq. (4.37) derived from the
variational method and Eq. (4.50) derived from the Galerkin method that the element
stiffness matrix can be derived from the shape functions, as recapped in the
following equation.
Z
Z 1
Z L
EAL 1 T
½BT ½Bdx ¼ EAj½Jj
½BT ½Bdx ¼
½B ½Bdx;
½k ¼ EA
2
0
1
1
where ½B ¼ d½N
dx . Also, Eqs. (4.39) and (4.51) show that the element nodal load
vectors derived from two separate weak formulations are
Z L Z 1 N1
N1
ffe g ¼
q
q
dx ¼
j½Jjdx.
N
N
0
1
2
2
In Section 4.4.2, we manually integrate the equations for [k] and [ fe] using
knowledge gathered, and learning the calculus methods needed to identify the
element shape functions and nodal load vectors. The integration for a bar element
is very simple. However, when the [B] matrix becomes more sophisticated, such
as what would be needed for a 2D, 4-node plane element, it becomes more difficult
to integrate the equation analytically. Because it is problematic to create computer
programs for sophisticated analytic calculus, and the FE method is itself a set of numerical procedures, it is only reasonable that the integration to identify the element
stiffness matrix is carried out numerically.
The purpose of integration of a function is, in essence, to find the area underneath
a curve. For example, integration of a linear velocityetime history function is the
distance traveled. Therefore numerical integration is basically a set of numerical
procedures designed to find integration values. Many available numerical integration
methods are reported, such as the Riemann integral (Riemann sum), trapezoidal rule,
and Simpson’s rule. However, these methods are only suitable for integrating
tabulated data with many data points, such as accelerationetime histories with a
very small delta t (time interval) between two adjacent data points. Using these
methods, the computational cost would be too high to afford when trying to create
the element stiffness matrix for all elements. There is a need to use a method that
requires considerably fewer data points to complete the integration with comparable
accuracy. Additionally, this method needs to accommodate integrations over the 2D
area or 3D volume needed to formulate 2D or 3D element types.
One such method is the Gaussian (or Gauss) quadrature rule, named after
German mathematician (Johann) Carl Friedrich Gauss (Apr. 1777eFeb. 1855),
4.5 Derive Element Stiffness Matrix From Shape Functions
according to the MacTutor History of Mathematics (O’Connor and Robertson).
Historically speaking, the mathematical term “quadrature” is used to describe the
determination of area, and therefore numerical integration is known as quadrature
or numerical quadrature. Like other numerical methods, the Gauss quadrature rule
turns an integration problem into a simple multiplication and summation question.
We suspect that at an early age, the idea of turning a complex mathematical operation into a more easily solvable operation had been in Gauss’ mind. At the age of
seven, Gauss was able to find the sum of the series of integers from 1 to 100 quickly
by using a formula of (1 þ 100) 50 ¼ 5050. His idea was to list these numbers
twice, one in a forward array (from 1 to n) and the other in a reversed order array
(from n to 1), as shown below:
Forward
1
2
3
: : :
99
100
Backward
Sum
100
101
99
101
98
101
: : :
: : :
2
101
1 .
101
By adding the two arrays together to form the third row, he found that each of the
pairs of the forward and backward data resulted in a constant value of n þ 1.
Because there are a total of n items and the array is repeated twice, the summation
of the series of numbers from 1 to n becomes
n
X
i¼1
1
i ¼ ½n ðn þ 1Þ.
2
Consider a definite integral I from 1 to 1 for a function f(x) that has the form
Z 1
I¼
f ðxÞdx.
(4.52)
1
Using the middle Riemann sum method studied in calculus, an integral can be
approximated by 2 f(0) using one division, and 0.25 [ f(0.75) þ f(0.25) þ
f(0.25) þ f(0.75)] if four divisions are used. To obtain reasonable accuracy, more
complex functions require more divisions. The Gauss quadrature rule has a similar
flavor to Gauss’ summation method in that the complex operation can be done using
simple operations, and the rule is based on a variation of the Riemann sum method.
For this rule, multiplication and summation can be used to find the integral, as shown
in the equation below:
Z 1
m
X
I¼
f ðxÞdxz
Wi f ðxi Þ;
(4.53)
1
i¼1
where Wi is the weight at the integration point i, and f(xi) is the function evaluated at
the same integration point i. As long as the integration points and associated weights
Wi are identified, the definite integral can be found. Clearly, the difference between
the Riemann sum and Gauss quadrature are (1) the Riemann sum method applies
equal weight to each division while the Gauss quadrature utilizes different weighting
185
186
CHAPTER 4 Element Stiffness Matrix
factors for different divisions, and (2) the division for Gauss quadrature does not
need to be equal in size. More explanations will follow in the selection of Gauss integration points.
For a linear function f(x) ¼ a1 þ a2x, the exact solution when integrated from 1
to 1 is
Z 1
Z 1
f ðxÞdx ¼
ða1 þ a2 xÞdx ¼ 2a1 .
(4.54)
1
1
The Gauss quadrature rule mandates that for a polynomial of 2m 1 order, a total of m integration points and m weights must be used. Based on this rule, this linear
polynomial integral needs to be evaluated at only one integration point, x1
ð2m 1 ¼ 10m ¼ 1Þ, where f(x1) and the integral can be written as
f ðx1 Þ ¼ a1 þ a2 x1 and
Z
1
1
f ðxÞdxz
1
X
Wi f ðxi Þ ¼ a1 W1 þ a2 W1 x1 .
(4.55)
(4.56)
i¼1
For Eq. (4.56) to be identical to Eq. (4.54), we must have W1 ¼ 2 and x1 ¼ 0.
Therefore the integration point (or Gauss point) for integrating a linear function is
located at x ¼ 0 and has a weight of 2.
Next, we assume a third order polynomial function f(x) ¼
a1 þ a2x þ a3x2 þ a4x3. The Gauss quadrature rule requires the use of two integration points x1 and x2 ð3 ¼ 2m 10m ¼ 2Þ, with two weights W1 and W2.
The exact solution has the form
Z 1
Z 1
2
a1 þ a2 x þ a3 x2 þ a4 x3 dx ¼ 2a1 þ a3 .
f ðxÞdx ¼
(4.57)
3
1
1
The Gauss quadrature has the form
Z
1
1
f ðxÞdxz
2
X
Wi f ðxi Þ ¼ W1 f ðx1 Þ þ W2 f ðx2 Þ
i¼1
¼ a1 ðW1 þ W2 Þ þ a2 ðW1 x1 þ W2 x2 Þ þ a3 W1 x1 2 þ W2 x2 2 þ a4 W1 x1 3 W2 x2 3 .
(4.58)
Comparing Eqs. (4.57) and (4.58), we must have
W1 þ W2 ¼ 2;
W1 x1 þ W2 x2 ¼ 0;
2
W1 x1 2 þ W2 x2 2 ¼ ; and
3
W1 x1 3 þ W2 x2 3 ¼ 0:
4.5 Derive Element Stiffness Matrix From Shape Functions
Table 4.1 Location and Weighting Factors for 1- and 2-Point Gauss
Quadratures
No. of Gauss Point
Location of Gauss Point
1
2
x¼0
x¼
p1ffiffiz0:57735
3
Weighting Factor
2
1
1ffiffi, and x ¼ p1ffiffi. Note
By solving these four equations, we have W1 ¼ W2 ¼ 1, x1 ¼ p
2
3
3
that p1ffiffiz0:57735, and these two representations are used interchangeably during nu3
merical evaluations.
So, the integration points (or Gauss points) for integrating a third-degree polynomial function are located at x ¼ p1ffiffi, each with a weight of 1. Because only the
3
simplest element types are described in the early chapters of this book, Table 4.1 lists
the Gauss points and corresponding weighting factors for up to a 2-point Gauss
quadrature. Formulas that need up to five points are considered in Section 8.2.1,
where Table 8.1 contains a longer list.
Example R4.4
1
Find I ¼ 1 2x2 þ x3 dx using 1- and 2-point Gauss quadrature.
Solution
R
xnþ1 . Hence, the exact solution is
Recall that the indefinite integral xn dx ¼ nþ1
1
R1 I ¼ 1 2x2 þ x3 dx ¼ 23x3 þ 14x4
¼ 43. This solution is only provided for
1
checking the accuracy of the Gauss integration.
For 1-point Gauss quadrature, we evaluate the integral at x ¼ 0 with a
weighting factor of 2, in accordance with Table 4.1. Hence, f ðxÞx¼0 ¼
2
2x þ x3 x¼0 ¼ 0. Therefore the resulting integral is I z W1 f(x ¼ 0) ¼
2 f(x) ¼ 2 0 ¼ 0, which is obviously inadequate. The largest exponent is 3,
therefore similar to the previous example, 3 ¼ 2m 10m ¼ 2. We need a 2point Gauss quadrature.
For 2-point Gauss quadrature, both weighting factors W1 ¼ W2 ¼ 1, and
x ¼ p1ffiffi, the integral is
3
1
1
IzW1 f x ¼ pffiffiffi þ W2 f x ¼ pffiffiffi ¼
3
3
2 3 !
1
1
þ
2 pffiffiffi pffiffiffi
3
3
3 !
1
þ pffiffiffi
;
3
2
1
2 pffiffiffi
3
187
188
CHAPTER 4 Element Stiffness Matrix
which simplifies to
Iz
3 !
2
1
þ
pffiffiffi
3
3
3 !
2
1
4
¼ ;
þ pffiffiffi
3
3
3
as expected.
4.5.2 1D ELEMENT STIFFNESS MATRIX USING GAUSS QUADRATURE
4.5.2.1 Stiffness Matrix for a Bar Element
For a bar element with a length L, the element stiffness matrix is
R1
T
½k ¼ EAL
1 ½B ½Bdx, where
2
dN dN dx 2 d½ð1 xÞ=2 d½ð1 þ xÞ=2
1 1
¼
¼
½B ¼
and (4.59)
¼
dx
dx dx L
dx
dx
L L
2 1 1 3
Z 1
Z 1
6 L2 L2 7
EAL
EAL
6
7dx.
(4.60)
½k ¼
½BT ½Bdx ¼
4
2
2
1 5
1
1 1
L2 L 2
R1
1 1
EA
Because L is constant, Eq. (4.60) can be rearranged as ½k ¼ 2L 1
dx.
1 1
R1
R1
For 1-point Gauss integration (x1 ¼ 0, W1 ¼ 2), 1 1 dxz2 1 ¼ 2 and 1 ð1Þ
1 1
dxz2 ð1Þ ¼ 2 Hence, ½k ¼ EA
: This result demonstrates that
L
1 1
using Gauss quadrature to find the element stiffness matrix requires no calculus
background, and it is very easy to program this procedure into a computer.
4.5.2.2 Stiffness Matrix of a Beam Element
To find the element stiffness matrix of a beam element, we can use the straine
displacement matrix [B] in Eq. (2.43) or (3.19), as shown below:
6 12x
4 6x
6 12x
2 6x
½B14 ¼ 2 þ 3 þ 2
3 þ 2
L L
L L
L
L
L2
L
6x 3x 1 6x 3x þ 1
½B ¼
L
L
L2
L2
Eq. (4.20) shows the strain energy function, and Eq. (2.42) expresses the strain
within a beam. Both equations are repeated as follows:
Z
1
U¼
sT εdV
2 V
4.5 Derive Element Stiffness Matrix From Shape Functions
du
d2 w
d2
¼ z 2 ¼ z 2 ½ N1 N2
dx
dx
dx
¼ z½Bf w1 q1 w2 q2 gT
εxx ¼
N3
N4 f w1
q1
w2
q 2 gT
If we use the generalized displacement {w} to represent f w1 q1 w2 q2 gT , the
strain energy function for a linear isotropic beam can be written as
Z
Z
Z
1
1
1 T
T
T
T
T
2
s εdV ¼
Ezfwg ½B z½BfwgdV ¼ ½w
½B E½Bz dV fwg.
U¼
2 V
2 V
2
V
(4.61)
For a beam with the axis coinciding with the x-axis and the cross section lying
on the yez plane, the area moment of inertia, or second moment of inertia, can
Rh
Rh
be expressed as Iyy ¼ 2 h z2 dA ¼ 2 h z2 dydz, where h is the height of the beam,
2
2
dA ¼ dydz, and dV ¼ dAdx. Because the axis of the beam element coincides
with the x-axis, [B] matrix is a function of x only, and hence is not affected by
R
R
RL
R
integration over dA. Thus, V ½BdV ¼ 0 ½Bdx and V z2 dV ¼ V z2 dAdx ¼
R L R h2
0
h2
z2 dydz dx ¼
RL
0 Iyy dx.
Therefore the strain energy function shown in
Eq. (4.61) can be written as
2
3
Z L
1
½BT EI½Bdx5fwg;
U ¼ fwgT 4
2
0
(4.62)
where Iyy above is simplified as I. Because {w} represents the generalized displacement for a beam element, the strain energy of the beam has the form
U ¼ 12fwgT ½kfwg. Comparing this equation with Eq. (4.62) yields the element stiffness matrix for this 2-node beam element as
Z 1
Z L
½BT EI½Bdx ¼
½BT EI½Bj½Jjdx.
(4.63)
½k ¼
0
1
T
Because ½B ½B is a second order polynomial, a 2-point Gauss integration needs
to be performed to obtain the integral. Recall from Eq. (4.53) that
R1
1ffiffi, x ¼ p1ffiffi, and W ¼ W ¼ 1 as
p
1
2
2
1 f ðxÞdxzW1 f ðx1 Þ þ W2 f ðx2 Þ, where x1 ¼
3
3
shown in Table 4.1. For a beam with a constant elastic modulus and area moment
of inertia, the derivation of the element stiffness matrix starts with
189
190
CHAPTER 4 Element Stiffness Matrix
2
6x
L2
6
6
6
6
6 3x 1
Z 1
Z 16
6
EIL
EIL
6 L
½k ¼
½BT ½Bdx ¼
6
2 1
2 1 6 6x
6
6 L2
6
6
6
4 3x þ 1
L
3
7
7
7
7
7
7
7 6x
7
7 2
7 L
7
7
7
7
7
5
3x 1
L
6x
L2
3x þ 1
dx.
L
Next, we multiply the 41 [B]T matrix with the 14 [B] matrix to obtain the 44
matrix
2
3
36x2
18x2 6x
36x2
18x2 þ 6x
6
7
6
7
L3
L3
L4
L4
6
7
6
7
2
2
2
2
18x þ 6x
9x 1 7
6 18x 6x ð3x 1Þ
7
Z 16
6
7
EIL
L3
L3
L2
L2
6
7 dx.
½k ¼
6
7
2 1 6 36x2
18x2 þ 6x
36x2
18x2 6x 7
6
7
6
7
L3
L3
L4
L4
6
7
6
7
2
2
2
4 18x2 þ 6x
9x 1
18x 6x ð3x þ 1Þ 5
L3
L2
L3
L2
Step-by-step derivations of the 2-point Gauss quadrature from this equation are
listed in the exercise section for reference. The final stiffness matrix for the beam
element calculated from Gauss quadrature is
2
3
12 6L
12 6L
6
7
6L 2L2 7
EI 6 6L 4L2
7.
½k ¼ 3 6
(4.64)
L 6
12 6L 7
4 12 6L
5
6L
2L2
6L
4L2
As expected, Eq. (4.64), which is based on Gauss quadrature, is the same as Eq.
(4.11). Obviously, the Gauss quadrature method is much easier than the direct
method mentioned in Section 4.2, and it can easily be programmed with computer
language.
4.5.3 GAUSS INTEGRATION POINTS FOR 2D AND 3D ELEMENTS
The natural coordinate system selected for defining a 3-node triangular element is
the area ratio coordinate system (Section 3.4). A right triangle is chosen as the
template for this isoparametric mapping. Fig. 4.7 shows mapping using 1- and
3-point Gauss integration, where the corresponding weights are 1 and 1/3.
4.5 Derive Element Stiffness Matrix From Shape Functions
FIGURE 4.7
Locations of the 1- and 3-point Gauss integration point(s) for a triangular element.
When using Gauss integration points for a 2D quadrilateral element, we need to
consider both axes. To account for the 2D nature of the element, the Gauss quadrature has the form
Z 1 Z 1
n X
m
X
(4.65)
f ðx; hÞdxdhz
Wi Wj f xi ; hj .
1
1
i¼1 j¼1
For a 3D, 1-point Gauss quadrature, the integration point is located at x ¼ 0,
h ¼ 0 and the corresponding weight is 4 (a weight of 2 in the x-axis and 2 in the
h-axis results in 2 2 ¼ 4 in total). Fig. 4.8 shows representations of 4-point
(2 2) and 9-point (3 3) Gauss integration. As shown in Eq. (4.65), the product
of the weights from both the x- and h-axes needs to be used. Similar to the 2D
FIGURE 4.8
Locations of Gauss integration points for a 2D, 4-node plane element with 2 2 and
3 3 integration schemes.
191
192
CHAPTER 4 Element Stiffness Matrix
quadrilateral element, the Gauss quadrature for the 3D solid element can be obtained
by simply expanding Eq. (4.65) to include the z-axis as
Z 1 Z 1 Z 1
n X
m X
r
X
f ðx; h; zÞdxdhdzz
Wi Wj Wk f xi ; hj ; zk .
1
1
1
i¼1 j¼1 k¼1
(4.66)
For a 3D, 1-point Gauss quadrature, the integration point is located at x ¼ 0,
h ¼ 0, and z ¼ 0, while the corresponding weight is 2 2 2 ¼ 8. A higher number of integration points include 8-point (2 2 2) and 27-point (3 3 3) integrations. Note that we always choose the same number of integration points for each
axis for demonstration purposes. While this is practical, in order to avoid hourglass
modes (to be discussed in Section 4.5.6), we may use a different number of integration points on different axes for some types of elements.
Example 4.5
R1 R1
Find the integral of I ¼ 1 1 3þ2x
5þh using 1-point (1 1) and 4-point (2 2)
Gauss quadrature.
Solution
The exact solution is
Z
I¼
1
1
Z
3 þ 2x
dxdh ¼
1 5 þ h
1
Z
1
6
dh ¼ 6 ln ð5 þ hÞ11 ¼ 6½ln 6 ln 4 ¼ 2:433:
5
þ
h
1
1-point integration: w ¼ 2 2 ¼ 4, x1 ¼ h1 ¼ 0, and
Iz4 3
¼ 2:4:
5
The weights for the 2D, 4-point integration are:
and the Gauss
wi ¼1 1 ¼ 1 points are located at: Gi ¼ ðxm ; hn Þ ¼
1ffiffi . Hence, the integral is
p1ffiffi; p
3
1ffiffi; p
1ffiffi
p
3 3
;
1ffiffi; p1ffiffi
p
3 3
;
p1ffiffi; p1ffiffi
3 3
;
3
3 2 0:577 3 2 0:577 3 þ 2 0:577 3 þ 2 0:577
þ
þ
þ
¼ 2:432:
Iz
5 0:577
5 þ 0:577
5 þ 0:577
5 0:577
4.5.4 2D AND 3D ELEMENT STIFFNESS MATRICES USING GAUSS
QUADRATURE
While analytically integrating the element shape functions to find the [B] matrix
and the stiffness matrix [k] for a bar element is simple, similar integrations to
find the [k] matrix in 2D and 3D could become very tedious. Two primary causes
of the tediousness are that the dimensions of the [B] matrix become larger, and the
4.5 Derive Element Stiffness Matrix From Shape Functions
elements may not be aligned with the global axes. As such, Gauss quadrature is
used to simplify this task. As mentioned previously, all element-related information needs to be outlined in the same global coordinate system for assembly into
a set of global forceedisplacement equations before the FE software can be
used to calculate the nodal displacements. In this section, we outline the procedures needed to formulate the [k] matrix from the [B] matrix prescribed in the natural coordinate system and the nodal coordinates provided in the global coordinate
system.
4.5.4.1 2D Plane Element Stiffness Matrix
Assume a quadrilateral plane element has a constant thickness of t. Because this
element has a total of eight DOFs, the element stiffness matrix must have dimensions of 88. From the strain energy formula, the element stiffness matrix can be
written as
Z
ZZ
Z 1 Z 1
½k88 ¼ ½BT ½E½BdV ¼
½BT ½E½BtdA ¼ t
½BT ½E½Bj½Jjdxdh;
V
A
1
1
(4.67)
where [B] and j[J]j are both evaluated at (x, h). We know from Example 4.4 that a
2D, 4-point (2 2) Gauss quadrature of a function f(x, h) has the form
Iz
2 X
2
X
Wi Wj f xi ; hj
i¼1 j¼1
¼ W1 W1 f ðx1 ; h1 Þ þ W1 W2 f ðx1 ; h2 Þ þ W2 W1 f ðx2 ; h1 Þ þ W2 W2 f ðx2 ; h2 Þ.
(4.68)
Because W1 ¼ W2 ¼ 1 for a (2 2) Gauss quadrature, we can bypass the weights
in the previous equation and write
½k88 ¼ ½Bðx1 ; h1 ÞT ½E½Bðx1 ; h1 Þj½Jðx1 ; h1 Þj þ ½Bðx1 ; h2 ÞT ½E½Bðx1 ; h2 Þj
½Jðx1 ; h2 Þj þ ½Bðx2 ; h1 ÞT ½E½Bðx2 ; h1 Þj½Jðx2 ; h1 Þj
þ ½Bðx2 ; h2 ÞT ½E½Bðx2 ; h2 Þj½Jðx2 ; h2 Þj;
(4.69)
where (x1, h1) ¼ (0.577, 0.577), (x1, h2) ¼ (0.577, 0.577), (x2, h2) ¼ (0.577,
0.577), and (x2, h1) ¼ (0.577, 0.577). The following sections describe the way
to determine each component needed to calculate [k] from Eq. (4.69).
4.5.4.1.1 The [E] Matrix
In Eq. (4.69), the simplest term is the [E]33 matrix, which comes directly from
the corresponding material law. For a plane stress element with a constant
elastic modulus E and Poisson’s ratio y, the [E] matrix can be determined from
the Young’s modulus and Poisson’s ratio using
193
194
CHAPTER 4 Element Stiffness Matrix
2
8
9
1
>
6
< sxx >
=
E 6
6y
syy ¼
26
>
>
:
; 1y 4
sxy
0
y
1
0
3
2
8
9
1
7>
6
< εxx >
=
7
6
E
0 7 εyy 0½E ¼
6y
7>
26
>
1
y
:
;
5
4
1y
gxy
0
2
0
y
1
0
0
3
7
0 7
7
7
1 y5
2
(4.70)
4.5.4.1.2 The Jacobian
The second easiest term to compute in Eq. (4.69) is the Jacobian. Recall that the
3
2
vx vy
6 vx vx 7
7
6
Jacobian matrix for a 2D plane element is ½Jðx; hÞ ¼ 6
7. Because we
4 vx vy 5
vh vh
P4
P4
know that x ¼ 1 Ni xi and y ¼ 1 Ni yi , all four entries of the Jacobian matrix
can be calculated from the known nodal coordinates xi and yi, as shown in the
following four equations:
vx
v
¼ ðN1 x1 þ N2 x2 þ N3 x3 þ N4 x4 Þ
vx vx
1
¼ ½ ð1 hÞx1 þ ð1 hÞx2 þ ð1 þ hÞx3 ð1 þ hÞx4 4
vy
v
¼ ðN1 y1 þ N2 y2 þ N3 y3 þ N4 y4 Þ
vx vx
1
¼ ½ ð1 hÞy1 þ ð1 hÞy2 þ ð1 þ hÞy3 ð1 þ hÞy4 4
vx
v
¼
ðN1 x1 þ N2 x2 þ N3 x3 þ N4 x4 Þ
vh vh
1
¼ ½ ð1 xÞx1 ð1 þ xÞx2 þ ð1 þ xÞx3 þ ð1 xÞx4 4
vy
v
¼
ðN1 y1 þ N2 y2 þ N3 y3 þ N4 y4 Þ
vh vh
1
¼ ½ ð1 xÞy1 ð1 þ xÞy2 þ ð1 þ xÞy3 þ ð1 xÞy4 4
We can find the determinant of the Jacobian matrix, or the Jacobian, by using the
vy
vy
vx
equation j½Jj ¼ vx
vx vh vh vx. Using this equation, we can calculate the four Jacobians from the Gauss integration points, j½Jðx1 ; h1 Þj; j½Jðx1 ; h2 Þj; j½Jðx2 ; h1 Þj;
4.5 Derive Element Stiffness Matrix From Shape Functions
and j½Jðx2 ; h2 Þj. This exercise has been previously performed in Example 3.7 for
finding the Jacobian ratio. To find the element stiffness matrix, the results of the
four Jacobians need to be substituted into Eq. (4.69). This concept is further demonstrated in the following numerical example.
Example 4.6
Consider the previous example for a four-node plane element with nodal coordinates P1(0, 0), P2(6, 0), P3(6, 4.2), and P4 (0, 4). Calculate the four Jacobians using a (2 2) Gauss quadrature rule.
Solution
This is a problem that should be solved with a numerical tool, not through
manual calculation. These numerical tools can range from simple tools like
spreadsheets to more sophisticated ones like MATLAB (MathWorks, Natick,
MA) or Mathcad (PTC, Needham, MA). Here, manual calculation for the first
term is listed to provide additional aid regarding the process. To find
j½Jðx1 ; h1 Þj, we first determine the four entries of the Jacobian matrix by
substituting (x, h) as (x1,h1) ¼ (0.577, 0.577) into the equations that follow,
and then the Jacobian at (x1, h1) can be obtained. The final step, which is left to
you, is to sum all the Jacobians.
P
vx vð Ni xi Þ 1
¼
¼ ½ ð1 hÞx1 þ ð1 hÞx2 þ ð1 þ hÞx3 ð1 þ hÞx4 ¼ 3
vx
vx
4
P
vy vð Ni yi Þ 1
¼
¼ ½ ð1 hÞy1 þ ð1 hÞy2 þ ð1 þ hÞy3 ð1 þ hÞy4 ¼ 0:02115
vx
vx
4
P
vx vð Ni xi Þ 1
¼
¼ ½ ð1 xÞx1 ð1 þ xÞx2 þ ð1 þ xÞx3 þ ð1 xÞx4 ¼ 0
vh
vh
4
P
vy vð Ni yi Þ 1
¼
¼ ½ ð1 xÞy1 ð1 þ xÞy2 þ ð1 þ xÞy3 þ ð1 xÞy4 ¼ 2:02115
vh
vh
4
j½Jðx1 ; h1 Þj ¼
vx vy vy vx
¼ 6:06345
vx vh vx vh
j½Jðx1 ; h2 Þj ¼ 6:06345; j½Jðx2 ; h2 Þj ¼ 6:23655; j½Jðx2 ; h1 Þj ¼ 6:23655
P
Note that 4i¼1 det½Ji ¼ 24:6, which is the same as the area of the trapezoid
defined by the given coordinates for this example. We can easily understand that
the Jacobian calculated at each Gauss integration point is related to the area ratio,
that is, the ratio of the area represented by each individual Jacobian to the area of
the total trapezoid. The sum of the Jacobians for all four integration points is the
total area of this plane element.
195
196
CHAPTER 4 Element Stiffness Matrix
4.5.4.1.3 The [B] Matrix
The most cumbersome terms to compute in Eq. (4.69) are the [B]T and [B] matrices.
From Eq. (3.43), the previously shown [B] matrix is now expanded as
2
3
vy v vy v
0
6
7
6 vh vx vx vh
7
6
7
6
7"
#
6
N
0 N2 : : 0
1 6
vx v
vx v 7
7 1
0
½B38 ¼
6
7
vx vh vh vx 7 0 N
j½Jj 6
0 : : N4 28
1
6
7
6
7
6 vx v
7
vx v vy v vy v 5
4
vx vh vh vx vh vx vx vh
3
2
6 vy vN1 vy vN1
6 vh vx vx vh
6
6
1 6
6
0
¼
6
j½Jj 6
6
6 vx vN1 vx vN1
6
4 vx vh vh vx
: :
0
vx vN1
vx vh
vx
vh
vN1
vx
: :
vy vN1
vh vx
vy
vx
vN1
vh
: :
vy vN4
vh vx
vy
vx
vN4
vh
0
vx vN4
vx vh
vx
vh
vN4
vx
0
vx vN4
vx vh
vx
vh
vy vN4
vh vx
vy
vx
7
7
7
7
vN4 7
7
vx 7
7
7
7
vN4 7
vh 5
.
38
(4.71)
Because all four entries
vx
vy vx vy vx; vx; vh; vh
in the Jacobian matrix have already been
computed at each Gauss point, as shown in Example 4.5, the only items left to
vNi
i
compute are vN
vx , and vh .
vN1 ð1 hÞ vN2 ð1 hÞ vN3 ð1 þ hÞ vN4 ð1 þ hÞ
;
;
;
¼
¼
¼
¼
4
4
4
4
vx
vx
vx
vx
vN1 ð1 xÞ vN2 ð1 þ xÞ vN3 ð1 þ xÞ vN1 ð1 xÞ
;
;
;
¼
¼
¼
¼
4
4
4
4
vh
vh
vh
vh
It is easy to make a mistake when doing manual calculations for this type of
work. Hence, a computer program should be written. However, to demonstrate the
concept, we calculate B11 in Eq. (4.71) using the same nodal coordinates as those
in Example 4.5.
1 vy vN1 vy vN1
B11 ðx1 ; h1 Þ ¼ B11 ð0:577; 0:577Þ ¼
j½Jj vh vx
vx vh
ðx1 ;h1 Þ
1
½2:02115 ð0:39425Þ 0:02115 ð0:39425Þ ¼ 0:13
6:06345
Similarly, we can use the same set of procedures to determine B12, B13, ., B37, B38,
which are the remaining components of the [B(x1, h1)] matrix. We then use the same
procedures to determine [B(x1, h2)], [B(x2, h2)], and [B(x2, h1)] for the other three
¼
4.5 Derive Element Stiffness Matrix From Shape Functions
Gauss integration points. Now, recall Eq. (4.69), where the element stiffness matrix
[k] is written as
½k88 ¼ ½Bðx1 ; h1 ÞT ½E½Bðx1 ; h1 Þj½Jðx1 ; h1 Þj þ ½Bðx1 ; h2 ÞT ½E½Bðx1 ; h2 Þj
½Jðx1 ; h2 Þj þ ½Bðx2 ; h1 ÞT ½E½Bðx2 ; h1 Þj½Jðx2 ; h1 Þj
þ ½Bðx2 ; h2 ÞT ½E½Bðx2 ; h2 Þj½Jðx2 ; h2 Þj
We have all the numerical values needed to complete the calculation; therefore as
long as all the nodal coordinates, Young’s modulus, and Poisson’s ratio are available,
we can determine all numerical entries of the element stiffness matrix. Obviously, a
computer is needed for this task.
4.5.4.2 3D Element Stiffness Matrix
A 3D hexahedral element has a total of 24 DOFs. Hence, the dimension of the
element stiffness matrix must be 2424. Like a 2D element, the stiffness matrix
for a 3D element can be written as
ZZZ
Z 1 Z 1 Z 1
T
½B ½E66 ½Bdx dy dz ¼
½BT ½E½Bj½Jjdx dh dz.
½k2424 ¼
1
1
1
(4.72)
In a similar manner to previous examples, we will use the Gauss quadrature
method to numerically integrate Eq. (4.72) to determine the Jacobians and element
stiffness matrix. Taking two Gauss points for each direction yields a total of eight
integration points. These eight Gauss points are located at
ð 0:577; 0:577; 0:577Þ; ð0:577; 0:577; 0:577Þ; ð0:577; 0:577; 0:577Þ;
ð0:577; 0:577; 0:577Þ; ð0:577; 0:577; 0:577Þ; ð0:577; 0:577; 0:577Þ;
ð0:577; 0:577; 0:577Þ; and ð0:577; 0:577; 0:577Þ.
Let f ðx; h; zÞ ¼ ½BT ½E½Bj½Jj.
(4.73)
The element stiffness matrix can be obtained using the following equation:
Z 1 Z 1 Z 1
2 X
2 X
2
X
½k2424 ¼
(4.74)
fdx dh dzz
Wi Wj Wk f xi ; hj ; zk
1
1
1
i¼1 j¼1 k¼1
4.5.5 FULL AND REDUCED INTEGRATION
From several aforementioned examples, it is clearly noted that Gauss quadrature using fewer Gauss points may result in less accurate results. The minimum required
number of Gauss points can be determined from the order of polynomials. If the
Gauss points selected meet this minimum requirement, the integration is called
full integration. If a lower number of Gauss point is chosen, this integration scheme
is called reduced integration. It seems obvious that full integration should always be
197
198
CHAPTER 4 Element Stiffness Matrix
used for the best accuracy of model predictions. Unfortunately, this is not always the
case. For this reason, we need to discuss different integration schemes.
As P
seen in Example
P 4.6, the four
P entries in the
P Jacobian matrix are
v
N
x
v
N
y
v
N
x
v
Ni y i Þ
ð
Þ
ð
Þ
ð
Þ
ð
i
i
i
i
i
i
vy
vy
vx ¼
vx ¼
; vx ¼
; vh
; vh ¼
. Hence, the [B]
vx
vx
vx
vh
vh
matrix, which is calculated from the Jacobian shown in Eq. (4.71), consists of
only linear polynomial terms. However, the function f(x, h, z) in Eq. (4.73) involves
[B]2. For this reason, it is more accurate to use two Gauss points for integration along
each direction, that is, full integration of a 3D brick element requires 8 Gauss
points, 2 Gauss points along x-, h-, and z-directions each for a 2 2 2 ¼ 8point integration. Another integration scheme, termed “reduced integration,”
uses only 1-point for integrating the 1D, 2-node bar element; 2D, 4-node quadrilateral element; and 3D, 8-node solid element. In between the 8-point fullintegration scheme and 1-point reduced-integration scheme for a 2D plane or 3D
brick element, there is a selective-reduced-integration scheme with a selective
reduction from 2- to 1-point integration along one (for 2D or 3D problems) or
two (for 3D problems) preselected direction(s). In the event that a 3D model requires a higher accuracy along the x-axis while lower accuracy is sufficient for
the other two axes, a 2-point Gauss integration along the x-axis and only 1-point
Gauss integration along the h- and z-axes may be used. In this case, the total number of Gauss points is 2 1 1 ¼ 2.
From the previous exercise used to identify the numerical entries for the [B] matrix, we can easily imagine that the computational cost would be significantly lower
when the reduced-integration scheme is adopted. In the reduced-integration scheme,
the integration point is located at the center of the element. Hence, the strains/
stresses are only calculated at the center of the element, unless analytical methods
are used.
Knowing that using more Gauss points tends to improve the integration accuracy of polynomial equations, it seems that an FE model based on the reducedintegration scheme would provide less accurate results. However, this is not the
case, because the additional Gauss points used in the full-integration scheme provide some resistance to those deformation modes not supported by the reducedintegration scheme. In other words, an FE model that is developed based on the
full-integration scheme usually has a stiffer performance than that calculated
from the analytical solution. This phenomenon of producing a much smaller
displacement field (sometimes a reduction by several orders of magnitude) due
to the artificial stiffening effect induced by the full-integration scheme is called
“mesh locking.” Many scholarly articles have been devoted to this subject. Interested readers should consider reading these articles in order to get a better understanding of this phenomenon.
4.5.6 ZERO-ENERGY MODE
With better accuracy and lower computational costs, it would seem evident that the
reduced-integration scheme should always be used. Unfortunately, there is an
4.5 Derive Element Stiffness Matrix From Shape Functions
undesirable effect associated with this scheme. If neglected or improper constraints
(boundary conditions) are applied to the FE model, there may be instances where
all nodes have identical displacements, and therefore there are no strains within the
element. This situation, where there are some obvious nodal displacements but no
actual deformation within the element(s), and in which zero strain (hence zero energy)
is yield, is known as the zero-energy mode. Fig. 4.9 shows patterns of “rigid body
deformation,” in a 2D plane element. The deformations include translations along
the x- or y-axis, and rotation about the z-axis. In these modes of deformation, the dimensions and shape of the element remain unchanged. Because all nodes have identical displacements, there are no strains within the element. As mentioned earlier,
neglecting or using improper constraints for the model causes the zero-energy
mode. By correcting the boundary conditions, the problem should disappear.
There are other deformation modes that produce no strain energy when using the
reduced-integration scheme for 4-node plane and 8-node solid elements, as shown in
Fig. 4.10. In general, the strains calculated at the Gauss points are reasonably accurate. However, no such assertion can be made for areas away from the integration
points. Additionally, when the reduced 1-point integration scheme is applied, the
single Gauss point may not be able to resist the “bending-like” deformation
mode, which is described as one side of the element being in compression, while
the other side is in tension. In this case, there will be no strain at the single Gauss
point (center of the element), and hence these deformation modes produce zero
strain energy. This phenomenon is often described with other terms, including
spurious mode, hourglass mode, and instability.
It is termed spurious mode since the deformation mode is physically impossible.
When a group of elements undergoes bending-like deformation, the combined
deformed shape looks like a series of hourglasses linked to each other, hence it is
named the hourglass mode. Fig. 4.11 demonstrates the hourglass mode when a femur
is loaded in a 3-point bending configuration. The loading configuration in this case
exhibits a style of point loading, which is known to trigger the occurrence of hourglass modes much more easily than what would occur from distributed loading.
Because of the way a triangular (2D) or tetrahedral (3D) element is formulated,
no hourglassing occurs for these element types. To prevent the impractical hourglass
deformation modes from happening in other element types, proper hourglass control
FIGURE 4.9
Zero strain energy produced when the element has only rigid body motions. Left:
Translation along the y-axis. Middle: Translation along the x-axis. Right: Rotation about the
z-axis.
199
200
CHAPTER 4 Element Stiffness Matrix
FIGURE 4.10
Top: Two bending-like deformation modes produce no strain energy in a 4-node plane
element. Bottom: Two exemplary zero-energy deformation modes for a solid element.
FIGURE 4.11
A femur loaded in a 3-point bending configuration. Without a proper hourglass control,
the deformed shape looks like a series of hourglasses linked together.
is needed. The two most popular ways of reducing the chances of hourglassing are
adding a small elastic stiffness and viscous damping. You should consult the user
manual provided by the software vendor to select the most suitable hourglass control
method to avoid such undesirable deformations. Obviously, additional energy is
needed to prevent hourglassing, and this needs to be monitored closely. In general,
the ratio of hourglass energy to the peak internal energy should be evaluated to determine if the energy spent to control hourglassing is reasonable. Ratios less than 10%
for the entire FE model and less than 10% for each part are deemed acceptable.
Considering both ratios is to prevent some special cases where a small part within
the entire model may have produced a very high percentage of hourglassing, but
this locally seen high hourglass energy would not have contributed a significant
portion in the total hourglass energy of the overall model, due to its small volume.
4.6 Method of Superposition
4.6 METHOD OF SUPERPOSITION
The bar, truss, and beam elements discussed so far are idealized structures not
frequently found in the real world. For example, a real-world frame can resist an
axial load in ways that are not accounted for by a beam element. Furthermore, a
real-world beam resists not only vertical deflection (that produces rotation of the
neutral axis), but also the torsional load (rotation about the axis of the beam). The
first step in determining element stiffness matrices for these real-world structures
is to formulate individual isoparametric element shape functions [N] from the natural coordinate system. The strainedisplacement matrix [B] can then be derived from
the shape functions. By integrating the ½BT ½E½B matrices, we calculate the element
stiffness matrix [k]. The formulation processes are discussed in previous sections. As
an example, the isoparametric element shape functions of any element type can be
derived by setting one DOF to unity at a time while all other DOFs are zero. This
work is very tedious. An application of the superposition principle is a simpler
way to achieve the same goal.
Within the superposition principle is the implication that the overall response of a
structure subjected to multiple loading conditions can be obtained by summing all
the responses calculated from each individual loading. For example, the deformation
of a beam subjected to force and moment can be calculated by combining the deflections due to force alone and moment alone. For application of this principle to result
in adequate accuracy, all individual systems involved must be linear.
4.6.1 STIFFNESS MATRIX OF A 2D FRAME ELEMENT
An example in which this principle is used is the development of the stiffness matrix
for a pseudo-3D, 2-node frame element, which consists of the combined properties of
a beam and a bar. Although this element type is actually a 1D element, it can be positioned in any orientation for connection with other elements within the 3D structure.
As mentioned previously, a 1D, 2-node bar element has two uniaxial translational DOFs and a 1D, 2-node beam element has four DOFs, two vertical deflections
and two rotations. Fig. 4.12 shows a 1D, 2-node frame element in which a
FIGURE 4.12
Left: A 1D, 2-node frame element with two axial displacements, two vertical deflections,
and two rotations for a total of six DOFs. Right: For a 3D, 2-node frame element, the
torsional DOFs need to be added, making a total of 12 DOFs (3 translations and 3
rotations per node) for the 3D, 2-node frame element.
201
202
CHAPTER 4 Element Stiffness Matrix
beam and a bar element are combined, for a total of six DOFs, such that each
of the nodes can undergo translation, deflection, and rotation. Because we already
know the element stiffness matrices for both constituent element types, we can
easily determine the element stiffness matrix corresponding to
f u1 w1 q1 u2 w2 q2 gT . The stiffness matrix of a 1D, 2-node frame
element can be written as
2
3
AE
AE
6
0
0
0
0 7
6 L
7
L
6
7
6
12EI
6EI
12EI 6EI 7
6 0
7
0
6
7
6
L3
L2
L3
L2 7
6
7
6
6EI
4EI
6EI
2EI 7
6 0
7
0
6
L
L 7
L2
L2
6
7
½kframe ¼ ½kbar þ ½kbeam ¼ 6
7.
6 AE
7
AE
6
0
0
0
0 7
6 L
7
L
6
7
6
12EI 6EI
12EI 6EI 7
6
7
0
6 0
3
2
3
2 7
6
7
L
L
L
L
6
7
6
6EI
2EI
6EI
4EI 7
4 0
5
0
L
L
L2
L2
(4.75)
4.6.2 STIFFNESS MATRIX OF A 2-NODE, PSEUDO-3D FRAME
ELEMENT
Compared to a 1D frame element, which has a total of six DOFs, a 2-node pseudo3D frame element, in essence, is a 1D element positioned in a 3D space. This
element type can resist all three translational and three rotational deformations,
and hence has a total of 12 DOFs. Here, the three rotational deformations include
the two deflections and a torsion. In an analogous manner to a bar subjected to axial
loading, the governing equation due to torsional stiffness and the torsional element
stiffness matrix can be written as
Z 1
dqx
GIx 1 1
Mx ¼ GIx
and ½k ¼
½BT GIx ½Bj½Jjdx ¼
;
(4.76)
dx
L 1 1
1
where Mx is the twisting moment about the x-axis, G is the shear modulus, Ix
is the torsional moment of inertia with Ix ¼ Iy þ Iz, and qx is the torsion. By
applying the principle of superposition, the element stiffness matrix of a 3D frame
element in correspondence to the 12 nodal DOFs of f u1 v1 w1 qx1
qy1 qz1 u2 v2 w2 qx2 qy2 qz2 gT can be written as
4.7 Coordinate Transformation
2
6 EA
6
6
6 L
6
6
6 0
6
6
6
6
6 0
6
6
6
6
6 0
6
6
6
6 0
6
6
6
6
6 0
6
6
½k ¼ 6
6 EA
6
6 L
6
6
6
6 0
6
6
6
6 0
6
6
6
6
6 0
6
6
6
6
6 0
6
6
6
6
6 0
4
203
3
0
0
0
0
0
EA
L
0
0
0
0
12EIz
L3
0
0
0
6EIz
L2
0
12EIz
L3
0
0
0
0
12EIy
L3
0
6EIy
L2
0
0
0
12EIy
L3
0
6EIy
L2
0
0
GIx
L
0
0
0
0
0
GIx
L
0
0
6EIy
L2
0
4EIy
L
0
0
0
6EIy
L2
0
2EIy
L
6EIz
L2
0
0
0
4EIz
L
0
6EIz
L2
0
0
0
0
0
0
0
0
EA
L
0
0
0
0
12EIz
L3
0
0
0
6EIz
L2
0
12EIz
L3
0
0
0
0
12EIy
L3
0
6EIy
L2
0
0
0
12EIy
L3
0
6EIy
L2
0
0
GIx
L
0
0
0
0
0
GIx
L
0
0
6EIy
L2
0
2EIy
L
0
0
0
6EIy
L2
0
4EIy
L
6EIz
L2
0
0
0
2EIz
L
0
6EIz
L2
0
0
0
7
7
7
7
7
6EIz 7
7
L2 7
7
7
7
0 7
7
7
7
7
0 7
7
7
7
0 7
7
7
7
2EIz 7
7
L 7
7
7.
7
0 7
7
7
6EIz 7
7
7
L2 7
7
7
0 7
7
7
7
7
0 7
7
7
7
7
0 7
7
7
7
4EIz 7
7
L 5
0
(4.77)
4.7 COORDINATE TRANSFORMATION
Except for the spring element discussed in Chapter 1, all element types discussed so
far are intentionally made to align with the x- or y-axis. This geometric arrangement
does not happen frequently when solving real-world problems. In case an element
does not line up with the global Cartesian coordinate system, we can transfer the
element stiffness matrix already derived in the local natural coordinate system to
the global coordinate system, to ease the burden of deriving a new stiffness matrix.
Consider the bar element P1eP2 lying on the xey plane but not aligned with
either the x- or y-axis, as shown in Fig. 4.13. Assume the coordinates of P1 (x1,
y1) and P2 (x2, y2), elastic modulus E, and cross-sectional area A of the bar are all
provided in the input data deck. Additionally, the forceedisplacement equation,
in the form of a nodal load vector, element stiffness matrix, and nodal displacements,
is defined in the natural coordinate xeh system as shown in Eq. (4.78).
u1x
f1x
EA 1 1
¼
(4.78)
L 1 1
f2x
u2x
204
CHAPTER 4 Element Stiffness Matrix
FIGURE 4.13
Left: A bar element in the xeh coordinate system translated by (x0, y0) and rotated by a
counterclockwise angle of q. Right: The geometric relationships between (Vx and Vh) with
respect to (Vx and Vy) for a vector V.
Before we can assemble all element stiffness matrices to form the global stiffness
matrix, all individual element stiffness matrices need to be based on the same
coordinate system. All nodal load vectors must be transferred into the same global
coordinate system, as well. After these steps, we can assemble the global forcee
displacement equations. By solving the global forceedisplacement equation, we
can determine the nodal displacements, which are also prescribed in the same global
coordinate system. To achieve this goal, the methods to transfer the load vectors
and stiffness matrix from the local xeh system to the global xey system are
introduced.
4.7.1 2D TRANSFORMATION OF VECTOR
!
In a 2D space, a vector V can be decomposed in the xey coordinate system as V ¼
!
!
! !
!
V x þ V y and in the xeh coordinate system as V ¼ V x þ V h . Note that x and h are
simply use to indicate a different coordinate system, which is not bounded by the
definition of natural coordinate system in which both x and h range from 1 to 1.
From the geometric relationships shown in Fig. 4.13, the magnitudes of Vx and Vy
in terms of Vx and Vh can be written as
Vx ¼ Vx cos q Vh sin q and
(4.79)
Vy ¼ Vx sin q þ Vh cos q.
(4.80)
We can write Eqs. (4.79) and (4.80) in the matrix form as
Vx
Vx
cos q sin q
¼
.
Vy
Vh
sin q cos q
(4.81)
We shall now define [T ]1, called the inverse transformation matrix or inverse
rotation matrix, as the matrix dedicated to transfer a 2D vector from the local
4.7 Coordinate Transformation
coordinate system xeh with a counterclockwise rotation angle of q to the global coordinate system x-y, where
cos q sin q
.
½T1 ¼
sin q cos q
Because ½T½T1 ¼ ½T1 ½T ¼ ½I, where [I] is identity matrix, we can easily calculate [T ] as
cos q sin q
½T ¼
sin q cos q
From linear algebra, we can determine the transpose matrix of [T ], [T ]T, by turning
all the rows of a given matrix into columns as
cos q sin q
½TT ¼
sin q cos q
From these calculations, we notice that [T]1 ¼ [T ]T. In linear algebra, a matrix with
this characteristic is called an orthogonal matrix. We now multiply [T] to both sides
of Eq. (4.81):
Vx
Vx
Vx
cos q sin q
cos q sin q cos q sin q
¼
¼
.
Vy
Vh
Vh
sin q cos q
sin q cos q sin q cos q
(4.82)
Reversing the order of Eq. (4.82), we have
Vx
Vx
Vx
cos q sin q
¼
¼ ½T
.
Vy
Vy
Vh
sin q cos q
(4.83)
The matrix [T ] is called the transformation matrix or rotation matrix, which allows the transformation of a 2D vector from the global coordinate system to the local
coordinate system. For easier memorization, Eqs. (4.81) and (4.83) are collectively
written as listed in Table 4.2.
As an example of how to use Table 4.2 to find a component in the local coordinate system Vx or Vh, multiply the entries in the column under the local component
of interest with the corresponding global component vectors Vx and Vy, and then sum
them. For example, Vx ¼ Vxcos q þ Vysin q. For finding a global component Vx or Vy,
multiply the entries in the row of interest by the corresponding local component vectors Vx and Vh, and then sum them. For example, Vx ¼ Vxcos q Vhsin q.
Table 4.2 Transformation Matrix for a Vector From
the Local to the Global Coordinate Systems and Vice
Versa
Vx
Vy
Vx
Vh
cos q
sin q
sin q
cos q
205
206
CHAPTER 4 Element Stiffness Matrix
4.7.2 2D TRANSFORMATION OF STIFFNESS MATRIX
4.7.2.1 Rotation of a 2D Bar Element Stiffness Matrix
As vector requires coordinate transformation, the element stiffness matrix needs to
first be transferred from the local to the global coordinate system. For a 1D, 2-node
bar element not aligned with the x- or y-axis, the 1D forceedisplacement equation in
the x coordinate can be expanded into the 2D xeh plane, as shown in the following
equation:
9
8
f1x >
>
>
>
>
>
>
=
<f >
f1x
u1x
EA 1 1
1h
¼
0
>
> f2x >
L 1 1
f2x
u2x
>
>
>
>
>
;
:
f2h
9
38
2
u1x >
1 0
1 0 >
>
>
>
>
>
7>
6
<
=
7
6
v
0 07
EA 6 0 0
1h
.
(4.84)
¼
L 6
1 0 7
u2x >
>
>
5>
4 1 0
>
>
>
>
:
;
0 0
0 0
v2h
From Table 4.2, we can write the nodal displacements described in the local coordinate system in terms of the global coordinate system. Letting C represent cos q
and S represent sin q, the relationship between the local and global coordinate systems for the nodal displacements can be expressed as
8
9 2
9
38
u1x >
u1x >
C S
0 0 >
>
>
>
>
>
>
>
>
>
>
>
7>
<v >
= 6
7< v1y =
6 S C
0
0
1h
7
6
¼6
.
(4.85)
>
C S 7
u2x >
u2x >
>
>
>
>
5>
4 0 0
>
>
>
>
>
>
>
>
:
:
;
;
0 0
S C
v2h
v2y
Note that the transformation matrix
2
C S 0
6 S C 0
6
6
4 0 0 C
0
0
S
shown in Eq. (4.85) can be expressed as
3
0
½T22 ½022
07
7
¼
7
½022 ½T22
S5
C
to highlight the fact that this is a 2D transformation
Similarly, the relationship between the local and global coordinate systems for
the nodal load vectors can be expressed as
8
9 2
9
38
f1x >
f1x >
C S
0 0 >
>
>
>
>
>
>
>
>
>
>
> 6
>
7>
<
7< f1h =
f1y = 6 S C
0
0
7
¼6
.
(4.86)
6 0 0
>
C S 7
f2x >
f2x >
>
>
>
>
5>
4
>
>
>
>
>
>
>
>
:
:
;
;
0 0
S C
f2y
f2h
4.7 Coordinate Transformation
By substituting Eqs. (4.84) and (4.85) into Eq. (4.86), we have
8
9 2
9
3
38
2
f1x >
u1x >
C S
0 0
1 0
1 0 >
>
>
>
>
>
>
>
>
>
7
6
7>
6
>
>
>
>
>
>
7
>
6
7>
>
>
>
6
>
>
>
>
>
>
>
>
7
6
7
6
>
>
>
f
v
S
C
0
0
0
0
0
0
< 1y = 6
=
7 EA 6
7< 1h >
7
6
7
6
¼6
7
7
6
>
>
>
L 6 1 0
> f2x >
> 6
C S 7
1 0 7
u2x >
>
>
>
>
7
6 0 0
7>
6
>
>
>
>
>
>
>
>
7
6
7
6
>
>
>
>
>
>
>
>
5
4
5
4
>
>
>
>
>
>
>
>
:
:
;
;
0 0
S C
0 0
0 0
f2y
v2h
3
3
2
32
2
C S
0 0
C S
0 0
1 0
1 0
7
7
6
76
6
7
7
6
76
6
7
7
6
76
6
0 0 7 EA 6 0 0
0 0 7
0 0 76 S C
6 S C
7
7
6
76
6
¼6
7
7
76
6
7
6 0 0
6
7
6
L
C S 7
C S 7
1 0 76 0 0
7
6
6 1 0
7
7
6
76
6
5
5
4
54
4
0 0
S C
0 0
S C
0 0
0 0
8
9
u1x >
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
v
< 1y >
=
>
>
>
> u2x >
>
>
>
>
>
>
>
>
>
>
>
>
>
:
;
v2y
9
3 8
2
u1x >
1 0
1 0
>
>
>
>
7 >
6
>
>
>
>
7 >
>
6
>
>
>
>
7 >
6
v1y >
0
0
0
0
<
=
7
6
EA
7
6
T
0fkgglobal ¼ ½TT ½klocal ½T.
¼ ½T
7½T
6
>
>
7
L 6 1 0
>
>
1 0 7 >
u2x >
6
>
>
>
7 >
6
>
>
>
5 >
4
>
>
>
>
:
;
0 0
0 0
v2y
(4.87)
Eq. (4.87) shows that the element stiffness matrix for a 2-node bar element in the
global coordinate system can be determined using the rotation matrix [T ] and the
stiffness matrix expressed in the local coordinate system as
2
3
C 2 CS
C 2 CS
6
7
CS S2
CS S2 7
EA 6
T
6
7.
½kglobal ¼ ½T ½klocal ½T ¼
(4.88)
7
2
L 6
C 2 CS 5
4 C CS
CS
S2
CS
S2
207
208
CHAPTER 4 Element Stiffness Matrix
Eq. (4.88) is identical to Eq. (1.46) in which the direct method is used to derive
the stiffness matrix of a 1D bar element with its axis not in line with the x-axis. Obviously, the matrix rotation method is easier to program with computer code as
compared to the direct method.
4.7.2.2 Rotation of 2D Beam Element Stiffness Matrix
Like a 1D bar element rotated from a 1D domain into a 2D plane, the stiffness matrix
of a beam element can be calculated using Eq. (4.87). Recall that the element stiffness matrix of a 2-node beam element is
2
3
12 6L
12 6L
6
7
6L 2L2 7
EI 6 6L 4L2
7.
½klocal ¼ 3 6
L 6
12 6L 7
4 12 6L
5
6L
2L2
6L
4L2
Consider a 2-node beam element that is rotated in a counterclockwise direction for an angle of q, as shown in Fig. 4.14. Conventionally, a beam element is set
to be along the x-axis. This element has two DOFs for each node, a vertical deflection (in the z-direction) and a rotation (about the h-axis). From the geometric
relationship shown in the figure, the vertical deflection and rotation of the 2node beam element can be written as Eq. (4.89), where C represents cos q and S
represents sin q.
FIGURE 4.14
A beam element rotated in a counterclockwise direction for an angle of q. A subscript z
added for the vertical deflection of both points to indicate that these two displacements
are in the local coordinate system. From the geometric relationship, we can determine
!
!
!
that w 1z ¼ w 1x þ w 1z ¼ w1x sin q þ w1z cos q.
4.7 Coordinate Transformation
w1z
w2z
"
¼
S
C
0
0
0
S
2
S C 0
6
6 0 0 1
¼6
6 0 0 0
4
0
0
8
w1z
>
>
>
>
< q
1h
0
>
>
w
>
>
> 2z
>
>
:
; >
q2h
8
9
w
>
> 1x >
>
>
>
>
3>
>
>
>
w1z >
>
>
>
>
>
>
>
7>
<
7 q1h =
7
7> w >
2x >
5>
>
>
>
>
>
>
>
>
>
>
w
2z
>
>
>
>
>
>
:
;
q2h
8
> w1x
>
#>
>
0 < w1z
w2x
C >
>
>
>
:
w2z
0
0
0
0 0 0
S C 0
0
0
0
1
9
>
>
>
>
=
209
9
>
>
>
>
=
>
>
>
>
;
(4.89)
From Eq. (4.87), the element stiffness matrix in the global coordinate system can
be calculated from the stiffness matrix derived from the local coordinate system,
shown as follows:
2
½kglobal
6
6
6
6
6
6
6
6
T
¼ ½T ½klocal ½T ¼ 6
6
6
6
6
6
6
4
S
0
C
0
0
1
0
0
0
0
0
2
0
6
6
6
6
6
6
6
EI 6
¼ 36
L 6
6
6
6
6
6
6
4
3
7
7 2
0 7
7
7 6
7 6
6
0 0 7
7 EI 6
7 36
7
S 0 7 L 6
6
7 4
7
C 07
7
5
0
0
0
6L
6L
12
6L 4L2
6L
12
12
6L
12
6L
2L2
6L
1
12S 6SL
6
6
6 12C
6
6
6
EI 6
6 6L
¼ 3 6
6 12S
L
6
6
6
6 12C
6
4
2
0
12S
6CL
12C
4L2
6L
6SL
12S
6CL
12C
2L2
6L
6SL
6L
4L2
0 0
7
72
6CL 7
7 S C 0
76
6
2 7
6
2L 7
76 0 0 1
76
7
6SL 76
6 0 0 0
74
7
6CL 7
7 0 0 0
5
4L2
12S2
12CS
12SC
12C 2
6CL
12CS
12C 2
6SL
6CL
4L2
6SL
6CL
12SC
6SL
2
12S
12SC
12CS
12C 2
6CL
12CS
12C 2
6SL
6CL
2L2
6SL
6CL
12S
0
0
0
0
0
S
0
0
0
6SL
0 0
0
0
3
7
7
0 7
7
7
S C 0 7
7
5
0 0
0
0
3
7
7
0 7
7
7
C 07
7
5
3
6SL
2
S C
76
76
6
2L2 7
76 0 0 1
76
7
6L 76
6 0 0 0
54
12SC
12S
2
32
1
3
7
7
6CL 7
7
7
7
2 7
2L 7
7.
7
6SL 7
7
7
7
6CL 7
7
5
2
4L
(4.90)
1
210
CHAPTER 4 Element Stiffness Matrix
In Eq. (4.89), w1x and w2x are the respective x-components of the two vertical
deflections when the axis of the beam is not aligned with the x-axis of the
global coordinate system. We need to be careful to not confuse these terms with
the axial displacements u1 and u2. If we want to include u1 and u2 in a frame
element, the stiffness due to axial displacements needs to be superimposed into
Eq. (4.90).
These two examples show that for finding the element stiffness matrix in the
global coordinate system, we must first correctly identify the rotation matrix.
Once the rotation matrix is obtained, the element stiffness matrix derived from
the local coordinate system can be easily transferred to the global coordinate system.
In this way, all element-related information can be stored in the same global coordinate system in preparation for using the FE model to solve for nodal
displacements.
4.7.3 2D TRANSFORMATION OF INCLINED BOUNDARY CONDITION
Sometimes, boundary conditions are not necessarily prescribed in accordance to the
global Cartesian coordinate directions. For example, Fig. 4.15 shows a boundary
condition, in which an a degrees inclined roller support is applied at P2 of a 4node, plane stress element. One of the ways to resolve such an anomalous boundary
condition is to form a new element stiffness matrix [k0 ] by altering part of the original element stiffness matrix [k] according to the boundary conditions (Griffiths,
1990). Upon completion, the nodal displacements can be calculated using Gauss
elimination or other methods.
The element shown in Fig. 4.15 allows two DOFs (u and v) per node, for a total of
eight DOFs. Under the Cartesian coordinate system, the forceedisplacement equation for this element has the form
FIGURE 4.15
A 2D, 4-node plane stress element with a fixed boundary condition at P1 and a roller
support along an inclined plane at P2.
4.7 Coordinate Transformation
2
6
6
6
6
6
6
6
6
6
6
6
6
6
4
k11
k12
k13
k14
k15
k16
k17
k22
k23
k24
k25
k26
k27
k33
k34
k44
k35
k45
k36
k46
k37
k47
k55
k56
k66
k57
k67
S
Y
M
M
k77
38 9 8
9
k18 > u1 > > F1H >
>
>
>
>
>
>
>
>
> v1 >
>
>
>
>
k28 7
F1V >
7>
>
>
>
>
>
>
>
>
7> >
>
>
>
>
>
>
>
>
>
k38 7>
u
F
2
2H
>
>
>
>
> >
> >
>
7>
>
>
<
<
=
7
k48 7 v2
F2V =
¼
;
7
>
k58 7>
u3 >
F3H >
>
>
>
>
>
>
>
7>
>
>
>
>v >
>
> F3V >
> >
>
k68 7
>
>
7>
>
> 3>
> >
>
>
>
>
7>
>
>
>
>
>
>
k78 5>
u
F
> 4H >
> 4>
> >
>
>
>
: ; :
;
k88
v4
F4V
(4.91)
where H indicates the horizontal component and V indicates the vertical component.
We can describe the boundary conditions at P1 in the global coordinate system as
u1 ¼ v1 ¼ 0; however, it is much easier to describe the roller support condition at P2
in a local coordinate systems as h2 ¼ 0. From the geometric relationship, we relate
the transformed and untransformed DOFs as
u2 ¼ x2 cos a h2 sin a
v2 ¼ x2 sin a þ h2 cos a.
(4.92)
Note that Eq. (4.92) has the same form as Eq. (4.81), with the exception that the
amplitude of the angle is changed from q to a, because the geometric relationships
between the two coordinate systems are identically set up. Of course, the same
boundary condition can be applied at any of the nodes as needed. Hence, we can
rewrite Eq. (4.92) without denoting the specific node P2 in the equation as
u ¼ x cos a h sin a
v ¼ x sin a þ h cos a
(4.93)
The boundary conditions based on the new local coordinate system are applied at
P2 only, and therefore the only entries in the original stiffness matrix [k] that are
affected are those related to DOFs associated with P2 (i.e., the third and fourth
rows and columns). According to Griffiths, we write the altered stiffness matrix
[k0 ] as
38 9 8
2
9
0
0
k11 k12 k13
k14
k15 k16 k17 k18 > u1 > > F1H >
>
>
>
>
>
>
0
0
7>
6 k
>
> v1 >
>
>
>
>
F1V >
6 21 k22 k23 k24 k25 k26 k27 k28 7>
>
>
>
>
>
>
>
>
6 0
>
> >
> >
>
0
0
0
0
0
0
0 7
>
>
>
>
>
7>
6 k31 k32
x
F
k33
k34
k35
k36
k37
k38
2x
2
>
>
>
>
>
>
>
7>
6
>
>
>
>
<
=
=
0
0
0
0
0
0
0 7<
6 k0
h
F
k
k
k
k
k
k
k
2h
6 41
2
42
43
44
45
46
47
48 7
¼
. (4.94)
7
6
0
0
>
> u3 >
6
F3H >
k53
k54
k55 k56 k57 k58 7
>
>
>
>
>
>
>
7>
6
>
> >
> >
>
>
>
6 SYM
0
0
> >
>
k63
k64
k66 k67 k68 7
>
7>
6
> F3V >
> v3 >
> >
>
>
>
>
>
7
6
>
>
>
0
0
>
>
4
u4 >
F4H >
k73 k74
k77 k78 5>
>
>
>
>
>
>
>
>
: ; :
;
0
0
v4
F4V
k83 k84
SYM
k88
211
212
CHAPTER 4 Element Stiffness Matrix
In Eq. (4.94), for i ¼ 1, 2, 5, 6, 7, 8, we have
0
0
¼ k3i
¼ ki3 cos a þ ki4 sin a
ki3
(4.95)
0
0
ki4
¼ k4i
¼ ki3 sin a þ ki4 cos a.
For the remaining (i ¼ 3, 4), we have
0
0
¼ k43
¼ ðk44 k33 Þsin a cos a þ k34 cos 2 a
k34
0
k33
¼ k33 cos2 a þ k34 sin 2 a þ k44 sin2 a
0
k44
(4.96)
¼ k33 sin a k34 sin 2 a þ k44 cos a.
2
2
After these transformations, the inclined roller support can be treated just like a
typical roller in the x- or y-direction. Thus, the zero-displacement boundary condition h2 ¼ 0 can be managed by eliminating the fourth row and column from further
calculations. Eqs. (4.95) and (4.96) are written to transfer the third and fourth DOFs
(corresponding to P2) from xey to xeh coordinates. In general, an inclined support
associated with the mth and nth DOFs within a qq stiffness matrix is transferred by
replacing the entries associated with the mth and nth DOFs as
0
0
kim
¼ kmi
¼ kim cos a þ kin sin a
(4.97)
0
0
kin
¼ kni
¼ kim sin a þ kin cos a;
where i ¼ 1, 2, ., q; i s m; and i s n. Additionally,
0
0
¼ knm
¼ ðknn kmm Þsin a cos a þ kmn cos 2 a
kmn
0
kmm ¼ kmm cos2 a þ kmn sin 2 a þ knn sin2 a
0
knn
(4.98)
¼ kmm sin a kmn sin 2 a þ knn cos a.
2
2
When more than one node has similar inclined boundary conditions, the same
procedures can be applied sequentially. Upon completion of transferring all inclined
boundary conditions, the resulting forceedisplacement equation can be solved for
finding solutions. Example 4.7 demonstrates the application of Griffiths’s method.
Example 4.7
A 3-node, 3-element truss structure with a constant EA ¼ 60 for each element is
pinned at P1, supported on a roller at P2, and loaded at P3 with a force F ¼ 100 as
shown in Fig. 4.16. Calculate the nodal displacements at P2 and P3.
Solution
In this 2D truss problem, each node has two pseudo-DOFs for a total of six DOFs.
For element 1, the corresponding four DOFs are u1, v1, u2, and v2. The element
stiffness matrix is written as
2
1
EA 6
6 0
½k1 ¼
6
L1 4 1
0
3 2
15
0 1 0
6
0 0 07
7 6 0
7¼6
0 1 0 5 4 15
0
0 0 0
0
0
0
0
15
0
15
0
3
0
07
7
7.
05
0
4.7 Coordinate Transformation
FIGURE 4.16
A 3-truss structure represented by three nodes and three elements.
For element 2, the corresponding four DOFs are u2, v2, u3, and v3, and the
element stiffness matrix is written as
2
0 0
6
EA 6 0 1
½k2 ¼
6
L2 4 0 0
0 1
0
0
0
0
3 2
0
0
7
6
1 7 6 0
7¼6
0 5 40
0
20
0
0 20
1
3
0
0
0 20 7
7
7.
0
0 5
0 20
For element 3, cos q ¼ 0.8 and sin q ¼ 0.6, and the element stiffness matrix,
corresponding to u1, v1, u3, and v3, is written as
2
½k3 ¼
C2
EA 6
6 CS
6
L3 4 C2
CS
S2
CS
CS S2
C2
CS
C2
CS
CS
3
2
7:68
6 5:76
S2 7
7 6
7¼6
CS 5 4 7:68
5:76
S2
3
5:76 7:68 5:76
4:32 5:76 4:32 7
7
7.
5:76 7:68
5:76 5
4:32 5:76
4:32
We assemble these three element matrices to form the global stiffness matrix as
2
15 þ 7:68
6 5:76
6
6
6 15
½K ¼ 6
6
0
6
6
4 7:68
5:76
15
0
7:68
5:76
4:32
0
0
5:76
0
15
0
0
0
0
20
0
5:76
0
0
7:68
4:32
0
20
5:76
5:76
4:32
0
20 5:76 20 þ 4:32
3
22:68
5:76 15
0
7:68 5:76
6 5:76
4:32
0
0
5:76 4:32 7
7
6
7
6
6 15
0
15
0
0
0 7
7.
¼6
6 0
0
0
20
0
20 7
7
6
7
6
4 7:68 5:76
0
0
7:68
5:76 5
5:76 4:32
0
20 5:76
24:32
2
3
7
7
7
7
7
7
7
7
5
213
214
CHAPTER 4 Element Stiffness Matrix
Because of the inclined boundary condition h2 ¼ 0 at P2, we need to transfer u2
and v2 from the global coordinate system to the local coordinate system. Hence,
the third and fourth DOFs u2 and v2 in [K] need to be changed to [K0 ] corresponding to x2 and h2. We use Eqs. (4.97) and (4.98) with m ¼ 3, n ¼ 4, cos
q ¼ 0.8, and sin q ¼ 0.6, and the following entries that correspond to the third and
fourth rows and columns of [K0 ] are
0
0
k13
¼ k31
¼ 0:8k13 þ 0:6k14 ¼ 0:8ð15Þ þ 0:6ð0Þ ¼ 12
0
k23
0
k53
0
k63
0
k14
0
k24
0
k54
0
k64
0
k34
0
k33
0
¼ k32
0
¼ k35
0
¼ k36
0
¼ k41
¼ 0:8k23 þ 0:6k24 ¼ 0:8ð0Þ þ 0:6ð0Þ ¼ 0
¼ 0:8k53 þ 0:6k54 ¼ 0:8ð0Þ þ 0:6ð0Þ ¼ 0
¼ 0:8k63 þ 0:6k64 ¼ 0:8ð0Þ þ 0:6ð20Þ ¼ 12
¼ 0:6k13 þ 0:8k14 ¼ 0:6ð15Þ þ 0:8ð0Þ ¼ 9
0
¼ k42
¼ 0:6k23 þ 0:8k24 ¼ 0:6ð0Þ þ 0:8ð0Þ ¼ 0
0
¼ k45
¼ 0:6k53 þ 0:8k54 ¼ 0:6ð0Þ þ 0:8ð0Þ ¼ 0
0
¼ k46 ¼ 0:6k63 þ 0:8k64 ¼ 0:6ð0Þ þ 0:8ð20Þ ¼ 16
0
¼ k43
¼ 0:48ðk44 k33 Þ þ 0:28k34 ¼ 0:48ð20 15Þ þ 0:28ð0Þ ¼ 2:4
¼ 0:64k33 þ 0:96k34 þ 0:36k44 ¼ 0:64ð15Þ þ 0:96ð0Þ þ 0:36ð20Þ ¼ 16:8
0
¼ 0:36k33 0:96k34 þ 0:64k44 ¼ 0:36ð15Þ 0:96ð0Þ þ 0:64ð20Þ ¼ 18:2
k44
Replacing these corresponding values to the third and fourth rows and columns of
[K ], the global forceedisplacement equation of the system is changed to
2
22:68
5:76
12
9
6 5:76
4:32
0
0
6
6
6 12
0
16:8 2:4
6
6 9
0
2:4 18:2
6
6
4 7:68 5:76
0
0
5:76 4:32 12 16
7:68
5:76
0
0
7:68
5:76
9
38 9 8
5:76 > u1 > > 0 >
>
>
>
>
>
>
>
>
>v >
>
>
>
4:32 7
0 >
1>
>
>
>
>
7>
>
>
>
>
>
7> > >
12 7< x2 = < 0 =
7
¼
.
> >
>
16 7
>
7>
> >
> h2 >
> 0 >
>
>
>
>
7>
> >
> u3 >
> 100 >
>
5:76 5>
>
>
>
>
>
>
; >
: >
:
;
v3
24:32
0
The boundary conditions in this set of simultaneous equations are
u1 ¼ v1 ¼ h2 ¼ 0; thus, we eliminate the first, second, and fourth rows and columns from further calculations, as shown below:
2
16:8
6
4 0
12
9
38 9 8
0
12 >
< x2 >
= >
< 0 >
=
7
7:68 5:76 5 u3 ¼ 100
>
>
: >
; >
:
;
5:76 24:32
v3
0
Using Gauss elimination, the nodal displacements are:
9
8 9 8
>
=
< x2 >
= >
< 4:6875 >
u3 ¼ 17:9427 .
>
>
;
: >
; >
:
6:5625
v3
4.7 Coordinate Transformation
4.7.4 3D ROTATION
Consider a vector V in a 3D space where the angles between the vector V and the x-,
y-, and z-axes are a, b, and g, respectvely. The direction cosines of the vector are
defined as the cosines of these three angles. Similarly, the direction cosines of the
axes of two coordinate systems are defined as the cosines of the nine angles formed
by x, y, z versus x, h, z axes, respectively, as shown in Table 4.3.
Table 4.3 Direction Cosines of the Axes
x-axis
h-axis
z-axis
x-axis
l1 ¼ cosð:xxÞ
l1 ¼ cosð:xhÞ
l1 ¼ cosð:xzÞ
y-axis
m1 ¼ cosð:yxÞ
m1 ¼ cosð:yhÞ
m1 ¼ cosð:yzÞ
z-axis
n1 ¼ cosð:zxÞ
n1 ¼ cosð:zhÞ
n1 ¼ cosð:zzÞ
Example 4.8
In Fig. 4.13, identify the direction cosines of the axes.
Solution
The nine angles formed between two axes in the xeyez global coordinate system
and xehez coordinate system, respectively, are
:xx ¼ q; :xh ¼
P
P
þ q; :xz ¼ ;
2
2
:yx ¼
P
P
q; :yh ¼ q; :yz ¼ ; and
2
2
:zx ¼
P
P
; :zh ¼ ; :zz ¼ 0:
2
2
So, the direction cosines of the axes are listed in the following table:
x-axis
x-axis
y-axis
z-axis
h-axis
l1 ¼ cos q
m1 ¼ cos
n1 ¼ cos
l1 ¼ cos
Pq
2
P
2
¼0
¼ sin q
z-axis
Pþq
2
¼ sin q
m1 ¼ cos q
n1 ¼ cos
P
2
l1 ¼ cos
m1 ¼ cos
¼0
P
2
P
2
¼0
¼0
n1 ¼ cos(0) ¼ 1
As expected, results in Example 4.6 have the same form as those shown in
Table 4.2. Using Table 4.3, we can easily write the relationship between the nodal displacements in the local coordinate system in terms of the global coordinate systems as
8
9 2
9
8
9
38
l1 m1 n1 >
>
>
< uðx; h; zÞ >
=
< uðx; y; zÞ >
=
< uðx; y; zÞ >
=
6
7
vðx; h; zÞ ¼ 4 l2 m2 n2 5 vðx; y; zÞ ¼ ½T vðx; y; zÞ ;
(4.99)
>
>
>
>
>
>
:
;
:
;
:
;
wðx; h; zÞ
wðx; y; zÞ
wðx; y; zÞ
l3 m3 n3
215
216
CHAPTER 4 Element Stiffness Matrix
where [T ] is the rotation matrix or transformation matrix, and l1, l2, l3, m1, m2, m3,
n1, n2, and n3 are direction cosines of the axes. Since [T ] is an orthogonal matrix, we
can also write
9 2
9
9
8
8
38
l1 l2 l3 >
>
>
=
=
=
< uðx; h; zÞ >
< uðx; h; zÞ >
< uðx; y; zÞ >
6
7
vðx; y; zÞ ¼ ½TT vðx; h; zÞ ¼ 4 m1 m2 m3 5 vðx; h; zÞ . (4.100)
>
>
>
>
>
>
;
;
;
:
:
:
wðx; h; zÞ
wðx; h; zÞ
wðx; y; zÞ
n1 n2 n 3
Equations with the same form as Eqs. (4.99) and (4.100) can be used for transferring all other vectors, such as the nodal coordinates and nodal forces and movements, from one coordinate system to the other. In a manner similar to forming an
element stiffness matrix in the 2D global coordinate system from the element stiffness matrix derived in the local coordinate system, we can describe the 3D element
stiffness matrix in the global coordinate system as fkgglobal ¼ ½TT ½klocal ½T, which
is just like that shown in Eq. (4.87).
4.8 CHAPTER SUMMARY USING A NUMERICAL EXAMPLE
With all the explanations presented so far, you must wonder how we can use these
equations to write an FE software program to find solution for a simple FE problem.
Example 4.8 summarizes the concepts of determining the [B] and [k] matrices from
the element shape functions [N ] using the Gauss quadrature. Once all element stiffness matrices are found, they can be assembled into a global stiffness matrix. By
applying the boundary and loading conditions, the nodal displacements can be
solved. Even with a very simple problem, we soon realize that going through the procedures is rather tedious, especially when calculations are done manually. The
advantage of using the FE method is clear once these procedures are programed
into a computer.
Example 4.9
Consider a 4-node, 2D plane stress element that has an edge P1eP4 aligned with
the local t-axis as shown in Fig. 4.17A. Also, the constants in a set of consistent
units are E ¼ 30,000,000, n ¼ 0.3 and thickness t ¼ 0.1. The element has fixed
boundary conditions at P1 and P4 and is loaded at P2 along the negative t-axis
and at P3 along the s-axis.
1. Find the Jacobian matrix and the Jacobian (determinant of the Jacobian
matrix).
2. Establish the four [B] matrices from the global coordinate system at each of
the four Gauss integration points.
3. Determine the [k] matrix from [B] matrices.
4. Assemble the structure stiffness matrix [K ].
4.8 Chapter Summary Using a Numerical Example
(B)
(A)
’( 0,20)
’( 10,15)
’( 10,5)
’( 0,0)
FIGURE 4.17
(A) A quadrilateral element loaded along the negative t-axis at P2 and positive s-axis
at P3 and (B) the element rotated until the edge P10 eP40 is aligned with the y-axis.
After rotation, the loading is along the negative t-axis at P2 and positive s-axis at P3.
5. Calculate the nodal load vectors.
6. Apply the boundary conditions and calculate the nodal displacements
u2, v2, u3 and v3 in the global coordinate system.
Solutions
We solve the problem outlined in Fig. 4.17A based on isoparametric formulations. Because we apply the Jacobian transformation, the formulations are turned
into the global xey coordinate system. Consequently, all nodal displacements
calculated are in the global xey coordinate system as well.
Step 1: Find the four Jacobians at the four Gauss points
The isoparametric coordinate systemebased shape functions for a 4-node, 2D
plane stress element are
ð1 xÞð1 hÞ
ð1 þ xÞð1 hÞ
ð1 þ xÞð1 þ hÞ
; N2 ¼
; N3 ¼
; N4
4
4
4
ð1 xÞð1 þ hÞ
.
¼
4
N1 ¼
(4.101)
From these four shape functions, any physical quantity anywhere within the
element can be expressed as
4¼
4
X
i¼1
Ni 4i
(4.102)
217
218
CHAPTER 4 Element Stiffness Matrix
We now derive the formula to calculate v4
vx and
v4
vh
as
v4 vN1
vN2
vN3
vN4
4 þ
4 þ
4 þ
4
¼
vx 1
vx 2
vx 3
vx 4
vx
1
¼ ½ð1 þ hÞ41 þ ð1 hÞ42 þ ð1 þ hÞ43 ð1 þ hÞ44 4
1
¼ ½ð 41 þ 42 þ 43 44 Þ þ hð41 42 þ 43 44 Þ
4
v4 vN1
vN2
vN3
vN4
¼
4 þ
4 þ
4 þ
4
vh
vh 1
vh 2
vh 3
vh 4
(4.103)
1
¼ ½ð1 þ xÞ41 ð1 þ xÞ42 þ ð1 þ xÞ43 þ ð1 xÞ44 4
1
¼ ½ð 41 42 þ 43 þ 44 Þ þ xð41 42 þ 43 44 Þ.
4
vy
vx
To find the Jacobian matrix and Jacobian, we first need to find vx
vx; vx; vh, and
Inserting x1 ¼ 0, x2 ¼ 5, x3 ¼ 1, and x4 ¼ 12 as well as y1 ¼ 0, y2 ¼ 10,
y3 ¼ 18, and y4 ¼ 16 to the two equations listed above, we have
vy
vh.
vx 1
1
¼ ½ð x1 þ x2 þ x3 x4 Þ þ hðx1 x2 þ x3 x4 Þ ¼ ð16 þ 6hÞ ¼ 4 þ 1:5h.
vx 4
4
(4.104)
Similarly,
vy
vx
vy
¼ 3 2h;
¼ 4:5 þ 1:5x;
¼ 6 2x.
vx
vh
vh
(4.105)
Inserting these four entries in the Jacobian matrix, we have the Jacobian matrix
and Jacobian as
3
vx vy
6 vx vx 7 4 þ 1:5h
7
6
½J ¼ 6
7¼
4 vx vy 5
4:5 þ 1:5x
vh vh
2
3 2h
; j½Jj ¼ 37:5 12:5x.
6 2x
(4.106)
In this 4-node plane stress element, we need four Gauss integration points to
conduct the full Gauss quadrature. These four Gauss points have the natural coordinates of
G1 ð0:5773; 0:5773Þ; G2 ð0:5773; 0:5773Þ; G3 ð0:5773; 0:5773Þ;
G4 ð0:5773; 0:5773Þ.
(4.107)
The four Jacobians in correspondence to the four Gauss points G1, G2, G3, and
G4 can be calculated by inserting the x and h values either in the Jacobian matrix
4.8 Chapter Summary Using a Numerical Example
219
or in the determinant of Jacobian matrix. For demonstration purpose, we calculate
J(G1). In real-world practice, a computer program should be written to calculate
the Jacobians.
4 þ 1:5h
3 2h 4 þ 1:5 ð0:5773Þ
3 2 ð0:5773Þ ½JG1 ¼ ¼
4:5 þ 1:5x 6 2x 4:5 þ 1:5 ð0:5773Þ 6 2 ð0:5773Þ 3:13405 4:1546 ¼
¼ 44:71625;
5:36595 7:1546 or j½Jj ¼ 37:5 12:5x ¼ 37:5 12:5 ð0:5773Þ ¼ 44:71625
(4.108)
Applying the same procedures, we have J(G2) ¼ 44.71625, J(G3) ¼ 30.28375,
and J(G4) ¼ 30.28375.
Step 2: Find the four [B] matrices associated with the four Gauss integration points
From Eq. (4.71), the 38 [B] matrix for a 2D, 4-node isoparametric plane
stress element is written as
3
vy v vy v
0
7
6
7
6 vh vx vx vh
7
6
7
6
1 6
vx v
vx v 7
7
6
0
¼
6
vx vh vh vx 7
j½Jj 6
7
7
6
6 vx v
vx v vy v vy v 7
5
4
vx vh vh vx vh vx vx vh
2
½B3x8
"
N1
0
N2
,
,
0
0
N1
0
,
,
N4
#
2x8
3x2
½B3x8 ¼
1
1
4 j½Jj
2
vy
vy
6 ðh 1Þ ðx 1Þ
6 vh
vx
6
6
6
6
6
0
6
6
6
6
6 vx
vx
4
ðx 1Þ ðh 1Þ
vx
vh
0
vy
vy
ð1 hÞ ð1 xÞ
vh
vx
0
vy
vy
ð1 þ hÞ ð1 þ xÞ
vh
vx
0
vy
vy
ð1 hÞ ð1 xÞ
vh
vx
vx
vx
ðx 1Þ ðh 1Þ
vx
vh
0
vx
vx
ð1 xÞ ð1 hÞ
vx
vh
0
vx
vx
ð1 þ xÞ ð1 þ hÞ
vx
vh
0
vy
vy
ðh 1Þ ðx 1Þ
vh
vx
vx
vx
ð1 xÞ ð1 hÞ
vx
vh
vy
vy
ð1 hÞ ð1 xÞ
vh
vx
vx
vx
ð1 þ xÞ ð1 þ hÞ
vx
vh
vy
vy
ð1 þ hÞ ð1 þ xÞ
vh
vx
vx
vx
ð1 xÞ ðh 1Þ
vx
vh
(4.109)
Because vx
vx;
vy vx
vy
vx ; vh, and vh are already calculated in Step 1, the [B] matrix can
be determined from the above equation using a computer program. Here, we
calculate B11 for demonstration purpose as
B11 ¼
1
1
vy
vy
1
ðh 1Þ ðx 1Þ ¼
½ð6 2xÞðh 1Þ ð3 2hÞðx 1Þ
4 j½Jj
vh
vx
4j½Jj
B11 ðG1 Þ ¼
1
f½6 2ð0:5773Þ½ð0:5773Þ 1 ½3 2ð0:5773Þ½ð0:5773Þ 1g ¼ 0:026455
4j½JjG1
B11 ðG2 Þ ¼
1
f½6 2ð0:5773Þ½ð0:5773Þ 1 ½3 2ð0:5773Þ½ð0:5773Þ 1g ¼ 0:000635
4j½JjG2
B11 ðG3 Þ ¼
1
f½6 2ð0:5773Þ½ð0:5773Þ 1 ½3 2ð0:5773Þ½ð0:5773Þ 1g ¼ 0:010468
4j½JjG3
B11 ðG4 Þ ¼
1
f½6 2ð0:5773Þ½ð0:5773Þ 1 ½3 2ð0:5773Þ½ð0:5773Þ 1g ¼ 0:048595
4j½JjG4
3
7
7
7
7
7
7
vx
vx
7
ð1 xÞ ðh 1Þ 7
vx
vh
7
7
7
7
vy
vy
5
ð1 hÞ ð1 xÞ
vh
vx
0
220
CHAPTER 4 Element Stiffness Matrix
After completing all other entries, the resulting four [B] matrices, based on
four different Gauss points, are provided here for readers to check their works.
2
0:0265 0:0000 0:0729 0:0000
6
½BðG1 Þ ¼ 4 0:0000 0:0750 0:0000 0:0399
0:0750 0:0265 0:0399 0:0729
2
0:0006 0:0000 0:0213 0:0000
6
½BðG2 Þ ¼ 4 0:0000 0:0556 0:0000 0:0012
0:0556 0:0006 0:0012 0:0213
2
0:0105
6
½BðG3 Þ ¼ 4 0:0000
0:0000
0:0297
0:0297
0:0409
0:0000
0:0105 0:0507
0:0000
0:0507
0:0409
0:0071 0:0000 0:0535
0:0000 0:0201 0:0000
0:0201 0:0071
0:0150
3
0:0000
7
0:0150 5
0:0535
3
0:0587 0:0000 0:0794 0:0000
7
0:0000 0:0588 0:0000 0:0044 5
0:0588 0:0587 0:0044 0:0794
3
0:0391 0:0000 0:0695 0:0000
7
0:0000 0:1107 0:0000 0:0303 5
0:1107 0:0391 0:0303 0:0695
2
3
0:0486 0:0000 0:1172 0:0000 0:0372 0:0000 0:0314 0:0000
6
7
½BðG4 Þ ¼ 4 0:0000 0:0583 0:0000 0:0065 0:0000
0:0535
0:0000 0:0017 5
0:0583 0:0486 0:0065 0:1172 0:0535 0:0372 0:0017 0:0314
Step 3: Find element stiffness matrix [k]
From Eq. (4.67), the element stiffness matrix for a 2D, 4-node plane stress
element is approximated by Gauss quadrature as
Z
½k88 ¼
V
Z
¼
ZZ
½BT ½E½BdV ¼
1
1
Z
½BT ½E½BtdA
A
1
1
½BT t½E½Bj½Jjdxdhz
4
X
(4.110)
½BðGi ÞT t½E½BðGi Þj½JðGi Þj.
i¼1
T
We can easily find [B(Gi)] from [B(Gi)] by changing from the row to column.
Additionally, the equation representing the elastic constants for a plane stress
element previously shown in Eq. (1.22) is repeated here for convenience.
3
2
8
9
9
8
1 n
0
εxx >
>
7>
6
< sxx >
=
=
<
E 6
0 7
7 εyy
6n 1
syy ¼
7
6
2
>
>
:
; 1n 4
;
:g >
1 n 5>
sxy
xy
0 0
2
4.8 Chapter Summary Using a Numerical Example
For this problem, we have E ¼ 30 106, t ¼ 0.1, and y ¼ 0.3, that is,
2
3
3
2
1 0:3
0
7
6
6 0:1
7
Et 6
30
10
7
6
0 7¼
6n 1
0 5
t½E ¼
4 0:3 1
7
1 n2 6
1 0:09
4
1 n5
0
0 0:35
0 0
2
2
3
300 90
0
6
7
¼ 106 4 90 300 0 5
1
0
n
0
0
105
Taking the first Gauss point G1 as an example, the [k(G1)] is calculated using
Eq. (4.110) as
½kðG1 Þ ¼ ½BðG1 ÞT t½E½BðG1 Þj½JðG1 Þj
2
3
0:0265
0
0:0750
6
7
6
0
0:0750 0:0265 7
6
7
6
7
6
7
6 0:0729
0
0:0399 7 2
3
6
7
300 90
0
6
7
6
7 6
7
0
0:0399
0:0729
6
7 6
7 6 90 300 0 7
¼ 106 6
7
6
7 4
5
6 0:0071
0
0:0201 7
6
7
0
0 105
6
7
6
0
0:0201
0:0071 7
6
7
6
7
6
7
6 0:0535
7
0
0:0150
4
5
2
0
0:0265
0:0150
0:0000
0:0535
0:0729 0:0000 0:0071 0:0000 0:0535
6
6
6 0:0000 0:0750 0:0000 0:0399
4
0:0750 0:0265 0:0399 0:0729
2
0:3931
0:1900 0:4387
6
6 0:1900
0:8644 0:2962
6
6
6
6 0:4387 0:2962 0:8658
6
6
6 0:3287 0:5405 0:2788
6
6
¼ 10 6
6
6 0:1053 0:0509 0:1176
6
6
6 0:0509 0:2316 0:0794
6
6
6
6 0:1510
0:1571 0:5447
4
0:1896
0:0000 0:0201
0:0000
0:0201 0:0071
0:0150
0:0000
3
7
7
0:0150 7 ½JðG1 Þ
5
0:0535
0:3287 0:1053 0:0509
0:1510
0:5405 0:0509 0:2316
0:1571
0:2788
0:1176
0:0794
0:5447
0:5091
0:0881
0:1449
0:0383
0:0881
0:0282
0:0136
0:0405
0:1449
0:0136
0:0621
0:0421
0:0383 0:0405 0:0421
0:0922 0:0620 0:1134 0:0508
0:0247
0:4342
0:0767
0:1896
3
7
0:0922 7
7
7
7
0:0620 7
7
7
0:1134 7
7
7
7
0:0508 7
7
7
0:0247 7
7
7
7
0:0767 7
5
0:1809
(4.111)
The same procedures are repeated to find [k(G2)], [k(G3)], and [k(G4)]. Once these
calculations are completed, we sum up the four [k(Gi)] matrices to determine the
221
222
CHAPTER 4 Element Stiffness Matrix
element stiffness matrix [k]. It is obvious that a computer program should be used
to do the calculations. Results listed below are for readers to check their works.
2
0:1595
6 0:0034
6
6
6 0:0054
6
6 0:0610
6
½kðG2 Þ ¼ 106 6
6 0:1742
6
6 0:1701
6
6
4 0:0201
0:2278
0:0034
0:0054
0:0610
0:1742
0:1701
0:0201
0:4556
0:0523
0:0523
0:0668
0:0104
0:0024
0:1463
0:1877
0:4839
0:0589
0:1953
0:2491
0:0104
0:1463
0:0024
0:1877
0:0235
0:0676
0:0676
0:6870
0:0747
0:3310
0:0090
0:7005
0:4839
0:1953
0:0387
0:0589
0:2491
0:0090
0:0747
0:0090
0:0879
0:3310
0:7005
0:2523
0:6880
0:2198
0:2787
0:2198
0:9296
0:0335
0:0201
0:0097
0:0265
0:1555
0:0752
0:1041
0:0917
0:0178
0:0178
0:2571
0:1351
0:1346
0:0752
0:0363
0:3420
0:0665
0:0729
0:2304
0:1351
0:0752
0:1346
0:0363
0:3150
0:0990
0:0990
0:5804
0:5041
0:2806
0:0621
0:3885
0:3420
0:0729
0:1153
0:0665
0:2304
0:0859
0:5041
0:0621
0:0540
0:2806
0:3885
0:3044
1:2763
0:2719
0:4302
0:2719
0:5148
0:1369
2
0:0417
6 0:0201
6
6
6 0:0097
6
6 0:0265
6
½kðG3 Þ ¼ 106 6
6 0:1555
6
6 0:0752
6
6
4 0:1041
0:0816
2
0:3543
6 0:1837
6
6
6 0:5818
6
6 0:2480
6
6
½kðG4 Þ ¼ 10 6
6 0:0715
6
6 0:0021
6
6
4 0:1559
0:0665
½k88 z
4
X
0:1837
0:5818
0:2480
0:0715
0:0021
0:1559
0:4213
0:2155
0:2155
1:3726
0:2369
0:0495
0:0259
0:4229
0:2479
0:1793
0:0578
0:3678
0:2369
0:0259
0:0495
0:4229
0:4841
0:2118
0:2118
0:2381
0:2479
0:0578
0:0635
0:1793
0:3678
0:0133
0:1175
0:0133
0:1297
0:1291
0:1133
0:0568
0:1175 0:0133
0:1291 0:1133
0:3340
0:0480
0:0315
0:0480
0:0986
0:0036
0:2278
3
0:0387 7
7
7
0:0090 7
7
0:0879 7
7
7
0:2523 7
7
0:2787 7
7
7
0:0335 5
0:3279
0:0816
3
7
7
7
7
7
0:0540 7
7
7
0:3044 7
7
0:4302 7
7
7
0:1369 5
0:1153
0:0859
0:2608
0:0665
3
0:0635 7
7
7
0:0133 7
7
0:1297 7
7
7
0:0568 7
7
0:0315 7
7
7
0:0036 5
0:0348
½BðGi ÞT t½E½BðGi Þj½JðGi Þj
i¼1
2
0:9486
6
6 0:3972
6
6
6 1:0161
6
6
6 0:6643
6
6
¼ 10 6
6 0:3635
6
6
6 0:2984
6
6
6 0:4311
4
0:5654
0:3972
1:0161
1:8329
0:5819
0:5819
2:5622
0:6527
0:1961
0:2984 0:1540
1:3056
0:3841
0:4830
1:3921
0:1253
0:0016
0:6643 0:3635 0:2984
0:4311
0:6527 0:2984 1:3056
0:4830
0:5654
3
7
0:1253 7
7
7
0:1961 0:1540 0:3841 1:3921 0:0016 7
7
7
1:3317
0:4665 0:4021 0:0016 0:2770 7
7
7
0:4665
1:5337
0:4961 1:0161 0:6643 7
7
7
0:4021 0:4961
2:3603 0:5819 0:6527 7
7
7
0:0016 1:0161 0:5819 1:9772
0:0972 7
5
0:2770 0:6643 0:6527 0:0972
0:8044
(4.112)
Step 4: Find structure stiffness matrix [K ]
In this problem, the entire structure consists of only one element. Thus, the
element stiffness matrix [k] is the same as the structure stiffness matrix [K ].
4.8 Chapter Summary Using a Numerical Example
For problems involving more than one element, we need to use the method
explained in Section 1.3 to assemble all element stiffness matrices into [K].
Step 5: Find nodal load vectors { f }
Because the loading conditions are prescribed in the local set coordinate system, we first need to transfer these loads into the global xey coordinate system.
Based on the coordinates of P1 and P4, the counterclockwise rotation from the
y-axis to t-axis is q ¼ tan112
16 ¼ 36:87 degrees. Thus, sin q ¼ 0.6 and cos
q ¼ 0.8. Using Table 4.2 in Section 4.7.1, we know that
fx ¼ fs cos q ft sin q
fy ¼ fs sin q þ ft cos q.
(4.113)
At P2, the nodal load vector along the negative t-axis is expressed in the global
coordinate system as
f2x ¼ 0:8fs 0:6ft ¼ 0:8ð0Þ 0:6ð50000Þ ¼ 30000
f2y ¼ 0:6fs þ 0:8ft ¼ 0:6ð0Þ þ 0:8ð50000Þ ¼ 40000
At P3, the nodal load vector along the s-axis is expressed in the global coordinate
system as
f3x ¼ 0:8fs 0:6ft ¼ 0:8ð50000Þ 0:6ð0Þ ¼ 40000
f3y ¼ 0:6fs þ 0:8ft ¼ 0:6ð50000Þ þ 0:8ð0Þ ¼ 30000
Step 6: Apply boundary conditions and calculate nodal displacements
In this problem, the element/structure is fixed at P1 and P4. Thus, the first, second, seventh, and eighth DOFs are set to zero. In terms of mathematics, either the
elimination method or penalty method can be used to determine the nodal displacements. For illustration purposes, we remove the first, second, seventh, and
eighth rows and columns. The reduced forceedisplacement equation and nodal
displacements based on Gauss elimination are
2
2:5622
6 0:1961
6
106 6
4 0:1540
0:3841
9
38 9 8
0:1961 0:1540 0:3841 > u2 > > 30000 >
>
>
>
>
>
>
>
< = < 40000 >
=
1:3317
0:4665 0:4021 7
7 v2
¼
7
0:4665
1:5337
0:4961 5>
u3 >
>
> >
> 40000 >
>
>
>
: >
; >
:
;
0:4021 0:4961
2:3603
v3
30000
9
8 9 8
u2 > > 0:0199 >
>
>
>
>
>
>
>
>
=
< v = < 0:0522 >
2
¼
.
(4.114)
>
> >
> 0:0470 >
>
> u3 >
>
>
; >
:
;
: >
v3
0:0093
Discussion: We can solve the same problem using a different approach. In this
new approach, the element stiffness matrix [k] is formed in the local coordinate
system, in contrast to the global coordinate system used earlier. Fig. 4.17B shows
that the t-axis is rotated to overlap the y-axis. Using the method shown in
223
224
CHAPTER 4 Element Stiffness Matrix
Section 4.7.2, we transfer this [k] in the local coordinate system to [k0 ] in the
global coordinate system. If all calculations are properly done, [k0 ] should be
identical to that shown in Eq. (4.112).
The first step is to determine the nodal coordinates in the local set coordinate
system. Using Table 4.2 in Section 4.7.1, we first identify the equations needed to
transfer the nodal coordinates from the global (xey) to local coordinate system as
Vs ¼ Vx cos q þ Vy sin q
(4.115)
Vt ¼ Vx sin q þ Vy cos q.
For P1 (0, 0) in the global coordinate system, the coordinates in the local coordinate system P10 is also (0, 0). P2 (5, 10) in the global coordinate system is
transferred to P20 s ¼ 0.8 5 þ 0.6 10 ¼ 10 and P20 t ¼ 0.6 5 þ 0.8 10 ¼ 5. Also, P3 (1, 8) in the global coordinate system is transferred to
P30 s ¼ 0.8 (1) þ 0.6 18 ¼ 10 and P30 t ¼ 0.6 (1) þ 0.8 18 ¼ 15.
Finally, P4 (12, 16) in the global coordinate system is transferred to
P40 s ¼ 0.8 (12) þ 0.6 16 ¼ 0 and P40 t ¼ 0.6 (12) þ 0.8 16 ¼ 20.
We will now use these four local coordinates to form the element stiffness matrix
[k0 ] using Steps 1 to 3 shown previously. We assume that readers already have
written programs to complete these steps, and therefore only some key results
are listed for accuracy checking.
Step 1: The Jacobians
½J ¼
5
2:5h
0 7:5 2:5x
JðG1 Þ ¼ 44:7163; JðG2 Þ ¼ 44:7163; JðG3 Þ ¼ 30:2838; JðG4 Þ ¼ 30:2838
Step 2: The [B] matrices
2
0:0661 0:0000
0:0823
0:0000 0:0177 0:0000 0:0339
6
BðG1 Þ ¼ 4 0:0000 0:0441 0:0000 0:0118 0:0000 0:0118 0:0000
0:0441 0:0661 0:0118 0:0823 0:0118 0:0177 0:0441
2
0:0339
0:0000
0:0177
0:0000
0:0823 0:0000 0:0661
6
BðG2 Þ ¼ 4 0:0000 0:0441 0:0000 0:0118 0:0000 0:0118
0:0441 0:0339 0:0118 0:0177 0:0118 0:0823
0:0000
0:0441
2
0:0262 0:0000
0:0023
0:0000 0:0977 0:0000 0:0738
6
BðG3 Þ ¼ 4 0:0000 0:0174 0:0000 0:0651 0:0000 0:0651 0:0000
0:0174 0:0262 0:0651 0:0023 0:0651 0:0977 0:0174
2
0:0738 0:0000
0:0977
0:0000 0:0023 0:0000 0:0262
6
BðG4 Þ ¼ 4 0:0000 0:0174 0:0000 0:0651 0:0000 0:0651 0:0000
0:0174 0:0738 0:0651 0:0977 0:0651 0:0023 0:0174
0:0000
3
7
0:0441 5
0:0339
0:0000
3
7
0:0441 5
0:0661
0:0000
3
7
0:0174 5
0:0738
3
0:0000
7
0:0174 5
0:0262
4.8 Chapter Summary Using a Numerical Example
Step 3: The [k] matrix
2
0:7451
6 0:2794
6
6
6 0:7753
6
6 0:1526
6
½kðG1 Þ ¼ 106 6
6 0:1997
6
6 0:0749
6
6
4 0:2298
0:0519
2
0:2693
6 0:1431
6
6
6 0:0616
6
6 0:0226
6
6
½kðG2 Þ ¼ 10 6
6 0:4376
6
6 0:2049
6
6
4 0:2298
2
0:2794
0:5123
0:1201
0:7753
0:1201
1:0051
0:1526
0:2040
0:0932
0:1997
0:0749
0:2078
0:0749
0:1373
0:0322
0:2298
0:0844
0:4376
0:2040
0:0749
0:0932
0:2078
0:3698
0:0409
0:0409
0:0535
0:0547
0:0201
0:2049
0:0616
0:1373
0:0844
0:0322
0:4376
0:0547
0:2049
0:0201
0:0616
0:0368
0:0226
0:0226
0:2693
0:1710
0:1811
0:2206
0:0139
0:0458
0:1431
0:1431
0:0616 0:0226
0:4376
0:2049
0:2298
0:3458
0:0139
0:0458
0:1811
0:2206
0:0519
0:0139
0:0535
0:0201
0:2078
0:0409
0:1997
0:0322
1:0051
0:0547
0:0932
0:0749
0:7753
0:0458 0:0201 0:0368
0:1811 0:2078 0:0322
3
0:0519
0:1710 7
7
7
0:1811 7
7
0:2206 7
7
7
0:0139 7
7
0:0458 7
7
7
0:1431 5
0:3458
0:0844
0:2206 0:0409
0:0547
0:0932
0:3698
0:1526
0:0519
0:1997
0:0749
0:7753
0:1526
0:7451
0:0844
0:1710
0:0749
0:1373
0:1201
0:2040
0:2794
0:5123
0:0790
0:0296
0:0336
0:0496
0:2949
0:1106
0:1823
0:0313
0:0543
0:0583
0:0583
0:1487
0:1113
0:0099
0:1106
0:1253
0:2027
0:2176
0:0226
0:0570
0:1113
0:0099
0:4234
0:1851
0:4152
0:1454
1:1003
0:4126
0:6801
6 0:0296
6
6
6 0:0336
6
6 0:0496
6
6
½kðG3 Þ ¼ 10 6
6 0:2949
6
6 0:1106
6
6
4 0:1823
0:0313
2
0:5548
6 0:0836
6
6
6 0:6801
6
6 0:0844
6
½kðG4 Þ ¼ 106 6
6 0:0570
6
6 0:1454
6
6
4 0:1823
0:0226
2
1:6483
6 0:5357
6
6
6 1:4834
6
6 0:0412
6
½k ¼ 106 6
6 0:9891
6
6 0:5357
6
6
4 0:8242
0:0412
0:1106 0:1253 0:1851
0:2027 0:2176 0:4152
0:4126
0:7564
0:0844
0:0226
0:0371
0:0570 0:1454
0:1692 0:1194
0:6801
0:1169
0:0844
0:1385
0:5548
0:0836
0:0836
0:2209
0:6801 0:0844
0:1169 0:1385
0:0570
0:1692
0:1454
0:1194
0:1823
0:0313
0:4126
0:1253
0:1851
0:2949
0:7564
0:2176
0:4152
0:1106
0:1487
0:0099
0:0099
0:4234
0:0336
0:0496
0:1106
0:0336
0:0496
0:0790
0:2027
0:0583
0:1113
0:0296
0:1169
1:1003
0:1385 0:4126
0:1692 0:1253 0:2176
0:1194 0:1851 0:4152
0:0313 0:2949
0:0371
0:5357
0:1106
1:4834
0:0412
0:9891
0:5357
1:1332
0:0412
0:1854
0:5357
0:6800
0:0412
0:1854
2:3075
0:5357
0:5357
1:5864
0:1650
0:0412
0:0412
0:7210
0:5357
0:1650
0:0412
2:3075
0:5357
0:6800
0:0412
0:7210
0:5357
1:5864
0:0412
0:9891
0:5357
1:4834
0:0412
0:2678
0:5357
0:6800
0:0412
0:1854
0:8242
3
0:1710 7
7
7
0:0749 7
7
0:1373 7
7
7
0:1201 7
7
0:2040 7
7
7
0:2794 5
3
7
7
7
7
7
0:1194 7
7
7
0:1169 7
7
0:1385 7
7
7
0:0836 5
0:2209
0:0371
0:1692
3
0:0226
0:0371 7
7
7
0:1106 7
7
0:2027 7
7
7
0:0583 7
7
0:1113 7
7
7
0:0296 5
0:0543
0:0412
3
0:0412 0:2678 7
7
7
0:9891 0:5357 7
7
0:5357 0:6800 7
7
7
1:4834 0:0412 7
7
0:0412 0:1854 7
7
7
1:6483 0:5357 5
0:5357
1:1332
After adding the loading conditions and eliminating the rows and columns
associated with the fixed boundary conditions at P1 and P4, we have
225
226
CHAPTER 4 Element Stiffness Matrix
2
2:3075
6 0:5357
6
6
10 6
4 0:1650
0:5357
0:1650
1:5864
0:0412
0:0412
2:3075
0:0412
0:7210
0:5357
9
38 9 8
0
0:0412 > s2 > >
>
>
>
>
>
>
>
>
>
<
=
<
7
50000 =
0:7210 7 t2
¼
7
0:5357 5>
>
> 50000 >
> >
> s3 >
>
>
;
:
; >
: >
0
t3
1:5864
The solutions for this equation are
8 9 8
9
s2 > > 0:0154 >
>
>
>
>
>
>
>
>
< t = < 0:0537 >
=
2
¼
>
> s3 >
> >
> 0:0320 >
>
>
>
: >
; >
:
;
t3
0:0356
Using Eq. (4.113), we have
u2 ¼ cos q s2 sin q t2 ¼ 0:8 ð0:0154Þ 0:6 ð0:0537Þ ¼ 0:0199
v2 ¼ sin q s2 þ cos q t2 ¼ 0:6 ð0:0154Þ þ 0:8 ð0:0537Þ ¼ 0:0522
u3 ¼ cos q s3 sin q t3 ¼ 0:8 ð0:0320Þ 0:6 ð0:0356Þ ¼ 0:0470
v3 ¼ sin q s3 þ cos q t3 ¼ 0:6 ð0:0320Þ þ 0:8 ð0:0356Þ ¼ 0:0093
As expected, these answers are identical to those shown in Eq. (4.114). We
encourage you to use Eq. (4.116) to transfer the stiffness matrix in the local
coordinate system to the global coordinate system as a practice. Of course, the
answers were already provided in Eq. (4.112).
fkgglobal ¼ ½TT ½klocal ½T;
(4.116)
where
2
C S
6 S C
6
6
6 0 0
6
6 0 0
6
½T ¼ 6
6 0 0
6
6 0 0
6
6
4 0 0
0 0
0 0
0 0
C S
S C
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
C S
S C
0 0
0 0
0
0
0
0
3
0
07
7
7
07
7
07
7
7.
07
7
07
7
7
S5
0
0
C
S C
4.8 Chapter Summary Using a Numerical Example
EXERCISES
1. Write a simple program to determine the four Jacobians based on the 2 2
integration scheme and the sum of these four Jacobians for a quadrilateral
element with coordinates P1(0, 0), P2(6, 0), P3(6.3, 4.2), and P4(0, 4).
2. Derive the element stiffness matrix for a 2D frame element that provides
stiffness to resist axial displacement, vertical deflection, and rotation. Here,
the frame element is rotated in a counterclockwise direction for an angle of
q.
R1
3. Solve the function y ¼ 1 x sin x2 þ cosð xÞ dx to compare the 2 2
Gauss quadrature approach with the 2-point, midpoint Riemann Sum.
4. Develop the stiffness matrix using the direct method of a set of three bars, see
Fig. 4.1. However, the elements are arranged as equilateral triangles rather
than right triangles.
5. Find the stiffness matrix of a 3-node, 2D element with points at P1 (0, 0), P2
(2, 1), and P3 (1, 4). Assume a linear elastic material with elastic modulus of
20, thickness of 0.5, and a Poisson’s ratio of 0.23.
6. Use the variational method (principle of minimum potential energy) to find
the energy equation of a beam of length L with the origin at the beginning of
the element. Assume no body forces and no surface traction forces.
7. Using the big matrix above Eq. (4.64), derive the stiffness matrix of a beam
element in the natural coordinate system using 2-point Gauss quadrature. It
is best to write a program to calculate all the points, but the first one is
calculated for you.
8. Write a program that can rotate any bar element stiffness matrix from a local
coordinate system to a global coordinate system given an angle and stiffness
matrix.
9. A 2D element has the orientation as shown in Fig. 4.15. Change the slanted
boundary condition to apply to P3 instead of P2. Given the stiffness matrix,
also as shown below, find the reduced stiffness matrix. a ¼ 25 degrees
2
6
6
6
6
6
6
6
6
6
6
6
6
6
4
47 18
61
19
2
23 18
2
15
18 30
2
47
18
4
23
61
2
47
Symm
4
2
3
46 7
7
7
18 7
7
46 18 19 7
7
7
18 19 2 7
7
61
2
15 7
7
7
47 18 5
2
61
227
228
CHAPTER 4 Element Stiffness Matrix
10. Write a program to calculate the reduced global stiffness matrix for any four
points in a 4-node element. Elastic modulus, Poisson’s ratio, thickness, and
the applied forces are given. Assume node 1 is constrained in x and y and
node 2 has a rotated boundary condition of q1 degrees and constraint in the x
direction. Use 2-point Gauss quadrature to calculate the integrals (four
Gauss points). Include a step to rotate the whole stiffness matrix by an angle
q2 degrees.
11. Show step-by-step derivation of the stiffness matrix for a beam element.
Solution: From Section 4.5.2, we found that the stiffness matrix can be
described as
2
6x
6
6 L2
6
6
6 3x 1
Z 1
Z 16
6
EIL
EIL
6 L
½k ¼
½BT ½Bdx ¼
6
2 1
2 1 6 6x
6
6 L2
6
6
6
4 3x þ 1
L
and multiplying this out gives us
2
36x2
L4
6
6
6
6
6 18x2 6x
Z 16
6
EIL
L3
6
½k ¼
2 1 6
6 36x2
6
6
L4
6
6
4 18x2 þ 6x
L3
2
3
7
7
7
7
7
7
7 6x
7
7 2
7 L
7
7
7
7
7
5
3x 1
L
6x
L2
3x þ 1
dx;
L
3
18x2 þ 6x
7
7
L3
7
7
ð3x 1Þ2
18x2 þ 6x
9x2 1 7
7
7
L3
L2
L2
7dx
7
2
2
2
18x þ 6x
36x
18x 6x 7
7
7
L3
L3
L4
7
7
2 5
2
2
9x 1
18x 6x ð3x þ 1Þ
L2
L3
L2
18x2 6x
L3
36x2
L4
(4.117)
To avoid denominators within the matrix, L14 is pulled out for the next step. Notice that there is already an L
EI . Using
in the component to the left-hand side of the matrix, and therefore this component becomes 2L
3
x ¼ p1ffiffi, the matrix is
3
2
1
6
ð36Þ pffiffiffi
6
3
6
6
6
2
6
6 ð18Þ p1ffiffiffi ð6Þ p1ffiffiffi
6
6
3
3
6
6
2
6
1
6
ð36Þ pffiffiffi
6
6
3
6
6
2
6
6
1
1
4 ð18Þ pffiffiffi þ ð6Þ pffiffiffi
3
3
2
1
1
18 ð6Þ pffiffiffi L
ð36Þ pffiffiffi
3
3
2
2
1
1
1
ð18Þ pffiffiffi þ ð6Þ pffiffiffi
ð3Þ pffiffiffi 1 L2
3
3
3
2
2
!
1
1
1
L
ð36Þ pffiffiffi
ð18Þ pffiffiffi þ ð6Þ pffiffiffi
3
3
3
!
2
2
1
1
1
ð9Þ pffiffiffi 1 L2
ð18Þ pffiffiffi ð6Þ pffiffiffi
3
3
3
3
2
!
1
1
L 7
ð18Þ pffiffiffi þ ð6Þ pffiffiffi
7
3
3
7
7
!
7
2
7
1
7
2
ð9Þ pffiffiffi 1 L
7
7
3
7
2
! 7
7
1
1
7
L7
ð18Þ pffiffiffi ð6Þ pffiffiffi
7
3
3
7
7
2
7
7
1
2
5
ð3Þ pffiffiffi þ 1 L
3
4.8 Chapter Summary Using a Numerical Example
229
1ffiffi, the matrix is
Using x ¼ p
2
3
1 2
6
ð36Þ pffiffiffi
6
3
6
6
6
2
6
6 ð18Þ p1ffiffiffi ð6Þ p1ffiffiffi
6
6
3
3
6
6
6
1 2
6
6
ð36Þ pffiffiffi
6
3
6
6
2
6
6
1
1
4 ð18Þ pffiffiffi þ ð6Þ pffiffiffi
3
3
!
1 2
1
L
ð18Þ pffiffiffi ð6Þ pffiffiffi
3
3
2
1
ð3Þ pffiffiffi 1 L2
3
!
1 2
1
ð18Þ pffiffiffi þ ð6Þ pffiffiffi
L
3
3
!
2
1
ð9Þ pffiffiffi 1 L2
3
1 2
ð36Þ pffiffiffi
3
2
1
1
ð18Þ pffiffiffi þ ð6Þ pffiffiffi
3
3
1 2
ð36Þ pffiffiffi
3
2
1
1
ð18Þ pffiffiffi ð6Þ pffiffiffi
3
3
Simplifying the two matrices results in
2
6
12
6
6
6
6
6
6 6 pffiffiffi L
6
3
6
6
6
6
12
6
6
6
6
4 6 þ p6ffiffiffi L
3
6
6 pffiffiffi L
3
6
4 pffiffiffi L2
3
6
6 þ pffiffiffi L
3
2L2
12
6
6 þ pffiffiffi L
3
12
6
6 pffiffiffi L
3
6
6 þ pffiffiffi L
3
3
7
7
7
7
7
2
7
2L
7
7
7 and
7
6
6 pffiffiffi L 7
7
3 7
7
7
6
4 þ pffiffiffi L2 5
3
2
3
6
6
6
p
ffiffi
ffi
p
ffiffi
ffi
L
12
6
L 7
12
6þ
6
7
3
3
6
7
6
7
6
7
6
6
6
2
2
6 6 þ pffiffiffi L
7
p
ffiffi
ffi
p
ffiffi
ffi
4
þ
L
L
2L
6
6
7
3
3
3
6
7
6
7
6
7
6
6
6
12
6 þ pffiffiffi L 7
12
6 pffiffiffi L
6
7
3
3
6
7
6
7
6
7
6
6
2
2 5
4 6 p6ffiffiffi L
6 þ pffiffiffi L
2L
4 pffiffiffi L
3
3
3
Summing the two matrices produces
2
3
24
12L
24
12L
6 12L
8L2 12L 4L2 7
7
6
7
6
4 24 12L
24
12L 5
12L
4L2
12L
8L2
Placing this matrix back in Eq. (4.117) yields
2
3
24
12L
24
12L
2
2 7
6
12L 4L 7
8L
EI 6 12L
½k ¼ 3 6
7
2L 4 24 12L
24
12L 5
12L
4L2
12L
8L2
3
!
1 2
1
L 7
ð18Þ pffiffiffi þ ð6Þ pffiffiffi
7
3
3
7
7
!
7
2
7
1
7
2
ð9Þ pffiffiffi 1 L
7
7
3
7
2
! 7
7
1
1
7
ð18Þ pffiffiffi ð6Þ pffiffiffi
L7
7
3
3
7
7
2
7
7
1
2
5
ð3Þ pffiffiffi þ 1 L
3
230
CHAPTER 4 Element Stiffness Matrix
Dividing through by 2 provides the final equation as
2
12 6L
6
EI 6 6L 4L2
½k ¼ 3 6
L 6
4 12 6L
6L
2L2
12
6L
3
7
6L 2L2 7
7.
12 6L 7
5
6L
(4.118)
4L2
REFERENCES
Griffiths, D.V., 1990. Treatment of skew boundary conditions in finite element analysis.
Computers & Structures 36 (6), 1009e1012.
Turner, M.J., 1959. The direct stiffness method of structural analysis. In: Structural and
Materials Panel Paper, AGARD Meeting, Aachen, Germany.
CHAPTER
Material Laws and
Properties
5
King H. Yang
Wayne State University, Detroit, Michigan, United States
5.1 MATERIAL LAWS
By now, you are well aware of how to derive element shape functions, and how to
use these functions and Gauss quadrature to develop the element stiffness matrix
for an isotropic linear elastic material. In this chapter, some commonly used material
laws are discussed along with the methodology including how to measure associated
properties.
Our world comprises a large variety of materials. Typical engineering materials,
such as steel and aluminum, are well characterized, but there are also numerous
ceramic materials, biological tissues, synthetic composite materials, and metal
alloys (e.g., magnesium alloy) that are not well described in scientific literature.
Most commercially available FEA software packages contain a substantial library
of material laws from which users can choose. For example, the material library
in the LS-DYNA package (LSTC, Livermore, CA) has more than 100 material
laws, with provisions for users to implement their own material laws to supplement
any deficiencies in the provided package.
For many materials, responses vary based on the speed with which it is lengthened
or compressed. As such, a single material law without loading-rate dependency
is not adequate for all simulation analyses. Thus, it is conceivable that for
different types of engineering analysis, different material laws are needed to
obtain acceptably accurate results. For example, a linear elastic material law
would probably suffice for determining the efficacy of a structural design or
for estimating the risk of a structural failure under load. However, an elastice
plastic law is needed to investigate the energy-absorbing capability of a car
crashing into a wall. In most of the cases, for the simulation of high-speed
impact responses of soft biological tissues (e.g., a response to a baseball player
being struck by a fast ball), a strain-rate dependent, viscoelastic material law
needs to be implemented.
As mentioned previously, a constitutive equation provides the characteristics
of the material in terms of the stressestrain relationship. We understand from
Chapter 4 that a stressestrain relationship can be identified by differentiating
the strain energy density with respect to strain (Eq. 4.19 s ¼ 12 sε ¼ 12 Eε2 ).
However, only experimental data can provide the stressestrain relationships
Basic Finite Element Method as Applied to Injury Biomechanics. http://dx.doi.org/10.1016/B978-0-12-809831-8.00005-2
Copyright © 2018 Elsevier Inc. All rights reserved.
231
232
CHAPTER 5 Material Laws and Properties
needed to derive constitutive laws. Additionally, experimental data are needed to
deduce the necessary constants in the constitutive equations to fully define the
material behaviors. This chapter addresses several material laws that are
commonly used to represent engineering materials and biological tissues. In
addition to the information on the basic material laws covered here, you are
encouraged to read about the many other material laws that are described in various
FEA software manuals.
5.1.1 LINEAR ELASTIC MATERIAL
An elastic material will return to its original shape once the external load is
removed. Additionally, the stressestrain curves due to loading and unloading
are identical in an elastic material. For a linear elastic material, the relationship
between the stress and strain is linear, whereas a nonlinear elastic material exhibits
a nonlinear stressestrain relationship. Although materials in the real world are
mostly nonlinear and inelastic, the linear elastic constitutive equation is the
most widely used to represent any solid material that is subjected to infinitesimal
strains.
There are no universally accepted definitions for an “inelastic” material. Some
suggest that any material that does not follow the definition of an elastic material
is considered an inelastic material. Hence, rigid body, plastic, incompressible, strain
hardening, strain softening, and viscoelastic materials can all be classified as inelastic material. In using the term “inelastic” our intention is to show that many
real-world materials are too complex. As such, they are simplified and approximated
as elastic material when conducting FEA.
In Section 2.1, we show the difference between engineering strain (DL/L0) and
true strain (DL/L) for a 1D material. As long as the deformation (DL) is “small,” the
difference between these two strain measures is negligible. The question is: how
small is considered small? To partially answer this question, we must return to
the original definition of strain, which is the ratio of deformation to length.
The generalized deformation {D} can be expressed in terms of generalized
displacements {u} as {D} ¼ {u(x þ dx) u(x)}. Expanding {u(x þ dx)} using
2
Taylor series, we write fuðx þ dxÞgzuðxÞ þ vu
vx dx þ O dx , where “O” represents
all other second order terms. We can neglect all second order terms, when dx is infinitesimally small and dx2 becomes negligible. For this reason, the term “infinitesimal” deformation is also known as “linearized” deformation. Following this
linearized deformation, the infinitesimal strain is the infinitesimal deformation
divided by small length dx. Thus, the infinitesimal strain for the 1D element is
fεxx g ¼ du
dx, and that for 3D element is
5.1 Material Laws
2
8
εxx
>
>
>
>
>
>
εyy
>
>
>
>
< ε
zz
>
g
> xy
>
>
>
>
>
gyz
>
>
>
:
gzx
8
εxx
>
>
>
>
>
>
εyy
>
>
>
>
< ε
zz
¼
>
>
2ε
> xy
> >
>
>
>
>
> 2ε
> >
>
>
>
>
> yz
>
>
:
; >
2εzx
9
>
>
>
>
>
>
>
>
>
>
=
6
6
6
6
6
6
9 6
6
6
>
>
6
>
>
6
>
>
6
>
>
>
6
>
= 6
6
¼6
>
6
>
>
6
>
>
> 6
>
6
>
>
6
>
; 6
6
6
6
6
6
6
6
4
3
v
vx
0
0
v
vy
0
v
vz
0 7
7
7
7
7
7
v
0 7
7
vy
7
7
7
8 9
v 7
7> u >
0
7
vz 7< =
7 v ;
7>
v
;
: >
0 7
w
7
vx
7
7
7
v v 7
7
vz vy 7
7
7
7
v 7
5
0
vx
0
as shown in Eqs. (1.9)e(1.11).
Many researchers suggest that a “few” percentage points of strain can apply
infinitesimal strain theory without defining the magnitude explicitly. In our own laboratory, we allow up to 5% of stretch to be considered small on special occasions.
Note here that at 5% stretch, the engineering strain is 5%, stepwise true strain is
4.762%, and the natural log of the true strain is 4.879%. In all commercially available software packages, postprocessors allow users to output any strain measures.
Hence, we recommend that you properly select the material law when developing
your FE models and choose responsible variables in accordance with the material
law selections.
Fig. 5.1 shows an idealized engineering stressestrain curve for typical mild steel
under tension. Although the curve is rather complex, a linear region of the curve
exists where the strain is relatively small. In this region, the slope of the curve is
called Young’s modulus, and it represents the stiffness of the material. This material
property has been used in all examples presented so far. In problems where the risk
of failure is of major concern, the linear elastic material model is generally used to
identify the peak stress under a foreseeable peak-loading condition. If the ratio of the
peak stress to yield stress is higher than the preestablished safety factor, the structure
is considered safe. Here the term “safety factor,” also known as “factor of safety” in
structural mechanics, is defined as the ratio of the design stress to yield stress. Generally speaking, all engine components require a safety factor of 6e8, bridge components 5e7, structural steelworks 4e6, etc. For nuclear power plants and space
vehicles, the required safety factors are even higher. Please note that even though
all components are designed with a safety factor of n, the safety factor for the entire
structure may be less than n, because the assembly processes may reduce the safety
factor.
233
234
CHAPTER 5 Material Laws and Properties
FIGURE 5.1
An idealized stressestrain diagram of a slender bar made of mild steel subjected to tensile
loading. Initially, the stressestrain curve behaves in a linear manner (linear region) until
the elastic limit. The slope of the linear region (E ) is known as Young’s modulus. At the
point marked as “yield stress,” the strain magnitude is around 1.25% for a mild steel.
After passing the “yield stress (and the corresponding yield strain)” point before entering
into the “strain hardening” region, this idealized material behaves in a perfectly plastic
manner. After passing the point marked as “ultimate stress,” the diameter of the test
sample begins to shrink, and hence this region is called the “necking” of the specimen.
The linear elastic material law is also very useful for modeling the hard tissue in a
biological system that consists of multiple tissue types. For example, the response of
a femur or tibia during sports activity can be modeled as a linear elastic material,
because the strains in these long bones are well below their respective yield strains.
Aside from Young’s modulus, Poisson’s ratio is also needed to fully characterize
a linear elastic material. When an elastic specimen is subjected to uniaxial loading, it
deforms not only in the direction of loading but also distorts in directions perpendicular to the load. By definition, Poisson’s ratio is the negative ratio of transverse strain
to axial strain, n ¼ εlateral/εaxial. Because most of the materials expand sideways
when subjected to axial compression, a negative sign will generate a positive
Poisson’s ratio. By this definition, almost all materials have a positive Poisson’s ratio
ranging from 0 to 0.5. An exception was found by Professor Roderic Lakes, who
reported that a negative Poisson’s ratio could be found in certain manufactured
foam materials (Lakes, 1987).
Examples of materials with near-zero Poisson’s ratio, in which no sideways
expansion or contraction is displayed upon axial loading (εlateral ¼ 0), include
low-density foam, cork, and highly osteoporotic trabecular bone. A practical application for a material with a near-zero Poisson’s ratio is the cork seal for a wine bottle.
The cork must allow easy insertion into and removal from the bottle and at the same
time can resist the internal pressure created as wine develops and matures.
5.1 Material Laws
For rubber materials and brain tissues, which display a nearly incompressible
behavior, Poisson’s ratio is very close to 0.5. Many beginners think steel is incompressible and rubber is compressible. This common misconception is due to the fact that an
incompressible material is classically defined as a material that exhibits no volume
change upon loading. Despite large shape changes upon loading, a rubber material
is considered an incompressible material as long as the volume is conserved. Volumetric strain is defined as the change in volume divided by the original volume, i.e.,
DV/V. For a cube with dimensions of a b c, we express the volumetric strain as
DV ða þ DaÞðb þ DbÞðc þ DcÞ abc
¼
V
abc
¼ ð1 þ εxx Þð1 þ εyy Þð1 þ εzz Þ 1 z εxx þ εyy þ εzz :
Thus an incompressible material must satisfy εxx þ εyy þ εzz ¼ 0. Assume that
the cube is axially loaded along the z-axis and εxx ¼ εyy, we deduce that
εxx ¼ 0.5εzz. Based on the definition, the Poisson’s ratio for an incompressible material is calculated as
n¼
lateral strain
0:5εzz
¼
¼ 0:5:
axial strain
εzz
A Poisson’s ratio of 0.5 simply indicates that the volume is conserved. Unlike
cork, rubber could not be used as a wine bottle sealer, because the compressive force
needed to insert the sealer would cause a large sideway expansion and jam the bottle.
To measure Poisson’s ratio, we can simply attach two strain gages on an axially
loaded specimen, with the first gage aligned in the axial direction and the second
gage aligned perpendicular to the first. There are many other methods reported in
scientific literature to measure this ratio, and we recommend that you study these
measurement methods on your own.
For an isotropic linear elastic material, only two material constants, Young’s
modulus and Poisson’s ratio, are needed to fully describe the material. All other
constants, such as shear modulus, bulk modulus, can be calculated from these two
constants. For an orthotropic linear elastic material, nine material constants are
required to fully describe this material. They are the three Young’s moduli: Exx,
Eyy, and Ezz, three Poisson’s ratios: nyz, nzx, and nxy, and three shear moduli: Gyz,
Gzx, and Gxy. Orthotropic materials are described in more detail in Section 5.1.5.
5.1.2 ELASTICePLASTIC MATERIAL
In a crashworthy vehicle, many components are designed to serve double purposes.
For example, while the chassis is designed to provide the main structural support and
attachment points for other components of the car, it also serves as the leading safety
component to absorb crash energy, and thus to lower the severity of the impact. As
discussed in Chapter 4, it is known that strain energy can be calculated from the area
under the forceedeformation curve. Because the yield strain (the strain magnitude at
the yield point) is very low in magnitude, as shown in Fig. 5.1, the elastic portion of
235
236
CHAPTER 5 Material Laws and Properties
σ
σ
ε
ε
FIGURE 5.2
(Left): Characteristic stressestrain responses for a mild steel and an aluminum alloy.
(Right): An idealized elasticeplastic material law used to represent these materials, which
are capable of sustaining large plastic deformations that absorb greater energy. The
elastic modulus E and tangent modulus Et, shown in the figure are provided by model
developers before FE simulations can commence.
the forceedeformation curve can absorb only a very limited portion of the crash
energy. To lower the enormous amount of kinetic energy during a crash, the plastic
portion of the forceedeformation curve is far more significant than the small elastic
region. To model this kind of problem, the material law needs to accommodate the
plastic region in addition to the elastic portion. Fig. 5.2 (left) shows idealized stresse
strain responses for a mild steel and an aluminum alloy. These material characteristics can be idealized and simplified as elasticeplastic materials, as shown in Fig. 5.2
(right). In addition to Young’s modulus and Poisson’s ratio, we need the yield strain
(stress) and the tangent modulus Et for each material to fully define the elastice
plastic behaviors.
5.1.3 HYPERELASTIC MATERIAL
An ideal hyperelastic material can sustain a very large deformation and then recover
fully once the load is removed (Fig. 5.3). Exemplary materials that follow the
hyperelastic constitutive law are rubber and sponge. During the vulcanization processes of rubber, crosslinks are formed among the polymer chains, which in turn
make it possible for the material to recover fully from deformations. This type of
material is frequently used in modeling tires, engine mounts, and some energyabsorbing foams used in automobiles for reducing impact severities.
Several hyperelastic models based on different formulation methods are available in most software packages. Examples are the ArrudaeBoyce hyperelastic
rubber model (Arruda and Boyce, 1993), BlatzeKo hyperelastic rubber model
(Blatz and Ko, 1962), MooneyeRivlin incompressible hyperelastic rubber (Mooney,
1940; Rivlin, 1948), and hyper-viscoelastic rubber (Ogden, 1984). Because hyperelastic materials are nearly incompressible in nature, the bulk modulus of the material
is much higher than the shear modulus. Hence, the most critical parameter required
5.1 Material Laws
FIGURE 5.3
A typical stressestretch diagram for a hyperelastic material. The stiffness of the material
is usually higher under axial compression than under tension.
is the shear modulus as a function of time or loading rate. You are encouraged to read
descriptions of these material laws and try them out before deciding which one is
best suited for your rubber modeling project.
5.1.4 VISCOELASTIC MATERIAL
A viscoelastic material possesses both viscous and elastic characteristics under
loading. A vehicular shock absorber, made of a spring and a damper, is a typical
structural component that behaves in a viscoelastic manner. The damper reduces
the shock and absorbs some of the energy, while the spring serves to return the
component back to its original position. There are three main characteristics of
viscoelastic materials: creep, stress relaxation, and hysteresis. The creep phenomenon is used to describe the continued deformation of a viscoelastic material after the
load has reached a constant state (Fig. 5.4A). Under a constant deformation, the
stress relaxation describes the continued reduction in stress inside a viscoelastic material (Fig. 5.4B). A hysteresis loop, the shaded area shown in Fig. 5.4C, describes
the differences in loading and unloading curves in a viscoelastic material as well as
the energy dissipated. Based on these three properties, it is easily understood that a
material loaded at a faster rate will have a higher peak stress than a material loaded at
a lower rate (Fig. 5.4D). The higher the loading rate lesser the stress relaxation, and
hence a higher peak stress.
Just as a shock absorber is made of a spring and a damper, the behaviors of a
viscoelastic material can be approximated by a proper arrangement of springs and
dampers. Some commonly used springedamper models are the KelvineVoigt
model (a spring and a damper in parallel) and Maxwell model (a spring and a
damper in series). A commonly used viscoelastic material law is based on the linear
237
238
CHAPTER 5 Material Laws and Properties
(A)
(B)
(C)
(D)
FIGURE 5.4
Characteristics of a viscoelastic material include: (A) creep, (B) stress relaxation, and
(C) hysteresis. (D) The peak stress due to a high loading rate applied to a viscoelastic
material is higher than that of a low loading rate, because a low loading rate affords a
longer time for the material to relax during loading.
viscoelastic properties reported by Herrmann and Peterson (1968). This material is
formulated by superposition of a linearly viscoelastic material and an elastic material under hydrostatic pressure. Like a hyperelastic material, a viscoelastic material
has a very high bulk modulus in comparison to its shear modulus. Hence, the
behavior of a viscoelastic material is best described through its shear deformation,
with the time-dependent shear relaxation modulus defined as
GðtÞ ¼ GN þ ðG0 GN Þebt ;
(5.1)
where G0 is the short-time shear modulus, GN is the long-time shear modulus, and b
is the decay constant.
Because brain tissues have a very high water content, brain has been classified as
a nearly incompressible material. Both the hyperelastic and viscoelastic material
laws have been used to model responses of the brain. Brain tissues also exhibit a
higher stiffness due to axial compression as compared to that in tension
(Jin et al., 2013; Miller and Chinzei, 2003). Additionally, Jin et al. (2013) reported
that brain tissues are loading-rate-dependent. To mimic the complex brain material
properties, a linear viscoelastic material law (e.g., Mao et al., 2013), a hyperelasticviscoelastic material law (e.g., Kleiven, 2007), and a heterogeneous anisotropic
hyperelastic-viscoelastic material law (Sahoo et al., 2016) have all been used in
the modeling of the human brain.
5.1 Material Laws
5.1.5 ORTHOTROPIC MATERIAL
An orthotropic material has different material properties along the three mutually
perpendicular axes. Typical orthotropic materials include wood and continuous
fiber-reinforced composites. In order to fully describe this class of material, a total
of nine material properties is needed: three elastic moduli (Exx, Eyy, Ezz), three shear
moduli (Gxy, Gyz, Gzx), and three Poisson’s ratios (nxy, nyz, nzx). The constitutive
equations for an orthotropic material can be written as
2
3
6 1 yyz nzy yyx þ yzx nyz yzx þ yyx nzy
0
0
0 7
78
8
9 6
9
Eyy Ezz D
Eyy Ezz D
6 Eyy Ezz D
7
εxx >
s
6
7>
>
>
xx
>
>
>
>
6
7
>
>
>
>
>
>
>
6 yxy þ yxz nzy 1 yzx nxz yzy þ yzx nxy
7>
>
>
>
>
>
>
>
>
6
7
>
>
>
>
0
0
0
ε
s
>
>
>
>
yy
yy
6
>
>
>
>
7
E
E
D
E
E
D
E
E
D
>
>
>
>
zz xx
zz xx
zz xx
>
>
>
>
6
7
>
>
>
>
>
>
> 6
>
7<
<
=
szz
6 yxz þ yxy nyz yyz þ yxz nyx 1 yxy nyx
7 εzz =
7
¼6
;
0
0
0 7>
6 E E D
>
gxy >
sxy >
Exx Eyy D
Exx Eyy D
>
>
>
6 xx yy
7>
>
>
>
>
>
>
>
6
7>
>
>
>
>
>
>
>
6
7>
>
>
>
>
>
>
>
>
6
7
0
0
0
G
0
0
g
xy
>
> syz >
> 6
>
yz
>
>
>
7>
>
>
>
>
>
>
>
6
7
>
>
> 6
>
:
:
;
7
0
0
0
0 Gyz 0 7 gzx ;
szx
6
6
7
4
5
0
0
0
0
0 Gzx
(5.2)
where
D¼
1 nxy nyx nyz nzy nzx nxz 2nxy nyz nzx
; and
Exx Eyy Ezz
(5.3)
Eyy
Ezz
Exx
yxy ; yzy ¼
yyz ; yxz ¼
yzx .
Exx
Eyy
Ezz
(5.4)
yyx ¼
A subset of orthotropic material is the transversely isotropic material. In this
class of material, the two elastic moduli in the transverse plane are identical, while
the elastic modulus along the axial direction is different. Typical transversely
isotropic materials include unidirectional fiber-reinforced composite materials,
rolled steel rod, and long bones in human legs. According to Wolff’s law, bone
will be deposited along the direction of gravity (Wolff, 1986), and this is the explanation for why long bones in legs are transversely isotropic. Understandably then,
human femurs and tibiae tend to have higher elastic moduli along the axial direction
than in the two directions in the transverse plane. As such, human long bones are
frequently modeled as transversely isotropic materials.
Assume a transversely isotropic material has the same elastic modulus in the yez
plane (Eyy ¼ Ezz) and a higher elastic modulus in the axial direction (Exx). We can
use Eq. (5.4) to determine that nyz ¼ nzy. Also, nyx is smaller than nxy because Exx
239
240
CHAPTER 5 Material Laws and Properties
is greater than Eyy. Furthermore, the isotropy in the transverse plane requires that
Eyy
. These values can be substituted into Eq.
nxy ¼ nxz, Gxy ¼ Gxz, and Gyz ¼ 2ð1þn
yz Þ
(5.2) to identify the constitutive equations for a transversely isotropic material.
5.1.6 FOAM MATERIAL
Foam materials are frequently used to absorb compression-induced energy. However, this class of material is not stiff when subjected to tensile or shear loading.
Critically, the properties of materials in this class are frequently dependent upon
the rate of loading. The stressestrain curves of typical foam materials are extremely
complex and substantially vary in a manner dependent upon choices of materials and
the extent of air (void) spaces. For example, a stressestrain curve of polyurethane
foam has an initial elastic-yield region followed by a plateau compaction region
and then a densification region (Fig. 5.5). In contrast, polyethylene foam has an
initial low stiffness (toe) region, followed by a densification region, where the
stiffness is much higher than in the toe region. The characteristic behavior of ligaments is similar to that just described for polyethylene. A third foam material is
polystyrene, which has an initial stiffness that is very high and becomes lower after
passing a certain strain magnitude.
Polyethylene and polyurethane foams are the most common forms of foam
material. Designers of vehicle interiors typically use polyurethane foam as a filler
material to reduce the severity of head impacts against the interior header/pillar
during automobile accidents. Polyurethane foam is also frequently used to make
consumer products, such as bedding, sofas, and carpet underlay. Polystyrene foams,
on the other hand, are frequently used as filler materials in car bumpers and knee
bolsters to absorb substantial magnitudes of impact energy.
σ
ε
FIGURE 5.5
Typical stressestrain curves for polystyrene, polyethylene, and polyurethane foams. The
tensile behavior of human trabecular bone is similar to a polyurethane foam. Many
biological soft tissues behave like a polyethylene foam, with a toe region followed by linear
region prior to the point of failure.
5.2 Material Test Strategy and Associated Property
There are many other engineering materials that can be formed into special
cellular shapes to absorb a great amount of energy at nearly constant stress during
deformation. For example, paper and aluminum honeycombs are manufactured
through extrusion of 2D hexagonal cells into 3D (Gibson and Ashby, 1997). On
the other hand, Skydex are 3D cellular lightweight material, which consists of layers
of periodic twin-hemispherical microstructures made of thermoplastic polyurethane.
Zhu et al. (2013) tested panels of Skydex material at strain rates of 0.01e10 s1,
developed a 3D FE model, and validated the model against experimental data.
Despite the highly nonlinear feature of the stressestrain curve, the model was
able to capture the behaviors at low strain rate. Human cancellous bone, which
has a cell-like structure, can also be modeled using constitutive equations developed
for various foam materials. In particular, the tensile stressestrain curve for cancellous bone, reported by Carter et al. (1980), has a shape that is nearly identical to that
of a typical polyurethane foam.
Fig. 5.5 depicts that there is no such thing as a typical stressestrain curve for a
foam material. Because a set of constitutive equations can only be derived from
experimental results, we can easily understand that each different class of foam material requires a specific set of constitutive equations to fully describe their behaviors. From this understanding, we can easily deduce that material constants
needed to model a foam material are highly dependent upon the choices of material
laws. As the stressestrain curve is extremely complex, some software packages provide curve-fitting subroutines to allow the users to input experimental data, which is
then used in the software interpolation routines to determine which properties to
adopt. It is important to carefully read through the user manual before deciding
which material law best suits the purpose.
5.1.7 MATERIAL DEFINED BY EQUATION OF STATE
When modeling the deployment phase of an airbag or a blast event, the expansion of
the gas volume happens quickly and it is impossible to model such a quick event
using solid elements. In such scenarios, the gas volume is determined from an equation of state (EOS), which relates material parameters such as pressure and temperature to gas volume and gas density. Due to the enormous numbers of causalities
during recent conflicts in Iraq and Afghanistan, Chapter 18 is devoted to methods
related to simulating blast events based on applying the EOS.
5.2 MATERIAL TEST STRATEGY AND ASSOCIATED PROPERTY
The American Society for Testing and Materials (ASTM) International prescribes
numerous material testing protocols for identifying different properties of materials.
The standards for testing engineering materials are fairly complete. Thus material
properties for common structural materials are readily available. Table 5.1 lists
the elastic moduli and Poisson’s ratios for some commonly used engineering materials compiled from various engineering handbooks.
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Table 5.1 List of Young’s Modulus and Poisson’s Ratio for Commonly Used
Engineering Materials
Material
Young’s Modulus (GPa)
Poisson’s Ratio
Wood
Concrete
Magnesium
Aluminum
Cast Iron
Brass
Bronze
Titanium alloy
Fiber-reinforced composite
Mild steel
Tungsten
10e20 along fiber direction
15e30
40e45
65e70
100e130
100e125
95e130
100e120
70e400
200
400
0.3e0.45
0.1e0.2
0.28e0.35
0.32e0.35
0.2e0.3
0.33e0.36
0.34e0.36
0.32
0.2e0.35
0.27e0.3
0.25e0.3
Unlike engineering materials, standards for testing biological tissues are far from
perfect. Biological tissues tend to behave nonlinearly and are known to be age- and
loading-rate dependent. More critically, biological tissues are difficult to acquire due
to ethical concerns, and it is challenging to cut the tissues into specific shapes, such
as the dog boneeshaped specimen required in many ASTM testing specifications.
Compared to the stressestrain curves for engineering materials, biological
tissues tend to behave in a manner that is exceedingly dependent on test conditions.
For example, a confined compression test produces a distinctly different result than a
nonconfined compression test, and a friction-free compression test generates disparate results from tests with friction present between the specimen and the test bed.
Additionally, biological tissues contain hierarchical structures that need to be
considered before testing is conducted. For example, when measuring the mechanical properties of a vertebral body, the size of the specimen cannot be too small; if it
were, the continuum mechanics assumptions, upon which the FE method is based,
would be violated. Understandably, there are wide ranges of properties reported in
literature for biological tissues. The following sections describe some commonly
used test methodologies for testing biological tissues.
5.2.1 EXPERIMENTAL TYPES FOR BIOLOGICAL TISSUE TESTING
Generally speaking, experiments conducted to determine mechanical properties can
be separated into two major categories: (1) component, subsystem, or whole system
tests, and (2) material tests. A component, subsystem, or whole system test is usually
designed to identify structural responses and integrity, while a material test tends to
be used to obtain stressestrain curves of specimens with simple geometries.
A sled system can be used for dynamic testing of a whole system subjected to a great
amount of energy (Fig. 5.6). Accelerometers, load cells, and other instrumentations
5.2 Material Test Strategy and Associated Property
FIGURE 5.6
Evaluation of NASCAR restraint system on a pneumatically driven sled testing system.
in conjunction with high-speed videos are used to acquire the test results. For testing
subsystems or components, a drop stand based on gravitational force or a pneumatically driven linear impactor is sufficient (Fig. 5.7).
Most of the material tests are conducted using servo hydraulic universal testing
systems capable of uniaxial (tension/compression) or biaxial (tension/compression
and torsion) loading. Such a system usually comes with a force/moment load cell
FIGURE 5.7
(Left) A double-wire drop test system for testing the efficacy of a football helmet. (Right) A
pneumatically driven linear impactor system.
(Left) Photo courtesy of Professor Liying Zhang, Wayne State University, Detroit, MI.
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CHAPTER 5 Material Laws and Properties
FIGURE 5.8
A universal material testing system capable of conducting tension/compression and
torsion tests.
and linear and rotational displacement measurement devices (Fig. 5.8). Forcee
displacement (F-d ) and momenteangle (M-q) diagrams can be created from data
obtained from such a system. If the specimen is of uniform geometry, such as a
dog boneeshaped specimen, the stressestrain curve can be calculated from the
forceedeformation data. Many other material properties, such as Young’s modulus,
shear modulus, bulk modulus, and Poisson’s ratio, can be gathered from the use of a
universal testing system.
For quasi-static testing, a very slow loading rate is used to acquire test data from
a universal testing system. This mode of loading is particularly useful in the observation of post-yield deformation behavior as well as failure mode. However, the
effects of inertial and damping forces are excluded in quasi-static testing. There
are specially designed machines that allow material testing at high speeds. However,
as the speed increases, it becomes more difficult to stop the loading ram in a short
distance. Therefore only tensile testing is allowed, because a high-speed compression test may not stop in time, and this may result in damage to the machine.
Many soft tissues are too compliant to be cut into a specimen of a specific shape
for material testing. In such a case, an indentation test is a good choice for obtaining
the material properties (Fig. 5.9). In this type of tests, a small probe (indentor) is
pushed into the specimen to a preset depth while the displacement and force are
recorded. To obtain creep responses, the probe may be held in place for a certain
period of time. The hardness and modulus can be calculated from the unloading
curve in accordance with the shape of the probe, such as spherical, conical,
5.2 Material Test Strategy and Associated Property
Spherical
Indenter
Brain Tissue
Sample
FIGURE 5.9
Microindentation test of rat brain tissue.
Courtesy of Professor Kurosh Darvish at Temple University, Philadelphia, PA.
paraboloid, flat, etc. (Oliver and Pharr, 2004). This experimental method is quite
useful for obtaining in vivo material properties in cases where cutting a piece of
specimen from the subject is prohibited.
To obtain the shear modulus, a simple shear test can be conducted on a universal
testing system. Additionally, a dynamic shear rheometer or a rotational rheometer for
an oscillatory mode can be used to acquire the storage modulus G0 and loss modulus
G00 . For a more liquid-like material, the loss modulus is larger than the storage
modulus, while the opposite is true for a more gel-like material. Temperature control
in these experiments is important, because the shear property varies with changing
temperature. Instead of reporting these two parameters, a combined parameter named
complex modulus (G*) is conventionally reported for shear properties, where
G ¼ G0 þ i G00 :
(5.5)
Note that the storage and loss moduli described here are different from the shortand long-term moduli described in Section 5.1.4. Depending on the frequency (u)
with which the material is tested, the approximate relationships between these
two descriptions of shear material properties are expressed as
G0 ¼ GN þ ðG0 GN Þ
u2
u 2 þ b2
ub
G ¼ ðG0 GN Þ 2
;
u þ b2
00
where b is the decay constant as shown in Eq. (5.1).
(5.6)
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CHAPTER 5 Material Laws and Properties
In certain materials, mechanical properties obtained from loading along different
axes are substantially different. For example, fiber-reinforced composite material
and human aorta both have properties that are different along the axial and transverse directions. For this class of material, multiple tests need to be conducted under
uniaxial testing conditions to fully characterize the material behaviors. Biaxial
testing apparatus may be useful for determining the properties of such materials.
Examples of biaxial testing equipment include an equal biaxial tester, for which a
cruciform-shaped sample is used; a bubble inflation test device; and an equal biaxial
tester, which is the best choice for testing tissue samples grown on a foundation
membrane.
A high-speed equal biaxial tester, reported by Mason et al. (2005), uses a pneumatically driven cam mechanism to stretch the clamped cruciform samples at equal
speeds along each axis (Fig. 5.10A). A cruciform-shaped sample allows all major
FIGURE 5.10
(Top Left): A pneumatically propelled, cam-driven testing device that provides equal
biaxial loading. Each loading ram has a clamp for the attachment of the specimen and a
load cell to measure the force. (Top Right): A cruciform-shaped human aorta sample
clamped in the test apparatus. (Bottom Left): An equal biaxial tester for stretching a
membrane with three test tissue samples adhered to the membrane. During the test, the
well moves downward while the indenter stays at the same place to stretch the membrane
(green colored, dark gray in print version). (Bottom Right): As the well moves downward,
equal biaxial stretches are applied to the membrane and the tissue samples.
Bottom figures courtesy of Professor Barclay Morrison III at the Columbia University, New York, NY.
5.2 Material Test Strategy and Associated Property
deformations to occur at the central region, which in turn allows easy calculation of the
in-plane stressestrain responses. For 3D deformation, laser measuring devices can be
added to determine the amount of thinning of the specimen in the normal direction.
A bubble inflation on circular samples is performed by clamping the specimen
between a bottom metal plate and a top metal ring. By injecting high-pressure air
into the space between the bottom plate and the specimen, the specimen will inflate
into a dome shape. High-speed videos are used to collect the deformation time
histories so that the corresponding stressestrain relationships can be calculated.
Use of a device based on this principle requires the calculation of 3D deformation
patterns, and hence the data processing can be somewhat difficult.
Another test device based on a similar principle is shown in the bottom of
Fig. 5.10. A substrate membrane is attached to an external circular well, and pulling
down the well causes the membrane (and the tissue samples adhere to it) to stretch
equally on the surface. Again, a high-speed video is used to obtain the deformation
patterns from which the stressestrain relationship can be obtained. This device eliminates the need to calculate 3D deformation patterns, and hence it is easier than the
bubble inflation device in terms of data processing.
Because of the low stiffness values, large deformations are observed in most soft
material/tissue testing. Hence, it is important that true stresses and true strains are
measured during these tests.
The aforementioned test methodologies are satisfactory for measuring the
mechanical behaviors from quasi-static to medium loading rates, such as those
seen in contact sports and car crashes. In recent military conflicts, explosions caused
by improvised explosive devices (IEDs) have resulted in substantial numbers of casualties to military personnel and equipment. These types of explosions produce
high-rate impacts. Behaviors of materials exposed to explosion are quite different
from those of materials exposed to more typical impacts under medium loading
rates. To determine these high-rate material properties, many researchers have
used a split Hopkinson pressure bar (SHPB), which is based on the principle
reported by Bertram Hopkinson (January 1874eAugust 1918) and later enhanced
by Herbert (Harry) Kolsky (September 1916eMay 1992).
Briefly, an SHPB has an incident and a transmitted bar, with a thin specimen
placed in between the 2 bars. A pressurized gas gun is used to generate an elastic
pressure pulse in the incident bar (Fig. 5.11). At the interface between the incident
bar and the specimen, the elastic stress wave is partially reflected and partially transmitted to the thin specimen, and plastic deformations are induced. At the next
interface, which is between the specimen and the transmitted bar, the stress wave
is again partially reflected and partially transmitted. The material properties of the
specimen can be deduced from strain histories obtained from strain gages that detect
strains from the incident and transmitted bars.
5.2.2 REVERSE ENGINEERING METHODOLOGY
While traditional engineering materials, in general, are homogeneously distributed
and exhibit no directional dependence, directional dependence is obvious in many
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FIGURE 5.11
A split pressure Hopkinson bar system used for high-rate material property testing.
synthetic materials, such as polymers and fiber-reinforced composites. Testing of
these modern materials requires different protocols from those used for traditional
materials. However, these materials can readily be cut into specific shapes that are
necessary for acquired samples. The same statement cannot be made regarding
acquisition of material properties from biological tissues.
There are many difficulties associated with acquiring samples of biological
tissues. First, it is difficult to acquire biological tissues, especially human. This is
partially due to ethical concerns. Second, biological tissues are age, gender, directional, and strain-rate dependent. Hence, a very large sample size is required in order
to depict the relative contribution of each of these factors. Third, biological tissues
are extremely complex in hierarchical structure, and different regions of the same
class of material have distinct behaviors. For example, the hierarchical composition
of a ligament or tendon consists of several fascicles, each containing a number of
fibrils that consist of a number of subfibrils, which are further divided into numerous
microfibrils. Another example is the regional differences in articular cartilage, which
consists of the superficial zone, middle zone, deep zone, and calcified zone, with
each zone containing different arrangements of cellular structures. Although each
substructure of a ligament or each zone of cartilage plays a role in the overall
mechanical response of the ligament or cartilage, the small dimensions make it
extremely difficult to test the properties of each substructure or zone. Additionally,
extreme needs for computational power would render it unfeasible to explicitly
model the hierarchical or layered structures of such materials. Fourth, many biological soft tissues, such as adipose (fat) or brain tissues, cannot withstand their own
weight when separated from the whole body, thus making it very difficult to determine the magnitudes of stress and strain.
A typical universal material testing system measures only the force/moment and
deformation/torsional angle. If a test sample cannot be cut into an accurate geometric
shape, stress and strain cannot be determined from the force and displacement time
histories. To overcome the inability to have a uniformly shaped material sample,
5.2 Material Test Strategy and Associated Property
Zhu et al. (2010) used a reverse engineering approach with sample-specific FE
models created from laser scanning of each test sample. In this approach, a set of optimization procedures was used to systematically adjust the material properties in the
model until the model calculated forceedeflection curves simulating low and highspeed loading conditions matched those obtained experimentally for the same two
loading conditions. For the final step, the optimized properties obtained from the first
two series of test data were implemented into the model to simulate a third series of
tests conducted at a medium loading rate. If the model predicted forceedisplacement
data well matched with that reported for the third test series, the material properties
obtained from the optimization procedures were considered accurate.
Since no cutting techniques exist for accurate dissection of soft tissues, geometric
variations are quite common. Variations that result from testing samples of different
shapes largely confound experimental data obtained from biological tissues. Using
sample-specific reverse engineering methodologies with protocols similar to those
described above can reduce this large variability, thus removing the uncertainty
due to geometric variations, so that any discrepancies seen between the test and simulation can be attributed only to the properties of the material and not the geometric
variation.
5.2.3 LIST OF COMMON MATERIAL PROPERTIES OF BIOLOGICAL
TISSUES
Because a large quantity of biological tissue samples is expensive and difficult to
obtain, few studies report differences in properties attributed to age and gender.
For those studies involving the age- and gender-dependent properties, the sample
size is typically on the low side. However, Kalra et al. (2015) tested 278 isolated
rib samples taken from 53 males and 29 females, with ages ranging from 21 to
87 years. They reported cortical bone thickness and other biomechanical properties
as functions of age, gender, height, and weight. Even though the sample size was
reasonably large, the biomechanical data obtained from this study was still limited
to a quasi-static loading scheme, and results may not be applicable for simulating
high-speed sports collisions or automotive accidents in the real world. This fact
highlights the need for organized systematic investigations of human tissue properties that are based on age, gender, and loading rate. This consideration is particularly
true since personalized protective equipment appears to be the trend for future
protection of humans in sports activities.
Listing the material properties of biological tissues is not practical due to
different conditions employed in the test as well as variations in age and gender.
However, Table 5.2 lists the material laws and associated properties used in a
50th percentile male human model sponsored by the Global Human Body Modeling
Consortium, developed jointly by the University of Virginia, University of Waterloo,
Wayne State University, and Wake Forest University. While this set of data represents state-of-the-art selections, the material laws and associated properties chosen
may need to be updated as further research data becomes available.
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CHAPTER 5 Material Laws and Properties
Table 5.2 Exemplary Material Laws and Corresponding Properties Used in
the Global Human Body Modeling Consortium 50th Percentile Male Model
Components
Material Laws
Properties
Cervical cortical shell
Power law plasticity
Femur/tibia
Strain-rate dependent
plasticity
Rib cortical shell
Piecewise linear plastic
E ¼ 16.7 GPa, n ¼ 0.3,
k ¼ 0.45, N ¼ 0.3
E ¼ 15.5 GPa, n ¼ 0.3,
separate curves are used to
define Young’s modulus
and yield stress based on
effective strain rate
E ¼ 10.2 GPa, n ¼ 0.3,
Etan ¼ 2.3 GPa,
sy ¼ 65.3 MPa
Cowper Symond coeff.,
C ¼ 2.5 (1/s), P ¼ 7
Compact Bone
Trabecular Bone
Linear elastic
Plastic kinematic
E ¼ 0. 02 GPa, n ¼ 0.3
E ¼ 0.445 GPa, n ¼ 0.3,
sy ¼ 5.3 MPa
Linear elastic
Linear elastic
Linear elastic
E ¼ 0.4 GPa, n ¼ 0.2
E ¼ 0.05 GPa, n ¼ 0.4
E ¼ 0.2 GPa, n ¼ 0.2
Cervical ALL, PLL
Elastic discrete beam
ACL/PCL
Mat soft tissue
Patella tendon
Linear elastic
Based on user defined
curves
Ci’s ¼ 0.0068, 0, 4.2e4,
58.9, 0.2793
B ¼ 4.315 GPa, L ¼ 1.06
Si’s ¼ 0.153, 0.026, 0.348
E ¼ 1.2 GPa, n ¼ 0.3
Lumbar vertebrae
Femur/Tibia
Cartilage
Cervical endplates
Intercostal cartilage
Femur/tibia cartilage
Ligament/Tendon
Internal Organ Parenchyma
Heart tissue
Heart
Lung tissue
Lung
Liver
Hyperelastic rubber
Kidney
Ogden rubber
P ¼ 2.4825 GPa,
Hi’s ¼ 24.26, 40.52, 1.63
B ¼ 0.00266 GPa, d ¼ 0.1,
a ¼ 0.213, b ¼ 0.343,
Li’s ¼ 1.002e6, 2.04
n ¼ 0.49, Vi’s ¼ 3.5e6,
2.8e6
Mi’s ¼ 4.0e6, 2.4e06
Di’s ¼ 9.7e4, 1.6e04
n ¼ 0.49, R ¼ 2.5e4 GPa
m1 ¼ 1.5e4, a1 ¼ 0.18
m2 ¼ 5.26e4, a2 ¼ 0.24
5.3 Building Laboratory-Specific Material Property Library
Table 5.2 Exemplary Material Laws and Corresponding Properties Used in the Global
Human Body Modeling Consortium 50th Percentile Male Modeldcont’d
Components
Material Laws
Properties
Spleen
Viscous foam
Ei ¼ 9.8e5 GPa,
N1 ¼ 3.0, V2 ¼ 0.015,
Ev ¼ 8.5e5 GPa,
N2 ¼ 0.2, n ¼ 0.45
Psoas, erector
Viscoelastic
Leg/thigh muscles
Foot flesh
Elastic
Soft tissue visco
K ¼ 0.0325 GPa,
G0 ¼ 51 kPa, Gf ¼ 26 kPa
E ¼ 0.2 GPa, n ¼ 0.2
B ¼ 0.02 GPa, C1 ¼ 1.2e7,
C2 ¼ 2.5e7
S1 ¼ 1.162, T1 ¼ 23.48,
S2 ¼ 0.808, T2 ¼ 63.25
Passive Muscles
B, bulk modulus; Ci’s, Hyperelastic coefficients; Di’s, decay constants ( ith/ jth terms); Ei, initial Young’s
modulus; Ev, elastic modulus for viscosity; Hi’s, material coefficients for heart model; K, stiffness coefficient; k, strength coefficient; L, stretch ratio at which fibers are straightened; Li’s, material
coefficients for lung model; Mi’s, shear relaxation moduli ( ith/ jth terms); N, hardening exponent; N1,
exponent for power law in Young’s modulus; N2, exponent for power law in viscosity; R, shear relaxation
modulus; S, spectral strengths for Prony series relaxation kernel; Si’s, spectral strengths for Prony
series relaxation kernel; T, characteristic times for Prony series relaxation kernel; Vi’s, viscoelastic
constants; a, exponent term; m, shear modulus.
5.3 BUILDING LABORATORY-SPECIFIC MATERIAL PROPERTY
LIBRARY
System-level FE models can only be reliably used to predict system behaviors if validations have been conducted at the material level and subsystem first. Engineering
materials, such as mild steel and aluminum, are well described in scientific literature.
The constitutive laws and associated properties of such materials are well investigated and published. Despite abundant information on material and subsystem level
validations, we still see variations in material properties. These are most likely due
to differences in manufacturing processes. For example, sheet metal and plate metal
are known to have different properties. A well-managed FE modeling laboratory
should have a material property library in which well-documented material laws
and associated properties are categorized for laboratory members to select from,
and as new materials become available, the library needs to be augmented.
In some cases, no existing material laws can be used to fully describe the behaviors of new materials, such as specific alloys. For example, stressestrain responses
of die cast magnesium alloy AM60B, designed to reduce vehicular weight while
maintaining high material strength during crash events, do not fit any existing
material laws. As previously mentioned, constitutive laws can only be developed
from experimental data. Some materials, such as AM60B, exhibit changes in
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constitutive behaviors as the strain rate changes. Developing suitable constitutive
laws for such materials requires dynamic testing in a variety of environments.
This produces a great amount of data that must be evaluated. This extensive testing
and data evaluation is extremely time consuming. During a time when fuel economy
standards were increased, there was an immediate need to integrate AM60B into
production of cars. Because the material is strain-rate dependent, and it had not
yet been tested in various crash environments, alternative solutions were needed
to reasonably replicate the behaviors of AM60B in whole car computational models.
Zhu et al. (2012) investigated MATs 24, 88, 99, and 107, available in the LS-DYNA
material law library, to determine if one of these material laws can be used with reasonable accuracy to replicate the behavior of AM60B. All these laws are equipped with
varying degrees of strain-rate effects and failure-simulation capabilities. Physical tests,
including coupon testing from quasi-static to a strain rate of 800 s1, four-point bending,
and crush testing of thin-walled structural components, were conducted. Design optimization procedures were used to find the best fit material laws and associated properties.
The authors conclusion is that MAT 99 is a well-suited material law for modeling
AM60B subjected to static and dynamic loading, and the associated material properties
are now stored in our material property library for future applications.
Biological tissues are similar to alloys such as AM60B in that constitutive laws
and associated properties are often not readily available. Therefore, choosing material
properties usually requires making many assumptions. For most of the engineering
materials, constitutive laws and associated properties can be readily obtained since
specimens are abundant and can be machined into specific sizes for testing under
various loading conditions. Biological tissues are not easily obtained and they exhibit
greater variations due to differences in age, gender, and ethnicity, among other things.
One good example pertaining to a material that is difficult to characterize is the
skull of the adult rat, for which current literature is limited despite the fact that adult
rats are the most frequently used animal in the study of traumatic brain injury.
Because the brain is encased in skull, properties of the skull affect the impact
responses of the brain. Mao et al. (2011) tested skull samples in three-point bending
at loading velocities of 0.02 and 200 mm/s. Using classical beam theory, the elastic
modulus, energy absorbed to failure, energy density, and bending stress were calculated. Results demonstrated that bending velocity (a parameter related to the strain
rate) had a significant effect on elastic modulus and bending stress, but not on energy
and energy density. Since the properties reported by the Mao study were based on
classical homogeneous beam theory, there was no accounting for the variations
seen in the diploe layer sandwiched between the outer and inner tables. To account
for specimen-to-specimen variations, Guan et al. (2011) conducted microCT scans
on skull samples. The authors then applied an optimization methodology to establish
material properties of rat skull. These two examples demonstrate that material properties can be affected by mesh size. As the mesh size becomes smaller, more detailed
properties are needed to properly represent the structure of interest.
For animal tissues, it is typically much more difficult to determine the properties
of soft as compared to hard tissues, which are easier to procure and manipulate.
5.3 Building Laboratory-Specific Material Property Library
Specimen
preparation
Laser
scanning
Geometry
Material
Material
library
FE
simulation
Mechanical
test
Optimization
Optimized
parameters
FIGURE 5.12
Procedures to identify accurate material properties of very compliant biological tissues.
High-resolution laser, microCT, or microMRI scanning is used to identify specimenspecific geometry, while reverse engineering method is used in conjunction with
optimization procedures to determine high-fidelity material properties.
Figure reproduced from Zhu, F., Jin, X., Guan, F., Zhang, L., Mao, H., Yang, K.H., King, A.I., 2010. Identifying
the properties of ultrasoft materials using a new methodology of combined specimen-specific finite element
model and optimization techniques. Materials and Design 31, 4704e4712 with permission.
Indentation testing is a good approach for obtaining elastic moduli of soft tissues.
However, this methodology requires many assumptions and cannot be used to
directly obtain shear modulus, which is needed for very compliant materials. Due
to the compliant nature, dissecting soft tissues into predefined shapes is very difficult;
the shape changes under its own weight. As such, extremely large variations are
observed and reported in scientific literature regarding properties of such soft tissues.
To reduce such variabilities, a new approach based on reverse engineering was
used to identify the material properties of ultrasoft human brain tissues (Zhu et al.,
2010). In this approach, each brain sample was scanned by a high-resolution laser,
and specimen-specific FE models were developed (Fig. 5.12). By using samplespecific models, effects of geometry were removed; only the material properties
could affect the test results. Because large deformations occur when testing soft
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tissues, the more reliable forceedeflection curves (instead of stressestrain curves)
were used for optimizations. Two sets of experimental data at two different loading
rates were used to identify material properties, while a third set was used for model
validation. This new approach of identifying material constants for ultrasoft biological tissues and engineering materials (e.g., silicone gels and rubbers) is recommended for building material properties of soft tissues.
In some FE preprocessing software packages, properties for many alloys made
with iron, nickel, aluminum, copper, magnesium, and titanium as the major components are listed in the material library databases. Even with such databases, it is still
the best practice to validate the recommended properties prior to subsystem model
validation. Yang and Chou (2015) outlined the procedures recommended for the
validations of the properties before integrating the values into the material library
of a laboratory. This kind of behind-the-scene efforts could dictate the accuracy
of FE models created in a laboratory.
EXERCISES
1. Download the LS-DYNA theory manual from http://www.lstc.com/pdf/lsdyna_theory_manual_2006.pdf published by Dr. John Hallquist. In Section
19 of this manual, identify which material models are covered in this
textbook.
2. Describe how you would test the strength of an armadillo shell.
3. Why does cortical bone have different properties at different locations of the
same body?
4. Explain what happens with a material that has a negative Poisson’s ratio.
5. From the left graph of Fig. 5.2, identify the following properties for both mild
steel and the aluminum alloy: elastic modulus, yield tensile strength,
ultimate tensile strength, and failure strain on the figure.
6. Explain why the bulk modulus is much higher than the shear modulus for an
incompressible material.
7. Explain why viscoelastic materials are time-dependent.
8. Write the constitutive equation (i.e., stressestrain relationship) for a
transverse isotropic material.
9. What is the difference between a viscoelastic material and a foam material?
10. Calculate the complex (or dynamic) modulus of a material with the following
properties: GN ¼ 10 MPa, G0 ¼ 40 MPa, and u ¼ 10 Hz. How does b affect
the complex modulus?
References
REFERENCES
Arruda, E.M., Boyce, M.C., 1993. A three-dimensional model for the large stretch behavior of
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Blatz, P.J., Ko, W.L., 1962. Application of finite element theory to the deformation of rubbery
materials. Transactions of the Society of Rheology 6, 223e251.
Carter, D.R., Schwab, G.H., Spengler, D.M., 1980. Tensile fracture of cancellous bone. Acta
Orthopaedica Scandinavica 51, 733e741.
Gibson, L.J., Ashby, M.F., 1997. Cellular Solids: Structure and Properties, second ed.
Cambridge University Press, Cambridge, ISBN 0521499119.
Guan, F., Mao, H., Han, X., Wagner, C., Yeni, Y.N., Yang, K.H., 2011. Application of optimization methodology and specimen-specific finite element models for investigating
material properties of rat skull. Annals of Biomedical Engineering 39, 85e95.
Herrmann, L.R., Peterson, F.E., 1968. A numerical procedure for viscoelastic stress analysis.
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CHAPTER
Prescribing Boundary and
Loading Conditions to
Corresponding Nodes
6
King H. Yang
Wayne State University, Detroit, Michigan, United States
6.1 ESSENTIAL AND NATURAL BOUNDARY CONDITIONS
In the FE method, boundary conditions are defined as a set of prescribed values for
the nodal DOFs or a set of equations, which specify the behavior at the boundary of
the FE model. Examples for prescribed boundary values include a fixed or prescribed displacement and rotation in structural mechanics, a preset surface temperature in heat transfer mechanics, and a steady fluid flow in fluid mechanics.
As briefly discussed in Chapter 4, any prescribed conditions are classified as the
essential boundary conditions.
In the explanation of weak formulations in Section 4.4, the derivative of the
displacement function at the boundary is considered a natural boundary condition. In other words, the natural boundary condition is used to impose a specific
rate of change for a variable. An example of a natural boundary condition in
structural mechanics is surface traction. Other examples of natural boundary conditions include the zero gradient of fluid flow at the outlet in fluid mechanics and
an insulated surface in heat transfer mechanics. When setting up an FE solution, an
essential boundary condition has a higher priority than a natural boundary
condition.
The essential and natural boundary conditions employed in the FE method are
related to the Dirichlet and Neumann boundary conditions used in mathematics
for solving differential equations. The Dirichlet boundary condition, named after
Johann Peter Gustav Lejeune Dirichlet (Feb. 1805eMay 1859), designates a fixed
value at the boundary. The Neumann boundary condition, named after Carl Gottfried
Neumann (May 1832eMar. 1925), defines the derivative of a function on the boundary. These explanations seem to suggest that the Dirichlet boundary condition is
comparable to an essential boundary condition, while the Neumann condition is
equivalent to a natural one. There are special cases in which the above equivalencies
are not necessarily true, but this topic is beyond the scope of this book.
In structural mechanics and impact biomechanics, the most pertinent boundary
conditions are related to the kinematics of the structure, such as body forces, displacement constraints, initial velocities, and prescribed accelerations. Recall from Section
4.7 that in order to solve an FE problem, all nodal coordinates, boundary conditions,
Basic Finite Element Method as Applied to Injury Biomechanics. http://dx.doi.org/10.1016/B978-0-12-809831-8.00006-4
Copyright © 2018 Elsevier Inc. All rights reserved.
257
258
CHAPTER 6 Prescribing Boundary and Loading Conditions
and loading conditions need to be described in the same global coordinate system.
Because boundary conditions are typically designated in a local coordinate system,
a proper transformation from each local to global coordinate system needs to be
performed within the FE solver prior to calculating the nodal displacements.
6.2 NODAL CONSTRAINT AND PRESCRIBED DISPLACEMENT
In the FE method, a boundary or loading condition can only be applied to the proper
DOFs of the corresponding nodes. To assist users in the development of FE models,
some computer programs with graphical user interfaces are designed to calculate
nodal loads from loads that do not necessarily coincide with nodal locations. These
programs are called FE preprocessors. As described in Section 6.1, essential boundary conditions are equivalent to nodal constraints, which is a term used in many FE
software packages. On the other hand, natural boundary conditions are termed as
loading conditions.
6.2.1 NODAL CONSTRAINT
There are two types of nodal constraints: single-point and multipoint. The singlepoint constraint restricts one or several DOFs from specific movements (e.g., zero
displacement in any direction, zero vertical motion but free to move horizontally,
etc.) at a node. Multipoint constraint allows the user to define the movement of a
group of nodes controlled by a prescribed equation or by motions of a control
node. In most FE solvers, nodal constraints need to be uniquely defined. For
example, a node cannot have a single-point constraint while at the same time being
defined as part of a set of multipoint constraints.
For a single-point constraint, the defined value can be zero or nonzero. A zero
displacement constraint is also known as a homogeneous constraint, while a nonzero
constraint is known as a prescribed motion or nonhomogeneous constraint. For the
static forceedisplacement equations shown in Eq. (6.1), if u1 is a fixed DOF (i.e.,
zero displacement), the first row and first column are removed from further
calculation.
2
38 9 8 9
k11 k12 : : k1n > u1 > > f1 >
>
> >
> >
> >
>
6
7>
>
>
>
>
>
>
>
>
>
6 k21 k22 : : k2n 7>
>
>
u2 >
f2 >
>
>
>
>
6
7>
>
>
>
=
6
7< = < >
6
7
¼
: : : : 7 :
:
6 :
> >
>
6
7>
> >
> >
> >
>
6
7>
>
>
>
>
>
>
>
>
6 :
7
: : : : 5>
>:>
> : >
> >
>
4
>
>
>
>
>
>
>
: ; : >
;
(6.1)
kn1 kn2 : : knn
un
fn
2
38 9 8 9
f2 >
k22 : : k2n >
>
>
> u2 >
>
>
>
>
>
6
>
>
>
>
7>
>
>
>
>
>
>
>
6 : : : : 7< : = < : >
=
6
7
6
7
06
¼
7 > > > >.
6 : : : : 7>
: >
>:>
> >
>
>
4
5>
> >
> >
> >
>
>
>
: >
: ; >
;
kn2 : : knn
un
fn
6.2 Nodal Constraint and Prescribed Displacement
FIGURE 6.1
Different boundary conditions are used to represent a fully restrained beam and a
propped beam.
Similarly, for a fixed displacement boundary condition um ¼ 0, where
1 m n, before nodal displacements u1 through um1 and umþ1 to un are calculated, the mth row and mth column should be removed from the forceedisplacement
equation shown in Eq. (6.1). For this boundary condition, fm becomes an unknown
reaction force, which can be calculated by using Eq. (6.2) after all DOFs are identified. Eq. (6.2) represents the equilibrium condition of a node at which the displacement is constrained to zero. Hence, this equation is called the constraint equation.
fm ¼ km1 u1 þ km2 u2 þ . þ kmn un .
(6.2)
Fig. 6.1 illustrates the way to address certain boundary conditions. In order to
represent boundary conditions for a fully restrained beam, all displacements (u, v,
and w) and rotations (Rx, Ry, and Rz) at both ends need to be set to zero. For a propped
beam, no displacements or rotations are allowed at the fixed end, and the only
boundary condition is a zero vertical (w) deflection at the roller support. Another
example of a nodal constraint is a pinned joint, which allows no displacements in
either vertical (w) or horizontal (u) directions, but allows free rotation about the
y-axis. This pinned joint constraint is frequently used in simplified human models
for the idealization of the motion of the knee joint.
6.2.2 PRESCRIBED DISPLACEMENT
To impose a prescribed displacement boundary condition (e.g., u2 ¼ c) for the static
problem stated for Eq. (6.1), the second row of the equation needs to be rewritten, as
shown in Eq. (6.3). In other words, if a specific displacement is prescribed at the mth
DOF, we need to set kmm ¼ 1 and kmi ¼ 0 (where m s i). We then solve this new
equation to obtain all other nodal displacements. The following example shows a
cantilever beam subjected to a prescribed displacement.
259
260
CHAPTER 6 Prescribing Boundary and Loading Conditions
2
k11
6
6 0
6
6 :
6
6
4 :
kn1
k12
:
:
1
0
:
:
:
:
:
:
:
kn2
:
:
38 9 8 9
f1 >
k1n >
>
>
> u1 >
>
>
>
>
>
>
>
>
>
7>
>
>
>
>
0 7>
u
c
>
>
>
2
7< = < =
¼
: 7
f
:
7> > > 3 >.
7>
>:>
> : >
> >
>
: 5>
>
> >
>
> >
>
>
: >
: >
; >
;
un
fn
knn
(6.3)
Example 6.1
Consider an isotropic linear elastic cantilever beam with a length of
100 in., cross-sectional moment of inertia 100 in4., and Young’s modulus of
10 106 psi. This beam is loaded at the midpoint with a prescribed downward
displacement of 0.00417 in. (Fig. 6.2). Find the end deflection and rotation
angle.
FIGURE 6.2
A cantilever beam is loaded with a prescribed displacement.
Solution
Examples 6.1 and 4.1 have the same geometric configuration but different
loading conditions. Taking the global stiffness equation already derived for
Example 4.1, we eliminate the first and second rows and columns because of
the fixed boundary conditions at P1. The resulting global forceedisplacement
equation has the form
9
38 9 8
2
w2 > > F2 >
24
0
12
300
>
>
>
>
>
>
> =
> <
>
<
6 0
20; 000 300 5000 7
M2 =
7 q2
6
¼
.
80006
7
>
4 12 300
12
300 5>
w3 >
F3 >
>
>
>
>
>
>
>
>
: ; :
;
300
5000 300 10; 000
q3
M3
Because the only loading condition is a prescribed boundary displacement of
w2 ¼ 0.00417, we replace the first row of the stiffness matrix as described for
6.2 Nodal Constraint and Prescribed Displacement
Eq. (6.3). Additionally, we know that no other forces or moments are applied, that
is, M2 ¼ F3 ¼ M3 ¼ 0. We rewrite the above equation as
3
2
1
6
0
0
0 78 w2 9 8 0:00417 9
>
7>
6 8000
>
>
>
> >
>
>
7>
6
=
<
<q >
= >
7
6 0
0
2
20;
000
300
5000
7
.
¼
80006
7> w > >
6
>
0
3
>
>
>
>
7
6 12 300
>
>
> ;
> :
12
300 7:
;
6
5 q3
4
0
300
5000 300 10; 000
Notice that the entry in the first row and column is 1 when it is multiplied by
the constant outside of the matrix. Solving this equation yields the same answers
as previously described in Example 4.1 (i.e., f w2 q2 w3 q3 gT ¼ f 0:00417 0:000125 0:01042 0:000125 gT ). Notice that the only difference between the two examples is the loading condition. A downward force of
100 lb is applied in Example 4.1, while the current example is loaded with a
downward displacement of 0.00417 in. This example illustrates that either force
or displacement can be applied as the boundary/loading condition used to find the
remaining nodal DOFs.
6.2.3 PENALTY METHOD
Removing one or more rows and columns from the global stiffness equation based
on prescribed boundary conditions is easier said than done. This statement is especially true when the order of the global stiffness matrix is large and the task is performed by hand. Even if such work is accomplished by a computer, additional
computational resources are needed to attain the goal of rearranging the old global
stiffness matrix into a new one with reduced order. A technique to ease this burden is
called the penalty method.
We will use an example to explain the concept of the penalty method. Assume
u2 ¼ 0 (the boundary condition at the 2nd DOF is fixed) in Eq. (6.1). Instead of
removing the 2nd row and 2nd column from the global stiffness matrix, we multiply
a very large stiffness value (i.e., penalty factor) to k22 before carrying out the Gauss
elimination. In other words, except for the large penalty multiplied to k22, nothing is
changed for Eq. (6.1). When a suitable penalty factor is selected, the resulting u2 is
very close to, but definitely not, zero. The penalty method can also be used for more
sophisticated nodal constraints. For example, if the displacements of nodes in a
group are linearly related, penalty factors proportional to the displacements can
be added to the proper entries within the stiffness matrix. Due to the limited scope
of this book, further discussion on this topic will not be provided.
261
262
CHAPTER 6 Prescribing Boundary and Loading Conditions
Example 6.2
A cantilever beam with the same geometric configuration, E and I as used in
Example 6.1 is shown in Fig. 6.2. Instead of the constant displacement of
0.00417 in. applied at P2, the node is loaded with a concentrated force of
100 lb in the negative y-direction. Calculate the generalized displacements at
P2 and P3 using the penalty method.
Solution
Both elements 1 and 2 are 50 in. long; thus EI
L3 ¼ 8000. The stiffness matrices for
both elements are identical and have the form of
3
2
12
300
12
300
6 300 10; 000 300 5000 7
7
6
½k1 or 2 ¼ 8000 6
7.
4 12 300
12
300 5
300
5000
300 10; 000
By assembling the two element stiffness matrices, we obtain the structure
stiffness matrix as
3
2
12
300
12
300
0
0
6 300 10; 000 300 5000
0
0 7
7
6
7
6
6 12 300
24
0
12
300 7
7.
6
½K ¼ 8000 6
5000
0
20; 000 300 5000 7
7
6 300
7
6
4 0
0
12
300
12
300 5
0
0
300
5000
300
10; 000
In this problem, we use the penalty method to address the boundary conditions of
w1 ¼ 0 and q1 ¼ 0. From the structure stiffness matrix, the largest entry has a
magnitude of 1.6 108. Thus it should be adequate to use a factor of 1010 for
multiplication to both k11 and k22, the corresponding diagonal terms for the nodal
DOFs where the two boundary conditions applied. All other entries should remain
unchanged. The resulting forceedisplacement matrix becomes
2
6
6
6
6
6
6
8000 6
6
6
6
6
4
12 1010
300
300
10; 000 1010
12
300
300
5000
0
0
0
0
38 9 8
9
w1 >
0 >
>
>
>
>
>
>
>
>
>
>
7>
>
>
>
>
>
>
>
>
>
>
>
q
0
300 5000
0
0 7
>
>
>
>
1
7>
> >
> >
>
>
>
>
>
>
7>
>
>
>
>
<
=
<
=
7
w
100
24
0
12
300 7
2
¼
.
7
> q2 >
>
0 >
>
>
>
0
20; 000 300 5000 7
>
>
>
>
7>
>
>
>
>
> >
>
7>
>
>
>
> w3 >
>
>
>
>
>
0 >
12
300
12
300 7
>
>
>
>
5>
>
>
>
>
>
>
: >
; >
:
;
q3
0
300
5000 300 10; 000
12
300
0
0
6.2 Nodal Constraint and Prescribed Displacement
Finally, the generalized nodal displacements are
8 9 8
9
w1 > > 1:042 1013 >
>
>
>
>
>
> >
> >
>
>
>
15 >
>
>
>
q1 >
6:250
10
>
>
>
>
>
>
>
>
>
>
>
< w = < 4:167 103 >
=
2
¼
.
>
>
q2 >
1:25 104 >
>
>
>
>
>
>
>
>
>
> >
> >
>
>
2 >
>
> >
>
>
> 1:042 10 >
> w3 >
> >
>
:
:
;
;
4
q3
1:25 10
As seen in this set of solutions, w1 and q1 are not zero, but the magnitudes are so
small that they can be considered as zeros. You can readily see the advantage of
using the penalty method is that there is no need to rearrange rows and columns in
the [ K] matrix, which is necessary if the elimination method is used. The
disadvantage is that the generalized displacements at the boundary nodal DOFs
are not zero. Our other finding from this example is that the 100 lb downward
force has the same effect as the 0.00417 in. of downward displacements applied at
P2. In other words, these two “loading” conditions are identical, and the FE
method allows the “loading” condition to be either displacement or force.
6.2.4 SYMMETRIC FINITE ELEMENT MODELING THROUGH NODAL
CONSTRAINT
When a symmetric FE model is loaded symmetrically, running a submodel by cutting the original model at the symmetry line(s) with appropriate nodal constraints
can achieve the same results as running the whole model. The major advantages
of running a submodel are related to saving computational resources and shortening
the simulation time.
Fig. 6.3 shows a thin rectangular plate that has a circular hole located in the center
and is loaded in tension. It is clear that this plate has two symmetric lines, the x- and
y-axes. Additionally, the tensile loading is symmetric about the y-axis. Thus, only a
quarter of the whole plate needs to be modeled in order to simulate the model
responses of the whole plate (Fig. 6.3 right). Intuitively we understand that since
the forces are equal and opposite along the two midlines, no x-displacement will
occur on the y-axis and no y-displacement will occur on the x-axis. Similarly, any
nodes located on the x-axis will have no y-displacements. These nodal constraints
are equivalent to properly positioned roller supports, as shown in Fig. 6.3. The comparable loading condition for the quarter model becomes half of the original tensile
load because of the y-symmetry.
Although a symmetric structure is frequently seen in the real world, symmetric
loading conditions are not as prevalent. Thus, FE simulations using submodels with
263
264
CHAPTER 6 Prescribing Boundary and Loading Conditions
FIGURE 6.3
A thin symmetric rectangular plate is loaded symmetrically by tensile load P. This problem
can be simulated using a quarter plate with appropriate nodal constraints.
suitable nodal constraints due to symmetry are not commonly encountered. However, FE simulations using a symmetric half model or quarter model are very useful in
debugging the “goodness” of a model. When a symmetrically loaded symmetric FE
model fails to show symmetric results, the model developers must be aware that
there are some mistakes that exist in the model. The term “goodness” used here is
not related to the mesh quality previously discussed in Section 3.8. For example,
if the left and right hemispheres of a human head is symmetrically modeled and
loaded, the intracranial responses should be symmetric about the midsagittal plane.
Any deviations from symmetric responses indicate that the model was not properly
developed.
6.3 NATURAL BOUNDARY/LOADING CONDITIONS
All examples thus far involve a concentrated force or displacement that is directly
applied to the corresponding node(s), as required by the FE method. In real-world
problems, concentrated loads are not common, and they may not be located at the
nodes. Additionally, distributed loads (such as water pressure at the wall surface
of a dam, cumulated snow on top of a roof, or wind load directed at a building)
are more common in real-world problems. This section discusses how to mathematically handle these natural boundary/loading conditions.
From Chapters 2 and 3, we know that element shape functions have many applications in the FE method. For example, shape functions can be used with nodal
values for interpolating physical quantities not located at the nodes. For instance,
temperature at any location within an element can be found by using the temperatures at the nodes. Using this application in a reverse manner, the element shape
functions can be used to redistribute the load applied at a non-nodal location within
an element to the appropriate nodal DOFs of neighboring nodes within the same
element. We will use some examples to demonstrate this procedure.
6.3 Natural Boundary/Loading Conditions
6.3.1 CONCENTRATED LOADS
6.3.1.1 Bar Problem
Recall that the shape functions of a 2-node linear element in the natural coordinate
system are
1x
1þx
and N2 ¼
.
(6.4)
2
2
For a physical quantity 4 located within this 2-node element, the magnitude can
be determined from the element shape functions and nodal quantities 41 and 42, as
shown in Eq. (6.5). The following examples demonstrate usage of this equation in a
reverse manner to identify nodal loads.
N1 ¼
4 ¼ N1 41 þ N2 42
(6.5)
Example 6.3
Fig. 6.4 shows a concentrated load F applied at x ¼ 2 on a 2-node bar element.
Determine the nodal load vectors at P1 and P2.
2
4
FIGURE 6.4
A linear bar element subjected to a concentrated axial load at x ¼ 2.
Solution
Because we are using the shape functions derived from the natural coordinate system, the first step is to change the x coordinateebased geometric information into
x-coordinate values, ranging between 1 and 1. Using Eq. (3.11) or directly mapping x ¼ 0 / x ¼ 1 and x ¼ 6 / x ¼ 1, the linear transformation mapping
L . Based on this simple coordinate transformation, the
function is x ¼ 2ð1þxÞ
xex relationship for this example is x ¼ 1 þ 13 x. At x ¼ 2, the corresponding
x coordinate is x ¼ 13.
By inserting this value into the element shape functions, we find
1þx 1
2
1
N1 x ¼ 13 ¼ 1x
2 ¼ 3 and N2 x ¼ 3 ¼ 2 ¼ 3.
For concentrated loads, the nodal load vectors (also known as consistent load
forces), are the products of the shape functions (evaluated at the location of the
265
266
CHAPTER 6 Prescribing Boundary and Loading Conditions
applied load, i.e., x ¼ 2 or x ¼ 1/3) and the magnitudes of the loads. To determine nodal forces f1 and f2, we apply this concept and directly calculate the nodal
load vectors, as shown in Eq. (6.6).
9
8
2 >
>
>
>
F
=
<
f1
N1 ðxÞ
3
(6.6)
F¼
¼
f fe g ¼
>
>
N2 ðxÞ x¼2 or x¼1
f2
1
>
>
: F;
3
3
6.3.1.2 Beam Problem
A beam element has two DOFs (vertical deflection and rotation) at each node. The
shape functions of the 2-node beam element from Eq. (3.17) are repeated here as Eq.
(6.7). As shown in the example below, concentrated and distributed loads that are not
placed in nodal locations can be redistributed and expressed in terms of nodal load
vectors.
2 3x þ x3 ð1 xÞ2 ð2 þ xÞ
¼
4
4
2
3
L 1xx þx
Lð1 xÞ2 ð1 þ xÞ
N2 ¼
¼
8
8
N1 ¼
2 þ 3x x3 ð1 þ xÞ2 ð2 xÞ
N3 ¼
¼
4
4
2
3
L 1xþx þx
Lð1 þ xÞ2 ðx 1Þ
N4 ¼
¼
8
8
(6.7)
Example 6.4
As previously shown in Example 4.1, a concentrated load P, located at x ¼ L2,
needs to be redistributed to the corresponding nodal DOFs in order for the FE
solver to find the solution. Show that the nodal load vectors displayed in
Fig. 6.5 are correctly stated.
FIGURE 6.5
A concentrated load P, applied at the center of the beam, decomposed into four
nodal load vectors.
6.3 Natural Boundary/Loading Conditions
Solutions
Because P is a concentrated load in the downward direction, the nodal load vectors can be derived from the shape functions by directly inserting the proper xvalue into Eq. (6.8). At the center of the element, the natural coordinate x ¼ 0.
Thus the nodal load vectors can be found by multiplying each of the shape functions by the magnitude of the load P (the negative sign is used to designate a
downward direction):
2
3
9
8
2 3x þ x3
>
>
> P >
6
7
>
>
>
>
6
7
>
>
4
>
>
2
6
7
>
>
8
9
>
>
6
>
>
7
2
3
>
>
V1 >
L
1
x
x
þ
x
>
>
>
6
7
PL
>
>
>
>
>
>
>
>
6
7
=
<
<
=
6
7
M1
8
8
7
.
(6.8)
¼ ðPÞ6
¼
ffe g ¼
6
7
>
>
P >
V2 >
>
>
>
>
6
7
2 þ 3x x3
>
>
>
>
>
>
:
;
6
7
>
>
>
2 >
6
7
M2
4
>
>
>
>
6 7
>
>
>
>
6
7
PL
>
>
2
3
>
>
4L 1 x þ x þ x 5
>
>
:
8 ;
8
x¼0
P A quick
P check on the calculated nodal load vectors should be conducted on
F and M to ensure that
were made doing the calculaP no careless mistakes
P
tions. We can see that
F ¼ P and
M ¼ 0, just as the original loading
condition.
Example 6.5
Determine the nodal load vectors for a concentrated moment M applied at the
center of a 2-node beam element as shown in Fig. 6.6.
Solution
2
Recall from Chapter 3 that dx
dx ¼ L when mapping an element of length L to the
natural coordinate system. At the midpoint of the element, the natural coordinate
is x ¼ 0. As such, the nodal load vectors for a concentrated moment can be found
by multiplying the derivatives of the shape functions by the magnitude of the
concentrated moment (a positive sign indicates that the moment is applied in a
counterclockwise direction), as shown in Eq. (6.9).
267
268
CHAPTER 6 Prescribing Boundary and Loading Conditions
FIGURE 6.6
A concentrated moment applied at the midpoint of a beam element distributed to
four nodal DOFs. The corresponding nodal load vectors are displayed at the
bottom.
8
9
V1 >
>
>
>
>
>
T
<
M1 =
dN
vN vx T
¼ ðMÞ
¼ ðMÞ
f fe g ¼
>
dx x¼0
vx vx x¼0
> V2 >
>
>
>
:
;
M2
3
2
9
8
3 þ 3x2
>
>
3M
>
>
7
6
>
>
>
>
7
6
>
>
4
>
>
2L
7
6 >
>
>
>
7
6
>
>
2
>
>
>
6 L 1 2x þ 3x 7
M >
>
>
>
>
7
6
=
<
7
2 6
8
4
7
6
¼ ðMÞ
¼
7
6
>
>
L 6
>
7
3 3x2
>
> 3M >
>
>
7
6
>
>
>
7
6
4
>
> 2L >
>
>
7
6 >
>
>
7
6
M >
>
>
>
>
4 L 1 þ 2x þ 3x2 5
>
;
: 4 >
8
x¼0
(6.9)
6.3 Natural Boundary/Loading Conditions
Example 6.6
The concentrated load F is applied to the top edge of a 4-node bilinear element as
shown in Fig. 6.7. Find the nodal load vectors.
FIGURE 6.7
The concentrated load F applied to the top edge of a 4-node bilinear element at x ¼ 2,
y ¼ 4. Note that the origin of the global coordinate system is located at P1, while the origin
of the local (natural) coordinate system is located at the center of the element.
Solution
In this example, the force is applied on the perimeter of the element. For this kind
of edge load, the nodal load vectors involve only the line segment between P4 and
P3. As such, only the 2-node element shape functions, as opposed to the 4-node
element shape functions, are appropriate for use in the determination of the nodal
load vectors. Using the same procedures outlined in Example 6.3, the nodal load
vectors are expressed in Eq. (6.10).
9
8
2 >
>
>
F>
=
<
f4y
N1 ðxÞ
3
(6.10)
¼
¼ ðFÞ
ffe g ¼
>
f3y
N2 ðxÞ x¼1 >
>
;
: 1 F >
3
3
where the subscript y shows that the loading is along the vertical direction.
269
270
CHAPTER 6 Prescribing Boundary and Loading Conditions
In the event that the concentrated load is applied within (inside) a 4-node bilinear
element, the nodal load vectors have to be calculated from the bilinear element shape
functions. We first repeat the shape functions of the 4-node bilinear element defined
in Chapter 3, as shown in Eq. (6.11).
1
N1 ¼ ð1 xÞð1 hÞ
4
1
N2 ¼ ð1 þ xÞð1 hÞ
4
1
N3 ¼ ð1 þ xÞð1 þ hÞ
4
1
N4 ¼ ð1 xÞð1 þ hÞ
4
(6.11)
Example 6.7
!
The concentrated load F ¼ 10!
x 20!
y is applied at x ¼ 4.5, y ¼ 3 to a 4-node
bilinear element, as shown in Fig. 6.8. Find the nodal load vectors.
FIGURE 6.8
A 4-node bilinear element subjected to a single-point, concentrated load at
x ¼ 4.5, y ¼ 3.
Solution
As in all isoparametric elements, the same shape functions are used to find the
nodal load vectors that result from horizontally oriented and vertically oriented
loads. A simple linear transformation yields x ¼ 13 x 1 and h ¼ 12 y 1. At the
6.3 Natural Boundary/Loading Conditions
point of the applied load, the natural coordinates are (0.5, 0.5). We can find the
nodal load vectors as shown in Eq. (6.12).
2
3
ð1 xÞð1 hÞ
6
7
6
7
4
9
6
7
8
9
8
6
7
f1x >
0:625 >
>
>
6
7
>
>
>
>
>
>
>
>
>
6 ð1 þ xÞð1 hÞ 7
>
>
>
>
>
>
>
>
>
>
>
6
7
=
< f2x =
<
1:875
4
6
7
6
7
¼ ð10Þ6
¼
ffex g ¼
7
>
>
>
>
>
>
6 ð1 þ xÞð1 þ hÞ 7
5:625 >
f3x >
>
>
>
>
>
>
>
>
6
7
>
>
>
>
>
>
>
>
6
7
;
:
;
:
4
6
7
1:875
f4x
6
7
6
7
4 ð1 xÞð1 þ hÞ 5
4
x¼0:5;h¼0:5
2
3
ð1 xÞð1 hÞ
6
7
6
7
4
6
7
9
9
8
8
6
7
f
1y
>
>
>
> 1:25 >
6
7
>
>
>
>
>
>
>
>
>
>
>
6 ð1 þ xÞð1 hÞ 7
>
>
>
>
>
>
>
>
6
7
=
< f2y =
<
3:75
4
6
7
6
7
¼ ð20Þ6
¼
ffey g ¼
7
>
>
>
>
>
>
>
>
6 ð1 þ xÞð1 þ hÞ 7
>
>
>
>
> f3y >
> 11:25 >
6
7
>
>
>
>
>
>
>
>
6
7
;
;
:
:
4
6
7
f4y
3:75
6
7
6
7
4 ð1 xÞð1 þ hÞ 5
4
x¼0:5;h¼0:5
(6.12)
6.3.2 DISTRIBUTED LOAD
6.3.2.1 Body Force
The most common type of distributed load is the volume load, also known as the
body force, which is measured in force per unit volume. Examples of volume
load include the force due to gravitational loading, inertial loading, prestress, and
centrifugal forces. Consider a body force with a constant intensity q on a line
element (measured in force per unit length). The nodal load vectors for a 2-node
bar element are described in Eq. (4.39) and are repeated here as Eq. (6.13). This
equation is used to distribute the body force load into nodal load vectors for bar
elements.
Z 1 f1
N1
ðqÞ
¼
(6.13)
ffe g ¼
j½Jjdx
f2
N2
1
271
272
CHAPTER 6 Prescribing Boundary and Loading Conditions
Example 6.8
A tapered bar with a mass density of 500 kg/m3 is subjected to its own weight.
This tapered bar is idealized into four bar elements as shown in Fig. 6.9. Find
the width of each element and the nodal load vectors at P2, P3, P4, and P5.
1
2
3
4
FIGURE 6.9
A tapered bar with a base width of 10 m reduced to 4 m at the end. The length of
the tapered bar is 40 m, while the constant thickness is 0.1 m. This tapered bar is
idealized into four equal-length bar elements (right).
Solution
Because the width (w) of the tapered bar reduces linearly as the z-coordinate increases, we can apply the same concept of linear transformation to find the relationship between w and z. In other words, we are required to map w ¼ 10 at z ¼ 0
and w ¼ 4 at z ¼ 40. Hence, the linear transformation equation is w ¼ 10 0.15z. Next we divide the tapered bar into four taper elements of equal length
and calculate the respective volume of each element by integrating the product
of the width (w) and thickness (t) over the element length, as shown in Eq. (6.14).
6.3 Natural Boundary/Loading Conditions
V1
V2
V3
V4
0:15 2 10
z
¼
tð10 0:15zÞdz ¼ 0:1 10z 2
0
0
20
Z 20
0:15 2
z
¼
tð10 0:15zÞdz ¼ 0:1 10z 2
10
10
30
Z 30
0:15 2
z
¼
tð10 0:15zÞdz ¼ 0:1 10z 2
20
20
40
Z 40
0:15 2
z
¼
tð10 0:15zÞdz ¼ 0:1 10z 2
30
30
Z
10
¼ 9:25
¼ 7:75
(6.14)
¼ 6:25
¼ 4:75
If we sum up the results of Eq. (6.14), we have 28 m3. A quick check for the total
volume using analytical methods reveals the same 28 m3 for the whole tapered
bar. However, we still have not explicitly calculated the widths of the four rectangular elements. It is helpful to understand what the values from Eq. (6.14)
represent. If we find volumes of the rectangular bar elements by using the widths
at the beginning of each of the elements, which are 10, 8.5, 7, and 5.5 respectively
at z ¼ 0, 10, 20, and 30, then we end up with 31 m3. However, choosing the
widths of the tapered bar at the vertical centers (i.e., z ¼ 5, 15, 25, and 35 m)
yields widths of 9.25, 7.75, 6.25, and 4.75 m, respectively. Summing these latter
volumes results in 28 m3, which is the same as the volume found from summing
the values from Eq. (6.14).
Because the mass (i.e., inertia of the body, a 1 kg mass would weigh 1 kg on the
earth) of each rectangular-shaped element is the density times the volume, we can
calculate the mass for element 1 as m1 ¼ 500 9.25 (width) 10 (length) 0.1
(thickness) ¼ 4625 kg. Similarly, m2 ¼ 3875, m3 ¼ 3125, and m4 ¼ 2375 kg. If acceleration due to gravity is rounded to 10 m/s2 and the element mass is equally
distributed between the two nodes forming the element, such that each force related
to an element accounts for half of the total force, then f1(elem 1) ¼ f2(elem 1) ¼
23,125 N for element 1, f2(elem 2) ¼ f3(elem 2) ¼ 19,375 N for element 2, f3(elem
3) ¼ f4(elem 3) ¼ 15,625 N for element 3, and f4(elem 4) ¼ f5(elem 4) ¼ 11,875 N
for element 4, where f1, f2, f3, f4, and f5 are nodal forces contributed by each element
at nodes P1, P2, P3, P4, and P5, respectively. By assembling these nodal force vectors that were calculated for the four elements, we have
f1 ðelem 1Þ ¼ 23; 125; f2 ðelem 1 þ elem 2Þ ¼ 42; 500; f3 ðelem 2 þ elem 3Þ
¼ 35; 000; f4 ðelem 3 þ elem 4Þ ¼ 27; 500; and f5 ðelem 4Þ
¼ 11; 875; where all units are in N.
Again, we check the differences between the nodal load vectors calculated
from the four rectangular-shaped elements and those of the four equal-length
tapered bar elements. We start by finding the Jacobian (the length ratio in a 1D
element) for all four elements. Recall from Section 3.7 that the Jacobian is
273
274
CHAPTER 6 Prescribing Boundary and Loading Conditions
dz ¼ L, which in this case is 5, because each element has the same length
j½Jj ¼ dx
2
of 10 m. Next, we use Eq. (3.50) to transform the coordinates. For the first shape
element we find that z ¼ 5 þ 5x. We plug this representation of z into the equation
for width to find width as a function of x: w ¼ 10 0.15z ¼ 10
0.15(5 þ 5x) ¼ 9.25 0.75x. Finally, we use Eq. (6.13) to determine the
element load vectors:
2 3
Z 1
N1
ffe gelement 1 ¼
rgt ð9:25 0:75xÞ4 5 5dx
1
N2
3
2
1x
6 2 7
Z 1
7
6
7dx
¼ 5rgt ð9:25 0:75xÞ6
7
6
1
41 þ x5
2
2
3
Z 1
4:625 5x þ 0:375x2
4
5dx
¼ 5 500 10 0:1 2
1
4:625 þ 4:25x 0:375x
2
3
23; 750
5
¼4
22; 500
For element 2, z ¼ 15 þ 5x, w ¼ 10 0.15(15 þ 5x) ¼ 7.75 0.75x
2 3
Z 1
N1
rgt ð7:75 0:75xÞ4 5 5dx
ffe gelement 2 ¼
1
N2
3
2
1x
6 2 7
Z 1
7
6
7dx
¼ 5rgt ð7:75 0:75xÞ6
7
6
1
41 þ x5
2
Z
¼ 5 500 10 0:1 2
¼4
20; 000
2
1
1
4
3:875 4:25x þ 0:375x2
3:875 þ 3:5x 0:375x
3
5dx
2
3
5
18; 750
16; 250
12; 500
and f fe gelement 4 ¼
.
Similarly, f fe gelement 3 ¼
15; 000
11; 250
When we assemble the nodal load vectors for all elements, we have
f1 ¼ 23,750, f2 ¼ 42,500, f3 ¼ 35,000, f4 ¼ 27,500, and f5 ¼ 11,250. Comparing
6.3 Natural Boundary/Loading Conditions
these calculated results with those found from individual rectangular elements
reveals that the only differences are at the edge nodes f1 and f5. Despite these differences, the total force remains the same when calculated using either methods.
Obviously, it is much easier to use the simplified rectangular elements.
6.3.2.2 Distributed Load
Example 6.9
A uniformly distributed load with an intensity of q is applied to a 2-node beam
element with a length of L, as shown in Fig. 6.10. Find the corresponding nodal
load vectors.
FIGURE 6.10
A 2-node beam element subjected to a uniformly distributed vertical load of intensity q.
Solution
The nodal load vectors (shear force and bending moment) are obtained from the
shape functions of a 2-node beam element, as shown in Eq. (6.15). Here the 1D
dx ¼ L.
Jacobian, or the length ratio, is dx
2
9
8
V
1 >
>
>
>
>
>
>
>
> Z
>
>
Z 1
=
< M1 >
1
dx
¼
dx
q N T dx ¼
q NT
ffe g ¼
>
>
dx
1
1
>
>
V
2
>
>
>
>
>
>
>
>
;
:
M2
2
3
8
9
3
>
qL >
>
>
2
3x
þ
x
>
>
6
7
>
>
>
>
6
7
>
>
2
4
>
>
6
7
>
>
(6.15)
>
>
6
7
>
>
>
>
6 L
7
>
>
2
> qL >
>
7
>
6
>
>
1 x x2 þ x3 7
>
>
>
>
Z 16
<
=
6
7
8
12
qL
6
7
¼
6
7dx ¼
>
7
2 1 6
>
>
qL >
2 þ 3x x3
>
>
6
7
>
>
>
>
6
7
>
>
>
>
2
6
7
4
>
>
>
>
6
7
>
>
>
>
6 7
>
>
>
2
6
7
2
3
>
>
qL >
>
>
4L 1 x þ x þ x 5
>
>
>
: 12 >
;
8
275
276
CHAPTER 6 Prescribing Boundary and Loading Conditions
Example 6.10
A bilinear 4-node plane element is subjected to a linearly varying edge load on the
P4eP3 edge, as shown in Fig. 6.11. Determine the nodal load vectors.
Solution
Because the linearly distributed load is perpendicular to the P4eP3 edge, there
is no horizontal component for the redistribution of the nodal load vectors.
With only vertical loading, we can deduce an important concept regarding
this type of problem: we only need to use the 2-node linear shape functions
along the x-direction to redistribute the edge load, because the loading is
applied at the perimeter of the 4-node element. For the edge P4eP3, the load
intensity is q ¼ q4 at x ¼ 0 and q ¼ q3 at x ¼ L. Hence, we can write the equa4
tion of mapping as q ¼ q4 þ q3 q
L x, where x is between 0 and L. We use linear
transformation to find x ¼ L2 ð1 þ xÞ. The nodal load vectors along the vertical
(y) direction, are
3
2
f1y
Z L
Z L
q3 q4
7
6
x N T dx
q N T dx ¼
q4 þ
ffe g ¼ 4
5¼
L
0
0
f2y
Z L
q3 q4
¼
x
q4 þ
L
0
N T J dx
2
3
1x
7
6
Z 1
6 2 7L
q3 q4 L
6
7
ð1 þ xÞ 6
¼
dx
q4 þ
7
6
7 2
2
L
1
41 þ x5
2
2
3
q3 2q4
2
3
þ
63
q3 þ 2q4
3 7
6
7
L
5
7 ¼ L4
¼ 6
7 6
26
4 2q
5
2q3 þ q4
q4
3
þ
3
3
6.3 Natural Boundary/Loading Conditions
FIGURE 6.11
A bilinear plane, 4-node element loaded on the P4eP3 edge with linearly
distributed vertical loading.
6.3.3 INITIAL VELOCITY AND ACCELERATION
In contact-impact analyses, such as a car crashing into a rigid wall or a bird striking
the windshield of an aircraft, the loading condition mentioned in the previous section
is frequently replaced by an initial condition (e.g., an initial preimpact velocity) or
an accelerationetime history (e.g., a crash pulse). In these dynamic simulations,
velocity profiles are obtained by integrating acceleration-time histories or differentiating displacementetime histories. The velocity profile is then imposed on the
model to run the simulations. More discussion related to dynamic simulations will
be provided in Chapter 8.
EXERCISES
1. Consider a beam element subjected to a concentrated load P and a
concentrated moment M at the center. Find the nodal load vectors.
2. Describe and provide a specific example (not given in the book) of natural and
essential boundary conditions in your own words.
3. A 3-node propped beam (see Fig. 6.1) has a 10 kN downward force applied to
the middle node, which is located at the center of the beam. Another 1.7 kN
force is applied at the propped end to pull the beam away from the wall. The
length of the beam is 100 m, the elastic modulus is 35 106 MPa, the area is
277
278
CHAPTER 6 Prescribing Boundary and Loading Conditions
constant at 5 m2, and I is constant at 50 m4. Note: use a 2D frame element to
properly solve this problem.
4. Below is a 3-element beam model. Assume P3 is fixed in the x- and
y-directions and P1 is supported in the y-direction. A displacement of
0.01768 m is applied at P2 along the x-direction. L2, and L3 are 50 m. E for
all the elements is 70 MPa and A is 0.15 m2. Find the reaction forces and u1.
5. A truss model is shown below. Form the global stiffness matrix and apply the
boundary conditions using the penalty method. Each truss member has a
cross-sectional dimension of 0.1 0.1 m, a Young’s modulus of 200 GPa,
and an ultimate strength of 400 MPa. Once you find the displacements,
calculate the reaction forces.
6. For a 3D box loaded symmetrically, how many planes of symmetry can be
used, and what fraction of the model is formed? What is the maximum
number of cuts can you make on a model?
6.3 Natural Boundary/Loading Conditions
7. Write a function or create an Excel spreadsheet that uses the shape functions
of bar elements to determine the forces on the nodes for any force at any
location.
8. A tapered beam (shown below) is 60 m in length. The beginning width is 12 m,
the ending width is 3 m, and the thickness is 0.6 m. An electromagnetic force
is applied from a magnet above the tapered beam. It provides a 0.1 kN/m2
downward force. Divide the beam into four elements and apply the forces at
these nodes. Find the forces and moments applied at each node.
9. In the model below, q4 is 10 kN and q3 is 7 kN. Assume the element is in the
natural coordinate system; L is 2 m. E is 500 MPa, v is 0.26, and the
thickness is 0.05 m. Assume P1 is fixed and P2 is constrained in the y
direction. Find the nodal displacements.
279
280
CHAPTER 6 Prescribing Boundary and Loading Conditions
10. The length of the beam model below is 75 m, I is 4 m4, and E is 70 GPa.
The load is expressed by the equation q(x) ¼ 25 tan(x) with the tangent
calculated in degrees. With P1 fixed, find the deflection of P2.
CHAPTER
Stepping Through Finite
Element Analysis
7
King H. Yang
Wayne State University, Detroit, Michigan, United States
7.1 INTRODUCTION
Thus far we have learned how to idealize a structure into an FE model, form an
element stiffness matrix [k] from shape functions [N ] under a specific material
law, assemble a global (structure) stiffness matrix [K ] from individual element stiffness matrices, determine nodal load vectors { f }, and set up boundary conditions.
With this information, numerical procedures, such as the Gauss elimination method
(described in Chapter 1), are used to calculate the nodal displacements {u} in static
problems [K](u) ¼ { f }, and the strainedisplacement [B] matrix is used to find the
strains from nodal displacements, followed by using constitutive equations to find
the stresses. In the next section, numerical procedures related to static analyses
are presented.
In general, numerical procedures that are routinely used for solving a linear
system of equations, such as [K ]{u} ¼ { f }, can be classified as direct and indirect
methods. There are a number of direct methods, such as the Gauss elimination
method. The advantages and disadvantages among these direct methods are mostly
related to the algorithms utilized to allocate computer memory (in-core or out-ofcore) and the resulting computational speed. In this chapter, two indirect methods
based on numerical iterations are introduced.
7.2 ITERATIVE PROCEDURES VERSUS GAUSSIAN
ELIMINATION
In Section 1.3.7, we discussed the use of Gaussian (or Gauss) elimination method to
find nodal displacements from a set of static forceedisplacement equations [K ]
{u} ¼ {f }. This method provides the exact solutions if the problem is properly set
up and well defined, but there is a trade-off between using this method to get exact
solutions and using methods that require lesser computational resources. This is
especially true when a large system of equations is involved. When using the direct
method, the requirement for computer in-core memory increases as the number of
DOFs increases. For example, the memory necessary to store an n by n matrix
Basic Finite Element Method as Applied to Injury Biomechanics. http://dx.doi.org/10.1016/B978-0-12-809831-8.00007-6
Copyright © 2018 Elsevier Inc. All rights reserved.
281
282
CHAPTER 7 Stepping Through Finite Element Analysis
requires a total of n2 cells, each needing at least 8 bytes of memory to store numbers
in a double-precision, floating-point format. Assuming we have a structure of
120,000 DOFs (about 40,000 nodes, each with three DOFs), we would need a total
of 115 gigabytes (GB) of in-core memory just to store the [K ] matrix. The requirement for in-core memory doubles if a quadruple-precision (16 bytes, or 128 bits)
computer is used. Without considering resources needed for doing matrix manipulations, the requirement for large amounts of memory simply to store the [K ] matrix
would be problematic even for a modern computer. Strategies for reducing the
necessary amount of in-core memory have been reported. Most of the entries in
the [K ] matrix are zero, as seen in a number of examples we have illustrated thus
far. Data storage formats in these reported strategies include, but are not limited
to, the half-bandwidth, skyline, compressed sparse row, and compressed sparse
column formats. Because these storage methods are hidden from users for most
FE solvers, no further discussion on this topic is provided.
In addition to the requirement for a substantial amount of computer memory,
Gauss elimination algorithm is computationally inefficient. In this method, all of
the processes need to be carried out to the end before the first answer can be determined. During the execution of these processes, updates to the entire stiffness matrix
need to be made at each and every step when forming an upper triangular matrix,
which is computationally expensive. As an alternative, iterative procedures, or indirect methods, are proposed to find approximate solutions. If the convergence criterion is not stringently required, only a limited number of iterations is needed before a
set of approximate solutions are obtained. This low convergence criterion approach
is useful in the early stage of model development, the stage in which the main purpose is to get a quick estimate of the possible trend. In late stage modeling, many
more iterations are used to ensure convergence to the exact solutions. One possible
drawback to using the iteration method is that some problems may be very slow in
convergence, while others may not converge at all.
We will demonstrate a couple of indirect methods using the following example of
a 4-node, 3-element bar problem. Assume each element has a stiffness of EA
L ¼5
units, as shown in Fig. 7.1. After eliminating the first row and column associated
1
2
3
FIGURE 7.1
A 4-node, 3-element bar fixed at the left-hand side and loaded at the right-hand side with
a force of 15 units.
7.2 Iterative Procedures Versus Gaussian Elimination
with the boundary condition u1 ¼ 0, the global forceedisplacement equation
becomes
2
38 9 8 9 8 9
10 5 0 >
=
= >
< F2 >
= >
< 0 >
< u2 >
6
7
0 .
(7.1)
¼
¼
5
10
5
u
F
4
5
3
3
>
;
; >
: >
; >
: >
: >
15
0 5 5
u4
F4
The exact solutions of ½ 3 6 9 T units can be easily identified using knowledge gained in algebra or by simply applying the Gauss elimination method. Two
iterative procedures are illustrated in the following subsections to demonstrate the
nature of approaching the solution gradually.
7.2.1 JACOBI OR SIMULTANEOUS DISPLACEMENT METHOD
The Jacobi method is named after Carl Gustav Jakob Jacobi (Dec. 1804eFeb. 1851).
The first step (iteration) of this method is to rearrange Eq. (7.1) into three new
equations: (1) express the first unknown u2 as a function of the rest of the unknown
displacements (u3 and u4 in this case); (2) express the second unknown u3 as a function of the rest of the unknown displacements (u2 and u4 in this case); and (3) express
the third unknown u4 as a function of the rest of the unknown displacements (u2 and
u3 in this case).
10u2 5u3 ¼ 00ðu2 Þstep
1
¼
ð5u3 Þ
¼ 0:5u3
10
5u2 þ 10u3 5u4 ¼ 00ðu3 Þstep
1
¼
ð5u2 þ 5u4 Þ
¼ 0:5ðu2 þ u4 Þ
10
(7.2)
(7.3)
ð15 þ 5u3 Þ
¼ 3 þ u3
(7.4)
5
For the second iteration, the values obtained from the first iteration are inserted into
Eqs. (7.2)e(7.4) as follows:
5u3 þ 5u4 ¼ 150ðu4 Þstep
1
¼
ðu2 Þstep 2 ¼ 0:5ðu3 Þstep 1
h
i
ðu3 Þstep 2 ¼ 0:5 ðu2 Þstep 1 þ ðu4 Þstep 1
ðu4 Þstep 2 ¼ 3 þ ðu3 Þstep 1
Because all displacements are updated at the end of each iteration, the Jacobi
method is also known as the simultaneous displacement method. The easiest way
to start the iteration is to assume all three unknown displacements u2, u3, u4 are 0,
because we have no way of knowing what the nodal displacements should be. These
processes are iteratively moved forward until convergence occurs.
Step 1:
u2 ¼ u3 ¼ u4 ¼ 0
ðu2 Þstep 1 ¼
ð5u3 Þ
¼ 0:5u3 ¼ 0
10
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CHAPTER 7 Stepping Through Finite Element Analysis
ðu3 Þstep 1 ¼
ð5u2 þ 5u4 Þ
¼ 0:5ðu2 þ u4 Þ ¼ 0
10
ðu4 Þstep 1 ¼
ð15 þ 5u3 Þ
¼ 3 þ u3 ¼ 3
5
Step 2:
ðu2 Þstep 2 ¼ 0:5ðu3 Þstep 1 ¼ 0
h
i
ðu3 Þstep 2 ¼ 0:5 ðu2 Þstep 1 þ ðu4 Þstep 1 ¼ 1:5
ðu4 Þstep 2 ¼ 3 þ ðu3 Þstep 1 ¼ 3
These iterative processes are quite tedious when solving manually, but all tedious
jobs are well suited for computers. A simple spreadsheet program, or a program
in any numerical software package, can be utilized to do the calculation. Table 7.1
lists the calculated nodal displacements using the Jacobi iterative approach. If a 5%
error is allowed, it takes 23 iterations to achieve a reasonable convergence, while a
total of 36 iterations are needed for the solution to be within 1% of the exact solution.
Fig. 7.2 graphically displays the convergence processes.
In real-world problems, we cannot use a percentage error to decide at which iteration the calculations should stop, because we have no way of knowing the exact solution. Instead, the total difference in displacement between the present step and the
predecessor step is calculated for each iteration. The calculated total difference will
steadily decrease, with some occasional zigzag patterns, until an asymptote state is
reached. A predetermined convergence value can be set to decide when the calculations stop.
7.2.2 GAUSSeSEIDEL OR SUCCESSIVE DISPLACEMENT METHOD
GausseSeidel method is an improved form of Jacobi method, also known as the
successive displacement method. This method is named after Carl Friedrich Gauss
(Apr. 1777eFeb. 1855) and Philipp Ludwig von Seidel (Oct. 1821eAug. 1896).
Again, we assume that the starting values are u2 ¼ u3 ¼ u4 ¼ 0. The difference
between the GausseSeidel and Jacobi methods is that the Jacobi method uses the
values obtained from the previous step while the GausseSeidel method always
applies the latest updated values during the iterative procedures, as demonstrated
in Table 7.2. The reason the GausseSeidel method is commonly known as the
successive displacement method is because the second unknown is determined
from the first unknown in the current iteration, the third unknown is determined
from the first and second unknowns, etc.
Although the three resulting values for both methods are identical in the first step,
you should be able to notice the subtle differences between the two methods. In the
Jacobi method, no updates are applied until the next step. For the GausseSeidel
method, the new u3 is calculated from the new u2 in the first equation, and the
Table 7.1 Iterative Results From the Jacobi Simultaneous Displacement Method
Iteration
0
0
0.75
0.75
1.3125
1.3125
1.734375
1.734375
2.050781
2.050781
2.288086
2.288086
2.466064
2.466064
2.599548
2.599548
2.699661
2.699661
2.774746
2.774746
2.831059
2.831059
2.873295
2.873295
2.904971
U3
0
1.5
1.5
2.625
2.625
3.46875
3.46875
4.101563
4.101563
4.576172
4.576172
4.932129
4.932129
5.199097
5.199097
5.399323
5.399323
5.549492
5.549492
5.662119
5.662119
5.746589
5.746589
5.809942
5.809942
U4
3
3
4.5
4.5
5.625
5.625
6.46875
6.46875
7.101563
7.101563
7.576172
7.576172
7.932129
7.932129
8.199097
8.199097
8.399323
8.399323
8.549492
8.549492
8.662119
8.662119
8.746589
8.746589
8.809942
Iteration
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
U2
2.904971
2.928728
2.928728
2.946546
2.946546
2.95991
2.95991
2.969932
2.969932
2.977449
2.977449
2.983087
2.983087
2.987315
2.987315
2.990486
2.990486
2.992865
2.992865
2.994649
2.994649
2.995986
2.995986
2.99699
2.99699
U3
5.857456
5.857456
5.893092
5.893092
5.919819
5.919819
5.939864
5.939864
5.954898
5.954898
5.966174
5.966174
5.97463
5.97463
5.980973
5.980973
5.98573
5.98573
5.989297
5.989297
5.991973
5.991973
5.99398
5.99398
5.995485
U4
8.809942
8.857456
8.857456
8.893092
8.893092
8.919819
8.919819
8.939864
8.939864
8.954898
8.954898
8.966174
8.966174
8.97463
8.97463
8.980973
8.980973
8.98573
8.98573
8.989297
8.989297
8.991973
8.991973
8.99398
8.99398
7.2 Iterative Procedures Versus Gaussian Elimination
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
U2
285
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CHAPTER 7 Stepping Through Finite Element Analysis
FIGURE 7.2
Convergence processes of using the Jacobi iterative procedures for a 4-node, 3-element
bar problem.
Table 7.2 Difference Between the Jacobi and GausseSeidel Iterative
Procedures Assuming the Initial Values u2 ¼ u3 ¼ u4 ¼ 0 for the Problem
Outlined in Fig. 7.1
Jacobi Method
GausseSeidel Method
ðu2 Þstep 1 ¼
ð5u3 Þ
10
ðu3 Þstep 1 ¼
ð5u2 þ5u4 Þ
10
¼ 0:5u3 ¼ 0
¼ 0:5ðu2 þ u4 Þ ¼ 0
3Þ
ðu2 Þstep 1 ¼ ð5u
10 ¼ 0:5u3 ¼ 0
ðu3 Þstep 1 ¼
½5ðu2 Þstep 1þ5u4 10
¼ 0:5ðu2 Þstep 1
þ0:5u4 ¼ 0
3Þ
¼ 3 þ u3 ¼ 3
ðu4 Þstep 1 ¼ ð15þ5u
5
ðu4 Þstep 1 ¼
½15þ5ðu3 Þstep 1
5
¼ 3 þ ðu3 Þstep 1 ¼ 3
new u4 is calculated from the new u2 and u3 in the first and second equations. Note
that while u2 also needs to be updated in the third equation, it just happens that u2 is
not present in the third equation for this particular case. Table 7.3 and Fig. 7.3 show
the iterative results and convergence steps of the GausseSeidel method for the same
4-node, 3-element problem used for the Jacobi method.
Comparing results obtained from the Jacobi and GausseSeidel methods for this
particular example problem, we observed that the convergence occurs much quicker
for the GausseSeidel method. Although this is true in most problems, some special
cases may have opposite results. In terms of computational efficiency, the simultaneous displacement (Jacobi) method is perfectly designed for parallel computing,
because none of the variables within each iteration change until the iteration is
completed. As such, all variables need to be stored in memory until the iteration
is finished. On the other hand, the GausseSeidel method can replace each variable
as soon as a new update becomes available.
7.2 Iterative Procedures Versus Gaussian Elimination
Table 7.3 Iterative Results From the GausseSeidel Successive
Displacement Method
Iteration
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
U2
0
0
0.75
1.3125
1.734375
2.050781
2.288086
2.466064
2.599548
2.699661
2.774746
2.831059
2.873295
2.904971
2.928728
2.946546
2.95991
2.969932
2.977449
2.983087
U3
0
1.5
2.625
3.46875
4.101563
4.576172
4.932129
5.199097
5.399323
5.549492
5.662119
5.746589
5.809942
5.857456
5.893092
5.919819
5.939864
5.954898
5.966174
5.97463
U4
3
4.5
5.625
6.46875
7.101563
7.576172
7.932129
8.199097
8.399323
8.549492
8.662119
8.746589
8.809942
8.857456
8.893092
8.919819
8.939864
8.954898
8.966174
8.97463
FIGURE 7.3
Convergence processes of using the GausseSeidel iterative procedures for the 4-node,
3-element bar problem.
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CHAPTER 7 Stepping Through Finite Element Analysis
Other iterative procedures apply different and yet conceptually similar
approaches. Thus, no further discussion is made regarding other iterative solvers.
As seen in the two iterative procedures shown above, iterative methods slowly reach
the final solution rather than a large final step, as seen in the backward substitution
procedures of the Gauss elimination.
In summary, the direct method requires more in-core computer memory, but the
solutions are accurate. On the other hand, the indirect method reaches the final
solution gradually. However, as the level of convergence can be set by the users, a
lower precision may be desired in order to detect the potential trend of the changing
design variables much quicker than the direct method can provide.
7.3 VERIFICATION AND VALIDATION
The FE method is considered a branch in the field of computational mechanics. A
structural FE model is based on idealization of a real-world structure through a system of mathematical equations. By solving these equations numerically, we find
strains and stresses within the structure when subjected to loads and boundary conditions. As long as the geometry is adequately represented with a high-quality mesh
(see Section 3.8), element types and constitutive laws are properly selected, and
associated material properties are used, the model should provide approximate solutions acceptable to engineers. As the modern-day computer becomes so incredibly
powerful, many FE models of numerous components based on very sophisticated
constitutive laws are now routinely employed in engineering analyses. In addition
to using FE models for routine engineering analyses and design iterations, these
complicated FE models are frequently used to mimic situations where experiments
are too costly (e.g., a car-to-car crash) or nearly impossible to conduct (e.g., a battlefield blast).
As an example, simulating a severe car crash requires the FE solver to be equipped with capabilities for solving problems involving large deformations and large
strains, complex material laws, and nonlinear material properties. Additionally,
the software package needs to be able to handle sophisticated contact impact algorithms, including external contact between the car and an external structure (e.g.,
another car or a tree) and self-contact (e.g., folding of an energy absorbing tube/
box that involves outer-surface-to-outer-surface and inner-surface-to-inner-surface
contacts). A software package with this much sophistication needs to be verified
before it is used. Also, the model-predicted results need to be validated, whenever
experimental data are available, so that sufficient confidence for using such a software package can be established.
7.3.1 HISTORICAL ASPECTS
Harlow Shapley (Nov. 1885eOct. 1972), an American astronomer, once said, “No
one trusts a model except the man who wrote it; everyone trusts an observation
7.3 Verification and Validation
except the man who made it.” Professor George E.P. Box (Oct. 1919eMar. 2013), a
statistician at the University of WisconsineMadison, is credited with coining a common saying that evolved into “all models are wrong, but some are still useful” (Box,
1976). FE models have become more sophisticated than ever, so the famous sayings
by Shapley and Box must come into modelers’ minds. In particular, FE models
related to large systems of nonlinear, multiphysics problems with multiple, multiscale material compositions, where numerous assumptions are made during the
model formulation processes, could become problematic. To ensure that the
model-predicted responses match those that actually occur, careful validation of
the simulation results is necessary.
Since large software development projects are complex and the resulting products must be accurate, validation and verification (V&V) requirements will most
likely be initiated in the computer software industry. Before V&V processes are
required, little or no documentation is written down for future software maintenance
purposes. The V&V processes demand computer programmers to detail the software
architecture and any changes made at each and every step during the development
and modification processes. Software developed following these processes must
meet the requirements set forth by end users (validation) and must result in correct
products (verification). In other words, validation is a set of processes related to
building the precise software product, while verification is related to building the
software product correctly. As early as 1979, the Institute of Electrical and Electronic Engineers (IEEE) had already established a standard called IEEE 730. This
standard prescribed guidelines for the initiation, planning, controlling, and execution
of software quality assurance (SQA) in software development and maintenance processes. The current update, which is called IEEE 730-2014, greatly expands the previous versions, and the changes are so significant that it is more like a brand new
standard than the revision of an old one (IEEE, 2014).
Because different fields call for different requirements, the SQA guidelines
listed by IEEE are not likely to include all the details necessary to cover the requirements for different branches of computational engineering. Of the various areas of
engineering, the V&V processes probably began in the field of nuclear engineering,
partly because of the high safety requirements mandated in this field. In 1985, the
American Nuclear Society published a set of guidelines related to V&V for the
nuclear industry (ANS, 1985). This work was published subsequently by the American National Standards Institute (ANSI, 1987). In the following year, Roache
published a comprehensive book that covers topics from V&V of computer code
and quantification of uncertainty to quality assurance of computer code, using
computational solid mechanics and fluid dynamics problems to illustrate these
topics (Roache, 1998).
In 2009 the Department of Defense (DoD) updated its Instruction 5000.61, previously published in 2003. This update is intended to ensure that all modeling and
simulation (M&S) activities within DoD follow the Verification, Validation, and
Accreditation (VV&A) procedures, practices, and requirements for accreditation
documentation. Subsequently, all four branches of the U.S. armed services (Army,
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CHAPTER 7 Stepping Through Finite Element Analysis
Navy, Air Force, and Marines) and some DoD subagencies have developed their own
guidelines to meet the VV&A processes prescribed by the DoD.
In other government agencies and professional organizations, setting up a standard to determine if a model is accredited seems like a task that is too difficult to
prescribe in a general sense. As such, the emphasis becomes the verification and
validation processes, without accreditation. Following the disaster of the space shuttle Columbia, which occurred on Feb. 1, 2003, the National Aeronautics and Space
Administration (NASA) issued a standard called STD-7009 in 2008, which provides
requirements for processes, procedures, practices, and methods for dealing with
modeling and simulations (NASA, 2008). In this document, the verification process
is defined as the steps needed to address the questions: “were the models implemented correctly, and what was the numerical error/uncertainty?” For the validation
process, the modelers need to document “how M&S results compare favorably to the
referent data and how close is the referent to the real-world system”?
The definitions of “verification” and “validation” in this NASA documentation
differ from those defined in the IEEE standard, and are closer to those used in the
fields of computational solid mechanics and impact biomechanics. Notice that the
earlier IEEE standard uses the sequence of “validationeverification,” while standards related to engineering fields follow a reverse order of “verificatione
validation.” The confusion caused by this reverse order is obvious; it is too difficult
to define exactly what verification is and what the validation is, especially given that
according to http://www.thesaurus.com/, these two words are synonymous.
7.3.2 VERIFICATION
The American Society of Mechanical Engineers (ASME) uses a definition that is
similar to that of NASA. In the Performance Test Codes (PTC) 60/V&V 10
(ASME, 2006), the ASME defines verification as “the process of determining that
a computational model accurately represents the underlying mathematical model
and its solution.” Likewise, the Los Alamos National Laboratory (LANL) declares
verification is “concerned with identifying and removing errors in the model by
comparing numerical solutions to analytical or highly accurate benchmark solutions.”
According to ASME, the verification processes can be separated into two parts:
verification of the code and verification of the calculation. Although it is the software vendor’s responsibility to guarantee the correctness and accuracy of the product, it is in your best interest, as the user, to ensure that this is indeed the case. In
particular, some material laws in an updated version of the FE solver may have defects that are unknown to the software developer, even though several older versions
of the same software may have been used successfully for numerous problems.
Additionally, any newly added material laws need to be carefully verified before
they are implemented into FE models.
Some example problems the author has encountered are: (1) a material law
worked for a simple model of a perfect mesh (i.e., all elements confirmed to prescribed idealized shapes) but failed in a real-world problem with a less than perfect
7.3 Verification and Validation
mesh; (2) the instructions provided by the software vendor were too vague to follow
exactly; (3) the software was misused, because the same parameter-name defined in
one software package was contrarily defined in another software package (e.g., the
decay constants for the linear viscoelastic material in LS-DYNA and PAM-CRASH
were reciprocal to each other); and (4) an FE model that ran flawlessly in an earlier
version was not executable in the new version.
Issues related to the problem mentioned in (4) are particularly frustrating. Does
the newfound error mean that the previously working FE model was incorrectly
developed? In a worse scenario, does it mean that the previous version of the FE
solver was wrongly coded? If so, shall you retract all simulation results acquired
from the earlier version? In some cases, a simple operation system update can
resolve the problem. In other cases, lengthy discussions with technical support
from the software vendor are required.
7.3.2.1 Verification of Code
In some commercial software packages, a verification manual is provided by the
vendor. Such a manual frequently outlines some exemplary cases in which analytical
solutions are compared with FE simulation results. For FE solvers without such a
manual, or for applications that are beyond the scope considered by the software
vendor (e.g., using a gross FE solver to solve a biological tissue at the cellular level),
we recommend users setting up their own verification cases to ensure each material
law to be used in the model is coded correctly. In other words, if ten different material laws are needed to represent the overall structure, all ten laws need to be verified with appropriate material-specific FE models prior to use with the overall
structure. In a well-established FE modeling laboratory, a collection of problems
with known analytical solutions, or with repeatable experimental results, is used
to evaluate the accuracy of the code implemented in the software.
The issue of convergence for an FE mesh discussed earlier is worthy of additional comment here. For a material model needed as part of an overall structure,
the minimal mesh density that is required to achieve convergence during the verification of the material model needs to be maintained in the overall structural mesh.
7.3.2.2 Verification of Calculation
Binary code underlies all computer languages, which means numeric values are
stored using only the digits 0 and 1, representing “Off” or “On,” respectively.
With the exception of software used in the financial world (where only twodecimal precision is needed to represent pennies, and values rarely exceed 100 trillion), numbers are typically stored in scientific notation, and more specifically, in a
format referred to as floating point. Table 7.4 lists the number of bits used to store a
floating-point number in computers ranging from 8-bit to 128-bit.
In this table, the “sign” (1 for negative, 0 for positive) and “exponent” bits are selfexplanatory, “mantissa” represents the fractional part, and “exponent excess” represents an offset for the exponent, which is defined as one half the range of integers
that can fit into the exponent bits. For example, the 8-bit computer has an exponent
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CHAPTER 7 Stepping Through Finite Element Analysis
Table 7.4 The Number of Bits Allocated to Store a Floating-Point Number
for Computers Ranging From 8-bit to 128-bit
No. of Bits
Sign Bits
Exponent Bits
Mantissa Bits
Exponent Excess
8
16
32
64
128
1
1
1
1
1
4
6
8
11
15
3
9
23
52
112
7
31
127
1023
16,383
excess of 7, because the largest integer for the four exponent bits in the 8-bit computer
is 15 (24 1). The exponent excess is needed to allow for negative exponents. From
this table, it is easily understandable that a computer equipped with a higher number
of bits allows for more significant digits, and hence a higher precision.
When verifying computer code, we must be aware of two potential sources of
error: roundoff error and truncation error. The roundoff error comes from the limited
precision associated with a digital computer. As shown in Table 7.4, a higher bit
computer allocates more bits for the fractional part, and hence it is more accurate.
Let us suppose we are using a computer in which numbers are stored and calculated
in base 10, as opposed to binary, and there is an accuracy of three decimal places. To
find the sum of 13 þ 13 þ 13, we must first convert each entry as 0.333 in this computer.
As a result, the final answer will be 0.999 instead of 1.
The truncation error frequently comes from code development as well as the
calculation. Sometimes, a code developer intentionally eliminates certain higher
order terms within an equation, in order to increase the overall computational speed;
this is done at the price of sacrificing the accuracy. The calculation error comes from
using a limited number of (time) steps to acquire the result. For example, the analytical solution to a differential equation arises from a solution that is based on an infinite number of steps. If only a limited number of steps are used, errors will take place
at each iteration. In the explanation of the Jacobi method versus the GausseSeidel
method, we can easily understand the effects of different numerical schemes on the
overall solutions. Additional magnitudes of truncation error come from accumulation of all errors through all iterations. This issue is particularly true when solving
long-duration, dynamic problems. There are no specific methods available to avoid
truncation error. However, finer meshes and smaller time steps usually reduce the
potential of having such errors.
7.3.3 VALIDATION
According to ASME, validation of the FE method is the quantification of the accuracy of the predictions made by an FE model as compared to real-world experimental
data. This statement is somewhat ambiguous; if we can run experiments, why do we
need a computer model? Many FE practitioners may tell you that a validated model
7.3 Verification and Validation
allows the application engineers to save time and money by not having to conduct
expensive experiments. These statements imply that while some experiments are conducted, and results obtained from these experiments are used to validate the FE
model, many applications of the model have no experimental data to sustain accuracy. Additionally, FE models are frequently used for situations that are too costly,
or when it would be impossible to conduct experiments. Thus, validation of these
kinds of models is not practical. As such, most (if not all) FE models can only be
considered partially validated, while total validation is (nearly) impossible.
One noteworthy issue is that experiments tend to provide only limited measurements (e.g., the failure strain, maximum force, maximum deflection, and/or peak
acceleration at certain locations), while an FE model permits a large set of response
variables, including stress that is not measurable using experimental means, and at
any location within the model. Again, the inability to measure the stress experimentally is probably the ultimate reason to run an FE model. This statement alone illustrates that total validation of an FE model is not practical.
Another issue is verbiage. The term “validation” is a very strong and assertive
word in the English language. As such, some researchers suggest the use of “confirmation” instead of “validation” to better reflect the actual situation (Oreskes et al.,
1994). For example, one may want to use the wording “the model predicted strain
at the mid-shaft of the femur confirmed with that measured by strain gauge attached
to the same location” instead of saying “the model is validated against strain data
measured experimentally.” Although the author agrees with this notion, only a handful
of researchers use this term. Hence, the term “validation” will be used in this chapter.
In computational mechanics, large degrees of model validation are slightly easier
to obtain if only man-made materials are involved. This is simply because test specimens needed to conduct carefully designed experiments are easy to procure. For
biological tissues, test specimens are difficult to acquire due to ethical concerns,
and mechanical properties of the specimens are, in general, age- and genderdependent, and may be affected by the strain-rate employed in the experiments. A
consensus among FE biological tissue/system model developers is that all validation
cases associated with a model need to be clearly documented. This way potential
users of the previously developed model can decide if they want to commit the
resources (both computational and monetary) needed to conduct additional experiments in order to validate the model for loading conditions not previously validated.
Depending on the number of available experimental datasets, model validation
processes can be classified into three levels. The simplest one is to compare a set
of predictions made by a deterministic model against experimental data. This first
level of validation is suitable for FE models that are designed to identify material
properties where highly repeatable experimental data are available for model validation. The second level is comparing one set of deterministic model predictions
against the corridors or distribution functions obtained from several sets of experimental data. With this level of validation, it is assumed that the material properties,
loading conditions, and boundary conditions are well defined, but that the experimental data are not highly repeatable. The third level is matching the cumulative
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distribution function (CDF) obtained experimentally to that predicted by an FE
model that is based on probabilistic approach (e.g., including the variations due to
uncertainties in material properties, boundary conditions, and loading conditions
in the FE analysis). Obviously, the cost becomes exponentially higher with requirements for high-level model validation.
For modeling a complex system (e.g., a human body or a detailed car FE model
subjected to impact), model validations should be conducted in a hierarchical
fashion, from the component level (e.g., a femur in a person or a suspension system
in a car), to a subsystem level (e.g., a lower extremity or a collapsible steering
column and airbag protection system), to the complete system level (e.g., a whole
body or an entire car). Again, validating a car model using the hierarchical approach
is a lot easier than validating a human model, because an identical car can always be
produced for the purpose of model validation. Conversely, no two human beings are
exactly alike, and testing live humans to the extent of injury is unethical. While it is
very expensive to crash a car for evaluating the crash safety performance and
providing data for model validation, car manufacturers are willing to commit the
needed resources in order to meet legal obligations and corporate commitments.
In contrast, regardless of the financial resources to which any person may have
access, it is not possible to conduct sufficient experiments to generate impactresponse data on humans, covering all combinations of gender, age, body regions,
and suitable loading rates.
According to most published literature, variations seen in experiments can be
attributed to differences in age, gender, anthropometries, and material properties,
which are reflections of the extent of physical activities. With the exception of
age and gender, such information is unfortunately not available, unless the experiments are specifically conducted for the purpose of acquiring data for model validations. As an example, when a fractured femur was discovered in the posttest autopsy
after a whole-body impact experiment, researchers did not routinely report some
fracture-associated dimensions, such as the diameter at the midshaft, the total length
of the femur, and the quality of the bone. Thus, while some experimental data are
available in scientific literature, most data do not provide sufficient specificities
for the purpose of validating FE models.
In consideration of the modeling side, as reported in scientific literature, FE
models that were developed to mimic experimental conditions tended to be deterministic. In other words, a single model was used to simulate all tests under the
same experimental protocol, despite the fact that there were differences in age,
gender, and the extent of physical activities among all test samples. In order for
the model-predicted responses to match those measured experimentally, material
properties (or other associated parameters) were adjusted manually (trial and error)
or through the use of optimization techniques until reasonable matches were
achieved. A model developed using this approach of “tuning” may become questionable when the loading conditions change. Only those FE models validated using
hierarchical approaches can add sufficient confidence for running the model beyond
the loading and boundary conditions for which they were validated.
7.3 Verification and Validation
Although design and test engineers are well aware of the shortcomings related to
the deficiency of model validations, lack of financial resources tends to prohibit
more experiments from being conducted. Without clear indication of the model
fidelity, a higher factor of safety (FoS) is usually assigned to ensure the integrity
of the structure. For areas where permissible error is extremely low, such as in
NASA or nuclear power plant designs, all numerical models used in designing those
critical systems need to have the highest level of validation and FoS.
7.3.4 QUANTIFYING THE EXTENT OF VALIDATION
For numerical models to be considered reliable and predictive, they need to be
assessed through a series of rigorous validation processes as described in the previous
section. The extent to which models are validated against experimental data needs to
be assessed quantitatively rather than qualitatively. In many publications, qualitative
assessments are seen with phrases such as “the model is well validated” or “the model
is in good agreement with experimental data.” A numerical value would afford users
of the model the ability to determine the extent of model validation before the model
is applied to develop protection equipment, among other applications.
Several government agencies, such as the United States Department of Defense,
American Institute of Aeronautics and Astronautics, American Society of Mechanical Engineers, and Advanced Simulation and Computing (ASC) of the United
States of Department of Energy have investigated fundamental concepts/methodologies for validation of large-scale numerical models. Unfortunately, there are
currently no universally accepted quantification methods to determine the degree
of model validation against experimental observations. This problem is due, in
part, to the large variations and uncertainties seen in experimental data, and many
experimental studies do not provide detailed test conditions that were used while
the experiments were being conducted.
Generally speaking, modern experiments use multiple electronic sensors, instrumented at different locations within the structure or human body, to generate timehistory data. These sensors, which include accelerometers, load cells, pressure
gauges, displacement transducers, etc., are connected to signal conditioners that
turn the measured voltages into physical units, such as g’s, kN (or lb), kPa (or
psi), mm (or inches). Additionally, high-speed videos may be used to record the
kinematics of an event. Some publications compare simulated results against experimental data, based on still photographs and time-history figures, for subjective
engineering judgments to determine the extent of model validation. Such subjective
graphical comparisons cannot be considered quantitative when determining the
reliability and predictive capability of numerical models. Quantitative assessment
of model validity uses a single numerical value to reckon the discrepancy between
model-predicted and experimental data. This assessment method removes or minimizes subjectivity from the evaluation processes, and it will be further explained
in the next sections.
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7.3.4.1 Assessment of Variations in Magnitude
In analyzing structural mechanics and impact biomechanics responses, frequently
asked questions are related to issues such as whether or not the peak response exceeds the maximum allowable response. As such, the simplest quantitative assessment is to report the percentage difference between the peak values predicted by
the FE model and the average value obtained experimentally. This simple comparison is useful, especially for cases related to the determination of thresholds for
structural damage or bone fracture.
Because both positive and negative values exist in most time-history signals,
some researchers consider taking the absolute values to compare the difference in
the overall magnitude. Geers (1984) defined the following three equations based
on the integral of the squared response and the product of experimental and
modeling values in order to quantify the difference in magnitude between values
predicted by the model and those obtained experimentally:
Z t2
Iexp ¼
½expðtÞ2 dt
t1
Z
Imod ¼
t2
t1
½modðt þ sÞ2 dt
Z
Iexpmod ¼
t2
(7.5)
½expðtÞ½modðt þ sÞdt
t1
where “exp” represents the experimental values, “mod” represents the modelcalculated magnitudes, s is the delay between starting times for the two responses,
and t1 and t2 are the beginning and ending times of the interval of interest. If exp(t) is
the same as mod(t), it is easily understood that there is no shift in time between the
two responses. Further, Geers defines a parameter G to describe the difference in
magnitude, based on the ratio of the areas under the model-predicted and experimental curves
sffiffiffiffiffiffiffiffiffi
Imod
.
(7.6)
Gðt; sÞ ¼
Iexp
In this equation, a value of 1 indicates that the two areas underneath the modelpredicted and experimentally obtained curves are identical. A subsequent modified
version reported by Sprague and Geers (2004) retains the essence of the original idea
by Geers (1984), but uses the average value instead of the overall integration value,
as shown in Eq. (7.7). The parameter G remains the same as shown in Eq. (7.6).
Z t2
1
Iexp ¼
½expðtÞ2 dt
t2 t1 t1
Z t2
1
Imod ¼
½modðt þ sÞ2 dt
(7.7)
t2 t1 t1
Z t2
1
Iexpmod ¼
½expðtÞ½modðt þ sÞdt
t2 t1 t1
7.3 Verification and Validation
The third equation in each of the sets listed as Eqs. (7.5) and (7.7) are related to a
time shift, and will be discussed in the next section. The assessment tool used by
Professor Geers et al. considers the difference in magnitude over the entire interval
of interest. This assessment measure is not suitable for events where the magnitude
at one single time point or a group of several time points is more critical than the
overall response.
Deb et al. (2010) defined a gross correlation index (GCI) by including three measurements, the peak load, mean load, and energy absorption (as listed in Eq. 7.8) for
assessing the validity of an FE model representing a top-hat section.
( "
#)1=2
1 ðPmod Pexp Þ2 ðMmod Mexp Þ2 ðEmod Eexp Þ2
GCI ¼ 1 þ
þ
;
2
2
3
P2exp
Mexp
Eexp
(7.8)
where Pmod and Pexp are the model-predicted and experimentally obtained peak
loads, Mmod and Mexp are the model-predicted and experimentally obtained mean
loads, and Emod and Eexp are the model-predicted and experimentally obtained energy absorptions. From this definition, a GCI value of zero indicates that there is
no correlation at all, while a GCI value of one (1.0) implies a perfect correlation between the numerical model prediction and experimental results.
A modified gross correlation index (MGCI) was proposed later by Zhu et al.
(2012). Because there are no suitable constitutive laws available to represent a die
cast AM60B magnesium alloy, optimization-based methodologies based on existing
constitutive equations were used to produce a set of material properties for this alloy.
Several FE models have been created to mimic experiments designed to study slowspeed axial crushing, high-speed axial crushing, and quasi-static, four-point bending
responses of thin-walled double top-hat beam components. By selecting the peak
load, displacement at the peak load, and energy absorption as the assessment metrics, the correlation between the model-predicted results and experimental data is
made using the MGCI, as listed in Eq. (7.9), to evaluate the goodness-of-fit.
2
ðPmod Pexp Þ2
ðDmod Dexp Þ2
MGCI ¼ 1 4a þ
b
P2exp
D2exp
31=2
ðEmod Eexp Þ2 5
þc ;
(7.9)
2
Eexp
where a, b, and c are the weighting factors of the parameters P, D, and E, respectively, and a þ b þ c ¼ 1. The weighting factors a, b, and c can be judiciously chosen based on relative importance of the associated parameters chosen by the user.
7.3.4.2 Assessment of Variations in Phase and Shape
In addition to variations in magnitude between the model-predicted and experimentally obtained results, differences in time histories include the time-shift and shape
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FIGURE 7.4
Comparison of a model-predicted acceleration time history (thick solid line) versus
experimental data represented by the average value (thin solid line) and 1 standard
deviation (dashed lines).
difference. Fig. 7.4 shows a set of hypothetical, experimentally obtained acceleration time histories presented in terms of the average and corridors formed by
plus/minus one standard deviation. The figure also displays the FE modelpredicted response subjected to the same loading condition, with the time zero
shifted to coincide with experimental data at the origin of the curves.
In Eqs. (7.5) and (7.7), the parameter “s” is used to adjust the delay between
starting times for the model and experimental responses. By visual observation of
the curves shown in Fig. 7.4, in which the starting times were adjusted, the
model-predicted responses seem to reflect the phenomenon observed in the physical
tests as having double peaks. However, the times at which these two peaks are
reached during experiments seem to be different from that predicted by the model.
Such a difference in time to reach the peak magnitude is considered another type of
phase shift. As such, it has been quite problematic for researchers to decide whether
the phase shift should be adjusted for the time at which a significant increase in the
magnitude is observed or at which the peak value is reached. Additionally, debates
continue regarding whether or not to normalize the duration of impact in order to
compare model-predicted and experimental results more realistically.
In Eqs. (7.5) and (7.7), the integral Iexpmod described by Geers et al. is used to
quantify the phase discrepancy between the curve predicted by the model and that
obtained experimentally. They propose a second parameter H, shown in Eq.
(7.10), for this purpose.
sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
Iexpmod
Hðt; sÞ ¼
(7.10)
ðImod Iexp Þ1=2
7.3 Verification and Validation
If Imod ¼ Iexp, it can be understood that there is no difference in the starting time
and hence no phase shift when H ¼ 1. With parameter G (Eq. 7.6) and H, Geers
further defines a combined comprehensive error factor Ec from these two parameters
using Eq. (7.11).
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
(7.11)
Ec ðsÞ ¼ ðG 1Þ2 þ ð1 HÞ2
In this equation, a value of 1 is added to the parameter G so that a zero value
indicates that the two areas underneath the model-predicted and experimentally
obtained curves are identical. Similarly, 1 H is used because a unit H value indicates no phase shift. Note that there is no physical meaning presented in Eq. (7.11). It
is simply a way to combine two parameters selected to identify the differences in
magnitude and phase shift into one. Other combinations have been reported, but
they are beyond the scope of this book.
The International Organization for Standardization (ISO) Technical Committee
(TC) 22 aims to develop voluntary, consensus-based, market-relevant International
Standards that support innovation and provide solutions to global challenges related
to road vehicles. The subcommittees (SCs) 10 and 12 virtual testing workgroup
(WG) 4 was to “provide a validated metric to calculate the level of correlation
between two non-ambiguous signals (e.g., time-history signals) obtained from a
physical test and a virtual computational model of the same test. The defined metric
shall be primarily aimed at vehicle safety applications” (Barbat et al., 2013). In this
study, the four metrics evaluated are: the CORrelation and Analysis (CORA) metric
(Gehre et al., 2009), enhanced error assessment of response time histories
(EEARTH) metric (Zhan et al., 2011a), model reliability metric (Zhan et al.,
2012), and Bayesian confidence metric (Zhan et al., 2011b). The website https://
www.iso.org/obp/ui/#iso:std:iso:tr:16250:ed-1:v1:en provides detailed descriptions
of various rating systems. Also available is an objective signal rating system software package for users to calculate scores of different rating systems. In this
book, only the high-level concepts of these metrics are introduced.
Briefly, the CORA method combines a corridor rating and a cross-correlation
rating to determine the correlation of two response curves obtained from simulation
results and experimental data into one single value. The software comes with a set of
default values to calculate the fit between two curves, but this default set of parameters may not be suitable for all potential situations. You can download the CORA
method, user manual, and associated information from http://www.pdb-org.com/en/
information/18-cora-download.html, which is maintained by the Partnership for
Dummy Technology and Biomechanics (pdb).
The EEARTH metric provides three independent error measures: phase, magnitude, and topological differences. Deviations in magnitude and phase are similar to
other evaluation tools, such as those defined by Geers et al. For the deviation in topology, a time-warping technique is used to align the peaks and valleys by shortening and
lengthening the time axis before evaluating the topological variance. All three errors
are then combined into one parameter for evaluating the fit between the two curves.
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The model reliability metric uses a predefined reliability target to determine if a
computational model is adequate. The reliability target makes use of a threshold factor, which is defined according to the experience of experts, and the probability of fit,
which is determined based on the number of data points that are within the lower and
upper bounds of the threshold interval. A typical Bayesian metric is aimed to quantitatively assess the quality of a multivariate system. By adding the probabilistic
principal component analysis (PPCA) to reduce the dimensions for data analysis,
the Bayesian confidence metric can be used to quantitatively assess the model
validity.
Both the model reliability metric and the Bayesian confidence metric provide
simple and easily understandable overall scores for the model validation, but they
cannot distinguish crucial differences, such as phase, magnitude, and slope. As
such, Barbat et al. (2013) further identified the key shortcoming of the Bayesian confidence metric and CORA method. As the leader of the WG 4, Barbat and his colleagues who volunteered for ISO TC22, SC10, and SC12 recommended the adoption
of a combined ISO metric, which was based on variations in the corridor, phase,
magnitude, and slope.
Before using this ISO recommended rating system, all data should be digitized at
a sampling rate of 10 kHz, properly filtered, and then evaluated at each relevant data
point. Three different rating zones, a narrow zone within the inner corridors, a wider
zone between the inner and outer corridors, and a zone outside the outer corridors are
identified to determine the score for the corridor submetric. A data point within the
inner corridors is given a numeric rating of “1,” while a point outside the outer corridors is rated “0.” For other points between the inner and outer corridors, a quadratic
function based on how far the data point is away from the inner corridors is used to
score the point with a value from 1 to 0. Finally, the average score of all data points is
used to represent the corridor submetric on this curve.
To determine the score for the phase submetric, the relevant interval on the time
axis should be preselected and then assigned a value of 100%. The maximum allowable percentage of time shift is then assigned. By shifting the model-predicted curve
left and right on the time axis, cross-correlations between the shifted curve and the
experimentally obtained curve are determined. A score of “1” is given if there is no
need to shift the time axis, and a score of “0” is assigned if the shift exceeds the
maximum allowable percentage. If the score is between these two extremes, a
regression method is used to calculate the phase score.
After the phase shift is minimized, the magnitude submetric can be determined.
Similar to the phase submetric, a maximum allowable magnitude threshold needs to
be preselected. For a data point on the model-predicted curve that exhibits no difference against the experimentally obtained magnitude, a score of “1” is assigned. A
score of “0” indicates that the error is equal or greater than the maximum allowable
error. In between, the score for the magnitude submetric is calculated based on a
regression method. Similarly, the slope submetric is calculated after minimizing
the time shift. A score of “1” is assigned to the slope submetric if no difference is
found between the measured and model-predicted data. A score of “0” is assigned
7.3 Verification and Validation
if the difference in slopes exceeds the maximum allowable difference. In between,
the score for the slope submetric is calculated based on a regression method.
The combined ISO metric recommends the weighting factors of 0.4, 0.2, 0.2, and
0.2, respectively, for the four submetrics. As such, the total combined rating would
range from “0” to “1.” According to Barbat et al. (2013), total ISO ratings of
0.94e1.0, 0.80e0.94, 0.58e0.80, and 0.0e0.58 are considered to be excellent,
good, fair, and poor, respectively.
Each quantification system described above has limitations. In particular, no
evaluation methods have the capability to deal with multiple sets of data obtained
from experiments conducted under identical test conditions but still having variations in test results. Therefore, sound engineering judgment is still crucial when
determining the extent of model validation.
7.3.5 UNCERTAINTY QUALIFICATION
As described earlier, Harlow Shapley once said that “everyone trusts an observation
except the man who made it,” because an experiment provides real data and hence
they must be correct. For people who have prepared a sophisticated experiment, they
know well that it is impossible to place sensors at identical spots, calibrate all sensors just before the test, position the test subject (e.g., a crash dummy) at a specific
location (e.g., in a car seat), and use the same instrumentation at all times. However,
quantifying experimental uncertainties is beyond the scope of this book and will not
be further discussed.
Uncertainties in FE modeling arise mostly from uncertainties in material properties. Variations in the mechanical properties of engineering materials do exist, but
the discrepancies are usually minimal. For biological materials, the properties
vary greatly with age, gender, extent of exercise, etc. For these reasons, a deterministic approach (i.e., one-model only) for most FE modeling may be doubtful. As
such, probabilistic approaches based on sound statistical methods should be considered as part of the model validation.
Uncertainty can also come from numerical procedures. Some examples are: uncertainty arisen from discretization of a structure into mesh, truncation of significant
digits due to limited precision, selection of the time step (Dt), and added energy into
FE modeling to stabilize instability (hourglass mode). A mesh with a large element
size will not provide a high-stress concentration area at a corner or around a small
hole. This reason alone is sufficient to check all FE models for mesh convergence.
Because of more recent advancements in computing technology, truncation is not a
substantial problem. The default minimum time step, usually calculated by taking
90% of the time needed for a sound wave to pass through the element, may not
be sufficiently small to ensure numerical stability. Finally, the ratio of the hourglass
energy to internal energy should be reported to ensure that hourglass mode does
not occur or does not significantly affect the simulation results.
As mentioned in Section 7.3.3, FE models can never be fully validated because
they are confirmed to a limited number of experimentally obtained datasets.
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Additionally, these models are used for cases where experiments are too expensive
or not possible to conduct. In other words, no validation datasets exist in these cases.
Thus, we do not trust any validation metric blindly without sound engineering judgment. Better yet, we highly recommend that model developers intimately participate
in the experiments, so that uncertainties due to experimental procedures are
minimized.
Lastly, test and FE analysis teams need to work together to ensure that an integrated V&V plan is developed; this may be a cultural challenge. Development of
more generally accepted and easily applied Uncertainty Quantification in FEA is
required.
7.4 RESPONSE VARIABLES
Outcome/output variables of interest in an experiment or a simulation are called
response or dependent variables. Independent variables, also known as explanatory
variables or predictors, are those that can be set during experiments or simulations to
obtain outputs. As an example, the response variable in the statement “age is a predictor of bone strength” is “bone strength.”
In the field of structural mechanics or impact biomechanics, the FE method is in
essence a method for stress analysis. In general, response variables selected are those
related to failure of the material, because the purpose of stress analyses is to decide
whether or not a material will fail. The stressestrain response shown in Fig. 5.1 is
for a typical ductile material. This material can sustain large plastic deformation,
and hence a great amount of energy can be absorbed before fracture occurs. As
such, assessment of failure in a ductile material is usually derived from yield stress.
On the contrary, a typical brittle material exhibits (very) little plastic deformation
before fracture, and only little energy is absorbed. Hence, assessing the failure of
a brittle material is usually derived from the maximum stress. To illustrate the difference between these two types of materialda typical steel has a 0.2% yield strain
(based on the conventional definition) and a failure strain ranging from 12% to 20%,
and is known as a ductile material. Human cortical bone has a failure strain of about
2% and is considered as brittle material.
7.4.1 PRINCIPAL STRESS
In Section 1.2, we discussed the six stress components (three axial stresses sxx, syy,
and szz and three shear stresses sxy, syz, and szx). As the coordinate system selected
to formulate an FE model can be arbitrary, it is easily understandable that a different
set of stress values will be present when a different coordinate system is chosen.
Because the material does not care which coordinate system it was based on, a
more scientific way to evaluate the stress components is to determine the maximum
tensile stress, maximum compressive stress, and maximum shear stress. All these
values can be obtained by rotating the coordinate system until the shear stress
7.4 Response Variables
FIGURE 7.5
A plane stress element is rotated with an angle qP so that the shear stress on the plane
vanishes. As such, the only stress components left are the maximum and minimum
principal stresses.
components become zero, and the resulting stress components are known as the
principal stresses.
Fig. 7.5 shows a plane stress element with normal stresses sxx and syy as well as a
shear stress sxy. After the coordinate system is properly rotated, the only stress components left are the maximum principal stress (smax) and minimum principal stress
(smin).
As upper-level undergraduate and beginning graduate students already know the
concept behind the principal stresses in 2D from previous classes, only the relevant
equations are provided here, for the sake of easy referencing. In order to eliminate
the shear stress, the rotation angle qP must satisfy Eq. (7.12).
2sxy
2sxy
1
tan 2qP ¼
or qP ¼ tan1
(7.12)
2
sxx syy
sxx syy
And the corresponding maximum and minimum principal stresses are shown in
Eq. (7.13).
ffi
sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
sxx þ syy
sxx syy 2
smax ¼
þ
þ s2xy
2
2
(7.13)
sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
sxx þ syy
sxx syy 2
smin ¼
þ s2xy
2
2
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The principal stresses in 3D are, in essence, the eigenvalues of the 3D stress
tensor. More information regarding the calculations of eigenvalues and eigenvectors
is provided in Section 8.3. A second-order tensor, such as a stress or strain tensor,
possesses three invariants, where the term “invariant” indicates that the value does
not change when viewing from different coordinate systems. In accordance with
concepts in linear algebra, the trace (tr) of a tensor is defined as the sum of the three
diagonal terms, from upper left to lower right. The three stress invariants are shown
in Eq. (7.14).
2
3
sxx sxy sxz
6
7
6
7
½s ¼ 6 syx syy syz 7
4
5
szx szy szz
I1 ¼ trðsÞ ¼ sxx þ syy þ szz
(7.14)
1 I2 ¼ tr s2 ¼ sxx syy þ syy szz þ szz sxx s2xy s2yz s2zx
2
1 I3 ¼ tr s3 ¼ sxx syy szz sxx s2yz syy s2zx szz s2xy þ 2sxy syz szx
3
The corresponding maximum principal stress (s1), minimum principal stress
(s3), and intermediate principal stress (s2) are written as shown in Eq. (7.15).
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi I1 2
smax ¼ s1 ¼ þ
I 21 3I2 cos j
3 3
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi I1 2
4p
2
(7.15)
smin ¼ s3 ¼ þ
I 1 3I2 cos j 3
3 3
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi I1 2
2p
2
sint ¼ s2 ¼ þ
I 1 3I2 cos j 3
3 3
where j ¼
1cos1
3
2I13 9I1 I2 þ27I3
2ðI12 3I2 Þ
3=2
.
Eqs. (7.14) and (7.15) can be used to calculate the principal stresses directly from
the stress tensor as observed from the xeyez coordinate system. In other textbooks,
the three stress invariants may be defined by taking the stress tensor that has been
prerotated to coincide with the principal axes. That is, the (principal) stress tensor
is a diagonal matrix with a form of:
2
3
smax 0
0
6
7
½s ¼ 4 0
sint
0 5;
0
0 smin
7.4 Response Variables
where smax, sint, and smin are the three eigenvalues. In this case, the three invariants
are defined as
I1 ¼ smax þ sint þ smin
I2 ¼ smax sint þ sint smin þ smin smax
I3 ¼ smax sint smin
You may notice that these stress invariants are identical to the three invariants listed
in Eq. (7.14), if sxy, syz and szx are considered zero. Also, the third invariant is the
determinant of the stress tensor.
Failure criteria related to the maximum principal stress are credited to
Charles-Augustin de Coulomb (Jun. 1736eAug. 1806) and William John Macquorn Rankine (Jul. 1820eDec. 1872), who stated that a brittle material will
fail when the maximum principal stress exceeds its threshold. Hence, the principal stresses are frequently used as the design criteria to prevent brittle material
from failure.
7.4.2 MAXIMUM SHEAR STRESS
For ductile material, Henri Édouard Tresca (Oct. 1814eJun. 1885) proposed the use
of the maximum shear stress as the yielding criterion. Similar to the principal
stresses, the angle of rotation qS for a plane stress problem at which the maximum
shear stress is presented is written in Eq. (7.16).
sxx syy
sxx syy
1 1
or qS ¼ tan
tan 2qS ¼ (7.16)
2
2sxy
2sxy
And the corresponding maximum shear stress is shown in Eq. (7.17).
sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
sxx syy 2
ðsxy Þmax ¼
þ s2xy
2
(7.17)
7.4.3 VON MISES STRESS
Another yielding criterion for ductile material is proposed by Richard Edler von
Mises (Apr. 1883eJul. 1953). This yield criterion is based on the distortional energy stored in the material, and hence it is an effective stress method. Yielding is to
occur when the von Mises stress, as calculated in Eq. (7.18), reaches the yield
stress.
rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
sVM ¼ s2xx þ s2yy þ s2zz sxx syy syy szz szz sxx þ 3 s2xy þ s2yz þ s2zx
(7.18)
305
306
CHAPTER 7 Stepping Through Finite Element Analysis
If the principal stresses are already calculated, the alternate form of von Mises
stress is written as shown in Eq. (7.19).
sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
ðs1 s2 Þ2 þ ðs2 s3 Þ2 þ ðs3 s1 Þ2
sVM ¼
(7.19)
2
In most of the FEA software packages, all stress components discussed above
(the principal stresses, maximum shear stress, and von Mises stress), are calculated.
It is for users to decide which failure or injury criteria to use for their particular problem. You should be aware that numerous other yield or fracture criteria have been
presented in scientific literature. If none of these simple criteria suit your needs,
additional research should be conducted to identify the best criterion to use for
your problem.
EXERCISES
1. Interpret in your own words the quotation given in Section 7.3.1: “No one trusts
a model except the man who wrote it; everyone trusts an observation except
the man who made it.”
2. Why is setting up validation and verification standards so important?
3. Create an Excel spread sheet that performs the Jacobi method for solving
Exercise 6.3, where the final equation is
2
0:7
0
0
6
0:0336 0
6 0
6
10 6 0
0
28
6
0
0
4 0:35
0
0:42
7
0:35
0
66
0
0:35
0
9
9 8
38 9 8
F2x >
0
u2 >
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
> >
7> >
>
>
10000 >
w2 >
F2z >
0:42 7>
>
>
>
>
>
>
=
<
<
=
=
<
7
7
1700
7 7 q2 ¼ M2 ¼
>
> >
>
> >
>
>
>
> >
7>
>
>
0
u >
0 5>
>
> F3x >
>
> >
>
>
>
>
>
>
>
>
> >
>
> 3>
;
:
:
;
; >
:
0
q3
M3
14
0
4. Create another Excel spreadsheet that uses the GausseSeidel method to solve
Exercise 6.5, where the final equation is:
2
10
6
6 0
6
½K ¼ 10 6 5
6
4 0
0
5
0
6:67
0
0
0
3
7
6:67 7
7
0
6:44 1:44 1:92 7
7
7
0
1:44
4
0 5
0 6:67 1:92
0
10:67
9
8
9
8 9 8
F
0
u
>
>
>
>
2x >
2>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
> >
>
>
0
>
>
>
=
< F2y >
<
= >
=
< v2 >
0
½Kfug ¼ fFg ½K u3 ¼ F3x ¼
>
>
>
>
>
>
>
>
> >
>
>
> F4x >
>
>
>
20000 >
u4 >
>
>
>
>
>
>
>
> >
>
>
>
>
>
>
;
:
:
; >
;
: >
F4y
50000
v4
86
References
FIGURE 7.6
Facial impact forceetime response corridors.
From Biomechanical Response Requirements of the THOR NHTSA Advanced Frontal Dummy, GESAC Report
No. 05-03. Available at: https://www.nhtsa.gov/sites/nhtsa.dot.gov/files/thorbio05_1.pdf.
5. Fig. 7.6 shows the forceetime corridors due to rigid bar facial impacts
reconstructed by Nyquist et al. (1986) from experimental data obtained by
Allsop et al. (1988). (1) Construct the forceetime equation that represents
the average response of rigid bar facial impact, and (2) Assume that the
forceetime history for the face FE model of a crash dummy has the form of
F ¼ 33.93 t3 þ 321.4 t2 158.9 t, where F is in N and t is in ms. Calculate
all I’s, G, H, and Ec based on Geer’s work.
6. Given that sxx ¼ 45 MPa, syy ¼ 34 MPa, and sxy ¼ 16 MPa, calculate the
rotation angle to eliminate shear stress, maximum and minimum principal
stresses, the three stress invariants, the rotation angle of maximum shear
stress, the maximum shear stress, and von Mises stress. Remember any stress
in the z direction is assumed zero.
REFERENCES
Allsop, D., Warner, C., Wille, M., Schneider, D., Nahum, A., 1988. Facial impact response e
A comparison of the hybrid III dummy and human cadaver. In: Proceedings of the 32nd
Stapp Car Crash Conference. http://dx.doi.org/10.4271/881719. SAE Technical Paper
881719.
American National Standards Institute (ANSI), 1987. American National Standard Guidelines
for the Verification and Validation of Scientific and Engineering Computer Programs for
the Nuclear Industry, vol. 10. The Society.
307
308
CHAPTER 7 Stepping Through Finite Element Analysis
American Nuclear Society (ANS), 1985. Guidelines for the Verification and Validation of Scientific and Engineering Computer Programs for the Nuclear Industry: ANSI/ANS-10.4.
American Nuclear Society, La Grange Park, IL.
American Society of Mechanical Engineers (ASME), 2006. Guide for Verification and Validation in Computational Solid Mechanics. ASME, New York, New York.
Barbat, S., Fu, Y., Zhan, Z., Yang, R.-J., Gehre, C., 2013. Objective rating metric for dynamic
systems. Paper Number 13-0448. In: Technical Session: Testing and Modeling of Structural Performance in Frontal Crashes, in the Proceedings of the 23rd Enhanced Safety
of Vehicles Conference, Seoul, Korea May 27e30.
Box, G.E.P., 1976. Science and statistics. Journal of the American Statistical Association 71
(356), 791e799. http://dx.doi.org/10.1080/01621459.1976.10480949.
Deb, A., Haorongbam, B., Chou, C.C., 2010. Efficient Approximate Methods for Predicting
Behaviors of Steel Hat Section under Impact Axial Loading. SAE Paper No. 2010-011015.
Department of Defense (DoD), 2009. DoD Modeling and Simulation (M&S) Verification,
Validation, and Accreditation (VV&A). DoD Instruction 5000.61.
Geers, T.L., June 1984. An objective error message for the comparison of calculated and
measured transient response histories. The Shock and Vibration Bulletin 54, 99e107.
Gehre, C., Gades, H., Wernicke, P., 2009. Objective rating of signals using test and simulation
responses. Paper Number 09-0407. In: 21st ESV Conference; Stuttgart; Germany.
Institute of Electrical and Electronic Engineers (IEEE), 2014. 730-2014-IEEE Standard for
Software Quality Assurance Processes. http://dx.doi.org/10.1109/IEEESTD.2014.
6835311. ISBN:978-0-7381-9168-3.
National Aeronautics and Space Administration (NASA), 2008. NASA Technical Standard,
Standard for Models and Simulations. NASA-STD-7009.
Nyquist, G., Cavanaugh, J., Goldberg, S., King, A., 1986. Facial impact tolerance and
response. In: Proceedings of the 30th Stapp Car Crash Conference. http://dx.doi.org/
10.4271/861896. SAE Technical Paper 861896.
Oreskes, N., Shrader-Frechette, K., Belitz, K., 1994. Verification, validation, and confirmation
of numerical models in the earth sciences. Science 264, 641e646.
Roache, P.J., 1998. Verification and Validation in Computational Science and Engineering.
Hermosa Publisher, Albuquerque, NM. ISBN:10: 0913478083.
Sprague, M.A., Geers, T.L., 2004. A spectral-element method for modeling cavitation in transient fluidestructure interaction. International Journal for Numerical Methods in Engineering 60 (15), 2467e2499.
Zhan, Z., Fu, Y., Yang, R.-J., 2011a. Enhanced error assessment of response time histories
(EEARTH) metric and calibration process. In: SAE 2011 World Congress; SAE 201101-0245; Detroit; MI; USA.
Zhan, Z., Fu, Y., Yang, R.-J., Peng, Y., 2011b. An enhanced Bayesian based model validation
method for dynamic systems. ASME Journal of Mechanical Design 133 (4), 041005.
Zhan, Z., Fu, Y., Yang, R.-J., Peng, Y., 2012. Development and application of a reliabilitybased multivariate model validation method. International Journal of Vehicle Design 60
(3/4), 194e205.
Zhu, F., Chou, C.C., Yang, K.H., Chen, X., Wagner, D., Bilkhu, S., 2012. Obtaining material
parameters for die cast AM60B magnesium alloy using optimization techniques. International Journal of Vehicle Safety 6 (2), 178e190.
CHAPTER
Modal and Transient
Dynamic Analysis
8
King H. Yang
Wayne State University, Detroit, Michigan, United States
8.1 INTRODUCTION
As it has been discussed so far, the FE method is frequently used to solve static structural or biomechanical problems that contain complex geometry, multiple material
compositions, and complicated boundary and loading conditions. Likewise, the
FE method is often used to solve dynamic problems of similar nature. There are
two common dynamic analysis types involved: the modal analysis and transient
dynamic analysis.
Modal analysis method involves the determination of natural/resonant frequencies and associated mode shapes (vibration modes) of a component or structure
under free (unforced) vibration. Mathematically speaking, the aim of modal analysis
is to identify the eigenvalues and eigenvectors. We can deduce natural frequencies
and associated characteristic eigenvectors with these two parameters, which in
turn is used to calculate nodal displacements, velocities, and accelerations. When
analytical solution methods are used for modal analysis, it is difficult to visually
display the vibration modes unless the structure of interest is very simple. On the
other hand, mode shapes can be graphically displayed with ease through the use
of an FE postprocessor. Considering that the FE method can handle complex geometry much easier than the analytical method, it is understandable that the FE method
is the preferred choice for performing modal analysis in many circumstances.
Transient dynamic analysis (also known as time-history analysis) aims at finding
dynamic responses of a structure under arbitrary time-dependent loads. After
completing FE calculations, time-dependent nodal displacements, velocities, accelerations, strains, stresses, forces, etc. within a structure can be graphically displayed.
Because of the high computational cost associated with solving a large system transient dynamic analysis, the total duration of simulation period is generally limited.
Fig. 8.1 shows a deformed rib cage of an elderly female at a particular instant to
demonstrate the result of a typical transient dynamic analysis. Computer animation
can be generated during postprocessing to visualize the processes of vehicular structural deformation, occupant kinematics, rib cage deformation, internal organ
rupture, among others.
Basic Finite Element Method as Applied to Injury Biomechanics. http://dx.doi.org/10.1016/B978-0-12-809831-8.00008-8
Copyright © 2018 Elsevier Inc. All rights reserved.
309
310
CHAPTER 8 Modal and Transient Dynamic Analysis
FIGURE 8.1
Deformed rib cage of a 70-year-old female occupant subjected to a medium-speed rightlateral impact (Kalra, 2016).
8.2 ELEMENT MASS MATRIX
In static analysis, the governing forceedisplacement equilibrium equations are
½Kfug ¼ ff g, where [K] is the structural (global) stiffness matrix, {u} are the nodal
displacements, and {f} are the nodal load vectors. Using this equation, a single
spring system of length l in equilibrium has the form k Dl ¼ F, where k is the
spring constant, Dl is the change in length, and F is the applied force. Obviously,
this relationship was derived through Hooke’s law, a law that was developed more
than four centuries ago and was named after Robert Hooke (Jul. 1635eMar.
1703). Because mass and damping play no role in determining the structural
response due to static loading, we have not yet introduced these two terms.
For dynamic equilibrium problems, the governing equation is different from
static equilibrium problems due to the need for inertial terms. From previous courses
in dynamics or vibration, we know that the equation of motion for an unforced,
single massespring system has the following form
€ þ kxðtÞ ¼ 0;
mxðtÞ
(8.1)
€ indicates double differentiation of the displacement function with respect
where ðxÞ
to time. Obviously, displacements, velocities, and accelerations in a moving system
€ as functions of
all need to be functions of time. Explicitly denoting both x(t) and xðtÞ
time is aimed at highlighting the difference between static and dynamic problems.
8.2 Element Mass Matrix
As familiarity grows with the concept that time is inherent in dynamic problems, we
may omit explicit denotations.
Eq. (8.1) is also known as the equation of motion or the dynamic equilibrium
equation for an undamped (i.e., the damping coefficient is zero) and free (i.e., no
external force involved) response (or vibration) of a single massespring system.
We can expand this equation to represent a system of masses and springs subjected
to external loading as
€
½MfxðtÞg
þ ½KfxðtÞg ¼ ff ðtÞg;
(8.2)
where [M] is the structure mass matrix, [K] is the structure stiffness matrix, and a
nonzero {f(t)} represents a forced vibration. If {f(t)} ¼ 0, it is a free vibration.
We have emphasized previously that the FE method requires all boundary and
loading conditions be prescribed at the corresponding nodal DOFs. We easily understand that the overall mass of the structure needs to be individually prescribed to the
corresponding nodal DOFs, as well. Because we have used the element shape functions Ni to interpolate coordinates and displacements anywhere within an element, it
is only natural to think about using element shape functions to distribute the element
mass to those nodes that form the element.
The most logical approach is to apply the same set of shape functions [N], which
are used to identify the element strain displacement matrix [B] and element stiffness
matrix [k], to determine the element mass matrix [m]. The resulting element mass
matrix, based on choosing the same set of shape functions, is called the stiffness
consistent mass matrix, or consistent mass matrix for short. Once all element
mass matrices are determined, they can be assembled into a global mass matrix
[M]. Again, only the simplest types of elements, the 2-node bar and beam elements;
3-node plane elements; 4-node plane-stress, plane-strain, and plate elements; and
8-node solid elements are discussed, because these element types produce more
accurate answers than other higher order element types when a great number of
elements are used to represent the structure.
Note that the term “mass” refers to a scalar while “weight” is a vector. In order
for a scalar mass to reveal its weight, an acceleration (such as the earth gravitational
acceleration with a magnitude of 9.8 m/s2 along the vertical direction) needs to be
multiplied to the quantity of mass. Additionally, [M] needs to be a square matrix,
similar to that of the [K] matrix.
8.2.1 CONSISTENT MASS MATRIX
Using the principle of variational formulation described in Section 4.4, the kinetic
energy of an element is written as
Z
1
ðK:E:Þe ¼
rfve gT fve gdV;
(8.3)
2
Ve
where e represents a specific element, r is the density, K.E. is the kinetic energy, V is
the domain volume, and v is the velocity. For most engineering problems, the density
311
CHAPTER 8 Modal and Transient Dynamic Analysis
r remains constant throughout the entire dynamic domain. Hence, we will assume
density is constant for the rest of this section. We express the element velocity field
(ve) by differentiating the element displacement field (ue) with respect to time as
ve ¼
n
X
Ni vi ¼
i¼1
n
X
i¼1
Ni
dui
;
dt
(8.4)
where i indicates the node numbers that form the element and n is the number of
nodes that form an element.
By definition, static problems are not dependent upon time. For instance, nodal
displacements {ui} are constant and nodal velocities {vi} are zero for static problems. In contrast, dynamic values are dependent upon time, and specifically {ve}
and {ue} are functions of time. Importantly, these values are not dependent upon
spatial geometry (x, y, z), and therefore they can be removed from volume integrals.
Conversely, element shape functions Ni are dependent only on spatial geometry, and
not time. Inserting ve shown in Eq. (8.4) into Eq. (8.3) and moving all {ue}-related
terms outside the integration with respect to spatial geometry results in Eq. (8.5).
Z
1
ðK.E.Þe ¼
rfve gT fve gdV
2 Ve
(8.5)
Z
duj
1 dui T
T
¼
rfNi g fNj gdV
2 dt
dt
Ve
By comparing Eq. (8.5) to the definition of kinetic energy derived from Newton’s
second law of motion, KE ¼ 1 2 mv2 , it is obvious that the element mass matrix [m]e
can be written as shown in Eq. (8.6). This equation forms the basis for calculating the
consistent element mass matrix.
Z
½me ¼
(8.6)
rfNi gT fNj gdV
=
312
Ve
8.2.1.1 Element Mass Matrix for a 2-Node Bar Element
As mentioned previously, a 2-node bar element requires only C0 continuity while a
2-node beam element needs a C1 continuity. Hence, the element shape functions
derived for a bar/spring element and a beam element are different. We can use either
of the shape functions shown in Eq. (2.16) (based on a Cartesian coordinate system
with the origin located at the first node of the element) or Eq. (3.6) (based on a natural coordinate system) to calculate the consistent element mass matrix.
We will use Eq. (2.16) to demonstrate the process, and you are encouraged to go
through this process on your own using Eq. (3.6). Keep in mind that the shape functions for Eq. (3.6) are based on the natural coordinate system, and you must ensure
that the proper Jacobian is applied to account for the length ratio.
Consider a 2-node, 1D bar element with constant density r, constant crosssectional area A, and length L. For this element, the derivation processes for
8.2 Element Mass Matrix
obtaining the consistent element mass matrix with respect to the corresponding
DOFs f u1 u2 gT are
Z
½me ¼
rfNi gT fNj gdV
Ve
¼r
9
8
>
L x>
>
>
>
>
Z L>
=
< L >
Lx
0
>
>
>
>
:
x
L
>
>
>
>
;
L
2
L2 2Lx þ x2
6
Z L6
L2
6
¼ rA
6
0 6
4
Lx x2
L2
2
3
2 1
rAL 4
5
¼
6
1 2
x
Adx
L
3
Lx x2
7
L2 7
7
7 dx
7
2
5
x
2
L
(8.7)
Based on the basic information defined for this element, the total mass is calculated as m ¼ rAL. As a quick check, the sum of all entries in the consistent element
mass matrix is also rAL. This simple check confirms that the calculation is likely to
be correct, and indeed it is correct.
As previously discussed regarding static problems, which have no kinematic
motion, each node has only one DOF. The only concern for a static bar element is
axial deformation. If the bar was allowed to move in both the x- and z-directions,
as shown in Fig. 8.2, kinematic motions would pose as DOFs, and therefore, unlike
static problems, each node would have two DOFs. For a bar element with possible
motions in a 2D plane, the consistent mass matrix corresponding to the four DOFs
f u1 w1 u2 w2 gT is written as
3
2
2 0 1 0
7
rAL 6
60 2 0 17
(8.8)
½me ¼
7.
6
6 41 0 2 05
0 1
0
2
As we can see from Eq. (8.8), the sum of all entries for this 2-node, 1D bar
element is twice that of the static bar element; the sum is 2rAL rather than rAL.
This discrepancy is due to the inclusion of inertial mass in each direction; we
need a mass of rAL in the x-direction and a mass of rAL in the z-direction.
313
314
CHAPTER 8 Modal and Transient Dynamic Analysis
FIGURE 8.2
A 2-node bar element considered as a two DOFs per node element, because the element
is allowed to move in both x- and z-directions due to dynamic motions.
8.2.1.2 Element Mass Matrix for a 2-Node Beam Element
To derive the consistent mass matrix for a 2-node, 1D beam element of length L, we
use shape functions based on the natural coordinate system, as shown in Eq. (3.17).
In this beam element, the Jacobian (also known as the determinant of the Jacobian
dx ¼ L, that is, dx ¼ L dx. Again, a constant density is assumed for
matrix) is jJj ¼ dx
2
2
this element, and the consistent element mass matrix derived in a step-by-step
process is
L 1 ξ
2 − 3ξ + ξ
2 − 3ξ + ξ
4
L 1− ξ − ξ + ξ
×
ξ
ξ
2 + 3ξ − ξ
L 1− ξ − ξ + ξ
1 ξ
ξ
ξ
ξ
ξ
8
×
4
L
1 ξ
L 1− ξ − ξ + ξ
8
4
×
L
2 − 3ξ + ξ
2 + 3ξ − ξ
×
4
4
8
8
4
1 ξ
ξ
2 − 3ξ + ξ
4
4
L
ξ
8
×
L 1 ξ
×
L 1− ξ − ξ + ξ
4
4
ξ +ξ
8
L −1 − ξ + ξ + ξ
8
2 − 3ξ + ξ
×
4
8
×
4
×
2 + 3ξ − ξ
4
L −1 − ξ + ξ + ξ
8
)
dξ
8
L −1 − ξ + ξ + ξ
8
(8.9)
where the associated DOFs are f w1 q1 w2 q2 gT .
Note that the highest order of the polynomial in Eq. (8.9) is 6. Based on the Gauss
quadrature rule, a polynomial equation with the highest order of 2n 1 requires a
minimum of n Gauss points to acquire the exact solution. For a sixth order polynomial, n ¼ 3.5. Hence, a 4-point Gauss integration is required to obtain the exact solution. Gauss points and associated weighting factors for up to the fifth order Gauss
quadrature are listed in Table 8.1.
8.2 Element Mass Matrix
Table 8.1 Locations and Weighting Factors for up to 5-Points Gauss
Quadrature
No. of Gauss
Points
Locations of Gauss
Point
Weighting
Factors
Suitable Order
of Polynomial
1
2
3
0
0.57735
0.77460
0
0.86113
0.33998
0
0.53846
0.90617
2
1
0.55556
0.88889
0.34785
0.65214
0.56889
0.47863
0.23693
1st
2nd and 3rd
4th and 5th
4
5
6th and 7th
8th and 9th
To avoid making arithmetic mistakes, a simple computer program should be used
to incorporate the locations of the Gauss points and associated weighting factors.
The program could be used to determine the consistent mass matrix, one entry at
a time. For the purpose of demonstration, we calculate m11 and m12(¼m21) using
the 4-point Gauss quadrature as follows:
Z 1
2
rAL 1 2 3x þ x3
m11 ¼
16
1 2
Z
rAL 1 4 12x þ 9x2 þ 4x3 6x4 þ x6 dx
¼
(8.10)
32 1
z
4
X
wi fi ¼ 0:37143rAL
i¼1
m12 ¼ m21 ¼
z
4
X
rAL L
2
32
Z
1
1
2 5x þ x2 þ 6x3 4x4 x5 þ x6 dx
wi fi ¼ 0:05238rAL2
i¼1
Note that for Eq. (8.10) we expand (2 3x þ x3)2 prior to inserting the x values
for Gauss integration. Since definite integrals of odd functions are zero when evaluations are between {a to a}, and since some polynomial terms are odd functions
while others are even, there is a likelihood of making a mistake if Gauss integration
is applied prior to expanding the squared operation.
The entire consistent mass matrix for a 2-node beam element is listed in
Eq. (8.11), which you can use to check your computer program. Note that the entries
associated with the two rotational DOFs (q1 and q2, corresponding to the second and
fourth rows as well as second and fourth columns) of the matrix shown in Eq. (8.11)
315
316
CHAPTER 8 Modal and Transient Dynamic Analysis
are unrelated to the total mass of the beam element. By summing all other values
(m11, m13, m31, and m33), we can see that the total mass remains as rAL.
3
2
0:37143
0:05238L
0:12857
0:03095L
7
6
6 0:05238L
0:00952L2
0:03095L 0:00714L2 7
7 (8.11)
6
½me ¼ rAL6
0:03095L
0:37143
0:05238L 7
5
4 0:12857
2
2
0:03095L 0:00714L 0:05238L 0:00952L
8.2.1.2.1 Element Mass Matrix for a 3-Node, 2D Triangular Element
For a 3-node, 2D constant-strain, triangular element of constant thickness t, there are
a total of six DOF’s, three horizontal displacements and three vertical displacements.
The three area coordinates introduced in Section 3.4.1 are used as the element shape
functions. To integrate these shape functions that are based on area coordinates, we
need to know that
Z
n!p!q!
;
(8.12)
x1 n x2 p x3 q dA ¼ 2A
ð2 þ n þ p þ qÞ!
A
where x1, x2, and x3 are the three area coordinates, A is the area of the triangle, n, p,
and q are exponents of the respective area coordinates,
and the symbol “!” denotes
R
the factorial. For example, if we need to find A x1 2 dA, we can see from Eq. (8.12)
that n ¼ 2 and p ¼ q ¼ 0. As such, we have
Z
n!
2!
A
¼ 2A
¼ .
x1 2 dA ¼ 2A
ð2 þ nÞ!
ð2 þ 2Þ! 6
A
R
Similarly, to find A x1 x2 dA, we need to have n ¼ 1, p ¼ 1, and q ¼ 0. From
Eq. (8.12), we can find
Z
n!p!
1
A
¼ 2A
¼ .
x1 x2 dA ¼ 2A
ð2 þ n þ pÞ!
ð2 þ 1 þ 1Þ! 12
A
Using this integration formula shown in Eq. (8.12) of determining the consistent
mass matrix of the triangular element gives us
2 3
x1
Z 6 7
6 7
7
½me ¼ r6
6 x2 7½ x1 x2 x3 dV
V 4
5
x3
(8.13)
2
3
2 2
3
2 1 1
x1 x1 x2 x1 x3
6
7
7
Z 6
7
6
7
rAt 6
2
6
7
6
7
¼ rt 6 x1 x2 x2 x2 x3 7dA ¼
6 1 2 1 7.
12
A4
4
5
5
2
1 1 2
x1 x3 x2 x3 x3
8.2 Element Mass Matrix
Obviously, Eq. (8.13) covers only one of the two DOFs for each node. In order to
accommodate all six DOFs of this triangular element, the consistent mass matrix,
corresponding to f u1 v1 u2 v2 u3 v3 gT needs to be repeated as shown
in Eq. (8.14). As a quick check, the total mass along the horizontal and vertical directions are both rAt.
3
2
2 0 1 0 1 0
60 2 0 1 0 17
7
6
7
6
1 0 2 0 1 07
rAt 6
7
6
(8.14)
½me ¼
7
12 6
60 1 0 2 0 17
7
6
41 0 1 0 2 05
0
1 0
1
0
2
In the event that integration is conducted along the first side of a triangular
element (i.e., x1 ¼ 0), Eq. (8.12) must be modified to accommodate this
special case, as denoted in Eq. (8.15). Similar attention is needed when x2 ¼ 0 or
x3 ¼ 0.
Z
m!n!
(8.15)
x2 m x3 n dA ¼ L
ð1
þ
m þ nÞ!
L
8.2.1.2.2 Element Mass Matrix for a 4-Node, 2D Plane Element
For a bilinear 4-node, 2D rectangular plane-stress or plane-strain element of constant
thickness t, length 2a in the x-direction, and width 2b in the y-direction, we use the
isoparametric shape functions described in Section 3.4.2 to calculate the consistent
mass matrix of the element. Eq. (8.16) shows the derivation of the consistent mass
matrix just before carrying out the double integrations.
8
9
>
>
>
> ð1 xÞð1 hÞ >
>
>
>
>
>
>
>
4
>
>
>
>
>
>
>
>
>
>
>
> ð1 þ xÞð1 hÞ >
>
>
>
>
>
>
>
Z <
=
4
ð1 xÞð1 hÞ ð1 þ xÞð1 hÞ ð1 þ xÞð1 þ hÞ ð1 xÞð1 þ hÞ
t dx dy
½me ¼
r
>
4
4
4
4
Ve >
>
> ð1 þ xÞð1 þ hÞ >
>
>
>
>
>
>
>
>
>
4
>
>
>
>
>
>
>
>
>
>
>
>
>
>
ð1
xÞð1
þ
hÞ
>
>
>
>
:
;
4
2
3
1 x2 ð1 hÞ2
1 x2 1 h2
ð1 xÞ2 1 h2
ð1 xÞ2 ð1 hÞ2
6
7
6
7
2
2 7
Z 1 Z 1 6 1 x2 ð1 hÞ2 ð1 þ xÞ2 ð1 hÞ2 ð1 þ xÞ2 1 h2
1
x
1
h
6
7
rabt
6
7dxdh
¼
7
16 1 1 6
6 1 x2 1 h 2
ð1 þ xÞ2 1 h2
ð1 þ xÞ2 ð1 þ hÞ2
1 x2 ð1 þ hÞ2 7
6
7
4
5
2
2
2
2
2
2
2
2
ð1 xÞ 1 h
ð1 xÞ ð1 þ hÞ
1x 1h
1 x ð1 þ hÞ
(8.16)
317
318
CHAPTER 8 Modal and Transient Dynamic Analysis
Because the highest order of polynomials for both x and h is 2, it is sufficient to
use the 2 2 Gauss quadrature rule to integrate Eq. (8.16). That is, the four Gauss
points needed to carry out the integration are (x1 ¼ 0.57735, h1 ¼ 0.57735),
(x2 ¼ 0.57735, h2 ¼ 0.57735), (x3 ¼ 0.57735, h3 ¼ 0.57735), (x4 ¼ 0.57735,
h4 ¼ 0.57735). With all weighting factors along both axes being 1, the combined
weighting factor for all four Gauss points is 1, as well. Additionally, the area of the
element is A ¼ 4ab. Eq. (8.17) shows the consistent mass matrix for the 2D, 4-node
bilinear element. Again, the sum of all entries is rAt. Because eight DOFs need to be
considered for 2D motions, Eq. (8.17) needs to be repeated and expanded to an 88
matrix.
3
2
0:11111 0:05556 0:02778 0:05556
6 0:05556 0:11111 0:05556 0:02778 7
7
6
(8.17)
½me ¼ rAt6
7
4 0:02778 0:05556 0:11111 0:05556 5
0:05556
0:02778
0:05556
0:11111
8.2.1.2.3 Element Mass Matrix for a 3D, 8-Node Solid Element
For an 8-node, 3D brick element of constant density r and dimensions
2a 2b 2c, the shape functions described in Section 3.5.2 are used to calculate
the consistent mass matrix. The size of this mass matrix is 88 for each of the three
axes. Because detailed derivations are quite lengthy, only the results are provided for
reference in Eq. (8.18). In this equation, V represents the volume of the element, and
V ¼ 8abc. Of course, the complete consistent mass matrix for a 24-DOFs, 8-node,
solid element is a 2424 matrix.
2
6
6
6
6
6
6
6
6
6
6
½me ¼ rV 6
6
6
6
6
6
6
6
6
4
0:037037 0:018519 0:009259 0:018519 0:018519 0:009259
0:00463
0:037037 0:018519 0:009259 0:009259 0:018519 0:009259
0:037037 0:018519
0:00463
0:037037 0:009259
0:009259 0:018519
0:00463
0:009259
0:037037 0:018519 0:009259
S
Y
M
M
0:037037 0:018519
0:037037
0:009259
3
7
0:00463 7
7
7
0:009259 7
7
7
7
0:018519 7
7
7
0:018519 7
7
7
7
0:009259 7
7
7
0:018519 7
5
0:037037
(8.18)
8.2.2 LUMPED MASS MATRIX
Compared to static analysis, one substantially different aspect in dynamic analysis is
the calculation of time histories for displacement, velocity, acceleration, stress, etc.
The consistent mass matrix described above has both diagonal and off-diagonal terms.
Solving the dynamic equilibrium differential Eq. (8.1) with these off-diagonal terms is
8.2 Element Mass Matrix
highly computationally intensive. To mitigate this deficiency, diagonalization of the
mass matrix is desired. There are several advantages of having a diagonalized mass
matrix. First, a diagonalized matrix can be stored as a vector, and thus storage space
is saved. Second, assembling all diagonalized element mass matrices result in a diagonalized structure mass matrix [M] is very easy. Third, finding [M]1 of a diagonal
matrix [M] is very simple, straight forward, and results in another diagonalized matrix.
€
When multiplying [M]1 to both sides of Eq. (8.1), the first term involves only fxg,
which can be easily integrated using an implicit or explicit integration scheme (discussed in Section 8.5).
The method used to formulate a consistent mass matrix into a diagonalized mass
matrix is called lumping. There are several lumping methods reported in the literature for the generation of a lumped mass matrix. Some commercially available software packages allow users to determine which specific lumping method is the
preferred choice of the user. Unfortunately, there are no universal rules in determining which lumping method is best suited for what condition. For this reason,
only the HintoneRockeZienkiewicz (HRZ) method (Hinton et al., 1976) and the
very simple row sum method are described to show the essence of mass lumping.
8.2.2.1 HRZ Lumping Method
The HRZ mass lumping method involves two steps for each direction of the coordinate system:
1. Sum all diagonal terms related to the translational DOFs of the element mass
matrix [m]e and call it “S.” In other words, any DOFs associated with rotational
DOFs should be excluded in this calculation. For example, m22 and m44 in Eq.
(8.11) for a 2-node beam element are associated with rotational DOFs and
should not be added.
2. For each diagonal term mii related to translational DOFs, calculate mii/S.
Example 8.1
Calculate the lumped mass matrix of a 2-node beam element using the HRZ
lumping method.
Solutions
The corresponding DOFs for a 2-node beam element are f w1 q1 w2 q2 gT .
From Eq. (8.11), the sum of the translational DOFs-related diagonal items (m11
and m33) is S ¼ 0.74286rAL. Therefore, the lumped mass matrix based on the
HRZ method is found by dividing the four diagonal terms by 0.74286 as
3
2
0:5
0
0
0
7
6
7
6 0 0:0128L2 0
0
7.
6
(8.19)
½me ¼ rAL6
7
0
0
0:5
0
5
4
0
0
0 0:0128L2
319
320
CHAPTER 8 Modal and Transient Dynamic Analysis
From Eq. (8.19), the total calculated mass exceeds the mass of the beam
element (i.e., rAL), because the steps used for diagonalization of the mass
matrix may change some fundamental physical properties, such as the rotational
moment of inertia, of the element. Again, as long as the solutions are reasonable
and acceptable by engineers, and no other methods are better than using the FE
method to find the solution, this simplification will be continuously used for
solving dynamic problems.
8.2.2.2 Row Sum Lumping Method
Another commonly used lumping method is the row sum method, achieved by
simply summing all terms in the consistent mass matrix onto the diagonal term.
For example, the 4-node, 2D-plane element of area A and thickness t can be calculated from Eq. (8.17) as shown in Eq. (8.20). As can be seen, summation of all four
diagonal terms yields the same total mass as the element.
3
2
0:25
0
0
0
6 0
0:25
0
0 7
7
6
(8.20)
½me ¼ rAt6
7
4 0
0
0:25
0 5
0
0
0
0:25
8.3 MODAL ANALYSIS
Modal analysis, in essence, is the method to identify the natural frequencies and associated mode shapes. The concept of modal analysis was introduced by Sir John William Strutt (Nov. 1842eJun. 1919), who became Lord (Baron) Rayleigh in 1873, for
studying the vibration of linear dynamic problems without damping (Rayleigh, 1877).
Once the resonant frequencies and their respective mode shapes are known, the time
histories of the vibrating structures can be directly constructed. As such, the main
reason for using modal analysis instead of transient dynamic analysis is to reduce
computational costs. For an FE model of n DOFs, this system also has n mode shapes.
However, only a few mode shapes associated with the lowest few natural frequencies
are critical in structural analysis. If all or most mode shapes are required to capture the
response of the FE model, the cost-saving effect will diminish fairly rapidly. Note that
eigenvalues applied to other fields may utilize a few highest eigenvalues and their
corresponding mode shapes. In this book, the emphasis is placed on problems that
are related to structural analysis with eigenvalues on the lower end side.
In mechanics, the identifications of natural frequencies and mode shapes due to
vibrations are identical to calculating the eigenvalues and eigenvectors in mathematical problems. A note sung by a professionally trained singer or produced by a music
instrument may hit the resonant frequency, causing significant transfer of mechanical energy from sound to a wine glass, overstraining it, and forcing the wine glass to
8.3 Modal Analysis
shatter. An out-of-balance car tire can induce unwanted vibration when driving at a
certain speed. Also, bridges and skyscrapers can swing due to wind load or earthquake. If unattended, these structures may swing violently and cause catastrophic
failure. As such, analysis of eigenvalues and eigenvectors is a very important subject
in dynamic analysis.
We shall begin discussing the analytical background of modal analysis with a
review of fundamental knowledge regarding the determinations of eigenvalues
and eigenvectors in simple massespring systems subjected to free vibration. We
then discuss motions due to forced vibration. Using numerical methods derived
from these fundamental theories, the natural frequencies and mode shapes of a structure can be calculated from FE models.
8.3.1 FREE VIBRATION OF A MASSeSPRING SYSTEM
For a single massespring element (i.e., no damping, c ¼ 0) under free (i.e.,
unforced) vibration, the governing differential equation of motion of this system is
mx€ þ kx ¼ 0:
(8.21)
lt
Assuming that the solution to this differential equation has a form of x ¼ e , then
x€ ¼ l2 elt and the characteristic equation for this system and the solutions of the
characteristic equation are
rffiffiffiffi
k
2
ml þ k ¼ 0 and l ¼ i
;
(8.22)
m
respectively, where i is the imaginary unit (defined as i2 ¼ 1), the two l’s are the
eigenvalues, and the positive solution of l is related to the natural frequency of the
system (to be further illustrated in Example 8.2). A free-vibration system vibrates at
the natural frequency and will vibrate perpetually in a harmonic, sinusoidal manner.
A two-mass, two-spring system without external loads (including gravity) is
shown in Fig. 8.3. This system is represented by a 3-node, 2-element FE model,
where the location of mass m1 coincides with node P2 and mass m2 coincides
with node P3.
In this configuration, Eq. (8.21) for a single DOF system needs to be expanded to
reflect the increased DOFs as
€ þ ½Kfxg ¼ f0g.
½Mfxg
(8.23)
The global mass matrix [M] is formed by the two concentrated masses m1 and m2 as
m1 0
½M ¼
;
(8.24)
0 m2
and the global stiffness matrix [K], after removing the first row and column corresponding to the fixed boundary condition at P1 from further calculation, is
k1 þ k2 k2
½K ¼
.
(8.25)
k2
k2
321
322
CHAPTER 8 Modal and Transient Dynamic Analysis
FIGURE 8.3
A two-spring, two-mass system and the corresponding FE model.
Finally, the governing differential equation of motion for this system becomes
m1 0
k1 þ k2 k2
x2
x€2
þ
¼ 0:
(8.26)
x€3
0 m2
k2
k2
x3
Differential equations with the form of Eqs. (8.23) or (8.26) are mathematically
known as eigenvalue problems. With each eigenvalue (characteristic value), there is
an associated eigenvector (characteristic vector). In mechanics, the positive eigenvalues are related to the natural frequencies of the system and the associated eigenvectors describe the shapes of vibration, which are also known as mode shapes,
natural modes, normal modes, or principle modes of vibration.
Note that the global mass matrix seen in Eq. (8.26) must be of diagonal form
without rotational inertia (i.e., m12 ¼ m21 ¼ 0). Otherwise, finding analytical or numerical solutions for the eigenvalues would be more difficult. As such, the methods
for finding lumped element mass matrix described in Section 8.2.2 is well suited for
forming numerical methods to calculate eigenvalues, to be described in Section
8.3.3. Example 8.2 is provided for gaining a better understanding of the analytical
solution processes in determining the eigenvalues and eigenvectors.
Example 8.2
A two-spring, two-mass system with m1 ¼ m2 ¼ 1, k1 ¼ 3, and k2 ¼ 2 is shown in
Fig. 8.3. Determine the two natural frequencies and two vibration mode shapes.
Solutions
Assume the solution has the form {x} ¼ {X elt}; then by inserting the values of
m1, m2, k1, and k2 into Eq. (8.26), we have
1 0
3 þ 2 2
x2
x€2
þ
¼0
x€3
x3
0 1
2
2
"
l2
0
#
0
l2
5 2
þ
2 2
!
X2
X3
¼0
(8.27)
8.3 Modal Analysis
Because X2 and X3 need not be zero, Eq. (8.27) can only be valid if the determinant
of the characteristic matrix, shown in Eq. (8.28), is equal to zero.
l2 þ 5
2
2
l þ2
2
¼0
(8.28)
Expanding Eq. (8.28) gives us
2
l þ 5 l2 þ 2 4 ¼ 0
l4 þ 7l2 þ 6 ¼ 0
2
l þ 1 l2 þ 6 ¼ 0:
(8.29)
Solving Eq. (8.29) results in the eigenvalues l2 ¼ 1 and 6. If we equate l to
the frequency u as
l ¼ iu
(8.30)
l2 ¼ u2 ;
then the natural frequency is u, which has a dimension of rad/s. This is the reason
we say that the eigenvalues are “related to” natural (resonant) frequencies, but are
not natural frequencies. For a quick note, 2p rad around a circle is equal to 1
cycle, or 2p circular frequency (rad/s) is equal 1 cyclic frequency (cycle/s, or Hz).
Thus, 1 rad/s z 0.159 Hz. In this problem, the two natural frequencies are 1 and
pffiffiffiffi
6 rad/s. After obtaining the eigenvalues, we can calculate the eigenvectors {V}
associated with each of the eigenvalues determined from Eq. (8.29) as
" 2
)
#(
lj þ 5
2
VjxðP2 Þ
¼ 0;
(8.31)
VjxðP3 Þ
2
l2j þ 2
where j represents the jth eigenvalue, {V} represents the eigenvectors, and x(P2)
and x(P3) denote the x-direction vectors (motion), which are associated with P2
and P3 DOFs, respectively. For j ¼ 1, the first eigenvalue l2 ¼ 1, Eq. (8.31)
becomes
)
(
V1xðP2 Þ
4 2
¼ 0:
(8.32)
V1xðP3 Þ
2 1
Note that the determinant of the eigenequation (i.e., the matrix shown in Eq.
8.32) must be equal to zero to have a nontrivial solution. In other words, an eigenequation is a singular matrix. Thus, there are multiple combinations of V1xðP2 Þ
and V1xðP3 Þ that could satisfy Eq. (8.32). We will take the first equation
4V1xðP2 Þ 2V1xðP3 Þ ¼ 0 and describe the first eigenvector as V1xðP3 Þ ¼ 2V1xðP2 Þ.
In plain English, this eigenvector designates that for each unit movement in
the x-direction at P2, there are movements of two units at P3 in the same
direction (see Fig. 8.4, top). We can derive the eigenvector using the second
eigenequation 2V1xðP2 Þ þ V1xðP3 Þ ¼ 0, shown in Eq. (8.32). Obviously, the first
323
324
CHAPTER 8 Modal and Transient Dynamic Analysis
and second eigenequations result in the same eigenvector. In vector form, the
first eigenvector {V1} is expressed as
1
fV1 g ¼
.
(8.33)
2
As previously mentioned, there are many eigenvectors that could satisfy Eq.
(8.33), as long as the ratio between V1xðP2 Þ and V1xðP3 Þ remains the same. To avoid
the confusion of having multiple eigenvectors associated with the same eigenvalue, other textbooks prefer the use of “unit eigenvector,” i.e., an eigenvector
of unit length. Because the length of the vector shown in Eq. (8.33) is
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffi
12 þ 22 ¼ 5, the unit eigenvector becomes
9
8
1 >
>
>
>
p
ffiffi
ffi
>
>
<
5=
.
(8.34)
fV1 gunit ¼
>
2 >
>
>
>
>
p
ffiffi
ffi
;
:
5
In the FE method, forcing the eigenvector to be of unit length is not critical,
because we frequently enlarge the displacement field during postprocessing to
better visualize the outcome of calculations. Additionally, the absolute motion
of the two masses depends on the initial conditions. Hence, we will exclude
the derivation of the unit eigenvector from further discussion.
For j ¼ 2, the second eigenvalue l2 ¼ 6, and Eq. (8.31) becomes
)
(
V2xðP2 Þ
1 2
¼ 0:
(8.35)
V2xðP3 Þ
2 4
For this second set of eigenvalues, the associated eigenvector {V2} is
2
.
fV2 g ¼
1
(8.36)
Using the method of superposition and the two eigenvectors shown in Eqs.
(8.33) and (8.36), the nodal displacement of this example problem is
pffiffi
1 it
x2 ðtÞ
2
¼
e þ
ei 6t .
(8.37)
2
x3 ðtÞ
1
As we can see in Eq. (8.37), we can construct the entire displacementetime history as long as the eigenvalues and eigenvectors are known. Hence, modal analysis
is an efficient method compared to the transient dynamics method. Depending on
the initial condition, the system could vibrate completely
pffiffiffiffi in mode 1 (natural frequency of 1 rad/s), in mode 2 (natural frequency of 6 rad=s), or a combination
of the two modes using superposition. In mode 1, the associated eigenvector indicates that mass 2 is moving at twice the speed of mass 1 and in the same direction.
In mode 2, mass 1 is moving at twice the speed of mass 2, but in opposite directions.
Fig. 8.4 shows the time histories for mode 1, mode 2, and the superposition of the
two modes, respectively.
8.3 Modal Analysis
FIGURE 8.4
The mode shapes of the two-mass, two-spring system, which is rotated
90 degrees to the vertical orientation for demonstration purposes. Top: Mode
shapes of p
the
ffiffiffi harmonic motion at 1 rad/s. Middle: Mode shapes of the harmonic
motion at 6 rad=s. Bottom: Mode shapes of combined mode 1 and mode 2. In
all three plots, the horizontal axis represents time, and the vertical axis represents
the movements of m1 and m2 in the x-direction as shown in Fig. 8.3.
325
326
CHAPTER 8 Modal and Transient Dynamic Analysis
Here is a quick note on the characteristics of eigenvectors. Recall that the
dot product of two vectors is zero if they
perpendicular
(orthogonal)
n! are
! !
!o
to each other. Similarly, a set of vectors V 1 ; V 2 ; V 3 ; .; V n are mutually
orthogonal if every pair of vectors is orthogonal. We now take the dot product
of the two eigenvectors calculated from Example 8.2 and find that
f 1 2 gT $f 2 1 gT ¼ 0. Thus, we can state that the two eigenvectors in
this example are orthogonal to each other. This statement can be extended to
a set of n eigenvectors calculated from a system with n DOFs, as long as the
eigenmatrix is symmetric. Because proving that eigenvectors of a real symmetric matrix are orthogonal requires lengthy work, it will not be included in this
book.
8.3.1.1 Confusing Nomenclature
In mechanical systems, we use eigenvalues to relate the resonant frequencies
and eigenvectors to show the shapes of different vibration modes. Such information
can be directly used to reduce or avoid unwanted vibration. Additionally, “ringing”
generated by striking a structural component at one of its natural frequencies can be
used as a tool to examine the integrity of a structure. For example, an undamaged
beam will ring similar to the original beam while a cracked or deformed beam
will not, because the natural frequency has changed.
In other fields, concepts around eigenvalues and eigenvectors are used for
various purposes. For example, eigenvalues are used in the field of control theory
to determine if a system is stable, and eigenvector-based principal component
analysis (PCA) is used in the field of industrial engineering to identify critical
components. PCA is also used in image processing to design algorithms to reduce
image sizes. Additionally, eigenvalues are the key measures for determining the
sharpness of a photo. Even the Google search engine uses eigenvector-based
algorithms; they are used to rank the popularity of websites. Because eigenvalues
and eigenvectors are used in multiple fields, nomenclature for associated terms
varies. In this section, we discuss a few terms that may be confusing due to the
varying nomenclature.
For Example 8.2, the governing equation of motion, based on Eq. (8.26), is
1 0
5 2
x2
x€2
þ
¼ 0:
x€3
0 1
2 2
x3
By substituting {x}"¼ {Xelt},# where l ¼ iu in this equation, we establish that
5 2
x2
x2
l2 0
€ ¼ l2 x and
þ
¼ 0. For a nontrivial
fxg
2
2 2
x3
x3
0 l
solution, the determinant of the characteristic matrix must equal zero, as shown
below.
8.3 Modal Analysis
l2 þ 5
2
2
l2 þ 2
¼0
In other textbooks, x may be represented by fxg ¼ fX sin ltg,
fxg ¼ fX cos ltg, or the combination of both terms. In all of these cases,
€ ¼ l2 x and the determinant of the characteristics matrix would become
fxg
5 l2
2
¼ 0. As such, l2 ¼ 1 or 6 is found for Example 8.2 and
2
2 l2
l2 ¼ 1 or 6 is the result of the second case. Despite the difference in signs, we
take only the positive values for eigenvalues; hence, there is no confusion that
the eigenvalues (l2) are 1 and 6. Whether l2 is positive or negative, the good thing
is that the natural frequencies calculated from both assumptions of {x} remain 1
pffiffiffiffi
and 6 rad/s.
In other approaches used for calculating eigenvalues, the equation of motion
€ þ ½Kfxg ¼ 0, shown in Eq. (8.23), is
for a multiple DOFs system ½Mfxg
multiplied by [M]1 to both sides of the equation and the equation of motion
becomes
€ þ ½Afxg ¼ 0;
€ þ ½M1 ½Kfxg ¼ ½Ifxg
½Ifxg
where [A] ¼ [M]1[K]. The characteristic equation for the above equation is [AlI]
{X} ¼ 0. Assuming fxg ¼ fX sin utg, we can deduce that l ¼ u2, which is
different from l2 ¼ u2 shown in Eq. (8.30). Thus, the eigenvalue could be
l or l2, depending on which modified version of the equation of motion,
shown in Eq. (8.2), is chosen. Because a different assumption of the trial
solution results in a different “understanding” of eigenvalues, we prefer the
reference of natural frequency, instead of eigenvalue, to avoid unnecessary
confusion.
For a mechanical system, the first mode with the smallest eigenvalue is
called the fundamental eigenvalue, and the largest eigenvalue is called
the principal eigenvalue or dominant eigenvalue. In this example, the two
eigenvalues are 1 and 6. Therefore, the fundamental eigenvalue accounts
for 14.29% of the sum of the two eigenvalues, and the largest (second in
this case) accounts for the largest portion (85.71%) of the sum. Thus, the
principle or dominant eigenvalue has the highest frequency. A point that
needs to be stressed is that the words fundamental, principal, and dominant can
represent similar concepts, and hence it is not always clear which one has the
smallest or largest absolute value. Throughout my teaching career, students
have expressed confusion regarding this point, and that is why it is explicitly noted
here for clarity.
327
328
CHAPTER 8 Modal and Transient Dynamic Analysis
8.3.2 FORCED VIBRATION
When an external force is applied at the same frequency as one of the natural
frequencies of a structure, and that force is applied to the structure in the same
direction as the associated mode shape, resonance occurs. This is a positive
feedback loop, where motion of the structure due to the external force amplifies
that force. Without damping, the magnitude of the motion of the structure
eventually approaches infinity. During mechanical resonance, the magnitude of
the vibration is amplified over time and may result in structural failure. The
dramatic collapse of the Tacoma Narrows Bridge into Puget Sound, Washington,
on November 7, 1940, was one of the most noted examples related to forced
resonance-induced failure of a bridge (Billah and Scanlan, 1991). In structure
mechanics, we are only interested in a few of the lowest natural frequencies,
because lower frequency loading requires the least amount of mechanical
energy to initiate resonance. Also, higher frequency loadings are very difficult
to generate and even more difficult to impose on the system. Hence, higher frequency vibration modes are not usually investigated when studying structural
vibrations.
A two-spring, two-mass system subjected to forced vibration without gravity is
represented by a 3-node, 2-element FE model as shown in Fig. 8.5, where the
external force f2 is applied at P2. If we neglect the damping, the matrix form of
the equation of motion can be directly modified from Eq. (8.26) as
m1
0
0
m2
x€2
x€3
þ
k1 þ k2
k2
k2
k2
x2
x3
¼
f2
0
.
(8.38)
Again, analytical methods are used to illustrate the solution procedures. These
methods are directly applicable in the development of numerical methods for finding
eigenvalues and eigenvectors.
FIGURE 8.5
A two-spring, two-mass system is represented by a 3-node, 2-element FE model. A force
f2 is applied externally at P2.
8.3 Modal Analysis
Example 8.3
In Fig. 8.5, assume m1 ¼ m2 ¼ 1, k1 ¼ 3, and k2 ¼ 2. We further assume that the
external force is a harmonic forcing function with the form of f2 ¼ F0 sin ut.
Show the motion of m1 and m2.
Solution
Eq. (8.38) can be solved by finding the homogeneous solution and particular
solution separately. For the homogeneous solution, this example is identical to
that described in Example 8.2 for free vibration. As such, the
pffiffiffiffieigenvalues are
l2 ¼ 1 and 6. Also, the two natural frequencies are 1 and 6 rad/s, because
l ¼ iu.
For the particular solution, assume that x2 ¼ X2 sin ut and x3 ¼ X3 sin ut are
the trial solutions which satisfy Eq. (8.38). Inserting these two trial solutions into
Eq. (8.38) gives us
)
#( ) (
"
X2
F0 sin ut
k2
k1 þ k2 m1 u2
0
¼
sin ut
X3
0
k2
k2 m2 u2
(8.39)
"
#( ) ( )
X2
F0
k1 þ k2 m1 u2
k2
.
¼
X3
0
k2
k2 m2 u2
We can solve Eq. (8.39) by applying Cramer’s rule as
8
9
k2 m2 u2 F0
>
>
>
>
>
>
>
>
<
ðk1 þ k2 m1 u2 k2 m2 u2 k2 2 =
X2
.
¼
>
>
X3
k2 F 0
>
>
>
>
>
>
:
;
ðk1 þ k2 m1 u2 k2 m2 u2 k2 2
(8.40)
Because X2 and X3 are the exact solutions for Eq. (8.39), this set of trial solutions
must be the valid solutions for the equation of motion shown as Eq. (8.38). We
also note that the magnitude of X2 and X3 shown in Eq. (8.40) would approach
infinity (i.e., the state of resonance), if the denominator were to approach zero as
shown below.
k1 þ k2 m1 u2 k2 m2 u2 k22 ¼ 0
(8.41)
By inserting the values for m1, m2, k1, and k2 into Eq. (8.41), we have
5 u2 2 u2 4 ¼ 00u4 7u2 þ 6 ¼ 00 u2 1 u2 6 ¼ 0: (8.42)
329
CHAPTER 8 Modal and Transient Dynamic Analysis
Relative magnitude versus ω
400
300
200
X Fo
330
100
0
0.95
0.96
0.97
0.98
0.99
1
1.01
1.02
1.03
1.04
1.05
-100
-200
Radians/Second
FIGURE 8.6
The movement of m1 as a function of u. We can see that as u approaches 1 rad/s,
the first natural frequency, the magnitude of the motion draws to near infinity.
Eq. (8.42) is consistent with Eq. (8.28), with the two natural frequencies of 1 and
pffiffiffiffi
6 rad/s. Fig. 8.6 shows the magnitude of X2 as a function of u ranging from 0.95
to 1.05 rad/s, where X2 is calculated directly from Eq. (8.40). We see that the
magnitude increases significantly when u approaches 1 rad/s, which is the lower
of the two natural frequencies. Although not plotted, we can deduce that a similar
pffiffiffi
sharp rise in magnitude happens when u approaches 6z2:449 rad=s.
8.3.3 NUMERICAL METHODS FOR FINDING EIGENVALUES AND
EIGENVECTORS
Examples described for free (Section 8.3.1) and forced (Section 8.3.2) vibrations
that have only two DOFs can be easily solved by using analytical methods. As
the number of DOFs increases, determination of all resonant frequencies and associated mode shapes becomes difficult, if not impossible, when using analytical
methods. As such, numerical schemes designed to find approximate answers of a
few of the lowest natural frequencies are desired. Again, the lowest few resonant frequencies are more important than the others in mechanical systems, because these
frequencies are easier to impose on the structure. We will start with an example
by rewriting Eq. (8.23) for an undamped system of n DOFs as
€ n1 þ ½Knn fxgn1 ¼ f0gn1 .
½Mnn fxg
(8.43)
In this system, there are n eigenvalues (li, i ¼ 1 to n) and n associated
eigenvectors (Vi, i ¼ 1 to n) with 0<l1 < l2 < . < lp < . < ln, where p is the
first few vibration modes we want to solve and l ¼ iu or l2 ¼ u2.
8.3 Modal Analysis
By assuming x ¼ elt and l ¼ iu, we rewrite Eq. (8.43) for the overall governing
equation and the equation for the jth vibration mode is written as
½K u2 ½M fVg ¼ 0; and
(8.44)
½K uj 2 ½M (8.45)
Vj ¼ 0;
where {V} is the set of eigenvectors, {Vj} is the jth eigenvector, and uj is the jth
eigenvalue. Because it is trivial for the eigenvector to be zero, we must have
½K u2 ½M ¼ 0, where j j stands for the determinant. We now describe some
commonly used numerical methods for determining the eigenvalues and eigenvectors. Because a number of textbooks are devoted to this subject, only a couple of
basic numerical methods are illustrated in this book.
8.3.3.1 Rayleigh Quotient Iteration
We first multiply {Vj}T to Eq. (8.45) and then switch [M] related terms to the righthand side. The resulting equation is
fVj gT ½KfVj g ¼ uj 2 fVj gT ½MfVj g.
(8.46)
Based on the descriptions of Parlett (1974) regarding Lord Rayleigh’s work on
the theory of sound, we define the strain energy (potential energy) and kinetic energy
to have the forms of
1
1
_ 2.
S.E. ¼ ½Kfxg2 and K.E. ¼ ½Mfxg
2
2
From Eq. (8.47), the strain energy is
1
S.E. ¼ fVj gT ½KfVj g;
2
(8.47)
(8.48)
and the kinetic energy is
1
K.E. ¼ uj 2 fVj gT ½MfVj g.
(8.49)
2
Parlett (1974) also stated that “Lord Rayleigh showed that the frequency can be
written in terms of the mode shape” and the ratio of the two energies. Thus, we
rewrite Eq. (8.46) as
uj 2 ¼
fVj gT ½KfVj g
fVj gT ½MfVj g
.
(8.50)
Because the eigenvector {Vj} in Eq. (8.50) has not been determined (i.e., it is unknown), we replace {Vj} by an arbitrary vector {A}. Now, the generalized Rayleigh
quotient RQ is written as
RQ ¼
fAgT ½KfAg
fAgT ½MfAg
.
(8.51)
331
332
CHAPTER 8 Modal and Transient Dynamic Analysis
In Eq. (8.51), if the arbitrarily selected vector {A} were matched exactly to one of the
mode shapes, then RQ would be identical to the corresponding eigenvalue, just like
that indicated in Eq. (8.50). In an event that the vector {A} is close but not matching
exactly to one of the eigenvectors, it can be shown that the Rayleigh quotient represents a value that is close to the corresponding eigenvalue. Due to the limited scope,
we will not prove this statement in this basic FE method book. The proof can be found
in a study by Parlett (1974), and there are other references available for more details.
It is important to note that the accuracy of the Rayleigh quotient method is better
than it may seem. A 20% error in the selection of the eigenvector will result in only
a 4% variation in the eigenvalue, while a 10% error in selecting eigenvectors creates
only a 1% inaccuracy. Additionally, the magnitude of the Rayleigh quotient will always
be between the smallest and largest eigenvalues (l1 < RQ < ln). As such, the Rayleigh
quotient method can easily be used to compute the lowest or highest eigenvalues with
great confidence. In a conservative mechanical system, the Rayleigh principle asserts
that the Rayleigh quotient has a relatively stationary value in the neighborhood of
resonant vibration modes. Thus, other eigenvalues can be found from Rayleigh quotients versus the selected component of the eigenvector curve at the local minimum
point, inflection point (a point on a curve at which the second derivative of the curve
changes from positive to negative or vice versa), and local maximum point. We can
use these properties to iteratively calculate the eigenvalues through optimization techniques or randomly selected vectors using the following steps.
Steps for identifying natural frequencies using the Rayleigh quotient procedures:
1.
2.
3.
4.
5.
Randomly select a vector {A} as an estimated eigenvector.
Compute the corresponding strain energy of the system.
Compute the corresponding kinetic energy of the system
S.E. .
Compute RQ ¼ K.E.
Identify which values of RQ are near the eigenvalues of the system. Note that the
highest RQ approximates the highest eigenvalue and the lowest RQ approximates
the lowest eigenvalue.
Example 8.4
Using the Rayleigh quotient method, compute the natural frequency for the freevibration problem shown in Example 8.2.
Solution
Assuming a unit (1) displacement vector is applied at P2, we estimate that the
displacement vector at P3 is also 1. Thus, the first trial vector is
1
. By inserting the values for k1 and k2 into
fAgT ¼ f 1 1 g and fAg ¼
1
Eq. (8.25) for [K], and values for m1 and m2 into Eq. (8.24) for [M], the strain
energy and kinetic energy of the system are calculated as
8.3 Modal Analysis
1
S:E. ¼ f 1
2
"
5
2
#( )
1
¼ 1:5 and
2 2
1
"
#( )
1 0
1
1
K:E. ¼ f 1 1 g
¼ 1:
2
0 1
1
1g
From these two energies, we compute the Rayleigh quotient as RQ ¼ 1.5. Note
that although the calculations related to the 22 matrix in this example is rather
simple, it would be impractical, considering both accuracy and time resources, to
calculate values for a larger system manually. This would require the use of computer program, even if you needed to write the program, yourself. Table 8.2 lists
10 calculated Rayleigh quotients based on randomly selected vectors representing
the estimated mode shapes.
Table 8.2 Results of Rayleigh Quotient Based on Estimated Eigenvectors
Case No.
1
2
3
4
5
6
7
8
9
10
AP2
1
1
1
1
4
3
2
0.6
0.5
0.4
AP3
1
2
L0.5
3
1
1
1
1
1
1
Strain Energy
1.5
2.5
3.75
5.5
33
17.5
7
0.7
0.625
0.6
Kinetic
Energy
1
2.5
0.625
5
8.5
5
2.5
0.68
0.625
0.58
Rayleigh
Quotient
1.5
1
6
1.1
3.882
3.5
2.8
1.029
1
1.034
Where AP2 and AP3 are displacement vectors applied at nodes P2 and P3, respectively.
From Table 8.2, we observe that Cases 2 and 9 have the lowest Rayleigh quotients on these limited selected sample vectors. In fact, the estimated eigenvectors
are identical in Cases 2 and 9 since the ratios of AP2 over AP3 are of the same
magnitude. Also, Case 3 has the highest Rayleigh quotients. You may wish to
generate more cases to partially prove that 1 and 6 are the minimum and
maximum, respectively, of Rayleigh quotients for this example. Because we
know that l1 < RQ < ln, the vector f 1 2 gT , with the lowest RQ value, must
approximate the eigenvector associated with the first mode shape. Analogously,
the vector f 1 0:5 gT with the highest RQ value must approximate the eigenvector associated with the mode corresponding to the highest magnitude of natural frequency. Indeed, these two cases are not only approximately the same as
the eigenvalues; they are exactly the same as those determined in Example 8.2.
333
334
CHAPTER 8 Modal and Transient Dynamic Analysis
Finally, the lowest and highest natural
pffiffiffiffi frequencies, determined using the Rayleigh quotient method, are 1 and 6 rad/s, respectively.
As a final note, there are at least two reasons that the Rayleigh quotient
method is useful. First, the method is sensitive to the errors in the estimated
mode shapes. For example, if we select the vector f 1 2:1 gT or f 1 1:9 gT
instead of f 1 2 gT , shown in Case 2, we can see that the energies changed
greatly, and the Rayleigh quotients for both cases are higher than the case with
the exact eigenvector of the system. Second, an experienced engineer can usually
guess the correct mode shapes, which makes it easy to use the Rayleigh quotient
method to determine the natural frequency of the system.
8.3.3.2 Matrix Iteration Method
The matrix iteration method originated from Dunkerley’s equation, based on the
work by professor Stanley Dunkerley (Apr. 1870eSep. 1912). In his experimental
works related to a long shaft loaded with multiple pulleys in a machine shop, he
discovered a simple way to approximate the critical rotational speeds before vibration initiated. Although his method dealt with only the lowest critical speeds, it was
exactly what was needed to solve vibration problems in machine shops of that time
period. As the Rayleigh quotient method provides an “upper bound” estimation of
natural frequency, the Dunkerley method complements it by providing a “lower
bound” approximation of resonances. Here the upper bound means that the value
of the actual natural frequency will be lower than or equal to the approximate value
that is found when using the Rayleigh quotient method, while the lower bound
means that the actual natural frequency will be higher than or equal to the value
calculated using the Dunkerley method. When both methods are used, we can
quickly establish a band within which the actual natural frequencies lie.
We begin the discussion of Dunkerley’s method by assuming that a shaft vibrates
at a frequency (or rotational speed) of u1 due to an isolated loading condition 1, at a
frequency of u2 when subjected to an isolated loading condition 2, and at a frequency of u3 owing to an isolated loading condition 3. Based on his experimental
observations, Dunkerley found that the reciprocal rule worked quite well for finding
the resulting frequency of vibration due to simultaneous application of all three
loads. In other words, the combined vibration frequency can be estimated by
1
1
1
1
z
þ
þ
.
u 2 u 1 2 u2 2 u3 2
(8.52)
Eq. (8.52) can be expanded to identify the vibration frequency due to application of
more loading modes at the same time. To use the Dunkerley’s method within the FE
method, we first multiply [K]1, known as the flexibility matrix (the inverse of stiffness is flexibility), to Eq. (8.23) as
€ þ ½K1 ½Kfxg ¼ 0:
½K1 ½Mfxg
(8.53)
8.3 Modal Analysis
We now define the matrix [D] ¼ [K]1[M], which is known as the system matrix
or dynamic matrix, because the dynamic properties of the system are well preserved
by this matrix. We can also determine [D]1 as
½D ¼ ½K1 ½M
½D1 ¼ ½M1 ½K.
(8.54)
We define [D]1 so that we can also multiply [M]1 to Eq. (8.23) before
applying Dunkerley’s method. For now, we use only the [D] matrix for the remaining discussion of this section. Calculating the double derivative of a trial solution
€ ¼ u2 Xeiut
or
({x} ¼ {Xeiut} or fxg ¼ fX sin utg) gives us fxg
2
€ ¼ u fX sin utg. By inserting the results into Eq. (8.53), we have
fxg
€ þ ½Ifxg ¼ f0g
½Dfxg
u2 ½DfXg þ ½IfXg ¼ f0g
(8.55)
1
½D 2 ½I fXg ¼ f0g;
u
1 0
where [I] is identity matrix, ½I ¼
.
0 1
To numerically find the first eigenvalue and associated eigenvector, we start by
setting the determinant of the third line of Eq. (8.55) equal to zero. With the exception
that a new system (dynamic) matrix [D] is introduced, so far there are no differences
between the Dunkerley method and what we did with analytical equations used for
finding the natural frequencies of the free-vibration system shown in Section 8.3.1.
To describe the next step, we use a 2-DOFs system in which the structural mass
matrix [M] is a lumped matrix (i.e., diagonal matrix, as discussed in Section 8.2.2).
The same approach can be used for a larger system. Because we use a 2-DOFs system to illustrate Dunkerley’s method, the sizes of [K], [K]1, and [M] matrices are
22. As such, the size of the [D] matrix is also 22. We assume that the entries for
[M] and [K]1 have the forms
"
#
m1 0
½M ¼
0 m2
#
"
(8.56)
a11 a12
1
;
½K ¼
a21 a22
where we use a randomly selected letter a to avoid the use of k, which was taken
previously to represent stiffness. The matrix [D] can be found by multiplying
[K]1 and [M] as
a11 m1 a12 m2
1
½D ¼ ½K ½M ¼
.
(8.57)
a21 m1 a22 m2
335
336
CHAPTER 8 Modal and Transient Dynamic Analysis
By inserting the entries in Eq. (8.57) to Eq. (8.55), and designating the determinant
to be zero for nontrivial solution, we have
a11 m1 1
u2
a12 m2
¼0
1
(8.58)
a21 m1
a22 m2 2
u
1
1
ða11 m1 þ a22 m2 Þ 2 þ m1 m2 ða11 a22 a12 a21 Þ ¼ 0:
4
u
u
Eq. (8.58) has four roots, two positive and two negative. However, the natural
frequencies of a mechanical system cannot have negative values. Thus, we consider
only the two positive roots or two natural frequencies (u1, u2) for this 2-DOFs
system. We can write the eigenvalues lj ¼ u1j 2 in matrix form as
3
2
1
0 7
6 2
7
6 u1
7.
6
(8.59)
6
1 7
5
4 0
u2 2
Eq. (8.59) is a diagonal matrix that satisfies the characteristics of the [D] matrix
as shown in Eq. (8.56) (i.e., determinant is zero). According to the first invariant
of a matrix, the trace (i.e., summation of all diagonal terms) of the natural frequency shown in Eq. (8.59) and the trace of [D] matrix should be identical.
Thus, we have
1
1
þ
¼ a11 m1 þ a22 m2
u 1 2 u2 2
(8.60)
As previously mentioned, the simple 2-DOFs system can be expanded to cover the
eigenvalues for a system with n DOFs. Since the traces for Eqs. (8.59) and (8.60) are
linear combinations of diagonal terms, we can rewrite Eq. (8.60) in an analogous
manner for a system with n DOFs as
1
1
1
þ
þ / þ 2 ¼ a11 m1 þ a22 m2 þ / þ ann mn .
u1 2 u2 2
un
(8.61)
Although there is no need to follow a particular order, it is common practice to
place the lowest to the highest natural frequencies from left to right, respectively
(u1 < u2 < . < un). There is a further assumption that all natural frequencies
are sufficiently separated in the frequency domain, and the first natural frequency
is much lower than the rest of the frequencies (i.e., u1 << u2 < . < un ). For
this reason, we can assume
8.3 Modal Analysis
1
1
1
þ
þ / þ 2 z 0:
un
u22 u23
(8.62)
Inserting Eq. (8.62) to Eq. (8.51) gives us
1
z a11 m1 þ a22 m2 þ / þ ann mn .
u1 2
(8.63)
Eq. (8.63) depicts two things. First, only the first natural frequency u1 can be obtained. Secondly, each mass makes some contribution to the first natural frequency.
Because [K] and [M] matrices are known, the first natural frequency can be found.
Though other books may define this differently, here we declare that the “first” or
“lowest” natural frequency is the fundamental frequency or the resonant frequency
of the lowest magnitude. Because we did not take u12 ; u12 ; .; u12 into consideration (i.e., the terms associn
2
3
ated with the second and higher modes are neglected when calculating the first natural frequency), Dunkerley’s method tends to overestimate the magnitude of the
natural frequency, and hence it is considered the lower bound estimation of the first
natural frequency.
The matrix iteration method is based on Dunkerley’s method (Dunkerley,
1893). It is a set of iterative procedures that allow users to determine the first
natural frequency of vibration of a system. The method requires an assumed set
of trial vectors, which are iteratively updated until the error is less than a preset
value, that is, the solution converges. To find the natural frequency for the second
and higher modes, the principle of orthogonality needs to be applied to form a new
matrix that is free from the first mode of vibration. The method to apply this
orthogonality principle is illustrated numerically using Example 8.5 in Section
8.3.3.3. After extraction of the first mode from further consideration, the
matrix iteration method is then used to find the second natural frequency. The
same set of procedures can be repeated until all desired modes of vibrations are
obtained.
To use the matrix iteration method, the derivations provided above from Eqs.
(8.52) to (8.63) are shortened into seven steps:
Set up the flexibility matrix [K]1.
Set up the [D] matrix as [K]1[M].
Select a trial vector {x1}.
Multiply the trial vector {x1} by the [D] matrix to form the next trial vector {x2}.
Normalize the resulting vector {x2}. The factor used for the normalization is the
estimated eigenvalue. This step is based on rearranging Eq. (8.53) as
½Dfxg ¼ u12 fxg. Again, the actual values for entries in an eigenvector are not
important. As long as the ratio between the two entries is correct, the eigenvector is correct.
6. Calculate the vibration frequency from the normalized vector {x2} and determine
the difference in frequency Du.
1.
2.
3.
4.
5.
337
338
CHAPTER 8 Modal and Transient Dynamic Analysis
7. If Du is smaller than an accepted range of error (i.e., u converges), the iteration
completes. Otherwise, repeat the procedures for determining {x3}, {x4}, .,
{xn} until u converges.
Example 8.5
Using the matrix iteration method, compute the first natural frequency for the
free-vibration problem shown in Example 8.2.
Solution
5 2
From Eq. (8.27), the [K] and [M] matrices of the system are ½K ¼
and
2 2
1 0
0:33333 0:33333
1
1
½M ¼
. We calculate the [K] matrix as ½K ¼
.
0 1
0:33333 0:83333
0:33333 0:33333
From these values, the [D] matrix is calculated as ½D ¼
0:33333 0:83333
0
0:33333 0:33333
¼
.
1
0:33333 0:83333
1
, we calculate {x2} through {xn}
Assuming that the trial vector is fx1 g ¼
1
using the iterative procedures as shown below. Once the difference between two
consecutive estimated frequencies is lower than a preset threshold, the iteration
can be stopped, and the last estimated frequency will be very close to the first
mode of the natural frequency.
Iteration 1: we calculate {x2} from the product of [D] and {x1} as
0:33333 0:33333
1
0:66667
¼
.
fx2 g ¼ ½Dfx1 g ¼
0:33333 0:83333
1
1:16667
1
0
We select the value for the first entry as the normalization factor. The goal of
normalization is to make the first entry of the vector equal 1. To accomplish this,
we pull the 0.66667 out of the vector and end up with fx2 g ¼
1:00000
0:66667
. That is, u12 ¼ 0:666670u ¼ 1:22474 and the updated
1:75000
1:00000
.
fx2 g ¼
1:75000
Iteration 2:
0:33333 0:33333
1:00
0:91667
¼
½Dfx
¼
¼
fx3 g
2g
0:33333 0:83333
1:75
1:79167
8.3 Modal Analysis
Normalizing {x3} results in fx3 g ¼ 0:91667
1:00000
. That is, u12 ¼
1:00000
0:916670u ¼ 1:04447; Du ¼ 0:18027, and the updated fx3 g ¼
.
1:95454
1:95454
Iteration 3:
0:33333 0:33333
fx4 g ¼ ½Dfx3 g ¼
0:33333 0:83333
1:00
1:95454
Normalizing {x4} results in fx4 g ¼ 0:98483
¼
1:00000
0:98483
1:96211
. That is, u12 ¼
1:00000
.
0:984830u ¼ 1:00767; Du ¼ 0:03680 and the updated fx4 g ¼
1:99233
Iteration 4:
0:33333 0:33333
1:00000
0:99743
¼
fx5 g ¼ ½Dfx4 g ¼
0:33333 0:83333
1:99233
1:99360
1:99233
Normalizing {x5} results in fx5 g ¼ 0:99743
1:00000
. That is, u12 ¼
1:00000
.
0:997430u ¼ 1:00129; Du ¼ 0:00638, and the updated fx5 g ¼
1:99874
Iteration 5:
0:33333 0:33333
1:00000
0:99957
¼
fx6 g ¼ ½Dfx5 g ¼
0:33333 0:83333
1:99874
1:99894
1:99874
Normalizing {x5} results in fx6 g ¼ 0:99957
1:00000
. That is, u12 ¼
1:00000
0:999570u ¼ 1:00022; Du ¼ 0:00107, and the updated fx6 g ¼
.
1:99980
We can see that the eigenvector identified at the fifth iteration (Du < 0.005) is
almost the same as that shown in Eq. (8.33) in Section 8.3.1. Using the matrix
iteration method, the first natural frequency can be obtained fairly accurately
(Du < 0.02) in four iterations. As expected, the calculated natural frequency at
the fifth iteration (u ¼ 1.00022 rad/s) is slightly higher than the 1 rad/s calculated
using the analytical method discussed in Section 8.3.1. Again, this is why Dunkerley’s method is considered a lower bound method.
1:99980
339
340
CHAPTER 8 Modal and Transient Dynamic Analysis
8.3.3.3 Second Natural Frequencies
Upon convergence of the matrix iteration method, the resulting frequency and iteration vector must be very close to the first eigenvalue and associated eigenvector. As
mentioned in Section 8.3.1, all eigenvectors are orthogonal to each other. Thus, if
we use a new trial iteration vector orthogonal to the eigenvector associated with
the first eigenvalue, we can exclude the first eigenvector from the matrix. By using
this principle, we can calculate the second lowest natural frequency and associated
mode shape.
We start by assuming that a new trial iteration vector {y1} is a linear combination
of all eigenvectors {V1}, {V2}, ., {Vn}, of an n-DOFs system, that is
fy1 g ¼ c1 fV1 g þ c2 fV2 g þ / þ cn fVn g.
(8.64)
T
Multiplying both sides of Eq. (8.64) by {V1} [M] results in
fV1 gT ½Mfy1 g ¼ c1 fV1 gT ½MfV1 g þ c2 fV1 gT ½MfV2 g þ / þ cn fV1 gT ½MfVn g.
(8.65)
Because {V1}, the first eigenvector, is orthogonal to the rest of the eigenvectors, we
can write fV1 g$fVj g ¼ fV1 gT fVj g ¼ 0, where j ¼ 2 to n. Thus, Eq. (8.65) can be
rewritten as
fV1 gT ½Mfy1 g ¼ c1 fV1 gT ½MfV1 g
c1 ¼
fV1 gT ½Mfy1 g
T
fV1 g ½MfV1 g
(8.66)
.
Now, we can take away the effect of the first eigenvector from the trial vector by
subtracting c1{V1} from the trial vector as
fy1 g c1 fV1 g ¼ fy1 g fV1 gfV1 gT ½Mfy1 g
¼ ½Ify1 g fV1 gT ½MfV1 g
fV1 gfV1 gT ½M
T
fV1 g ½MfV1 g
fy1 g ¼
½I fV1 gfV1 gT ½M
fV1 gT ½MfV1 g
fy1 g.
(8.67)
The process of taking away a mode of vibration is called sweeping. For Eq. (8.67),
we “sweep” the effect of the first mode from the original equation so we can apply the
iteration procedures to solve for the second natural frequency and the associated mode
shapes. Hence, the method of excluding one mode of vibration is also called the
sweeping method. After sweeping, the updated system dynamic matrix becomes
½D2 fy1 g ¼ ½D1 ½I fV1 gfV1 gT ½M
fV1 gT ½MfV1 g
fy1 g.
(8.68)
Eq. (8.68) can be used to find the second natural frequency and associated mode
shapes. Again, we will use a numerical example to find the second natural frequency
and mode shape.
8.3 Modal Analysis
Example 8.6
Find the second natural frequency and mode shape for the same problem listed in
Example 8.5.
Solution
From Example 8.5, we know that
1:0000
0:33333 0:33333
1 0
.
½ D1 ¼
; ½M ¼
; and fV1 g ¼
1:9998
0:33333 0:83333
0 1
From Eq. (8.68), we calculate the following components for further data processing as
1:0000
1 0
1
1:9998
T
¼
fV1 gfV1 g ½M ¼
f1:0000 1:9998g
1:9998
0 1
1:9998 3:9992
"
#(
)
1 0
1:0000
¼ 4:9992
fV1 gT ½MfV1 g ¼ f1:0000 1:9998g
0 1
1:9998
"
# "
# "
#!
0:33333 0:33333
1 0
0:20003 0:40002
½D2 fy1 g ¼
fy1 g
0:33333 0:83333
0 1
0:40002
0:8
"
#"
#
0:33333 0:33333
0:79997 0:40002
¼
fy1 g
0:33333 0:83333 0:40002
0:2
"
#
0:13332 0:06667
¼
fy1 g
0:06669 0:03333
1
into the iteration as
Iteration 1: Now, insert the trial vector fy1 g ¼
"
#( ) 1 (
)
0:13332 0:0667
1
0:06665
¼
½D2 fy1 g ¼
0:06669 0:03333
1
0:03336
(
)
1
¼ 0:06665
0:50053
Iteration 2:
0:13332 0:0667
0:06669 0:03333
1
0:50053
¼
0:16669
0:08337
¼ 0:16669
¼ 0:16667
1
0:50016
Iteration 3:
0:13332 0:0667
0:06669 0:03333
1
0:50016
¼
0:16667
0:08336
1
0:50016
341
342
CHAPTER 8 Modal and Transient Dynamic Analysis
Because this example has only two DOFs, once the first mode is removed,
the second mode converges quickly. As shown in Eq. (8.59), the eigenvalue we
employ in Sections 8.3.3.2 and 8.3.3.3 is lj ¼ u1j 2 . Thus, the second natural
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffi frequency is
1=0:16667 z 6 and the corresponding eigenvector is
1
. These results are the same as those shown in Example 8.2. In a
0:50016
system with more than two DOFs, this set of procedures can be repeated to
find the natural frequencies of third, fourth, etc., until all the modes of interest
are resolved.
8.3.3.4 Jacobi Method
Several numerical methods are published for determining eigenvalues. Due to the
limited scope of this book, most of these methods and algorithms are not discussed.
For a more in-depth study in this subject, we recommend looking into the power forward iterative method introduced separately by Von Luigi Vianello and Aurel Stodola. Vianello’s method was initially developed to calculate critical buckling
loads of columns due to bending (Von Vianello, 1898). In separated efforts, Stodola
discovered the same iterative procedures while studying vibrating shafts (Stodola
and Loewenstein, 1906). Another popular method is Holzer’s method (Holzer,
1921; Thomson, 1981), which is based on a tabulated trial-and-error scheme for
finding the natural frequency.
The methods we have introduced thus far (the Rayleigh quotient, matrix
iteration method, and exclusion of a particular mode of vibration based on Dunkerley’s scheme) are designed to solve one mode of vibration at a time. As such,
we feel compelled to mention the Jacobi method (Jacobi, 1846) through which
users can find multiple eigenvalues simultaneously. This method was proposed
by Carl Gustav Jacob Jacobi (1846), the same scientist we discussed in Section
7.2.1 for his iterative procedures designed for solving simultaneous linear
equations.
Finding eigenvalues of a matrix [A] involves finding the characteristic values li that
represent the essence of the original matrix. In order to do so, we mandate that the
determinant of the characteristic equation jA lIj ¼ 0. Because we multiply the eigenvalues to an identity matrix, li are arranged in diagonal terms. For a real, symmetric matrix [D], the Jacobi method calls for sequential rotations of the matrix until it is
diagonalized. To achieve this goal, a pair of off-diagonal terms (Dij and Dji) are forced
to zero first by rotating the matrix [D] to ½D1 ¼ ½TT ½DfTg, where [T] is the rotation
matrix. If the same procedures are successively applied with rotation matrices [T1],
[T2], ., [Tn], the final updated matrix [Dn] will be very close to a diagonal matrix,
and the diagonal entries will correspond to eigenvalues of [D]. Finally, the eigenvectors are listed in individual columns of the overall rotation matrix [T1][T2].[Tn]. A
8.3 Modal Analysis
44 real, symmetric matrix is used to explain the procedures for finding the rotation
angle needed to force an off-diagonal pair Dij ¼ Dji to zeros.
We start by assuming that we are to find the eigenvalues and eigenvectors for a
structure with a system dynamic matrix [D] of
2
3
D11 D12 D13 D14
6
D22 D23 D24 7
6
7
½D ¼ 6
(8.69)
7.
4 SYM
D33 D34 5
D44
The most effective way to diagonalize this matrix is to start with forcing the
largest off-diagonal term to zero. Let D34 be the entry with the largest magnitude
of all off-diagonal terms. To make D34 ¼ 0, we need a rotation matrix [T] similar
to that used in 2D-coordinate transformations discussed in Section 4.7 as
3
2
1 0 0 0
60 1 0 0 7
7
6
(8.70)
½T ¼ 6
7;
4 0 0 c s 5
0 0
s
c
where c represents cos q and s represents sin q. We now use Eqs. (8.69) and (8.70) to
update from [D] to [D1] as
2
1
6
60
6
½D1 ¼ ½T ½D½T ¼ 6
60
4
T
0
2
0
1
0
0
c
0 s
0
32
D11
76
6
07
76 D12
76
6
s7
54 D13
c
D14
D14
32
D13
D22
D23
D23
D33
76
6
D24 7
76 0
76
6
D34 7
54 0
D24
D34
D44
D13
D14
D22
D23 c þ D24 s
D23
D33 c þ D34 s
D24
D34 c þ D44 s
D13 s þ D14 c
D23 s þ D24 c
D33 s þ D34 c
D34 s þ D44 c
D11
D12
D12
D22
6
6
6
6
6
¼ 6 D13 c þ D14 s
6
6
6
4
D13 s þ D14 c
D23 c þ D24 s
D23 s þ D24 c
1
D12
D12
D11
6
D12
6
¼6
4 D13 c þ D14 s
2
0
D13 c þ D14 s
D23 c þ D24 s
!
D33 c2 þ D34 cs
þD34 cs þ D44 s2
D33 cs þ D34 c2
D34 s2 þ D44 cs
!
0 0
1 0
0
7
0 7
7
7
s 7
5
c
0
c
0
0
s
32
1
0 0
76 0
76
76
54 0
0
1 0
0 c
0
3
s
0
3
0 7
7
7
s 5
c
3
D13 s þ D14 c
7
D23 s þ D24 c
!7
7
2
7
D33 cs D34 s
7
7.
2
þD34 c þ D44 cs 7
!7
7
D33 s2 D34 cs
5
D34 cs þ D44 c2
(8.71)
In Eq. (8.71), the entries ½D1 34 ¼ ½D1 43 ¼ ðD44 D33 Þcs þ D34
. Since
the Jacobi method calls for the diagonalization of the [D1] matrix, the next step is
c2
s2
343
344
CHAPTER 8 Modal and Transient Dynamic Analysis
to force these two off-diagonal terms to be zero. From trigonometry, we know that
sin 2q ¼ 2 sin q cos q and cos 2q ¼ cos q2 sin q2 . Hence, we can write
1
½D1 34 ¼ ½D1 43 ¼ ðD44 D33 Þsinð2qÞ þ D34 cosð2qÞ ¼ 0:
2
Rearranging Eq. (8.72), we have
1
ðD44 D33 Þsinð2qÞ ¼ D34 cosð2qÞ
2
2D34
.
tanð2qÞ ¼
D33 D44
(8.72)
(8.73)
In this example, we assume that D34 is the “largest” entry of all off-diagonal terms
and hence we select rows and columns 3 and 4 to calculate the rotation angle. Note
that the “largest” entry may be the entry with the largest value (e.g., a value of
0.8 > 1.4) or the largest absolute value (e.g., j1:4j > 0.8). The decision regarding
which off-diagonal term to make zero only affects the number of iterations needed to
converge the [D] matrix to a diagonalized matrix. Because choosing D34 is only for
illustration purpose, a more general term Dmn may have the largest magnitude of all
off-diagonal terms. So, we rewrite Eq. (8.73) as
tanð2qÞ ¼
2Dmn
;
Dmm Dnn
(8.74)
which is the general equation needed to identify the rotation angle and to form a
rotation matrix that will force [D1]mn to zero. In our demonstration matrix shown
in Eq. (8.69), we found that a properly chosen angle q can force
[D1]34 ¼ [D1]43 ¼ 0. If we apply the same procedure to columns and rows 1 and
2, we can deduce that [D1]12 ¼ [D1]21 ¼ 0. However, a different angle q will be
needed, and this may make the previously updated [D1]34 or [D1]43 become nonzero.
As such, the rotation matrix needs to be successively applied so that the final updated
matrix {Dn} is very close to a diagonal matrix. To write this concept in equation
form, the second update [D2] has the form of
½D2 ¼ ½T2 T ½D1 ½T2 ¼ ½T2 T ½T1 T ½D½T1 ½T2 .
(8.75)
The same procedures shown as Eq. (8.75) can be successively applied for a total
of n iterations, shown as Eq. (8.76). This will cause all off-diagonal terms to
converge to zero.
½Dn ¼ ½Tn T .½T2 T ½T1 T ½D½T1 ½T2 .½Tn (8.76)
Once the final updated matrix is “almost” diagonalized, all natural frequencies
are listed in the diagonalized matrix [Dn], and the corresponding eigenvectors can
be found from individual columns of the final [T] matrix, where
½T ¼ ½T1 ½T2 .½Tn .
(8.77)
8.3 Modal Analysis
Example 8.7
Using the Jacobi method, compute the natural frequencies for the free-vibration
problem shown in Example 8.2.
Solution
0:33333 0:33333
From Example 8.5, we have ½D ¼
. Because there are only
0:33333 0:83333
two rows and columns, the rotation angle can be found by zeroing the off-diagonal term D12 with
tan 2q ¼
2D12
2 0:33333
¼ 1:33332
¼
D11 D22 0:33333 0:83333
1
q ¼ tan1 ð1:33332Þ ¼ 26:565 degrees.
2
The rotation matrix
0:89443 0:44721
.
0:44721 0:89443
is then
calculated to be ½T ¼
c s
¼
s c
We now update from [D] to [D1] matrix as
"
#"
#
0:89443 0:44721 0:33333 0:33333
T
½D1 ¼ ½T ½D½T ¼
0:44721 0:89443
0:33333 0:83333
"
#
0:89443 0:44721
0:44721 0:89443
"
#"
#
0:14907 0:07453
0:89443 0:44721
¼
0:44721 0:89442
0:44721 0:89443
"
#
0:16666
0
¼
0
0:99999
Because [D1] is already diagonalized, no further rotation is needed. The two
natural frequencies can be calculated from the two diagonal entries as
1
1
u2 ¼ 0:16666 and u2 ¼ 0:99999. A quick note to add here is that natural frequencies calculated using the Jacobi method may not be arranged by the magnitude, as in other methods. In this example, the first (lowest) natural frequency u1
is located at [D1]22 while the second natural frequency u2 is located at [D1]11.
Thus, pthe
ffiffiffi natural frequencies should be sorted so that u1 ¼ 1 and
u2 ¼ 6 rad/s. The first eigenvector associated with u1 is associated with the
second column of [T] as
0:44721
1
¼ 0:44721
.
fV1 g ¼
0:89443
2
345
346
CHAPTER 8 Modal and Transient Dynamic Analysis
The second eigenvector associated with u2 is associated with the first column
of [T] as
0:89443
2
fV2 g ¼
¼ 0:44721
.
0:44721
1
Note that the normalization of the second eigenvector is with respect to the
first eigenvector, that is, for both eigenvectors the value pulled out of the vector
is {V11}.
As we have previously stated, as long as the ratio is correct, the actual values
for the eigenvectors are not important. Therefore, we again show that the eigenvalues and eigenvectors obtained from the Jacobi method are identical to those
calculated using the analytical method shown in Section 8.3.1. Example 8.7 requires only one iteration, and it is very rare that this situation exists. The
following example illustrates that the Jacobi method is better. As mistakes are
easily made when doing these types of calculations manually, we highly recommend writing a program to perform the calculations. Values are listed for you to
check your results.
2
3
Example 8.8
6
Assume that the system dynamic matrix [D] has the form ½D ¼ 4 2
4
3
2 4
7
9 2 5.
2 3
Using the Jacobi method, determine the three natural frequencies and three
eigenvectors.
Solution
The largest off-diagonal term is D13. However, we cannot choose rows and columns 1 and 3, since D11 ¼ D33. This would result in division by zero when calculating the angle of rotation. Instead, we choose rows and columns 1 and 2 to start
the iteration.
Iteration 1: The angle of rotation is calculated using Eq. (8.74) as
tan 2q1 ¼
2D12
22
¼
¼ 0:66667
D11 D22 3 9
1
q1 ¼ tan1 ð0:66667Þ ¼ 16:845 degrees
2
At this angle, the rotation
2
c
6
½T1 ¼ 4 s
0
matrix is
3 2
3
s 0
0:95709 0:28978 0
7 6
7
c 0 5 ¼ 4 0:28978 0:95709 0 5
0 1
0
0
1
8.3 Modal Analysis
We now update from [D] to [D1] using Eq. (8.76) as
2
32
3
0:95709 0:28978 0
3 2 4
6
76
7
6
76
7
6
76
7
T
½D1 ¼ ½T1 ½D½T1 ¼ 6 0:28978 0:95709 0 76 2 9 2 7
6
76
7
4
54
5
0
0
1
4 2 3
2
0:95709
0:28978 0
3
6
7
6
7
6
7
6 0:28978 0:95709 0 7
6
7
4
5
0
0
1
2
2:29171 0:69384 3:2488
6
6
6
¼ 6 2:78352
6
4
4
2
9:19337
0:95709
2
3
7
7
7
3:0733 7
7
5
3
0:28978 0
3
6
7
6
7
6
7
6 0:28978 0:95709 0 7
6
7
4
5
0
0
1
2
2:39443 0:00002 3:2488
3
6
7
6
7
6
7
¼ 6 0:00002 9:60549 3:0733 7
6
7
4
5
3:2488 3:0733
3
Iteration 2: The largest off-diagonal term is D13
tan 2q2 ¼
2D13
2 3:2488
¼ 10:72973
¼
D11 D33 2:39443 3
1
q2 ¼ tan1 ð10:72973Þ ¼ 42:33774 degrees
2
3 2
3
2
c 0 s
0:73919 0 0:6735
7 6
7
6
7 6
7
½T2 ¼ 6
¼6
0
1
0
7
40 1 0 7
5 4
5
s 0 c
0:6735 0 0:73919
347
348
CHAPTER 8 Modal and Transient Dynamic Analysis
2
6
6
6
½D2 ¼ ½T2 ½D1 ½T2 ¼ 6
6
4
0:73919 0 0:6735
T
0
1
0:6735
0
2
6
6
6
6
6
4
0
0:6735
0
1
0
3
7
7
7
7
7
5
0 0:73919
0:41813 2:06985 0:38098
6
6
6
¼ 6 0:00002
6
4
4:01413
2
6
6
6
6
6
4
9:60549
2:27177
0:73919
0
0:6735
0
1
0
0:6735
2
2:39443 0:00002
76
76
76
0
76 0:00002 9:60549
76
54
0:73919
3:2488
3:0733
0:73919
0:6735
2
32
7
7
7
3:0733 7
7
5
4:40564
3
7
7
7
7
7
5
0 0:73919
0:56567 2:06985 0:00001
6
6
6
¼ 6 2:06985
6
4
0:00001
3
9:60549
2:27177
3
7
7
7
2:27177 7
7
5
5:96012
Iteration 3:
tan 2q3 ¼
2D23
2 2:27177
¼ 1:24639
¼
D22 D33 9:60549 5:96012
1
q3 ¼ tan1 ð1:24639Þ ¼ 25:62967 degrees
2
2
3 2
3
1 0 0
1
0
0
6
7 6
7
6
7 6
7
½T3 ¼ 6 0 c s 7 ¼ 6 0 0:90161 0:43255 7
4
5 4
5
0 s c
0 0:43255 0:90161
3:2488
3
7
7
7
3:0733 7
7
5
3
8.3 Modal Analysis
2
1
0
6
6
6
½D3 ¼ ½T3 ½D2 ½T3 ¼ 6
60
6
4
0
T
2
0:90161
0:43255
1
0
6
6
6
6
6 0 0:90161
6
4
0 0:43255
2
0:56567
6
6
6
¼6
6 1:86619
6
4
0:89532
2
32
0
0:56567
6
6
6
¼6
6 1:86619
6
4
0:89532
0:56567
76
76
76
6
0:43255 7
76 2:06985
76
54
0:00001
0:90161
0
2:06985
9:60549
2:27177
3
7
7
7
2:27177 7
7
7
5
5:96012
3
7
7
7
0:43255 7
7
7
5
0:90161
2:06985 0:00001
9:64306
2:1066
0:00001
1
76
76
76
6
4:6263 7
76 0
76
54
4:39105
0
1:86619 0:89532
10:69539
32
0
0
3
7
7
7
0:90161 0:43255 7
7
7
5
0:43255 0:90161
3
7
7
7
0:00001 7
7
7
5
4:87023
Iteration 4:
tan 2q4 ¼
0:00001
2D12
2 ð1:86619Þ
¼ 0:33144
¼
D11 D22 0:56567 10:69539
1
q4 ¼ tan1 ð0:33144Þ ¼ 9:16863 degrees
2
3
2
3 2
c s 0
0:98722 0:15934 0
7
6
7 6
7 6 0:15934 0:98722 0 7
½T4 ¼ 6
7
4 s c 05 ¼ 6
4
5
0 0 1
0
0
1
349
350
CHAPTER 8 Modal and Transient Dynamic Analysis
2
0:98722
0:15934
6
6
6
½D4 ¼ ½T4 ½D3 ½T4 ¼ 6 0:15934 0:98722
6
4
T
0
0
0
32
0:56567 1:86619
76
76
76
0 76 1:86619
76
54
0:89532
1
2
0:89532
3
10:69539
7
7
7
0:00001 7
7
5
0:00001
4:87023
3
0:98722 0:15934 0
6
7
6
7
6
7
6 0:15934 0:98722 0 7
6
7
4
5
0
2
0
0:8558
6
6
6
¼ 6 1:75221
6
4
0:89531
2
1
0:13814
10:85606
0:00001
0:86687 0:00001
6
6
6
¼ 6 0:00001
6
4
10:99652
0:14265
0:88388
0:88388
32
0:98722 0:15934
76
76
76
0:14265 76 0:15934
76
54
4:87023
0:88388
0
3
7
7
7
0:14265 7
7
5
4:87023
Iteration 5:
tan 2q5 ¼
2D13
2 ð0:88388Þ
¼ 0:30813
¼
D11 D33 0:86687 4:87023
1
q5 ¼ tan1 ð0:30813Þ ¼ 8:56282 degrees
2
2
3 2
3
c 0 s
0:98885 0 0:14889
6
7 6
7
7 6
7
½T5 ¼ 6
0
1
0
40 1 0 5 ¼ 4
5
s
0
c
0:14889 0 0:98885
0:98722
0
0
3
7
7
7
07
7
5
1
8.3 Modal Analysis
2
6
6
6
½D5 ¼ ½T5 ½D4 ½T5 ¼ 6
6
4
0 0:14889
0:98885
T
0
1
0
0:14889
0
0:98885
2
6
6
6
6
6
4
0:14889
0
1
0
6
6
6
¼6
6 0:00001
6
4
2
0:88388
0
0:98881
0:86687
76
76
76
76 0:00001
76
54
0:98885
0:14889 0
2
32
0:00001
10:99652
0:14265
0:02123
10:99652
0:14890
0:99995
0:02123
0:00001
10:99652
0:00001
0:14106
32
76
76
76
6
0:14265 7
76
76
54
4:94752
0:98885
0
0:14889
3
7
7
7
0:14106 7
7
7
5
5:00328
Iteration 6:
2D23
2 ð0:14106Þ
¼
¼ 0:04707
D22 D33 10:99652 5:00328
1
q6 ¼ tan1 ð0:04707Þ ¼ 1:34746 degrees
2
2
3 2
3
1 0 0
1
0
0
6
7 6
7
7 6
7
½T6 ¼ 6
4 0 c s 5 ¼ 4 0 0:99972 0:02352 5
c
4:87023
7
7
7
7
7
5
0:14106
s
7
7
7
0:14265 7
7
5
3
0:74496
0
3
0:98885
6
6
6
¼6
6 0:02123
6
4
tan 2q6 ¼
0:88388
0 0:02352 0:99972
0 0:14889
1
0
0 0:98885
3
7
7
7
7
7
7
5
351
352
CHAPTER 8 Modal and Transient Dynamic Analysis
2
1
0
32
0
0:99995
76
6
76
6
76
6
½D6 ¼ ½T6 ½D5 ½T6 ¼ 6 0 0:99972 0:02352 76 0:02123
76
6
54
4
0:00001
0 0:02352 0:99972
T
2
1
0
6
6
6
6 0 0:99972
6
4
0 0:02352
2
0:99995
1
0:99995
6
6
6
¼ 6 0:02122
6
4
0:00051
0:14106
3
0:00001
0
0
3
3
7
7
7
0:02352 7
7
5
0:99972
0:02122
0:00051
3
7
7
7
0:14106 7
7
5
5:00328
7
7
7
10:99676 0:2587 7
7
5
0:11762 4:99856
6
6
6
6 0 0:99972
6
4
0 0:02352
2
10:99652
0:00001
7
7
7
0:02352 7
7
5
0:99972
0:02123
6
6
6
¼ 6 0:02122
6
4
0:00051
2
0
0:02123
3
7
7
7
10:99976 0:00002 7
7
5
0:00002 4:99993
If we consider 0.02122 as sufficiently small, then the updated matrix [D6] is
considered converged. In this case, the three eigenvalues (i.e., the three diagonal
entries) are l1 ¼ 0.99995, l2 ¼ 10.99976, and l3 ¼ 4.99993, which are very
close to the 1, 11, and 5, respectively, calculated using analytical methods.
The Jacobi method also allows the calculations of all eigenvalues simultaneously.
Also, the eigenvalues obtained are not arranged in any particular order and
need to be sorted. To find the eigenvectors, we need to find the overall rotation
matrix as
8.3 Modal Analysis
2
0:95709
0:28978 0
32
6
76
6
76
76
½T1 ½T2 ½T3 ½T4 ½T5 ½T6 ¼ 6
0:28978
0:95709
0
6
76
4
54
0:73919
0
0:6735
0
1
0
3
7
7
7
7
5
0:6735 0 0:73919
2
32
3
1
0
0
0:98722 0:15934 0
6
76
7
6
76
7
76 0:15934 0:98722 0 7
6
0
0:90161
0:43255
6
76
7
4
54
5
0
0
0 0:43255
2
6
6
6
6
4
0:90161
0:98885
0
0:14889
0
1
0
0:14889 0
2
1
0:70787
6
6
¼6
6 0:00139
4
0:98885
0
32
0
1
76
76
76 0
76
54
0
0:40699
0:57728
0:70633 0:40950
0:57741
0
0:99972
1
0
3
7
7
0:02352 7
7
5
0:02352 0:99972
3
7
7
0:81649 0:57735 7
7.
5
The three eigenvectors are listed in the three columns of the [T] matrix associated
with the three eigenvalues. As expected, the three eigenvectors listed below are
very close to the three eigenvectors calculated analytically, as listed on the far
right of each column vector.
8
9
8
9
8
9
>
>
>
< 0:70787 >
=
< 1 >
=
< 1 >
=
Ti1 ¼
0:00139
¼ 0:70787 0:002 z0:70787 0
>
>
>
>
>
>
:
;
:
;
:
;
0:70633
0:998
1
8
9
8
9
8 9
>
>
>
< 0:40699 >
=
< 1 >
=
<1>
=
Ti2 ¼ 0:81649 ¼ 0:40699 2:006 z0:40699 2
>
>
>
>
>
:
;
:
;
: >
;
0:40950
1:006
1
8
9
8
9
8
9
1
>
>
>
>
< 0:57728 >
=
<
=
< 1 >
=
Ti3 ¼ 0:57735 ¼ 0:57728 1:0001 z0:57728 1
>
>
>
>
>
>
:
;
:
;
:
;
0:57741
1:0002
1
353
354
CHAPTER 8 Modal and Transient Dynamic Analysis
Examples shown in Section 8.3 represent a type of dynamic analysis in which a
kinematicsetime history can be constructed from natural frequencies and mode
shapes using modal analysis. You may note that modal analysis is a type of dynamic
analysis, and the most important variable in dynamic analysis is “time.” Yet, none of
the calculations involve time when performing modal analysis. Because there is no
need for explicit integration or differentiation over time, using modal analysis for
computing structural kinematics affords great computational cost savings when
compared to using direct integration methods, to be discussed in Section 8.5.
You may also note that the Jacobi method sequentially rotates a matrix until all
off-diagonal terms approach zero. As such, the Jacobi method can also be used to
calculate the principal stresses described in Section 7.4.1.
8.3.4 SECTION REMARKS
The effects from unwanted vibrations can range from excessive rattle or shortened
fatigue life to catastrophic failure of a component or a whole structure. As such, one
important responsibility of product and structure engineers is to reduce the magnitude or isolate the source of vibration. Modal analysis is a computationally efficient
way for analyzing a dynamic system through the characteristic eigenvalues and eigenvectors. When conducting modal analysis using FE solvers, we must be very
careful about the low stiffness hourglass modes, which are energies numerically
added to stabilize the computation. These hourglass modes may create unrealistic
mode shapes associated with low eigenvalues.
By knowing the natural frequencies of a system, we have the opportunity to
reduce these vibration modes from being activated. From Eq. (8.39) in Section
8.3.2, we know that by changing the mass and stiffness values, the corresponding
natural frequency can be altered. Based on this principle, a properly constructed
springemass device, known as the dynamic vibration absorber, was used for centuries to reduce the possibility of resonance without damping.
Besides external loading, vibration-inducing internal sources in rotary machines
require attention. For example, an out-of-balance flywheel or tire may provoke significant vibration. Although rebalancing the flywheel or tire would take care of the unwanted problem of vibration, a short-term solution is to adjust the operating speed
so the frequency generated by the motions of the flywheel or tire is nowhere near
the resonance frequency. Additionally, various methods to isolate the source of vibration are used in factory and car designs to reduce the vibration-induced noise and to
slow down wear and tear of components. The key to isolating vibration sources is to
reduce the transmission of vibration from one component to the supporting structure
or to another component. For example, engine mounts are carefully designed and machine tools in factories are strategically arranged to reduce transmission of vibration to
other parts. Modal analysis can provide needed information to work on these problems.
There are situations where the direction and magnitude of external loading to a
structure are unpredictable. A couple of examples are the shaking of a high rise
building due to earthquake and twisting of a suspended bridge due to gusting
8.4 Damping
wind. In these cases, the way to dissipate energy is through structure damping, which
is the topic of the next section.
8.4 DAMPING
Before discussing structure damping, we must first declare that damping described
in this section is conceptually related to, but not directly associated with, “viscous
damping” discussed in Section 4.5.6. For the latter, a set of numerical procedures
is used to control the unrealistic/artificial hourglass energy. Damping as discussed
in this section is related to the behavior of the material itself or adding stiffness
or mass to the system.
Damping converts mechanical energy into thermal energy (heat), which in turn
reduces the magnitude of vibration. Frequently used methods for controlling unwanted vibrations include balancing the vibration produced by the rotary element
of a machine, reducing or increasing the speed to temporarily avoid unwanted vibration, or isolating the vibrating component from others. Important topics associated
with the same purpose include the application of damping components (such as a
dashpot) or making use of damping materials (such as a foam pad). Damping can
be accomplished by addressing internal or external sources.
Internal damping is related to energy dissipation in cyclic loaded components,
such as those used as structural components in the aviation or automotive industry.
The internal damping in an airplane or automobile structural component may come
from microstructural inhomogeneity within metal due to relative motions among
crystal grains. Also, fluid motion in cartilage can be a source of internal damping.
Sources for external damping frequently occur at the boundary, such as the friction
at the joint between two truss members and the pin used to connect them. The motion of a child on a swing will diminish unless the adult standing next to the child
keeps on pushing. External damping can also come from two cars crashing into
each other. In these cases, Coulomb damping (friction) is commonly used to represent this effect. Because kinetic energy is converted into heat for both internal and
external damping, these processes are irreversible.
Aside from Coulomb damping, dashpots are commonly used to provide external
damping. Fig. 8.7 shows the 6 m in diameter, 728-ton tuned mass damper installed in
Taipei 101, a 509 m tall, 101 story skyscraper located in Taipei, Taiwan. This mass
damper is suspended at the 92nd floor with eight steel cables, and is anchored at the
87th floor by eight viscous dampers. The tuned mass sphere can move up to a distance of 1.5 m in any direction to protect the building from the strongest earthquakes
or gale winds up to 216 km/h.
8.4.1 COULOMB DAMPING
Effects of damping can come from friction. For example, friction causes a car or a
swing to slow down if no additional energy enters. Consider that a simple masse
355
356
CHAPTER 8 Modal and Transient Dynamic Analysis
FIGURE 8.7
The 728-ton tuned mass damper of Taipei 101 designed to withstand strong winds and
earthquakes.
spring system without externally applied force is sliding on a tabletop as shown in
Fig. 8.8. Without friction, the mass will oscillate perpetually in a harmonic motion at
the natural frequency, as discussed in Section 8.3.1.
If friction exists, the oscillating mass will come to a full stop once all energy is
expended through friction-generated heat. This type of frictional force between the
mass and table is termed Coulomb friction, named after Charles Augustin de
Coulomb (Jun. 1736eAug. 1806). The damping induced by Coulomb friction is
FIGURE 8.8
A single springemass system oscillating on a tabletop.
8.4 Damping
termed Coulomb damping. According to Coulomb, the direction of frictional force is
opposite to the velocity of the moving mass, and the magnitude is determined by
_ m g;
F ¼ sgnðxÞm
(8.78)
_ indicates the sign of velocity, m is the coefficient of friction, which may
where sgnðxÞ
either be static or kinetic coefficient of friction. For example, a tire with a pressure of
275.8 kPa (40 psi) on pavement has a static coefficient of friction of 0.9, but the kinetic coefficient of friction is 0.85 on a dry surface and 0.69 on a wet surface. The
negative sign in Eq. (8.78) is intended to highlight the fact that Coulomb friction is
opposite to the direction of motion. With Coulomb friction, the equation of motion
for this system becomes
_ m g.
mx€ þ kx ¼ sgnðxÞm
(8.79)
In a special event when the mass is moving in only one direction, no sign change
needs to be considered. The solution for the nonhomogeneous differential equation
shown in Eq. (8.79) can be expressed as
rffiffiffiffi
rffiffiffiffi
k
k
mmg
xðtÞ ¼ A cos
.
(8.80)
t þ B sin
t
m
m
k
However, the nonlinear term related to the sign change makes it very difficult to
solve Eq. (8.79), where the direction of velocity could be positive or negative. At
each end of the oscillation, where the velocity is zero, the direction of frictional force
will change sign instantaneously. As such, there is a large jump in frictional force at
the end of oscillation from negative to positive, or vice versa. To overcome this problem, many FE solution packages treat Coulomb friction as a contact problem, based
on the penalty method or Lagrange multiplier, to remove the effect due to force
discontinuity. For detailed descriptions of various methods used for modeling
Coulomb friction, you may want to refer Weylera et al. (2012). To properly use software packages to model Coulomb damping, it is imperative that you carefully read
the user’s manual provided by the software vendor.
8.4.2 VISCOUS DAMPING
Example 8.2 in Section 8.3.1 describes an unforced vibration situation, in which the
peak magnitudes of oscillation for both m1 and m2 remain the same with time, as
demonstrated in Fig. 8.4. If an external cyclic force is applied to this two-spring,
two-mass system at one of the two natural frequencies as shown in Section 8.3.2,
the system will begin resonance and the peak magnitude will continue to increase
until failure occurs. When such an undesired situation is expected to arise, the structure designer needs to modify (or control) the structural behavior to avoid catastrophic failure.
In cars, shock absorbers (dampers or dashpots) operate in combination with the
suspension system to absorb (dissipate) some of the impact energy and dampen
shocks when driving on poor pavement. A typical shock absorber converts the
357
358
CHAPTER 8 Modal and Transient Dynamic Analysis
kinetic energy caused by the bumps into heat (through friction) to reduce the effects
of impact. It also limits the amplitude at resonance. By adding dampers to an undamped spring system, the new springedamper system becomes a damped vibration
problem.
Materials that exhibit similar effects of damping include rubber, foam, thermoplastics, composites, etc. Damping can also occur in structural components in which
fluid or fluid-like materials are contained. Typical materials that contain fluid
include soils and biological tissues. Damping is an important topic in studying structure responses due to earthquakes or in studying biomechanics for injury prevention
of biological tissues.
The governing equations of motion for undamped and damped systems under
forced vibration are expressed as
€
½MfxðtÞg
þ ½KfxðtÞg ¼ ff ðtÞg
€
_
½MfxðtÞg
þ ½CfxðtÞg
þ ½KfxðtÞg ¼ ff ðtÞg;
(8.81)
where [M] is the structural mass matrix, [K] is the global stiffness matrix, [C] is the
_ are the velocity
global damping matrix, {x} represents the displacement vectors, fxg
€ are the acceleration vectors, and {f} are the nodal force vectors, which
vectors, fxg
can all be constant or functions of time. Upon impact, a dashpot or viscous component generates resistance force that is inversely proportional to the velocity at which
the damper is impacted. It can be deduced from Eq. (8.81) that the dimension for the
global damping matrix [C] is N s/m.
Previously, we have described the methods used to determine the global mass
matrix [M] and stiffness matrix [K]. Similar to the [M] and [K] matrices, the [C] matrix needs to be a square matrix. As the element mass matrix [m] and stiffness matrix
[k] can be directly derived from element shape functions, we may think that the same
shape functions can be used to derive the element damping matrix. Unfortunately,
damping properties, especially those due to internal damping, of most materials
are not well established and cannot be measured from static tests. Even if these coefficients could be measured, solving Eq. (8.81) for a complex structure is not a trivial task. In this section, we discuss the method used to eliminate the need for defining
[C] in Eq. (8.81) by first reviewing the characteristics of damping coefficients and
then discussing an alternative Rayleigh damping method.
8.4.2.1 Damping Constant and Damping Ratio
A single mass, spring, and damper system, subjected to unforced vibration, is first
used to review the effect of damping. In this simple system, the governing differential equation has the form of
€ þ cxðtÞ
_ þ kxðtÞ ¼ 0;
mxðtÞ
(8.82)
lt
where c is called the damping constant. Assuming x ¼ e , we have x_ ¼ lelt and x€ ¼
l2 elt . Thus, the characteristic equation for Eq. (8.82) is
ml2 þ cl þ k ¼ 0:
(8.83)
8.4 Damping
pffiffiffiffiffiffiffiffiffiffiffiffi
c2 4mk.
The solution for the quadratic equation listed in Eq. (8.83) is l ¼ c 2m
Depending on the magnitude of c with respect to m and k, the solution for this characteristic equation reveals whether a system will vibrate or not. For example, a critical damping condition occurs when c2 ¼ 4mk. In this configuration, motion of the
mass returns to the resting positon in the shortest amount of time. Shock absorbers in
a car are designed to critically damp out the vibration. When c2 > 4 m k, the system
is overdamped. The damping mechanism in a door closer is designed this way to
limit the speed at which the door closes. In other words, the time needed to return
the mass to its resting position in an overdamped system is longer than that needed
for a critically damped system. For an underdamped situation to occur, we must have
c2 < 4 m k. In this case, vibrations continue but gradually tapper off before reaching
a stationary position. In order to easily distinguish these three conditions, the amount
of damping is usually described by a parameter called the damping ratio (z), which
is defined as the ratio of the damping constant c to the critical damping constant cc. A
system with a damping ratio <1 will oscillate, while a damping ratio >1 indicates
nonoscillatory behavior.
8.4.2.2 Rayleigh Damping
From Eq. (8.22), Section 8.3.1, we knowpthat
ffiffiffiffiffiffiffiffiffi the natural frequency of a single
springemass system is expressed as u ¼ k=m. At critical damping, the damping
constant has a value of cc 2 ¼ 4mk. From these two relationships, we write the critical damping constant as a function of m and u as
pffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
cc ¼ 4mk ¼ 4m mu2 ¼ 2mu.
(8.84)
We can declare that the damping constant can be represented by linear combinations of m and k, that is
c ¼ am þ bk;
(8.85)
where a is the mass damping component and b is the structural- or stiffness-damping
component. We declare this relationship because the magnitudes for m and k are usually larger than the magnitude of c. Also, direct modeling of damping is very complex and is considered more of an art than a science. From Eq. (8.85), the damping
ratio is calculated as
z¼
c
a
bu
þ
.
¼
cc 2u
2
(8.86)
Eqs. (8.85) and (8.86) are related to a single massespringedamper system. We can
analogously proclaim that the same relationship holds true for a large system of
masses, springs, and dampers as
½C ¼ a½M þ b½K
(8.87)
Eq. (8.87) is called the equation of Rayleigh damping, named after Lord Rayleigh as described in Section 8.3.3. Although the rheological or physical association
359
CHAPTER 8 Modal and Transient Dynamic Analysis
α damping
0.6
0.5
Damping Ratio
360
0.4
0.3
0.2
0.1
0.0
1
2.5
4
5.5
7
Frequency (Radians/Second)
8.5
10
FIGURE 8.9
Effects of mass damping. The calculated damping ratios represent fixed a values of 0.2
(the purple solid-double dotted line (dark gray in print versions) with the lowest
magnitude), 0.4, 0.6, 0.8, and 1.0 (the red solid line (darkest gray in print versions) with
the highest magnitude) as a function of frequency ranges from 1 to 10 rad/s.
of this equation is not clear, acquiring the damping matrix through this method and
using it to solve transient responses was found to be quite successful.
If we consider only the mass damping (i.e., neglect stiffness damping), the damping ratios can be calculated from a and u using Eq. (8.86) (see results shown in
Fig. 8.9). We can see from this figure that the damping ratio z is inversely proportional to the frequency. Additionally, as the frequency becomes higher, the effect
of mass damping diminishes. Thus, mass damping is most suitable for damping
out oscillations due to low frequency and high amplitude. Because only a single a
value is allowed as an input parameter to an FE model, the specific frequency that
dominates the impact responses should be the one used to calculate the a value to
damp out unwanted vibrations of the system.
If we consider only the structural damping (i.e., neglect mass damping), the
damping ratios can be calculated from b and u in Eq. (8.86), and results are shown
in Fig. 8.10. It can be seen from this figure that the damping ratio z is directly proportional to the frequency. Additionally, as the frequency becomes higher, the effect
of damping increases linearly. Thus, stiffness damping is good for damping out oscillations due to high frequency and low amplitude. Again, to damp out unwanted
vibrations, responses due to the most dominant frequency should be used to determine a single b value as the input parameter to an FE model.
In real-world problems, both mass and stiffness damping are desired to damp out
vibrations from both low and high frequencies. Assuming that we want to have a
8.4 Damping
β damping
6
Damping Ratio
5
4
3
2
1
0
1
2.5
4
5.5
7
Frequency (Radians/Second)
8.5
10
FIGURE 8.10
Effects of structural damping. The calculated damping ratios represent fixed b values of
0.2 (the purple double dotted line (dark gray in print versions) with the lowest magnitude),
0.4, 0.6, 0.8, and 1.0 (the red solid line (darkest gray in print versions) with the highest
magnitude) as a function of frequency ranges from 1 to 10 rad/s.
constant damping ratio of 0.5 for frequencies ranging from u1 to u2 rad/s. From Eq.
(8.86), we can write
a
bu1
¼ 0:5
þ
2u1
2
a
bu2
¼ 0:5:
þ
2u2
2
(8.88)
We further assume that u1 ¼ 1 and u2 ¼ 10. Eq. (8.88) represents two equations for
solving two unknowns. We find that a ¼ 0.909 and b ¼ 0.0909. Inserting these
values into Eq. (8.86) gives us the calculated damping ratios are shown in
Fig. 8.11 as functions of frequency. It can be seen that mass damping dominates
the effect in the low-frequency segment, while stiffness damping dictates the
high-frequency portion of the response.
In an underdamped system, the peak magnitude of each oscillation decreases
logarithmically over time. The damping ratio of the system can be determined
from experimental data. Assuming the period of a damped vibration is T ¼ 2p
u for
one cycle within which there are two consecutive peaks with peak values of x1
and x2. The logarithmic decrement D between these two consecutive peaks is calculated as
x1
D ¼ ln .
(8.89)
x2
The damping ratio can be calculated as
D
z ¼ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi .
ð2pÞ2 þ D2
(8.90)
361
CHAPTER 8 Modal and Transient Dynamic Analysis
α,β, and combined damping
0.6
0.5
Damping Ratio
362
0.4
0.3
0.2
0.1
0.0
1
2.5
4
5.5
7
Frequency (Radians/Second)
8.5
10
FIGURE 8.11
Damping ratios due to individual and combined effects of mass (a ¼ 0.909) and stiffness
(b ¼ 0.0909) damping. Blue solid-double dotted line (light gray in print versions)
represents mass damping only, red dashed line (dark gray in print versions) shows only
the effect of stiffness damping, and gray solid line shows combined effects.
Once the damping ratio is determined from the test data obtained from the most
dominate frequency, we can determine the values of a and b using Eq. (8.88). These
values are then inserted into an FE model to predict the effects due to Rayleigh
damping by calculating the decay time and other transient dynamic responses.
As the number of DOFs of a system increases, there are more mode shapes to be
considered. As such, finding a set of reasonable a and b values (which may or may
not be functions of time) for Rayleigh damping becomes a rather difficult task. In
most occasions, trial and error is the only way to find proper combinations of these
values for the model-predicted responses to match those obtained experimentally.
Due to the limited scope of the current book, we will not discuss this issue any
further. For further exploration of the various procedures used for estimating the
values of a and b, study by Adhikari and Phani (2007), the book by Bathe and Wilson (1976), and the dissertation by Pilkey (1998) are recommended.
8.5 DIRECT INTEGRATION METHODS
In Section 7.2 we discussed the benefit of reducing computational costs when
comparing the usage of iterative methods versus Gaussian elimination. If the stiffness matrix involves some extent of nonlinearity (for example, the stiffness may increase due to deformation), then the final solutions can only be achieved through
iterative procedures. However, solutions obtained through iterative methods may
not be exact.
8.5 Direct Integration Methods
Of all problems discussed thus far in this chapter, the term “time” has not
been a factor, despite the fact that the governing differential equations of a dynamic system involves time-dependent displacement, velocity, and acceleration
terms. Even for the forced vibration cases, where time-dependent harmonic
forces are applied to force the system to oscillate, we do not use time in the analysis. The reason we do not need to consider time is because we can solve for the
natural frequencies and mode shapes through modal analysis, which in turn can
be used to construct time-dependent parameters, such as the displacementetime
history. We can say that the solution schemes used in modal analysis are “time
implicit” approaches.
In this section, we discuss two methods commonly used to obtain dynamic responses of a structure through direct integration, the most general technique available for solving dynamic problems in the FE method. Consider that the governing
equations of motion, also known as the dynamic equilibrium equation, have the
form of
€
_
½MfxðtÞg
þ ½CfxðtÞg
þ ½KfxðtÞg ¼ ff ðtÞg.
(8.91)
Because [C] is very difficult to deduce and usually small in magnitude, methods
described in the Rayleigh damping section allow a convenient way for distributing
the effects of damping through mass and stiffness damping. If we consider [M] and
[K] matrices as premodified to include mass and stiffness damping or we simply
neglect the [C] matrix, Eq. (8.91) can be rewritten as
€
½MfxðtÞg
þ ½KfxðtÞg ¼ ff ðtÞg.
(8.92)
1
By multiplying [M]
to both sides of the equation and rearranging terms, we have
€ i Þg ¼ ½M1 ½Kfxðti Þg ff ðti Þg .
(8.93)
fxðt
If acceleration is known from initial or other conditions, we can directly integrate
Eq. (8.93) once to obtain the velocity, and twice to obtain the displacement. Likewise, we can differentiate displacement once to obtain velocity, and twice to obtain
acceleration. With the direct integration method, we apply this principle in a stepby-step manner to calculate entire time histories of displacement, velocity, and acceleration, provided we have initial values. In this section, two numerical methods
are presented for solving dynamic problems.
Before discussing various direct integration methods that are reported, we assume that the initial time is zero (t0 ¼ 0), and the time increment (Dt) at each
time step is constant throughout the entire integration process for all methods to
be discussed. Thus, the discrete time points of interest are 0, Dt, 2Dt, ., nDt.
Compared to analytical solutions, which provide accurate solutions at any time t,
we can only claim that solutions from direct integration methods are reasonably accurate at any of these discrete time points, and furthermore, this accuracy can be
assumed only if the time step is properly chosen.
363
364
CHAPTER 8 Modal and Transient Dynamic Analysis
8.5.1 CENTRAL DIFFERENCE METHOD
Because velocity is the time derivative of displacement, and acceleration is the time
derivative of velocity, we can approximate the velocity and acceleration using the
trapezoidal rule as
fxðtnþ1 Þg fxðtn Þg
Dt
(8.94)
_ nþ1 Þg fxðt
_ n Þg
fxðt
€ n Þg ¼
;
fxðt
Dt
th
where the subscript n indicates the n time step. Instead of differentiation, we can
also apply the trapezoidal rule to acquire velocity by integrating acceleration and
to obtain displacement by integrating velocity as
_ n Þg ¼
fxðt
Dt
€ n Þg þ fxðt
€ nþ1 ÞgÞ
ðfxðt
2
(8.95)
Dt
_ n Þg þ fxðt
_ nþ1 ÞgÞ.
fxðtnþ1 Þg ¼ fxðtn Þg þ ðfxðt
2
However, using two consecutive steps by the trapezoidal rule to determine the
velocity and acceleration can result in a large error when analyzing dynamic problems with high-frequency contents, such as those seen in car crashes. A more accurate way to approximate the solution is to find velocity and acceleration from three
consecutive time steps. Because the derivative and integration for the middle time
step are directly calculated from the two data points at one step ahead and one
step behind, this method is called the central difference method, also known as
the centered difference method.
_ nþ1 Þg ¼ fxðt
_ n Þg þ
fxðt
8.5.1.1 Derivation
Fig. 8.12 shows three consecutive steps of the displacementetime history. We can
easily approximate the velocity from Fig. 8.12 as
1
ðfxðtnþ1 Þg fxðtn1 ÞgÞ.
(8.96)
2Dt
To approximate the acceleration at tn, we first calculate the velocity at tn 12Dt and
tn þ 12Dt as
o
n 1
¼ ðfxðtn Þg fxðtn1 ÞgÞ
x_ tnDt=2
Dt
(8.97)
n o
1
x_ tnþDt=2
¼ ðfxðtnþ1 Þg fxðtn ÞgÞ.
Dt
From Fig. 8.12 and Eq. (8.97), we approximate the acceleration as
o n o
1 n € n Þg ¼
x_ tnþDt=2
x_ tnDt=2
fxðt
Dt
(8.98)
1
¼
ðfxðt
Þg
2fxðt
Þg
þ
ÞgÞ.
fxðt
n
n1
nþ1
ðDtÞ2
_ n Þg ¼
fxðt
8.5 Direct Integration Methods
FIGURE 8.12
Displacementetime histories of three discrete, consecutive time steps.
As shown in Eqs. (8.96) and (8.98), the velocity and acceleration are expressed by
_ in Eq. (8.96) and
the displacements at only two consecutive steps. By inserting fxg
€ in Eq. (8.98) into Eq. (8.91), we obtain
fxg
€ n Þg þ ½Cfxðt
_ n Þg þ ½Kfxðtn Þg ¼ ff ðtn Þg
½Mfxðt
½M
1
ðDtÞ
2
ðfxðtn1 Þg 2fxðtn Þg þ fxðtnþ1 ÞgÞ þ ½C
1
ðfxðtnþ1 Þg
2Dt
fxðtn1 ÞgÞ þ ½Kfxðtn Þg ¼ ff ðtn Þg
!
!
½M
½C
2½M
þ
fxðtnþ1 Þg ¼ ff ðtn Þg ½K fxðtn Þg
ðDtÞ2 2Dt
ðDtÞ2
(8.99)
!
½C
þ
fxðtn1 Þg.
ðDtÞ2 2Dt
½M
Note that the top line of Eq. (8.99) involves the dynamic equilibrium equation only
at the time tn. This central difference method allows us to find {x(tnþ1)} from the
previous two steps {x(tn1)} and {x(tn)}, as shown in the third line of Eq. (8.99), using only the equilibrium condition at tn. As such, this integration scheme is called the
explicit integration method or explicit integration scheme. In contrast, the implicit
integration method is defined for the integration scheme that requires a dynamic
equilibrium equation at time tnþ1. We also notice from Eq. (8.99) that [K] and
[M] do not need to be updated through all time steps. This characteristic makes
the central difference method computationally efficient.
365
366
CHAPTER 8 Modal and Transient Dynamic Analysis
The time step Dt needs to be properly selected so that the integration is reasonably accurate. For the implicit integration scheme, the selected Dt must be small
enough to cover the frequency spectrum associated with the problem. For a system
with the first natural frequency of H Hz (i.e., Dt ¼ 1/H s), a general rule of thumb
is to select a time step of no greater than 1/pH s. For the explicit integration
scheme, the critical time step needs to be smaller than the time needed for the
acoustic wave to propagate through the element. In most FE solvers, the default
time step is made to be (e.g., 90%) smaller than the critical time step to ensure accurate model predictions.
To write Eq. (8.99) in a less cumbersome way, we first calculate six constants,
namely
a1 ¼
1
ðDtÞ
; a2 ¼
2
1
2
a5 ðDtÞ2
2
; a3 ¼
¼
.
;
a
¼
2Dt;
a
¼
ðDtÞ
;
a
¼
4
6
5
2Dt
2
2
ðDtÞ2
(8.100)
Using the first three of these constants and inserting them appropriately into Eq.
(8.99) gives us
ða1 ½M þ a2 ½CÞfxðtnþ1 Þg ¼ ff ðtÞg ð½K a3 ½MÞfxðtn Þg
ða1 ½M þ a2 ½CÞfxðtn1 Þg.
(8.101)
For the central difference method to perform efficiently, the [M] and [C] matrices
need to be diagonalized so that the left-hand side of Eq. (8.101) can be easily calcu_ 0 Þg, and fxðt
€ 0 Þg
lated. We then calculate {x(t1)} from initial conditions fxðt0 Þg; fxðt
by inserting n ¼ 0 into Eq. (8.101) as
ða1 ½M þ a2 ½CÞfxðt1 Þg ¼ ff ðtÞg ð½K a3 ½MÞfxðt0 Þg
ða1 ½M þ a2 ½CÞfxðt1 Þg.
(8.102)
We quickly realize from Eq. (8.102) that we need to know {x(t1)} before finding
{x(t1)}. To calculate {x(t1)}, we assign n ¼ 0 to Eqs. (8.96) and (8.98) and obtain
_ 0 Þg ¼
fxðt
1
ðfxðt1 Þg fxðt1 ÞgÞ0
2Dt
(8.103)
_ 0 Þg þ fxðt1 Þg; and
fxðt1 Þg ¼ a4 fxðt
€ 0 Þg ¼
fxðt
1
ðDtÞ2
ðfxðt1 Þg 2fxðt0 Þg þ fxðt1 ÞgÞ0
€ 0 Þg þ 2fxðt0 Þg fxðt1 Þg.
fxðt1 Þg ¼ a5 fxðt
(8.104)
8.5 Direct Integration Methods
If we subtract Eq. (8.104) from Eq. (8.103), we have
_ 0 Þg þ fxðt1 Þg a5 fxðt
€ 0 Þg 2fxðt0 Þg þ fxðt1 Þg ¼ 0
a4 fxðt
€ 0 Þg þ a4 fxðt
_ 0 Þg 2fxðt0 Þg ¼ 0
2fxðt1 Þg a5 fxðt
(8.105)
€ 0 Þg Dtfxðt
_ 0 Þg þ fxðt0 Þg.
fxðt1 Þg ¼ a6 fxðt
By inserting {x(t1)} as derived in Eq. (8.105) into Eqs. (8.103) and (8.104), we
_ 0 Þg and fxðt
€ 0 Þg. These descriptions outline the procedures needed to
can find fxðt
_ 0 Þg, and fxðt
€ 0 Þg.
find {x(t1)}, fxðt
Once {x(t1)} is known, we will be able to calculate {x(t2)}. Similarly, we can
determine {x(t3)}, {x(t4)}, ., {x(tn)} in a step-by-step manner. If velocitye and
accelerationetime histories are also needed, we can compute the velocity
_ 1 Þg; fxðt
_ 2 Þg; .; fxðt
_ n Þg and acceleration fxðt
€ 1 Þg; fxðt
€ 2 Þg; .; fxðt
€ n Þg time hisfxðt
tories from {x(t0)}, {x(t1)}, ., {x(tn)}.
8.5.1.2 Application
To summarize, the central difference integration method can be carried out using the
following steps. Again, a computer program is recommended to perform this task.
1. Properly select Dt, calculate, and store the constants a1 through a6
a1 ¼
1
ðDtÞ
2
; a2 ¼
1
2
ðDtÞ2
; a3 ¼
; a4 ¼ 2Dt; a5 ¼ ðDtÞ2 ; a6 ¼
2
2Dt
2
ðDtÞ
2. Identify initial conditions fx0 g; fx_0 g, and fx€0 g from the equilibrium equation
€
_
½MfxðtÞg
þ ½CfxðtÞg
þ ½KfxðtÞg ¼ ff ðtÞg
3. Calculate {x(t1)} using Eq. (8.105)
€ 0 Þg Dtfxðt
_ 0 Þg þ fxðt0 Þg
fxðt1 Þg ¼ a6 fxðt
4. Calculate {x(t1)} using Eq. (8.102), which is listed below
ða1 ½M þ a2 ½CÞfxðt1 Þg ¼ ff ðtÞg ð½K a3 ½MÞfxðt0 Þg ða1 ½M þ a2 ½CÞ
fxðt1 Þg
As noted earlier, [M] and [C] need to be diagonal matrices for easier calculations. In
such a case, the vector on the left-hand side of this equation becomes
f ða1 M11 þ a2 C11 Þx1 ðt1 Þ
ða1 M22 þ a2 C22 Þx2 ðt1 Þ .
ða1 Mnn þ a2 Cnn Þxn ðt1 Þ gT .
Thus, the nodal displacement xi can be expressed as
xi ðt1 Þ ¼
f f ðtÞg ð½K a3 ½M Þf xðt0 Þg ða1 ½M þ a2 ½C Þf xðt1 Þg
;
ða1 Mii þ a2 Cii Þ
where ii is used to indicate the diagonal entries, and not related to index notation.
367
368
CHAPTER 8 Modal and Transient Dynamic Analysis
5. Calculate {x(t2)} from {x(t1)} and {x(t0)} using Eq. (8.101). For diagonalized
[M] and [C], the nodal DOFs are expressed as
xi ðt2 Þ ¼
ff ðtÞg ð½K a3 ½MÞfxðt1 Þg ða1 ½M þ a2 ½CÞfxðt0 Þg
ða1 Mii þ a2 Cii Þ
_ i Þg using Eq. (8.96)
6. If velocity is needed, determine fxðt
_ i Þg ¼ a2 ðfxðtiþ1 Þg fxðti1 ÞgÞ
fxðt
€ i Þg using Eq. (8.98)
7. If acceleration is needed, determine fxðt
€ i Þg ¼ a1 ðfxðti1 Þg 2fxðti Þg þ fxðtiþ1 ÞgÞ
fxðt
8. Similar to Step 5, calculate {x(tiþ1)} from {x(ti)} and {x(ti1)}, where i ¼ 2 to
n, using Eq. (8.101)
_ i Þg and fxðt
€ i Þg using Eqs. (8.96) and
9. Similar to Steps 6 and 7, determine fxðt
(8.98), if needed
10. Repeat Steps 8 and 9 for all remaining time steps (i ¼ 3 to n)
Example 8.9
Determine the displacementetime histories for a forced vibration problem of the
2-spring, 2-mass system described in Example 8.2 using the central difference
_ [ 0Þg [ 0, and
method. Here the initial conditions are {x(t [ 0)} ¼ {0}, fxðt
the loading condition is f2(t) [ 10.
Solution
With nodal DOFs denoted in the subscript, the dynamic equilibrium equation
directly taken from Example 8.2 is
5 2
x2 ðtÞ
10
1 0
x€2 ðtÞ
þ
¼
.
€
ðtÞ
x
2 2
0
x3 ðtÞ
0 1
3
_
€
First step: find fxð1Þg; fxð0Þg,
and fxð0Þg.
The initial conditions assigned are x2(t ¼ 0) ¼ x3(t ¼ 0) ¼ 0 and
x_2 ðt ¼ 0Þ ¼ x_3 ðt ¼ 0Þ ¼ 0. Using the equilibrium equation listed above and
initial conditions provided, we find x€2 ðt ¼ 0Þ ¼ 10 and x€3 ðt ¼ 0Þ ¼ 0. We will
sidestep the writing of “t ¼ ” in parentheses to save space from this point on.
First, we select a time step of Dt ¼ 0.25 s. We then calculate the six constants as
a1 ¼
1
ðDtÞ
2
¼ 16; a2 ¼
1
2
¼ 32; a4 ¼ 2Dt ¼ 0:5;
¼ 2; a3 ¼
2Dt
ðDtÞ2
a5 ¼ ðDtÞ2 ¼ 0:0625; a6 ¼
ðDtÞ2
¼ 0:03125
2
8.5 Direct Integration Methods
Second, we identify {x(t1)}. By inserting the initial displacements, velocities, and accelerations into Eq. (8.105), we have
)
)
(
(
) (
)
(
x2 ðt1 Þ
x2 ðt0 Þ
x€2 ðt0 Þ
x_2 ðt0 Þ
¼ a6
þ
Dt
x€3 ðt0 Þ
x_3 ðt0 Þ
x3 ðt1 Þ
x2 ðt0 Þ
)
( ) ( ) (
)
( )
(
0:3125
0
0
10
x2 ðt1 Þ
¼
þ
0:25
¼ 0:03125
0
0
0
0
x3 ðt1 Þ
Third, we calculate {x(t1)} using Eq. (8.101) as
) ( ) 0"
#
"
#1( )
"
#(
10
5 2
1 0
0
1 0
x2 ðt1 Þ
A
¼
@
32
16
0
2 2
0 1
0
x3 ðt1 Þ
0 1
"
#(
)
1 0
0:3125
16
0 1
0
) (
)
(
0:3125
x2 ðt1 Þ
¼
0
x3 ðt1 Þ
Fourth, we calculate {x(t2)} by letting n ¼ 1 in Eq. (8.101) as
) ( ) 0"
#
"
#1(
"
#(
)
10
5 2
1 0
1 0
x2 ðt2 Þ
0:3125
@
A
¼
32
16
0
2 2
0 1
x3 ðt2 Þ
0 1
0
"
#( )
1 0
0
16
0 1
0
) ( )
"
#!(
) (
) (
)
(
10
27 2
0:3125
x2 ðt2 Þ
1:15234
x2 ðt2 Þ
¼
0
¼
16
0
2 30
0
0:03906
x3 ðt2 Þ
x3 ðt2 Þ
_ 1 Þg and fxðt
€ 1 Þg from Eqs. (8.96) and (8.98), respectively, as
Fifth, we find fxðt
! 2:30469
1:15234
0
_ 1 Þg ¼ a2 ðfxðt2 Þg fxðt0 ÞgÞ ¼ 2
fxðt
¼
0:07813
0:03906
0
€ 1 Þg ¼ a1 ðfxðt0 Þg 2fxðt1 Þg þ fxðt2 ÞgÞ
fxðt
! 0
0:3125
1:1523
8:4368
¼ 16
2
þ
¼
0
0
0:0391
0:6256
These processes can be carried out iteratively until all time steps are
completed. For validation of your computer program, Table 8.3 lists the calculated values for the first nine time steps.
369
370
CHAPTER 8 Modal and Transient Dynamic Analysis
Table 8.3 Displacement, Velocity, and Acceleration for the Second and Third DOFs Under
Forced Motion Solved by the Central Difference Method
Dt
2Dt
3Dt
4Dt
5Dt
6Dt
7Dt
8Dt
9Dt
x2
x3
x_2
0.313
0.000
2.305
1.152
0.039
3.899
2.262
0.217
4.329
3.317
0.651
3.559
4.042
1.418
1.978
4.306
2.513
0.243
4.163
3.832
0.963
3.824
5.193
1.199
3.564
6.382
0.423
x_3
x€2
0.078
0.435
1.224
2.402
3.724
4.828
5.359
5.100
4.053
8.438
4.316
0.875
5.282
7.372
6.502
3.152
1.265
4.946
0.625
2.227
4.089
5.332
5.247
3.585
0.662
2.737
5.637
x€3
8.5.2 THE NEWMARK METHOD
The Newmark method is named after Professor Nathan M. Newmark (Sep.
1910eJan. 1981) based on an assumption of average acceleration. It is a numerical
integration method used to solve differential equations, such as the dynamic equilibrium equations shown as Eq. (8.91). Details of this method can be found in Newmark
(1959).
8.5.2.1 Derivation
Consider that Taylor’s expansion of any function f(t) is written as
f ðtnþ1 Þ ¼ f ðtnþDt Þ
df ðtn Þ 1 2 d2 f ðtn Þ
Dts ðsÞ
þ Dt
f ðtn Þ
¼ f ðtn Þ þ Dt
þ
/
þ
dt
2
dt2
s!
Z tnþ1
1
þ
f ðsþ1Þ ðsÞðtnþ1 sÞs ds;
s! tn
(8.106)
where f(s) represents the sth time derivative of the function f. To find the velocity at
_ nþ1 Þ and s ¼ 0 as
time tnþ1, we let f ðtnþ1 Þ ¼ xðt
Z tnþ1
_ nþ1 Þg ¼ fxðt
_ nþDt Þg ¼ fxðt
_ n Þg þ
€
fxðt
fxðsÞgds.
(8.107)
tn
Similarly, to find the displacement at time tnþ1, we let f ðtnþ1 Þ ¼ xðtnþ1 Þ and s ¼ 1 as
Z tnþ1
€
_ n Þg þ
fxðtnþ1 Þg ¼ fxðtnþDt Þg ¼ fxðtn Þg þ Dtfxðt
fxðsÞgðt
nþ1 sÞds.
tn
(8.108)
€
To carry out the integration in Eqs. (8.107) and (8.108), we need to find fxðsÞg
first. Because s is between tn and tnþ1, we attain from the Taylor’s expansions
€ n Þg and f ðtnþ1 Þ ¼ fxðt
€ nþ1 Þg,
expressed in Eq. (8.106) by assuming f ðtn Þ ¼ fxðt
respectively, and s ¼ 2 as
€ n Þg ¼ fxðsÞg
€
þ xð3Þ ðsÞðtn sÞ þ xð4Þ ðsÞ
fxðt
ðtn sÞ2
þ .;
2
(8.109)
8.5 Direct Integration Methods
371
and
ðtnþ1 sÞ2
þ .;
(8.110)
2
€
where “.” represents higher order terms not explicitly written. To extract fxðsÞg,
which is used in Eqs. (8.107) and (8.108), we multiply (1 g) to Eq. (8.109) and
g to Eq. (8.110), where 0 < g < 1. We then sum up the two results as
€ nþ1 Þg ¼ fxðsÞg
€
þ xð3Þ ðsÞðtnþ1 sÞ þ xð4Þ ðsÞ
fxðt
€ n Þg þ gfxðt
€ nþ1 Þg ¼ ð1 g þ gÞfxðsÞg
€
ð1 gÞfxðt
þ xð3Þ ðsÞ½ð1 gÞðtn sÞ
þ gðtnþ1 sÞ þ /
€
€ n Þg þ gfxðt
€ nþ1 Þg þ xð3Þ ðsÞðs Dtg tn Þ þ /
¼ ð1 gÞfxðt
fxðsÞg
As part of Eq. (8.107), we find
as
Z
tnþ1
Z
€
¼
fxðsÞgds
tn
R
(8.111)
€
by applying the mean value theorem
fxðsÞgds
i
€ nþ1 Þg þ xð3Þ ðsÞðs gDt tn Þ þ . ds
€ n Þg þ gfxðt
ð1 gÞfxðt
tnþ1 h
tn
ðs gDt tn Þ tnþ1
€ nþ1 Þg þ xð3Þ ðsÞ
€ n Þg þ Dtgfxðt
þ .;
¼ Dtð1 gÞfxðt
2
tn
€ n Þg þ Dtgfxðt
€ nþ1 Þg þ ðDtÞ2
¼ Dtð1 gÞfxðt
1
g xð3Þ ðesÞ þ .
2
(8.112)
where tn < es < tnþ1 . By inserting this integral to Eq. (8.107) and then ignoring
higher order terms, we conclude that
_ nþ1 Þgzfxðt
_ n Þg þ Dt½ð1 gÞfxðt
€ n Þg þ gfxðt
€ nþ1 Þg.
fxðt
(8.113)
€
Similarly, by replacing g by 2b in Eq. (8.111) to extract fxðsÞg,
we have
€
€ n Þg þ 2bfxðt
€ nþ1 Þg þ xð3Þ ðsÞðs 2Dtb tn Þ þ .. (8.114)
2bÞfxðt
fxðsÞgzð1
To find {x(tnþ1)} in Eq. (8.108), we first find the integral
Z
tnþ1
tn
Z
¼
€
fxðsÞgðt
nþ1 sÞds
i
€ nþ1 Þg þ xð3Þ ðsÞðs 2bDt tn Þ þ . ðtnþ1 sÞds
€ n Þg þ 2bfxðt
ð1 2bÞfxðt
tnþ1 h
tn
¼ ðDtÞ2
1
1
€ n Þg þ ðDtÞ2 bfxðt
€ nþ1 Þg þ ðDtÞ3
b fxðt
b xð3Þ ðesÞ þ ..
2
6
(8.115)
372
CHAPTER 8 Modal and Transient Dynamic Analysis
By inserting results of this integral into Eq. (8.108) and neglecting all higher order
terms, we have
1
€ n Þg þ bfxðt
_ n Þg þ ðDtÞ2
€ nþ1 Þg .
b fxðt
fxðtnþ1 Þgzfxðtn Þg þ Dtfxðt
2
(8.116)
This set of basic equations can be used to solve dynamic equilibrium problems
numerically. Of course, proper selections of the time step, g, and b values will affect
the integration results.
8.5.2.2 Application
According to the Newmark method, Eqs. (8.113) and (8.116) are approximations for
the velocity and displacement at time tnþ1 ¼ tn þ Dt. These two equations are
relisted below for easy reference.
_ nþ1 Þgzfxðt
_ n Þg þ Dt½ð1 gÞfxðt
€ n Þg þ gfxðt
€ nþ1 Þg
fxðt
1
2
€
_
€
xðt
b
Þgzfxðt
Þg
þ
Dtf
xðt
Þg
þ
ðDtÞ
Þg
þ
bf
xðt
Þg
f n
fxðtnþ1
n
n
nþ1
2
As we can see, we cannot use these two equations directly to find the displacement and velocity at tnþ1 unless the acceleration at tnþ1 is available. The first step in
finding the acceleration is to rearrange Eq. (8.113) as
_ nþ1 Þg fxðt
_ n Þg ¼ Dt½ð1 gÞfxðt
€ n Þg þ gfxðt
€ nþ1 Þg
fxðt
€ nþ1 Þg fxðt
€ n Þg.
€ n Þg þ g½fxðt
¼ Dt½fxðt
(8.117)
We now define the difference or increment between two consecutive time steps
tnþ1 and tn in terms of displacement, velocity, acceleration, and force in the four
equations listed in Eq. (8.118).
Dfxðtn Þg ¼ fxðtnþ1 Þg fxðtn Þg; Dfx_ðtn Þg ¼ fx_ðtnþ1 Þg fx_ðtn Þg
Dfx€ðtn Þg ¼ ½fx€ðtnþ1 Þg fx€ðtn Þg; Df f ðtn Þg ¼ f f ðtnþ1 Þg f f ðtn Þg;
(8.118)
where f represents the force as a function of time. Using these definitions of increment, we rewrite Eq. (8.117) as
€ n Þg þ ðgDtÞ½Dfxðt
€ n Þg.
_ n Þg ¼ ðDtÞfxðt
½Dfxðt
(8.119)
Similarly, we rearrange Eq. (8.116) and apply the definitions in Eq. (8.118) as
h
i
ðDtÞ2
€ n Þg þ bðDtÞ2 ½Dfxðt
€ n Þg.
(8.120)
fxðt
2
€ n Þg by multiplying 1/b(Dt2) to both sides of Eq. (8.120) as
We now solve ½Dfxðt
_ n Þg þ
½Dfxðtn Þg ¼ Dtfxðt
€ n Þg ¼
½Dfxðt
½Dfxðtn Þg
bðDtÞ
2
1
1
_ n Þg fxðt
€ n Þg.
fxðt
bDt
2b
(8.121)
8.5 Direct Integration Methods
€ n Þg found in Eq. (8.121) into Eq.
By inserting the difference in acceleration ½Dfxðt
_ n Þg as
(8.119), we obtain the difference in velocity ½Dfxðt
"
#
½Dfxðtn Þg
1
1
_ n Þg fxðt
€ n Þg
€ n Þg þ ðgDtÞ
_ n Þg ¼ ðDtÞfxðt
½Dfxðt
fxðt
bDt
2b
bðDtÞ2
¼
g
g
g
_ n Þg þ Dt 1 € n Þg.
½Dfxðtn Þg fxðt
fxðt
bDt
b
2b
(8.122)
For time points tn and tnþ1, we replace “t” in the dynamic equilibrium equation
shown in Eq. (8.91) with tn and tnþ1 as
€ n Þg þ ½Cfxðt
_ n Þg þ ½Kfxðtn Þg ¼ ff ðtn Þg
½Mfxðt
€ nþ1 Þg þ ½Cfxðt
_ nþ1 Þg þ ½Kfxðtnþ1 Þg ¼ ff ðtnþ1 Þg.
½Mfxðt
(8.123)
Subtracting the two dynamic equilibrium equations in Eq. (8.123) yields
_ n Þg þ ½K½Dfxðtn Þg ¼ fDf ðtn Þg.
€ n Þg þ ½C½Dfxðt
½M½Dfxðt
(8.124)
By inserting the increment in displacement [D{x(tn)}] shown in Eq. (8.120), accel€ n Þg shown in Eq. (8.121), and velocity ½Dfxðt
_ n Þg shown into Eq.
eration ½Dfxðt
(8.122) in Eq. (8.124), we obtain
#
"
½Dfxðtn Þg
1
1
g
g
_ n Þg fxðt
€ n Þg þ ½C
_ n Þg
½Dfxðtn Þg fxðt
½M
fxðt
2
bDt
2b
bDt
b
bðDtÞ
g
€ n Þg þ ½K½Dfxðtn Þg ¼ fDf ðtn Þg0
fxðt
2b
#
"
½M
g½C
½M g½C
_ n Þg
þ
½K
½Dfxðt
þ
þ
Þg
¼
ðt
Þg
þ
fxðt
fDf
n
n
bDt
b
bðDtÞ2 bDt
þ Dt 1 ½M
g
€ n Þg.
þ Dt
1 ½C fxðt
þ
2b
2b
(8.125)
Eq. (8.125) allows us to find the increment/difference in displacement D{x(tn)}
from which we can calculate {x(tnþ1)}. The aforementioned procedures are repeated
until the displacement, velocity, and acceleration for all time steps are determined.
Before we start working on an example problem, we summarize the steps commonly
taken to apply the Newmark method as follows:
€ 0 Þg from initial displacement {x(t0)} and
1. Determine the initial acceleration fxðt
_ 0 Þg using the dynamic equilibrium equation
velocity fxðt
€ 0 Þg ¼ ff ðt0 Þg ½Kfxðt0 Þg ½Cfxðt
_ 0 Þg.
½Mfxðt
373
374
CHAPTER 8 Modal and Transient Dynamic Analysis
2. Properly select Dt, g, and b. In general, choosing g ¼ 0.5 and 0.167 b 0.25
yields reasonable answers.
3. Calculate
and vectors as derived in Eq. (8.125)
" the following matrices
#
½M
g½C
Kb ¼
þ ½K ; Fb ¼ f f ðtnþ1 Þg f f ðtn Þg
þ
2
bDt
bðDtÞ
.
½M
g½C
½M
g
Ab ¼
þ
þ Dt
1 ½C
; Bb ¼
bDt
b
2b
2b
4. Solve for [D{x(tn)}] from initial velocity and acceleration by assigning n ¼ 0 in
_ n Þg þ Bb fxðt
€ n Þg. Note that as Kb is not
Kb ½Dfxðtn Þg ¼ Fb þ Ab fxðt
1
necessarily a diagonalized matrix, we can find Kb
first, then multiply the
inverse matrix to both sides of the equation to make calculations easier. We only
need to perform this task once, because there is no need to update Kb and
1
Kb
after the first time. Alternatively, we can diagonalize it or make it an
upper triangular matrix in the calculations.
5. Calculate the displacement for the next time step {x(tnþ1)} by using
fxðtnþ1 Þg ¼ fxðtn Þg þ ½Dfxðtn Þg.
6. Calculate the difference in acceleration by using Eq. (8.121) as
1 fxðt
1 fxðt
€ n Þg ¼ ½Dfxðtn Þg
_ n Þg 2b
€ n Þg.
½Dfxðt
bDt
2
bðDtÞ
7. Calculate
the
difference
in
velocity
using Eq.
g
g
€ n Þg.
_ n Þg þ Dt 1 2b
_ n Þg ¼ bDt
½Dfxðtn Þg gb fxðt
½Dfxðt
fxðt
by
(8.122)
as
8. Repeat steps 4 to 7 by sequentially assigning n ¼ 1, 2, ., n 1, to complete the
computation of displacement, velocity, and accelerationetime histories.
Example 8.10
Using the Newmark method, determine the displacementetime histories for the
forced vibration problem of a 2-spring, 2-mass system described in Example
_ ¼ 0Þg ¼ 0, and the
8.2. The initial conditions are fxðt ¼ 0Þg ¼ f0g; fxðt
loading condition is f2(t) ¼ 10. Also, the time step is 0.25 s, g is assumed to be
0.5, and b is assumed to be 0.25.
Solution
€ 0 Þg are f 10 0 gT . We
As shown in Example 8.9, the
fxðt
accelerations
initial
b
calculate the three matrices Kb ; A , Bb , and the vector Fb as
0
2 0
16 0
69 2
; Fb ¼
; Bb ¼
Kb ¼
; Ab ¼
0
0 2
0 16
2 66
8.5 Direct Integration Methods
375
Assign n ¼ 0 and apply the matrices and vector listed above to
_ n Þg þ Bb fxðt
€ n Þg. Note that we need to
Kb ½Dfxðtn Þg ¼ Fb þ Ab fxðt
_ n Þg and by the initial
multiply through by the initial velocity to obtain Ab fxðt
b
€ n Þg before further calculations can be carried
acceleration to determine B fxðt
1
out. We then multiply Kb
to both sides of the equation in order to obtain
Dx2 ðt0 Þ
0:2901
the difference in displacement as
¼
. Because the initial
Dx3 ðt0 Þ
0:0088
displacements are zero, the displacements at t1 are the same as the difference, that
x2 ðt1 Þ
0:2901
is,
¼
.
x3 ðt1 Þ
0:0088
With known displacements at time t1, we can determine the difference in
1:433
Dx€2 ðt1 Þ
acceleration and velocity, also at time t1, as
¼
and
Dx€3 ðt1 Þ
0:563
2:321
Dx_2 ðt1 Þ
¼
. From these values, acceleration and velocity at time
Dx_3 ðt1 Þ
0:070
t1 can be calculated. Table 8.4 lists the values for you to compare your results
using the Newmark method.
There are several special values related to g and b that are worthy of some quick
notes. The Newmark method is unstable when g < 0.5 and conditionally stable
when g 0.5. For g ¼ 0.5, the Newmark method is accurate at least for the second
order. For all other selections, the method is only accurate for the first order.
If we choose g ¼ 0.5 and b ¼ 0.25, the acceleration within the time interval tn
and tnþ1 is constant. In undamped cases, this choice leads to an unconditionally stable time integration with good accuracy. Choosing g ¼ 0.5 and b ¼ 1/6 results in the
linear acceleration method, because the third time derivative of x is zero. Lastly, with
Table 8.4 Displacement, Velocity, and Acceleration for the Second and Third DOFs Under
Forced Motion Solved by the Newmark Method
Dt
2Dt
3Dt
4Dt
5Dt
6Dt
7Dt
8Dt
9Dt
x2
x3
x_2
0.290
0.009
2.321
1.078
0.067
3.985
2.145
0.256
4.551
3.206
0.674
3.938
4.004
1.391
2.448
4.392
2.412
0.651
4.372
3.661
0.809
4.091
4.988
1.436
3.780
6.206
1.057
x_3
x€2
x€3
0.070
0.394
1.119
2.224
3.511
4.659
5.332
5.286
4.455
8.567
4.742
0.214
4.684
7.241
7.135
4.538
0.481
3.512
0.563
2.023
3.779
5.065
5.228
3.960
1.422
1.793
4.852
376
CHAPTER 8 Modal and Transient Dynamic Analysis
g ¼ 0.5 and b ¼ 0, the method is, in essence, the same as the central difference
method.
Many other direct numerical integration methods are available for solving dynamic equilibrium equations. For example, the Houbolt implicit method is used to
approximate the velocity and acceleration in terms of displacement in a similar
manner to the central difference method (Houbolt, 1950). The difference between
the two methods is that the equilibrium is considered at tnþ1 instead of tn used in
the central difference method. Thus, the Houbolt method is an implicit method
while the central difference is an explicit method. Also, we did not discuss the Bathe
method, which combines the Newmark and Euler backward methods (Bathe, 2007).
We did not speak about the Wilson-q method, which is a modified version of the
Newmark method based on the linear acceleration assumption (Wilson et al.,
1973). Neither did we talk about the HilbereHugheseTaylor (HHT) method, which
is an implicit method that allows for second order accuracy (Hughes, 1983). These
methods are important and offer many advantages, but they are beyond the scope of
this textbook.
Importantly, we did not discuss numerical stability. While this is a critical topic,
stability is highly dependent upon the particular problems being solved, and the
topic is too complex to be covered in this book on basics. Research studies specifically designed to explore the intricacies of this topic would be good sources for
more information. Despite the complexity of specifics on this topic, we can discuss
basic concepts. An integration method is considered “unconditionally stable” if the
solution does not grow without bound for any time-step size. The method is deemed
“conditionally stable” if the step size needs to be within a certain limit. By choosing
different magnitudes for various parameters used in numerical integration schemes,
results could be widely different in cases where instability exists. Even if solutions
are stable, we observe slightly different results between the central difference and
Newmark methods, as shown in Tables 8.3 and 8.4.
8.6 IMPLICIT AND EXPLICIT SOLVERS
In static and quasi-static engineering analysis problems, in which the global stiffness
matrix is not altered due to increasing displacement, the equilibrium equation [K]
{x} ¼ {f} can be solved by multiplying [K]1 to both sides of the equation, using
the Gauss elimination method, or applying iterative procedures as discussed in Section 7.2. In these cases, the final nodal displacements can be directly calculated. For
nonlinear problems, we can gradually increase the loading intensity and calculate
intermediate nodal displacements. For dynamic problems containing only low frequencies, the displacement at each time instant (t1, t2, ., tn) is considered constant;
that is, not a function of time. For these problems, the velocity and acceleration are
zero, and we can neglect the effects due to mass and damping. These types of problems can be solved by using implicit schemes. Most commercially available FE
8.6 Implicit and Explicit Solvers
software packages, such as ANSYS, LS-DYNA Implicit, NASTRAN, and OptiStruct, provide implicit solvers.
In high-speed vehicular crashes, contact penetration, and ballistic and blast problems, large displacements occur in extremely brief periods of time. For these scenarios, transient dynamic analysis is required. Velocity and acceleration are no
longer zero and need to be explicitly calculated. Commonly used explicit FE solvers
include Abaqus, LS-DYNA, Pam-crash, Radioss, etc.
The terms “explicit” and “implicit,” as used in relationship to the FE method, are
specifically designated to be associated with numerical schemes used for integration
over time. They reference the process of determining displacement, velocity, and acceleration through iterations in which the solution at a prior time step is used to
obtain the solution at the next step. For implicit integration schemes, the dynamic
equilibrium equation is assessed at the next time step tnþ1, while explicit integration
methods evaluate the equilibrium equation at the current time step tn. Lastly, the implicit solver often employs the Newmark, Newton Raphson, Wilson-q methods, etc.,
while the explicit solver commonly uses the central difference method. In this section, subtle differences between the implicit and explicit iteration schemes are
briefly discussed.
8.6.1 IMPLICIT SOLVER
An implicit solver requires the formation of a global stiffness matrix [K]. In static problems, the equilibrium is maintained when [K]1 is multiplied to both sides of the static
equilibrium equation. After that, nodal displacements can be easily calculated. One
disadvantage of carrying out this operation is that large sizes of storage space and
in-core memory are needed to find [K]1. Additionally, nonlinearity causes [K] to
become a function of displacement. As such, updating or factorizing the global stiffness matrix [K] before inverting it to [K]1 for computing the nodal displacements
needed before the next time step can be proceed is computationally intensive.
For dynamic analysis, we need the global [M] and [C] matrices. However, [C]
cannot be easily obtained, and hence equivalent Rayleigh damping or ignoring
damping are commonly used to solve the problem. For the implicit solution, the current quantities (displacement, velocity, or acceleration) are determined at time tnþ1,
and these quantities are calculated from the same quantities already computed in the
previous time step tn. For example, the Newmark method shown in Section 8.5.2 is
used to calculate Dx(tn) (the displacement increment between two time steps) prior
to computing the displacement at the next step x(tnþ1) from x(tn). Again, factorizing
the global stiffness matrix [K] before inverting it to [K]1 for computing the next
time step is computationally intensive.
In the implicit time integration scheme, the solution in each discrete time step is
unconditionally stable because global equilibrium is achieved at each iteration.
Since equilibrium is maintained at all time, the step size may be chosen without
many restrictions. As such, implicit solution schemes typically involve a relatively
small number of time steps, but results are rather accurate.
377
378
CHAPTER 8 Modal and Transient Dynamic Analysis
8.6.2 EXPLICIT SOLVER
_ and external
For explicit solvers, both the element internal force ([k]{x} and ½cfxg)
force {f} are summed in the right-hand side of the dynamic equilibrium equation, as
discussed in Section 8.5.1, before the computation is initiated. Because the mass matrix is diagonalized, accelerations can be easily computed by dividing the sum of the
forces acting on the nodal mass (mii). In doing so, the explicit integration scheme
unavoidably includes inertial effects, and hence this scheme is good for transient dynamic analysis. The acceleration obtained is then used to acquire the displacement
by using methods such as the central difference method.
Because the global stiffness matrix is not required, the maximum time-step size
must satisfy the CouranteFriedricheseLevy condition to ensure that dynamic equilibrium is satisfied. As there is a maximal allowable requirement in the time-step
size, all explicit integration methods are “conditionally
stable.” For a 1D problem,
pffiffiffiffiffiffiffiffi
the wave propagation velocity is known to be c ¼ r=E, where r is the density
and E is Young’s modulus. Hence, the time needed for the wave to propagate through
the 1D element of length L is Dt ¼ L=c. The same idea is applied for the determination of the time-step size for 2D or 3D elements. After time-step sizes for all elements in an FE models are determined, the shortest time step would be the time step
of the entire model. To ensure acceptable accuracy, the time step is usually set to
90% of the shortest time step.
Because the size of the time step needs to be extremely small (about 1 ms in
typical car-crash simulations), an explicit integration algorithm typically requires
numerous time steps. Despite the fact that the calculation within each time step is
inexpensive, use of this type of algorithm is computationally expensive because
the computation needs to be carried through all elements. Running simulations on
models with a large number of elements is especially expensive. Also, while it is
possible to solve transient dynamic problems on a computer with limited core memory, as there is no need to create the global stiffness matrix, this is often not practical.
Since numerous time steps are needed (even for solving a transient dynamic problem
with an extremely brief time period), and there is a large number of elements in a
detailed FE model, a massively parallel-processing computer is frequently required.
Instead of solving for the displacement {x(tn)} in implicit integration schemes,
€ n Þg is computed in explicit integration schemes. As there is no
acceleration fxðt
more need to invert the stiffness matrix, inverting the diagonalized mass matrix
[M] is an easy, trivial task. However, using explicit modeling for solving a transient
dynamic problem requires far more understanding of the FE method as compared to
implicit modeling. It is highly advantageous to become familiar with implicit solvers
prior to using an explicit solver.
8.6.3 USING FE SOLVERS
Some questions are frequently raised regarding why students need to learn the
fundamental aspects of FE formulations. As learning how to use FE software
8.6 Implicit and Explicit Solvers
packages often leads to increased career opportunities, some students feel that there
is no need to learn the background information regarding how FE solvers are written.
Regrettably, creating an FE model without proper knowledge may lead to choosing
an incorrect element type, creating a poor-quality mesh, selecting a wrong option for
the material law, picking an erroneous time iteration scheme, etc. Inputting such a
questionable FE model into a “black-box,” such as an FE software package, can
lead to undesirable “garbage in, garbage out” results.
Although damping was deliberated, we did not discuss different material laws
used for modeling viscoelastic materials. All materials deform under load. Upon
removal of the force, an elastic material will return to its original shape and size.
For a viscous material, the material property is determined by the resistance to
flow. For example, honey is more resistant flowing than water, and as such, the viscosity of honey (approximately 10,000 cP at room temperature) is much higher than
that of water (1 cP). As a result of resistance to flow, it takes a very long time for a
material with high viscosity to recover the original shape after loading is removed.
A viscoelastic material possesses both elastic and viscous characteristics when
undergoing deformation, and it possesses three basic phenomena: creep, stress relaxation, and hysteresis. As water accounts for up to 65% of an adult human, most human tissues and organs are characterized as viscoelastic material. Upon loading,
water in intervertebral discs and articular cartilages moves out of the tissues. These
tissues act like shock absorbers when loaded quickly and can slowly deform due to
static loading. Similarly, many types of rubber, foam, and polymers possess viscoelastic properties. Taking advantages of their energy absorbing capabilities, viscoelastic materials are used to dampen the magnitude of noise, isolate sources of
vibration, reduce impact to the head protected in foam padded helmet or in vehicle
interior trims made of energy absorption polymers, absorb shock when a car bumper
hits a telephone pole, protect a building from an earthquake shock, etc.
Modeling viscoelastic materials is one of the most difficult tasks in dynamic
FEA. Unfortunately, no universal guidelines are available to direct software users
on how to model viscoelastic material. For many viscoelastic materials, the stresse
strain relationships are so complex that simple constitutive equations cannot be
easily derived from experimental results. In these cases, various numerical approximation schemes are reported for describing the material laws. Sometimes, tabulated
experimental data are used directly in software instead of constitutive equations.
Detailed discussion of different ways to model viscoelastic materials is out of the
scope of this current book.
Choosing a proper integration scheme is not an easy task. As mentioned, implicit
analysis is recommended for static and low-frequency dynamic analyses, while an
explicit solver is recommended for car-crash and blast analyses. Sometimes, a combination of both implicit and explicit modeling on the same structure is necessary.
For example, an implicit code is used to determine the static crush strength of a
car roof, while an explicit code is applied to simulate a rollover event, in which
the roof is dynamically deformed. If we need one model each for implicit and
379
380
CHAPTER 8 Modal and Transient Dynamic Analysis
explicit analysis, the model development cost would be high and there would be no
guarantee that the two models would be consistent with each other. For example, to
avoid the need for extraordinary requirements in computer memory and storage
space, a coarse mesh FE model is desired when using the implicit scheme. On the
other hand, to better capture deformation modes, a detailed model is commonly
used for explicit solutions. Extra effort is needed to develop FE models that can
be adequately used with both implicit and explicit solvers.
Finally, we need to stress again that explicit analysis requires a lot more understanding than implicit analysis. For explicit analysis, issues related to the minimum
time step alone require a lot more knowledge than we can describe in this limitedscope book. For example, can we increase the time step by mass scaling or reduction of the stiffness without losing accuracy? Is there an appropriate occasion to
alter the time step? For these reasons, it is once again highly recommended that
you understand the use of implicit solvers before attempting to perform explicit
analysis.
EXERCISES
1. In Section 8.2.1, the integration of Eqs. (8.8) and (8.9) is done by using
4-point Gauss quadrature. Redo the entire integration using 1-, 2-, 3-, and
5-point Gauss quadrature and report any discrepancies observed.
2. Create lumped sum matrices using both the HRZ method and Row Sum
method for (1) 1D, 2-node bar element; (2) 2-node beam element (3) 3-node,
2D triangular element; (4) 4-node, 2d plane element; and (5) 3D, 8-node
solid element.
3. A spring system has three springs and two masses. Spring 1 connects to a wall
and Mass 1. Spring 2 connects the same wall and Mass 2. Spring 3 connects
Mass 2 to Mass 1. Mass 1 is 1 and Mass 2 is 2. The stiffnesses of Springs 1, 2,
and 3 are 3, 6, and 9, respectively. Analytically, find the smallest natural
frequency and eigenvector of the system for free vibration.
4. Use the situation in Problem 3, but apply a 5 cosðutÞ force to Mass 1. If the
applied force has a frequency of 7 Hz, what is the maximum displacement of
each node?
5. Create an Excel sheet or other program to calculate the Rayleigh quotient for
any system with up to five nodes. Check your answer with the values in
Table 8.2.
6. Write a program to perform the matrix iteration method to find the first and
second natural frequencies and associated eigenvectors.
References
7. Using the Jacobi method, find the first three natural frequencies and
eigenvectors of the dynamic matrix below.
2
5
4 6
6
2 4
6
½D ¼ 4 4
3
7
8 25
8. The time between the two consecutive peaks is 0.125 s of a damped vibration
and peak values are 475 and 264. Find the Rayleigh damping coefficients
best suited for this experimental system for damping between u1 ¼ 0.25 and
u2 ¼ 17.3. Also, plot the damping ratios over the frequencies.
9. Using the central difference method, write a program to calculate the
displacementetime histories of a system. To simplify the problem, apply a
constant force. Check your program with Example 8.9.
10. Write a program or create an Excel sheet that performs the Newmark method
to calculate the displacements, velocities, and accelerations of any dynamic
equilibrium equation. Check your answer with Example 8.10.
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Wilson, E.L., Farhoomand, I., Bathe, K.J., 1973. Nonlinear dynamic analysis of complex
structures. Earthquake Engineering and Structural Dynamics 1, 241e252.
PART
Modeling Human
Body for Injury
Biomechanics
Analysis
2
Saving lives is the noblest thing a safety engineer could ever do. Finite element
human modeling could help you get there.
Edited by
King-Hay Yang, PhD
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Introduction
Worldwide, road traffic incidents claim the lives and cause injuries and disabilities to
staggering numbers of people. According to the World Health Organization (WHO),
there were 1.25 million such deaths and more than 20 million injuries or disabilities
in 2013. Bhalla et al. (2011) reported that road traffic crashes contributed more than
2% to global mortality and morbidity. Despite much smaller vehicle fleets in most
low- and middle-income countries, road traffic death rates were 2.1e7.5 times those
seen in the Netherlands, Sweden, and the United Kingdom, as shown in Figure II-1.
Among high-income countries, the death rate in the United States in 2010 was
more than double the rates seen in the three countries with the best road-safety
performance, according to the Bhalla report. Tangible costs, which included
lost productivity, workplace losses, legal and court expenses, medical costs,
emergency medical services, insurance administration costs, congestion costs,
and property damage costs, were estimated to exceed $242 billion in the year
2010 (the most recent year for which cost data is available) in the United States
(NHTSA, 2017). When intangible valuations, such as those related to quality of
life, were taken into consideration, the total societal HARM1 resulting from
motor-vehicle crashes reached $836 billion. Reducing the human and economic
costs associated with road traffic injuries is a crucial need that calls for exceptional
intervention.
Crash-test dummies or anthropomorphic test devices (ATDs) are human
surrogates designed to mimic the behaviors of human bodies in motor-vehicle
collisions. In ongoing safety testing and simulation paradigms that are based on
crash-test dummies seated in or standing outside of vehicles, all available dummies
are designed only for testing unidirectional responses. These dummies have no
internal organs for directly estimating risks of injury to these vital components.
The crash dummies are also limited in terms of age, gender, and size representations.
Accelerometers, displacement transducers, and load cells can be mounted inside the
crash dummies for measuring head and spinal accelerations, chest deflections along
the anterioreposterior and lateral directions, and forces in the femur, pubic symphysis, etc. Additionally, deformation patterns in the rib cage can be measured by using
a chest band, which consists of multiple optical fiber sensors or strain gauges, to
depict deformation contours during impact. From these electronic data, the risks
of injury to different body regions, not individual organs, can be estimated.
In contrast, finite element (FE) human models could potentially afford
predictions of location-specific bony fractures, ligamentous injuries, and internal
1
Here the metric HARM was initially proposed by Malliaris et al. (1982) as a tool for accounting the
societal costs by considering the number of injury incidents, types and severities of injuries, and cost
of each injury. Using this metric, each AIS code has prescribed medical and indirect costs (such as
loss of wages).
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Introduction
FIGURE II-1
Composite plot of road traffic death rates in 18 countries compared to the Netherlands,
Sweden, and the United Kingdom.
Figure reprinted from Bhalla, K., Sharaz, S., Abraham, J., Bartels, D., Yeh, P.-H., 2011. Road injuries
in 18 countries. Department of Global Health and Population, Harvard School of Public Heath, Boston,
MA, USA, with permission.
organ contusions or ruptures that might occur anywhere within the body. If properly
developed and validated, human modeling using the FE method may shift the
paradigm of advancements in safety from dummy-based crash-tests to numerically
based, human-centered crash simulations. Such validated models may be morphed
to exemplify different ages, genders, and anthropometries to represent the entire
population for omnidirectional impact simulations. Because FE models employ
tissue-level injury tolerances to calculate risks of injury, the locations and intensities
of injuries could be more accurately predicted. With better understanding of human
impact responses through the use of FE human models, safety engineers may
develop new countermeasures to better protect people who use roadsdfrom
pedestrians, pedal cyclists, motorcyclists, to vehicular occupants.
As crash dummies are used worldwide in the automotive industry as assessment
tools for evaluating automotive safety since early 1970s, transitions to a new paradigm of relying totally on computer simulations would be painful and costly. Many
evaluation tools and guidelines have already been developed by each original equipment manufacturer (OEM) in accordance with these dummies, so evaluating dummy
performances could be done easily and rapidly. By switching to numerical modeling,
OEMs would need to develop a new set of tools and would face increasing workloads, as there are many more response variables generated by human body models
than the handful of injury metrics provided by crash dummies. Despite that, the age
of simulation-based automotive safety rating is coming sooner than most people had
predicted. For example, the EuroNCAP is now accepting numerical simulations for
estimating the risks of pedestrian injuries.
Introduction
Considering that the future of car safety is not for dummies, in Part II of this book
we describe procedures needed to develop human models, common practices for
model validation, and limitations for models of various regions. Also included in
Part II are descriptions of modeling techniques related to predicting precrash muscle
responses, using automated procedures for morphing standard models into
population-specific models (parametric modeling), and modeling of vulnerable subjects. Finally, with increasing need for understanding combat-related injuries, fundamentals of blast modeling are presented.
REFERENCES
Bhalla, K., Sharaz, S., Abraham, J., Bartels, D., Yeh, P.-H., 2011. Road Injuries in 18
Countries. Department of Global Health and Population, Harvard School of Public Heath,
Boston, MA, USA.
Malliaris, A.C., Hitchcock, R., Hedlund, J., 1982. A search for priorities in crash protection.
In: Crash Protection, SAE SP-513, Society of Automotive Engineers, SAE Technical
Paper 820242, pp. 1e33. http://dx.doi.org/10.4271/820242.
NHTSA, 2017. Traffic Safety Facts, Summary of Motor Vehicle Crashes. DOT HS 812 376.
US National Highway Traffic Safety Administration, 1200 New Jersey Ave. S.E.,
Washington, DC 20590.
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CHAPTER
Developing FE Human
Models From Medical
Images
9
Anil Kalra
Ford Motor Company, Dearborn, Michigan, United States
9.1 INTRODUCTION
This chapter discusses the general methods used for developing high-quality finite
element (FE) meshes for accurate modeling of biological systems. Just like any FE
models, accurate geometric information and mechanical properties are needed to
ensure that the FE models developed to represent a biological system are biofidelic.
With the advancement of high-resolution medical imaging techniques, it is now
possible to capture accurate geometric details of most different human body parts
without meticulous dissections. Different image scanning modalities, such as
computed tomography (CT or CAT) or magnetic resonance imaging (MRI), enable
the acquisition of two-dimensional (2D) images of thin slices throughout the regions
of interest (ROIs) in a body. Local adaptive thresholds can be applied to each 2D slice
for segmentation of different body organs. Multiple images from the adjacent slices
can be rendered together to get a 3D volume of the segmented region.
This chapter begins with a brief introduction for different medical imaging
modalities including X-ray, CT, MRI, PET (positron emission tomography), and
ultrasound imaging. This section also discusses the rationale for using different
modalities for different applications and the relative importance of using each
type of modality for segmentation of ROIs.
Different procedures involved in generating a human body model (HBM) can be
divided into three broad categories: pre-mesh, mesh, and post-mesh planning. The
pre-mesh planning includes steps which require coregistration of medical images,
image segmentation, surface smoothing, and 3D computer-aided design (CAD) surface extraction of different body parts or ROIs. Also, the physics behind segmentation of medical images from different medical modalities along with all these
processes involved during pre-meshing phase are explained in detail. An overview
of different kinds of free and commercially available software packages is provided
along with their capabilities to perform medical image analysis.
Once the highly accurate CAD models representing different body parts or
regions are available, further steps to generate high-quality meshes of human
body regions are explained in the next phase. The FE mesh for CAD geometries
or for surface entities such as NURBS, B-rep, STL, or point cloud can be
Basic Finite Element Method as Applied to Injury Biomechanics. http://dx.doi.org/10.1016/B978-0-12-809831-8.00009-X
Copyright © 2018 Elsevier Inc. All rights reserved.
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constructed by means of well-devised and widely tested computer programs. Other
types of mesh-generation approaches (George, 1992; Ho-Le, 1988; Lo, 2014;
Young et al., 2008) using different algorithms (Delaunay triangulation, BowerWatson, etc.) have been developed based upon the type and size of elements
(tetrahedral, brick, 2D quad, or triangular) required in the model. For 3D volume
meshing, the generation of tetrahedral elements is relatively easy, as it requires
connecting different vertices to form gross elements. Although, brick elements
are computationally more accurate, they require more efforts to acquire the
topology of the complicated CAD surface before such elements can be formed.
The difference between these types of meshes and the pros and cons of generating
these meshes directly from segmented CAD surfaces rendered from medical images
are also explained in this chapter. The most widely used technique in creating highquality hexahedra mesh for advanced HBMs called “multiblock meshing” (Armstrong et al., 2015) is further explained with suitable examples as well.
After generating a preliminary mesh, it is important to perform element quality
checks so that a robust solution can be guaranteed from the FE model. The chapter
explains some of the common diagnostic checks for good quality mesh and the suggested threshold for each quality check parameter. Further assignment of accurate
material properties as per constitutive material laws for the representative tissue
type, such as bones, soft tissues, and ligaments, should be followed to exhibit stiffness characteristics for these tissue meshes. Additionally, the contact algorithms
used to transfer the momentum or the force between unmerged Lagrangian meshes
are also discussed. Further suggestions are included to perform some pre-checks
while assigning the different contact definitions between these unmerged mesh components which represent different tissue behaviors during finite element analysis. To
conclude, an example using different procedures and tools for generating a full-body
FE model of a 70-year-old female is provided.
9.2 BIOMEDICAL IMAGES FOR FINITE ELEMENT MESH
DEVELOPMENT
9.2.1 X-RAY IMAGING
X-ray imaging is one of the oldest transmission-based techniques which uses ionizing
radiation to take 2D images of exposed tissues by sending X-ray’s beams through
ROIs. These X-ray beams are absorbed in different amounts depending on the density
of the material through which it is passing. These densities are expressed using the
Hounsfield (HU) scale values, named after Sir Godfrey Newbold Hounsfield (Aug.
1919eAug. 2004), as shown in Fig. 9.1. The values are measured based on a scale
where radio density of water at standard pressure and temperature is defined as zero
HU, while radio density of air on the same scale is defined with 1000 HU. The formula for calculating HU value based on this scale is shown in Eq. (9.1).
m mwater
HU ¼ 1000 ;
(9.1)
mwater mair
9.2 Biomedical Images for Finite Element Mesh Development
FIGURE 9.1
Hounsfield number for different human tissues.
where m ¼ original linear attenuation coefficient of substance, mwater ¼ linear attenuation coefficient of water, and mair ¼ linear attenuation of air. In a typical
transmission-based imaging technique, the emitted X-rays pass through ROIs, which
affect the level of energy reaching to the detector, depending upon the type of tissue
or organs encountered along the path. Tissues having low Hounsfield scale values
darken the film such as lungs and fat, while tissues having higher Hounsfield scale
values lighten the film and provide white spots such as hard bones. Therefore, the
denser the tissue, the brighter the image will be as the detector will return weak signals. Similar tissues can provide different X-ray images depending on the hardness
or penetrating ability of X-rays, which is adjusted by selecting the voltage for the
emitter. But longer emissions of these X-rays can be harmful as they ionize the biological tissues. The most common clinical uses of X-rays are detecting fractured
bones, dental cavities, swallowed objects, and breast mammography. Conventional
X-ray techniques provide flattened 2D images which cannot be used to generate 3D
segmented regions of tissues.
9.2.2 COMPUTED TOMOGRAPHY IMAGING
CT, or computer axial tomography (CAT), uses multiple X-ray projections taken
from different angles to produce detailed cross-sectional images of ROIs. Similar
to X-ray imaging, when these beams pass through different dense tissues, it gets
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weaker after absorption and the final intensity is measured by the detectors providing
different contrast imaging of tissues. Since the same cross section is scanned at
different angles, it can be used to reconstruct a 3D image of the tissues. By adding
more than one detector, the processes can be faster and more accurate. CT imaging is
commonly used to diagnose presence of tumors, colon cancer, bone injuries, and internal bleeding.
9.2.3 MAGNETIC RESONANCE IMAGING
Magnetic resonance imaging (MRI) generates cross-sectional images of the tissues
by using a strong magnetic field to magnetize protons within the tissue. The basic
principle of MRI imaging is to align the nuclei of atoms which have a spin and
exhibit magnetic moment. Magnetic field intensity lies between 0.1 and 3.0 T, a
standard unit of magnetic flux density named after Nikola Tesla (Jul. 1856eJan.
1943). The image contrasts are achieved by using different pulse sequences and
by changing the imaging parameters relative to longitudinal relaxation time (T1)
and transverse relaxation time (T2). The signal intensities on the T1- and T2weighted images correspond to specific tissue characteristics. Irrespective of the difference between T1- and T2-weighted images, proton density weighing is also used
to get contrast between soft tissue images. The major difference between MRI and
other transmission-based techniques, such as CT and X-ray, is that MRI is based on
signals sent by the tissue while the others use an external source such as X-rays to get
the contrasts. With CT, detailed anatomical details can be achieved, while with the
MRI tissues having different biological functions can be distinguished more clearly.
MRI imaging is generally used to evaluate abnormal tissues, spinal injuries, brain
abnormalities, tendon or ligaments tears, etc.
9.2.4 POSITRON EMISSION TOMOGRAPHY
Positron emission tomography (PET) is a nuclear imaging technique which uses a
dye-like substance having radioactive tracers to identify cellular level changes in
the tissue. Computer analysis of tracer concentration in the tissue along with a CT
scan helps in generating a 3D image. PET is primarily used in clinical oncology
for medical imaging of tumors, and in the diagnosis of neurological diseases, e.g.,
Alzheimer’s and multiple sclerosis, as it is able to collect diagnostic information
which cannot be acquired by other methods. This imaging modality is sparingly
used in FE model development as it is quite expensive.
9.2.5 ULTRASOUND IMAGING
The principle behind ultrasound imaging is to record the reflection of sound waves
penetrating through different tissues and reflecting back from boundaries of structures
having different densities and velocities of sound-wave propagation. The velocities of
ultrasound waves vary in different substances as shown in Fig. 9.2. The contrast
9.2 Biomedical Images for Finite Element Mesh Development
FIGURE 9.2
Speed of ultrasound waves in different tissues.
images of different tissues are generated based on these sound-wave propagations.
Depending on the reflection-echo quality, ultrasound images may be noisy and
have spatial deformities. There are different methods to get these contrast 2D
images such as B (brightness)-mode, M (motion)-mode, or D (Doppler)-mode.
Ultrasound imaging is mainly used in fetal scans during pregnancy, and to evaluate
symptoms of pain, swelling, or infections. The contact between the transducer and
the adjoining surface is very crucial and refined by using gel-like substances or
water-filled plastic bags.
Although large numbers of medical modalities are available, the task of segmenting the detailed geometry of all human organs, such as bones, soft tissues, muscles,
and veins, through medical images provided by different modalities is not straightforward. One single imaging technique cannot be used to acquire all 3D geometry
information due to the limitations associated with each technique. For example,
CT images can be used mainly for segmenting bone, since they have a unique spectrum in terms of Hounsfield scale values which provides white shades during a scan.
The similar attenuation of the signal is not provided by soft tissues, which will show
up as several overlapping grey spectrums of fat, skin, muscles, etc. The CT, or
Hounsfield, number described as the density are assigned to a voxel in a CT scan
on an arbitrary scale on which water has density 0, air 1000, and compact
bone þ1000 (Fig. 9.1). Most of the soft tissues have a CT number closer to water;
therefore the boundaries of adjacent soft tissues cannot be easily separated in CT
scan.
MRI technique can provide excellent soft-tissue contrast as it uses a magnetic
field which aligns orientations of protons (nuclei of hydrogen atoms) and is abundant
in water and fat. Therefore, in general, 3D geometry of bones is segmented through
CT scan images, while the soft tissues, such as heart, lungs, and abdominal organs,
are segmented with the help of MRI images where the distinction between the
boundaries of various soft tissues can be made relatively sharper. Other techniques
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such as PET, single-photon emission computed tomography (SPECT), contrast CT,
or contrast MRI can also be used to distinguish the boundaries of adjacent soft tissues more clearly, if required.
9.3 PHYSICS BEHIND 3D SEGMENTATION OF MEDICAL
IMAGES
In a broad sense, segmentation can mean to divide the image into a group or patch of
information related to intensity, color, or other attributes, which are homogenous in
nature. These attributes or elements are called pixels (where “pix” stands for picture
and “els” for elements) in 2D space and voxels (where “vox” stands for volume and
“els” for elements) in 3D space as shown in Fig. 9.3.
Medical images can be considered a cluster of voxels in 3D space where each
voxel has a certain attribute, such as CT value or HU (Hounsfield units) value,
associated with it. In a typical image-segmentation procedure, whether manual or
using an algorithm, voxels can be subgrouped for a range of values which will
provide the 3D representation for that range. A preset threshold value can be
assigned to generate a volumetric region in different software packages, which are
used to segment 3D images obtained from medical scans. This preset threshold value
can be different for bones and soft tissues.
FIGURE 9.3
Elements of a brain based on voxel images taken from MRI scan.
Reproduced from Despotovic, I., Goossens, B., Philips, W., 2015. MRI segmentation of the human brain:
challenges, methods, and applications. Computational and Mathematical Methods in Medicine, an open source
literature.
9.4 Meshing Human Body
In a human body, there are different types of organs (e.g., bone, muscles, ligaments, internal soft tissues) which have different attributes from macroscopic (material, structure, bone architecture, etc.) to microscopic (molecular compositions,
collagens, fibers, etc.) levels. Different medical imaging modalities use different approaches (based on physics) to capture these variations to provide distinguishing
voxel-based images for each organ. The efficiency of each technique can be categorized based on its applications. For the current application, it is important to capture
the details regarding the geometry of different body regions which makes it mandatory to separate the boundary layers of two adjacent tissues. However, it is very difficult to distinguish the relative difference of two adjacent pixels or voxels, especially
at the interface of two adjoining tissues. As a result, many pixels may contain one or
more tissue elements associated with it, which is known as a partial volume effect.
Also very often the noise-to-signal ratio is approximately 10% in medical images,
which makes differentiating the barrier between two adjacent tissues even more
difficult. The complexity is exacerbated by the uneven shapes of ROIs.
The pixel or voxel assignment for two neighboring candidates is always probabilistic and uncertain. The probability of wrong tissue assignment increases when
lower resolution values are used for scanning the ROIs. Therefore for detailed and
accurate geometry ROIs should be scanned with relatively higher resolution, such
as capturing of trabecular bone architecture requires microcomputed tomography
(microCT) scanning. Segmentation of 3D CAD surfaces with similar minute details
for whole body skeletal requires use of microCT scanning and it is a very difficult
and time-consuming task. Although such scanning, capturing miniature surface topology and the corresponding FE model generation using very small element sizes is
possible, it will limit its applications for analysis of individual body organs rather
than for full human body FE analysis. For generating CAD images during FE model
development of the human body, CT and MRI images are used most commonly for
bone and soft-tissue segmentation, respectively, with best possible resolution in the
range of 1 1 1 mm.
9.4 MESHING HUMAN BODY
Fig. 9.4 shows a schematic diagram of different procedures involved when developing an FE HBM. Based on the different nature of the work, the entire model
development process is divided into pre-mesh, mesh, and post-mesh segments. In
the pre-mesh segment, emphasis is placed on the image-processing technique. For
the mesh segment, the techniques for developing FE meshing from CAD surfaces
processed through medical images are explained. For the post-mesh segment,
considerations of mesh quality and completions of input data deck are discussed.
Some early FE human models took model geometry from commercial databases,
such as those available in the Viewpoint Data Lab. However, these geometric data
are collected for the purposes of 3D animations in movies and video games and
hence lack the anatomical details for injury simulations. In 1986, the National Library of Medicine (NLM) initiated a Visible Human Project, through a willed
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CHAPTER 9 Developing FE Human Models From Medical Images
FIGURE 9.4
Procedures in the development of a human body FE model.
body program. Anatomically detailed, 3D CT and magnetic resonance imaging
(MRI) images of a male and a female whole body were acquired. Additionally, cryosectional images sectioned at 1 mm intervals for male, and one-third of a millimeter interval for female were attained. These Visible Human data were used in
the creation of several FE human models. In more recent human models, clinical
CT and MRI data are utilized for segmenting bony and soft-tissue geometries for human model development. The following section explains each step for these most
recent model generations in three different segments.
9.4.1 PRE-MESH SEGMENT
This section explains the procedures needed for capturing accurate geometry of
different tissues or ROIs. Most of the time, FE modeling of geometric entities requires a closed volume or a surface, further divided into nodes and elements, which
is called mesh. Meshing of ideal geometric entities such as cube, cylinder, and
sphere, can be directly created in preprocessing software used for completing
mesh-generation tasks. However, meshing of nonuniform 3D volumes such as human tissue requires extra effort so that the generated elements can capture the
required shape or geometry of these tissues. Therefore, retrieving the 3D CAD surfaces or volumes having geometric details and accurate shape of human tissues is the
most important task. This task for retrieving geometric surfaces from scanned medical images is the “pre-mesh segment” of the complete set of procedures used in
developing a HBM. The different steps involved in this pre-mesh segment are
described in detail in following subsections.
9.4 Meshing Human Body
9.4.1.1 Image Registration
The first step of the procedure is to import and register the medical images available in different formats, most commonly in the digital imaging and communications in medicine (Dicom) format, in the image-processing software. In a broad
sense, during the image-registration process, the images are aligned in the
image-processing software based on features/landmarks or voxel attributes in
different slices of medical scans. The 2D images/slices are taken on planes which
have distance/thickness associated with them. Therefore, the pixels of each slice
have “A” attributes need to be mapped to the pixels on the next slice which
also have the same “A” attributes. These attributes are unique characteristics of
different human tissues. Along with these similarities, the minimized distance
depending on the slice thickness, which also guides the image-registration process
to find the similar spatial attributes in the adjacent images. The main aim of the
image-registration process is to configure corresponding anatomical locations in
adjacent images of series of medical scans of a modality. Although, the process
can be applied to images retrieved through different modalities, multimodal image
registration, its measurement applications are limited to single-modality registration especially in 3D CAD model retrieval applications for FE mesh generations.
The major application of image registration in medical modalities is to find the
suspected abnormalities in the images in terms of unusual spatial intensities,
such as tumor, cancer, or other inflammatory diseases.
There are different methods, techniques, and algorithm features used in imageprocessing software to register the images from similar modalities (Brown, 1992;
Maintz and Viergever, 1998; Wyawahare et al., 2009). Registration is generally
achieved by an optimized similarity function such as a correlation coefficient or
by minimized absolute difference between two spatial properties. There are several
ways to register the images based on the level of automation (manual, interactive,
semiautomatic, and automatic). As suggested, manual methods allow the user to
orient the images based on some landmarks or control points, while interactive
methods make the problem less complex by performing some operations but still
need user input to finalize the registration. Semiautomatic method provides automated registration of images but still requires user approval to finalize it. Automatic
registration method completes the registration by itself with no user interaction.
When more than one modality gets involved during the analysis of images, the automation level decreases and more user-dependent interactions are needed to orient the
images properly. With the advancement of computer algorithms and imageprocessing techniques, the image-registration process has been fully automated in
most of the software used to segment medical images.
9.4.1.2 Image Segmentation
After registering the images, segmentation is the most important step for generating
an accurate 3D CAD model from medical scan images. Generally, segmentation
means dividing into separate parts or segments. During medical image segmentation, different tissues having similar image properties (threshold, contrast, or HU
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CHAPTER 9 Developing FE Human Models From Medical Images
FIGURE 9.5
Bone threshold used during segmentation procedures to generate 3D human skull from
CT images.
values) are divided into different segments having similar volume in 3D space as of
that tissue. For example, in a CT scan, the bones will have separate thresholds and
can be segmented from medical images, and further, can be stacked together on top
of each other in order to retrieve 3D volume for that bone. Fig. 9.5 shows an example
for generation of a 3D CAD model of a human skull using bone threshold through
CT scans.
Due to currently available advanced algorithms, the overall segmentation process
is automated in most of the software programs used for analyzing medical images
for various applications. Although the spatial elements and resolutions are different
for images retrieved through different modalities, a wide variety of segmentation
techniques are available to tackle these variances. These techniques can be classified
into different categories, such as measurement spaceeguided spatial-clustering,
single-linkage region growing schemes, hybrid-linkage region growing schemes,
centroid-linkage region growing schemes, spatial-clustering schemes, and splitand-merge schemes (Haralick and Shapiro, 1985; Wyawahare et al., 2009). These
schemes and techniques can be further used in different methods available in
image-segmentation software and can be additionally classified as:
•
•
•
manual segmentation
intensity-based methods (including thresholding, region growing, classification,
and clustering)
atlas-based methods
9.4 Meshing Human Body
•
•
surface-based methods (including active contours and surfaces, and multiphase
active contours)
hybrid segmentation methods
Further review and details of these methods are explained in Pham et al. (2000),
Zhang (1996), and Despotovic et al. (2015). These techniques used different algorithms, such as edge relaxation, Hough transform, or artificial neural network
(ANN)-based algorithms to capture the details in terms of different tissue’s properties. All these algorithms provide the common objective of subgrouping voxels having similar spatial attributes together.
As discussed earlier, there may be artifacts related to segmentation of different tissue types depending on the modalities used for scanning the tissue. For example, the
requirements for segmentation and artifacts related to the brain and thorax tissues
might be different. The problems such as partial volume effects while segmenting adjacent soft-tissue boundaries are more prominent in the brain than the thorax, while motion artifacts will be more prominent in the thorax tissue scan due to air inflation
breathing cycles. Therefore, the efficacy of segmentation algorithms depends on the
ability to tackle different types of artifacts, such as motion artifacts, partial volume artifacts, RF noise, and ring artifacts, present in medical images achieved through medical modalities. To tackle these problems and fully automate the segmentation process,
different smoothing algorithms or filters are applied for removing the noise from the
observed noisy images provided by different medical modalities.
Although the segmentation process has been fully automated with the recent
advancement in segmentation techniques, some manual efforts might still be needed
depending on the application for which segmentation is done. To create 3D CAD
images for FE model developments, manual adjustments are usually made to capture
the adjacent tissue boundaries and subgrouping ROIs. It refers to a process where a
human operator segments (paints) and labels the medical image by hand. The
manual task of segmenting or painting ROIs is also called as masking. Generally,
a “semiautomatic” approach is followed where initially a rough threshold is applied
on registered images, and then, manual painting or masking operations can be performed to further improve the segmentation on ROIs.
Many segmentation methods have been reported (Kang et al., 2003; Pham et al.,
2000; Zhang, 1996) for processing the images retrieved through medical scans for
skeletal, vessel, and brain. With the advanced imaging technique, it is possible to
scan different body regions with higher resolution which helps in extracting the geometric details of ROIs with higher accuracy. Further, these segmented slices are used
in developing 3D models of ROIs for different applications, such as computer-aided
diagnosis system, implant design, assistance in robotic surgery, and finite element
mesh developments for different ROIs. Various commercial software packages,
such as Mimics (Materialise, Leuven, Belgium), Simpleware (Synopsys, Mountain
View, USA), and 3D slicer (Open-source software available at www.slicer.org), are
available to retrieve these 3D models or computer-aided drawing models for ROIs of
the human body.
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FIGURE 9.6
(Left) Detailed outer surface of the human brain (segmented with Mimics 12.0
(Materialise, Leuven, Belgium)) and (Right) smoothened brain outer surface
(smoothened with 3-matic software (Materialise, Leuven, Belgium)).
9.4.1.3 3D Surface Smoothing
For a user-defined confidence level, which varied for different applications, the
voxel surface is smoothed to the STereoLithography (STL) file format (also known
as “Standard Triangle Language” or “Standard Tessellation Language” format) and
can be exported for mesh generation in preprocessing software. Depending on the
type of mesh required, the exported CAD surface may need to be smoothed to get
a feature-free geometry. For example, complete details regarding the shape of the
outer surface of a 3D segmented model of a brain scan can still be captured by
smoothing the finer irregularities (Fig. 9.6 left). This type of enclosed volume
with surface irregularities can be meshed with small-sized tetrahedral elements to
capture topology details. To mesh such an enclosed volume with hexahedral
elements will require very small element size to align the faces of the hexahedral
elements along the surface topology, which will yield very small element size resulting into larger simulation time in explicit solvers. Therefore smoothing algorithms
are applied before exporting, which can provide a simplified feature-free surface
making it relatively easy to mesh with a bigger element size. Different algorithms
(Özsa
glam and Çunkaş, 2015) are available to smooth surfaces, such as Laplacian
smoothing (Vollmer et al., 1999) and fuzzy vectorebased smoothing (Shen and
Barner, 2004). Many of these algorithms are directly available in imagesegmentation software which can be applied to achieve simplified feature-free surfaces without losing too many of the geometrical details, i.e., overall volume and
shape. Fig. 9.6 (right) shows an example of a smoothed surface of the human brain
retrieved through 3-matic software (Materialise, Leuven, Belgium).
9.4.1.4 Image Analysis Software Packages
Numerous software packages are developed by several research groups (Jansen et al.,
2005; Kwon et al., 2009; Pieper et al., 2004; Shattuck and Leahy, 2002; Udupa et al.,
9.4 Meshing Human Body
1992) for analyzing medical images and further, segmenting the images for generation of 3D models. The most commonly used file extension format from medical
images is .dicom. Various free as well as commercially available software tools are
available to read different file formats. There are even web-based software tools
which can be used to analyze the medical images. These tools are becoming
very popular because they process the data quickly and easily. Some of these software programs are 3D-Doctor (Able Software Corp, Lixington, MA, USA), eFilm
Workstation, PACSPlus Viewer (Medical Standard, Korea), AMIDE (http://amide.
sourceforge.net/), Simpleware (Synopsys, Mountain View, USA), Imaris (Bitplane, Zurich, Switzerland), Mimics (Materialise, Leuven, Belgium), Analyze
(http://analyzedirect.com/), Vitrea 2-Fusion7D (MediMark Europe), Medx,
3DVIEWNIX (http://www.mipg.upenn.edu/Vnews/index.html), 3D Slicer (https://
www.slicer.org/), OsiriX (Pixmeo, Bernex, Switzerland), BrainSuite (http://
brainsuite.org/), MIPAV (https://mipav.cit.nih.gov/), ITK (Insight Toolkit)
(https://itk.org/), and MRIcro (http://www.mccauslandcenter.sc.edu/crnl/mricro).
These programs have different functions ranged from image registration to viewing
or processing the medical images depending on the application.
Some commonly used software packages for segmenting 3D CAD surfaces
from medical images are Mimics, Simpleware (commercially available), ITK,
and 3D Slicer (open-source free software). These software packages are userfriendly and can be customized for user operations as well. For example, ITK is
a Cþþ cross-platform software and uses built-in environment CMake to manage
platform-specific project generation and compilation process in a platformindependent way. Developers around the world can use, debug, maintain, and
extend the software capabilities to perform user-specific operations in terms of
registration, filtering, or segmentation of medical images. Because of the recent
advancements in imaging techniques as well as advanced algorithm development
for segmentation of 3D images from medical scans, the quality of retrieved 3D
surface models are well-refined, and detailed, as well as efficient. Using ScanIP
and the þFE module from the Simpleware software platform (Synopsys, Mountain View, USA), a high fidelity, multidomain (33 part), water-tight volumetric
FE mesh of the human head was generated directly from the segmented images
of a series of MRI scans and is shown in Fig. 9.7.
However, there are still problems associated with registering and segmenting
poorer quality images, as well as medical images with missing voxels. Additionally,
as explained earlier, the voxel distribution cannot be accurately predicted between
adjacent soft-tissue boundaries, and depending on the algorithm used for segmenting
the image, the problems might be different in nature. For example, the most
commonly used algorithm for segmenting medical images is based on a simpleregion growing approach which uses global threshold criteria to distribute the voxels
in a section which yields inefficient results at the boundaries. An example for a
similar problem called partial volume effect is shown in Fig. 9.8. To deal with
this issue, a local threshold can be applied at the adjoining region which might fix
the problem. Similarly, the problem of missing pixels during 2D imaging can be
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CHAPTER 9 Developing FE Human Models From Medical Images
FIGURE 9.7
Detailed 3D FE model of a human headeneck segmented from MRI images capturing
detailed tissues, such as: (A) ventricles, (B) cerebral hemispheres, (C) skull with neck
vertebrae, (D) white matter, and (E) complete brain with bridging veins and sinuses.
Courtesy of Simpleware (Synopsys, Mountain View, USA).
overcome by applying smoothing filters to or by manually painting (masking operation) the problem area.
9.4.2 FINITE ELEMENT MESH DEVELOPMENT
Once the required 3D CAD surface is ready for all ROIs, the next step is to generate a
high-quality mesh. There are different methods which can be used to develop FE meshes
from CAD surfaces depending on the type of mesh needed for the analysis. The different
types of mesh and the association with different types of surfaces retrieved through medical images can be divided into three categories as shown in Fig. 9.9.
The unstructured mesh made of mainly tetrahedral elements can be achieved
directly from voxel matrix or 3D CAD enclosed volumes for ROIs retrieved through
9.4 Meshing Human Body
FIGURE 9.8
Effect of partial volume during segmentation of CT scan images.
FIGURE 9.9
Mesh pattern based on medical scan segmentation: unstructured and mixed
meshesdgrid-based approaches; structured meshdglobal human body modeling
consortium (GHBMC) model.
Unstructured and mixed meshes: photos courtesy of Simpleware (Synopsys, Mountain View, CA, USA).
medical scans. Different algorithms, such as advancing front or Delaunay meshing
algorithms, are used to generate these meshes with the help of different software
packages. Some advanced algorithms and grid-based methods (voxel and volumetric
marching cube meshing) can be used directly to convert voxels from 3D medical
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CHAPTER 9 Developing FE Human Models From Medical Images
FIGURE 9.10
(A) Voxel-based mesh and (B) diamond-shaped quad surface mesh with hexagonal
inside elements.
image to unstructured tetrahedral or hexahedral mesh. However, hexahedral meshes
which are generated using the original voxel cannot capture the detailed outer geometry of the surface due to the cubic nature of mesh advancing as shown in Fig. 9.10A
and provides stepped interfaces at the mesh boundaries with a “Lego brick” appearance which overestimates the surface area and can yield unrealistic results during
physics-based numerical calculations. This problem can be solved by shrink wrapping the outer edges for voxel mesh on the surface, and the inside volume can still be
meshed for hexahedral cubic meshes as shown in Fig. 9.10B. Both of these meshed
surfaces can be achieved by a fully automated process, but have drawbacks in terms
of providing a robust numerical solution.
Mixed meshes, which consist of both tetrahedral and hexahedral elements, can
be achieved with fully automated, newly developed advanced algorithms such as
the enhanced volumetric marching cubes (EVoMaCs) approach (Cotton et al.,
2016; Young et al., 2008). These sorts of mixed meshes can also be manually generated by combining structured and unstructured mesh approaches.
The voxel-based approach can be used to capture very minute details of the complex human body shapes, including bridging vein structures in brain segmentations,
heart valves in different chambers, and trabecular architecture in detailed scan of
human bones, etc. The accuracy of meshing these detailed geometries depends
mainly on resolution and segmentation efficiency of the algorithms. However, the
main disadvantages of creating fully connected tetrahedral meshes using voxelbased approaches are: (1) an uncontrolled mesh size which requires a large number
9.4 Meshing Human Body
FIGURE 9.11
(Left) A human head FE model having 3.2 million tetrahedral elements generated by
directly converting segmented voxels to mesh: (left) isometric view and (right) lateral view.
Photos courtesy of Simpleware (Synopsys, Mountain View, CA, USA).
of elements, and (2) relatively more difficult to handle than a structured hexagonal
mesh of larger element size and lesser number of elements. Cotton et al. (2016) presented such an FE model of the human head meshed using these approaches to capture detailed segmented brain tissues consisting of 3.72 million tetrahedral elements
as shown in Fig. 9.11.
On the other hand, structured meshes, which mainly consist of continuous hexahedral elements, are preferable as they satisfy the basic requirements and theory
behind the FE-based approach for solving physics-related problems. To date, there
is no fully robust, automatic, continuous structured hex meshing algorithm available, and therefore the hexahedral meshing of complex assemblies, with components having nonuniform geometries such as HBMs, is quite time-consuming,
laborious, and costly. Many constraints and challenges are associated with automated hexahedral meshing such as rigid connectivity between assemblies and
complex shapes of the geometries. These and other challenges are discussed by
Blacker (2001) in detail.
9.4.2.1 Multiblock Approach
The most common approach used for generating high-quality hexahedral elements
is the “multiblock meshing” technique. The 3D enclosed volumes retrieved
through medical images of ROIs are divided into blocks which consist of vertices
and edges. The vertices can be placed on the outer surface of geometry. A set of
step-by-step procedures for generating a high-quality hexahedral mesh is shown
in Fig. 9.12.
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CHAPTER 9 Developing FE Human Models From Medical Images
FIGURE 9.12
Example showing structured hexahedral mesh generation with segmented CAD models
using “Multiblock” technique, where N, number of nodes on each edge of single block; n,
arbitrary number. (A) 3D segmented CAD surface, (B) multiple blocks representing the
vertebral body, (C) mesh with element size ¼ i/4, and (D) mesh with element size ¼ i.
First, the 3D CAD surface (.STL file extension) of T7 Vertebrae (half vertebral
surface was used due to its symmetrical nature) segmented through CT scan images
with the help of Mimics 12.0 (Materialise, Leuven, Belgium) software package was
developed. This .STL file was imported into commercially available software ICEM
CFD 12.1 (Ansys Inc., PA, USA) for multiblock mesh generation. The vertices and
faces are projected in a multistep process on the outer edge of the closed surface.
Once all vertices and edges are in place on the outer surface, the internal vertices
and edges were arranged to get a suitable mesh. The proper connectivity of the
blocks was assured during these operations. A high-quality continuous-structured
hexahedral element mesh was the final product. Using a multiblock approach, the
mesh controls can be assigned manually on the edges to predefine mesh size restriction. Further, by changing the number of nodes through the vertices of each block,
the corresponding size, as well as the number of elements, was changed. Figs. 9.12C
and D prove this, as the latter had the number of nodes on each edge doubled, which
resulted in mesh with smaller elements and higher number of elements.
The major advantage of the multiblock meshing technique is the ease of use, and
its usefulness for conducting the mesh sensitivity study. The optimized size of
elements can be achieved based on a convergence study using different numbers
of elements. This approach is relatively easier than manual procedure, which is
9.4 Meshing Human Body
FIGURE 9.13
Hexagonal (structured) mesh of human head using block meshing technique. The mesh
is generated from the model reported by Mao et al. (2013).
based on traditional operations for generating hexahedral meshing, i.e., solid mapping, sweeping, etc. Also, the nonuniform geometric shape can be meshed at the curvature boundaries more efficiently with the multiblock approach. Mao et al. (2013)
presented the development of high-quality hexahedral human brain meshes
(Fig. 9.13) using a feature-based multiblock approach and recommended it to
develop high-quality FE models to enhance the acceptance and application of numerical simulations.
9.4.3 POST-MESH SEGMENT
9.4.3.1 Mesh Quality Check
The final output retrieved from preprocessing software is considered to be preliminary
mesh, which requires further manual rectifications or modifications to meet certain
mesh quality criteria. The diagnostic checks used for justifying the quality of the
mesh depends on certain parameters, including mesh element size, Jacobian, warpage,
skew, etc. Although a mesh sensitivity study is usually done to verify the size of the
element needed to obtain an optimized solution using FE for any given problem, it
can be a very difficult task to conduct in a full HBM due to the high number of elements involved. In general, the current generations of HBMs have an average element
size of 1e5 mm for different body regions so that an affordable time step can be
achieved during the explicit solution for a model involving higher number of elements.
In terms of achieving a robust solution from FE meshes, an element having a perfect Jacobian value of 1, such as a cube or a cuboid, may be the ultimate goal, but
such elements are not possible to achieve due to the complex shapes of human organs. Different parameters of elements, such as Jacobian, warpage, skew, and aspect
ratio, are considered to be parts of quality check for a mesh. When generating a highquality mesh it is important to avoid sharp angles, distortion in the elements, and
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extraordinarily small elements. There might be smaller elements near small features
which capture the geometry or shape, but the size of smaller elements needs to be
optimized to avoid a smaller time step during explicit solution. The total number
of elements may also be optimized through mesh sensitivity study as increasing
number of elements increases the solution time.
Some of the most common parameters designated to check the mesh quality
depend on the types of elements selected. These parameters are previously explained
in Section 3.8. Different FE preprocessing software packages may choose different
names to represent these parameters. Hence, readers are advised to check the user
manuals before using any software package.
9.4.3.2 Mesh Quality Improvement
Improving the mesh quality is an iterative process and for a large part depends on the
skill level of the operator. There are diagnostic checks available while performing
“multiblock” operation, but the mesh generation will depend on the expertise of
the user and the surface topology of ROIs. The quality of poor elements can be
improved by two ways:
1. Automated process: Previously, mesh quality improvement has not been fully
automated. Some preprocessing software programs have algorithms available to
automate some of the mesh refinement processes, but use of these tools might
require special attention as it can be parameter-specific improvement, which
might worsen other unconsidered parameters, including the quality.
2. Manual adjustment: The most adaptive way of refining the quality is by translating
or dragging the poor quality mesh nodes, or by remeshing the poor quality mesh
regions. This task is time-consuming and dependent on the expertise of the user.
In addition to these mesh refinements needed to meet certain parameters, there
are other checks which should be performed before proceeding further. For example,
there should be no free (or floating) nodes between shared edges of elements, nor
should the mesh have a nodal connection within a single volume. No duplicate elements or nodes should be present. Proper care should be taken to ensure consistent
orientation of the normal direction in shell elements. If tetrahedral meshes are present, there should be no T-connections between them. The element connectivity issues should be resolved before proceeding to the next step.
Table 9.1 presents suggested values for these parameters as a diagnostic check
for a good quality mesh for the human body FE model. The corresponding value
of each parameter recommended below is a quantitative measure to infer the difference between a poor or a good quality mesh, and is adopted through user experience
and Altair Hyperworks’ good quality mesh guide, but the value of these parameters
can be varied based on user’s choice or preference depending on different applications (Hyperworks, 2016).
After it is assured that a high-quality mesh has been developed that captures structural boundaries for different body regions, the next step involves assigning material
properties, element types, boundary conditions, contacts, and loading conditions.
9.4 Meshing Human Body
Table 9.1 Suggested Values for Mesh Quality Check Parameters in a Human
Body FE Model
No.
Parameter
Requirement
1
2
3
4
5
6
Jacobian
Warpage
Skew
Aspect ratio
Minimum length
Maximum length
7
8
9
10
Time step
Duplicate elements
Free edges/nodes
Tet collapse
>0.4
<50
<60
<5
1 mm
170% of mean
mesh size
0.1 e6 ms
None
None
None
9.4.3.3 Material Law and Property
Various material laws have been used in literature (Yang et al., 2006) to characterize
the stressestrain behavior (relationship) of different human body parts, such as
elastic or elasticeplastic constitutive material laws primarily for bones, and viscoelastic or hyperelastic formulations mainly for soft tissues. Parameters of each material model were set based on experimental data in order to better represent the
mechanical behavior of various human organs. The skeletal meshes are modeled
with elastic or elasticeplastic constitutive relations, and the cortical bones are
modeled with shell elements covering solid elements which represent trabecular
bone elements. The soft tissues are represented with viscoelastic material properties.
9.4.3.4 Articular Joints and Contact Algorithms
Once material properties have been assigned, boundary conditions and interactions
between parts need to be defined. Articular joints are used to represent kinematic
behavior of body regions. Appropriate contacts between adjacent body regions
and active/passive muscle responses (see Chapter 11) are defined to capture accurate
biomechanical behavior of the human body.
The contacts and physical joints between adjacent tissues are assigned where
element connectivity is not available. The bones in the human body are joined to
each other for the sake of motion of different body parts. For example, fingers are
joined with each other and then to the ulna and radius, which are then joined to
the humerus, which is in turn joined to the scapula. These bone connections are
simulated using mechanical joints available in finite element codes, such as spherical and revolute joints, and additionally, stiffness characteristics are applied to these
joints to get the desired range of motion and kinematics of human bone joints.
During large deformation or impact problems, different tissues make contact
with each other inside the human body. Bones, ligaments, muscles, and connected
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soft tissues are assembled into a human body to perform different kinematic and biological functions, but it is difficult to accurately model these tissue connections in
full-body finite element models. The finite element solution is based on continuum
mechanics which requires the continuous flow of material within a subject. Once the
material is discrete, the laws of continuum mechanics do not hold. Therefore the kinematics or force transfer between two adjacent and distinct bodies is governed by
contact mechanics, which is based on empirical relations, not continuum. These
empirical equations, which define the transfer of physical conditions from one object
to another, depend on numerous factors, such as contact boundary conditions, material properties, contact detection methods, and contact geometry. The contact
problems become more complicated in advanced applications, such as metal forming or fluid structure interactions. In FE HBMs, different types of contacts are established between different tissues or organ meshes. Along with high-quality nodal
connected mesh, the contact algorithms play an important role in providing accurate
modeling solutions during large deformation problems.
In a typical FE solver, each tissue is represented by the mesh having elements
and nodes, and is referred to as “part.” Each part has its own material (based on
different constitutive material models) and element (formulation of element, shell,
or solid, etc.) properties associated with it. Additionally, each part, node, or element
has a unique ID (identification number) assigned. These parts, elements, or nodes
may also be grouped as “sets.” A contact is defined between parts, part sets, segment
(outer surface of elements) sets, or node sets. With the help of contact definitions,
these unmerged Lagrangian meshes can interact with each other. The contacts can
also be used to tie unmerged mesh components to each other, such as constraint
of flesh with bones. The entities between which contact is assigned are categorized
further as “master” or “slave” elements. The definition of master or slave is not fully
categorical in FE solvers, but in general recommendations:
•
•
•
The slave surface should be the surface having fine mesh.
For similar mesh densities, the slave surface should be made from the softer
material.
For similar mesh densities and material, the slave surface should have a curved
surface.
Different types of contact algorithms are available in FE solvers, including kinematic constraint, Lagrange multiplier, penalty, or distributed parameter (LS-DYNA
theory manual). The most commonly used method in an explicit solver such as LSDYNA is the penalty method which is further divided into three types, based on the
algorithms:
1. Standard penalty formulation
2. Soft constraint penalty formulation
3. Segment-based penalty formulation
In the standard penalty formulation, the stiffness of the contact interface should
be similar in magnitude to the stiffness of the interface element. The soft constraint
9.4 Meshing Human Body
penalty formulation is implemented to treat contact between bodies with dissimilar
material properties, such as boneeflesh interface. These formulations generally use
“the slave node-master segment” approach, whereas the segment-based penalty formulations use “the slave segment-master segment” approach. Other types of contact
algorithms are available to treat complex unacceptable penetrations such as “slidingonly contact” option to treat shockwaveestructure interactions. The most commonly
used contact types used in HBM development are “automatic surface to surface,”
“automatic single surface,” and “tied surface to surface.”
The contact occurrence is identified in a solver by first checking the potential penetration of a slave node through master segment with the help of different algorithms at
each integration point. For penalty-based contacts, the resistant force is proportional to
the penetration. The type of contact, such as one-way (slave-master), both-ways
(slave-master and master-slave), or tied treatment, also affects the execution of the
contact algorithms. In one-way contacts, only the user-defined slave nodes are checked
for penetration of the master segments, whereas for both-ways contacts, the penetration check is done for both slave-master and master-slave segments. One-way contacts
are generally used when the master is a rigid body, or when a relatively fine mesh
(slave) contacts a coarser mesh (master). In tied contact types, the slave nodes are constrained to the master surface and can be defined with different translational and rotational degrees of freedom. It is generally recommended to define contact regions by
node/segment sets, rather than part/part sets (Bala, 2001).
Further, the stiffness of the contact can be represented by linear springs between
slave nodes and the master segments and can be specified by the user, either by
giving an estimated value of contact thickness or by giving a scaling factor
for the existing contact thickness. The penalty-based approach for modeling
contacts generally uses a default method based on contact segment size and material
properties to determine the stiffness, whereas the soft-constraint-based approach
uses contacting nodal masses and the global time step size to calculate this stiffness.
The latter is suited for defining contact between dissimilar material properties due to
its independence on contacting elements material characteristics. Along with thickness offset to define contact stiffness, other factors such as static or dynamic friction
values (based on Coulomb formulation of friction), viscous damping values, contact
energy, etc., play a major role in the success of the defined contact. A majority of the
solvers provide options to define different contact parameters to successfully execute
the contact between unmerged Lagrangian meshes.
Special attention is needed while assigning different types of contact algorithm such
as proper frictional coefficient assignment, no physical penetration between two bodies,
and type of contact used (such as segment or soft-based penalty formulations).
Some common rules, which can be applied to resolve contact-related problems,
based on user experience are:
1. There should be no initial penetration between contact surfaces. Initial
penetration between the meshes may create instability in the calculation. Even
though the solvers try to autocorrect the geometry by moving interpenetration
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CHAPTER 9 Developing FE Human Models From Medical Images
2.
3.
4.
5.
nodes to the master surface on the first cycle, it may not be done accurately. This
may also cause interpenetrations in another contact surface which can lead to
nonphysical contact behavior.
Segment normal should be checked for their orientation, if “nonautomatic” types
of contacts are provided. The problem can be solved if “automatic” types of
contacts are used as it looks in both directions of segment. In general, the
contact thickness is equal to the shell thickness for single surface contacts, but
in “automatic” type of contacts, the contact surfaces are defined from shell-mid
plane, a distance equal to one half of contact thickness.
Redundant contacts between two surfaces should be avoided.
Mesh sizes of the contact interfaces should be as similar as possible.
The adjacent surfaces should be adequately offset to account for shell thickness
during mesh generation, as contact surfaces are offset from midplanes of shell
elements.
9.4.3.5 Boundary and Loading Condition
After assigning appropriate material properties, element formulations, and different
boundary conditions to the mesh components representing different body regions,
the behavior of different body regions should be validated. The loading and boundary conditions as well as data sampling rates should be consistent with those used in
the experiments. Different experimental tests taken from the literature can be used to
validate the behavior of HBMs at a component level, in addition to whole body simulations. The biomechanical results of forceedeflection, stressestrain histories, and
one-to-one kinematics comparison between simulations and experiments correlate
to assure the biofidelic behavior of developed FE models.
9.5 EXEMPLARY WHOLE BODY FE MESH DEVELOPMENT
So far, details regarding different procedures used in the development of human
body FE models are presented. Various algorithms and software packages were
mentioned in this chapter, all of which can be used to complete the aforementioned
procedures. An example of the software used for the development of a full-body FE
model for a representative elderly female (collaborative human advanced research
models, CHARM 70F) is shown in Fig. 9.14 (Kalra, 2017). The figure explains
the processes followed as well as the software or tool which was used to accomplish
the task. Image registration and segmentation was done using Mimics 12.0 (Materialise, Leuven, Belgium) and the segmented 3D surfaces for ROIs were further
smoothed using 3-matic (Materialise, Leuven, Belgium). These surfaces were
further meshed with the “multiblock” technique using ICEM CFD 12.1 (Ansys
Inc., PA, USA). The diagnostic check for preliminary mesh quality as well as the
refinement was done using Hypermesh 12.0 (Altair engineering Inc., Troy, MI,
USA). Additionally, the material properties, loading conditions, contact algorithms,
etc., were assigned using LS-Prepost 4.1 (LSTC, Livermore, CA, USA), and finally
the problems for different loading conditions were solved using LS-DYNA (LSTC,
Livermore, CA, USA) explicit solver.
References
FIGURE 9.14
Example of tools and processes used in FE model development of an elderly female
(CHARM 70F) in context with concepts explained in this chapter. More information
regarding the CHARM-70F model can be found in Chapter 17.
ACKNOWLEDGMENTS
The author likes to thank Ms. Kathlein Endlein, Research Analyst from Ford Motor Company,
for her valuable editorial comments on the manuscript.
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Conference. Stapp Car Crash Journal 50, 429.
References
Young, P., Beresford-West, T., Coward, S., Notarberardino, B., Walker, B., Abdul-Aziz, A.,
2008. An efficient approach to converting three-dimensional image data into highly accurate computational models. Philosophical Transactions of the Royal Society of London A:
Mathematical, Physical and Engineering Sciences 366 (1878), 3155e3173.
Zhang, Y.J., 1996. A survey on evaluation methods for image segmentation. Pattern Recognition 29 (8), 1335e1346.
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CHAPTER
Parametric Human
Modeling
10
Jingwen Hu
University of Michigan, Ann Arbor, Michigan, United States
10.1 INTRODUCTION
10.1.1 WHAT IS A PARAMETRIC HUMAN MODEL?
The phrase “parametric human model” refers to a set of models that can be generated
through an automated procedure using parameters, or measurements, obtained from
humans. The parameters typically used for such models are related to geometry or
material properties, thus they can be used to account for morphological and biomechanical variations among the population. Because the material properties in FE
models are generally easy to change, this chapter will focus on how to quantify
the geometric variations in the human body and how to automate the development
of models with different geometry targets.
10.1.2 WHY ARE PARAMETRIC HUMAN MODELS NEEDED?
10.1.2.1 Increased Injury Risks Among Vulnerable Populations
It is well documented in scientific literature that among the adult population, small
females, elderlies, and obese people are at an increased risk of death and serious injuries in motor-vehicle crashes (MVCs) as compared with midsized, young, male occupants. Newgard and McConnell (2008) found that the effectiveness of vehicle
airbag deployment with regard to injuries was lower for smaller occupants than
for midsized men. Kent et al. (2009) reported that if the injury risks for people of
all ages were the same as for people at age 20, the number of occupants injured every
year would lessen by 1.13e1.32 million in the United States alone. This is nearly
half the number of total annual injuries in MVCs. Field data analyses have also
shown that obese occupants are at a higher risk of fatality and injury in frontal
crashes, as compared to normal-weight individuals (Carter et al., 2014; Cormier,
2008; Ma et al., 2011; Ryb et al., 2010; Tagliaferri et al., 2009; Viano et al.,
2008; Zhu et al., 2006). All the above findings highlight the potential benefit of
safety systems specifically optimized for vulnerable populations.
In addition to overall risks of injury, it has been well documented that injury risks
to various body regions of people involved in MVCs are associated with age, sex,
and body mass index (BMI), a parameter measuring obesity level. Increased age
in adults is associated with increased serious injury risks to almost every region
Basic Finite Element Method as Applied to Injury Biomechanics. http://dx.doi.org/10.1016/B978-0-12-809831-8.00010-6
Copyright © 2018 Elsevier Inc. All rights reserved.
417
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CHAPTER 10 Parametric Human Modeling
of the body and in every crash mode. Among all the body regions, thoracic injuries
are disproportionately common for occupants in the older population (Kent et al.,
2005a; Morris et al., 2002, 2003), while lower extremity injuries are affected by
age as well as sex and BMI (Moran et al., 2003; Ridella et al., 2012). Rupp and Flannagan (2011) did a comprehensive study by analyzing the effects of age, sex, and
BMI on injuries in different body regions, as well as the specific types of injuries.
It was further confirmed that aging increases the risk of injury to every region of
the body, and the body regions for which the effects of age are most prevalent in
frontal crashes are the thorax and lower extremities. Among all thorax and lower extremity injuries, age is mostly associated with injuries to the ribs and the knee, thigh,
and hip (KTH) complex. Despite the large effects of age on these risks, the effects of
BMI and sex on risk of injury to the thorax and lower extremities should not be
neglected when optimizing safety systems. Field data was used from several
studies to determine trends for increased or decreased risk of injury for different
body regions in terms of effects of obesity (Fig. 10.1). It is clear that the thorax
(Boulanger et al., 1992; Cormier, 2008; Mock et al., 2002; Moran et al., 2002; Reiff
et al., 2004) and lower extremities (Arbabi et al., 2003; Boulanger et al., 1992;
Rupp et al., 2013; Ryb and Dischinger, 2008; Zarzaur and Marshall, 2008) are
more likely to be injured in obese than nonobese occupants.
10.1.2.2 Increasing Proportion of Older and Obese Populations
Due to increasing life expectancy and decreasing birth rates, the rate of growth of the
older population in the United States, Japan, China, and many other countries is
increasing, and is expected to continue for the next several decades. By 2030, 20%
of the US population will be 65 years of age or older (http://www.census.gov).
Similarly, China will have 285 million people over the age of 60 by 2025, and the
projected proportion of China’s population over 65 years will be more than 23%
in 2050. According to the World Health Organization (WHO), the proportion of
FIGURE 10.1
Effects of obesity on risks of injury to different body regions.
10.1 Introduction
people who are obese has increased significantly worldwide since the 1980s. In 2014,
also worldwide, 39% of adults aged 18 years and over were overweight, and 13%
were obese. In the United States, the prevalence of overweightness and obesity
were 68.8 and 35.7%, respectively in 2009e10, compared with 55.9 and 22.9% in
1988e94 (Flegal et al., 2012). In a study by Finkelstein et al. (2012), it was predicted
that the prevalence of obesity would be up to 42% in the United States by 2030. The
documented evidence that age and obesity are strongly related to risks of injury in
MVCs, together with the projected increase in older and obese populations, is ample
motivation for developing advanced injury assessment tools to evaluate vehicle-safety
designs aimed at mitigating injuries in these vulnerable populations.
10.1.2.3 Factors Associated With Increased Injury Risks for Vulnerable
Populations
While any combination of small female, elderly, or obese increases the risks of thorax and lower extremity injuries in MVCs, the exact mechanisms and factors associated with these increases vary. In general, factors affecting risks of injury can be
grouped into three categories of characteristics (Kent et al., 2005b): geometric,
compositional, and material (Fig. 10.2). Geometric characteristics include the 3D
external body contour as well as size, shape, and orientation of the bones; compositional characteristics include the cross-sectional areas of the cortical bones and soft
tissues; and material characteristics include the mechanical properties of cortical and
cancellous bones as well as soft tissues.
FIGURE 10.2
Factors affecting injury risks for female, older, and obese populations (Feik et al., 1997;
Nalla et al., 2004; Shi et al., 2014).
419
420
CHAPTER 10 Parametric Human Modeling
10.1.2.4 Age-Related Factors
It is well established that injury tolerance decreases in the thorax and lower extremities with aging (Laituri et al., 2005; Zhou et al., 1996), and such decreases are
caused by changes in factors of all three characteristics categories (i.e., geometric,
compositional, and material). Aging causes several geometric changes in the thorax
region, including an increased kyphosis of the thoracic spine (Goh et al., 2000;
Puche et al., 1995) and, due to more horizontally oriented rib angles in the older population, an increase in the depth of rib cage (Gayzik et al., 2008; Kent et al., 2005b).
These geometric changes may affect both the force required to deflect the thorax and
the distribution of strain occurring within the thorax when a load is applied. In addition to the geometric changes, compositional changes in cortical bone crosssectional areas with aging also affect injury tolerance. For example, a review of
scientific literature reveals that the cross-sectional area of a rib may decrease
approximately 0.19 mm2/year after age 25 due to progressive circumendosteal
resorption (Stein, 1976). Similarly the reduction in cross-sectional area of cortical
bone in an aged proximal femoral metaphysis also significantly influences the propensity to fracture (Holzer et al., 2009; Verhulp et al., 2008). In addition to geometric and compositional changes, both cortical and cancellous bones exhibit a decrease
in Young’s modulus with aging, mainly due to the decrease in bone mineral density.
The fracture toughness (failure strain/stress) of cortical bone also shows significant
deterioration with aging (Nalla et al., 2004).
10.1.2.5 Obesity-Related Factors
Compared to effects of aging, the effects of obesity on injury risks in MVCs are relatively simple, with most effects coming from geometric changes related to adipose
tissue. Field data analyses (Boulanger et al., 1992; Viano et al., 2008), cadaver tests
(Forman et al., 2009a, 2009b; Kent et al., 2010; Michaelson et al., 2008), and
computational studies (Turkovich et al., 2013; Zhu et al., 2010) have all shown
that the increased mass of an obese individual can cause that individual to move
further forward in a frontal crash, thereby increasing the contact loading and causing
increased risks of thorax and KTH injuries. It was also found that obesity effectively
introduces slack in the seat belt system by routing the belt further away from the
skeleton (Cormier, 2008), and due to the increased amount of adipose tissue around
the pelvic and abdominal areas, the interaction between the lap belt and pelvis is
postponed (Reed et al., 2012). Such a belt fit is expected to adversely affect occupant
kinematics in a frontal crash, causing increased risks for thorax and lower extremity
injuries.
10.1.2.6 Sex-Related Factors
Women generally have lower injury tolerance than men, but effects due to sex are often
complicated by a correlation with stature. In addition to the effect of stature, the difference due to sex can be reflected in both geometric and material differences between
males and females. For example, differences in pelvic-bone anatomy and shape could
explain differences in the risks for some lower extremity injuries. Wang et al. (2004)
10.2 Current State-of-The-Art FE, Whole-Body, Human Models
found that the female hip socket tends to face more forward than that of a male. As a
result, a greater proportion of the surface area of the femoral head is engaged during
frontal-impact in females, when there is loading through the KTH complex. This factor
could decrease the risk of hip fracture for females, but increase the possibility of a knee
or thigh injury in frontal crashes. In addition, elderly women often have increased bone
porosity and decreased bone mineral density, and therefore are at a greater risk of fracture than are elderly men (Riggs et al., 2004). However, it should be noted that there is
an interaction effect between age and sex.
10.1.3 NEED FOR PARAMETRIC HUMAN MODEL
Development and validation of increasingly sophisticated models of only a few sizes
and shapes is the primary focus of FE modelers over the past two decades. Yet, the
greatest potential for human models in injury biomechanics and vehicle development lies in representing the large range of human variability that is not represented
by the traditional human models and anthropomorphic test devices (ATDs). Based
on discussion in the previous section, vulnerable populations are at higher risk of
injury because of changes in geometric, compositional, and material characteristics,
all shown to be associated with injury occurrence. These characteristics affect directions and magnitudes of loading to the human body in collisions. The relative
contributions from the effects of age, sex, stature, and BMI on injury risks in crashes
can best be assessed using FE models that are customized with specific parameters
related to individual populations. Such models can be generated automatically,
efficiently overcoming limitations in existing methods for safety designs in which
there is not adequate consideration for geometrical and biomechanical variability.
10.2 CURRENT STATE-OF-THE-ART FE, WHOLE-BODY,
HUMAN MODELS
Several FE, whole-body, human models have been reported in scientific literature
(Table 10.1), including HUMOS (Robin, 2001), H-model (Haug et al., 2004),
Ford Human Body Model (Ruan et al., 2003, 2005), WSU Human Model
(Kim et al., 2005; Shah et al., 2001), THUMS model (Hayashi et al., 2008; Iwamoto
et al., 2002), GHBMC model (Shin et al., 2012; Untaroiu et al., 2013; Vavalle et al.,
2013), and the most recent CHARM-70F model. Due to the time-consuming development process and the desire to compare predictions between FE models and ATD
models, the FE models have typically been constructed to simulate the same small
number of body sizes (midsized male, small female, and large male) currently represented by physical ATDs. As a result, the traditional FE human models are limited in
the same way that adult ATDs are: they are not able to represent variability in skeleton
and body shapes among diverse populations.
In the HUMOS2 project, one of the first existing parametric, whole-body, FE, human models was developed. In this model, parametric anthropometry is used to scale
421
Table 10.1 Overview of Recent Whole-Body Human FE Models for Injury Prediction
Model
HUMOS
H-Model
Ford Model
WSU Model
THUMS
GHBMC
CHARM70F
Reference
Robin (2001)
Vezin and
Verriest (2005)
Haug et al.
(2004)
Ito et al. (2012)
Shah et al. (2001)
Kim et al. (2005)
To be
published
Radioss
PAM-CRASH
Midsize male
Small female
Large male
Scale to others
No
No
Seated
PAM-CRASH
LS-DYNA
Iwamoto et al.
(2002)
Hayashi et al.
(2008)
LS-DYNA
Shin et al. (2012)
Vavalle et al. (2013)
Software
Ruan et al.
(2003)
El-Jawahri et al.
(2010)
LS-DYNA
LS-DYNA
LS-DYNA
Midsize male
Small female
Midsize male
Midsize male
Midsize male
Small female
Large male
Midsize male
Small female
Large male
Midsize
female
35, 75
No
Seated
35, 55, 75
No
Seated
No
No
Seated
No
No
Seated and
standing
No (65)
No
Seated and standing
70
No
Seated and
standing
Visible Human
Project
Visible Human
Project
Visible Human
Project
CT scans
CT scans
Not Stated
w120 k
w120 k
w1.8 million
MRI, UMRI, CT,
external
anthropometry
w2.2 million
Size
Age
Obesity
Posture
Figure (most
recent
version)
Geometry
Elements
External
anthropometry,
X-ray
w70 k
w1.5 million
10.3 How to Build a Parametric Human Model
an FE model of a midsized male (Vezin and Verriest, 2005) into different statures.
However, the HUMOS2 is based on only a small number of landmark locations
from mostly young, nonobese subjects, and it does not capture variability on compositional and material levels. Furthermore, HUMOS2 models do not include variability
in external body geometry, which is important to study when modeling obese occupants. More recently, researchers in a few studies have tried to turn the H-model and
the Ford Human Body Model into models representing the aging population by
changing the bone geometry and material properties (Dokko et al., 2009;
El-Jawahri et al., 2010; Ito et al., 2009, 2012). However, the models used in these attempts represented only the midsized male population, and while the overall shape of
the rib cage and cortical bone thickness were varied, a systematic change of the crosssectional geometry of each rib is necessary to adequately study this group. Moreover,
using the traditional approach to turn an FE model of a young male into an older male
is still time-consuming, which has limited the number of models that could be generated for the older population. Shi et al. (2015) developed four human models representing occupants with four BMI levels (25/30/35/40) by morphing the THUMS v4,
midsized, male model, which is the first study in which the effects of body shape were
considered for developing human models to be used in injury prediction. However, in
that study stature was held constant. More recently, Schoell et al. (2015) developed a
65-year-old, midsized male model by morphing the GHBMC midsized male model.
The geometries of the brain, head, rib cage, pelvis, femur, and tibia were predicted by
statistical-geometry models, and the material properties of the head, thorax, pelvis,
and lower extremities were adjusted based on scientific literature. In the previous
studies on building human models representing populations other than young,
midsized men, the process of mesh construction or mesh morphing was not fully
automated, and its time-consuming nature limited the number of human models
that could be developed within necessary time constraints.
10.3 HOW TO BUILD A PARAMETRIC HUMAN MODEL
10.3.1 METHOD OVERVIEW
The framework for developing a parametric, whole-body, human FE model for crash
simulations is shown in Fig. 10.3 (Hu et al., 2012, 2016; Hwang et al., 2016b). Statistical models of human geometry are the foundation of parametric models. These statistical models describe morphological variations as functions of occupant parameters
(age, sex, stature, and/or BMI), and these parameters are used in a method designed to
rapidly morph the mesh of a baseline model into other geometries while maintaining
high accuracy of the geometry and good mesh quality. The statistical-geometry model
shown in Fig. 10.3 includes only the skeleton models of the rib cage and lower extremities and the model of the external body surface. However, geometry models representing other bones and internal organs can be included as well. These models can
be integrated together in a manner that incorporates a driving posture based on
423
424
CHAPTER 10 Parametric Human Modeling
FIGURE 10.3
Method for developing parametric human models representing a diverse population.
Figure based on Hwang, E., Hu, J., Chen, C., Klein, K.F., Miller, C.S., Reed, M.P., Rupp, J.R., Hallman, J.J., 2016b. Development, evaluation, and sensitivity analysis of
parametric finite element whole-body human models in side impacts. Stapp Car Crash Journal 60, 473e508, and reproduced with permission from The Stapp Association.
10.3 How to Build a Parametric Human Model
volunteer data. The baseline model can be any contemporary, whole-body, human FE
model (e.g., GHBMC-OS model in Fig. 10.3), which can be morphed into different
geometric targets through an automated mesh-morphing process.
10.3.2 STATISTICAL MODELS OF HUMAN GEOMETRY
Although the method for developing a statistical model of human geometry may vary
slightly from study to study, the general approach involves four steps, as illustrated in
Fig. 10.4. First, clinical CT/MRI scans or body scans performed with lasers are obtained using a protocol approved by an institutional review board (IRB) to ensure that
the procedures for the acquisition of the data are conducted in accordance with all
federal, institutional, and ethical guidelines. CT scans are reviewed by a radiologist
to ensure no abnormalities exist in the anatomical regions of interest (ROIs). Second,
data extraction from images or body-surface scans is conducted, and possibly image
processing and segmentation, data cleaning, landmark identification, and/or template
mesh mapping are included. The goal of the data extraction is to obtain a homologous
set of landmarks for every subject. These sets are used to define the geometry of the
skeleton or the external body shape. Because landmarks at corresponding locations
should be identified in all the sampled subjects, this step can be time-consuming,
although semiautomated methods have been developed in recent studies. Third, a
series of statistical analyses are performed to develop the predictive geometry model.
These methods include the generalized Procrustes alignment (GPA), principal
component analysis (PCA), and multivariate regression analysis. Lastly, the final statistical model can be represented by a set of coefficient matrices, and with any given
set of human parameters (e.g., age, sex, stature, and BMI), the landmark locations or
the nodal coordinates of the template mesh can be predicted.
10.3.2.1 Generalized Procrustes Alignment
GPA is used to align the landmarks/meshes from different subjects, so further statistical analyses can be performed. GPA has been widely used in morphometric studies
of anthropology and anatomy (Dijksterhuis and Gower, 1992; Slice, 2005; Stegmann
and Gomez, 2002), which involve the following steps:
1. To represent the geometry for each subject, construct a matrix xi with a
dimension of n 3, where n is the landmark number, and each row of the
matrix represents the landmark 3D coordinate.
2. Calculate the mean shape x of all the (m) subjects xi:
x¼
m
1 X
xi .
m i¼1
(10.1)
3. Use the Procrustes superimposition to align the remaining shapes to the mean
shape:
x0i ¼
1
ðxi CÞ½T;
CS
(10.2)
425
426
CHAPTER 10 Parametric Human Modeling
FIGURE 10.4
Method for developing statistical models of human geometry.
10.3 How to Build a Parametric Human Model
where [T] is a 3 3 matrix that includes the orthogonal rotation and reflection
components, C is the translation component along the x-, y-, and z-directions, and
CS is the centroid size (CS), which is the square root of the sum-of-the-squared
coordinate values in all dimensions, as shown in Eq. (10.3).
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
CS ¼
tr xTi xi ;
(10.3)
where “tr” is the trace of xTi xi , a matrix with a dimension of 3 3.
4. Recalculate the mean shape x0 of all the aligned shapes x0i :
x0 ¼
m
1 X
x0 .
m i¼1 i
(10.4)
5. If the difference between x and x0 is greater than a threshold (e.g., 1 106 mm),
return to step 3. After the GPAs, the geometry data of all the subjects are
aligned, they are separated into two components, the CS and the shape vector
(xi C)[T].
10.3.2.2 Principal Component Analysis and Regression Analysis
PCA is used to express the geometry data on an orthogonal basis and to quantify the
data variance in an efficient way. Geometrically, the first principal component (PC)
is the direction in the space of the data with the highest geometric variance, the second PC is in the direction orthogonal to the first PC with the second highest variance,
and so on. Multivariate regression analysis is used to predict the variation of PC
scores, associated with the PCs generated by PCA, with age, sex, height, BMI,
and/or other occupant parameters, which in turn is used to predict detailed body
geometry. In addition, if a random component with a standard deviation given by
the residual vectors is added to the regression model, the variations of the body
geometry with the same set of subject parameters can be predicted.
The PCA method introduced here follows the method reported by Li et al. (2011).
The coordinates of the landmarks were rigidly aligned using Procrustes alignment
and rescaling (Slice, 2007). Three coordinates at each of the landmarks formed a
geometry vector (denoted as g for one subject) with a length of l (which can be calculated as the product of the number of landmarks and 3, the number of coordinates per
landmark). The geometry vectors for all subjects were joined together to construct a
geometry matrix G1. To make the PCA method work properly, the geometry matrix
G1 was centered by subtracting the mean g from the individual geometry vectors gi.
The resulting matrix is called the data centered matrix G. The PCA was computed by
calculating the eigenvalues and eigenvectors of the covariance matrix, which is from
the centered geometry matrix G. G was deduced as follows:
G ¼ SP
(10.5)
S ¼ GPT ;
(10.6)
and
427
428
CHAPTER 10 Parametric Human Modeling
where S is an N l matrix called principal component (PC) scores and P is the
eigenvectors of G, which is an l l-normalized matrix. Any subject’s nodal coordinates or thicknesses could be obtained based on Eq. (10.7),
gi ¼ g þ PTN STNi ;
(10.7)
where SNi is the row of matrix SN corresponding to the ith subject’s PC scores.
To use parameters such as age, BMI, and bone length to predict PC scores (Sk), and
in turn, to predict detailed geometry, a regression analysis was performed. A regression
model was generated following the procedure used in Reed and Parkinson (2008):
STk ¼ CF þ εT ;
(10.8)
where F is the feature matrix, C is the coefficient matrix, and ε is a vector of zero
mean and normally distributed residuals.
T
10.3.2.3 Examples of Statistical Models of Human Geometry
As reported in scientific literature, many statistical models of human geometry have
been developed for the skeleton and internal organs, such as the skull (Urban et al.,
2014), rib cage (Gayzik et al., 2008; Holcombe et al., 2016; Shi et al., 2014; Wang
et al., 2016; Weaver et al., 2014a, 2014b), femur (Bredbenner and Nicolella, 2008;
Bryan et al., 2009; Klein et al., 2015; Zhu and Li, 2011), tibia (Baka et al., 2014;
Bredbenner et al., 2010), brain (Danelson et al., 2008), liver (Lamecker et al.,
2004; Lu and Untaroiu, 2014), and spleen (Yates et al., 2016). Statistical models
of external body surfaces are also available (Park et al., 2015; Park and Reed,
2015; Reed and Parkinson, 2008). Because the thorax and lower extremities are
the two body regions most affected by age and obesity in MVCs, specific examples
of rib cage, pelvis, and femur models are introduced below.
In two consecutive studies by Shi et al. (2014) and Wang et al. (2016), a statistical model of the adult rib cage was developed and improved. In Wang’s study,
anonymous clinical rib-cage CT scans (n ¼ 101) were obtained from the University
of Michigan Health System, using a protocol approved by an IRB at the University
of Michigan. All subjects were adult female (n ¼ 47) or male (n ¼ 54) patients
without skeletal pathology. The rib-cage geometry for each subject was collected
through a series of image analyses, including threshold-based image segmentation,
landmark identification on each rib, sternum and spine, landmark reprocessing
through B-spline, and landmark symmetry adjustment. A total of 1016 landmarks
on the rib cage, including 960 landmarks on the ribs, 32 landmarks on the sternum,
and 24 landmarks on the spine, were identified for each subject. After the landmarks
were identified, a template rib-cage model was mapped into the geometry of each
subject, using a mesh-morphing method. GPA, PCA, and regression analyses
were then used to develop a parametric model that used age, sex, stature, and
BMI to predict nodal locations of the FE rib-cage mesh.
Fig. 10.5 shows the effects of age, sex, stature, and BMI on rib-cage geometry.
Age affects the rib angle and rib-cage depth, and except for ribs 11 and 12, an
increase in age is associated with a raise in the front edges of the ribs. This, in
10.3 How to Build a Parametric Human Model
FIGURE 10.5
The effects of age, sex, stature, and BMI on rib-cage geometries.
Reprinted from article by Shi, X., Cao, L., Reed, M.P., Rupp, J.D., Hoff, C.N., Hu, J., 2014. A statistical human
rib cage geometry model accounting for variations by age, sex, stature and body mass index. Journal of
Biomechanics 47, 2277e2285, with permission from Elsevier.
turn, results in greater overall rib-cage depth. However, this increase in depth is associated with a decrease in width, especially in the middle of the rib cage. The effects of
sex are significant for rib angles and depths of rib cages, with men having flatter angles and greater depths than women with the same stature. Interestingly, in contrast to
the effects of age, the widths of the rib cages were greater in men than in women with
the same stature, indicating that men’s rib cages have greater volume. In addition, the
effects due to sex are consistent at different statures. The effects of an increase in BMI
on the rib-cage geometry are similar to those of an increase in age, both of which are
associated with a flatter rib angle and an increase in depth and decrease in width of
the rib cage. However, across the range in the current sample, the effects due to BMI
are greater than the effects due to age. Increases in stature are associated with
increased depth, width, and height of the rib cage, but decreased rib angle. More
detailed results can be found in the studies by Shi et al. (2014) and Wang et al. (2016).
Using a similar method to the one described for building the rib-cage model, statistical models of lower extremity bones were developed by Klein et al. (2015), based
on CT scans from more than 100 subjects between the ages of 17 and 89 years, with
heights of 1.5e2 m, and BMIs of 15e46 kg/m2. The unique aspect of this study is
that during the data-extraction step, landmarks representing both the bone shape
and the cortical thickness were collected. Fig. 10.6 shows the results of the femur geometries after GPA, PCA, and regression analysis reflecting the effects of age, femur
length, BMI, and sex were completed. These femur models were created by varying
429
430
CHAPTER 10 Parametric Human Modeling
FIGURE 10.6
The effects of age, BMI, femur length, and sex on femur geometry predicted by the
parametric models.
Klein (2015) University of Michigan, PhD Dissertation; adapted with permission.
one parameter at a time and holding the other parameters constant. The cross sections
for five evenly spaced locations along the shaft are also shown for comparison. As
shown, the femur length has the most significant effects on femur geometry, but
age, sex, and BMI all affect the femur shape and cortical bone thickness. Similarly,
the effects of age, bispinous breadth, BMI, and sex on pelvis geometry predicted by
the pelvis model are shown in Fig. 10.7. The sex effect is strong in determining the
pelvis shape, but the stature, age, and BMI effects are not as significant.
10.3.3 MESH MORPHING
Due to the time-consuming process of building a human, whole-body FE model,
current models are only in a few body sizes, and thus are not able to account
for the variability in human geometry. Although there is a large variability
10.3 How to Build a Parametric Human Model
FIGURE 10.7
The effects of age, BMI, bispinous breadth, and sex on pelvis geometry predicted by the
parametric models.
Klein (2015) University of Michigan, PhD Dissertation; adapted with permission.
in geometry, we should recognize that human bodies are anatomically similar. It is
possible that the mesh from a model can be changed smoothly into other geometries
without the need for developing new meshes. Therefore, the basic concept for developing a parametric model is to morph a baseline model into different geometries using an automated mesh-morphing method. In this way, multiple models with
different combinations of human characteristics can be generated rapidly.
Although mesh-morphing methods were introduced in scientific literature in the
late 1990s, the concept of parametric modeling became popular only in recent years.
Table 10.2 lists some of the studies on the development of parametric human models.
Note that all these models are at the component level; a parametric, whole-body, human FE model is rare. Although methods for morphing mesh vary among studies,
they can be divided into two types: landmark-based and surface matchingebased.
Because the nature of the statistical-geometry model is based on landmarks,
landmark-based mesh morphing is more suitable for linking statistical-geometry
models to baseline FE models.
Among landmark-based mesh-morphing methods, radial basis functions (RBFs)
are among the most popular. RBFs have been widely used in image processing and
neural networks (Bennink et al., 2007; Carr et al., 2001). To use RBFs for mesh
morphing, corresponding landmarks need to be identified in both the statisticalgeometry model and the baseline FE model, so that the nodal displacement at
each landmark location can be calculated. Using RBFs, a 3D displacement field
throughout the entire space of the geometry can be calculated based on the landmark
displacements. By applying this displacement field to the baseline FE mesh, a new
model with new geometry can be obtained. In terms of geometry accuracy and mesh
quality, the thin-plate spline function and multiquadratic function are the most
431
432
Besnault
et al. (1998)
Bryan
(2010)
Grassi
et al.
(2011)
Grosland
et al. (2009)
Bucki et al.
(2010)
Couteau
et al. (2000)
O’Reilly
and Whyne
(2008)
Li et al.
(2011)
Body
region
Morphing
method
Pelvis
Femur
Femur
Phalanx
Face
Femur
Spine
Head
Kriging
Radial basis
function
Deformable
registration
algorithm
Meshmatching
algorithm
Elastic
volumetric
registration
Landmarkbased
parametric
meshing
Radial basis
function
Morphing
type
Landmarkbased
Surface
matching
and
Laplace
smoothing
Surface
matching
Landmarkbased
Surface
matching
Surface
matching
Surface
matching
Landmarkbased
Landmarkbased
Reference
Figure
CHAPTER 10 Parametric Human Modeling
Table 10.2 Recent Parametric Human FE Models
10.3 How to Build a Parametric Human Model
suitable RBFs available for mesh morphing (Li et al., 2012). It should be noted that
RBFs are not only used for FE meshing, but can also be used to systematically
change the information associated with each node, such as the values of the cortical
bone thickness and material properties.
To use meshes morphed with landmark-based RBFs, landmarks must be identified at locations on the baseline FE model corresponding to the locations predicted
by the statistical-geometry models. The basic concepts and formulas of the RBF
interpolation are previously described (Bennink et al., 2007; Carr et al., 1997).
The applications of this method for developing parametric FE models have also
been well documented (Hu et al., 2012, 2016; Hwang et al., 2016a, 2016b; Li
et al., 2011, 2012; Shi et al., 2015).
In general, an RBF takes the form
sðxÞ ¼ pðxÞ þ
n
X
li 4ðkx xi kÞ.
(10.9)
i¼1
where p(x) is a low-order polynomial, li is the weighting coefficient, 4 is the basis
function, and kx xik is the Euclidean norm (the distance between x and xi).
A first-order polynomial is selected for p(x), and the thin-plate spline function
4(r) ¼ r2logr is selected as the basis function because it generally results in smooth
mesh (Li et al., 2012). In this study, because of the 3D coordinates of the nodes, the
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
2
Euclidean norm is defined as rij ¼ ðxi xj Þ2 þ yi yj þ ðzi zj Þ2 .
We assume that the numbers of landmarks on the baseline model and the target
geometry are both n. To compute the weighting coefficient li, and the coefficients of
the polynomial function p(x), RBF can be written in matrix form as Eq. (10.10),
by applying interpolation requirements and boundary conditions:
A þ aI B
l
T
¼
;
(10.10)
BT
0
c
0
where A is a basis function 4 in the n n matrix, in which A0ij ¼ 4ðrij Þ ¼ rij2 logðrij Þ;
2
3
1 x1 y1 z 1
6
7
B¼ 4« «
«
« 5 , where x, y, and z are the coordinates of the baseline
1
xn
yn zn nx4
3
2
X1 Y1 Z 1
7
6
«
« 5 , where X, Y, and Z are the coordinates of the
landmarks; T ¼ 4 «
Xn Yn Zn nx3
target landmarks; and a is a smoothing factor, but in this study a ¼ 0 is assumed,
as the exact interpolation is to be represented; and c is constant.
Once l and c are determined by solving Eq. (10.10), with the assumption that the
number of nodes in the baseline model is N, the coordinates of the nodes in the
morphed model (T 0 ) can be calculated by Eq. (10.11):
0
0 l
A þ aI B0
T
¼
;
(10.11)
c
B0T
0
0
433
434
CHAPTER 10 Parametric Human Modeling
where A0 is an N n matrix in which Aij ¼ 4(rij) ¼ rij2 logðrij Þ, i is the number of
nodes 2in the baseline model,
and j is the number of the baseline landmarks.
3
1 x1 y1 z 1
7
6
B0 ¼ 4 « «
«
« 5 , where x, y, and z are the coordinates of the nodes
1 xN yN zN Nx4
3
2
X 1 Y 1 Z1
7
6
«
« 5 , where X, Y, and Z are the coordiin the baseline model; T 0 ¼ 4 «
XN YN ZN Nx3
nates of the nodes in the morphed model; and a ¼ 0 is assumed to represent an exact
interpolation.
10.3.4 EXAMPLE OF PARAMETRIC, WHOLE-BODY, HUMAN MODEL
To build a parametric, whole-body, human model, the statistical models of the skeleton and external body surfaces need to be integrated. Because these models were
typically developed based on different groups of subjects, it is important to follow
the defined procedures rigorously when performing the integration. Hwang et al.
(2016a) used limited-body landmarks and joint centers that were available in the
body-shape model to position the rib cage, pelvis, femur, and tibia models.
Fig. 10.8 shows the reference points with respect to each bone model for a typical
FIGURE 10.8
Reference points for positioning each statistical skeleton model.
10.3 How to Build a Parametric Human Model
bone-positioning process. From the model of the external body surface, a set of joint
center points and body-surface landmarks are chosen as the reference points, while
corresponding landmarks on the bones are also identified. Singular value decomposition is used to minimize the sum-of-the-square spatial distances between the corresponding landmark coordinates from the external body surface and the bones. The
statistical-geometry model of the skeleton may include only the bones (rib cage,
pelvis, femur, and tibia) that are most likely to be injured in MVCs; therefore, all
other bony structures can be scaled/morphed based on already registered bones
and landmarks associated with the model of the external body shape.
Once the geometry models are integrated, morphing the mesh of the whole
body can be conducted. Because the computational cost of mesh morphing increases substantially as the number of landmarks increase, and because the human model has over 20,000 landmarks, the model can be divided into several
regions, and the regions can be morphed separately, prior to combining the regions together to form the entire model. When the process of morphing the whole
body is done on a contemporary computer, it typically requires less than 20 min,
even for the most complex models, such as THUMS v4 or GHBMC models.
Examples of morphed human models with a wide range of stature and BMI are
shown in Fig. 10.9.
FIGURE 10.9
Six morphed, human body models with a wide range of stature and BMIs.
435
436
CHAPTER 10 Parametric Human Modeling
10.3.5 TISSUE MATERIAL PROPERTIES FOR A PARAMETRIC,
HUMAN MODEL
Once the FE mesh is generated, material properties suitable for occupants with
different characteristics, such as age, sex, and BMI, need to be assigned to different
body components. The effects of age and sex on bone material properties are widely
reported in scientific literature (Burstein et al., 1976; Kemper et al., 2005, 2007;
Lobdell et al., 1973; Takahashi et al., 2000; Wall et al., 1979; Yamada, 1970), but
similar effects on properties of soft tissue are not well understood. Furthermore,
large variations generally exist in material properties of human tissues, even in populations of people of the same age and sex. Thus, stochastic material models,
including not only the means but also the standard deviations of the material parameters, are needed for a parametric, whole-body, human model. Such models are not
yet available for human tissues from different body regions, but the method for
creating such models has been developed and demonstrated in a study by
Hu et al. (2011), in which a stochastic visco-hyperelastic model of human placenta
tissue was developed using a combination of tensile testing, specimen-specific FE
modeling, and stochastic optimization methods.
10.4 HOW TO VALIDATE A PARAMETRIC HUMAN MODEL
Although cadaver tests were routinely used to validate human FE models, the validations were limited to a few body sizes (i.e., midsized male, small female, and large
male) with the same body shape. Effects of age and obesity have rarely been considered in model-validation processes. The major difference between validating a single FE model and a parametric one is that the parametric model can be morphed into
geometries representing specific cadavers. Consequently, more accurate, subjectspecific model validations can be conducted, leading to a better understanding of
the relationship between material properties and human impact responses.
The validation of a parametric, human FE model should involve morphing the
whole model to represent groups of individual cadavers, and morphing segments
of the model to represent cadaver bone specimens used in biomechanical tests to
characterize responses and tolerances, reconstruct loading conditions applied to
each of these subjects/specimens, and compare model-predicted responses to those
measured experimentally for different test subjects. The goal of this validation
process is to match the overall trends in measured responses for each group of
test subjects considering effects of age and obesity rather than matching only cases
of single responses.
Table 10.3 lists studies focused on the effects of age and obesity that can be used
for model validation. Cadaver studies with whole-body CT scans, which are the
most suitable for parametric model validation, are still largely lacking. For cadaver
component tests with CT scans available, subject-specific FE models should be
generated as part of the test reconstruction process, while for whole-body cadaver
10.4 How to Validate a Parametric Human Model
Table 10.3 Studies Can Be Used in Validating the Parametric Human FE
Model
Study References
Validation Data
Ivarsson et al. (2009)
Rupp et al. (2003)
Femur compression/bending
Pelvis, femur, knee impact
response
Pelvis, femur, knee impact
response
Rib bending
Isolated rib-cage impact
Thorax pendulum impact
Various thorax impact conditions
Whole-body sled test
Abdomen belt loading
Abdomen belt loading
Rupp et al. (2008)
Charpail et al. (2005)
Vezin and Berthet (2009)
Kroell and Schneider
(1971)
Kent and Patrie (2005)
Kent et al. (2010)
Foster et al. (2006)
Lamielle et al. (2008)
Body
Region
Effects
Femur
KTH
Age
Age
KTH
Age
Ribs
Rib cage
Thorax
Age
Age
Age
Thorax
Whole body
Abdomen
Abdomen
Age
Obesity
Obesity
Obesity
tests, the subject-specific FE model should be generated to not only represent accurate geometry of the skeleton and body shape, but also the sitting posture of the
cadaver in the test. During the model-validation process, material properties and
boundary conditions can be tuned using optimization methods, so that the model responses best fit the cadaver responses. Compared to the traditional model-validation
method, in which the responses of a single human model are compared with a testing
corridor, without considering the geometry variations among the cadavers, subjectspecific models would significantly reduce the geometry and posture differences between the model and the cadaver. Consequently, the material properties of the model
could be tuned more accurately, and the errors in the final-impact responses could be
significantly reduced. However, it should be noted that to achieve this level of parametric model validation, additional cadaveric studies with whole-body CT scans
focusing on the effects of age, sex, and obesity effects, are necessary.
In scientific literature, studies on validating a parametric whole-body, human
model against cadaver test results are scarce. Shi et al. (2015) compared simulation
results from human models with different BMI levels to data from cadaver sled tests
in frontal-crash conditions. Only the effects of obesity on the increased percentage
of body excursion (head, shoulder, hip, and knee) were used to evaluate the model
accuracy, while the statures of the models were not associated with the cadavers.
More recently, Hwang et al. (2016b) conducted subject-specific model evaluations
against four cadaver side-impact tests using two cadavers with notably different stature and BMI levels. The simulation results showed a preliminary trend that the correlations between the impact responses predicted by the parametric model and those
measured in the four cadaver tests were better than the correlations found between
the original midsized male model and the normalized cadaver responses. Although
437
438
CHAPTER 10 Parametric Human Modeling
Hwang’s study is limited by the number of cadavers, it demonstrated the advantages
of using a parametric human model to predict human impact responses accounting
for various human characteristics.
10.5 CHAPTER CONCLUSION
For this chapter, a literature review was conducted on the effects of age, sex, and
obesity on MVC-induced injuries and the recent developments in human modeling
technologies for investigating the impact responses for various vulnerable populations. Recent studies on statistical, human, geometry models; mesh morphing; human tissue tests; and whole-body cadaver tests all converge nicely toward a
parametric, human FE model, which can represent a diverse population. The method
for building the parametric, human model will enable population-based simulations
for future safety design optimizations targeting various vulnerable populations that
cannot be represented with current injury assessment tools.
ACKNOWLEDGMENTS
The material in this chapter is based on the research studies funded by National Highway
Traffic Safety Administration, the National Science Foundation, the National Institute of Justice, Toyota, General Motors, and Ford Motor Company. The author would like to thank Dr.
Matthew Reed, Dr. Jonathan Rupp, Dr. Byong-Keon Park, Dr. Monica Jones, and Dr. Eunjoo
Hwang from the University of Michigan Transportation Research Institute (UMTRI) for their
tremendous support on these projects. The author would also like to thank Dr. Katlyn Hunter,
Dr. Zhigang Li, Dr. Xiangnan Shi, and Dr. Yulong Wang for their Ph.D. dissertation research
at UMTRI related to parametric human modeling.
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CHAPTER
Modeling Passive and
Active Muscles
11
Masami Iwamoto
Toyota Central Research & Development Laboratories, Inc., Nagakute-city, Japan
11.1 INTRODUCTION
In real-world automotive accidents, drivers often engage in protective activity to
avoid the crash itself or subsequent damage to their bodies. Researchers have
investigated drivers’ performances before upcoming impacts. According to data
from the Institute for Traffic Accident Research and Data Analysis (ITARDA,
2005), more than 60% of drivers involved in traffic accidents in Japan made
evasive maneuvers, such as braking, steering, or both. Hault-Dubrulle et al.
(2009) investigated drivers’ performances in a series of volunteer tests in which
a driving simulator was used to determine activity that might ensue after the driver
sees a vehicle approaching from the front. The volunteers included 80 adult subjects, both male and female. Sixty-seven percent of the volunteers extended their
right legs to step on the brake pedal and their arms to push on the steering wheel
to brace their bodies for the upcoming crash. These findings suggest that drivers
could activate their muscles for upcoming crashes. In another study, Ejima et al.
(2009) conducted a series of volunteer tests using adult males to investigate differences in occupant behaviors during decelerations of 0.8 G in both relaxed and
tensed states. Their results demonstrated that head extrusions of tensed occupants
were significantly different from those of relaxed occupants. Results of these
studies suggest that muscle activity could have a significant influence on occupant
kinematics during low-speed impacts.
Kallieris et al. (1995) compared injury outcomes between cadaver tests and realworld accidents in frontal crashes with a consistent level of impact velocity at
50 km/h, and concluded that more rib fractures were observed in the cadaver tests
and more lower or upper extremity injuries in the real-world accidents. Cadavers
have no muscle activity, while people involved in the real-world accidents use a variety of muscle activity, and it is hypothesized that muscle activity could affect injury
outcomes. Hence, tests with cadavers are not adequate for predicting injury outcomes, and muscle modeling with passive and active properties is critical for investigating differences in occupant motions and injury outcomes between cadaver tests
and real-world accidents.
Basic Finite Element Method as Applied to Injury Biomechanics. http://dx.doi.org/10.1016/B978-0-12-809831-8.00011-8
Copyright © 2018 Elsevier Inc. All rights reserved.
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CHAPTER 11 Modeling Passive and Active Muscles
FIGURE 11.1
Structure models of muscles (left lower arm). (A) Anatomical text (Gray’s anatomy),
(B) truss model, and (C) hybrid (truss þ solid).
Macroscopic structures and mechanical properties of muscles are predominant in
impact situations, and therefore we will not cover microscopic anatomical structures
of human skeletal muscle, especially microlevel structures such as actin or myosin.
From a macroscopic view, a muscle has volume that includes muscle fibers that lie
along the muscle action line. In addition, a muscle is connected with bones via tendons, and contraction of the muscle controls displacement and rotation of the bones
attached to it. Muscles are activated to generate or suppress force, supply heat, support joints and body posture, and provide a protective cushion to keep internal body
structures safe from external forces. In this chapter, simplified anatomical and mechanical models are introduced and used to reproduce such macroscopic muscle
functions. Fig. 11.1 shows two types of muscle models for the left human arm
and an anatomical diagram for comparison. In Fig. 11.1B, muscles are modeled
as truss elements, and the tendons that connect the muscles to the bones are modeled
as seat-belt elements. In Fig. 11.1C, muscles are modeled as hybrid elements with a
combination of truss elements with active-muscle properties and solid elements with
passive-muscle properties. The tendons are modeled as shell or bar elements with
nonlinear elastic properties.
Typical mechanical responses of muscles can be described by using the Hill-type
muscle model, which according to Zajac (1989) is often represented by the
11.1 Introduction
FIGURE 11.2
Hill-type muscle model.
schematic shown in Fig. 11.2. The passive and active properties of a muscle are
described in the following equation:
n
M
M o
M
$FCE V LM ; L_
F M ¼ PCSAM aM $sM
þ sM
max $FCE L L
max $FPE L
(11.1)
From this equation, we can see that a muscle force is the sum of the active and passive forces of the muscle. The variable aM , which ranges from 0 to 1, represents the
activation level of the skeletal muscle M, and PCSAM represents the physiological
cross-sectional area of this muscle. The constant sM
max stands for the variable of
maximum contraction force per unit cross-sectional area of muscle M. According
2
to Gans (1982), sM
max is 5.5 kgf/cm for humans.
Fig. 11.3 shows typical function curves representing relationships between active
force and length, between active force and velocity, and between passive force and
elongation of a muscle. The active force is proportional to the active forceelength
relationship multiplied by the active forceevelocity relationship. The active forcee
FIGURE 11.3
Typical function curves of Hill-type muscle model. (A) Active forceelength, (B) active
forceevelocity, and (C) passive forceelength.
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CHAPTER 11 Modeling Passive and Active Muscles
length relationship was reported by Gordon et al. (1966) as an important feature of
the muscle. They conducted isometric contraction tests using frogs’ muscles and obtained the lengthetension relationship. The active forceevelocity relationship is
also an important feature. Hill (1938) conducted isotonic contraction tests using
frogs’ muscles to obtain the velocityetension relationship. The forceeelongation relationships pertaining to passive characteristics were obtained from tensile tests data
in which human or animal muscles were used (Yamada, 1970).
The Hill-type muscle model is used to simulate macroscopic forces generated by
the muscle. Major software packages for simulating human motion include SIMM
(Software for Interactive Musculoskeletal Modeling, MusculoGraphics, Inc.,
USA) and AnyBody (AnyBody Technology, Denmark). Software packages such
as MADYMO (MAthematical DYnamic MOdels, TASS International, The
Netherlands) are also capable of multi-body modeling. All of these software packages make use of the Hill-type muscle model. It is worth noting that explicit finite
element codes LS-DYNA (LSTC, USA) and VPS (Virtual Performance Solution,
ESI Group, France) also make use of the Hill-type muscle model.
In 1986, the US National Library of Medicine initiated a Visible Human Project,
through which high-resolution MRI and CT scans and thin-layer cryosectioning of
two cadaveric bodies were conducted. The donors for this project were a 38-year-old
male with a height of 180 cm and weight of 90 kg, and a 59-year-old female with a
height of 167 cm and weight of 72 kg. The PCSAs of these two subjects were determined from the high-resolution photographs of the relevant cryosections. The
PCSAs of lower extremity muscles were larger than the ones in the arms and
neck. This indicates that lower extremity muscles generate larger forces while standing or walking (Table 11.1).
11.2 METHODS FOR MODELING PASSIVE MUSCLE
The passive properties of muscles are derived from data gathered from tensile tests
performed along the direction of the muscle fiber and compressive or impact tests
performed in an orthogonal direction. Series of tensile tests were conducted on muscles from human cadavers by Yamada (1970) and on dog muscles by Gras et al.
(2012). Iwamoto et al. (2009, 2012) conducted indentation tests for biceps brachii
muscles on human volunteers, and Loocke et al. (2008) conducted compressive tests
using porcine muscles.
Östh et al. (2012) used truss and bar elements to develop a human body model containing models of one-dimensional muscles in which the passive properties were
described as nonlinear elastic material. The stressestrain curves that were used in
these models were based on experimental data. These one-dimensional muscle
models can be used to simulate tensile properties of the muscles, but not compressive
properties orthogonal to the direction of the muscle fibers. Other researchers have
developed a human body model containing muscle models with three-dimensional
geometry (Behr et al., 2006; Hedenstierna et al., 2008; Iwamoto et al., 2011). A
11.2 Methods for Modeling Passive Muscle
Table 11.1 Physiological Cross-sectional Areas of Human Muscles
PCSA (mm2)
Body Part
Muscle Name
Neck
Sternocleidomastoid
Longus colli
Splenius capitis
Semispinalis capitis
Levator scapulae
Trapezius
Deltoid
Pectoralis major
Biceps brachii
Brachialis
Flexor carpi ulnaris
Extensor digitorum
Flexor carpi radialis
Rectus abdominis
External oblique
Gluteus medius
Gluteus Maximus
Sartorius
Vastus medialis
Vastus lateralis
Tibialis anterior
Flexor digitorum longus
Extensor digitorum Longus
Shoulder
Upper arm
Lower arm
Abdomen
Hip
Upper leg
Lower leg
Young
Male Sym
166.00
39.30
99.00
95.20
993.00
2323.00
2282.00
1179.00
319.00
881.00
557.00
430.00
310.00
658.00
685.00
1710.00
1813.00
320.00
3237.00
4063.00
1277.00
1897.00
667.00
Ratio
Elder
Female Sef
131.50
25.90
75.40
50.40
125.70
868.50
1455.40
872.08
174.67
247.67
129.17
81.48
92.59
372.48
439.86
1308.00
1458.00
268.00
1560.00
1648.00
848.00
586.00
404.00
Sef/Sym (%)
79.22
65.90
76.16
52.94
12.66
37.39
63.78
73.97
54.76
28.11
23.19
18.95
29.87
56.61
64.21
76.49
80.42
83.75
48.19
40.56
66.41
30.89
60.57
hybrid combination of truss elements with active-muscle properties and solid elements with passive-muscle properties were used for this latter model. Rubber-like
materials were used to emulate the passive properties, and the necessary stressestrain
curves were based on experimental data. This type of muscle modeling can be used to
simulate both tensile and compressive properties of muscles.
In 2009, Iwamoto et al. used an indentation machine and an electromyography
(EMG) machine to simultaneously measure forceedeflection curves and muscle activity in the biceps brachii of a human male volunteer. The 33-year-old subject, with
75 kg weight and 179 cm height, held his posture in a supine position while keeping
his elbow at 90 , once with and once without a 5-kg load attached to his right wrist.
Forceedeflection curves were measured with the circular head of the indentation
machine, which the subject himself pushed into the largest part of his biceps brachii.
The experimental setup is shown in Fig. 11.4.
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CHAPTER 11 Modeling Passive and Active Muscles
Indentation probe
Indentation machine
muscular power
Indentation machine
Skin
Displacement
90°
Electromyograph
Biceps brachii
Weight
(5kg, 0kg)
Diameter = 7 mm
Sectional image of indentation probe
FIGURE 11.4
Experimental setup for indentation tests.
Fig. 11.5A shows a simulation setup for indentation tests used to validate properties from compression orthogonal to the direction of the muscle fibers. In the simulation, the simple boundary conditions were reproduced for the indentation tests
with the position of the humerus bone fixed, based on the assumption that the arm
posture changed little during the tests. In this simulation, the biceps brachii muscle
was represented with the hybrid model mentioned above, in which truss elements
were merged with solid elements by shared nodes. An impactor was pushed into
the middle of the whole muscle while both ends of the muscle were fixed to a rigid
wall, which represented a bone. The data for time history curves of displacements
with and without the weight were obtained from the tests and were entered into
the indentation head model. EMG data recorded during voluntary isometric contraction were used to normalize the EMG data that were recorded when the impactor
was pushed into the muscle. The muscle-activation levels with and without the
weight were set as constant values of 5% and 0.16%, respectively.
Fig. 11.5B, which shows the deformations in the muscle model with and without
the weight, demonstrates that a larger deformation is produced without the weight.
Fig. 11.6 shows a comparison between the model prediction and the test data. The
predicted forceedisplacement curves well agreed with test data both with and
FIGURE 11.5
Simulation setup for validation of muscle stiffness change. (A) Simulation condition, (B)
simulation results on muscle deformation.
11.3 Methods for Modeling Muscular Activation
FIGURE 11.6
Comparison of forceedisplacement curves between model prediction and test data.
without the weight. These simulation results indicate that the hybrid model of the
biceps brachii muscle reproduced the stiffness of passive properties in the direction
orthogonal to the muscle fiber as well as the stiffness change with increasing muscle
activation.
11.3 METHODS FOR MODELING MUSCULAR ACTIVATION
Active muscle properties include activation level, forceelength curve, and forcee
velocity curve, according to Eq. (11.1). The forceelength and forceevelocity curves
can be obtained only from Hill-type material descriptions (for an example see
Thelen, 2003). For this reason, settings for forceelength and forceevelocity curves
are nearly identical among models of active human muscles. Since the only property
that substantially changes is the activation level, this property is obviously important
for depicting the differences in contributions from each of the modeled muscles.
There are particular methods that can be used to estimate the activation levels of
individual muscles. Normalized EMG data of surface muscles that have been
measured in experimental tests can be used directly, while the activation levels of
the other muscles are estimated by considering the role of each muscle (Iwamoto
et al., 2012). Muscle activations can be estimated with a PID controller (Iwamoto
et al., 2015; Östh et al., 2012; Rooij, 2011) or through reinforcement learning
(Iwamoto et al., 2012). Both methods are used to measure error levels of feedback
and adjust the model to address the errors. Detailed description of each method is
presented below.
11.3.1 ESTIMATION OF MUSCLE ACTIVATION BASED ON EMG DATA
EMG data obtained from experimental tests on surface muscles have been used as
estimates for muscle activations (Chang et al., 2009; Choi et al., 2005; Iwamoto
et al., 2012). Fig. 11.7 shows a diagram of an experimental setup to obtain EMG
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CHAPTER 11 Modeling Passive and Active Muscles
24 muscles for EMG measurement
Muscles (Lower ex.) Muscles (Upper ex.)
Soleus
Pectoralis Major
Deltoid Ant.
Tibialis Ant.
Gastrocnem Med.
Deltoid Lat.
Gastrocnem Lat.
Vastus Med.
Vastus Lat.
Rectus Femoris
Adductor Longus
Biceps Femoris
Semitendinosus
Gluteus Maximus
Deltoid Post.
Infraspinatus
Biceps Brachii
Brachialis
Triceps Brachii
Flexor Carpi Radialis
Flexor Carpi Ulnaris
Brachioradialis
Extensor Carpi Radialis
Extensor Carpi Ulnaris
3D motion analysis system
(3D motion of subjects) Camera (x 12)
Six axis load cell (steering reaction force)
Reflective marker
Force plate
(reaction force on seat)
Seat sensor
(pressure distribution on seat)
EMG measurement system (24ch)
(electromyography)
Six axis load cell
(pedal force)
FIGURE 11.7
Diagram of a measuring system for bracing motion.
data for muscles in the upper and lower extremities of human volunteers engaging in
bracing motions (Iwamoto et al., 2012). In the test, the same human male subject
involved in the indentation test (shown in Fig. 11.4) performed a bracing posture,
where he used maximal voluntary force to extend his right leg to step on the brake
pedal and his arms to push on the steering wheel. This posture was based on a previous study performed by Hault-Dubrulle et al. (2009). Several data sets, such as kinematic data of 3D motions of the whole body, EMG data from 24 skeletal muscles
of upper and lower extremities, pressure distributions on seats, and reaction forces
on the pedal, steering wheel, and seats, were obtained from this test and were
analyzed.
Muscle-activation levels were determined from the normalized EMG data obtained from the tests. Some of the results from tests on the lower and upper extremities are shown in Fig. 11.8. Fig. 11.9A shows the initial posture of a human body FE
FIGURE 11.8
Activation level time histories of some of the muscles in lower and upper extremities.
11.3 Methods for Modeling Muscular Activation
FIGURE 11.9
Driver’s motion predicted by model for bracing condition.
model with each muscle modeled with hybrid, 3D geometry elements. The simulation setup reproduced the volunteer test setup. The FE model was set to a sitting position on rigid seats with the right foot positioned on the brake pedal and the hands
positioned as gripping the steering wheel. The muscle-activation levels from the
EMG data of the 24 lower and upper extremity muscles were directly entered into
the corresponding muscle models, while those of the other muscles were estimated
by considering the actions of the muscles, such as flexion, extension, inversion, and
eversion of the corresponding joint, and whether the muscle was primary or supplementary. Fig. 11.9B shows a bracing posture predicted at 300 ms. From the time of
initial posture to the one shown in this figure, the hip displaced upward, the right leg
displaced forward and downward, and the head rotated rearward. This predicted
posture at 300 ms was similar to that observed in the volunteer test. Fig. 11.10 shows
FIGURE 11.10
Results of comparisons between tests and simulations for reaction forces in bracing
conditions with maximal force.
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CHAPTER 11 Modeling Passive and Active Muscles
a comparison of reaction forces between simulation results and volunteer test data
for the bracing condition with the maximal braking force. Predicted forces of the
pedal, steering wheel, and seat back showed good agreement with those from the
test data. The reaction force of the seat cushion was zero in the simulation, while
the force was 100 N in the test. This inconsistency is because the hip was completely
apart from the seat cushion in the bracing motion of the simulation.
11.3.2 ESTIMATION OF MUSCLE ACTIVATION USING PID
CONTROLLER
A PID (proportional, integral, derivative) controller is often used to predict occupant
motions during decelerations due to braking or low-speed impact (Iwamoto et al.,
2015; Östh et al., 2012; Rooij, 2011). Fig. 11.11A shows the head and neck portion
of an FE model of the human body. Each muscle in this region is represented with
hybrid, 3D geometry elements. This model was used for predicting head and neck
motions during a low-speed rear impact. Fig. 11.12 shows a diagram of a muscle
control system using a PID controller for an anatomical joint. A PID controller is
a generic control-loop feedback mechanism widely used in industrial control systems. It calculates an “error” value as the difference not only between a measured
process and a desired trajectory, but also the rate at which the model is approaching
the desired state. The controller attempts to minimize the error by adjusting the
process control inputs prior to passing the set point (Johnson and Moradi, 2005).
The first step in using the muscle control system is to use LS-DYNA to calculate
the occupant’s motions during deceleration caused by impact. The current angle q
of each joint, which is set as shown in Fig. 11.11B, is calculated based on the
X-, Y-, Z-coordinates and displacements of the three nodes associated with the
angle in time t. As referenced in scientific literature (for example, Johnson and
Moradi, 2005), a PID controller provides feedback regarding what the current
FIGURE 11.11
Headeneck FE model used for study with PID controller. (A) Headeneck model with
muscles and (B) definition of head rotational angle.
11.3 Methods for Modeling Muscular Activation
User input data
Reaction Time Delay
(option)
Current joint
angle θ
Target joint
angle θT
PID gains
Kp, Ki, Kd
Difference
e = θ– θ
PID Control
Neural Delay
(option)
CCR
(Co-Contraction
Ratio)
Activation of
each muscle
Simulation using
human FE model
Feedback
FIGURE 11.12
Feedback loop for muscle control with incorporation of PID controller.
angle q should be in current time in order to have the best opportunity of reaching the
predetermined target angle qs at the desired time. The manipulator MV is used
to adjust model parameters in accordance with the assessed deviation as follows:
Z
de
MVðtÞ ¼ KP e þ KI edt þ KD
(11.2)
dt
where e is the difference between the current and target joint angles; KP, KI, and KD
are the gains for P, I, and D controls, respectively. According to Koike and Kawato
(2000), muscle activation is related to the firing rate of motor neurons, which is
described by a sigmoid function (Dayna and Abbott, 2001). Therefore, the activation
level AM ðtÞ of a muscle M associated with a joint can be described by the following
sigmoid function in time t, using the manipulator MVðtÞ,
1
M
A ðtÞ ¼ CCR$
(11.3)
1 þ expðS$BðAM ðt DtÞ þ MV ðtÞÞ þ SÞ
where CCR is the co-contraction ratio of flexors or extensors relative to total contraction area, S is used to distinguish activity of flexors from extensors with 1.0 for
flexors and 1.0 for extensors, B is a constant of 5.0 for the sigmoid function, and
Dt is the time-step. Because the deviations eX, eY, and eZ for the X, Y, and Z axes,
respectively, exist, MVX, MVY, and MVZ also exist; therefore AX (t), AY (t), and
AZ (t) must also exist. It is hypothesized that muscle-activation patterns depend on
percentages of contributions from the muscles used for specific joint motions,
such as flexion, extension, inversion, eversion, internal, and external rotation. The
activation level ALM ðtÞ of muscle M at time t is defined as:
M
M
M
M
M
M
ALM ðtÞ ¼ ALM
0 þ cmX $AX ðtÞ þ cmY $AY ðtÞ þ cmZ $AZ ðtÞ
(11.4)
M
where ALM
0 ðtÞ is the initial activation level of muscle M, and cmX is the contribution
M
percentages of inversion and eversion about the X-axis, cmY is that of flexion and
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CHAPTER 11 Modeling Passive and Active Muscles
Table 11.2 Percentages of Neck Muscle Contributions
Name of
Muscles
Longus colli
Scalenus
anterior
Semispinalis
capitis
Splenius
capitis
Flexion
Extension
Inversion
Int.
Rotation
Eversion
Ext.
Rotation
0.663
0.139
0.0
0.0
0.048
0.188
0.0
0.0
0.29
0.673
0.0
0.0
0.0
0.479
0.0
0.229
0.0
0.292
0.0
0.529
0.0
0.163
0.0
0.308
extension along the Y-axis, and cmM
Z is that of internal or external rotation about the
Z-axis. The contribution percentages for each of the neck muscles were obtained
from scientific literature (Rooij, 2011), as shown in Table 11.2. Translational and
rotational accelerations of T1, obtained from experimental test data (Ono et al.,
1997), were entered into the program for the center of gravity (CG) of T1. Both
PMHSs (White et al., 2009) and volunteers (Ono et al., 1997), were used in these
experiments with the initial velocity of 8 km/h.
Fig. 11.13 shows comparison curves between simulation results and PMHS test
data of head angles relative to time in low-speed rear impacts of 8 km/h. For the
simulation, gains for P, I, and D were all set to 0 to represent cadaveric responses
without muscle activation. Simulation results without muscle activation showed
good agreement with the PMHS test data.
(A)
(B)
PMHS test 1
Model (P=0,I=0,D=0）
PMHS test 2
20
Head rotational angle (deg)
458
0
-20
0
0.1
0.2
0.3
0.4
-40
-60
-80
-100
Time (sec)
FIGURE 11.13
Comparison between model response and PMHS tests for head angles in low-speed, rear
impact at 8 km/h. (A) Maximum head rotation (0.35 s after impact) and (B) comparison of
head angles between model and PMHS tests.
11.3 Methods for Modeling Muscular Activation
(A)
(B)
Model (CCR=0.7)
20
0
0
-20
0.1
0.2
0.3
0.4
-40
-60
-80
-100
Time (sec)
Volunteer test
Head rotational angle (deg)
Head rotational angle (deg)
Volunteer test
Model (CCR=0.4)
20
0
-20
-40
-60
-80
-100
Time (sec)
FIGURE 11.14
Comparison between model responses and volunteer tests for head angles in low-speed,
rear impact at 8 km/h. (A) CCR for flexors ¼ 0.7 and (B) CCR for flexors ¼ 0.4.
Fig. 11.14 shows curves for time histories of head angles for two different CCR
values, with each graph containing the same results from experimental data. When
the CCR for the flexor muscles was set to 0.7, the flexor muscles constrained head
rotation by about 20 degrees relative to the head rotation of the volunteers. At 0.4,
the maximum head angle predicted by the model was almost the same as that
of the volunteers. This indicates that the CCR is a significant factor in controlling
the rotational motion. Fig. 11.15 shows comparisons of activation levels of the longus colli and scalenus anterior flexors, and the semispinalis capitis and splenius
(A)
(B)
Longus Colli
Semispinalis Capitis
Scalenus Anterior
Splenius Capitis
1
Muscle Activation Level
Muscle Activation Level
Longus Colli
Semispinalis Capitis
0.8
0.6
0.4
0.2
0
Time (sec)
Scalenus Anterior
Splenius Capitis
1
0.8
0.6
0.4
0.2
0
Time (sec)
FIGURE 11.15
Predicted activation levels of some muscles during low-speed rear impact at 8 km/h. (A)
CCR for flexors ¼ 0.7 and (B) CCR for flexors ¼ 0.4.
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CHAPTER 11 Modeling Passive and Active Muscles
capitis extensors for the two CCR values. The solid lines show the flexor muscles
while the dashed lines show the extensors. For both CCRs, the activation patterns
show almost the same oscillating behaviors. When the activation levels of flexors
increased, those of extensors decreased.
11.3.3 ESTIMATION OF MUSCLE ACTIVATION USING
REINFORCEMENT LEARNING
Reinforcement learning (RL) is rarely used for prediction of occupant motions during decelerations due to braking or low-speed impact. To date, Iwamoto et al. (2012)
were the only researchers to report having used this method. This is probably
because the RL method requires substantial computational resources due to the
repetitive nature of trial-and-error estimation. To reduce the overall computational
cost, it is necessary to reduce the cost of each cycle. Fig. 11.17 shows a musculoskeletal FE model of the human head and neck created with simple structures, with
bones modeled as rigid bodies and each muscle as the 1D model mentioned previously. No other soft tissues, such as fat or skin, were represented in the model. The
rotational angle of the head can be calculated as a change in angle about the center of
gravity (CG) of the head. Fig. 11.16 shows a muscle control system using reinforcement learning for maintenance of the head and neck posture. Reinforcement learning
is a process used for developing habits, and it involves the basal ganglia in the brain.
Briefly, a decision is made, and the basal ganglia receive feedback that enhances the
circuit if the feedback is rewarding or reduces the circuit if not (Yin and Knowlton,
2006). This circuitry that involves the basal ganglia also plays an important role in
posture maintenance (Takakusaki et al., 2003, 2004). This process can also be used
for learning the actions of a process for which the underlying parameters are unknown, and there have been several mathematical models (Doya, 2000). The system
produces a control function for the contribution of each muscle around a predetermined articulated joint. The functions are then used to hold the posture in a target
position.
FIGURE 11.16
Muscle control system using reinforcement learning.
11.3 Methods for Modeling Muscular Activation
FIGURE 11.17
Outline of occupant posture prediction using optimal muscle control functions.
Three steps are necessary to perform the muscle control simulation in a normal
environment of 1 G (Fig. 11.17). (1) A predetermined target joint angle is set to
0 degrees, and the range of the joint angle with respect to the target point is set
from 40 to 40 degrees. The CG of the head is provided by the system along
with a boundary to the prescribed motion for rotation of the head around the CG.
The inputs are used for the creation of a database that includes positions of the
model. (2) The initial angle is randomly determined from the range of available joint
angles, and an agent of the RL process transfers random muscle-activation levels
through an interface to the muscle model within the musculoskeletal FE model.
The joint motions are then set according to the activation levels of the muscles
associated with the joint, and the current position data is then sent back through
the interface to the RL agent. For each muscle, the agent evaluates the validity of
the muscle-activation level based on the difference between the current and target
positions, and outputs a control function of the muscle. The control function shows
a relationship of the activation level of the muscle with the joint angle and joint
angular velocity (cf. Fig. 11.18). This routine is repeated until an optimal control
function is achieved for each muscle. (3) Joint motion of the musculoskeletal FE
model is predicted under an impact deceleration condition from the optimal control
functions of the muscles. Fig. 11.18 shows muscle control functions of longus colli
(flexor) and splenius capitis (extensor), which were obtained after 660 trials by the
reinforcement learning.
Fig. 11.19A shows a comparison of simulation results and experimental test data
for the time histories of the head angle about the CG. The results from the simulation
in which there was no muscle activation is in good agreement with cadaver test data.
Similarly, the results from the simulation in which there was muscle activation
generally agree with the test data from the live volunteers. Fig. 11.19B shows the
predicted time histories of activation levels for the primary muscles, including the
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CHAPTER 11 Modeling Passive and Active Muscles
FIGURE 11.18
Muscle control functions obtained from reinforcement learning (trial number 660).
(A) longus colli (flexor muscle) and (B) splenius capitis (extensor muscle).
FIGURE 11.19
Simulation results using reinforcement learning. (A) Head angle time history and
(B) activation level time history.
sternocleidomastoid (STCLM), longus colli (LUCL), splenius capitis (SPCP), and
semispinalis capitis (SMCP), with the time history of the head angle. The sternocleidomastoid and semispinalis capitis are superficial muscles, while the longus colli
and splenius capitis are deep; the sternocleidomastoid and longus colli are flexor
muscles, and the semispinalis capitis and splenius capitis are extensors. As shown
in Fig. 11.19B, the deep muscles were activated at the onset of the head rotation,
while the superficial muscles were activated about 40 ms later. Additionally, the
deep muscles had larger activation levels than the superficial muscles.
11.4 Application of Muscle Models
11.3.4 DISCUSSION FOR BETTER ESTIMATION OF MUSCLE
ACTIVATION
In this section, three methods to estimate muscle activations currently used for
impact biomechanics are introduced. There are advantages and disadvantages for
each. The advantage of the EMG-based method is that measurements are more reliable than estimates. However, it is not possible to measure the EMG data of all muscles (more than 200), simultaneously. Estimations are needed for muscles that are
not measured. Additionally, the raw data obtained from the EMG measurements
cannot be used directly; they must be normalized with data collected for maximum
voluntary force (MVF). The manner in which the MVF is determined obviously affects the validity of the final results. The method in which the PID controller is used
has the advantage that activation levels of all muscles in the body can be estimated
with predetermined dynamic scenarios of deceleration due to impact or braking.
However, some of the parameters, such as PID gains, should be determined for
each dynamic situation, and it is not confirmed that the activation levels predicted
by the PID-controller method are reliable until the predicted activation levels are
validated using EMG measurement data. Reinforcement learning has the advantage
that activation levels of all muscles can be estimated under various dynamic situations. However, high computational costs of learning through trial and error are
needed to obtain the control function of each muscle, and like the PID-controller
method, it is not confirmed that the predicted activation levels are reliable until
they are validated with EMG measurement data. Further studies are needed for better
estimation of muscle activations in the future.
11.4 APPLICATION OF MUSCLE MODELS
This section shows an application result using a human body FE model containing
muscle models. In the simulations, a frontal crash with precrash braking was
assumed as an impact situation. Fig. 11.20 shows a simulation condition in which
a male adult driver steps on the brake pedal and grasps the steering wheel while
sitting on a flat rigid seat. There is a 3-point seat belt equipped with pretension
and force limits of 4 kN, and there are no airbags. The model that was used as the
driver was a human, whole body FE model with 3D geometry muscles, as shown
in Fig. 11.9. Two simulations were performed, one with and one without muscle activation. The sled model used for the simulation in which there was no muscle activation included seats, a steering wheel, and a brake pedal. This model provided
braking deceleration of 0.7 G for 600 ms and then an impact deceleration corresponding to 50 km/h. In the simulation that included muscle activation, the braced
condition shown previously (cf. Fig. 11.9) was held from the beginning to the end
of the simulation. Fig. 11.21 shows comparisons of injury outcomes with and
without muscle activation. More head injuries and rib fractures were predicted in
the case without muscle activation, while more upper and lower extremity injuries
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CHAPTER 11 Modeling Passive and Active Muscles
FIGURE 11.20
Simulation setup of frontal impact with brake G. (A) Simulation condition and (B) inputted
acceleration for sled model.
FIGURE 11.21
Comparison of injury outcomes with and without muscle activation. (A) Without activation
(four rib fractures) and (B) with activation (one rib fracture).
11.5 Chapter Conclusion
FIGURE 11.22
Comparisons of injury trends between cadaver tests and real-world accidents.
were predicted in the case with muscle activation. Fig. 11.22A summarizes the
numbers of fractures in the ribs and in the legs and arms for the two simulations.
Fig. 11.22B shows a comparison of the mean numbers for fractured ribs or arms
and legs between cadaver tests and real-world accidents that occurred in almost
the same impact conditions as the cadaver tests. These data were obtained from
Kallieris et al. (1995). More rib fractures were observed in cadaver tests, while
more fractures in the legs and arms were observed in real-world accidents.
Comparing Fig. 11.22A with Fig. 11.22B, we can see that injury outcomes predicted
by the simulation without muscle activation correspond to those observed in cadaver
tests, while injury outcomes predicted by the case with muscle activation correspond
to those observed in real-world accidents. These results clearly demonstrate that
there is a distinct difference between injuries that occur with and without muscle
activation, and that these differences can be well captured in simulations. A human
body model with muscle-activation capabilities is crucial for work in predicting injuries. Therefore, there is a critical need for more detailed investigation of occupant
injury mechanisms in real-world accidents.
11.5 CHAPTER CONCLUSION
This chapter describes how anatomical structures and physiological functions of
passive and active muscles can be modeled for impact biomechanics. Advanced
technology for active safety and autonomous driving could reduce impact velocity,
and muscle activation could have significant effects on occupant motions and injury
outcomes in these low-speed impacts. Therefore, the modeling of passive and active
muscles is imperative in impact biomechanics. Compared to 3D modeling, it is relatively easy to use 1D modeling of muscles to estimate muscle-activation levels. It is
possible to use 1D models to simulate occupant motions relative to muscle
465
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CHAPTER 11 Modeling Passive and Active Muscles
activations during decelerations due to impacts or braking. However, 1D models
cannot be used to simulate muscular stiffness changes that occur with increasing
levels of muscle activation. This could affect injury outcomes from impacts. For
example, when a driver is in a braced position, thoracic muscles such as the pectoralis major could become stiff enough that the number of rib fractures caused by a
shoulder seat belt would be reduced. Changes in muscular stiffness can be reproduced with 3D modeling.
There are two different types of control functions that can be used for estimating
occupant motions when the muscles are activated during deceleration due to impact or
braking. One makes use of a PID controller, and the other uses reinforcement learning.
Of the two, the PID-controller method is easier to use. However, parameters such as
the PID gains and percentages of muscle contributions must be adjusted for each
case of premeasured deceleration time history curves. In the real world, accidents
with decelerations that are similar to the scope of assumptions do not necessarily
occur. Although reinforcement learning is computationally costly due to the repetitions
necessary for the trial-and-error method for obtaining an optimal muscle control function, it can be used in dynamic conditions for predicting occupant motions during muscle activation. Further research is needed to reduce computation time for reinforcement
learning applied to 3D muscle models. These modeling approaches are promising in
accident injury reconstruction using a human body FE model.
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CHAPTER
Modeling the Head
for Impact Scenarios
12
Haojie Mao
Western University, London, Ontario, Canada
12.1 WHY IS NUMERICAL MODELING OF THE HUMAN HEAD
ESSENTIAL?
The human head is composed of complex structures, including one of the hardest
materials of our bodies (the skull) and the softest ones (the brain). Unlike other
body organs where materials can deform without damage, such as the heart and muscles, in the head, the soft brain does not deform much under normal conditions and is
shielded by other soft tissues. The various skull foramina allow the soft spinal cord,
nerves, and vessels to communicate with the brain. This isolation, or protection, of
the brain causes difficulty in observing brain responses such as impact-induced deformations. Hence, both numerical human-brain models and advanced noninvasive
techniques are critical tools needed for understanding how the brain reacts to
external loadings.
In addition to the brain, the head hosts the eyes, nose, and ears, all with complex
structures of materials that may absorb and dissipate energy during trauma, which
affects brain responses. Meanwhile, these components are vulnerable to blunt forces
and can get injured. In addition, the facial muscles that are integral to our daily
expressions comprise a complex system that is vulnerable to external impacts.
For understanding head and brain injuries and predicting these injuries for
improved head protection, numerical human head models have unlimited potential.
First, numerical models can be used to predict mechanics-related head damage. This
includes acute damage such as scalp and muscle tearing, skull and facial fractures,
brain laceration, and acute vessel breakage such as subdural hematoma. Meanwhile,
chronic brain damage, such as ventricular hypertrophy or Chiari I malformation, in
which brain tissues are deformed by disease-related stresses, can be investigated by
mechanical parameters calculated from computational models. Furthermore, with
the improvement of computational power and model resolution, numerical head
models can be used to map out comprehensive stress/strain responses at each region
of the head, while physical measurements are limited to several locations, because
too many instrumented sensors will affect the physics of the head. Second, numerical models can assist in understanding of initial mechanical changes that take place
after trauma, which, once coupled with subsequent biological processes, will help in
understanding and predicting brain neuronal damage such as axonal injury, dendrite
Basic Finite Element Method as Applied to Injury Biomechanics. http://dx.doi.org/10.1016/B978-0-12-809831-8.00012-X
Copyright © 2018 Elsevier Inc. All rights reserved.
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activation, and cell damage. Such prediction is critical because the neuronal damage
that happens inside the brain is responsible for various symptoms that patients suffer.
Some of the most challenging symptoms are cognitive, such as memory loss,
executive function disorder, and lack of emotional control. Unfortunately, the biological consequences from impact (brain mechanics) to symptoms remain largely
unknown. Hence, there is a strong need for high-quality human head models that
can offer the opportunity of “seeing” into the mechanical world of the brain,
providing a unique way to understand the connections between the mechanical
and biological worlds, and to predict various types of brain damage.
12.2 INTRODUCTION OF CORRESPONDING ANATOMY
12.2.1 UNDERSTANDING HUMAN HEAD ANATOMY WHILE KEEPING
HEAD BIOMECHANICS AND INJURY IN MIND
The human head has complex anatomical features, making it a challenge to develop a
comprehensive, high-quality human head model. Not only do head modelers need to
understand the anatomy of the head, they must also think through injuries that need
to be addressed. They must understand research questions, and based on these questions, they must make decisions such as how detailed a model should be, and what
main characteristics the model should represent. For example, skull deformations are
not expected to cause diffuse brain deformations in sports-related traumatic events,
and therefore, a simplified, rigid-skull head model is acceptable for simulations. However, to study blunt or blast-induced head injuries, where either head-acceleration measurements are not readily available or mechanisms such as skull deformation and wave
propagation need to be considered, a comprehensive human head model including a
deformable skull is needed. The prevalence of brain injury further justifies the need
for comprehensive head models.
Despite the brain being encased in the protective skull, on which there are only a
few foramina, the human brain is injured frequently, affecting approximately 1.7
million people annually in the United States. Further, the consequences of brain injury
can be detrimental, in some cases lasting for the rest of a patient’s life. In addition,
modern neuroscience research has revealed that a traditionally categorized subtle
blow to the head, which might have been previously overlooked, can induce longterm damage to the brain. Hence, it is crucial to understand how mechanical forces
affect the brain. With this understanding, we can develop more effective strategies
to prevent, diagnose, and treat these brain injuries. To this end, we need to understand
the anatomy of the human head together with head injuries and biomechanics.
12.2.2 ANATOMY OF THE HUMAN HEAD
12.2.2.1 The Brain
12.2.2.1.1 Gray and White Matter
The soft tissues in the brain include gray and white matter (Fig. 12.1). Gray matter is
composed of neurons that are nerve cells, glial cells (support neurons), capillaries
12.2 Introduction of Corresponding Anatomy
FIGURE 12.1
Distribution of gray and white matter of the brain from a sagittal section view.
Gray H., Wikimedia Commons.
(body’s smallest blood vessels), neuropil (formed from the unmyelinated axons),
and dendrites of neurons, along with branches (dendrites) of glial cells. White matter
is composed of myelinated axons, named because of the whitish color of myelin.
Because the material properties are different for these two types of tissues, we
need to represent them as two different types of material in a head model.
12.2.2.1.2 Brain Components
The brain consists of various structures (Fig. 12.2). The main components are the
cerebrum, cerebellum, and brainstem. The cerebellum is usually modeled as one
component. The brainstem is made of the midbrain, pons, and medulla. The cerebrum contains several types of the brain tissues and has features such as the cortex,
hippocampus, thalamus, corpus callosum, and other smaller components. The
average material properties of gray matter can be used for these small components
when developing computational models representing the biomechanics of the head.
However, anatomical locations must be noted for explaining damage to specific
brain regions.
12.2.2.1.3 Ventricles
The ventricle system comprises a series of caverns through which cerebrospinal fluid
(CSF) flows. Included in the system are two lateral ventricles, a third ventricle, a
fourth ventricle, and a cerebral aqueduct that connects the third ventricle to the
fourth (Fig. 12.3). Since these ventricles host CSF, they need to be represented in
a model and treated as a fluid-like material.
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FIGURE 12.2
Brain components including cerebrum, corpus callosum, thalamus, middle brain, pons,
medulla, and cerebellum.
Gray H., Wikimedia Commons.
FIGURE 12.3
Brain ventricle system.
Gray H., Wikimedia Commons.
12.2.2.1.4 BraineSkull Complex and Superior Sinuses
Generally speaking, the CSF fills in a space between the brain and skull (Fig. 12.4,
top and bottom). To be specific, the CSF runs between the pia that tightly covers the
brain and the arachnoid that connects to the dura. The dura layer fixes tightly to the
12.2 Introduction of Corresponding Anatomy
FIGURE 12.4
Top: Braineskull interface. Bottom: Illustration of details of the braineskull interface.
Gray H., Wikimedia Commons.
interior surface of the cranium. It should be emphasized that there are multiple
trabeculae and vessels existing between the pia and arachnoid, thus providing shear
resistance. Hence, modeling the brain CSF layer as a fluid, which could not provide
structural resistance to shearing, is not supported by the anatomy. Furthermore, there
are arachnoid border cells and dura border cells between the arachnoid and dura
membranes. The breakage of cerebral vessels in the region of border cells may
lead to acute subdural hematoma (Fig. 12.4, bottom). Overall, the commonly
referred “braineskull interface” is a complex structure (Fig. 12.4, bottom) that needs
to be considered during modeling. For more specific information regarding the
braineskull interface, readers may wish to read the study reported by Haines
et al. (1993).
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In addition to the thin CSF layer, the superior sagittal sinus runs along the top of
the brain (Fig. 12.4, top and bottom). Modeling such a structure is usually required
because of the fluid-like material property and considerable volume of the superior
sagittal sinus.
12.2.2.1.5 Bridging Veins and Brain Vasculature
Bridging veins pierce through the meninges to connect cerebral veins to the superior
sagittal sinus. The anatomy of bridging veins is thoroughly studied through postmortem
dissection. Although the numbers of bridging veins vary in the literature, on average,
there are 11 pairs in the superior sagittal space, connecting the soft brain to the stiff
dura/skull. These veins provide constraining effects on brain motions during impacts,
and hence, they need to be simulated in computational human head models.
Besides the bridging vein, both brain arteries and veins surround and go deep into
the brain. The inlets of the brain vasculature network start from the circle of Willis,
which connects to the two internal carotid arteries (left and right) and one basilar
artery. The outlets are veins that draw blood out of the brain. Hence, the vasculature
system introduces two levels of challenges in terms of modeling. First, the vessels
have much more complex geometry (Fig. 12.5) than the brain does in terms of developing FE meshes. Second, the blood inside the vessels is non-Newtonian fluid,
which may generate irritable flows during trauma loading. Due to the complexity,
simulating the entire 3D, brain vasculature network in a head model remains challenging. So far, most 3D head models do not explicitly model the vasculature. However, these vessels are hundreds of orders stiffer than the brain tissue (more details in
Section 12.4), and are expected to act as reinforcement fibers or networks that could
induce heterogeneous brain deformations.
FIGURE 12.5
The brain vasculature from the bottom view.
Gray H., Wikimedia Commons.
12.2 Introduction of Corresponding Anatomy
12.2.2.1.6 Falx and Tentorium
Both the falx and tentorium play roles in partitioning the brain (Fig. 12.6). The falx
separates the left and right hemispheres, while the tentorium separates the cerebrum
and cerebellum. Both membranes are hundreds of times stiffer than the brain, and
they are tightly connected to the stiff skull. As such, these structures provide significant constraints on brain displacements during events with large head rotations, and
they must to be simulated in a computational human head model.
12.2.2.2 The Skull
The skull is made of 22 bones, 8 bones for the cranium and 14 bones for the face
(Fig. 12.7). Between bones there are fissures, which are strongly fused during adulthood. Usually, these fissures are not needed in adult human head models because of
the strong fusion that makes the fissures as strong as the bone. However, these
fissures need to be simulated for head models of children, especially infants who
have unfused, soft, and ductile fissures. In adult cranial bones, the inner and outer
surfaces are made of dense cortical bones, and the medium is made of trabecular
bones with porosities. Modern imaging techniques can be used to convert imaging
signals into stiffness, and this allows distinctions of specific material properties
for each skull element. However, until researchers in this field can reach a consensus
FIGURE 12.6
The falx and tentorium.
Gray H., Wikimedia Commons.
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FIGURE 12.7
Side view of the human skull.
Gray H., Wikimedia Commons.
on the accuracy of correlating image signals and bone stiffness, classifying skull
bones into two general categories, cortical and trabecular, will be sufficient.
12.2.2.3 Head Sinuses/Cavities
Modelers tend to overlook sinuses and cavities due to the time constraint and difficulty in creating complex hexahedral meshes. There are four sinuses in the head:
frontal, ethmoidal, sphenoidal, and maxillary (Fig. 12.8). These sinuses are actually
needed for studying impacts to the face and forehead. For example, in simulating a
horizontal impact to the forehead, the frontal sinus plays a critical role when simulating how impact energy is distributed through the bone materials. Also, if a person
was exposed to loading from a blast, the cavity of sinuses would affect the stresswave transmission. Without representing these sinuses, the skull would become
too stiff in an FE model, and the predicted brain responses would be inaccurate.
12.2.2.4 Foramina and Canals
The brain is mostly sealed by the skull. However, the foramina and canals in the
skull are needed to allow the brain to communicate with other organs (Fig. 12.9).
The most prominent foramen, foramen magnum, is where the brainstem connects
to the spinal cord. This biggest opening of the skull needs to be simulated, because
it affects the way the brainstem is restrained. However, the other small openings of
the skull might be neglected if research interests are not directly focused on these
12.2 Introduction of Corresponding Anatomy
FIGURE 12.8
Sinuses of the head.
Gray H., Wikimedia Commons.
FIGURE 12.9
Foramina and canals.
Wikimedia Commons.
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CHAPTER 12 Modeling the Head for Impact Scenarios
regions. In scenarios where optical nerve damage or basilar skull fracture may be of
concern, a detailed model of the corresponding foramina will be needed.
12.2.2.5 The Extra-skull Soft Tissue, Eye, and Ear
12.2.2.5.1 Facial Muscle
Connected to the skull are various muscles (Fig. 12.10). Simply put, these muscles
could be grouped together and simulated as a chunk of soft tissue. However, in
scenarios where cranial injuries need to be studied, such as facial surgery, a refined
model with passive and active properties of each of the muscles will be needed.
12.2.2.5.2 The Eye
In studying skull or brain responses, the eye could be simplified as a chunk of soft
material providing adequate inertial effects during an impact. However, in conditions such as a direct blow or blast wave impacting upon the eye, a detailed eye
model is preferred. To create such a model, major components of the eye, including
the aqueous humor, ciliary body, cornea, choroid, iris, lens, optic nerve, conjunctiva,
pupil, retina, sclera, vitreous humor, and zonules, need to be simulated (Fig. 12.11).
Also, the boundary conditions of the eye, which include the muscles as well as bony
sockets, need to be represented. Pioneering work from Stitzel et al. (2002) is a valuable reference for developing high-quality eye models.
12.2.2.5.3 The Ear
Modeling a detailed ear is usually neglected when studying brain responses. However, with tinnitus remaining as a major concern, it is desirable to add an ear model
FIGURE 12.10
Facial muscles.
Gray H., Wikimedia Commons.
12.2 Introduction of Corresponding Anatomy
FIGURE 12.11
The eye. Left: A, aqueous humor; B, ciliary body; C, cornea; D, choroid; I, iris; L, lens; N,
optic nerve; O, conjunctiva; P, pupil; R, retina; S, sclera; V, vitreous humor; and Z, zonules
(Stitzel et al., 2002). Right: Eye muscles.
Gray H., Wikimedia Commons.
to an available human head model. The detailed cavity and geometry of the ear
(Fig. 12.12) could be collected through medical imaging. However, the membranes
and cells that sense sound waves remain very challenging to simulate. A 3D head
model with a detailed ear simulating blunt impacts is yet to be reported.
FIGURE 12.12
The ear.
Gray H., Wikimedia Commons.
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12.3 INJURY MECHANISM
12.3.1 FOCUSING ON BRAIN INJURY
The challenge in understanding the mechanisms of traumatic brain injury (TBI)
partially lies in the “invisibility” of the brain. In a dynamic, high-rate event, there
are very limited methods that are effective and valid for looking into the brain.
Hence, little is known on how the brain responds during an impact event. As an alternative, a popular approach is to measure the kinematics of the head, including linear
and rotational motions of the skull. Researchers have heavily studied the correlations
between head kinematic data, which are measured immediately during the event,
and brain injuries that are observed at a later time.
12.3.1.1 Linear Acceleration
Dr. Gurdjian and Professor Lissner from Wayne State University are pioneers in
measuring head kinematics in a laboratory setting. Since 1939, they conducted a series of experiments in which they dropped human cadavers from various heights to
the ground, and measured linear head accelerations during the impacts. Besides
measuring head accelerations, they investigated head injuries such as skull fractures
and brain contusions. After decades of effort by these pioneers and their colleagues,
the Wayne State tolerance curve (WSTC) was proposed, in which the risk of head
injury was reportedly determined by both the duration and magnitude of head linear
acceleration. In other words, the head can tolerate higher magnitudes of acceleration
when the impact duration is short, but tolerates lower magnitudes of acceleration if
the impact duration is long (Fig. 12.13).
Based on the WSTC and subsequent efforts from the impact biomechanics community, the head-injury criterion (HIC) was proposed (Eq. 12.1). Both HIC36 and
FIGURE 12.13
Wayne State tolerance curve (WSTC) for head injury.
12.3 Injury Mechanism
HIC15 have been used, with 36 and 15 corresponding to delta t (t2t1), representing
36 and 15 ms, respectively.
2
2
32:5 3
Z t2
6
6 1
7 7
HIC ¼ 4ðt2 t1 Þ4
adt5 5
(12.1)
ðt2 t1 Þ t1
MAX
12.3.1.2 Rotational Acceleration
Challenging the linear-accelerationebased injury mechanisms and criteria, researchers
made strong arguments regarding the effects of rotational acceleration. Among them,
researchers from the University of Pennsylvania conducted a series of animal
experiments and demonstrated that, without any linear acceleration, head rotation
alone would cause significant brain damage, such as axonal injury and subdural
hematoma (Gennarelli et al., 1982; Meaney, 1991; Ommaya and Gennarelli, 1974).
Hence, it was demonstrated that both linear and rotational accelerations could cause
brain damage.
12.3.2 SO, HOW DO THESE LINEAR AND ROTATIONAL
ACCELERATIONS OF THE HEAD AFFECT THE BRAIN?
Now, we look deeper into injury mechanismsdhow do these linear and rotational
accelerations, which describe how the head moves during an impact, affect the brain,
which is well protected by the skull?
12.3.2.1 Understanding Linear Acceleration
Linear accelerations of the head are calculated based on accelerometers instrumented
on the skull. During an impact, the entire skull moves immediately because of the high
stiffness of the skull. This movement is what is measured through accelerometers.
However, the brain, due to inertia, tends to remain in its original position. Hence,
this disparity of movement between the skull and the brain induces compressive forces
at the brain region where the skull is moving toward the brain, and induces tensile
forces at the brain region where the skull is moving away from the brain
(Fig. 12.14). These compressive and tensile forces are expected to induce positive
and negative pressures to brain tissues, leading to brain damage. These are typical
linear-accelerationeinduced coup and contrecoup injury mechanisms.
12.3.2.2 Understanding Rotational Acceleration
During rotational motion, the brain tends to remain in its original position due to
inertia, while the loaded skull shears nearby brain tissues through the skullebrain
interaction. The loaded brain tissues then shear nearby materials. These loads induce
brain deformations (Fig. 12.15). In a dynamic trauma event in which the head experiences complex boundary conditions and may rotate around various rotational axes,
it is expected that the shearing will distribute all over the brain. Hence, rotational
loading is reported to induce diffuse injuries.
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FIGURE 12.14
Illustration of the effect of linear motion.
FIGURE 12.15
Illustration of the effects of rotational motion. The brain motion traces on the right refer to
the work of Parnaik et al. (2004).
12.3.3 ANYTHING ELSE BESIDES LINEAR AND ROTATIONAL
ACCELERATIONS?
There are laboratory settings in which both the linear and rotational accelerations are
zero or minimized, yet the brain is injured. One of the classic examples is controlled
cortical impact (CCI), in which either a craniotomy is performed and an impactor is
driven to directly deform brain surfaces, or an impact is applied to the skull to
deform the brain (Fig. 12.16). CCI may induce brain damage in the cortex, hippocampus, thalamus, and corpus callosum. The brain damage induced by CCI demonstrates a limitation of using head acceleration or other kinematics as injury metrics,
because in CCI where the brain damage is induced, none of the acceleration-based
injury metrics will be able to predict the injury. This, again, highlights the importance of investigating braineintracranial biomechanics using FE modeling. Animal
12.3 Injury Mechanism
FIGURE 12.16
Illustration of head impact without linear or rotational head motions.
experiments, such as brain injury on rats, demonstrate that contusion development is
consistent with high strains as predicted by a ratbrain model (Mao et al., 2006).
12.3.4 INSTEAD OF ACCELERATIONS, DESCRIBING BRAIN
TISSUE-LEVEL RESPONSE IS KEY IN UNDERSTANDING
INJURY MECHANISMS
All linear and rotational head motions and direct impacts upon the head induce
tissue-level responses of the brain. These responses (strain/stress), even in their
mild forms, can affect brain cells that lead to brain dysfunction. Hence, a key benefit
of an FE model is its capability to accurately describe these strain/stress responses,
which, once being combined with tissue-level injury tolerances, will help explain
brain injuries that occur at specific brain regions.
12.3.4.1 Tissue-Level Injury Tolerances
Traditional head-injury tolerances are built upon head kinematics or impact forces.
For example, HIC specifies the magnitude and duration of linear accelerations that
the head can tolerate. Aside from HIC, there are still many experimental studies
investigating the correlation between head accelerations/blunt forces and head injuries. One recent focus has been related to finding the correlations between
linear/rotational accelerations (or velocities) and concussion. These kinematicsbased injury tolerances provide guidance for engineering design, such as designing
helmets that can mitigate concussion. A limitation of traditional head-injury tolerances is that we do not understand how the brain is affected.
With techniques that directly stretch live brain tissues, we can now prove that
brain tissues get injured when they experience certain levels of stress/strain. These
levels (or limits) are called tissue-level injury tolerances. These tolerances, combined with accurate FE models of the human head that are capable of predicting
stress/strain for the entire brain, have the potential to guide better design of safety
countermeasures that reduce stresses and strains of the brain.
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CHAPTER 12 Modeling the Head for Impact Scenarios
There are different methods to experimentally measure tissue-level injury
tolerances of the brain. One of such methods is to culture and test in vitro brain tissues. Morrison et al. (2011) conducted a comprehensive review over in vitro models
in the field. Briefly, researchers culture brain slices on thin membranes, which are
subjected to mechanical stretch (unilateral, bilateral, or circumferential). Then,
the correlations between cell damage and mechanical stretch (strain) are calculated
as tissue-level and the corresponding injury tolerances can then be determined. For
example, Cater et al. (2006) described the correlations between strain, post-injury
time, and cell-death percentages in the rat hippocampus as shown in (Eq. 12.2).
Cell DeathCA ¼ 0:0389ð0:0011Þ Strain0:3663ð0:0029Þ Time2:0150ð0:0216Þ ;
Cell DeathDG ¼ 0:0323ð0:0017Þ Strain0:3721ð0:0056Þ Time1:8209ð0:0407Þ ;
(12.2)
where CA represents the cornu ammonis and DG stands for the dentate gyrus.
Complementing in vitro models, FE animal head models play a unique role in
calculating tissue-level tolerances. The logic of this approach is straightforward. FE
models of animal heads are used to simulate laboratory experiments and provide mechanical response maps of the brain, while histological staining or imaging techniques
provide injury maps of the brain. Then, tissue-level injury tolerances can be calculated
by comparing mechanical maps to injury maps. However, there are two main
challenges of using this FE approach. First, head models must be of very high quality
and provide accurate predictions, which this chapter will help. Second, researchers
need to be able to understand the observed brain injuries, distinguishing the injuries
primarily caused by mechanical forces from the injuries caused by biochemical
processes, which may or may not be related to the initial mechanical loading.
12.4 MATERIAL MODELS
The challenges of defining head materials mostly are related to assigning properties for
the brain, which behaves as a nonlinear, viscoelastic material. Hence, this section introduces several classical material models that have been used for representing the brain.
In assigning brain material models, the first thing a head modeler needs to do is to
make justifications between complexity and simplicity, as well as representation of
all physics and improvement of efficiency/stability (Fig. 12.17). For example, an FE
model with complex material properties but with poor stability may provide more
troubles than benefits for an industrial user who needs to frequently exercise the
FE model under extreme conditions.
12.4.1 BRAIN MATERIAL MODELS
Linear viscoelastic models expressed in Prony series can be used to represent viscosity of the brain. During experiments, a tension, compression, shear, or bulk compression is applied. Shear testing is the most used experimental mode, because it is
12.4 Material Models
FIGURE 12.17
Using complex or simple material models.
directly related to the brain’s critical shear deformation. The immediate stress and
stress relaxing over time are measured. The experimental shear-stress time history
can be fitted with the following Prony series:
GðtÞ ¼ GN þ N
N
X
Gi expð t=si Þ
(12.3)
i¼1
where GN denotes long-term shear modulus, and Gi and si denote shear moduli and
the associated relaxation constant, respectively.
Nonlinearity of the brain can be represented through several constitutive models.
Several basic concepts need to be introduced before understanding these models.
The left CauchyeGreen tensor, B, a product of deformation gradient and transverse deformation gradient, has three invariants denoted as I. l is stretch. J is the
determinant of the deformation gradient.
I1 ¼ tr ðBÞ ¼ l21 þ l22 þ l23
i
1h
I2 ¼ ðtrBÞ2 tr B2 ¼ l21 l22 þ l22 l23 þ l23 l21
2
(12.4)
I3 ¼ det B ¼ J 2 ¼ l21 l22 l23
In representing the brain as an isotropic, incompressible material, l1, l2, and l3
can be written as follows:
l1 l2 l3 ¼ 1
l1 ¼ l
.pffiffiffi
l
l2 ¼ l3 ¼ 1
(12.5)
For compression and tension, invariants can be written as follows:
I1 ¼ l2 þ 2l1
I2 ¼ l2 þ 2l
(12.6)
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CHAPTER 12 Modeling the Head for Impact Scenarios
For shear, where l is equal to 1, invariants can be written as follows:
I1 ¼ I2 ¼ 3 þ g2
(12.7)
where g is nominal shear strain.
To represent nonlinearity, various constitution models have been proposed in the
literature. What we need to do is to derive the equations that express engineering
stress for compression/tension and shear modes. Once we get the equations of
engineering stress, we will be able to fit testing data, which are usually expressed
in the form of engineering stress and strain, to calculate material parameters.
12.4.1.1 Gent (1996)
According to Gent (1996), the strain energy function W is written as Eq. (12.8a). In
the uniaxial compression/tension mode, engineering stress T is written as Eq.
(12.8b), related to stretch parameter l. In the shear mode, engineering stress is written as Eq. (12.8c), related to nominal shear strain g. m, and Jm are two material parameters to be calculated.
m
I1 3
W ¼ Jm ln 1 (12.8a)
2
Jm
T¼
mJm
l l2
1
Jm l 2l þ 3
(12.8b)
mJm g
Jm g2
(12.8c)
2
Ts ¼
12.4.1.2 Fung (1967)
The Fung model is written with the shear modulus m, constant b, and constant e
(Eq. 12.9a). In the uniaxial compression/tension mode, engineering stress T is
written as Eq. (12.9b), related to stretch parameter l. In the shear mode, engineering
stress is written as Eq. (12.9c), related to shear nominal strain g.
i
m h bðI1 3Þ
W¼
1
(12.9a)
e
2b
2
1
T ¼ mebðl þ2l 3Þ l l2
(12.9b)
Ts ¼ mgebg
2
(12.9c)
12.4.1.3 Mooney (1940)
The MooneyeRivlin model, the strain energy W is written with components I1, I2, C1,
and C2 (Eqs. 12.10a and 12.10b). In the uniaxial compression/tension mode, engineering stress T is written as Eq. (12.10c), related to stretch parameter l. In the shear
mode, engineering stress is written as Eq. (12.10d), related to shear nominal strain g.
WðI1 ; I2 Þ ¼ C1 ðI1 3Þ þ C2 ðI2 3Þ
(12.10a)
12.5 Material Properties
m ¼ 2ðC1 þ C2 Þ
T ¼ C1 2l 2l2 þ C2 2 2l3
(12.10b)
Ts ¼ 2ðC1 þ C2 Þg
(12.10d)
(12.10c)
12.4.1.4 Ogden (1984)
The strain energy function in the Ogden model is written with components l, shear
modulus m, and constant a (Eq. 12.11a). In the uniaxial compression/tension mode,
engineering stress T is written as Eq. (12.11b), related to stretch parameter l. In the
shear mode, engineering stress is written as Eq. (12.11c), related to shear parameter g.
2m a
(12.11a)
l1 þ la2 þ la3 3
2
a
a2þ1 i
2m h a1
l
T¼
l
(12.11b)
a
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi a qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi a
m
1
g
g
þ 1 þ ðg2 =4Þ
þ 1 þ ðg2 =4Þ
Ts ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
a 1 þ ðg2 =4Þ 2
2
(12.11c)
W¼
12.4.2 SKULL, FLESH, AND SCALP MATERIAL MODELS
A simple elastic model can be used to define the skull where the deformation is small
and does not reach the yielding point, as we see in most mild TBI cases. In the cases
when plastic deformation and damage of the skull is involved, plasticity could be
defined by specifying yield stress. The post-initial-yield stress could be defined
via a tangent modulus, which is approached differently in different software packages. In LS-DYNA, the tangent modulus is the slope of the post-yield stressestrain
curve. In ABAQUS, the tabular inputs of two or more yield stresses and plastic
strains are defined, and the tangent modulus can be calculated by the difference
of two stresses being divided by the difference of two strains. Also, there are options
in software packages, such as LS-DYNA or ABAQUS, to specify different responses
under compression or tension loading modes. Finally, the failure of bone can be
defined by specifying a failure strain.
Flesh and scalp materials could be generally considered as incompressible soft
materials. That being said, the material models for the brain could be applied here.
12.5 MATERIAL PROPERTIES
Decades of experimental measurements of head tissues, especially brain tissues,
have provided a large pool of data for modelers to use. However, the growing
pool of data provided by different groups contain large variances on brain material
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CHAPTER 12 Modeling the Head for Impact Scenarios
properties. For example, the elastic moduli of brain tissues range from several kPa to
hundreds of kPa. In 2010, Chatelin et al. comprehensively summarized 50 years of
brain material testing.
Material testing for the skull bones and flesh come with more consistent results.
Usually, the elastic moduli of cortical bone, the inner and outer table of skull bones,
vary from several GPa’s to tens of GPa’s. The elastic moduli of trabecular bone,
middle portion of skull bones, vary from tens of MPa’s to hundreds of MPa’s.
An efficient and effective approach for head modelers to start is to refer to
material properties used in several high-quality human head models. A comprehensive review of human head models, conducted in 2011, is recommended to readers
(Yang et al., 2011). Table 12.1 provides parameters of head material properties as an
example. In the example, the linear viscoelastic material model is used for the brain,
which has been proven to be numerically stable and robust under extreme impact
conditions (Mao et al., 2013b). To define the brain as a nonlinear and viscoelastic
material, parameters from human head, FE models, such as Kleiven’s, can be
referred to Kleiven (2002).
12.6 TEST DATA AVAILABLE FOR MODEL VALIDATION
Experiments using postmortem human subjects (PMHSs) have generated highquality data for model validation. Below is a summary of these experiments.
12.6.1 BRAIN PRESSURE
Nahum et al. (1977) subjected the foreheads of re-pressurized PMHSs to impacts
from a rigid impactor covered by various padding materials. These impacts were
in an angled direction, with impact velocities changing from 4.36 to 12.95 m/s.
The detailed contact-force and head-acceleration time histories were reported for
one case (no. 37) but not for other cases. However, the acceleration from case no.
37 could be chosen as a baseline which could be morphed to match peak accelerations reached in other cases, providing loading conditions for the FE head model.
The impact force, head acceleration, and brain pressures are shown in Fig. 12.18.
Trosseille et al. (1992) conducted PMHS experiments in which brain pressures
were measured. In the study, the re-pressurized PMHS heads were instrumented
with a 12-accelerometer array to measure the entire 3D kinematics of the head.
Miniature pressure transducers were placed in the subarachnoid space and in the
ventricular system to measure intracranial and ventricular pressures. The PHMSs
were suspended in a sitting position and were subjected to impacts to the facial
regions by a 23.4 kg impactor at a velocity of 7 m/s in the anteroposterior direction.
The typical brain pressures at coup and contrecoup for Case MS428-2 are shown in
Fig. 12.19.
Table 12.1 Material Properties for the Head
Viscoelastic
Component
Density
(kg/m3)
Bulk Modulus
(GPa)
Short-Time Shear
Modulus (GPa)
Long-Time Shear
Modulus (GPa)
Decay Constant
Facial tissue
Scalp
1100
1100
0.005
0.02
0.00034
0.0017
0.00014
0.00068
0.00003
0.00003
Elastic
Density (kg/m3)
Young’s Modulus (GPa)
Poison’s Ratio
Membrane
Skin
Dura
Falx, pia
Arachnoid
Tentorium
Maxillary, sphenoid, and ethmoidal
sinus
1100
1100
1100
1100
1100
1100
1000
0.0315
0.01
0.0315
0.0125
0.012
0.0315
0.001
0.315
0.45
0.35
0.35
0.35
0.3
0.3
Linear Elastic Plasticity
Component
Young’s Modulus
(GPa)
Poisson’s
Ratio
Yield Stress
(GPa)
Tangent
Modulus (GPa)
Plastic Strain
Failure
1130
1500
0.03
10
0.48
0.25
0.00413
0.0122
0.25
1500
6
0.25
1000
1
0.22
1000
0.6
0.22
489
Bridging vein
Cranial cortical
bone
Facial cortical
bone
Cranial
trabecular bone
Facial trabecular
bone
Density
(kg/m3)
12.6 Test Data Available for Model Validation
Component
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CHAPTER 12 Modeling the Head for Impact Scenarios
FIGURE 12.18
Impact force, head acceleration, and brain pressures from experiments reported by
Nahum et al. (1977).
FIGURE 12.19
Brain pressures at coup and contrecoup regions due to frontal impact from experiments
reported by Trosseille et al. (1992).
12.6 Test Data Available for Model Validation
12.6.2 BRAIN MOTION
Hardy et al. (2001, 2007) performed a series of comprehensive impact tests with
PMHS heads monitored with a unique high-speed, biplanar X-ray system combined
with radio-opaque, reduced-density, or neutral-density targets (NDTs). This setting
allowed the researchers to measure relative motion between the skull and brain at
different regions. Over the years, Hardy and colleagues have collected and analyzed
test results from sagittal, coronal, and horizontal impacts aligned to the center of
gravity of the head, or off center. The data that were collected are critical for validating FE models of the human head. With the data, modelers can compare node
displacements predicted by the model to those found in experiments (Fig. 12.20).
For presenting comprehensive validation results, modelers can compare both
model-predicted traces to experimental data and model-predicted displacemente
time histories to experimenta