tl5Fb'%59i@i&d,miI9'g;.4NW+ i!%SmcWEHrtj'lf-LI!!S&SW3 ft , tp!.ti!&!!%IE?ii5k i SHuuS! $i-'*". ... f.., "b ..Fi iiiU4i , JR* j!FUN AMEN L FINITEELE ENT ANALY ISND APPllC I NSWith Mathematica and MATLABComputationsM. ASGHAR

WILEYJOHN WILEY 8t SONS, INC.METU LIBRARYMathematica is a registeredtrademarkof WolframResearch, Inc.MATLAB is a registeredtrademarkof The MathWorks, Inc.ANSYSis a registered trademarkof ANSYS, Inc.ABAQUS is a registeredtrademarkof ABAQUS, Inc.This book is printed on acid-freepaper. eCopyright 2005 by John Wiley & Sims, Inc. All rights reserved.Published by John Wiley& Sons, Inc., Hoboken,NewJerseyPublishedsimultaneouslyin Canada.No part of this publicationmay be reproduced, stored in a retrieval systemor transmittedin any formor by anymeans, electronic, mechanical, photocopying,recording,scanningor otherwise, except as permittedunderSection 107or 108of the 1976UnitedStates CopyrightAct, without either the prior writtenpermissionof thePublisher,or authorizationthroughpayment of the appropriateper-copyfee to the CopyrightClearanceCenter,222 RosewoodDrive, Danvers, MA01923, (978) 750-8400, fax (978) 750-4470, or on the web atwww.copyright.com. 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Neither the publisher/norauthor shall be liable for any loss of profitor any othercommercial damages, includingbut not limited to special, incidental, consequential,or other damages.For general informationon our other products and services or for technical support, please contact our CustomerCare Department within the UnitedStates at (800) 762-2974, outside the UnitedStates at (317) 572-3993or fax(317) 572-4002.Wileyalso publishesits books in a varietyof electronicformats. Some content that appears in print may not beavailablein electronic books. For more informationabout Wileyproducts, visit our website at www.wiley.com.Library of CongressCataloging-in-Publication DataBhatti, M. AsgharFundamental finiteelement analysis and applications: withMathematicaand Matlab computations/ M. Asghar Bhatti.p. cm.Includes index.ISBN 0,471-64808-61. Structural analysis (Engineering) 2. Finite element method, J, Title.TA646.B56 2005620' .001' 51825-dc22Printed in the United States of America1098765432Vr/\. ff (;".+ CONTENTSCONTENTS OF THE BOOKWEB SITEPREFACExixiii1 FINITE ELEMENT METHOD: THE BIG PICTURE 11.1 Discretization and Element Equations / 21.1.1 Plane Truss Element / 41.1.2 Triangular Element for Two-Dimensional Heat Flow/ 71.1.3 General Remarks on Finite Element Discretization/ 141.1.4 Triangular Element for Two-Dimensional Stress Analysis / 161.2 Assembly of Element Equations -/ 211.3 Boundary Conditions and Nodal Solution/ 361.3.1 Essential Boundary Conditions by Rearranging Equations / 371.3.2 Essential Boundary Conditions by Modifying Equations / 391.3.3 Approximate Treatment of Essential Boundary Conditions / 401.3.4 Computation of Reactions to Verify Overall Equilibrium/ 411.4 Element Solutions and Model Validity / 491.4.1 Plane Truss Element / 491.4.2 Triangular Element for Two-Dimensional Heat Flow/ 511.4.3 Triangular Element for Two-Dimensional Stress Analysis/ 541.5 Solution of Linear Equations / 581.5.1 Solution Using Choleski Decomposition/ 581.5.2 ConjugateGradientMethod/ 62vvi CONTENTS1.6 Multipoint Constraints / 721.6.1 Solution Using Lagrange Multipliers / 751.6.2 Solution Using Penalty Function / 791.7 Units / 832 MATHEMATICAL FOUNDATION OF THEFINITE ELEMENT METHOD 982.1 Axial Deformation of Bars / 992.1.1 Differential Equation for Axial Deformations I 992.1.2 Exact Solutions of Some Axial Deformation Problems / 1012.2 Axial Deformation of Bars Using Galerkin Method/ 1042.2.1 Weak Form for Axial Deformations / 1052.2.2 Uniform Bar Subjected to Linearly Varying Axial Load / 1092.2.3 Tapered Bar Subjected to Linearly Varying Axial Load / 1132.3 One-Dimensional BVJ;>Using.Galerkin Method / 115_... -,-2.3.1 Overall Solution Procedure Using GalerkinMethod/ 1152.3.2 Highet Order Boundary Value Problems / 1192.4 Rayleigh-Ritz Method/ 1282.4.1 Potential Energy for Axial Deformation of Bars / 1292.4.2 Overall Solution Procedure Using the Rayleigh-Ritz Method/ 1302.4.3 Uniform Bar Subjected to Linearly Varying Axial Load I 1312.4.4 Tapered Bar Subjected to Linearly Varying Axial Load / 1332.5 Comments on Galerkinand Rayleigh-Ritz Methods / 1352.5.1 Admissible AssumedS ~ l u t i o n / 1352.5.2 Solution Convergence-the Completeness Requirement / 1362.5.3 Galerkinversus Rayleigh-Ritz / 1382.6 Finite Element Form of Assumed Solutions / 1382.6.1 LinearInterpolation Functions for Second-Order Problems / 1392.6.2 Lagrange Interpolation/ 1422.6.3 Galerkin Weighting Functions in Finite Element Form/ 1432.9.4 Hermite Interpolation for Fourth-Order Problems / 1442.7 Finite Element Solution of Axial Deformation Problems / 1502.7.1 Two-Node Uniform Bar Element for Axial Deformations / 1502.7.2 Numerical Examples / 1553 ONE-DIMENSIONAL BOUNDARYVALUEPROBLEM3.1 Selected Applications of 1D BVP/ 1743.1.1 Steady-State Heat Conduction/ 1743.1.2 Heat Flow through Thin Fins / 175173CONTENTS3.1.3 Viscous Fluid Flowbetween Parallel Plates-LubricationProblem/ 1763.1.4 Slider Bearing/ 1773.1.5 Axial Deformation of Bars / 1783.1.6 Elastic Buckling of Long Slender Bars/ 1783.2 Finite Element Formulation for Second-Order ID BVP/ 1803.2.1 Complete Solution Procedure/ 1863.3 Steady-State Heat Conduction/ 1883.4 Steady-State Heat Conduction and Convection/ 1903.5 Viscous Fluid FlowBetween Parallel Plates/ 1983.6 Elastic Buckling of Bars / 2023.7 Solution of Second-Order 1DBVP/ 2083.8 A Closer Look at the Interelement DerivativeTerms / 2144 TRUSSES, BEAMS, AND FRAMES 2224.1 Plane Trusses/ 2234.2 Space Trusses/ 2274.3 Temperature Changes and Initial Strains in Trusses/ 2314.4 Spring Elements/ 2334.5 Transverse Deformation of Beams / 2364.5.1 Differential Equation for Beam Bending/ 2364.5.2 Boundary Conditions for Beams / 2384.5.3 Shear Stressesin Beams/ 2404.5.4 Potential Energy for Beam Bending/ 2404.5.5 Transverse Deformation of a Uniform Beam/ 2414.5.6 Transverse Deformation of a TaperedBeam Fixed atBoth Ends/ 2424.6 Two-Node Beam Element / 2444.6.1 Cubic Assumed Solution/ 2454.6.2 Element Equations Using Rayleigh-Ritz Method/ 2464.7 Uniform Beams Subjected to Distributed Loads / 2594.8 Plane Frames / 2664.9 Space Frames/ 2794.9.1 Element Equations in Local Coordinate System/ 2814.9.2 Local-to-Global Transformation/ 2854.9.3 Element Solution/ 2894.10Frames in Multistory Buildings/ 293viiviii CONTENTS5 TWO-DIMENSIONALELEMENTS5.1 Selected Applications of the 2D BVP/ 3135.1.1 Two-Dimensional Potential Flow/ 3135.1.2 Steady-State Heat Flow/ 3165.1.3 Bars Subjected to Torsion/ 3175.1.4 Waveguidesin Electromagnetics/ 3195.2 Integration by Parts in Higher Dimensions/ 3205.3 Finite Element Equations Using the Galerkin Method/ 3255.4 Rectangular Finite Elements/ 3295.4.1 Four-Node Rectangular Element / 3295.4.2 Eight-Node Rectangular Element / 3465.4.3 Lagrange Interpolation for Rectangular Elements / 3505.5 Triangular Finite Elements/ 3575.5.1 Three-Node Triangular Element / 3585.5.2 Higher Order Triangular Elements/ 3713116 MAPPEDELEMENTS 3816) Integration Using Change of Variables / 3826.1.1 One-Dimensional Integrals/ 3826.1.2 Two-Dimensional Area Integrals/ 3836.1.3 Three-Dimensional VolumeIntegrals / 3866.2 Mapping Quadrilaterals Using Interpolation Functions / 3876.2.1 Mapping Lines / 3876.2.2 MappingAreas/ 3926.2.3 Mapped Mesh/ 4056.3 Numerical Integration Using Gauss Quadrature/ 4086.3.1 Gauss Quadrature for One-Dimensional Integrals/ 4096.3.2 Gauss Quadrature for Area Integrals/ 4146.3.3 Gauss Quadrature for VolumeIntegrals/ 4176.4 Finite Element Computations InvolvingMapped Elements / 4206.4.1 Assumed Solution/ 4216.4.2 Derivatives of the Assumed Solution/ 4226.4.3 Evaluation of Area Integrals/ 4286.4.4 Evaluation of Boundary Integrals/ 4366.5 Complete Mathematica and MATLABSolutions of 2D BVP InvolvingMapped Elements / 4416.6 Triangular Elements by Collapsing Quadrilaterals/ 4516.7 InfiniteElements / 4526.7.1 One-DirnensionalBVP/ 4526.7.2 Two-Dimensional BVP/ 458CONTENTS7 ANALYSIS OF ELASTICSOLIDS 4677.1 Fundamental Concepts in Elasticity/ 4677.1.1 Stresses/ 4677.1.2 Stress Failure Criteria/ 4727.1.3 Strains/ 4757.1.4 ConstitutiveEquations/ 4787.1.5 TemperatureEffects and Initial Strains/ 4807.2 GoverningDifferential Equations/ 4807.2.1 Stress EquilibriumEquations/ 4817.2.2 GoverningDifferential Equations in Terms of Displacements / 4827.3 General Form of Finite Element Equations/ 4847.3.1 Potential Energy Functional / 4847.3.2 WeakForm/ 4857.3.3 Finite Element Equations / 4867.3,4 Finite Element Equations in the Presence of Initial Strains/ 4897.4 Plane Stress and Plane Strain/ 4907.4.1 Plane Stress Problem/ 4927.4.2 Plane Strain Problem/ 4937.4.3 Finite Element Equations / 4957.4.4 Three-Node Triangular Element / 4977.4.5 Mapped Quadrilateral Elements/ 5087.5 Planar Finite Element Models/ 5177.5.1 Pressure Vessels / 5177.5.2 Rotating Disks and Flywheels/ 5247.5.3 Residual Stresses Due to Welding / 5307.5.4 Crack Tip Singularity/ 5318 TRANSIENT PROBLEMS 5458.1 TransientField Problems/ ,5458.1.1 Finite Element Equations/ 5468.1.2 Triangular Element / 5498.1.3 Transient Heat Flow/ 5518.2 Elastic Solids Subjected to Dynamic Loads/ 5578.2.1 Finite Element Equations/ 5598.2.2 Mass Matrices for Common Structural Elements/ 5618.2.3 Free-VibrationAnalysis/ 5678.2.4 Transient Response Examples/ 573lxx CONTENTS9 p-FORMULATION 5869.1 p-Formulation for Second-Order 1DBVP/ 5869.1.1 Assumed Solution Using Legendre Polynomials/ 5879.1.2 Element Equations/ 5919.1.3 Numerical Examples / 5939.2 p-Formulation for Second-Order 2D BVP/ 6049.2.1 p-Mode Assumed Solution/ 6059.2.2 Finite Element Equations / 6089.2.3 Assembly of Element Equations/ 6179.2.4 Incorporating Essential Boundary Conditions/ 6209.2.5 Applications/ 624A USE OF COMMERCIAL FEA SOFTWARE 641A.1 ANSYS Applications/ 642A.1.1 General Steps/ 643A.1.2 Truss Analysis/ 648A.1.3 Steady-State Heat Flow/ 651A.1.4 Plane Stress Analysis/ 655A.2 Optimizing Design Using ANSYS/ 659A.2.1 General Steps/ 659A.2.2 Heat FlowExample/ 660A.3 ABAQUSApplications/ 663A.3.1 Execution Procedure/ 663A.3.2 Truss Analysis/ 66'5A.3.3 Steady-State Heat Flow/ 666A.3.4 Plane Stress Analysis/ 671B VARIATIONAL FORMFOR BOUNDARYVALUEPROBLEMS 676B.1 Basic Concept of Variationof a Function/ 676B.2 Derivation of Equivalent Variational Form/ 679B.3 Boundary ValueProblem Corresponding to a GivenFunctional / 683BIBLIOGRAPHY 687INDEX 695CONTEN'TS OF THE BOOK WEBS ~ T E(www.wiley.com/go/bhatti)ABAQUS ApplicationsAbaqusUse\AbaqusExecutionProcedure.pdfAbaqusUse\HeatFlowAbaqusUse\PlaneStressAbaqusUse\TmssAnalysisANSYSApplicationsAnsysUse\AppendixA. AnsysUse\Chap5AnsysUse\Chap7AnsysUse\Chap8AnsysUse\GeneralProcedure.pdfFull Detail Text ExamplesFull Detail Text Examples\ChaplExarnples.pdfFull Detail Text Examples\Chap2Examples.pdfFull Detail Text Examples\Chap3Examples.pdf .Full Detail Text Exarnples\Chap4Exarnples.pdfFull Detail Text Exarnples\Chap5Exarnples.pdfFull Detail Text Examples\Chap6Examples.pdfFull Detail Text Examples\Chap7Examples.pdfxixii CONTENTS OFTHE BOOKWEBSITEFull Detail Text Examples\Chap8Examples.pdfFullDetail Text Examples\Chap9Examples.pdfMathematica ApplicationsMathematicaUse\MathChapl.nbMathematicaUse\MathChap2.nbMathematicaUse\MathChap3.nbMathematicaUse\MathChap4.nbMathematicaUse\MathChap5.nbMathematicaUse\MathChap6.nbMathematicaUse\MathChap7.nbMathematicaUse\MathChap8.nbMathematicaUse\Mathematica Introduction.nbMATLAB ApplicationsMatlabFiles\Chap IMatlabFiles\Chap2MatlabFiles\Chap3MatlabFiles\Chap4MatlabFiles\Chap5MatlabFiles\Chap6MatlabFiles\Chap7MatlabFiles\Chap8MatlabFiles\Common ISample CourseOutlines, Lectures, and ExaminationsSupplementary Material and CorrectionsPREFACELarge numbers of books have been written on the finite element method. However, effectiveteaching of the method using most existing books is a difficult task. The vast majorityofcurrent books present the finite element method as an extension of the conventional matrixstructuralanalysis methods. Using this approach,one can teach the mechanical aspects ofthe finite elementmethod fairlywell, but thereare nosatisfactoryexplanationsfor eventhe simplest theoreticalquestions. Whyare rotational degrees of freedomdefined for thebeamand plateelements but not for the planestressandtruss elements?Whatis wrongwith connecting corner nodes of a planar four-nodeelement tothe rnidsidenodesofaneight-nodeelement? The applicationof the methodto nonstructural problems is possibleonly if one can interpret problem parameters in terms of their structural counterparts. Forexample, one can solve heat transfer problemsbecausetemperaturecan be interpreted asdisplacement in a structural problem.More recently, several new textbooks on finite elements have appeared that emphasizethe mathematical basis of the finite element method. Usingsomeof these books, thefi-nite element methodcan be presentedas a methodfor .obtaining approximate solutionofordinary and partial differential equations. Thechoice of appropriate degrees of freedom,boundaryconditions, trialsolutions, etc., cannow befullyexplainedwiththistheoreti-cal background. However, the vast majority of these bookstend to be too theoreticalanddo notpresent enoughcomputational detailsandexamplesto beof value, especiallytoundergraduate and first-year graduate students in engineering.The finite element coursesface one more hurdle. One needs to performcomputationsin orderto effectively learn the finite element techniques. However, typical finite elementcalculations arevery longandtedious, especiallythoseinvolvingmappedelements. Infact, some of these calculations are essentially impossible to perform by hand. To alleviatethis situation, instructors generally rely on programswritten in FORTRAN or some otherxiiixiv PREFACEconventional programming language. In fact, there are several books available that includethese types of programs with them. However, realistically, in a typical one-semester course,most students cannot be expected to fully understand these programs. At best they use themas black boxes, which obviously does not help in learning the concepts.Inadditiontotraditional research-orientedstudents, effectivefiniteelement coursesmust also cater to the needs and expectations of practicing engineersand others interestedonly in the finite element applications. Knowing the theoreticaldetails alone does not helpin creatingappropriate models for practical, and often complex, engineering systems.This bookisintendedtostrikean appropriatebalanceamongthetheory, generality,and practical applications of the finite element method. The method is presented as a fairlystraightforward extension of the classical weighted residual and the Rayleigh-Ritz methodsfor approximate solutionof differential equations. The theoretical details are presented inan informal style appealing to the reader's intuition rather than mathematical rigor. To makethe concepts clear, all computational details are fully explained and numerous examples areincluded showing all calculations. To overcomethe tedious natureof calculationsassoci-ated with finite elements, extensive use of MATLABand Mathematicd'' is made in theboole. All finite element proceduresare implemented in the form of interactive Mathemat-ica notebooksand easy-to-follow MATLAB code. All necessarycomputations are readilyapparent fromthese implementations. Finally, to addressthe practical applicationsof thefinite elementmethod, the bookintegratesa series of computer laboratories and projectsthat involve modelingand solution using commercialfinite element software. Short tuto-rials and carefully chosen sampleapplications of ANSYSand ABAQUS are containedinthe book.Thebookis organizedin sucha way that it can be usedvery effectively ina lecture/computer laboratory(lab)format. Inover20yearsofteachingfiniteelements, usingavarietyof approaches, theauthor hasfoundthatpresentingthematerial inatwo-hourlectureand one-hour lab per week isi ~ e a l l y suited for the first finite element course. Thelecturepart develops suitable theoretical backgroundwhile the lab portiongives studentsexperience in finite element modeling and actual applications. Both parts should be taughtinparallel. Of course, it takes timeto developthe appropriate theoretical background inthe lecturepart. The lab part, therefore, is ahead of the lecturesand, in the initial stages,students are usingthefinite element softwareessentiallyasablackbox. However, thisapproach has two main advantages. The first is that students have some time to get familiarwiththeparticular computer systemandthefinite element packagebeingutilized. Thesecond, and more significant, advantage is that it raises students' curiosity in learning moreabout why things must be done in a certain way. During early labs students often encountererrorssuchas "negativepivot found"or "zeroor negative Jacobian forelement."When,during the lecture part,they find out mathematical reasonsfor sucherrors, it makesthemappreciatethe importanceof learningtheory in orderto becomebetterusers of the finiteelement technology.The author also feels strongly that the labs must utilize one of the several commerciallyavailable packages, instead of relyingon simple home-grownprograms. Use of commer-cial programsexposesstudentsto at least one state-of-the-art finite element package withitsbuilt-inor associatedpre-andpostprocessors. Sincethegeneral proceduresareverysimilar among different programs, it is relatively easy to learn a different package after thisPREFACEexposure. Most commercial prol$nims also include analysismodules for linear and nonlin-ear staticand dynamic analysis, buclding, fluid flow, optimization, and fatigue. Thuswiththesepackagesstudentscanbeexposedtoa variety offiniteelement applications, eventhough theregenerally is not enough timeto develop theoretical detailsof all thesetopicsinone finite element course. With more applications, students also perceive the course asmore practical and seem to put moreeffort into learning.TOPICS COVEREDThebook coversthe fundamental concepts and is designed fora first course on finiteele-mentssuitable for upper division undergraduate students and first-year graduate students.It presents the finite element method asa tool to find approximate solution of differentialequations and thus can be used by students froma variety of disciplines. Applications cov-eredinclude heat flow, stress analysis, fluidflow, andanalysis ofstructural frameworks.The material is presented in nine chapters and two appendixes as follows.1. Finite Element Method:The Big Picture. Thischapter presentsanoverviewof thefinite element method. To give a clear idea of the solution process, the finite element equa-tions for a few simple elements (plane truss, heat flow, and plane stress) are presented in thischapter. A few general remarks on modeling and discretization are also included. Importantsteps of assembly, handling boundary conditions, and solutions for nodal unknowns and el-ement quantities are explained in detail in this chapter. These steps are fairly mechanical innatureand do not require complex theoretical development. Theyare, however, central toactually obtaining a finite element solution for a given problem. The chapter includes briefdescriptions of both direct and iterative methods for solution oflinear systems of equations.Treatment of linear constraints through Lagrange multipliers and penalty functions is alsoincluded.Thischapter gives enough background to students so that theycanquickly start usingavailablecommercial finite element packages effectively. It playsan important roleinthelecture/lab format advocated-for the first finite element course.2. Mathematical Foundationsof theFiniteElement Method. Froma mathematicalpoint ofviewthefinite element methodis aspecial formof the well-knownGalerkinandRayleigh-Ritzmethodsfor findingapproximatesolutionsof differential equations.Thebasicconceptsareexplained inthis chapter withreferencetotheproblemofaxialdeformation of bars. The derivation of thegoverning differential equation is included forcompleteness. Approximate solutions using theclassical formof Galerkin andRayleigh-Ritzmethods arepresented. Finally, themethodsarecast intotheformthat is suitablefordevelopingfinite element equations. Lagrange and Hermitian interpolation functions,commonly employed in derivation of finite element equations, are presented in this chapter.3. One-Dimensional BoundaryValue Problem. A largehumber of practical problemsare governed by a one-dimensional boundary value problem of the formd ( dU(X))dx k ( x ) ~ + p(x) u(x) +q(x) =0xvxvi PREFACEFinite element formulationand solutions of selected applications that are governed by thedifferential equation of this form are presented in this chapter.4. Trusses, Beams, andFrames. Manystructural systemsusedinpractice consist oflongslendermembersof variousshapesusedintrusses, beams, and frames. Thischap-ter presentsfinite element equationsfor these elements. The chapter is important for civiland mechanical engineeringstudentsinterestedin structures. It alsocoverstypical mod-elingtechniquesemployedin framedstructures, suchas rigidendzonesandrigidfloordiaphragms. Thosenot interestedin these applicationscan skip thischapter without anyloss in continuity.5. Two-Dimensional Elements. Inthis chapterthe basic finite element conceptsare il--lustratedwith referenceto the following partial differential equation defined over an arbi-trary two-dimensional region:The equation can easily be recognizedas a generalizationof the one-dimensional bound-ary value problem considered in Chapter 3. Steady-state heat flow, a variety of fluid flow,and thetorsion of planarsectionsare some of the commonengineering applications thatare governed by the differential equationsthat are special cases. of this general boundaryvalue problem. Solutionsof these problemsusing rectangular and triangular elementsarepresented in this chapter.6. Mapped Elements. Quadrilateral elementsand other elementsthat canhave curvedsidesare much more usefulin accuratelymodelingarbitrary shapes. Successful develop-ment of these elementsis basedon the key conceptof mapping.Theseconcepts are dis-cussed in this chapter. Derivation of the Gaussian quadrature used to evaluate equations formapped elements is presented. Four-sand eight-nodequadrilateral elementsare presentedfor solutionof two-dimensional boundaryvalue problems. The chapter also includespro-cedures for forming triangles by collapsing quadrilateralsand for developingthe so-calledinfinite elements to handle far-field boundary conditions.7. Analysis ofElastic Solids. The problem of determining stresses and strains in elasticsolidssubjectedtoloadingandtemperaturechangesisconsideredinthischapter. Thefundamental conceptsfromelasticityare reviewed.Usingthese concepts, thegoverningdifferentialequationsin termsof stressesand displacementsare derived followedbythegeneralformoffinite element equationsforanalysisof elasticsolids. Specificelementsforanalysisof planestressand planestrainproblemsare presented inthischapter. Theso-called singularity elements, designed to capturea singular stress field neara crack tip,are discussed. Thischapter is important for those interestedin stressanalysis. Thosenotinterested in these applications can skip this chapter without any loss in continuity.8. Transient Problems. Thischapterconsidersanalysis of transient problemsusingfi-nite elements. Formulations for both the transient field problems and the structural dynam-ics problems are presented in this chapter.9. p-Formulation. In conventional finite element formulation, each element is based ona specific set of interpolation functions. After choosingan elementtype, the onlyway toPREFACEobtaina better solutionisto refihe the model. Thisformulationiscalled h-formulation,whereh indicates thegenericsizeofanelement. Analternativeformulation, calledthep-formulation, is presented in this chapter. In this formulation, the elements are basedoninterpolation functions that mayinvolve very highorderterms. Theinitial finite elementmodel is fairlycoarseand is basedprimarilyon geometricconsiderations. Refinedsolu-tions are obtained by increasing the order of the interpolation functions used in the formu-lation. Efficient interpolation functionshave been developed so that higher order solutionscan be obtained in a hierarchical manner from the lower order solutions.10. Appendix A: Use of Commercial FEA Software. This appendix introduces studentsto two commonly used commercial finite element programs, ANSYSand ABAQUS. Con-cise instructions for solution of structural frameworks,heat flow, and stress analysis prob-lems are given for both programs.11. Appendix B: VariationalFormfor Boundary Value Problems. The main body of thetext employsthe Galerkin approach for solutionof general boundary value problems andthe variational approach(usingpotential energy)forstructural problems.Thederivationof the variational functional requiresfamiliaritywith the calculusof variations. In the au-thor'sexperience, giventhat onlylimitedtime isavailable,most undergraduate studentshave difficultyfullycomprehendingthistopic. Forthisreason, andsincethederivationisnot central tothefiniteelement development, thematerial ondevelopingvariationalfunctionalsismovedtothisappendix. If desired, thismaterial canbecoveredwiththediscussion of the Rayleigh-Ritz method in Chapter 2.To keepthe bookto a reasonable lengthand to make it suitablefora wideraudience,important structural oriented topics, such as axisymmetric and three-dimensional elasticity,plates and shells, material and geometric nonlinearity, mixed and hybrid formulations, andcontact problemsarenot coveredinthisbook. These topics arecoveredindetail inacompanion textbook by the author entitled Advanced Topics in FiniteElement Analysis ofStructures:With Mathematico'"and lvIATLABComputations, John Wiley, 2006.UNIQUE FEATURES(i) All key. ideas are introduced in chaptersthat emphasize themethodasa waytofind approximate solution of boundary value problems. Thus the book can be usedeffectively for students from a variety of disciplines..(ii) The "big picture" chapter gives readersan overview of all the mechanical detailsof the finite element method very quickly. Thisenables instructors to start usingcommercial finiteelement softwareearlyinthe semester; thus allowingplentyof opportunity to bring practicalmodelingissues into the classroom. Theauthorisnot awareofanyother bookthat startsout inthismanner. Fewbooks thatactuallytrytodothisdo so bytaldngdiscretespringand bar elements. Inmyexperience this does not work very well because studentsdo not see actual finiteelement applications. Also, this approachdoes not makesenseto thosewho arenot interested in structural applications.xvllxviii PREFACE(iii) Chapters2and3introducefundamental finiteelementconcepts throughone-dimensional examples. Theaxial deformation problem is used for a gentle intro-ductiontothesubject. This allowsfor parameterstobeinterpreted inphysicalterms. Thederivation ofthegoverning equationsand simple techniques forob-tainingexact solutions are included to helpthosewhomaynotbe familiar withthe structural terminology. Chapter 3 alsoincludessolutionofone-dimensionalboundary valueproblems without reference to any physical application fornon-structural readers.(iv) Chapter 4, on structural frameworks, is quite unique for bookson finite elements.No current textbook that approaches finite elements froma differential equationpoint of view also has a complete coverage of structural frames, especially in threedimensions. In fact, even most books specifically devoted to structural analysis donot have as satisfactory a coverage of the subject as provided in this chapter.(v) Chapters5and6aretwoimportant chaptersthat introduce keyfiniteelementconceptsinthecontext of two-dimensional boundary valueproblems. To keeptheintegrationanddifferentiationissues fromcloudingthebasicideas, Chap-ter 5 startswith rectangular elements and presents complete examples using suchelements. Thetriangular elementsarepresented next. Bythetimethemappedelements are presented in Chapter 6, thereare no real finite element-related con-ceptsleft. It isall just calculus. This clear distinction between the fundamentalconcepts and calculus-related issues gives instructors flexibilityin presenting thematerial to students with a widevarietyof mathematics background.(vi) Chapter 9, on p-formulation, is unique. Noother book geared towardthe first fi-nite element course even mentions this important formulation. Several ideas pre-sented inthischapter are used inrecent development oftheso-called meshlessmethods.(vii) Mathematica and MATLAB ,implementations are included to showhow calcula-tions can be organized using' a computer algebra system. These implementationsrequireonlytheverybasicunderstandingof thesesystems. Detailed examplesare presented in Chapter 1 showing how to generate andassemble element equa-tions, reorganize matrices to account for boundary conditions, and then solve forprimary and secondary unknowns. These steps remain exactly the same for all im-plementations. Most of the other implementations are nothing more than elementmatrices written using Mathematica or MATLABsyntax.(viii) Numerous numerical examples are included to clearly showall computations in-volved.(ix) All chapters containproblems for homeworkassignment. Most chapters alsocontainproblems suitablefor computerlabs andprojects. Theaccompanyingwebsite(www.wiley.com/go/bhatti)containsall text examples, MATLABandMathematicafunctions, andANSYSandABAQUSfilesinelectronic form. Tokeeptheprinted book to a reasonable length most examplesskipsomecompu-tations. Theweb site contains fullcomputational detailsof-these examples. Alsothe book generally alternates between showing examples done with MathematicaandMATLAB. Thewebsitecontainsimplementationsofall examplesinbothMathematica and MATLAB.PREFACEtVPICAL COURSESThe book can be used to develop a number of courses suitable for different audiences.FirstFiniteElement Coursefor Engineering Students About 32 hoursof lec-tures and 12 hours of labs (selected materials from indicated chapters):Chapter l: Finite element procedure, discretization, element equations, assembly,boundaryconditions, solution of primary unknownsand element quantities, reactions,solution validity (4 hr)Chapter 2: Weakformfor approximatesolutionof differential equations, Galerkinmethod, approximatesolutionsusingRayleigh-Ritzmethod, comparisonof GalerkinandRayleigh-Ritzmethods, Lagrange andHermite interpolation, axial deformationelement using Rayleigh-Ritz and Galerkin methods(6 hr)Chapter 3: IDBVP, FEA solution ofBVP, ID BVP applications (3 hr)Chapter 4: Finite element for beam bending, beam applications, structural frames (3 hr)Chapter 5: Finite elements for 2Dand3Dproblems, linear triangular element forsecond-order 2D BVP, 2D fluid flowand torsion problems(4 hr)Chapter 6: 2D Lagrangeand serendipity shape functions, mapped elements, evaluationof area integralsfor2D mapped elements, evaluationof lineintegralsfor 2D mappedelements (4 hr)Chapter 7: Stressesand strains in solids, finite element analysisof elastic solids, CSTand isoparametric elements for plane elasticity (4 hr)Chapter 8: Transient problems (2 hr)Review, exams (2 hr)About 12 hours of labs (some sections from the indicated chapters supplemented by docu-mentation of the chosen commercial software):Appendix: Introduction to Mathematica and/or MATLAB (2 hr)Chapters1 and 4: Software documentation, basic finite element procedure using com-mercial software, truss and frame problems (2 hr)Chapters1 and5: Softwaredocumentation, 2Dmeshgeneration, heat flow problems(2 hr)Chapters 1 and 7: 2D, axisymmetric, and 3D stress analysis problems (2 hr)Chapter 8: Transient problems (2 hr)Software documentation: Constraints, design optimization (2 hr)First FiniteElement Coursefor StudentsNotInterested inStructural Appli-cations Skip Chapters4 and 7. Spend more time on applications in Chapters5 and 6.IntroduceChapter9: p-Formulation. Inthe labs replacetruss, frame, and stressanalysisproblems with appropriate applications.Finite Element Course for Practicing Engineers Fromthe current book: Chapters1, 2, 6, and 7. Fromthecompanionadvancedbook: Chapters 1, 2, and5andselectedmaterial from Chapters6, 7, and 8.xixxx PREFACEFiniteElementModelingand Applications For a short courseonfinite elementmodelingor self-study, itis suggestedto coverthe first chapter in detail and thenmoveon to Appendix A for specific examples of usingcommercial finite element packages forsolution of practical problems.ACKNOWLEDGMENTSMost of the material presented in the book has becomepart of the standard finite elementliterature, and henceit is difficult to acknowledgecontributions of specific individuals. Iam indebtedto the pioneersin the field and the authorsof all existing booksand journalpapers on the subject. I have obviously benefited from their contributionsand have used agood number of them in my over 20 years of teaching the subject.I wrote the first draft of the book in early1990. However, the printedversion has prac-tically nothing in common with that first draft. Primarilyas a resultof questions from mystudents, I have had to makeextensive revisionsalmostevery year. Overthe lastcoupleof yearsthe processbeganto showsigns of convergenceandthe result iswhat youseenow. Thus I would like to acknowledge all direct and indirect contributionsof my formerstudents. Theirquestionshopefullyled meto explainthingsin waysthat makesensetomost readers. (A note to future students and readers: Please keep the questions coming.)Iwanttothankmyformer graduatestudent RyanVignes, whoreadthroughseveraldrafts of the book and provided valuable feedback. ProfessorsJia Liu and Xiao Shaopingused early versions of the book when they taught finite elements. Theirsuggestionshavehelped a great deat in improving the book. My colleagues Professors Ray P.S. Han, HosinDavidLee, and Ralph Stephenshave helpedby sharingtheirteaching philosophyand bykeeping me in shape through heated games of badminton and tennis.Finally, I would like to acknowledge the editorial staff of John Wiley for doing a greatjobintheproduction of thebook. 1'/amespeciallyindebtedto JimHarper, who, fromour first meeting in Seattle in 2003, has been in constant communication and has kept theprocess going smoothly. Contributions of senior production editor Bob Hilbert and editorialassistant Naomi Rothwellare gratefully acknowledged.CHAPTER ONE5FINITEELEMENT METHOD:THEBIG PICTUREApplication of physical principles, suchas massbalance, energyconservation, and equi-librium, naturallyleadsmanyengineeringanalysis situations intodifferential equations.Methodshave beendeveloped for obtaining exactsolutionsforvariousclassesofdiffer-ential equations. However, thesemethods donot applytomanypractical problemsbe-causeeither theirgoverningdifferential equationsdonot fall intotheseclasses ortheyinvolve complex geometries. Finding analytical solutionsthat also satisfy boundary condi-tions specified over arbitrary two- and three-dimensional regionsbecomesa very difficulttask. Numerical methods are therefore widely used for solution of practical problems in allbranches of engineering.The finite element method is one of the numerical methodsforobtainingapproximatesolution of ordinary and partial differential equations. It is especially powerful when deal-ing with boundary conditions defined over complex geometries that are common in practi-cal applications. Other numericalmethodssuchas finitedifferenceand boundary elementmethods maybecompetitiveorevensuperiortothefiniteelementmethodfor certainclasses of problems. However, becauseof its versatility in handling arbitrary domainsandavailability of sophisticated commercial finite element software,over the last few decades,the finite element methodhas become the preferred method forsolutionof manypracti-cal problems. Only the finite element method is considered in detail in this book. Readersinterested in other methodsshouldconsult appropriate references, Books by Zienkiewiczand Morgan[45], Celia and Gray [32], and Lapidus and Pinder [37] are particularly usefulfor those interested in a comparisonof different methods.Theapplication of the finite element method to a given problem involves the followingsix steps:2 FINITEELEMENTMETHOD:THE BIGPICTURE1. Development of element equations2. Discretization of solution domain into a finite element mesh3. Assembly of element equations4. Introduction of boundary conditions5. Solution for nodal unknowns6. Computation of solution and related quantities over each elementThekey ideaofthe finite element methodisto discretizethe solution domainintoanumber ofsimpler domainscalledelements. Anapproximate solutionisassumedoveran element intermsofsolutionsat selectedpointscallednodes. To give a clearideaofthe overall finite element solution process, the finite element equationsfora few simpleelements are presentedinSection1.1. Obviouslyat thisstageitisnot possibletogivederivations of these equations. The derivations must wait until later chapters after we havedeveloped enough theoretical background. Few general remarkson discretization are alsomadeinSection1.1. Morespecific commentson modelingarepresented in laterchap-ters when discussing various applications. Important steps of assembly, handling boundaryconditions, andsolutionsfornodalunknownsandelement quantities remainessentiallyunchanged for any finite element analysis. Thus these procedures are explained in detail inSections1.2, 1.3, and104. Thesesteps are fairly mechanical in natureand do not requirecomplextheoretical development. They are, however, central to actuallyobtaininga finiteelement solution for a given problem. Therefore, it is importantto fully master these stepsbefore proceedingto the remaining chapters in the book.The finite element process results in a large system of equations that must be solved fordetermining nodal unknowns. Several methods are available for efficient solution of theselargeand relativelysparsesystemsof equations. Abriefintroductiontotwo commonlyemployedmethodsis given in Section1.5. In somefinite elementmodeling situationsitbecomes necessarytointroduceconstraintsinthefiniteelement equations. Section 1.6presents examples of few such situations and discusses two different methods for handlingtheseso-calledmultipoint constraints.A brief sectiononappropriateuseof unitsin nu-merical calculations concludesthis chapter.1.1 DISCRETIZATION AND ELEMENT EQUATIONSEachanalysissituationthatisdescribedintermsofoneormoredifferential equationsrequiresan appropriateset of elementequations. Even for the samesystemof governingequations, several elementswith different shapesand characteristics maybeavailable. Itis crucial to chooseanappropriateelementtype for the applicationbeingconsidered. Aproper choicerequires knowledgeofall detailsofelement formulationandathoroughunderstanding of approximations introduced during its development.A key step in the derivation of element equations is an assumption regarding the solutionof the goveming differential equation over an element. Several practical elements are avail-able that assume a simple linear solution. Other elements use more sophisticated functionsto describe solution over elements. The assumed element solutionsare written in terms ofunknown solutions at selected points called nodes. The unknown solutions at the nodes areDISCRETIZATION ANDELEMENT EQUATIONSgenerally referred to as the nodal degrees offreedom, a terminologythat dates back to theearly development of the method by structuralengineers. The appropriatechoice of nodaldegrees of freedomdependson the governingdifferential equationand will be discussedin the following chapters.The geometry of an element depends on the type of the governing differential equation.For problems defined by one-dimensional ordinary differential equations, the elementsarestraight or curved line elements. For problems governed by two-dimensional partial differ-ential equationsthe elementsare usuallyof triangular or quadrilateral shape. Theelementsides may be straight or curved. Elementswith curved sides are useful for accurately mod-eling complex geometries common in applications such as shell structuresand automobilebodies. Three-dimensionalproblemsrequire tetrahedralorsolidbrick-shapedelements.Typical element shapes for one-, two-, andthree-dimensional (lD, 2D, and 3D) problemsare shown in Figure1.1. The nodes on the elementsare shown as dark circles.Element equationsexpressa relationship betweenthe physical parameters in the gov-erning differential equations and the nodal degrees of freedom. Since the number of equa-tions for some of the elementscan be very large, the element equationsare almostalwayswritten using a matrix notation. The computations are organized in two phases. In the firstphase (the element derivation phase), the element matricesare developed for a typical ele-ment that is representative of all elementsin the problem. Computations are performed ina symbolic form without using actual numerical values for a specific element. The goal isto develop general formulas for element matricesthat can later be used for solution of anynumericalproblembelonging to that class. Inthe secondphase, the general formulas areused to write specific numerical matrices for each element.Oneofthemainreasons for thepopularityofthefiniteelement methodisthewideavailability of general-purpose finite element analysis software. This software developmentispossiblebecause general element equations canbeprogrammedinsuchawaythat,given nodal coordinatesand other physical parameters for an element, the program returnsnumerical equations forthat element. Commercial finite element programs containa largelibrary of elementssuitable for solution of a wide variety of practical problems.3IDElements2D Elements3D ElementsFigure 1.1. Typical finite element shapes4 FINITEELEMENTMETHOD:THE BIGPICTURETo give a clear pictureof the overall finite element solution procedure, the general fi-niteelement equationsforfewcommonlyusedelements aregiven below. Thedetailedderivations of these equationsare presented in later chapters.1.1.1 Plane Truss ElementManystructuralsystemsused in practiceconsistof longslendershapesof variouscrosssections. Systems in which the shapes are arranged so that each member primarily resistsaxial forces are usually known as trusses. Common examples are roof trusses, bridge sup-ports, crane booms, and antennatowers. Figure1.2 showsa transmissiontower that canbe modeledeffectivelyas a planetruss. For modelingpurposesall members are consid-ered pin jointed. The loadsare appliedat the joints. Theanalysis problem is to find jointdisplacements, axial forces, and axial stresses in different members of the truss."Clearlythe basicelement to analyzeany planetruss structureisa two-nodestraight-lineelement orientedarbitrarilyinatwo-dimensional x-yplane, as shownin the Figure1.3. The element end nodal coordinatesare indicated by(Xl' YI)and (x2' Y2). The elementaxis s runs from the first node of the elementto the second node. The angle a defines theorientationof the element with respectto a global x-ycoordinatesystem. Eachnode hastwo displacement degreesof freedom, u indicatingdisplacement in theXdirection and vindicatingdisplacement in the y direction.The elementcan be subjected to loadsonlyatits ends.Usingthese elements, the finite element modelof thetransmissiontower is as shownin Figure1.4. The modelconsistsof16 nodesand 29 planetruss elements. Theelementnumbers and node numbers are assigned arbitrarily for identification purposes.60057054048042010001b10001b300180o300 180 96 o 6096 180 300 inFigure 1.2. Transmission toweryDISCRETIZATION ANDELEMENT EQUATIONS 5Nodal dof End loadsFigure 1.3. Plane truss elementxElement numbersFigure 1.4. Planetruss element model of the transmission towerUsingproceduresdiscussedinlater chapters, it can beshownthat thefiniteelementequations for a plane truss-dement are as follows:IslnsIn;-lslns-In;-1;-Is Insz2sIslnswhere E = elastic modulus of the material(Young's modulus), A = area of cross section ofthe element, L =lengthof the element, andIs. Insare the directioncosines of the elementaxis (line from element node1 to 2). Here, Is is the cosine of angle a between the elementaxis and the x axis (measured 'counterclockwise) and Ins is the cosine of angle between theelement axis and the y axis. Interms of element nodal coordinates,6 FINITEELEMENTMETHOD:THE BIGPICTUREIn the element equationsthe left-hand-side coefficient matrix is usually calledthe stiffnessmatrix and the right-hand-side vector as the nodal load vector. Note that once the elementend coordinates, material property, cross-sectionalarea, and element loading are specified,the only unknowns in the element equations are the nodal displacements.It is important to recognize that the element equations refer to an isolatedelement,Wecannotsolve forthenodaldegreesof freedom forthe entirestructureby simplysolvingthe equations for one element.We must consider contributions of all elements, loads,andsupport conditions before solving for the nodal unknowns. These proceduresare discussedin detail in later sections of this chapter.Example 1.1 Write finiteelement equations for element number14 in the finite elementmodel of thetransmissiontower shownin Figure1.4. Thetower ismadeofsteel (!i..=-29 x 106Ib/in2) angle sections. The area of cross section of element14 is 1.73 in2.The element is connected between nodes 7 and 9. We can choosefirstnode of the element. Choosingnode 7 as the first node establishes the element s axis asgoing from node 7 toward 9. The origin of the global x-y coordinate system can be placedat any convenientlocation. Choosingthe centerlineof thetoweras the origin, thenodalcoordinates for the element14 are as follows:First node (node 7) = (-60, 420) in;Second node (node 9) = (-180, 480) in;XI=-60;x2 =-180;YI = 420Y2 = 480Using these coordinates, the element lengthand the direction cosines can easily be calcu-lated as follows:Element length:Element direction cosines:-------L =-xll + (Y2- YI)2 =60-{5 in;' _x2- XI _ 2. 112 =Y2- YI =_l_Is - -L-- --{5' s L -{5From the given material and section properties,E =29000000Ib/in2; E: =373945. lb/inUsingthese values, the element stiffnessmatrix(the left-hand sideof the element equa-tions) can easily be written as follows:[z2k =EA L -I;-lm,Isms112;-Isms-112;-I;-Isms12sIsms-Isms] [299156.-m; _-149578.Isms - "':299156.112; 149578.-149578.74789.149578.-74789.-299156.149578.299156.'-149578.149578.]-74789.-149578.74789.Theright-hand-side vectorofelement equationsrepresents appliedloadsattheelementends. There are no loads applied at node 7. The applied load of 1000 lb at node 9 is sharedby elements 14, 16,23, and 24. Theportiontaken byelement 14 cannotbedeterminedDISCRETIZATION ANDELEMENT EQUATIONSwithout detailed analysis of the tower, which is exactly what we are attempting to do in thefirst place. Fortunately, to proceed with the analysis, it is not necessary to know the portionof the load resisted by different elements meeting at a common node. As will become clearin the next section, in which we consider the assembly of element equations, our goal is togenerate a global system of equations applicable to the entire structure. As far as the entirestructure is concerned, node 9 has an applied load of 1000 lb in the -y direction. Thus, it isimmaterial how we assign nodalloadsto the elements as longas the total loadat the nodeis equal to the applied load.Keepingthis in mind, whencomputing element equations, wecansimplyignoreconcentrated loadsappliedatthe nodesandapplythemdirectlyto theglobal equations at the start of the assemblyprocess. Detailsof this process are presentedin a following section. nod'!JJQ.ads are ..14ar:...f9JIRWJ/;,.7[299156. -149578. -299156.-149578. 74789. 149578.-299156. 149578. 299156.149578. -74789. -149578.149578.] [U7] [0] -74789. v7_ 0-149578. u9' - 0 '74789. v90 MathematicafMATLAB Implementation:n..l on the Book Web Site:Plane truss element equations1.1.2 Triangular Element for Two-DimensionalHeat FlowConsider the problem of finding steady-state temperature distribution in long chimneylikestructures. Assuming no temperature gradient in the longitudinal direction, we can talce aunit sliceof sucha structure and modelit as a two-dimensional problem to determine thetemperature T(x, y). Using conservation of energyon a differential volume, the followinggoverningdifferential equation can easily be established.:_. a (aT) a( aT)axkx ax + ay ky ay + Q=0where kx and kyare thermal conductivities in the x and y directions and Q(x, y) is specifiedheat generation per unit volume. Typical units for k are W/m- C or Btu/hr ft OF and thosefor Qare W1m3or Btu/hr . ft3. The possible boundaryconditions are as follows:(i) Knowntemperature along a boundary:T =To specified(ii) Specifiedheat flux alonga boundary:8 FINITEELEMENTMETHOD: THE BIGPICTUREy (m)0.030.015 qooToon0.03 0.06x (m)nFigure 1.5. Heat flow throughan L-shaped solid: solution domainand unit normalswhere nx and ny are the x andY, componentsof the outer unit normal vector to theboundary (see Figure1.5 for an example):Inl =~ n; + n; =1On an insulatedboundaryor acrossa line of symmetrythere is no heat flow andthus qo =O. The sign convention for heat flow is that heat flowing intoa body ispositive and that flowing out of the body is negative.(iii) Heat loss due to convection along a boundary:st (aT aT)-k an == - kx ax nx + ky ay ny =h(T - Too)where h is the convection coefficient, T is the unknown temperature at the bound-ary, andToois the knowntemperature of the surrounding fluid. Typical units for hare W/m2 C and Btulhr ft2 "P,As aspecificexample, consider two-dimensional heat flowover anL-shapedbodyshown in Figure1.5. Thethermal conductivity in both directionsis the same,kx = ky =DISCRETIZATION ANDELEMENTEQUATIONS45 Wlm . C. The bottom is maintained at a temperature ofTo = 110C. Convection heatlosstakes placeon thetop wheretheambient air temperature is 20C andthe convectionheattransfer coefficient is h =55 W/m2 C. Theright sideis insulated. Theleftsideissubjected to heat flux at a uniform rate of qo =8000 W/m2. Heat is generated in the bodyat a rate of Q =5 X 106W1m3.Substitutingthegivendataintothegoverningdifferentialequationandtheboundaryconditions, we see that the temperature distribution over this body mustsatisfythe follow-ing conditions:9Over the entire L-shaped regionOn the left side (lix = -1, ny = 0)On the bottom of the regionOn the right side(nx =1, ny=0)On the horizontal portions ofthe top side(nx =0, ny=1)On the vertical portion of thetop side (nx =1, l1y=0)45 (a2; + a2; ) +5 X 106=0ax ay_ (45 aT (-1) =8000 =>aT= 8000 along x =0ax ax 45T =110 along y =0aT =0 along x =0.06ax(aT ) ei 55- 45 ay (1) =55(T - 20) => ay =- 45 (T- 20)(aT ) et 55- 45 ax (1) =55(T - 20) => ax=- 45 (T- 20)Clearlythereislittle hopeof findinga simplefunctionT(x, y) that satisfiesall thesere-quirements. We must resort to various numerical techniques. In the finite element method,the domain is discretized into a collection of elements, each one of them being of a simplegeometry, suchas a triangle, a rectangle, or a quadrilateral.A triangular element for solution of steady-state heatflow overtwo-dimensional bod-iesis shownin Figure1.6. Theelement canbeusedforfindingtemperature distributiony------xFigure 1.6. Triangular element for heat flow10 FINITEELEMENTMETHOD: THE BIGPICTUREoveranytwo-dimensional bodysubjectedto conductionand convection. Theelement isdefined by threenodes withnodal coordinatesindicatedby(xI' YI)' (Xz' Yz), and(x3'Y3)'The starting node of the triangle is arbitrary, but we must move counterclockwise aroundthe triangle to define the other two nodes. The nodal degrees of freedom are the unknowntemperatures at each node TpTz' and 13.For the truss model considered in the previous section, the structure was discrete to startwith, and thus there was only one possibility for a finite element model. This is not the casefor the two-dimensional regions. There are many possibilities in which a two-dimensionaldomaincan be discretizedusingtriangularelements. One mustdecideon the numberofelementsand their arrangement.In general, the accuracy of the solutionimprovesas thenumberof elements is increased. The computational effort, however, increases rapidlyaswell. Concentrating more elements in regions where rapid changes in solution are expectedproduces finite element discretizations that giveexcellent results withreasonablecom-putational effort. Some generalremarksonconstructinggoodfinite element meshesarepresentedin a following section. For the L-shapedsolid a very coarsefinite element dis-cretization is as shown in Figure 1.7 for illustration. To get results that are meaningful froman actual design point of view, a much finer mesh, one with perhaps100 to 200 elements,would be required.Thefinite element equations for a triangular elementfor two-dimensional steady-stateheat flow are derived in Chapter5. Theequationsare basedon the assumptionof linearyElement numbers0.030.0250.020.0150.010.0050x0 0.01 .0.02 0.03 0.04 0.05 0.06yNodenumbers0.030.0250.020.015210.010.005 2001 6 11 16 19x0 0.01 0.02 0.03 0.04 0.05 0.06Figure 1.7. Triangular elementmesh for heat flow throughan L-shaped solidDISCRETIZATION ANDELEMENT EQUATIONStemperature distributionover the element. In terms of nodal temperatures, the temperaturedistributionover a typical element is written as follows:whereThe quantities Ni, i =1, 2, 3, are known as interpolation or shape functions. The.superscriptTover Nindicatesmatrixtranspose. Thevector d is the vectorof nodal unknowns. Theterms bl, cI' ... depend on element coordinatesand are defined as follows:11CI=X3-X2;II =XiY3 - x3Y2 ;C2 =xI-X3;12 =X3YI -XIY3;b3 =YI - Y2C3=X2- Xj13 = XIY2- X2YIThe area of the triangle A can be computed from the following equation:where det indicatesdeterminant of the matrix.A note on the notation employed for vectors and matricesin this book is in order here.As an easy-to-remember convention, all vectors are considered column vectors and aredenoted by boldface italic characters. When an expression needsa row vector, a super-script T is used to indicate that it is the transposeof a column vector. Matricesare alsodenotedby boldfaceitaliccharacters. Thenumbersof rowsandcolumnsina matrixshould be carefullynotedin the initialdefinition. Remember that, for matrixmultipli-cation to make sense, the numberof columns in the first matrix should be equalto thenumber of rows in the second matrix. Since large column vectors occupy lot of space ona page, occasionally vector elements may be displayed in arow to save space. However,for matrix operations, they are still treated as column vectors.As shown in Chapter 5, the finite element equations for this element are as follows:12 FINITEELEMENTMETHOD:THE BIGPICTUREwhere kx =heat conduction coefficient in the xdirection, ky =heat conductioncoefficientin the y direction, and Q =heatgeneratedper unit volume over the element. The matrixJe"and the vector r"take intoaccount any specified heat loss due to convection along oneor moresidesofthe element. If the convection heat loss is specified alongside1 oftheelement, then we haveConvection along side 1:[2 1 0)k =hL121 2 0 ."6 '000- hTooL12[11)r" - --2- .owhereh = convectionheat flow coefficient, Too = temperatureoftheairorotherfluidsurrounding the body, and L12 =length of side 1 of the element.For convection heat flowalong sides 2 or 3, the matrices are as follows:J. 0nr" = 0)Convection along side 2:e" ' 620 1, 0

- hTooL31

Convection along side 3:e" 60 r,,- --?-I' 10 - 1whereL23 andL31 are lengths ofsides2and3 ofthe element. Thevector rqisduetopossible heat flux q applied along one or more sides of the element:Applied flux along side 1:Applied flux along side 2:Applied flux along side 3:r, UJ qi'[!j qi' mIf convectionor heat flux isspecified onmorethanonesideof an element, appropriatematrices are written for each side and then added together. For an insulated boundary q = 0,and hence insulated boundaries do not contribute anything to the element equations.DISCRETIZATION ANDELEMENT EQUATIONSAs mentioned in the previous section, we cannot solve for nodal temperatures by simplysolvingthe equations for one eiement. We must consider contributions of all elements andspecifiedboundary conditions beforesolvingforthenodal unknowns. Theseproceduresare discussed in detail in later sectionsin this chapter.Example 1.2 Writefinite element equations for element number 20 in the finite elementmodel of the heat flow throughthe L-shaped solid shown in Figure 1.7.Theelement issituated between nodes4, 10,and5. Wecanchooseanyof thethreenodes as the first node of the element and define the other two by moving counterclockwisearound the element. Choosing node 4 as the first node establishes line 4-10 as the first sideof the element, line10-5 as the second side,and line 5-4 as thethird side. Theorigin ofthe global x-y coordinate system can be placed at any convenient location. Choosing node1 as the origin, the coordinates of the element end nodesare as follows:13Node1 (global node 4) =(O., 0.0225) m;Node2 (global node10) =(O.015, 0.03) m;Node 3 (global node 5) =(O., 0.03) m;XI =0.;x2 =0.015;x3 =0.;YI =0.0225Y2 = 0.03Y3 =0.03Usingthesecoordinates, theconstantsbi' ci, and I, andtheelementareacaneasilybecomputed as follows:b, =0.;cI= -0.015;II =0.00045;b2 =0.0075;c2 =0.;12 =0.;b3=-0.0075c3=0.01513 =-0.0003375Element Area.= 0.00005625From thegivendatathethermal conductivitiesandheat generated overthesolidareasfollows:Q=5000000Substituting these numerical values into the element equation expressions, the matriceslckand rQ can easily be written as follows:There is an applied heat flux on side 3 (line 5-4) of the element. The lengthof this side ofthe element is 0.0075 m and Withq =8000 (a positive value since heat is flowing into thebody)the rq vector for the element is as follows:Heat flux on side 3 with coordinates ({O., 0.0225) (O., 0.03)),L =0.0075; q =8000(45.lck = O.-45.(93.75]"a = 93.7593.7514 FINITEELEMENTMETHOD:THE BIGPICTURE[30.)rq = a ,30.Theside 2 ofthe elementissubjected toheat lossbyconvection. Theconvectiontermgenerates a matrix khand a vector rho Substituting the numerical values into the formulas,these contributions are as follows:Convection on side 2 with coordinates ((0.015, 0.03) (0., 0.03}),L =0.015; h =55;Too =20 a 0.1375 0.275 8.25Adding matrices kkand kh and vectors rQ, rq,and rh, the complete element equations areas follows:[45.O.-45.O. -45. )[T4) [123.75)11.525 -11.1125 TIO= 102.-11.1125 56.525 Ts132. MathematicalMATLAB Implementation 1.2 on theBook Web Site:Triangularelement for heat flow1.1.3 General Remarks on Finite Element DiscretizationTheaccuracyof afinite element analysisdependsonthenumber ofelements usedinthe model andthe arrangement of elements. In general, the accuracy of the solution im-provesasthenumber ofelementsis' increased. Thecomputational effort, however, in-creases rapidly as well. Concentratingmore elementsin regionswhere rapidchanges insolutionareexpectedproducesfiniteelement discretizations that giveexcellent resultswith reasonablecomputational effort. Some general remarkson constructing good finiteelement meshes follow.1. Physical Geometry of the Domain. Enoughelementsmust beusedtomodel thephysicaldomainas accuratelyas possible. For example,when a curveddomainis to bediscretized by using elements with straight edges, one must use a reasonably large numberof elements; otherwise there will be a large discrepancy in the actual geometry and the dis-cretized geometry used in the model. Figure1.8 illustrates error in the approximation of acurved boundary for a two-dimensional domain discretized using triangular elements. Us-ing more elements along the boundary will obviously reduce this discrepancy. If available,a better option is to use elements that allow curved sides.2. Desired Accuracy. Generally, using more elements produces more accurate results.3. Element Formulation. Someelement formulations producemoreaccurateresultsthan others, andthus formulationemployedina particular elementinfluencesthe num-ber of elements needed in the model for a desired accuracy.DISCRETIZATION ANDELEMENTEQUATIONSActual boundaryFigure 1.8. Discrepancy in the actual physical boundaryand the triangular element model geometry15Valid meshxInvalid meshFigure 1.9. Valid and invalid mesh for four-nodeelements4. Special SolutionCharacteristics. Regionsover whichthe solutionchangesrapidlygenerally require a large number of elements to accurately capture high solution gradients.A good modelingpracticeisto start witha relativelycoarsemeshto get an ideaof thesolution and then proceed with more refined models. The results from the coarse model areused to guide the mesh refinement process.5. Available Computational Resources. Models with more elements require more com-putational resources in terms of memory, disk space, and computer processor.6. Element Interfaces. are joinedtogetherat nodes (typicallyshown as darkcircles on the finite element meshes). The solutions at these nodes are the primary variablesin the finite element procedure. For reasons that will become clear after studyingthe nextfew chapters, it is important to create meshes in which the adjacent elementsare alwaysconnected fromcomer tocomer. Figure1.9 showsan exampleof a valid andan invalidmesh when empioying four-node quadrilateral elements. The reason why the three-elementmesh on the right is invalid is becausenode 4 that formsa comer of elements2 and 3 isnot attached to one of the four comersof element 1.7. Symmetry. For many practical problems, solution domains and boundary conditionsare symmetric, and henceone canexpect symmetryin thesolutionas well. It is impor-tant to recognizesuch symmetryand to model only the symmetric portionof the solutiondomain that gives information for the entire model. One commonsituationis illustratedin the modeling of a notched-beam problem in the following section. Besidesthe obviousadvantage of reducingthe model size, bytakingadvantageof symmetry, one isguaran-teed to obtaina symmetricsolutionforthe problem. Dueto the numerical natureof the16 FINITEELEMENT METHOD: THEBIGPICTUREFigure 1.10. Unsymmetrical finite element mesh for a symmetric notchedbeam501b/in2Figure 1.11. Notchedbeamfinite element methodand the unique characteristics of elementsemployed, modelingtheentiresymmetricregion may in fact produceresults thatare notsymmetric. Asa simpleillustration, considerthetriangularelement meshshown inFigure 1.10 thatmodels theentire notched beam of Figure1.11. The actual solution should be symmetricwith respectto the centerlineof the beam.However, the computedfinite element solutionwill notbeentirely symmetric becausethe arrangement of the triangular elements in the model is notsymmetricwith respect to the midplane.Ageneralrule of thumbto followina finite element analysisisto startwitha fairlycoarse mesh. The number and arrangement of elements should be just enough to get a goodapproximationof the geometry, loading, and other physical characteristics of the problem.Fromtheresultsofthiscoarsemodel, select regionsinwhichthesolutionischangingrapidlyforfurther refinement. Toseesolutionconvergence, select oneormorecriticalpointsinthe model and monitor thesolutionatthese pointsas thenumber ofelements(or the total number of degrees of freedom)in the model is increased. Initially, when themeshesare relativelycoarse, thereshouldbe significantchangeinthesolution at thesepoints from one mesh to the other. The solution should begin to stabilize after the numberof elements used in the model has reached a reasonable level.1.1.4 Triangular Element for Two-Dimensional Stress AnalysisAs a final example of the element equations, consider the problem of finding stresses in thenotched beam of rectangular cross section shown in Figure 1.11. The beam is 4 in thick inthe direction perpendicular to the plane of paper and is made of concretewith modulusofelasticity E =3 x1061b/in2and Poisson's ratio v =0.2.Since the beam thickness is small as compared to the other dimensions, it is reasonableto consider the analysis as a plane stress situation in which the stress changes in the thick-ness directionare ignored.Furthermore, we recognizethat the loadingandthe geometryare symmetricwith respectto the plane passingthrough the midspan.Thusthe displace-ments must be symmetric and the points on the plane passingthrough the midspan do notexperience any displacement in the horizontal direction. Taking advantage of these simpli-DISCRETIZATION ANDELEMENTEQUATIONS 17xxsoso4040 3030 2020Element numbers1010y1210864200y1210864200Figure 1.12. Finiteelement model of the notchedbeamfications, we need to construct a two-dimensional plane stress finite element model of onlyhalf of the beam.As an illustration, a coarse finite element model of the right half of thebeam using triangular elements is shown in Figure 1.12. All nodes on the right end are fixedagainst displacement because of the given boundary condition. The left end of the modelis on the symmetry plane, and thus nodes on the left end cannot displacein the horizontaldirection. Onceagain, in an actualstressanalysisa muchfiner finite element mesh willbe needed to get accurate values of stresses and displacements. Even in the coarse modelnotice that relatively small elements are employed in the notched region where high stressgradients are expected.A typical triangular element for the solution of the two-dimensional stress analysis prob-lem is shown in Figure 1.13. The element is defined by three nodes with nodal coordinatesindicated by(XI' YI)' (x2' Y2)' and (x3' Y3)' The starting node of the triangleis arbitrary, butwe must movecounterclockwise aroundthetriangleto definethe other two nodes. Thenodaldegreesof freedomare thedisplacementsintheXand Y directions, indicatedbyu and v. Onone or moresidesof the element, uniformlydistributedloadin thenormaldirection qn and that in the tangential directionqrcan be specified.The element is basedon theassumptionof linear displacementsoverthe element. Interms of nodal degreesof freedom, the displacementsover an element can bewrittenasfollows:u(x, y) = NIul +N2u2+N3u3vex, y) =NI VI +N2v2 +N3v3ul~ JVI( u(x, y) ) =(NI0 N20 N3u2=NTdvex, y) 0 NI0 N20 v2u3v318 FINITEELEMENTMETHOD:THEBIGPICTUREy-------------xFigure 1.13. Planestress triangular elementwhere the Ni , i =1, 2, 3, are the samelinear triangle interpolation functionsas thoseusedfor the heat flow element:Cj=X3-XZ;II=XZY3 - X3Yz;Cz=XI -X3;Iz=X3Yl- XlY3;C3 = Xz-XlI3=XlYz- XZYIThe element area A can be computed as follows:Using these assumed displacements, the element strains can be written as follows:o bz0 b3o Cz0Czbzc3where Ex and Y1 =~lYI + -{ITh =6 ===:> h=3m2Yl +0Yz + ..;7Y3 =7===:> Y3 =3 ~lYI + (--{IT)y:i + (-..;7)Y3 + {iY4 =8===:> Y4 =3 ~Using the LTmatrix, the solution for actual variables (backward substitution) is as follows:53"133 {IT113-17433-{i62 FINITEELEMENTMETHOD: THE BIGPICTURERow 4:Row 3:Row 2:Row 1:- (;;2d -.-:!1. ---- d - 11-yL. 4- 3-12 --.' 4- 6- '7d + (-- '7)d =.l.L====> d = 323-y I 3 vI 4 30 3 4Z{fidz + Od3 + (-{fi)d4 =3 ~ ====> dZ =':rt3dl + 1dZ + 2d3 + Id4 =~ ====> d, = - 7 i ~1.5.2 Conjugate Gradient MethodThe well-knownconjugategradient methodfor solutionofunconstrainedoptimizationproblems canbeusedto developaniterativemethodforsolutionofa linearsystemofequations. Themethodhasbeen usedeffectivelyforsolutionof nonlinear finiteelementproblems that requirerepeated solutionsoflargesystemsof equations.Consider a sym-metric system of n X n linear equationsexpressed in matrix form as follows:Kd=R['''kJZk'"r] ["]kZJkF10.11 dz_ rz":,1"liZ.. cL I ~ I IFinding the solutionof this systemof equationsis equivalent to minimizing the followingquadratic function:minf(d) =4dTKd - dTR,/Thiscan easilybeseen byusingthe necessaryconditions forthe minimum, namely, thegradient of the function must be zero:whereSince Kis a symmetric matrix, the transposeof the first term ineachrow issameas thesecondterm. Thus the gradient can be expressed as follows:SOLUTION OF LINEAREQUATIONSNoting thatref] [1 0 ... 0]'eI 010 ...: =: : -. : =identity matnxeT 0 0 ... 1IIwe see that the gradient of the quadratic function isg(d) =Kd-RA condition of zero gradient implies Kd- R =0, which clearly is the given linear system ofequations. Thus solution of a linear system of equations is equivalent to finding a minimumof the quadratic function f.Basic Conjugate Gradient Method In the conjugate gradient method, the minimumof f is located bystartingfroman assumedsolutiond(O) (say withall variablesequal tozero) and performing iterations as follows:63k =0, 1,...wherea(k) is knownas the step lengthandh(k) is the search direction. At the kth iterationthe search directionh(k) is determined as follows:Thescalar multiplier fJdeterminestheportionoftheprevious directiontobeaddedtodetermine the new direction.According to the Fletcher-Reeves formula,After establishing the search direction, the minimization problem reducesto finding a(k) inorder toFor the quadratic function f we have64 FINITEELEMENTMETHOD:THEBIGPICTUREExpanding,!Cd(k) +(P)h(k)) =Kd(k) + Kh(k)+ Kd(k) + Kh(k) - (d(k){R - a(kJch(k){RFor the minimum thederivative ofthisexpressionwithrespect toa(k) must beequal tozero, giving the following equation for the step length:Noting the symmetry of K, the first two terms can be combined and moved to the right-handside to givegivingSince f3 is usually small, h(k) "" _g(k), and thus for computational efficiency the step lengthexpression is slightlymodified as follows: "BasicConjugateGradientAlgorithm Thecomputationalstepscanbeorganizedinto the followingalgorithm.Choose astartingpoint d(O). Computeg(O)=Kd(O) - R. Set h(O) =_g(O). ComputereO) =(g(O){g(O). Choose a convergence tolerance parameter E. Set k O.1. If -,J-k) .sE, stop;d(k) is the desired solution. Otherwise, continue.2."Compute the step length:3. Next point:SOLUTION OF LINEAREQUATIONS4. Update the quantities for the next iteration:g(k+l) =K(d(k) +o:(k)/z(k) _R==g(k) + o:(k)Z(k),.(k+l) =(g(k+l){g(k+l)fP) =,.(k+l)h(k)/z(k+1) =-s"'" +(3(k)/z(k)5. Set k =k: + 1 and go to step 1.Example 1.14 Findthe solutionofthe followingsystemofequationsusingthe basicconjugate gradient method:Starting point,d(O) =(0,0,0, O)g(O) =Kd(O) -R =(-5., -6., -7., -8.)/z(0) =_g(O) =(5.,6.,7., 8.),.(0)=g(O)Tg(O) =174.;(3(0) =0Iteration 1:z(O) =Klz(O) :::; {129., 21., 79.,88.)Step length, 0:(0) =,.(O)/h(O)Tz(O) =0.0857988Next point, d(1) =d(O) +0:(0)/z(0) == (0.428994,0.514793,0.600592,0.686391)g(l) =g(O) + o:(O)z(O) =(6:06805,-4.19822, -0.221893, -0.449704),.(1) =g(I)Tg(l) =54.6978(3(1) =,.(1)/,.(0) =0.314355/z(l) =-g(l) +(3(0)/z(0) =(':4.49627,6.08435,2.42238, 2.96454)Iteration 2:z(l) =KIz(l) =(1.21451,34.7228, -2.98549, -24.1888)Step length,o:m=,.(I)/Iz(l)Tz(l) =0.431153Next point,d(2) =d(1) +o:(I)/z(l) = (-1.50959, 3.13808, 1.64501, 1.96456)g(2) =g(l) + o:(I)Z(I) =(6.59169, 10.7726, -1.50909, -10.8788),.(2) =gC2)TgC2) =280.125(3(2) =r(2)/,.(I) =5.121321z(2) =_g(2) +(3(l}Jz(l) =(-29,6185, 20.3873,13.9149,26.0612)Iteration 3:Z(2) =KIz(2)=(-43.7321,-76.9897,-114.179, 184.981)Step length, 0:(2) =,.(2)//z(2)Tz(2) =0.09470986566 FINITEELEMENTMETHOD:THE BIGPICTURENext point, d(3) ==d(2) +OP)h(2) == (-4.31475,5.06896,2.96288, 4.43281)g(3) ==g(2) +(1'(2)Z(2) ==(2.44982, 3.48091, -12.323, 6.64076)r(3) == g(3)Tg t =-xSubstituting these, the bounding curves areCj: s= 2; C2: l =tx =x =>t =1; C3: s =4;Thus, in terms of (s, t)the integration region is a square, as shown in the figure.From the given change of variables we' can solve for x and y to getS2!3x'=-'{i'y = Y;{iThe Jacobian matrix and the Jacobian of the transformation are as follows.(axJ=&~asax) [_2Ft = 3fSfi~ fial 3?iJs2fJ ] -31413fS 'J?i31detJ =3t(1.1, 1.82}(2.52, 1.59}x(4,3}(4, I}---s(1.59, 1.26}Figure 6.1. Two-dimensional area in original x-y and transformed s-t coordinatesINTEGRATIONUSINGCHANGE OF VARIABLESThusthe integral can be evaluated as follows:If(x6) If (sV3j{i)6 If s3l dAxy = ({S{i? detl ds dt = 3t4dsdt~ ~ ~ ,L4(f3S3) L426s3520= 4 dt ds = --ds =-s=2 1=1 3t s=2 243 81Derivation of dAxy =del Jdsdt An important relationshipused in the computationwith mappingis that dA.ry= detl ds dt, For the derivation of this result,assume that in thes, t coordinates the differential area dAslis a small rectangle, as shown in Figure6.2. ThepointsPi' i =0, ... , 3, denotethe verticesof this rectangle. Withthe lengthsof the sidesdenoted by ds and dt, the coordinatesof the vertices are expressed as follows:385P2(so + ds, to+ dt);Thecorresponding pointsin the x, y domainare denotedbyQi'i =0, ... ,3,and are ob-tainedfromthemappinggivingthedifferential areadA.ryasshowninFigure6.2. Thecoordinates of pointQoare determinedas follows:Similarly, the coordinates of point QIare determined as follows:In general, the coordinates xI' YI are complicated functionsof ds. However, sincewe aredealing with small differential lengths, we can write these coordinates using a Taylor seriesas follows:YIyi--ds ---ITdt1--+--------s--'-t--------xFigure 6.2. Differential area in mapped and original coordinates386 MAPPEDELEMENTSIn a similarmannerthe coordinatesof the othertwo points can be expressedin terms ofderivatives of the mapping functions. Thusthe coordinatesof pointsQi' i =0, ... , 3, canbe written as follows:These four points form a parallelogram whose area can be determined by taking the crossproduct of vectorsQOQ1and QOQ3 as follows:( ax ay )Vector QOQ1=Q1- Qo=as ds, as ds( ax ay )Vector QOQ3 = Q3- Qo= at dt, at dtax ay ay axdAxy =QOQ3 x QOQ1=as ds at dt- as ds at dt == detJ dsdtwheredetJ = Iax ay._ax ay Ias at atas,/is the Jacobianas defined earlier. Thus, withthe changeof variables,the area integralistransformedas follows:II!(x,y)dAxy =II!(x(s,t),y(s,t))detJdsdtAX)' A"Note that, since ds dtis the area dAs/' the ratio of the areas Ax!As/ is equal to the absolutevalue of the Jacobian.6.1.3 Three-Dimensional Volume IntegralsConsider a function lex,y, z) of three variables x, y, and z that needs to be integrated overa three-dimensional regionIIIlex, y, z) dVxyz~ y . :MAPPINGQUADRILATERALSUSINGINTERPOLATIONFUNCTIONSThesubscript xyz is used on Vto indicatethat the integration is over a volume defined interms' of x, y, and z. Consider a change of variables to 1', S, t given by the mapping functionsas follows:387x =x(r, s, t); y =y(r, s, t); z =z(r, s, t)Following the same reasoningas forthetwo-dimensional case, it can be shownthat therelationship betweenthe differentialvolumedVStarting point of liney(1)=: Yz =>Ending point of liney(1)=: (y] +Yz)12 =>Midpoint of lineMAPPINGQUADRILATERALS USINGINTERPOLATION FUNCTIONSMapping aCurve Thesameideacanbeusedtowriteamapping betweenaspacecurve andatwo-unit-longlineintheScoordinate. If acurveisdefinedinterms of npoints (x"YP z]), (xz'Yz, zz), ... , then we choosea two-unit-long master line element with11 equallyspacednodes. Forthemaster element wewritethe (n - I)-order Lagrange in-terpolation functionsNi(s),i =1, 2, ... , 11. Themapping of x and Y coordinates is thenasfollows:xeS) =N;(s)xj+Nz(s)xz+ +N,,(s)xlIyes) =N] (s)Yj +Nz(s)yz + +NII(s)YIIz(s) =N] (s)Zj +Nz(s)zz+ +N/s)zlIAsan example, considera quadraticcurveinthex, Y planedescribedinterms ofcoor-dinatesat three pointsas shown in Figure 6.4. Themaster element is defined: with nodesat S =-1, 0, 1. Forinterpolating betweenthethreedatapoints, wewritethequadratics Lagrange interpolation functions for the master lineas follows:389.N3 =!s(l +s)The mappingof x and y coordinates is then as follows:xeS) =!(-1 + s)sx] + (1 - sZ)xz+ !s(l +s)x3yes) =!(-1 + s)sYj + (1 - sZ)Yz + !s(l +S)Y3We can clearlysee thatxes) and yes) mapthethreegiven pointson the curveto thethreenodes on the master line by evaluating the coordinates of the curve at the corresponding svalues as follows:Ats = -1:Ats =0:Ats =1 :x(-l)=xj;x(O) = xz;x(l) = x3;y( -1) =Yj ~ Starting point of lineyeO) =Yz ~ Second point on liney(l) =Y3 ~ Ending point of lineIn order to see thatthe intermediate pointsare mapped correctlyas well, we consider thefollowing numerical example.Threenode master lineNode-1Noe 2 Node 3sy Arc.L{X3, Y3)I(xz, yz)(Xj, yllxFigure 6.4. Master line in s-t and a quadratic curve line in the x-y coordinates390 MAPPEDELEMENTSExample 6.3 Givenaquadraticcurve that passes throughpoints II, I}, 12,712}, 14, 5},develop an appropriate mapping to map this curve to a straight line two units long. Numer-icallydemonstratethat all points onthecurvearemappeduniquelyto thepointsonthestraight line.In this example we haveYJ =1; Y- 12- 2'Substituting the given coordinates intothemapping of x and Y coordinates, we havexes) =1Ws - I)s) + -I)(s + 1)) ++ 1)) = ++ 2yes) =1- I)s) + H-(s - I)(s + 1)) + 5+1)) =+ 2s +Using this mapping, we evaluate x and y coordinates for several values of s. The computa-tions are summarized in the following table:s xes) yes)-1 1 1-0.75 1.15625 1.71875-0.5 1.375 2.375-0.25 1.65625 2.96875O. 2. 3.50.25 2.40625 3.968750.5 2.875 4.3750.75 3.40625 4.718751. 4. 5.-rThe table clearly demonstrates that the given points are mapped into the nodes ofthe masterelement and all pointsin between are mapped appropriately.Restrictions on Mapping of Lines For the mapping to be useful in finite element ap-plications, it must be one to one. That is, each point on the master element must map into aunique point on the given curve. Unfortunately, the mapping based on the Lagrange inter-polation doesnot alwayssatisfy thisrequirement. As an example, consider a line passingthrough the following three data points. These points are the same as those used in Example6.3 except for the x coordinate of the second point.Coordinates:(1, IJ, (4, 5JUsingthesecoordinateswiththequadratic Lagrange interpolation functions, weget thefollowing mapping:xes) =1Ws-I)s) + H-(s -1)(s+ 1))+ 1)) = ++yes) =1- I)s) + H-(s - I)(s + 1)) + 5+ 1)) =+ 2s +MAPPINGQUADRILATERALS USINGINTERPOLATION FUNCTIONS:rhe followingtableshows several z, ypointsevaluatedfromthismappingfordifferentsvalues. Theyvaluesareall-reasonable. However, someof the xvaluesbetweens =-1, 0 are not reasonable. We would expect all these values to be between1 and i =1.25.However, the computed values for s =-0.8, -0.6, -0.4 are all less than1. Also differentvalues of s are getting mappedto the same point in x. For example, x= 1 is obtained fromboth s =-1and s =-0.2. This shows that the mapping is not one to one.s x(s) y(s)-1 1 1-0.8 0.85 1.58-0.6 0.8 2.12-0.4 0.85 2.62-0.2 1. 3.080 1.25 3.50.2 1.6 3.880.4 2.05 4.220.6 2.6 4.520.8 3.25 4.781. 4. 5.From this example it is clearthat for mappingto be one to one the functionsx(s)and y(s)shouldbeeither increasingordecreasingfunctions overtheentirerange-1 -s s dsFor the numerical example we havedy > 0dsfor all -1ss s I.dx 5s 3-- ds =2+ 2;dy- =2-sdsThese derivatives are plotted in Figure 6.5. We can clearly see that dxlds is negative froms =-1 to s =-0,6. Therefore the mapping for the x coordinate is not good over this range.The mapping for the y coordinate is good for the entire range.Usingthepositive-derivative requirement for themappingfunction, itispossibletowrite a generalrequirement for placement of the middle pointfor a quadratic curve. Dif-ferentiatingthe general formulafor mapping of a quadratic curve, we havedx 1( )- =-(-1 +2s)xl- 4sx2-+(l +2s)x3ds 2 .Setting L; =x3- xl' where L; is the horizontallength of the line, we have392 MAPPEDELEMENTS-- dx/ds-- dy/dssFigure 6.5. Derivatives for the mappingAts =-1Ats =13L 3L----' + 2x - 2x > 0=:} x < x +----'2 1 2 2 1 4Thederivative will be positive overthe entire range for s between -1 and1ifIn otherwords, themappingof aquadraticcurve isfineaslongasthesecondpointischosen anywhere over the middle half of the curve. MathematicalMATLAB Implementation 6.1 on the Book Web Site:Mapping lines6.2.2 Mapping Quadrilateral AreasFor mapping two-dimensional areasthe master area is a 2 x 2 square in the s, t coordinates.The interpolation functions are written by substituting numerical values of the nodal coor-dinates of the master area into the expressions given for rectangular elements in Chapter 5.Themapping between x and y coordinates is obtained by treating the coordinates of the ac-tualelement as the data for Lagrange interpolation. The idea is first explained bydefiningmapping fora quadrilateral withstraight sides. It is then generalized to quadrilaterals withcurved sides.Quadrilaterals with Straight Sides Consider a 2 x 2 square master area and an arbi-traryquadrilateral areaas shown in Figure 6.6. Theinterpolation functions forthemasterareacaneasilybewritten bytakingproductsoflinear Lagrange interpolation functions MAPPINGQUADRILATERALS USINGINTERPOLATION FUNCTIONS 393(xz, yz)xY Quadrilateral area(X3, Y3)4I21Master area.t1----2---1Figure 6.6. Four-nodemaster area and actual quadrilateral areainthe s and t directionsas discussedin Chapter 5. Usingthe numerical values of the co-ordinates at thefour nodesofthemaster area, thefollowinginterpolationfunctions areobtained:N=!(l - s)(l - t)!(s + l)(l- t)!(s + l)(t + 1)!(l - s)(t + 1)Multiplying these interpolation functionswithnodalcoordinatesof the quadrilateral, themapping for the quadrilateral is then as follows:Xes, t) = !(l - s)(l - t)xj+-!(s + 1)(1 - t)xz+ !(s + 1)(t + l)x3 + !(l - s)(t + l)x4yes, t) =!(l - s)(l - t)Yj +!(s + 1)(1 - t)yz+ !(s + 1)(t + 1)Y3 + !(l - s)(t + 1)Y4To seethat xes, t) and yes,t) represent the appropriate mapping, consider the sidesofthequadrilateral. The side (xl,'Yj) to (xz' Yz) of the quadrilateral should be mapped to the masterarea side 1-2. To verify this, we evaluate the coordinates of the quadrilateral at t =-1andselected s values as follows:At (s,t) = (-1, -1): x(-I, -1) =xj; y(-1,-:-: 1) = Yj =>Starting point of side1-2At (s, t)=(0, -1):At (s, t) =(l, -1):X =&(xj +Xz); Y =&(Yj +Yz) => Midpoint of side 1-2x =Xz; y.=Yz => Endpoint of side 1-2Similarly we can verifythat each pointon the master area maps to a uniquepoint on thequadrilateral. The origin of the area (0, 0) corresponds to the point with x, Y coordinatesasthe average of the coordinatesof the four comers:(s, t) =(0,0) =>{!(xi +Xz+x3+x4), !(Yj +Yz +Y3 +Y4)}394 MAPPEDELEMENTSRestrictions on Mapping of Areas The mapping must be one to one to be usefulinthe finite element computations. Formapping of lines it was shownthatthis requirementis fulfilled if the derivative of the mapping is positive over the range of the master line. Forthe mapping of areas, a similar criteria can be developed. Themapping now is a functionof two variables;therefore, usingthe chain rule, we write its total derivatives as follows:OX oxdx= -ds+ -dtOs otoy oydy =-ds+ -dtos otWriting the two differentials in a matrixform, we have(dX)=(fsdy !!x.as~ ) ( d S )!!x. dtatForgiven ds and dt valuesthis equation determines a corresponding value of dxand dy.Formapping to be oneto one, we should beableget uniquevaluesof dsand dt for anygiven values of dx and dy. That is, we should be able to get an inverse relationship( )(axds asdt = ~ax) ( )-7ft dxfu dyaswhere Jis the 2 x2 Jacobian matrixand detJ is its determinant and is simplycalled theJacobian:(axJ= as!!x.as~ ) ; / detJ =ox oy_ox oy!!x. , os ot otosatIt is evident that detJ mustnotbezeroanywhere over -1::; s, t ::; 1 forthe mapping tobeinvertible. Thisrequirement issatisfiedas longasdetJiseither positiveornegativeover the entire master element. By adopting a convention of defining areas by moving in acounterclockwise direction aroundthe boundaries, detJ should alwaysbe positive. Thus,for the mappingto be valid, we have the following criteria:Mapping is valid if detJ > 0 for all - 1 ::; s, t s 1Example 6.4 GoodMapping Determine the mapping to a 2 x 2 square for a quadrilat-eral with the followingcorner coordinates:xY1 0.5 02 2.25 1.33 3.1 2.74 1.1 2MAPPINGQUADRILATERALS USINGINTERPOLATION FUNCTIONS 395Substituting these coordinates into the equationsfor mapping, we getxes, t)=- s)(1- t)0.5 + + 1)(1 - t)2.25 ++ 1)(t + 1)3.1 +- s)(t + 1)1.1yes, t) = - s)(1- t)O+ + 1)(1 - t)1.3 ++ 1)(t + 1)2.7 + s)(t + 1)2These expressions simplify toxes, t) =0.0625ts + 0.9375s + 0.3625t + 1.7375=.,..0. 15ts + 0.5s + 0.85t + 1.5Differentiating these expressions, the Jacobian matrix and the Jacobian of the mapping areas follows:1 =(0.0625t + 0.9375 0.0625s + 0.3625)0.5 - 0.15t 0.85 - 0.15sdetl =-0. 171875s +0.1075t +0.615625By solving the equation detl = 0 for s, we find the line over which the Jacobian is zero asfollows:s =-5.81818(-0.1075t - 0.615625)Figure 6.7 shows the line along which the Jacobian is zero. Thisline is clearly outside ofthe master area. Thus detl *' 0 over -1:;; s, t:;; 1 and hence the mapping is good.Since detl is a simple function for this example, it is easy to set detl =0 and solve fors to draw the line along which deEl is zero. When detl is a complicated function, this willnot be possible. Thus, to generate a zero Jacobian line, we need a tool to draw contour plotof functions of two variables. Both MATLAB and Mathematica have functions for drawingcontour plots. The following Mathematica commands are used to draw Figure 6.7.o -1-2DetJ=O4 3 2 o-3__'-":" -.J-1sFigure 6.7. Zero Jacobian lineand the master area396 MAPPEDELEMENTSdetJ = -0.171875s +0.1075t +0.615625;Block[{$DisplayFunction =Identity},zeroContour =ContourPlot[detJ, Is, -4, 4}, It, -3, 3}, Contours -. {O},ContourShading -. False, FrameLabel -.{"s", "t"},AspectRatio -. Automatic]];Show[{zeroContour, Graphics[{GrayLevel[0.7], Rectangle[(-1, -1}, (1, 1)],GrayLevel[O], Text["DetJ =0", {1.5, -2.3}]}]}];Figure6.8 providesagraphical illustration ofthe mapping. A regulargrid ofpointsischosen on the master area. For each pair of s, tvalues, the corresponding x, y values arecomputed from the mapping. All points are clearly mapped properly.s xes, t) Yes, t)-1 -1 0.5 O.-1 -0.5 0.65 0.5-1 O. 0.8 1.-1 0.5 0.95 1.5-1 1. 1.1 2.-0.5 -1 0.9375 0.325-0.5 -0.5 1.10313 0.7875-0.5 O. 1.26875 1.25-0.5 0.5 1.43438 1.7125-0.5 1. 1.6 2.175O. -1 1.375 0.65O. -0.5 1.55625 1.075O. O. 1.7375 1.5O. 0.5 /1.91875 1.925O. 1. 2.1 2.35xY QuadrilateralG""Master area,,-'J : :L-::------.Figure 6.8. Four-node master area and actual quadrilateral areaMAPPINGQUADRILATERALS USINGINTERPOLATION FUNCTIONS 397s x(s, t) y(s, t)0.5 -1 1.8125 0.9750.5 -0.5 2.00938 1.36250.5 O. 2.20625 1.750.5 0.5 2.40312 2.13750.5 1. 2.6 2.5251. -1 2.25 1.31. -0.5 2.4625 1.651. O. 2.675 2.1. 0.5 2.8875 2.351. 1. 3.1 2.7Example 6.5 BadMapping Determine the mapping to a 2x2 square for a quadrilateralwith the followingcorner coordinates:xY1 0.5 02 1.2 1.33 3.1 2.74 1.1 2Substituting these coordinates into the equations for mapping, we getx(s, t) =0.325ts + 0.675s + 0.625t + 1.475y(s, t) =-0. 15ts + 0.5s + 0.85t + 1.5The Jacobian matrix and the Jacobian of the mapping are as follows:J =(0.325t +0.675 0.325s+ 0.625);0.5 - 0.15t 0.85 - 0.15sdetJ =-0.26375s +0.37t + 0.26125By solvingthe equation det1 =0 for s, we find the line over which the Jacobian is zero asfollows:s = -3.79147(-0.37t - 0.26125)10.5o... -0.5-1-1.5- 2 ~ ..:J-1 0 2sFigure 6.9. Zero Jacobian lineand the masterarea398 MAPPEDELEMENTSIILMaster area"oe"yQuadrilateralFigure 6.10. Four-nodemaster area and actual quadrilateral areaFigure6.9 shows the linealongwhich the Jacobianis zero. Thisline overlapsthe masterarea and hence the mapping is not one to one. Figure 6.10 providesa graphical illustrationof the mapping. A regulargrid of pointsis chosen on the master area. For each pair of s, tvalues,the corresponding x, y values are computed from the mapping. It is clear fromthefigure that some points are mappedoutside of the quadrilateral.S xes, t) yes, t)-1 -1 0.5 O.-1 -0.5 0.65 0.5-1 O. 0.8 1.-1 0.5 0.95 1.5-1 1. 1.1 2.-0.5 -1 /0.675 0.325-0.5 -0.5 0.90625 0.7875-0.5 O. 1.1375 1.25-0.5 0.5 1.36875 1.7125-0.5 1. 1.6 2.175O. -1 0.85 0.65O. -0.5 1.1625 1.075O. O. 1.475 1.5O. 0.5 1.7875 1.925O. 1. 2.1 2.350.5 -1 1.025 0.9750.5 -0.5 1.41875 1.36250.5 O. 1.8125 1.750.5 0.5 2.20625 2.13750.5 1. 2.6 2.5251. -1 1.2 1.31. -0.5 1.675 1.651. O. 2.15 2.1. 0.5 2.625 2.351. 1. 3.1 2.7MAPPINGQUADRILATERALS USINGINTERPOLATION FUNCTIONSQuadrilateralswithCurvedSides Usingappropriateinterpolation functions, it ispossibletomapquadrilateralswithoneormoresidescurved. Themaster element sidethat correspondstoa curvewillhaveasmanyequallyspacednodesasthoseneededtodefinethe curveontheactual area. For example, fora siderepresented byaquadraticcurve, there willbethreenodes onthemaster element placedat- i, 0, and1. Thus, ifall sides of an elementare quadraticcurves, then there will be a total of eight nodes. Theinterpolation functions for this' eight-nodemaster element aretheserendipityfunctionswrittenfor rectangular elements inChapter 5. If the two oppositesidesare curves, thenthe interpolation functionsare written using the product Lagrange formula. If one, two, orthree adjacent sides of an element ar