MATLAB Codes for Finite Element Analysis in Solid Mechanics

Solid Mechanics and Its Applications Antonio J M Ferreira Nicholas Fantuzzi MATLAB Codes for Finite Element Analysis Solids and Structures Second Edition Solid Mechanics and Its Applications Volume 157 Founding Editor G M L Gladwell University of Waterloo Waterloo ON Canada Series Editors J R Barber Department of Mechanical Engineering University of Michigan Ann Arbor MI USA Anders Klarbring Mechanical Engineering Linköping University Linköping Sweden The fundamental questions arising in mechanics are Why How and How much The aim of this series is to provide lucid accounts written by authoritative researchers giving vision and insight in answering these questions on the subject of mechanics as it relates to solids The scope of the series covers the entire spectrum of solid mechanics Thus it includes the foundation of mechanics variational formulations computational mechanics statics kinematics and dynamics of rigid and elastic bodies vibrations of solids and structures dynamical systems and chaos the theories of elasticity plasticity and viscoelasticity composite materials rods beams shells and membranes structural control and stability soils rocks and geomechanics fracture tribology experimental mechanics biomechanics and machine design The median level of presentation is the first year graduate student Some texts are monographs defining the current state of the field others are accessible to final year undergraduates but essentially the emphasis is on readability and clarity Springer and Professors Barber and Klarbring welcome book ideas from authors Potential authors who wish to submit a book proposal should contact Dr Mayra Castro Senior Editor Springer Heidelberg Germany email mayracastrospringercom Indexed by SCOPUS Ei Compendex EBSCO Discovery Service OCLC ProQuest Summon Google Scholar and SpringerLink More information about this series at httpwwwspringercomseries6557 Antonio J M Ferreira Nicholas Fantuzzi MATLAB Codes for Finite Element Analysis Solids and Structures Second Edition 123 Antonio J M Ferreira Engenharia Mecânica Universidade do Porto Porto Portugal Nicholas Fantuzzi DICAM Department University of Bologna Bologna Italy ISSN 09250042 ISSN 22147764 electronic Solid Mechanics and Its Applications ISBN 9783030479510 ISBN 9783030479527 eBook httpsdoiorg1010079783030479527 1st edition Springer ScienceBusiness Media BV 2009 2nd edition The Editors if applicable and The Authors under exclusive license to Springer Nature Switzerland AG 2020 This work is subject to copyright All rights are solely and exclusively licensed by the Publisher whether the whole or part of the material is concerned specifically the rights of translation reprinting reuse of illustrations recitation broadcasting reproduction on microfilms or in any other physical way and transmission or information storage and retrieval electronic adaptation computer software or by similar or dissimilar methodology now known or hereafter developed The use of general descriptive names registered names trademarks service marks etc in this publication does not imply even in the absence of a specific statement that such names are exempt from the relevant protective laws and regulations and therefore free for general use The publisher the authors and the editors are safe to assume that the advice and information in this book are believed to be true and accurate at the date of publication Neither the publisher nor the authors or the editors give a warranty express or implied with respect to the material contained herein or for any errors or omissions that may have been made The publisher remains neutral with regard to jurisdictional claims in published maps and institutional affiliations This Springer imprint is published by the registered company Springer Nature Switzerland AG The registered company address is Gewerbestrasse 11 6330 Cham Switzerland I dedicate this book to Sara with love António Ferreira To Ilaria Nina and Lena Amor gignit amorem Nicholas Fantuzzi Preface to the Second Edition This new edition comes 10 years after the first publication The main reason is due to some physiological changes into MATLAB programming and tools The aim of the book is to present finite element programming with the help of MATLAB easy implementation style Codes are not optimized to get best performances but to enhance clarity to readers Finite element programming is presented via classical examples from structural mechanics Readers can easily start from the given codes and modify them according to their needs In this book most common problems for 1D and 2D structures are presented such as static free vibration buckling and linear time history analyses Not all the given analyses are presented and solved for all the given structural models However readers can easily use theories and codes presented in order to extend the given codes to problems not given in the book Major modifications to the first edition are listed below Reviewed and improved MATLAB introductory chapter with more samples and programming details General finite element code review and cleaning Removal of MATLAB struct implementations only plain MATLAB codes are used Expanded theory and codes for the free vibration analysis of 2D and 3D trusses Expanded theory and codes for the free vibration analysis of 2D and 3D Bernoulli frames Expanded theory and codes for the buckling problem of Bernoulli beams Enhanced graphical output using Hermite interpolation for Bernoulli beams and frames Improved theoretical background of Timoshenko beam theory Expanded theory and codes for the free vibration analysis of 2D plane stress problems Expanded theory and codes of Q8 and Q9 elements for plane stress New codes for stress extrapolation and interelement averaging for 2D plane stress vii New codes for bending of Kirchhoff plates with conforming and notconforming elements Improved theory and new codes of Q8 and Q9 elements for Mindlin and lam inated FSDT plates Expanded theory and codes for buckling of laminated FSDT plates New chapter for the bending and free vibration solutions of functionally graded Timoshenko beams New chapter for the bending and free vibration solutions of functionally graded Mindlin plates New chapter on linear time transient analysis for Timoshenko beams New chapter on linear time transient analysis for Mindlin plates The authors do not guarantee that the codes are errorfree although a major effort was taken to verify all of them The given codes have been tested under MATLAB R2019a therefore users should use this version or greater ones when running these codes Any suggestions or corrections are welcomed by an email to ferreirafeuppt Porto Portugal Antonio J M Ferreira Bologna Italy 2019 Nicholas Fantuzzi viii Preface to the Second Edition Preface to the First Edition This book intend to supply readers with some MATLAB codes for finite element analysis of solids and structures After a short introduction to MATLAB the book illustrates the finite element implementation of some problems by simple scripts and functions The following problems are discussed discrete systems such as springs and bars beams and frames in bending in 2D and 3D plane stress problems plates in bending free vibration of Timoshenko beams and Mindlin plates including laminated composites buckling of Timoshenko beams and Mindlin plates The book does not intend to give a deep insight into the finite element details just the basic equations so that the user can modify the codes The book was prepared for undergraduate science and engineering students although it may be useful for graduate students The MATLAB codes of this book are included in the disk Readers are wel comed to use them freely The author does not guarantee that the codes are errorfree although a major effort was taken to verify all of them Users should use MATLAB 70 or greater when running these codes Any suggestions or corrections are welcomed by an email to ferreirafeuppt Porto Portugal 2008 Antonio Ferreira ix Contents 1 Short Introduction to MATLAB 1 11 Introduction 1 12 Getting Started 2 13 Matrices 3 131 Operating with Matrices 4 132 Statements 5 133 Matrix Functions 5 134 Inverse 6 135 Component Operations 6 136 Colon Notation and Submatrices 7 14 Loops and Repetitive Actions 10 141 Conditionals if and Switch 10 142 Loops For and While 11 143 Relations and Logical Operators 12 144 Logical Indexing 13 15 Library and User Defined Functions 14 151 Standard Library 14 152 Vector Functions 15 153 Matrix Functions 15 154 Scripting and Users Defined Functions 16 155 Debug Mode 19 16 Linear Algebra 19 17 Graphics 20 171 2D Linear Plots 20 172 3D Linear Plots 21 173 3D Surface Plots 23 174 Patch Plots 23 References 25 xi 2 Discrete Systems 27 21 Introduction 27 22 Springs and Bars 27 23 Equilibrium at Nodes 29 24 Some Basic Steps 29 25 First Problem and First MATLAB Code 30 References 36 3 Bars or Trusses 37 31 Introduction 37 32 A Bar Element 38 33 Postcomputation of Stress 43 34 Numerical Integration 43 35 Isoparametric Bar Under Uniform Load 44 36 Fixed Bar with Spring Support 48 37 Bar in Free Vibrations 52 References 56 4 Trusses in 2D Space 57 41 Introduction 57 42 2D Trusses 57 43 Stiffness Matrix 59 44 Mass Matrix 59 45 Postcomputation of Stress 60 46 First 2D Truss Problem 61 47 Second 2D Truss Problem 66 48 2D Truss with Spring 69 49 2D Truss in Free Vibrations 72 Reference 75 5 Trusses in 3D Space 77 51 Introduction 77 52 Basic Formulation 77 53 First 3D Truss Problem 79 54 Second 3D Truss Example 83 55 3D Truss Problem in Free Vibrations 86 Reference 88 6 Bernoulli Beams 89 61 Introduction 89 62 Bernoulli Beam 89 63 Bernoulli Beam Problem 93 64 Bernoulli Beam with Spring 97 xii Contents 65 Bernoulli Beam Free Vibrations 99 66 Stability of Bernoulli Beam 101 References 104 7 Bernoulli 2D Frames 105 71 Introduction 105 72 2D Frame Element 105 73 First 2D Frame Problem 107 74 Second 2D Frame Problem 111 75 2D Frame in Free Vibrations 118 8 Bernoulli 3D Frames 123 81 Introduction 123 82 Matrix Transformation in 3D Space 123 83 Stiffness Matrix and Vector of Equivalent Nodal Forces 126 84 Mass Matrix 127 85 First 3D Frame Problem 128 86 Second 3D Frame Problem 131 87 3D Frame in Free Vibrations 136 9 Grids 141 91 Introduction 141 92 First Grid Problem 143 93 Second Grid Problem 147 10 Timoshenko Beams 151 101 Introduction 151 102 Static Analysis 151 103 Free Vibrations 159 104 Buckling Analysis 165 References 170 11 Plane Stress 171 111 Introduction 171 112 Displacements Strains and Stresses 171 113 Boundary Conditions 173 114 Hamilton Principle 173 115 Finite Element Discretization 174 116 Interpolation of Displacements 174 117 Element Energy 175 1171 Quadrilateral Element Q4 176 1172 Quadrilateral Elements Q8 and Q9 179 118 Postprocessing 181 1181 Stress Extrapolation 182 1182 Interelement Averaging 184 Contents xiii 119 Plate in Traction 184 1110 2D Beam in Bending 197 1111 2D Beam in Free Vibrations 202 Reference 205 12 Kirchhoff Plates 207 121 Introduction 207 122 Mathematical Background 208 123 Finite Element Approximation 209 1231 Interpolation Functions 209 1232 Stiffness Matrix 212 124 Isotropic Square Plate in Bending 215 125 Orthotropic Square Plate in Bending 226 References 227 13 Mindlin Plates 229 131 Introduction 229 132 The Mindlin Plate Theory 229 1321 Displacement Field 229 1322 Strains 230 1323 Stresses 231 1324 Hamiltons Principle 232 133 Finite Element Discretization 233 134 Stress Recovery 235 135 Square Mindlin Plate in Bending 235 136 Free Vibrations of Mindlin Plates 244 137 Stability of Mindlin Plates 253 References 267 14 Laminated Plates 269 141 Introduction 269 142 Displacement Field 269 143 Strains 270 144 Stresses 271 145 Hamiltons Principle 273 146 Finite Element Approximation 275 1461 StrainDisplacement Matrices 276 1462 Stiffness Matrix 277 1463 Load Vector 278 1464 Mass Matrix 278 147 Stress Recovery 278 148 Static Analysis 279 149 Free Vibrations 293 xiv Contents 1410 Buckling Analysis 300 14101 Buckling of Cross and AnglePly Laminates 305 References 310 15 Functionally Graded Structures 313 151 Introduction 313 152 Functionally Graded Materials 313 153 Timoshenko Beam 314 1531 Finite Element Approximation 317 1532 Bending of MicroBeams 318 1533 Free Vibrations of MicroBeams 322 154 Mindlin Plate 325 1541 Bending of MicroPlates 327 1542 Free Vibrations of MicroPlates 331 References 334 16 Time Transient Analysis 335 161 Introduction 335 162 Numerical Time Integration 335 163 Clamped Timoshenko Beam 337 164 SimplySupported Laminated Plate 340 References 345 Index 347 Contents xv Chapter 1 Short Introduction to MATLAB Abstract This chapter introduces MATLAB by presenting programs that investigate elementary mathematical problems The primarily objective is to learn quickly the first steps The emphasis here is learning by doing Therefore the best way to learn is by trying it yourself Working through the examples will give you a feel for the way that MATLAB operates In this introduction we will describe how MATLAB handles simple numerical expressions and mathematical formulas 11 Introduction MATLAB is a commercial software and a trademark of The MathWorks Inc USA The name MATLAB stands for MATrix LABoratory MATLAB was written origi nally to provide easy access to matrix software developed by the LINPACK linear system package and EISPACK Eigen system package projects It is an integrated programming system including graphical interfaces and a large number of special ized toolboxes MATLAB is getting increasingly popular in all fields of science and engineering due to its simple programming very close to linear algebra and powerful and easy to use Integrated Development Environment IDE MATLAB started as an interactive program for doing matrix calculations and has now grown to a high level mathematical language that can solve integrals and differential equations numerically and plot a wide variety of two and three dimen sional graphs In this subject you will mostly use it interactively and also create MATLAB scripts that carry out a sequence of commands MATLAB also contains a programming language that is rather like Pascal It is a highperformance language for technical computing It integrates com putation visualization and programming environment Furthermore MATLAB is a modern programming language environment it has sophisticated data structures contains builtin editing and debugging tools and supports objectoriented program ming These factors make MATLAB an excellent tool for teaching and research MATLAB has many advantages compared to conventional computer languages eg C Fortran for solving technical problems MATLAB is an interactive sys tem whose basic data element is an array that does not require dimensioning The The Editors if applicable and The Authors under exclusive license to Springer Nature Switzerland AG 2020 A J M Ferreira and N Fantuzzi MATLAB Codes for Finite Element Analysis Solid Mechanics and Its Applications 157 httpsdoiorg10100797830304795271 1 2 1 Short Introduction to MATLAB first version of MATLAB was produced in the mid 1970s as a teaching tool The soft ware package has been commercially available since 1984 and is now considered as a standard tool at most universities and industries worldwide A deeper study of MATLAB can be obtained from many MATLAB books 1 2 and the very useful help of MATLAB This chapter introduces MATLAB by presenting programs that investigate ele mentary but interesting mathematical problems The primarily objective is to learn quickly the first steps The emphasis here is learning by doing Therefore the best way to learn is by trying it yourself Working through the examples will give you a feel for the way that MATLAB operates In this introduction we will describe how MATLAB handles simple numerical expressions and mathematical formulas 12 Getting Started When you start MATLAB a special window called the MATLAB desktop appears The desktop is a window that contains other windows The major tools within or accessible from the desktop are Command window Command history Current directory Workspace The command window is a white plain window in which it is possible to edit and run commands in order to see directly the effects of MATLAB syntax The command history collects all the commands that have been inserted The current directory is the current working folder in which the program is working this folder defines the root folder of your project The workspace collects all the variables memory that are introduced Note that while MATLAB is running is filling the workspace which represents the RAM of the machine Simple example of interactive calculation is given just by typing the expression in the command window We want to calculate the expression 5 3 thus we type at the prompt command and obtain immediately 8 You will have noticed that if you do not specify an output variable MATLAB uses a default variable ans short for answer to store the results of the current calculation Note that the variable ans is created or overwritten if it already exists and added to the workspace To avoid this we may assign a value to a variable or output argument name x 5 3 This variable name can always be used to refer to the results of the previous computations Therefore computing 4x results in ans 32 If a complex operation has to be computed ie 5 432 MATLAB works according to the priorities The contents of all parentheses are evaluated first starting from the innermost parentheses and working outward All exponentials are evaluated working from left to right 12 Getting Started 3 All multiplications and divisions are evaluated working from left to right All additions and subtractions are evaluated starting from left to right Thus the earlier calculation was for 5 432 by priority 3 Typing pi the number π 3141592 is shown in the command window If you type PI an error appears due to the keysensitive MATLAB property It is important to pay attention calling variables with capitals or lowercase letters MATLAB has also some built it functions for example typing exp1 the natural exponent appears e 271828 The usage of comments is fundamental while a program is developed Comments are very useful also if you open a program that you made months before and do not remember its structure and purpose Comments in MATLAB are introduced with symbol at the beginning of a line Moreover a double symbol at a beginning of a line defines a section which is highlighted by MATLAB editor and can help during code debugging While the instructions are written can be useful to maintain code line alignments It is helpful because makes the program easy to read and also the searching for any error Every variable can be declared in every part of the program it is not needed to declare all the variables in the initialization part Another very important thing about MATLAB declaration is that each variable might not be declared compulsorily this makes MATLAB very practical and easy to use 13 Matrices Matrices are the fundamental object of MATLAB and are particularly important in this book Matrices can be created in MATLAB in many ways the simplest one obtained by the commands A1 2 34 5 67 8 9 A 1 2 3 4 5 6 7 8 9 Note the semicolon at the end of each matrix line We can also generate matrices by predefined functions such as random matrices rand3 ans 08147 09134 02785 09058 06324 05469 01270 00975 09575 Rectangular matrices can be obtained by specification of the number of rows and columns as in 4 1 Short Introduction to MATLAB rand23 ans 09649 09706 04854 01576 09572 08003 131 Operating with Matrices We can add subtract multiply and transpose matrices For example we can obtain a matrix cab by the following commands arand4 a 02769 06948 04387 01869 00462 03171 03816 04898 00971 09502 07655 04456 08235 00344 07952 06463 brand4 b 07094 06551 09597 07513 07547 01626 03404 02551 02760 01190 05853 05060 06797 04984 02238 06991 cab c 09863 13499 13985 09381 08009 04797 07219 07449 03732 10692 13508 09515 15032 05328 10190 13454 The matrices can be multiplied for example ead as shown in the following example drand41 d 08909 09593 05472 01386 ead e 11792 06220 14787 12914 13 Matrices 5 The transpose of a matrix is given by the apostrophe as arand32 a 01493 02543 02575 08143 08407 02435 a ans 01493 02575 08407 02543 08143 02435 132 Statements Statements are operators functions and variables always producing a matrix which can be used later Some examples of statements a3 a 3 ba3 b 9 eye3 ans 1 0 0 0 1 0 0 0 1 If one wants to cancel the echo of the input a semicolon at the end of the statement suffices It is recalled that MATLAB is casesensitive variables a and A being different objects We can erase variables from the workspace by using clear A given object can be erased such as clear A 133 Matrix Functions Some useful matrix functions are given in Table 11 Some examples of such functions are given in the following commands here we build matrices by blocks 6 1 Short Introduction to MATLAB Table 11 Some useful functions for matrices eye Identity matrix zeros A matrix of zeros ones A matrix of ones diag Creates or extract diagonals rand Random matrix eye3diageye3rand32 ans 10000 0 0 10000 08147 09134 0 10000 0 10000 09058 06324 0 0 10000 10000 01270 00975 Another example of matrices built from blocks Arand3 A 05497 07572 05678 09172 07537 00759 02858 03804 00540 B A zeros32 zeros23 ones2 B 05497 07572 05678 0 0 09172 07537 00759 0 0 02858 03804 00540 0 0 0 0 0 10000 10000 0 0 0 10000 10000 134 Inverse Given a square matrix A the inverse matrix is given by invA The main use of matrices and vectors is in solving sets of linear equations This kind of systems can be implemented in MATLAB using their matrix form Ax b In order to solve these systems the inverse matrix has to be used x A1b MATLAB has an improved algorithm backslash that compute this problem x Ab which works better than x invAb 135 Component Operations Sometimes it is useful to do some operation component by component between vectors or scalars To do so a dot must be added as a prefix of the operator In order to 13 Matrices 7 do a generic exponent n of each component of a vector v11 5 2 the command window code for n 3 should be v13 and the command window output is ans 1 125 8 The dot symbol underlines that the operation in general a product a division and an exponent should be done over each component of the vectormatrix 136 Colon Notation and Submatrices The colon is a shortcut for calling back vectors and matrices components For example typing u 03 it gives u 0 1 2 3 if the step is different from 1 it becomes v 0042 Generally a b c produces a vector of entries starting with the value a incrementing by the value b until it gets to c it will not produce a value beyond c It should be noted that the colon substitutes the for loop see loop section below With the colon it is possible to extract bits of a vectormatrix Considering the following vector v1 126 026 which means v1 1 3 5 0 2 4 6 To get from 2nd to 5th entries v125 ans 3 5 0 2 or to get alternate entries v1127 and get ans 1 5 2 6 In MATLAB it is possible to manipulate matrices in order to make code more compact or more efficient For example using the colon we can generate vectors as in x18 x 1 2 3 4 5 6 7 8 or using increments x120537 x 12000 17000 22000 27000 32000 37000 8 1 Short Introduction to MATLAB This sort of vectorization programming is quite efficient no forend cycles are used This efficiency can be seen in the generation of a table of sines x0pi22pi x 0 15708 31416 47124 62832 bsinx b 0 10000 00000 10000 00000 x b ans 0 0 15708 10000 31416 00000 47124 10000 62832 00000 The colon can also be used to access one or more elements from a matrix where each dimension is given a single index or vector of indices A block is then extracted from the matrix as illustrated next arand34 a 06551 04984 05853 02551 01626 09597 02238 05060 01190 03404 07513 06991 a23 ans 02238 a1223 ans 04984 05853 09597 02238 a1end ans 02551 a1 ans 06551 04984 05853 02551 a3 ans 05853 02238 07513 It is interesting to note that arrays are stored linearly in memory from the first dimension second and so on So we can in fact access vectors by a single index as show below 13 Matrices 9 a1 2 34 5 6 9 8 7 a 1 2 3 4 5 6 9 8 7 a3 ans 9 a7 ans 3 a1 2 3 4 ans 1 4 9 2 a ans 1 4 9 2 5 8 3 6 7 Subscript referencing can also be used in both sides a a 1 2 3 4 5 6 9 8 7 b b 1 2 3 4 5 6 b1a1 b 1 2 3 4 5 6 b1a2 b 4 5 6 4 5 6 b2 b 4 6 4 6 10 1 Short Introduction to MATLAB a30 a 1 2 3 4 5 6 0 0 0 b3120 b 4 6 4 6 20 0 We can insert one element in matrix b and MATLAB automatically resizes the matrix Note that vectormatrix indexing in MATLAB starts from 1 and not from 0 zero as in other programming languages eg Python C etc so if a vector is generated as v16 its size is 6 as the last index used for the creation 14 Loops and Repetitive Actions Every programming language MATLAB included has at least three structures for sequential alternative and repetitive computing These structures are fundamental from basic to advanced programming Conditionals and loops are the most common and wide used structures for basic programming which are shown below 141 Conditionals if and Switch Often a function needs to branch based on runtime conditions MATLAB offers structures for this as in most programming languages Here is an example illustrating most of the features of if x1 if x0 dispBad input elseif maxx 0 y x1 else y xˆ2 end If there are many options it may better to use switch instead For instance switch units case length dispmeters case volume 14 Loops and Repetitive Actions 11 dispcubic meters case time disphours otherwise dispnot interested end 142 Loops For and While Many programs require iteration or repetitive execution of a block of statements Again MATLAB is similar to other languages here This code for calculating the first 10 Fibonacci numbers illustrates the most common type of forend loop f1 2 f 1 2 for i310fifi1fi2end f f 1 2 3 5 8 13 21 34 55 89 It is sometimes necessary to repeat statements based on a condition rather than a fixed number of times This is done with while x10while x 1 x x2end x 5 x 25000 x 12500 x 06250 Other examples of forend loops x for i 14 xxiˆ2 end x 1 x 1 4 x 1 4 9 x 1 4 9 16 and in inverse form 12 1 Short Introduction to MATLAB x for i 411 xxiˆ2 end x 16 x 16 9 x 16 9 4 x 16 9 4 1 Note the initial values of x as empty vectormatrix and the possibility of decreasing cycles 143 Relations and Logical Operators Relations in MATLAB are shown in Table 12 Note the difference between and logical equal The logical operators are given in Table 13 The result if either 0 or 1 as in 353535 ans 1 ans 0 ans 0 The same is obtained for matrices as in a rand5 b triua a b a Table 12 Some relation operators Less than Greater than Less or equal than Greater or equal than Equal to Not equal Table 13 Logical operators and or not 14 Loops and Repetitive Actions 13 01419 06557 07577 07060 08235 04218 00357 07431 00318 06948 09157 08491 03922 02769 03171 07922 09340 06555 00462 09502 09595 06787 01712 00971 00344 b 01419 06557 07577 07060 08235 0 00357 07431 00318 06948 0 0 03922 02769 03171 0 0 0 00462 09502 0 0 0 0 00344 ans 1 1 1 1 1 0 1 1 1 1 0 0 1 1 1 0 0 0 1 1 0 0 0 0 1 144 Logical Indexing Logical indexing arise from logical relations resulting in a logical array with ele ments 0 or 1 a a 1 2 3 4 5 6 0 0 0 a2 ans 0 0 1 1 1 1 0 0 0 Then we can use such array as a mask to modify the original matrix as shown next aans20 a 1 2 20 20 20 20 0 0 0 This will be very useful in finite element calculations particularly when imposing boundary conditions 14 1 Short Introduction to MATLAB 15 Library and User Defined Functions MATLAB has a lot of builtin functions like sinx cosx absx expx etc which can be applied to vectors and matrices In addition users can create their own customized functions for several purposes like avoiding code repetitions and re use the same routines for different computer programs The use and implementation of user defined functions is highly recommended because it helps in reducing code errors and code clarity 151 Standard Library The sine of a value is mathematically written as y sin x y is the sine of the generic number x Since MATLAB has only matrices and vectors the expression above means each component of vector x has a sine y following the relation y sin x so x 0pi65pi y sinx gives y 0 05000 08660 10000 A not complete list of functions are given in Table 14 Another example with matrices is given below arand34 a 04387 07952 04456 07547 03816 01869 06463 02760 07655 04898 07094 06797 bsina b 04248 07140 04310 06851 03724 01858 06022 02725 06929 04704 06514 06286 csqrtb c 06518 08450 06565 08277 06102 04310 07760 05220 08324 06859 08071 07928 15 Library and User Defined Functions 15 Table 14 Scalar functions sin asin exp abs round cos acos log sqrt floor tan atan rem sign ceil Table 15 Vector functions max sum median any min prod mean all 152 Vector Functions Some MATLAB functions operate well basically on vectors only This definition is not general because these functions work also with matrices but need more complex definition A not complete list is illustrated in Table 15 Consider for example vector x110 The sum mean and maximum values are evaluated as x110 x 1 2 3 4 5 6 7 8 9 10 sumx ans 55 meanx ans 55000 maxx ans 10 153 Matrix Functions Some important matrix functions which are used for matrix structured data are listed in Table 16 In some cases such functions may use more than one output argument as in Arand3 A 08147 09134 02785 09058 06324 05469 01270 00975 09575 16 1 Short Introduction to MATLAB Table 16 Matrix functions eig Eigenvalues and eigenvectors chol Cholesky factorization inv Inverse lu LU decomposition qr QR factorization schur Schur decomposition poly Characteristic polinomial det Determinant size Size of a matrix norm 1norm 2norm Fnorm norm cond Conditioning number of 2norm rank Rank of a matrix yeigA y 01879 17527 08399 where we wish to obtain the eigenvalues only or in VDeigA V 06752 07134 05420 07375 06727 02587 00120 01964 07996 D 01879 0 0 0 17527 0 0 0 08399 where we obtain the eigenvectors and the eigenvalues of matrix A 154 Scripting and Users Defined Functions A Mfile is a plain text file with MATLAB comands saved with extension m The Mfiles can be scripts of functions By using the editor of MATLAB we can insert commentsorstatementsandthensaveorcompilethemfileNotethatthepercentsign represents a comment No statement after this sign will be executed Comments are quite useful for documenting the file Mfiles are useful when the number of statements is large or when you want to execute it at a later stage or frequently or even to run it in background 15 Library and User Defined Functions 17 A simple example of a script is given below program 1 programmer Antonio Ferreira date 20080530 purpose show how Mfiles are built data a matrix of numbers b matrix with sines of a arand34 bsina Users can create their own functions Generally it is computationally convenient to divide the whole program in subprograms in which the code has different purposes This makes the code more readable afterwards Functions act like subroutines in FORTRAN where a particular set of tasks is performed For example a rectangle area calculus is shown The input data are the 2 rectangle dimensions a b and the outputs are the area A the perimeter p and the diagonal d The first line we should name the function and give the input parameters ab in parenthesis and the output parameters Apd in square parenthesis function Apd rectab A ab p 2a b d sqrtaˆ2 bˆ2 end The function has been defined with the sintax function Apd rect ab where the input data are ab written in round brackets and the output data in square brackets Apd It is noted that MATLAB does not mix the letters A and a up because it is a keysensitive code This function must be saved with the name rectm and it can be recalled in other mfile or in the command window directly area perim diag rect23 and it gives ans 60000 100000 36056 Another MATLAB function sample is given below 18 1 Short Introduction to MATLAB function abc antoniomnp a hilbm b magicn c eyemp end We then call this function as abcantonio234 producing abcantonio234 a 10000 05000 05000 03333 b 8 1 6 3 5 7 4 9 2 c 1 0 0 0 0 1 0 0 It is possible to use only some output parameters by not typing the last symbols of the function definition abantonio234 a 10000 05000 05000 03333 b 8 1 6 3 5 7 4 9 2 or by removing some of them with symbol as a cantonio234 a 10000 05000 05000 03333 c 1 0 0 0 0 1 0 0 15 Library and User Defined Functions 19 155 Debug Mode Most of the time we work on MATLAB scripts in the MATLAB editor MATLAB itself identifies possible code problems as warning or errors However MATLAB has powerful debugging features that help us checking the code while it is running line by line Of all debugging tools the breakpoints are the most practical ones Each runnable line of a MATLAB script has an hyphen on the left side of the MATLAB editor It is sufficient to press on the hyphen to see a red dot If the MATLAB script is run the code will stop running at the specific line showing a green pointing arrow From now on it is possible to check workspace status variable values and even execute code line by line or continue the script run It is suggested to the reader to check MATLAB documentation for the latest debugging features and get familiar with MATLAB debugger 16 Linear Algebra In our finite element calculations we typically need to solve systems of equations or obtain the eigenvalues of a matrix MATLAB has a large number of functions for linear algebra Only the most relevant for finite element analysis are here presented Consider a linear system axb where arand3 a 08909 01386 08407 09593 01493 02543 05472 02575 08143 brand31 b 02435 09293 03500 The solution vector x can be easily evaluated by using the backslash command xab x 07837 29335 10246 Consider two matrices for example a stiffness matrix and a mass matrix for which we wish to calculate the generalized eigenproblem arand4 a 01966 03517 09172 03804 20 1 Short Introduction to MATLAB 02511 08308 02858 05678 06160 05853 07572 00759 04733 05497 07537 00540 brand4 b 05308 05688 01622 01656 07792 04694 07943 06020 09340 00119 03112 02630 01299 03371 05285 06541 vdeigab v 01886 00955 10000 09100 00180 10000 05159 04044 10000 02492 02340 00394 09522 08833 06731 10000 d 48305 0 0 0 0 06993 0 0 0 0 01822 0 0 0 0 07628 The MATLAB function eig can be applied to the generalized eigenproblem pro ducing matrix v each column containing an eigenvector and matrix d containing the eigenvalues at its diagonal If the matrices are the stiffness and the mass matrices then the eigenvectors will be the modes of vibration and the eigenvalues will be the square roots of the natural frequencies of the system 17 Graphics MATLAB allows to produce graphics in a simple way either 2D or 3D plots Basic implemented graphical functions such as plot plot3 surf and patch are shown in the present section 171 2D Linear Plots Using the command plot we can produce simple 2D plots in a figure using two vectors with x and y coordinates A simple example x 2pipi1002pi y sinx plotxy producing the plot of Fig11 We can insert a title legend modify axes etc as shown in Table 17 By using hold on we can produce several plots in the same figure We can also modify colors of curves or points as in 17 Graphics 21 8 6 4 2 0 2 4 6 8 1 08 06 04 02 0 02 04 06 08 1 Fig 11 Sample of 2D line plot of a sine Table 17 Some graphics commands Title Title xlabel xaxis legend ylabel yaxis legend axisxminxmaxyminymax Sets limits to axis axis auto Automatic limits axis square Same scale for both axis axis equal Same scale for both axis axis off Removes scale axis on Scales again x0pi1002pi y1sinx y2sin2x y3sin4x plotxy1xy2xy3 producing the plot of Fig 12 172 3D Linear Plots As for 2D plots we can produce 3D plots with plot3 using x y and z vectors For example 22 1 Short Introduction to MATLAB 0 1 2 3 4 5 6 7 1 08 06 04 02 0 02 04 06 08 1 Fig 12 Sample of application of line colors and markers 1 05 0 05 1 1 05 0 05 1 0 05 1 15 2 25 x 10 5 Fig 13 Sample of 3D linear plot t010120pi xcost ysint ztˆ3 plot3xyz produces the plot illustrated in Fig13 17 Graphics 23 Fig 14 Sample of 3D surface plot 173 3D Surface Plots A simple example of surface plot is given below xy meshgrid10510120 z sinx2 cosy3 surfxyz and the graphical result is depicted in Fig14 174 Patch Plots For creating mesh filled polygons where the filling represents a generic field such as displacement or stress patch MATLAB command can be used Create a single polygon by specifying the x y coordinates of each vertex Then add two more polygons to the figure Create a red square with vertices at 0 0 1 0 1 1 and 0 1 Specify x as the xcoordinates of the vertices and y as the ycoordinates patch automatically connects the last x y coordinate with the first x y coor dinate x 02 25 34 0 y 01 05 2 15 patchxyred 24 1 Short Introduction to MATLAB 0 1 2 3 4 5 6 7 8 0 1 2 3 4 5 6 7 8 Fig 15 Specifying patch coordinates Create two polygons by specifying x and y as twocolumn matrices Each column defines the coordinates for one of the polygons patch adds the polygons to the current axes without clearing the axes x2 2 5 2 5 7 7 8 8 y2 4 0 8 2 7 3 4 0 patchx2y2green graphical result is given in Fig15 Different polygon color faces can be set and in particular interpolated polygon face colors can be created Create two polygons and use a different color for each polygon vertex Use a colorbar to show how the colors map into the colormap Create the polygons using matrices x and y Interpolate colors across polygon faces by specifying a color at each polygon vertex and use a colorbar to show how the colors map into the colormap Matrix c must be a matrix of the same size as x and y defining one color per vertex and add a colorbar x 2 5 2 5 7 7 8 8 y 4 0 8 2 7 3 4 0 c 0 05 08 03 08 02 05 1 figure patchxyc colorbar result is plot in Fig16 References 25 2 3 4 5 6 7 8 0 1 2 3 4 5 6 7 8 0 01 02 03 04 05 06 07 08 09 1 Fig 16 Different polygon color faces References 1 B Hahn DT Valentine Essential MATLAB for Engineers and Scientists 6th edn Academic Press Cambridge MA USA 2016 2 S Attaway MATLAB A Practical Introduction to Programming and Problem Solving ButterworthHeinemann Oxford UK 2018 Chapter 2 Discrete Systems Abstract In this chapter some basic concepts of the finite element method are illus trated by solving basic discrete systems built from springs and bars Generation of element stiffness matrix and assembly for the global system is performed First basic steps on finite element programs are described 21 Introduction The finite element method is nowadays the most used computational tool in science and engineering applications The finite element method had its origin around 1950 with reference works of Courant 1 Argyris 2 and Clough 3 Many finite element books are available such as the books by Reddy 4 Oñate 5 Zienkiewicz 6 Hughes 7 Hinton 8 just to name a few Some recent books deal with the finite element analysis with MATLAB codes 9 10 The programming approach in these books is quite different from the one presented in this book In this chapter some basic concepts are illustrated by solving discrete systems built from springs and bars 22 Springs and Bars Consider a bar or spring element with 2 nodes 2 degrees of freedom corresponding to 2 axial displacements ue 1 ue 2 where the superscript e refers to a generic finite element as illustrated in Fig21 We suppose an element of length L constant cross section with area A and modulus of elasticity E The element supports axial forces only The deformation in the bar is obtained as ϵ ue 2 ue 1 Le 21 The Editors if applicable and The Authors under exclusive license to Springer Nature Switzerland AG 2020 A J M Ferreira and N Fantuzzi MATLAB Codes for Finite Element Analysis Solid Mechanics and Its Applications 157 httpsdoiorg10100797830304795272 27 28 2 Discrete Systems Fig 21 Spring or bar finite uf 1 e 2 ul element with 2 nodes Ro RS L while the stress in the bar is given by the Hookes law as e e 6 por 0 Be EO 22 The axial resultant force is obtained by integration of stresses on the crosssection area of the bar as 1 y N Ao A Ee EAO At ao 1 23 Taking into account the static equilibrium of the axial forces RO and RY according to Fig21 such nodal forces can be expressed as EA e RY R N uy uy 24 we can write the equations in the form taking k 4 RY 11 fui e 1 fF 1 Ka q Re k 11 ui Ka 25 where K is the stiffness matrix of the bar spring element a is the displacement vector and q represents the vector of nodal forces In case a bar element is con sidered the element might undergo the action of uniformly distributed forces thus it is necessary to transform those forces into nodal forces by 11 fui bly 1 q KZ Ly oD Kg 26 1 1 U5 2 1 with f being the vector of nodal forces equivalent to distributed forces b More details regarding this aspect will be given in the following chapter 23 Equilibrium at Nodes 29 23 Equilibrium at Nodes In Eq 26 we show the equilibrium relation for one element but we also need to obtain the equations of equilibrium for the structure Therefore we need to assemble the contribution of all elements so that a global system of equations can be obtained To do that we recall that in each node the sum of all forces arising from various adjacent elements equals the applied load at that node We then obtain Ne SiR fj 27 e1 where n represents the number of elements in the structure producing a global system of equations in the form Ky Ki2 Kin uy fi Ko Ky Kon uo hr Knit Ki2 ste Kann Un Sh or in a more compact form Kaf 28 Here K represents the system or structure stiffness matrix a is the system displace ment vector and f represents the system force vector 24 Some Basic Steps In any finite element problem some calculation steps are typical e define a set of elements connected at nodes e for each element compute stiffness matrix K and force vector f e assemble the contribution of all elements into the global system Ka f e modify the global system by imposing essential displacements boundary condi tions e solve the global system and obtain the global displacements a e for each element evaluate the strains and stresses postprocessing 30 2 Discrete Systems 25 First Problem and First MATLAB Code To illustrate some of the basic concepts and introduce the first MATLAB code we consider a problem illustrated in Fig 22 where the central bar is defined as rigid Our problem has 3 finite elements and 4 nodes Three nodes are clamped being the boundary conditions defined as uj u3 u4 0 In order to solve this problem we set k 1 for all springs and the external applied load at node 2 to be P 10 We can write for each element in turn the local equilibrium equation Spring 1 RY KO 1 l1 u RS 11 us Spring 2 R KE 1 l u RY 1 1 us Spring 3 R 1 l1 u RY 1 1 us We then consider the compatibility conditions to relate local element and global structure displacements as u u us u u U2 us U3 u U2 us U4 29 By expressing equilibrium of forces at nodes to 4 we can write 3 Node 1 R Fi Ri F 210 e1 i 2 3 1 2 P P10 Op ss a 2 4 ul uz ua O Rigid bar Fig 22 Problem 1 a spring problem 25 First Problem and First MATLAB Code 31 3 Node 2 R P RS R R P 211 e1 3 Node 3 R Fy RS Fy 212 e1 3 Node 4 R Fy RS Fy 213 e1 and then obtain the static global equilibrium equations in the form ky k 0 O uy Fy k ky ky kz ky k ud P 0 k ko 0 U3 F3 214 0 k 0 k3 ug F4 Taking into account the boundary conditions u u3 u4 0 we may write ky k 0 O 0 F ky kj ko k3 ko k ur P 0 k kn O Of 215 0 k3 0 ks 0 F4 At this stage we can compute the reactions F F3 Fy only after the computation of the global displacements We can remove lines and columns of the system corre sponding to uw u3 u4 0 and reduce the global system to one equation ki ko k3u2 P The reactions can then be obtained by kiu2 Fi kou2 F3 k3u2 Fy Note that the stiffness matrix was obtained by summing the contributions of each element at the correct lines and columns corresponding to each element degrees of freedom For instance the degrees of freedom of element are 1 and 2 andthe 2 x 2 stiffness matrix of this element is placed at the corresponding lines and columns of the global stiffness matrix ky k00 k k 00 dd 1 Ky KY 0 0 00 216 0 O 00 32 2 Discrete Systems For element 2 the global degrees of freedom are 2 and 3 and the 2 2 stiffness matrix of this element is placed at the corresponding lines and columns of the global stiffness matrix K2 0 0 0 0 0 k2 k2 0 0 k2 k2 0 0 0 0 0 217 For element 3 the global degrees of freedom are 2 and 4 and the 2 2 stiffness matrix of this element is placed at the corresponding lines and columns of the global stiffness matrix K3 0 0 0 0 0 k3 0 k3 0 0 0 0 0 k3 0 k3 218 A first MATLAB code problem1m is introduced to solve the problem illustrated in Fig22 Many of the concepts used later on more complex elements are already given in this code MATLAB codes for Finite Element Analysis problem1m AJM Ferreira N Fantuzzi 2019 clear memory clear elementNodes connections at elements elementNodes 1 22 32 4 numberElements number of Elements numberElements sizeelementNodes1 numberNodes number of nodes numberNodes 4 for structure displacements displacement vector force force vector stiffness stiffness matrix displacements zerosnumberNodes1 force zerosnumberNodes1 stiffness zerosnumberNodes applied load at node 2 force2 100 25 First Problem and First MATLAB Code 33 computation of the system stiffness matrix for e 1numberElements elementDof element degrees of freedom Dof elementDof elementNodese stiffnesselementDofelementDof stiffnesselementDofelementDof 1 11 1 end boundary conditions and solution prescribed dofs prescribedDof 134 free Dof activeDof activeDof setdiff1numberNodesprescribedDof solution displacementsactiveDof stiffnessactiveDofactiveDofforceactiveDof output displacementsreactions outputDisplacementsReactionsdisplacementsstiffness numberNodesprescribedDof We discuss some of the programming steps The workspace is deleted by clear In matrix elementNodes we define the connections left and right nodes at each element elementNodes 1 22 32 4 In the first line of this matrix we place 1 and 2 corresponding to nodes 1 and 2 and proceed to the other lines in a similar way By using the MATLAB function size that returns the number of lines and columns of a rectangular matrix we can detect the number of elements by inspecting the number of lines of matrix elementNodes numberElements sizeelementNodes1 Note that in this problem the number of nodes is 4 numberNodes 4 In this problem the number of nodes is the same as the number of degrees of freedom which is not the case in many other examples Because the stiffness matrix is the result of an assembly process involving summing of contributions it is important to initialize it It is also a good programming practice in MATLAB to increase the speed of for loops Using MATLAB function zeros we initialize the global displacement vector dis placements the global force vector force and the global stiffness matrix stiffness respectively 34 2 Discrete Systems displacements zerosnumberNodes1 force zerosnumberNodes1 stiffness zerosnumberNodes It is remarked that the initiation of the displacements vector is optional in the present problem because the same vector is carried out from MATLAB computation at the solution section of the code We now place the applied force at the corresponding degree of freedom force2 100 We compute now the stiffness matrix for each element in turn and then assemble it in the global stiffness matrix for e 1numberElements elementDof element degrees of freedom Dof elementDof elementNodese stiffnesselementDofelementDof stiffnesselementDofelementDof 1 11 1 end In the first line of the loop we inspect the degrees of freedom at each element in a vector elementDof For example for element 1 elementDof 12 for element 2 elementDof 2 3 and so on elementDof elementNodese Next we state that the stiffness matrix for each element is constant and then we perform the assembly process by spreading this 2 2 matrix at the corresponding lines and columns defined by elementDof stiffnesselementDofelementDof stiffnesselementDofelementDof 1 11 1 The line stiffnesselementDofelementDof 1 11 1 of the code can be interpreted as stiffness1 21 2 stiffness1 21 2 1 11 1 for element 1 stiffness2 32 3 stiffness2 32 3 1 11 1 for element 2 and stiffness2 42 4 stiffness2 42 4 1 11 1 25 First Problem and First MATLAB Code 35 for element 3 This sort of coding allows a quick and compact assembly This global system of equations cannot be solved at this stage We need to impose essential boundary conditions before solving the system Ka f The lines and columns of the prescribed degrees of freedom as well as the lines of the force vector will be eliminated at this stage First we define vector prescribedDof corresponding to the prescribed degrees of freedom Then we define a vector containing all activeDof degrees of freedom by setting up the difference between all degrees of freedom and the prescribed ones The MATLAB function setdiff allows this operation prescribed dofs prescribedDof 134 free Dof activeDof activeDof setdiff1numberNodesprescribedDof Note that the solution is performed with the active lines and columns only by using a mask displacementsactiveDof stiffnessactiveDofactiveDofforceactiveDof We then call function outputDisplacementsReactionsm to output displacements and reactions as function outputDisplacementsReactions displacementsstiffnessGDofprescribedDof output of displacements and reactions in tabular form GDof total number of degrees of freedom of the problem displacements dispDisplacements jj 1GDof format jj displacements reactions F stiffnessdisplacements reactions FprescribedDof dispreactions prescribedDof reactions end Reactions are computed by evaluating the total force vector as f Ka Because we only need reactions forces at prescribed degrees of freedom we then use reactions F stiffnessdisplacements reactions FprescribedDof 36 2 Discrete Systems When running this code we obtain detailed information on matrices or results depending on the user needs for example displacements and reactions Displacements ans 10000 0 20000 33333 30000 0 40000 0 reactions ans 10000 33333 30000 33333 40000 33333 References 1 R Courant Variational methods for the solution of problems of equilibrium and vibration Bull Am Math Soc 49 123 1943 2 JH Argyris Matrix displacement analysis of anisotropic shells by triangular elements J Roy Aero Soc 69 801805 1965 3 RW Clough The finite element method in plane stress analysis inProceedings of 2nd ASCE Conference in Electronic Computation Pittsburgh PA 1960 4 JN Reddy An Introduction to the Finite Element Method McGrawHill International Editions New York 1993 5 E Onate Calculo de estruturas por el metodo de elementos finitos CIMNE Barcelona 1995 6 OC Zienkiewicz The Finite Element Method McGrawHill 1991 7 TJR Hughes The Finite Element MethodLinear Static and Dynamic Finite Element Analysis Dover Publications New York 2000 8 E Hinton Numerical Methods and Software for Dynamic Analysis of Plates and Shells Piner idge Press 1988 9 W Young Kwon and Hyochoong Bang Finite Element Method Using MATLAB CRC Press Inc Boca Raton FL USA 1996 10 PI Kattan MATLAB Guide to Finite Elements An Interactive Approach 2nd edn Springer 2007 Chapter 3 Bars or Trusses Abstract In this chapter we analyze axially loaded structural elements termed bars or trusses A truss is connected to other elements only through pins which are con nections that do not constrain rotations Trusses are modeled as discrete elements or springs because only axial force traction or compression and elongation is evalu ated In the present chapter an isoparametric finite element formulation is considered for the bartruss problem 31 Introduction In this chapter we analyze axially loaded structural elements termed bars or trusses A truss is connected to other elements only through pins which are connections that do not constrain rotations Trusses are modeled as discrete elements or springs because only axial force traction or compression and elongation is evaluated In general a finite element is formulated in a reference or parent domain thus a coordinate transformation is accomplished that regards both geometry and dependent variables Interpolation functions are used for both transformations known also as mapping According to the degree of approximation of both geometry and dependent variables finite element formulation are classified as 1 Superparametric the approximation used for the geometry is higher order than that used for the dependent variable 2 Isoparametric equal degree of approximation is used for both geometry and dependent variables 3 Subparametric the approximation used for the geometry is lower order than that used for the dependent variable In the present chapter an isoparametric finite element formulation is considered for the bartruss problem The Editors if applicable and The Authors under exclusive license to Springer Nature Switzerland AG 2020 A J M Ferreira and N Fantuzzi MATLAB Codes for Finite Element Analysis Solid Mechanics and Its Applications 157 httpsdoiorg10100797830304795273 37 38 3 Bars or Trusses 32 A Bar Element Consider the twonode bar finite element shown in Fig31 with constant cross section area A and length L 2a The bar element can undergo only axial stresses ox which are uniform in every crosssection The equilibrium of the bar can be expressed according to the Hamiltons Principle 1 as to dK 5U dW dt 0 31 t where 6 represents the variation K is the kinetic energy U the internal strain energy and W the external work due to applied loads It is recalled that JT U W represents the total potential energy of the system The kinetic energy can be expressed as 1 du pA Ou K dV d 32 5 ha of where V indicates the volume of the bar and p its density By evaluating the variation of the kinetic energy and by integrating by parts the following expression is carried out a 2 u 0K pA Diz oe dx 33 The internal work done or strain energy stored by the bar element is u avesfe d 34 5 one a Oxx AX Straindisplacement relation is Ou 35 Ox By assuming a linear elastic behaviour of the bar material we can write Fig 31 A bar element in its U local coordinate system dx px 1 2 OO xra xra L2a 32 A Bar Element 39 0 o Ee ES 36 Ox where E is the modulus of elasticity vSf ou 4 37 2 J ax the variation of the internal energy is derived as Ou Oou dU EA d 38 Ox Ox 38 If we consider p as the applied forces by unit length the virtual external work at each element is bW p ou dx 39 Finally the equilibrium of the bar is given by Pu Ou Oou 4 pa Sroudrrea Nar f pouds0 310 which is called also weak or variational form of the bar problem Lets consider now a twonoded finite element as illustrated in Fig 32 The axial displacements can be interpolated as u Ni uy Nou2 311 where the shape functions are defined as 1 1 Nig 39 N2 7 9 312 in the natural coordinate system 11 The interpolation 311 can be defined in matrix form as Fig 32 A twonode bar é1 é1 element ea ea 1 2 fa OO OU L2 L2a 40 3 Bars or Trusses uy e u N No Nu 313 The element strain energy can be carried out in the natural system after coordinate transformation x ag dx ad as dN dN 6U utes dx uv 314 q dx dx d ddg ld by recalling that anddx adE as y g that T Gedx ade mde ads su saute EN Na dew 004 f NTNUEw G15 ou ra dE uw bu u 1 a dé d a 1 dN where N and dg dU du Ku 316 The element stiffness matrix K is given by EA Ke NNdé 317 a J1 The integral is evaluated in the natural system by considering the stretch formulation x a which is equivalent to a geometric transformation mapping Classically the Jacobian matrix is introduced for such transformation which in the present problem is J a For 1D problems the procedure can be done without introduc ing formally the Jacobian However many books and references use formally this notation while evaluating the integrals for finite element analysis 1 In this element the derivatives of the shape functions are dN 1 dN 1 7 318 dé 2 d 2 In this case the stiffness matrix can be given in explicit form as 1 t EA 2 EA1 1 K t 1 dé 319 af eae a ft G19 2 By using L 2a we obtain the same stiffness matrix as in the direct method presented in the previous chapter The virtual work done by the external forces is defined as 32 A Bar Element 41 Fig 33 Bar discretized into 1 2 3 4 5 Nodes 4 elements oooo0 tews L L L L a 1 1 OWe p du dx p ou adé surTa f pN dé 320 a 1 l1 or OW ou f 321 where the vector of nodal forces that are equivalent to distributed forces is given only if the distributed forces are uniformly distributed by raaf nrag f t 8 ae ap 322 4 P of frefs The bar mass matrix is derived from the variation of the kinetic energy as a 1 M pa NNdx pa NNa dé 323 a 1 by including the shape function vector definition the socalled consistent mass matrix takes the form 1 e PA 1 pA 21 m feel é 1éad a 12 324 It is possible to avoid the integration for the mass matrix by considering the lumped mass matrix as eo 10 M pAa E 4 325 which is the bar total mass divided by 2 as 2 node element has been considered For a system of bars the contribution of each element must be assembled For example in the bar of Fig 33 we consider 5 nodes and 4 elements In this case the structure vector of displacements is given by uw uy Un U3 Ug us 326 42 3 Bars or Trusses Summing the contribution of all elements we obtain the strain energy the energy done by the external forces and the kinetic energy as 4 6U du Ku du Ku 327 e1 4 6W du f ou 328 e1 4 5K 6u Mii du Mii 329 e1 where K M and f are the structure stiffness matrix mass matrix and the force vector respectively The stiffness matrix is then assembled as 1 1000 00 000 110 0 0 EA 1 1000 0 1 100 EA 1 210 0 K 0 000001 1 007j0 1 2 1 0 0 0000 Jo0 0 00 lo 012 1 0 0000 00 000 0 0 01 1 Ns element 1 element 2 330 whereas the vector of equivalent forces is given by 1 2 fap 2 331 2 1 similar assembly procedure follows for the mass matrix also We then obtain a global system of equations as Mu Kuf 332 to be solved after the imposition of the boundary conditions as explained before The algebraic problem 332 can be used to consider the static problem when M 0 or the free vibrations problem when f 0 or timehistory analysis via Newmarks method In order to carry out the free vibration problem the solution has to be sought in the form u de thus the final algebraic problem becomes 32 A Bar Element 43 K wM i0 333 where i represent the eigenvector and w the eigenvalue In the present text only static and free vibration problem is considered in the following 33 Postcomputation of Stress The stress in the generic element is defined by Eq 36 By including the finite element approximation and using the coordinate transformation it leads dN E dN E oy Ee Ewu u u5 u 334 dx a dé 2a where uw and uz are the nodal displacements of the generic element Note that using linear interpolation the stress at the element is constant 34 Numerical Integration The integrals arising from the variational formulation can be solved by numeri cal integration for example by Gauss quadrature In this section we present the Gauss method for the solution of one dimensional integrals We consider a function fx x 1 1 In the Gauss method the integral 1 I St xdx 335 1 is replaced by a sum of p Gauss points in which the function at those points is multiplied by some weights as in 1 P I fxdx 0 fi Wi 336 1 il where W is the ith point weight In Table 31 the coordinates and weights of the Gauss technique are presented This technique is exact whenever fx is a polynomial of degree p by employing 5p 1 integration points 2 When p 1 is odd the nearest larger integer should be selected For quadratic functions fx p 2 two integration points represent exact integration over the element For linear p 1 or constant p 0 functions fx one integration point is exact such integration is called reduced integration with respect to the twopoints integration 44 3 Bars or Trusses Table 31 Coordinates and weights for Gauss integration n xi Wi 1 00 20 2 05773502692 10 3 07745966692 00 05555555556 08888888889 4 08611363116 03399810436 03478548451 06521451549 Fig 34 One dimensional Gauss quadrature for two and one integration points 1 2 1 2 ξ ξ ξ1 1 3 ξ2 1 3 ξ1 0 Inthepresentcaselinearinterpolationfunctionsareconsideredthustheintegrand of Ke 319 is constant requiring only onepoint quadrature whereas the integrand of Me 324 is quadratic requiring twopoints quadrature The integral of fe 322 is evaluated exactly by only onepoint integration because the integrand function is linear In Fig34 point location for Gauss integration are illustrated 35 Isoparametric Bar Under Uniform Load MATLAB code problem2m solves the bar problem illustrated in Fig35 in which the modulus of elasticity is E 30 106 and the area of the crosssection is A 1 The bar is subjected to a uniform constant axial load p 50 Isoparametric element means that the unknown field u in this case is approximated with the same shape functions used for approximating the geometry of the finite element This concept will be more clear later when twodimensional structures are presented Fig 35 Clamped bar subjected to distributed load p problem2m 1 2 3 4 1 2 3 30 30 30 p 50 35 Isoparametric Bar Under Uniform Load 45 The present problem has exact solution in terms of displacements and stresses as pLx x pLl x u1 o 337 2EA L A2 L The code problem2m solves the present problem Bc ee ee ee ee ee eee ee ee eee eee eee eee MATLAB codes for Finite Element Analysis problem2m AJM Ferreira N Fantuzzi 2019 SS clear memory clear close all E modulus of elasticity A area of cross section L length of bar E 30e6 A 1 EA EA L 90 p 50 generation of coordinates and connectivities numberElements number of elements numberElements 3 generation equal spaced coordinates nodeCoordinates linspace0LnumberElements1 xx nodeCoordinates numberNodes number of nodes numberNodes sizenodeCoordinates 2 elementNodes connections at elements ii 1numberElements elementNodes1 ii elementNodes2 ii1 for structure z displacements displacement vector force force vector stiffness stiffness matrix displacements zeros numberNodes 1 force zeros numberNodes 1 stiffness zeros numberNodesnumberNodes computation of the system stiffness matrix and force vector for e 1numberElements elementDof element degrees of freedom Dof elementDof elementNodese nn lengthelementDof lengthelement nodeCoordinateselementDof2 nodeCoordinates elementDof 1 detJacobian lengthelement2 invJacobian 1detJacobian central Gauss point xi0 weight W2 46 3 Bars or Trusses shapenaturalDerivatives shapeFunctionL200 Xderivatives naturalDerivativesinvJacobian B matrix B zeros1nn B1nn Xderivatives stiffnesselementDofelementDof stiffnesselementDofelementDof BB2detJacobianEA forceelementDof1 forceelementDof1 2shapepdetJacobian end prescribed dofs prescribedDof findxxminnodeCoordinates xxmaxnodeCoordinates free Dof activeDof activeDof setdiff1numberNodesprescribedDof solution GDof numberNodes displacements solutionGDofprescribedDofstiffnessforce output displacementsreactions outputDisplacementsReactionsdisplacementsstiffness numberNodesprescribedDof stresses at elements sigma zerosnumberElements1 for e 1numberElements elementDof element degrees of freedom Dof elementDof elementNodese nn lengthelementDof lengthelement nodeCoordinateselementDof2 nodeCoordinateselementDof1 sigmae Elengthelement1 1displacementselementDof end drawing nodal displacements figure axes1 axes hold on box on displacements figure plotaxes1nodeCoordinatesdisplacements okmarkersize8linewidth15 figure axes2 axes hold on box on stresses figure graphical representation with interpolation for each element interpNodes 10 for e 1numberElements nodeA elementNodese1 nodeB elementNodese2 XX linspacenodeCoordinatesnodeAnodeCoordinatesnodeB interpNodes ll nodeCoordinatesnodeBnodeCoordinatesnodeA dimensionless coordinate xi XX nodeCoordinatesnodeA2ll 1 linear shape function phi1 051 xi phi2 051 xi displacement at the element u phi1displacementsnodeA phi2displacementsnodeB plotaxes1XXuklinewidth15 35 Isoparametric Bar Under Uniform Load 47 plotaxes1XXpLXX2EA1 XXLblinewidth15 stress at the element sigma Ell ones1interpNodes displacementsnodeB displacementsnodeA plotaxes2XXsigmaklinewidth15 plotaxes2XXpLA05 XXLblinewidth15 end setaxes1fontsize18 setaxes2fontsize18 xlimaxes10 L xlimaxes20 L The nodal coordinates are obtained by an equalspaced division of the domain using linspace generation of coordinates and connectivities numberElements number of elements numberElements 3 generation equal spaced coordinates nodeCoordinates linspace0LnumberElements1 The connectivities are obtained by a vectorized cycle elementNodes connections at elements ii 1numberElements elementNodes1 ii elementNodes2 ii1 The evaluation of the stiffness matrix involves the integral 319 We use a Gauss quadrature with one central point ξ 0 and weight 2 see Table31 thus the inte gration of the stiffness matrix described by constant functions is computed exactly So the stiffness matrix and its global assembly becomes stiffnesselementDofelementDof stiffnesselementDofelementDof BB2detJacobianEA where B is a matrix with the derivatives of the shape functions B matrix B zeros1nn B1nn Xderivatives The shape functions and their derivatives with respect to natural coordinates are computed in function shapeFunctionL2m function shapenaturalDerivatives shapeFunctionL2xi shape function and derivatives for L2 elements shape Shape functions naturalDerivatives derivatives wrt xi xi natural coordinates 1 1 shape 1xi1xi2 48 3 Bars or Trusses naturalDerivatives 112 end end function shapeFunctionL2 The function solutionm will be used in the remaining of the book This function computes the displacements of any FE system in the forthcoming problems function displacementssolutionGDofprescribedDofstiffnessforce function to find solution in terms of global displacements GDof number of degree of freedom prescribedDof bounded boundary dofs stiffness stiffness matrix force force vector activeDof setdiff1GDof prescribedDof U stiffnessactiveDofactiveDofforceactiveDof displacements zerosGDof1 displacementsactiveDof U The postcomputation is performed by following Eq334 using the nodal dis placements and analytical derivatives of shape functions sigmae Elengthelement1 1displacementselementDof matrix of derivatives of shape functions B matrix can be used instead without changing the result Representation of results in terms of displacements and stresses is given in Fig36 where it is clear that the displacement numerical solution is exact in the nodes and approximated by linear interpolation in the elements Numerical stress is constant in the elements but it should be linear A better solution in terms of displacements and stresses can be achieved by increasing the number of finite elements or by increasing the order of approximation eg using quadratic shape functions 36 Fixed Bar with Spring Support Another problem involving bars and springs is illustrated in Fig37 The MATLAB code for this problem is problem3m using direct stiffness method In other words the stiffness matrix of the element is computed exactly according to Eq319 not using Gauss quadrature 36 Fixed Bar with Spring Support 49 0 20 40 60 80 0 05 1 15 2 103 0 20 40 60 80 3000 2000 1000 0 1000 2000 3000 Fig 36 Deformed shape top and stress plot bottom of a fixed bar under constant load compar ison between numerical solid line and exact dash line solutions 1 2 3 4 1 2 3 2 m 2 m 8 kN E 70000 MPa A 200 mm2 k 2000 Nmm k Fig 37 Illustration of problem 3 problem3m 50 3 Bars or Trusses MATLAB codes for Finite Element Analysis problem3m ref D Logan A first course in the finite element method third Edition page 121 exercise P310 direct stiffness method AJM Ferreira N Fantuzzi 2019 clear memory clear E modulus of elasticity A area of cross section L length of bar k spring stiffness E 70000 A 200 k 2000 generation of coordinates and connectivities numberElements number of elements numberElements 3 numberNodes 4 elementNodes 1 2 2 3 3 4 nodeCoordinates 0 2000 4000 4000 xx nodeCoordinates for structure displacements displacement vector force force vector stiffness stiffness matrix displacements zerosnumberNodes1 force zerosnumberNodes1 stiffness zerosnumberNodesnumberNodes applied load at node 2 force2 8000 computation of the system stiffness matrix ea zeros1numberElements for e 1numberElements elementDof element degrees of freedom Dof elementDof elementNodese L nodeCoordinateselementDof2 nodeCoordinateselementDof1 if e 3 eae EAL else eae k end stiffnesselementDofelementDof stiffnesselementDofelementDof eae1 11 1 end boundary conditions and solution 36 Fixed Bar with Spring Support 51 prescribed dofs prescribedDof 14 free Dof activeDof activeDof setdiff1numberNodesprescribedDof solution displacements solutionnumberNodesprescribedDofstiffnessforce output displacementsreactions outputDisplacementsReactionsdisplacementsstiffness numberNodesprescribedDof The isoparametric version for the problem illustrated in Fig37 is given in prob lem3am Thus Gauss quadrature is used in this code for computing the stiffness matrix of the element MATLAB codes for Finite Element Analysis problem3am ref D Logan A first course in the finite element method third Edition page 121 exercise P310 with isoparametric formulation AJM Ferreira N Fantuzzi 2019 clear memory clear E modulus of elasticity A area of cross section L length of bar E 70000 A 200 EA EA k 2000 generation of coordinates and connectivities numberElements 3 numberNodes 4 elementNodes 1 2 2 3 3 4 nodeCoordinates 0 2000 4000 4000 xx nodeCoordinates for structure displacements displacement vector force force vector stiffness stiffness matrix displacements zerosnumberNodes1 force zerosnumberNodes1 stiffness zerosnumberNodesnumberNodes applied load at node 2 force2 80000 computation of the system stiffness matrix 52 3 Bars or Trusses ea zeros1numberElements for e 1numberElements elementDof element degrees of freedom Dof elementDof elementNodese if e 3 bar elements nn lengthelementDof lengthelement nodeCoordinateselementDof2 nodeCoordinateselementDof1 detJacobian lengthelement2 invJacobian 1detJacobian central Gauss point xi0 weight W2 shapenaturalDerivatives shapeFunctionL200 Xderivatives naturalDerivativesinvJacobian B matrix B zeros1nn B1nn Xderivatives eae EA stiffnesselementDofelementDof stiffnesselementDofelementDof BB2detJacobianeae else spring element stiffnesselementDofelementDof stiffnesselementDofelementDof k1 11 1 end end boundary conditions and solution prescribedDof 14 solution displacements solutionnumberNodesprescribedDofstiffnessforce output displacementsreactions outputDisplacementsReactionsdisplacementsstiffness numberNodesprescribedDof Both codes give the same solution and matches the analytical solution presented in Logan 3 The displacements at nodes 2 and 3 are 0935mm and 0727mm respectively The reactions at the supports 1 and 4 are 6546kN and 1455kN respectively 37 Bar in Free Vibrations The following problem involves the free vibration problem of the structure given in Fig38 The MATLAB code for this problem is problem3vibm using isoparametric elements and four methods for computing the mass matrix consistent lumped full and reduced integration 37 Bar in Free Vibrations 53 1 2 3 4 1 2 3 2 m 2 m E 70000 MPa A 200 mm2 k EA4000 ρ 1000 tonmm3 k Fig 38 Illustration of problem 3 vibrations problem3vibm MATLAB codes for Finite Element Analysis problem3vibm ref JN Reddy An introduction to the Finite Element Method third Edition page 86 example 254 AJM Ferreira N Fantuzzi 2019 clear memory clear E modulus of elasticity A area of cross section L length of bar rho density E 70000 A 200 EA EA k EA4000 rho 1000 generation of coordinates and connectivities numberElements 3 numberNodes 4 elementNodes 1 2 2 3 3 4 nodeCoordinates 0 2000 4000 4000 xx nodeCoordinates for structure displacements displacement vector force force vector stiffness stiffness matrix mass mass matrix displacements zerosnumberNodes1 force zerosnumberNodes1 stiffness zerosnumberNodesnumberNodes mass zerosnumberNodesnumberNodes computation of the system stiffness matrix ea zeros1numberElements for e 1numberElements elementDof element degrees of freedom Dof elementDof elementNodese if e 3 bar elements 54 3 Bars or Trusses nn lengthelementDof lengthelement nodeCoordinateselementDof2 nodeCoordinateselementDof1 detJacobian lengthelement2 invJacobian 1detJacobian stiffness matrix Central Gauss point xi0 weight W2 shapenaturalDerivatives shapeFunctionL200 Xderivatives naturalDerivativesinvJacobian B matrix B zeros1nn B1nn Xderivatives eae EA stiffnesselementDofelementDof stiffnesselementDofelementDof BB2detJacobianeae mass matrix exact integration masselementDofelementDof masselementDofelementDof 2 11 2detJacobianrhoA3 lumped mass matrix masselementDofelementDof masselementDofelementDof 1 00 1detJacobianrhoA Gauss quadrature calculation twopoints integration coincides with exact gaussLocations 0577350269189626 0577350269189626 gaussWeights ones21 onepoint integration reduced integration gaussLocations 00 gaussWeights 2 for q 1sizegaussWeights1 shape shapeFunctionL2gaussLocationsq masselementDofelementDof masselementDofelementDof shapeshapegaussWeightsqdetJacobianrhoA end else spring element stiffnesselementDofelementDof stiffnesselementDofelementDof k1 11 1 end end boundary conditions and solution prescribedDof 14 free vibration problem modeseigenvalues eigenvaluenumberNodesprescribedDof stiffnessmass0 omega sqrteigenvaluessqrtrhoE4000 37 Bar in Free Vibrations 55 The structure of the code given follows the one of problem3m where the static solution is substituted by eigenvalue solver function eigenvalue function modeseigenvalues eigenvalueGDofprescribedDof stiffnessmassmaxEigenvalues function to find solution in terms of global displacements GDof number of degree of freedom prescribedDof bounded boundary dofs stiffness stiffness matrix mass mass matrix maxEigenvalues maximum eigenvalues to be computed If 0 all the eigenvalues are requested suggested for beam structures activeDof setdiff1GDof prescribedDof if maxEigenvalues 0 VD eigstiffnessactiveDofactiveDof massactiveDofactiveDof else VD eigsstiffnessactiveDofactiveDof massactiveDofactiveDofmaxEigenvaluessmallestabs end eigenvalues diagD modes zerosGDoflengtheigenvalues modesactiveDof V end the generalized eigenvalue problem is solved with the help of the MATLAB function eig and eigenfrequencies and eigenmodes are collected in the two vectors eigenval ues and modes Note that the eigenvalues are the frequencies squared according to Eq333 The function provided is able to calculate all the eigenvalues and eigen vectors of the problem by setting maxEigenvalues0 or a certain number of eigenvalues by defining a number for the aforementioned variable This feature will be useful for twodimensional problems where eigenvalue problems might be large The exact solution provided by Reddy 2 is ω1 202874 ω2 491318 where ω ωLρE Four implementations of the mass matrix of the element are given By default the full Gauss quadrature formula is applied with 2 points because linear shape functions have to be computed The reader can comment and uncomment the lines needed to carried out the results listed in Table32 Consistent refers to the exact integration of the mass matrix in Eq324 and lumped to the lumped mass matrix in Eq324 Full and reduced refer to the two points and onepoint Gauss integration rule respectively Since full integration is exact the same result is achieved as consistent mass matrix case whereas reduced integration and lumped mass matrix are not because they come from different math ematical procedures 56 3 Bars or Trusses Table 32 First two vibration frequencies of the bar in problem3vibm ω Exact 2 Consistent Lumped Full Gauss Reduced Gauss 1 202875 211896 200000 211896 218518 2 491318 605416 400000 605416 1035495 The error on the first frequency is small with respect to the exact one On the contrary the errors are larger for all computations for the second frequency due to reduced number of finite elements used By increasing the number of finite elements accuracy improves References 1 JN Reddy Energy Principles and Variational Methods in Applied Mechanics 3rd edn Wiley Hoboken 2017 2 JN Reddy An Introduction to the Finite Element Method 3rd edn McGrawHill International Editions New York 2005 3 DL Logan A First Course in the Finite Element Method BrooksCole Pacific Grove 2002 Chapter 4 Trusses in 2D Space Abstract This chapter deals with the static and free vibration analyses of two dimen sional trusses which are basically bars oriented in two dimensional Cartesian sys tems A transformation of coordinate basis is necessary to translate the local element matrices into the structural coordinate system Trusses support compressive and ten sile forces only as in bars All forces are applied at the nodes After the presentation of the element formulation some examples are solved by MATLAB codes 41 Introduction This chapter deals with the static and free vibration analyses of two dimensional trusses which are basically bars oriented in two dimensional Cartesian systems A transformation of coordinate basis is necessary to translate the local element matrices stiffness matrix mass matrix and force vector into the structural global coordinate system Trusses support compressive and tensile forces only as in bars All forces are applied at the nodes After the presentation of the element formulation some examples are solved by MATLAB codes 42 2D Trusses In Fig 41 we consider a typical 2D truss in global x y plane The local system of coordinates x y defines the local displacements u 1 u 2 The element possesses 2 degrees of freedom in the local setting uT u 1 u 2 41 while in the global coordinate system the element is defined by 4 degrees of freedom uT u1 u2 u3 u4 42 The Editors if applicable and The Authors under exclusive license to Springer Nature Switzerland AG 2020 A J M Ferreira and N Fantuzzi MATLAB Codes for Finite Element Analysis Solid Mechanics and Its Applications 157 httpsdoiorg10100797830304795274 57 58 4 Trusses in 2D Space UA os 2 2 y Fig 41 2D truss element local and global degrees of freedom The relation between both local and global displacements is given by u u cos uz sind 43 uy u3C0SO 4 u4 sind 44 where is the angle between local axis x and global axis x or in matrix form as u Lu 45 being matrix L defined as pats The J m elements of matrix LZ can be defined by the nodal coordinates as lcosé m sing 2 47 Le Le being L the length of the element Le V x2 x1 92 1 48 43 Stiffness Matrix 59 43 Stiffness Matrix In the local coordinate system the stiffness matrix of the 2D truss element is given by the bar stiffness as before EAT 1 1 K Le I 49 In the local coordinate system the strain energy of this element is given by e 1 Tyr US 5 Ku 410 Replacing u Lu in 410 we obtain 1 Ue su ILKLu 411 It is now possible to express the global stiffness matrix as KLKL 412 or P lm P lm EA Im m lm m K L lm 2 Im 413 lm m Im m 44 Mass Matrix The mass matrix has to be also converted in the global Cartesian system The con sistent Mc and lumped M mass matrices for the truss in local basis are pa 2 1 Mc G be I 414 pA 1 0 M GZ be lo 4 415 respectively Starting from the kinetic energy the definition of the mass matrix in the global reference system can be derived as 60 4 Trusses in 2D Space e 1 IT NA gy 1 T Tra K 5u Mu 5u L MLu 416 It is now possible to express the global mass matrix as ML7ML 417 for the consistent mass matrix 2 2am P lm pa 2im 2m Im m M 6 P Im 27 2m 418 lm m2 2lm 2m and for the lumped mass matrix 2 Im O 0 pa Im m 0O 0 M be 0 0 2 Im 419 0 0 Im m In contrast to stiffness translational masses never vanish thus all translational masses must be retained in the local mass matrix In other words by setting Jm 0 and 1 otherwise in the definitions 418 and 419 as 2 0 1 0 pA 10 2 0 1 M er el1 0 2 0 420 0 1 0 2 and for the lumped mass matrix 1 0 0 0 pA 10 1 0 0 0 0 0 1 45 Postcomputation of Stress In the local coordinate system the stresses are defined as o Fe Taking into account the definition of strain in the bar we obtain us Uj E ui E y E I1 1 I1 1 422 nape io te 422 45 Postcomputation of Stress 61 1 2 4 3 10 000 1 3 2 120 120 x y E 30 106 A 2 Fig 42 First 2D truss problem problem4m By transformation of local to global coordinates we obtain stresses as function of the displacements as σx E Le 1 1Lu E Le l m l mu 423 46 First 2D Truss Problem In a first 2D truss problem illustrated in Fig42 we consider a downward point force 10000 applied at node 1 The modulus of elasticity is E 30 106 and all elements are supposed to have constant crosssection area A 2 The supports are located in nodes 2 3 and 4 Structure degrees of freedom are shown in Fig 43 The code problem4m listing is as MATLAB codes for Finite Element Analysis problem4m AJM Ferreira N Fantuzzi 2019 clear memory clear 62 4 Trusses in 2D Space E modulus of elasticity A area of cross section E 30e6 A 2 EA EA generation of coordinates and connectivities numberElements 3 numberNodes 4 elementNodes 1 21 31 4 nodeCoordinates 0 00 120120 120120 0 xx nodeCoordinates1 yy nodeCoordinates2 for structure displacements displacement vector force force vector stiffness stiffness matrix GDof 2numberNodes GDof total number of degrees of freedom displacements zerosGDof1 force zerosGDof1 applied load at node 2 force2 100000 computation of the system stiffness matrix stiffness formStiffness2DtrussGDofnumberElements elementNodesnumberNodesnodeCoordinatesxxyyEA boundary conditions and solution prescribedDof 38 solution displacements solutionGDofprescribedDofstiffnessforce drawing displacements us 122numberNodes1 vs 222numberNodes figure L xx2xx1 XX displacementsus YY displacementsvs dispNorm maxsqrtXXˆ2YYˆ2 scaleFact 15000dispNorm hold on drawingMeshnodeCoordinatesscaleFactXX YYelementNodes L2k drawingMeshnodeCoordinateselementNodesL2k axis equal setgcafontsize18 stresses at elements stresses2DtrussnumberElementselementNodes xxyydisplacementsE output displacementsreactions outputDisplacementsReactionsdisplacementsstiffness GDofprescribedDof 46 First 2D Truss Problem 63 1 2 3 4 6 5 8 7 x y Fig 43 First 2D truss problem degrees of freedom Note that this code calls some new functions The first function formStiffness2 Dtrussm computes the stiffness matrix of the 2D truss twonode element function stiffness formStiffness2DtrussGDofnumberElements elementNodesnumberNodesnodeCoordinatesxxyyEA stiffnesszerosGDof computation of the system stiffness matrix for e 1numberElements elementDof element degrees of freedom Dof indice elementNodese elementDof indice121 indice12 indice221 indice22 xa xxindice2xxindice1 ya yyindice2yyindice1 lengthelement sqrtxaxayaya C xalengthelement S yalengthelement k1 EAlengthelement CC CS CC CS CS SS CS SS CC CS CC CSCS SS CS SS stiffnesselementDofelementDof 64 4 Trusses in 2D Space stiffnesselementDofelementDofk1 end end The function stresses2Dtrussm computes stresses of the 2D truss elements Both functions used the expressions shown in the beginning of this chapter function stresses2DtrussnumberElementselementNodes xxyydisplacementsE stresses at elements for e 1numberElements indice elementNodese elementDof indice121 indice12 indice221 indice22 xa xxindice2xxindice1 ya yyindice2yyindice1 lengthelement sqrtxaxayaya C xalengthelement S yalengthelement sigmae Elengthelement C S C SdisplacementselementDof end dispstresses sigma end The code problem4m is therefore easier to read by using functions that can also be used for other 2D truss problems Displacements reactions and stresses are in full agreement with analytical results by Logan 1 Displacements ans 10000 00041 20000 00159 30000 0 40000 0 50000 0 60000 0 70000 0 80000 0 reactions 46 First 2D Truss Problem 65 ans 10e03 00030 0 00040 79289 00050 20711 00060 20711 00070 20711 00080 0 stresses ans 10e03 39645 14645 10355 The deformation of the structure is illustrated in Fig44 We use a drawing routine drawingMesh for the purpose This routine needs the input of nodal coordinates and elements connectivities and draws either undeformed and deformed meshes Moreover element type is required for the correct representation as well as type of line needed for the plot which can be given according to MATLAB plot function Fig 44 Deformed shape of 2D truss 0 20 40 60 80 100 120 0 20 40 60 80 100 120 66 4 Trusses in 2D Space 47 Second 2D Truss Problem The next problem is illustrated in Fig45 The degrees of freedom are illustrated in Fig46 The MATLAB code is problem5m The analytical solution of this problem is presented in 1 The results of this code agree well with the analytical solution although the analytical solution considered only half of the structure 1 2 3 5 4 6 50 kN 100 kN 50 kN x y 2 1 6 11 10 4 9 3 8 5 7 3 m 3 m 3 m E 70000 MPa A 300 mm2 Fig 45 A second truss problem problem5m 1 2 5 6 9 10 3 4 7 8 11 12 x y Fig 46 A second truss problem degrees of freedom 47 Second 2D Truss Problem 67 MATLAB codes for Finite Element Analysis problem5m AJM Ferreira N Fantuzzi 2019 clear memory clear E modulus of elasticity A area of cross section E 70000 A300 EAEA generation of coordinates and connectivities elementNodes 1 21 32 32 41 43 43 64 54 63 55 6 nodeCoordinates 0 00 30003000 03000 30006000 06000 3000 numberElements sizeelementNodes1 numberNodes sizenodeCoordinates1 xx nodeCoordinates1 yy nodeCoordinates2 for structure displacements displacement vector force force vector stiffness stiffness matrix GDof 2numberNodes U zerosGDof1 force zerosGDof1 applied load at node 2 force4 50000 force8 100000 force12 50000 computation of the system stiffness matrix stiffness formStiffness2DtrussGDofnumberElements elementNodesnumberNodesnodeCoordinatesxxyyEA boundary conditions and solution prescribedDof 1 2 10 solution displacementssolutionGDofprescribedDofstiffnessforce us 122numberNodes1 vs 222numberNodes drawing displacements figure L xx2xx1 XX displacementsus YY displacementsvs dispNorm maxsqrtXXˆ2YYˆ2 scaleFact 2dispNorm hold on drawingMeshnodeCoordinatesscaleFactXX YY elementNodesL2k drawingMeshnodeCoordinateselementNodesL2k 68 4 Trusses in 2D Space axis equal setgcafontsize18 output displacementsreactions outputDisplacementsReactionsdisplacementsstiffness GDofprescribedDof stresses at elements stresses2DtrussnumberElementselementNodes xxyydisplacementsE Results are the following Displacements ans 10000 0 20000 0 30000 71429 40000 90386 50000 52471 60000 162965 70000 52471 80000 200881 90000 104942 100000 0 110000 33513 120000 90386 reactions ans 10e05 00000 00000 00000 10000 00001 10000 stresses ans 2109015 1224318 625575 47 Second 2D Truss Problem 69 Fig 47 Deformed shape problem 5 0 1000 2000 3000 4000 5000 6000 1000 0 1000 2000 3000 442349 1731447 884697 625575 1731447 442349 1224318 2109015 The deformed shape of this problem is shown in Fig47 48 2D Truss with Spring In Fig 48 we consider a structure that is built from two truss elements and one spring element For the truss elements the modulus of elasticity is E 210000 MPa and the crosssection area is A 500 mm2 This problem is modeled with four points and three elements Figure49 illustrates the degrees of freedom according to our finite element discretization The listing of the code problem6m is presented MATLAB codes for Finite Element Analysis problem6m ref D Logan A first course in the finite element method third Edition mixing trusses with springs AJM Ferreira N Fantuzzi 2019 70 4 Trusses in 2D Space clear memory clear E modulus of elasticity A area of cross section E 210000 A 500 EA EA generation of coordinates and connectivities nodeCoordinates 0 05000cospi4 5000sinpi4 10000 0 elementNodes 1 21 31 4 numberElements sizeelementNodes1 numberNodes sizenodeCoordinates11 spring added xx nodeCoordinates1 yy nodeCoordinates2 for structure displacements displacement vector force force vector stiffness stiffness matrix GDof 2numberNodes U zerosGDof1 force zerosGDof1 stiffness zerosGDof applied load at node 2 force2 25000 computation of the system stiffness matrix stiffness formStiffness2DtrussGDofnumberElements1 elementNodesnumberNodesnodeCoordinatesxxyyEA spring stiffness in global Dof stiffness2 72 7 stiffness2 72 7 20001 11 1 boundary conditions and solution prescribedDof 38 solution displacements solutionGDofprescribedDofstiffnessforce output displacementsreactions outputDisplacementsReactionsdisplacementsstiffness GDofprescribedDof stresses at elements stresses2DtrussnumberElements1elementNodes xxyydisplacementsE The functions for forming the stiffness matrix formStiffness2Dtruss and computing the stresses in each truss stresses2Dtruss are used with a reduced number of elements numberElements1 due to the presence of the spring 48 2D Truss with Spring 71 25 kN 2 4 3 1 1 3 2 k 10 m 5 cos 45m E 210000 MPa A 500 mm2 k 2000 Nmm x y 45 Fig 48 Mixing 2D truss elements with spring elements problem6m Fig 49 Mixing 2D truss elements with spring elements degrees of freedom 2 3 4 7 5 6 1 x y In fact the spring stiffness is added to global degrees of freedom 2 and 7 corre sponding to vertical displacements at nodes 1 and 4 after the assembly of the truss structure Displacements reactions and stresses are listed below Displacements are exactly the same as the analytical solution 1 Stresses in bars show that bar 1 is under tension and bar 2 is under compression 72 4 Trusses in 2D Space Displacements ans 10000 17241 20000 34483 30000 0 40000 0 50000 0 60000 0 70000 0 80000 0 reactions ans 10e04 00003 18103 00004 18103 00005 18103 00006 0 00007 06897 00008 0 stresses ans 512043 362069 49 2D Truss in Free Vibrations The structure in Fig45 without the applied loads is now studied in free vibrations considering a constant density for each member of ρ 1000 tonmm3 The listing of the code problem5vibm follows 49 2D Truss in Free Vibrations 73 MATLAB codes for Finite Element Analysis problem5vibm AJM Ferreira N Fantuzzi 2019 clear memory clear E modulus of elasticity A area of cross section E 70000 A 300 EA EA rho 1000 rhoA rhoA generation of coordinates and connectivities elementNodes 1 21 32 32 41 43 43 64 54 63 55 6 nodeCoordinates 0 00 30003000 03000 30006000 06000 3000 numberElements sizeelementNodes1 numberNodes sizenodeCoordinates1 xx nodeCoordinates1 yy nodeCoordinates2 for structure displacements displacement vector force force vector stiffness stiffness matrix GDof 2numberNodes U zerosGDof1 computation of the system stiffness matrix stiffness formStiffness2DtrussGDofnumberElements elementNodesnumberNodesnodeCoordinatesxxyyEA computation of the system stiffness matrix mass formMass2DtrussGDofnumberElements elementNodesnumberNodesnodeCoordinatesxxyyrhoA boundary conditions and solution prescribedDof 1 2 10 free vibration problem modeseigenvalues eigenvalueGDofprescribedDof stiffnessmass0 us 122numberNodes1 vs 222numberNodes modeNumber 1 drawing displacements figure L xx2xx1 XX modesusmodeNumber YY modesvsmodeNumber dispNorm maxsqrtXXˆ2YYˆ2 scaleFact 1e12dispNorm hold on drawingMeshnodeCoordinatesscaleFactXX YY 74 4 Trusses in 2D Space elementNodesL2k drawingMeshnodeCoordinateselementNodesL2k axis equal setgcafontsize18 omega sqrteigenvalues with respect to the correspondent static problem a function for the assembly of the mass matrix is given function mass formMass2DtrussGDofnumberElements elementNodesnumberNodesnodeCoordinatesxxyyrhoA masszerosGDof computation of the system stiffness matrix for e 1numberElements elementDof element degrees of freedom Dof indice elementNodese elementDof indice121 indice12 indice221 indice22 xa xxindice2xxindice1 ya yyindice2yyindice1 lengthelement sqrtxaxayaya consistent mass matrix k1 rhoAlengthelement6 2 0 1 0 0 2 0 1 1 0 2 0 0 1 0 2 lumped mass matrix k1 rhoAlengthelement2eye4 masselementDofelementDof masselementDofelementDofk1 end end where the selection of consistent and lumped mass matrices can be selected respec tively by commenting and uncommenting the correspondent lines of the code The eigenvalue function is used to obtain eigenfrequencies and eigenmodes The drawingMesh used in the static simulation is used here to plot one mode shape at a time according to the variable modeNumber The first mode shape of the structure is given in Fig410 The frequencies are compared to the ones obtained by the same problem studied with a commercial FE software and listed in Table 41 Reference 75 0 1000 2000 3000 4000 5000 6000 500 0 500 1000 1500 2000 2500 3000 3500 Fig 410 First mode shape problem 5 vibrations Table 41 First three natural frequencies ω of the structure in problem5vib ω 104 Ref Lumped Consistent 1 79193 79193 82926 2 116238 116238 130233 3 178650 178650 227597 Reference 1 DL Logan A First Course in the Finite Element Method BrooksCole 2002 Chapter 5 Trusses in 3D Space oe Abstract The present chapter generalizes the 2D truss model of the previous chapter as trusses in 3D Cartesian space Static and free vibration problems are solved trans forming the local stiffness into global 3D quantities Some simple problems are solved in MATLAB and verified with reference codes 51 Introduction The present chapter generalizes the 2D truss model of the previous chapter as trusses in 3D Cartesian space Static and free vibration problems are solved transforming the local stiffness mass matrices and load vector into global 3D quantities Some simple problems are solved in MATLAB and verified with reference codes 52 Basic Formulation We consider now trusses in 3D space A typical twonoded 3D truss element is illustrated in Fig 51 Each node has three global degrees of freedom The displacement vector in local coordinates does not change with respect to the one in the previous chapter 51 On the contrary the displacements in global coordinates projected from node and node 2 are uw u ur uz ug Us U6 51 The relationship between local and global coordinates is due to the direction cosines matrix 45 as lL l10 00 L7 4 2 I 00h 1 32 The Editors if applicable and The Authors under exclusive 77 license to Springer Nature Switzerland AG 2020 A J M Ferreira and N Fantuzzi MATLAB Codes for Finite Element Analysis Solid Mechanics and Its Applications 157 httpsdoiorg10100797830304795275 78 5 Trusses in 3D Space u1 u2 u3 u1 u2 u3 u4 u5 u6 u4 u5 u6 x y z y x z Fig 51 Trusses in 3D coordinates local and global coordinate sets where the cosines are obtained as lx x2 x1 Le ly y2 y1 Le lz z2 z1 Le The stiffness matrix in global coordinates is given by K LT KeL E A Le l2 x lxly lxlz l2 x lxly lxlz l2 y lylz lxly l2 y lylz l2 z lxlz lylz l2 z l2 x lxly lxlz l2 y lylz sym l2 z 53 Analogous transformation is performed for the global mass matrix The consistent mass matrix becomes M LT MeL ρA 6 Le 2 0 0 1 0 0 2 0 0 1 0 2 0 0 1 2 0 0 2 0 sym 2 54 52 Basic Formulation 79 The lumped mass matrix is M LT MeL ρA 2 Le 1 0 0 0 0 0 1 0 0 0 0 1 0 0 0 1 0 0 1 0 sym 1 55 We then perform a standard assembly procedure to obtain the stiffness and mass matrices and the global vector of equivalent nodal forces for the complete system as we did for 2D trusses 53 First 3D Truss Problem We consider the 3D truss problem illustrated in Fig52 The MATLAB code problem7m is used to evaluate displacements reactions and forces at elements The Cartesian coordinates of the nodes P1 P2 P3 P4 are listed in Fig52 Bound ary conditions are indicated by vector Ui ui vi wi for i 1 2 3 4 in Fig52 In this problem the displacement at node 1 along y v1 0 is fixed as well as all displacements of nodes 2 3 and 4 The area of the members are indicated by Ai for i 1 2 3 MATLAB codes for Finite Element Analysis problem7m ref D Logan A first course in the finite element method third Edition A 3D truss example AJM Ferreira N Fantuzzi 2019 clear memory clear E modulus of elasticity A area of cross section E 12e6 A 030207290187 area for various sections generation of coordinates and connectivities nodeCoordinates 72 0 0 0 36 0 0 36 72 0 0 48 elementNodes 1 21 31 4 numberElements sizeelementNodes1 numberNodes sizenodeCoordinates1 xx nodeCoordinates1 yy nodeCoordinates2 80 5 Trusses in 3D Space 1 000 x y z 4 3 2 1 E 12 106 P1 72 0 0 P2 0 36 0 P3 0 36 72 P4 0 0 48 U2 U3 U4 0 0 0 v1 0 A1 0302 A2 0729 A3 0187 1 2 3 Fig 52 A 3D truss problem geometry mesh loads and boundary nodes problem7m for structure displacements displacement vector force force vector stiffness stiffness matrix GDof global number of degrees of freedom GDof 3numberNodes U zerosGDof1 force zerosGDof1 applied load at node 2 force3 1000 stiffness matrix stiffness formStiffness3DtrussGDofnumberElements elementNodesnumberNodesnodeCoordinatesEA boundary conditions and solution prescribedDof 2 412 solution displacements solutionGDofprescribedDofstiffnessforce output displacementsreactions outputDisplacementsReactionsdisplacementsstiffness GDofprescribedDof 53 First 3D Truss Problem 81 stresses at elements stresses3DtrussnumberElementselementNodesnodeCoordinates displacementsE The code is supported by formStiffness3Dtrussm for the assembly and generation of the stiffness matrix in the global coordinates and illustrated above function stiffness formStiffness3DtrussGDofnumberElements elementNodesnumberNodesnodeCoordinatesEA stiffnesszerosGDof computation of the system stiffness matrix for e 1numberElements elementDof element degrees of freedom Dof indice elementNodese elementDof 3indice12 3indice11 3indice1 3indice22 3indice21 3indice2 x1 nodeCoordinatesindice11 y1 nodeCoordinatesindice12 z1 nodeCoordinatesindice13 x2 nodeCoordinatesindice21 y2 nodeCoordinatesindice22 z2 nodeCoordinatesindice23 L sqrtx2x1x2x1 y2y1y2y1 z2z1z2z1 CXx x2x1L CYx y2y1L CZx z2z1L T CXxCXx CXxCYx CXxCZx CYxCXx CYxCYx CYxCZx CZxCXx CZxCYx CZxCZx stiffnesselementDofelementDof stiffnesselementDofelementDof EAeLT T T T end end The stress calculation for each member in stresses3Dtruss function stresses3DtrussnumberElementselementNodes nodeCoordinatesdisplacementsE stresses in 3D truss elements fprintfStresses in elements format for e 1numberElements elementDof element degrees of freedom Dof indice elementNodese elementDof 3indice12 3indice11 3indice1 3indice22 3indice21 3indice2 x1 nodeCoordinatesindice11 y1 nodeCoordinatesindice12 82 5 Trusses in 3D Space z1 nodeCoordinatesindice13 x2 nodeCoordinatesindice21 y2 nodeCoordinatesindice22 z2 nodeCoordinatesindice23 L sqrtx2x1x2x1 y2y1y2y1 z2z1z2z1 CXx x2x1L CYx y2y1L CZx z2z1L u displacementselementDof memberstresse ELCXx CYx CZx CXx CYx CZxu fprintf3d 128f e memberstresse end The results are in excellent agreement with analytical solution in 1 Displacements ans 10000 00711 20000 0 30000 02662 40000 0 50000 0 60000 0 70000 0 80000 0 90000 0 100000 0 110000 0 120000 0 reactions ans 20000 2231632 40000 2561226 50000 1280613 60000 0 70000 7024491 80000 3512245 90000 7024491 100000 4463264 110000 0 120000 2975509 53 First 3D Truss Problem 83 Stresses in elements 1 94819142387 2 144536842298 3 286854330060 54 Second 3D Truss Example In Fig53 a second example of a 3D truss is illustrated MATLAB codes for Finite Element Analysis problem8m ref D Logan A first course in the finite element method third Edition A second 3D truss example AJM Ferreira N Fantuzzi 2019 clear memory clear E modulus of elasticity A area of cross section E 210000 A 100 100 100 100 area for various sections 14 4 3 58 1 1 44 0 3 20 4 0 30 4 6 4 2 3 1 10 000 x y z E 210000 MPa A 100 mm2 U2 U3 U4 U5 0 0 0 Fig 53 Second 3D problem problem8m 84 5 Trusses in 3D Space generation of coordinates and connectivities nodeCoordinates 4000 4000 3000 0 4000 0 0 4000 6000 4000 0 3000 8000 1000 1000 elementNodes 1 21 31 41 5 numberElements sizeelementNodes1 numberNodes sizenodeCoordinates1 xx nodeCoordinates1 yy nodeCoordinates2 for structure displacements displacement vector force force vector stiffness stiffness matrix GDof global number of degrees of freedom GDof 3numberNodes U zerosGDof1 force zerosGDof1 applied load at node 2 force2 10000 stiffness matrix stiffness formStiffness3DtrussGDofnumberElements elementNodesnumberNodesnodeCoordinatesEA boundary conditions and solution prescribedDof 415 solution displacements solutionGDofprescribedDofstiffnessforce output displacementsreactions outputDisplacementsReactionsdisplacementsstiffness GDofprescribedDof stresses at elements stresses3DtrussnumberElementselementNodesnodeCoordinates displacementsE The results are in excellent agreement with analytical solution in 1 Displacements ans 10000 03024 20000 15177 30000 02688 40000 0 50000 0 60000 0 54 Second 3D Truss Example 85 70000 0 80000 0 90000 0 100000 0 110000 0 120000 0 130000 0 140000 0 150000 0 reactions ans 10e03 00040 02709 00050 0 00060 02032 00070 13546 00080 0 00090 10160 00100 0 00110 79681 00120 0 00130 16255 00140 20319 00150 08128 Stresses in elements 1 338652236 2 1693261180 3 7968086584 4 2726097914 Code problem8m call functions formStiffness3Dtrussm for stiffness computa tion and function stresses3Dtrussm for computation of stresses at 3D trusses introduced for problem7m 86 5 Trusses in 3D Space 55 3D Truss Problem in Free Vibrations We consider the 3D truss geometry presented in Sect53 for introducing the free vibration problem of 3D trusses The density of all members have been considered as unitary ρ 1 The problem is listed in problem7vibm MATLAB codes for Finite Element Analysis problem7vibm A 3D truss example in free vibrations AJM Ferreira N Fantuzzi 2019 clear memory clear E modulus of elasticity A area of cross section E 12e6 A 030207290187 area for various sections rho 111 density for various sections generation of coordinates and connectivities nodeCoordinates 72 0 0 0 36 0 0 36 72 0 0 48 elementNodes 1 21 31 4 numberElements sizeelementNodes1 numberNodes sizenodeCoordinates1 xx nodeCoordinates1 yy nodeCoordinates2 for structure displacements displacement vector force force vector stiffness stiffness matrix GDof global number of degrees of freedom GDof 3numberNodes U zerosGDof1 force zerosGDof1 stiffness matrix stiffness formStiffness3DtrussGDofnumberElements elementNodesnumberNodesnodeCoordinatesEA mass matrix mass formMass3DtrussGDofnumberElements elementNodesnumberNodesnodeCoordinatesrhoA boundary conditions and solution prescribedDof 2 412 free vibration problem modeseigenvalues eigenvalueGDofprescribedDof stiffnessmass0 55 3D Truss Problem in Free Vibrations 87 us 133numberNodes2 vs 233numberNodes1 ws 333numberNodes modeNumber 1 drawing displacements figure L xx2xx1 XX modesusmodeNumber YY modesvsmodeNumber ZZ modeswsmodeNumber dispNorm maxsqrtXXˆ2YYˆ2ZZˆ2 scaleFact 1e3dispNorm hold on drawingMeshnodeCoordinatesscaleFactXX YY ZZ elementNodesL3k drawingMeshnodeCoordinateselementNodesL3k axis equal setgcafontsize18 view4545 omega sqrteigenvalues The consistent 54 and lumped 55 mass matrices are carried out in Code formMass3Dtrussm function stiffness formMass3DtrussGDofnumberElements elementNodesnumberNodesnodeCoordinatesrhoA stiffnesszerosGDof computation of the system stiffness matrix for e 1numberElements elementDof element degrees of freedom Dof indice elementNodese elementDof 3indice12 3indice11 3indice1 3indice22 3indice21 3indice2 x1 nodeCoordinatesindice11 y1 nodeCoordinatesindice12 z1 nodeCoordinatesindice13 x2 nodeCoordinatesindice21 y2 nodeCoordinatesindice22 z2 nodeCoordinatesindice23 L sqrtx2x1x2x1 y2y1y2y1 z2z1z2z1 CXx x2x1L CYx y2y1L CZx z2z1L T CXxCXx CXxCYx CXxCZx CYxCXx CYxCYx CYxCZx CZxCXx CZxCYx CZxCZx consistent mass matrix Mass 162 0 0 1 0 0 0 2 0 0 1 0 0 0 2 0 0 1 1 0 0 2 0 0 0 1 0 0 2 0 0 0 1 0 0 2 lumped mass matrix Mass 12eye6 88 5 Trusses in 3D Space Table 51 Vibration frequencies for 3D truss ω Consistent Lumped Ref 1 92104 75203 752028 2 158813 129670 1296705 stiffnesselementDofelementDof stiffnesselementDofelementDof rhoeAeLMass end end The reader can easily switch from consistent to lumped mass matrix by comment ing and uncommenting related code lines Results of the present analysis are listed in Table51 where consistent and lumped mass matrix are used Since the structure has only two free motions of node 1 two frequencies are carried out The reference solution has been carried out with a commercial finite element code with same number of finite elements Excellent agreement is shown and clearly the selected commercial code considers only lumped mass matrix for truss elements Reference 1 DL Logan A First Course in the Finite Element Method BrooksCole Pacific Grove 2002 Chapter 6 Bernoulli Beams Abstract Bernoulli theory is a classical beam theory where the transverse shear deformation is neglected and the deflection of the beam indicated by w is the only degree of freedom of the model and the inplane rotation si given by the derivative of the transverse deflection with respect to the beam axis In this chapter we perform the static vibration and buckling analysis of Bernoulli beams in bending configuration Results will be compared to analytical and reference results from the literature 61 Introduction Bernoulli theory is a classical beam theory where the transverse shear deformation is neglected and the deflection of the beam indicated by w is the only degree of freedom of the model and the inplane rotation si given by the derivative of the transverse deflection with respect to the beam axis The classical solution of the present problem considers an approximation using the wellknown Hermite interpolation functions which are able to give exact nodal solution in the generic finite element such formu lation is known as superconvergent element In this chapter we perform the static vibration and buckling analysis of Bernoulli beams in bending configuration Results will be compared to analytical and reference results from the literature 62 Bernoulli Beam The beam is defined in the x z plane with constant crosssection area A Fig61 The Bernoulli beam theory assumes that undeformed plane sections remain plane under deformation The axial displacement u at a distance z of the beam middle axis is given by The Editors if applicable and The Authors under exclusive license to Springer Nature Switzerland AG 2020 A J M Ferreira and N Fantuzzi MATLAB Codes for Finite Element Analysis Solid Mechanics and Its Applications 157 httpsdoiorg10100797830304795276 89 90 6 Bernoulli Beams Zz Ww Pp LETTE r a v2 a Fig 61 Bernoulli beam element with 2 nodes ow zZ7 61 Ox 61 where w is the transverse displacement Thus the movement of the beam is totally described by the vertical displacement Strains are defined as ou aw du Ow x 70 p we t 0 62 Ox 9x2 Yar 97 ax 62 The elastic strain deformation is obtained as 1 1 4 U oyexdV EezdV 63 2 Jy 2 Jy Taking dV dAdx and integrating upon the area A we obtain 1 aw U 2 7 El Qx2 dx 64 where J is the second moment of area of the beam crosssection The kinetic energy is obtained as K5 oi pijav5 ply ob paw dx 65 2Jy 2Ja ax where dot identifies time derivative and the first term indicates the rotary inertia and the second one is the vertical bulk inertia of the beams cross section For thin beams rotary inertia can be neglected as it is done in the following for the sake of simplicity Interested readers can easily include the rotary inertia contribution in the mass matrix following 1 The external work for this element by considering the transverse pressure p and the axial load N that accounts for nonlinear Von Karman strain is given by 62 Bernoulli Beam 91 4 a dw 06 ow p dw ax noe or dx 66 a a Ox Ox With kinetic strain energies and external work Hamiltons Principle can be formu lated At each node we consider 2 degrees of freedom w and gw the transverse dis placement and rotation of the crosssection 0 0 wt E OM yy 627 Ox Ox The transverse displacement is interpolated by Hermite shape functions 1 as w Néw 68 being the shape functions defined as 1 3 M1 7236 8 69 a 2 3 No 7 610 1 N3 y2 38 611 a 2 3 NaS FClL 8 8 612 where xa identifies the dimensionless axial coordinate These functions known as Hermite approximation functions can be carried out from the elastic solution of a cantilever beam by enforcing alternatively unitary displacements and rotations at the boundaries The strain energy is obtained as 1 aw 1 pl El wy u5 EI ax5 2 92 ade 2 Jaa ax 2J at ag 1 El f swt NNdéw 613 2 a 1 nN dN where N de The element stiffness matrix is then obtained as 3 3a 3 3a e Ely nT wo El 3a 4a 3a 2a K N Nod 73 33g 3 3a 614 3a 2a 3a 4a 92 6 Bernoulli Beams The kinetic energy takes the form 1 2 1 2 1 eT T e K 5 pAwdx 5 pAwadé aw pAaN Ndéw 615 a 1 1 The mass matrix is clearly identified by 156 44a 54 26a 1 2 2 e T pAa 44a 16a 26a 12a M pAaN Nde 9 54 26a 156 44a 616 26a 12a 44a 16a The work done by the distributed forces is defined as a 1 1 éWe p dw dx p dw adé wa pNdé 617 a l 1 The vector of nodal forces equivalent to distributed uniform p forces is obtained as 3 1 fe ap Nodg 4 618 1 3 3 a The work done by the axial force N is defined as dw ddw N dw ddw éWs N d d 2 dx ox ee ag 145 N 1 r dw N Ndéw 619 a J1 The stability matrix is defined as 36 6a 36 6a el flare 1 3L 16a 6a 4a G 7N Ndé oa 36 6a 36 6a 620 6a 4a 6a 16a After assembly the algebraic solving system is Mii Ku NGu f 621 62 Bernoulli Beam 93 The expression 621 allows to solve the static free vibration and buckling problem of Bernoulli beam Note that the buckling and free vibration problem are both solved in the form of an eigenvalue problem Thus the codes of the two problems will be very similar interchanging the mass matrix with the stability matrix 63 Bernoulli Beam Problem In Figs62 and 63 we consider a simplysupported and clamped Bernoulli beam in bending under uniform load For the sake of simplicity unitary values of the stiffness E I 1 beam length L 1 and applied load p 1 are considered Thus in terms of central displacements the exact solution for the simply supported beam is δexact 5384 00130208333 and for the clamped case δexact 1384 00026041667 Code problem9m solves these problems for both boundary conditions The user can define the number of ele L 1 p 1 z w x u EI 1 exact 5 384 pL4 EI Fig 62 Simplysupported Bernoulli problem under uniform load problem9m L 1 p 1 z w x u EI 1 exact 1 384 pL4 EI Fig 63 Clamped Bernoulli problem under uniform load problem9m 94 6 Bernoulli Beams Fig 64 Deformed shape for simplysupported and clamped beams 0 02 04 06 08 1 0014 0012 001 0008 0006 0004 0002 0 0 02 04 06 08 1 3 25 2 15 1 05 0 103 ments however the present formulation is exact in the nodes and approximated in the elements through Hermite polynomials Thus a minimum number of 2 finite element is required in order to obtain exact solution at beam central section The maximum transverse displacement wmax for simplysupported beam and clamped beam match the analytical solution as illustrated in Fig64 with 2 finite elements Interpolation is performed using 50 points in order to have a good display of the deformed shape see drawInterpolatedBeamm code given however in case sev eral finite elements are set a reduction of interpolation points is suggested in order to increase code speed while displaying final deformed shape 63 Bernoulli Beam Problem 95 MATLAB codes for Finite Element Analysis problem9m AJM Ferreira N Fantuzzi 2019 clear memory clear E modulus of elasticity I second moment of area L length of bar E 1 I 1 EIEI generation of coordinates and connectivities numberElements 2 nodeCoordinates linspace01numberElements1 L maxnodeCoordinates numberNodes sizenodeCoordinates1 xx nodeCoordinates1 elementNodes zerosnumberElements2 for i 1numberElements elementNodesi1i elementNodesi2i1 end distributed load P 1 for structure displacements displacement vector force force vector stiffness stiffness matrix GDof global number of degrees of freedom GDof 2numberNodes stiffess matrix and force vector stiffnessforce formStiffnessBernoulliBeamGDofnumberElements elementNodesnumberNodesxxEIP boundary conditions and solution clampedclamped fixedNodeU 1 2numberElements1 fixedNodeV 2 2numberElements2 simply supportedsimply supported fixedNodeU 1 2numberElements1 fixedNodeV clamped at x0 fixedNodeU 1 fixedNodeV 2 prescribedDof fixedNodeUfixedNodeV solution displacements solutionGDofprescribedDofstiffnessforce output displacementsreactions outputDisplacementsReactionsdisplacementsstiffness 96 6 Bernoulli Beams GDofprescribedDof reordering displacements and rotations W displacements122numberNodes R displacements222numberNodes drawing nodal displacements figure plotnodeCoordinatesWokmarkersize8linewidth15 setgcafontsize18 graphical representation with interpolation for each element drawInterpolatedBeam This code calls function formStiffnessBernoulliBeamm for the computation of the stiffness matrix and the force vector for the Bernoulli beam 2node element Note that the stiffness matrix is computed exactly without applying Gauss quadrature The solutions given match the exact solutions because the Hermite approximation polynomials are derived by following the solution of the elastic line of the Bernoulli beam 1 function stiffnessforce formStiffnessBernoulliBeamGDofnumberElements elementNodesnumberNodesxxEIP force zerosGDof1 stiffness zerosGDof calculation of the system stiffness matrix and force vector for e 1numberElements elementDof element degrees of freedom Dof indice elementNodese elementDof 2indice111 2indice21 2indice211 2indice212 length of element LElem xxindice2xxindice1 k1 EILElemˆ312 6LElem 12 6LElem 6LElem 4LElemˆ2 6LElem 2LElemˆ2 12 6LElem 12 6LElem 6LElem 2LElemˆ2 6LElem 4LElemˆ2 f1 PLElem2 PLElemLElem12 PLElem2 PLElemLElem12 equivalent force vector forceelementDof forceelementDof f1 stiffness matrix stiffnesselementDofelementDof stiffnesselementDofelementDofk1 end end 64 Bernoulli Beam with Spring 97 p 103 L 10 t L103 k 10 E 106 Fig 65 Bernoulli beam with spring under uniform load problem9am 64 Bernoulli Beam with Spring Figure65 illustrates a beam in bending clamped at one end and supported by a spring at the other end The beam width is considered as unitary Code problem9am illustrates the use of MATLAB for solving this problem MATLAB codes for Finite Element Analysis problem9am AJM Ferreira N Fantuzzi 2019 clear memory clear E modulus of elasticity I second moment of area L length of bar E 1e6 L10 tL1000 I1tˆ312 EIEI generation of coordinates and connectivities numberElements 3 nodeCoordinates linspace0LnumberElements1 L maxnodeCoordinates numberNodes sizenodeCoordinates1 xx nodeCoordinates1 elementNodes zerosnumberElements2 for i 1numberElements elementNodesi1 i elementNodesi2 i1 end 98 6 Bernoulli Beams distributed force P 1000 for structure displacements displacement vector force force vector stiffness stiffness matrix GDof global number of degrees of freedom GDof 2numberNodes stiffnessSpring zerosGDof1 forceSpring zerosGDof11 stiffess matrix and force vector stiffnessforce formStiffnessBernoulliBeamGDofnumberElements elementNodesnumberNodesxxEIP spring added stiffnessSpring1GDof1GDof stiffness forceSpring1GDof force k 10 stiffnessSpringGDof1 GDof1GDof1 GDof1 stiffnessSpringGDof1 GDof1GDof1 GDof1 k kk k boundary conditions and solution fixedNodeU 1 fixedNodeV 2 prescribedDof fixedNodeUfixedNodeVGDof1 solution displacements solutionGDof1prescribedDof stiffnessSpringforceSpring displacements dispDisplacements jj 1GDof1 format jj displacements reordering displacements and rotations W displacements122numberNodes R displacements222numberNodes drawing nodal displacements figure plotnodeCoordinatesWokmarkersize8linewidth15 setgcafontsize18 graphical representation with interpolation for each element drawInterpolatedBeam exact solution by Bathe Solutions Manual of Procedures load LP3LP3LP6 K EILˆ3189 108 27108 135 5427 54 27kLˆ3EI X Kload plot0 33333 66667 1000000 Xxb markersize8linewidth15 64 Bernoulli Beam with Spring 99 Results are compared with finite element solution by Bathe 2 in his Solution Man ual The transverse displacement at the right end of the beam is 3749906 for the MATLAB solution while Bathe presents 3947275 using three finite elements The relative error between the two solutions is about 5 The present implementation coincides with exact solution available in the book by Reddy 3 given by 1 pLt kL3 wL 1 8EI 3EI 622 3 3 3 71 dw pL I kL L kL dx 6EI 24EI 3EI 65 Bernoulli Beam Free Vibrations With references to Fig 62 by removing the applied transverse loads we consider a simplysupported Bernoulli beam in free vibrations The reference code problem9vibm is given below which has the same structure of problem9m where the static loads are removed and substituted by the mass matrix and eigenvalue problem Bee eececceeeen eee eet eee eee rete e eee eeee MATLAB codes for Finite Element Analysis problem9vibm AJM Ferreira N Fantuzzi 2019 clear memory clear E modulus of elasticity I second moment of area L length of bar E1 I 1 EI EI rho 1 A 23 rhoA rhoA generation of coordinates and connectivities numberElements 64 nodeCoordinates linspace01numberElements1 L maxnodeCoordinates numberNodes sizenodeCoordinates1 xx nodeCoordinates1 elementNodes zeros numberElements 2 for i 1numberElements elementNodes i1 i elementNodes i 2 i1 end for structure z displacements displacement vector 100 6 Bernoulli Beams stiffness stiffness matrix mass mass matrix GDof global number of degrees of freedom GDof 2numberNodes stiffess matrix stiffness formStiffnessBernoulliBeamGDofnumberElements elementNodesnumberNodesxxEI1 stiffess matrix mass formMassBernoulliBeamGDofnumberElements elementNodesnumberNodesxxrhoA boundary conditions and solution clampedclamped fixedNodeU 1 2numberElements1 fixedNodeV 2 2numberElements2 simply supportedsimply supported fixedNodeU 1 2numberElements1 fixedNodeV clamped at x0 fixedNodeU 1 fixedNodeV 2 prescribedDof fixedNodeUfixedNodeV free vibration problem modeseigenvalues eigenvalueGDofprescribedDof stiffnessmass0 natural frequencies omega sqrteigenvalues exact frequencies omegaexact11 piˆ2sqrtEIrhoALˆ4 omegaexact21 4piˆ2sqrtEIrhoALˆ4 omegaexact31 9piˆ2sqrtEIrhoALˆ4 The function formMassBernoulliBeamm computes the mass matrix and it is listed below function mass formMassBernoulliBeamGDofnumberElements elementNodesnumberNodesxxrhoA mass zerosGDof calculation of the system mass matrix for e 1numberElements elementDof element degrees of freedom Dof indice elementNodese elementDof 2indice111 2indice21 2indice211 2indice212 length of element LElem xxindice2xxindice1 k1 rhoALElem420156 22LElem 54 13LElem 65 Bernoulli Beam Free Vibrations 101 22LElem 4LElem2 13LElem 3LElem2 54 13LElem 156 22LElem 13LElem 3LElem2 22LElem 4LElem2 6 mass matrix masselementDofelementDof mass elementDofelementDof k1 end end The convergence of the present numerical solution is shown in Table 61 compared to the exact frequencies for simply supported beams 52 EI QO nn for n12 623 pAL4 The first three frequencies converge with 64 finite elements NV 64 whereas the first frequency converges with 16 N 16 66 Stability of Bernoulli Beam A simply supported beam under axial load only is considered Such problem leads to the wellknown Euler buckling loads for the beam which exact solution is No n for n 12 624 n SNM T5 or n12 624 The reference code problem9bukm given below solves the linear buckling problem of Bernoulli beam As previously stated this problem is analogous to the free vibration one eigenvalue problem wherein the mass matrix is substituted by the stability matrix Table 61 First three natural frequencies of simply supported beam 1 w2 3 Present N2 65335 288926 726239 N4 65095 261340 596406 N8 65079 260381 586458 N16 65078 260317 585753 N 32 65078 260313 585707 N 64 65078 260313 585704 Exact 65078 260313 585704 102 6 Bernoulli Beams MATLAB codes for Finite Element Analysis problem9bukm AJM Ferreira N Fantuzzi 2019 clear memory clear E modulus of elasticity I second moment of area L length of bar E 1 I 1 EIEI generation of coordinates and connectivities numberElements 64 nodeCoordinates linspace01numberElements1 L maxnodeCoordinates numberNodes sizenodeCoordinates1 xx nodeCoordinates1 elementNodes zerosnumberElements2 for i 1numberElements elementNodesi1i elementNodesi2i1 end for structure displacements displacement vector stiffness stiffness matrix stability geometric matrix GDof global number of degrees of freedom GDof 2numberNodes stiffess matrix stiffness formStiffnessBernoulliBeamGDofnumberElements elementNodesnumberNodesxxEI1 stability matrix stability formStabilityBernoulliBeamGDofnumberElements elementNodesnumberNodesxx boundary conditions and solution clampedclamped fixedNodeU 1 2numberElements1 fixedNodeV 2 2numberElements2 simply supportedsimply supported fixedNodeU 1 2numberElements1 fixedNodeV clamped at x0 fixedNodeU 1 fixedNodeV 2 prescribedDof fixedNodeUfixedNodeV free vibration problem modeseigenvalues eigenvalueGDofprescribedDof stiffnessstability0 66 Stability of Bernoulli Beam 103 natural frequencies N0 eigenvalues exact frequencies simplysupported beam N0exact11 piˆ2EILˆ2 N0exact21 4piˆ2EILˆ2 N0exact31 9piˆ2EILˆ2 The function formStabilityBernoulliBeamm computes the stability matrix and it is listed below function stability formStabilityBernoulliBeamGDofnumberElements elementNodesnumberNodesxx stability zerosGDof calculation of the system mass matrix for e 1numberElements elementDof element degrees of freedom Dof indice elementNodese elementDof 2indice111 2indice21 2indice211 2indice212 length of element LElem xxindice2xxindice1 k1 130LElem36 3LElem 36 3LElem 3LElem 4LElemˆ2 3LElem LElemˆ2 36 3LElem 36 3LElem 3LElem LElemˆ2 3LElem 4LElemˆ2 stability matrix stabilityelementDofelementDof stabilityelementDofelementDof k1 end end The convergence of the numerical solution is given in Table 62 compared to the exact buckling loads for simply supported beams The solution converges up to the third buckling load with 64 elements N 64 the critical load is obtained with 16 elements N 16 104 6 Bernoulli Beams Table 62 First three buckling loads of simply supported beam N 0 1 N 0 2 N 0 3 Present N 2 99438 480000 1287228 N 4 98747 397754 917847 N 8 98699 394986 890484 N 16 98696 394797 888410 N 32 98696 394785 888274 N 64 98696 394784 888265 Exact 98696 394784 888264 References 1 JN Reddy An introduction to the finite element method 3rd edn McGrawHill International Editions New York 2005 2 KJ Bathe Finite element procedures in engineering analysis Prentice Hall 1982 3 JN Reddy Energy principles and variational methods in applied mechanics 3rd edn Wiley Hoboken NJ USA 2017 Chapter 7 Bernoulli 2D Frames Abstract In this chapter twodimensional frames under static loading and free vibrations are analyzed The present formulation is a generalization of the previ ous Bernoulli beam in local coordinates The stiffness and mass matrices are given by transformation of the same matrices in local coordinates by a matrix of rotation which is a function of the beam slope with respect to the horizontal axis 71 Introduction In this chapter twodimensional frames under static loading and free vibrations are analyzed The present formulation is a generalization of the previous Bernoulli beam in local coordinates The stiffness and mass matrices are given by transformation of the same matrices in local coordinates by a matrix of rotation which is a function of the beam slope with respect to the horizontal axis 72 2D Frame Element In Fig71 we show the twonoded Bernoulli beam element Each node has three global degrees of freedom two displacements in global axes and one rotation The vector of global displacements is given by uT u1 u4 u2 u5 u3 u6 71 Note that the new numbering of global degrees of freedom with respect to the 2D truss problem presented in the previous chapters to exploit MATLAB programming strengths In order to match the ordering of degrees of freedom the stiffness matrix has to be rearranged as shown in the code listing We define a local basis with cosines l m with respect to θ the angle between x and x In this local coordinate set the displacements are detailed as The Editors if applicable and The Authors under exclusive license to Springer Nature Switzerland AG 2020 A J M Ferreira and N Fantuzzi MATLAB Codes for Finite Element Analysis Solid Mechanics and Its Applications 157 httpsdoiorg10100797830304795277 105 106 7 Bernoulli 2D Frames u1 u2 u3 u3 u1 u2 u4 u5 u6 u6 u4 u5 x y Fig 71 A 2D frame element uT u 1 u 4 u 2 u 5 u 3 u 6 72 Noting that u 3 u3 u 6 u6 we derive a localglobal transformation matrix in the form u Lu 73 where L l 0 m 0 0 0 0 l 0 m 0 0 m 0 l 0 0 0 0 m 0 l 0 0 0 0 0 0 1 0 0 0 0 0 0 1 74 In local coordinates the stiffness matrix of the frame element is obtained by com bination of the stiffness of the bar element and the Bernoulli beam element in the form Ke E L3 AL2 AL2 0 0 0 0 AL2 0 0 0 0 12I 12I 6I L 6I L 12I 6I L 6I L 4I L2 2I L2 sym 4I L2 75 72 2D Frame Element 107 In global coordinates the strain energy is given by e 1 Tyr I TT TK T U 5u Ku 5uLKLuu Ku 76 where KLKL 77 In local coordinates the mass matrix of the frame element is obtained by combination of the mass of the bar element and the Bernoulli beam element in the form 140 70 0 O O 0 140 0 0 O 0 ie PAL 156 54 22L 13L MW 720 156 13L 22L 78 4L 3L sym 4L In global coordinates the kinetic energy is given by e 1 T Je 1 T1T It T K 5u Mu jv LD Miliu Mu 79 where M LML 710 The load vector for the present problem has to be defined according to the global Cartesian reference system and the order of the degrees of freedom reported in Eq 71 as the following scheme F Foi Fen Fy es Fyn Mi My 711 where F F and M are the horizontal and vertical concentrated forces and moments applied at the nodes The number of nodes in the mesh is indicated with n Static and free vibration with F 0 problems are shown in the following through codes The static problem involves stiffness matrix inversion and free vibration prob lem is carried out as eigenvalue problem 73 First 2D Frame Problem Consider the twodimensional frame illustrated in Fig 72 The code for solving this problem is problem10m The degrees of freedom are shown in Fig 73 108 7 Bernoulli 2D Frames 1 2 3 4 10 kN 10 kN 5 kNm 5 kNm 1 2 3 3 2 m x y 45 3 m 3 2 m 3 m Fig 72 A 2D frame example problem10m 1 5 9 2 6 10 3 7 11 4 8 12 x y Fig 73 Degrees of freedom for problem 10 MATLAB codes for Finite Element Analysis problem10m AJM Ferreira N Fantuzzi 2019 clear memory clear E modulus of elasticity I second moment of area E 210000 A 100 I 2e8 EA EA EI EI generation of coordinates and connectivities numberElements 3 p1 30001cospi4 nodeCoordinates 0 3000sqrt23000 3000sqrt2p1 0p13000 0 elementNodes zerosnumberElements2 for i 1numberElements elementNodesi1 i 73 First 2D Frame Problem 109 elementNodesi2 i1 end numberNodes sizenodeCoordinates1 xx nodeCoordinates1 yy nodeCoordinates2 for structure displacements displacement vector force force vector stiffness stiffness matrix GDof global number of degrees of freedom GDof 3numberNodes force zerosGDof1 force vector force6 10000 force7 10000 force10 5e6 force11 5e6 stiffness matrix stiffness formStiffness2DframeGDofnumberElements elementNodesnumberNodesxxyyEIEA boundary conditions and solution prescribedDof 1 4 5 8 9 12 solution displacements solutionGDofprescribedDofstiffnessforce output displacementsreactions outputDisplacementsReactionsdisplacementsstiffness GDofprescribedDof drawing undeformed and deformed meshes figure XX displacements1numberNodes YY displacementsnumberNodes12numberNodes dispNorm maxsqrtXXˆ2YYˆ2 scaleFact 500dispNorm hold on drawingMeshnodeCoordinatesscaleFactXX YYelementNodes L2k drawingMeshnodeCoordinateselementNodesL2k axis equal setgcafontsize18 plot interpolated deformed shape acoording to Lagrange and Hermite shape functions drawInterpolatedFrame2D Code problem10m calls function formStiffness2Dframem to compute the stiff ness matrix of twodimensional frame elements 110 7 Bernoulli 2D Frames function stiffness formStiffness2DframeGDofnumberElements elementNodesnumberNodesxxyyEIEA stiffnesszerosGDof computation of the system stiffness matrix for e 1numberElements elementDof element degrees of freedom Dof indice elementNodese elementDof indice indicenumberNodes indice2numberNodes nn lengthindice xa xxindice2xxindice1 ya yyindice2yyindice1 lengthelement sqrtxaxayaya cosa xalengthelement sena yalengthelement ll lengthelement L cosaeye2 senaeye2 zeros2 senaeye2 cosaeye2 zeros2 zeros24 eye2 oneu 1 11 1 oneu2 1 11 1 oneu3 1 11 1 oneu4 4 22 4 k1 EAlloneu zeros24 zeros2 12EIllˆ3oneu 6EIllˆ2oneu3 zeros2 6EIllˆ2oneu2 EIlloneu4 stiffnesselementDofelementDof stiffnesselementDofelementDof Lk1L end end Results are given as Displacements ans 10000 0 20000 00000 30000 00000 40000 0 50000 0 60000 13496 70000 13496 80000 0 90000 0 100000 00005 73 First 2D Frame Problem 111 Fig 74 Deformed shape of problem 10 0 2000 4000 6000 8000 2000 1000 0 1000 2000 3000 110000 00005 120000 0 reactions ans 10e07 00000 00000 00000 00000 00000 00010 00000 00010 00000 22596 00000 22596 The deformed shape is given in Fig74 The interpolation of the deformed shape according to the calculated nodal variables is carried out in the script drawInter polatedFrame2Dm which is given but not reported in the text for the sake of conciseness Notethateventhoughonlyoneelementpersegmentisselectedclampedboundary conditions in terms of boundary rotations are satisfied zero slope due to the Hermite interpolation By increasing the finite elements by dividing the given 3 elements accuracy improves as it is shown in the following examples 74 Second 2D Frame Problem Consider the frame illustrated in Fig75 The code for solving this problem is prob lem11m The degrees of freedom are shown in Fig76 112 7 Bernoulli 2D Frames 1 2 3 4 15 kN 10 kNm 6 m 6 m E 210000 MPa A 200 mm2 I 2 108 mm4 x y Fig 75 A 2D frame example geometry materials and loads problem11m 1 2 4 3 5 8 6 7 9 11 10 12 x y Fig 76 A 2D frame example degree of freedom ordering 74 Second 2D Frame Problem 113 MATLAB codes for Finite Element Analysis problem11m 2D frame AJM Ferreira N Fantuzzi 2019 clear memory clear E modulus of elasticity I second moment of area E 210000 A 200 I 2e8 EA EA EI EI generation of coordinates and connectivities numberElements 3 nodeCoordinates 0 00 60006000 60006000 0 elementNodes zerosnumberElements2 for i 1numberElements elementNodesi1 i elementNodesi2 i1 end numberNodes sizenodeCoordinates1 xx nodeCoordinates1 yy nodeCoordinates2 for structure displacements displacement vector force force vector stiffness stiffness matrix GDof global number of degrees of freedom GDof 3numberNodes force zerosGDof1 force vector force2 15000 force10 10e6 stiffness matrix stiffness formStiffness2DframeGDofnumberElements elementNodesnumberNodesxxyyEIEA boundary conditions and solution prescribedDof 1 4 5 8 9 12 solution displacements solutionGDofprescribedDofstiffnessforce output displacementsreactions outputDisplacementsReactionsdisplacementsstiffness GDofprescribedDof drawing mesh and deformed shape 114 7 Bernoulli 2D Frames figure XX displacements1numberNodes YY displacementsnumberNodes12numberNodes dispNorm maxsqrtXXˆ2YYˆ2 scaleFact 50dispNorm hold on drawingMeshnodeCoordinatesscaleFactXX YYelementNodes L2k drawingMeshnodeCoordinateselementNodesL2k axis equal setgcafontsize18 plot interpolated deformed shape acoording to Lagrange and Hermite shape functions drawInterpolatedFrame2D Results are listed below Displacements ans 10000 0 20000 52843 30000 44052 40000 0 50000 0 60000 06522 70000 06522 80000 0 90000 0 100000 00005 110000 00006 120000 0 reactions ans 10e07 00000 00009 00000 00006 00000 00005 00000 00005 00000 30022 00000 22586 The deformed shape of this problem is illustrated in Fig77 As it was aformentioned the Bernoulli frame element is exact at the nodes super convergent element and interpolated in the domain By increasing the number of elements will give a better approximation more exact nodal values in other parts of the structure It is generally suggested to place a node where numerical value is needed rather than interpolate it The following Code problem11bm considers 12 elements 4 divisions for each segment in the same frame structure 74 Second 2D Frame Problem 115 Fig 77 Deformed shape of problem 11 0 1000 2000 3000 4000 5000 6000 7000 0 1000 2000 3000 4000 5000 6000 MATLAB codes for Finite Element Analysis problem11bm 2D frame AJM Ferreira N Fantuzzi 2019 clear memory clear E modulus of elasticity I second moment of area E 210000 A 200 I 2e8 EA EA EI EI generation of coordinates and connectivities numberElements 12 nodeCoordinates 0 00 15000 30000 4500 0 60001500 60003000 6000 4500 60006000 60006000 4500 6000 30006000 15006000 0 elementNodes zerosnumberElements2 for i 1numberElements elementNodesi1 i elementNodesi2 i1 end numberNodes sizenodeCoordinates1 xx nodeCoordinates1 yy nodeCoordinates2 for structure displacements displacement vector force force vector stiffness stiffness matrix GDof global number of degrees of freedom 116 7 Bernoulli 2D Frames GDof 3numberNodes force zerosGDof1 stiffness zerosGDof force vector force5 15000 force31 10e6 stiffness matrix stiffness formStiffness2DframeGDofnumberElements elementNodesnumberNodesxxyyEIEA boundary conditions and solution prescribedDof 1 13 14 26 27 39 solution displacements solutionGDofprescribedDofstiffnessforce output displacementsreactions outputDisplacementsReactionsdisplacementsstiffness GDofprescribedDof drawing mesh and deformed shape figure XX displacements1numberNodes YY displacementsnumberNodes12numberNodes dispNorm maxsqrtXXˆ2YYˆ2 scaleFact 50dispNorm hold on drawingMeshnodeCoordinatesscaleFactXX YYelementNodes L2k drawingMeshnodeCoordinateselementNodesL2k axis equal setgcafontsize18 plot interpolated deformed shape acoording to Lagrange and Hermite shape functions drawInterpolatedFrame2D Generation of nodal coordinates and element nodes is fundamental for force vector and boundary conditions definitions The order of degrees of freedom follows the same rule considered before as u degrees of freedom displacement along x then the v degrees of freedom displacement along y and finally rotations Results are given below Displacements ans 10000 0 20000 06857 30000 22689 40000 40387 74 Second 2D Frame Problem 117 50000 52843 60000 50645 70000 48447 80000 46249 90000 44052 100000 32197 110000 17606 120000 05226 130000 0 140000 0 150000 01630 160000 03261 170000 04891 180000 06522 190000 01942 200000 00687 210000 00912 220000 06522 230000 04891 240000 03261 250000 01630 260000 0 270000 0 280000 00008 290000 00012 300000 00011 310000 00005 320000 00002 330000 00001 340000 00002 350000 00006 360000 00009 370000 00010 380000 00006 390000 0 reactions ans 10e07 00000 00009 00000 00006 00000 00005 00000 00005 00000 30022 00000 22586 The deformed shape of this problem is illustrated in Fig78 118 7 Bernoulli 2D Frames Fig 78 Deformed shape of problem 11b 0 1000 2000 3000 4000 5000 6000 7000 0 1000 2000 3000 4000 5000 6000 75 2D Frame in Free Vibrations The free vibrations of the 2D frame structure of the previous example is shown in problem11bvibm The material density has been considered as ρ 805 109 tonmm3 The code is given below MATLAB codes for Finite Element Analysis problem11bvibm 2D frame AJM Ferreira N Fantuzzi 2019 clear memory clear close all E modulus of elasticity I second moment of area E 210000 A 200 I 2e8 rho 805e9 EA EA EI EI rhoA rhoA generation of coordinates and connectivities numberElements 12 nodeCoordinates 0 00 15000 30000 4500 0 60001500 60003000 6000 4500 60006000 60006000 4500 6000 30006000 15006000 0 elementNodes zerosnumberElements2 for i 1numberElements elementNodesi1 i elementNodesi2 i1 75 2D Frame in Free Vibrations 119 end numberNodes sizenodeCoordinates1 xx nodeCoordinates1 yy nodeCoordinates2 for structure displacements displacement vector stiffness stiffness matrix GDof global number of degrees of freedom GDof 3numberNodes U zerosGDof1 stiffness matrix stiffness formStiffness2DframeGDofnumberElements elementNodesnumberNodesxxyyEIEA mass matrix mass formMass2DframeGDofnumberElements elementNodesnumberNodesxxyyrhoA boundary conditions and solution prescribedDof 1 13 14 26 27 39 solution modeseigenvalues eigenvalueGDofprescribedDof stiffnessmass0 omega sqrteigenvalues drawing mesh and deformed shape modeNumber 3 U modesmodeNumber figure XX U1numberNodes YY UnumberNodes12numberNodes dispNorm maxsqrtXXˆ2YYˆ2 scaleFact 20dispNorm hold on drawingMeshnodeCoordinatesscaleFactXX YYelementNodes L2k drawingMeshnodeCoordinateselementNodesL2k axis equal setgcafontsize18 plot interpolated deformed shape acoording to Lagrange and Hermite shape functions displacements U drawInterpolatedFrame2D 120 7 Bernoulli 2D Frames Code problem11bvibm calls function formMass2Dframem to compute the mass matrix of twodimensional frame elements function mass formMass2DframeGDofnumberElements elementNodesnumberNodesxxyyrhoA mass zerosGDof computation of the system stiffness matrix for e 1numberElements elementDof element degrees of freedom Dof indice elementNodese elementDof indice indicenumberNodes indice2numberNodes xa xxindice2xxindice1 ya yyindice2yyindice1 lengthelement sqrtxaxayaya cosa xalengthelement sena yalengthelement ll lengthelement L cosaeye2 senaeye2 zeros2 senaeye2 cosaeye2 zeros2 zeros24 eye2 oneu 162 11 2 oneu2 17026 9 9 26 oneu3 ll42022 13 13 22 oneu4 llˆ24204 3 3 4 m1 rhoAlloneu zeros24 zeros2 oneu2 oneu3 zeros2 oneu3 oneu4 masselementDofelementDof masselementDofelementDof Lm1L end 75 2D Frame in Free Vibrations 121 1000 0 1000 2000 3000 4000 5000 6000 0 1000 2000 3000 4000 5000 6000 0 2000 4000 6000 0 1000 2000 3000 4000 5000 6000 7000 0 2000 4000 6000 0 1000 2000 3000 4000 5000 6000 7000 0 1000 2000 3000 4000 5000 6000 0 1000 2000 3000 4000 5000 6000 Fig 79 First four mode shapes of problem11bvibm Table 71 First four natural frequencies of the 2D frame in problem11bvibm ω1 ω2 ω3 ω4 Ref 4142418 8698700 11816826 15773839 Present 4223818 8733047 12373850 16394477 Error 197 039 471 393 The results are verified with another finite element code with the same number of elements and degrees of freedom per node The first four mode shapes are shown in Fig79 and the correspondent frequencies are listed in Table 71 where a good agreement is observed between the two solutions Chapter 8 Bernoulli 3D Frames Abstract The analysis of three dimensional frames is quite similar to the analysis of 2D frames In the 2node 3D frame finite element we now consider in each node three displacements and three rotations with respect to the three global cartesian axes However the complexity in such structures is due to the orientation of the beam in space other than in 2D plane Before introducing the stiffness and mass matrices in the global reference system rotation matrices for vectors in 3D space are firstly introduced 81 Introduction The analysis of three dimensional frames is quite similar to the analysis of 2D frames In the 2node 3D frame finite element we now consider in each node three displace ments and three rotations with respect to the three global cartesian axes However the complexity in such structures is due to the orientation of the beam in space other than in 2D plane Before introducing the stiffness and mass matrices in the global reference system rotation matrices for vectors in 3D space are firstly introduced 82 Matrix Transformation in 3D Space It is assumed that the local axis x is aligned with the beam major axis as shown in Fig81 For 2D frames one single rotation is sufficient for the definition of the beam in the x y plane on the contrary at least three rotations are needed in the 3D space The definition of the direction cosines according to axis x is straightforward and follows the presentation given in the 2D frames chapter as Cxx x2 x1 Le Cyx y2 y1 Le Czx z2 z1 Le 81 The Editors if applicable and The Authors under exclusive license to Springer Nature Switzerland AG 2020 A J M Ferreira and N Fantuzzi MATLAB Codes for Finite Element Analysis Solid Mechanics and Its Applications 157 httpsdoiorg10100797830304795278 123 124 8 Bernoulli 3D Frames U11 wah Uio v 4 U5 ug us ur U10 Us ur U2 ug ug ub fo 12 U U Uh 7 U1 Ua U3 Us y U6 UG x 2 z Fig 81 A 3D frame element and its local and global reference systems where x y and z refer to the global reference system and x y and 2 is the local beam reference system and Le VX2 x1 92 v1 2 21 i is the beam length The vector rotation matrix can be represented in matrix form as Cy Cyy Czy r Cyy Cyy Coy 83 Cyy Cyz Cry where the definitions of the last two rows is shown below The vector rotation matrix in 3D space 83 is given by the product of three rotation matrices as r RRgR 84 where a 3 and are the rotation angles about x y and z axes respectively Rotation about z axis R is given by cosy siny 0 R siny cosy 0 85 0 0 1 82 Matrix Transformation in 3D Space 125 where cos y CyyCyy sin y CyxCyy and Cyy C2 C Rotation about y axis Rg is given by cos O sin Z R3 0 1 0 86 sin G0 cos 3 where cos 3 C and sin C Combining the rotation about y and z axis the vector rotation matrix is Cry Cyx Cry Cyx Cyy 0 r RgR Cyy Cry 87 Cxy Czy Cyx Czy oO Cry Cyy Cry Note that when 3 90 or G 270 global coordinates of the 2node beam element change only along z thus the vector rotation matrix takes a special form as 0 O0Cy r 0 10 88 Cy 0 0 for z2 z or 3 90 C 1 otherwise for z z2 or G 270 C 1 If the extra beam rotation a is included its rotation matrix is 1 0 0 Ry Ocosa sina 89 0 sina cosa Finally the vector rotation matrix becomes r RRgR Cyy Cyy Coy Cyyr Czy Sina Cyy Cosa CyxrC zy Sina Cyx COS Cyy sin Cry Cry Cyy Czy Cosa Cy Sina CyxCz COSA Cyy SIN a TT rome MC oT Cy cos Cry Cry 810 Special case of vertical members 3 90 and 3 270 can be derived for the present case also as 0 0 Cy r Cz sina cosa 0 811 C cosa sina 0 126 8 Bernoulli 3D Frames obviously for α 0 the previous case 88 is obtained For the sake of simplicity in the following members without α orientation are considered eg beams of circular cross section The interested reader can easily extend the codes using the aforementioned rotation matrices taking into account the rotation α 83 Stiffness Matrix and Vector of Equivalent Nodal Forces In the local coordinate system the stiffness matrix is given by K E A L 0 0 0 0 0 E A L 0 0 0 0 0 12E Iz L3 0 0 0 6E Iz L2 0 12E Iz L3 0 0 0 6E Iz L2 12E Iy L3 0 6E Iy L2 0 0 0 12E Iy L3 0 6E Iy L2 0 G J L 0 0 0 0 0 G J L 0 0 4E Iy L 0 0 0 6E Iy L2 0 2E Iy L 0 4E Iz L 0 6E Iz L2 0 0 0 2E Iz L E A L 0 0 0 0 0 12E Iz L3 0 0 0 6E Iz L2 12E Iy L3 0 6E Iy L2 0 G J L 0 0 4E Iy L 0 sym 4E Iz L 812 After transformation to the global axes the stiffness matrix in global coordinates is obtained as K RT KR where the rotation matrix R is defined as R r 0 0 0 0 r 0 0 0 0 r 0 0 0 0 r 813 being r as indicated in 87 or 810 The latter has not been coded in the present book for the sake of simplicity α 0 but it can be easily introduced by the reader The twonode 3D frame element has six degrees of freedom per node Once the static problem is solved nodal displacements carried out it is possible to calculate the reactions at the supports by F KU 814 where K and U are the structure stiffness matrix and the vector of nodal displacement respectively The element nodal forces can be evaluated by axes transformation as 83 Stiffness Matrix and Vector of Equivalent Nodal Forces 127 fe KRUe 815 where fe is the element nodal forces vector and Ue is the global vector of displace ments referred to the element e 84 Mass Matrix In case of free vibration analysis the consistent mass matrix in the local coordinate system is defined as M ρAL 420 140 0 0 0 0 0 70 0 0 0 0 0 156 0 0 0 22L 0 54 0 0 0 13L 156 0 22L 0 0 0 54 0 13L 0 140r2x 0 0 0 0 0 70r2x 0 0 4L2 0 0 0 13L 0 3L2 0 4L2 0 13L 0 0 0 3L2 140 0 0 0 0 0 156 0 0 0 22L 156 0 22L 0 140r2x 0 0 4L2 0 sym 4L2 816 where r2 x I y I zA where I y and I z are the second moment of area of the cross section about the principal y and z axes Whereas the lumped mass matrix is given by M ρAL 2 1 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 r2 x 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 1 0 0 0 r2 x 0 0 0 0 sym 0 817 The mass matrix in the global coordinate system takes the form M RT MR 818 where the rotation matrix R is defined in 813 128 8 Bernoulli 3D Frames 1 3 2 4 10 kN 15 kN 4 m 3 m 3 m x y z Fig 82 A 3D frame example problem12m 85 First 3D Frame Problem The first 3D frame example is illustrated in Fig82 We consider E 210 GPa G 84 GPa A 2 102 m2 Iy 1 104 m4 Iz 2 104 m4 J 05 104 m4 Code problem12m solves this problem and calls function formStiffness 3Dframem that computes the stiffness matrix of the 3D frame element MATLAB codes for Finite Element Analysis problem12m AJM Ferreira N Fantuzzi 2019 clear memory clear E modulus of elasticity I second moment of area J polar moment of inertia G shear modulus E 210e9 A 002 Iy 10e5 Iz 20e5J 5e5 G 84e9 generation of coordinates and connectivities nodeCoordinates 0 0 0 3 0 0 0 0 3 0 4 0 xx nodeCoordinates1 yy nodeCoordinates2 zz nodeCoordinates3 elementNodes 1 21 31 4 numberNodes sizenodeCoordinates1 numberElements sizeelementNodes1 85 First 3D Frame Problem 129 for structure displacements displacement vector force force vector stiffness stiffness matrix GDof global number of degrees of freedom GDof 6numberNodes force zerosGDof1 stiffness zerosGDofGDof force vector force1 10e3 force3 15e3 stiffness matrix stiffness formStiffness3DframeGDofnumberElements elementNodesnumberNodesnodeCoordinatesEAIzIyGJ boundary conditions prescribedDof 724 solution displacements solutionGDofprescribedDofstiffnessforce displacements dispDisplacements jj 1GDof format long f jj displacements fprintfnode U fprintf3d 128f f Listing of formStiffness3Dframem function stiffness formStiffness3DframeGDofnumberElements elementNodesnumberNodesnodeCoordinatesEAIzIyGJ stiffness zerosGDof computation of the system stiffness matrix for e 1numberElements elementDof element degrees of freedom Dof indice elementNodese elementDof 6indice15 6indice14 6indice13 6indice12 6indice11 6indice1 6indice25 6indice24 6indice23 6indice22 6indice21 6indice2 x1 nodeCoordinatesindice11 y1 nodeCoordinatesindice12 z1 nodeCoordinatesindice13 x2 nodeCoordinatesindice21 y2 nodeCoordinatesindice22 z2 nodeCoordinatesindice23 L sqrtx2x1x2x1 y2y1y2y1 130 8 Bernoulli 3D Frames z2z1z2z1 k1 EAL k2 12EIzLLL k3 6EIzLL k4 4EIzL k5 2EIzL k6 12EIyLLL k7 6EIyLL k8 4EIyL k9 2EIyL k10 GJL k k1 0 0 0 0 0 k1 0 0 0 0 0 0 k2 0 0 0 k3 0 k2 0 0 0 k3 0 0 k6 0 k7 0 0 0 k6 0 k7 0 0 0 0 k10 0 0 0 0 0 k10 0 0 0 0 k7 0 k8 0 0 0 k7 0 k9 0 0 k3 0 0 0 k4 0 k3 0 0 0 k5 k1 0 0 0 0 0 k1 0 0 0 0 0 0 k2 0 0 0 k3 0 k2 0 0 0 k3 0 0 k6 0 k7 0 0 0 k6 0 k7 0 0 0 0 k10 0 0 0 0 0 k10 0 0 0 0 k7 0 k9 0 0 0 k7 0 k8 0 0 k3 0 0 0 k5 0 k3 0 0 0 k4 if x1 x2 y1 y2 if z2 z1 Lambda 0 0 1 0 1 0 1 0 0 else Lambda 0 0 1 0 1 0 1 0 0 end else CXx x2x1L CYx y2y1L CZx z2z1L D sqrtCXxCXx CYxCYx CXy CYxD CYy CXxD CZy 0 CXz CXxCZxD CYz CYxCZxD CZz D Lambda CXx CYx CZx CXy CYy CZy CXz CYz CZz end R Lambda zeros39 zeros3 Lambda zeros36 zeros36 Lambda zeros3zeros39 Lambda stiffnesselementDofelementDof stiffnesselementDofelementDof RkR end end 85 First 3D Frame Problem 131 Results are listed as follows Displacements node U 1 000000706 2 000000006 3 000001063 4 000000109 5 000000088 6 000000113 7 000000000 8 000000000 9 000000000 10 000000000 11 000000000 12 000000000 13 000000000 14 000000000 15 000000000 16 000000000 17 000000000 18 000000000 19 000000000 20 000000000 21 000000000 22 000000000 23 000000000 24 000000000 86 Second 3D Frame Problem The next 3D problem is illustrated in Fig83 and considers E 210 GPa G 84 GPa A 2 102 m2 Iy 1 104 m4 Iz 2 104 m4 J 05 104 m4 The MATLAB code for this problem is problem13m MATLAB codes for Finite Element Analysis problem13m AJM Ferreira N Fantuzzi 2019 clear memory clear 132 8 Bernoulli 3D Frames E modulus of elasticity I second moments of area J polar moment of inertia G shear modulus E 210e9 A 002 Iy 10e5 Iz 20e5 J 5e5 G 84e9 generation of coordinates and connectivities nodeCoordinates 0 0 0 0 0 4 4 0 4 4 0 0 0 5 0 0 5 4 4 5 4 4 5 0 xx nodeCoordinates1 yy nodeCoordinates2 zz nodeCoordinates3 elementNodes 1 52 63 7 4 8 5 6 6 7 7 8 8 5 numberNodes sizenodeCoordinates1 numberElements sizeelementNodes1 for structure displacements displacement vector force force vector stiffness stiffness matrix GDof global number of degrees of freedom GDof 6numberNodes force zerosGDof1 stiffness zerosGDof force vector force37 15e3 stiffness matrix stiffness formStiffness3DframeGDofnumberElements elementNodesnumberNodesnodeCoordinatesEAIzIyGJ boundary conditions prescribedDof 124 solution displacements solutionGDofprescribedDofstiffnessforce displacements dispDisplacements jj 1GDof format long f jj displacements fprintfnode U fprintf3d 128f f drawing mesh and deformed shape U displacements figure XX U166numberNodes YY U266numberNodes 86 Second 3D frame problem 133 ZZ U366numberNodes scaleFact 500 hold on drawingMeshnodeCoordinatesscaleFactXX YY ZZelementNodes L2k drawingMeshnodeCoordinateselementNodesL2k axis equal setgcafontsize18 view17045 plot interpolated deformed shape acoording to Lagrange and Hermite shape functions drawInterpolatedFrame3D Results are obtained as Displacements node U 1 000000000 2 000000000 3 000000000 4 000000000 5 000000000 6 000000000 7 000000000 8 000000000 9 000000000 10 000000000 11 000000000 1 3 4 2 5 7 8 6 15 kN 5 m 4 m 4 m x y z Fig 83 A second 3D frame example problem13m 134 8 Bernoulli 3D Frames 12 000000000 13 000000000 14 000000000 15 000000000 16 000000000 17 000000000 18 000000000 19 000000000 20 000000000 21 000000000 22 000000000 23 000000000 24 000000000 25 000039898 26 000000298 27 000058935 28 000003552 29 000035809 30 000004453 31 000212492 32 000000684 33 000058935 34 000003552 35 000035809 36 000022176 37 000213205 38 000000684 39 000058935 40 000003552 41 000035940 42 000022305 43 000039898 44 000000298 45 000058935 46 000003552 47 000035940 48 000004455 The deformed shape of this structure is depicted in Fig84 The results for the present case are compared to the solution carried out by a commercial finite element code and listed in Table 81 where u1 u2 and u3 are the translational displacements along x y and z respectively while u4 u5 and u6 are the rotations in radians about x y and z respectively Good agreement is observed 86 Second 3D frame problem 135 Fig 84 Deformed shape for problem 13 Table 81 Comparison in terms of displacements and rotations all multiplied by 104 of the 3D frame in problem13m Node 5 Node 6 Node 7 Node 8 Ref Present Ref Present Ref Present Ref Present u1 39898 39898 212492 212492 213205 213205 39898 39898 u2 00298 00298 00684 00684 00684 00684 00298 00298 u3 58935 58935 58935 58935 58935 58935 58935 58935 u4 03552 03552 03552 03552 03552 03552 03552 03552 u5 35809 35809 35809 35809 35940 35940 35940 35940 u6 04453 04453 22176 22176 22305 22305 04455 04455 between the two solutions It is recalled that the solution is exact in the nodes and approximated through interpolation functions Lagrange for axial displacements and Hermite for transverse displacements along the beams 136 8 Bernoulli 3D Frames 87 3D Frame in Free Vibrations The present problem considers the structure in Fig83 without external applied forces and considers E 210 GPa G 84 GPa A 2 102 m2 Iy 1 104 m4 Iz 2 104 m4 J 05 104 m4 and ρ 7850 kgm3 The MATLAB code for this problem is problem13vibm MATLAB codes for Finite Element Analysis problem13vibm AJM Ferreira N Fantuzzi 2019 clear memory clear E modulus of elasticity I second moments of area J polar moment of inertia G shear modulus E 210e9 A 002 rho 7850 Iy 10e5 Iz 20e5 J 5e5 G 84e9 generation of coordinates and connectivities nodeCoordinates 0 0 0 0 0 4 4 0 4 4 0 0 0 5 0 0 5 4 4 5 4 4 5 0 xx nodeCoordinates1 yy nodeCoordinates2 zz nodeCoordinates3 elementNodes 1 52 63 7 4 8 5 6 6 7 7 8 8 5 numberNodes sizenodeCoordinates1 numberElements sizeelementNodes1 for structure displacements displacement vector stiffness stiffness matrix mass mass matrix GDof global number of degrees of freedom GDof 6numberNodes U zerosGDof1 stiffness matrix stiffness formStiffness3DframeGDofnumberElements elementNodesnumberNodesnodeCoordinatesEAIzIyGJ mass matrix mass 87 3D Frame in Free Vibrations 137 formMass3DframeGDofnumberElements elementNodesnumberNodesnodeCoordinatesrhoAIzIy boundary conditions prescribedDof 124 solution modeseigenvalues eigenvalueGDofprescribedDof stiffnessmass0 omega sqrteigenvalues drawing mesh and deformed shape modeNumber 1 U modesmodeNumber figure XX U166numberNodes YY U266numberNodes ZZ U366numberNodes scaleFact 20 hold on drawingMeshnodeCoordinatesscaleFactXX YY ZZelementNodes L2k drawingMeshnodeCoordinateselementNodesL2k axis equal setgcafontsize18 view17045 plot interpolated deformed shape acoording to Lagrange and Hermite shape functions drawInterpolatedFrame3D Listing of formMass3Dframem is given below function mass formMass3DframeGDofnumberElements elementNodesnumberNodesnodeCoordinatesrhoAIzIy mass zerosGDof computation of the system mass matrix for e 1numberElements elementDof element degrees of freedom Dof indice elementNodese elementDof 6indice15 6indice14 6indice13 6indice12 6indice11 6indice1 6indice25 6indice24 6indice23 6indice22 6indice21 6indice2 x1 nodeCoordinatesindice11 y1 nodeCoordinatesindice12 z1 nodeCoordinatesindice13 x2 nodeCoordinatesindice21 y2 nodeCoordinatesindice22 z2 nodeCoordinatesindice23 138 8 Bernoulli 3D Frames L sqrtx2x1x2x1 y2y1y2y1 z2z1z2z1 p IzIyA lumped mass matrix m rhoAL2diag1 1 1 p 0 0 1 1 1 p 0 0 consistent mass matrix m rhoAL420140 0 0 0 0 0 70 0 0 0 0 0 0 156 0 0 0 22L 0 54 0 0 0 13L 0 0 156 0 22L 0 0 0 54 0 13L 0 0 0 0 140p 0 0 0 0 0 70p 0 0 0 0 22L 0 4Lˆ2 0 0 0 13L 0 3Lˆ2 0 0 22L 0 0 0 4Lˆ2 0 13L 0 0 0 3Lˆ2 70 0 0 0 0 0 140 0 0 0 0 0 0 54 0 0 0 13L 0 156 0 0 0 22L 0 0 54 0 13L 0 0 0 156 0 22L 0 0 0 0 70p 0 0 0 0 0 140p 0 0 0 0 13L 0 3Lˆ2 0 0 0 22L 0 4Lˆ2 0 0 13L 0 0 0 3Lˆ2 0 22L 0 0 0 4Lˆ2 if x1 x2 y1 y2 if z2 z1 Lambda 0 0 1 0 1 0 1 0 0 else Lambda 0 0 1 0 1 0 1 0 0 end else CXx x2x1L CYx y2y1L CZx z2z1L D sqrtCXxCXx CYxCYx CXy CYxD CYy CXxD CZy 0 CXz CXxCZxD CYz CYxCZxD CZz D Lambda CXx CYx CZx CXy CYy CZy CXz CYz CZz end R Lambda zeros39 zeros3 Lambda zeros36 zeros36 Lambda zeros3zeros39 Lambda masselementDofelementDof masselementDofelementDof RmR end end Note that both consistent and lumped mass matrices are given they have to be commented and uncommented according to the readers needs The first four natural frequencies are compared to the ones given by a commercial code and listed in Table 82 As expected by using a lumped mass matrix the error on the natural frequency is larger because more finite elements has to be used in the computation 87 3D Frame in Free Vibrations 139 Table 82 First four natural frequencies of the 3D frame in problem13vibm Ref Consistent Error Lumped Error ω1 433481 434457 023 407301 604 ω2 577971 579639 029 485147 1606 ω3 591716 593490 030 539221 887 ω4 1175250 1186686 097 1002990 1466 Fig 85 First four mode shapes for problem13vibm The first four mode shapes for a consistent mass matrix of this structure are shown in Fig85 Chapter 9 Grids Abstract In this chapter we perform the static analysis of grids which are planar structures where forces are applied normal to the grid plane In other words the grid element is analogous to the 2D frame element where the axial stiffness is replaced by the torsional one 91 Introduction In this chapter we perform the static analysis of grids which are planar structures where forces are applied normal to the grid plane In other words the grid element is analogous to the 2D frame element where the axial stiffness is replaced by the torsional one At each node a transverse displacement and two rotations are assigned The stiffness matrix in local cartesian axes is given by K 12E I L3e 0 6E I L2e 12E I L3e 0 6E I L2e 0 G J Le 0 0 G J Le 0 6E I L2e 0 4E I Le 6E I L2e 0 2E I Le 12E I L3e 0 6E I L2e 12E I L3e 0 6E I L2e 0 G J Le 0 0 G J Le 0 6E I L2e 0 2E I Le 6E I L2e 0 4E I Le 91 where E is the modulus of elasticity I is the second moment of area J the polar moment of inertia and G the shear modulus The element length is denoted by Le The Editors if applicable and The Authors under exclusive license to Springer Nature Switzerland AG 2020 A J M Ferreira and N Fantuzzi MATLAB Codes for Finite Element Analysis Solid Mechanics and Its Applications 157 httpsdoiorg10100797830304795279 141 142 9 Grids 1 2 w1 w2 θz1 θx1 θz2 θx2 L x y z Fig 91 A typical 2node grid element We consider direction cosines C cos θ and S sin θ being θ the angle between global axis x and local axis x The rotation matrix is defined as R 1 0 0 0 0 0 0 C S 0 0 0 0 S C 0 0 0 0 0 0 1 0 0 0 0 0 0 C S 0 0 0 0 S C 92 The stiffness matrix in global cartesian axes is obtained as K RT KR 93 Six degrees of freedom are linked to every grid element as illustrated in Fig91 After computing displacements in global coordinate set we compute reactions by F KU 94 where K and U is the stiffness matrix and the vector of nodal displacements of the structure respectively Element forces are also possible to compute by transformation fe KRUe 95 where fe is the element nodal forces vector and Ue is the global vector of displace ments referred to the element e 92 First Grid Problem 143 3 1 2 10kN z x 3 2 1 3m 3m 4m x y z Fig 92 A first grid example problem14m 92 First Grid Problem The first grid problem is illustrated in Fig92 The grid is built from two elements as illustrated Given E 210 GPa G 84 GPa I 20 105 m4 J 5 105 m4 the MATLAB problem14m computes displacements reactions and stresses MATLAB codes for Finite Element Analysis problem14m AJM Ferreira N Fantuzzi 2019 clear memory clear E modulus of elasticity I second moments of area J polar moment of inertia G shear modulus E 210e9 G 84e9 I 20e5 J 5e5 generation of coordinates and connectivities nodeCoordinates 4 0 0 3 0 3 xx nodeCoordinates1 yy nodeCoordinates2 elementNodes 1 2 3 1 numberNodes sizenodeCoordinates1 numberElements sizeelementNodes1 GDof global number of degrees of freedom GDof 3numberNodes force zerosGDof1 force vector force1 10e3 144 9 Grids stiffness matrix stiffness formStiffnessGridGDofnumberElements elementNodesxxyyEIGJ boundary conditions prescribedDof 49 solution displacements solutionGDofprescribedDofstiffnessforce output displacementsreactions outputDisplacementsReactionsdisplacementsstiffness GDofprescribedDof forces in elements dispforces in elements EF forcesInElementGridnumberElementselementNodes xxyyEIGJdisplacements The code for computing the stiffness matrix of the grid element is listed below function stiffness formStiffnessGridGDof numberElementselementNodesxxyyEIGJ function to form global stiffness for grid element stiffness zerosGDof for e 1numberElements elementDof element degrees of freedom Dof indice elementNodese elementDof indice1131 indice1132 indice1133 indice2131 indice2132 indice2133 xa xxindice2xxindice1 ya yyindice2yyindice1 L sqrtxaxayaya C xaL S yaL a1 12EILLL a2 6EILL a3 GJL a4 4EIL a5 2EIL stiffness in local axes k a1 0 a2 a1 0 a2 0 a3 0 0 a3 0 a2 0 a4 a2 0 a5 a1 0 a2 a1 0 a2 0 a3 0 0 a3 0 a2 0 a5 a2 0 a4 transformation matrix a 1 0 0 0 C S0 S C 92 First Grid Problem 145 R a zeros3zeros3 a stiffnesselementDofelementDof stiffnesselementDofelementDof RkR end end The code for computing the forces in elements is listed below function EF forcesInElementGridnumberElements elementNodesxxyyEIGJdisplacements forces in elements EF zeros6numberElements for e 1numberElements elementDof element degrees of freedom Dof indice elementNodese elementDof indice1131 indice1132 indice1133 indice2131 indice2132 indice2133 xa xxindice2xxindice1 ya yyindice2yyindice1 L sqrtxaxayaya C xaL S yaL a1 12EILLL a2 6EILL a3 GJL a4 4EIL a5 2EIL stiffness in local axes k a1 0 a2 a1 0 a2 0 a3 0 0 a3 0 a2 0 a4 a2 0 a5 a1 0 a2 a1 0 a2 0 a3 0 0 a3 0 a2 0 a5 a2 0 a4 transformation matrix a 1 0 0 0 C S0 S C R a zeros3zeros3 a forces in element EFe kRdisplacementselementDof end end Results for displacements reactions and forces in elements are listed below 146 9 Grids Displacements ans 10000 00048 20000 0 30000 00018 40000 0 50000 0 60000 0 70000 0 80000 0 90000 0 reactions ans 10e04 00004 05000 00005 13891 00006 20000 00007 05000 00008 13891 00009 20000 forces in elements EF 10e04 05000 05000 00888 00888 00666 24334 05000 05000 00888 00888 24334 00666 Comparison in terms of displacements at node 1 with a commercial finite element code gives the displacements listed in Table 91 93 Second Grid Problem 147 Table 91 Comparison in terms of displacements and rotations all multiplied by 103 of the grid in problem14m Node 1 Ref Present w 47622 47622 θx 00000 00000 θz 17611 17611 2 3 4 1 20 kN x z 3 2 4 1 4 m 4 m x y z Fig 93 A second grid example problem15m 93 Second Grid Problem The second grid problem is illustrated in Fig93 The grid is built from three elements as illustrated Given E 210 GPa G 84 GPa I 20 105 m4 J 5 105 m4 the MATLAB problem15m computes displacements reactions and stresses MATLAB codes for Finite Element Analysis problem15m AJM Ferreira N Fantuzzi 2019 clear memory clear E modulus of elasticity 148 9 Grids I second moments of area J polar moment of inertia G shear modulus E 210e9 G 84e9 I 20e5 J 5e5 generation of coordinates and connectivities nodeCoordinates 4 4 0 4 0 0 4 0 xx nodeCoordinates1 yy nodeCoordinates2 elementNodes 1 2 3 1 4 1 numberNodes sizenodeCoordinates1 numberElements sizeelementNodes1 GDof global number of degrees of freedom GDof 3numberNodes force vector force zerosGDof1 force1 20e3 stiffness matrix stiffness formStiffnessGridGDofnumberElements elementNodesxxyyEIGJ boundary conditions prescribedDof 412 solution displacements solutionGDofprescribedDofstiffnessforce output displacementsreactions outputDisplacementsReactionsdisplacementsstiffness GDofprescribedDof forces in elements dispforces in elements EF forcesInElementGridnumberElementselementNodes xxyyEIGJdisplacements Results for displacements reactions and forces in elements are listed below Displacements ans 10000 00033 20000 00010 30000 00010 40000 0 50000 0 60000 0 70000 0 80000 0 90000 0 93 Second Grid Problem 149 100000 0 110000 0 120000 0 reactions ans 10e04 00004 10794 00005 01019 00006 31776 00007 01587 00008 04030 00009 04030 00010 10794 00011 31776 00012 01019 forces in elements EF 10e04 10794 01587 10794 01019 00000 01019 11398 05699 31776 10794 01587 10794 01019 00000 01019 31776 14679 11398 Chapter 10 Timoshenko Beams Abstract Unlike the Bernoulli beam formulation the Timoshenko beam formula tion accounts for transverse shear deformation It is therefore capable of modeling thin or thick beams In this chapter we perform the analysis of Timoshenko beams in static bending free vibrations and buckling We present the basic formulation and show how a MATLAB code can accurately solve this problem 101 Introduction Unlike the Bernoulli beam formulation the Timoshenko beam formulation accounts for transverse shear deformation It is therefore capable of modeling thin or thick beams In this chapter we perform the analysis of Timoshenko beams in static bend ing free vibrations and buckling We present the basic formulation and show how a MATLAB code can accurately solve this problem 102 Static Analysis The Timoshenko theory assumes the deformed crosssection planes to remain plane but not normal to the middle axis If the beam lays in the x z plane the displacement field is defined as u1x z t zθyx t u3x z t wx t 101 where u1 and u3 are the axial and transverse displacements of the threedimensional fibers of the beam w and θy denote the kinematic parameters of the theory as constant transverse displacement and rotation of the crosssection plane about a normal to the middle axis x The Editors if applicable and The Authors under exclusive license to Springer Nature Switzerland AG 2020 A J M Ferreira and N Fantuzzi MATLAB Codes for Finite Element Analysis Solid Mechanics and Its Applications 157 httpsdoiorg101007978303047952710 151 152 10 Timoshenko Beams Normal and transverse shear strains are defined as Ou 06 z 102 ax Ox 102 Ou Ou3 Ow we 8 4 103 Yxz az ax y az The strain energy considers both bending and shear contributions 1 1 U oxesdV TrzYxed V 104 2 Jv 2 Jy where the normal stress is obtained by the Hookes law as Oy Ee 105 and the transverse shear stress is obtained as Txz kG xz 106 being G the shear modulus G E 107 21 v and k the shear correction factor This factor is dependent on the crosssection and on the type of problem Generally it is considered as 56 and this value will be used in the computations Considering dV dA dx and integrating in the crosssection we obtain the strain energy in terms of the generalized displacements 1 2 1 2 U 7 EegdVt ez kGydV 2 Jv 2 Jv sf 1 2 4 5 KGa ow 9 a x x 2J Ox 2 Ja Ox 108 Each node of this 2node element considers one transverse displacement w and one rotation as illustrated in Fig 101 Thus the displacement vector is u w we Oy 2 109 102 Static Analysis 153 2 Oy1 Oyo W1 W2 z L Fig 101 Timoshenko beam element degrees of freedom of the twonoded element In opposition to Bernoulli beams here the interpolation of displacements is inde pendent for both w and 6 w Nw 1010 0 NO 1011 So the displacement vector becomes u w 65 1012 where the shape functions are defined as N509 3049 1013 in natural coordinates 1 1 We can now compute the stiffness matrix as u 6 ELNTN dx 6 3 fo Fb x O 1 f eny en Nee e 5 kGA wN OSN Nw NOS dx 1014 aN where N ae Coordinate transformation is applied to have the integrals in natural x coordinates as 1 Ely U U U 6 NN adé 06 2 J y if 1 Al kGA wN 0SN Nw N adé 1015 2 I41 a a 154 10 Timoshenko Beams Uy is the bending part first term of the stiffness matrix and it can be easily computed from 1015 On the contrary the shear term U should be reordered as lf eget LENT pine W U 5 kealw 0 E NN ba 1 in suf kGA 4 iN Nadéu 1016 2 N 4 1 r 1 LNTN INN 5u kGA Fr NN ad u Finally the strain energy becomes 1 fELfO 0 kGANN aN7N U 3 0 NN a aN N a2NN dg u 1017 Therefore the stiffness matrix for a generic element size 4 x 4 is EL TO 0 kGA NTN aNN Ke E NN da NTN enn dé 1018 The exact integration of the linear element stiffness matrix is strongly not recom mended due to shear locking in thin beams 1 Established suggestion is to the compute the bending stiffness exactly via 2 points Gauss quadrature and the shear part is calculated using reduce integration single point Gauss quadrature in this context 24 This is only one possible solution An alternative is to employ shape functions for the transverse displacement of higher order with respect to the rota tions eg quadratic shape functions for the displacements and linear for the rotations the correspondent stiffness matrix is 5 x 5 Note that superconvergent Timoshenko beam element exact solution in the nodal points and approximated elsewhere is given by cubic polynomials for the transverse displacement and quadratic for the rotation the correspondent stiffness matrix for this case is 7 x 7 To carry out static analysis the external work should be derived We pw dx 1019 by including finite element approximation it leads a a T 1 T We wtf pN dx uf 0 Jo uf 0 eas 1020 a a l so the force vector is given by 102 Static Analysis 155 1 T N f P adé 1021 iL 0 Code problem16m compute the displacements of Timoshenko beams in bending The code considers unitary beam width b so that the second moment of inertia is J bh 12 h12 and elastic modulus E 108 and Poisson ratio v 03 Bc ee ee ee ee ee eee ee ee eee eee eee eee MATLAB codes for Finite Element Analysis problem16m Timoshenko beam in bending AJM Ferreira N Fantuzzi 2019 BB clear memory clear E modulus of elasticity G shear modulus I second moments of area L length of beam thickness thickness of beam E 1e8 poisson 030 L 1 thickness 0001 I thickness312 EI EI kapa 56 P 1 uniform pressure constitutive matrix G E21poisson C EI 0 0 kapathicknessG mesh numberElements 40 nodeCoordinates linspace0LnumberElements1 xx nodeCoordinates elementNodes zeros sizenodeCoordinates212 for i 1sizenodeCoordinates 2 1 elementNodes i1 i elementNodes i 2 i1 end generation of coordinates and connectivities numberNodes sizexx1 GDof global number of degrees of freedom GDof 2numberNodes stiffness matrix and force vector stiffness force formStiffnessMassTimoshenkoBeamGDofnumberElements elementNodes numberNodesxxCP11Ithickness boundary conditions simplysupported at both ends fixedNodew 1 numberNodes fixedNodeTX 156 10 Timoshenko Beams boundary conditions clamped at both ends fixedNodeW 1 numberNodes fixedNodeTX fixedNodeW boundary conditions cantilever fixedNodeW 1 fixedNodeTX 1 prescribedDof fixedNodeW fixedNodeTXnumberNodes solution displacements solutionGDofprescribedDofstiffnessforce output displacementsreactions outputDisplacementsReactionsdisplacementsstiffness GDofprescribedDof U displacements ws 1numberNodes max displacement dispmax displacement minUws The code calls one function formStiffnessMassTimoshenkoBeamm which com putes the stiffness matrix the force vector and the mass matrix of the 2node Timo shenko beam the computation of the mass matrix relevant for free vibrations will be discussed later in this chapter function stiffnessforcemass formStiffnessMassTimoshenkoBeamGDofnumberElements elementNodesnumberNodesxxCPrhoIthickness computation of stiffness mass matrices and force vector for Timoshenko beam element stiffness zerosGDof mass zerosGDof force zerosGDof1 2x2 Gauss quadrature gaussLocations 05773502691896260577350269189626 gaussWeights ones21 bending contribution for matrices for e 1numberElements indice elementNodese elementDof indice indicenumberNodes indiceMass indicenumberNodes ndof lengthindice lengthelement xxindice2xxindice1 detJacobian lengthelement2 invJacobian1detJacobian for q 1sizegaussWeights1 pt gaussLocationsq shapenaturalDerivatives shapeFunctionL2pt1 102 Static Analysis 157 Xderivatives naturalDerivativesinvJacobian B matrix B zeros22ndof B1ndof12ndof Xderivatives stiffness matrix stiffnesselementDofelementDof stiffnesselementDofelementDof BBgaussWeightsqdetJacobianC11 force vector forceindice forceindice shapePdetJacobiangaussWeightsq mass matrix massindiceMassindiceMass massindiceMassindiceMass shapeshapegaussWeightsqIrhodetJacobian massindiceindice massindiceindice shapeshape gaussWeightsqthicknessrhodetJacobian end end shear contribution for the matrices gaussLocations 0 gaussWeights 2 for e 1numberElements indice elementNodese elementDof indice indicenumberNodes ndof lengthindice lengthelement xxindice2xxindice1 detJ0 lengthelement2 invJ0 1detJ0 for q 1sizegaussWeights1 pt gaussLocationsq shapenaturalDerivatives shapeFunctionL2pt1 Xderivatives naturalDerivativesinvJacobian B B zeros22ndof B21ndof Xderivatives B2ndof12ndof shape stiffness matrix stiffnesselementDofelementDof stiffnesselementDofelementDof BBgaussWeightsqdetJacobianC22 end end end Timoshenko codes also call function shapeFunctionL2m which computes shape functions and derivatives with respect to ξ see Sect35 for further details Distributed load p 1 is uniform The code is ready for simplysupported clamped conditions at both ends or cantilever boundary configurations The user can easily introduce new essential boundary conditions A simplysupported Timoshenko beam 158 10 Timoshenko Beams Zz Ww p1 xu L1 Fig 102 Simplysupported Timoshenko problem under uniform load problem16m with reference symbols and geometry is depicted in Fig 102 Timoshenko model is able to analyze both thick or thin beams The present code is compared with exact solutions based on assumed first order shear deformation theory 5 The analytical solution for simplysupported SS Tim oshenko beam is x PLA x 2x3 4 x4 4 PL x x3 1022 x 2 w dad L 13 14 28 LB being S kGA the shear stiffness and D a the flexural stiffness The analytical solution for cantilever CF Timoshenko beam is x PL 62 4x3 x4 PL 5x x 1023 ea PL B14 os VL Pe The maximum displacements for simplysupported SS Bernoulli beam is 5 pLt max 1024 max 384 El ao and for cantilever beam is 1 pL Wmax Spy 1025 8 EI In Table 101 we compare the present solution obtained by MATLAB code and the exact solutions by previous equations 5 for the maximum displacement of the given structures We consider 40 elements and analyze various hL ratios From the table is clear that deflections of the cantilever beam do not depend on the shear deformation because the Bernoulli and Timoshenko solutions coincide 103 Free Vibrations 159 Table 101 Comparation of maximum displacement for Timoshenko beam hL Exact 5 Present SS Bernoulli 15584 0001 15625 15609 001 15631 103 15613 103 01 16210 10 15999 106 CF Bernoulli 150 0001 150 150 001 00150 00150 01 15156 10 15156 109 103 Free Vibrations The kinetic energy considers two parts one related with translations 9A and one related to rotations rotary inertia p in the form 1 a 1 a K pAwdx pl0edx 1026 2 a 2 a By applying the aforementioned linear interpolation 6 and by introducing the coordinate transformation in order to evaluate the integrals in natural coordinates the kinetic energy becomes 1 1 K sw pANN ad w 50 plyNN adé 6 1027 1 l By collecting the terms of the displacement vector u it leads Kalu Aa NNO plya oo déu 1028 3 fe 0 00NN The element mass matrix can be written as mea PaaNN 0 dé 1029 sy 0 plyaNN The stiffness matrix of the twonoded element has been carried out in the previous Sect 102 The first problem considers a thin ZL 1 h 0001 cantilever beam The non dimensional natural frequencies are given by 160 10 Timoshenko Beams Table 102 Comparing natural frequencies for cantilever isotropic thin beam using code prob lem16vibrationsm Mode Present Exact 6 1 elem 2 elem 5 elem 10 elem 50 elem 1 3464 3592 3532 3520 3516 3516 2 5883488 40407 24313 22583 22056 22035 pA owL 1030 El Results for this clamped thin beam are presented in Table 102 Results are in excellent agreement with exact solution 6 Bc ee ee ee ee ee eee ee ee eee eee eee eee MATLAB codes for Finite Element Analysis probleml6vibrationsm Timoshenko beam in free vibrations AJM Ferreira N Fantuzzi 2019 B clear memory clear E modulus of elasticity G shear modulus I second moments of area L length of beam thickness thickness of beam E 1e8 poisson 030 L 1 thickness 0001 rho 1 I thickness312 EI EI A 1thickness kapa 56 constitutive matrix G E21poisson C EI 0 0 kapathicknessG mesh numberElements 50 nodeCoordinates linspace0LnumberElements1 xx nodeCoordinates x xx elementNodes zeros sizenodeCoordinates212 for i 1sizenodeCoordinates 2 1 elementNodesi1 i elementNodesi2 i1 end generation of coordinates and connectivities numberNodes sizexx1 GDof global number of degrees of freedom GDof 2numberNodes 103 Free Vibrations 161 computation of the system stiffness force mass stiffnessforcemass formStiffnessMassTimoshenkoBeamGDofnumberElements elementNodes numberNodesxxC0rhoIthickness boundary conditions simplysupported at both ends fixedNodew 1 numberNodes fixedNodeTX boundary conditions clamped at both ends fixedNodew 1 numberNodes fixedNodeTX fixedNodeW boundary conditions cantilever fixedNodew 1 fixedNodeTX 1 prescribedDof fixedNodeW fixedNodeTXnumberNodes free vibration problem modeseigenvalues eigenvalueGDofprescribedDof stiffnessmass0 omega sqrt eigenvalues LLsqrt rhoAET display first 2 dimensionless frequencies omega 12 drawing mesh and deformed shape modeNumber 4 V1 modes1modeNumber drawing eigenmodes figure drawEigenmodes1D modeNumber numberNodes V1x Fig 103 illustrates the first four modes of vibration for this beam with vy 03 as computed by code problem16vibrationsm using 40 elements This code calls function formStiffnessMassTimoshenkom already presented in this chapter The code calls function drawEigenmodes1Dm which draws eigenmodes for this case The next example computes natural frequencies of a system suggested by Lee and Schultz 7 The shear correction factor is taken as 56 We consider beams clamped or simplysupported at the ends The nondimensional frequencies are listed according to L which is due to the exact natural frequency calculation by EI EI w L 1031 4 pA pAL where XL takes the following forms according to the boundary conditions of the beam Cantilever beam L 12n 12 simply supported beam L nz clamped beam L 72n 12 where n represents the mode number Results are listed in Tables 103 and 104 and show excellent agreement with those of Lee and Schultz 7 Figures 104 and 105 illustrate the modes of vibration for beams clamped or simplysupported at both ends with v 03 using 40 twonoded elements 162 10 Timoshenko Beams 0 01 02 03 04 05 06 07 08 09 1 50 0 0 01 02 03 04 05 06 07 08 09 1 60 40 20 0 20 40 0 01 02 03 04 05 06 07 08 09 1 60 40 20 0 20 40 0 01 02 03 04 05 06 07 08 09 1 60 40 20 0 20 40 Fig 103 First 4 modes of vibration for a cantilever beam Table 103 Nondimensional natural frequencies λnL for a Timoshenko beam clamped at both ends ν 03 k 56 number of elements N 40 Mode Ref 7 hL 0002 001 01 1 473004 47345 47330 45835 2 785320 78736 78675 73468 3 109956 110504 110351 98924 4 141372 142526 142218 122118 5 172788 174888 174342 143386 6 204204 207670 206783 163046 7 235619 240955 239600 181375 8 267035 274833 272857 198593 9 298451 309398 306616 214875 10 329867 344748 340944 230358 11 361283 380993 375907 245141 12 392699 418249 411574 259179 13 424115 456642 448016 262929 14 455531 496312 485306 268419 15 486947 537410 523517 273449 103 Free Vibrations 163 Table 104 Nondimensional natural frequencies λnL for a Timoshenko beam simplysupported at both ends ν 03 k 56 number of elements N 40 Mode Ref 7 hL 0002 001 01 1 314159 31428 31425 31169 2 628319 62928 62908 60993 3 942478 94573 94503 88668 4 125664 126437 126271 113984 5 157080 158596 158267 137089 6 188496 191127 190552 158266 7 219911 224113 223186 177811 8 251327 257638 256231 195991 9 282743 291793 289749 213030 10 314159 326672 323806 229117 11 345575 362379 358467 244404 12 376991 399022 393803 259017 13 408407 436721 429883 260647 14 439823 475605 466780 262782 15 471239 515816 504566 268779 0 01 02 03 04 05 06 07 08 09 1 0 5 0 01 02 03 04 05 06 07 08 09 1 5 0 5 0 01 02 03 04 05 06 07 08 09 1 5 0 5 0 01 02 03 04 05 06 07 08 09 1 5 0 5 Fig 104 First 4 modes of vibration for a beam clamped at both ends 164 10 Timoshenko Beams 0 01 02 03 04 05 06 07 08 09 1 4 2 0 0 01 02 03 04 05 06 07 08 09 1 5 0 5 0 01 02 03 04 05 06 07 08 09 1 5 0 5 0 01 02 03 04 05 06 07 08 09 1 5 0 5 Fig 105 First 4 modes of vibration for a beam simplysupported at both ends Code problem16vibrationsSchultzm considers a number of boundary condi tions the user should change according to the problem MATLAB codes for Finite Element Analysis problem16vibrationsSchultzm Timoshenko beam in free vibrations LeeSchultz problem AJM Ferreira N Fantuzzi 2019 clear memory clear E modulus of elasticity G shear modulus I second moments of area L length of beam thickness thickness of beam E 10e7 poisson 030 L 1 thickness 0002 rho 1 I thicknessˆ312 EI EI A 1thickness kapa 56 constitutive matrix G E21poisson C EI 0 0 kapathicknessG mesh numberElements 40 104 Buckling Analysis 165 nodeCoordinates linspace0LnumberElements1 xxnodeCoordinates x xx elementNodes zeros sizenodeCoordinates212 for i 1sizenodeCoordinates 2 1 elementNodesi1 i elementNodesi2 i1 end generation of coordinates and connectivities numberNodes sizexx1 GDof global number of degrees of freedom GDof 2numberNodes computation of the system stiffness force mass stiffnessforcemass formStiffnessMassTimoshenkoBeamGDofnumberElements elementNodes numberNodesxxC0rhoIthickness boundary conditions CC fixedNodeW findxxminnodeCoordinates xxmax nodeCoordinates fixedNodeTX fixedNodew prescribedDof fixedNodeW fixedNodeTXnumberNodes boundary conditions SS fixedNodeW findxxminnodeCoordinates xxmaxnodeCoordinates prescribedDof fixedNodeW free vibration problem modeseigenvalues eigenvalueGDofprescribedDof stiffnessmass0 omega sqrt eigenvalues sqrt rhoAL4ET display first 15 dimensionless frequencies sqrt omega115 drawing mesh and deformed shape modeNumber 4 V1 modes1modeNumber drawing eigenmodes figure drawEigenmodes1D modeNumber numberNodes V1 x 104 Buckling Analysis The work energy due to the applied compression load is 2 1 Ow W P dx 1032 2 Ja Ox 166 10 Timoshenko Beams Table 105 Critical loads using 40 elements Lh Ss cc Exact 8 Present Exact 8 Present 10 80138 80218 29766 29877 100 8223 8231 32864 32999 1000 00082 00082 00329 00330 The finite element approximation is applied 1 eT P IT Wy e W zw NNadéw 1033 2 1 a The relation is written in terms of the displacement vector 1 PNN0 Wy 5u el 0 0 déu 1034 Thus the geometric stiffness matrix is 1 P N7N 0 K 0 0 as 1035 The buckling analysis of Timoshenko beams considers the solution of the eigen problem K AK xX0 1036 where are the critical loads and X the buckling modes We now consider simply supported SS and clamped CC beams The exact solution 8 is p Tt 1 4 El 1037 Lee LigkGA where Leg is the effective beam length For pinnedpinned beams Leg L and for fixedfixed beams Leg L2 Table 105 shows the buckling loads for SS and CC Timoshenko beams Results are in excellent agreement with those of Bazant and Cedolin 8 Code problem16Bucklingm is listed below and calls function formStiffness BucklingTimoshenkoBeamm to compute the stiffness matrix and the geometric stiffness matrix 104 Buckling Analysis 167 MATLAB codes for Finite Element Analysis problem16Bucklingm Timoshenko beam under buckling loads P AJM Ferreira N Fantuzzi 2019 clear memory clear E modulus of elasticity G shear modulus I second moments of area L length of beam thickness thickness of beam E 10e6 poisson 0333 L 1 thickness 01 I thicknessˆ312 EI EI A 1thickness kapa 56 constitutive matrix G E21poisson C EI 0 0 kapathicknessG mesh numberElements 40 nodeCoordinates linspace0LnumberElements1 xx nodeCoordinates x xx elementNodes zerossizenodeCoordinates212 for i 1sizenodeCoordinates21 elementNodesi1 i elementNodesi2 i1 end generation of coordinates and connectivities numberNodes sizexx1 GDof global number of degrees of freedom GDof 2numberNodes computation of the system stiffness Kg stiffnessKg formStiffnessBucklingTimoshenkoBeamGDofnumberElements elementNodesnumberNodesxxCIthickness boundary conditions CC fixedNodeW findxxminnodeCoordinates xxmaxnodeCoordinates fixedNodeTX fixedNodeW prescribedDof fixedNodeW fixedNodeTXnumberNodes boundary conditions SS fixedNodeW findxxminnodeCoordinates xxmaxnodeCoordinates prescribedDof fixedNodeW buckling problem modeseigenvalues eigenvalueGDofprescribedDof stiffnessKg0 168 10 Timoshenko Beams reordering eigenvalues eigenvaluesii sorteigenvalues modes modesii Bazant Cedolin solution for SS and CC PcrSS pipiEILˆ211pipiEILLkapaGA PcrCC pipiEIL2ˆ211pipiEILL4kapaGA drawing mesh and deformed shape modeNumber 4 V1 modes1modeNumber drawing eigenmodes figure drawEigenmodes1DmodeNumbernumberNodesV1x Code formStiffnessBucklingTimoshenkoBeamm follows next function stiffnessKg formStiffnessBucklingTimoshenkoBeamGDofnumberElements elementNodesnumberNodesxxCIthickness computation of stiffness matrix and geometric stiffness for Timoshenko beam element stiffness zerosGDof Kg zerosGDof 2x2 Gauss quadrature gaussLocations 05773502691896260577350269189626 gaussWeights ones21 bending contribution for stiffness matrix for e 1numberElements indice elementNodese elementDof indice indicenumberNodes ndof lengthindice lengthelement xxindice2xxindice1 detJacobian lengthelement2invJacobian1detJacobian for q 1sizegaussWeights1 pt gaussLocationsq shapenaturalDerivatives shapeFunctionL2pt1 Xderivatives naturalDerivativesinvJacobian B matrix B zeros22ndof B1ndof12ndof Xderivatives K stiffnesselementDofelementDof stiffnesselementDofelementDof BBgaussWeightsqdetJacobianC11 Kgindiceindice Kgindiceindice XderivativesXderivativesgaussWeightsqdetJacobian end end shear contribution for stiffness matrix 104 Buckling Analysis 169 gaussLocations 0 gaussWeights 2 for e 1numberElements indice elementNodese elementDof indice indicenumberNodes ndof lengthindice lengthelement xxindice2xxindice1 detJ0 lengthelement2 invJ01detJ0 for q 1sizegaussWeights1 pt gaussLocationsq shapenaturalDerivatives shapeFunctionL2pt1 Xderivatives naturalDerivativesinvJacobian B B zeros22ndof B21ndof Xderivatives B2ndof12ndof shape K stiffnesselementDofelementDof stiffnesselementDofelementDof BBgaussWeightsqdetJacobianC22 end end end Figures106 and 107 illustrate the first four buckling loads for simplysupported and clamped Timoshenko beams respectively Both beams consider Lh 10 and ν 03 0 01 02 03 04 05 06 07 08 09 1 04 02 0 0 01 02 03 04 05 06 07 08 09 1 02 0 02 0 01 02 03 04 05 06 07 08 09 1 01 0 01 0 01 02 03 04 05 06 07 08 09 1 01 0 01 Fig 106 First 4 modes of buckling for simply supported Timoshenko beam 170 10 Timoshenko Beams 0 01 02 03 04 05 06 07 08 09 1 04 02 0 0 01 02 03 04 05 06 07 08 09 1 02 0 02 0 01 02 03 04 05 06 07 08 09 1 02 01 0 0 01 02 03 04 05 06 07 08 09 1 02 0 02 Fig 107 First 4 modes of buckling for clamped Timoshenko beam References 1 JN Reddy An Introduction to the Finite Element Method 3rd edn McGrawHill International Editions New York 2005 2 KJ Bathe Finite Element Procedures in Engineering Analysis Prentice Hall Upper Saddle River 1982 3 E Onate Calculo de estruturas por el metodo de elementos finitos CIMNE Barcelona 1995 4 RD Cook DS Malkus ME Plesha RJ Witt Concepts and Applications of Finite Element Analysis Wiley New York 2002 5 CM Wang JN Reddy KH Lee Shear Deformable Beams and Plates Elsevier Amsterdam 2000 6 M Petyt Introduction to Finite Element Vibration Analysis Cambridge University Press Cam bridge 1990 7 J Lee WW Schultz Eigenvalue analysis of timoshenko beams and axisymmetric mindlin plates by the pseudospectral method J Sound Vib 26934 609621 2004 8 ZP Bazant L Cedolin Stability of Structures Oxford University Press New York 1991 Chapter 11 Plane Stress Abstract This chapter deals with the static and dynamic analysis of 2D solids par ticularly in plane stress Plane stress analysis refers to problems where the thickness is quite small when compared to other dimensions in the reference plane xy The loads and boundary conditions are applied at the reference or middle plane of the structure In this chapter we consider isotropic homogeneous materials fournode Q4 eightnode Q8 and ninenode Q9 quadrilateral elements 111 Introduction This chapter deals with the static and dynamic analysis of 2D solids particularly in plane stress Plane stress analysis refers to problems where the thickness is quite small when compared to other dimensions in the reference plane x y The loads and boundary conditions are applied at the reference or middle plane of the struc ture Displacements are computed at the reference plane The stresses related with z coordinates are assumed to be very small and not considered in the formulation The plane strain is analogous to plane stress where the solid is considered as indefinitely long Between the two problems only the constitutive matrix is different therefore for the sake of brevity plane strain is not presented here In this chapter we consider isotropic homogeneous materials fournode Q4 eightnode Q8 and ninenode Q9 quadrilateral elements The problem is defined in a convex domain Ω bounded by Γ as illustrated in Fig111 112 Displacements Strains and Stresses The plane stress problem considers two global displacements u and v defined in global directions x and y respectively The Editors if applicable and The Authors under exclusive license to Springer Nature Switzerland AG 2020 A J M Ferreira and N Fantuzzi MATLAB Codes for Finite Element Analysis Solid Mechanics and Its Applications 157 httpsdoiorg101007978303047952711 171 172 11 Plane Stress Fig 111 Plane stress r illustration of the 2D domain 2 and its boundary I y v ux y ux y 111 x y han 111 Strains are obtained by derivation of displacements Ou Ox x Ov eX y dy 112 Yxy Ou 4 Ov Oy Ox By assuming a linear elastic material we obtain stresses as E VE 0 11 11 Ox Ex o o Ce vE E 0 y 113 Txy 1 py 1 py Yxy 0 0 G where F is the modulus of elasticity v the Poissons coefficient and G x 4y V is the shear modulus C the elastic constitutive matrix The static equilibrium equations are defined as Oo OTxy 445b0 114 Ox Oy 114 OTxy Ooy z 54b0 115 Ox Oy 115 where b b are body forces 113 Boundary Conditions 173 113 Boundary Conditions Essential or displacement boundary conditions are applied on the boundary displace ment part I as uu 116 Natural or force boundary conditions are applied on so that ot 117 where t is the surface traction per unit area and o the normal vector to the plate boundary If necessary o can be computed according to Cartesian components of stress by OxNy Txyn ny On Ox XEOX xyly x y on ee 4 a 0 ny Oy 118 Try where n and n are the Cartesian components of the unit normal vectors to the surface 114 Hamilton Principle The total potential energy can be defined as MWUVe 119 where U is the elastic strain deformation 1 T 1 T U heodQ he Ced2 1110 2 Je 2 Je The energy produced by the external domain and boundary forces is given by VeWe nu de hut dl 1111 2 where is the unitary thickness of the 2D domain The kinetic energy is defined as 1 T 1 Tos K pwudVh pwudQ 1112 2Jy 2 Jo 174 11 Plane Stress The Hamilton principle reads bh th hpoa adQ hde Ce dQ f hou bd f houtdr 0 Q Q 2 r 1113 115 Finite Element Discretization Given a domain denoted by 2 and a boundary denoted by I the nnoded finite element displacement vector is defined by 2n degrees of freedom we ur ur etn V1 2 2 1114 where n is due to the number of grid nodes per element used in the approximation For instance for Q4 element the finite element displacement vector is uo u ua Uz Ug Vy V2 V3 vs 1115 116 Interpolation of Displacements The displacement vector in each element is interpolated by the nodal displacements as u Niu v Nfv 1116 il il where N denote the element shape functions This can also be expressed in matrix form as Ni NyNp 0 0 0 y w 4 0 0 Meng weft Nu 1117 where N is the matrix of the shape functions which is needed for computing the stress at each element and for evaluating boundary tractions as shown below The strain vector can be obtained by derivation of the displacements according to Eq 112 as 116 Interpolation of Displacements 175 ON ONS ONS 0 0 O Ox Ox Ox ON ONS ONS 0 0 0 S u Bu dy Oy ay u Bu 1118 ON ONS ON ONY ONS ONS Oy Oy Oy Ox Ox Ox where B is the straindisplacement matrix This matrix is needed for computation of the stiffness matrix 117 Element Energy The total potential energy can be defined at each element by MN UV 1119 where the strain energy is defined as e 1 T e 1 eT T ewe Uf he CedQ u AB CB dQQu 1120 2 Jae 2 ne where is the plate thickness and the element stiffness matrix is obtained as K hB CB d 1121 2 The energy produced external work by body forces and boundary tractions is given by ViWE ut hNb dQ ut f ANt dr 1122 Qe re where the vector of nodal forces is obtained as f AN b dQ ANt dr 1123 Qe re We can introduce these expressions into the total potential energy as e 1 eT ize ye eT ge l 5u Ku uf 1124 The kinetic energy defines the element mass matrix as 176 11 Plane Stress 1 eT T ewe 1 e we Ku hpNN dQQuf uMu 1125 2 Qe 2 where the mass matrix is M hpN N dx 1126 2 Thus the Hamiltons Principle can be used to carry out dynamic equilibrium equa tions for the present 2D solid 1171 Quadrilateral Element Q4 We consider a quadrilateral element illustrated in Fig 112 The element is defined by 4 nodes in natural coordinates 7 The coordinates are interpolated as 4 4 x oMx y Nii 1127 il il where N are the Lagrange shape functions given by 1 ME hi ga 9d 0 1128 1 N2 7 nQlim gad 9d 7 1129 1 1 4 3 4 3 g pe g 1 2 1 2 Fig 112 Quadrilateral Q4 element in natural coordinates with single point integration 7 0 0 and two points integration 7 1 V3 1 V3 117 Element Energy 177 1 N3 7 1hQla qu 6 7 1130 1 NaS 4 Qb gad 9d 0 1131 Note that this is the geometrical approximation of the domain This is an important aspect because the same shape functions are used here to approximate the unknown field u and v as well as geometry This finite element is known as isoparametric How ever it will be discussed in the following that it is possible to implement geometry approximation and unknown field variables with different shape functions Displacements are interpolated as 4 4 u So Nui vo Nv 1132 il il where uv are the displacements at any point in the element and u v for i 1 23 4 are the nodal displacements 0 O Derivatives in natural coordinates can be found as Og On 0 Ox Oy 0 ag d BE dx 1133 O Ox Oy oO On On On Oy In matrix form we can write relations 1133 as 0 0 jJ 1134 0g J Ox where J is the Jacobian operator relating natural and global coordinates The deriva tives with respect to the global coordinates can be found as 0 0 jy 1135 Ox OE Note that in very distorted elements the Jacobian inverse J may not exist In other words matrix inversion becomes inaccurate because of matrix high conditioning number Recalling the definition of the stiffness matrix for the generic element d2 det Jddn where det J is the determinant of the Jacobian matrix The stiffness matrix is then obtained by 178 11 Plane Stress 1 pl 1 pl K hB CB det Jdédn h Fdédn 1136 1J1 1J1 where F B CB det J Note that B by definition 1118 depends on the Cartesian coordinates x y but the element stiffness matrix 1 136 has to be integrated in natural coordinates 77 The integral in the stiffness matrix is computed numerically by Gauss quadrature in two dimensions Each integral is transformed as a weighted sum by the product of weights and value of the function at the given nodes 1 F has to be written as a function of the natural points 7 with coordinate transformation Integration points and integration weights depend on the type of integration the user wishes to perform In the 4node element we can use a 2 x 2 numerical integration for exact integration and a single point integration for the reduced integration Thus the element stiffness matrix with exact integration is 1 1 2 2 Ke nf B CB det J dédn h BCB det Jujw 1137 lv1 i1 jl All Gauss points have unitary values in this integration rule The element stiffness matrix with reduced integration matrices are evaluated in 0 0 is 1 pl K nf B CB det J dédn 4hB CB det J 1138 1J1 recalling that for reduced integration the weight is w 2 Integration points are depicted for the present element in Fig 112 The force vector due to body forces can be carried out using the aforementioned procedure as 1 1 2 2 f n Nb detJ dédnh 5 Nb det Jww 1139 lel il jl for the full integration case For the sake of simplicity natural boundary conditions in the present book are derived upon integration of the boundary stress applied According to Eq 1111 the boundary tractions should be computed and applied at nodes as f Nn6dr 1140 re where linear shape functions are needed on each element side for computing the stress because Q4 element is considered Using the local coordinate s starting from corner node on each edge these shape functions take the form 117 Element Energy 179 s s viG1 ynls 1141 a a where a is the edge length Thus for an edge parallel to the axis y and a tension along x constant p boundary forces f f f become no a pa f vipas 1142 Ae pa fe vspas 0 The element mass matrix for Q4 element computed with full 2 x 2 Gauss quadra ture takes the form 1 pl 2 2 M hpNN det J ddn Y Y hpNN det Jwiw 1143 vlesl i1 jl whereas reduced single point integration leads matrices are evaluated in 0 0 1 pl M hpNN det J ddn 4hpNN det J 1144 1J1 1172 Quadrilateral Elements Q8 and Q9 For the sake of conciseness the complete formulation for 8 nodes and 9 nodes ele ments is not repeated but only main expressions are reported The reader can refer to the previous section for missing details in the present one The Lagrange shape functions N for 8 node elements are given by Nif7 0251 Hd mdém N2E 0251 md N3E 0251 d md NaE 0251 Hdmdém 4s Nsm 051 1 9 No 051 9 N7Em 0511 1 0 Nsm 051 1 9 The shape function for Q9 are 180 11 Plane Stress 1 1 4 7 3 4 7 3 a pe 1 5 2 1 5 2 Fig 113 Quadrilateral Q8 element in natural coordinates with two points reduced integration 9 1 V3 1 3 and quadrilateral Q9 element in natural coordinates with three points integration 7 0 0 7 35 35 NiE 9 025n Din 1 N29 0258nE Din 1D N3 9 0258nE Dn VD Na 9 025 Din VD Ns 051 n 1 1146 No 05E DU 17 N7E0 050 n Ns Em 05E 1 17 No 11 17 The element stiffness matrix Pq K hSSBCBedet Jujw 1147 i1 jl where pq are the number of points for the integration 3 x 3 full integration and 2 x 2 reduced integration see Fig 113 for details Generally full integration is used for the stiffness matrix computation and reduced integration points 2 x 2 are used for the stress recovery in the postcomputation 1 Body forces for Q8 and Q9 elements can be easily carried out following the pro cedure discussed for Q4 On the contrary the procedure for getting natural boundary 117 Element Energy 181 conditions for Q8 and Q9 is discussed below For such elements quadratic shape functions are needed on each element side using the local coordinate s these shape functions take the form Ss 2s tis 15 1 a a das 41 1148 a a Ss Ss uas 125 a a where a is the edge length Thus for an edge parallel to the axis y and a tension along x constant p boundary forces f f f f become no a pa fi vip ds 6 0 Ae 4 2pa fy Yop ds 1149 0 3 no a pa fi u3p ds 6 0 The element mass matrix for Q8 and Q9 element takes the form 1 pl Pq M hpNN det J ddn YY hpNN det Jwiw 1150 vtesl i1 jl where pq are the number of points for the integration 3 x 3 full integration and 2 x 2 reduced integration 118 Postprocessing Postprocessing technique is fundamental for the stress recovery since the formu lation is based on displacements Stresses should be recovered in order to perform structural design Stresses are carried out from computed displacements thus they are derived quantities The accuracy of such quantities is generally lower than pri mary variables displacements For an accuracy of the displacements of 1 the stresses might be accurate at 10 or lower at the boundaries 1 To calculate the strain and stresses a loop over all the elements is performed For the eth element the strains can be defined as e Bu 1151 182 11 Plane Stress and the stresses are σ Cϵ CBue 1152 The stresses are evaluated in the integration points of the elements It is a good practice to carry out stresses using 2 2 Gauss integration for all elements Q4 Q8 and Q9 Note that such post computation does not involve Gauss integration this solution is used for the practical way of computing stresses in 2D finite elements In the following element nodal point stresses are evaluated Such stresses are not generally the same among adjacent elements because stresses are not required to be continuous in the finite element method In the applications it is of interest to evaluate and report these stresses at the element nodal points located on the corners and possibly midpoints of the element These are called element nodal point stresses Therefore stress averaging is applied in order to improve stress accuracy Three approaches can be followed in order to recover stresses in finite elements 1 Direct evaluation stresses are carried out directly by substituting element nodal locations in shape functions 2 Stress extrapolation stresses are evaluated at integration points and an extrap olation technique is used to carry out stresses at the nodal points 3 Patch recovery stress at a nodal point is assumed to be a polynomial expansion of the same complete order of the shape functions used over an element patch surrounding the current node The first approach wont be discussed because it is straightforward The second one is given below 1181 Stress Extrapolation Consider ξ η as natural coordinates of the current parent element and ˆξ ˆη are the coordinates of the same element defined by the four integration points of the 2 2 integration the relationship between these sets of coordinates is ξ η ˆξ ˆη 3 or ˆξ ˆη ξ η 3 1153 Stresses σx σy and τxy at any point P termed σP can be obtained as classical interpolation using shape functions which are evaluated at the coordinates of point P as 118 Postprocessing 183 4 9G1 a a a a a or Moc MiE A Na MEM NEM PE CL54 il OG4 where og are the stresses evaluated at the integration points and N 7 are the shape functions evaluated in the reference system of the integration points The extrapolation can be applied by replacing each corner coordinates in the integration points natural system using relations 1153 For instance the first shape function N at the first integration point 3 3 becomes 1 M 7 J31 V3n 1155 such shape function evaluated at the four corners of a parent Q4 element leads 1 053 05 1 053 05 1156 this vector represents the extrapolation of the stress og at the four corner points of the parent Q4 element If this operation is done for each integration point the following relationship applies oO 140573 05 1053 05 Tcl 02 05 14053 05 10573 ca 1157 03 105V3 05 14053 05 0G3 4 05 1053 05 140573 L7c4 The same procedure shown in 1154 can be extended to Q8 and Q9 elements shape functions do not change because stresses are evaluated using 2 x 2 integration grid For instance for Q8 elements it leads o1 140573 05 1053 05 oD 05 14053 05 1053 03 1053 05 14053 05 O61 4 05 110573 05 140573 og 1158 Os 1 34 V34 1 V34 34 203 06 1 V34 1 V34 1 V34 Cl V34 Loe o7 1 34 1 34 1 V34 1 34 78 1 34 1 34 1 V34 1 V34 and for Q9 becomes 184 11 Plane Stress σ1 σ2 σ3 σ4 σ5 σ6 σ7 σ8 σ9 1 05 3 05 1 05 3 05 05 1 05 3 05 1 05 3 1 05 3 05 1 05 3 05 05 1 05 3 05 1 05 3 1 34 1 34 1 34 1 34 1 34 1 34 1 34 1 34 1 34 1 34 1 34 1 34 1 34 1 34 1 34 1 34 025 025 025 025 σG1 σG2 σG3 σG4 1159 1182 Interelement Averaging The previous approach shows jumps among elements Therefore a general smooth ing should be applied for the whole mesh Obviously the aforementioned stresses should be averaged at nodal stresses in the following two ways 1 Unweighted averaging the same weight is assigned to all elements that share a node 2 Weighted averaging a weight is assigned to each element according to the stress component element geometry and when applicable element type Stress extrapolation and unweighted averaging are shown below for the simple case of cantilever wall beam 119 Plate in Traction We consider a thin plate under uniform traction forces at its extremes The problem is illustrated in Fig114 Isotropic material properties are given as E 108 Poisson ratio ν 03 and applied pressure p 106 Plate length and width are indicated in Fig114 as L 10 2 10 p p Fig 114 Thin plate in traction problem17m 119 Plate in Traction 185 and W 2 Due to the symmetry of the problem only onefourth of the plate is modeled Analytically the maximum displacement expected should be umax pL E A 005 1160 Three MATLAB codes are provided for this problem using Q4 problem17m Q8 problem17am and Q9 problem17bm elements MATLAB codes for Finite Element Analysis problem17m 2D problem thin plate in tension using Q4 elements AJM Ferreira N Fantuzzi 2019 clear memory clear close all material properties E 10e7 poisson 030 matrix C C E1poissonˆ21 poisson 0poisson 1 00 0 1poisson2 load P 1e6 mesh generation Lx 5 Ly 1 numberElementsX 10 numberElementsY 5 numberElements numberElementsXnumberElementsY nodeCoordinates elementNodes rectangularMeshLxLynumberElementsXnumberElementsYQ4 xx nodeCoordinates1 yy nodeCoordinates2 figure drawingMeshnodeCoordinateselementNodesQ4 axis equal numberNodes sizexx1 GDof global number of degrees of freedom GDof 2numberNodes calculation of the system stiffness matrix stiffness formStiffnessMass2DGDofnumberElements elementNodesnumberNodesnodeCoordinatesC11Q4complete boundary conditions fixedNodeX findnodeCoordinates10 fixed in XX fixedNodeY findnodeCoordinates20 fixed in YY prescribedDof fixedNodeX fixedNodeYnumberNodes force vector distributed load applied at xxLx force zerosGDof1 rightBord findnodeCoordinates1Lx forcerightBord PLynumberElementsY 186 11 Plane Stress forcerightBord1 PLynumberElementsY2 forcerightBordend PLynumberElementsY2 solution displacements solutionGDofprescribedDofstiffnessforce displacements dispDisplacements jj 1GDof format f jj displacements fprintfnode U fprintf3d 128f f UX displacements1numberNodes UY displacementsnumberNodes1GDof scaleFactor 10 deformed shape figure drawingFieldnodeCoordinatesscaleFactorUX UY elementNodesQ4UXU XX hold on drawingMeshnodeCoordinatesscaleFactorUX UY elementNodesQ4 drawingMeshnodeCoordinateselementNodesQ4 colorbar titleDisplacement field ux on deformed shape axis off stresses at nodes stressstrain stresses2DGDofnumberElements elementNodesnumberNodesnodeCoordinatesdisplacements CQ4complete drawing stress fields on deformed shape figure hold on drawingField2nodeCoordinateselementNodes scaleFactorUXUYQ4stress1 axis equal drawingMeshnodeCoordinateselementNodesQ4 colorbar titleStress field sigmaxx on deformed shape axis off MATLAB codes for Finite Element Analysis problem17am 2D problem thin plate in tension using Q8 elements AJM Ferreira N Fantuzzi 2019 clear memory clear close all material properties E 10e7 poisson 030 matrix C C E1poissonˆ21 poisson 0poisson 1 00 0 1poisson2 119 Plate in Traction 187 load P 1e6 mesh generation Lx 5 Ly 1 numberElementsX 10 numberElementsY 5 numberElements numberElementsXnumberElementsY nodeCoordinates elementNodes rectangularMeshLxLynumberElementsXnumberElementsYQ8 xx nodeCoordinates1 yy nodeCoordinates2 figure drawingMeshnodeCoordinateselementNodesQ8 axis equal numberNodes sizexx1 GDof global number of degrees of freedom GDof 2numberNodes calculation of the system stiffness matrix stiffness formStiffnessMass2DGDofnumberElements elementNodesnumberNodesnodeCoordinatesC11Q8complete boundary conditions fixedNodeX findnodeCoordinates10 fixed in XX fixedNodeY findnodeCoordinates20 fixed in YY prescribedDof fixedNodeX fixedNodeYnumberNodes force vector distributed load applied at xxLx force zerosGDof1 rightBord findnodeCoordinates1Lx forcerightBord12end PLynumberElementsY3 forcerightBord22end PLynumberElementsY23 forcerightBord1 PLynumberElementsY6 forcerightBordend PLynumberElementsY6 solution displacements solutionGDofprescribedDofstiffnessforce displacements dispDisplacements jj 1GDof format f jj displacements fprintfnode U fprintf3d 128f f UX displacements1numberNodes UY displacementsnumberNodes1GDof scaleFactor 10 deformed shape figure drawingFieldnodeCoordinatesscaleFactorUX UY elementNodesQ9UXU XX hold on drawingMeshnodeCoordinatesscaleFactorUX UY 188 11 Plane Stress elementNodesQ9 drawingMeshnodeCoordinateselementNodesQ9 colorbar titleDisplacement field ux on deformed shape axis off stresses at nodes stressstrain stresses2DGDofnumberElements elementNodesnumberNodesnodeCoordinatesdisplacements CQ8complete drawing stress fields on deformed shape figure hold on drawingField2nodeCoordinateselementNodes scaleFactorUXUYQ4stress1 sigma XX axis equal drawingMeshnodeCoordinateselementNodesQ8 colorbar titleStress field sigmaxx on deformed shape axis off MATLAB codes for Finite Element Analysis problem17bm 2D problem thin plate in tension using Q9 elements AJM Ferreira N Fantuzzi 2019 clear memory clear close all material properties E 10e7 poisson 030 matrix C C E1poissonˆ21 poisson 0poisson 1 00 0 1poisson2 load P 1e6 mesh generation Lx 5 Ly 1 numberElementsX 10 numberElementsY 5 numberElements numberElementsXnumberElementsY nodeCoordinates elementNodes rectangularMeshLxLynumberElementsXnumberElementsYQ9 xx nodeCoordinates1 yy nodeCoordinates2 figure drawingMeshnodeCoordinateselementNodesQ9 axis equal 119 Plate in Traction 189 numberNodes sizexx1 GDof global number of degrees of freedom GDof 2numberNodes calculation of the system stiffness matrix stiffness formStiffnessMass2DGDofnumberElements elementNodesnumberNodesnodeCoordinatesC11Q9complete boundary conditions fixedNodeX findnodeCoordinates10 fixed in XX fixedNodeY findnodeCoordinates20 fixed in YY prescribedDof fixedNodeX fixedNodeYnumberNodes force vector distributed load applied at xxLx force zerosGDof1 rightBord findnodeCoordinates1Lx forcerightBord12end PLynumberElementsY3 forcerightBord22end PLynumberElementsY23 forcerightBord1 PLynumberElementsY6 forcerightBordend PLynumberElementsY6 solution displacements solutionGDofprescribedDofstiffnessforce displacements dispDisplacements jj 1GDof format f jj displacements fprintfnode U fprintf3d 128f f UX displacements1numberNodes UY displacementsnumberNodes1GDof scaleFactor 10 deformed shape figure drawingFieldnodeCoordinatesscaleFactorUX UY elementNodesQ9UXU XX hold on drawingMeshnodeCoordinatesscaleFactorUX UY elementNodesQ9 drawingMeshnodeCoordinateselementNodesQ9 colorbar titleDisplacement field ux on deformed shape axis off stresses at nodes stressstrain stresses2DGDofnumberElements elementNodesnumberNodesnodeCoordinatesdisplacements CQ9complete drawing stress fields on deformed shape figure hold on drawingField2nodeCoordinateselementNodes scaleFactorUXUYQ4stress1 sigma XX axis equal drawingMeshnodeCoordinateselementNodesQ8 190 11 Plane Stress colorbar titleStress field sigmaxx on deformed shape axis off These codes have several supporting functions for stiffness matrix formation and post computation analysis Function formStiffnessMass2Dm forms the finite ele ment matrix according to the typology selected Q4 Q8 or Q9 and Gauss integration required function stiffnessmass formStiffnessMass2DGDof numberElements elementNodesnumberNodesnodeCoordinates CrhothicknesselemType quadType compute stiffness and mass matrix for plane stress quadrilateral elements stiffness zerosGDof mass zerosGDof quadrature according to quadType gaussWeightsgaussLocations gaussQuadraturequadType for e 1numberElements indice elementNodese elementDof indice indicenumberNodes ndof lengthindice cycle for Gauss point for q 1sizegaussWeights1 GaussPoint gaussLocationsq xi GaussPoint1 eta GaussPoint2 shape functions and derivatives shapeFunctionnaturalDerivatives shapeFunctionsQxietaelemType Jacobian matrix inverse of Jacobian derivatives wrt xy JacobinvJacobianXYderivatives JacobiannodeCoordinatesindicenaturalDerivatives B matrix B zeros32ndof B11ndof XYderivatives1 B2ndof12ndof XYderivatives2 B31ndof XYderivatives2 B3ndof12ndof XYderivatives1 stiffness matrix stiffnesselementDofelementDof stiffnesselementDofelementDof BCthicknessBgaussWeightsqdetJacob mass matrix 119 Plate in Traction 191 massindiceindicemassindiceindice shapeFunctionshapeFunction rhothicknessgaussWeightsqdetJacob massindicenumberNodesindicenumberNodes massindicenumberNodesindicenumberNodes shapeFunctionshapeFunction rhothicknessgaussWeightsqdetJacob end end end Function shapeFunctionsQm computes the shape functions and their deriva tives with respect to natural ξ η coordinates for Q4 Q8 and Q9 elements Function Jacobianm computes the Jacobian matrix and its inverse The computation of Gauss point locations and weights is made in function gaussQuadraturem The listing of these functions is given below function shapenaturalDerivatives shapeFunctionsQxietaelemType shape function and derivatives for Q4 Q8 and Q9 elements shape Shape functions naturalDerivatives derivatives wrt xi and eta xi eta natural coordinates 1 1 switch elemType case Q4 Q4 element shape 141xi1eta 1xi1eta 1xi1eta 1xi1eta naturalDerivatives 14 1eta 1xi 1eta 1xi 1eta 1xi 1eta 1xi case Q8 Q8 element shape 141xi1eta1xieta 1xi1eta1xieta 1xi1eta1xieta 1xi1eta1xieta 21xixi1eta 21xi1etaeta 21xixi1eta 21xi1etaeta naturalDerivatives 14 eta2xieta1 2etaxixi1 eta2xieta1 2etaxixi1 eta2xieta1 2etaxixi1 eta2xieta1 xi12etaxi 4xieta1 2xiˆ21 21etaˆ2 4etaxi1 4xieta1 21xiˆ2 2etaˆ21 4etaxi1 192 11 Plane Stress case Q9 Q9 element shape 14xietaxi1eta1 xietaxi1eta1 xietaxi1eta1 xietaxi1eta1 2etaxixi1eta1 2xixi1etaeta1 2etaxixi1eta1 2xixi1etaeta1 4xixi1etaeta1 naturalDerivatives 14 eta2xi1eta1xixi12eta1 eta2xi1eta1xixi12eta1 eta2xi1eta1xixi12eta1 eta2xi1eta1xixi12eta1 4xietaeta1 2xi1xi12eta1 22xi1eta1eta14xietaxi1 4xietaeta1 2xi1xi12eta1 22xi1eta1eta14xietaxi1 8xietaˆ21 8etaxiˆ21 end end end function function JacobianMatrixinvJacobianXYDerivatives JacobiannodeCoordinatesnaturalDerivatives JacobianMatrix Jacobian matrix invJacobian inverse of Jacobian Matrix XYDerivatives derivatives wrt x and y naturalDerivatives derivatives wrt xi and eta nodeCoordinates nodal coordinates at element level JacobianMatrix nodeCoordinatesnaturalDerivatives invJacobian invJacobianMatrix XYDerivatives naturalDerivativesinvJacobian end end function Jacobian function weightslocations gaussQuadratureoption Gauss quadrature for 2D elements option third 3x3 option complete 2x2 option reduced 1x1 locations Gauss point locations weights Gauss point weights switch option case third 119 Plate in Traction 193 locations 0774596669241483 0774596669241483 0 0774596669241483 0774596669241483 0774596669241483 0774596669241483 0 0 0 0774596669241483 0 0774596669241483 0774596669241483 0 0774596669241483 0774596669241483 0774596669241483 weights 05555555555555560555555555555556 05555555555555560888888888888889 05555555555555560555555555555556 08888888888888890555555555555556 08888888888888890888888888888889 05555555555555560888888888888889 05555555555555560555555555555556 05555555555555560888888888888889 05555555555555560555555555555556 case complete locations 0577350269189626 0577350269189626 0577350269189626 0577350269189626 0577350269189626 0577350269189626 0577350269189626 0577350269189626 weights 1111 case reduced locations 0 0 weights 4 end end end function gaussQuadrature The post computation of the stress field is carried out in stresses2Dm which is listed below function stressstrain stresses2DGDofnumberElements elementNodesnumberNodesnodeCoordinates displacementsCelemTypequadType quadrature according to quadType gaussWeightsgaussLocations gaussQuadraturequadType stresses at nodes stress zerosnumberElementssizegaussLocations13 stressPoints 1 11 11 11 1 for e 1numberElements indice elementNodese elementDof indice indicenumberNodes nn lengthindice for q 1sizegaussWeights1 pt gaussLocationsq wt gaussWeightsq xi pt1 194 11 Plane Stress eta pt2 shape functions and derivatives shapeFunctionnaturalDerivatives shapeFunctionsQxietaelemType Jacobian matrix inverse of Jacobian derivatives wrt xy JacobinvJacobianXYderivatives JacobiannodeCoordinatesindicenaturalDerivatives B matrix B zeros32nn B11nn XYderivatives1 B2nn12nn XYderivatives2 B31nn XYderivatives2 B3nn12nn XYderivatives1 element deformation strain BdisplacementselementDof stresseq Cstrain end end end end function stresses2D Some functions included in the codes above will be used also for plate problems In Fig115 we show the finite element mesh considering 10 5 elements In Fig116 the deformed shape of the problem is illustrated using Q4 Q8 and Q9 and in Fig117 the stress distribution along the x axis is shown Due to the constant state of stress in the beam the normal stress σxx field is constant in all the points of the beam Note that stiffness matrix calculation for Q8 and Q9 have been carried out using reduced 2 2 Gauss integration The same integration is used for Q4 which results in exact integration Small differences are shown in terms of displacements and stresses according to the finite element used However these differences are small among each other and can be reduced by applying mesh refinement Fig 115 Finite element mesh for a thin plate in tension 119 Plate in Traction 195 Fig 116 Plate in traction using Q4 Q8 and Q9 displacement field ux 196 11 Plane Stress Fig 117 Plate in traction using Q4 Q8 and Q9 stress field σxx 1110 2D Beam in Bending 197 Fig 118 Thin plate in bending problem18m 1 5 p 1110 2D Beam in Bending We show in this example code problem18m a 2D beam in bending Fig118 Note some of the differences to problem17m Fixed boundary conditions are considered on the left edge x 0 of the plate for both u and v The constant applied force is in the y direction so care must be taken to ensure that degrees of freedom are properly assigned as well as lumped nodal forces on such elements MATLAB codes for Finite Element Analysis problem18m 2D problem beam in bending using Q4 elements AJM Ferreira N Fantuzzi 2019 clear memory clear close all materials E 10e7 poisson 030 matrix C C E1poissonˆ21 poisson 0poisson 1 00 0 1poisson2 load P 1e6 mesh generation Lx 5 Ly 1 numberElementsX 20 numberElementsY 10 numberElements numberElementsXnumberElementsY nodeCoordinates elementNodes rectangularMeshLxLynumberElementsXnumberElementsYQ4 xx nodeCoordinates1 yy nodeCoordinates2 figure drawingMeshnodeCoordinateselementNodesQ4 198 11 Plane Stress axis equal numberNodes sizexx1 GDof global number of degrees of freedom GDof 2numberNodes calculation of the system stiffness matrix stiffness formStiffnessMass2DGDofnumberElements elementNodesnumberNodesnodeCoordinatesC11Q4complete boundary conditions fixedNodeX findnodeCoordinates10 fixed in XX fixedNodeY findnodeCoordinates10 fixed in YY prescribedDof fixedNodeX fixedNodeYnumberNodes force vector distributed load applied at xxLx force zerosGDof1 rightBord findnodeCoordinates1Lx forcerightBordnumberNodes PLynumberElementsY forcerightBord1numberNodes PLynumberElementsY2 forcerightBordendnumberNodes PLynumberElementsY2 solution displacements solutionGDofprescribedDofstiffnessforce displacements and deformed shape dispDisplacements jj 1GDof format f jj displacements fprintfnode U fprintf3d 128f f UX displacements1numberNodes UY displacementsnumberNodes1GDof scaleFactor 01 deformed shape figure drawingFieldnodeCoordinatesscaleFactorUX UY elementNodesQ4UXU XX hold on drawingMeshnodeCoordinatesscaleFactorUX UY elementNodesQ4 drawingMeshnodeCoordinateselementNodesQ4 colorbar titleDisplacement field ux on deformed shape axis off stresses at nodes stressstrain stresses2DGDofnumberElements elementNodesnumberNodesnodeCoordinatesdisplacements CQ4complete drawing stress fields on deformed shape figure hold on drawingField2nodeCoordinateselementNodes scaleFactorUXUYQ4stress1 axis equal axis off drawingMeshnodeCoordinateselementNodesQ4 colorbar titleStress field sigmaxx on deformed shape stress extrapolation stressExtr zerosnumberElements43 for e 1numberElements for i 13 1110 2D Beam in Bending 199 stressExtrei 105sqrt3 05 105sqrt3 05 05 105sqrt3 05 105sqrt3 105sqrt3 05 105sqrt3 05 05 105sqrt3 05 105sqrt3 stresse1istresse2istresse3istresse4i end end stress averaging at nodes stressAvg zerosnumberNodes3 for i 13 currentStress stressExtri for n 1numberNodes idx findnelementNodes stressAvgni sumcurrentStressidx lengthcurrentStressidx end end surface representation figure hold on for k 1sizeelementNodes1 patchnodeCoordinateselementNodesk141 nodeCoordinateselementNodesk142 nodeCoordinateselementNodesk1410 stressAvgelementNodesk141 end axis equal axis off colorbar titleAveraged nodal stress field sigmaxx In Fig119 we show the displacements field of ux on top of the deformed shape of the beam If the user wishes to plot another displacement just change the displacement component upon calling drawingFieldm In Fig1110 we show the stress field of σx in the beam according to its values in the integration points If the user wishes to plot another stress just change the number of the stress component upon calling drawingField2m The analogous codes for Q8 and Q9 are not listed for the sake of conciseness The reader can inspect given codes problem18am and problem18bm for Q8 and Q9 element implementations Thestressextrapolationandstressaveragingisgiveninthelastpartofthecodesfor Q4 Q8andQ9elements Thesefinal routines implement directlythetheorypresented inthischapterForcomparisonTable111listmaximumσxx inplanestressevaluated at the integration points and after extrapolation procedure As expected the numerical values of the extrapolated stresses are higher than the same at integration points 200 11 Plane Stress Fig 119 Plate in bending using Q4 Q8 and Q9 displacement field ux 1110 2D Beam in Bending 201 Fig 1110 Plate in bending using Q4 Q8 and Q9 stress field σxx 202 11 Plane Stress Table 111 Maximum normal σxx 107 stresses evaluated at the integration points and after extrap olation Q4 Q8 Q9 Integration points 29617 31072 30720 Extrapolation 31947 35688 35131 1111 2D Beam in Free Vibrations The present example shows the free vibration analysis of cantilever beam The same geometry of the previous example is considered Fig118 without applied forces and ρ 1000 The implementation using Q4 elements is listed in code prob lem18vibm MATLAB codes for Finite Element Analysis problem18vibm 2D problem beam in free vibrations using Q4 elements AJM Ferreira N Fantuzzi 2019 clear memory clear close all materials E 10e7 poisson 030 rho 1000 matrix C C E1poissonˆ21 poisson 0poisson 1 00 0 1poisson2 mesh generation Lx 5 Ly 1 numberElementsX 20 numberElementsY 10 numberElements numberElementsXnumberElementsY nodeCoordinates elementNodes rectangularMeshLxLynumberElementsXnumberElementsYQ4 xx nodeCoordinates1 yy nodeCoordinates2 figure drawingMeshnodeCoordinateselementNodesQ4 axis equal numberNodes sizexx1 GDof global number of degrees of freedom GDof 2numberNodes stiffness and mass matrices stiffnessmass formStiffnessMass2DGDofnumberElements elementNodesnumberNodesnodeCoordinatesCrho1 Q4complete boundary conditions 1111 2D Beam in Free Vibrations 203 fixedNodeX findnodeCoordinates10 fixed in XX fixedNodeY findnodeCoordinates10 fixed in YY prescribedDof fixedNodeX fixedNodeYnumberNodes solution modeseigenvalues eigenvalueGDofprescribedDof stiffnessmass15 omega sqrteigenvalues sort out eigenvalues omegaii sortomega modes modesii drawing mesh and deformed shape modeNumber 3 displacements modesmodeNumber displacements and deformed shape UX displacements1numberNodes UY displacementsnumberNodes1GDof scaleFactor 05 deformed shape figure drawingFieldnodeCoordinatesscaleFactorUX UY elementNodesQ4UXU XX hold on drawingMeshnodeCoordinatesscaleFactorUX UY elementNodesQ4 drawingMeshnodeCoordinateselementNodesQ4 colorbar titleDisplacement field ux on deformed shape axis off The mass matrix is carried out in function formStiffnessMass2Dm The eigen value problem is solved using eigenvaluem used in 1D beam problems The desired mode shape can be represented by changing the value of the variable modeNumber The first three mode shapes of the beam modelled using Q4 elements are shown in Fig1111 Note that the first two modes are flexural and the third one is axial Codes for Q8 and Q9 elements are given by problem18avibm and problem18bvibm wherein functions call are modified for such elements These codes are not listed for the sake of conciseness 204 11 Plane Stress Fig 1111 First three mode shapes of a plate in free vibrations using Q4 elements Displacement field ux 1111 2D Beam in Free Vibrations 205 Table 112 First five natural frequencies of cantilever beam ω Q4 Q8 Q9 1 126548 124847 124811 2 686069 674494 674243 3 996379 995644 995550 4 1648584 1610925 1610303 5 2769458 2683379 2682412 The comparison in terms of natural frequencies using the three implementations is given in Table 112 Reference 1 JN Reddy An introduction to the finite element method 3rd edn McGrawHill International Editions New York 2005 Chapter 12 Kirchhoff Plates Abstract In the present chapter finite element implementation of Kirchhoff plates in bending is discussed using the socalled conforming and not conforming Her mite shape functions Note that Hermite shape functions other than more common Lagrangefunctionsthatconsidernodalparametersonlyusemorekinematicparame ters than the ones representing the displacement field of the mathematical differential problem that is currently in use 121 Introduction The present Kirchhoff plate theory considers thin plates made of orthotropic materi als the theory is valid also for isotropic ones In the present chapter finite element implementation of Kirchhoff plates in bending is discussed using the socalled con forming and not conforming Hermite shape functions Note that Hermite shape func tions other than more common Lagrange functions that consider nodal parameters only use more kinematic parameters than the ones representing the displacement field of the mathematical differential problem that is currently in use The classical and easier beam problem directly comes into the mind As a matter of fact Bernoulli beam in bending has displacement and rotation as dwdx parameters at each bound ary node If the beam has 2 nodes this results in a finite element with 4 degrees of freedom where the continuity among the elements is ensured up to the first order derivative due to the presence of the rotation in terms of first order derivative The same can be done for the Kirchhoff plates however kinematic approximation and interelement continuity should be assured according to 2 Cartesian directions x and y For this reason two approximations for the finite element analysis are generally introduced a not conforming with 3 degrees of freedom dof and a conforming one with 4 dofs per node when the element is considered with 4 nodes More details regarding these implementations will be given below The Editors if applicable and The Authors under exclusive license to Springer Nature Switzerland AG 2020 A J M Ferreira and N Fantuzzi MATLAB Codes for Finite Element Analysis Solid Mechanics and Its Applications 157 httpsdoiorg101007978303047952712 207 208 12 Kirchhoff Plates 122 Mathematical Background Kirchhoff plate in bending only is based on the following displacement field ow ow uy x y Z t zZ UyX y Z t zZ u3x yZt w 121 ox oy where wx y t is the transverse displacement parameter No axial displacement is considered due to uncoupling phenomena between bending and axial behaviors for orthotropic plates Straindisplacement relations are 02 Ej aay a2 z az we zDw 122 02 and 3 4 5 O due to Kirchhoff assumptions Note that VoigtKelvin notation is used for strain definitions eg 11 122 233 323 413 512 6 Constitutive law is indicated as oO Q1 Qo O é o2Qn2 Qn 0 3 123 56 0 O Q6 LE where E Vi2Ey Ey On 9n On 066 Giz 124 1 vi2v21 1 vj2v21 1 vj2v21 The orthotropic properties of the lamina should be given Ej Ey vj2 Giz and v2 V12EE applies In case of isotropic materials two material properties are needed as E and v moreover Q1 Q22 Qi2 vQ1 and Qe G apply with 2G E1 v It is convenient to write constitutive equations in matrix form by including the straindisplacement relations 122 as o Qe zQDw 125 Equilibrium equations are carried out using the Hamiltons Principle The strain energy for the plate is 1 1 T U of 0282 06 dV ao edV 2 V 2 Vv 1 1 1 eQe dV DpwzQDw dV DwDWwd2 2Jy 2 Jy 2 Je 126 122 Mathematical Background 209 where D is the bending stiffness matrix Di Dp 0 DDiy Dy 0 127 0 0 Dee and Dj Qih 12 for ij 1 26 Potential work done by transverse loads is V ow yw dxdy 128 A For the static analysis Hamiltons Principle becomes the Principle of minimum total potential energy or principle of virtual displacements thus 6U6V0 129 which means the equilibrium At this stage the finite element approximation should be introduced in order to get the solution in weak form 123 Finite Element Approximation The finite element approximation for the Kirchhoff plate theory using Hermite shape functions starts with the classical polynomial approximation wx yt Njx yd Nd 1210 jl where d t are the parameters related to w and its derivatives at the nodes and Nj y the interpolation functions Subscript identifies that quantities are defined within a generic finite element and n is a function of kinematic parameters chosen for each element node Vectors N and d collect element shape functions and generalized displacement parameters respectively Definitions of interpolation functions and their derivations according to nodal parameters are given below 1231 Interpolation Functions Approximation polynomials for the present finite elements should be taken consid ering the following polynomial expansion scheme 210 12 Kirchhoff Plates lx x y xy wy xy 1211 y2 xy x2y x3 y xy x2y3 3353 by considering all these aforementioned terms the polynomial is complete up to the third order and it is valid when 4 generalized displacements per node are considered as such w dwdx dwdy and d7wdxdy note that IwIx Iwdy represent rotation of the fiber at the point In this case the Hermite approximation is called conforming A not conforming approximation with w dwdx dwdy can be obtained by removing the terms xy xy xy3 x3 given in Eq 1211 Thus the generation of the shape functions for the not conforming element starts from the approximation wa aox a3y agxy asx ay apxy 1212 agxy ax ayy ayxy ayoxy Since the shape functions will represent both w and its derivatives First order deriva tives should be carried out as ow 2 2 2 3 dz aay 2a5x 2azxy agy 3dox 3a41xy ai2y ox aw 1213 by a3 agx 2agy ayx 2agxy 3aioy ayyx B3a2xy All these three approximations are valid in the 4 nodal corners of the finite element defined by the coordinates x y for i 1 2 3 4 However since finite element mapping has to be considered for transforming each element of general shape into the one in a parent space 7 the approximation should be written in such space which nodal corner values are 1 71 1 1 2 m2 1 1 33 CU D and 4 74 1 1 By defining the vector ordering as T d w Wo ws W4 Wort Wix2 Wx3 Word Wy Wy Wy3 Wye 1214 where x and y represent partial derivatives with respect tox and y Equations 1212 1213 have to written for each corner and collected in matrix form as d Aa 1215 where A is a known matrix of coefficients which are function of the node coordinates ni for i 1 2 3 4 Thus the vector a is the one including coefficients a for j1 12 coefficients can be carried out by matrix inversion as aAd 1216 123 Finite Element Approximation 211 The solution obtained is substituted into the initial approximation function 1212 and each term which multiplies d represents the shape function associated with that degree of freedom In compact form they can be written as Ni 81 G 1 2 3 4 N gp 5678 1217 Ni gp G 9 10 11 12 where gj 0125 1 1 m0 2 10 1 8i2 01258 E 1 1 no 1 0 1218 gis 0125 Mo 1 1 0 1 0 and no nn for i 1 2 3 4 The master element is considered in 7 coordinates of side length 2 as most standard finite element procedures The conforming element is based on the approximation as 2 2 2 2 3 W a ax asy agxy asx dey a7xy agxy dox 1219 ayy tax y apxy ai3xy ayaxy aisxy ayoxy with derivatives dw 2 2 2 Ox a2 aay 2asx 2azxy agy 3dox 3ay1xy ayy 2ay3xy Bay4xy 2aisxy B3aioxy ow a3 44x 2agy anx 2agxy 3a10y ayx dy 1220 3ay2xy 2a3xy Qay4xry B3asxy B3ay6xy w 4 a4 2azx 2agy 3a11x axdy Bay 4aj3xy 6ay4xy 6aisxy 9ayoxy The same aforementioned procedure can be followed letting to the following shape functions associated with the displacement parameter vector d wy Wo W3 Wa Wirt Wix2 Wix3 Word Way Way Wy Wy T 1221 Wxyl W xy2 W xy3 Wxy4 Shape functions result to be 212 12 Kirchhoff Plates Nj j j 1234 N j j 5 67 8 il gin 1222 Ni g3 G 910 1112 Niegi G 13 14 15 16 where 5 5 8j1 90625E Eo 2n 1 N0 2 Bj 006256 E C1 Go ni no 2 1223 gs 00625 1Go 2n ni no gia 006258 ni E L 0 m1 no and no nn for i 1 23 4 The master element is considered in n coordinates of side length 2 as most standard finite element procedures 1232 Stiffness Matrix Once the interpolation functions are defined according to the chosen selected degrees of freedom per node stiffness matrix creation should be performed The terms that have to be integrated are given by the strain energy Eq 126 once the approximation 1210 is introduced in it However strain energy is written in Cartesian coordinates x y and it has to be mapped into reference ones 7 This procedure involves derivatives up to at least second order thus Jacobian matrix transformation is required The first order derivatives of an arbitrary function defined in the Cartesian x y plane with respect to x and y are given by the Jacobian matrix definition as from Eq 1133 0 0 0 ax y Nx 0 J 0 12 24 fs E fs a 1224 dy an an The above 2 x 2 matrix denoted by J is the inverse of Jacobian matrix of the transformation J Note that the first order derivatives of and n with respect to x and y are indicated as nx y ny respectively It is possible to obtain the inverse matrix of Jacobian as J der 2 a det J xy xnYe 1225 TX XE where det J is the determinant of the Jacobian and comparing the inverse matrix of Jacobian in 1225 with that in 1224 the following relationships are obtained Jn n é é Gy Tay deta Gets 1226 123 Finite Element Approximation 213 The substitution of 1226 into 1224 yields 0 0 0 det J na 2 ox 05 on 1227 0 det J 0 4x 0 ae x ay ag an The second order derivatives of a function can be derived from 1224 as a 5 0 5 0 a 0 0 ri t s 27 xa ax a2 af x pp Ex dE On g aE n an a7 5 0 5 0 a 0 0 az Fay thes t ma hy et I 1228 dy 5 a2 oP Sy ny dE dn 5yy aE On a a a a 0 0 Doan SxS ap OT My a OF Sey Sy Nx SRT Ot Say 5S OT ay Oxdy Sib 553 1 rr Ny Seay by 5g hy 7 Then the second order derivatives of with respect to x and y can be expressed as Ex det I Yn en y det IW det Je YeYnn YeVn det J det J 1229 Ey det I xpxeq x7 det JT det Je xe Xn Xex det J det J In a similar manner the second order derivatives of n with respect to x and y can also be obtained ax det J ynyee ynye det J det Je yeven yz det J det J 1230 Nyy det J XpXee XpXe det J det Jg xexgy xz det J detJ where det J and det J are the first order derivatives of the determinant of Jacobian with respect to and n respectively Differentiation of det J in 1225 leads to det J x VeXen VyXee X é EVEn YEXEn T Vn XGE nYEE 1231 det Jn XnYen Yn Xen VEXnn XEYny and finally the mixed derivatives of and 7 with respect to x and y are given by Exy det J yyxXen YyXn det J det Je yexnn Yexy det J det Jr Ny det J yexen YnXe det Jo det Je ynxee yexe det J det J 1232 The above general coordinate transformation formulation is valid for any transforma tion mapping Herein a4 node linear element is considered for geometrical mapping thus the finite element is not isoparametric but subparametric because the num ber of parameters used for the geometry approximation are less then the ones used 214 12 Kirchhoff Plates for interpolating the strain field within each element leading to a higher continuity among the elements This assumption is sufficient for the present scope The strain energy for Kirchhoff plate problem with the present finite element approximation 1210 takes the form 1 T 1 eT T e U 5 Dw DDw d2 54 DN DDN d2 d a e 1233 sa f BDB d2 d 2 Ia where B DN includes the derivatives of the shape functions with respect to the Cartesian system as B B B Bs 1234 that have to be mapped according to the transformation of coordinates as B EPNes neNun 2 NNéEn Ey Ne NxNy Bo ENee 0Ny 28myNen yNe tyyNy 1235 B3 2 EE Nee nxNyNay Exny nxNen Ey Nez NxyNn Derivatives of the shape functions are used and indicated as aN aN aN ON oN Nee a Nmaay Nn ae7 Neap NrnqQ 236 aé an adn 0 an The stiffness matrix of the generic finite element can be carried out from equation 1233 by rewriting the area integral in the parent domain using Jacobian matrix as 1 pl Ké BDB det Jddn 1237 1J1 Integration of 1237 is carried out by Gauss integration Summarizing geometrical element mapping is due to Q4 Lagrange shape functions whereas finite element approximation is carried out with Hermite interpolation functions The load vector is given by direct substitution of the finite element approximation within the potential definition 128 as 1 pl f pN det Jdédn 1238 1J1 After assembly of the stiffness matrix and load vector of the generic element the static problem can be solved using classical Gauss elimination method 124 Isotropic Square Plate in Bending 215 124 Isotropic Square Plate in Bending We consider a simplysupported SSSS and clamped CCCC square plate side a b 1 under uniform transverse pressure p 1 and thickness h The modulus of elasticity is taken E 109201 and the Poissons ratio is taken as ν 03 The nondimensional transverse displacement is set as w w D pa4 1239 where the bending stiffness D is taken as D Eh3 121 ν2 1240 The code for solving the present problem is listed in problemKm and given below MATLAB codes for Finite Element Analysis problemKm Kirchhoff plate in bending AJM Ferreira N Fantuzzi 2019 clear memory clear close all isotropic material E 10920 poisson 030 thickness 001 I thicknessˆ312 D11 Ethicknessˆ3121poissonˆ2 D22 D11 D12 poissonD11 D66 1poissonD112 orthotropic material E1 318e6 E2 102e6 poisson12 031 G12 096e6 poisson21 poisson12E2E1 thickness 001 I thicknessˆ312 D11 E1I1poisson12poisson21 D22 E2I1poisson12poisson21 D12 poisson12D22 D66 G12I matrix C bending part Cbending D11 D12 0 D12 D22 0 0 0 D66 1The reader may be curious about the reason for this particular value of E With a 1 thickness h 01 and the mentioned values for E and ν we obtain a flexural stiffness of 1 This is only a practical convenience for nondimensional results not really a meaningful value 216 12 Kirchhoff Plates load P 1 3 nonconforming 4 node element 4 conforming 4 node element dofpernode 3 number of kinematic parameters per node mesh generation L 1 numberElementsX 20 numberElementsY numberElementsX numberElements numberElementsXnumberElementsY nodeCoordinates elementNodes rectangularMeshLLnumberElementsXnumberElementsYQ4 xx nodeCoordinates1 yy nodeCoordinates2 figure drawingMeshnodeCoordinateselementNodesQ4 axis equal numberNodes sizexx1 GDof global number of degrees of freedom GDof dofpernodenumberNodes stiffness formStiffnessMatrixKGDofnumberElements elementNodesnumberNodesnodeCoordinates Cbendingcompletedofpernode force formForceVectorKGDofnumberElementselementNodes numberNodesnodeCoordinatesPcompletedofpernode boundary conditions prescribedDofactiveDof EssentialBCssssGDofxxyynodeCoordinatesnumberNodes solution displacements solutionGDofprescribedDofstiffnessforce displacements dispDisplacements jj 1GDof format f jj displacements fprintfnode U fprintf3d 128f f format long isotropic dimensionless deflection D1 Ethicknessˆ3121poissonˆ2 mindisplacements1numberNodesD1Lˆ4 orthotropic dimensionless deflection mindisplacements1numberNodesD122D66PLˆ4 surface representation figure hold on for k 1sizeelementNodes1 patchnodeCoordinateselementNodesk141 124 Isotropic Square Plate in Bending 217 nodeCoordinateselementNodesk142 displacementselementNodesk14 displacementselementNodesk14 end setgcafontsize18 view4545 The mesh is generated automatically using rectangularMeshm code The boundary conditions for the plate are assigned using EssentialBCm function Some predefined boundary condition configurations are given such as clamped CCCC and simplysupported SSSS others can be easily implemented The aforementioned code is listed below function prescribedDofactiveDof EssentialBCtypeBCGDofxxyynodeCoordinatesnumberNodes essential boundary conditions for rectangular plates W transverse displamcent TX rotation about y axis TY rotation about x axis switch typeBC case ssss simply supported plate fixedNodeW findxxmaxnodeCoordinates1 xxminnodeCoordinates1 yyminnodeCoordinates2 yymaxnodeCoordinates2 fixedNodeTX findyymaxnodeCoordinates2 yyminnodeCoordinates2 fixedNodeTY findxxmaxnodeCoordinates1 xxminnodeCoordinates1 case cccc clamped plate fixedNodeW findxxmaxnodeCoordinates1 xxminnodeCoordinates1 yyminnodeCoordinates2 yymaxnodeCoordinates2 fixedNodeTX fixedNodeW fixedNodeTY fixedNodeTX case scsc fixedNodeW findxxmaxnodeCoordinates1 xxminnodeCoordinates1 yyminnodeCoordinates2 yymaxnodeCoordinates2 fixedNodeTX findxxmaxnodeCoordinates2 xxminnodeCoordinates2 fixedNodeTY case cccf fixedNodeW findxxminnodeCoordinates1 yyminnodeCoordinates2 yymaxnodeCoordinates2 218 12 Kirchhoff Plates fixedNodeTX fixedNodeW fixedNodeTY fixedNodeTX end prescribedDof fixedNodeW fixedNodeTXnumberNodes fixedNodeTY2numberNodes activeDof setdiff1GDofprescribedDof end This script has several supporting functions for stiffness matrix formStiffness MatrixKm and force vector formForceVectorKm generation function K formStiffnessMatrixKGDofnumberElements elementNodesnumberNodesnodeCoordinates CbendingquadTypedofpernode computation of stiffness matrix for Kirchhoff plate element K stiffness matrix K zerosGDof Gauss quadrature for bending part gaussWeightsgaussLocations gaussQuadraturequadType cycle for element for e 1numberElements indice nodal connectivities for each element elementDof element degrees of freedom indice elementNodese if dofpernode 3 elementDof indice indicenumberNodes indice2numberNodes else 4 dof elementDof indice indicenumberNodes indice2numberNodes indice3numberNodes end ndof lengthelementDof cycle for Gauss point for q 1sizegaussWeights1 GaussPoint gaussLocationsq xi GaussPoint1 eta GaussPoint2 part related to the mapping shape functions and derivatives natDerQ4 shapeFunctionKQ4xieta if dofpernode 3 naturalDerivatives shapeFunctionK12xieta else 4 dof naturalDerivatives shapeFunctionK16xieta end 124 Isotropic Square Plate in Bending 219 Jacobian matrix inverse of Jacobian derivatives wrt xy JacobinvJacobianXYDerQ4 JacobianKnodeCoordinatesindicenatDerQ4 detJ detJacob detJxi XYDerQ411XYDerQ425 XYDerQ421XYDerQ415 XYDerQ422XYDerQ413 XYDerQ412XYDerQ423 detJeta XYDerQ412XYDerQ425 XYDerQ422XYDerQ415 XYDerQ421XYDerQ414 XYDerQ411XYDerQ424 xix invJacobian11 xiy invJacobian21 etax invJacobian12 etay invJacobian22 xixx detJˆ2XYDerQ422XYDerQ425 XYDerQ422ˆ2detJdetJxi XYDerQ421XYDerQ424 XYDerQ421XYDerQ422detJdetJeta xiyy detJˆ2XYDerQ412XYDerQ415 XYDerQ412ˆ2detJdetJxi XYDerQ411XYDerQ414 XYDerQ411XYDerQ412detJdetJeta etaxx detJˆ2XYDerQ422XYDerQ423 XYDerQ422XYDerQ421detJdetJxi XYDerQ421XYDerQ425 XYDerQ421ˆ2detJdetJeta etayy detJˆ2XYDerQ412XYDerQ413 XYDerQ412XYDerQ411detJdetJxi XYDerQ411XYDerQ415 XYDerQ411ˆ2detJdetJeta xixy detJˆ2XYDerQ422XYDerQ415 XYDerQ422XYDerQ412detJdetJxi XYDerQ421XYDerQ414 XYDerQ421XYDerQ412detJdetJeta etaxy detJˆ2XYDerQ421XYDerQ415 XYDerQ422XYDerQ411detJdetJxi XYDerQ422XYDerQ413 XYDerQ421XYDerQ411detJdetJeta B matrix bending Bb zeros3ndof Bb11ndof xixˆ2naturalDerivatives3 etaxˆ2naturalDerivatives4 2xixetaxnaturalDerivatives5 xixxnaturalDerivatives1 etaxxnaturalDerivatives2 Bb21ndof xiyˆ2naturalDerivatives3 etayˆ2naturalDerivatives4 2xiyetaynaturalDerivatives5 xiyynaturalDerivatives1 220 12 Kirchhoff Plates etayynaturalDerivatives2 Bb31ndof 2xixxiynaturalDerivatives3 etaxetaynaturalDerivatives4 xixetay xiyetaxnaturalDerivatives5 xixynaturalDerivatives1 etaxynaturalDerivatives2 stiffness matrix bending KelementDofelementDof KelementDofelementDof BbCbendingBbgaussWeightsqdetJacob end Gauss point end element end function force formForceVectorKGDofnumberElementselementNodes numberNodesnodeCoordinatesPquadTypedofpernode computation of force vector for Kirchhoff plate element force force vector force zerosGDof1 Gauss quadrature for bending part gaussWeightsgaussLocations gaussQuadraturequadType cycle for element for e 1numberElements indice nodal connectivities for each element indice elementNodese if dofpernode 3 elementDof indice indicenumberNodes indice2numberNodes else 4 dof elementDof indice indicenumberNodes indice2numberNodes indice3numberNodes end ndof lengthelementDof cycle for Gauss point for q 1sizegaussWeights1 GaussPoint gaussLocationsq GaussWeight gaussWeightsq xi GaussPoint1 eta GaussPoint2 part related to the mapping shape functions and derivatives natDerQ4 shapeFunctionKQ4xieta if dofpernode 3 shapeFunction shapeFunctionK12xieta else 4 dof 124 Isotropic Square Plate in Bending 221 shapeFunction shapeFunctionK16xieta end Jacobian matrix inverse of Jacobian derivatives wrt xy Jacob JacobianKnodeCoordinatesindicenatDerQ4 force vector forceelementDof forceelementDof shapeFunctionPdetJacobGaussWeight end end Gauss point loop end end element loop end For the generation of stiffness matrix and force vector shape functions and their derivatives are needed Thus specific functions are given for not conform ing shapeFunctionK12m and conforming shapeFunctionK16m elements as well as Jacobian matrix calculation JacobianKm All these functions are listed below function shapenaturalDerivatives shapeFunctionK12xieta shape function and derivatives for not conforming elements shape Shape functions naturalDerivatives derivatives wrt xi and eta xi eta natural coordinates 1 1 natural derivatives order ddx ddy dˆ2dxˆ2 dˆ2dyˆ2 dˆ2dxdy shape eta 1xi 1etaˆ2 eta xiˆ2 xi 28 eta 1xi 1etaˆ2 eta xiˆ2 xi 28 eta 1xi 1 etaˆ2 eta xiˆ2 xi 28 eta 1xi 1etaˆ2 eta xiˆ2 xi 28 eta 1xi 1ˆ2xi 18 eta 1xi 1xi 1ˆ28 eta 1xi 1xi 1ˆ28 eta 1xi 1ˆ2xi 18 eta 1ˆ2eta 1xi 18 eta 1ˆ2eta 1xi 18 eta 1eta 1ˆ2xi 18 eta 1eta 1ˆ2xi 18 naturalDerivatives1 eta 1etaˆ2 eta 3xiˆ2 38 eta 1etaˆ2 eta 3xiˆ2 38 eta 1 etaˆ2 eta 3xiˆ2 38 eta 1 etaˆ2 eta 3xiˆ2 38 eta 1 3xiˆ2 2xi 18 eta 13xiˆ2 2xi 18 eta 13xiˆ2 2xi 18 eta 1 3xiˆ2 2xi 18 eta 1ˆ2eta 18 eta 1ˆ2eta 18 eta 1eta 1ˆ28 222 12 Kirchhoff Plates eta 1eta 1ˆ28 naturalDerivatives2 xi 13etaˆ2 xiˆ2 xi 38 xi 1 3etaˆ2 xiˆ2 xi 38 xi 1 3etaˆ2 xiˆ2 xi 38 xi 13etaˆ2 xiˆ2 xi 38 xi 1ˆ2xi 18 xi 1xi 1ˆ28 xi 1xi 1ˆ28 xi 1ˆ2xi 18 xi 1 3etaˆ2 2eta 18 xi 1 3etaˆ2 2eta 18 xi 13etaˆ2 2eta 18 xi 13etaˆ2 2eta 18 naturalDerivatives3 3xieta 14 3xieta 14 3xieta 14 3xieta 14 3xi 1eta 14 3xi 1eta 14 3xi 1eta 14 3xi 1eta 14 0 0 0 0 naturalDerivatives4 3etaxi 14 3etaxi 14 3etaxi 14 3etaxi 14 0 0 0 0 3eta 1xi 14 3eta 1xi 14 3eta 1xi 14 3eta 1xi 14 naturalDerivatives5 12 3xiˆ28 3etaˆ28 3etaˆ28 3xiˆ28 12 12 3xiˆ28 3etaˆ28 3etaˆ28 3xiˆ28 12 xi4 3xiˆ28 18 18 3xiˆ28 xi4 3xiˆ28 xi4 18 3xiˆ28 xi4 18 eta4 3etaˆ28 18 3etaˆ28 eta4 18 3etaˆ28 eta4 18 18 3etaˆ28 eta4 end 124 Isotropic Square Plate in Bending 223 function shapenaturalDerivatives shapeFunctionK16xieta shape function and derivatives for conforming elements shape Shape functions naturalDerivatives derivatives wrt xi and eta xi eta natural coordinates 1 1 natural derivatives order ddx ddy dˆ2dxˆ2 dˆ2dyˆ2 dˆ2dxdy shape eta 1ˆ2eta 2xi 1ˆ2xi 216 eta 1ˆ2eta 2xi 1ˆ2xi 216 eta 1ˆ2eta 2xi 1ˆ2xi 216 eta 1ˆ2eta 2xi 1ˆ2xi 216 eta 1ˆ2eta 2xi 1ˆ2xi 116 eta 1ˆ2eta 2xi 1xi 1ˆ216 eta 1ˆ2eta 2xi 1xi 1ˆ216 eta 1ˆ2eta 2xi 1ˆ2xi 116 eta 1ˆ2eta 1xi 1ˆ2xi 216 eta 1ˆ2eta 1xi 1ˆ2xi 216 eta 1eta 1ˆ2xi 1ˆ2xi 216 eta 1eta 1ˆ2xi 1ˆ2xi 216 eta 1ˆ2eta 1xi 1ˆ2xi 116 eta 1ˆ2eta 1xi 1xi 1ˆ216 eta 1eta 1ˆ2xi 1xi 1ˆ216 eta 1eta 1ˆ2xi 1ˆ2xi 116 naturalDerivatives1 3xiˆ2 1eta 1ˆ2eta 216 3xiˆ2 1eta 1ˆ2eta 216 3xiˆ2 1eta 1ˆ2eta 216 3xiˆ2 1eta 1ˆ2eta 216 eta 1ˆ2eta 2 3xiˆ2 2xi 116 eta 1ˆ2eta 23xiˆ2 2xi 116 eta 1ˆ2eta 23xiˆ2 2xi 116 eta 1ˆ2eta 2 3xiˆ2 2xi 116 3xiˆ2 1eta 1ˆ2eta 116 3xiˆ2 1eta 1ˆ2eta 116 3xiˆ2 1eta 1eta 1ˆ216 3xiˆ2 1eta 1eta 1ˆ216 eta 1ˆ2eta 1 3xiˆ2 2xi 116 eta 1ˆ2eta 13xiˆ2 2xi 116 eta 1eta 1ˆ23xiˆ2 2xi 116 eta 1eta 1ˆ2 3xiˆ2 2xi 116 naturalDerivatives2 3etaˆ2 1xi 1ˆ2xi 216 3etaˆ2 1xi 1ˆ2xi 216 3etaˆ2 1xi 1ˆ2xi 216 3etaˆ2 1xi 1ˆ2xi 216 3etaˆ2 1xi 1ˆ2xi 116 3etaˆ2 1xi 1xi 1ˆ216 3etaˆ2 1xi 1xi 1ˆ216 3etaˆ2 1xi 1ˆ2xi 116 xi 1ˆ2xi 2 3etaˆ2 2eta 116 xi 1ˆ2xi 2 3etaˆ2 2eta 116 xi 1ˆ2xi 23etaˆ2 2eta 116 xi 1ˆ2xi 23etaˆ2 2eta 116 xi 1ˆ2xi 1 3etaˆ2 2eta 116 xi 1xi 1ˆ2 3etaˆ2 2eta 116 xi 1xi 1ˆ23etaˆ2 2eta 116 224 12 Kirchhoff Plates xi 1ˆ2xi 13etaˆ2 2eta 116 naturalDerivatives3 3xieta 1ˆ2eta 28 3xieta 1ˆ2eta 28 3xieta 1ˆ2eta 28 3xieta 1ˆ2eta 28 3xi 1eta 1ˆ2eta 28 3xi 1eta 1ˆ2eta 28 3xi 1eta 1ˆ2eta 28 3xi 1eta 1ˆ2eta 28 3xieta 1ˆ2eta 18 3xieta 1ˆ2eta 18 3xieta 1eta 1ˆ28 3xieta 1eta 1ˆ28 3xi 1eta 1ˆ2eta 18 3xi 1eta 1ˆ2eta 18 3xi 1eta 1eta 1ˆ28 3xi 1eta 1eta 1ˆ28 naturalDerivatives4 3etaxi 1ˆ2xi 28 3etaxi 1ˆ2xi 28 3etaxi 1ˆ2xi 28 3etaxi 1ˆ2xi 28 3etaxi 1ˆ2xi 18 3etaxi 1xi 1ˆ28 3etaxi 1xi 1ˆ28 3etaxi 1ˆ2xi 18 3eta 1xi 1ˆ2xi 28 3eta 1xi 1ˆ2xi 28 3eta 1xi 1ˆ2xi 28 3eta 1xi 1ˆ2xi 28 3eta 1xi 1ˆ2xi 18 3eta 1xi 1xi 1ˆ28 3eta 1xi 1xi 1ˆ28 3eta 1xi 1ˆ2xi 18 naturalDerivatives5 9etaˆ2 1xiˆ2 116 9etaˆ2 1xiˆ2 116 9etaˆ2 1xiˆ2 116 9etaˆ2 1xiˆ2 116 3etaˆ2 1 3xiˆ2 2xi 116 3etaˆ2 13xiˆ2 2xi 116 3etaˆ2 13xiˆ2 2xi 116 3etaˆ2 1 3xiˆ2 2xi 116 3xiˆ2 1 3etaˆ2 2eta 116 3xiˆ2 1 3etaˆ2 2eta 116 3xiˆ2 13etaˆ2 2eta 116 3xiˆ2 13etaˆ2 2eta 116 3etaˆ2 2eta 1 3xiˆ2 2xi 116 3etaˆ2 2eta 13xiˆ2 2xi 116 3etaˆ2 2eta 13xiˆ2 2xi 116 3etaˆ2 2eta 1 3xiˆ2 2xi 116 end 124 Isotropic Square Plate in Bending 225 function JacobianMatrixinvJacobianXYDerivatives JacobianKnodeCoordinatesnaturalDerivatives JacobianMatrix Jacobian matrix invJacobian inverse of Jacobian Matrix XYDerivatives derivatives wrt x and y naturalDerivatives derivatives wrt xi and eta nodeCoordinates nodal coordinates at element level JacobianMatrix nodeCoordinatesnaturalDerivatives12 invJacobian invJacobianMatrix XYDerivatives nodeCoordinatesnaturalDerivatives end end function Jacobian Moreover for geometric approximation the code shapeFunctionKQ4m has been considered This code is different from the one used in the previous chapter because higher order derivatives of the shape functions are needed in the present problem shapeFunctionKQ4m is listed below function shapenaturalDerivatives shapeFunctionKQ4xieta shape function and derivatives for Q4 elements shape Shape functions naturalDerivatives derivatives wrt xi and eta xi eta natural coordinates 1 1 shape14 1xi1eta1xi1eta 1xi1eta1xi1eta natural derivatives order ddx ddy dˆ2dxˆ2 dˆ2dyˆ2 dˆ2dxdy naturalDerivatives 141eta 1xi1eta 1xi 1eta 1xi1eta 1xi naturalDerivatives514 14 14 14 end Notethatdifferentshapefunctionsareconsideredforthegeometricapproximation and the displacement field approximation In Table 121 we present nondimensional transverse displacement results obtained by the code problemKm for various boundary conditions Conforming elements have a faster convergence than not conforming ones however this results is not valid in general for any Kirchhoff plate problem Therefore more degrees of freedom does not mean in this case that fast convergence is observed In Fig121 we show the deformed shape of a clamped and simplysupported plate using a 20 20 Q4 mesh 226 12 Kirchhoff Plates Table 121 Dimensionless deflection w 102 for square isotropic plate CCCC SSSS NC C NC C 5 5 01180 01115 03976 03821 10 10 01290 01268 04136 04093 20 20 01272 01266 04081 04070 30 30 01268 01266 04071 04066 Exact 1 0126 04062 Fig 121 Deformed shape of a clamped and simplysupported square plate meshed by 20 20 Q4 elements 125 Orthotropic Square Plate in Bending An orthotropic square plate under uniform load is considered in the present section with simplysupported boundary conditions The problem has been taken from the book 2 where the following material properties are selected E1 318 Mpsi E2 102 Mpsi ν12 031 G12 096 Mpsi Dimensionless central deflections are considered as w wD12 2D66 pa4 1241 The same code shown above can be used to carried out the following calculations The reader needs to comment and uncomment lines relative to material properties anddimensionlessformulaforthetransversedisplacementResultsusingconforming and not conforming elements are listed in Table 122 Note that for the present case the not conforming element has a faster convergence with respect to the conforming one References 227 Table 122 Dimensionless deflection w 103 for square orthotropic plate SSSS NC C 5 5 09014 08541 10 10 09317 09200 20 20 09229 09200 30 30 09213 09200 Exact 2 09225 References 1 SP Timoshenko S WoinowskyKrieger Theory of Plates and Shells 2nd edn McGrawHill International Student Edition Tokyo 1959 2 JN Reddy An Introduction to the Finite Element Method 3rd edn McGrawHill International Editions New York 2005 Chapter 13 Mindlin Plates Abstract This chapter considers the static free vibration and buckling problem of Mindlin plates in bending Many implementation codes will be taken from the previ ous chapters such as mesh generation Gauss integration and field representation The theory of Mindlin plates is firstly presented and several applications are described 131 Introduction This chapter considers the static free vibration and buckling problem of Mindlin plates in bending Many implementation codes will be taken from the previous chap ters such as mesh generation Gauss integration and field representation The theory of Mindlin plates is firstly presented 132 The Mindlin Plate Theory The Mindlin plate theory or firstorder shear deformation theory for plates includes the effect of transverse shear deformations 1 It may be considered an extension of the Timoshenko theory for beams in bending The main difference from the thin Kirchhofftype theory is that in the Mindlin theory the normals to the undeformed middle plane of the plate remain straight but not normal to the deformed middle surface 1321 Displacement Field If only transverse loads are applied Mindlin plate is subjected to bending and shear deformations only no axial deformation occurs Thus the assumed displacement The Editors if applicable and The Authors under exclusive license to Springer Nature Switzerland AG 2020 A J M Ferreira and N Fantuzzi MATLAB Codes for Finite Element Analysis Solid Mechanics and Its Applications 157 httpsdoiorg101007978303047952713 229 230 13 Mindlin Plates y ts b Oy x Fig 131 Mindlin plate illustration of geometry and rotational degrees of freedom field for a thick plate of thickness can be defined without axial displacements as usx y xs t ZOxx y t Uzx yZt 20 x y 1 131 u3x y xs t wx y t where 6 6 are the rotations of the normal to the middle plane with respect to axes y and x respectively and w is the uniform transverse displacement of the plate Physical meaning of the kinematic parameters is shown in Fig 131 1322 Strains Straindisplacement relations are carried out which give the relations between strains and degrees of freedom w 6 and 0 Bending flexural strains are obtained as Ou 1 00 x i Ox Ox Our 06 y 132 y SS EO Our 4 Our 00 4 06 Tay Oy dx ax Oy 132 The Mindlin Plate Theory 231 while the transverse shear deformations are obtained as γxz u3 x u1 z w x θx γyz u3 y u2 z w y θy 133 that in matrix form become ϵb zϵ1 ϵs γ0 134 where ϵb ϵx ϵy γxyT ϵs γxz γyzT 135 Note that inplane strains are linear through the thickness and they will be involved inbendingwhereastransversesheardeformationsareconstantthroughthethickness The latter needs a shear correction factor for the transverse stress quantities 1323 Stresses The linear elastic stressstrain relations in bending are defined for a homogeneous isotropic material as σb Qbϵb 136 where σb σx σy τxyT 137 are the bending stresses and strains and Qb is defined as Qb E 1 ν2 1 ν 0 ν 1 0 0 0 1 ν 2 138 whereas the linear elastic stressstrain relations in transverse shear are defined as σs Qsϵs 139 where σs τxz τyzT 1310 are the transverse shear stresses and strains and Qs is defined as 232 13 Mindlin Plates 10 a651 1311 where G is the shear modulus 1324 Hamiltons Principle The strain energy of the Mindlin plate is given as 1 2 1 T k T U oepdV oedV 1312 2 Jy 2Jy The k parameter also known as the shear correction factor can be taken as 56 3 as also introduced for Timoshenko beams in Chap 10 Introducing the stresses 136 and 139 into the strain energy 1312 we obtain 1 T k T U Qhe dV Qsés dV 2 Jv 2 Jv 1 k EMT Q 6 dV TQ yO dv 1313 2Jy 2Jv De dQ OT Ay dQ 2 Jo 2 Jo where D and A are the bending and shear stiffness matrices in the form lv O Eh p ft 0 fg Kean 1314 121 v lvy 01 00 2 The potential can be defined as VeWe pwdQ 1315 2 where p is the transverse pressure applied on the plate which works for the transverse displacement only Other types of load are neglected in the present investigation The kinetic energy of the plate is defined by 1 K p w 26 6 dV 1 v 1316 mor m6 m20y Jaa 2 Ja 132 The Mindlin Plate Theory 233 where my and m2 are termed main and rotary inertias respectively h2 h2 h3 2 p mo pdzph ma pz dz 1317 h2 h2 12 133 Finite Element Discretization The generalized displacements are independently interpolated using the same shape functions w SIME nwi O YONEMOi O NMEMi 1318 il il il wheren 4 for Q4n 8 for Q8 andn 9 for Q9 N 7 identify the Lagrangian shape functions according to element choice The finite element approximation 1318 can be conveniently written in matrix form as w N00 Ww u60N0 6 Nd 1319 9y 00N LO where W wi Lee wr 6 Ax1 Lee Oxn 6 1 Lee Ayn dé collects all the degrees of freedom of the generic element in vector form and N is the matrix of the shape functions Strains are defined as ZBd Bd 1320 The straindisplacement matrices for bending and shear contributions are obtained by derivation of the shape functions by ON ON 00 0 0 Ox Ox OM ON B00 0 O 1321 Oy Oy 00 OM ON ON ON OD I Be Oy ON ONn oe FE Mo Nn 0 0 x x B aN aN 1322 00 Mm MN Oy dy We then obtain the plate strain energy 1313 as 234 13 Mindlin Plates e 1 eT 2pT e ge US d ZB QBy dzd2 d 2 2 Jz 1 dk BQB dzdQ d 1323 2 2 Jz The stiffness matrix of the Mindlin plate is then obtained as Ke B DB dQ B AB dQ 1324 Qe Qe The external work potential 1315 with the finite element approximation becomes ViWe a N pdx 1325 Re where p po o only transverse loads are considered thus the force vector for the Mindlin plate is given by f N pd 1326 Re Finally the kinetic energy 1316 takes the form e I eT T eae K d N IN dQQd 1327 2 92 where I is the inertia matrix given by mo 0 0 I 0 m 0 1328 0 O m2 where mm represents the rotary inertia that for thin plates is generally negligible Mass matrix is given by Mé NIN do 1329 2 Geometric mapping is applied in order to get integrals in natural coordinates Such transformation is achieved with the determinant of the Jacobian matrix det J as done for the plane stress case The element stiffness matrix is 1 pl 1 pl K B DB det J ddn B AB det J ddn 1330 1J1 lJl 133 Finite Element Discretization 235 The vector of nodal forces is 1 pl f N pdet J dédn 1331 1J1 and the mass matrix in natural coordinates is 1 pl M NIN det J dédn 1332 1J1 All element matrices are computed by Gauss integration Mindlin theory as Timo shenko one has demonstrated to suffer from shear locking It has been demonstrated that the simplest remedy to this numerical behavior is to perform reduced integra tion of the shear component For instance the stiffness integral for Q4 element is solved by considering 2 x 2 Gauss integration exact for the bending contribution and single point quadrature reduced for the shear contribution 4 5 134 Stress Recovery Once the nodal solution is carried out d stresses can be recovered from constitutive equations as 1 oy Qe Gee QBpd 1333 Os Qss QBd It is noted that a and o are evaluated at the integration points GaussLegendre points Since the bending stresses are linear through the plate thickness in the fol lowing they will be computed at the top layer of the plate z h2 On the contrary shear stresses are constant through the thickness thus they are independent on z Values for the element corner points can be obtained by extrapolation as illustrated in the Sect 1181 Accurate values of the transverse shear stresses can be carried out by solving the 3D equilibrium equations with a as known functions 135 Square Mindlin Plate in Bending We consider a simplysupported and clamped square plate side a b 1 under uniform transverse pressure p 1 and thickness h The modulus of elasticity is taken E 10920 and the Poissons ratio is taken as vy 03 The nondimensional transverse displacement is set as The reader may be curious about the reason for this particular value of E With a 1 thickness h 01 and the mentioned values for E and v we obtain a flexural stiffness of 1 This is only a practical convenience for nondimensional results not really a meaningful value 236 13 Mindlin Plates Table 131 Nondimensional transverse displacement of a square plate under uniform pressure simplysupported SSSS boundary conditions ah Mesh Q4 Q8 Q9 Exact 10 2 2 0003545 0004039 0004408 6 6 0004245 0004272 0004274 10 10 0004263 0004273 0004273 20 20 0004270 0004273 0004273 30 30 0004271 0004273 0004273 0004270 10000 2 2 0003188 0001541 0004194 6 6 0004024 0000801 0004064 10 10 0004049 0003797 0004063 20 20 0004059 0004061 0004062 30 30 0004060 0004062 0004062 0004060 w w D pa4 1334 where the bending stiffness D is taken as D Eh3 121 ν2 1335 In Tables 131 and 132 we present nondimensional transverse displacement results obtained by the code problem19m for various thickness values and boundary con ditions In Fig132 we show the deformed shape of a simplysupported plate using a 20 20 Q4 mesh MATLAB codes for Finite Element Analysis problem19m Mindlin plate in bending Q4 elements AJM Ferreira N Fantuzzi 2019 clear memory clear close all materials E 10920 poisson 030 kapa 56 thickness 01 I thicknessˆ312 constitutive matrix bending part Cbending IE1poissonˆ2 1 poisson 0poisson 1 00 0 1poisson2 shear part 135 Square Mindlin Plate in Bending 237 Cshear kapathicknessE21poissoneye2 load P 1 mesh generation L 1 numberElementsX 20 numberElementsY 20 numberElements numberElementsXnumberElementsY nodeCoordinates elementNodes rectangularMeshLLnumberElementsXnumberElementsYQ4 xx nodeCoordinates1 yy nodeCoordinates2 figure drawingMeshnodeCoordinateselementNodesQ4 axis equal numberNodes sizexx1 GDof global number of degrees of freedom GDof 3numberNodes computation of the system stiffness matrix and force vector stiffness formStiffnessMatrixMindlinGDofnumberElements elementNodesnumberNodesnodeCoordinatesCshear CbendingQ4completereduced force formForceVectorMindlinGDofnumberElements elementNodesnumberNodesnodeCoordinatesPQ4reduced boundary conditions prescribedDofactiveDof EssentialBCssssGDofxxyynodeCoordinatesnumberNodes solution displacements solutionGDofprescribedDofstiffnessforce displacements dispDisplacements jj 1GDof format f jj displacements fprintfnode U fprintf3d 128f f format long D1 Ethicknessˆ3121poissonˆ2 mindisplacements1numberNodesD1Lˆ4 surface representation figure hold on for k 1sizeelementNodes1 patchnodeCoordinateselementNodesk141 nodeCoordinateselementNodesk142 displacementselementNodesk14 238 13 Mindlin Plates displacementselementNodesk14 end setgcafontsize18 view4545 postcomputation stressshear MindlinStressGDofnumberElements elementNodesnumberNodesnodeCoordinatesdisplacements CshearCbendingthicknessQ4completereduced This MATLAB code calls functions formStiffnessMatrixMindlinm for computa tion of stiffness matrix and formForceVectorMindlinm for computation of the force Fig 132 Mesh of 20 20 Q4 elements and deformed shape 135 Square Mindlin Plate in Bending 239 Table 132 Nondimensional transverse displacement of a square plate under uniform pressure clamped CCCC boundary conditions ah Mesh Q4 Q8 Q9 Exact 10 2 2 0000357 0001730 0001757 6 6 0001486 0001505 0001507 10 10 0001498 0001505 0001505 20 20 0001503 0001505 0001505 30 30 0001503 0001505 0001505 10000 2 2 351010 0001541 0001541 6 6 0001239 0000173 0001267 10 10 0001255 0000199 0001266 20 20 0001262 0001142 0001265 30 30 0001264 0001254 0001265 0001260 vector Such functions can be used also for the finite element computation with Q8 and Q9 elements function K formStiffnessMatrixMindlinGDofnumberElements elementNodesnumberNodesnodeCoordinatesCshear CbendingelemTypequadTypeBquadTypeS elemType type of element Q4 Q8 Q9 quadTypeB type of quadrature for bending quadTypeS type of quadrature for shear computation of stiffness matrix for Mindlin plate element K stiffness matrix K zerosGDof Gauss quadrature for bending part gaussWeightsgaussLocations gaussQuadraturequadTypeB cycle for element for e 1numberElements indice nodal connectivities for each element elementDof element degrees of freedom indice elementNodese elementDof indice indicenumberNodes indice2numberNodes ndof lengthindice cycle for Gauss point for q 1sizegaussWeights1 GaussPoint gaussLocationsq xi GaussPoint1 eta GaussPoint2 shape functions and derivatives shapeFunctionnaturalDerivatives 240 13 Mindlin Plates shapeFunctionsQxietaelemType Jacobian matrix inverse of Jacobian derivatives wrt xy JacobinvJacobianXYderivatives JacobiannodeCoordinatesindicenaturalDerivatives B matrix bending Bb zeros33ndof Bb1ndof12ndof XYderivatives1 Bb22ndof13ndof XYderivatives2 Bb3ndof12ndof XYderivatives2 Bb32ndof13ndof XYderivatives1 stiffness matrix bending KelementDofelementDof KelementDofelementDof BbCbendingBbgaussWeightsqdetJacob end Gauss point end element shear stiffness matrix Gauss quadrature for shear part gaussWeightsgaussLocations gaussQuadraturequadTypeS cycle for element for e 1numberElements indice nodal connectivities for each element elementDof element degrees of freedom indice elementNodese elementDof indice indicenumberNodes indice2numberNodes ndof lengthindice cycle for Gauss point for q 1sizegaussWeights1 GaussPoint gaussLocationsq xi GaussPoint1 eta GaussPoint2 shape functions and derivatives shapeFunctionnaturalDerivatives shapeFunctionsQxietaelemType Jacobian matrix inverse of Jacobian derivatives wrt xy JacobinvJacobianXYderivatives JacobiannodeCoordinatesindicenaturalDerivatives B matrix shear Bs zeros23ndof Bs11ndof XYderivatives1 Bs21ndof XYderivatives2 Bs1ndof12ndof shapeFunction Bs22ndof13ndof shapeFunction stiffness matrix shear KelementDofelementDof KelementDofelementDof 135 Square Mindlin Plate in Bending 241 BsCshearBsgaussWeightsqdetJacob end gauss point end element end function force formForceVectorKGDofnumberElementselementNodes numberNodesnodeCoordinatesPquadTypedofpernode computation of force vector for Kirchhoff plate element force force vector force zerosGDof1 Gauss quadrature for bending part gaussWeightsgaussLocations gaussQuadraturequadType cycle for element for e 1numberElements indice nodal connectivities for each element indice elementNodese if dofpernode 3 elementDof indice indicenumberNodes indice2numberNodes else 4 dof elementDof indice indicenumberNodes indice2numberNodes indice3numberNodes end ndof lengthelementDof cycle for Gauss point for q 1sizegaussWeights1 GaussPoint gaussLocationsq GaussWeight gaussWeightsq xi GaussPoint1 eta GaussPoint2 part related to the mapping shape functions and derivatives natDerQ4 shapeFunctionKQ4xieta if dofpernode 3 shapeFunction shapeFunctionK12xieta else 4 dof shapeFunction shapeFunctionK16xieta end Jacobian matrix inverse of Jacobian derivatives wrt xy Jacob JacobianKnodeCoordinatesindicenatDerQ4 force vector forceelementDof forceelementDof shapeFunctionPdetJacobGaussWeight 242 13 Mindlin Plates end end Gauss point loop end end element loop end The imposition of the essential boundary conditions is made in function Essen tialBCm which has been introduced in the previous chapter for the study of Kirch hoff plates In the last part of the main code the postcomputation of the stresses is given in MindlinStressm Postcomputation implementation is given in the code below function stressshear MindlinStressGDofnumberElements elementNodesnumberNodesnodeCoordinatesdisplacements CshearCbendinghelemTypequadTypeBquadTypeS Mindlin Stress computes normal and shear stresses according to Mindlin theory note that transverse shear stresses are not corrected normal stresses 1 sigmaxx 2 sigmayy 3 tauxy stress zerosnumberElements43 Gauss quadrature for bending part gaussWeightsgaussLocations gaussQuadraturequadTypeB cycle for element for e 1numberElements indice nodal connectivities for each element indiceB element degrees of freedom indice elementNodese indiceB indice indicenumberNodes indice2numberNodes nn lengthindice cycle for Gauss point for q 1sizegaussWeights1 pt gaussLocationsq wt gaussWeightsq xi pt1 eta pt2 shape functions and derivatives shapeFunctionnaturalDerivatives shapeFunctionsQxietaelemType Jacobian matrix inverse of Jacobian derivatives wrt xy JacobinvJacobianXYderivatives JacobiannodeCoordinatesindicenaturalDerivatives 135 Square Mindlin Plate in Bending 243 B matrix bending Bb zeros33nn Bb1nn12nn XYderivatives1 Bb22nn13nn XYderivatives2 Bb3nn12nn XYderivatives2 Bb32nn13nn XYderivatives1 stresses strain h2BbdisplacementsindiceB stresseq Cbendingstrain end Gauss point end element shear stresses 1 tauxz 2 tauyz by constitutive equations shear zerosnumberElements12 Gauss quadrature for shear part gaussWeightsgaussLocations gaussQuadraturequadTypeS cycle for element for e 1numberElements indice nodal connectivities for each element indiceB element degrees of freedom indice elementNodese indiceB indice indicenumberNodes indice2numberNodes nn lengthindice cycle for Gauss point for q 1sizegaussWeights1 pt gaussLocationsq wt gaussWeightsq xi pt1 eta pt2 shape functions and derivatives shapeFunctionnaturalDerivatives shapeFunctionsQxietaelemType Jacobian matrix inverse of Jacobian derivatives wrt xy JacobinvJacobianXYderivatives JacobiannodeCoordinatesindicenaturalDerivatives B matrix shear Bs zeros23nn Bs11nn XYderivatives1 Bs21nn XYderivatives2 Bs1nn12nn shapeFunction Bs22nn13nn shapeFunction sliding BsdisplacementsindiceB sheareq Cshearsliding 244 13 Mindlin Plates end end gauss point loop end end element loop end MATLAB codes for solving the static problem for Mindlin plates with Q8 and Q9 elements are not reported for the sake of conciseness but given in codes prob lem19am and problem19bm 136 Free Vibrations of Mindlin Plates By using the Hamilton Principle 2 we may express the equations of motion of Mindlin plates as Mu Kuf 1336 where M Kf are the system mass and stiffness matrices and the force vector respectively and U u are the accelerations and displacements Assuming a harmonic motion we obtain the natural frequencies and the modes of vibration by solving the generalized eigenproblem 6 K wM X 0 1337 where w is the natural frequency and X the mode of vibration By using the mass 1329 and stiffness 1324 matrices defined before the free vibration problem can be solved after assembly We consider a square plate side a with thicknesstoside ratio ha 001 and ha 01 The nondimensional natural frequency is given by P W Wmn4 G where p is the material density G the shear modulus G E21 v E the modulus of elasticity and v the Poissons ratio Indices m and n are the vibration halfwaves along x and y axes In this problem we consider simplysupported SSSS and clamped CCCC plates as well as SCSC and CCCF plates where F means free side For CCCC and CCCF we use a shear correction factor k 08601 while for SCSC plates we use k 0822 For SSSS plates we consider k 56 136 Free Vibrations of Mindlin Plates 245 Table 133 Convergence of natural frequency ω for CCCC plate with k 08601 ν 03 ha 001 Mesh Q4 Q8 Q9 Ref 7 10 10 01800 01756 01754 15 15 01774 01754 01754 20 20 01765 01754 01754 25 25 01761 01754 01754 01754 ha 01 Mesh Q4 Q8 Q9 Ref 7 10 10 16259 15911 15911 15 15 16063 15911 15911 20 20 15996 15910 15910 25 25 15965 15910 15910 15940 Table 134 Convergence of natural frequency ω for SSSS plate with k 08333 ν 03 ha 001 Mesh Q4 Q8 Q9 Ref 7 10 10 00973 00963 00963 15 15 00968 00963 00963 20 20 00965 00963 00963 25 25 00965 00963 00963 00963 ha 001 Mesh Q4 Q8 Q9 Ref 7 10 10 09399 09303 09303 15 15 09346 09303 09303 20 20 09327 09303 09303 25 25 09318 09303 09303 0930 In Table 133 we show the convergence of the fundamental frequency for CCCC plate with k 08601 ν 03 and two thickness to width ratios ha 001 and ha 01 Q4 Q8 and Q9 finite elements are employed We obtain quite good agreement with the analytical solution 7 In Table 134 we show the convergence of the fundamental frequency for SSSS plate with k 08333 ν 03 for two thickness to width ratios ha 001 and ha 01 Again we obtain quite good agreement with a analytical solution 7 Tables 135 and 136 list the natural frequencies of a SSSS plate with ha 01 and ha 01 being k 0833 ν 03 Our finite element solution agrees with the tridimensional solution and analytical solution given by Mindlin 6 Tables 137 and 138 compare natural frequencies with the RayleighRitz solution 6 and a solution by Liew et al 8 246 13 Mindlin Plates Table 135 Natural frequencies of a SSSS plate with ha 01 k 0833 ν 03 using a mesh 15 15 Mode m n Q4 Q8 Q9 3D Mindlin 1 1 1 09346 09303 09303 0932 0930 2 2 1 22545 22194 22194 2226 2219 3 1 2 22545 22194 22194 2226 2219 4 2 2 34592 34058 34058 3421 3406 5 3 1 43031 41504 41504 4171 4149 6 1 3 43031 41504 41504 4171 4149 7 3 2 53535 52065 52065 5239 5206 8 2 3 53535 52065 52065 5239 5206 9 4 1 69413 65246 65246 6520 10 1 4 69413 65246 65246 6520 11 3 3 70318 68354 68354 6889 6834 12 4 2 78261 74506 74506 7511 7446 13 2 4 78261 74506 74506 7511 7446 Analytical solution Table 136 Natural frequencies of a SSSS plate with ha 001 k 0833 ν 03 using a mesh 20 20 Mode m n Q4 Q8 Q9 Mindlin 1 1 1 00965 00963 00963 00963 2 2 1 02430 02406 02406 02406 3 1 2 02430 02406 02406 02406 4 2 2 03890 03847 03847 03847 5 3 1 04928 04808 04808 04807 6 1 3 04928 04808 04808 04807 7 3 2 06380 06246 06246 06246 8 2 3 06380 06246 06246 06246 9 4 1 08550 08164 08164 08156 10 1 4 08550 08164 08164 08156 11 3 3 08857 08641 08641 08640 12 4 2 09991 09599 09599 09592 13 2 4 09991 09599 09599 09592 Analytical solution 136 Free Vibrations of Mindlin Plates 247 Table 137 Natural frequencies of a CCCC plate with ha 01 k 08601 ν 03 using a mesh 20 20 Mode m n Q4 Q8 Q9 Rayleigh Ritz 7 Liew et al 8 1 1 1 15955 15910 15910 15940 15582 2 2 1 30662 30390 30390 30390 30182 3 1 2 30662 30390 30390 30390 30182 4 2 2 42924 42626 42626 42650 41711 5 3 1 51232 50253 50253 50350 51218 6 1 3 51730 50729 50729 50780 51594 7 3 2 61587 60803 60803 60178 8 2 3 61587 60803 60803 60178 9 4 1 76554 74142 74142 75169 10 1 4 76554 74142 74142 75169 11 3 3 77703 76805 76805 77288 12 4 2 84555 82618 82617 83985 13 2 4 85378 83371 83370 83985 Table 138 Natural frequencies of a CCCC plate with ha 001 k 08601 ν 03 using a mesh 20 20 Mode m n Q4 Q8 Q9 Rayleigh Ritz 7 Liew et al 8 1 1 1 0175 01754 01754 01754 01743 2 2 1 03635 03574 03574 03576 03576 3 1 2 03635 03574 03574 03576 03576 4 2 2 05358 05266 05266 05274 05240 5 3 1 06634 06400 06400 06402 06465 6 1 3 06665 06431 06431 06432 06505 7 3 2 08266 08020 08020 08015 8 2 3 08266 08020 08020 08015 9 4 1 10875 10227 10227 10426 10 1 4 10875 10227 10227 10426 11 3 3 11049 10683 10683 10628 12 4 2 12392 11754 11754 11823 13 2 4 12446 11804 11804 11823 248 13 Mindlin Plates Table 139 Natural frequencies for SCSC plate with ha 01 k 0822 ν 03 using a mesh 15 15 Mode m n Q4 Q8 Q9 Mindlin 6 1 1 1 12940 12837 12837 1302 2 2 1 23971 23641 23641 2398 3 1 2 29290 28595 28595 2888 4 2 2 38394 37735 37735 3852 5 3 1 43475 42021 42021 4237 6 1 3 51354 49095 49095 4936 7 3 2 55094 53737 53736 8 2 3 58974 57075 57075 9 4 1 69384 65325 65324 10 1 4 72939 70968 70967 11 3 3 77968 72776 72776 12 4 2 78516 75033 75032 13 2 4 84308 79849 79847 Table 1310 Natural frequencies for SCSC plate with ha 001 k 0822 ν 03 using a mesh 15 15 Mode m n Q4 Q8 Q9 Mindlin 6 1 1 1 01424 01410 01410 01411 2 2 1 02710 02664 02664 02668 3 1 2 03484 03374 03374 03377 4 2 2 04722 04597 04596 04608 5 3 1 05191 04975 04974 04979 6 1 3 06710 06279 06279 06279 7 3 2 07080 06811 06809 8 2 3 07944 07517 07516 9 4 1 08988 08289 08288 10 1 4 10228 09695 09690 11 3 3 10758 10040 10036 12 4 2 11339 10129 10129 13 2 4 12570 11387 11385 Tables 139 and 1310 compare natural frequencies for SCSC plate with ha 01 and ha 01 being k 0822 ν 03 respectively Sides located at x 0 L are simplysupported Tables 1311 and 1312 compare natural frequencies for CCCF plates with ha 01 and ha 001 respectively being k 0822 ν 03 Side located at x L is free 136 Free Vibrations of Mindlin Plates 249 Table 1311 Natural frequencies for CCCF plate with ha 01 k 08601 ν 03 using a mesh 15 15 Mode m n Q4 Q8 Q9 Mindlin 6 1 1 1 10923 10803 10802 1089 2 2 1 17566 17428 17428 1758 3 1 2 27337 26557 26555 2673 4 2 2 32591 31953 31953 3216 5 3 1 33541 32882 32881 3318 6 1 3 46395 45554 45553 4615 7 3 2 49746 47287 47285 8 2 3 54620 52428 52427 9 4 1 55245 53091 53090 10 1 4 65865 63901 63899 11 3 3 66347 64428 64426 12 4 2 76904 71330 71328 13 2 4 81626 76590 76589 Table 1312 Natural frequencies for CCCF plate with ha 001 k 08601 ν 03 Mode m n Q4 Q8 Q9 Mindlin 6 1 1 1 01180 01166 01166 01171 2 2 1 01967 01947 01947 01951 3 1 2 03193 03079 03078 03093 4 2 2 03830 03733 03733 03740 5 3 1 04031 03921 03920 03931 6 1 3 05839 05671 05669 05695 7 3 2 06387 05948 05947 8 2 3 07243 06541 06539 9 4 1 08817 06820 06818 10 1 4 09046 08399 08393 11 3 3 10994 08590 08584 12 4 2 11407 09771 09769 13 2 4 11853 10338 10335 Figure 133 shows the modes of vibration for a CCCC plate with ha 01 using 20 20 Q4 elements 250 13 Mindlin Plates Figure 134 shows the modes of vibration for a SSSS plate with ha 01 using 20 20 Q4 elements Figure 135 shows the modes of vibration for a SCSC plate with ha 001 using 20 20 Q4 elements Figure 136 shows the modes of vibration for a CCCF plate with ha 01 using 20 20 Q4 elements TheMATLABcodeproblem19Vibrationsmsolvesthefreevibrationproblem of Mindlin plates The user is requested to change input details according to the problem MATLAB codes for Finite Element Analysis problem19aVibrationsm Mindlin plate in free vibrations Q8 elements AJM Ferreira N Fantuzzi 2019 clear memory clear close all materials E 10920 poisson 030 G E21poisson thickness 01 rho 1 I thicknessˆ312 kapa 08601 cccc cccf case kapa 0822 scsc case kapa 56 ssss case constitutive matrix bending part Cbending IE1poissonˆ2 1 poisson 0poisson 1 00 0 1poisson2 shear part Cshear kapathicknessE21poissoneye2 mesh generation L 1 numberElementsX 20 numberElementsY 20 numberElements numberElementsXnumberElementsY nodeCoordinates elementNodes rectangularMeshLLnumberElementsXnumberElementsYQ8 xx nodeCoordinates1 yy nodeCoordinates2 figure drawingMeshnodeCoordinateselementNodesQ8 axis equal numberNodes sizexx1 GDof global number of degrees of freedom GDof 3numberNodes 136 Free Vibrations of Mindlin Plates 251 computation of the system stiffness and mass matrices stiffness formStiffnessMatrixMindlinGDofnumberElements elementNodesnumberNodesnodeCoordinatesCshear CbendingQ8thirdcomplete mass formMassMatrixMindlinGDofnumberElements elementNodesnumberNodesnodeCoordinatesthickness rhoIQ8third boundary conditions prescribedDofactiveDof EssentialBCssssGDofxxyynodeCoordinatesnumberNodes free vibrations modeseigenvalues eigenvalueGDofprescribedDof stiffnessmass15 omega sqrteigenvalues sort out eigenvalues omegaii sortomega modes modesii dimensionless omega omegabar omegaLsqrtrhoG drawing mesh and deformed shape modeNumber 1 displacements modesmodeNumber surface representation figure hold on for k 1sizeelementNodes1 patchnodeCoordinateselementNodesk141 nodeCoordinateselementNodesk142 displacementselementNodesk14 displacementselementNodesk14 end setgcafontsize18 view4545 This code calls function formMassMatrixMindlinm which computes the mass matrices of the Mindlin Q4 Q8 and Q9 elements The code for computing the stiff ness matrix has already been presented 252 13 Mindlin Plates function mass formMassMatrixMindlinGDofnumberElements elementNodesnumberNodesnodeCoordinatesthickness rhoIelemTypequadType computation of mass matrix for Mindlin plate element mass mass matrix mass zerosGDof Gauss quadrature for bending part gaussWeightsgaussLocations gaussQuadraturequadType cycle for element for e 1numberElements indice nodal connectivities for each element indice elementNodese ndof lengthindice cycle for Gauss point for q 1sizegaussWeights1 GaussPoint gaussLocationsq xi GaussPoint1 eta GaussPoint2 shape functions and derivatives shapeFunctionnaturalDerivatives shapeFunctionsQxietaelemType Jacobian matrix inverse of Jacobian derivatives wrt xy JacobinvJacobianXYderivatives JacobiannodeCoordinatesindicenaturalDerivatives mass matrix massindiceindice massindiceindice shapeFunctionshapeFunctionthickness rhogaussWeightsqdetJacob massindicenumberNodesindicenumberNodes massindicenumberNodesindicenumberNodes shapeFunctionshapeFunctionI rhogaussWeightsqdetJacob massindice2numberNodesindice2numberNodes massindice2numberNodesindice2numberNodes shapeFunctionshapeFunctionI rhogaussWeightsqdetJacob end end Gauss point loop end end element loop end Codes for solving the free vibration problem with Q8 and Q9 elements are not shown for the sake of conciseness 137 Stability of Mindlin Plates 253 w1 1036663 1 w2 1995017 08 08 06 06 04 04 02 02 0 0 0 02 04 06 08 1 0 02 04 06 08 1 1 w3 1995017 w4 27951 08 08 06 06 04 04 02 02 0 0 0 02 04 06 08 1 0 02 04 06 08 1 Fig 133 Modes of vibration for a CCCC plate with ha 01 using 20 x 20 Q4 elements 137 Stability of Mindlin Plates In this section we formulate and implement the buckling analysis of Mindlin plates After presenting the basic finite element formulation we present a MATLAB code for buckling analysis of a simplysupported isotropic square plate under uniaxial initial stress In order to study the buckling problem of Mindlin plates the potential of initially stresses plates has to be considered Initial stress works for the nonlinear gradient of displacements as 6 T VO 0 en dV 1338 Vv 254 13 Mindlin Plates 1 w1 604444 4 w2 1451034 08 08 06 06 04 04 02 02 0 0 0 02 04 06 08 1 0 02 04 06 08 1 1 w3 1451034 4 w4 2226577 08 08 06 06 04 04 02 02 0 0 0 02 04 06 08 d 0 02 04 06 08 1 Fig 134 Modes of vibration for a SSSS plate with ha 01 using 20 x 20 Q4 elements T where o o oF TY and yz represents the nonlinear strains also known as Von Karman nonlinear strains as Ou Our Ouz3 2 Ox Ox Ox 1 Ou 2 Our Ouz3 2 nL 13 1339 2 Oy Oy Oy Ou Ou Our Ou Ou3 Ou gt eee ee Ox Oy Ox Oy Ox Oy 137 Stability of Mindlin Plates 255 1 w1 9186 4 w2 17432 08 08 06 06 04 04 02 02 0 0 0 02 04 06 08 1 0 02 04 06 08 1 w3 222583 1 w4 302419 08 08 06 06 04 04 02 02 0 0 0 02 04 06 08 1 0 02 04 06 08 1 Fig 135 Modes of vibration for a SCSC plate with ha 001 using 20 x 20 Q4 elements by including the Mindlin displacement field 131 the nonlinear strains take the form 2 2 2 z 06x z 065 Ow Ox Ox Ox 1 00 00 Ow ex 2 42 4 1340 2 Oy Oy Oy y 22 Oe 20M By dw dw Ox Oy Ox Oy Ox Oy the potential of second order displacements can be rewritten as 256 13 Mindlin Plates 1 w1 70446 1 w2 1134491 08 08 06 06 04 04 02 02 0 0 0 02 04 06 08 1 0 02 04 06 08 1 1 w3 1749087 1 w4 2093844 08 08 06 06 04 04 02 02 0 0 0 02 04 06 08 1 0 02 04 06 08 1 Fig 136 Modes of vibration for a CCCF plate with ha 01 using 20 x 20 Q4 elements 1 30 00 Aw 2 0 2 x 2 y Ve 5 za Dn sf fe 2 e 2 2 G2 2 2 2 of 2 OY 4 2 OH 4 Oe 1341 dy dy dy 4279 700 OO 4 790 06 4 ow Ow dv 7 x at ay TY 4 TT wy 5 By Oy ax Oy Ox Oy Rearranging the nonlinear terms we have 1 VO vu 6m 2VOTGVO Pvaléve av 1342 Vv 137 Stability of Mindlin Plates 257 where V 00x 00x is the gradient operator and 00 A0 Fx Try o Ki 3 1343 and finally 2 1 T 70 he T 70 he T 20 e Vi s hVw ao Vwt VO6 VO VO a VO JdQ 1344 2 Jae 12 12 by collecting all the terms in matrix form as 1 Vw VO 5 Vw Vor V0T8 VO dee 1345 ae VO where S is a banded 6 x 6 matrix as no 0 0 h 26 s 9 Go 0 1346 12 3 0 0 6 12 where 0 is a2 x 2 matrix of zeros Since the scope is to introduce the finite element approximation 1318 it is convenient to convert the vector of gradients as Vw V00 w Vd 0V06 Vu 1347 VO 00V 4 where 0 is a2 x matrix of zeros and V is a6 x 3 operator including partial deriva tives with respect to x and y Finally the second order potential 1345 can be rewritten in matrix form and the finite element approximation 1318 can be included as g T 60 e Ve 5 Vu SVudsx2 ao 1 1348 d7 VNSVNdQd a GSGdQd 2 Qe 2 Qe Thus the geometric stiffness matrix K is defined Ké GSGd2 1349 Re 258 13 Mindlin Plates where G is a6 x 3n matrix with the following structure Nix Nox NnxO O 0 0 O 0 Nyy Noy NnyO9 OO 0 0 O 0 G 0 O 0 Mix Nox NnxO OO 0 0 O 0 My No NnyO OO 0 0 O 0 0 O 0 My Nox Nnx 0 O 0 0 OO 0 My Noy Nay mee 1350 N 0 0 N 0 0 ON 0 ON 0 0 ON 0 ON where N and Nj fori 12n are the partial derivatives of the shape functions and N Nix Nox Nnx Ny Ni Noy Nny Due to the banded structure of G matrix two contributions can be identified so the geometric stiffness matrix Kg may be written as 6 KG KG KG 1351 The first term involves the derivatives of w and that is the conventional buckling term associated with the classical plate theory On the other hand the remaining parts socalled curvature terms become significant for moderately thick plates and play a role akin to the rotary inertia in the free vibration problem The bending contribution Kg in natural coordinates is given by 1 pl K G6Gyh det J dédn 1352 1J1 where N 00 G IN 0 0 1353 137 Stability of Mindlin Plates 259 Table 1313 Buckling factors an D for a simply supported square plate under uniaxial initial stress v 03 using 10 x 10 mesh ah Exact 6 Q4 Q8 Q9 1000 4000 40897 40033 40001 20 3944 40153 39287 39288 10 3786 38097 37315 37315 5 3264 31813 31255 31256 The shear contribution Kg is given by 1 1 h3 Ke G7 6G1 det Jdédn 1J1 12 1 1 r 20 h3 f Gio Gaz det J ddn 1354 where Gs lo x o Go lo Nn 1355 All the geometric stiffness matrix components should be carried out using reduced integration single point for Q4 and 2 x 2 for Q8 and Q9 elements This selection has demonstrated to have higher accuracy of the finite element solution The stability problem involves the solution of the eigenproblem K AKcga 0 1356 where K is the global stiffness matrix Kg is the global geometric matrix and is a constant by which the inplane loads must be multiplied to cause buckling Vector a represents the buckling mode correspondent to the buckling load factor A By solving the generalized eigevalue problem 1356 buckling loads and buckling modes can be carried out Table 1313 summarizes results for simply supported square plates of various thicknesses under uniaxial o initial stress We consider a 10 x 10 mesh Fig 138 and compare present finite element formulation with closed form solution 6 The schematic geometry loads and boundary conditions are illustrated in Fig 137 In Figs 139 1310 and 1311 the eigenmodes are illustrated for ah 10 Table 1314 lists the results for simplysupported plates of various thicknesses under uniaxial 0 and biaxial 7 initial stresses where y oy o using a 30 x 30 mesh The results are compared to the ones given in 9 The MATLAB code problem19Bucklingm computes the problem of a Mindlin plate under compressive uniaxial and biaxial loads 260 13 Mindlin Plates a a σ0 x σ0 x Simplysupported x y Fig 137 Buckling problem a Mindlin plate under uniaxial initial stress Fig 138 Buckling of Mindlin Plate 10 10 mesh This code calls function formGeometricStiffnessMindlinm for the computation of the geometric stiffness matrix MATLAB codes for Finite Element Analysis problem19Bucklingm Buckling analysis of Q4 Mindlin plates AJM Ferreira N Fantuzzi 2019 clear memory 137 Stability of Mindlin Plates 261 clear close all material E 10920 poisson 030 kapa 56 thickness 0001 I thicknessˆ312 constitutive matrix bending part Cbending IE1poissonˆ21 poisson 0 poisson 1 0 0 0 1poisson2 shear part Cshear kapathicknessE21poissoneye2 initial stress matrix sigmaX 1thickness sigmaXY 0 sigmaY 0 sigmaMatrix sigmaX sigmaXY sigmaXY sigmaY mesh generation L 1 numberElementsX number of elements in x numberElementsY number of elements in y numberElementsX 20 numberElementsY 20 number of elements numberElements numberElementsXnumberElementsY nodeCoordinates elementNodes rectangularMeshLLnumberElementsXnumberElementsYQ4 xx nodeCoordinates1 yy nodeCoordinates2 figure drawingMeshnodeCoordinateselementNodesQ4 axis equal numberNodes sizexx1 number of nodes GDof 3numberNodes total number of DOFs stiffness and geometric stiffness matrices stiffness formStiffnessMatrixMindlinGDofnumberElements elementNodesnumberNodesnodeCoordinatesCshear CbendingQ4completereduced geometric formGeometricStiffnessMindlinGDofnumberElements elementNodesnumberNodesnodeCoordinatessigmaMatrix thicknessQ4reducedreduced Essential boundary conditions prescribedDofactiveDof EssentialBCssssGDofxxyynodeCoordinatesnumberNodes buckling analysis modeslambda eigenvalueGDofprescribedDof stiffnessgeometric15 262 13 Mindlin Plates sort out eigenvalues lambdaii sortlambda modes modesii dimensionless omega lambdabar lambdaLLpipiCbending11 drawing mesh and deformed shape modeNumber 1 displacements modesmodeNumber surface representation figure hold on for k 1sizeelementNodes1 patchnodeCoordinateselementNodesk141 nodeCoordinateselementNodesk142 displacementselementNodesk14 displacementselementNodesk14 end setgcafontsize18 view4545 plot of the first 4 eigenmodes for k 14 modeNumber k displacements modesmodeNumber contourFieldnumberElementselementNodesxxyy displacements1numberNodesInfInf11221 titlelambdanum2strk num2strlambdak colorbar off box on setgcalinewidth5fontsize18 end function KG formGeometricStiffnessMindlinGDofnumberElements elementNodesnumberNodesnodeCoordinatessigmaMatrix thicknesselemTypequadTypeBquadTypeS computation of geometric stiffness for Mindlin plate element KG geometric matrix KG zerosGDof Gauss quadrature for bending part gaussWeightsgaussLocations gaussQuadraturequadTypeB cycle for element for e 1numberElements indice nodal connectivities for each element elementDof element degrees of freedom indice elementNodese elementDof indice indicenumberNodes indice2numberNodes ndof lengthindice 137 Stability of Mindlin Plates 263 cycle for Gauss point for q 1sizegaussWeights1 GaussPoint gaussLocationsq xi GaussPoint1 eta GaussPoint2 shape functions and derivatives shapeFunctionnaturalDerivatives shapeFunctionsQxietaelemType Jacobian matrix inverse of Jacobian derivatives wrt xy JacobinvJacobianXYderivatives JacobiannodeCoordinatesindicenaturalDerivatives geometric matrix Gb zeros23ndof Gb11ndof XYderivatives1 Gb21ndof XYderivatives2 KGelementDofelementDof KGelementDofelementDof GbsigmaMatrixthicknessGb gaussWeightsqdetJacob end end Gauss point loop end end element loop Gauss quadrature for shear part gaussWeightsgaussLocations gaussQuadraturequadTypeS cycle for element for e 1numberElements indice nodal connectivities for each element elementDof element degrees of freedom indice elementNodese elementDof indice indicenumberNodes indice2numberNodes ndof lengthindice cycle for Gauss point for q 1sizegaussWeights1 GaussPoint gaussLocationsq xi GaussPoint1 eta GaussPoint2 shape functions and derivatives shapeFunctionnaturalDerivatives shapeFunctionsQxietaelemType Jacobian matrix inverse of Jacobian derivatives wrt xy JacobinvJacobianXYderivatives JacobiannodeCoordinatesindicenaturalDerivatives Geometric matrix Gs1 zeros23ndof Gs11ndof12ndof XYderivatives1 Gs12ndof12ndof XYderivatives2 264 13 Mindlin Plates KGelementDofelementDof KGelementDofelementDof Gs1sigmaMatrixthicknessˆ312Gs1 gaussWeightsqdetJacob Gs2 zeros23ndof Gs212ndof13ndof XYderivatives1 Gs222ndof13ndof XYderivatives2 KGelementDofelementDof KGelementDofelementDof Gs2sigmaMatrixthicknessˆ312Gs2 gaussWeightsqdetJacob end end Gauss point loop end end element loop end Fig 139 Buckling modes 14 for a SSSS plate with ha 0001 using 20 20 Q4 elements 137 Stability of Mindlin Plates 265 Fig 1310 Buckling modes 58 for a SSSS plate with ha 0001 using 20 20 Q4 elements Codes problem19aBucklingm and problem19bBucklingm solve Q8 and Q9 buckling Mindlin problem They are not shown for the sake of conciseness and they can be simply derived changing the call to the respective functions Table1314 266 13 Mindlin Plates Fig 1311 Buckling modes 912 for a SSSS plate with ha 0001 using 20 20 Q4 elements Table 1314 Buckling factors λ λb2π2D for simply supported plates under uniaxial γ 0 and biaxial γ 1 initial stress using 30 30 mesh ab 05 1 γ bh Ref 9 Q4 Q8 Q9 Ref 9 Q4 Q8 Q9 0 10 5523 53209 53092 53092 3800 37400 37314 37314 20 6051 59945 59798 59798 3948 39381 39287 39287 100 6242 62546 62386 62386 3998 40069 39971 39971 1000 6250 62659 62499 62499 4000 40098 40000 40000 1 10 4418 42568 42474 42474 1900 18700 18657 18657 20 4841 47956 47839 47839 1974 19691 19643 19643 100 4993 50037 49909 49909 1999 20034 19985 19985 1000 5000 50127 49999 49999 2000 20049 20000 20000 Kirchhoff theory or Classical Plate theory CPT 9 References 267 References 1 JN Reddy Mechanics of laminated composite plates CRC Press New York 1997 2 M Petyt Introduction to finite element vibration analysis Cambridge University Press 1990 3 JN Reddy An introduction to the finite element method McGrawHill International Editions New York 1993 4 KJ Bathe Finite element procedures in engineering analysis Prentice Hall 1982 5 JN Reddy An introduction to the finite element method 3rd edn McGrawHill International Editions New York 2005 6 E Hinton Numerical methods and software for dynamic analysis of plates and shells Pineridge Press 1988 7 DJ Dawe OL Roufaeil Rayleighritz vibration analysis of mindlin plates J Sound Vib 693 345359 1980 8 KM Liew J Wang TY Ng MJ Tan Free vibration and buckling analyses of sheardeformable plates based on fsdt meshfree method J Sound Vib 276 9971017 2004 9 JN Reddy Energy principles and variational methods in applied mechanics 3rd edn Wiley Hoboken NJ USA 2017 Chapter 14 Laminated Plates Sack og Abstract In this chapter we consider a first order shear deformation theory for the static free vibration and buckling analysis of laminated plates We introduce a computation of the shear correction factor and solve some examples with MAT LAB codes The main difference between the present chapter and the previous one related to Mindlin plates is that due to lamination there might be a coupling between membrane and bending behaviors 141 Introduction Here we consider a first order shear deformation theory for the static free vibration and buckling analysis of laminated plates We introduce a computation of the shear correction factor and solve some examples with MATLAB codes The main differ ence between the present chapter and the previous one related to Mindlin plates is that due to lamination there might be a coupling between membrane and bending behaviors 142 Displacement Field In the first order shear deformation theory displacements are the same as in Mindlin plate theory plus the inplane displacements as in plane stress analysis as ux ys a t ux y t 20x x y t ux yZt vx yt z0 y t 141 u3x y 20 we y0 the displacement vector can be defined as u u vw 0 Oy The Editors if applicable and The Authors under exclusive 269 license to Springer Nature Switzerland AG 2020 A J M Ferreira and N Fantuzzi MATLAB Codes for Finite Element Analysis Solid Mechanics and Its Applications 157 httpsdoiorg101007978303047952714 270 14 Laminated Plates 143 Strains Straindisplacement relations by neglecting normal transverse strain according to the first order assumptions 1 with Von Karman strains are obtained by derivation as Ou 0c 1 dw Ox ax 2 ax dv 8y 1 dw éy z Oy Oy 2 dy Ou Ov 00 O06 Ow Ow My 5t5t27442 142 Oy Ox Oy Ox Ox Oy Ow xz 6 Yxe Ox Ow Yyz Oy ye y Oy or in matrix form as du 1 2 00 ax ey Au e Ox 2 0x a ee6 au z 5 3 2 By 0 1 y 2d0y Yet Ly wey du Ov dw aw 0 OBy Oy Ox Ox dy Oy Ox 143 Ow 0 0 y 35 ay 144 Vxz Vez 6 ow Ox Note that membrane strains are linear through the thickness whereas shear strains are constant Thus strains can be conveniently collected as e 6 e y y 6 6 c He z 2 0 rs 0 O40 se Ie WO Lb T T T where VO 7 WW 7 and e e YO 143 Strains 271 Strain characteristics and e can be written according to the displace ment parameters of the model giving the definition of three differential operators as 0 0 000 Ox u 0 VU O 0 000 W Du Oy Ox as 0 goo Oy Ox 000 0 0 Ox u 9 v 146 Y 000 0 na W Du dy Ox a a L 000 Oy Ox O u 00 10 Ox v 7 w Du 0 0oo1 Oy 6 144 Stresses Laminated composite plates are considered made of several orthotropic plies Normal stress a is neglected so reduced elastic constants should be used The stressstrain relations in laminate coordinates 1 are k 4 k o Q11 Qi2 Vio Ex Oy Qir Qo Or y 147 Oxy O16 Q26 Qe Yay k A A k 2 ie ss 148 Oxz Qs Oss Yxz where k identifies the generic lamina and O jt j 124 5 6 are the reduced elastic coefficients 1 272 14 Laminated Plates In matrix form the constitutive equations for the generic ply become TI LO QI Ly by including the straindisplacement relations 146 Cn QnE zOne T Qy 1410 Due to constant shear stresses given by the model the shear correction factors K1 K should be included Being K K the shear correction factors in both directions At each layer inter face the transverse shear continuity must be guaranteed The equilibrium equation in x direction is written as Ooy OTxy OT xz 424 90 1411 Ox Oy Oz Assume cylindrical bending é Oox OM D4z Ox Qx Txz sz dz sy edz Dy zzdz gz i Ox nj2 Ox Ry Ry Jn2 Ri 1412 where e Q is the shear force in xz plane h2 eR D zzdz represents the plate stiffness in x direction h2 e zis the thickness coordinate Zz egz Dzzdz represents the shear shape h2 Function gz which represents the shear stress diagram becomes parabolic for homogeneous sections gz Dh81 Azh The strain energy is given by A2 72 2 ph2 42 Ws gz 8 4 1413 nj2 G13Z Ry Jnj2 G13 2 being G3z is the transverse shear modulus in xz plane For a constant transverse shear deformation the strain is given by h2 Q2 OQ Ws Vx2Gi3ZVxdzZ hG 1414 WG hG 144 Stresses 273 where h2 hG G13zdz 1415 h2 and 7 is the mean value for transverse shear strain It is now possible to obtain the shear correction factor K as Ws R2 Kj ey 1416 Ws 5 WG fe Gntedz h2 To obtain the second shear factor K2 we proceed in a similar way 2 145 Hamiltons Principle Governing equations of the present theory are derived using the Hamiltons Principle Strain energy is given by 1 for lf oro T 4 27 0 U oedV o720 7 y dV 1417 2 Jy 2 Jv by including the stress definitions 1410 where the subscript has been removed for the sake of simplicity 1 e 06 yO OqyOdV 1418 since the plate is considered made of several plies the volume integral can take the form ne Zk41 ave fajae dzd2 1419 Vv 2 z DQ N pay PR where nc is the number of plies in the stacking sequence of the laminate Stiffness constants can be defined as 274 14 Laminated Plates Fig 141 Laminated plate x organization of layers in the thickness direction ek1 Zk ne Ondz SOP Gra A é kl A Ie k 52 2 ZQn dz 2 as Zit z B Zz 1420 1 2 k 53 3 Qn dz 5 YO yi st D é kl ne x0 dz KQ Zk41 Zk A é k1 where K Kz K is the shear correction factor Figure 141 illustrates the position of the z coordinates across the thickness direction In conclusion the strain energy becomes U OT ACO 4 OTRO 4 OTB 2 Jo ODE 4 YOTA Jac 1421 The potential is Ve pwdV up dV 1422 Vv Vv 145 Hamiltons Principle 275 where p 0 0 pd oy thus it is assumed that only transverse loads p are applied to the plate The kinetic energy takes the form 1 K 5 o 20 205 wav 1423 Vv performing multiplications and by introducing the inertia terms ne Zk1 I y pzidz for i012 1424 kl where nc indicates the number of layers the kinetic energy becomes 1 K5 sa 0 w 21 U0 105 H Jae 1425 Q in matrix form it can be written as lf wray K wMudQ 1426 2 Ja where u i bw Oy 6 and the inertia matrix is given by 00 0 0OhHh00h M0010 0 1427 00h0 Ondvd0h 146 Finite Element Approximation The displacement parameters are independently interpolated using the same func tions as in the Mindlin problem wo YO NilEmui vo D NiEmvi wo D NiE nwi il il il 0 ONE MOxi Oy D NiEmOyi 1428 il il where n depend on the order of the finite element used Such approximation can be easily rewritten in matrix form by introducing the shape function matrix as 276 14 Laminated Plates Uo No000 Ta Uo 0ON000 v uwo00N00 W Nad 1429 A 000NO0 4 Oy 0000NL where N is a matrix of size 5 x 5n which includes the shape functions N for each T a T T A degree of freedom u ui Lee Un v v1 Lee vn w wi tee Wn O Oct Orn Oy1 Ayn 1461 StrainDisplacement Matrices The straindisplacement matrices B B and B are derived by including the finite element approximation in the Eqs 146 as Du DNd Bd e Dyu DNd Bid 1430 y Du DNd BO d The membrane component B of size 3 x 5 is given by oN 0 000 Ox B 0 ON 900 for i12n 1431 dy ON ON 000 Oy Ox the bending component BY of size 3 x 5n is oN 000 0 Ox BY 000 0 ON for i12n 1432 Oy ON ON 000 Oy Ox 146 Finite Element Approximation 277 and the shear component B is ON 00 NO Ox Bo for i127 1433 ON o 00ON Oy 1462 Stiffness Matrix After the introduction of the straindisplacement matrices B BY and B the strain energy for the finite element becomes e 1 eT T A e eT e eT e US 54 o By AB B BB B BB BY DB BOTABO Jane d 1434 Thus the following matrices are given Ko fo By AB d2 Kj Jo By BBO aa Kn Jae BT BB dQ 1435 K fo By DB d2 KO BOTA BOdQ The element stiffness matrix is defined by the sum of all these components as K Ky Ky King Kip KY 1436 Note that to avoid shear locking the shear component of the stiffness matrix is com puted using reduced integration as it was introduced in the previous chapter All the integrals are evaluated in the natural coordinate system for instance one term of the stiffness matrix 1436 is given by 1 pl Ko BY AB det Jddn 1437 1J1 thus Gauss integration can be easily applied 278 14 Laminated Plates 1463 Load Vector The potential of the external loads with the finite element approximation becomes Ve d N pdQ d f 1438 2 where f are the equivalent nodal values of the finite element due to the load In natural coordinates the force vector takes the form 1 pl f N pdet J dédn 1439 1J1 1464 Mass Matrix The mass matrix can be carried out by including the finite element approximation into the kinetic energy 1426 as e eT T e qe Kf 54 N MN dX d 1440 2 and the mass matrix can be carried out as M N MN dx 1441 Re Finally the mass matrix can be written in natural coordinates as 1 pl M N MN det J dédn 1442 1J1 147 Stress Recovery Once the nodal solution is carried out d stresses can be recovered from constitutive equations please note that dependency has been omitted for the sake of simplicity as O 0 0 d om Que Qné Qnze 1443 QnBn Q1zB d T Q7 QBd 1444 147 Stress Recovery 279 It is noted that σm and τ are only evaluated at the integration points Gauss Legendre points Values for the element corner points can be obtained by extrapo lation as shown in the previous chapters 148 Static Analysis We analyze a laminated sandwich 3layer square plate a b simplysupported on all sides under uniform pressure This is known as the Srinivas problem 3 with the following core properties Qcore 0999781 0231192 0 0 0 0231192 0524886 0 0 0 0 0 0262931 0 0 0 0 0 0266810 0 0 0 0 0 0159914 The material properties for the skins are obtained from those of the core and a multiplying factor R Qskin RQcore The thickness of the skins is h10 and the one of the core is 4h5 In this example we present transverse displacement and stresses in dimensionless form w 0999781 w a 2 a 2 0 hq σ1 x σ1 x a 2 a 2 h 2 q σ2 x σ1 x a 2 a 2 2h 5 q σ3 x σ2 x a 2 a 2 2h 5 q σ1 y σ1 y a 2 a 2 h 2 q σ2 y σ1 y a 2 a 2 2h 5 q σ3 y σ2 y a 2 a 2 2h 5 q τ 1 xz τ 2 xz 0 a 2 0 q τ 2 xz τ 2 xz 0 a 2 2h 5 q For various values of R we compare results with thirdorder theory of Pandya 4 and various finite element and meshless results by Ferreira 5 6 Results are quite good with the exception of the transverse shear stresses that should be further corrected 2 τ cor xz G13γxz gz g 1445 280 14 Laminated Plates where h2 g gzdz 1446 h2 A viable alternative for the computation of the transverse shear stresses is to solve the 3D equilibrium equation by considering o 7yy and Ty from the solution of the 2D problem Tables 141 142 and 143 list all the present results for Q4 Q8 and Q9 elements The codes of the Q8 and Q9 elements are not shown here for the sake of brevity Code problem20m solves this problem Bc ee ee ee ee ee eee ee ee eee eee eee eee MATLAB codes for Finite Element Analysis problem20m laminated plate Srinivas problem using Q4 elements S Srinivas A refined analysis of composite laminates J Sound and Vibration 30 1973 495507 AJM Ferreira N Fantuzzi 2019 6 clear memory clear close all materials thickness 01 load P 1 mesh generation L 1 numberElementsX 10 numberElementsY 10 numberElements numberElementsxXnumberElementsY nodeCoordinates elementNodes rectangularMeshLLnumberElementsxXnumberElementsyY Q4 xx nodeCoordinates1 yy nodeCoordinates2 figure drawingMesh nodeCoordinates elementNodes Q4 axis equal numberNodes sizexx1 GDof global number of degrees of freedom GDof 5numberNodes computation of the system stiffness matrix the shear correction factors are automatically calculted for any laminate 148 Static Analysis 281 AMatrixBMatrixDMatrixSMatrixqbarra srinivasMaterialthickness stiffness formStiffnessMatrixMindlinlaminated5dof GDofnumberElements elementNodesnumberNodesnodeCoordinatesAMatrix BMatrixDMatrixSMatrixQ4completereduced computation of the system force vector force formForceVectorMindlin5dofGDofnumberElements elementNodesnumberNodesnodeCoordinatesPQ4reduced boundary conditions prescribedDofactiveDof EssentialBC5dofssssGDofxxyynodeCoordinatesnumberNodes solution U solutionGDofprescribedDofstiffnessforce drawing deformed shape and normalize results to compare with Srinivas ws 1numberNodes dispmaximum displacement absminUws0999781thickness surface representation figure hold on for k 1sizeelementNodes1 patchnodeCoordinateselementNodesk141 nodeCoordinateselementNodesk142 UelementNodesk14 UelementNodesk14 end setgcafontsize18 view4545 stress computation Srinivas only dispstress computation Srinivas only stresslayer1stresslayer2 stresslayer3shearlayer1 shearlayer2 SrinivasStressGDofnumberElements elementNodesnumberNodesnodeCoordinates qbarraUthicknessQ4completereduced normalized stresses look for table in the book format absminstresslayer331 absminstresslayer231 absminstresslayer131 maxshearlayer211 maxshearlayer111 282 14 Laminated Plates Table 141 Square sandwich plate under uniform pressure R 5 Method w σ1 x σ2 x σ3 x τ1 xz τ2 xz HSDT 4 25613 6238 4691 9382 3089 2566 FSDT 4 23610 6187 4950 9899 3313 2444 CLT 21694 61141 48623 9783 45899 3386 Ferreira 5 25874 5921 4561 9122 3593 3593 Ferreira N 15 6 25738 58725 46980 9396 3848 2839 Analytical 3 25897 60353 46623 9340 43641 32675 HSDT 7 N 11 2536710 596447 464292 92858 38449 19650 HSDT 7 N 15 2562387 601834 468581 93716 42768 22227 HSDT 7 N 21 2571100 603660 470028 94006 45481 23910 Present 4 4 Q4 2600321 546108 436887 87377 23922 119608 Present 10 10 Q4 2593004 584403 467523 93505 29841 149207 Present 20 20 Q4 2592797 589507 471606 94321 31980 159902 Present 4 4 Q8 2590307 580208 464167 92833 29432 147159 Present 10 10 Q8 2592715 591667 473334 94667 32269 161347 Present 20 20 Q8 2592778 590739 472591 94518 33254 166272 Present 4 4 Q9 2596740 596249 476999 95400 29332 146661 Present 10 10 Q9 2592875 591653 473323 94665 32269 161347 Present 20 20 Q9 2592788 591302 473042 94608 33254 166272 148 Static Analysis 283 Table 142 Square sandwich plate under uniform pressure R 10 Method w σ1 x σ2 x σ3 x τ1 xz τ2 xz HSDT 4 15233 6465 5131 5131 3147 2587 FSDT 4 131095 6780 5424 4424 3152 2676 CLT 11887 65332 48857 5356 43666 37075 Ferreira 5 159402 6416 4772 4772 3518 3518 Ferreira N 15 6 15855 62723 5016 501 3596 3053 Analytical 3 15938 65332 48857 4903 40959 35154 HSDT 7 N 11 1530084 647415 494716 49472 27780 18207 HSDT 7 N 15 1542490 652223 498488 49849 31925 21360 HSDT 7 N 21 1546581 653809 499729 49973 35280 23984 Present 4 4 Q4 1622395 581236 464989 46499 15126 151261 Present 10 10 Q4 1599120 623765 499012 49901 18995 189954 Present 20 20 Q4 1596820 629474 503580 50358 20371 203713 Present 4 4 Q8 1594510 619570 495656 49566 18721 187207 Present 10 10 Q8 1596065 629404 503523 50352 20557 205571 Present 20 20 Q8 1596108 630874 504699 50470 21190 211903 Present 4 4 Q9 1598469 619773 495818 49582 18674 186735 Present 10 10 Q9 1596166 629403 503523 50352 20557 205571 Present 20 20 Q9 1596114 630874 504699 50470 21190 211903 284 14 Laminated Plates Table 143 Square sandwich plate under uniform pressure R 15 Method w σ1 x σ2 x σ3 x τ1 xz τ2 xz HSDT 4 11043 6662 5197 3465 3035 2691 FSDT 4 9085 7004 5603 3753 3091 2764 CLT 81768 69135 55308 3687 42825 38287 Ferreira 5 121821 65650 4709 3140 3466 3466 Ferreira N 15 6 121184 63214 50571 3371 3466 3099 Analytical 3 12172 66787 48299 3238 39638 35768 HSDT 7 N 11 1135941 663646 498957 33264 21686 15578 HSDT 7 N 15 1143874 667830 502175 33478 26115 19271 HSDT 7 N 21 1146442 669196 503230 33549 30213 22750 Present 4 4 Q4 1252176 584574 467659 31177 10975 164621 Present 10 10 Q4 1223318 628602 502881 33525 13857 207849 Present 20 20 Q4 1220283 634574 507659 33844 14872 223084 Present 4 4 Q8 1218046 624614 499691 33313 13653 204795 Present 10 10 Q8 1219292 634574 507660 33844 15010 225146 Present 20 20 Q8 1219327 636058 508847 33923 15476 232142 Present 4 4 Q9 1221077 624785 499828 33322 13624 204362 Present 10 10 Q9 1219371 634574 507659 33844 15010 225146 Present 20 20 Q9 1219332 636058 508847 33923 15476 232142 148 Static Analysis 285 The computation of the material constitutive matrices and shear correction compu tation is made in function srinivasMaterialm Since the lamination is symmetrical computation of the coupling constitutive matrix B is not needed The implementation of membranebending coupling stiffnesses are left to the reader for generalizing the present code function AMatrixBMatrixDMatrixSMatrixqbarra srinivasMaterialthickness SRINIVAS EXAMPLE multiplying factor for skins rf 15 plate thickness h thickness matrix D inplane dmat 0999781 0231192 0 0231192 0524886 00 0 0262931 shear dm 026681 0 0 0159914 nc number of layers nc 3 layers angles ttt 0 ttt1 0 th1 ttt th2 ttt1 th3 ttt1 coordinates z1 upper and z2 lower for each layer z12h5 2h5 h2 z2h2 2h5 2h5 thickness for each layer thick1nc z11nc z21nc coefe shear correction factors k1 and k2 coefe12 00 gbarf12 00 rfact12 00 sumla12 00 trlow12 00 upter12 00 middle axis position bending dsumm 00 for ilayr 1nc dzeta z1ilayr z2ilayr zheig dsumm thickilayr2 dindx1 rfdmat11 dindx2 dmat22 upter12 upter12 dindx12zheigthickilayr trlow12 trlow12 dindx12thickilayr dsumm dsumm thickilayr end zeta212 upter12trlow12 shear correction factors calculation for ilayr1nc diff1z1ilayrz2ilayr d1rfdmat11 d2rfdmat22 d3rfdm11 d4rfdm22 ifilayr2 286 14 Laminated Plates d1dmat11 d3dm11 d4dm22 d2dmat22 end index10 for i12 zeta1izeta2i zeta2izeta1idiff1 diff2izeta2iˆ2zeta1iˆ2 diff3izeta2iˆ3zeta1iˆ3 diff5izeta2iˆ5zeta1iˆ5 dindxd1d2 gindxd3d4 gbarfigbarfigindxidiff120 rfactirfactidindxidiff3i30 term1 sumlaisumlaidiff1 term2 dindxizeta1iˆ4diff140 term3 dindxidiff5i200 term4 dindxizeta1izeta1idiff3i60 term5 sumlaizeta1izeta1idiff1 term6 sumlaidiff3i30 coefei coefeiterm1dindxi term2term3term4term5term6gindxi index index1 sumlai sumlaidindxidiff2i20 end end coefe12 rfact12rfact1220gbarf12coefe12 kapa coefe1 constitutive matrix membrane bending and shear a11 0 a22 0 a12 0 a33 0 dd zeros2 d zeros3 for i 1nc theta thi q11 rfdmat11 q12 rfdmat12 q22 rfdmat22 q33 rfdmat33 cs costheta ss sintheta ss11 rfdm11kapa ss22 rfdm22kapa if i 2 core layer q11 dmat11 q12 dmat12 q22 dmat22 q33 dmat33 cs costheta ss sintheta ss11 dm11kapa ss22 dm22kapa end dd11 dd11 ss11costhetaˆ2 ss22sinthetaˆ2z1iz2i dd22 dd22 ss11sinthetaˆ2 ss22costhetaˆ2z1iz2i d11 d11 q11csˆ42q122q33sssscscs q22ssˆ4z1iˆ3z2iˆ33 d22 d22 q11ssˆ42q122q33sssscscs q22csˆ4z1iˆ3z2iˆ33 148 Static Analysis 287 d12 d12 q11q224q33sssscscs q12ssˆ4csˆ4z1iˆ3z2iˆ33 d33 d33q11q222q122q33sssscscs q33ssˆ4csˆ4z1iˆ3z2iˆ33 a11 a11 q11thicki a22 a22 q22thicki a33 a22 q33thicki a12 a12 q12thicki reduced stiffness coefficients qbarra11i q11 qbarra12i q12 qbarra22i q22 qbarra33i q33 qbarra44i ss11 qbarra55i ss22 end nc equivalent membrane bending and coupling stiffness monoclinic coeff are not present due to symmetric lamination scheme A44 dd22 A55 dd11 D11 d11 D12 d12 D22 d22 D66 d33 A11 a11 A12 a12 A66 a33 A22 a22 AMatrix A11A120A12A22000A66 srinivas case symmetric BMatrix zeros3 BMatrix B11B120B12B22000B66 DMatrix D11D120D12D22000D66 SMatrix A4400A55 end Because this plate element has 5 degrees of freedom instead of 3 degrees of freedom as in Mindlin plates some changes were introduced and new functions are needed Function formStiffnessMatrixMindlinlaminated5dofm computes the stiffness matrix for the Q4 Q8 and Q9 Mindlin plate with 5 DOFs function K formStiffnessMatrixMindlinlaminated5dofGDofnumberElements elementNodesnumberNodesnodeCoordinatesAMatrix BMatrixDMatrixSMatrixelemTypequadTypeBquadTypeS computation of stiffness matrix for laminated plate element 288 14 Laminated Plates K stiffness matrix K zerosGDof Gauss quadrature for bending part gaussWeightsgaussLocations gaussQuadraturequadTypeB cycle for element for e 1numberElements indice nodal connectivities for each element elementDof element degrees of freedom indice elementNodese elementDof indice indicenumberNodes indice2numberNodes indice3numberNodes indice4numberNodes ndof lengthindice cycle for Gauss point for q 1sizegaussWeights1 GaussPoint gaussLocationsq xi GaussPoint1 eta GaussPoint2 shape functions and derivatives shapeFunctionnaturalDerivatives shapeFunctionsQxietaelemType Jacobian matrix inverse of Jacobian derivatives wrt xy JacobinvJacobianXYderivatives JacobiannodeCoordinatesindicenaturalDerivatives B matrix bending Bb zeros35ndof Bb1ndof12ndof XYderivatives1 Bb22ndof13ndof XYderivatives2 Bb3ndof12ndof XYderivatives2 Bb32ndof13ndof XYderivatives1 B matrix membrane Bm zeros35ndof Bm13ndof14ndof XYderivatives1 Bm24ndof15ndof XYderivatives2 Bm33ndof14ndof XYderivatives2 Bm34ndof15ndof XYderivatives1 stiffness matrix bendingbending KelementDofelementDof KelementDofelementDof BbDMatrixBbgaussWeightsqdetJacob membranemembrane KelementDofelementDof KelementDofelementDof BmAMatrixBmgaussWeightsqdetJacob membranebending KelementDofelementDof KelementDofelementDof BmBMatrixBbgaussWeightsqdetJacob bendingmembrane KelementDofelementDof KelementDofelementDof BbBMatrixBmgaussWeightsqdetJacob end Gauss point 148 Static Analysis 289 end element shear stiffness matrix Gauss quadrature for shear part gaussWeightsgaussLocations gaussQuadraturequadTypeS cycle for element for e 1numberElements indice nodal connectivities for each element elementDof element degrees of freedom indice elementNodese elementDof indice indicenumberNodes indice2numberNodes indice3numberNodes indice4numberNodes ndof lengthindice cycle for Gauss point for q 1sizegaussWeights1 GaussPoint gaussLocationsq xi GaussPoint1 eta GaussPoint2 shape functions and derivatives shapeFunctionnaturalDerivatives shapeFunctionsQxietaelemType Jacobian matrix inverse of Jacobian derivatives wrt xy JacobinvJacobianXYderivatives JacobiannodeCoordinatesindicenaturalDerivatives B matrix shear Bs zeros25ndof Bs11ndof XYderivatives1 Bs21ndof XYderivatives2 Bs1ndof12ndof shapeFunction Bs22ndof13ndof shapeFunction stiffness matrix shear KelementDofelementDof KelementDofelementDof BsSMatrixBsgaussWeightsqdetJacob end end gauss point loop end end element loop end Function formForceVectorMindlin5dofm computes the corresponding force vector function force formForceVectorMindlin5dofGDofnumberElements elementNodesnumberNodesnodeCoordinatesP elemTypequadType 290 14 Laminated Plates computation of force vector for laminated plate element force force vector force zerosGDof1 Gauss quadrature for bending part gaussWeightsgaussLocations gaussQuadraturequadType cycle for element for e 1numberElements indice nodal connectivities for each element indice elementNodese cycle for Gauss point for q 1sizegaussWeights1 GaussPoint gaussLocationsq GaussWeight gaussWeightsq xi GaussPoint1 eta GaussPoint2 shape functions and derivatives shapeFunctionnaturalDerivatives shapeFunctionsQxietaelemType Jacobian matrix inverse of Jacobian derivatives wrt xy JacobinvJacobianXYderivatives JacobiannodeCoordinatesindicenaturalDerivatives force vector forceindice forceindice shapeFunctionPdetJacobGaussWeight end end Gauss point loop end end element loop end Function EssentialBC5dofm defines the constrained degrees of freedom in vector form according to the selected condition function prescribedDofactiveDoffixedNodeW EssentialBC5doftypeBCGDofxxyynodeCoordinatesnumberNodes essentialBoundary conditions for recatngular plates 5Dof switch typeBC case ssss fixedNodeW findxxmaxnodeCoordinates1 xxminnodeCoordinates1 yyminnodeCoordinates2 yymaxnodeCoordinates2 fixedNodeTX findyymaxnodeCoordinates2 yyminnodeCoordinates2 fixedNodeTY findxxmaxnodeCoordinates1 148 Static Analysis 291 xxminnodeCoordinates1 fixedNodeU findxxminnodeCoordinates1 fixedNodeV findyyminnodeCoordinates2 fixedNodeU findyymaxnodeCoordinates2 yyminnodeCoordinates2 fixedNodeV findxxmaxnodeCoordinates1 xxminnodeCoordinates1 case ssss2 fixedNodeW findxxmaxnodeCoordinates1 xxminnodeCoordinates1 yyminnodeCoordinates2 yymaxnodeCoordinates2 fixedNodeTX findyymaxnodeCoordinates2 yyminnodeCoordinates2 fixedNodeTY findxxmaxnodeCoordinates1 xxminnodeCoordinates1 fixedNodeU findxxmaxnodeCoordinates1 xxminnodeCoordinates1 fixedNodeV findyymaxnodeCoordinates2 yyminnodeCoordinates2 case cccc fixedNodeW findxxmaxnodeCoordinates1 xxminnodeCoordinates1 yyminnodeCoordinates2 yymaxnodeCoordinates2 fixedNodeTX fixedNodeW fixedNodeTY fixedNodeTX fixedNodeU fixedNodeTX fixedNodeV fixedNodeTX end prescribedDoffixedNodeWfixedNodeTXnumberNodes fixedNodeTY2numberNodes fixedNodeU3numberNodesfixedNodeV4numberNodes activeDofsetdiff1GDofprescribedDof For the Srinivas example stresses are postprocessed in function SrinivasStressm function stresslayer1stresslayer2stresslayer3 shearlayer1shearlayer2 SrinivasStressGDof numberElementselementNodesnumberNodesnodeCoordinates qbarraUhelemTypequadTypeBquadTypeS computes normal and shear stresses for Srinivas case note that transverse shear stresses are not corrected normal stresses in each layer stresslayer1 zerosnumberElements43 stresslayer2 zerosnumberElements43 stresslayer3 zerosnumberElements43 292 14 Laminated Plates Gauss quadrature for bending part gaussWeightsgaussLocations gaussQuadraturequadTypeB cycle for element for e 1numberElements indice nodal connectivities for each element indiceB element degrees of freedom indice elementNodese indiceB indice indicenumberNodes indice2numberNodes indice3numberNodes indice4numberNodes nn lengthindice cycle for Gauss point for q 1sizegaussWeights1 pt gaussLocationsq wt gaussWeightsq xi pt1 eta pt2 shape functions and derivatives shapeFunctionnaturalDerivatives shapeFunctionsQxietaelemType Jacobian matrix inverse of Jacobian derivatives wrt xy JacobinvJacobianXYderivatives JacobiannodeCoordinatesindicenaturalDerivatives B matrix bending Bbzeros35nn Bb1nn12nn XYderivatives1 Bb22nn13nn XYderivatives2 Bb3nn12nn XYderivatives2 Bb32nn13nn XYderivatives1 B matrix membrane Bmzeros35nn Bm13nn14nn XYderivatives1 Bm24nn15nn XYderivatives2 Bm33nn14nn XYderivatives2 Bm34nn15nn XYderivatives1 stresses stresslayer1eq 2h5qbarra13132BbUindiceB qbarra13132BmUindiceB stresslayer2eq 2h5qbarra13133BbUindiceB qbarra13133BmUindiceB stresslayer3eq h2qbarra13133BbUindiceB qbarra13133BmUindiceB end end Gauss point loop end end element loop shear stresses in each layer by constitutive equations shearlayer1 zerosnumberElements12 148 Static Analysis 293 shearlayer2 zerosnumberElements12 shearlayer3 zerosnumberElements12 Gauss quadrature for shear part gaussWeightsgaussLocations gaussQuadraturequadTypeS cycle for element for e 1numberElements indice nodal connectivities for each element indiceB element degrees of freedom indice elementNodese indiceB indice indicenumberNodes indice2numberNodes indice3numberNodes indice4numberNodes nn lengthindice cycle for Gauss point for q 1sizegaussWeights1 pt gaussLocationsq wt gaussWeightsq xi pt1 eta pt2 shape functions and derivatives shapeFunctionnaturalDerivatives shapeFunctionsQxietaelemType Jacobian matrix inverse of Jacobian derivatives wrt xy JacobinvJacobianXYderivatives JacobiannodeCoordinatesindicenaturalDerivatives B matrix shear Bs zeros25nn Bs11nn XYderivatives1 Bs21nn XYderivatives2 Bs1nn12nn shapeFunction Bs22nn13nn shapeFunction shearlayer1eq qbarra45451BsUindiceB shearlayer2eq qbarra45452BsUindiceB end end gauss point loop end end element loop end 149 Free Vibrations The free vibration problem of laminated plates follows the same procedure as for Mindlin plates The stiffness matrix is as previously computed and the mass matrix is obtained according to Eq1442 and coded in formMassMatrixMindlinlami nated5dofm which is shown below 294 14 Laminated Plates We consider crossply stacking sequences boundary conditions and thickness toside ratios according to Liew 8 Both square and rectangular plates are studied Eigenvalues are expressed in terms of the nondimensional frequency parameter w defined as wh ph w mW Do where Exh Do 121 42V21 Also a constant shear correction factor K 1712 is used for all computations The examples considered here are limited to thick symmetric crossply laminates with layers of equal thickness The material properties for all layers of the laminates are identical as E1Ex2 40 G23 05E2 G13 Gi2 06E 2 Vj2 025 v2 000625 We consider SSSS simply supported on all sides and CCCC clamped on all sides boundary conditions for their practical interest The convergence study of frequency parameters w for threeply 0900 sim ply supported SSSS rectangular laminates is performed in Table 144 while the corresponding convergence study for CCCC rectangular laminate is performed in Table 145 It can be seen that a faster convergence is obtained for higher tb ratios irrespective of ab ratios In both SSSS and CCCC cases the results converge well to Liew 8 results Q8 and Q9 have a faster convergence compared to Q4 as expected Note that Liew 8 considers only bending vibrations whereas the present finite element code is able to calculate membrane and bending vibrations The MATLAB code problem21m for this case is listed next Bee eececceeeen eee eet eee eee rete e eee eeee MATLAB codes for Finite Element Analysis problem21m free vibrations of laminated plates using Q4 elements See reference K M Liew Journal of Sound and Vibration Solving the vibration of thick symmetric laminates by ReissnerMindlin plate theory and the pRitz method Vol 198 Number 3 Pages 343360 1996 AJM Ferreira N Fantuzzi 2019 149 Free Vibrations 295 clear memory clear close all materials h 0001 rho 1 I hˆ312 AMatrixBMatrixDMatrixSMatrixQ liewMaterialh mesh generation L 1 numberElementsX 10 numberElementsY 10 numberElementsnumberElementsXnumberElementsY nodeCoordinates elementNodes rectangularMeshLLnumberElementsXnumberElementsYQ4 xxnodeCoordinates1 yynodeCoordinates2 figure drawingMeshnodeCoordinateselementNodesQ4 axis equal numberNodes sizexx1 GDof global number of degrees of freedom GDof 5numberNodes stiffness and mass matrices stiffness formStiffnessMatrixMindlinlaminated5dof GDofnumberElements elementNodesnumberNodesnodeCoordinatesAMatrix BMatrixDMatrixSMatrixQ4completereduced mass formMassMatrixMindlinlaminated5dofGDof numberElementselementNodesnumberNodesnodeCoordinates rhohIQ4complete boundary conditions prescribedDofactiveDof EssentialBC5dofccccGDofxxyynodeCoordinatesnumberNodes eigenproblem free vibrations numberOfModes 12 modeseigenvalues eigenvalueGDofprescribedDof stiffnessmassnumberOfModes omega sqrteigenvalues Liew pRitz D0 Q22hˆ312e2hˆ3121miu12miu21 dimensionless omega omegabar omegaLLpipisqrtrhohD0 sort out eigenvalues omegaii sortomega modes modesii drawing mesh and deformed shape 296 14 Laminated Plates modeNumber 1 displacements modesmodeNumber surface representation figure hold on for k 1sizeelementNodes1 patchnodeCoordinateselementNodesk141 nodeCoordinateselementNodesk142 displacementselementNodesk14 displacementselementNodesk14 end setgcafontsize18 view4545 Table 144 Convergence study of frequency parameters ω ωb2π2ρhD0 for threeply 0900 simply supported SSSS rectangular laminates Modes ab hb Mesh 1 2 3 4 5 6 1 0001 5 5 Q4 69607 107831 251919 308932 325750 408649 10 10 67066 97430 178158 263878 278395 323408 20 20 66454 95190 165801 254234 268237 279061 5 5 Q8 67298 118761 255129 263293 409942 431296 10 10 66257 94594 162887 251265 266098 269812 20 20 66252 94472 162067 251149 264989 266657 5 5 Q9 66273 94709 165004 252341 266244 282859 10 10 66253 94486 162254 251223 265063 267742 20 20 66252 94471 162064 251149 264985 266650 Liew 8 66252 94470 162051 251146 264982 266572 020 5 5 Q4 35913 62812 76261 88475 113313 121324 10 10 35479 58947 73163 86545 97538 112835 20 20 35367 58036 72366 85856 93768 109971 5 5 Q8 35333 57852 72220 85732 93562 109857 10 10 35329 57745 72107 85619 92617 109076 20 20 35329 57738 72100 85613 92551 109022 5 5 Q9 35333 57850 72218 85703 93553 109848 10 10 35329 57745 72107 85619 92617 109076 20 20 35329 57738 72100 85613 92551 109022 Liew 8 35939 57691 73972 86876 91451 112080 continued 149 Free Vibrations 297 Table 144 continued Modes ab hb Mesh 1 2 3 4 5 6 2 0001 5 5 Q4 24798 80538 81040 115816 235944 237622 10 10 23905 69399 69817 99192 159748 160852 20 20 23689 67016 67415 95617 146808 147815 5 5 Q8 25395 81685 93337 190756 208629 230502 10 10 23625 66406 66728 96330 143779 144307 20 20 23618 66254 66647 94479 142886 143861 5 5 Q9 23625 66543 66939 94916 146031 147032 10 10 23619 66271 66665 94500 143087 144066 20 20 23618 66253 66647 94472 142883 143860 Liew 8 23618 66252 66845 94470 142869 163846 020 5 5 Q4 20006 37932 55767 61626 62479 74516 10 10 19504 35985 50782 55720 59030 72281 20 20 19379 35493 49610 54132 58064 71031 5 5 Q8 19342 35396 49379 54064 57885 71097 10 10 19338 35334 49237 53636 57746 70604 20 20 19338 35329 49227 53606 57738 70584 5 5 Q9 19342 35394 49378 54048 57857 70860 10 10 19338 35333 49237 53636 57746 70603 20 20 19338 35329 49227 53606 57738 70584 Liew 8 19393 35939 48755 54855 57691 71177 Function formMassMatrixMindlinlaminated5dofm computes the corresponding mass matrix function M formMassMatrixMindlinlaminated5dofGDofnumberElements elementNodesnumberNodesnodeCoordinatesrhothicknessI elemTypequadType computation of mass matrix for Mindlin plate element M zerosGDof Gauss quadrature for bending part gaussWeightsgaussLocations gaussQuadraturequadType cycle for element for e1numberElements indice nodal connectivities for each element indiceelementNodese ndoflengthindice 298 14 Laminated Plates cycle for Gauss point for q1sizegaussWeights1 GaussPointgaussLocationsq xiGaussPoint1 etaGaussPoint2 shape functions and derivatives shapeFunctionnaturalDerivatives shapeFunctionsQxietaelemType Jacobian matrix inverse of Jacobian derivatives wrt xy JacobinvJacobianXYderivatives JacobiannodeCoordinatesindicenaturalDerivatives mass matrix Mindiceindice Mindiceindice shapeFunctionshapeFunctionthicknessrho gaussWeightsqdetJacob MindicenumberNodesindicenumberNodes MindicenumberNodesindicenumberNodes shapeFunctionshapeFunctionIrho gaussWeightsqdetJacob Mindice2numberNodesindice2numberNodes Mindice2numberNodesindice2numberNodes shapeFunctionshapeFunctionIrho gaussWeightsqdetJacob Mindice3numberNodesindice3numberNodes Mindice3numberNodesindice3numberNodes shapeFunctionshapeFunctionthicknessrho gaussWeightsqdetJacob Mindice4numberNodesindice4numberNodes Mindice4numberNodesindice4numberNodes shapeFunctionshapeFunctionthicknessrho gaussWeightsqdetJacob end end Gauss point loop end end element loop end Codes problem21am and problem21bm which use Q8 and Q9 are not listed but they can be easily obtained by setting proper parameters 149 Free Vibrations 299 Table 145 Convergence study of frequency parameters ω ωb2π2ρhD0 for threeply 0900 clamped CCCC rectangular laminates Modes ab hb Mesh 1 2 3 4 5 6 1 0001 5 5 Q4 166943 220807 510268 577064 599352 765002 10 10 151249 184938 276970 426545 443895 455585 20 20 147776 178233 252187 375788 399809 416217 5 5 Q8 206595 331178 434331 512849 537652 641788 10 10 147825 183351 265149 394917 395065 429600 20 20 146665 176205 245328 355836 391616 407921 5 5 Q9 146889 176999 252296 395279 396226 412499 10 10 146668 176191 245570 357569 391863 407984 20 20 146655 176140 245143 355465 391582 407695 Liew 8 146655 176138 245114 355318 391572 407685 020 5 5 Q4 45013 73324 79268 94186 119311 123170 10 10 44145 68373 76291 92078 103964 114680 20 20 43931 67178 75509 91264 100084 111927 5 5 Q8 43873 66999 75364 91167 100065 111835 10 10 43860 66799 75254 90984 98899 111063 20 20 43860 66786 75247 90975 98820 111010 5 5 Q9 43873 66996 75357 91128 100055 111801 10 10 43860 66798 75254 90984 98899 111062 20 20 43860 66786 75247 90975 98820 111010 Liew 8 44468 66419 76996 91852 97378 113991 2 0001 5 5 Q4 57995 150869 151794 209280 477268 480954 10 10 52624 113863 114494 155733 230606 232217 20 20 51435 107292 107872 146188 203457 204857 5 5 Q8 99892 169918 179324 243426 273170 291237 10 10 52978 111633 112456 174494 208141 211924 20 20 51070 105337 105904 143631 195855 197203 5 5 Q9 51130 106374 106948 144893 204132 205539 10 10 51056 105336 105900 143346 196218 197561 20 20 51051 105269 105833 143248 195708 197047 Liew 8 23618 66252 66845 94470 142869 163846 020 5 5 Q4 32165 44538 64677 66996 70825 78720 10 10 30946 42705 59857 60821 67327 78602 20 20 30646 42207 58436 59323 66221 76291 5 5 Q8 30568 42108 58399 59093 66072 76816 10 10 30548 42043 57985 58847 65852 75264 20 20 30547 42039 57959 58831 65840 75157 5 5 Q9 30568 42101 58352 59091 66039 76623 10 10 30548 42043 57985 58847 65852 75263 20 20 30547 42039 57959 58831 65840 75157 Liew 8 30452 42481 57916 59042 65347 76885 300 14 Laminated Plates 1410 Buckling Analysis Here we perform the buckling analysis of some rectangular laminated plates using the laminated plate formulation presented before First of all the second order poten tial energy has to be carried out by including the laminated FSDT plate displacement field 142 in the nonlinear Von Karman strains Thus following the mathematical steps illustrated for Mindlin plates the second order potential energy becomes 1 VP 5 nvuravn hVv GV hVw eV 2 1447 h 0 3 a0 5 VOC VO VOLE vo Jae that rewritten in matrix form becomes Vu 1 Vv VO Vu Vol Vw vo Ver S8 Vw d2 1448 2 Jae v6 V0 where S is a banded 10 x 10 matrix as no 0 0 0 0 0 n 0 0 0 0 076 0 0 S A 1449 0 0 0 Dp 0 0 0 0 0 6 127 where 0 is a2 x 2 matrix of zeros Since the scope is to introduce the finite element approximation 1428 it is convenient to convert the vector of gradients as Vu V0000 Uu Vu 0v000 v Vw00V00 wVu 1450 VOx 000V01 4 VO 0000V 14 where 0 is a 2 x 1 matrix of zeros and V is a 10 x 5 operator including partial derivatives with respect to x and y Finally the second order potential 1448 can be rewritten in matrix form and the finite element approximation 1428 can be included as 1410 Buckling Analysis 301 2 1 T 0 e Veo x Vu SVudQ Jae 1 1451 d7 VNSVNdQd d GSGdQd 2 Qe 2 Qe Thus the geometric stiffness matrix K is defined Ke GSGd2 1452 Re where G is a 10 x 5n matrix with the following structure N 0 0 0 O N 0 0 0 0 ON 0 0 0 ON 0 0 0 0 ON 0 0 G 0 0N 0 0 1453 0 0 ON 0 0 0 ON 0 0 0 0 0N 000 ON where N and N fori 12n are the partial derivatives of the shape functions and N Nix Nox Nnx Ny Niy Noy Nyy Due to the banded structure of G matrix three contributions can be identified so the geometric stiffness matrix Kg may be written as 9 KG KG KG KG 1454 The first term involves the derivatives of u which gives a strong contribution in the buckling load calculation only for anisotropic plates The second term involves the derivatives of w and that is the conventional buckling term associated with the classical plate theory The third socalled curvature terms becomes significant for moderately thick plates and play a role akin to the rotary inertia in the free vibration problem The membrane contribution Kg in natural coordinates is given by 1 pl Kon GG 6Gmih det J ddn 1J1 1 pl G1 6Gnoh det J ddn 1455 1J1 302 14 Laminated Plates where N 0000 ON000 Gm IN 000 o Gna lo N 00 o 1456 The bending contribution Kg in natural coordinates is given by 1 pl Ki GP 6Gyh det J dédn 1457 1J1 where 00N 00 G lo ON 0 o 1458 The shear contribution Kg is given by 1 1 h3 K G16G1 det J dédn 1J1 12 1 1 h3 GL Go5 detJddn 1459 1J1 where 000N 90 0000N Ga lo 00N o Ga lo 000 Nn 1460 Alternatively the second order potential can be written in the following compact matrix form Q1f 0 Nx Ve a JohNx Nyé Ral d2Qa 47 nN NeN a2 6 2 wee Ny A a0 Nx A w i hNxNyé N dQw 1461 ar n3 a0 Ny a 6 fon Nx Ny o N dX2 0 aT 3 a0 N aA 6 fo B Nx Nye N dQ 0 All the geometric stiffness matrix components should be carried out using reduced integration single point for Q4 and 2 x 2 for Q8 and Q9 elements This selection has demonstrated to have higher accuracy of the finite element solution 1410 Buckling Analysis 303 Table 146 Buckling of square and rectangular plates N Ncrb2π2D22 with four antisymmet ric crossplies 090090 simplysupported SSSS with uniaxial load k 0 E1E2 ab 05 5 10 20 25 40 5 5 Q4 50167 44374 40895 40143 38976 10 10 47797 42237 38900 38179 37061 20 20 47235 41730 38427 37714 36608 5 5 Q8 48876 42582 38820 38009 36751 10 10 47059 41568 38275 37564 36461 20 20 47050 41563 38272 37561 36458 5 5 Q9 47079 41589 38296 37584 36481 10 10 47052 41564 38273 37562 36460 20 20 47050 41563 38272 37561 36458 Exact 1 4705 4157 3828 3757 3647 ab 10 5 5 Q4 28111 23333 20526 19926 18999 10 10 26835 22238 19541 18964 18074 20 20 26531 21978 19307 18736 17854 5 5 Q8 29355 23505 20083 19353 18225 10 10 26443 21899 19234 18664 17784 20 20 26432 21893 19230 18661 17782 5 5 Q9 26447 21906 19242 18672 17793 10 10 26432 21893 19231 18661 17783 20 20 26432 21892 19230 18661 17782 Exact 1 2643 2189 1923 1866 1778 ab 15 5 5 Q4 34918 29827 26814 26167 25167 10 10 30790 25993 23163 22557 21620 20 20 29846 25136 22361 21767 20849 5 5 Q8 37476 30560 26485 25612 24263 10 10 29775 24993 22179 21576 20645 20 20 29551 24868 22110 21520 20607 5 5 Q9 29654 24963 22200 21608 20694 10 10 29556 24874 22116 21525 20613 20 20 29550 24868 22110 21519 20607 Exact 1 2955 2487 2211 2152 2061 304 14 Laminated Plates Table 147 Buckling of square plate N Ncrb2E22h3 with two antisymmetric angleplies 45 45 simplysupported SSSS with uniaxial load E1E2 ah 10 10 25 40 5 5 Q4 83135 128766 165941 10 10 78489 122307 157329 20 20 77394 120775 154926 5 5 Q8 77073 120324 154347 10 10 77037 120274 154160 20 20 77035 120271 154148 5 5 Q9 77072 120323 154337 10 10 77037 120274 154160 20 20 77035 120271 154148 Exact 1 7847 12231 15774 ah 20 5 5 Q4 94536 15657 213698 10 10 88612 147192 201326 20 20 87229 144996 198421 5 5 Q8 86825 144353 197570 10 10 86779 144280 197475 20 20 86776 144276 197469 5 5 Q9 86822 144349 197566 10 10 86779 144280 197475 20 20 86776 144276 197469 Exact 1 8727 14513 19861 ah 100 5 5 Q4 98958 168424 235794 10 10 92497 157641 220850 20 20 90993 155129 217366 5 5 Q8 90625 154464 216416 10 10 90505 154312 216233 20 20 90501 154307 216226 5 5 Q9 90551 154391 216342 10 10 90504 154312 216233 20 20 90501 154307 216226 Exact 1 9052 15435 21628 1410 Buckling Analysis 305 14101 Buckling of Cross and AnglePly Laminates Antisymmetric cross090090 and angleply 4545 simplysupported lami nates have been tested below according to the studies of Reddy 1 The orthotropic material properties are E11E22 10 G23 02E22 G13 G12 05E22 ν12 025 ν21 0025 with a shear correction factor Ks 56 and the layers have the same thickness according to the number of plies considered Results for various mesh sizes and elements are listed in Tables146 and 147 Excellent agreement is observed among the solutions in comparison with the exact solution given by Reddy 1 Q8 and Q9 finite elements have a faster convergence as expected with respect to Q4 elements All results have been listed in dimensionless form according to N Ncr b2 π2D22 1462 for the crossply case and N Ncr b2 E22h3 1463 for the angleply case The MATLAB code problem20Bucklingm for this case is listed next MATLAB codes for Finite Element Analysis problem20Bucklingm buckling laminated plate using Q4 elements AJM Ferreira N Fantuzzi 2019 clear memory clear close all materials thickness 0001 initial stress matrix sigmaX 1thickness sigmaXY 0 sigmaY 0 sigmaMatrix sigmaX sigmaXY sigmaXY sigmaY 306 14 Laminated Plates Mesh generation Lx 1 Ly 1 numberElementsX 10 numberElementsY 10 numberElements numberElementsXnumberElementsY nodeCoordinates elementNodes rectangularMeshLxLynumberElementsXnumberElementsYQ4 xx nodeCoordinates1 yy nodeCoordinates2 figure drawingMeshnodeCoordinateselementNodesQ4 axis equal numberNodes sizexx1 GDof global number of degrees of freedom GDof 5numberNodes computation of the laminate stiffness matrix AMatrixBMatrixDMatrixSMatrix reddyLaminateMaterialBukthickness computation of the system stiffness matrix stiffness formStiffnessMatrixMindlinlaminated5dof GDofnumberElementselementNodesnumberNodes nodeCoordinatesAMatrix BMatrixDMatrixSMatrix Q4completereduced computation of the system force vector geometric formGeometricStiffnessMindlinlaminated5dof GDofnumberElementselementNodesnumberNodes nodeCoordinatessigmaMatrixthickness Q4reducedreduced boundary conditions prescribedDofactiveDof EssentialBC5dofssssGDofxxyynodeCoordinatesnumberNodes solution buckling analysis modeslambda eigenvalueGDofprescribedDof stiffnessgeometric15 sort out eigenvalues lambdaii sortlambda modes modesii dimensionless omega see tables in the book lambdabar lambdaLyˆ2pipiDMatrix22 lambdabar lambdaLyˆ2thicknessˆ3 lambdabar1 drawing mesh and deformed shape modeNumber 1 displacements modesmodeNumber 1410 Buckling Analysis 307 surface representation figure hold on for k 1sizeelementNodes1 patchnodeCoordinateselementNodesk141 nodeCoordinateselementNodesk142 displacementselementNodesk14 displacementselementNodesk14 end setgcafontsize18 view4545 Function formGeometricStiffnessMindlinlaminated5dofm computes the corre sponding geometric stiffness matrix function KG formGeometricStiffnessMindlinlaminated5dofGDof numberElementselementNodesnumberNodes nodeCoordinatessigmaMatrixthicknesselemType quadTypeBquadTypeS computation of geometric stiffness for laminated plate element KG geometric matrix KG zerosGDof Gauss quadrature for bending part gaussWeightsgaussLocations gaussQuadraturequadTypeB cycle for element for e 1numberElements indice nodal connectivities for each element elementDof element degrees of freedom indice elementNodese elementDof indice indicenumberNodes indice2numberNodes indice3numberNodes indice4numberNodes ndof lengthindice cycle for Gauss point for q 1sizegaussWeights1 GaussPoint gaussLocationsq xi GaussPoint1 eta GaussPoint2 shape functions and derivatives shapeFunctionnaturalDerivatives shapeFunctionsQxietaelemType Jacobian matrix inverse of Jacobian derivatives wrt xy JacobinvJacobianXYderivatives JacobiannodeCoordinatesindicenaturalDerivatives geometric matrix w Gb zeros25ndof Gb11ndof XYderivatives1 308 14 Laminated Plates Gb21ndof XYderivatives2 KGelementDofelementDof KGelementDofelementDof GbsigmaMatrixthicknessGb gaussWeightsqdetJacob geometric matrix u Ga1 zeros25ndof Ga113ndof14ndof XYderivatives1 Ga123ndof14ndof XYderivatives2 KGelementDofelementDof KGelementDofelementDof Ga1sigmaMatrixthicknessGa1 gaussWeightsqdetJacob geometric matrix v Ga2 zeros25ndof Ga214ndof15ndof XYderivatives1 Ga224ndof15ndof XYderivatives2 KGelementDofelementDof KGelementDofelementDof Ga2sigmaMatrixthicknessGa2 gaussWeightsqdetJacob end Gauss point end element shear stiffness matrix Gauss quadrature for shear part gaussWeightsgaussLocations gaussQuadraturequadTypeS cycle for element for e 1numberElements indice nodal condofectivities for each element elementDof element degrees of freedom indice elementNodese elementDof indice indicenumberNodes indice2numberNodes indice3numberNodes indice4numberNodes ndof lengthindice for q 1sizegaussWeights1 GaussPoint gaussLocationsq xi GaussPoint1 eta GaussPoint2 shape functions and derivatives shapeFunctionnaturalDerivatives shapeFunctionsQxietaelemType Jacobian matrix inverse of Jacobian derivatives wrt xy JacobinvJacobianXYderivatives JacobiannodeCoordinatesindicenaturalDerivatives Geometric matrix Gs1 zeros25ndof Gs11ndof12ndof XYderivatives1 Gs12ndof12ndof XYderivatives2 KGelementDofelementDof KGelementDofelementDof 1410 Buckling Analysis 309 Gs1sigmaMatrixthicknessˆ312Gs1 gaussWeightsqdetJacob Gs2 zeros25ndof Gs212ndof13ndof XYderivatives1 Gs222ndof13ndof XYderivatives2 KGelementDofelementDof KGelementDofelementDof Gs2sigmaMatrixthicknessˆ312Gs2 gaussWeightsqdetJacob end gauss point end element end The material considered in the present applications is taken from Reddy 1 and code is listed below function AMatrixBMatrixDMatrixSMatrixbarQ reddyLaminateMaterialBukthickness REDDY PLATE BUCKLING EXAMPLE plate thickness h thickness stack 0 90 0 90 antisymmetric crossply stack 45 45 antisymmetric angleply nlam lengthstack hk hnlam reddy orthotropic properties E2 1 E1 10E2 G12 05E2 G13 05E2 G23 02E2 nu12 025 nu21 nu12E2E1 kapa 56 Reduced stiffness constants Q11 E11nu12nu21 Q12 nu12E21nu12nu21 Q22 E21nu12nu21 Q66 G12 Q44 G23 Q55 G13 A zeros6 B zeros6 D zeros6 barQ zeros66nlam for k 1lengthstack theta stackk barQk effectivepropsQtheta zk h2 hkk zk1 zk h2 hkk1 zk 310 14 Laminated Plates A A zk zk barQk B B 12zkˆ2 zkˆ2barQk D D 13zkˆ3 zkˆ3barQk end AMatrix A11A12A16 A12A22A26 A16A26A66 BMatrix B11B12B16 B12B22B26 B16B26B66 DMatrix D11D12D16 D12D22D26 D16D26D66 SMatrix kapaA44kapaA45 kapaA45kapaA55 end effective properties according to the orientation theta function barQ effectivepropsQthetak theta deg2radthetak cc costheta ss sintheta barQ11 Q11ccˆ4 2Q12 2Q66ccˆ2ssˆ2 Q22ssˆ4 barQ12 Q11Q224Q66ccˆ2ssˆ2 Q12ccˆ4 ssˆ4 barQ22 Q11ssˆ4 2Q12 2Q66ccˆ2ssˆ2 Q22ccˆ4 barQ16 Q11 Q12 2Q66ccˆ3ss Q12Q222Q66ccssˆ3 barQ26 Q11Q12 2Q66ccssˆ3 Q12Q222Q66ccˆ3ss barQ66 Q11Q222Q122Q66ccˆ2ssˆ2 Q66ccˆ4ssˆ4 barQ44 Q44ccˆ2Q55ssˆ2 barQ45 Q55Q44ccss barQ55 Q55ccˆ2Q44ssˆ2 end Codes problem20aBucklingm and problem20bBucklingm which use Q8 and Q9 are not listed but they can be easily obtained by setting proper parameters References 1 JN Reddy Mechanics of Laminated Composite Plates and Shells CRC Press Boca Raton 2004 2 JA Figueiras Ultimate load analysis of anisotropic and reinforced concrete plates and shells University of Wales 1983 3 S Srinivas A refined analysis of composite laminates J Sound Vib 30 495507 1973 4 BN Pandya T Kant Higherorder shear deformable theories for flexure of sandwich plates finite element evaluations Int J Solids Struct 24 419451 1988 References 311 5 AJM Ferreira Analysis of composite plates and shells by degenerated shell elements FEUP 1997 6 AJM Ferreira A formulation of the multiquadric radial basis function method for the analysis of laminated composite plates Compos Struct 59 385392 2003 7 AJM Ferreira CMC Roque PALS Martins Analysis of composite plates using higher order shear deformation theory and a finite point formulation based on the multiquadric radial basis function method Compos Part B 34 627636 2003 8 KM Liew Solving the vibration of thick symmetric laminates by reissnermindlin plate theory and the pritz method J Sound Vib 1983 343360 1996 9 E Hinton Numerical methods and software for dynamic analysis of plates and shells Pineridge Press Swansea 1988 Chapter 15 Functionally Graded Structures Abstract InthepresentchapterfunctionallygradedmaterialsFGMsandstructures are presented In particular the static and free vibration problems of Timoshenko beams and Mindlin plates are studied The buckling problem for both structures can be developed following analogous problems presented in Chap 10 for Timoshenko beams and in Chap 14 for laminated FSDT plates 151 Introduction In the present chapter functionally graded materials FGMs and structures are pre sented In particular the static and free vibration problems of Timoshenko beams with 3 degrees of freedom per node and Mindlin plates with 5 degrees of freedom per node are studied The buckling problem for both structures can be developed following analogous problems presented in Chap 10 for Timoshenko beams and in Chap 14 for laminated FSDT plates A short introduction to functionally graded materials and their implementation in the constitutive model is given Since the FGMs are introduced only at the constitutive level of a model short theoretical background is given The reader should refer to the theoretical backgrounds of Timoshenko and Mindlin plates given in the previous chapters 152 Functionally Graded Materials Functionally graded materials FGMs are a new class of composite materials that have a gradual variation along a given direction These materials have been proposed as thermal barrier for coating applications They are isotropic but not homogenous along one direction For beams and plates the direction of homogeneity is the thick ness direction In the most common applications the material is made of two con stituents such as metal and ceramic These materials are used on one hand as thermal barrier ceramic and ductility metal FGMs are mathematically presented as a con The Editors if applicable and The Authors under exclusive license to Springer Nature Switzerland AG 2020 A J M Ferreira and N Fantuzzi MATLAB Codes for Finite Element Analysis Solid Mechanics and Its Applications 157 httpsdoiorg101007978303047952715 313 314 15 Functionally Graded Structures tinuous variation of the mechanical properties though the thickness direction Thus such model involves the definition of the stress resultant for beams and reduced elastic coefficients for plates The most wide used formula is the powerlaw distribution which is valid for elastic modulus E and material density p 1 2 Given two material properties P and P as the material at the top material 1 and bottom material 2 of the two faces of the beam or the plate the powerlaw distribution is given by Pz Pi Pa fZ P2 151 where poo 242 152 LOSNOT and n is the powerlaw index Note that forn 0 P P andn w P P3 Thus material properties and structural behavior can be tailored by the index n The following integrals will be used in the codes below for the computation of the structural properties h2 h z dz loamy h2 nh2 zz dz 153 hp 2nnh3 fz 2dz Atntn yh h2 4in lIn2n 3 FGMs are now applied to the study of beams and plates For functionally graded structures the bending and axial behaviors are generally coupled so they cannot be treated separately Bending and stretching are uncoupled only if fz is symmetric with respect to the 153 Timoshenko Beam The displacement field of Timoshenko beams considering both axial and bending behavior is ux zt uxt 20 xt 1 uxt x t 154 u3x Zs t wx t 153 Timoshenko Beam 315 Note that axial displacements are here introduced because FGM constitutive law couples axial and bending behaviors of the beam Straindisplacement relationship is 2 cp Oy Om LL WY O ze Ox Ox 2 0x 155 ow 6 7 Vaz By Oe Vez all the other strains are zero for this theory In matrix form such relations become a 10 0 Ox 20x u M 9 0 I lw 156 WN Ox Ox 0 0 1 Ox and in matrix form becomes Du The constitutive equations are Oy EZ Tez GZ Vez 157 Note that the elastic and shear moduli have a variation along the beam height The beam strain energy is 1 U 5 Oxy TxzYxz JUV 158 2 Jy By introducing the definitions Axx Byx D 1 Zz 2 Ez d Sy a Ezd92Q 159 XX XX9 XX 2 9A XZ 201 2 where K is the shear correction factor and using the integrals of fz given in 153 the following definitions apply M M1 Axx E Ay By EByp ln 211 n2n 6 3n 3n7M 4 8n 3n n3 Dyx Eyl a 6 11n 6n2 n3 KE Ao M bh S Sto FT bh By bh2 Ip 1510 21v ln 12 316 15 Functionally Graded Structures where M EE is the ratio of the two elastic constituents These elastic coeffi cients can be collected in the matrix Axx Byx 0 C By Dy O 1511 0 0 Sy The strain energy in matrix form becomes 1 ft 1 ft U a oedx e Ce dx 1512 2 Jo 2 Jo the straindisplacement formulae 156 can be included into the strain energy as 1 L U Du CDu dx 1513 0 The kinetic energy of the beam is I 2 42 I 2 2 K pujtu3dV putz0x pw dV 1514 2 Jv 2Jv the following inertia definitions apply AO py mp2 ao y pp mo np2 my 2 0 Itn Pl p2 1 20 n2 n Pl p2 643 3n 8 3n 3 ny 19 On tM Bn 30 FMD O45 15 6 11n 6n2 n3 and the kinetic energy can be written in matrix form as 1 L K a a Tu dx 1516 2 Jo where the inertia matrix I is Mo 0 my I0 m O 1517 m O mp 153 Timoshenko Beam 317 1531 Finite Element Approximation The finite element approximation uy Un N 0 0 u0 N 0O Nd 1518 0 0 NI w Ox ben is introduced in the strain energy 1513 in order to obtain the strain energy for the element N is the matrix of the shape functions linear shape functions are considered for simplicity and d is the vector of nodal displacements The strain energy for the Timoshenko beam is e 1 eT T e 1 eT T e US 54 DN CDN dx dé 54 B CB dx d 1519 where B DN is the matrix of the derivative of the shape functions as ON 0 0 Ox B9 9 2 1520 OX ON 0 N Ox The element stiffness matrix is a 1 K B CB dx B CB det Jdé 1521 a l where the integral has been transformed into natural coordinates Gauss integration is applied for obtaining the stiffness matrix To avoid shear locking reduced integration has to be applied to the shear component of the matrix thus 1 1 K B CB det J dé B CB det J dé Ki K 1522 1 1 318 15 Functionally Graded Structures where ON 9 9 0 0 0 ox 0 0 0 B 0 0 ON Bs on 1523 Ox 0 N 0 0 0 Ox full integration is used for Ky and reduced integration for K The kinetic energy 1516 for the element is e 1 eT a T e 1 eT T e Ke 54 N INdxd 54 N INdet Jdé d 1524 a 1 the mass matrix is 1 M NIN det Jdé 1525 1 1532 Bending of MicroBeams Simplysupported functionally graded microbeams are considered according to the example provided by Reddy 2 The beams have the following mechanical properties 54yv E 144GPa E E10 038 K 2 Bil v 6 5v h8810m b2h L20h p1Nm 1526 The numerical results are presented in terms of the central deflection in dimen sionless form as w wL2E2IopL Uniform and point loads are considered for simplysupported beams with different powerlaw exponent n Consider the point load as pL applied at the beam central point The finite element code which solves the present problem is listed in code problem16fgmm The code is an extension of the static problem of Timoshenko beams presented in Chap 10 Bee eececceeeen eee eet eee eee rete e eee eeee MATLAB codes for Finite Element Analysis probleml6éfgmm Functionally graded Timoshenko beam in bending under uniform and point loads AJM Ferreira N Fantuzzi 2019 clear memory 153 Timoshenko Beam 319 clear E1 modulus of elasticity of material 1 E2 modulus of elasticity of material 2 L length of beam thickness height of crosssection width width of the crosssection E1 144e9 E2 144e9 rho1 1 rho2 1 poisson 038 thickness 88e6 width 2thickness L 20thickness n 5 FGM powerlaw index M E1E2 A0 widththickness B0 widththicknessˆ2 I0 widththicknessˆ312 Axx E2A0Mn1n Bxx E2B0nM121n2n Dxx E2I063n3nˆ2M 8n3nˆ2nˆ3611n6nˆ2nˆ3 kapa 51poisson65poisson Sxz kapaE2A021poissonMnn1 P 1 uniform pressure constitutive matrix C Axx Bxx 0 Bxx Dxx 0 0 0 Sxz m0 A0rho1 nrho2n1 m1 B0nrho1rho22n1n2 m2 I063n3nˆ2rho1 8n3nˆ2nˆ3rho2 611n6nˆ2nˆ3 inertia matrix I m0 0 m1 0 m0 0 m1 0 m2 mesh numberElements 40 nodeCoordinates linspace0LnumberElements1 xx nodeCoordinates elementNodes zerossizenodeCoordinates212 for i 1sizenodeCoordinates21 elementNodesi1i elementNodesi2i1 end generation of coordinates and connectivities numberNodes sizexx1 GDof global number of degrees of freedom GDof 3numberNodes computation of the system stiffness matrix stiffnessforce 320 15 Functionally Graded Structures formStiffnessMassTimoshenkoFgmBeamGDofnumberElements elementNodesnumberNodesxxCPIthickness uncomment to apply the point load force force0 forceroundnumberNodes2numberNodes PL boundary conditions simplysupported at both bords fixedNodeU fixedNodeW 1 numberNodes fixedNodeTX prescribedDof fixedNodeU fixedNodeWnumberNodes fixedNodeTX2numberNodes solution displacements solutionGDofprescribedDofstiffnessforce output displacementsreactions outputDisplacementsReactionsdisplacementsstiffness GDofprescribedDof U displacements ws 1numberNodes max displacement dispmax displacement minUwsnumberNodes wbar Uroundlengthws2numberNodesE2I0PLˆ4100 Anewcode formStiffnessMassTimoshenkoFgmBeammisgivenforthecom putation of the stiffness mass and force vector It is recalled that reduced integration is applied for the shear part of the stiffness matrix whereas full integration is con sidered for the mass matrix and force vector Such code is listed below function stiffnessforcemass formStiffnessMassTimoshenkoFgmBeamGDofnumberElements elementNodesnumberNodesxxCPIthickness computation of stiffness mass matrices and force vector for Timoshenko beam element stiffness zerosGDof mass zerosGDof force zerosGDof1 2x2 Gauss quadrature gaussLocations 05773502691896260577350269189626 gaussWeights ones21 bending contribution for matrices for e 1numberElements indice elementNodese elementDof indice indicenumberNodes indice2numberNodes indiceMass indicenumberNodes 153 Timoshenko Beam 321 ndof lengthindice lengthelement xxindice2xxindice1 detJacobian lengthelement2 invJacobian1detJacobian for q 1sizegaussWeights1 pt gaussLocationsq shapenaturalDerivatives shapeFunctionL2pt1 Xderivatives naturalDerivativesinvJacobian B matrix B zeros33ndof B11ndof Xderivatives B22ndof13ndof Xderivatives stiffness matrix stiffnesselementDofelementDof stiffnesselementDofelementDof BCBgaussWeightsqdetJacobian force vector forceindiceMass forceindiceMass shapePdetJacobiangaussWeightsq B matrix B zeros33ndof B11ndof shape B21ndof2ndof shape B312ndof3ndof shape mass matrix masselementDofelementDof masselementDofelementDof BIBgaussWeightsqdetJacobian end end shear contribution for the matrices gaussLocations 0 gaussWeights 2 for e 1numberElements indice elementNodese elementDof indice indicenumberNodes indice2numberNodes ndof lengthindice lengthelement xxindice2xxindice1 detJ0 lengthelement2 invJ0 1detJ0 for q 1sizegaussWeights1 pt gaussLocationsq shapenaturalDerivatives shapeFunctionL2pt1 Xderivatives naturalDerivativesinvJacobian B B zeros33ndof B3ndof12ndof Xderivatives B32ndof13ndof shape stiffness matrix stiffnesselementDofelementDof 322 15 Functionally Graded Structures Table 151 Center deflections w 102 of simplysupported FGM microbeams n Uniform ref 2 Present Point load ref 2 Present 0 01310 01309 02100 02098 1 03062 03016 04906 04787 5 05968 05962 09562 09556 10 06571 06565 10532 10526 100 10610 10599 17006 16996 stiffnesselementDofelementDof BCBgaussWeightsqdetJacobian end end end The results given by the present code with 40 finite element are listed in Table 151 compared to the same results given by the semianalytical Navier method presented by Reddy 2 For the uniform load case the force vector has to be used in the form given by formStiffnessMassTimoshenkoFgmBeamm For the point load the force vector has to be zero except for the force applied at the central point thus comments should be removed from lines uncomment to apply the point load force force0 forceroundnumberNodes2numberNodes PL The code automatically applies the point load in the central node of the finite element mesh Very good match can be observed varying the powerlaw exponent n Note that when n 1 the beam is isotropic made of material 1 on the contrary for n material 2 is the constituent of the isotropic beam 1533 Free Vibrations of MicroBeams For the free vibration problem the shear correction factor width and beam length are the same as the static case other parameters are given below h 176 106 m ρ1 122 103 kgm ρ2 122 103 kgm 1527 153 Timoshenko Beam 323 The finite element code which solves the present problem is listed in prob lem16fgmVibm The code is an extension of the free vibration problem of Timo shenko beams presented in Chap 10 MATLAB codes for Finite Element Analysis problem16fgmVibm Functionally graded Timoshenko beam in free vibrations AJM Ferreira N Fantuzzi 2019 clear memory clear close all E1 modulus of elasticity of material 1 E2 modulus of elasticity of material 2 rho1 density of material 1 rho2 density of material 2 L length of beam thickness height of crosssection width width of the crosssection E1 144e9 E2 144e9 rho1 122e3 rho2 122e3 poisson 038 thickness 176e6 width 2thickness L 20thickness n 10 FGM powerlaw index M E1E2 A0 widththickness B0 widththicknessˆ2 I0 widththicknessˆ312 Axx E2A0Mn1n Bxx E2B0nM121n2n Dxx E2I063n3nˆ2M 8n3nˆ2nˆ3611n6nˆ2nˆ3 kapa 51poisson65poisson Sxz kapaE2A021poissonMnn1 constitutive matrix C Axx Bxx 0 Bxx Dxx 0 0 0 Sxz m0 A0rho1 nrho2n1 m1 B0nrho1rho22n1n2 m2 I063n3nˆ2rho1 8n3nˆ2nˆ3rho2 611n6nˆ2nˆ3 inertia matrix I m0 0 m1 0 m0 0 m1 0 m2 mesh numberElements 40 nodeCoordinates linspace0LnumberElements1 324 15 Functionally Graded Structures xx nodeCoordinates elementNodes zerossizenodeCoordinates212 for i 1sizenodeCoordinates21 elementNodesi1i elementNodesi2i1 end generation of coordinates and connectivities numberNodes sizexx1 GDof global number of degrees of freedom GDof 3numberNodes computation of the system stiffness matrix stiffnessmass formStiffnessMassTimoshenkoFgmBeamGDofnumberElements elementNodesnumberNodesxxC1Ithickness boundary conditions simplysupported at both bords fixedNodeU 1 fixedNodeW 1 numberNodes fixedNodeTX prescribedDof fixedNodeU fixedNodeWnumberNodes fixedNodeTX2numberNodes free vibrations modeseigenvalues eigenvalueGDofprescribedDof stiffnessmass0 omega sqrteigenvaluesLLsqrtrho2A0E2I0 display first 2 dimensionless frequencies omega13 drawing mesh and deformed shape modeNumber 4 V1 modes1modeNumber drawing eigenmodes figure drawEigenmodes1DmodeNumbernumberNodes V11numberNodes2numberNodesxx The mass matrix is computed by the function formStiffnessMassTimoshenko FgmBeamm Finite element analysis with 40 elements is carried out Results in terms of the first three natural frequencies ωn ωnL2ρ2 A0E2I0 are listed in Table 152 and compared to the same given by Reddy 2 The first four mode shapes are depicted in Fig151 Good agreement is observed between the two solutions which proof the validity of the present code 154 Mindlin Plate 325 Table 152 First three natural frequencies ωn n 1 2 3 of simplysupported FGM microbeams ω1 ω2 ω3 n ref 2 Present ref 2 Present ref 2 Present 0 983 98353 3882 389412 8563 862024 1 867 86730 3429 343826 7579 760537 10 1028 102898 4047 405756 8880 882976 0 05 1 15 2 25 3 35 4 104 5 0 104 0 05 1 15 2 25 3 35 4 104 5 0 5 104 0 05 1 15 2 25 3 35 4 104 5 0 5 104 0 05 1 15 2 25 3 35 4 104 2 0 2 104 Fig 151 First 4 modes of vibration for a simplysupported FGM microbeam 154 Mindlin Plate Due to the coupling between bending and membrane plate behavior the implemen tation of functionally graded Mindlin plates follows the theoretical background pre sented for laminated FSDT plates in Chap 14 where 5 dofs per node have been considered for the plate For simplicity plates made of a single FGM ply are con sidered instead of laminated composites because the generalization is simple by following the present discussion and the one already given for orthotropic laminated plates Note that in the present section integral for evaluating the mechanical coefficients are performed analytically this is made possible since powerlaw is relatively simple to treat and Poisson ratio is considered constant Numerical integration through the plate thickness can be carried out or the FGM ply can be seen as an equivalent laminate made of several isotropic plies where each ply is made of a fraction of Ez according to the abscissa z 326 15 Functionally Graded Structures The constitutive equation for FGM plates made of a single ply is Ox Q1 Qn O 0 0 ex Oy Qn On O 0 0 y Try 0 0 O66 0 0 Vey 1528 Txz 0 0 0 K Q66 0 Yxz Tyz 0 0 0 0 Ks Q66 YVyz where K is the shear correction factor and Ez VEz Ez lv Qu Ta G2 Tapa Vu Qos Xin 2 O11 1529 The stiffness coefficients for the Mindlin plate can be calculated by integration 2 as h2 hI2 Bz An Ondz s az h2 njy2 lv l h2 a eofe bd 11 Jap E h2 3 wvre tia lv Jin Eh M fon Mrn 1530 11 l h2 h2 YEz An QO az s dz vA 1531 h2 ayy lv h2 hl Ez lv Aw O66 dz dz A 1532 h2 nj2 20 V 20 where M EE The other coefficients are given by h2 h2 2B z B ZO a ZO 4 h2 njy lv 7 Ey Ba foe Exzdz EM In Ta Jing ht I EE TP WHA E DN 2 1533 154 Mindlin Plate 327 h2 h2 2B z Dis 2 Oy dz cE dz h2 nj2 lv l h2 77 Ey Er f zz Exz dz lv Jin Exh M12n n 1 1 1744 Dm2n 3 12 1534 lvp lv By VBiy Boo a Bu DypvDi1 Doo az Pu 1535 These elastic properties are included into the strain energy definition 1421 for obtaining the stiffness matrix of the FGM plate The inertia terms needed for the free vibration problem takes the form h2 h2 f odas mfomd de 01536 h2 h2 where p and 2 are the densities of the two functionally graded constituents By carrying out the integrals the inertia terms become poh p2nh ly p1 p2 t ph I p1 p2 n1 2n 1n 2 pyhi2n n h b pi p2h dpe 1537 4n 1n 2n 3 12 such terms are used in the inertia matrix definition 1427 for carrying out the mass matrix of the plate 1541 Bending of MicroPlates The finite element modelling of the present problem has the same structure of the one reported for laminated plates in Chap 14 The only difference is in the coding of the stiffness coefficients which substitute the laminated composite configuration The code problem20Fgmm solves the bending of functionally graded plates under uniform loads The comparison is performed in terms of maximum deflection at the plate center for simplysupported conditions as shown 2 for different FGM powerlaws n Plate properties are 328 15 Functionally Graded Structures E1 144 GPa E2 E110 h 176 μm a b 20h p 1 Nm2 1538 where a b are the plate dimensions and h its thickness Uniform applied transverse load is p The main code problem20Fgmm given below MATLAB codes for Finite Element Analysis problem20Fgmm functionally graded plate using Q4 elements AJM Ferreira N Fantuzzi 2019 clear memory clear close all materials thickness 176e6 n 5 powerlaw index load P 1 mesh generation L 20thickness numberElementsX 10 numberElementsY 10 numberElements numberElementsXnumberElementsY nodeCoordinates elementNodes rectangularMeshLLnumberElementsXnumberElementsYQ4 xx nodeCoordinates1 yy nodeCoordinates2 figure drawingMeshnodeCoordinateselementNodesQ4 axis equal numberNodes sizexx1 GDof global number of degrees of freedom GDof 5numberNodes computation of the system stiffness matrix the shear correction factors are automatically calculted for any laminate AMatrixBMatrixDMatrixSMatrix reddyFgmMaterialthicknessn stiffness formStiffnessMatrixMindlinlaminated5dof GDofnumberElements elementNodesnumberNodesnodeCoordinatesAMatrix BMatrixDMatrixSMatrixQ4completereduced computation of the system force vector force 154 Mindlin Plate 329 formForceVectorMindlin5dofGDofnumberElements elementNodesnumberNodesnodeCoordinatesPQ4reduced boundary conditions prescribedDofactiveDof EssentialBC5dofssssGDofxxyynodeCoordinatesnumberNodes solution U solutionGDofprescribedDofstiffnessforce drawing deformed shape and normalize results to compare with Srinivas ws 1numberNodes dispmaximum displacement absminUws144e9thicknessˆ3PLˆ4 surface representation figure hold on for k 1sizeelementNodes1 patchnodeCoordinateselementNodesk141 nodeCoordinateselementNodesk142 UelementNodesk14 UelementNodesk14 end setgcafontsize18 view4545 Material properties and stiffness matrices A B D are calculated in code red dyFgmMaterialm which is given below function AMatrixBMatrixDMatrixSMatrixInertia reddyFgmMaterialthicknessn REDDY FGM MATERIAL plate thickness h thickness elastic moduli E1 144e9 E2 144e9 rho1 122e3 rho2 122e3 poisson 038 kapa 51poisson65poisson c1 11poissonˆ2 c2 poisson1poissonˆ2 c3 121poisson stiffness calculation A11 c1E1E2hn1 hE2 A12 c2E1E2hn1 hE2 A66 c3E1E2hn1 hE2 A44 kapac3E1E2hn1 hE2 A55 A44 330 15 Functionally Graded Structures Table 153 Center deflections w of simplysupported FGM microplates with 10 10 mesh n Ref 2 Q4 Q8 Q9 0 00044 00042 00042 00042 05 00071 00070 00070 00070 1 00100 00098 00099 00099 5 00194 00192 00192 00192 10 00214 00212 00212 00212 B11 c1E1E2nhˆ22n1n2 B12 c2E1E2nhˆ22n1n2 B66 c3E1E2nhˆ22n1n2 D11 c1E1E2hˆ32nnˆ24n1n2n3 E2hˆ312 D12 c2E1E2hˆ32nnˆ24n1n2n3 E2hˆ312 D66 c3E1E2hˆ32nnˆ24n1n2n3 E2hˆ312 inertia calculation I0 rho1rho2hn1 rho2h I1 rho1rho2nhˆ22n1n2 I2 rho1rho2hˆ32nnˆ24n1n2n3 rho2hˆ312 AMatrix A11A120A12A11000A66 BMatrix B11B120B12B11000B66 DMatrix D11D120D12D11000D66 SMatrix A4400A55 Inertia I0 0 0 0 0 0 I2 0 I1 0 0 0 I2 0 I1 0 I1 0 I0 0 0 0 I1 0 I0 end The modelling is presented for Q4 Q8 and Q9 elements with a mesh of 10 10 Codes problem20aFgmm and problem20bFgmm are not shown for the sake of conciseness Table153 lists the maximum plate displacement according to n power law index The deformed shape of the plate considering a 10 10 mesh and simplysupported edges is shown in Fig152 Since the reference uses a higher order theory shear correction factor is not specified for this reason the shear correction factor of the beam case has been considered as Ks 51 ν 6 5ν Very good agreement is observed between our solution and the one given in the literature Stress postcomputation is not performed However the reader can consider the previous implementation given for laminated plates to carrying them out 154 Mindlin Plate 331 Fig 152 Deformed shape of a simplysupported microplate 1542 Free Vibrations of MicroPlates Free vibrations of the same plates presented in the previous sections are shown Den sities of the two materials are ρ1 122 103 kgm3 and ρ2 ρ110 All the other parameters geometrical and mechanical are the same as in the previous example The main code is listed in problem21Fgmm and given below MATLAB codes for Finite Element Analysis problem21Fgmm free vibrations of FGM plates using Q4 elements AJM Ferreira N Fantuzzi 2019 clear memory clear close all materials thickness 176e6 n 10 powerlaw index AMatrixBMatrixDMatrixSMatrixInertia reddyFgmMaterialthicknessn mesh generation L 20thickness numberElementsX 10 numberElementsY 10 numberElementsnumberElementsXnumberElementsY nodeCoordinates elementNodes rectangularMeshLLnumberElementsXnumberElementsYQ4 xx nodeCoordinates1 yy nodeCoordinates2 figure drawingMeshnodeCoordinateselementNodesQ4 332 15 Functionally Graded Structures axis equal numberNodes sizexx1 GDof global number of degrees of freedom GDof 5numberNodes stiffness and mass matrices stiffness formStiffnessMatrixMindlinlaminated5dof GDofnumberElements elementNodesnumberNodesnodeCoordinatesAMatrix BMatrixDMatrixSMatrixQ4completereduced mass formMassMatrixFgmPlate5dofGDofnumberElements elementNodesnumberNodesnodeCoordinatesInertia Q4complete boundary conditions prescribedDofactiveDof EssentialBC5dofssssGDofxxyynodeCoordinatesnumberNodes eigenproblem free vibrations numberOfModes 12 modeseigenvalues eigenvalueGDofprescribedDof stiffnessmassnumberOfModes omega sqrteigenvalues sort out eigenvalues omegaii sortomega modes modesii dimensionless omega omega1sqrt122e3Lˆ4144e9thicknessˆ2 drawing mesh and deformed shape modeNumber 1 displacements modesmodeNumber surface representation figure hold on for k 1sizeelementNodes1 patchnodeCoordinateselementNodesk141 nodeCoordinateselementNodesk142 displacementselementNodesk14 displacementselementNodesk14 end setgcafontsize18 view4545 In order to account for the mass matrix with all inertia contributions bulk and rotary inertias the new code for the mass matrix generation is given in formMass MatrixFgmPlate5dofm and listed below 154 Mindlin Plate 333 function M formMassMatrixFgmPlate5dofGDofnumberElements elementNodesnumberNodesnodeCoordinatesI elemTypequadType computation of mass matrix for Mindlin plate element M zerosGDof Gauss quadrature for bending part gaussWeightsgaussLocations gaussQuadraturequadType cycle for element for e1numberElements indice nodal connectivities for each element elementDof element degrees of freedom indiceelementNodese elementDof indice indicenumberNodes indice2numberNodes indice3numberNodes indice4numberNodes ndoflengthindice cycle for Gauss point for q1sizegaussWeights1 GaussPointgaussLocationsq xiGaussPoint1 etaGaussPoint2 shape functions and derivatives shapeFunctionnaturalDerivatives shapeFunctionsQxietaelemType Jacobian matrix inverse of Jacobian derivatives wrt xy JacobinvJacobianXYderivatives JacobiannodeCoordinatesindicenaturalDerivatives N matrix N zeros55ndof N11ndof shapeFunction N2ndof12ndof shapeFunction N32ndof13ndof shapeFunction N43ndof14ndof shapeFunction N54ndof15ndof shapeFunction mass matrix MelementDofelementDof MelementDofelementDof NINgaussWeightsqdetJacob end end Gauss point loop end end element loop end The fundamental frequency of square microplates are given in Table 154 by varying the powerlaw index n for a 10 10 Q4 Q8 and Q9 elements and com 334 15 Functionally Graded Structures Table 154 Fundamental frequency ω of simplysupported FGM microplates with 10 10 mesh Powerlaw index n 0 1 2 3 4 5 6 7 8 9 10 Ref 2 610 539 522 532 551 571 588 604 617 627 636 Q4 6168 5448 5277 5387 5583 5785 5967 6124 6255 6365 6455 Q8 6102 5390 5220 5329 5523 5723 5903 6058 6188 6297 6386 Q9 6102 53890 5220 5329 5523 5723 5903 6058 6188 6297 6386 pared with the results presented in 2 where a higher order shear deformation theory has been considered Very good agreement is observed between the two solutions which proofs the validity of the present code Codes problem21aFgmm and prob lem21bFgmm for Q8 and Q9 elements are not explicitly shown the reader just need to change parameters in the present code accordingly in order to get such elements References 1 Y Miyamoto WA Kaysser BH Rabin A Kawasaki RG Ford Functionally Graded Mate rials Design Processing and Applications Materials Technology Series Springer US 2013 2 JN Reddy Energy Principles and Variational Methods in Applied Mechanics 3rd edn Wiley Hoboken NJ USA 2017 Chapter 16 Time Transient Analysis Abstract In the present chapter time transient analysis is presented for Timoshenko beams and laminated FSDT plates The theoretical background mainly focuses on how to implement linear time transient analysis in numerical methods therefore the reader should refer to chapters 10 and 14 for the beam and plate theories and implementation respectively 161 Introduction In the present chapter time transient analysis is presented for Timoshenko beams and laminated FSDT plates The theoretical background mainly focuses on how to implement linear time transient analysis in numerical methods therefore the reader should refer to Chaps10 and 14 for the beam and plate theories and implementation respectively 162 Numerical Time Integration Newmarks time integration method for second order differential equations is briefly described below In the Newmark method functions of time and their derivatives are approximated using Taylors series truncated up to the second order derivative Time increment is indicated as dt ts1 ts where s1 and s indicate the forward and backward time integration points If Δ indicates the generalized global vector of kine matic displacements of the finite element discrete model velocity and acceleration vectors can be carried by Δs1 Δs a1 Δs a2 Δs1 Δs1 a3Δs1 Δs a4 Δs a5 Δs 161 The Editors if applicable and The Authors under exclusive license to Springer Nature Switzerland AG 2020 A J M Ferreira and N Fantuzzi MATLAB Codes for Finite Element Analysis Solid Mechanics and Its Applications 157 httpsdoiorg101007978303047952716 335 336 16 Time Transient Analysis where a1 1 αdt a2 α dt a3 2 γdt2 a4 a3 dt a5 1 γ γ 162 The parameters α and γ are selected according to the time integration method imple mented Their choice is related to the approximation error introduced by the time integration method These methods are stable when the introduced error is bounded eg limited or conditionally stable when the error is bounded only according to a stability condition such as dt dtcr 1 2ωmax α γ12 163 where ωmax is the maximum eigenvalue computed with the linear eigenvalue problem used for the free vibration analysis The Newmarks method contains the following methods α 1 2 γ 1 2 constant average acceleration method stable α 1 2 γ 1 3 linear acceleration method conditionally stable α 1 2 γ 1 6 FoxGoodwin scheme conditionally stable α 1 2 γ 0 central difference method conditionally stable α 3 2 γ 8 5 Galerkin method stable α 3 2 γ 2 backward difference method stable The algebraic system of equations at the time ts1 takes the form ˆKΔs1 ˆF Δs1 ˆK1 ˆF 164 where ˆK Ks1 a3Ms1 ˆF Fs1 Ms1a3Δs a4 Δs a5 Δs 165 Note that all the Δ quantities at ts are known as well as time evolution of stiffness Ks1 and mass Ms1 matrices and force vector Fs1 at ts1 In the following applications only the external force vector will change whereas stiffness and mass matrices will remain constant 162 Numerical Time Integration 337 Definition 162 fail for central difference scheme with γ 0 thus an alternative form of the Eq164 should be considered K Δs1 F Δs1 K1 F 166 where K Ms1 1 a3 Ks1 F Fs1 Ks1Δs a4 a3 Δs a5 a3 Δs 167 so the problem is solved in terms of accelerations instead of displacements Both algebraic systems 164 and 166 need starting values to be initiated eg initial conditions If displacement Δ0 and velocity Δ0 vectors should be known at the initial time step t0 the acceleration vector Δ0 is not known However it should be carried out as Δ0 M1 0 F0 K0Δ0 168 Numerical implementation follows the following steps 1 Define time vector and time step dt 2 Identify best time integration method for the list given and set α and γ 3 Initialize displacement Δ0 and velocity Δ0 vectors 4 Carry out acceleration vector Δ0 with expression 168 5 Use 164 or 166 for evaluating the solution at the generic time step ts 6 Calculate acceleration Δs1 and velocity Δs1 vectors or displacement Δs1 and velocity Δs1 vectors according to the previous selection 7 Iterate the last two steps up to the end of the time frame For simplicity in the present applications structural damping C Δs is not included in the discrete model However it can be easily included without losing generality 1 For instance the wellknown Rayleigh damping can be used for including damping effects in the numerical model 163 Clamped Timoshenko Beam This example has been taken from the book by Reddy 1 which considers the trans verse motion of a beam with initial configuration in free motion eg no exiting force is applied The beam is of unitary length L 1 with initial conditions wx 0 sin πx πx1 x θx 0 π cos πx π1 2x 169 338 16 Time Transient Analysis Fig 161 Central transverse deflection of a clamped beam in free motion with dt 0005 and α γ 05 0 01 02 03 04 05 time 02 015 01 005 0 005 01 015 02 central point motion beam stiffness EI 1 and ρA 1 then kG A 4E IH 2 where H is the cross sectionheightassumingcrosssectionrectangularMomentofinertia I BH 312 poisson ratio ν 025 and shear correction factor k 56 A stable method with α γ 05 is considered dt 0005 for a total time ttot 05 The code problem16timeReddym is listed below and gives the time history of the central point of the clamped beam under study depicted in Fig161 The present result closely matches the one provided by Reddy 1 MATLAB codes for Finite Element Analysis problem16timeReddym Timoshenko beam time transient analysis ref JN Reddy an introduction to Finite Element Method 3rd Ed Example 622 page 332 AJM Ferreira N Fantuzzi 2019 clear memory clear E modulus of elasticity G shear modulus I second moments of area L length of beam thickness thickness of beam poisson 025 L 1 thickness 001 I thicknessˆ312 E 1I rho 100 EI EI kapa 56 constitutive matrix G E21poisson C EI 0 0 kapathicknessG mesh 163 Clamped Timoshenko Beam 339 numberElements 40 nodeCoordinates linspace0LnumberElements1 xx nodeCoordinates elementNodes zerossizenodeCoordinates212 for i 1sizenodeCoordinates21 elementNodesi1i elementNodesi2i1 end generation of coordinates and connectivities numberNodes sizexx1 GDof global number of degrees of freedom GDof 2numberNodes computation of the system stiffness matrix stiffnessforcemass formStiffnessMassTimoshenkoBeamGDofnumberElements elementNodesnumberNodesxxC0rhoIthickness boundary conditions simplysupported at both bords fixedNodeW 1 numberNodes fixedNodeTX boundary conditions clamped at both bords fixedNodeW 1 numberNodes fixedNodeTX fixedNodeW boundary conditions cantilever fixedNodeW 1 fixedNodeTX 1 prescribedDof fixedNodeW fixedNodeTXnumberNodes Time transient simulation timeStep 0005 totalTime 05 time timeSteptimeSteptotalTime Newtons parameters alpha 12 gamma 12 a1 1alphatimeStep a2 alphatimeStep a3 2gammatimeStepˆ2 a4 a3timeStep a5 1gammagamma initialization displacementsTime zerosGDoflengthtime velocitiesTime zerosGDoflengthtime accelerationsTime zerosGDoflengthtime initial conditions ws 1numberNodes displacementsTimews1 sinpixx pixx1xx displacementsTimewsnumberNodes1 picospixx pi12xx accelerationsTime1 massforce stiffnessdisplacementsTime1 for i 2lengthtime forceHat force massa3displacementsTimei1 340 16 Time Transient Analysis a4velocitiesTimei1 a5accelerationsTimei1 stiffnessHat stiffness a3mass displacementsTimei solutionGDofprescribedDof stiffnessHatforceHat accelerationsTimei a3displacementsTimei displacementsTimei1 a4velocitiesTimei1 a5accelerationsTimei1 velocitiesTimei velocitiesTimei1 a1accelerationsTimei1 a2accelerationsTimei end central point vs time figure plottimedisplacementsTimeroundnumberNodes2 linewidth2markersize16 xlabeltime ylabelcentral point motion ylim024 024 setgcalinewidth2fontsize14 grid on box on 164 SimplySupported Laminated Plate Transient solution of antisymmetric crossply laminate 090 using FSDT is consid ered below Q4 10 10 mesh is considered with material and geometric properties as E1 25E2 E2 21 106 Ncm2 G12 G13 05E2 G23 02E2 ν12 025 ν21 001 ρ 8 106 Ns2cm4 a b 25 cm 1610 Two values of the sidetothickness ratios are considered ah 10 and ah 25 The plate is under uniformly distributed load p q01 cosω0t where q0 1 Ncm2 and ω0 00185 µHz Time scale is in μs microseconds and the plate is simplysupported Newtons parameters are chosen as stable with α 12 and γ 12 The code problem20timeReddym lists the program for the present problem which has been taken from 2 of Sect76 some data are not given thus have been selected by the authors MATLAB codes for Finite Element Analysis problem20timeReddym laminated plate time transient using Q4 elements 164 SimplySupported Laminated Plate 341 ref JN Reddy Mechanics of Laminated Composite Plates and Shells 2nd Ed Section 674 page 364 AJM Ferreira N Fantuzzi 2019 clear memory clear close all materials thickness 25 Mesh generation L 25 numberElementsX 10 numberElementsY 10 numberElements numberElementsXnumberElementsY nodeCoordinates elementNodes rectangularMeshLLnumberElementsXnumberElementsYQ4 xx nodeCoordinates1 yy nodeCoordinates2 figure drawingMeshnodeCoordinateselementNodesQ4 axis equal numberNodes sizexx1 GDof global number of degrees of freedom GDof 5numberNodes computation of the system stiffness matrix the shear correction factors are automatically calculted for any laminate AMatrixBMatrixDMatrixSMatrixInertia reddyLaminateMaterialthickness stiffness formStiffnessMatrixMindlinlaminated5dof GDofnumberElements elementNodesnumberNodesnodeCoordinatesAMatrix BMatrixDMatrixSMatrixQ4completereduced computation of the system force vector force formForceVectorMindlin5dofGDofnumberElements elementNodesnumberNodesnodeCoordinates0Q4complete mass formMassMatrixFgmPlate5dofGDofnumberElements elementNodesnumberNodesnodeCoordinatesInertia Q4complete boundary conditions prescribedDofactiveDof EssentialBC5dofssssGDofxxyynodeCoordinatesnumberNodes Time transient simulation timeStep 5 342 16 Time Transient Analysis totalTime 1000 time timeSteptimeSteptotalTime Newtons parameters alpha 12 gamma 12 a1 1alphatimeStep a2 alphatimeStep a3 2gammatimeStepˆ2 a4 a3timeStep a5 1gammagamma initialization displacementsTime zerosGDoflengthtime velocitiesTime zerosGDoflengthtime accelerationsTime zerosGDoflengthtime initial conditions accelerationsTime1 massforce stiffnessdisplacementsTime1 q0 1 P q01 cos00185time for i 2lengthtime force formForceVectorMindlin5dofGDofnumberElements elementNodesnumberNodesnodeCoordinatesPiQ4complete forceHat force massa3displacementsTimei1 a4velocitiesTimei1 a5accelerationsTimei1 stiffnessHat stiffness a3mass displacementsTimei solutionGDofprescribedDof stiffnessHatforceHat accelerationsTimei a3displacementsTimei displacementsTimei1 a4velocitiesTimei1 a5accelerationsTimei1 velocitiesTimei velocitiesTimei1 a1accelerationsTimei1 a2accelerationsTimei end dimensionless transverse displacement vs time centralPt findxxL2 yyL2 wbar displacementsTimecentralPt21e6thicknessˆ3q0Lˆ41e2 figure hold on plottimewbarlinewidth2markersize16 xlabeltime ylabelcentral point motion ylim02 42 setgcalinewidth2fontsize14 grid on box on 164 SimplySupported Laminated Plate 343 The stiffness matrices for the present material configuration are carried out in reddyLaminateMaterialm listed below Note that the code computes A B and D matrices of the consitutive law and inertia matrix I Since for the present lamination scheme I1 0 previous implementation of the mass matrix formation could be used which needs material density ρ and inertia h312 However the present implemen tation that takes some snippets from the FGM codes is more general and it is valid for anisotropic lamination schemes also wherein inertia matrix has I1 0 function AMatrixBMatrixDMatrixSMatrixInertia reddyLaminateMaterialthickness REDDY TIME TRANSIENT EXAMPLE plate thickness h thickness stack 0 90 antisymmetric crossply nlam lengthstack hk hnlam reddy orthotropic properties E2 21e6 E1 25E2 G12 05E2 G13 05E2 G23 02E2 nu12 025 nu21 nu12E2E1 rho0 8e6 kapa 56 Reduced stiffness constants Q11 E11nu12nu21 Q12 nu12E21nu12nu21 Q22 E21nu12nu21 Q66 G12 Q44 G23 Q55 G13 A zeros6 B zeros6 D zeros6 I0 0 I1 0 I2 0 inertias barQ zeros66nlam for k 1lengthstack theta stackk barQk effectivepropsQtheta zk h2 hkk zk1 zk h2 hkk1 zk A A zk zk barQk B B 12zkˆ2 zkˆ2barQk D D 13zkˆ3 zkˆ3barQk I0 I0 zk zk rho0 I1 I1 12zkˆ2 zkˆ2rho0 I2 I2 13zkˆ3 zkˆ3rho0 end 344 16 Time Transient Analysis AMatrix A11A12A16 A12A22A26 A16A26A66 BMatrix B11B12B16 B12B22B26 B16B26B66 DMatrix D11D12D16 D12D22D26 D16D26D66 SMatrix kapaA44kapaA45 kapaA45kapaA55 Inertia I0 0 0 0 0 0 I2 0 I1 0 0 0 I2 0 I1 0 I1 0 I0 0 0 0 I1 0 I0 end effective properties according to the orientation theta function barQ effectivepropsQthetak theta deg2radthetak cc costheta ss sintheta barQ11 Q11ccˆ4 2Q122Q66ccˆ2ssˆ2 Q22ssˆ4 barQ12 Q11 Q22 4Q66ccˆ2ssˆ2 Q12ccˆ4 ssˆ4 barQ22 Q11ssˆ4 2Q122Q66ccˆ2ssˆ2 Q22ccˆ4 barQ16 Q11 Q12 2Q66ccˆ3ss Q12 Q22 2Q66ccssˆ3 barQ26 Q11 Q12 2Q66ccssˆ3 Q12 Q22 2Q66ccˆ3ss barQ66 Q11 Q22 2Q12 2Q66ccˆ2ssˆ2 Q66ccˆ4 ssˆ4 barQ44 Q44ccˆ2 Q55ssˆ2 barQ45 Q55 Q44ccss barQ55 Q55ccˆ2 Q44ssˆ2 end The ratio between static and dynamic response for the present plate iswdws 2000 the value indicated by Reddy 2 is 2049 Deflection is shown in dimensionless form as w w0 0 tE2h3q0a4 102 in Fig162 for different values of ah Figure163 shows that the chosen time integration method α γ 05 is stable for any dt chosen in fact the solutions do not change by changing dt Further calculations can be carried out for instance by plotting the transient stresses after postcomputation is applied For implementing postcomputation the reader can refer to the bending of laminated FSDT plates In order to see the effects due to another possible implementation is to introduce evolution of the stiffness and mass matrices also not only external force for modeling viscous material behavior The present section does not consider Q8 and Q9 elements the reader can easily obtain these codes by setting proper parameters to the given one References 345 Fig 162 Central transverse deflection of a simplysupported 090 plate with dt 5 µs and α γ 05 0 200 400 600 800 1000 time 0 05 1 15 2 25 3 35 4 central point motion ah25 ah10 Fig 163 Central transverse deflection of a simplysupported 090 plate with dt 25 10 5 µs and α γ 05 0 200 400 600 800 1000 time 0 05 1 15 2 25 3 35 4 central point motion dt5 s dt10 s dt25 s References 1 JN Reddy An Introduction to the Finite Element Method 3rd edn McGrawHill International Editions New York 2005 2 JN Reddy Mechanics of Laminated Composite Plates and Shells CRC Press 2004 Index A Assembly of stiffness matrix 31 34 Axes transformation 126 Axial stresses 38 B Bar element 27 Bending stiffness 215 236 Bending stiffness matrix 277 Bending strains 231 Bending stresses 231 Bernoulli beam 89 Bernoulli beam free vibrations 99 Bernoulli beam problem 93 Bernoulli beam with spring 97 B matrix 47 175 233 bending 276 membrane 276 shear 277 Boundary conditions 242 Buckling analysis of Mindlin plates 253 Buckling analysis of Timoshenko beams 165 C Constitutive matrix 172 Coordinate transformation 126 142 Crossply laminates 294 Cylindrical bending 272 D Determinant of Jacobian matrix 234 Distributed forces 28 E Eigenproblem 166 259 Equations of motion of Mindlin plates 244 Essential boundary conditions 35 173 Benoulli beam 89 Exact Gauss quadrature 178 235 External forces 173 External work 39 90 F Finite element steps 29 Force vector 3D frame 126 grids 142 Mindlin plate 235 plane stress 175 Free vibrations of laminated plates 293 Free vibrations of Mindlin plates 244 Free vibrations of Timoshenko beams 159 Functionally graded materials 313 Fundamental frequency 245 G Gauss quadrature 43 154 178 Generalized eigenproblem 244 259 Geometric stiffness matrix The Editors if applicable and The Authors under exclusive license to Springer Nature Switzerland AG 2020 A J M Ferreira and N Fantuzzi MATLAB Codes for Finite Element Analysis Solid Mechanics and Its Applications 157 httpsdoiorg1010079783030479527 347 348 Index Mindlin plate 258 301 Grid example 143 147 Grids 141 stiffness matrix 141 H Hamilton principle 244 Hermite shape functions 91 Hookes law 38 I Initially stressed Mindlin plate 253 Integration points 178 Interpolation of displacements 153 Inverse of Jacobian 177 J Jacobian 40 177 K Kinetic energy 159 bar element 42 2D frame 107 Kirchhoff plates 207 L Lagrange shape functions 176 179 Laminated plates 269 Local coordinate system 59 105 M Mass matrix Mindlin plate 235 2D frame 107 MATLAB codes EssentialBCm 217 EssentialBC5dofm 290 eigenvaluem 55 forcesInElementGridm 145 formForceVectorKm 218 formForceVectorMindlinm 239 formForceVectorMindlin5dofm 289 formGeometricStiffnessMindlinm 265 formGeometricStiffnessMindlin laminated5dofm 307 formMass2Dframem 120 formMass3Dframem 137 formMass3Dtrussm 88 formMassMatrixFgmPlate5dofm 332 formMassMatrixMindlinlaminated 5dofm 297 formMassMatrixMindlinm 252 formStabilityBernoulliBeamm 103 formStiffness2Dframem 110 formStiffness2Dtrussm 63 formStiffness3Dframem 129 formStiffness3Dtrussm 81 formStiffnessBernoulliBeamm 96 formStiffnessBucklingTimoshenko Beamm 169 formStiffnessGridm 144 formStiffnessMassTimoshenko Beamm 157 formStiffnessMassTimoshenkoFgm Beamm 320 formStiffnessMass2Dm 190 formStiffnessMatrixKm 218 formStiffnessMatrixMindlinlaminat ed5dofm 287 formStiffnessMatrixMindlinm 239 gaussQuadraturem 193 JacobianKm 221 Jacobianm 191 MindlinStressm 242 244 problem1m 32 problem2m 45 problem3am 51 problem3m 48 problem3vibm 52 problem4m 61 problem5m 66 problem5vibm 72 problem6m 69 problem7m 79 problem7vibm 87 problem8m 84 problem9m 94 problem9am 97 problem9bukm 103 problem9vibm 99 problemKm 215 problem10m 107 109 problem11m 111 114 problem11bm 114 problem11bvibm 118 problem12m 128 problem13m 131 problem13vibm 136 problem14m 143 144 Index 349 problem15m 147 problem16Bucklingm 166 problem16fgmm 318 problem16fgmVibm 323 problem16m 156 problem16timeReddym 338 problem16vibrationsm 161 problem16vibrationsSchultzm 164 problem17am 185 problem17bm 185 problem17m 184 185 problem18am 199 problem18bm 199 problem18m 197 problem18vibm 202 problem19Bucklingm 260 problem19m 236 problem19Vibrationsm 250 problem20Bucklingm 305 problem20Fgmm 328 problem20aFgmm 330 problem20bFgmm 330 problem20m 285 problem20timeReddym 340 problem21Fgmm 331 problem21m 294 reddyFgmMaterialm 329 reddyLaminateMaterialBukm 309 shapeFunctionK12m 221 shapeFunctionK16m 221 shapeFunctionsQm 191 shapeFunctionKQ4m 225 SrinivasStressm 291 solutionm 48 srinivasMaterialm 287 stresses2Dm 191 stresses2Dtrussm 64 stresses3Dtrussm 82 outputDisplacementsReactionsm 35 Mindlin plate theory 269 Mindlin plates 229 Modes of vibration 161 244 N Natural boundary conditions 173 Natural coordinate system 39 Natural frequencies 159 244 Nodal point stresses 182 P Plane stress 171 Potential energy 173 Prescribed degrees of freedom 35 Problem 14 144 Problem 16 vib 161 Problem 17 185 Problem 17a 185 Problem 17b 185 Problem 18 197 Problems Essential BC 217 formMass3Dtruss 88 formStabilityBernoulliBeamm 103 problem K 215 problem 1 32 problem 2 45 problem 3 48 problem 4 61 problem 5 66 problem 5 vibrations 72 problem 6 69 problem 7 79 87 problem 8 84 problem 9 94 problem 9a 97 problem9bukm 103 problem 9vib 99 problem 10 107 109 problem 11 111 114 problem 11b 114 problem 11bvib 118 problem 11vib 118 problem 12 128 problem 13 131 problem 13vib 136 problem 14 143 problem 15 147 problem 16 156 problem16Buckling 166 problem 16 FGM 318 problem 16 FGM vibrations 323 problem 16 time transient Reddy 338 problem16vibrations 161 problem16vibrationsSchultz 164 problem 17 184 problem 18 197 problem 18a 199 problem 18b 199 problem 18vib 202 problem 19 236 problem19Buckling 260 problem19Vibrations 250 problem20 285 problem20Buckling 305 problem 20 FGM 328 problem 20 time transient Reddy 340 350 Index problem21 294 problem 21 FGM 331 Q Q4 element 233 Q8 element 233 Q9 element 233 Quadrilateral element Q4 176 233 Quadrilateral element Q8 233 Quadrilateral element Q9 233 R Reactions 35 Reduced Gauss quadrature 235 Rotation matrix 126 142 S Shape functions 39 40 91 153 174 176 179 233 Shear correction factor 152 232 272 294 305 Shear deformations theories Mindlin theory 269 Shear locking 154 235 Spring element 27 Stiffness matrix 40 assembly process 42 bar element 28 40 Bernoulli beam 91 grids 141 Mindlin plate 234 plane stress 175 Timoshenko beams 153 3D frame 126 3D truss 78 2D frame 106 2D truss element 59 Strain energy 42 59 317 bar element 38 42 Bernoulli beam 90 Mindlin plate 232 233 plane stress 173 Timoshenko beams 152 2D frame 107 Stresses 2D truss 60 Stress recovery 181 Stressstrain relations 231 Surface tractions 175 T Timoshenko beams 151 Transformation 106 Transverse shear stresses 152 231 Transverse strains 231 3D frame 123 stiffness matrix 126 3D frame problem 128 3D truss 77 stiffness matrix 78 stresses 85 3D truss problem 79 83 2D frame 105 mass matrix 107 120 stiffness matrix 106 109 2D frame problem 107 111 2D frame problem free vibrations 118 2D truss 57 stiffness matrix 63 stresses 64 2D truss problem 61 66 2D truss problem with spring 69 Twonode bar finite element 38 Twonode element 159 V Vector of equivalent forces 42