Introduction to Finite Element Analysis Using MATLAB and Abaqus

w w w c r c p r e s s c o m K16894 Amar Khennane Introduction to Finite Element Analysis Using MATLAB and Abaqus Khennane Introduction to Finite Element Analysis Using MATLAB and Abaqus Introduction to Finite Element Analysis Using MATLAB and Abaqus A very good introduction to the finite element method with a balanced treatment of theory and implementation F Albermani Reader in Structural Engineering The University of Queensland Australia There are some books that target the theory of the finite element while others focus on the programming side of things Introduction to Finite Element Analysis Using MATLAB and Abaqus accomplishes both This book teaches the first principles of the finite element method It presents the theory of the finite element method while maintaining a balance between its mathematical formulation programming implemen tation and application using commercial software The computer implementation is carried out using MATLAB while the practical applications are carried out in both MATLAB and Abaqus MATLAB is a highlevel language specially designed for dealing with matrices making it particularly suited for programming the finite element meth od while Abaqus is a suite of commercial finite element software Introduction to Finite Element Analysis Using MATLAB and Abaqus introduces and explains theory in each chapter and provides corresponding examples It offers introductory notes and provides matrix structural analysis for trusses beams and frames The book examines the theories of stress and strain and the relationships be tween them The author then covers weighted residual methods and finite element ap proximation and numerical integration He presents the finite element formulation for plane stressstrain problems introduces axisymmetric problems and highlights the theory of plates The text supplies stepbystep procedures for solving problems with Abaqus interactive and keyword editions The described procedures are implemented as MATLAB codes and Abaqus files can be found on the CRC Press website Mathematics 6000 Broken Sound Parkway NW Suite 300 Boca Raton FL 33487 711 Third Avenue New York NY 10017 2 Park Square Milton Park Abingdon Oxon OX14 4RN UK an informa business w w w c r c p r e s s c o m Introduction to Finite Element Analysis Using MATLAB and Abaqus 2013 by Taylor Francis Group LLC 2013 by Taylor Francis Group LLC Boca Raton London New York CRC Press is an imprint of the Taylor Francis Group an informa business Amar Khennane Introduction to Finite Element Analysis Using MATLAB and Abaqus 2013 by Taylor Francis Group LLC MATLAB is a trademark of The MathWorks Inc and is used with permission The MathWorks does not warrant the accuracy of the text or exercises in this book This books use or discussion of MATLAB software or related products does not constitute endorsement or sponsorship by The MathWorks of a particular pedagogical approach or particular use of the MATLAB software CRC Press Taylor Francis Group 6000 Broken Sound Parkway NW Suite 300 Boca Raton FL 334872742 2013 by Taylor Francis Group LLC CRC Press is an imprint of Taylor Francis Group an Informa business No claim to original US Government works Version Date 20130220 International Standard Book Number13 9781466580213 eBook PDF This book contains information obtained from authentic and highly regarded sources Reasonable efforts have been made to publish reliable data and information but the author and publisher cannot assume responsibility for the valid ity of all materials or the consequences of their use The authors and publishers have attempted to trace the copyright holders of all material reproduced in this publication and apologize to copyright holders if permission to publish in this form has not been obtained If any copyright material has not been acknowledged please write and let us know so we may rectify in any future reprint Except as permitted under US Copyright Law no part of this book may be reprinted reproduced transmitted 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Contents List of Figures xiii List of Tables xxv Preface xxvii Author xxix Chapter 1 Introduction 1 11 Prologue 1 12 Finite Element Analysis and the User 1 13 Aim of the Book 2 14 Book Organization 2 Chapter 2 Bar Element 5 21 Introduction 5 22 OneDimensional Truss Element 5 221 Formulation of the Stiffness Matrix The Direct Approach 5 222 TwoDimensional Truss Element 7 23 Global Stiffness Matrix Assembly 9 231 Discretization 9 232 Elements Stiffness Matrices in Local Coordinates 9 233 Elements Stiffness Matrices in Global Coordinates 10 2331 Element 1 11 2332 Element 2 11 2333 Element 3 12 234 Global Matrix Assembly 12 2341 Only Element 1 Is Present 13 2342 Only Element 2 Is Present 13 2343 Only Element 3 Is Present 13 235 Global Force Vector Assembly 14 24 Boundary Conditions 15 241 General Case 15 25 Solution of the System of Equations 16 26 Support Reactions 17 27 Members Forces 18 28 Computer Code trussm 19 281 Data Preparation 20 2811 Nodes Coordinates 20 2812 Element Connectivity 20 2813 Material and Geometrical Properties 20 2814 Boundary Conditions 20 2815 Loading 21 282 Element Matrices 21 2821 Stiffness Matrix in Local Coordinates 21 2822 Transformation Matrix 22 v 2013 by Taylor Francis Group LLC vi Contents 2823 Stiffness Matrix in Global Coordinates 22 2824 Steering Vector 22 283 Assembly of the Global Stiffness Matrix 23 284 Assembly of the Global Force Vector 23 285 Solution of the Global System of Equations 23 286 Nodal Displacements 23 287 Element Forces 23 288 Program Scripts 24 29 Problems 27 291 Problem 21 27 292 Problem 22 32 210 Analysis of a Simple Truss with Abaqus 35 2101 Overview of Abaqus 35 2102 Analysis of a Truss with Abaqus Interactive Edition 36 21021 Modeling 36 21022 Analysis 51 2103 Analysis of a Truss with Abaqus Keyword Edition 57 Chapter 3 Beam Element 63 31 Introduction 63 32 Stiffness Matrix 63 33 Uniformly Distributed Loading 67 34 Internal Hinge 71 35 Computer Code beamm 73 351 Data Preparation 73 3511 Nodes Coordinates 73 3512 Element Connectivity 74 3513 Material and Geometrical Properties 74 3514 Boundary Conditions 74 3515 Internal Hinges 74 3516 Loading 75 3517 Stiffness Matrix 76 352 Assembly and Solution of the Global System of Equations 76 353 Nodal Displacements 76 354 Element Forces 77 36 Problems 81 361 Problem 31 81 362 Problem 32 84 363 Problem 33 87 37 Analysis of a Simple Beam with Abaqus 90 371 Interactive Edition 90 372 Analysis of a Beam with Abaqus Keyword Edition 103 Chapter 4 Rigid Jointed Frames 107 41 Introduction 107 42 Stiffness Matrix of a BeamColumn Element 107 43 Stiffness Matrix of a BeamColumn Element in the Presence of Hinged End 107 2013 by Taylor Francis Group LLC Contents vii 44 Global and Local Coordinate Systems 108 45 Global Stiffness Matrix Assembly and Solution for Unknown Displacements 109 46 Computer Code framem 109 461 Data Preparation 109 4611 Nodes Coordinates 110 4612 Element Connectivity 110 4613 Material and Geometrical Properties 110 4614 Boundary Conditions 110 4615 Internal Hinges 111 4616 Loading 111 462 Element Matrices 112 4621 Stiffness Matrix in Local Coordinates 112 4622 Transformation Matrix 113 4623 Stiffness Matrix in Global Coordinates 113 4624 Steering Vector 113 4625 Element Loads 113 463 Assembly of the Global Stiffness Matrix 113 464 Solution of the Global System of Equations 114 465 Nodal Displacements 114 466 Element Forces 114 47 Analysis of a Simple Frame with Abaqus 124 471 Interactive Edition 124 472 Keyword Edition 132 Chapter 5 Stress and Strain Analysis 135 51 Introduction 135 52 Stress Tensor 135 521 Definition 135 522 Stress TensorStress Vector Relationships 137 523 Transformation of the Stress Tensor 139 524 Equilibrium Equations 139 525 Principal Stresses 140 526 von Mises Stress 141 527 Normal and Tangential Components of the Stress Vector 141 528 Mohrs Circles for Stress 143 529 Engineering Representation of Stress 144 53 Deformation and Strain 144 531 Definition 144 532 Lagrangian and Eulerian Descriptions 145 533 Displacement 146 534 Displacement and Deformation Gradients 147 535 Green Lagrange Strain Matrix 148 536 Small Deformation Theory 149 5361 Infinitesimal Strain 149 5362 Geometrical Interpretation of the Terms of the Strain Tensor 150 5363 Compatibility Conditions 152 537 Principal Strains 152 2013 by Taylor Francis Group LLC viii Contents 538 Transformation of the Strain Tensor 153 539 Engineering Representation of Strain 153 54 StressStrain Constitutive Relations 154 541 Generalized Hookes Law 154 542 Material Symmetries 155 5421 Symmetry with respect to a Plane 155 5422 Symmetry with respect to Three Orthogonal Planes 157 5423 Symmetry of Rotation with respect to One Axis 157 543 Isotropic Material 158 5431 Modulus of Elasticity 160 5432 Poissons Ratio 160 5433 Shear Modulus 160 5434 Bulk Modulus 160 544 Plane Stress and Plane Strain 162 55 Solved Problems 163 551 Problem 51 163 552 Problem 52 164 553 Problem 53 167 554 Problem 54 168 555 Problem 55 170 556 Problem 56 171 557 Problem 57 172 558 Problem 58 174 Chapter 6 Weighted Residual Methods 175 61 Introduction 175 62 General Formulation 175 63 Galerkin Method 176 64 Weak Form 178 65 Integrating by Part over Two and Three Dimensions Green Theorem 179 66 Rayleigh Ritz Method 183 661 Definition 183 662 Functional Associated with an Integral Form 183 663 Rayleigh Ritz Method 183 664 Example of a Natural Functional 185 Chapter 7 Finite Element Approximation 191 71 Introduction 191 72 General and Nodal Approximations 191 73 Finite Element Approximation 193 74 Basic Principles for the Construction of Trial Functions 195 741 Compatibility Principle 195 742 Completeness Principle 196 75 TwoDimensional Finite Element Approximation 197 751 Plane Linear Triangular Element for C0 Problems 197 7511 Shape Functions 197 7512 Reference Element 199 7513 Area Coordinates 202 752 Linear Quadrilateral Element for C0 Problems 203 2013 by Taylor Francis Group LLC Contents ix 7521 Geometrical Transformation 203 7522 Construction of a Trial Function over a Linear Quadrilateral Element 206 76 Shape Functions of Some Classical Elements for C0 Problems 207 761 OneDimensional Elements 207 7611 TwoNodded Linear Element 207 7612 ThreeNodded Quadratic Element 207 762 TwoDimensional Elements 207 7621 FourNodded Bilinear Quadrilateral 207 7622 EightNodded Quadratic Quadrilateral 208 7623 ThreeNodded Linear Triangle 208 7624 SixNodded Quadratic Triangle 208 763 ThreeDimensional Elements 208 7631 FourNodded Linear Tetrahedra 208 7632 TenNodded Quadratic Tetrahedra 209 7633 EightNodded Linear Brick Element 209 7634 TwentyNodded Quadratic Brick Element 210 Chapter 8 Numerical Integration 211 81 Introduction 211 82 Gauss Quadrature 211 821 Integration over an Arbitrary Interval a b 214 822 Integration in Two and Three Dimensions 215 83 Integration over a Reference Element 216 84 Integration over a Triangular Element 217 841 Simple Formulas 217 842 Numerical Integration over a Triangular Element 218 85 Solved Problems 219 851 Problem 81 219 852 Problem 82 221 853 Problem 83 226 Chapter 9 Plane Problems 231 91 Introduction 231 92 Finite Element Formulation for Plane Problems 231 93 Spatial Discretization 234 94 Constant Strain Triangle 235 941 Displacement Field 236 942 Strain Matrix 237 943 Stiffness Matrix 237 944 Element Force Vector 237 9441 Body Forces 238 9442 Traction Forces 238 9443 Concentrated Forces 239 945 Computer Codes Using the Constant Strain Triangle 240 9451 Data Preparation 241 9452 Nodes Coordinates 243 9453 Element Connectivity 243 9454 Material Properties 243 2013 by Taylor Francis Group LLC x Contents 9455 Boundary Conditions 243 9456 Loading 243 9457 Main Program 243 9458 Element Stiffness Matrix 245 9459 Assembly of the Global Stiffness Matrix 246 94510 Solution of the Global System of Equations 246 94511 Nodal Displacements 246 94512 Element Stresses and Strains 246 94513 Results and Discussion 247 94514 Program with Automatic Mesh Generation 249 946 Analysis with Abaqus Using the CST 253 9461 Interactive Edition 253 9462 Keyword Edition 260 95 Linear Strain Triangle 263 951 Displacement Field 264 952 Strain Matrix 265 953 Stiffness Matrix 266 954 Computer Code LSTPLANESTRESSMESHm 266 9541 Numerical Integration of the Stiffness Matrix 270 9542 Computation of the Stresses and Strains 271 955 Analysis with Abaqus Using the LST 272 9551 Interactive Edition 272 9552 Keyword Edition 278 96 The Bilinear Quadrilateral 279 961 Displacement Field 280 962 Strain Matrix 281 963 Stiffness Matrix 282 964 Element Force Vector 282 965 Computer Code Q4PLANESTRESSm 284 9651 Data Preparation 284 9652 Main Program 287 9653 Integration of the Stiffness Matrix 289 9654 Computation of the Stresses and Strains 290 9655 Program with Automatic Mesh Generation 291 966 Analysis with Abaqus Using the Q4 Quadrilateral 295 9661 Interactive Edition 295 9662 Keyword Edition 302 97 The 8Node Quadrilateral 304 971 Formulation 304 972 Equivalent Nodal Forces 307 973 Program Q8PLANESTRESSm 307 9731 Data Preparation 307 9732 Main Program 311 9733 Integration of the Stiffness Matrix 314 9734 Results with the Coarse Mesh 314 9735 Program with Automatic Mesh Generation 315 974 Analysis with Abaqus Using the Q8 Quadrilateral 321 98 Solved Problem with MATLAB 326 2013 by Taylor Francis Group LLC Contents xi 981 Strip Footing with the CST Element 326 982 Strip Footing with the LST Element 331 983 Bridge Pier with the Q8 Element 336 Chapter 10 Axisymmetric Problems 353 101 Definition 353 102 StrainDisplacement Relationship 353 103 StressStrain Relations 354 104 Finite Element Formulation 355 1041 Displacement Field 355 1042 Strain Matrix 355 1043 Stiffness Matrix 356 1044 Nodal Force Vectors 356 10441 Body Forces 356 10442 Surface Forces Vector 356 10443 Concentrated Loads 357 10444 Example 357 105 Programming 358 1051 Computer Code AXISYMT6m 359 10511 Numerical Integration of the Stiffness Matrix 362 10512 Results 363 1052 Computer Code AXISYMQ8m 365 10521 Numerical Integration of the Stiffness Matrix 368 10522 Results 370 106 Analysis with Abaqus Using the 8Node Quadrilateral 372 Chapter 11 Thin and Thick Plates 379 111 Introduction 379 112 Thin Plates 379 1121 Differential Equation of Plates Loaded in Bending 379 1122 Governing Equation in terms of Displacement Variables 382 113 Thick Plate Theory or Mindlin Plate Theory 383 1131 StressStrain Relationship 384 114 Linear Elastic Finite Element Analysis of Plates 385 1141 Finite Element Formulation for Thin Plates 385 11411 Triangular Element 385 11412 Rectangular Element 387 1142 Finite Element Formulation for Thick Plates 388 115 Boundary Conditions 389 1151 Simply Supported Edge 389 1152 Builtin or Clamped Edge 390 1153 Free Edge 390 116 Computer Program for Thick Plates Using the 8Node Quadrilateral 390 2013 by Taylor Francis Group LLC xii Contents 1161 Main Program ThickplateQ8m 390 1162 Data Preparation 395 11621 Stiffness Matrices 395 11622 Boundary Conditions 395 11623 Loading 396 11624 Numerical Integration of the Stiffness Matrix 397 1163 Results 398 11631 Determination of the Resulting Moments and Shear Forces 398 11632 Contour Plots 399 117 Analysis with Abaqus 400 1171 Preliminary 400 11711 ThreeDimensional Shell Elements 401 11712 Axisymmetric Shell Elements 401 11713 Thick versus Thin Conventional Shell 401 1172 Simply Supported Plate 401 1173 ThreeDimensional Shells 406 Appendix A List of MATLAB Modules and Functions 419 Appendix B Statically Equivalent Nodal Forces 445 Appendix C Index Notation and Transformation Laws for Tensors 447 References and Bibliography 453 Index 455 2013 by Taylor Francis Group LLC List of Figures FIGURE 21 Truss structure 6 FIGURE 22 Bar element 6 FIGURE 23 Degrees of freedom of a rod element in a twodimensional space 7 FIGURE 24 Truss element oriented at an arbitrary angle θ 8 FIGURE 25 Model of a truss structure 10 FIGURE 26 Free body diagram of the truss 14 FIGURE 27 Free body diagram of element 3 18 FIGURE 28 Equilibrium of node 3 19 FIGURE 29 Model of Problem 21 28 FIGURE 210 Model of Problem 22 32 FIGURE 211 Abaqus documentation 36 FIGURE 212 Starting Abaqus 36 FIGURE 213 Abaqus CAE main user interface 37 FIGURE 214 Creating a part 37 FIGURE 215 Choosing the geometry of the part 37 FIGURE 216 Fitting the sketcher to the screen 38 FIGURE 217 Drawing using the connected line button 38 FIGURE 218 Drawing the truss geometry 38 FIGURE 219 Finished part 38 FIGURE 220 Material definition 39 FIGURE 221 Material properties 39 FIGURE 222 Create section window 40 FIGURE 223 Edit material window 40 FIGURE 224 Section assignment 40 FIGURE 225 Regions to be assigned a section 41 FIGURE 226 Edit section assignment 41 FIGURE 227 Loading the meshing menu 41 FIGURE 228 Selecting regions to be assigned element type 42 xiii 2013 by Taylor Francis Group LLC xiv List of Figures FIGURE 229 Selecting element type 42 FIGURE 230 Mesh 43 FIGURE 231 Assembling the model 43 FIGURE 232 Creating instances 44 FIGURE 233 Numbering of the degrees of freedom 44 FIGURE 234 Creating boundary conditions 45 FIGURE 235 Type of boundary conditions 45 FIGURE 236 Selecting a region to be assigned boundary conditions 46 FIGURE 237 Edit boundary condition dialog box for pinned support 46 FIGURE 238 Edit boundary condition dialog box for roller support 47 FIGURE 239 Creating a step for load application 47 FIGURE 240 Create step dialog box 48 FIGURE 241 Edit step dialog box 48 FIGURE 242 Creating a load 49 FIGURE 243 Creating a concentrated load 49 FIGURE 244 Selecting a joint for load application 50 FIGURE 245 Entering the magnitude of a joint force 50 FIGURE 246 Loaded truss 50 FIGURE 247 Creating a job 51 FIGURE 248 Naming a job 51 FIGURE 249 Editing a job 52 FIGURE 250 Submitting a job 52 FIGURE 251 Monitoring of a job 52 FIGURE 252 Opening the visualization module 53 FIGURE 253 Common plot options 53 FIGURE 254 Elements and nodes numbering 53 FIGURE 255 Deformed shape 54 FIGURE 256 Field output dialog box 54 FIGURE 257 Contour plot of the vertical displacement U2 55 FIGURE 258 Viewport annotations options 55 FIGURE 259 Normal stresses in the bars 55 FIGURE 260 Selecting variables to print to a report 56 FIGURE 261 Choosing a directory and the file name to which to write the report 56 FIGURE 262 Running Abaqus from the command line 61 2013 by Taylor Francis Group LLC List of Figures xv FIGURE 31 Beam element 64 FIGURE 32 Differential element of a beam 64 FIGURE 33 Nodal degrees of freedom 65 FIGURE 34 Statically equivalent nodal loads 68 FIGURE 35 Loading bending moment and shear force diagrams 68 FIGURE 36 Support reactions for individual members 71 FIGURE 37 Beam with an internal hinge 71 FIGURE 38 Beam elements with a hinge 73 FIGURE 39 Example of a continuous beam 73 FIGURE 310 Example 1 Continuous beam results 81 FIGURE 311 Problem 31 81 FIGURE 312 Problem 32 and equivalent nodal loads for elements 3 and 4 84 FIGURE 313 Problem 33 87 FIGURE 314 Continuous beam 90 FIGURE 315 Beam cross section dimensions are in mm 90 FIGURE 316 Creating the BeamPart 91 FIGURE 317 Drawing using the connected line icon 91 FIGURE 318 Material definition 91 FIGURE 319 Creating a beam profile 92 FIGURE 320 Entering the dimensions of a profile 92 FIGURE 321 Creating a section 93 FIGURE 322 Editing a beam section 93 FIGURE 323 Editing section assignments 94 FIGURE 324 Beam orientation 94 FIGURE 325 Assigning beam orientation 94 FIGURE 326 Rendering beam profile 95 FIGURE 327 Rendered beam 95 FIGURE 328 Selecting a beam element 96 FIGURE 329 Seeding a mesh by size 96 FIGURE 330 Node and element labels 97 FIGURE 331 Creating a node set 97 FIGURE 332 Selecting multiple nodes 98 FIGURE 333 Creating element sets 98 2013 by Taylor Francis Group LLC xvi List of Figures FIGURE 334 Imposing BC using created sets 98 FIGURE 335 Selecting a node set for boundary conditions 99 FIGURE 336 Editing boundary conditions 99 FIGURE 337 Imposing BC using created sets 100 FIGURE 338 Imposing a concentrated load using a node set 100 FIGURE 339 Imposing a line load on an element set 101 FIGURE 340 Field output 101 FIGURE 341 Submitting a job in Abaqus CAE 101 FIGURE 342 Plotting stresses in the bottom fiber 102 FIGURE 41 Beam column element with six degrees of freedom 108 FIGURE 42 Example 1 Portal frame 110 FIGURE 43 Frame with an internal hinge 119 FIGURE 44 Finite element discretization 119 FIGURE 45 Statically equivalent nodal loads 120 FIGURE 46 Portal frame 124 FIGURE 47 Profiles sections dimensions are in mm 125 FIGURE 48 Creating the Portalframe part 125 FIGURE 49 Material and profiles definitions 126 FIGURE 410 Creating sections 126 FIGURE 411 Editing section assignments 127 FIGURE 412 Assigning beam orientation 127 FIGURE 413 Rendering beam profile 127 FIGURE 414 Seeding by number 128 FIGURE 415 Mesh 128 FIGURE 416 Creating the element set Rafters 129 FIGURE 417 Imposing BC using created sets 129 FIGURE 418 Imposing a line load in global coordinates 130 FIGURE 419 Imposing a line load in local coordinates 130 FIGURE 420 Analyzing a job in Abaqus CAE 131 FIGURE 421 Plotting stresses in the bottom fiber interactive edition 131 FIGURE 422 Plotting stresses in the bottom fiber keyword edition 134 FIGURE 51 Internal force components 136 FIGURE 52 Stress components at a point 136 FIGURE 53 Stress components on a tetrahedron 137 2013 by Taylor Francis Group LLC List of Figures xvii FIGURE 54 Equilibrium of an infinitesimal cube 139 FIGURE 55 Principal directions of a stress tensor 141 FIGURE 56 Tangential and normal components of the stress vector 142 FIGURE 57 Mohrs circles 143 FIGURE 58 Schematic representation of the deformation of a solid body 145 FIGURE 59 Reference and current configurations 146 FIGURE 510 Deformations of an infinitesimal element 147 FIGURE 511 Geometrical representation of the components of strain at a point 151 FIGURE 512 Monoclinic material 155 FIGURE 513 Symmetry of rotation 157 FIGURE 514 A state of plane stress 162 FIGURE 515 State of plane strain 163 FIGURE 516 Change of basis 165 FIGURE 517 Displacement field Problem 53 167 FIGURE 518 Displacement field Problem 55 170 FIGURE 519 Strain rosette 171 FIGURE 520 Problem 57 172 FIGURE 521 Displacements without the rigid walls 173 FIGURE 61 Graphical comparison of exact and approximate solution 178 FIGURE 62 Integration by parts in two and three dimensions 180 FIGURE 63 Infinitesimal element of the boundary 180 FIGURE 64 Graphical comparison of the exact and approximate solutions 186 FIGURE 71 Thick wall with embedded thermocouples 192 FIGURE 72 Finite element discretization 193 FIGURE 73 Finite element approximation 195 FIGURE 74 Geometrical illustration of the compatibility principle 195 FIGURE 75 Linear triangle 197 FIGURE 76 Geometrical transformation for a triangular element 200 FIGURE 77 Threenode triangular element with an arbitrary point O 202 FIGURE 78 Threenode triangular reference element 204 FIGURE 79 Geometrical transformation 204 FIGURE 710 Onedimensional elements 207 FIGURE 711 Twodimensional quadrilateral elements 207 2013 by Taylor Francis Group LLC xviii List of Figures FIGURE 712 Twodimensional triangular elements 208 FIGURE 713 Threedimensional tetrahedric elements 209 FIGURE 714 Threedimensional brick elements 210 FIGURE 81 Positions of the sampling points for a triangle Orders 1 2 and 3 219 FIGURE 82 Gauss quadrature over an arbitrary area 219 FIGURE 83 Double change of variables 220 FIGURE 84 Coarse mesh of two 8nodded elements 221 FIGURE 85 Eight elements finite element approximation with two 8nodded elements 222 FIGURE 86 Estimation of rainfall using finite element approximation 226 FIGURE 91 Discretization error involving overlapping 234 FIGURE 92 Discretization error involving holes between elements 235 FIGURE 93 Plane elements with shape distortions 235 FIGURE 94 Geometrical discretization error 235 FIGURE 95 Linear triangular element 236 FIGURE 96 Element nodal forces 239 FIGURE 97 Analysis of a cantilever beam in plane stress 240 FIGURE 98 Finite element discretization with linear triangular elements 241 FIGURE 99 Deflection of the cantilever beam 248 FIGURE 910 Stresses along the xaxis 249 FIGURE 911 Automatic mesh generation with the CST element 252 FIGURE 912 Deflection of the cantilever beam obtained with the fine mesh 253 FIGURE 913 Stresses along the xaxis obtained with the fine mesh 253 FIGURE 914 Creating the BeamCST Part 254 FIGURE 915 Drawing using the createlines rectangle icon 254 FIGURE 916 Creating a partition 255 FIGURE 917 Creating a plane stress section 255 FIGURE 918 Editing section assignments 255 FIGURE 919 Mesh controls 256 FIGURE 920 Selecting element type 256 FIGURE 921 Seeding part by size 256 FIGURE 922 Mesh 257 FIGURE 923 Imposing BC using geometry 257 FIGURE 924 Imposing a concentrated force using geometry 257 FIGURE 925 Analyzing a job in Abaqus CAE 258 2013 by Taylor Francis Group LLC List of Figures xix FIGURE 926 Plotting displacements on deformed and undeformed shapes 258 FIGURE 927 Generating a mesh manually in Abaqus 261 FIGURE 928 Displacement contour 263 FIGURE 929 Linear strain triangular element 263 FIGURE 930 Automatic mesh generation with the LST element 271 FIGURE 931 Deflection of the cantilever beam obtained with the LST element 272 FIGURE 932 Stresses along the xdirection obtained with the LST element 273 FIGURE 933 Aluminum plate with a hole 273 FIGURE 934 Making use of symmetry 273 FIGURE 935 Creating the PlateLST Part 274 FIGURE 936 Creating a plane stress section 274 FIGURE 937 Editing section assignments 275 FIGURE 938 Mesh controls 275 FIGURE 939 Seeding edge by size and simple bias 276 FIGURE 940 Creating a node set 276 FIGURE 941 Creating a surface 277 FIGURE 942 Imposing BC using node sets 277 FIGURE 943 Imposing a pressure load on a surface 278 FIGURE 944 Plotting the maximum inplane principal stress under tension 279 FIGURE 945 Plotting the maximum inplane principal stress under compression 279 FIGURE 946 Linear quadrilateral element 280 FIGURE 947 Element loading 283 FIGURE 948 Equivalent nodal loading 284 FIGURE 949 Finite element discretization with 4nodded quadrilateral elements 285 FIGURE 950 Contour of the vertical displacement v2 290 FIGURE 951 Contour of the stress σxx 291 FIGURE 952 Automatic mesh generation with the Q4 element 295 FIGURE 953 Contour of the vertical displacement v2 295 FIGURE 954 Contour of the stresses along the xaxis σxx 295 FIGURE 955 Creating the BeamQ4 Part 296 FIGURE 956 Creating a partition 296 FIGURE 957 Creating a plane stress section 297 FIGURE 958 Editing section assignments 297 FIGURE 959 Mesh controls 297 2013 by Taylor Francis Group LLC xx List of Figures FIGURE 960 Selecting element type 298 FIGURE 961 Seeding part by size 298 FIGURE 962 Mesh 298 FIGURE 963 Imposing BC using geometry 299 FIGURE 964 Imposing a concentrated force using geometry 299 FIGURE 965 Plotting displacements on deformed and undeformed shapes 300 FIGURE 966 Generating a mesh manually in Abaqus 302 FIGURE 967 Mesh generated with the keyword edition 304 FIGURE 968 Displacement contour 305 FIGURE 969 Eightnodded isoparametric element 305 FIGURE 970 Equivalent nodal loads 307 FIGURE 971 Geometry and loading 307 FIGURE 972 Coarse mesh 308 FIGURE 973 Contour of the vertical displacement v2 314 FIGURE 974 Contour of the stress σxx 314 FIGURE 975 Contour of the stress τxy 315 FIGURE 976 Slender beam under 4point bending 315 FIGURE 977 Automatic mesh generation with the Q8 element 319 FIGURE 978 Contour of the vertical displacement v2 320 FIGURE 979 Contour of the stress σxx 320 FIGURE 980 Contour of the stress τxy 320 FIGURE 981 Creating the DeepBeamQ8 Part 321 FIGURE 982 Creating a plane stress section 321 FIGURE 983 Editing section assignments 322 FIGURE 984 Mesh controls and element type 322 FIGURE 985 Mesh 323 FIGURE 986 Creating the node set Loadednode 323 FIGURE 987 Creating the node set Centerline 324 FIGURE 988 Creating the node set Support 324 FIGURE 989 Imposing BC using a node set 325 FIGURE 990 BC and loads 325 FIGURE 991 Contour of the vertical displacement 326 FIGURE 992 Contour of the horizontal stress σxx 326 FIGURE 993 Strip footing 327 2013 by Taylor Francis Group LLC List of Figures xxi FIGURE 994 Strip footing model 328 FIGURE 995 Mesh with the CST element 328 FIGURE 996 Computed result with the CST element 332 FIGURE 997 Mesh with the LST element 332 FIGURE 998 Statically equivalent loads for the LST element 333 FIGURE 999 Computed result with the LST element 336 FIGURE 9100 Bridge pier 337 FIGURE 9101 Bridge pier model 338 FIGURE 9102 Element internal node numbering 338 FIGURE 9103 Finite element discretization of the pier model 339 FIGURE 9104 Contour of the vertical displacement 350 FIGURE 9105 Contour of the maximum principal stress σ1 350 FIGURE 9106 Contour of the minimum principal stress σ2 351 FIGURE 101 Typical axisymmetric problem 354 FIGURE 102 Strains and corresponding stresses in an axisymmetric solid 354 FIGURE 103 Tangential strain 354 FIGURE 104 Axisymmetric equivalent nodal loads 356 FIGURE 105 Typical quadrilateral element on which axisymmetric loads are applied 357 FIGURE 106 Circular footing on a sandy soil 358 FIGURE 107 Geometrical model for the circular footing 358 FIGURE 108 Finite element mesh using the 6node triangle 362 FIGURE 109 Contour plot of the vertical displacement 363 FIGURE 1010 Contour plot of the radial stress 364 FIGURE 1011 Contour plot of the vertical stress 364 FIGURE 1012 Contour plot of the shear stress 365 FIGURE 1013 Finite element mesh using the 8node quadrilateral 369 FIGURE 1014 Contour plot of the vertical displacement 370 FIGURE 1015 Contour plot of the radial stress 370 FIGURE 1016 Contour plot of the vertical stress 371 FIGURE 1017 Contour plot of the shear stress 371 FIGURE 1018 Creating the FOOTINGQ8 Part 372 FIGURE 1019 Creating an axisymmetric section 372 FIGURE 1020 Editing section assignments 373 FIGURE 1021 Edge partition 373 2013 by Taylor Francis Group LLC xxii List of Figures FIGURE 1022 Mesh controls and element type 374 FIGURE 1023 Mesh 374 FIGURE 1024 Imposing BC using geometry 375 FIGURE 1025 Imposing loads using geometry 375 FIGURE 1026 Contour of the vertical displacement 376 FIGURE 1027 Contour of the vertical stress σyy 376 FIGURE 111 Deformed configuration of a thin plate in bending 380 FIGURE 112 Internal stresses in a thin plate Moments and shear forces due to internal stresses in a thin plate 380 FIGURE 113 Moments and shear forces due to inernal stresses in a thin plate 380 FIGURE 114 Free body diagram of a plate element 382 FIGURE 115 Deformed configuration of a thick plate in bending 383 FIGURE 116 Threenode triangular plate bending element 386 FIGURE 117 Fournode rectangular plate bending element 387 FIGURE 118 Plate boundary conditions 390 FIGURE 119 Simply supported plate on all edges 391 FIGURE 1110 Finite element mesh of one quadrant of the simply supported plate 395 FIGURE 1111 Contour plot of the vertical displacement 399 FIGURE 1112 Contour plot of the moment Mxx 400 FIGURE 1113 Contour plot of the moment Mxy 400 FIGURE 1114 Lifting of corners of a plate 401 FIGURE 1115 Creating the SlabS4R Part 402 FIGURE 1116 Sketching the SlabS4R Part 402 FIGURE 1117 Creating a homogeneous shell section 402 FIGURE 1118 Editing section assignments 403 FIGURE 1119 Mesh controls and element type 403 FIGURE 1120 Mesh 404 FIGURE 1121 Creating a node set 404 FIGURE 1122 Imposing BC EdgeX0 using geometry 404 FIGURE 1123 Imposing BC EdgeZ18 using geometry 405 FIGURE 1124 Imposing BC EdgeZ0 using geometry 405 FIGURE 1125 Imposing BC EdgeX9 using geometry 405 FIGURE 1126 Imposing a concentrated force using a node set 406 FIGURE 1127 Plotting displacements on deformed shape 407 2013 by Taylor Francis Group LLC List of Figures xxiii FIGURE 1128 Castellated beam 407 FIGURE 1129 Base profile 407 FIGURE 1130 Castellated beam profile 408 FIGURE 1131 Geometrical details of the castellated beam 408 FIGURE 1132 Loading and boundary conditions 408 FIGURE 1133 Sketching the I profile 409 FIGURE 1134 Adding dimensions 409 FIGURE 1135 Finishing dimensioning the profile 410 FIGURE 1136 Editing shell extrusion 410 FIGURE 1137 Selecting a plane for an extruded cut 410 FIGURE 1138 Magnify View tool 411 FIGURE 1139 Sketching a hexagon 411 FIGURE 1140 Delete tool 412 FIGURE 1141 Dimension tool 412 FIGURE 1142 Linear pattern tool 413 FIGURE 1143 Editing a linear pattern 413 FIGURE 1144 Edit cut extrusion 414 FIGURE 1145 Creating a shell section 414 FIGURE 1146 Editing section assignments 415 FIGURE 1147 Mesh controls and element type 415 FIGURE 1148 Element type 416 FIGURE 1149 Mesh 416 FIGURE 1150 Imposing BC using geometry 417 FIGURE 1151 Applying a pressure load on a shell surface 417 FIGURE 1152 Contour of the vertical displacement 418 FIGURE 1153 Contour plot of the von Mises stress 418 FIGURE B1 Common beam loadings 445 FIGURE C1 Transformation of coordinates 449 FIGURE C2 Rotation around the third axis 450 2013 by Taylor Francis Group LLC 2013 by Taylor Francis Group LLC List of Tables TABLE 51 Relationships between the Coefficients of Elasticity 161 TABLE 81 Abscissa and Weights for Gauss Quadrature 213 TABLE 82 Abscissae and Weights for a Triangle 218 TABLE 83 Coordinates of Rain Gages and Precipitations 227 xxv 2013 by Taylor Francis Group LLC 2013 by Taylor Francis Group LLC Preface The advent of the digital computer has revolutionized engineering curricula In this day and age the analysis of all but the simplest problem is carried out with the aid of a computer program that not only speeds up calculations but also allows the display of results in fancy graphics For instance when graduate engineers enter the design office they encounter advanced commercial finite element software whose capabilities and the theories behind their development are far more superior to the training they have received during their university studies These packages also come with a graphical user interface GUI Most of the time this is the only component the user will interact with and learning how to use the software is often a matter of trial and error assisted by the documentation that accompanies the software However proficiency in using the GUI is by no means related to the accuracy of the results The latter depends very much on a deep understanding of the mathematics governing the theory So what is to be taught This is the challenge facing experts and educators in engineering Should only the theory be taught with the practical aspects to be picked up later Or on the other hand should the emphasis be on more handson applications using computer software at the expense of theory The many textbooks that describe the theory of the finite element andor its engineering applications fall into one of the following two categories those that deal with the theory assuming that the reader has access to some sort of software and those that deal with the programming aspect assuming that the reader has some theoretical knowledge of the method The theoretical approach is beneficial to students in the long term as it provides them with a deeper understanding of the mathematics behind the development of the finite element method It also helps them prepare for postgraduate studies However it leaves very little time for practical applications and as such it is not favored by employers as they have to provide extra training for graduates in solving reallife problems In addition from my personal experience it is often less attractive to students as it involves a lot of mathematics such as differential equations matrix algebra and advanced calculus Indeed finite element analysis subjects are usually taught in the two later years of the engineering syllabus and at these later stages in their degree most students expect that they have completed their studies in mathematics in the first two years The handson approach on the other hand makes extensive use of the availability of computer facilities Reallife problems are usually used as examples It is very popular with students as it helps them solve problems quickly and efficiently with the results presented in attractive graphics Students become experts at using the pre and postprocessor abilities of the software and usually claim competency with a given computer package which employers look well upon However this approach gives students a false sense of achievement When faced with a novel problem they usually do not know how to choose a suitable model and how to check the accuracy and the validity of the answers In addition modern packages have abilities beyond the student knowledge and experience This is a serious cause for concern In addition given the many available computer software it is also very unlikely that after graduating a student will use the same package on which he or she was trained The aim of this book therefore is to bridge this gap It introduces the theory of the finite element method while keeping a balanced approach between its mathematical formulation programming implementation and as its application using commercial software The computer implementation is carried out using MATLAB while the practical applications are carried out in both MATLAB and Abaqus MATLAB is a highlevel language specially designed for dealing with matrices making it particularly suited for programming the finite element method In addition it also allows the reader to focus on the finite element method by alleviating the programming burden Experience has shown that books that include programming examples that can be implemented are of benefit to beginners This book also includes detailed stepbystep procedures for solving problems with Abaqus interactive and keyword editions Abaqus is one of the leading finite element packages and xxvii 2013 by Taylor Francis Group LLC xxviii Preface has much operational and verification experience to back it up notwithstanding the quality of the pre and postprocessing capabilities Finally if you want to understand the introductory theory of the finite element method to program it in MATLAB andor to get started with Abaqus then this book is for you ABAQUS is a registered trade mark of Dassault Systèmes For product information please contact Web www3dscom MATLAB is a registered trademark of The MathWorks Inc For product information please contact The MathWorks Inc 3 Apple Hill Drive Natick MA 017602098 USA Tel 5086477000 Fax 5086477001 Email infomathworkscom Web wwwmathworkscom 2013 by Taylor Francis Group LLC Author Dr Amar Khennane is a senior lecturer in the School of Engineering and Information Technology at the University of New South Wales Canberra Australian Capital Territory Australia He earned his PhD in civil engineering from the University of Queensland Australia a master of science in structural engineering from Heriot Watt University United Kingdom and a bachelors degree in civil engineering from the University of TiziOuzou Algeria His teaching experience spans 20 years and 2 continents He has taught structural analysis structural mechanics and the finite element method at various universities xxix 2013 by Taylor Francis Group LLC 2013 by Taylor Francis Group LLC 1 Introduction 11 PROLOGUE Undoubtedly the finite element method represents one of the most significant achievements in the field of computational methods in the last century Historically it has its roots in the analysis of weightcritical framed aerospace structures These framed structures were treated as an assemblage of onedimensional members for which the exact solutions to the differential equations for each member were well known These solutions were cast in the form of a matrix relationship between the forces and displacements at the ends of the member Hence the method was initially termed matrix analysis of structures Later it was extended to include the analysis of continuum structures Since continuum structures have complex geometries they had to be subdivided into simple components or elements interconnected at nodes It was at this stage in the development of the method that the term finite element appeared However unlike framed structures closed form solutions to the differential equations governing the behavior of continuum elements were not available Energy prin ciples such as the theorem of virtual work or the principle of minimum potential energy which were well known combined with a piecewise polynomial interpolation of the unknown displacement were used to establish the matrix relationship between the forces and the interpolated displacements at the nodes numerically In the late 1960s when the method was recognized as being equivalent to a minimization process it was reformulated in the form of weighted residuals and variational calculus and expanded to the simulation of nonstructural problems in fluids thermomechanics and electromagnetics More recently the method is extended to cover multiphysics applications where for example it is possible to study the effects of temperature on electromagnetic properties that might affect the performance of electric motors 12 FINITE ELEMENT ANALYSIS AND THE USER Nowadays in structural design the analysis of all but simple structures is carried out using the finite element method When graduate structural engineers enter the design office they will encounter advanced commercial finite element software whose capabilities and the theories behind its devel opment are far superior to the training they have received during their undergraduate studies Indeed current commercial finite element software is capable of simulating nonlinearity whether material or geometrical contact structural interaction with fluids metal forming crash simulations and so on Commercial software also come with advanced pre and postprocessing abilities Most of the time these are the only components the user will interact with and learning how to use them is often a matter of trial and error assisted by the documentation accompanying the software However proficiency in using the pre and postprocessors is by no means related to the accuracy of the results The preprocessor is just a means of facilitating the data input since the finite element method requires a large amount of data while the postprocessor is another means for presenting the results in the form of contour maps The user must realize that the core of the analysis is what happens in between the two processes To achieve proficiency in finite element analysis the user must understand what happens in this essential part often referred to as the black box This will only come after many years of highlevel exposure to the fields that comprise FEA technology differential equations numerical analysis and vector calculus A formal training in numerical procedures and matrix algebra as applied in the finite element method would be helpful to the user particularly if heshe is one of the many design engineers applying finite element techniques in their work without a prior training in numerical procedures 1 2013 by Taylor Francis Group LLC 2 Introduction to Finite Element Analysis Using MATLAB and Abaqus 13 AIM OF THE BOOK The many textbooks that describe the theory of the finite element andor its engineering applications can be split into two categories those that deal with the theory assuming that the reader has access to some sort of software and those that deal with the programming aspect assuming that the reader has some theoretical knowledge of the method The aim of this book is to bridge this gap It introduces the theory of the finite element method while keeping a balanced approach between its mathematical formulation programming implementation and its application using commercial software The key steps are presented in sufficient details The computer implementation is carried out using MATLAB while the practical applications are carried out in both MATLAB and Abaqus MATLAB is a highlevel language specially designed for dealing with matrices This makes it particularly suited for programming the finite element method In addition MATLAB will allow the reader to focus on the finite element method by alleviating the programming burden Experience has shown that books that include programming examples are of benefit to beginners It should be pointed out however that this book is not about writing software to solve a particular problem It is about teaching the first principles of the finite element method If the reader wishes to solve reallife problems heshe will be better off using commercial software such as Abaqus rather than writing hisher own code Homewritten software may have serious bugs that can compromise the results of the analysis while commercial software has much operational and verification experience to back it up notwithstanding the quality of the pre and postprocessing abilities For this purpose detailed stepbystep procedures for solving problems with Abaqus interactive and keyword editions are given in this book Abaqus is a suite of commercial finite element codes It consists of Abaqus Standard which is a general purpose finite element software and Abaqus Explicit for dynamic analysis It is now owned by Dassault Systèms and is part of the SIMULIA range of products httpwwwsimuliacomproductsunifiedfeahtml Data input for a finite element analysis with Abaqus can be done either through AbaqusCAE or CATIA which are intuitive graphic user interfaces They also allow monitoring and viewing of results Data can be entered in or using an input file prepared with a text editor and executed through the command line or using a script prepared with Python Python is an objectoriented programming language and is included in Abaqus as Abaqus Python The latter is an advanced option reserved for experienced users and will not be covered in this book 14 BOOK ORGANIZATION The organization of the book contents follows the historical development of the finite element method After some introductory notes in Chapter 1 Chapters 2 through 4 introduce matrix struc tural analysis for trusses beams and frames The matrix relationships between the forces and nodal displacements for each element type are derived using the direct approaches from structural mechanics Using a truss as an example in Chapter 1 the different steps required in a finite ele ment code such as describing loads supports material and mesh preparation matrix manipulation introduction of boundary condition and equation solving are described succinctly Indeed a truss offers all the attributes necessary to illustrate the coding of a finite element code Similar codes are developed for beams and rigid jointed frames in Chapters 3 and 4 respectively The described procedures are implemented as MATLAB codes at the end of each chapter In addition detailed stepbystep procedures for solving similar problems with both the Abaqus interactive and keyword editions are provided at the end of each chapter Chapter 5 marks the change of philosophy between matrix structural analysis and finite element analysis of a continuum In matrix analysis there is only one dominant stress which is the lon gitudinal stress In a continuum on the other hand there are many stresses and strains at a point Chapter 5 introduces the theories of stress and strain and the relationships between them It also includes many solved problems that would help the reader understand the developed theories 2013 by Taylor Francis Group LLC Introduction 3 Chapters 6 and 7 introduce respectively the weighted residual methods and finite element approximation which include the various types of continuum elements and the different techniques used to construct the piecewise polynomial interpolations of the unknown quantities These methods are necessary to establish the matrix relationships between forces and nodal displacements for continuum elements of complicated geometry and whose behavior is governed by differential equations for which closed form solutions cannot be easily established Chapter 8 is entirely devoted to numerical integration using the Gauss Legendre and Hammer formulae with many examples at the end of the chapter Indeed during the implementation of the finite element method many integrals arise as will be seen in Chapters 9 through 11 When the number of elements is large andor their geometrical shape is general as is the case in most applications the use of analytical integration is quite cumbersome and ill suited for computer coding In Chapter 9 the finite element formulation for plane stressstrain problems is presented The stiffness matrices for the triangular and quadrilateral families of elements are developed in detail enabling the reader to solve a wide variety of problems The chapters also include a wide variety of solved problems with MATLAB and Abaqus Chapter 10 introduces axisymmetric problems while Chapter 11 is devoted to the theory of plates The stiffness matrices for the most common elements are developed in detail and numerous examples are solved at the end of each chapter using both MATLAB and Abaqus The appendices and httpwwwcrcpresscomproductisbn9781466580206 contain all the MATLAB codes used in the examples 2013 by Taylor Francis Group LLC 2013 by Taylor Francis Group LLC 2 Bar Element 21 INTRODUCTION There is no better way of illustrating the steps involved in a finite element analysis than by analyzing a simple truss Indeed a truss is the first structural system introduced into the cursus of engineering studies As early as the first year the student becomes acquainted with a truss in engineering statics A truss offers all the attributes needed to illustrate a finite analysis without the need to resort to advanced mathematical tools such as numerical integration and geometrical transformations that are required in the analysis of complicated structures A truss is a structure that consists of axial members connected by pin joints as shown in Figure 21 The loads on a truss are assumed to be concentrated at the joints The members of a truss support the external load through axial force as they do not undergo bending deformation Therefore no bending moments are present in truss members 22 ONEDIMENSIONAL TRUSS ELEMENT 221 FORMULATION OF THE STIFFNESS MATRIX THE DIRECT APPROACH A member of a truss is the simplest solid element namely an elastic rod with ends 1 and 2 referred to hereafter as nodes Consider an element of length L cross section A and made of a linear elastic material having a Youngs modulus E as represented in Figure 22a If we apply a normal force N1 at node 1 and at the same time maintaining node 2 fixed in space the bar shortens by an amount u1 as represented in Figure 22b The force N1 is related to the displacement u1 through the spring constant N1 AE L u1 21 In virtue of Newtons third law there must be a reaction force R2 at node 2 equal in magnitude and opposite in direction to the force N1 that is R2 AE L u1 22 Similarly if we apply a normal force N2 at node 2 and at the same time maintaining node 1 fixed in space the bar lengthens by an amount u2 as represented in Figure 22c In the same fashion the force N2 is related to the displacement u2 through the spring constant N2 AE L u2 23 Again in virtue of Newtons third law there must be a reaction force R1 at node 1 equal in magnitude and opposite in direction to the force N2 that is R1 AE L u2 24 5 2013 by Taylor Francis Group LLC Introduction to Finite Element Analysis Using MATLAB and Abaqus where the vector u is the vector of nodal displacements the vector F is the vector of nodal forces The matrix K is called the stiffness matrix it relates the nodal displacements to the nodal forces Knowing the forces F1 and F2 one may be tempted to solve the system of Equation 26 to obtain the displacements u1 and u2 This is not possible at least in a unique sense Indeed taking a closer look at the matrix K it can be seen that its determinant is equal to zero that is detK AEL2 AEL2 0 That is any set of displacements u1 and u2 is a solution to the system As odd as it may appear at this stage this actually makes a lot of physical sense In Figure 22d the bar is subject to the forces F1 and F2 Under the action of these forces the bar will experience a rigid body movement since it is not restrained in space There will be many sets of displacements u1 and u2 that are solutions to the system 26 To obtain a unique solution the bar must be restrained in space against rigid body movement The state of restraints of the bar or the structure in general is introduced in the form of boundary conditions This will be covered in detail in Section 24 222 TWODIMENSIONAL TRUSS ELEMENT As shown in Figure 21 a plane truss structure consists of axial members with different orientations A longitudinal force in one member may act at a right angle to another member For example the force F in Figure 21 acts at right angle to member a and therefore causing it to displace in its transversal direction The nodal degrees of freedom nodal displacements of the rod element become four as represented in Figure 23 and they are given as δd u1 v1 u2 v2T The corresponding stiffness matrix becomes K AEL 0 AEL 0 0 0 0 0 AEL 0 AEL 0 0 0 0 Note that the second and fourth columns and rows associated with the transversal displacements are null since the truss member has axial deformation only 8 Introduction to Finite Element Analysis Using MATLAB and Abaqus X X θ θ Y Y y y x x u1 v1 v2 V1 U1 fx1 Fx1 Fx2 fx2 fy1 fy2 Fy1 Fy2 U2 u2 V2 a b FIGURE 24 Truss element oriented at an arbitrary angle θ a Nodal displacements b Nodal forces Another problem that arises from the fact that all truss members do not have the same orientation is that when it comes to assemble the global stiffness we need to have the element degrees of freedom nodal displacements given in terms of the common reference axes of the truss Figure 24 shows a truss element oriented at an arbitrary angle θ with respect to the horizontal axis X Y of the structure It also shows two sets of nodal displacements The first set u v is given in terms of the local set of axis x y associated with the element while the second set of displacements U V is associated with the global set of axis X Y The element stiffness matrix is expressed in terms of the local displacements u and v In order to be assembled with the stiffness matrices of the other elements to form the global stiffness matrix of the whole structure it should be transformed such that it is expressed in terms of the global displacements U and V If we consider node 1 it can be seen that the displacements U1 and V1 can be written in terms of u1 and v1 as U1 u1 cos θ v1 sin θ V1 u1 sin θ v1 cos θ 211 In a similar fashion U2 and V2 can be expressed in terms of u2 and v2 as U2 u2 cos θ v2 sin θ V2 u2 sin θ v2 cos θ 212 Grouping Equations 211 and 212 yields U1 V1 U2 V2 cos θ sin θ 0 0 sin θ cos θ 0 0 0 0 cos θ sin θ 0 0 sin θ cos θ u1 v1 u2 v2 213 or in a more compact form as de Cde 214 2013 by Taylor Francis Group LLC Bar Element 9 The matrix C is called the transformation matrix It is an orthonormal matrix with a determinant equal to one Its inverse is simply equal to its transpose that is C1 CT 215 The vector of the global nodal forces fe Fx1 Fy1 Fx2 Fy2T may be also obtained from the vector of local nodal forces fe fx1 fy1 fx2 fy2T as fe Cfe 216 In the local coordinate system the forcedisplacement relation is given as Ke de fe 217 Using de CTde and fe CTfe and substituting in 217 yields KeCTde CTfe 218 Premultiplying both sides by C yields CKeCTde fe 219 which can be rewritten as Kede fe 220 with Ke CKeCT 221 The matrix Ke is called the element stiffness matrix in the global coordinate system it relates the global nodal displacements to the global nodal forces 23 GLOBAL STIFFNESS MATRIX ASSEMBLY 231 DISCRETIZATION To illustrate how the elements stiffness matrices are put together to form the global stiffness matrix we proceed with a very simple example Consider the truss represented in Figure 25 First we number all the elements and the nodes as well as identifying the nodal degrees of freedom global displacement as shown in Figure 25 In total there are three nodes three elements and six degrees of freedom U1 V1 U2 V2 U3 V3 232 ELEMENTS STIFFNESS MATRICES IN LOCAL COORDINATES Referring to Equation 210 it can be seen that the element stiffness matrix is a function of the material properties through the elastic modulus E the crosssectional area A of the element and its length L The elastic modulus refers to the material used to build the truss If we assume that all the members of the truss are made of steel with an elastic modulus of 200000 MPa and all the elements have the same crosssectional area say 2300 mm2 then it is possible to evaluate each element stiffness matrix 2013 by Taylor Francis Group LLC 10 Introduction to Finite Element Analysis Using MATLAB and Abaqus 3 V3 U3 V2 V1 Y x x y y y x 1 2 3 2 X U1 1 4 m 12 kN 6 m U2 FIGURE 25 Model of a truss structure Element 1 has a length of 4000 mm Substituting in Equation 210 its stiffness matrix in its local coordinates is obtained as K1L 115000 0 115000 0 0 0 0 0 115000 0 115000 0 0 0 0 0 222 Element 2 has a length of 6000 mm Its stiffness matrix in its local coordinates is obtained as K2L 7666667 0 7666667 0 0 0 0 0 7666667 0 7666667 0 0 0 0 0 223 Element 3 has a length of 7211 mm which can be calculated with the wellknown Pythagoras formula Its stiffness matrix in its local coordinates is obtained as K3L 6379143 0 6379143 0 0 0 0 0 6379143 0 6379143 0 0 0 0 0 224 233 ELEMENTS STIFFNESS MATRICES IN GLOBAL COORDINATES The elements stiffness matrices as respectively given by Equations 222 through 224 cannot be assembled into the global stiffness matrix of the truss because they are formulated in their respective 2013 by Taylor Francis Group LLC Bar Element 11 local coordinate systems In order to do so they need to be transformed from their local coordinate systems x y to the global coordinate system X Y 2331 Element 1 The local axis x of element 1 makes an angle of 0 with the global X axis of the structure In virtue of Equation 213 its transformation matrix C is given as C1 cos0 sin0 0 0 sin0 cos0 0 0 0 0 cos0 sin0 0 0 sin0 cos0 1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1 225 The transformation matrix C1 of element 1 is an identity matrix Therefore as per Equation 219 premultiplying the matrix K1L by C1 and postmultiplying it by C1T that is C1K1LC1T would not change anything the reason being that the local axes of element 1 are colinear with global axes X Y of the structure Therefore the stiffness matrix of element 1 K1G in the global coordinates system remains unchanged that is K1G U1u1 V1v1 U2u2 V2v2 U1u1 115000 0 115000 0 V1v1 0 0 0 0 U2u2 115000 0 115000 0 V2v2 0 0 0 0 226 In its local coordinates x y it has the degrees of freedom u1 v1 u2 v2 while in the global coordinates as shown in Figure 25 it has the global degrees of freedom U1 V1 U2 V2 The top row and the left column outside the matrix show the correspondence between the local and the global degrees of freedom 2332 Element 2 The local axis x of element 2 makes an angle of 90 with the global X axis of the structure In virtue of Equation 213 its transformation matrix C is given as C2 0 1 0 0 1 0 0 0 0 0 0 1 0 0 1 0 227 Premultiplying the matrix K2L by C2 and postmultiplying it by C2T yields the stiffness matrix K2G C2K2LC2T of element 2 in the global system of axes K2G U2u1 V2v1 U3u2 V3v2 U2u1 0 0 0 0 V2v1 7666667 0 7666667 0 U3u2 0 0 0 0 V3v2 7666667 0 7666667 0 228 2013 by Taylor Francis Group LLC 12 Introduction to Finite Element Analysis Using MATLAB and Abaqus In its local coordinates x y it has the degrees of freedom u1 v1 u2 v2 while in the global coordinates as shown in Figure 25 it has the global degrees of freedom U2 V2 U3 V3 2333 Element 3 The local axis x of element 3 makes an angle of θ tan164 5631 with the global X axis of the structure Using Equation 213 its transformation matrix C3 is given as C3 0554699 0832051 0 0 0832051 0554699 0 0 0 0 0554699 0832051 0 0 0832051 0554699 229 Premultiplying the matrix K3L by C3 and postmultiplying it by C3T yields the stiffness matrix K3G C3K3LC3T of element 3 in the global system of axes K3G U1u1 V1v1 U3u2 V3v2 U1u1 19628 29442 19628 29442 V1v1 29442 44163 29442 44163 U3u2 19628 29442 19628 29442 V3v2 29442 44163 29442 44163 230 In its local coordinates x y it has the degrees of freedom u1 v1 u2 v2 while in the global coordinates as shown in Figure 25 it has the global degrees of freedom U1 V1 U3 V3 234 GLOBAL MATRIX ASSEMBLY As shown in Figure 25 the truss has six degrees of freedom U1 V1 U2 V2 U3 V3 that is two degrees of freedom per node Its stiffness matrix must therefore have six lines and six columns each corresponding to a degree of freedom K U1 V1 U2 V2 U3 V3 U1 0 0 0 0 0 0 V1 0 0 0 0 0 0 U2 0 0 0 0 0 0 V2 0 0 0 0 0 0 U3 0 0 0 0 0 0 V3 0 0 0 0 0 0 231 2013 by Taylor Francis Group LLC Bar Element 13 To populate the global stiffness matrix imagine three hypothetical states First only element 1 is present Second only element 2 is present Third only element 3 is present 2341 Only Element 1 Is Present K U1 V1 U2 V2 U3 V3 U1 115000 0 115000 0 0 0 V1 0 0 0 0 0 0 U2 115000 0 115000 0 0 0 V2 0 0 0 0 0 0 U3 0 0 0 0 0 0 V3 0 0 0 0 0 0 232 Notice that only the cases corresponding to the global degrees of freedom of element 1 are populated 2342 Only Element 2 Is Present K U1 V1 U2 V2 U3 V3 U1 0 0 0 0 0 0 V1 0 0 0 0 0 0 U2 0 0 0 0 0 0 V2 0 0 0 7666667 0 7666667 U3 0 0 0 0 0 0 V3 0 0 0 7666667 0 7666667 233 2343 Only Element 3 Is Present K U1 V1 U2 V2 U3 V3 U1 19628 29442 0 0 19628 29442 V1 29442 44163 0 0 29442 44163 U2 0 0 0 0 0 0 V2 0 0 0 0 0 0 U3 19628 29442 0 0 19628 29442 V3 29442 44163 0 0 29442 44163 234 2013 by Taylor Francis Group LLC 14 Introduction to Finite Element Analysis Using MATLAB and Abaqus By direct addition of the preceding matrices the global structure stiffness matrix is obtained as K U1 V1 U2 V2 U3 V3 U1 115000 19628 29442 115000 0 19628 29442 V1 29442 44163 0 0 29442 44163 U2 115000 0 115000 0 0 0 V2 0 0 0 7666667 0 7666667 U3 19628 29442 0 0 19628 29442 V3 29442 44163 0 7666667 29442 44163 7666667 235 235 GLOBAL FORCE VECTOR ASSEMBLY Figure 26 shows a free body diagram where all the external forces acting on the truss are represented At node 1 which is pinned there are two reaction forces RX1 and RY1 At node 2 which is a roller support there is one reaction force RY2 Node 3 is free but there is an external force of 12000 N acting in the positive xdirection The external forces can be grouped in the global force vector as F RX1 RY1 0 RY2 12000 0 236 Y 1 1 4 m 2 2 3 3 12 kN 6 m X RX1 RY1 RY2 FIGURE 26 Free body diagram of the truss 2013 by Taylor Francis Group LLC Bar Element 15 24 BOUNDARY CONDITIONS 241 GENERAL CASE Once the global stiffness matrix and the global force vector are assembled the equilibrium equations of the truss are written as follows 134628 29442 115000 0 19628 29442 29442 44163 0 0 29442 44163 115000 0 115000 0 0 0 0 0 0 7666667 0 7666667 19628 29442 0 0 19628 29442 29442 44163 0 7666667 29442 12082967 U1 V1 U2 V2 U3 V3 RX1 RY1 0 RY2 12000 0 or in more compact form as Kδ F 237 As given by Equation 237 the system of equations cannot be solved in a unique fashion since the matrix K is singular Indeed it is assembled from the elements stiffness matrices which are singular In addition the righthandside vector contains the unknown support reactions To solve the system of equations it is necessary to partition the matrix K according to known and unknown quantities The vector of displacements δ can be partitioned into known and unknown quantities Node 1 is a pinned support therefore the displacements U1 and V1 are both equal to zero Node 2 is a roller support therefore the displacement V2 is also equal to zero It follows therefore that the vector δ can be partitioned as follows δ U1 0 V1 0 V2 0 U2 U3 V3 238 Similarly the righthandside vector of global forces can be partitioned accordingly F RX1 RY1 RY2 0 12000 0 239 Note that unknown displacements correspond to known forces and known displacements correspond to unknown forces 2013 by Taylor Francis Group LLC 16 Introduction to Finite Element Analysis Using MATLAB and Abaqus Finally the matrix K is partitioned as 134628 29442 0 115000 19628 29442 29442 44163 0 0 29442 44163 0 0 7666667 0 0 7666667 115000 0 0 115000 0 0 19628 29442 0 0 19628 29442 29442 44163 7666667 0 29442 12082967 U1 0 V1 0 V2 0 U2 U3 V3 RX1 RY1 RY2 0 12000 0 As a result of the position of V2 being interchanged with that of U2 in the vector δ column 3 and line 3 have also been respectively interchanged with column 4 and line 4 in the matrix K Finally the partitioned system of equations can be rewritten in a compact form as KPP KPF KFP KFF δP δF FP FF 240 where The subscripts P and F refer respectively to the prescribed and free degrees of freedom δPT 0 0 0 the vector of the known prescribed displacements δFT U2 U3 V3 the vector of the unknown free displacements FPT RX1 RY1 RY2 the vector of the unknown reaction forces corresponding to the prescribed displacements FFT 0 12000 0 the vector of the known applied external forces 25 SOLUTION OF THE SYSTEM OF EQUATIONS Equation 240 can be expanded to yield KPP δP KPF δF FP 241 KFP δP KFF δF FF 242 Since δP and FF are known quantities it is then possible to obtain from Equation 242 the vector δP as δF KFF1 FF KFP δP 243 2013 by Taylor Francis Group LLC However since δpT 0 0 0 Equation 243 reduces to δF KFF1 F which is simply equivalent to eliminating the lines and the columns corresponding to the restrained degrees of freedom in the global matrix that is U2 U3 115000 0 0 0 19628 29442 0 29442 12082967 1 0 12000 Solving the system of equations yields δF U2 U3 0 09635 02348 mm In summary the vector of global displacements can be obtained as δ U1 0 V1 0 U2 0 V2 0 U3 09635 V3 02348 26 SUPPORT REACTIONS Once δF is known it is possible to obtain from Equation 241 the vector of the unknown reaction forces F R1 RY1 RY2 Since δpT 0 0 0 the vector F is obtained as F KFF δF That is R1 RY1 RY2 115000 19628 29442 0 29442 44163 0 0 7666667 0 12 18 kN Considering vertical equilibrium yields ΣY RY2 RY1 0 RY1 18 kN Considering horizontal equilibrium yields ΣX 12 R1 0 R1 12 kN 18 Introduction to Finite Element Analysis Using MATLAB and Abaqus 27 MEMBERS FORCES Once all the displacements are known the member forces can be easily obtained For example ele ment 3 has the following vector of global displacements d3 extracted from the global displacements vector δ Equation 245 d3 U1 0 V1 0 U3 09635 V3 02348 The vector of displacements in local coordinates d3 is obtained using the inverse transformation d3 C3Td3 that is d3 0554699 0832051 0 0 0832051 0554699 0 0 0 0 0554699 0832051 0 0 0832051 0554699 0 0 09635 02348 0 0 03391 09319 Multiplying the local stiffness matrix of element 3 K3L by the local displacement vector d3 yields the local vector of forces f3 that is f3 6379143 0 6379143 0 0 0 0 0 6379143 0 6379143 0 0 0 0 0 0 0 03391 09319 21631 0 21631 0 kN The forces on the bar element are represented graphically in Figure 27 It can be seen that the member is under a tensile force of 21631 kN This result can be checked using the method of joints Y 3 3 x y 1 4 m 6 m X 21631 kN 21631 kN FIGURE 27 Free body diagram of element 3 2013 by Taylor Francis Group LLC FIGURE 28 Equilibrium of node 3 Consider the free body diagram of node joint 3 as shown in Figure 28 The equilibrium of the joint in the x direction requires ΣX 12 F3 sin3369 0 F3 21633 kN This confirms the obtained result with the finite element method Remark The preceding sections illustrate the steps required in a finite element analysis As can be noticed even for a small number of elements in this case 3 the calculations are rather involved For very large structures with a large number of elements the calculation effort is so intensive that it is virtually impossible to carry out without the help of a digital computer However it can be also noticed that the calculations involve matrix algebra and the steps are quite repetitive which makes them ideally suited for programming on a digital computer 20 Introduction to Finite Element Analysis Using MATLAB and Abaqus These steps are best illustrated by means of an example Let us consider the truss represented in Figure 25 The main program is given in the Mfile trussm 281 DATA PREPARATION Since the basic building block in MATLAB is a matrix the data will be prepared in the form of tables whenever possible as they are very easily translated into matrices Although there are many ways of reading data in MATLAB in what follows we will use an Mfile truss1datam to read the data relevant to the truss Note that a consistent set of units is required in any finite element analysis In this case mm are used for length and N for forces The input data for this structure consist of nnd 3 number of nodes nel 3 number of elements nne 2 number of nodes per element nodof 2 number of degrees of freedom per node 2811 Nodes Coordinates The coordinates x and y of the nodes are given in the form of a matrix geomnnd 2 geom 0 0 4000 0 4000 6000 2812 Element Connectivity The table of connectivity describes how the elements are connected to each other The nodal coordinates are given in the matrix connecnel 2 connec 1 2 2 3 1 3 2813 Material and Geometrical Properties The material and geometrical properties are given in the matrix propnel 2 prop 200000 2300 200000 2300 200000 2300 2814 Boundary Conditions Boundary conditions give information on how the structure is restrained in space against any rigid body movement Without the introduction of boundary conditions the global stiffness matrix is singular To solve the equilibrium equations we need to know how the nodes are restrained in space In what follows we adopt the following convention A restrained degree of freedom is assigned the digit 0 A free degree of freedom is assigned the digit 1 The information on the boundary conditions is given in the matrix nfnnd nodof This matrix is called the matrix of nodal freedom matrix It is first initialized to 1 then the degrees of freedom are read 2013 by Taylor Francis Group LLC Bar Element 21 nf 0 0 1 0 1 1 The free degrees of freedom different from zero are then counted and their rank assigned back into the matrix nfnnd nodof nf 0 0 1 0 2 3 In this case the total number of active degrees of freedom is obtained as n 3 At this stage it is possible to initialize the global matrix KKn n 0 and the global force vector Fn 0 KK 0 0 0 0 0 0 0 0 0 and F 0 0 0 Note that we will only assemble the quantities corresponding to the active degrees of freedom that is the lines and the columns in the matrix KK corresponding respectively to the active degrees of freedom 1 2 and 3 As to the restrained degrees of freedom with a number equal to 0 they will be simply eliminated 2815 Loading Finally to be able to solve for the unknown displacements we need to know how the structure is loaded The information about the loading is also given in the form of a matrix loadnnd 2 load 0 0 0 0 1200 0 The data are stored in the Mfile truss1datam At this level in the main program trussm the model data are written to the file truss1resultstxt using the Mfile printtrussmodelm This is not necessary however it is always helpful to write the data to a file because it is easier to check for errors 282 ELEMENT MATRICES 2821 Stiffness Matrix in Local Coordinates For each element from 1 to nel we set up the local stiffness and transformation matrices Once the stiffness matrix kl is set up in local coordinates it is transformed into global coordinates kg through the transformation matrix C and then assembled to the global stiffness matrix KK For any element i we retrieve its first and second node from the connectivity matrix node1 conneci 1 node2 conneci 2 Then using the values of the nodes we retrieve their x and y coordinates from the geometry matrix x1 geomnode1 1 y1 geomnode1 2 x2 geomnode2 1 y2 geomnode2 2 2013 by Taylor Francis Group LLC Introduction to Finite Element Analysis Using MATLAB and Abaqus 22 Next using Pythagoras theorem we evaluate the length of the element L x2 x12 y2 y12 Finally we retrieve the material and geometrical property of the section E propi 1 A propi 2 before evaluating the matrix kl using Equation 210 The MATLAB script for evaluating the matrix kl is given in Appendix A in the Mfile trussklm 2822 Transformation Matrix Once the nodal coordinates are retrieved it is also possible to evaluate the angle θ that the element makes with the global X axis θ tan1 y2 y1x2 x1 However care should be taken when the element is at right angle with the global axis X as x2x1 0 The matrix C is evaluated using Equation 225 The MATLAB script is given in Appendix A in the Mfile trussCm 2823 Stiffness Matrix in Global Coordinates The element stiffness matrix kg is obtained using Equation 221 kg C kl CT 2824 Steering Vector Once the matrix kg is formed we retrieve the steering vector g containing the number of degrees of freedom of the nodes of the element g nfnode1 1 nfnode1 2 nfnode2 1 nfnode2 2 For example for element 1 the vector g will look like g 0 0 1 0 The only nonzero component in the vector g is located in the third position and its value is equal to 1 That is only the element corresponding to the third line and third column in the matrix kg33 will be assembled and it will occupy the position KK1 1 in the global matrix The MATLAB script for constructing the steering vector g is given in Appendix A in the Mfile trussgm Bar Element 23 283 ASSEMBLY OF THE GLOBAL STIFFNESS MATRIX The global stiffness matrix KK is assembled using a double loop over the components of the vector g loop i for any gi 0 loop j for any gj 0 add kgi j to KKgi gj end loop j end loop i The script is given in Appendix A in the Mfile formKKm 284 ASSEMBLY OF THE GLOBAL FORCE VECTOR A loop is carried over all the nodes If a degree of freedom j of a node i is free that is nfi j 0 then it is susceptible of carrying an external force Fnfi j loadi j The global force vector is formed in Appendix A in the Mfile formtrussFm 285 SOLUTION OF THE GLOBAL SYSTEM OF EQUATIONS In MATLAB it is very easy to solve a system of linear equations one statement does it all In this case the global displacements vector delta is obtained as delta KKF The backslash symbol is used to divide a matrix by a vector 286 NODAL DISPLACEMENTS Once the global displacements vector delta is obtained it is possible to retrieve any nodal displace ments A loop is carried over all the nodes If a degree of freedom j of a node i is free that is nfi j 0 then it could have a displacement different from zero The value of the displacement is extracted from the global displacements vector delta nodedispi j deltanfi j 287 ELEMENT FORCES To obtain the member forces a loop is carried over all the elements 1 Form element stiffness matrix kl in local xy 2 Form element transformation matrix C 3 Transform the element matrix from local to global coordinates kg C kl CT 4 Form element steering vector g a Loop over the degrees of freedom of the element to obtain element displacements vector edg in global coordinates b If gj 0 then the degree of freedom is restrained edgj 0 c Otherwise edgj deltagj 2013 by Taylor Francis Group LLC 24 Introduction to Finite Element Analysis Using MATLAB and Abaqus 5 Obtain element force vector in global XY coordinates fg kg edg 6 Transform element force vector to local coordinates fl CT fg 7 For each element store the third component of fl If the component is positive the element is under tension otherwise it is under compression The results of the analysis are written to the file trussresultstxt using the Mfile print1resultsm given in Appendix A A copy of the file trussresultstxt is included within Section 288 288 PROGRAM SCRIPTS Filetrussm trussm LINEAR STATIC ANALYSIS OF A TRUSS STRUCTURE clc Clear screen clear Clear all variables in memory Make these variables global so they can be shared by other functions global nnd nel nne nodof eldof n global geom connec prop nf load dispExecuting trussm ALTER THE NEXT LINES TO CHOOSE AN OUTPUT FILE FOR THE RESULTS Open file for output of results fid fopentruss1resultstxtw dispResults printed in file truss1resultstxt ALTER THE NEXT LINE TO CHOOSE AN INPUT FILE truss1data Load the input file printtrussmodel Print model data KK zerosn Initialize global stiffness matrix to zero Fzerosn1 Initialize global force vector to zero for i1nel kltrusskli Form element matrix in local xy C trussCi Form transformation matrix kgCklC Transform the element matrix from local to global coordinates gtrussgi Retrieve the element steering vector KK formKKKK kg g assemble global stiffness matrix end 2013 by Taylor Francis Group LLC Bar Element 25 F formtrussFF Form global force vector End of assembly delta KKF solve for unknown displacements Extract nodal displacements for i1nnd for j1nodof nodedispij 0 if nfij 0 nodedispij deltanfij end end end Calculate the forces acting on each element in local coordinates and store them in the vector force for i1nel kltrusskli Form element matrix in local xy C trussCi Form transformation matrix kgCklC Transform the element matrix from local to global coordinates gtrussgi Retrieve the element steering vector for j1eldof if gj 0 edgj0 displacement 0 for restrained freedom else edgj deltagj end end fg kgedg Element force vector in global XY flCfg Element force vector in local xy forcei fl3 end printtrussresults fclosefid Filetruss1datam File truss1datam The following variables are declared as global in order to be used by all the functions Mfiles constituting the program global nnd nel nne nodof eldof n global geom connec prop nf load format short e Beginning of data input nnd 3 Number of nodes nel 3 Number of elements nne 2 Number of nodes per element nodof 2 Number of degrees of freedom per node eldof nnenodof Number of degrees of freedom per element 2013 by Taylor Francis Group LLC 26 Introduction to Finite Element Analysis Using MATLAB and Abaqus Nodes coordinates X and Y geomzerosnnd2 geom 0 0 X and Y coord node 1 4000 0 X and Y coord node 2 4000 6000 X and X coord node 3 Element connectivity conneczerosnel2 connec 1 2 1st and 2nd node of element 1 2 3 1st and 2nd node of element 2 1 3 1st and 2nd node of element 3 Geometrical properties prop11 E prop12 A propzerosnel2 prop 200000 2300 E and A of element 1 200000 2300 E and A of element 2 200000 2300 E and A of element 3 Boundary conditions nf onesnnd nodof Initialize the matrix nf to 1 nf11 0 nf12 0 Prescribed nodal freedom of node 1 nf22 0 Prescribed nodal freedom of node 3 Counting of the free degrees of freedom n0 for i1nnd for j1nodof if nfij 0 nn1 nfijn end end end loading load zerosnnd 2 load31200 0 forces in X and Y directions at node 3 End of input Filetruss1resultstxt PRINTING MODEL DATA Number of nodes 3 Number of elements 3 Number of nodes per element 2 Number of degrees of freedom per node 2 Number of degrees of freedom per element 4 Node X Y 1 000000 000000 2 400000 000000 3 400000 600000 Element Node1 Node2 1 1 2 2013 by Taylor Francis Group LLC Bar Element 27 2 2 3 3 1 3 Element E A 1 200000 2300 2 200000 2300 3 200000 2300 Node dispU dispV 1 0 0 2 1 0 3 2 3 Node loadX loadY 1 000000 000000 2 000000 000000 3 120000 000000 Total number of active degrees of freedom n 3 PRINTING ANALYSIS RESULTS Global force vector F 0 1200 0 Displacement solution vector delta 000000 009635 002348 Nodal displacements Node dispX dispY 1 000000 000000 2 000000 000000 3 009635 002348 Members actions element force action 1 000 Compression 2 180000 Compression 3 216333 Tension 29 PROBLEMS Prepare a data file for the trusses shown next and carry out the analysis using the code trussm 291 PROBLEM 21 FIGURE 29 Input file File trussproblem1datam The following variables are declared as global in order 2013 by Taylor Francis Group LLC 28 Introduction to Finite Element Analysis Using MATLAB and Abaqus 4 4 8 6 12 8 2 m 2 1 1 2 3 5 7 9 9 11 13 15 14 10 5 7 6 3 32m 42m Horizontal members Diagonal members E30106 kNm2 A 0045 m2 E 30 106 kNm2 A 002 m2 7 kN 15 kN 5 kN 10 kN FIGURE 29 Model of Problem 21 to be used by all the functions Mfiles constituting the program global nnd nel nne nodof eldof n global geom connec prop nf load format short e Beginning of data input nnd 9 Number of nodes nel 15 Number of elements nne 2 Number of nodes per element nodof 2 Number of degrees of freedom per node eldof nnenodof Number of degrees of freedom per element Nodes coordinates X and Y geomzerosnnd2 geom 0 0 X and Y coord node 1 1 2 X and Y coord node 2 2 0 X and Y coord node 3 3 2 X and Y coord node 4 4 0 X and Y coord node 5 5 2 X and Y coord node 6 6 0 X and Y coord node 7 7 2 X and Y coord node 8 8 0 X and Y coord node 9 Element connectivity conneczerosnel2 connec 1 2 1st and 2nd node of element 1 1 3 1st and 2nd node of element 2 2 3 1st and 2nd node of element 3 2 4 1st and 2nd node of element 4 3 4 1st and 2nd node of element 5 3 5 1st and 2nd node of element 6 4 5 1st and 2nd node of element 7 4 6 1st and 2nd node of element 8 2013 by Taylor Francis Group LLC Bar Element 29 5 6 1st and 2nd node of element 9 5 7 1st and 2nd node of element 10 6 7 1st and 2nd node of element 11 6 8 1st and 2nd node of element 12 7 8 1st and 2nd node of element 13 7 9 1st and 2nd node of element 14 8 9 1st and 2nd node of element 15 Geometrical properties prop11 E prop12 A propzerosnel2 prop 30e6 002 E and A of element 1 30e6 0045 E and A of element 2 30e6 002 E and A of element 3 30e6 0045 E and A of element 4 30e6 002 E and A of element 5 30e6 0045 E and A of element 6 30e6 002 E and A of element 7 30e6 0045 E and A of element 8 30e6 002 E and A of element 9 30e6 0045 E and A of element 10 30e6 002 E and A of element 11 30e6 0045 E and A of element 12 30e6 002 E and A of element 13 30e6 0045 E and A of element 14 30e6 002 E and A of element 15 Boundary conditions nf onesnnd nodof Initialize the matrix nf to 1 nf11 0 nf12 0 Prescribed nodal freedom of node 1 nf92 0 Prescribed nodal freedom of node 3 Counting of the free degrees of freedom n0 for i1nnd for j1nodof if nfij 0 nn1 nfijn end end end loading load zerosnnd 2 load215 0 forces in X and Y directions at node 2 load30 5 forces in X and Y directions at node 3 load40 7 forces in X and Y directions at node 4 load70 10 forces in X and Y directions at node 7 End of input Results file PRINTING MODEL DATA Number of nodes 9 Number of elements 15 Number of nodes per element 2 2013 by Taylor Francis Group LLC 30 Introduction to Finite Element Analysis Using MATLAB and Abaqus Number of degrees of freedom per node 2 Number of degrees of freedom per element 4 Node X Y 1 000000 000000 2 000100 000200 3 000200 000000 4 000300 000200 5 000400 000000 6 000500 000200 7 000600 000000 8 000700 000200 9 000800 000000 Element Node1 Node2 1 1 2 2 1 3 3 2 3 4 2 4 5 3 4 6 3 5 7 4 5 8 4 6 9 5 6 10 5 7 11 6 7 12 6 8 13 7 8 14 7 9 15 8 9 Element E A 1 3e007 002 2 3e007 0045 3 3e007 002 4 3e007 0045 5 3e007 002 6 3e007 0045 7 3e007 002 8 3e007 0045 9 3e007 002 10 3e007 0045 11 3e007 002 12 3e007 0045 13 3e007 002 14 3e007 0045 15 3e007 002 Node dispU dispV 1 0 0 2 1 2 3 3 4 4 5 6 5 7 8 6 9 10 7 11 12 8 13 14 9 15 0 Node loadX loadY 1 000000 000000 2 001500 000000 2013 by Taylor Francis Group LLC Bar Element 31 3 000000 00500 4 000000 00700 5 000000 000000 6 000000 000000 7 000000 01000 8 000000 000000 9 000000 000000 Total number of active degrees of freedom n 15 PRINTING ANALYSIS RESULTS Global force vector F 15 0 0 5 0 7 0 0 0 0 0 10 0 0 0 Displacement solution vector delta 000014 000010 000003 000019 000010 000023 000006 000023 000007 000021 000009 000018 000005 000009 000010 Nodal displacements Node dispX dispY 1 000000 000000 2 000014 000010 3 000003 000019 4 000010 000023 5 000006 000023 6 000007 000021 7 000009 000018 8 000005 000009 9 000010 000000 2013 by Taylor Francis Group LLC 32 Introduction to Finite Element Analysis Using MATLAB and Abaqus Members actions element force action 1 769 Compression 2 1844 Tension 3 769 Tension 4 2187 Compression 5 210 Compression 6 2281 Tension 7 573 Compression 8 2025 Compression 9 573 Tension 10 1769 Tension 11 573 Compression 12 1512 Compression 13 1691 Tension 14 756 Tension 15 1691 Compression 292 PROBLEM 22 FIGURE 210 Input file File trussproblem2datam The following variables are declared as global in order to be used by all the functions Mfiles constituting the program global nnd nel nne nodof eldof n global geom connec prop nf load format short e 75 kN 3 6 5 5 7 9 18 m 18 m 18 m 24 m 24 m 24 m 4 10 kN E 30000 MPa A 20000 mm2 10 kN 4 2 2 3 1 1 8 6 FIGURE 210 Model of Problem 22 2013 by Taylor Francis Group LLC Bar Element 33 Beginning of data input nnd 6 Number of nodes nel 9 Number of elements nne 2 Number of nodes per element nodof 2 Number of degrees of freedom per node eldof nnenodof Number of degrees of freedom per element Nodes coordinates X and Y geomzerosnnd2 geom 0 0 X and Y coord node 1 2400 1800 X and Y coord node 2 2400 5400 X and Y coord node 3 4800 3600 X and Y coord node 4 4800 5400 X and Y coord node 5 7200 5400 X and Y coord node 6 Element connectivity conneczerosnel2 connec 1 2 1st and 2nd node of element 1 1 3 1st and 2nd node of element 2 2 3 1st and 2nd node of element 3 2 4 1st and 2nd node of element 4 3 4 1st and 2nd node of element 5 3 5 1st and 2nd node of element 6 4 5 1st and 2nd node of element 7 5 6 1st and 2nd node of element 8 4 6 1st and 2nd node of element 9 Geometrical properties prop11 E prop12 A propzerosnel2 prop 30000 20000 E and A of element 1 30000 20000 E and A of element 2 30000 20000 E and A of element 3 30000 20000 E and A of element 4 30000 20000 E and A of element 5 30000 20000 E and A of element 6 30000 20000 E and A of element 7 30000 20000 E and A of element 8 30000 20000 E and A of element 9 Boundary conditions nf onesnnd nodof Initialize the matrix nf to 1 nf12 0 Prescribed nodal freedom of node 1 nf61 0 nf62 0 Prescribed nodal freedom of node 6 Counting of the free degrees of freedom n0 for i1nnd for j1nodof if nfij 0 nn1 nfijn end end end loading load zerosnnd 2 load20 10000 forces in X and Y directions at node 2 2013 by Taylor Francis Group LLC 34 Introduction to Finite Element Analysis Using MATLAB and Abaqus load37500 0 forces in X and Y directions at node 3 load40 10000 forces in X and Y directions at node 4 End of input Results file PRINTING MODEL DATA Number of nodes 6 Number of elements 9 Number of nodes per element 2 Number of degrees of freedom per node 2 Number of degrees of freedom per element 4 Node X Y 1 000000 000000 2 240000 180000 3 240000 540000 4 480000 460000 5 480000 540000 6 720000 540000 Element Node1 Node2 1 1 2 2 1 3 3 2 3 4 2 4 5 3 4 6 3 5 7 4 5 8 5 6 9 4 6 Element E A 1 30000 20000 2 30000 20000 3 30000 20000 4 30000 20000 5 30000 20000 6 30000 20000 7 30000 20000 8 30000 20000 9 30000 20000 Node dispU dispV 1 1 0 2 2 3 3 4 5 4 6 7 5 8 9 6 0 0 Node loadX loadY 1 000000 000000 2 000000 1000000 3 750000 000000 4 000000 1000000 2013 by Taylor Francis Group LLC Bar Element 35 5 000000 000000 6 000000 000000 Total number of active degrees of freedom n 9 PRINTING ANALYSIS RESULTS Global force vector F 0 0 10000 7500 0 0 10000 0 0 Displacement solution vector delta 080865 026183 065965 018000 061631 017710 095294 009000 095294 Nodal displacements Node dispX dispY 1 080865 000000 2 026183 065965 3 018000 061631 4 017710 095294 5 009000 095294 6 000000 000000 Members actions element force action 1 833333 Tension 2 1641476 Compression 3 722222 Tension 4 1024394 Tension 5 2459549 Tension 6 2250000 Compression 7 000 Compression 8 2250000 Compression 9 3162278 Tension 210 ANALYSIS OF A SIMPLE TRUSS WITH ABAQUS 2101 OVERVIEW OF ABAQUS Abaqus is a suite of commercial finite element software It consists of Abaqus Standard which is a general purpose finite element software and Abaqus Explicit for dynamic analy sis It is now owned by Dassault Systèms and is part of the SIMULIA range of products httpwwwsimuliacomproductsunifiedfeahtml 2013 by Taylor Francis Group LLC 36 Introduction to Finite Element Analysis Using MATLAB and Abaqus Modeling and Visualization Reference Programming Interfaces Version 68 Update Information Abaqus Release Notes Abaqus Keywords Reference Manual Abaqus Theory Manual Abaqus Verification Manual Abaqus User Subroutines Reference Manual Abaqus Glossary Manual Abaqus Scripting Users Manual Abaqus Scripting Reference Manual Abaqus GUI Toolkit Users Manual Abaqus GUI Toolkit Reference Manual Abaqus Interface for MSCADAMS Users Manual Abaqus Interface for Moldflow Users Manual Analysis Examples Tutorials Information Installation and Licensing Abaqus Installation and Licensing Guide Using Abaqus Online Documentation AbaqusCAE Users Manual Abaqus Analysis Users Manual Abaqus Example Problems Manual Abaqus Benchmarks Manual Getting Started with Abaqus Interactive Edition Getting Started with Abaqus Keywords Edition FIGURE 211 Abaqus documentation Data input for a finite element analysis with Abaqus can be done either through AbaqusCAE which is an intuitive graphic user interface that also allows monitoring and viewing the results or through an input file prepared with a text editor and executed through the command line or finally using a script prepared with Python which is an objectoriented programming language Python is included in Abaqus as Abaqus Python The latter is an advanced option reserved for experienced users and will not be covered in this book Note that Python is free to use even for commercial products because of its OSIapproved opensource license httpwwwpythonorg Abaqus also comes with an integrated user manual Abaqus Documentation that can be opened in a browser see Figure 211 New users usually prefer using the graphic interface and they can start with the tutorial provided in the documentation Getting started with Abaqus Interactive edition This tutorial takes the user through all the steps required to build a finite element model analyze it and visualize the results There are also many tutorials available on the web Students can join the SIMULIA Learning Community and they may be eligible for a free copy of Abaqus Student Version httpwwwsimuliacomacademicspurchasehtml 2102 ANALYSIS OF A TRUSS WITH ABAQUS INTERACTIVE EDITION 21021 Modeling In this section we will analyze the truss shown in Figure 29 with the Abaqus interactive edition Click Start All Programs and locate Abaqus as shown in Figure 212 FIGURE 212 Starting Abaqus 2013 by Taylor Francis Group LLC Bar Element 37 Double click onAbaqus CAE to reveal the main user inter face Click on Create Model Database to start a new anal ysis On the main menu click on File and Set Work Direc tory to choose your working directory Click on Save As and name the file Trusscae Figure 213 FIGURE 213 Abaqus CAE main user interface On the lefthandside menu click on Part to begin creating the model Figure 214 FIGURE 214 Creating a part The creating part window shown in Figure 215 appears on the screen Name the part Trusspart and check 2D Pla nar as this is a planar truss check on Deformable in the type Choose Wire as the base feature Enter an approximate size of 10 m and click on Continue WARNING There are no predefined system of units within Abaqus so the user is responsible for ensuring that the correct units are specified FIGURE 215 Choosing the geometry of the part 2013 by Taylor Francis Group LLC 38 Introduction to Finite Element Analysis Using MATLAB and Abaqus Click on Autofit View to fit the view of the sketcher to the screen You can also place the cursor on the center of the sketcher and zoom in and out using the middle mouse button Figure 216 FIGURE 216 Fitting the sketcher to the screen In the sketcher menu choose the CreateLines Connected button to begin drawing the geometry of the truss Figure 217 FIGURE 217 Drawing using the connected line button Begin drawing the truss The coordinates of the cursor are given in the topleft corner You could also enter them using the Pick a point or enter XY coordinates in the box situated in the bottomleft corner Once finished click on Done in the bottomleft corner to exit the sketcher Figure 218 FIGURE 218 Drawing the truss geometry The finished part should appear as shown in Figure 219 FIGURE 219 Finished part 2013 by Taylor Francis Group LLC Bar Element 39 Next under the model tree click on Materials to create a material for the truss Since all the members of the truss are made of the same mate rial we will only define one material which we will name Trussmaterial Then click on Mechanical then Elastic ity and Elastic Figure 220 FIGURE 220 Material definition Enter 30e6 kNm2 for the elas tic modulus and 03 for Poissons ratio even though it is not applica ble for a truss Figure 221 FIGURE 221 Material properties 2013 by Taylor Francis Group LLC 40 Introduction to Finite Element Analysis Using MATLAB and Abaqus The longitudinal members of the truss have a cross area of 0045 m2 and the diagonal members have a cross area of 002 m2 To input this data we need to define two sections Figure 222 Under the Model tree click on Sections and the Create Section window appears Name the section Longi tudinal In the Category check Beam and in the Type choose Truss Click on Continue FIGURE 222 Create section window Next the Edit Section win dow appears Scroll through Material and choose the already created material Trussmaterial to assign it to the section In Cross sectional area enter 0045 m2 and click OK Figure 223 Follow exactly the same procedure to create another section named Diagonal and enter 002 m2 for the cross area FIGURE 223 Edit material window Next we assign the defined sections to the corresponding members Expand the menu under Trusspart and click on Section assignment The message Select the regions to be assigned a section should appear on the bottom left corner of the main win dow Figure 224 FIGURE 224 Section assignment 2013 by Taylor Francis Group LLC Bar Element 41 Keep the Shift key down and with the mouse select the horizontal members Once a member is selected it changes color Click on done in the bottomleft corner next to the message Sel ect the regions to be ass igned a section The Edit Section Assignment win dow appears Figure 225 FIGURE 225 Regions to be assigned a section In Section scroll to Longitudinal and click on OK Figure 226 Repeat the same thing for the diagonal members Keep the Shift key down and with the mouse select the diagonal members Click on done in the bottom left corner next to the message Select the regions to be assigned a section The Edit Section Assignment window appears In Section scroll to Diagonal and click on OK FIGURE 226 Edit section assignment In the next step we will define the elements Expand the menu under Trusspart and click on Meshempty to load the meshing menu Figure 227 FIGURE 227 Loading the meshing menu 2013 by Taylor Francis Group LLC 42 Introduction to Finite Element Analysis Using MATLAB and Abaqus On the main menu click on Mesh and then on Element Type and with the mouse select the whole truss Click on Done in the bottomleft corner of the main window Figure 228 FIGURE 228 Selecting regions to be assigned element type The element type dialog box appears In Element Library click on Standard In Ele ment family scroll down and choose Truss In Geomet ric order choose Linear The message T2D2 A 2 node linear 2D truss should appear in the dialog box Figure 229 FIGURE 229 Selecting element type 2013 by Taylor Francis Group LLC Bar Element 43 On the main menu click on Seed then on Edge by number and select the whole truss Enter 1 in the bottomleft corner of the main window and press Enter The seeding on the truss should look like Figure 230 On the main menu click on Mesh again and then on Part to mesh the truss Once meshed the truss changes color to blue FIGURE 230 Mesh Expand the menu under Assembly and double click on instances In Abaqus you can create many parts and assemble them together to form a model You can also create many instances from one part For example in a bridge you do not have to draw all the gird ers If they are similar draw ing one is enough The others are created as instances of the first one Figure 231 FIGURE 231 Assembling the model 2013 by Taylor Francis Group LLC 44 Introduction to Finite Element Analysis Using MATLAB and Abaqus The create instance dia log box appears In this case we have only one part Trusspart Select it and click OK Figure 232 FIGURE 232 Creating instances Before introducing the boundary con ditions we need to understand how the degrees of freedom are numbered The translations along the axes x y and z are respectively numbered 1 2 and 3 The rotations around these axes are respectively numbered 4 5 and 6 Figure 233 Y 2 5 3 4 1 X 6 Z FIGURE 233 Numbering of the degrees of freedom 2013 by Taylor Francis Group LLC Bar Element 45 Expand the menu under Steps and Initial click on BC to introduce the boundary con ditions Figure 234 FIGURE 234 Creating boundary conditions The Create Boundary Condition dialog box appears Name the boundary con dition Pinnedsupport Choose SymmetryAntisym metryEncastré and click on Continue Figure 235 FIGURE 235 Type of boundary conditions 2013 by Taylor Francis Group LLC 46 Introduction to Finite Element Analysis Using MATLAB and Abaqus Select the leftside support and click on Done Figure 236 FIGURE 236 Selecting a region to be assigned boundary conditions The Edit Boundary Condition dia log box appears Select PINNEDU1 U2 U3 0 and click on OK Figure 237 FIGURE 237 Edit boundary condition dialog box for pinned support 2013 by Taylor Francis Group LLC Bar Element 47 Under Steps and Initial click on BC to create the boundary conditions for the roller In the Create Boundary Con dition dialog box name the boundary condition RollerSupport Choose SymmetryAntisymmetryEncastré and click on Continue Select the right support and click on Done In the Edit Boundary Condition dialog box select XASYMMU2 U3 UR1 0 and click on OK Figure 238 FIGURE 238 Edit boundary condition dialog box for roller support In the lefthandside menu right click on Steps to crate another step for applying the loads Click on Continue Figure 239 FIGURE 239 Creating a step for load application 2013 by Taylor Francis Group LLC 48 Introduction to Finite Element Analysis Using MATLAB and Abaqus In the Create Step dialog box name the step ApplyLoads select Static General and click on Continue Figure 240 FIGURE 240 Create step dialog box In the Edit step dialog box although it is not necessary you can still provide a descrip tion such as applying joint loads Leave all the other details as they are and click on OK Figure 241 FIGURE 241 Edit step dialog box 2013 by Taylor Francis Group LLC Bar Element 49 In the lefthandside menu under Steps and ApplyLoads click on Loads as shown in Figure 242 FIGURE 242 Creating a load In the Create load dialog box name the load Horizontal 15 kN force In Step scroll to ApplyLoads which means that the load will be applied in this step In Category choose Mechanical and in Type choose Concentrated Force Click on Continue Figure 243 FIGURE 243 Creating a concentrated load 2013 by Taylor Francis Group LLC 50 Introduction to Finite Element Analysis Using MATLAB and Abaqus With the mouse select the topleft joint as shown in Figure 244 and click on done in the bottomleft corner of the same window FIGURE 244 Selecting a joint for load application In the Edit Load dialog box enter 15 for CF1 and click on OK Figure 245 FIGURE 245 Entering the magnitude of a joint force Repeat the same procedure for the other joint loads Since they are vertical loads point ing in opposite direction to the axis y their magnitude should be entered in CF2 as nega tive Once finished the loaded truss should look like the one shown in Figure 246 FIGURE 246 Loaded truss 2013 by Taylor Francis Group LLC Bar Element 51 21022 Analysis Under Analysis right click on Jobs and then click on Create Figure 247 FIGURE 247 Creating a job The Create Job dialog box appears Name the job TrussProblem1 and click on Continue Figure 248 FIGURE 248 Naming a job 2013 by Taylor Francis Group LLC 52 Introduction to Finite Element Analysis Using MATLAB and Abaqus The Edit Job dialog box appears Enter a description for the job Check Full analysis and choose to run the job in Background and check to start it immediately Click OK Figure 249 FIGURE 249 Editing a job Expand the tree under Jobs right click on TrussProblem1 Then click on Submit Figure 250 FIGURE 250 Submitting a job If you get the following mes sage Job TrussProblem1 completed successfully in the bottom window then your job is free of errors and was exe cuted properly Now it is time to view the analysis results Figure 251 FIGURE 251 Monitoring of a job 2013 by Taylor Francis Group LLC Bar Element 53 Under the top menu in the Module scroll to Visualization and click to load Abaqus Viewer On the main menu under File click Open navigate to your working directory and open the file TrussProblem1odb It should have the same name as the job you submitted Figure 252 FIGURE 252 Opening the visualization module Click on the Common options icon to display the Common Plot options dia log box Under labels check Show Ele ment labels and Show Node labels to display elements and nodes numbering Figure 253 FIGURE 253 Common plot options You may obtain a dif ferent nodes and ele ments numbering to the one shown in Figure 254 However you must ensure that there are 15 elements and 9 nodes only FIGURE 254 Elements and nodes numbering 2013 by Taylor Francis Group LLC 54 Introduction to Finite Element Analysis Using MATLAB and Abaqus Click on the icon Plot Deformed Shape to dis play the deformed shape of the truss Figure 255 FIGURE 255 Deformed shape On the main menu click on Results then on Field Output to open the Field Output dialog box Choose U Spatial dis placements at nodes For component choose U2 to plot the vertical displace ment Figure 256 FIGURE 256 Field output dialog box 2013 by Taylor Francis Group LLC Bar Element 55 Figure 257 shows the contour plot of the vertical displacement U2 as well as the legend block 0000e00 U U2 1945e05 3890e05 5835e05 7780e05 9726e05 1167e04 1362e04 1556e04 1751e04 1945e04 2140e04 2334e04 14 15 11 7 13 10 3 9 6 5 1 5 4 2 8 12 FIGURE 257 Contour plot of the vertical displacement U2 If you cannot read the dis placements values in the leg end block on the main menu click on Viewport Annota tion Options Under Legend click on Set font and enter a bigger font Figure 258 FIGURE 258 Viewport annotations options On the main menu click on Results then on Field Output to open the Field Output dialog box Choose S Stress components at integration points For component choose S11 to plot the stresses in the bars Note that Abaqus does not plot the normal forces in the bars Figure 259 FIGURE 259 Normal stresses in the bars 2013 by Taylor Francis Group LLC 56 Introduction to Finite Element Analysis Using MATLAB and Abaqus To create a text file containing the reac tion forces and nodal displacements in the menu bar click on Report and Field Output In the Report Field Output dialog box for Position select Unique nodal check RF1 and RF2 for RF Reaction force and check U1 and U2 for U Spatial displacement Then click on click on Set up Figure 260 FIGURE 260 Selecting variables to print to a report Click on Select to nav igate to your working directory Name the file TrussProblem1rpt Uncheck Append to file and click OK Figure 261 FIGURE 261 Choosing a directory and the file name to which to write the report 2013 by Taylor Francis Group LLC Bar Element 57 Open your working directory and take a look at the files generated by Abaqus You can happily ignore most of them However you should keep the inp file as it contains all the information about the model We will write a similar file in the next section The odb is binary and contains all the information about the model and the results of the analysis It is used by the visualization module to view the results The dat contains written output such as results Most importantly it contains any errors made during the setting up of the model The msg file that appears as an outlook item in Windows contains any error that arises during the analysis It is particularly useful in nonlinear problems The rest of the files you can ignore them for the time being Use your favorite text editor and open the file TrussProblem1rpt Field Output Report written Fri Apr 01 091709 2011 Source 1 ODB FTRAVAILNEWBOOKAbaqusexamplestrussproblem1odb Step Loadstep Frame Increment 1 Step Time 1000 Loc 1 Nodal values from source 1 Output sorted by column Node Label Field Output reported at nodes for part TRUSS1 Node RFRF1 RFRF2 UU1 UU2 Label Loc 1 Loc 1 Loc 1 Loc 1 1 0 0 736886E06 213242E06 2 0 0 512812E06 940781E06 3 0 15125 985185E06 15125E36 4 0 0 103689E06 233414E06 5 0 0 873148E06 182554E06 6 0 0 136096E06 100075E06 7 0 0 611111E06 230828E06 8 0 0 273148E06 186493E06 9 15 6875 15E36 6875E36 Note that at node 9 the horizontal reaction is equal to 15 kN and the vertical reaction is equal to 6875 The horizontal and vertical displacements at node 7 are respectively equal to 611111e 06 230828e06 m which are the same as previously obtained with the MATLAB code trussm node 5 000006 000023 m 2103 ANALYSIS OF A TRUSS WITH ABAQUS KEYWORD EDITION In Abaqus you can create a complete finite element model by simply using a text editor The input file must have the extension inp It contains Abaqus commands in the format of Keywords A keyword starts with a In the Abaqus Documentation click on the Abaqus Keywords Reference Manual to find the meaning and usage of all the Abaqus keywords They are organized in an alphabetical order In this section we will prepare an input file for the truss shown in Figure 29 We will keep the same node and element numbering The problem at hand is very simple therefore the file should be very easy to understand Using a text editor create a file and save it as trussproblem1keywordinp Before creating the model make sure you adhere to the following rules 2013 by Taylor Francis Group LLC 58 Introduction to Finite Element Analysis Using MATLAB and Abaqus Any line that starts with two stars represents a comment that will be ignored by Abaqus Any line that starts with only one represents a command and Abaqus will attempt to execute it If it is not a proper keyword an error will result Any line that does not start with or represents data Do not leave blank lines instead use two stars HEADING Example TrussProblem1KeywordEdition the HEADING Example TrussProblem1KeywordEdition will appear on any output files created by Abaqus Geometry definition Enter the nodal coordinates of the nodes Node Nset allnodes 1 0 0 2 1 2 3 2 0 4 3 2 5 4 0 6 5 2 7 6 0 8 7 2 9 8 0 Define node sets to be used for BC and applying loads Nset nsetPinnedsupport 1 Nset nsetRollersupport 9 Nset nsetHF15 2 Nset nsetVF5 3 Nset nsetVF7 4 Nset nsetVF10 7 Select element type as T2D2 planar truss element and define element connectivity Element typeT2D2 1 1 2 2 1 3 3 2 3 4 2 4 5 3 4 6 3 5 7 4 5 8 4 6 9 5 6 10 5 7 11 6 7 12 6 8 13 7 8 14 7 9 15 8 9 2013 by Taylor Francis Group LLC Bar Element 59 Create two element sets one for the horizontal elements named Horizontal and one for the diagonal elements named Diagonal elset elset Horizontal 2 4 6 8 10 12 14 elset elset Diagonal 1 3 5 7 9 11 13 15 Define material and name it Mymaterial Material nameMymaterial Elastic 3e07 Define a section for the horizontal members Solid Section elset Horizontal materialMymaterial 0045 Define a section for the diagonal members Solid Section elsetDiagonal materialMymaterial 002 Define Boundary Conditions Boundary Rollersupport XASYMM Pinnedsupport PINNED Define step and name it Loadstep Step nameLoadstep Static 1 1 1e05 1 Apply the loads as concentrated forces Cload HF15 1 15 VF5 2 5 VF10 2 10 VF7 2 7 OUTPUT REQUESTS FIELD OUTPUT Only request the default field output Output field variablePRESELECT HISTORY OUTPUT Only request the default History output Output history variablePRESELECT End Step The file starts with the keyword HEADING Below in the data line put any text you want to describe the model The text will appear on any output files created by Abaqus Next define the geometry of the nodes using the keyword node You can group all the nodes in a node set named allnodes In the data line below the keyword enter the node number followed 2013 by Taylor Francis Group LLC 60 Introduction to Finite Element Analysis Using MATLAB and Abaqus by its x and y coordinates Use one line per node and make sure you separate the entered values by commas Otherwise you will get an error Once all the nodes are defined create node sets that will be used later for imposing the boundary conditions and applying the loads Nset nsetPinnedsupport creates a node set named Pinnedsupport that contains the node 1 entered in the data line Nset nsetRollersupport creates a node set named Rollersupport containing node 9 Nset nsetHF15 creates a node set named HF15 containing node 2 Nset nsetVF5 creates a node set named VF5 containing node 3 Nset nsetVF7 creates a node set named VF7 containing node 4 Nset nsetVF10 creates a node set named VF10 containing node 7 Next using the keyword elset create two elements sets one for the horizontal members named Horizontal and one for the diagonal members named Diagonal Using the keyword Material create a material named Mymaterial The created material is elastic and has a Youngs modulus of 3e 07 given in the data line of the keyword Elastic Using the keyword Solid Section create a section for the horizontal members with the element set Horizontal and Mymaterial for material Enter the cross section of 0045 in the data line Create another one for the diagonal members using the element set Diagonal and the same material This time enter 002 for the cross section Using the keyword Boundary apply the boundary condition We assign YSYMM symmetry about a plane Y constant to node set Rollersupport It means the degrees of freedom 2 4 and 6 are suppressed In the next data line we assign PINNED to node set Pinnedsupport It means the degrees of freedom 1 3 and 3 are suppressed Next using the keyword step create a step and name it Loadstep The keyword static indicates that it will be a general static analysis It is important to note that there are four values in the data line of the keyword static These values represents pseudotime in Abaqus Standard that is a mapping between time and load The first value equal to 1 represents the initial time increment In other words Abaqus will initially try to apply the total load as one increment The second value also equal to 1 is the total time period of the step The third value corresponds to the minimum time increment This particularly happens in nonlinear analysis If Abaqus cannot apply the load as a whole it keeps reducing the increment until it reaches this minimum value The fourth and last value is the maximum time increment allowed The keyword cload indicates that the loads will be applied as concentrated loads In the data lines HF15 1 15 indicates that a positive 15 kN load is applied in the direction 1 x direction to node set HF15 defined previously VF5 2 5 indicates that a negative 5 kN load is applied in the direction 2 y direction to node set VF5 defined previously VF10 2 10 indicates that a negative 10 kN load is applied in the direction 2 Y direction to node set VF10 defined previously VF7 2 7 indicates that a negative 7 kN load is applied in the direction 2 Y direction to node set VF7 defined previously You can request outputs that will be written to the database file odb using the keyword output There are two types of outputs field and history When the variable is set equal to PRESELECT only the default variables will be printed Field output is intended for infrequent requests for a large portion of the model and can be used to generate contour plots animations and so on History output on the other hand is intended for relatively frequent output requests for small 2013 by Taylor Francis Group LLC Bar Element 61 FIGURE 262 Running Abaqus from the command line portions of the model and is displayed in XY data plots For example if we want to monitor the displacement of a node with load this is the type of output that needs to be requested You can create many steps in Abaqus but each one of them must end with the keyword end step If your operatingsystem is Windows in Start Menu clickon Accessories and then onCommand prompt to open a DOS shell Using DOS commands navigate to your working directory At the command line type Abaqus jobtrussproblem1keyword inter followed by Return The outcome should be similar to the one shown in Figure 262 If you get an error open the file with extension dat to see what type of error To load the visualization model type Abaqus Viewer at the command line 2013 by Taylor Francis Group LLC 2013 by Taylor Francis Group LLC 3 Beam Element 31 INTRODUCTION A beam constitutes the simplest way of spanning a gap between two objects As structural elements beams are prominent in both civil and mechanical engineering They are used as supports for floors in buildings decks in bridges wings in aircraft or axles for cars A beam is generally slender and carries loadings applied perpendicular to its longitudinal axis In matrix structural analysis or finite element for that matter a beam is regarded as an element with a node at each end When the element is loaded as shown in Figure 31a each node will undergo a vertical displacement w and a rotation θ as shown in Figure 31b The end nodes 1 and 2 are subject to shear forces and moments which result in vertical translations and rotations Each node therefore has two degrees of freedom In total the element has four degrees of freedom The nodal forces and displacements can be expressed in vector form as Fe F1 M1 F2 M2T 31 de w1 θ1 w2 θ2T 32 The differential equations describing the behavior of a beam element are well known They are referred to as the EulerBernoulli theory of bending or simply known as the engineering beam theory For a differential element dx of the beam as shown in Figure 32 the relationships between deflection slope load shear and moment are given in the form of differential equations as d2w dx2 M EI 33 d3w dx3 1 EI dM dx S EI 34 d4w dx4 1 EI dS dx qx EI 35 where w M S EI and qx represent respectively the deflection moment shear force stiffness and uniformly distributed load 32 STIFFNESS MATRIX It is possible to develop the matrix relationship between the nodal forces F1 M1 F2 M2T and the nodal displacements w1 θ1 w2 θ2T by integrating the differential equations 33 through 35 The integration produces constants of integration that can be identified by considering the boundary conditions of the element A simpler way of establishing the matrix relationship is to operate as for the bar element see Section 221 It consists in placing simple supports at each end of the beam then set the degrees of freedom to unity one at a time and calculate the nodal forces needed to produce the deformed state The reactions at the supports resulting from the imposition of unit displacementsrotations at the nodes are called stiffness influence coefficients To obtain these coefficients we will use the theorem of Castigliano 63 2013 by Taylor Francis Group LLC 64 Introduction to Finite Element Analysis Using MATLAB and Abaqus E I L E I L a F1 w1 w2 F2 y y M1 θ1 θ2 M2 x x b FIGURE 31 Beam element a Forces and b displacements qx qxdqx Mx dMx SxdSx Mx Sx dx y x FIGURE 32 Differential element of a beam ROTATION θ2 Consider the beam element shown in Figure 33a The member is initially straight If we try to rotate node 2 by an amount θ2 then reaction forces will be developed at nodes 1 and 2 Considering vertical equilibrium yields Fy1 Fy2 0 36 Taking moments around z with respect to node 2 gives M1 M2 Fy1L 0 37 Taking moments around z with respect to x as shown yields Mx M1 Fy1x 38 The moment Mx may also be written as a function of M2 Mx Fy1L x M2 39 2013 by Taylor Francis Group LLC Beam Element 65 FIGURE 33 Nodal degrees of freedom a Rotation θ2 b rotation θ1 c displacement w2 and d displacement w1 The strain energy of a beam in bending is given as Π L 0 Mx2 2EI dx 1 2EI L 0 Fy1L x M22dx 1 2EI F2 y1 L3 3 M2 2L Fy1M2L2 310 Using the theorem of Castigliano and taking the derivative with respect to Fy1 yields Π Fy1 2 3 L3Fy1 M2L2 w1 0 311 and Π M2 1 2EI 2M2L Fy1L2 θ2 312 Solving for M2 and Fy1 using Equations 311 and 312 yields M2 4EI L θ2 313 Fy1 6EI L2 θ2 314 Since M1 M2 Fy1 0 we also have M1 2EI L θ2 315 ROTATION Θ1 By simply transposing the suffixes similar expressions can be obtained for M1 M2 and Fy2 when considering a rotation θ1 Figure 33b that is M1 4EIL θ1 316 M2 2EIL θ2 317 Fy2 6EIL2 θ1 318 DISPLACEMENT w2 The initially straight member is now given a vertical displacement w2 as represented in Figure 33c The bending moment at a distance x is obtained as Mx M1 Fyx 319 or as Mx M1 Fyx 320 Substituting in the expression of the bending energy yields Π L0 Mx22EI dx 12EI L0 M1 Fy22dx 12EI M12L F2y2 L33 Fy2M1L2 321 Using the theorem of Castigliano we obtain ΠM1 12EI 2ML Fy2L2 θ1 0 322 and ΠFy2 12EI 2Fy2L33 M1L2 w2 323 Solving for M1 and Fy2 using Equations 322 and 323 yields M1 6EIL2 w2 324 Fy2 12EIL3 w2 325 From equilibrium of the moments we obtain M2 as M2 6EIL2 w2 326 Beam Element 67 DISPLACEMENT w1 Again by simply transposing the suffices similar expressions can be obtained for M1 M2 and Fy1 when considering a displacement w1 Figure 33d that is M1 6EI L2 w1 327 M2 6EI L2 w1 328 Fy1 12EI L3 w1 329 The preceding results can be grouped in a matrix form Fy1 M1 Fy2 M2 12EIL3 6EIL2 12EIL3 6EIL2 6EIL2 4EIL 6EIL2 2EIL 12EIL3 6EIL2 12EIL3 6EIL2 6EIL2 2EIL 6EIL2 4EIL w1 θ1 w2 θ2 330 or simply as fe Keδe 331 where Ke is the stiffness matrix that relates the nodal displacements to the nodal forces 33 UNIFORMLY DISTRIBUTED LOADING The stiffness matrix for a beam element was developed for loadings applied only at its nodes Quite often however beams support uniformly distributed loading along or part of their length This requires modification in order to be used in an analysis The distributed loading is replaced by a system of statically equivalent nodal forces that are always of opposite sign from the fixed end reactions as shown in Figure 34 Figure B1 in Appendix B shows the equivalent nodal loads for the most common loadings on beams The displacements computed using equivalent nodal loads are exact in a finite element sense however the internal reactions computed in individual elements using the relation Fe Kede are not Instead to obtain the correct internal reactions the following relation must be used Fe Kede F0 332 where F0 represents the vector of equivalent nodal forces at element level To illustrate the computation of the reaction forces let us consider a beam for which a solution can be easily obtained Such a beam is presented in Figure 35 together with the bending moment and shear force diagrams which have been obtained with the method of moment distribution From the shear force diagram the support reactions at A B and C are respectively given as RA 16 kN RB 118 kN RC 18 kN 333 2013 by Taylor Francis Group LLC 68 Introduction to Finite Element Analysis Using MATLAB and Abaqus L q Loading Equivalent nodal loads qL2 12 qL 2 qL 2 qL2 12 qL2 Fixed end reactions 12 qL2 12 qL 2 qL 2 FIGURE 34 Statically equivalent nodal loads IAB 120106 mm4 IBC 240106 mm4 6 kNm B C 3 m 32 102 SF kN 16 16 BM kNm 104 138 4 m A FIGURE 35 Loading bending moment and shear force diagrams 2013 by Taylor Francis Group LLC Beam Element 69 From the bending moment diagrams the support moments are obtained as MA 16 kNm MB 32 kNm MC 104 kNm 334 Using the finite element method let us calculate these support reactions Element AB Considering that the beam is made of steel with an elastic modulus of 200 106 kNm2 and using a consistent set of units kN and m from Equation 330 the stiffness matrix of element AB is obtained as KAB 10667 16000 10667 16000 16000 32000 16000 16000 10667 16000 10667 16000 16000 16000 16000 32000 335 Element AB is not subjected to any external loading FAB 0 0 0 0 336 Element BC KBC 9000 18000 9000 18000 18000 48000 18000 24000 9000 18000 9000 18000 18000 24000 18000 48000 337 The applied uniformly distributed load is transformed into equivalent static loads as shown in Figure 34 FAB qL2 12 kN qL212 8 kNm qL2 12 kN qL212 8 kNm 338 Assembling the global stiffness matrix and force vector results in 10667 16000 10667 16000 0 0 16000 32000 16000 16000 0 0 10667 16000 19667 2000 9000 18000 16000 16000 2000 80000 18000 24000 0 0 9000 18000 9000 18000 0 0 18000 24000 18000 48000 wA θA wB θB wC θC 0 0 12 8 12 8 339 2013 by Taylor Francis Group LLC 70 Introduction to Finite Element Analysis Using MATLAB and Abaqus The boundary conditions for the beam are given as wA θA wB wC θC 0 340 Eliminating the lines and columns corresponding to these degrees of freedom results in one single equation 80000 θB 8 θB 00001rd 341 The results for each span will be computed individually The nodal displacements of element AB are obtained as dAB wA 0 θA 0 wB 0 θB 00001 342 The final reactions for element AB are caused by the rotation of joint B VA MA VB1 MB 10667 16000 10667 16000 16000 32000 16000 16000 10667 16000 10667 16000 16000 16000 16000 32000 0 0 0 00001 16 16 16 32 343 It can be noticed that VA RA 16 kN MA 16 kNm MB 32 kNm As to the notation VB1 it means that only the end shear at point B is considered The total reaction at B is equal to the end shear from element AB plus the end shear at point B from element BC that is RB VB1 VB2 Similarly the final reactions for element BC are caused by joint B rotation minus the equivalent nodal loads that replaced the uniformly distributed load that is VB2 MB VC MC 9000 18000 9000 18000 18000 48000 18000 24000 9000 18000 9000 18000 18000 24000 18000 48000 0 00001 0 0 12 8 12 8 102 32 138 104 344 Finally we obtain RB VB1 VB2 16 102 118 kN The final results shown in Figure 36 are exactly the same as the ones shown in Figure 35 2013 by Taylor Francis Group LLC Beam Element 71 16 kNm 16 kN 16 kN 102 kN 6 kNm 138 kN 104 kNm 32 kNm FIGURE 36 Support reactions for individual members 34 INTERNAL HINGE In some cases a beam may contain an internal hinge which results in a discontinuity in the slope of the deflection curve as well as a zero value of the bending moment If we are to analyze the beam shown in Figure 37 using the finite element method we will discretize the beam using two elements The hinge should be accounted for only once either associated with element 1 or with element 2 If the beam is discretized with two elements one with a hinge at its right end and the other with a hinge at its left the result will be a singular stiffness matrix Using Equation 330 the forcedisplacement relationship for element 1 is written as 12EIL3 6EIL2 12EIL3 6EIL2 6EIL2 4EIL 6EIL2 2EIL 12EIL3 6EIL2 12EIL3 6EIL2 6EIL2 2EIL 6EIL2 4EIL w11 θ11 w12 θ12 F11 M11 F12 M12 0 345 To eliminate the moment M12 which is equal to zero we partition the system of equations as follows 12EIL3 6EIL2 12EIL3 6EIL2 6EIL2 4EIL 6EIL2 2EIL 12EIL3 6EIL2 12EIL3 6EIL2 6EIL2 2EIL 6EIL2 4EIL w11 θ11 w12 θ12 F11 M11 F12 M12 0 346 L Internal hinge FIGURE 37 Beam with an internal hinge 2013 by Taylor Francis Group LLC or in a more compact form as k11 k12 0 k21 k22 d θ12 F M12 0 347 Expanding Equation 347 yields k11d k12θ12 F 348 k21d k22θ12 M12 Solving for θ12 using the second equation of 348 yields θ12 k221M12 k21d 349 Substituting for θ12 in the first equation of 348 and rearranging yields k11 k12k221k21 d F k12k221M12 350 or in a more compact form as Kcd Fc 351 where Kc is a condensed matrix When the partitioned parts of Equation 348 are substituted in Equation 351 the condensed matrix becomes Kc 3EIL3 3EIL2 3EIL3 3EIL3 EIL2 3EIL 3EIL2 3EIL2 3EIL3 3EIL2 3EIL3 352 It is true that moment M12 is equal to zero at the hinge but not the rotation θ12 and as such it should not have been eliminated from Equation 351 To include the rotation θ12 we expand Equation 351 as follows 3EIL3 3EIL2 3EIL2 0 w11 θ11 F11 M11 353 3EIL2 3EIL3 3EIL2 0 w12 θ12 or for element 2 with a hinge at its left end Equation 353 is rewritten as 3EIL3 0 3EIL3 3EIL2 w21 θ21 F21 M21 354 0 0 0 0 0 0 0 0 3EIL2 0 3EIL2 3EIL Beam Element 73 M11 θ11 θ12 θ21 θ22 M22 M12 M21 0 Element 1 Element 2 F22 W22 W21 F21 W12 F12 F11 W11 FIGURE 38 Beam elements with a hinge 35 COMPUTER CODE BEAMm Except for slight differences that need to be taken into account writing a MATLAB code for the analysis of slender beams is not much different from that for a truss structure First the elements stiffness do not need to be transformed from local to global coordinates Second each element will have two types of loading one that consists of the external forces directly applied to the nodes and another that only consists of the statically equivalent nodal loads Therefore in the development of the program BEAMm we will follow the same style as that used in the program TRUSSm Let us consider the beam shown in Figure 39 351 DATA PREPARATION To read the data we will use the Mfile beam1datam Again we will use a consistent set of units mm for length and N for force The input data for this beam consist of the following nnd 4 number of nodes nel 3 number of elements nne 2 number of nodes per element nodof 2 number of degrees of freedom per node 3511 Nodes Coordinates The abscissae x of the nodes are given in the form of a vector geomnnd 1 geom 0 4000 9000 16000 1 1 2 m 20 kN 4 m 5 m 7 m 4 kNm 2 2 3 3 4 E 200 000 MPa I 200 106 mm4 FIGURE 39 Example of a continuous beam 2013 by Taylor Francis Group LLC 74 Introduction to Finite Element Analysis Using MATLAB and Abaqus 3512 Element Connectivity The table of connectivity describes how the elements are connected to each other The element connectivity is given in the matrix connecnel 2 connec 1 2 2 3 3 4 3513 Material and Geometrical Properties The material and geometrical properties are given in the matrix propnel 2 The first column represents the Youngs modulus while the second represents the second moment of inertia of the cross section prop 200000 200e 6 200000 200e 6 200000 200e 6 3514 Boundary Conditions In the same fashion as for the truss a restrained degree of freedom is assigned the digit 0 while a free degree of freedom is assigned the digit 1 As previously explained a node in a beam element has two degrees of freedom a vertical translation along the axis y and a rotation around the axis z perpendicular to the plan xy As shown in Figure 39 nodes 1 and 4 are fully fixed encastré Their degrees of freedom are all assigned the digit 0 Nodes 2 and 3 are simple supports They are restrained vertically but are free to rotate Therefore their degrees of freedom w and θ are respectively assigned the digits 0 and 1 The information on the boundary conditions is given in the matrix nfnnd nodof nf 0 0 0 1 0 1 0 0 The free degrees of freedom different from zero are then counted and their rank assigned back into the matrix nfnnd nodof nf 0 0 0 1 0 2 0 0 In this case the total number of active degrees of freedom is obtained as n 2 3515 Internal Hinges To account for internal hinges we create a vector Hingenel 2 that we initialize to 1 If a particular element k has a hinge at its left end then we assign it the digit 0 at the position of its first node that is Hingek 1 0 2013 by Taylor Francis Group LLC A hinge must be considered for one element only 3516 Loading When it comes to loading a beam element differs from a rod element As previously explained a beam element can have two types of loading loads applied directly at the nodes and statically equivalent nodal loads A good computer code should cater for both loadings To distinguish between the two loading systems we will use two matrices Jointloadsnnd2 and Elementloadsnel4 There are no loads applied directly at the nodes Therefore the matrix Jointloadsnnd2 is empty Jointloads 0 0 0 0 0 0 0 0 0 0 Elements 1 and 2 have loads applied along their length which need to be transformed to statically equivalent nodal loads as shown in Figure 39 Element Fy1 My F2y2 M2 1 104 107 104 107 2 104 8333 x 106 104 8333 x 106 3 0 0 0 0 These data are stored in the Mfile beam1datam in the matrix Elementloads The two systems of loads are added to form the global force vector Fn This is carried out in the Mfile formbeamFm as follows Joint loads To assemble the nodal loads we create a loop over the nodes If a degree of freedom nfij is not restrained then it is susceptible of carrying a load That load is assembled into the global force vector at the position Fnfij Element loads To assemble the statically equivalent nodal loads we create a loop over the elements Since the loads are element based we need the steering vector g containing the number of the degrees of freedom of the nodes of the element It is formed in the same way as in the program trussm The script is given in the Mfile beamgm Then we create a loop over the degrees of freedom of the element If a degree of freedom nfij is not restrained then it is susceptible of carrying a load That load is assembled into global force vector at the position Fgj The data preparation is now complete and the model data are written to the file beam1resultstxt using the Mfile printbeammodelm At this stage it is possible to initialize the global matrix KKnn 0 KK 0 0 0 0 76 Introduction to Finite Element Analysis Using MATLAB and Abaqus Again we will only assemble the quantities corresponding to the active degrees of freedom that is the lines and the columns in the matrix KK corresponding respectively to the active degrees of freedom 1 and 2 The restrained degrees of freedom with a number equal to 0 will be eliminated 3517 Stiffness Matrix For a beam element there is no need to transform the element stiffness matrix from local to global coordinates since both sets of axes are colinear Therefore for each element from 1 to nel we set up the local stiffness matrix and directly assemble it into the global stiffness matrix KK For any element i we retrieve its first and second nodes from the connectivity matrix node1 conneci 1 node2 conneci 2 Then using the values of the nodes we retrieve their x coordinates from the geometry matrix x1 geomnode1 x2 geomnode2 Next we evaluate the length of the element as L x2 x1 Finally we retrieve the material and geometrical property of the section E propi 1 I propi 2 Depending on whether nodes 1 or 2 are internal hinges we evaluate the stiffness matrix kl as follows if Hingei node1 0 evaluate the matrix kl using Equation 353 if Hingei node2 0 evaluate the matrix kl using Equation 352 else evaluate the matrix kl using Equation 330 The MATLAB script for evaluating the matrix kl is given in the Mfile beamkm 352 ASSEMBLY AND SOLUTION OF THE GLOBAL SYSTEM OF EQUATIONS The global stiffness matrix KK is assembled using the same script formKKm as in the program trussm The global displacements vector delta is obtained as delta KKF 353 NODAL DISPLACEMENTS To retrieve the nodal displacements a loop is carried over all the nodes If a degree of freedom j of a node i is free that is nfi j 0 then it could have a displacement different from zero The value of the displacement is extracted from the global displacements vector delta nodedispi j deltanfi j 2013 by Taylor Francis Group LLC Beam Element 77 354 ELEMENT FORCES To obtain the member forces a loop is carried over all the elements 1 Form element stiffness matrix kl 2 Form element steering vector g a Loop over the degrees of freedom of the element to obtain element displacements vector edg b If gj 0 then the degree of freedom is restrained edj 0 c Otherwise edj deltagj 3 Obtain element force vector due to joint loads as fl kl ed 4 Obtain element equivalent nodal forces as f0 Elementloadsi 5 Obtain element forces as forcei fl f0 The results of the analysis are written to the file beam1resultstxt using the Mfile printbeamresultsm A copy of the file beam1resultstxt is included within the section Program scripts Filebeamm beamm LINEAR STATIC ANALYSIS OF A CONTINUOUS BEAM clc Clear screen clear Clear all variables in memory Make these variables global so they can be shared by other functions global nnd nel nne nodof eldof n geom connec F prop nf Elementloads Jointloads force Hinge dispExecuting beamm Open file for output of results ALTER THE NEXT LINES TO CHOOSE AN OUTPUT FILE FOR THE RESULTS dispResults printed to file beam1resultstxt fid fopenbeam1resultstxtw ALTER THE NEXT LINE TO CHOOSE AN INPUT FILE beam1data Load the input file KK zerosn Initialize global stiffness matrix to zero Fzerosn1 Initialize global force vector to zero F formbeamFF Form global force vector printbeammodel Print model data for i1nel 2013 by Taylor Francis Group LLC 78 Introduction to Finite Element Analysis Using MATLAB and Abaqus klbeamki Form element matrix gbeamgi Retrieve the element steering vector KK formKKKK kl g assemble global stiffness matrix end End of assembly delta KKF solve for unknown displacements Extract nodal displacements for i1nnd for j1nodof nodedispij 0 if nfij 0 nodedispij deltanfij end end end Calculate the forces acting on each element in local coordinates and store them in the vector force for i1nel klbeamki Form element matrix gbeamgi Retrieve the element steering vector for j1eldof if gj 0 edj0 displacement 0 for restrained freedom else edj deltagj end end fl kled Element force vector in global XY f0 Elementloadsi forcei flf0 end printbeamresults fclosefid Filebeam1datam File Beam1datam The following variables are declared as global in order to be used by all the functions Mfiles constituting the program global nnd nel nne nodof eldof n geom connec prop nf Elementloads Jointloads Hinge format short e Beginning of data input 2013 by Taylor Francis Group LLC Beam Element 79 nnd 4 Number of nodes nel 3 Number of elements nne 2 Number of nodes per element nodof 2 Number of degrees of freedom per node eldof nnenodof Number of degrees of freedom per element Nodes coordinates X and Y geomzerosnnd1 geom 0 X coord node 1 4000 X coord node 2 9000 X coord node 3 16000 X coord node 4 Element connectivity conneczerosnel2 connec 1 2 1st and 2nd node of element 1 2 3 1st and 2nd node of element 2 3 4 1st and 2nd node of element 3 Geometrical properties prop11 E prop12 I propzerosnel2 prop 200000 200e6 E and I of element 1 200000 200e6 E and I of element 2 200000 200e6 E and I of element 3 Boundary conditions nf onesnnd nodof Initialize the matrix nf to 1 nf11 0 nf12 0 Prescribed nodal freedom of node 1 nf21 0 Prescribed nodal freedom of node 2 nf31 0 Prescribed nodal freedom of node 3 nf41 0 nf42 0 Prescribed nodal freedom of node 4 Counting of the free degrees of freedom n0 for i1nnd for j1nodof if nfij 0 nn1 nfijn end end end Internal Hinges Hinge onesnel 2 loading Jointloads zerosnnd 2 Enter here the forces in X and Y directions at node i Elementloads zerosnel 4 Elementloads1 1e4 1e7 1e4 1e7 Elementloads2 1e4 8333e6 1e4 8333e6 End of input 2013 by Taylor Francis Group LLC 80 Introduction to Finite Element Analysis Using MATLAB and Abaqus Filebeam1resultsm PRINTING MODEL DATA Number of nodes 4 Number of elements 3 Number of nodes per element 2 Number of degrees of freedom per node 2 Number of degrees of freedom per element 4 Node X 1 000000 2 400000 3 900000 4 1600000 Element Node1 Node2 1 1 2 2 2 3 3 3 4 Element E I 1 200000 2e008 2 200000 2e008 3 200000 2e008 Nodal freedom Node dispw Rotation 1 0 0 2 0 1 3 0 2 4 0 0 Applied Nodal Loads Node loadY Moment 1 000000 000000 2 000000 166700000 3 000000 833300000 4 000000 000000 Total number of active degrees of freedom n 2 PRINTING ANALYSIS RESULTS Global force vector F 1667e006 8333e006 Displacement solution vector delta 000001 000016 2013 by Taylor Francis Group LLC Beam Element 81 977 kNm 177 354 354 4 kNm 104 104 20 kN 98 kN 1017 1138 862 076 076 FIGURE 310 Example 1 Continuous beam results Nodal displacements Node dispy rotation 1 000000 000000 2 000000 000001 3 000000 000016 4 000000 000000 Members actions element fy1 M1 Fy2 M2 1 982992 977323020 1017008 1045353960 2 1138117 1045353960 861883 354767327 3 76022 354767327 76022 177383663 The results are shown graphically for each element in Figure 310 36 PROBLEMS Prepare a data file for the beams shown next and carry out the analysis using the code beamm 361 PROBLEM 31 FIGURE 311 Input file File Beamproblem1datam The following variables are declared as global in order to be used by all the functions Mfiles constituting the program global nnd nel nne nodof eldof n geom connec prop nf Elementloads Jointloads Hinge format short e 4 m 1 1 2 2 3 3 4 4 5 6 m 4 m 8 m 20 kN 5 kNm 5 kNm E 200000 MPa I 200 106 mm4 FIGURE 311 Problem 31 2013 by Taylor Francis Group LLC 82 Introduction to Finite Element Analysis Using MATLAB and Abaqus Beginning of data input nnd 5 Number of nodes nel 4 Number of elements nne 2 Number of nodes per element nodof 2 Number of degrees of freedom per node eldof nnenodof Number of degrees of freedom per element Nodes coordinates X and Y geomzerosnnd1 geom 0 X coord node 1 4000 X coord node 2 10000 X coord node 3 14000 X coord node 4 22000 X coord node 5 Element connectivity conneczerosnel2 connec 1 2 1st and 2nd node of element 1 2 3 1st and 2nd node of element 2 3 4 1st and 2nd node of element 3 4 5 1st and 2nd node of element 4 Geometrical properties propzerosnel2 prop 200000 200e6 E and I of element 1 200000 200e6 E and I of element 2 200000 200e6 E and I of element 3 200000 200e6 E and I of element 4 Boundary conditions nf onesnnd nodof Initialize the matrix nf to 1 nf11 0 nf12 0 Prescribed nodal freedom of node 1 nf31 0 Prescribed nodal freedom of node 3 Counting of the free degrees of freedom n0 for i1nnd for j1nodof if nfij 0 nn1 nfijn end end end Internal Hinges Hinge onesnel 2 loading Jointloads zerosnnd 2 Jointloads220000 5e6 Elementloads zerosnel 4 Elementloads4 2e4 266666e7 2e4 266666e7 End of input 2013 by Taylor Francis Group LLC Beam Element 83 Results file PRINTING MODEL DATA Number of nodes 5 Number of elements 4 Number of nodes per element 2 Number of degrees of freedom per node 2 Number of degrees of freedom per element 4 Node X 1 000000 2 400000 3 1000000 4 1400000 5 2200000 Element Node1 Node2 1 1 2 2 2 3 3 3 4 4 4 5 Element E I 1 200000 2e008 2 200000 2e008 3 200000 2e008 4 200000 2e008 Nodal freedom Node dispw Rotation 1 0 0 2 1 2 3 0 3 4 4 5 5 6 7 Applied Nodal Loads Node loadY Moment 1 000000 000000 2 2000000 500000000 3 000000 000000 4 2000000 2666660000 5 2000000 2666660000 Total number of active degrees of freedom n 7 PRINTING ANALYSIS RESULTS Global force vector F 20000 5e006 0 20000 266666e007 20000 266666e007 2013 by Taylor Francis Group LLC 84 Introduction to Finite Element Analysis Using MATLAB and Abaqus Displacement solution vector delta 1557600 000561 001870 12813333 004270 53373339 005337 Nodal displacements Node dispy rotation 1 000000 000000 2 1557600 000561 3 000000 001870 4 12813333 004270 5 53373339 005337 Members actions element fy1 M1 Fy2 M2 1 3264000 12140000000 3264000 916000000 2 5264000 416000000 5264000 32000000000 3 4000000 32000000000 4000000 16000000000 4 4000000 16000000000 000 000 362 PROBLEM 32 FIGURE 312 Input file File Beamproblem2datam The following variables are declared as global in order to be used by all the functions Mfiles constituting the program global nnd nel nne nodof eldof n geom connec prop nf Elementloads Jointloads Hinge 75 75 4375 72916 10416 1875 375 375 Equivalent nodal loads 2 m 20 kN 5 kNm 2 m 3 m 25 m 1 1 2 2 3 3 4 4 5 E200106 kNm2 I600106 m4 E 200 106 kNm2 I 300 106 m4 Element 3 Element 4 FIGURE 312 Problem 32 and equivalent nodal loads for elements 3 and 4 2013 by Taylor Francis Group LLC Beam Element 85 format short e Beginning of data input nnd 5 Number of nodes nel 4 Number of elements nne 2 Number of nodes per element nodof 2 Number of degrees of freedom per node eldof nnenodof Number of degrees of freedom per element Nodes coordinates X and Y geomzerosnnd1 geom 0 X coord node 1 2 X coord node 2 4 X coord node 3 7 X coord node 4 95 X coord node 5 Element connectivity conneczerosnel2 connec 1 2 1st and 2nd node of element 1 2 3 1st and 2nd node of element 2 3 4 1st and 2nd node of element 3 4 5 1st and 2nd node of element 4 Geometrical properties prop11 E prop12 I propzerosnel2 prop 200e6 600e6 E and I of element 1 200e6 600e6 E and I of element 2 200e6 300e6 E and I of element 3 200e6 300e6 E and I of element 4 Boundary conditions nf onesnnd nodof Initialize the matrix nf to 1 nf11 0 Prescribed nodal freedom of node 1 nf41 0 Prescribed nodal freedom of node 3 Counting of the free degrees of freedom n0 for i1nnd for j1nodof if nfij 0 nn1 nfijn end end end Internal Hinges Hinge onesnel 2 loading Jointloads zerosnnd 2 Jointloads220 0 Elementloads zerosnel 4 Elementloads3 75 375 75 375 2013 by Taylor Francis Group LLC 86 Introduction to Finite Element Analysis Using MATLAB and Abaqus Elementloads4 4375 72916 1875 10416 End of input Results file PRINTING MODEL DATA Number of nodes 5 Number of elements 4 Number of nodes per element 2 Number of degrees of freedom per node 2 Number of degrees of freedom per element 4 Node X 1 000000 2 000200 3 000400 4 000700 5 000950 Element Node1 Node2 1 1 2 2 2 3 3 3 4 4 4 5 Element E I 1 2e008 00006 2 2e008 00006 3 2e008 00003 4 2e008 00003 Nodal freedom Node dispw Rotation 1 0 1 2 2 3 3 4 5 4 0 6 5 7 8 Applied Nodal Loads Node loadY Moment 1 000000 000000 2 02000 000000 3 00750 00375 4 000000 00354 5 00188 000104 Total number of active degrees of freedom n 8 PRINTING ANALYSIS RESULTS 2013 by Taylor Francis Group LLC Beam Element 87 Global force vector F 0 20 0 75 375 35416 1875 10416 Displacement solution vector delta 000065 000113 000039 000142 000008 000058 000135 000053 Nodal displacements Node dispy rotation 1 000000 000065 2 000113 000039 3 000142 000008 4 000000 000058 5 000135 000053 Members actions element fy1 M1 Fy2 M2 1 1594 000 1594 3188 2 406 3187 406 2375 3 406 2375 1906 1094 4 625 1094 000 000 363 PROBLEM 33 FIGURE 313 Input file File beamproblem3datam The following variables are declared as global in order to be used by all the functions Mfiles constituting the program global nnd nel nne nodof eldof n geom connec prop nf Elementloads Jointloads Hinge 4 m 4 m 6 m 4 3 3 2 2 1 1 12 kN Internal hinge E 200 106 kNm2 I 600106 m4 FIGURE 313 Problem 33 2013 by Taylor Francis Group LLC 88 Introduction to Finite Element Analysis Using MATLAB and Abaqus format short e Beginning of data input nnd 4 Number of nodes nel 3 Number of elements nne 2 Number of nodes per element nodof 2 Number of degrees of freedom per node eldof nnenodof Number of degrees of freedom per element Nodes coordinates X and Y geomzerosnnd1 geom 0 X coord node 1 4 X coord node 2 8 X coord node 3 14 X coord node 4 Element connectivity conneczerosnel2 connec 1 2 1st and 2nd node of element 1 2 3 1st and 2nd node of element 2 3 4 1st and 2nd node of element 3 Geometrical properties prop11 E prop12 I propzerosnel2 prop 200e6 600e6 E and I of element 1 200e6 600e6 E and I of element 2 200e6 600e6 E and I of element 3 Boundary conditions nf onesnnd nodof Initialize the matrix nf to 1 nf11 0 nf120 Prescribed nodal freedom of node 1 nf41 0 nf420 Prescribed nodal freedom of node 4 Counting of the free degrees of freedom n0 for i1nnd for j1nodof if nfij 0 nn1 nfijn end end end Internal Hinges Hinge onesnel 2 Hinge22 0 loading Jointloads zerosnnd 2 Jointloads212 0 Elementloads zerosnel 4 End of input 2013 by Taylor Francis Group LLC Beam Element 89 Results file PRINTING MODEL DATA Number of nodes 4 Number of elements 3 Number of nodes per element 2 Number of degrees of freedom per node 2 Number of degrees of freedom per element 4 Node X 1 000000 2 000400 3 000800 4 001400 Element Node1 Node2 1 1 2 2 2 3 3 3 4 Element E I 1 2e008 00006 2 2e008 00006 3 2e008 00006 Nodal freedom Node dispw Rotation 1 0 0 2 1 2 3 3 4 4 0 0 Applied Nodal Loads Node loadY Moment 1 000000 000000 2 01200 000000 3 000000 000000 4 000000 000000 Total number of active degrees of freedom n 4 PRINTING ANALYSIS RESULTS Global force vector F 12 0 0 0 Displacement solution vector delta 000096 000027 2013 by Taylor Francis Group LLC 90 Introduction to Finite Element Analysis Using MATLAB and Abaqus 000158 000040 Nodal displacements Node dispy rotation 1 000000 000000 2 000096 000027 3 000158 000040 4 000000 000000 Members actions element fy1 M1 Fy2 M2 1 936 2690 936 1055 2 264 1055 264 000 3 264 000 264 1582 37 ANALYSIS OF A SIMPLE BEAM WITH ABAQUS 371 INTERACTIVE EDITION In this section we will analyze the continuous beam shown in Figure 314 with the Abaqus interactive edition The cross section of the beam is shown in Figure 315 The material is steel with an elastic modulus of 200 GPa 4 m 5 m 2 m 20 kN 4 kNm 10 kNm 7 m FIGURE 314 Continuous beam 172 13 333 359 8 FIGURE 315 Beam cross section dimensions are in mm 2013 by Taylor Francis Group LLC Beam Element 91 Start Abaqus CAE Click on Create Model Database On the main menu click on File and Set Work Directory to choose your work ing directory Click on Save As and name the file Beamcae On the lefthandside menu click on Part to begin creating the model Name the part BeamPart check 2D Planar and check Deformable in the type Choose Wire as the base feature Enter an approximate size of 20 m and click on Continue Figure 316 FIGURE 316 Creating the BeamPart In the sketcher menu choose the CreateLines Connected icon to begin drawing the geometry of the beam Click on Done in the bottomleft corner of the viewport win dow Figure 317 FIGURE 317 Drawing using the connected line icon Under the model tree click on material to create a material and name it Steel Click on Mechanical then Elasticity and Elastic Enter 200e6 kNm2 for the elastic modulus and 03 for Poissons ratio Figure 318 FIGURE 318 Material definition 2013 by Taylor Francis Group LLC 92 Introduction to Finite Element Analysis Using MATLAB and Abaqus Under the model tree click on Profiles to create a profile and name it BeamProfile Click on Continue Figure 319 FIGURE 319 Creating a beam profile In the Edit Profile dialog box enter the dimensions of the profile Make sure you enter them in meters to keep a con sistent set of units Click on OK Figure 320 FIGURE 320 Entering the dimensions of a profile 2013 by Taylor Francis Group LLC Beam Element 93 Under the model tree click on Sections to create a section and name it Beamsection In the Category check Beam and in the Type choose Beam Click on Continue Figure 321 FIGURE 321 Creating a section In the Edit Section dialog box in the Profile name choose BeamProfile and in Material choose Steel Leave the Poissons ratio as zero Click on OK Figure 322 FIGURE 322 Editing a beam section 2013 by Taylor Francis Group LLC 94 Introduction to Finite Element Analysis Using MATLAB and Abaqus Expand the menu under BeamPart and double click on Section Assignments With the mouse select the whole beam in the drawing area and click on Done in the left bottom corner In the Edit Section Assign ments dialog box make sure that Beamsection appears in the section Click on OK Figure 323 FIGURE 323 Editing section assignments In Abaqus a beam element must have an orientation in space The default orientation is the one shown in Figure 324 The axis n1 is in opposite direction to the global axis Z For beams in a plane the n1 direction is always 00 00 10 that is normal to the plane in which the motion occurs Therefore pla nar beams can bend only about the first beamsection axis Y X Z n2 n1 t FIGURE 324 Beam orientation Change the Module to Prop erty Click on the Assign Beam Orientation icon and select the entire geometry from the viewport In the prompt in the leftbottom corner of the viewport accept 00 00 10 as the direction for n1 and click Return Click OK to confirm Figure 325 FIGURE 325 Assigning beam orientation 2013 by Taylor Francis Group LLC Beam Element 95 In the menu bar select View then Part Display Options In the Part Display Options in Idealizations check Ren der beam profiles Click Apply Figure 326 FIGURE 326 Rendering beam profile Using the Rotate View icon you can rotate the beam to appear as shown in Figure 327 If you are happy with what you see go back and uncheck Render beam profiles FIGURE 327 Rendered beam 2013 by Taylor Francis Group LLC 96 Introduction to Finite Element Analysis Using MATLAB and Abaqus In the model tree double click on Mesh under the BeamPart and in the main menu under Mesh click on Element Type With the mouse highlight all members in the viewport and select Done In the dialog box select Standard for element type Linear for geometric order and beam for family The name of the element B21 and its description are given below the element controls Click on OK Figure 328 FIGURE 328 Selecting a beam element In the main menu under Seed click on Edge by size With the mouse highlight all the beam in the viewport In the prompt area of the viewport enter 10 that is each ele ment will have a length of 1 m Click on Return then click Done Figure 329 FIGURE 329 Seeding a mesh by size 2013 by Taylor Francis Group LLC Beam Element 97 In the main menu under Mesh click on Part In the prompt area of the viewport click on Yes In the menu bar select View then Part Dis playOptions In thePartDis play Options under Mesh check Show node labels and Show element labels Click Apply The element and node labels will appear in the viewport Figure 330 FIGURE 330 Node and element labels In the model tree under BeamPart double click on Sets In the dialog box name the set FixedSupport check Node in type and click on Continue With the mouse highlight node 1 which is the fixed support and click on Done in the prompt area of the viewport Figure 331 FIGURE 331 Creating a node set 2013 by Taylor Francis Group LLC 98 Introduction to Finite Element Analysis Using MATLAB and Abaqus FIGURE 332 Selecting multiple nodes FIGURE 333 Creating element sets Again double click on Sets In the dialog box name the set RollerSupports check Node in type and click on Continue While keeping the SHIFT key down with the mouse highlight nodes 2 3 and 4 When selected they change color as shown in Figure 332 Click on Done in the prompt area of the viewport Again double click on Sets Name the set LoadedNode check Node in type and click on Continue With the mouse highlight node 6 Click on Done in the prompt area of the viewport Next create two element sets one for the elements subject to the 4 kNm load and the other for the elements subject to 10 kNm Double click on Sets In the dialog box name the set UDL4 check Element in type and click on Continue While keeping the SHIFT key down with the mouse highlight elements 5 6 7 8 and 9 When selected they change color as shown in Figure 333 Click on Done in the prompt area of the viewport Create another element set named UDL10 and select elements 10 to 16 In the model tree expand the Assembly and double click on Instances Select Dependent for the instance type and click OK In the model tree expand Steps and Initial and double click on BC Name the boundary con dition fixed select Dis placementRotation for the type and click on Continue In the rightbottom corner of the viewport you can see Sets Figure 334 Double click on it FIGURE 334 Imposing BC using created sets 2013 by Taylor Francis Group LLC Beam Element 99 In the dialog box that appears select BeamPart1 FixedSupport and check Highlight selections in viewport Click on Continue Figure 335 FIGURE 335 Selecting a node set for boundary conditions Fill up the Edit Boundary Conditions in the dialog box as shown by restricting all the degrees of freedom Click on OK Figure 336 FIGURE 336 Editing boundary conditions 2013 by Taylor Francis Group LLC 100 Introduction to Finite Element Analysis Using MATLAB and Abaqus Click on BC again Name the boundary condition Rollers select Displacement Rotation for the type and click on Continue Double click on Sets Select BeamPart 1RollerSupports Fill up the Edit Boundary Conditions by restrict ing only U2 Click on OK Figure 337 FIGURE 337 Imposing BC using created sets In the model tree double click on Steps Name the step ApplyLoads Set the procedure to General and select Static General Click on Continue Give the step a description and click OK In the model tree under steps and under ApplyLoads click on Loads Name the load Concentrated load and select Concentrated force as the type Click on Continue In the Region Selection dia log box select BeamPart 1Loadednode Click on Continue In the Edit Load dialog box enter 20 for CF2 Click OK Figure 338 FIGURE 338 Imposing a concentrated load using a node set 2013 by Taylor Francis Group LLC Beam Element 101 Click on Loads again Name the load UDL4 and select Line load as the type Click on Continue In the Region Selection dialog box select BeamPart1UDL4 Click on Continue In the Edit Load dialog box enter 4 for Component 2 Click OK Repeat the procedure again to create the 10 kNm distributed load over element set BeamPart1UDL10 Figure 339 FIGURE 339 Imposing a line load on an element set In the model tree expand the Field Output Requests and then double click on F Output1 FOutput1 is the default and is automatically generated when creating the step Uncheck the variables Contact and select any other variable you wish to add to the field output Click on OK Figure 340 FIGURE 340 Field output Under Analysis right click on Jobs and then click on Create In the Create Job dialog box name the job BEAMProblem and click on Continue In the Edit Job dialog box enter a description for the job Check Full analysis select to run the job in Background and check to start it immediately Click OK Expand the tree under Jobs right click on BEAMProblem Then click on Submit If you get the following message BEAMProblem completed successfully in the bottom window then your job is free of errors and was executed properly Figure 341 FIGURE 341 Submitting a job in Abaqus CAE 2013 by Taylor Francis Group LLC 102 Introduction to Finite Element Analysis Using MATLAB and Abaqus FIGURE 342 Plotting stresses in the bottom fiber Under the top menu in the Module scroll to Visualization and click to load Abaqus Viewer On the main menu under File click Open navigate to your working directory and open the file BeamProblemodb It should have the same name as the job you submitted Click on the Common options icon to display the Common Plot options dialog box Under labels check Show Element labels and Show Node labels to display elements and nodes numbering Click on the icon Plot Deformed Shape to display the deformed shape of the beam On the main menu click on Results then on Field Output to open the Field Output dialog box Choose S Stress components at integration points For component choose S11 to plot the stresses in the bars Figure 342 Click on Section points to open the section point dialog box Check bottom to plot the stresses in the lower fiber or Top for the stresses in the top fiber In the menu bar click on Report and Field Output In the Report Field Output dialog box for Position select Unique nodal check RF2 and RM3 for RF Reaction force and check U2 and UR3 for U Spatial displacement Then click on Set up Click on Select to navigate to your working directory Name the file BeamProblemrpt Uncheck Append to file and click OK Use your favorite text editor and open the file BeamProblemrpt which should be the same as the one listed next Field Output Report written Mon Apr 11 115508 2011 Source 1 ODB CAbaqusWorking DirectoryBeamProblemodb Step ApplyLoads Frame Increment 1 Step Time 1000 Loc 1 Nodal values from source 1 Output sorted by column Node Label Field Output reported at nodes for part BEAMPART1 Node RFRF2 RM3 UU2 UR3 Label Loc 1 Loc 1 Loc 1 Loc 1 1 137126 146329 137126E36 146329E36 2 835087 0 835087E36 352684E06 3 585745 0 565745E36 148255E03 4 293621 0 243621E36 299728E03 5 0 0 232859E06 245559E06 2013 by Taylor Francis Group LLC Beam Element 103 6 0 0 494778E06 581202E06 7 0 0 35009E06 24655E06 8 0 0 356109E06 360552E06 9 0 0 72162E06 307269E06 10 0 0 972231E06 66529E06 11 0 0 85733E06 487975E06 12 0 0 21104E03 216607E03 13 0 0 434032E03 188215E03 14 0 0 588021E03 94655E06 15 0 0 623626E03 324961E06 16 0 0 523046E03 161662E03 17 0 0 300053E03 261264E03 Minimum 0 0 623626E03 216607E03 At Node 17 17 15 12 Maximum 585745 146329 972231E06 299728E03 At Node 3 1 10 4 Total 110000 146329 249686E03 161611E03 372 ANALYSIS OF A BEAM WITH ABAQUS KEYWORD EDITION In this section we will prepare an input file for the beam shown in Figures 314 and 315 We will use the same number of elements and nodes as earlier The file is named BeamProblemKeywordinp and is listed next Heading BeamProblem Model keyword edition Preprint echoNo modelNO historyNO Define the end nodes Node 1 9 0 17 6 0 Generate the remaining nodes Ngen 1171 Define element 1 Element typeB21 112 Generate the elements Elgen elset allelements 116 1 1 Nset nsetFixedsupport 1 Nset nsetRollersupports 5 10 17 Nset nsetLoadednode 3 Elset elsetUDL4 generate 5 9 1 2013 by Taylor Francis Group LLC 104 Introduction to Finite Element Analysis Using MATLAB and Abaqus Elset elsetUDL10 generate 10 16 1 Section Beamsection Profile BeamProfile Beam Section elsetallelements materialSteel sectionI 01795 0359 0172 0172 0013 0013 0008 001 MATERIALS Material nameSteel Elastic 2e08 03 BOUNDARY CONDITIONS Boundary Fixedsupport encastre Rollersupports 2 2 STEP ApplyLoads Step nameApplyLoads Static 1 1 1e05 1 LOADS Cload Loadednode 2 20 Dload UDL4 PY 4 UDL10 PY 10 OUTPUT REQUESTS Output field Node Output CF RF RM U Element Output S Output history variablePRESELECT End Step The file starts with the keyword HEADING which in this case is entered as BeamProblem Model keyword edition Using the keyword node we define the two extreme nodes 1 and 17 and give their coordinates x and y Using the keyword ngen which stands for node generate we generate all the remaining nodes from 1 to 17 in an increment of 1 Using the keyword Element typeB21 representing a beam element in the plane In the data line we enter 1 as the element number with nodes 1 and 2 all separated by Next we generate the elements using the keyword elgen We group the elements in a set named allelements In the data line we enter the master element that has been previously defined that is element 1 then the number of elements to be generated 16 followed by the increment in node numbers of corresponding nodes from element to element which in this case is 1 then the increment in element numbers which is again 1 2013 by Taylor Francis Group LLC Beam Element 105 Once all the elements and nodes are defined using the keyword nset we create the following node sets Fixedsupport which contains node 1 Rollersupports which contains nodes 5 10 and 17 and Loadednode which contains node 3 Next with the keyword elset and the parameter generate we create element sets UDL4 and UDL10 containing respectively elements 5 to 9 and 10 to 16 When the parameter generate is included each data line should give a first element a last element and the increment in element numbers between these elements If it is not included then all the elements forming the set must be listed in the data lines With the keyword Beam Section we define a section for the elements contained in the set allelements the material is Steel and the section is the form of I In the first data line we enter the dimensions of the section and in the second its orientation with respect to the global coordinates Using the keyword Material we create a material named Steel The material is elastic and its properties are given in the data line of the keyword elastic Using the created node sets we impose the boundary conditions with the keyword Bound ary We fully fix the node set Fixedsupport by using encastre All the nodes in the node set Rollersupports are fixed in the direction 2 Next using the keyword step we create a step named ApplyLoads The keyword static indicates that it will be a general static analysis Using the keyword cload we apply a concentrated load of 20 kN in the direction 2 to the node in node set Loadednode Using the keyword dload for distributed load we apply line loads of 4 and 10 kNm to the elements contained respectively in element sets UDL4 and UDL10 Using the keywords Output field and Node Output we request the nodal variables CF concentrated force RF reaction force RM reaction moment and displacements U to be written to the database file odb With Element Output we also add the stresses S to the database file Output history variable PRESELECT requests the default variables for history output Finally we end the step and the file with End Step At the command line type Abaqus jobBeamProblemKeyword inter followed by Return If you get an error open the file with extension dat to see what type of error To load the visualization model type Abaqus Viewer at the command line 2013 by Taylor Francis Group LLC 2013 by Taylor Francis Group LLC 4 Rigid Jointed Frames 41 INTRODUCTION Rigid jointed frames are often used in buildings They resist the combined effects of horizontal and vertical loads They derive their strength from the moment interactions between the beams and the columns at the rigid joints As a result the elements are subjected not only to bending but also to axial force Such elements are referred to as beamcolumn elements Their nodal displacements include both translations and rotation u v θ as shown in Figure 41 In total there are six degrees of freedom de u1 v1 θ1 u2 v2 θ2T 41 corresponding to six nodal loads Fe Fx1 Fy1 M1 Fx2 Fy2 M2T 42 42 STIFFNESS MATRIX OF A BEAMCOLUMN ELEMENT If we assume that the deformations are infinitesimally small and the material is linear elastic then the axial displacements of the beamcolumn element do not interact with the bending deformations Consequently the principle of superposition applies and the displacements forces and stiffness matrix of the beamcolumn element can be obtained by simply adding the respective matrices of a truss element Equation 210 and that of a beam element Equation 330 Ke AEL 0 0 AEL 0 0 0 12EIL3 6EIL2 0 12EIL3 6EIL2 0 6EIL2 4EIL 0 6EIL2 2EIL AEL 0 0 AEL 0 0 0 12EIL3 6EIL2 0 12EIL3 6EIL2 0 6EIL2 2EIL 0 6EIL2 4EIL 43 43 STIFFNESS MATRIX OF A BEAMCOLUMN ELEMENT IN THE PRESENCE OF HINGED END Sometimes a designer may specify an internal hinge in a frame which results in a zero value for the bending moment To account for the presence of a hinge the stiffness matrix can be obtained by superimposing the respective matrices of a truss element Equation 210 and that of a beam element with a hinge at its right end Equation 352 or a hinge at its left end Equation 353 107 2013 by Taylor Francis Group LLC 108 Introduction to Finite Element Analysis Using MATLAB and Abaqus u1 u2 v1 Fy1 Fy2 Fx2 Fx1 M2 v2 Y X y y Displacements E A I θ Forces x x θ1 θ2 M1 FIGURE 41 Beam column element with six degrees of freedom It follows that the stiffness matrix of a beamcolumn element with a hinge at its right end is given as Ke AEL 0 0 AEL 0 0 0 3EIL3 3EIL2 0 3EIL3 0 0 3EIL2 3EIL 0 3EIL2 0 AEL 0 0 AEL 0 0 0 3EIL3 3EIL2 0 3EIL3 0 0 0 0 0 0 0 44 and with a hinge at its left end as Ke AEL 0 0 AEL 0 0 0 3EIL3 0 0 3EIL3 3EIL2 0 0 0 0 0 0 AEL 0 0 AEL 0 0 0 3EIL3 0 0 3EIL3 3EIL2 0 3EIL2 0 0 3EIL2 3EIL 45 As with a beam system a hinge should be associated only with one element 44 GLOBAL AND LOCAL COORDINATE SYSTEMS Like for a truss member beamcolumn or frame elements do not all have the same orientation in space Similarly when it comes to assembling the global stiffness we need to have the element degrees of freedom nodal displacements given in terms of the common reference axes of the 2013 by Taylor Francis Group LLC Rigid Jointed Frames 109 frame The transformation is similar to that of a bar element except that the transformation matrix is given as cos θ sin θ 0 0 0 0 sin θ cos θ 0 0 0 0 0 0 1 0 0 0 0 0 0 cos θ sin θ 0 0 0 0 sin θ cos θ 0 0 0 0 0 0 1 46 The transformation is carried out as follows Ke CKeCT 47 where Ke represents the element stiffness matrix in the global coordinate system 45 GLOBAL STIFFNESS MATRIX ASSEMBLY AND SOLUTION FOR UNKNOWN DISPLACEMENTS The assembly of the global stiffness matrix is similar to that of a truss detailed in Section 234 except that a beam column element has six degrees of freedom The introduction of the boundary conditions also follows the same principle The only difference is that a node possesses three degrees of freedom two translations and a rotation Any of these degrees of freedom can be free or restrained When all the degrees of freedom at a node are restrained the node is sometimes referred to as encastré Distributed loads along a beamcolumn element are also treated in the same fashion as for a beam element Section 33 46 COMPUTER CODE framem Writing a MATLAB code for the analysis of a frame is merely a combination of the codes previously written for a truss and a beam structure The only differences reside in the matrices dimensions Similar to a truss structure the elements stiffness matrices need to be transformed from local to global coordinates Likewise to a beam structure each beamcolumn element will have two types of loading one that consists of the external forces directly applied to the nodes and another that only consists of the statically equivalent nodal loads Therefore in the development of the program framem we will borrow the same style as that used in the programs trussm and beamm Let us consider the portal frame shown in Figure 42 461 DATA PREPARATION To read the data we will use the Mfile frameproblem1datam Again we will use a consistent set of units mm for length and N for force The input data for this beam consist of nnd 5 number of nodes nel 4 number of elements nne 2 number of nodes per element nodof 3 number of degrees of freedom per node Note that a beamcolumn element has three degrees of freedom per node nodof 3 2013 by Taylor Francis Group LLC 110 Introduction to Finite Element Analysis Using MATLAB and Abaqus 2 5 m 1 m 2 12 kNm 12 kNm 3 All the elements have the same cross section second moment of inertia and made from the same material A5210 mm2 I864e6 mm4 E 200 GPa 3 1 12 m 5 4 4 1 FIGURE 42 Example 1 Portal frame 4611 Nodes Coordinates The coordinates x and y of the nodes are given in the form of a matrix geomnnd 2 geom 0 0 0 5000 6000 6000 12000 5000 12000 0 4612 Element Connectivity The element connectivity is given in the matrix connecnel 2 connec 1 2 2 3 3 4 4 5 4613 Material and Geometrical Properties The material and geometrical properties are given in the matrix propnel 3 The first column represents the Youngs modulus the second represents the crosssectional area and the third the second moment of inertia of the cross section prop 200000 5310 864e 6 200000 5210 864e 6 200000 5210 864e 6 4614 Boundary Conditions In the same fashion as for the truss and the beam a restrained degree of freedom is assigned the digit 0 while a free degree of freedom is assigned the digit 1 As previously explained a node in 2013 by Taylor Francis Group LLC Rigid Jointed Frames 111 a beamcolumn element has three degrees of freedom a horizontal translation along the axis X a vertical translation along the axis Y and a rotation around the axis Z perpendicular to the plan XY As shown in Figure 42 nodes 1 and 5 are fully fixed encastré Their degrees of freedom are all assigned the digit 0 Nodes 2 3 and 4 are free Their degrees of freedom u v and θ are assigned the digit 1 The information on the boundary conditions is given in the matrix nfnnd nodof nf 0 0 0 1 1 1 1 1 1 1 1 1 0 0 0 The free degrees of freedom different from zero are then counted and their rank assigned back into the matrix nfnnd nodof nf 0 0 0 1 2 3 4 5 6 7 8 9 0 0 0 In this case the total number of active degrees of freedom is obtained as n 9 4615 Internal Hinges To account for internal hinges we create a vector Hingenel 2 that we initialize to 1 If a particular element k has a hinge at its left end then we assign it the digit 0 at the position of its first node that is Hingek 1 0 On the other hand the hinge may be accounted for with the element j having it at its right In that case we assign it the digit 0 at the position of its second node that is Hingej 2 0 A hinge must be considered for one element only 4616 Loading A beamcolumn element can have two types of loading loads applied directly at the nodes and statically equivalent nodal loads A good computer code should cater for both loadings To dis tinguish between the two loading systems we will use two matrices Jointloadsnnd 3 and Elementloadsnel 4 There are no loads applied directly at the nodes Therefore the matrix Jointloadsnnd 3 is empty Jointloads 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 2013 by Taylor Francis Group LLC Elements 2 and 3 have loads applied along their length which need to be transformed to statically equivalent nodal loads as shown in Figure 34 The MATLAB script for evaluating the matrix kl is given in the Mfile beamcolumnkm 114 Introduction to Finite Element Analysis Using MATLAB and Abaqus 464 SOLUTION OF THE GLOBAL SYSTEM OF EQUATIONS The solution of the global system of equations is obtained with one statement delta KKF The backslash symbol is used to divide a matrix by a vector 465 NODAL DISPLACEMENTS Once the global displacements vector delta is obtained it is possible to retrieve any nodal displace ments A loop is carried over all the nodes If a degree of freedom j of a node i is free that is nfi j 0 then it could have a displacement different from zero The value of the displacement is extracted from the global displacements vector delta nodedispi j deltanfi j 466 ELEMENT FORCES To obtain the member forces a loop is carried over all the elements 1 Form element stiffness matrix kl in local xy 2 Form element transformation matrix C 3 Transform the element matrix from local to global coordinates kg C kl CT 4 Form element steering vector g a Loop over the degrees of freedom of the element to obtain element displacements vector edg in global coordinates b If gj 0 then the degree of freedom is restrained edgj 0 c Otherwise edgj deltagj 5 Obtain element force vector in global XY coordinates fg kg edg 6 Transform element force vector to local coordinates fl CT fg 7 Retrieve the element statically equivalents loads f0 Elementloadsi if any 8 Obtain the elements internal forces as forcei fl f0 The results of the analysis are written to the file frameproblem1resultstxt using the Mfile printframeresultsm A copy of the file frameproblem1resultstxt is included within Fileframem PROGRAM framem LINEAR STATIC ANALYSIS OF A RIGID JOINTED FRAME Make these variables global so they can be shared by other functions clc clear all global nnd nel nne nodof eldof n geom connec F global prop nf Elementloads Jointloads force Hinge format short e dispExecuting framem Open file for output of results 2013 by Taylor Francis Group LLC Rigid Jointed Frames 115 ALTER NEXT LINES TO CHOOSE OUTPUT FILES fid fopenframeproblem1resultstxtw dispResults printed to file frameproblem1resultstxt Beginning of data input frameproblem1data Load the input file F zerosn1 Initialize global force vector to zero F AssemJointLoadsF Assemble joint loads to global force vector printframemodel Print model data KK zerosn n Initialize the global stiffness matrix to zero for i1nel klbeamcolumnki Form element matrix in local xy C beamcolumnCi Form transformation matrix kgCklC Transform the element matrix from local to global coordinates fl Elementloadsi Retrieve element equivalent nodal forces in local xy fgCfl Transform the element force vector from local to global coordinates gbeamcolumngi Retrieve the element degrees of freedom KK formkkKK kg g assemble global stiffness matrix F AssemElemloadsF fg g assemble global force vector end End of assembly delta KKF solve for unknown displacements Extract nodal displacements for i1nnd for j1nodof nodedispij 0 if nfij 0 nodedispij deltanfij end end end for i1nel klbeamcolumnki Form element matrix in local xy C beamcolumnCi Form transformation matrix kgCklC Transform the element matrix from local to global coordinates gbeamcolumngi Retrieve the element degrees of freedom for j1eldof if gj 0 edgj0 displacement 0 for restrained freedom else edgj deltagj end end fg kgedg Element force vector in global XY fl Cfg Element force vector in local xy f0 Elementloadsi Equivalent nodal loads 2013 by Taylor Francis Group LLC 116 Introduction to Finite Element Analysis Using MATLAB and Abaqus forceli flf0 forcegi Cflf0 end printframeresults fclosefid Fileframeproblem1datam File frameproblem1datam The following variables are declared as global in order to be used by all the functions Mfiles constituting the program global nnd nel nne nodof eldof n geom connec F prop nf Elementloads Jointloads force Hinge format short e nnd 5 Number of nodes nel 4 Number of elements nne 2 Number of nodes per element nodof 3 Number of degrees of freedom per node eldof nnenodof Number of degrees of freedom per element Nodes coordinates x and y geomzerosnnd2 geom110 geom12 0 x and y coordinates of node 1 geom210 geom22 5000 x and y coordinates of node 2 geom316000 geom32 6000 x and y coordinates of node 3 geom4112000 geom42 5000 x and y coordinates of node 4 geom5112000 geom52 0 x and y coordinates of node 4 Element connectivity conneczerosnel2 connec11 1 connec12 2 First and second node of element 1 connec21 2 connec22 3 First and second node of element 2 connec31 3 connec32 4 First and second node of element 3 connec41 4 connec42 5 First and second node of element 4 Geometrical properties propzerosnel3 prop1120e5 prop125210 prop13864e6 EA and I element 1 prop2120e5 prop225210 prop23864e6 EA and I element 2 prop3120e5 prop325210 prop33864e6 EA and I element 3 prop4120e5 prop425210 prop43864e6 EA and I element 4 Boundary conditions nf onesnnd nodof Initialize the matrix nf to 1 nf11 0 nf12 0 nf13 0 Prescribed nodal freedom of node 1 nf51 0 nf52 0 nf53 0 Prescribed nodal freedom of node 5 Counting of the free degrees of freedom n0 for i1nnd for j1nodof if nfij 0 nn1 nfijn 2013 by Taylor Francis Group LLC Rigid Jointed Frames 117 end end end Internal Hinges Hinge onesnel2 loading Jointloads zerosnnd 3 Joint loads are usually entered in global coordinates Enter here the forces in X and Y directions and any concentrated moment at node i Staticaly equivalent loads are entered in local coordinates of the element Elementloads zerosnel 6 Elementloads2 0 364965e3 37e6 0 364965e3 37e6 Elementloads3 0 364965e3 37e6 0 364965e3 37e6 End of input Fileframeproblem1resultstxt PRINTING MODEL DATA Number of nodes 5 Number of elements 4 Number of nodes per element 2 Number of degrees of freedom per node 3 Number of degrees of freedom per element 6 Node X Y 1 000000 000000 2 000000 500000 3 600000 600000 4 1200000 500000 5 1200000 000000 Element Node1 Node2 1 1 2 2 2 3 3 3 4 4 4 5 Element E A I 1 200000 5210 864e007 2 200000 5210 864e007 3 200000 5210 864e007 4 200000 5210 864e007 Nodal freedom Node dispu dispu Rotation 1 0 0 0 2 1 2 3 3 4 5 6 4 7 8 9 5 0 0 0 2013 by Taylor Francis Group LLC 118 Introduction to Finite Element Analysis Using MATLAB and Abaqus Applied joint Loads Node loadX loadY Moment 1 000000 000000 000000 2 599999 3599993 3700000000 3 1199998 000000 7400000000 4 599999 3599993 3700000000 5 000000 000000 000000 Total number of active degrees of freedom n 9 PRINTING ANALYSIS RESULTS Global force vector F 599999 359999 37e007 12000 0 74e007 599999 359999 37e007 Displacement solution vector delta 2503159 016363 000712 2504119 000000 000686 2503159 016363 000712 Nodal displacements Node dispx dispy rotation 1 000000e000 000000e000 000000e000 2 250316e001 163630e001 711912e003 3 250412e001 798515e015 685508e003 4 250316e001 163630e001 711912e003 5 000000e000 000000e000 000000e000 Members actions in local coordinates element fx1 fy1 M1 fx2 Fy2 M2 1 341005529 119999753 546036264780 341005529 119999753 53962499239 2 62306063 356093620 53962499239 62306063 373836380 00000 3 62306063 373836380 00000 62306063 356093620 53962499239 4 341005529 119999753 53962499239 341005529 119999753 546036264780 Members actions in global coordinates element fx1 fy1 M1 fx2 Fy2 M2 1 119999753 341005529 546036264780 119999753 341005529 53962499239 2 119999753 341005529 53962499239 00000 378992988 00000 3 00000 378992988 00000 119999753 341005529 53962499239 4 119999753 341005529 53962499239 119999753 341005529 546036264780 2013 by Taylor Francis Group LLC Rigid Jointed Frames 119 E 70e 6 Nm2 A 01 m2 I 1333e 3 m4 E 70e 6 Nm2 A 01 m2 I 1333e3 m4 E35e6 Nm2 A016 m2 I21333e 3 m4 E35e6 Nm2 A016 m2 I21333e 3 m4 3 m 20 kN 20 kN 20 kNm 3 m 3 m 5 m 5 m FIGURE 43 Frame with an internal hinge Figure 43 shows a twolevel frame with an internal hinge in the top beam The frame is made from two different materials The columns have an elastic modulus of 35e6 kNm2 a cross area of 016 m2 and second moment of inertia of 21333e 3 m4 The beams have an elastic modulus of 70e6 kNm2 a cross area of 01 m2 and second moment of inertia of 1333e 3 m4 In addition two concentrated loads are applied along the lower beam Instead of considering the lower beam as one element with the concentrated loads transformed into statically equivalent loads we will simply discretize the beam into three elements such that the two concentrated loads are applied at joints The finite element discretization is shown in Figure 44 2 2 3 3 4 4 5 5 6 6 7 7 8 8 9 9 1 1 FIGURE 44 Finite element discretization 2013 by Taylor Francis Group LLC 120 Introduction to Finite Element Analysis Using MATLAB and Abaqus Element 4 Element 6 q q L L 5qL 5qL 3qL 3qL qL2 8 qL2 8 8 8 8 8 FIGURE 45 Statically equivalent nodal loads Elements 4 and 6 are both subject to a uniformly distributed load that needs to be transformed to statically equivalent nodal loads However the elements are joined by a hinge at node 5 In such a case the statically equivalent nodal loads are the reactions of a propped cantilever Figure 45 Input File File frameproblem2datam The following variables are declared as global in order to be used by all the functions Mfiles constituting the program global nnd nel nne nodof eldof n geom connec F prop nf Elementloads Jointloads force Hinge format short e nnd 9 Number of nodes nel 9 Number of elements nne 2 Number of nodes per element nodof 3 Number of degrees of freedom per node eldof nnenodof Number of degrees of freedom per element Nodes coordinates x and y geomzerosnnd2 geom110 geom12 0 x and y coordinates of node 1 geom210 geom22 5 x and y coordinates of node 2 geom310 geom32 10 x and y coordinates of node 3 geom413 geom42 5 x and y coordinates of node 4 geom5145 geom52 10 x and y coordinates of node 5 geom616 geom62 5 x and y coordinates of node 6 geom719 geom72 10 x and y coordinates of node 7 geom819 geom82 5 x and y coordinates of node 8 geom919 geom92 0 x and y coordinates of node 9 Element connectivity 2013 by Taylor Francis Group LLC Rigid Jointed Frames 121 conneczerosnel2 connec11 1 connec12 2 First and second node of element 1 connec21 2 connec22 3 First and second node of element 2 connec31 2 connec32 4 First and second node of element 3 connec41 3 connec42 5 First and second node of element 4 connec51 4 connec52 6 First and second node of element 5 connec61 5 connec62 7 First and second node of element 6 connec71 6 connec72 8 First and second node of element 7 connec81 7 connec82 8 First and second node of element 8 connec91 8 connec92 9 First and second node of element 9 Geometrical properties propzerosnel3 prop1135e6 prop12016 prop1321333e3 EA and I of element 1 prop2135e6 prop22016 prop2321333e3 EA and I of element 2 prop3170e6 prop3201 prop3313333e3 EA and I of element 3 prop4170e6 prop4201 prop4313333e3 EA and I of element 4 prop5170e6 prop5201 prop5313333e3 EA and I of element 5 prop6170e6 prop6201 prop6313333e3 EA and I of element 6 prop7170e6 prop7201 prop7313333e3 EA and I of element 7 prop8135e6 prop82016 prop8321333e3 EA and I of element 8 prop9135e6 prop92016 prop9321333e3 EA and I of element 9 Boundary conditions nf onesnnd nodof Initialize the matrix nf to 1 nf11 0 nf12 0 nf13 0 Prescribed nodal freedom of node 1 nf91 0 nf92 0 nf93 0 Prescribed nodal freedom of node 9 Counting of the free degrees of freedom n0 for i1nnd for j1nodof if nfij 0 nn1 nfijn end end end Internal Hinges Hinge onesnel 2 Hinge42 0 Hinge accounted with element 4 loading Joint loads are usually entered in global coordinates Enter here the forces in X and Y directions and any concentrated moment at node i Jointloads zerosnnd 3 Jointloads40 20 0 Jointloads60 20 0 Staticaly equivalent loads are entered in local coordinates of the element Elementloads zerosnel 6 Elementloads4 0 5625 50625 0 3375 0 Elementloads6 0 3375 0 0 5625 50625 End of input 2013 by Taylor Francis Group LLC 122 Introduction to Finite Element Analysis Using MATLAB and Abaqus Results File PRINTING MODEL DATA Number of nodes 9 Number of elements 9 Number of nodes per element 2 Number of degrees of freedom per node 3 Number of degrees of freedom per element 6 Node X Y 1 000000 000000 2 000000 000500 3 000000 001000 4 000300 000500 5 000450 001000 6 000600 000500 7 000900 001000 8 000900 000500 9 000900 000000 Element Node1 Node2 1 1 2 2 2 3 3 2 4 4 3 5 5 4 6 6 5 7 7 6 8 8 7 8 9 8 9 Element E A I 1 35e007 016 00021333 2 35e007 016 00021333 3 7e007 01 00013333 4 7e007 01 00013333 5 7e007 01 00013333 6 7e007 01 00013333 7 7e007 01 00013333 8 35e007 016 00021333 9 35e007 016 00021333 Nodal freedom Node dispu dispu Rotation 1 0 0 0 2 1 2 3 3 4 5 6 4 7 8 9 5 10 11 12 6 13 14 15 7 16 17 18 8 19 20 21 9 0 0 0 Applied joint Loads Node loadX loadY Moment 1 000000 000000 000000 2 000000 000000 000000 3 000000 05625 05063 4 000000 02000 000000 2013 by Taylor Francis Group LLC Rigid Jointed Frames 123 5 000000 06750 000000 6 000000 02000 000000 7 000000 05625 005063 8 000000 000000 000000 9 000000 000000 000000 Total number of active degrees of freedom n 21 PRINTING ANALYSIS RESULTS Global force vector F 0 0 0 0 5625 50625 0 20 0 0 675 0 0 20 0 0 5625 50625 0 0 0 Displacement solution vector delta 000004 000010 000049 000004 000018 000366 000001 000008 000016 000000 002762 000732 000001 000008 000016 000004 000018 000366 000004 000010 000049 Nodal displacements Node dispx dispy rotation 1 000000e000 000000e000 000000e000 2013 by Taylor Francis Group LLC 124 Introduction to Finite Element Analysis Using MATLAB and Abaqus 2 415908e005 982143e005 489326e004 3 361455e005 178571e004 365810e003 4 138636e005 838724e005 158328e004 5 898622e018 276241e002 731947e003 6 138636e005 838724e005 158328e004 7 361455e005 178571e004 365810e003 8 415908e005 982143e005 489326e004 9 000000e000 000000e000 000000e000 Members actions in local coordinates element fx1 fy1 M1 fx2 Fy2 M2 1 1100000 84705 138690 1100000 84705 284833 2 900000 562264 786320 900000 562264 2025000 3 646969 200000 501487 646969 200000 98513 4 562264 900000 2025000 562264 00000 00000 5 646969 00000 98513 646969 00000 98513 6 562264 00000 00000 562264 900000 2025000 7 646969 200000 98513 646969 200000 501487 8 900000 562264 2025000 900000 562264 786320 9 1100000 84705 284833 1100000 84705 138690 Members actions in global coordinates element fx1 fy1 M1 fx2 Fy2 M2 1 84705 1100000 138690 84705 1100000 284833 2 562264 900000 786320 562264 900000 2025000 3 646969 200000 501487 646969 200000 98513 4 562264 900000 2025000 562264 00000 00000 5 646969 00000 98513 646969 00000 98513 6 562264 00000 00000 562264 900000 2025000 7 646969 200000 98513 646969 200000 501487 8 562264 900000 2025000 562264 900000 786320 9 84705 1100000 284833 84705 1100000 138690 47 ANALYSIS OF A SIMPLE FRAME WITH ABAQUS 471 INTERACTIVE EDITION In this section we will analyze the portal frame shown in Figure 46 with the Abaqus interactive edition The cross sections of the profiles used are shown in Figure 47 The material is steel with an elastic modulus of 200 GPa 15 m 6 m 14 m 12 kNm 12 kNm 15 kNm 10 kNm FIGURE 46 Portal frame 2013 by Taylor Francis Group LLC Rigid Jointed Frames 125 326 14 8 300 359 333 13 13 172 172 Columns Rafters FIGURE 47 Profiles sections dimensions are in mm Start Abaqus CAE Click on Create Model Database On the main menu click on File and Set Work Directory to choose your working directory Click on Save As and name the file Portalframecae On the lefthandside menu click on Part to begin creating the model Name the part PortalFrame check 2D Planar check Deformable in the type Choose Wire as the base feature Enter an approximate size of 20 m and click on Continue In the sketcher menu choose the CreateLines Connected icon to begin drawing the geometry of the frame Click on Done in the bottomleft corner of the viewport win dow Figure 48 FIGURE 48 Creating the Portalframe part 2013 by Taylor Francis Group LLC 126 Introduction to Finite Element Analysis Using MATLAB and Abaqus Under the model tree click on material to create a material and name it Steel Click on Mechanical then Elasticity and Elastic Enter 200e6 kNm2 for the elastic modulus and 03 for Poissons ratio Next click on Profiles to cre ate a profile and name it ColumnProfile Click on Continue Enter the dimen sions of the profile section Repeat the procedure to create another profile for the rafters which will be named RafterProfile Figure 49 FIGURE 49 Material and profiles definitions Under the model tree click on Sections to create a section and name it Columnsection In the Category check Beam and in the Type choose Beam Click on Continue In the Edit Section dialog box in the Profile name select ColumnProfile and in Material choose Steel Leave the Poissons ratio as zero Click on OK Repeat the procedure to create a section for the rafters named Columnsection using the profile RafterProfile Figure 410 FIGURE 410 Creating sections 2013 by Taylor Francis Group LLC Rigid Jointed Frames 127 Expand the menu under Parts and PortalFrame and dou ble click on Section Assign ments By keeping the Shift key down click on the columns in the viewport area Click on Done in the left bottom corner In the Edit Section Assignments dialog box select Columnsection and click on OK Repeat the procedure by selecting this time the rafters and in the Edit Section Assign ments dialog box select Raftersection Click on OK to finish Figure 411 FIGURE 411 Editing section assignments To check the beam orien tations change the Module to Property Click on the Assign Beam Orientation icon and select the entire geometry from the viewport In the prompt in the left bottom corner of the view port accept 00 00 10 as the direction for n1 and click Return Click OK to confirm Figure 412 FIGURE 412 Assigning beam orientation In the menu bar select View then Part Display Options In the Part Display Options in Idealizations check Ren der beam profiles Click Apply Figure 413 FIGURE 413 Rendering beam profile 2013 by Taylor Francis Group LLC 128 Introduction to Finite Element Analysis Using MATLAB and Abaqus In the model tree double click on Mesh under the PortalFrame and in the main menu under Mesh click on Element Type With the mouse highlight all members in the viewport and select Done In the dialog box select Standard for element type Linear for geometric order and beam for family Click on OK In the main menu under Seed click on Edges With the mouse high light all the frame in the view port In the dialog box select edge by number and enter 4 Click on Apply and on OK Under mesh click on Part and Yes in the prompt area Figure 414 FIGURE 414 Seeding by number In the menu bar select View then Part Display Options In the Part Display Options under Mesh check Show node labels and Show ele ment labels Click Apply The element and node labels will appear in the viewport Figure 415 FIGURE 415 Mesh 2013 by Taylor Francis Group LLC Rigid Jointed Frames 129 In the model tree under PortalFrame double click on Sets In the dia log box name the set PinnedSupports check Node in type and click on Continue With the mouse highlight the nodes forming the supports Click on Done in the prompt area of the viewport Make sure you select Element for Type repeat the procedure to create the following ele ment sets Leftcolumn Rightcolumn Columns LeftRafter RightRafter and Rafters Figure 416 FIGURE 416 Creating the element set Rafters In the model tree expand the Assembly and double click on Instances Select Portalframe for Parts and click OK In the model tree expand Steps and Initial and double click on BC Name the boundary condition Pinned select Symmetry AntisymmetryEncastre for the type and click on Continue In the right bottom corner of the viewport click on Sets and select PortalFrame 1Pinnedsupports In the Edit Boundary Condition select PINNED Click OK Figure 417 FIGURE 417 Imposing BC using created sets 2013 by Taylor Francis Group LLC 130 Introduction to Finite Element Analysis Using MATLAB and Abaqus In the model tree double click on Steps Name the step ApplyLoads Set the procedure to General and select Static General Click on Continue Give the step a description and click OK In the model tree under steps and under ApplyLoads click on Loads Name the load W15 and select Line load as the type Click on Continue In the Region Selection dialog box select PortalFrame1LeftColumn Click on Continue In the Edit Load dialog box select Global for System and enter 15 for Component 1 Click OK Repeat exactly the same proce dure for the right column name the load W10 and enter 10 for the magnitude Figure 418 FIGURE 418 Imposing a line load in global coordi nates In the model tree under steps and under ApplyLoads click on Loads Name the load DOWN12 and select Line load as the type Click on Continue In the Region Selection dialog box select PortalFrame1LeftRafter Click on Continue In the Edit Load dialog box select Local for System and enter 12 for Component 2 Click OK Repeat exactly the same procedure for the right rafter name the load UP12 and enter 12 for the magnitude Figure 419 FIGURE 419 Imposing a line load in local coordi nates In the model tree expand the Field Output Requests and then double click on FOutput1 FOutput1 is the default and is automatically generated when creating the step Uncheck the variables Contact and select any other variable you wish to add to the field output Click on OK Under Analysis right click on Jobs and then click on Create In the Create Job dialog box name the job Portalframe and click on Continue In the Edit Job dialog box enter a description for the job Check Full analysis select to run the job in 2013 by Taylor Francis Group LLC Rigid Jointed Frames 131 FIGURE 420 Analyzing a job in Abaqus CAE Background and check to start it immediately Click OK Expand the tree under Jobs right click on Portalframe Then click on Submit If you get the following message Portalframe completed successfully in the bottom window then your job is free of errors and was exe cuted properly Notice that Abaqus has generated an input file for the job Portalframeinp Figure 420 Open it with your preferred text editor and compare it with the one given in Section 472 Under the top menu in the Module scroll to Visualization and click to load Abaqus Viewer On the main menu under File click Open navigate to your working directory and open the file Portalframeodb It should have the same name as the job you submitted Click on the Common options icon to display the Common Plot options dialog box Under labels check Show Element labels and Show Node labels to display elements and nodes numbering Click on the icon Plot Deformed Shape to display the deformed shape of the beam On the main menu click on Results then on Field Output to open the Field Output dialog box Choose S Stress components at integration points For component choose S11 to plot the stresses in the elements Click on Section points to open the section point dialog box Check bottom to plot the stresses in the lower fiber or Top for the stresses in the top fiber Figure 421 In the menu bar click on Report and Field Output In the Report Field Output dialog box for Position select Unique nodal check RF1 RF2 and RM3 for RF Reaction force and check U1 U2 and UR3 for U Spatial displacement Then click on Set up Click on Select to navigate to your working directory Name the file PortalFramerpt Uncheck Append to file and click OK Use your favorite text editor and open the file PortalFramerpt which should be the same as the one listed next FIGURE 421 Plotting stresses in the bottom fiber interactive edition 2013 by Taylor Francis Group LLC 132 Introduction to Finite Element Analysis Using MATLAB and Abaqus Field Output Report written Sun May 01 145307 2011 Source 1 ODB FTRAVAILNEWBOOKAbaqusexamplesPortalframeodb Step Applyloads Frame Increment 1 Step Time 1000 Loc 1 Nodal values from source 1 Output sorted by column Node Label Field Output reported at nodes for part PORTALFRAME1 Node RFRF1 RFRF2 RM3 UU1 UU2 UR3 Label Loc 1 Loc 1 Loc 1 Loc 1 Loc 1 Loc 1 1 100845 75 0 895948E36 750000E36 748338E03 2 0 0 0 352883E03 259456E06 301636E03 3 0 0 0 352071E03 410271E03 175904E03 4 0 0 0 351052E03 259456E06 316545E03 5 851552 75 0 776552E36 750000E36 728888E03 6 0 0 0 111346E03 648639E06 712556E03 7 0 0 0 212186E03 129728E06 614198E03 8 0 0 0 294479E03 194592E06 471234E03 9 0 0 0 360710E03 364285E03 120266E03 10 0 0 0 362105E03 428683E03 342622E03 11 0 0 0 358483E03 258914E03 140226E03 12 0 0 0 358292E03 332892E03 131950E03 13 0 0 0 361470E03 482755E03 207421E03 14 0 0 0 359490E03 391929E03 135997E03 15 0 0 0 290684E03 194592E06 479722E03 16 0 0 0 207946E03 129728E06 610826E03 17 0 0 0 108549E03 648639E06 697875E03 472 KEYWORD EDITION In this section we will prepare an input file for the portal frame shown in Figures 46 and 47 The file named FrameProblemKeywordinp is listed next Heading FrameProblem Model keyword edition Preprint echoNo modelNO historyNO Define the end nodes Node 1 0 0 5 0 6 9 7 75 13 14 6 17 14 0 Generate the remaining nodes Ngen 151 591 9131 13171 2013 by Taylor Francis Group LLC Rigid Jointed Frames 133 Define element 1 Element typeB21 112 Generate the elements Elgen elset allelements 116 1 1 Nset nsetPinnedsupports 1 17 Elset elsetLeftColumn generate 1 4 1 Elset elsetRightColumn generate 13 16 1 Elset elsetColumns LeftColumn RightColumn Elset elsetLeftRafter generate 5 8 1 Elset elsetRightRafter generate 9 12 1 Elset elsetRafters LeftRafterRightRafter Section Beamsection Profile RafterProfile Beam Section elsetRafters materialSteel sectionI 01795 0359 0172 0172 0013 0013 0008 001 Section Beamsection Profile ColumnProfile Beam Section elsetColumns materialSteel sectionI 0163 0326 0172 0172 0013 0013 0014 001 MATERIALS Material nameSteel Elastic 2e08 03 BOUNDARY CONDITIONS Boundary PinnedsupportsPINNED STEP ApplyLoads Step nameApplyLoads Static 1 1 1e05 1 LOADS Dload Rightcolumn PX 10 Leftcolumn PX 15 Leftrafter P2 12 Rightrafter P2 12 2013 by Taylor Francis Group LLC 134 Introduction to Finite Element Analysis Using MATLAB and Abaqus OUTPUT REQUESTS Output field Node Output CF RF RM U Element Output S Output history variablePRESELECT End Step At the command line type Abaqus jobFrameProblemKeyword inter followed by Return If you get an error open the file with extension dat to see what type of error To load the visualization model type Abaqus Viewer at the command line On the main menu under File click Open navigate to your working directory and open the file FrameProblemKeywordodb It should have the same name as the job you submitted Click on the Common options icon to display the Common Plot options dialog box Under labels check Show Element labels and Show Node labels to display elements and nodes numbering Click on the icon Plot Deformed Shape to display the deformed shape of the beam On the main menu click on Results then on Field Output to open the Field Output dialog box Choose S Stress components at integration points For component choose S11 to plot the stresses in the elements Figure 422 Click on Section points to open the section point dialog box Check bottom to plot the stresses in the lower fiber or Top for the stresses in the top fiber Notice that the stress contour is exactly the same as obtained previously except that the node and element numbering is different FIGURE 422 Plotting stresses in the bottom fiber keyword edition 2013 by Taylor Francis Group LLC 5 Stress and Strain Analysis 51 INTRODUCTION This chapter deals with the notions of stressstrain and straindisplacements relation which are quite essential for understanding the remaining developments in the book It marks the change of philosophy between matrix structural analysis and finite element analysis of a continuum In the previous Chapters 2 through 4 we only considered structural elements whose behavior can be formulated as a function of a single variable x which is the longitudinal direction of the element This is of course possible because of the geometry where two dimensions are insignificant compared to the third one The only stress of interest therefore is the longitudinal stress σx along the dominant dimension Yet in a threedimensional solid where all the dimensions are of the same size this assumption is not valid anymore When a threedimensional solid is subjected to external forces andor displacements and at the same time is restrained against rigid body movement internal forces are induced and these result in more than one stress at a point Additionally these external forces result in material points within the body being displaced When there is a change in distance between two points straining has taken place Again there is more than one strain at a point As will be shown in the Sections 533 and 534 a segment of infinitesimal length not only experiences a change in length but also a change in direction As stresses and strains are interrelated we will also consider the relations between them Such relations are called constitutive equations since they describe the macroscopic behavior resulting from the internal constitution of the material Materials however exhibit different behaviors over their entire range of deformations As such it is not possible to write one set of mathematical equations to describe these behaviors Yet for many engineering applications the theory of linear elasticity offers a useful and reliable model for analysis 52 STRESS TENSOR 521 DEFINITION Let us consider a body in equilibrium under external forces as represented in Figure 51 Let us take a cut through the body as represented by the plane Σ and denote by dA an infinitesimal element of the internal cross section A force dF is exerted on this small area It represents the influence of the right section on the left section of the body The vector dF can be expressed in terms of its normal and tangential components dFn and dFt to the surface dA The stresses acting on the surface are then given as σn lim dA0 dFn dA 51 σi lim dA0 dFi dA 52 It can be seen that dFt has also two components on the plane of the surface dA In total therefore there are three stress components one normal and two tangential However as the infinitesimal element dA shrinks to a point there will be an infinite number of planes passing through that point It would be impossible therefore to consider all of them However if we choose three mutually perpendicular planes as represented in Figure 52 the stresses can be written for all 135 136 Introduction to Finite Element Analysis Using MATLAB and Abaqus dA dA dFt1 dFt dFt2 dF dFn FIGURE 51 Internal force components 3 σ31 σ33 σ32 σ23 σ22 σ21 σ12 σ11 σ13 2 1 FIGURE 52 Stress components at a point them It should be emphasized however that the parallelepiped represented in Figure 52 is not a block of material cut from the body but a simple yet convenient schematic device to repre sent the stresses acting at a point It can be seen that there are nine components of stress acting at a point The nine stresses are arranged in a stress tensor as σ σ11 σ12 σ13 σ21 σ22 σ23 σ31 σ32 σ33 53 It should be noted that a stress component has two indices the first index indicates the direction of the normal to the plane on which it acts and the second refers to the direction of the stress component A stress component is positive if it acts on a positive face in the positive direction or on a negative face in the negative direction 2013 by Taylor Francis Group LLC The stress tensor is symmetric since by taking moments about the axis passing through the point or the center of the cube as shown it can be shown that σ12 σ21 σ13 σ31 σ23 σ32 54 In general σij σji 55 This shows that the stress tensor contains only six independent components 522 STRESS TENSORSTRESS VECTOR RELATIONSHIPS In order to study the transformation of stress let us isolate an infinitesimal tetrahedron as shown in Figure 53 The plane ABC is perpendicular to an arbitraryoriented normal n written in vector matrix notation as n n1 n2 n3T 56 The components ni are the direction cosines of the normal n If ΔS is the area of the surface ABC then the areas of the other surfaces can be expressed as for COB ΔS1 n1ΔS for AOC ΔS2 n2ΔS for BOA ΔS3 n3ΔS 57 These expressions denote that the faces are the projections of the oblique face onto the coordinates planes Figure 53 also shows the stress vectors Ti T2 T3 which are the components of the T on the three mutually orthogonal planes These stress vectors are resolved along the coordinate axes 1 2 and 3 as σ11 σ12 σ13 for T1 σ21 σ22 σ23 for T2 σ31 σ32 σ33 for T3 58 The asterisk indicates that we are dealing with average values Remember that a stress vector is a point quantity The body force vector q that acts throughout the body is also shown Resolution of q and T into the directions of the coordinate axis yields q qi q2 q3T 59 T T1 T2 T3T 510 Equilibrium requires the vector sum of all forces acting on the tetrahedron to be zero To obtain the forces acting on the tetrahedron the stress components must be multiplied by the respective areas on which they act Requiring equilibrium in the xdirection yields T1ΔS q1ΔV σn11n1ΔS σn21n2ΔS σn31n3ΔS 0 511 The volume of the tetrahedron can be written as ΔV 1 3 hΔS 512 where h is the perpendicular distance from point O to the base ABC Substituting for V and dividing by S in the equilibrium equation yields T1 q1 1 h σn11n1 σn21n2 σn31n3 0 513 Now let the tetrahedron shrink to a point by taking the limit as h 0 and noting in this process that the starred average quantities take on the actual values of those quantities at a point results in T1 σ11n1 σ21n2 σ31n3 514 Similarly we obtain for the y and z directions T2 σ12n1 σ22n2 σ32n3 515 T3 σ13n1 σ23n2 σ33n3 516 These expressions can be grouped in a matrix form as follows T1 T2 T3 σ11 σ21 σ31 σ12 σ22 σ32 σ13 σ23 σ33 n1 n2 n3 517 or simply as T σTn or in index notation as Ti σij nj 518 Stress and Strain Analysis 139 523 TRANSFORMATION OF THE STRESS TENSOR If the components of the stress tensor σ in the basis e1 e2 e3 are known let us find the components of the same tensor in another basis e1 e2 e3 obtained from a rotation of axes around the origin Since the stress tensor is a secondorder tensor it obeys the same transformation laws for second order tensor as detailed in Appendix C Therefore the components of the stress tensor in the new basis are obtained respectively in index and matrix notations as σ km lkilmjσij 519 σ QσQT 520 The components lij or Qij are the cosines of the angles formed by the unit vectors ei ej The inverse transformations are obtained as σij lkilmjσ km σ QTσQ 521 524 EQUILIBRIUM EQUATIONS Equilibrium of a small cube of material that is removed from a larger body subject to external forces requires that the resultant force and moment acting on the cube must be equal to zero In Figure 54 the components of stress acting on the positive faces of the element are shown The components acting on the negative faces are omitted for the sake of clarity of the figure The omitted components are σ11 σ12 σ13 on face 1 σ21 σ22 σ23 on face 2 and σ31 σ32 σ33 on face 3 The stresses vary throughout the body and it is assumed that their components and derivatives are continuous functions of the coordinates To express this variation the wellknown rules of differential calculus can be used σ11x dx σ11x σ11 x 522 In addition to the stress components acting on the body body forces such as the ones due to gravity are also present and have intensities bx by and bz or simply bi When the stress components are σ33 σ31 σ13 σ11 σ12 2 3 σ13 σ11 σ12 σ33 σ31 z x z x y dx dy dz dx dz σ21 σ22 σ23 σ32 σ21 σ22 σ23 σ32 x y z y dx dy dz dy 1 FIGURE 54 Equilibrium of an infinitesimal cube 2013 by Taylor Francis Group LLC multiplied by the area on which they act force components are obtained Requiring equilibrium in the x direction leads to σ11dydz σ11 σ11x dx dy dz σ21dxdz σ21 σ21x dx dz σ31dxdy σ31 σ31x dx dx dy bxdxdydz 0 After rearranging Equation 523 becomes σ11x σ21y σ31z bx 0 Requiring equilibrium in y and z directions as well results in σ12x σ22y σ32z by 0 σ13x σ23y σ33z bz 0 Noticing that x y and z are actually the first second and third directions Equations 524 and 525 can be simply written as σji bj 0 or because of the symmetry of the stress tensor as σij bi 0 The comma in expressions 526 and 527 indicates derivative with respect to a direction designated by the index following the comma 525 PRINCIPAL STRESSES Since the stress tensor is a secondorder tensor the calculation of the principal stress values and their associated principal directions is exactly the same as for a general secondorder tensor detailed in Appendix C In other words in the basis e1 e2 e3 the stress vector T σn on the cutting plane Pn is not parallel to the normal n the problem is to find the cutting plane Pn whose normal n is parallel to T such that T σn λn where λ is a scalar This plane together with two other planes which are all mutually perpendicular forms a basis called the principal basis of the tensor Figure 55 This new basis is made of the principal directions of the tensor In this basis the tensor reduces to its diagonal form σ σ1 0 0 0 σ2 0 0 0 σ3 Stress and Strain Analysis 141 n Pn n T Pn FIGURE 55 Principal directions of a stress tensor where σ1 σ2 and σ3 are the principal stresses and roots of the characteristic equation of the tensor σ3 I1σ2 I2σ I3 0 529 where I1 I2 and I3 are the stress invariants which are independent of the coordinates system They are obtained as I1 σii I2 1 2σiiσjj σijσij 530 I3 σij detσ These invariants can also be expressed in terms of σ1 σ2 and σ3 which are invariants themselves I1 σ1 σ2 σ3 I2 σ1σ2 σ2σ3 σ3σ1 531 I3 σ1σ2σ3 526 VON MISES STRESS What is referred to as von Mises stress is another form of invariant of the stress tensor As the reader will find out in subsequent chapters Abaqus by default plots a contour of the von Mises stress This quantity is very useful when plastic yielding of a material is present Indeed it is possible for a material to yield under a given combination of the principal stresses even though none of them exceeds the yield stress of the material The von Mises stress is a formula combining the principal stresses into an equivalent stress that can be compared to the yield stress of the material and it is given as σ1 σ22 σ2 σ32 σ3 σ12 2σ2 e 532 527 NORMAL AND TANGENTIAL COMPONENTS OF THE STRESS VECTOR In a basis formed by the principal directions of the stress tensor the stress vector may be resolved into a normal and tangential component as shown in Figure 56 2013 by Taylor Francis Group LLC The same can be said about points M and N MN ON OM dOM d s Substituting 559 and 560 expression 558 is rewritten as d s d s d uM Introducing the Cartesian components of the vectors d s dX1 e1 dX2 e2 dX3 e3 d s d x1 e1 d x2 e2 d x3 e3 d u d u1 e1 d u2 e2 d u3 e3 Equation 562 becomes d xi d X i u iX j dX j δji uijdX j FijdX j or in matrix notation as d X d x u iX jd X u1X1 u1X2 u1X3 u2X1 u2X2 u2X3 u3X1 u3X2 u3X3 d X and in a more compact form as d x I u Id X d x d x 142 Introduction to Finite Element Analysis Using MATLAB and Abaqus t T tt tn σ3 σ2 n σ1 FIGURE 56 Tangential and normal components of the stress vector By definition the stress vector is expressed respectively in vector index and matrix notations as T σn Ti σijnj 533 T σn The normal is the scalar or dot product of T with n written in vector index and matrix notations as σn T n σn ninjσij 534 σn nTσn In the principal basis the components of the stress vector can also be expressed as T1 σ1n1 T2 σ2n2 535 T3 σ3n3 Substituting Equations 535 in any equation of 534 yields σn σ1n2 1 σ2n2 2 σ3n2 3 536 Using Pythagoras theorem gives the tangential or shear component as σ2 s TiTi σ2 n 537 Notice that the term TiTi represents the modulus of the stress vector T it is actually the scalar product of T by itself Substituting 535 and 536 in 537 yields σ2 s σ2 1n2 1 σ2 2n2 2 σ2 3n2 3 σ1n2 1 σ2n2 2 σ3n2 3 538 2013 by Taylor Francis Group LLC d s2 d d s2 d s 2 d OM 2 d O M 2 d s 2 d S2 d s 2 d S2 d s d u d s d u d T d u d u d S2 d s d s 2 2Eij dX dX Eij frac12 left fracpartial u ipartial X j fracpartial u jpartial X i fracpartial u kpartial X j fracpartial u kpartial X i right Equations 570 and 571 can be written in matrix notation as d s 2 d S2 d X T abla uTd x d X T abla uTd x d x T abla uT abla uTd x d X T abla uT abla uT abla uT abla ud x 2d XTmathitEd x with E frac12 abla u abla uT abla uT abla u Stress and Strain Analysis 143 When the principal stresses are ordered according to σ1 σ2 σ3 the maximum shear stress is given as σs 1 2σ1 σ3 539 Combining Equations 538 and 537 with the identity n2 1n2 2n2 3 1 and solving for the direction cosines ni we obtain n2 1 σn σ2σn σ3 σ2 s σ1 σ2σ1 σ3 n2 2 σn σ1σn σ3 σ2 s σ2 σ1σ2 σ3 540 n2 3 σn σ1σn σ2 σ2 s σ3 σ1σ3 σ2 These equations serve as the basis for Mohrs circle of stress 528 MOHRS CIRCLES FOR STRESS Mohrs circles provide a convenient graphical twodimensional representation of the three dimensional state of stress Mohrs circles are drawn in the σn σs stress space Given the ordering σ1 σ2 σ3 it can be seen that the numerator of the righthandside of Equation 540 is positive that is σn σ2σn σ3 σ2 s 0 541 This equation represents stress points in the σn σs stress space that are on or outside the circle C1 shown in Figure 57 which has for equation σn σ2 σ322 σ2 s σ2 σ322 542 σs σn σ2 σ3 2 σ1 σ2 2 σ1 σ3 2 C3 C1 C2 FIGURE 57 Mohrs circles 2013 by Taylor Francis Group LLC 144 Introduction to Finite Element Analysis Using MATLAB and Abaqus The same approach can be used to draw two other circles C2 and C3 represented by the following equations σn σ3 σ122 σ2 s σ3 σ122 543 σn σ1 σ222 σ2 s σ1 σ222 544 529 ENGINEERING REPRESENTATION OF STRESS Previously it was shown that the stress tensor is symmetric and therefore possesses only six inde pendent components For this reason engineers more often write the stress tensor as a vector with six components σ1 σ2 σ3 σ4 σ5 σ6 σ11 σ22 σ33 σ12 σ23 σ13 σxx σyy σzz σxy σyz σxz 545 With this notation the transformation law for stress in the case of a rotation around the axis 3 or axis z by an angle ψ is written as σ 11 σ 22 σ 33 σ 12 σ 23 σ 13 cos2 ψ sin ψ 0 2 sin ψ cos ψ 0 0 sin2 ψ cos2 ψ 0 2 sin ψ cos ψ 0 0 0 0 1 0 0 0 sin ψ cos ψ sin ψ cos ψ cos2 ψ sin2 ψ 0 0 0 0 0 0 0 cos ψ sin ψ 0 0 0 0 sin ψ cos ψ σ11 σ22 σ33 σ12 σ23 σ13 546 53 DEFORMATION AND STRAIN 531 DEFINITION The term deformation refers to a change in shape of the body between some initial undeformed configuration and some final deformed configuration as represented in Figure 58 After deformation point M moves to M and point N moves to N The segment MN not only undergoes a change in length but also a change in its direction Most often deformation is not just a function of the spatial 2013 by Taylor Francis Group LLC Stress and Strain Analysis 145 Before deformation M N M N After deformation FIGURE 58 Schematic representation of the deformation of a solid body coordinates but it is also a function of time Some deformation processes such as creep in concrete or relaxation in prestressing tendons occur over very long periods of time Often the configuration at time t 0 is chosen as the reference configuration and the current configuration refers to the configuration which the body occupies at current time t 532 LAGRANGIAN AND EULERIAN DESCRIPTIONS During deformation the particles of a body move along various paths Relative to a Cartesian coordi nate system a particle that originally occupied a position X Y Z in the undeformed configuration occupies the position x y z in the deformed configuration This motion may be expressed by the equations x xX Y Z t y yX Y Z t 547 z zX Y Z t or more compactly in index notation as xi xiX1 X2 X3 t 548 Equations 547 and 548 can be thought of as a mapping of the initial configuration into the current configuration This description of motion is known as the Lagrangian description For instance when a body undergoes deformation a quantity associated with a particle such as temperature changes with time Such changes in temperature can be expressed according to Equation 548 as θ θX1 X2 X3 t 549 The Lagrangian description is also known as the material description or reference description On the other hand the motion may be given in the form Xi Xix1 x2 x3 t 550 Given the current position of a particle this description can be thought of as one that provides a tracing to the original position of the particle This description is known as the Eulerian description The triples X Y Z and x y z are also known respectively as material and spatial coordinates The Lagrangian description seems the most suitable in solid mechanics since in these problems there is usually an easy way to identify a reference configuration for which all information is known 2013 by Taylor Francis Group LLC 146 Introduction to Finite Element Analysis Using MATLAB and Abaqus However it is of little use in fluid mechanics because in nonsteady flow the reference position at time t 0 of a particle is generally not known In this book since we are primarily dealing with solid mechanics we will use the Lagrangian description The coordinates of a particle in the initial configuration are labeled X Y Z 533 DISPLACEMENT Relative to a Cartesian coordinate system let e1 e2 e3 be the unit vectors in the directions of the superposed coordinates X1 X2 X3 and x1 x2 x3 The position of the particle M at time t 0 can be described by the vector OM as shown in Figure 59 OM X1 e1 X2 e2 X3 e3 551 The particle originally at M moves to M in the current configuration at time t Its new position is described by the vector OM OM x1 e1 x2 e2 x3 e3 552 The equation xi xiX1 X2 X3 t describes the path of the particle which at time t 0 is located at M The vector MM is the displacement vector from the reference to the current configuration obtained as MM OM OM 553 which after substitution of Equations 551 and 552 becomes MM x1 X1 e1 x2 X2 e2 x3 X3 e3 554 This equation is normally written as MM u u1 e1 u2 e2 u3 e3 555 X2 x2 X1 x1 e2 e3 e1 O M M X3 x3 FIGURE 59 Reference and current configurations 2013 by Taylor Francis Group LLC Stress and Strain Analysis 147 The terms ui are the components of the displacement vector and they are assumed to be continuous functions of the coordinates Xi ui uiX1 X2 X3 t 534 DISPLACEMENT AND DEFORMATION GRADIENTS Once more let us consider the deformed and undeformed configuration in a Cartesian coordinate system where the unit vectors e1 e2 e3 are the directions of the superposed coordinates X1 X2 X3 and x1 x2 x3 Figure 510 represents the deformation process of an infinitesimal element originally at MN in the undeformed configuration which moves to the position MN in the deformed configuration During deformation point M moves to M and its new position is given by OM OM MM 556 The vector MM represents the displacement of point M and is noted uM Point N also moves to N and its new position is given by the vector position ON ON NN 557 Again the vector NN represents the displacement of point N noted uN The relative position between points N and M after deformation is expressed as MN ON OM ON OM uN uM 558 Since points M and N are very close to each other MN is an infinitesimal element with length dS It follows therefore that MN ON OM d OM dS 559 O X2 x2 X1 x1 X3 x3 e2 e3 e1 M uM uN N dS ds M N FIGURE 510 Deformations of an infinitesimal element 2013 by Taylor Francis Group LLC negligible and equal to zero As a result the Green Lagrange strain tensor reduces to the infinitesimal strain tensor which is written in both index and matrix notations as eij 12 uixj ujxi 576 e 12 VuI VuIT 577 Within the context of small deformation theory Equation 570 is rewritten as dℓ32 dℓ32 2eijdxidxj 578 Further assuming that dS ds for small deformations this equation may be put in the form ds dS ds eij dXi ds dXj ds 579 or in matrix form as ds dS ds 1dS2 dX1 dX2 dX3 T e dX1 dX2 dX3 580 The lefthandside of Equations 579 or 580 is recognized as the change in length of the differential element and is called the normal strain for the element originally having direction cosines dXdS Introducing the direction cosines αi dXidS Equations 579 and 580 become ds dS ds eijαiαj 581 1dS2 α1 α2 α3 T e α1 α2 α3 582 In addition in small deformation theory there is very little difference between the material X1 X2 X3 and spatial x1 x2 x3 coordinates Hence it is immaterial whether the infinitesimal strain tensor is written as 12 uixj ujxi or 12 uixj ujxi 5362 Geometrical Interpretation of the Terms of the Strain Tensor In the context of small deformation theory let us consider the deformation behavior of two orthogonal infinitesimal elements MN and ML respectively parallel to the axis x1 and x2 as shown in Figure 511 MN dxi e1 ui u1x1 e1 uz u2x1 e2 dx1 u1x1 dx1 e1 u2x1 e2 584 The term u1x1dx1 e11dx1 represents the change of length of the infinitesimal element MN in the direction x1 It follows therefore that e11 represents the straining at point M in the direction x1 and the term u2x1dx1 e21dx1 represents its distortion in the direction x2 The same reasoning can be carried out for the infinitesimal element ML ML dx2 e2 uz u2x2 e2 ui u1x2 e1 u1x2 e1 dx2 u2x2 e2 585 Again the term u2x2dx2 e22dx2 represents the change in length of the infinitesimal element ML in the direction x2 It follows therefore that e22 represents the straining at point M in the direction x2 and the term u1x2dx2 e12dx2 represents its distortion in the direction x1 The angle γ1 between the directions MN and MN after deformation is such that tan γ1 u2x1 dx1 u2x1dx1 u1x1 u2x1 586 The same can be said for the angle γ2 between the directions ML and ML tan γ2 u1x2 587 It follows therefore that the angle between MN and ML that was equal to π2 before deformation reduces by γ1 γ2 after deformation has taken place The angle γ12 is called the engineering shear strain at point M written as γ12 2e12 588 5363 Compatibility Conditions The strain tensor contains six independent components Integration of these six components should lead to the three displacement components u1 u2 u3 However the solution is not unique unless the six components of strain verify the following compatibility equations 2eijxkxj 2eijxixk 2 2eijxjxk for i j 589 2eiixjxk eikxj eijxk for i j k 590 537 PRINCIPAL STRAINS In terms of components the strain tensors E and e bear some resemblance to the stress tensor Therefore the entire development for principal strains principal strain directions and strain invariants may be carried out exactly as done for the stress tensor In particular in the basis made of the principal directions the strain tensor reduces to its diagonal form e e1 0 0 0 e2 0 0 0 e3 591 where e1 e2 and e3 are the principal stresses and roots of the characteristic equation of the tensor e3 I1e2 I2e I3 0 592 where I1 eii I2 12 eik eij eij eik 593 I3 eij dete These invariants can also be expressed in terms of e1 e2 and e3 which are invariants themselves as I1 e1 e2 e3 I2 e1 e2 e2 e3 e3 e1 594 Stress and Strain Analysis 153 538 TRANSFORMATION OF THE STRAIN TENSOR Like the stress tensor the strain tensor transforms according to the transformation law of second order tensors If the components of the strain tensor ϵ are known in the basis e1 e2 e3 then its components in the basis e1 e2 e3 are obtained in both index and matrix notations as ϵ km lkilmjϵij 595 ϵ QϵQT 596 The components lij or Qij are the cosines of the angles formed by the unit vectors ei ej The inverse transformations are obtained as ϵij lkilmjϵ km ϵ QTϵQ 597 539 ENGINEERING REPRESENTATION OF STRAIN Like the stress tensor the strain tensor is symmetric and therefore possesses only six independent components Engineers also prefer to substitute for the shear strains the engineering shear strains as ϵ1 ϵ2 ϵ3 ϵ4 ϵ5 ϵ6 ϵ11 ϵ22 ϵ33 γ12 γ23 γ13 ϵxx ϵyy ϵzz γxy γyz γxz ϵxx ϵyy ϵzz 2ϵxy 2ϵyz 2ϵxz 598 With this notation the transformation law for strain in the case of a rotation around the axis 3 or axis z is written as ϵ 11 ϵ 22 ϵ 33 γ 12 γ 23 γ 13 cos2 ψ sin ψ 0 sin ψ cos ψ 0 0 sin ψ cos2 ψ 0 sin ψ cos ψ 0 0 0 0 1 0 0 0 2 sin ψ cos ψ 2 sin ψ cos ψ cos2 ψ sin2 ψ 0 0 0 0 0 0 0 cos ψ sin ψ 0 0 0 0 sin ψ cos ψ ϵ11 ϵ22 ϵ33 γ12 γ23 γ13 599 2013 by Taylor Francis Group LLC 154 Introduction to Finite Element Analysis Using MATLAB and Abaqus 54 STRESSSTRAIN CONSTITUTIVE RELATIONS 541 GENERALIZED HOOKES LAW The stress tensor is related to the strain tensor through the generalized Hookes law which is given in index notation as σij Dijklϵkl 5100 where Dijkl is the stiffness tensor This is a fourthorder tensor with 81 components Equation 5100 represents actually nine equations of which the first one is given as σ11 D1111ϵ11 D1112ϵ12 D1113ϵ13 D1121ϵ21 D1122ϵ22 D1123ϵ23 D1131ϵ31 D1132ϵ32 D1133ϵ33 5101 Luckily in practice the equations are much simpler and not all the 81 components are indepen dent The symmetry of both the stress and strain tensors introduces some simplifications into the constitutive equations Dijkl Dijlk Djikl Djilk 5102 In addition the assumption of linear elastic material behavior implies the existence of a strain energy density function Omitting the proof this energy density function is given as dU σijϵij σ11ϵ11 σ22ϵ22 σ33ϵ33 σ12ϵ12 σ23ϵ23 σ13ϵ13 5103 According to Equations 5101 and 5103 it follows that U ϵ11 σ11 D1111ϵ11 D1112ϵ12 D1113ϵ13 D1121ϵ21 D1122ϵ22 D1123ϵ23 D1131ϵ31 D1132ϵ32 D1133ϵ33 5104 and U ϵ22 σ22 D2211ϵ11 D2212ϵ12 D2213ϵ13 D2221ϵ21 D2222ϵ22 D2223ϵ23 D2231ϵ31 D2232ϵ32 D2233ϵ33 5105 Hence 2U ϵ11ϵ22 D1122 D2211 5106 and in general 2U ϵklϵmn Dklmn Dmnkl 5107 Equation 5107 shows that the fourthorder tensor Dijkl is symmetric In other words the number of independent elastic coefficients is reduced from 36 to 21 2013 by Taylor Francis Group LLC Stress and Strain Analysis 155 The generalized Hookes law for an anisotropic material can now be written using engineering matrix notation as σ11 σ22 σ33 σ12 σ23 σ13 D1111 D1122 D1133 D1112 D1123 D1113 D2211 D2222 D2233 D2212 D2223 D2213 D3311 D3322 D3333 D3312 D3323 D3313 D1211 D1222 D1233 D1212 D1223 D1213 D2311 D2322 D2333 D2312 D2323 D2313 D1311 D1322 D1333 D1312 D1323 D1313 ϵ11 ϵ22 ϵ33 γ12 2ϵ12 γ23 2ϵ23 γ13 2ϵ13 5108 with Dklmn Dmnkl In practice it is sometimes more useful to express observed strains in terms of applied stresses using the compliance tensor obtained by inverting 5108 ϵ11 ϵ22 ϵ33 γ12 γ23 γ13 C1111 C1122 C1133 C1112 C1123 C1113 C2211 C2222 C2233 C2212 C2223 C2213 C3311 C3322 C3333 C3312 C3323 C3313 C1211 C1222 C1233 C1212 C1223 C1213 C2311 C2322 C2333 C2312 C2323 C2313 C1311 C1322 C1333 C1312 C1323 C1313 σ11 σ22 σ33 σ12 σ23 σ13 5109 Further simplifications in the number of constants can be achieved if certain symmetries exist in the material But before investigating these material symmetries it is important to know how a fourth order tensor is transformed Since the components of stress and strain are functions of the system of reference axes the elastic coefficients in Equation 5108 are also functions of this orientation If the components of the stiffness tensor Dijkl in the basis e1 e2 e3 are known its components in the basis e 1 e 2 e 3 are obtained according to the following transformation rule D prst lpilrjlskltlDijkl 5110 542 MATERIAL SYMMETRIES 5421 Symmetry with respect to a Plane A material that exhibits symmetry of its elastic properties to one plane is called a monoclinic material This symmetry is expressed by the requirement that the material constants do not change under a change from the basis e1 e2 e3 to e 1 e 2 e 3 such as the one represented in Figure 512 e2 e2 e2 e2 e3 e1 e1 e1 e1 e3 e3 e3 FIGURE 512 Monoclinic material 2013 by Taylor Francis Group LLC 156 Introduction to Finite Element Analysis Using MATLAB and Abaqus The direction cosines of the primed axis with respect to the unprimed axis are given as lij cose i ej 1 0 0 0 1 0 0 0 1 5111 It follows therefore that D 1111 l1kl1ll1ml1nDklmn l11l11l11l11D1111 D1111 5112 In a similar way for this type of symmetry it follows that when e3 and e 3 are in the same direction we obtain D 1113 l1kl1ll1ml3nDklmn l11l11l11l33D1113D1113 D1113 5113 and when they are in the opposite direction we obtain D 1113 l1kl1ll1ml3nDklmn l11l11l11l33D1113D1113 D1113 5114 which is impossible It follows therefore that D1113 0 In a similar fashion it can be shown that the number of elements is reduced from 21 to 13 that is the elastic matrix is written as follows D D1111 D1122 D1133 D1112 0 0 D2211 D2222 D2233 D2212 0 0 D3311 D3322 D3333 D3312 0 0 D1211 D1222 D1233 D1212 0 0 0 0 0 0 D2323 D2313 0 0 0 0 D1323 D1313 5115 Similarly the compliance matrix becomes C C1111 C1122 C1133 C1112 0 0 C2211 C2222 C2233 C2212 0 0 C3311 C3322 C3333 C3312 0 0 C1211 C1222 C1233 C1212 0 0 0 0 0 0 C2323 C2313 0 0 0 0 C1323 C1313 5116 2013 by Taylor Francis Group LLC Stress and Strain Analysis 157 5422 Symmetry with respect to Three Orthogonal Planes A material that exhibits symmetry of its elastic planes with respect to three orthogonal planes is called an orthotropic material Following the same reasoning as for the symmetry with respect to a single plane and equating terms to zero where contradictions arise the elastic matrix reduces from 13 terms to 9 D D1111 D1122 D1133 0 0 0 D2211 D2222 D2233 0 0 0 D3311 D3322 D3333 0 0 0 0 0 0 D1212 0 0 0 0 0 0 D2323 D2313 0 0 0 0 D1323 D1313 5117 5423 Symmetry of Rotation with respect to One Axis A material that posseses an axis of symmetry in the sense that all rays at right angle to this axis have the same elastic properties is called a transversely isotropic material If this axis is for example e3 as shown in Figure 513 then a change of basis obtained by rotation around e3 will leave the elastic properties unaltered Making use of this property leads to D1111 D2222 D2323 D1313 D1133 D2233 D1212 1 2D1111 D1122 5118 e3 e2 e1 FIGURE 513 Symmetry of rotation 2013 by Taylor Francis Group LLC 158 Introduction to Finite Element Analysis Using MATLAB and Abaqus The number of independent coefficient in the elastic matrix is now reduced to 5 D D1111 D1122 D1133 0 0 0 D1122 D1111 D1133 0 0 0 D1133 D1133 D3333 0 0 0 0 0 0 1 2D1111 D1122 0 0 0 0 0 0 D1313 0 0 0 0 0 0 D1313 5119 The compliance matrix is obtained as C C1111 C1122 C1133 0 0 0 C1122 C1111 C1133 0 0 0 C1133 C1133 C3333 0 0 0 0 0 0 2C1111 C1122 0 0 0 0 0 0 C1313 0 0 0 0 0 0 C1313 5120 543 ISOTROPIC MATERIAL A material is isotropic if its elastic properties are the same in any direction and therefore do not depend on the choice of the coordinates system The elastic and compliance matrices remain unaltered by any change of orthonormal basis The use of these properties leads to D1313 1 2D1111 D1122 5121 D3333 D1111 D1133 D1122 5122 The elastic matrix is written as D D1111 D1122 D1122 0 0 0 D1122 D1111 D1122 0 0 0 D1122 D1122 D1111 0 0 0 0 0 0 1 2D1111 D1122 0 0 0 0 0 0 1 2D1111 D1122 0 0 0 0 0 0 1 2D1111 D1122 5123 2013 by Taylor Francis Group LLC Stress and Strain Analysis 159 and the compliance matrix as C C1111 C1122 C1122 0 0 0 C1122 C1111 C1122 0 0 0 C1122 C1122 C1111 0 0 0 0 0 0 2C1111 C1122 0 0 0 0 0 0 2C1111 C1122 0 0 0 0 0 0 2C1111 C1122 5124 In fact the elastic matrix possesses only two independent components Introducing the elastic properties λ and μ known as the Lamés constants the stressstrain relations for an isotropic material become σ11 σ22 σ33 σ12 σ23 σ13 λ 2μ λ λ 0 0 0 λ λ 2μ λ 0 0 0 λ λ λ 2μ 0 0 0 0 0 0 μ 0 0 0 0 0 0 μ 0 0 0 0 0 0 μ ϵ11 ϵ22 ϵ33 γ12 γ23 γ13 5125 In index notation the previous relationship is written as σij λδijϵkk 2μϵij 5126 The compliance matrix is given as ϵ11 ϵ22 ϵ33 γ12 γ23 γ13 λ 2μ μ3λ 2μ λ 2μ3λ 2μ λ 2μ3λ 2μ 0 0 0 λ 2μ3λ 2μ λ 2μ μ3λ 2μ λ 2μ3λ 2μ 0 0 0 λ 2μ3λ 2μ λ 2μ3λ 2μ λ 2μ μ3λ 2μ 0 0 0 0 0 0 1 μ 0 0 0 0 0 0 1 μ 0 0 0 0 0 0 1 μ σ11 σ22 σ33 σ12 σ23 σ13 5127 which can also be written in index notation as ϵij λδij 2μ3λ 2μσnn 1 2μσij 5128 Notice that in index notation the engineering shear strain γij is not used 2013 by Taylor Francis Group LLC 160 Introduction to Finite Element Analysis Using MATLAB and Abaqus 5431 Modulus of Elasticity Let us consider a uniaxial tension or compression test In this case the only stress that is different from zero is σ11 From Equation 5127 it can be seen that all the shear strains γij are equal to zero The strain in the direction of the test is given as ϵ11 λδij 2μ3λ 2μσ11 1 2μσ11 5129 This relation can be rearranged to give σ11 μ3λ 2μ λ μ ϵ11 Eϵ11 5130 which is the wellknown Hookes law Equation 5130 shows the relationship between the elastic modulus E and the Lamé constants λ and μ 5432 Poissons Ratio From Equation 5127 when only σ11 is different from zero it can also be seen that the strains in the directions 2 and 3 are given as ϵ22 ϵ33 λ 2μ3λ 2μσ11 λ 2λ μϵ11 νϵ11 E νσ11 5131 The coefficient ν λ2λμ is called Poissons ratio Equation 5131 gives the relationships between Poissons ratio and the Lamé constants 5433 Shear Modulus Let us consider a pure shear test in the plane e1 e2 made by the directions 1 and 2 The only stress that is different from zero is σ12 τ The stressstrain relations can be written as σ12 2μϵ12 μγ12 5132 The coefficient μ is called the shear modulus It is much better known as G 5434 Bulk Modulus Another test to consider is the application of hydrostatic compression or tension σ12 σ23 σ13 0 In this test σ11 σ22 σ33 1 3σii p 5133 where p stands for hydrostatic pressure As a result of this test the strains are also spherical ϵ11 ϵ22 ϵ33 1 3ϵii ϵv 5134 where ϵv stands for volumetric strain 2013 by Taylor Francis Group LLC TABLE 51 Relationships between the Coefficients of Elasticity λ μ E ν λ Eν 1 ν1 2ν μ μ E 21 ν G E μ3λ 2μ λ μ E E λ 2λ μ ν λ 2 3 μ K λ 2 3 μ E 31 2ν 3G E It follows that p λ 2 3 μ εv Kεv 5135 The coefficient K is called the bulk modulus or the compressibility modulus Table 51 gives the relationships between the coefficients of elasticity Finally the stressstrain relationships for an isotropic material can be written in terms of E and ν as σ11 σ22 σ33 σ12 σ23 σ13 E 1 ν1 2ν 1 ν ν ν 0 0 0 ν 1 ν ν 0 0 0 ν ν 1 ν 0 0 0 0 0 0 1 2ν 2 0 0 0 0 0 0 1 2ν 2 0 0 0 0 0 1 2ν 2 5136 for the elastic matrix and as ε11 ε22 ε33 γ12 γ23 γ13 1 E 1 ν ν 0 0 0 ν 1 ν 0 0 0 ν ν 1 0 0 0 0 0 0 21 ν 0 0 0 0 21 ν 0 0 0 0 0 21 ν σ11 σ22 σ33 σ12 σ23 σ13 5137 162 Introduction to Finite Element Analysis Using MATLAB and Abaqus 544 PLANE STRESS AND PLANE STRAIN In reality all solids are three dimensional Fortunately for many problems that are of practical interest some simplifying assumptions can be made regarding the stress or strain distributions and solutions can be carried out in a relatively simpler manner A solid with one dimension relatively small compared to the two others and loaded in its plane can be analyzed using the plane stress approach The surfaces of the beam shown in Figure 514 z t2 are free of forces and therefore the stress components σ33 σ13 and σ23 are equal to zero If the beam is thin it can be reasonably assumed that these components are zero throughout the thickness of the beam and the other stress components σ11 σ22 and σ12 remain practically constant The nonzero stresses are σ11 σ22 and σ12 Therefore Equation 5137 becomes ϵ11 ϵ22 γ12 1 E 1 ν 0 ν 1 0 0 0 21 ν σ11 σ22 σ12 5138 Inverting expression 5138 yields σ11 σ22 σ12 E 1 ν2 1 ν 0 ν 1 0 0 0 1 ν 2 ϵ11 ϵ22 γ12 5139 It should be pointed out that in plane stress ϵ33 is not equal to zero and is given as ϵ33 ν E σ11 σ22 5140 Plane strain on the other hand occurs in a threedimensional solid subject to a uniform loading acting constantly along its length A typical example is a very long strip footing subject to a uniformly distributed load as shown in Figure 515 In these conditions change of thickness is y x H t L z FIGURE 514 A state of plane stress 2013 by Taylor Francis Group LLC Stress and Strain Analysis 163 FIGURE 515 State of plane strain prevented Therefore the ends of the footing are prevented from moving in the zdirection that is the displacement w of each face in the zdirection is equal to zero By symmetry at the midsection of the footing w must be also equal to zero In such a case the components of strain ϵ33 γ13 and γ23 are equal to zero The nonzero stresses are ϵ11 ϵ22 and γ12 Therefore Equation 5136 becomes σ11 σ22 σ12 E 1 ν1 2ν 1 ν ν 0 ν 1 ν 0 0 0 1 2ν 2 ϵ11 ϵ22 γ12 5141 Inverting expression 5141 yields ϵ11 ϵ22 γ12 1 ν E 1 ν ν 0 ν 1 ν 0 0 0 2 σ11 σ22 σ12 5142 Note also that in a state of plane strain σ33 is not equal to zero but it is given as σ33 νϵ11 ϵ22 5143 55 SOLVED PROBLEMS 551 PROBLEM 51 The stress tensor at a point P is given as σ 2 4 3 4 0 0 3 0 1 Find the stress vector on a plane that passes through P and is parallel to the plane x2y2z6 0 2013 by Taylor Francis Group LLC 164 Introduction to Finite Element Analysis Using MATLAB and Abaqus Solution The function defining the surface of the plane can be written as fxyz x 2y 2z 6 0 The vector normal to the plane V is obtained as V f x e1 f y e2 f z e3 1 e1 2 e2 2 e3 The normal unit vector n to the plane is therefore obtained as n 1 e1 2 e2 2 e3 V 1 3 e1 2 3 e2 2 3 e3 Hence T1 σ11n1 σ21n2 σ31n3 2 1 3 4 2 3 3 2 3 16 3 T2 σ12n1 σ22n2 σ23n3 4 1 3 0 2 3 0 2 3 4 3 T3 σ13n1 σ23n2 σ33n3 3 1 3 0 2 3 1 2 3 1 3 552 PROBLEM 52 The state of stress at point is given with respect to the Cartesian axes o x y z by the stress matrix σ 2 2 0 2 2 0 0 0 2 Determine the stress tensor σ in the Cartesian axes o x y z obtained by rotating the axes o x y z around z by 45 anticlockwise Check the result using the engineering notation of stress Solution Index and matrix notations The basis e 1 e 2 e 3 is obtained from the basis e1 e2 e3 by a rotation of 45 around e3 as shown in Figure 516 The transformation tensor lij or matrix Q are respectively given as lij cose i ej 22 22 0 22 22 0 0 0 1 2013 by Taylor Francis Group LLC Stress and Strain Analysis 165 e3 e3 e2 e2 45 e1 e1 FIGURE 516 Change of basis and Q cose i ej 22 22 0 22 22 0 0 0 1 The stress tensor σ in the basis e 1 e 2 e 3 is obtained as σ km lkilmjσij Let us consider the first component σ 11 It is obtained as σ 11 l1il1jσij l11l11σ11 l11l12σ12 l11l13σ13 l12l11σ21 l12l12σ22 l12l13σ23 l13l11σ31 l13l12σ32 l13l13σ33 σ 11 2 2 2 2 2 2 2 2 2 2 2 2 0 0 2 2 2 2 2 2 2 2 2 2 2 2 0 0 0 2 2 0 0 2 2 0 0 0 2 2 2 1 Repeating the same process for all the other terms we obtain σ 2 2 1 2 2 1 0 2 2 1 2 2 3 0 0 0 2 2013 by Taylor Francis Group LLC 166 Introduction to Finite Element Analysis Using MATLAB and Abaqus In matrix notation the transformation is carried out as σ QσQT 22 22 0 22 22 0 0 0 1 2 2 0 2 2 0 0 0 2 22 22 0 22 22 0 0 0 1 2 2 1 2 2 1 0 2 2 1 2 2 3 0 0 0 2 Engineering notation According to the engineering notation the stress transformation law is given as σ 11 σ 22 σ 33 σ 12 σ 23 σ 13 cos2 ψ sin ψ 0 2 sin ψ cos ψ 0 0 sin ψ cos2 ψ 0 2 sin ψ cos ψ 0 0 0 0 1 0 0 0 sin ψ cos ψ sin ψ cos ψ cos2 ψ sin2 ψ 0 0 0 0 0 0 0 cos ψ sin ψ 0 0 0 0 sin ψ cos ψ σ11 σ22 σ33 σ12 σ23 σ13 Introducing the numerical values we obtain σ 11 σ 22 σ 33 σ 12 σ 23 σ 13 05 05 0 1 0 0 05 05 0 1 0 0 0 0 1 0 0 0 05 05 0 0 0 0 0 0 0 0 2 2 2 2 0 0 0 0 2 2 2 2 2 2 2 2 0 0 2 2 1 2 2 3 2 2 2 1 0 0 The results compare very well 2013 by Taylor Francis Group LLC 553 PROBLEM 53 The Lagrangian description of the deformation of a body is given by x1 X1 x2 X2 02X3 x3 X3 02X2 Determine the deformation gradients F and the Green Lagrange strain matrix E Calculate the change in squared length of the lines OA AC and the diagonal OC for the small undeformed rectangle shown in Figure 517 Solution The deformation gradient is given by Fij xi Xj F x1 X1 x1 X2 x1 X3 x2 X1 x2 X2 x2 X3 x3 X1 x3 X2 x3 X3 The Green Lagrange strain matrix is given as E 1 2 FTF I 1 2 1 0 0 0 1 02 0 02 1 1 0 0 0 1 02 0 02 1 1 0 0 0 1 0 0 0 1 0 0 0 0 002 02 02 02 002 554 PROBLEM 54 Assuming small strain theory determine the linear strain tensor ε for the displacement field given by u x1 x32e1 x2 x3e2 x1x2e3 At point P0 2 1 determine The engineering normal strain in the direction 8e1 1e2 4e3 The change in right angle between v1 8e1 1e2 4e3 and v2 4e1 2e2 7e3 Solution The linear strain tensor is given as εij 1 2 ui xj uj xi or in matrix form as ε 1 2 u uT The displacement gradient is given as uijVudu1x1u1x2u1x3u2x1u2x2u2x3u3x1u3x2u3x32x1x302x1x30 2x2x32x2x3x2x1 At P021 the strain tensor is given as e2012 0211210 The unit vector in the direction v18e11e24e3 is given by v1v18g11g24g3 The engineering normal strain in this direction is given as e8919492012002106891 49 The unit vector in the direction v24e14e27e3 is given by v2v24g149g27g3 The change of right angle between v1 and v2 is given as γ12289194920120021031881 555 PROBLEM 55 A twodimensional solid is deformed as shown in Figure 518 Under the restriction of small deformation theory determine the linear strain tensor The solid lines represent the undeformed state Deduce the engineering form of the strain tensor The dimensions are given in mm Solution Comparing with Figure 511 it can be clearly seen that e11u1x100202001 e22u2x2003630012 γ1u2x1001020005 γ2u1x2001230004 γ12γ1γ20009 e12γ12200045 The strain tensor is given as e0010000450004500012 Using engineering notation the strain tensor is given in a vector form as εe11e22γ12γxy001 0012 0009 556 PROBLEM 56 A 45 strain rosette measures longitudinal strain along the axes shown in Figure 519 The following readings are obtained at point P ε110005ε110004 and ε220007 mmmm Determine the shear strain γ11 at the point Solution The unit vector in the direction X1 is given as n22 e122 e20e3 The stretch or engineering normal strain in the direction X1 is given as e222200005ε12 0ε12 0007000000 It follows therefore that 242ε1200120004γ120004 557 PROBLEM 57 Consider a cube of an isotropic linear elastic body whose edges are 10 mm long sitting in a rigid mold with a gap of 002 mm between the faces of the mold and that of the cube Figure 520 Determine the pressure on the lateral faces and the maximum shearing stress in the cube when the uniform pressure applied in the zdirection reaches 1200 MPa Take E 60000 MPa and ν 03 Solution First check that the pressure is causing enough lateral strains for the cube to reach the wall If we assume that the walls are inexistent then only nonzero stress is σ33 It follows from the strainstress relations that ε 1 ν ν 0 0 0 ν 1 ν 0 0 0 ν ν 1 0 0 0 0 0 21 ν 0 0 0 0 21 ν 0 0 0 0 21 ν which yields ε11 ε22 μE σ33 0360000 1200 0006 The displacement of the lateral faces is ΔL 0006 10 006 mm which is the total displacement Therefore each face is displaced by 003 mm which is greater than the 002 mm gap As a result lateral forces will develop preventing the lateral faces from expanding more than 002 mm Figure 521 At contact the lateral strains in the cube will be equal to ε11 ε22 00410 0004 Since the loading is in the principal directions the only nonzero strains and stresses are ε11 ε22 ε33 σ11 σ22 σ33 of which ε11 ε22 σ33 are known and σ11 σ22 ε33 are the unknowns Stress and Strain Analysis 173 003 mm 003 mm FIGURE 521 Displacements without the rigid walls The stressstrain relations can therefore be written as 0004 0004 ϵ33 0 0 0 1 E 1 ν ν 0 0 0 ν 1 ν 0 0 0 ν ν 1 0 0 0 0 0 0 21 ν 0 0 0 0 0 0 21 ν 0 0 0 0 0 0 21 ν σ11 σ22 1200 0 0 0 which yield three equations 0004 1 60000σ11 03σ22 03 1200 0004 1 6000003σ11 σ22 03 1200 ϵ33 1 6000003σ11 03σ22 1200 Solving the system of equations yields σ11 σ22 17143 MPa and ϵ33 0018 The maximum shear stress is given as σs 1 2σ11 σ33 1 217143 1200 5143 MPa 2013 by Taylor Francis Group LLC 558 PROBLEM 58 The displacement field of a circular bar that is being twisted by equal and opposite end moments is given by u1 0 u2 2 105x3 u3 2 105x1x3 The length of the bar is 2000 mm and the diameter is 400 mm If the bar is made of an isotropic linear elastic material with E 21 105 MPa and ν 03 using small deformation theory determine the state of stress in the points 2000 100 100 and 1000 100 100 What can be concluded about the variation of the stress along the length of the beam Solution The displacement gradient is given as u 0 0 0 2 105x3 0 0 0 2 105x2 2 105x1 0 The strain tensor small deformations is given as ε 12 u uᵀ 0 1 105x3 1 105x2 1 105x3 0 0 0 1 105x2 0 It can be seen that the strain tensor is not a function of the x1 coordinate ε2000 100 100 ε1000 100 100 0 1 103 1 103 1 103 0 0 1 103 0 The stresses do not vary along the length of the beam σ 0 16154 16154 16154 0 0 16154 0 MPa 6 Weighted Residual Methods 61 INTRODUCTION In Chapters 2 and 3 we used wellknown methods of structural analysis to develop the stiffness matrices of the bar and beam elements The reason being that these elements are onedimensional and the exact solutions of the differential equations governing their behaviors are well known For other structural problems in two and three dimensions such direct approaches are inexistent for the obvious reason that it is not possible to find analytical solutions to the differential equations governing their behavior except in the case of very simple geometries The alternative is to replace the differential equations by approximate algebraic equations This is achieved by using weighted residual methods 62 GENERAL FORMULATION Given a physical problem be it structural or not whose behavior is governed by a set of differential equations Bu 0 on Ω 61 where B represents a linear differential operator u the unknown function Ω is the geometrical domain Since the variable u is unknown we may try to substitute for it a trial or approximate function of our choosing say u given as a polynomial function u i1 to n αiPix 62 where the coefficients αi are general parameters Pix is a polynomial base Substituting u for u will not in general satisfy the differential equation 61 and will result in a residual over the whole domain Ω that is Bu 0 on Ω 63 The essence of the weighted residual methods is to force the residual to zero in some average over the whole domain Ω To do so we multiply the residual by a weighting function ψ and force the integral of the weighted residual to zero over the whole domain that is W Ω ψBu dΩ 0 64 There is a variety of residual methods such as collocation method subdomain method leastsquares method method of moments and Galerkin method They all differ in the choice of the weighting function ψ The most popular however is the Galerkin method and it is the only one described in this chapter This differential equation has an exact solution given by ux 1 sinx sin1 Given the following differential equation Bux d2ux dx2 ux x 0 on Ω 0 1 Equation 624 is an integral form of the differential Equation 620 and its natural boundary condition Integration by parts in two and three dimensions In the same manner if we integrate along the direction y we obtain intintOmega Phi fracpartial Psipartial y dy dx intintOmega fracpartial Phipartial y Psi dx dy intGamma Phi Psi nd Gamma 662 FUNCTIONAL ASSOCIATED WITH AN INTEGRAL FORM Consider Equation 625 If we adopt the method of Galerkin and substituting δux for ψ we obtain 66 RAYLEIGH RITZ METHOD 661 DEFINITION A functional Π is a function of a set of functions and their derivatives Π Πu ux ²ux² Taking the first variation of Π1 we obtain δΠ1 Π1 α1 δα1 0 6 20 α1 1 12 663 The trial function can therefore be written as u1x 5 18 xx 1 665 The total potential energy of a structure or solid in equilibrium is defined as the sum of the internal energy strain energy and the external energy the potential energy of the externally applied forces that is Π Ui Ue 670 Carrying out the matrix multiplication yields mathbfwx N1x N2x N3x N4x beginbmatrix w1 heta1 w2 heta2 endbmatrix ag683 B left beginarraycc left frac6l2 frac12xl3 right left frac4l frac6xl2 right left frac6l2 frac12xl3 right left frac2l frac6xl2 right endarray right ag691 int0fracl2 beginbmatrix N1x N2x N3x N4x endbmatrix qx dx beginbmatrix fracqL2 fracqL212 fracqL2 fracqL212 endbmatrix ag696 7 Finite Element Approximation Tx coincide with the exact solution at the five data points xi called nodes It follows or simply as Tx N T Nodal values V1 V2 Vn1 Vn Elements Ω1 x1x2 Ω2 x2x3 Ωn1 xn1xn Finite Element Approximation 196 Introduction to Finite Element Analysis Using MATLAB and Abaqus the displacement ux is a linear function of x As a result it is possible to construct a trial function ux for the displacement using a linear polynomial ux α1 α2x 720 The parameters α1 and α2 are identified using the two end nodal values U1 and U2 The bar problem is classified as a C0 problem The trial solution must be continuous and its derivative must exist Now let us consider the beam problem Under the applied uniformly distributed loading every cross section of the beam is subject to a vertical displacement wx and a rotation θx From the engineering beam theory the rotation θx is obtained as the first derivative of the deflection wx that is θx dwxdx The slope θx must be continuous otherwise the beam would develop kinks in its deflected shape Therefore if we are about to construct a trial function wx for the deflection then both the trial function and its first derivative must be continuous The second derivative which represents the curvature of the beam must exist A suitable trial function that satisfies these requirements would be wx α1 α2 x α3 x2 α4 x3 721 The four parameters α1 α2 α3 and α4 can be identified using the two end nodal values for the deflection w1 w2 and the two end values for the slope θ1 and θ2 The beam problem is classified as a C1 problem The trial solution and its first derivative must be continuous the second derivative must exist In general the compatibility principle can be formulated as follows For a class C0 problem continuity C0 the trial solution must be continuous across the boundary of the elements but not necessarily its derivatives For a class C1 problem continuity C1 both the trial solution and its firstorder derivatives must be continuous across the boundary of the elements but not necessarily its secondorder derivatives For a class Cn problem continuity Cn the trial solution and its n 1th order derivatives must be continuous across the boundary of the elements but not necessarily its nth order derivatives 742 COMPLETENESS PRINCIPLE Again consider the bar problem in Figure 74a If the applied force P is different from zero then the displacement ux has a finite value different from zero at any point x belonging to the bar except at x 0 where a displacement equal to zero is imposed boundary condition If we choose to discretize the bar with a linear twonodded element then the adopted trial function given in Equation 720 will make a suitable choice since if the size of the elements shrinks to zero that is limx0 ux α1 which is a constant representing the actual value of the displacement at that point However if the trial function did not contain a constant term limx0 ux will be equal to zero which actually does not represent the real case Furthermore the constant term is necessary for the trial function to be able to represent a rigid body motion In this case all points must have the same displacement ux α In addition we have duxdx α2 which represents the real case of the bar with a constant deformation This leads to the definition of the completeness principle which can be stated as follows When the size of the element shrinks to zero the trial function must be able to represent For a class C0 problem continuity C0 a constant value of the exact function as well as constant values of its firstorder derivatives 2013 by Taylor Francis Group LLC For a class C1 problem continuity C1 a constant value of the exact function as well as constant values of its first and secondorder derivatives For a class C2 problem continuity C2 a constant value of the exact function as well as constant values of its derivatives up to the nth order Expression 730 is a nodal approximation as opposed to 722 which is a general approximation The shape functions Nᵢx y i 1 2 3 are obtained as N₁x y 12Ay₂ y₂x₂ x x₃ x₂y₂ y 731 The trial function expression 722 can be rewritten in the form Ux y 1 x yᵀ a b c 723 The nodes of the parent element correspond to the nodes of the reference element and inversely The parameters αi can be easily obtained by solving the system 744 The inverse of the matrix A is obtained as A¹ 1 0 0 1 1 0 1 0 1 745 and the parameters αi as α1 α2 α3 1 0 0 1 1 0 1 0 1 x1 x2 x3 746 Substituting for the parameters αi in Equation 741 yields xξ η 1 ξ η 1 0 0 1 1 0 1 0 1 x1 x2 x3 747 Expanding and rearranging Equation 747 yields xξ η τ1ξ ηxi τ2ξ ηx2 τ3ξ ηx3 748 with τ1ξ η 1 ξ η τ2ξ η ξ τ3ξ η η 749 Following exactly the same process for the variable y yields yξ η τ1ξ ηyi τ2ξ ηy2 τ3ξ ηy3 750 Expressions 748 and 750 represent well and truly a linear geometrical transformation This can be easily checked The x coordinate of the midpoint between node 1 and node 2 of the parent element is given as x x1 x22 The ξ η coordinates of the corresponding point on the reference element are given as 12 0 Substituting these values in expression 749 and in expressions 748 yields x 1 05 0xi 05x2 0x3 x1 x2 2 The Jacobian of the transformation is given by J xξ xη yξ yη 751 After deriving and rearranging the Jacobian is written in the form of a product of two matrices J 1 1 0 1 0 1 x1 y1 x2 y2 x3 y3 752 202 Introduction to Finite Element Analysis Using MATLAB and Abaqus Since the geometrical transformation is well defined we will construct the trial function Ux y for an unknown function Fx y over the reference element The unknown function defined over the parent element is of class C0 with nodal values F1 F2 and F3 Since it is of the same class as the coordinates x and y we will reuse the same trial function that is Uξ η α1 α2ξ α3η 753 Following exactly the same procedure as previously and replacing x1 x2 and x3 respectively with the nodal values F1 F2 and F3 we end up with Uξ η N1ξ ηF1 N2ξ ηF2 N3ξ ηF3 754 with N1ξ η 1 ξ η N2ξ η ξ N3ξ η η 755 Remark The shape functions Niξ η are exactly the same as the functions τiξ η of the geometrical transformation This is due to the fact that the function Uξ η is of the same class as the coordinates x and y and most importantly the geometrical nodes the nodes used to define the geometry of the element are the same as the interpolation nodes the nodes used to define the nodal values of the unknown function Such an element is called an isoparametric same parameters element since it uses the same nodes to define both the geometry and interpolate the function 7513 Area Coordinates Let us consider an arbitrary point O of the triangular element shown in Figure 77 The area coordinates L1 L2 and L3 are defined as L1 AreaO23 Area123 756 3 O 1 2 FIGURE 77 Threenode triangular element with an arbitrary point O 2013 by Taylor Francis Group LLC Finite Element Approximation 203 L 2 AreaO13 Area123 757 L3 AreaO12 Area123 758 From these definitions it follows that L1 L 2 L3 1 759 It is also obvious that When point O coincides with node 1 L1 1 L 2 0 and L3 0 When point O coincides with node 2 L1 0 L 2 1 and L3 0 When point O coincides with node 3 L1 0 L 2 0 and L3 1 In addition moving point O in any direction will result in a linear variation of the area coordinates L1 L2 and L3 in terms of x and y Therefore it should be clear to the reader that the area coordinates L1 L2 and L3 are indeed the same as the shape functions N1 N2 and N3 given in Equation 731 that is L1 N1x y L2 N2x y L3 N3x y 760 In the case of a reference triangular element as shown in Figure 78 the area coordinates are expressed in terms of the coordinates ξ η as follows L1 N1ξ η 1 ξ η L2 N2ξ η ξ L3 N3ξ η η 761 752 LINEAR QUADRILATERAL ELEMENT FOR C0 PROBLEMS 7521 Geometrical Transformation In the quadrilateral family of elements except for the square or the rectangle it is impossible to construct the shape functions directly in terms of x and y as we did for the triangle The only way to construct these functions is to use a reference element which is a square of side 2 units as represented in Figure 79 To define the geometrical transformation we will assume that the coordinates x y of an arbitrary point of the parent element are the unknown functions defined over the domain represented by the reference element in its local coordinate system ξ η Notice that both the variables x and y belong to the C0 class of functions since they are continuous and their first derivatives are constant equal to 1 Therefore we start by constructing a general approximation for x in terms of ξ and η x α1 α2ξ α3η α4ξη 762 2013 by Taylor Francis Group LLC or in a matrix form as x 1 ξ η ξη α1 α2 α3 α4 763 Then we will transform the general approximation Equation 762 to a nodal approximation by using the nodal values x1 x2 x3 and x4 respectively at nodes 1 2 3 and 4 Notice also that the couple ξ η takes on the values of 1 1 1 1 1 1 and 1 1 respectively at nodes 1 2 3 and 4 It follows x1 α1 α2 α3 α4 x2 α1 α2 α3 α4 x3 α1 α2 α3 α4 x4 α1 α2 α3 α4 764 which when rewritten in a matrix form yields x1 x2 x3 x4 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 α1 α2 α3 α4 765 or in a more compact form as X Aα 766 The parameters αi can be obtained easily by solving the system 765 It can be noticed that the columns of the matrix A are actually orthogonal vectors of norm 4 Hence the inverse of the matrix A is obtained as A¹ 14 Aᵀ 14 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 767 and the parameters αi as α1 α2 α3 α4 14 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 x1 x2 x3 x4 768 Substituting for the parameters αi in Equation 763 yields xξ η 1 ξ η ξη 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 4 x1 x2 x3 x4 769 Expanding and rearranging Equation 769 leads to xξ η τ1ξ ηxi τ2ξ ηx2 τ3ξ ηx3 τ4ξ ηx4 770 with τ1ξ η 0251 ξ η ξη τ2ξ η 0251 ξ η ξη τ3ξ η 0251 ξ η ξη τ4ξ η 0251 ξ η ξη 771 Following exactly the same process for the variable y we obtain yξ η τ₁ξ ηy₁ τ₂ξ ηy₂ τ₃ξ ηy₃ τ₄ξ ηy₄ 772 Expressions 770 and 772 represent well and truly a linear geometrical transformation This can be easily checked as follows The center of the reference square is given by ξ η 0 0 Substituting these values in expression 771 and then in expressions 770 and 772 yields x 14 x₁ x₂ x₃ x₄ y 14 y₁ y₂ y₃ y₄ which are the coordinates of the center of the parent element in the x y coordinate system The Jacobian of the transformation is given by J xξ yξ xη yη ⁴ᵢ₁ xᵢξ ⁴ᵢ₁ yᵢξ ⁴ᵢ₁ xᵢη ⁴ᵢ₁ yᵢη 773 After deriving and rearranging the Jacobian is written in the form of a product of two matrices J 14 1 η 1 ξ 1 η 1 η 1 ξ 1 ξ 1 ξ 1 ξ 774 7522 Construction of a Trial Function over a Linear Quadrilateral Element Now let us construct a trial function Ux y for an unknown function Fx y of class C⁰ with nodal values F₁ F₂ F₃ and F₄ defined over the parent element Since the geometrical transformation is well defined we will construct the trial function over the reference element The function Fx y is of the same class as the coordinates x and y we will use the same trial function that is Uξ η α₁ α₂ξ α₃η α₄ξη Following exactly the same procedure as previously described we end up with Uξ η 1 ξ η ξη 1 1 1 1 1 1 1 1 1 1 1 1 F₁ F₂ F₃ F₄ 776 which after expanding and rearranging becomes Uξ η N₁ξ ηF₁ N₂ξ ηF₂ N₃ξ ηF₃ N₄ξ ηF₄ 777 with N₁ξ η 0251 ξ1 η ξη N₂ξ η 0251 ξ1 η N₃ξ η 0251 ξ η ξη N₄ξ η 0251 ξ η ξη 778 208 Introduction to Finite Element Analysis Using MATLAB and Abaqus Threenodded linear Sixnodded quadratic η η 3 4 6 5 1 2 ξ 1 2 3 ξ FIGURE 712 Twodimensional triangular elements 7622 EightNodded Quadratic Quadrilateral N1ξ η N2ξ η N3ξ η N4ξ η N5ξ η N6ξ η N7ξ η N8ξ η 0251 ξ1 η1 ξ η 0501 ξ21 η 0251 ξ1 η1 ξ η 0501 ξ1 η2 0251 ξ1 η1 ξ η 0501 ξ21 η 0251 ξ1 η1 ξ η 0501 ξ1 η2 782 7623 ThreeNodded Linear Triangle Figure 712 N1ξ η N2ξ η N3ξ η 1 ξ η ξ η 783 7624 SixNodded Quadratic Triangle N1ξ η N2ξ η N3ξ η N4ξ η N5ξ η N6ξ η 1 ξ η1 21 ξ η 4ξ1 ξ η ξ1 2ξ 4ξη η1 2η 4η1 ξ η 784 763 THREEDIMENSIONAL ELEMENTS 7631 FourNodded Linear Tetrahedra N1ξ η ζ N2ξ η ζ N3ξ η ζ N4ξ η ζ 1 ξ η ζ ξ η ζ 785 2013 by Taylor Francis Group LLC Finite Element Approximation 209 Fournodded linear Tennodded quadratic 2 1 4 10 7 9 5 η 4 2 8 1 6 ζ ζ 3 3 η ξ ξ FIGURE 713 Threedimensional tetrahedric elements 7632 TenNodded Quadratic Tetrahedra Figure 713 N1ξ η ζ N2ξ η ζ N3ξ η ζ N4ξ η ζ N5ξ η ζ N6ξ η ζ N7ξ η ζ N8ξ η ζ N9ξ η ζ N10ξ η ζ 1 ξ η ζ1 21 ξ η ζ 4ξ1 ξ η ζ ξ1 2ξ 4ξη η1 2η 4η1 ξ η ζ 4ζ1 ξ η ζ 4ξζ 4ηζ ζ1 2ζ 786 7633 EightNodded Linear Brick Element N1ξ η ζ N2ξ η ζ N3ξ η ζ N4ξ η ζ N5ξ η ζ N6ξ η ζ N7ξ η ζ N8ξ η ζ 1 8 1 ξ1 η1 ζ 1 ξ1 η1 ζ 1 ξ1 η1 ζ 1 ξ1 η1 ζ 1 ξ1 η1 ζ 1 ξ1 η1 ζ 1 ξ1 η1 ζ 1 ξ1 η1 ζ 787 2013 by Taylor Francis Group LLC 210 Introduction to Finite Element Analysis Using MATLAB and Abaqus 7634 TwentyNodded Quadratic Brick Element Figure 714 N1ξ η ζ N2ξ η ζ N3ξ η ζ N4ξ η ζ N5ξ η ζ N6ξ η ζ N7ξ η ζ N8ξ η ζ N9ξ η ζ N10ξ η ζ N11ξ η ζ N12ξ η ζ N13ξ η ζ N14ξ η ζ N15ξ η ζ N16ξ η ζ N17ξ η ζ N18ξ η ζ N19ξ η ζ N20ξ η ζ 1 81 ξ1 η1 ζ2 ξ η ζ 1 41 ξ21 η1 ζ 1 81 ξ1 η1 ζ2 ξ η ζ 1 41 ξ1 η21 ζ 1 81 ξ1 η1 ζ2 ξ η ζ 1 41 ξ21 η1 ζ 1 81 ξ1 η1 ζ2 ξ η ζ 1 41 ξ1 η21 ζ 1 41 ξ1 η1 ζ2 1 41 ξ1 η1 ζ2 1 41 ξ1 η1 ζ2 1 41 ξ1 η1 ζ2 1 81 ξ1 η1 ζ2 ξ η ζ 1 41 ξ21 η1 ζ 1 81 ξ1 η1 ζ2 ξ η ζ 1 41 ξ1 η21 ζ 1 81 ξ1 η1 ζ2 ξ η ζ 1 41 ξ21 η1 ζ 1 81 ξ1 η1 ζ2 ξ η ζ 1 41 ξ1 η21 ζ 788 Eightnodded linear Twentynodded quadratic ζ ζ 2 8 11 17 18 20 12 19 13 14 15 10 9 16 1 1 4 7 7 5 8 6 6 η η ξ ξ 4 3 5 3 2 FIGURE 714 Threedimensional brick elements 2013 by Taylor Francis Group LLC 81 INTRODUCTION In Section 664 analytical integration was used to integrate the expression of the theorem of virtual work during the evaluation of the stiffness matrix of the beam element That was relatively easy because a beam element is unidimensional However when the number of elements is large andor their geometrical shape is general as is the case in most finite element applications the use of analytical integration is quite cumbersome and illsuited for computer coding The alternative is to use numerical integration There exist many numerical methods for evaluating a definite integral Simpsons rule Newton Cotes and Gauss quadrature are examples of such methods The basic idea of numerical integration is to replace the continuous integral with a series of finite sums b a fx dx n i1 Aifxᵢ error 81 The parameters Aᵢ are called the weights of the integration In finite element application Gauss quadrature also called the GaussLegendre method is the most widely used as it is the most precise For Equation 84 to be identically satisfied for all αi we must have the following equalities TABLE 81 Abscissa and Weights for Gauss Quadrature Using only six significant digits for the abscissa and the weights the integral becomes I 0 to π 0 to 3 x² x sin y dx dy 224 Introduction to Finite Element Analysis Using MATLAB and Abaqus eta sampj1 WJ sampj2 derfun fmquadsamp ij Form the vector of the shape functions and the matrix of their derivatives JAC dercoord Evaluate the Jacobian DET detJAC Evaluate determinant of Jacobian matrix Ixx Ixx dotfunY2WIWJDET end end end Ixx TwoQ8 Input module TwoQ8m Two elements mesh global geom connec nel nne nnd RI RE nnd 13 Number of nodes The matrix geom contains the x and y coordinates of the nodes geom RI 0 node 1 RIcospi8 RIsinpi8 node 2 RIcospi4 RIsinpi4 node 3 RIcos3pi8 RIsin3pi8 node 4 RIcospi2 RIsinpi2 node 5 RIRE2 0 node 6 RIRE2cospi4 RIRE2sinpi4 node 7 RIRE2cospi2 RIRE2sinpi2 node 8 RE 0 node 9 REcospi8 REsinpi8 node 10 REcospi4 REsinpi4 node 11 REcos3pi8 REsin3pi8 node 12 REcospi2 REsinpi2 node 13 nel 2 Number of elements nne 8 Number of nodes per element The matrix connec contains the connectivity of the elements connec 1 6 9 10 11 7 3 2 Element 1 3 7 11 12 13 8 5 4 Element 2 End of input module TwoQ8m EightQ8m Eight elements mesh global geom connec nel nne nnd RI RE nnd 37 Number of nodes The matrix geom contains the x and y coordinates of the nodes geom RI 0 node 1 RIRERI4 0 node 2 RIRERI2 0 node 3 RI3RERI4 0 node 4 RE 0 node 5 RIcospi16 RIsinpi16 node 6 RIRERI2cospi16 RIRERI2sinpi16 node 7 REcospi16 REsinpi16 node 8 2013 by Taylor Francis Group LLC Numerical Integration 225 RIcospi8 RIsinpi8 node 9 RIRERI4cospi8 RIRERI4sinpi8 node 10 RIRERI2cospi8 RIRERI2sinpi8 node 11 RI3RERI4cospi8 RI3RERI4sinpi8 node 12 REcospi8 REsinpi8 node 13 RIcos3pi16 RIsin3pi16 node 14 RIRERI2cos3pi16 RIRERI2sin3pi16 node 15 REcos3pi16 REsin3pi16 node 16 RIcospi4 RIsinpi4 node 17 RIRERI4cospi4 RIRERI4sinpi4 node 18 RIRERI2cospi4 RIRERI2sinpi4 node 19 RI3RERI4cospi4 RI3RERI4sinpi4 node 20 REcospi4 REsinpi4 node 21 RIcos5pi16 RIsin5pi16 node 22 RIRERI2cos5pi16 RIRERI2sin5pi16 node 23 REcos5pi16 REsin5pi16 node 24 RIcos6pi16 RIsin6pi16 node 25 RIRERI4cos6pi16 RIRERI4sin6pi16 node 26 RIRERI2cos6pi16 RIRERI2sin6pi16 node 27 RI3RERI4cos6pi16 RI3RERI4sin6pi16 node 28 REcos6pi16 REsin6pi16 node 29 RIcos7pi16 RIsin7pi16 node 30 RIRERI2cos7pi16 RIRERI2sin7pi16 node 31 REcos7pi16 REsin7pi16 node 32 RIcospi2 RIsinpi2 node 33 RIRERI4cospi2 RIRERI4sinpi2 node 34 RIRERI2cospi2 RIRERI2sinpi2 node 35 RI3RERI4cospi2 RI3RERI4sinpi2 node 36 REcospi2 REsinpi2 node 37 nel 8 Number of elements nne 8 Number of nodes per element The matrix connec contains the connectivity of the elements connec 1 2 3 7 11 10 9 6 Element 1 3 4 5 8 13 12 11 7 Element 2 9 10 11 15 19 18 17 14 Element 3 11 12 13 16 21 20 19 15 Element 4 17 18 19 23 27 26 25 22 Element 5 19 20 21 24 29 28 27 23 Element 6 25 26 27 31 35 34 33 30 Element 7 27 28 29 32 37 36 35 31 Element 8 End script EightQ8m Next we provide the abscissae and weights necessary to perform a gauss quadrature These are given in the script gaussm listed in Appendix A which is a function that returns the matrix sampngp 2 The first column contains the abscissa and the second column the weights For each element we retrieve the coordinates of its nodes using the script coordq8m also listed in Appendix A The double sum is evaluated using three Gauss points ngp 3 The shape functions Niξi ηj given in the vector funnne as well as their derivatives returned in the matrix der2 nne are all evaluated at the Gauss points using the script fmquadm listed in Appendix A The Jacobian is simply evaluated as jac der coord The second moment of area is obtained as a sum of all the terms Ixx Ixx dotfun Y2 WI WJ DET with Ixx being previously initialized to zero After execution of the code the second moment of area obtained with the coarse mesh is Ixx 4205104 mm4 2013 by Taylor Francis Group LLC 230 Introduction to Finite Element Analysis Using MATLAB and Abaqus 65 105 Node 5 40 60 Node 6 Precipitations recorded by the rain gauges q 20 15 10 20 30 25 Connectivity connec 1 2 3 4 5 6 Element 1 AT 0 Initialize total area to zero QT 0 Initialize total rainfall to zero for i1nel for each element retrieve the vector qe containing the precipitations at its nodes as well as the matrix coord containing the x and y coordinates of the nodes for k1 nne qek qconnecik for j12 coordkjgeomconnecikj end end for ig 1npt WI sampig3 derfun fmT6quadsamp ig JAC dercoord DET detJAC calculate its area AT AT WIDET Estimate quantity of rain over its area QT QT WIdotfunqeDET end end AT QT After execution of the code we obtain exactly the same results as with the linear triangular elements that is A 3775 km2 Q 75500 mm km2 2013 by Taylor Francis Group LLC 9 Plane Problems 91 INTRODUCTION By now it should have become clear to the reader that in any finite element analysis we are not analyzing the actual physical problem but a mathematical model of it As a result we introduce some simplifications and hence some modeling errors In reality all solids are threedimensional Fortunately for many problems which are of practical interest some simplifying assumptions can be made regarding the stress or strain distributions For example in Chapters 2 through 4 dealing with skeletal structures linetype elements were used because of the predominance of the longitudinal stress In Section 544 we have also seen that when the loading andor geometry permit it a solid can be analyzed as a plane stress or plane strain problem There are also other simplifications for solids that posses a symmetry of revolution in both geometry and loading and for flat solids loaded perpendicular to their plane These will be dealt respectively in Chapters 10 and 11 However unlike skeletal structures whose discretization into an assembly of elements is relatively easy the connecting joints naturally constitute the nodes such an intuitive approach does not exist for a two or threedimensional continuum There are no joints to be used as nodes or cleavage lines to be used as elements edges Hence the discretization becomes a process that requires an understanding of the physical problem at hand It should be also added that the more physical details one tries to capture the more complex the model becomes In particular the user has to decide on the choice of element type and size These depend on the physical makeup of the body the loading and on how close to the actual behavior the user wants the results to be Heshe also has to decide whether the model can be simplified And how could the results be checked There are of course no definite answers to these questions In this chapter dealing with plane problems and in Chapters 10 and 11 we will formulate the finite element method and in the process attempt to answer some of these questions The user however is reminded that only practice makes perfect 92 FINITE ELEMENT FORMULATION FOR PLANE PROBLEMS The stressstrain relationships for a plane problem see Section 544 are given for plane stress as σxx σyy τxy E 1 ν2 1 ν 0 ν 1 0 0 0 1 ν 2 ϵxx ϵyy γxz 91 and for plane strain as σxx σyy τxy E 1 ν1 2ν 1 ν ν 0 ν 1 ν 0 0 0 1 2ν 2 ϵxx ϵyy γxy 92 Whether it is a state of plane stress or plane strain a material point can only move in the direc tions x and y Therefore the two displacement variables that play a role are ux y and vx y 231 2013 by Taylor Francis Group LLC Plane Problems 235 FIGURE 92 Discretization error involving holes between elements Elongated element Near triangular Highly skewed Near triangular FIGURE 93 Plane elements with shape distortions FIGURE 94 Geometrical discretization error 94 CONSTANT STRAIN TRIANGLE The linear triangular element shown in Figure 95 is perhaps the earliest finite element It has three nodes and each node has two degrees of freedom Its shape functions have already been obtained in Chapter 7 and they are given as N1x y m11 m12x m13y N2x y m21 m22x m23y 922 N3x y m31 m32x m33y 2013 by Taylor Francis Group LLC Plane Problems 237 or more compactly as U Na 928 942 STRAIN MATRIX Substituting for the displacements u and v in Equation 96 using Equation 927 the strain vector is obtained as ϵ Ba 929 with B N1 x 0 N2 x 0 N3 x 0 0 N1 y 0 N2 y 0 N3 y N1 y N1 x N2 y N2 x N3 y N3 x 930 Substituting Equations 922 and 923 in 930 the matrix B becomes B m12 0 m22 0 m32 0 0 m13 0 m23 0 m33 m13 m12 m23 m22 m33 m32 931 Remark The matrix B is independent of the Cartesian coordinates x and y It is a function of the nodal coordinates only and it is constant all over the element It follows therefore that the strain vector is constant over the element That is the reason why the element is termed constant strain triangle 943 STIFFNESS MATRIX The stiffness matrix of the element is given by Equation 921 Since both the matrices B and D are constant the stiffness matrix becomes Ke BTDBtAe 932 where Ae represents the area of the element and is given by Equation 924 944 ELEMENT FORCE VECTOR The element force vector is given by Equation 922 2013 by Taylor Francis Group LLC Plane Problems 241 3 1 kN 3 2 2 1 1 4 5 5 6 8 8 11 11 10 9 12 15 16 14 19 20 23 24 22 21 16 19 18 17 17 14 13 13 10 60 mm 20 mm 12 15 18 21 20 7 7 4 6 9 FIGURE 98 Finite element discretization with linear triangular elements 9451 Data Preparation To read the data we will use the Mfile CSTCOARSEMESHDATAm listed next FILE CSTCOARSEMESHDATAm File CSTCOARSEMESHDATAm The following variables are declared as global in order to be used by all the functions Mfiles constituting the program global nnd nel nne nodof eldof n global geom connec dee nf Nodalloads format short e nnd 21 Number of nodes nel 24 Number of elements nne 3 Number of nodes per element nodof 2 Number of degrees of freedom per node eldof nnenodof Number of degrees of freedom per element Nodes coordinates x and y geom zerosnnd2 geom 0 10 Node 1 0 0 Node 2 0 10 Node 3 10 10 Node 4 10 0 Node 5 10 10 Node 6 20 10 Node 7 20 0 Node 8 20 10 Node 9 30 10 Node 10 30 0 Node 11 30 10 Node 12 40 10 Node 13 40 0 Node 14 40 10 Node 15 50 10 Node 16 50 0 Node 17 50 10 Node 18 60 10 Node 19 2013 by Taylor Francis Group LLC 242 Introduction to Finite Element Analysis Using MATLAB and Abaqus 60 0 Node 20 60 10 Node 21 Element connectivity conneczerosnel3 connec 1 4 2 Element 1 4 5 2 Element 2 2 5 3 Element 3 5 6 3 Element 4 4 7 5 Element 5 7 8 5 Element 6 5 8 6 Element 7 8 9 6 Element 8 7 10 8 Element 9 10 11 8 Element 10 8 11 9 Element 11 11 12 9 Element 12 10 13 11 Element 13 13 14 11 Element 14 11 14 12 Element 15 14 15 12 Element 16 13 16 14 Element 17 16 17 14 Element 18 14 17 15 Element 19 17 18 15 Element 20 16 19 17 Element 21 19 20 17 Element 22 17 20 18 Element 23 20 21 18 Element 24 Material E 200000 Elastic modulus in MPa vu 03 Poissons ratio thick 5 Beam thickness in mm Form the elastic matrix for plane stress dee formdsigEvu Boundary conditions nf onesnnd nodof Initialize the matrix nf to 1 nf191 0 nf192 0 Prescribed nodal freedom of node 19 nf201 0 nf202 0 Prescribed nodal freedom of node 20 nf211 0 nf212 0 Prescribed nodal freedom of node 21 Counting of the free degrees of freedom n0 for i1nnd for j1nodof if nfij 0 nn1 nfijn end end end loading Nodalloads zerosnnd 2 Nodalloads21 0 Nodalloads22 1000 Node 2 End of input 2013 by Taylor Francis Group LLC Plane Problems 243 The input data for this beam consist of nnd 21 number of nodes nel 24 number of elements nne 3 number of nodes per element nodof 2 number of degrees of freedom per node The thickness of the beam which is a geometrical property is given as thick 5 9452 Nodes Coordinates The coordinates x and y of the nodes are given in the form of a matrix geomnnd 2 9453 Element Connectivity The element connectivity is given in the matrix connecnel 3 Note that the internal numbering of the nodes is anticlockwise 9454 Material Properties The material properties namely elastic modulus and Poissons ratio are given in the variables E 200000 and vu 03 With these properties we form the elastic matrix for plane stress using the function formdsigm listed in Appendix A which returns the matrix dee 9455 Boundary Conditions In the same fashion as for a truss or a beam a restrained degree of freedom is assigned the digit 0 while a free degree of freedom is assigned the digit 1 As previously explained a node has two degrees of freedom a horizontal translation along the axis X and a vertical translation along the axis Y As shown in Figure 98 nodes 19 20 and 21 represent the fixed end of the cantilever which is fully fixed The prescribed degrees of freedom of these nodes are assigned the digit 0 All the degrees of freedom of all the other nodes which are free are assigned the digits 1 The information on the boundary conditions is given in the matrix nfnnd nodof 9456 Loading The concentrated force of 1000 N is applied at node 2 The force will be assembled into the global force vector fg in the main program 9457 Main Program The main program CSTPLANESTRESSm is listed next THIS PROGRAM USES AN 3NODE LINEAR TRIANGULAR ELEMENT FOR THE LINEAR ELASTIC STATIC ANALYSIS OF A TWO DIMENSIONAL PROBLEM clear all clc Make these variables global so they can be shared by other functions global nnd nel nne nodof eldof n global geom connec dee nf Nodalloads format long g ALTER NEXT LINES TO CHOOSE THE NAME OF THE OUTPUT FILE fid fopenCSTCOARSEMESHRESULTStxtw 2013 by Taylor Francis Group LLC 244 Introduction to Finite Element Analysis Using MATLAB and Abaqus To change the size of the problem or change elastic properties supply another input file CSTCOARSEMESHDATA End of input Assemble the global force vector fgzerosn1 for i1 nnd if nfi1 0 fgnfi1 Nodalloadsi1 end if nfi2 0 fgnfi2 Nodalloadsi2 end end Assembly of the global stiffness matrix initialize the global stiffness matrix to zero kk zerosn n for i1nel beegA elemT3i Form strain matrix and steering vector kethickAbeedeebee Compute stiffness matrix kkformkkkkke g assemble global stiffness matrix end End of assembly delta kkfg solve for unknown displacements nodedispzerosnnd2 for i1 nnd if nfi1 0 xdisp 0 else xdisp deltanfi1 end if nfi2 0 ydisp 0 else ydisp deltanfi2 end nodedispi xdisp ydisp end Retrieve the xcoord and ydisp of the nodes located on the neutral axis k 0 for i1nnd if geomi2 0 kk1 xcoordk geomi1 verticaldispknodedispi2 end end 2013 by Taylor Francis Group LLC Plane Problems 245 for i1nel beegA elemT3i Form strain matrix and steering vector eldzeroseldof1 Initialize element displacement to zero for m1eldof if gm0 eldm0 else eldmdeltagm Retrieve element displacement end end epsbeeeld Compute strains EPSieps Store strains for all elements sigmadeeeps Compute stresses SIGMAisigma Store strains for all elements end Print results to file printCSTresults Plot the stresses in the xdirection xstress SIGMA1 cmin minxstress cmax maxxstress caxiscmin cmax patchFaces connec Vertices geom FaceVertexCDataxstress FacecolorflatMarkero colorbar plottools After declaring the global variables that will be used by the functions and the naming of the output results file CSTCOARSEMESHRESULTStxt the program starts by uploading the data file and assembling the global force vector fg The elements stiffness matrices the assembly of the global stiffness matrix the solution of the global equations and the computation of stresses and strains are obtained as follows 9458 Element Stiffness Matrix For each element from 1 to nel we set up its strain matrix bee its steering vector g and calculate its area A This is achieved in the function elemT3m which can be found in Appendix A For any element i retrieve the coordinates x and y of its nodes x1 geomconneci 1 1 y1 geomconneci 1 2 x2 geomconneci 2 1 y2 geomconneci 2 2 x3 geomconneci 3 1 y3 geomconneci 3 2 Calculate the area of the element using Equation 734 and the coefficients mjk j k 1 2 3 using Equation 736 Using the coefficients mjk assemble the matrix bee using Equation 931 2013 by Taylor Francis Group LLC 246 Introduction to Finite Element Analysis Using MATLAB and Abaqus Using the matrix of nodal freedom nf in combination with the connectivity matrix retrieve the steering vector g for the element g nfconnec1 1 1 nfconnec1 1 2 nfconnec2 1 1 nfconnec2 1 2 nfconnec3 1 1 nfconnec3 1 2 Once the matrix bee is formed the element stiffness matrix ke is obtained as ke thick A beeT dee bee 9459 Assembly of the Global Stiffness Matrix As shown in Figure 95 a linear triangular element has in total 6 degrees of freedom The global stiffness matrix KK is assembled using a double loop over the components of the vector g The script is exactly the same as the one used in the codes Trussm Beamm and Framem It is given in the function formKKm listed in Appendix A 94510 Solution of the Global System of Equations The solution of the global system of equations is obtained with one statement delta KKfg 94511 Nodal Displacements Once the global displacements vector delta is obtained it is possible to retrieve any nodal displace ments A loop is carried over all the nodes If a degree of freedom j of a node i is free that is nfi j 0 then it could have a displacement different from zero The value of the displacement is extracted from the global displacements vector delta nodedispi j deltanfi j 94512 Element Stresses and Strains To obtain the element stresses and strains a loop is carried over all the elements 1 Form element strain matrix bee and steering vector g a Loop over the degrees of freedom of the element to obtain element displacements vector edg b If gj 0 then the degree of freedom is restrained edgj 0 c Otherwise edgj deltagj 2 Obtain element strain vector eps bee edg 3 Obtain element stress vector sigma dee bee edg 4 Store the strains for all the elements EPSi eps for printing to result file 5 Store the stresses for all the elements SIGMAi sigma for printing to result file 2013 by Taylor Francis Group LLC Plane Problems 247 94513 Results and Discussion After running the program CSTPLANESTRESSm the results are written to the text file CSTCOARSEMESHRESULTStxt listed next CSTCOARSEMESHRESULTStxt PRINTING ANALYSIS RESULTS Nodal displacements Node dispx dispy 1 145081e002 649329e002 2 328049e004 652078e002 3 142385e002 647141e002 4 142332e002 497317e002 5 182950e004 494530e002 6 138358e002 494091e002 7 129745e002 350495e002 8 137982e004 346630e002 9 126721e002 347556e002 10 109224e002 219922e002 11 895233e005 214870e002 12 107002e002 216958e002 13 808085e003 113485e002 14 256420e005 107261e002 15 790991e003 110480e002 16 446383e003 388383e003 17 663586e005 319069e003 18 426507e003 366370e003 19 000000e000 000000e000 20 000000e000 000000e000 21 000000e000 000000e000 Element stresses element sigmaxx sigmayy tauxy 1 78546e000 78546e000 78546e000 2 13515e000 51683e000 13112e001 3 66118e002 98937e000 91400e000 4 91400e000 36192e000 98937e000 5 25827e001 21744e000 48607e000 6 15601e000 81980e000 15027e001 7 69913e001 66741e001 59323e000 8 24966e001 56374e000 14180e001 9 42552e001 50356e000 16983e000 10 22662e000 10785e001 18024e001 11 16757e000 23552e000 28152e000 12 41961e001 84119e000 17462e001 13 59121e001 76315e000 14550e000 14 26997e000 13258e001 20813e001 15 27809e000 50108e000 22163e001 16 59202e001 11322e001 20864e001 17 75391e001 10170e001 45429e000 18 25481e000 14627e001 23117e001 19 41445e000 76816e000 30783e000 20 76988e001 13636e001 24504e001 21 93536e001 14198e001 49720e000 22 14584e000 43753e001 24544e001 23 16603e000 99582e000 77540e000 24 93738e001 28121e001 28182e001 Element strains element epsilonxx epsilonyy gammaxy 1 27491e005 27491e005 10211e004 2013 by Taylor Francis Group LLC 248 Introduction to Finite Element Analysis Using MATLAB and Abaqus 2 14510e005 27869e005 17045e004 3 14510e005 49369e005 11882e004 4 40271e005 43858e006 12862e004 5 12587e004 27869e005 63189e005 6 44967e006 38650e005 19535e004 7 44967e006 43858e006 77120e005 8 11637e004 92623e006 18434e004 9 20521e004 38650e005 22078e005 10 48459e006 50524e005 23431e004 11 48459e006 92623e006 36597e005 12 19719e004 20883e005 22701e004 13 28416e004 50524e005 18915e005 14 63881e006 62239e005 27057e004 15 63881e006 20883e005 28812e006 16 27903e004 32191e005 27123e004 17 36170e004 62239e005 59058e005 18 92001e006 69314e005 30052e004 19 92001e006 32191e005 40018e005 20 36448e004 47301e005 31856e004 21 44638e004 69314e005 64636e005 22 66359e006 00000e000 31907e004 23 66359e006 47301e005 10080e004 24 42651e004 00000e000 36637e004 Once the calculations are done the first thing that needs to be checked is whether the results are reasonable or not This task is even more difficult when inhouse software is used as is the case here The results as shown earlier are in the form of numbers hence difficult to interpret The first thing we can do is to check whether the deflected shape is correct For this we plot the vertical displacement of the nodes situated along the neutral axis of the cantilever as shown in Figure 99 As it appears the shape is acceptable however the computed values are just over half those obtained with the analytical solution Equation 941 Next we plot a contour of the longitudinal stress σxx using the MATLAB patch function as shown in Figure 910 The elements above the neutral axis are in tension while those below the neutral axis are in compression which is obviously correct Most importantly the stress value is constant over each element However the neutral axis should be stress free and that is not the case As they are the results are not satisfactory Indeed we are asking too much of the constant strain stress triangle that is to model a stress gradient when evidently it cannot do so We have also used a coarse mesh without sufficient refinement to model the stress gradient 0 002 004 006 Vertical deflection mm 008 01 012 0 10 20 30 40 50 60 Numerical Analytical Length mm FIGURE 99 Deflection of the cantilever beam 2013 by Taylor Francis Group LLC Plane Problems 249 10 σxx kNmm2 80 60 40 20 20 40 60 80 0 8 6 4 2 0 2 4 6 8 10 0 10 20 30 40 50 60 FIGURE 910 Stresses along the xaxis 94514 Program with Automatic Mesh Generation To better model the stress gradient with a triangular element we need to refine the mesh However this will require many elements and nodes which is not easy to prepare by hand as we did for the coarse mesh In the new program named CSTPLANESTRESSMESHm listed next the mesh is automatically created by calling the function T3meshm This function prepares the elements connectivity and nodal geometry matrices and is listed after the main program CSTPLANESTRESSMESHm THIS PROGRAM USES AN 3NODE LINEAR TRIANGULAR ELEMENT FOR THE LINEAR ELASTIC STATIC ANALYSIS OF A TWO DIMENSIONAL PROBLEM IT INCLUDES AN AUTOMATIC MESH GENERATION Make these variables global so they can be shared by other functions clear all clc global nnd nel nne nodof eldof n global geom dee nf Nodalloads global Length Width NXE NYE Xorigin Yorigin format long g To change the size of the problem or change elastic properties supply another input file Length 60 Length of the model Width 20 Width NXE 24 Number of rows in the x direction NYE 10 Number of rows in the y direction dhx LengthNXE Element size in the x direction dhy WidthNYE Element size in the x direction Xorigin 0 X origin of the global coordinate system Yorigin Width2 Y origin of the global coordinate system nne 3 nodof 2 eldof nnenodof T3mesh Generate the mesh Material 2013 by Taylor Francis Group LLC 250 Introduction to Finite Element Analysis Using MATLAB and Abaqus E 200000 Elastic modulus in MPa vu 03 Poissons ratio thick 5 Beam thickness in mm Form the elastic matrix for plane stress dee formdsigEvu Boundary conditions nf onesnnd nodof Initialize the matrix nf to 1 Restrain in all directions the nodes situated x Length for i1nnd if geomi1 Length nfi 0 0 end end Counting of the free degrees of freedom n0 for i1nnd for j1nodof if nfij 0 nn1 nfijn end end end loading Nodalloads zerosnnd 2 Initialize the matrix of nodal loads to 0 Apply the load as a concentrated load on the node having coordinate X Y 0 Force 1000 N for i1nnd if geomi1 0 geomi2 0 Nodalloadsi 0 Force end end End of input Assemble the global force vector fgzerosn1 for i1 nnd if nfi1 0 fgnfi1 Nodalloadsi1 end if nfi2 0 fgnfi2 Nodalloadsi2 end end Assembly of the global stiffness matrix initialize the global stiffness matrix to zero kk zerosn n 2013 by Taylor Francis Group LLC Plane Problems 251 for i1nel beegA elemT3i Form strain matrix and steering vector kethickAbeedeebee Compute stiffness matrix kkformkkkkke g assemble global stiffness matrix end End of assembly delta kkfg solve for unknown displacements for i1 nnd if nfi1 0 xdisp 0 else xdisp deltanfi1 end if nfi2 0 ydisp 0 else ydisp deltanfi2 end nodedispi xdisp ydisp end Retrieve the xcoord and ydisp of the nodes located on the neutral axis k 0 verticaldispzeros1NXE1 for i1nnd if geomi2 0 kk1 xcoordk geomi1 verticaldispknodedispi2 end end for i1nel beegA elemT3i Form strain matrix and steering vector eldzeroseldof1 Initialize element displacement to zero for m1eldof if gm0 eldm0 else eldmdeltagm Retrieve element displacement end end epsbeeeld Compute strains EPSieps Store strains for all elements sigmadeeeps Compute stresses SIGMAisigma Store stresses for all elements end Plot stresses in the xdirection xstress SIGMA1 cmin minxstress cmax maxxstress caxiscmin cmax patchFaces connec Vertices geom FaceVertexCDataxstress FacecolorflatMarkero 2013 by Taylor Francis Group LLC 252 Introduction to Finite Element Analysis Using MATLAB and Abaqus colorbar plottools T3meshm This function generates a mesh of triangular elements global nnd nel nne nodof eldof n global geom connec dee nf Nodalloads global Length Width NXE NYE Xorigin Yorigin dhx dhy nnd 0 k 0 for i 1NXE for j1NYE k k 1 n1 j i1NYE 1 geomn1 i1dhx Xorigin j1dhy Yorigin n2 j iNYE1 geomn2 idhx Xorigin j1dhy Yorigin n3 n1 1 geomn3 i1dhx Xorigin jdhy Yorigin n4 n2 1 geomn4 idhx Xorigin jdhy Yorigin nel 2k m nel 1 connecm n1 n2 n3 connecnel n2 n4 n3 nnd n4 end end The variables NXE and NYE represent respectively the number of intervals along the x and y directions as shown in Figure 911 For each interval i and j four nodes n1 n2 n3 and n4 and two elements are created The first element has nodes n1 n2 n3 while the second element has nodes n2 n4 n3 In total the number of elements and nodes created are respectively equal to n3 n1 dhx dhy i1 iNXE j 1 j NYE n2 n4 FIGURE 911 Automatic mesh generation with the CST element 2013 by Taylor Francis Group LLC Plane Problems 253 0 002 004 006 Vertical displacement mm 008 01 012 0 10 20 30 40 Numerical Analytical 50 60 Length mm FIGURE 912 Deflection of the cantilever beam obtained with the fine mesh 10 8 6 4 2 0 2 4 6 8 10 150 150 100 100 50 50 0 σxx Nmm2 10 20 30 40 50 60 0 FIGURE 913 Stresses along the xaxis obtained with the fine mesh nel 2 NXE NYE and nnd NXE 1 NYE 1 The module also returns the matrices geomnnd 2 and connecnel nne The results obtained with the fine mesh are displayed in Figures 912 and 913 Figure 912 shows the deflection of the nodes situated along the center line neutral axis It can be clearly seen that the solution matches closely the analytical solution Figure 913 displays a contour of the stresses in the xdirection The stress gradient can be clearly seen even though each element displays a constant stress Those elements within the vicinity of the neutral axis display stress values close to zero 946 ANALYSIS WITH ABAQUS USING THE CST 9461 Interactive Edition In this section we will analyze the cantilever beam shown in Figure 97 with the Abaqus interactive edition We keep the same geometrical properties C 10 mm L 60 mm t 5 mm the same mechanical properties a Youngs modulus of 200000 MPa and a Poissons ratio of 03 and the same loading a concentrated force P of 1000 N 2013 by Taylor Francis Group LLC 254 Introduction to Finite Element Analysis Using MATLAB and Abaqus Start Abaqus CAE Click on Create Model Database On the main menu click on File and set Set Work Directory to choose your work ing directory Click on Save As and name the file BEAMCSTcae On the lefthandside menu click on Part to begin creating the model Name the part BeamCST check 2D Planar check Deformable in the type Choose Shell as the base feature Enter an approximate size of 100 mm and click on Continue Figure 914 FIGURE 914 Creating the BeamCST Part In the sketcher menu choose the CreateLines Rectangle icon to begin drawing the geometry of the beam Click on Done in the bottomleft corner of the viewport window Figure 915 FIGURE 915 Drawing using the createlines rectangle icon 2013 by Taylor Francis Group LLC Plane Problems 255 If we want to make sure that we will have nodes lying on the neu tral axis of the beam it is advisable to partition the beam along the neutral axis On the main menu click on Tools then on Partition In the dia log box check Face in Type and Use shortest path between 2 points in Method Select the two end points as shown in Figure 916 and in the prompt area click on Create partition FIGURE 916 Creating a partition Define a material named steel with an elastic modulus of 200000 MPa and a Poissons ratio of 03 Next click on Sections to create a section named Beamsection In the Category check Solid and in the Type check Homoge neous Click on Continue In the Edit Section dialog box check Plane stressstrain thickness and enter 5 mm as the thickness Click on OK Figure 917 FIGURE 917 Creating a plane stress section Expand the menu under Parts and BEAMCST and dou ble click on Section Assign ments With the mouse select the whole part In the Edit Section Assignments dialog box select Beamsection and click on OK Figure 918 FIGURE 918 Editing section assignments 2013 by Taylor Francis Group LLC 256 Introduction to Finite Element Analysis Using MATLAB and Abaqus In the model tree double click on Mesh under the BEAMCST In the main menu under Mesh click on Mesh Controls In the dia log box check Tri for Ele ment shape and Structured for Technique Click on OK Figure 919 FIGURE 919 Mesh controls In the main menu under Mesh click on Element Type With the mouse select all the part in the view port In the dialog box select Standard for element library Linear for geometric order The description of the element CPS3 A 3node linear plane stress triangle can be seen in the dialog box Click on OK Figure 920 FIGURE 920 Selecting element type In the main menu under Seed click on Part In the dialog box enter 5 for Approximate global size Click on OK and on Done Figure 921 FIGURE 921 Seeding part by size 2013 by Taylor Francis Group LLC Plane Problems 257 In the main menu under Mesh click on Part In the prompt area click on Yes In the main menu select View then Part Display Options In the Part Display Options under Mesh check Show node labels and Show ele ment labels Click Apply The element and node labels will appear in the viewport Figure 922 FIGURE 922 Mesh In the model tree expand the Assembly and double click on Instances Select BEAMCST for Parts and click OK In the model tree expand Steps and Ini tial and double click on BC Name the boundary condition FIXED select Symmetry AntisymmetryEncastre for the type and click on Continue Keep the shift key down and with the mouse select the right edge and click on Done in the prompt area In the Edit Boundary Con dition check ENCASTRE Click OK Figure 923 FIGURE 923 Imposing BC using geometry In the model tree double click on Steps Name the step Applyloads Set the proce dure to General and select Static General Click on Continue Give the step a description and click OK In the model tree under steps and under Applyloads click on Loads Name the load PointLoad and select Con centrated Force as the type Click on Continue Using the mouse click on the middle of the left edge and click on Done in the prompt area In the Edit Load dialog box enter 1000 for CF2 Click OK Figure 924 FIGURE 924 Imposing a concentrated force using geometry 2013 by Taylor Francis Group LLC 258 Introduction to Finite Element Analysis Using MATLAB and Abaqus In the model tree expand the Field Output Requests and then double click on FOutput1 FOutput1 is the default and is automatically generated when creating the step Uncheck the variables Contact and select any other variable you wish to add to the field output Click on OK Under Analysis right click on Jobs and then click on Create In the Create Job dialog box name the job BEAMCST and click on Continue In the Edit Job dialog box enter a description for the job Check Full analysis select to run the job in Background and check to start it immediately Click OK Expand the tree under Jobs right click on BEAMCST Then click on Submit If you get the following message BEAMCST completed successfully in the bottom window then your job is free of errors and was executed properly Figure 925 Notice that Abaqus has generated an input file for the job BEAMCSTinp which you can open with your preferred text editor Under the top menu in the Module scroll to Visualization and click to load Abaqus Viewer On the main menu under File click Open navigate to your working directory and open the file BEAMCSTodb It should have the same name as the job you submitted Click on the Common options icon to display the Common Plot options dialog box Under labels check Show Element labels and Show Node labels to display elements and nodes numbering Click on the icon Plot Contours on both shapes to display the deformed shape of the beam Under the main menu select U and U2 to plot the vertical displacement It can be seen that the displacement of the left edge is equal to 0965 mm which is almost similar with the analytical solution and the results obtained with the MATLAB code Figure 926 In the menu bar click on Report and Field Output In the Report Field Output dialog box for Position select Unique nodal check U1 and U2 under U Spatial FIGURE 925 Analyzing a job in Abaqus CAE FIGURE 926 Plotting displacements on deformed and undeformed shapes 2013 by Taylor Francis Group LLC Plane Problems 259 displacement Then click on click on Set up Click on Select to navigate to your working directory Name the file BEAMCSTrpt Uncheck Append to file and click OK Use your favorite text editor and open the file BEAMCSTrpt which should be the same as the one listed next Field Output Report written Wed May 11 011514 2011 Source 1 ODB CAbaqusFILESBEAMCSTodb Step Applyloads Frame Increment 1 Step Time 1000 Loc 1 Nodal values from source 1 Output sorted by column Node Label Field Output reported at nodes for part BEAMCST1 Node UU1 UU2 Label Loc 1 Loc 1 1 2157E06 9656E03 2 2659E36 8537E36 3 2141E33 2628E36 4 2210E03 9595E03 5 2202E03 9590E03 6 2269E33 9544E36 7 7791E06 8447E03 8 7086E06 7289E03 9 6846E06 6163E03 10 6312E06 5084E03 11 5586E06 4066E03 12 4725E06 3125E03 13 3683E06 2276E03 14 2280E06 1536E03 15 2085E06 9187E03 16 2590E06 4408E03 17 3564E06 1232E03 18 1641E33 2734E36 19 3622E03 2115E03 20 6811E03 5371E03 21 9712E03 1008E03 22 1232E03 1616E03 23 1464E03 2347E03 24 1665E03 3186E03 25 1835E03 4118E03 26 1975E03 5126E03 27 2084E03 6197E03 28 2161E03 7314E03 29 2202E03 8456E03 30 1099E03 9603E03 31 1085E03 9600E03 32 2187E03 8445E03 33 2144E03 7301E03 34 2068E03 6185E03 35 1962E03 5115E03 36 1824E03 4106E03 37 1655E03 3175E03 38 1456E03 2335E03 39 1225E03 1603E03 40 9620E03 9948E03 41 6673E03 5248E03 42 3418E03 2033E03 43 1539E33 1045E36 2013 by Taylor Francis Group LLC 260 Introduction to Finite Element Analysis Using MATLAB and Abaqus 44 1098E03 8451E03 45 1070E03 7296E03 46 1029E03 6173E03 47 9734E03 5096E03 48 9029E03 4080E03 49 8172E03 3142E03 50 7160E03 2296E03 51 5992E03 1558E03 52 4668E03 9433E03 53 3184E03 4664E03 54 1547E03 1399E03 55 1601E03 1416E03 56 3183E03 4601E03 57 4635E03 9356E03 58 5934E03 1551E03 59 7081E03 2289E03 60 8075E03 3136E03 61 8916E03 4075E03 62 9603E03 5090E03 63 1014E03 6167E03 64 1052E03 7290E03 65 1077E03 8442E03 9462 Keyword Edition In this section we will use a text editor to prepare an input file for the cantilever beam shown in Figure 97 The file is named BEAMCSTKeywordinp and is listed next Heading Analysis of cantilever beam as a plane stress problem Preprint echoYES Node generation NODE 1 0 0 5 0 20 61 60 0 65 60 20 NGENNSETLeftEdge 15 NGENNSETRightEdge 6165 NFILL LeftEdgeRightEdge125 NSET NSET Loadednode 3 Element generation ELEMENTTYPECPS3 1 1 6 7 ELGEN ELSET ODD 1 4 1 2 12 5 8 ELEMENTTYPECPS3 2 1 7 2 ELGENELSET EVEN 2 4 1 2 12 5 8 ELSET ELSET AllElements EVEN ODD MATERIAL NAME STEEL ELASTIC 200000 03 2013 by Taylor Francis Group LLC Plane Problems 261 SOLID SECTION ELSET AllElements MATERIAL STEEL 5 BOUNDARY CONDITIONS Boundary RightEdge encastre STEP ApplyLoads Step nameApplyLoads Static 1 1 1e05 1 LOADS Cload Loadednode 2 1000 OUTPUT REQUESTS Output field variablePRESELECT Output history variablePRESELECT End Step 1 The input file always starts with the keyword HEADING which in this case is entered as Analysis of cantilever beam as a plane stress problem 2 Using Preprint echoYES will allow to print an echo of the input file to the file with an extension dat 3 Using the keyword Node we define the four corner nodes 1 5 61 and 65 as shown in Figure 927 4 Using the keyword NGEN we generate the nodes located on the left edge In the data line we enter the number of the first end node 1 which has been previously defined then the number of the second end node 5 which also must have been previously defined followed by the increment in the numbers between each node along the line which in this case is the default 1 We then group the nodes in a set named LeftEdge 5 Using the keyword NGEN again we generate the nodes located on the right edge and group them in a set named RightEdge 6 Using the keyword NFILL we generate all the remaining nodes by filling in nodes between two bounds In the data line we enter first the node sets LeftEdge and 6 60 mm 61 65 20 mm 1 21 2 3 5 10 8 7 7 1 kN FIGURE 927 Generating a mesh manually in Abaqus 2013 by Taylor Francis Group LLC 262 Introduction to Finite Element Analysis Using MATLAB and Abaqus RightEdge followed by the number of intervals along each line between bounding nodes in this case 12 and the increment in node numbers from the node number at the first bound set end which in this case is 5 as shown in Figure 927 7 Using the keyword NSET NSET Loadednode we create a node set containing node 3 This will be used to apply the concentrated load of 1000 N 8 Using the keyword ELEMENT and Type CPS3 which stands for a continuum plane stress three node triangle we define elements 1 and 2 as well as their connectivity 9 Using the keyword ELGEN we generate all the elements having an odd number which we group in the set ODD The keyword ELGEN requires in its data line a Master element number b Number of elements to be defined in the first row generated including the master element c Increment in node numbers of corresponding nodes from element to element in the row The default is 1 d Increment in element numbers in the row The default is 1 e If necessary copy this newly created master row to define a layer of elements f Number of rows to be defined including the master row The default is 1 g Increment in node numbers of corresponding nodes from row to row h Increment in element numbers of corresponding elements from row to row i If necessary copy this newly created master layer to define a block of elements only necessary for a 3D mesh j Number of layers to be defined including the master layer The default is 1 k Increment in node numbers of corresponding nodes from layer to layer l Increment in element numbers of corresponding elements from layer to layer 10 Using the same procedure we generate all the elements having an even number which we group in the set EVEN 11 Next we use the keyword elset to group all the elements in an element set named AllElements consisting of element sets ODD and EVEN listed in the data line 12 Using the keywords Material and elastic we define a material named steel having an elastic modulus of 200000 MPa and a Poissons ratio of 03 13 Using the keyword solid section we assign the material steel to all the elements and in the data line we enter the thickness of the domain which in this case is 5 mm 14 Using the created node sets we impose the boundary conditions with the keyword Boundary We fully fix the node set RightEdge by using encastre 15 Next using the keyword step we create a step named ApplyLoads The keyword static indicates that it will be a general static analysis 16 Using the keyword cload we apply a concentrated load of 1000 N in the direction 2 to the node in node set Loadednode 17 Using the keywords Output field variablePRESELECT and Output history variablePRESELECT we request the default variables for both field and history outputs 18 Finally we end the step and the file with End Step At the command line type Abaqus jobBEAMCSTKeyword inter followed by Return If you get an error open the file with extension dat to see what type of error To load the visualization model type Abaqus Viewer at the command line On the main menu under File click Open navigate to your working directory and open the file BEAMCSTKeywordodb Click on the Common options icon to display the Common Plot options dialog box Under labels check Show Element labels and Show Node labels to display elements and nodes numbering Click on the icon Plot Deformed Shape to display the deformed shape of the beam On the main menu click on Results then on Field Output to open the Field Output dialog box Choose U Spatial displacements at nodes For component choose U2 to plot 2013 by Taylor Francis Group LLC Plane Problems 263 FIGURE 928 Displacement contour the vertical displacement Notice that the displacements contour is exactly the same as obtained previously except that the node and element numbering is different Figure 928 95 LINEAR STRAIN TRIANGLE A more versatile element in the triangular family is the linear strain triangle shown in Figure 929 It has six nodes The sides can be straight or curved It can be used to mesh domains with curved boundaries Its shape functions have already been defined in Chapter 7 and they are given as N1ξ η N2ξ η N3ξ η N4ξ η N5ξ η N6ξ η λ1 2λ 4ξλ ξ1 2ξ 4ξη η1 2η 4ηλ 942 with λ 1 ξ η y x 1 2 3 4 6 5 η ξ 01 u3 u4 u5 u6 u1 u2 v3 v4 v5 v6 v1 v2 2 1 6 5 4 3 00 10 FIGURE 929 Linear strain triangular element 2013 by Taylor Francis Group LLC Plane Problems 267 To change the size of the problem or change elastic properties supply another input file Length 60 Length of the model Width 20 Width NXE 12 Number of rows in the x direction NYE 5 Number of rows in the y direction XIG zeros2NXE11 YIGzeros2NYE11 Vectors holding grid coordinates dhx LengthNXE Element size in the x direction dhy WidthNYE Element size in the x direction Xorigin 0 X origin of the global coordinate system Yorigin Width2 Y origin of the global coordinate system nne 6 nodof 2 eldof nnenodof T6mesh Generate the mesh Material E 200000 Elastic modulus in MPa vu 03 Poissons ratio thick 5 Beam thickness in mm nhp 3 Number of sampling points Form the elastic matrix for plane stress dee formdsigEvu Boundary conditions nf onesnnd nodof Initialize the matrix nf to 1 Restrain in all directions the nodes situated x Length for i1nnd if geomi1 Length nfi 0 0 end end Counting of the free degrees of freedom n0 for i1nnd for j1nodof if nfij 0 nn1 nfijn end end end loading Nodalloads zerosnnd 2 Initialize the matrix of nodal loads to 0 Apply an equivalent nodal load of Pressurethickdhx to the central node located at x0 and y 0 Force 1000 N 2013 by Taylor Francis Group LLC 268 Introduction to Finite Element Analysis Using MATLAB and Abaqus for i1nnd if geomi1 0 geomi2 0 Nodalloadsi 0 Force end end End of input Assemble the global force vector fgzerosn1 for i1 nnd if nfi1 0 fgnfi1 Nodalloadsi1 end if nfi2 0 fgnfi2 Nodalloadsi2 end end Assembly of the global stiffness matrix Form the matrix containing the abscissas and the weights of Hammer points samphammernhp initialize the global stiffness matrix to zero kk zerosn n for i1nel coordg elemT6i Form strain matrix and steering vector kezeroseldofeldof Initialize the element stiffness matrix to zero for ig 1nhp wi sampig3 derfun fmT6quadsamp ig jac dercoord d detjac jac1invjac Compute inverse of the Jacobian derivjac1der Derivative of shape functions in global coordinates beeformbeederivnneeldof Form matrix B keke dthickwibeedeebee Integrate stiffness matrix end kkformkkkkke g assemble global stiffness matrix end End of assembly delta kkfg solve for unknown displacements for i1 nnd if nfi1 0 xdisp 0 else xdisp deltanfi1 end if nfi2 0 ydisp 0 else ydisp deltanfi2 end nodedispi xdisp ydisp 2013 by Taylor Francis Group LLC Plane Problems 269 end Retrieve the xcoord and ydisp of the nodes located on the neutral axis k 0 for i1nnd if geomi2 0 kk1 xcoordk geomi1 verticaldispknodedispi2 end end nhp 1 Calculate stresses at the centroid of the element samphammernhp for i1nel coordg elemT6i Retrieve coordinates and steering vector eldzeroseldof1 Initialize element displacement to zero for m1eldof if gm0 eldm0 else eldmdeltagm Retrieve element displacement from the global displacement vector end end for ig1 nhp derfun fmT6quadsamp ig Derivative of shape functions in local coordinates jacdercoord Compute Jacobian matrix jac1invjac Compute inverse of the Jacobian derivjac1der Derivative of shape functions in global coordinates beeformbeederivnneeldof Form matrix B epsbeeeld Compute strains sigmadeeeps Compute stresses end Compute stresses SIGMAisigma Store stresses for all elements end Prepare stresses for plotting ZX ZY ZT Z1 Z2preparecontourdataSIGMA Plot mesh using patches patchFaces connec Vertices geom FaceVertexCDatahsvnel FacecolornoneMarkero Plot stresses in the xdirection Ch contourfXIGYIGZX40 clabelCh colorbar plottools T6meshm This function generates a mesh of the linear strain triangular element global nnd nel geom connec XIG YIG global Length Width NXE NYE Xorigin Yorigin dhx dhy 2013 by Taylor Francis Group LLC Plane Problems 271 n3 n1 n2 dhy dhx i1 j1 jNYE i NXE n4 n5 n6 n9 n8 n7 FIGURE 930 Automatic mesh generation with the LST element 1 For every element i 1 to nel 2 Retrieve the coordinates of its nodes coordnne 2 and its steering vector geldof using the function elemt6m 3 Initialize the stiffness matrix to zero a Loop over the Hammer points ig 1 to nhp b Retrieve the weight wi as sampig 3 c Use the function fmT6quadm to compute the shape functions vector fun and their local derivatives der at the local coordinates ξ sampig 1 and η sampig 2 d Evaluate the Jacobian jac der coord e Evaluate the determinant of the Jacobian as d detjac f Compute the inverse of the Jacobian as jac1 invjac g Compute the derivatives of the shape functions with respect to the global coordinates x and y as deriv jac1 der h Use the function formbeem to form the strain matrix bee i Compute the stiffness matrix as ke ke d thick wi bee dee bee 4 Assemble the stiffness matrix ke into the global matrix kk The abscissa and weights for the Hammer formula are listed in Table 82 and given by the function hammerm listed in Appendix A 9542 Computation of the Stresses and Strains Once the global system of equations is solved we will compute the stresses at the centroid of the elements For this we set nhp 1 Then for each element 1 Retrieve the coordinates of its nodes coordnne 2 and its steering vector geldof using the function elemt6m 2 Retrieve its nodal displacements eldeldof from the global vector of displacements deltan 2013 by Taylor Francis Group LLC 272 Introduction to Finite Element Analysis Using MATLAB and Abaqus 0 002 004 006 Vertical displacement 008 01 012 0 10 20 30 40 50 60 Numerical Analytical Length mm FIGURE 931 Deflection of the cantilever beam obtained with the LST element a Loop over the Hammer points ig 1 to nhp b Use the function fmT6quadm to compute the shape functions vector fun and their local derivatives der at the local coordinates ξ sampig 1 and η sampig 2 c Evaluate the Jacobian jac der coord d Evaluate the determinant of the Jacobian as d detjac e Compute the inverse of the Jacobian as jac1 invjac f Compute the derivatives of the shape functions with respect to the global coordinates x and y as deriv jac1 der g Use the function formbeem to form the strain matrix bee h Compute the strains as eps bee eld i Compute the stresses as sigma dee eps 3 Store the stresses in the matrix SIGMAnel 3 The stresses computed at the centers of the elements are reorganized in a format suitable for plotting with the MATLAB graphic functions In the present case the stresses stored in the array SIGMAnel 3 are fed to the function preparecontourdatam listed in Appendix A For every node the function locates all the elements surrounding it Then the stresses are aver aged and assigned to the node and stored in the matrices ZX ZY ZT Z1 and Z2 corresponding respectively to σxx σyy and τxy and the principal stresses σ1 and σ2 In this particular case the matrix ZX and the vectors XIG and YIG are used in the MATLAB function contourf to produce a plot of the stresses σxx The results of the analysis are displayed in Figures 931 and 932 Figure 931 shows the deflection of the nodes situated along the center line neutral axis It can be clearly seen that the solution matches closely the analytical solution Figure 932 displays a contour plot the stresses in the xdirection The stress gradient can be clearly seen The stresses along the neutral axis are equal to zero 955 ANALYSIS WITH ABAQUS USING THE LST 9551 Interactive Edition In this section we will analyze the plate with a hole shown in Figure 933 using the linear strain triangle The plate is made of aluminum with an elastic modulus of 70 GPa and a Poissons ratio of 033 The plate is 5 mm thick and subject to a uniform pressure on both sides of 50 MPa Since the plate presents two planes of symmetry in both geometry and loading we will analyze a quarter only as shown in Figure 934 Indeed whenever possible always take advantage of symmetry to simplify the model 2013 by Taylor Francis Group LLC Plane Problems 273 10 Y 5 5 10 0 10 20 30 40 50 60 150 100 50 50 100 σxx Nmm2 0 0 X Height mm Length mm FIGURE 932 Stresses along the xdirection obtained with the LST element 50 MPa 20 140 mm 70 mm 70 mm FIGURE 933 Aluminum plate with a hole 45 mm 10 70 mm Fixed in the xdirection 50 MPa Fixed in the ydirection FIGURE 934 Making use of symmetry 2013 by Taylor Francis Group LLC 274 Introduction to Finite Element Analysis Using MATLAB and Abaqus Start Abaqus CAE Click on Create Model Database On the main menu click on File and set Set Work Directory to choose your working directory Click on Save As and name the file PlateLSTcae On the left handside menu click on Part to begin creating the model Name the part PlateLST check 2D Planar and check Deformable in the type Choose Shell as the base fea ture Enter an approximate size of 100 mm and sketch a quarter of the part as shown In the sketcher menu choose the Create arc center and 2 end points icon to cre ate the arc and CreateLines Rectangle icon to create the edges When finished click on Done in the bottomleft corner of the viewport win dow Figure 935 FIGURE 935 Creating the PlateLST Part Define a material named Aluminum with an elastic modulus of 70000 MPa and a Poissons ratio of 032 Next click on Sections to create a section named Platesection In the Category check Solid and in the Type check Homo geneous Click on Con tinue In the Edit Section dialog box check Plane stressstrain thickness and enter 5 mm as the thickness Click on OK Figure 936 FIGURE 936 Creating a plane stress section 2013 by Taylor Francis Group LLC Plane Problems 275 Expand the menu under Parts and PlateLST and dou ble click on Section Assign ments With the mouse select the whole part In the Edit Section Assignments dialog box select Platesection and click on OK Figure 937 FIGURE 937 Editing section assignments In the model tree dou ble click on Mesh under the PlateLST In the main menu under Mesh click on Mesh Controls In the dialog box check Tri for Element shape and Struc tured for Technique Click on OK In the main menu under Mesh click on Ele ment Type In the dialog box select Standard for ele ment library Quadratic for geometric order The descrip tion of the element CPS6M 6node modified quadratic plane stress triangle can be seen in the dialog box Click on OK Figure 938 FIGURE 938 Mesh controls 2013 by Taylor Francis Group LLC 276 Introduction to Finite Element Analysis Using MATLAB and Abaqus In the main menu under Seed click on Edges Select the arc first In the Local seeds select by number and enter 15 Click on the vertical left edge enter 20 and select sim ple for bias The idea of this is to refine the mesh in the vicin ity of the hole Do the same for the other edges When fin ished click on OK and on Done Under Mesh click on Part and then Yes to mesh the part Figure 939 FIGURE 939 Seeding edge by size and simple bias Expand the menu under PlateLST and click on Sets In the Create set dialog box name the set LeftEdge and check Node Click on Con tinue and with the mouse select the nodes as shown in Figure 940 Repeat the procedure to create another node set that you will name BottomEdge FIGURE 940 Creating a node set 2013 by Taylor Francis Group LLC Plane Problems 277 Expand the menu under PlateLST and click on Sur faces In the Create Sur face dialog box name the set LoadedSurface and check Geometry Click on Con tinue and with the mouse select the left edge shown in Figure 941 FIGURE 941 Creating a surface In the model tree expand the Assembly and double click on Instances Select PlateLST for Parts and click OK In the model tree expand Steps and Initial and double click on BC Name the boundary con dition Leftside select Dis placementRotation for the type and click on Con tinue In the bottomright cor ner of the viewport click on sets and in the dia log box select PlateLST 1LeftEdge Click on Con tinue In the Edit Boundary Condition check U1 Click OK Double click again on BC Name the boundary con dition Bottomside select DisplacementRotation for the type and click on Con tinue In the bottomright cor ner of the viewport click on sets and in the dia log box select PlateLST 1BottomEdge Click on Continue In the Edit Bound ary Condition check U2 Click OK Figure 942 FIGURE 942 Imposing BC using node sets 2013 by Taylor Francis Group LLC 278 Introduction to Finite Element Analysis Using MATLAB and Abaqus In the model tree double click on Steps Name the step Applyloads Set the proce dure to General and select Static General Click on Continue Give the step a description and click OK In the model tree under steps and under Applyloads click on Loads Name the load Pressure and select Pressure as the type Click on Con tinue In the rightbottom cor ner of the viewport click on Surfaces In the dialog box select loadedSurface and click on Continue In the new dialog box enter 50 MPa Figure 943 FIGURE 943 Imposing a pressure load on a surface In the model tree expand the Field Output Requests and then double click on FOutput1 FOutput1 is the default and is automatically generated when creating the step Uncheck the variables Contact and select any other variable you wish to add to the field output Click on OK Under Analysis right click on Jobs and then click on Create In the Create Job dialog box name the job PlateLST and click on Continue In the Edit Job dialog box enter a description for the job Check Full analysis select to run the job in Background and check to start it immediately Click OK Expand the tree under Jobs right click on PlateLST Then click on Submit If you get the following message PlateLST completed successfully in the bottom window then your job is free of errors and was executed properly Under the top menu in the Module scroll to Visualization and click to load Abaqus Viewer On the main menu under File click Open navigate to your working directory and open the file PlateLSTodb Click on the Common options icon to display the Common Plot options dialog box Under labels check Show Element labels and Show Node labels if you wish to display elements and nodes numbering Click on the icon Plot Contours on deformed shape to display the deformed shape of the beam Under the main menu select S and MaxInPlane Principal to plot the first principal stress as shown in Figure 944 9552 Keyword Edition Except for simple geometries it is very difficult to generate a mesh using keywords as we did previ ously Hence in this example instead of writing an input file we will simply open the one generated previously by Abaqus Navigate into the working directory and locate the file PlateLSTinp and open it with your preferred text editor It is a very long file as it lists all the nodes their coordinates and all the elements with their connectivity Note that the two node sets created are present as well as the surface Scroll to the end of the file and locate Name Pressure Type Pressure Dsload LoadedSurface P 50 2013 by Taylor Francis Group LLC Plane Problems 279 FIGURE 944 Plotting the maximum inplane principal stress under tension FIGURE 945 Plotting the maximum inplane principal stress under compression Change the value of 50 to 50 to apply a compressive pressure Rename the file PlateLSTKeywordinp Submit the job through the command line CWorkingDirectoryAbaqus jobPlateLSTKeyword inter When the job is successfully completed start Abaqus viewer and open the file PlateLSTKeywordodb Click on the Common options icon to display the Common Plot options dialog box Under labels check Show Element labels and Show Node labels if you wish to display elements and nodes numbering Click on the icon Plot Contours on deformed shape to display the deformed shape of the beam Under the main menu select S and MaxInPlane Principal to plot the first principal stress Figure 945 Now compare with Figure 944 96 THE BILINEAR QUADRILATERAL The linear strain quadrilateral has four nodes and straight edges as shown in Figure 946 Its shape functions have already been obtained in Chapter 7 and they are also given here 2013 by Taylor Francis Group LLC Plane Problems 285 6 3 2 2 3 4 4 6 5 7 7 8 10 10 60 mm 13 16 19 9 9 11 15 18 21 20 mm 12 12 20 17 14 11 8 5 1 1 1 kN FIGURE 949 Finite element discretization with 4nodded quadrilateral elements To change the size of the mesh alter the next statements nnd 21 Number of nodes nel 12 Number of elements nne 4 Number of nodes per element nodof 2 Number of degrees of freedom per node ngp 2 number of Gauss points eldof nnenodof Number of degrees of freedom per element Nodes coordinates x and y geom 0 100 x and y coordinates of node 1 00 00 x and y coordinates of node 2 00 100 x and y coordinates of node 3 100 100 x and y coordinates of node 4 100 00 x and y coordinates of node 5 100 100 x and y coordinates of node 6 200 100 x and y coordinates of node 7 200 00 x and y coordinates of node 8 200 100 x and y coordinates of node 9 300 100 x and y coordinates of node 10 300 00 x and y coordinates of node 11 300 100 x and y coordinates of node 12 400 100 x and y coordinates of node 13 400 00 x and y coordinates of node 14 400 100 x and y coordinates of node 15 500 100 x and y coordinates of node 16 500 00 x and y coordinates of node 17 500 100 x and y coordinates of node 18 600 100 x and y coordinates of node 19 600 00 x and y coordinates of node 20 600 100 x and y coordinates of node 21 disp Nodes XY coordinates geom Element connectivity connec 1 4 5 2 Element 1 2 5 6 3 Element 2 4 7 8 5 Element 3 5 8 9 6 Element 4 7 10 11 8 Element 5 8 11 12 9 Element 6 10 13 14 11 Element 7 11 14 15 12 Element 8 2013 by Taylor Francis Group LLC 286 Introduction to Finite Element Analysis Using MATLAB and Abaqus 13 16 17 14 Element 9 14 17 18 15 Element 10 16 19 20 17 Element 11 17 20 21 18 Element 12 disp Elements connectivity connec E 200000 Elastic modulus in MPa vu 03 Poissons ratio thick 5 Beam thickness in mm Form the elastic matrix for plane stress dee formdsigEvu Boundary conditions nf onesnnd nodof Initialize the matrix nf to 1 nf19 0 0 Node 19 is restrained in the x and y directions nf20 0 0 Node 20 is restrained in the x and y directions nf21 0 0 Node 21 is restrained in the x and y directions Counting of the free degrees of freedom n0 for i1nnd for j1nodof if nfij 0 nn1 nfijn end end end loading Nodalloads zerosnnd 2 Initialize the matrix of nodal loads to 0 Apply a concentrated at the node having x 0 and y 0 Force 1000 N Nodalloads1 0 Force The input data for this beam consist of nnd 21 number of nodes nel 12 number of elements nne 4 number of nodes per element nodof 2 number of degrees of freedom per node The coordinates x and y of the nodes are given in the form of a matrix geomnnd 2 The element connectivity is given in the matrix connecnel 4 Note that the internal numbering of the nodes is anticlockwise As shown in Figure 949 nodes 19 20 and 21 represent the fixed end of the cantilever which is fully fixed The prescribed degrees of freedom of these nodes are assigned the digit 0 All the degrees of freedom of all the other nodes which are free are assigned the digit 1 The information 2013 by Taylor Francis Group LLC Plane Problems 287 on the boundary conditions is given in the matrix nfnnd nodof The concentrated force of 1000 N is applied at node 2 The force will be assembled into the global force vector fg in the main program 9652 Main Program The main program Q4PLANESTRESSm is listed next THIS PROGRAM USES AN 4NODDED QUADRILATERAL ELEMENT FOR THE LINEAR ELASTIC STATIC ANALYSIS OF A TWO DIMENSIONAL PROBLEM Make these variables global so they can be shared by other functions clc clear all global nnd nel nne nodof eldof n ngp global geom connec dee nf Nodalloads format long g To change the size of the problem or change the elastic properties supply another input file Q4COARSEMESHDATA End of input Assemble the global force vector fgzerosn1 for i1 nnd if nfi1 0 fgnfi1 Nodalloadsi1 end if nfi2 0 fgnfi2 Nodalloadsi2 end end Form the matrix containing the abscissas and the weights of Gauss points ngp 2 sampgaussngp Numerical integration and assembly of the global stiffness matrix initialize the global stiffness matrix to zero kk zerosn n for i1nel coordg elemq4i coordinates of the nodes of element i and its steering vector kezeroseldofeldof Initialize the element stiffness matrix to zero for ig1 ngp wi sampig2 for jg1 ngp wjsampjg2 derfun fmlinsamp igjg Derivative of shape functions in local coordinates jacdercoord Compute Jacobian matrix ddetjac Compute determinant of Jacobian matrix jac1invjac Compute inverse of the Jacobian derivjac1der Derivative of shape functions 2013 by Taylor Francis Group LLC 288 Introduction to Finite Element Analysis Using MATLAB and Abaqus in global coordinates beeformbeederivnneeldof Form matrix B keke dthickwiwjbeedeebee Integrate stiffness matrix end end kkformkkkkke g assemble global stiffness matrix end End of assembly delta kkfg solve for unknown displacements dispnode xdisp ydisp for i1 nnd if nfi1 0 xdisp 0 else xdisp deltanfi1 end if nfi2 0 ydisp 0 else ydisp deltanfi2 end dispi xdisp ydisp Display displacements of each node DISPi xdisp ydisp end ngp1 Calculate stresses and strains at the center of each element sampgaussngp for i1nel coordg elemq4i coordinates of the nodes of element i and its steering vector eldzeroseldof1 Initialize element displacement to zero for m1eldof if gm0 eldm0 else eldmdeltagm Retrieve element displacement from the global displacement vector end end for ig1 ngp wi sampig2 for jg1 ngp wjsampjg2 derfun fmlinsamp igjg Derivative of shape functions in local coordinates jacdercoord Compute Jacobian matrix jac1invjac Compute inverse of the Jacobian derivjac1der Derivative of shape functions in global coordinates beeformbeederivnneeldof Form matrix B epsbeeeld Compute strains sigmadeeeps Compute stresses end end SIGMAisigma Store stresses for all elements end 2013 by Taylor Francis Group LLC Plane Problems 289 Average stresses at nodes ZX ZY ZT Z1 Z2stressesatnodesQ4SIGMA Plot stresses in the xdirection U2 DISP2 cmin minU2 cmax maxU2 caxiscmin cmax patchFaces connec Vertices geom FaceVertexCDataU2 FacecolorinterpMarker colorbar 9653 Integration of the Stiffness Matrix The stiffness matrix of the element is given by Equation 974 For each element it is evaluated as follows 1 For every element i 1 to nel 2 Retrieve the coordinates of its nodes coordnne 2 and its steering vector geldof using the function elemQ4m 3 Initialize the stiffness matrix to zero a Loop over the Gauss points ig 1 to ngp b Retrieve the weight wi as sampig 2 i Loop over the Gauss points jg 1 to ngp ii Retrieve the weight wj as sampjg 2 iii Use the function fmlinm to compute the shape functions vector fun and their derivatives matrix der in local coordinates ξ sampig 1 and η sampjg 1 iv Evaluate the Jacobian jac der coord v Evaluate the determinant of the Jacobian as d detjac vi Compute the inverse of the Jacobian as jac1 invjac vii Compute the derivatives of the shape functions with respect to the global coordinates x and y as deriv jac1 der viii Use the function formbeem to form the strain matrix bee ix Computethestiffnessmatrixas ke ke d thick wi wj bee dee bee 4 Assemble the stiffness matrix ke into the global matrix kk The evaluation of the stiffness matrix requires the use of Gauss quadrature To do so the abscissas and the weight of the corresponding Gauss points need to be made available to the program These are arranged in the array sampngp 2 organized as follows ξi sampi 1 and Wi sampi 2 981 The MATLAB function gaussm is listed in Appendix A and can be used for up to ngp 4 The function elemq4m is also listed in Appendix A It returns the coordinates of the nodes of each element as well as its steering vector textbfg The function fmlinm also listed in Appendix A returns the shape functions vector fun and their derivatives matrix der in local coordinates 2013 by Taylor Francis Group LLC 290 Introduction to Finite Element Analysis Using MATLAB and Abaqus 9654 Computation of the Stresses and Strains Once the global system of equations is solved we will compute the stresses at the centroid of the elements For this we set ngp 1 1 For each element 2 Retrieve the coordinates of its nodes coordnne 2 and its steering vector geldof using the function elemQ4m 3 Retrieve its nodal displacements eldeldof from the global vector of displacements deltan a Loop over the Gauss points ig 1 to ngp b Loop over the Gauss points jg 1 to ngp c Use the function fmlinm to compute the shape functions vector fun and their local derivatives der at the local coordinates ξ sampig 1 and η sampjg 1 d Evaluate the Jacobian jac der coord e Evaluate the determinant of the Jacobian as d detjac f Compute the inverse of the Jacobian as jac1 invjac g Compute the derivatives of the shape functions with respect to the global coordinates x and y as deriv jac1 der h Use the function formbeem to form the strain matrix bee i Compute the strains as eps bee eld j Compute the stresses as sigma dee eps 4 Store the stresses in the matrix SIGMAnel 3 The stresses computed at the centers of the elements are averaged at the nodes using the function StressesatnodesQ4m listed in Appendix A which returns σx σx τx σ1 and σ2 In the present case we can either feed any of the stresses or the displacements of the nodes to the MATLAB function patch with the argument interp to interpolate between the values at the nodes and get contour plots Figures 950 and 951 show respectively the contours of the vertical displacement v2 and of the stress σxx It can be clearly seen that the displacement of the tip equal to 0104 mm is very close to the exact displacement equal to 1108 mm obtained with Equation 941 On the other hand the stresses are not correct This is not a problem with the element but rather with the calculations of the stresses in the program Indeed in the program the stresses are calculated at the center of the elements then averaged at the nodes The maximum stress of about 75 MPa represents the value at the center of the element 10 Height mm 5 5 10 10 20 30 40 50 60 01042 00521 0 Vertical displacement mm 0 0 Length mm FIGURE 950 Contour of the vertical displacement v2 2013 by Taylor Francis Group LLC Plane Problems 291 10 5 0 5 10 Height mm 10 20 30 40 50 60 50 0 50 σxx MPa 0 Length mm FIGURE 951 Contour of the stress σxx 9655 Program with Automatic Mesh Generation To better model the stress gradient we need to refine the mesh In the new program named Q4PLANESTRESSMESHm listed next the mesh is automatically created by calling the func tion Q4meshm This function prepares the elements connectivity and nodal geometry matrices and is listed after the main program THIS PROGRAM USES AN 4NODDED QUADRILATERAL ELEMENT FOR THE LINEAR ELASTIC STATIC ANALYSIS OF A TWO DIMENSIONAL PROBLEM Make these variables global so they can be shared by other functions clc clear all global nnd nel nne nodof eldof n ngp global geom connec dee nf Nodalloads global Length Width NXE NYE Xorigin Yorigin dhx dhy format long g To change the size of the mesh alter the next statements Length 60 Length of the model Width 20 Width NXE 24 Number of rows in the x direction NYE 8 Number of rows in the y direction dhx LengthNXE Element size in the x direction dhy WidthNYE Element size in the x direction Xorigin 0 X origin of the global coordinate system Yorigin Width2 Y origin of the global coordinate system nne 4 nodof 2 eldof nnenodof Q4mesh Generate the mesh E 200000 Elastic modulus in MPa vu 03 Poissons ratio thick 5 Beam thickness in mm Form the elastic matrix for plane stress dee formdsigEvu Boundary conditions 2013 by Taylor Francis Group LLC 292 Introduction to Finite Element Analysis Using MATLAB and Abaqus nf onesnnd nodof Initialize the matrix nf to 1 Restrain in all directions the nodes situated x Length for i1nnd if geomi1 Length nfi 0 0 end end Counting of the free degrees of freedom n0 for i1nnd for j1nodof if nfij 0 nn1 nfijn end end end loading Nodalloads zerosnnd 2 Initialize the matrix of nodal loads to 0 Apply a concentrated at the node having x 0 and y 0 Force 1000 N for i1nnd if geomi1 0 geomi2 0 Nodalloadsi 0 Force end end End of input Assemble the global force vector fgzerosn1 for i1 nnd if nfi1 0 fgnfi1 Nodalloadsi1 end if nfi2 0 fgnfi2 Nodalloadsi2 end end Form the matrix containing the abscissas and the weights of Gauss points ngp 2 sampgaussngp Numerical integration and assembly of the global stiffness matrix initialize the global stiffness matrix to zero kk zerosn n for i1nel coordg elemq4i coordinates of the nodes of element i and its steering vector kezeroseldofeldof Initialize the element stiffness matrix to zero for ig1 ngp 2013 by Taylor Francis Group LLC Plane Problems 293 wi sampig2 for jg1 ngp wjsampjg2 derfun fmlinsamp igjg Derivative of shape functions in local coordinates jacdercoord Compute Jacobian matrix ddetjac Compute determinant of Jacobian matrix jac1invjac Compute inverse of the Jacobian derivjac1der Derivative of shape functions in global coordinates beeformbeederivnneeldof Form matrix B keke dthickwiwjbeedeebee Integrate stiffness matrix end end kkformkkkkke g assemble global stiffness matrix end End of assembly delta kkfg solve for unknown displacements dispnode xdisp ydisp for i1 nnd if nfi1 0 xdisp 0 else xdisp deltanfi1 end if nfi2 0 ydisp 0 else ydisp deltanfi2 end dispi xdisp ydisp Display displacements of each node DISPi xdisp ydisp end ngp1 Calculate stresses and strains at the center of each element sampgaussngp for i1nel coordg elemq4i coordinates of the nodes of element i and its steering vector eldzeroseldof1 Initialize element displacement to zero for m1eldof if gm0 eldm0 else eldmdeltagm Retrieve element displacement from the global displacement vector end end for ig1 ngp wi sampig2 for jg1 ngp wjsampjg2 derfun fmlinsamp igjg Derivative of shape functions in local coordinates jacdercoord Compute Jacobian matrix 2013 by Taylor Francis Group LLC 294 Introduction to Finite Element Analysis Using MATLAB and Abaqus jac1invjac Compute inverse of the Jacobian derivjac1der Derivative of shape functions in global coordinates beeformbeederivnneeldof Form matrix B epsbeeeld Compute strains sigmadeeeps Compute stresses end end SIGMAisigma Store stresses for all elements end Average stresses at nodes ZX ZY ZT Z1 Z2stressesatnodesQ4SIGMA Plot stresses in the xdirection U2 DISP2 cmin minU2 cmax maxU2 caxiscmin cmax patchFaces connec Vertices geom FaceVertexCDataU2 FacecolorinterpMarker colorbar Q4meshm This module generates a mesh of linear quadrilateral elements global nnd nel nne nodof eldof n global geom connec dee nf Nodalloads global Length Width NXE NYE Xorigin Yorigin dhx dhy nnd 0 k 0 for i 1NXE for j1NYE k k 1 n1 j i1NYE 1 geomn1 i1dhx Xorigin j1dhy Yorigin n2 j iNYE1 geomn2 idhx Xorigin j1dhy Yorigin n3 n1 1 geomn3 i1dhx Xorigin jdhy Yorigin n4 n2 1 geomn4 idhx Xorigin jdhy Yorigin nel k connecnel n1 n2 n4 n3 nnd n4 end end The variables NXE and NYE represent respectively the number of intervals along the x and y directions as shown in Figure 952 For each interval i and j four nodes n1 n2 n3 and n4 and one element are created The element has nodes n1 n2 n4 n3 In total the number of elements and nodes created are respectively equal to nel NXENYE and nnd NXE1NYE1 The module also returns the matrices geomnnd 2 and connecnel nne The results obtained with the fine mesh are displayed in Figures 953 and 954 respectively as contour plots of the vertical displacement v2 and the stress σxx The stress values are more accurate They are very similar to those obtained with the linear strain triangular element shown in Figure 913 2013 by Taylor Francis Group LLC Plane Problems 295 NXE 30 NYE10 Height mm 10 8 6 4 2 2 4 6 8 10 0 10 20 30 40 50 60 0 Length mm FIGURE 952 Automatic mesh generation with the Q4 element 0 0 5 5 10 10 20 30 40 50 60 10 Height mm 01 005 0 Vertical displacement mm Length mm FIGURE 953 Contour of the vertical displacement v2 5 10 10 Height mm 0 5 0 10 20 30 40 50 60 100 100 0 Stress σxx MPa Length mm FIGURE 954 Contour of the stresses along the xaxis σxx 966 ANALYSIS WITH ABAQUS USING THE Q4 QUADRILATERAL 9661 Interactive Edition In this section we will analyze the cantilever beam shown in Figure 97 with the Abaqus interactive edition We keep the same geometrical properties C 10 mm L 60 mm t 5 mm the same mechanical properties a Youngs modulus of 200000 MPa and a Poissons ratio of 03 and the same loading a concentrated force P of 1000 N 2013 by Taylor Francis Group LLC 296 Introduction to Finite Element Analysis Using MATLAB and Abaqus Start Abaqus CAE Click on Create Model Database On the main menu click on File and set Set Work Directory to choose your working direc tory Click on Save As and name the file BEAMQ4cae On the lefthandside menu click on Part to begin creat ing the model Name the part BeamQ4 check 2D Pla nar and check Deformable in the type Choose Shell as the base feature Enter an approximate size of 100 mm and click on Continue In the sketcher menu choose the CreateLines Rectangle icon to begin drawing the geom etry of the beam Click on Done in the bottomleft cor ner of the viewport window Figure 955 FIGURE 955 Creating the BeamQ4 Part If we want to make sure that we will have nodes lying on the neutral axis of the beam it is advisable to partition the beam along the neutral axis On the main menu click on Tools then on Partition In the dialog box check Face in Type and Use shortest path between 2 points in Method Select the two end points as shown in Figure 956 and in the prompt area click on Create partition FIGURE 956 Creating a partition 2013 by Taylor Francis Group LLC Plane Problems 297 Define a material named steel with an elastic modulus of 200000 MPa and a Poissons ratio of 03 Next click on Sections to create a section named BeamsectionQ4 In the Category check Solid and in the Type check Homo geneous Click on Con tinue In the Edit Section dialog box check Plane stressstrain thickness and enter 5 mm as the thickness Click on OK Figure 957 FIGURE 957 Creating a plane stress section Expand the menu under Parts and BEAMQ4 and dou ble click on Section Assign ments With the mouse select the whole part In the Edit Section Assign ments dialog box select BeamsectionQ4 and click on OK Figure 958 FIGURE 958 Editing section assignments In the model tree dou ble click on Mesh under the BEAMQ4 In the main menu under Mesh click on Mesh Controls In the dialog box check Quad for Ele ment shape and Structured for Technique Click on OK Figure 959 FIGURE 959 Mesh controls 2013 by Taylor Francis Group LLC 298 Introduction to Finite Element Analysis Using MATLAB and Abaqus In the main menu under Mesh click on Element Type With the mouse select all the part in the viewport In the dialog box select Stan dard for element library Lin ear for geometric order In Quad check Reduced inte gration The description of the element CPS4R A 4 node bilinear plane stress quadrilateral reduced inte gration hourglass control can be seen in the dialog box Click on OK Figure 960 FIGURE 960 Selecting element type In the main menu under Seed click on Part In the dialog box enter 5 for Approximate global size Click on OK and on Done Figure 961 FIGURE 961 Seeding part by size In the main menu under Mesh click on Part In the prompt area click on Yes In the main menu select View then Part Display Options In the Part Display Options under Mesh check Show node labels and Show ele ment labels Click Apply The element and node labels will appear in the viewport Figure 962 FIGURE 962 Mesh 2013 by Taylor Francis Group LLC Plane Problems 299 In the model tree expand the Assembly and double click on Instances Select BEAMQ4 for Parts and click OK In the model tree expand Steps and Initial and double click on BC Name the boundary condi tion FIXED select Symmetry AntisymmetryEncastre for the type and click on Continue Keep the shift key down and with the mouse select the right edge and click on Done in the prompt area In the Edit Boundary Condition check ENCASTRE Click OK Figure 963 FIGURE 963 Imposing BC using geometry In the model tree double click on Steps Name the step Applyloads Set the proce dure to General and select Static General Click on Continue Give the step a description and click OK In the model tree under steps and under Applyloads click on Loads Name the load PointLoad and select Con centrated Force as the type Click on Continue Using the mouse click on the middle of the left edge and click on Done in the prompt area In the Edit Load dialog box enter 1000 for CF2 Click OK Figure 964 FIGURE 964 Imposing a concentrated force using geometry In the model tree expand the Field Output Requests and then double click on FOutput1 FOutput1 is the default and is automatically generated when creating the step Uncheck the variables Contact and select any other variable you wish to add to the field output Click on OK Under Analysis right click on Jobs and then click on Create In the Create Job dialog box name the job BEAMQ4 and click on Continue In the Edit Job dialog box enter a description for the job Check Full analysis select to run the job in Background and check to start it immediately Click OK Expand the tree under Jobs right click on BEAMQ4 Then click on Submit If you get the following message BEAMQ4 completed successfully in the bottom window then your job is free of errors and was executed properly Under the top menu in the Module scroll to Visualization and click to load Abaqus Viewer On the main menu under File click Open navigate to your working directory and open the file BEAMQ4odb It should 2013 by Taylor Francis Group LLC 300 Introduction to Finite Element Analysis Using MATLAB and Abaqus FIGURE 965 Plotting displacements on deformed and undeformed shapes have the same name as the job you submitted Click on the Common options icon to display the Common Plot options dialog box Under labels check Show Element labels and Show Node labels to display elements and nodes numbering Click on the icon Plot Contours on both shapes to display the deformed shape of the beam Under the main menu select U and U2 to plot the vertical displacement It can be seen that the displacement of the left edge is equal to 01263 mm which is almost similar to the analytical solution and the results obtained with the MATLAB code Figure 965 In the menu bar click on Report and Field Output In the Report Field Output dialog box for Position select Unique nodal check U1 and U2 under U Spatial displacement Then click on Set up Click on Select to navigate to your working directory Name the file BEAMQ4rpt Uncheck Append to file and click OK Use your favorite text editor and open the file BEAMQ4rpt which should be the same as the one listed next Field Output Report written Tue Jun 07 141655 2011 Source 1 ODB CABAQUSFILESBEAMQ4odb Step Applyloads Frame Increment 1 Step Time 1000 Loc 1 Nodal values from source 1 Output sorted by column Node Label Field Output reported at nodes for part BEAMQ41 Node UU1 UU2 Label Loc 1 Loc 1 1 329597E18 126304E03 2 0 106912E36 3 190702E33 399459E36 4 282845E03 124280E03 5 282845E03 124280E03 6 190702E33 399459E36 7 208167E18 105542E03 8 176942E18 951550E03 2013 by Taylor Francis Group LLC Plane Problems 301 9 602816E18 771076E03 10 637511E18 662657E03 11 188276E18 506013E03 12 205998E18 407127E03 13 372966E18 280818E03 14 142030E18 199475E03 15 359684E18 111797E03 16 986624E18 559480E03 17 116891E18 124520E03 18 218595E33 470851E36 19 478542E03 279632E03 20 888433E03 686775E03 21 125353E03 122002E03 22 161799E03 209244E03 23 189209E03 289212E03 24 217734E03 414322E03 25 237241E03 511911E03 26 257807E03 667828E03 27 268659E03 774101E03 28 283401E03 953045E03 29 283412E03 106447E03 30 145331E03 120582E03 31 145331E03 120582E03 32 283412E03 106447E03 33 283401E03 953045E03 34 268659E03 774101E03 35 257807E03 667828E03 36 237241E03 511911E03 37 217734E03 414322E03 38 189209E03 289212E03 39 161799E03 209244E03 40 125353E03 122002E03 41 888433E03 686775E03 42 478542E03 279632E03 43 218595E33 470851E36 44 143650E03 110266E03 45 135899E03 913903E03 46 133946E03 804582E03 47 125300E03 635695E03 48 117534E03 531450E03 49 105325E03 389201E03 50 935962E03 298439E03 51 772590E03 190591E03 52 615508E03 121474E03 53 420098E03 559630E03 54 192738E03 167195E03 55 192738E03 167195E03 56 420098E03 559630E03 57 615508E03 121474E03 58 772590E03 190591E03 59 935962E03 298439E03 60 105325E03 389201E03 61 117534E03 531450E03 62 125300E03 635695E03 63 133946E03 804582E03 64 135899E03 913903E03 65 143650E03 110266E03 Minimum 283412E03 126304E03 At Node 29 1 Maximum 283412E03 470851E36 At Node 32 43 Total 388578E18 315015 2013 by Taylor Francis Group LLC 302 Introduction to Finite Element Analysis Using MATLAB and Abaqus 22 11 10 1 kN 6 1 1 7 331 342 12 60 mm 20 mm FIGURE 966 Generating a mesh manually in Abaqus 9662 Keyword Edition In this section we will use a text editor to prepare an input file for the cantilever beam We will refine the mesh by using 10 elements along the yaxis and 30 elements along the longitudinal direction In total there will be 300 elements and 342 nodes The corner nodes are shown in Figure 966 The file is named BEAMQ4Keywordinp and is listed next Heading Analysis of cantilever beam as a plane stress problem using the 4node bilinear quadrilateral Preprint echoYES Node generation NODE 1 0 0 11 0 20 331 60 0 342 60 20 NGENNSETLeftEdge 111 NGENNSETRightEdge 331342 NFILL LeftEdgeRightEdge3011 NSET NSET Loadednode 6 Element generation ELEMENTTYPECPS4R 1 1 12 13 2 ELGEN ELSET AllElements 1 10 1 1 30 11 10 MATERIAL NAME STEEL ELASTIC 200000 03 SOLID SECTION ELSET AllElements MATERIAL STEEL 5 BOUNDARY CONDITIONS Boundary 2013 by Taylor Francis Group LLC Plane Problems 303 RightEdge encastre STEP ApplyLoads Step nameApplyLoads Static 1 1 1e05 1 LOADS Cload Loadednode 2 1000 OUTPUT REQUESTS Output field variablePRESELECT Output history variablePRESELECT End Step 1 The input file always starts with the keyword HEADING which in this case is entered as Analysis of cantilever beam as a plane stress problem using the 4node bilinear quadrilateral 2 Using Preprint echoYES will allow to print an echo of the input file to the file with an extension dat 3 Using the keyword Node we define the four corner nodes 1 11 331 and 342 as shown in Figure 966 4 Using the keyword NGEN we generate the nodes located on the left edge In the data line we enter the number of the first end node 1 which has been previously defined then the number of the second end node 11 which also must have been previously defined followed by the increment in the numbers between each node along the line which in this case is the default 1 We then group the nodes in a set named LeftEdge 5 Using the keyword NGEN again we generate the nodes located on the right edge and group them in a set named RightEdge 6 Using the keyword NFILL we generate all the remaining nodes by filling in nodes between two bounds In the data line we enter first the node sets LeftEdge and RightEdge followed by the number of intervals along each line between bounding nodes in this case 30 and the increment in node numbers from the node number at the first bound set end which in this case is 11 7 Using the keyword NSET NSET Loadednode we create a node set containing node 6 This will be used to apply the concentrated load of 1000 N 8 Using the keyword ELEMENT and Type CPS4R which stands for a continuum plane stress four node quadrilateral we define element 1 as well as its connectivity 9 Using the keyword ELGEN we generate all the elements that we group in the set Allelements The keyword ELGEN requires in its data line a Master element number b Number of elements to be defined in the first row generated including the master element c Increment in node numbers of corresponding nodes from element to element in the row The default is 1 d Increment in element numbers in the row The default is 1 e If necessary copy this newly created master row to define a layer of elements f Number of rows to be defined including the master row The default is 1 g Increment in node numbers of corresponding nodes from row to row 2013 by Taylor Francis Group LLC 304 Introduction to Finite Element Analysis Using MATLAB and Abaqus FIGURE 967 Mesh generated with the keyword edition h Increment in element numbers of corresponding elements from row to row i If necessary copy this newly created master layer to define a block of elements only necessary for a 3D mesh j Number of layers to be defined including the master layer The default is 1 k Increment in node numbers of corresponding nodes from layer to layer l Increment in element numbers of corresponding elements from layer to layer 10 Using the keywords Material and elastic we define a material named steel having an elastic modulus of 200000 MPa and a Poissons ratio of 03 11 Using the keyword solid section we assign the material steel to all the elements and in the data line we enter the thickness of the domain which in this case is 5 mm 12 Using the created node sets we impose the boundary conditions with the keyword Boundary We fully fix the node set RightEdge by using encastre 13 Next using the keyword step we create a step named ApplyLoads The keyword static indicates that it will be a general static analysis 14 Using the keyword cload we apply a concentrated load of 1000 N in the direction 2 to the node in node set Loadednode 15 Using the keywords Output field variablePRESELECT and Output history variablePRESELECT we request the default variables for both field and history outputs 16 Finally we end the step and the file with End Step At the command line type Abaqus jobBEAMQ4Keyword inter is followed by Return If you get an error open the file with extension dat to see what type of error To load the visualization model type Abaqus Viewer at the command line Figure 967 On the main menu under File click Open navigate to your working directory and open the file BEAMQ4Keywordodb Click on the Common options icon to display the Common Plot options dialog box Under labels check Show Element labels and Show Node labels to display the mesh generated Uncheck Show Element labels and Show Node labels then click on the icon Plot Deformed Shape to display the deformed shape of the beam On the main menu click on Results then on Field Output to open the Field Output dialog box Choose U Spatial displacements at nodes For component choose U2 to plot the vertical displacement Figure 968 97 THE 8NODE QUADRILATERAL 971 FORMULATION The 8nodded quadrilateral element has curved sides which makes it very useful in modeling structures with curved edges Figure 969 The element shape functions are given as 2013 by Taylor Francis Group LLC Plane Problems 305 FIGURE 968 Displacement contour u7 v7 v6 5 4 3 2 1 v5 η u6 u5 u4 u3 u2 u1 u8 v8 v2 v1 v4 ξ v3 7 8 6 y x FIGURE 969 Eightnodded isoparametric element N1ξ η N2ξ η N3ξ η N4ξ η N5ξ η N6ξ η N7ξ η N8ξ η 0251 ξ1 η1 ξ η 0501 ξ21 η 0251 ξ1 η1 ξ η 0501 ξ1 η2 0251 ξ1 η1 ξ η 0501 ξ21 η 0251 ξ1 η1 ξ η 0501 ξ1 η2 982 The displacement field over the element is approximated as u N1u1 N2u2 N3u3 N4u4 N5u5 N6u6 N7u7 N8u8 983 v N1v1 N2v2 N3v3 N4v4 N5v5 N6v6 N7v7 N8v8 984 2013 by Taylor Francis Group LLC Plane Problems 307 L q qL qL qL 1 6 1 6 2 3 FIGURE 970 Equivalent nodal loads 500 Reinforced concrete E 40000 MPa ν 017 Depth100 mm 100 700 400 300 30 kN 30 kN FIGURE 971 Geometry and loading 972 EQUIVALENT NODAL FORCES When the shape of the loading on an element edge is complicated the integration process detailed in Section 964 should be used However if the loads are uniformly distributed then the equivalent nodal loads shown in Figure 970 can be used 973 PROGRAM Q8PLANESTRESSm The program is virtually identical to its predecessor Q4PLANESTRESSm except that some of the arrays have slightly bigger dimensions because of the increased number of degrees of freedom per element In order to assess the performance of the element we will analyze the simply supported deep beam subject to fourpoint bending shown in Figure 971 Taking advantage of symmetry only half the model is analyzed We will use 32 elements to discretize the domain as shown in Figure 972 The nodes numbered 113121 represent the midspan These nodes are allowed to displace vertically but not horizontally The program is listed next 9731 Data Preparation To read the data we will use the Mfile Q8COARSEMESHDATAm listed next FILEQ8COARSEMESHDATAm Beginning of data input global nnd nel nne nodof eldof n ngp global geom connec dee nf Nodalloads nnd 121 Number of nodes 2013 by Taylor Francis Group LLC 308 Introduction to Finite Element Analysis Using MATLAB and Abaqus 500 100 1 1 2 3 4 5 6 7 8 9 4 79 121 32 29 99 113 85 71 57 43 29 15 700 x 400 y 300 30 kN FIGURE 972 Coarse mesh nel 32 Number of elements nne 8 Number of nodes per element nodof 2 Number of degrees of freedom per node ngp 2 number of Gauss points eldof nnenodof Number of degrees of freedom per element Thickness of the domain thick 100 Nodes coordinates x and y geom 0 0 x and y coordinates of node 1 0 50 0 100 0 150 0 200 0 250 0 300 0 350 0 400 50 0 50 100 50 200 50 300 50 400 100 0 100 50 100 100 100 150 100 200 100 250 100 300 100 350 100 400 150 0 150 100 150 200 150 300 150 400 200 0 200 50 2013 by Taylor Francis Group LLC Plane Problems 309 200 100 200 150 200 200 200 250 200 300 200 350 200 400 250 0 250 100 250 200 250 300 250 400 300 0 300 50 300 100 300 150 300 200 300 250 300 300 300 350 300 400 350 0 350 100 350 200 350 300 350 400 400 0 400 50 400 100 400 150 400 200 400 250 400 300 400 350 400 400 450 0 450 100 450 200 450 300 450 400 500 0 500 50 500 100 500 150 500 200 500 250 500 300 500 350 500 400 550 0 550 100 550 200 550 300 550 400 600 0 600 50 600 100 600 150 600 200 600 250 600 300 600 350 600 400 650 0 650 100 650 200 650 300 2013 by Taylor Francis Group LLC 310 Introduction to Finite Element Analysis Using MATLAB and Abaqus 650 400 700 0 700 50 700 100 700 150 700 200 700 250 700 300 700 350 700 400 750 0 750 100 750 200 750 300 750 400 800 0 800 50 800 100 800 150 800 200 800 250 800 300 800 350 800 400 x and y coordinates of node 121 Element connectivity connec 1 10 15 16 17 11 3 2 Element 1 3 11 17 18 19 12 5 4 5 12 19 20 21 13 7 6 7 13 21 22 23 14 9 8 15 24 29 30 31 25 17 16 17 25 31 32 33 26 19 18 19 26 33 34 35 27 21 20 21 27 35 36 37 28 23 22 29 38 43 44 45 39 31 30 31 39 45 46 47 40 33 32 33 40 47 48 49 41 35 34 35 41 49 50 51 42 37 36 43 52 57 58 59 53 45 44 45 53 59 60 61 54 47 46 47 54 61 62 63 55 49 48 49 55 63 64 65 56 51 50 57 66 71 72 73 67 59 58 59 67 73 74 75 68 61 60 61 68 75 76 77 69 63 62 63 69 77 78 79 70 65 64 71 80 85 86 87 81 73 72 73 81 87 88 89 82 75 74 75 82 89 90 91 83 77 76 77 83 91 92 93 84 79 78 85 94 99 100 101 95 87 86 87 95 101 102 103 96 89 88 89 96 103 104 105 97 91 90 91 97 105 106 107 98 93 92 99 108 113 114 115 109 101 100 101 109 115 116 117 110 103 102 103 110 117 118 119 111 105 104 105 111 119 120 121 112 107 106 Element 32 Material properties E40000 vu017 Youngs modulus and Poissons ratio Form the matrix of elastic properties 2013 by Taylor Francis Group LLC Plane Problems 311 deeformdsigEvu Matrix of elastic properties for plane stress Boundary conditions nf onesnnd nodof Initialize the matrix nf to 1 for i1nnd if geomi1 800 nfi 0 1 end if geomi1 100 geomi2 0 nfi 1 0 end end Counting of the free degrees of freedom n0 for i1nnd for j1nodof if nfij 0 nn1 nfijn end end end disp Nodal freedom nf disp Total number of active degrees of freedom n loading Nodalloads zerosnnd 2 Nodalloads79230000 Vertical load on node 79 End input The input data for this beam consist of nnd 121 number of nodes nel 32 number of elements nne 8 number of nodes per element nodof 2 number of degrees of freedom per node The coordinates x and y of the nodes are given in the form of a matrix geomnnd 2 The element connectivity is given in the matrix connecnel 8 Note that the internal numbering of the nodes is anticlockwise As shown in Figure 972 nodes 113121 are fixed in the xdirection only Node 15 which represents the simple support is fixed in the ydirection only The information on the boundary conditions is given in the matrix nfnnd nodof The concentrated force of 30000 N is applied at node 79 Notice the negative sign to indicate that the force acts in the negative ydirection The force will be assembled into the global force vector fg in the main program 9732 Main Program The main program Q8PLANESTRESSm is listed next THIS PROGRAM USES AN 8NODDED QUADRILATERAL ELEMENT FOR THE LINEAR ELASTIC STATIC ANALYSIS OF A TWO DIMENSIONAL PROBLEM 2013 by Taylor Francis Group LLC 312 Introduction to Finite Element Analysis Using MATLAB and Abaqus Make these variables global so they can be shared by other functions clc clear all global nnd nel nne nodof eldof n ngp global geom connec dee nf Nodalloads format long g This is where the to input the data in the form of a file with an extension m Q8coarsemeshdata End of input Assemble the global force vector fgzerosn1 for i1 nnd if nfi1 0 fgnfi1 Nodalloadsi1 end if nfi2 0 fgnfi2 Nodalloadsi2 end end Form the matrix containing the abscissas and the weights of Gauss points sampgaussngp Numerical integration and assembly of the global stiffness matrix initialize the global stiffness matrix to zero kk zerosn n for i1nel coordg elemq8i coordinates of the nodes of element i and its steering vector kezeroseldofeldof Initialize the element stiffness matrix to zero for ig1 ngp wi sampig2 for jg1 ngp wjsampjg2 derfun fmquadsamp igjg Derivative of shape functions in local coordinates jacdercoord Compute Jacobian matrix ddetjac Compute determinant of Jacobian matrix jac1invjac Compute inverse of the Jacobian derivjac1der Derivative of shape functions in global coordinates beeformbeederivnneeldof Form matrix B keke dthickwiwjbeedeebee Integrate stiffness matrix end end kkformkkkkke g assemble global stiffness matrix end End of assembly 2013 by Taylor Francis Group LLC Plane Problems 313 delta kkfg solve for unknown displacements dispnode xdisp ydisp for i1 nnd if nfi1 0 xdisp 0 else xdisp deltanfi1 end if nfi2 0 ydisp 0 else ydisp deltanfi2 end dispi xdisp ydisp Display displacements of each node DISPi xdisp ydisp end ngp1 Calculate stresses and strains at the center of each element sampgaussngp for i1nel coordg elemq8i coordinates of the nodes of element i and its steering vector eldzeroseldof1 Initialize element displacement to zero for m1eldof if gm0 eldm0 else eldmdeltagm Retrieve element displacement from the global displacement vector end end for ig1 ngp wi sampig2 for jg1 ngp wjsampjg2 derfun fmquadsamp igjg Derivative of shape functions in local coordinates jacdercoord Compute Jacobian matrix jac1invjac Compute inverse of the Jacobian derivjac1der Derivative of shape functions in global coordinates beeformbeederivnneeldof Form matrix B epsbeeeld Compute strains sigmadeeeps Compute stresses end end SIGMAisigma Store stresses for all elements end ZX ZY ZT Z1 Z2stressesatnodesQ8SIGMA U2 DISP2 Choose one the quantities U2 ZX ZY ZT Z1 Z2 to plot cmin minZT cmax maxZT caxiscmin cmax patchFaces connec Vertices geom FaceVertexCDataZT FacecolorinterpMarker colorbar 2013 by Taylor Francis Group LLC 314 Introduction to Finite Element Analysis Using MATLAB and Abaqus 9733 Integration of the Stiffness Matrix The computation of the stiffness matrix is carried out in the same fashion as for the linear quadrilateral element except that the function elemQ4m is replaced by elemQ8m and fmlinm by fmquadm The exact integration of the stiffness matrix requires 3 Gauss points in each direction 9734 Results with the Coarse Mesh Figures 973 through 975 show respectively the contours of the vertical displacement v2 the stress σxx and the shear stress τxy The stresses are calculated at the centers of the elements and averaged at the nodes The program predicts a displacement at midspan equal to 015 mm To check whether this result is accurate consider the present deep beam as a slender beam and use the engineering beam theory to calculate the midspan deflection For the slender beam with a stiffness EI shown in Figure 976 the midspan deflection is obtained as δmax Pa3L2 4a2 24EI 991 400 350 300 250 200 Height mm 150 100 50 0 0 100 200 300 400 500 600 700 800 015 01 005 0 Vertical displacement mm Length mm FIGURE 973 Contour of the vertical displacement v2 400 350 300 250 200 Height mm 150 100 100 200 300 400 500 600 700 800 3 2 1 0 1 2 3 Horizontal stress σxx MPa 50 0 0 Length mm FIGURE 974 Contour of the stress σxx 2013 by Taylor Francis Group LLC Plane Problems 315 400 350 300 250 200 Height mm 150 100 50 0 100 200 300 400 500 600 700 800 0 1 06 08 04 02 0 02 04 Shear stress τxy MPa Length mm FIGURE 975 Contour of the stress τxy a a P P L El FIGURE 976 Slender beam under 4point bending According to Equation 991 the midspan displacement is equal to 012 mm which is less than the value of 015 mm obtained with the program This is somewhat logical since the engineering beam theory on which the analytical formula is based does not take into account the extra shear deflections that develop in deep beams We can therefore confidently affirm that the displacement obtained with the program is as good as can be obtained with a coarse mesh The contour of the horizontal stress σxx in Figure 974 looks acceptable compression at the top and tension at the bottom with the neutral axis is free of any stress The contour of the shear stress τxy is also acceptable A shear band can be seen between the support and the load application point Elsewhere the shear stresses are quite negligible 9735 Program with Automatic Mesh Generation In the program PlaneQ8MESHm the mesh is automatically generated with the module Q8meshm This module prepares the elements connectivity and nodal geometry matrices and is listed next after the main program PlaneQ8meshm THIS PROGRAM USES AN 8NODDED QUADRILATERAL ELEMENT FOR THE LINEAR ELASTIC STATIC ANALYSIS OF A TWO DIMENSIONAL PROBLEM IT CONTAINS AN AUTOMATIC MESH GENERATION MODULE Q8meshm Make these variables global so they can be shared by other functions clc clear all global nnd nel nne nodof eldof n ngp global geom connec dee nf Nodalloads 2013 by Taylor Francis Group LLC 316 Introduction to Finite Element Analysis Using MATLAB and Abaqus global Length Width NXE NYE Xorigin Yorigin dhx dhy format long g To change the size of the problem alter the next lines Length 800 Length of the model Width 400 Width NXE 32 Number of rows in the x direction NYE 16 Number of rows in the y direction dhx LengthNXE Element size in the x direction dhy WidthNYE Element size in the x direction Xorigin 0 X origin of the global coordinate system Yorigin 0 Y origin of the global coordinate system nne 8 nodof 2 eldof nnenodof ngp 2 Q8mesh Generate the mesh E 40000 Elastic modulus in MPa vu 017 Poissons ratio thick 100 Beam thickness in mm Form the elastic matrix for plane stress dee formdsigEvu Boundary conditions nf onesnnd nodof Initialize the matrix nf to 1 Restrain in all directions the nodes situated x Length for i1nnd if geomi1 Length nfi 0 1 end if geomi1 100 geomi2 0 nfi 1 0 end end Counting of the free degrees of freedom n0 for i1nnd for j1nodof if nfij 0 nn1 nfijn end end end loading Nodalloads zerosnnd 2 Initialize the matrix of nodal loads to 0 Apply a concentrated at the node having x 0 and y 0 2013 by Taylor Francis Group LLC Plane Problems 317 Force 30000 N for i1nnd if geomi1 500 geomi2 400 Nodalloadsi 0 Force Force acting in negative direction end end End of input Assemble the global force vector fgzerosn1 for i1 nnd if nfi1 0 fgnfi1 Nodalloadsi1 end if nfi2 0 fgnfi2 Nodalloadsi2 end end Form the matrix containing the abscissas and the weights of Gauss points sampgaussngp Numerical integration and assembly of the global stiffness matrix initialize the global stiffness matrix to zero kk zerosn n for i1nel coordg elemq8i coordinates of the nodes of element i and its steering vector kezeroseldofeldof Initialize the element stiffness matrix to zero for ig1 ngp wi sampig2 for jg1 ngp wjsampjg2 derfun fmquadsamp igjg Derivative of shape functions in local coordinates jacdercoord Compute Jacobian matrix ddetjac Compute determinant of Jacobian matrix jac1invjac Compute inverse of the Jacobian derivjac1der Derivative of shape functions in global coordinates beeformbeederivnneeldof Form matrix B keke dthickwiwjbeedeebee Integrate stiffness matrix end end kkformkkkkke g assemble global stiffness matrix end End of assembly delta kkfg solve for unknown displacements dispnode xdisp ydisp for i1 nnd if nfi1 0 xdisp 0 else xdisp deltanfi1 end 2013 by Taylor Francis Group LLC 318 Introduction to Finite Element Analysis Using MATLAB and Abaqus if nfi2 0 ydisp 0 else ydisp deltanfi2 end dispi xdisp ydisp Display displacements of each node DISPi xdisp ydisp end ngp1 Calculate stresses and strains at the center of each element sampgaussngp for i1nel coordg elemq8i coordinates of the nodes of element i and its steering vector eldzeroseldof1 Initialize element displacement to zero for m1eldof if gm0 eldm0 else eldmdeltagm Retrieve element displacement from the global displacement vector end end for ig1 ngp wi sampig2 for jg1 ngp wjsampjg2 derfun fmquadsamp igjg Derivative of shape functions in local coordinates jacdercoord Compute Jacobian matrix jac1invjac Compute inverse of the Jacobian derivjac1der Derivative of shape functions in global coordinates beeformbeederivnneeldof Form matrix B epsbeeeld Compute strains sigmadeeeps Compute stresses end end SIGMAisigma Store stresses for all elements end ZX ZY ZT Z1 Z2stressesatnodesQ8SIGMA Plot stresses in the xdirection U2 DISP2 cmin minU2 cmax maxU2 caxiscmin cmax patchFaces connec Vertices geom FaceVertexCDataU2 FacecolorinterpMarker colorbar Q8meshm This module generates a mesh of 8nodded quadrilateral elements global nnd nel nne nodof eldof n 2013 by Taylor Francis Group LLC Plane Problems 319 global geom connec dee nf Nodalloads global Length Width NXE NYE Xorigin Yorigin dhx dhy nnd 0 k 0 for i 1NXE for j1NYE k k 1 n1 i13NYE22j 1 n2 i3NYE2j NYE 1 n3 i3NYE22j1 n4 n3 1 n5 n3 2 n6 n2 1 n7 n1 2 n8 n1 1 geomn1 i1dhx Xorigin j1dhy Yorigin geomn3 idhx Xorigin j1dhy Yorigin geomn2 geomn11geomn312 geomn12geomn322 geomn5 idhx Xorigin jdhy Yorigin geomn4 geomn31 geomn512 geomn32 geomn522 geomn7 i1dhx Xorigin jdhy Yorigin geomn6 geomn51 geomn712 geomn52 geomn722 geomn8 geomn11 geomn712 geomn12 geomn722 nel k nnd n5 conneck n1 n2 n3 n4 n5 n6 n7 n8 end end The variables NXE and NYE represent respectively the number of intervals along the x and y directions as shown in Figure 977 For each interval i and j one element with nodes n1 n8 is created The module returns the matrices geomnnd 2 and connecnel nne as well as the number of elements nel and the number of nodes nnd The results obtained with the fine mesh NXE 32 n5 n6 n8 n4 n3 n2 n1 n7 y dhx dhy x j 1 to NYE i1 to NXE FIGURE 977 Automatic mesh generation with the Q8 element 2013 by Taylor Francis Group LLC 320 Introduction to Finite Element Analysis Using MATLAB and Abaqus and NYE 16 are displayed in Figures 978 through 980 respectively as contour plots of the vertical displacement v2 the stress σxx and the shear stress τxy The stresses are calculated at the centers of the elements and averaged at the nodes More details can be obtained with a finer mesh for example notice the stress concentration at the load application point and the shape of the shear band 400 300 200 Height mm 100 100 200 300 400 500 600 700 800 015 005 0 Vertical displacement mm 01 0 0 Length mm FIGURE 978 Contour of the vertical displacement v2 400 300 200 Height mm 100 0 100 200 300 400 500 600 700 800 4 4 Horizontal stress σxx MPa 2 2 0 0 Length mm FIGURE 979 Contour of the stress σxx 400 300 200 Height mm 100 0 100 200 300 400 500 600 700 800 0 1 1 2 2 0 Shear stress τxy MPa Length mm FIGURE 980 Contour of the stress τxy 2013 by Taylor Francis Group LLC Plane Problems 321 974 ANALYSIS WITH ABAQUS USING THE Q8 QUADRILATERAL In this section we will analyze the simply supported deep beam subject to fourpoint bending shown in Figure 971 Taking advantage of symmetry only half the model is analyzed We will use an element size of 25 mm so we could compare the results with those obtained previously Start Abaqus CAE Click on Create Model Database On the main menu click on File and set Set Work Directory to choose your working directory Click on Save As and name the file DeepBeamQ8cae On the lefthandside menu click on Part to begin creating the model Name the part DEEPBeamQ8 check 2D Planar and check Deformable in the type Choose Shell as the base feature Enter an approximate size of 1000 mm and click on Continue In the sketcher menu choose the Create Lines Rectangle icon to begin drawing the geometry of the beam Click on Done in the bottomleft corner of the viewport window Figure 981 FIGURE 981 Creating the DeepBeamQ8 Part Define a material named RConcrete with an elas tic modulus of 40000 MPa and a Poissons ratio of 017 Next click on Sections to create a section named BeamsectionQ8 In the Category check Solid and in the Type check Homoge neous Click on Continue In the Edit Section dialog box check Plane stressstrain thickness and enter 100 mm as the thickness Click on OK Figure 982 FIGURE 982 Creating a plane stress section 2013 by Taylor Francis Group LLC 322 Introduction to Finite Element Analysis Using MATLAB and Abaqus Expand the menu under Parts and DeepBeamQ8 and double click on Section Assignments With the mouse select the whole part In the Edit Section Assignments dialog box select BeamsectionQ8 and click on OK Figure 983 FIGURE 983 Editing section assignments In the model tree double click on Mesh under the DeepBeamQ8 In the main menu under Mesh click on Mesh Controls In the dialog box check Quad for Ele ment shape and Structured for Technique Click on OK Under Mesh click on Ele ment Type In the dialog box select Standard for ele ment library Quadratic for geometric order In Quad check Reduced integration The description of the ele ment CPS8R A 8node biquadratic plane stress quadrilateral reduced inte gration can be seen in the dialog box Click on OK Figure 984 FIGURE 984 Mesh controls and element type 2013 by Taylor Francis Group LLC Plane Problems 323 In the main menu under Seed click on Part In the dialog box enter 25 for Approxi mate global size Click on OK and on Done In the main menu under Mesh click on Part In the prompt area click on Yes Figure 985 FIGURE 985 Mesh Under Part in the lefthand side menu click on Sets In the dialog box name the set Loadednode and check Node for Type Click on Con tinue In the viewport locate the node situated at 300 mm from the right edge which is the centerline of the beam Click on Done Figure 986 FIGURE 986 Creating the node set Loadednode 2013 by Taylor Francis Group LLC 324 Introduction to Finite Element Analysis Using MATLAB and Abaqus Repeat the procedure and this time name the node set Cen terline In the viewport locate all the nodes situated on the right edge Click on Done Figure 987 FIGURE 987 Creating the node set Centerline Repeat the procedure and this time name the node set Sup port In the viewport locate the node situated at 100 mm from the left bottom corner Click on Done Figure 988 FIGURE 988 Creating the node set Support 2013 by Taylor Francis Group LLC Plane Problems 325 In the model tree expand the Assembly and double click on Instances Select DeepBeamQ8 for Parts and click OK In the model tree expand Steps and Initial and double click on BC Name the boundary condition Roller select DisplacementRotation for the type and click on Continue In the bottom right corner of the viewport click on Sets and select DeepBeamQ8 1Support and click on Continue In the Edit Boundary Condition check U2 Click OK Repeat the procedure again this time select the set DeepBeamQ8 1Centerline and click on Continue In the Edit Boundary Condition check U1 Click OK Figure 989 FIGURE 989 Imposing BC using a node set In the model tree double click on Steps Name the step Applyloads Set the proce dure to General and select Static General Click on Continue Click on OK In the model tree under steps and under Applyloads click on Loads Name the load PointLoad and select Concentrated force as the type Click on Continue In the bottomright corner of the viewport click on sets and select DeepBeam Q81Loaded node In the Edit Load dialog box enter 30000 for CF2 Click OK Figure 990 FIGURE 990 BC and loads Under Analysis right click on Jobs and then click on Create In the Create Job dialog box name the job DeepBeamQ8 and click on Continue In the Edit Job dialog box enter a description for the job Check Full analysis select to run the job in Background and check to start it immediately Click OK Expand the tree under Jobs right click on DeepBeamQ8 Then click on Submit If you get the following message DeepBeamQ8 2013 by Taylor Francis Group LLC 326 Introduction to Finite Element Analysis Using MATLAB and Abaqus FIGURE 991 Contour of the vertical displacement FIGURE 992 Contour of the horizontal stress σxx completed successfully in the bottom window then your job is free of errors and was executed properly Under the top menu in the Module scroll to Visualization and click to load Abaqus Viewer On the main menu under File click Open navigate to your working directory and open the file DeepBeamQ8odb It should have the same name as the job you submitted Click on the icon Plot on Undeformed shape Under the main menu select U and U2 to plot the vertical displacement Figure 991 It can be seen that the displacement contour is exactly the same as that obtained with the MATLAB code Figure 978 Under the main menu select S and S11 to plot σxx Figure 992 Again the contour is very similar to that shown in Figure 979 98 SOLVED PROBLEM WITH MATLAB 981 STRIP FOOTING WITH THE CST ELEMENT Figure 993 represents a strip footing on a sandy soil with an elastic modulus E 105 kNm2 and a Poissons ratio μ 03 The footing is 2 m wide and supports a uniformly distributed load of 5 kNm2 Five meters beneath the footing the soil is made up of a solid rock formation that can be 2013 by Taylor Francis Group LLC Plane Problems 327 2 m 6 m 5 m 5 kNm2 E 105 kNm2 ν 03 Rock substratum FIGURE 993 Strip footing considered very stiff In addition assume that 6 m away from the center of the footing the horizontal displacement of the soil is negligible Consider an element length of 05 m analyze the footing using both the CST and LST elements Plot the vertical deflection of the center line as a function of depth Produce a contour of the second principal stress σ2 The finite strip footing is a threedimensional solid However the longitudinal direction is very important which therefore prevents thickness change The ends of the strip foundations are prevented from moving in the zdirection then the displacement w is equal to zero At the midspan of the footing by symmetry w must be also equal to zero Therefore we assume that w is zero everywhere and the displacements u and v are functions of x and y only Such a state is characterized by ϵzz ϵxz ϵyz 0 and it is a state of plane strain The function formdepsm is used to generate the matrix of the elastic properties In addition the geometry of the footing is symmetrical therefore only the right half is discretized as shown in Figure 994 The domain is discretized using 12 intervals along the xdirection NXE 12 and 10 along the ydirection NYE 10 These give an element size of 05 m in both directions as shown in Figure 995 The boundary conditions of restrained nodes are generated using their coordinates as follows The nodes directly beneath the center of the footing x 0 and the nodes situated on the right boundary x Length are restrained in the xdirection if geomi1 0 geomi1 Length nfi 0 1 end The nodes situated on the rocky substratum y 0 are restrained in all directions if geomi2 0 nfi 0 0 end 2013 by Taylor Francis Group LLC 328 Introduction to Finite Element Analysis Using MATLAB and Abaqus 5 m Y X Displacement along X restrained Displacement along X restrained Displacements along X and Y directions are restrained 6 m 1 m 5 kNm FIGURE 994 Strip footing model 125 kN 11 22 33 25 kN 125 kN Y X FIGURE 995 Mesh with the CST element The mesh generating function T3meshm does not actually generate the loading This was added manually to the figure Indeed since T3meshm numbers the nodes in the ydirection it is not difficult to see in Figure 995 that nodes 11 22 and 33 are the loaded nodes Equivalent statically concentrated loads are applied to these nodes as follows Nodalloads11 0 125 Nodalloads22 0 250 Nodalloads33 0 125 2013 by Taylor Francis Group LLC Plane Problems 329 CSTSTRIPFOOTINGm THIS PROGRAM USES AN 3NODE LINEAR TRIANGULAR ELEMENT FOR THE LINEAR ELASTIC STATIC ANALYSIS OF A TWO DIMENSIONAL PROBLEM IT INCLUDES AN AUTOMATIC MESH GENERATION Make these variables global so they can be shared by other functions clear all clc global nnd nel nne nodof eldof n global geom dee nf Nodalloads global Length Width NXE NYE Xorigin Yorigin format long g To change the size of the problem or change elastic properties supply another input file Length 6 Length of the model Width 5 Width NXE 12 Number of rows in the x direction NYE 10 Number of rows in the y direction dhx LengthNXE Element size in the x direction dhy WidthNYE Element size in the x direction Xorigin 0 X origin of the global coordinate system Yorigin 0 Y origin of the global coordinate system nne 3 nodof 2 eldof nnenodof T3mesh Generate the mesh Material E 100000 Elastic modulus in MPa vu 03 Poissons ratio thick 1 Beam thickness in mm Form the elastic matrix for plane strain dee formdepsEvu Boundary conditions nf onesnnd nodof Initialize the matrix nf to 1 Restrain in the xdirection the nodes situated x 0 or x Length for i1nnd if geomi1 0 geomi1 Length nfi 0 1 end end Restrain in all directions the nodes situated y 0 for i1nnd if geomi2 0 nfi 0 0 end end 2013 by Taylor Francis Group LLC 330 Introduction to Finite Element Analysis Using MATLAB and Abaqus Counting of the free degrees of freedom n0 for i1nnd for j1nodof if nfij 0 nn1 nfijn end end end loading Nodalloads zerosnnd 2 Initialize the matrix of nodal loads to 0 Apply equivalent concentrated loads on nodes 11 22 and 33 in the ydirection Nodalloads11 0 125 Nodalloads22 0 250 Nodalloads33 0 125 End of input Assemble the global force vector fgzerosn1 for i1 nnd if nfi1 0 fgnfi1 Nodalloadsi1 end if nfi2 0 fgnfi2 Nodalloadsi2 end end Assembly of the global stiffness matrix initialize the global stiffness matrix to zero kk zerosn n for i1nel beegA elemT3i Form strain matrix and stering vector kethickAbeedeebee Compute stiffness matrix kkformkkkkke g assemble global stiffness matrix end End of assembly delta kkfg solve for unknown displacements for i1 nnd if nfi1 0 xdisp 0 else xdisp deltanfi1 end if nfi2 0 ydisp 0 else ydisp deltanfi2 2013 by Taylor Francis Group LLC Plane Problems 331 end nodedispi xdisp ydisp end Retrieve the ydisp of the nodes located on center line beneath the footing k 0 verticaldispzeros1NYE1 for i1nnd if geomi1 0 kk1 ycoordk geomi2 verticaldispknodedispi2 end end for i1nel beegA elemT3i Form strain matrix and stering vector eldzeroseldof1 Initialize element displacement to zero for m1eldof if gm0 eldm0 else eldmdeltagm Retrieve element displacement end end epsbeeeld Compute strains EPSieps Store strains for all elements sigmadeeeps Compute stresses SIGMAisigma Store stresses for all elements end Calculate the principal stresses SIG1zerosnel1 SIG2zerosnel1 for i 1nel SIG1iSIGMAi1SIGMAi22 sqrtSIGMAi1SIGMAi222 SIGMAi32 SIG2iSIGMAi1SIGMAi22 sqrtSIGMAi1SIGMAi222 SIGMAi32 end cmin minSIG2 cmax maxSIG2 caxiscmin cmax patchFaces connec Vertices geom FaceVertexCDataSIG2 FacecolorflatMarkero colorbar plottools The computed results are shown in Figure 996 A patch plot of the principal stress σ2 as well as the vertical displacements of the nodes situated just below the center of the footing are shown Both the maximum displacement and maximum stress respectively equal to 012 mm and 8 kNm2 occur just below the footing 982 STRIP FOOTING WITH THE LST ELEMENT The domain is also discretized using 12 intervals along the xdirection NXE 12 and 10 along the ydirection NYE 10 These give an element size of 05 m in both directions as shown in 2013 by Taylor Francis Group LLC 332 Introduction to Finite Element Analysis Using MATLAB and Abaqus Patch plot of the principal stress σ2 kNm2 5 4 3 2 1 0 5 4 3 Depth m 2 1 0 0 02 04 06 08 1 12 0 1 2 3 4 5 6 1 2 3 4 5 6 7 8 Vertical displacement m 104 FIGURE 996 Computed result with the CST element Y 0416 1666 1666 0833 0416 105 42 63 84 21 X FIGURE 997 Mesh with the LST element Figure 997 The boundary conditions of restrained nodes are entered in the same way as done previously using the nodal coordinates The mesh generating function T6meshm does not generate the loading This was added man ually as shown in Figure 997 The function T6meshm numbers the nodes in the ydirection therefore it is not difficult to see in Figure 997 that nodes 21 42 63 84 and 105 are the loaded nodes The equivalent statically concentrated loads are calculated as shown in Figure 998 and they are entered as follows Nodalloads21 0 0416 Nodalloads42 0 1666 Nodalloads63 0 0833 Nodalloads84 0 1666 Nodalloads105 0 0416 2013 by Taylor Francis Group LLC Plane Problems 333 q L qL 2qL 3 6 qL 6 Statically equivalent loads Actual loads FIGURE 998 Statically equivalent loads for the LST element LSTSTRIPFOOTINGm THIS PROGRAM USES A 6NODE LINEAR TRIANGULAR ELEMENT FOR THE LINEAR ELASTIC STATIC ANALYSIS OF A TWO DIMENSIONAL PROBLEM IT INCLUDES AN AUTOMATIC MESH GENERATION Make these variables global so they can be shared by other functions clear all clc global nnd nel nne nodof eldof n global connec geom dee nf Nodalloads XIG YIG global Length Width NXE NYE Xorigin Yorigin format long g To change the size of the problem or change elastic properties supply another input file Length 6 Length of the model Width 5 Width NXE 12 Number of rows in the x direction NYE 10 Number of rows in the y direction XIG zeros2NXE11 YIGzeros2NYE11 Vectors holding grid coordinates dhx LengthNXE Element size in the x direction dhy WidthNYE Element size in the x direction Xorigin 0 X origin of the global coordinate system Yorigin 0 Y origin of the global coordinate system nne 6 nodof 2 eldof nnenodof T6mesh Generate the mesh Material E 100000 Elastic modulus in MPa vu 03 Poissons ratio thick 1 Beam thickness in mm nhp 3 Number of sampling points Form the elastic matrix for plane stress dee formdepsEvu Boundary conditions nf onesnnd nodof Initialize the matrix nf to 1 2013 by Taylor Francis Group LLC 334 Introduction to Finite Element Analysis Using MATLAB and Abaqus Restrain in the xdirection the nodes situated x 0 or x Length for i1nnd if geomi1 0 geomi1 Length nfi 0 1 end end Restrain in all directions the nodes situated y 0 for i1nnd if geomi2 0 nfi 0 0 end end Counting of the free degrees of freedom n0 for i1nnd for j1nodof if nfij 0 nn1 nfijn end end end loading Nodalloads zerosnnd 2 Initialize the matrix of nodal loads to 0 Apply equivalent concentrated loads on nodes 21 42 63 84 and 105 in the ydirection Nodalloads21 0 0416 Nodalloads42 0 1666 Nodalloads63 0 0833 Nodalloads84 0 1666 Nodalloads105 0 0416 End of input Assemble the global force vector fgzerosn1 for i1 nnd if nfi1 0 fgnfi1 Nodalloadsi1 end if nfi2 0 fgnfi2 Nodalloadsi2 end end Assembly of the global stiffness matrix Form the matrix containing the abscissas and the weights of Hammer points samphammernhp 2013 by Taylor Francis Group LLC Plane Problems 335 initialize the global stiffness matrix to zero kk zerosn n for i1nel coordg elemT6i Form strain matrix and stering vector kezeroseldofeldof Initialize the element stiffness matrix to zero for ig 1nhp wi sampig3 derfun fmT6quadsamp ig jac dercoord d detjac jac1invjac Compute inverse of the Jacobian derivjac1der Derivative of shape functions in global coordinates beeformbeederivnneeldof Form matrix B keke dthickwibeedeebee Integrate stiffness matrix end kkformkkkkke g assemble global stiffness matrix end End of assembly delta kkfg solve for unknown displacements for i1 nnd if nfi1 0 xdisp 0 else xdisp deltanfi1 end if nfi2 0 ydisp 0 else ydisp deltanfi2 end nodedispi xdisp ydisp end Retrieve the xcoord and ydisp of the nodes located on the neutral axis k 0 for i1nnd if geomi1 0 kk1 ycoordk geomi2 verticaldispknodedispi2 end end nhp 1 Calculate stresses at the centroid of the element samphammernhp for i1nel coordg elemT6i Retrieve coordinates and stering vector eldzeroseldof1 Initialize element displacement to zero for m1eldof if gm0 eldm0 else eldmdeltagm Retrieve element displacement from the global displacement vector 2013 by Taylor Francis Group LLC 336 Introduction to Finite Element Analysis Using MATLAB and Abaqus end end for ig1 nhp derfun fmT6quadsampig Derivative of shape functions in local coordinates jacdercoord Compute Jacobian matrix jac1invjac Compute inverse of the Jacobian derivjac1der Derivative of shape functions in global coordinates beeformbeederivnneeldof Form matrix B epsbeeeld Compute strains sigmadeeeps Compute stresses end SIGMAisigma Store stresses for all elements end Prepare stresses for plotting ZX ZY ZT Z1 Z2preparecontourdataSIGMA Plot mesh using patches patchFacesconnecVerticesgeomFaceVertexCDatahsvnel FacecolornoneMarkero Plot stresses in the xdirection Ch contourfXIGYIGZ240 clabelCh colorbar plottools The computed results with the LST element are shown in Figure 999 A contour plot of the principal stress σ2 as well as the vertical displacements of the nodes situated just below the center of the footing are shown Like with the CST element both the maximum displacement and maximum stress respectively equal to 012 mm and 8 kNm2 occur just below the footing 983 BRIDGE PIER WITH THE Q8 ELEMENT Using the code Q8PLANESTRESSm analyze the bridge pier shown in Figure 9100 It is subject to six concentrated loads of 170 kN each The material is reinforced concrete with an elastic 0 5 4 3 2 1 0 5 4 3 2 1 0 1 2 3 4 5 6 2 4 6 8 Depth m 0 02 04 06 08 1 12 14 Contour plot of the principal stress σ2 kNm2 Vertical displacement m 104 FIGURE 999 Computed result with the LST element 2013 by Taylor Francis Group LLC Plane Problems 337 170 kN 1 m 2 m 2 m 1 m 1 m 10 m 070 m 2 m 25 m 25 m 15 m FIGURE 9100 Bridge pier modulus of 50000 MPa and a Poissons ratio of 017 Assume the base support as extremely rigid The first step consists of finding ways of simplifying the model The thickness of the pier is equal to 070 m which is relatively small compared to the horizontal and vertical dimensions The pier can therefore be analyzed as a plane stress problem In addition both the loading and geometry are symmetrical therefore only half the pier can be analyzed The second step consists of identifying the boundary conditions on the model as shown in Figure 9101 The third step consists of constructing an appropriate mesh of the domain Indeed the quality of the numerical results depends very much on the quality of the mesh However this is probably the most difficult and timeconsuming task in any finite element analysis specially when complex geometries are considered Like in the present case the domain is not regular therefore the mesh generation routine Q8meshm presented previously cannot be used since it was written for regular rectangular domains To mesh the present domain the Abaqus interactive edition was used As seen previously Abaqus generates an input file The nodal coordinates and elements connectivity are imported into MATLAB However this is not a straightforward procedure Indeed as shown in Figure 9102 within the 8node quadrilateral element Abaqus numbers differently the nodes The following in an excerpt of the Abaqus input file pierinp which lists the connectivity of elements 1 to 10 Element typeCPS8R 1 65 67 117 64 173 174 175 176 2 67 65 66 116 173 177 178 179 3 62 20 114 68 180 181 182 183 4 118 113 69 120 184 185 186 187 5 145 158 161 156 188 189 190 191 6 70 69 2 1 192 193 194 195 7 141 78 79 152 196 197 198 199 8 120 69 70 119 186 192 200 201 9 112 73 60 61 202 203 204 205 10 81 137 121 167 206 207 208 209 2013 by Taylor Francis Group LLC 338 Introduction to Finite Element Analysis Using MATLAB and Abaqus Edge fixed in xdirection and free in ydirection Edge fully fixed in both directions 170 kN 170 kN 170 kN FIGURE 9101 Bridge pier model 5 1 5 2 6 6 3 1 2 3 7 7 4 4 8 8 Abaqus numbering Present code numbering FIGURE 9102 Element internal node numbering For these data to be used in the MATLAB code Q8PLANESTRESSm they are rearranged as follows connec 65 173 67 174 117 175 64 176 Element 1 67 173 65 177 66 178 116 179 62 180 20 181 114 182 68 183 118 184 113 185 69 186 120 187 145 188 158 189 161 190 156 191 70 192 69 193 2 194 1 195 141 196 78 197 79 198 152 199 120 186 69 192 70 200 119 201 112 202 73 203 60 204 61 205 81 206 137 207 121 208 167 209 2013 by Taylor Francis Group LLC Plane Problems 339 0 0 2 4 6 2 2 Meters 4 6 8 4 Meters FIGURE 9103 Finite element discretization of the pier model Note how the columns are swapped to comply with the MATLAB code numbering scheme This can be achieved by importing the input file to Microsoft Excel and rearranging the columns manually or by a writing a MATLAB code that reads the input file and rearranges the columns The data consist of 138 elements and 481 nodes The details are given in the file PIERQ8datam and the actual mesh is shown in Figure 9103 Note that a consistent set of units is used dimensions in meters forces in kN and Youngs modulus in kNm2 All the nodes situated at x 0 are fixed in the xdirections and all the nodes situated at y 75 forming the base are fixed in both directions Nodes 18 19 and 20 situated respectively at x 5 m y 35 m x 3 m y 35 m and x 1 m y 35 m are each subject to a vertical force of 170 kN PIERQ8datam Beginning of data input Data file for the bridge pier analysis using 8nodded quadrilaterals global nnd nel nne nodof eldof n ngp global geom connec dee nf Nodalloads nnd 481 Number of nodes nel 138 Number of elements nne 8 Number of nodes per element nodof 2 Number of degrees of freedom per node ngp 2 number of Gauss points eldof nnenodof Number of degrees of freedom per element Thickness of the domain 2013 by Taylor Francis Group LLC 340 Introduction to Finite Element Analysis Using MATLAB and Abaqus thick 07 Thickness in meters Material properties E50e6 vu017 Youngs modulus kNm2and Poissons ratio Form the matrix of elastic properties deeformdsigEvu Matrix of elastic properties plane strain Nodes coordinates x and y geom 14489 03882 x and y coordinates of node 481 12990 07500 10607 10607 07500 12990 03882 14489 15000 00000 00000 15000 15000 75000 35000 75000 35000 00000 35480 04877 39213 13889 42322 17678 46111 20787 55123 24520 60000 25000 60000 35000 50000 35000 30000 35000 10000 35000 00000 35000 15000 05000 15000 10000 15000 15000 15000 20000 15000 25000 15000 30000 15000 35000 15000 40000 15000 45000 15000 50000 15000 55000 15000 60000 15000 65000 15000 70000 20000 75000 25000 75000 30000 75000 35000 70000 35000 65000 35000 60000 35000 55000 35000 50000 35000 45000 35000 40000 35000 35000 35000 30000 35000 25000 35000 20000 35000 15000 35000 10000 35000 05000 36903 09567 50433 23097 60000 30000 2013 by Taylor Francis Group LLC Plane Problems 341 55000 35000 45000 35000 40000 35000 35000 35000 25000 35000 20000 35000 15000 35000 05000 35000 00000 30000 00000 25000 00000 20000 05279 25104 14959 30626 15745 08916 17401 04784 30678 00527 32060 05869 24709 30574 36579 16005 44579 23344 45316 29797 49797 31580 18298 00390 19024 04486 19821 09532 39824 20741 29942 19923 20045 25012 30014 35050 20045 40063 30035 50060 20042 55061 30033 60056 20026 60047 30027 65043 20013 65036 30016 70020 20002 70015 30035 55056 20042 50056 20038 45049 30033 45053 30027 40051 20039 35054 20034 30033 29993 30030 29970 24998 20057 19916 20044 14742 29993 14785 30327 09731 30998 04957 33175 11463 39957 30076 34702 29078 29566 30118 19863 30733 13008 12396 10014 30384 09828 15734 05929 19562 05039 30118 15590 14193 20705 05842 19288 10906 36925 25511 48666 26297 2013 by Taylor Francis Group LLC 342 Introduction to Finite Element Analysis Using MATLAB and Abaqus 25004 24996 25046 40070 25049 55071 25027 65046 25035 60055 25047 50066 25043 45058 25031 35058 25017 30033 24994 19860 25006 14524 25053 08546 27641 04931 28417 11161 39934 25661 13049 17687 28367 06671 31567 15168 21675 01239 25011 70019 26413 01504 24210 26375 28474 25634 19773 26633 15193 26440 10320 25819 20271 15224 24001 10934 24406 06514 22846 03356 28123 15885 19850 22736 11200 21537 32207 24235 15744 22513 27330 21697 23705 22499 16578 18873 30443 20121 23874 15056 20083 18949 23454 18741 53776 29409 43237 26067 35630 22212 25059 02226 33638 18329 17419 15990 27962 03080 26233 18785 02640 25052 05159 27611 02520 30059 00000 27500 00000 22500 02964 19781 05604 22333 12500 35000 10007 32692 12487 30505 14980 32813 14299 13294 14377 10656 17516 09911 17439 12549 27902 23665 28887 20909 2013 by Taylor Francis Group LLC Plane Problems 343 31325 22178 30341 24934 16573 06850 14368 08208 13858 05740 15945 04333 19987 00815 18661 02048 20935 03921 22260 01058 19053 05313 19996 08374 22286 30653 24854 32787 22500 35000 19932 32866 39879 23201 38430 25586 36277 23861 37727 21477 51221 27853 49550 24697 52743 23924 54449 26964 46623 24821 45345 22065 48215 22048 54388 32205 52500 35000 49898 33290 51786 30494 60000 27500 56888 29705 57550 24880 16649 00195 15000 02500 17012 04743 32999 04979 30663 07344 32664 09866 35000 07500 19423 07009 15000 07500 17411 09766 42202 22043 41073 19210 44140 19325 32496 14892 35000 12500 30160 12258 29956 22460 27468 19891 24999 22428 27487 24997 32497 30015 35000 27500 32485 24999 29981 27514 30020 37550 27522 35054 25038 37564 27536 40061 32517 45026 35000 42500 32513 40026 30030 42552 30035 52558 2013 by Taylor Francis Group LLC 344 Introduction to Finite Element Analysis Using MATLAB and Abaqus 27541 50063 25048 52569 27542 55063 22531 60051 20019 62541 22520 65041 25031 62551 22507 70017 25019 67533 20007 67525 32517 60028 30030 62549 32514 65021 35000 62500 27527 65045 27534 60056 25042 57563 30034 57556 30022 67531 32508 70010 35000 67500 20001 72507 17501 70007 15000 72500 17500 75000 17506 65018 15000 67500 32500 75000 35000 72500 30008 72510 32517 50030 32517 55028 35000 52500 22545 55066 22545 50061 20042 52559 25045 47562 30034 47556 27538 45055 22541 45054 20040 47553 32507 35025 35000 37500 22546 40067 22535 35056 20042 37558 25024 32546 30004 32540 27505 30031 22525 30033 20036 32544 32471 19961 35000 22500 22524 25004 22525 19888 20051 22464 25000 17192 22525 14633 20051 17329 25030 11535 22437 09039 19933 12137 26347 06739 27690 09138 29320 04944 32839 00264 30838 02215 2013 by Taylor Francis Group LLC Plane Problems 345 35000 02500 19780 13065 21645 10920 23938 12995 22072 15140 47557 30688 47500 35000 45158 32399 60000 32500 57500 35000 32134 29598 34851 32039 32500 35000 29783 32559 27138 30346 27500 35000 17411 30679 17500 35000 11807 11501 11900 09131 05020 32559 02500 35000 00000 32500 11418 14065 08664 14362 09131 11900 08565 20550 07878 17648 11438 16711 12125 19612 01958 14872 04905 17026 00000 17500 14319 15940 07800 25461 10167 28101 07527 30251 37896 14947 34877 13734 35039 10515 37952 11785 14872 01958 17850 02587 33770 05373 31369 03198 35120 02450 36076 07257 32618 08666 28392 08916 26209 11048 24204 08724 26386 06593 38201 18373 34634 20270 35109 17167 40675 15860 42636 29936 42500 35000 39978 32538 46991 28047 27027 03218 28545 00488 15000 12500 17522 14871 37330 29577 37500 35000 17528 19958 2013 by Taylor Francis Group LLC 346 Introduction to Finite Element Analysis Using MATLAB and Abaqus 15000 22500 17522 25006 35000 17500 29968 17354 15000 27500 17517 30017 20039 27523 17519 35027 15000 37500 17523 40032 35000 32500 15000 42500 17519 45025 20041 42556 17521 50028 15000 52500 17521 55031 35000 47500 15000 57500 17513 60023 20034 57554 35000 57500 15000 62500 27514 70020 25006 72509 27500 75000 22500 75000 25045 42564 15000 47500 25010 27515 15000 32500 27500 14655 15000 17500 24732 04370 26511 02653 28164 04876 23950 05951 24629 02430 21768 18845 20177 17087 23664 16899 35814 27294 39945 27868 30796 11312 32371 13315 29845 15527 28270 13523 24459 28474 26342 26004 29020 27876 15076 28533 17483 26537 19818 28683 15468 24477 17797 22624 19812 24685 07500 35000 12757 26130 05740 13858 21992 26504 21190 03541 23367 01733 22555 06178 33455 26656 41586 25864 44277 27932 25736 00361 2013 by Taylor Francis Group LLC Plane Problems 347 30213 06270 32603 16748 43908 24706 10760 23678 13472 22025 29320 01804 29283 18003 32041 19225 23579 20620 21778 22618 19967 20842 25517 22098 24844 18763 26782 20241 14813 18280 16161 20693 18845 15607 16504 15091 25999 15471 27178 17335 23957 24437 16998 17431 33919 23224 18330 18911 x and y coordinates of node 481 Element connectivity connec 65 173 67 174 117 175 64 176 Element 1 67 173 65 177 66 178 116 179 62 180 20 181 114 182 68 183 118 184 113 185 69 186 120 187 145 188 158 189 161 190 156 191 70 192 69 193 2 194 1 195 141 196 78 197 79 198 152 199 120 186 69 192 70 200 119 201 112 202 73 203 60 204 61 205 81 206 137 207 121 208 167 209 165 210 122 211 54 212 15 213 54 211 122 214 75 215 14 216 165 217 56 218 18 219 77 220 16 221 55 222 165 213 15 223 79 197 78 224 6 225 22 226 52 227 107 228 106 229 51 230 80 231 79 226 22 232 23 233 14 215 75 234 81 235 13 236 105 237 50 238 51 229 106 239 102 240 82 241 132 242 123 243 101 244 47 245 48 246 102 247 98 248 84 249 130 250 124 251 97 252 44 253 45 254 98 255 94 256 86 257 128 258 125 259 127 260 89 261 91 262 126 263 93 264 142 265 126 262 91 266 41 267 88 268 90 269 40 270 127 263 126 271 90 268 88 272 125 273 127 272 88 274 94 259 40 269 90 275 92 276 39 277 36 278 93 279 35 280 8 281 93 266 91 282 34 283 35 279 38 284 9 285 39 276 92 286 43 287 86 256 94 288 42 289 87 290 125 258 128 291 95 292 129 293 128 257 86 294 97 295 95 291 128 293 129 296 96 297 46 298 84 248 98 254 45 299 85 300 124 250 130 301 99 302 131 303 130 249 84 304 101 305 2013 by Taylor Francis Group LLC 348 Introduction to Finite Element Analysis Using MATLAB and Abaqus 99 301 130 303 131 306 100 307 49 308 82 240 102 246 48 309 83 310 123 242 132 311 103 312 103 311 132 313 133 314 104 315 133 316 134 317 80 318 104 314 135 319 134 320 106 228 107 321 10 322 71 323 107 227 52 324 149 325 120 326 150 327 162 328 76 329 77 219 18 330 57 331 55 332 17 333 56 217 165 222 111 334 110 335 59 336 19 337 73 338 111 337 19 339 60 203 68 340 112 205 61 341 62 183 69 185 113 342 3 343 2 193 64 175 117 344 63 345 21 346 113 347 115 348 4 349 3 342 155 350 116 351 115 352 138 353 7 354 5 355 116 178 66 356 113 184 118 357 138 352 115 347 117 174 67 358 148 359 114 360 12 361 74 362 108 363 53 364 1 365 6 224 78 366 70 195 11 367 72 368 71 322 10 369 72 367 11 370 53 363 108 371 139 372 136 373 150 374 151 375 74 376 81 209 167 377 169 378 74 361 12 379 13 235 81 376 109 380 76 331 57 381 58 382 77 329 76 383 122 210 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4 348 115 351 116 355 112 443 146 450 144 438 73 202 70 366 78 196 141 451 119 200 141 452 168 424 151 453 119 451 119 453 151 374 150 326 120 201 2013 by Taylor Francis Group LLC Plane Problems 349 156 454 110 334 111 440 145 191 137 455 166 456 76 380 109 433 123 310 83 397 100 306 131 420 124 300 85 404 96 296 129 418 125 290 87 411 89 260 127 273 92 275 90 271 126 265 142 414 141 199 152 428 143 457 168 452 108 434 136 372 139 458 72 371 140 435 108 362 74 378 169 459 166 460 75 214 122 383 76 456 148 461 155 462 157 444 147 448 72 458 139 426 171 463 71 368 161 464 153 436 140 459 169 465 163 429 164 466 159 467 154 468 158 469 159 466 164 470 172 471 138 472 160 473 157 462 155 353 155 461 148 358 67 179 116 350 120 325 149 474 170 475 118 187 136 437 153 476 162 327 150 373 152 198 79 231 80 317 134 427 161 189 158 471 172 477 153 464 146 446 154 467 159 478 144 450 118 475 170 479 160 472 138 357 121 432 110 454 156 480 167 208 158 188 145 439 144 478 159 469 154 445 157 473 160 481 163 468 161 465 169 377 167 480 156 190 162 476 153 477 172 470 164 431 75 460 166 455 137 206 81 234 171 425 168 457 143 385 71 463 170 474 149 430 163 481 160 479 Element 138 Boundary conditions nf onesnnd nodof Initialize the matrix nf to 1 for i1nnd if geomi1 0 nfi 0 1 end if geomi2 75 nfi 0 0 end end Counting of the free degrees of freedom n0 for i1nnd for j1nodof if nfij 0 nn1 nfijn end end end disp Nodal freedom nf disp Total number of active degrees of freedom n loading Nodalloads zerosnnd 2 Nodalloads182170 Vertical load on node 18 Nodalloads192170 Vertical load on node 19 2013 by Taylor Francis Group LLC 350 Introduction to Finite Element Analysis Using MATLAB and Abaqus 2 Meters 0 2 4 Vertical displacement m 4 6 8 0 1 2 3 4 5 6 15 1 05 0 104 Meters FIGURE 9104 Contour of the vertical displacement 2 0 2 4 Principal stress σ1 kNm2 4 6 8 50 100 150 200 250 0 1 2 3 4 5 6 FIGURE 9105 Contour of the maximum principal stress σ1 2013 by Taylor Francis Group LLC Plane Problems 351 100 200 300 400 500 600 2 0 2 4 4 6 8 Principal stress σ2 kNm2 0 2 4 6 FIGURE 9106 Contour of the minimum principal stress σ2 Nodalloads202170 Vertical load on node 20 End input To run this example in the program Q8PLANESTRESSm replace Q8coarsemeshdata with PIERQ8datam The obtained results are displayed in Figures 9104 through 9106 respectively as contour plots of the vertical displacement v2 the first principal stress σ1 and the second principal stress σ2 The contours of the principal stresses may not be very accurate since they are calculated at the centers of the elements and averaged at the nodes More details can be obtained with a finer mesh 2013 by Taylor Francis Group LLC 354 Introduction to Finite Element Analysis Using MATLAB and Abaqus r r θ v u z FIGURE 101 Typical axisymmetric problem z v z v r u r u εθ σθ εrr εzz σzz σrr τzr γzr θ θ FIGURE 102 Strains and corresponding stresses in an axisymmetric solid r B θ dθ A u u u r dr x y FIGURE 103 Tangential strain 103 STRESSSTRAIN RELATIONS In three dimensions the generalized Hookes law for an isotropic material with a modulus of elasticity E and a Poissons ratio ν is given in terms of the elasticity matrix by Equation 5136 and in terms of the compliance matrix by Equation 5137 In an axisymmetric problem the shear strains γrθ and γzθ and the shear stresses τrθ and τzθ all vanish because of the radial symmetry Hence Equation 5136 is rewritten only in terms of the four stresses σrr σzz σθ and τzr and the 2013 by Taylor Francis Group LLC 358 Introduction to Finite Element Analysis Using MATLAB and Abaqus 105 PROGRAMMING Figure 106 represents a circular footing on a sandy soil with an elastic modulus E 105 kNm2 and a Poissons ratio ν 03 The footing is 2 m in radius and supports a load of 200 kN Nine meters beneath the footing the soil is made up of a solid rock formation that can be considered very stiff Assume that 7 m away from the centerline of the footing the horizontal displacement of the soil is negligible Consider an element length of 05 m analyze the footing using both the 6node triangle and the 8node quadrilateral elements Figure 107 shows the geometrical domain and the boundary conditions Because of symmetry only half the domain will be discretized Nodes on the centerline will only displace in the vertical direction Idem for the nodes placed at a 7 m radius because the horizontal movement of the soil at this distance is assumed negligible The nodes placed at a depth of 9 m are fixed in all directions because the rock substratum is assumed indeformable The 200 kN is also transformed into an equivalent uniformly distributed load of 63662 kNm2 200 kN 2 m E105kNm2 ν035 9 m R7 m Rock substratum FIGURE 106 Circular footing on a sandy soil Y 1 m 63662 kNm2 R 7 m X 9 m FIGURE 107 Geometrical model for the circular footing 2013 by Taylor Francis Group LLC Axisymmetric Problems 359 1051 COMPUTER CODE AXISYMT6m The following program AXISYMT6m is an adaptation of the plane stressstrain program LSTPLANESTRESSMESHm to axisymmetric conditions The program is listed next and includes the automatic mesh generation function T6meshm THIS PROGRAM USES A 6NODE TRIANGULAR ELEMENT FOR THE LINEAR ELASTIC STATIC ANALYSIS OF AN AXISYMMETRIC PROBLEM IT INCLUDES AN AUTOMATIC MESH GENERATION Make these variables global so they can be shared by other functions clear all clc global nnd nel nne nodof eldof n global connec geom dee nf Nodalloads global Length Width NXE NYE Xorigin Yorigin format long g To change the size of the problem or change elastic properties supply another input file Length 7 Length of the model Width 9 Width NXE 14 Number of rows in the x direction NYE 18 Number of rows in the y direction dhx LengthNXE Element size in the r direction dhy WidthNYE Element size in the z direction Xorigin 0 r origin of the global coordinate system Yorigin 0 z origin of the global coordinate system nne 6 nodof 2 eldof nnenodof T6mesh Generate the mesh Material E 100000 Elastic modulus in kNm2 vu 035 Poissons ratio nhp 3 Number of sampling points Form the elastic matrix for plane stress dee formdaxEvu Boundary conditions nf onesnnd nodof Initialize the matrix nf to 1 for i1nnd if geomi1 0 geomi1 Length nfi 0 1 Restrain in direction r the nodes situated x 0 and x Length end if geomi2 0 nfi 0 0 Restrain in all directions the nodes situated y 0 Rock substratum end 2013 by Taylor Francis Group LLC 360 Introduction to Finite Element Analysis Using MATLAB and Abaqus end Counting of the free degrees of freedom n0 for i1nnd for j1nodof if nfij 0 nn1 nfijn end end end loading Nodalloads zerosnnd 2 Initialize the matrix of nodal loads to 0 Apply an equivalent nodal load to the nodes located at r 0 z 9 r 025 z 9 and r 05 z 9 r 75 z 9 r 1 z 9 pressure 63662 kNm2 for i1nnd if geomi1 0 geomi2 9 Nodalloadsi pressure0 0 elseif geomi1 025 geomi2 9 Nodalloadsi pressure0 00833 elseif geomi1 05 geomi2 9 Nodalloadsi pressure0 0083300833 elseif geomi1 075 geomi2 9 Nodalloadsi pressure0 025 elseif geomi1 1 geomi2 9 Nodalloadsi pressure0 00833 end end End of input Assemble the global force vector fgzerosn1 for i1 nnd if nfi1 0 fgnfi1 Nodalloadsi1 end if nfi2 0 fgnfi2 Nodalloadsi2 end end Assembly of the global stiffness matrix Form the matrix containing the abscissas and the weights of Hammer points samphammernhp initialize the global stiffness matrix to zero kk zerosn n for i1nel coordg elemT6i Form strain matrix and steering vector kezeroseldofeldof Initialize the element stiffness matrix to zero for ig 1nhp wi sampig3 2013 by Taylor Francis Group LLC Axisymmetric Problems 361 derfun fmT6quadsamp ig jac dercoord d detjac jac1invjac Compute inverse of the Jacobian derivjac1der Derivative of shape functions in global coordinates beeradiusformbeeaxiderivnnefun coordeldof Form matrix B keke dwibeedeebeeradius Integrate stiffness matrix end kkformkkkkke g assemble global stiffness matrix end End of assembly delta kkfg solve for unknown displacements for i1 nnd if nfi1 0 xdisp 0 else xdisp deltanfi1 end if nfi2 0 ydisp 0 else ydisp deltanfi2 end DISPi xdisp ydisp end nhp 1 Calculate stresses at the centroid of the element samphammernhp for i1nel coordg elemT6i Retrieve coordinates and steering vector eldzeroseldof1 Initialize element displacement to zero for m1eldof if gm0 eldm0 else eldmdeltagm Retrieve element displacement from the global displacement vector end end for ig1 nhp derfun fmT6quadsampig Derivative of shape functions in local coordinates jacdercoord Compute Jacobian matrix jac1invjac Compute inverse of the Jacobian derivjac1der Derivative of shape functions in global coordinates beeradiusformbeeaxiderivnnefun coordeldof Form matrix B epsbeeeld Compute strains sigmadeeeps Compute stresses end SIGMAisigma Store stresses for all elements end ZX ZY ZTHETA ZT StressesatnodesaxiSIGMA U2 DISP2 Plot stresses in the xdirection cmin minZT cmax maxZT 2013 by Taylor Francis Group LLC 364 Introduction to Finite Element Analysis Using MATLAB and Abaqus Contour plot of the radial stress σrr kNm2 Vertical direction m 0 2 4 6 9 8 7 6 5 4 3 2 1 0 5 10 15 20 25 30 35 40 45 0 Radial direction m FIGURE 1010 Contour plot of the radial stress 0 2 4 6 Vertical direction m 9 8 7 6 5 4 3 2 1 0 70 60 50 40 30 20 10 0 Contour plot of the vertical stress σzz kNm2 Radial direction m FIGURE 1011 Contour plot of the vertical stress 2013 by Taylor Francis Group LLC Axisymmetric Problems 365 10 5 0 15 0 2 4 6 9 8 7 6 5 4 3 2 1 0 Vertical direction m Contour of the shear stress τrz kNm2 Radial direction m FIGURE 1012 Contour plot of the shear stress 1052 COMPUTER CODE AXISYMQ8m The following program AXISYMQ8m is an adaptation of the plane stressstrain program PLANEQ8MESHm to axisymmetric conditions The program is listed next and includes the automatic mesh generation function Q8meshm THIS PROGRAM USES AN 8NODDED QUADRILATERAL ELEMENT FOR THE LINEAR ELASTIC STATIC ANALYSIS OF AN AXISYMMETRIC PROBLEM IT CONTAINS AN AUTOMATIC MESH GENERATION MODULE Q8meshm Make these variables global so they can be shared by other functions clc clear all global nnd nel nne nodof eldof n ngp global geom connec dee nf Nodalloads global Length Width NXE NYE Xorigin Yorigin dhx dhy format long g To change the size of the problem or change elastic properties ALTER the q8inputmodulem Length 7 Length of the model Width 9 Width NXE 14 Number of rows in the x direction NYE 18 Number of rows in the y direction dhx LengthNXE Element size in the r direction dhy WidthNYE Element size in the z direction Xorigin 0 r origin of the global coordinate system Yorigin 0 z origin of the global coordinate system nne 8 nodof 2 eldof nnenodof ngp 3 2013 by Taylor Francis Group LLC 366 Introduction to Finite Element Analysis Using MATLAB and Abaqus Q8mesh Generate the mesh E 100000 Elastic modulus in kNm2 vu 035 Poissons ratio Form the elastic matrix for plane stress dee formdaxEvu Boundary conditions nf onesnnd nodof Initialize the matrix nf to 1 Restrain in all directions the nodes situated x Length for i1nnd if geomi1 0 geomi1 Length nfi 0 1 Restrain in direction r the nodes situated x 0 and x Length end if geomi2 0 nfi 0 0 Restrain in all directions the nodes situated y 0 Rock substratum end end Counting of the free degrees of freedom n0 for i1nnd for j1nodof if nfij 0 nn1 nfijn end end end loading Nodalloads zerosnnd 2 Initialize the matrix of nodal loads to 0 Apply an equivalent nodal load to the nodes located at r 0 z 9 r 025 z 9 and r 05 z 9 r 75 z 9 r 1 z 9 pressure 63662 kNm2 for i1nnd if geomi1 0 geomi2 9 Nodalloadsi pressure0 0 elseif geomi1 025 geomi2 9 Nodalloadsi pressure0 00833 elseif geomi1 05 geomi2 9 Nodalloadsi pressure0 0083300833 elseif geomi1 075 geomi2 9 Nodalloadsi pressure0 025 elseif geomi1 1 geomi2 9 Nodalloadsi pressure0 00833 end end End of input 2013 by Taylor Francis Group LLC Axisymmetric Problems 367 Assemble the global force vector fgzerosn1 for i1 nnd if nfi1 0 fgnfi1 Nodalloadsi1 end if nfi2 0 fgnfi2 Nodalloadsi2 end end Form the matrix containing the abscissas and the weights of Gauss points sampgaussngp Numerical integration and assembly of the global stiffness matrix initialize the global stiffness matrix to zero kk zerosn n for i1nel coordg elemq8i coordinates of the nodes of element i and its steering vector kezeroseldofeldof Initialize the element stiffness matrix to zero for ig1 ngp wi sampig2 for jg1 ngp wjsampjg2 derfun fmquadsamp igjg Derivative of shape functions in local coordinates jacdercoord Compute Jacobian matrix ddetjac Compute determinant of Jacobian matrix jac1invjac Compute inverse of the Jacobian derivjac1der Derivative of shape functions in global coordinates beeradiusformbeeaxiderivnnefun coordeldof Form matrix B keke dwiwjbeedeebeeradius Integrate stiffness matrix end end kkformkkkkke g assemble global stiffness matrix end End of assembly delta kkfg solve for unknown displacements dispnode xdisp ydisp for i1 nnd if nfi1 0 xdisp 0 else xdisp deltanfi1 end if nfi2 0 ydisp 0 else ydisp deltanfi2 end dispi xdisp ydisp Display displacements of each node DISPi xdisp ydisp end 2013 by Taylor Francis Group LLC 370 Introduction to Finite Element Analysis Using MATLAB and Abaqus 10522 Results Figures 1014 through 1017 show respectively the contours of the vertical displacement v the radial stress σrr the vertical stress σzz and the shear stress τrz obtained with the 6node triangle element Contour of the vertical displacement m 9 8 7 6 6 8 6 4 2 104 5 4 4 3 2 2 Vertical direction m 1 0 0 Radial direction m FIGURE 1014 Contour plot of the vertical displacement 0 2 4 6 35 30 25 20 15 10 5 0 9 8 7 6 5 4 3 2 Vertical direction m 1 0 Contour of the radial stress σrr kNm2 Radial direction m FIGURE 1015 Contour plot of the radial stress 2013 by Taylor Francis Group LLC Axisymmetric Problems 371 9 8 7 6 5 4 3 2 Vertical direction m 1 0 0 2 4 6 Contour plot of the vertical stress σzz kNm2 30 40 50 60 70 20 10 Radial direction m FIGURE 1016 Contour plot of the vertical stress 0 2 4 6 9 8 7 6 5 4 3 2 2 4 6 8 10 12 14 16 1 0 Vertical direction m Contour of the shear stress τrz kNm2 Radial direction m FIGURE 1017 Contour plot of the shear stress 2013 by Taylor Francis Group LLC 372 Introduction to Finite Element Analysis Using MATLAB and Abaqus 106 ANALYSIS WITH ABAQUS USING THE 8NODE QUADRILATERAL In this section we will analyze the circular footing shown in Figure 106 using the Abaqus interactive edition Taking advantage of symmetry only half the model is analyzed We will use an element size of 05 m so that we could compare the results with those obtained previously Start Abaqus CAE Click on Create Model Database On the main menu click on File and set Set Work Directory to choose your working directory Click on Save As and name the file FOOTINGQ8cae On the lefthandside menu click on Part to begin creating the model Name the part FOOT INGQ8 check Axisymmet ric check Deformable in the type Choose Shell as the base feature Enter an approx imate size of 20 m and click on Continue In the sketcher menu choose the Create Lines Rectangle icon to begin drawing the geometry of the footing Make sure that the sketch is to the right or to the left of the centerline Click on Done in the bottomleft corner of the viewport win dow Figure 1018 FIGURE 1018 Creating the FOOTINGQ8 Part Define a material named Dirt with an elastic modulus of 100000 kNm2 and a Poissons ratio of 035 Next click on Sections to create a section named FootingsectionQ8 In the Category check Solid and in the Type check Homogeneous Click on Continue In the Edit Section dialog box uncheck Plane stressstrain thickness Click on OK Figure 1019 FIGURE 1019 Creating an axisymmetric section 2013 by Taylor Francis Group LLC Axisymmetric Problems 373 Expand the menu under Parts and FOOTINGQ8 and dou ble click on Section Assign ments With the mouse select the whole part In the Edit Section Assignments dialog box select Foot ingsectionQ8 and click on OK Figure 1020 FIGURE 1020 Editing section assignments It will be useful to parti tion the top edge so that we could apply the pressure load over a length of 2 m Therefore under the main menu expand Tools and click on Partition In the partition dialog box select Edge for Type and Enter parameter for Method In the command line of the viewport enter 0714285714285 57 as shown in Figure 1021 Click on Create partition FIGURE 1021 Edge partition 2013 by Taylor Francis Group LLC 374 Introduction to Finite Element Analysis Using MATLAB and Abaqus In the model tree double click on Mesh under the FOOTINGQ8 In the main menu under Mesh click on Mesh Controls In the dialog box check Quad for Ele ment shape and Structured for Technique Click on OK Under Mesh click on Ele ment Type In the dialog box select Standard for ele ment library Quadratic for geometric order In Quad check Reduced integration The description of the ele ment CAX8R A 8node biquadratic axisymmetric quadrilateral reduced inte gration can be seen in the dialog box Click on OK Figure 1022 FIGURE 1022 Mesh controls and element type In the main menu under Seed click on Part In the dialog box enter 05 for Approx imate global size Click on OK and on Done In the main menu under Mesh click on Part In the prompt area click on Yes Figure 1023 FIGURE 1023 Mesh 2013 by Taylor Francis Group LLC Axisymmetric Problems 375 In the model tree expand the Assembly and double click on Instances Select FOOTINGQ8 for Parts and click OK In the model tree expand Steps and Ini tial and double click on BC Name the boundary condition Centerline select Displace mentRotation for the type and click on Continue In the viewport with the mouse select the centerline and click on Continue In the Edit Boundary Condition check U1 Click OK Repeat the pro cedure again this time select the right edge and click on Continue In the Edit Bound ary Condition check U1 Click OK Repeat the proce dure again this time select the bottom edge and click on Continue In the Edit Bound ary Condition check U1 and U2 Click OK Figure 1024 FIGURE 1024 Imposing BC using geometry In the model tree double click on Steps Name the step Applyloads Set the proce dure to General and select Static General Click on Continue Give the step a description and click OK In the model tree under steps and under Applyloads click on Loads Name the load Pressure and select Pres sure as the type Click on Continue In the viewport with the mouse select the left part of the partitioned top edge In the Edit Load dia log box enter 63662 kNm2 Click OK Figure 1025 FIGURE 1025 Imposing loads using geometry 2013 by Taylor Francis Group LLC 376 Introduction to Finite Element Analysis Using MATLAB and Abaqus FIGURE 1026 Contour of the vertical displacement FIGURE 1027 Contour of the vertical stress σyy 2013 by Taylor Francis Group LLC Axisymmetric Problems 377 Under Analysis right click on Jobs and then click on Create In the Create Job dialog box name the job FOOTINGQ8 and click on Continue In the Edit Job dialog box enter a description for the job Check Full analysis select to run the job in Background and check to start it immediately Click OK Expand the tree under Jobs right click on FOOTINGQ8 Then click on Submit If you get the following message FOOTINGQ8 completed successfully in the bottom window then your job is free of errors and was executed properly Under the top menu in the Module scroll to Visualization and click to load Abaqus Viewer On the main menu under File click Open navigate to your working directory and open the file FOOTINGQ8odb It should have the same name as the job you submitted Click on the icon Plot on Undeformed shape Under the main menu select U and U2 to plot the vertical displacement Figure 1026 It can be seen that the displacement contour is similar to that obtained with the MATLAB code Figures 109 and 1014 Under the main menu select S and S22 to plot σyy Figure 1027 Again the contour is very similar to that shown in Figure 1016 2013 by Taylor Francis Group LLC 11 Thin and Thick Plates 111 INTRODUCTION Plates are very important structural elements They are mainly used as slabs in buildings and bridge decks They are structural elements that are bound by two lateral surfaces The dimensions of the lateral surfaces are very large compared to the thickness of the plate A plate may be thought of as the twodimensional equivalent of a beam Plates are also generally subject to loads normal to their plane 112 THIN PLATES 1121 DIFFERENTIAL EQUATION OF PLATES LOADED IN BENDING The small deflection theory of plates attributed to Kirchhoff is based on the following assumptions 1 The xy plane coincides with the middle plane of the plate in the undeformed geometry 2 The lateral dimension of the plate is at least 10 times its thickness 3 The vertical displacement of any point of the plate can be taken equal to that of the point below or above it in the middle plane 4 A vertical element of the plate before bending remains perpendicular to the middle surface of the plate after bending 5 Strains are small deflections are less than the order of 1100 of the span length 6 The strain of the middle surface is zero or negligible Considering the plate element shown in Figure 111 the inplane displacements u and v respectively in the directions x and y can be expressed as u zw x 111 v zw y 112 where w represents the vertical displacement of the plate midplane Because of the assumption number 4 that is a vertical element of the plate before bend ing remains perpendicular to the middle surface of the plate after bending the transverse shear deformation is negligible The inplane strains can therefore be written in terms of the displacements as ϵxx ϵyy γxy u x v y u y v x z2w x2 z2w y2 2z 2w xy z χx χy χxy 113 The vector χ χx χy χxyT is called the vector of curvature or generalized strain 379 2013 by Taylor Francis Group LLC 382 Introduction to Finite Element Analysis Using MATLAB and Abaqus qxy Mx Mxy Mx dx x Qx Qx dx x Qy Qx Qy Mx My Mxy Mxy Qy dy Mxy Mxy dy y My dx dy z y My dy y Mxy dx x x FIGURE 114 Free body diagram of a plate element Moment equilibrium about the yaxis leads to Mxy y Mxx x Qx 1116 Substituting 1115 and 1116 in 1114 results in the governing equation 2Mxx x2 2Mxy xy 2Myy y2 qx y 0 1117 Since no relations regarding material behavior have entered Equation 1117 it is valid for all types of materials 1122 GOVERNING EQUATION IN TERMS OF DISPLACEMENT VARIABLES Substitution of Equation 1112 into the equilibrium equation 1117 leads to the general differential equation of simple rectangular plates 4w x4 2 4w 2x2y 4w y4 qx y Dr 1118 which is often written as 4w q Dr 1119 with Dr Eh3 121 ν2 1120 The solution of a simple rectangular plate in bending requires finding a function wx y that satisfies Equation 1118 and also the boundary conditions of the specific problem 2013 by Taylor Francis Group LLC Thin and Thick Plates 391 z 1000 lb 44482 N H 025 in 635 mm Simply supported 36 in 9144 mm 36 in 9144 mm Simply supported Simply supported Simply supported x y FIGURE 119 Simply supported plate on all edges 44482 N at the center The size of the plate is 36 36 in2 9144 9144 mm2 and the thickness is 025 in 635 mm It is made of steel E 30 106 psi 206843 MPa and ν 03 The main program ThickplateQ8m is listed next THIS PROGRAM USES AN 8NODDED QUADRILATERAL ELEMENT FOR THE LINEAR ELASTIC STATIC ANALYSIS OF A THICK PLATE IN BENDING Make these variables global so they can be shared by other functions clc clear all global nnd nel nne nodof eldof n ngpb ngps global geom connec deeb dees nf load dim format long g To cchange the size of the problem or change the elastic properties ALTER the PlateQ8inputmodulem dim 2 nne 8 nodof 3 eldof nnenodof PlateQ8inputmodule Length 18 Length of the in xdirection Width 18 Width of the model in ydirection NXE 9 Number of rows in the x direction NYE 9 Number of rows in the y direction dhx LengthNXE Element size in the x direction dhy WidthNYE Element size in the y direction Xorigin 0 x origin of the global coordinate system Yorigin 0 y origin of the global coordinate system thick 025 Thickness of plate ngpb 3 number of Gauss points bending ngps 2 number of Gauss points for shear Q8mesh Generate the mesh E 30e6 Elastic modulus in kNm2 vu 03 Poissons ratio Form the matrix of elastic properties 2013 by Taylor Francis Group LLC 392 Introduction to Finite Element Analysis Using MATLAB and Abaqus deebformdeebEvuthick Matrix of elastic properties for plate bending deesformdeesEvuthick Matrix of elastic properties for plate shear Boundary conditions nf onesnnd nodof Initialize the matrix nf to 1 for i1nnd if geomi1 0 nfi1 0 Restrain in direction w nfi3 0 Restrain rotation thetay around x elseif geomi2 0 nfi1 0 Restrain displacement w nfi2 0 Restrain rotation thetax around y elseif geomi1 Length nfi2 0 Restrain rotation thetax around y elseif geomi2 Width nfi3 0 Restrain rotation thetay around x end end Counting of the free degrees of freedom n0 for i1nnd for j1nodof if nfij 0 nn1 nfijn end end end disp Nodal freedom nf disp Total number of active degrees of freedom n loading load zerosnnd 3 for i1nnd if geomi1 Length geomi2 Width loadi1 10004 Vertical load of 250 lb on the center node end end End of input Assemble the global force vector fgzerosn1 for i1 nnd for j1nodof if nfij 0 fgnfij loadij end end end Form the matrix containing the abscissas and the weights of Gauss points sampbgaussngpb sampsgaussngps 2013 by Taylor Francis Group LLC Thin and Thick Plates 393 Numerical integration and assembly of the global stiffness matrix initialize the global stiffness matrix to zero kk zerosn n for i1nel coordg platelemq8i coordinates of the nodes of element i and its steering vector kebzeroseldofeldof Initialize the element bending stiffness matrix to zero keszeroseldofeldof Initialize the element Shear stiffness matrix to zero Integrate element bending stiffness and assemble it in global matrix for ig1 ngpb wi sampbig2 for jg1 ngpb wjsampbjg2 derfun fmquadsampb igjg Derivative of shape functions in local coordinates jacdercoord Compute Jacobian matrix ddetjac Compute the determinant of Jacobian matrix jac1invjac Compute inverse of the Jacobian derivjac1der Derivative of shape functions in global coordinates beebformbeebderivnneeldof Form matrix B kebkeb dwiwjbeebdeebbeeb Integrate stiffness matrix end end kkformkkkkkeb g assemble global stiffness matrix Integrate element Shear stiffness and assemble it in global matrix for ig1 ngps wi sampsig2 for jg1 ngps wjsampsjg2 derfun fmquadsamps igjg Derivative of shape functions in local coordinates jacdercoord Compute Jacobian matrix ddetjac Compute determinant of Jacobian matrix jac1invjac Compute inverse of the Jacobian derivjac1der Derivative of shape functions in global coordinates beesformbeesderivfunnneeldof Form matrix B keskes 56dwiwjbeesdeesbees Integrate stiffness matrix end end kkformkkkkkes g assemble global stiffness matrix end End of assembly delta kkfg solve for unknown displacements format short e dispnode wdisp xslope yslope for i1 nnd if nfi1 0 wdisp 0 else 2013 by Taylor Francis Group LLC 394 Introduction to Finite Element Analysis Using MATLAB and Abaqus wdisp deltanfi1 end if nfi2 0 xslope 0 else xslope deltanfi2 end if nfi3 0 yslope 0 else yslope deltanfi3 end dispi wdisp xslope yslope Display displacements of each node DISPi wdisp xslope yslope end ngp1 Calculate moments and shear forces the center of each element sampgaussngp for i1nel coordg platelemq8i coordinates of the nodes of element i and its steering vector eldzeroseldof1 Initialize element displacement to zero for m1eldof if gm0 eldm0 else eldmdeltagm Retrieve element displacement from the global displacement vector end end for ig1 ngp wi sampig2 for jg1 ngp wjsampjg2 derfun fmquadsamp igjg Derivative of shape functions in local coordinates jacdercoord Compute Jacobian matrix ddetjac Compute the determinant of Jacobian matrix jac1invjac Compute inverse of the Jacobian derivjac1der Derivative of shape functions in global coordinates beebformbeebderivnneeldof Form matrix Bb chib beebeld compute bending curvatures Moment deebchib Compute moments beesformbeesderivfunnneeldof Form matrix Bs chis beeseld compute shear curvatures Shear deeschis Compute shera forces end end ElementForcesiMoment Shear end W DISP1 MX MY MXY QX QY ForcesatnodesplateElementForces cmin minW cmax maxW 2013 by Taylor Francis Group LLC Thin and Thick Plates 395 y x 280 262 73 81 9 19 29 2 1 1 20 30 2 3 18 in 4572 mm 18 in 4572 mm Symmetry Symmetry Simply supported edge Simply supported edge FIGURE 1110 Finite element mesh of one quadrant of the simply supported plate caxiscmin cmax patchFaces connec Vertices geom FaceVertexCDataW FacecolorinterpMarker colorbar Because of symmetry only one quadrant of the plate is discretized The finite element mesh is shown in Figure 1110 and generated using the function meshQ8m Both the nodes and the elements are numbered in the ydirection In total there are 282 nodes and 81 elements 1162 DATA PREPARATION 11621 Stiffness Matrices Note two different integrations schemes are used one consisting of a 3 3 rule ngpb 3 to integrate the flexural matrix deeb and the other consisting of a 2 2 rule ngps 2 to integrate the shear stiffness matrix dees The matrices are respectively formed with the functions formdeebm and formdeesm listed in Appendix A 11622 Boundary Conditions The boundary conditions are given as follows Edge x 0 w 0 θy 0 Edge x 18 in θx 0 Edge y 0 w 0 θx 0 Edge y 18 in θy 0 They are introduced as follows For all the nodes located at x 0 restrain the degree of freedom No 1 corresponding to the vertical translation and the degree of freedom No 3 corresponding to the rotation θy around the axis x 2013 by Taylor Francis Group LLC 396 Introduction to Finite Element Analysis Using MATLAB and Abaqus For all the nodes located at y 0 restrain the degree of freedom No 1 corresponding to the vertical translation and the degree of freedom No 2 corresponding to the rotation θx around the axis y For all the nodes located at x Length the length of the quarter plate that is restrain the degree of freedom No 2 corresponding to the rotation θx around the axis y For all the nodes located at y Width the width of the quarter plate that is restrain the degree of freedom No 3 corresponding to the rotation θy around the axis x 11623 Loading A quarter of the 1000 lb load is applied at node 282 in the opposite zdirection This node is located by its coordinates File PlateQ8inputmodulem Beginning of data input global nnd nel nne nodof eldof n ngpb ngps global geom connec deeb dees nf load dim dim2 Dimension nnd 21 Number of nodes nel 4 Number of elements nne 8 Number of nodes per element nodof 3 Number of degrees of freedom per node ngpb 3 number of Gauss points bending ngps 2 number of Gauss points shear eldof nnenodof Number of degrees of freedom per element Thickness of the domain thick 025 Nodes coordinates x and y geom 00 18 x and y coordinates of node 1 00 135 x and y coordinates of node 2 00 9 x and y coordinates of node 3 00 45 x and y coordinates of node 4 00 0 x and y coordinates of node 5 45 18 x and y coordinates of node 6 45 9 x and y coordinates of node 7 45 0 x and y coordinates of node 8 9 18 x and y coordinates of node 9 9 135 x and y coordinates of node 10 9 9 x and y coordinates of node 11 9 45 x and y coordinates of node 12 9 0 x and y coordinates of node 13 135 18 x and y coordinates of node 14 135 9 x and y coordinates of node 15 135 0 x and y coordinates of node 16 18 18 x and y coordinates of node 17 18 135 x and y coordinates of node 18 18 9 x and y coordinates of node 19 18 45 x and y coordinates of node 20 18 0 x and y coordinates of node 21 disp Nodes XY coordinates geom Element connectivity connec 1 2 3 7 11 10 9 6 Element 1 3 4 5 8 13 12 11 7 Element 2 9 10 11 15 19 18 17 14 Element 3 11 12 13 16 21 20 19 15 Element 4 2013 by Taylor Francis Group LLC Thin and Thick Plates 397 disp Elements connectivity connec Material properties E30e6 vu03 Youngs modulus and Poissons ration Form the matrix of elastic properties deebformdeebEvuthick Matrix of elastic properties for plate bending deesformdeesEvuthick Matrix of elastic properties for plate shear Boundary conditions nf onesnnd nodof Initialize the matrix nf to 1 nf11 0 nf130 nf21 0 nf230 nf31 0 nf330 nf41 0 nf430 nf51 0 nf520 nf530 nf630 nf810 nf820 nf930 nf1310 nf1320 nf1430 nf1610 nf1620 nf1720nf1730 nf1820 nf1920 nf2020 nf2110nf2120 Counting of the free degrees of freedom n0 for i1nnd for j1nodof if nfij 0 nn1 nfijn end end end disp Nodal freedom nf disp Total number of active degrees of freedom n loading load zerosnnd 3 load171 10004 Vertical load of 250 lb on node 17 End input 11624 Numerical Integration of the Stiffness Matrix The stiffness matrix is given by Equation 1166 For each element it is computed as follows 1 For every element i 1 to nel 2 Retrieve the coordinates of its nodes coordnne 2 and its steering vector geldof using the function platelemq8m 3 Initialize the stiffness matrices to zero 2013 by Taylor Francis Group LLC 398 Introduction to Finite Element Analysis Using MATLAB and Abaqus a Loop over the Gauss points ig 1 to ngpb b Retrieve the weight wi as sampbig 2 i Loop over the Gauss points jg 1 to ngpb ii Retrieve the weight wj as sampbjg 2 iii Use the function fmquadm to compute the shape functions vector fun and their derivatives matrix der in local coordinates ξ sampbig 1 and η sampbjg 1 iv Evaluate the Jacobian jac der coord v Evaluate the determinant of the Jacobian as d detjac vi Compute the inverse of the Jacobian as jac1 invjac vii Compute the derivatives of the shape functions with respect to the global coordinates x and y as deriv jac1 der viii Use the function formbeebm to form the strain matrix beeb ix Compute the stiffness matrix as keb keb d wi wj beeb deeb beeb 4 Assemble the stiffness matrix keb into the global matrix kk a Loop over the Gauss points ig 1 to ngps b Retrieve the weight wi as sampsig 2 i Loop over the Gauss points jg 1 to ngps ii Retrieve the weight wj as sampsjg 2 iii Use the function fmquadm to compute the shape functions vector fun and their derivatives matrix der in local coordinates ξ sampsig 1 and η sampsjg 1 iv Evaluate the Jacobian jac der coord v Evaluate the determinant of the Jacobian as d detjac vi Compute the inverse of the Jacobian as jac1 invjac vii Compute the derivatives of the shape functions with respect to the global coordinates x and y as deriv jac1 der viii Use the function formbeesm to form the strain matrix bees ix Compute the stiffness matrix as kes kes d wi wj bees dees bees 5 Assemble the stiffness matrix kes into the global matrix kk The functions formbeebm and formbeesm which form the flexural and shear strain matrices are listed in Appendix A 1163 RESULTS 11631 Determination of the Resulting Moments and Shear Forces Once the global equations are solved or the global displacement for each element we retrieve its nodal displacements and calculate the resulting moments and shear forces at its center For such we use only one Gauss point as detailed next 1 For every element i 1 to nel 2 Retrieve the coordinates of its nodes coordnne 2 and its steering vector geldof using the function platelemq8m 3 Retrieve its vector of nodal displacements eldeldof a Loop over the Gauss points ig 1 to ngp b Retrieve the weight wi as sampig 2 i Loop over the Gauss points jg 1 to ngp ii Retrieve the weight wj as sampjg 2 2013 by Taylor Francis Group LLC Thin and Thick Plates 399 iii Use the function fmquadm to compute the shape functions vector fun and their derivatives matrix der in local coordinates ξ sampig 1 and η sampjg 1 iv Evaluate the Jacobian jac der coord v Evaluate the determinant of the Jacobian as d detjac vi Compute the inverse of the Jacobian as jac1 invjac vii Compute the derivatives of the shape functions with respect to the global coordinates x and y as deriv jac1 der viii Use the function formbeebm to form the strain matrix beeb ix Compute the flexural curvature χb beeb eld and the corresponding moments as Moment deeb χb x Use the function formbeesm to form the strain matrix bees xi Compute the shear curvature χs bees eld and the corresponding shear forces as Shear dees χs 4 Store the moments and shear forces in the array ElementForcesnel 5 Using the data stored in the array ElementForcesnel 5 the function Forcesatnodesplatem calculates the moments and shear forces at the nodes and returns them as arrays for plotting using the MATLAB function patch 11632 Contour Plots Figure 1111 shows the contour plot of the vertical displacement The program predicts a vertical displacement of 035239 in at node 282 which is the center of the plate that is very close to the exact solution of 035022 in Figures 1112 and 1113 show the contour plots of the moments Mxx and Mxy It is very interesting to note in Figure 1113 that the corner of the plate tends to rise Indeed it is well known that the corners of a flat plate under transverse load have the tendency to rise when upward displacements are not restricted as shown in Figure 1114 Vertical displacement w in 0 005 01 015 02 025 03 035 0 0 2 4 6 8 10 12 14 16 18 5 10 15 Width in Length in FIGURE 1111 Contour plot of the vertical displacement 2013 by Taylor Francis Group LLC 400 Introduction to Finite Element Analysis Using MATLAB and Abaqus Moment Mxx per unit length lb in 250 200 150 100 50 0 15 10 Width in 5 0 0 2 4 6 8 10 12 14 16 18 Length in FIGURE 1112 Contour plot of the moment Mxx Moment Mxy per unit length lb in 10 15 20 25 30 35 40 45 50 55 60 0 0 2 4 6 8 10 12 14 16 18 5 10 15 Width in Length in FIGURE 1113 Contour plot of the moment Mxy 117 ANALYSIS WITH ABAQUS 1171 PRELIMINARY Abaqus does not have plate elements as such Instead it uses shell elements In Abaqus a plate is merely considered as a flat shell A shell element can be considered as a sophisticated version of 2013 by Taylor Francis Group LLC Thin and Thick Plates 401 FIGURE 1114 Lifting of corners of a plate a plate element that can carry inplane forces Abaqus offers two types of threedimensional shell elements conventional shell elements and continuum shell elements Detailed descriptions of these elements can be found in the Abaqus manual and in Ref 6 In Abaqus shell elements are named as in the following sections 11711 ThreeDimensional Shell Elements S8R5W S conventional stressdisplacement shell SC continuum stressdisplacement shell STRI triangular stressdisplacement thin shell DS heat transfer shell 8 number of nodes R reduced integration optional 5 number of degrees of freedom per node optional W warping considered in smallstrain formulation 11712 Axisymmetric Shell Elements SAX2T S stressdisplacement shell DS heat transfer shell AX axisymmetric AXA axisymmetric with nonlinear asymmetric deformation 2 order of interpolation T coupled temperature displacement 11713 Thick versus Thin Conventional Shell Before choosing a shell element in Abaqus it is worthwhile to check whether it is suitable for thin shells only thick shells only or both The following elements are suitable for both S3 S3R S3RS S4 S4R S4RSW SAX1 SAX2 SAX2T SC6R and SC8R They include the transverse shear deformation which becomes very small as the shell thickness decreases The following elements S8R and S8RT are only for use in thick shell problems Elements STRI3 S4R5 STRI65 S8R S9R5 SAXA1n and SAXA2n should not be used for thick shells where transverse shear deformation is important 1172 SIMPLY SUPPORTED PLATE In this section we will analyze the simply supported square plate shown in Figure 119 As before we will only analyze a quarter for reasons of symmetry in both geometry and loading We will use the S4R element which is suitable for both thin and thick shells 2013 by Taylor Francis Group LLC 402 Introduction to Finite Element Analysis Using MATLAB and Abaqus Start Abaqus CAE Click on Create Model Database On the main menu click on File and set Set Work Direc tory to choose your working directory Click on Save As and name the file SLABS4Rcae On the lefthandside menu click on Part to begin creating the model Name the part SLABS4R check 3D check Deformable in the type Choose Shell as the shape and Extrusion for type Enter an approx imate size of 20 in and click on Continue Figure 1115 FIGURE 1115 Creating the SlabS4R Part In the sketcher menu choose the CreateLines connected icon to draw a straight line 18 in long In the prompt area in the bottomleft corner of the viewport window click on Sketch the section for the shell extrusion In the Edit base extrusion dialog box enter 18 in for depth and click OK Figure 1116 FIGURE 1116 Sketching the SlabS4R Part Define a material named steel with an elastic modulus of 30000000 psi and a Poissons ratio of 03 Next click on Sections to create a section named SlabsectionS4R In the Category check Shell and in the Type check Homogeneous Click on Continue In the Edit Section dialog box enter 025 in as the thickness Check Simpson for thickness integration rule Click on OK Figure 1117 FIGURE 1117 Creating a homogeneous shell section 2013 by Taylor Francis Group LLC Thin and Thick Plates 403 Expand the menu under Parts and SLABS4R and dou ble click on Section Assign ments With the mouse select the whole part In the Edit Section Assign ments dialog box select SlabsectionS4R and click on OK Figure 1118 FIGURE 1118 Editing section assignments In the model tree double click on Mesh under the SLABS4R In the main menu under Mesh click on Mesh Controls In the dia log box check Quad for Element shape and Struc tured for Technique Click on OK Under Mesh click on Element Type In the dia log box select Standard for element library Linear for geometric order In Quad check Reduced integration The description of the ele ment S4R A 4node doubly curved thin or thick shell reduced integration hour glass control finite mem brane strains can be seen in the dialog box Click on OK Figure 1119 FIGURE 1119 Mesh Controls and element type 2013 by Taylor Francis Group LLC 404 Introduction to Finite Element Analysis Using MATLAB and Abaqus In the main menu under Seed click on Part In the dialog box enter 2 in for Approx imate global size Click on OK and on Done In the main menu under Mesh click on Part In the prompt area click on Yes Figure 1120 FIGURE 1120 Mesh Under Part in the lefthand side menu click on Sets In the dialog box name the set Loadednode and check Node for Type Click on Con tinue In the viewport locate the central node as shown in Figure 1121 Click on Done FIGURE 1121 Creating a node set In the model tree expand the Assembly and double click on Instances Select SLABS4R for Parts and click OK In the model tree expand Steps and Initial and double click on BC Name the boundary condition EdgeX0 select DisplacementRotation for the type and click on Continue With the mouse select edge hav ing X 9 in as shown in Figure 1122 and click on Done in the prompt area In the Edit Boundary Condition check U2 UR1 UR2 no displacement is allowed along Y and no rota tions are allowed around X and Y Click OK FIGURE 1122 Imposing BC EdgeX0 using geometry 2013 by Taylor Francis Group LLC Thin and Thick Plates 405 Repeat the procedure and this time name the bound ary condition EdgeZ18 select Displacement Rotation for the type and click on Continue With the mouse select edge having Z 18 in as shown in Figure 1123 and click on Done in the prompt area In the Edit Boundary Condi tion check U2 UR2 UR3 no displacement is allowed along Y and no rotations are allowed around Y and Z Click OK FIGURE 1123 Imposing BC EdgeZ18 using geometry Repeat the procedure and this time name the bound ary condition EdgeZ0 select DisplacementRotation for the type and click on Continue With the mouse select edge having Z 0 in as shown in Figure 1124 and click on Done in the prompt area In the Edit Boundary Condition check U3 UR1 UR2 because of symmetry no displacement is allowed along Z and no rotations are allowed around X and Y Click OK FIGURE 1124 Imposing BC EdgeZ0 using geometry Repeat the procedure and this time name the boundary con dition EdgeX9 select Dis placementRotation for the type and click on Continue With the mouse select edge having X 9 in as shown in Figure 1125 and click on Done in the prompt area In the Edit Boundary Condi tion check U1 UR2 UR3 because of symmetry no dis placement is allowed along X and no rotations are allowed around Y and Z Click OK FIGURE 1125 Imposing BC EdgeX9 using geometry 2013 by Taylor Francis Group LLC 406 Introduction to Finite Element Analysis Using MATLAB and Abaqus In the model tree double click on Steps Name the step Applyloads Set the procedure to General and select Static General Click on Continue Give the step a description and click OK In the model tree under steps and under Applyloads click on Loads Name the load PointLoad and select Concentrated Force as the type Click on Continue In the bottomright corner of the viewport click on Sets and select SLABS4R 1Loadednode In the Edit load dialog box enter 250 a quarter of the load for CF2 Click OK Figure 1126 FIGURE 1126 Imposing a concentrated force using a node set In the model tree expand the Field Output Requests and then double click on FOutput1 FOutput1 is the default and is automatically generated when creating the step Uncheck the variables Contact and select any other variable you wish to add to the field output Click on OK Under Analysis right click on Jobs and then click on Create In the Create Job dialog box name the job SLABS4R and click on Continue In the Edit Job dialog box enter a description for the job Check Full analysis select to run the job in Background and check to start it immediately Click OK Expand the tree under Jobs right click on SLABS4R Then click on Submit If you get the following message SLABS4R completed successfully in the bottom window then your job is free of errors and was executed properly Under the top menu in the Module scroll to Visualization and click to load Abaqus Viewer On the main menu under File click Open navigate to your working directory and open the file SLABS4Rodb It should have the same name as the job you submitted Click on the Common options icon to display the Common Plot options dialog box Under labels check Show Element labels and Show Node labels to display elements and nodes numbering Click on the icon Plot Contours on both shapes to display the deformed shape of the beam Under the main menu select U and U2 to plot the vertical displacement It can be seen that the displacement of center of the plate is equal to 0351 in which is very close to the analytical solution Figure 1127 1173 THREEDIMENSIONAL SHELLS In this section we will show some more features of modeling with Abaqus We will analyze a castellated beam as an assembly of threedimensional shell elements Castellated beams such as the one shown in Figure 1128 are widely used in the steel construction industry They are fabricated from standard universal beam sections The beam is initially split along its length in a zigzag cut The two halves of the beam are then separated displaced by one profile to join the peaks and welded together to increase the depth of the beam From a universal beam section such as the one shown in Figure 1129 we will make a castellated beam whose cross section is shown in Figure 1130 Notice that we will only model 2013 by Taylor Francis Group LLC Thin and Thick Plates 407 FIGURE 1127 Plotting displacements on deformed shape H15 h w 05 h h w H FIGURE 1128 Castellated beam 178 78 406 128 FIGURE 1129 Base profile the middle plane as the behavior of a conventional shell element is described by that of its middle plane Figure 1131 shows the castellated beam over a length of 12 m There are 19 hexagons through the length spaced at 203 mm The beam will be fixed at both ends and subject to uniformly distributed load of 178 kNm as shown in Figure 1132 2013 by Taylor Francis Group LLC 408 Introduction to Finite Element Analysis Using MATLAB and Abaqus 178 5706 FIGURE 1130 Castellated beam profile 609 316 316 203 203 12000 406 406 FIGURE 1131 Geometrical details of the castellated beam 178 kNm 12 m FIGURE 1132 Loading and boundary conditions 2013 by Taylor Francis Group LLC Thin and Thick Plates 409 Start Abaqus CAE Click on Create Model Database On the main menu click on File and set Set Work Directory to choose your working directory Click on Save As and name the file Castellatedbeamcae On the lefthandside menu click on Part to begin creating the model Name the part Castel latedbeam check 3D check Deformable in the type Choose Shell as the shape and Extrusion for type Enter an approximate size of 1000 mm and click on Continue In the sketcher menu choose the Create Lines con nected icon to begin drawing the profile of the beam Draw an I profile as shown in Figure 1133 without paying too much atten tion to the dimensions FIGURE 1133 Sketching the I profile Click on the Add Dimension icon With the mouse click on the first vertice of the flange and on the second vertice rep resenting the middle of the flange as shown In the com mand line of the viewport enter 89 mm as shown Click on Return Figure 1134 FIGURE 1134 Adding dimensions 2013 by Taylor Francis Group LLC 410 Introduction to Finite Element Analysis Using MATLAB and Abaqus Repeat the operation for the parts of the flanges and enter 5962 mm for the web Click on Return The result should look like the one shown in Figure 1135 FIGURE 1135 Finishing dimensioning the profile When finished the Add dimension too and click on Done in the prompt area to sketch the section for the shell extrusion In the Edit base extrusion enter 12000 mm as shown in Figure 1136 and click OK FIGURE 1136 Editing shell extrusion Under the main menu click on Shape Cut and Extrude Select the web as the plane for the extruded cut Next select the right hand end of the beam as the edge or the axis that will appear vertical on the right Figure 1137 FIGURE 1137 Selecting a plane for an extruded cut 2013 by Taylor Francis Group LLC Thin and Thick Plates 411 The sketcher is loaded again This time we will use it to sketch the hexagon Use the Magnify View tool to increase the size of the sketch Figure 1138 FIGURE 1138 Magnify view tool Draw a circle and using the Add dimension tool enter its radius as 203 mm Then draw two other circles as shown in Figure 1139 each having a radius of 203 mm Then using the Create Lines con nected tool join the intersect ing points as shown to create a perfect hexagon FIGURE 1139 Sketching a hexagon 2013 by Taylor Francis Group LLC 412 Introduction to Finite Element Analysis Using MATLAB and Abaqus Select the Delete tool By keeping the Shift key down select all the cir cles When finished click in the prompt area on Done All that is left is a hexagon Figure 1140 FIGURE 1140 Delete tool Next we need to position the hexagon at exactly 316 mm from the edge Using the Add dimen sion tool enter 316 as the distance from the left vertex to the edge Figure 1141 FIGURE 1141 Dimension tool 2013 by Taylor Francis Group LLC Thin and Thick Plates 413 Next we need to copy the hexagon along the length of the beam Click on the Lin ear Pattern tool and select the hexagon Click on Done Figure 1142 FIGURE 1142 Linear pattern tool In the Edit Linear Pattern dialog box enter 19 for direc tion 1 and 1 for direction 2 Enter the distance from ver tice to vertice as 609 mm Click on OK Figure 1143 FIGURE 1143 Editing a linear pattern 2013 by Taylor Francis Group LLC 414 Introduction to Finite Element Analysis Using MATLAB and Abaqus Then in the prompt area of the viewport click on Sketch the section for the extruded cut In the Edit Cut Extrusion dialog box select Through all for the type and click OK The result should be an image of a castellated beam Figure 1144 FIGURE 1144 Edit cut extrusion Define a material named Steel with an elastic modulus of 200000 MPa and a Pois sons ratio of 027 Next click on Sections to create two sections one for the web and the other for the flanges Name the first one Websection In the Cate gory check Shell and in the Type check Homogeneous Click on Continue In the Edit Section dialog box enter the web thickness as 78 mm and the material as steel Click on OK Create another section named Flangesection Enter the shell thickness as 128 mm as shown in Figure 1145 Click on OK FIGURE 1145 Creating a shell section 2013 by Taylor Francis Group LLC Thin and Thick Plates 415 Expand the menu under Parts and Castellatedbeam and double click on Section Assignments With the mouse select the web In the Edit Section Assign ments dialog box select Websection and click on OK Double click on Section Assignments again select the flanges In the Edit Section Assignments dialog box select Flangesection and click on OK In the prompt area click on Done Figure 1146 FIGURE 1146 Editing section assignments In the model tree double click on Mesh under the Castel latedbeam In the main menu under Mesh click on Mesh Controls select all the regions and click on Done In the dialog box check Quad for Element shape and Struc tured for Technique A pop up will appear stating that the web is too complex to be meshed with a structured technique As a result select the flanges only for a struc tured mesh and the web on its own for a free mesh Figure 1147 FIGURE 1147 Mesh controls and element type 2013 by Taylor Francis Group LLC 416 Introduction to Finite Element Analysis Using MATLAB and Abaqus Under Mesh click on Ele ment Type In the dialog box select Standard for element library Linear for geomet ric order In Quad check Reduced integration The description of the element S4R can be seen Click on OK Figure 1148 FIGURE 1148 Element type In the main menu under Seed click on Part In the dialog box enter 40 for Approxi mate global size Click on OK and on Done In the main menu under Mesh click on Part In the prompt area click on Yes Figure 1149 FIGURE 1149 Mesh 2013 by Taylor Francis Group LLC Thin and Thick Plates 417 In the model tree expand the Assembly and double click on Instances Select Castellatedbeam for Parts and click OK In the model tree expand Steps and Initial and double click on BC Name the boundary condition FIXED select Symme tryAntisymmetryEncastre for the type and click on Continue In the viewport select the two ends of the beam and click on Continue In the Edit Boundary Con dition check Encastre Click OK Figure 1150 FIGURE 1150 Imposing BC using geometry In the model tree double click on Steps Name the step Applyloads Set the proce dure to General and select Static General Click on Continue Click on OK In the model tree under steps and under Applyloads click on Loads Name the load Pressure and select Pres sure as the type Click on Continue In the viewport select the two top surfaces If any of the surface appears brown select it and flip the color to purple in the prompt area In the Edit Load dialog box enter 1 Nmm2 for magnitude Click OK Figure 1151 FIGURE 1151 Applying a pressure load on a shell surface 2013 by Taylor Francis Group LLC 418 Introduction to Finite Element Analysis Using MATLAB and Abaqus FIGURE 1152 Contour of the vertical displacement Under Analysis right click on Jobs and then click on Create In the Create Job dialog box name the job CastellatedBeam and click on Continue In the Edit Job dialog box enter a description for the job Check Full analysis select to run the job in Background and check to start it immediately Click OK Expand the tree under Jobs right click on CastellatedBeam Then click on Submit If you get the following message CastellatedBeam completed successfully in the bottom window then your job is free of errors and was executed properly Under the top menu in the Module scroll to Visualization and click to load Abaqus Viewer On the main menu under File click Open navigate to your working directory and open the file CastellatedBeam It should have the same name as the job you submitted Click on the icon Plot on Undeformed shape Under the main menu select U and U2 to plot the vertical displacement Figure 1152 Under the main menu select S and Mises to plot the von Mises stress Figure 1153 FIGURE 1153 Contour plot of the von Mises stress 2013 by Taylor Francis Group LLC Appendix A List of MATLAB Modules and Functions A1 AssemElemloadsm functionF AssemElemloadsF fg g This function assemble the global force vector global eldof This function assembles the global force vector for idof1eldof if gidof 0 Fgidof Fgidof fgidof end end end function AssemElemloads A2 AssemJointLoadsm functionF AssemJointLoadsF This function assembles the joints loads to the global force vector global nnd nodof global nf Jointloads for i1nnd for j1nodof if nfij 0 Fnfij Jointloadsij end end end end End function formbeamF A3 beamcolumnCm functionC beamcolumnCi This function forms the transformation between local and global coordinates global nnd nel nne nodof eldof global geom connec prop nf load retrieve the nodes of element i node1conneci1 node2conneci2 Retrieve the x and y coordinates of nodes 1 and 2 419 2013 by Taylor Francis Group LLC 420 Appendix A List of MATLAB Modules and Functions x1geomnode11 y1geomnode12 x2geomnode21 y2geomnode22 Evaluate the angle that the member makes with the global axis X ifx2x10 ify2y1 theta2atan1 else theta2atan1 end else thetaatany2y1x2x1 end Construct the transformation matrix C costheta sintheta 0 0 0 0 sintheta costheta 0 0 0 0 0 0 1 0 0 0 0 0 0 costheta sintheta 0 0 0 0 sintheta costheta 0 0 0 0 0 0 1 end function beamcolumnC A4 beamcolumngm functiong beamcolumngi This function forms the steering vector for element i global nnd nel nne nodof eldof global geom connec prop nf load retrieve the nodes of element i node1conneci1 node2conneci2 Retrieve the element degrees of freedom to be stored in the steering vector gnfnode11 nfnode12 nfnode13 nfnode21 nfnode22 nfnode23 end function beamcolumng A5 beamcolumnkm functionkl beamcolumnki This function forms the beamcolumn element stiffness in local coordinates global nnd nel nne nodof eldof global geom connec prop nf load Hinge retrieve the nodes of element i node1conneci1 node2conneci2 2013 by Taylor Francis Group LLC Appendix A List of MATLAB Modules and Functions 421 Retrieve the x and y coordinates of nodes 1 and 2 x1geomnode11 y1geomnode12 x2geomnode21 y2geomnode22 Evaluate length of element i L sqrtx2x12 y2y12 Retrieve section properties of element i E propi1 A propi2 I propi3 EAEA EIEI Calculate element stiffness matrix in its local coordinates if Hingei1 0 klEAL 0 0 EAL 0 0 0 3EIL3 0 0 3EIL3 3EIL2 0 0 0 0 0 0 EAL 0 0 EAL 0 0 0 3EIL3 0 0 3EIL3 3EIL2 0 3EIL2 0 0 3EIL2 3EIL elseif Hingei2 0 klEAL 0 0 EAL 0 0 0 3EIL3 3EIL2 0 3EIL3 0 0 3EIL2 3EIL 0 3EIL2 0 EAL 0 0 EAL 0 0 0 3EIL3 3EIL2 0 3EIL3 0 0 0 0 0 0 0 else klEAL 0 0 EAL 0 0 0 12EIL3 6EIL2 0 12EIL3 6EIL2 0 6EIL2 4EIL 0 6EIL2 2EIL EAL 0 0 EAL 0 0 0 12EIL3 6EIL2 0 12EIL3 6EIL2 0 6EIL2 2EIL 0 6EIL2 4EIL end End function beamcolumnk A6 beamgm functiong beamgi This function forms the steering vector for element i global connec nf retrieve the nodes of element i node1conneci1 node2conneci2 Form the steering vector from elements degrees of freedom gnfnode11 nfnode12 nfnode21nfnode22 end function beamg 2013 by Taylor Francis Group LLC 422 Appendix A List of MATLAB Modules and Functions A7 beamkm functionkl beamki This function forms the element stiffness in local coordinates global nnd nel nne nodof eldof global geom connec prop nf load Hinge retrieve the nodes of element i node1conneci1 node2conneci2 Retrieve the x and y coordinates of nodes 1 and 2 x1geomnode1 x2geomnode2 Evaluate length of element i L absx2x1 Retrieve section properties of element i EI propi1propi2 Calculate element stiffness matrix in its local coordinates if Hingei 1 0 kl 3EIL3 0 3EIL3 3EIL2 0 0 0 0 3EIL3 0 3EIL3 3EIL2 3EIL2 0 3EIL2 3EIL elseif Hingei 2 0 kl 3EIL3 3EIL2 3EIL3 0 3EIL2 3EIL 3EIL2 0 3EIL3 3EIL2 3EIL3 0 0 0 0 0 else kl 12EIL3 6EIL2 12EIL3 6EIL2 6EIL2 4EIL 6EIL2 2EIL 12EIL3 6EIL2 12EIL3 6EIL2 6EIL2 2EIL 6EIL2 4EIL end End function beamk A8 coordq8m functioncoord coordq8knne geom connec This function returns the coordinates of the nodes of element k coordzerosnne2 for i1 nne coordigeomconnecki end End function coordq8 2013 by Taylor Francis Group LLC Appendix A List of MATLAB Modules and Functions 423 A9 elemq4m functioncoordg elemq4i This function returns the coordinates of the nodes of element i and its steering vector g global nnd nel nne nodof eldof n ngp global geom connec dee nf load l0 coordzerosnnenodof for k1 nne for j1nodof coordkjgeomconnecikj ll1 glnfconnecikj end end End function elemq4 A10 Elemq8m functioncoordg elemq8i This function returns the coordinates of the nodes of element i and its steering vector global nnd nel nne nodof eldof n ngp global geom connec dee nf load l0 coordzerosnnenodof for k1 nne for j1nodof coordkjgeomconnecikj ll1 glnfconnecikj end end End function elemq8 A11 elemT3m functionbeegA elemT3i This function returns the coordinates of the nodes of element i and its steering vector global nnd nel nne nodof eldof n global geom connec dee nf load x1 geomconneci11 y1 geomconneci12 x2 geomconneci21 y2 geomconneci22 x3 geomconneci31 y3 geomconneci32 A 05det1 x1 y1 1 x2 y2 1 x3 y3 m11 x2y3 x3y22A m21 x3y1 x1y32A 2013 by Taylor Francis Group LLC 424 Appendix A List of MATLAB Modules and Functions m31 x1y2 y1x22A m12 y2 y32A m22 y3 y12A m32 y1 y22A m13 x3 x22A m23 x1 x32A m33 x2 x12A bee m12 0 m22 0 m32 0 0 m13 0 m23 0 m33 m13 m12 m23 m22 m33 m32 l0 for k1nne for j1nodof ll1 glnfconnecikj end end End function elemT3 A12 elemT6m functioncoordg elemT6i This function returns the coordinates of the nodes of element i and its steering vector global nnd nel nne nodof eldof n global geom connec dee nf load l0 coordzerosnnenodof for k1 nne for j1nodof coordkjgeomconnecikj ll1 glnfconnecikj end end End function elemT6 A13 fmlinm functionderfun fmlinsamp igjg This function returns the vector of the shape function and their derivatives with respect to xi and eta xisampig1 etasampjg1 fun 0251 xi eta xieta 1 xi eta xieta 1 xi eta xieta 1 xi eta xieta der 0251eta 1eta 1eta 1eta 1xi 1xi 1xi 1xi end function fmlin 2013 by Taylor Francis Group LLC Appendix A List of MATLAB Modules and Functions 425 A14 fmquadm functionderfun fmquadsamp igjg This function returns the vector of the shape function and their derivatives with respect to xi and eta at the gauss points for an 8nodded quadrilateral xisampig1 etasampjg1 etam1eta etap1eta xim1xi xip1xi fun1 025ximetam1 xi eta fun2 051 xi2etam fun3 025xipetam1 xi eta fun4 05xip1 eta2 fun5 025xipetap1 xi eta fun6 051 xi2etap fun7 025ximetap1 xi eta fun8 05xim1 eta2 der11025etam2xi eta der121etamxi der13025etam2xieta der14051eta2 der15025etap2xieta der161etapxi der17025etap2xieta der18051eta2 der21025xim2etaxi der22051 xi2 der23025xipxi2eta der241xipeta der25025xipxi2eta der26051xi2 der27025ximxi2eta der281ximeta end function fmquad A15 fmT6quadm functionderfun fmT6quadsamp ig This function returns the vector of the shape function and their derivatives with respect to xi and eta at the gauss points for an 8nodded quadrilateral xisampig1 etasampig2 lambda 1 xi eta fun1 lambda12lambda fun2 4xilambda fun3 xi12xi fun4 4xieta fun5 eta12eta fun6 4etalambda der1114lambda der124lambdaxi der1314xi der144eta der150 der164eta der2114lambda der224xi der230 der244xi der2514eta der264lambdaeta end function fmT6quad 2013 by Taylor Francis Group LLC 426 Appendix A List of MATLAB Modules and Functions A16 Forcesatnodesplatem functionMX MY MXY QX QYForcesatnodesplateElementForces This function averages the stresses at the nodes global nnd nel nne connec for k 1nnd mx 0 my 0 mxy 0 qx 0 qy 0 ne 0 for iel 1nel for jel1nne if connecieljel k nene1 mx mx ElementForcesiel1 my my ElementForcesiel2 mxy mxy ElementForcesiel3 qx qx ElementForcesiel4 qy qy ElementForcesiel5 end end end MXk1 mxne MYk1 myne MXYk1 mxyne QXk1 qxne QYk1 qyne end A17 FileformbeamFm functionF formbeamFF This function forms the global force vector global nnd nodof nel eldof global nf Elementloads Jointloads for i1nnd for j1nodof if nfij 0 Fnfij Jointloadsij end end end for i1nel gbeamgi Retrieve the element steering vector for j1eldof if gj 0 Fgj Fgj Elementloadsij end end end End function formbeamF A18 Fileformffm functionffformfffffg g This function assemble the global force vector global nodof nne eldof 2013 by Taylor Francis Group LLC Appendix A List of MATLAB Modules and Functions 427 This function assembles the global force vector for idof1eldof if gidof 0 ffgidof ffgidof fgidof end end end function formff A19 FileformKKm functionKKformKKKK kg g This function assembles the global stiffness matrix global eldof This function assembles the global stiffness matrix for i1eldof if gi 0 for j1 eldof if gj 0 KKgigj KKgigj kgij end end end end end function formKK A20 formtrussFm functionF formtrussFF This function forms the global force vector global nnd nodof global nf load for i1nnd for j1nodof if nfij 0 Fnfij loadij end end end End function formtrussF A21 formbeem functionbee formbeederivnneeldof This function assembles the matrix bee from the derivatives of the shape functions in global coordinates beezeros3eldof for m1nne k2m lk1 xderiv1m bee1lx bee3kx yderiv2m 2013 by Taylor Francis Group LLC 428 Appendix A List of MATLAB Modules and Functions bee2ky bee3ly end End function formbee A22 formbeeaxim functionbee radius formbeeaxiderivnnefun coordeldof This function assembles the matrix bee for an axisymmetric problem from the derivatives of the shape functions in global coordinates beezeros4eldof radius dotfuncoord1 for m1nne k2m lk1 xderiv1m bee1lx bee4kx yderiv2m bee2ky bee4ly bee3l funmradius end End function formbeeaxi A23 formbeebm functionbeeb formbeebderivnneeldof This function assembles the matrix beeb from the derivatives of the shape functions in global coordinates for a thick plate element bending action beebzeros3eldof for m1nne k3m jk1 xderiv1m beeb1jx beeb3kx yderiv2m beeb2ky beeb3jy end End function formbeeb A24 formbeesm functionbees formbeesderivfun nneeldof This function assembles the matrix bees from the derivatives of the shape functions in global coordinates for the shear action in a plate element beeszeros2eldof for m1nne k3m 2013 by Taylor Francis Group LLC Appendix A List of MATLAB Modules and Functions 429 jk1 ik2 xderiv1m yderiv2m bees2ix bees1iy bees1k funm bees2j funm end End function formbees A25 formdaxm functiondee formdaxEvu This function forms the elasticity matrix for a plane stress problem v1 1 vu c E1 vu1 2vu dee cv1 vu vu 0 vu v1 vu 0 vu vu v1 0 0 0 0 51vu end function fromdeps A26 formdeebm functiondeeb formdeebEvuthick This function forms the elasticity matrix for a bending action in a plate element DR Ethick3121vuvu deebDR1 vu 0 vu 1 0 0 0 1vu2 end function fromdeeb A27 formdeesm functiondees formdeesEvuthick This function forms the elasticity matrix for the shear action in a thick plate element G E21vu deesGthick 0 0 thick end function fromdees A28 formdepsm functiondee formdepsEvu This function forms the elasticity matrix for a plane strain problem v11vu cE1vu12vu 2013 by Taylor Francis Group LLC 430 Appendix A List of MATLAB Modules and Functions deev1c vuc 0 vuc v1c 0 0 0 5c12vu end function fromdeps A29 formdsigm functiondee formdsigEvu This function forms the elasticity matrix for a plane stress problem cE1vuvu deec1 vu 0 vu 1 0 0 0 51vu end function formdsig A30 gaussm functionsampgaussngp This function returns the abscissas and weights of the Gauss points for ngp equal up to 4 sampzerosngp2 if ngp1 samp0 2 elseif ngp2 samp1sqrt3 1 1sqrt3 1 elseif ngp3 samp 2sqrt15 59 0 89 2sqrt15 59 elseif ngp4 samp 0861136311594053 0347854845137454 0339981043584856 0652145154862546 0339981043584856 0652145154862546 0861136311594053 0347854845137454 end End function Gauss A31 hammerm functionsamphammernpt This function returns the abscissae and weights of the integration points for npt equal up to 7 sampzerosnpt3 if npt1 samp13 13 12 elseif npt2 npt3 npt3 samp16 16 16 23 16 16 2013 by Taylor Francis Group LLC Appendix A List of MATLAB Modules and Functions 431 16 23 16 elseif npt4 npt5 npt4 samp 13 13 2796 15 15 2596 35 15 2596 15 35 2596 elseif npt6 a 0445948490915965 b 0091576213509771 samp a a 0111690794839005 12a a 0111690794839005 a 12a 0111690794839005 b b 0054975871827661 12b b 0054975871827661 b 12b 0054975871827661 elseif npt7 a 6sqrt1521 b 47 a A 155sqrt152400 B 31240 A samp 13 13 980 a a A 12a a A a 12a A b b B 12b b B b 12b B end End function hammer A32 platelemq8m functioncoordg platelemq8i This function returns the coordinates of the nodes of element i and its steering vector global nne nodof geom connec nf dim coordzerosnnedim for k1 nne for j1dim coordkjgeomconnecikj end end l0 for k1 nne for j1nodof ll1 glnfconnecikj end end End function platelemq8 A33 preparecontourdatam functionZX ZY ZT Z1 Z2preparecontourdataSIGMA This function averages the stresses at the nodes and rearrange the values in the matrices Z for contouring global nnd nel nne geom connec XIG YIG NXE NYE for k 1nnd sigx 0 sigy 0 tau 0 2013 by Taylor Francis Group LLC 432 Appendix A List of MATLAB Modules and Functions ne 0 for iel 1nel for jel1nne if connecieljel k nene1 sigx sigxSIGMAiel1 sigy sigy SIGMAiel2 tau tau SIGMAiel3 end end end xc geomk1 yc geomk2 for i 12NXE1 for j12NYE 1 if xc XIGi yc YIGj ZXji sigxne ZYji sigyne ZTjitaune Z1ji sigxsigy2 sqrtsigxsigy22 tau2ne Z2ji sigxsigy2 sqrtsigxsigy22 tau2ne end end end end A34 printbeammodelm fprintffid PRINTING MODEL DATA Print Nodal coordinates fprintffid fprintffid Number of nodes g nnd fprintffid Number of elements g nel fprintffid Number of nodes per element g nne fprintffid Number of degrees of freedom per node g nodof fprintffid Number of degrees of freedom per element g eldof fprintffid fprintffid Node X for i1nnd fprintffid g 072f i geomi end fprintffid Print element connectivity fprintffid fprintffid Element Node1 Node2 for i1nel fprintffid g g g i conneci1 conneci2 end fprintffid Print element property fprintffid fprintffid Element E I for i1nel fprintffid g g g i propi1 propi2 end fprintffid Print Nodal freedom 2013 by Taylor Francis Group LLC Appendix A List of MATLAB Modules and Functions 433 fprintffid fprintffid Nodal freedom fprintffid Node dispw Rotation for i1nnd fprintffid g g g i nfi1 nfi2 end fprintffid Print Nodal loads fprintffid fprintffid Applied Nodal Loads fprintffid Node loadY Moment for i1nnd for j1nodof nodeforceij 0 if nfij 0 nodeforceij Fnfij end end fprintffid g 072f 072f i nodeforcei1 nodeforcei2 end fprintffid fprintffid fprintffidTotal number of active degrees of freedom n g n fprintffid A35 printbeamresultsm fprintffid fprintffid PRINTING ANALYSIS RESULTS Print global force vector fprintffid fprintffidGlobal force vector F fprintffid g F fprintffid Print Displacement solution vector fprintffid fprintffidDisplacement solution vector delta fprintffid 85f delta fprintffid Print nodal displacements fprintffid fprintffid Nodal displacements fprintffid Node dispy rotation for i1nnd fprintffid g 85f 85f i nodedispi1 nodedispi2 end fprintffid Print Members actions fprintffid fprintffid Members actions fprintffid element fy1 M1 Fy2 M2 for i1nel 2013 by Taylor Francis Group LLC 434 Appendix A List of MATLAB Modules and Functions fprintffid g 92f 92f 92f 92f i forcei1forcei2forcei3forcei4 end A36 printCSTresultsm fprintffid fprintffid PRINTING ANALYSIS RESULTS Print nodal displacements fprintffid fprintffid Nodal displacements fprintffid Node dispx dispy for i1nnd fprintffid g 85e 85e i nodedispi1 nodedispi2 end fprintffid Print element stresses fprintffid fprintffid Element stresses fprintffid element sigmaxx sigmayy tauxy for i1nel fprintffid g 74e 74e 74e i SIGMAi1SIGMAi2SIGMAi3 end Print element strains fprintffid fprintffid Element strains fprintffid element epsilonxx epsilonyy gammaxy for i1nel fprintffid g 74e 74e 74e i EPSi1EPSi2EPSi3 end A37 printframemodelm fprintffid PRINTING MODEL DATA Print Nodal coordinates fprintffid fprintffid Number of nodes g nnd fprintffid Number of elements g nel fprintffid Number of nodes per element g nne fprintffid Number of degrees of freedom per node g nodof fprintffid Number of degrees of freedom per element g eldof fprintffid fprintffid Node X Y for i1nnd fprintffid g 072f 072f i geomi1 geomi2 end 2013 by Taylor Francis Group LLC Appendix A List of MATLAB Modules and Functions 435 fprintffid Print element connectivity fprintffid fprintffid Element Node1 Node2 for i1nel fprintffid g g g i conneci1 conneci2 end fprintffid Print element property fprintffid fprintffid Element E A I for i1nel fprintffid g g g g i propi1 propi2 propi3 end fprintffid Print Nodal freedom fprintffid fprintffid Nodal freedom fprintffid Node dispu dispu Rotation for i1nnd fprintffid g g g g i nfi1 nfi2nfi3 end fprintffid Print joint loads fprintffid fprintffid Applied joint Loads fprintffid Node loadX loadY Moment for i1nnd for j1nodof nodeforceij 0 if nfij 0 nodeforceij Fnfij end end fprintffid g 072f 072f 072f i nodeforcei1 nodeforcei2 nodeforcei3 end fprintffid fprintffid fprintffidTotal number of active degrees of freedom n g n fprintffid A38 printframeresultsm fprintffid fprintffid PRINTING ANALYSIS RESULTS Print global force vector fprintffid fprintffidGlobal force vector F fprintffid g F fprintffid 2013 by Taylor Francis Group LLC 436 Appendix A List of MATLAB Modules and Functions Print Displacement solution vector fprintffid fprintffidDisplacement solution vector delta fprintffid 85f delta fprintffid Print nodal displacements fprintffid fprintffid Nodal displacements fprintffid Node dispx dispy rotation for i1nnd fprintffid g 85e 85e 85e i nodedispi1 nodedispi2nodedispi3 end fprintffid Print Members actions fprintffid fprintffid Members actions in local coordinates fprintffid element fx1 fy1 M1 fx2 Fy2 M2 for i1nel fprintffid g 74f 74f 74f 74f 74f 94f i forceli1forceli2forceli3forceli4forceli5forceli6 end fprintffid fprintffid Members actions in global coordinates fprintffid element fx1 fy1 M1 fx2 Fy2 M2 for i1nel fprintffid g 74f 74f 74f 74f 74f 94f i forcegi1forcegi2forcegi3forcegi4forcegi5forcegi6 end A39 printtrussmodelm fprintffid PRINTING MODEL DATA Print Nodal coordinates fprintffid fprintffid Number of nodes g nnd fprintffid Number of elements g nel fprintffid Number of nodes per element g nne fprintffid Number of degrees of freedom per node g nodof fprintffid Number of degrees of freedom per element g eldof fprintffid fprintffid Node X Y for i1nnd fprintffid g 072f 072f i geomi1 geomi2 end fprintffid Print element connectivity fprintffid fprintffid Element Node1 Node2 for i1nel fprintffid g g g i conneci1 conneci2 end 2013 by Taylor Francis Group LLC Appendix A List of MATLAB Modules and Functions 437 fprintffid Print element property fprintffid fprintffid Element E A for i1nel fprintffid g g g i propi1 propi2 end fprintffid Print Nodal freedom fprintffid fprintffid Node dispU dispV for i1nnd fprintffid g g g i nfi1 nfi2 end fprintffid Print Nodal loads fprintffid fprintffid Node loadX loadY for i1nnd fprintffid g 072f 072f i loadi1 loadi2 end fprintffid fprintffid fprintffidTotal number of active degrees of freedom n g n fprintffid A40 printtrussresultsm fprintffid fprintffid PRINTING ANALYSIS RESULTS Print global force vector fprintffid fprintffidGlobal force vector F fprintffid g F fprintffid Print Displacement solution vector fprintffid fprintffidDisplacement solution vector delta fprintffid 85f delta fprintffid Print nodal displacements fprintffid fprintffid Nodal displacements fprintffid Node dispX dispY for i1nnd fprintffid g 85f 85f i nodedispi1 nodedispi2 end fprintffid 2013 by Taylor Francis Group LLC 438 Appendix A List of MATLAB Modules and Functions Print Members actions fprintffid fprintffid Members actions fprintffid element force action for i1nel if forcei 0 fprintffid g 92f s i forcei Tension else fprintffid g 92f s i forcei Compression end end A41 Q4meshm This module generates a mesh of linear quadrilateral elements global nnd nel nne nodof eldof n global geom connec dee nf Nodalloads global Length Width NXE NYE Xorigin Yorigin dhx dhy nnd 0 k 0 for i 1NXE for j1NYE k k 1 n1 j i1NYE 1 geomn1 i1dhx Xorigin j1dhy Yorigin n2 j iNYE1 geomn2 idhx Xorigin j1dhy Yorigin n3 n1 1 geomn3 i1dhx Xorigin jdhy Yorigin n4 n2 1 geomn4 idhx Xorigin jdhy Yorigin nel k connecnel n1 n2 n4 n3 nnd n4 end end A42 Q8meshm This function module a mesh of 8nodded quadrilateral elements global nnd nel nne nodof eldof n global geom connec dee nf Nodalloads global Length Width NXE NYE Xorigin Yorigin dhx dhy nnd 0 k 0 for i 1NXE for j1NYE k k 1 n1 i13NYE22j 1 n2 i3NYE2j NYE 1 n3 i3NYE22j1 n4 n3 1 n5 n3 2 n6 n2 1 n7 n1 2 n8 n1 1 2013 by Taylor Francis Group LLC Appendix A List of MATLAB Modules and Functions 439 geomn1 i1dhx Xorigin j1dhy Yorigin geomn3 idhx Xorigin j1dhy Yorigin geomn2 geomn11geomn312 geomn12geomn322 geomn5 idhx Xorigin jdhy Yorigin geomn4 geomn31 geomn512 geomn32 geomn522 geomn7 i1dhx Xorigin jdhy Yorigin geomn6 geomn51 geomn712 geomn52 geomn722 geomn8 geomn11 geomn712 geomn12 geomn722 nel k nnd n5 conneck n1 n2 n3 n4 n5 n6 n7 n8 end A43 Stressesatnodesaxim functionZX ZY ZTHETA ZTStressesatnodesaxiSIGMA This function averages the stresses at the nodes global nnd nel nne geom connec for k 1nnd sigx 0 sigy 0 sigtheta 0 tau 0 ne 0 for iel 1nel for jel1nne if connecieljel k nene1 sigx sigxSIGMAiel1 sigy sigy SIGMAiel2 sigtheta sigtheta SIGMAiel3 tau tau SIGMAiel4 end end end ZXk1 sigxne ZYk1 sigyne ZTk1taune ZTHETAk1 sigthetane end A44 StressesatnodesQ4m functionZX ZY ZT Z1 Z2StressesatnodesQ4SIGMA This function averages the stresses at the nodes global nnd nel nne geom connec XIG YIG NXE NYE for k 1nnd sigx 0 sigy 0 tau 0 ne 0 for iel 1nel for jel1nne if connecieljel k nene1 sigx sigxSIGMAiel1 sigy sigy SIGMAiel2 tau tau SIGMAiel3 end end end ZXk1 sigxne ZYk1 sigyne 2013 by Taylor Francis Group LLC 440 Appendix A List of MATLAB Modules and Functions ZTk1taune Z1k1 sigxsigy2 sqrtsigxsigy22 tau2ne Z2k1 sigxsigy2 sqrtsigxsigy22 tau2ne end A45 StressesatnodesQ8m functionZX ZY ZT Z1 Z2StressesatnodesQ8SIGMA This function averages the stresses at the nodes global nnd nel nne geom connec for k 1nnd sigx 0 sigy 0 tau 0 ne 0 for iel 1nel for jel1nne if connecieljel k nene1 sigx sigxSIGMAiel1 sigy sigy SIGMAiel2 tau tau SIGMAiel3 end end end ZXk1 sigxne ZYk1 sigyne ZTk1taune Z1k1 sigxsigy2 sqrtsigxsigy22 tau2ne Z2k1 sigxsigy2 sqrtsigxsigy22 tau2ne end A46 T3meshm This module generates a mesh of triangular elements global nnd nel nne nodof eldof n global geom connec dee nf Nodalloads global Length Width NXE NYE Xorigin Yorigin dhx dhy nnd 0 k 0 for i 1NXE for j1NYE k k 1 n1 j i1NYE 1 geomn1 i1dhx Xorigin j1dhy Yorigin n2 j iNYE1 geomn2 idhx Xorigin j1dhy Yorigin n3 n1 1 geomn3 i1dhx Xorigin jdhy Yorigin n4 n2 1 geomn4 idhx Xorigin jdhy Yorigin nel 2k m nel 1 connecm n1 n2 n3 connecnel n2 n4 n3 nnd n4 end end for i1nel x geomconneci11 geomconneci21 geomconneci31 y geomconneci12 geomconneci22 geomconneci32 end 2013 by Taylor Francis Group LLC Appendix A List of MATLAB Modules and Functions 441 A47 T6meshm This module generates a mesh of the linear strain triangular element global nnd nel geom connec XIG YIG global Length Width NXE NYE Xorigin Yorigin dhx dhy nnd 0 k 0 for i 1NXE for j1NYE k k 1 n1 2j1 2i22NYE1 n2 2j1 2i12NYE1 n3 2j1 2i2NYE1 n4 n1 1 n5 n2 1 n6 n3 1 n7 n1 2 n8 n2 2 n9 n3 2 geomn1 i1dhx Xorigin j1dhy Yorigin geomn2 2i12dhx Xorigin j1dhy Yorigin geomn3 idhx Xorigin j1dhy Yorigin geomn4 i1dhx Xorigin 2j12dhy Yorigin geomn5 2i12dhx Xorigin 2j12dhy Yorigin geomn6 idhx Xorigin 2j12dhy Yorigin geomn7 i1dhx Xorigin jdhy Yorigin geomn8 2i12dhx Xorigin jdhy Yorigin geomn9 idhx Xorigin jdhy Yorigin nel 2k m nel 1 connecm n1 n2 n3 n5 n7 n4 connecnel n3 n6 n9 n8 n7 n5 maxn maxn1 n2 n3 n4 n5 n6 n7 n8 n9 ifnnd maxn nnd maxn end XIN and YIN are two vectors that holds the coordinates X and Y of the grid necessary for the function contourf XINYIN stress XIG2i1 geomn11 XIG2i geomn21 XIG2i1 geomn31 YIG2j1 geomn12 YIG2j geomn42 YIG2j1 geomn72 end end A48 trussCm functionC trussCi This function forms the transformation between local and global coordinates global geom connec retrieve the nodes of element i node1conneci1 node2conneci2 Retrieve the x and y coordinates of nodes 1 and 2 x1geomnode11 y1geomnode12 2013 by Taylor Francis Group LLC 442 Appendix A List of MATLAB Modules and Functions x2geomnode21 y2geomnode22 Evaluate the angle that the member makes with the global axis X ifx2x10 ify2y1 theta2atan1 else theta2atan1 end else thetaatany2y1x2x1 end Construct the transformation matrix C costheta sintheta 0 0 sintheta costheta 0 0 0 0 costheta sintheta 0 0 sintheta costheta end function trussC A49 trussgm functiong trussgi This function forms the steering vector for element i global connec nf retrieve the nodes of element i node1conneci1 node2conneci2 Form the steering vector from elements degrees of freedom gnfnode11 nfnode12 nfnode21nfnode22 end function trussg A50 trussklm functionkl trusskli This function forms the element stiffness matrix in local coordinates global geom connec prop retrieve the nodes of element i node1conneci1 node2conneci2 Retrieve the x and y coordinates of nodes 1 and 2 x1geomnode11 y1geomnode12 x2geomnode21 y2geomnode22 Evaluate length of element i 2013 by Taylor Francis Group LLC Appendix A List of MATLAB Modules and Functions 443 L sqrtx2x12 y2y12 Retrieve section properties of element i E propi1 Apropi2 Calculate element stiffness matrix in its local coordinates klEAL 0 EAL 0 0 0 0 0 EAL 0 EAL 0 0 0 0 0 End function trusskl 2013 by Taylor Francis Group LLC Appendix B Statically Equivalent Nodal Forces Actual load Statically equivalent nodal loads P P P PL PL qL2 qL2 qL2 3qL 5qL 3qL qL2 7qL qL2 qL qL 2 2 12 30 8 8 20 20 20 12 2 8 8 2 L q q q 8 FIGURE B1 Common beam loadings 445 2013 by Taylor Francis Group LLC Appendix C Index Notation and Transformation Laws for Tensors C1 INDEX NOTATION FOR VECTORS AND TENSORS C11 VECTOR AND TENSOR COMPONENTS Operations on Cartesian components of vectors and tensors can be expressed very efficiently and clearly using index notation The index notation refers to vectors or tensors by their general term with the indices ranging over the dimensions of the vector or the tensor Let u be a vector and a a secondorder tensor defined in a Cartesian basis Using matrix notation they can be represented by their Cartesian components as u u1 u2 u3 a a11 a12 a13 a21 a22 a23 a31 a32 a33 C1 Using index notation the vector u and the tensor a can be expressed in a compact manner as u ui a aij C2 C12 EINSTEIN SUMMATION CONVENTIONS Under the rules of index notation if an index is repeated in a product of vectors or tensors summation is implied over the range of the repeated index For example for a range from 1 to 3 the following expressions can be developed as aibi a1b1 a2b2 a3b3 C3 ci aikxk a11x1 a12x2 a13x3 a21x1 a22x2 a23x3 a31x1 a32x2 a33x3 C4 λ aijbij a1b1 a1b2 a1b3 a2b1 a2b2 a2b3 a3b1 a3b2 a3b3 C5 cij aikbkj ai1b1j ai2b2j ai3b3j C A B C6 aij bji A BT C7 Expression C6 is equivalent to the product of two matrices 447 2013 by Taylor Francis Group LLC Appendix C Index Notation and Transformation Laws for Tensors 451 The matrix Q is an orthonormal matrix and has the following properties QT Q1 C27 In index notation the relationship C27 is given as likljk δij C28 Note that in index notation and by analogy to matrix notation you cannot write C28 as ljilij δij This is completely erroneous in index notation since the repeated indices in the first term imply summation therefore the first term is a scalar and the second a tensor Given an arbitrary vector v represented in the base e1 e2 e3 as v v1 e1 v2 e2 v3 e3 vj ej C29 The same vector can also be represented in the base e 1 e 2 e 3 as v v 1 e 1 v 2 e 2 v 3 e 3 v i e i C30 Using Equation C19 Equation C30 is rewritten as v v ilij ej C31 Comparing Equations C29 and C31 reveal that the vector components in the primed and unprimed basis are related by vj v ilij lijv i C32 in matrix notation v QTv C33 The inverse transformation is defined as v i vilij C34 or in matrix notation as v Qv C35 C22 TRANSFORMATION LAWS FOR TENSORS Given two arbitrary vectors u and v represented in the base e1 e2 e3 respectively as u u1 e1 u2 e2 u3 e3 C36 v v1 e1 v2 e2 v3 e3 C37 Now suppose the existence of a linear application between the two vectors defined by u fu and expressed in index notation as ui aijvj C38 2013 by Taylor Francis Group LLC 452 Appendix C Index Notation and Transformation Laws for Tensors or in matrix notation as u av C39 In another base say e 1 e 2 e 3 the vectors u and v are expressed as u u 1 e 1 u 2 e 2 u 3 e 3 C40 v v 1 e 1 v 2 e 2 v3 e 3 C41 and the relationship u fu is expressed in index notation as u i a ijv j C42 and in matrix notation as u av C43 The problem is to find a relationship between the tensors a and a Using C32 Equation C38 is rewritten as ui aijlmjv m lmjaijv m C44 Substituting in C44 for ui using C32 leads to lkiu k aijlmjv m lmjaijv m C45 Multiplying both sides of the equations by lni and noting that lnilki δnk Equation C46 becomes δnku k lkilmjaijv m C46 That is u k lkilmjaijv m C47 Comparing C42 and C47 it follows a km lkilmjaij C48 Using matrix notation and after substituting Equation C33 Equation C39 becomes u aQTv C49 Replacing the vector u by QTu Equation C49 becomes QTu aQTv C50 and premultiplying both sides of the equation by the matrix Q and noting QQT I yields the result u QaQTv C51 Comparing Equations C43 and C51 yields the result a QaQT C52 The inverse relation is expressed as a QTaQ C53 2013 by Taylor Francis Group LLC References and Bibliography In the course of writing this present work many books on matrix structural analysis and the theories of elasticity and finite element methods have been consulted Some of these books have been explicitly cited while others not An exhaustive list of all the books consulted is given below REFERENCES 1 Dhatt G and Touzot G Une Présentation de la Méthode des Éléments Finis Deuxième édition Maloine SA Editeurs Paris France 1984 2 Hammer PC Marlowe OJ and Stroud AH Mathematical tables and other aids to computation American Mathematical Society 1055 130136 1956 3 Smith IM and Griffiths DV Programming the Finite Element Method 2nd edn Wiley Chichester UK 1988 4 Timoshenko S and Goodier J Theory of Elasticity 3rd edn McGrawHill New York 1970 5 Timoshenko S and WoinowskyKrieger S Theory of Plates and Shells McGrawHill New York 1959 6 Zienkiewicz OC The Finite Element Method 3rd edn McGrawHill York London 1977 BIBLIOGRAPHY Chandrupatla TR and Belegundu AD Introduction to Finite Elements in Engineering 3rd edn PrenticeHall Upper Saddle River NJ 2002 Cook RD Finite Element Modeling for Stress Analysis Wiley New York 1995 Kwon YW and Bang H The Finite Element Method Using Matlab 2nd edn CRC Press London UK 2000 Logan DL A First Course in the Finite Element Method Using Algor 2nd edn BrooksCole Thompson Learning Pacific Groove CA 2001 Mase GE Schaums Outline Series Theory and Problems of Continuum Mechanics McGrawHill New York 1970 McGuire M Gallagher GH and Ziemian RD Matrix Structural Analysis 2nd edn Wiley New York 2000 Meek JL Computer Methods in Structural Analysis E FN SPON London UK 1991 Reddy JN An Introduction to the Finite Element Method 3rd edn McGrawHill New York 2006 Saada AS Elasticity Theory and Applications 2nd edn Krieger Publishing Melbourne FL 1993 453 2013 by Taylor Francis Group LLC w w w c r c p r e s s c o m K16894 Amar Khennane Introduction to Finite Element Analysis Using MATLAB and Abaqus Khennane Introduction to Finite Element Analysis Using MATLAB and Abaqus Introduction to Finite Element Analysis Using MATLAB and Abaqus A very good introduction to the finite element method with a balanced treatment of theory and implementation F Albermani Reader in Structural Engineering The University of Queensland Australia There are some books that target the theory of the finite element while others focus on the programming side of things Introduction to Finite Element Analysis Using MATLAB and Abaqus accomplishes both This book teaches the first principles of the finite element method It presents the theory of the finite element method while maintaining a balance between its mathematical formulation programming implemen tation and application using commercial software The computer implementation is carried out using MATLAB while the practical applications are carried out in both MATLAB and Abaqus MATLAB is a highlevel language specially designed for dealing with matrices making it particularly suited for programming the finite element meth od while Abaqus is a suite of commercial finite element software Introduction to Finite Element Analysis Using MATLAB and Abaqus introduces and explains theory in each chapter and provides corresponding examples It offers introductory notes and provides matrix structural analysis for trusses beams and frames The book examines the theories of stress and strain and the relationships be tween them The author then covers weighted residual methods and finite element ap proximation and numerical integration He presents the finite element formulation for plane stressstrain problems introduces axisymmetric problems and highlights the theory of plates The text supplies stepbystep procedures for solving problems with Abaqus interactive and keyword editions The described procedures are implemented as MATLAB codes and Abaqus files can be found on the CRC Press website Mathematics 6000 Broken Sound Parkway NW Suite 300 Boca Raton FL 33487 711 Third Avenue New York NY 10017 2 Park Square Milton Park Abingdon Oxon OX14 4RN UK an informa 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