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This 13 page Class Notes was uploaded by Jackeline Wuckert on Friday October 23, 2015. The Class Notes belongs to ME512 at University of Louisville taught by RogerBradshaw in Fall. Since its upload, it has received 46 views. For similar materials see /class/228364/me512-university-of-louisville in Mechanical Engineering at University of Louisville.
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Date Created: 10/23/15
ME 512 Finite Elements in ME Design IVSTR L CTOR Prof Roger Bradshaw O ire H can 393 203 Sackett Hall 31 W F 111001 145 502 8526099 M WF 300330 rogerbrad shaw louisvilleedu By appointment email or stop by gt Goals for ME 512 Gain a good understanding of nite element analysis FEA works Understand important concepts of nite element modeling Ability to solve course problems and apply to future work in career Familiarity and competence with ANSYS ME 512 is not an FEA theory course 7 focus on applications A couple of images from past FEA work I have been involved in coo 6116 mm waded Syllabus and Computer Accounts gt Roster gt Review syllabus gt ME computer lab account for those without accounts ME 512 FINITE ELEMENTS IN ME DESIGN W 500 730 Sackett Hall 208 Fall 2009 INSTRUCTOR Prof Roger Bradshaw OtZIL Hours 203 Sackett Hall M F 300400 502 852 6099 Tu Th 10001100 rogerbradshawlouisvilleedu By appointment email or stop by CATALOG INFORMATION Matrix analysis of static and dynamic structural systems and steadystate heat transfer Computer aided design of trusses frames plane stress structures as well as one and two dimensional thermal systems including conduction and convection Prerequisite ME 422 TEXTBOOK The textbook for the course is Concepts and Applications of Finite Element Analysis Fourth Edition R D Cook D S Malkus M E Plesha and R J Witt John Wiley 8 Sons 2002 Please let me know ASAP if you are not able to obtain a copy Finite Element Analysis Overview gt What is finite element analysis FEA Amethod of calculating a field quantity over the domain of the body being modeled Two typical field quantities for FEA StressAnalysis Displacement V 7 Thermal Analysis Temperature 1 gt Overview Consider a body under various loads 2 and subjected to various displacement KW constraints example is for stress analysis case FEA is a method to aggroximate the deformation of the body at eve goim in the body FEA is a numerical solution for a given problem 7 Different body 7 different solution FEA result is usually approximate unless problem is very easy 7 In this case analytical solutionslikely exist Finite Element Definition V Field quantity solved at every point in body dx Key pomt 7 keep this in mind V Infinitesimal elements dy Break domain into series of elements Exact solution 7 reduce elements to infinitesimal size Calculus approach Exact solution dx dy gt 0 gt Finite elements Ax Break domain into series of elements Example 7 square Ax 7 Ay Ay many other choices possible in FEA 130 notirdeducle to 1nf1mtes1mal Size Approx solution nstea eve op an approx1mate method of solution most cases Ax Ay 66 0 Apply approach throughout domain Nodes and Elements gt Example from text 7 2D model of gear tooth Triangular elements 6 nodes per element V m Nudes Points where eld quantities are obtained These determine the eld quantity at all locations in the body A gure 111 A two dimensional model of a gear tooth All nodes and elements he in the plane of the gure Supports are not shmvu Nodes and Elements gt Nodes and elements Divide structure into a finite number of pieces elements Nodes 16 shown Elements connected at common points nodes ments 9 shown gt FEA nodeelement behavior Elements are completely joined together 7 The nodes are NOT like a series of spot welds gluing the elements together FEA solves for the field quantity at the nodes 7 Stress analysis 7 nodal displacement 7 Thermal analysis 7 nodal temperature Nodal solution known gt can predict the field quantity everywhere in the model Can utilize this field quantity to calcuate other items 7 Stress analysis 7 stress strain 7 Thermal analysis 7 heat flux Nature of FEA Method gt FEA solution approach Solution occurs at nodes Nodal solutions describe eld everywhere 7 how gt Represent field quantity as a piecewise polynomial Consider the solution of a v i 39 J i O 39 function y fx using a series V r y of parabolic segments l 391 39 f l SolVe fx at points nodes i i 2 For eVery 3 points elements mummwwx the associated parabola is known WWH J pm raw k r hex Lljv 49 x 2 aquot39 6x gt Seems silly 7 just curve fitting right No echoice of polynomial is key to Step 1 Based on that choice you are able to solve for nodal Values Nodal Values known gt can solve for Values of fx everywhere Benefits of FEA gt From textbook p 12 EA has advantages over most other numerical analysis methods including versatility and physical appeal FEA is applicable to any eld problem heat transfer stress analysis magnetic elds and to on We Thcbud 39 LP Boundary conditions and loading are not restricted For example in stress analysis any portion 039 y W v Lquot quot 39L J he applied to any other portion 0 Material mm um L a 39 l0 mother or even within an element 39 Components that have different behaviors and different mathematical descriptions Lari be combined Thus a single FE model might Contain bar beam plate cable and friction elements An FE structure closely resembles the actual body or region to be analyzed The approximation is easily improved by grading the mesh so that more elements appear where eld gradients are high and more resolution is required Examples gt From textbook Simplification 7 2D axisymmetric solid of revolution Note several materials can be utilized 2m dimIII G m Iilimml 6 PM Fab Steel mu quot i p a 7 pomnn and possihle nite Element mesh right portion from 11 Examples gt From textbook Other fields can be studied 7 magnetic flux in this case Simplification 7 break into segments assume static History of FEA gt Various numerical methods have been used throughout history for structural analysis thermal analysis etc V 39 1943 19505 1960 19605 19705 19805 7 Nonlinear models advanced material models element types etc Finite element analysis is a fairly recent development Courant proposed FEA as a method to solve torsion problem method impractical without computers to automate Application with wing structures stress analysis thin wings not suitable for classical methods Name nite elementquot coined by Clough Application of FEA to other eld problems beyond stress analysis thermal electrical magnetic etc Development of large FEA software packages limited to solution on mainframe computers FEA becomes available on personal computers I first used ANSYS 31 on a Compaq 38620 in 1988 Continuing development Example Simple FEA Model gt Consider a tapered post under constant axial loading gt Several simpli cations before FEA model 39 Support conditions 7 should you model the ground 39 Discretize 7 series of constant area sections 39 NodesElements 7 use 3 1D bar elements 4 nodes load P 7 Number of elements affects solution more elements a better result Figure 121 Steps in modeling and FE analysis of a tapered support post Stmcture Model Discretized models P 1 Ar A1 A2 1 5 An Rigid Finite element support representation representation Example Simple FEA Model gt Consider previous example with tensile load gt Bar tapers from area of 7A to A over its length gt Break into elements 39 US6 constant area 6161116118 Assume area to be that of actual structure at element center gt Boundary conditions and loads Fix the model at the left end gt 111 0 Apply load P as nodal load gt F4 P Mathematical model Finite elemenl model g 1 2 3 ii g 6A 7 4A 2A 7 39 l gt s L gt s s l L r l 3531 L T 1 Figure 13 A tapered bar discretized by three uniform twonode elements ME 323 Deformation 5 Under Axial Load gt Consider bar under axial load P Length of bar is L B B Constant crosssectional area A Modulus of elasticity E Stretched by amount 5 L gt Calculate stress and strain for bar P 6 039 8 A L C gt Assume linear elastic behavior A C P 039E8 Fig 222 039 gt gt Can solve for 5 under this load A key fomlula from ME 323 P L know material property E gt 5 a P 5 and dimensions L A of bar A L A E L can nd 5 for a given load P 114 Example Simple FEA Model Finite element model gt Apply ME 323 solution method a 1 2 a 4 p Result will be identical to that from FEA 6A L M L M Displ varies linearly lt gt constant stress PL 12L PL 1 0 M2 m 3 2m 4 quot3 m 1393 3 Note these terms are axial strain in AL u u v u u 2 3 2 4 3 each element 6EEE E 0 E T E LS2 L 17172 L 2 3 L 34 L u r I Axral displacement I Axial mess I 39I FEA Exam x 0 115 Figure 131 A tapered bar discretized by three uniform twonode elements FEA Fundamental Equation and Steps gt Fundamental equation of FEA a I A 4 Lqu uac cr 39539 K K L Jlspiawn w Veda 39 J 1 6 54149 Mezz v x f hand written notation a underneath letter means vector or matrix gt Steps in performing FEA common to all packages Preprocessor Build model apply loads Solver Solve equation K d r for d Postprocessor Obtain results from solution lists plots contours etc gt ANSYS does all three steps while some codes only do prepostprocessing and rely on other software for solver Linear Algebra Review Linear algebra is a key skill for FEA Review of linear algebra basics 7 handout Will post optional reading Kreyszig if need additional review V Solve equation K d r for d Matrix Vector Mathematics Handout M 512 Fall 2009 ln nm handout we mu mum mamx mm mm nmimnmncai concepts mm mu hc inch as c gtMESan i i quot 39 A Iif 39 Mun n Auybm Consider an in lt 3 matrix A bold in typed work iiiidemlde A in hand Written Work By de nition this inatiix has In rm u and u column and Jas the form A An 1 A A3 A22 Ala Am Ayn 39 39 Am The sun A is 11 Value in the t lUW End j column 1D Finite Elements Bars and Beams gt ME 512 begin study with simplest elements gt 1D springs and 1D bar elements Axial loads mechanics of materials 7 axial bar two force member Used in earlier example of tapered column Mamemalical model Finite elemenl model 1 1 2 a 6A 4A L 2A 4P gt 1D beam elements Transverse loads moments mechanics of materials 7 beam theory Simple example 7 cantilever beam under end loading Flu and Landed End l O M z 4 o o 4quot IO My 1 2 Node 1D Spring Elements gt Consider a simple spring element V 1D 7 in X direction 2 nodes 1 2 2 DOFs ul uz F 2 nodal forces F1 F2 Spring stiffness k Element stiffness matrix K Relationship between nodal forces and displacements DOFs 2 nodal forces 2 DOFs gt K is 2gtlt2 matix assuming linear spring K r N m K L U Milli l i FL J TW39 4 Km We can find K by a series of thought experiments Note that a linear relationship between N DOFs and N nodal forces can always be written in this way where K is an NgtltN matrix DOF degree of freedom 22 1D Spring Determination of K quotfif s Ti s u Am f ii391xi U r 17 J gt Thought experiment 1 7 set u2 0 fixed at node 2 K M F lt10 4 2 FL 1 V is my i3 Fl ul 5 K k re eg uullbrlum 5 a 55 K2 k gt Thought experiment 2 7 set 111 0 fixed at node 1 kl 1quot F kLzMLFz 1 2 FW Lf a 2 41 F2 gt K22k 5374 eg ubnum Fl FL 1 J gt All terms in K known 7 stiffness matrix obtained 23 1D Spring Determination of K gt K determination procedure repeated in handout Review quickly see me if questions H pm L v vquot39 5 a ti Ifrind M h lui Fl it 7 K h M 139 E H Km 44 a J Fwd Kquot Kn i 17 KlL F79 0642 A 94 22 gt Men E an m to 67 gt MMJ F Km ru 0 L a C 14 mi 5 50M irr lj Mlq n ski2 7 if 4 Ea k z i rim out we 4 have F s Kn 2 4 ka oMnmmi r 1 1D Spring Stiffness Matrix K p I L 77 gt m E Vii G 2 gt Stiffness matrix K is now known for 1D spring Consider source for each item below from previous slide 0u207gt u2 0u107gt 1St column ofK f f 2nd column ofK I I k k k 7k u F gt I 1 1 l I k k I kj 1quot2 F 2 gt Features of stiffness matrix K for 1D spring Singular 7 det K O k27 k2 0 System can undergo rigid body motion 7 consider u1 u2 5 Jz 4amp7 f g 0 uail l a Lx K Requires no nodal loads 7 value of5 is arbitrary 7 110 unique solution Direct Method To Find Stiffness Matrix K gt Two standard approaches to find stiffness matrix K Approach used for spring 7 direct method Formal approach 7 formal method derives from energy method gt Direct method 7 start with fundamental equation of FEA m K a i E t J Nodal DOFs 7 set n h DOF equal to 1 all others equal to 0 Determine nodal loads required to cause this set of DOFs These values are the rim column of K d 1 gt Km 1 gt Km 1 Repeat process for all DOFs associated with element gt Limitations for direct method Nodal loads known for a given displacement case Works well for springs bars and beams simple elements Not possible for more complicated elements Formal method must be used 7 will review in a little while H 5 aquot FL va 4 5 3 m m matting 2 Node 1D Bar Elements gt Consider a simple bar element r oi 11 1D 7 in X direction F V r F 1 2 2 nodes 1 2 Q 2 2 DOFs u1u2 MFA 2 nodal forces F1 F2 Element length L constant cross sectional area A and modulus E gt Use mechanics of materials to calculate stiffness of element lllffamlml F k L g F AL f a F 4L 4 4 SOME gt Stiffness matrix same as spring gt K AE 1 1 39 Changekto AEL L 1 1 Displacement Field 2 Node 1D Bar Element gt Previous work determined the nodal displacements What is displacement of any point in the bar element gt Begin with nodal displacements at nodes 1 and 2 ul uz Assume linear behavior over element domain LL lt F W39 I ab UL J 552762 gt7lt LL quot 7C ilk 4 1 L E x 7 A 2 degrees of freedom 111 and uz 7 eld quantity values that can vary 7 Using nodal values as DOFs is the standard FEA practice Can also express as ux given as below 31 and 32 to be found 7 This approach Will be used to highlight certain element behaviors in X r th 51mg mew 28
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