FIN ELEM MTH MECH DES I
FIN ELEM MTH MECH DES I ME 512
U of L
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This 16 page Class Notes was uploaded by Maud Armstrong on Friday October 23, 2015. The Class Notes belongs to ME 512 at University of Louisville taught by Roger Bradshaw in Fall. Since its upload, it has received 34 views. For similar materials see /class/228366/me-512-university-of-louisville in Mechanical Engineering at University of Louisville.
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Date Created: 10/23/15
Shape Functions Demonstration gt Demonstrate nature of shape functions Begin with 1D bar presented in weeks 12 End with CST from Problem l33c 7 Similar to what you will do in Homework 4 Demonstration of Shape Functions Roger Bradshaw ME 512 Shape functions are a cnucal step to understand for nite element analysis As described in class I the shape functions relate the eld quantity over the domain of the element to the nodal values F ii I i ame eldand uj r 47 i from it can be evaluated anywhere In the element Simply stated a shape function needs to satisfy the element behavior i e linear in x for the 10 bar h elenienL linear in x and y for the constant strain triangle etc and be l at Its associated node and 0 at all other nodes Lets demonstrate this rst for the 1D bar and then for the constant strain triangle from Problem l33Cl similar to what you will do in Homework 4 a gtlt 1 10 Bar Element H quot Assume that the bar ranges from x 0 node 1to x l i riode 2 you can easin recreate this 5 p Shape function l0 needs to be I at node I and D at node 2 Shape function 2 needs to be I at node 2 and 0 at node I These are given by PrOblem 1393393 N1x i r x Nllx x We can de ne the eld quantity in terms of the nodal values uI and u2 this provides a description or39ulx I i L A 39 r d i L 39 quotvalues CST Displacement Along Edges gt Key feature of F EA consider behavior along element edge Nodes 12 6A9 3 0 vi LIquot 1qu3 W 1quotquot11 Nodesl3 Egg L We e a ogiu gas Nodes 23 59 3 4 b re El ll 2 l g 3M5 D gt Key Resul displacement along an edge depends 0 1112 on the nodal displacements located on that edge gt Now consider group of elements Same finding for adjacent elements Displacements along edges match Continuous displacement field throughout entire model StrainDisplacement Matrix gt Recall strain displacement relationship write in matrix form a wax go 2 W 3 0 a r I r 93 j 5 ngjL 63 93 6 v 3 J J 23 abk 9 1 quotJ 39 33 ex 5 w 3 UL gt Now express displacement eld in terms of nodal values 0 Straindisplacement matrix B 3 u zg 5 Can now evaluate stress as well 0 Constitutive matrix E 7 relates stress and strain vectors 6 E 2 gt 6 E B d 0 Strain energy density W go B dT B d dT BTE Bd V 3 21 Example Constant Strain Triangle CST gt Constant strain triangle CST ug V 0 3 nodes plane element 0 2 DOFs per node 0 0 Calculate straindisplacement matrix B U j 2 a I u 7 See Eq 3410 use in HW problem 9x7 3 an d 9 72 um Kg 39 f 33 my 3 V j y 7 093 r A5047 0 M100 0 mad 0 39quot H17 Vbr Vt J 0 No7 a mo 0 may 1 1 7 11 i 0 3w ls y j V j WW 3 L3H 5x1 1quot 3X2 24 0 x 5 lt3 4 j 2 E 1 4 J J xtzwys 54s quot 3 22 Constant Strain Triangle CST gt Constant strain triangle 3 Nodes l 2 3 at xi yi W 8 Book 7 special element case where x1 y1 y2 0 Straindisplacement matrix 1 quot2 gt xu B given below 3911 02 quot1 1 o l o o o 1 8X x2 x2 1 x x x 1 2 By o 3 2 0 3 0 1410 x2 3 3 23393 3 3 2 7w x2 1 22 1 1 o quota 12 x2 323 3 x2 3 v3 B Constant Strain Triangle CST gt Constant strain triangle for the general case Nodes l 2 3 at xi yi quot2 Relationship 5 below From RD Cook Finite Element quot1 2 WM 5 I13 1 The strain eld obtained from the shape functions 1 in Fig 321 A constant strain triangle m the form 2 Bd39 ls quot1 Its six nodal dof are shown v 5 1 Y2 0 hi 0 in 0 u 2 0 x 0 x 0 x 32 3 8y 2A 32 n 21 v2 ny 132 Yza 13 Vat x21 Y u 3 where xi and yi are nodal coordinates i 1 2 3 xi x x and yij y y ij 1 2 3 and 2A is twice the area of the mangle 24 my 13012 Node numbers are arbitrary except that the sequence 123 must go counterclockwise around the element if A is to be positive 324 Compatibility gt Consider any displacement eld 0 Can always differentiate to get associated strain field gt Nature of strain fields 0 Not all strain elds are acceptable admissible 0 Must not lead to gaps overlaps or other unrealistic behavior gt Compatibility 0 Strain eld is acceptable if satisfy compatibility equations 0 3D 7 6 equations to be satis ed 0 2D 7 1 equation to be satis ed below 925 9726 9753 079 9X1 Q 22 gt Polynomial displacement elds satisfy compatibility FEA is based upon polynomial displacement elds 0 Thus FEA displacement fields lead to compatible strain elds Equilibrium gt Equilibrium gt no acceleration forces balance 0 Consider at a point in a continuum 0 3D 7 stress eld must satisfy 3 differential equations 0 2D 7 stress field was satisfy 2 differential equations 97quot 97139 Body forces X I Z f 5 0 3x 6 lt90 at J 7 33 o 3 9x 0 Any problem that satisfies equilibrium everywhere is exact gt Finite element analysis satisfy equilibrium approximately fo 9 5 E 3953 8 o m mum LII 54 nvi i39 m 0 wake2 Exact Case Cantilever With End Moment gt Consider two cantilever beam cases with exact solutions Will compare various element results to these cases quot1 iiZea L Rue Ben t39g Caml39ll Lr NEW Shut gt Pure bending case from beam theory Vl X1 4 also lawn I zEI M e 39 4 1 C7 2 5310 E u3tave539e 73 5x3 0 We 7 Ext I V XL 16 Exact Case Cantilever With End Load gt End shear load Beam theory shear stress analysis P a Vjo rxa 54x gam 9Ljf u9 4 0Lj399 Ii 1 91 2x1 2Lx3 0quotquot 1E2 2 3 LLFX39LgXJ V 5x51x1 gt The stress state is also fully known TX FL U I M 0 3 0 13 s 3 11 txg th Hgt 1 gt Two cases now lets consider various element types 3 28 Comparison of Various Elements Handout l llmlcu k quotNut uhicctcri 039 H1 ml Limd I39ll II CI H i39icmcnl I 0Ill391 i In 1 t N rerl I matII umlmuri II I M1 kml Munmn h i iglmumul i Hint I39IIIc Ihmlr In 11 r I I Ilinl i39lwm Hm IIII iuiIIIuII U I hm u III39umI I Mr I 11 u n 1 39 I ITquot7ii I Mm mi ilw mum rivmmh H uI quJumun i It i Ince ii It I wwhw A i i Ind I lenizi39 iiw mm mIIJIrII In Him mummy Ii i lii39 oicliml iii 1I i vimlinitln i MINl mm Iu In 4 I u min mm mmImn lIv mmrim it I it In Mum hi mm It 739 III 4 WW II I III Immu i II I MIMI mm t Hi m I I JHH UCVI It39IAI K uIIIiouI Illl I 39 39 In HIM Iu xgmg u no mmr iiri 9 H HiI H AHH l u 3 u I gIII H H utmm mummy m rm H I V m m H I i V H V H r I my H m uu mm InI II pH gt Hi i In I H il n unniJht IIIIJ 7 4n n tii i IIII m I I c Ivil v m H n I i mum Iil i Hi I iiI n V w I EX Mummy Hixvlnnil Itldnli39 Wm M I 2I iH MI W W M WNW I m it i I II 329 Constant Strain Triangle CST Congratin 3qu TVIM LL r 39 3 hadn 39r b lkhjLL RAeri L 1115ka mix 39hriknjus MSHAPE Ooumd 5 f F 331 4 Wm rowers ILLUJI awexrlwl 3 6x a a ex COKShLF 35nd quot AWSL1 QQ 1L3 Constant Strain Triangle CST gt Imagine CST elements making up cantilever beam 1 element through the beam thickness CST Doe s Worf Lima Paye 5414 am 0r siww39x k4 est J PM Elma I Tvaw Canhicutr39 a I 031 xtj 4 an XL 1 30933 QCX X 3033 x3 6 newsth a 3 fny 7 5 6 mwmi D a Xx onwai awi o 11 gt Consistent with ANSYS warnings do not use CST Linear Strain Triangle LST Linglur gincork Ttrcamski LS 49 huh hnutatc Ana s PLANE 81 at quotVJD h tq39wacjv C HSRAPQ amnime l a 54 pix k qxt gkj bvt Stumisu be V g SivaMS x 63 3 Y3 ladi f KILIS 9 90 L Z5Ygt gtj C 79093 Simiimv 920903 VUf ire bu Linear Strain Triangle LST gt Compare findings to exact cases 5 A ffllxlggxquot 33x3 V 16 D quotX PUMA 95 F xy 5 3 given 0 a 33 Fltgt95 o K011 gt Linear strain triangle 0 Solves pure bending case exactly 0 Shear load case not exact but very good result gt If triangular elements needed LST is preferred choice Bilinear Quadrilateral Q4 FLANF47 0 3 no ethdmflacmwfv shape 473 1 4 Elllnaur Quoad C Q4 4 Mo e AWSOP 125011 Mimi39th z huts gt Amrian I t waular lTCU A quot4 E W wirwr x 57 l y relaholnstup e D 171 39lquot 1 a g LL VB l39 j k f x3 39 CIC1gtCC3 Bilinear Quadrilateral Q4 gt Compare findings to exact cases A Ewe Eewlmi I Tray Gunfile u 43095 Xy CCXB V 4 X3 X1 V go sz x7 3X C x13 x3 6 193 9 37 2ny 3 a 4amp0 o o Xxj ram 0 3911 gt Looks close for pure bending strain 8X especially It actually has some problems due to strain interrelationships quot quot Note mlw x 5i 73 y relal39 mstup e3 m 236 g 65 06 4 158 Improved Bilinear Quadrilateral 06 gt Consider beam in pure bending actual vs Q4 element Moments Mb in actual case M61 in nite element model Angles between faces of 9b and 961 in actual and nite element cases robv 4 a M 39 Exquot SM 21b 1 139 uf ytquot Bel 1 V2 L x i l 91 1 v a 2 quot quot l Tb lt 2a gt 24 a b Figure 362 a Deformation mode of a rectangular block of material in pure bending b Defamation mode of the QA clement under bending load gt Set Mb Me1 how are 9b and 961 related Ratio 961 9b gt 0 as a b gt 00 FE model much stiffer than actual gt Refer to this phenomenon as locking Keep aspect ratio ab close to l 7 better results for Q4 Q4 becomes excessively stiff if aspect ratio becomes large Improved Bilinear Quadrilateral 06 I Awa d Edimar Quail Q 4 Ma and I W quot 4 01M tsPlLuWLnl skau AMIIs VLJoSE 41 wz l k 50 1 l 13M tyingwa skac Calm L dz 4 gt Element canhave any shape M 39 Consider this square 2x2 element for demonstration UL mm Ll W I 71 t gp h 0 9 nnivzs Plankth Few bew ma EmaHt xi reki39am dfov 3737 Improved BIIInear Quad rIIateraI 06 gt Downside of Q6 element 39 Nonconforming element 39 Element edges do not stay connected together 39 Book refers to this as an incompatibility between elements F 153 F F r A I I I m I 2r J31 2r 2r I 2 W 1 4 a b r I F r I F I Cl Figun1101aDisplaccmnmodcsu l 7 m and u I 7 Equot in the Q6 clunan m Incompatibilin helwecn adjacent 06 elements a No inmmpaljbi ly belween adjaognl Q4 lemma gt Effect is acceptable gt convergence as element size reduces 3738 Improved Bilinear Quadrilateral QB gt Q6 elements are computationally intensive Start offwith 12 DOFs 8 for Q4 4 extra displacement shapes Reduce back to 8 DOFs Via static condensation see Sect 67 Stiffness matrix K has same size as Q4 element but different values gt Consider displacement and strain terms assume rectangular Exact in pure bending if rectangular Very good results for shear loaded cantilever case Q6 L 6 C5Xj9zltquot 41quot V p 5 5 mg 391 x P 0 3 e 3 quot1quot 7quot Fm Viix QB Static Condensation gt Q6 8 external and 4 internal DOFs Internal DOFs a1 7 a4 in image at right Static condensation 7 eliminate 4 internal DOFs Result is 8X8 K matrix as in Q4 case 0M5 element Flgun 671 1 Let dof d be partitioned d d where 1 are boundary 10139 to be retained and dc are internal or noduless dof to be condensed Loads 1 which axe o en zero are similarly partitioned Thus equations kd r become 1er er dr rr chr kcc at rt 39 674 The lower panition is solved for dc which is hen substituted into the upper partition 1 tkr lttkiiairi 572 kl knllkccl llkuhidri rl kmllk lil irci 673 condensed k condensed r QB Quadratic Quadrilateral 831335 Quade cricer i 8 W04 5va AwSLTS Fame 81 KF 3qx 531 343 r5yx3 p aL P ixzj 4 p5x71 Q g Tvaw Cahili L Cx j lex 71 x7 777x 023 4 x3 x1 C a quot3 3m W13 3 0 217 5 E91 93 F057 3 25 gt97 X7 x1 0 a r 0 7 x3 52 71 o r1011 QB Quadratic Quadrilateral gt Q8 provides excellent performance Solves pure bending case exactly Shear loaded cantilever still not exact but very close gt Q8 is referred to as a serendipity element Mathematically a quadratic element with center node is more natural 7 n pi i 2 3 6713mm Elem a Q 7 ND uej conquotNM 4 fr39acj7 1 It turns out that you simply do not need the center node 7 leave it out serendipity ser en dip i ty sgr39en cITp39T te seren dip i ties The faculty of making fortunate discoveries by accident uhe fact or occurrence of such discoveriei An Instance of making such a discovery 12 Elements Not Discussed gt Several other types of plane elements gt Drilling DOF Suppose we add nodal rotation to each node 3 DOFs node u v G 7 last is referred to as drilling DOF Applicable to triangles and quadrilaterals Read text if interested not required This topic will come u in the context of plates and shells Figure 3101 c gt Q9 7 quadrilateral with midside and center nodes Lagrange biquadratic element multiplication of 2 quadratic functions Read text if interested not required glf tlmhc v Clem 6 Us Jej co Mm M 1 4 frag r 343 I lmucm l39mnpdrumh In mlxluwr mum uluucled I a lull Munmn Ml amp clmxmm lnl39rul A I39m M liclmn In I 11 l I Jul1 Insxt I39nm Hm l lnxu lw tlu39mnlmHIH H Nc m Inlmmi u hul lmwnl quot mmmu mm m r I mm m u M mm mm NH mum H ml v i i v 5 33 5 1l n V H 311 f 1H4 w m Wm M H In 1 ml y In 5 u I n u um m v Wm m Lu o v A 0 mm mm m xx Him nun I Mmmn umhwm Immw lumpm39 mm 39 M quotquot l My rm 1m clmwmw l quot quot 344 Comparison of Element Results gt Compare several elements under common analysis mite ElementModeling or StressAnalysis by RD Cook Cantilever with end shear load previously discussed Exact result at left includes shear de ection 7 about 3 of total Two models at right 1 element through thickness in all cases nu 5104 v03 Lalo hi aBeam theory gives PL 6 PL 0000 0 ac 3H 5AG l 031 a 39 300 a J FE supports and loads 10 d csr 1 5 Q4 3 a as m 2 LST l2 15 2 oz A N Comparison of Element Results One Znnda beam element an a a 71 u 0264 DOFS 395 c 24 24 c LST C 30 a z 200 u 0693 a a 254 0c 0357 24 o4r 16 LST C 30 quot aquot 1 90 ac osoz v 250 u 0924 24 061 I I L I C as T 14 26 quot am 270 uc L016 quot a3oo 05 1023 24 as C QB C 26 a 231 MC 0930 a 27a ac L035 Reaction Solution gt Finite element model 0 Nodal DOF 7 u V 9 etc 0 Corresponds to nodal load 7 Fx Fy M etc gt You always specify one or the other for every nodal DOF Specify nodal DOF 7 there is a force that causes this to occur 0 Specify nodal load 7 use this to find nodal DOF Specify nothing 7 same as saying the nodal load is zero gt Reaction solution have asked for this in preVious homeworks Nodal DOF specified by you 0 The load to cause t comes from outside of the model 0 This is referred to as a reaction force 0 Sum up all reaction forces 7 total load to keep model in equilibrium gt List reaction forces in AN SYS 0 List Results gt Reaction Solution 0 Can use this as a check for your model validity 7 Do the reaction solutions make sense Stress Terms gt Recall general stress state at a point 6 terms 0 Typically Xy refer to global axes verify if using a new element 0 Can refer to other axes Via results coordinate system CS discuss later ALL Ty s NH 7x V3 VT MDIMI 3X 37 5 1 311 77x Shear Sham 5x7 5 7 e S xi gt Principal stress state CS where shear stress is zero 0 General stress state has 3 principal stresses 4 m A 0 7 lLJk I X lo PIA 55w 1 030 yaw m Hus39Uqu9 7 gt61 gt 0 0 r nrf tu may 939 Stress Terms gt Recall Mohr s circle for general 3D stress state 0 Plot normal stress versus shear stress for all possible CS 0 Maximum shear stress rm is often a useful designvalue 0 Can obtain in AN SYS as a term called stress intensit twice rm 6 1u a New 7quot Es 0 3 r2 3 r3 Prmapql SW5 6 0 3 SM vainbe 039 3 L E S r s 2 3 395 3M7 349