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# FINELEMMTHMECHDESI ME512

U of L

GPA 3.61

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This 5 page Class Notes was uploaded by Maud Armstrong 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 39 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

Isoparametric Elements gt Up until now we have considered elements with simple shapes Triangles and rectangles A Shape functions can be easily obtained Stiffness matrices can be easily obtained gt We need more varied shapes for general models How can we consider more complex element shapes Example 7 8 node quad with curved boundary l H 3M r f 7 k g 8 Lx Md boundla Seek to express the displacement fields Shape functions Nixy required 8 7 Need to vary linearly quadratically etc x9 y ZN x y u 7 This may not be possible directly 11 Element in Natural Coordinate System gt Isoparametric element formulation Use a square element to consider general element This has a natural or intrinsic coordinate system st or n Shape functions known in the natural coordinate system CS 7 Can similarly obtain for other elements 8 node quad etc exam 4 M9 M Isff f Uquot a 00 N M I svcure N3 59 rsXI e Ma Q 5 quot 7397 gt Map general element to this square element Use this mapping to evaluate various quantities 4 LlA QIQ t L S Xl t Mapping Element To Natural CS gt Can express any property p in the natural CS Nist known 0 Assume that we know nodal values pi 0 Evaluate p at any point st in element 0 p can be any element property no restnctron gt Now consider mapping of an actual element 0 Edges form lines of constant s ort CDCX I13 gt Map point st to actual point Xy 0 Simply setp to be X or y 0 Dif cult to do reverse Xy gt st xst 2N1 SJ x1 yst Emmy pwgt ZNXs Op Displacements in Natural CS gt We know how to identify a given point 0 Relate st to a given value Xy 0 Can describe entire element geometry 7 Use A rocedure for everJ A oint in st gt Similar approach for displacement fields 0 Nodal displacements ui vi known 0 Predict ust and vst for any st 0 Can describe entire displacement field 7 Use procedure for every point in 50 gt lsoparametric element formulation xst 2N1 SJ x1 yst ZNisty ust ZNist ui vst ZNistvi Same parameters descrrbe element geometry and drsplacements 09133 Example Q4 lsopametrlc Element Spreadsheet gt Consider a Q4 element odes 17 4 each located at X1 y for i 1 7 gt Microso Excel spreadshe t Tool I develo ed for ME 512 to lot an iso arametric 4 element Shows lines of constant n to demonstrate elem t ape Also maps point to in thatyou specify to associated xy Value gt Consider demonstations for nodes 141 7 use 11 075 025 11 21 22 12 7 simple square 451 34 152 105 7gen al shape 451 34 64 105 7 what is wrong with this shape Strains in lsoparametric Elements gt We need a way to relate displacements to strains For simplicity consider only X Similar treatment for other terms gt If shape functions known in xy it is easy Example 7 see constant strain triangle discussed previously We y Z N bay 6M 6N 5x2 1xyy I 6x 6x gt We know ermetr and dis lacement elds in st xw 2mm um Emma yVJZNVJM VGJFZNlVJN How can we determine the strain elds Use calculus chain rule to accomplish Chain Rule gt We can easily express derivatives in terms of st amp a 9 Du sot t N z 9 quot 9x 5 39339 m E MX mt ZMM y 1 g a g EJ 1stZNxstux 93 at vltstZNstv gt Rewrite this in matrix form 9 1 25 9 5 Jun A Qze 9V 5quot 9t 94 mi J Kit w lt4 Tm snrn g an aalude Sena u by c Vtwk SM 7 a xc u and gag kw Lt sit gt We can now evaluate ex Bu 6X Solution for Strain Terms gt Use the inverse of the Jacobian J J39l to nd strains a 3 9 gt Recover exas 6 55 x J39j Q r If gt Every item above is known in terms of Nist Can always make this evaluation Can similarly evaluate entire strain state gt After careful rearranging we can ultimately write 5 se 5 Eli c W wad cal AnaA1 lng uwg Ci sueJ aw N m M 034 amt gt Known mapping st gt Xy so we can say exst gt exXy Demonstration of Jacobian Matrix gt Simple example to demonstrate Jacobian I Consider single point A in 211 coordinates I Consider step A constant 11 and nd associated Ax and Ay I Consider step An constant lt2 and find associated Ax and Ay I Show that the relationship between At An Ax Ay is Jacobian J Demunsrmxlon Examgle nl Jacoblan Matrix Roger Bradshaw ME 5i2 In ims handout we constder an example oi me Jacobian mainx and what it represents Begin with a 1 domain 0mm Qiiwm ckmmi39 a Gama inward 79 u in Sv 5 km 391 a I39 Q 9amp0 177 3 2 q 4019 M X M x iii i HUv Lilt ili 39 Caquot 6 x 5m MaOx It 479 Determination of Stiffness Matrices gt The last step is to find the stiffness matrix K for element gt Recall that we previously wrote this as an integral equation JV IA x gt It can be shown that this is equivalent to 2D elements I N K r r39 J detJ 5 J 1 g Lye434 ll I L Vm m x t C ayc s an gt Similar expressions for distributed loads pressures etc I Recall earlier integrals for workequivalent nodal loads gt Note that in general iJi varies over element I Exception constant for rectangles and parallelograrns

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