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by: Maud Armstrong


Maud Armstrong
U of L
GPA 3.61


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Class Notes
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This 9 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 59 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
Input Files For ANSYS Analysis gt In the lab we have created input files to analyze problems Solve the problem completely Edit the log le to create an input le Read that input le 7 ANSYS solves the problem gt This feature is an extremely powerful part of AN SYS Useful for considering variations on models using parameters Can incorporate DO loops other programming methods Can also ask user for parameters ASK command gt Questions see APDL in Help APDL ANSYS Parametric Design Language input les Minor differences since Steps Used To Make Input File On 101603 In Lab quotquot5 handout was me Roger Bradshaw ME 512 Fall 2003 is different model in lab Otherwise to use copy the log le a le called filenamejnp where lename is whatever name you want to V 39 r 39 139 39 quot quotwungsforANSYS 2 Make a backup copy of your log le so if you mess it up you can always start again Finite Element Models In 3D gt Up until now our models have been 2D Expressed solely in the Xy domain Bars and beams 7 1D line elements Plane analysis 7 2D elements AXisymmetric analysis 7 2D elements Each case was a simpli cation of a 3D problem gt Now we consider modeling a 3D problem completely Model can be considered in Xyz domain Cartesian coordinates Each node contains 3 degrees of freedom Z 7 u movement in X direction 7 V movement in y direction J y 7 w movement in z direction X W gt Solid modeling now requires a new entity Volumes 7 mesh to make the 3D model Volumes consist ofa series ofjoined areas 11 Others we already know areas from lines lines from keypoints gt Lets now review several types of elements gt gt Linear Strain Tetrahedron Constant and Linear Strain Tetrahedra Constant Strain Tetrahedron Element consists of4 nodes 7 Simplest element to use in 3D modeling Dis lacement eld is limited 4 DOFs dir 7 Strain elds are constant at all points Analogous to constant strain triangle in 2D Generally poor results 7 do not use this element Element consists of 10 nodes 7 4 corner nodes 6 Inidside nodes Displacement eld is quadratic 7 Strain elds are linear in element Analogous to linear strain triangle in 2D Leads to excellent results This is the element that you should use if you cannot use brick elements 17 Inidside nodes mean that the element edges can become curved parabolic 613 gt Trilinear Hexahedron 8 node brick V Quadratic Hexahedron 20 node brick Brick Hexahedra Elements Element consists of 8 comer nodes Displacement eld is linear in all 3 dirs 7 Strain elds are not constant in element Analogous to Q4 element in 2D Shear locking is a problem with this element 7 Can add extra displacement shapes as in Q6 7 This is the ANSYS default condition Element consists of 20 nodes 7 8 corner nodes 12 midside nodes Displacement eld is quadratic 7 Strain elds are linear in element Analogous to Q8 in 2D Leads to excellent results Serendipity element 7 natural element has 27 nodes above 6 midface 1 center 17 Inidside nodes mean that the element edges can become curved parabolic 614 V Additional Elements Prisms 6 node prism 7 6 corner nodes 15 node prism 7 6 corner 9 midside nodes CST or LST extruded out of plane 7 Common use 7 mesh area with 2D triangles and extrude for 3D model gt Pyramids 13 nodes 7 5 corner 8 midside nodes There is not a 5 node pyramid Used for transition regions going from bricks to tetrahedra aFini1e element model b EXDIDdEd View 039 pyramids gt SOLID45 7 8 node biick V SOLID95 7 20 node biick ANSYS 3D Elements SOLID45 Without extra displacement shapes trilinear hexahedron With extra displacement shapes eliminate shear locking similar to Q6 for 2D Degenerate into other shapes M 4 node tetrahedron v M3 6 node prism I R K L 3 SOLID95 0 20 node brick element Degenerate into other shapes lO node tetrahedron l5 node prism 13 node pyramid ANSYS 3D Meshing gt Recall from 2D that tllere were two types ofmeslling Mapped mesh 7 3 or 4 sided areas esh with quadrilaterials also triangles but not common Free mesh 7 irregular areas consisting ofmany sides 7 M could consist oftriangles quads or a combination gt Meshing in 3D is much more dif cult Mapped mesh 7 6 to 8 sided volumes 7 Mesh with brick elements Free mesh 7 irregular volumes consisting ofmany sides 7 It is not possible to generally mesh with brick elements 7 Thus mesh can only consist oftetIahedIa gt General strategy for 3D meshing Try and create volumes that can be meshed using mapped meshing Mesh mapped regions rst then free regions 47 7 l 3D FEA Numerical Integration RBM gt As always integrals are solved using numerical integration Stiffness matrix K distributed loads 7 nodal loads Requires 3D quadrature method gt Three approaches can be used in ANSYS Specialized 14 point quadrature default oids hourglass instability re 7 Restrictions to avoid instability 7 Fast but not as accurate gt Rigid body motion RBM Must provide suf cient BCs to prevent RBM Much more complicated in 3D than in 2D Six possible RBMs 7 3 translation 3 rotation Elements In ANSYS gt ANSYS contains many elements We have discussed and used several 1D LINK1BEAM3 2D PLANE42 PLANESZ 3D SOLID45 SOLID95 We will use others as we nish the semester gt Reference 7 all elements in ANSYS listed in handout Many are designed for advanced material models behavior Do not use an element if you do not understand what it does 7 Read ANSYS Element Reference Theory for any element new to you N 8 Elements le512er Bradshaw 243 ngwequancy Ouazl laleral Sohd 37D Tehahezlval HighrFrequenc 370 anew de HmhrFreuenc 243 47Node Mixed up Hyperelastlc Solid 37D Mode xed up Hyperelastlc Solid 27D Mode Mixed Up Hyperelastlc Solid Inertial Forces gt Inertia forces Due to acceleration and mass of body Bicycle problem 7 applied gravity to nd weight gt Units are important for inertial forces Bicycle problem 7 two ways were provided to apply units Leave density in lbs and apply acceleration as 1 1 g Convert density to mass and apply g 3864 ins2 f gt Correct approach for inclusion of inertial forces Recall length and force units provide the stress units Length time and mass units provide force units F m a These need to be internally consistent 7 provides necessary mass units 7 Will need to use in upcoming homework with rotating disk 1lm11lull m In m Bu I39rucliccs for Crush Modeling and n Simuluuu Path Plots gt Report results in ANSYS several ways Contour plots Vector plots List results gt Another powerful tool is path plots Describe path in ANSYS Plotlist various quantities along path l mh l lms Ruin this sccllull ifA lL Kliun Plate and Shell Finite Element Models gt Simplifications of 3D structures 0 Beams and bars 0 Plane stress plane strain analysis axisymmetric analysis 0 Plates and shells are another simplification 7 Consider 3D structure as a 2D element gt Plates 0 A flat planar body whose thickness outof plane is relatively small compared to other inplane dimensions gt Shells 0 Similar definition as plates except curved 0 We will refer generically to plates but will encompass shells as well N gt K K L w u L te i K e PM Sm Metal a ems TL l 4 LIW b54515 aw 71 Plates Beams Analogy gt Recall how beam elements were used 0 Simplified 3D structure as a 1D axial element Aci ml F M1 gt Plates are analogous structures to beams 0 Evaluate a 3D plate as a 2D sheet element 0 Consider transverse loading momenm first then add in plane loads w 9193 21gt Thin Plate Theory Kirchhoff39s Hypothesis gt Plates simplify 3D structures to 2D elemenm 39 This is done via an approximation of the actual deformation gt Kirchhost hypothesis for thin plates 18405 39 Approach dates to 1840s Plate is thin and initially at 39 Midplane exists at xy plane 2 is out ofplane 39 Line AOB is perpedicular to midplane prior to deformation gt Consider shape a er defamation u 7 X din w 7 2 dir 39 Line AOB remans perpendicular after deformation 39 Line AOB does not change length i raml Pm m V quot L A015 J xlv hwr w lnl PIA W new Mr M 9 int Thin Plate Theory Kirchhoff39s Hypothesis gt Consider deformed plate shape more closely 39 Look at line AOB and see WhatWe can determine 39 Displacement of midplane point 0 is uO and WO in x and 2 dirs 39 Call the angle ofmidplane when deformed as e 39 By similar triangles e is angle of AOB with the vertical axis gt Consider an arbitrary point p along line AOB 39 Can express displacement up 311de using Kirchhoff assumptions Ztm9 UP bio 2 39MS 1 e w 29 r W u tanHrdW 7 2 are ob 3de wp mwa b 6 pos as shown Thin Plate Theory Kirchhoff39s Hypothesis gt Can repeat analysis in y direction deformation v I Similar nding as that for displacement in x direction up I For transverse loading we need only consider displacement W gt Bene t of Kirchoff s hypothesis I Displacement of every point known in terms of midplane deformation I Reduces 3D problem whole body to 2D problem midplane only I From this assumption strain state everywhere follows w wO 9amp1 91W 3 7x3 22 9x23 2 31w QJ ex 2 e 7 gt Use of Kirchhofs hypothesis in FEA I Bending response only requires outofplane deformation w I Use nodal DOFs to express this value as well as nodal rotation angles 6 a 0 WM 6 3320 weeks he


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