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This 12 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 25 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
Stress Terms gt Von Mises stress 7 will use this a great deal 0 Requires some setup to fully understand gt Consider the stress state as a 2 tensor 0 Proper form in terms of continuum mechanics Errms Tag 039 quotMW re 7 1LkL a 4 39 cm 73 ConMann manhunt f gt Rotate stress state from Xyz CS to another abc CS no restriction 0 Is there anything that is the same about the stress state in all abc coordinate systems gt Stress invariants 7 same in all CS 0 There are a total of 3 distinct values 7 0 Any combination is also invariant Stress Invariants gt lSt invariant 7 hydrostatic pressure 0 Corresponds to equal pressure in all directions 6H 0 This stress state leads to a change in volume of the cube 7H Uquot W l7JVH E z a H H Z7114 gt Any stress state 7 break into two parts 0 Hydrostatic part 7 change in volume of cube 0 Deviatoric part 7 change in shape of cube O xr 6 H u 2quot Q 01 E 074 I W 4 Olamg SM Aydmlr ikc she44 quotckwyz rave 0 6nlt7 quotJ gt We now seek to characterize the deviatoric part quotcutfu mm n 5913 399 Von Mises Stress gt Deviatoric stress has its own invariants g 0139 I gt gt The von Mises stress is one of these 0 Given by the following expression d w utquot C 3734441 0 Can also be written in terms of principal stresses 1 WM a 73 03 v393 a lt71 1sa31 fm iM 2 gt Bene ts of von Mises stress 0 Characterizes stress state by a single scalar quantity 0 Can be used in various failure yield theories more later M Ty Kin5 CTiM VBW Mtge SM Importance of Stress State gt Several quantities of stress available gt Why do we care about these values 0 Failure prediction maximum load capability 0 Identify areas of high stress redesign 0 Load for yielding ermanent deformation 0 Analysis of fatigue effects cycles to failure 0 Fracture mechanics effect of a crack gt Review uniaXial tensile test result before proceeding airMk MM 01 Distortion Energy Theorem gt Prediction of yielding Via stress results Ductile material with yield stress 0y Obtained from uniaxial tension test Vi H gt Linear analysis W ME 512 7 limited to linear analysis 03 Model is valid up to yielding linear elastic material model 39 Beyond that stressstrain response M 7quot IL is nonlinear need nonlinear model gt Finite element analysis Predict general stress state at all points in model domain Can we predict when permanent deformation occurs gt Distortion energy theorem Yielding occurs when the von Mises stress exceeds yield stress Benefit 7 can use von Mises stress to predict yielding Failure Prediction Using FEA gt What about prediction of failure Need a failure theory to relate stress state to failure gt Several types of failure theories Maximum stress Maximum strain Maximum shear stress Distortion energy theorem gt Handout 7 optional reading on failure theories Will be posted to Blackboard no paper handout 42 Funn by View runu mum Whether because a material inhomagcneiiy or nonuniforniity nl regions of high men may he prescni in which localized yielding occurs As increases the inelastic action beau1 mo 39 p rsniing eventually in a slate of gaimzl yielding The rapidity wilh which 1 min mm m In general yielding occurs is dependent upon the service candilions as well as Lil distribul39mu of mass and the properties of he male nls Alumni the various service mndidons mmncnxure morch a unlimith simi unt fancy Linear FEA and Failure Analysis gt Consider a simple example of linear analysis and failure Stress state for plate with hole in tension under far eld stress 50 Multiply the far eld stress 50 similarly scale omax 0 T l Eat gm UL Ana 4 30 3 9 9 f a la gt Suppose stressstrain curve 0 for material is given at right 5g Elasticperfectly plastic K Now compare the model prediction to FEA prediction Linear FEA and Failure Analysis gt Consider stress in model along horizontal line y 0 Farfield stress 50 30 ksi r Perform linear FEA Predicts omax 7 N Mi This is clearly not possible gt What happens in practice Localized yielding around the hole nonlinear analysis needed Be cautious of trying to predict failure using linear FEA 90 MPA PW a I 60 J JA Wquot j r V eJul l tc jIUk 44351 w 39 RJLS Inn rm 7 I I r I 2b 2 a flaw awnMahm 39ttf Bi 2 E A 5 i e gt Thermal Stresses gt Consider stresses caused by temperature in a body Assume homogeneous and isotropic Reference temperature Tef 7 strainfree at Tef when unloaded V Change in temperature AT T 7 Tef Body will expand as below 7 no stress induced Same is true if temperature eld varies linearly in xyz 047 139 ex F777 7 21 Plan X AT SM Thermal stresses Body is restrained andor temperature varies over the domain of the body in a form other than linearly in xyz Suppose body 1s at unltorm temperature 39139 and unrestra1ned Coef cient of thermal expansion on l F or l C 7 material property V V Thermal Stresses How do we consider cases where thermal stresses occur Break strain eld into three parts Total strain E 7 strain observed by measuring part what you see Thermal strain 51 7 strain that would exist if unrestrained Mechanical strain EM 7 strain caused by stress state Now de ne relationships Mechanical strain EM a 7 31 Stress state 039 E 8M E 87 51 Unrestrained 039 0 7 e 51 Fully restrained E 0 7 039 7E 61 Expand this equation Stress state 039 E s 7 E er First part 7 stresses calculated as if no thermal strain Second part 7 stress caused by fully restrained thermal strain Subtract second from rst to get actual stress state Thermal Stress in FEA gt Description of thermal stress in textbook 0 Identical to that described on previous slide Thermal Stms Problem Stresses in an elastic FE model produced by a temperature eld are calculated as follows The procedure is not restricted to bar and beam elements It is used for all nite elements that are based on displacement elds and use nodal displace ments as dof The following steps are carried out automatically by software 1 In each element compute initial stresses which are stresses produced by the tem perature eld when all displacements are prohibited Also compute loads applied to nodes by stresses See Sections 31 and 33 for a general treatment of initial stresses 0390 2 Assemble the elements andloads calculated in Step 1 The result is a stnpcture whose nodes as yet have no displacements but are loaded by initial stresses produced by temperature changes Solve for nodid dof produced by loads of Step 2 and compute element strains and stresses they produce Superpose on these stresses the initial stresses calculated in Step I If mechanical loads are also applied they may be superposed on thermal loads in Step 2 U Distributed Loads Plane Analysis gt Several types of distributed loads for plane analysis ed e 39 1AM Maw xes 0 Lucia NaeL AM a silly ex5 are 2th as ka quot2quot 5 51 eesI V y j atan mm Q gt gt As before convert to nodal loads using work equivalence I vquot jif evum raee qcf39 f ft MIA J e htyoc Alter1m r 17 4 Jam sane lama1Q MILy 134 y 3 7 fj mel N 57 JV l V lithium km AltM 6 361 a ijQMf Distributed Loads Example gt Consider example from textbook Pressure over surface ofan element it 25 kwhell w m 5 712 swim 2 bLuv ab 1 De w s 0 an a ta 1 AI whwf 4 VIAVa daQc m LA Ar AM 62 6 F GWM 5 53 71quot h39ci tax44 aF n 324 5 5 9 VL39 Hit a V MA 6 V1 z gt Coupling l Constraint Equations gt Models thus far have considered various boundary conditions Nodal loads Nodal displacements Distributed loads pressure gravity gt work equivalent nodal loads Symmetry antisymmetry gt various combinations of DOF restraint gt Consider two other forms of model conditions Coupling Constraint equations Coupling and Column Equation viqu V M a M a 41 r I lam mde symmelry bed wm elements You an anth you to Im dcyza armadan u waI am chunk em Ema 2 r mm DoFs MW gt003 1 39fnlte on cantata but u it Elements of General Shape gt Up until now focused on triangles and rectangles 0 General quadrilaterals 7 will address with isoparametric elements gt A few points to keep in mind 0 Odd shaped elements are usually detrimental to accuracy Cl Mei Curved boundaries are best accomplished with midside nodes 7 Both cases look good if suf cient number of elemenw do swjhi 124 H U 1 569wa I a L A cums Lc39a g 4 i as 8 was ab 4 m cuffed ELK j 0 Straight edges are best in interior of model Element Connectivity gt Follow proper procedures in connecting elements 0 Bad practices shown in Figure 3102 ANSYS will not permit these if using solid modeling gt Consider two types of bad element connectivity R I L cLu 711d yaamp 145 lung auf m Sloan I r l An 1 arqcr untilI I or 45 04 F W 139 magma 43ch V PArAbGIVL byrk PW AP IL 03 fhr mun4439 bv 4 41241 FrgU S gt Elements with midside nodes ANSYS can remove any all of g the midside nodes as requested 0 refJ 39ve J O or he a 355 Elements With Removed Midside Nodes gt When would we want to remove midside nodes Midside node elements 7 regions of high stress gradients Elements Without midside nodes 7 regions of low stress gradients gt Consider plate With a hole 7 one possible model More likely to use this approach with 3D models to reduce DOFs I ML grinii imamu quottquot397 lquot 4 nub thawt 01a Ni k S vdut 7r CAM L4 2 no1 ebmemquot ralfrrxd gt Note that order matters for meshing Mesh regions Without midside nodes then midside node regions 9 quot MW AW kmw 0 I CVHDVC l AS mC de VLIAJQ39 gt Demonstration in ANSYS Order in which you mesh matters Elements Without midside nodes rst then those with midside nodes Element Connectivity 4 numu Ilm h lnn two men 39 cimmm Hm I cm in do but lmc m 1 u n thoughquot mdu whi o4v c qu mlrl mun hue Helm h Mt He ll m xim submuan m in t 39 mkwm Itrt c walect l the mud etemcnh Ital on Element Shape Warnings V V Element aspect ratio Essentially ratio of lengths of element edges longest shortest Can be complicated see AN SYS help for discussion Element edge angle Angle between adjacent edges 2D or faces 3D ANSYS wamings during solution Warning if aspect ratio is greater than 20 Warning if interior angle is greater than 1500 Note that warnings do not necessarily mean bad results Check using siml le analJ ses Example 7 cantilever beam with end load Homework problem 7 increasing aspect ratio with Q4 elements Shear locking occurs 7 AN SYS warns when aspect ratio exceeds 20 Same warning if we use Q8 elements Is there a problem in this case Element Aspect Ratio V Cantilever beam problem from homework Solve using Q4 7 compare to exact case shear locking Repeat using Q8 elements 7 very accurate results Lecklng with 045 Aspect v 14 v Exam Error Y us o o a a o 6 2 o n o E u c r5 9 2 D 496 u 6 b6393 39 343 7996MB 3974203 000 2 3176mm SHEEDUG 000 run in 7 JIFGM noon Fun annm Chuck 11 late mm NOTE M41 CF 3783 NE 135725 Present tint B is has then DI equal tn the vrcvious tine fine quotin Ilri null lu z 4 ll RNING N 3983 HE 13525 d rlm KP Qumdrilatma Element 1 In an up ratio at 35 lllich Axum g Imninu Unit 05 28 Point Loads in FEA gt Consider a point load P applied to a model assume 2D 0 Distribute load over height h 0 0 Equivalent pressure p Ph 0 Along line 6x p equilibrium r I lt z r h r gt Cons1der l1m1t case 3 0 Let h a 0 2 point load l 3 Pressurep oo 0 Stress 6x a 00 Infinite stress theoretically exists under a point load Same result for 3D analysis gt Finite element model 0 Stress can never go to in nity except for certain special elements not discussed 0 Instead stresses increase with mesh refinement 0 Consider a 2D example to demonstrate this effect 1quot Point Loads in FEA gt Consider following distributed load over 2 elements 0 Determine associated workequivalent nodal load 0 cu m 01 g v 3 PM s L w Leaked quot gt Finite element model stress at point load is G 2 P L 0 At a point load stress increases with re nement 0 Re ne mesh Lelement 0 means 6 a so gt Lesson for nite element modeling 0 Think about how load is introduced into structure 0 Distribute load appropriately over structure 0 Do not typically introduce a large load at a point element Stress Concentrations gt Stress concentrations 0 Items such as holes notches and changes in dimension 0 Increased stress relative to farfield stress values gt Concern 7 model with sharp reentrant comers 0 Stress concentration is theoretically infinite FEA 7 refine mesh smaller and smaller and stress 7 oo 5N nb u Lb Fmme ewm 47 L7Xign3mgv 7 77 ii 523 E 7e a m 32mm W N a g Symm Symm gt Important lessons 0 Re ect actual shape if concerned with stresses at a particular location 0 Refine mesh in regions of high stress levels esp at a point 0 If stresses go up and up likely an infinite stress concetration point Flexible Boundary Conditions gt Boundary conditions so far have been in nitely stiff 0 Example 7 UY 0 means node cannot move regardless of force gt Often useful to have exible boundary conditions 0 Re ect stiffness of underlying structure 0 Can accomplish using springs or other exible elemenm 2A 2k A mmmmnnem gt Spr1ng supports 0 Place springs over desired width Stiffness to match support structure 5 0 Preferred approach is to not use a 4k 2 midside nodes for such supports
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