COMPUTER AIDED DESIGN
COMPUTER AIDED DESIGN MECH 403
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Auxiliary View Sketching Assignment Mech 403 Rice University Often a part has planar features that do not lie in one of the standard viewing planes To be able to dimension such features in true shape it is necessary to provide auxiliary views that show the part from a line of sight which is perpendicular to the auxiliary surface C J Inclined plane D lt Inclined plane A39 B Edge View of A B inclined plane D C D C 9 A 0 I I New 0 55 u nt 7 39g A a i V a A ltr 0m v 1eW DC D C Page 1 of 3 Auxiliary View Sketching Assignment Mech 403 Rice University In manual sketching true shape features in the auxiliary view can be projected back to the standard views to locate the corresponding points in the nontrue shape display Auxiliary views usually show only the true shape features in the auxiliary plane Features behind that plane are usually omitted f included the view is called a full auxiliary view below They are usually cluttered so most auxiliary views are just partial auxiliary views or details that contain dimensions to features cot available elsewhere in true shape Page 2 of 3 Auxiliary View Sketching Assignment Mech 403 Rice University Full Auxiliary Partial Auxiliaiy For the part shown below prepare a hand sketch of three standard views and one auxiliary view of the inclined bottom surface feature Include dimensions only on the partial auxiliary view to save time Page 3 of 3 Crossing Pipes Discussion Draft 2 92106 Chapter 7 of 4 outlines a steady state thermal analysis of an assumed pipeline junction Here alternate points of view and additional postprocessing features will be presented The rst difference is to recognize that the geometry material properties and boundary conditions have a plane of symmetry Therefore a half model can be employed That lets the full mesh be ef ciently applied to nonredundant results The half model is seen in Figure l where each pipe hot end surface has been color coded light red the original interior is green and the material eXposed on the symmetry plane is in yellow This is one of those problems where the temperature solution depends only on the geometry and is independent of the material used The actual material is brass and its thermal properties from the material library les are listed in Figure 2 Of course the heat uX vectors and the thermal reactions will always depend on the thermal conductivity k Figure 1 Us Mm 94455M aquothybrmutt w on wk lm Coppe SIGYLD Yield strength 2SBBBSBDUB meAZ cammercialamr ALPX Thermal expansion cos 139005 JKelvin i la Copper Kgtlt Thermal conductivity 110 WmKLS 1 I3 Copper CohaIt B C Spectflcheat 390 Ja kgKj quot r6 0 39quotquotvquotquotv 39VA 39 mmrw quotV smpacawvwmvmxgmw o 39 39 vz w ramsx Chromium 00pm SIGYLD Yield strength 3476395 psi 2 39 a Commercial er ALP X Thermal expansion DUE 19005 Fahrenheit i a Copper KX Thermal conductivity 0001412225 ElTUfinsFJ 39uF3tCnnnerCnhaltB x CW S1i2iiqlegt 99331 113t2mEMQREL r Figure 2 Most experimental properties are known only to two or three signi cant gures However the library table values can be misleading because the properties sometimes were measured in a different set of units and multiplied by a conversion factor and incorrectly displayed to 7 or 8 signi cant gures Here the conductivity of brass is displayed Figure 2 top in its experimentally measured units as k 110 WmK but had you used English units it would be converted and displayed Figure 2 bottom to 8 signi cant gures instead of a realistic value of about k 147e3 BTUinsF Page 1 of 10 Copyright J E Akin All rights reserved Thispipejunctionhasfour f 39 oi 39 conditions I 39 assigning it a given value of4oo c The other three ends are treated in asimilar my The first quotFilre3le f 39 L L L 39 been replaced with more meaningful ones That practice often saves time later when aproblem has to be reviewed lnlual temperature mltaem lt3 I 5 Temperature QC a QASludy 1 rDefaullr sold a M LoadRestraint I7 Show review 39 p R m I Design Scenario E contactGaps Global r Figures The boundary condition L quot r39 39 f the 39 39 39 39 r39 f 39 39 That is there is no heat ow normal to the plane qw 2 a That is the natural boundary condition in afinite element 39 39 quot 39 quot satisfied quotL 39 39 quot 39 39 sneci call addressed lyeithei39 39 39 39 39 a mine They both 39 A f 39 39 39 A 39 39 39 and at p nan 39 39 39 39 conditions there will be considered later in the appendix All that remains is to generate amesh mpute the temperatures and postprocess them e automatic mesh generator does a good job When the solution is run the default temperature plot in Figure 4 shows two hot pipe regions and two cooler ones However in the author s opinion the sc s t r ereforeyin s The newinnerand outer e it a little easierto check that the J t plauc ruiu quotcan u I39 Figure 5 the plot settings are changed to first show discrete colorband surface temperature displays are seen in Figure 6 They m temperatures uuuiei 39 39 L 39 C Figure 7 You can 39 39 39 39 or com 39 39 39 39 or contours lines displayed in eithermode pagezoflo Copyright JE Akin Allrights reserved Temp Celsius 4 DUEP002 3 739002 3 47392002 3 2094002 2 3324002 2 5794002 2 405002 2 1354002 1 Weill 1 BDEHJUE 1 3324002 1 EVE002 30024001 l Temp Celsius B DUE001 Figure 4 Figure 5 1 e 5 Edit De nition Animatequot Section Clipping Iso Capping Chart Options W ya u r g 5 a VA g i 1 A AW AV A 1 v A 5 VA VA A AV 5 7 VA l Mendel name Cvussedplpes Sludy name Sludyl Meshlype Sum mesh Temp Celsius 11 00214102 3 73mm 3 4e002 3 20202 2 9324032 2 B7E002 2 4024002 2 138002 1 0724002 1 6094002 1 EBB002 1 0724002 0 0094001 l Figure 6 Page 3 of 10 Copyright 1E Akin All rights reserved It is common to make technical reports more speci c by graphing the results along selected lines or curves That is done by utilizing the List Selected option a er the temperature has been plotted Then you select an edge of model split line pick update and then plot The selected line at an inner edge adjacent to the hottest inlet and seen previously in Figure 6 is seen in Figure 8 The resulting temperature gmph versus the nondimensional position along the line is given on the le in Figure 9 While its continuation along the curved cylinder intersection cmve is shown on the right Since there is no convection heat genemtion or nonzero heat ux conditions the tempemture results and contours are the same for all materials However the heat ux does change magnitude With different materials Temp Celsius 4 muzmuz E Buyout Figure 3 Page 4 of 10 Copyright JE Akin All rights reserved Study name Study 1 Study name Study 1 Plot type Thermal Plan le type Thermal Plan Temp Celsms Temp Celslus 020 040 060 080 l00 020 040 060 080 100 Parametrlestance Pavamemc Distance Figure 9 The heat ux magnitude contours are given in Figure 10 left The contour lines are less smooth than the previous ones for the temperature because the heat ux is always less accurate than e temperatures Had these contours shown larger wiggles it would be a signal that a ner mesh should be utilized Since the heat ux is a vector quantity it should also be plotted as a vector That setting change is also shown on the right in Figure 11 The resulting vector plot is shown in Figure 11 a er changes with Vector Plot Options The heat ux vectors show that heat ows into the junction at the 400 C surface and out at the 80 and 100 C ends The 250 C end has some in ow and some heat out ow Figure 10 HFquN Wm 2 5 2454005 Mums E WmAZ 595mm E Fringe v 522240113 a L Vector er WWWm 5 7321005 5 215005 4 7024005 41924005 a E72Uu5 3 1531005 7 2 5554005 I 2 1324005 l 5294305 I HFLUXN1Raultant hear flu v Page 5 of 10 Copyright JE Akin All rights reserved HFFUXN Wmquot2 62424005 57324005 5 2124005 4 7024005 4 1324005 35724005 31624005 25524005 2 1324005 1 5224005 1 1124005 5 5524004 32221003 Figure 11 Here on can 39 L 39 r 39 39 39 I ain speci ed temperatures To do that use the L25 selected option and select each ofthe four pipe ends in order At each one I 39 L list and sum quot L I quot quot 39 mu 39g re 17 39 39 They w ofpower in at the two hottest 39 39 39 rr A 39 m u that ure 8212mm v2 2v2n22 8212m2d121212m22 Summary NA NA 8212mm ttsms 1 Face 821212d ttsms 8212m2d121212m22 NA w 8212mm ttsms 2 aawaso Wm 2 4 unhas wwz 2 007524 wwz 3 035524 wwz 100 C End Figure 12 Page6of10 Copyright JE Akin Allrights reserved needs to be doubled to determine the required input power of about 1700 W to maintain the specified temperature restraints If the current example were changed to steel with a conductivity of about k 519 WmK then the heat ux magnitudes and required power would drop by about a factor of two while the temperature distribution would be unchanged References l Akin JE Finite Element Analysis with Error Estimators ElseVier 2006 Chapman AJ Fundamentals of Heat Transfer Macmillan 1987 Kurowski PM Finite Element Analysis for Design Engineers SAE International 2004 Kurowski PM Engineering Analysis with CosmosWorks Professional 2006 SDC 2006 59 Page 7 of 10 Copyright JE Akin All rights reserved Appendix Adding convection Convection conditions or known heat ow on the boundaries Will change the temperature distribution in the original problem To illustrate this point assume the outer surface convects to air at a temperature of 30 C 303 K With a convection coef cient of about h 5 WmA2 Also let the pipe interiors convect to oil at 70 C 343 K With an assumed convection coef cient of about 600 WmA2 You simply have to apply two convection conditions seen in Figure 13 and recomputed the solution Q Crossedpipes iirarameters allstudv 1 eDefaulte 39 50M I7 Show prewaw a mun ME ShawAI ninsmm3ml39l iquot I7 Show preview gure 13 Page 8 of10 Copyright JE Akin All rights reserved The resulting discrete temperatures are shown in Figure 14 There are steeper temperature gradients than seen in Figure 6 of the original problem Likewise the original graph of temperatures in Figure 9 shows less temperature drop than the new graph in Figure 15 The new heat ux vectors are given in Figure 16 There you note that the location of the maximum heat ux has changed Ifyou again compute the reaction heat ow given in Figure 17 you need 2 2070 Ws 4140 ws Temp Celsius Temp cam 4 We 4 cinema 3 732mm 3 73mm 3 47mm 3 472mm 3 2D2UE2 3 202mm 2 932 2 2 saawuz 2 E72UE2 2 E7E 2 2 4D2 2 2 mama 2 132UE2 2 l ei DZ l E7EUE2 i 57mm 1 ED2UE2 l EUEWZ l 332UE2 1 335mm 1 U72UE2 l U7EVOU2 E unw m a mammal Figure 14 St dy name Study 1 Plot type Thermal Plot l Temp Celsius c c c M c o iuu I 000 020 040 060 080 100 Parametric Distance Figure 15 Page 9 of 10 Copyright JE Akin All rights reserved HFquN WIm39QJ 6349005 5279005 57094005 5 1394005 4559605 40094005 3 439005 2889005 2299005 1 7294005 1189005 58890011 2 019003 Figure 16 Sawamaararararrca Summary Sawamaarararanca Summary Va ue Umts NA Tma H93 VTl 4 W Tma H93 3 W Saxamaarararanca Summary Saxamaarararanaa Summary NA Figure 17 Page 10 of10 Copyright JE Akin Allrights reserved ANSYS Tutorial Release 9 L Lawren oe Kent Mechamca and Aemspaoe Engmeermg Unwersny of Texas atArhngmn Schro Deve opmenl Cowman on wwwschrn cnm wwwschm euruvecum Plane Stress Plane Strain Lesson 2 Plane Stress Plane Strain lrl OVERVIEW Plane stress anol plane stxam problems are an rmportant subelass of general threer dlmenslonal problems The tutonals ln thrs lesson olemonstrate o Solvlng planar stress eoneentraaon problems oEyaluaang potenaal rnaeeuraeres ln the solutrons oUsrng the yanous ANSYS 2D element formulatlons 24 INTRODUCTION It ls posslble for an object s ueh as the one e m nts of stress when subjected to arbltxary threerdlmenslonal loadln s referencedto a Cartesran coordlnate system these eomponents of stress are on the eoyer of thrs book to have slx g When Normal Stresses ax try 392 shear Stresses try ryz fzr Figure lrl Stresses m 3 dlmenslons Ingeneral b l b A 4 ln Lesson 4 However Wordlmenslonal moolels are often easrer to develop easrer to solve anol ean be employe ln many srtuaaons if they ean aeeurately represent the behayror ofthe object unoler loadlng L2 Plane Stress Plane Strain A state of Plane Stress exists in a thin object loaded in the plane of its largest dimensions Let theXYplane be the plane of analysis The nonzero stresses trx Ty 1W lie in the XY plane and do not vary in the z direction Further the other stresses and 61192 and Tn are all zero for this kind of geometry and loading A thin beam loaded 39 39 r 4 put ear ANSYS provides a anode planar triangular element along with 4node and 8node quadrilateral elements for use in the development of plane stress models We will use L3 PLA39IE WITH CENTRAL HOLE To start off let s solve a problem with a known solution so that we can check our m0 loaded thin plate with a central hole as shown in Figure 22 70w Figure L2 Plate with central hole Thel 39 antler al 39 39 It is made of steel with material properties elastic modulus E 207 x 10 Nm2 and Poisson s ratio y 029 We apply a horizontal tensile loading in the form ofa pressure p r o Nm2 along the vertical edges ofthe plate Because holes are necessary for fasteners such as bolts rivets etc the need to lmow stresses and deformations near them occurs very o en and has received a great deal of study The results of these studies are widely published and we can look up the stress ethodsy u rr r 1 enen mentall L A 39 A Plane Stress Plane Strain zes The unlfol39m homogeneous plate above ts symmetne about hortzontal axes tn both geome and loadlng Thts means that the state of stress and deformatlon belovv a honzontal eenterltne ts a mtrror tmage of that above the eenterltne and llkewlse f0 We ean take advantage of the symmetry and by applytng the eoneet boundary eondtttons use only a quarter of the plate for the ftntte element model For save plate potnts on the eenterltnes wlll move along the eenterltnes but not perpendlcular to them Thts tndteates the approprtate dtsplaeement eondtuons to use as shown belovv A A A A Figure 273 Quadrant used for analysts 1n Tutortal 2A vve wlll use ANSYS to determtne the maxlmum honzontal stress tn the plate and eompare the eomputed results vvtth the maxlmum value that ean be ealeulated be used to formulate and solve the problem 274 TUTORIAL 2A 7 PLATE Ohieetive Flnd the maximum axial seress tn the plate vvtth a eentral hole and eompare PREPROCESING 1 1quot NW 39 39 you vvtth thts problem Also set the Jnhname to Tutnriale or somethtng memorable and provtde a Title Ifyou vvantto make ehanges tn the Jobname Worklng Dtreetory or Tttle atteryou39ve started ANSYS use File gt Change Jnhname or Direetnry or Title Select the x node tnangular elementto use for the soluuon ofthts problem L4 Plane Stress Plane Strain Figure L4 Sixnode triangle enu gt Preprocessor gt Elunen 2 Main M Type gt AddEdilDelele gtAdd gt Structural Solid gt Triange 6 node 2 gt 0K rary n1 Elzmznt mm Llhvaw hr Element was Shell anstvalnt HYDEVelastl rnanqle Shade 2 Element tvve Vefevence numhev 4 0K nut mt Help Figure L5 Element selection Select the option where you de ne the plate thickness 3 Options Element behavior K3 gt Plane strs wthk gt OK gt Close Plane Stress Plane Strain mm element tyne ununns ovum fav mun E amer me up Nu Elnmant behavan 4 m Extra stress WINK K5 ND extra 011le V Eelmnt nutvut K6 Bas eleme l 39 OK and Heb Figure 245 Elanent opuons 4 Main Menu gt Prepmcessm gt Real Cunstants gt AddEditDelem gt Add gt 0K Iemennyperumgj 2 Du md ReaKnnsxam in Most ahmml ty am up 0K m Figure 277 Real Constants Ema the plate thlclmess of 0 01 m gtEnta 001 gt OK gt Class Rea mm rm mane Stvess Mm mews szopvaa mm 55 P lum OK W cm Hep Figure Ls Ema the plane uuckness Enter the makenal propemes zrs Plane Stress Plane Strain 5 Main Menu gt Preprncessnr gt Material Prnps gt Material Mnclels Matenal Model Number 1 Double cllck Structural gt Linear gt Elastic gt Isntrnpic T Fm M V n lM d l Behavlor Window create the geometry for the upper nglnt quadrant of the plate by subtraeung a 0 2 rn drarneter errele from a 0 5 x 0 2 rn rectangle Generate the rectangle rst 5 Main Menu gt Preprncessnr gt Mncleling gt Create gt Area gt Rectangle gt By 2 Cnrners Enter lower lelt comer WP X 00 WP Y 00 and Width 05 Height 02 gt 0K 7 Main Menu gt Preprncessnr gt Mmleling gt Create gt Areas gt Circle gt Snlid Circle Enter WPX 00 WP Y 00 and Radius 01 gt 0K errauqle by 2 51 a re r r npick me a Pick r quotnpick u e we x 7 Clubs x e u e u e Glnha x e z e u e MP X 39u 2 width n5 we r in Height M nauins 01 r ox apply on apply Reset Cancel Reset Cancel Help Help Figure 23 create areas Plane Stress Plane Strain Figure 240 Feetangle and clrcle bottom of the sereen as neeessary a Main Menu gt Freprneeeenr gt Nindeling gt Operate gt Bnnleane gt Subtract gt Area gt Plck the reetangle gt OK then plck the cliche gt 0K Figure zen Geometry for quadrant of plate create amesh oftxlangular elements over the quadrant area 9 Main Menu gt Freprneeeenr gt Meshing gt Mesh gt Area gt Free Pick the quadrant gt 0K NINA Figure 242 Tn angular elernent mesh WV V 39 m1 Apply the deplaeernentboundary eondtlons andloads 10 Main Menu gt Freprneeeenr gt Land gt De ne ands gt Apply gt Structural gt Dieplaeernentgt 0n Lineeplek the lett edge ofthe quadrant gt OK gt ux o gt 0K zes Plane Stress 11ane Strain 11 Marn Menu gt Preprneessnr gt Land gt De ne ands gt Apply gt Struenrra1 gt D1sp1aeernentgt 0n Linesprek the bottom edge ofthe quadrant gt OK gt 11v o gt 0K 121VIain Menu gt Preprneessnr gt Land gt De ne ands gt Apply gt Struenrra1 gt Pressure gt On Lines Trek the rrght edge ofthe quadrant gt OK gt Pressure 10 gt 0K The 39emuwy 15 y again Figure 213 Model wrth loadmg ano1 dASpIacement boundary condmons pressure 15 shown as a smgle arrow A Th A1 p rtrtt be safe save the moo1e1 13 Utility Menu gt File gt Save as Jnhnannenh or Save as use a new narne soumm The rnteraetrve so1utron proeeeds as 1llustxated1n the tutona1s ofLesson 1 141v1a1n Menu gt Snlntinn gt Salve gt Current LS gt 0K The STATUS Cnmmand wmdow duplays the prob1ern parameters and the Salve Current Luz Step wmdow 15 shown Check the so1uuon optrons m the STATUS wmdow and1f an 15 OK se1eetF11egt c1nse 5e1eet 0K elnse the Snllltinn isDnne wmdow POSTPROCESING Fr 1 examme the dernrnned shape 15nr M Undef gt 0K Plane Stress Plane Strain 279 msmcmm s x V a 7 AAAAAAAA A A A A Figure 244 Plot ofDeformed shape The deformed shape looks correct The undeformed shape is mddcated by the dashed lmes d n 39 Note that the element edges on the erreular are are represented by straight hnes Thxs is an amfact wd F u d you wru edge T mum ddsplacement is shown on the graph legend as 032ee11 whxch seems reasonable The umts of ddsplacement are matters be ause we employedmeters and Nrn2 m the problem fonnulaaon Now plot the stress m thederecnon 15 M M r r Stressgt Xedireeu39nn SX gt 0K pre and canvas as set to Arrnws to dsplay the pressure 1oads 2710 Plane Stress 11ane Strain ELEMENT SDLU39KIDquot STEPJ muAth azuxrii 3na7s seem 775775 1 ese 2 557 A m5 23555 m 2 m 3 5m A 55 Figure 2715 Element 5X stresses The minimum SMN and maximum SMX stresses as well as the color bar legend give overall evaluation of the sx stress state We are interested in the maximum stress at the hole Use the Znnm to focus on the area Wth highest stress Figure 2716 SX stress detail Plane Stress Plane Strain zen Stress yanatrons ln the aetual rsotropre homogeneous plate should be smooth and that the number of elements used ln thrs model ls too few to aeeurately ealeulate the st es yalues near the hole beeause of the stress gradlents there We will nnt accept thrs stress solutron More slxrnode elements are needed ln the reglon near the hole to flnd ues of he stress n the otherhand ln the rrght half of the model away from the stress rrser the ealeulated stress eontours are smooth and SK would seem to be aeeurately determrned there ln lo 1 t l nu l u contour all mwum TF um It for problems llke the one ln thrs tutonal the stress values wlll be averaged before plottrng and any eontour dlscontlnultles and thus errors wlll be hldden If you plot nodal solutron stresses you wlll always see smooth eontours dlrecdy from solld meehanres studses and there ls no apprommatron ln the solutrons for truss struetures formulated and solved ln the ways dlscussedln Lesson 1 The cnntinlulm elements sueh as the ones for plane stress and plane stratn on the other hand are normally deyeloped uslng dssplaeement funetrons of a polynomlal type to m r V r the aeeuraey The ANSYS sun ode trrangle uses a quadratre polynomlal andls eapable ofrepresenung lrnear stress and strarn yanatrons wrthrn an element Near stress eoneentratrons the stress gradlents yary qurte sharply To eapture thrs yanatron the number of elements near the stress eoneentratrons must be rnereased proportronately 17 Main Menu gt Preprncessnr gt Meshing gt Mullin Mesh gt Re ne At gt All Select ll V VAVAV fe exm tm gt A VABAVXQEXAVAWAVAV Figure 247 Global mesh re nement We wlll also re ne the mesh seleetlvely nearthe hole zelz Plane Stress 11ane Strain Pr prneeeenr gt Meshing gt Mudin Mesh gt Re ne At gt Nudes Selectthe unree nodes shown gt 0K Seleenne Level nrre nernem 1 gt 0K VAVAVAVAVA 4quot V un l l we VAV VAVAV A W 1 AAV gt1 VA A Figure zels Seleeuve re nement at nodes Note agam Nowrepeatthe solunon andreplot the stress sx 191VIain Menu gt Snllltinn gt Salve gt Current Ls gt 0K Stressgt Xedireeu39nn gt SX gt 0K Plane stress Plane Strain VAVAV AVAVAVAVAV V b yquot AVMVQQV 39 AV Maw rs a m W Fidmv y4n quot W E EE AV gt 4 AVE VAVAVAVAVAVAYAVA 1523 2545 3573 3 as 4 ans 7225253 795527 2335 mm 13 Figure 719 SX stress contour after mesh re nement Figure 720 SX stress detatl contour after mesh re nement smo and NW stress legend shows amaxlmum value of4 38Pa T eck Lhts result nd the stress Concentration factor for Lhts problem In a text or reference boo or horn a web stte such as www etbx corn For the geometry of thts 2714 Plane Stress Plane Strain example we nd K 217 We can compute the maximum stress using Kloadnet cross sectional area Using the pressure p 10 Pa we obtain QM 217p0400104702001 434Pa The computed maximum value is 433 Pu which is less than one per cent in error assuming that the value of1ltis exact The result from the rst mesh ms 58 in error 275 THE APPROXIMATE NATURE OF FEM As mentioned above the stiffness matrix for the truss elements of Lesson 1 can 39 d be developed directly and simply from element soli mechanics principles For continuum problems in two and threedimensional stress this is generally no longer m ihle AL 39 39 39 quot dev pedby 39 quot 39 39 39 39 within an element Ordinarily this is done by spec39 ying the highest degree ofthe polynomial that governs the displacement distribution with39 an ele For hemahod elements the polynomial at the nodes are called shape runcu39ons In ANSYS modeled with sixenode triangles rourenode quadrilaterals or eighlrnode quadrilaterals Figure 2721 Triangular and quadrilateral elements The greater the number ofnodes the higher the order ofthe polynomial and the greater the accuracy in describing displacements stresses and strains within the element If the stress is constant throughout aregion avery simple model is suf cient to describe the stress state perhaps only one or two elements If there are gradients in the stress distributions within a region highdegree displacement polynomials andor many elements are required to accurately analyze the situation Plane Stress Plane Strain 2r15 of elements were used to solve the problem m the prevlous tutorral and why the englneer must carefully pre are a model start wth small models grow the models as understandlng of the problem develops and carefully rnterpret the ealeulated results The ease wth whreh models ean be prepared and solved sometrmes leads to eareless eyaluatron othe eomputedresults zrs ANSYS GEOMIETRY and ls separate from the th flnlte element model el the truss examples of The flnlte element model eonsrsts of elements and nodes eome on eh rt may be based 1t ls posslble to bulld e wrthout eonsrderatron of any underlymg geometry as was don n Lesson 1 but m many eases development of the geometry ls the rst task Twordlmenslonal geometry m ANSYS ls burlt from keypnlnts lines stratght ares spllnes and areas These geometne rtems are asslgned numbers and ean be lrsted numbered manrpulated and plotted The keypornts 23456 llnes 2352210 and area 3 forTutorral 2A are shown below Figure 242 Keypornts llnes and areas The flnlte element model developed prevlously for thrs part used the area A3 for deyelopme t the eelement FEM mesh The loads dlsplacement boun condluons and pressures were applredto the geometry llnes en e solutron step was exeeuted the loa w re transferred from the llnes to the FEM model nodes Applylng boundary eondrtrons and loads to e geometry facllltates remeshlng the problem The geometry does not ehange only the number and loeatron of nodes and elements and at solutron trme the loads are transferredto the new mesh ean be rt ean be ereated by readmg a text le For example the geometry of Tutorral 2A ean be mg text flle uslng the File gt Read Input rrnm eommand those sho generated wn above wth the follow sequence Thekeyporntlrne ete numbers wlll be dlfferentfrom 2716 Plane Stress 1lane Strain FILNAHGeum t1t1e Stress cententretmn Geometry 1 Exam 1e of ereetmg geometry ue1ng keypmnte 11nee area 1 xeypmnt 1 s at 00 00 1 L1ne from keypumte 2 to 3 1 ere rren keynmnt 2 tn 5 eenter kp 1 radlus e1 LARC 2 s 1 e1 AL 1 2 3 4 5 1 Area deilned by 111125 12a 15 Geometry for FEM analysls also tan be treated wrth solld modelmg CAD or other H neutral le ls a eommon format usedto exchange geometry between eomputer programs Tutorral 2 demonstrates ths opaon forANSYS geometry development 277 TUTORIAL zBe SEATBELT COMIFON39ENT Ohieetjye Determme the stresses and deformatlon of the prototype seatbelt eomponent shown m the gure below rfrt ls subjected to tensrle load of 1000 lhf F1 gure zezs Seatbelt eomponent steel has an over all length of about 2 5 mehes andls part was developed m a CAD system orted mto ANSYS or analysls For a The seatbelt eomponent ls made of 332 0 09375 mehes thck A solld model of the and exported as an GES le The le ls lmp p or torrgu Plane Stress Plane Strain 2717 Figure 224 Seatbelt tongue39 PREPROCESING 1 Start ANSYS Rnn Interactive set inhnome and working directnry create the top half of the geometry above The latth retention slot 15 0 375 x 0 8125 inthes andls lotated 0 375 inth from the nght edge Kyou are not using an IGES me to define the geometry for this enertise you can treate the geometry directly in ANSYS Wth key points lines arcs by seietong File gt Read Input from to read in the text the given below and skipping the IGES import steps 2 3 FILNAHSEarhEJr t1t1e seetheit Geometry 1 Example of treoting geometry using kevpoints iines orts 1 creote geometry ir 1 ou oo KEYFDIUE 1 is ot ou oo ir 2 o75 oo ir a 1125 oo ir 1 15 oo ir 5 1 5 o5 ir s 1 25 o75 ir 7 o u o75 ir a 1 125 o 375 ir 9 1 c9375 o aoszs ir 1n n o125 aoszs ir 1 o 75 375 ir 1 125 o 5 ir zels Plane Stress 11ane Strain t Llne from keypemee 1 ee 2 u 2 t at from keypeme 5 ee 5 center kp 12 radlus n25 eee 5 n25 mac s12 mac a 9 13 uu3125 mac m 11 14 uu525 ALyall t Use all 111125 ee create ehe area LAltemzu39v e1yuse z snlid mndeler tn create the tap half nf the enmpnhent Shawn zhnve in the XX plane and expm an IGES me nf the part To Import the IGES le 3 Utility Menu gt File gt Imp m gt IGES Seleet the IGES le you created earher Aeeept the ANSYS import default settmgs If you have trouble Wth the Import seleet the a1temate optmhs and try agam Defezulring t n process to remove mechststehetes that may extst m the IGES me for example lmes that because of the modelmg or the le translation process do not qutte Josh mum mas r nuxxs10pmouttansrmxeisxmpmt 1925 mm ohm r Na eeteemtne r oereememaeet MERGE Mevqe mnndenl k2vvt57 17 V25 50on vests sahd tr awhiah e 17 V25 SMALL oeteze smaH 512557 17 V25 0K mt Met Figure 245 IGES Import plane Plane stress Plane Strain 2719 4 Utility Mun gt PlotCtrls gt Pan Zoorn Rotate gt Back or use the sidebar icon Figure 2726 Seatbelt solid front and back 5 Main Meml gt Preprocessu gt Element Type gt AddEditmdete gt Add gt Solid gt Quad snode 133 gt 0K Use the Ssnode qusdnlstenrl element for thrs problem 6 Options gt Plane strs wthkgt OK gt Close Enter the threkness 7 M enn gt Preprocessor gt Real Constants gt AddE itDelete gt Add gt Type 1 Plane 183 gt OK gtEnter 009375 gt OK gt Close Enter the material plopertles 8 Main Menu gt Preprocessu gt Material Props gt Material Models Mitten n Modpl Nuth I 39 39 r39 in I n 7 Model Rehnvlot wlndow Now mesh the xy plane area m area numbers on mt helps 9 Main Menu gt Preprocessor gt Meshing gt Mesh gt Areas gt Free Rck the XVY planar area gt OK Important note The mesh below was developed from an IGES geometry le Using the If o use the Modify Mesh re nement tools to ohtrnn s mesh density whreh produces results wrth accuracies zezo Plane Stress 1lane Strain Figure 247 Quad 8 mesh The IGES solld rnodel ls not needed any longer and slnce lts llnes and areas may lnterfere Wlth subsequent rnodehng operaaons delete ltfrorn the sesslon 10 Main Menu gt Preprucessur gt Mudellnggt Deletegt Vulume and Beluw Don39t be 11 Utility Menu gtPlntCtrls gt Fan Znnnn Rntate gt Frnnt If necessary to see the front slde ofnnesh Figure 248 Mesh front vlew W left edge and zero UY along bottonn edge enu gt Preprncessnr gt Land gt De ne ands gt Apply gt Structural gt Displacernentgt On Line Plck the left edge gt x 0 gt 0K enu gt Preprncessnr gt Land gt De ne ands gt Apply gt Structural gt Displacernentgt On Line Plck the lower edge gt UY o gt 0K The vemcal lnslde edge of the lateh retentlon slot 10001bf0 09375 ln n 0 75 n 1 enu gt Preprncessnr gt Land gt De ne ands gt Apply gt Structural gt Pressure gt On Line Plane stress 1gtlane strain 721 Select the lnslde line and setpressure 14000 gt 0K Figure 729 Applled displacement and pressure Condlhons Solve the equatlons SOLUTION 15 Main MEnugt Sulutiun gt Salve gt Current LS gt OK POSTPROCESSING vun Mises stress wth the material yleld stress ls an accepted way of Comparlng the evaluating yleldlng for ductile memls com lned stress state so we enter the postproeessor and plot the element solution ofvoh Mlses stress SEQV 16 39 r Stress gt scroll down vim Mises gt SEQV gt OK oom In on the small fillet where the maxlmum stresses occur The element solutroh stress contours are reasonably smooth and the maxlmum voh Mlses stress ls around 113000 psi Further mesh refinement glv es a stress value a llLLle over 120 000 psi Figure 2430 Von MlSeS stresses Plane Stress 1lane Strain uh u at If your model has a 12111 radius nvteh your flnlterslze elements wlll show a very hlgh lnflnlty The flnlte elerheht teehruque heeessanly desehbes flmte quanuues and earth t dlrecdy treat an lnflnlt stress at a sin ar paint so out chase a srhgularrty39 Kyou do not eare what happens at the hoteh statre load dueule materral ete do not worry about th5 loeatroh but look at the other reglons If you really are eoheerhed about the maxrrhurh stress here fatlgue loads or buttle matehal then use the aetual part hoteh radlus however small 132 for ths tutohal do sure the mesh ls sumerehtly re ned hear the hoteh If a eraek up ls the object of the you should look at fracture rheeharues approaehes to the problem See AN h analys s help topres on fracture rhee arues The engrheer39s respohsrbrhty ls not only to bulld useful models but also to rhterpret the results of sueh rhodels rh rhtelhgeht and meanlngful ways Thls ear often get overlooked m the rush to get answers 17 M M t Strainrtntalgt 1n prin gt 0K m M M t r DOF snlutinn gt Translatinn UXgt 0K Figure 231 UK drsplaeerhehts The maxrrhurh de eetroh m the erdrectlon ls about 000145 inches and oeeurs as expeeted at the eehter ofthe nghtrhand edge ofthe lateh retehuoh slot Plane Stress Plane Strain zrzs zrs MAPPED MExiUNG Quadniaterai rnesnes ean also be ereated by mapping a square With a regular array of eeus onto a general quadrriaterai or triangular region To illustrate nus delete the last une uueu mm the textme above so that e areais nut created justthe lines and readrtrnto ANSYS Use FlntCtrlsto turn Keypuint Numbering 0n Then use 1 Main Menu gt Preprueeeeur gt Mudeiing gt Create gt Line gt Line gt Straight Line Figure 242 Lines added to geometry 2 Main Menu gt Preprueeeeur gt Mudeiing gt Create gt Area gt Arbitrary gt By Lines quadniaterai areas gt Applygt 0K Figure 243 Quadniaterairn angular regions 3 Main Menu gt Preprueeeeur gt Mudeiing gt Operate gt Buuieane gt Glue gt Area gt Pick Au meshing 2724 Plane Stress 1lane Strain 4 IVIain Menu gt Preprueessur gt Meshing gt She Cntrls gt Manualslxe gt Lines gt All Lines Enter 4 far N39DIV Nn element divisinns gt OK All hnes Wlll be dlvrdedmto four segments formesh ereatdon d ues LESIZE Element slzes an chked hnes 5sz Elementedqelenqth NDW ND af element dlvlslans NDW ts used duly t 5sz ts hlankav 12m mow sleNDlv an be hanged specs Seerud tend ANGle Dlvlslan av deqveei H W use ANGle duly t numhev af dlvlslani NDW and element edde lendm sle ave hlankav 12m deal attached 51255 and valumes 1 ND 0K mt ester net Figure 2734 Element slze on preked hnes 51VIain Menu gt Preprneessnr gt Element Type gt AddEdiUDelete gt Add gt Send gt Quad annde 183 gt 0K Use the Ernode quadnlateral element for the mesh 61VIainM2nugt Preprneessnr gt Meshinggt Meshgt Areas gt IVLapped gt 3 An 4 sided gt Pick All nnl w h lmd rv v M r t poor mesh qushty Notree the long and narrow quads near the pornt of maxlmum stress We need more elements and they need to be better shaped wth smaller aspeet miles to obtam satlsfactory results Plane stress Plane strain 2725 Figure 735 Mapped mesh and Von Mlses results whlch llnes Thls allows much better control over the quallty of the mesh and an ecample ofuslng ths approach 15 desmbed m Lesson4 79 CONVERGENCE The goal of flnlte element analysls as dlscussed m thls lesson 15 to amve at computed enough terms are summed occur These constralnts are relaxed as the element polynomlal IS lncreased or as more elements are used Thus your computed dlsplacements should converge smoothly from below to flxedvalues stralns are the x andor y denvatwes of the dlsplacements and thus depend on the m an enatrc way as the mesh 15 re ned rst smaller than the ultrmate computed values than larger etc Not all elements are developed uslng the ldeas dlscussed above and some wlll glve See Lesson 6 Lo Computed Al M mail and Varlatlons as quot the solutlon ofa problem 210 TWOVDHVJJENSIONAL ELEMENT OPTIONS The analysls optrons for Wordlmenslonal elements are Plane Stress Axisymmetric Plan train 39 39 L p mun l u othe part 2 26 Plane Stress Plane Strain The first analysis option Plane Stress is the ANSYS default and provides an analysis for a part with unit thickness If you are working on a design problem in which the thickness is not yet known you may wish to use this option and then select the thickness based upon the stress strain and de ection distributions found for a unit thickness The second option Axisymmetric analysis is covered in detail in Lesson 3 Plane Strain occurs in a problem such as a cylindrical roller bearing caged against axial motion and uniformly loaded in a direction normal to the cylindrical surface Because there is no axial motion there is no axial strain Each slice through the cylinder behaves like every other and the problem can be conveniently analyzed with a planar model Another plane strain example is that of a long retaining wall restrained at each end and loaded uniformly by soil pressure on one or both faces 2 11 SUMMARY Problems of stress concentration in plates subject to inplane loadings were used to illustrate ANSYS analysis of plane stress problems Free triangular and quadrilateral element meshes were developed and analyzed Mapped meshing with quads was also presented Similar methods are used for solving problems involving plane strain one only has to choose the appropriate option during element selection The approach is also applicable to axisymmetric geometries as discussed in the next lesson 2 12 PROBLEMS In the problems below use triangular andor quadrilateral elements as desired Triangles may produce more regular shaped element meshes with free meshing The sixnode triangles and eightnode quads can approximate curved surface geometries and when stress gradients are present give much better results than the fournode elements 2 1 Find the maximum stress in the aluminum plate shown below Use tabulated stress concentration factors to independently calculate the maximum stress Compare the two results by determining the percent difference in the two answers 2727 Plane Stress Plane Strain 600 mm 75 mm dwa O H WZKN a ummum p ate 10 mm thwck 300 mm Figure P271 n pun u m the centerlme and top edge as shown now need to model half of the plate instead ofjust one quarter and properly restxam vemcal ngxd body nnouon one way to o this to an one keypomt along the centerlme from UY dupla t 600 mm 0 75 mm dwa E H 5 8 12m 10mmtmck Figure P272 Plane Stress 1lane Strain us 24 An alumlnum square 10 lnches on a slde has a semen drameterlnole at the eenter The object ls ln a s at anl p ssure f 5 lnch dla 1500 psl Determrne tlne magmtuole anol loeatron of Lh r al stress h pnnerpal strarn and he maxlmum yon Mses stress No thlckness ls requlred for plane strarn analysls lo x lo m Figure P23 24 Repeat 2 for a steel plate one men tlnrekrn a state ofplane stress 25 See lf you ean reoluee tlne maxlmum stress for the plate of problem 24 by addlng noles as shown below Seleet a hole slze anol loeatron that you thnk wlll smootln out tlne stxess ow39 eauseol by tlne loaol transmrssron through the plate 600 mm 75mmdla o O o m WZKN alumlnum plate 10 mm hlck 300 mm Figure P25 zes Repeat 24 but the object ls aplate wrtln notelnes or wrtln a step m the geometry Seleet m V rl l quot 1 d tlnat tlnere ls a unlfol39m state of axlal stress at ertner enol relaayely far from noteln or hole create your geometry accordlng y Plane Stress Plane Strain 249 Figure P24 277 Determrne the stresses and de eeuons ln an object at hand39 sueh as a seatbelt tongue or retarmng wall whose geometry and loadln make rt surtable for plane stress or plane strarn analysls Do all th eeessary modellng ofgeometry use a CAD system lf Y e ou wrsh matenals andloadrngs H A eanuleyer beam wth a umt yndth reetangular eross seeuon ls loaded wth a unlform pressure along rts upper surface Model the beam as aproblem ln plane stress Compute the nd de eeuon and the maxlmum stress at the anuleyer support Compare yourresults tothose you would flnd uslng elementary beam theory Figure PM Restxaln UK along the eanuleyer support lrne but restrarn UY at only one keypornt along tlus hne Otherwrse the strarn ln the Y drreeuon due to the Polsson effect ls preyented ereated Try xlng all node pornts ln UK and UY and see whathappens Seleet your own dlmenslons matenals and pressure Try a beam that39s long and slender d one that39s short and thck The effect of shear loadng must be rneluded ln the de eetron analysls as the slenderness deereases 2 30 Plane Stress Plane Strain NOTES