# Introduction to Finite Element Modeling

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1 Introduction to Finit Elmnt Modling Enginring analysis of mchanical systms hav bn addrssd by driving diffrntial quations rlating th variabls of through basic physical principls such as quilibrium, consrvation of nrgy, consrvation of mass, th laws of thrmodynamics, Maxwll's quations and Nwton's laws of motion. Howvr, onc formulatd, solving th rsulting mathmatical modls is oftn impossibl, spcially whn th rsulting modls ar nonlinar partial diffrntial quations. Only vry simpl problms of rgular gomtry such as a rctangular of a circl with th simplst boundary conditions wr tractabl. h finit lmnt mthod (FEM) is th dominant discrtization tchniqu in structural mchanics. h basic concpt in th physical intrprtation of th FEM is th subdivision of th mathmatical modl into disjoint (non-ovrlapping) componnts of simpl gomtry calld finit lmnts or lmnts for short. h rspons of ach lmnt is xprssd in trms of a finit numbr of dgrs of frdom charactrizd as th valu of an unknown function, or functions, at a st of nodal points. h rspons of th mathmatical modl is thn considrd to b approximatd by that of th discrt modl obtaind by conncting or assmbling th collction of all lmnts. h disconnction-assmbly concpt occurs naturally whn xamining many artificial and natural systms. For xampl, it is asy to visualiz an ngin, bridg, building, airplan, or sklton as fabricatd from simplr componnts. Unlik finit diffrnc modls, finit lmnts do not ovrlap in spac. Objctivs of FEM in this Cours Undrstand th fundamntal idas of th FEM Know th bhavior and usag of ach typ of lmnts covrd in this cours B abl to prpar a suitabl FE modl for structural mchanical analysis problms Can intrprt and valuat th quality of th rsults (know th physics of th problms) B awar of th limitations of th FEM (don't misus th FEM - a numrical tool) Finit Elmnt Analysis A typical finit lmnt analysis on a softwar systm rquirs th following information: 1. Nodal point spatial locations (gomtry). Elmnts conncting th nodal points 3. Mass proprtis 4. Boundary conditions or rstraints 5. Loading or forcing function dtails 6. Analysis options

2 Bcaus FEM is a discrtization mthod, th numbr of dgrs of frdom of a FEM modl is ncssarily finit. hy ar collctd in a column vctor calld u. his vctor is gnrally calld th DOF vctor or stat vctor. h trm nodal displacmnt vctor for u is rsrvd to mchanical applications. FEM Solution Procss Procdurs 1. Divid structur into pics (lmnts with nods) (discrtization/mshing). Connct (assmbl) th lmnts at th nods to form an approximat systm of quations for th whol structur (forming lmnt matrics) 3. Solv th systm of quations involving unknown quantitis at th nods (.g., displacmnts) 4. Calculat dsird quantitis (.g., strains and strsss) at slctd lmnts Basic hory h way finit lmnt analysis obtains th tmpraturs, strsss, flows, or othr dsird unknown paramtrs in th finit lmnt modl ar by minimizing an nrgy functional. An nrgy functional consists of all th nrgis associatd with th particular finit lmnt modl. Basd on th law of consrvation of nrgy, th finit lmnt nrgy functional must qual zro. h finit lmnt mthod obtains th corrct solution for any finit lmnt modl by minimizing th nrgy functional. h minimum of th functional is found by stting th drivativ of th functional with rspct to th unknown grid point potntial for zro. F hus, th basic quation for finit lmnt analysis is = 0 p whr F is th nrgy functional and p is th unknown grid point potntial (In mchanics, th potntial is displacmnt.) to b calculatd. his is basd on th principl of virtual work, which stats that if a particl is undr quilibrium, undr a st of a systm of forcs, thn for any displacmnt, th virtual work is zro. Each finit lmnt will hav its own uniqu nrgy functional. As an xampl, in strss analysis, th govrning quations for a continuous rigid body can b obtaind by minimizing th total potntial nrgy of th systm. h total 1 potntial nrgy Π can b xprssd as: Π = σ εdv d bdv d qds whr s Ω Ω and ar th vctors of th strss and strain componnts at any point, rspctivly, d is th vctor of displacmnt at any point, b is th vctor of body forc componnts pr unit volum, and q is th vctor of applid surfac traction componnts at any surfac point. h volum and surfac intgrals ar dfind ovr th ntir rgion of th structur Ω and that part of its boundary subjct to load. h first trm on th right hand sid of this

3 quation rprsnts th intrnal strain nrgy and th scond and third trms ar, rspctivly, th potntial nrgy contributions of th body forc loads and distributd surfac loads. In th finit lmnt displacmnt mthod, th displacmnt is assumd to hav unknown valus only at th nodal points, so that th variation within th lmnt is dscribd in trms of th nodal valus by mans of intrpolation functions. hus, within any on lmnt, d = N u whr N is th matrix of intrpolation functions trmd shap functions and u is th vctor of unknown nodal displacmnts. (u is quivalnt to p in th basic quation for finit lmnt analysis.) h strains within th lmnt can b xprssd in trms of th lmnt nodal displacmnts as = B u whr B is th strain displacmnt matrix. Finally, th strsss may b rlatd to th strains by us of an lasticity matrix (.g., Young s modulus) as s = E. h total potntial nrgy of th discrtizd structur will b th sum of th nrgy contributions of ach individual lmnt. hus, Π = Π whr Π rprsnts th total potntial nrgy of an individual lmnt. Π 1 u ( B EB) udv u N pdv = u N qds = 0 Ω Ω. Π 1 aking th drivativ = ( B EB) udv N udv u Ω Ω lmnt quilibrium quation k u f = 0 whr f = N udv + k = ( B EB) Ω udv and k is known as th lmnt stiffnss matrix. Ω N qds = 0 N qds on gts th and h physical significanc of th vctors u and f varis according to th application bing modld. Application Problm Structurs and solid mchanics Stat (DOF) vctor d rprsnts Displacmnt Forcing vctor f rprsnts Mchanical forc Hat conduction mpratur Hat flux Acoustic fluid Displacmnt potntial Particl vlocity Potntial flows Prssur Particl vlocity Gnral flows Vlocity Fluxs Elctrostatics Elctric potntial Charg dnsity Magntostatics Magnticpotntial Magntic intnsity

5 3-D two-nod bam spcify momnts of inrtia in both local X and local Y 6 DOF Mx My Mz A typical bam 3-D 4-nod plat Modling of structurs whr bnding (out of plan) and/or mmbran (in-plan) strss play qually important rols in th bhavior of that particular structur Each nod has 6 DOF Must spcify plat thicknss -D (3)4-nod solid "isoparamtric four-nod solid" Common in -D strss problms and natural frquncy analysis for solid structur. "hin" in that strss magnitud in third dirction is considrd constant ovr th lmnt thicknss. translational DOF and no rotational DOF -D -nod truss No bnding loads uniaxial tnsion-comprssion Straight bar with uniform proprtis from nd to nd DOF, no rotation -D -nod bam 3 DOF - translation x, y and rotation about z Any cross sction for which momnt of inrtia can b computd

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