Middle East Technical University Department of Mechanical Engineering ME 413 Introduction to Finite Element Analysis
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1 Middl East Tchnical Univrsity Dpartmnt of Mchanical Enginring ME 43 Introduction to Finit Elmnt Analysis Chaptr 3 Computr Implmntation of D FEM Ths nots ar prpard by Dr. Cünyt Srt csrt@mtu.du.tr Ths nots ar prpard with th hop to b usful to thos who want to larn and tach FEM. You ar fr to us thm. Plas snd fdbacks to th abov mail addrss. 3-
2 Summary of Chaptr In Chaptr Ritz mthod was improvd into FEM. Solution is not global anymor. Solution ovr ach lmnt is simpl. Complicatd D and 3D solutions on complx domains can b obtaind, by using ncssary numbr of lmnts. Problms with multipl matrials can b solvd. Approximation function slction is vry wll dfind and indpndnt of BCs. But th procdur of Chaptr still hav difficultis. Writing approximation functions on-by-on for ach nod is difficult. In D and 3D it ll b vn mor difficult. Approximation functions chang whn msh changs. Symbolic math is limitd and costly. Exrcis : Modify Exampl_v.m cod so that it asks th usr to ntr NE and works automatically. Using tic & toc commands, masur computation tim for NE = 5,, 5,,. Improv th spd of th cod in any way you can. METU Dpt. of Mchanical Enginring ME 43 Int. to Finit Elmnt Analysis Lctur Nots of Dr. Srt 3-
3 What Is This Chaptr About? W ll improv th solution procdur of Chaptr so that it is almost compltly msh indpndnt approximation function calculations ar asy symbolic calculations ar avoidd solution is fficint and vry algorithmic To achiv ths w ll us Elmntal wak form Mastr lmnt concpt Gauss Quadratur numrical intgration METU Dpt. of Mchanical Enginring ME 43 Int. to Finit Elmnt Analysis Lctur Nots of Dr. Srt 3-3
4 Nod Basd Intgral Calculation of Chaptr In FEM φ s hav local support. For linar lmnts, φ s ar nonzro ovr at most two lmnts. φ i i i- i- i i+ + i+ i+3 x Intgral of th i th qn. contains φ i in all its trms (s slid -). But φ i is nonzro only ovr lmnts - and. Thrfor intgral calculations simplify as follows Intgral of qn. i : I i = f φ i dx = f φ i dx + f φ i dx Ω Ω Ω Ovr th whol problm domain Ovr lmnt - only Ovr lmnt only METU Dpt. of Mchanical Enginring ME 43 Int. to Finit Elmnt Analysis Lctur Nots of Dr. Srt 3-4
5 Nod Basd Intgral Calculation of Chaptr (cont d) For a 5 nod msh φ φ φ 3 φ 4 φ 5 All intgrals ar i= i= i=3 i=4 i=5 x I = f φ Ω dx I = f φ Ω dx + f φ Ω dx I 3 = f φ 3 Ω dx + f φ 3 Ω 3 dx I 4 = f φ 4 Ω 3 dx + f φ 4 Ω 4 dx I 5 = f φ 5 Ω 4 dx This is nod basd thinking. W valuat th intgral of ach quation, which ar associatd with on nod and on φ. In FEM cods w prfr lmnt basd oprations. METU Dpt. of Mchanical Enginring ME 43 Int. to Finit Elmnt Analysis Lctur Nots of Dr. Srt 3-5
6 Nw Elmnt Basd Intgral Calculation Instad of thinking about ach quation individually, concntrat on lmnts and dtrmin th contribution of ach lmnt to ach quation. I = Ω f φ dx = contributs to I and I = contributs to I and I 3 = = =3 =4 x I = Ω f φ dx + Ω f φ dx =3 contributs to I 3 and I 4 I 3 = Ω f φ 3 dx + Ω 3 f φ 3 dx =4 contributs to I 4 and I 5 I 4 = Ω 3 f φ 4 dx + Ω 4 f φ 4 dx I 5 = Ω 4 f φ 5 dx METU Dpt. of Mchanical Enginring ME 43 Int. to Finit Elmnt Analysis Lctur Nots of Dr. Srt 3-6
7 For th following modl DE wak form is d dx du a dx Elmntal Thinking du + b dx + cu = f, < x < L L a du dw dx dx du + bw Lwf dx + cwu dx = dx + wa du dx L Q L In Chaptr FE solution of this rsultd in a NN NN global systm wa du dx Q K u = F + Q Th nw ida is to writ wak form ovr ach lmnt individually obtain lmntal systms add thm up to gt th global systm METU Dpt. of Mchanical Enginring ME 43 Int. to Finit Elmnt Analysis Lctur Nots of Dr. Srt 3-7
8 Local Nod Numbring and Elmntal Wak Form Considr th following linar lmnt Global nod numbrs Local nod numbrs i x i i+ x i+ x x x x Elmntal wak form of th modl DE is x x a du dw dx dx + bw du dx + cwu dx = x x wf dx + w du dx x w du dx x which will rsult in th following NN NN lmntal systm K u = F + Q METU Dpt. of Mchanical Enginring ME 43 Int. to Finit Elmnt Analysis Lctur Nots of Dr. Srt 3-8
9 Elmntal Systm K u = F + Q Elmntal systms ar spars. On a msh of 4 linar lmnts = = =3 =4 x For = = + For = = + For =3 = + For =4 = + METU Dpt. of Mchanical Enginring ME 43 Int. to Finit Elmnt Analysis Lctur Nots of Dr. Srt 3-9
10 Small Elmntal Systms of Siz NEN x NEN Each lmntal systm contributs to only qns of th global systm. It is bttr to think of lmntal systms as NEN NEN, instad of NN NN whr NEN is th numbr of lmnt s nods (= for linar lmnts) K u = F + Q Elmntal stiffnss matrix NEN NEN Elmntal unknown vctor NEN Elmntal forc vctor NEN Elmntal boundary trm vctor NEN For xampl for =3, small lmntal systm is 3 3 K K 3 3 K K u u = F 3 F + Q 3 Q Assmbly opration = + METU Dpt. of Mchanical Enginring ME 43 Int. to Finit Elmnt Analysis Lctur Nots of Dr. Srt 3-
11 From Approximation Functions to Shap Functions φ i φ i+ Nod basd thinking i- i- i i+ + i+ i+3 x Elmnt basd thinking i i+ x S S Shap functions x h = x i+ x i = x x METU Dpt. of Mchanical Enginring ME 43 Int. to Finit Elmnt Analysis Lctur Nots of Dr. Srt 3-
12 Shap Functions Similar to φ s, shap functions also hav th Kronckr-dlta proprty S S S i x j = if i = j if i j x x x For a linar lmnt shap functions ar FE solution ovr lmnt is u S u u S S = x x h, S = x x h x u = NEN j= u j S j Numbr of lmnt s nods (= for linar lmnts) Nodal unknown at lmnt s j th nod METU Dpt. of Mchanical Enginring ME 43 Int. to Finit Elmnt Analysis Lctur Nots of Dr. Srt 3-
13 Shap Functions (cont d) Difficultis with ths shap functions ar For ach lmnt thy will b diffrnt functions of x. Intgration ovr an lmnt will hav limits of x and x, which ar not appropriat for Gauss Quadratur intgration. S x S x x Th cur is to us th concpt of mastr lmnt. METU Dpt. of Mchanical Enginring ME 43 Int. to Finit Elmnt Analysis Lctur Nots of Dr. Srt 3-3
14 Mastr Elmnt D, linar mastr lmnt is dfind using th local coordinat ξ (ksi). S (ξ) S (ξ) S = ξ Suprscript is NOT ncssary hr ξ Nod Nod ξ = - ξ = S = + ξ For all linar lmnts in a D msh, thr is only a singl mastr lmnt. Mastr lmnt has a lngth of. End points ar ξ = and ξ =, which ar suitabl for Gauss Quadratur. METU Dpt. of Mchanical Enginring ME 43 Int. to Finit Elmnt Analysis Lctur Nots of Dr. Srt 3-4
15 Mapping Btwn an Actual Elmnt & Mastr Elmnt If a msh has only linar lmnts, than w only nd to dfin shap functions. This is a grat simplification, but it coms with a pric. In ordr to xprss vrything in th intgrals in trms of ξ, w nd to obtain th rlation btwn th global x coordinat and th local ξ coordinat. This rlation is linar as shown x x An actual lmnt of lngth h x x = Aξ + B x(ξ) =? Using th fact that nd points of th actual lmnt coincid with thos of th mastr lmnt, w gt x = h ξ + x + x This rlation is diffrnt for ach and vry lmnt ξ = ξ = ξ Mastr lmnt of lngth METU Dpt. of Mchanical Enginring ME 43 Int. to Finit Elmnt Analysis Lctur Nots of Dr. Srt 3-5
16 Jacobian of an Elmnt In th intgrals of th wak form w hav th first drivativ of u. u = NEN j= u j S j du NEN dx = j= u j ds j dx Mastr lmnt shap functions ar writtn in trms of ξ. Thrfor x drivativs should b xprssd in trms of ξ drivativs. ds j dx = ds j dξ dξ dx ds dξ =.5, ds dξ =.5 = h (using th boxd quation of th prvious slid. J = dx = h dξ is th Jacobian of lmnt. It is th ratio of actual lmnt s lngth to th lngth of th mastr lmnt. METU Dpt. of Mchanical Enginring ME 43 Int. to Finit Elmnt Analysis Lctur Nots of Dr. Srt 3-6
17 Exampl 3..g. Exampl 3. Solv th following problm using a uniform msh of 4 linar lmnts of lngth h =.5. Elmntal wak form is d u dx u = x, < x < u =, u = x x du dx dw dx wu dx = x x wx dx + w du dx x w du dx x To gt x lmntal systm of qns, substitut th following approximat solution into th lmntal wak form NEN u = u j S j j= METU Dpt. of Mchanical Enginring ME 43 Int. to Finit Elmnt Analysis Lctur Nots of Dr. Srt 3-7
18 Exampl 3. (cont d) Ω d dx u j S j dw dx w u j S j dx = Ω wx dx + w du dx x w du dx x Elmntal systm is x and w nd wight functions to gt it. In GFEM w = S, w = S Eqn : Ω d dx u j S j ds dx S u j S j dx = Ω S x dx + S du dx x S du dx x Q Eqn : Ω d dx u j S j ds dx S u j S j dx = Ω S x dx + S du dx x S du dx x Q In gnral th i th qn of lmnt is obtaind by using w = S i Eqn i : Ω u j ds j dx ds i dx S i u j S j dx = Ω S i x dx + Q i METU Dpt. of Mchanical Enginring ME 43 Int. to Finit Elmnt Analysis Lctur Nots of Dr. Srt 3-8
19 Exampl 3. (cont d) Chang th intgration paramtr from x to ξ (rfr to slid 3-6) dx = J dξ Eqn i : u j ds j dx ds i dx S i u j S j dx = S i x dx + Q i ds dx = ds dξ J x = h ξ + x + x Eqn i : u j ds j dξ J ds i dξ J S i u j S j J dξ = S i f ξ J dξ + Q i Tak th summation sign outsid th intgral and tak th intgrand into u j paranthsis. ds i ds j Eqn i : dξ J dξ J S is j J dξ u j = S i f ξ J dξ + Q i METU Dpt. of Mchanical Enginring ME 43 Int. to Finit Elmnt Analysis Lctur Nots of Dr. Srt 3-9
20 Exampl 3. (cont d) Eqn i : ds i dξ ds j J dξ K ij J S is j J dξ u j = S i f ξ J dξ + Q i F i Summation sign is ovr j =,. i indx also gos from to. i = givs th first quation, i = givs th scond quation. x lmntal systm is K u = F + Q W don t nd to do any calculations for Q i valus (Dtails will com). METU Dpt. of Mchanical Enginring ME 43 Int. to Finit Elmnt Analysis Lctur Nots of Dr. Srt 3-
21 Exampl 3. (cont d) x = x For ach lmnt h =.5. Jacobian for ach lmnt is J = h / =.5 All lmnts ar linar. Shap functions and thir drivativs ar S = ξ, S = + ξ ds dξ =.5, ds dξ =.5 W nd vrything to valut th ntris of K and F on-by-on for ach lmnt. METU Dpt. of Mchanical Enginring ME 43 Int. to Finit Elmnt Analysis Lctur Nots of Dr. Srt 3-
22 Exampl 3. (cont d) K ij = ds i dξ ds j J dξ J S is j J dξ For = K = ds dξ ds J dξ J S S J dξ = 47 K = ds dξ ds J dξ J S S J dξ = 97 4 K = K ([K ] is symmtric. Intrchang i & j and s) K = K = ds dξ ds J dξ J S S J dξ = 47 METU Dpt. of Mchanical Enginring ME 43 Int. to Finit Elmnt Analysis Lctur Nots of Dr. Srt 3-
23 No nd to calculat [K ], [K 3 ] or [K 4 ]. Exampl 3. (cont d) Thy will all b qual to [K ]. This is a spcial cas for this problm. Can you s why? Lt s start {F } calculations. F i = S i f ξ J dξ f ξ = x ξ For = : f = h ξ + x + x = ξ + 8 F = S f ξ J dξ = 768, F = S f ξ J dξ = METU Dpt. of Mchanical Enginring ME 43 Int. to Finit Elmnt Analysis Lctur Nots of Dr. Srt 3-3
24 Exampl 3. (cont d) For = : f = ξ+3 8 F = S f ξ J dξ = 768, F = S f ξ J dξ = For =3 : f =? (find yourslf) F 3 = S f ξ J dξ = 33 S f ξ J dξ = , F 3 = 768 For =4 : f =? (find yourslf) F 4 = S f ξ J dξ = 67 S f ξ J dξ = 8 768, F 4 = 768 METU Dpt. of Mchanical Enginring ME 43 Int. to Finit Elmnt Analysis Lctur Nots of Dr. Srt 3-4
25 Exampl 3. (cont d) Four lmntal systms ar = = =3 =4 x For = : u u = Q Q For = : u u = Q Q For =3 : u 3 u = Q Q 3 For =4 : u 4 u = Q Q 4 METU Dpt. of Mchanical Enginring ME 43 Int. to Finit Elmnt Analysis Lctur Nots of Dr. Srt 3-5
26 Exampl 3. (cont d) Assmbl lmntal systms into 5x5 global systm (s slid 3-9). = = =3 =4 x K K K K + K K K K 3 + K 3 K 3 K K + K K 4 4 K 4 4 K u u u 3 u 4 u 5 = F F + F F + F 3 F 3 + F 4 F 4 + Q Q + Q Q + Q 3 Q 3 + Q 4 Q 4 METU Dpt. of Mchanical Enginring ME 43 Int. to Finit Elmnt Analysis Lctur Nots of Dr. Srt 3-6
27 Exampl 3. (cont d) Put th numbrs in to gt = = =3 =4 x u u u 3 u 4 u 5 = Q Q + Q Q + Q 3 Q 3 + Q 4 Q 4 Balanc of scondary variabls : Q + Q = Q + Q 3 = Q 3 + Q 4 = du dx x du dx x du dx x 3 + du dx x + du dx x 3 + du dx x 4 = = = METU Dpt. of Mchanical Enginring ME 43 Int. to Finit Elmnt Analysis Lctur Nots of Dr. Srt 3-7
28 Exampl 3. (cont d) Global systm is u u u 3 u 4 u 5 = Q Q 5 u and u 5 ar known. Rduc th systm by dropping th st and 5 th quations u u 3 u 4 = FE solution is u u 3 u 4 = METU Dpt. of Mchanical Enginring ME 43 Int. to Finit Elmnt Analysis Lctur Nots of Dr. Srt 3-8
29 Apply EBCs without Rduction Rduction is not asy to implmnt in a computr cod. A simplr tchniqu is to kp th st and 5 th qns, but modify thm as follows K K K 3 K 4 K 5 K 3 K 3 K 33 K 34 K 35 K 4 K 4 K 43 K 44 K 45 u u u 3 u 4 u 5 = U U 5 + Equat diagonal ntris to, and non-diagonal ntris to zro Ths ar th spcifid valus of u and u 5 Equat unknown Q s to zro. Disadvantags ar symmtry of [K] is lost. an unncssarily larg systm is solvd. METU Dpt. of Mchanical Enginring ME 43 Int. to Finit Elmnt Analysis Lctur Nots of Dr. Srt 3-9
30 Apply EBCs without Rduction (cont d) A third altrnativ for EBCs modifis st and 5 th qns as follows L K K K 3 K 4 K 5 K K K 3 K 4 K 5 K 3 K 3 K 33 K 34 K 35 K 4 K 4 K 43 K 44 K 45 K 5 K 5 K 53 K 54 L K 55 u u u 3 u 4 u 5 = L K U F F 3 F 4 L K 55 U 5 + whr L is larg nough numbr. If L is larg nough th st and 5 th qns simplify to LK u + Ngligibly small trms = LK U u = U LK 55 u 5 + Ngligibly small trms = LK 55 U 5 u 5 = U 5 This tchniqu prsrvs possibl symmtry of [K]. METU Dpt. of Mchanical Enginring ME 43 Int. to Finit Elmnt Analysis Lctur Nots of Dr. Srt 3-3
31 NBCs If a NBC is providd, th spcifid Q valu is usd in th global systm. Similar to th Ritz mthod, NBCs ar satisfid not xactly, but approximatly. B carful in dtrmining th SV corrctly. If a hat conduction problm is formulatd starting from d dx dt ka dx = thn Q = ka dt dx and Q NN = ka dt dx L SV is hat in Watts If in th sam problm ka is constant and droppd from th DE d dx thn Q = dt dx and Q NN = dt dx L dt dx = SV is tmpratur gradint in K/m METU Dpt. of Mchanical Enginring ME 43 Int. to Finit Elmnt Analysis Lctur Nots of Dr. Srt 3-3
32 MBCs Put th givn mixd BC into th form SV = αpv + β whr α and β ar known valus. Us αpv + β in th propr plac of th Q vctor. Transfr αpv to th [K] matrix and lav β on th RHS of th global systm. If a mixd BC is givn at th 5 th (last) nod of a 4 lmnt msh K K K 3 K 4 K 5 K K K 3 K 4 K 5 K 3 K 3 K 33 K 34 K 35 K 4 K 4 K 43 K 44 K 45 K 5 K 5 K 53 K 54 K 55 u u u 3 u 4 u 5 = F F F 3 F 4 F 5 + Q αu 5 + β Modify K 55 as K 55 α METU Dpt. of Mchanical Enginring ME 43 Int. to Finit Elmnt Analysis Lctur Nots of Dr. Srt 3-3
33 Gauss Quadratur (GQ) Intgration In FEM intgrals similar to th following ons nd to b valuatd K ij = ds i dξ ds j J dξ J S is j J dξ, F i = S i f ξ J dξ Th limits [-,] ar suitabl for GQ intgration, which convrts an intgral into a summation Numbr of GQ points I = g ξ dξ = NGP k= Coordinats of GQ points g(ξ k )W k GQ wights METU Dpt. of Mchanical Enginring ME 43 Int. to Finit Elmnt Analysis Lctur Nots of Dr. Srt 3-33
34 GQ Intgration (cont d) GQ points and wights for diffrnt NGP valus ar NGP ξ k W k / 3 = / 3 = = = /9 = /9 = /9 = NGP point GQ intgration can valuat ( NGP ) ordr polynomial functions xactly. METU Dpt. of Mchanical Enginring ME 43 Int. to Finit Elmnt Analysis Lctur Nots of Dr. Srt 3-34
35 .g. Exampl 3. Exampl 3. Evaluat K and F of Exampl 3. using GQ intgration. K = ds dξ ds J dξ J S S J dξ = (.5).5 (.5).5 ξ ξ.5 dξ = ξ + ξ g(ξ) dξ Using point GQ : K = g = Using point GQ : K = g + g = Using 3 point GQ : K = 5 g g + 5 g.6 = Both ar xact METU Dpt. of Mchanical Enginring ME 43 Int. to Finit Elmnt Analysis Lctur Nots of Dr. Srt 3-35
36 Exampl 3. (cont d) F = S f ξ J dξ = ξ ξ dξ = ξ 3 + ξ ξ 4 g(ξ) dξ Using point GQ : F = g =.953 Using point GQ : F = g + g = Both ar Using 3 point GQ : F = 5 g g + 5 g.6 =.3 xact METU Dpt. of Mchanical Enginring ME 43 Int. to Finit Elmnt Analysis Lctur Nots of Dr. Srt 3-36
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