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 Introduction to Finit Elmnt Analysis Chaptr 5 Two-Dimnsional Formulation 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. 5-

2 What Ar W Going to Larn? Compard to D, major diffrncs in D FEM formulation ar application of IBP. mastr lmnts and shap functions for triangular and quadrilatral lmnts. Jacobian transformation. boundary intgral valuation. Triangular lmnt Quadrilatral lmnt METU Dpt. of Mchanical Enginring ME Int. to Finit Elmnt Analysis Lctur Nots of Dr. Srt 5-

3 Modl DE in D Poisson quation in D is a u = f whr a(x, y) and f(x, y) ar known functions and u(x, y) is th unknown. For a problm in th xy plan of th Cartsian coordinat systm, gradint oprator is = i x + j y In th xy plan Poisson quation bcoms i x + j y u u a i + a x y j a u = f x u a x y a u y = f METU Dpt. of Mchanical Enginring ME Int. to Finit Elmnt Analysis Lctur Nots of Dr. Srt 5-

4 Modl DE in D If function a is constant ovr th problm domain, Poisson qn. bcoms a u = f u = g u x + u x = g Homognous form of this quation is calld th Laplac s quation u = 0 u x + u x = 0 Poisson quation modls many physical phnomna such as potntial flow hat conduction groundwatr flow transvrs dflction of plats lctrostatics and magntostatics METU Dpt. of Mchanical Enginring ME Int. to Finit Elmnt Analysis Lctur Nots of Dr. Srt 5-

5 Obtaining Wak Form in D Modl DE : x a u x y a u y = f Wightd rsidual intgral statmnt of this DE is Ω w x u a x y a u y f dω = 0 nd ordr drivativs of u can b rducd to st ordr using th following gnral quations Ω Ω w F x dω w F y dω = F w x dω Ω = F w y dω Ω + wf n x dγ Γ + wfn y dγ Γ METU Dpt. of Mchanical Enginring ME Int. to Finit Elmnt Analysis Lctur Nots of Dr. Srt 5-5

6 Obtaining Wak Form in D Ω w x u a x w y a u y wf dω = 0 Ω a u w x x dω Γ wa u x n x dγ Ω a u w y y dω Γ wa u y n y dγ Elmntal wak form is a u w x x + u w dω = wf dω + w a u Ω y y Ω x n x + a u y n y Γ dγ q n : SV of th problm whr n x and n y ar th Cartsian componnts of th unit outward normal of Γ. METU Dpt. of Mchanical Enginring ME Int. to Finit Elmnt Analysis Lctur Nots of Dr. Srt 5-6

7 Approximat solution ovr an lmnt is D Formulation (cont d) u = NEN j= u j S j (x, y) For linar triangular and quadratic lmnts NEN is and, rspctivly. -nod triangular lmnt NEN = -nod quadrilatral lmnt NEN = METU Dpt. of Mchanical Enginring ME Int. to Finit Elmnt Analysis Lctur Nots of Dr. Srt 5-7

8 To gt th i th quation of lmnt D Formulation (cont d) substitut approximat u into th lmntal wak form and slct w = S i NEN a u x j S j x + u y j S j y dω = S i f dω + S i q n dγ Ω Ω Γ j= S i NEN j= S i Arrang to gt NEN j= Ω a S j x S i x + S j y S i y dω u j = S i f dω + S i q n dγ Ω Γ K ij F i Q i METU Dpt. of Mchanical Enginring ME Int. to Finit Elmnt Analysis Lctur Nots of Dr. Srt 5-8

9 NEN NEN lmntal systm is D Formulation (cont d) K u = F + {Q } K ij = a S j Ω x S i x + S j y S i y dω F i = S i f dω Q i = S i q n dγ Ω Γ To valuat ths intgrals triangular and quadrilatral mastr lmnts will b introducd. shap functions will b writtn in mastr lmnt coordinats. D Jacobian transformation will b usd. GQ intgration will b usd. METU Dpt. of Mchanical Enginring ME Int. to Finit Elmnt Analysis Lctur Nots of Dr. Srt 5-9

10 D Quadrilatral Mastr Elmnt Actual quadrilatral lmnt Mastr quadrilatral lmnt y η (-,) (,) ξ x (-,-) (,-) Mastr quadrilatral lmnt is a squar of siz x. Its nods ar always numbrd in a CCW ordr starting with (-,-) cornr. Nods of th actual lmnt ar also numbrd in CCW ordr. It dos NOT mattr which nod is slctd as th first on. METU Dpt. of Mchanical Enginring ME Int. to Finit Elmnt Analysis Lctur Nots of Dr. Srt 5-0

11 Shap Functions of D Quadrilatral Mastr Elmnt Gnral form of Lagrang typ D shap functions ovr a -nod quadrilatral lmnt is S = A + Bξ + Cη + Dξη Unknown constants A, B, C and D can b found using th fact that shap functions satisfy th Kronckr-Dlta proprty if i = j S j ξ i, η i = 0 if i j S = ( ξ)( η) η Shap functions ar S = ( + ξ)( η) S = ( + ξ)( + η) S = ( ξ)( + η) ξ METU Dpt. of Mchanical Enginring ME Int. to Finit Elmnt Analysis Lctur Nots of Dr. Srt 5-

12 D Triangular Mastr Elmnt Actual triangular lmnt Mastr triangular lmnt η y (0,) x (0,0) (,0) ξ Mastr triangular lmnt is a right triangl with an ara of 0.5. Its nods ar always numbrd in a CCW ordr, starting with (0,0) cornr. METU Dpt. of Mchanical Enginring ME Int. to Finit Elmnt Analysis Lctur Nots of Dr. Srt 5-

13 Shap Functions of D Triangular Mastr Elmnt Gnral form of Lagrang typ D shap functions ovr a -nod triangular lmnt is S = A + Bξ + Cη Unknown constants A, B and C can b found using th Kronckr-Dlta proprty of th shap functions Shap functions ar S = ξ η S = ξ S = η η ξ METU Dpt. of Mchanical Enginring ME Int. to Finit Elmnt Analysis Lctur Nots of Dr. Srt 5-

14 Jacobian Transformation in D S/ x and S/ y drivativs appar in th intgrals of Slid 5-9. Ths drivativs nd to b xprssd in trms of S/ ξ and S/ η drivativs. This rquirs th transformation btwn (x, y) and (ξ, η) coordinats. Actual lmnt in (x, y) y x = x(ξ, η) y = y(ξ, η) Mastr lmnt in (ξ, η) η ξ x METU Dpt. of Mchanical Enginring ME Int. to Finit Elmnt Analysis Lctur Nots of Dr. Srt 5-

15 Jacobian Transformation in D (cont d) Rmmbr that in D x(ξ) rlation was x = h ξ + x + x This rlation can also b xprssd as x = NEN j= x j S j x = ξ x + + ξ x This works du to th Kronckr-Dlta proprty of th shap functions ξ = is mappd to x = x ξ = is mappd to x = x Sam logic can also b usd in D to gt x(ξ, η) and x(ξ, η) rlations. METU Dpt. of Mchanical Enginring ME Int. to Finit Elmnt Analysis Lctur Nots of Dr. Srt 5-5

16 Jacobian Transformation in D (cont d) NEN x = x j S j j= NEN and y = y j S j j= Ths can b usd for both quadrilatral and triangular lmnts. x j and y j ar th coordinats of th cornr points of th lmnts..g. Exampl 5.: Obtain x(ξ, η) and y(ξ, η) rlations for th following lmnt. y x Cornr coordinats ar Cornr : (5, 6) Cornr : (0, 7) Cornr : (, 0) METU Dpt. of Mchanical Enginring ME Int. to Finit Elmnt Analysis Lctur Nots of Dr. Srt 5-6

17 Exampl 5. (cont d) x = y = j= x j S j y j S j = 5 ξ η + 0 ξ + η = 5 5ξ η = 6 ξ η + 7 ξ + 0 η = 6 + ξ 6η j= η Evry point on th mastr lmnt can b mappd to a point on th actual lmnt using ths rlations. P(0.5,0.5) For xampl point P with ξ, η = (0.5, 0.5) maps to x = = y ξ y = =.5 Both points P and Q ar th mid-points of th Q(,.5) facs opposit to nod. x METU Dpt. of Mchanical Enginring ME Int. to Finit Elmnt Analysis Lctur Nots of Dr. Srt 5-7

18 Jacobian Transformation in D (cont d) With th link btwn (x, y) and (ξ, η) coordinats, S/ x and S/ y drivativs can b linkd to S/ ξ and S/ η. S ξ = S x x ξ + S y y ξ S η = S x x η + S y y η S ξ S η = x ξ x η y ξ y η Jacobian matrix [J ] S x S y In D Jacobian was J = dx dξ. In D Jacobian is a matrix. In gnral [J ] is diffrnt for ach lmnt of th FE msh. METU Dpt. of Mchanical Enginring ME Int. to Finit Elmnt Analysis Lctur Nots of Dr. Srt 5-8

19 Jacobian Transformation in D (cont d) For th intgrals of slid 5-9 what w actually nd is S x S y = ξ x ξ y η x η y Invrs of th Jacobian matrix J S ξ S η Sinc w know x and y as a function of ξ and η, but not th othr way, it is NOT practical to calculat J dirctly. Instad w first calculat J and thn tak its invrs. METU Dpt. of Mchanical Enginring ME Int. to Finit Elmnt Analysis Lctur Nots of Dr. Srt 5-9

20 Exampl 5..g. Exampl 5.: Obtain [J ] and J of th lmnt that w studid in xrcis 5.. y x Cornr coordinats ar Cornr : (5, 6) Cornr : (0, 7) Cornr : (, 0) J = x ξ x η y ξ y η = x j S j ξ x j S j η y j S j ξ y j S j η = S ξ S η S ξ S η S ξ S η x x x y y y This gnral J calculation formula applis to both triangular and quadrilatral lmnts. METU Dpt. of Mchanical Enginring ME Int. to Finit Elmnt Analysis Lctur Nots of Dr. Srt 5-0

21 For a triangular lmnt shap functions ar S = ξ η, S = ξ, S = η For a triangular lmnt drivativs of th shap functions ar S S S ξ ξ ξ = 0 S S S 0 η η η Jacobian of th lmnt is Exampl 5. (cont d) J = = 5 6 METU Dpt. of Mchanical Enginring ME Int. to Finit Elmnt Analysis Lctur Nots of Dr. Srt 5-

22 Exampl 5. (cont d) Invrs of th Jacobian matrix is J = J J J J J J = J J J J = 5 6 = J = 6 5 = Not that in D J was qual to th ratio of actual lmnt s lngth to mastr lmnt s lngth. Similarly for a -nod triangular lmnt J is qual to th ratio of actual lmnt s ara to mastr lmnt s ara. In this xrcis th ara ratio is 6.5/(0.5) =. METU Dpt. of Mchanical Enginring ME Int. to Finit Elmnt Analysis Lctur Nots of Dr. Srt 5-

23 Exampl 5..g. Exampl 5.: Obtain [J ] and J of th lmnt shown blow. y x Cornr coordinats ar Cornr : (5, 6) Cornr : (0, 7) Cornr : (0, 0) Cornr : (, 0) J = S ξ S η S ξ S η S ξ S η S ξ S η x x x x y y y y = η ξ η ξ η + ξ + η ξ METU Dpt. of Mchanical Enginring ME Int. to Finit Elmnt Analysis Lctur Nots of Dr. Srt 5-

24 Exampl 5. (cont d) J = η 7 η + ξ ξ Dtrminant of J is J = J J J J = 8 ξ η + 7 Not that this tim both th Jacobian matrix and its dtrminant ar functions of ξ and η. Intgral of J ovr th mastr lmnt will giv th ara of th actual lmnt. η= ξ= 8 ξ η + 7 J dξ dη =.5 This can b gnralizd as follows which will b usd in GQ intgration Actual lmnt s ara dxdy Ω = J dξdη Ω (Tru for both triangular and quadrilatral lmnts) METU Dpt. of Mchanical Enginring ME Int. to Finit Elmnt Analysis Lctur Nots of Dr. Srt 5-

25 Calculation of Intgrals Ovr a Mastr Elmnt Now th intgrals of Slid 5-9 can b valuatd on a mastr lmnt using GQ. Us x = x j S j and y = y j S j to convrt x and y of function a to ξ and η. Sam for f of F i. K ij = a S j S i x x + S j S i dω, F Ω y y i = S i f dω Ω dxdy = J dξdη Ω Ω mastr S x = S ξ ξ x + S η η x = S ξ J + S η J S y = S ξ ξ y + S η η y = S ξ J + S η J METU Dpt. of Mchanical Enginring ME Int. to Finit Elmnt Analysis Lctur Nots of Dr. Srt 5-5

26 Gauss Quadratur Ovr Quadrilatral Elmnts For a quadrilatral mastr lmnt both ξ and η chang btwn - and. [-, ] ar th limits usd in D GQ intgration. Thrfor for D quadrilatral lmnts D GQ tabls can b usd. Considr th valuation of th following intgral using NGP points in both ξ and η dirctions. I = η= ξ= g dξ dη In D thr ar NGP GQ points I = NGP n= NGP m= g ξ m, η n W m W n NGP or I = g ξ k, η k W k k= Sum for η Sum for ξ Combind sum for ξ and η W k = W m W n METU Dpt. of Mchanical Enginring ME Int. to Finit Elmnt Analysis Lctur Nots of Dr. Srt 5-6

27 Gauss Quadratur Ovr Quadrilatral Elmnts (cont d) D GQ Intgration NGP ξ k W k point GQ ovr a quad. lmnt / / η /9 8/9 5/9 ξ D GQ Intgration Ovr Quads NGP ξ k η k W k / / / / / / / / /8 0/8 5/8 0/8 6/8 0/8 5/8 0/8 5/8 METU Dpt. of Mchanical Enginring ME Int. to Finit Elmnt Analysis Lctur Nots of Dr. Srt 5-7

28 Gauss Quadratur Ovr Triangular Elmnts For a triangular mastr lmnt limits for ξ and η ar [0, ] and [0, -ξ], rspctivly. ξ f dξdη = f dηdξ Ω ξ=0 η=0 Thrfor a nw GQ tabl is ncssary. D GQ Intgration Ovr Triangls NGP ξ k η k W k / / 0.5 η / / /6 /6 /6-7/96 5/96 5/96 5/96 point GQ ovr a triangular lmnt ξ METU Dpt. of Mchanical Enginring ME Int. to Finit Elmnt Analysis Lctur Nots of Dr. Srt 5-8

29 Exampl 5..g. Exampl 5.: Calculat th first ntry of th following lmntal forc vctor F i = x S i dxdy Ω ovr th following lmnt using point GQ intgration. y Cornr coordinats ar x Cornr : (0, 0) Cornr : (5, 0) Cornr : (, ) Cornr : (0, ) METU Dpt. of Mchanical Enginring ME Int. to Finit Elmnt Analysis Lctur Nots of Dr. Srt 5-9

30 W nd to calculat Exampl 5. (cont d) F = x S dxdy Ω Switching to mastr lmnt coordinats th intgral bcoms F = x ξ, η ( ξ)( η) J dξdη S dxdy W first nd x as a function of ξ and η. x = x j S j = 0 S + 5 S + S + 0 S j= x = 5 + ξ η + + ξ + η x = ( + ξ η ξη) METU Dpt. of Mchanical Enginring ME Int. to Finit Elmnt Analysis Lctur Nots of Dr. Srt 5-0

31 Exampl 5. (cont d) Nxt w nd to calculat th Jacobian and its dtrminant (similar to Slid 5-) J = η ξ η ξ η + ξ + η ξ = η ξ 0 J = η ξ 0 = η Th intgral bcoms F = + ξ η ξη g(ξ,η) ξ η η dξdη F = g dξdη METU Dpt. of Mchanical Enginring ME Int. to Finit Elmnt Analysis Lctur Nots of Dr. Srt 5-

32 Exampl 5. (cont d) point GQ ovr th quadrilatral mastr lmnt will b F = g ξ, η W + g ξ, η W + g ξ, η W + g ξ, η W whr points and wights ar providd in Slid 5-7 Th rsult will b Nots: F = = In gnral J is a function of ξ and η and it nds to b valuatd at GQ points. In this xampl w did not nd y(ξ, η) bacaus f was not a function of y. In this xampl w did not calculat th invrs of J bcaus forc vctor dos not contain shap function drivativs. Stiffnss matrix calculation will nd it. Is th abov rsult xact? What is th xact valu? What will point and 9 point intgrations giv? METU Dpt. of Mchanical Enginring ME Int. to Finit Elmnt Analysis Lctur Nots of Dr. Srt 5-

33 Calculation of Q {Q} intgrals nd to b valuatd only for th ral boundary facs whr NBC or MBC is spcifid. Considr th following problm with a msh of lmnts and 6 nods. 6 NBC = 5 EBC = = NBC METU Dpt. of Mchanical Enginring ME Int. to Finit Elmnt Analysis Lctur Nots of Dr. Srt 5-

34 Calculation of Q (cont d) Thr ar lmnts facs at NBC or MBC boundaris. Th assmbld global {Q} will b (first local nod of ach lmnt is shown, th othrs ar locatd in a CCW ordr) NBC 6 EBC 5 Q = Q Q Q Q Q 5 Q 6 = Q Q + Q Q Q + Q Q + Q + Q Q NBC Q, Q 5 and Q 6 ar not ncssary bcaus PVs ar known at ths nods. Only th circld ons ar ncssary. Not : In this xampl w do not hav an intrnal nod (a nod that is not locatd at a boundary), but for thos nods sum of Q i s will b zro. METU Dpt. of Mchanical Enginring ME Int. to Finit Elmnt Analysis Lctur Nots of Dr. Srt 5-

35 Calculation of Q (cont d) For th Poisson quation {Q } is calculatd as (Slid 5-9) Q i = S i q n ds Γ q n = a u x n x + a u y n y Γ is th boundary of th lmnt and it is composd of NEN straight lins. For a triangular lmnt th intgral can b dcomposd into parts. Q i = S i q n ds f + S i q n ds f + S i q n ds f f : Fac = s + + s s METU Dpt. of Mchanical Enginring ME Int. to Finit Elmnt Analysis Lctur Nots of Dr. Srt 5-5

36 Calculation of Q (cont d) Considr th lmnt problm of Slid 5-. W nd Q, Q, Q, Q, Q = : Q = S q n ds f + S q n ds f + S q n ds f S is zro on facs and NBC + S q n ds f 6 EBC 5 NBC Q = S q n ds f + S q n ds f No nd (f is intrnal) + S q n ds f + S q n ds f S is zro on facs and Q = S q n ds f + S q n ds f S is zro on facs and + S q n ds f + S q n ds f No nd (f is intrnal) METU Dpt. of Mchanical Enginring ME Int. to Finit Elmnt Analysis Lctur Nots of Dr. Srt 5-6

37 Calculation of Q (cont d) = : 6 Q = = : Q = S q n ds f S q n ds f + S q n ds f S is zro on fac + S q n ds f + S q n ds f + S q n ds f No nd (f is intrnal) NBC EBC 5 NBC No nd (f is intrnal) S is zro on fac Conclusion : Boundary intgrals nd to b calculatd only for ral boundary facs whr NBC or MBC is providd. METU Dpt. of Mchanical Enginring ME Int. to Finit Elmnt Analysis Lctur Nots of Dr. Srt 5-7

38 Calculation of Q (cont d) Considr th common and simpl cas of q n = constant. Lt s study th calculation of Q = S q n ds f + S q n ds f S is a D shap function but it rducs to a first ordr function ovr facs and of lmnt. NBC q n = q L 6 EBC 5 NBC, q n = q B S = s f L f s L f f f S = s L f s f L f METU Dpt. of Mchanical Enginring ME Int. to Finit Elmnt Analysis Lctur Nots of Dr. Srt 5-8

39 Calculation of Q (cont d) Q = L f s=0 S q n ds f Q = s q B ds + L f + S q n ds f L f s=0 s q L ds L f Q = q B L f + q L L f Sam procdur can b followd to calculat Q. Q = Q = S q n ds f s q B ds f L f Q = q B L f S = s L f s f L f f f METU Dpt. of Mchanical Enginring ME Int. to Finit Elmnt Analysis Lctur Nots of Dr. Srt 5-9

40 Calculation of Q (cont d) Calculation of Q, Q and Q follow th sam procdur. Q = q L L f, Q = q B L f, Q = q L L f Summary : From th bottom fac of =, amount of providd SV is q B L f and it is dividd qually to Q and Q. From th bottom fac of =, amount of providd SV is q B L f and it is dividd qually to Q and Q. From th lft fac of =, amount of providd SV is q L L f and it is dividd qually to Q and Q. From th lft fac of =, amount of providd SV is q L L f and it is dividd qually to Q and Q. q L L f q L L f 6 5 EBC q B L f q B L f METU Dpt. of Mchanical Enginring ME Int. to Finit Elmnt Analysis Lctur Nots of Dr. Srt 5-0

41 Assmbld {Q} vctor is Calculation of Q (cont d) Q = Q Q Q Q Q 5 Q 6 = q B L f + q L L f q B L f + q B L f Q q L L f + q L L f Q 5 Q 6 NBC q n = q L 6 EBC 5 NBC, q n = q B Not that it is not possibl to valuat Q, Q 5 and Q 6 xactly, and ths ar not ncssary du to givn EBC for u, u 5 and u 6. METU Dpt. of Mchanical Enginring ME Int. to Finit Elmnt Analysis Lctur Nots of Dr. Srt 5-

42 Calculation of Q (cont d) Qustion : What if q n is not constant at an NBC boundary? Answr : Just valuat th lin intgrals with th givn variabl q n. Qustion : What if th BC is not NBC but MBC? Considr th following cas whr bottom BC is MBC with constant α and β. 6 NBC q n = q L EBC 5 MBC, q n = αu + β METU Dpt. of Mchanical Enginring ME Int. to Finit Elmnt Analysis Lctur Nots of Dr. Srt 5-

43 Calculation of Q (cont d) Q = S q n ds f + S q n ds f f Sam as bfor = q L L f s αu L f + β ds f NBC q n = q L 6 EBC 5 MBC, q n = αu + β u u f S s f u = s u + s u f u f L f = u + u u s L f L u f L f METU Dpt. of Mchanical Enginring ME Int. to Finit Elmnt Analysis Lctur Nots of Dr. Srt 5-

44 Calculation of Q (cont d) L f Q = s s=0 α u + u u s + β ds L f L f + q L L f Q = β L f + αl f u + αl f 6 u + Contribution of th MBC of th bottom fac q L L f Contribution of th NBC of th lft fac NBC q n = q L 6 EBC 5 To calculat Q a similar intgral is valuatd but this tim with S. L f Q s = α u + u u s + β ds s=0 L f L f MBC, q n = αu + β Q = β L f + αl f 6 u + αl f u METU Dpt. of Mchanical Enginring ME Int. to Finit Elmnt Analysis Lctur Nots of Dr. Srt 5-

45 Calculation of Q is just th sam as Q. Q = β L f + αl f u + αl f 6 u Assmbld {Q} is Calculation of Q (cont d) NBC q n = q L 6 EBC 5 MBC, q n = αu + β Q = Q Q Q Q Q 5 Q 6 = β L f β L f + αl f + αl f u + αl f 6 u + q L L f 6 u + αl f u + β L f + αl f u + αl f 6 u Q q L L f + q L L f Q 5 Q 6 Circld trms nd to b transfrrd to th [K] matrix. So it is bttr to writ thm using global indics. METU Dpt. of Mchanical Enginring ME Int. to Finit Elmnt Analysis Lctur Nots of Dr. Srt 5-5

46 Exampl 5.5.g. Exampl 5.5: Dtrmin th tmprtur distribution ovr th following D gomtry. Obtain unknown nodal tmpraturs. Thrmal conductivity of th mdium is. W (mk). First local cornrs of th lmnts ar shown with s insid th lmnts. Insulatd y 6 5 T = 00 k dt dy = h(t T ) h = 5 W m K, T = 0 x Nod coordinats [m] Nod : (0, 0) Nod : (0.5, 0) Nod : (, 0) Nod : (0, 0.5) Nod 5 : (0.5, 0.5) Nod 6 : (0, ) METU Dpt. of Mchanical Enginring ME Int. to Finit Elmnt Analysis Lctur Nots of Dr. Srt 5-6

47 Exampl 5.5 (cont d) Govrning DE is k T = 0 Elmntal wak form (Slid 5-6) is k T w x x + T w y y dω = wf dω + wq n dγ Ω Ω Γ K ij = k S i Ω x S j x + S i y S j y dω q n = k T x n x + k T y n y F i = fs i dω Ω W can start by calculating th Jacobian matrix of ach lmnt. METU Dpt. of Mchanical Enginring ME Int. to Finit Elmnt Analysis Lctur Nots of Dr. Srt 5-7

48 Exampl 5.5 (cont d) = : J = 0 0 J = 0.5 J = 0 0 = : J = 0 0 J = 0.5 J = = = y x Elmnts and hav th sam shap and siz as lmnt and thir first local nod is at th right angl cornr. Thrfor thir Jacobian matrics ar th sam. J = J = J METU Dpt. of Mchanical Enginring ME Int. to Finit Elmnt Analysis Lctur Nots of Dr. Srt 5-8

49 Exampl 5.5 (cont d) Elmntal systms can now b calculatd. K ij = k S i Ω x S j x + S i y S j y dω K ij = Ω + k S i ξ J + S i η J S i ξ J + S i η J S j ξ J + S j η J S j ξ J + j η J J dω F i = 0 = : K = METU Dpt. of Mchanical Enginring ME Int. to Finit Elmnt Analysis Lctur Nots of Dr. Srt 5-9

50 Exampl 5.5 (cont d) = : K = = and : K = K = [K ] Now th {Q} vctor should b calculatd. Only contribution will com from th two MBC facs at th bottom. NBC q n = 0 y 6 5 T = 00 q n = k dt dy = h(t T ) α = 5, β = 00 x METU Dpt. of Mchanical Enginring ME Int. to Finit Elmnt Analysis Lctur Nots of Dr. Srt 5-50

51 Exampl 5.5 (cont d) Contributions of th MBC at fac of = to Q and Q. βl f + αl f T + αl f 6 T βl f + αl f 6 T + αl f T Contribution of th MBC at fac of = to Q and Q. Th contribution to Q is not rquird bcaus nod is an EBC nod. βl f + αl f T + αl f 6 T No nd METU Dpt. of Mchanical Enginring ME Int. to Finit Elmnt Analysis Lctur Nots of Dr. Srt 5-5

52 Exampl 5.5 (cont d) Q = (0.5) (0.5) T + 5(0.5) T + 00 Q 0 Q 5 T + 5(0.5) T 6 5(0.5) Pay attntion T + 5(0.5) 6 T Q 6 Q = 5 0.8T 0.67T T.6667T 0.67T Q 0 Q 5 Q 6 METU Dpt. of Mchanical Enginring ME Int. to Finit Elmnt Analysis Lctur Nots of Dr. Srt 5-5

53 Exampl 5.5 (cont d) Global systm is sym T T T T T 5 T 6 = 5 0.8T 0.67T T.6667T 0.67T Q 0 Q 5 Q 6 Tak th unknonws du to MBC from th {Q} vctor into th [K] matrix T T T T T 5 T 6 = 5 50 Q 0 Q 5 Q 6 METU Dpt. of Mchanical Enginring ME Int. to Finit Elmnt Analysis Lctur Nots of Dr. Srt 5-5

Exampl 5.5 (cont d) Apply rduction for th known T, T 5 and T 6.. 0. 0.65 0..667 0 0.65 0.6 T T T = 5 50 + 0. 00 +.(00) 0 +. 00 + +0.65(00) As sn MBC s do not dstroy th symmtry of th rducd systm.

54 Exampl 5.5 (cont d) Apply rduction for th known T, T 5 and T T T T = (00) (00) As sn MBC s do not dstroy th symmtry of th rducd systm. Solv for th unknown primary variabls T T T = Constant T lins should b paralll to th EBC boundary and thy should b prpndicular to th insulatd boundary. METU Dpt. of Mchanical Enginring ME Int. to Finit Elmnt Analysis Lctur Nots of Dr. Srt 5-5

Middle East Technical University Department of Mechanical Engineering ME 413 Introduction to Finite Element Analysis

Middle East Technical University Department of Mechanical Engineering ME 413 Introduction to Finite Element Analysis 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 http://www.m.mtu.du.tr/popl/cunyt