MAE4700/5700 Finite Element Analysis for Mechanical and Aerospace Design

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1 MAE4700/5700 Finit Elmnt Analysis for Mchanical and Arospac Dsign Cornll Univrsity, Fall 2009 Nicholas Zabaras Matrials Procss Dsign and Control Laboratory Sibly School of Mchanical and Arospac Enginring 101 Rhods Hall Cornll Univrsity Ithaca, NY MAE 4700 FE Analysis for Mchanical & Arospac Dsign U N I V E R S I T Y 1

2 Many nginring structurs consist of straight mmbrs connctd at thir nds by bolts, pins or wlding. Truss analysis Th truss lmnt forms th basic lmnt of such structurs. It can only tak forcs along its span (no momnts). MAE 4700 FE Analysis for Mchanical & Arospac Dsign U N I V E R S I T Y 2

3 Truss analysis Intrnal forcs in a truss lmnt act along '( ) '( ) th mmbr F1y = F2y = 0 Howvr, displacmnts at th nods can hav both componnts (x - and y -dirctions, in local coordinats). This is du to rotation MAE 4700 FE Analysis for Mchanical & Arospac Dsign U N I V E R S I T Y 3

4 Truss analysis To analyz a truss lmnt in th global coordinats x and y, you nd to account for both componnts of displacmnt: ( ) ( ) ( ) ( ) u, u, u, u 1y 2x 2y Also not that th cross sction of truss lmnts can vary as shown. MAE 4700 FE Analysis for Mchanical & Arospac Dsign U N I V E R S I T Y 4

5 Stiffnss of a truss lmnt ( ) ( ) F 1 ( ) ( ) ( ) ( ) F 1 k k u 1 ( ) = ( ) ( ) ( ) F2 k k u2 k = A E L Th intrnal forc p in th F2 truss is givn (s fr body diagram) as: ( ) ( ) p = F2 = F1 = Aσ Assuming lastic dformations: ( ) ( ) p = F = F = A E ε Th (small) strain is givn as: u2 u1 ε = L Finally: 2 1 ( ) ( ) AE F2 = F1 = ( u2 u1) = L = k ( u u ) 2 1 MAE 4700 FE Analysis for Mchanical & Arospac Dsign U N I V E R S I T Y 5

6 Truss lmnt stiffnss in local coordinats '( ) ( ) ( ) '( ) F k k u '( ) = ( ) ( ) '( ) F2x k k u2x Notic (as it should b) that: F '( ) '( ) 1y = F2y = 0 W can r-writ th lmnt stiffnss quations as: '( ) '( ) F u '( ) '( ) F 1 ( ) y u1y '( ) = k '( ) F x u2x '( ) '( ) F 2y u 2y ( ) ( ) [ K ' ] ( ) { F' } { d' } { F } = [ K' ] d { } '( ) ( ) '( ) MAE 4700 FE Analysis for Mchanical & Arospac Dsign U N I V E R S I T Y 6

7 Elmnt stiffnss in global coordinats Th angl φ is masurd anti-clockwis from x to x W nd to b abl to transform displacmnts from th x and y axs to displacmnts along th x and y axs. W start with th rvrs: '( ) ( ) u cosφ sinφ 0 0 u '( ) ( ) u1y sinφ cosφ 0 0 u1y '( ) = ( ) u 2x 0 0 cosφ sinφ u2x '( ) ( ) u 0 0 sinφ cosφ 2y u 2y ( ) { d ' } T ( ) ( ) { d } Transformation matrix T () MAE 4700 FE Analysis for Mchanical & Arospac Dsign U N I V E R S I T Y 7

8 Coordinat transformation { d' } = [ T ]{ d } '( ) ( ) u cosφ sinφ 0 0 u '( ) ( ) u1y sinφ cosφ 0 0 u1y '( ) = ( ) u 2x 0 0 cosφ sinφ u2x '( ) ( ) u 0 0 sinφ cosφ 2y u 2y ( ) ( ) [ ] ( ) { d ' } T Not that : { d } [ ] T T [ T ] = [ I] { d } = [ T ] T { d' } MAE 4700 FE Analysis for Mchanical & Arospac Dsign U N I V E R S I T Y 8

9 Coordinat transformation cosφ sinφ 0 0 cosφ sinφ 0 0 sinφ cosφ 0 0 sinφ cosφ 0 0 = 0 0 cosφ sinφ 0 0 cosφ sinφ 0 0 sinφ cosφ 0 0 sinφ cosφ I Vrify that: [ ] T T [ T ] = [ I] 4 4 T T MAE 4700 FE Analysis for Mchanical & Arospac Dsign U N I V E R S I T Y 9 x T

10 Stiffnss of a truss lmnt Similarly for th forcs: { F ' } = [ T ]{ F } '( ) ( ) F cosφ sinφ 0 0 F '( ) ( ) F1y sinφ cosφ 0 0 F1y '( ) = ( ) F 2x 0 0 cosφ sinφ F2x '( ) ( ) F 0 0 sinφ cosφ 2y F 2y { ' } [ ] F T { F } ( ) { F } = [ T ] T { F' } MAE 4700 FE Analysis for Mchanical & Arospac Dsign U N I V E R S I T Y 10

11 Stiffnss of a truss lmnt { } { } ( ) ( ) ( ) ( ) ( ) [ T ] F = [ K' ][ T ] d { } '( ) ( ) '( ) { F } = [ K' ] d '( ) ( ) Using { '( ) { F } = [ K' ] d } and th transformation quations, { F } = [ T ] F w can writ th stiffnss in th x,y systm as follows: ( ) ( ) ( ) ( ) [ K ] = [ T ] T [ K' ][ T ] { } '( ) ( ) ( ) { d } = [ T ] d { } '( ) ( ) ( ) { ( ) } ( ) T ( ) ( ) { ( ) F [ T ] [ K' ][ T ] d } = MAE 4700 FE Analysis for Mchanical & Arospac Dsign U N I V E R S I T Y 11 ( ) [ K ]

12 K = k ( ) ( ) [ ' ], Truss lmnt stiffnss ( ) ( ) T ( ) ( ) [ K ] = [ T ] [ K' ][ T ] T ( ) [ ] cosφ sinφ 0 0 sinφ cosφ 0 0 = 0 0 cosφ sinφ 0 0 sinφ cosφ [ K ] = k ( ) ( ) 2 2 cos φ sinφ cosφ cos φ sinφ cosφ 2 2 sinφ cosφ sin φ sinφ cosφ sin φ 2 2 cos φ sinφ cosφ cos φ sinφ cosφ 2 2 sinφ cosφ sin φ sinφ cosφ sin φ MAE 4700 FE Analysis for Mchanical & Arospac Dsign U N I V E R S I T Y 12

13 Truss lmnt stiffnss F ( ) 2 2 F1 y ( ) sinφ cosφ sin φ sinφ cosφ sin φ u k = F cos φ sinφ cosφ cos φ sinφ cosφ u u ( ) 2 2 ( ) cos φ sinφ cosφ cos φ sinφ cosφ u ( ) 2 2 2x ( ) 2 2 F sinφ cosφ sin φ sinφ cosφ sin φ 2 y Not th 2x2 symmtric submatrix structur This implis that you can rvrs th numbring of nods (1 and 2) without any changs in th lmnt stiffnss. F ( ) 2 2 F2 y ( ) sin cos sin sin cos sin u k φ φ φ φ φ φ ( ) = 2 2 F cos φ sinφ cosφ cos φ sinφ cosφ u ( ) 2 2 F sinφ cosφ sin φ sinφ cosφ sin φ 1y u ( ) 2 2 ( ) 2x cos φ sinφ cosφ cos φ sinφ cosφ u2x MAE 4700 FE Analysis for Mchanical & Arospac Dsign U N I V E R S I T Y 13 ( ) 2x ( ) ( ) 1y ( ) ( ) 2x ( ) 2 y

14 Assmbly procss Th assmbly procss is idntical to th on discussd for `spring structurs and it will not b rpatd hr in its gnral form (no nd to show at this point complicatd looking matrix oprations). W will howvr provid soon a simpl xampl dmonstrating this assmbly procss. MAE 4700 FE Analysis for Mchanical & Arospac Dsign U N I V E R S I T Y 14

15 Gnralizing th application of ssntial BCs You alrady hav sn through an xampl how ssntial boundary conditions ar applid to th global systm of qs: [ K]{ d} = { F} In ssnc, w partition th stiffnss matrix in a way that sparats known from unknown dgrs of frdom as follows: d E df ff : : : Known displacmnts Unknown displacmnts Applid (known) forcs KE KEF d E f E T = KEF KF df ff MAE 4700 FE Analysis for Mchanical & Arospac Dsign U N I V E R S I T Y 15 f E : Unknown raction forcs corrsponding to nods/dirctions with prscribd displacmnt

16 Gnralizing th application of ssntial BCs K K f E EF d E E T = KEF KF df ff K d + K d = f E E EF F E K d + K d = f T EF E F F F Th unknown displacmnts ar obtaind from th 2 nd q. as: T 1 T K d E + K d = f d = K ( f K d E ) EF F F F F F F EF d F With known, w can rturn to th 1 st q. to comput th raction forcs: f = K d + K d E K F E E EF F Not that th matrix is symmtric and positiv dfinit, so a solution for always xists! d F MAE 4700 FE Analysis for Mchanical & Arospac Dsign U N I V E R S I T Y 16

17 A truss xampl 1,2 3,4 5,6 A A = A = 2A 3 A = A Not that point D is fr to mov in th x dirction E = 10 7 Pa Construct th global stiffnss matrix and load vctor Partition th matrics and solv for th unknown displacmnts at point B, and displacmnt in x dirction at point D. Find th strsss in th thr bars Find th ractions at C, D and F 7,8 MAE 4700 FE Analysis for Mchanical & Arospac Dsign U N I V E R S I T Y 17

18 A truss xampl: Elmnt 1 (1) 0 φ =135 (1) [ K ] EA = /2 1/2 1/2 1/2 1/2 1/2 1/2 1/2 1/2 1/2 1/2 1/2 1/2 1/2 1/2 1/2 Not: Rcall that you can numbr th corrsponding global nods in th (1) squnc without any changs in [ K ]. MAE 4700 FE Analysis for Mchanical & Arospac Dsign U N I V E R S I T Y

19 A truss xampl: Elmnt 2 3,4 A 2 = 2A 0 φ = 90 (2) [ K ] EA = MAE 4700 FE Analysis for Mchanical & Arospac Dsign U N I V E R S I T Y

20 A truss xampl: Elmnt 3 3 A = A 0 φ = 45 (3) [ K ] = EA /2 1/2 1/2 1/2 1/2 1/2 1/2 1/2 1/2 1/2 1/2 1/2 1/2 1/2 1/2 1/ MAE 4700 FE Analysis for Mchanical & Arospac Dsign U N I V E R S I T Y 20

21 A truss xampl: Assmbly (lmnt 1) (1) [ K ] = /2 1/2 1/2 1/2 EA 1/2 1/2 1/2 1/2 2 1/2 1/2 1/2 1/2 1/2 1/2 1/2 1/ [ K] = /2 1/ /2 1/2 1/2 1/ /2 1/ EA /2 1/ /2 1/2 1/2 1/ /2 1/ MAE 4700 FE Analysis for Mchanical & Arospac Dsign U N I V E R S I T Y 21

22 A truss xampl: Assmbly (lmnt 2) (2) [ K ] = EA [ K] = /2 1/ /2 1/2 1/2 1/ /2 1/ EA /2 1/ /2+ 0 1/2+ 0 1/2 1/ /2 1/ MAE 4700 FE Analysis for Mchanical & Arospac Dsign U N I V E R S I T Y 22

23 A truss xampl: Assmbly (lmnt 3) (3) [ ] K = /2 1/2 1/2 1/2 EA 1/2 1/2 1/2 1/2 2 1/2 1/2 1/2 1/2 1/2 1/2 1/2 1/ [ K] = /2 1/ /2 1/2 1/2 1/ /2 1/ EA /2 1/2 1/2 1/ /2 1/2 1/2 1/2 1/2 1/ /2 1/2 1/ /2 1/ /2 1/2 1/ /2 1/2 1/2+ 1/2 1/ / MAE 4700 FE Analysis for Mchanical & Arospac Dsign U N I V E R S I T Y 23

24 A truss xampl: Assmbly [ K] = /2 1/ /2 1/2 1/2 1/ /2 1/ EA /2 1/2 1/2 1/ /2 1/2 1/2 1/2 1/2 1/ /2 1/ /2 1/ /2 1/ MAE 4700 FE Analysis for Mchanical & Arospac Dsign U N I V E R S I T Y 24

25 A truss xampl: Partition and BCs 1/2 1/ /2 1/2 d1 = 0 r1 1/2 1/ /2 1/2 d2 = 0 r d3 = 0 r3 EA d4 = 0r /2 1/2 1/2 1/2 d /2 1/2 1/2 1/2 d6 = 0r 6 1/2 1/ /2 1/ d7 10 Nt 1/2 1/ /2 1/ d 8 0 MAE 4700 FE Analysis for Mchanical & Arospac Dsign U N I V E R S I T Y 25

26 A truss xampl: Partition and BCs d E = 0 [ K E ] [ K EF ] f E 1/2 1/ /2 1/2 d1 = 0 r1 1/2 1/ /2 1/2 d2 = 0 r d3 = 0 r3 EA d4 = 0r /2 1/2 1/2 1/2 d6 = 0 r /2 1/2 1/2 1/2 d 5 0 1/2 1/ /2 1/ d7 10 Nt 1/2 1/ /2 1/ d 8 0 T [ K ] EF MAE 4700 FE Analysis for Mchanical & Arospac Dsign U N I V E R S I T Y 26 [ K F ] d F f F

27 A truss xampl: Partition and BCs 1/2 1/2 1/2 d 0 5 EA 3 1/2 1 0 d7 = 10 Nt 2 1/ d K F d m d7 = m d 0.005m 8 d F f F MAE 4700 FE Analysis for Mchanical & Arospac Dsign U N I V E R S I T Y 27

28 A truss xampl: Raction calculation f = K d + K d E E E EF F d E = 0 [ K E ] [ K EF ] f E 1/2 1/ /2 1/2 d1 = 0 r1 1/2 1/ /2 1/2 d2 = 0 r d3 = 0 r3 EA d4 = 0r /2 1/2 1/2 1/2 d6 = 0 r /2 1/2 1/2 1/2 d 5 0 1/2 1/ /2 1/ d7 10 Nt 1/2 1/ /2 1/ d 8 0 T [ K ] EF MAE 4700 FE Analysis for Mchanical & Arospac Dsign U N I V E R S I T Y 28 [ K F ] d F f F

29 A truss xampl: Raction calculation r1 0 1/2 1/2 r 2 0 1/2 1/2 d5 EA r3 = d7 2 r d 8 r df 6 1/2 1/2 1/2 f E K EF r1 RFx 1000Nt r R 2 Fy 1000Nt r3 RCx = 0 r R 1000Nt 4 Cy r 0 6 R Dy MAE 4700 FE Analysis for Mchanical & Arospac Dsign U N I V E R S I T Y 29

30 A truss xampl: Comput th strsss Combining th 2 Eqs givs: u u u u L L ' ' ' ' 2x 2x = σ = E E σ = cosφ sinφ cosφ sin φ { d } L ε Howvr: '( ) ( ) u cosφ sinφ 0 0 u '( ) ( ) u1y sinφ cosφ 0 0 u1y '( ) = ( ) u 2x 0 0 cosφ sinφ u2x '( ) ( ) u 2y 0 0 sinφ cosφ u 2y { d } MAE 4700 FE Analysis for Mchanical & Arospac Dsign U N I V E R S I T Y 30

31 A truss xampl: Comput th strsss E σ = cosφ sinφ cosφ sin φ { d } L Applying this to ach lmnt, w hav: σ (1) 0 E = = kPa m 0.005m 0 (2) 0 σ = E[ ] = 50kPa m 0.005m m (3) E σ = = 0kPa m 0.005m MAE 4700 FE Analysis for Mchanical & Arospac Dsign U N I V E R S I T Y 31

32 Thr-dimnsional (spac) truss structurs Th local stiffnss qs ar xactly as bfor, i.. '( ) ( ) ( ) '( ) Howvr, w now hav to includ displacmnts and forcs in th x, y and z axs. F k k u, '( ) = ( ) ( ) '( ) F2x k k u2x MAE 4700 FE Analysis for Mchanical & Arospac Dsign U N I V E R S I T Y 32 k ( ) = AE L

33 Thr-dimnsional (spac) truss structurs '( ) ( ) ( ) '( ) F k k u '( ) = ( ) ( ) '( ) F2x k k u2x A unit vctor along th dirction x of a 3D truss lmnt has th componnts (dirction cosins of th axs btwn x and x,y,z, rspctivly): x x y y z z l =, m =, n = s s s L L L L = ( x x ) + ( y y ) + ( z z ) whr ( x, y, z ) and ( x ar th nodal , y2, z ) 2 coordinats in th (x,y,z) systm. MAE 4700 FE Analysis for Mchanical & Arospac Dsign U N I V E R S I T Y 33

34 Thr-dimnsional (spac) truss structurs '( ) ( ) ( ) '( ) F k k u '( ) = ( ) ( ) '( ) F2x k k u2x Th displacmnt transformation thn taks th form: Similar transformation is applid for th forcs: ( ) u ( ) u1 y '( ) ( ) u l m n s s s u1z [ ]{ } '( ) = ( ) T d u2x l m n u s s s 2x ( ) [ T ] u 2 y ( ) u 2z '( ) F { d } [ T ]{ F } '( ) = F 2x MAE 4700 FE Analysis for Mchanical & Arospac Dsign U N I V E R S I T Y 34

35 Thr-dimnsional (spac) truss structurs '( ) ( ) ( ) '( ) F k k u '( ) = ( ) ( ) '( ) F2x k k u2x Similarly to th drivation for 2D trusss, th stiffnss matrix in global coordinats is thn: T [ K] [ T ] { K } [ T ] 6x6 6x2 2x2 2x6 MAE 4700 FE Analysis for Mchanical & Arospac Dsign U N I V E R S I T Y 35

36 Stiffnss of a spac lmnt ( ) [ ] K = l m l n l l m l n l s s s s s s s s s s ml m mn ml m mn 2 2 nl mn n nl mn n s s s s s s s s s s 2 2 L l m l n l l m l n l s s s s s s s s s s ml m mn ml m mn s s s s s s s s s s 2 2 nl mn n n l m n n s s s s s s s s s s EA s s s s s s s s s s x x y y z z whr : l =, m =, n = s s s L L L L = ( x x ) + ( y y ) + ( z z ) MAE 4700 FE Analysis for Mchanical & Arospac Dsign U N I V E R S I T Y 36

37 Computing th strsss in a spac truss lmnt ' ' u2x u ε = L u u E σ L L σ E = l m n l m n { d } s s s s s s L ' ' 2x ' ' = E = u 2x u ( ) ( ) u ( ) Howvr: u1 y '( ) ( ) u l m n s s s u 1z [ ]{ } '( ) = ( ) T d u2x l m n u s s s 2x ( ) u 2 y Combining th 2 Eqs givs: ( ) u 2z MAE 4700 FE Analysis for Mchanical & Arospac Dsign U N I V E R S I T Y 37

38 Accounting for thrmal ffcts in truss analysis Considr a truss structur that is hatd. W nd to account for thrmal xpansion ffcts. Not: total strain Hook s law is now modifid as follows (using th x coordinat systm): u2 u1 σ = E εlastic = E ( ε ε ) ( ) thrmal = E α ΔT L total strain u u F = F = p = A = A E ( Δ T ) = ( ) ( ) σ L α ε = k ( u u ) A E ε, k = ε = ε + ε lastic thrmal AE F 1 k k u F k k u ( ) ( ) ( ) ( ) AEε ( ) 0 = ( ) ( ) ( ) Thrma lnodal vctor MAE 4700 FE Analysis for Mchanical & Arospac Dsign U N I V E R S I T Y 38 L 0

39 Elmnt quations with thrmal ffcts Expanding ths quations to includ nodal displacmnts in th y axis givs: '( ) '( ) F u '( ) '( ) F 1 0 ( ) y u1y '( ) + AEε0 = k '( ) F x u2x '( ) '( ) F 2y u 2y ( ) ( ) { F' thrmal } [ K' ] ( ) { F' } { d' } W nd to transform this to x and y displacmnts (our dgrs of frdom for this lmnt) { d' } = [ T ]{ d } { F ' } = [ T ]{ F } MAE 4700 FE Analysis for Mchanical & Arospac Dsign U N I V E R S I T Y 39

40 Elmnt quations with thrmal ffcts '( ) '( ) F u '( ) '( ) F 1 0 ( ) y u1y '( ) + AEε0 = k '( ) F x u2x '( ) '( ) F 2y u 2y ( ) ( ) { F' thrmal } [ K' ] ( ) { F' } { d' } { d' } = [ T ]{ d } { F ' } = [ T ]{ F } W can transform ths lmnt quations as follows: From ths quations, w conclud that: ( ) [ T ]{ F } + { F' } [ ' ][ ]{ thrmal = K T d } T T ( ) { F } [ T ] { F' thrmal} [ T ] [ K' ][ T ]{ d } + = thrmal { F } MAE 4700 FE Analysis for Mchanical & Arospac Dsign U N I V E R S I T Y 40 ( ) [ K ]

41 Elmnt quations with thrmal ffcts T T ( ) { F } [ T ] { F' thrmal} [ T ] [ K' ][ T ]{ d } + = thrmal { F } ( ) [ K ] Us: Finally w obtain: T ( ) [ ] cosφ sinφ 0 0 sinφ cosφ 0 0 = 0 0 cosφ sinφ 0 0 sinφ cosφ (for 2D trusss) ( ) 2 2 ( ) F cosφ cos φ sinφ cosφ cos φ sinφ cosφ u ( ) 2 2 ( ) F1 y sinφ ( ) sinφ cosφ sin φ sinφ cosφ sin φ + AE ( ) ε 0 = k u1 y 2 F2 x cosφ 2 ( ) cos φ sinφ cosφ cos φ sinφ cosφ u2x ( ) F sin 2 2 ( ) 2 y φ sin cos sin sin cos sin u φ φ φ φ φ φ 2 y ( ) { F } thrmal { F } ( ) ( ) [ K ] d MAE 4700 FE Analysis for Mchanical & Arospac Dsign U N I V E R S I T Y 41

42 Elmnt quations with thrmal ffcts What do you nd to do to account for thrmal ffcts in truss analysis? For ach truss lmnt that is hatd, simply add to th lmnt forc, th following xtra trm cosφ sinφ A E whr T cosφ sinφ ε0, ε 0 = α Δ thrmal { F } You will nd to dfin at which truss lmnts thrmal ffcts tak plac and for ach of thm rad th valus α and ΔT. MAE 4700 FE Analysis for Mchanical & Arospac Dsign U N I V E R S I T Y 42

43 Principl of minimum potntial nrgy An altrnativ quivalnt approach to solving many structural problms is th principl of minimum potntial nrgy. From all th possibl compatibl displacmnts of a structur, th on that minimizs th total potntial nrgy is th xact solution. Potntial nrgy for givn displacmnts = Strain nrgy for ths givn displacmnts - Work don by xtrnal loads on ths givn displacmnts MAE 4700 FE Analysis for Mchanical & Arospac Dsign U N I V E R S I T Y 43

44 Principl of minimum potntial nrgy Lt us s this principl applid to th truss problms discussd arlir. Assmbly procss min { d} PE, PE = U W 1 '( ) '( ) '( ) '( ) PE = σε dv ( F u + F2x u2x ) 2 Ω Elastic strain nrgy dnsity ( work/ volum) Extrnal Work MAE 4700 FE Analysis for Mchanical & Arospac Dsign U N I V E R S I T Y 44

45 Principl of minimum potntial nrgy Lts apply this principl to on truss lmnt. W nd to minimiz with rspct to th nodal displacmnts (local coordinats) '( ) '( u ), u2x. Rcall that: u u ε ', σ ε L ' 2x = = E 1 PE = E ε dv F u F u = 2 '( ) '( ) '( ) '( ) 2x 2x 2 Ω Ω ' 1 u2x u' E ( ) dv F u F u 2 L 2 '( ) '( ) '( ) '( ) 2x 2x Adx' MAE 4700 FE Analysis for Mchanical & Arospac Dsign U N I V E R S I T Y 45

46 Principl of minimum potntial nrgy ' 1 min ( 2x ) 2 '( ) '( ) '( ) '( ) PE = E A L F u F2x u2x u u u u 2 L min ' ', ' 2x, ' 2x u u' Tak partial drivativs of PE wrt u, u ' : PE E A ' '( ) = 0 ( ' 2x) 0 u u F = u' L PE E A ' '( ) = 0 ( 2x ' ) 2x 0 u u F = u' L 2x MAE 4700 FE Analysis for Mchanical & Arospac Dsign U N I V E R S I T Y 46 ' 2x '( ) ( ) ( ) '( ) F k k u '( ) = ( ) ( ) '( ) F2x k k u2x Ths ar th sam Eqs as thos obtaind with th dirct mthod!

47 Principl of minimum potntial nrgy In gnral (not just for mchanics problms!), th principl of minimum potntial nrgy taks th following form: 1 min { ( ) } T [ ( ) ]{ ( ) } { } T PE = d K d d { F } { d} { d} 2 min or aftr assmbly: min Not that th mimimization givs us th familiar solution: [ K]{ d} = { F} PE { d} { d} 1 min { } T [ ]{ } { } T = d K d d { F } 2 MAE 4700 FE Analysis for Mchanical & Arospac Dsign U N I V E R S I T Y 47

48 Principl of minimum potntial nrgy W will not us this mthod to rpat th truss calculations. Howvr, it will b our starting point for computing th stiffnss of bam lmnts (lctur 4). Th mthod of minimum potntial nrgy can b applid to many problms not rlatd to mchanics howvr thr ar many problms whr this tchniqu is not applicabl. Aftr discussing bam bnding problms, w will nd to look for mor powrful (`unfortunatly also mor mathmatical) mthods (wak (Galrkin) formulations). MAE 4700 FE Analysis for Mchanical & Arospac Dsign U N I V E R S I T Y 48

49 Rvisiting th 2-nod truss lmnt Up to now w usd th dirct mthod to xprss th nodal loads vs. nodal displacmnts for th 2-nod truss lmnt. '( ) Lt us linarly intrpolat th displacmnt u x of any point x in th lmnt in trms of th nodal displacmnts: '( ) '( ) x' '( ) x' '( ) x' x' u ( ) ( ) ux = (1 ) u + u2 x = [1, ] [ '( ) ] { } = N d L L L L u2x basis functions ( ) ( ) N matrix 1 N2 lmnt basis functions Nodal displacmnts ' dux Th strain in this 2-nod lmnt ε = can now dx' b computd as follows: ' ( ) ( ) ( ) '( ) ( ) ( ) dux dn [ ]{ d } dn ( ) ( ) ( ) ux = [ N ]{ d } ε = = = [ ]{ d } [ B ]{ d } dx' dx' dx' ( ) ( ) dn1 dn u u u [ B ] = [, ] = [, ] ε = [, ] '( ) = dx' dx' L L L L u L 2x '( ) '( ) '( ) ( ) 2x Gradint of basis functions matrix This is xactly th sam approximation w usd bfor (constant strain lmnt) MAE 4700 FE Analysis for Mchanical & Arospac Dsign U N I V E R S I T Y 49

50 Rvisiting th 2-nod truss lmnt '( ) ( ) ( ) ( ) ( ) Lt us us ths intrpolation formulas u = [ N ]{ d }, ε = [ B ]{ d } to xprss th potntial nrgy in a format that will bcom vry familiar as w procd forward in this cours. 1 2 '( ) '( ) '( ) '( ) PE = E ε dv F u F2x u2x = 2 For now ths calculations ar idntical to our arlir approach! Indd: Ω Ω 1 T { d } [ B ] E [ B ]{ d } dv F u F u = 2 ( ) T ( ) '( ) '( ) '( ) '( ) 2x 2x Adx' T ε ε 1 ( ) T T ( ) = { d } [ B ] E [ B ] dv { d } 2 F u F u Ω [ K ] Elmnt stiffnss matrix '( ) '( ) '( ) '( ) 2x 2x 1 L L T T L 1 1 AE 1 1 [ K ] = [ B ] E [ B ] dv = [ B ] E [ B ] A dx' = E [, ] A dx' = 1 L L 0 0 L 1 1 Ω k L MAE 4700 FE Analysis for Mchanical & Arospac Dsign U N I V E R S I T Y 50

51 Truss analysis with displacmnt constraints Up to this point, w imposd ssntial boundary conditions in trms of prscribd x- or y- nodal displacmnts. How about if th support is inclind as in th figur blow: For this problm, th constraint is that thr is no normal displacmnt at th support 1 Hr, w don t know th displacmnts at nod 1 but w know th rlation btwn u and u 1y. In gnral w writ ths constrains on our nodal dgrs of frdom as: Cd=q. MAE 4700 FE Analysis for Mchanical & Arospac Dsign U N I V E R S I T Y 51

52 Truss analysis with displacmnt constraints Not that at nod 1 w don t hav ssntial boundary conditions w hav a displacmnt constraint. To solv this problm w us th principl of minimum potntial nrgy with th constraint Cd=q: Hr, w nforc th constraint using Lagrang multiplirs. Find d and λ such that 1 T T T min L= { d} [ K]{ d} { d} { F} + { { } λ} ([ C]{ d} { q}) d 2 Potntial nrgy of unconstraind systm Lagrang multiplir nforcing th constraint Constraint MAE 4700 FE Analysis for Mchanical & Arospac Dsign U N I V E R S I T Y 52

53 Truss analysis with displacmnt constraints Find d and λ such that 1 T T T min L = { d} [ K]{ d} { d} { F} + { { } λ} ([ C]{ d} { q}) d 2 Potntial nrgy of unconstraind systm Lagrang multiplir nforcing th constraint Constraint λ is th Lagrang multiplir that nforcs th constraint it is nothing ls but th raction forc at nod 1 (normal to th support!) Minimization is now prformd with rspct to both d and λ. K C T d F = C 0 q λ Apply ssntial boundary conditions and thn solv for {d F } and λ (hr, raction forc at nod 1) MAE 4700 FE Analysis for Mchanical & Arospac Dsign U N I V E R S I T Y 53

54 Displacmnt constraints: Implmntation How do you implmnt this in th MatLab libraris of HW1? u sin 30 + u cos30 = 0 x1 y1 Introduc th constraints in th InputData.m and thn modify th stiffnss and load vctors in th NodalSoln.m. Apply th ssntial boundary conditions first bfor you augmnt th rducd global quations (K f ) with th Lagrang multiplir. InputData.m % Rad information for constraints C = zros(1,nq-lngth(dbc)); %Th dimnsion of C is nq minus th % prscribd DOF via ssntial BCs % Hr thr is only on constraint C(1) = sin(pi/6); C(2) = cos(pi/6); q = 0; MAE 4700 FE Analysis for Mchanical & Arospac Dsign U N I V E R S I T Y 54

55 Displacmnt constraints: NodalSoln.m function [d, rf, lambda] = NodalSoln(K, R, dbc, bcvals, C, q) % K=global stiffnss, R=global forc, dbc=dgrs of frdom with spcifid valus, bcvals=spcifid displacmnts dof = lngth(r); % Extract th total dgrss-of-frdom df = stdiff(1:dof, dbc); % Sts th diffrnc btwn two sts of indics, i.. th global dgrs of frdom minus th % dgrs of frdom with prscribd ssntial boundary conditions Kf = K(df, df); % Rmov qs. corrsponding to prscribd displacmnts Rf = R(df) - K(df, dbc)*bcvals; % Modify th rmaining load vctor to account for th ssntial boundary conditions [m n] = siz(c); % Extract numbr of constraints Kf = [Kf C'; % Augmnt global quations with th Lagrang multiplirs C zros(m)]; Rf = [Rf;q]; % Augmnt load vctor K C T d F = C 0 λ q dfvals = Kf\Rf; % Solv th linar systm of quations. Hr for simplicity, w us Gauss limination. d = zros(dof,1); % Rstor th solution vctor (i.. includ back th nods with prscribd displacmnts). d(dbc) = bcvals; % Us th originally stablishd ordring of th nods. d(df) = dfvals(1:(lngth(dfvals)-m)); rf = K(dbc,:)*d - R(dbc); % Calculat th raction vctor at nods with prscribd displacmnts lambda = dfvals((lngth(dfvals)-m+1):lngth(dfvals)); % Calculat Langrang multiplirs (ractions at nods with constraints) MAE 4700 FE Analysis for Mchanical & Arospac Dsign U N I V E R S I T Y 55

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