TRANSVERSE VIBRATION OF A BEAM VIA THE FINITE ELEMENT METHOD Revision E

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1 RANSVERSE VIBRAION OF A BEAM VIA HE FINIE ELEMEN MEHOD Revision E B om Irvine Emil: tomirvine@ol.com November 8 8 Introuction Mn structures re too complex for nlsis vi clssicl meto. Close-form solutions re tus unvilble for tese structures. For exmple structure m be compose of severl ifferent mterils. Some of te mterils m be nisotropic. Furtermore te structure m be n ssembl of pltes bems n oter components. Consier tree exmples:. A circuit bor s numerous cips crstl oscilltors ioes connectors cpcitors ump wires n oter piece prts.. A lrge ircrft consisting of fuselge wing sections til section engines etc.. A builing s plies fountion bems floor sections n lo-bering wlls. e finite element meto is numericl meto tt cn be use to nlze complex structures suc s te tree exmples. e purpose of tis tutoril is to erive for meto for nlzing bem vibrtion using te finite element meto. e meto is bse on Reference. eor Consier bem suc s te cntilever bem in Figure. L Figure. were E I is te moulus of elsticit. is te re moment of inerti.

2 L is te lengt. is mss per lengt. e prouct is te bening stiffness. e vibrtion moes of te cntilever bem cn be foun b clssicl metos. Specificll te funmentl frequenc is 8. L () is problem presents goo opportunit to compre te ccurc of te finite element meto to te clssicl solution. Let (xt) represent te isplcement of te bem s function of spce n time. e free trnsverse vibrtion of te bem is governe b te eqution: x x ( x) (x t) ( x) (x t) t () Eqution () neglects rotr inerti n ser eformtion. Note tt it is lso inepenent of te bounr conitions wic re pplie s constrint equtions. Assume tt te solution of eqution () is seprble in time n spce. (xt) Y(x)f (t) () x x ( x) Y(x)f (t) ( x) Y(x)f (t) t () f (t) x x ( x) Y(x) Y(x) ( x) f (t) t (b) e prtil erivtives cnge to orinr erivtives. f (t) Y(x) Y(x) ( x) f (t) t ()

3 Y(x) Y(x) f (t) f (t) ( x) t () e left-n sie of eqution () epens on x onl. e rigt n sie epens on t onl. Bot x n t re inepenent vribles. us eqution () onl s solution if bot sies re constnt. Let be te constnt. Y(x) ( x) Y(x) f (t) f (t) t () Eqution () iels two inepenent equtions. (x) Y(x) (x) Y(x) (8) f (t) f (t) (9) t Eqution (8) is omogeneous fort orer orinr ifferentil eqution. e weigte resiul meto is pplie to eqution (8). is meto is suitble for bounr vlue problems. An lterntive meto woul be te energ meto. e energ meto is introuce in Appenix A. ere re numerous tecniques for ppling te weigte resiul meto. Specificll te Glerkin pproc is use in tis tutoril. e ifferentil eqution (8) is multiplie b test function φ (x). Note tt te test function φ(x) must stisf te omogeneous essentil bounr conitions. e essentil bounr conitions re te prescribe vlues of Y n its first erivtive. e test function is not require to stisf te ifferentil eqution owever. e prouct of te test function n te ifferentil eqution is integrte over te omin. e integrl is set eqution to zero. (x) (x) Y(x) (x) φ Y(x) ()

4 e test function φ(x) cn be regre s virtul isplcement. e ifferentil eqution in te brckets represents n internl force. is term is lso regre s te resiul. us te integrl represents virtul work wic soul vnis t te equilibrium conition. Define te omin over te limits from to b. ese limits represent te bounr points of te entire bem. b (x) (x) Y(x) (x) φ Y(x) () b φ(x) (x) Y(x) Integrte te first integrl b prts. b φ(x) { (x) Y(x) } () b φ(x) (x) Y(x) b φ(x) (x) Y(x) b φ(x) { (x) Y(x) } () φ(x) (x) Y(x) b b φ(x) (x) Y(x) b φ(x) { (x) Y(x) } ()

5 { } Y(x) (x) (x) Y(x) (x) (x) Y(x) (x) (x) b b b φ φ φ () Integrte b prts gin. { } Y(x) (x) (x) Y(x) (x) (x) Y(x) (x) (x) Y(x) (x) (x) b b b b φ φ φ φ () { } Y(x) (x) (x) Y(x) (x) (x) Y(x) (x) (x) Y(x) (x) (x) b b b b φ φ φ φ () e essentil bounr conitions for cntilever bem re () Y (8) Y x (9)

6 us te test functions must stisf φ ( ) () φ x () e nturl bounr conitions re (x) Y(x) x b () (x) Y(x) x b () Eqution () requires (x) φ(x) x b () Appl equtions () () n () to eqution (). e result is b φ(x) (x) Y(x) b φ(x) { (x) Y(x) } () Note tt eqution () woul lso be obtine for oter simple bounr conition cses. Now consier tt te bem consists of number of segments or elements. e elements re rrnge geometricll in series form. Furtermore te enpoints of ec element re clle noes.

7 e following eqution must be stisfie for ec element. (x) (x) Y(x) φ(x) φ { (x) Y(x) } Furtermore consier tt te stiffness n mss properties re constnt for given element. () φ(x) Y(x) φ(x) Y(x) () Now express te isplcement function Y(x) in terms of nol isplcements n s well s te rottions n. Y(x) L L L L ( ) < x < (8) Note tt is te element lengt. In ition ec L coefficients is function of x. Now introuce nonimensionl nturl coorinte. x / (9) Note tt is te segment lengt. e isplcement function becomes. Y( ) L L L L < < () e slope eqution is Y '( ) L ' L' L' L' < < () e isplcement function is represente terms of nturl coorintes in Figure.

8 Y(x) (-) x Figure. Represent ec L coefficient in terms of cubic polnomil. Li c i ci ci ci i Y( ) { c c c c } { c c c c } { c c c c } { c c c c } < < () () 8

9 Y '( ) { c c c } { c c c } { c c c } { c c c } < < () Solve for te coefficients c i. e constrint equtions re Y () () Y() () Y '() () Y '() (8) Evlute te isplcement t. Y() { c } { c} { c} { c} (9) Bounr conition () requires { c } { c} { c} { c} (9) c () c () c () c () 9

10 e isplcement equtions becomes Y( ) { c c c } { c c c } { c c c } { c c c } < < () e slope equtions becomes Y '( ) { c c c } { c c c } { c c c } { c c c } < < () Evlute te slope t. { c } { c } { c } { c } Y '() () Bounr conition () requires. { c } { c } { c } { c } () c (8) c (9) c () c ()

11 e isplcement equtions becomes Y( ) { c c } { c c } { c c } { c c } < < e slope equtions becomes Y'( ) { c c } { c c } { c c } { c c } < < () { c c } { c c } { c c } { c c } Y() () () Bounr conition () requires { c c } { c c } { c c } { c c } () c c () c c () c c (8) c c (9)

12 c c () c c () c c () c c () e isplcement eqution becomes Y( ) {[ c ] c } [ c ] c [ c ] c [ c ] c { } { } { } < < e slope eqution becomes Y '( ) { [ c ] c } { c c } { [ c ] c } { [ c ] c } < < () ()

13 e slope eqution becomes Y '() { [ c ] c } { c c } { [ c] c } { [ c ] c } () Bounr conition (8) requires { [ c ] c } { c c } { [ c] c } { [ c ] c } { [ c ] c } { c c } { [ c] c } { [ c ] c } () (8) c (9) c () c () c () c ()

14 c () c () e isplcement eqution becomes Y( ) {[ ( ) ] } [ ( ) ] { } { [ ] } { [ ( ) ] ( ) } < < Y( ) { } { } { } { } < < () () Recll x / (8) us / (9) (9b) / (8) Note (8)

15 { } { } { } { } Y(x) ( ) x x / < < (8) { / } {[ ] [ ] [ ] [ ] } Y(x) ( ) x x / < < (8) { / } {[ ] [ ] Y(x) [ ] [ ] } ( ) x x / < < (8) Now Let Y(x) L ( ) x x / (8) were L (8) L (8) L (88) L (89)

16 [ ] (9) e erivtive terms re x / x ) ( L' Y(x) (9) x / x ) ( L" Y(x) (9) Note tt primes inicte erivtives wit respect to. In summr. L (9) ' L (9) " L (9)

17 Recll φ(x) Y(x) φ(x) Y(x) (9) e essence of te Glerkin meto is tt te test function is cosen s φ ( x) Y(x) (9) us Y(x) Y(x) [ Y(x) ] (9) Cnge te integrtion vrible using eqution (9b). Also ppl te integrtion limits. Y(x) Y(x) [ Y(x) ] (9) L" L" [ L ][ L ] (9)

18 { [ L" ] [ L" ] } [ L ][ L ] (98) { [ L" ][ L" ] } [ L ][ L ] (99) { L" L" } { L L } () { L" L" } { L L } () { L" L" } { L L } () For sstem of n elements K M... n () were K { L" L" } () 8

19 9 { } L L M () [ ] L" L" () ( )( ) ( )( ) ( )( ) ( )( ) ( )( ) ( )( ) ( )( ) ( )( ) ( )( ) ( )( ) L" L" () Note tt onl te upper tringulr components re sown ue to smmetr. ( )( ) ( )( ) ( )( ) ( )( ) ( )( ) ( )( ) ( )( ) ( )( ) ( )( ) ( )( ) L" L" (8)

20 L" L" ( )( ) ( )( ) ( )( ) ( )( ) ( )( ) ( )( ) ( )( ) ( )( ) ( )( ) ( )( ) (9) L" L" () K ()

21 K () K () K ()

22 [ ] L L () ( ) ( )( ) ( )( ) ( )( ) ( ) ( )( ) ( )( ) ( ) ( )( ) ( ) L L () L L () 9 (8) (9) 9 ()

23 8 () () () () 9 () 8 () () Recll { } L L M (8) { } 9 M (9) 9 M () 9 M () M () M ()

24 { } M () M () M () M () M (8) { } 9 M (9) 9 M () 9 M () 9 M () M ()

25 { } 8 M (9) 8 M () 8 M () M () { } M () M () M () M () M () { } M (8)

26 M (9) M () M () { } M () M () M () M () { } 9 M () 9 M () 9 M (8)

27 M (9) { } 8 M () 8 M () 8 M () M () { } M () M () M () M ()

28 Recll K { L" L" } (8) M { L L } (9) K () M () Exmple Moel te cntilever bem in Figure s single element using te mss n stiffness mtrices in equtions () n (). e moel consists of one element n two noes s sown in Figure. E N N Figure. Note tt L. 8

29 9 e eigen problem is L L () () were L () L (b) e bounr conitions t noe re () () e first two columns n te first two rows of ec mtrix in eqution () cn tus be struck out.

30 e resulting eigen eqution is tus e eigenvlues re foun using te meto in Reference () (8)..98 (8b) e finite element results for te nturl frequencies re tus L..98 (9) L..8 (8) e finite element results re compre to te clssicl results in ble. ble. Nturl Frequenc Comprison Element Inex Finite Element Moel L Clssicl Solution L...8.

31 e clssicl results re tken from Reference. e finite element results tus overpreicte te nturl frequencies. Neverteless goo greement is obtine for te first frequenc. Exmple Moel te cntilever bem in Figure wit two elements using te mss n stiffness mtrices in equtions () n (). Let ec element ve equl lengt. e moel consists of two elements n tree noes s sown in Figure. E E N N N Figure. ere re severl kes to tis problem. One is tt L/. e oter is tt noe N receives mss n stiffness contributions from bot elements E n E. us te resulting globl mtrices ve imension x prior to te ppliction of te bounr conitions. e locl stiffness mtrix for element is ( L / ) e isplcement vector is lso sown in eqution (8) for reference. (8)

32 e locl stiffness mtrix for element is ( ) L / (8) e locl mss mtrix for element is ( ) / L (8) e locl mss mtrix for element is ( ) / L (8) e globl eigen problem ssemble from te locl mtrices is ( ) ( ) 8 L / 8 L / (8)

33 Agin te bounr conitions t noe re (8) (8) e first two columns n te first two rows of ec mtrix in eqution (8) cn tus be struck out. ( L / ) 8 ( L / ) 8 (88) Let ( L / ) ( ) L / (89) ( L / )( L / ) (9) L (9) L (9b)

34 e eigen problem in eqution (88) becomes 8 8 (9) e eigenvlues re foun using te meto in Reference. Eqution (9) iels four eigenvlues (9) (9b) e finite element results for te first two nturl frequencies re tus..9 L (9).. L (9)

35 e finite element results re compre to te clssicl results in ble. ble. Nturl Frequenc Comprison Elements Finite Element Clssicl Moel Solution Inex L L.... Excellent greement is obtine for te first two roots. e next step woul be to solve for te eigenvectors wic represent te moe spes. A greter number of elements woul be require to obtin ccurte moe spes owever. Exmple Repet exmple wit elements. Let ec element ve equl lengt. e globl stiffness n mss mtrix re omitte for brevit. e eigenvlue scle fctor is ( ) L (9) e moel iels eigenvlues. e first four re.9e -.9e -.8. (9)

36 (98) ( L ) (99) L () e finite element results re compre to te clssicl results in ble. ble. Nturl Frequenc Comprison Elements Finite Element Clssicl Moel Solution Inex L L Excellent greement is obtine for te first four roots. e corresponing moe spes re sown in Figures troug 8. Note tt ec moe spe is multiplie b n rbitrr mplitue scle fctor. e bsolute mplitue scle is tus omitte from te plots.

37 CANILEVER BEAM ELEMEN MODEL MODE Displcement L Figure. x CANILEVER BEAM ELEMEN MODEL MODE Displcement L Figure. x

38 CANILEVER BEAM ELEMEN MODEL MODE Displcement L Figure. x CANILEVER BEAM ELEMEN MODEL MODE Displcement L Figure 8. x 8

39 Aitionl exmples re given in Appenix C. Reference. L. Meirovitc Computtionl Metos in Structurl Dnmics Sitoff & Nooroff e Neterlns Irvine e Generlize Eigenvlue Problem W. omson eor of Vibrtion wit Applictions Secon Eition Prentice-Hll New Jerse 98.. K. Bte Finite Element Proceures in Engineering Anlsis Prentice-Hll New Jerse 98. 9

40 APPENDIX A Energ Meto e totl strin energ P of bem is L P (A-) e totl kinetric energ of bem is L [ ] n (A-) Agin let Y(x) L ( ) x x / (A-) Y(x) L' ( ) x x / (A-) Y(x) L" ( ) x x / (A-) / (A-) Assume constnt mss ensit n stiffness. e strin energ is converte to loclize stiffness mtrix s K { L" L" } (A-)

41 e kinetic energ is converte to loclize mss mtrix s M { L L } (A-8) e totl strin energ is set equl to te totl kinetic energ per te Rleig meto. e result is generlize eigenvlue problem. For sstem of n elements K M... n (A-9) were K (A-) M (A-)

42 APPENDIX B Bem Bening - Alternte Mtrix Formt e isplcement mtrix for bem bening is (B-) e stiffness mtrix for bem bening is K (B-) e mss mtrix for bem bening is M (B-)

43 APPENDIX C Free-Free Bem Repet exmple from te min text wit single element but wit free-free bounr conitions. e eigen problem is L L (C-) (C-) were L (C-) L (C-)

44 e eigenvlues re foun using te meto in Reference. Eqution (C-) iels four eigenvlues.. (C-)..9 (C-) e finite element results for te first four nturl frequencies re tus..9 L (C-) 9..8 L (C-8)

45 e finite element results re compre to te clssicl results in ble C-. ble C-. Nturl Frequenc Comprison Elements Finite Element Clssicl Moel Solution Inex L L