MEEN 617 Handout #12 The FEM in Vibrations A brief introduction to the finite element method for modeling of mechanical structures

Transcription

1 MEEN 67 Handout # T FEM in Vibrations A brif introduction to t finit lmnt mtod for modling of mcanical structurs T finit lmnt mtod (FEM) is a picwis application of a variational mtod. Hr I provid you wit a fundamntal introduction to t FE mtod congrunt wit prior analysis of mcanical systms and t assumd mods mtod. For a complt and most lucid introduction to t FEM plas rad Rddy, J.N., Introduction to t Finit Elmnt Mtod, Jon Wily Pubs. T bauty of t FEM mtod lis on its simplicity sinc its formulation is indpndnt of t actual rspons of t systm. Tat is, littl knowldg about an pctd answr is rquird a-priori. Lt s rviw t Assumd Mods Mtod. In any mcanical systm, t Hamiltonian t T V Wt dt () t is t fundamntal principl of mcanics from wic t laws of motion ar drivd, i.., Nwton s Laws and/or Lagrangian Mcanics. In gnral, t kintic nrgy (T) and t strain nrgy (V) of a mcanical systm ar functions of t displacmnt vctor ( ) and its tim drivativs, i.., MEEN 67 HD Introduction to FEM in Vibrations 3 Dr. L. San Andrés

2 T T v, v,matrialproprtis () V V v, v,matrialproprtis wr ( ) = d/dt andv= vi + vy j + vzk of a matrial point in t domain of intrst Ω= Ω( i ) is t vctor of displacmnts X3 Domain (Xi) v X X Fig. Domain for analysis In t assumd mods mtod an approimation to t displacmnt function in a continuous systm is prssd as n i v j i i (3) ( t ) wr ac ψ i(ω) dscribs a dflctd sap ovr t ntir domain. Eq. (3) is a linar combination of t basis functions st {ψ i(ω) } As sown in past lcturs, substitution of Eq. (3) into Eq. () lads to an N-DOF matmatical modl of t mcanical systm, wos quations of motion ar: MEEN 67 HD Introduction to FEM in Vibrations 3 Dr. L. San Andrés

3 Mv+Kv=Q (4) wr M = M T and K = K T ar t N N matrics of mass and stiffnss cofficints, rspctivly, and Q is t vctor of gnralizd forcs. T cofficints in M and K ar dtrmind from rlations of t sap functions and its drivativs. For ampl: For an lastic bar subjctd to aial motions: M M A d, ij ji i j L L d d d d i j Kij K ji EA d u,a (5.a) wil, for an lastic bam undr transvrs or latral dformations, M M A d, ij ji i j L L d d i j Kij K ji EI d d d v,e,i (5.b) Abov, t st of sap functions i i =,,.. must satisfy b an N admissibl st (wic mans): * i must b a linarly indpndnt st, * satisfy t ssntial boundary conditions, * b sufficintly diffrntiabl as rquird by t strain nrgy function. MEEN 67 HD Introduction to FEM in Vibrations 3 Dr. L. San Andrés 3

4 Howvr, tr ar a numbr of problms associatd wit ts rquirmnts: a) a compl systm gomtry rquirs of compl sap functions, i.., a difficult task for t inprincd usr; b) i ar dfind ovr t ntir domain () and tus, ty lad to a igly coupld systm of quations; ar rlatd to a particular problm; and consquntly, not gnral. c) i T FEM ovrcoms ts difficultis and provids a sound basis for t analysis of vibrations of mcanical systms. T FEM can b nvisiond in t prsnt contt as an application of t assumd mods mtod wrin t sap functions rprsnt dflction ovr just a portion (finit lmnt) of t i structur, wit t lmnts bing assmbld to form t structural systm. MEEN 67 HD Introduction to FEM in Vibrations 3 Dr. L. San Andrés 4

5 A FE modl for aial dformations of an lastic bar Figur sows an lastic bar (on fid nd) subjctd to aial loads (P, f) (t) tat caus aial displacmnts or lastic dformations u (,t).,a u(,t) f L P=AE du/d Fig. Elastic bar undr aial displacmnts T first stp in t FEM is to divid t domain Ω into a sris of finit lmnts {Ω } and tn constructing a finit lmnt ms, as sown in Fig. 3. lmnt lngt N N N+ N nod or joint Fig 3. Discrtization of bar into N finit lmnts MEEN 67 HD Introduction to FEM in Vibrations 3 Dr. L. San Andrés 5

6 Drivation of EOM for on finit lmnt A typical lmnt Ω = ( A, B ) is isolatd from t ms. =B-A A =-A Figur 4 dpicts a fr body diagram of t lmnt, wr u and u ar t (nodal) aial displacmnts at t tips of t lmnt, and P and P ar t (nod) aial forcs arising from t ractions wit t nigboring lmnts. Not f rprsnts a distributd forc/unit lngt acting on t bar. P u f Distributd forc/lngt u,a P u P EA u P EA ; Fig 4. Fr body diagram for a finit lmnt in aial bar T nodal aial forcs P ar u u P EA ; P EA MEEN 67 HD Introduction to FEM in Vibrations 3 Dr. L. San Andrés 6

7 T kintic nrgy (T ) and potntial (strain) nrgy (V ) for t finit lmnt Ω ar: u T A d, t u V EA d and t virtual work from t trnal forcs ovr t lmnt is t (6) ( A) ( B) (7) W f u d u P u P T Hamiltonian Principl applis ovr t wol systm; and nc, it also olds ovr t lmnt (Ω ); tus t T V W dt t (8) t In Ω = ( A, B ), lt t displacmnt u b rprsntd as u t, i u i () t (9) i wr u u,, A t u ub, tar t nodal displacmnts and and ar sap functions admissibl to t problm. Not tat on prsums to know u, u. MEEN 67 HD Introduction to FEM in Vibrations 3 Dr. L. San Andrés 7

8 Substitution of Eq. (9) into Eq. (6) givs: ij i j, ij i j () T M uu V K uu i j i j wr t cofficints of t lmnt mass and lmnt stiffnss matrics ar: T sap functions in Eq. (9) satisfy: ij ji i j M M A d, d d i j K K E A d ij ji d d i ( t) i t, i At A or, u u u u, (, t ) () u u u u must ( ) ( ) At or, B, (3a) ( ) ( ) u u u u (, t) ( ) ( ), (3b) ( ) ( ) MEEN 67 HD Introduction to FEM in Vibrations 3 Dr. L. San Andrés 8

9 Not tat from Eq. (), t sap functions nd to b at last onc C. diffrntiabl ovr t lmnt Ω, i.., i i,.. n Not n > in Eq. (9) is also a coic. Howvr, igr ordr lmnts ar not discussd r for brvity. Work now wit t virtual work prformd by t trnal forcs. Substitution of Eq. (9) into Eq. (7) lads to: Wt f i dui u P () u P ( ) Wt Fi ui up up (4) Sinc (). AbovF ( ) i f i d is a vctor of distributd forcs, and rcall tat t nodal raction aial forcs. u u P EA ; P EA Substitution of Eqs. (, 4) into t Hamiltonian Principl in Eq. (8) lads to t systm of quations for ONE lmnt: ar or in matri form n n M ijuj Kijuj Fi Pi j j (5a) Mu +Ku=F +P (5b) MEEN 67 HD Introduction to FEM in Vibrations 3 Dr. L. San Andrés 9

10 From Eqs. (3) t sap functions i must satisfy t ssntial i =, boundary conditions, ;, and C, i i =, Slct, ; (6) Fig 5. Sap functions for a finit lmnt in aial bar Not tat a) ψ and ψ ar linar combinations of t linarly indpndnt complt st, t sap functions ar a,. In addition, partition of unity (Tis mans t sap functions i i =, ar abl to modl rigid body displacmnts). is diffrnt from zro only on Ω and lswr is b) t st i i =, zro. Tis quality is calld a local support and it is trmly important MEEN 67 HD Introduction to FEM in Vibrations 3 Dr. L. San Andrés

11 to mak t global M and K matrics bandd, i.., wit a small numbr of non-zro valus. T drivation of t sap (or intrpolation) functions i i =, dos not dpnd on t problm. T functions do dpnd on t typ of lmnt (gomtry, numbr of nods or joints, and t numbr of primary unknowns). Substitution of t sap functions, Eq. (6), into t lmnt mass and stiffnss matrics givs: A AE f M ; ; 6 K F (7) for a bar wit uniform cross sctional ara (A ) and matrial proprtis (ρ, E) witin t lmnt. Not tat K is a singular matri. T nt stp constructs (in t computr) t matrics (M, K ) =,,..N for ac lmnt in t domain of intrst. Nt, on prforms t intrconnction or assmbly of t lmnt EOMs. For t sak of discussion, suppos t domain Ω = (, L) is dividd into 3 lmnts of possibly unqual lngts, as sown blow 3 P Global nods MEEN 67 HD Introduction to FEM in Vibrations 3 Dr. L. San Andrés

12 Dfin t global displacmnts U={U, U, U 3, U 4 } U U U3 U4 3 4 Global displacmnts wil t (local) nodal displacmnts in ac lmnt () ar: u u 3 3 u u u u 3 4 lmnt displacmnts 3 T lmnts ar connctd at global nods () and (3); sinc t displacmnt U nds to b continuous (i.., witout cracks, fracturs, tc.). Tn, t intrlmnt continuity conditions ar U u, u U u u U u u U MEEN 67 HD Introduction to FEM in Vibrations 3 Dr. L. San Andrés,, (8) A BOOLEAN or CONNECTIVITY ARRAY (B) stats t corrspondnc btwn local nods in an lmnt and t global nods. B ij t global nod numbr corrsponding to t j-t nod of lmnt i i =,....N : numbr lmnts on ms. j =,....N n : numbr of nods pr lmnt.

13 Prsntly, B 3 34 Rptition of a numbr in B indicats tat t cofficints of K and M associatd wit t nod numbr will add. In a computr implmntation of t FE scm, t connctivity array is usd tnsivly for t automatic assmbly of t global systm of quations. For t tr lmnts in t ampl, t lmnt Eqns. (7) ar writtn in global coordinats or nods as: =: =: =3: U U F P U A AE U F P 6 U U A 6 A U 4 U 4 U U U U F P AE U 3 U 3 F P U 4 U 4 U U U AE U U 3 U3 F P 3 3 U 4 U4 F P MEEN 67 HD Introduction to FEM in Vibrations 3 Dr. L. San Andrés 3

14 T global EOMS for t wol systm ar obtaind by suprposition (addition) of t quations abov: MU KUF+P F P F F P P 3 3 F F P P 3 3 F P (8) wr N ; M M K K (9) N ar t global matrics of mass and stiffnss cofficints, and N ; F f P P () N ar t vctors of distributd forcs and nodal forcs, rspctivly. Not abov K is singular sinc t boundary condition (fid nd at =, i.. U =) is yt to b applid. Imposition of boundary conditions In gnral, du to continuity (action=raction), at t intrnal nods P P, P P 3 Or P P Q, P P Q, 3 3 P Q () 3 4 wit Q, Q 3 and Q 4 as spcifid nodal (aial) forcs acting on t bar. Incidntally, U = is spcifid, wil t wall raction forc an unknown. P? is MEEN 67 HD Introduction to FEM in Vibrations 3 Dr. L. San Andrés 4

15 In gnral, lt Ua U=, wr U a is a vctor containing t activ U d DOF (dgrs of frdom) and U d ar t spcifid (tim invariant) DOF. Tn, t global quations of motion ar partitiond as Maa Mad U Kaa Kad Ua S a + = a Mda Mdd Ud Kda Kdd Ud Sd () wr Mad Mda, Kda K ad. S a =F+Q is a vctor of (known) applid forcs (distributd and nodal) and S d is a vctor of unknown raction forcs (to b calculatd) Sinc U d, pand Eqs. () as M U K U K U S M U K U K U S aa a aa a ad d a da a da a dd d d Solv for U a from t first quation abov, M U K U S K U (3) aa a aa a a ad d and nt, find t intrnal forcs S d from t nd Equation S M U K U K U (4) d da a da a dd d For t lastic bar, t systm of qns. (3) is tridiagonal and tus, its solution can b obtaind quickly witout a matri invrsion. Not tat t ssntial BCs is a spcifid U = wil t raction forc P is unknown. Hnc, t final EOMS ar: MEEN 67 HD Introduction to FEM in Vibrations 3 Dr. L. San Andrés 5

16 M, M,3 M,4U K, K,3 K,4 U M,3 M3,3 M3,4 U 3 K,3 K3,3 K3,4 U3 M 4, M3,4 M 4,4 U 4 K4, K3,4 K 4,4 U 4 P ( t ) (5) and onc t quation abov is solvd, M U K U M U K U M U K U P (6),,,3 3,3 3,4 4,4 4 In t ampl configuration, t satisfaction of t ssntial constant U = rmovs t singularity of t stiffnss matri (i.., rmovs t rigid body mod). For t ampl cas, considring lmnts of qual lngt, = L/3, t quations of motion ar: 4 U U AL 3 4 AE U 3 U3 8 L U U P t 4 4 ( ) (7) MEEN 67 HD Introduction to FEM in Vibrations 3 Dr. L. San Andrés 6

17 Us MODAL ANALYSIS to solv MU KU F (8) +Initial Conditions To tis nd, first solv t omognous EOMs MU KU and mak t modal matri, i K i M (9)... N (3) wr N is t activ dgrs of frdom in t systm. Using t modal transformation U () t () t, t pysical EOMS bcom in modal coordinats: m m m (3) M K Q M M K K, and T m m T wr ; T Qm F (3) ar t modal mass and stiffnss matrics (diagonal) and t modal forc vctor, rspctivly. By now, you do know ow to solv t uncoupld st of N ODES and tn rturn to pysical coordinats. MEEN 67 HD Introduction to FEM in Vibrations 3 Dr. L. San Andrés 7

18 Eampls How good is a FE modl? Rcall t EOM for on-lmnt M U K U F (a) T wr U U ; F F U F (a) T and M AL 6 AE, K L (a3) wit L as t bar lngt wit cross-sction A and matrial proprtis (,E). Bar wit on nd fid: Lt s find its fundamntal nat. frquncy T St U U U ; F F? F EOM (a) rducs to AL AE F 6 U L U AL AE 3 L Or U U T Tn _ FEM 3 E L / and / E (act) L HD4 T ratio of natural frquncis FEM/act =. MEEN 67 HD Introduction to FEM in Vibrations 3 Dr. L. San Andrés 8

19 Bar wit two fr nds: Find its fundamntal lastic mod natural frquncy T St U U U ; F F F EOM (a) rducs to AL U AE U 6 U L U Procss DOF quation abov to obtain E, L _ FEM _ FEM / T / E, (act) L Handout 4 T ratio of lastic mod nat. frquncis FEM/act=. ONE FE modl dlivrs clos prdictions for t lastic natural frquncis. A qustion to pondr: Wy dos t FE modl producs LARGER natural frquncis tan t act (analytical) ons? MEEN 67 HD Introduction to FEM in Vibrations 3 Dr. L. San Andrés 9

20 A FE modl for bnding of lastic bams Figur 6 dpicts an lastic cantilvr bam wit latral forcs (F) and momnts (M) acting on it tat produc t bam latral dflction or bnding v (,t). F o and M o ar a latral forc and momnt applid at t nd of t bam (=L), and f () is a distributd forc/unit lngt. T bam as matrial dnsity () and lastic modulus (E) and gomtric proprtis of ara (A) and ara momnt of inrtia (I).,E,I F v(,t) f o f M o Fig 6. Elastic bam undr latral (bnding) displacmnts For a finit lmnt ( ) or pic of t bam wit widt, not t sign convntion: f V L M M V lmnt lngt Fig 7. Notation for sar forcs (V) and momnts (M) Lt V and M dnot t sar forc and bnding momnt. For an Eulrtyp bam MEEN 67 HD Introduction to FEM in Vibrations 3 Dr. L. San Andrés

21 v M EI ; V M (33) In t FEM, discrtiz t bam () into a collction of finit lmnts, as sown blow in Fig. 8. lmnt lngt N N N+ N nod or joint Fig 8. Discrtization of bam into N finit lmnts Lt v and v b t latral displacmnt and bam rotation (primary and scondary variabls). Fig. 9 sows a fr body diagram in t lmnt Ω, wit, MEEN 67 HD Introduction to FEM in Vibrations 3 Dr. L. San Andrés

22 V f V3 f: Distributd forc/lngt Displacmnt V and rotation,e,i lmnt lngt Q Q + Q3 Q4 Latral forcs and momnts Fig 9. Fr body diagram for a finit lmnt in a lastic bam T sar forcs at t nds of t lmnt v Q V ( ) EI Q V EI ( ) 3 And t bnding momnts at t nds ar v Q M( ) EI 6 v Q4 M( ) EI ; v (34) (35) MEEN 67 HD Introduction to FEM in Vibrations 3 Dr. L. San Andrés

23 Lt V v ; V v V 3, t, t ; V 4, t, t (36) V V 4 V V3 = = Assum tat t bam latral displacmnt and rotation (angl) in ar givn by t approimation v 4 ( t, ) i V i( t ) i (, t) MEEN 67 HD Introduction to FEM in Vibrations 3 Dr. L. San Andrés 3 4 i d d i V i ( t ) (37) From a prior discussion on t assumd mods mtod, t lmnts of t mass and stiffnss matrics for ar ij ji i j M M A d, d d i j K K E I d ij ji d d (38)

24 Not tat C. T sap functions satisfy t ssntial i i,..4 boundary conditions, v v, t, t V ; 3 4 () () () () V ; ( ) ( ) ( ) ( ) (39a) v, t, t 3 () () 3() 4() d d d d d d d d V ; 4( ) ( ) ( ) 3( ) d d d d d d d d V ; (39b) T lowst ordr polynomial tat satisfis t conditions abov is tird ordr (and is twic diffrntiabl), i.., 3 vt, c c c c3 c c 3c t, 3 3 (4) Lading to t following conditions, : V c, V : V c c c c, c 3 3 c c 3c 3 V4 Solving t st of four Eqs. (4) givs (4) MEEN 67 HD Introduction to FEM in Vibrations 3 Dr. L. San Andrés 4

25 ( ) ( ) 3 ( ) 4 ( ) (4) Figur dpicts t sap functions. Not tat ( ) ( ) ( ) ( ) Fig. Sap functions for structural lmnt undr bnding. MEEN 67 HD Introduction to FEM in Vibrations 3 Dr. L. San Andrés 5

26 Substitution of Eq. (4) into Eq. (38), rndrs t following lmnt mass and stiffnss matrics A T M M (43) 6 6 E I T K 3 K (44) and t vctor of gnralizd forcs {F i } is: i F f d (45) t, it, Assuming a constant distributd forc f (,t) ovr t lmnt, obtain: F f f f f (45b) and t constraint nodal forcs ar obtaind from t virtual work prformd: W Q V Q V Q V Q V (46) Q (46b) Hnc, Q Q Q3 Q4 Tn, t systm of quations for t lmnt ar: T T MEEN 67 HD Introduction to FEM in Vibrations 3 Dr. L. San Andrés 6

27 4 4 M ijvj KijVj Fi Qi (47) j j or (47) MV +KV =F +Q wr V V V V3 V4 Assmbly of t lmnt matrics to produc t global systm of quations is asily don as mplifid bfor. B carful to kp continuity of t displacmnts at t nods (joints) and also t continuity of constraint forcs at t nods. T global systm of quations bcoms MV+KV=F+Q (48) wr t global vctor of displacmnts is V VV,, VV,,... V, V (49) 3 4 wit N as t global numbr of nods); N ; N T N N N M M K K (5) ar t global matrics of mass and stiffnss cofficints, and N ; F f Q Q (5) N ar t vctors of distributd forcs and nodal forcs, rspctivly. Not abov K is singular sinc t boundary conditions ar yt to b applid. T MEEN 67 HD Introduction to FEM in Vibrations 3 Dr. L. San Andrés 7

28 T global FEM vctor of forcs Q will in gnral av zro componnts at t intrnal nods. T lmnts of tis vctor ar of t form, s Fig. : Hr k dnots t global nod numbr. Q Q Q ; Q Q Q (5) k 3 k 4 V3 = V = Q3 + Q4 + Q Q Fig. Balanc of forcs at a nod T qns. abov constitut a statmnt of quilibrium of forcs at t nodal intrfac (boundary) of an lmnt. Rcall from Eqns. (34, 35) tat v v Q Q EI EI 3 v v Q4 Q EI EI If no trnal forcs or momnts ar applid at t nod, tn MEEN 67 HD Introduction to FEM in Vibrations 3 Dr. L. San Andrés 8

29 Q Q Q, k 3 Q Q Q k 4 (53) Howvr, if an trnal nodal sar forc or momnt is applid at t joint, tn t componnts of t forc vctor Q will not b zro. Considr, for ampl, t cas of a support spring wit stiffnss k s conncting t bam to ground, i.., Q Q F spring 3 ; Q Q Q F kv (54) V = = 3 k 3 spring s V V + Q3 + Q Q ks + kv s F spring Fig. Balanc of forcs at a nod wit a spring connction Not tat tis spring rstoring forc dpnds on t lmnt latral dflction, and trfor, it is unknown. Hnc, on nds to modify t global stiffnss matri and add t contribution of t support stiffnss k s.. MEEN 67 HD Introduction to FEM in Vibrations 3 Dr. L. San Andrés 9

30 Coordinat transformation of lmnt matrics: A plan lastic fram lmnt combins t lastic proprtis of a bar and a bam; and nc, it as tr displacmnt coordinats at ac nd (two ortogonal displacmnts and on rotation). As sown in Figur 3, tis lmnt as a local coordinat systm (,y) wr t - coordinat aligns wit t major ais (lngt) of t bar-bam. U 5 U 6 y U 4 U U 3 U Fig 3. Arbitrary fram lmnt and displacmnts T -ais is tiltd angl rlativ to a global (inrtial) coordinat systm (X, Y) to wic all lmnts in t structur will b rlatd. In t (X,Y) coordinat systm, t displacmnts ar dfind as sown in Fig. 4. X MEEN 67 HD Introduction to FEM in Vibrations 3 Dr. L. San Andrés 3

31 y Y U 3 U 5 U 6 U 4 U U Fig 4. Arbitrary fram lmnt and displacmnts in (X,Y) as X in t T transformation btwn t local displacmnts i i,...6 (X,Y) to t displacmntsv i in t local coordinat systm i,...6 (,y) is givn by t (coordinat transformation) quation V =TV (55) V wr T cos sin sin cos cos sin sin cos (56) T lmnt EOMs in t (,y) coordinat systm ar (57) MV +KV =F +Q MEEN 67 HD Introduction to FEM in Vibrations 3 Dr. L. San Andrés 3

32 Substitution of (Eq. 55) into (57) and pr-multiplying by T T givs t following: T M T V +T K T V =T F +T Q (58) T T T T T T T T Lt: M T M T, K T K T, F T F, Q T Q (59) and writ t EOM as: MV +KV =F +Q (6) Not tat t rsulting mass and stiffnss matrics ar still symmtric. T assmbly of t lmnt matrics procds in t usual way to obtain t global systm of quations. Constraints and rduction of dgrs of frdom: Tus far, t analysis taks all gnralizd displacmnts as indpndnt of ac otr. T assumption lad to t global systm of quations: MV+KV=F+Q (6) Frquntly, tr ariss a nd to spcify rlationsips amongst systm displacmnt coordinats. Tis is quivalnt to spcifying N displacmnts wic ar linarly indpndnt and t N +, N +,....N dpnd on t N displacmnts. rst T discussion prsntly rlats to a constraint quation of t form: f V, V,... V g V, V,... V,..., i N N N i N i N (6) N T quation abov can b writtn in matri form as: Va RV= R da R dd = Vd (63) MEEN 67 HD Introduction to FEM in Vibrations 3 Dr. L. San Andrés 3

33 wr V d is t vctor of N d dpndnt coordinats or displacmnts, and V a is t vctor of N Na = N a indpndnt or ACTIVE coordinats or displacmnts. Not N + N d= N a+ N d= N. From q. (63) RdaVa RddVd find V =T V = -R R V (64) - d da a dd da a wr T da is t (N d N a ) matri transformation btwn activ to dpndnt dgrs of frdom. Now, t total global displacmnt vctor can b writtn as: Va Iaa V = Va TV Vd Tda wr I aa is a (N a N a ) unit matri. Substitution of (64) into t EOM (6) and pr-multiplying by givs: T T a a T T or T M TV +T K TV =T F+T Q MaV a +Ka V a =F a +Qa a (65) T T (66) wr t mass and stiffnss matrics ar rducd to (N a N a ) activ DOF. Not tat, T T T T M T MT, K T KT, F T F, Q T Q (67) a a a a T systm of quations (66) accounts only for t activ dgrs of frdom N a. MEEN 67 HD Introduction to FEM in Vibrations 3 Dr. L. San Andrés 33

34 EXAMPLE: Using ONE finit lmnt dtrmin t first natural frquncy for t bam configurations (pin-pin nds, fid nd-fr nd, fid nd-fid nd) and compar t rsults wit avialabl closd form formula. Sow t prcntag rror. T bam as lngt L, cross sctional ara A, inrtia Ip and lastic modulus E. T gnralizd mass and stiffnss matrics for bam bnding ar: LSA ME67 56 L 54 3L L L 4L 3L 3L EI 6L M = K = L 56 L L 3 3L 3L L 4L 6L 6L 4L 6L L 6L 6L 6L L 6L 4L 3 4 displacmnt & slop (lft nd) displacmnt & slop (rigt nd) Pin-Pin Ends: No displacmnt and momnt at ac nd st boundary conditions: Momnt Momnt 3 L 4 56 L 54 3L L 4L 3L 3L 54 3L 56 L 4L 3L 3L L 3 4 EIp L 3 6L 6L 6L 4L 6L L 6L 6L 6L L 6L 4L 3 4 = Forc Momnt Forc Momnt L 4 56 L 54 3L L 4L 3L 3L 54 3L 56 L 3L 3L L 4L 4 EIp L 3 6L 6L 6L 4L 6L L 6L 6L 6L L 6L 4L 4 = Forc Forc wic rducs to a by systm of quations for fr vibration L 4 4L 3L 3L 4L 4 EIp L 3 4L L L 4L 4 = t natural frquncis can b dtrmind by t ignvalu analysis ORIGIN I M K = M K = 4 L 7L 4 4L 3L sort( ignvals( A) ) = L EIp 3L 4L act first natural frquncy i EIp L 3 rror 4L L i L 4L i 3 L 4 EIp L 4 EIp 3 L 4 EIp L 4 EIp rror = L % EIp A 3 3

35 Fid nd-fr nd: zro displacmnt and slop at t fid nd, and zro sar and momnt at t fr nd. st boundary conditions: Forc Momnt L 4 56 L 54 3L L 4L 3L 3L 54 3L 56 L 4L 3L 3L L 3 4 EIp L 3 6L 6L 6L 4L 6L L 6L 6L 6L L 6L 4L 3 4 = Forc Momnt Forc Momnt L 4 56 L 54 3L L 4L 3L 3L 54 3L 56 L 3L 3L L 3 4L 4 EIp L 3 6L 6L 6L 4L 6L L 6L 6L 6L L 6L 4L 3 4 = Forc Momnt wic rducs to a by systm of quations for fr vibration L 4 56 L L 4L 3 4 EIp L 3 6L 6L 4L 3 4 = Now t natural frquncis can b dtrmind by t ignvalu analysis I M K = M K 4 L 4L 4L L L 56 EI L 3 6L 6L 4L 5 L 4 EI 6 EI L 5 9 L 3 EI 476 L 4 EI A i sort( ignvals( A) ) i i = L EIp act first natural frquncy = L EIp rror rror.483 %

36 Fid End - Fid End: Rduc bam into two parts, ac of lngt L/. T boundaty conditions bcom zro displacmnt and slop at t fid nd, and zro slop at t otr nd of t / bam. st boundary conditions: 4 Forc L o L = wr Lo is t original lngt of t bam L 4 56 L 54 3L L 4L 3L 3L 54 3L 56 L 4L 3L 3L L 3 4 EIp L 3 6L 6L 6L 4L 6L L 6L 6L 6L L 6L 4L 3 4 = Forc Momnt Forc Momnt L 4 56 L 54 3L L 4L 3L 3L 54 3L 56 L 3L 3L L 4L 3 EIp L 3 6L 6L 6L 4L 6L L 6L 6L 6L L 6L 4L 3 = Forc Momnt Momnt wic rducs to just on quation for fr vibration of t dflction at midspan L EIp L 3 = 3 Now t natural frquncy is = 4 4 L o 4 56 EIp.736 = L o EIp = L o EIp act first natural frquncy rror rror.6 %