MU+CU+KU=F MU+CU+KU=0

Transcription

1 MEEN 67 Handout # MODAL ANALYSIS OF MDOF Systems with VISCOUS DAMPING ^ Symmetic he motion of a n-dof linea system is descibed by the set of 2 nd ode diffeential equations MU+CU+KU=F t whee U (t) and F (t) ae n ows vectos of displacements and extenal foces, espectively. M, K, C ae the system (nxn) matices of mass, stiffness, and viscous damping coefficients. hese matices ae symmetic, i.e. M=M, K=K, C=C. he solution to Eq. () is detemined uniquely if vectos of initial displacements U o and initial velocities dt ( ) t () V d U o ae specified. Fo fee vibations, the foce vecto F (t) =, and Eq. () is MU+CU+KU= A solution to Eq. (2) is of the fom (2) U e t ψ (3) whee in geneal is a complex numbe. Substitution of Eq. (3) into Eq. (2) leads to the following chaacteistic equation: 2 M C K Ψ f Ψ (4) MEEN 67 HD Modal Analysis of MDOF Systems with Viscous Damping L. San Andés 23

2 whee f is a nxn squae matix. he system of homogeneous equations (4) has a nontivial solution only if the deteminant of the system of equation equals zeo, i.e n f c c c2 c3... c2n (5) he oots of the chaacteistic polynomial given by Eq. (5) can be of thee types: a) Real and negative,, coesponding to ove damped modes. b) Puely imaginay, i, fo undamped modes. c) Complex conjugate pais of the fom, id, fo unde damped modes. Clealy if the eal pat of any, it means the system is unstable. he constituent solution, Eq. (3), supeposition of the solution oots satisfying Eq. (4), i.e., U e t ψ can be witten as the e and its associated vectos ψ U 2n t t C Ψ e Ψ Ψ Ψ... Ψ nx2 n n (7) o letting 2 2 (6) wite Eq. (6) as Only if the system is defined by symmetic matices. Othewise, the complex oots may NO BE complex conjugate pais. MEEN 67 HD Modal Analysis of MDOF Systems with Viscous Damping L. San Andés 23 2

3 U Ψ (8) Ce t t Howeve, a tansfomation of the fom, t 2 nx Unx Ψ q (9) is not possible since this implies the existence of 2n- modal coodinates which is not physically appaent when the numbe of physical coodinates is only n. o ovecome this appaent difficulty, efomulate the poblem in a slightly diffeent fom. Let Y be a 2n- ows vecto composed of the physical velocities and displacements, i.e. U Y= U, and Q= F (t) () be a modified foce vecto. hen wite MU+CU+KU=F t as M U -M U + = M C U K U F o whee (.a) AY+BY=Q M -M A=, B= M C K (.b) (2) MEEN 67 HD Modal Analysis of MDOF Systems with Viscous Damping L. San Andés 23 3

4 A and B ae 2nx2n matices, symmetic if the M, C, K matices ae also symmetic. Fo fee vibations, Q=, and a solution to Eq. (.b) is sought of the fom: U U =Y=Φ Substitution of Eq. (3) into Eq. (.b) gives: e t (3) A+B Φ (4) which can be witten in the familia fom: DΦ Φ (5) whee - - M M D=-B A= -, o -K M C I D= - - -K M -K C (6) with I as the nxn identity matix. Fom Eq. (5) wite D Ι Φ f Φ (7) he eigenvalue poblem has a nontivial solution if f (8) MEEN 67 HD Modal Analysis of MDOF Systems with Viscous Damping L. San Andés 23 4

5 Fom Eq. (8) detemine 2n eigenvalues eigenvectos Φ, =, 2.., 2n and associated. Each eigenvecto must satisfy the elationship: DΦ Φ (9) and can be witten as Ψ Φ = 2 Ψ whee Ψ is a nx vecto satisfying: I Ψ Ψ = K M -K CΨ Ψ fom the fist ow of Eq. (2) detemine that: 2 I Ψ Ψ o (2) 2 Ψ Ψ (2) and fom the second ow of Eq. (2), with substitution of the elationship in Eq. (2), obtain -K - MΨ K CΨ Ψ o K M - K C -I Ψ = (23) fo =, 2,.2n. Note that multiplying Eq. (23) by (- K) gives 2 2 M C K Ψ (4) i.e., the oiginal eigenvalue poblem. MEEN 67 HD Modal Analysis of MDOF Systems with Viscous Damping L. San Andés 23 5

6 Solution of Eq. (9), DΦ Φ, delives the 2n-eigenpais ; Φ Ψ Ψ (24),2,..2n In geneal, the j-components of the eigenvectos numbes witten as Ψ ae complex i a i b e Ψ j j j j j=,2 n j whee and φ denote the magnitude and the phase angle. Note: fo viscous damped systems, not only the amplitudes but also the phase angles ae abitay. Howeve, the atios of amplitudes and diffeences in phase angles ae constant fo each of the elements in the eigenvecto Ψ. hat is, / const and const fo j, k =, 2,..N j k jk j k jk A constituent solution of the homogeneous equation (fee vibation poblem) is then given as: U 2n U t Y= C e Φ (25) Let the (oots) be witten in the fom (when unde-damped) i (26) d and wite Eq. (25) as MEEN 67 HD Modal Analysis of MDOF Systems with Viscous Damping L. San Andés 23 6

7 2n U Y= U Φ i C e d t (27) and since Ψ Φ Ψ, the vecto of displacements is just U 2n i C Ψ e d t (27) ORHOGONALIY OF DAMPED MODES Each eigenvalue and its coesponding eigenvecto Φ satisfy the equation: ΑΦ B Φ (28) Conside two diffeent eigenvalues (not complex conjugates): s; sand q; q Φ Φ, then if is easy to demonstate that: and infe and Α = Α B=B (a symmetic system), it s q Φ A Φ Φs AΦ q= ; fo s q s q s q Φ ΒΦ = (29) Now, constuct a modal damped matix Φ (2nx2n) fomed by the columns of the modal vectosφ, i.e. Φ Φ Φ... Φ.. Φ Φ (3) 2 n 2n 2n MEEN 67 HD Modal Analysis of MDOF Systems with Viscous Damping L. San Andés 23 7

8 And wite the othogonality popety as: ΦΑΦ=σ Φ ΒΦ=β (3) Whee σ and β ae (2nx2n) diagonal matices. Now, ecall that the equations of motion in physical coodinates ae : MU+CU+KU=F t ( ) () With the definition U Y U, Eqs. () ae conveted into 2n fist ode diffeential equations: Α Y+Β Y=Q F () t M -M whee A=, B= M C K o uncouple the set of 2n fist-ode Eqs. (32), a solution of the following fom is assumed: 2n Y Φ z ΦZ t t t (32) (2) (33) Substitution of Eq. (33) into Eq. (32) gives: ΑΦΖ+ ΒΦΖ=Q (34) MEEN 67 HD Modal Analysis of MDOF Systems with Viscous Damping L. San Andés 23 8

9 Pemultiply this equation by the damped modes to get: Φ Α Φ Ζ+ ΦΒΦΖ= Φ Q σζ+β Ζ=G o Φ and use the othogonality popety 2 of Φ Q Eq. (36) epesents a set of 2n uncoupled fist ode equations: z z g t z z g t ( ) ( ) (35) (36) (37) z z g 2N 2N 2N 2N 2N t ( ) Φ ΑΦ Φ B Φ, =, 2..2N whee ; / (38) since ΑΦ Β Φ. In addition, g Φ Q (39) () t () t Initial conditions ae also detemined fom Y U o o Uo with the tansfomation Y Φ Z t σζ =Φ o ΑYo (4.a) 2 he esult below is only valid fo symmetic systems, i.e. with M, K and C as symmetic matices. Fo the moe geneal case (non symmetic system), see the textbook of Meiovitch to find a discussion on LEF and RIGH eigenvectos. MEEN 67 HD Modal Analysis of MDOF Systems with Viscous Damping L. San Andés 23 9

10 o o Φ ΑY o =, 2, 2n (4.b) he geneal solution of the fist ode equation z z g t, with initial condition z z t o, is deived fom the Convolution integal with / t t t (4) z z e g e d o Once each of the coodinates to obtain: z () t solutions is obtained, then etun to the physical 2n U Y Φ z ΦZ U t t t (33=43) and since given by: Ψ Φ Ψ, the physical displacement dynamic esponse is U t 2n Ψ z t (44) and the velocity vecto is coespondingly equal to: U t 2n Ψ z t (45) MEEN 67 HD Modal Analysis of MDOF Systems with Viscous Damping L. San Andés 23

11 Read/study the accompanying MAHCAD woksheet with a detailed example fo discussion in class. MEEN 67 HD Modal Analysis of MDOF Systems with Viscous Damping L. San Andés 23

12 MODAL ANALYSIS of MDOF linea systems with viscous damping Oiginal by D. Luis San Andes fo MEEN 67 class / SP 8, 2 he equations of motion ae: M d 2 U/dt 2 + C du/dt + K U = F(t) whee M,C,K ae nxn SYMMERIC matices of inetia, viscous damping and stiffness coefficients, and U, du/dt, d 2 U/dt 2,and ae the nx vectos of displacements, velocity and acceleations. F(t) is the nx vecto of genealized foces. Eq () is solved with appopiate initial conditions, at t=, Uo,Vo=dU/dt () ======================================================================== Define elements of inetia, stiffness, and damping matices: m k. 7 c 5 m 2 kg N/m N.s/m k 2. 7 c 2 2 m 3 5 k c 3 n 3 # of DOF Make matices: M m m 2 m 3 K k k 2 k 2 k 2 k 2 k 3 k 3 Initial conditions in displacement and velocity: k 3 k 3 C c c 2 c 2 c 2 c 2 c 3 c 3 c 3 c 3 Uo Vo. PLOs - only N 24# fo time steps natual feqs - undamped damped eigenvals ===================================================================== 2. Evaluate the damped eigenvalues: Rewite Eq (), as A dy/dt + B Y = Q(t), whee Y = [ du/dt, U ], and Q=[, F(t)] ae 2n ow vectos of (velocity, displacements ) and genealized foces; and initial conditions Yo = [ Vo, Uo ] M A = B = M C M K ae 2nx2n Symmetic matices

13 2. define the A & B augmented matices: zeo identity( n) identity( n) the (nxn) null matix A stack( augment( zeo M) augment( M C) ) B stack( augment( M zeo) augment( zeo K) ) 2.2 Use MAHCAD function to calculate eigenvectos and eigenvalues of the genealized eigenvalue poblem, M X = N X. In vibations poblems we set the poblem as: + B =, hence M=-B, N=A to use popely the MAHCAD functions genvecs & genvals genvals( B A) ad/s D genvecs( B A) i i i i i i Recall the undamped natual fequencies note that the (damped) eigenvalues ae complex conjugates, with the same eal pat and +/- imaginay pats D.35.2i.6.2i.64.23i i i i i.6.2i.64.23i i i i i.7.2i.22.44i i i i 4 Note that the eigenvectos ae conjugate pais, i.e. they show the same eal pat and +/- imaginay pat. In addition the fist n-ows of an eigenvecto ae popotional to the 2nd n-ows. he popotionality constant is the damped eingenvalue.

14 2.2. Fom "damped" modal matices using the othogonality popeties: D A D D B D.54.75i.54.75i.26.26i.26.26i.2.i.2.i i i i i i i i i i i i i 4 9.2i i i i i Note off-diagonal tems ae small - but NO zeo (as they should) Check the othogonality popety: compae / atios to eigenvalues jj jj i i i i i i i i i i i i undamped:

15 fo undedamped systems only damping atios: eal pat: imaginay pat: n.5 d = n 2 j 2n j Im j Re j 2.5 nj Re j j dj natual fequencies Im j damped natual feqs. damped eigenvals ( ) damping atios. n d ( natual ) feqs. ad/s damped natual feqs. ( ) ( ) fom undamped modal analysis damped modal esponse

16 3. he 2n fist ode equations A dy/dt + B Y = Q(t) with the tansfomation Y= d Z become 2n equations of the fom: i dz i /dt + i Z i = G i, i=,2,3,...2n whee G= d Q(t) and initial conditions Zo= d A Yo: 3. solve the 2n-fist ode diffeential equations fo the Fee Response to initial conditions, F(t)=: Fom initial conditions in displacement and velocity, set Yo stack( Vo Uo) and in damped modal space: Zo D A Yo p N t p ( p ) t numbe of data points time sequence m 2n get the modal esponse: Z jp Zo j e jt p andback into the physical coodinates: Y= d Z FREE RESPONSE using DAMPED MODES s 2n m Y sp q Ds q Z qp j 2n damped modal esponse

17 Plot the displacements (last n ows of Y vecto): RESPONSE: U ecall: Uo U damped undamped time(secs) RESPONSE: U2.5 U2.5 RESPONSE: U Damped Undamped time(secs) U Damped Undamped time(secs) ( )

18 4. Foced esponse to step load F(t)=Fo and initial conditions Set initial conditions in displacement and velocity: Uo Vo SEP FORCE Fo O Step load esponse hen set: Yo stack( Vo Uo) Qo stack( O Fo) and tansfom to damped modal space: Zo D A Yo Go D Qo max 8 secs f p N and in physical coodinates: SEP RESPONSE using DAMPED MODES t p ( p ) t s m m Y= d Z 2n =================================================================== 6. Fo completeness, obtain also the undamped foced esponse: o Mm M Uo o Mm M Vo Fo t to get the modal esponse max N Z jp j 2n Zo j e jt p Go j e jt p jj m Y sp Ds q q Z qp j n jp o j cos j t p o j sin j t p j j Km jj cos j t p and into physical coodinates: s n SEP RESPONSE using UNDAMPED MODES U ================================================================== Plot the displacements (last n ows of Y vecto): Step load esponse n U sp q sq qp static esponse - check U s K Fo

19 p p RESPONSE: U. U s U damped undamped time(secs) RESPONSE: U2..5 U RESPONSE: U3 Damped Undamped time(secs).5 U Damped Undamped time(secs) ( )

20 5. Peiodic esponse to loading, F(t)=Fo sin(t): Response due to initial conditions vanishes afte a long time because of damping: feq. of excitation Fo fhz 8 2 fhz ecall the undamped fequencies Plot the displacements (last n ows of Y vecto): RESPONSE: U.2 ecall: j U RESPONSE: U2. damped undamped time(secs) U Damped Undamped time(secs) RESPONSE: U3..5 U3.5

21 6) FRF Response to peiodic loading, F=Fo cos(t) UNDAMPED CASE modal foce magntidue: SE and the modal "S-S" esponse is modes. i n Fo i j n max 3 n kmin kmax Fo MODAL esponse k kmin kmax q jk ad/sec <=== focing fequency k k kmax max kmax x cos(t) kmin 2.6 j Km jj k. FRF in physical plane: 2 j 2 s n 2 j k j n U sk i i si q ik x cos(t) hee i=imaginay unit DAMPED CASE set: Qo stack O Fo ( ) focing vecto Go D Qo k kmin kmax k k kmax max ad/s <=== focing fequency j 2n modal esponse Z jk Go j jj k i j x cos(t) and back into the physical plane:

22 s 2n m 2n PERIODIC RESPONSE using DAMPED MODES m Y sk q Ds q Z qk x cos(t) Plot magnitude of esponse in physical space

23 DAMPED Amplitude FRF. Damped physical esponse. Amplitude fequency (ad/s) U U2 U3 UNDAMPED Amplitude FRF UnDamped physical esponse. Amplitude fequency (ad/s) U U2 U3

24 select coodinate to displate physical esponse - damped & undamped jj 2 n 3 DOFs fequency (ad/s) damped undamped LINEAR vetical scale undamped: ( ) fequency (ad/s) damped undamped