4.3 - Modal Aalyss Physcal coordates are ot always the easest to work Egevectors provde a coveet trasformato to modal coordates Modal coordates are lear combato of physcal coordates Say we have physcal coordates x ad wat to trasform to some other coordates u u = x + 3x u u = x 3x u = 3 3 x x Revew of the Egevalue Problem Start wth M x + Kx= where x s a vector ad M ad K are matrces. Assume tal codtos x ad x. Rewrte as M M M ( q x+ Kx= ad let M x= q x= q (coord. tras. #) Egeproblem (cot) Premultply by M to get M MM q + M KM q= q + K q= (4.55) Now we have a symmetrc, real matrx Guaratees real egevalues ad dstct, mutually orthogoal egevectors Egevectors = Mode Shapes? Mode shapes are solutos to Mω u =Ku physcal coordates. Egevetors are characterstcs of matrces. he two are related by a smple trasformato, but they are ot syoymous. 3 4
Egevectors vs. Mode Shapes he egevectors of the symmetrc PD matrx K are orthoormal,.e., P P = I. Are the mode shapes orthoormal? Usg the trasformato x = M q, the modes shapes U = M P P = M U. Now, P P = U M M U = U MU = I. hus, the mode shapes are orthogoal oly w.r.t. the mass matrx. Smlarly, U KU = P M KM K he Matrx of egevectors ca be used to decouple the equatos of moto If P orthoormal (utary), PP= I P = P P =Λ coordates r(t) 5 6 hus, PKP =Λ= dagoal matrx of egevalues. Back to q+ K q=. Make the addtoal coordate trasformato q= Pr ad premultply by P r+ r= r+λ r= PP PKP I (4.59) Now we have decoupled the EOM.e., we have depedet d-order systems modal Wrtg out equato (4.59) yelds rt ( ) ω rt ( ) (4.6) rt () + = ω rt () rt () + ω rt () = (4.6) rt () + ω rt () = (4.63) We must also trasform the tal codtos r() r / r = = = P () = P M () (4.64) r () r q x r () r / r = = = P () = P M () (4.65) r () r q x 7 hs trasformato takes the problem from couple equatos the physcal coordate system to decoupled equatos the modal coordates x k m k Physcal Coordates. Coupled equatos x m x = M P r r (ω ) r (ω ) Modal Coordates. Ucoupled equatos Fgure 4.5 8
Modal rasforms to SDOF he modal trasformato P M trasforms our DOF to SDOF systems hs allows us to solve the two decoupled SDOF systems depedetly usg the methods of chapter he we ca recombe usg the verse trasformato to obta the soluto terms of the physcal coordates. he free respose s calculated for each mode depedetly usg the formulas from chapter : r rt t r t () = sω + cos ω, =, ω or (see Wdow 4.3 for a remder) r ω r rt () = r+ s( t+ ta ), =, ω ω r Note, the above assumes ether frequecy s zero 9 Oce the soluto modal coordates s determed (r ) the the respose Physcal Coordates s computed: Wth DOFs these trasformatos are: x (t) = S r (t) where S = M P x x Steps solvg va modal aalyss (Wdow 4.4) (where = the prevous sldes)
Modal Coordates: Idepedet oscllators Modal ad Physcal Resposes 4 - Free respose modal coordates r r Secto 4.4 More the Degrees of Freedom = ω = λ -4 3 4 5 6 7 8 9 = π = 4. 44sec, λ = 4 ω = 4 = πsec Physcal Coordates: Coupled oscllators Note IC - π sec 4.44 sec Free respose physcal coordates -4 3 4 5 6 7 8 9 me (s) 3 x x Extedg prevous secto to ay umber of degrees of freedom Fg 4.8 Fg 4.74 A FBD of the system of fgure 4.8 yelds the equatos of moto o the form: m + k ( x x ) k + ( x x )=, =,, 3L (4.83) Wrtg all of these equatos ad castg them matrx form yelds: where: M x + Kx =, (4.8) 5 the relevat matrces ad vectors are: k + k k m k k+ k3 k3 m K k 3 M=, = (4.83) k + k k m k k x() t x() t x() t x() t x() t =, x() t = x() t x() t 6
For such systems as fgure 4.7 ad 4.8 the process stays the same just more modal equatos result: Process stays the same as secto 4.3 r ( t) + ω r r ( t) + ω r + ω 3 r + r 3r3 r ω = = = = Just get more modal equatos, oe for each degree of freedom ( s the umber of dof) he Mode Summato Approach Based o the dea that ay possble tme respose s just a lear combato of the egevectors Startg wth q + K q = (4.88) j λ t j λ t ( ) let q= q = ae + be v = = two learly depedet solutos for each term. = ( ω φ ) ca also wrte ths as q = d s t+ v (4.9) See example 4.4. for detals7 8 Mode Summato Approach (cot) Fd the costats d ad φ from the I.C. q() = d s φ v ad q () = dω cosφ v = = Assumg egevectors ormalzed such that v v j ()= jd s = d s ( j ) = dj s j = = Smlarly for the tal veloctes, v q ()= d ω cosφ = δ j j v q v φ v φ v v φ j j j j Mode Summato Approach (cot) Solve for d ad φ from the two IC equatos v () () q ω v q d = ad φ = ta s φ v q () IMPORAN NOE about q()= f you just crak t through the above expressos you mght coclude that d =,.e., the trval sol. Be careful wth q () = as well. 9
Mode Summato Approach for zero tal dsplacemet If q() =, the retur to q() = d s φ v = ad realze that φ = stead of d =. he compute d from the velocty expresso v q ()= ω d cosφ Mode Summato Approach wth rgd body modes (ω = ) f λ =, q (t) = ( a e j t +b e j t )v = ( a +b )v does ot gve two learly depedet solutos. Now we must use the expaso ( ) q(t) = ( a + b t)v + a e j λ t +b e j λ t = ad adjust calculato of the costats from the tal codtos accordgly. Note that the uderle term s a traslatoal moto v Example 4.3. solved by the mode summato method / 3 From before, we have M = ad = V 3 / / Approprate IC are q= M x=, q= = M v π ω () () φ v q ω v q φ = ta = ta = () φ π v q v q() d 3 d = = sφ d 3 Example 4.3. costructg the summato of modes q () t 3 3 s π t s π t q () t = + the frst mode the secod mode rasformg back to the physcal coordates yelds: x(t) = M / q = 3 s t π 3 + 3 s t π 3 = 3 s t π 3 + 3 s t π 3 3 4
Example 4.3. a comparso of the two soluto methods shows they yeld detcal results.5 -.5 free respose of DOF modal MSA - 3 4 5 6 7 8 9 tme (sec) 3 - - free respose of DOF modal MSA -3 3 4 5 6 7 8 9 tme (sec) Steps for Computg the Respose By Mode Summato. Wrte the equatos of moto matrx form, detfy M ad K. Calculate M -/ (or L) 3. Calculate K = M KM 4. Compute the egevalue problem for the matrx K ad get ω ad v 5. rasform the tal codtos to q(t) q( ) = M x() ad q () = M x () 5 6 Summary of Mode Summato Cotued 6. Calculate the modal expaso coeffcets ad phase costats ω () () φ ta v q v q =, d = v q() sφ 7. Assemble the tme respose for q q(t) = = d s( ω t + φ )v 8. rasform the soluto to physcal coordates x(t) = M q(t) = d s( ω t + φ )u = 7 Nodes of a Mode Shape Examato of the mode shapes Example 4.4.3 shows that the thrd etry of the secod mode shape s zero! Zero elemets a mode shape are called odes. A ode of a mode meas there s o moto of the mass or (coordate) correspodg to that etry at the frequecy assocated wth that mode. 8
he secod mode shape of Example 4.4.3 has a ode Note that for more the DOF, a mode shape may have a zero valued etry hs s called a ode of a mode..887.887 u = ode.887 hey make great moutg pots maches 9 A rgd body mode s the mode assocated wth a zero frequecy Fg 4. Note that the system Fg 4. s ot costraed ad ca move as a rgd body Physcally f ths system s dsplaced we would expect t to move off the page whlst the two masses oscllate back ad forth 3 Example 4.4.4 Rgd body moto he free body dagram of fgure 4. yelds mx = k( x x) ad mx =k( x x) m x x k m + = x x Solve for the free respose gve: m = kg, m = 4 kg, k = 4 N subject to x =. m ad v = 3 Followg the steps of Wdow 4.4 /. M = 4 / /. K = M KM = 4 = 4λ 3. det( K λi) = det ( λ 5λ) λ = = λ = ad λ = 5 ω =, ω =.36 rad/s Idcates a rgd body moto 3
Now calculate the egevectors ad ote partcular that they caot be zero eve f the egevalue s zero 4 v λ = 4 v v v = =.447 v= or after ormalzg v=.8944.8944.447.8944 Lkewse: v = P =.447.8944.447 As a check ote that PP= I ad PKP = dag 5 33 5. Calculate the matrx of mode shapes.447.8944.447.8944 / S= M P= /.8944.447 =.447.36.447.7889 S =.8944.8944 7. Calculate the modal tal codtos:.447.7889..447 r() = S x =.8944.8944 =.8944 r() = S x = 34 7. Now compute the soluto modal coordates ad ote what happes to the frst mode. Sce ω = the frst modal equato s r + ()r = r (t) = a + bt Ad the secod modal equato s r (t) + 5r (t) = Rgd body traslato r (t) = a cos 5t Oscllato 35 Applyg the modal tal codtos to these two soluto forms yelds: r () = a=.447 r () = b=. rt ( ) =.4 as the past problems the tal codtos for r yeld rt ( ) =.89cos 5t.4 r() t =.89cos 5t 36
8. rasform the modal soluto to the physcal coordate system.447.8944.45 x(t) = Sr(t) =.447.36.89 cos 5t x(t) = x (t). + 7.6 cos 5t x (t) = 3 m..99 cos 5t Each mass s moved a costat dstace ad the oscllates at a sgle frequecy. Order the frequeces It s coveto to call the lowest frequecy ω so that ω < ω < ω 3 < Order the modes (or egevectors) accordgly It really does ot make a dfferece computg the tme respose However: Whe we measurg frequeces, they appear lowest to hghest Physcally the frequeces respod wth the hghest eergy the lowest mode (mportat flutter calculatos, ru up rotatg maches, etc.) 37 38 he system of Example 4..5 solved by Mode Summato From Example 4..6 we have: ω =, u = 3, ω =, u = 3 Use the followg tal codtos ad ote that oly oe mode should be excted (why?) 3 x() =, x () = rasform coordates M = 9 M = 3 ad M = /3 hus the tal codtos become 3 3 q() = M x() = = 3 () M q = x () = = 39 4
rasform Mode Shapes to Egevectors v = M u = 3 3 = v = M u = 3 3 = egevectors Note that ulke the mode shapes, the egevectors are orthogoal: Note that v v = [ ] =, but u u = 3 3 = 3 Normalzg yelds: v = ad v = 4 From Equato (4.9): = d s( t + φ ) v q = = = q ω d ω cos( ω t + φ ) v v Set t= ad multply by v : q () = dω cosφv = = d cosφ v + d cosφ v = d cosφ φ = π / Or drectly from Eq. (4.97) 4 From the tal dsplacemet: d = v q() s(π /) = [ ] d = v q() s(π /) = [ ] thus q(t) = d s(ω t +φ )v = = (4.98) = Egevector = cos = cos Egevector 43 rasformg Back to Physcal Coordates: x(t) = M q(t) = 3 t) cos( cos t = 3 cos t x (t) = 3 cos t ad x (t) = cos t So, the tal codtos geerated moto oly the frst mode (as expected) 44
Alterate Path to Symmetrc Sgle-Matrx Egeproblem Square root of matrx coceptually easy, but computatoally expesve q+ q= q+ q= M MM M KM K More effcet to decompose M to product of upper ad lower tragular matrces (Cholesky decomposto) 45 Cholesky Decomposto Let M= U U where U s upper tragular Itroduce the coordate trasformato Ux= q x= U q U U x+ K x = q U q premultply by U to get q+ q= q+ q= I U KU K ote that: [ ] [ ] U KU = U K U = U KU 46