Lecture 20 - Wave Propagation Response

Transcription

1 .09/.093 Fiite Eleet Aalysis of Solids & Fluids I Fall 09 Lecture 0 - Wave Propagatio Respose Prof. K. J. Bathe MIT OpeCourseWare Quiz #: Closed book, 6 pages of otes, o calculators. Covers all aterials icludig this week s lectures. L w C = tw (C depeds o aterial properties) For this syste, C is the wave speed (give), L w is the critical wavelegth to be represeted, t w is the total tie for this wave to travel past a poit, L e is the effective legth of a fiite eleet, ad is equivalet to Lw (give). To solve, we should use = L e C. Mesh L e should be saller tha the shortest wave legth we wat to pick up. To establish a esh, we use low-order eleets (4-ode eleets i D, 8-ode eleets i 3D). We use the cetral differece ethod, which requires stability. We eed to esure that cr = ω. Recall that i oliear aalysis, the wave speed chages. We kow that { } () ω ax ω { } () where ω is the largest frequecy of a assebled fiite eleet esh ad ax ω is the largest eleet frequecy of all eleets i the esh. The we ca use ad coservatively, we use a slightly saller value. = { } () ax ω 1

2 Lecture 0 Wave Propagatio Respose.09/.093, Fall 09 () How to Fid ω () For lower-order eleets, we have bouds for ω. (See Sectios 9.3/9.4 ad Table 9.5 that gives forulae for ω ().) Proof: We kow ( ) φ T ΣK () φ ω = ( ) (a) φ T ΣM () φ ω Σ () = U (b) Σ J () U () = φ T K () φ J () = φ T M () φ Note that K () is of the sae size as K. Also, for a eleet (), sice φ is ot a eigevector for K () : φ T K () φ ( ) ω () φ T M () φ ( ) U () J () ω () (c) Plug (c) ito (b) ad we have ( ) () () Σ ω { ( ) } ω J J ax ω () () Σ I practice, wave propagatio probles are very difficult to solve, due to reflectios, absorptios, ad the ay differet wave types (shear waves, etc.). As etioed, the cetral differece ethod is alost always used for wave propagatio solutios, because wave propagatio odelig eeds fie eshes. I beas, rotatioal DOFs result i high frequecies, so rotatioal ass is frequetly set to zero. If we have zero asses i the odel, we eed to use static codesatio. [ ] [ ] [ ] [ ] K aa K ab φ a = ω M a 0 φ a K ba K bb φ b 0 0 φ b No ass o b-dofs Use the d equatio to eliiate φ b fro the 1st equatio. K ba φ a + K bb φ b = 0 φ b = K 1 bb K baφ a We obtai ( ) K aa K ab K 1 K ba φ a = ω M a φ a bb It is the sae as Gauss eliiatio o b-dofs. Hece, if there are zero asses, we use static codesatio prior to the use of the cetral differece ethod. All we have discussed regardig trasiet/dyaic aalysis is also applicable (with odificatios!) i heat trasfer & fluid flow aalysis.

3 Lecture 0 Wave Propagatio Respose.09/.093, Fall 09 Trasiet Heat Trasfer Aalysis The goverig fiite eleet equatio is Cθ + Kθ = Q (give 0 θ) (A) where C is the heat capacity atrix, K is the coductivity atrix, ad Q is the heat flow iput vector. I. Mode Superpositio Frequetly, heat trasfer solutios ca be obtaied with coarser eshes. Let s exaie the syste Assue Cθ + Kθ = 0 (o heat flow iput) θ = e λt φ λ e λt Cφ + K e λt φ = 0 K φ = λcφ The eigevalues are 0 λ 1 λ... λ, ad the eigevectors are φ 1, φ,..., φ. For costat teperature over the esh, we have λ 1 = 0. Also: φ T i Cφ j = δ ij (B) We obtai the decoupled equatios. φ T i Kφ j = λ i δ ij η i + λ i η i = q i (i = 1,..., ) ; 0 η i (C) θ = Σ φ i η i i=1 [ ] Φ T CΦ = I ; Φ = φ 1... φ θ = Φη 0 η = Φ T C 0 θ Φ T KΦ =... λ i... q = Φ T Q q i = φ i T Q To solve Eq. (C) or perfor direct itegratio o Eq. (A), we ca use the Euler backward ethod or the Euler forward ethod. II. The Euler Backward Method Readig assiget: Sectio 9.6 ( ) t+ θ t+ θ t θ = Use Eq. (A) at tie t + to solve for t+ θ. This is a iplicit ethod, ad is ucoditioally stable. oly eeds to be selected for accuracy. 3

4 Lecture 0 Wave Propagatio Respose.09/.093, Fall 09 III. The Euler Forward Method The, use Eq. (A) at tie t. t θ = t+ θ t θ C t θ + K t θ = t Q 1 ( ) 1 C t+ θ = C t θ K t θ + t Q If C is a diagoal atrix, the o factorizatio is ivolved. The Euler forward ethod results i: 1 C t+ θ = t Qˆ The ethod is coditioally stable, ad ust satisfy Recall: Hece, Kφ = ω Mφ is the eigevalue proble. cr = λ MU + KU = 0 U = φ si ω (t τ) U = ω φ si ω (t τ ) Mω φ si ω (t τ ) + Kφ si ω (t τ ) = 0 η i + λ i η i = q i cr = λ ẍ i + ω i x i = r i cr = ω The explicit ethod uses the goverig equatios at tie t to obtai the solutio at t +. The iplicit ethod uses the goverig equatios at tie t + to obtai the solutio at t +. 4

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