TWO-STAGE ISOLATION FOR HARMONIC BASE EXCITATION Revision A. By Tom Irvine February 25, 2008

Transcription

1 TWO-STAGE ISOLATION FOR HARMONIC BASE EXCITATION Revision A By Tom Irvine tomirvine@aol.com February 5, 008 Introduction Consider a base plate mass m and an avionics mass m modeled as two-degree-of-freedom. Evaluate the benefits and drawbacks of this two-stage isolation scheme. m x k m x k & y& Figure. The system also has damping, but it is modeled as modal damping. A free-body diagram of mass is given in Figure. A free-body diagram of mass is given in Figure 3.

2 k ( x - x ) m x k ( x - y ) Figure. Determine the equation of motion for mass. Σ F = m & x () ( ) ( y) m & x = k x x k x () ( ) y m & x k x k x x = k (3) ( ) y m & x k x k x x = k (4) ( ) y m & x k k x k x = k (5) m x k (x -x ) Figure 3.

3 Derive the equation of motion for mass. Σ F = m & x (6) m ( ) & x = k x x (7) ( ) 0 m & x k x x = (8) m & x k x k x = 0 (9) Assemble the equations in matrix form. m 0 0 && x k k m && x k k x k = y k x 0 (0) Define a relative displacement z such that x = z y () x = z y () Substitute equations () and () into (0). m 0 0 && z && y k k m && z && y k k z y k = y k z y 0 (3) m 0 0 && z m && y k k m && z m && y k k z k k k z k k y k = y k y 0 (4) m 0 0 && z m && y k k m && z my && k k z ky = k y k z 0 0 (5) 3

4 m 0 0 && z k k m && z k k z m = && y k z my && (6) Decoupling Equation (6) is coupled via the stiffness matrix. An intermediate goal is to decouple the equation. Simplify, where M & z K z = F (7) m 0 M = (8) 0 m k k k K = (9) k k z z = (0) z m = && y F m && y () Consider the homogeneous form of equation (7). M & z K z = 0 () Seek a solution of the form ( jt) z = q exp (3) The q vector is the generalized coordinate vector. 4

5 Note that ( jt) z& = j q exp (4) & z = q exp (5) ( jt) Substitute equations (3) through (5) into equation (). ( jt) K q exp( jt) = 0 M q exp (6) { M q K q} exp( jt) = 0 (7) n M q K q = 0 (8) { M K} q = 0 { K M } q = 0 (9) (30) Equation (30) is an example of a generalized eigenvalue problem. The eigenvalues can be found by setting the determinant equal to zero. det { K M } = 0 (3) k k k m 0 det = 0 k k 0 m (3) ( ) k k m det k = 0 k k m (33) k m k m k (34) ( k ) = 0 4 m [ m ( k k )] k m m k = 0 (35) 5

6 The eigenvalues are the roots of the polynomial. b b 4ac = (36) a where b b 4ac = (37) a a = m m (38) [ ( )] b = mk3 m k k (39) c = k (40) The eigenvectors are found via the following equations. { M } q = 0 K (4) { M } q = 0 K (4) where q q = (34) q q q = (44) q An eigenvector matrix Q can be formed. The eigenvectors are inserted in column format. [ q ] Q = q (45) 6

7 q q Q = (46) q q The eigenvectors represent orthogonal mode shapes. Each eigenvector can be multiplied by an arbitrary scale factor. A mass-normalized eigenvector matrix Qˆ can be obtained such that the following orthogonality relations are obtained. Qˆ T M Qˆ = I (47) and Qˆ T K Qˆ = Ω (48) where superscript T represents transpose I is the identity matrix Ω is a diagonal matrix of eigenvalues Note that qˆ qˆ Qˆ = qˆ qˆ (49a) T qˆ qˆ Qˆ = qˆ qˆ (49b) Rigorous proof of the orthogonality relationships is beyond the scope of this tutorial. Further discussion is given in References 5 and 6. Nevertheless, the orthogonality relationships are demonstrated by an example in this tutorial. Now define a modal coordinate η (t) such that z = Qˆ η (50a) z = qˆ η qˆ η (50b) 7

8 z = qˆ η qˆ η (50c) Recall x = z y (5a) x = z y (5b) The displacement terms are x = y qˆ η qˆ η (5a) x = y qˆ η qˆ η (5b) The velocity terms are x& = y& qˆ η& qˆ η& (53a) x& = y& qˆ η& qˆ η& (53b) The acceleration terms are & x = && y qˆ η&& qˆ η& (54a) & x = && y qˆ η&& qˆ η& (54b) Substitute equation (50a) into the equation of motion, equation (7). M Qˆ η& & K Qˆ η = F (55) Premultiply by the transpose of the normalized eigenvector matrix. Qˆ T M Qˆ T T η& & Qˆ K Qˆ η = Qˆ F (56) The orthogonality relationships yield I η& & Ω η = Qˆ T F (57) For the sample problem, equation (57) becomes 8

9 0 0 η&& η&& 0 0 η = qˆ η qˆ qˆ m && y qˆ m && y (58) Note that the two equations are decoupled in terms of the modal coordinate. Now assume modal damping by adding an uncoupled damping matrix. 0 0 η&& η&& ξ 0 0 ξ η& η& 0 0 η = qˆ η qˆ qˆ m && y qˆ m && y (59) Equation (59) yields two equations η & ξ η = [ qˆm qˆ m ]y & (60) η & ξ η = [ qˆm qˆ m ]y & (6) Now assume a harmonic base input. ( jt) & y = A exp (6) Assume a harmonic modal displacement. ( jt) η i= ψ i exp (63) ( jt) η& i = j i ψ i exp (64) η& & i = i ψ i exp( jt) (65) By substitution, { } j ξ ψ exp ( jt) = [ qˆ m qˆ m ] A exp( jt) { } j ξ ψ exp ( jt) = [ qˆ m qˆ m ] A exp( jt) (66) (67) 9

10 j ξ ψ exp ( jt) = [ qˆ m qˆ m ] A exp( jt) (68) j ξ ψ exp( jt) = [ qˆm qˆ m ] A exp( jt) (69) [ ] qˆ η ( ) qˆm m = ψ exp jt = A exp( jt) (70) j ξ [ ] qˆ η ( ) qˆm m = ψ exp jt = A exp( jt) (7) j ξ The modal velocity is [ ] j qˆ η& qˆm m = A exp( jt) (7) j ξ [ ] j qˆ η& qˆm m = A exp( jt) (73) j ξ The modal acceleration is [ qˆ m qˆ ] η& & m = A exp( jt) (74) j ξ 0

11 [ qˆ m qˆ ] η& & m = A exp( jt) (75) j ξ Recall & x = && y qˆ η&& qˆ η& (76) & x = && y qˆ η&& qˆ η& (77) && x(t) = qˆ qˆ [ qˆ ] [ qˆ ] qˆm j m ξ qˆm j m ξ A exp ( jt) (78) && x(t) = qˆ qˆ [ qˆ ] [ qˆ ] qˆm j m ξ qˆm j m ξ A exp ( jt) (79) The Fourier transform equation is Xˆ i (f) [ jt ] = & x (t) exp - dt - i (80)

12 Take the Fourier transform of each side of equations (78) and (79). Xˆ (f) / A = qˆ qˆ [ qˆ ] [ qˆ ] qˆm j m ξ qˆm j m ξ (8) Xˆ (f) / A = qˆ qˆ [ qˆ ] [ qˆ ] qˆm j m ξ qˆm j m ξ (8) References. T. Irvine, An Introduction to the Shock Response Spectrum Revision P, Vibrationdata, 00.. T. Irvine, Response of a Single-degree-of-freedom System Subjected to a Classical Pulse Base Excitation, Revision A, Vibrationdata, R. Cook, Finite Element Modeling for Stress Analysis, Wiley, New York, NE/Nastran User s Manual, Version 8, Noran Engineering, Los Alamitos, CA, Bathe, Finite Element Procedures in Engineering Analysis, Prentice-Hall, New Jersey, Weaver and Johnston, Structural Dynamics by Finite Elements, Prentice-Hall, New Jersey, L. Meirovitch, Analytical Methods in Vibrations, Macmillan, New York, 967.

13 APPENDIX A EXAMPLE Normal Modes Analysis x m k m x k & y& Figure A-. A 5-lbm avionics component (m ) is mounted on a -lbm base plate (m ). Each spring stiffness is 4.6e04 lbf/in. Analyze the energy transmitted to the avionics mass with and without the base plate stage. Table A-. Parameters Variable m m k k Value lbm 5 lbm 4.6e04 lbf/in 4.6e04 lbf/in 3

14 Furthermore, assume that each mode has a damping value of 5%. m 0 0 && z k k m && z k k z my && = k z my && (A-) Solve for the acceleration response time histories. The homogeneous, undamped problem is / && z 9.e 04 5/386 && z 4.6e e 04 z 0 = 4.6e 04 z 0 (A-) The natural frequencies are f = 0.3 Hz (A-3) f = Hz (A-4) 4

15 FRF Analysis TRANSFER MAGNITUDE EXAMPLE 00 Base Plate Avionics Mass 0 TRANS (G/G) Figure A FREQUENCY (Hz) 5

16 TRANSFER MAGNITUDE AVIONICS MASS EXAMPLE 00 Single-Stage Two-Stage 0 TRANS (G/G) FREQUENCY (Hz) Figure A-3. The Single-Stage curve represents the avionics mass and its spring by themselves. The results are mixed. The optimum design depends on the base excitation frequency. 6

17 APPENDIX B EXAMPLE Repeat the example from Appendix A but with the base plate and avionics mass both at 5 lbm. Table B-. Parameters Variable m m k k Value 5 lbm 5 lbm 4.6e04 lbf/in 4.6e04 lbf/in The natural frequencies are f = 85.4 Hz (B-) f = Hz (B-) 7

18 TRANSFER MAGNITUDE EXAMPLE 00 Base Plate Avionics Mass 0 TRANS (G/G) Figure B-. FREQUENCY (Hz) 8

19 TRANSFER MAGNITUDE AVIONICS MASS EXAMPLE 00 Single-Stage Two-Stage 0 TRANS (G/G) Figure B-3. FREQUENCY (Hz) The Single-Stage curve represents the avionics mass and its spring by themselves. Again, the results are mixed. The optimum design depends on the base excitation frequency. 9

20 APPENDIX C EXAMPLE 3 Repeat the example from Appendix A but with the base plate at 5 lbm. Table C-. Parameters Variable m m k k Value 5 lbm 5 lbm 4.6e04 lbf/in 4.6e04 lbf/in The natural frequencies are f = 44.6 Hz (C-) f = 359. Hz (C-) 0

21 TRANSFER MAGNITUDE EXAMPLE 3 00 Avionics Mass Base Plate 0 TRANS (G/G) FREQUENCY (Hz) Figure C-.

22 TRANSFER MAGNITUDE AVIONICS MASS EXAMPLE 3 00 Single-Stage Two-Stage 0 TRANS (G/G) FREQUENCY (Hz) Figure C-. The Single-Stage curve represents the avionics mass and its spring by themselves. The Two-Stage design provides greater attenuation above an excitation frequency of 00 Hz.