SHOCK RESPONSE OF MULTI-DEGREE-OF-FREEDOM SYSTEMS Revision F By Tom Irvine May 24, 2010

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1 SHOCK RESPONSE OF MULTI-DEGREE-OF-FREEDOM SYSTEMS Revision F By Tom Irvine tomirvine@aol.com May 4, 010 Introduction The primary purpose of this tutorial is to present the Modal Transient method for calculating the response of a multi-degree-of-freedom system to an arbitrary base input. A secondary purpose is to compare the results of this method to simplified methods for multi-degree-of-freedom shock response. Shock Response Spectrum The shock response spectrum is inherently a single-degree-of-freedom concept, as discussed in Reference 1. Nevertheless, there is an occasional need to determine the response of a multi-degree-of-freedom system to shock response spectrum input for engineering purposes. The modal transient method is the formal method for carrying out this calculation, but it may require intensive calculations depending on the complexity of the model and the base input time history. Over the years, a number of simplified modal combination methods have been proposed for estimating the multi-degree-of-freedom response. Each requires the eigenvalues, eigenvectors, and the modal participation factors of the system, or at least estimates of these parameters. Equation of Motion The governing equation of motion for the displacement η i of mode i within a multi-degreeof-freedom system subjected to base excitation is & η i + ξi ωi η& i + ωi η i = Γ i& y (1) Γ i is the modal participation factor for mode i, as given in Appendix C & y& ω i ξ i is the base excitation acceleration is the natural frequency is the viscous damping ratio 1

2 The response of a physical coordinate in a multiple-degree-of-freedom system can be approximated as the summation response of a series of single-degree-of-freedom systems. The method assumes that the modal frequencies are well-separated. The corresponding single-degree-of-freedom response for the relative displacement D j is D & j+ ξi ωi D& j + ωi D j = & y () The subscript j denotes the connection of D with a particular mode. Thus, η j(t) = Γ j D j(t) (3) The advantage of this approach is that D j can be calculated with relative ease, thereby providing an indirect solution for η j. On the other hand, a direct calculation of η j requires intensive calculations. Absolute Sum (ABSSUM) The absolute sum method is a conservative approach because it assumes that the maxima of all modes appear at the same instant of time. The maximum relative displacement ( z i ) max is calculated from the modal coordinates as N z max qˆ i j η j max (4) j= 1 ( i ) where qˆ i j is the mass-normalized eigenvector coefficient for coordinate i and mode j. The corresponding ABSSUM method is N (5) j = 1 ( zi ) max Γj qˆ i j D j, max

3 Equation (5) is taken from Reference, equation (7.53). Again, the maximum relative displacement D j is taken as the single-degree-of-freedom response. Square Root of the Sum of the Squares (SRSS) The square root of the sum of the squares equation is N z i max i j j, max (6) j= 1 ( ) [ qˆ η ] The SRSS method with the single-degree-of-freedom response D j is N z i max j i j D j,max (7) j= 1 ( ) [ Γ qˆ ] Naval Research Laboratories (NRL) Method The maximum relative displacement is calculated from the modal coordinates as N z max (qˆ ik )( η k, max ) (8) k = 1, k j ( i ) = qˆ i j η j, max + where the j-th mode is the mode that has the largest magnitude of the product qˆ i j η j max. Note that the NRL method is the same as the ABSSUM method for N=. The simplified formula is N z max (qˆ ik )( Γ k D k, max ) (9), max k = 1, k j ( i ) = ( qˆ i j)( Γ j D j) + 3

4 Acceleration Response Equations (6) through (7) can be extended for the absolute acceleration response, by making the appropriate substitutions. References 1. T. Irvine, An Introduction to the Shock Response Spectrum Revision P Vibrationdata, 00.. R. Kelly and G. Richman, Principles and Techniques of Shock Data Analysis, SVM- 5; The Shock and Vibration Information Center, United States Department of Defense, Washington D.C., A. Chopra, Dynamics of Structures, Prentice-Hall, New Jersey, Bathe, Finite Element Procedures in Engineering Analysis, Prentice-Hall, New Jersey, Weaver and Johnston, Structural Dynamics by Finite Elements, Prentice-Hall, New Jersey, T. Irvine, Response of a Single-degree-of-freedom System Subjected to a Classical Pulse Base Excitation, Revision A, Vibrationdata,

5 APPENDIX A Two-degree-of-freedom System, Modal Analysis The method of generalized coordinates is demonstrated by an example. Consider the system in Figure A-1. m x k m 1 x 1 k 1 & y& Figure A-1. The system also has damping, but it is modeled as modal damping. A free-body diagram of mass 1 is given in Figure A-. A free-body diagram of mass is given in Figure A-3. k 3 ( x - x 1 ) m 1 x 1 k 1 ( x 1 - y ) Figure A-. Determine the equation of motion for mass 1. 5

6 Σ F = m 1 & x 1 (A-1) ( ) ( y) m1& x 1 = k 3 x x 1 k 1 x 1 (A-) ( ) y m 1& x 1 + k 1 x 1 k 3 x x 1 = k 1 (A-3) ( ) y m 1& x1 + k1 x1 + k3 x1 x = k1 (A-4) ( + ) y m 1& x1 + k1 k3 x1 k3x = k1 (A-5) k (x - y) m x k 3 (x -x 1 ) Figure A-3. Derive the equation of motion for mass. Σ F = m & x (A-6) m ( ) ( y) & x = k 3 x x 1 k x (A-7) ( ) y m & x + k x + k 3 x x 1 = k (A-8) ( + ) y m & x + k k 3 x k 3 x 1 = k (A-9) Assemble the equations in matrix form. m1 0 0 && x1 k k 3 m && x k3 k 3 x1 = k1 y + k k3 x k y (A-10) 6

7 Define a relative displacement z such that x 1 = z1 + y (A-11) x = z + y (A-1) Substitute equations (A-11) and (A-1) into (A-10). m1 0 0 && z1 + && y k1 + k3 + m && z + && y k3 k3 z1 + y k1 y = k + k3 z + y k y (A-13) m1 0 0 && z1 m1y && k k 3 m && z my && k3 k 3 z1 k k 3 + k k3 z k3 k 3 y = k1 y + k k3 y k y (A-14) m && z 1 m && y k k m && z m && y k3 k 3 z 1 k + 1 y k = 1 y + k k3 z k y k y (A-15) m && z 1 k k 3 m && z k3 k 3 z 1 m = 1 && y + k k3 z m && y (A-16) Decoupling Equation (A-16) is coupled via the stiffness matrix. An intermediate goal is to decouple the equation. Simplify, M & z + K z = F (A-17) 7

8 where m1 0 M = (A-18) 0 m k1 + k3 k3 K = (A-19) k3 k + k3 z = z1 (A-0) z m1y && F = my && (A-1) Consider the homogeneous form of equation (A-17). M & z + K z = 0 (A-) Seek a solution of the form ( jωt) z = q exp (A-3) The q vector is the generalized coordinate vector. Note that ( jωt) z& = jω q exp (A-4) & z = ω q exp (A-5) ( jωt) Substitute equations (A-3) through (A-5) into equation (A-). ( jωt) + K q exp( jωt) = 0 ω M q exp (A-6) { ω M q + K q} exp( jωt) = 0 (A-7) ω n M q + K q = 0 (A-8) 8

9 { ω M + K} q = 0 { K ω M } q = 0 (A-9) (A-30) Equation (A-30) is an example of a generalized eigenvalue problem. The eigenvalues can be found by setting the determinant equal to zero. det { K ω M } = 0 (A-31) k1 + k3 det k3 k3 m1 ω k + k = 0 m (A-3) det ( k + k ) 1 3 ω m1 k 3 = 0 k ( k k ) ω m (A-33) [( k k ) 3 ω m1]( k + k3 ) [ ω m ] k = (A-34) [ m ( k + k ) + m ( k + k )] k = 0 ω 4 m m 1 ω (A-35) The eigenvalues are the roots of the polynomial. b b 4ac ω 1 = (A-36) a where b b + 4ac ω = (A-37) a a = m 1 m (A-38) [ m ( k + k ) + m ( k )] b = k3 (A-39) 9

10 c = k 3 (A-40) The eigenvectors are found via the following equations. { ω 1 M } q = 0 K 1 (A-41) { ω M } q = 0 K (A-4) where q11 q 1 = (A-34) q1 q1 q = (A-44) q An eigenvector matrix Q can be formed. The eigenvectors are inserted in column format. [ q ] Q = 1 q (A-45) q11 q1 Q = (A-46) q1 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 (A-47) and Qˆ T K Qˆ = Ω (A-48) where superscript T represents transpose 10

11 I is the identity matrix Ω is a diagonal matrix of eigenvalues Note that qˆ 11 qˆ 1 Qˆ = qˆ 1 qˆ (A-49a) T qˆ 11 qˆ 1 Qˆ = qˆ 1 qˆ (A-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 (A-50a) z1 = qˆ 11 η1 + qˆ 1 η (A-50b) z = qˆ 1 η1 + qˆ η (A-50c) Recall x 1 = z1 + y (A-51a) x = z + y (A-51b) The displacement terms are x 1 = y + qˆ 11 η1 + qˆ 1 η (A-51a) x = y + qˆ 1 η1 + qˆ η (A-5b) The velocity terms are x& 1 = y& + qˆ 11 η& 1 + qˆ 1 η& (A-53a) x& = y& + qˆ 1 η& 1 + qˆ η& (A-53b) 11

12 The acceleration terms are & x 1 = && y + qˆ 11 η&& 1 + qˆ 1 η& (A-54a) & x = && y + qˆ 1 η&& 1 + qˆ η& (A-54b) Substitute equation (50a) into the equation of motion, equation (A-17). M Qˆ & η& + K Qˆ η = F (A-55) Premultiply by the transpose of the normalized eigenvector matrix. T Qˆ M Qˆ T T η& & + Qˆ K Qˆ η = Qˆ F (A-56) The orthogonality relationships yield I η& & + Ω η = Qˆ T F (A-57) For the sample problem, equation (A-57) becomes && η1 η&& + ω ω η1 = qˆ11 η qˆ1 qˆ 1 m1y && qˆ my && (A-58) Note that the two equations are decoupled in terms of the modal coordinate. Now assume modal damping by adding an uncoupled damping matrix η&& 1 η&& ξ ω ξ ω η& 1 + η& ω ω η1 = qˆ11 η qˆ1 qˆ 1 m1y && qˆ my && Now consider the initial conditions. Recall (A-59) z = Qˆ η (A-60) Thus, z ( 0) Qˆ η( 0) = (A-61) 1

13 Premultiply by Qˆ T M. Qˆ T M z = T (A-6) ( 0) Qˆ M Qˆ η( 0) Recall Qˆ T M Qˆ = I (A-63) Qˆ T Qˆ T M z M z ( 0) I η( 0) = (A-64) ( 0) η( 0) = (A-65) Finally, the transformed initial displacement is η T (A-66) ( 0 ) = Qˆ M z( 0) Similarly, the transformed initial velocity is η & = T (A-67) ( 0 ) Qˆ M z& ( 0) The product of the first two matrices on the left side of equation (A-59) equals a vector of participation factors. Γ1 = Γ qˆ11 qˆ1 qˆ 1 m 11 qˆ m (A-68) η&& 1 η&& + ξ1 ω1 0 0 ξ ω η& 1 + η& ω ω η1 Γ = 1&& y η Γ&& y (A-69) Equation (A-69) can be solved in terms of Laplace transforms, or some other differential equation solution method. As an example, the solution for a half-sine input is given in Reference 6. 13

14 APPENDIX B EXAMPLE 1 Normal Modes Analysis m x k 3 k x 1 m 1 k 1 y Figure B-1. Consider the system in Figure B-1. Assign the values in Table B-1. Table B-1. Parameters Variable m 1 m k 1 k k 3 Value 3.0 kg.0 kg 400,000 N/m 300,000 N/m 100,000 N/m Furthermore, assume 1. Each mode has a damping value of 5%.. Zero initial conditions Next, assume that the base input function is a half-sine pulse. 14

15 BASE INPUT 50 G, 11 msec HALF-SINE PULSE ACCEL (G) TIME (SEC) Figure B-. & y&(t) πt Asin, 0 t T T = 0, t > T (B-1) Assign A = 50 G and T = seconds. 15

16 Figure B-3. Figure B-4. 16

17 Solve for the acceleration response time histories. The homogeneous, undamped problem is m1 0 0 && z1 k k 3 m && z k 3 k 3 z1 m1y && = k + k 3 z my && (B-) && z1 500,000 + && z 100, ,000 z1 0 = 400,000 z 0 (B-3) The eigenvalue problem is 500,000 ω 100, ,000 q 1 0 = 400,000 ω q 0 (B-4) Set the determinant equal to zero 500,000 ω 100,000 det = 0 100, ,000 ω (B-5) The roots of the polynomial are ω 1 = rad /sec (B-6) ω = rad /sec (B-7) f 1 = 59.4 Hz (B-8) f = 75.9 Hz (B-9) The frequencies are rather close together, which is a concern relative to an assumption behind the modal combination methods. 17

18 The corresponding eigenvector matrix is Q = q11 q1 (B-10) q1 q Q = (B-11) 1 1 The next goal is to obtain a normalized eigenvector matrix Qˆ such that Qˆ T M Qˆ = I (B-1) The normalized eigenvector matrix is Qˆ = (B-13a) Qˆ T = (B-13b) Note that the eigenvector in the first column has a uniform polarity. Thus, the two masses vibrate in phase for the first mode. The eigenvector in the second column has two components with opposite polarity. The two masses vibration 180 degrees out of phase for the second mode. Γ1 = Γ qˆ11 qˆ1 qˆ 1 m 1 qˆ m 1 (B-14) Γ1 = Γ qˆ11 qˆ1 qˆ 1 m 1 qˆ m (B-15) 18

19 Γ = Γ (B-16) Γ1 = Γ (B-17) Modal Transient Analysis The combination of equations (B-1) and (B-6) are solved using the method in Reference 6, as modified for a multi-degree-of-freedom system. The modifications are made via equations (A-69) and (A-50a). The resulting relative displacement maxima are given in Table B-. The time history responses are plotted in Figure B-5. Table B-. Modal Transient Analysis, Relative Displacement Parameter Mass 1 (inch) Mass (inch) Maximum Absolute

20 Figure B-5. SDOF Response The peak acceleration value for a single-degree-of-freedom can be calculated per the method in Reference 6. The resulting maxima are shown in Table B-3. Table B-3. SDOF Response Values, Q=10 Mode Natural Frequency (Hz) Peak Relative Displacement (inch) Note that the coordinates in Table B-3 occur on the response spectrum curve in Figure B-4. 0

21 SRSS Method Again, the SRSS equation is N z i max j i j D j,max (B-18) j= 1 ( ) [ Γ qˆ ] Equation (B-18) is solved using 1. The eigenvalues from equation (B-13a). The participation factors from equation (B-17) 3. The displacement values from Table B-3 The results are given in Table B-4. Table B-4. Comparison of SRSS Results with Modal Transient Analysis, Relative Displacement Mass Modal Transient Analysis (inch) SRSS Analysis (inch) Error % % Sample Calculation z1 max 1 11 D1, max qˆ1 D, max (B-19a) ( ) Γ qˆ + [ Γ ] ( z ) [(.64)(0.479)(0.) ] + [(0.079)( 0.30)(0.140) ] 1 max (B-19b) ( ) z1 max inch (B-19c) 1

22 The acceleration results are Table B-5. Comparison of SRSS Results with Modal Transient Analysis, Absolute Acceleration Mass Modal Transient Analysis (G) SRSS Analysis (G) Error % % ABSSUM Method Again, the ABSSUM equation is N (B-0) j = 1 ( zi ) max Γj qˆ i j D j, max Equation (B-0) is solved using 1. The eigenvalues from equation (B-13a). The participation factors from equation (B-17) 3. The displacement values from Table B-3 The results are given in Table B-6. Table B-6. Comparison of the Sum of the Absolute Magnitudes with Modal Transient Analysis, Relative Displacement Mass Modal Transient Analysis (inch) ABSSUM (inch) Error % %

23 ( z1 ) max Γ1 qˆ 11 D 1, max + Γ qˆ 1 D, max (B-1a) ( z ) (.64)(0.479)(0.) + (0.079)( 0.30)(0.140) 1 max (B-1b) ( ) z1 max inch (B-1c) The acceleration results are: Table B-7. Comparison of ABSSUM Results with Modal Transient Analysis, Absolute Acceleration Mass Modal Transient Analysis (G) ABSSUM Analysis (G) Error % % 3

24 APPENDIX C Modal Participation Factor Consider a discrete dynamic system governed by the following equation where M K & x& M & x + K x = F (C-1) is the mass matrix is the stiffness matrix is the acceleration vector x is the displacement vector F is the forcing function or base excitation function A solution to the homogeneous form of equation (1) can be found in terms of eigenvalues and eigenvectors. The eigenvectors represent vibration modes. Let φ be the eigenvector matrix. The system s generalized mass matrix mˆ is given by mˆ = φ T M φ (C-) Let r be the influence vector which represents the displacements of the masses resulting from static application of a unit ground displacement. Define a coefficient vector L as The modal participation factor matrix L = φ T M r (C-3) Γ i for mode i is Γ L i i = mˆ (C-4) ii 4

25 The effective modal mass m eff, i for mode i is L i m eff, i= (C-5) mˆ ii 5

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

TWO-STAGE ISOLATION FOR HARMONIC BASE EXCITATION Revision A By Tom Irvine Email: tomirvine@aol.com February 5, 008 Introduction Consider a base plate mass m and an avionics mass m modeled as two-degree-of-freedom.