EFFECTIVE MODA MASS & MODA PARTICIPATION FACTORS Revision F By Tom Irvine Email: tomirvine@aol.com March 9, 1 Introduction The effective modal mass provides a method for judging the significance of a vibration mode. Modes with relatively high effective masses can be readily excited by base excitation. On the other hand, modes with low effective masses cannot be readily excited in this manner. Consider a modal transient or frequency response function analysis via the finite element method. Also consider that the system is a multi-degree-of-freedom system. For brevity, only a limited number of modes should be included in the analysis. How many modes should be included in the analysis? Perhaps the number should be enough so that the total effective modal mass of the model is at least 9% of the actual mass. Definitions The equation definitions in this section are taken from Reference 1. Consider a discrete dynamic system governed by the following equation where M is the mass matrix K x M x K x F (1) is the stiffness matrix is the acceleration vector x is the displacement vector F is the forcing function or base excitation function 1
A solution to the homogeneous form of equation (1) can be found in terms of eigenvalues and eigenvectors. The eigenvectors represent vibration modes. et be the eigenvector matrix. The system s generalized mass matrix mˆ is given by mˆ T M () et 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 as The modal participation factor matrix T M r (3) i for mode i is The effective modal mass i i (4) mˆ ii m eff, i for mode i is eff, i i mˆ m (5) i i mˆ Note that ii = 1 for each index if the eigenvectors have been normalized with respect to the mass matrix. Furthermore, the off-diagonal modal mass ( mˆ, i j ) terms are zero regardless of the i j normalization and even if the physical mass matrix M has distributed mass. This is due to the orthogonality of the eigenvectors. The off-diagonal modal mass terms do not appear in equation (5), however.
Example Consider the two-degree-of-freedom system shown in Figure 1, with the parameters shown in Table 1. m x k 3 k m 1 x 1 k 1 y Figure 1. Table 1. Parameters Variable m 1 m k 1 k k 3. kg 1. kg Value 1 N/m N/m 3 N/m The homogeneous equation of motion is m 1 x 1 k 1 k 3 m x k 3 k 3 x 1 k k 3 x (6) The mass matrix is M kg 1 (7) 3
The stiffness matrix is 4 3 K N / m 3 5 (8) The eigenvalues and eigenvectors can be found using the method in Reference. The eigenvalues are the roots of the following equation. det K M (9) The eigenvalues are 91.9 rad / sec 1 (1) 1 3.3 rad /sec (11) f 1 4.78 Hz (1) 698 rad /sec (13) 78.9 rad /sec (14) f 1.4 Hz (15) The eigenvector matrix is.68.4597.351.8881 (16) The eigenvectors were previously normalized so that the generalized mass is the identity matrix. mˆ T M (17) 4
.68.4597.68.351 mˆ (18).351.8881 1.4597.8881.68.4597 1.56.65 mˆ (19).351.8881.4597.8881 1 mˆ () 1 Again, r is the influence vector which represents the displacements of the masses resulting from static application of a unit ground displacement. For this example, each mass simply has the same static displacement as the ground displacement. 1 r (1) 1 The coefficient vector is T M r ().68.4597 1 (3).351.8881 1 1.68.4597 (4).351.8881 1 1.7157 kg.379 (5) 5
The modal participation factor i for mode i is i i (6) mˆ ii The modal participation vector is thus 1.7157.379 (7) The coefficient vector and the modal participation vector are identical in this example because the generalized mass matrix is the identity matrix. The effective modal mass For mode 1, m eff, i for mode i is i m eff, i (8) mˆ ii 1.7157 kg meff,1 (9) 1 kg meff,1.944 kg (3) For mode, meff,.379 kg (31) 1 kg meff,.56 kg (3) Note that meff,1 meff,.944 kg.56 kg (33) 6
meff,1 meff, 3 kg (34) Thus, the sum of the effective masses equals the total system mass. Also, note that the first mode has a much higher effective mass than the second mode. Thus, the first mode can be readily excited by base excitation. On the other hand, the second mode is negligible in this sense. From another viewpoint, the center of gravity of the first mode experiences a significant translation when the first mode is excited. On the other hand, the center of gravity of the second mode remains nearly stationary when the second mode is excited. Each degree-of-freedom in the previous example was a translation in the X-axis. This characteristic simplified the effective modal mass calculation. In general, a system will have at least one translation degree-of-freedom in each of three orthogonal axes. ikewise, it will have at least one rotational degree-of-freedom about each of three orthogonal axes. The effective modal mass calculation for a general system is shown by the example in Appendix A. The example is from a real-world problem. Aside An alternate definition of the participation factor is given in Appendix B. References 1. M. Papadrakakis, N. agaros, V. Plevris; Optimum Design of Structures under Seismic oading, European Congress on Computational Methods in Applied Sciences and Engineering, Barcelona,.. T. Irvine, The Generalized Coordinate Method For Discrete Systems, Vibrationdata,. 3. W. Thomson, Theory of Vibration with Applications nd Edition, Prentice Hall, New Jersey, 1981. 4. T. Irvine, Bending Frequencies of Beams, Rods, and Pipes, Rev M, Vibrationdata, 1. 5. T. Irvine, Rod Response to ongitudinal Base Excitation, Steady-State and Transient, Rev B, Vibrationdata, 9. 7
APPENDIX A Equation of Motion y z x kz1 m, J kx1 kz ky1 kx ky kz3 kz4 kx3 ky3 kx4 ky4 Figure A-1. Isolated Avionics Component Model The mass and inertia are represented at a point with the circle symbol. Each isolator is modeled by three orthogonal DOF springs. The springs are mounted at each corner. The springs are shown with an offset from the corners for clarity. The triangles indicate fixed constraints. indicates the origin. 8
y x z a1 a C. G. b c1 c Figure A-. Isolated Avionics Component Model with Dimensions All dimensions are positive as long as the C.G. is inside the box. At least one dimension will be negative otherwise. 9
The mass and stiffness matrices are shown in upper triangular form due to symmetry. m M m m Jx Jy Jz (A-1) 4kx K = 4ky 4k z 4k z b ky kx c1 c 4kxb c1 c ky a1 a 4k z b k z a 1 a b k y c 1 c k z 4kxc a 1 a kz a1 a k y a 1 a c 1 c kx c1 cb k a a y 1 4k x b (A-) 1
The equation of motion is M x x y y z z K (A-3) The variables α, β, and θ represent rotations about the X, Y, and Z axes, respectively. Example A mass is mounted to a surface with four isolators. The system has the following properties. M Jx Jy Jz kx ky kz a1 a b c1 c = 4.8 lbm = 44.9 lbm in^ = 39.9 lbm in^ = 18.8 lbm in^ = 8 lbf/in = 8 lbf/in = 8 lbf/in = 6.18 in = -.68 in = 3.85 in = 3. in = 3. in 11
et r be the influence matrix which represents the displacements of the masses resulting from static application of unit ground displacements and rotations. The influence matrix for this example is the identity matrix provided that the C.G is the reference point. 1 r 1 1 1 1 1 (A-4) The coefficient matrix is T M r (A-5) The modal participation factor matrix i for mode i at dof j is i j i j (A-6) mˆ ii Each mˆ ii coefficient is 1 if the eigenvectors have been normalized with respect to the mass matrix. The effective modal mass m eff, i vector for mode i and dof j is i j meff, i j (A-7) mˆ ii The natural frequency results for the sample problem are calculated using the program: six_dof_iso.m. The results are given in the next pages. 1
six_dof_iso.m ver 1. March 31, 5 by Tom Irvine Email: tomirvine@aol.com This program finds the eigenvalues and eigenvectors for a six-degree-of-freedom system. Refer to six_dof_isolated.pdf for a diagram. The equation of motion is: M (d^x/dt^) + K x = Enter m (lbm) 4.8 Enter Jx (lbm in^) 44.9 Enter Jy (lbm in^) 39.9 Enter Jz (lbm in^) 18.8 Note that the stiffness values are for individual springs Enter kx (lbf/in) 8 Enter ky (lbf/in) 8 Enter kz (lbf/in) 8 Enter a1 (in) 6.18 Enter a (in) -.68 Enter b (in) 3.85 Enter c1 (in) 3 13
Enter c (in) 3 The mass matrix is m =.111.111.111.1163.134.487 The stiffness matrix is k = 1.e+4 *.3.13.3 -.1418.3 -.13.1418 -.13.763 -.5458.1418 -.5458 1.14.13 -.1418 1.3 Eigenvalues lambda = 1.e+5 *.13.57.886.98 1.5699.7318 14
Natural Frequencies = 1. 7.338 Hz. 1. Hz 3. 7.4 Hz 4. 7.47 Hz 5. 63.6 Hz 6. 83.19 Hz Modes Shapes (rows represent modes) x y z alpha beta theta 1. 5.91-6.81-1.4. 8.69.954 -.744 3. 7.17 6.3 4. 1.4 -.6-1.95 5. -3.69 1.61 -.3 6. 1.96 -.5 4.3 Participation Factors (rows represent modes) x y z alpha beta theta 1..656 -.755 -.693..963.111 -.769 3..795.691 4..115 -.63 -. 5. -.49.187 -.38 6..17 -.5.1 Effective Modal Mass (rows represent modes) x y z alpha beta theta 1..43.569.48..98.13.59 3..63.477 4..133.69.48 5..168.35.566 6..471.63.439 Total Modal Mass.111.111.111.116.13.487 15
APPENDIX B The following definition is taken from Reference 3. Note that the mode shape functions are unscaled. Hence, the participation factor is unscaled. Modal Participation Factor Consider a beam of length loaded by a distributed force p(x,t). Consider that the loading per unit length is separable in the form The modal participation factor where p(x, t) Po p(x)f (t) (B-1) i for mode i is defined as 1 i p(x) i (x)dx (B-) i (x) is the normal mode shape for mode i 16
APPENDIX C The following convention appears to be more useful than that given in Appendix B. Modal Participation Factor for a Beam et Y n(x) = mass-normalized eigenvectors m(x) = mass per length The participation factor is n m(x) Yn(x)dx (C-1) The effective modal mass is m(x) Yn(x)dx m eff, n (C-) m(x) Yn(x) dx The eigenvectors should be normalized such that (x) Yn(x) dx 1 m (C-3) Thus, meff, n n m(x) Yn(x)dx (C-4) 17
APPENDIX D Effective Modal Mass Values for Bernoulli-Euler Beams The results are calculated using formulas from Reference 4. The variables are E I = is the modulus of elasticity = is the area moment of inertia = is the length = is (mass/length) Table D-1. Bending Vibration, Beam Simply-Supported at Both Ends Mode Natural Frequency n Participation Factor Effective Modal Mass 1 8 4 3 9 3 8 9 4 16 5 5 5 8 5 6 36 7 49 7 8 49 95% of the total mass is accounted for using the first seven modes. 18
Table D-. Bending Vibration, Fixed-Free Beam Mode Natural Frequency n Participation Factor Effective Modal Mass 1 1.8751.783. 6131 4.6949.4339.1883 3 5.544.6474 4 7.1818.336 9% of the total mass is accounted for using the first four modes. 19
APPENDIX E Rod, ongitudinal Vibration The results are taken from Reference 5. Table E-1. ongitudinal Vibration of a Rod, Fixed-Free Mode Natural Frequency n Participation Factor 1.5 c / 1.5 c / 3 3.5 c / 5 Effective Modal Mass 8 8 9 8 5 The longitudinal wave speed c is c E (E-1) 93% of the total mass is accounted for by using the first three modes.