January, 2000Page 1 of 1 MODAL MASS, STIFFNESS AND DAMPING Mark H. Richardson Vibrant Technology, Inc. Jamestown, CA INTRODUCTION For classically damped structures, modal mass, stiffness and damping can be defined directly from formulas that relate the full mass, stiffness and damping matrices to the transfer function matrix. The modal mass, stiffness, and damping definitions are derived in a previous paper [1], and are re- stated here for convenience. The transfer function is defined over the complex Laplace plane, as a function of the variable ) j s ( ω + σ · . Experi- mentally, the values of a transfer function are measured only along the ω j -axis in the s-plane, that is for ) j s ( ω · . These values are referred to as the Frequency Response Function (FRF). CLASSICALLY DAMPED STRUCTURE A classically damped structure is one where the modal damping is much smaller than the damped natural fre- quency of each mode (it is lightly damped), and the mode shapes are primarily real valued (they approximate normal modes). Light Damping: A structure is lightly damped if the damp- ing coefficient ) ( k σ of each mode (k) is much less than the damped natural frequency ) ( k ω . That is, k k ω << σ (1) Normal Mode Shapes: If the imaginary part of each mode shape vector } u { k is much less than the real part, that is if, ( ) ( ) } u { Re } u { Im k k << (2) where, ( ) ( ) } u { Im j } u { Re } u { k k k + · (3) the structure's mode shapes approximate normal modes. Both of these assumptions are satisfied by a large variety of real structures from which experimental modal data has been acquired. MODAL MASS MATRIX When the mass matrix is post-multiplied by the mode shape matrix and pre-multiplied by its transpose, the result is a diagonal matrix, shown in equation (4). This is a definition of modal mass. [ ] ] ] ] ω · · φ φ O O O O A 1 m ] [ ] M [ ] [ t (4) where, · ] M [ (n by n) mass matrix. [ ] · · φ } u { } u { } u { ] [ m 2 1 K (n by m) mode shape ma- trix. · } u { k n-dimensional mode shape vector for the th k mode, k = 1 to m. m = number of modes of vibration. n = number of DOFs of the structure model. [ ] ] ] ] ω · O O O O A 1 m = (m by m) modal mass matrix. The modal mass of each mode (k) is a diagonal element of the modal mass matrix, Modal mass: k k k A 1 m ω · k = 1 to m (5) · k p · ω + σ − k k j pole location for the th k mode. · σ k damping coefficient of the th k mode. · ω k damped natural frequency of the th k mode. · k A a scaling constant for the th k mode. MODAL STIFFNESS MATRIX When the stiffness matrix is post-multiplied by the mode shape matrix and pre-multiplied by its transpose, the result is a diagonal matrix, shown in equation (6). This is a defini- tion of modal stiffness. [ ] ] ] ] ] ω ω + σ · · φ φ O O O O A k ] ][ K [ ] [ 2 2 t (6) where, January, 2000 Page 2 of 2 · ] K [ (n by n) stiffness matrix. [ ] ] ] ] ] ω ω + σ · O O O O A k 2 2 = (m by m) modal stiffness matrix. The modal stiffness of each mode (k) is a diagonal element of the modal stiffness matrix, Modal stiffness: k k 2 k 2 k k A k ω ω + σ · k = 1 to m (7) MODAL DAMPING MATRIX When the damping matrix is post-multiplied by the mode shape matrix and pre-multiplied by its transpose, the result is a diagonal matrix, shown in equation (8). This is a defini- tion of modal damping. [ ] ] ] ] ω σ · · φ φ O O O O A 2 c ] [ ] C [ ] [ t (8) where, · ] C [ (n by n) damping matrix. [ ] ] ] ] ω σ · O O O O A 2 c = (m by m) modal damping ma- trix. The modal damping of each mode (k) is a diagonal element of the modal damping matrix, Modal damping: k k k k A 2 c ω σ · k = 1 to m (9) SDOF RELATIONSHIPS The familiar single degree-of-freedom (SDOF) relationships follow from the definitions of modal mass, stiffness, and damping for multiple DOF systems, ) ( m k 2 k 2 k k k ω + σ · k = 1 to m (10) ) 2 ( m c k k k σ · k = 1 to m (11) SCALING MODE SHAPES TO UNIT MODAL MASSES Mode shapes are called "shapes" because they are unique in shape, but not in value. That is, the mode shape vector } u { k for each mode (k) does not have unique values. It can be arbitrarily scaled to any set of values, but the rela- tionship of one shape component to any other is unique. In other words, the "shape" of } u { k is unique, but its values are not. A mode shape is also called an eigenvector, which means that its "shape" is unique, but its values are arbitrary. Notice also, that each of the modal mass, stiffness, and damping matrix definitions (5), (7), and (9) includes a scal- ing constant ) A ( k . This constant is necessary because the mode shapes are not unique in value, and therefore can be arbitrarily scaled. Unit Modal Masses One of the common ways to scale mode shapes is to scale them so that the modal masses are one (unity). Normally, if the mass matrix [ ] M were available, the mode vectors would simply be scaled such that when the triple product [ ] [ ][ ] U M U t was formed, the resulting modal mass matrix would equal an identity matrix. However, when the modal data is obtained from experimental transfer function meas- urements (FRFs), no mass matrix is available for scaling in this way. Even without the mass matrix however, experimental mode shapes can still be scaled to unit modal masses by using the relationship between residues and mode shapes. t k k k } u }{ u { A )] k ( r [ · k = 1 to m (12) where, · )] k ( r [ (n by n) residue matrix for the th k mode. Residues are the constant numerators of the transfer function matrix when it is written in partial fraction form, ∑ · − − − · m 1 k * k * k ) p s ( j 2 )] k ( r [ ) p s ( j 2 )] k ( r [ )] s ( H [ (13) * -denotes the complex conjugate. Residues have unique values, and have engineering units. Since the transfer functions typically have units of (motion / force), and the denominators have units of Hz or (radi- ans/second), residues have units of (motion / force) (Hz). Equation (12) can be written for the th j column (or row) of the residue matrix and for mode (k) as, January, 2000 Page 3 of 3 ( ) ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ' ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ' ¹ ⋅ ⋅ ⋅ · ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ' ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ' ¹ ⋅ ⋅ ⋅ · ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ' ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ' ¹ ⋅ ⋅ ⋅ nk jk k 2 k 1 jk k nk nk 2 jk jk k 2 jk k 1 k nj jj j 2 j 1 u u u u u A u u u u u u u A ) k ( r ) k ( r ) k ( r ) k ( r (14) Unique Variable k=1,…, m The importance of this relationship is that residues are unique in value and reflect the unique physical properties of the structure, while the mode shapes aren't unique in value and can therefore be scaled in any manner desired. The scaling constant k A must always be chosen so that equation (14) remains valid. The value of k A can be chosen first, and the mode shapes scaled accordingly so that equa- tion (14) is satisfied. Or, the mode shapes can be scaled first and k A computed so that equation (14) is still satisfied. In order to obtain mode shapes scaled to unit modal masses, we simply set the modal mass to one (1) and solve equation (5) for k A , k k 1 A ω · k=1 to m (15) Driving Point Measurement The unit modal mass scaled mode shape vectors are ob- tained from the th j column (or row) of the residue matrix by substituting equation (15) into equation (14), ( ) ( ) ( ) ( ) ( ) ( ) ( ) ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ' ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ' ¹ ⋅ ⋅ ⋅ ω · ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ' ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ' ¹ ⋅ ⋅ ⋅ · ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ' ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ' ¹ ⋅ ⋅ ⋅ k r k r k r k r k r k r k r u A 1 u u u nj j 2 j 1 jj k nj j 2 j 1 jk k nk k 2 k 1 (16) UMM k=1,…, m Notice that the driving point residue ( ) k r jj (where the row index(j) equals the column index(j)), plays an important role in this scaling process. Therefore, the driving point residue for each mode(k) is required in order to use equation (16). Triangular Measurement For cases where the driving point measurement cannot be made, an alternative set of measurements can be used to provide the driving point mode shape component jk u . From equation (14) we can write, ) k ( r ) k ( r ) k ( r A u pq jq jp k jk · k=1 to m (17) Equation (17) can be substituted for jk u in equation (16) to yield mode shapes scaled to unit modal masses. Equation (17) says that as an alternative to making a driving point measurement, three other measurements can be made in- volving DOF(p), DOF(q), and DOF(j). DOF(j) is the reference (fixed) DOF for the th j column (or row) of transfer function measurements, so the two meas- urements jp H and jq H would normally be made. In addi- tion, one extra measurement pq H is also required in order to solve equation (17). Since the measurements jp H , jq H , and pq H form a triangle in the transfer function matrix, they are called a triangular measurement. CONVERTING RESIDUES TO DISPLACEMENT UNITS Vibration measurements are often made using accelerome- ters to measure acceleration response, or vibrometers to measure velocity. Excitation forces are typically measured with a load cell. Therefore, transfer function measurements made with these transducers will have units of either (accel- eration/force) or (velocity/force). Modal residues always carry the units of the transfer func- tion multiplied by (radians/second). Therefore, residues taken from transfer functions with units of (accelera- tion/force) will have units of (acceleration/force-sec). Likewise, residues taken from measurements with units of (velocity/force) would have units of (velocity/force-sec). Similarly, residues taken from measurements with units of (displacement/force-sec) would have units of (displace- ment/force-sec). Since the modal mass, stiffness, and damping equations (4), (6), and (8) assume units of (displacement/force), residues with units of (acceleration/force-sec) or (velocity/force-sec) must be "integrated" to units of (displacement/force-sec) units before performing mode shape scaling. January, 2000 Page 4 of 4 Integration of a time domain function has an equivalent op- eration in the frequency domain. Integration of a transfer function is done by dividing it by the Laplace variable(s), 2 a v d s )] s ( H [ s )] s ( H [ )] s ( H [ · · (18) where, )] s ( H [ d = transfer matrix in (displacement/force) units. )] s ( H [ v = transfer matrix in (velocity/force) units. )] s ( H [ a = transfer matrix in (acceleration/force) units. Since residues are the result of a partial fraction expansion of a transfer function, residues can be "integrated" directly as if they were obtained from an integrated transfer function using the formula, 2 k a k v d ) p ( )] k ( r [ p )] k ( r [ )] k ( r [ · · k=1 to m (19) where, )] k ( r [ d = residue matrix in (displacement/force) units. )] k ( r [ v = residue matrix in (velocity/force) units. )] k ( r [ a = residue matrix in (acceleration/force) units. · k p · ω + σ − k k j pole location for the th k mode. Since we are assuming that damping is light and the mode shapes are normal, equation (19) can be simplified to, )] k ( r [ ) F ( )] k ( r [ F )] k ( r [ a 2 k v k d · · k=1 to m (20) where, ) ( F 2 k 2 k k k ω + σ ω · k=1 to m (21) Equations (20) and (21) can be summarized in the following table. To change transfer function units Multiple residues From To By FORCE ON ACCELERATI FORCE NT DISPLACEME 2 F FORCE VELOCITY FORCE NT DISPLACEME F Table 1. Residue Scale Factors. Where: ) ( F 2 2 ω + σ ω · (seconds) EXAMPLE OF UNIT MODAL MASS SCALING Suppose that we have the following data for a single mode of vibration, Frequency = 10.0 Hz. Damping = 1.0 % Residue Vector = ¹ ¹ ¹ ' ¹ ¹ ¹ ¹ ' ¹ + + − 5 . 0 0 . 2 1 . 0 Also, suppose that the measurements from which this data was obtained have units of (Gs/Lbf). Also assume that the driving point is at the second DOF of the structure. Hence the driving point residue = 2.0. Converting the frequency and damping into units of radi- ans/second, Frequency = 62.83 Rad/Sec Damping = 0.628 Rad/Sec The residues always carry the units of the transfer function measurement multiplied by (radians/second). Therefore, for this case the units of the residues are, Residue Units = Gs/(Lbf-Sec) = 386.4 Inches/(Lbf-Sec 3 ) Therefore, the residues become, Residue Vector = ¹ ¹ ¹ ' ¹ ¹ ¹ ¹ ' ¹ + + − 2 . 93 1 8 . 72 7 64 . 8 3 Inches/(Lbf-Sec 3 ) Since the modal mass, stiffness, and damping equations (4), (6), and (8) assume units of (displacement/force), the above residues with units of (acceleration/force) have to be con- verted to (displacement/force) units. This is done by using the appropriate scale factor from Table 1. For this case: 000253 . 0 83 . 62 1 F 2 2 · , ` . | ≅ (Seconds 2 ) Multiplying the residues by 2 F gives, January, 2000 Page 5 of 5 Residue Vector = ¹ ¹ ¹ ' ¹ ¹ ¹ ¹ ' ¹ + + − 0.0488 0.1955 0.00977 Inches/(Lbf-Sec) Finally, to obtain a mode shape scaled to unit modal mass, Equation (18) is used. The mode shape of residues must be multiplied by the scale factor, 927 . 17 0.1955 83 . 62 r SF jj · + · ω · to obtain the unit modal mass mode shape, UMM Mode Shape = ¹ ¹ ¹ ' ¹ ¹ ¹ ¹ ' ¹ + + − 875 . 0 505 . 3 175 . 0 Inches/(Lbf-Sec) REFERENCES [1] Richardson, M.H. "Derivation of Mass, Stiffness and Damping Parameters From Experimental Modal Data" Hew- lett Packard Company, Santa Clara Division, June, 1977. [2] Potter, R. and Richardson, M.H. "Mass, Stiffness and Damping Matrices from Measured Modal Parame- ters",.I.S.A. International Instrumentation - Automation Conference, New York, New York, October 1974 kk 2 2 = (σ k + ωk ) mk ck = ( 2σ k ) mk k = 1 to m (10) [H(s )] = ∑ [r(k )] [r(k )]* − 2 j (s − p * ) k =1 2 j (s − p k ) k m (13) * -denotes the complex conjugate. and damping for multiple DOF systems. Residues are the constant numerators of the transfer function matrix when it is written in partial fraction form. trix. Unit Modal Masses One of the common ways to scale mode shapes is to scale them so that the modal masses are one (unity). Even without the mass matrix however. if the mass matrix M were available. stiffness. and (9) includes a scaling constant ( A k ) . the mode vectors would simply be scaled such that when the triple product matrix. but the relationship of one shape component to any other is unique. th column (or row) of Page 2 of 2 . The modal damping of each mode (k) is a diagonal element of the modal damping matrix. [φ]t [C] [φ] = where. which means that its "shape" is unique. k = 1 to m (12) [r(k )] = (n by n) residue matrix for the k th mode. 2000 [K ] = (n by n) stiffness matrix. (7). the resulting modal mass matrix [ O 2σ cO = = (m by m) modal damping maAω O ] O would equal an identity matrix. 2 2 σ k + ωk Modal stiffness: k k = A k ωk k = 1 to m (7) MODAL DAMPING MATRIX When the damping matrix is post-multiplied by the mode shape matrix and pre-multiplied by its transpose. no mass matrix is available for scaling in this way. the "shape" of {u k } is unique. In other words. However. but its values are not. when the modal data is obtained from experimental transfer function measurements (FRFs). The modal stiffness of each mode (k) is a diagonal element of the modal stiffness matrix. This is a definition of modal damping. and have engineering units. shown in equation (8). It can be arbitrarily scaled to any set of values. the mode shape vector {u k } for each mode (k) does not have unique values. A mode shape is also called an eigenvector. stiffness. k = 1 to m (11) Residues have unique values. the result is a diagonal matrix. but its values are arbitrary. and the denominators have units of Hz or (radians/second). and therefore can be arbitrarily scaled. Normally. residues have units of (motion / force) (Hz). The familiar single degree-of-freedom (SDOF) relationships follow from the definitions of modal mass. but not in value. [ O kO ] O σ 2 + ω2 = = (m by m) modal stiffness Aω O SCALING MODE SHAPES TO UNIT MODAL MASSES Mode shapes are called "shapes" because they are unique in shape. 2σ k Modal damping: c k = A k ωk SDOF RELATIONSHIPS k = 1 to m (9) [r(k )] = A k {u k }{u k }t where. that each of the modal mass. Since the transfer functions typically have units of (motion / force). That is. This constant is necessary because the mode shapes are not unique in value. [ O 2σ cO = Aω O ] O (8) [ ] [C] = (n by n) damping matrix.January. [U] t [M ][U] was formed. Notice also. and damping matrix definitions (5). Equation (12) can be written for the j the residue matrix and for mode (k) as. experimental mode shapes can still be scaled to unit modal masses by using the relationship between residues and mode shapes. three other measurements can be made involving DOF(p). one extra measurement H pq is also required in order to solve equation (17). In addition. and DOF(j). u jk = A k rjp (k ) rjq (k ) rpq (k ) k=1 to m (17) k=1.…. Since the modal mass. yield mode shapes scaled to unit modal masses. Similarly. Or. (14) From equation (14) we can write.January. plays an important role in this scaling process. Therefore. Modal residues always carry the units of the transfer function multiplied by (radians/second). th and H pq form a triangle in the transfer function matrix. CONVERTING RESIDUES TO DISPLACEMENT UNITS Vibration measurements are often made using accelerometers to measure acceleration response. In order to obtain mode shapes scaled to unit modal masses. residues taken from measurements with units of (displacement/force-sec) would have units of (displacement/force-sec). H jq . they are called a triangular measurement. we simply set the modal mass to one (1) and solve equation (5) for A k . so the two measurements H jp and H jq would normally be made. and the mode shapes scaled accordingly so that equation (14) is satisfied. while the mode shapes aren't unique in value and can therefore be scaled in any manner desired. th u 1k u 2k ⋅ 1 = ⋅ A k u jk ⋅ u nk UMM r1 j (k ) r (k ) 2j ⋅ = ⋅ ⋅ rnj (k ) ωk rjj (k ) r1 j (k ) r (k ) 2j ⋅ (16) ⋅ ⋅ rnj (k ) k=1. an alternative set of measurements can be used to provide the driving point mode shape component u jk . Likewise. Excitation forces are typically measured with a load cell. The value of A k can be chosen first. DOF(q). Ak = 1 ωk k=1 to m (15) Driving Point Measurement The unit modal mass scaled mode shape vectors are obtained from the j column (or row) of the residue matrix by substituting equation (15) into equation (14). Since the measurements H jp . the driving point residue for each mode(k) is required in order to use equation (16). residues taken from transfer functions with units of (acceleration/force) will have units of (acceleration/force-sec). residues with units of (acceleration/force-sec) or (velocity/force-sec) must be "integrated" to units of (displacement/force-sec) units before performing mode shape scaling. and (8) assume units of (displacement/force). transfer function measurements made with these transducers will have units of either (acceleration/force) or (velocity/force). Page 3 of 3 . Therefore. DOF(j) is the reference (fixed) DOF for the j column (or row) of transfer function measurements. stiffness. 2000 r1 j (k ) u 1k u jk u 1k r (k ) u u u 2j 2k jk 2k ⋅ ⋅ ⋅ ⋅ = A k ⋅ = A k u jk ⋅ r (k ) (u ) 2 u jj jk jk ⋅ ⋅ ⋅ r (k ) u nk u nk u nk nj Unique Variable Triangular Measurement For cases where the driving point measurement cannot be made. The scaling constant A k must always be chosen so that equation (14) remains valid. and damping equations (4). m Notice that the driving point residue rjj (k ) (where the row index(j) equals the column index(j)).…. (6). Therefore. m Equation (17) can be substituted for u jk in equation (16) to The importance of this relationship is that residues are unique in value and reflect the unique physical properties of the structure. Equation (17) says that as an alternative to making a driving point measurement. or vibrometers to measure velocity. the mode shapes can be scaled first and A k computed so that equation (14) is still satisfied. residues taken from measurements with units of (velocity/force) would have units of (velocity/force-sec). 0. To change transfer function units From ACCELERATION FORCE VELOCITY FORCE Multiple residues To DISPLACEMENT FORCE DISPLACEMENT FORCE By F2 F 1 F2 ≅ = 0. For this case: Equations (20) and (21) can be summarized in the following table.January.1 Residue Vector = + 2.8 Inches/(Lbf-Sec3) + 193.000253 62. Table 1. Integration of a transfer function is done by dividing it by the Laplace variable(s). p k = − σ k + jω k = pole location for the k th mode. [ra (k )] = residue matrix in (acceleration/force) units. [H v (s )] = transfer matrix in (velocity/force) units. [H v (s )] [H a (s )] = s s2 (18) EXAMPLE OF UNIT MODAL MASS SCALING Suppose that we have the following data for a single mode of vibration. Residue Scale Factors.83 Multiplying the residues by 2 (Seconds2) F 2 gives.2 Since the modal mass. − 0. Frequency = 62. equation (19) can be simplified to. Since residues are the result of a partial fraction expansion of a transfer function.628 Rad/Sec The residues always carry the units of the transfer function measurement multiplied by (radians/second).83 Rad/Sec Damping = 0.4 Inches/(Lbf-Sec3) Therefore. and (8) assume units of (displacement/force). the residues become. [rd (k )] = where. Therefore. Frequency = 10. [rv (k )] = residue matrix in (velocity/force) units. Also assume that the driving point is at the second DOF of the structure. the above residues with units of (acceleration/force) have to be converted to (displacement/force) units. and damping equations (4). Hence the driving point residue = 2.5 Also. Page 4 of 4 .0 % [H d (s )] = transfer matrix in (displacement/force) units. [rv (k )] [ra (k )] = pk (p k ) 2 k=1 to m (19) [rd (k )] = residue matrix in (displacement/force) units.64 Residue Vector = + 772. [H a (s )] = transfer matrix in (acceleration/force) units. Since we are assuming that damping is light and the mode shapes are normal. Residue Units = Gs/(Lbf-Sec) = 386. stiffness. Damping = 1. for this case the units of the residues are.0 Hz. This is done by using the appropriate scale factor from Table 1. residues can be "integrated" directly as if they were obtained from an integrated transfer function using the formula.0 + 0. Where: F= ω (σ + ω2 ) 2 (seconds) [H d (s )] = where. Converting the frequency and damping into units of radians/second. 2000 Integration of a time domain function has an equivalent operation in the frequency domain. Fk = ωk 2 (σ + ωk ) 2 k k=1 to m (21) − 38. suppose that the measurements from which this data was obtained have units of (Gs/Lbf). [rd (k )] = Fk [rv (k )] = (Fk ) 2 [ra (k )] k=1 to m (20) where. (6). 875 REFERENCES [1] Richardson. Stiffness and Damping Parameters From Experimental Modal Data" Hewlett Packard Company. to obtain a mode shape scaled to unit modal mass. International Instrumentation . SF = ω = rjj 62. "Mass..S.505 Inches/(Lbf-Sec) + 0.A.175 UMM Mode Shape = + 3. Santa Clara Division. R.I.83 = 17.1955 Inches/(Lbf-Sec) + 0. and Richardson. June. M.1955 to obtain the unit modal mass mode shape. The mode shape of residues must be multiplied by the scale factor.00977 Residue Vector = + 0.H.0488 Finally.H. 2000 − 0. "Derivation of Mass. New York. 1977. Stiffness and Damping Matrices from Measured Modal Parameters". Equation (18) is used.January.Automation Conference. New York. M. October 1974 Page 5 of 5 . [2] Potter. − 0.927 + 0.