Theoretical Basis of Modal Analysis

American Journal of Mechanical Engineering, 03, Vol., No. 7, 73-79 Available online at http://pubs.sciepub.com/ajme//7/4 Science and Education Publishing DOI:0.69/ajme--7-4 heoretical Basis of Modal Analysis Pavol Lengvarský *, Jozef Bocko Department of Applied Mechanics and Mechatronics, echnical University of Košice, Košice, Slovakia * Corresponding author: pavol.lengvarsky@tuke.sk Received October 09, 03; Revised October 8, 03; Accepted November 3, 03 Abstract here is given the theoretical background of the modal analysis in this paper. In the first part modal analysis, its theoretical and experimental procedure of vibration analysis is defined. In document one degree of freedom system and its behavior, equations and next modification as system of free vibration, harmonic excitation, structural damping is defined. Also, there is defined important property of the modal model - orthogonal property. In the next step is discussed multi degrees of freedom system, its properties, damping and solution. Keywords: modal analysis, vibration, damping, orthogonal properties, excitation Cite his Article: Pavol Lengvarský, and Jozef Bocko, heoretical Basis of Modal Analysis. American Journal of Mechanical Engineering, no. 7 (03): 73-79. doi: 0.69/ajme--7-4.. Introduction Using of modal analysis we can determine the natural frequencies, the damping at natural frequencies and the mode shapes at natural frequencies [3]. Figure shows the theoretical procedure of vibration analysis. It displays the three phase procedure of the theoretical vibration analysis. We are starting with a description of the physical properties of the structures, usually as their mass, stiffness and damping characteristics []. deduce the modal model properties. his method is called the experimental procedure of vibration analysis and is shows in Figure []. In Figure 3 the general procedure for linking simulation and experimental analysis in modal analysis is shown [6]. Figure. heoretical procedure of vibration analysis [] he theoretical modal analysis of solid model leads to a description of the behavior of the structure as modes of vibration so-called the modal model. his model is defined as a set of natural frequencies with damping factor and natural modes vibration. his solution describes various methods of the structure vibration []. he response of model is the next part of the theoretical procedure analysis, which corresponds to the excitation and its amplitude. Model describes a set of frequency response functions []. Figure. Experimental aim [] It is possible to perform the analysis in the opposite direction, from the response properties, which we can Figure 3. General procedure for linking simulation and experimental analysis in modal analysis []. One Degree of Freedom System In Figure 4 one degree of freedom system is shown. he system is composed of a perfectly rigid body of mass m, spring with stiffness k and viscous damper with damping ratio d. he force in the spring is directly proportional to the displacement x and the damping force to the velocity x. he spring and damper can be considered as linear []. he equation of motion has the form

74 American Journal of Mechanical Engineering mx + dx + kx = F( t), () where Ft () is the excitation force and x is the displacement of the body of the equilibrium position. δ t jω t j t ( ) [ d ω xt = e Ce d + Ce ], () where C and C are constants. hese ones must have real value to both sides of the equation. he condition is valid if C = ( A jb), C = ( A + jb), (3) where A and B are real constants. Substituting the equation (3) into the equation () we obtain the displacement in the form δ t ( ) = ( cosω + sinω ) xt e A dt B dt ( ω t φ) δ t = xe ˆ sin d + (4) where amplitude ˆx and phase angle φ are xˆ = A + B, (5) Figure 4. One degree of freedom system [3].. Free Vibration Substituting Ft () = 0 to equation () we get the equation of motion in the form mx + dx + kx = 0. () Dividing the equation () of mass m, we obtain x+ δx + ω n x = 0, (3) where δ is the damping constant and ω n is the natural angular frequency If we assume the result in the form we obtain the equation d δ =, (4) m k ω n =. (5) m x() t = Ce λt (6) ( ) λ + δλ + ω n C = 0. (7) For C 0 has the characteristic equation form + + = 0. (8) λ δλ ω n he roots of the characteristic equation (8) are *, = ±, (9) λλ δ δ ω n where δ is damping constant. If 0 < δ < ωn we have damping system where *, = ± n = ± j d, (0) λλ δ ω δ δ ω d = n. () ω ω δ With the general solution of damping system we have equation in the form A φ = arctg. (6) B (6) For the case that damping constant is δ = 0 we get ( ) ω ω ˆ ( ω φ) xt = Acos t+ Bsin t= xsin t+. (7) n n n he function in Figure 5 shows the free vibration and consists of the harmonic displacement x. he value ω n is natural angular frequency of the system without damping and with period π and the time of period n and thus where f n is the natural frequency. ωnn = π, (8) π n =, (9) ωn ωn fn = =, (0) n π Figure 5. Free vibration without damping [3] Constants A and B are determined from the initial x t 0 x xt = 0 = x. hus conditions ( = ) = 0 and ( ) 0 0 0, x A= x B =. () ω n In Figure 6 is the course of function e δ t decreasing in time and the harmonic value corresponding of the damping harmonic vibration is shown []. Proportion of damping constant δ we can express the value of the damping the coefficient attenuation D, or the logarithmic decrement ϑ. he damping factor has the form

American Journal of Mechanical Engineering 75 Logarithmic decrement has the form ϑ D = δ. () ω n xn = ln, (3) xn+ where x n is the value of the displacement from the equation (4) at the time t 0 and x n + is the value of the displacement at the time t 0 + d, where π d =. (4) ωd d is the time period of damping vibration. ϑ δ = d = and for D << can write π D D (5) ϑ πd. (6) Figure 6. Damping vibration [3].. Harmonic Excitation After description of free vibration we consider forced vibration with harmonic excitation. We consider excitation force in the form ( ) ˆ cos ω. F t = F t (7) he solution of equation of motion () is represented by a complex value of the displacement the excitation force F(t) has the form z = x + jy, (8) ( ) ˆ ( ω ω ) cos sin ˆ j t F t = F t+ j t = Fe ω. (9) From the equation () we obtain and solution where constant C is ˆ j t mz + dz + kz= Fe ω (30) j t ( ) Ce ω, z t = (3) Fˆ C =. (3) k mω + jdω he displacement xt () Re { zt ()} with amplitude and phase angle xˆ = = leads to xt ( ) = xˆ cos( ωt ε) (33) Fˆ ( k mω ) + ( dω) dω ε = arctg k m ω. (34) (35) he mass performs harmonic motion during harmonic excitation with frequency ω. he displacement pursute the force with phase angle ε. For static force is ω = 0. he displacement is then Fˆ xs =. (36) k Amplitude ˆx is divided by magnification factor x S with a constant ratio of frequencies, thus coefficient of tuning is η = (37) ω n and damping factor D from equation () is substituted into (34), thus function of magnification is obtained ω xˆ V = = xs ( η ) + ( Dη) and from the equation (35) we obtain Dη ε = arctg. η (38) (39) he values V and ε are shown in Figure 7 and Figure 8 as a ratio of the function force relative frequency and damping factor. he function of magnification is also known as the resonance curve. Each of the resonance curves starts at point and decreases to 0 for large values η. For D / = 0.707 curve has maximum in relative frequency Vmax = D D (40) η = (4) max D. For small damping D << is obtained Vmax D (4) ηmax. (43) he form /D is called resonance sharpness or Q factor xˆ max Q = Vmax. D = x (44) s

76 American Journal of Mechanical Engineering replaced by meω 0. Using the equations (33) and (34) the solution is in the form ( ) ( ω ε) xt = xˆ cos t, (48) Figure 7. Function of magnification V [3] Figure 8. he phase angle [3] From Figure 8 is clear that phase angle lies between the values 0 and π. For η = is ω = ωn and is equal to π /, independent of the damping coefficient. Figure 7 indicates that a resonance curve is wider with growth of damping. For D << is =, (45) ωmax ωmax D ω ω ω where ω and ω are angular frequencies at x ˆ = x ˆ max / 0.707x ˆ max (obr.9). he system from Figure 4 is supplemented of the source excitation with imbalance (Figure 0). he result is the excitation force in the form ( ) ω F t = me 0 cos ωt. (46) With a total mass m= m+ m0 we obtain the equation of motion mx + d x + k x = m0 eω cos ωt. (47) he partial results follow directly from the results of amplitude with constant excitation F ˆ, which will be Figure 9. Curve of resonance [3] where amplitude is xˆ = me 0 ω ( k mω ) + ( dω) or m0 xˆ = Ve, m+ m0 where the value magnification is V = η ( η ) + ( dη). (49) (50) (5) Figure 0. Excited system [3] he amplitude of constant force is equal to the phase angle ε. From equation (40) is obvious, that V max is identical with the maximum value V while the function of magnification V starts at 0 and ends at. he maximum value V is for the case η max =. D (5)

American Journal of Mechanical Engineering 77 Figure. Function of magnification V [3] In Figure is shown the system with excitation from the ground. he equation of motion has the form Substituting my + d y + k y= k z+ d z. (53) y = x z (54) we obtain the equation of motion in the form my + d y + k y = m z( t). (55) field, reducing the value of the growth of vibration and thus also the noise when passing through the resonance area. Structural damping is generally smaller than other types of damping []. All materials exhibit some degree of damping due to their hysteresis properties. he typical example of this effect is the graph of the time dependence of the displacement on force shown in Figure 3a, in which the area contained by loop represents lost of energy in one cycle vibration between illustrated boundaries []. Another common source of energy absorption in real structures and also the damping is the friction, which exists in the connections between the components of the structure. If these effects are macro slip between adjacent parts, or rather micro slip on connection between them can be described by a simple model of dry friction shown in Figure 3b with its corresponding dependence displacement on force []. Viscous damper is shown in Figure 3c, while it is necessary to assume the harmonic motion at the frequency ω, for the purpose of constructing a dependence displacement on force. he problem that arises with a model viscous damping is that the frequency dependence of the energy loss per cycle, while the dry friction device is unaffected frequency of load []. Figure 3. he time dependence of the displacement on force [] (a) Hysteresis material, (b) Dry friction, (c) Viscous damper 3. Undamped Multidegrees of Freedom Systems Figure. System with excitation from the ground [3] For the harmonic motion zt ˆ( ) = zcosω t has the equation of motion the form my + d y + ky = mzˆ ω cos ωt, (56) where right-hand sites of the equations (56), (47) are similar, thus ( ) ( ω ε) yt = Vzˆ cos t, (57) where the function of magnification V is obtained from the equation (5) []... Structural Damping Structural damping causes faster damping of free vibration, reducing of amplitude vibration in the resonance 3.. Free Vibration - Modal Properties For undamped system MDOF with N degrees of freedom, we can write the equations of motion in the matrix notation Mx () t + Kx() t = f(), t (58) where M and K are NxN matrices of mass and stiffness, x ( t) and f ( t) are Nx vectors of displacement and forces dependent on time []. We will consider the first solution of free damping (to determine the normal and modal properties) that We assume the solution in the form f( t ) = 0. (59) ( t) = e iω t, x X (60) where X is Nx vector of time-independent amplitudes. For this case it is clear that x = ω Xe ω. (6) Substituting the equations (59), (60) and (6) in the equation of motion (58), leads to the equation

78 American Journal of Mechanical Engineering ( ω ) K M Xe ω = 0, (6) for which the only non-trivial solution satisfies or K ω M = (63) det 0, N N N + N + 0 = 0, (64) α ω α ω α from which we can find N values of ω = ω, ω,..., ωr,..., ωn, undamped natural angular ( ) frequencies of the system. 3.. Orthogonal Property of Modal Model he most important property of the modal model is orthogonal property. he equation of motion for free vibration we can written in the form ( ω ) K M Xe ω = 0. (65) he equation (65) for r -th mode shape we multiply by the transposed vector ψ s and we get ( ψskψr) ωr ( ψsmψ r) = 0. (66) hen for the s -th mode shape we multiply by the transposed vector ψ r, we get the equation in the form ( ψrkψs) ωs ( ψrmψ s) = 0. (67) Since the stiffness matrix K and mass matrix M are symmetrical matrices, we can the equation (67) transpose and written in the form ( ψskψr) ωs ( ψsmψ r) = 0. (68) Subtracting the equation (68) from the equation (66) we get expression ( ω ω ) s r s r if it is satisfied ωr ωs, that ψ Mψ = 0, (69) ψ s M ψ r = 0; r s. (70) From the equation (70) follows that mode shapes r and s are orthogonal with respect to the mass matrix M. Substituting the equation (70) to the equation (66) we get expression ψ s K ψ r = 0; r s. (7) From the equation (7) follows that mode shapes r and s are orthogonal with respect to the stiffness matrix K. In the special case when r = s or ωr = ωs, equations (70) and (7) are not valid. For this special case we can write the equation (68) in the form if ω r = kr / mr, then ( ψr K ψr) = ωr ( ψr K ψ r), (7) ψr K ψr = mr a ψr K ψ r = kr. (73) 4. Multidegrees of Freedom Systems with Structural Damping 4.. Solution of Free Vibration- Complex Modal Properties In this section we will examine properties of the general elements of hysteric damping. he general equation of motion for the system with multi degrees of freedom with hysteric damping and harmonic excitation has the form Mx + Kx + idx= Fe ω. (74) We will considered the case without excitation and we choose a solution in the form x = X e λ. (75) Substituting to the equation (74), the solution (75) leads to eigen problem, which solution is in the form of two matrices containing eigenvalues and eigenvectors. In this case each natural frequency and each mode shape is described by complex quantities such as ( i ) λr = ωr + ηr, (76) where ω r is natural angular frequency and η r is the damping loss factor. It is not necessary that natural angular frequency ω r equal to natural angular frequency of the undamped system ω r. his solution can be considered as the same type of orthogonal properties which have been demonstrated for undamped system and can be defined by equations Ψ M Ψ = mr ; Ψ ( K+ id) Ψ = k r. (77) 5. Conclusion his paper is only general entry to the modal analysis. It define the theoretical background of modal analysis while other models of modal analysis exist, for example models with different damping, excitation and other examples of calculation of modal analysis. Acknowledgement his article was created with support of VEGA grant project VEGA /05/ Numerical modeling of mechatronic systems. References [] EWINS, D.J., Modal esting: heory, Practice and Application, England: Wiley, 000. Second Edition. [] REBUŇA, F. ŠIMČÁK, F.: Príručka experimentálnej mechaniky, SjF U, Košice, 007. [3] MEAD, D.: Passive Vibration Control, University of Southampton,UK: Wiley, 000.

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