DYNAMICS OF MACHINES 41614

DYNAMICS OF MACHINES 464 PART I INTRODUCTION TO THE BASIC TOOLS OF MODELLING, SIMULATION, ANALYSIS & EXPERIMENTAL VALIDATION 6 x 4 Angular Velocity: rpm Mode: Nat. Freq.: 4.788 Hz 4 4 3 x 4 Signal (a) in Time Domain (b) in Frequency Domain 6.5 5 (a) Amplitude [m/s ].5 x 7 8 6 4 Angular Speed: Mode: Nat. Freq.: 5.685 Hz 4 3 5 5 5 3 x 5 4 3 (b) Amplitude [m/s ] 5 5 5 frequency [Hz] 3 4 3 4 5 5 5.5 x 3.5 4 6 8 Dr.-Ing. Ilmar Ferreira Santos, Associate Professor Department of Mechanical Engineering Technical University of Denmark Building 358, Room 59 8 Lyngby Denmark Phone: +45 45 5 6 69 Fax: +45 45 88 4 5 E-Mail: ifs@mek.dtu.dk Kompendie 34 February 4

Contents Introduction to Dynamical Modelling of Machines and Structures and Experimental Analysis of Mechanical Vibrations based on the Human Senses 3. Summary........................................ 3. Description of the Test Facilities........................... 3.3 Data of the Mechanical System............................ 5.4 Calculating Equivalent Stiffness Coefficients Beam Theory........... 5.5 Calculating Stiffness Matrices Beam Theory................... 7.6 Mechanical Systems with D.O.F........................... 9.6. Physical System and Mechanical Model................... 9.6. Mathematical Model.............................. 9.6.3 Analytical and Numerical Solution of the Equation of Motion........6.4 Analytical and Numerical Solution of Equation of Motion using Matlab. 5.6.5 Comparison between the Analytical and Numerical Solutions of Equation of Motion.................................... 6.6.6 Homogeneous Solution or Free-Vibrations or Transient Response - Experimental Analysis................................ 9.6.7 Natural Frequency ω n [rad/s] or f n [Hz].................. 9.6.8 Damping Factor ξ or Logarithmic Decrement β............... 9.6.9 Forced Vibrations or Steady-State Response................ 4.6. Resonance and Phase............................. 4.6. Superposition of Transient and Forced Vibrations in Time Domain (Simulation)..................................... 7.6. Resonance Experimental Analysis in Time Domain............ 8.7 Mechanical Systems with D.O.F........................... 3.7. Physical System and Mechanical Model................... 3.7. Mathematical Model.............................. 3.7.3 Analytical and Numerical Solution of System of Differential Linear Equations...................................... 3.7.4 Modal Analysis using Matlab eig-function [u, w] = eig( B, A)....... 37.7.5 Analytical and Numerical Solutions of Equation of Motion using Matlab. 4.7.6 Analytical and Numerical Results of the System of Equations of Motion. 4.7.7 Programming in Matlab Frequency Response Analysis.......... 4.7.8 Understanding Resonances and Mode Shapes using your Eyes and Fingers 43.7.9 Resonance Experimental Analysis in Time Domain............ 48.8 Mechanical Systems with 3 D.O.F........................... 49.8. Physical System and Mechanical Model................... 49.8. Mathematical Model.............................. 49.8.3 Programming in Matlab Theoretical Parameter Studies and Experimental Validation.................................. 5.8.4 Theoretical Frequency Response Function.................. 55.8.5 Experimental Natural Frequencies..................... 56.8.6 Experimental Resonances and Mode Shapes................ 56.9 Exercises........................................ 58. Project Identification of Model Parameters (An Example)........... 6. Project Modal Analysis & Validation of Models................. 66

Introduction to Dynamical Modelling of Machines and Structures and Experimental Analysis of Mechanical Vibrations based on the Human Senses. Summary The aim of this study is to show some theoretical and experimental examples to facilitate the understanding of the physical meaning of the main topics and definitions used in relation to vibrations in machines. Theoretical and experimental studies are led side-by-side, clarifying the definitions of stiffness, flexibility, natural frequency, damping factor, logarithmic decrement, resonance, phase, beating, unbalance, natural mode shape, modal node and so on. The experimental investigations are always carried out in low frequency ranges, aiming at making easy the visualization of the movements by the human eyes and the understanding of the mechanical vibrations without necessarily having to use sensors and electronic devices.. Description of the Test Facilities Figures and show the simple elements used during the experimental investigations: a flexible beam (ruler), concentrated masses or magnets, a support, an accelerometer, a signal amplifier, a shaker and a signal analyzer. The beam or ruler already has a scale, enabling it to easily achieve the information about the position (length) where the magnets will be mounted. At each position more than one magnet can be mounted, allowing changes in the values of the concentrated masses. The masses or magnets can be easily moved and attached to different positions along the ruler, aiming at investigating changes in the natural frequencies of the magnet-ruler system (mass-spring system). The ruler is very flexible in one plane only due to the characteristic of its cross-section (moment of inertia of area). The beam can easily be mounted with different boundary conditions, as free-free, clamped-free, simply supported in both ends, etc. allowing an investigation of the influence of the boundary conditions on its flexibility, and consequently on the natural frequencies of the system. Regarding rotating machines some analogies can be made between the mass-spring system presented here and a centrifugal compressor, while comparing the ruler (or flexible beam) to the flexible shaft, and the magnets (or concentrated masses) to the impellers or rigid discs. Changes in the montage of shaft into the bearings (boundary conditions) or changes in the positioning of the impellers along the shaft will lead to different critical speeds and mode shapes. As mentioned above, the mass-spring system was designed to have a very flexible behavior with very low natural frequencies. This allows the detection of natural frequencies, modes shapes and resonance using the human eyes as sensors (or sighting senses). Moreover, it is also made possible to excite the structure with human fingers, aiming at understanding the 9 degrees phase between displacement and excitation (while operating around resonance conditions) by means of tactile senses. Accelerometer, amplifier, shaker and signal analyzer are used aiming at confirming what the human senses detect. 3

By understanding the topics related to mechanical vibration in low frequency ranges (high flexibility and slow motions detectable by human senses), it gets easier to understand mechanical vibration in high frequency ranges where one has faster motions with small amplitudes, just detectable by sensors and electronic devices. Figure : Experimental investigation of a mechanical continuous system with concentrated masses modelled as equivalent spring-mass systems with, and 3 degrees of freedom (D.O.F.). Figure : Signal analyzer and shaker used for inducing and measuring mechanical vibrations while analyzing the behavior of the spring-mass systems with D.O.F., D.O.F. and 3 D.O.F. 4

.3 Data of the Mechanical System ρ material density of the beam 7, 8 Kg/m 3 E elasticity modulus N/m L total length of the beam.6 m b width of the beam.3 m h thickness. m m i concentrated mass (i =,...,6).9 Kg Table : Data of the mass-spring system A. ρ material density of the beam 7, 8 Kg/m 3 E elasticity modulus N/m L total length of the beam.3 m b width of the beam.5 m h thickness. m ρ material density (steel) 7, 8 Kg/m 3 m i concentrated mass (i =,...,6).9 Kg Table : Data of the mass-spring system B..4 Calculating Equivalent Stiffness Coefficients Beam Theory (a) (b) Figure 3: (a) Flexible beam clamped-free boundary condition case with force applied to the end L; (b) clamped-free boundary condition case with force applied to a general position L ; By applying a vertical force F at the end of the beam as shown in figure 3(a) and using Beam Theory, one can write: EI d4 y(x) dx 4 = () Integrating in X once, one has: EI d3 y(x) dx 3 = F(x) = C () 5

Integrating twice in X, it gives: EI d y(x) dx = M(x) = C x + C (3) Integrating again in X, one gets the rotation angle of the beam: EI dy(x) dx = EIΘ(x) = C x + C x + C 3 (4) And finally, integrating the last time in X, one achieves the equation responsible for describing the deflexion of the beam: EI y(x) = C x 3 6 + C x + C 3x + C 4 (5) The boundary conditions for the clamped-free beam are: y(x = ) = Θ(x = ) = M(x = L) = F(x = L) = F (reaction) (6) After calculating the constants C i using the boundary condition for the clamped-free beam case, one gets y(x) = F E I ( x 3 6 L x ) (7) dy(x) dx = Θ(x) = F E I ( ) x L x (8) Using the relationship between applied force F and the induced deflexion at a given point along the beam length, x = L for instance, one gets the equivalent stiffness as: K = F y(l) = 3 EI L 3 (9) Suggestion (I): Change the beam boundary conditions, for example bi-supported at both ends, and calculate the equivalent stiffness in the new case. Suggestion (II): Change the beam boundary conditions, for example clamped-clamped at both ends, and calculate the equivalent stiffness in the new case. 6

.5 Calculating Stiffness Matrices Beam Theory Two Different Lengths for Applying Forces To facilitate the understanding of steps which will be presented, one can introduce the follow nomenclature (see figure 3(b)): L = L or L = L length where the force F is applied. x = L or x = L length where the displacement is measured. Taking into account two different points for applying the forces and measuring the displacements of the beam, one works with the following set of equations x [, L ] y(x) = F E I dy(x) dx = F E I ( ) x 3 6 L x ( ) x L x () and x [L, L] y(x) = y(l ) + dy(x) dx dy(x) dx = dy(x) dx x=l (x L ) x=l () which are responsible for describing the deflection of the beam, considering the loading on different coordinates. Let us introduce an example of a system with two points of force application. Assuming in case (I) the force F is applied to the first coordinate L = L. One can measure and/or calculate the beam deflection at the coordinates x = L and x = L through equations () and (): y = y(l ) = F L3 3 EI and y = y(l ) = F 6EI (L3 + 3L (L L )) (3) () Assuming in case (II) that the force F is applied to the second coordinate L = L, one can measure and/or calculate the follow beam displacements at the coordinates x = L and x = L through the equations () and (): y = y(l ) = F 6EI (L3 + 3L (L L )) (4) and y = y(l ) = F L3 3 EI (5) 7

In order to calculate the stiffness matrix (k, k, k, k ) of the d.o.f. system, Ky = f (6) where [ k k K = k k ] y = { y y } T f = { f f } T (7) one has to solve the following system of equations for case (I) and case (II): [ k k k k ] { y y } = { f f } (8) Case (I): f = { F } T [ k k k k ] { F F L 3 3 EI 6EI (L3 + 3L (L L )) } = { F } (9) Case (II): f = { F } T [ k k k k ] { F 6EI (L3 + 3L (L L )) F L 3 3 EI } = { F } () It leads to a system of 4 equations, which can be written in a matrix form: F L 3 3 EI F 6EI (L3 + 3L (L L )) F L 3 3 EI F 6EI (L3 + 3L F L (L L )) 3 3 EI F 6EI (L3 + 3L (L L )) k k k k = F F F 6EI (L3 + 3L (L L )) F L 3 3 EI () Solving this matrix system of order 4 by using the software MATHEMATICA, or by using Cramer s rule, one achieves the stiffness matrix: K = EI (4L L )(L L ) (L /L ) 3 6(L 3L )/L 6(L 3L )/L () Similar procedure can be made in order to get the stiffness matrix of the 3 d.o.f system. This is the motivation of an exercise later on. The results (stiffness matrix with 9 stiffness coefficient) will be presented in the section describing mechanical systems with 3 d.o.f. 8

.6 Mechanical Systems with D.O.F..6. Physical System and Mechanical Model (a) (b) (c) Figure 4: (a) Real mechanical system composed of a turbine attached to an airplane flexible wing; (b) Laboratory prototype built by a lumped mass attached to a flexible beam); (c) Equivalent mechanical model with D.O.F. for a lumped mass attached to a flexible beam..6. Mathematical Model It is important to point out, that the equations of motion in Dynamics of Machinery will frequently have the form of second order differential equations: ÿ(t) = F(y(t), ẏ(t)). Such equations can generally be linearized around an operational position of a physical system, leading to second order linear differential equations. It means that the coefficients which are multiplying the variables ÿ(t), ẏ(t), y(t) (co-ordinate chosen for describing the motion of the physical system) do not depend on the variables themselves. In the mechanical model presented in figure 4 these coefficients are constants: m, d and k. One of the aims of the course Dynamics of Machinery is to help the students to properly find these coefficients so that the equations of motion can really describe the movement of the physical system. The coefficients can be predicted using theoretical or experimental approaches. 9

After having created the mechanical model for the physical system, the next step is to derive the equation of motion based on the mechanical model. The mechanical model is composed of a lumped mass m (assumption!!!), spring with equivalent stiffness coefficient k (calculated using beam theory) and damper with equivalent viscous coefficient d (obtained experimentally). While creating the mechanical model and assuming that the mass is a particle, the equation of motion can be derived, for example, using Newton s second law: m ÿ (t) + d ẏ (t) + k y (t) = f (t) (3) ÿ (t) + d m ẏ (t) + k m y (t) = f m (t) (4) ÿ (t) + ξω n ẏ (t) + ω ny (t) = f m (t) (5).6.3 Analytical and Numerical Solution of the Equation of Motion After having created the mechanical model (step ) and derived the mathematical model (step ) for this mechanical model, the next step is to solve the equation of motion (step 3), aiming at analyzing (step 4) the dynamical behavior of the physical system. Here the analytical and numerical solutions of linear differential equations are presented and compared. Later on you can choose the most convenient way to solve the equations and analyze the dynamical behavior of physical systems. It is important to mention that the numerical procedure is very simple, an integrator of first-order. Other integrators can be used depending on the characteristics of the mechanical models, for example, Runge-Kutta of higher order (third, fourth, etc.) among others. Homogeneous Solution or Transient Solution and Transient Analysis The homogeneous solution of a linear differential equation is also called transient solution. The homogeneous differential equation is achieved when the right side of the equation is set zero (see eq.(6), or in other words, when no force acts on the system. All analyzes and conclusions obtained from the homogeneous solution are called transient analyzes, and provide information about the dynamical behavior of the system while perturbations of displacement and velocities (in case of second order differential equations of motion) are introduced into the system. ÿ(t) + ξω n ẏ(t) + ω ny(t) = (6) y h (t) = Ce λt, (assumption) (7) ẏ h (t) = λce λt ÿ h (t) = λ Ce λt The assumption (7) and its derivatives are introduced into the differential equation (6), leading to λ Ce λt + ξω n λce λt + ω nce λt = (λ + ξω n λ + ω n)ce λt = (8)

Demanding (λ + ξω n λ + ω n) =, because Ce λt in eq.(8), one gets two values of λ, or two roots for the equation (λ + ξω n λ + ω n) = : λ = ξω n ω n ξ i λ = ξω n + ω n ξ i (9) It is important to highlight that all analyzes carried out here will be concentrated in cases where ξ < (sub-critically damped system). Cases where ξ = or ξ > are called critically or super-critically damped systems respectively. In these cases no problem related to amplification of vibration amplitudes can be found while crossing resonances. Using the roots λ and λ the homogenous solution is: y h (t) = C e λ t + C e λ t where C and C are defined as a function of the initial condition of the movement when t = : initial displacement y() = y ini initial movement ẏ() = v ini It is important to point out that these constants have to be calculated by using the general solution, which will be presented later. Permanent Solution and Steady-State Analysis The permanent solution of a linear differential equation is also called steady-state solution. The complete differential equation is achieved when the right side of the equation is completed with the excitation (see eq.(3)), or in other words, when static or dynamic forces act on the system. All analyzes and conclusions obtained from the permanent solution are called steady-state analyzes, and provide information about the dynamical behavior of the system while excitation forces are introduced into the system. Let us introduce an excitation force which value oscillates in time with frequency ω [rad/s]: ÿ(t) + ξω n ẏ(t) + ω ny(t) = f m eiωt (3) y h (t) = Ae iωt, (assumption) (3) ẏ h (t) = jωae iωt ÿ h (t) = (jω) Ae iωt = (ω) Ae iωt The assumption (3) and its derivatives are introduced into the differential equation (3), leading to: (ω) Ae iωt + ξω n (jωae jωt ) + ω nae iωt = f m eiωt ( ω + ω n + iξω n ω)a = f m (3)

The amplitude A eq.(33) and the particular solution y p (t) eq.(34) are derived by eq.(3) and (3): A = f/m ω + ω n + iξω n ω (33) [ y p (t) = A e iωt = f/m ω + ω n + iξω n ω ] e iωt (34) General Solution = Transient Solution + Steady-State Solution The general solution of a linear differential equation is achieved by adding the homogenous and the permanent solutions, and sequentially by defining the initial conditions of the movement. This solution will provide information about the transient and steady-state response of the mechanical model. Considering that the order of the mechanical model is correct (in this case, one degree-of-freedom system), the solution of the linear differential equation will be useful for predicting the dynamical behavior of the physical system, if the coefficients of the differential equation (m, d and k, or ω n and ξ) are properly chosen, using either the theoretical or experimental information or a combination of both. The general solution of the differential equation of motion is given by: y(t) = C e λ t + C e λ t + Ae iωt (35) where y(t) is the displacement of the mass-damping-spring system. For achieving the velocity of the mass-damping-spring system, eq.(35) has to be differentiated in time: ẏ(t) = λ C e λ t + λ C e λ t + iωae iωt (36) Introducing the initial conditions of displacement and velocity into eq.(35) and (36), when t =, one gets: y() =C e λ + C e λ + Ae iω y ini = C + C + A (37) ẏ() =λ C e λ + λ C e λ + iωae iω v ini = λ C + λ C + iωa (38) Rewriting eq.(37) and eq.(38) in matrix form, one gets: [ λ λ ] { C C } = { yini A v ini jωa } (39) Solving the linear system eq.(39) using Cramer s rule, the constants C and C are calculated: C = [ ] yini A det v ini A λ [ ] = λ (y ini A) (v ini iωa) λ λ det λ λ

det C = det [ ] yini A λ v ini A [ λ λ ] = λ (y ini A) + (v ini iωa) λ λ Summarizing, below is the analytical solution of second order differential equation, which is responsible for describing the movements of the mass-damping-spring system in time domain, as a function of the excitation force and initial condition of displacement and velocity: y(t) = C e λ t + C e λ t + Ae iωt (4) where λ = ξω n ω n ξ i λ = ξω n + ω n ξ i A = f/m ω + ω n + iξω n ω C = λ (y ini A) (v ini iωa) λ λ C = λ (y ini A) + (v ini iωa) λ λ Numerical Solution According to Taylor s expansion, an equation can be approximated by: f(t) f(t ) + df (t t dt ) + d f dt (t t )... + dn f dt n (t t ) (4) t=t t=t t=t Assuming a very small time step t t, the higher order terms of eq.(4) can be neglected. It turns: f(t) f(t ) + df (t t dt ) (4) t=t Knowing the initial conditions of the movement when t =, y() = y = y ini ẏ() = ẏ = v ini and the equation of motion, which has to be solved, ÿ(t) = ξω n ẏ(t) ω ny(t) + f m eiωt (43) 3

one can get the initial acceleration, when t =, on the basis of initial conditions: t = ẏ y ÿ = ξω n ẏ ω ny + f m eiωt The first predicted values of displacement, velocity and acceleration in time t = t, using the approximation given by eq.(4), are: t = t ẏ = ẏ + ÿ t y = y + ẏ t ÿ = ξω n ẏ ω ny + f m eiωt The second predicted values of displacement, velocity and acceleration in time t = t + t, using the approximation given by eq.(4), are: t = t ẏ = ẏ + ÿ t y = y + ẏ t ÿ = ξω n ẏ ω ny + f m eiωt The N-th predicted values of displacement, velocity and acceleration in time t N = t N + t, using the approximation given by eq.(4), are: t N = N t ẏ N = ẏ N + ÿ N t y N = y N + ẏ N t ÿ N = ξω n ẏ N ω ny N + f m eiωt N (44) Plotting the points [y, y, y 3,..., y N ] versus [t, t, t 3,..., t N ], one can observe the numerical solution of the differential equation, which describes the displacement of the mass-dampingspring system in time domain. Plotting the points [ẏ, ẏ, ẏ 3,...,ẏ N ] versus [t, t, t 3,..., t N ] or [ÿ, ÿ, ÿ 3,...,ÿ N ] versus [t, t, t 3,..., t N ] one can also observe velocity and acceleration of the mass-damping-spring system in time domain. The analytical and numerical solutions eq.(43) of the second order differential equation are illustrated using a Matlab code. 4

.6.4 Analytical and Numerical Solution of Equation of Motion using Matlab %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % DYNAMICS OF MACHINERY LECTURES (435) % % MEK - DEPARTMENT OF MECHANICAL ENGINEERING % % DTU - TECHNICAL UNIVERSITY OF DENMARK % % % % Copenhagen, October 3th, 3 % % % % IFS % % % % D.O.F. SYSTEM - EXACT AND NUMERICAL SOLUTION % %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% clear all; close all; %Concentred Masses m=.9; %[Kg] m=.9; %[Kg] m3=.9; %[Kg] m4=.9; %[Kg] m5=.9; %[Kg] m6=.9; %[Kg] %Elastic Properties of the Beam of 6 [mm] E = e; %elasticity modulus [N/m^] b =.3 ; %width [m] h =. ; %thickness [m] Iz= (b*h^3)/; %area moment of inertia [m^4] %Mass-Spring-Damping System Properties L=.6; %beam length K= 3*E*Iz/L^3; %stiffness coefficient M=m+m; %mass coefficient xi=.5; %damping factor [no-dimension] D=*xi*sqrt(K*M); %damping coefficient wn=sqrt(k/m); %natural frequency [rad/s] fn=wn//pi %natural frequency [Hz] fnexp=.875; %measured natural frequency [Hz] dif=(fn-fnexp)/fnexp;%error between calculated %and measured frequencies % %Initial Condition y_ini= -. % beam initial deflection [m] v_ini= -. % beam initial velocity [m/s] freq_exc=.95 % excitation frequency [Hz] force=-. % excitation force [N] time_max=3.; % integration % % %EXACT SOLUTION % EQUATION (4) n=6; % number of points for plotting j=sqrt(-); % complex number w=*pi*freq_exc; % excitation frequency [rad/s] lambda=-xi*wn+j*wn*sqrt(-xi*xi); lambda=-xi*wn-j*wn*sqrt(-xi*xi); AA=(force/M)/(wn*wn-w*w + j**xi*wn*w); C=( lambda*(y_ini-aa)-(v_ini-j*w*aa))/(lambda-lambda); C=(-lambda*(y_ini-AA)+(v_ini-j*w*AA))/(lambda-lambda); for i=:n, t(i)=(i-)/n*time_max; y_exact(i)=c*exp(lambda*t(i)) +... C*exp(lambda*t(i)) +... AA*exp(j*w*t(i)); end % %NUMERICAL SOLUTION % EQUATION (44) % trying different time steps to observe convergence % deltat=.365; % time step [s] % deltat=.3; % time step [s] % deltat=.; % time step [s] % deltat=.5; % time step [s] deltat=.; % time step [s] n_integ=time_max/deltat; % Initial Conditions y_approx() = y_ini; yp_approx() = v_ini; % number of points (integration) % beam initial deflection [m] % beam initial velocity [m/s] for i=:n_integ, t_integ(i)=(i-)*deltat; ypp_approx(i) =-(wn*wn)*y_approx(i)... -(*xi*wn)*yp_approx(i)... +(force/m)*exp(j*w*t_integ(i)); yp_approx(i+)= yp_approx(i) + ypp_approx(i)*deltat; y_approx(i+) = y_approx(i) + yp_approx(i+)*deltat; end % %Graphical Results title( Simulation of D.O.F System in Time Domain ) subplot(3,,), plot(t,real(y_exact), b ) title( (a) Exact Solution - (b) Numerical Solution (delta T =. s) - (c) Comparison ) xlabel( ) ylabel( (a) y(t) [m] ) grid subplot(3,,), plot(t_integ(:n_integ),real(y_approx(:n_integ)), r ) xlabel( ) ylabel( (b) y(t) [m] ) grid subplot(3,,3), plot(t,real(y_exact), b,t_integ(:n_integ), real(y_approx(:n_integ)), r ) xlabel( ) ylabel( (c) y(t) [m] ) grid 5

.6.5 Comparison between the Analytical and Numerical Solutions of Equation of Motion. (a) Exact Solution (b) Numerical Solution (delta T =.365 s) (c) Comparison (a) y(t) [m].5.5. 3 4 5 6 7 8 9.4 (b) y(t) [m] (c) y(t) [m]...4 3 4 5 6 7 8 9.4...4 3 4 5 6 7 8 9 Figure 5: Comparison between the analytical and numerical solutions when a time step of.365 [s] is used Divergence and numerical instability.. (a) Exact Solution (b) Numerical Solution (delta T =.3 s) (c) Comparison (a) y(t) [m] (b) y(t) [m] (c) y(t) [m].5.5. 3 4 5 6 7 8 9.... 3 4 5 6 7 8 9.... 3 4 5 6 7 8 9 Figure 6: Comparison between the analytical and numerical solutions when a time step of.3 [s] is used Divergence between the numerical and analytical solution due to a roof time step. 6

. (a) Exact Solution (b) Numerical Solution (delta T =. s) (c) Comparison (a) y(t) [m] (b) y(t) [m] (c) y(t) [m].5.5. 3 4 5 6 7 8 9..5.5. 3 4 5 6 7 8 9..5.5. 3 4 5 6 7 8 9 Figure 7: Comparison between the analytical and numerical solutions when a time step of. [s] is used Accumulation of errors with the number of iterations and divergence between the numerical and analytical solutions.. (a) Exact Solution (b) Numerical Solution (delta T =.5 s) (c) Comparison (a) y(t) [m] (b) y(t) [m] (c) y(t) [m].5.5. 3 4 5 6 7 8 9..5.5. 3 4 5 6 7 8 9..5.5. 3 4 5 6 7 8 9 Figure 8: Comparison between the analytical and numerical solutions when a time step of.5 [s] is used Accumulation of errors with the number of iterations and divergence between the numerical and analytical solutions. 7

. (a) Exact Solution (b) Numerical Solution (delta T =. s) (c) Comparison (a) y(t) [m] (b) y(t) [m] (c) y(t) [m].5.5. 3 4 5 6 7 8 9..5.5. 3 4 5 6 7 8 9..5.5. 3 4 5 6 7 8 9 Figure 9: Comparison between the analytical and numerical solutions when a time step of. [s] is used Good agreement between the numerical and analytical solutions in the total range of time. (a) Exact Solution (b) Numerical Solution (delta T =. s) (c) Comparison (a) y(t) [m] (b) y(t) [m] (c) y(t) [m].5.5 5 5 5 3.5.5 5 5 5 3.5.5 5 5 5 3 Figure : Comparison between the analytical and numerical solutions when a time step of. [s] is used Good agreement between the numerical and analytical solutions in the total range of time. System behavior when excited by a harmonic force with a frequency near the natural frequency. 8

.6.6 Homogeneous Solution or Free-Vibrations or Transient Response - Experimental Analysis ÿ (t) + ξω n ẏ (t) + ω ny (t) = (45).6.7 Natural Frequency ω n [rad/s] or f n [Hz] f n = k = π m π 3EI L 3 mi [Hz] number Length f n f e xp f e xp of masses [m] [Hz] [Hz] [Hz] (theor.) (*) (**).6.3 / =..5.6.87 8.5/ =.85.875 3.6.7 7/ =.7.75 4.6.6 6/ =.6.65 Table 3: Measuring the natural frequency of the mass-spring system A with. d.o.f, (*) using the human eyes and a watch, and (**) using an accelerometer attached to the mass, and making a comparison to the theoretical mathematical model. number Length f e xp of masses [m] [Hz] (*).6.875.3.875 Table 4: Measuring the natural frequency of the mass-spring system A with. d.o.f, (*) using the human eyes and a watch, when the equivalent stiffness of the system is changed, by modifying the position (length) where the lumped mass is attached to the beam..6.8 Damping Factor ξ or Logarithmic Decrement β Experimental identification without using sensors (OBS: Experiment carried out using a watch, light and shadow, a calculator, and the mass-spring system oscillating from an initial condition of displacement y o = y ini until half the initial amplitude y N = y ini /.) ξ = ( ) πn ln yo y N + [ πn ln ( yo y N )] or β = π ξ 9

Using the information of figure (signal in time domain), where y o = 3. 5 [m/s ] (first peak of signal in time domain, figure ) y N = y 6 =. 5 [m/s ] (after 6 peaks) N = 6, one gets Damping Factor of the system A (ξ) ξ = ( ) π 6 ln 3. 5. 5 [ ( )] = + π 6 ln 3. 5. 5.755.58. Log Dec β = π ξ = π. =.68 Equivalent Viscous Damping (d) d = ξ ω n m =. (.87 π).38.6 [N s/m] 4 x 5 Signal (a) in Time Domain (b) in Frequency Domain (a) Amplitude [m/s ] 4 5 5 5 3 x 5 (b) Amplitude [m/s ].8.6.4. 5 5 5 frequency [Hz] Figure : Transient Vibration Acceleration of the clamped-free flexible beam when two concentrated masses m = m +m =.38 Kg are attached at its free end (L =.6 m) Natural frequency of the mass-spring system A :.87 Hz.

5 x 5 Signal (a) in Time Domain (b) in Frequency Domain (a) Amplitude [m/s ] 5 5 5 5 3 x 5 (b) Amplitude [m/s ] 5 5 5 frequency [Hz] Figure : Transient Vibration Acceleration of the clamped-free flexible beam when two concentrated masses m = m +m =.38 Kg are attached at its free end (L =.3 m) Natural frequency of the mass-spring system B :.75 Hz. Damping Factor of the system B From fig., one gets: y o = 4.6 5 [m/s ], y N = y 54 =. 5 [m/s ] and N = 54: ξ = ( ) π 54 ln 4.6 5. 5 [ ( )] = + π 54 ln 4.6 5. 5.4498.5 (46). Equivalent Viscous Damping (d) d = ξ ω n m =.5 (.75 π).38.4 [N s/m]

x 4 Signal (a) in Time Domain (b) in Frequency Domain (a) Amplitude [m/s ].5.5 5 5 5 3.5 x 5 (b) Amplitude [m/s ].5.5 5 5 5 frequency [Hz] Figure 3: Free vibration Spring-mass systems with D.O.F. Two masses m = m + m =.38 Kg fixed at the middle of the beam L =.55 m, resulting in a system B natural frequency of 3.8 Hz Damping Factor of the system B From fig.3, one gets: y o =.95 4 [m/s ], y N = y 34 =.5 4 [m/s ] and N = 34: ξ = ( ) π 34 ln.95 4.5 4 [ ( )] = + π 34 ln.95 4.5 4.39.5.3 Equivalent Viscous Damping (d) d = ξ ω n m =.3 (3.8 π).38.5 [N s/m]

x 4 Signal (a) in Time Domain (b) in Frequency Domain (a) Amplitude [m/s ].5.5 5 5 5 3 x 5 (b) Amplitude [m/s ].8.6.4. 5 5 5 frequency [Hz] Figure 4: Free vibration Spring-mass systems with D.O.F. One mass m = m =.9 Kg fixed at the middle of the beam L =.55 m, resulting in a system B natural frequency of 4.94 Hz Damping Factor of the system B From fig.3, one gets: y o =.95 4 [m/s ], y N = y 4 =.5 4 [m/s ] and N = 4: ξ = ( ) π 4 ln.95 4.5 4 [ ( )] = + π 4 ln.95 4.5 4.796.6.7 Equivalent Viscous Damping (d) d = ξ ω n m =.7 (4.94 π).9.8 [N s/m] IMPORTANT CONCLUSION: THE DAMPING FACTOR IS A CHARACTER- ISTIC OF THE GLOBAL MECHANICAL SYSTEM AND SIMULTANEOUSLY DE- PENDS ON MASS m, STIFFNESS k AND DAMPING d COEFFICIENTS, NOT ONLY ON THE DAMPING COEFFICIENT, AS YOU CAN SEE IN THE DEFINI- TION: ξ = d m ω n = d m k IT IS POSSIBLE TO INCREASE THE DAMPING FACTOR OF A MECHANICAL SYSTEM EITHER BY DECREASING THE MASS m, OR BY DECREASING THE STIFFNESS k OR BY INCREASING THE DAMPING COEFFICIENT d. THE DAMPING FACTOR IS A VERY USEFUL PARAMETER FOR DEFINING THE RESERVE OF STABILITY IN MACHINERY DYNAMICS. 3

.6.9 Forced Vibrations or Steady-State Response The two most frequent ways of representing the frequency response function of mechanical systems are presented in figure 5: (a) and (b) real and imaginary parts as a function of the excitation frequency; (c) and (d) magnitude and phase as a function of the excitation frequency. Other alternative ways are presented in figures 6 and 7. (a) Frequency Response Function (Real Part) 5 ξ=.5 ξ=.5 Real(A(ω)) [m/n] 5.5.5 Frequency [Hz] (b) Frequency Response Function (Imaginary Part) ξ=.5 ξ=.5 Imag(A(ω)) [m/n] 4 6 8 Phase(A(ω)) [ o] (c) Frequency Response Function (Amplitude) ξ=.5 8 ξ=.5 A(ω) [m/n] 6 4 5.5.5 Frequency [Hz] (d) Frequency Response Function (Phase) 5 ξ=.5 ξ=.5.5.5 Frequency [Hz].5.5 Frequency [Hz] Figure 5: Steady state response in the frequency domain or Frequency Response Function f/m (FRF): (a) and (b) illustrate the real and imaginary part of A = ω +ωn +iξωnω; (c) and (d) illustrate the magnitude and phase of the complex function A = f/m ω +ω n+iξω nω..6. Resonance and Phase In order to understand the 9 degree phase while in resonance use the tactile senses Remember the experiments in the classroom using the mass-beam system and forces applied by your finger, and outside building 44, using a car and a tree and the forces applied by your hands (synchronization). Complex Vector Diagram of Resonance and Phase 4

Frequency Response Function (Real Part) ξ=.5 ξ=.5 Imag(A(ω)) [m/n] 3 4 5 6 7 8 9 5 4 3 3 4 5 Real(A(ω)) [m/n] Figure 6: Steady state response in the frequency domain or Frequency Response Function (FRF) illustrated as the real versus the imaginary part of A = f/m ω +ω n +iξωnω. Frequency Response Function ξ=.5 Imag(A(ω)) [m/n].5.5.5.5 Real(A(ω)) [m/n].5.5 Frequency [Hz] Figure 7: Steady state response in the frequency domain or Frequency Response Function (FRF) f/m illustrated in a 3D-plot: the real and imaginary parts of A = ω +ωn+iξω nω as a function of the frequency. 5

(a) (b) (c) Figure 8: Complex vector diagram using an excitation force of constant magnitude: (a) the frequency of the excitation force is lower than the natural frequency; (b) the frequency of the excitation force is coincident with the natural frequency, characterizing a resonance case where the phase between the force and the displacement is 9 degrees, i.e. the phase between the force and the velocity is degrees, meaning a synchronization between force and velocity; (c) the frequency of the excitation force is bigger than the natural frequency. 6

.6. Superposition of Transient and Forced Vibrations in Time Domain (Simulation).5 (a) Excitation Frequency :. Hz.5 (a) Excitation Frequency :.7 Hz.5.5.5.5 y(t) [m] y(t) [m].5.5.5.5 (a).5 5 5 5 3 (b).5 5 5 5 3.5 (a) Excitation Frequency :.8 Hz.5 (a) Excitation Frequency :.87 Hz.5.5.5.5 y(t) [m] y(t) [m].5.5.5.5 (c).5 5 5 5 3 (d).5 5 5 5 3.5 (a) Excitation Frequency :.9 Hz.5 (a) Excitation Frequency :. Hz.5.5.5.5 y(t) [m] y(t) [m].5.5.5.5 (e).5 5 5 5 3 (f).5 5 5 5 3 Figure 9: Time response of the system with D.O.F. excited by forces with different frequencies (a) ω =. Hz (before resonance); (b) ω =.7 Hz; (before resonance); (c) ω =.8 Hz (before resonance but close beating); (d) ω =.87 Hz (at resonance); (e) ω =.9 Hz (after resonance but close beating); (f) ω =. Hz (after resonance). 7

.6. Resonance Experimental Analysis in Time Domain 3 x 5 Signal (a) in Time Domain (b) in Frequency Domain (a) Amplitude [m/s ] 3 5 5 5 3.5 x 5 (b) Amplitude [m/s ].5 5 5 5 frequency [Hz] x 5 Signal (a) in Time Domain (b) in Frequency Domain x 5 Signal (a) in Time Domain (b) in Frequency Domain 3 3 (a) Amplitude [m/s ] (a) Amplitude [m/s ] 3 3 5 5 5 3 5 5 5 3.5 x 5.5 x 5 (b) Amplitude [m/s ].5 (b) Amplitude [m/s ].5 5 5 5 frequency [Hz] 5 5 5 frequency [Hz] Figure : Resonance phenomena due to force excitations with frequency around the natural frequency of the mass-spring system: D.O.F. system with natural frequency of.87 Hz, excited by the shaker with frequencies of.8 Hz,.87 Hz and.9 Hz Spring-mass system (A) with two masses m = m + m =.38 Kg fixed at the beam (A) length L =.6 m, resulting in a natural frequency of.87 Hz. 8

5 x 5 Signal (a) in Time Domain (b) in Frequency Domain (a) Amplitude [m/s ] 5 5 5 5 3.5 x 5 (b) Amplitude [m/s ].5.5 5 5 5 frequency [Hz] Figure : Beating phenomena with two transient responses Two spring-mass systems with D.O.F. each, vibrating with very similar natural frequencies (transient responses), resulting in beating phenomena: Spring-mass system A with three masses m = m +m +m 3 =.576 Kg fixed at the beam length L =.85 m, resulting in a natural frequency near.75 Hz; Springmass system B with masses m = m 4 + m 5 =.38 Kg fixed at the end of the beam L =.3 m, resulting in a natural frequency of.75 Hz 9

(a) Amplitude [m/s ] 6 4 4 6 x 6 Signal (a) in Time Domain (b) in Frequency Domain 8 5 5 5 3.5 x 6 (a) Amplitude [m/s ].5 x 5 Signal (a) in Time Domain (b) in Frequency Domain.5.5.5 5 5 5 3 8 x 6 (b) Amplitude [m/s ].5.5 (b) Amplitude [m/s ] 6 4 5 5 5 frequency [Hz] 5 5 5 frequency [Hz] Figure : Beating phenomena due to transient (low damping factor) and forced vibrations with similar frequencies: D.O.F. system A with natural frequency of.87 Hz, excited by a shaker with frequencies of.8 Hz and.9 Hz - Spring-mass system A with two masses m = m +m =.38 Kg fixed at the beam length L =.6 m, resulting in a natural frequency of.87 Hz. 3

.7 Mechanical Systems with D.O.F..7. Physical System and Mechanical Model (a) (b) (c) Figure 3: (a) Real mechanical system built by two turbines attached to an airplane flexible wing; (b) Laboratory prototype built by two lumped masses attached to a flexible beam); (c) Equivalent mechanical model with D.O.F. for the two lumped masses attached to the flexible beam..7. Mathematical Model Mÿ(t) + Dẏ(t) + Ky(t) = f(t) (47) [ m m m m ]{ ÿ ÿ } [ d d + d d ] { ẏ ẏ } [ k k + k k ] { y y } = { f f } (48) 3

where m = m + m m = m = m = m 3 + m 4 (49) d = ξ k m d = d = d = ξ k m (Approximation of damping coefficients using the exp. damping factor!) (5) EI k = (4L L )(L L ) (L /L ) 3 6EI k = (4L L )(L L ) (L 3L )/L 6EI k = (4L L )(L L ) (L 3L )/L EI k = (4L L )(L L ) (5) The coefficients of the mass matrix can easily be achieved. Each of the single masses has m = m = m 3 = m 4 = m 5 = m 6 =.9 Kg. The stiffness coefficients k ij were calculated in the section.5, using the geometry of the beam (I, L, L ) and its material properties (E). The damping matrix can be approximated by using proportional damping, for example, D = αm + βk. The coefficients α and β can be chosen, so that the damping factor ξ of the first resonance is of the same order as the one in the previous section. Please, note that this is just an approximation which be verified using Modal Analysis Techniques. Another way of approximating the damping coefficients is given by eq.(5)..7.3 Analytical and Numerical Solution of System of Differential Linear Equations System of Equation of motion [ ]{ } [ m m ÿ d d + m m ÿ d d ]{ ẏ ẏ } [ k k + k k ]{ y y } { f = f } e jωt Mÿ + Dẏ + Ky = fe jωt (5) Differential Equations nd order st order [ ] { } [ ] { } { M D ÿ K ẏ f + = M ẏ M y } e jωt Aż(t) + Bz(t) = fe jωt (53) 3

{ ẏ(t) z(t) = y(t) } = ẏ (t) ẏ (t) y (t) y (t) velocity velocity displacement displacement (54) The analytical solution can be divided into three steps: (I) homogeneous solution (transient analysis); (II) permanent solution (steady-state analysis) and (III) general solution (homogeneous + permanent). Homogeneous Solution and Transient Analysis The homogeneous differential equation is achieved when the right side of the equation is set zero (see eq.(55)), or in other words, when no force is acting on the system. Aż h (t) + Bz h (t) = (55) z h (t) = ue λt, (assumption) (56) ż h (t) = λue λt The assumption (56) and its derivative are introduced into the differential equation (55), leading to an eigenvalue-eigenvector problem: [λa + B]ue λt = [λa + B]u = (57) Eigenvalues λ i can be calculated by using eq.(58): determinant(λa + B) = λ, λ, λ 3, λ 4 (58) Eigenvectors u i can be calculated by using eq.(59): λ Au = Bu u λ Au = Bu u λ 3 Au = Bu u 3 (59) λ 4 Au = Bu u 4 The homogeneous solution can be written as: z h (t) = C u e λ t + C u e λ t + C 3 u 3 e λ 3t + C 4 u 4 e λ 4t where C, C, C 3 and C 4 are constants depending on the initial displacement and velocities of the coordinates y and y, when t =. 33

Permanent Solution and Steady-State Analysis The permanent solution takes into account the right side of the differential equation (see eq.(6)), or in other words, the effect of the force on the system. In case of harmonic excitation, one can write: Aż p (t) + Bz p (t) = fe jωt (6) z p (t) = Ae jωt, (assumption!) (6) ż p (t) = jωae jωt The assumption adopted in eq.(6) and its derivative are introduced into the differential equation (6), leading to: [jωa + B]Ae jωt = fe jωt A = [jωa + B] f (6) The permanent solution of the equation of motion is given by: z p (t) = Ae jωt z p (t) = [jωa + B] fe jωt (63) General Solution The general solution of a linear differential equation is achieved by adding the homogenous and the permanent solutions, and by sequentially defining the initial conditions of the movement. This solution will provide information about the transient and steady-state response of the mechanical model. Considering that the order of the mechanical model is correct (in this case, the two degree-of-freedom system), the solution of the linear differential equation will be useful for predicting the dynamical behavior of the physical system, if the coefficients of the differential equation (M, D and K, or A and B) are properly chosen, either by using theoretical or experimental information or both. The general solution of the differential equation of motion is given by: z(t) = C u e λ t + C u e λ t + C 3 u 3 e λ 3t + C 4 u 4 e λ 4t + Ae iωt (64) where z(t) gives information about the displacement and velocity of the coordinates y and y. Introducing the initial conditions of displacement and velocity z ini = { v ini v ini y ini y ini } T (65) into eq.(64), when t =, one obtains z() = z ini = C u e λ + C u e λ + C 3 u 3 e λ 3 + C 4 u 4 e λ 4 + Ae iω (66) or z ini = C u + C u + C 3 u 3 + C 4 u 4 + A = [ u u u 3 u 4 ] C C C 3 C 4 + A (67) Solving the linear system by inverting the modal matrix U = [ u u u 3 u 4 ] one achieves the vector c = { C C C 3 C 4 } T : 34

z ini = U c + A c = U {(z ini A)} (68) Summarizing, below is the analytical solution of a second order differential equation, which is responsible for describing the displacements and velocities of the coordinates y (t) and y (t) in time domain, as a function of the excitation force and the initial condition of displacement and velocity: z(t) = C u e λ t + C u e λ t + C 3 u 3 e λ 3t + C 4 u 4 e λ 4t + Ae iωt (69) where λ = ξ ω n ω n ξ i and u λ = ξ ω n + ω n ξ i and u λ 3 = ξ ω n ω n ξ i and u 3 λ 4 = ξ ω n + ω n ξ i and u 4 C C C 3 C 4 A = [jωa + B] f = [ u u u 3 u 4 ] { z ini A} Numerical Solution The numerical solution of the system of differential equations can be found by using the approximation according to Taylor s expansion. Thus, one equation can be approximated by: f(t) f(t ) + df (t t dt ) + d f dt (t t )... + dn f dt n (t t ) (7) t=t t=t t=t Assuming a very small time step t t, the higher order terms of eq.(7) can be neglected. It turns: f(t) f(t ) + df (t t dt ) (7) t=t Knowing the initial conditions of the movement, when t = t =, y() = y { y () y () } = { yini y ini } (7) 35

and ẏ() = ẏ { ẏ () ẏ () } = { vini v ini } (73) and the equation of motion eq.(5), which has to be solved, one can calculate the acceleration, when t = t = : ÿ = M { Dẏ + Ky fe jωt } The first predicted values of displacement, velocity and acceleration in time t = t, using the approximation given by eq.(7), are: t = t ẏ = ẏ + ÿ t y = y + ẏ t ÿ = M { Dẏ + Ky fe jωt } The second predicted values of displacement, velocity and acceleration in time t = t + t, using the approximation given by eq.(7), are: t = t ẏ = ẏ + ÿ t y = y + ẏ t ÿ = M { Dẏ + Ky fe jωt } The N-th predicted values of displacement, velocity and acceleration in time t N = t N + t, using the approximation given by eq.(74), are: t N = N t ẏ N = ẏ N + ÿ N t y N = y N + ẏ N t ÿ N = M { Dẏ N + Ky N fe jωt N } (74) Plotting the points [y,y,y 3,...,y N ] versus [t, t, t 3,..., t N ], one can observe the numerical solution of the differential equation, which describes the displacements of the mass-dampingspring system in time domain. Plotting the points [ẏ,ẏ,ẏ 3,...,ẏ N ] versus [t, t, t 3,..., t N ] or [ÿ,ÿ,ÿ 3,...,ÿ N ] versus [t, t, t 3,..., t N ] one can also observe the velocity and acceleration of the mass-damping-spring system in time domain. The analytical and numerical solutions of the second order differential equation, eq.(5), are illustrated using a Matlab code. 36

.7.4 Modal Analysis using Matlab eig-function [u, w] = eig( B, A) %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % MACHINERY DYNAMICS LECTURES (464) % % MEK - DEPARTMENT OF MECHANICAL ENGINEERING % % DTU - TECHNICAL UNIVERSITY OF DENMARK % % % % Copenhagen, February th, % % IFS % % % % MODAL ANALYSIS % % % % D.O.F. SYSTEMS - MODAL ANALYSIS - 3 EXPERIMENTAL CASES % %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% clear all; close all; %Concentred Masses m=.9; %[Kg] m=.9; %[Kg] m3=.9; %[Kg] m4=.9; %[Kg] m5=.9; %[Kg] m6=.9; %[Kg] %Elastic Properties of the Beam of 6 mm E= e; %elasticity modulus [N/m^] b=.3 ; %width [m] h=. ; %thickness [m] I= (b*h^3)/; %area moment of inertia [m^4] % (.CASE) Data for the mass-spring system % M=m; %concentrated mass [Kg] M=m; %concentrated mass [Kg] L=.3; %length for positioning M [m] L=.6; %length for positioning M [m] % % Coefficients of the Stiffness Matrix LL=(L-4*L)*(L-L)^; K= -*(E*I/LL)*L^3/L^3; %equivalent Stiffness [N/m] K= -6*(E*I/LL)*(L-3*L)/L; %equivalent Stiffness [N/m] K= -6*(E*I/LL)*(L-3*L)/L; %equivalent Stiffness [N/m] K= -*(E*I/LL); %equivalent Stiffness [N/m] %Mass Matrix M= [M ; M]; %Stiffness Matrix K= [K K; K K]; %Damping Matrix D= [ ; ]; %State Matrices A= [ M D ; zeros(size(m)) M ] ; B= [ zeros(size(m)) K ; -M zeros(size(m))]; %Dynamical Properties of the Mass-Spring System [u,w]=eig(-b,a); % Eigenvectors u % Eigenvalues w [rad/s] w u pause; w_vector=diag(w); [natfreq,indice]=sort(w_vector); % Sorting the natural Frequencies % Natural Frequencies (Hz) w=abs(imag(w_vector(indice())))//pi % First natural frequency w=abs(imag(w_vector(indice(3))))//pi % Second natural frequency mass_position() = ; mass_position() = L; mass_position(3) = L; % first mode shape uu()=; uu()=u(,indice()); uu(3)=u(,indice()); figure() plot(uu,mass_position, r*-.,-uu,mass_position, r*-., LineWidth,.5) grid f=numstr(w); set(gca, FontSize,, FontAngle, oblique ); title([ First Mode Shape - Freq.:,f, Hz ]) % second mode shape uu()=; uu()=u(,indice(3)); uu(3)=u(,indice(3)); figure() plot(uu,mass_position, r*-.,-uu,mass_position, r*-., LineWidth,.5) grid f=numstr(w); set(gca, FontSize,, FontAngle, oblique ); title([ Second Mode Shape - Freq.:,f, Hz ]) pause; % (.CASE) Increasing the Mass Values % M=m+m; %concentrated mass [Kg] M=m3+m4; %concentrated mass [Kg] L=.3; %length for positioning M [m] L=.6; %length for positioning M [m] % % Coefficients of the Stiffness Matrix LL=(L-4*L)*(L-L)^; K= -*(E*I/LL)*L^3/L^3; %equivalent Stiffness [N/m] K= -6*(E*I/LL)*(L-3*L)/L; %equivalent Stiffness [N/m] K= -6*(E*I/LL)*(L-3*L)/L; %equivalent Stiffness [N/m] K= -*(E*I/LL); %equivalent Stiffness [N/m] %Mass Matrix M= [M ; M]; %Stiffness Matrix K= [K K; K K]; %Damping Matrix D= [ ; ]; %State Matrices A= [ M D ; zeros(size(m)) M ] ; B= [ zeros(size(m)) K ; -M zeros(size(m))]; %Dynamical Properties of the Mass-Spring System [u,w]=eig(-b,a); % Eigenvectors u % Eigenvalues w [rad/s] w u pause; w_vector=diag(w); [natfreq,indice]=sort(w_vector); % Sorting the natural Frequencies % Natural Frequencies (Hz) w=abs(imag(w_vector(indice())))//pi % First natural frequency w=abs(imag(w_vector(indice(3))))//pi % Second natural frequency mass_position() = ; mass_position() = L; mass_position(3) = L; % first mode shape uu()=; uu()=u(,indice()); uu(3)=u(,indice()); figure(3) plot(uu,mass_position, b*-.,-uu,mass_position, b*-., LineWidth,.5) grid f=numstr(w); set(gca, FontSize,, FontAngle, oblique ); title([ First Mode Shape - Freq.:,f, Hz ]) % second mode shape uu()=; uu()=u(,indice(3)); uu(3)=u(,indice(3)); figure(4) plot(uu,mass_position, b*-.,-uu,mass_position, b*-., LineWidth,.5) grid f=numstr(w); set(gca, FontSize,, FontAngle, oblique ); title([ Second Mode Shape - Freq.:,f, Hz ]) pause; % (3.CASE) Increasing the Mass Values % M=m+m+m3; %concentrated mass [Kg] M=m4+m5+m6; %concentrated mass [Kg] L=.3; %length for positioning M [m] L=.6; %length for positioning M [m] %... 37

%Dynamical Properties of the Mass-Spring System [u,w]=eig(-b,a); % Eigenvectors u % Eigenvalues w [rad/s] (.CASE) w = +48.857i -48.857i + 7.357i - 7.357i u =...33.33 -.33 -.33.. -.355i +.355i -.778i +.778i +.7i -.7i -.356i +.356i (.CASE) w = +34.544i -34.544i + 5.985i - 5.985i u =.. -.33 -.33 -.33 -.33 -. -. -.89i +.89i +.635i -.635i +.96i -.96i +.94i -.94i (3.CASE) u =.. -.33 -.33 -.33 -.33 -. -. -.5i +.5i +.449i -.449i +.68i -.68i +.36i -.36i w = +8.5i -8.5i + 4.445i - 4.445i 38

.7 First Mode Shape Freq.:.7 Hz.7 Second Mode Shape Freq.: 7.775 Hz.6.6.5.5.4.4.3.3.... (a).8.6.4...4.6.8.8.6.4...4.6.8.7 First Mode Shape Freq.:.8736 Hz.7 Second Mode Shape Freq.: 5.4979 Hz.6.6.5.5.4.4.3.3.... (b).8.6.4...4.6.8.8.6.4...4.6.8.7 First Mode Shape Freq.:.67554 Hz.7 Second Mode Shape Freq.: 4.489 Hz.6.6.5.5.4.4.3.3.... (c).8.6.4...4.6.8.8.6.4...4.6.8 Figure 4: First and second mode shapes of the mechanical system modelled with D.O.F. (a) (.CASE) one mass attached to each of the two coordinates; (b) (.CASE) two masses attached to each one of the two coordinates; (c) (3.CASE) three masses attached to each one of the two coordinates. 39

.7.5 Analytical and Numerical Solutions of Equation of Motion using Matlab %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % DYNAMICS OF MACHINERY LECTURES (73) % % MEK - DEPARTMENT OF MECHANICAL ENGINEERING % % DTU - TECHNICAL UNIVERSITY OF DENMARK % % % % Copenhagen, February th, % % % % IFS % % % % D.O.F - EXACT AND NUMERICAL SOLUTION % %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% clear all; close all; %Concentred Masses m=.9; %[Kg] m=.9; %[Kg] m3=.9; %[Kg] m4=.9; %[Kg] m5=.9; %[Kg] m6=.9; %[Kg] %Elastic Properties of the Beam of 6 [mm] E = e; %elasticity modulus [N/m^] b =.3 ; %width [m] h =. ; %thickness [m] Iz= (b*h^3)/; %area moment of inertia [m^4] % (.CASE) Data for the mass-spring system % M=m+m; %concentrated mass [Kg] M=m3+m4; %concentrated mass [Kg] L=.3; %length for positioning M [m] L=.6; %length for positioning M [m] % % Coefficients of the Stiffness Matrix LL=(L-4*L)*(L-L)^; K= -*(E*Iz/LL)*L^3/L^3; % Stiffness coeff.[n/m] K= -6*(E*Iz/LL)*(L-3*L)/L; % Stiffness coeff.[n/m] K= -6*(E*Iz/LL)*(L-3*L)/L; % Stiffness coeff.[n/m] K= -*(E*Iz/LL); % Stiffness coeff.[n/m] % Coefficients of the Damping Matrix % (damping factor xi=.5) D= *.5*sqrt(M/K); % Damping coeff.[ns/m] D= ; % Damping coeff.[ns/m] D= ; % Damping coeff.[ns/m] D= *.5*sqrt(M/K); % Damping coeff.[ns/m] %Mass Matrix M= [M ; M]; %Damping Matrix D=[D D; D D]; %Stiffness Matrix K= [K K; K K]; %State Matrices A & B % EQUATION (5) A= [ M D ; zeros(size(m)) M ] ; B= [ zeros(size(m)) K ; -M zeros(size(m))]; %Dynamical Properties of the Mass-Spring System [u,w]=eig(-b,a); %eigenvectors u %eigenvalues w w=abs(imag(w(3,3)))//pi; %first natural freq.[hz] w=abs(imag(w(,)))//pi; %second natural freq.[hz] wexp=.785; %measured natural freq.[hz] wexp=5.563; %measured natural freq.[hz] dif=(w-wexp)/wexp; %error calculated and measured freq. dif=(w-wexp)/wexp; %error calculated and measured freq. % %Initial Condition y_ini = -. % beam initial deflection [m] y_ini = -. % beam initial deflection [m] v_ini = -. % beam initial velocity [m/s] v_ini = -. % beam initial velocity [m/s] freq_exc =. % excitation frequency [Hz] force = -. % excitation force [N] force = -. % excitation force [N] time_max = 3.; % integration % % %EXACT SOLUTION % EQUATION (68) n=; % number of points for plotting j=sqrt(-); % complex number w_exc=*pi*freq_exc; % excitation frequency [rad/s] z_ini = [v_ini v_ini y_ini y_ini] ; force_exc = [force force ] ; vec_aux = z_ini - inv((j*w_exc*a + B))*force_exc; lambda=w(,); lambda=w(,); lambda3=w(3,3); lambda4=w(4,4); u=u(:4,); u=u(:4,); u3=u(:4,3); u4=u(:4,4); C=inv(u)*(vec_aux); c=c(); c=c(); c3=c(3); c4=c(4); for i=:n, t(i)=(i-)/n*time_max; y_exact=c*u*exp(lambda*t(i)) +... c*u*exp(lambda*t(i)) +... c3*u3*exp(lambda3*t(i)) +... c4*u4*exp(lambda4*t(i)) +... inv((j*w_exc*a + B))*force_exc*exp(j*w_exc*t(i)); end y_exact(i) = y_exact(3); y_exact(i) = y_exact(4); figure() title( Simulation of D.O.F System in Time Domain ) subplot(,,), plot(t,real(y_exact), b ) title( Exact Solution ) xlabel( ) ylabel( y_{exact}(t) [m] ) grid subplot(,,), plot(t,real(y_exact), b ) xlabel( ) ylabel( y_{exact}(t) [m] ) grid pause; % %NUMERICAL SOLUTION % EQUATION (73) % deltat=.365; % time step [s] % deltat=.3; % time step [s] % deltat=.; % time step [s] % deltat=.5; % time step [s] % deltat=.; % time step [s] deltat=.5; % time step [s] n_integ=time_max/deltat; % Initial Conditions y_approx() = y_ini; y_approx() = y_ini; yp_approx() = v_ini; yp_approx() = v_ini; for i=:n_integ, t_integ(i)=(i-)*deltat; % number of points (integration) % beam initial deflection [m] % beam initial deflection [m] % beam initial velocity [m/s] % beam initial velocity [m/s] ypp_approx(i)=-/m*(k*y_approx(i)+k*y_approx(i)... + D*yp_approx(i)+D*yp_approx(i)... -(force)*exp(j*w_exc*t_integ(i))); ypp_approx(i)=-/m*(k*y_approx(i)+k*y_approx(i)... +D*yp_approx(i)+D*yp_approx(i)... -(force)*exp(j*w_exc*t_integ(i))); yp_approx(i+)=yp_approx(i) + ypp_approx(i)*deltat; yp_approx(i+)=yp_approx(i) + ypp_approx(i)*deltat; y_approx(i+)=y_approx(i)+yp_approx(i+)*deltat; y_approx(i+)=y_approx(i)+yp_approx(i+)*deltat; end % 4

% %Graphical Results figure() title( Simulation of D.O.F System in Time Domain ) subplot(,,), plot(t_integ(:n_integ), real(y_approx(:n_integ)), r ) title( Numerical Solution (delta T =.5 s) ) xlabel( ) ylabel( y_{approx}(t) [m] ) grid subplot(,,), plot(t_integ(:n_integ), real(y_approx(:n_integ)), r ) xlabel( ) ylabel( y_{approx}(t) [m] ) grid % %Graphical Results (Comparison Exact vs. Numerical) figure(3) subplot(,,), plot(t,real(y_exact), b, t_integ(:n_integ),real(y_approx(:n_integ)), r ) title( Simulation of D.O.F System in Time Domain - Exact Solution vs. Numerical Solution (delta T =.5 s) ) xlabel( ) ylabel( y_{approx}(t) [m] ) grid subplot(,,), plot(t,real(y_exact), b, t_integ(:n_integ),real(y_approx(:n_integ)), r ) xlabel( ) ylabel( y_{approx}(t) [m] ) grid.7.6 Analytical and Numerical Results of the System of Equations of Motion 4 x 5 Exact Solution y exact (t) [m] 4 6 5 5 5 3 y exact (t) [m] 4 4 6 6 x 5 8 5 5 5 3 5 x 5 Numerical Solution (delta T =.5 s) y approx (t) [m] 5 5 5 5 3 y approx (t) [m] 4 4 6 6 x 5 8 5 5 5 3 Figure 5: Analytical and Numerical Solutions (a) Analytical solution with initial velocity condition at ẏ () = mm/s, ẏ () = mm/s, y () = mm and y () = mm; (b) Numerical solution (time step of.5 [s]) with the same initial conditions Transient Analysis. 4

.7.7 Programming in Matlab Frequency Response Analysis %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % MACHINERY DYNAMICS LECTURES (73) % % IKS - DEPARTMENT OF CONTROL ENGINEERING DESIGN % % DTU - TECHNICAL UNIVERSITY OF DENMARK % % % % Copenhagen, February th, % % IFS % % % % D.O.F. SYSTEMS - FRF (FREQUENCY RESPONSE FUNCTION) % %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %Concentred Masses m=.9; %[Kg] m=.9; %[Kg] m3=.9; %[Kg] m4=.9; %[Kg] m5=.9; %[Kg] m6=.9; %[Kg] %Elastic Properties of the Beam of 6 [mm] E= e; %elasticity modulus [N/m^] b=.3 ; %width [m] h=. ; %thickness [m] I= (b*h^3)/; %area moment of inertia [m^4] % (.CASE) Data for the mass-spring system % M=m+m+m5; %concentrated mass [Kg] M=m3+m4+m6; %concentrated mass [Kg] L=.3; %length for positioning M [m] L=.6; %length for positioning M [m] % % Coefficients of the Stiffness Matrix LL=(L-4*L)*(L-L)^; K= -*(E*I/LL)*L^3/L^3; %equivalent Stiffness [N/m] K= -6*(E*I/LL)*(L-3*L)/L; %equivalent Stiffness [N/m] K= -6*(E*I/LL)*(L-3*L)/L; %equivalent Stiffness [N/m] K= -*(E*I/LL); %equivalent Stiffness [N/m] % Coefficients of the Damping Matrix D= *.*(*pi*5.)*m; %equivalent Damping [N/m] D= ; %equivalent Damping [N/m] D= ; %equivalent Damping [N/m] D= *.*(*pi*.)*m; %equivalent Damping [N/m] %Mass Matrix M= [M ; M]; %Damping Matrix D=[D D; D D]; %Stiffness Matrix K= [K K; K K]; %State Matrices A= [ M D ; zeros(size(m)) M ] ; B= [ zeros(size(m)) K ; -M zeros(size(m))]; %Dynamical Properties of the Mass-Spring System [u,w]=eig(-b,a); %natural frequency [rad/s] %Dynamical Properties of the Mass-Spring System w=sort(diag(abs(w)))//pi %natural frequency [rad/s] w=w() %first natural frequency [Hz] w=w(3) %second natural frequency [Hz] wexp=.65 %measured natural frequency [Hz] wexp=4.45 %measured natural frequency [Hz] dif=(w-wexp)/wexp %error between calculated and measured freq. dif=(w-wexp)/wexp %error between calculated and measured freq. % FRF -- FREQUENCY RESPONSE FUNCTION N = 8 ; % number of points for plotting (see HP-Analyzer) N_factor = ; % Given Excitation Function acting on the the point fo = ; % [N] force amplitude acting on the point of the beam fo = ; % [N] force amplitude acting on the point of the beam for i=:n; F(i)= fo; F(i)= fo; end; % Calculation of the displacement, velocity and acceleration responses for i=:n; w(i) = *pi*i/n_factor; j = sqrt(-); AA = [(-M*w(i)*w(i)+K)+D*w(i)*j]; % Dynamical Stiffness Matrix x = inv(aa)*[f(i) F(i)] ; % Displacement (complex) x(i) = x(); % rail vibration displacement x(i) = x(); % sleeves displacement end; % Given Excitation Function acting on the the point fo = ; % [N] force amplitude acting on the point of the beam fo = ; % [N] force amplitude acting on the point of the beam for i=:n; F(i)= fo; F(i)= fo; end; % Calculation of the displacement, velocity and acceleration responses for i=:n; w(i) = *pi*i/n_factor; j = sqrt(-); AA = [(-M*w(i)*w(i)+K)+D*w(i)*j]; % Dynamical Stiffness Matrix x = inv(aa)*[f(i) F(i)] ; % Displacement (complex) x(i) = x(); % rail vibration displacement x(i) = x(); % sleeves displacement end; % Plotting the results figure() subplot(,,),plot(w//pi,abs(x)) title( Excitation on Point and Response of Point ) xlabel( Frequency [Hz] ) ylabel( (a) y [m/n] ) grid on subplot(,,),plot(w//pi,abs(x)) title( Excitation on Point and Response of Point ) xlabel( Frequency [Hz] ) ylabel( (b) y[m/n] ) grid on subplot(,,3),plot(w//pi,abs(x)) title( Excitation on Point and Response of Point ) xlabel( Frequency [Hz] ) ylabel( (c) y [m/n] ) grid on subplot(,,4),plot(w//pi,abs(x)) title( Excitation on Point and Response of Point ) xlabel( Frequency [Hz] ) ylabel( (d) y [m/n] ) grid on 4

Excitation on Point and Response of Point.5 Excitation on Point and Response of Point.8 y (ω) [m/n]..5..5 y (ω) [m/n].6.4. 4 6 8 Frequency [Hz] 4 6 8 Frequency [Hz] Phase [ o] 5 5 5 3 35 4 6 8 Frequency [Hz] Phase [ o] 5 5 5 3 35 4 6 8 Frequency [Hz] Figure 6: Forced Vibration Theoretical Frequency Response Function (FRF) of the clampedfree flexible beam when two concentrated masses m = m + m =.38 Kg are attached at its free end (L =.6 m) and two additional masses m = m + m =.38 Kg are attached at its middle (L =.3 m) Natural frequencies of the. D.O.F. mass-spring system A :.8 Hz and 5.56 Hz..7.8 Understanding Resonances and Mode Shapes using your Eyes and Fingers Understanding the 9 Degree Phase between excitation force and displacement response while in Resonance using tactile senses. In other words, understanding Zero Degree Phase between the excitation force and velocity response while in resonance using tactile senses. Visualization of the participation of modes shapes in the transient response Visualization using your eyes! Transient motion of the physical system excited with different initial conditions by using your fingers! 43

Excitation on Point and Response of Point.8 Excitation on Point and Response of Point.5 y (ω) [m/n].6.4. y (ω) [m/n].5.5 4 6 8 Frequency [Hz] 4 6 8 Frequency [Hz] Phase [ o] 5 5 5 3 35 4 6 8 Frequency [Hz] Phase [ o] 5 5 5 3 35 4 6 8 Frequency [Hz] Figure 7: Forced Vibration Theoretical Frequency Response Function (FRF) of the clampedfree flexible beam when two concentrated masses m = m + m =.38 Kg are attached at its free end (L =.6 m) and two additional masses m = m + m =.38 Kg are attached at its middle (L =.3 m) Natural frequencies of the. D.O.F. mass-spring system A :.8 Hz and 5.56 Hz. 44

y i (ω) [m/n].8.6.4. Excitation on Point point point y i (ω) [m/n].5.5.5 Excitation on Point point point Phase [ o] 5 5 3 4 6 8 Frequency [Hz] 5 point point 35 4 6 8 Frequency [Hz] Phase [ o] 5 5 3 4 6 8 Frequency [Hz] 5 point point 35 4 6 8 Frequency [Hz] Figure 8: Forced Vibration Theoretical Frequency Response Function (FRF) of the clampedfree flexible beam when two concentrated masses m = m + m =.38 Kg are attached at its free end (L =.6 m) and two additional masses m = m + m =.38 Kg are attached at its middle (L =.3 m) Natural frequencies of the. D.O.F. mass-spring system A :.8 Hz and 5.56 Hz. 45

. FRF Excitation on Point point point Imag(y i (ω)/f (ω)) (i=,) [m/n]...3.4.5.6.7.4.3.....3.4 Real(y i (ω)/f (ω)) (i=,) [m/n] Figure 9: Forced Vibration Theoretical Frequency Response Function (FRF) of the clampedfree flexible beam when two concentrated masses m = m + m =.38 Kg are attached at its free end (L =.6 m) and two additional masses m = m + m =.38 Kg are attached at its middle (L =.3 m) Natural frequencies of the. D.O.F. mass-spring system A :.8 Hz and 5.56 Hz. 46

FRF Excitation on Point point point.5 Imag(y i (ω)/f (ω)) (i=,) [m/n].5.5.5 Real(y i (ω)/f (ω)) (i=,) [m/n].5 4 6 Frequency [Hz] 8 Figure 3: Forced Vibration Theoretical Frequency Response Function (FRF) of the clampedfree flexible beam when two concentrated masses m = m + m =.38 Kg are attached at its free end (L =.6 m) and two additional masses m = m + m =.38 Kg are attached at its middle (L =.3 m) Natural frequencies of the. D.O.F. mass-spring system A :.8 Hz and 5.56 Hz. 47

.7.9 Resonance Experimental Analysis in Time Domain x 5 Signal (a) in Time Domain (b) in Frequency Domain x 5 Signal (a) in Time Domain (b) in Frequency Domain (a) Amplitude [m/s ] (a) Amplitude [m/s ] 5 5 5 3 4 x 6 5 5 5 3 8 x 6 (b) Amplitude [m/s ] 3 (b) Amplitude [m/s ] 6 4 5 5 5 frequency [Hz] 5 5 5 frequency [Hz] Figure 3: Resonance phenomena due to the excitation force with frequency around the natural frequency of the mass-spring system: D.O.F. system with the natural frequencies of.6 Hz and 4.59, excited by the shaker Spring-mass system A with three masses m = m +m +m 3 =.573 Kg fixed at the beam length L =.6 m and three additional masses m = m 4 +m 5 +m 6 =.573 Kg fixed at the middle L =.3 m resulting in two natural frequencies of.6 Hz and 4.6 Hz. 48

.8 Mechanical Systems with 3 D.O.F..8. Physical System and Mechanical Model (a) (b) (c) Figure 3: (a) Real mechanical system built by three turbines attached to an airplane flexible wing; (b) Laboratory prototype built by three lumped masses attached to a flexible beam); (c) Equivalent mechanical model with 3 D.O.F. for the three lumped masses attached to a flexible beam..8. Mathematical Model It is important to point out again, that the equations of motion in Dynamics of Machinery will frequently have the form of a second order differential equation: ÿ(t) = F(y(t), ẏ(t)). Such equations can generally be linearized around an operational position of the physical system, leading to second order linear differential equations. It means that the coefficients which are multiplying the variables ÿ (t), ẏ (t), y (t), ÿ (t), ẏ (t), y (t), ÿ 3 (t), ẏ 3 (t), y 3 (t) (coordinates chosen to describe the motion of the physical system) do not depend on the variables themselves. In the case of the mechanical model presented in figure 3, these coefficients are constants: m, m and m 3 are related to the masses; d, d, d 3, d, d, d 3, d 3, d 3 and d 33 are related to the equivalent viscous damping; and k, k, k 3, k, k, k 3, k 3, k 3 and k 33 are related to equivalent stiffness. One of the goals of the course (Dynamics of Machines) is to present theoretical or experimental approaches to properly find these coefficients, so that the 49

equations of motion can really describe the movement of the physical system. After creating the mechanical model for the physical system, the next step is to derive the equation of motion based on the mechanical model. The mechanical model is built by lumped masses m, m, m (assumption!!!), springs with equivalent stiffness coefficient (calculated using Beam Theory) and dampers with equivalent viscous coefficient (obtained experimentally). While creating the mechanical model and assuming that the mass is a particle, the equation of motion can be derived using Newton s or Lagrange axioms. For the 3 D.O.F system one can write: Mÿ(t) + Dẏ(t) + Ky(t) = f(t) (75) or m m m 3 m m m 3 m 3 m 3 m 33 The mass coefficients m = m + m m = m 3 = m = m = m 3 + m 4 m 3 = m 3 = m 3 = m 33 = m 5 + m 6 ÿ ÿ ÿ 3 + d d d 3 d d d 3 d 3 d 3 d 33 + ẏ ẏ ẏ 3 + k k k 3 k k k 3 k 3 k 3 k 33 y y y 3 = f f f 3 ejωt (76) can easily be achieved either by measuring the masses or by having the material density and mass dimensions. The equivalent damping coefficients can be approximated by d = ξ k m d = d 3 = d = d = ξ k m (Approximation!!!) (78) d 3 = d 3 = d 3 = d 33 = ξ k 33 m 33 or by assuming, for example, proportional damping D = αm + βk. The coefficients α and β can be chosen, so that the damping factor ξ of the first resonance is of the same order as the damping factor achieved in the previous section. Please, note that this is just an approximation (77) 5

which could be verified using Modal Analysis Techniques. Another way of approximating the damping coefficients is given by eq.(78), which should also be verified through experiments! The stiffness coefficients k = 3EIL 3 (L 4L 3 ) L 3 (L L ) (L L + L + L L 3 4L L 3 ) k = 3EI( 3L (L L 3 )L 3 + L (L L L 3 L 3 )) L (L L ) (L L 3 )(L L + L + L L 3 4L L 3 ) k 3 = 9EIL L (L L )(L L 3 )(L L + L + L L 3 4L L 3 ) k = 3EI( 3L (L L 3 )L 3 + L (L L L 3 L 3 )) L (L L ) (L L 3 )(L L + L + L L 3 4L L 3 ) k = k 3 = k 3 = k 3 = k 33 = 3EI(L 4L 3 )(L L 3 ) (L L ) (L L 3 ) (L L + L + L L 3 4L L 3 ) 3EI(L L L L 3L L 3 + 6L L 3 ) (L L )(L L 3 ) (L L + L + L L 3 4L L 3 ) 9EIL L (L L )(L L 3 )(L L + L + L L 3 4L L 3 ) 3EI(L L L L 3L L 3 + 6L L 3 ) (L L )(L L 3 ) (L L + L + L L 3 4L L 3 ) 3EI(L 4L ) (L L 3 ) (L L + L + L L 3 4L L 3 ) (79) can be calculated using the geometry of the beam (I, L, L, L 3 ) and its material properties (E), according to section.3. You can try to get these coefficients using the information presented in section.5. Differential Equations nd order st order [ ] { } [ ] { } { M D ÿ K ẏ f + = M ẏ M y } e jωt Aż(t) + Bz(t) = fe jωt (8) z(t) = { ẏ(t) y(t) } = ẏ (t) ẏ (t) ẏ 3 (t) y (t) y (t) y 3 (t) velocity velocity velocity displacement displacement displacement (8) 5

The analytical solution can be divided into three steps: (I) homogeneous solution (transient analysis); (II) permanent solution (steady-state analysis) and (III) general solution (homogeneous + permanent), as mentioned in section.7.3. Introducing the initial conditions of displacement and velocity z ini = { v ini v ini v 3ini y ini y ini y 3ini } T (8) one gets z(t) = C u e λ t + C u e λ t + C 3 u 3 e λ 3t + C 4 u 4 e λ 4t + C 5 u 5 e λ 5t + C 6 u 6 e λ 6t + Ae iωt (83) where λ = ξ ω n ω n ξ i and u λ = ξ ω n + ω n ξ i and u λ 3 = ξ ω n ω n ξ i and u 3 λ 4 = ξ ω n + ω n ξ i and u 4 λ 5 = ξ 3 ω n3 ω n3 ξ3 i and u 5 λ 6 = ξ 3 ω n3 + ω n3 ξ3 i and u 6 C C C 3 C 4 C 5 C 6 A = [jωa + B] f = [ u u u 3 u 4 u 5 u 6 ] { z ini A}.8.3 Programming in Matlab Theoretical Parameter Studies and Experimental Validation 5

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % MACHINERY DYNAMICS LECTURES (73) % % IKS - DEPARTMENT OF CONTROL ENGINEERING DESIGN % % DTU - TECHNICAL UNIVERSITY OF DENMARK % % % % Copenhagen, February th, % % IFS % % % % 3 D.O.F. SYSTEMS - 4 DIFFERENT EXPERIMENTAL CASES % %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %Concentred Masses Values m=.9; %[Kg] m=.9; %[Kg] m3=.9; %[Kg] m4=.9; %[Kg] m5=.9; %[Kg] m6=.9; %[Kg] %Elastic Properties of the Beam of 6 [mm] E=.7e; %elasticity modulus [N/m^] b=.3 ; %width [m] h=. ; %thickness [m] Iz= (b*h^3)/; %area moment of inertia [m^4] % (.CASE) Data for the mass-spring system % M=m; %concentrated mass [Kg] M=m; %concentrated mass [Kg] M3=m3; %concentrated mass [Kg] L=.3; %length for positioning M [m] L=.46; %length for positioning M [m] L3=.6; %length for positioning M3 [m] % % Coefficients of the Stiffness Matrix [N/m] K= (3*E*Iz*L^3*(L - 4*L3))/(L^3*(L - L)^*(... *L*L + L^ + L*L3-4*L*L3)); K= (-3*E*Iz*(-3*L*(L - *L3)*L3 + L*(L^ -... *L*L3 - *L3^)))/(L*(L - L)^*(L -... L3)*(*L*L + L^ + L*L3-4*L*L3)); K3= (-9*E*Iz*L^)/(L*(L - L)*(L - L3)*(... *L*L + L^ + L*L3-4*L*L3)); K= (-3*E*Iz*(-3*L*(L - *L3)*L3 + L*(L^ -... *L*L3 - *L3^)))/(L*(L - L)^*(L -... L3)*(*L*L + L^ + L*L3-4*L*L3)); K= (3*E*Iz*(L - 4*L3)*(L - L3)^)/((L -... L)^*(L - L3)^*(*L*L + L^ +... L*L3-4*L*L3)); K3= (-3*E*Iz*(L^ - *L*L - *L^ - 3*L*L3 +... 6*L*L3))/((L - L)*(L - L3)^*(*L*L +... L^ + L*L3-4*L*L3)); K3= (-9*E*Iz*L^)/(L*(L - L)*(L - L3)*(*L*L... + L^ + L*L3-4*L*L3)); K3= (-3*E*Iz*(L^ - *L*L - *L^ - 3*L*L3 +... 6*L*L3))/((L - L)*(L - L3)^*(*L*L +... L^ + L*L3-4*L*L3)); K33= (3*E*Iz*(L - 4*L))/((L - L3)^*(*L*L +... L^ + L*L3-4*L*L3)); %Mass Matrix M= [M ; M ; M3]; %Stiffness Matrix K= [K K K3; K K K3; K3 K3 K33]; %Damping Matrix D= [ ; ; ]; %State Matrices A= [ M D ; zeros(size(m)) M ] ; B= [ zeros(size(m)) K ; -M zeros(size(m))]; %Dynamical Properties of the Mass-Spring System [u,w]=eig(-b,a); %natural frequency [rad/s] %Dynamical Properties of the Mass-Spring System w=sort(diag(abs(w)))//pi %natural frequency [rad/s] w=w(); %first natural frequency [Hz] w=w(3); %second natural frequency [Hz] w3=w(5); %third natural frequency [Hz] wexp=.3 wexp=7. wexp3=9.3 dif=(w-wexp)/wexp dif=(w-wexp)/wexp dif3=(w3-wexp3)/wexp3 pause; % (.CASE) Increasing the Mass Values %measured natural frequency [Hz] %IMPORTANT: Freq resolution 4 lines %measured natural frequency [Hz] %IMPORTANT: Freq resolution 4 lines %measured natural frequency [Hz] %IMPORTANT: Freq resolution 4 lines %error between calculated and measured freq. %error between calculated and measured freq. %error between calculated and measured freq. % Data for the mass-spring system % M=m+m4; %concentrated mass [Kg] M=m+m5; %concentrated mass [Kg] M3=m3+m6; %concentrated mass [Kg] L=.3; %length for positioning M [m] L=.46; %length for positioning M [m] L3=.6; %length for positioning M3 [m] % % Coefficients of the Stiffness Matrix [N/m] K= (3*E*Iz*L^3*(L - 4*L3))/(L^3*(L - L)^*(... *L*L + L^ + L*L3-4*L*L3)); K= (-3*E*Iz*(-3*L*(L - *L3)*L3 + L*(L^ -... *L*L3 - *L3^)))/(L*(L - L)^*(L -... L3)*(*L*L + L^ + L*L3-4*L*L3)); K3= (-9*E*Iz*L^)/(L*(L - L)*(L - L3)*(... *L*L + L^ + L*L3-4*L*L3)); K= (-3*E*Iz*(-3*L*(L - *L3)*L3 + L*(L^ -... *L*L3 - *L3^)))/(L*(L - L)^*(L -... L3)*(*L*L + L^ + L*L3-4*L*L3)); K= (3*E*Iz*(L - 4*L3)*(L - L3)^)/((L -... L)^*(L - L3)^*(*L*L + L^ +... L*L3-4*L*L3)); K3= (-3*E*Iz*(L^ - *L*L - *L^ - 3*L*L3 +... 6*L*L3))/((L - L)*(L - L3)^*(*L*L +... L^ + L*L3-4*L*L3)); K3= (-9*E*Iz*L^)/(L*(L - L)*(L - L3)*(*L*L... + L^ + L*L3-4*L*L3)); K3= (-3*E*Iz*(L^ - *L*L - *L^ - 3*L*L3 +... 6*L*L3))/((L - L)*(L - L3)^*(*L*L +... L^ + L*L3-4*L*L3)); K33= (3*E*Iz*(L - 4*L))/((L - L3)^*(*L*L +... L^ + L*L3-4*L*L3)); %Mass Matrix M= [M ; M ; M3]; %Stiffness Matrix K= [K K K3; K K K3; K3 K3 K33]; %Damping Matrix D= [ ; ; ]; %State Matrices A= [ M D ; zeros(size(m)) M ] ; B= [ zeros(size(m)) K ; -M zeros(size(m))]; %Dynamical Properties of the Mass-Spring System [u,w]=eig(-b,a); %natural frequency [rad/s] %Dynamical Properties of the Mass-Spring System w=sort(diag(abs(w)))//pi %natural frequency [rad/s] w=w(); %first natural frequency [Hz] w=w(3); %second natural frequency [Hz] w3=w(5); %third natural frequency [Hz] wexp=.7875 %measured natural frequency [Hz] %IMPORTANT: Freq resolution 4 lines wexp=5.5 %measured natural frequency [Hz] %IMPORTANT: Freq resolution 4 lines wexp3=4.3 %measured natural frequency [Hz] %IMPORTANT: Freq resolution 4 lines dif=(w-wexp)/wexp %error between calculated and measured freq. dif=(w-wexp)/wexp %error between calculated and measured freq. dif3=(w3-wexp3)/wexp3 %error between calculated and measured freq. pause; 53

% (3.CASE) Changing the Position of the Concentrated Masses % Data for the mass-spring system % M=m+m; %concentrated mass [Kg] M=m3+m4; %concentrated mass [Kg] M3=m5+m6; %concentrated mass [Kg] L=.5; %length for positioning M [m] L=.3; %length for positioning M [m] L3=.45; %length for positioning M3 [m] % % Coefficients of the Stiffness Matrix [N/m] K= (3*E*Iz*L^3*(L - 4*L3))/(L^3*(L - L)^*(... *L*L + L^ + L*L3-4*L*L3)); K= (-3*E*Iz*(-3*L*(L - *L3)*L3 + L*(L^ -... *L*L3 - *L3^)))/(L*(L - L)^*(L -... L3)*(*L*L + L^ + L*L3-4*L*L3)); K3= (-9*E*Iz*L^)/(L*(L - L)*(L - L3)*(... *L*L + L^ + L*L3-4*L*L3)); K= (-3*E*Iz*(-3*L*(L - *L3)*L3 + L*(L^ -... *L*L3 - *L3^)))/(L*(L - L)^*(L -... L3)*(*L*L + L^ + L*L3-4*L*L3)); K= (3*E*Iz*(L - 4*L3)*(L - L3)^)/((L -... L)^*(L - L3)^*(*L*L + L^ +... L*L3-4*L*L3)); K3= (-3*E*Iz*(L^ - *L*L - *L^ - 3*L*L3 +... 6*L*L3))/((L - L)*(L - L3)^*(*L*L +... L^ + L*L3-4*L*L3)); K3= (-9*E*Iz*L^)/(L*(L - L)*(L - L3)*(*L*L... + L^ + L*L3-4*L*L3)); K3= (-3*E*Iz*(L^ - *L*L - *L^ - 3*L*L3 +... 6*L*L3))/((L - L)*(L - L3)^*(*L*L +... L^ + L*L3-4*L*L3)); K33= (3*E*Iz*(L - 4*L))/((L - L3)^*(*L*L +... L^ + L*L3-4*L*L3)); %Mass Matrix M= [M ; M ; M3]; %Stiffness Matrix K= [K K K3; K K K3; K3 K3 K33]; %Damping Matrix D= [ ; ; ]; %State Matrices A= [ M D ; zeros(size(m)) M ] ; B= [ zeros(size(m)) K ; -M zeros(size(m))]; %Dynamical Properties of the Mass-Spring System [u,w]=eig(-b,a); %natural frequency [rad/s] %Dynamical Properties of the Mass-Spring System w=sort(diag(abs(w)))//pi %natural frequency [rad/s] w=w(); %first natural frequency [Hz] w=w(3); %second natural frequency [Hz] w3=w(5); %third natural frequency [Hz] wexp=.94 %measured natural frequency [Hz] %IMPORTANT: Freq resolution 4 lines wexp=7.88 %measured natural frequency [Hz] %IMPORTANT: Freq resolution 4 lines wexp3=.5 %measured natural frequency [Hz] %IMPORTANT: Freq resolution 4 lines dif=(w-wexp)/wexp %error between calculated and measured freq. dif=(w-wexp)/wexp %error between calculated and measured freq. dif3=(w3-wexp3)/wexp3 %error between calculated and measured freq. pause; % (4.CASE) Changing the Position and the Values of the Concentrated Masses % Data for the mass-spring system % M=m+m4+m5; %concentrated mass [Kg] M=m+m6; %concentrated mass [Kg] M3=m3; %concentrated mass [Kg] L=.5; %length for positioning M [m] L=.3; %length for positioning M [m] L3=.45; %length for positioning M3 [m] % % Coefficients of the Stiffness Matrix [N/m] K= (3*E*Iz*L^3*(L - 4*L3))/(L^3*(L - L)^*(... *L*L + L^ + L*L3-4*L*L3)); K= (-3*E*Iz*(-3*L*(L - *L3)*L3 + L*(L^ -... *L*L3 - *L3^)))/(L*(L - L)^*(L -... L3)*(*L*L + L^ + L*L3-4*L*L3)); K3= (-9*E*Iz*L^)/(L*(L - L)*(L - L3)*(... *L*L + L^ + L*L3-4*L*L3)); K= (-3*E*Iz*(-3*L*(L - *L3)*L3 + L*(L^ -... *L*L3 - *L3^)))/(L*(L - L)^*(L -... L3)*(*L*L + L^ + L*L3-4*L*L3)); K= (3*E*Iz*(L - 4*L3)*(L - L3)^)/((L -... L)^*(L - L3)^*(*L*L + L^ +... L*L3-4*L*L3)); K3= (-3*E*Iz*(L^ - *L*L - *L^ - 3*L*L3 +... 6*L*L3))/((L - L)*(L - L3)^*(*L*L +... L^ + L*L3-4*L*L3)); K3= (-9*E*Iz*L^)/(L*(L - L)*(L - L3)*(*L*L... + L^ + L*L3-4*L*L3)); K3= (-3*E*Iz*(L^ - *L*L - *L^ - 3*L*L3 +... 6*L*L3))/((L - L)*(L - L3)^*(*L*L +... L^ + L*L3-4*L*L3)); K33= (3*E*Iz*(L - 4*L))/((L - L3)^*(*L*L +... L^ + L*L3-4*L*L3)); %Mass Matrix M= [M ; M ; M3]; %Stiffness Matrix K= [K K K3; K K K3; K3 K3 K33]; %Damping Matrix D= [ ; ; ]; %State Matrices A= [ M D ; zeros(size(m)) M ] ; B= [ zeros(size(m)) K ; -M zeros(size(m))]; %Dynamical Properties of the Mass-Spring System [u,w]=eig(-b,a); %natural frequency [rad/s] %Dynamical Properties of the Mass-Spring System w=sort(diag(abs(w)))//pi %natural frequency [rad/s] w=w(); %first natural frequency [Hz] w=w(3); %second natural frequency [Hz] w3=w(5); %third natural frequency [Hz] exp=.3 %measured natural frequency [Hz] %IMPORTANT: Freq resolution 4 lines wexp=7.9 %measured natural frequency [Hz] %IMPORTANT: Freq resolution 4 lines wexp3=8. %measured natural frequency [Hz] %IMPORTANT: Freq resolution 4 lines dif=(w-wexp)/wexp %error between calculated and measured freq. dif=(w-wexp)/wexp %error between calculated and measured freq. dif3=(w3-wexp3)/wexp3 %error between calculated and measured freq. pause; 54

.8.4 Theoretical Frequency Response Function. Excitation on Point.4 Excitation on Point.8 Excitation on Point 3 (a) y [m/n].5 (b) y [m/n].3.. (c) y3 [m/n].6.4. 3 Frequency [Hz].4 3 Frequency [Hz].8 3 Frequency [Hz] (d) y [m/n].3.. (a) y [m/n].6.4. (b) y3 [m/n].5.5 3 Frequency [Hz].8 3 Frequency [Hz] 3 Frequency [Hz] 3 (c) y3 [m/n].6.4. (d) y3 [m/n].5.5 (d) y33 [m/n] 3 Frequency [Hz] 3 Frequency [Hz] 3 Frequency [Hz] Figure 33: Forced Vibration Theoretical Frequency Response Function (FRF) of the clampedfree flexible beam when two concentrated masses m 33 = m + m =.38 Kg are attached at its free end (L =.6 m), two additional masses m = m 3 + m 4 =.38 Kg are attached at L =.4 m and two additional masses m = m 5 + m 6 =.38 Kg are attached at L =. m Natural frequencies of the 3 D.O.F. mass-spring system A :.3 Hz, 7. Hz and 9.3 Hz. 55

.8.5 Experimental Natural Frequencies (a) Amplitude [m/s ] 3 x Signal (a) in Time Domain (b) in Frequency Domain 4 5 5 5 3 x 5 (b) Amplitude [m/s ] 5 5 5 frequency [Hz] Figure 34: Transient Vibration Acceleration of the clamped-free flexible beam when two concentrated masses m = m + m =.38 Kg are attached at its free end (L 3 =.6 m), two additional masses m = m + m =.38 Kg are attached at its length (L =.4 m) and two additional masses m = m +m =.38 Kg are attached at its length (L =. m) Natural frequencies of the 3 D.O.F. mass-spring system A :.3 Hz, 7. Hz and 9.3 Hz..8.6 Experimental Resonances and Mode Shapes Visualization of the participation of modes shapes in the transient response Visualization using your eyes! Transient motion of the physical system excited with different initial conditions by using your fingers! Applying an oscillatory excitation by using your finger at the 3 different points of the physical system (co-ordinates of the mechanical model) and detecting the participation of the mode shapes in the permanent solution or steady-state response by using your eyes. 56

x 5 Signal (a) in Time Domain (b) in Frequency Domain (a) Amplitude [m/s ] 5 5 5 3 x 5 (b) Amplitude [m/s ].5.5 5 5 5 frequency [Hz] x 5 Signal (a) in Time Domain (b) in Frequency Domain (a) Amplitude [m/s ] 5 5 5 3 x 5 (b) Amplitude [m/s ].5.5 5 5 5 frequency [Hz] x 5 Signal (a) in Time Domain (b) in Frequency Domain (a) Amplitude [m/s ] 5 5 5 3 x 5 (b) Amplitude [m/s ].5.5 5 5 5 frequency [Hz] Figure 35: Resonance phenomena due to the excitation force with frequencies around the natural frequencies of the mass-spring system: 3 D.O.F. system with the natural frequencies of.75 Hz, 5. Hz and 4.68 Hz, excited by the shaker Spring-mass system A with two masses m = m + m =.38 Kg fixed at the beam length L 3 =.6 m, two additional masses fixed at L =.4 m and two more at L =. m. 57

.9 Exercises Answer the questions using the Matlab program dof-integration.m:. Vary the cross section parameters of the beam (b,h) while exciting the mass-spring system with just an initial velocity (initial displacement and excitation force are set zero). (a) Explain what happens with the natural frequency of the mass-spring system; (b) What happens with the maximum vibration amplitude of the system?. Vary the beam length (L) while exciting the mass-spring system with just an initial velocity (initial displacement and excitation force are set zero). (a) Explain what happens with the natural frequency of the mass-spring system; (b) What happens with the maximum vibration amplitude of the system? 3. Vary the number of masses attached to the beam while exciting the mass-spring system with just an initial velocity (initial displacement and excitation force are set zero). (a) Explain what happens with the natural frequency of the mass-spring system; (b) What happens with the maximum vibration amplitude of the system? 4. Vary the damping factor ξ while exciting the mass-spring-damping system with just an initial velocity (initial displacement and excitation force are set zero). (a) Explain what happens with the natural frequency w n of the mass-spring system and the damped natural frequency w d ; (b) What happens with the maximum vibration amplitude of the system? 5. Set the damping factor ξ =.5 while exciting the mass-spring-damping system with just an excitation force of f =. e j w t [N] and initial velocity (initial displacement is set zero). Explain the vibration behavior of the system in terms of amplitudes and frequencies when: (a) w = %w n ; (b) w = 5%w n ; (c) w = 9%w n ; (d) w = w n ; (e) w = %w n ; (f) w = 5%w n and (g) w = %w n. 6. Set the damping factor at times more than before, ξ =.5, while exciting the massspring-damping system with just an excitation force of f =. e j w t [N] and initial velocity (initial displacement is set zero). Explain the vibration behavior of the system in terms of amplitudes and frequencies when: (a) w = %w n ; (b) w = 5%w n ; (c) w = 9%w n ; (d) w = w n ; (e) w = %w n ; (f) w = 5%w n and (g) w = %w n. 7. Explain how this parameters variation could be useful in the case of a real machine? Answer the questions using the Matlab program dof-integration.m:. Excite the mass-spring system just with an initial velocity at the first coordinate (ẏ ini ) (initial displacements and excitation forces are set zero). Describe the vibration behavior of points y and y.. Excite the mass-spring system just with an initial velocity at the second coordinate (ẏ ini ) (initial displacements and excitation forces are set zero). Describe the vibration behavior of points y and y. 3. Compare the two last simulations. Why is the transient behavior so different when the system is perturbed with initial velocity at point y or at point y? 58

4. Vary the number of masses attached to the first coordinate y the beam while exciting the mass-spring system with just an initial velocity at the first coordinate (ẏ ini ) (initial displacements and excitation forces are set zero). (a) Explain what happens with the natural frequencies of the system; (b) How many natural frequencies change when you change the mass in just one point of the structure? Explain. 5. Vary the number of masses attached to the second coordinate of the beam, y, while exciting the mass-spring system with just an initial velocity at the first coordinate (ẏ ini ) (initial displacements and excitation forces are set zero). (a) Explain what happens with the natural frequencies of the system; (b) How many natural frequencies change when you change the mass in just one point of the structure? Explain. 6. Set the damping factor ξ =.5, while exciting the mass-spring-damping system with just an excitation force of f =. e j w t [N] (initial velocities and initial displacements are set zero). Explain the vibration behavior of the system in terms of amplitudes and frequencies, when: (a) w = %w n ; (b) w = 5%w n ; (c) w = 9%w n ; (d) w = w n ; (e) w = %w n ; (f) w = 9%w n ; (g) w = w n ; (h) w = %w n ; (i) w = %w n. 7. Set the damping factor ξ =.5, while exciting the mass-spring-damping system with just an excitation force of f =. e j w t [N] (initial velocities and initial displacements are set zero). Explain the vibration behavior of the system in terms of amplitudes and frequencies, when: (a) w = %w n ; (b) w = 5%w n ; (c) w = 9%w n ; (d) w = w n ; (e) w = %w n ; (f) w = 9%w n ; (g) w = w n ; (h) w = %w n ; (i) w = %w n. 8. Set the damping factor ξ =.5, while exciting the mass-spring-damping system with just an excitation force of f =. e j w t [N] (initial velocities and initial displacements are set zero). Explain the vibration behavior of the system in terms of amplitudes and frequencies, when: (a) w = %w n ; (b) w = 5%w n ; (c) w = 9%w n ; (d) w = w n ; (e) w = %w n ; (f) w = 9%w n ; (g) w = w n ; (h) w = %w n ; (i) w = %w n. 9. Set the damping factor ξ =.5, while exciting the mass-spring-damping system with just an excitation force of f =. e j w t [N] (initial velocities and initial displacements are set zero). Explain the vibration behavior of the system in terms of amplitudes and frequencies, when: (a) w = %w n ; (b) w = 5%w n ; (c) w = 9%w n ; (d) w = w n ; (e) w = %w n ; (f) w = 9%w n ; (g) w = w n ; (h) w = %w n ; (i) w = %w n ;. Explain how the variation of such parameters could be useful in a case with a real machine? Create a program dof3-integration.m based on dof-integration.m and answer the following questions:. Make use of the beam theory, and show how to get the 9 stiffness coefficients k, k, k 3, k, k, k 3, k 3, k 3 and k 33.. Excite the mass-spring system with just an initial velocity at the first coordinate (ẏ ini ) (initial displacements and excitation forces are set zero). Describe the vibration behavior of the points y, y and y 3. 3. Excite the mass-spring system with just an initial velocity at the second coordinate (ẏ ini ) (initial displacements and excitation forces are set zero). Describe the vibration behavior of the points y, y and y 3. 59

4. Excite the mass-spring system with just an initial velocity at the third coordinate (ẏ 3ini ) (initial displacements and excitation forces are set zero). Describe the vibration behavior of the points y, y and y 3. 5. Compare the three last simulations. Why is the transient behavior so different when the system is perturbed with initial velocity at point y, y or y 3. 6. Vary the number of masses attached to the first coordinate of the beam, y, while exciting the mass-spring system with just an initial velocity at the first coordinate (ẏ ini ) (initial displacements and excitation forces are set zero). (a) Explain what happens with the natural frequencies of the system; (b) How many natural frequencies change when you change the mass in just one point of the structure? Explain. 7. Vary the number of masses attached to the second coordinate of the beam, y, while exciting the mass-spring system with just an initial velocity at the first coordinate (ẏ ini ) (initial displacements and excitation forces are set zero). (a) Explain what happens with the natural frequencies of the system; (b) How many natural frequencies change when you change the mass in just one point of the structure? Explain. 8. Set the damping factor ξ =.5, while exciting the mass-spring-damping system with just an excitation force of f =. e j w t [N] (initial velocities and initial displacements are set zero). Explain the vibration behavior of the system in terms of amplitudes and frequencies, when: (a) w = %w n ; (b) w = 5%w n ; (c) w = 9%w n ; (d) w = w n ; (e) w = %w n ; (f) w = 9%w n ; (g) w = w n ; (h) w = %w n ; (i) w = %w n ; (j) w = 9%w n3 ; (k) w = w n3 ; (l) w = %w n3 ; (m) w = %w n3. 9. Set the damping factor ξ =.5, while exciting the mass-spring-damping system with just an excitation force of f =. e j w t [N] (initial velocities and initial displacements are set zero). Explain the vibration behavior of the system in terms of amplitudes and frequencies, when: (a) w = %w n ; (b) w = 5%w n ; (c) w = 9%w n ; (d) w = w n ; (e) w = %w n ; (f) w = 9%w n ; (g) w = w n ; (h) w = %w n ; (i) w = %w n ; (j) w = 9%w n3 ; (k) w = w n3 ; (l) w = %w n3 ; (m) w = %w n3.. Set the damping factor ξ =.5, while exciting the mass-spring-damping system with just an excitation force of f 3 =. e j w t [N] (initial velocities and initial displacements are set zero). Explain the vibration behavior of the system in terms of amplitudes and frequencies, when: (a) w = %w n ; (b) w = 5%w n ; (c) w = 9%w n ; (d) w = w n ; (e) w = %w n ; (f) w = 9%w n ; (g) w = w n ; (h) w = %w n ; (i) w = %w n ; (j) w = 9%w n3 ; (k) w = w n3 ; (l) w = %w n3 ; (m) w = %w n3.. Set the damping factor ξ =.5, while exciting the mass-spring-damping system with just an excitation force of f 3 =. e j w t [N] (initial velocities and initial displacements are set zero). Explain the vibration behavior of the system in terms of amplitudes and frequencies, when: (a) w = %w n ; (b) w = 5%w n ; (c) w = 9%w n ; (d) w = w n ; (e) w = %w n ; (f) w = 9%w n ; (g) w = w n ; (h) w = %w n ; (i) w = %w n ; (j) w = 9%w n3 ; (k) w = w n3 ; (l) w = %w n3 ; (m) w = %w n3 ;. Explain how such a variation of parameters could be useful in a case with a real machine? 6

. Project Identification of Model Parameters (An Example) GOAL With the first project the student will face a practical problem of the real life: how to properly choose the coefficients of linear differential equations of second order, aiming at achieving a reliable mathematical model, which can predict the machine dynamics? (a) (b) Figure 36: (a) Offshore platform http : //www.civl.port.ac.uk/comp prog/offshore platforms; (b) Laboratory prototype composed of one concentrated mass (foundation and rotor) attached to four flexible beams An equivalent model of D.O.F. system for analyzing the platform s linear vibration in the horizontal direction. To represent the D-movements of the offshore platform shown in figure 36(a) a laboratory prototype was built, as it can be seen in figure 36(b). This simplified test rig is composed of one concentrated mass (foundation and rotor) attached to four flexible beams. An equivalent model of D.O.F. system can be created with the purpose of analyzing the platform s linear vibration in the horizontal direction. m.8 [kg] platform mass L.5 [m] beam length b.5 [m] beam width h. [m] beam thickness E.9 [N/m ] steel elastic modulus Table 5: Main parameters of the test rig (platform). Create a mechanical model of one-degree-of-freedom for describing the horizontal vibration of the test rig. Use Newton s law and equivalent coefficients of mass m [Kg], viscous damping d [N/(m/s)] and linear stiffness k [N/m].. There are two different ways of experimentally obtaining the forced vibration response of the platform in the frequency domain, i.e. its frequency response functions F RF(ω), namely by means of H(ω) and H(ω) functions. Detail about how to experimentally obtain H(ω) and H(ω) will be given in the second part of manuscript. Anyway, for now, it is important to relate such experimental functions to the frequency response functions 6

presented in section.6.9, to learn how to deal with experimental data and how to extract important information. H(ω) as well as H(ω) are obtained when an excitation force is horizontally acting on the platform mass and its value is measured by using a force transducer, and simultaneously the acceleration response of the platform is measured using a acceleration sensor. The experimental setup can be seen in figure 36(b). The most common ways of representing the frequency response functions H(ω) and H(ω) are in the form of amplitude and phase, or real and imaginary. In table 6 such values are presented in the real and imaginary forms and in figure 37 they are plotted. Please, plot H(ω) and H(ω) in terms of magnitude and phase. Frequency - H - H COHERENCE [Hz] [(m/s )/N] [(m/s )/N].65.49 -.4i.595 -.49i.836.75.6 -.5i.698 -.58i.8745.875.75 -.7i.855 -.8i.8776..86 -.68i.945 -.78i.8744.5.974 -.5i.93 -.8i.897.5.4 +.5i.4 +.6i.9335.375.3 -.8i.393 -.9i.943.5.4 -.i.477 -.5i.9555.65.54 -.i.6 -.6i.96.75.87 -.4i.889 -.4i.969.875.3 -.45i.84 -.5i.9677 3..474 -.7i.5 -.76i.989 3.5.676 -.3i.76 -.34i.987 3.5.36 -.364i.38 -.37i.979 3.375.346 -.487i.3536 -.497i.9787 3.5.47 -.467i.473 -.479i.9758 3.65.46 -.673i.4697 -.687i.9795 3.75.57 -.93i.565 -.94i.9887 3.875.5894 -.39i.5974 -.56i.9866 4..686 -.47i.6994 -.454i.98 4.5.8 -.959i.88 -.987i.9857 4.5.938 -.84i.9539 -.87i.9834 4.375.3 -.4i.49 -.46i.973 4.5.39 -.5385i.37 -.5556i.9693 4.65.5586 -.83i.63 -.845i.9555 4.75.8764 -.4666i.9844 -.559i.9456 4.875.883 -.95i.664 -.938i.94 5..59-3.7i.36-3.373i.9 5.5.98-3.6778i.43-4.3i.8694 5.5 -.665-3.734i -.7564-3.5568i.93 5.375 -.54 -.3597i -.336 -.4844i.9498 5.5 -.657 -.4778i -.476 -.5337i.9635 5.65 -.989 -.57i -.9659 -.37i.98 5.75 -.758 -.84i -.778 -.88i.9873 5.875 -.55 -.63i -.5665 -.6384i.99 6. -.4389 -.56i -.4438 -.533i.9966 6.5 -.35 -.495i -.3559 -.437i.997 6.5 -.59 -.348i -.6 -.3436i.9976 6.375 -.843 -.3i -.866 -.36i.998 6.5 -.4 -.668i -.64 -.673i.998 6.65 -.76 -.3i -.75 -.34i.999 6.75 -.6 -.i -.6 -.3i.999 6.875 -.975 -.98i -.97 -.98i.9993 7. -.9376 -.766i -.9384 -.767i.999 7.5 -.973 -.63i -.977 -.64i.9995 7.5 -.8793 -.58i -.8797 -.59i.9995 7.375 -.856 -.459i -.859 -.459i.9996 7.5 -.87 -.377i -.873 -.377i.9998 7.65 -.87 -.59i -.89 -.6i.9997 7.75 -.7879 -.4i -.788 -.4i.9997 7.875 -.7695 -.7i -.7696 -.7i.9998 8. -.753 -.98i -.755 -.98i.9998 8.5 -.7374 -.58i -.7376 -.59i.9998 8.5 -.74 -.38i -.74 -.38i.9998 8.375 -.7 -.994i -.73 -.994i.9998 8.5 -.6963 -.949i -.6964 -.95i.9998 8.65 -.686 -.93i -.686 -.93i.9999 8.75 -.679 -.897i -.6793 -.897i.9998 8.875 -.675 -.886i -.676 -.886i.9998 9. -.666 -.847i -.667 -.847i.9999 9.5 -.6544 -.85i -.6545 -.85i.9998 9.5 -.6463 -.793i -.6465 -.794i.9997 9.375 -.6365 -.765i -.6367 -.765i.9997 9.5 -.687 -.774i -.689 -.774i.9997 9.65 -.65 -.76i -.653 -.76i.9996 9.75 -.67 -.7i -.69 -.7i.9997 9.875 -.66 -.7i -.66 -.7i.9997. -.68 -.74i -.683 -.74i.9998 Table 6: Two experimental frequency response functions H(ω) and H(ω) of the test rig. 3. Identification of model parameters Suppose you have no access to the values of mass 6

REAL(Acc/force) [(m/s )/N] 5 5 5 Experimental Frequency Response Function (with damper) measured 5 3 4 5 6 7 8 9 Frequency [Hz] (a) IMAG(Acc/force) [(m/s )/N] 3 4 measured 3 4 5 6 7 8 9 Frequency [Hz] REAL(Acc/force) [(m/s )/N] 5 5 5 Experimental Frequency Response Function (with damper) measured 5 3 4 5 6 7 8 9 Frequency [Hz] IMAG(Acc/force) [(m/s )/N] 3 4 measured (b) 3 4 5 6 7 8 9 Frequency [Hz] Figure 37: Two experimental frequency response function F RF(ω) of the test rig obtained by means of H(ω) and H(ω) functions. 63

m [kg], damping d [N/(m/s)] and stiffness k [N/m] of the platform and then can not be calculated. However, you can measure the forces applied to the platform and its acceleration response (steady-state response) and are able to experimentally obtain the frequency response function presented in table 6 and plotted in figure 37. Remember that the frequency response function is acceleration/force in this case, or FRF(ω) = ω /m ( ω + ωn + ξ ω n ω i) = ω ( m ω + d ω i + k) Elaborate a simple parameter identification procedures based on the Least-Square Method, assuming that the frequency response functions FRF(ω) are known, and identify simultaneously the three parameter, i.e. mass, stiffness and damping: (84) ω ω ω3...... ωn { m k } = ( REAL ( REAL ( REAL (... REAL ω FRF(ω ) ω FRF(ω ) ω3 FRF(ω 3 ) ω N FRF(ω N ) ) ) ) ) = A x = b (85) x = ( A T A ) AT b (86) ω ω ω 3... ω N { } d = ( IMAG ( IMAG ( IMAG (... IMAG ω FRF(ω ) ω FRF(ω ) ω3 FRF(ω 3 ) ω N FRF(ω N ) ) ) ) ) = Ā x = b (87) x = ( Ā T Ā) Ā T b (88) Implement the identification procedure using MAPLE, or MATHEMATICA or MATLAB or MATCAD or another software. Use H(ω) as well as H(ω) for identifying the coefficients of mass m [kg], stiffness k [N/m] and damping d [N/(m/s)]. 4. Find theoretical and experimental ways of checking the identified values of mass, stiffness and damping, in order to assure that such values are really the correct mass, stiffness and damping coefficients of the real system. The more checking procedures you can find, the better! Explain them all in details. 5. Model application On the platform top a rotating machine is mounted, as it can be seen in figure.36(b). Its characteristics are delivered by the manufacturer. The maximum angular velocity is, [rpm] ( [Hz]). It is also known that the machine unbalance (m unb ε) is. [Kg m]. By using your mathematical model, plot the vibration amplitude of the platform as a function of the machine angular velocity. Determine the maximum vibration amplitude of the platform. It is important to notice that the values of H(ω) and H(ω) presented in table 6 are H(ω) and H(ω). When using the values of such functions to identify the model parameters, they have to be multiplied by -. 64

6. Model application As explained in the last item, the motor characteristics are delivered by the manufacturer. The maximum angular velocity is, [rpm] ( [Hz]). It is also known that the machine unbalance (m unb ε) is. [Kg m]. Based on the dynamic equilibrium between motor torque, and torques associated to inertia, losses and external loading, the motor start-up curve shows a linear variation of angular velocity from to, [rpm] in a period of 4 [s]. Based on your mathematical model, please simulate computationally the vibration behavior of the platform during the period of 4 [s], knowing that, when the motor starts, the platform is already deformed due to a constant lateral wind. The platform static deformation is. [m]. Plot the platform displacement in the time domain, considering two cases: (a) considering the static wind force acting on the platform the whole time; (b) considering the lateral wind force suddenly disappears 3 [s] after the motor start-up. Analyze and discuss the behavior of the plots. What is the maximum vibration amplitude of the platform in both cases? 7. Question Explain why the test rig can be modelled as an one-degree-of-freedom system in the range of to Hz, if one knows that a rigid body in the space (platform mass of the test rig) should be represented by a mechanical model of six-degrees-of-freedom, i.e. three linear and three angular displacements. 8. Write your final conclusions. (No Technical report!) 65

. Project Modal Analysis & Validation of Models GOAL To get familiar with the dynamic interaction between machine and structure, the elaboration of mechanical and mathematical models for representing rotor-structure vibrations in D, implementation and vibration analysis using Matlab, visualization of natural frequencies and mode shapes. To test the accuracy of an analytical mathematical model proposed for describing the system dynamic behavior, i.e. natural frequencies and mode shapes. Remember, if the measured frequencies and mode shapes agree with those predicted by the analytical mathematical model, the model is verified and can be useful for design proposes and vibration predictions with some confidence. Otherwise, the analytical models are useless. (a) (b) Figure 38: Machine-structure dynamical interaction (a) Offshore platform http : //www.oil gas.uwa.edu.au/troll A Graphics.htm; (b) Equivalent laboratory prototype composed of four concentrated masses attached to four flexible beams: - mass on the first floor, - mass on the second floor, 3- mass on the third floor, 4- mass on the fourth floor, 5- motor-disk with unbalanced masses for simulating a rotating machine with unbalance problems, 6- acceleration sensor attached to the second mass, 7- acceleration sensor attached to the third mass, 8- magnetic actuator for simulating wave excitation, 9- magnetic actuator for simulating waves excitation. Figures 38(a) and (b) illustrate an offshore platform and an equivalent laboratory prototype, where the students can carry on measurements and vibration analyzes. The laboratory prototype is composed of four concentrated masses attached to four flexible beams. Elements,,3 and 4 are the four masses connected by means of flexible beams. Element 5 is a motor-disk with a 66