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# CU-Boulder ASEN 5022 - Vibration of Two Degrees of Freedom Systems

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ASEN5022: Dynamics of Aerospace StructuresLecture 6: Vibration of Two Degrees of Freedom SystemsThis lecture covers:• Begin with a 2-dof spring-mass-damper model system.• Obtain the frequency response functions.• Learn the modes and mode shapes and their physical meanings.• Study the properties of 2-dof frequency response functions.16.1 Equations of MotionConsider a two-DOF model as shown in Fig. 1. Summing the forces acting on each mass, the equationsof motion for the coupled two-mass-spring-damper system can be written asM1¨x1= f1(t) − K1x1+ k2(x2− x1) + c2(˙x2−˙x1)m2¨x2= −k2(x2− x1) − c2( ˙x2− ˙x1)(6.1)As the preceding equation involves two displacements, x1and x2, its general solution involves complexmatrix differential algebra. For design considerations, however, important insight can be gained byconsidering the special case of forcing function given byf1(t) = F1ejω t, f2(t) = F2ejω t(6.2)so that the solution assumes the form ofx1(t) = X1ejω t, x2(t) = X2ejω t(6.3)2Fig. 1. Two DOF Vertical Motion ModelSubstituting (6.2) and (6.3) into (6.1), we obtain−ω2M1X1= F1− K1X1+ k2(X2− X1) + jωc2(X2− X1)−ω2m2X2= F2− k2(X2− X1) − jωc2(X2− X1)(6.4)In order to solve for X1and X2, let’s rearrange the preceding equation to read(−ω2M1+ jωc2+ K1+ k2) X1− (k2+ jωc2) X2= F1−(k2+ jωc2)X1+ (−ω2m2+ jωc2+ k2) X2= F2(6.5)3Solving for X1and X2we obtainX1=H2(ω)F1+ H12(ω)F2H1(ω) H2(ω) − H12(ω)2X2=H1(ω)F2+ H12(ω)F1H1(ω)H2(ω) − H12(ω)2H1(ω) = (−ω2M1+ jωc2+ K1+ k2)H2(ω) = (−ω2m2+ jωc2+ k2)H12(ω) = (k2+ jωc2)(6.6)Although (6.6) appears to be very complex, several simplifications are possible to aid vibrationdesigners. This is studied below.46.2 What Are Vibration Modes and Mode Shape?It turns out that the motions of X1and X2given by (6.6) are not randomly independent as thesolution may suggest. They are interlinked by the property called mode shapes. Understanding thephysical properties of mode shapes is important if one is tasked to design structures subject tovibrations.To this end, let us consider the following special 2-DOF case:m1= m2= 1, k1= k2= 2.618π2, c2= 0, f1(t) = 0 (6.7)For this model, we apply two different excitations:f2(t) =0.2 × sin (0.98 ∗ πt)sin (0.98 ∗ 2.618πt)(6.8)Figure 2 illustrates the two responses subject to the two excitations specified in (6.8). Observethat, for the case of f2(t) = 0.2 × sin (0.98 ∗ πt), both mass m1and mass m2are moving inphase, that is, they move in the same direction in time. However, when the system is subjected tof2(t) = sin (0.98 ∗ 2.618πt), mass m1and mass m2are moving out phase. That is, they move in the5Fig. 2. Motions of two masses under two different excitationsopposite directions in time. In other words, depending on the excitation frequency, the motions of thetwo masses are drastically different. To understand this strange phenomena, one has to understandthe roles of modes and mode shapes. To this end, let us recast (6.1) in a matrix form with c = 0:m100 m2¨x1¨x2+k1+ k2−k2−k2k2x1x2=f1(t)f2(t)(6.9)6The characteristic equation of the above coupled 2-dof differential equation(6.9) can be obtained asfollows. First, we assume the solution of their homogeneous equations in the formx1x2=¯x1¯x2ejωt(6.10)Second, substituting this into (6.9) with f1= f2= 0 yields:−ω2m100 m2+k1+ k2−k2−k2k2¯x1¯x2= 0. (6.11)Hence, the characteristic equation is obtained by requiring that the above equation has a nontrivialsolution:det |(k1+ k2− ω2m1) −k2−k2(k2− ω2m2)| = 0 ⇒ ω4− (k1+ k2m1+k2m2)ω2+k1k2m1m2= 0 (6.12)Note that, with m1= m2= 1, k1= k2= 2.618π2, the two roots of the above characteristic equationare given by7ω21= π2, ω22= (2.618π)2. (6.13)These two values are called characteristic values whose square roots are called the natural frequen-cies or vibration modes of the system.The corresponding eigenvectors can be computed from the second row of (6.11):−ω2¯x2− k2¯x1+ k2¯x2= 0 ⇒¯x1¯x2=k2− ω2k2= 1 −ω2k2. (6.14)Note that for the two modes (frequencies) computed in (6.13), we have two different expressions:For ω = π :¯x1¯x2= 1 −π22.618π2= 0.618For ω = 2.618π :¯x1¯x2= 1 −2.6182π22.618π2= −1.618. (6.15)The ratios of these eigenvector sets are plotted in Figure 3.Observe from Fig. 3 that for the case of the first mode when mass 1 moves in the same directionwith its amplitude of 0.616 while mass 2 moves a unit amplitude. In other words, the two massesmove in phase as illustrated in Figure 2. As for the second mode, mass 1 moves in the oppositedirection by -1.618 while mass 2 moves a unit amplitude, which is also illustraed in Fig. 3. From these8Fig. 3. Mode Shapes of 2-DOF Example Problemobservations we conclude that mode shapes indicate how the system will deform when subjected toa harmonic excitation whose frequency is close to one of the natural frequencies (or vibration modes)of the system. Thus, mode-shape information is useful in designing structures subjected to harmonic9excitations. Examples of such systems include propelered airplanes, ships, motor vehicles, and manyother machinery equipment.6.3 Vibration analysis of 2-DOF Spring-Mass Problem by MATLABFor eigenvalue analysis of the 2-dof example problem in Matlab, we invoke the following routine:[X, D] = eig(K, M);where X is the eigenvector, D is the eigenvalues.%% 2 dof eigenvalue analysis%%stiffness matrixK = [2*2.618*pi^2 -2.618*pi^2;-2.618*pi^2 2.618*pi^2];% mass matrixM = [1 0;100 1];% call eigenvalue routine[X, D] = eig(K, M);% write eigenvector and eigenvalueslambda = diag(D);disp([’Eigenvalues of 2-dof system ’ num2str(lambda’) ]);disp(’ ’ );disp([’Eigenvectors of 2-dof system ’ num2str(X(1,:)) ] );disp([’ ’ num2str(X(2,:)) ] );% compute the frequenciesfreq = diag(D); % extract the two diagonal entries of D matrixdisp(’ ’ );freq = sqrt(freq);disp([’Frequencies of 2-dof system ’ num2str(freq’)]);11After executing the preceding Matlab code, we find the following results:Eigenvalues of 2-dof system 67.6464 9.86948Eigenvectors of 2-dof system 0.85065 0.52573-0.52573 0.85065Frequencies of 2-dof system 8.2247 3.1416Note that MATLAB prints out the highest mode first. Hence, the mode shapes are given byFor the

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