Session 2: MDOF systems
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1 BRUFACE Vibrations and Acoustics MA Academic year 7-8 Cédric Dumoulin Arnaud Deraemaeker Session : MDOF systems Exercise : Multiple DOFs System Consider the following three degrees-of-freedom model: k c m k c m k c m x x x f Figure -DOFs system A) Write the equations of motion in the time domain in a matrix form. m x c + c c x k + k k m x m + c c c x c c + k k k k k x x x x x = f B) For the numerical values m = kg, k = k = 6 N/m, c = c =. Ns/m, compute the eigenfrequencies and the modeshapes of the conservative system. Represent the modeshapes, and check the orthogonality conditions.. Solve the problem: ( K ω i M ) ψ i = The orthogonality conditions are given by ψ i Kψ T i = ω i ψ i Mψ T i = ω i µ i
2 clear all; close all;clc; % Build matrices of the system m=; k=6; k=6; c=.; c=.; K=[k+k k ; k *k k ; k k]; M=[m ; m ; m]; C=[c+c c ; c *c c ; c c]; % compute and represent modeshapes [V,D]=eig(K,M); f=/(*pi)*sqrt(diag(d)); % frequencies in Hz figure; subplot(,,); barh(v(:,)); title(['mode / ' numstr(f()) ' Hz']) subplot(,,); barh(v(:,)); title(['mode / ' numstr(f()) ' Hz']) subplot(,,); barh(v(:,)); title(['mode / ' numstr(f()) ' Hz']) % Check orthogonality MU=diag(V'*M*V) % mass normalized WMU=diag(V'*K*V) disp(mu) disp(wmu) disp([wmu./mu diag(d)]); mode /.8 Hz mode /.7985 Hz.5.5 mode /.47 Hz.5.5 C) Compute the mode shapes and the eigenfrequencies when k =. Represent the mode shapes. Comment %% now with k=; k=; Kd=[k+k k ; k *k k ; k k]; M=[m ; m ; m]; C=[c+c c ; c *c c ; c c];
3 % compute and represent modeshapes [Vd,Dd]=eig(Kd,M); fd=/(*pi)*sqrt(diag(dd)); % frequencies in Htz figure; subplot(,,); barh(vd(:,)); title(['mode / ' numstr(fd()) '... Hz']) subplot(,,); barh(vd(:,));title(['mode / ' numstr(fd()) '... Hz']) subplot(,,); barh(vd(:,)); title(['mode / ' numstr(fd()) '... Hz']) mode / 6.65e 9 Hz mode /.666 Hz.5.5 mode /.7 Hz.5.5 The first mode shape has a frequency of Hz. This is because when k =, the three masses are free to float. The mode shape corresponds to a translation of all masses in phase and with the same amplitude. This is called a rigid body mode, because there is no strain energy in the system (the springs are not compressed). D) For the value of k = k, compute the impulse response for x by projecting the equations of motion in the modal basis. Represent the Bode diagram for the same coordinate x, and for the acceleration ẍ. Is the modal damping hypothesis valid. Multiply by a factor 5 the damping coefficient of mode and plot the Bode diagram for x on the same graph as with the initial value of the damping. Comment. The projection in the modal basis results in three indepent equations of the form µ i ( z i + ξ i ω i ż i + ω i z i ) = F i
4 The force F i exciting each mode is given by (the force f is only acting on x ): T V V V F = V V V V V V V = V V Therefore we see that: F i = V i For one mode, the impulse response corresponds to the impulse response of a DOF system with h i (t) = e ξ iω i t µ i ω di sin(ω di t) ω di = ω i ξ i The impulse response at point x is therefore given by (one must multiply the impulse response both by F i and the amplitude of the respective mode at x ) h(t) = V i F i h i (t) = i= i= V e ξ iω i t i sin(ω di t) µ i ω di Similarly, the response in the frequency domain is given by: H (ω) = i= V i ω ω i + jξ i ωω i Here is the corresponding Matlab code: %% Projection in the modal basis mui=diag(v'*m*v); wi=sqrt(diag(v'*k*v))./mui; xi=(./(*mui.*wi)).*diag(v'*c*v); wd=wi.*(sqrt( xi.^)); F=V'*[;;]; %% Impulse response t=linspace(,,); u=zeros(,length(t)); for i=: u=u+v(,i)*(f(i)*exp( xi(i)*wi(i)*t)/(mui(i)*wd(i))).*sin(wd(i)*t); figure; set(gca,'fontsize',5) plot(t,u); xlabel('time (sec)'); ylabel('amplitude (m)') title('impulse response x_') 4
5 .4 impulse response x.. Amplitude (m) time (sec) %% Bode diagram w=linspace(,,); U=zeros(,length(w)); for i=: U=U+ (F(i)*V(,i))./(w.^ wi(i)^ + *j*xi(i)*wi(i)*w); figure; set(gca,'fontsize',5) plot(w,*log(abs(u)),'k','linewidth',); hold on xlabel('freq(rad/s)'); ylabel('ampl (db)') xi()=xi()*5; U=zeros(,length(w)); for i=: U=U+ (F(i)*V(,i))./(w.^ wi(i)^ + *j*xi(i)*wi(i)*w); plot(w,*log(abs(u)),'r','linewidth',) leg('initial','5 times damp mode ') title('bode diagram displacement x_') 4 Bode diagram displacement x initial 5 times damp mode Ampl (db) Freq(rad/s) 5
6 % Acceleration A=( w.^).*u; figure; set(gca,'fontsize',5) plot(w,*log(abs(a)),'k','linewidth',); xlabel('freq(rad/s)'); ylabel('ampl (db)') title('bode diagram acceleration x_') 4 Bode diagram acceleration x Ampl (db) Freq(rad/s) The modal damping hypothesis is valid because the damping matrix is proportional to the stiffness matrix. When multiplying by 5 the damping factor of mode, the frequency response function at x is only affected around the second natural frequency of the system. E) Consider the case when c =. Is the modal damping hypothesis still verified? Draw the Bode diagram for x using the full system of equations (solve frequency by frequency). Compare with the modal approach in which the coupling is neglected. Comment c=; C=[c+c c ; c *c c ; c c]; % Direct reference method U=zeros(,length(w)); for i=:length(w) U(:,i)=(K w(i)^*m + j*w(i)*c)\[ ]'; figure; set(gca,'fontsize',5) plot(w,*log(abs(u(,:))),'k','linewidth',); hold on xlabel('freq(rad/s)'); ylabel('ampl (db)') % Modal damping hypothesis xi=(./(*mui.*wi)).*diag(v'*c*v); U=zeros(,length(w)); for i=: U=U+ (F(i)*V(,i))./(w.^ wi(i)^ + *j*xi(i)*wi(i)*w); plot(w,*log(abs(u)),'r','linewidth',) leg('direct method','modal approach') title('bode diagram displacement x_') 6
7 4 Bode diagram displacement x direct method modal approach Ampl (db) Freq(rad/s) The damping matrix is not proportional to the stiffness matrix anymore. Nevertheless, as the damping is quite small (only a few percents for each mode, see values of ξ i in Matlab), the approximation is very good and there is almost no difference between the direct exact method and the decoupled modal approach. Note: Use the eig function to compute the eigenfrequencies and mode shapes. 7
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