Dynamics of Machines - Part II - IFS.pdf

1.7.4 Modal Analysis using Matlab eig-function [u, w] = eig(−B, A)
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% MACHINERY DYNAMICS LECTURES (41614) %
% MEK - DEPARTMENT OF MECHANICAL ENGINEERING %
% DTU - TECHNICAL UNIVERSITY OF DENMARK %
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% Copenhagen, February 11th, 2000 %
% IFS %
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% MODAL ANALYSIS %
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% 2 D.O.F. SYSTEMS - MODAL ANALYSIS - 3 EXPERIMENTAL CASES %
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clear all; close all;
%Concentred Masses
m1= 0.191; %[Kg]
m2= 0.191; %[Kg]
m3= 0.191; %[Kg]
m4= 0.191; %[Kg]
m5= 0.191; %[Kg]
m6= 0.191; %[Kg]
%Elastic Properties of the Beam of 600 mm
E= 2e11; %elasticity modulus [N/m^2]
b= 0.030 ; %width [m]
h= 0.0012 ; %thickness [m]
I= (b*h^3)/12; %area moment of inertia [m^4]
% (1.CASE) Data for the mass-spring system
%__________________________________________________
M1=m1; %concentrated mass [Kg] |
M2=m2; %concentrated mass [Kg] |
L1= 0.310; %length for positioning M1 [m] |
L2= 0.610; %length for positioning M2 [m] |
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% Coefficients of the Stiffness Matrix
LL=(L1-4*L2)*(L1-L2)^2;
K11= -12*(E*I/LL)*L2^3/L1^3; %equivalent Stiffness [N/m]
K12= -6*(E*I/LL)*(L1-3*L2)/L1; %equivalent Stiffness [N/m]
K21= -6*(E*I/LL)*(L1-3*L2)/L1; %equivalent Stiffness [N/m]
K22= -12*(E*I/LL); %equivalent Stiffness [N/m]
%Mass Matrix
M= [M1 0; 0 M2];
%Stiffness Matrix
K= [K11 K12; K21 K22];
%Damping Matrix
D= [0 0; 0 0];
%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)
w1=abs(imag(w_vector(indice(1))))/2/pi % First natural frequency
w2=abs(imag(w_vector(indice(3))))/2/pi % Second natural frequency
mass_position(1) = 0; mass_position(2) = L1; mass_position(3) =
L2;
% first mode shape
uu1(1)=0; uu1(2)=u(1,indice(1)); uu1(3)=u(2,indice(1)); figure(1)
plot(uu1,mass_position,’r*-.’,-uu1,mass_position,’r*-.’,’LineWidth’,1.5)
grid f1=num2str(w1); set(gca,’FontSize’,12,’FontAngle’,’oblique’);
title([’First Mode Shape - Freq.: ’,f1,’ Hz’])
37
% second mode shape
uu2(1)=0; uu2(2)=u(1,indice(3)); uu2(3)=u(2,indice(3)); figure(2)
plot(uu2,mass_position,’r*-.’,-uu2,mass_position,’r*-.’,’LineWidth’,1.5)
grid f2=num2str(w2); set(gca,’FontSize’,12,’FontAngle’,’oblique’);
title([’Second Mode Shape - Freq.: ’,f2,’ Hz’]) pause;
% (2.CASE) Increasing the Mass Values
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M1=m1+m2; %concentrated mass [Kg] |
M2=m3+m4; %concentrated mass [Kg] |
L1= 0.310; %length for positioning M1 [m] |
L2= 0.610; %length for positioning M2 [m] |
%__________________________________________________|
% Coefficients of the Stiffness Matrix
LL=(L1-4*L2)*(L1-L2)^2;
K11= -12*(E*I/LL)*L2^3/L1^3; %equivalent Stiffness [N/m]
K12= -6*(E*I/LL)*(L1-3*L2)/L1; %equivalent Stiffness [N/m]
K21= -6*(E*I/LL)*(L1-3*L2)/L1; %equivalent Stiffness [N/m]
K22= -12*(E*I/LL); %equivalent Stiffness [N/m]
%Mass Matrix
M= [M1 0; 0 M2];
%Stiffness Matrix
K= [K11 K12; K21 K22];
%Damping Matrix
D= [0 0; 0 0];
%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)
w1=abs(imag(w_vector(indice(1))))/2/pi % First natural frequency
w2=abs(imag(w_vector(indice(3))))/2/pi % Second natural frequency
mass_position(1) = 0; mass_position(2) = L1; mass_position(3) =
L2;
% first mode shape
uu1(1)=0; uu1(2)=u(1,indice(1)); uu1(3)=u(2,indice(1)); figure(3)
plot(uu1,mass_position,’b*-.’,-uu1,mass_position,’b*-.’,’LineWidth’,1.5)
grid f1=num2str(w1); set(gca,’FontSize’,12,’FontAngle’,’oblique’);
title([’First Mode Shape - Freq.: ’,f1,’ Hz’])
% second mode shape
uu2(1)=0; uu2(2)=u(1,indice(3)); uu2(3)=u(2,indice(3)); figure(4)
plot(uu2,mass_position,’b*-.’,-uu2,mass_position,’b*-.’,’LineWidth’,1.5)
grid f2=num2str(w2); set(gca,’FontSize’,12,’FontAngle’,’oblique’);
title([’Second Mode Shape - Freq.: ’,f2,’ Hz’]) pause;
% (3.CASE) Increasing the Mass Values
%__________________________________________________
M1=m1+m2+m3; %concentrated mass [Kg] |
M2=m4+m5+m6; %concentrated mass [Kg] |
L1= 0.310; %length for positioning M1 [m] |
L2= 0.610; %length for positioning M2 [m] |
%__________________________________________________|
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