Dynamics of Machines - Part II - IFS.pdf

1.7.7 Programming in Matlab – Frequency Response Analysis
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% MACHINERY DYNAMICS LECTURES (72213) %
% IKS - DEPARTMENT OF CONTROL ENGINEERING DESIGN %
% DTU - TECHNICAL UNIVERSITY OF DENMARK %
% %
% Copenhagen, February 11th, 2000 %
% IFS %
% %
% 2 D.O.F. SYSTEMS - FRF (FREQUENCY RESPONSE FUNCTION) %
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%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
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M1=m1+m2+m5; %concentrated mass [Kg] |
M2=m3+m4+m6; %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]
% Coefficients of the Damping Matrix
D11= 2*0.01*(2*pi*5.0)*M1; %equivalent Damping [N/m]
D12= 0; %equivalent Damping [N/m]
D21= 0; %equivalent Damping [N/m]
D22= 2*0.01*(2*pi*1.0)*M2; %equivalent Damping [N/m]
%Mass Matrix
M= [M1 0; 0 M2];
%Damping Matrix
D=[D11 D12; D21 D22];
%Stiffness Matrix
K= [K11 K12; K21 K22];
%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)))/2/pi %natural frequency [rad/s]
w1=w(1) %first natural frequency [Hz]
w2=w(3) %second natural frequency [Hz]
wexp1=0.625 %measured natural frequency [Hz]
wexp2=4.405 %measured natural frequency [Hz]
dif1=(w1-wexp1)/wexp1 %error between calculated and measured freq.
dif2=(w2-wexp2)/wexp2 %error between calculated and measured freq.
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% FRF -- FREQUENCY RESPONSE FUNCTION
N = 800 ; % number of points for plotting (see HP-Analyzer)
N_factor = 100 ;
% Given Excitation Function acting on the the point 1
fo1 = 1 ; % [N] force amplitude acting on the point 1 of the beam
fo2 = 0 ; % [N] force amplitude acting on the point 2 of the beam
for i=1:N;
F1(i)= fo1;
F2(i)= fo2;
end;
% Calculation of the displacement, velocity and acceleration responses
for i=1:N;
w(i) = 2*pi*i/N_factor;
j = sqrt(-1);
AA = [(-M*w(i)*w(i)+K)+D*w(i)*j]; % Dynamical Stiffness Matrix
x = inv(AA)*[F1(i) F2(i)]’; % Displacement (complex)
x11(i) = x(1); % rail vibration displacement
x21(i) = x(2); % sleeves displacement
end;
% Given Excitation Function acting on the the point 2
fo1 = 0 ; % [N] force amplitude acting on the point 1 of the beam
fo2 = 1 ; % [N] force amplitude acting on the point 2 of the beam
for i=1:N;
F1(i)= fo1;
F2(i)= fo2;
end;
% Calculation of the displacement, velocity and acceleration responses
for i=1:N;
w(i) = 2*pi*i/N_factor;
j = sqrt(-1);
AA = [(-M*w(i)*w(i)+K)+D*w(i)*j]; % Dynamical Stiffness Matrix
x = inv(AA)*[F1(i) F2(i)]’; % Displacement (complex)
x12(i) = x(1); % rail vibration displacement
x22(i) = x(2); % sleeves displacement
end;
% Plotting the results
figure(1)
subplot(2,2,1),plot(w/2/pi,abs(x11))
title(’Excitation on Point 1 and Response of Point 1’)
xlabel(’Frequency [Hz]’)
ylabel(’(a) y11 [m/N]’)
grid on
subplot(2,2,2),plot(w/2/pi,abs(x12))
title(’Excitation on Point 2and Response of Point 1’)
xlabel(’Frequency [Hz]’)
ylabel(’(b) y12[m/N]’)
grid on
subplot(2,2,3),plot(w/2/pi,abs(x21))
title(’Excitation on Point 1 and Response of Point 2’)
xlabel(’Frequency [Hz]’)
ylabel(’(c) y21 [m/N]’)
grid on
subplot(2,2,4),plot(w/2/pi,abs(x22))
title(’Excitation on Point 2 and Response of Point 2’)
xlabel(’Frequency [Hz]’)
ylabel(’(d) y22 [m/N]’)
grid on