The simplest multi-degree of freedom system is, of course, the two degree of freedom system, for which we will consider free vibration in this example. As-sume that the system is as pictured in Figure 14.1. Note that we need two coordinates to de…ne this system’s motion. Applying Newton’s second law to each mass results in two governing equations
m1
x::1+ (k1+ k2)x1 k2x2= 0 (14.1)
m2x::2 k2x1+ (k2+ k3) x2= 0: (14.2)
The motion of the masses are coupled. To solve these equations, it would be advantageous if we could …nd a coordinate system in which the equations were not coupled. Then, the problem would reduce to two single degree of freedom
Figure 14.1: Typical two degree of freedom free oscillator.
101
102 CHAPTER 14 FREE VIBRATION OF MDOF SYSTEMS systems, which we have already solved. The procedure of decoupling can be shown in matrix notation:
2
;: Let us denote the matrix multiplying x:: by M (the mass matrix) and the matrix multiplying x by K. The modal analysis procedure may then be summarized as follows:
1. Find the eigenvalues and matrix of eigenvectors for the matrix K M . Let this matrix of eigenvectors be U , and denote its columns (the eigenvectors) by fugi:
2. Normalize the eigenvectors with respect to the mass matrix through the products fugTi M fugi = 1; fugTi K fugi = !2i = i; where !i is the natural frequency of mode i:
3. From the solution of the motion in the decoupled modal coordinates, we arrive at: The main bene…t of this procedure (from a MATLAB standpoint) is that it is easily extended to n degrees of freedom by changing the limit of the summation in Equation 14.4 from 2 to n. Of course, we would also have n natural frequencies and eigenvectors, but the procedures for …nding them (i.e., solving the eigenvalue problem) and normalizing the eigenvectors with respect to the mass matrix are unchanged. The following program, modfree.m, is capable of solving n degree of freedom free vibration through the modal analysis procedure outlined above.
The following …ve …gures show the results of di¤erent masses, sti¤nesses, and initial conditions on the motion. Recall that this is free vibration; zero initial conditions will lead to zero motion. For the …rst three …gures, m1= 1; m2= 4;
k1= k3= 10; and k2= 2. Since the masses and sti¤nesses are not changing, the natural frequencies for these cases will be identical: !1= 3:480; and !2= 1:700:
For Figure 14.2, the initial displacement of each degree of freedom is 1. Notice how the second mode of vibration seems to show little evidence of the …rst mode; it looks like a single sinusoid, instead of a sum of sinusoids. This is due to the di¤erence in mass between the two degrees of freedom; remember that the second mass is four times the …rst.
As validation for this assertion, look at Figure 14.3. This …gure has zero initial velocity as well, but the initial displacements are now 3 and –2, respec-tively. Again, the second mode is vibrating along a single sinusoid; since the
103 mass ratio is unchanged, the mode behavior is as well. If we eliminated an initial displacement from the second degree of freedom, we would expect the system to vibrate in both modes, at least in the second degree of freedom. As Figure 14.4 shows, this is the case. Notice that the …rst mode only vibrates in one frequency this time; this is because the displacement of the second mass and its velocity are too small to impart any appreciable force on the …rst mass.
As a …nal proof of the e¤ects of the mass on the mode shapes, look at Figure 14.5. This …gure was created by interchanging the masses; the mass of 4 units is in the …rst degree of freedom, and the mass of 1 unit is in the second. Notice how the …rst mode shape is now a single sinusoid, and the second is a sum. A peculiarity to this case is that the system is symmetric; that is, the response of the …rst mode in Figure 14.2 matches that of the second mode in Figure 14.5, and vice versa. This is due to the symmetric sti¤ness matrix. If we change the value of either k1 or k3, the symmetric behavior would no longer exist.
Figure 14.6 shows the e¤ects of adding an initial velocity to the system.
As expected from a single degree of freedom system, the initial velocity a¤ects the slope of the response curve at the starting point, but has no e¤ect on the qualitative aspects of the mode shapes, or on the natural frequencies. Also, the maximum amplitude of vibration is increased for both modes.
Figure 14.7 shows the result of changing the mass matrix to m = 2 43 0
0 9 3 5 : As a result, !1= 2:014; and !2= 1:131: These reduced natural frequencies are re‡ected in the …gure; when compared to Figure 14.2, the curves of Figure 14.7 have a lower frequency. The curves look a little more “spread out” than those in Figure 14.2. However, the second mode shape is still a single sinusoid, not a sum like the …rst shape. Note that both initial displacements were again 1.
Finally, Figure 14.8 shows the result of changing k1 and k3 to 1 unit, and k2 to 3 units. The natural frequencies become !1 = 2:150; and !2 = 0:6154:
Again, the di¤erence between Figure 14.2 and Figure 14.8 is readily apparent.
A peculiarity that develops is that the …rst mass’s vibration comes in more distinct peaks; one peak reaches 1, and the next is about 0.7. In Figure 14.2, the di¤erence is 1 to less than 0.5.
104 CHAPTER 14 FREE VIBRATION OF MDOF SYSTEMS Program 13-1: modfree.m
% This code solves n-degree of freedom free
% vibration using the modal analysis techniques
% of Section 7.6.1.
% clear;
M=input(’Enter the mass matrix: ’);
[n,o]=size(M);
if n~=o then
error(’M matrix must be square!’);
end
K=input(’Enter the sti¤ness matrix: ’);
[n,o]=size(K);
if n~=o then
error(’K matrix must be square!’);
end qu=0;
[u,l]=eig(K,M);
% Using “eig” in this way allows us to subtract M*w^2
% from K, instead of I*w^2 (where I is the n by n identity
% matrix).
% The output from “eig” gives unit-length eigenvectors.
% We need to scale them with respect to M.
% for s=1:n
alfa=sqrt(u(:,s)’*M*u(:,s));
u(:,s)=u(:,s)/alfa;
end
x0=input(’Enter the initial displacement column vector: ’);
xd0=input(’Enter the initial velocity column vector: ’);
tf=input(’Enter the …nal time: ’);
t=0:0.1:tf;
q=tf/0.1;
x=zeros(size(n,q));
% Applying Equation 7.183.
%
% Plotting the modes in a subplot format.
% Note that, for more than 3 or 4 degrees
% of freedom, the plots will become nearly
% unreadable.
105
% for r=1:n subplot(n,1,r) plot(t,x(r,:))
xlabel(’Time, seconds’);
ylabel([’Response x’,num2str(r)]);
end
106 CHAPTER 14 FREE VIBRATION OF MDOF SYSTEMS
0 5 10 15 20 25 30
-1 -0.5 0 0.5 1
Displacement, m
Response x1
0 5 10 15 20 25 30
-2 -1 0 1 2
Time, seconds
Displacement, m
Response x2
Figure 14.2: Response to initial displacement vector of [1 1].
107
0 5 10 15 20 25 30
-4 -2 0 2 4
Displacement, m
Response x1
0 5 10 15 20 25 30
-2 -1 0 1 2
Time, seconds
Displacement, m
Response x2
Figure 14.3: Response after changing initial displacement to [3 –2].
108 CHAPTER 14 FREE VIBRATION OF MDOF SYSTEMS
0 5 10 15 20 25 30
-1 -0.5 0 0.5 1
Displacement, m
Response x1
0 5 10 15 20 25 30
-0.2 -0.1 0 0.1 0.2
Time, seconds
Displacement, m
Response x2
Figure 14.4: Response after changing initial displacement to [1 0].
109
0 5 10 15 20 25 30
-1 -0.5 0 0.5 1
Time, seconds
Displacement, m
Response x2
0 5 10 15 20 25 30
-2 -1 0 1 2
Displacement, m
Response x1
Figure 14.5: Response after interchanging system masses.
110 CHAPTER 14 FREE VIBRATION OF MDOF SYSTEMS
0 5 10 15 20 25 30
-2 -1 0 1 2
Displacement, m
Response x1
0 5 10 15 20 25 30
-2 -1 0 1 2
Time, seconds
Displacement, m
Response x2
Figure 14.6: Response with initial velocity [4 2] and initial displacement [1 1].
111
0 5 10 15 20 25 30
-1 -0.5 0 0.5 1
Displacement, m
Response x1
0 5 10 15 20 25 30
-2 -1 0 1 2
Time, seconds
Displacement, m
Response x2
Figure 14.7: Response with increased system masses.
112 CHAPTER 14 FREE VIBRATION OF MDOF SYSTEMS
0 5 10 15 20 25 30
-1 -0.5 0 0.5 1
Displacement, m
Response x1
0 5 10 15 20 25 30
-2 -1 0 1 2
Time, seconds
Displacement, m
Response x2
Figure 14.8: Response with reduced system sti¤nesses.