ASEN 5022 - Spring 2006

Dynamics of Aerospace Structures

Lecture 02: 19 January 2006

Vibration of Two-DOF Systems

1. Equations of Motion

Consider a two-DOF model as shown in Fig. 1. The free-body diagrams for massesM1 and m2 are also shown in Fig. 1. Note that the inertia forces for both massesare associated with minus sign, for the inertia forces can be considered as resistingforces (cf.,

∑f − ma = 0). Summing the forces acting on each mass, the equations

of motion for the coupled two-mass-spring-damper system can be written as

For M1 :∑

f = f1(t) − M1 x1 + {k2(x2 − x1) + c2(x2 − x1) − K1x1} = 0

For m2 :∑

f = f2(t) − m2 x2 − {k2(x2 − x1) + c2(x2 − x1)} = 0

⇓M1 x1 = f1(t) − K1 x1 + k2 (x2 − x1) + c2 (x2 − x1)

m2 x2 = −k2 (x2 − x1) − c2 (x2 − x1)

(1)

TwoDOFSpring-M

ass-DamperModel

k

f

C

x 1

m2

1

2

2

x 2

1M

K

1

FreeBodyDiagram

forTwoDOFSpring-M

ass-DamperModel

1M

f 1(t

)K

1x1

m2

M1

-()

m2x2

-()

c 2(x

2x 1

)-

k2 c 2(x

2x 1

)-

k2(

x2x

1)

-

(x

2x1)

-

f 2

f 2

x1

Fig. 1 Two DOF Simplified Vertical Motion Model

As the preceding equation involves two displacements, x1 and x2, its general solutioninvolves complex matrix differential algebra. For design considerations, however,important insight can be gained by considering the special case of forcing functiongiven by

f1(t) = F1e jω t , f2(t) = F2e jω t (2)

so that the solution assumes the form of

x1(t) = X1e jω t

x2(t) = X2e jω t(3)

Substituting (2) and (3) into (1), we obtain

−ω2 M1 X1 = F1 − K1 X1 + k2 (X2 − X1) + jωc2 (X2 − X1)

−ω2m2 X2 = F2 − k2 (X2 − X1) − jωc2 (X2 − X1)(4)

In order to solve for X1 and X2, let’s rearrange the preceding equation to read

(−ω2 M1 + jωc2 + K1 + k2) X1 = F1 + (k2 + jωc2) X2

(−ω2m2 + jωc2 + k2) X2 = F2 + (k2 + jωc2)X1

(5)

Solving for X1 and X2 we obtain

X1 = H2(ω)F1 + H12(ω)F2

H1(ω) H2(ω) − H 212(ω)

X2 = H1(ω)F2 + H12(ω)F1

H1(ω)H2(ω) − H 212(ω)

H1(ω) = (−ω2 M1 + jωc2 + K1 + k2)

H2(ω) = (−ω2m2 + jωc2 + k2)

H12(ω) = (k2 + jωc2)

(6)

Although (6) appears to be very complex, several simplifications are possible to aidvibration designers. This is studied below.

2. What Are Vibration Modes and Mode Shape?

It turns out that the motions of X1 and X2 given by (6) are not randomly independentas the solution may suggest. They are interlinked by the property called mode shapes.Understanding the physical properties of mode shapes is important if one is tasked todesign structures subject to vibrations.

As a motivation, let us consider the following special 2-DOF case:

m1 = m2 = 1, k1 = k2 = 2.618π2, c2 = 0, f1(t) = 0 (7)

For this model, we apply two different excitations:

f2(t) ={ 0.2 × sin (0.98 ∗ π t)

sin (0.98 ∗ 2.618π t)(8)

Figure 2 illustrates the two responses subject to the two excitations specified in (8).Observe that, for the case of f2(t) = 0.2 × sin (0.98 ∗π t), both mass m1 and mass m2

are moving in phase, that is, they move in the same direction in time. However, whenthe system is subjected to f2(t) = sin (0.98 ∗ 2.618π t), mass m1 and mass m2 aremoving out phase, that is, they move in the opposite directions in time. In other words,depending on the excitation frequency, the motions of the two masses are drasticallydifferent. To understand this strange phenomena, one has to understand the roles ofmodes and mode shapes. To this end, let us recast (1) in a matrix form with c = 0:

[m1 00 m2

] {x1

x2

}+

[k1 + k2 −k2

−k2 k2

] {x1

x2

}=

{f1(t)f2(t)

}(9)

The characteristic equation of the above coupled 2-dof differential equation(9) can beobtained as follows. First, we assume the solution of their homogeneous equations inthe form

{x1

x2

}=

{x1

x2

}e jωt (10)

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2

0

0.5

1

1.5

2

2.5

MassPositionsforMode1

Time

2-dof system response due to harmonic input

at mass 2 with omega= 0.98*omega1

Position of

Mass 1

m1

m2Position of

Mass 2

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2

0

0.5

1

1.5

2

2.5

MassPositionsforMode2

2-dof system response due to harmonic input

at mass 2 with omega= 0.98*omega2

Time

Position of

Mass 1

m1

m2

Position of

Mass 2

k1

k2

Figure 2 Motions of two masses under two different excitations

Second, substituting this into (9) with f1 = f2 = 0 yields:

[−ω2

[m1 00 m2

]+

[k1 + k2 −k2

−k2 k2

]]

{x1

x2

}= 0 (11)

Hence, the characteristic equation is obtained by requiring that the above equation hasa nontrivial solution:

det

[k1 + k2 − ω2m1 −k2

−k2 k2 − ω2m2

]= 0 ⇒ ω4−(

k1 + k2

m1+ k2

m2)ω2+ k1k2

m1m2= 0

(12)

Note that, with m1 = m2 = 1, k1 = k2 = 2.618π2, the two roots of the abovecharacteristic equation are given by

ωn1 = π, ωn2 = 2.618π (13)

These two values are called characteristic values whose square roots are called thenatural frequencies or vibration modes of the system.

The corresponding eigenvectors can be computed from the second row of (11):

−ω2 x2 − k2 x1 + k2 x2 = 0 ⇒ x1

x2= k2 − ω2

k2= 1 − ω2

k2(14)

Note that for the two modes computed in (13), we have two different expressions:

For ω = π,x1

x2= 1 − π2

2.618π2= 0.618

For ω = 2.618π,x1

x2= 1 − 2.6182π2

2.618π2= −1.618

(15)

The ratios of these eigenvector sets are plotted in Figure 3.

Modal Amplitude for Mode 1

MassPoint

Fixed

end

1

2

0.6 1.00.0

Mode Shape for Mode 1

0 1.0-1.6

Modal Amplitude for Mode 2

MassPoint

1

2

Mode Shape for Mode 2

Fixed

end

Figure 3 Mode Shapes of 2-DOF Example Problem

Observe from Figure 3 that for the case of the first mode when mass 1 moves in thesame direction with its amplitude of 0.616 while mass 2 moves a unit amplitude. Inother words, the two masses move in phase as illustrated in Figure 2. As for the secondmode, mass 1 moves in the opposite direction by -1.618 while mass 2 moves a unitamplitude, which is also illustraed in Figure 2. From these observations we concludethat mode shapes indicate how the system will deform when subjected to a harmonicexcitation whose frequency is close to one of the natural frequencies (or vibrationmodes) of the system. Thus, mode-shape information is useful in designing structuressubjected to harmonic excitations. Examples of such systems include propelleredairplanes, ships, motor vehicles, and many other machinery equipment.

3. Vibration analysis of 2-DOF Spring-Mass Problem by MATLAB

In Matlab we invoke the following routine:

[X, D] = eig(K, M);where X is the eigenvector, D is the eigenvalues.

For example, the eigenproblem of the 2-DOF system studied in the preceding sectionscan now be analyzed by MATLAB with

K =[

2 ∗ 2.618π2 −2.618π2

−2.618π2 2.618π2

], M =

[1 00 1

](16)

Upon using the following MATLAB code, we find the following results:

%% 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;0 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 frequencies

freq = diag(D); % extract the two diagonal entries of D matrix

disp(’ ’ );freq = sqrt(freq);disp([’Frequencies of 2-dof system ’ num2str(freq’)]);

we find the following results:

Eigenvalues of 2-dof system 67.6464 9.86948

Eigenvectors of 2-dof system 0.85065 0.52573-0.52573 0.85065

Frequencies of 2-dof system 8.2247 3.1416

Note that MATLAB prints out the highest mode first. Hence, the mode shapes aregiven by

For the second mode with ω2 = 2.618π = 8.2247 : X2 ={

x21

x22

}=

{0.85065−0.5257

}

For the first mode with ω1 = π = 3.14157 : X1 ={

x11

x12

}=

{0.5257310.850650

}(17)

The mode shapes computed by MATLAB is normalized so that their vectorial lengthis unity. If the modal amplitude at mass 2, x22, for the second mode shape is scaledto be unity from MATLAB computed value x22 = −0.52573, then we would have

x21 = −1.618 as given by (15). The same is true for the modal amplitude for mode 1at mass 1 and mass 2 points. Finally, one may express X2 as

For the second mode with ω2 = 2.618π = 8.22474 : X2 ={

x21

x22

}=

{ −0.850650.52573

}(18)

which indicates that X2 and −X2 are equally valid representation.

4. Den Hartog-Ormondroyd Oscillators

An important application of the two-dof vibration model is to study shock isolationdesign of a suspension system as shown Figure 1. If Figure 1 is rearranged as shownin Figure 4, the same model can be used as a resonator model whose generic charac-teristics can be used for the design of accelerometers and more recently as filters ofmicroelectro-mechanical systems (MEMS). In either case, the governing equation (1)is applicable.

Let us parameetrize the model as

ω1 =√

K1/M1, ω2 =√

k2/m2, c = 2ζm2ω2, µ = m2/M1 (19)

and normalize the frequence response X1(ω) given in (6) by its static displacementwith F2 = 0:

Two DOF Spring-Mass-Damper Suspension Modelm

2

x 1C2

x 2

1M

1

f 2f

k2/2

1K /2

Figure 4: 2-DOF Suspension Model

xst (1) = F1/K2 (20)

such that from (7) we have

H1 = X1(ω)

xst (1)= K1 H22(ω)

H11(ω) H22(ω) − H 212(ω)

(21)

Dividing both the nominator and the denominator by M1 m2 and utilizing the pa-rameters introduced in (20), the frequency response function H1(ω) at mass 1 can beexpressed as

H1 = H22(ω)

H11(ω) H22(ω) − µH 212(ω)

H11(ω) = −ω2 + j2ζµω2ω + ω21 + µω2

2

H22(ω) = −ω2 + j2ζω2ω + ω22

H12(ω) = ( j2ζω2ω + ω22)

(22)

10110-1

100

101

Frequency (Hz)

Amplitude

Invariant points

Frequency Response Functions for Different Damping Ratios

of a 2-DOF Suspension Model

Figure 5: Frequency Response Function at Mass 1

Figure 5 illustrates the frequency response function (FRF) for different damping ratioswith

ω1 = ω2 = 10 H z, µ = 0.1 (23)

Note that there are two points that all FRF curves pass through two invariant points.It is these invariant points that were first discovered by den Hartog and Ormondroydin 1928 who subsequently utilized their properties for improved design of mechanicalshock attenuation.

In their design study by using Figure 1, den Hartog and Ormondroyd observed thatif the magnitudes of the two invariant points are made to be the same by varying themass ratio and the damping ratio, a near optimum design goal can be achieved.

For the design of accelerometers using Figure 4, the objective is to maximize theamplitude of the frequency response of mass 2 in order to decrease the signal to noiseratio while realizing a flat plateau of frequency interval of interest.

For the design of resonators using Figure 4, in addition to maximizing the amplitudeof FRF at mass 2, the gap between the two invariant points should be minimized.

MATLAB code that produced Figure 5 islisted below.

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

% hartog_oscillator.m

% by k c park 15 january 2002

clear

% Define some useful numbers:

dtr=pi/180;

Hz2rps=2*pi; rps2Hz=1/2/pi; % Conversions to/from Hz and rad/s

% Natural frequencies of individual masses: 10Hz

w1=10*Hz2rps;

w2=w1;

mu = 0.1;

for zeta =0:0.05:0.8 %zeta is the damping ratio

% Define a system in transfer function form

a0=1; a1 =2*zeta*w2; a2 = w2^2;

b0=1;

b1 =2*zeta*(1+mu)*w2;

b2 = w1^2+w2^2 + mu*(1+4*zeta^2)*w2^2 - 4*mu*zeta^2*w2^2;

b3 = 2*zeta*(mu*w2^3 + w2*(w1^2+mu*w2^2)) - 4*zeta*mu*w2^3;

b4 = w2^2*(w1^2+mu*w2^2)- mu*w2^4;

sys = tf ( [a0 a1 a2], [b0 b1 b2 b3 b4]);

% Pick a range of frequencies to look at from 0 Hz to 100 Hz

wf= logspace(0.5,1.5,400)*Hz2rps;

% Find the frequency response

H=freqresp(sys,wf); % This ends up being a 3 dimensional array!

H=reshape(H,size(wf)); % this makes it the same size as wf

% Plot the amplitude as a function of frequency in Hz

loglog(wf*rps2Hz,w1^2*abs(H)), grid

xlabel(’Frequency (Hz)’)

ylabel(’Amplitude’)

hold on;

end

axis([5 30 0.1 30]);

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

References

1. R. E. D. Bishop, Vibration, Cambridge University Press, 1979.

2. J. P. Den Hartog, Mechanical Vibrations, Dover Pub., 1984.

3. Maurice Roseau, Vibrations in Mechanical Systems, Springer-Verlag, 1984.