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5 Vibration of Linear Multiple-Degree-of-
Freedom Systems
F (t)1
F (t)2
F (t)3
x2
k1
c1
x1
x3
k2
k3
m1
m3
m2
=0
=0
x1x1x1x1x1
F (t)1m1
FC1
FK1FK3
FK2
FK3
FK2 m2
F (t)3m3
F (t)2
x1x1x1 x2
x3
Fig. 5.1: Multi-Degree-of-Freedom System with Free Body Diagram
5.1 Equation of Motion
The equation of motion can be derived by using the principles we have learned such as
Newtons/Eulers laws or Lagranges equation of motion. For a general linear system mdof
system we found that we can write in matrix form
FxNKxGCxM =++++ &&& (5.1.1)
with the matrices
M: Mass matrix (symmetric)T
MM =
C: Damping matrix (symmetric)T
CC=
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K: Stiffness matrix (symmetric)T
KK=
G : Gyroskopic matrix (skew-symmetric)T
GG =
N: Matrix of non-conservative forces (skew-symmetric)T
NN =
F: External forces
Note:A general matrixA can be decomposed into the symmetric part and the skew-symmetric part
by the following manipulation:
( ) ( )
4444 34444 21
4342143421
part
symmetricskew
T
symmetric
TAAAAA
++=2
1
2
1
In the standard case that we have no gyroscopic forces and no non-conservative displacement
dependent forces but only inertial forces, damping forces and elastic forces the last equation
reduces to
FxKxCxM =++ &&& (5.1.2)
Example:
The system shown in fig. 5.1, where the masses can slide without friction ( = 0), has thefollowing equation of motion.
=
++
+
+
)(
)(
)(
0
0
000
000
00
00
00
00
3
2
1
3
2
1
33
22
32321
3
2
11
3
2
1
3
2
1
tF
tF
tF
x
x
x
kk
kk
kkkkk
x
x
xc
x
x
x
m
m
m
&
&
&
&&
&&
&&
(5.1.3)
5.2 Influence of the Weight Forces and Static Equilibrium
The static equilibrium displacements are calculated by ( 0== statstat xx &&& ):
statstat FxK = (5.2.1)
which in the case of the example shown in fig. 5.2:
==
gm
gmFxK statstat
2
1
The dynamic problem for this example is
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( )
{
mequilibriustatictheabout
vibrationthedescribingmotiontheofpart
dynstat
forcesstatic
stat
xxx
tFFxKxM
+=
+=+321
&&
(5.2.2)
g
k2
k1
m1
m1
m2
m2
k1
k2
2
1
Fig. 5.2: Static equilibrium position of a two dof system
From the last equation also follows that
dyndyn xxxx &&&&&& ==
so that
( )tFFxxKxM statstatdyndyn +=++ )(&& (5.2.3)
and after rearrangement
( )tFxKFxKxM statstatdyndyn +=+=
44 344 21&&
0
(5.2.4)
)(tFxKxM dyndyn =+&& (5.2.5)
As can be seen the static forces and static displacements can be eliminated and the equation of
motion describes the dynamic process about the static equilibrium position.
Note:
In cases where the weight forces influences the dynamic behavior a simple elimination of the
static forces and displacements is not possible. In the example of an inverted pendulum shown
in fig. 5.3 the restoring moment is mglsin, where lis the length of the pendulum.
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k k
mg
m
Fig. 5.3: Case where the static force also influences the dynamics
5.3 Ground Excitation
Fig. 5.4 shows a mdof system
k2
k1
m1
m2
c1
x2
x1
xo
f2
f1
Fig. 5.4: 2dof System with excitation by ground motionx0
Without ground motionx0 = 0 the equation of motion is
=
+
+
+
2
1
2
1
22
221
2
11
2
1
2
1
00
0
0
0
f
f
x
x
kk
kkk
x
xc
x
x
m
m
&
&
&&
&&
(5.3.1)
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Now, if we include the ground motion, the differences (x1-x0) and the relative velocity
d(x1-x0)/dtdetermine the elastic and the damping force, respectively at the lower mass. This
can be expressed by addingx0 to the last equation in the following manner
)(0)(000
0
0
0
0
1
0
1
2
1
2
1
22
221
2
11
2
1
2
1
tx
c
tx
k
f
f
x
x
kk
kkk
x
xc
x
x
m
m
&&
&
&&
&&
+
+
=
+
+
+
(5.3.2)
The dynamic forcef0 of the vibrating system on the foundation is
(5.3.3)( ) ( 0110110 xxcxxkf && += )
Fig. 5.5: Example for ground motion excitation of a building structure (earthquake excitation)
Fig. 5.6: Excitation of a vehicle by rough surface
5.4 Free Undamped Vibrations of the Multiple-Degree-of-Freedom
System
5.4.1 Eigensolution, Natural Frequencies and Mode Shapes of the System
The equation of motion of the undamped system is
0=+ xKxM && (5.4.1)
To find the solution of the homogeneous differential equation, we make the harmonic solution
approach as in the sdof case. However, now we have to consider a distribution of the
individual amplitudes for each coordinate. This is done by introducing (the unknown) vector:
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ti
ti
ex
ex
=
=
&&(5.4.2)
Putting this into eqn. (5.4.1) yields
) 0 = MK (5.4.3)
This is a homogeneous equation with unknown scalarand vector . If we set we see
that this is a general matrix eigenvalue problem
2=1:
) 0= MK (5.4.4)
where is the eigenvalue and is the eigenvector. Because the dimension of the matrices isf
byfwe get f pairs of eigenvalues and eigenvectors:
fiii ...,2,1......... == (5.4.5)
i is the i-th natural circular frequency and
ithe i-th eigenvector which has the physical meaning of a vibration mode shape
The solution of the characteristic equation
0det = MK (5.4.6)
yields the eigenvalues and natural circular frequencies , respectively. The natural
frequencies are:
2=
2
iif = (5.4.7)
The natural frequencies are the resonant frequencies of the structure.
The eigenvectors can be normalized arbitrarily, because they only represent a vibration mode
shape, no absolute values. Commonly used normalizations are
1) Normalizei
so that 1=i
2) Normalizei
so that the maximum component is 1.
3) Normalizei
so that the modal mass (the generalized mass) is 1.
Generalized mass or modal mass:i
T
iiMM = (5.4.8)
1The well-known special eigenvalue problem has the form 0)( = xIA , where I is the identity matrix, x
the eigenvector and the eigenvalue.
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Generalized stiffness:i
T
iiKK = (5.4.9)
1:
=
==
i
Ti
Mif
KK iii(5.4.10)
The so-calledRayleigh ratio is
ii
T
iM
K
Tii = (5.4.11)
It allows the calculation of the frequency if the vectors are already known.
5.4.2 Modal Matrix, Orthogonality of the Mode Shape Vectors
If we order the natural frequencies so that
f ...321
and put the corresponding eigenvectors columnwise in a matrix, the so-called modal matrix,
we get
Modal Matrix: [ ]
==
ffff
f
f
f
...
............
...
...
,...,,
21
22221
11211
21(5.4.12)
The first subscript of the matrix elements denotes the no. of the vector component while the
second subscript characterizes the number of the eigenvector.
The eigenvectors are linearly independent and moreover they are orthogonal. This can be
shown by a pairi andj
( ) 02 =ii
MK and ( ) 02 =jj
MK (5.4.13)
Premultplying by the transposed eigenvector with indexj and i respectively:
02 =ii
Tj MK and 0
2 =jj
Ti MK (5.4.14)
If we take the transpose of the second equation:
02 = iTjTTj MK (5.4.15)
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and consider the symmetry of the matrices:T
MM = and TKK= and subtract this equation
( ) 02 =ij
Tj MK from the first equation (5.4.14) we get
( ) ( )022 =
i
T
jijM (5.4.16)
which means that if the eigenvalues are distinct ji for ji the second scalar product
expression must be equal to zero:
0=i
Tj M (5.4.17)
That means that the two distinct eigenvectors ji are orthogonal with respect to the massmatrix. For all combinations we can write:
SymbolKroneckerji
ji
for
for
MK
MM
ij
iiiji
T
j
iiji
T
j
==
=
=.............
0
1
(5.4.18)
or with the modal matrix:
{ }
{ }
==
==
ff
iiT
f
iT
M
M
M
MdiagK
M
M
M
MdiagM
...0
0
...0
0
22
11
2
1
(5.4.19)
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Example: Mode shapes and natural frequencies of a two storey structure
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5.4.3 Free Vibrations, Initial Conditions
The free motion of the undamped systemx(t)is a superposition of the modes vibrating with
the corresponding natural frequency:
( ) (=
+=f
iisiiciitAtAtx
1
sincos ) (5.4.20)
Each mode is weighted by a coefficientAci andAsi which depend on the initial displacement
shape and the velocities. In order to get these coefficients, we premultiply (5.4.20) by the
transposedj-th eigenvector:
( ) ( ) ( )tAtAMtAtAMtxM jsjjcjM
j
T
j
jifr
f
iisiicii
T
j
T
j
j
sincossincos
0.
1
+=+=
=
=
43421444444 3444444 21
(5.4.21)
All but one of the summation terms are equal to zero due to the orthogonality conditions.
With the initial conditions fort= 0 we can derive the coefficients:
( )
cjjT
jAMxM
xtx
t
=
==
=
0
00
0
j
T
jcj
M
xMA
0= (5.4.22)
( ) ( )=
+=f
i
isiiciiitAtAtx
1
cossin &
( )
sjjjT
jAMxM
vtx
t
=
==
=
0
00
0
&
& jj
T
jsj
M
vMA
0= (5.4.23)
which we have to calculate for modesj .
5.4.4 Rigid Body Modes
x1 x2
m1 m2k
Fig. 5.7: A two-dof oscillator which can perform rigid body motion
As learned earlier the constraints reduce the dofs of the rigid body motion. If the number of
constraints is not sufficient to suppress rigid body motion the system has also zero
eigenvalues. The number of zero eigenvalues corresponds directly to the number of rigid body
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modes. In the example shown in Fig. 5.7 the two masses which are connected with a spring
can move with a fixed distance as a rigid system. This mode is the rigid body mode, while the
vibration of the two masses is a deformation mode.
The equation of motion of this system is
=
+
0
0
0
0
2
1
2
1
2
1
x
x
kk
kk
x
x
m
m
&&
&&
The corresponding eigenvalue problem is
=
0
0
2
1
2
1
mkk
kmk
The eigenvalues follow from the determinant which is set equal to zero:
[ ] 0))((det 221 == kmkmk L
( ) 021212 =+ kmkmmm
Obviously, this quadratic equation has the solution
0211 == and
21
21222
mm
mmk +==
The corresponding (unnormalized) eigenvectors are
=
1
1
1
which is the rigid body mode: both masses have the same displacement, no potential energy is
stored in the spring and hence no vibration occurs. The second eigenvector, the deformation
mode is
=
2
12
1
m
m
which is a vibration of the two masses. Other examples for systems with rigid body modes are
shown in the following figures.
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Fig. 5.8: Examples for systems with torsional and transverse bending motion with rigid body
motion
Fig. 5.9: Flying airplane (Airbus A318) as a system with 6 rigid body modes and deformation
modes
Fig. 5.10: Commercial communication satellite system (EADS) with 6 rigid body modes and
deformation modes
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5.5 Forced Vibrations of the Undamped Oscillator under Harmonic
Excitation
k1
k2
m2
m1
k2
x1
x2f (t)2
f (t)1
2 2
Fig. 5.11: Example for a system under forced excitation
The equation of motion for this type of system is
( )tFxKxM =+&& (5.5.1)
For a harmonic excitation we can make an exponential approach to solve the problem as we
did with the sdof system
{
ti
vectorAmplitudecomplex
eFtF )( = (5.5.2)
We make a complex harmonic approach for the displacements with as the excitation
frequency:
tieXx = (5.5.3)
The acceleration vector is the second derivativetieXx = && (5.5.4)
Putting both into the equation of motion and eliminating the exp-function yields
( ) FXMK = (5.5.5)
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which is a complex linear equation system that can be solved by hand for a small number of
dofs or numerically. The formal solution is
( ) FMKX 1= (5.5.6)
which can be solved if determinant of the coefficient matrix : 0det MK
If the excitation frequency coincides with one of the natural frequencies i we getresonance of the system with infinitely large amplitudes (in the undamped case)
Resonance: ( ) iMK == 0det
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