multiple degrees of freedom
TRANSCRIPT
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CHAPTER V5.1 Multi-Degree-of-Freedom Systems
The discrete and lumped mass systems have more than one coordinate todescribe their motion. In general, have fnite numbers o degrees o reedom.
Such systems are called multi-degree o reedom systems. For example,
there is one Euation o motion or each degree o reedom! i generali"ed
coordinates are used! there is one generali"ed coordinate or each degree o
reedom. The Euations o motion can be obtained rom #e$ton%s second la$
o motion or by using the in&uence coe'cients. (o$ever, it is oten more
convenient to derive the Euations o motion o a multi-degree o reedom
system by using )agrange%s Euations. There are n natural reuencies, each
associated $ith its o$n mode shape, or a system having n degrees o
reedom. The method o determining the natural reuencies rom the
characteristic Euation obtained by euating the determinant to "ero also
applies to these systems. (o$ever, as the number o degrees o reedom
increases, the solution o the characteristic Euation becomes more
complex. The mode shapes exhibit a property *no$n as orthogonality, $hich
oten enables us to simpliy the analysis o multidegree o reedom systems.
Some examples o systems having multi-degrees o reedom are sho$n.
+ particular structure can be analy"ed by considering the euivalent system
sho$n belo$. To isolate a structure rom the vibration generated by a
machine, the machine is mounted on a large bloc*. The bloc* is supported on
springs as sho$n in Figure . Figure sho$s a multiple degrees o
reedom simplifed model o a motor vehicle. +bridge structure can modeled
by a simply supported beam $ith lumped masses at di/erent stations as
sho$n in Figure . The aircrat $ing can also be modeled as cantilevered
beam $ith lumped masses that replaces the pylons, control suraces, uel
tan*s, engines, and $ing stations, see Figure 0.
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Figure 1. Three 2egree o Freedom 3odels o Structures
Figure . 3odel o )arge 4loc* Foundation o the 3achine
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Figure . + simplifed model or a motor vehicle
Figure . 3odeling o 4ridge structure
Figure 0. Simplifed 3odel o +ircrat 5ing $ith )umped 3asses
5.2 Derii!g t"e Di#ere!ti$l E.o.M of Multi%le D.o.F
Systems
Example 1
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The t$o degree o reedom system sho$n consists o a pulley, a mass m, t$o
springs, a dashpot, and a cable connecting them as sho$n. The pulley%s
centroidal mass moment o inertia is I, and the dashpot has a coe'cient o c.
There is a su'cient riction bet$een the cable and the pulley to prevent the
cable rom slipping. 2etermine the di/erential Euations o motion o the
system $hen the pulley is sub6ected to a time varying moment 3 7t8 as
sho$n.
Figure 9. T$o 2egrees o Freedom System
Solution
The generali"ed coordinates : and x are selected
1. Pulley
MO=I
k1r2
+k2 (xr ) r+M(t)=I
I+( k1+k2) r2
k2 rx=M( t) 7a8
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2. M$ss
Fx=m x
k2 (xr)c x=m x
m x+cx+k2xk
2r=0 7b8
Euations 7a8 and 7b8 are the di/erential Euations o motion o the system.
Example 2
5rite the di/erential Euations o Example 78 in matrix orm
[I 00 m ]{x}+[0 00 c ]{x}+[(k1+k2)r2 k
2r
k2 r k2] {x}={M( t)0 }
Mass, Stifness, and Damping Matrices
The matrix orm o Euations $ill generally involve a mass matrix 3, a
sti/ness matrix ;, and a damping matrix
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in $hich x, x0, xnare the generali"ed coordinates,
Fis a column matrix involving excitation orces and=or moments.
For undamped ree vibration
MX+KX=0(3)
in $hich &is a null column matrix
Notes
. For linear systems, the mass, sti/ness, and damping matrices are all
symmetric, that is, mi6> m6i, ci6> c6i, and *i6> *6i,0. The mass matrix is usually a diagonal matrix 7in the dynamic coupling,
the mass matrix is non-diagonal8.9. In general, many o the elements o the sti/ness matrix are "ero
every$here, except or those $hich are on the diagonal and
immediately above and belo$ it. The sti/ness matrix is thus reerred to
as a '$!ded m$tri(.
Example 3
2etermine the di/erential Euations o motion in matrix orm or the
undamped ree vibration o the three-story building sho$n. +ssume that the
mass distribution o the building can be represented by the lumped masses
at di/erent levels, considering the columns as springs in parallel.
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Figure ?. @ndamped 3odel o Three Story 4uilding
Solution
+ssume x9A x0A x
k1x
1+k
2(x2x1)=m1 x1
k2 (x2x1 )+k3 (x3x2 )=m2 x2
k3 (x3x2 )=m3 x3
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[m
1 0 0
0 m2
0
0 0 m3
]{x
1
x2
x3
}+[k
1+k
2 k
2 0
k2
k2+k
3 k
3
0 k3
k3
]{x
1
x2
x3
}={000}or,
MX+KX=0
5.) *!+ue!,e Coe,ie!ts
(Stifness, Flexibility, Damping and Inertia)
The di/erential Euations o motion can be $ritten in terms o a &exibility
matrix +, $hich is the inverse o the sti/ness matrix ;, that is ;-> + or +->
;.
The elements *i6, ai6, ci6 and mi6 o the sti/ness, &exibility, damping and
inertia matrices, respectively, are reerred to as in&uence coe'cients.
The di/erential Euations o motion and the Euations or determining the
natural reuencies can be $ritten by simply inspecting the system and
applying the defnition o the in&uence coe'cient used.
Stifness Inuence Coecients
2esignating the coordinates , 0 , B, n o an n-degree-o-reedom
system as the generali"ed coordinates that defne linear and=or angular
displacements.
The sti/ness coe'cient *i6 is the orce or moment reuired to hold a
particular coordinate ifxed $hen a coordinate 6is given a unit linear 7orangular8 displacement, $ith all other coordinates held fxed.
The sti/ness coe'cient *ii 7i>68 is the static orce or moment reuired to
give the coordinate ia unit linear or angular displacement.
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The sti/ness coe'cient *6, *06,B, *n6in the matrix ; are the elements o
the 6thcolumn. I a sti/ness coe'cient 7orce or moment8 acts in the same
sense assumed positive or the associated generali"ed coordinate, it $ill be
positive. I it acts in the opposite sense, it $ill be negative.
Example 4
It is desired to determine the sti/ness matrix ; or the three-story building o
Example 798.
Figure C. 3odel o Three Story 4uilding
Solution
*>*D *0 *0> -*0 *9>
*0> - *0 *00> *0D *9 *09> - *9
*9> *90> -*9 *99> *9
K=[k
1+k
2 k
2 0
k2
k2+k
3 k
3
0 k3
k3
]stiffness
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KX=M X
Example 5
+ shat $ith three dis*s attached to it, is fxed at one end. 2etermine
a. The stiffness matrix,
b. The differential Equations of motion.
Figure . Shat $ith Three 2is*s
Solution
K=[2 k k 0k 2 k k
0 k k]stiffness
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[I
1 0 0
0 I2 0
0 0 I3]{
1
2
3
}+[2 k k 0k 2 k k0 k k]{
1
2
3
}={000}Example 7
For the six-degree-o-reedom system sho$n, fve-story building $ith a spring
and mass system attached to the ourth &oor. 2etermine the sti/ness matrix.
Solution
In determining the ourth column o ;, $e visuali"e the hori"ontal orces
reuired to maintain the confguration o x? > , $ith all other masses
7stations8 held fxed.
Example 7
T$o identical bars o length l, mass,
and moment o inertia I, , 0,
and 9 are the generali"ed
coordinates. 2etermine;.
Figure G. 3ultiple 2egrees o
Freedom System
Solution
* is the orce applied at the mass
center to give it a unit translation
7> 8.
*0is the moment reuired to *eep
the upper mass rom rotating 70
>8 $hen *is applied.
Figure G. 3odel o Five Story
4uilding
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*9is the moment reuired to *eep the lo$er bar rom rotating 79>8 $hen
*is applied.
Note: The *i6 has been given a sense the same as the sense assumed
positive.
FV=0=k116 k
MG=0=k214 k l
2+2 k
l
2
MO=0=k31+2 k
l
2k l
2
the first column of K is then
6 k kl k l
2T
[k11 k21 k31 ]T=
K=k[ 6 l
l2
l 3
2l
2 34
l2
l2
34
l2 3
4l
2 ]Example 8T$o identical bars, each has a length l, a mass m, and a mass moment o
inertia Ioabout the pinned end. 2etermine the sti/ness matrix and Euations
o motion.
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Figure 1. 2ouble Hendulum System
Solution
The spring orce*lsho$n is based upon :is small so that
kl sin 1kl
1=kl
MA=0=k11mgl
2k l2
MB=0=k21+k l2
which
k11=k l2+mg
l
2
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k21=k l2
MA=0=k12+k l2
MB=0=k22mgl
2k l2
which
k12=k l2
k22=k l2+mgl2
Thus the ifferenti!l "#u!tions of motion isgi$en %&
[IO 00 IO]{
1
2}+
[
k l2+mg
l
2k l2
k l2 k l2+mgl
2
]{1
2}={00}
5. Fle(i'ility *!+ue!,e Coe,ie!ts
The &exibility coe'cient ai6is the linear 7or angular8 displacement that occurs
or a particular coordinate i$hen a unit orce or moment is applied at the
location o a coordinate 6$ith all other coordinates ree to displace.
In using &exibility coe'cients to obtain the di/erential Euations o motion,
let us use the concept o dynamic euilibrium. The inertia e/ects are eual in
magnitude to the orces that the various masses exerts, and can thus be
treated as i they $ere orces acting on the system. #ote that the total
displacement o a particular mass is eual to the sum o the displacements
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caused by each inertia e/ect 7orce8, and remembering the defnition o a
&exibility coe'cient.
The total de&ection at station iis given by
x i='=0
n
!i' f'
x1=!
11m
1x
1!
12m
2x
2+(!
1 nmnxn
x2=!
21m
1x
1!
22m
2x
2+(!
2 n mn xn
( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( (
xn=!n 1 m1 x1!n 2 m2 x2+(!nnmn xn
in matrix orm!
{x
1
x2
:
xn}+
[!
11
!21
.
!n1
!12
!22
.
!n 2
..
.
.
..
.
.
..
.
.
!1 n
!2 n
.
!nn][m
11
m21
.
mn 1
m12
m22
.
mn 2
..
.
.
..
.
.
..
.
.
m1 n
m2 n
.
mnn]{x
1
x2
:
xn}={
00
:
0 }(4)or more simply!
X+AM X=0(5)
+is the &exibility matrix,
3ultiplying by +-
MX+A1X=0(6)
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it can be seen that +-> ; or ;-> +
Thus, the inverse o the &exibility matrix +is the sti/ness matrix ;.
+ is symmetric matrix, since the inverse o a symmetric matrix is also
symmetric.
Example 9
2etermine the &exibility matrix o the three-story building sho$n.
Figure . Three Story 4uilding
Solution
(! )F1=( 3 k) !
11!
11=
1
3 k
Since the hori"ontal orces on the columns supporting the upper t$o masses
are "ero!!
21=!
31=!
11
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7b8 The columns in each o the frst t$o stories are sub6ected to the unit orce
F0, thus!
!12=
1
3 k
!22=
1
3 k+
1
2 k=
5
6 k
!32=!22= 5
6 k
7c8 Similarly, the columns are sub6ected to the orce F9,
!13=
1
3 k
!23= 1
3 k+
1
2k=
5
6 k
!33= 1
3 k+
1
2 k+
1
k=
11
6 k
The di/erential Euations o motion are
{x
1
x2
x3
}+ 16 k[2 2 22 5 52 5 11]{m
1x
1
m2x
2
m3x
3
}={000}Example 10
+ dis* o mass m and centroidal moment o area I about a "-axis is attached
to the end o a cantilever beam o length l and negligible mass, sti/ness
actor EI, the dis* has general plane motion 7it both translates and rotates8
as sho$n, so that it has t$o degrees o reedom that are defned by the
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generali"ed coordinates > yand 0>:. Sho$ that +;>I, in $hich I is
the identity matrix.
Figure a. + 8,
&=!11= l
3
3"I=!
21= l
2
2"I
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$e next apply unit moment 73>8
&=!12=
l2
2"I*=!
22=
l
"I
Thus, the &exibility matrix,
A= l
6"I[2 l2
3 l
3 l 6]
In determining the sti/ness matrix ;!
Figure b. +
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)=k12=6"I
l2 M=k
22=
4"I
l
The sti/ness matrix is thus
K=6"I
l [ 2
l2
1l
1l
2
3]
AK= l
6"I
6"I
l
[2l
23 l
3 l 6
][ 2
l2
1l
1l
23
]=
[1 0
0 1
]=I
5.5 C"oi,e of Sti#!ess or Fle(i'ility M$tri(
In some problems it is easier to determine the &exibility coe'cients a i6than
the sti/ness coe'cients *i6, $hile in others the reverse is true. For example
. The lumped-mass model o an overhanging beam sho$n 7a8 becomes
statically indeterminate to the second degree 7b8 $hen sti/ness
coe'cients are considered as *0and *9act the same as un*no$n pin
reactions in *eeping m0and m9fxed $hen *is applied. It is easy to
determine the &exibility coe'cients, since the beam sho$n 7c8 or
determining the frst column o + is or a statically determinate beam.
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Figure . Extended 4eam $ith )umped 3asses0. In the case sho$n 7d8 and 7e8, the sti/ness coe'cients can be
determined by simple inspection. For example, i F> *is applied to
station 7mass 8 so that x > , and m0 and m9 are held fxed by *0
and *9, respectively, a simple inspection, and a consideration o
statics, sho$s that
k11=k
1+k
2* k
21=k
2* k
31=0
5hile, i a unit orce 7F>8 is applied in station , $ith m, m0, and
m9ree to displace, it is ound necessary to $rite three Euations o
euilibrium, $hich must solved simultaneously to obtain the &exibility
coe'cients a, a0, and a9.
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Figure 0a. Three 2egree o Freedom Systems
9. 5hen systems are unconstrained 7ree-ree8 as sho$n 78 and 7g8, the
&exibility coe'cients are infnite. For example, , i a unit orce 7F>8
is applied to m o the system 78, the displacements 7the a i6%s8 become
a> a0> a9> ,since there are no other external orces to resist
this statically applied orce. These ree-ree systems are reerred to as
semidefnite systems, and there Euations o motion must be
determined using sti/ness coe'cients rather than &exibilitycoe'cients. Jne o the natural reuencies o such systems $ill be
"ero, corresponding to the "ero root o the reuency Euation $hen
the system moves as a rigid body.
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Figure 0b. Three 2egree o Freedom Systems
Example 11
The &exible mass-less cantilever beam sho$n in Figure is carrying three
masses at stations 78, 708, and 798. The beam has a uniorm cross section
$ith property KEIL. 2etermine the in&uence coe'cients. +lso, determine the
Euations o motion.
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Figure 9. Flexible 3ass-less )L, then
a11=IE3
L3
a21=IE3
L3
+ L y|x = L=IE6
L5 3
a31=IE3
L3
+ 2 L y|x = L=IE3
L4 3
Hlacing a unit orce at station 708, Ka > 0 )L, then
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a12=IE6
L5 3
a22= IE3
L8 3
a32=IE3
L8 3
+ L y|x = 2 L=IE3
L14 3
Hlacing a unit orce at station 798, Ka > 9 )L, then
a13=IE3
L4 3
a23=IE3
L14 3
a33=IE3
L27 3
It is clear that Kai6> a6iL. The &exibility matrix is $ritten as
27144
1485.2
45.21
EI3
L3
I $e consider the $hole system, then
=
3
2
1
333231
232221
131211
3
2
1
f
f
f
aaa
aaa
aaa
x
x
x
KL, K0L and K9L are the inertia orces acting on the masses. I there are
other orces acting on one o the masses, such as external or damping
orces, they should be included. The orces are given by
> - m1y , 0 > - m0
2y , 9> - m93y
Thereore,
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3
2
1
y
y
y
=
3
2
1
3
2
13
y
y
y
m00
0m0
00m
27144
1485.2
45.21
EI3
L
The above Euations ta*e the orm
0
y
y
y
y
y
y
m27144
14m852
452m
EI3
L
3
2
1
3
2
1
3
2
13
=
+
.
.
5./ D$m%i!g *!+ue!,e Coe,ie!ts
The damping coe'cient ci6 is the orce or moment reuired to hold a
particular coordinate ifxed $hen a coordinate 6is given a unit linear 7orangular8 velocity, $ith all other coordinates held fxed. ci6> c6i
Example 12
It is desired to determine the damping matrix or the three-story building
sho$n.
Solution
Figure ?. Three Story 4uilding $ith 2amping
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c> cD c0 c0> -c0 c9>
c0> -c0 c00> c0D c9 c09> -c9
c9> c90> -c9 c99> c9
MX+CX+KX=0
There is no practical means o determining accurate values or ci6elements o
the damping matrix or structural systems.
5.0 *!erti$ *!+ue!,e Coe,ie!ts
The elements o the mass matrix, mi6, are *no$n as the inertia in&uence
coe'cients. +lthough it is more convenient to derive the inertia in&uence
coe'cients rom the expression or *inetic energy o the system 7next
section8, the coe'cients mi6can be computed using the impulse-momentum
relations. The inertia in&uence coe'cients mi6 are defned as the set o
impulses applied at points 7i=1,2,,n8, respectively, to produce a unit
velocity at point 6 and "ero velocity at every other point. Thus, or a
multidegree o reedom system, the total impulse at point i, Fi , can be
ound by summing up the impulses causing the velocitiesx' 7j = 1, 2, ,
n8 as
Fi='=1
n
mi' x'
In matrix orm!
F=[ m ] x
$herexF are the velocity and impulse vectors given by
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x=
{
x1
x2.
.
.xn
}*F=
{
F1
F2.
.
.Fn
}and NmOis the mass matrixFor linear system the inertia in&uence coe'cients are symmetric, that is m i6> m6i. The ollo$ing procedure can be used to derive the inertia in&uence
coe'cients,
. +ssume that a set o impulses fi' are applied at various points i
7i>,0,B,n8 so as to produce a unit velocity at point6 7 x' >$ith
j=1to start $ith8 and a "ero velocity at all other points. 4y defnition,
the set o impulsesfi' denote the inertia in&uence coe'cients mij.
0. +ter step orj=1, the procedure is repeated or6 > 0, 9, B, n.
#ote that ixjdenote an angular coordinate, thenx' represents an angular
velocity andF' indicate an angular impulse.
Example 13
Find the inertia in&uence coe'cients o the system sho$n
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Figure C. 3ultiple 2egrees o Freedom o Trailer
Solution
)et x7t8 and :7t8denote the coordinates to defne the linear and angular
positions o the trailer 738 and the compound pendulum 7m, l8. To derive the
inertia in&uence coe'cients, frst, impulses o magnitudes mand m0are
applied along the directions x7t8 and :7t8 to results in the velocities
x=1=0 . Then the linear impulse-linear momentum Euation gives
m> 73 D m878
and the angular impulse-angular momentum Euation 7about J8 yields
m21=m(1)l
2
#ext, impulses o magnitudes m0and m00are applied along the directions
x7t8 and :7t8 to obtain the velocities x=0=1 . Then the linear impulse-
linear momentum Euation provides
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m12=m(1)
l
2
and the angular impulse-angular momentum Euation 7about J8 gives
m22=
m l2
3(1)
Thus, the mass or inertia matrix o the system is given by
[ m ]=
[M+m
ml
2
ml
2
ml2
3
]5. Pote!ti$l $!d i!eti, E!ergy E(%ressio!s i! M$tri(
Form
)et xidenote the displacement o mass m iand Fi the orce applied in the
direction o xiat mass miin an n degree o reedom system similar to the one
sho$n in Figure. The elastic potential energy 7also *no$n as strain energy or
energy o deormation8 o the ithspring is given by
Figure . 3ultiple 2egrees o Freedom SystemVi=
1
2Fix i(7)
The total potential energy can be expressed as
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V=i=1
n
Vi=1
2i=1
n
Fixi
Since Fi=i=1
n
ki'x '
V=1
2i=1
n
(i=1
n
ki'x')x i=1
2i=1
n
i=1
n
ki'x'x i
The last Euation can also be $ritten in matrix orm as
V=1
2 xT
[ k]x (8)
$here x is the displacement vector and [k] is the sti/ness matrix.
The *inetic energy associated $ith mass miis, by defnition, eual to
Ti=1
2mi x i
2(9)
The total *inetic energy o the system can be expressed as
T=i=1
n
Ti=1
2i=1
n
mi x i2
$hich can be $ritten in matrix orm as
T=1
2xT [m ]x (10)
5herex is the velocity vector and NmOis the mass matrix.
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It can be seen that the potential energy is a uadratic unction o the
displacements, and the *inetic energy is a uadratic unction o the
velocities. (ence they are said to be in uadratic orm. Since *inetic energy,
by defnition, cannot be negative and vanishes only $hen all the velocities
vanish, T is called positive defnite uadratic orms and the mass matrix NmO
is called a positive defnite matrix. Jn the other hand, the potential energy
expression, P is a positive defnite uadratic orm, but the matrix N*O is
positive defnite only i the system is a stable one. There are systems or
$hich the potential energy is "ero $ithout the displacements or coordinates
x, x0, B, xn being "ero. In these cases the potential energy $ill be a
positive uadratic unction rather than positive defnite! correspondingly, the
matrix N*O is said to be positive. + system or $hich N*O is positive and NmO is
positive defnite is called a semi-defnite system.
5.3 Solutio! of t"e E4u$tio!s of Motio!
The Euations o motion are usually $ritten in the matrix orm
fyKyCyM =++
708
5here KML is the mass matrix, KCLis damping coe'cients matrix, and KLis
the sti/ness matrix. +ll orces have the same reuency. + complete analysis
o a system $ith several degrees o reedom reuires the determination o
the natural reuencies, the damping reuencies and damping actors, and
the orced response. These analyses are summari"ed as ollo$s.
5.1&Determi!$tio! of t"e $tur$l Fre4ue!,ies $!d Mode
S"$%es
(!igen"#alues and !igen"#ectors)$$
** The term Eigen-value comes from the German word "Eigenwert" in which
Eigen means "Characteristic" and Wert means "Value".
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+n n-degree-o-reedom system has n natural reuencies, and or each
natural reuency there is a corresponding normal mode shape that defnes a
relationship bet$een the amplitudes o the generali"ed coordinates or that
mode.
The suare o the natural reuencies and sets o coordinate value describing
the normal mode shape are reerred to as eigenvalues and eigenvectors
respectively.
It $as sho$n that the damped natural reuency
+=+1,2
and that they are the same in magnitude or many real systems that have
damping o less than 0Q 7RM.08. For this reason, the eigenvalues and
eigenvectors o multi-degree-o-reedom systems, are usually determined
considering undamped ree vibration.
The Euations o motion are in the orm
MX+KX=0
Hremultiply by 3-7the inverse o 38 to obtain
X+M1KX=0(13)
+ssume x i=Xi ei+t
x i=+2Xi e
i+t
Substituting into Euation 798 $e obtain
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+2X+M1KX=0
[M1K+2I]X=0
[-I]X=0
The dynamic matrix KDL or its inverse KD L is ormed as
D = M 1K
D- 1= K 1M
= 2
The Eigenvalues o KDL give the natural reuencies and the Eigenvectors
are the corresponding modal vectors. KD-L gives the inverse o natural
reuencies and the corresponding modal vectors.
Example 15
For the system sho$n determine
a. The eigenvalues o the system,
b. The natural reuencies in 7("8,
c. The eigenvectors.
l > cm, d > 9.0C cm,
> 1?.? Ha, I > .91 ;g.m0
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Figure G. T$o 2egrees o Freedom System
Solution
M=[I 00 I] *
K=[ k kk 2 k]
-=M1K=[ k
I
kI
ki
2k
I]
|-I|=0
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|k
I
kI
ki
2 k
I |=0
2
3 k
I +(kI)
2
=0
1=+
1
2=k
I[ 352 ] * 2=+22=kI[ 3+52 ]
k=G/
l =
84.4
10
9
032 ( 0.031125)4
0.6=13170.21 .m /r!
1=+12=3645.34,2=+2
2=25000
f1=
+1
2 0=9.6123 * f
2=
+2
2 0=25.1523
[k
I
kI
ki
2 k
I ]{4142}={00}
( kI)41kI4 2=0 ( ! )
kI
41+( 2 kI)42=0(%)
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The eigenvectors or Uand U0can be determined rom either Euation 7a8 or
7b8, using Euation 7a8
42
41
=
k
I
.
k
I
421
411=0.62 (first moe )
422
412
=1.62 ( secon moe )
Assuming ! $!lue of unit&one of the eigen$ectors com5onents
(e . g . 41=1)
{
41
42
}1
=
{
1
0.62
}eigen$ector for first moe (in5h!se )
{4142}
2
={ 11.62}eigen$ector for secon moe(180o out of 5h!se)
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Figure 1. raphical Vepresentation o 3ode Shapes
Example 16
For the semidefnite system sho$n determine!
a. the natural reuencies,b. the normal mode shapes 7eigenvectors8.
Figure . Example o Semidefnite System
Solution
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K=[ k k 0k 2 k k
0 k k]M=[I 0 0
0 I 0
0 0 2I]
[( kI) kI 0kI ( 2 kI ) kI0
k2I ( k2I)]{
414
2
43
}={000}
"x5!ning the etermin!nt *sim5lif&ing * gi$es the fre#uenc& e#u!tion !s
3
7
2 ( kI)2+2( kI)2
=0
2
7
2
k
I
+2
(
k
I
)
2
=0
the roots!re
1=0 f
1=0
2=0.72k
I
f2= 1
20
0.72
k
I
3=2.78k
I f3=
1
2 02.78 kI
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%Thefirstthir e#u!tions !re
( kI)41kI4 2=06 k2I4 2+( k2I)43=0
4
2
41
=
k
I
k
I
43
41
=
k
I
2( k2I)
(42
41 )1=1,(
43
41 )1=1 (rigi %o& motion)
( 4241)
2
=0.28,( 4341)
2
=0.64 (1st $i%r!tionmoe )
(
42
41
)3
=1.78,
(
43
41
)3
=0.39 (2n $i%r!tion moe )
{4
1
42
43
}1
={1
1
1}*{
41
42
43
}2
={ 1
0.28
0.64}{4
1
42
43
}3
={ 1
1.780.39
}
WW The components o the eigenvectors are said to be normali"ed $.r.t.4
1 .
They can also be normali"ed $.r.t. either4
2 or4
3 . The magnitude o the
reerencing components is reerred to as the normali"ed actor. The
eigenvector can be normali"ed by giving it a unit magnitude, in $hich case
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the normali"ed actor is the suare root o the sum o the suares o the
eigenvector components.
Figure . raphical Vepresentation o 3ode Shapes
Example 17
2etermine the natural reuencies and corresponding normal mode
confgurations o the three-story building sho$n.
m > 1WC*g, * > ?W1#=m
Solution
|-|=|(
5 k
3 m) 2 k
3I 0
k2 m (3 k2 m) k2m
0 k
m ( km)|=0
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"x5!ning the etermin!nt *sim5lif&ing * gi$es the fre#uenc& e#u!tion !s
3
25
6(km )2+276( km )2
( km )3
=0
.320.83(10)2.2+11.25(10)5.12.50(10)7=0
The roots can be determined by incremental-search method.
1=149.60 f
1=1.95023
2=652.6 f2=4.0723
3=1280.9 f
3=5.723
[(5 k3 m) 2k3 m 0
k2
m
(
3 k
2m
)
k2
m0
km ( km)
]{X1X
2
X3
}=
{
0
0
0
}Thefirstthir e#u!tions !re
(5 k3 m)X1 2 k3 mX2=06 kmX2+( km)X3=0
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X
2
X1
=
5 k
3 m
2 k
3 m
X3
X1
=( 32 )5 k
3 m
(k
m)
(X2X1)
1
=2.05,(X3X1)
1
=2.93 (1st$i%r!tion moe )
(X2X1)
2
=0.54,(X3X1)
2
=1.78 ( 2n $i%r!tionmoe )
(X2X1)
3
=1.34,(X3X1)
3
=0.86 (3r $i%r!tionmoe )
I $e normali"ed these relationships $.r.t. Xby assuming that X>
{X
1
X2X
3
}1
=
{ 1
2.05
2.93
}*
{X
1
X2X
3
}2
=
{ 1
0.54
1.78
}
{X
1
X2X
3
}3
=
{ 1
1.340.86
}
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Figure 0. First Three 3odes o Pibration