Multiple Degrees of Freedom

8/9/2019 Multiple Degrees of Freedom

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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


34/44

+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


35/44

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


36/44

|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(%)


37/44

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)


38/44

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


39/44

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


40/44

%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


41/44

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


42/44

"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


43/44

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

}


44/44

Figure 0. First Three 3odes o Pibration

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Multiple Degrees of Freedom

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