# vibraciones de dos grados de libertad

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3he vibrations o syst ms having more than one

degree o freedom

Many real systems can e represented y a single degree of freedom model. However,most actual systems have several bodies and several restraints and therefore severaldegrees of freedom. The number of degrees of freedom that a system possesses isequal to the number of independent coordina es necessary to describe the motion ofthe system. Since no body is completely rigid, and no spring is without mass, everyreal system has more than one degree of freedom, and sometimes it is not sufficientlyrealistic to approximate a system by a single degree of freedom model. Thus, it isnecessary to study the vibration o systems with more than one degree of freedom.

Each flexibly connected body in a multi-degree of freedom system can moveindependently of the other bodies, and only under certain conditions will all bodiesundergo an harmonic motion at the same frequency. Since all bodies move with thesame frequency, they all attain their amplitudes at the same time even if they do notall move in the same direction. When such motion occurs the frequency is called a

n tur lfrequency of the system, and the motion is a principal mode of vibration: thenumber o natural frequencies and principal modes that a system possesses is equalto the number of degrees of freedom of that system. The deployment of the systemat its lowest or first natural frequency is called its first mode, at the next highest orsecond natural frequency it is called the second mode and so on.

A two degree of freedom system will e considered initially. This is because theaddition of more degrees of freedom increases the labour of the sohion procedurebut does not introduce any new analytical principies.

Initially, we will obtain the equations ofmotion for a two degree offreedom model,

and from these find the natural frequencies and corresponding mode shapes.Sorne examples of two degree of freedom models of vibrating systems are shownin Figs 3.l aHh).

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The vibration of systems with two egrees of freedom 9

MFig. 3.l a) Linear undamped system, horizontal motion. Coordinates x 1 and x 2 .

m,

Fig. 3.l b) Linear undamped system, vertical motion. Coordinates y and y2

Fig. 3.1 c) Torsional undamped system. Coordinates e, and 8 2 .

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9 he vibratons of syst ms having more than one degree of freedom Ch . 3

Fig. 3.l d) Coupled pendula . Coordinates 1 and 2

,--....em /

G

X

Figt. 3 . l e) System with combined translation and rotation . Coordinates x and }

Fig. 3.1(f) Shear frame. Coordinates x 1 and x 2

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Sec 3 1] The vibration o systems with two degrees of freedom 9

Fig. 3.2 g) Two degree of freedom model, rotation plus translation. Coordinates y nd 8

antilever spring

Fig. 3.1 h) Two degree of freedom model, trans lation vibration. Coordinates x 1 nd x 2

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9 he vibrations o systems having more than one degree of freedom [Ch. 3

3 1 THE VIBR TION O SYSTEMS WITH TWO DEGREES O FREEDOM

3 1 1 Free vibration of an undamped systemOf the examples of two degree of freedom models shown in Figs. 3.1 a)- h),consider the system shown in Fig. 3.l a). I f x

1> x 2 the FBDs are as shown in

Figure 3.2.

k 1x 1 Q k x 1 2 )

k x 1 2 D k 2 x 2

a) b)

Fig. 3.2. a) Applied forces; b) effective forces.

The equations of motion are therefore,

for body 1, 3.1)

and

for body 2. 3.2)

The same equations are obtained if x 1 < x 2 is assumed because the direction ofthe central spring force is then reversed.

Equations 3.1) and 3.2) can be solved for the natural frequencies and correspondingmode shapes by assuming a solution of the form

x1

= A1sin wt

+t/1 and x

2= A

2sin wt

+t/J .

This assumes that x 1 and x 2 oscillate with the same frequency w and are either inphase or n out of phase. This is a sufficient condition to make w a natural frequency.

Substituting these solutions into the equations of motion gives

m1 A 1 w

2 sin wt + t/J = 1 A 1 sin wt + t/J)- k A 1 A 2 )sin wt + t/1and

m2 A 2 w

2 sin wt + t/1 = k A 1 - A 2 )sin wt + t/1)- k 2 A 2 sin wt + t/1 .

Since these solutions are true for all values of t,A 1 k + k 1 - m 1w 2 ) + A 2 - k = O

and

3.3)

3.4)

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Sec 3.1] The vibration of systems with two degrees. of freedom 93

A 1 and 2 can be eliminated by writing

\

k k 1 m 1w2

- k \ = O- k k k 2 - m 2 w

23.5)

This is the characteristic or frequency equation. Alternatively, we may write

A 1/A 2 = k/ k k 1 - m1 w2

) from 3.3)

and

3.6)

A/A 2 = k 2 k - m2 w2 )/k from 3.4)

Thus

k/ k k 1 - m1w2

) = k 2 k - m2 w2 )/k

and

k+ k 1 - m 1w2 ) k 2 k - m2 w

2 k 2 =O 3.7)

This result is the frequency equation and could also be obtained by multiplying outthe above determinant, equation 3.5).

The solutions to equation 3.7) give the natural frequencies of free vibration for

the system in Fig. 3.1 a). The corresponding mode shapes are found by substitutingthese frequencies, in turn, into either of equations 3.6).

Consider the case when k 1 = k 2 =k , and m 1 = m2 =m. The frequency equation is2 k - mw 2 k2 =O; that is,

m2 w 4 - 4mkw 2 3k 2 = O or mw 2 - k) mw 2 - 3k) = O.

Therefore, either mw 2 - k = O or mw 2 - 3k = O.Thus

w 1 = .j k/m)rad/s and w 2 = .j 3k/m) rad/s.

f

w = .j k/m)rad/s,

and if

w = .j 3k/m)rad/s, from 3.6)

This gives the mode shapes corresponding to the frequencies w1

and w2

Thus, thefirst mode of free vibration occurs a t a frequency 1/2n).j k/m)Hz and A 1 /A 2 )

1 = 1,that is, the bodies move in phase with each other and with the same amplitude as ifconnected by a rigid link, Fig. 3.3. The second mode of free vibration occurs at afrequency 1/2n).j 3k/m)Hz and A/A 2 )

11 = - 1 that is, the bodies move exactly outof phase with each other, but with the same amplitude, see Fig. 3.3.

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94 The vibrations o systems having more than one degree o freedom [Ch. 3

l v ~x

1st mode

1/ .= Hz1 27T

.

t t

Fig . 3.3 Natural frequencies and mode shapes for two degree of freedom translation vibration system .Bodies of equal mass and springs of equal stiffne ss.

3 1 2 Free motion

The two modes of vibration can be written

and

where the ratio A 1 /A 2 is specified for each mode.Since each solution satisfies the equation of motion the general solution is

where A 1 A 2 1 2 are found from the initial conditions.

A2 y -A2

A1 y +A1

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9 he vibrations of syst ms having more than one degree of freedom Ch. 3

3 1 3 Coordinate coupliogIn sorne systems the motion is such that the coordinates are coupled in the equationsof motion. Consider the system shown in Fig. 3.l e); only motion in the plane of thefigure is considered, horizontal motion being neglected because the lateral stiffnessof the springs is assumed to be negligible. The coordinates of rotation, O, andtranslation, x are coupled as shown in Fig. 3.4. G is the centre of mass of the rigidbeam of mass m and moment of inertia 1 about G.

Fig. 3.4. Two degree of freedom model, rotation plus translation.

The FBDs are shown in Fig. 3.5; since the weight of the beam is supported by thesprings, both the initial spring forces and the beam weight may be omitted.

la l lb

m i

Fig. 3.5. (a) Applied forces. (b) Effective force and moment.

For small amplitudes of oscillation (so that sin ~ O the equations of motion are

x = k 1 x L 10 k 2 x L 2 0

and

that is,

F en x

M en z

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Sec. 3.1) The vibration of systems with two degrees of freedom 9

andX 2 2l u - k 1L 1 - k 2 L 2 )x + k 1 L 1 + k 2 L 2 )0 =O.

l t will be noticed that these equations can be uncoupled by making k 1L 1 = k 2 L 2 ;if this is arranged, translation x motion) and rotation 8 motion) can take placeindependently. Otherwise translation and rotation occur simultaneously.

Assuming x 1 = A 1 sin wt + t/1 and } = A 2 sin wt + t/1 , substituting into theequations of motion gives

-m w 2 A 1 + k 1 + k 2)A 1 - k 1L 1 - k 2 L 2 A 2 =O,and

- l w 2 A 2 - k 1 L 1 - k2 L 2 )A 1 + k 1 Li + k 2 L; A 2 =O.That is,

A 1 k 1 + k 2 - mw 2 ) + A 2 k 1L 1 - k 2 L 2 =O,and

A 1 k 1 L 1 - k 2 L 2 + A 2 k 1 Li + k 2 L ; - 2 ) =O.Hence the frequency equation is

\

k 1 k 2 - m w2

k 1L 1 2 L 2 ) 12 2 2 =O.

k 14 - k 2 L 2 ) k 1L 1 + k 2 L 2 - IwFor each natural frequency, there is a corresponding mode shape, given by A 1 /A 2

Example 8A system is modelled as a straight link AB 1m long of mass 10 kg, supportedhorizontally by springs A and B of stiffness k 1 and k 2 respectively. The moment ofinertia of AB about its centre of mass Gis kgm 2 , and Gis located t distance afrom A and b from B

Find the relationship between k 1 , k2 , a, and b so that one mode of free vibration

shall be translational motion on

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