Ch 4 - The Vibrations of Systems Having TDOF

4

The Vibrations of Systems Having Two Degree of Freedom

4.1. Introduction Many real systems can be represented by a single degree of freedom model. However, most actual systems have several bodies and several restraints and therefore several degrees of freedom. The number of degrees of freedom that a system possesses is equal to the number of independent coordinates necessary to describe the motion of the system. Since no body is completely rigid, and no spring is without mass, every real system has more than one degree of freedom, and sometimes it is not sufficiently realistic to approximate a system by a single degree of freedom model. Thus, it is necessary to study the vibration of systems with more than one degree of freedom. Each flexibly connected body in a multi-degree of freedom system can move independently of the other bodies, and only under certain conditions will all bodies undergo an harmonic motion at the same frequency. Since all bodies move with the same frequency, they all attain their amplitudes at the same time even if they do not all move in the same direction. When such motion occurs the frequency is called a natural frequency of the system, and the motion is a principal mode of vibration: the number of natural frequencies and principal modes that a system possesses is equal to the number of degrees of freedom of that system. The deployment of the system at its lowest or first natural frequency is called its first mode, at the next highest or second natural frequency it is called the second mode and so on. A two degree of freedom system will be considered initially. This is because the addition of more degrees of freedom increases the labour of the solution procedure but does not introduce any new analytical principles. Initially, we will obtain the equations of motion for a two degree of freedom model, and from these find the natural frequencies and corresponding mode shapes. Some examples of two degree of freedom models of vibrating systems are shown in Figures (4.1a-h).

Chapter 4 The Vibrations of Systems Having Two Degree of Freedom

Fig. 4.1a: Linear undamped system, horizontal motion. Coordinates x1

and x2.

Fig. 4.1b: Linear undamped system, vertical motion. Coordinates y1 and

y2.

Fig. 4.1c: Torsional undamped system. Coordinates 1 and 2.


Fig. 4.1d: Coupled pendulum. Coordinates 1 and 2.

Fig. 4.1e: system with combined translation and rotation. Coordinates x

and .

Fig. 4.1f: Shear frame. Coordinates x1 and x2.


Fig. 4.1g: Two degree of freedom model, rotation plus translation.

Coordinates y and .

Fig. 4.1g: Two degree of freedom model, translation vibration. Coordinates x1 and x2.

4. 1. Free Vibration of an Undamped System Of the examples of two degree of freedom models shown in Figures (4.1a-h), consider the system shown in Figure (4.1a). If x1 > x2 the FBDs are as shown in Figure (4.2).


1m11xk 21 xxk

2m 22xk 21 xxk 2m

22xm

1m11xm

(b) (a)

Fig. 4.2: (a) Applied forces, (b) effective forces.

The equations of motion are therefore,

)( 211111 xxkxkxm for body 1,........ (4.1) and

222122 )( xkxxkxm for body 2,.......... (4.2)

The same equations are obtained if xl < x2 is assumed because the direction of the central spring force is then reversed. Equations (4.1) and (4.2) can be solved for the natural frequencies and corresponding mode shapes by assuming a solution of the form

tAx sin11 and tAx sin22 This assumes that xl and x2 oscillate with the same frequency and are either in phase or out of phase. This is a sufficient condition to make a natural frequency. Substituting these solutions into the equations of motion gives

tAAktAktAm sin)(sinsin 2111211 and

tAktAAktAm sinsin)(sin 2221222

Since these solutions are true for all values of t,

0)( 22111 kAmkkA ...... (4.3) and

0)( 22221 mkkAkA ......... (4.4)

A1 and A2 can be eliminated by writing

02

22

211

mkkk

kmkk.......... (4.5)

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


)4.4(/)(

)3.4()(/

22221

21121

fromkmkkAA

and

frommkkkAA

..... (4.6)

Thus

kmkkmkkk )()( 2222

11

and

0))(( 22222

11 kmkkmkk ...... (4.7) This result is the frequency equation and could also be obtained by multiplying out the above determinant, equation (4.5). The solutions to equation (4.7) give the natural frequencies of free vibration for the system in Figure (4.1a). The corresponding mode shapes are found by substituting these frequencies, in turn, into either of Eqs. (4.6). Consider the case when k1 = k2 = k, and ml = m2 = m. The frequency equation is 0)2( 222 kmk ; that is,

034 2242 kmkm or 0322 kmkm

Therefore, either 02 km , or 032 km Thus

sradm

k/1 and sradm

k/

32

If

sradm

k/ , 1/ 21

m

kAA

and if

sradm

k/

3 , 1/ 321

m

kAA

This gives the mode shapes corresponding to the frequencies 1 and 2. Thus, the first mode of free vibration occurs at a frequency mk2/1 Hz and (A1/A2)

I = 1,that is, the bodies move in phase with each other and with the same amplitude as if connected by a rigid link, Figure (4.3). The


second mode of free vibration occurs at a frequency mk32/1 Hz and (A1/A2)

II = -1 , that is, the bodies move exactly out of phase with each other, but with the same amplitude, see Figure (4.3).

Fig. 4.3: Natural frequencies and mode shapes for two degree of freedom translation vibration system. Bodies of equal mass and springs of equal

stiffness.

4. 2. Free Motion The two modes of vibration can be written

)sin( 112

1

2

1

tA

A

x

xII

,

and

)sin( 222

1

2

1

tA

A

x

xIIII

,

where the ratio A1/A2 is specified for each mode. Since each solution satisfies the equation of motion, the general solution is


)sin()sin( 222

111

2

1

2

1

tA

At

A

A

x

xIII

,

where A1, A2, 1 , 2 are found from the initial conditions.

For example, for the system considered above, if one body is displaced a distance X and released,

Xx )0(1 and 0)0()0()0( 212 xxx

where )0(1x means the value of x1 when t = 0, and similarly for )0(2x ,

)0(1x and )0(2x .

Remembering that in this system mk1 , mk32 , and

11

2

1

A

A and 12

2

1

A

A

we can write 211 3sinsin tmktmkx ,

and

212 3sinsin tmktmkx Substituting the initial conditions xl(0) = X and x2(0) = 0 gives

21 sinsin X ,

and

21 sinsin0

that is, 2/sinsin 21 X

The remaining conditions give 0coscos 21 .

Hence

tmkXtmkXx 3cos)2/(cos)2/(1 ,

and tmkXtmkXx 3cos)2/(cos)2/(2

That is, both natural frequencies are excited and the motion of each body has two harmonic components.


4. 3. Coordinate Coupling In some systems the motion is such that the coordinates are coupled in the equations of motion. Consider the system shown in Figure (4.1e); only motion in the plane of the figure is considered, horizontal motion being neglected because the lateral stiffness of the springs is assumed to be negligible. The coordinates of rotation, , and translation, x, are coupled as shown in Figure (4.4). G is the centre of mass of the rigid beam of mass m and moment of inertia I about G.

Fig. 4.4: Two degree of freedom model, rotation plus translation.

The free body diagrams are shown in Figure (4.5); since the weight of the beam is supported by the springs, both the initial spring forces and the beam weight may be omitted.

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

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

)()( 2211 LxkLxkxm ,

and

222111 )()( LLxkLLxkI


that is, 0)()( 221121 LkLkxkkxm ,

and

0)()( 2222112211 LkLkxLkLkI

It will be noticed that these equations can be uncoupled by making k1L1 = k2L2; if this is arranged, translation (x motion) and rotation ( motion) can take place independently. Otherwise translation and rotation occur simultaneously. Assuming tAx sin11 and tA sin2 , substituting into the equations of motion gives

0)()( 2221112112 ALkLkAkkAm ,

and

0)()( 2222

211122112

2 ALkLkALkLkAI

that is,

0221122211 LkLkAmkkA ,

and

0)( 2222211222111 ILkLkALkLkA

Hence the frequency equation is

022

222112211

22112

21

ILkLkLkLk

LkLkmkk

4. 4. Forced Vibration Harmonic excitation of vibration in a system may be generated in a number of ways, for example by unbalanced rotating or reciprocating machinery, or it may arise from periodic excitation containing a troublesome harmonic component. A two degree of freedom model of a dynamic system excited by an harmonic force F sin t is shown in Figure (4.6). Damping is assumed to be negligible. The force has a constant amplitude F and a frequency /2 Hz.


tF sin

Fig. 4.6: Two degree of freedom model with forced excitation.

The equations of motion are tFxxkxkxm sin)( 211111

and

222122 )( xkxxkxm Since there is zero damping, the motions are either in phase or out of phase with the driving force, so that the following solutions may be assumed:

tAx sin11 and tAx sin22

Substituting these solutions into the equations of motion gives

FkAmkkA 22111

and

022221 mkkAkA

Thus

2

221

mkkFA ,

and

FkA2 ,

where

022211222 kmkkmkk and = 0 is the frequency equation.

Hence the response of the system to the exciting force is determined.


4. 5. The Undamped Dynamic Vibration Absorber If a single degree of freedom system or mode of a multi-degree of freedom system is excited into resonance, large amplitudes of vibration result with accompanying high dynamic stresses and noise and fatigue problems. In most mechanical systems this is not acceptable. If neither the excitation frequency nor the natural frequency can conveniently be altered, this resonance condition can often be successfully controlled by adding a further single degree of freedom system. Consider the model of the system shown in Figure (4.7), where k1 and m1 are the effective stiffness and mass of the primary system when vibrating in the troublesome mode. The absorber is represented by the system with parameters k2 and m2. From section 4.1.4 it can be seen that the equations of motion are

tFxxkxkxm sin)( 2121111 , for the primary system and )( 21222 xxkxm , for the vibration absorber. Substituting

tXx sin11 and tXx sin22 gives

FkXmkkX 2221211 ,

tF sin

tXx sin11

1k

2k

1m

2m

tXx sin22

Fig. 4.7: System with undamped vibration absorber.


and 0222221 mkXkX

Thus

2

221

mkFX and

22

FkX ,

where 222121222 kmkkmk and = 0 is the frequency equation.

It can be seen that not only does the system now possess two natural frequencies, n1 and n2 instead of one, but by arranging for 2mk , Xo can be made zero.

Thus if 1122 mkmk , the response of the primary system at its original resonance frequency can be made zero. This is the usual tuning arrangement for an undamped absorber because the resonance problem in the primary system is only severe when 11 mk rad/s. This is shown in Figure (4.8).

1n 2n

1

0

k

F

1X

Fig. 4.8: Amplitude-frequency response for system with and without

tuned absorber.


When 0X , 22 / kFX , so that the force in the absorber spring, 22 Xk is

F ; thus the absorber applies a force to the primary system which is equal and opposite to the exciting force. Hence the body in the primary system has a net zero exciting force acting on it and therefore zero vibration amplitude. If an absorber is correctly tuned 1122

2 mkmk , and if the mass

ratio m2/m1, the frequency equation = 0 is

0122

1

2

4

nn m

m

This is a quadratic equation in 2

n . Hence

212

1

212

2

221

mm

m

mmm

n ....... (4.8)

Eq. (4.8) gives us the frequency of tuned system by finding the roots of the equation with m2/m1 as the parameter. There are two resonant frequencies the steady state response of the system is just like shown in Figure (4.8). Figure (4.9) gives the plot of mass ratio m2/m1 versus resonance frequency.

2n

1n

Fre

quen

cy r

atio

n/

Frequency ratio m2/m1

Fig. 4.9: Effect of absorber mass ratio on natural frequencies.


4. 6. Semidefinite Systems Semidefinite systems are also known as unrestrained or degenerate systems. Two examples of such systems are shown in Figure (4.10). The arrangement in Figure (4.10a) may be considered to represent two railway cars of masses m1 and m2 with a coupling spring k. The arrangement in Fig. (4.10c) may be considered to represent two rotors of mass moments of inertia J1 and J2 connected by a shaft of torsional stiffness kt.

(a)

(b) (c)

Fig. 4.10: Semidefinite systems.

In a railway train, the rail cars can be modeled as lumped masses and the couplings between the cars as springs. A train rolling down the track can be considered as a system having rigid-body, unrestrained, translational motion. At the same time, the rail cars can vibrate relative to one another. The presence of an unrestrained degree of freedom in the equation of motion changes the analysis. The stiffness matrix of an unrestrained system will be singular. One of the natural frequencies of an unrestrained two-degree-of-freedom system will be zero. For such a system, the motion is composed of translation and vibration. The analysis of unrestrained systems is presented by considering the system shown in Figure (4.10a). The equations of motion of the system can be written as (Figure (4.10b)):

0)( 2111 xxkxm

0)( 1222 xxkxm . (4.9)

For free vibration, we assume the motion to be harmonic: tXx sin11 and tXx sin22 ... (4.10)


Substitution of Eq. (4.10) into Eq. (4.9) gives 02121 kXXkm

02221 XkmkX ... (4.11)

By equating the determinant of the coefficients of X1and X2 to zero, we obtain the frequency equation as

0212212 mmkmm ... (4.12)

from which the natural frequencies can be obtained:

01 n and

21

212 mm

mmkn

.... (4.13)

As stated earlier, Eq. (4.13) shows that one of the natural frequencies of the system is zero, which means that the system is not oscillating. In other words, the system moves as a whole without any relative motion between the two masses (rigid-body translation). Such systems, which have one of the natural frequencies equal to zero, are called semidefinite systems. We can verify, by substituting into Eq. (4.11), that and are opposite in phase. There would thus be a node at the middle of the spring.

4. 6. Lagrange's Method We have learnt in section 4. 3 that if a system has n degree of freedom, it can be specified by a set of n generalized coordinates. So, the equations of motion of a vibrating system can often be derived in a simple manner in terms of generalized coordinates by the use of Lagrange's method. Lagrange's equations can be stated, for an n degree-of-freedom system, as

niQq

U

q

EK

q

EK

dt

di

iii

........,,2,1,..

. (4.13)

where tqq ii is the generalized velocity and Qi is the nonconservative

generalized force corresponding to the generalized coordinate qi. The forces represented by Qi may be dissipative (damping) forces or other external forces that are not derivable from a potential function. For example, if Fxk, Fxyk and Fzk represent the external forces acting on the kth mass of the system in the x, y, and z directions, respectively, then the generalized force can be computed as follows:


k i

kzk

i

kyk

i

kxki q

zF

q

yF

q

xFQ ... (4.14)

where xk, yk and zk are the displacements of the kth mass in the x, y, and z directions, respectively. Note that for a torsional system, the force Fxk, for example, is to be replaced by the moment acting about the x-axis (Mxk), and the displacement by the angular displacement about the x-axis (xk) in Eq. (4.14). For a conservative system, Qi = 0, so Eq. (4.13) takes the form

niq

U

q

EK

q

EK

dt

d

iii

........,,2,1,0..

... (4.15)

Eqs (4.13) or (4.15) represent a system of n differential equations, one corresponding to each of the n generalized coordinates. Thus the equations of motion of the vibrating system can be derived, provided the energy expressions are available.

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Ch 4 - The Vibrations of Systems Having TDOF