c. r. yang, ntnu mt chapter 5 two degree freedom …compute the eigenvalues or natural frequencies...

台灣師範大學機電科技學系

C. R. Yang, NTNU MT

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Chapter 5Two Degree Freedom Systems

5台灣師範大學機電科技學系

C. R. Yang, NTNU MT

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Learning Objectives

Formulate (用公式表示) the equations of motion of two-degree-of-

freedom systems

Identify the mass, damping, and stiffness matrices from theequations of motion

Compute the eigenvalues or natural frequencies of vibration and themodal (形態上的) vectors

Determine the free-vibration solution using the known initialconditions

Understand the concepts of coordinate coupling and principalcoordinates

Determine the forced-vibration solutions under harmonic forces

Understand the concepts of self-excitation and stability of thesystem

Use the Laplace transform approach for solution of two-DOF systems

Solve two-DOF free- and forced-vibration problems using MATLAB


C. R. Yang, NTNU MT

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Chapter Outline

5.1 Introduction

5.2 Equations of Motion for Forced Vibration

5.3 Free Vibration Analysis of an Undamped System

5.4 Torsional System

5.5 Coordinate Coupling and Principal Coordinates

5.6 Forced-Vibration Analysis

5.7 Semidefinite Systems

5.8 Self-Excitation and Stability Analysis

5.9 Transfer-Function Approach

5.10 Solutions Using Laplace Transform

5.11 Solutions Using Frequency Transfer Functions


C. R. Yang, NTNU MT

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5.1Introduction

5.1


C. R. Yang, NTNU MT

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5.1 Introduction

• Two-degree-of-freedom systems are defined as systems that

require two independent coordinates to describe their motion,

as introduced in Fig. 1.12.

Fig. 1.12 Two-degree-of-freedom systems


C. R. Yang, NTNU MT

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X1(t) X2(t)

(t)

X (t)

隆起團塊的


C. R. Yang, NTNU MT

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Fig. 5.2 Automobile

X1(t)X2(t)


C. R. Yang, NTNU MT

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Fig. 5.3 Multistory building

subjected to an earthquake

土地

多層的


C. R. Yang, NTNU MT

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Fig. 5.4 Packaging of an instrument

Fig. 5.4(a) Illustrates the packaging of an instrument of mass m.Assuming that the motion of the instrument is confined to the xy-plane, the system can be modeled as a mass m supported by springsin the x and y directions, as shown in Fig. 5.4(b). Thus the systemhas one point mass m and two degrees of freedom, because themass has two possible types of motion (translations along the x andy directions).


C. R. Yang, NTNU MT

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5.1 Introduction

• The general rule for the computation of the number of degrees of freedom can be stated as follows:

• There are two equations of motion for a two-DOF system, one for each mass (more precisely, for each degree of freedom). They are generally in the form of coupled differential equations—that is, each equation involves all the coordinates.

• If a harmonic solution is assumed for each coordinate, the equations of motion lead to a frequency equation that gives two natural frequencies for the system. If we give suitable initial excitation, the system vibrates at one of these natural frequencies.

No. of degrees of freedom of the system

No. of masses in the system x No. of possible types of motion of each mass

=


C. R. Yang, NTNU MT

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5.1 Introduction

• During free vibration at one of the natural frequencies, the amplitudes of the two-DOFs (coordinates) are related in a specific manner and the configuration is called a normal mode, principal mode, or natural mode of vibration. Thus a two-DOF system has two normal modes of vibration corresponding to the two natural frequencies.

• If we give an arbitrary initial excitation to the system, the resulting free vibration will be a superposition of the two normal modes of vibration. However, if the system vibrates under the action of an external harmonic force, the resulting forced harmonic vibration take place at the frequency of the applied force.

• Under harmonic excitation, resonance occurs (i.e., the amplitudes of the two coordinates will be maximum) when the forcing frequency is equal to one of the natural frequencies of the system.


C. R. Yang, NTNU MT

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C. R. Yang, NTNU MT

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5.2Equations of Motion for Forced Vibration

5.2台灣師範大學機電科技學系

C. R. Yang, NTNU MT

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• Consider a viscously damped two-degree-of-freedom spring-mass

system, shown in the figure below

f f

Fig. 5.5 A two-DOF spring-mass-damper system

k2=c2=0 uncouple


C. R. Yang, NTNU MT

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• The application of Newton’s second law of motion to each of the masses gives the equations of motion:

1

1 2 2 1 2 2 1 1 1 1 1 1 1

2

2 2 2 1 2 2 1 3 2 3 2 2 2

:

( ) ( )

:

( ) ( )

F ma

For mass m

f k x x c x x k x c x m x

For mass m

f k x x c x x k x c x m x

1 1 1 2 1 2 2 1 2 1 2 2 1

2 2 2 1 2 3 2 2 1 2 3 2 2

( ) ( ) (5.1)

( ) ( ) (5.2)

m x c c x c x k k x k x f

m x c x c c x k x k k x f


C. R. Yang, NTNU MT

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• To rewrite the equations of motion as:

• It can be seen that Eq. (5.1) contains terms involving x2

(i.e., ), whereas Eq. (5.2) contains terms involving x1

(i.e., ). Hence they represent a system of two coupled

second-order differential equations. We can therefore expect that the motion of the mass m1 will influence the motion of the mass m2, and vice versa.

1 1 1 2 1 2 2 1 2 1 2 2 1

2 2 2 1 2 3 2 2 1 2 3 2 2

( ) ( ) (5.1)

( ) ( ) (5.2)

m x c c x c x k k x k x f

m x c x c c x k x k k x f

2 2 2 2 c x and k x

2 1 2 1 c x and k x


C. R. Yang, NTNU MT

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• Both equations (5.1 and 5.2) can be written in matrix form as

where [m], [c], and [k] are called the mass, damping, and stiffness matrices, respectively, and are given by

• It can be seen that the matrices [m], [c], and [k] are symmetric:

where the superscript T denotes the transpose of the matrix.

• And the displacement and force vectors are given respectively:

[ ] ( ) [ ] ( ) [ ] ( ) ( ) (5.3)m x t c x t k x t f t

1 2 2 1 2 21

2 2 3 2 2 32

0[ ] [ ] [ ]

0

c c c k k kmm c k

c c c k k km

1 1

2 2

( ) ( )( ) , ( )

( ) ( )

x t f tx t f t

x t f t

][][],[][],[][ kkccmm TTT


C. R. Yang, NTNU MT

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C. R. Yang, NTNU MT

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5.3Free-Vibration Analysis of an Undamped System


C. R. Yang, NTNU MT

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5.3 Free-Vibration Analysis of an Undamped System

• The solution of Eqs.(5.1) and (5.2) involves four constants of

integration (two for each equation). We shall first consider the free

vibration solution of Eqs.(5.1) and (5.2).

• By setting f1(t) = f2(t) = 0, and damping disregarded, i.e., c1 = c2 =

c3=0, and the equation of motion is reduced to:

• Assuming that it is possible to have harmonic motion of m1 and m2

at the same frequency ω and the same phase angle Φ, but with

different amplitude, we take the solutions as

)5.5(0)()()()(

)4.5(0)()()()(

2321222

2212111

txkktxktxm

txktxkktxm

)6.5()cos()(

)cos()(

22

11

tXtx

tXtx

Eq. (5.1)

Eq. (5.2) f1(t)=f2(t)= 0

c1= c2=c3=0


C. R. Yang, NTNU MT

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• Substituting Eq. (5.6) into Eqs.(5.4) and (5.5), we obtain

• Since Eq.(5.7)must be satisfied for all values of the time t, the

terms between brackets must be zero. Thus,

which represent two simultaneous homogenous algebraic equations

in the unknown X1 and X2.

2

1 1 2 1 2 2

2

2 1 2 2 3 2

( ) cos( ) 0

( ) cos( ) 0 (5.7)

m k k X k X t

k X m k k X t

2

2 2 1 1 2

1 1 2 1 2 2

1 2

2 2 2

2 1 2 2 3 2 2

1 2 2 3

( )( ) 0

( ) 0 (5.8) ( )

X m k km k k X k X

X k

X kk X m k k X

X m k k


C. R. Yang, NTNU MT

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• For trivial solution, i.e., X1 = X2 = 0, which implies that there is no

vibration. For a nontrivial solution, the determinant of the

coefficients of X1 and X2 must be zero:

or

Eq. (5.9) is called the frequency or characteristic equation, because

its solution yields the frequencies or the characteristic values of the

system.

0)(

)(det

21

2

12

221

2

1

kkmk

kkkm

)9.5(0))((

)()()(

2

23221

132221

4

21

kkkkk

mkkmkkmm


C. R. Yang, NTNU MT

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• The roots of Eq. (5.9) are given by:

• The roots 1 and 2 are called natural frequencies of the system.

2 2 1 2 2 2 3 1

1 2

1 2

2

1 2 2 2 3 1

1 2

2

1/21 2 2 3 2

1 2

( ) ( )1,

2

( ) ( )1[

2

( )( )4 ] (5.10)

k k m k k m

m m

k k m k k m

m m

k k k k k

m m

2 21 21 2

1 2

, k k

m m


C. R. Yang, NTNU MT

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• To determine the values of X1 and X2, these values depend on the natural

frequencies 1 and 2. We shall denote the values of X1 and X2 corresponding

to 1 as X1(1) and X2

(1) and those corresponding to 2 as X1(2) and X2

(2). From

Eq. (5.8) gives

• The normal modes of vibration corresponding to ω12 and ω2

2 can be

expressed, respectively, as

• The vectors , which denote the normal modes of vibration, are

known as the modal vectors of the system.

2 1 1 2 2

1

1 2 2 2 3

2 1 1 2 2

2

1 2 2 2 3

(2) 2

(1) 2

1

(1) 2

2

2

1

(2)

2

( )

( )

( )(5.11)

( )

X m k k kr

X k m k k

X m k k kr

X k m k k

)12.5( and )2(

12

)2(

1

)2(

2

)2(

1)2(

)1(

11

)1(

1

)1(

2

)1(

1)1(

Xr

X

X

XX

Xr

X

X

XX

(1) (2) X and X


C. R. Yang, NTNU MT

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• The free-vibration solution or the motion in time can be expressed,

using Eq. (5.6), as

Where the constants are determined by the initial

conditions.

(1) (1)

1 1 1(1)

(1) (1)

2 1 1 1

(2) (2)

1 1 2(2)

(2) (2)

2 2 1

1

2 2

1

2

( ) cos( )( )

( ) cos( )

( ) cos( )( ) (5.13)

(second m

first mo

o) cos( )

de

de

x t X tx t

x t r X t

x t X tx t

x t r X t

(1) (2)

1 1 1 2, , , X X and

1 1 1

2 2 2

( ) cos( ) for

( ) cos( ) for (5.6)

x t X t m

x t X t m

(1)

(1)

(2

2

1

1

2

2

)

(2)

1

(5.11)

Xr

X

Xr

X


C. R. Yang, NTNU MT

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• Initial conditions

As stated in Section 5.1, the system can be made to vibrate in its i-th

normal mode (i=1, 2) by subjecting it to the specific initial conditions.

However, for any other general initial conditions, both modes will be

excited. The resulting motion, which is given by the general solution of Eqs.

(5.4) and (5.5), can be obtained by a linear superposition of the two

normal modes, Eq.(5.13):

where c1 and c2 are constants. Since already involve the

unknown constants (see Eq. (5.13)), we can choose c1=c2=1

with no loss of generality.

1 1 1

2 1 2

( )

( )

( 0) some constant, ( 0) 0,

( 0) , ( 0) 0

i

i

i

x t X x t

x t r X x t

1 1 2 2( ) ( ) ( ) (5.14)x t c x t c x t

(1) (2)( ) ( )x t and x t

(1) (2)

1 1 X and X


C. R. Yang, NTNU MT

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• Thus, the components of the vector can be expressed, using

Eq. (5.14) with c1=c2=1 and Eq. (5.13), as

• The unknown constants can be determined from

the initial conditions:

(1) (2)

2 2

(1) (2)

1

(1) (2) (1) (2)

1 1 1 1 1 1 1 2

1 2 1

2

(1) (2)

2 2 2 1 1 2 2

1 1 2 2

( ) ( ) ( ) cos( ) cos( )

( ) ( ) ( ) cos( ) cos( )

cos( ) cos( ) (5.15)

X X

r X r

x t x t x t X t X t

x t x t x t t t

t tX

1 1 1 1

2 2 2 2

( 0) (0), ( 0) (0),

( 0) (0), ( 0) (0) (5.16)

x t x x t x

x t x x t x

( )x t

(1) (2)

1 1 1 2, , , X X and


C. R. Yang, NTNU MT

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• Substituting Eq. (5.16) into Eq.(5.15) leads to

• The solution of Eq. (5.17) can be expressed as

)17.5(sinsin)0(

coscos)0(

sinsin)0(

coscos)0(

2

)2(

1221

)1(

1112

2

)2(

121

)1(

112

2

)2(

121

)1(

111

2

)2(

11

)1(

11

XrXrx

XrXrx

XXx

XXx

)(

)0()0(sin,

)(

)0()0(sin

)0()0(cos,

)0()0(cos

122

2112

)2(

1

121

2121

)1(

1

12

2112

)2(

1

12

2121

)1(

1

rr

xxrX

rr

xxrX

rr

xxrX

rr

xxrX

Four algebraic equations

in the unknowns

(1)

1 1

(2)

1 2

(1)

1 1

(2)

1 2

cos

cos

sin

sin

X

X

X

X


C. R. Yang, NTNU MT

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• We can obtain the desired solution as

1/22 2

(1) (1) (1)

1 1 1 1 1

1/22 2

(2) (2) (2)

1 1

1/22

2 2 1 2

2 1 2 2

2 1 1

1/22

2 1 1 2

1 1

2 1 2

(1)

1 1

1

2 2

2 1 2

(0) (0)1(0) (0)

( )

(0) (0

cos sin

cos si

)1(0)

n

sta

)

n

(0)(

X X X

X X

r x xr x x

r r

r x xr x x

r r

X

X

1

(1)

1 1

(2)

1 1 2

2 (2)

1

1 2 1 2

1 2 1 2

1 1 1 2

2 1 1 22

in

cos

sinta

(0) (0)tan

[ (0) (0)

(0) (0)tan (5.18)

[ (0) (0n

c s )o

r x x

r x x

r x x

r x x

X

X

X


C. R. Yang, NTNU MT

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Example 5.1 Frequencies of Spring-Mass System


C. R. Yang, NTNU MT

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Fig. 5.6 Two-DOF system

2 21 21 2

1 2

, k k

m m


C. R. Yang, NTNU MT

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(1) (1)

2 1 1X r X

(2) (2)

2 2 1X r X

for m1 vibration

for m2 vibration

for m1 vibration

for m2 vibration

under 1

under 2

superposition

superposition


C. R. Yang, NTNU MT

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for m1 vibration

for m2 vibration


C. R. Yang, NTNU MT

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Example 5.3

Free-Vibration Response of a Two-Degree-of-Freedom System

Find the free-vibration response of the system shown in Fig.5.5(a)

with k1 = 30, k2 = 5, k3 = 0, m1 = 10, m2 = 1 and c1 = c2 = c3 = 0 for

the initial conditions .1 1 2 2(0) 1, (0) (0) (0) 0x x x x


C. R. Yang, NTNU MT

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Example 5.3

Free-Vibration Response of a Two Degree of Freedom System

Solution

For the given data, the eigenvalue problem, Eq.(5.8), becomes

By setting the determinant of the coefficient matrix in Eq.(E.1) to zero,

we obtain the frequency equation,

(E.1)0

0

55-

5 3510

0

0

2

1

2

2

2

1

32

2

22

221

2

1

X

X

X

X

kkmk

kkkm

4 210 85 150 0 (E.2)


C. R. Yang, NTNU MT

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Example 5.3


Solution

The natural frequencies can be found as

The substitution of in Eq. (E.1) leads to ,

while yields . Thus the normal modes

(or eigenvectors) are given by

2 2

1 2 1 22 5 6 0 1 5811 2 4495 or E.3). , . . , . (

1 2

1 11 1 2 2

1 11 2

2 2

1 1

2 5

X X X X X X E.4) E.

X5

X)

( ) ( )

( ) ( ) ( ) ( )

( ) ( )( (

2 2

1 2 5. 1 1

2 12X X( ) ( )

2 2

2 6 0. 2 2

2 15X X( ) ( )

(或1與2代入Eq. (5.11)求得r1與r2)

(1與2代入 (E.1)求得X1=rX2)


C. R. Yang, NTNU MT

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Example 5.3


Solution

The free-vibration responses of the masses m1 and m2 are given by

(see Eq.5.15):

By using the given initial conditions in Eqs.(E.6) and (E.7), we obtain

(E.7))4495.2cos(5)5811.1cos(2)(

(E.6))4495.2cos()5811.1cos()(

2

)2(

11

)1(

12

2

)2(

11

)1(

11

tXtXtx

tXtXtx

(1) (2)

1 1 1 1 2

(1) (2)

2 1 1 1 2

(1) (2)

1 1 1 1 2

(1) (2)

2 1 1 1 2

( 0) 1 cos cos (E.8)

( 0) 0 2 cos 5 cos (E.9)

( 0) 0 1.5811 sin 2.4495 sin (E.10)

( 0) 0 3.1622 sin 12.2475 sin (E.11)

x t X X

x t X X

x t X X

x t X X

The unknown constants

can be determined from

the initial conditions:

X X and(1) (2)

1 1 1 2, , ,


C. R. Yang, NTNU MT

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Example 5.3


Solution

The solution of Eqs.(E.8) and (E.9) yields

The solution of Eqs.(E.10) and (E.11) leads to

Equations (E.12) and (E.13) gives

(E.12)7

2cos;

7

5cos 2

)2(

11

)1(

1 XX

(1) (2)

1 1 1 2 1 2sin 0, sin 0 sin 0, sin 0 (E.13) X X

(E.14)0,0,7

2,

7

521

)2(

1

)1(

1 XX


C. R. Yang, NTNU MT

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Example 5.3


Solution

Thus the free vibration responses of m1 and m2 are given by

(E.16)4495.2cos7

105811.1cos

7

10)(

(E.15)4495.2cos7

25811.1cos

7

5)(

2

1

tttx

tttx


C. R. Yang, NTNU MT

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5.4Torsional System

5.4


C. R. Yang, NTNU MT

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• Consider a torsional system as shown in Fig.5.8. The three segments of the

shaft have rotational spring constants kt1, kt2, and kt3, The mass moments of

inertia of discs are J1 and J2, the applied torques Mt1 and Mt2, and the

rotational degrees of freedom 1 and 2.

• The differential equations of rotational motion for the discs can be derived as

1 1 1 1 2 2 1 1

2 2 2 2 1 3 2 2

( )

( )

t t t

t t t

J k k M

J k k M

Fig. 5.8 Torsional system with discs mounted on a shaft

M J


C. R. Yang, NTNU MT

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• Upon rearrangement become

• For the free vibration analysis of the

system, Eq.(5.19) reduces to

• Note that Eq. (5.20) is similar to Eqs. (5.4) and (5.5). In fact, Eq.

(5.20) can be obtained by substituting 1, 2, J1, J2, kt1, kt2, and kt3

for x1, x2, m1, m2, k1, k2, and k3 , respectively. Thus the analysis

presented in Section 5.3 is also applicable to torsional systems.

)20.5(0)(

0)(

2321222

2212111

ttt

ttt

kkkJ

kkkJ

)19.5()(

)(

22321222

12212111

tttt

tttt

MkkkJ

MkkkJ


C. R. Yang, NTNU MT

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Example 5.4

Natural Frequencies of a Torsional System

Find the natural frequencies and mode shapes for the torsional system

shown in the figure below for J1 = J0 , J2 = 2J0 and kt1 = kt2 = kt .


C. R. Yang, NTNU MT

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Example 5.4


Solution

The differential equations of motion, Eq.(5.20), reduce to (with kt3 = 0,

kt1 = kt2 = kt, J1 = J0 and J2 = 2J0):

Rearranging and substituting the harmonic solution:

(E.1) 02

02

2120

2110

tt

tt

kkJ

kkJ

( ) cos( ); 1,2 (E.2)i i

t t i

1 1 1 2 1 2 2

2 2 2 1 2 3 2

( ) 0

( ) 0 (5.20)

t t t

t t t

J k k k

J k k k

Matrix

form


C. R. Yang, NTNU MT

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Example 5.4


Solution

This gives the frequency equation of

The solution of Eq.(E.3) gives the natural frequencies

4 2 2 2 2 2

0 0 1 2

0 0

(5 17) (5 17)2 5 0 (E.3) ,

4 4

t t

t t

k kJ J k k

J J

1 2

0 0

(5 17) and (5 17) (E.4)4 4

t tk k

J J


C. R. Yang, NTNU MT

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Example 5.4


Solution

The amplitude ratios are given by

Equations (E.4) and (E.5) can also be obtained by substituting the

following values into Eqs.(5.10) and (5.11).

(E.5)4

)175(2

4

)175(2

)2(

1

)2(

22

)1(

1

)1(

21

r

r

0and2,

,,

3022011

2211

kJJmJJm

kkkkkk tttt

Similar to Eq. (5.13), we can obtain-First mode and Second mode


C. R. Yang, NTNU MT

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Similar to Eq. (5.13), we can obtain-First mode and Second mode

Example 5.4


Solution

Where the constants are determined by the initial

conditions.

(1) (1)

1 1 1(1)

(1) (1)

2 1 1 1

(2) (2)

1 1 2(2)

(2) (2)

2 2 1

1

2 2

1

2

( ) cos( )( )

( ) cos( )

( ) cos( )( ) (5.13)

( ) cos(second mode

first od

)

m et t

tt r t

t tt

t r t

(1) (2)

1 1 1 2, , , and


C. R. Yang, NTNU MT

-48-

1 2

0 0

5.4

(5 17) and (5 17) (E.4)4 4

t t

Example

k k

J J


C. R. Yang, NTNU MT

-49-

5.5Coordinate Coupling and Principal Coordinates


C. R. Yang, NTNU MT

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(一般化座標)

尾座

頭座


C. R. Yang, NTNU MT

-51-

Fig. 5.11 Lathe

尾座頭座


C. R. Yang, NTNU MT

-52-

Fig. 5.12 Modeling of a lathe

Generalized coordinates

有四種表示方式：

1 2

1

( ), ( ); ( ), ( )

( ), ( ); ( ), ( )

x t x t x t t

x t t y t t


C. R. Yang, NTNU MT

-53-


Equations of Motion using x(t) and (t) (支點在C.G.):

From the free-body diagram shown in Fig. 5.12 (a), with the

positive values of the motion variables as indicated, the force

equilibrium equation in the vertical direction can be written as

And the moment equation about C.G. can be expressed as

Eqs.(5.21) and (5.22) can be rearranged and written in matrix form

as

1 1 2 2 1 2 2 2 1 1(5.2( ) ( ) ( ) 01) ( )mx k x l k x l mx k l k lk k x

2 2

2 2

0 1 1 1 2 2 2 0 1 11 1 2 2(5( ) ( ) ( ).22) 0( )J k x l l k x l k l kl J k l k ll x

)23.5(0

0

)( )(

)( )(

0

0 2

2

2

12211

221121

0 21

x

lklklklk

lklkkkx

J

m

(向下為正)

(順時針為正)


C. R. Yang, NTNU MT

-54-

Equations of motion Using y(t) and θ(t) (支點在P):

From Fig. 5.12(b), where y(t) and θ(t) are used as the generalized

coordinates of the system, the equations of motion for translation and

rotation can be written as

1 1 2 2 1 2 2 2 1 1 0my k y l k y l me my me k k y k l k l( ) ( ) ( ) ( )

2 2

1 1 1 2 2 2 1 1 2 2 1 1 2 2 0P PJ k y l l k y l l mey J mey k l k l y k l k l( ) ( ) ( ) ( )

(向下為正)

(順時針為正)

作用在C.G.點的轉矩

作用在C.G.點的力


C. R. Yang, NTNU MT

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These equations can be rearranged and written in matrix form as

)25.5(0

0

)()(

)()(

2

2

2

112211

112221

2

y

lklklklk

lklkkky

Jme

mem

P

me mey即有和不等於零項

=mass (m) × tangential acceleration ( )

=force ( ) × moment arm (e)my

e


C. R. Yang, NTNU MT

-56-

固有的


C. R. Yang, NTNU MT

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Example 5.6

Principal Coordinates of Spring-

Mass System

Determine the principal coordinates

for the spring-mass system shown in

the figure 5.6.

Fig. 5.6 Two-DOF system


C. R. Yang, NTNU MT

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Example 5.6 Principal Coordinates of Spring-Mass SystemSolution

Approach: Define two independent solutions as principal coordinates and express them in terms of the solutions x1(t) and x2(t).

The general motion of the system shown in Fig. 5.6 is given by Eq. (E.10) of Example 5.1:

Where are constants.

(E.1)3

coscos)(

3coscos)(

22112

22111

tm

kBt

m

kBtx

tm

kBt

m

kBtx

(1) (2)

1 1 2 1 1 2, , , B X B X and


C. R. Yang, NTNU MT

-59-


Example 5.6 Principal Coordinates of Spring-Mass System

Solution

We define a new set of coordinates q1(t) and q2(t) such that

Since q1(t) and q2(t) are harmonic functions, their corresponding

equations of motion can be written as

1 1 1

2 2 2

( ) cos

3( ) cos (E.2)

kq t B t

m

kq t B t

m

1 1

2 2

0

30 (E.3)

kq q

m

kq q

m

These equations represent a two-DOF system

whose natural frequencies are

.

Because there is neither static nor dynamic

coupling in the equations of motion (E.3), q1(t) and q2(t) are principal coordinates.

1 2 3k m and k m

Note that the equation of motion corresponding to the solution q=Bcos(t+) is given by 2 0q q


C. R. Yang, NTNU MT

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Example 5.6 Principal Coordinates of Spring-Mass System

Solution

From Eqs.(E.1) and (E.2), we can write

The solution of Eqs.(E.4) gives the principal coordinates:

(E.4))()()(

)()()(

212

211

tqtqtx

tqtqtx

(E.5))]()([2

1)(

)]()([2

1)(

212

211

txtxtq

txtxtq


C. R. Yang, NTNU MT

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5.6Forced-Vibration Analysis


C. R. Yang, NTNU MT

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• The equations of motion of a general two-degree-of-freedom system under external forces can be written as

Eqs. (5.1) and (5.2) can be seen to be special cases of Eq. (5.27), with m11=m1, m22=m2, and m12=0.

• Consider the external forces to be harmonic:

where ω is the forcing frequency.

• We can write the steady-state solutions as

Where X1 and X2 are, in general, complex quantities that depend on and the system parameters.

)27.5(

2

1

2

1

2221

1211

2

1

2221

1211

2

1

2212

1211

F

F

x

x

kk

kk

x

x

cc

cc

x

x

mm

mm

0 1 2 5 28i t

j jF t F e j( ) , , ( . )

)29.5(2,1,)( jeXtx ti

jj


C. R. Yang, NTNU MT

-63-

• Substitution of Eqs. (5.28) and (5.29) into Eq. (5.27) leads to

as in Section 3.5 (Page 278), we define the mechanical impedance Zrs(i) as

2 2

11 11 11 12 12 12 1

2 2212 12 12 22 22 22

10

20

( ) ( )

( ) ( )

(5.30)

m i c k m i c k X

Xm i c k m i c k

F

F

rs rs rs rsZ i m i c k r s2

( ) , 1,2 (5.31)



C. R. Yang, NTNU MT

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• We can write Eq.(5.30) as:

where

• Eq.(5.32) can be solved to obtain:

)32.5()( 0FXiZ

20

10

0

2

1

2212

1211matrix Impedance

)( )(

)( )()(

F

FF

X

XX

iZiZ

iZiZiZ

1

05 33

( ) ( . )X FZ i


C. R. Yang, NTNU MT

-65-


• The inverse of the impedance matrix is given by

• Eqs.(5.33) and (5.34) lead to the solution

• By substituting Eq. (5.35) into Eq. (5.29), we can find the complete

solution, x1(t) and x2(t).

1 22 12

2

11 22 12 12 11

1 Z i - Z iZ i

Z i Z i Z i Z i Z i

( ) ( )( )

( ) ( ) ( ) ( ) ( )

22 10 12 201 2

11 22 12

12 10 11 202 2

11 22 12

( ) ( )( )

( ) ( ) ( )

( ) ( )( ) (5.35)

( ) ( ) ( )

Z i F Z i FX i

Z i Z i Z i

Z i F Z i FX i

Z i Z i Z i

22 121

21 11

1

det

a a

a aA

A

1

0 5 33( .( ) )X Z i F

( ) , 1,2 (5.29)i t

j jx t X e j


C. R. Yang, NTNU MT

-66-


Example 5.8

Steady-State Response of Spring-

Mass System

Find the steady-state response of system

shown in Fig.5.15 when the mass m1 is

excited by the force F1(t) = F10 cos ωt.

Also, plot its frequency response curve.

Solution

The equations of motion of the system

can be expressed as (from Eq. (5.3))

Fig. 5.15 A two-mass system subjected to harmonic force.

1 1 10

2 2

cos 0 2 (E.1)

0 2 0

x x F tm k - k

m x -k k x


C. R. Yang, NTNU MT

-67-


Example 5.8 Steady-State Response of Spring-Mass System

Solution

Comparison of Eq. (E.1) with Eq. (5.27) shows that

We assume the solution to be as follows

Eq.(5.31) gives

( ) cos ; 1,2 (E.2)j jx t X t j

2

11 22 12( ) ( ) 2 , ( ) (E.3)Z Z m k Z k

m m m m c c c

k k k k k F F t F

11 22 12 11 12 22

11 22 12 1 10 2

, 0, 0,

2 , , cos , 0

Since F10cost=Re(F10eit), we

shall assume the solution also to

be xj=Re(Xjeit)=Xjcost, j=1, 2.


C. R. Yang, NTNU MT

-68-


Example 5.8 Steady-State Response of Spring-Mass System

Solution

Hence X1 and X2 are given by Eq. (5.35)

By defining , Eqs.(E.4) and (E.5) can be expressed

as (E.6) and (E.7).

2 2

10 101 2 2 2 2 2

10 102 2 2 2 2 2

( 2 ) ( 2 )( ) (E.4)

( 2 ) ( 3 )( )

( ) (E.5)( 2 ) ( 3 )( )

m k F m k FX

m k k m k m k

kF kFX

m k k m k m k

k m and k m2 2

1 2 3


C. R. Yang, NTNU MT

-69-

E.7)(

1

)(2

1

2

1

2

1

2

102

k

FX

2

10

1

1 2 2 2

2

1 1 1

2

( ) (E.6)

1

F

X

k


C. R. Yang, NTNU MT

-70-

Fig. 5.16 Frequency-response curves of example 5.8


C. R. Yang, NTNU MT

-71-

5.7Semidefinite Systems


C. R. Yang, NTNU MT

-72-

5.7 Semidefinite (半定) Systems

退化未拘束

奇異的(行列式=0)


C. R. Yang, NTNU MT

-73-


Fig. 5.17 Semidefinite systems with two-DOF台灣師範大學機電科技學系

C. R. Yang, NTNU MT

-74-


• For Fig. 5.17 (a) and (b), the equations of motion can be written as

• For free vibration, we assume the motion to be harmonic:

• Substituting Eq.(5.37) into Eq.(5.36) gives

1 1 2 1 1 1 1 2

2 2 2 1 2 2 2 1

( ) ( ) 0

( ) ( ) 0 (5.36)

m x k x x m x k x x

m x k x x m x k x x

)37.5(2,1),cos()( jtXtx jjj

)38.5(0)(

0)(

2

2

21

21

2

1

XkmkX

kXXkm


C. R. Yang, NTNU MT

-75-


• By equating the determinant of the coefficients of X1 and X2 to zero,

we obtain the frequency equation as

• From which the natural frequencies can be obtained:

)39.5(0)]([ 21

2

21

2 mmkmm

)40.5()(

and 021

2121

mm

mmk


C. R. Yang, NTNU MT

-76-

Example 5.9 Free Vibration of an Understrained System


C. R. Yang, NTNU MT

-77- 台灣師範大學機電科技學系

C. R. Yang, NTNU MT

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C. R. Yang, NTNU MT


C. R. Yang, NTNU MT

-80-

5.8Self-Excitation and Stability Analysis

5.8


C. R. Yang, NTNU MT

-81-

5.8 Self-Excitation and Stability Analysis


C. R. Yang, NTNU MT

-82-


C. R. Yang, NTNU MT

-83-

四次方的

羅斯 - 赫維茲


C. R. Yang, NTNU MT

-84-

The same with Eq. (5.46)

(羅斯 -赫維茲準則)Routh-Hurwitz criterion


C. R. Yang, NTNU MT

-85-

(羅斯 -赫維茲準則)Routh-Hurwitz criterion


C. R. Yang, NTNU MT

-86-

0 2

1 3

1

a a

a aA

a

0 4

1 5

1

a a

a aB

a

0 6

1

6

1

0

a a

aa

a

1 3

a a

A BC

A

1 5

6

a a

A aD

A

1 0

00

a

A

A


C. R. Yang, NTNU MT


C. R. Yang, NTNU MT

-88-

5.9Transfer-Function Approach

5.9


C. R. Yang, NTNU MT

-89-


• For two-DOF system shown in Fig. 5.5, the equations of motion are

(Eqs. (5.1) and (5.2)):

• By taking the Laplace transforms of Eqs. (5.50) and (5.51),

assuming zero initial conditions, we obtain

5.51

5.50

2122321223222

1221212212111

fxkxkkxcxccxm

fxkxkkxcxccxm

5.53

5.52

212232122322

2

2

122121221211

2

1

sFsXksXkkssXcssXccsXsm

sFsXksXkkssXcssXccsXsm


C. R. Yang, NTNU MT

-90-

• Eqs. (5.52) and (5.53) can be rearranged to obtain

• Eqs. (5.54) and (5.55) indicate two simultaneous linear algebraic

equation in X1(s) and X2(s).These can be solved using Cramer’s rule

as

2

1 1 2 1 2 2 2 1

2

2 2

1

2

2

1 2 3 3 22 2

[ ] ( ) 5.54

( ) [ ) 5.53

m s c c s k k c s k F s

c s k m s c c s k

X s X s

X s sk sX F


1 2

1

2

2

1 5.56 and

(t) (

5.57

)Apply inverse Laplace transfor

D s D sX s X

ms x and x

sD s D s

t


C. R. Yang, NTNU MT

-91-


In Eqs. (5.56) and (5.57)


C. R. Yang, NTNU MT

-92-


• Note that

1. The denominator, D(s), in the expressions of X1(s) and X2(s)

given by Eq. (5.60), is a fourth-order polynomial in s and

denotes the characteristic polynomial of the system. Because

the characteristic polynomial is of order four, the model (or

system) is said to be a fourth-order model (or system).

2. Equations (5.56) and (5.57) permit us to apply inverse Laplace

transforms to obtain the fourth-order differential equations for

x1(t) and x2(t) , respectively.

3. Equations (5.56) and (5.57) can be used to derive the transfer

functions of x1(t) and x2(t) corresponding to any specified

forcing function.

1 2

1 2 : and X s X s

Transfer function H s H sF s F s


C. R. Yang, NTNU MT

-93-

5.10Solutions Using Laplace Transform

5.10

The computation of response of two-degree-of-freedom systems using the Laplace transform.


C. R. Yang, NTNU MT

-94-

5.10Solutions Using Laplace Transform

振動問題

解析

建模聯立微分方程式

(ODEs)

(t的函數)

代數方程式(s的函數)

(僅需加減乘除運算，可簡化求解的程序)

求解1

X1(s)與X2(s)

求解2

x1(t)與x2(t)

解釋物理現象

Cramer’s rule

求解

取£

取£-1

部份分式展開轉換公式應用

£[x]

£[x]


C. R. Yang, NTNU MT


C. R. Yang, NTNU MT

-96-


Example 5.12

Response Under Impulse Using Laplace Transform Method

Two railway cars, of masses m1 = M and m2 = m are connected by a

spring of stiffness k, as shown in the figure. If the car of mass M is

subjected to an impulse , determine the time responses of the

cars using the Laplace transform method.

tF 0


C. R. Yang, NTNU MT

-97-


Example 5.12


Solution

The responses of the cars can be determined using either of the

following approaches:

a. Consider the system to be undergoing free vibration due to the

initial velocity caused by the impulse applied to car M.

b. Consider the system to be undergoing forced vibration due to the

force applied to car M (with the displacements and velocities

of cars M and m considered to be zero initially). tF 0


C. R. Yang, NTNU MT

-98-


Example 5.12


Solution

Using the second approach, the equations of motion of the cars can be

expressed from Eq. (5.36)

Using the Laplace transforms, Eqs. (E.1) and (E.2) can be written as

1 1 2 0

2 2 1

E.1

0 E.2

Mx k x x F t

mx k x x

E.4 0

E.3

2

2

1

021

2

sXkmsskX

FskXsXkMs


C. R. Yang, NTNU MT

-99-


Example 5.12


Solution

Equations (E.3) and (E.4) can be solved by Cramer’s rule

E.6

E.5

22

02

22

2

01

mMkMmss

kFsX

mMkMmss

kmsFsX


C. R. Yang, NTNU MT

-100-


Example 5.12


Solution

Using partial fractions, Eqs. (E.5) and (E.6) can be rewritten as

where

E.8 11

E.7 1

222

02

222

01

ws

w

wsmM

FsX

ws

w

wM

m

smM

FsX

E.9 112

mMkw


C. R. Yang, NTNU MT

-101-


Example 5.12


Solution

The inverse transforms of Eqs. (E.7) and (E.8), using the results of

Appendix D, yield the time responses of the cars as

01

02

sin E.10

1sin E.11

F mx s t wt

M m wM

Fx s t wt

M m w


C. R. Yang, NTNU MT

-102-

5.11Solutions Using Frequency Transfer Functions

5.11

The frequency transfer function can be obtained by substituting i in place of s in the general transfer function of the system.


C. R. Yang, NTNU MT

-103-


Example 5.13

Derivation of Frequency

Transfer Functions

Derive the frequency transfer

functions of x1(t) and x2(t) for

the system shown in figure.


C. R. Yang, NTNU MT

-104-


Example 5.13

Derivation of Frequency Transfer Functions

Solution

From the free-body diagrams of the masses, the equations

of motion of the system is

E.2 0

E.1 sin

212212222

01212212111111

pxxkxxcxm

wtPpxxkxxcxkxcxm

(向下為正)


C. R. Yang, NTNU MT

-105-


Example 5.13


Solution

By taking the Laplace transforms of Eqs. (E.1) and (E.2), assuming

zero initial conditions,

E.4 0

E.3

12212222

121221211111

2

1

sXsXksXsXcsXm

sPsXsXksXsXcsXkssXcsXsm


C. R. Yang, NTNU MT

-106-


Example 5.13


Solution

The solutions X1(s) and X2(s) of Eqs. (E.3) and (E.4) are

where

E.6 and E.5 22

11

sD

sDsX

sD

sDsX

2

1 2 2 2

12 2

1

2

E.7

E.8

D s m s c s k

D s

P s

c s k P s


C. R. Yang, NTNU MT

-107-


Example 5.13


Solution

We have

The general transfer functions is

E.9 211221

2

21221221

3

221221

4

22

kkskckc

scckmkmkmscmcmcmsmmsD

2

21 22

1 1

2 2 2 E.9 and E.10X s X s

P s P

m s c s k c s k

D s D ss


C. R. Yang, NTNU MT

-108-


Example 5.13


Solution

The frequency transfer functions (s → i) of x1(t) and x2(t) is

where

E.13 and E.12 22

1

222

2

2

1

1

iwD

kiwc

iwP

iwX

iwD

kiwcwm

iwP

iwX

21122121221221

2

221221

4

21

4

kkkckciwcckmkmkmw

cmcmcmiwwmmwiwD


C. R. Yang, NTNU MT

-109-

5.12 Examples using MATLAB

•To practice by yourself from Ex. 5.15 to Ex.5.21

•The source codes of all MATLAB programs are given at the companion website

c. r. yang, ntnu mt chapter 5 two degree freedom …compute the eigenvalues or natural frequencies...

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