week 15 vibration

TWO DEGREE OF FREEDOM SYSTEM

INTRODUCTION Systems that require two independent coordinates

to describe their motion; Two masses in the system X two possible types of

motion of each mass. Example: motor pump system.

There are two equations of motion for a 2DOF system, one for each mass (more precisely, for each DOF).

They are generally in the form of couple differential equation that is, each equation involves all the coordinates.

EQUATION OF MOTION FOR FORCED VIBRATION Consider a viscously damped two degree

of freedom spring-mass system, shown in Fig.5.3.

Figure 5.3: A two degree of freedom spring-mass-damper system

4

Both equations can be written in matrix form as

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

EQUATIONS OF MOTION FOR FORCED VIBRATION

)2.5()()()1.5()()(

2232122321222

1221212212111

FxkkxkxccxcxmFxkxkkxcxccxm

)3.5( )()(][)(][)(][ tFtxktxctxm

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

5

And the displacement and force vectors are given respectively:


322

221

322

221

2

1

][

][ 0

0 ][

kkkkkk

k

cccccc

cm

mm

)()(

)()()(

)(2

1

2

1

tFtF

tFtxtx

tx

It can be seen that the matrices [m], [c], and [k] are all 2 x 2 matrices whose elements are known masses, damping coefficient and stiffnesses of the system, respectively.

6

where the superscript T denotes the transpose of the matrix.


oThe solution of Eqs.(5.1) and (5.2) involves four constants of integration (two for each equation). Usually the initial displacements and velocities of the two masses are specified as

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

oFurther, these matrices can be seen to be symmetric, so that,

x1(t = 0) = x1(0) and 1( t = 0) = 1(0), x2(t = 0) = x2(0) and 2 (t = 0) = 2(0).

xx x

x

7

Assuming that it is possible to have harmonic motion of m1 and m2 at the same frequency ω and the same phase angle Φ, we take the solutions as

FREE VIBRATION ANALYSIS OF AN UNDAMPED SYSTEM

)5.5(0)()()()()4.5(0)()()()(

2321222

2212111

txkktxktxmtxktxkktxm

By setting F1(t) = F2(t) = 0, and damping disregarded, i.e., c1 = c2 = c3 = 0, and the equation of motion is reduced to:

)6.5()cos()()cos()(

22

11

tXtxtXtx

8

Since Eq.(5.7)must be satisfied for all values of the time t, the terms between brackets must be zero. Thus,


)7.5( 0)cos()(

0)cos()(

2322

212

221212

1

tXkkmXk

tXkXkkm

Substituting into Eqs.(5.4) and (5.5),

)8.5(0)(

0)(

2322

212

221212

1

XkkmXk

XkXkkm

9

or

0

)(

)(det

212

12

2212

1

kkmk

kkkm

which represent two simultaneous homogenous algebraic equations in the unknown X1 and X2. For trivial solution, i.e., X1 = X2 = 0, there is no solution. For a nontrivial solution, the determinant of the coefficients of X1 and X2 must be zero:

)9.5(0))((

)()()(223221

1322214

21

kkkkk

mkkmkkmm


10

The roots are called natural frequencies of the system.

which is called the frequency or characteristic equation. Hence the roots are:

)10.5())((4

)()(21

)()(21,

2/1

21

223221

2

21

132221

21

13222122

21

mmkkkkk

mmmkkmkk

mmmkkmkk


11

The normal modes of vibration corresponding to ω1

2 and ω22 can be

expressed, respectively, as

)11.5()(

)(

)()(

32222

2

2

21221

)2(1

)2(2

2

32212

2

2

21211

)1(1

)1(2

1

kkmk

kkkm

XXr

kkmk

kkkm

XXr

To determine the values of X1 and X2, given ratio

)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

which are known as the modal vectors of the system.


12

Where the constants , , and are determined by the initial conditions. The initial conditions are

(5.17)mode second)cos(

)cos(

)(

)()(

modefirst )cos(

)cos(

)(

)()(

22)2(

12

22)2(

1

)2(2

)2(1)2(

11)1(

11

11)1(

1

)1(2

)1(1)1(

tXr

tX

tx

txtx

tXr

tX

tx

txtx

The free vibration solution or the motion in time can be expressed itself as

0)0(,)0(

,0)0(constant, some )0(

2)(

12

1)(

11

txXrtx

txXtxi

i

i


)1(1X

)2(1X 1 2

13

Thus the components of the vector can be expressed as

)14.5()()()( 2211 txctxctx

The resulting motion can be obtained by a linear superposition of the two normal modes, Eq.(5.13)

)15.5()cos()cos(

)()()(

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

22)2(

1211)1(

11

)2(2

)1(22

22)2(

111)1(

1)2(

1)1(

11

tXrtXr

txtxtx

tXtXtxtxtx

where the unknown constants can be determined from the initial conditions:


14

Substituting into Eq.(5.15) leads to)16.5()0()0(),0()0(

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

2222

1111

xtxxtxxtxxtx

)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

The solution can be expressed as

)()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

rrxxrX

rrxxrX

rrxxrX

rrxxrX


15

from which we obtain the desired solution

)18.5()0()0([

)0()0(tancossintan

)0()0([)0()0(tan

cossintan

)0()0()0()0()(

1

sincos

)0()0()0()0()(

1

sincos

2112

2111

2)2(

1

2)2(

112

2121

2121

1)1(

1

1)1(

111

2/1

22

22112

21112

2/122

)2(1

22

)2(1

)2(1

2/1

21

22122

21212

2/121

)1(1

21

)1(1

)1(1

xxrxxr

XX

xxrxxr

XX

xxrxxrrr

XXX

xxrxxrrr

XXX


16

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

EXAMPLE 5.3:FREE VIBRATION RESPONSE OF A TWO DEGREE OF FREEDOM SYSTEM

).0()0()0( ,1)0( 2211 xxxx

Find the free vibration response of the system shown in Fig.5.3(a) with k1 = 30, k2 = 5, k3 = 0, m1 = 10, m2 = 1 and c1 = c2 = c3 = 0 for the initial conditions

(E.1)00

55-

5 3510

00

2

1

2

2

2

1

322

22

2212

1

XX

XX

kkmk

kkkm

or

17

from which the natural frequencies can be found as

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

EXAMPLE 5.3 SOLUTION

(E.2)01508510 24

E.3)(4495.2,5811.10.6,5.2

21

22

21

The normal modes (or eigenvectors) are given by

E.5)(5

1

E.4)(21

)2(1)2(

2

)2(1)2(

)1(1)1(

2

)1(1)1(

XX

XX

XX

XX

18

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

The free vibration responses of the masses m1 and m2 are given by (see Eq.5.15):

(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

(E.11)sin2475.121622.3)0(

(E.10)sin4495.2sin5811.10)0(

(E.9)cos5cos20)0(

(E.8)coscos1)0(

2)2(

1)1(

12

2)2(

11)1(

11

2)2(

11)1(

12

2)2(

11)1(

11

XXtx

XXtx

XXtx

XXtx


19

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

The solution of Eqs.(E.8) and (E.9) yields(E.12)

72cos;

75cos 2

)2(11

)1(1 XX

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

11)1(

1 XX

Equations (E.12) and (E.13) give(E.14)0,0,

72,

75

21)2(

1)1(

1 XX


20

EXAMPLE 5.3 SOLUTIONThus the free vibration responses of m1 and m2 are given by

(E.16)4495.2cos7

105811.1cos7

10)(

(E.15)4495.2cos725811.1cos

75)(

2

1

tttx

tttx

21

Figure 5.6: Torsional system with discs mounted on a shaft

TORSIONAL SYSTEM

Consider a torsional system as shown in Fig.5.6. The differential equations of rotational motion for the discs can be derived as

22

which upon rearrangement become

TORSIONAL SYSTEM

22312222

11221111

)(

)(

ttt

ttt

MkkJ

MkkJ

)19.5()(

)(

22321222

12212111

tttt

tttt

MkkkJ

MkkkJ

For the free vibration analysis of the system, Eq.(5.19) reduces to

)20.5(0)(

0)(

2321222

2212111

ttt

ttt

kkkJ

kkkJ

23

Find the natural frequencies and mode shapes for the torsional system shown in Fig.5.7 for J1 = J0 , J2 = 2J0 and kt1 = kt2 = kt .

Solution: The differential equations of motion,

Eq.(5.20), reduce to (with kt3 = 0, kt1 = kt2 = kt, J1 = J0 and J2 = 2J0):

EXAMPLE 5.4:NATURAL FREQUENCIES OF TORSIONAL SYSTEM

(E.1) 02

02

2120

2110

tt

tt

kkJ

kkJ

Fig.5.7:

Torsional system

24

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

gives the frequency equation:


(E.2)2,1);cos()( itt ii

(E.3)052 20

220

4 tt kkJJ

(E.4))175(4

and)175(4 0

20

1 Jk

Jk tt

Rearranging and substituting the harmonic solution:

25

Equations (E.4) and (E.5) can also be obtained by substituting the following in Eqs.(5.10) and (5.11).

The amplitude ratios are given by

(E.5)4

)175(2

4)175(2

)2(1

)2(2

2

)1(1

)1(2

1

r

r

0and2,,,

3022011

2211

kJJmJJmkkkkkk tttt


26

Generalized coordinates are sets of n coordinates used to describe the configuration of the system.•Equations of motion Using x(t) and θ(t).

COORDINATE COUPLING AND PRINCIPAL COORDINATES

27

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

)21.5()()( 2211 lxklxkxm

From the free-body diagram shown in Fig.5.10a, with the positive values of the motion variables as indicated, the force equilibrium equation in the vertical direction can be written as

)22.5()()( 2221110 llxkllxkJ

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


28

The lathe rotates in the vertical plane and has vertical motion as well, unless k1l1 = k2l2. This is known as elastic or static coupling.

From Fig.5.10b, the equations of motion for translation and rotation can be written as

)23.5(00

)( )(

)( )(

00

22

212211

221121

0 21

x

lklklklk

lklkkkxJ

m

•Equations of motion Using y(t) and θ(t).

melyklykym )()( 2211


29

These equations can be rearranged and written in matrix form as

)24.5()()( 222111 ymellykllykJP

)25.5(00

)()(

)()(

22

2112211

112221

2

y

lklklklk

lklkkkyJmemem

P

If , the system will have dynamic or inertia coupling only.

2211 lklk

Note the following characteristics of these systems:


30

1.In the most general case, a viscously damped two degree of freedom system has the equations of motions in the form:

)26.5(00

2

1

2221

1211

2

1

2221

1211

2

1

2221

1211

xx

kkkk

xx

cccc

xx

mmmm

2.The system vibrates in its own natural way regardless of the coordinates used. The choice of the coordinates is a mere convenience.

3.Principal or natural coordinates are defined as system of coordinates which give equations of motion that are uncoupled both statically and dynamically.


31

EXAMPLE 5.6:PRINCIPAL COORDINATES OF SPRING-MASS SYSTEM

Determine the principal coordinates for the spring-mass system shown in Fig.5.4.

32

We define a new set of coordinates such that

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 is


(E.1)3coscos)(

3coscos)(

22112

22111

tmkBt

mkBtx

tmkBt

mkBtx

33

Since the coordinates are harmonic functions, their corresponding equations of motion can be written as

(E.2)3cos)(

cos)(

222

111

tmkBtq

tmkBtq

(E.3)03

0

22

11

qmkq

qmkq


34

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

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

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

212

211

tqtqtxtqtqtx

(E.5))]()([21)(

)]()([21)(

212

211

txtxtq

txtxtq


35

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

Consider the external forces to be harmonic:

FORCED VIBRATION ANALYSIS

)27.5(

2

1

2

1

2221

1211

2

1

2221

1211

2

1

2212

1211

FF

xx

kkkk

xx

cccc

xx

mmmm

)28.5(2,1,)( 0 jeFtF tijj

where ω is the forcing frequency. We can write the steady-state solutions as

)29.5(2,1,)( jeXtx tijj

36

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

)30.5(

)()(

)()(

20

10

2

1

22222221212122

1212122

1111112

FF

XX

kcimkcim

kcimkcim

We defined as in section 3.5 the mechanical impedance Zre(iω) as

)31.5(2,1,,)( 2 srkcimiZ rsrsrsrs


37

And write Eq.(5.30) as:

)32.5()( 0FXiZ

Where,

20

100

2

1

2212

1211 matrix Impedance)( )()( )(

)(

FF

F

XX

X

iZiZiZiZ

iZ


38

where the inverse of the impedance matrix is given

)33.5()( 01FiZX

)34.5()( )()( )(

)()()(1)(

1112

12222

122211

1

iZiZi-ZiZ

iZiZiZiZ

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

)35.5()()()(

)()()(

)()()()()()(

2122211

201110122

2122211

201210221

iZiZiZFiZFiZiX

iZiZiZFiZFiZiX

FORCED VIBRATION ANALYSISEq.(5.32) can be solved to obtain:

39

Find the steady-state response of system shown in Fig.5.13 when the mass m1 is excited by the force F1(t) = F10 cos ωt. Also, plot its frequency response curve.

EXAMPLE 5.8:STEADY-STATE RESPONSE OF SPRING-MASS SYSTEM

40

The equations of motion of the system can be expressed as


(E.1)0

cos2

2 0

0 10

2

1

2

1

tFxx

k-k-kk

xx

mm

E.2)(2,1;cos)( jtXtx jj

We assume the solution to be as follows.

Eq.(5.31) gives

(E.3))(,2)()( 122

2211 kZkmZZ

41

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

)(

(E.4)))(3(

)2()2(

)2()(

2210

22210

2

2210

2

22210

2

1

kmkmkF

kkmkFX

kmkmFkm

kkmFkmX

Eqs.(E.4) and (E.5) can be expressed as

Hence,

E.6)(

1

2

)(2

1

2

1

2

1

2

10

2

1

1

k

F

X


42Fig.5.14: Frequency response curves

E.7)(

1

)(2

1

2

1

2

1

2

102

k

FX


week 15 vibration

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