vibrations 2dof ppt

MDPN471 & MDP410

Mechanical Vibrations CHAPTER

5 Two-Degree-Of-Freedom

Vibration

2-DOF Vibration 5.1 Introduction

• Two degree of freedom systems are defined as systems that

require two independent coordinates to describe their motion.

Consider the Automobile

Suspension system.

2-DOF Vibration 5.1 Introduction

• Consider the motor-pump system.

– The vertical displacement of the system and the angular

coordinate denoting the rotation of the mass about its C.G. make

up two independent coordinates.

Motor-pump system

2-DOF Vibration

No. of degrees of freedom of the system =

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

Packaging of an instrument

5.1 Introduction

2-DOF Vibration 5.2 Eqs of Motion for Forced Vibration

Consider a viscously damped two degree of freedom

spring-mass system.

A two degree of freedom spring-mass-damper system


)2.5()()(

)1.5()()(

2232122321222

1221212212111

Fxkkxkxccxcxm

Fxkxkkxcxccxm

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

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:

where [m], [c], and [k] are called the mass, damping,

and stiffness matrices, respectively, and are given by


322

221

322

221

2

1

][

][

0

0 ][

kkk

kkkk

ccc

cccc

m

mm

)(

)()(

)(

)()(

2

1

2

1

tF

tFtF

tx

txtx

And the displacement and force vectors are given

respectively:

It can be seen that the matrices [m], [c], and [k]

are symmetric:

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


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

•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

It can be seen that the matrices [m], [c], and [k]

are symmetric:

2-DOF Vibration 5.3 Free Vib. Analysis of Undamped System

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

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

2321222

2212111

txkktxktxm

txktxkktxm

)6.5()cos()( ),cos()( 2211 tXtxtXtx

Assuming solutions are harmonic motion of m1

and m2 at the same frequency ω and the same

phase angle Φ, we take the solutions as

By setting F1(t) = F2(t) = 0, and damping disregarded,

i.e., c1 = c2 = c3 = 0, and the equation of motion is

reduced to:

)2.5()()(

)1.5()()(

2232122321222

1221212212111

Fxkkxkxccxcxm

Fxkxkkxcxccxm


)7.5( 0)cos()(

0)cos()(

232

2

212

22121

2

1

tXkkmXk

tXkXkkm

2

1 1 2 1 2 2

2

2 1 2 2 3 2

( ) 0

( ) 0 (5.8)

m k k X k X

k X m k k X

For non-trivial solution, thus,

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


0)(

)(det

21

2

22

221

2

1

kkmk

kkkm

)9.5(0))((

)()()(

2

23221

132221

4

21

kkkkk

mkkmkkmmor

• These 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. Therefore,


)10.5())((

4

)()(

2

1

)()(

2

1,

2/1

21

2

23221

2

21

132221

21

1322212

)2(

2

)1(

mm

kkkkk

mm

mkkmkk

mm

mkkmkk

The roots are called natural frequencies of the system.

which is called the frequency or characteristic equation.

The roots are:


Remember that:

To find x1(t) and x2(t), we need to find X1, X2, and .

But at which ??

)cos()( ),cos()( 2211 tXtxtXtx

)()()(

)()()(

)2(

2

)1(

22

)2(

1

)1(

11

txtxtx

txtxtx

By a linear superposition:


)2(

1

)2(

2)2(

)1(

1

)1(

2)1(

X

Xr

X

Xr

Substituting the natural frequencies into Eq 5.8,

)8.5(0)(

0)(

232

2

212

22121

2

1

XkkmXk

XkXkkm

we can determine the values of X1 and X2 at each mode,

)11.5()(

)(

)(

)(

32

2

22

2

2

21

2

21

32

2

12

2

2

21

2

11

kkm

k

k

kkm

kkm

k

k

kkm


)12.5( and )2(

1)2(

)2(

1

)2(

2

)2(

1)2(

)1(

1)1(

)1(

1

)1(

2

)1(

1)1(

Xr

X

X

XX

Xr

X

X

XX

The normal modes of vibration corresponding to ω(1)2

and ω(2)2 can be expressed, respectively, as

which are known as the modal vectors of the system.


(5.13)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

)14.5()()()( )2()1( txtxtx iii

The resulting motion can be obtained by a linear

superposition of the two normal modes, Eq.(5.13)


Thus the components of the vector can be expressed as

where the 4 unknown constants can be

determined from the 4 initial conditions:

)15.5()cos()cos(

)()()(

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

)2()2(

)2(

1)2()1()1(

)1(

1)1(

)2(

2

)1(

22

)2()2(

)2(

1)1()1(

)1(

1

)2(

1

)1(

11

tXrtXr

txtxtx

tXtXtxtxtx

√ √ √ √


)16.5()0()0(),0()0(

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

2222

1111

xtxxtx

xtxxtx

)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

Substituting into Eq.(5.15) leads to

The solution can be expressed as


1/22 2

(1) (1) (1)

1 1 1 1 1

1/22

2 2 1 2

2 1 2 2

2 1 1

1/22 2

(2) (2) (2)

1 1 2 1 2

1/22

2 1 1 2

1 1 2 2

2 1 2

(1)1 1

1

cos sin

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

( )

cos sin

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

( )

stan

X X X

r x xr x x

r r

X X X

r x xr x x

r r

X 11 2 1 2

(1)

1 1 1 2 1 2

(2)1 11 2 1 1 2

2 (2)

1 2 2 1 1 2

in (0) (0)tan

cos [ (0) (0)

sin (0) (0)tan tan (5.18)

cos [ (0) (0)

r x x

X r x x

X r x x

X r x x

from which we obtain the desired solution

2-DOF Vibration Example 5.3 Free Vibration Response of a 2DOF

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

0

0

2

1

32

2

22

221

2

1

X

X

kkmk

kkkm

Solution:

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

2-DOF Vibration

(E.2)01508510 24

E.3)(4495.2,5811.1

0.6,5.2

21

2

2

2

1

E.5)(5

1

E.4)(2

1

)2(

1)2(

2

)2(

1)2(

)1(

1)1(

2

)1(

1)1(

XX

XX

XX

XX

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,

The normal modes (or eigenvectors) are given by

Example 5.3 Free Vibration Response of a 2DOF

2-DOF Vibration

(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

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):


2-DOF Vibration

(E.12)7

2cos;

7

5cos 2

)2(

11

)1(

1 XX

(E.13)0sin,0sin 2

)2(

11

)1(

1 XX

(E.14)0,0,7

2,

7

521

)2(

1

)1(

1 XX

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

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

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


2-DOF Vibration

(E.16)4495.2cos7

105811.1cos

7

10)(

(E.15)4495.2cos7

25811.1cos

7

5)(

2

1

tttx

tttx

Thus the free vibration responses of m1 and m2 are

given by


2-DOF Vibration 5.4 Torsional System

22312222

11221111

)(

)(

ttt

ttt

MkkJ

MkkJ

)19.5()(

)(

22321222

12212111

tttt

tttt

MkkkJ

MkkkJ

which upon rearrangement become

Consider a torsional system as shown in Fig.5.6. The

differential equations of rotational motion for the discs

can be derived as

For the free vibration analysis of the system,

Eq.(5.19) reduces to

2-DOF Vibration 5.4 Torsional System

)20.5(0)(

0)(

2321222

2212111

ttt

ttt

kkkJ

kkkJ

Figure 5.6: Torsional system with discs mounted on a shaft

2-DOF Vibration

(E.1) 02

02

2120

2110

tt

tt

kkJ

kkJ

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):

Fig.5.7:

Torsional system

Example 5.4 Natural Frequencies of Torsional System

2-DOF Vibration

28

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

(E.3)052 2

0

22

0

4

tt kkJJ

(E.4))175(4

and)175(4 0

2

0

1J

k

J

k tt

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

gives the frequency equation:

Rearranging and substituting the harmonic solution:


2-DOF Vibration

29

(E.5)4

)175(2

4

)175(2

)2(

1

)2(

22

)1(

1

)1(

21

r

r

0and2,

,,

3022011

2211

kJJmJJm

kkkkkk tttt

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


vibrations 2dof ppt

Documents

Transcript of vibrations 2dof ppt