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Two-Degree of Freedom Systems

Introduction

Equation of Motion

Free Vibration of an Undamped System

Torsional System

Coordinate Coupling

Forced Vibration Analysis

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Introduction

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Importance of the Study of

Vibration

Why study vibration? Vibrations can lead to excessive deflections

and failure on the machines and structures

To reduce vibration: proper design ofmachines and their mountings

To utilize profitably in several consumer andindustrial applications

To improve the efficiency of certain machining,

casting, forging & welding processes

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Basic Concepts of Vibration

Vibration = any motion that repeatsitself after an interval of time

Vibratory System consists of:1) spring or elasticity

2) mass or inertia

3) damper

Involves transfer of potential energy tokinetic energy and vice versa

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Degree of Freedom (d.o.f.): min. no. of independent

coordinates required to determine completely the

positions of all parts of a system at any instant of time

Examples of single degree-of-freedom systems:

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Examples of single degree-of-freedom

systems:

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Stiffness and Mass

Vibration is cause by the interaction of two different forces

one related to position (stiffness) and one related to

acceleration (mass).

m

k

x

Displacement

Mass Spring

Stiffness (k )

Mass (m )

)(t kx F k

)()( t x mt maF m

Proportional to displacement

Proportional to acceleration

statics

dynamics

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Equation of Motion

From Newton’s Law for this simple mass-spring systemthe two forces must be equal i.e. F M = F k .

M

k x

Displacement

Mass Spring

0

)()(

)()(

t kx t x m

t kx t x m

or

This is a 2nd order differentialequation and all phenomena that

have differential equations of this

type for their equation of motion

will exhibit oscillatory behavior.

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Examples of Single-Degree-of-Freedom

Systems

m

l =length

Pendulum

0 )()( t l

g t q q

Gravity g

Shaft and Disk

0 )()( t k t J q q

TorsionalStiffness

k Moment

of inertia

J

NEWTON’s Law

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Examples of Two degree-of-freedom

systems:

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Examples of Three degree of freedom

systems:

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Example of Infinite-number-of-degrees-of-freedom system:

Infinite number of degrees of freedom system

are termed continuous or distributed systems

Finite number of degrees of freedom are termeddiscrete or lumped parameter systems

More accurate results obtained by increasing

number of degrees of freedom

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Classification of Vibration

Free Vibration: A system is left to vibrate on its own after an

initial disturbance and no external force acts on

the system. E.g. simple pendulum

Forced Vibration: A system that is subjected to a repeating

external force. E.g. oscillation arises from diesel

engines

Resonance occurs when the frequency of the

external force coincides with one of the

natural frequencies of the system

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Undamped Vibration:When no energy is lost or dissipated in friction

or other resistance during oscillations

Damped Vibration:

When any energy is lost or dissipated in friction

or other resistance during oscillations

Linear Vibration:

When all basic components of a vibratorysystem, i.e. the spring, the mass and the damper

behave linearly

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Nonlinear Vibration:If any of the components behave nonlinearly

Deterministic Vibration:

If the value or magnitude of the excitation (force

or motion) acting on a vibratory system is known

at any given time

Nondeterministic or random Vibration:

When the value of the excitation at a given timecannot be predicted

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Examples of deterministic and randomexcitation:

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Vibration Analysis Procedure

Step 1: Mathematical Modeling

Step 2: Derivation of Governing

EquationsStep 3: Solution of the Governing

Equations

Step 4: Interpretation of the Results

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Vibration Analysis Procedure

Example of the modeling of a forginghammer:

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Using mathematical model to representthe actual vibrating system

E.g. In figure below, the mass and damping

of the beam can be disregarded; the system

can thus be modeled as a spring-mass

system as shown.

Mass or Inertia Elements

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20

30.1sinsin t A A x q

31.1cos t Adt

dx

32.1sin 22

2

2

xt Adt

xd

Harmonic Motion

Periodic Motion: motion repeated after equal intervals

of time.

Harmonic Motion: simplest type of periodic motion.

Displacement ( x): (on horizontal axis)

Velocity:

Acceleration:

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Definitions of Terminology:

Amplitude ( A) is the maximum displacement

of a vibrating body from its equilibrium

position

Period of oscillation (T ) is time taken to

complete one cycle of motion

Frequency of oscillation ( f ) is the no. of

cycles per unit time

59.1

2

T

60.12

1

T f

Harmonic Motion

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Summary of simple harmonic motion

t

x(t)

0 x

Slope

here is v 0

n

Period

n

T

2

Amplitude

A

Maximum

Velocity

An

Hzs

cycles

rad/cycle

rad/s

222

nnnnf

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Initial Conditions

If a system is vibrating then we must assume that somethingmust have (in the past) transferred energy into to the system

and caused it to move. For example the mass could have

been:

•moved a distance x 0 and then released at t=0 (i.e. given

Potential energy) or

•given an initial velocity v 0 (i.e. given Kinetic energy) or

•Some combination of the two above cases

From our earlier solution we know that:

))

))

cos(cos()(

sin(sin()(

A A x v

A A x x

nnn

n

00

00

0

0

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Harmonic Motion

Definitions of Terminology:

Natural frequency is the frequency which a system oscillates

without external forces

Phase angle ( ) is the angular difference between two

synchronous harmonic motions

62.1sin

61.1sin

22

11

t A x

t A x

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62.1sin61.1sin

22

11

t A x

t A x

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•Complex Fourier Series:

The Fourier series can also be represented in

terms of complex numbers.

Harmonic Analysis

)79.1(sincos)78.1(sincos

t it e

t it e

t i

t i

Also,

)81.1(2

sin

)80.1(2cos

i

eet

eet

t it i

t it i

and

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x(t)

t

undampedoverdamped

underdamped

Critically

damped

d

2

nd

1

0

n

2

1

1

x0

Comparison of motion with different types of damping

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OTHER EXAMPLE

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Two-Degree of Freedom Systems

Equation of Motion

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Two degree of freedom systems are defined as

systems that require two independentcoordinates to describe their motion.

Consider the motor-pump system shown in

Fig.5.1. 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.

Figure 5.1: Motor-pump system

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Consider the system shown in Fig.5.2.

The general rule for the computation of the

number of degrees of freedom can be stated as

follows:

No. of degrees No. of masses in the system

of freedom of = x no. of possible types of the system motion of each mass

Figure 5.2: Packaging of an instrument

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0

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0

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OTHER EXAMPLE

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Deriving 2-DOF mathematical model

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Deriving 2 DOF mathematical model

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Free Vibration of an Undamped 2-DOF

System

These are coupled eqns of motion. For normal mode of oscillation, each

mass undergoes harmonic motion of same freq passing through theequilibrium position.

In matrix form, (1)

)1(0)(

0)(

1223222

2212111

xk xk k xm

xk xk k xm

0)(

)(

0

0

2

1

322

221

2

1

2

1

x

x

k k k

k k k

x

x

m

m

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Free Vibration of an Undamped 2-DOF

System

Assume harmonic motion. Let,

)sin((

2

)sin(

2

2

1

2

1

2

1

2

1

t X X

x x

t X

X

x

x

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Subst. (2) into (1)

)3(0)(

(

0)(

)(

0

0

2

1

2

2322

2

2

121

2

1

322

221

2

1

2

2

2

1

X

X

mk k k

k mk k

or

X

X

k k k

k k k

X

X

m

m

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For non-trivial solution,

0))(())((

0

':

12212211

2221

1211

A A A A

A A

A A

Rule sCramer Note

0)(

)(det

32

2

22

221

2

1

k k mk

k k k m

or

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or

0))((

)()()(

2

23221

132221

4

21

k k k k k

mk k mk k mm

A

AC B B x

c Bx Ax

x

Substitute

mm

k k k k k k

m

k k

m

k k

2

4

0

,

0)()()(

2

2

2

21

1332212

2

32

1

214

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To determine the values of X1 and X2,

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The normal modes of vibration corresponding to ω12 and ω22 canbe expressed, respectively, as

)11.5()(

)(

)()(

)3(

32

2

22

2

2

21

2

21

)2(

1

)2(

22

32

2

12

2

2

21

2

11)1(

1

)1(

21

k k m

k

k

k k m

X

X r

k k mk

k k k m

X X r

From

1 2,

)12.5( and )2(

12

)2(

1

)2(

2

)2(

1)2(

)1(

11

)1(

1

)1(

2

)1(

1)1(

X r

X

X

X X

X r

X

X

X X

Known as the modal vectors of the system.

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Modal vector

First mode

Second mode

.)2(

2

)2(

12

)1(

2

)1(

1121

X and X asat and

X and X astocorrespond X and X of Values

The free vibration solution or the motion in time can be

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In i t ial condi t ions . The initial conditions are

(5.13)modesecond

)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(

t X r

t X

t x

t xt x

t X r t X

t xt xt x

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

t x X r t x

t x X t x

i

i

i

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Substituting into Eq.(5.15) leads to

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

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

2222

1111

xt x xt x

xt x xt x

)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

X r X r x

X r X r x

X X x X X x

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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

r r x xr X

r r x xr X

r r

x xr X

r r

x xr X

from which we obtain the desired solution

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)18.5()0()0([

)0()0(tan

cos

sintan

)0()0([

)0()0(

tancos

sin

tan

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

)(

1

sincos

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

)(

1

sincos

2112

2111

2

)2(

1

2

)2(

11

2

2121

2121

1)1(

1

1

)1(

11

1

2/1

2

2

2

2112

211

12

2/12

2

)2(

1

2

2

)2(

1

)2(

1

2/1

2

1

2

2122

212

12

2/12

1

)1(

1

2

1

)1(

1

)1(

1

x xr

x xr

X

X

x xr

x xr

X

X

x xr x xr

r r

X X X

x xr x xr

r r

X X X

Example 5.3

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99

p

Find the free vibration response of the system

shown with k1 = 30, k2 = 5, k3 = 0, m1 = 10, andm2 = 1 for the initial conditions

).0()0()0( ,1)0( 2211 x x x x

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Solut ion : For the given data, the eigenvalue problem,

Eq.(5.8), becomes

(E.1)0

0

55-

5 3510

0

0

2

1

2

2

2

1

322

22

221

2

1

X

X

X

X

k k mk

k k k m

or

By setting the determinant of the coefficient matrix in

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101

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,

(E.2)01508510 24

)2(

1

)2(

2

2

2

2

)1(

1

)1(

2

2

1

2

21

22

21

5)1(0.6

2)1(5.2Subst

E.3)(4495.2,5811.1

0.6,5.2

X X toleads E in

X X toleads E in

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102

The normal modes (or eigenvectors) are given by

E.5)(5

1

E.4)(2

1

)2(

1)2(

2

)2(1)2(

)1(

1)1(

2

)1(

1)1(

X X

X X

X X

X X

The free vibration responses of the masses m1 and m2

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103

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

t X t X t x

t X t X t x

(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

X X t x

X X t x

X X t x

X X t x

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

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

(E.12)7

2cos;7

5cos 2

)2(

11

)1(

1 X X

(E.13)0sin,0sin 2

)2(

11

)1(

1 X X

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

(E.14)0,0,7

2

,7

521

)2(

1

)1(

1 X X

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Thus the free vibration responses of m1 and m2 are

given by

(E.16)4495.2cos7

105811.1cos7

10)(

(E.15)4495.2cos7

25811.1cos

7

5)(

2

1

t t t x

t t t x

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1 1 1 2 2 1

2 2 2 2 2

0 ( ) 0

0 0

m x k k k x

m x k k x

1 1

2 2

sin( )sin( )

x A t x A t

2

1 1

2

2 2

sin( )

sin( )

x A t

x A t

2

1 1 2 2 11

22 2 2 22

( ) 00

00

A k k k Am

A k k Am

2

1 2 1 2 1

222 2 2

( ) 0

0

k k m k A

Ak k m

Can be solvedonly if

2

1 2 1 2

2

2 2 2

( )0

k k m k

k k m

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21 2 1 2

2

2 2 2

4 2 2

1 2 1 2 2 2 1 1 2 2 2

( )0

( ) ( ) ( ) 0

k k m k

k k m

m m k k m k m k k k k

2 2

1 2 2 2 1 1 2 2 2 1 1 2 1 2 2 22 2

1 2

1 2

( ) ( ) 4( ) ( ),

2( )

k k m k m k k m k m m m k k k k

m m

1 1

2 2

first natural frequency

first natural frequency

n

n

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k

k

x 1

x 2

m

m

For a case

2 4 2 2( ) 3 0m km k

2 2 2 2 2 2

21 2 2

2

2

1

2

3 9 4 3 52 2

3 5 3 5

2 4 2 4

1.618

0.618

n

n

n

km k m m k km m k m m

k k k k

m m m m

k

m

k

m

4 2 21 2 1 2 2 2 1 1 2 2 2( ) ( ) ( ) 0m m k k m k m k k k k

m1= m2

= m

k 1= k

2= k

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Mode shapes

(1) 2

2

(1)

1

(2 ) (1.618) ( / )0.618

A k m k m

A k

(1) 2

2

(1)

1

(2 ) (0.618) ( / )1.618

A k m k m

A k

1.0

-0.618

1.0

1.618

1

1.0

0.618

2

1.0

1.618

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However for any other general initial conditions both

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However, for any other general initial conditions, bothmodes will be excited. The resulting motion can beobtained by a linear superposition of the two normal

modes

)()()( 2211 t xct xct x

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

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

22

)2(

1211

)1(

11

)2(

2

)1(

22

22)2(111)1(1)2(1)1(11

t Ar t Ar t xt xt x

t At At xt xt x

asexpressedbecanvectortheof componentstheThus

.generalityof lossnowithchoosecanweand

constantsunknowntheinvolvealreadyandSince

constants.areandwhere

)(

1,

)()(

21

)2(

1

)1(

1

)2()1(

21

t x

cc A A

t xt x

cc

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:conditionsinitialthefrom

determinebecanandconstantsunknownthewhere 21

)2(

1

)1(

1 ,, A A

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

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

2222

1111

xt x xt x

xt x xt x

2

)2(

1221

)1(

1112

2

)2(

121

)1(

112

2

)2(

121

)1(

111

2

)2(

11

)1(

11

sinsin)0(

coscos)0(

sinsin)0(

coscos)0(

Ar Ar x

Ar Ar x

A A x

A A x

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Example

115

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Example

119

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