Mechanical Vibrations - Suranaree University of...

Mechanical

Vibrations

Singiresu S. Rao

SI Edition

Free Vibration of

SDOF

© 2005 Pearson Education South Asia Pte Ltd.2

2.1 Introduction

2.2 Free Vibration of an Undamped

Translational System

2.3 Free Vibration of an Undamped

Torsional System

2.4 Stability Conditions

2.5 Free Vibration with Viscous Damping

Chapter Outline


2.1 Introduction

• Free Vibration occurs when a system oscillates

only under an initial disturbance with no external

forces acting after the initial disturbance

• Undamped vibrations result when amplitude of

motion remains constant with time (e.g. in a

vacuum)

• Damped vibrations occur when the amplitude of

free vibration diminishes gradually overtime, due

to resistance offered by the surrounding medium

(e.g. air)


2.1 Introduction

• Several mechanical and structural

systems can be idealized as single

degree of freedom systems, for example,

the mass and stiffness of a system

© 2005 Pearson Education South Asia Pte Ltd.

Equation of motion

• Newton’s second law

• Principle of Conservation of

Energy

• Rayleigh method

• Lagrange equation

• D’Alembert principle

5


If mass m is displaced a distance when acted

upon by a resultant force in the same direction,

Free Vibration of an Undamped Translational

System

)(tx

)(tF

dt

txdm

dt

dtF

)()(

If mass m is constant, this equation reduces to

2

2

( )( )

d x tF t m mx

dt

where is the acceleration of the mass2

2 )(

dt

txdx


Free Vibration of an Undamped Rotational

System

For a rigid body undergoing rotational motion,

Newton’s Law gives

( )M t J

where is the resultant moment acting on the

body and and are the resulting

angular displacement and angular acceleration,

respectively.

M

22 /)( dttd


Newton’s second law: SDOF

8

( )F t mxUndamped Translational System



9

( )M t JUndamped Rotational System



10

( )F t mx

kx mx

0mx kx

Equation of motion(free vibration of SDOF)


Undamped free vibration of SDOF

11

Laplace transform

Natural frequency of n k m



12

where

Inverse Laplace, Solution of the governing equation

or

(1)

(2)


A system is said to be conservative if no energy is

lost due to friction or energy-dissipating nonelastic

members.

Principle of Conservation of Energy

If no work is done on the conservative system by

external forces, the total energy of the system

remains constant. Thus the principle of

conservation of energy can be expressed as:

constant

( ) 0

T U

dT U

dt

or


Principle of Conservation of Energy:

SDOF

The kinetic and potential energies are given by:

21

2T mx and 21

2U kx

2 2

constant

1 1 constant

2 2

T U

mx kx

( ) 0

1 12 2 0

2 2

dT U

dt

m x x k x x

0mx kx



Free Vibration of an Undamped Torsinal

System

Torsional Spring Constant:4

32t

Gdk

l

( )M t J

For a rigid body undergoing

rotational motion, Newton’s Law gives

tk J

0tJ k



Example 2.8

Effect of Mass on ωn of a Spring

Determine the effect of the mass of the spring on

the natural frequency of the spring-mass system

shown in the figure below.


Example 2.8 Solution

(E.1)2

12

l

xydy

l

mdT s

s

The kinetic energy of the spring element of length

dy is

where ms is the mass of the spring. The total

kinetic energy of the system can be expressed as

2 22

20

2 2

kinetic energy of mass ( ) kinetic energy of spring ( )

1 1

2 2

1 1(E.2)

2 2 3

m s

ls

y

s

T T T

m y xmx dy

l l

mmx x



(E.3)2

1 2kxU

The total potential energy of the system is given by

2 2 2

constant

1 1 1 constant

2 2 3 2

s

T U

mmx x kx

( ) 0

03

03

s

s

dT U

dt

mmx x kx

mm x kx



we obtain the expression for the natural

frequency:

(E.7)

3

2/1

s

n mm

k

Thus the effect of the mass of spring can be

accounted for by adding one-third of its mass to

the main mass.


where c = damping constant

Free Vibration with Viscous Damping: SDOF

• Damping force: F cx

Newton’s law yields the equation of motion:

0

mx cx kx

mx cx kx

or

2

2

0

4

2

ms cs k

c c mks

m

Characteristic equation



• Critical Damping Constant and Damping Ratio:

2 4 0

2 2 2c n

c mk

kc mk m m

m

The critical damping cc is defined as the value of

the damping constant c for which the radical in

becomes zero:

or

The damping ratio ζ is defined as:

c

c

c


Case1.


(0 1 or )cc c

Assuming that ζ ≠ 0, consider the following 3 cases:

For this condition, Poles is negative and the roots

are:2

1

2

2

1

1

n n n d

n n n d

s i i

s i i

Damped natural frequency

21d n

1i



where (C’1,C’2), (X,Φ), and (X0, Φ0) are arbitrary

constants to be determined from initial conditions.

1 2

1 2

2

0 0

( )

cos sin

sin

cos 1 (2.70)

n d n d

n

n

n

i t i t

t

d d

t

d

t

n

x t C e C e

e C t C t

Xe t

X e t

and the solution can be written in different forms:



)71.2(1

and 2

00201

n

nxxCxC

0 00

2( ) cos sin

1

nt nd d

n

x xx t e x t t

and hence the solution becomes

For the initial conditions at t = 0,

Eq. describes a damped harmonic motion. Its

amplitude decreases exponentially with time, as

shown in the figure below.



Underdamped Solution

0 00

2( ) cos sin

1

nt nd d

n

x xx t e x t t


Due to repeated roots, the solution of Eq.(2.59) is

given by


1 22

cn

cs s

m

)78.2()()( 21

tnetCCtx

)79.2( and 00201 xxCxC n

In this case, the two roots(Poles) are:

Case2. )/2or or 1( mkmc/cc c

Application of initial conditions gives:

Thus the solution becomes:

)80.2()( 000

t

nnetxxxtx


Due to repeated roots, the solution of critical

damped system is given by


1 22

cn

cs s

m

1 2( ) ( ) ntx t C C t e

1 0 2 0 0and nC x C x x

In this case, the two roots are:

Case2.

( 1 or )cc c

Application of initial conditions gives:

Thus the solution becomes:

0 0 0( ) nt

nx t x x x t e


In this case, the solution Eq.(2.69) is given by:

The roots are real and distinct and are given by:


2

1

2

2

1 0

1 0

n n

n n

s

s

2 21 1

1 2( )n n n nt t

x t C e C e

For the initial conditions at t = 0,

Case3. )/2or or 1( mkmc/cc c

2

0 0

12

2

0 0

12

1

2 1

1

2 1

n

n

n

n

x xC

x xC


Since , the motion will eventually

diminish to zero, as indicated in the figure below.


tetn as 0


1

2

1

1

0 1 01

2 0 2 0

cos( )

cos( )

n

n

n

n d

n d

t

d

t

d

t

t

X e tx

x X e t

ee

e

Solution of Underdamped free vibration:


32

The logarithmic decrement can be obtained

1

22

2 2ln

21n d n

d

x c

x m

2 22

Hence,


Home work - 1

33

Determine the differential equation of motion and

the equivalent stiffness for the system (N/m) of Figure


Home work - 2

34

For free vibration of underdamped system( = 0.8) as shown in Figure.

Determine stiffness of system


Home work - 3

35

The free response of the undamped system of Figure.

Determine

- Natural frequency, Initial velocity and Initial displacement


Home work - 4

36

The free response of the underdamped system in Figure.

Calculate the damping ratio, logarithm decrement


Chapter summaries

• Equation of motion for SDOF undamped

system

• Equation of motion for SDOF damped system

37


Chapter summaries

• Undamped free vibration response

• Overdamped free vibration response

• Critically damped free vibration response

38


Chapter summaries

• Underdamped free vibration response

• Logarithm decrement

39

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