Chapter 9 Vibration

Topic 9 (Chapter 22) :

Vibrations

EME1076 Applied Dynamics

This lecture note is taken and modified from Engineering Mechanics-Dynamics, R.C.Hibbeler, Prentice Hall which copyright belongs to Pearson Education South Asia Pte Ltd.

Trimester2, 2014/15

2

Chapter Objectives

To discuss undamped and viscous damped one-degree-of-freedom vibration of a rigid

body using the equation of motion method.

To study the analysis of undamped forced vibration and viscous damped forced

vibration.

3

Undamped Free Vibration

Undamped Forced Vibration

Viscous Damped Free Vibration

Viscous Damped Forced Vibration

Chapter Outline

4

22.1 Undamped Free Vibration

A vibration is a periodic motion of a body or system of connected bodies displaced from a position of equilibrium.

Free vibration occurs when the motion is maintained by gravitational or elastic restoring forces

Forced vibration is caused by an external periodic or intermittent force applied to the system

Both of these types of vibration may be either damped or undamped

5

Undamped vibrations can continue indefinitely because frictional effects are neglected in the

analysis

In reality, motion of all vibrating bodies is actually damped because of frictional forces present.

Example of a undamped free vibration is as shown


6

The block has a mass m and is attached to a spring having a stiffness k

When released from a displaced position, x, the spring pulls on the block, causing a

vibrating motion

The block will attain a velocity such that it will proceed to move out of equilibrium when x = 0, and provided the supporting surface is

smooth, oscillation will continue indefinitely.


7

The time-dependent path of motion of the block may be determine by applying equation of

motion to the block when it is in the displaced

position x

The elastic restoring force F = kx is always directed toward the equilibrium position,

whereas the acceleration a is assumed to act in

the direction of positive displacement


8

From the figure, we have

The acceleration is proportional to the blocks displacement and such motion is called

simple harmonic motion

Rearranging,

,

xmxkmaF xx

;

02 xx nm

kn


9

The constant n is called the natural frequency, expressed in rad/s,

is a homogeneous, second-order, linear, differential equation with constant

coefficients, and the general solution is

m

kn

02 xx n

tBtAx nn cossin


10

The general solution may also be expressed in terms of simple sinusoidal motion.

Let

where C and are new constant to be determine in place of A and B

Therefore,

sincos CBCA

tCtCx nn cossinsincos


11

Since sin ( + ) = sin cos + cos sin ,

If this equation is plotted on an x-versus-nt axis, the graph shown in figure below is obtained

)sin( tCx n


12

The maximum displacement of the block from its equilibrium position is defined as the amplitude of vibration

From the figure, the amplitude is C and angle is the phase angle since it represents the amount by which the curve is displaced from the origin when t = 0.

A

B

BAC

1

22

tan


13

Note that the sine curve, completes one cycle in time when

The length of time is call a period, which may be represented as

t

n

2

k

m 2


14

The frequency f is defined as the number of cycles completed per unit of time, which is

the reciprocal of the period:

The frequency is expressed in cycles/s. This ratio of units is called a hertz (Hz)

m

kf n

2

1

2

1


15

When a body or system of connected bodies is given an initial displacement from its

equilibrium position and released, it will

vibrate with the natural frequency, n.

Provided the body has a single degree of freedom, then the vibrating motion of the

body will have the same characteristics as

the simple harmonic motion of the block and

spring just presented.


16

Consequently, the bodys motion is described by a differential equation of the same standard form as

Hence, if the natural frequency of the body is known, the period of vibration, natural frequency

and other vibrating characteristics of the body

can be established with the equations.

02 xx n


17

EXAMPLE 22.1

Determine the period of vibration for the simple

pendulum shown. The bob has a mass m and is

attached to a cord of length l.

18

Free-Body Diagram

Motion of the system will be related to the position coordinate (q =) .

When the bob is displaced by an angle , the restoring force acting on the bob is created by the weight

component mg sin .

at acts in the direction of increasing s (or )

EXAMPLE 22.1

19

Equation of Motion

Applying the equation of motion in the tangential direction, since it involves the

restoring force, yields

Kinematics

at = d2s/dt2. furthermore, s may be related to

by the equation s = l, so that

ttt mamgmaF sin;

lat

(1)

EXAMPLE 22.1

20

Hence, Eq. 1 reduces to

The solution of this equation involves the use of an elliptic integral.

For small displacements, sin , in which case

0sin l

g

0 l

g

EXAMPLE 22.1

21

Comparing this equation with

It can be seen that n = (g/l)

The period of time required for the bob to make one complete swing is therefore

02 xx n

g

l

n

2

2

EXAMPLE 22.1

22

EXAMPLE 22.2

The 10-kg rectangle plate shown is suspended at

its center from a rod having torsional stiffness k =

1.5 N.m/rad. Determine the natural period of

vibration of the plate when it is given a small

angular displacement in the plane of the plate.

23

Free-Body Diagram

Since the plate is displaced in its own plane, the torsional restoring moment created by the rod is M = k.

This moment acts in the direction opposite to the angular displacement

The angular acceleration acts in the direction of positive

EXAMPLE 22.2

24

Equation of Motion

or

Since this equation is in the standard form, the natural frequency is

n = (k/IO)

0

;

O

OOO

I

k

IkIM

EXAMPLE 22.2

25

The moment of inertia of the plate about an axis coincident with the rod is given by

The natural period of vibration is

222

22

108.0)3.0()2.0()10(12

1

)(12

1

mkg

bamIO

sk

IO

n

69.15.1

108.022

2

EXAMPLE 22.2

26

The bent rod has a negligible mass and supports

a 5-kg collar at its end. Determine the natural

period of vibration for the system.

EXAMPLE 22.3

27

Free-Body and Kinetic Diagrams

The rod is displaced by a small amount from the equilibrium position.

Since the spring is subjected to an initial compression of xst for equilibrium, then when the

displacement x > xst, the spring exerts a force of Fs = kx kxst on the rod

EXAMPLE 22.3

28

To obtain the standard form, 5ay acts upward, which is in accordance with positive displacement

Equation of Motion

Moments will be summed about point B to eliminate the unknown reaction at this point. Since is small,

)2.0()5()2.0(05.49)1.0()1.0(

;)(

yst

BkB

akxkx

M

EXAMPLE 22.3

29

The second term on the left side, -kxst(0.1), represents the moment created by the spring

force which is necessary to hold the collar in

equilibrium, i.e. x = 0

Since this moment is equal and opposite to the moment 49.05(0.2) created by the weight

of the collar, these two terms cancel in the

previous equation, so that

)2.0(5)1.0( yakx (1)

EXAMPLE 22.3

30

Kinematics

The positions of the spring and the collar may be related to the angle .

Since is small, x = (0.1 m) and

y = (0.2 m),

Substituting into Eq.1 yields

2.0 yay

2.0)2.0(51.0)1.0(400

EXAMPLE 22.3

31

Rewriting this equation in the standard form gives,

Compared with , we have

The natural period of vibration is therefore,

020

02 xx n

sradnn /47.4202

sn

40.147.4

22

EXAMPLE 22.3

32

A 5-kg block is suspended from a cord that

passes over a 7.5-kg disk. The spring has a

stiffness k = 3500 N/m. Determine the natural

period of vibration for the system.

EXAMPLE 22.4

33

Free-body and Kinetic Diagrams

The system consists of the disk, which undergoes a rotation defined by the angle , and the block, which translates by an amount s

The vector acts in the direction of positive , and consequently mBas acts downward in the direction of positive s

OI

EXAMPLE 22.4

34

Equation of Motion

Summing moments about point O to eliminate the reactions Ox and Oy, realising

that IO = mr2, yields

) 25 . 0 ( ) 5 ( ) 25 . 0 )( 5 . 7 ( 2

1

) 25 . 0 ( ) 25 . 0 ( ) 81 . 9 ( 5

; ) (

2 s

s

B k O

a

F

M

(1)

EXAMPLE 22.4

35

Kinematics

As shown, a small positive displacement of the disk causes the block to lower by

an amount s = 0.25

When = 0, the spring force required for

equilibrium of the disk is 5(9.81) N, acting to the right

EXAMPLE 22.4

36

For position , the spring force is

Fs = (3500 N/m)(0.25 m) + 5(9.81)N

Substituting these results into Eq.1 and simplifying yields,

Hence, sradnn /20400

0400

2

EXAMPLE 22.4

Therefore, the natural period of vibration is

sTn

314.020

22

37

22.3 Undamped Forced Vibration

Undamped forced vibration is one of the most important types of vibrating motion in

engineering work

The principles which describe the nature of this motion may be used to analyze the force

which cause vibration in many types of

machines and structures.

38

Periodic Force

The block and spring shown provide a convenient model which represents the

vibrational characteristics of a system

subjected to a periodic force F = F0 sin 0t

This force has an amplitude of F0 and a forcing frequency 0


39

The free-body diagram for the block when it is displaced a distance x is shown

Applying the equation of motion yields

or

tm

Fx

m

kx

xmkxtFmaF xx

0

0

0

sin

sin;


40

This equation is referred to as a nonhomogeneous second-order differential

equation

The general solution consists of a complimentary solution, xc, plus a particular solution, xp

The complimentary solution is determined by setting the term on the right side equal to

zero and solving the resulting homogenous

equation


41

The solution is

where n is the natural frequency, n = (k/m)

Since the motion is periodic, the particular solution may be determined by assuming a solution of the form

where C is a constant.

tBtAx nnc cossin

tCxp 0sin


42

Taking the second time derivative and substituting into the differential equation,

Factoring out sin t and solving for C,

tm

FtC

m

ktC 0

000

2 sin)sin(sin

2

0

2

0

)/(1

/

)/(

/

n

kF

mk

mFC


43

Therefore we will obtain the particular solution

The general solution is therefore

tkF

xn

p 02

0 sin)/(1

/

tkF

tBtAxn

nn 02

0 sin)/(1

/cossin


44

Here x described two types of vibrating motion of the block

The complimentary solution xc defines the free vibration, which depends on the circular frequency n =(k/m) and the constants A and B


45

The particular solution xp describes the forced vibration of the block caused by applied force F = F0 sin t


46

The resultant vibration x is as shown

Since all vibrating systems are subjected to friction, the free vibration, xc, will in time dampen out


47

For this reason, the free vibration is referred to as transient, and the forced vibration is

called steady-state, since it is the only vibration that remains


48

The amplitude of forced vibration depends on the frequency ratio /n

If the magnification factor MF is defined as the ratio of the amplitude of steady-state

vibration, (xp)max, to the static deflection F0/k, which would be produced by the amplitude of

the periodic force F0, therefore,

20

max

)/(1

1

/

)(

n

p

kF

x

MF


49

Note that for 0 0, the MF 1, because of the very low frequency

50

Resonance occurs when the force F is applied with a frequency close to the natural

frequency of the system, / 1, therefore the amplitude become extremely large

because the block follows the motion of the

block.


51

Periodic Support Displacement

Forced vibrations can also arise from the periodic excitation of the support of a system.

The model shown represents the periodic vibration of a block which is caused by harmonic movement of the support t00 sin


52

As seen from the FBD, the coordinate x is measured from the point of zero displacement of

the support

Therefore, general displacement of the spring is

Applying the equation of motion yields,

or

)sin( 0 tx

tm

kx

m

kx

xmtxkmaF xx

sin

)sin(;

0

0

+


53

EXAMPLE 22.7

The instrument is rigidly attached to a platform P, which in

turn is supported by four springs, each having a stiffness k

= 800 N/m. Initially the platform is at rest when the floor is

subjected to a displacement = 10 sin (8t) mm, where t is

in seconds.

If the instrument is constrained to move vertically and the

total mass of the instrument and platform is 20 kg,

determine the vertical displacement y of the platform as a

function of time, measured from the equilibrium position.

What floor vibration is required to cause resonance?

54

Since the induced vibration is caused by the displacement of the supports,

ttBtAy On

nn

sin

)/(1cossin

2

0

(1)

View Free Body Diagram

EXAMPLE 22.7

55

Here mm, so that

The amplitude of vibration caused by floor displacement is

)8sin(10sin0 tt

sradm

k

sradmm

n /6.1220

)800(4

/810 00

mmyn

n7.16

)]6.12/()8[(1

10

)/(1)(

22

0max

(2)

EXAMPLE 22.7

56

Hence, Eq.1 and its time derivative become

Evaluating for A and B, the vibrating motion is described by the equation,

Resonance will occur when the amplitude of vibration approaches infinity. From Eq.2, it is

)8sin(7.16)6.12sin(5.10

)8cos(3.133)6.12sin()6.12()6.12cos()6.12(

)8sin(7.16)6.12cos()6.12sin(

tty

ttBtAy

ttBtAy

sradn /6.12

EXAMPLE 22.7

57

22.4 Viscous Damped Free Vibration*

In many cases damping is attributed to the resistance created by the substance, such as water, air, oil, or in which the system vibrates

Provided the body moves slowly through this substance, the resistance to motion is directly proportional to the bodys speed

This type of force developed under these conditions is called a viscous damping force, which is expressed by

xcF

58


The constant c, is the coefficient of viscous damping and has units of N.s/m

As shown, the effect of damping is provided by the dashpot of coefficient of damping c connected to the block on the right side.

Damping occurs when the piston P moves to the right or left within the enclosed cylinder

59

If block is displaced a distance x from its equilibrium position, both the spring force kx and the damping force cx oppose the forward motion of the block.

Applying the equation of motion yields

or

.

;

0

x xF ma kx cx mx

mx cx kx

+


60

This linear, second-order, homogeneous, differential equation has solutions of form

where e is the base of the natural logarithm and is a constant

The value of may be obtained by differentiating the above equation twice and

substituting back into the equation of motion

tex


61

Thus we obtained two values of ,

There are three possible combinations of 1, 2 which must be considered

m

k

m

c

m

c

m

k

m

c

m

c

2

2

2

1

22

22


62

The critical damping coefficient cc which will make the radical in the previous equation

equal to zero, i.e.,

or

nc

c

mm

kmc

m

k

m

c

22

02

2


63

Overdamped System

When c > cc, the roots 1 and 2 are both real

Motion corresponding to this solution is nonvibrating

The effect of damping is so strong that when the block is displaced and released, it simply creeps back to its original position without oscillating.

ttBeAex 21


64

Critically Damped System

If c = cc, then 1 = 2 = - cc/2m = n, it represents a condition when c has the

smallest value necessary to cause the

system to be nonvibrating

tneBAx )(


65

Underdamped System

Cases where c < cc The roots 1 and 2 are complex numbers and

the general solution can be written as

Where D and are constants which can be determined from the initial conditions of the

problem

)sin()2/( teDx dtmc


66

The constant d is called the damped natural frequency of the system, and has a value of

c /cc is called the damping factor

22

12

c

ndc

c

m

c

m

k


67

As shown on the graph, the initial limit of motion, D, diminishes with each cycle of vibration, since motion is confined within the bounds of the exponential curve


68

Using the damped natural frequency d, the period of damped vibration may be written as

dd

2


69

22.5 Viscous Damped Forced Vibration*

The most general case of single-degree-of-freedom vibrating motion occurs when the

system includes the effects of forced motion

and induced damping

As shown from diagram, the differential equation which describes the motion is

tFkxxcxm sin0

70


The particular solution will be of the form

Solving for constant A and B,

tBtAxp cossin

22222

0

2222

22

0

/

/

/

/

mc

mcFB

mc

mFA

n

n

n

71

It is also possible to express the particular solution in the form of

where

)sin( tCxp

2220

)/)(/(2)/(1

/

ncn cc

kFC

2

1

)/(1

)/)(/(2tan

n

nccc


72

The angle represent the phase difference between the applied force and the resulting

steady-state vibration of the damped system.

The magnification factor MF is

2220 )/)(/(2)/(11

/MF

ncccnkF

C


73

From the graph, the magnification of the

amplitude increases as the

damping factor decreases

Resonance occurs only when the damping is zero

and the frequency ratio

equal 1


74

EXAMPLE 22.8

The 30-kg electric motor is

supported by four springs, each

having a stiffness of 200 N/m. if

the rotor R is unbalanced such

that its effect is equivalent to a 4-

kg mass located 60 mm from the

axis of rotation, determine the

amplitude of vibration when the

rotor is turning at = 10 rad/s. The damping ratio is c/cc = 0.15

75

The periodic force which causes the motor to vibrate is the centrifugal force due to the

unbalanced rotor

This force has a constant magnitude of

Since F = F0 sin t, where = 10 rad/s, then

NmrmaF n 242)10)(06.0(42

0

tF 10sin24

View Free Body Diagram

EXAMPLE 22.8

76

The stiffness of the entire system of four springs is k = 4(200) = 800 N/m

Therefore the natural frequency of vibration is

Since the damping factor is known, the steady-state amplitude is

sradm

kn /16.5

30

800

mm7.10

)/)(/(2)/(1

/

222

0

ncn cc

kFC

EXAMPLE 22.8

77

CHAPTER REVIEW

Undamped Free Vibration

A body has free vibration provided gravitational or elastic restoring forces cause

the motion.

This motion is undamped when friction forces are neglected.

The periodic motion of an undamped, freely vibrating body can be studied by displacing

the body from the equilibrium position and

then applying the equation of motion along

the path.

78

For a one-degree-of-freedom system, the resulting differential equation can be written

in the form

The frequency, or number of cycles completed per unit of time is

02 xx n

2

1

2

n

n

f

CHAPTER REVIEW

79

Undamped Forced Vibration

When the equation of motion is applied to the body, which is subjected to a periodic force

or support displacement having a frequency

, then the displacement consists of a complimentary solution and a particular

solution.

The complementary solution is caused by the free vibration and can be neglected

CHAPTER REVIEW

80

The particular solution is caused by the forced vibration

Resonance will occur if the natural period of vibration, n, is equal to the forcing frequency,

This should be avoided since the motion will become unbounded.

t

kFx O

nO

Op

sin

/1

/2

CHAPTER REVIEW

81


A viscous damping force is caused by a fluid drag on the system as it vibrates

If the motion is slow, this drag force is then proportional to the velocity, that is

C is the coefficient of viscous damping

xcF

CHAPTER REVIEW

82


By comparison its value to the critical damping coefficient cc = 2mn, we can specify the type of vibration that occurs

If c > cc, it is an overdamped system

If c = cc, it is a critically damped system

If c < cc, it is an undamped system

CHAPTER REVIEW

83

Viscous Damped Force Vibration

The most general type of motion for a one-degree-freedom system occurs when the system

is damped and subjected to periodic forced

motion.

The solution provides inside as to how the damping factor, c/cc, and the frequency ratio, /n, influence the vibration

Resonance is avoided provided c/cc, 0 and /n, 0

CHAPTER REVIEW

Chapter 9 Vibration

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