ERT 452 VIBRATION MUNIRA MOHAMED NAZARI SCHOOL OF BIOPROCESS UNIVERSITI MALAYSIA PERLIS 1ERT 452 SESION 2011/2012

ERT 452 SESION 2011/2012 1

Lecture 2Free Vibration of Single Degree

of Freedom Systems

ERT 452VIBRATION

MUNIRA MOHAMED NAZARISCHOOL OF BIOPROCESSUNIVERSITI MALAYSIA PERLIS

ERT 452 SESION 2011/2012 2

CO 2

Ability to DEVELOP and PLAN the solutions to vibration problems that contain free and forced-vibration analysis of one degree of

freedoms system.

COURSE OUTCOME

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2.1 Introduction2.2 Free Vibration of an Undamped

Translational System2.3 Free Vibration of an Undamped

Torsional System

COURSE OUTLINE

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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

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A spring-mass system in horizontal position.

2.1Introduction

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Several mechanical and structural systems can be idealized as single degree of freedom systems, for example, the mass and stiffness of a system.

2.1Introduction

Equivalent spring-mass system for the cam

follower system.

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Modeling of tall structure as spring-mass system.

2.1Introduction

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Procedure◦ Select a suitable coordinate to describe the position

of the mass of rigid body in the system (linear or angular).

◦ Determine the static equilibrium configuration of the system and measure the displacement of the mass or rigid body from its static equilibrium position.

◦ Draw free body diagram.

◦ Apply Newton’s second law of motion.

2.2 Free Vibration of an Undamped Translational System

Equation of Motion Using Newton’s Second Law of Motion:

9


If mass m is displaced a distance when acted upon by a resultant force in the same direction,

)(tx

)(tF

dt

txdm

dt

dtF

)()(

If mass m is constant, this equation reduces to

(2.1))(

)(2

2

xmdt

txdmtF

where

2

2 )(

dt

txdx

is the acceleration of the

mass.

10


where

2

2 )(

dt

txdx

is the acceleration of the

mass.For a rigid body undergoing rotational motion, Newton’s Law gives

)2.2()( JtM

where is the resultant moment acting on the body and and are the resulting angular displacement and angular acceleration, respectively.

M

22 /)( dttd

11


For undamped single degree of freedom system, the application of Eq. (2.1) to mass m yields the equation of motion:

)3.2(0

or

)(

kxxm

xmkxtF

12


Equation of Motion Using Other Methods:

(2.4b) 0)(

)2.4a( 0)(

JtM

xmtF

The application of D’Alembert’s principle to the system shown in Fig.(c) yields the equation of motion:

)3.2(0or 0 kxxmxmkx

1)D’Alembert’s Principle.The equations of motion, Eqs. (2.1) & (2.2) can be rewritten as

13

2)Principle of Virtual Displacements.“If a system that is in equilibrium under the action of a set of forces is subjected to a virtual displacement, then the total virtual work done by the forces will be zero.”


Consider spring-mass system as shown in figure, the virtual work done by each force can be computed as:

14

Since the virtual displacement can have an arbitrary value, , Eq.(2.5) gives the equation of motion of the spring-mass system as

When the total virtual work done by all the forces is set equal to zero, we obtain


xxmW

xkxW

i

S

)( force inertia by the done work Virtual

)( force spring by the done work Virtual

)5.2(0 xkxxxm

0x

)3.2(0 kxxm

15

3)Principle of Conservation of Energy.A system is said to be conservative if no energy is lost due to friction or energy-dissipating nonelastic members.


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:

)6.2(0)(

constant

UTdt

d

UTor

16


The kinetic and potential energies are given by:

)7.2(2

1 2xmT

or )8.2(2

1 2kxU

Substitution of Eqs. (2.7) & (2.8) into Eq. (2.6) yields the desired equation

)3.2(0 kxxm

17


Equation of Motion of a Spring-Mass System in Vertical Position:

18


For static equilibrium, )9.2(stkmgW

where W = weight of mass m, = static deflection g = acceleration due to gravity

st

The application of Newton’s second law of motion to mass m gives

Wxkxm st )(

and since , we obtain Wk st

)10.2(0 kxxm

19

This indicates that when a mass moves in a vertical direction, we can ignore its weight, provided we measure x from its static equilibrium position.


Notice that Eqs. (2.3) and (2.10) are identical.

20


The solution of Eq. (2.3) can be found by assuming

)11.2()( stCetx

Where C and s are constants to be determined. Substitution of Eq. (2.11) into Eq. (2.3) gives

)12.2()( 2 kmsC

Since C ≠ 0, we have

)13.2(02 kms

21


And hence,

)13.2()( 2/1nim

ks

Where i = (-1) and1/2

)14.2()( 2/1

m

kn

Roots of characteristic equation or known as eigenvalues of the problem.

22


)15.2()( 21titi nn eCeCtx

Hence, the general solution of Eq. (2.3) can be expressed as

where C1 and C2 are constants. By using the identities

)16.2(sincos)( 21 tAtAtx nn

where A1 and A2 are new constants.

tite ti sincos

23


Hence, . Thus the solution of Eq. (2.3) subject to the initial conditions of Eq. (2.17) is given by

)17.2()0(

)0(

02

01

xAtx

xAtx

n

nxAxA / and 0201

)18.2( sincos)( 00 t

xtxtx n

nn

24


Harmonic Motion:

)23.2()sin()( 00 tAtx n

where A0 and are new constants, amplitude and phase angle respectively:

0

)24.2(

2/12

0200

n

xxAA

and

)25.2(tan0

010

x

x n

Eqs.(2.15),(2.16) & (2.18) are harmonic functions of time. Eq. (2.16) can also be expressed as:

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1) Circular natural frequency:


Note the following aspects of spring-mass systems:

)26.2(2/1

m

kn

Spring constant, k:

)27.2(stst

mgWk

Hence,

)28.2(2/1

stn

g

26

)29.2(2

12/1

stn

gf


)30.2(21

2/1

gfst

nn

Hence, natural frequency in cycles per second:

and, the natural period:

27

2) Velocity and the acceleration of the

mass m at time t can be obtained as:


)(tx )(tx

)31.2()cos()cos()()(

)2

cos()sin()()(

222

2

tAtAtdt

xdtx

tAtAtdt

dxtx

nnn

nnnn

n

3) If initial displacement is zero, 0x

)32.2(sin2

cos)( 00 tx

tx

tx nn

nn

If initial velocity is zero, 0x

)33.2(cos)( 0 txtx n

28

4) The response of a single degree of freedom system can be represented in the state space or phase plane:


)35.2()sin(

)34.2()cos(

A

y

A

xt

A

xt

nn

n

nxy /Where,

By squaring and adding Eqs. (2.34) & (2.35)

)36.2(1

1)(sin)(cos

2

2

2

2

22

A

y

A

x

tt nn

29


Phase plane representation of an undamped system

30

Example 2.2Free Vibration Response Due to Impact

A cantilever beam carries a mass M at the free end as shown in the figure. A mass m falls from a height h on to the mass M and adheres to it without rebounding. Determine the resulting transverse vibration of the beam.

31

The initial conditions of the problem can be stated:

Example 2.2 Solution

(E.1)2

)(

0

0

ghmM

mv

mM

mx

xmMmv

m

m

(E.2) 2, 00 ghmM

mx

k

mgx

Using the principle of conservation of momentum:

or

Thus the resulting free transverse vibration of the beam can be expressed as:

)cos()( tAtx n

32


where

)(

3

tan

3

0

01

2/12

020

mMl

EI

mM

k

x

x

xxA

n

n

n

33

Example 2.5 Natural Frequency of Pulley System

Determine the natural frequency of the system shown in the figure. Assume the pulleys to be frictionless and of negligible mass.

34

21

222

k

W

k

W


(E.1))(4

)(4114

mass theofnt displacemeNet constant spring Equivalent

mass theofWeight

21

21

21

21

21

kk

kkk

kk

kkW

kkW

k

W

eq

eq

The total movement of the mass m (point O) is:

The equivalent spring constant of the system:

35


(E.2)0 xkxm eq

(E.3)rad/sec)(

2/1

21

21

2/1

kkm

kk

m

keqn

By displacing mass m from the static equilibrium position by x, the equation of motion of the mass can be written as

Hence, the natural frequency is given by:

(E.4)cycles/sec)(4

1

2

2/1

21

21

kkm

kkf nn

36

Free Vibration of an Undamped Torsinal System

37

l

GIM t

0

2.3 Free Vibration of an Undamped Torsinal System

From the theory of torsion of circular shafts, we have the relation:

Shear modulus

Polar moment of inertia of cross section of shaft

Length shaftTorque

38

)38.2(32

4

0

dI

2.3 Free Vibration of an Undamped Torsional System

)39.2(32

40

l

Gd

l

GIMk tt

Polar Moment of Inertia:

Torsional Spring Constant:

39


Equation of Motion:

)40.2(00 tkJ Applying Newton’s Second Law of Motion,

Thus, the natural circular frequency:

)41.2(2/1

0

J

ktn

The period and frequency of vibration in cycles per second are:

)43.2(2

1

)42.2(2

2/1

0

2/1

0

J

kf

k

J

tn

tn

40


Note the following aspects of this system:1)If the cross section of the shaft supporting the

disc is not circular, an appropriate torsional spring constant is to be used.

2)The polar mass moment of inertia of a disc is given by:

3)An important application: in a mechanical clock

g

WDDhJ

832

44

0

where ρ is the mass density h is the thickness D is the diameter W is the weight of the disc

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where ωn is given by Eq. (2.41) and A1 and A2 can be determined from the initial conditions. If


General solution of Eq. (2.40) can be obtained:

)44.2(sincos)( 21 tAtAt nn

)45.2()0()0( and )0( 00 tdt

dtt

The constants A1 and A2 can be found:

)46.2(/02

01

nA

A

Eq. (2.44) can also represent a simple harmonic motion.

42

Example 2.6Natural Frequency of Compound Pendulum

Any rigid body pivoted at a point other than its center of mass will oscillate about the pivot point under its own gravitational force. Such a system is known as a compound pendulum (shown in Figure). Find the natural frequency of such a system.

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For a displacement θ, the restoring torque (due to the weight of the body W ) is (Wd sin θ ) and the equation of motion is


E.1)(0sin0 WdJ

E.2)(00 WdJ Hence, approximated by linear equation:

The natural frequency of the compound pendulum:

(E.3)2/1

0

2/1

0

J

mgd

J

Wdn

44

Comparing with natural frequency, the length of equivalent simple pendulum:


E.4)(0

md

Jl

If J0 is replaced by mk02, where k0 is the radius of

gyration of the body about O,

(E.6)

(E.5)

2

2/1

2

0

0

d

kl

k

gdn

45

If kG denotes the radius of gyration of the body about G, we have:


(E.8)

E.7)( 2

2220

dd

kl

dkk

G

G

(E.9)2

d

kGA G

If the line OG is extended to point A such that

and

Eq.(E.8) becomes

(E.10)OAdGAl

46

Hence, from Eq.(E.5), ωn is given by


E.11)( /

2/12/12/1

20

OA

g

l

g

dk

gn

This equation shows that, no matter whether the body is pivoted from O or A, its natural frequency is the same. The point A is called the center of percussion.

47

Text book ProblemsLet’s try!!!

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A simple pendulum is set into oscillation from its rest position by giving it an angular velocity of 1 rad/s. It is found to oscillate with an amplitude of 0.5 rad. Find the natural frequency and length of the pendulum.

Problem 2.64

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Derive an expression for the natural frequency of the simple pendulum shown in Fig. 1.10. Determine the period of oscillation of a simple pendulum having a mass, m = 5 kg and a length l = 0.5 m.

Problem 2.66

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A uniform circular disc is pivoted at point O, as shown in Fig. 2.99. Find the natural frequency of the system. Also find the maximum frequency of the system by varying the value of b.

Problem 2.77

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Derive the equation of motion of the system shown in Fig. 2.100, by using Newton’s second law of motion method.

Problem 2.78

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ERT 452 VIBRATION MUNIRA MOHAMED NAZARI SCHOOL OF BIOPROCESS UNIVERSITI MALAYSIA PERLIS 1ERT 452 SESION 2011/2012