Ch19Mechanical Vibrations

7/30/2019 Ch19Mechanical Vibrations

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VECTOR MECHANICS FOR ENGINEERS:

DYNAMICS

NinthEdition

Ferdinand P. Beer

E. Russell Johnston, Jr.

CHAPTER

2010 The McGraw-Hill Companies, Inc. All rights reserved.

19Mechanical Vibrations


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Vector Mechanics for Engineers: DynamicsNinth

Edition

Contents

19 - 2

IntroductionFree Vibrations of Particles. Simple

Harmonic Motion

Simple Pendulum (Approximate

Solution)

Simple Pendulum (Exact Solution)

Sample Problem 19.1

Free Vibrations of Rigid Bodies

Sample Problem 19.2

Sample Problem 19.3

Principle of Conservation of Energy

Sample Problem 19.4Forced Vibrations

Sample Problem 19.5

Damped Free Vibrations

Damped Forced Vibrations

Electrical Analogues


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Edition

Introduction

19 - 3

Mechanical vibration is the motion of a particle or body which

oscillates about a position of equilibrium. Most vibrations in

machines and structures are undesirable due to increased stresses

and energy losses.

Time interval required for a system to complete a full cycle of the

motion is theperiodof the vibration.

Number of cycles per unit time defines thefrequency of the vibrations.

Maximum displacement of the system from the equilibrium position is

the amplitude of the vibration.

When the motion is maintained by the restoring forces only, the

vibration is described asfree vibration. When a periodic force is applied

to the system, the motion is described asforced vibration.

When the frictional dissipation of energy is neglected, the motion

is said to be undamped. Actually, all vibrations are dampedto

some degree.


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Edition

Free Vibrations of Particles. Simple Harmonic Motion

19 - 4

If a particle is displaced through a distancexm from its

equilibrium position and released with no velocity, the

particle will undergo simple harmonic motion,

0

kxxm

kxxkWFma st

General solution is the sum of twoparticular solutions,

tCtC

tm

kCt

m

kCx

nn cossin

cossin

21

21

x is aperiodic function and n is the natural circular

frequency of the motion.

C1 and C2 are determined by the initial conditions:

tCtCx nn cossin 21 02 xC

nvC 01 tCtCxv nnnn sincos 21

NE


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Edition


19 - 5

txx nm sin

n

n

2period

2

1 n

n

nf natural frequency

202

0 xvx nm amplitude

nxv 001

tan phase angle

Displacement is equivalent to thex component of the sum of two vectors

which rotate with constant angular velocity21 CC

.n

02

01

xC

vCn

NE


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Edition


19 - 6

txx nm sin

Velocity-time and acceleration-time curves can be

represented by sine curves of the same period as thedisplacement-time curve but different phase angles.

2sin

cos

tx

tx

xv

nnm

nnm

tx

tx

xa

nnm

nnm

sin

sin

2

2

NE


7/32 2010 The McGraw-Hill Companies, Inc. All rights reserved.


Edition

Simple Pendulum (Approximate Solution)

19 - 7

Results obtained for the spring-mass system can be

applied whenever the resultant force on a particle is

proportional to the displacement and directed towards

the equilibrium position.

for small angles,

g

l

t

lg

nn

nm

22

sin

0

:tt maF

Consider tangential components of acceleration and

force for a simple pendulum,

0sin

sin

l

gmlW

NE




Edition

Simple Pendulum (Exact Solution)

19 - 8

0sin

l

gAn exact solution for

leads to

2

022 sin2sin1

4

m

nd

g

l

which requires numerical solution.

g

lKn

2

2

V M h i f E i D iNE




Edition

Sample Problem 19.1

19 - 9

A 50-kg block moves between vertical

guides as shown. The block is pulled

40mm down from its equilibrium

position and released.

For each spring arrangement, determine

a) the period of the vibration, b) the

maximum velocity of the block, and c)

the maximum acceleration of the block.

SOLUTION:

For each spring arrangement, determine

the spring constant for a single

equivalent spring.

Apply the approximate relations for the

harmonic motion of a spring-masssystem.

V M h i f E i D iNE




Edition

Sample Problem 19.1

19 - 10

mkN6mkN4 21 kkSOLUTION:

Springs in parallel:

- determine the spring constant for equivalent spring

mN10mkN104

21

21

kkP

k

kkP

- apply the approximate relations for the harmonic motion

of a spring-mass system

nn

n m

k

2

srad14.14kg20

N/m104

s444.0n

srad4.141m040.0

nmm xv

sm566.0mv

2sm00.8ma 2

2

srad4.141m040.0

nmm axa

V t M h i f E i D iNE




Edition

Sample Problem 19.1

19 - 11

mkN6mkN4 21 kk Springs in series:

- determine the spring constant for equivalent spring

- apply the approximate relations for the harmonic motion

of a spring-mass system

nn

nm

k

2

srad93.6kg20

400N/m2

s907.0n

srad.936m040.0

nmm xv

sm277.0mv

2sm920.1ma 2

2

srad.936m040.0

nmm axa

mN10mkN104

21

21

kkP

k

kkP





Edition

Free Vibrations of Rigid Bodies

19 - 12

If an equation of motion takes the form

0or022

nnxx the corresponding motion may be considered

as simple harmonic motion.

Analysis objective is to determine n.

mgWmbbbmI ,22but 23222

121

0

5

3sin

5

3

b

g

b

g

g

b

b

g

nnn

3

52

2,

5

3then

For an equivalent simple pendulum,

35bl

Consider the oscillations of a square plate

ImbbW sin





Edition

Sample Problem 19.2

19 - 13

k

A cylinder of weight Wis suspended

as shown.

Determine the period and naturalfrequency of vibrations of the cylinder.

SOLUTION:

From the kinematics of the system, relatethe linear displacement and acceleration

to the rotation of the cylinder.

Based on a free-body-diagram equation

for the equivalence of the external and

effective forces, write the equation ofmotion.

Substitute the kinematic relations to arrive

at an equation involving only the angular

displacement and acceleration.





Edition

Sample Problem 19.2

19 - 14

SOLUTION:

From the kinematics of the system, relate the linear

displacement and acceleration to the rotation of the cylinder.rx rx 22

rra

ra

Based on a free-body-diagram equation for the equivalence of

the external and effective forces, write the equation of motion.

: effAA MM IramrTWr 22

rkWkTT 2but21

02

Substitute the kinematic relations to arrive at an equation

involving only the angular displacement and acceleration.

0

3

8

22 22121

m

k

mrrrmrkrWWr

m

kn

3

8

k

m

n

n

8

32

2

m

kf nn

3

8

2

1

2





Edition

Sample Problem 19.3

19 - 15

s13.1

lb20

n

W

s93.1n

The disk and gear undergo torsional

vibration with the periods shown.

Assume that the moment exerted by the

wire is proportional to the twist angle.

Determine a) the wire torsional spring

constant, b) the centroidal moment of

inertia of the gear, and c) the maximum

angular velocity of the gear if rotated

through 90o and released.

SOLUTION:

Using the free-body-diagram equation forthe equivalence of the external and

effective moments, write the equation of

motion for the disk/gear and wire.

With the natural frequency and moment

of inertia for the disk known, calculatethe torsional spring constant.

With natural frequency and spring

constant known, calculate the moment of

inertia for the gear.

Apply the relations for simple harmonic

motion to calculate the maximum gear

velocity.

V t M h i f E i D iN

Ed




Edition

Sample Problem 19.3

19 - 16

s13.1

lb20

n

W

s93.1n

SOLUTION:

Using the free-body-diagram equation for the equivalence

of the external and effective moments, write the equation ofmotion for the disk/gear and wire.

: effOO MM

0

I

K

IK

K

I

I

K

nnn

2

2

With the natural frequency and moment of inertia for the

disk known, calculate the torsional spring constant.

22

2

21 sftlb138.0

12

8

2.32

20

2

1

mrI

K

138.0213.1 radftlb27.4 K

Vector Mechanics for Engineers D namicsNi

Ed




Edition

Sample Problem 19.3

19 - 17

s13.1

lb20

n

W

s93.1n

radftlb27.4 K

K

I

I

K

nnn

2

2

With natural frequency and spring constant known,

calculate the moment of inertia for the gear.

27.4293.1

I 2sftlb403.0 I

Apply the relations for simple harmonic motion to

calculate the maximum gear velocity.

nmmnnmnm tt sinsin

rad571.190 m

s93.1

2rad571.1

2

nmm

srad11.5m

Vector Mechanics for Engineers DynamicsNi

Ed




dition

Principle of Conservation of Energy

19 - 18

Resultant force on a mass in simple harmonic motion

is conservative - total energy is conserved.

constantVT

222

2212

21 constant

xx

kxxm

n

Consider simple harmonic motion of the square plate,

01 T 2

21

21 2sin2cos1

m

m

Wb

WbWbV

22352122

32

212

21

2

212

21

2

m

mm

mm

mb

mbbm

IvmT

02 V

00 2223

5

2

12

2

1

2211

nmm mbWb

VTVT

bgn 53

Vector Mechanics for Engineers: DynamicsNi

Ed



Vector Mechanics for Engineers: Dynamicsinth

dition

Sample Problem 19.4

19 - 19

Determine the period of small

oscillations of a cylinder which rolls

without slipping inside a curved

surface.

SOLUTION:

Apply the principle of conservation ofenergy between the positions of maximum

and minimum potential energy.

Solve the energy equation for the natural

frequency of the oscillations.


Ed



Vector Mechanics for Engineers: Dynamicsnthdition

Sample Problem 19.4

19 - 20

01 T

2

cos1

2

1

mrRW

rRWWhV

22

43

22

2

2

1

2

12

2

1

2212

21

2

m

mm

mm

rRm

r

rR

mrrRm

IvmT

02 V

SOLUTION:

Apply the principle of conservation of energy between the

positions of maximum and minimum potential energy.

2211 VTVT


Ed




Sample Problem 19.4

19 - 21

2211 VTVT

02

022

43

2

mm rRmrRW

22

43

2

2 mnmm

rRmrRmg

rR

gn

3

22

g

rR

nn

2

32

2

01 T 22

1 mrRWV

2243

2 mrRmT 02 V

Solve the energy equation for the natural frequency of the

oscillations.

Vector Mechanics for Engineers: DynamicsNin

Ed




Forced Vibrations

19 - 22

:maF

xmxkWtP stfm sin

tPkxxm fm sin

xmtxkW fmst sin

tkkxxm fm sin

Forced vibrations - Occurwhen a system is subjected to

a periodic force or a periodic

displacement of a support.

f forced frequency


Ed




Forced Vibrations

19 - 23

txtCtC

xxx

fmnn

particulararycomplement

sincossin 21

222

11 nf

m

nf

m

f

mm

kP

mk

Px

tkkxxm fm sin

tPkxxm fm sin

At f = n, forcing input is in

resonance with the system.

tPtkxtxm fmfmfmf sinsinsin2

Substituting particular solution into governing equation,


Ed




Sample Problem 19.5

19 - 24

A motor weighing 350 lb is supported

by four springs, each having a constant

750 lb/in. The unbalance of the motor is

equivalent to a weight of 1 oz located 6in. from the axis of rotation.

Determine a) speed in rpm at which

resonance will occur, and b) amplitude

of the vibration at 1200 rpm.

SOLUTION:

The resonant frequency is equal to thenatural frequency of the system.

Evaluate the magnitude of the periodic

force due to the motor unbalance.

Determine the vibration amplitude

from the frequency ratio at 1200 rpm.


Ed


25/32


Vector Mechanics for Engineers: Dynamicsnthition

Sample Problem 19.5

19 - 25

SOLUTION:

The resonant frequency is equal to the natural frequency of

the system.

ftslb87.102.32

350 2m

ftlb000,36

inlb30007504

k

W= 350 lb

k= 4(350 lb/in)

rpm549rad/s5.57

87.10

000,36

m

kn

Resonance speed = 549 rpm


Ed


26/32


Vector Mechanics for Engineers: Dynamicsnth

ition

Sample Problem 19.5

19 - 26

W= 350 lb

k= 4(350 lb/in)

rad/s5.57n

Evaluate the magnitude of the periodic force due to the

motor unbalance. Determine the vibration amplitude from

the frequency ratio at 1200 rpm.

ftslb001941.0sft2.32

1

oz16

lb1oz1

rad/s125.7rpm1200

2

2

m

f

lb33.157.125001941.0 2126

2

mrmaP nm

in001352.0

5.577.1251

300033.15

1 22

nf

mm

kPx

xm = 0.001352 in. (out of phase)


Edi


27/32



ition


19 - 27

With viscous damping due to fluid friction,

:maF

0

kxxcxm

xmxcxkW st

Substitutingx = eltand dividing through by elt

yields the characteristic equation,

m

k

m

c

m

ckcm

2

2

22

0 lll

Define the critical damping coefficient such that

ncc m

m

kmc

m

k

m

c220

2

2

All vibrations are damped to some degree by

forces due to dry friction,fluid friction, or

internal friction.


Edi


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ition


19 - 28

Characteristic equation,

m

k

m

c

m

ckcm

2

2220 lll

nc mc 2 critical damping coefficient

Heavy damping: c > cctt

eCeCx 21

21

ll - negative roots- nonvibratory motion

Critical damping: c = cc

tnetCCx 21 - double roots- nonvibratory motion

Light damping: c < cc

tCtCex ddtmc cossin 21

2

2

1c

ndc

c damped frequency


Edi


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ition

Damped Forced Vibrations

19 - 29

2

222

1

2tan

21

1

nf

nfc

nfcnf

m

m

m

cc

cc

x

kP

x

magnification

factor

phase difference between forcing and steady

state response

tPkxxcxm fm sin particulararycomplement xxx


Edi


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tion


19 - 30

Consider an electrical circuit consisting of an inductor,

resistor and capacitor with a source of alternating voltage

0sin C

qRi

dt

diLtE fm

Oscillations of the electrical system are analogous todamped forced vibrations of a mechanical system.

tEqC

qRqL fm sin1


Edit


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Vector Mechanics for Engineers: Dynamicsthtion


19 - 31

The analogy between electrical and mechanical

systems also applies to transient as well as steady-state oscillations.

With a charge q = q0 on the capacitor, closing the

switch is analogous to releasing the mass of the

mechanical system with no initial velocity atx = x0.

If the circuit includes a battery with constant voltage

E, closing the switch is analogous to suddenly

applying a force of constant magnitude P to the

mass of the mechanical system.

Vector Mechanics for Engineers: DynamicsNint

Edit


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Vector Mechanics for Engineers: Dynamicsthtion

Electrical Analogues The electrical system analogy provides a means of

experimentally determining the characteristics of a given

mechanical system.

For the mechanical system,

0212112121111 xxkxkxxcxcxm

tPxxkxxcxm fm sin12212222 For the electrical system,

02

21

1

121111

C

qq

C

qqqRqL

tEC

qqqqRqL fm sin2

1212222

The governing equations are equivalent. The characteristics

of the vibrations of the mechanical system may be inferred

from the oscillations of the electrical system.

Ch19Mechanical Vibrations

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