CHAP19 Mechanical Vibrations

VECTOR MECHANICS FOR ENGINEERS:

DYNAMICS

Seventh Edition

Ferdinand P. Beer

E. Russell Johnston, Jr.

Lecture Notes:

J. Walt Oler

Texas Tech University

CHAPTER

© 2003 The McGraw-Hill Companies, Inc. All rights reserved.

19Mechanical Vibrations


Vector Mechanics for Engineers: Dynamics

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Contents

Introduction

Free 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 19.4

Forced Vibrations

Sample Problem 19.5

Damped Free Vibrations

Damped Forced Vibrations

Electrical Analogues



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Introduction• 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 the period of the vibration.

• Number of cycles per unit time defines the frequency 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 as free vibration. When a periodic force is applied to the system,

the motion is described as forced vibration.

• When the frictional dissipation of energy is neglected, the motion is

said to be undamped. Actually, all vibrations are damped to some

degree.



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Free Vibrations of Particles. Simple Harmonic Motion• If a particle is displaced through a distance xm from its

equilibrium position and released with no velocity, the

particle will undergo simple harmonic motion,

( )

0=+

−=+−==

kxxm

kxxkWFma st

��

δ

• General solution is the sum of two particular solutions,

( ) ( )tCtC

tm

kCt

m

kCx

nn ωω cossin

cossin

21

21

+=

+

=

• x is a periodic function and ωn is the natural circular

frequency of the motion.

• C1 and C2 are determined by the initial conditions:

( ) ( )tCtCx nn ωω cossin 21 += 02 xC =

nvC ω01 =( ) ( )tCtCxv nnnn ωωωω sincos 21 −== �



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Free Vibrations of Particles. Simple Harmonic Motion

( )φω += txx nm sin

==n

nω

πτ

2period

===π

ω

τ 2

1 n

nnf natural frequency

( ) =+= 20

20 xvx nm ω amplitude

( ) == −nxv ωφ 00

1tan phase angle

• Displacement is equivalent to the x component of the sum of two vectors

which rotate with constant angular velocity 21 CC��

+

.nω

02

01

xC

vC

n

=

=ω



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Free Vibrations of Particles. Simple Harmonic Motion

( )φω += txx nm sin

• Velocity-time and acceleration-time curves can be

represented by sine curves of the same period as the

displacement-time curve but different phase angles.

( )

( )2sin

cos

πφωω

φωω

++=

+=

=

tx

tx

xv

nnm

nnm

�

( )

( )πφωω

φωω

++=

+−=

=

tx

tx

xa

nnm

nnm

sin

sin

2

2

��



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Simple Pendulum (Approximate Solution)• 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

l

g

nn

nm

πω

πτ

φωθθ

θθ

22

sin

0

==

+=

=+��

:tt maF =∑

• Consider tangential components of acceleration and force

for a simple pendulum,

0sin

sin

=+

=−

θθ

θθ

l

g

mlW

��

��



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Simple Pendulum (Exact Solution)

0sin =+ θθl

g��An exact solution for

leads to ( )

∫−

=2

022 sin2sin1

4π

φθ

φτ

m

nd

g

l

which requires numerical solution.

=

g

lKn π

πτ 2

2



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Sample Problem 19.1

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-mass system.



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Sample Problem 19.1

mkN6mkN4 21 == kkSOLUTION:

• Springs in parallel:

- determine the spring constant for equivalent spring

mN10mkN104

21

21

==

+==

+=

kkP

k

kkP

δ

δδ

- apply the approximate relations for the harmonic motion of a

spring-mass system

nn

nm

k

ω

πτ

ω

2

srad14.14kg20

N/m104

=

===

s 444.0=nτ

( )( )srad 4.141m 040.0=

= nmm xv ω

sm566.0=mv

2sm00.8=ma( )( )2

2

srad 4.141m 040.0=

= nmm axa



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Sample Problem 19.1

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

ω

πτ

ω

2

srad93.6kg20

400N/m2

=

===

s 907.0=nτ

( )( )srad .936m 040.0=

= nmm xv ω

sm277.0=mv

2sm920.1=ma( )( )2

2

srad .936m 040.0=

= nmm axa

mN10mkN104

21

21

==

+==

+=

kkP

k

kkP

δ

δδ



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Free Vibrations of Rigid Bodies

• If an equation of motion takes the form

0or0 22 =+=+ θωθω nn xx ��

the corresponding motion may be considered as

simple harmonic motion.

• Analysis objective is to determine ωn.

( ) ( )[ ] mgWmbbbmI ==+= ,22but 23222

121

05

3sin

5

3=+≅+ θθθθ

b

g

b

g��

g

b

b

g

nnn

3

52

2,

5

3then π

ω

πτω ===

• For an equivalent simple pendulum,

35bl =

• Consider the oscillations of a square plate

( ) ( ) θθθ �� ImbbW +=− sin



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Sample Problem 19.2

k

A cylinder of weight W is suspended as

shown.

Determine the period and natural

frequency of vibrations of the cylinder.

SOLUTION:

• From the kinematics of the system, relate

the 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 of motion.

• Substitute the kinematic relations to arrive at

an equation involving only the angular

displacement and acceleration.



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Sample Problem 19.2SOLUTION:

• From the kinematics of the system, relate the linear displacement

and acceleration to the rotation of the cylinder.

θrx = θδ rx 22 ==

θα ��rra ==θα ��

= θ��

ra =

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

external and effective forces, write the equation of motion.

( ) :∑∑ = effAA MM ( ) αIramrTWr +=− 22

( )θδ rkWkTT 2but21

02 +=+=

• Substitute the kinematic relations to arrive at an equation

involving only the angular displacement and acceleration.

( )( ) ( )

03

8

22 2

21

21

=+

+=+−

θθ

θθθ

m

k

mrrrmrkrWWr

��

��

m

kn

3

8=ω

k

m

nn

8

32

2π

ω

πτ ==

m

kf nn

3

8

2

1

2 ππ

ω==



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Sample Problem 19.3

s13.1

lb20

n =

=

τ

W

s93.1=nτ

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 for

the 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, calculate the

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.



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Sample Problem 19.3

s13.1

lb20

n =

=

τ

W

s93.1=nτ

SOLUTION:

• Using the free-body-diagram equation for the equivalence of

the external and effective moments, write the equation of

motion for the disk/gear and wire.

( ) :∑∑ =effOO MM

0=+

−=+

θθ

θθ

I

K

IK

��

��

K

I

I

K

nnn π

ω

πτω 2

2===

• With the natural frequency and moment of inertia for the disk

known, calculate the torsional spring constant.

22

2

21 sftlb 138.0

12

8

2.32

20

2

1⋅⋅=

== mrI

K

138.0213.1 π= radftlb27.4 ⋅=K



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Sample Problem 19.3

s13.1

lb20

n =

=

τ

W

s93.1=nτ

radftlb27.4 ⋅=K

K

I

I

K

nnn π

ω

πτω 2

2===

• With natural frequency and spring constant known,

calculate the moment of inertia for the gear.

27.4293.1

Iπ= 2

sftlb 403.0 ⋅⋅=I

• Apply the relations for simple harmonic motion to calculate

the maximum gear velocity.

nmmnnmnm tt ωθωωωθωωθθ === sinsin

rad 571.190 =°=mθ

( )

=

=

s 93.1

2rad 571.1

2 π

τ

πθω

nmm

srad11.5=mω



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Principle of Conservation of Energy• Resultant force on a mass in simple harmonic motion is

conservative - total energy is conserved.

constant=+ VT

=+

=+

222

2212

21 constant

xx

kxxm

nω�

�

• Consider simple harmonic motion of the square plate,

01 =T ( ) ( )[ ]2

21

21 2sin2cos1

m

m

Wb

WbWbV

θ

θθ

≅

=−=

( ) ( )( ) 22

35

21

22

32

212

21

2

212

21

2

m

mm

mm

mb

mbbm

IvmT

θ

ωθ

ω

�

�

=

+=

+= 02 =V

( ) 00222

35

212

21

2211

+=+

+=+

nmm mbWb

VTVT

ωθθ bgn 53=ω



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Sample Problem 19.4

Determine the period of small

oscillations of a cylinder which rolls

without slipping inside a curved

surface.

SOLUTION:

• Apply the principle of conservation of energy

between the positions of maximum and

minimum potential energy.

• Solve the energy equation for the natural

frequency of the oscillations.



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Sample Problem 19.4

01 =T ( )( )

( )( )2

cos1

2

1

mrRW

rRWWhV

θ

θ

−≅

−−==

( ) ( )

( ) 22

43

22

221

212

21

2212

21

2

m

mm

mm

rRm

r

rRmrrRm

IvmT

θ

θθ

ω

�

��

−=

−+−=

+= 02 =V

SOLUTION:

• Apply the principle of conservation of energy between the

positions of maximum and minimum potential energy.

2211 VTVT +=+



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Sample Problem 19.4

2211 VTVT +=+

( ) ( ) 02

0 22

43

2

+−=−+ mm rRmrRW θ

θ�

( )( ) ( ) ( )22

43

2

2 mnmm rRmrRmg ωθ

θ−=−

rR

gn

−=

3

22ωg

rR

nn

−==

2

32

2π

ω

πτ

01 =T ( )( )221 mrRWV θ−≅

( ) 22

43

2 mrRmT θ�−= 02 =V

• Solve the energy equation for the natural frequency of the

oscillations.



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Forced Vibrations

:maF =∑

( ) xmxkWtP stfm ��=+−+ δωsin

tPkxxm fm ωsin=+��

( ) xmtxkW fmst ��=−+− ωδδ sin

tkkxxm fm ωδ sin=+��

Forced vibrations - Occur when

a system is subjected to a

periodic force or a periodic

displacement of a support.

=fω forced frequency



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Forced Vibrations

[ ] txtCtC

xxx

fmnn

particulararycomplement

ωωω sincossin 21 ++=

+=

( ) ( )22211 nf

m

nf

m

f

mm

kP

mk

Px

ωω

δ

ωωω −=

−=

−=

tkkxxm fm ωδ sin=+��

tPkxxm fm ωsin=+��

At ωf = ωn, forcing input is in

resonance with the system.

tPtkxtxm fmfmfmf ωωωω sinsinsin2 =+−

Substituting particular solution into governing equation,



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Sample Problem 19.5

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 6 in.

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 the

natural 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.



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Sample Problem 19.5SOLUTION:

• The resonant frequency is equal to the natural frequency of the

system.

ftslb87.102.32

350 2⋅==m

( )ftlb000,36

inlb30007504

=

==k

W = 350 lb

k = 4(350 lb/in)

rpm 549rad/s 5.57

87.10

000,36

==

==m

knω

Resonance speed = 549 rpm



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Sample Problem 19.5

W = 350 lb

k = 4(350 lb/in)

rad/s 5.57=nω

• 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

oz 16

lb 1oz 1

rad/s 125.7 rpm 1200

2

2⋅=

=

===

m

f ωω

( )( )( ) lb 33.157.125001941.02

126

2

==

== ωmrmaP nm

( ) ( )

in 001352.0

5.577.1251

300033.15

122

−=

−=

−=

nf

mm

kPx

ωω

xm = 0.001352 in. (out of phase)



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Damped Free Vibrations

• With viscous damping due to fluid friction,

:maF =∑ ( )

0=++

=−+−

kxxcxm

xmxcxkW st

��

��δ

• Substituting x = eλt and dividing through by eλt yields

the characteristic equation,

m

k

m

c

m

ckcm −

±−==++

22

220 λλλ

• Define the critical damping coefficient such that

ncc m

m

kmc

m

k

m

cω220

2

2

===−

• All vibrations are damped to some degree by forces

due to dry friction, fluid friction, or internal friction.



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Damped Free Vibrations• Characteristic equation,

m

k

m

c

m

ckcm −

±−==++

22

220 λλλ

== nc mc ω2 critical damping coefficient

• Heavy damping: c > cc

tteCeCx 21

21λλ += - negative roots

- nonvibratory motion

• Critical damping: c = cc

( ) tnetCCxω−+= 21 - double roots

- nonvibratory motion

• Light damping: c < cc

( ) ( )tCtCex ddtmc ωω cossin 21

2 += −

=

−=

2

1c

ndc

cωω damped frequency



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Damped Forced Vibrations

( )[ ] ( )( )[ ]( )( )

( )=

−=

=

+−

==

2

222

1

2tan

21

1

nf

nfc

nfcnf

m

m

m

cc

cc

x

kP

x

ωω

ωωφ

ωωωωδ

magnification

factor

phase difference between forcing and steady state

response

tPkxxcxm fm ωsin=++ �� particulararycomplement xxx +=



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Electrical Analogues• Consider an electrical circuit consisting of an inductor, resistor

and capacitor with a source of alternating voltage

0sin =−−−C

qRi

dt

diLtE fm ω

• Oscillations of the electrical system are analogous to damped

forced vibrations of a mechanical system.

tEqC

qRqL fm ωsin1

=++ ��



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Electrical Analogues

• 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 at x = 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.



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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 ωsin12212222 =−+−+ ��

• For the electrical system,

( ) 02

21

1

121111 =

−++−+

C

qq

C

qqqRqL ��

( ) tEC

qqqqRqL fm ωsin

2

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.

CHAP19 Mechanical Vibrations

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Transcript of CHAP19 Mechanical Vibrations