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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
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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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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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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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Free Vibrations of Particles. Simple Harmonic Motion
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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Free Vibrations of Particles. Simple Harmonic Motion
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
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Vector Mechanics for Engineers: DynamicsNinth
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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
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Vector Mechanics for Engineers: DynamicsNinth
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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
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Vector Mechanics for Engineers: DynamicsNinth
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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
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Vector Mechanics for Engineers: DynamicsNinth
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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
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Vector Mechanics for Engineers: DynamicsNinth
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
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Vector Mechanics for Engineers: DynamicsNinth
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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
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Vector Mechanics for Engineers: DynamicsNinth
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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.
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Vector Mechanics for Engineers: DynamicsNinth
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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
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Vector Mechanics for Engineers: DynamicsNinth
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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
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Vector Mechanics for Engineers: DynamicsNinth
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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
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Vector Mechanics for Engineers: DynamicsNinth
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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
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Vector Mechanics for Engineers: DynamicsNinth
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
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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.
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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
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Vector Mechanics for Engineers: Dynamicsnthdition
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.
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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
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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,
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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.
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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
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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)
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Damped Free Vibrations
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.
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Damped Free Vibrations
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
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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
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Electrical Analogues
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
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Electrical Analogues
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.
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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 sin12212222 For the electrical system,
02
21
1
121111
C
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.