dynamics - chapter 19 [beer7]
TRANSCRIPT
VECTOR MECHANICS FOR ENGINEERS: DYNAMICS
Seventh Edition
Ferdinand P. Beer
E. Russell Johnston, Jr.
Lecture Notes:
J. Walt Oler
Texas Tech University
CHAPTER
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19Mechanical Vibrations
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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
• General solution is the sum of two particular solutions,
tCtC
tm
kCt
m
kCx
nn cossin
cossin
21
21
• x is a periodic 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
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Free Vibrations of Particles. Simple Harmonic Motion
txx nm sin
n
n 2
period
2
1 n
nnf natural frequency
20
20 xvx nm amplitude
nxv 00
1tan phase angle
• Displacement is equivalent to the x component of the sum of two vectorswhich rotate with constant angular velocity
21 CC
.n
02
01
xC
vC
n
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Free Vibrations of Particles. Simple Harmonic Motion
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
tx
tx
xv
nnm
nnm
tx
tx
xa
nnm
nnm
sin
sin2
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
tl
g
nn
nm
22
sin
0
:tt maF
• Consider tangential components of acceleration and force for a simple pendulum,
0sin
sin
l
g
mlW
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Simple Pendulum (Exact Solution)
0sin l
gAn exact solution for
leads to
2
022 sin2sin1
4
m
nd
g
l
which requires numerical solution.
g
lKn
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 19.1mkN6mkN4 21 kk SOLUTION:
• Springs in parallel:
- determine the spring constant for equivalent spring
mN10mkN10 4
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
s 444.0n
srad 4.141m 040.0 nmm xv
sm566.0mv
2sm00.8ma 22
srad 4.141m 040.0
nmm axa
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Sample Problem 19.1mkN6mkN4 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
n m
k
2
srad93.6kg20
400N/m2
s 907.0n
srad .936m 040.0 nmm xv
sm277.0mv
2sm920.1ma 22
srad .936m 040.0
nmm axa
mN10mkN10 4
21
21
kkP
k
kkP
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Free Vibrations of Rigid Bodies• If an equation of motion takes the form
0or0 22 nn xx the corresponding motion may be considered as simple harmonic motion.
• Analysis objective is to determine n.
mgWmbbbmI ,22but 23222
121
05
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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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.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.
03
8
22 221
21
m
k
mrrrmrkrWWr
m
kn 3
8
k
m
nn 8
32
2
m
kf nn 3
8
2
1
2
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Sample Problem 19.3
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 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
W
s93.1n
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
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
221 sftlb 138.0
12
8
2.32
20
2
1
mrI
K
138.0213.1 radftlb27.4 K
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Sample Problem 19.3
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 2sftlb 403.0 I
• Apply the relations for simple harmonic motion to calculate the maximum gear velocity.
nmmnnmnm tt sinsin
rad 571.190 m
s 93.1
2rad 571.1
2 n
mm
srad11.5m
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Principle of Conservation of Energy• 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
221
21 2sin2cos1
m
m
Wb
WbWbV
2235
21
2232
212
21
2212
21
2
m
mm
mm
mb
mbbm
IvmT
02 V
00 22235
212
21
2211
nmm mbWb
VTVT
bgn 53
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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
cos12
1
mrRW
rRWWhV
22
43
22
221
212
21
2212
21
2
m
mm
mm
rRm
r
rRmrrRm
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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Sample Problem 19.4
2211 VTVT
02
0 2243
2 m
m rRmrRW
2243
2
2 mnmm rRmrRmg
rR
gn
3
22g
rR
nn
2
32
2
01 T 221 mrRWV
2243
2 mrRmT 02 V
• Solve the energy equation for the natural frequency of the oscillations.
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Forced Vibrations
:maF
xmxkWtP stfm sin
tPkxxm fm sin
xmtxkW fmst sin
tkkxxm fm sin
Forced vibrations - Occur when a system is subjected to a periodic force or a periodic displacement of a support.
f forced frequency
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Forced Vibrations
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
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 2m
ftlb000,36
inlb30007504
k
W = 350 lb
k = 4(350 lb/in)
rpm 549rad/s 5.5787.10
000,36
m
kn
Resonance speed = 549 rpm
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Sample Problem 19.5
W = 350 lb
k = 4(350 lb/in)
rad/s 5.57n
• 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
oz 16
lb 1oz 1
rad/s 125.7 rpm 1200
22
m
f
lb 33.157.125001941.0 2126
2
mrmaP nm
in 001352.0
5.577.1251
300033.15
1 22
nf
mm
kPx
xm = 0.001352 in. (out of phase)
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Damped Free Vibrations
• With viscous damping due to fluid friction,
:maF 0
kxxcxm
xmxcxkW st
• Substituting x = et and dividing through by et yields the characteristic equation,
m
k
m
c
m
ckcm
22
220
• Define the critical damping coefficient such that
ncc m
m
kmc
m
k
m
c 2202
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
nc mc 2 critical damping coefficient
• Heavy damping: c > cc
tt eCeCx 2121
- 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
nd c
c 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
magnification
factor
phase difference between forcing and steady state response
tPkxxcxm fm sin 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
• Oscillations of the electrical system are analogous to damped forced vibrations of a mechanical system.
tEqC
qRqL fm sin1
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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 sin12212222
• For the electrical system,
02
21
1
121111
C
C
qqqRqL
tEC
qqqqRqL fm sin
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.