Introduction to Dynamics - Purdue University
Transcript of Introduction to Dynamics - Purdue University
Introduction to Dynamics (N. Zabaras)
Mechanical VibrationsProf. Nicholas Zabaras
Warwick Centre for Predictive Modelling
University of Warwick
Coventry CV4 7AL
United Kingdom
Email: [email protected]
URL: http://www.zabaras.com/
April 10, 2016
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Introduction to Dynamics (N. Zabaras)
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 (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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Closely following the Vector Mechanics for Engineers, Beer and Johnston (Chapter 19), and Engineering Mechanics:
Dynamics Hibbeler (Chapter 21)
Introduction to Dynamics (N. Zabaras)
Since all forces
are
conservative:
Differentiate (with respect to time):
The mass-spring system
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Introduction to Dynamics (N. Zabaras)
Divide through by :
(The Governing
Equation)
Consider new coordinate system:
Where, (from equilibrium)
Substitution into the governing equation, gives:
steady
The mass-spring system
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Introduction to Dynamics (N. Zabaras)
Assume a solution of
the form:
A)
B)
C)
D) Substitution of the solution gives:
The natural frequency is then:
steady
The mass-spring system
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Introduction to Dynamics (N. Zabaras)
Since all forces are
conservative:
Treat as fixed-axis rotation, so:
If is small ℎ = 𝑟𝜃𝑠𝑖𝑛𝜃
2≈ 𝑟𝜃2
1
2, so:
And differentiate:
⇒
Now divide by to get:
(The Governing Equation)
The Pendulum
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𝐾. 𝐸. =1
2𝑚𝑣2 =
1
2𝑚 𝑟 𝜃
2=1
2𝑚𝑟2 𝜃2
Introduction to Dynamics (N. Zabaras)
mass-spring system
Governing equation:
Natural frequency:
The Pendulum
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Introduction to Dynamics (N. Zabaras)
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
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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Introduction to Dynamics (N. Zabaras)
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 9
Introduction to Dynamics (N. Zabaras)
Simple Harmonic Motion
txx nm sin
n
n
2period
2
1 n
nnf natural frequency
20
20 xvx nm amplitude
1
0 0tan nx v phase angle
• Displacement is equivalent to the x component of the sum of two vectors
which rotate with constant angular velocity 21 CC
.n
02
01
xC
vC
n
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Introduction to Dynamics (N. Zabaras)
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
sin
2
2
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Introduction to Dynamics (N. Zabaras)
Sample Problem
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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Introduction to Dynamics (N. Zabaras)
Sample Problem
mkN6mkN4 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
410 N/m14.14 rad s
50kg
2
n
n
n
k
m
s 444.0n
srad 4.141m 040.0
nmm xv
sm566.0mv
2sm00.8ma
2
20.040 m 14.14 rad s
m m na x
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Introduction to Dynamics (N. Zabaras)
Sample Problem
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
2400N/m6.93rad s
50 kg
2
n
n
n
k
m
s 907.0n
srad .936m 040.0
nmm xv
sm277.0mv
2sm920.1ma
2
20.040 m 6.93 rad s
m m na x
1 2
1 2 1 2
1 2
1 1
1 1 1 1 12.4 /
4 6
P PP
k k k k
k kN mk P k k
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Introduction to Dynamics (N. Zabaras)
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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Introduction to Dynamics (N. Zabaras)
Principle of Conservation of Energy
• Resultant force on a mass in simple harmonic motion
is conservative - total energy is conserved.
constantVT2 21 1
2 2
2 2 2
constant
constantn
mx kx
x x
• Consider simple harmonic motion of the square plate,
01 T 2
1
212
1 cos 2sin 2m m
m
V Wb Wb
Wb
2 21 12 2 2
22 21 1 2
2 2 3
2 2 2 25 51 12 3 2 3
m m
m m
m n
T mv I
m b mb
mb mb
02 V
1 1 2 2
2 2 251 12 2 3
0 0m n
T V T V
Wb mb
bgn 53
16
Note that the max
velocity is m n
12
sin 2m m For small angles:
1: Position of max displacement
2: Position of equilibrium (also maximum
Velocity)
Introduction to Dynamics (N. Zabaras)
Sample Problem
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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Introduction to Dynamics (N. Zabaras)
Sample Problem
01 T
2
cos1
2
1
mrRW
rRWWhV
2 21 12 2 2
22 2 2 21 1 1
2 2 2
2 234
m m
m m
I
m
T mv I
R rm R r mr
r
m R r
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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𝑐𝑜𝑠𝜃 = 1 −𝜃2
2!+𝜃4
4!− ⋯ ≈ 1 −
𝜃2
2𝑓𝑜𝑟 𝜃 ≪ 1
Motion of the centre of mass:
Rotation around O.
Rotation of the disk. To compute
the angular velocity of the disk
use
m m mv R r r
Instanteneous
centre of rotation
m mv r
Introduction to Dynamics (N. Zabaras)
Sample Problem
2211 VTVT
02
0 22
43
2
mm rRmrRW
22
43
2
2 mnmm rRmrRmg
rR
gn
3
22g
rR
nn
2
32
2
01 T 221 mrRWV
22
43
2 mrRmT 02 V
• Solve the energy equation for the natural frequency
of the oscillations.
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Introduction to Dynamics (N. Zabaras)
Sample Problem
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Consider a 3 kg mass, A, suspended from the centre of a uniform pulley, B. A light
inextensible flexible cord passes around the periphery of the pulley and is
attached to a fixed upper surface, on one side via a spring of stiffness k =50 N/m.
The pulley has a mass of 2 kg, a radius of 60 mm, and a moment of inertia I = 3.6
×10-4 kg m2.
Using the principle of work and energy, derive the second-order differential
equation describing the motion of A, and calculate the frequency of vibration.
Introduction to Dynamics (N. Zabaras)
Sample Problem
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What is the kinetic energy of the system?
1
2mAv 2
1
2mBv 2
1
2I
v
r
2
2.55v 2
What is the potential energy of the system?
P.E.=
(mA mb )gx 1
2k(2x)2 49.1x 100x2
Consider conservation of mechanical energy and take the time derivative =0.
2.55v2 49.1x 100x2 const
2.55 2vdv
dt
49.1
dx
dt100 2x
dx
dt
0
5.1dx
dt
d2x
dt2 49.1
dx
dt 200x
dx
dt 0
5.1d2x
dt2 200x 49.1
d2x
dt2 39.22x 0
39.22 6.26
Principle of work and energy: P.E. + K.E. = const. Therefore,
K.E. =
𝑥 = 𝑟𝜃, v = rω = 𝑟 𝜃Displacement of spring =2x
Introduction to Dynamics (N. Zabaras)
The damping force is:
where c is the damping constant (N⋅s/m)
Damping is a forcenon-conservative
Damping – The Dashpot
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Introduction to Dynamics (N. Zabaras)
Non-Conservative Forces
Non-conservative forces are not a function of
position only (e.g. air resistance, dry friction)
If a system has non-conservative forces, then:
work done on
system/mass by non-
conservative forces
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Introduction to Dynamics (N. Zabaras)
Since damping force is non-
conservative, we use:
U is work done by damping:
(at x=0, spring extension=0)
⇒ Differentiate (with respect to time):
The mass-spring damper system
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Introduction to Dynamics (N. Zabaras)
Divide through by :
If we redefine x as distance from
equilibrium:
(a damped 2nd-order
system) What happens if we move mass and release?
(step response)
The mass-spring damper system
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Introduction to Dynamics (N. Zabaras)
A) B)
C) D)
What is the step response?
t t
x
x
it depends on the particular combination of m, c and k
2nd Order System: Step Response
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Introduction to Dynamics (N. Zabaras)
A) B)
C) D)
What is the form of the step response?
t t
x
x
Solution has a component of the form: What is the form of the step response if ω is...
purely imaginary: D)C)
§
A) & B)purely real: complex (both):
2nd Order System: Step Response
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Introduction to Dynamics (N. Zabaras)
Trial
solution:
So:
Now use quadratic formula:
What happens to the step response if:
over damped
under damped
critically damped
2nd Order System: Step Response
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Introduction to Dynamics (N. Zabaras)
Natural frequency is that with no
damping, i.e.:
For convenience, we define the
damping factor :
This gives:
over damped
under damped
critically damped
2nd Order System: Step Response
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Introduction to Dynamics (N. Zabaras)
Damped Free Vibrations
• With viscous damping due to fluid friction,
:maF
0
kxxcxm
xmxcxkW st
• Substituting x = elt and dividing through by elt
yields the characteristic equation,
m
k
m
c
m
ckcm
22
220 lll
• Define the critical damping coefficient s.t.
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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Introduction to Dynamics (N. Zabaras)
Damped Free Vibrations• Characteristic equation,
m
k
m
c
m
ckcm
22
220 lll
nc mc 2 critical damping coefficient
• Heavy damping: c > cc
tteCeCx 21
21ll
- 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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Introduction to Dynamics (N. Zabaras)
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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Introduction to Dynamics (N. Zabaras)
Forced Vibrations
txtCtC
xxx
fmnn
particulararycomplement
sincossin 21
22211 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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Introduction to Dynamics (N. Zabaras)
Sample Problem
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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Introduction to Dynamics (N. Zabaras)
Sample Problem
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)
rpm 549rad/s 5.57
87.10
000,36
m
kn
Resonance speed = 549 rpm
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Introduction to Dynamics (N. Zabaras)
Sample Problem
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
2
2
m
f
lb 33.157.125001941.02
126
2
mrmaP nm
in 001352.0
5.577.1251
300033.15
122
nf
mm
kPx
xm = 0.001352 in. (out of phase)
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Introduction to Dynamics (N. Zabaras)
Unforced governing equation:
Often we are interested in an
oscillatory external force (e.g.
vibration)
Treat external force in the
same way as we did for the
non-conservative damping
force
Governing equation with external
forcing is then:
The response is known as frequency response
Forced 2nd Order System
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Introduction to Dynamics (N. Zabaras)
Substitution gives:
Governing equation with oscillatory forcing:
After initial transient response,
solution of the form:
Rearrange for normalized response:
when frequency is very low
(quasi-steady), G=1
Frequency Response Amplitude
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Introduction to Dynamics (N. Zabaras)
Natural frequency:
Damping factor:
The frequency response can now
be written:
The amplitude is: The phase lag:
(between force and response)
Frequency Response: Amplitude
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Introduction to Dynamics (N. Zabaras) 40
Introduction to Dynamics (N. Zabaras)
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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Introduction to Dynamics (N. Zabaras)
Helicopter Ground Resonance
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Introduction to Dynamics (N. Zabaras)
Summary
The 2nd-order equation describing damped
oscillation is:
over damped
under damped
critically damped
The response to step inputs or external forces
is characterised by the degree of damping:
If system is under damped ( ) driving near to
the natural frequency causes resonance
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