NOTES OF VIBRATION SEARCH

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Vibration of single degree of
freedom systems
y
Assoc. Prof. Dr. Pelin Gundes Bakir
[email protected]
Course Schedule
•
A short
h review
i on the
h d
dynamic
i b
behaviour
h i
off the
h single
i l d
degree off ffreedom
d
systems
•
A short review on the dynamic behaviour of multi‐degree of freedom structures
•
Objectives for vibration monitoring
•
Fourier Series Expansion, Fourier Transforms, Discrete Fourier Transform
•
Digital signal processing, problems associated with analog‐to‐digital conversion, sampling,
aliasing,
l
leakage,
l k
windowing,
d
filters
fl
•
Steps in instrumenting a structure, selection and installation of instruments,maintenance,
vibration
ib ti iinstrumentation,
t
t ti
exciters,
it
ttransducers,
d
performance
f
specification,
ifi ti
d
data
t acquisition
i iti
systems, strong‐motion data processing
Course Schedule
•
Random
R
d
variables,
i bl stochastic
h i processes, statistical
i i l analysis,
l i correlation
l i and
d convolution,
l i
coherence, time and frequency domain representation of random dynamic loads
•
Dynamic response of single and multi degree of freedom systems to random loads
•
Modal analysis
•
Applications in bridges, buildings, mechanical engineering and aerospace structures
•
MATLAB exercises
•
Term Projects
References
Modal
M
d l analysis:
l i
• Heylen W., Lammens S. And Sas P., ‘Modal Analysis Theory and Testing’, Katholieke
Universiteit Leuven, 1997.
• Ewins D
D.J.,
J ‘Modal Testing
Testing, Theory
Theory, Practice
Practice, and Application’ (Mechanical Engineering
Research Studies Engineering Design Series), Research Studies Pre; 2 edition (August 2001)
ISBN‐13: 978‐0863802188
• Maia, N. M. M. and Silva, J. M. M.Theoretical and Experimental Modal Analysis, Research
Studies Press Ltd,, Hertfordshire, 1997, 488 pp.,ISBN 0863802087
Signal processing:
• Blackburn,, James A,, Modern instrumentation ffor scientists and engineers,
g
, New York :
Springer, 2001
•
Stearns S. D. and David, R. A., Signal Processing Algorithms in Matlab, Prentice‐Hall Inc, 1996
• Mitra S.K., ‘Digital Signal Processing’, A Computer based approach, Mc‐Graw Hill, 3rd Edition,
2006.
• Heylen W., Lammens S. And Sas P., ‘Modal Analysis Theory and Testing’, Katholieke
Universiteit Leuven, 1997.
• Keith Worden ‘Signal Processing and Instrumentation’, Lecture Notes,
http://www.dynamics.group.shef.ac.uk/people/keith/mec409.htm
References
Signal processing:
•
Lynn, P. A. Introductory Digital Signal processing With Computer Applications. John Wiley &
Sons, 1994.
•
Stearns D. D. and David, R. A., Signal Processing Algorithms in Matlab, Prentice‐Hall Inc, 1996
•
Ifeachor E.C. and Jervis B.W. Digital Signal Processing: A Practical Approach, Addison‐Wesley,
1997
General vibration theory
•
Rao S.S., ‘Mechanical vibrations’, Pearson, Prentice Hall, 2004.
•
Inman D.J., ‘Engineering Vibration’, Prentice Hall, 1994.
•
Meirovitch L., ‘Fundamentals of vibrations’, McGrawHill, 2001.
References
R d
Random
vibrations:
ib i
•
Bendat J.S. and Piersol A.G., ‘Random data analysis and measurement procedures’, Wiley Series in
Probabilityy and Statistics,, 3rd Edition,, 2004.
•
Lutes L.D. and Sarkani S., ‘Random Vibrations: Analysis of structural and mechanical systems’, Elsevier, 631
pp, 2004.
•
Newland D.E., ’An introduction to random vibrations, spectral and wavelet analysis’, Longman,
1975/1984/1993.
•
Soong T.T. and Grigoriu, ‘Random vibration of mechanical and structural systems’, Prentice Hall, 1993.
•
Wirsching P.H. and Paez T.L. and Ortiz K. ‘Random vibrations: Theory and Practice’, John Wiley and sons,
1995.
199
•
Bendat J.S. and Piersol A.G. ‘Engineering applications of correlation and spectral analysis’, John Wiley and
Sons,, 2nd Edition,, 1993.
References
Vib i Instrumentation:
Vibration
I
i
•
Vibration, monitoring, testing and instrumentation handbook, CRC Press, Taylor and Francis,
Edited by
by: Clarence W
W. De Silva
Silva.
•
Aszkler C., ‘Acceleration, shock and vibration sensors’, Sensors handbook, Chapter 5, pages
137 159
137‐159.
•
McConnell K.G., ‘Vibration testing, theory and practice’, John Wiley and Sons, 1995.
•
Prerequisites: Basic knowledge on structural analysis.
Vibration based health monitoring
B i Information:
Basic
I f
ti
•
Instructor: Assoc. Prof. Dr. Pelin Gundes Bakir
(http://atlas.cc.itu.edu.tr/~gundes
•
Email: [email protected]
•
Office hours TBD by email appointment
• Website:
W b it
http://atlas.cc.itu.edu.tr/~gundes/lectures
•
Lecture time: Wednesday 14.00
14 00‐17
17.00
00
•
Lecture venue: NH 404
Vibration based health monitoring
‘Vibration based str
structural
ct ral health monitoring’ is
i a multidisciplinary
ltidi i li
research topic. The course is suitable both for undergraduate and
graduate students as well as the following departments:
• Civil engineering
• Earthquake engineering
• Mechanical engineering
• Aerospace engineering
• Electrical and electronic engineering
Vibration based health monitoring
Basic Information:
•
70 % attendance is required.
Grading:
• Quiz+homeworks: 35%
• Mid‐term project:25%
• Final project:40%
Introduction
•Concepts
Concepts from vibrations
•Degrees of freedom
•Classification of vibration
Concepts from vibrations
NEWTON’S
’ LAWS
First law:
If there are no forces acting upon a particle, then the particle will move in
a straight line with constant velocity.
Second law:
A particle acted upon by a force moves so that the force vector is equal to
the time rate of change of the linear momentum vector.
Third law:
When two particles exert forces upon one another, the forces lie along the
line joining the particles and the corresponding force vectors are the
negative of each other.
Definition
•
The
h minimum number
b off independent
d
d
coordinates
d
required
d to d
determine
completely the positions of all parts of a system at any instant of time
defines the degree of freedom of the system. A single degree of freedom
system requires only one coordinate to describe its position at any instant
of time.
Single degree of freedom system
•
FFor the
th simple
i l pendulum
d l
in
i th
the fi
figure, the
th motion
ti can be
b stated
t t d either
ith in
i terms
t
off θ or x and
d y. If th
the
coordinates x and y are used to describe the motion, it must be recognized that these coordinates are not
independent. They are related to each other through the relation
x2 + y2 = l 2
where l is the constant length of the pendulum. Thus any one coordinate can describe the motion of the
pendulum. In this example, we find that the choice of θ as the independent coordinate will be more
convenient than the choice of x and y.
Two degree of freedom system
•
SSome examples
l off ttwo d
degree off freedom
f d
systems
t
are shown
h
iin th
the fi
figure. Th
The fi
firstt fi
figure shows
h
a ttwo
mass – two spring system that is described by two linear coordinates x1 and x2. The second figure denotes
a two rotor system whose motion can be specified in terms of θ1 and θ2. The motion of the system in the
third figure can be described completely either by X and θ or by x,y and X.
Discrete and continuous systems
•
A large
l
number
b off practical
ti l systems
t
can be
b d
described
ib d using
i a finite
fi it number
b off d
degrees off ffreedom,
d
such
h as
the simple system shown in the previous slides.
•
Some systems, especially those involving continuous elastic members, have an infinite number of degrees
of freedom as shown in the figure. Since the beam in the figure has an infinite number of mass points, we
need an infinite number of coordinates to specify its deflected configuration. The infinite number of
coordinates defines its elastic deflection curve. Thus, the cantilever beam has infinite number of degrees
of freedom.
Discrete and continuous systems
•
SSystems
t
with
ith a fi
finite
it number
b off d
degrees off ffreedom
d
are called
ll d discrete
di
t or
lumped parameter systems, and those with an infinite number of degrees
of freedom are called continuous or distributed systems.
•
Most of the time, continuous systems are approximated as discrete
systems, and solutions are obtained in a simple manner. Although
treatment of a system as continuous gives exact results, the analytical
methods available for dealing with continuous systems are limited to a
narrow selection of problems, such as uniform beams, slender rods and
thin
hi plates.
l
•
Hence,, most of the p
practical systems
y
are studied byy treatingg them as finite
lumped masses, springs and dampers. In general, more accurate results
are obtained by increasing the number of masses, springs and dampers‐
that is by increasing the number of degrees of freedom.
Classification of vibration
• Free vibration:
ibration If a system,
t
after
ft an iinitial
iti l disturbance
di t b
iis lleft
ft tto
vibrate on its own, the ensuing vibration is known as free vibration.
No external force acts on the system. The oscillation of a simple
pendulum is an example of free vibration.
vibration
• Forced vibration: If a system is subjected to an external force (often
a repeating
ti type
t
off force),
f
) the
th resulting
lti vibration
ib ti is
i known
k
as forced
f
d
vibration.
– If the frequency of the external force coincides with one of the natural
frequencies of the system, a condition known as resonance occurs,
and the system undergoes dangerously large oscillations. Failures of
such structures as buildings
buildings, bridges
bridges, turbines
turbines, and airplane wings
have been assoicated with then occurrence of resonance.
Classification of vibration
•
•
•
Undamped
d
d vibration:
ib i
Iff no energy is lost
l or d
dissipated
d in ffriction or other
h
resistance during oscillation, the vibration is known as undamped vibration.
If any energy is lost in this way however, it is called damped vibration.
While the spring
p g forms a physical
p y
model for storingg kinetic energy
gy and hence
causing vibration, the dashpot, or damper, forms the physical model for
dissipating energy and damping the response of a mechanical system. A
dashpot
p consists of a piston
p
fit into a cylinder
y
filled with oil. This piston
p
is
perforated with holes so that motion of the piston in the oil is possible. The
laminar flow of the oil through the perforations as the piston moves causes a
damping
p g force on the piston.
p
Classification of vibration
•
Linear vibration:
Li
ib ti
If allll th
the b
basic
i components
t
of a vibratory system‐the spring, the mass,
and the damper, behave linearly, the
resulting vibration is known as linear
vibration. The differential equations that
govern the behaviour of vibratory linear
y
are linear. Therefore,, the principle
p
p of
systems
superposition holds.
•
Nonlinear vibration: If however, any of the
basic components behave nonlinearly, the
vibration is called ‘nonlinear vibration’. The
differential equations that govern the
behaviour of vibratory non‐linear systems
are non‐linear. Therefore, the principle of
superposition does not hold.
Classification of vibration
Linear and
Li
d nonlinear
li
vibrations
ib ti
contd:
td
• The nature of the spring force can be
deduced by performing a simple static
experiment With no mass attached
experiment.
attached, the
spring stretches to a position labeled as xo=0
in the figure.
• As successively more mass is attached to the
spring, the force of gravity causes the spring
to stretch further. If the value of the mass is
recorded, along with the value of the
displacement of the end of the spring each
time more mass is added, the plot of the
force (mass denoted by m, times the
acceleration due to gravity, denoted by g),
versus this displacement denoted by x, yields
a curve similar to that shown in the figure.
Classification of vibration
Linear and
d nonlinear
l
vibrations
b
contd:
d
• Note that in the region of values for x
between 0 and about 20 mm,, the curve is
a straight line. This indicates that for
deflections less than 20 mm and forces
less than 1000 N,
N the force that is applied
by the spring to the mass is proportional
to the stretch of the spring.
•
The constant of proportionality is the slope
of the straight
g line.
Classification of vibration
•
Deterministic
i i i vibration:
ib i
Iff the
h value
l or magnitude
i d off the
h excitation
i i (f
(force or
motion) acting on a vibratory system is known at any given time, the excitation is
called ‘deterministic’. The resulting vibration is known as ‘deterministic vibration’.
•
Nondeterministic vibration: In some cases, the excitation is non‐deterministic or
random; the value of excitation at a given time cannot be predicted. In these
cases, a large collection of records of the excitation may exhibit some statistical
regularity. It is possible to estimate averages such as the mean and mean square
values of the excitation.
Classification of vibration
•
•
Examples
l off random
d
excitations are wind
d velocity,
l
road
d roughness,
h
and
d
ground motion during earthquakes.
If the excitation is random,, the resultingg vibration is called random
vibration. In the case of random vibration, the vibratory response of the
system is also random: it can be described only in terms of statistical
quantities.
quantities
Mathematical background
•Homogeneous
Homogeneous linear ODEs with
constant coefficients
•Nonhomogeneous
N h
ODEs
ODE
Introduction
• Th
The dynamic
d
i behaviour
b h i
off mechanical
h i l systems
t
iis d
described
ib d b
by what
h t
we call second order Ordinary Differential Equations.
• The input to the mechanical structure appears on the right hand
side of the equation and is the Force and the solution of the
equation gives the output which is usually the displacement.
• In order to be able solve these equations, it is imperative to have a
solid background
g
on the solution of homogeneous
g
and
nonhomogeneous Ordinary Differential Equations.
• Homogeneous Ordinary Differential Equations represent the ‘Free
Free
Vibrations’ and the non‐homogeneous Ordinary Differential
Equations represent ‘Forced Vibrations’.
Homogeneous linear ODEs with
constant coefficients
•
We shall
h ll now consider
d second‐order
d d h
homogeneous llinear ODEs whose
h
coefficients a and b are constant.
y′′ + ay′ + byy = 0
•
The solution of a first order linear ODE:
y′ + ky = 0
•
•
By separating variables and integrating, we obtain:
dy
= −kdx
y
ln y = − ∫ kdx + c *
Taking exponents on both sides:
y ( x) = ce ∫
− kdx
•
= ce − kx
Let’s tryy the above solution in the first equation.
q
Usingg a constant
coefficient k: λx
y=e
Substituti
Subs
u ngg itss de
derivative
v ves : y′ = λe λx
(λ2 + aλ + b)e λx = 0
andd y′′ = λ2 e λx
Homogeneous linear ODEs with
constant coefficients
•
Hence iff λ is a solution
l
off the
h important characteristic
h
equation ((or
auxiliary equation)
λ2 + aλ + b = 0
•
Then the exponential solution y = e λx is a solution of the
y′′ + ay′ + byy = 0
•
Now from elementary algebra we recall that the roots of this quadratic
equation are:
1
λ = (−a + a 2 − 4b)
1
2
1
λ2 = (−a − a 2 − 4b)
2
•
The functions below are solutions to y′′ + ay′ + by = 0
y1 = e λ1x and
y 2 = e λ2 x
Homogeneous linear ODEs with
constant coefficients
•
From algebra
l b we know
k
that
h the
h quadratic
d
equation below
b l
may have
h
three
h
kinds of roots:
λ2 + aλ + b = 0
•
•
•
Case I: Two real roots if a 2 − 4b > 0
Case II: A real double root if a 2 − 4b = 0
Case III: Complex conjugate roots if a 2 − 4b < 0
•
CASE I: In this case
case, a basis of solutions of y′′ + ay′ + by = 0
λx
λ x
in any interval is: y1 = e and y2 = e
because y1 and y2 are defined and real for all x and their q
quotient is not
constant. The corresponding general solution is:
1
2
y = c1e λ1x + c2 e λ2 x
Homogeneous linear ODEs with
constant coefficients
CASE II: Reall double
d bl root λ=‐a/2
λ /
• If the discriminant a 2 − 4b is zero, we see from
1
2
1
λ2 = (−a − a 2 − 4b)
2
λ1 = (−a + a 2 − 4b)
that we get only one root:
λ = λ1 = λ2 = − a / 2,
hence only one solution :
y1 = e −( a / 2) x
TTo obtain
b i a second
d independent
i d
d
solution
l i y2 needed
d d for
f a basis,
b i we use the
h
method of order of reduction. Setting
y2 = uy1 , Substituting this and its derivatives y 2 ' = u ' y1 + uy1 ' and y2 ' ' into
y′′ + ay′ + by = 0
Homogeneous linear ODEs with
constant coefficients
CASE II: Reall double
d bl root λ=‐a/2
λ /
• We have : (u′′y1 + 2u′y1′ + uy1′′) + a(u′y1 + uy1′ ) + buy1 = 0
• Collecting terms u′′y + u′(2 y′ + ay ) + u( y′′ + ay′ + by ) = 0
• This expression in the last paranthesis is zero, since y1 is a solution of
1
1
1
1
1
1
y′′ + ay′ + by = 0
•
•
2 y′ = − ae
The expression in the second paranthesis is zero too since
We are thus left with
1
u ′′y1 = 0
Hence
u ′′ = 0
By two integrations
u = c1 x + c2
− ax / 2
= −ay1
Homogeneous linear ODEs with
constant coefficients
CASE II: Reall double
d bl root λ=‐a/2
λ /
To get a second independent solution y2=uy1, we can simply choose c1=1
and c2=0 and take u=x. Then y2=xyy1. Since these solutions are not
proportional, they form a basis. Hence in the case of a double root of
λ2 + aλ + b = 0
a basis
b i off solutions
l i
off
on any interval is:
•
y′′ + ay′ + by
b =0
e − ax / 2 , xe − ax / 2
The corresponding general solution is: y = (c1 + c2 x)e − ax / 2
Homogeneous linear ODEs with
constant coefficients
CASE III: Complex roots –a/2+iω
a/2+iω and –a/2‐iω
a/2 iω
•
This case occurs if the discriminant of the characteristic equation
λ2 + aλ + b = 0
is negative. In this case, the roots of the above equation and thus the solutions of the ODE
come at first out complex. However, we show that from them we can obtain a basis of real solutions:
y′′ + ay′ + by = 0
y1 = e − ax / 2 cos ω x ,
where
y 2 = e − ax / 2 sin
i ωx
1
4
ω 2 = b − a2
•
This is proved in the next slides. It can be verified by substitution that these are solutions in the present
case. They form a basis on any interval since their quotient cotωx is not constant. Hence, a real general
solution
l ti iin C
Case III iis:
y = e − ax / 2 ( A cos ω x + B sin ω x )
( A , B arbitrary )
Homogeneous linear ODEs with
constant coefficients: Proof
•
Complex
l number
b representation off h
harmonic motion: Since
r
r
X = OP
this vector can be represented as a complex number:
r
X = a + ib
r
where i = − 1 and a and b denote x and y components of X . Components
a and b are also called the real and the imaginary parts of the vector X. If A
denotes the modulus or the absolute value of the vector X, and θ denotes
r
the argument
g
of the angle
g between the vector and the x‐axis,, then X can
also be expressed as:
Homogeneous linear ODEs with constant coefficients:
P f
Proof
Complex number representation of harmonic motion
r
r
X = OP
r
X = a + ib
r
X
Homogeneous linear ODEs with
constant coefficients: Proof
•
A apparentt we have
As
h
two
t complex
l roots.
t These
Th
are:
1
2
1
2
λ1 = a + iω and λ1 = a − iω
•
We know from basic mathematics that a complex exponential function can
be expressed as:
r + it
r it
r
e
•
= e e = e (cos t + i sin t )
Thus the roots of the second order Ordinary Differential Equation can be
e λ x = e − ( a / 2) x +iωx = e − ( a / 2) x (cos ωx + i sin ωx)
expressed as:
1
e λ2 x = e −( a / 2 ) x −iωx = e −( a / 2) x (cos ωx − i sin ωx)
•
•
We now add these two lines and multiply the result by ½. This gives:
y1 = e − ax / 2 cos ω x
Then we subtract the second line from the first and multiply the result by
1/2i. This gives:
y 2 = e − ax / 2 sin ω x
Homogeneous linear ODEs with
constant coefficients
Case
Roots
I
Distinct real
Basis
General solution
e λ1x , e λ2 x
y = c1e λ1x + c2 e λ2 x
λ1, λ2
II
Real double root
λ=‐a/2
III
e
− ax / 2
, xe
− ax / 2
y = (c1 + c2 x)e − ax / 2
Complex
conjugate
1
2
1
λ2 = − a − i ω
2
λ1 = − a + iω
y1 = e − ax / 2 cos ωx
y2 = e − ax / 2 sin ωx
y = e − ax / 2 ( A cos ω x + B sin ω x )
Nonhomogeneous ODEs
•
In this
h section, we proceed
d from
f
homogeneous
h
to nonhomogeneous
h
ODEs.
′′
′
y + p ( x) y + q ( x) y = r ( x)
•
The general solution consists of two parts:
y ( x) = yh ( x) + y p ( x)
where yh = c1 y1 + c2 y2 is a general solution of the homogeneous ODE.
ODE
Term in r(x)
Choice for yp(x)
keγx
Ceγx
kx n (n = 0,1,2,...)
K n x n + K n −1 x n −1 + .... + K1 x + K 0
k cos ωx
k sin ωx
keαx cos ωx
keαx sin ωx
K cos ω x + M sin ω x
eαx ( K cos ωx + M sin ωx)
Nonhomogeneous ODEs
Choice rules for the method of undetermined coefficients
a)
Basic rule: If r(x) is one of the functions in the first column in the Table,
choose yp in the same line and determine its undetermined coefficients
by substituting yp and its derivatives into
y′′ + p ( x) y′ + q ( x) y = r ( x)
b) Modification rule: If a term in your choice for yp happens to be a solution
of the homogeneous ODE corresponding to the above equation, multiply
yyour choice of yp byy x ((or x^2 if this solution corresponds
p
to a double root
of the characteristic equation of the homogeneous ODE)
c) Sum rule: If r(x) is a sum of functions in the first column of the table,
choose for yp the sum of the functions in the corresponding lines of the
second column.
Free Vibration of Single Degree
of Freedom Systems
y
•Harmonic Motion
•Free vibration of undamped SDOF systems
•Free vibration of damped
p SDOF systems
y
Harmonic motion
• O
Oscillatory
ill t
motion
ti may repeatt it
itself
lf regularly,
l l as
in the case of a simple pendulum, or it may
display considerable irregularity
irregularity, as in the case of
ground motion during an earthquake.
• If the motion is repeated after equal intervals of
time, it is called periodic motion.
• The simplest type of periodic motion is harmonic
motion.
Harmonic motion
•
Shown in the figure is a vector
OP that rotates
counterclockwise with
g
velocityy ω.
constant angular
•
At any time t, the angle that
OP makes with the horizontal
is θ=ωt.
• Let y be the projection of
OP on the vertical axis. Then
y=A sin ωt. Here y, a
function of time is plotted
p
versus ωt.
Harmonic motion
•
A particle that experiences this
motion is said to have harmonic
motion.
•
The maximum displacement of a
vibrating body from its equilibrium
position is called the amplitude of
vibration Amplitude A is shown in
vibration.
the figure.
•
Range 2A is the peak to peak
displacement.
•
Now consider the units of θ. Let C
be the circumference of the circle
shown in the figure.
Harmonic motion
•
Thus C=2πA
C=2πA. Or we can write C=Aθ,
C=Aθ
where θ=2π for one revolution. Thus
defined, θ is said to be in radians and is
equivalent to 360°. Therefore, one
radian is approximately equal to 58.3
58.3°..
•
In general, for any arc length,
s=A θ , where θ is in radians. It follows
that ω in the figure would be in radians
per second.
•
As seen in the figure, the vectorial
method of representing harmonic
motion requires the description of both
the horizontal and vertical components.
•
The ti
Th
time ttaken
k to
t complete
l t one cycle
l
of motion is known as the period of
oscillation or time period and is
denoted by τ. The period is the time for
the motion to repeat (the value of τ in
the figure).
Harmonic motion
•
Note that ω τ=2 π where ω denotes the
angular velocity of the cyclic motion. The
angular velocity ω is also called the
circular
i l frequency.
f
•
The movement of a vibrating body from
its undisturbed or equilibrium position to
its extreme position in one direction,
then to the equilibrium position, then to
its extreme position in the other
direction, and back to equilibrium
position
i i is
i called
ll d a cycle
l off vibration.
ib i
•
One revolution (i.e., angular
displacement
p
of 2π radians)) of the p
pin P
in the figure or one revolution of the
vector OP in the figure constitutes a
cycle. Cycle is the motion in one period,
as shown in the figure.
Harmonic motion
•
Frequency is the
h number
b off cycles
l per unit time.
•
The most common unit of time used in vibration analysis is seconds.
seconds
Cycles per second is called Hertz.
•
The time the cycle takes to repat itself is the period T. In terms of the
period, the frequency is:
1
f =
τ
ω
• The frequency f is related to ω: f =
2π
ω = 2πf
Harmonic motion
•
Phase
h
angle:
l Consider
d two vibratory
b
motions denoted
d
d by:
b
x1 = A1 sin ωt
x2 = A2 sin(ωt + φ )
•
These two harmonic motions are called synchronous because they have
the same frequency or angular velocity ω. Two synchronous oscillations
need not have the same amplitude, and they need not attain their
g
maximum values at the same time as shown in the figure.
Harmonic motion
•
In this
h figure,
f
the
h second
d vector OP2 leads
l d the
h first
f
one OP1 by
b an angle
l φ
known as the phase angle. This means that the maximum of the second
vector would occur φ radians earlier than that of the first vector. These
two vectors are said to have a phase difference of φ.
Harmonic motion
•
•
From introductory
d
physics
h
and
d dynamics,
d
the
h fundamental
f d
l kinematical
k
l
quantities used to describe the motion of a particle are displacement,
velocity and acceleration vectors.
The acceleration of a particle is given by:
dv d 2 x
a=
= 2 = &x&
dt dt
•
Thus, displacement
Thus
displacement, velocity
velocity, and acceleration have the following
relationships in harmonic motion:
x = A sin ωt
v = x& = Aω cos ωt
a = &x& = − Aω 2 sin ωt
Operations on harmonic functions
•
r
Using complex
l number
b representation, the
h rotating vector X can b
be
written as:
r
where ω denotes the circular frequency (rad/sec) of rotation of the vector X
in counterclockwise direction. The differentiation of the harmonics given
b the
by
h above
b
equation
i gives:
i
•
Thus the displacement, velocity and acceleration can be expressed as:
Operations on harmonic functions
•
It can be
b seen that
h the
h
acceleration vector leads the
velocity vector by 90 degrees
and the velocity vector leads
the displacement vector by 90
degrees.
g
Harmonic motion
•
•
Naturall frequency:
f
Iff a system, after
f an initiall disturbance,
d
b
is left
l f to
vibrate on its own, the frequency with which it oscillates without external
forces is known as its natural frequency. As will be seen, a vibratory
system having n degrees of freedom will have, in general, n distinct
natural frequencies of vibration.
Beats: When two harmonic motions,
motions with frequencies close to one
another, are added, the resulting motion exhibits a phenomenon known
as beats. For example if:
x1 (t ) = X cos ωt
x2 (t ) = X cos(ω + δ )t
where δ is a small quantity.
The addition of these two motions yield:
x(t ) = x1 (t ) + x2 (t ) = X [cos ωt + cos(ω + δ )t ]
Harmonic motion
Beats:
x(t ) = x1 (t ) + x2 (t ) = X [cos ωt + cos(ω + δ )t ]
Using the relation
⎛ A+ B ⎞ ⎛ A− B ⎞
cos A + cos B = 2 cos⎜
⎟ cos⎜
⎟
2
2
⎝
⎠ ⎝
⎠
The first equation can be written as:
x(t ) = 2 X cos
δt
δ
cos(ω + )t
2
2
Harmonic motion
Beats:
• It can be seen that the resulting motion x(t) represents a cosine wave
δ
ω
+
with frequency
q
y
which is approximately
pp
y equal
q to ω and with a varying
y g
2
δt
amplitude 2 X cos 2 . Whenever, the amplitude reaches a maximum it is
called a beat.
• In machines and in structures,the
structures the beating phenomenon occurs when the
forcing frequency is close to the natural frequency of the system. We will
later return to this topic.
Harmonic motion
• O
Octave:
t
Wh the
When
th maximum
i
value
l off a range off
frequency is twice its minimum value, it is known
as an octave band.
band
• For example
example, each of the ranges 75‐150
75 150 Hz,
Hz 150‐
150
300 Hz, and 300‐600Hz can be called an octave
band.
• In each case
case, the maximum and minimum values
of frequency, which have a ratio of 2:1, are said to
differ byy an octave.
Free vibration of undamped SDOF
systems
•
Consider
d the
h single‐degree‐of‐freedom
l d
ff d
((SDOF)) system shown
h
in the
h ffigure.
The spring is originally in the unstretched position as shown. It is assumed
that the spring obeys Hooke’s law. The force in the spring is proportional
to displacement with the proportionality constant (spring constant) equal
to k.
Free vibration of undamped SDOF
systems
•
The
h stiffness
ff
in a spring can b
be related
l d more directly
d
l to materiall and
d
geometric properties of the spring. A spring like behaviour results from a
variety of configurations, including longitudinal motion (vibration in the
direction of the length), transverse motion (vibration perpendicular to the
length), and torsional motion(vibration rotating around the length).
Free vibration of undamped SDOF
systems
•
A spring is generally
ll made
d off an elastic
l
material.l For a slender
l d elastic
l
material of length l, cross‐sectional area A and elastic modulus E (or
Young’s modulus), the stiffness of the bar for vibration along its length is
given by:
EA
k=
•
l
The modulus E has the units of Pascal (denoted Pa) which are N/m2
N/m2.
Free vibration of undamped SDOF
systems
•
When
h the
h mass m (weight
(
h W)) is applied,
l d the
h spring willll deflect
d fl to a static
equilibrium position δst.
•
At this position, we find that:
W = mg = kδ st
•
If the mass is perturbed and allowed to move dynamically, the
displacement x, measured from the equilibrium position, will be a function
of time. Here,, x(t)
( ) is the absolute motion of the mass and the force in the
spring can be expresssed as:
− k ( x + δ st )
•
TTo d
determine
i the
h position
i i as a ffunction
i off time,
i
the
h equations
i
off motion
i
are employed; the free body diagrams are drawn as shown in the figure.
Note that x is measured positive downward.
Free vibration of undamped SDOF
systems
•
Applying
l
Newton’s
’ second
d law,
l
W − k ( x + δ st ) = m&x&
•
But from the static condition
condition, note that W=kδst.
st Thus,
Thus the equation of
motion becomes:
m&x& + kx = 0
•
With the standard form of:
•
Since in the above equation:
k
x=0
m
• This is Case III that has complex roots where the general solution was
computed as:
x = e − ax / 2 ( A cos ω t + B sin ω t )
a=0
&x& +
1
4
ω 2 = b − a2
Free vibration of undamped SDOF
systems
•
It is shown
h
that
h
x(t ) = A cos ωnt + B sin ωnt
where A and B are constants of integration and
k
(rad / sec)
m
Here ωn defines the natural frequency of the mass. This is the frequency at
which the mass will move regardless of the amplitude of the motion as
long as the spring in the system continues to obey Hooke
Hooke’ss law
law. The
natural frequency in Hertz is:
ωn =
fn =
1
2π
k
m
( Hz )
Free vibration of undamped SDOF
systems
•
The
h initiall conditions
d
x = xo at t=0
0 andd x& = x& o at t = 0 are usedd to evaluate
l
the constants of integration A and B. When substituted into the equation
x(t ) = A cos ωnt + B sin ωnt
we get:
A = xo and B =
•
x&o
ωn
and consequently x(t ) = xo cos ω n t +
The sum in the equation
x& o
ωn
sin ω n t
x(t ) = A cos ωnt + B sin ωnt
C = A2 + B 2
can also be combined to a p
phase shifted cosine with amplitude
p
and phase angle φ=arctan(B/A). For this purpose let:
A = C cos φ and B = C sin φ
Free vibration of undamped SDOF
systems
•
Introducing
d
the
h new values
l
off A and
d B into
x(t ) = A cos ωnt + B sin ωnt
we get:
x(t ) = C cos φ cos ωnt + C sin φ sin ωn t
•
Since
•
x(t)
( ) can b
be expressed
d as:
•
Where φ
φ=arctan(B/A)
arctan(B/A) and consequently:
cos( x − y ) = cos x cos y + sin x sin y
x(t ) = C cos(ωnt − φ )
⎛ x&o ⎞
⎛ x&o ⎞
2
2
2
⎜⎜ ⎟⎟
⎜
⎟
=
+
=
+
C
A
B
x
φ = arctan⎜
and
o
⎟
⎝ ωn xo ⎠
⎝ ωn ⎠
2
Free vibration of undamped SDOF
systems
•
The
h equation x(t ) = xo cos ω n t +
x& o
ωn
sin
i ω n t is a hharmonic ffunction off
time. Thus
time
Thus, the spring‐mass
spring mass system is called a harmonic oscillator.
oscillator The
nature of harmonic oscillation is shown in the figure. If C denotes a vector
of magnitude C, which makes an angle ωnt‐φ with respect to the vertical x
axis then the solution x(t ) = C cos(ωnt − φ ) can be seen to be the
axis,
projection of vector C on the x axis.
Damping
•
•
•
•
Undamped
U
d
d and
dd
damped
d vibration:
ib ti
Th response off a spring‐mass
The
i
model
d l predicts
di t
that the system will oscillate indefinitely. However, everyday observation indicates
that most freely oscillating systems eventually die out and reduce to zero motion.
Th choice
The
h i off representative
i model
d l ffor the
h observed
b
dd
decay iin an oscillating
ill i system
is based partially on physical observation and partially on mathematical
convenience. The theory of differential equations suggests that adding a term to
equation m&x&(t ) + kx(t ) = 0 of the form cx& , where c is a constant,
constant will result in a
solution x(t) that dies out.
Physical observation agrees fair well with this model and it is used very
successfully
f ll to
t model
d l th
the d
damping
i or d
decay iin a variety
i t off mechanical
h i l systems.
t
This type of damping is called the viscous damping.
Damping
The
h llaminar fl
flow off the
h oill through
h
h the
h perforations
f
as the
h piston moves
causes a damping force on the piston.
• The force is p
proportional
p
to the velocityy of the p
piston,, in a direction
opposite that of the piston motion. This damping force has the form:
f c = cx& (t )
where
h
c iis a constant off proportionality
i
li related
l d to the
h oilil viscosity.
i
i The
h
constant c, called the damping coefficient, has units of Ns/m, or kg/s.
•
Damped free vibration of SDOF system
•
Consider
d the
h spring‐mass system with
h an energy dissipating
d
mechanism
h
described by the damping force as shown in the figure. It is assumed that
the damping force FD is proportional to the velocity of the mass, as shown;
the damping coefficient is c. When Newton’s second law is applied, this
model for the damping force leads to a linear differential equation,
m&x& + cx& + kx
k =0
Damped free vibration of SDOF system
•
The
h ODE is homogeneous
h
llinear and
d has
h constant coefficients.
ff
The
h
characteristic equation is found by dividing the below equation by m:
s2 +
•
c
k
s+ =0
m
m
By the roots of a quadratic equation, we obtain:
s1 = −α + β , s2 = −α − β ,
where
c
α=
2m
•
and β =
1
c 2 − 4mk
2m
It is now most interesting that depending on the amount of damping
(much, medium or little) there will be three types of motion
corresponding to the three cases I, II and III.
Damped free vibration of SDOF system
C
Case
I
Roots
c > 4mk
2
Definition
fi i i
Distinct real roots
Overdamping
s1, s2
II
c 2 = 4mk
Real double root
Critical damping
p g
III
c 2 < 4mk
Complex conjugate roots
Underdamping
Damped free vibration of SDOF system
•
Define
f the
h criticall d
damping coefficient
ff
cc as that
h value
l off c that
h makes
k the
h
radical equal to zero,
cc = 2m
•
Define the damping factor as:
ξ=
•
k
= 2mωn
m
c
c
=
cc 2mωn
Introducing the above equation into
2
s1, 2
•
•
k
c
⎛ c ⎞
=−
± ⎜
⎟ −
2m
⎝ 2m ⎠ m
(
)
We find:
s1, 2 = − ξ ± ξ 2 − 1 ωn
Then the solution can be written as:
x(t ) = Ae
⎛⎜ −ζ + ζ 2 −1 ⎞⎟ω t
⎝
⎠ n
+ Be
⎛⎜ −ζ − ζ 2 −1 ⎞⎟ω t
⎝
⎠ n
Three cases of damping
• Heavy damping when c > cc
• Critical damping c = cc
• Light damping 0 < c < cc
Heavy damping (c > cc or ζ>1)
•
The
h roots are b
both
h real.l The
h solution
l
to the
h differential
d ff
l equation is:
x(t ) = Ae s1t + Be s2t
where A and B are the constants of integration.
integration Both s1 and s2 will be
2
2
2
negative because α > 0, β > 0, and β = α − k / m < α . Since
s1 = −α + β , s2 = −α − β , where α =
c
2m
and β =
1
c 2 − 4mk
2m
Thus, given any initial displacement, the mass will decay to the
equilibrium position without vibratory motion. An overdamped system
does not oscillate but rather returns to its rest p
position exponentially.
p
y
2
2
x(t ) = Ae ( −ζ + ζ −1 )ωnt + Be ( −ζ − ζ −1 )ωnt
Critical damping (c = cc, or ζ=1)
•
•
1
c 2 − 4mk is zero in this
Since β =
h case, s1=s2=‐α=‐cc/2m=‐ω
/
n.
2m
−ω t
Both roots are equal and the general solution is: x(t ) = ( A + Bt )e
.
Substitutingg the initial conditions,, x = xo at t=0 and x& = x& o at t = 0
A = xo and B = x&o + ωn xo
n
and the solution becomes :
x(t ) = [xo + ( x&o + ωn xo )t ]e −ωnt
•
The motion is again not vibratory and decays to the equilibrium position.
Light damping (0 < c < cc or ζ<1)
•
This
h case occurs iff the
h damping
d
constant c is so smallll that
h
c 2 < 4mk
imaginary
• Then β is no longer real but pure imaginary.
1
k
c2
2
4mk − c =
β = iω * where ω* =
−
2m
m 4m 2
•
The roots of the characteristic equation are now complex conjugate:
s1 = −α + iω*,
s2 = −α − iω *
with
α=
•
c
2m
Hence the corresponding general solution is:
x = e − α t ( A cos ω * t + B sin ω * t ) = Ce − α t cos( ω * t − φ o )
where
C 2 = A 2 + B 2 and tan φ o = B/A
Light damping (0 < c < cc )
•
The
h solution
l
can also
l be
b expressed
d as:
(
x(t ) = e −ζω nt A cos 1 − ζ 2 ωnt + B sin 1 − ζ 2 ωnt
•
)
The roots are complex. It is easily shown, using Euler’s formula that the
−ξω t
general solution is: x(t ) = [C cos(ωd t − φo )]e
where C and φ are the constants of integration. The damped natural
2
frequency ωd is given by ωd = ωn 1 − ξ
n
Light damping (0 < c < cc )
•
For the
h initiall conditions
d
x(t = 0) = xo
x& (t = 0) = x&o
•
The equation
(
x(t ) = e −ζω nt A cos 1 − ζ 2 ωnt + B sin 1 − ζ 2 ωnt
can be expressed as:
x(t ) = e
where
−ζω n t
)
⎛
⎞
x&o + ζω
ζ n xo
2
2
⎜⎜ xo cos 1 − ζ ωnt +
sin 1 − ζ ωnt ⎟⎟
ωd
⎝
⎠
ζ =
c
2mωn
Nature of the roots in the complex
plane
•
FFor ζ=0,
ζ 0 we obtain
bt i th
the iimaginary
i
roots
t
iωn and ‐iωn and a solution of
x(t ) = A cos ωnt + B sin ωnt
•
For 0<ζ<1, the roots are complex
conjugate and are located symmetrically
about the real axis.
•
As the value of ζ approaches 1,
both roots approach the point
‐ωn on the real axis.
axis
•
If ζ is greater than 1, both roots lie on the
real axis, one increasing and the other decreasing.
Logarithmic decrement
•
The
h logarithmic
l
h
d
decrement represents the
h rate at which
h h the
h amplitude
l d off
a free damped vibration decreases. It is defined as the natural logarithm
of the ratio of any two successive amplitudes.
•
Let t1 and t2 denote the times corresponding to two consecutive
amplitudes (displacements) measured one cycle apart for an
underdamped system as shown in the figure.
Logarithmic decrement
•
ξ t
Using x(t ) = [A cos(ωd t − φ )]e −ξω
, we can form
f
the
h ratio:
n
x1 X o e −ζω nt1 cos(ωd t1 − φo )
=
x2 X o e −ζω nt2 cos(ωd t 2 − φo )
•
But t2=t1+τd , hence
Logarithmic decrement
•
The
h llogarithmic
h
d
decrement can b
be ffound
d ffrom:
•
For small damping, the above equation can be approximated as:
Logarithmic decrement
•
Figure shows
h
the
h variation off the
h
logarithmic decrement δ with ζ
as shown in the equations:
•
It can be noticed that for values up to
ζ = 0.3, the two curves are difficult to
distinguish.
Logarithmic decrement
•
The logarithmic decrement is dimensionless and is actually another form
of the dimensionless damping ratio ζ. Once δ is known, ζ can be found by
solving:
g
•
If we use
instead of
we have:
Logarithmic decrement
•
Iff the
h damping
d
in the
h given system is not known,
k
we can determine
d
it
experimentally by measuring any two consecutive displacements x1 and
x2. By taking the natural logarithm of the ratio x1 and x2, we obtain δ. By
using
we can compute the damping ratio ζ.
• In fact the damping
p g ratio ζ can also be found byy measuringg two
displacements separated by any number of complete cycles. If x1 and xm+1
denote the amplitudes corresponding to times t1 and tm+1=t1+mτd where m
is an integer
integer, we obtain:
Logarithmic decrement
•
Since any two successive displacements
d l
separated
d by
b one cycle
l satisfy
f the
h
equation:
the equation
becomes:
•
The above equation yields
which can be substituted into the either of the equations
q
to obtain the
viscous damping ratio ζ:
Forced Vibration
•Harmonic
Harmonic excitation
•Base excitation
Harmonically excited vibration
•
•
•
•
A mechanical
h
l or structurall system is said
d to undergo
d
f
forced
d vibration
b
whenever external energy is supplied to the system during vibration.
External energy can be supplied to the system through either an applied
force or an imposed displacement excitation.
The applied force or displacement excitation may be harmonic,
nonharmonic but periodic,
periodic nonperiodic or random in nature.
nature The response
of a system to harmonic excitation is called harmonic response.
The nonperiodic excitations may have a long or short duration. The
response off a d
dynamic
i system to suddenly
dd l applied
li d nonperiodic
i di excitations
i i
is called transient response.
In this p
part of the course,, we shall consider the dynamic
y
response
p
of a
single degree of freedom system under harmonic excitations of the form
F (t ) = Fo ei (ωt +φ ) or F(t) = Fo cos (ωt + φ) or F(t) = Fo sin (ωt + φ)
h
where
Fo is the amplitude, ω is the frequency, and φ is the phase angle of the harmonic excitation.
Forced‐Excited System: Harmonic
Excitation
•
Some single
l degree
d
off freedom
f d
(SDOF)
(
) systems with
h an externall force
f
are
shown in the figure. Force can be applied both as an external force F(t), or
as a base motion y(t), as shown. The coordinate x(t) is the absolute motion
of the mass. The forces W and kδst are ignored in the free‐body diagrams
as we know they will add to zero in the equation of motion.
Forced‐Excited System: Harmonic
Excitation
•
•
•
Consider
d the
h force‐excited
f
d system off the
h figure,
f
where
h
the
h applied
l d force
f
is
F (t ) = Fo sin ωt
harmonic,
Applying
pp y g Newton’s second law,, the equation
q
of motion becomes:
m&x& + cx& + kx = Fo sin ωt
The general solution for this second‐order nonhomogeneous linear
diff
differential
i l equation
i is
i x(t ) = xh (t ) + x p (t )
where xh is the complementary solution or solution to the homogeneous
equation.
q
But this solution dies out soon . Our interest focuses on xp, the
particular solution. In vibration theory, the particular solution is also called
the steady‐state solution.
Forced‐Excited System: Harmonic
Excitation
•
•
•
The
h variations off h
homogeneous, particular,
l and
d generall solutions
l
with
h
time for a typical case are shown in the figure.
( ) becomes xp((t)) after some time ((τ
It can be seen that xh((t)) dies out and x(t)
in the figure).
The part of the motion that dies out due to damping (the free vibration
part) is called transient.
transient The particular solution represents the steady
state vibration and is present as long as the forcing function is present.
Forced‐Excited System: Harmonic
Excitation
•
The
h verticall motions off a mass‐spring system subjected
b
d to an externall
force r(t) can be expressed as:
mx′′ + cx′ + kx = r ((t )
•
•
•
Mechanically, this means that at each time instant t, the resultant of the
internal forces is in equilibrium with r(t). The resulting motion is called a
forced motion with forcing function r(t)
r(t), which is also known as the input
force or the driving force, and the solution x(t) to be obtained is called the
output or the response of the system to the driving force.
Of special interest are periodic external forces, and we shall consider a
driving force of the form:
r (t ) = Fo cos ωt
Then we have the nonhomogeneous ODE:
m x ′′ + c x ′ + kx = Fo cos ω t
Forced‐Excited System: Harmonic
Excitation
Solving the nonhomogeneous ODE
To find yp, we use the method of undetermined coefficients:
x p (t ) = a cos ω t + b sin
i ωt
x ′p (t ) = −ω a sin ω t + ω b cos ω t
x ′p′ (t ) = −ω 2 a cos ω t − ω 2 b sin ω t
Substituting the above equations into
m x ′′ + c x ′ + kx = Fo cos ω t
And collecting the cosine and the sine terms, we get
[( k − m ω 2 ) a + ω cbb ] cos ω t +[ −ω ca + ( k − m ω 2 )b ] sin ω t = Fo cos ω t
The cosine terms on both sides must be equal, and the coefficient of the
sine term on the left must be zero since there is no sine term on the right.
Forced‐Excited System: Harmonic
Excitation
•
This
h gives the
h two equations:
(k − mω 2 )a + ωcb = Fo
− ωca + (k − mω 2 )b = 0
for determining the unknown coefficients a and b.
b This is a linear system.
system
We can solve it by elimination to find:
k
= ωo
m
, we obtain:
k − mω 2
a = Fo
(k − mω 2 ) 2 + ω 2 c 2
ωc
b = Fo
(k − mω 2 ) 2 + ω 2 c 2
m(ωo2 − ω 2 )
a = Fo 2 2
m (ωo − ω 2 ) 2 + ω 2 c 2
ωc
b = Fo 2 2
m (ωo − ω 2 ) 2 + ω 2 c 2
•
If we set
•
We thus obtain the general solution of the nonhomogeneous ODE in the
form:
x(t ) = xh (t ) + x p (t )
Forced‐Excited System: Harmonic
Excitation
Case I: Undamped
d
d forced
f
d oscillations:
ll
• If the damping of the physical system is so small that its effect can be
neglected
g
over the time interval considered,, we can set c=0. Then
a = Fo
m(ωo2 − ω 2 )
m 2 (ωo2 − ω 2 ) 2 + ω 2 c 2
b = Fo
ωc
m (ω − ω 2 ) 2 + ω 2 c 2
reduces to
2
a=
2
o
Fo
m(ωo2 − ω 2 )
b=0
Hence
becomes
x p (t ) = a cos ω t + b sin ω t
x p (t ) =
Fo
Fo
=
cos
cos ωt
ω
t
ω 2
m(ωo2 − ω 2 )
k[1 − ( ) ]
ωo
Forced‐Excited System: Harmonic
Excitation
•
We thus
h have
h
the
h generall solution
l
off the
h undamped
d
d system as:
x(t ) = C cos(ωo t − δ ) +
•
Fo
cos ωt
m(ωo2 − ω 2 )
We see that this output is a superposition of two harmonic oscillations of
the natural frequency ωo/2π [cycles/sec] of the system, which is the
f
frequency
off the
h undamped
d
d motion
i and
d the
h ffrequency ω/2
/ π [cycles/sec]
[ l / ]
of the driving force.
Harmonic motion
Beats:
• As mentioned before, if the frequency of the forcing function and the
frequency of the system are very close to each other, then again beating effect
should be expected.
•
If we for example take the particular soluton:
x(t ) =
Fo
(cos ωt − cos ωot )
2
2
m(ωo − ω )
(ω ≠ ωo )
which can be rewritten as :
x(t ) =
•
2 Fo
⎛ ωo + ω ⎞ ⎛ ωo − ω ⎞
t ⎟ sin ⎜
t⎟
sin
⎜
m(ωo2 − ω 2 ) ⎝ 2
2
⎠ ⎝
⎠
Since ω is close to ωo, the difference ωo‐ ω is small. Hence the period of the last
sine function is large. This is because the greater the quantity under the sine,
the smaller the period is.
Harmonic motion
Beats:
Tb =
2TTo
4π
=
ωo − ω T − To
4π
ω + ωo
x(t ) =
Fo
⎛ ωo + ω ⎞ ⎛ ωo − ω ⎞
sin
t ⎟ sin ⎜
t⎟
⎜
2
2
m(ωo − ω ) ⎝ 2
⎠ ⎝ 2
⎠
Harmonic motion
Beats:
This phenomenon is also frequently observed in lightly damped sytems with close coupling of the torsional
and translational frequencies.
A typical example is the thirteen storey steel framed Santa Clara County Office Building as reported in :
Celebi and Liu, ‘Before and after retrofit‐ response of a building during ambient and strong motions’,
Journal of Wind Engineering and Industrial Aerodynamics, 77&78 (1998) 259‐268.
The proximity of the torsional frequency at 0.57 Hz to the translational frequency at 0.45 Hz causes the
observed coupling and beating effect in this structure.
Harmonic motion
Comparison of the frequencies of the building
Thirteen storey steel framed Santa Clara County Office Building as reported in :
Celebi and Liu, ‘Before and after retrofit‐ response of a building during ambient and strong motions’,
Journal of Wind Engineering and Industrial Aerodynamics
Aerodynamics, 77&78 (1998) 259
259‐268.
268
Resonance
•
If the
th damping
d
i off th
the system
t
is
i so smallll that
th t its
it effect
ff t can be
b neglected
l t d
over the time interval considered, we can set c=0. Then the particular
response can be expressed by:
xp =
•
Fo
Fo
cos ωt =
cos ωt
2
m(ω o − ω )
⎡ ⎛ ω ⎞2 ⎤
k ⎢1 − ⎜⎜ ⎟⎟ ⎥
⎢⎣ ⎝ ωo ⎠ ⎥⎦
2
Puttingg cosωt=1,, we see that the maximum amplitude
p
of the p
particular
solution is:
F
ao =
o
k
ρ
where ρ =
If
•
1
2
⎛ω ⎞
1 - ⎜⎜ ⎟⎟
⎝ ωo ⎠
ω → ωo , then ρ and ao tend to infinity.
This excitation of large oscillations by matching input and natural
frequencies is called resonance.
Resonance
F
In the
h case off resonance and
d no d
damping, the
h ODE &x& + ωo2 x = o cos ωot
m
becomes: m&x& + cx& + kx = Fo cos ωt
• Then from the modification rule,
rule the particular solution becomes:
•
x p (t ) = t (a cos ωot + b sin ωot )
•
By substituting this into the second equation, we find:
x p (t ) =
•
Fo
t sin ωot
2mωo
We see that because of the factor t, the amplitude becomes larger and
larger Practically speaking,
larger.
speaking systems with very little damping may undergo
large vibrations that can destroy the system.
Forced‐Excited System: Harmonic
Excitation
Case II: Damped
d forced
f
d oscillations:
ll
Iff the
h damping
d
off the
h mass‐spring
system is not negligibly small, we have c>0 and a damping term cx’ in
m&x& + cx& + kx = r ((t )
m&x& + cx& + kx = 0
Then the general solution yh of the homogeneous ODE approaches zero as
t goes to infinity. Practically, it is zero after a sufficiently long time. Hence
the transient solution given by
y (t ) = yh (t ) + y p (t )
approaches
pp
the steadyy state solution yp. This p
proves the following:
g
Steady state solution: After a sufficiently long time, the output of a
d
damped
d vibrating
ib i system under
d a purely
l sinusoidal
i
id l driving
d i i force
f
will
ill
practically be a harmonic oscillation whose frequency is that of the input.
Response of a damped system under
harmonic force
•
Iff the
h forcing
f
function
f
is given by
b
becomes:
the
h equation off motion
•
The particular solution is also expected to be harmonic; we assume it in
the following form:
where X and φ are constants to be determined that denote the amplitude
phase angle
g of the response,
p
, respectively.
p
y Byy substitutingg the
and the p
second equation into the first:
•
U i the
Using
h trigonometric
i
i relations
l i
b
below
l iin the
h above
b
equation
i
Response of a damped system under
harmonic force
•
We obtain:
b
•
If we solve the above equation
equation, we find:
•
If we insert the above into the
particular solution. Using:
, we find the
r=
ω
ωn
Response of a damped system under
harmonic force
•
We obtain:
b
Forced‐Excited System: Harmonic
Excitation
•
The
h quantity M=X/δ
/δst is known
k
as the
h magnification
f
factor,
f
amplification
lf
factor or the amplitude ratio. The amplitude of the forced vibration
becomes smaller with increasing values of the forcing frequency (that is,
M
0 as r
∞)
Xo
1
=
Fo / k
(1 − r 2 ) 2 + (2ξr ) 2
Forced‐Excited System: Harmonic
Excitation
•
A fact
f that
h creates d
difficulty
ff l ffor d
designers is that
h the
h response can b
become
large when r is close to 1 or when ω is close to ωn. This condition is called
resonance. The reduction in M in the presence of damping is very
significant at or near resonance.
Forced‐Excited System: Harmonic
Excitation
•
For an undamped
d
d system (ζ
(ζ=0),
) the
h phase
h
angle
l is zero for
f 0<r<1 and
d 180
degrees for r>1. This implies that the excitation and response are in phase
for 0<r<1 and out of phase for r>1 when ζ=0.
⎛ 2ζr ⎞
2 ⎟
⎝1− r ⎠
φ = arctan⎜
Forced‐Excited System: Harmonic
Excitation
•
For ζ>0
ζ and
d 0<r<1, the
h phase
h
angle
l is given by
b 0<φ<90,
φ
implying
l
that
h the
h
response lags the excitation.
⎛ 2ζr ⎞
2 ⎟
⎝1− r ⎠
φ = arctan⎜
Forced‐Excited System: Harmonic
Excitation
•
For ζ>0
ζ and
d r>1, the
h phase
h
angle
l is given by
b 90<φ<180,
φ
implying
l
that
h the
h
response leads the excitation.
⎛ 2ζr ⎞
2 ⎟
⎝1− r ⎠
φ = arctan⎜
Response
p
of a damped
p system
y
under
F (t ) = Fo e iωt
•
Using complex
l algebra,
l b let
l the
h harmonic
h
force
f
be:
b
F (t ) = Fo eiωt
where Fo is a real constant and i is the imaginary unit.
unit Assume that the
response has the same frequency as the force, but is, in general, out of
phase with the force
~
x(t ) = X o e i (ωt +φ ) = Xe iωt
~
where Xo is the amplitude of the displacement and X is the complex
displacement,
p
,
~
iφ
X = X oe
Substituting this into the differential equation of motion
~
(−mω 2 + icω + k ) Xeiωt = Fo eiωt
Forced‐Excited System: Harmonic
Excitation
•
Define
f the
h transfer
f function
f
((or ffrequency response function)
f
) H(ω)
( ) as the
h
complex displacement due to a force of unit magnitude (Fo=1). Thus,
H (ω ) =
•
1
(k − mω 2 ) + icω
Rationalizing, the transfer function becomes:
(k − mω 2 ) − icω
H (ω ) =
(k − mω 2 ) 2 + (cω ) 2
•
This is also the ratio between the complex
p displacement
p
response
p
and the
complex input forcing function.
Forced‐Excited System: Harmonic
Excitation
•
Define
f the
h gain function
f
as the
h modulus
d l off the
h transfer
f function
f
H (ω ) = H (ω ) H * (ω ) = (Re H ) 2 + (Im H ) 2
where H* is the complex conjugate. For the force excited system under
consideration,
H (ω ) =
•
1
(k − mω 2 ) 2 + (cω ) 2
The gain function is the amplitude of the displacement for Fo=1. Thus,
Xo
= H (ω )
Fo
Forced‐Excited System: Harmonic
Excitation
•
It is convenient to develop
d l a nondimensional
d
l form
f
off the
h gain function.
f
First define the frequency ratio
ω
r=
ωn
Multiplying the equation
H (ω ) =
1
(k − mω 2 ) 2 + (cω ) 2
by k and employing the definitions for ζ, cc and ωn, it is easily shown that
Xo
1
=
Fo / k
(1 − r 2 ) 2 + (2ξr ) 2
and the p
phase angle
g is:
⎛ 2ζr ⎞
2 ⎟
⎝1− r ⎠
φ = arctan⎜
Forced‐Excited System: Harmonic
Excitation
•
TTransfer
f ffunctions
ti
expressing
i the
th velocity
l it and
d acceleration
l ti responses can
be written based on equation
x(t ) = X o e
by multiplying H(ω) in equation
H (ω ) =
i (ωt +φ )
~ iω t
= Xe
1
(k − mω 2 ) 2 + (cω ) 2
by iω and (iω)^2=‐ω^2 respectively. The resulting gain function for the
velocity output would be derived by multiplying both sides of the above
equation and the equation
Xo
1
=
Fo / k
(1 − r 2 ) 2 + (2ξr ) 2
by ω and noting that ωXo is the amplitude of the velocity. Similarly, the
gain function for the acceleration can be obtained by multiplying both
sides by ω^2.
Quality factor and bandwidth
•
For smallll values
l
off d
damping we can take:
k
⎛ X
⎜⎜
⎝ δ st
⎞
⎛ X
⎟⎟
≅ ⎜⎜
⎠ max ⎝ δ st
⎞
1
⎟⎟
=
=Q
2
ζ
⎠ ω =ω n
where
X denotes the amplitude of the response and
F
δ st = o = deflection under the static force Fo
k
•
The value
Th
l off the
h amplitude
li d ratio
i at resonance is
i called
ll d the
h Q factor
f
or the
h
quality factor of the system. The points R1 and R2, where the amplification
factor falls to Q/√2, are called half power points because the power
absorbed (ΔW) by the damper (or by the resistor in an electrical circuit),
responding harmonically at a given frequency, is proportional to the
q
of the amplitude.
p
square
Quality factor and bandwidth
•
The
h d
differences
ff
b
between the
h frequencies
f
associated with the half power points R1
and R2 is called the bandwidth of the system.
system
•
To find the values of R1 and R2, we set
so that
or
•
From which, we can get:
Quality factor and bandwidth
•
For smallll values
l
off ζ,
ζ the
h roots
can be simplified as:
Then
Quality factor and bandwidth
•
Using the
h relation
l
in the
h equation
we find that the bandwidth Δω is given by:
Combining the above equation and the equation
⎛ X
⎜⎜
⎝ δ st
⎞
⎛ X
⎟⎟
≅ ⎜⎜
⎠ max ⎝ δ st
⎞
1
⎟⎟
=
=Q
2
ζ
⎠ ω =ω n
W obtain:
We
b i
•
It can be seen that the quality factor Q can be used for estimating the
equivalent viscous damping in a mechanical system.
Electrical systems
•
An electronic circuit is a closed path formed by the interconnection of
electronic components through which an electric current can flow.
•
We have just seen that linear ODEs have important applications in
mechanics. Similarly, they are models of electric circuits as they occur as
portions of large networks in computers and elsewhere.
•
The circuits we shall consider here are basic building blocks of such
networks.
•
They contain three kinds of components, namely, resistors, inductors and
capacitors.
•
Kirchhoff’s Voltage Law (KVL): The voltage (the electromotive force)
impressed on a closed loop is equal to the sum of the voltage drops across
p
the other elements of the loop.
Electrical systems
•
Figure shows
Fi
h
such
h a RLC circuit.
i i IIn iit a resistor
i
off resistance
i
R Ω (ohms), an inductor of inductance L H (Henrys) and a
capacitor of capacitance C F (farads) are wired in series as
shown,, and connected to an electromotive force E(t)
()
V(volts) (a generator for instance), sinusoidal as shown in
the figure or some other kind.
•
R, L, C, and E are given and we want to find the current I(t)
A(Amperes) in the circuit.
•
An ODE for the current I(t) in the RLC circuit in the figure is
obtained from the Kirchhoff’s Voltage Law.
•
In the figure, the circuit is a closed loop and the impressed
voltage E(t) equals the sum of the voltage drops across the
three elements R,L,C, of the loop.
Electrical systems
•
Voltage
l
drops:
d
Experiments show
h that
h the
h current I fl
flowing through
h
ha
resistor, inductor and capacitor causes a voltage drop (voltage difference,
measured in volts) at the two ends. These drops are:
•
•
•
RI (Ohm’s law) Voltage drop for a resistor of resistance R ohms Ω
dI
LI ′ = L
l
d
drop ffor an iinductor
d
off iinductance
d
Lh
henrys ((H))
dt Voltage
Q Voltage drop for a capacitor of capacitance C farads (F)
•
Here Q coulombs is the charge on the capacitor, related to the current by
C
I (t ) =
dQ
equivalently Q(t) = ∫ I(t)dt
dt
Electrical systems
•
Table
bl Elements
l
in an RLC circuit
Name
•
Symbol
Notation
Unit
Voltage
Drop
Ohm’s
resistor
Ohm’s
resistance, R
Ohms
(Ω)
RI
Inductor
Inductance, L
henrys
(H)
LdI/dt
Capacitor
p
Capacitance,
p
C farads
(F)
Q/C
According to Kirchhoff’ voltage law we thus have an RLC circuit with
electromotive force E(t)=E
( ) o sinωt ((Eo constant)) as a model for the ‘integro
g
differential equation’.
1
LI ′ + RI +
C∫
Idt = E (t ) = Eo sin ωt
Electrical systems
•
T gett rid
To
id off th
the iintegral
t
l iin
LI ′ + RI +
•
1
Idt = E (t ) = Eo sin ωt
C∫
We differentiate the above equation with repect to t, obtaining:
LI ′′ + RI ′ +
•
•
1
I = E ′(t ) = Eoω cos ωt
C
This shows that the current in an RLC circuit is obtained as the solution of
this nonhomogeneous second‐order ODE with constant coefficients.
1
Using LI ′ + RI + ∫ Idt = E (t ) = Eo sin ωt and noting that I=Q’ and I’=Q’’, we
C
have directly:
1
LQ′′ + RQ′ +
•
C
Q = Eo sin ωt
But in most practical problems, the current I(t) is more important than the
charge Q(t) and for this reason, we shall concentrate on the below
q
rather than the above.
equation
LI ′′ + RI ′ +
1
I = E ′(t ) = Eoω cos ωt
C
Solving ODE for the current
•
A general solution of
LI ′′ + RI ′ +
1
I = E ′(t ) = Eoω cos ωt
C
is the sum II=IIh+Ip , where Ih is a general solution of the homogeneous ODE
corresponding to the above equation and Ip is a particular solution. We first
determine Ip by the method of undetermined coefficients. We substitute:
I p = a cos ωt + b sin ωt
I ′p = ω (−a sin ωt + b cos ωt )
I ′p′ = ω 2 (−a cos ωt − b sin ωt )
into the first equation. Then we collect the cosine terms and equate them to
Eoωcosωt on the right, and we equate the sine terms into zero because there
i no sine
is
i term on the
h right.
i h
a
= Eoω
C
b
Lω 2 (−b) + Rω (− a) + = 0
C
Lω 2 (− a) + Rωb +
(Cosine terms)
(Sine
Si terms
t
)
Solving ODE for the current
•
TTo solve
l this
thi system
t
for
f a and
db
b, we fi
firstt iintroduce
t d
a combination
bi ti off L and
dC
C,
called the reactance:
1
S = ωL −
•
ωC
Dividing the previous two equations by ω, ordering them and substituting S
gives:
− Sa + Rb = E
o
− Ra − Sb = 0
•
We now eliminate b by multiplying the first equation by S and the second by R,
and adding. Then we eliminate a by multiplying the first equation by R and
second by –S, and adding. This gives:
a=
•
− Eo S
R2 + S 2
b=
Eo R
R2 + S 2
I p = a cos ωt + b sin ωt
I ′p = ω (−a sin ωt + b cos ωt )
I ′p′ = ω 2 (−a cos ωt − b sin ωt )
Equation for Ip with coefficients a and b as given above is the desired
particular solution of the nonhomogeneous ODE governing the current I in an
RLC circuit with sinusoidal electromotive force.
Solving ODE for the current
•
Using a =
− Eo S
R2 + S 2
b=
Eo R
R2 + S 2
we can write Ip in terms off physically
h
ll
visible quantities,
quantities namely,
namely amplitude Io and phase lag θ of the current
behind the electromotive force, that is,
I p (t ) = I o sin(ωt − θ )
where
Io =
a2 + b2 =
tan θ = −
Eo
R2 + S2
a
S
=
b
R
The q
quantity
y
R 2 + S 2 is called the impedance.
p
Our formula shows that the impedance
p
equals
q
the ratio
This is somewhat analogous to E/I = R (Ohm' s law)
Eo
.
Io
Solving ODE for the current
•
A generall solution
l
off the
h h
homogeneous equation corresponding
d to
LI ′′ + RI ′ +
is:
1
I = E ′(t ) = Eoω cos ωt
C
I h = c1e λ1t + c2 e λ2t
where λ1 and λ2 are the roots of the characteristic equation:
λ2 +
•
R
1
λ+
=0
L
LC
We can write the roots in the form λ1=‐α+β
α+β and λ2=‐α‐β,
α β, where
R
R2
1
1
4L
2
α=
, β=
−
=
−
R
2L
4 L2 LC 2 L
C
•
Now in an actual circuit, R is never zero (hence R>0). From this, it follows that
pp
zero,, theoreticallyy as t→∞,, but practically
p
y after a short time.
Ih approaches
Solving ODE for the current
•
Hence the
h transient current I=Ih+Ip tends
d to the
h steady
d state current Ip and
d
after some time the output will practically be a harmonic oscillation,
which is given by:
I p (t ) = I o sin(ωt − θ )
and whose frequency is that of the input (of the electromotive force)
Analogy of electrical and mechanical
quantities
•
EEntirely
ti l diff
differentt physical
h i l or other
th systems
t
may have
h
the
th same
mathematical model. For instance, the ODE of a mechanical system and
the ODE of an electric RLC circuit can be expressed by:
LQ′′ + RQ′ +
•
•
•
1
Q = Eo sin ωt
C
m y ′′ + c y ′ + ky = Fo cos ω t
The inductance L corresponds
p
to the mass,, and indeed an inductor
opposes a change in current, having an inertia effect similar to that of a
mass.
The resistance R corresponds to the damping constant c and a resistor
causes loss of energy, just as a damping dashpot does.
This analogy is strictly quantitative in the sense that to a given mechanical
system we can construct an electrical circuit whose current will give the
exact values of the displacement in the mechanical system when suitable
scale factors are used.
Analogy of electrical and mechanical
quantities
•
•
The
h practical
i l importance
i
off this
hi analogy
l
is
i almost
l
obvious.
b i
The
h analogy
l
may be
b
used for constructing an ‘electrical model’ of a given mechanical model, resulting
in substantial savings of time and money because electric circuits are easy to
assmeble,
bl and
d electric
l t i quantities
titi can be
b measured
d much
h more quickly
i kl and
d
accurately than mechanical ones.
1
m y ′′ + c y ′ + ky = Fo cos ω t
LQ′′ + RQ′ + Q = Eo sin ωt
C
Table: Analogy of electrical and mechanical quantities
Electrical system
Mechanical system
Inductance, L
Mass m
Resistance, R
Damping c
Reciprocal of capacitance
capacitance, 1/C
Spring modulus k
Electromotive force Eo sinωt
Driving force Focosωt
Current,, I(t)=dq/dt
( ) q/
Velocityy , v(t)=dy/dt
( ) y/
Charge, Q(t)
Displacement, y(t)
Base excited systems: absolute motion
•
Consider
d the
h base‐excited
b
d system
of the figure. The goal of the analysis
will be to determine the absolute
response x(t) (typically acceleration or
displacement of the mass) given the base
motion y(t).
•
FFrom the
h free‐body
f
b d diagram,
di
application
li i
of Newton’s second law leads directly to
the differential equation:
m&x& + cx& + kx = ky + cy&
Base excited systems: absolute motion
•
Assume that
h the
h base
b
motion is harmonic,
h
y (t ) = Yo e iωt
•
And assume that the response will be harmonic,
harmonic
~
x(t ) = Xe iωt
~
where X is the complex response. The transfer function and the gain
function are derived in the same manner as for the force excited system.
The transfer function is:
k + icω
H (ω ) =
(k − mω 2 ) + icω
•
The gain function is
Xo
k 2 + (cω ) 2
1 + (2ζr ) 2
= H (ω ) =
=
2
2 2
2
Yo
(k − mω ) + (cω )
(1 − r 2 ) 2 + (2ζr )
Base excited systems: absolute motion
•
•
Xo
1 + (2ξr ) 2
In nondimensional
d
l form
f
=
Yo
(1 − r 2 ) 2 + (2ξr ) 2
The gain function for the absolute displacement for the base‐excited
base excited
system is shown in the figure.
Base excited systems: absolute motion
•
•
•
•
The
h value
l off Td is
i unity
i at r=0
0 and
d close
l
to unity
i for
f smallll values
l
off r.
For an undamped system ζ=0, Td
∞ at resonance (r=1).
The value of Td is less than unity (Td <1) for values of r >√2 (for any amount of
damping ζ)
The value of Td is equal to unity (Td=1) for all values of ζ at r=√2
Xo
1 + (2ξr ) 2
=
Yo
(1 − r 2 ) 2 + (2ξr ) 2
Base excited systems: absolute motion
•
The
h equations
Xo
1 + (2ξr ) 2
=
Yo
(1 − r 2 ) 2 + (2ξr ) 2
Xo
k 2 + (cω ) 2
= H (ω ) =
Yo
(k − mω 2 ) 2 + (cω ) 2
can be interpreted as the gain functions for acceleration output given
acceleration input
input. And again note that the transfer function for velocity
and acceleration responses can be derived by multiplying the equation
k + icω
H (ω ) =
(k − mω 2 ) + icω
by iω and ‐ω^2, respectively.
Base excited system: Relative motion
Base‐excited
•
Consider
C
id th
the free
f
body
b d diagram
di
in
i the
th
figure. Now the response variable under
consideration will be the relative
displacement,
displacement
z (t ) = x(t ) − y (t )
•
In the model, the spring represents a
structural element. The stress in that
element will be proportional to z. Thus,
this problem would be relevant to
designers of structures subjected to base
motions, for example, earthquakes.
•
Letting z=x-y in the equation of motion
eads d
directly
ec y to
o m&z& + z& + kz = − m&y&
leads
Base excited system: Relative motion
Base‐excited
•
Assuming that
h the
h base
b
motion is harmonic,
h
y (t ) = Yo e iωt
•
And assuming the response is also harmonic,
~
z (t ) = Z eiωt
•
Following the procedure as described above, the transfer function is:
mω 2
H (ω ) =
(k − mω 2 ) + icω
• And the gain function is
Zo
mω 2
= H (ω ) =
Yo
(k − mω 2 ) 2 + (cω ) 2
Base excited system: Relative motion
Base‐excited
•
•
2
Z
r
o
In nondimensionless
d
l
fform,
=
Yo
(1 − r 2 ) 2 + (2ξr ) 2
The gain function for the relative motion for the base
base‐excited
excited system is
shown in the figure:
Base excited system: Relative motion
Base‐excited
•
Again note that
h the
h transfer
f function
f
for
f relative
l
velocity
l
and
d acceleration
l
responses can be derived by multiplying
mω 2
H (ω ) =
(k − mω 2 ) + icω
by iω and ‐ω^2, respectively.
• The gain functions for velocity and acceleration responses can be obtained
by multiplying both sides of the equations
Zo
mω 2
Zo
r2
= H (ω ) =
=
2
2
2
Yo
Yo
(k − mω ) + (cω )
(1 − r 2 ) 2 + (2ξr ) 2
by ω and ω^2, respectively.
Gain functions for force excited
systems
Gain functions for base‐excited
systems
Example
•
•
A fi
fixed
db
bottom
tt
offshore
ff h
structure
t t
is
i subjected
bj t d to
t oscillatory
ill t
storm
t
waves.
In a first approximation, it is estimated that the waves produce a harmonic
force F(t) having amplitude F=122 kN. The period of these waves is τ=8
sec The structure is modeled as having a lumped mass of 110 tons
sec.
concentrated in the deck. The weight of the structure itself is assumed to
be negligible. The natural period of the structure was measured as being
τn= 4
4.0
0 sec
sec. It is assumed that the damping factor is ζ=5%.
ζ=5% It is required to
determine the steady state amplitude of the response of the structure.
Solution: As modeled, this will be a force‐excited system, and the
response can be computed from the gain function of
Xo
1
=
Fo / k
(1 − r 2 ) 2 + (2ξr ) 2
The problem reduces to one of finding the frequency ratio r and the
stiffness k.
Example
•
Because r is the
h ratio off the
h forcing
f
frequency
f
to the
h naturall frequency,
f
it
follows that r will also be the ratio of the natural period to the forcing
period. Thus,
τ
4
r=
•
1
τn
= 0.25 Hz.
Then noting that the expression for the natural frequency is
fn =
•
= 0 .5
τ 8
To compute k, first note that the natural frequency is
fn =
•
=
n
We compute k=27667 N/m
1
2π
k
W /g
Example
•
Finally
ll substituting
b
into
Xo
1
=
Fo / k
(1 − r 2 ) 2 + (2ξr ) 2
•
Xo=5.87 m.
Resonance
• For resonance please download the Millenium
g and Tacoma Bridge
g videos in the
bridge
website of the course.
Background for response of
SDOF system
y
to random forces
•Response of SDOF system to impulsive forces
•Response of single degree of freedom system to arbitrary loading
•Relationship between the impulse response and the transfer
function
•Relationship
p between the Fourier transform of displacement
p
and
force
Response under a nonperiodic force
•
We will see that periodic forces of any general wave form can be represented by Fourier series as a
superposition of harmonic components of various frequencies.
•
The response of a linear sytem is then found by superposing the harmonic response to each of the
exciting forces
forces.
•
When the exciting force F(t) is nonperiodic, such as that due to blast from an explosion, a different
method of calculating the response is required.
•
Various methods can be used to find the response of the system to an arbitrary excitation as
follows:
–
Representing the excitation by a Fourier integral
–
Using the method of convolution integral
–
Using the method of Laplace transforms
–
First approximating F(t) by a suitable interpolation model and then using a numerical procedure
–
Numerically integrating the equations of motion.
Response of single degree of freedom
system to impulsive forces
•
A nonperiodic exciting force usually has a magnitude that varies with time
time;
it acts for a specified period of time and then stops.
•
The simplest form is the impulsive force‐
force a force that has a large
magnitude F and acts for a very short period of time Δt.
•
From dynamics
dynamics, we know that impulse can be measured by finding the
change in momentum of the system.
•
The unit impulse F acting at t=0 is also denoted by the Dirac delta
function, δ(t). The Dirac delta function at time t=τ, denoted as δ (t − τ )
has the properties
∞
∞
0
0
∫ δ (t − τ )dt = 1
∫ δ (t − τ ) F (t )dt = F (τ )
where 0<τ<∞. Thus an impulsive force acting at t=τ can be denoted as:
F (t ) = Fδ (t − τ )
Response of single degree of freedom
system to impulsive forces
•
Consider
d a SDOF system subjected
b
d to impulsive
l
lloading
d as shown
h
in ffigure.
The external force is:
F (t ) = Foδ (t )
where δ(t) is the Dirac delta function.
Response of single degree of freedom
system to impulsive forces
•
The
h equation off motion off the
h mass willll b
be similar
l to
m&x& + cx& + kx = Fo sin ωt
with the impulsive force of
F (t ) = Foδ (t )
on the right hand side. The unit impulse is defined as Fo=1. The response
x(t) to the unit impulse is denoted as h(t):
mh&& + ch& + kh = (1)δ (t )
•
Physically speaking
speaking, for t≈0,
t≈0 a radical change in the system motion takes
place when the short duration high amplitude force excites an initial
motion in the system. But for t>0, the response will be free vibration.
U i elementary
Using
l
mechanics,
h i F(Δt)=m(Δv)
F(Δ ) (Δ ) iit can b
be shown
h
that
h the
h
velocity of the system just after the impulse is:
1
h&(0 + ) =
m
Response of single degree of freedom
system to impulsive forces
•
U i th
Using
the iinitial
iti l conditions:
diti
1
h&(0 + ) =
m
h(0) = 0
•
⎞
x&o + ζω n xo
−ζω n t ⎛
2
2
⎜
⎟
=
−
+
−
x
(
t
)
e
x
cos
1
ζ
ω
t
sin
1
ζ
ω
t
In the equation:
n
n ⎟
⎜ o
ωd
⎝
⎠
•
The free vibration response is:
h(t ) =
•
1 −ξωnt
e
sin ωd t
mωd
t >0
Here h(t) is known as the force‐excited absolute displacement response,
impulse response function of the single‐degree‐of‐freedom system. Note that
h(t) characterizes a system just like the transfer function H(ω) does.
does The
velocity and acceleration impulse response functions can also be obtained as
derivatives of h(t).
Response of single degree of freedom
system to impulsive forces
•
If the magnitude of the impulse is F
instead of unity, the initial velocity x&o is
F/m and the response of the system
becomes:
x(t ) =
•
F −ξωnt
e
sin ωd t = Fg (t )
mωd
If the impulse F is applied at an arbitrary
time t=τ by an amount F/m as shown in
the figure, it will change the velocity at
t=τ by an amount F/m. Assuming that x=0
until the impulse is applied, the
displacement h at any subsequent time t,
caused by a change in the velocity at time
τ is given by the above equation with t
replaced by the time elapsed after the
application of the impulse, that is, t‐ τ. As
shown in Fig.b, we obtain
x(t ) = Fg (t − τ )
Response of Single‐Degree‐of‐
Freedom System to Arbitrary Loading
•
For a linear
l
system, the
h impulse
l response ffunction can be
b used
d to derive
d
the response of a system under an arbitrary loading history. Consider the
force shown in the figure
Response of Single‐Degree‐of‐
Freedom System to Arbitrary Loading
•
The
h iimpulse
l d
during
i Δτ
Δ is
i F(τ)
( ) Δτ.
Δ The
h
response to this impulse at any time
t>τ is approximately [F(τ) Δτ]h(t‐ τ).
Th the
Then
th response att t is
i the
th sum off
the responses due to a sequence of
impulses. In the limit as Δτ→0
t
x(t ) =
∫ F (τ )h(t − τ )dτ
−∞
where the input F(t) is accounted for
as t→‐∞, for example,F(t) could be
defined as zero for t<0. The expression
for x(t) is called the convolution
integral.
Response of Single‐Degree‐of‐
Freedom System to Arbitrary Loading
•
Note that h(t‐τ) =0 when τ>t. Thus,
we can expand the limits to the
interval (‐∞,∞):
∞
x(t ) =
∫ F (τ )h(t − τ )dτ
−∞
•
Another useful form is obtained by
letting θ=t‐τ:
t
x(t ) =
∫ F (t − θ )h(θ )dθ
−∞
Response of Single‐Degree‐of‐
Freedom System to Arbitrary Loading
•
B substituting
By
b tit ti th
the equation
ti
h(t ) =
into
1 −ξωnt
e
sin ωd t
mωd
t >0
∞
x(t ) =
∫ F (τ )h(t − τ )dτ
−∞
we obtain:
1
x(t ) =
mωd
t
∫ F (τ )e
−ζω n ( t −τ )
sin ωd (t − τ )dτ
0
which represents the response of an underdamped single degree of
freedom system to the arbitrary excitation F(t).
• Note that the above equation does not consider the effect of initial
conditions of the system.
• The integral in either of the two above equations is called the convolution
or Duhamel integral.
Relationship between h(t) and H(ω)
•
An important result
l ffrom Fourier transform
f
theory
h
is that
h h(t)
h( ) and
d H(ω)
( )
form a Fourier transform pair. This relationship is useful when deriving
responses of dynamic systems to random vibration inputs. Let,
F (t ) = eiωt
•
x(t ) = H (ω )e iωt
Then,
h
∞
= ∫ h(t − τ )e iωτ dτ
−∞
∞
= ∫ h(θ )e iω (t −θ ) dθ
−∞
=e
iωt
∞
∫ h(θ )e
−iωθ
dθ
−∞
which implies that h(t) and H(ω) form a Fourier transform pair.
∞
H (ω ) = ∫ h(θ )e
−∞
−iωθ
dθ
→
1
h(t ) =
2π
∞
∫ H (ω )e
−∞
iωt
dω
Relationship between h(t) and H(ω)
•
Based
B
d on th
the FFourier
i transform
t
f
off a convolution,
l ti
the
th expression
i for
f the
th
response to an arbitrary input in equation
∞
x(t ) =
and representation in
∫ F (τ )h(t − τ )dτ
−∞
∞
H (ω ) = ∫ h(θ )e −iωθ dθ
→
−∞
1
h(t ) =
2π
∞
∫ H (ω )e
i ωt
dω
−∞
it is clear that we can also express the response to an arbitrary input as:
1
x(t ) =
2π
∞
∫ F (ω ) H (ω )e
iωt
dω
−∞
where F(ω) is the Fourier transform of F(t). This expression is useful for the
analysis or numerical computation of system response or as the basis for
random vibration computations.
Relationship between X(ω) and F (ω)
•
The
h relationship
l
h b
between the
h Fourier transforms
f
off x(t)
( ) and
d F(t)
( ) is used
d to
derive responses of dynamic systems to random vibration input in random
vibration theory. Take the Fourier transform of both sides of
t
x(t ) =
∫ F (t − θ )h(θ )dθ
−∞
to find
fi d
⎡∞
⎤ − i ωt
F
t
θ
h
θ
d
θ
(
−
)
(
)
⎥ e dt
∫−∞⎢⎣−∫∞
⎦
Let τ = t − θ , dt = dτ
1
X (ω ) =
2π
∞
⎡∞
⎤ −iω (τ +θ )
F
(
τ
)
h
(
θ
)
d
θ
dτ
⎥e
∫−∞⎣⎢−∫∞
⎦
Rearranging :
1
X (ω ) =
2π
∞
∞
X(ω ) = ∫ h(θ )e
-∞
-iωθ
1
dθ
2π
∞
∫ F (τ )e
−∞
−iωτ
dτ
Relationship between X(ω) and F (ω)
•
From
∞
H (ω ) = ∫ h(θ )e −iωθ dθ
−∞
and the basic relationship of the Fourier transform:
1
u (τ ) =
2π
+∞
−iωt
(
)
g
t
e
dt
∫
−∞
it follows that
X(ω ) = H(ω )F(ω )
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