1 Single degree of freedom (SDOF) systems
James MW Brownjohn, University of Plymouth, January 2005
Contents
1.1 SDOF systems and their rationale
1.2 Motion of SDOF systems after initial disturbance
1.2.1 Undamped systems
1.2.2 Damped systems
1.3 Analysis of forced response of SDOF systems
1.3.1 Definition of a dynamic force
1.3.2 Types of dynamic force
1.3.3 Motion of SDOF systems under harmonic loading
1.3.3.1 Undamped system
1.3.3.2 Damped system
1.3.4 Visualising using rotating vectors
1-1
1.3.5 Periodic forcing
1.3.6 Rotating imblance
1.3.7 Impulse response function
1.3.8 Random forcing
1.4 Logarithmic decrement and damping
1.5 Base excitation
1.6 Transmissibility
References
1-2
1.1 SDOF systems
1.1.1 Definition and usage
Single-degree-of-freedom (SDOF) system is a system whose motion is
defined just by a single independent co-ordinate (or function) e.g. x which
is a function of time.
SDOF systems are often used as a very crude approximation for a generally
much more complex system.
However, behaviour of SDOF systems is probably the most important topic
to master in structural dynamics. This is because the behaviour of more
complex systems whose motion needs to be described by several
coordinates could be treated as if they are simply collections of several
SDOF systems hence understanding of SDOF systems is a prerequisite.
1-3
SDOF modelling of the vertically vibrating system shown in Figure 1 is
probably sufficient. However, sufficiently detailed description of the
motion of the horizontally vibrating system shown in Figure 2 would
require more than just one time-dependent function x(t).
Figure 1: SDOF physical system (after Smith, 1988).
1-4
Figure 2: SDOF mathematical modelling (after Smith, 1988).
1.1.2 Mathematical model of SDOF system
The aim of developing a SDOF mathematical model is to use it in order to
find the position x ( t ) of the moving mass m at any instant of time, also
x ( t ) . SDOF oscillators require a
often velocity x ( t ) and acceleration restoring force on the mass that increases as displacement (from a neutral
position) increases. Almost universally, a linear spring with constant k is
used giving restoring force –kx.
1-5
x (t )
displacement x
1
0
k
-1
m
Energy
1
PE
KE
PE+KE
0.5
0
0
0.1
0.2
0.3
0.4
0.5
0.6
time /seconds
0.7
0.8
0.9
1
Figure 3 Mathematical representation of undamped SDOF system and
energy balance
The system shown in Figure 3 is an idealised undamped SDOF system
which does not have a mechanism for removing energy from the vibrating
system. This energy consists of the potential energy stored in the spring:
(1)
1 2
U spring ( t ) = kx ( t ) ,
2
1-6
and kinetic energy stored in the moving mass:
(2)
1
T ( t ) = mx 2 ( t ) .
2
If such an idealised undamped SDOF system is set in motion by an initial
disturance, without any means of energy dissipating energy, the system
would continue to oscillate indefinitely. During this ideal process there is
an exchange between the potential and kinetic energies, as shown in Figure
3 but the total energy initially supplied to the system remains constant:
(3)
Etotal = U spring ( t ) + T ( t ) = const.
Damped systems have means of reducing the total vibration energy
supplied to the system. Our experience of vibrating systems is that, after
being set in motion to vibrate, the oscillations diminish and eventually
cease. Some form of damping is always present in real-life vibrating
systems. Essentially, damping has the ability to reduce to total vibration
energy within the system by some means of energy dissipation, such as
heating, radiation, etc.
1-7
1.2 Motion of SDOF systems after initial disturbances
Initial disturbances are applied at the beginning of vibration at time t = 0
and they are either initial displacement x (0) ≠ 0 or initial velocity x (0) ≠ 0
or both applied simultaneously at t = 0 .
1.2.1 Undamped systems
x (t )
mx( t )
x+ (t )
k
fk
m
m
Figure 4: Single spring-mass system with displacement x
1-8
Figure 5 shows an undamped SDOF system, featuring only mass and
stiffness where the positive direction of displacement x ( t ) , velocity x ( t )
x ( t ) and force are to the right.
and acceleration Note that we neglect friction, and gravity force is either irrelevant or
cancels from the analysis.
Via Newton’s third law, the acceleration resulting from spring force f is
given by
(4)
f = mx
The spring force in the positive direction is given by –kx hence
(5)
−kx ( t ) = mx( t )
or
− kx ( t ) − mx( t ) = 0
This can also be viewed (using d’Alambert’s principle) as equilibrium of
inertia and spring forces f I and fk, where f I ( t ) = −mx ( t ) is the inertia
1-9
force. The ‘-‘ sign indicates that it acts in the direction opposite of
acceleration, which is our experience of inertia. f k = −kx ( t ) is the spring
force (also acting in the direction opposite of displacement). Hence for
equilibrium of all forces acting on mass m:
(6)
fI (t ) + fs (t ) = 0
(7)
− mx ( t ) − kx ( t ) = 0
or
(8)
mx ( t ) + kx ( t ) = 0
This is an ordinary (time t is the only variable) second order (second
derivative is the highest derivative) linear (the only power is 1) differential
equation. Its solution can be assumed to be of a form:
(9)
x ( t ) = A sin(ω n t + φ )
This can be verified by calculating the first and second derivative of x ( t )
which are:
1-10
(10)
x ( t ) = ω n A cos (ω n t + φ ) ,
and
(11)
x ( t ) = −ω n2 A sin(ω n t + φ )
Substituting Equations (9)-(11) into equation of motion (8) yields
(12)
− mω n2 A sin (ω n t + φ ) = − kA sin (ω n t + φ )
Since the sine function is not necessarily zero and we are not interested in
zero amplitude (A=0) this equation can only be satisfied only if:
(13)
ω n2 =
k
.
m
In other words one particular type of harmonic oscillation, as assumed in
Equation (9), having frequency ω n given in Equation (13) is the solution.
The calculated frequency ω n is termed natural frequency or angular natural
frequency and is measured in radians per second. As shown in Figure 6, the
1-11
time for the cycle to repeat itself is the (natural) period , which is related to
ω n by:
(14)
2π rad 2π
s
T=
=
ω n rad/s ω n
Also known as the natural frequency is parameter f n defined as:
(15)
1 ωn
[Hz],
fn = =
T 2π
which indicates how many cycles of vibration are made per second.
To describe motion fully, it is necessary to find the two constants A and φ .
(Two constants are necessary because of the second order ordinary
differential equation). These constants can found from initial conditions
x ( 0 ) = x0 and x ( 0 ) = x0 ≡ v0 using equations (16) and (17):
(18)→
x ( 0 ) = x0 = A sin φ
and
1-12
(19)→
x ( 0 ) = v0 = ωn A cos φ
Combining the square of each equation,
x02 = A2 sin 2 φ
v 02 = ωn2 A2 cos 2 φ
(20)→
ω n2 x02 + v02
A=
ωn
(21)
and
⎛ ω n x0 ⎞
φ = tan ⎜
,
⎟
⎝ v0 ⎠
−1
leading to the final solution:
⎛
⎞
ω n2 x02 + v02
−1 ⎛ ω n x0 ⎞
sin ⎜ ω n t + tan ⎜
(22) x ( t ) =
⎟⎟
ωn
v
⎝ 0 ⎠ ⎠.
⎝
This solution is shown in Figure 6 and indicates that the vibration does not
diminish with time passing.
1-13
Figure 6: Summary
of description of
simple harmonic
motion
1-14
EXERCISE 1-1: SIMULATION OF EQUATION (22).
Simulation performed using interactive MATLAB software NDOF
System:
m=10kg, natural frequency f n = 10 Hz, damping: none
stiffness: k = ( 2π f n ) m =39,478N/m
2
Case 1: Initial displacement
Initial disturbance: x0 = 0.1m and v0 = 0
Peak displacement:
A=_______m
Peak velocity:
V=_______m/s
Case 2: Initial velocity
Initial disturbance: x0 = 0 and v0 = 10 m/sec
Peak displacement:
A=_______m
Peak velocity:
V=_______m/s
1-15
Case 3: Initial velocity and initial displacement
Initial disturbance: x0 = 0.1m and v0 = 10 m/sec
Peak displacement:
A=_______m
Peak velocity:
V=_______m/s
Results shown in Figure 6 and obtained in Exercise 1-1 demonstrate that,
as:
(23)
A= x +
2
0
vo2
ω
2
n
,
the peak displacement A of a SDOF system is a function not only of the
initial displacements and velocities, but also of its natural frequency ω n .
Similar considerations can be made for peak velocities (ω n A) and peak
accelerations (ω n2 A ).
1-16
1.2.2 Damped systems
Damping is a non-conservative force i.e. it dissipates energy. As a
mathematical convenience, the damping force is taken to be proportional to
and opposing velocity, which fits in with observations of ‘dashpots’ like
vehicle shock absorbers. Hence in addition to the stiffness term kx ( t ) that
acts in the opposite direction to deflection x we include a damping term
cx ( t ) that acts in the opposite direction to velocity x ( t ) .
Hence the term cx ( t ) is added to LHS of the equation of motion (8) so that:
(24)
mx ( t ) + cx ( t ) + kx ( t ) = 0 ,
The damping force will acts to dissipate energy by opposing motion (like
friction) so that any oscillations will die out or decay, provided that c is a
positive constant having units [N/s].
The cx ( t ) term represents a damping force fC ( t ) which acts in the
direction opposite of the direction of mass motion (velocity).
1-17
In other words, the force acts in the same direction as the stiffness and
inertia forces, as shown in Figure 7, and helps establish the following
equilibrium of all dynamic forces acting on the mass:
f I ( t ) + fc ( t ) + fk ( t ) = 0
(25)
x (t )
mx( t )
x+ (t )
k
c
fk
m
fc
m
Figure 7: Schematic of a SDOF system with viscous damping indicated by
a dashpot.
1-18
The solution of the differential equation for the undamped case was found
assuming that the response should be harmonic. In the same way, the
general solution for a second order differential equation is taken to be:
x ( t ) = ae λt ,
(26)
where a is a complex number featuring two constants in the same way as
two constants A and φ featured in the assumed solution in Equation (9).
x ( t ) , by inserting x ( t ) and its
After differentiation to obtain x ( t ) and derivatives into equation of motion of a damped system (Equation (24)),
(try yourself) the following equation is obtained:
(27)
( mλ
2
+ cλ + k ) ae λt = 0 .
As ae λt ≠ 0 , this leads to the following equation:
(28)
mλ 2 + cλ + k = 0 ,
known as the characteristic equation. The solution of this quadratic
equation is:
1-19
(29)
λ1,2 = −
c
1
c 2 − 4km .
±
m 2m
This leads us to three possible types of solution, depending on the sign of
c 2 − 4km . In examining the three cases, it is convenient to define: (1) a
critical damping coefficient as:
(30)
ccr = 2mω n = 2 km ,
and (2) damping ratio ζ as:
(31)
ζ =
c
.
ccr
Considering this new notation, and after dividing equation (24) by m , the
equation of motion can be re-written as:
(32)
x ( t ) + 2ζω n x ( t ) + ω n2 x ( t ) = 0 ,
which clearly shows how natural frequency ω n and damping ratioζ come
into the calculations of response.
1-20
1.2.2.1 CASE 1: UNDERDAMPED MOTION (ζ < 1)
In this case c 2 − 4km < 0 , so c < ccr i.e. ζ < 1 and the solutions λ1 and λ2
are:
(33)
λ1 = −ζω n − jω n 1 − ζ 2
and
(34)
λ2 = −ζω n + jω n 1 − ζ 2
where j = −1 .
Therefore, there are two solutions of Equation (32):
x1 ( t ) = a1e λ1t and x2 ( t ) = a2 e λ2t . The sum of the two solutions is also a
solution, so the overall solution is of the following form:
(35)
x ( t ) = a1e
−ζω n − jω n 1−ζ 2 t
+ a2 e
−ζω n + jω n 1−ζ 2 t
or
1-21
(36)
x (t ) = e
−ζω n
(a e
1
− jω n 1−ζ 2 t
+ a2 e
+ jω n 1−ζ 2 t
),
where a1 and a2 are complex-valued constants of integration to be
determined by initial conditions. Using Euler relations:
eφ j = cos φ + j sin φ ,
(37)
and
e −φ j = cos φ − j sin φ ,
(38)
Equation (36) can be written as
(39)
x ( t ) = Ae −ζωnt sin (ω d t + φ )
where: ω d = ω n 1 − ζ 2 is termed damped natural frequency. A and φ can
be expressed as a function of initial conditions:
(40)
A=
( v0 + ζωn x0 )
2
ω d2
+ ( x0ω d )
1-22
and
(41)
⎛ x0ω d ⎞
φ = tan ⎜
⎟
+
ζω
v
x
n 0 ⎠
⎝ 0
−1
A typical shape of the response to initial disturbances is shown in Figure 8
Figure 8: Response of an underdamped system
1-23
EXERCISE 1-2: SIMULATION OF EQUATION (39) –UNDERDAMPED MOTION
Simulation performed using NDOF
System 1: m=6500kg, natural frequency f n = 4.5Hz
k = (2π f n ) 2 m = 5200000N/m
Case 1: x0 = 0m , v0 = 0.1m/s and ζ = 1%
Peak displacement:
A=_______m
Case 1: x0 = 0m , v0 = 0.1m/s and ζ = 10%
Peak displacement:
A=_______m
1-24
System 2: m=6500kg, natural frequency f n = 0.08 Hz
stiffness: k = ( 2π f n ) m =1,621N/m
2
Case 1: x0 = 0m , v0 = 1m/s and ζ = 1%
Peak displacement:
A=_______m
Case 1: x0 = 0m , v0 = 1m/s and ζ = 10%
Peak displacement:
A=_______m
Conclusion from Exercise 1-2: damping has little influence on initial peak
displacement values.
1-25
1.2.2.2 CASE 2: OVERDAMPED MOTION (ζ > 1)
In this case:
(42)
x (t ) = e
−ζω n t
(a e
1
−ω n ζ 2 −1t
+ a2 e
+ ω n ζ 2 −1t
)
which is a non-oscillatory motion; examples are shown in Figure 9.
Figure 9: Response of an overdamped system
1-26
1.2.2.3 CASE 3: CRITICALLY DAMPED MOTION
When c = ccr = 2 km , then λ1 = λ2 = −ω n , so the solution takes form:
(43)
x ( t ) = ( a1 + a2 t ) e −ωnt .
From the form of Equation (43) it is clear that there is no oscillation (no sin
or cos terms). Similarly as before, constants a1 and a2 are determined from
initial conditions as, a1 = x0 , and a2 = v0 + ω n x0 , so the final solution for
critically damped system is:
(44)
x ( t ) = [ x0 + ( v0 + ω n x0 )t ] e −ωnt .
Examples of responses of a critically damped system are given in Figure
10.
1-27
Figure 10: Response of a critically damped system for three different initial
velocities. The system properties are: k = 225N/m , m = 100kg and ζ = 1.
Critical damping ccr can be understood in a number of ways:
1. it is the smallest value of damping that yields aperiodic motion (i.e.
removes oscillation as dynamic response of the system)
2. it can be shown (not shown here) that critical damping provides the
fastest return to zero without oscillation.
1-28
1.3 Analysis of forced response of SDOF systems
So far we have studied ‘free vibration’ where (for a given SDOF system)
the response depends on initial conditions of velocity and displacement and
no external forces are applied. We now consider the application of
‘dynamic forces’.
1.3.1 Definition of a dynamic force
Dynamic excitation force is a time-varying force which will engage
significant inertia, damping and stiffness forces in a structure to which it is
applied. Resonance is a condition where the applied force varies at a rate
that matches the natural frequency of the oscillator.
If the force variation is much slower than the natural oscillations, then the
oscillator behaves like a spring (governed by f=kx) in which case it is
‘stiffness-dominated’ and may be analysed by static analysis. A usual cutoff is r < 1/ 3.
Likewise if the force variation is much faster than the natural oscillations,
then the oscillator behaves like a mass, governed by f=ma.
1-29
1.3.2 Types of dynamic force
A variety of dynamic forces can be applied to a structural system which
will cause vibration. Typical examples of these forces are:
1.
harmonic (including rotating eccentric mass)
2.
periodic
3.
transient
4.
random
There are specific techniques best suited for each type of force.
1.3.3 Equation of motion for forced response of SDOF system
Figure 11 shows the damped SDOF oscillator with applied external force.
With the addition of the forcing function, equation (45) becomes
(46)
mx( t ) + cx ( t ) + kx ( t ) = f ( t )
Response of an undamped oscillator is analysed by deleting the cx ( t )
term.
1-30
x (t )
mx( t )
x+ (t )
k
c
fk
m
fc
m
f
Figure 11 Damped SDOF oscillator with external forcing f(t)
The following figures give examples of forcing functions.
1-31
Harmonic excitation and response (by rotating imbalance):
harmonic
1
ch 1 (N)
0.5
0
-0.5
-1
ch 4 (mm/sec2)
1
0.5
0
-0.5
-1
0
1
2
3
4
5
seconds
6
7
8
9
1-32
Periodic forcing and response: footfall forces (walking)
periodic
ch 1 (N)
1000
800
600
400
ch 4 (mm/sec2)
1000
500
0
-500
4
6
8
10
12
seconds
14
16
18
20
1-33
Transient excitation and response: earthquake
qk11
ch 1 (mm/sec2)
1
0.5
Input
0
(ground movement)
-0.5
-1
Response
2
ch 3 (mm/sec )
5
(building sway)
0
-5
500
550
600
650
700
750
seconds
800
850
900
950
1-34
Random forcing and response: e.g. wind
random
4
ch 1 (N)
2
0
-2
-4
ch 4 (mm/sec2)
10
5
0
-5
-10
0
5
10
15
20
seconds
1-35
1.3.4 Motion of SDOF systems under harmonic loading
Knowing structural/SDOF response to harmonic (i.e. sine) excitation is
important because:
1. It is a very common source of dynamic excitation, typically
applied by machinery
2. Periodic forces can be expressed as a sum of a series of harmonic
terms, so if we know how to calculate response to one term, by
superposition (which is applicable in the case of linear systems)
we can calculate total response to the periodic function as well.
3. Harmonic input-output (i.e. excitation-response) relationships
forms the foundation of vibration measurement.
4. Vibration isolation (to reduce transmitted forces or vibrations) is
based on the harmonic response analysis.
1-36
1.3.4.1 UNDAMPED SYSTEM
Under harmonic loading equation the forcing function for equation (46) is
f ( t ) = F0 cos (ωt ) where F0 is the amplitude and ω is the frequency in
radians per second.
The equation of motion is:
(47)
mx ( t ) + kx ( t ) = F0 cos (ω t )
or, after dividing by m :
(48)
x ( t ) + ω n2 x ( t ) = f 0 cos (ω t ) ,
where f 0 = F0 / m .
Differential equation (48) is linear and non-homogenous, which means that
its solution is a sum of homogenous and particular solutions:
(49)
x ( t ) = xh ( t ) + x p ( t )
1-37
Homogenous solution xh ( t ) has already been found (Equation (22)) and
corresponds to free vibrations of an initially disturbed undamped system,
which can be re-written as (Inman, 2001, p. 92):
(50)
xh ( t ) = A1 sin (ω n t ) + A2 cos (ω n t )
Particular solution x p ( t ) can be obtained by assuming that it is of the form
x p ( t ) = X cos (ω t ) , where X is an amplitude of the forced response. After
x p ( t ) into Equation (48),
double differentiation and inserting x p ( t ) and particular solution is obtained as:
(51)
f0
x p (t ) = 2
cos (ω t )
2
(ω n − ω )
The total solution (unless ω = ω n ) is therefore of the form:
1-38
x ( t ) = A1 sin (ω n t ) + A2 cos (ω n t )
(52)
f0
+ 2
cos (ω t )
2
(ω n − ω )
Finally, initial conditions are used to calculate A1 and A2 , so that the total
solution is of the form:
⎛
f0 ⎞
x (t ) =
sin (ω n t ) + ⎜ x0 − 2
cos (ω n t )
2 ⎟
ωn
ωn − ω ⎠
⎝
(53)
f0
cos (ω t )
+ 2
2
ωn − ω
v0
Two very important phenomena occur when the driving frequency becomes
close to the system’s natural frequency: beating and resonance. This is
illustrated in the following exercise.
1-39
EXERCISE 1-3: SIMULATION OF EQUATION (53).
Simulation performed using NDOF
System:
m=6500kg, natural frequency f n = 4.5Hz , damping: none
k = (2π f n ) 2 m = 5200000N/m
Case 1: Excitation frequency twice lower than the natural frequency
f = 2.25Hz leading to 14.173rad / sec
x0 = 0; v0 = 0; f ( t ) = F0 cos (ω t ) , F0 = 100kN
Peak displacement:
A=_______m
1-40
Case 2: Excitation frequency 90% of the natural frequency
f = 4.05Hz leading to ω = 25.447rad / sec
x0 = 0; v0 = 0; f ( t ) = F0 cos (ω t ) , F0 = 100kN
Case 3: Excitation frequency twice higher than the natural frequency
f = 9Hz leading to ω = 56.55rad / sec
x0 = 0; v0 = 0; f ( t ) = F0 cos (ω t ) , F0 = 100kN
Peak displacement:
A=_______m
Case 4: Excitation at natural frequency
f = 4.5 Hz leading to ω = 28.274rad / sec
x0 = 0; v0 = 0; f ( t ) = F0 cos (ω t ) , F0 = 100kN
Peak displacement:
A=_______m
1-41
In the case when ω = ω n the solution given in Equation (53) is no longer
valid (division by zero), and it can be shown (Inman, 2001, p. 96) that the
new solution is:
(54) x ( t ) =
v0
ω
sin (ω t ) + x0 cos (ω t ) +
f0
t sin (ω t ) ,
2ω
as shown in Figure 12. It can be seen that the response grows without
bound and this phenomenon is known as resonance.
1-42
Figure 12: Forced response of an undamped spring-mass system driven
harmonically at its natural frequency
1-43
1.3.4.2 DAMPED SYSTEM
Equation of motion of a damped system under harmonic loading is:
(55)
mx ( t ) + cx ( t ) + kx ( t ) = F0 cos (ω t )
or
(56)
x ( t ) + 2ζω n x ( t ) + ω n2 x ( t ) = f 0 cos (ω t ) ,
where all notation is the same as before.
Following exactly the same procedure as for the undamped case, the
solution of this differential equation can be presented as a sum of a
homogenous and particular solution, and it can be shown that:
(57) x ( t ) = Ae −ζωnt sin (ω d t + φ ) + X cos (ω t − θ ) ,
where:
(58)
X=
(ω
f0
2
n
− ω 2 ) + ( 2ζω nω )
2
,
1-44
(59)
⎛ 2ζω nω ⎞
θ = tan ⎜ 2
2 ⎟
ω
−
ω
⎝ n
⎠
−1
As before, constants A and φ are determined from initial conditions, but
the formulae are complex and are presented by Inman (2001, p. 99).
The structure of Equation (57) is such that the response has two parts: the
homogenous part, which decays and oscillates with the damped frequency
ω d , and homogenous part which is a harmonic function having the
frequency of the excitation force ω . This is why the homogeneous solution
is often called transient solution, whereas another term for particular
solution is steady-state solution. It should be stressed that even in the case
of homogenous (i.e. zero) initial conditions, transient response will exist as
a results of initial disturbance which the applied steady-state force imparts
on the structure at the beginning of excitation (Inman, 2001, Eq. (2.30),
p.99).
1-45
1.3.4.3 VISUALISING USING ROTATING VECTORS
A useful way to understand the components of equation (55) in steady state
after the transient part has decayed is by visualising the components as
rotating vectors (phasors) either ‘in phase’ or ‘in quadrature’.
Substituting the (steady state) solution x ( t ) = X cos (ωt − θ ) into equation
(55) leads to
(60) − mω 2 X cos (ωt − θ ) − cω X sin (ωt − θ ) + kX cos (ωt − θ ) = F0 cos (ωt )
The right hand side is a vector that rotates so that as ωt takes values 0,
π 2 , π , 3π 2 etc. the in phase component goes from F0 to 0 to - F0 to 0
while the quadrature term goes from 0 to F0 to 0 to - F0 . The terms in
equation (60) can be viewed as the in-phase components of vectors:
mω 2 X
cω X
kX
F0
at angles
π + ωt − θ
π 2 + ωt − θ
ωt − θ
ωt .
1-46
The terms in equation
− mω 2 X cos (ωt − θ ) − cω X sin (ωt − θ ) + kX cos (ωt − θ ) = F0 cos (ωt )
are thus added vectorially as follows:
quadrature
mω 2 X
ωcX
F0
ωt
kX
θ
F0 cos (ωt )
in phase
mω 2 X cos (ωt − θ )
cω X sin (ωt − θ )
kX cos (ωt − θ )
Figure 13 Vector representation of equation (60)
1-47
From this it can be seen that (e.g. by Pythagorus):
F0 = X
( k − mω ) + ( cω )
2 2
2
⎛ cω ⎞
and θ = tan ⎜
2 ⎟
ω
k
m
−
⎝
⎠
−1
Dividing left and right or top and bottom by m leads to equations (58) and
(59).
From Figure 13 can be seen that the mass ( mω 2 X ) and stiffness ( kX )
terms of equation (55) are similar in magnitude and opposite in sign and
both larger than the damping (cω X ) and forcing ( F0 ) terms. The phase
angle θ is approximately π 4 i.e. the response ‘lags’ the forcing by about
45°.
1-48
EXERCISE 1-4: SIMULATION OF EQUATION (57).
System:
m=6500kg, natural frequency f n = 4.5Hz
k = (2π f n ) 2 m = 5200000N/m
x0 = 0; v0 = 0; f ( t ) = F0 cos (ω t ) , F0 = 100kN
Case 1: Non-resonant forcing, Low-damping (ζ = 1% )
f = 9Hz leading to ω = 56.55rad / sec . Use T=10 seconds.
Transient peak displacement:
A=_______m
Steady-state peak displacement: A=_______m
Case 2: Resonant forcing, Low-damping (ζ = 1% )
f = 4.5 Hz leading to ω = 28.274rad / sec . Use T=10 seconds.
Steady-state peak displacement: A=_______m
1-49
Case 3: Non-resonant forcing, High-damping (ζ = 10% )
f = 9Hz leading to ω = 56.55rad / sec
Transient peak displacement:
A=_______m
Steady-state peak displacement: A=_______m
Case 4: Resonant forcing, High-damping (ζ = 10% )
f = 4.5 Hz leading to ω = 28.274rad / sec
Steady-state peak displacement: A=_______m
1-50
Conclusion from Exercise 1-4: Off resonance, steady state condition is
eventually established and damping has little influence on the peak values
of the steady state response. For resonant forcing, damping controls
maximum amplitude and the time taken to achieve it.
As transient part eventually dies out, it is often common to ignore the
transient part of the total solution and focus on the steady-state response:
(61)
x p ( t ) = X cos (ω t − θ )
With this in mind, it is of interest to consider magnitude X and phase θ
(relative to the force F0 cos (ω t ) ) of the steady-state response as a function
of the driving frequency ω . This is done by plotting the ratio of dynamic
amplitude X to static amplitude F0 / k , often referred to as the ‘dynamic
amplification factor’ or DAF and given as:
1-51
X ω n2
X
=
=
(62)
f0
⎛ F0 ⎞
⎜ ⎟
⎝ k ⎠
1
(1 − r
)
2 2
+ ( 2ζ r )
2
,
and
(63)
⎛ 2ζ r ⎞
,
2 ⎟
⎝1− r ⎠
θ = tan −1 ⎜
where r is the frequency ratio:
(64)
ω
r=
ωn
This relationship is shown in Figure 14.
Figure 14: Plot of a normalised
magnitude and its phase relative to the
excitation force.
1-52
The nature of resonance and the phase angles can be seen by revisiting
Figure 13 for the resonant condition with θ = π 2 .
The mass and stiffness terms are exactly equal and opposite. The forcing
term is also equal and opposite to the damping term. In fact at resonance
the driving force is acting only against the damping; the mass and stiffness
forces are perfectly balanced.
quadrature
mω 2 X
ωcX
F0
ωt
θ =π 2
kX
in phase
Figure 15 Vector sum of SDOF oscillator forces at resonance
1-53
Several important features of steady-state harmonic loading are visible in
Figure 14 and Figure 15:
1.
Well before resonance, the force and displacement response are
approximately in-phase i.e. the response follows the excitation
with a small delay indicated by the phase angle.
2.
Well after the resonance, the force and displacement are
approximately 1800 out-of-phase.
3.
At resonance the force and response is 900 behind the force and
Figure 15 shows that for a given (constant) displacement, this
leads to the smallest value of driving force.
4.
Hence there is a significant amplification of the response at the
resonance ( r = 1).
5.
Damping affects the response only for excitation frequencies
relatively close to the resonance.
1-54
EXERCISE 1-5: SIMULATION OF IN- AND OUT-OF-PHASE BEHAVIOUR
Simulation performed using NDOF
System:
m=6500kg, natural frequency f n = 4.5Hz
k = (2π f n ) 2 m = 5200000N/m , damping: ζ = 10%
x0 = 0; v0 = 0; f ( t ) = F0 cos (ω t ) , F0 = 100kN
Case 1: Low-frequency excitation
f = 2.25Hz leading to 14.173rad / sec
Compare timing of force and displacement
Case 2: High frequency excitation
f = 9Hz leading to ω = 56.55rad / sec
Compare timing of force and displacement
1-55
With regard to the last observation, when r = 1 the dynamic response can
be calculated from Equation (62) as:
(65)
F0 1
X=
k 2ζ
Therefore, at resonance, dynamic response is inversely proportional to
damping.
1-56
1.3.5 Periodic forcing
A function that is periodic repeats itself identically after one complete
period and using Fourier analysis can be decomposed into a sum of
sinusoidal functions.
For example a square wave repeating every 0.5 seconds (2Hz) can be
regenerated from an infinite sequence of sine waves with frequencies 2H,
6Hz, 10Hz, 14Hz etc..
The plot shows a sine wave and how the sum of 10 sinusoids, in phase at
time zero, approaches the square wave shape.
Using superposition, which applies for a linear system, the response to a
sum of sine (or cosine) waves is the sum of responses to the separate sine
(or cosine) waves.
1-57
5
0
-5
0
0.1
0.2
0.3
0.4
Fourier amplitude/ASD spectrum an
0.5
time/seconds
0.6
0.7
0.8
0.9
1
Fourier Phase spectrum φn
6
200
4
0
2
-200
0
0
20
40
frequency/Hz
60
80
0
20
40
frequency/Hz
60
80
1-58
EXERCISE 1-6: PERIODIC FORCING: WALKING ON A BRIDGE
f n = 4.25 Hz leading to f n = 26.7035rad / sec m = 6500kg
damping: ζ = 5%
x0 = 0; v0 = 0; f ( t ) = F0 cos (ωt )
Fundamental component: footfall rate
f = 2 Hz , F0 = 430 N leading to ω = 12.5664Hz
Second harmonic: twice footfall rate
f = 4 Hz F0 = 28 N leading to ω = 25.1327 Hz
Use T=2sec. Examine the acceleration trace and note peak values each
time.
Sum of fundamental and second harmonic
Run using ‘user .MAT’ and select file yh1p2.mat when prompted.
Compare with each of the two separate sinusoidal excitations.
1-59
1.3.6 Rotating imbalance
A common source of vibration problems is rotating machinery. A small
imbalance, characterised by a mass me at an eccentric distance re from the
shaft requires centripetal force Fr to maintain the rotation. The force is
proved by the mass to which it is attached. It is the reaction to this force,
-Fr, termed ‘centrifugal force’ that is a special case of harmonic excitation.
Figure 16 Harmonic excitation by rotating imbalance
k
me
ω
c
x(t)
1-60
The equation of motion for the system shown in Figure 16 is derived from
equation (47) as
(66)
mx( t ) + cx ( t ) + kx ( t ) = − Fr ( t )
The motion of the eccentric mass me at radius re is described as
xr = re cos (ωt )
and the centripetal force is
Fr = me xr = − me reω 2 cos (ωt )
so that equation (66) becomes
(67)
mx( t ) + cx ( t ) + kx ( t ) = me reω 2 cos (ωt ) .
The only difference from the solutions for harmonic excitation in section
1.3.4 is that the forcing function increases with frequency so that the
response is given by
x ( t ) = X cos (ωt − θ )
1-61
where
(68)
mr
X= ee
m
r2
(1 − r
)
2 2
+ ( 2ζ r )
2
with
⎛ 2ζ r ⎞
.
2 ⎟
⎝ 1− r ⎠
θ = tan −1 ⎜
Figure 17 Plot of equation (68): Response due to eccentric mass
1-62
1.3.7 Impulse response function
A common source of vibration is the sudden application of a short duration
force called an impulse. An impulse excitation is a force that is applied for
a very short, or infinitesimal, length of time. Shown in Figure 18 is the
force f ( t ) = 1 ε , with time duration ε starting at the time instant t = τ . As
ε approaches zero, the force amplitude becomes infinite, but the magnitude
of the impulse, defined by the time integral of f ( t ) remains equal to unity,
which is the area of the rectangle (1 ε ) ε = 1. Such a force in the limiting
case ε → 0 is called the unit impulse.
1-63
Unit ‘impulse’, height 5N, width
ε=1/5=0.2 seconds. This is an
approximation to δ ( 2 )
5
Force
4
ε
1/ε
τ=2=delay from time t=0.
3
τ
2
1
0
0
1
2
3
4
5
time
0.2
Damped response, h ( t − 2 )
1Hz, 1kg oscillator
0.1
displacement
Undamped response , h ( t − 2 )
1Hz, 1kg oscillator
0
Figure 18: Time history of an
impulse applied at time τ and the
corresponding response.
-0.1
-0.2
1-64
0
1
2
3
time
4
5
According to Newton’s second law of motion, if a force f ( t ) acts on a
body of mass m, the rate of momentum change of the body is equal to the
applied force, that is:
(69)
d
mx ( t ) ) = f ( t )
(
dt
Integrating both sides with respect to time t gives:
t2
(70)
m ( x ( t2 ) − x ( t1 ) ) = ∫ f ( t ) dt .
t1
The integral on the RHS of Equation (70) is the magnitude of the impulse.
In the case of a short unit impulse applied on a SDOF system: t1 = 0 and
t2 = ε , so the velocity at time t2 = ε can be calculated as:
t2
(71)
x ( ε ) =
∫ f ( t ) dt
t1
m
=
1
.
m
1-65
This is so because the duration of the impulse application is infinitesimally
short, so the SDOF spring and damper do not have the time to respond.
After the application of the impulse, the force is removed and the response
of the SDOF system can be calculated as for a freely vibrating system
1
+
disturbed only by initial velocity x ( 0 ) = . Due to the short duration of
m
the impulse it can be assumed that no displacement occurs, so x ( 0+ ) = 0
Response to an impulse
To calculate this response, Equations (39)-(41) can be used leading to
1
A=
and φ = 0 .
mω d
Therefore, the response of the system at to the unit impulse applied at time
t=0 can be calculated as:
(72)
1 −ζωnt
x(t ) =
e
sin ωd t
mωd
1-66
The impulse is denoted mathematically as δ ( t ) or ‘delta function’. When it
occurs at t=0 it is evaluated as δ ( 0 ) . The resulting response given by
equation (72) is termed the ‘impulse response function’ and denoted h ( t ) .
When the impulse occurs after a delay τ , the response at time t ( t > τ ) is
given by
(73)
1 −ζωn ( t −τ )
x(t ) =
e
sin ωd ( t − τ )
mωd
The delta function is then written δ (τ ) and the corresponding impulse
response function as h ( t − τ ) .
This is illustrated in Figure 18
Except for the infinitesimally short period of the impulse, the response is
entirely free vibration, with no more forcing.
1-67
EXERCISE 1-7: IMPULSE RESPONSE FUNCTION, EQUATION (73).
Simulation performed using NDOF
System:
m=1kg, natural frequency f n = 1Hz , damping 1%
k = ( 2π .1) m = 39.478 N / m
2
Case 1: Finite impulse, Damped response (ζ = 1% ), time step=0.01
x0 = 0; v0 = 0;
f(t)=0 until t=0, f(t)=10 until t=1 (set width=1);
Transient peak displacement:
=_______m
Transient peak velocity:
=_______m/sec
Examine the response up to t=1 sec.
1-68
Case 2: Infinitesimal impulse, Damped response (ζ = 1% ), time step0.01
x0 = 0; v0 = 0;
f(t)=0 until t=0, f(t)=100 until t=0.1;
Transient peak displacement:
=_______m
Transient peak velocity:
=_______m/sec
Examine the response up to t=1 sec.
1-69
1.3.8 Random forcing
Dynamic response of a SDOF system to a random signal can be found
using the method for response to an impulse. A random signal is not
periodic or harmonic as it never repeats itself and is not transient as it does
not go to zero after a short time.
1-70
The figures below represent the same data; a random signal with time
interval 0.32 seconds. In the left plot the data points are joined by smooth
continuously varying curves. In the right plot the data points are taken to be
constant at the sample time and dt/2 either side.
300
200
200
100
100
0
0
0
5
10
-100
-200
-100
-200
0
5
10
Figure 19 Random forcing function f(t)
1-71
Because a SDOF oscillator is linear, we can use superposition. Each bar in
The right hand plot is taken as an impulse scaled by 0.32×the value of the
data point i.e. by the area of each bar:
t
data value
bar area
equivalent impulse
impulse response
0
10.2133
3.268
3.268×δ (0)
3.268×h ( t )
0.32
11.0500
3.536
3.536×δ (0.32)
3.536×h ( t − 0.32 )
0.64
-12.2728
-3.927
-3.927×δ (0.64)
-3.927×h ( t − 0.64 )
0.96
30.6986
9.824
9.824×δ (0.96)
9.824×h ( t − 0.96 )
1.28
-20.9759
-6.712
-6.712×δ (1.28)
-6.712×h ( t − 1.28 )
1.60
51.0890
16.348
16.348×δ (1.60)
16.348×h ( t − 1.6 )
1.92
124.9847
39.995
39.995×δ (1.92)
39.995×h ( t − 1.92 )
2.24
42.5725
13.623
13.623×δ (2.24)
13.623×h ( t − 2.24 )
1-72
6
4
h(0)
h(0.32)
2
h(0.64)
The response to the random
forcing is the sum of the
impulse response functions.
h(0.96)
0
0
2
4
6
8
-2
h(1.28)
10
h(1.60)
h(1.92)
h(2.24)
Σ
-4
-6
Here the sum is taken for
functions only due to data up
to t=2.24.
6
5
4
3
2
1
0
-1 0
2
4
6
8
10
The operation of summing
shifted and scaled response
functions is called
‘convolution’.
-2
-3
-4
-5
Figure 20: Schematic explanation of a convolution integral.
1-73
Figure 21 shows the full response. Note that up to 2 seconds the result is
the same as for adding the impulse response functions.
random2
ch 1 (N)
500
0
-500
ch 2 (m)
20
0
-20
0
1
2
3
4
5
6
seconds
7
8
9
Figure 21 Calculated oscillator response to random forcing
1-74
Mathematically, to calculate a response of the system at time t j , it is
necessary to add up contributions of all impulses applied at times ti (for a
short period ∆t , assumed to be constant) before time t j . Therefore,
x ( t j ) = ∑ f ( ti ) ∆t × h ( t j − ti ) .
j
(74)
i =1
Finding the limit of this discrete sum as the number n of equal divisions ∆t
( ∆t = t n ) increases to infinity n → ∞ ( ∆t → dτ )yields:
t
(75)
x ( t ) = ∫ f (τ ) × h ( t − τ )dτ
0
Equation (75) is also known as a convolution or Duhamel integral.
1-75
Coursework: Exercise 1.8: Use spreadsheet ‘convit.xls’ to extend the
generation of response to random forcing represented by Figure 20 up to
the end of the sequence of random forcing data points i.e. about 10
seconds.
Repeat the exercise for a different damping ratio and then for a different
frequency and then use a forcing function of your choice (e.g. square or
triangular wave).
1-76
1.4 Logarithmic decrement and damping
Free vibration after initial disturbance ( x0 , v0 ) and impulse response
function due to impulsive forcing are both characterised by the decaying
exponential:
x(t ) = Ae −ζωnt sin (ωd t + φ )
When x0 ≠ 0, v0 = 0 , φ = 90° and the decay starts from the maximum
amplitude A.
When x0 = 0, v0 ≠ 0 or due to impulsive excitation, φ = 0° and the decay
starts from zero.
In either case the maximum values xn in successive cycles diminish
according to damping. If t0 is the time for the first maximum x0 then the
next maximum x1 will occur after an interval T = 2π ωd . Successive
maxima xn occur at times t0 + 2nπ ωd .
1-77
For the first maximum n=0, x0 = Ae −ζωnt0 sin (ωd t0 + φ )
For the send maximum n=1, x1 = Ae −ζωnt0 +T sin (ωd ( t0 + T ) + φ )
Ae −ζωnt0 sin (ωd t0 + φ )
x0
= −ζωnt0 +T
Then the ratio of these maxima is:
.
x1 Ae
sin (ωd ( t0 + T ) + φ )
Since ωd T = 2π the sine terms are identical and cancel as does A leaving:
(76)
x0
e −ζωnt0
= −ζω ( t +T ) = eζωnT
x1 e n 0
The ‘logarithmic decrement’ is the (natural) logarithm of this ratio i.e.
(77)
δ=
⎛ x0 ⎞
δ = ln ⎜ ⎟ = ζωnT
⎝ x1 ⎠
2πζωn
ωd
≈ 2πζ for small damping.
1-78
For maxima separated by m cycles, equation (76) becomes
(78)
−ζω t +iT
⎛ x (t ) ⎞
xi
e n( 0 )
ζωn mT
= −ζω ( t +iT +mT ) = e
→ ln ⎜
⎟ = mζωnT = mδ
n 0
xi +m e
⎝ x ( t = mT ) ⎠
Hence ‘log dec’ can also be calculated over several cylcels.
Logarithmic decrement (or ‘log dec’) has two uses.
First, the successive peak values of free oscillations can be found if one
peak value plus damping ratio are known.
Second, measurement of the ratio (or average ratio) of successive peak
allows for estimation of the (equivalent) damping ζ for a real system.
1-79
ω=1.6196Hz , ζ =0.56911%
0.4
0.3
0.2
ch 6
0.1
0
-0.1
-0.2
-0.3
-0.4
0
5
10
15
20
25
30
Figure 22 Free decay of
footbridge (right) and
measured damping
using ‘log dec’
1-80
EXERCISE 1-9: LOGARITHMIC DECREMENT.
System:
m=1kg, f n = 1Hz , k = ( 2π .1) m = 39.478 N / m
2
Case 1: x0 = 1m, v0 = 0 and ζ = 1%
Initial peak displacement:
A0=_______m at t0= _______seconds
Second peak displacement : A1=_______m at t1= _______seconds
Third peak displacement:
A2=_______m at t2= _______seconds
Calculate δ = ln ( A0 A1 ) , (1 2π ) δ . Also check ln ( A0 A2 ) , 1 ( t1 − t0 )
Case 2: x0 = 0, v0 = 1m / sec and ζ = 2%
First peak acceleration:
A0=_______m
Second peak acceleration :
A1=_______m
δ = ln ( A0 A1 ) _______
Third peak acceleration:
A2=_______m
δ = ln ( A1 A2 ) _______
1-81
1.5 Base excitation
As well as having a force applied directly to the mass of the oscillator, an
alternative and often more critical case is when the support of the structure
moves. Examples where understanding of base excitation is important
include earthquakes, suspension systems on vehicles and mounting of
sensitive machinery on a vibrating floor.
The analysis is the same if the movement is horizontal or vertical and
gravity is irrelevant. For consistency we consider horizontal movement (as
in a building moving in an earthquake).
1-82
xg ( t )
x (t )
mx( t )
x+ (t )
k
c
fk
m
fc
m
spring force
f k = k ( xg ( t ) − x ( t ) ) , (+ve when xg>x)
damping force
f c = k ( x g ( t ) − x ( t ) ) ,
inertia force
f I = −mx( t )
Figure 23 Damped SDOF system with base excitation
1-83
Using dynamic equilibrium, the equations of motion for mass m are:
(79)
f I ( t ) + fc ( t ) + fk ( t ) = 0 .
Substituting for fk etc. and dropping the (t),
(80)
− mx + c ( x g − x ) + k ( xg − x ) = 0
Note: using f=ma directly exchanges −mx on the left for + mx on the right.
It is the relative movement that gives rise to spring and damper forces,
hence we define relative displacement
(81)
xr = x − xg
so that we add mxg to both sides and equation (80) becomes:
(82)
m ( xg − x ) + c ( x g − x ) + k ( xg − x ) = mxg
i.e. (changing the sign as well)
1-84
(83)
mxr + cxr + kxr = − mxg
Equation (83) can be used for determining the relative response to any base
excitation after which the absolute response can be found by adding the
base response.
If we are interested in the internal stiffness and damping forces we can use
xr directly. Note that equation (83) is identical in form to equation(46):
(46)
mx( t ) + cx ( t ) + kx ( t ) = f ( t )
so all the methods described in section 1.3 for different forcing functions
can be applied. The one that interests us is harmonic base excitation.
1-85
1.5.1 Harmonic base excitation-absolute response
Starting with equation (82):
(82)
m ( xg − x ) + c ( x g − x ) + k ( xg − x ) = mxg
or
(84)
mx + cx + kx = −cx g − kxg
For harmonic base excitation
(85)
xg ( t ) = X g cos (ωt ) .
Assume that the transient part has decayed leaving a harmonic response:
(86)
x ( t ) = X cos (ωt − φ ) .
Hence equation (84) becomes
1-86
(87) − mω 2 X cos (ωt − φ ) − cω X sin (ωt − φ ) + kX cos (ωt − φ )
= −cω X g sin (ωt ) + kX g cos (ωt ) .
The terms in equation (87) are thus added vectorially as follows:
quadrature
mω 2 X
cω X g
2cω X
kX g
ωt
φ
kX
in phase
Figure 24 Vector representation of forces during base excitation
1-87
From this it can be seen that (e.g. by Pythagorus):
Xg k +ω c = X
2
2 2
( k − mω )
2 2
⎛ 2cω ⎞
−1 cω
- tan
.
+ ω c and φ = tan ⎜
2 ⎟
k
⎝ k − mω ⎠
−1
2 2
Using k = mωn2 and c = 2ζωn m
(88)
X
=
Xg
ωn2 + ( 2ζωnω )
(ω − ω
2
n
)
2 2
2
+ ( 2ζωnω )
⎛ 2ζωωn ⎞
−1
tan
, ∠ tan ⎜ 2
( 2ζωωn )
2 ⎟
⎝ ωn − ω ⎠
−1
2
Using frequency ratio r = ω ωn the dynamic amplification factor (DAF) is
(89)
X
=
Xg
1 + ( 2ζ r )
2
(1 − r ) + ( 2ζ r )
2
2
⎛ 2ζ r ⎞
−1
, ∠ tan −1 ⎜
tan
2ζ r .
2 ⎟
⎝ 1− r ⎠
Before studying the nature of this ratio, a different problem is studied,
having the same solution.
1-88
EXERCISE 1-10: SIMULATION OF IN- AND OUT-OF-PHASE BEHAVIOUR FOR
BASE EXCITATION: EQUATION (82)
Simulation performed using NDOF
System:
m=1kg, f n = 1Hz , k = ( 2π .1) m = 39.478 N / m , ζ = 10%
2
x0 = 0; v0 = 0; xg = 1.cos (ωt )
Use ‘base sine’, set dt=0.02.
Case 1: Low frequency base excitation, below resonance
f = 0.1Hz leading to ω = 0.62832Hz and T=50
Compare timing of base motion (blue trace) and displacement
Case 2: High frequency base excitation, above resonance
f = 2 Hz leading to ω = 12.5664 Hz and T=10
Compare timing of base motion (blue trace) and displacement.
1-89
1.6 Transmissibility
x (t )
mx( t )
x+ (t )
k
c
fk
m
fc
m
f
Recall that the oscillator response x ( t ) is given by the solution of
(46)
mx( t ) + cx ( t ) + kx ( t ) = f ( t )
We consider the case of harmonic forcing to investigate the ratio of force
transmitted to the (fixed) support or abutment to the external forcing.
1-90
The force fT transmitted to the support is the sum of the damping and
stiffness forces i.e. fT = f c + f k .
The ratio is defined as transmissibility ratio T or TR:
(90)
fT ( t )
f c + f k cx ( t ) + kx ( t )
T=
=
=
f (t )
f (t )
f (t )
For harmonic excitation f ( t ) = F0 cos (ωt ) the steady state part of the
solution (which we already know from equations (58) and (59) ) is
x ( t ) = X cos (ωt − θ ) .
Also, since (by equation (60)
F0 cos (ωt ) = − mω 2 X cos (ωt − θ ) − cω X sin (ωt − θ ) + kX cos (ωt − θ ) ,
T=
−cω X sin (ωt − θ ) + kX cos (ωt − θ )
− mω X cos (ωt − θ ) − cω X sin (ωt − θ ) + kX cos (ωt − θ )
2
.
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Once again the ratio of two vector quantities is best found via the force
polygon:
quadrature
mω 2 X
fT
F0
φ
ωt
ωcX
kX
in phase
Figure 25 Vector representation of transmissibility ratio T
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From Figure 25 it can be seen that
(91)
k + ( 2cω )
T=
( k − mω ) + ( cω )
2 2
ωn2 + ( 2ζωnω )
2
2
=
(ω
2
n
−ω
2
) + ( 2ζω ω )
2 2
2
n
The phase angle is less obvious,
⎛ 2ζωωn ⎞
−1
tan
but is ∠ tan ⎜ 2
( 2ζωωn )
2 ⎟
⎝ ωn − ω ⎠
−1
Using frequency ratio r = ω ωn
(92)
1 + ( 2ζ r )
2
⎛ 2ζ r ⎞
−1
tan
2ζ r
, ∠ tan ⎜
T=
2 ⎟
2
2
⎝ 1− r ⎠
(1 − r ) + ( 2ζ r )
−1
Equations (92) and (89) are identical so can be studied together.
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EXERCISE 1-11: SIMULATION OF IN- AND OUT-OF-PHASE BEHAVIOUR FOR
TRANSMISSIBILITY: EQUATION (91)
Simulation performed using NDOF
S ystem:
m=1kg, f n = 1Hz , k = ( 2π .1) m = 39.478 N / m , ζ = 10%
2
x0 = 0; v0 = 0; xg = 1.cos (ωt )
Use ‘sine’, set dt=0.02.
Case 1: Low frequency forcing (below resonance)
f = 0.1Hz leading to ω = 0.62832Hz T=50
Case 2: High frequency forcing (above resonance)
f = 2 Hz leading to ω = 12.5664 Hz T=10
Examine and compare forces ‘c+k’ and ‘external’
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1.6.1 Transmissibility Ratio Characteristics
Figure 26 Transmissibility ratio
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In both cases, as frequency ratio r tends to zero T=TR goes to unity
meaning that stiffness forces dominate: inertia forces goes to zero so that
for force transmissibility all the applied force is transmitted to the support,
while for base excitation, there is no relative motion caused by inertia.
In both cases for reasonable (small) damping ratios, phase angle passes
close to 90° at resonance.
An important feature shown in Figure 26 is that for r < 2 T increases with
lower damping whereas for r > 2 damping increases T.
1.6.2 Vibration isolation and base isolation
Vibration isolation aims to reduce transmission of forces generated by
rotating machinery or other sources of vibration excitation. Applications
include mountings for car engines, washing machine drums and
compressors.
Base isolation aims to reduce transmission of ground motion to a structure
or equipment by using a foundation or mounting comprising spring and
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damper elements. Applications include vehicle suspensions, protecting
buildings from earthquakes, preventing annoyance in hospitals and theatres
due to nearby railways and ensuring that very delicate or vibration sensitive
equipment experiences vibration levels that will not affect operation.
In both cases the isolation system is designed with r > 2 .
Figure 27 Vibration isolation of compressor and base isolation of building
For base isolation (e.g. of buildings) against earthquakes the stiffness of the
isolation mounting aims to reduce forces transmitted into the building that
have to be sustained by the structural elements, and these forces are
dominated by inertia, hence the need to isolate high frequency ground 1-97
motions. Design of such systems is complex partly because buildings are
not simple SDOF systems and earthquakes have energy at a range of
frequencies, but generally the damping levels designed to be high (e.g.
20%). If damping is too low and the frequency of the isolated building is
too low the structure will be wind-sensitive.
For vibration isolation (e.g. of pumps) the damping is kept low as the
system needs only to be effective for one frequency and for low damping
(above r > 2 ) T is kept very low. The compressor mounting in Figure 27
uses only springs having minimal damping.
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References and links
1. Inman, D. J. (2001). Engineering Vibration, 2nd Edition. Prentice Hall, New
Jersey. pp. 621.
2. Smith, J. W. (1987). Vibrations of Structures, Applications in Civil
Engineering Design. Chapman and Hall, pp. 338.
3. Neat animations from Prof Brian Stone, Australia’s top engineering lecturer:
http://www.mech.uwa.edu.au/bjs/Vibration/OneDOF/
4. Animation of base isolation:
http://widget.ecn.purdue.edu/~me563/Lectures/Forced/Isolation_Absorption/
Animations_01/animation01.html
5. Simulator (like NDOF, but as Java applet, single screen):
http://mathinsite.bmth.ac.uk/applet/msd/msd.html
6. EFUNDA formulae: http://www.efunda.com/formulae/vibrations
7. Lecture notes by Prof Dan Inman: www.cimss.vt.edu/me3504
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