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Dr. Millerjothi, BITS Pilani, Dubai Campus
�Forced system analysis
�Need for Vibration isolation and critical
speed and resonance
�Construction and functioning of
measuring instruments
Chapter 3
Dr. Millerjothi, BITS Pilani, Dubai Campus
• When a system is subjected to harmonic excitation, it is forced
to vibrate at the same frequency as that of the excitation.
• Common sources:
�Unbalance in rotating machines
�Forces produced by reciprocating machines
�The motion of the machine itself
• Resonance is to be avoided in most cases with dampers and
absorbers
Dr. Millerjothi, BITS Pilani, Dubai Campus
• Harmonic excitation may be in the form of a force or
displacement of some point in the system.
• Single DOF with viscous damping:
- Excited by
-
- Complementary (homogeneous)
and particular solutions
- The particular solution is a steady-state oscillation of
the same frequency as that of the excitation.
- The particular solution can be assumed to be
FORCED HARMONIC VIBRATION
tFkxxcxm ωsin0=++ &&&
( )φω −= tXx sin
Dr. Millerjothi, BITS Pilani, Dubai Campus
• X is the amplitude of oscillation and ϕ is the phase of the
displacement with respect to the exciting force.
• In harmonic motion, the phases of
the velocity and acceleration are ahead
of the displacement by 90o and 180o,
respectively, as in the figure given.
( ) ( )
k
m
k
c
k
c
k
m
k
F
mk
c
cmk
FX
222
2
0
2
1
222
0
1
tan ,
1
tan ,
ω
ω
φωω
ωω
φωω
−=
+
−
=
−=
+−= −
Dr. Millerjothi, BITS Pilani, Dubai Campus
In terms of the following quantities:
n
c
c
c
nc
n
k
c
c
c
k
c
factordampingc
c
dampingcriticalmc
nsoscillatioundampedoffrequencynaturalm
k
ωω
ζωω
ζ
ω
ω
2
2
==
==
==
==
Dr. Millerjothi, BITS Pilani, Dubai Campus
The non-dimensional expressions for the amplitude and
phase become
2
2220
1
2
tan
,
21
1
−
=
+
−
=
n
n
nn
F
Xk
ωω
ωω
ζφ
ωω
ζωω
Dr. Millerjothi, BITS Pilani, Dubai Campus
Xk/Fo and ϕ are functions
only of the frequency ratio
ω/ωn and the damping
ratio ζ. This can be plotted
as shown.
Dr. Millerjothi, BITS Pilani, Dubai Campus
For ω/ωn <<1, the impressed force is nearly equal to the
spring force, as in (a)
For ω/ωn =1, the phase angle is 90o and the force diagram is as in (b)
For ω/ωn >>1, ϕ approaches 180o and the impressed force is expended almost entirely in overcoming the large inertia force
Dr. Millerjothi, BITS Pilani, Dubai Campus
The differential equation and the complete solution
( )
( )1
2
1
222
0
02
1sin
21
)sin(
sin2
φωζ
ωω
ζωω
φω
ωωζω
ζω +−+
+
−
−=
=++
−teX
t
k
Ftx
tm
Fxxx
n
t
nn
nn
n
&&&
Dr. Millerjothi, BITS Pilani, Dubai Campus
Let harmonic force be represented by
The displacement can then be written as
Then
and
Complex frequency response
ti
oo eFtitF ωωω =+ )sin(cos
( ) ( )complexX
eXeXeXex titiiti
=
=== −− ,ωωφφω
( )
+
−
=
=++−
nn
i
k
F
X
FXkicm
ωζω
ωω
ωω
21
2
0
0
2
Dr. Millerjothi, BITS Pilani, Dubai Campus
The complex frequency response H(ω)
Often the factor 1/k is considered together with the force,
leaving the frequency response a non-dimensional quantity.
Thus H(ω) depends only on the frequency ratio and the
damping ratio.
( )
+
−
==
nn
i
k
F
XH
ωζω
ωω
ω2
1
1
2
0
Dr. Millerjothi, BITS Pilani, Dubai Campus
The real and imaginary parts of H(ω)
( ) ,
21
2
21
1
22
222
2
2
+
−
−
+
−
−
=
nn
n
nn
niH
ωζω
ωω
ωζω
ωζω
ωω
ωω
ω
Dr. Millerjothi, BITS Pilani, Dubai Campus
At resonance, the real part is zero and the response is
And the phase angle
( )ζ
ωωω2
1, iHn −==when
2
1
2
tan
−
=
n
n
ωω
ωζω
φ
Dr. Millerjothi, BITS Pilani, Dubai Campus
Unbalance in rotating machine:
� A spring-mass system constrained
to move in the vertical direction
� The unbalance is represented by
an eccentric mass m with eccentricity
e that is rotating with angular velocity ω
The displacement of the non-rotating mass (M – m) from the static
equilibrium position
ROTATING UNBALANCE
tex ωsin+
Dr. Millerjothi, BITS Pilani, Dubai Campus
The equation of motion
Since it is identical to tFkxxcxM
tmekxxcxM
xckxtexdt
dmxmM
ωωω
ω
sin
sin)(
0)sin()(
0
2
2
2
=++
=++
=++++−
&&&
&&&
&&&
222
2
02220
2222
0
]2[])(1[
)1
(
]2[])(1[
1
tan,)()(
nn
n
nn
F
MX
F
Xk
Mk
c
cMk
FX
ωω
ζωω
ω
ωω
ζωω
ωω
φωω
+−
=⇒
+−
=
−=
+−=
Dr. Millerjothi, BITS Pilani, Dubai Campus
Hence, the steady-state solution is
2222
2
2222
2
)(1
)(25
tan,
]2[])(1[
)(
tan,)()(
n
n
nn
n
me
MX
Mk
c
cMk
meX
ωωωω
φ
ωω
ζωω
ωω
ωω
φωω
ω
−=
+−
=
−=
+−=
Dr. Millerjothi, BITS Pilani, Dubai Campus
The complete solution is ( ) ( )
[ ])sin(
)(
1sin
222
2
1
2
1
φωωω
ω
φωζζω
−+−
+
+−= −
t
cMk
me
teXtx n
tn
me
MX
Dr. Millerjothi, BITS Pilani, Dubai Campus
Dr. Millerjothi, BITS Pilani, Dubai Campus
Dr. Millerjothi, BITS Pilani, Dubai Campus
Static unbalance:
When the unbalanced masses all lie in a single plane, as in the c
ase of a thin rotor disk, the resultant unbalance is a single radial
force.
The static unbalance can be detected by a static test
ROTOR UNBALANCE
System with static balance System with dynamic balance
Dr. Millerjothi, BITS Pilani, Dubai Campus
Dynamic unbalance:
When the unbalance appears in more than one plane, the
resultant is a force and a rocking moment, which is referred to as
dynamic unbalance
A static test may detect the resultant force, but the rocking
moment cannot be detected without spinning the rotor.
If the two unbalanced masses are equal and 180 apart, the rotor
will be statically balanced about the axis of the shaft. However,
when the rotor is spinning, each unbalanced disk would set up
a rotating centrifugal force, tending to rock the shaft on its
bearings.
Dr. Millerjothi, BITS Pilani, Dubai Campus
Machines to detect and correct the rotor unbalance
The balancing machine consists of supporting bearings
that are spring-mounted so as to detect the unbalanced
forces by their motion.
Balancing machines
Dr. Millerjothi, BITS Pilani, Dubai Campus
Whirling:
The rotating of the plane made by the bent shaft and the line of c
enters of the bearings.
Results from such various causes as mass unbalance, hysteresis
damping in the shaft, gyroscopic forces, fluid friction in bearings
and so on.
Can take place in the same or opposite direction as that of the
rotation of the shaft and the whirling speed may or may not be
equal to the rotating speed
WHIRLING OF ROTATING SHAFTS
Dr. Millerjothi, BITS Pilani, Dubai Campus
Consider a single disk of mass m symmetrically located on a sha
ft supported by two bearings. (see Figure)
The center of mass G of the disk is at a distance e, eccentricity
from the geometric center S of the disk.
Dr. Millerjothi, BITS Pilani, Dubai Campus
The centerline of the bearings intersects the plane of the disk O
and the shaft center is deflected by r = OS
Assume the shaft (i.e., the line e = SG) be rotating at a constant
speed ω, and in the general case, the line r = OS to be whirling
at speed dθ/dt, that is not equal to ω.
For the equation of motion, we can develop the acceleration of the
mass center as follows:
SGGG aaa
erOG
/+=
+=
Dr. Millerjothi, BITS Pilani, Dubai Campus
Resolving aG in the radial and tangential directions
[ ][ ] θθωωθθ
θωωθ
θωθθ
θωωθ
θωθω
&&&&&
&&&&
&&&&
&&&
crterrm
rckrterrm
jtewrr
iterra
jteiteirr
G
G
−=−−−
−−=−−−
⇓
−−++
−−−=
−+−+=
)sin()2(
)cos()(
)]sin()2[
)]cos()[(
)sin()cos(
2
22
2
22
Dr. Millerjothi, BITS Pilani, Dubai Campus
Which can be rearranged as
The general case of whirl comes under the classification of self-
excited motion, where the exciting forces inducing the motion
are controlled by the motion itself.
In the steady-state synchronous whirl, where and
the problem reduces to that of 1 DOF and 2
DOF.
)sin(2
)cos(
2
22
θωωθθ
θωωθ
−=
−+
−=
−++
terrm
cr
term
kr
m
cr
&&&&
&&&&
ωθ =&
0=== rr &&&&&θ
Dr. Millerjothi, BITS Pilani, Dubai Campus
Assume that and with integration , where φis the phase angle between e and r, which is now a constant.
With
For phase angle
Synchronous whirl:
ωθ =& φωθ −= t
0=== rr &&&&&θ
2
2
22
1
2
tan
sin
cos)(
−
=
=
=−
n
n
erm
c
erm
k
ωω
ωω
ζφ
φωω
φωω
Dr. Millerjothi, BITS Pilani, Dubai Campus
Noting from the vector triangle,
222
2
222
2
21
)()(
+
−
=
+−=
nn
n
e
cmk
mer
ωω
ζωω
ωω
ωω
ω
22
2
2
cos
+
−
−
=
ωω
ωωφ
m
c
m
k
k
Dr. Millerjothi, BITS Pilani, Dubai Campus
The eccentricity line e = SG leads the displacement line r = OS
by the phase angle φ.
When the rotation speed coincides with the critical speed
or the natural frequency of the shaft in lateral
vibration, a condition of resonance is encountered.
mkn =ω
Dr. Millerjothi, BITS Pilani, Dubai Campus
Figure shows the disk-shaft system under three different
conditions
At very high speed, the center of mass G tends to approach th
e fixed point O, and the shaft center S rotates about it in a
circle of radius e
Dr. Millerjothi, BITS Pilani, Dubai Campus
The equation of motion
SUPPORT MOTION
)sin(sin
,
0)()(
2 tYytYmymkzzczm
thenyxzLet
yxcyxkxm
ωωω ==−=++
−=
=−+−+
&&&&&
&&&&
Dr. Millerjothi, BITS Pilani, Dubai Campus
The solution can be written as
For absolute motion
2222
2
tan,)()(
)sin(
ωω
φωω
ω
φω
mk
c
cmk
YmZ
tZz
−=
+−=
−=
tiiti
tiiti
ti
eXeXex
eZeZeZ
Yey
yzxx
ωϕϕω
ωφφω
ω
−−
−−
==
==
=
+=
)(
)(
,
Dr. Millerjothi, BITS Pilani, Dubai Campus
ti
ti
tii
i
titii
titiitiitii
Yeicmk
ick
eYicmk
Ym
eYZex
icmk
YmZe
YemeZeicmk
YemeZekeZeiceZem
ω
ω
ωφ
φ
ωωφ
ωωφωφωφ
ωωω
ωωω
ωωω
ωωω
ωωωω
)(
)(
)(
)(
)()()(
2
2
2
2
2
22
222
+−+
=
++−
=
+=+−
=
=+−
=++−
−
−
−
−−−
since
Dr. Millerjothi, BITS Pilani, Dubai Campus
( )( ) ( )
( ) ( )22
3
22
2
2
222
22
tan
,
21
21
cmkk
mc
cmk
ck
Y
X
nn
n
ωωω
ϕ
ωζω
ωω
ωζω
ωω
ω
+−=
+
−
+
=+−
+=
Dr. Millerjothi, BITS Pilani, Dubai Campus
| X/Y | =1 at the frequency ω/ωn = √2
( ) ( )22
3
22
2
2
tan
21
21
cmkk
mc
Y
X
nn
n
ωωω
ϕ
ωζω
ωω
ωζω
+−=
+
−
+
=
Dr. Millerjothi, BITS Pilani, Dubai Campus
Isolation System
To protect a delicate object from excessive vibration transmitted to
it from its supporting structure, or
To prevent vibratory forces generated by machines from being
transmitted to its surroundings
The motion transmitted from the supporting structure to the mass
m is less than 1 when the ratio ω/ωn is greater than √2.
This can be done by using a soft spring.
Reducing the transmitted force:
VIBRATION ISOLATION
Dr. Millerjothi, BITS Pilani, Dubai Campus
The force to be isolated is transmitted through
the spring and damper, as shown in Fig.
12 <→>Y
X
nωω
when2
k
2
kc
F
Xcω
kX
φF
Xm 2ω( ) ( )
22
2
0
0
2
22
21
,sin
21
+
−
==
+=+=
nn
n
T
kF
XtFF
kXXckXF
ωςω
ωω
ω
ωςω
ω
when
Dr. Millerjothi, BITS Pilani, Dubai Campus
The transmissibility, defined as the ratio of the transmitted
force to that of the disturbing force.
( ) 12
1
1
1,
21
21
,
2
2
2
22
2
2
−∆
=→∆
==
−
=
+
−
+
==
2
gf
TRg
m
k
TR
F
FTR
n
n
n
n
D
T
πω
ωω
ωςω
ωω
ωςω
dampingNeglecting
bilityTransmissi
Dr. Millerjothi, BITS Pilani, Dubai Campus
Comparison of the preceding equation with the steady state
amplitude and phase
To reduce the amplitude X of the isolated mass m without
changing TR, m is often mounted on large mass M, as shown in
Figure. The stiffness k then be
increased to the ratio k/(m+M) constant.
The amplitude X is reduced because k
appears in the denominator
Y
X
F
FTR T ==
0
Dr. Millerjothi, BITS Pilani, Dubai Campus
Dr. Millerjothi, BITS Pilani, Dubai Campus
Dr. Millerjothi, BITS Pilani, Dubai Campus
Energy in a vibrating system is either dissipated into heat
or radiated away.
Energy dissipation:
• Usually determined under conditions of cyclic oscillations
• Force-Displacement relationship
• The force-displacement curve will enclose an area as
hysteresis loop
• Proportional to the energy lost per cycle
• The energy lost per cycle due to a damping force Fd
ENERGY DISSIPATED BY DAMPING
Dr. Millerjothi, BITS Pilani, Dubai Campus
Damping is present in all system
internal molecular friction – Structural damping
sliding friction – Coulomb damping
fluid resistance – Viscous damping
Since mathematical description of damping is quite complicate, simplified damping modal such as viscous has been developed to evaluate the system response.
Hysteresis loop : area enclosed by force-displacement curve.
proportional to the energy lost per cycle.
where is the energy lost per cycle due to a camping force
ENERGY DISSIPATED BY DAMPING
∫= dxFW dddW
dF
Dr. Millerjothi, BITS Pilani, Dubai Campus
In general, depends on many factors, such as temperature,
frequency, or amplitude.
Case of spring-mass system with viscous damping:
The steady-state displacement and velocity
The energy dissipated per cycle
dW
xcFd&=
( )φω −= tXx sin
( )φωω −= tXx cos&
22
0
222
2
)(cos XcdttXc
dtxcdxxcWd
ωπφωωωπ
∫
∫∫=−=
== &&
Dr. Millerjothi, BITS Pilani, Dubai Campus
Of particular interest is the energy dissipated in forced vibration at
resonance.
With and ,
Then
and the damping force becomes
By rearranging the foregoing equation to
mkn =ω kmc ζ2= 22 kXWd ζπ=
Dr. Millerjothi, BITS Pilani, Dubai Campus
The energy dissipated per cycle
is the area enclosed by the
ellipse.
If we add to the force kx,
the hysteresis loop is rotated as
shown in Fig ⇒Voight model
Specific Damping Capacity:
The energy loss per cycle divided by the peak potential energy
Loss Coefficient: The ratio of damping energy loss per radian
divided by the peak potential or strain energy
dF
π2/dW
Dr. Millerjothi, BITS Pilani, Dubai Campus
Equivalent damping Ceq: equate the energy dissipated by the
viscous damping to that of the nonviscous damping force
with assumed harmonic motion
Equivalent Viscous Damping
n
deqeqd
c
FX
X
WCXCW
ω
πωωπ
0
2
2
=
=→=
resonant&dampingviscous
Dr. Millerjothi, BITS Pilani, Dubai Campus
From experiments of most structural metals, damping is
independent of the frequency and proportional to the square
of the amplitude
Structural Damping
tFkxxxm
CXC
XW
eqeq
d
ωπωα
πωα
ωπ
α
sin0
2
2
=+
+
=→=
=
&&&
Dr. Millerjothi, BITS Pilani, Dubai Campus
Complex stiffness
( )( )
( ) kimk
FX
ikkeFxikxm
eFxikxm
xieXixeXx
ti
ti
titi
γω
γπα
γγ
πα
ωω
ω
ω
ωω
+−=
+
==++
=
++
===
−
20
0
0
:1
:,
,
stiffnesscomplex
factordampingstructural
motionharmonicassumestiffnesscomplex
&&
&&
&
Dr. Millerjothi, BITS Pilani, Dubai Campus
Compare to viscous damping at resonance
k
FX
γ0, =resonanceAt
ςφς
22
0 =∴⇒=k
FX
( ) ( )
( ) ( ) 222222
2
2
0
1,
1
1
1
1
,
γ
γ
γ
φ
ωω
+−
−=
+−
−=
+=+−
==
=
ry
r
rx
iyxirF
XrH
rn
where
FRF
Let
0
γ2
1
H(r)
-y
x
Dr. Millerjothi, BITS Pilani, Dubai Campus
The following equation is that of a circle with radius,
and centre,
( )γ
γ
γγ
irH
yxr
yx
−=
−===
=
++
1,0,1,
2
1
2
122
2
resonanceAt
Dr. Millerjothi, BITS Pilani, Dubai Campus
In forced vibration, there is a quantity Q related to damping that is
a measure of the sharpness of resonance.
Assume viscous damping
SHARPNESS OF RESONANCE
k
FX
F
Xk
tFkxxcxm
res
n
n
n
nn
ςωω
ωω
ωω
ςφ
ωω
ςωω
ω
2,1
1
2
tan,
21
1
sin
0
222
20
0
==
−
=
+
−
=
=++
when
&&&
Dr. Millerjothi, BITS Pilani, Dubai Campus
( )pointspowerhalfSidebands −⇒
= resres XXX
2
1707.0:
0X
X
1ω 2ω
ζ2
1
ζ2ζ22
1
nωω
Dr. Millerjothi, BITS Pilani, Dubai Campus
( ) 22
2
2
2
22
2
222
00
1221
418
21
1
22
1
2
ςςςωω
ωω
ςωω
ς
ωω
ςωω
ς
−±−=
+
−=⇒
+
−
===
n
nn
nn
res
F
kX
F
Xk
Dr. Millerjothi, BITS Pilani, Dubai Campus
n
nnn
n
n
n
ωωω
ωωω
ωωω
ωωω
ς
ςωω
ςωω
ςωω
ς
12
1212
2
2
1
2
2
2
2
2
1
2
2
4
21
21
21
1
−≅
+−
−=
−=
+=
−=
±=
<<Let
γ
ςωωω
1dampingstructuralFor
2
1
1212
=
=−
=−
=
Q
ff
fQ nn
Dr. Millerjothi, BITS Pilani, Dubai Campus
Seismic Unit of Fig
Displacement, velocity, or acceleration is indicated by the relative
motion of the suspended mass with respect to the case.
Vibration Measuring Instruments
m
k
c
x
y
tYmkzzczm
tYyyxz
yxkyxcxm
ωω
ω
sin
sin,
0)()(
2=++
=−=
=−+−+
&&&
&&&&
Let
motion suppot as Same
Dr. Millerjothi, BITS Pilani, Dubai Campus
In support motion x/y is of interest,
z/y is of interest for measuring instruments.
( ) ( ) ( )
−
=−
=
+
−
=+−
=
22
2
2
2
2
222
2
1
2
tan
21
n
n
nn
n
mk
cand
Y
cmk
YmZ
ωω
ωω
ζ
ωω
φ
ωω
ζωω
ωω
ωω
ω
Dr. Millerjothi, BITS Pilani, Dubai Campus
Response of a vibration-measuring instrument
Dr. Millerjothi, BITS Pilani, Dubai Campus
Seismometer is an instrument with low natural frequency
When the natural frequency is low in comparison to the vibration
frequency to be measured Z approaches Y regardless of the
damping ζ.
The mass m then remains stationary while the supporting case
moves with the vibrating body.
Such instruments are called seismometers
One of the disadvantages of the seismometer is its large size.
Seismometer
size large Reguire 1 , 2.5 r seismomete →≈
⟩=
y
z
nωω
Dr. Millerjothi, BITS Pilani, Dubai Campus
Accelerometer: instrument with high natural frequency
When the natural frequency of the instrument is high compared
to that of the vibration to be measured, the instrument
indicates acceleration.
The factor approaches unity for
ω/ωn → 0, so that
Accelerometer
Dr. Millerjothi, BITS Pilani, Dubai Campus
The useful range of the accelerometer can be seen from figure
below for various values of ζ
Dr. Millerjothi, BITS Pilani, Dubai Campus
The useful frequency range of the undamped accelerometer is
somewhat limited.
With ζ = 0.7, the useful frequency range is 0 ≤ ω/ωn ≤ 0.20
with a maximum error less than 0.01 percent. Thus, an instrument
with a natural frequency of 100 Hz has a useful frequency
range from 0 to 20Hz with negligible error.
Dr. Millerjothi, BITS Pilani, Dubai Campus
To reproduce a complex wave without changing its shape, the
phase of all harmonic components must remain unchanged with
respect to the fundamental.
This requires that the phase angle be zero or that all the
harmonic components must be shifted equally.
The first case of zero phase shift corresponds to ζ = 0 for
ω/ωn < 1.
The second case of an equal time- wise shift of all harmonics is
nearly satisfied for ζ = 0.7 for ω/ωn = 1.
When ζ = 0.7 for ω/ωn < 1 can be expressed as
Phase distortion