Download - Dynamics and Vibration, Wahab,2008-ch6
Dynamics & Vibration
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© John Wiley & Sons, Ltd
A harmonic motion is a periodic motion that varies in a sinusoidal way.
Displacement
Velocity
Acceleration
Simple harmonic motion
tAx nsin
tAdt
dxx nn cos.
tAdt
xdx nn sin2
2
2..
-0.15
-0.1
-0.05
0
0.05
0.1
0.15
0 2 4 6 8 10 12 14
Time (sec)D
isp
lace
men
t (m
)
Chapter 6: Free Vibration Of A Single Degree Of Freedom System (page 266)
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A cycle of vibration: is a measure of the motion of a vibrating system, from equilibrium (x=0) to a maximum, then to equilibrium again, then to a minimum and back to equilibrium.
Amplitude of vibration: is the maximum displacement of a vibrating system from its equilibrium position.
A period of oscillation: is the time taken to complete one cycle
A frequency of oscillation: is the number of cycles per unit time and is calculated as the inverse of the period of oscillation.
Definition
-0.15
-0.1
-0.05
0
0.05
0.1
0.15
0 2 4 6 8 10 12 14
Time (sec)
Dis
pla
cem
ent
(m)
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A natural frequency of a system: is the frequency at which the system is left to vibrate freely without the action of any external forces.
A phase angle of two harmonic motions: is the shift of the maximum values between two harmonic motions.
A Single Degree Of freedom (SDOF) system is the simplest vibrating system, which in general consists of a mass, a spring and a damper
Definition
x
k
c
f
m
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Free body diagram
Undamped free vibrationEquation of motion (page 268)
+x
kxm
x
k
m
..
xmFx ..
xmkx
Applying Newton’s 2nd Law
Rearrange
0..
kxxm 02..
xx n
Where the angular frequency of the system in rad/s is
m
kn
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Free body diagram
Undamped free vibrationEquation of motion
Applying Newton’s 2nd Law
From static equilibrium
Substituting in the equation of motion
x
k
m
mg
)( oxk
+x
m
mg
ok
+x
m
Static equilibrium
..
)( xmxkmg o
0 okmg o is the static deflection.
0....
kxxmxmkx
Which is the same equation of motion as before
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Undamped free vibrationEquation of motion using Energy method (page 270)
Total Energy in the general position
The total energy becomes
The static deflection is
Which is again the same equation of motion as before
x
k
m
k
m
(a) Equilibrium position (b) General
position
Datum
222.
2
1)(
2
1
2
1oo kxkmgxxmSEPEKE
22.
2
1
2
1kxxmSEPEKE
kmg
o
Conservation of energy
.333222111 constSEPEKESEPEKESEPEKE
02
1
2
1 22.
kxxm
dt
d00
...... kxxmxxkxxm
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The solution of the equation of motion is:
Time response (page 271)
)0( 2..
xx n
tBtAx nn sincos
Where A and B are constants
oxx oxx..
Initial conditions: at t=0,
tBtAx nnnn cossin
.
AxBAx oo 0sin0cos
n
o
nonnox
BBxBAx
.
..
0cos0sin
tx
txx nn
o
no
sincos
.
Which gives a final form of:
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The solution of the equation of motion is:
Time response)0( 2
..
xx n
tBtAx nn sincos
Where A and B are constants
oxx oxx..
Initial conditions: at t=0,
tBtAx nnnn cossin
.
AxBAx oo 0sin0cos
n
o
nonnox
BBxBAx
.
..
0cos0sin
tx
txx nn
o
no
sincos
.
Which gives a final form of: )sin( tCx nOr
2.
2
n
o
o
xxC
o
no
n
o
o
x
x
x
x.
1.
1 tantan
Where
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Time response
C is the maximum amplitude of vibration and
is the phase angle between the initial displacement and initial velocity
.
-1.5
-1
-0.5
0
0.5
1
1.5
0 0.2 0.4 0.6 0.8 1 1.2 1.4
Time (sec)
Am
plitu
de (m
m)
nn
2
nn
2 = period of motion in seconds (s)
2
1 n
nnf = natural frequency in Hz
(1 Hz = 1 cycle per second)
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Example 6.1: Undamped free vibration: wind turbine (page 274)
.
Wind turbine
m
k54m
Tower cross-section
0.3m0.5m
The rotor and hub mass is 15 tons
The tower is made of steel that has a Young’s modulus of 200 GPa.
Ignoring the tower own weight, determine:
a) Natural Frequency
b) The time response due to initial transverse displacement c) The maximum values of velocity and acceleration.
Solution:
The moment of inertia I is:
434444 m1067035.2)3.05.0(64
)(64
io DDI
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Example 6.1: Undamped free vibration: wind turbine
.
Wind turbine
m
k54m
Tower cross-section
0.3m0.5m
N/m1.1017554
1067035.210200333
39
3
L
EIk
The stiffness k of a cantilever beam is:
3
3
L
EI
u
Fk
rad/s 8236.01015
1.101753
m
kn
Hz131.02
nnf
The time response is: )sin( tCx n
m1.001.0 22
2.
2
n
o
o
xxC
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Example 6.1: Undamped free vibration: wind turbine
.
20
1.0tantan 1
.1
n
n
o
o
x
x
)2
8236.0sin(1.0
tx
-0.15
-0.1
-0.05
0
0.05
0.1
0.15
0 2 4 6 8 10 12 14
Time (sec)
Dis
pla
ce
me
nt
(m)
)2
8236.0cos(8236.01.0)cos(. ttCx nn
Velocity
Maximum velocity
m/s082.08236.01.0max
.
nCx Acceleration
)2
8236.0sin(8236.01.0)sin( 22.. ttCx nn
Maximum acceleration 222
max
..
m/s068.08236.01.0 nCx
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Example 6.2: Undamped free vibration : lift (page 277)
Natural frequency of the cage should not exceed 2.5 Hz x
E=200GPa
m=1100kg
L=50mA lift cage has a mass of 1100 kg
Starts its motion with an initial velocity of 3 m/s
.
Cable made of steel
Determine the minimum cross-section area of the cable and the amplitude of vibration
Solution:
A
P ELo
The stiffness k is
AA
L
EAPk
L
E
A
P
o
o 99
10450
10200
)(o
Pk
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Example 6.2: Undamped free vibration : lift
x
E=200GPa
m=1100kg
L=50m
.
69
1085.671100
10425.2
A
A
m
kn m2=67.85mm2
)sin( tCx n
time response
m19.025.2
30
22
2.2
n
oo
xxC
the maximum amplitude
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A pendulum (page 280)
mg
o
sinmg cosmg
L
+x
+y
The moment equivalent Newton’s 2nd law of motion ..
oo IM oI is the mass moment of inertia about ‘o’
..
Is the angular acceleration..
0sin ImgL
Rearranging and assuming small oscillation angle
0..
mgLIo Or 02..
n
Where
on I
mgL is the angular frequency in rad/s.
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A pendulum
mg
o
sinmg cosmg
L
+x
+y
The mass moment of inertia is 2
go mRI
where gR is the radius of Gyration
For a simple pendulum LRg
L
g
L
gLn
2
22gg
nR
gL
mR
mgL
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A hinged pin support at B
A mass m is attached at C
Write the equation of motion for the bell crank for small oscillation angle
Solution:
k
l2
l1
m
A
B
C
kxo
l2
l1
mg
A
Rx
C
Ry
B
Static equilibrium
From static equilibrium
012 lklmgM oB
1
2
l
l
k
mgo
Example 6.5: Undamped free vibration: bell crank (page 282)
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Example 6.5: Undamped free vibration: bell crank
l2
l1
A C
Vibrating system
l2
l1
mg
A
Rx
C
Ry
B
Free body diagram
)( 1 olk
From the free body diagram ..
112 )( BoB IllklmgM
..
11
212 )( BIl
l
l
k
mglklmg
Using 1
2
l
l
k
mgo
02
1..
BI
kl
Rearranging
2
2mlIB Where
m
k
l
l
l
l
m
knn
2
1
2
2
1..2
0
22
21
..
ml
kl
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A rotor on a fixed shaft (page 283)
k,G,IP
L
T,
J
The moment equation equivalent to Newton’s 2nd law of motion ..
JT T
..
Where is the sum of the torques
is the torsional angular acceleration.
J is the mass moment of inertia of the rotor (or disc)
L
GITk p
From static torsional analysis, the stiffness k is given by
Where G is the shear modulus of the shaft material and Ip is the polar moment of inertia of the shaft.
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A rotor on a fixed shaft
k,G,IP
L
T,
J
Where
The sum of the torques is ..
Jk
0..
kJ
Or
02..
n
J
kn is the angular frequency in rad/s.
LJ
GI pn
L
GIk pUsing gives
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Example 6.6: Undamped free vibration:torsional vibration (page 284)
4m
BrakeRotor
The rotor has a mass of 12 tons and a radius of gyration of 18m
A uniform hollow shaft of inner and outer diameters 0.2 m and 0.4 m
A brake is installed at a distance of 4m behind the rotor
Determine the period of free torsional vibration of the rotor when the break is stopping the rotor rotation
Solution:
The polar moment of inertia of the shaft is
434444 m10356.2)2.04.0(32
)(32
iop DDI
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Example 6.6: Undamped free vibration:torsional vibration
4m
BrakeRotor
The rotor mass moment of inertia J is
38880001812000 22 mRJ kg.m2
The angular frequency is
415.338880004
10356.21077 39
LJ
GI pn rad/s
The period of free vibration is:
84.1415.3
22 n
n s
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Viscous Damped Free VibrationEquation of motion (page 286)
The damping force produced by a viscous damper is equal to
.
xc
where c is the viscous damping coefficient in N.s/m.
+x
kx
.
xc
Free body diagram
From the free body diagram, Newton’s 2nd law is: ...
xmxckx
Rearranging
0...
kxxcxm Or 02 2...
xxx nn
Wherem
kn and is the damping ratio and is equal to:
nm
c
2
Defining the critical damping coefficient, cc, as nc mc 2
becomescc
c
x
k
m
c
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Time response (page 287)
+x
kx
.
xc
Free body diagram
The solution of the equation of motion is: tAex
Differentiating twice with respect to time and substitute in the equation of motion
02 22 nn
The roots are
)1( ),1( 22
21 nn
By adding the two solutions together
tt eAeAx 2121
Or tt nn eAeAx )1(2
)1(1
22
x
k
m
c
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Time response
tt nn eAeAx )1(2
)1(1
22 There are three categories of damped motion that can be defined a) Overdamped system 1
b) Critically damped system 1
c) Underdamped system 1
x
k
m
c
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Overdamped system (page 287)
1
1 2 and are real negative numbers
The time response will be in the form atAex
where a is a positive real number
tt nn eAeAx )1(2
)1(1
22
oxx oxx..
Applying initial conditions at t=0, and
21 AAxo
22
12
.
)1()1( AAx nno
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Overdamped system
Solving for A1 and A2 and substituting in the displacement time response equation
t
n
noot
n
noonn e
xxe
xxx
)1(
2
2.
)1(
2
2.
22
12
)1(
12
)1(
Displacement time response of an overdamped system
0
0.2
0.4
0.6
0.8
1
1.2
0 2 4 6 8 10 12
Time (sec)
Am
plit
ud
e (
mm
)
25
20The higher the damping ratio , the longer the time to decay
12
For heavily overdamped system, is much larger than 1,
)2
1(2
2
..
t
n
o
n
oo
nexx
xx
The displacement time response becomes
© John Wiley & Sons, Ltd
Critically damped system (page 289)
The displacement time response is the form
0
0.2
0.4
0.6
0.8
1
1.2
0 1 2 3 4 5 6
Time (sec)
Am
plit
ud
e (
mm
)
1
n 21
tnetAAx )( 21
oxx oxx..
Applying initial conditions at t=0, and
1Axo
12
.
AAx no
Solving for A1 and A2 and substituting in the displacement time response equation
tonoo
netxxxx ))((.
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Underdamped system (page 290)
1 012 and 22 11 i where 1i
The time response will be in the form baba eee Using the relationship
ttiti nnn eeAeAx
22 12
11
Defining the damped natural frequency d21 nd
The time response becomes
ttiti ndd eeAeAx 21
Using Euler formula xixe ix sincos
tdddd
netitAtitAx )sin(cos)sin(cos 21
tt nn eAeAx )1(2
)1(1
22
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Underdamped system
oxx oxx..
Applying initial conditions at t=0, and
Or tdd
netAtAx sincos 43
Where A3 and A4 are new constants
Another form for the equation
)sin( tDex dtn
Where 24
23 AAD
4
31tanA
A
3Axo
34
.
AAx ndo
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Underdamped system
Solving for A3 and A4 and substituting in the displacement time response equation
td
d
ono
donet
xxtxx
sincos
.
The period of motion in seconds is defined as
-0.8
-0.6
-0.4
-0.2
0
0.2
0.4
0.6
0.8
1
0 1 2 3 4 5 6
Time (s)
Dis
pla
cem
ent
(mm
)
1x2x
dd
2
)sin( tDex dtn
dd
2
The ratio of two successive amplitudes
))(sin(
)sin(
1)(
1
2
1
1
1
ddt
dt
tDe
tDe
x
xdn
n
dnex
x 2
1
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Underdamped system
-0.8
-0.6
-0.4
-0.2
0
0.2
0.4
0.6
0.8
1
0 1 2 3 4 5 6
Time (s)
Dis
pla
cem
ent
(mm
)
1x2x
dd
2
)sin( tDex dtn
Defining a logarithmic decrement
dnx
x 2
1ln
dd
2Using and
21 nd
22)2(
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Example 6.7: Damped free vibration I (page 293)
x
k
m
c
Mass = 300 kg
Damped period of vibration is 1.2s
The ratio between the two maximum amplitudes in two successive cycles is
15
1
1
2 x
x
Determine:a)The logarithmic decrement.b)The damping ratio.c)The damped angular frequency.d)The undamped angular frequency.e)The critical damping constant.f)The damping coefficient.g)The stiffness.
© John Wiley & Sons, Ltd
Example 6.7: Damped free vibration I
x
k
m
c
a) The logarithmic decrement
708.215lnln2
1 x
x
b) The damping ratio
3958.0708.2)2(
708.2
)2( 2222
c) The damped angular frequency.
236.5667.12.1
22.1
2
dd
d rad/s
d) The undamped angular frequency
7.53958.01
667.1
1 22
d
n rad/s
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Example 6.7: Damped free vibration I
x
k
m
c
f) The damping coefficient.
e) The critical damping constant.
N.s/m34207.530022 nc mc
N.s/m64.135334203958.0 ccc
g) The stiffness k
N/m97477.5300 22 nn mkm
k
© John Wiley & Sons, Ltd
Example 6.7: Damped free vibration I
0 1 2 3 4 5 6
Time
Dis
pla
cem
ent
1x
2x3x
Damped natural frequency is 1 Hz
The ratio between its maximum amplitude in the first cycle and that after two cycles is
81
3x
x
Find the damping ratio
Solution:
))2(sin(
)sin(
1)2(
1
3
1
1
1
ddt
dt
tDe
tDe
x
xdn
n
dnex
x 2
3
1 8ln2ln3
1 dnx
x
As dn 03972.12
8lnthus
The damping ratio is 163.003972.1)2(
03972.1
)2( 2222
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