Post on 05-Apr-2018
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Two degrees of freedom
A system with two degrees of freedom requires at least
two independent co ordinates to completely specify theconfiguration of the system.
A system with two degrees of freedom will have two
natural frequencies & the vibratory motion is a
combination of two harmonics of these two natural
frequencies.
The vibratory motion corresponding to the natural
frequencies are known asPrincipal modes of vibration. If amplitude of vibration of one of the masses is unity, it
is known as Normal mode of vibration.
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Determine the equations of motion & the natural
frequencies of the two degree of freedom spring mass
system shown in fig.
1k
1m
1x
2
k
2m
x2
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For the 2 d.o.f system shown in fig, the equations of
motion for free, undamped vibrations are given by;
Which may be rearranged as,
0)(
and0)(
12222
2121111
xxkxm
xxkxkxm
0
and
0)(
122222
2212111
xkxkxm
xkxkkxm
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The solutions of the above harmonic motions may
be given by
where A, B & are arbitrary constants
Substituting for x1 & x2 in the equations of motion,
& canceling out sin(t+)we get,
These are homogeneous linear algebraicequations in A & B, whose solution is obtained by
equating the determinant of the coefficients A &
B to zero.
)sin(&)sin( 21 tBxtAx
0)(
0)(
2222
2212
1
BmkAk
BkAkkAm
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The determinant of coefficients A & B will be
Expanding the determinant gives;
The two natural frequencies of the system are
obtained by solving the above equation.
02222
2
2
121
mkk
kmkk
021
212
2
2
1
214
mm
kkmk
mkk
21
21
2
2
2
1
21
2
2
1
212
4
1
22 mm
kk
m
k
m
kk
m
k
m
kk
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Thus, the general solution of the equations of
motion is composed of two harmonic motions of
frequency 1 & 2.
The amplitude ratios are;
)sin()sin(
)sin()sin(
2221112
2221111
tBtBx
tAtAx
22
2
222
2
2121
2
2
2
12
2
122
2
1121
2
1
1
1
1
k
mk
mkk
k
B
Ak
mk
mkk
k
B
A
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Hence the general solutions finally become
)sin()sin(
)sin()sin(
222211112
2221111
tAtAx
tAtAx
The constants A1, A2, 1 & 2 can be evaluated
by the four initial conditions
)0(and)0(),0(),0( 2211 xxxx
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Prob.1 For the system discussed before,
Determine the natural frequencies , Modal
vectors & mode shapes. Also locate the nodefor each mode of vibration. Given m1=2 kg,
m2=1 kg, k1=40N/m & k2=20 N/m.
Ans: 1=3.162 rad/sec, 2=6.325 rad/sec.
22
2
11
1
11modesecondfor theratioAmplitude
15.0modefirstfor theratioAmplitude
B
A
B
A
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First Mode Second Mode
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Modal Vector & Mode shapes:
The normal modes of vibration corresponding to
1 & 2 can be expressed respectively as;
1 1
1
1 1 1
2 2
2
2 2 2
0.5( )
1.01.0
( )1.0
A Ax
B AA A
xB A
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0.5
1.0
-1.0
1.0
l
l2
l1
First Mode Second Mode
For first mode both modal displacements positive. Hence there is nonode.
For the second mode, the node (Point where amplitude is zero) may befound from the geometry of the figure.
i.e. the node is located in the middle of the two masses.
12
2
1
2
1
2
21 5.02111
1ll
l
l
l
l
l
ll
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Prob 2. Set up the equations of motion and determine
the natural frequencies & mode shapes of the two d.o.f
system shown in fig.
m m
l l l
l l l
xx1
2
T
T
T
Tsin TsinTsin
Tsin
Left mass Right mass
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1 1 2 2
Assuming the string tension T to be constant for for small oscillations,
( )From the fig, sin = , sin = , sin =
Equation of equilibrium for the left mass is
x x x x
l l l
Natural Frequencies of vibration :
1 1 2
1
1 1 21
2
1 2 22
mx sin sin 0
( ). mx 0
Equation of equilibrium for the right mass is
mx sin sin 0
( ).
2mx 0 ...........
x
(i)
m
T
T T
x x xi e T T
l lT
x xl l
T T
x x xi e T T
l l
2 1 2
2mx 0 ........ )
0
...(iiT T
x xl l
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2 2
1 2
1 2
2
A and B are arbitrary constants.
Now x sin( ) & x sin( )
Substituting in (i) & (ii) we get
2m 0.........( )
x sin( ) and x sin( )Let A t B
where
A t B t
T TA B iii
t
l l
2
2
and
2m 0.........( )
i.e2
m
2
T TA B ivl l
T T
l l
T T
l l
For a non trivial solution of A & B, the determinant of the
coefficients of A & B must be zero.
2
0
m
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2 2
2
2 2
2
2
2 2
Cross multiplying, we get,
2m 0 or
2m . Taking square root,
2 m Taking +ve sign,
2m m
T T
l l
T T
l l
T Tl l
T T
l l
2
2
or
3Taking ve sign, m or
T
lT
l
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2
11
21 11
2
2
2
22 22
Using equations (iii) & (iv) we get,
2
12
2
12
Substitut
T Tm
A l lT TB
ml l
T Tm
A l lT TB
ml l
Mode shapes :
2 2
1 1 2
1 2
1 1 2 2
3ing or and we get
1 11 1
T T T
ml ml mlA A
B B
and
2
11
21 11
2
2
2
22 22
Using equations (iii) & (iv) we get,
2
12
2
12
Substitut
T Tm
A l lT TB
ml l
T Tm
A l lT TB
ml l
Mode shapes :
2 2
1 1 2
1 2
1 1 2 2
3ing or and we get
1 11 1
T T T
ml ml mlA A
B B
and
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1 2
1 1
1
1 1 1
2 2
2
2 2 2
The normal modes of vibration corresponding to &
can be expressed respectively as;1
( )1
1( )1
A Ax
B A
A AxB A
1 1
-1
1O
l l l
l1
First Mode
Second mode
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1 1
2 2
For the first mode, the two masses move in phase(i.e. both A and B are positive).Hence there is no node.
For the second mode, the two masses move out of phase.
(i.e. A and B are
Location of node :
11
1 1 1
opposite in sign) Hence node existssuch that
11 1 2 0.5
1
. the node is located at the middle of the two masses.
l l l ll l
l l l
i e
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Prob 3:
For the double pendulum shown in fig, obtain the
natural frequencies and mode shapes.
m
m
l
l
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m
m
l
l
1x
1x
2x2x -( )
mg
T2cos
T2
sinT1
T2
mg
cosT1
T2cos
T2
sin
sinT1
Equations of Motion :
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1
2
1 2 11 2 2 2 2
1 1 1 2 2
Let the tension in the upper string be T & that in the lower
string be T .From the free body diagrams of the masses,
sin , sin , T cos
cos T cos 2
x x xT mg
l lT T mg
Equations of Motion :
1 1 1 2 2
1 2 1
1
1 1 2
sin sin 0
2 0
30............( )
mg
mx T T
x x xmx mg mg
l l
mg mgmx x x i
l l
Equilibrium equation of top bob will be
Equilibrium equation of bottom bob will be
2 2 2
2 12
2 1 2
sin 0
i.e. 0
0............( )
mx T
x xmx mg
l
mg mgmx x x ii
l l
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1 2
2 2
1 2
2
Let x sin( ) and x sin( )
sin( ) and sin( )
substituting in the equations (i) & (ii), we get
30............( )
A t B t
x A t x B t
mg mgm A B iii
l l
mg mgA
l l
2
2
2
0............( )
To obtain the characteristic equation, equating the determinant of
the co efficients of A & B to zero,
3
0
m B iv
g gl l
g g
l l
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2
2 2
2
4
2
2
1
1
Expanding the determinant,
30
4
2 0
Using equations (iii) & (iv) we get,
3
g g g
l l l
g g
l l
Solving
g
A l
gBl
Mode shapes :
2 11
2
22 22
10.414
3 0.5858
12.414
3 3 3.414
g
l
g gl l
g g
A l lg g gB
l l l
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Modal Vector & Mode shapes:
The normal modes of vibration corresponding to1 & 2 can be
expressed respectively as;
1 1
1
1 1 1
2 2
22 2 2
0.414( )
1.0
2.414( )
1.0
A Ax
B A
A Ax
B A
0.414
1.0
-2.414
1.0
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Derive the frequency equation for the system shown in
fig. Assume the cord passing over the cylinder doesnot slip.
k
r
2k1
m2
m1
X
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1
1 1
2
2 1
For the mass m , the equilibrium equation is
0
Equation of equilibrium for the cylinder is
I ( ) 0
m x k x r
k r k r x r
Equations of motion :
2
2
1where I= , the mass moment of inertia of cylinder.
2m r
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2 2
2
1 1 1
2 2
1 1 2
)
Then x= sin( ) and = sin( )
substituting in (i) & (ii) and cancelling sin( ) through out,
( - ) 0..........( )
( ) 0...
A t B t
t
k m A k rB iii
k rA k k r I B
Assume x = Asin( t + & = Bsin ( t + )
2
1 1 1
2 2
1 1 2
2
2
.......( )
The frequency equation is obtained by equating the determinant
of the coefficients of A & B to zero.
( - )0
( )
1Simplifying & substituting I= , we get,
2
iv
k m k r
k r k k r I
m r
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Two simple pendulums are connected by a spring as
shown in fig. Determine the natural frequencies of the
pendulum.
m m
LL
k
a
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2 2
1 1 1 2
2 2 2
1 1 2
2 22 2 2 1
2
Taking moments about the hinged points, we have
For the left pendulum;
( ) 0( ) 0........( )
Similarly, for the right pendulum;
( ) 0
mL mgL kamL mgL ka ka i
mL mgL ka
mL
2 22 2 1
1 2
( ) 0........( )
sin( ) and sin( )
, we get
mgL ka ka ii
Let A t Let B t
Simplifying
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2 2 2 2
2 2 2 2
2 2 2 2
1
The non trivial solution for A & B may be obtained
by equating the determinant of coefficients of A & B to zero.
0
Taking +ve sign,
ra
mgL ka mL kaka mgL ka mL
mgL ka mL ka
g
L
2
2 2
d/sec Taking -ve sign,
2rad/sec.
g ka
L mL
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SEMI DEFINITE SYSTEMS
Sometimes if one of the roots of the characteristic
frequency equation is equal to zero, it indicates thatone of the natural frequencies of the system is zero.
Such systems are known as semi definite systems.
Practically this simply means that the system will moveas a rigid body without distortion of the springs that
connect the different masses of the system.
A vibrating system whose both ends are free will be a
semi definite system.
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Two blocks m1 and m2 are connected together by a spring of
constant K are resting on a smooth horizontal surface as
shown infig. Obtain an expression for the natural frequencies
for the system.
k
m1
m2
x x21
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For the 2 d.o.f system shown in fig, the equations of
motion for free, undamped vibrations are given by;
Which may be rearranged as,
1 1 1 2
2 2 2 1
( ) 0 and
( ) 0
m x k x x
m x k x x
1 1 1 2
2 2 1 2
0 and
0
m x kx kx
m x kx kx
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The solutions of the above harmonic motions may
be given by
where A, B & are arbitrary constants
Substituting for x1 & x2 in the equations of motion,
& canceling out sin(t+)we get,
These are homogeneous linear algebraicequations in A & B, whose solution is obtained by
equating the determinant of the coefficients A &
B to zero.
)sin(&)sin( 21
tBxtAx
2
2
1
2
( ) 0
( ) 0
k m A kB
kA k m B
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The determinant of coefficients A & B will be
Expanding the determinant we get;
2
1
22
0k m k
k k m
2 2 2
1 2
4 2 2 21 2 1 2
2 2
1 2 1 2
2
12 2
1 2 1 2
1 22
1 2
( )( ) =0
( ) 0
( ) 0
Hence either 0 0.
( ) 0
( )
k m k m k
m m m m k k k
m m m m k
or m m m m k
k m m
m m
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2
1 1 112
1 1 1 1
1
1 1
1 22
1 2
2
2 11
Amplitude ratio in first mode
1 Substituting 0,
11
0
k(m )Put = , we get the amplitude ratio in second mode
m
k(m
A k mkB k m k
A k k
B k k
m
m
A k
Bk m
Mode Shapes & Modal Vectors :
2
1
22 1
1 2 2
1 1
1)1 1
m
m
m
m m
m m
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Modal Vector & Mode shapes:
The normal modes of vibration corresponding to
1 & 2 can be expressed respectively as;
1 1
1 1 1
22 2
1
2 2 2
1
1
1
A A
B A
mA A
mB A
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1 1
-1
Node
First ModeSecond Mode
m2 m
1/
x
L
2
1 2
1
2 12 2
1 1
2
1
1 2
Let the node be at a distance x from the left end;
Taking ratio of corresponding sides from the two triangle,
1( )
( )
1
mm m
L x xmx L x
m mm m
mx
m m
L x xm m m
L
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For the Torsional system shown in fig,
determine the natural frequencies, modal
vectors & location of the node.2
II1
Kt
Solution: Let
and
be the angular twists in theshaft due to rotation of the rotors.
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For the 2 d.o.f system shown in fig, the equations of
motion for free, undamped vibrations are given by;
Which may be rearranged as,
1 1 1 2
2 2 2 1
( ) 0 and
( ) 0
t
t
I k
I k
1 1 1 2
2 2 2 1
0 and0
t t
t t
I k kI k k
Th l i f h b h i i
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The solutions of the above harmonic motions may
be given by
where A, B & are arbitrary constants
Substituting for 1 & 2 in the equations of
motion, & canceling out sin(t+)we get,
These are homogeneous linear algebraicequations in A & B, whose solution is obtained by
equating the determinant of the coefficients A &
B to zero.
1 2sin( ) & sin( )A t B t
2
2
1
2
( ) 0
( ) 0
t t
t t
k I A k B
k A k I B
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The determinant of coefficients A & B will be
Expanding the determinant we get;
2
1
22
0t t
t t
k I k
k k I
2 2 2
1 2
4 2 2 21 2 1 2
2 2
1 2 1 2
2
12 2
1 2 1 2
1 22
1 2
( )( ) =0
( ) 0
( ) 0
Hence either 0 0.
( ) 0
( )
t t t
t t t
t
t
k I k I k
I I I I k k k
I I I I k
or I I I I k
k I I
I I
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2
1 1112
1 1 1 1
1
1 1
1 22
1 2
2
2
Amplitude ratio in first mode
1 Substituting 0,
11
0
k (I )Put = , we get the amplitude ratio in second mode
I
t t
t t
t
t
t
t
t
k k IAB k I k
kA
B k
I
I
kA
Bk
Mode Shapes & ModalVectors :
2
1
21 2 11
1 2 2
1 1
1k (I )1 1
It
I
I
I II
I I
M d l V t & M d h
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Modal Vector & Mode shapes:
The normal modes of vibration corresponding to
1 & 2 can be expressed respectively as;
1 1
1 1 1
22 2
1
2 2 2
1
1
1
A A
B A
IA A
IB A
/I I
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1 1
-1
Node
First Mode Second Mode
/I2 I1
2
1 2
1
2 12 2
1 1
2
1
1 2
Let the node be at a distance x from the left end;
Taking ratio of corresponding sides from the two triangle,
1( )
( )
1
II I
L x xIx L x
I II I
Ix
I I
L x xI I I
L
G S
G S
G S
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Kt1
dA
dA
dB
IA
BIL
A
LB
Pinion
Gear
Rotor A
Rotor B
LA
LA LB'
dA
IA B'I
Geared SystemsActual system
Equivalent system
dA
dA
dB
IA
BIL
A
LB
Pinion
Gear
Rotor A
Rotor B
LA
LA LB'
dA
IA B'I
Geared SystemsActual system
Equivalent system
dA
dA
dB
IA
BIL
A
LB
Pinion
Gear
Rotor A
Rotor B
LA
LA LB'
dA
IA B'I
Geared SystemsActual system
Equivalent system
Kt1
Kt2
Kt2'
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1
The fig shows a geared system with a gear ratio = .
If the inertia of gears and shafts is negligible,
the second rotor system may be replaced by an
equivalent system B' (d =d ) such tha
A
B
B A
n
t:(i) The kinetic energy of the original system
is same as that of the equivalent system.
(ii) The strain energies of the original & equivalentsystem are the same.
(1) Equating the kinetic energies :
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2 ' '2 2
2
'
K.E of original system (B) =K.E of equivalent system (B')
1 1i.e. But '
2 2Mass moment of inertia of equivalent rotor
..
B B B B B A
BB B
A
I I
I I
(1) Equating the kinetic energies :
2
2 ' ' 2
2
2'
2 2 ' 2
.......( )
Strain energy of section L Strain energy of section L'
1 1But ,
2 2Hence, Torsional stiffness of equivalent shaft
k
t B
B
B B
t B
Bt t
i
k k t
k k
(2) Equating the kinetic energies :
2 2
2 2' 2 2........( )
B
B Bt t
A
k ii
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Problem
An engine drives a centrifugal pump through a2:1 speed reducer gear box. The mass momentof inertia of the engine flywheel and the pumpimpeller are 500 kg-m2 and 60 kg-m2 respectively.
The length and diameter of the engine shaft are250 mm & 50 mm respectively. The length &diameter of the pump impeller shaft are 150 mm& 40 mm respectively. The rigidity modulus of the
shaft material is 80 Gpa. Determine the naturalfrequencies and location of node.
G d S t
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dA
d
B
IA
BIL
A
LB
Pinion
Gear
Flywheel A
Impeller B
LA
LA LB'
dA
IA B'I
Geared SystemsActual system
Equivalent system
=0.05m
=0.04m
=0.25m
=0.15m
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2 2500 , 60 ,Gear ratio n=2
Dia of engine shaft d 50 0.05
Length of engine shaft L 250 0.25
Dia of engine shaft d 40 0.04
Length of engine shaft L 150
A B
A
A
B
B
I kgm I kgm
mm m
mm m
mm m
mm
4 47 4
1
4 47 4
2
0.15
Polar Moment of inertia of engine shaft
0.05J 6.136 10
32 32Polar Moment of inertia of pump shaft
0.04J 2.5133 10
32 32
A
B
m
dm
dm
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9 7
1
9 7
2
Torsional stiffness of engine shaft
80 10 6.136 10
0.25Torsional stiffness of impeller shaft
80 10 2.5133 10
0.15Mass moment of inertia of equ
A
B
GJ
L
GJ
L
196
1
' 2
ivalent impeller
Also, Torsional stiffness of equivalent impeller shaft
BI
15 kgm
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t eq
)
Equivalent system
500 kgm2
15 kgm2
K =28625.3 Nm/radt
L=1.7148m
0 As it is a semi definite system
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1
1 22
1 2
2
1 2
0
k (I ) 28625.3 (500 15)Also = = 44.34 / sec
I 500 15
Location of node on the equivalent shaft from left side rotor
15x = 1.7148
500 15
t I radI
IL
I I
As it is a semi definite system,
0.0499 0.05 .m
1 1
-1
Node
First Mode Second Mode
xL
/I2 I1
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Systems with two degree of freedom
Forced Vibration
m1
K1
m2
K2
x1
x2
F1 tsinFF 01
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Systems with two degree of freedom
Problem-1Governing equations
Newtons method
. Force equilibrium diagram
m1
K1x1
K2(x2-x1)
11xm
m2 22xm
m1
K1
m2
K2
x1
x2
F1
tsinFF 01
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Systems with two degree of freedom
Newtons method
. From Force equilibrium diagram of massm1
tsinF)x(xKxKxm 01221111
tsinFxK)xK(Kxm 02212111 1st Eqn.
of motion
. From Force equilibrium diagram of massm2
0)x(xKxm 12222
0xKxKxm 221222 2nd Eqn. of motion
Governing equations
m1
K1x1
K2(x2-x1)
11xm
m2 22xm
tsinFF
0
1
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Solution of governing equationsIt is possible to have pure harmonic free vibration for both themasses.Let us assume )t 1x Asin(
2x Bsin(t + )
Be the forced response of the system, where A and B are theamplitudes to be obtained
The above equations have to satisfy the governing equations
of motions
0xKxKxm 221222
1st Eqn. of motion
2nd Eqn. of motion
tsinFxK)xK(Kxm 02212111
where is the forcing
frequency
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Systems with two degree of freedom
Solution of governing equations
2
1 2 1 2 0(K K ) m Asin(t ) K Bsin(t ) F sint
22 2 2K Asin(t ) (K m )Bsin(t ) 0
0)sin( tIn above equations
B 2
1 2 1 2 0(K K ) m A K F sint
22 2 2K A (K m )B 0
The above equations reduces to: (characteristic equation)
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Systems with two degree of freedom
B 2
1 2 1 2 0(K K ) m A K F sint
22 2 2K A (K m )B 0
From above equations obtain A1 and A2 by Cramers rule
Solution of governing equations
20
22 2
KF........
0 K m A
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Systems with two degree of freedom
B
201 2 1
2
F(K K ) m
0K
0
mK
K........
K
m)K(K
2
22
2
2
2
121
where is the determinant of characteristic equations
Solution of governing equations
222222121 KmKm)K(K
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m1
K1
m2
K2
x1
x2
F1
Solution of governing equations
if is the natural frequency of system-I
1
11
m
K
if is the natural frequency of system-II
2
22
m
K
We have two resonant frequencies
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Dynamic vibration absorber
m2
K2
x2
x1
m1
K1
F1
Main system
having forcedvibration
Absorbersystem
The system can be used asDynamic vibration absorber
(Tuned damper)
Choose K2, m2 such thatvibrations of mass m1 isminimized
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Dynamic vibration absorber
If the system has to be used as Dynamic vibration absorber,the amplitude of vibration of mass m1, i.e. A=0
20
2
2 2
KF0 K m
A 0
20
2
2 2
KF0
0 K m
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0mKF 2220 )(
Dynamic vibration absorber
20
2
2 2
KF0
0 K m
0mK 222
2
22
m
K
2
2
mK
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2
2
m
K
Dynamic vibration absorber
Natural frequency of absorber system
2
2
2 m
K Excitation frequency of the system isequal to natural frequency of absorber
system
This shows that if excitation frequency of the system isequal to natural frequency of absorber system, thenvibration of the main system is zero
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Systems with two degree of freedom
Undesired maximum vibrations can occur only when mainsystem is under resonance, or near it i.e. 1
So, to reduce undesired vibration of main system mass m1,we choose K2, m2 in such a way that 2 1
2
2
1
1
m
K
m
K
2
2
1
1
m
K
m
KOR
Dynamic vibration absorber
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Dynamic vibration absorber
2
0 2 2
2 2 2
1 2 1 2 2 2
F (K m )A
(K K ) m K m K
20
2
2 2
KF
0 K m A
Dynamic vibration absorber
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y
To find the natural frequencies of the entire system, equatethe denominator of A to zero.
0mK
K........
K
m)K(K
2
22
2
2
2
121
1 2
2 2
2 1 2 2
1 1 2 1
k yields,
1 1 0k m m k k k k k
2 2 21 2 1 2 2 2 (K K ) m K m K 0
Dividing the entire equation by k
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Problem-1
Dynamic vibration Absorber
A reciprocating machine of mass M runs at aconstant speed of N rpm. After it was
installed, it was found that the forcingfrequency is too close to the naturalfrequency of the machine. What dynamicabsorber should be added, if the nearestnatural frequency of the system should be atleast 25% from the impressed frequency.
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Problem-1
Systems with two degree of freedom
Machine
Vibration absorber
Data Given
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Data Given
Natural frequencies of machine must be 25% away fromthe operating speed.Hence r=0.75
4 2
4 2
2 2
1 1
1 1
2 2
we have;
(2 ) 1 0
(0.75) (2 )(0.75) 1 0
0.34028. And =
where m , are mass and stiffness of the main systemand m , are mass and stiffness of the absorber system
Mass of abso
r r
m k
m k
kk
Hence
1
1
rber =0.34028 m 0.34028
Stiffness of absorber =0.34028k
M
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Problem-2
A machine part vibrates with large amplitudeswhen the compressor speed is 250 rpm. Tostudy the vibration, a springmass system is
suspended from the machine part to act as anabsorber. A 2kg absorber mass tuned to 250cpm resulted in two resonant frequencies of200 cpm and 300 cpm. Determine the
equivalent spring and mass constants for themachine part.
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