vibrations 2dof ppt
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vibrations 2dof pptTRANSCRIPT
MDPN471 & MDP410
Mechanical Vibrations CHAPTER
5 Two-Degree-Of-Freedom
Vibration
2-DOF Vibration 5.1 Introduction
• Two degree of freedom systems are defined as systems that
require two independent coordinates to describe their motion.
Consider the Automobile
Suspension system.
2-DOF Vibration 5.1 Introduction
• Consider the motor-pump system.
– The vertical displacement of the system and the angular
coordinate denoting the rotation of the mass about its C.G. make
up two independent coordinates.
Motor-pump system
2-DOF Vibration
No. of degrees of freedom of the system =
No. of masses in the system X No. of possible types of motion of each mass
Packaging of an instrument
5.1 Introduction
2-DOF Vibration 5.2 Eqs of Motion for Forced Vibration
Consider a viscously damped two degree of freedom
spring-mass system.
A two degree of freedom spring-mass-damper system
2-DOF Vibration 5.2 Eqs of Motion for Forced Vibration
)2.5()()(
)1.5()()(
2232122321222
1221212212111
Fxkkxkxccxcxm
Fxkxkkxcxccxm
)3.5( )()(][)(][)(][ tFtxktxctxm
Both equations can be written in matrix form as
The application of Newton’s second law of motion to
each of the masses gives the equations of motion:
where [m], [c], and [k] are called the mass, damping,
and stiffness matrices, respectively, and are given by
2-DOF Vibration 5.2 Eqs of Motion for Forced Vibration
322
221
322
221
2
1
][
][
0
0 ][
kkk
kkkk
ccc
cccc
m
mm
)(
)()(
)(
)()(
2
1
2
1
tF
tFtF
tx
txtx
And the displacement and force vectors are given
respectively:
It can be seen that the matrices [m], [c], and [k]
are symmetric:
)()(][)(][)(][ tFtxktxctxm
2-DOF Vibration 5.2 Eqs of Motion for Forced Vibration
][][],[][],[][ kkccmm TTT
•The solution of Eqs.(5.1) and (5.2) involves four
constants of integration (two for each equation).
•We shall first consider the free vibration solution
It can be seen that the matrices [m], [c], and [k]
are symmetric:
2-DOF Vibration 5.3 Free Vib. Analysis of Undamped System
)5.5(0)()()()(
)4.5(0)()()()(
2321222
2212111
txkktxktxm
txktxkktxm
)6.5()cos()( ),cos()( 2211 tXtxtXtx
Assuming solutions are harmonic motion of m1
and m2 at the same frequency ω and the same
phase angle Φ, we take the solutions as
By setting F1(t) = F2(t) = 0, and damping disregarded,
i.e., c1 = c2 = c3 = 0, and the equation of motion is
reduced to:
)2.5()()(
)1.5()()(
2232122321222
1221212212111
Fxkkxkxccxcxm
Fxkxkkxcxccxm
2-DOF Vibration 5.3 Free Vib. Analysis of Undamped System
)7.5( 0)cos()(
0)cos()(
232
2
212
22121
2
1
tXkkmXk
tXkXkkm
2
1 1 2 1 2 2
2
2 1 2 2 3 2
( ) 0
( ) 0 (5.8)
m k k X k X
k X m k k X
For non-trivial solution, thus,
Substituting into Eqs.(5.4) and (5.5),
2-DOF Vibration 5.3 Free Vib. Analysis of Undamped System
0)(
)(det
21
2
22
221
2
1
kkmk
kkkm
)9.5(0))((
)()()(
2
23221
132221
4
21
kkkkk
mkkmkkmmor
• These represent two simultaneous homogenous algebraic
equations in the unknown X1 and X2.
• For trivial solution, i.e., X1 = X2 = 0, there is no solution.
• For a nontrivial solution, the determinant of the coefficients of
X1 and X2 must be zero. Therefore,
2-DOF Vibration 5.3 Free Vib. Analysis of Undamped System
)10.5())((
4
)()(
2
1
)()(
2
1,
2/1
21
2
23221
2
21
132221
21
1322212
)2(
2
)1(
mm
kkkkk
mm
mkkmkk
mm
mkkmkk
The roots are called natural frequencies of the system.
which is called the frequency or characteristic equation.
The roots are:
2-DOF Vibration 5.3 Free Vib. Analysis of Undamped System
Remember that:
To find x1(t) and x2(t), we need to find X1, X2, and .
But at which ??
)cos()( ),cos()( 2211 tXtxtXtx
)()()(
)()()(
)2(
2
)1(
22
)2(
1
)1(
11
txtxtx
txtxtx
By a linear superposition:
2-DOF Vibration 5.3 Free Vib. Analysis of Undamped System
)2(
1
)2(
2)2(
)1(
1
)1(
2)1(
X
Xr
X
Xr
Substituting the natural frequencies into Eq 5.8,
)8.5(0)(
0)(
232
2
212
22121
2
1
XkkmXk
XkXkkm
we can determine the values of X1 and X2 at each mode,
)11.5()(
)(
)(
)(
32
2
22
2
2
21
2
21
32
2
12
2
2
21
2
11
kkm
k
k
kkm
kkm
k
k
kkm
2-DOF Vibration 5.3 Free Vib. Analysis of Undamped System
)12.5( and )2(
1)2(
)2(
1
)2(
2
)2(
1)2(
)1(
1)1(
)1(
1
)1(
2
)1(
1)1(
Xr
X
X
XX
Xr
X
X
XX
The normal modes of vibration corresponding to ω(1)2
and ω(2)2 can be expressed, respectively, as
which are known as the modal vectors of the system.
2-DOF Vibration 5.3 Free Vib. Analysis of Undamped System
(5.13)mode second)cos(
)cos(
)(
)()(
modefirst )cos(
)cos(
)(
)()(
22
)2(
12
22
)2(
1
)2(
2
)2(
1)2(
11
)1(
11
11
)1(
1
)1(
2
)1(
1)1(
tXr
tX
tx
txtx
tXr
tX
tx
txtx
The free vibration solution or the motion in time can be
expressed itself as
)14.5()()()( )2()1( txtxtx iii
The resulting motion can be obtained by a linear
superposition of the two normal modes, Eq.(5.13)
2-DOF Vibration 5.3 Free Vib. Analysis of Undamped System
Thus the components of the vector can be expressed as
where the 4 unknown constants can be
determined from the 4 initial conditions:
)15.5()cos()cos(
)()()(
)cos()cos()()()(
)2()2(
)2(
1)2()1()1(
)1(
1)1(
)2(
2
)1(
22
)2()2(
)2(
1)1()1(
)1(
1
)2(
1
)1(
11
tXrtXr
txtxtx
tXtXtxtxtx
√ √ √ √
2-DOF Vibration 5.3 Free Vib. Analysis of Undamped System
)16.5()0()0(),0()0(
),0()0(),0()0(
2222
1111
xtxxtx
xtxxtx
)17.5(sinsin)0(
coscos)0(
sinsin)0(
coscos)0(
2
)2(
1221
)1(
1112
2
)2(
121
)1(
112
2
)2(
121
)1(
111
2
)2(
11
)1(
11
XrXrx
XrXrx
XXx
XXx
)(
)0()0(sin,
)(
)0()0(sin
)0()0(cos,
)0()0(cos
122
2112
)2(
1
121
2121
)1(
1
12
2112
)2(
1
12
2121
)1(
1
rr
xxrX
rr
xxrX
rr
xxrX
rr
xxrX
Substituting into Eq.(5.15) leads to
The solution can be expressed as
2-DOF Vibration 5.3 Free Vib. Analysis of Undamped System
1/22 2
(1) (1) (1)
1 1 1 1 1
1/22
2 2 1 2
2 1 2 2
2 1 1
1/22 2
(2) (2) (2)
1 1 2 1 2
1/22
2 1 1 2
1 1 2 2
2 1 2
(1)1 1
1
cos sin
(0) (0)1(0) (0)
( )
cos sin
(0) (0)1(0) (0)
( )
stan
X X X
r x xr x x
r r
X X X
r x xr x x
r r
X 11 2 1 2
(1)
1 1 1 2 1 2
(2)1 11 2 1 1 2
2 (2)
1 2 2 1 1 2
in (0) (0)tan
cos [ (0) (0)
sin (0) (0)tan tan (5.18)
cos [ (0) (0)
r x x
X r x x
X r x x
X r x x
from which we obtain the desired solution
2-DOF Vibration Example 5.3 Free Vibration Response of a 2DOF
).0()0()0( ,1)0( 2211 xxxx
0
0
2
1
32
2
22
221
2
1
X
X
kkmk
kkkm
Solution:
Find the free vibration response of the system shown in
Fig.5.3(a) with k1 = 30, k2 = 5, k3 = 0, m1 = 10, m2 = 1
and c1 = c2 = c3 = 0 for the initial conditions
2-DOF Vibration
(E.2)01508510 24
E.3)(4495.2,5811.1
0.6,5.2
21
2
2
2
1
E.5)(5
1
E.4)(2
1
)2(
1)2(
2
)2(
1)2(
)1(
1)1(
2
)1(
1)1(
XX
XX
XX
XX
from which the natural frequencies can be found as
By setting the determinant of the coefficient matrix in
Eq.(E.1) to zero, we obtain the frequency equation,
The normal modes (or eigenvectors) are given by
Example 5.3 Free Vibration Response of a 2DOF
2-DOF Vibration
(E.7))4495.2cos(5)5811.1cos(2)(
(E.6))4495.2cos()5811.1cos()(
2
)2(
11
)1(
12
2
)2(
11
)1(
11
tXtXtx
tXtXtx
(E.11)sin2475.121622.3)0(
(E.10)sin4495.2sin5811.10)0(
(E.9)cos5cos20)0(
(E.8)coscos1)0(
2
)2(
1
)1(
12
2
)2(
11
)1(
11
2
)2(
11
)1(
12
2
)2(
11
)1(
11
XXtx
XXtx
XXtx
XXtx
By using the given initial conditions in Eqs.(E.6) and
(E.7), we obtain
The free vibration responses of the masses m1 and m2
are given by (see Eq.5.15):
Example 5.3 Free Vibration Response of a 2DOF
2-DOF Vibration
(E.12)7
2cos;
7
5cos 2
)2(
11
)1(
1 XX
(E.13)0sin,0sin 2
)2(
11
)1(
1 XX
(E.14)0,0,7
2,
7
521
)2(
1
)1(
1 XX
while the solution of Eqs.(E.10) and (E.11) leads to
The solution of Eqs.(E.8) and (E.9) yields
Equations (E.12) and (E.13) give
Example 5.3 Free Vibration Response of a 2DOF
2-DOF Vibration
(E.16)4495.2cos7
105811.1cos
7
10)(
(E.15)4495.2cos7
25811.1cos
7
5)(
2
1
tttx
tttx
Thus the free vibration responses of m1 and m2 are
given by
Example 5.3 Free Vibration Response of a 2DOF
2-DOF Vibration 5.4 Torsional System
22312222
11221111
)(
)(
ttt
ttt
MkkJ
MkkJ
)19.5()(
)(
22321222
12212111
tttt
tttt
MkkkJ
MkkkJ
which upon rearrangement become
Consider a torsional system as shown in Fig.5.6. The
differential equations of rotational motion for the discs
can be derived as
For the free vibration analysis of the system,
Eq.(5.19) reduces to
2-DOF Vibration 5.4 Torsional System
)20.5(0)(
0)(
2321222
2212111
ttt
ttt
kkkJ
kkkJ
Figure 5.6: Torsional system with discs mounted on a shaft
2-DOF Vibration
(E.1) 02
02
2120
2110
tt
tt
kkJ
kkJ
Find the natural frequencies and mode shapes for the
torsional system shown in Fig.5.7 for J1 = J0 , J2 = 2J0
and kt1 = kt2 = kt .
Solution:
The differential equations of motion,
Eq.(5.20), reduce to (with kt3 = 0,
kt1 = kt2 = kt, J1 = J0 and J2 = 2J0):
Fig.5.7:
Torsional system
Example 5.4 Natural Frequencies of Torsional System
2-DOF Vibration
28
(E.2)2,1);cos()( itt ii
(E.3)052 2
0
22
0
4
tt kkJJ
(E.4))175(4
and)175(4 0
2
0
1J
k
J
k tt
The solution of Eq.(E.3) gives the natural frequencies
gives the frequency equation:
Rearranging and substituting the harmonic solution:
Example 5.4 Natural Frequencies of Torsional System
2-DOF Vibration
29
(E.5)4
)175(2
4
)175(2
)2(
1
)2(
22
)1(
1
)1(
21
r
r
0and2,
,,
3022011
2211
kJJmJJm
kkkkkk tttt
Equations (E.4) and (E.5) can also be obtained by
substituting the following in Eqs.(5.10) and (5.11).
The amplitude ratios are given by
Example 5.4 Natural Frequencies of Torsional System