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NOISE AND VIBRATION
BDC 4013
Lecture 2 : Two Degree of FreedomLecture 2 : Two Degree of FreedomSystemsSystems
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1.Introduction
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2. 2-DOF System (a basic understanding
in multi-DOF)
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3.Deriving 2-DOF mathematical
model Equation of Motion
F1 F2
m1 m2
k1x1
1 1c x&
k3x2
3 2c x&
k2 (x2-x1)
( )2 2 1c x x& &
1x&&
2x&&
1F 2F
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m1k1x1
1 1c x&
k2 (x2-x1)( )2 2 1c x x& &
1x&&1F
m23 2
c x&
k2 (x2-x1)
( )2 2 1c x x& &
2x&&
2F
1 1 1 1 1 1 1 2 2 1 2 2 1
1 1 1 2 1 2 2 1 2 1 2 2 1
( ) ( )
( ) ( )
m x F k x c x k x x c x x
m x c c x c x k k x k x F
= + +
+ + + + =
&& & & &
&& & &
2 2 2 2 2 1 2 2 1 3 2 3 2
2 2 2 1 2 3 2 2 1 2 3 2 2
( ) ( )
( ) ( )
m x F k x x c x x k x c x
m x c x c c x k x k k x F
=
+ + + + =
&& & & &
&& & &
k3x2
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1 1 1 2 1 2 2 1 2 1 2 2 1( ) ( )m x c c x c x k k x k x F + + + + =&& & &
2 2 2 1 2 3 2 2 1 2 3 2 2( ) ( )m x c x c c x k x k k x F + + + + =&& & &
Rearrange in matrix notation
1 1 1 1
2 2 2 2
x x x F
x x x F
+ + =
&& &
&& &
( )( )
( )( )
1 2 2 1 2 21 1 1 1 1
2 2 3 2 2 32 2 2 2 2
00
c c c k k k m x x x F
c c c k k k m x x x F
+ + + + = + +
&& &
&& &
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General Multi-DOF Equation
[ ]{ [ ]{ } [ ]{ { }M x C x K x F+ + =&& &
[ ] [ ] [ ]
{ } { } { }
, and are symmetric matrices n x n
x , and are vectors n row
n is the number of the degrees of freedom
n is the number of natural frequencies
M C K
x x&& &
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4.Free Vibration of an Undamped 2-DOF
System
k2
k1
x1
x2
m1
m2
1 1 1 1 2 2 1 1
1 1 1 2 1 2 2
( ) 0
( ) 0
m x k x k x x F
m x k k x k x
+ = =
+ + =
&&
&&
2 2 2 2 1 2
2 2 2 1 2 2
( ) 0
0
m x k x x F
m x k x k x
+ = =
+ =
&&
&&
1 1 1 2 1 2 2
2 2 2 1 2 2
1 1 1 2 2 1
2 2 2 2 2
( ) 0
0
can be written in matrix
0 ( ) 0
0 0
m x k k x k x
m x k x k x
m x k k k x
m x k k x
+ + =
+ =
+ + =
&&
&&
&&
&&
1 1 2 2sin( ) sin( )x A t x A t = =
2 2
1 1 2 2sin( ) sin( )x A t x A t = = && &&
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1 1 1 2 2 1
2 2 2 2 2
0 ( ) 0
0 0
m x k k k x
m x k k x
+ + =
&&
&&
1 1
2 2
sin( )
sin( )
x A t
x A t
=
=
2
1 1
2
2 2
sin( )
sin( )
x A t
x A t
=
=
&&
&&2
1 1 2 2 11
2
2 2 2 22
( ) 00
00
A k k k Am
A k k Am
+ + =
2
1 2 1 2 1
222 2 2
( ) 0
0
k k m k A
Ak k m
+ =
Can be solvedonly if
2
1 2 1 2
2
2 2 2
( )0
k k m k
k k m
+ =
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[ ]
2
1 2 1 2
2
2 2 2
4 2 2
1 2 1 2 2 2 1 1 2 2 2
( )
0
( ) ( ) ( ) 0
k k m k
k k m
m m k k m k m k k k k
+ =
+ + + + =
[ ] [ ]2 2
1 2 2 2 1 1 2 2 2 1 1 2 1 2 2 22 2
1 2
1 2
( ) ( ) 4( ) ( ),
2( )
k k m k m k k m k m m m k k k k
m m
+ + + + + =
1 1
2 2
first natural frequency
first natural frequency
n
n
= =
= =
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2
1 2 1 1 2 1
22
2 2 2 1
( ) 0
0
n
n
k k m k A
Ak k m
+ =
1at first natural frequency = n
2 (1) (1)
1 2 1 1 1 2 2( ) 0nk k m A k A + =
2(1)
1 2 1 121 (1)
1 2
( )n
k k mAr
A k
+ = =
2 (2) (2)
1 2 1 2 1 2 2( ) 0nk k m A k A + =
2(2)1 2 1 22
2 (2)
1 2
( )n
k k mAr
A k
+ = =
Modal vector
1 1
1 1 1
1 1
2 1 1
A A
A r A
= =
( ) ( )
( )
( ) ( )A
(2) (2)
(2) 1 1
(2) (2)
2 2 1
A A
A r A
= =
A
First mode
Second mode
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k
k
x1
x2
m
m
Example 1;
Draw FBD, determine equation of motion and find thenatural frequencies of the system shown below
[ ]2 4 2 2( ) 3 0m km k + =
2 2 2 2 2 22
12 2
2
2
1
2
3 9 4 3 5
2 2
3 5 3 5
2 4 2 4
1.618
0.618
n
n
n
km k m m k km m k
m m
k k k k
m m m m
k
m
k
m
+ += =
= + = +
=
=
[ ]4 2 21 2 1 2 2 2 1 1 2 2 2( ) ( ) ( ) 0m m k k m k m k k k k + + + + =
m1 = m2 = m
k1 = k2 = k
Given;
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Free vibration with initial conditions
For a specific condition, the system can be made tovibrate in its ith normal mode (i = 1, 2) by subjecting it to
the specific initial conditions
.0)0(
,)0(
,0)0(
constant,some)0(
2
)(
12
1
)(
11
==
==
==
===
tx
Artx
tx
Atx
i
i
i
&
&
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However, for any other general initial conditions, bothmodes will be excited. The resulting motion can be
obtained by a linear superposition of the two normalmodes
)()()( 2211 txctxctx rrr
+=
)cos()cos()()()(
)cos()cos()()()(
22
)2(
1211
)1(
11
)2(
2
)1(
22
22
)2(
111
)1(
1
)2(
1
)1(
11
+++=+=
+++=+=
tArtArtxtxtx
tAtAtxtxtx
asexpressedbecanvectortheofcomponentstheThus
.generalityoflossnowithchoosecanweand
constantsunknowntheinvolvealreadyandSinceconstants.areandwhere
)(
1,
)()(
21
)2(
1
)1(
1
)2()1(
21
tx
ccAA
txtx
cc
r
rr
==
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:conditionsinitialthefrom
determinedbecanandconstantsunknownthewhere 21)2(
1
)1(
1 ,, AA
)0()0(),0()0(
)0()0(),0()0(
2222
1111
xtxxtx
xtxxtx
&&
&&
====
====
2
)2(
1221
)1(
1112
2
)2(
121
)1(
112
2
)2(
121
)1(
111
2
)2(
11
)1(
11
sinsin)0(
coscos)0(
sinsin)0(
coscos)0(
ArArx
ArArx
AAx
AAx
=
+=
=
+=
&
&
Four algebric equation
with unknowns..
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Example 2. Free Vibration Response of a Two
DOF System
).0()0()0(,1)0( 2211 xxxx && ===
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 andc1 = c2 = c3 = 0 for the initial conditions
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Solution: For the given data, the eigenvalue
problem, Eq.(5.8), becomes
(E.100
55-53510
00
2
1
2
2
2
1
32
2
22
221
2
1
=
+
+
=
++
++
X
X
X
X
kkmk
kkkm
or
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{ }
{ } )8.5(0)(0)(
2322
212
22121
2
1
=++
=++
XkkmXk
XkXkkm
{ }
{ } )9.5(0))((
)()()(
223221
132221
4
21
=+++
+++
kkkkk
mkkmkkmm
)15.5()cos()cos(
)()()(
)cos()cos()()()(
22
)2(
1211
)1(
11
)2(2
)1(22
22
)2(
111
)1(
1
)2(
1
)1(
11
+++=
+=
+++=+=
tXrtXr
txtxtx
tXtXtxtxtx
Related equations involved.
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Related equations involved.
)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
&&&&
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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,
Example 1 Solution
(E.2)01508510
24=+
E.3)(4495.2,5811.1
0.6,5.2
21
2
2
2
1
==
==
The normal modes (or eigenvectors) are given by
E.5)(5
1
E.4)(2
1
)2(
1)2(
2
)2(
1)2(
)1(
1)1(
2
)1(
1)1(
X
X
XX
XX
X
X
=
=
=
=
r
r
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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 1.Solution
(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
+==
===
===
+===
&
&
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while the solution of Eqs.(E.10) and (E.11) leads to
The solution of Eqs.(E.8) and (E.9) yields
Example 1. Solution
(E.12)
7
2cos;
7
5cos 2
)2(
11
)1(
1 == XX
(E.13)0sin,0sin 2)2(
11
)1(
1 == XXEquations (E.12) and (E.13) give
(E.14)0,0,7
2,7
521
)2(
1
)1(
1 ==== XX
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Example 1.Solution
Thus the free vibration responses of m1 and m2 are given
by
(E.16)4495.2cos7
105811.1cos7
10)(
(E.15)4495.2cos7
25811.1cos
7
5)(
2
1
tttx
tttx
=
+=
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