Bda31103 Lect03 -2 Dof Part2
66
7 Assume harmonic motion. Let, ( ) ( ) [ ] φ ω ω φ ω + − = ⇒ + = t A A x x t A A x x sin sin 2 2 1 2 1 2 1 2 1 & & & & .......(2) Substitute (2) into (1); 0 ) ( ) ( 0 0 2 1 3 2 2 2 2 1 2 1 2 2 2 1 = + − − + + − − A A k k k k k k A A m m ω ω
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2 DEGGREE OF FREEDOM
Transcript of Bda31103 Lect03 -2 Dof Part2
7
Assume harmonic motion. Let,
( )
( )[ ]φωω
φω
+−
=
⇒
+
=
tA
A
x
x
tA
A
x
x
sin
sin
2
2
1
2
1
2
1
2
1
&&
&&
.......(2)
Substitute (2) into (1);
0)(
)(
0
0
2
1
322
221
2
1
2
2
2
1 =
+−
−++
−
−
A
A
kkk
kkk
A
A
m
m
ωω
8
.......(3)
For non-trivial solution, matrix [ ] = 0
=
−+−
−−+
0
0
2
1
2
2322
2
2
121
A
A
mkkk
kmkk
ωω
02
2322
2
2
121 =
−+−
−−+
ωω
mkkk
kmkk
Note:
Cramer’s Rule; 02221
1211 =
AA
AA
( )( ) ( )( ) 012212211 =− AAAA
9
{ }{ } 0))((
)()()(
2
23221
132221
4
21
=−+++
+++−
kkkkk
mkkmkkmm ωω
A
ACBBx
cBxAx
x
Substitute
mm
kkkkkk
m
kk
m
kk
2
4
0
,
0)()()(
2
2
2
21
1332212
2
32
1
214
−±−=
=++
=
=
+++
++
+−
ω
ωω
10
The roots are;
2/12
21
2
23221
21
132221
21
1322212
2
2
1
))((4
)()(
2
1
)()(
2
1,
−++
−
+++
±
+++
=
mm
kkkkk
mm
mkkmkk
mm
mkkmkkωω
The roots are called natural frequencies of the
system.
22
11
frequency natural second
frequency naturalfirst
n
n
ωω
ωω
==
==
.......(4)
11
Equation (3) can be shown as below;
=
−+−
−−+
0
01
1
22
2322
2
2
121
AA
mkkk
kmkk
ωω
Replace, rA
A=
1
2
=
−+−
−−+
0
012
2322
2
2
121
rmkkk
kmkk
ωω
12
.......(5)
( ) ( )( ) 0)1(
0)1(
2
2
2322
12
2
121
=−++−
=−+−+
rmkkk
rkmkk
ω
ω
Amplitude ratio;
2
2232
22
2
2
11211
ω
ω
mkk
kr
k
mkkr
−+=
−+=
13
From equation (2), the solution will be (1 DOF),
)sin(1
)(
)sin(1
)(
22
2
22
11
1
11
φω
φω
+
=
+
=
tr
Atx
tr
Atx
.......(6)
)sin(1
)sin(1
)(
)(22
2
211
1
1
2
1 φωφω +
++
=
t
rAt
rA
tx
tx
)sin()sin()(
)sin()sin()(
222211112
2221111
φωφω
φωφω
+++=
+++=
trAtrAtx
tAtAtx .......(7)
Apply initial condition to get constant value;
2121 ,,, φφAA
The general solution is (2 DOF),
14
Example:
21 xx >
Find the free vibration response of the system
shown with k1 = k2 = k3 = k, m1 = m2 = m for the
initial conditions
assume0)0(,0)0(
0)0( ,1)0(
22
11
==
==
xx
xx
&
&
15
2221223
1121211
)(
)(
xmxxkxk
xmxxkxk
maF
&&
&&
=−+−
=−−−
=+→ ∑
FBD;
Rearrange;
0)(
0)(
1223222
2212111
=−++
=−++
xkxkkxm
xkxkkxm
&&
&&………...(1)
16
In matrix form;
=
+−
−++
0
0
)(
)(
0
0
2
1
322
221
2
1
2
1
x
x
kkk
kkk
x
x
m
m
&&
&&.......(1)
17
Assume harmonic motion. Let,
( )
( )[ ]φωω
φω
+−
=
⇒
+
=
tA
A
x
x
tA
A
x
x
sin
sin
2
2
1
2
1
2
1
2
1
&&
&&
.......(2)
Substitute (2) into (1);
0)(
)(
0
0
2
1
322
221
2
1
2
2
2
1 =
+−
−++
−
−
A
A
kkk
kkk
A
A
m
m
ωω
18
.......(3)
For non-trivial solution, matrix [ ] = 0
=
−+−
−−+
0
0
2
1
2
2322
2
2
121
A
A
mkkk
kmkk
ωω
02
2322
2
2
121 =
−+−
−−+
ωω
mkkk
kmkk
Note:
Cramer’s Rule; 02221
1211 =
AA
AA
( )( ) ( )( ) 012212211 =− AAAA
19
{ }{ } 0))((
)()()(
2
23221
132221
4
21
=−+++
+++−
kkkkk
mkkmkkmm ωω
A
ACBBx
cBxAx
x
Substitute
mm
kkkkkk
m
kk
m
kk
2
4
0
,
0)()()(
2
2
2
21
1332212
2
32
1
214
−±−=
=++
=
=
+++
++
+−
ω
ωω
20
The roots are;
2/1
21
2
23221
2
21
132221
21
1322212
2
2
1
))((4
)()(
2
1
)()(
2
1,
−++
−
+++
+++
=
mm
kkkkk
mm
mkkmkk
mm
mkkmkk
m
ωω
.......(4)
21
Given,
mmm
kkkk
==
===
21
321
From equation (4),
m
k
m
k±=
22ω
The two natural frequencies are;
m
k
m
k 3& 21 == ωω
22
Equation (3) can be shown as below;
=
−+−
−−+
0
01
1
22
2322
2
2
121
AA
mkkk
kmkk
ωω
Replace, rA
A=
1
2
=
−+−
−−+
0
012
2322
2
2
121
rmkkk
kmkk
ωω
23
.......(5)
( ) ( )( ) 0)1(
0)1(
2
2
2322
12
2
121
=−++−
=−+−+
rmkkk
rkmkk
ω
ω
Amplitude ratio;
2
2232
22
2
2
11211
ω
ω
mkk
kr
k
mkkr
−+=
−+=
24
To determine the mode shape, from (5),
atm
k=1ω
12
1 =−
=k
kkr
Motion at first mode,
First mode =
1
1
25
atm
k32 =ω
132
2 −=−
=kk
kr
Motion at second mode,
Second mode =
−1
1
26
From equation (2), the solution will be (1 DOF),
)sin(1
)(
)sin(1
)(
22
2
22
11
1
11
φω
φω
+
=
+
=
tr
Atx
tr
Atx
.......(6)
)sin(1
)sin(1
)(
)(22
2
211
1
1
2
1 φωφω +
++
=
t
rAt
rA
tx
tx
)sin()sin()(
)sin()sin()(
222211112
2221111
φωφω
φωφω
+++=
+++=
trAtrAtx
tAtAtx .......(7)
Apply initial condition to get constant value;
2121 ,,, φφAA
The general solution is (2 DOF),
27
Substitute value into equation (7),
)sin()sin()(
)sin()sin()(
222211112
2221111
φωφω
φωφω
+++=
+++=
trAtrAtx
tAtAtx
)sin()1()sin()1()(
)sin()sin()(
2221112
2221111
φωφω
φωφω
+−++=
+++=
tAtAtx
tAtAtx
21 & rr
)sin()sin()(
)sin()sin()(
2221112
2221111
φωφω
φωφω
+−+=
+++=
tAtAtx
tAtAtx.......(8)
28
Apply initial condition,
)sin()sin(0
)sin()sin()(
222111
2221112
φωφω
φωφω
+−+=
+−+=
tAtA
tAtAtx
0)0(,0)0(
0)0( ,1)0(
22
11
==
==
xx
xx
&
&
From equation (8),
)sin()sin(1
)sin()sin()(
222111
2221111
φωφω
φωφω
+++=
+++=
tAtA
tAtAtx
.......(9)
.......(10)
29
Equation (9) - (10),
1
1sin2
1
φ=A
From equation (8), the velocity equation & initial
condition,
.......(11)2
2sin2
1
φ=A
222111
222211111
coscos0
)cos()cos()(
φωφω
φωωφωω
AA
tAtAtx
+=
+++=&
.......(12)
.......(13)222111
222211112
coscos0
)cos()cos()(
φωφω
φωωφωω
AA
tAtAtx
−=
+−+=&
From equation (12) and (13),
0&0 2121 ≠≠≠≠ AAωω
30
So,
Where m & n are odd numbers.
From equation (11),
0coscos 21 == φφ
2&
221
πφ
πφ mn ==
2
121 == AA
31
Substitute value of into equation (8),
So,
)sin()sin()(
)sin()sin()(
2221112
2221111
φωφω
φωφω
+−+=
+++=
tAtAtx
tAtAtx
+−
+=
++
+=
2
3sin
2
1
2sin
2
1)(
2
3sin
2
1
2sin
2
1)(
2
1
ππ
ππ
tm
kt
m
ktx
tm
kt
m
ktx
+
−+
+
=
2sin
1
1
2
1
2sin
1
1
2
1
)(
)(
2
1 ππt
m
kt
m
k
tx
tx
Or,
21,2121 ,,,, φφωω AA
32
−
=
+
=
tm
kt
m
ktx
tm
kt
m
ktx
3cos
2
1cos
2
1)(
3cos
2
1cos
2
1)(
2
1
tm
kt
m
k
tx
tx 3cos
1
1
2
1cos
1
1
2
1
)(
)(
2
1
−+
=
Or,
Or,
OTHER EXAMPLE
33
34
Deriving 2-DOF mathematical model
35
2221223
1121211
)(
)(
xmxxkxk
xmxxkxk
maF
&&
&&
=−+−
=−−−
=∑
Free Vibration of an Undamped 2-DOF
System
36
Free Vibration of an Undamped 2-DOF
System
� These are coupled eqns of motion. For normal mode of oscillation, each mass undergoes harmonic motion of same freq passing through the equilibrium position.
� In matrix form, (1)
)1(0)(
0)(
1223222
2212111
=−++
=−++
xkxkkxm
xkxkkxm
&&
&&
0)(
)(
0
0
2
1
322
221
2
1
2
1 =
+−
−++
x
x
kkk
kkk
x
x
m
m
&&
&&
37
Free Vibration of an Undamped 2-DOF
System
� Assume harmonic motion. Let,
( )
)sin((
2
)sin(
2
2
1
2
1
2
1
2
1
φωω
φω
+−
=
+
=
tX
X
x
x
tX
X
x
x
&&
&&
38
Subst. (2) into (1)
)3(0)(
(
0)(
)(
0
0
2
1
2
2322
2
2
121
2
1
322
221
2
1
2
2
2
1
=
−+−
−−+
=
+−
−++
−
−
X
X
mkkk
kmkk
or
X
X
kkk
kkk
X
X
m
m
ω
ω
ω
ω
39
� For non-trivial solution,
0))(())((
0
':
12212211
2221
1211
=−
=
AAAA
AA
AA
RulesCramerNote
{ }{ } 0
)(
)(det
32
2
22
221
2
1 =
++−−
−++−
kkmk
kkkm
ω
ω
40
or{ }
{ } 0))((
)()()(
2
23221
132221
4
21
=−+++
+++−
kkkkk
mkkmkkmm ωω
A
ACBBx
cBxAx
x
Substitute
mm
kkkkkk
m
kk
m
kk
2
4
0
,
0)()()(
2
2
2
21
1332212
2
32
1
214
−±−=
=++
=
=
+++
++
+−
ω
ωω
41
which is called the frequency or characteristic
equation. Hence the roots are:
The roots are called natural frequencies of the system.
2/1
21
2
23221
2
21
132221
21
1322212
2
2
1
))((4
)()(
2
1
)()(
2
1,
−++
−
+++
+++
=
mm
kkkkk
mm
mkkmkk
mm
mkkmkk
m
ωω
22
11
frequency natural second
frequency naturalfirst
n
n
ωω
ωω
==
==
42
The normal modes of vibration corresponding to ω12 and ω2
2 can
be expressed, respectively, as
)(
)(
)(
)(
)3(
32
2
22
2
2
21
2
21
)2(
1
)2(
22
32
2
12
2
2
21
2
11
)1(
1
)1(
21
kkm
k
k
kkm
X
Xr
kkm
k
k
kkm
X
Xr
From
++−=
++−==
++−=
++−==
ωω
ωω
To determine the values of X1 and X2,
=
=
=
=)2(
12
)2(
1
)2(
2
)2(
1)2(
)1(
11
)1(
1
)1(
2
)1(
1)1( and Xr
X
X
XX
Xr
X
X
XX
rr
Known as the modal vectors of the system.
43
Modal vector
First mode
Second mode
.)2(
2
)2(
12
)1(
2
)1(
1121
XandXasatand
XandXastocorrespondXandXofValues
ω
ω
44
Initial conditions. The initial conditions are
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(
=
+
+=
=
=
+
+=
=
φω
φω
φω
φω
tXr
tX
tx
txtx
tXr
tX
tx
txtx
r
r
The free vibration solution or the motion in time can be
expressed itself as
0)0(,)0(
,0)0(constant, some )0(
2
)(
12
1
)(
11
====
=====
txXrtx
txXtx
i
i
i
&
&
45
Thus the components of the vector can be expressed
as
)cos()cos(
)()()(
)4()cos()cos(
)()()(
22
)2(
1211
)1(
11
)2(
2
)1(
22
22
)2(
111
)1(
1
)2(
1
)1(
11
φωφω
φωφω
+++=
+=
+++=
+=
tXrtXr
txtxtx
tXtX
txtxtx
where the unknown constants can be determined
from the initial conditions:
46
Substituting into Eq.(4) leads to
)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(
)5(sinsin)0(
coscos)0(
φωφω
φφ
φωφω
φφ
XrXrx
XrXrx
XXx
XXx
−−=
+=
−−=
+=
&
&
47
The solution can be expressed as
−−
=
−−−
=
−+−
=
−−
=
)(
)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
ωφ
ωφ
φφ
&&&&
48
from which we obtain the desired solution
{ }[ { } ]{ } { }
{ }[ { } ]{ } { }
+−−
=
=
−+−
=
=
−++−
−=
+=
−−+−
−=
+=
−−
−−
)0()0([
)0()0(tan
cos
sintan
)0()0([
)0()0(tan
cos
sintan
)0()0()0()0(
)(
1
sincos
)0()0()0()0(
)(
1
sincos
2112
2111
2
)2(
1
2
)2(
11
2
2121
2121
1
)1(
1
1
)1(
11
1
2/1
2
2
2
2112
211
12
2/12
2
)2(
1
2
2
)2(
1
)2(
1
2/1
2
1
2
2122
212
12
2/12
1
)1(
1
2
1
)1(
1
)1(
1
xxr
xxr
X
X
xxr
xxr
X
X
xxrxxr
rr
XXX
xxrxxr
rr
XXX
ωφφ
φ
ωφφ
φ
ω
φφ
ω
φφ
&&
&&
&&
&&
49
Example:
Find the free vibration response of the system
shown with k1 = 30, k2 = 5, k3 = 0, m1 = 10, and
m2 = 1 for the initial conditions
).0()0()0( ,1)0( 2211 xxxx && ===
50
2221223
1121211
)(
)(
xmxxkxk
xmxxkxk
maF
&&
&&
=−+−
=−−−
=+→ ∑
FBD;
Rearrange;
0)(
0)(
1223222
2212111
=−++
=−++
xkxkkxm
xkxkkxm
&&
&&………...(1)
51
In matrix form;
=
+−
−++
0
0
)(
)(
0
0
2
1
322
221
2
1
2
1
x
x
kkk
kkk
x
x
m
m
&&
&&.......(1)
Assume harmonic motion. Let,
( )
( )[ ]φωω
φω
+−
=
⇒
+
=
tA
A
x
x
tA
A
x
x
sin
sin
2
2
1
2
1
2
1
2
1
&&
&&
.......(2)
52
Assume harmonic motion. Let,
( )
( )[ ]φωω
φω
+−
=
⇒
+
=
tA
A
x
x
tA
A
x
x
sin
sin
2
2
1
2
1
2
1
2
1
&&
&&
.......(2)
Substitute (2) into (1);
0)(
)(
0
0
2
1
322
221
2
1
2
2
2
1 =
+−
−++
−
−
A
A
kkk
kkk
A
A
m
m
ωω
53
=
−+−
−−+
0
0
2
1
2
2322
2
2
121
A
A
mkkk
kmkk
ωω
.......(3)
=
−−
−−
0
0
55
51035
2
1
2
2
A
A
ωω
=
−−
−−
0
0
55
510352
2
ωω
0))(())((
0
':
12212211
2221
1211
=−
=
AAAA
AA
AA
RulesCramerNote
54
( ) ( )( )[ ] 025535503510 24 =−++− ωω
A
ACBBx
x
Substitute
2
4
01508510
,
2
2
2
−±−=
=+−
=
ωω
ω
01508510 24 =+− ωω
55
)10(2
)150)(10(4)85()85(,
2
2
2
2
1
−−±−−=ωω
01508510 2 =+− ωω
75.125.4,
20
35
20
85,
2
2
2
1
2
2
2
1
±=
±=
ωω
ωω
4494.2,5811.1
6,5.2
21
2
2
2
1
==
==
ωω
ωω
The two natural frequencies are;
56
Equation (3) can be shown as below;
=
−−
−−
0
01
55
510352
2
rωω
( ) ( )( ) 05)1(5
05)1(1035
2
2
1
2
=−+−
=−+−
r
r
ω
ω
57
.......(5)
( ) ( )( ) 05)1(5
05)1(1035
2
2
1
2
=−+−
=−+−
r
r
ω
ω
Amplitude ratio;
2
2
2
2
11
5
5
5
1035
ω
ω
−=
−=
r
r
58
To determine the mode shape, from (5),
at 5.22
1 =ω
25
)5.2(10351 =
−=r
Motion at first mode,
First mode =
2
1
59
at 62
2 =ω
565
52 −=
−=r
Motion at second mode,
Second mode =
−5
1
60
)sin()sin()(
)sin()sin()(
222211112
2221111
φωφω
φωφω
+++=
+++=
trAtrAtx
tAtAtx .......(7)
The general solution is (2 DOF),
Substitute value into equation (7),21 & rr
)sin()5()sin()2()(
)sin()sin()(
2221112
2221111
φωφω
φωφω
+−++=
+++=
tAtAtx
tAtAtx
)sin(5)sin(2)(
)sin()sin()(
2221112
2221111
φωφω
φωφω
+−+=
+++=
tAtAtx
tAtAtx.......(8)
61
Apply initial condition,
2211
2221112
sin5sin20
)sin()sin()(
φφ
φωφω
AA
tAtAtx
−=
+−+=
0)0(,0)0(
0)0( ,1)0(
22
11
==
==
xx
xx
&
&
From equation (8),
2211
2221111
sinsin1
)sin()sin()(
φφ
φωφω
++=
+++=
AA
tAtAtx
.......(9)
.......(10)
62
Equation (9) X 2 - (10),
1
1sin7
5
φ=A
From equation (8), the velocity equation & initial
condition,
.......(11)2
2sin7
2
φ=A
222111
222211111
coscos0
)cos()cos()(
φωφω
φωωφωω
AA
tAtAtx
+=
+++=&
.......(12)
.......(13)222111
222211112
cos5cos20
)cos(5)cos(2)(
φωφω
φωωφωω
AA
tAtAtx
−=
+−+=&
From equation (12) and (13),
0&0 2121 ≠≠≠≠ AAωω
63
So,
Where m & n are odd numbers.
From equation (11),
0coscos 21 == φφ
2&
221
πφ
πφ mn ==
7
2,
7
521 == AA
64
Substitute value of into equation (8),
So,
)sin(5)sin(2)(
)sin()sin()(
2221112
2221111
φωφω
φωφω
+−+=
+++=
tAtAtx
tAtAtx
+−
+=
++
+=
24494.2sin
7
10
25811.1sin
7
10)(
24494.2sin
7
2
25811.1sin
7
5)(
2
1
ππ
ππ
tttx
tttx
21,2121 ,,,, φφωω AA
65
Or,
Or,
( ) ( )
( ) ( )tttx
tttx
4494.2cos7
105811.1cos
7
10)(
4494.2cos7
25811.1cos
7
5)(
2
1
−=
+=
( ) ( )tttx
tx4494.2cos
5
1
7
25811.1cos
2
1
7
5
)(
)(
2
1
−+
=
+
−+
+
=
24494.2sin
5
1
7
2
25811.1sin
2
1
7
5
)(
)(
2
1 ππtt
tx
tx
Or,
QUESTION:
Determine the natural frequencies and mode shapes for
system,
66
m1 = m2 = m
k1 = k2 = k
)0()0( ),0()0(
)0()0( ),0()0(
2222
1111
xtxxtx
xtxxtx
&&
&&
====
====
67
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 tω ω= =
2 2
1 1 2 2sin( ) sin( )x A t x A tω ω ω ω= − = −&& &&
68
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
ω
ω
=
=2
1 1
2
2 2
sin( )
sin( )
x A t
x A t
ω ω
ω ω
= −
= −
&&
&&
21 1 2 2 11
22 2 2 22
( ) 00
00
A k k k Am
A k k Am
ωω
+ − − + = −−
2
1 2 1 2 1
222 2 2
( ) 0
0
k k m k A
Ak k m
ω
ω
+ − − = − −
Can be solved only if
2
1 2 1 2
2
2 2 2
( )0
k k m k
k k m
ω
ω
+ − − =
− −
69
[ ]
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
ω
ω
ω ω
+ − − =
− −
− + + + + − =
[ ] [ ]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 mω ω
+ + ± + + − + − =
1 1
2 2
first natural frequency
first natural frequency
n
n
ω ω
ω ω
= =
= =
70
2
1 2 1 1 2 1
222 2 2 1
( ) 0
0
n
n
k k m k A
Ak k m
ω
ω
+ − − = − −
1at first natural frequency = nω ω
2 (1) (1)
1 2 1 1 1 2 2( ) 0nk k m A k Aω + − − =
2(1)
1 2 1 121 (1)
1 2
( ) nk k mAr
A k
ω+ −= =
2 (2) (2)
1 2 1 2 1 2 2( ) 0nk k m A k Aω + − − =
2(2)
1 2 1 222 (2)
1 2
( ) nk k mAr
A k
ω+ −= =
Modal vector1 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
71
k
k
x1
x2
m
m
For a case
[ ]2 4 2 2( ) 3 0m km kω ω− + =
2 2 2 2 2 22
1 2 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
ω
ω
ω
+ − += =
= + = +
=
=
[ ]4 2 2
1 2 1 2 2 2 1 1 2 2 2( ) ( ) ( ) 0mm k k m k m k k k kω ω− + + + + − =m1 = m2 = m
k1 = k2 = k
72
Mode shapes
(1) 2
2
(1)
1
(2 ) (1.618) ( / )0.618
A k m k m
A k
−= = −
(1) 2
2
(1)
1
(2 ) (0.618) ( / )1.618
A k m k m
A k
−= =
1.0
-0.618
1.0
1.618
1
1.0
0.618φ
=
− 2
1.0
1.618φ
=
