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Mathcad - CMS (Component Mode Synthesis) Analysis.xmcd
Component Mode Synthesisby Julio C. Banks, P.E.
The treatment of large structural systems may be simplified by dividing the system into smallersystems called components . The components are related through the displacement, and force conditions at their junction points. Each component is represented by mode shapes (orfunctions) . The sum of the component mode shape functions allows the satisfaction of thedisplacement and force conditions at the junctions [1].
α 1 m 1 αL2
L1= L1 1 L2 αL1
Component 1: M1 11
5 M1 2
1
6
M2 1 M1 2 M2 21
7
Component 2: M3 3 1.0α M3 41
2α M3 5
1
5α
M4 3 M3 4 M4 41
3α M4 5
1
6α
M5 3 M3 5 M5 4 M4 5 M5 51
9α
Component 3: M6 6 α
Julio C. Banks, PE [email protected] page 1 of 11
Mathcad - CMS (Component Mode Synthesis) Analysis.xmcd
M
0.2000
0.1667
0.0000
0.0000
0.0000
0.0000
0.1667
0.1429
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
1.0000
0.5000
0.2000
0.0000
0.0000
0.0000
0.5000
0.3333
0.1667
0.0000
0.0000
0.0000
0.2000
0.1667
0.1111
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
1.0000
mL1
E 1 I 1 K1 1 4 K1 2 6 K2 1 K1 2
K2 2 12 K5 5 28.81
α3
K6 6 0
K
4.00
6.00
0.00
0.00
0.00
0.00
6.00
12.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
28.80
0.00
0.00
0.00
0.00
0.00
0.00
0.00
E I
L13
1
0
2
2
1
0
3
6
0
1
0
0
0
1
1
α
0
0
1
4
α
12
α2
1
0
0
0
p1
p2
p3
p4
P5
P6
0=
Apply the nonzero diagonal criterion to select the independent (generalized) coordinates, p.
1
0
2
2
0
1
0
0
0
1
1
α
0
0
1
4
α
12
α
p1
p3
p4
P5
1
0
3
6
1
0
0
0
p2
p6
=
Julio C. Banks, PE [email protected] page 2 of 11
Mathcad - CMS (Component Mode Synthesis) Analysis.xmcd
Or S p Qq=
where S
1
0
2
2
0
1
0
0
0
1
1
α
0
0
1
4
α
12
α2
= and Q
1
0
3
6
1
0
0
0
=
Let p2 q1= and p6 q2= therefore,
1
0
0
2
2
0
0
1
0
0
0
0
0
0
1
0
0
0
0
0
1
1
α
0
0
0
0
1
4
α
12
α2
0
0
0
0
0
0
1
p1
p2
p3
p4
P5
P6
1
1
0
3
6
0
1
0
0
0
0
1
q1
q2
= or S' p Q' q=
where S'
1
0
0
2
2
0
0
1
0
0
0
0
0
0
1
0
0
0
0
0
1
1
α
0
0
0
0
1
4
α
12
α2
0
0
0
0
0
0
1
and Q'
1
1
0
3
6
0
1
0
0
0
0
1
Therefore, p T q= Where T S'( )1Q'
1.000
1.000
2.000
2.333
0.333
0.000
1.000
0.000
2.500
2.667
0.167
1.000
Julio C. Banks, PE [email protected] page 3 of 11
Mathcad - CMS (Component Mode Synthesis) Analysis.xmcd
A TT
MT1.18
1.48
1.48
3.18
B TT
K T7.200
3.600
3.600
4.800
C A1B
11.40
4.19
2.80
0.20
λ sort eigenvals C( )( )1.37
10.23
ω λ
1.172
3.198
E I
mL14
Calculate the eigenvectors Nm length λ( ) 2
i 1 Nm
Φi eigenvec C λi
The mode shapes in normal coordinates is Φ0.2693
0.9631
0.9225
0.3859
Normalized mode shapes Φn Vnorm Φ( )0.280
1.000
1.000
0.418
First and Second Mode Shapes
Reference"Theory of Vibration with Applications, 5th Ed.", Thomson, W. T., and Marie Dillon Dahleh.1.Prentice Hall.ISBN 0-13-651068-X, Pp. 341 through 346.Julio C. Banks MSME Thesis - "Component Synthesis Methods for Vibrating Systems". Tufts2.University, Medford Massachusetts, May 1984.
Julio C. Banks, PE [email protected] page 4 of 11
Mathcad - CMS (Component Mode Synthesis) Analysis.xmcd
Appendix A
1
0
2
2
1
0
3
6
0
1
0
0
0
1
1
0
0
1
4
12
1
0
0
0
p1
p2
p3
p4
P5
P6
0=
Since the total number of coordinates used are six and there are four constraint equations, the number of generalized coordinates for the system is two (i.e., there are four superfluouscoordinates corresponding to the four constraint equations. We can thus choose any two (the first example uses the nonzero diagonal criterion) of the generalized coordinates, q. Let p1 = q1, and p6 = q6 be the generalized coordinates, and express p1 ..p6 in terms of q1, and q6according to the following steps:
1
0
3
6
0
1
0
0
0
1
1
α
0
0
1
4
α
12
α2
p1
p3
p4
P5
1
0
2
2
1
0
0
0
q2
q6
= or S p Qq=
Let p1 q1=
where S
1
0
3
6
0
1
0
0
0
1
1
α
0
0
1
4
α
12
α2
= and Q
1
0
2
2
1
0
0
0
=
and p6 q2=
or S' p Q' q=
1
0
0
3
6
0
0
1
0
0
0
0
0
0
1
0
0
0
0
0
1
1
α
0
0
0
0
1
4
α
12
α2
0
0
0
0
0
0
1
p1
p2
p3
p4
P5
P6
1
1
0
2
2
0
0
1
0
0
0
1
q1
q2
=
Julio C. Banks, PE [email protected] page 5 of 11
Mathcad - CMS (Component Mode Synthesis) Analysis.xmcd
For α 1.00
Where S'
1
0
0
0
0
0
0
1
0
3
6
0
0
0
1
0
0
0
0
0
1
1
α
0
0
0
0
1
4
α
12
α2
0
0
0
0
0
0
1
and Q'
1
1
0
2
2
0
0
1
0
0
0
1
p T q= Where T S'( )1Q'
1.00
1.00
2.00
2.33
0.33
0.00
0.00
1.00
4.50
5.00
0.50
1.00
A TT
MT1.1774
2.6614
2.6614
7.3206
B TT
K T7.200
10.800
10.800
19.200
C A1B
15.60
4.19
18.20
3.99
λ sort eigenvals C( )( )1.37
10.23
ω λ
1.172
3.198
E I
mL14
Julio C. Banks, PE [email protected] page 6 of 11
Mathcad - CMS (Component Mode Synthesis) Analysis.xmcd
Example
gc glb
lbf γ 0.280
lbf
in3
γ ρg
gc= ⇒ ρ γ
gc
g0.280
lb
in3
Do 1.315in Di 1.049in L1 10in E 30 106 psi
Area: Aπ
4Do
2Di
2 0.4939 in2
Moment of Inertia: Iπ
64Do
4Di
4 8.734 102 in
4
Thickness: tDo Di
20.133 in
V A L1 4.94 in3 m ρA 0.1383
lb
in
ω' ωE I
mL14
1.172
3.198
E I
mL14
fω'
2π
159.6
435.4
Hz
FEM Validation:
ANSYS FEA f'159.2
433.0
Hz error
f' ff
0.24
0.55
%
CAEFEM FEA f'165
470
Hz error
f' ff
3.4
8.0
%
Results CommentaryANSYS model uses 2D Beam Elements, while CAEFEM model utilized 3D Beam Elements (That is,ANSYS will most closely follow the closed-form solution since the latter is 2D. On the other hand,CAEFEM model had to be restrained in the out-of-plane dimension in order to emulate a 2D planeframe (CAEFEM has 3D beam elements only).
In general, the natural frequencies can be expressed as a function of = L2/L1 . The choices ofdependent, and independent coordinates follows those chosen in the reference. The results areidentical.
Julio C. Banks, PE [email protected] page 7 of 11
Mathcad - CMS (Component Mode Synthesis) Analysis.xmcd
The mass matrix: The Stiffness matrix:
M α( )
1
5
1
6
0
0
0
0
1
6
1
7
0
0
0
0
0
0
α
1
2α
1
5α
0
0
0
1
2α
1
3α
1
6α
0
0
0
1
5α
1
6α
1
9α
0
0
0
0
0
0
α
K α( )
4
6
0
0
0
0
6
12
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
28.8
α3
0
0
0
0
0
0
0
Recall S' α( )
1
0
0
2
2
0
0
1
0
0
0
0
0
0
1
0
0
0
0
0
1
1
α
0
0
0
0
1
4
α
12
α2
0
0
0
0
0
0
1
and Q'
1
1
0
3
6
0
1
0
0
0
0
1
p T q= Where T α( ) S' α( )( )1Q' T α( )
1.000
1.000
2.000
2.333
0.333
0.000
1.000
0.000
2.500
2.667
0.167
1.000
Julio C. Banks, PE [email protected] page 8 of 11
Mathcad - CMS (Component Mode Synthesis) Analysis.xmcd
α 1
A α( ) T α( )T
M α( ) T α( ) B α( ) T α( )T
K α( ) T α( )
A α( )1.1774
1.4840
1.4840
3.1753
B α( )7.2000
3.6000
3.6000
4.8000
C α( ) A α( )1
B α( )
C α( )11.40
4.19
2.80
0.20
λ sort eigenvals C α( )( )( )1.37
10.23
ω λ
E I
mL14
1.172
3.198
E I
mL14
fω
2π
159.6
435.4
Hz
Calculate the eigenvectors
Nm length λ( ) 2
i 1 Nm
Φi eigenvec C α( ) λi
The mode shapes in normal coordinates is Φ0.2693
0.9631
0.9225
0.3859
The mode shapes in physical coordinates is D T α( ) Φ
1.232
0.269
1.869
1.940
0.071
0.963
1.308
0.923
0.880
1.123
0.243
0.386
Julio C. Banks, PE [email protected] page 9 of 11
Mathcad - CMS (Component Mode Synthesis) Analysis.xmcd
Normalized mode shapes Dn Vnorm D( )
0.635
0.139
0.964
1.000
0.036
0.496
1.000
0.705
0.673
0.859
0.186
0.295
Julio C. Banks, PE [email protected] page 10 of 11
Mathcad - CMS (Component Mode Synthesis) Analysis.xmcd
Appendix BDefine two (2) procedures to unit-normalize the columns of a matrix. The first algorithm determinesthe maximum magnitude in each column. The second algorithm, expands the first algorithm to the unit-normalization phase of the solution
Vmax v( ) "Determine the maximum-magnitude"
"element in each column of a matrix"
Nr rows v( )
Nc cols v( )
max v1 j
max vi j vi j maxif
i 2 Nrfor
zj max
j 1 Ncfor
zreturn
Vnorm v( ) "Determine the maximum-magnitude"
"element in each column of a matrix"
Nr rows v( )
Nc cols v( )
max v1 j
max vi j vi j maxif
i 2 Nrfor
zj max
Vnj v j
zj
j 1 Ncfor
Vnreturn
Julio C. Banks, PE [email protected] page 11 of 11