me4213 mdof 2
DESCRIPTION
ME4213 MDOF 2TRANSCRIPT
ME 4213
Vibration of Multi-Degree-of-Freedom (MDOF)
Systems – eigenvalues and eigenvectors
H.P. LEEDepartment of Mechanical Engineering
EA-05-20Email: [email protected]
Semester 2 2014/2015
ME 4213
Matrices
M M T
A matrix M is defined to be symmetric if
A symmetric matrix M is positive definite if
xTMx 0 for all nonzero vectors x
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Matrices A symmetric positive definite matrix M can be factored
Here L is upper triangular, called a Cholesky matrix
M LLT
ME 4213
Matrices
The matrix square root is the matrix M 1/2 such that
M 1/2M 1/2 M
If M is diagonal, then the matrix square root is just the root
of the diagonal elements:
L M 1/2 m1 0
0 m2
(4.35)
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Matrices For a symmetric, positive
matrix M
11
2 2
111 1 1/ 2
1 12
1/ 2 1/ 2
1/ 2 1/ 2 1/ 2 1/ 2
identity symmetric
000, ,
00 0
Let ( ) ( ) and multiply by :
( ) ( ) (4.38)
mm
m m
I K
mM M M
m
t M t M
M MM t M KM t
x q
q q 0
1/ 2 1/ 2or ( ) ( ) where
is called the mass normalized stiffness and is similar to the scalar
used extensively in single degree of freedom analysis. The key here is that
i
t K t K M KM
kK
m
K
q q 0
s a SYMMETRIC matrix allowing the use of many nice properties and
computational tools
ME 4213
Matrix equation in terms of real symmetric matrix eigenvalue problem
2
2
vibration problem real symmetric eigenvalue problem
(4.40) (4.41)
Assume ( ) in ( ) ( )
, or
j t
j t j t
t e t K t
e K e
K K
q v q q 0
v v 0 v 0
v v v v v 0
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Observations
For a nxn matrix there are n eigenvalues
The eigenvalues are all real and positive
The matrix is similar to a diagonal matrix
The set of eigenvectors are orthogonal (we will
discuss this later)
The set of eigenvectors are independent (we will
discuss this later)
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Observation about the matrices
The original mass matrix is typically symmetric,
positive definite and up to now, diagonal.
The stiffness matrix is typically positive
semidefinite, which means that they may have a
zero eigenvalue.
The stiffness matrix as well as the mass
normalized stiffness matrix are symmetric.
ME 4213
Orthogonal and Normal Vectors
x
x1
M
xn
, y
y1
M
yn
, inner product is xTy xiyii1
n
x orthogonal to y if xTy 0
x is normal if xTx 1
if a the set of vectores is is both orthogonal and normal it
is called an orthonormal set
The norm of x is x xTx
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Normalization of a vector A vector can be normalized by dividing by its norm.
x
xTx
has norm of 1
x
xTx
xT
xTx
x
xTxxTx
xTx 1
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Recap on the previous example
1 13 31/ 2 1/ 2
2
2 2
1 1 2 2
0 27 3 0
0 1 3 3 0 1
3 1 so which is symmetric.
1 3
3- -1det( ) det 6 8 0
-1 3-
which has roots: 2 and 4
K M KM
K
K I
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Continue ,,,normalized eigenvectors
1 1
11
12
11 12 1
2 11 2
1
( )
3 2 1 0
1 3 2 0
10
1
(1 1) 1
11
12
K I
v
v
v v
v 0
v
v
v
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The second normalized eigenvector
v2 1
2
1
1
, v1
Tv2
1
2(11) 0
v1
Tv1
1
2(11) 1
v2
Tv2
1
2(1 (1)(1)) 1
v i are orthonormal
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Mode u and eigenvector v
u1 v1 and u2 v2
x M 1/2q u M 1/2
v
Note
M 1/2u1
3 0
0 1
13
1
1
1
v1
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Orthonormal set of vectors
1 2
1 1 1 2
2 1 2 2
1 2 1 1 2 2
1 2 21 1 1 2 1 2
1 2
21 2 1 2 2 2
1 0
0 1
0diag( , )
0
T T
T
T T
T T T
T T
T T
P
P P I
P KP P K K P
v v
v v v v
v v v v
v v v v
v v v v
v v v v
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Some terms that you need to know
P is known as the modal matrix
P is a orthogonal matrix
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To continue with the previous example
P v1 v1 1
2
1 1
1 1
PTP 1
2
1
2
1 1
1 1
1 1
1 1
1
2
11 11
11 11
1
2
2 0
0 2
I
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frequency
2
1
2
2
1 1 3 1 1 11 1
1 1 1 3 1 12 2
1 1 2 41
1 1 2 42
4 0 2 0 01
0 8 0 42 0
TP KP
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Conclusion
2diag diag( ) (4.48)T
i iP KP
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Another example
Equations of motion
1 1 1 2 1 2 2
2 2 2 1 2 3 2
( ) 0 (4.49)
( ) 0
m x k k x k x
m x k x k k x
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Matrix form for the equations of motion
1 2 21
2 2 32
00 (4.50)
0
k k km
k k km
x x
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Numerical example
m1 1 kg, m2 4 kg, k1 k3 10 N/m and k2 =2 N/m
1/ 2 1/ 2
2
1 2
1 2
1 0 12 2,
0 4 2 12
12 1
1 12
12 1det det 15 35 0
1 12
2.8902 and 12.1098
1.7 rad/s and 12.1098 ra
M K
K M KM
K I
d/s 3.48
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eigenvector
1
11
21
11 21
1
2 2 2 2 2
1 11 21 11 11
11
For equation (4.41 ) becomes:
12 - 2.8902 1 0
1 3- 2.8902
9.1089
Normalizing yields
1 (9.1089)
0.
v
v
v v
v v v v
v
v
v
21
1 2
1091, and 0.9940
0.1091 0.9940, likewise
0.9940 0.1091
v
v v
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Orthogonal matrix
1 2
0.1091 0.9940
0.9940 0.1091
0.1091 0.9940 12 1 0.1091 0.9940 2.8402 0
0.9940 0.1091 1 3 0.9940 0.1091 0 12.1098
0.1091 0.9940 0.1091 0.9940
0.9940 0.1091 0.9940 0.109
T
T
P
P KP
P P
v v
1 0
1 0 1
It shows that P is an orthogonal matrix
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Another note on v
In the previous section, we could have chosed v2 to be
v2 0.9940
0.1091
instead of v2
-0.9940
0.1091
because one can always multiple an eigenvector by a constant
and if the constant is -1 the result is still a normalized vector.
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Conclusion The procedure for finding the eigenvalues and
eigenvectors have been presented.
We have learnt the concept of modal vectors, normal vectors, modal matrix, normal matrix as well as their properties.
The most important property is that the modal or the normal matrix is an orthogonal matrix.
Or the modal vectors are orthogonal to each others.
The eigenvectors are independent.