Earthquake Engineering
MSc in Civil Engineering
Orthogonality and Normalization of Eigenvectors Orthogonality Consider the shear building with 2DOF which we have solved in the case of free vibration, that is, found the natural frequencies and mode shapes. The stiffness and mass matrix were given as:
10 − 5 N − 5 5 m
K =
and
5 0
M=
0
kg
5
The eigenvalues were found to be: λ 1 = 0.382 and λ 2 = 2.618 and the eigenvectors: X1
a 1 = 11 = a 21 1.618
and
X2
a 1 = 12 = a 22 − 0.618
It can be shown that: X1T M X2 = 0 and X1T K X2 = 0
(1)
In general, in a system with n DOF with eigenvectors X1, X2, …Xi, X j, …Xn , for any two different eigenvectors, XiT M X j = 0
and XiT K X j = 0 , for i≠j
(2)
This property of eigenvectors is known as the orthogonality. Normalization Previously, we normalized eigenvectors in the following way:
X1T ⋅ X1
=
1 + 1.618 2
= 1.902 1 / 1.902 0.526 ˆ = X = 1 1 . 618 / 1 . 902 0.851 X T2 ⋅ X 2
=
1 + 0.618 2
ˆ X 2
= 1.644 1 / 1.176 0.851 = = 0 . 618 / 1 . 176 − − 0.526
The matrix of normalised eigenvectors: 0.526 0.851 U= 0.851 − 0.526 Eigenvectors can also be normalized with respect to the mass matrix, M. Say:
Department Department of Engineering Systems
1
Earthquake Engineering
MSc in Civil Engineering
X1T M X1 = M1 Z1
=
(3a,b)
X1 M1
Here, M1 is known as the modal mass or generalized mass, and Z is the eigenvector normalized with respect to M. In our example: M1
= {1
Z1 =
5 0
1.618} ⋅
0
1 ⋅ = 18.09 5 1.618
18.09 = 0.235 1.618 0.380 18.09 1
It is easy to show that: Z1T M Z1 = 1
and (4a,b)
Z1T K Z1 = λ1 In the same way, the second eigenvector can be normalized so that the generalized mass, M2 = 1: X2T M X2 = M2 Z2
M2
=
X2 M2
5 = {1 − 0.618} ⋅ 0
0
1 ⋅ = 6.91 5 − 0.618
1 0.380 Z 2 = 6.91 = 1.618 − 0.235 6.91 Again: Z2T M Z2 = 1
and
Department of Engineering Systems
2
Earthquake Engineering
MSc in Civil Engineering
Z2T K Z2 = λ2 The matrix, Z, in which columns are the normalized eigenvectors, Zi, is known as the modal or modeshape matrix of the dynamic matrix M-1 K. For this example:
φ11 φ12 0.235 0.380 = 0.380 − 0.235 φ φ 21 22
Z=
(5)
It can be shown that: ZT M Z = I
and (6a,b)
ZT K Z = λ
λ 0 = 1 0 λ2
Here I is the unit matrix of order n, and l a diagonal matrix of order n, with eigenvalues on the diagonal; n is the number of degrees of freedom.
Forced Vibrations of MDOF Systems We shall continue to use the same example of a 2DOF system. The system is now undergoing forced vibrations (with the excitation force due to earthquake, wind, waves or other loading). F2(t) x2(t) m2 F1(t)
k 2 x1(t)
m1
k 1
The vector of external forces, F, is given as:
F1 ( t ) F ( t ) 2
F=
The equations of motion for this system are: M ⋅ x + K ⋅ x = F( t )
Department of Engineering Systems
(7)
3
Earthquake Engineering
MSc in Civil Engineering
or in a developed form: m1 ⋅ x 1 + ( k 1 + k 2 ) x 1 − k 2 x 2 m 2 ⋅ x 2
= F1 ( t )
(8a,b)
− k 2 x 1 + k 2 x 2 = F2 ( t )
These equations are coupled, that is, they both contain both of the unknown functions x1(t) and x2(t). In order to solve these equations, it is necessary to uncouple them in some way, so that each contains just one unknown function. We can substitute coordinates x by a new set of coordinates, q, such that: x=Zq
and
x = Z ⋅ q
(9)
Here, Z is the modeshape matrix for the system. After inserting Eq. 9 into Eq. 7, it becomes: M ⋅ Z ⋅ q + K ⋅ Z ⋅ q = F( t )
(10)
This is now pre-multiplied by ZT: Z T ⋅ M ⋅ Z ⋅ q + Z T ⋅ K ⋅ Z ⋅ q = Z T ⋅ F( t )
(11)
Using the expressions 6a,b we can simplify this equation into:
q + λ ⋅ q = Z T ⋅ F( t )
(12)
In the developed form, the matrix equation (12) can be written as: 2
q + ω ⋅ q = φ ⋅ F ( t ) + φ ⋅ F (t ) q 2 + ω 22 ⋅ q 2 = φ12⋅ F1 (t ) + φ 22⋅ F2 ( t ) 1
1
1
11
1
21
2
(13a,b)
These two equations are now independent and can be solved separately. The solution is found depending on the load, using any of the methods for SDOF systems.
Department of Engineering Systems
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