Chapter 4 Multiple Degree of Freedom Systems. The Millennium bridge required many degrees of freedom to model and design with. Extending the first 3 chapters to more then one degree of freedom. The first step in analyzing multiple degrees of freedom (DOF) is to look at 2 DOF.
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Chapter 4 Multiple Degree of Freedom Systems
The Millennium bridge required
many degrees of freedom to model
and design with.
Extending the first 3 chapters to more then one degree of freedom
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Fig 4.1
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A 2 degree of freedom system used to base
much of the analysis and conceptual
development of MDOF systems on.
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Figure 4.2
k2(x2 -x1)
m2
k1 x1
m1
x2
x1
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The two equations can be written in the form of a
single matrix equation (see pages 272-275 if matrices and vectors are a struggle for you) :
(4.4), (4.5)
(4.6), (4.9)
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IC’s can also be written in vector form
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For 1DOF we assumed the scalar solution aeltSimilarly, now we assume the vector form:
(4.15)
(4.16)
(4.17)
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The determinant results in 1 equation
in one unknown w(called the characteristic equation)
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(4.20)
(4.21)
Eq. (4.21) is quadratic inw2so four solutions result:
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This yields vector equation for each squared frequency:
Each of these matrix equations represents 2 equations in the 2 unknowns components of the vector, but the coefficient matrix is singular so each matrix equation results in only 1 independent equation. The following examples clarify this.
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Each value of w2 yields an expression or u:
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2 equations, 2 unknowns but DEPENDENT!
(the 2nd equation is -3 times the first)
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Note that the other equation is the same
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Several possibilities, here we just fix one element:
Choose:
Choose:
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Here we have introduce the name
mode shape to describe the vectors
u1 and u2. The origin of this name comes later
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We have computed four solutions:
(4.24)
Since linear, we can combine as:
(4.26)
Note that to go from the exponential
form to to sine requires Euler’s formula
for trig functions and uses up the
+/- sign on omega
determined by initial conditions.
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If IC’s correspond to mode 1 or 2, then the response is purely in mode 1 or mode 2.
x1
x2
k1
k2
m1
m2
Mode 1:
x2=A
x1=A/3
x1
x2
k1
k2
m1
m2
Mode 2:
x2=A
x1=-A/3
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At t=0 we have
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4 equations in 4 unknowns:
Yields:
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The final solution is:
(4.34)
These initial conditions gives a response that is a combination of modes. Both harmonic, but their summation is not.
Figure 4.3
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Determines how the second
frequency contributes to the
response
Determines how the first
frequency contributes to the
response
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Then M is said to be positive definite
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A matrix M is defined to be symmetric if
A symmetric matrix M is positive definite if
A symmetric positive definite matrix M can be factored
Here L is upper triangular, called a Cholesky matrix
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For a symmetric, positive definite matrix M:
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Window 4.1 page 285
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(4.43)
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(4.44)
To see this compute
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The first normalized eigenvector
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(4.37)
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called a matrix of eigenvectors (normalized)
P is called an orthogonal matrix
(4.47)
P is also called a modal matrix
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From the previous example:
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In general:
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The equations of motion:
Figure 4.4
In matrix form these become:
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Yes!
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Does this make any difference?
No! Try it in the previous example
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However, they can also be solved on
calculators with matrix functions and
with the codes listed in the last section
In fact, for more then two DOF one must
use a code to solve for the natural
frequencies and mode shapes.
Next we examine 3 other formulations for
solving for modal data
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From equation (4.17)
(4.53)
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Is the Generalized Symmetric Eigenvalue Problem
easy for hand computations, inefficient for computers
(ii) Is the Asymmetric Eigenvalue Problem
very expensive computationally
(iii) Is the Symmetric Eigenvalue Problem
the cheapest computationally
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