ME 440
Intermediate Vibrations
Th, March 26, 2009
Chapter 5: Vibration of 2DOF Systems
© Dan Negrut, 2009
ME440, UW-Madison
Before we get started…
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Last Time:
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Response to an arbitrary excitation: The total solution
Dynamic Load Factor
Response Spectrum
Today:
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HW Assigned (due April 2)
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Material Covered:
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5.1: Assume zero initial velocities
5.4: Assume small oscillations
Start Chapter 5: 2DOF systems
Next Time:
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Anonymously, please print out on a sheet of paper
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Two things that you disliked the most about the class
What you would do to improve this class (if you were teaching ME440)
2
Two Degree of Freedom Systems
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Number of Degrees of Freedom
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So far, we discussed one degree of freedom systems
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The number of generalized coordinates necessary to completely describe the
motion of a system
Recall that we had one natural frequency
Rule: for a n-degree of freedom system, one has n natural frequencies
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Associated with each natural frequency, there is a natural mode of vibration
The natural vibration modes turn out to be orthogonal (a concept a bit ahead
of its time…)
3
Multiple DOF Systems
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Probably the most important thing when trying
to derive the equations of motion associated
with a mechanical system is this:
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Make sure you understand how many degrees of
freedom you have
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You will have as many differential equations as
many degrees of freedom you have
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At left, the system has two degrees of freedom
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Choose 1 and 2 as the 2 DOFs
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Then x1, y1, x2, y2 are not independent gen.
coordinates, they’re derived based on 1 and 2
Note that you could select y1 and y2 to be the
independent generalized coordinates
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Then 1 and 2 become dependent coordinates
4
[Text]
Example
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Derive EOMs for system below
5
Matrix Notation
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Serves two purposes
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Most importantly, it brings sanity to the process of formulating the equations of motion
for large systems
It clearly shows the parallels that exist between the single and multiple DOF system in
relation to their EOMs
ém
ùìï &
ï
&ü
ê 1 0 úïí x 1 ïý +
ê 0 m úï x&
&
ïï
2ú
2
êë
ï
ûî þ
éc + c
ï &ü
ï
- c2 ùì
2
ê1
úïí x 1 ïý +
ê-c
úï x&ï
c
+
c
2
2
3
úï 2 þ
ï
ëê
ûî
ék + k
ï ü
ï
- k2 ùì
2
ê1
úïí x 1 ïý =
ê-k
úï x ï
c
+
k
2
2
3
úï 2 þ
ï
ëê
ûî
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In compact form, this equation assumes the form
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Observe the following notation convention:
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Matrices are in square brackets
Vectors are in curly brackets
ìï F (t )ü
ï 1 ïï
í
ý
ïï F2 (t )ïï
î
þ
6
Matrix Notation: Nomenclature
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Mass Matrix (symmetric!)
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Damping Matrix (symmetric!)
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Stiffness Matrix (symmetric!)
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Displacement Vector
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Force Vector
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Final Remarks, Matrix Notation
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The Mass Matrix
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Is symmetric
Is typically diagonal
It is not diagonal if there is dynamic coupling between the generalized
coordinates
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The Damping and Stiffness matrices
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This is the case for instance in Finite Element Analysis
Are symmetric – a consequence of Newton’s Third Law
Most of the time, they are not diagonal
Note: If [m], [c], and [k] are diagonal, we say that the equations of
motion are independent (they are decoupled)
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[Quick Review]
Matrix Algebra
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Definition: A matrix A is singular if its determinant is zero
det( A)  0  Matrix A is singular
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Examples:
 1 2 3
A   1 3 2
 0 2 2
1 7 
A

3
21


9
[Quick Review]
Inverse of a matrix A
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Result from Linear Algebra that we rely on heavily:
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A matrix A has an inverse (denoted by A-1) if, and only if, A is not a
singular matrix (that is, its determinant is not zero)
If A is nonsingular, that is, A-1 exists, then the solution of
the linear system Aa=b is simply a= A-1 b
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[Quick Review]
Dealing with a singular matrix
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Nomenclature
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For the linear system Aa=0, the vector a is called a nontrivial
solution if a satisfies the equation Aa=0 but a is not zero
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Example:
NOTE:
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3 6
A


1

2


 1 
a

 0.5
If A is nonsingular, then you cannot find a nontrivial solution a for
the problem Aa=0.
In other words, to find a nontrivial solution, the matrix A should be
singular:
det(A)=0
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[Quick Review]
On the number of trivial solutions
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As indicated, you can start looking for a nontrivial solution
provided the matrix A is singular
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Important observation:
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If a is a trivial solution, then so is 1.23a, 32.908a, -2.128a, etc.
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For any real number , if a is a nontrivial solution, then so is a
In other words, as soon as you find one trivial solution, you have as many
of them you wish
So you either don’t have any nontrivial solution at all, or have an infinite
number of them
Example of trivial solutions for Aa=0
3 6
A


1

2


 1 
a


0.5


2
a 
 1
 5 
a 
 2.5
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[Quick Review]
Finding a nontrivial solution
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Example: Find a nontrivial solution for Aa=0, given that
3 6
A


1

2


13
Free Vibration of Undamped Systems
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System at left leads to the following EOM
Assume a solution of the form:
Use the same old trick: substitute back into the EOM and
see what conditions A1, A2, , and  must satisfy so that
x1(t) and x2(t) verify the EOM
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[Cntd.]
Free Vibration of Undamped Systems
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Substituting back leads to the following relationship between A1 and A2
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In matrix form:
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This linear system has a nontrivial solution only if determinant of matrix is zero
IMPORTANT: This condition represents the characteristic equation (CE) associated
with our 2DOF system
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Recall from ME340: CE is the equation that provides the natural frequency of system
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[Cntd.]
Free Vibration of Undamped Systems
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Characteristic Equation:
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Characteristic Equation, after evaluating the determinant…
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Two real solutions that lead to two natural frequencies 1 and 2:
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[Cntd.]
Free Vibration of Undamped Systems
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A word on notation
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Why do I have m1 and n(1)? What’s the deal with the parentheses there?
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I want to emphasize the fact that the “1” in n(1) doesn’t have anything to
do with the “1” in m1
Rather, n(1) is a quantity that refers to the *entire* system
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Specifically, it indicates one of the natural frequencies of the *entire* system
It shows how both m1 *and* m2 move together in the first vibration mode
Note that n(2) is the other natural frequency at which the bodies move
*together* if left alone (free response)
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[Cntd.]
Free Vibration of Undamped Systems
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Recall the meaning of n(1) and n(2):
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Those values that zero out the determinant of the linear system (1) & (2)
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Consequently the linear system has an infinite number of solutions A1 and A2
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What is unique though, it’s the ration between A2/A1
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To find the first ratio (associated with n(1)), plug back value of n(1) to obtain
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To find second ratio (associated with n(2)), plug back value of n(2) to obtain
18
[Cntd.]
Free Vibration of Undamped Systems
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Remember what we assumed:
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In Matrix/Vector notation:
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NOTE: At this point, n(1) and r(1) are known, but not A1(1) and (1)
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Similarly, for the second natural frequency,
19
[Cntd.]
Free Vibration of Undamped Systems
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Notation – we call modal vectors the following quantities:
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Solution can be expressed now as
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Use ICs (two positions and two velocities) to find the following four unknowns:
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[Short Detour: Notation, FEM related]
Free Vibration of Undamped Systems
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Using again Matrix/Vector notation…
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Introduce the modal matrix, which is made up of the modal vectors
Important: the matrix [u] is constant (doesn’t change, an attribute of the m-k system!!)
Also, define {f(t)} as
Then, the solution can be expressed in matrix-vector notation (looks very similar
to what you have in FEM)
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Example, 2DOF:
m1=1kg, m2=2kg
k1=9N/m
k2=k3=18N/m
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 x (0)  3 x1 (0)  0
IC :  1
 x2 (0)  0 x2 (0)  9
Find modal vectors {u}(1), {u}(2)
Find mode ratios
Find natural frequencies
Find x(t)
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