ME 440Intermediate Vibrations

Th, March 26, 2009Chapter 5: Vibration of 2DOF Systems

© Dan Negrut, 2009ME440, UW-Madison

Before we get started…

Last Time: Response to an arbitrary excitation: The total solution Dynamic Load Factor Response Spectrum

Today: HW Assigned (due April 2)

5.1: Assume zero initial velocities 5.4: Assume small oscillations

Material Covered: Start Chapter 5: 2DOF systems

Next Time: Anonymously, please print out on a sheet of paper

Two things that you disliked the most about the class What you would do to improve this class (if you were teaching ME440)

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Two Degree of Freedom Systems

Number of Degrees of Freedom The number of generalized coordinates necessary to completely

describe the motion of a system

So far, we discussed one degree of freedom systems Recall that we had one natural frequency

Rule: for a n-degree of freedom system, one has n natural frequencies

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…)

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Multiple DOF Systems Probably the most important thing when

trying to derive the equations of motion associated with a mechanical system is this:

Make sure you understand how many degrees of freedom you have

You will have as many differential equations as many degrees of freedom you have

At left, the system has two degrees of freedom

Choose 1 and 2 as the 2 DOFs 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

Then 1 and 2 become dependent coordinates 4

[Text]

Example Derive EOMs for system below

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Matrix Notation Serves two purposes

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

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1 1 1 2 2 1 1 2 2 1 1

2 2 2 2 3 2 2 2 3 2 2

0 ( )

0 ( )

m x c c c x k k k x F t

m x c c c x k c k x F t

é ùì ü é ùì ü é ùì ü ì üï ï ï ï ï ï ï ï+ - + -ï ï ï ï ï ï ï ïê ú ê ú ê ú+ + =í ý í ý í ý í ýê ú ê ú ê úï ï ï ï ï ï ï ï- + - +ê ú ê ú ê úï ï ï ï ï ï ï ïë ûî þ ë ûî þ ë ûî þ î þ

&& &

&& &

In compact form, this equation assumes the form

Observe the following notation convention: Matrices are in square brackets Vectors are in curly brackets

Matrix Notation: Nomenclature

Mass Matrix (symmetric!)

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Damping Matrix (symmetric!)

Stiffness Matrix (symmetric!)

Displacement Vector

Force Vector

Final Remarks, Matrix Notation

The Mass Matrix Is symmetric Is typically diagonal It is not diagonal if there is dynamic coupling between the

generalized coordinates This is the case for instance in Finite Element Analysis

The Damping and Stiffness matrices 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 Definition: A matrix A is singular if its determinant is

zero

Examples:

det( ) 0 Matrix is singular A A

1 2 3

1 3 2

0 2 2

A1 7

3 21

A

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[Quick Review]

Inverse of a matrix A Result from Linear Algebra that we rely on heavily:

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

Nomenclature 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

Example:

NOTE: 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

3 6

1 2

A1

0.5

a

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[Quick Review]

On the number of trivial solutions

As indicated, you can start looking for a nontrivial solution provided the matrix A is singular

Important observation: If a is a trivial solution, then so is 1.23a, 32.908a, -2.128a,

etc. 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=03 6

1 2

A1

0.5

a

2

1

a

5

2.5

a

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[Quick Review]

Finding a nontrivial solution

Example: Find a nontrivial solution for Aa=0, given that

3 6

1 2

A

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Free Vibration of Undamped Systems

System at left leads to the following EOM

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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

[Cntd.]

Free Vibration of Undamped Systems

Substituting back leads to the following relationship between A1 and A2

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In matrix form:

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

Recall from ME340: CE is the equation that provides the natural frequency of system

[Cntd.]

Free Vibration of Undamped Systems

Characteristic Equation:

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Characteristic Equation, after evaluating the determinant…

Two real solutions that lead to two natural frequencies 1 and 2:

[Cntd.]

Free Vibration of Undamped Systems

A word on notation

Why do I have m1 and n(1)? What’s the deal with the parentheses there?

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 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) 17

Recall the meaning of n(1) and n(2): Those values that zero out the determinant of the linear system (1) & (2)

Consequently the linear system has an infinite number of solutions A1 and A2

What is unique though, it’s the ration between A2/A1

To find the first ratio (associated with n(1)), plug back value of n(1) to obtain

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[Cntd.]

Free Vibration of Undamped Systems

To find second ratio (associated with n(2)), plug back value of n(2) to obtain

Remember what we assumed:

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[Cntd.]

Free Vibration of Undamped Systems

In Matrix/Vector notation:

NOTE: At this point, n(1) and r(1) are known, but not A1(1) and (1)

Similarly, for the second natural frequency,

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[Cntd.]

Free Vibration of Undamped Systems

Notation – we call modal vectors the following quantities:

Solution can be expressed now as

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

Using again Matrix/Vector notation… 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!!)

Then, the solution can be expressed in matrix-vector notation (looks very similar to what you have in FEM)

Also, define {f(t)} as

Example, 2DOF:

m1=1kg, m2=2kg

k1=9N/m

k2=k3=18N/m

1 1

2 2

(0) 3 (0) 0:

(0) 0 (0) 9

x xIC

x x

Find modal vectors {u}(1), {u}(2)

Find mode ratios Find natural frequencies Find x(t)

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