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Lecture Notes: Introduction to Finite Element Method Chapter 7. Structural Vibration and Dynamics

1999 Yijun Liu, University of Cincinnati 163

II. Free Vibration

Study of the dynamic characteristics of a structure:

natural frequencies normal modes (shapes)

Let f(t) = 0 and C = 0(ignore damping) in the dynamic

equation (8) and obtain

0KuuM =+&& (12)

Assume that displacements vary harmonically with time,

that is,

),sin()(

),cos()(

),sin()(

2

tt

tt

tt

uu

uu

uu

=

==

&&

&

where u is the vector of nodal displacement amplitudes.

Eq. (12) yields,

[ ] 0uMK = 2 (13)

This is a generalized eigenvalue problem (EVP).

Solutions?

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Lecture Notes: Introduction to Finite Element Method Chapter 7. Structural Vibration and Dynamics

1999 Yijun Liu, University of Cincinnati 164

Trivial solution: 0u = for any values of (notinteresting).

Nontrivial solutions: 0u only if

02 = MK (14)

This is an n-th order polynomial of2, from which we canfind n solutions (roots) or eigenvalues i.

i (i = 1, 2, , n) are the natural frequencies (orcharacteristic frequencies) of the structure.

1 (the smallest one) is called the fundamentalfrequency.

For each i , Eq. (13) gives one solution (or eigen)vector

0uMK = ii2

.

iu (i=1,2,,n) are the normal modes (or natural

modes, mode shapes, etc.).

Properties of Normal Modes

0=jT

i uKu ,

0=jT

i uMu , for i j. (15)

That is, modes are orthogonal (or independent) to each

other with respect to Kand M matrices.

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Lecture Notes: Introduction to Finite Element Method Chapter 7. Structural Vibration and Dynamics

1999 Yijun Liu, University of Cincinnati 165

Normalize the modes:

.

,1

2ii

Ti

i

T

i

=

=

uKu

uMu

(16)

Note:

Magnitudes of displacements (modes) or stresses innormal mode analysis have no physical meaning.

For normal mode analysis, no support of the structure isnecessary.

i = 0 there are rigid body motions of the wholeor a part of the structure.

apply this to check the FEA model (check formechanism or free elements in the models).

Lower modes are more accurate than higher modes inthe FE calculations (less spatial variations in the lower

modes fewer elements/wave length are needed).

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Lecture Notes: Introduction to Finite Element Method Chapter 7. Structural Vibration and Dynamics

1999 Yijun Liu, University of Cincinnati 166

Example:

[ ]

.

422

22156

420

,

46

612

,0

0

223

2

22

=

=

=

LL

LAL

LL

L

L

EI

v

MK

MK

EVP:

in which EIAL 420/2= .

Solving the EVP, we obtain,

Exact solutions:

.03.22,516.32

1

42

21

41

=

=AL

EI

AL

EI

We can see that mode 1 is calculated much more accurately

than mode 2, with one beam element.

L

x1 2

v2

, A, EI 2

,044226

2261561222

=++

LLLL

LL

.62.7

1v,81.34

,38.1

1v,533.3

22

22

1

42

12

22

1

41

=

=

=

=

LAL

EI

LAL

EI

#1

#2#3