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

© 1999 Yijun Liu, University of Cincinnati

167

III. Damping

Two commonly used models for viscous damping.

A.

 

 Proportional Damping (Rayleigh Damping)

K

M

C

β

α

+

=

                             

(17)

where the constants 

α

 & 

β

 are found from

,

2

2

,

2

2

2

2

2

1

1

1

ω

β

αω

ξ

ω

β

αω

ξ

+

=

+

=

with 

2

1

2

1

 

&

 

 

,

 

,

ξ

ξ

ω

ω

 (damping ratio) being selected.

B. Modal Damping

Incorporate the viscous damping in modal equations.

Dam

p

ing

 r

a

ti

o

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

© 1999 Yijun Liu, University of Cincinnati

168

IV. Modal Equations

 

Use the normal modes (modal matrix) to transform the
coupled system of dynamic equations to uncoupled
system of equations.

We have

[

]

n

1,2,...,

     

,

2

=

=

i

i

i

0

u

M

K

ω

                  (18)

where the normal mode 

i

 satisfies:

=

=

,

0

,

0

j

T

i

j

T

i

u

M

u

u

K

u

 for 

 j,

and

=

=

,

,

1

2

i

i

T

i

i

T

i

ω

u

K

u

u

M

u

for i = 1, 2, …, n.

Form the modal matrix:

        

[

]

n

n

n

u

u

u

Ö

 

 

2

1

)

(

L

=

×

               

(19)

Can verify that

.

,

matrix)

 

Spectral

(

0

0

0

0

0

0

 

2

n

2

2

2

1

I

M

Ö

Ö

Ù

K

Ö

Ö

=

=

=

T

T

ω

ω

ω

L

O

M

M

L

  

(20)

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

© 1999 Yijun Liu, University of Cincinnati

169

Transformation for the displacement vector,

z

u

u

u

u

Φ

=

+

+

+

=

n

n

z

z

z

L

2

2

1

1

,     (21)

where

                







=

)

(

)

(

)

(

2

1

t

z

t

z

t

z

n

M

z

are called principal coordinates.

Substitute (21) into the dynamic equation:

Pre-multiply by 

Φ

T

, and apply (20):

),

t

p

z

z

C

z

=

+

+

&

&&

φ

                            

(22)

where 

+

=

β

α

φ

I

C

     (proportional damping),

   

)

t

T

f

p

Φ

=

.

Using Modal Damping

=

n

n

ω

ξ

ω

ξ

ω

ξ

φ

2

0

2

0

0

0

2

2

2

1

1

L

M

O

M

L

C

. (23)

).

t

f

z

K

z

C

z

M

=

Φ

+

Φ

+

Φ

&

&

&

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

© 1999 Yijun Liu, University of Cincinnati

170

Equation (22) becomes,

),

(

2

2

t

p

z

z

z

i

i

i

i

i

i

i

=

+

+

ω

ω

ξ

&

&&

 = 1,2,…,n. (24)

Equations in (22) or (24) are called modal equations.
These are uncoupled, second-order differential equations,
which are much easier to solve than the original dynamic
equation (coupled system).

To recover u from z, apply transformation (21) again, once
z is obtained from (24).

Notes:

 

Only the first few modes may be needed in constructing
the modal matrix 

Φ

 (i.e., 

Φ

 could be an n

×

m rectangular

matrix with m<n).  Thus, significant reduction in the
size of the system can be achieved.

 

Modal equations are best suited for problems in which
higher modes are not important (i.e., structural
vibrations, but not shock loading).

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

© 1999 Yijun Liu, University of Cincinnati

171

V. Frequency Response Analysis
     (Harmonic Response Analysis)

3

2

1

&

&&

loading

 

Harmonic

sin t

ω

F

Ku

u

C

u

M

=

+

+

                    (25)

Modal method:  Apply the modal equations,

 

,

sin

2

2

t

p

z

z

z

i

i

i

i

i

i

i

ω

ω

ω

ξ

=

+

+

&

&&

   i=1,2,…,m.      (26)

These are 1-D equations. Solutions are

                                   

),

sin(

)

2

(

)

1

(

)

(

2

2

2

2

i

i

i

i

i

i

i

t

p

t

z

θ

ω

η

ξ

η

ω

+

=

     (27)

                                   where



=

=

=

=

ratio

damping

 

,

2

,

angle

 

phase

 

,

1

2

arctan

i

2

i

i

c

i

i

i

i

i

i

i

m

c

c

c

ω

ξ

ω

ω

η

η

η

ξ

θ

   Recover u from (21).

Direct Method:   Solve Eq. (25) directly, that is, calculate

the inverse. With

t

i

e

ω

u

u

=

(complex notation), Eq. (25)

becomes

[

]

.

2

F

u

M

C

K

=

+

ω

ω

i

This equation is expensive to solve and matrix is ill-
conditioned if 

ω

 is close to any 

ω

i

.

z

i

ω

/

ω

i