MDOF SYSTEMS WITH
DAMPING
General case
Saeed Ziaei Rad
MDOF Systems with hysteretic
damping- general case
Free vibration solution:
[M ]{x}  ([K ]  i[ D]){x}  {0}
Assume a solution in the form of:
it
{x}  {X }e
Here  can be a complex number. The solution here is like
the undamped case. However, both eigenvalues and
Eigenvector matrices are complex.
The eigensolution has the orthogonal properties as:
[]T [M ][]  [mr ];
[]T [ K  iD][]  [kr ]
The modal mass and stiffness parameters are complex.
MDOF Systems with hysteretic
damping- general case
Again, the following relation is valid:
kr
2
 
  r (1  ir )
mr
2
r
A set of mass-normalized eigenvectors can be defined as:
{}r  (mr )
1/ 2
{ }r
What is the interpretation of complex mode shapes?
The phase angle in undamped is either 0 or 180.
Here the phase angle may take any value.
Numerical Example with
structural damping
x2
k4
k1
m2
m1
k2
k5
k3
m3
x1
k6
m1=0.5 Kg
m2=1.0 Kg
m3=1.5 Kg
k1=k2=k3=k4=k5=k6=1000 N/m
x3
Undamped
.5 0 0 


[M ]   0 1 0 
 0 0 1.5
 3  1  1


[K ]   1 3  1
 1  1 3 
Using command [V,D]=eig(k,M) in MATLAB
0
0 
950
.464  .218  1.318
[ r2 ]   0 3352 0  [ ]  .536  .782 .318 


 0
0
6698
.635 .493
.142 
Proportional Structural Damping
Assume proportional structural damping as:
d j  0.05k j ,
j  1,,6
0
0
950(1  .05i )


[2r ]  
0
3352(1  .05i )
0


0
0
6698(1  .05i )
.464(0 ) .218(180 ) 1.318(180 )





[]  .536(0 ) .782(180 )
.318(0 ) 

.635(0 ) .493(0 )

.
142
(
0
)


Non-Proportional Structural
Damping
Assume non-proportional structural damping as:
d1  0.3k1
d j  0,
j  2,,6
0
0
957(1  .067i )


[2r ]  
0
3354(1  .042i )
0


0
0
6690(1  .078i )
.463(5.5 ) .217(173 ) 1.321(181 ) 



 
[]   .537(0 )
.784(181 ) .316(6.7 )

 
 .636(0 )
.
492
(

1
.
3
)
.
142
(

3
.
1
)

Non-Proportional Structural
Damping
Each mode has a different damping factor.
All eigenvectors arguments for undamped and
proportional damp cases are either 0 or 180.
All eigenvectors arguments for non-proportional case
are within 10 degree of 0 or 180 (the modes are
almost real).
Exercise: Repeat the problem with
m1=1Kg, m2=0.95 Kg, m3=1.05 Kg
k1=k2=k3=k4=k5=k6=1000 N/m
FRF Characteristics
(Hysteretic Damping)
Again, one can write:
([K ]  i[ D]   2[M ]){X }eit  {F}eit
The receptance matrix can be found as:
H ( )  ([K ]  i[ D]   [M ])  [][   ][]
1
2
2
r
2
FRF elements can be extracted:
 jrkr
H jk ( )   2
2
2
r 1  r    ir r
N
or
N
H jk ( )  
r 1
r
Ajk
    ir
2
r
2
2
r
Modal Constant
T
MDOF Systems with viscous
damping- general case
The general equation of motion for this case can be
written as:
[M ]{x}  [C ]{x}  [ K ]{x}  { f }
Consider the zero excitation to determine the natural
frequencies and mode shapes of the system:
{x}  {X }e st
This leads to:
([M ]s 2  [C]s  [ K ]){X }  {0}
This is a complex eigenproblem. In this case, there are
2N eigenvalues but they are in complex conjugate pairs.
MDOF Systems with viscous
damping- general case
sr , sr*

*
{ }r , { }r
r  1,  , N
It is customary to express each eigenvalues as:
sr   r ( r  i 1   )
2
r
Next, consider the following equation:
(s [M ]  sr [C]  [ K ]){ }r  {0}
2
r
Then, pre-multiply by {}qH
:
{} (s [M ]  sr [C]  [K ]){}r  {0}
H
q
2
r
*
MDOF Systems with viscous
damping- general case
A similar expression can be written for { }q :
(s [M ]  sq [C]  [K ]){ }q  {0}
2
q
This can be transposed-conjugated and then multiply by{ }r
{} (s [M ]  sq [C]  [K ]){}r  {0}
H
q
2
q
**
Subtract equation * from **, to get:
(s  s ){} [M ]{}r  (sr  sq ){} [C]{}r  {0}
2
r
2
q
H
q
H
q
This leads to the first orthogonality equations:
(sr  sq ){} [M ]{}r  {} [C]{}r  {0}
H
q
H
q
(1)
MDOF Systems with viscous
damping- general case
Next, multiply equation (*) by sq and (**) by s r :
sr sq{} [M ]{}r {} [K ]{}r  {0}
H
q
H
q
(2)
Equations (1) and (2) are the orthogonality conditions:
If we use the fact that the modes are pair, then
sq   r ( r  i 1   r2 )
{ }q  { }
*
r
MDOF Systems with viscous
damping- general case
Inserting these two into equations (1) and (2):
{ }rH [C ]{ }r
cr
2 r r 

H
{ }r [ M ]{ }r mr
{ } [ K ]{ }r
kr
 

{ } [ M ]{ }r mr
2
r
H
r
H
r
Where mr , kr , c r are modal mass, stiffness and damping.
FRF Characteristics
(Viscous Damping)
The response solution is:
{X }  ([K ]  i[C]   2[M ])1{F}
We are seeking to a similar series expansion similar to the
undamped case.
To do this, we define a new vector {u}:
 x
{u}   
x 2 N 1
We write the equation of motion as:
[C : M ]N2 N {u}2 N1  [ K : 0]{u}  {0}N1
FRF Characteristics
(Viscous Damping)
This is N equations and 2N unknowns. We add an identity
Equation as:
[M : 0]{u}  [0 : M ]{u}  {0}
Now, we combine these two equations to get:
C
M

M
K
{u}  

0
0
0 
{u}  {0}

M
Which cab be simplified to:
[ A]{u}  [ B]{u}  {0}
3
FRF Characteristics
(Viscous Damping)
Equation (3) is in a standard eigenvalue form. Assuming a
trial solution in the form of {u}  {U}est
(sr [ A]  [ B]){ }r  {0}
r  1,,2 N
The orthogonality properties cab be stated as:
[ ] [ A][ ]  [ar ]
T
[ ] [ B][ ]  [br ]
T
With the usual characteristics:
br
sr 
ar
r  1,,2 N
FRF Characteristics
(Viscous Damping)
Let’s express the forcing vector as:
F 
{P}2 N 1   
0 
Now using the previous series expansion:
2N
X 
{ }Tr {P}{ }r



iX 2 N 1 r 1 ar (i  sr )
And because the eigenvalues and vectors occur in complex
conjugate pair:
N
X 
{ }Tr {P}{ }r { }rH {P}{ *}r

 *


*
i

X
a
(
i


s
)
a
(
i


s
)

2 N 1 r 1 r
r
r
r
FRF Characteristics
(Viscous Damping)
Now the receptance frequency response function H jk
Resulting from a single force Fk and response parameter X j
 jr kr
N
H jk ( )  
or
r 1
ar ( r r  i(   r 1   r2 )
N
H jk ( )  
Where:
r 1
r

 * *
jr
a*r ( r r  i(   r 1   r2 )
R jk  i( /  r )(r S jk )
 2   2  2ir r
r
{r Rk }  2( r Re{r Gk }  Im{r Gk } 1   )
2
r
{r S k }  2 Re{r Gk }
{r Gk }  ( kr / ar ){ }r
kr