Multi-degree of freedom systems – review
Free vibration analysis
x1  c11 c12   x1   k11 k12   x1   F1 
m11 m12   
m
  x   c
  x    k
x    F 
m
c
k
 21
22   2   21
22   2   21
22   2   2 
 No damping
Equation of motion:


[ M ]
x  [ K ]{x}  0
We know from theory of differential
A
equations that the solution is:  x  ( )e j t
B
Thus we need to find the frequency ω and
the amplitudes A and B.
Characteristic equation:
det([ K ]   2[ M ])  0
2 eigenvalues of [M]-1[K]
A
( )
B
eigenvectors of [M]-1[K].
: natural frequencies (number = number of
d.o.f.)
 A
 i  
 B

mode shapes (number = number of
d.o.f.).
General expression for free vibration:
 j t
 j t

1
x (t )  c1 1e
 c2 2e 2 


c1  1cos(1t   1)  c2  2cos(2t   2 )
Conclusions:
 Vibration=mode1*c1+mode2*c2
 Participation of each mode depends on
initial conditions
Steps in free vibration analysis
1. Find natural frequencies and mode shapes
2. Express displacements in terms of natural
frequencies and mode shapes:
 x1 (t ) 
A 
A 

  c1  1  cos(1t   1 )  c2  2  cos(2t   2 )
 x2 ( t ) 
 B1 
 B2 
3. Find constants and phase angles from
initial conditions
Example
x1
k
k
k
m
Second natural frequency:

x2
3k
( 2)  A   B
m
m
Mode shape for above frequency:
 A
  Usually assume first entry is 1.
  A
 1
Therefore, mode shape is:  
 1
Masses move in opposite directions
t
t
First natural frequency:

k
( 2)  A  B
m
 1
Mode shape for above frequency:  
 1
Masses move together.
t
t
Displacements of two masses are superposition
of displacements in the two modes:
 x1(t ) 
1
1
k
3k

  c1  cos(
t 1 )  c2   cos(
t  2 )
x
(
t
)
1

1
m
m
 
 
 2 
Find c1, c2 and 1 ,  2
from initial conditions
Orthogonality of mode shapes and
decoupling of equations of motion
Mode shapes are orthogonal wrt the mass
and stiffness matrices:
T



1 [ ]2  0
1T [ K ]2  0
Therefore:
[]T [M ][]  [M' ]
[]T [K ][]  [K ' ]
where [M'] and [K'] are diagonal matrices.
We can use the orthogonality property of the
mode shapes to uncouple the equations of
motion through the following
transformation:




x(t )  1(t ) p1(t )  2 (t ) p2 (t )  Φp(t )
This transformation expressed the unknown
displacements x (t ) in terms of the basis


vectors 1(t ) and 2 (t ) . If we substitute the
expression for the displacement into the
equations of motion then we have two new
equations wrt the unknown coordinates p (t ) .
M[Φ] p(t )  K [Φ] p (t )  0
These equations are equivalent to the
original system but they can be easily
decoupled.


Mp(t )  K  p(t )  0
where
M  [Φ]T [M][Φ]
K   [Φ]T [K ][Φ]
[M'] and [K'] are diagonal matrices, which
means that we have two uncoupled
equations with unknowns coordinates p1(t)
and p2(t).
Therefore, the equations of motion can be
written wrt to the new coordinate system can
be written as follows:
 p1(t )  k11
 p1(t )  0
m11
 p1(t )  k22
 p2 ( t )  0
m22
In some cases it is better to solve these two
uncoupled equations than the coupled
equations. To solve these equations we
need the initial conditions for coordinates
p1(t) and p2(t). These are obtained as
follows:
 p1(0) 
1  x1(0) 

  [Φ ] 

p
(
0
)
x
(
0
)
 2 
 2 
 p1(0) 
1  x1(0) 

[
Φ
]




 p 2 (0)
 x 2 (0)
Procedure for calculating free vibration
response by uncoupling the equations of
motion
1. Determine matrices:
M  [Φ]T [M][Φ]
K   [Φ]T [K ][Φ]
2. Determine initial values of coordinates
p1(t) and p2(t) and initial values of the
derivatives.
3. Solve uncoupled equations for p1(t) and
p2(t):
 p1(t )  k11
 p1(t )  0
m11
 p1(t )  k22
 p2 ( t )  0
m22
where m'ii and k'ii are the diagonal elements
of matrices M' and K', using the initial
conditions found in step 2.
4. Recover displacements x1(t) and x2(t)
using the following equations:




x(t )  1(t ) p1(t )  2 (t ) p2 (t )  Φp(t )