MIT - 16.20
Fall, 2002
Paul A. Lagace © 2001
Paul A. Lagace, Ph.D.
Professor of Aeronautics & Astronautics and Engineering Systems

MIT - 16.20 Fall, 2002
The difference is that it is a matrix equation:
m q˙˙
~
+ k q
~
= F (22-1)
~ = matrix
So apply the same solution technique as for a single degree-of-freedom system. Thus, first deal with…
Do this by again setting forces to zero:
F = 0
~ ~
m q˙˙
~
+ k q
~
= 0
Paul A. Lagace © 2001
(22-2)
Unit 22 - 2

MIT - 16.20 Fall, 2002
Again assume a solution which has harmonic motion. It now has multiple components: q t
~
= Ae
~ i ω t
(22-3) where ω are the natural frequencies of the system and:

M

 


A
M i


Substituting the assumed solution into the matrix set of governing equations:
⇒ − ω 2 m A e
~ ~ i ω t
+ k A e i ω t
= 0
To be true for all cases:
[ k − ω 2 m
]
A = 0 (22-4)
This is a standard eigenvalue problem.
Either:
A = 0 (trivial solution) or
Paul A. Lagace © 2001 Unit 22 - 3

MIT - 16.20 Fall, 2002
The determinant: k − ω 2 m = 0 (22-5)
There will be n eigenvalues for an n degree-of-freedom system.
In this case: eigenvalue = natural frequency
⇒ n degree-of-freedom system has n natural frequencies
Corresponding to each eigenvalue (natural frequency), there is an…
Eigenvector -- Natural Mode
• Place natural frequency ω r
into equation (22-4):
[ k − ω 2 r m
]
A =
~
• Since determinant = 0, there is one dependent equation, so one
~
~ by) one component
Paul A. Lagace © 2001 Unit 22 - 4

Fall, 2002 MIT - 16.20
• Say divide through by A n
:
[ k − ω i
2 ]

M

 A



 i
A
M
1 n





=
• Solve for A i
/ A n
for each ω r
• Call the eigenvector

M



A i
A
M n

 

 =

1


φ
~ i
• Do this for each eigenvalue
Indicates solution for ω r frequency: associated mode:
ω
1
, ω
2
……. ω n
φ
~ i
φ
~ i
φ
~ i
Paul A. Lagace © 2001 Unit 22 - 5

MIT - 16.20
For each eigenvalue, the homogeneous solution is:
~ q
i hom
= homogeneous
φ
~ i e i ω r t
= C
1
φ
~ sin ω r t + C
2
φ
~ i cos ω r t
Still an undetermined constant in each case (A n
) which can be determined from the Initial Conditions
Fall, 2002
• Each homogeneous solution physically represents a possible free vibration mode
• Arrange natural frequencies from lowest ( ω
1
) to highest ( ω n
)
• By superposition, any combinations of these is a valid solution
Example: Two mass system (from Unit 19)
Figure 22.1
Representation of dual spring-mass system
Paul A. Lagace © 2001 Unit 22 - 6

MIT - 16.20 Fall, 2002
The governing equation was:
 m
1

 0
0   q m
2
 
 
˙˙
1
2



+



( k
1
+ k
2
) − k
2
  q
1
  F
1

− k
2 k
2
  
  q
2
=  
 
F
2

Thus, from equation (22-5):
( k
1
+ k
2
) − ω 2 m
1
− k
2 k
2
−
− k
2
ω 2 m
2
= 0
This gives:
[ ( k
1
+ k
2
) − ω 2 m
1
][ k
2
− ω 2 m
2
]
− k
2
2 = 0
This leads to a quadratic equation in ω 2 . Solving gives two roots ( ω
1
2 and
ω
2
2 ) and the natural frequencies are ω
1
and ω
2
Find the associated eigenvectors in terms of A
2
(i.e. normalized by A
2
)
Paul A. Lagace © 2001 Unit 22 - 7

MIT - 16.20 Fall, 2002
Go back to equation (22-4) and divide through by A
2
:



( k
1
+ k
2
) − ω r
2 m
1
− k
2 k
2
−
− k
2
ω r
2 m
 
 
A
1

 = 0
2
  1 
Normalized constant
⇒ A
1
= k
1
+ k
2 k
2
−
Thus the eigenvectors are:
ω r
2 m
1 for ω r
mode
φ
~ i


k
1 = 


+ k
2 k
2
−
1
ω
1
2 m
1






φ = 




k
1
+ k
2 k
2
−
1
ω
2
2 m
1





For the case of Initial Conditions of 0, the cos term goes away and are left with… q t = φ
~ ~ sin ω r t
Physically the modes are:
Paul A. Lagace © 2001 Unit 22 - 8

MIT - 16.20
Figure 22.2
Representation of modes of spring-mass system
φφ (1)
Fall, 2002
ω
1
(lowest frequency) masses move in same direction
Paul A. Lagace © 2001
φ (2) masses move in opposite direction
ω
2
(higher frequency)
Unit 22 - 9

MIT - 16.20
General Rules for discrete systems:
• Can find various modes (without amplitudes) by considering combinations of positive and negative (relative) motion.
However, be careful of (-1) factor across entire mode.
For example, in two degree-of-freedom case
+ + + same mode same mode
- - +
Fall, 2002
• The more “reversals” in direction, the higher the mode (and the frequency)
• It is harder to excite higher modes
This can be better illustrated by considering the vibration of a beam. So look at:
Paul A. Lagace © 2001 Unit 22 - 10

MIT - 16.20
How?
• Lump mass into discrete locations with constraint that total mass be the same
Fall, 2002
• Connect masses by rigid connections with rotational springs at each mass
• Stiffnesses of connections are influence coefficients (dependent on locations of point masses)
• Forces applied to point masses
So:
Figure 22.3
Representation of cantilevered beam as single mass system
A B becomes for simplest case with torsional spring at joint where: m = ρ tw l length density thickness width
Paul A. Lagace © 2001 Unit 22 - 11

MIT - 16.20
Could also put mass at mid-point:
Figure 22.4
Representation of cantilevered beam as mid-point mass system
Fall, 2002
⇒ get a different representation
Consider the next complicated representation (simplest multimass/degree-of-freedom system)
Figure 22.5
Representation of cantilevered beam as dual spring-mass system
Paul A. Lagace © 2001 Unit 22 - 12

MIT - 16.20 each m is one half of total mass of beam for constant cross-section
Fall, 2002 case equation is:
 m
1

 0
0   q m
2
 
1
  k q
11
˙˙
2
+
 
 k
21 k
12 k
22
  q
  q
1
  
2

=
 F
1

 
 
F
2

Same form as before, so solution takes same form. For initial rest conditions: q t =
~
φ
~ sin ω r t
Have two eigenvalues (natural frequencies) and associated eigenvectors (modes)
⇒ Modes have clear physical interpretation here:
Paul A. Lagace © 2001 Unit 22 - 13

MIT - 16.20
Figure 22.6
Fall, 2002
Representation of deflection modes of cantilevered beam as dual spring-mass system q
2
(+) q
1
(+)
φ (1)
ω
1
(lowest frequency)
Paul A. Lagace © 2001
φ (2) q
1
(-) q
2
(+)
ω
2
(higher frequency)
Unit 22 - 14

MIT - 16.20
Can extend by dividing beam into more discrete masses
--> get better representation with more equations but same basic treatment/approach
Fall, 2002
In considering the modes that result from such an analysis, there is a key finding:
Orthogonality Relations
It can be shown that the modes of a system are orthogonal. That is: transpose
φ
~ m φ (s)
= 0 (22-6) for r ≠ s
If r = s, then a finite value results:
φ
~ m φ = M r
(22-7) some value
So the general relation for equations (22-6) and (22-7) can be written as:
Paul A. Lagace © 2001 Unit 22 - 15

MIT - 16.20
~ m
~
( s ) =
rs
M
δ rs
is the kronecker delta where:
δ rs
= 0 for r ≠ s
δ rs
= 1 for r = s r
(22-8)
This relation allows the transformation of the governing equation into a special set of equations based on the (normal) nodes…
Fall, 2002
These resulting equations are uncoupled and thus much easier to solve
The starting point is the eigenvectors (modes) and the orthogonality relations
One must also note that:
Paul A. Lagace © 2001
φ
~ k φ (s)
= δ rs
M r
ω r
2 (22-9)
Unit 22 - 16

MIT - 16.20
Thus:
Fall, 2002
(can show using equations (22-2) and (22-8) )
Have shown that the homogeneous solution to the general equation:
m q˙˙
~
+ k q = F is the sum of the eigenvectors (modes):
(22-1) q t
~
= n
φ ξ r
( ) r = 1 n = number of degrees of freedom
(22-10)
Where ξ r
(t) is basically a magnitude associated with the mode time t. The ξ r
become the “normalized coordinates”.
φ (r) at
 q
1
 





 q q q
M
2
3 n






=






φ
φ
1
2
M
M
( )
M
φ
1
M
M
M
(2)
M
φ
1
M
M
M
M
L
L
φ
φ
1 n
(n)
M
M 
M
 
 
 

 
ξ
ξ
ξ
ξ
1

2


3


 


M n


Paul A. Lagace © 2001 Unit 22 - 17

MIT - 16.20 which can be written as: q = φ ξ
~ ~ ~
Placing (22-11) into (22-1)
(22-11)
φ ξ
˙˙
+ k φ ξ
~ ~
= F
(22-12)
Now multiply this equation by the transpose of φ :
~
φ
~
T =





φ
φ
φ
1
2
 M
2
φ
L
L L 

L L L




φ
~
T m φ ξ
~ ~
+
~
φ T k φ ξ
~ ~
= φ
~
T
(22-13)
Notice that the terms of φ T
~
φ
~
and φ T k φ
~ ~
will result in most of the terms being zero due to the orthogonality relation (equation 22-8). Only the diagonal terms will remain.
Fall, 2002
Thus, (22-13) becomes a set of uncoupled equations: (via 22-8 and 22-9)
Paul A. Lagace © 2001 Unit 22 - 18

MIT - 16.20 Fall, 2002
M r
˙˙ r
+ M r
r
2
r
= Ξ r r = 1, 2 …n
(22-14)
That is:
M
1
ξ
1
+ M
1
ω
1
2 ξ
1
= Ξ
1
M
2
ξ
2
+ M
2
ω
2
2 ξ
2
= Ξ
2
M n
ξ n
+ M n
ω n
2 ξ n
= Ξ n where:
M r
= [ φ
1
φ
2
L ] m
~



φ
φ
1
2 r



 

M

= Generalized mass of rth mode and:
Paul A. Lagace © 2001 Unit 22 - 19

MIT - 16.20
Ξ r
= [ φ
1
φ
2
L ]
 F
1

 
 F
2
 = Generalized force of rth mode
Fall, 2002
ξ r
(t) = normal coordinates
The equations have been transformed to normal coordinates and are now uncoupled single degree-of-freedom systems
Implication: Each equation can be solved separately
The overall solution is then a superposition of the individual solutions
(normal nodes)
( Ξ = 0 )
--> solution…use same technique as before
• For any equation r:
ξ r
= a sin ω r t + b cos ω r t
Paul A. Lagace © 2001 Unit 22 - 20

MIT - 16.20
• Get a r
and b r
via transformed Initial Conditions
ξ r
0 =
1
φ
M ~ r m q i
0 = b r
ξ
˙ r
0 =
1
φ
M ~ r
m q˙ i
0 = a r
ω r
Notes:
• KEY SIMPLIFICATION is that often only first few (lowest) modes are excited so can solve only first few equations.
Can add more modes (equations) to improve solution if needed.
Fall, 2002
• This is a rigorous treatment -- no approximation made by going to normal coordinates.
But, this has all been based on the homogeneous case (free vibration), what about…
Paul A. Lagace © 2001 Unit 22 - 21

MIT - 16.20
( Ξ ≠ 0 )
Response is still made up of the natural modes. Solution is found using the same approach as for a single degree-of-freedom system…
• Break up each generalized force, Ξ r
, into a series of impulses
• Use Duhamel’s (convolution) integral to get response for each degree of freedom
• Stay in normalized coordinates
The solution for any mode will thus look like:
ξ r
( t ) =
1
M r
ω r
0 t
Ξ r
( τ ) sin ω r
( t − τ ) d τ and equation (22-11) then gives: q t i
( ) = n
φ i
ξ r
( ) r = 1
Again, use Initial Conditions to get constants
Fall, 2002
Paul A. Lagace © 2001 Unit 22 - 22

MIT - 16.20
Exact same procedure as single degree-of-freedom system. Do it multiple times and add up.
(Linear ⇒ Superposition)
Can therefore represent any system by discrete masses. As more and more discrete points are taken, get a better model of the actual behavior. Taking this to the limit will allow the full representation of the behavior of continuous systems .
Fall, 2002
Paul A. Lagace © 2001 Unit 22 - 23