Multiple Degree of Freedom Systems

Multiple Degree of Freedom

Systems

© D.J. Inman

2/58 Mechanical Engineering at Virginia Tech

The first step in analyzing multiple

degrees of freedom (DOF) is to look at 2

DOF

• DOF: Minimum number of coordinates to specify the position of a system

• Many systems have more than 1 DOF

• Examples of 2 DOF systems

– car with sprung and unsprung mass (both heave)

– elastic pendulum (radial and angular)

4.1 Two-Degree-of-Freedom

Model (Undamped)

A 2 degree of freedom system used to base

much of the analysis and conceptual

Free-Body Diagram of each mass

x

₁

x

₂

m

₁

m

2

k

₁

x

₁

Summing forces yields the

equations of motion:









1 1 1 1 2 2 1

2 2 2 2 1

1 1 1 2 1 2 2

2 2 2 1 2 2

( )

( )

( )

( )

(4.1)

( )

( )

( )

Rearranging terms:

( ) (

) ( )

( )

0

(4.2)

( )

( )

( )

0

m x t

k x t

k

x t

x t

m x t

k

x t

x t

m x t

k

k x t

k x t

m x t

k x t

k x t

 





 











Note that it is always the case

that

•

A 2 Degree-of-Freedom system has

– Two equations of motion!

The dynamics of a 2 DOF system consists

of 2 homogeneous and coupled equations

•

Free vibrations, so homogeneous eqs.

•

Equations are coupled:

– Both have x₁ and x₂.

– If only one mass moves, the other follows

– Example: pitch and heave of a car model

•

In this case the coupling is due to

k

₂

.

– Mathematically and Physically

Initial Conditions

•

Two coupled, second -order

, ordinary

differential equations with constant

coefficients

•

Needs

4 constants of integration

to solve

•

Thus

4 initial conditions

on positions and

velocities

1

(0)

10

, (0)

1 10

, (0)

2 20

, (0)

2 20

Solution by Matrix Methods

1 1 1

2 2 2

1 1 2 2

2 2 2

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 0 0

x t x t x t

t , t , t

x t x t x t

m k k k

M , K

m k k

M K        _{ }  _{ }  _{ }              _{ }  _{ }       

x x x

x x 0

0 ) ( ) ( ) ( 0 ) ( ) ( ) ( ) ( 2 2 1 2 2 2 2 2 1 2 1 1 1        t x k t x k t x m t x k t x k k t x m    

The two equations can be written in the form of a single matrix equation (see pages 272-275 if matrices and vectors are a struggle for you) :

(4.4), (4.5)

Initial Conditions

10 10

20 20

(0)

x

, and (0)

x

x

x





























x

x

The approach to a Solution:









2

2

Let ( )

1,

, , unknown

-j t

j t

t

e

j

M

K

e

M

K













 















x

u

u

0

u

u

0

u

0

For 1DOF we assumed the scalar solution aelt Similarly, now we assume the vector form:

(4.15)

(4.16)

This changes the differential

equation of motion into algebraic

vector equation:



2



1 2

-

(4.17)

This is two algebraic equation in 3 uknowns

( 1 vector of two elements and 1 scalar):

=

, and

M

K

u

u









 

 

 

u

0

The condition for solution of this matrix

equation requires that the the matrix

inverse does not exist:













2

1 2

2

If the inv - exists : which is the

static equilibrium position. For motion to occur

- does not exist

or det - (4.19)

M K

M K M K



 



  

_  

 

u 0

u 0

0

The determinant results in 1 equation

Back to our specific system: the

characteristic equation is defined as

฀ 

det -





2

M



K





0



det





2

m

₁



k

₁



k

₂



k

₂



k

₂





2

m

₂



k

₂













0



m

₁

m

₂



4



(

m

₁

k

₂



m

₂

k

₁



m

₂

k

₂

)



2



k

₁

k

₂



0

Eq. (4.21) is quadratic in



2



so four solutions result:

(4.20)

(4.21)

2 2

1

and

2 1

and

2

Once



is known, use equation (4.17) again to

calculate the corresponding vectors

u

₁

and

u

₂

2

1 1

2

2 2

( ) (4.22) and

( ) (4.23)

M K

M K 



  

  

u 0

u 0

This yields vector equation for each squared frequency:

Each of these matrix equations represents 2

equations in the 2 unknowns components of the vector, but the coefficient matrix is singular so

Examples 4.1.5 & 4.1.6:

calculating

u

and



•

m

₁

=9 kg,

m

₂

=1kg,

k

₁

=24 N/m and

k

₂

=3 N/m

•

The characteristic equation becomes



4

-6



2

+8=(



2

-2)(



2

-4)=0



2

= 2 and



2

=4 or



_1,3

 

2 rad/s,



_2,4

 

2 rad/s

Computing the vectors

u

11 2

1 1

12

2

1 1

11

12

11 12 11 12

For

=2, denote

then we have

(-

)

27 9(2)

3

0

3

3 (2)

0

9

3

0 and 3

0

u

u

M

K

u

u

u

u

u

u





 

  

 



 





 





 





 









 





 

 











u

u

0

2 equations, 2 unknowns but DEPENDENT!

11

11 12 12

2 1

1 1

results from both equations:

3 3

only the direction, not the magnitude can be determined!

This is because: det( ) 0.

The magnitude of the vector is arbitrary. To see this suppose

t

u

u u u

M K



  

  

1 2

1 1 1

2 2

1 1 1 1

hat satisfies

( ) , so does , arbitrary. So

( ) ( )

M K a a

M K a M K



 

  

      

u

u 0 u

u 0 u 0

Only the direction of vectors

u

can be

determined, not the magnitude as it

Likewise for the second value of



2

:

21 2

2 2

22

2 1

21

22

21 22 21 22

For

= 4, let

then we have

(-

)

27 9(4)

3

0

3

3 (4)

0

1

9

3

0 or

3

u

u

M

K

u

u

u

u

u

u









  







 













 

















 









 







 

u

u

0

What to do about the magnitude!

u

₁₂



1



u

₁



1 3

1













u

₂₂



1



u

₂



1 3

1













Several possibilities, here we just fix one element:

Choose:

Thus the solution to the algebraic

matrix equation is:



_{1, 3}

 

2, has mode shape

u

₁



1 3

1















_{2, 4}

 

2, has mode shape

u

₂



1 3

1













Here we have introduce the name

mode shape

to describe the vectors

Return now to the time response:



 



1 1 2 2

1 1 2 2

1 1 2 2

1 1 2 2

1 1 2 2

1 2

1 1 1 1 2 2 2 2

1 2 1 2

( ) , , ,

( )

( )

sin( ) sin( )

where , , , and are const

j t j t j t j t

j t j t j t j t

j t j t j t j t

t e e e e

t a e b e c e d e

t ae be ce de

A t A t

A A                                       

x u u u u

x u u u u

x u u

u u

ants of integration

We have computed four solutions:

Since linear, we can combine as:

determined by initial conditions.

(4.24)

(4.26)

Note that to go from the exponential

form to to sine requires Euler’s formula

Physical interpretation of all that

math!

• Each of the TWO masses is oscillating at TWO

natural frequencies ₁ and ₂

• The relative magnitude of each sine term, and

hence of the magnitude of oscillation of m₁ and m₂

is determined by the value of A₁u₁ and A₂u₂

• The vectors u₁ and u₂ are called mode shapes

What is a mode shape?

• First note that A₁, A₂, ₁ and ₂ are determined by the initial conditions

• Choose them so that A₂ = ₁ = ₂ =0

• Then:

• Thus each mass oscillates at (one) frequency ₁ with magnitudes proportional to u₁ the 1st mode shape

x(t)  x1(t) x₂(t)

 

 



  A₁ u11 u₁₂

 

  

A graphic look at mode shapes:

Mode 1:

k₁

m₁

x₁

m₂

x₂ k₂

Mode 2:

k₁

m₁

x₁

m₂

x₂ k₂

x₂=A x₂=A x₁=A/3

x₁=-A/3

u

₁



1 3

1













u

₂



1 3

1













If IC’s correspond to mode 1 or 2, then the response is purely in

Example 4.1.7

given the initial

conditions compute the time response

 

 

 

 

 

 

 

 





































































2 2 1 1 2 2 1 1 2 2 1 1 2 2 1 1

cos

2

cos

2

cos

3

2

cos

2

3

0

0

sin

sin

sin

3

sin

3

0

mm

1

















A

A

A

A

A

A

A

A

3



A

₁

sin

 



₁



A

₂

sin

 



₂

0



A

₁

sin

 



₁



A

₂

sin

 



₂

0



A

₁

2 cos

 



₁



A

₂

2 cos

 



₂

0



A

₁

2 cos

 



₁



A

₂

2 cos

 



₂

A

₁



1.5 mm,

A

₂

 

1.5 mm,



₁





₂





2

rad

4 equations in 4 unknowns:

The final solution is:

x

₁

(

t

)



0.5 cos 2

t



0.5 cos 2

t

x

₂

(

t

)



1.5 cos 2

t



1.5 cos 2

t

These initial conditions gives a response that is a combination of modes. Both harmonic, but their summation is not.

Figure 4.3

Solution as a sum of modes

฀ 

x

(

t

)



a

₁

u

₁

cos



₁

t



a

₂

u

₂

cos



₂

t

Determines how the first frequency contributes to the response

Things to note

• Two degrees of freedom implies two natural

frequencies

• Each mass oscillates at with these two frequencies

present in the response and beats could result

• Frequencies are not those of two component

systems

• The above is not the most efficient way to

calculate frequencies as the following describes

₁  2  k1

m₁  1.63,2  2 

k₂

Some matrix and vector reminders

1

2 2

1 2

1 2 2

1 1 2 2

2

1

0

0

0

0 for every value of except 0

T

T

T

a

b

d

b

A

A

c

c

ad

cb

c

a

x

x

m

M

M

m x

m x

m

M

M































































 



x x

x

x

x

x

x

4.2 Eigenvalues and Natural

Frequencies

•

Can connect the vibration problem with the

algebraic eigenvalue problem developed in

math

•

This will give us some powerful

computational skills

•

And some powerful theory

Some matrix results to help us use

available computational tools:

A matrix

M

is defined to be

symmetric

if

M



M

T

A symmetric matrix

M

is

positive definite

if

x

T

M

x



0 for all nonzero vectors

x

A symmetric positive definite matrix

M

can

be factored

M



LL

T

If the matrix

L

is diagonal, it defines

the

matrix square root

The matrix square root is the matrix M1/ 2 such that M1/ 2M1/ 2  M

If M is diagonal, then the matrix square root is just the root of the diagonal elements:

L  M1/ 2  m1 0

0 m₂ 

  



 

A change of coordinates is introduced to

capitalize on existing mathematics

1 1

2 ₂

1 1

1 1 1/ 2

1 1

2

1/ 2 1/ 2

1/ 2 1/ 2 1/ 2 1/ 2 identity symmetric

0 0 0 , , 0 0 0

Let ( ) ( ) and multiply by :

( ) ( ) (4.38)

m m

m _m

I K

m

M M M

m

t M t M

M MM t M KM t

               _{ }  _{ }          _{ }    x q

q q 0

1/ 2 1/ 2

or ( ) ( ) where

is called the mass normalized stiffness and is similar to the scalar

used extensively in single degree of freedom analysis. The key here is that i

t K t K M KM

k K m K     

q q 0

s a SYMMETRIC matrix allowing the use of many nice properties and computational tools

How the vibration problem relates to the

real symmetric eigenvalue problem

2

2

vibration problem real symmetric eigenvalue problem (4.40) (4.41)

Assume ( )

in ( )

( )

,

or

j t

j t j t

t

e

t

K

t

e

K e

K

K



 





l























q

v

q

q

0

v

v

0

v

0

v

v

v

v

v

0

2

Note that the martrix contains the same type of information

as does _n in the single degree of freedom case.

K

Important Properties of the

n

x

n

Real

Symmetric Eigenvalue Problem

•

There are

n

eigenvalues and they are all real

valued

•

There are

n

eigenvectors and they are all real

valued

•

The set of eigenvectors are orthogonal

•

The set of eigenvectors are linearly

independent

Square Matrix Review

• Let a_ik be the ikth element of A then A is symmetric if a_ik = a_ki denoted AT=A

• A is positive definite if xTAx > 0 for all nonzero x (also implies each l_i > 0)

• The stiffness matrix is usually symmetric and

positive semi definite (could have a zero eigenvalue)

Normal and orthogonal vectors

x  x₁ M x_n          

, y 

y₁ M y_n          

, inner product is xTy  x_iy_i

i1

n



x orthogonal to y if xTy  0 x is normal if xTx  1

   

if a the set of vectores is is both orthogonal and normal it is called an orthonormal set

Normalizing any vector can be

done by dividing it by its norm:

x

x

T

x

has norm of 1

To see this compute

(4.44)

x xTx

 xT

xTx

x xTx

 xTx

Examples 4.2.2 through 4.2.4

1 1

3 3

1/ 2 1/ 2

2

2 2

1 1 2 2

0

27

3

0

0

1

3

3

0

1

3

1

so

which is symmetric.

1

3

3-

-1

det(

)

det

6

8

0

-1

3-which has roots:

2

and

4

K

M

KM

K

K

l

I

l

l

l

l

l



l



 



 



 





 

 

 







 

 









 

























 





1 1

11 12

11 12 1

2 ₁

1 ₂

1

(

)

3 2

1

0

1

3 2

0

1

0

1

(1 1)

1

1

1

1

2

K

I

v

v

v

v

l









 





 





 





 









 





 

 

 



 

  

 



  



 



 

 

v

0

v

v

v

v

₂



1

2

1



1













,

v

₁T

v

₂



1

2

(1



1)



0

v

₁T

v

₁



1

2

(1



1)



1

v

T₂

v

₂



1

2

(1



(



1)(



1))



1



v

_i

are orthonormal

Likewise the second normalized eigenvector is

Modes

u

and Eigenvectors

v

are

different but related:

u

₁



v

₁

and

u

₂



v

₂

x



M

1/ 2

q



u



M

1/ 2

v

Note

M

1/ 2

u

₁



3

0

0

1













1

3

1



















1

1











 

v

₁

This orthonormal set of vectors is

used to form an

Orthogonal Matrix









1 2

1 1 1 2 2 1 2 2

1 2 1 1 2 2

1 2 2

1 1 1 2 1 2

1 2 2

1 2 1 2 2 2

1 0

0 1

0

diag( , )

0

T T T

T T

T T T

T T

T T

P

P P I

P KP P K K P l l

l l l   l l l   _{  }  _{  } _         _{ }       _{  } _        v v

v v v v

v v v v

v v v v

v v v v

v v v v

P _{is called an orthogonal matrix}

P _{is also called a modal matrix}

called a matrix of eigenvectors (normalized)

Example 4.2.3 compute

P

and show

that it is an orthogonal matrix

From the previous example:

P





v

₁

v

₁





1

2

1

1

1



1











 

P

T

P



1

2

1

2

1

1

1



1















1

1



1

1













1

2

1



1 1



1

1



1 1



1











 

1

2



2

₀

0

2









Example 4.2.4 Compute the square of

the frequencies by matrix manipulation

2 1

2 2

1 1 3 1 1 1

1 1

1 1 1 3 1 1

2 2

1 1 2 4

1

1 1 2 4

2

4 0 2 0 0

1

0 8 0 4

2 0 T P KP           _{     }                _   _             _{  } _{     }      





₁



2 rad/s and



₂



2 rad/s

In general:

 

2

diag

diag(

) (4.48)

T

i i

P KP

l



Example 4.2.5

Figure 4.4

The equations of motion:

1 1 1 2 1 2 2 2 2 2 1 2 3 2

(

)

0

(4.49)

(

)

0

m x

k

k x

k x

m x

k x

k

k x

















In matrix form these become:

1 2 2

1

2 2 3

2

0

0 (4.50) 0

k k k

m

k k k

m

 

 

 

 _{ } 

  _{ }

Next substitute numerical values

and compute

P

and



m₁  1 kg, m₂  4 kg, k₁  k₃  10 N/m and k₂=2 N/m





1/ 2 1/ 2

2

1 2

1 2

1 0 12 2

,

0 4 2 12

12 1 1 12

12 1

det det 15 35 0

1 12

2.8902 and 12.1098 1.7 rad/s and 12.1098 ra

M K

K M KM

K lI l l l

l l l            _{ }  _{ }           _{  }           _{ }           

Next compute the eigenvectors

1 11 21 11 21 1

2 2 2 2 2

1 11 21 11 11

11

For equation (4.41 ) becomes:

12 - 2.8902 1

0

1 3 - 2.8902

9.1089

Normalizing yields

1 (9.1089)

0.

v v v v

v v v v

v l          _               v v 21 1 2

1091, and 0.9940

0.1091 0.9940

, likewise

Next check the value of

P

to see

if it behaves as its suppose to:

 1 2

0.1091 0.9940

0.9940 0.1091

0.1091 0.9940 12 1 0.1091 0.9940 2.8402 0

0.9940 0.1091 1 3 0.9940 0.1091 0 12.1098

0.1091 0.9940 0.1091 0.9940

0.9940 0.1091 0.9940 0.109

T

T

P

P KP

P P



 

 _{  }

 

 

       

 _{      } _

 

       



 

 _ 

 

v v

1 0

1 0 1

   



   

   

A note on eigenvectors

In the previous section, we could have chosed v₂ to be

v₂  0.9940

0.1091

 

 



 instead of v₂  -0.9940

0.1091

 

 

 

because one can always multiple an eigenvector by a constant and if the constant is -1 the result is still a normalized vector.

Does this make any difference?

All of the previous examples can and should

be solved by “hand” to learn the methods

However, they can also be solved on

calculators with matrix functions and

with the codes listed in the last section

In fact, for more then two DOF one must

use a code to solve for the natural

frequencies and mode shapes.

Matlab commands

•

To compute the inverse of the square matrix

A:

inv(A)

or use

A\eye(n)

where n is the

size of the matrix

More commands

• To compute the matrix square root use sqrtm(A)

• To compute the Cholesky factor: L= chol(M)

• To compute the norm: norm(x)

• To compute the determinant det(A)

• To enter a matrix:

K=[27 -3;-3 3]; M=[9 0;0 1];

An alternate approach to normalizing

mode shapes

From equation (4.17)





M



2



K



u



0,

u



0

Now scale the mode shapes by computing



such that

 



_iu_i T M

 



_iu_i  1 



_i  1 uT_i u_i

2 2

is called

and it satisfies:

0

,

1, 2

i i i

T

i i i i i i

mass normalized

M

K

K

i













 





w

u

w

w

w

w

There are 3 approaches to computing

mode shapes and frequencies

(i) 2Mu  Ku (ii) 2u  M1Ku (iii) 2v  M12 KM12v

(i) Is the Generalized Symmetric Eigenvalue Problem easy for hand computations, inefficient for computers

(ii) Is the Asymmetric Eigenvalue Problem very expensive computationally