Multiple Degree of Freedom

Systems

The Millennium bridge required

many degrees of freedom to model

and design with.

© 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)

– motions of a ship (roll and pitch)

4.1 Two-Degree-of-Freedom

Model (Undamped)

A 2 degree of freedom system used to base

much of the analysis and conceptual

development of MDOF systems on.

Free-Body Diagram of each mass

x1 x2

m1 m2 k1 x1

k2(x2 -x1)

Figure 4.2

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!

– Two natural frequencies (as we shall see)!

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 k2.

– Mathematically and Physically

– If k2 = 0, no coupling occurs and can be solved

as two independent SDOF systems

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 10 1 10 2 20 2 20(0) , (0) , (0) , (0)x x x x x x x x

Solution by Matrix Methods

1 1 1

2 2 2

1 1 2 2

2 2 2

( ) ( ) ( )( ) ( ) ( )

( ) ( ) ( )

0

0

x t x t x tt , t , t

x t x t x t

m k k kM , K

m k k

M K

x x x

x x 0

0)()()(

0)()()()(

221222

2212111

txktxktxm

txktxkktxm

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)

(4.6), (4.9)

Initial Conditions

10 10

20 20

(0) , and (0)x x

x x

x x

IC’s can also be written in vector form

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)

(4.17)

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

u

The condition for solution of this matrix

equation requires that the the matrix

inverse does not exist:

2

12

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

in one unknown (called the characteristic equation)

Back to our specific system: the

characteristic equation is defined as

det -2M K 0

det2

m1 k1 k2 k2

k2 2m2 k2

0

m1m24 (m1k2 m2k1m2k2 )2 k1k2 0

Eq. (4.21) is quadratic in 2so four solutions result:

(4.20)

(4.21)

2 2

1 2 1 2 and and

Once is known, use equation (4.17) again to

calculate the corresponding vectors u1 and u2

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

each matrix equation results in only 1 independent

equation. The following examples clarify this.

Examples 4.1.5 & 4.1.6:calculating u

and

• m1=9 kg,m2=1kg, k1=24 N/m and k2=3 N/m

• The characteristic equation becomes

4-62+8=(2-2)(2-4)=0

2 = 2 and 2 =4 or

1,3 2 rad/s, 2,4 2 rad/s

Each value of 2 yields an expression or u:

Computing the vectors u

112

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! (the 2nd equation is -3 times the first)

1111 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

uu 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

remains arbitrary

Likewise for the second value of 2:

212

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

19 3 0 or

3

u

u

M K

u

u

u u u u

u

u 0

Note that the other equation is the same

What to do about the magnitude!

u12 1 u1 1

3

1

u22 1 u2 1

3

1

Several possibilities, here we just fix one element:

Choose:

Choose:

Thus the solution to the algebraic

matrix equation is:

1,3 2, has mode shape u1 1

3

1

2,4 2, has mode shape u2 1

3

1

Here we have introduce the name

mode shape to describe the vectors

u1 and u2. The origin of this name comes later

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

for trig functions and uses up the

+/- sign on omega

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

because the describe the relative magnitude of

oscillation between the two masses

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

sin1t A1u1 sin1t

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

u1 1

3

1

u2 1

3

1

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

mode 1 or mode 2.

Example 4.1.7 given the initial

conditions compute the time response

2211

22

11

2

1

2211

22

11

2

1

2cos22cos2

2cos23

2cos23

)(

)(

2sin2sin

2sin3

2sin3

)(

)(

0

0)0( mm,

0

1=(0)consider

tAtA

tA

tA

tx

tx

tAtA

tA

tA

tx

tx

xx

2211

22

11

2211

22

11

cos2cos2

cos3

2cos23

0

0

sinsin

sin3

sin3

0

mm 1

AA

AA

AA

AA

At t=0 we have

3 A1 sin 1 A2 sin 2

0 A1 sin 1 A2 sin 2

0 A1 2 cos 1 A2 2cos 2

0 A1 2 cos 1 A2 2cos 2

A1 1.5 mm,A2 1.5 mm,1 2

2 rad

4 equations in 4 unknowns:

Yields:

The final solution is: x1(t) 0.5cos 2t 0.5cos2t

x2(t) 1.5cos 2t 1.5cos2t

These initial conditions gives a response that is a combination

of modes. Both harmonic, but their summation is not.

Figure 4.3

(4.34)

Solution as a sum of modes

x(t) a1u1 cos1t a2u2 cos2t

Determines how the first

frequency contributes to the

response

Determines how the second

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 k1m1

1.63,2 2 k2

m2

1.732

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 bA A

c c c aad cb

x x

mM M m x m x

m

M M

x x

x x

x x x

Then M is said to be positive definite

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

• All the codes have eigen-solvers so these painful calculations can be automated

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

xTMx 0 for all nonzero vectors x

A symmetric positive definite matrix M can

be factored M LLT

Here L is upper triangular, called a Cholesky matrix

If the matrix L is diagonal, it defines

the matrix square root

The matrix square root is the matrix M 1/2 such that

M 1/2M 1/2 M

If M is diagonal, then the matrix square root is just the root

of the diagonal elements:

L M 1/2 m1 0

0 m2

(4.35)

A change of coordinates is introduced to

capitalize on existing mathematics

11

2 2

111 1 1/ 2

1 12

1/ 2 1/ 2

1/ 2 1/ 2 1/ 2 1/ 2

identity symmetric

000, ,

00 0

Let ( ) ( ) and multiply by :

( ) ( ) (4.38)

mm

m m

I K

mM M M

m

t M t M

M MM t M KM t

x q

q q 0

1/ 2 1/ 2or ( ) ( ) 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

kK

m

K

q q 0

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

computational tools

For a symmetric, positive definite matrix M:

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 in the single degree of freedom case.n

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

• The matrix is similar to a diagonal matrix

Square Matrix Review • Let aik be the ikth element of A then A is symmetric

if aik = aki denoted AT=A

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

• The stiffness matrix is usually symmetric and positive semi definite (could have a zero eigenvalue)

• The mass matrix is positive definite and symmetric (and so far, its diagonal)

Normal and orthogonal vectors

x

x1

M

xn

, y

y1

M

yn

, inner product is xTy xiyii1

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

The norm of x is x xTx (4.43)

Normalizing any vector can be

done by dividing it by its norm:

x

xTx

has norm of 1

To see this compute

(4.44)

x

xTx

xT

xTx

x

xTxxTx

xTx 1

Examples 4.2.2 through 4.2.4

1 13 31/ 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- -1det( ) det 6 8 0

-1 3-

which has roots: 2 and 4

K M KM

K

K Il

l l ll

l l

1 1

11

12

11 12 1

2 11 2

1

( )

3 2 1 0

1 3 2 0

10

1

(1 1) 1

11

12

K I

v

v

v v

l

v 0

v

v

vThe first normalized eigenvector

v2 1

2

1

1

, v1

Tv2

1

2(11) 0

v1

Tv1

1

2(11) 1

v2

Tv2

1

2(1 (1)(1)) 1

v i are orthonormal

Likewise the second normalized eigenvector is

computed and shown to be orthogonal to the first,

so that the set is orthonormal

Modes u and Eigenvectors v are

different but related:

u1 v1 and u2 v2

x M 1/2q u M 1/2

v

Note

M 1/2u1

3 0

0 1

13

1

1

1

v1

(4.37)

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 21 1 1 2 1 2

1 2

21 2 1 2 2 2

1 0

0 1

0diag( , )

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

ll l

ll 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)

(4.47)

Example 4.2.3 compute P and show

that it is an orthogonal matrix

From the previous example:

P v1 v1 1

2

1 1

1 1

PTP 1

2

1

2

1 1

1 1

1 1

1 1

1

2

11 11

11 11

1

2

2 0

0 2

I

Example 4.2.4 Compute the square of

the frequencies by matrix manipulation

2

1

2

2

1 1 3 1 1 11 1

1 1 1 3 1 12 2

1 1 2 41

1 1 2 42

4 0 2 0 01

0 8 0 42 0

TP KP

1 2 rad/s and 2 2 rad/s

In general: 2diag diag( ) (4.48)T

i iP 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 21

2 2 32

00 (4.50)

0

k k km

k k km

x x

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 1det 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 Il

l l ll

l l

d/s

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

0.9940 0.1091

v

v v

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

Yes!

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?

No! Try it in the previous example

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.

Next we examine 3 other formulations for

solving for modal data

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

• [P,D]=eig(A) computes the eigenvalues and

normalized eigenvectors (watch the order).

Stores them in the eigenvector matrix P and

the diagonal matrix D (D=)

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];

• To multiply: K*inv(chol(M))

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

iui TM iui 1 i

1

uiTui

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

(4.53)

There are 3 approaches to computing

mode shapes and frequencies

(i) 2 Mu Ku (ii) 2

u M 1Ku (iii) 2v M

12 KM

12 v

(i) Is the Generalized Symmetric Eigenvalue Problem

easy for hand computations, inefficient for computers

(ii) Is the Asymmetric Eigenvalue Problem

very expensive computationally

(iii) Is the Symmetric Eigenvalue Problem

the cheapest computationally