Chapter 4 Multiple Degree of Freedom Systems

© D.J. Inman1/58 Mechanical Engineering at Virginia Tech

Chapter 4 Multiple Degree of Freedom Systems

Extending the first 3 chapters to more then one degree of freedom

The Millennium bridge requiredmany degrees of freedom to modeland design with.


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)

Fig 4.1


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

m1m2k1 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 aet

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

det 2m1 k1 k2 k2

k2 2m2 k2

0

m1m24 (m1k2 m2k1 m2k2 )2 k1k2 0

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

(4.20)

(4.21)

2 21 2 1 2 and and


Once is known, use equation (4.17) again to

calculate the corresponding vectors u1 and u2

21 1

22 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 becomes4-62+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 u112

1 112

21 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

21

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

21 1 1

2 21 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

2 222

21

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 vectorsu1 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 exponentialform 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:

k1m1

x1

m2

x2k2

Mode 2:

k1

m1

x1

m2

x2k2

x2=A

x2=Ax1=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

sin30

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 2 cos 2 0 A1 2 cos 1 A2 2 cos 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) 0.5 cos 2t 0.5 cos2t

x2 (t) 1.5 cos 2t 1.5 cos2tThese 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 theresponse

Determines how the second frequency contributes to theresponse


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

m2

1.732


Some matrix and vector reminders

1

2 21 2

1 2 21 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

xT Mx 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

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

Window 4.1 page 285


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 i > 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 xT y xi yi

i1

n

x orthogonal to y if xT y 0

x is normal if xT x 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 xT x (4.43)


Normalizing any vector can be done by dividing it by its norm:

x

xT x has norm of 1

To see this compute

(4.44)

x

xT x

xT

xT x

x

xT x

xT x

xT x1


Examples 4.2.2 through 4.2.41 1

3 31/ 2 1/ 2

2

2 21 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 I


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

v 0

v

v

v

The first normalized eigenvector


v2 1

2

1

1

, v1

T v2 1

2(1 1) 0

v1T v1

1

2(11) 1

v2T v2

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 1/2v

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 21 2

21 2 1 2 2 2

1 0

0 1

0diag( , )

0

T TT

T T

T T T

T T

T T

P

P P I

P KP P K K P

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

PT P 1

2

1

2

1 1

1 1

1 1

1 1

1

2

11 1 1

1 1 11

1

2

2 0

0 2

I


Example 4.2.4 Compute the square of the frequencies by matrix manipulation

21

22

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


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 I

d/s


Next compute the eigenvectors1

11

21

11 21

1

2 2 2 2 21 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

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 eigenvectorsIn 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 methodsHowever, they can also be solved on calculators with matrix functions and with the codes listed in the last sectionIn 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 T M iui 1 i 1

uiT ui

2 2

is called and it satisfies:

0 , 1, 2

i i i

Ti 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) 2u M 1Ku (iii) 2v M

12 KM

12 v

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

(ii) Is the Asymmetric Eigenvalue Problemvery expensive computationally

(iii) Is the Symmetric Eigenvalue Problemthe cheapest computationally

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Chapter 4 Multiple Degree of Freedom Systems

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