Chapter 4 Multiple Degree of Freedom Systems

The Millennium bridge required

many degrees of freedom to model

and design with.

Extending the first 3 chapters to more

then one degree of freedom

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)

– Airplane roll, pitch and yaw

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

Figure 4.2

k2(x

2 -x

1)

k1 x

1

m1

m2

x1

x

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)!

• Thus some new phenomena arise in going from one to two

degrees of freedom

– Look for these as you proceed through the material

• Two instead of one natural frequency

– Leading to two possible resonance conditions

• The concept of a mode shape arises

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

0)()()(

0)()()()(

221222

2212111

txktxktxm

txktxkktxm

Solution by Matrix Methods

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) :

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

(4.4), (4.5)

(4.6), (4.9)

Initial Conditions (two sets needed one for each equation of

motion)

IC’s can also be written in vector form

10 10

20 20

(0) , and (0)x x

x x

x x

The approach to a Solution:

For 1DOF we assumed the scalar solution aeλt

Similarly, now

we assume the vector form:

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

(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 ω

(called the characteristic equation)

Back to our specific system: the characteristic equation is

defined as

2

2

1 1 2 2

2

2 2 2

4 2

1 2 1 2 2 1 2 2 1 2

det - 0

det 0

( ) 0

M K

m k k k

k m k

m m m k m k m k k k

Eq. (4.21) is quadratic in ω2 so four solutions result:

2 2

1 2 1 2 and and

Once ω is known, use equation (4.17) again to calculate

the corresponding vectors u1 and u2

This yields vector equation for each squared frequency:

2

1 1

2

2 2

( ) (4.22)

and

( ) (4.23)

M K

M K

u 0

u 0

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 ω

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

• The characteristic equation becomes

ω4-6ω2+8=(ω2-2)(ω2-4)=0

ω2 = 2 and ω2 =4 or

1,3 2,42 rad/s, 2 rad/s

Each value of ω2 yields an expression for 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)

Only the direction of vectors u can be determined, not

the magnitude as it remains arbitrary

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

Likewise for the second value of ω2

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!

Several possibilities, here we just fix one element:

13

12 1

13

22 2

1 1

1 1

u

u

u

u

Choose:

Choose:

Thus the solution to the algebraic matrix equation is:

13

1,3 1

13

2,4 2

2, has mode shape 1

2, has mode shape 1

u

u

Here we have introduce the name

mode shape to describe the vectors

u1 and u2. The origin of this name comes later

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

Return now to the time response:

(4.24)

(4.26)

We have computed four solutions:

Since linear, we can combine as:

determined by initial conditions.

Physical interpretation of all that math!

• Each of the TWO masses is oscillating at TWO natural

frequencies ω1and ω2

• 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

1 11

1 1 1 1 1

2 12

( )( ) sin sin

( )

x t ut A t A t

x t u

x u

What is a mode shape?

• First note that A1, A2, Φ1 and Φ2 are determined by

the initial conditions

• Choose them so that A2 = Φ1 = Φ2 =0

• Then:

• Thus each mass oscillates at (one) frequency w1 with

magnitudes proportional to u1 the 1st mode shape

13

21

u

13

11

u

A graphic look at mode shapes:

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

mode 1 or mode 2.

Mode 1:

Mode 2:

m1

m2

x1 x

2

x1=A/3

x2=A

m1

x1=-A/3 x

2=A

m2

x1 x

2

k1

k1

k2

k2

Example 4.1.7 given the initial conditions compute

the time response

1 21 2

1

21 1 2 2

1 21 2

1

21 1 2 2

1 0consider (0)= mm, (0)

0 0

sin 2 sin 2( )3 3

( )sin 2 sin 2

2 cos 2 2cos 2( )3 3

( )2 cos 2 2cos 2

A At tx t

x tA t A t

A At tx t

x tA t A t

x x

At t = 0 we have

2211

22

11

2211

22

11

cos2cos2

cos3

2cos23

0

0

sinsin

sin3

sin3

0

mm 1

AA

AA

AA

AA

4 equations in 4 unknowns:

1 1 2 2

1 1 2 2

1 1 2 2

1 1 2 2

1 2 1 2

3 sin sin

0 sin sin

0 2 cos 2cos

0 2 cos 2cos

1.5 mm, 1.5 mm, rad2

A A

A A

A A

A A

A A

Yields:

The final solution is:

1

2

( ) 0.5cos 2 0.5cos 2 mm

( ) 1.5cos 2 1.5cos 2 mm

x t t t

x t t t

These initial conditions gives a response that is a combination of modes.

Both harmonic, but their summation is not.

(4.34)

Figure 4.3a x1(t) Figure 4.3b x

2(t)

tatat 222111 coscos)( uux

Solution as a sum of modes

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

1 21 2

1 2

2 1.63, 2 1.732k k

m m

• The above is not the most efficient way to calculate

frequencies as the following describes

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:

TM M

0 for all nonzero vectors T M x x x

A symmetric positive definite matrix M can be factored

TM LL

Here L is upper triangular, called a Cholesky matrix

A matrix M is defined to be symmetric if

A symmetric matrix M is positive definite if

If the matrix L is diagonal, it defines the matrix square

root

1/2

1/2 1/2

The matrix square root is the matrix such that

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

of the diagonal elements:

M

M M M

M

11/2

2

0 (4.35)

0

mL M

m

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 diagonal, 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

Properties of the n x n Real Symmetric Matrix

• There are n eigenvalues and they are all real valued

• There are n eigenvectors and they are all real valued

• The eigenvalues are all positive if and only if the matrix is

positive definite

• The set of eigenvectors can be chosen to be orthogonal

• The set of eigenvectors are linearly independent

• The matrix is similar to a diagonal matrix

• Numerical schemes to compute the eigenvalues and

eigenvectors of symmetric matrix are faster and more efficient

Square n x n 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 λi > 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

11 1

, , inner product is

orthogonal to if 0

is normal if 1

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

i

nT

i i

i

n n

T

T

x yx y

x y

x y x y

x y x y

x x x

2

1

s called an set

The norm of is (4.43)n

T

i

i

orthonormal

x

x x x x

Normalizing any vector can be done by dividing it by its

norm:

has norm of 1T

x

x x

1T T

TT T T

x x x x x

x xx x x x x x

(4.44)

To see this compute

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 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

Likewise the second normalized eigenvector is computed

and shown to be orthogonal to the first, so that the set is

orthonormal

2 1 2

1 1

2 2

11 1, (1 1) 0

1 22

1(1 1) 1

2

1(1 ( 1)( 1)) 1

2

are orthonormal

T

T

T

i

v v v

v v

v v

v

Modes u and Eigenvectors v are different but related:

1 1 2 2

1/2 1/2

1/2

1 1

and

Note

13 0 13

0 1 11

M M

M

u v u v

x q u v

u v

(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

v v

v v v v

v v v v

v v v v

v v v v

v v v v

called a matrix of eigenvectors (normalized)

P is called an orthogonal matrix

P is also called a modal matrix

Example 4.2.4 compute P and show that it is an

orthogonal matrix

1 1

1 11

1 12

1 1 1 11 1

1 1 1 12 2

1 1 1 1 2 01 1

1 1 1 1 0 22 2

T

P

P P

I

v v

From the previous example:

Example 4.2.5 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 22 rad/s and 2 rad/s

2diag diag( ) (4.48)T

i iP KP

In general:

Example 4.2.6

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

1 2 21

2 2 32

00 (4.50)

0

k k km

k k km

x x

Figure 4.4 The equations of motion:

In matrix form these become:

Next substitute numerical values and compute P and Λ

1 2 1 3 21 kg, 4 kg, 10 N/m and =2 N/mm m k k k

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 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

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

2

2 2

In the previous section, we could have chosed to be

0.9940 -0.9940 instead of

0.1091 0.1091

because one can always multiple an eigenvector by a constant

and if the constant is -1 th

v

v v

e 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=L)

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))

2 0, 0M K u u

An alternate approach to normalizing mode shapes

Now scale the mode shapes by computing such that

1 1

T

i i i i iT

i i

M

u uu u

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

From equation (4.17)

(4.53)

1 12 22 2 1 2(i) (ii) (iii) M K M K M KM

u u u u v v

There are 3 approaches to computing mode shapes and

frequencies

(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

Some Review: Window 4.3

Orthonormal Vectors

similar to the unit vectors of statics and dynamics

. and of values all for

if be to set are vectors of set A

if 1

if 0

by dabbreviate is This

0xx if are and

1 and 1 if both are and

2T1

221121

ji

lorthonorman

ji

ji

orthogonal

normal

ijjTi

i

ijjTi

TT

xx

x

xx

xxxxxx

4.3 - Modal Analysis

• Physical coordinates are not always the easiest to work in

• Eigenvectors provide a convenient transformation to

modal coordinates

– Modal coordinates are linear combination of physical

coordinates

– Say we have physical coordinates x and want to

transform to some other coordinates u

212

211

3

3

xxu

xxu

1 1

2 2

1 3

1 3

u x

u x

Review of the Eigenvalue Problem

#1) trans. (coord.

let and

as Rewrite . and conditions initial

Assume matrices. are and and vector

a is where )( with Start

21

21

21

21

00

qxqx

0xx

xx

x0xx

q

MM

KMM

KM

KtM

Eigenproblem (cont)

0qqqq

K

KK

MKM

I

MM

M

T

~

~~

get to by yPremultipl

21

21

21

21

21

• Now we have a symmetric, real matrix

• Guarantees real eigenvalues and distinct, mutually

orthogonal eigenvectors

(4.55)

Eigenvectors = Mode Shapes?

.synonymous not are they but

tion,transforma simple a by related

are two The matrices. of sticscharacteri

are sEigenvetor s.coordinate physical in

= to solutions are shapes Mode 2 uu KM

Eigenvectors vs. Mode Shapes

P

K

KMMPKUU

IMUUUMMUPP

UMPPMU

M

IPP

K

TT

TTT

T

~

Similarly,

matrix. mass the w.r.t. only orthogonal are shapes

mode the Thus, .

Now, . shapes modes the

, tiontransforma the Using l?orthonorma

shapes mode the Are . i.e., l,orthonorma are

~ matrix PD symmetric the of rseigenvecto The

21

21

21

21

21

21

21

qx

The Matrix of eigenvectors can be used to decouple the

equations of motion

1If orthonormal (unitary), T TP P P I P P

Thus, diagonal matrix of eigenvalues.

Back to 0. Make the additional coordinate

transformation and premultiply by

Pr Pr 0 (4.59)

T

T

r

T T

P KP

q Kq

q P P

P P K Ir r

• Now we have decoupled the EOM i.e., we have n

independent 2nd-order systems in modal coordinates r(t)

Writing out equation (4.59) yields

21 11

22 22

2

1 1 1

2

2 2 2

( ) ( )1 0 00 (4.60)

( ) ( )0 1 00

( ) ( ) 0 (4.62)

( ) ( ) 0 (4.63)

r t r t

r t r t

r t r t

r t r t

We must also transform the initial conditions

101 1/2

0

202

101 1/2

0

202

(0)(0) (0) (4.64)

(0)

(0)(0) (0) (4.65)

(0)

T T

T T

rrP P M

rr

rrP P M

rr

r q x

r q x

rx PM 21

This transformation takes the problem from couple

equations in the physical coordinate system in to

decoupled equations in the modal coordinates

k1 k

2

1)2

1

r1

2)2

1

r2

x1

x2

m1 m2

Physical Coordinates. Coupled equations Modal Coordinates.

Uncoupled equations

21

MPT

Modal Transforms to SDOF

• The modal transformation

transforms our 2 DOF into 2 SDOF systems

• This allows us to solve the two decoupled SDOF systems

independently using the methods of chapter 1

• Then we can recombine using the inverse transformation

to obtain the solution in terms of the physical coordinates.

The free response is calculated for each mode

independently using the formulas from chapter 1:

00

22 10 00 2

0

( ) sin cos , 1,2

or (see Window 4.3 for a reminder)

( ) sin( tan ), 1,2

ii i i i

i

i i ii i i

i i

rr t t r t i

r rr t r t i

r

Note, the above assumes neither frequency is zero

Once the solution in modal coordinates is determined (ri)

then the response in Physical Coordinates is computed:

• With n DOFs these transformations are:

12

1 1

where nxn

(t) S (t)

n nn n

S M P

nxnn n

x r

(where n = 2 in the previous slides)

Steps in solving via modal analysis (Window 4.5)

Example 4.3.1 via MATLAB (see text for hand

calculations)

9 0 27 3 0 0, , (0) , (0)

0 1 3 3 1 0M K

x x

• Follow steps in Window 4.5 (page 337)

21

Calculate 1)

M 21

21~

Calculate 2)

MKMK

» Minv2 = inv(sqrt(M))

Minv2 =

0.3333 0

0 1.0000

»Kt =Minv2*K*Minv2

Kt =

3 -1

-1 3

Example 4.3.1 solved using MATLAB as a calcuator

% 3) Calculate the symmetric eigenvalue problem for K tilde [P,D] = eig(Kt);

[lambda,I]=sort(diag(D)); % just sorts smallest to largest

P=P(:,I); % reorder eigenvectors to match eigenvalues

»lambda =

2

4

P =

-0.7071 -0.7071

-0.7071 0.7071

Example 4.3.1 (cont)

% 4) Calculate S = M^(-1/2) * P and Sinv = P^T * M^(1/2)

S = Minv2 * P;

Sinv = inv(S);

% 5) Calculate the modal initial conditions

r0 = Sinv * x0;

rdot0 = Sinv * v0;

Example 4.3.1 (cont)

% 6) Find the free response in modal coordinates

tmax = 10;

numt = 1000;

t = linspace(0,tmax,numt);

[T,W]=meshgrid(t,lambda.^(1/2));

% Use Tony's trick

R0 = r0(:,ones(numt,1));

RDOT0 = rdot0(:,ones(numt,1));

r = RDOT0./W.*sin(W.*T) + R0.*cos(W.*T);

% 7) Transform back to physical space

x = S*r;

Example 4.3.1 (cont) % Plot results

figure

subplot(2,1,1)

plot(t,r(1,:),'-',t,r(2,:),'--')

title('free response in modal coordinates')

xlabel('time (sec)')

legend('r_1','r_2')

subplot(2,1,2)

plot(t,x(1,:),'-',t,x(2,:),'--')

title('free response in physical coordinates')

xlabel('time (sec)')

legend('x_1','x_2')

1 1

1

1

2 1

2

2 2

2

2 4.44sec,

4 2

sec

T

T

Modal and Physical Responses

Modal Coordinates:

Independent

oscillators

Physical

Coordinates:

Coupled oscillators

Note IC

Free response in modal coordinates

Free response in physical coordinates

0 1 2 3 4 5 6 7 8 9 10 -4

-2

0

2

4

Time (s)

x 1

x 2

0 1 2 3 4 5 6 7 8 9 10 -4

-2

0

2

4

r 1

r 2

sec sec

Section 4.4 More then 2 Degrees of Freedom

Fig 4.8

Extending previous section to

any number of degrees of

freedom

Fig 4.7

A FBD of the system of figure 4.8 yields the n equations

of motion o the form:

1 1 1 0, 1,2,3 (4.83)i i i i i i i im x k x x k x x i n

Writing all n of these equations and casting them in matrix

form yields:

( ) ( ) , (4.80)M t K t x x 0

where:

the relevant matrices and vectors are:

1 2 2

1

2 2 3 3

2

3

1

0 00 0

00 0

, 0 (4.83)

0 00 0

n n n

n

n n

k k km

k k k km

M K k

k k km

k k

1 1

2 2

( ) ( )

( ) ( )( ) , ( )

( ) ( )n n

x t x t

x t x tt t

x t x t

x x

For such systems as figure 4.7 and 4.8 the process stays

the same…just more modal equations result:

0)()(

0)()(

0)()(

0)()(

4.3 section as same the stays Process

2

3233

2222

1211

trtr

trtr

trtr

trtr

nnn

Just get more modal

equations, one for each

degree of freedom (n is

the number of dof)

See example 4.4.2 for details

The Mode Summation Approach

• Based on the idea that any possible time response is just a

linear combination of the eigenvectors

1 1

1

Starting with ( ) ( ) (4.88)

let ( )= ( )

two linearly independent solutions for each term.

can also write this as ( ) sin (4.92)

i i

n nj t j t

i i i i

i i

n

i i i i

i

t K t

t t a e b e

t d t

q q 0

q q v

q v

Mode Summation Approach (cont)

jjjTj

jj

n

ii

Tjii

n

iiii

Tj

Tj

ijiTj

n

iiiii

n

iiii

ii

d

ddd

dd

d

cos=)0( ,velocities initial the for Similarly

sinsin=sin=)0(

that such normalized rseigenvecto Assuming

cos)0( and sin)0(

I.C. the from and constants the Find

11

11

qv

vvvvqv

vv

vqvq

Mode Summation Approach (cont)

1

Solve for and from the two IC equations

(0) (0) and tan

sin (0)

about (0)=

if you just crank it through the above expressions

you might conclude that 0,

i i

T T

i i ii i T

i i

i

d

d

d

v q v q

v q

IMPORTANT NOTE q 0

i.e., the trivial soln.

Be careful with (0) = 0 as well. q

Mode Summation Approach for zero initial displacement

1

If (0) = 0, the return to

(0) sin

and realize that 0 instead of 0.

The compute from the velocity expression

(0)= cos

n

i i i

i

i i

i

T

i i i i

d

d

d

d

q

q v

v q

Mode Summation Approach with rigid body modes (ω1 =

0)

1

0 0

1 1 1 1 1

if 0,

( ) ( )v ( )v

does not give two linearly independent solutions.

Now we must use the expansion

j t j t

i iq t a e b e a b

1 1 1

2

( ) ( )v ( )v

and adjust calculation of the constants from the

initial conditions accordingly.

i i

nj t j t

i i i

i

q t a b t a e b e

Note that the underline term is a translational motion

Example 4.3.1 solved by the mode summation method

1/2

1/2 1/2

0 0 0 0

11 1

2

3 0 1 11From before, we have and =

0 1 1 12

3 0Appropriate IC are = = , = =

0 0

(0) (0) 2tan tan

(0) 0

2

(0)

T T

i i i ii T

i

T

ii

M V

M M

d

q x q v

v q v q

v q

v q 1

2

3 22

sin 3 22

i

d

d

Example 4.3.1 constructing the summation of modes

1

2

the first mode the second mode

( ) 1 13 2 1 3 2 1sin 2 sin 2

( ) 1 12 2 2 22 2

q tt t

q t

Transforming back to the physical coordinates yields:

1/21 10 01 13 2 1 3 2 13 3( ) sin 2 sin 2

1 12 2 2 22 20 1 0 1

1 13 2 1 3 2 1 sin 2 sin 2

1 12 2 2 23 2 3 2

t M t t

t t

x q

Example 4.3.1 a comparison of the two solution

methods shows they yield identical results

21

21~

KMMK

Steps for Computing the Response By Mode Summation

1. Write the equations of motion in matrix form, identify M

and K

2. Calculate M -1/2

(or L)

3. Calculate

4. Compute the eigenvalue problem for the matrix

iiK v and get and ~ 2

5. Transform the initial conditions to q(t)

1 12 2(0) (0) and (0) (0)M M q x q x

Summary of Mode Summation Continued

6. Calculate the modal expansion coefficients and phase

constants

i

Ti

iTi

Tii

i d

sin

)0( ,

)0(

)0(tan 1 qv

qv

qv

7. Assemble the time response for q

n

iiiii tdt

1

sin)( vq

8. Transform the solution to physical coordinates

1

2

1

( ) ( ) sinn

i i i i

i

t M t d t

x q u

Nodes of a Mode Shape

• Examination of the mode shapes in Example 4.4.3 shows

that the third entry of the second mode shape is zero!

• Zero elements in a mode shape are called nodes.

• A node of a mode means there is no motion of the mass or

(coordinate) corresponding to that entry at the frequency

associated with that mode.

2

0.2887

0.2887

0

0.2887

u

The second mode shape of Example 4.4.3 has a node

• Note that for more then 2 DOF, a mode shape may have a

zero valued entry

• This is called a node of a mode.

node

They make great mounting points in machines

A rigid body mode is the mode associated with a zero

frequency

• Note that the system in Fig 4.12 is not constrained and can

move as a rigid body

• Physically if this system is displaced we would expect it to move

off the page whilst the two masses oscillate back and forth

Fig 4.11

Example 4.4.4 Rigid body motion

1 1 2 1 2 2 2 1

1 1 1

2 2 2

( ) and ( )

0 1 1 0

0 1 1 0

m x k x x m x k x x

m x xk

m x x

1 2

0 0

1 kg, 4 kg, 400 N subject to

0.01 m and 0

0

m m k

x v

The free body diagram of figure 4.11 yields

Solve for the free response given:

Following the steps of Window 4.5

1/2

1/2 1/2

2

1 2 1 2

1 0

1. 10

2

1 0 1 01 1 400 200

2. 400 1 11 1 200 1000 0

2 2

4 23. det 100det 100 5 0

2 1

0 and 5 0, 2.236 rad/s

M

K M KM

K I

Indicates a rigid body motion

Now calculate the eigenvectors and note in particular that

they cannot be zero even if the eigenvalue is zero

11

11 21

21

1 1

2

4 0 2 00 100 4 2 0

2 1 0 0

1 0.4472 or after normalizing

2 0.8944

0.8944 0.4472 0.8944Likewise:

0.4472 0.8944 0.4472

vv v

v

P

v v

v

and diag 0 5T TP P I P KP

As a check note that

5. Calculate the matrix of mode shapes

1/2

1

1

0

1 0 0.4472 0.8944 0.4472 0.8944

0 1/ 2 0.8944 0.4472 0.4472 0.2236

0.4472 1.7889

0.8944 0.8944

7. Calculate the modal initial conditions:

0.4472 1.7889(0)

0.8944 0.894

S M P

S

S

r x

1

0

0.01 0.004472

4 0 0.008944

(0) 0S

r x

1 1

1

(0) 0

( )

r r

r t a bt

2 2

2 2

( ) 5 ( ) 0

( ) cos 5

r t r t

r t a t

7. Now compute the solution in modal coordinates and

note what happens to the first mode.

Since ω1 = 0 the first modal equation is

Rigid body translation

And the second modal equation is

Oscillation

Applying the modal initial conditions to these two

solution forms yields:

1

1

1

2

2

(0) 0.004472

(0) 0.0

( ) 0.0042

as in the past problems the initial conditions for yield

( ) 0.0089cos 5

0.0042( )

0.0089cos 5

r a

r b

r t

r

r t t

tt

r

8. Transform the modal solution to the physical

coordinate system

1 3

2

0.00450.4472 0.8944( ) ( )

0.4472 0.2236 0.0089cos 5

( ) 2.012 7.60cos 5 ( ) 10 m

( ) 2.012 1.990cos 5

t S tt

x t tt

x t t

x r

x

Each mass is moved a constant distance and then oscillates at a single

frequency.

Order the frequencies

• It is convention to call the lowest frequency ω1 so that ω1 < ω2 <

ω3 < …

• Order the modes (or eigenvectors) accordingly

• It really does not make a difference in computing the time

response

• However:

– When we measuring frequencies, they appear lowest to highest

– Physically the frequencies respond with the highest energy in

the lowest mode (important in flutter calculations, run up in

rotating machines, etc.)

The system of Example 4.1.5 solved by Mode Summation

1 1 2 2

1 13 32, , 2,

1 1

u u

1 03(0) , (0)

01

x x

From Example 4.1.6 we have:

Use the following initial conditions and note that only one mode

should be excited (why?)

Transform coordinates

1 12 2

9 0 3 0 1/ 3 0 and

0 1 0 1 0 1M M M

12

12

13 0 13(0) (0)

0 1 11

3 0 0 0(0) (0)

0 1 0 0

M

M

q x

q x

Thus the initial conditions become

1 2

1 11 1 and

1 12 2

v v

Transform Mode Shapes to Eigenvectors

12

1 1

12

2 2

13 0 13

0 1 11

13 0 13

0 1 11

M

M

v u

v u

1 2 1 2

11 21 3 Note that 1 1 0, but 1 031 31

T T

v v u u

eigenvectors

Note that unlike the mode shapes, the eigenvectors are orthogonal:

Normalizing yields:

From Equation (4.92):

iiiii

iiiii

i tdttdt vqvq )cos()()sin()(2

1

2

1

1

1

2

1cos2

1

1

2

1cos2

0

0

cos)0(

1221111

2

1

TTT

iiii

i

dd

d

vvv

vq

1 1 10 cos / 2d

Set t=0 and multiply by v1:

Or directly from Eq. (4.97)

2

1

( ) sin( )

1 11 2 cos( 2 ) cos( 2 )

1 12

i i i i

i

t d t

t t

q v

From the initial displacement:

11

22

1(0) 1 21 1

1sin( / 2) 2 2

1(0) 11 1 0

1sin( / 2) 2

T

T

d

d

v q

v q

(4.98)

Eigenvector 2

Eigenvector 1

thus

Transforming Back to Physical Coordinates:

12

1 0 13( ) ( ) cos( 2 )

10 1

1cos 2

3

cos 2

t M t t

t

t

x q

1 2

1( ) cos 2 and ( ) cos 2

3x t t x t t

So, the initial conditions generated motion only in the

first mode (as expected)

Alternate Path to Symmetric Single-Matrix Eigenproblem

• Square root of matrix conceptually easy, but

computationally expensive

0qqqq

KMKMMM~2

121

21

21

• More efficient to decompose M into product of upper and

lower triangular matrices (Cholesky decomposition)

Cholesky Decomposition

1

1

1

1 1

Let where is upper triangular

Introduce the coordinate transformation

0

premultiply by to get

note that:

T

T

T

T

T T TT T T

M U U U

U U U U K

U

U

I U K U K

U K U U K U U

x q x q x x

qq

q q q q 0

1T K U

Cholesky (cont)

1 12 2 M M M

TM U U

• sqrtm requires a singular value decomposition (SVD),

whereas Cholesky requires only simple operations

12 Note that = for diagonal M U M

• Is this really faster? Let’s ask MATLAB

• sqrtm requires a singular value decomposition

(SVD), whereas Cholesky requires only simple

operations

»M = [9 0 ; 0 1];

»flops(0); sqrtm(M); flops

ans = 65

»M = [9 0 ; 0 1];

»flops(0); chol(M); flops

ans = 5

Section 4.5 Systems with Viscous Damping

Extending the first 4 sections to included

the effects of viscous damping (dashpots)

Viscous Damping in MDOF Systems

• Two basic choices for including damping

– Modal Damping

• Attribute some amount to each mode based on

experience, i.e., an artful guess or

• Estimate damping due to viscoelasticity using some

approximation method

– Model the damping mechanism directly (hard and still

an area of research-good for physicists but engineers

need models that are correct enough).

Modal Damping Method

Solve the undamped vibration problem following Window 4.5

0rr0xx )()()()( ttItKtM

Here the mode shapes and eigenvectors are real valued and

form orthonormal sets, even for repeated natural frequencies

(known because 21

21~

KMMK is symmetric)

)cossin()( tBtAetr diidiit

iii

21di i i

Modal Damping (cont)

• Decouple system based on M and K, i.e., use the

“undamped” modes

• Attribute some zi (zeta) to each mode of the decoupled

system (a guess. Not known beforehand. Can be tested

with gross data like x):

idit

ii

iiiiii

teAtr

rrr

ii

sin)(

02 2

Alternately:

here

(4.106)

(4.107)

Transform Back to Get Physical Solution

• Use modal transform to obtain modal initial conditions and

compute Ai and Fi:

02

12

11

02

12

11

)0()0()0(

)0()0()0(

xxxr

xxxr

MPMPS

MPMPS

TT

TT

• With r(t) known, use the inverse transform to recover the

physical solution:

)()()()( 21

21

tStPMtMt rrqx

Modal Damping by Mode Summation

• Can also use mode summation approach

• Again, modes are from undamped system

• The higher the frequency, the smaller the effect (because of the

exponential term). So just few first modes are enough.

)0(+)0(

)0(tan and

sin

)0(

1 and ,

where sin)(

1

2221

21

1

qvqv

qvqv

vv

vq

Tiii

Ti

Tidi

i

i

Ti

i

iidiiii

n

iiidi

ti

d

KMM

tedt ii

Compute q(t), Transform back

• To get the proper initial conditions use:

)0()0( and ),0()0( 21

21

xqxq MM

• Use the above to compute q(t) and then:

)()( 21

tMt qx

the response in physical coordinates.

9 0 6 2

0 4 2 2

x x 0

0 0

1 0

0 0

x x

Example

1 20.01 and 0.1

Consider:

Subject to initial conditions:

Experiments do not give C. They provide zeta (in modal

coordinates) by the half power method.

Compute the solution assuming modal damping of:

Compute the modal decomposition L =sqrt(M)

615.0

788.0

788.0

615.0

615.0

788.0,947.0 and ,

788.0

615.0,240.0

~

500.0333.0

333.0667.0~ ,

20

03

2211

11

P

K

KLLKL

vvvv

0

365.2

846.1

308.0394.0

263.0205.0

0

01

01

r

xr

SPLS

Compute the modal initial conditions:

Compute the modal solutions:

958.0,963.0,49.0,49.0

,1.0,01.0

2211

21

dd

0.004896

1

0.096

2

( ) 4.208 sin(0.49 1.561)

( ) 3.346 sin(0.958 1.471)

t

t

r t e t

r t e t

Using eq (4.108) and (4.109) yields

Then use x(t)=Sr(t)

1

2

( )0.205 0.263( ) ( )

( )0.394 0.308

r tx t Sr t

r t

0.004896 0.096

1

0.004896 0.096

2

( ) 0.863 sin(0.49 1.561) 0.88 sin(0.958 1.471)

( ) 1.658 sin(0.49 1.561) 1.029 sin(0.958 1.471)

t t

t t

x t e t e t

x t e t e t

So, first separate solutions in the modal coordinates were

found and then the modes were assembled by the use of S.

The response in the physical coordinates is therefore a

combination of the modal responses just as in the undamped

case. See page 357 for an additional example.

Lumped Damping models

• In some cases (FEM, machine modeling), the damping

matrix is determined directly from the equations of motion.

• Then our analysis must start with:

0 0

( ) ( ) ( ) ,

subject to and

M t C t K t x x x 0

x x

1 1 1 1 2 2 1

1 1 2 2 1

2 2 2 2 1

2 2 1

( )

( )

( )

( )

m x c x c x x

k x k x x

m x c x x

k x x

Generic Example:

Fig 4.15

• If the damping

mechanisms are

known then

• Sum forces to find

the equations of

motion

Free Body Diagram:

Matrix form of Equations of Motion:

0

0

)(

)(

)(

)(

)(

)(

0

0

2

1

22

221

2

1

22

221

2

1

2

1

tx

tx

kk

kkk

tx

tx

cc

ccc

tx

tx

m

m

The C and K matrices have the same form.

It follows from the system itself that consisted damping and stiffness

elements in a similar manner.

A Question of matrix decoupling

• Can we decouple the system with the same coordinate

transformations as before?

0

? diagonal

0

21

21

rrr

xxx

PCMMPI

KCMT

• In general, these can not be decoupled since K and C can

not be diagonalized simultaneously

A Little Matrix Theory

• Two symmetric matrices have the same eigenvectors

if and only if the matrices commute

• Define 1 / 2 1 / 2C M C M

• Transform the damped equations of motion into:

0qqq )(ˆ)(~

)( tKtCtI

• Let P be the matrix of eigenvectors vi of and TK P KP

Then PCPT ~ will be diagonal if and only if transformed

K and C have same eigenvectors, i.e.

for all , so i i iC i CK KC v v

More Matrix Stuff and Normal Mode Systems

CKMKCM

CMKMMKMCMM

CMMKMMKMMCMM

CKKC

11

21

121

21

121

21

21

21

21

21

21

21

21

~~~~

• This does not require a matrix square root to check

• This informs us explicitly whether or not the equations of motion

can be decoupled

• If true, such systems are called “normal mode” systems or said

to possess “classical normal modes”

Happens if and only if CM-1

K is symmetric

Proportional Damping

• It turns out that CM-1

K = symmetric is a necessary and

sufficient condition for C to be diagonalizable by the

eigenvectors of the “undamped” system, i.e., those based on

M,K

• Best known example is “proportional”damping.

• The coefficients are obtained through experiments or just

by guess.

1 1 1

linear combination of and .

both symmetric

C M K M K

CM K M K M K K KM K

Proportional Damping (cont)

12

2

Write the system as

diagonal!

Thus, the damping ratios in the decoupled system are

22 2

ii i i i

i

M M K K

M I I K K

P I I

x x x 0

q x q q q 0

q r r r r 0

(4.124)

Generalized Proportional Damping

For any value of n up to the number of degrees of freedom:

1

1

1

0

0 1

ni

i

i

C K

C K K I K

For example for n = 2 we get the previous proportional

damping formulation:

Section 4.6 Modal Analysis of the Forced

Response

Extending the chapters 2 and 3 to more then one degree of

freedom

Forced Response: the response of an mdof system to a

forcing term

1 1 12 2 2

4

00 0 0 0

00 0 0 0( ) (4.126)

00 0 0 0

( )0 0 0 1

Assume diagonalizable for now, i.e.,

where

M C K B t

F t

C

C K M C M CM

x x x F

q q q F

x2

k2

x1

13

11

u

c2

m1

m2

k1

c1

F1 F2

If the system of equations decouple then the methods of

Chapters 2 and 3 can be applied

12

th 2

Decouple the system with the eigenvalues of

2

so the equation would be 2 ( )

T

i i i

i i i i i i i

K

I P M B

i r r r f t

r r r F

• Responses to harmonic, periodic, or general forces as in

Chapters 2 and 3

• Note that the modal forcing function is a linear combination of

many physical forces

(4.129)

(4.130)

With the modal equation in hand the general solution is

given

2

0

( ) 2 ( ) ( ) ( ) (4.130)

( ) sin

1 ( ) sin (4.131)

i i

i i i i

i i i i i i i

t

i i di i

t

t

i di

di

r t r t r t f t

r t d e t

e f t e t d

The applied force is distributed across the all of the

modes except in a special case.

12( ) ( ) for the decoupled EOM.Tt P M B t

f F

• An excitation on a single physical DOF may “spread” to all modal DOFs

(one F generates many f’s)

• It is actually possible to drive a MDOF system at one of its natural

frequencies and not experience resonant response (an unusual

circumstance)

th

Let ( ) ( ), where is some spatial vector

and ( ) is any fuction of time. What if happens

to be related to the mode shape by u ?i

t f t

f t

i M

F b b

b

b

Example 4.6.1

A 2-dof system

1

131/2 1/2

1 13 31/2 1/2

09 0 2.7 0.3 27 3

( )0 1 0.3 0.3 3 3

3 0 0,

0 1 0 1

0 2.7 0.3 0 0.3 0.1

0 1 0.3 0.3 0 1 0.1 0.3

F t

M M

C M CM

x x x

Figure 4.16

Compute the mass normalized stiffness matrix and its

eigen solution

1 13 31/2 1/2

1

2

0 27 3 0 3 1

0 1 3 3 0 1 1 3

2 1 1, 0.707

4 1 1

K M KM

K P

v v

From before:

Transform the damping matrix, the forcing function and

write down the modal equations

1/2

2

1 1 0.3 0.1 1 1 0.2 00.707 0.707

1 1 0.1 0.3 1 1 0 0.4

2 0

0 4

00.2357 0.7071( ) ( )

( )0.2357 0.7071

T

T

T

P CP

P KP

t P M B tF t

f F

1 1 1

2 2 2

( ) 0.2 ( ) 2 ( ) (0.7071)(3)cos 2 2.1213cos 2

( ) 0.4 ( ) 4 ( ) (0.7071)(3)cos 2 2.1213cos 2

r t r t r t t t

r t r t r t t t

From the above coefficients the modal equations

become (note that the force is distributed to each mode)

1

2

2

1 1 1

2

2 2 2

0.20.0707

2 2

0.40.1000

2(2)

1 1.41

1 1.99

d

d

Compute the modal values using the single degree of

freedom formulas

• The modal damping

ratios and damped

natural frequencies are

computed using the usual

formulas and the

coefficients from the

terms in the modal

equations:

Use SDOF formula for the particular solution given in

equation (2.36)

1

2

12

1

2

( ) 1.040cos(2 0.1974)

( ) 2.6516sin(2 )

1.040cos(2 0.1974)( )

2.6516sin(2 )

( ) 0.2451cos(2 0.1974) 0.6249sin 2

( ) 0.7354cos(2 0.1974) 1.8749sin 2

p

p

ss

r t t

r t t

tt M P

t

x t t t

x t t t

x

Now transform back to physical coordinates

Note that the force effects both degrees of freedom even though it is applied to one.

The Frequency Response of each mode is plotted:

• This graph shows the amplitude of each mode due to an input modal force f

1 and f

2.

• A force applied to mass # 2 F

2 will

contribute to both modal forces!

0 1 2 3 4 5 -30

-20

-10

0

10

20

Frequency ()

Amplitude (dB)

R 1 ( )/f

1 ( ))

R 2 ( )/f

2 ( ))

The frequency response of each degree of freedom is

plotted

• This graph shows

the amplitude of

each mass due to

an input force on

mass #2.

• Each mass is

excited by the

force on mass #2

• Both masses are

effected by both

modes

0 1 2 3 4 5 -50

-40

-30

-20

-10

0

10

Frequency ω

Amplitude (dB)

X 1 ( )/F

2 ( ))

X 2 ( )/F

2 ( ))

Resonance for multiple degree of freedom systems can

occur at each of the systems natural frequencies

• Note that the frequency response of the previous example

shows two peaks

• If in the odd case that b is orthogonal to one of the mode

shapes then resonance in that mode may not occur (see

example 4.6.2)

• If the modes are strongly coupled the resonant peaks may

combine (see X1/F2 in the previous slide) and be hard to

notice

Special cases:

9 0 27 3 3cos 2

0 1 3 3 1t

x x

Example: Illustrating the effect of the input force

allocation

1/2 1/21/ 3 0

0 1

1 0 3 1 1/ 3 0 3 1cos 2 cos 2

0 1 1 3 0 1 1 1

M M

t t

x q

q q

Consider:

Compute the modal equations and discuss resonance.

Solution:

1 11

1 12P

Calculating the natural frequencies and mode shapes yields:

1 22 and 2 rad/s

1 13 3

1 2, 1 1

u u

1 2

1 11 1,

1 12 2

v v

The mass normalized eigenvectors are:

Transform and compute the modal equations:

1 1

2 2

1 1

2 2

yields

1 0 2 0 1 1 11cos 2

0 1 0 4 1 1 12

2 2 cos 2

4 0

P

r rt

r r

r r t

r r

q r

2 2 , the driving frequency

No resonance even though

An example with three masses

1 1 2 2

2 2 2 3 3

3 3 3 4

0 0 0

0 0

0 0 0

m k k k

M m K k k k k

m k k k

k1

k3

k4

k2

x1 x

2 x

3

F1 F

2 F

3

c1 c

2 c

3 c

4 m

1 m

2 m

3

m1=m

2=m

3=2Kg k

1=k

2=k

3= k

1=3N/m C=0.02K

Solving a system with 3 masses is best done using a code.

Using Matlab we can calculate the eigenvectors and eigenvalues and

hence the mode shapes and natural frequencies.

1/2

0.707 0 0 0.5 0.707 0.5

0 0.707 0 0.707 0 -0.707

0 0 0.707 0.5 -0.707 0.5

M P

35405.0354.0

5.005.0

354.05.00.354

.

U 1 2 3

1 2 3

0.94 1.73 2.26

0.0094 0.017 0.0226

The frequency response of each mode computed

separately:

0 1 2 3 4 5

-30

-20

-10

0

10

20

30

40

50

Frequency ω

Amplitude (dB)

r 1 ( )/f 1

( ))

r 2 ( )/f 2

( ))

r 3 ( )/f 3

( ))

A comparison of the Frequency response between driving

mass #1 and driving mass #2

0 1 2 3 4 5

-80

-60

-40

-20

0

20

40

Frequency ω Frequency ω

0 1 2 3 4 5 -60

-40

-20

0

20

40

Amplitude (dB)

X 1 ( )/F

2 ( ))

X 2 ( )/F

2 ( ))

X 3 ( )/F

2 ( ))

X1 w( ) / F1 w( )

X2 w( ) / F1 w( )

X3 w( ) / F1 w( )

X1 w( ) / F2 w( )

X2 w( ) / F2 w( )

X3 w( ) / F2 w( )

( ) ( ) ( )M t K t t x x F

12ˆ( ) ( ) ( )t K t M t

q q F

Computing the forced response via the mode summation

technique

1

( ) sin (4.134)n

H i i i i

i

t d t

q v

Consider

Transform:

From eq. (4.92) the homogeneous solution in mode

summation form is

The total solution in mode summation form is:

1particular

homogenous

( ) [ sin cos ] ( ) (4.135)n

i i i i i p

i

t b t c t t

q v q

12( ) ( )p pt M tq x

1

2

1

( ) sin cos ( ) (4.136)n

i i i i i p

i

t b t c t M t

q v x

But

Next use the initial conditions and orthogonality to

evaluate the constants

12

12

12

12

(0) (4.138)

(0) (4.139)

(0)

1 (0)

T T

i i i p

T T

i i i i p

T T

i i i p

T T

i i i p

i

c M

b M

c M

b M

0

0

0

0

v q v x

v q v x

v q v x

v q v x

Substitution of the constants into Equation (4.136) and

multiplying by M-1/2 yields

1

( ) sin cos ( )n

i i i i i p

i

t d t c t t

x u x

(4.141)

1/2( ) ( )t M P tx r

Decoupled Forced EOM

Physical Co-ordinates. Coupled equations

Modal Co-ordinates. Uncoupled equations

k1

m1

x1

m2

x2

k2

F1

1

1

r2

f1

f2

w1

2

w 2

2

r1

4.7 Lagrange’s Equations

Defining work, energy and

virtual displacements and

work we will learn an

alternate method of deriving

equations of motion

Generalized coordinates: 2 not 4!

Recall equations (1.63) and (1.64)

Definitions (from Dynamics)

2

1

0

1

1 2

0

1 1Kinetic Energy:

2 2

Work Done by a force:

a reference position then the potential energy is

( )

TT m m

W d

V r d

r

r

r

r

r r r r

F r

r

F r

Strain Energy in a Spring

curve vs )( the under area the is which

2

1)()(

:spring a for

energy) potential (elastic energy Strain

200

xxF

kxdkdFxV

kxF

xx

F(x)

x

Strain energy in an elastic material

The variation of , denoted ( ) is given by

( , )( )= ( , )

( , )The axial stress is ( , ) ( , )

( )

dx dx

u x tdx dx x t dx

x

P x tx t E x t

A x

P(x,t) P(x,t) +Px(x,t) dx

dx +ux(x,t)dx

Example of a bar of cross section

A(x) elongated by force P(x,t) Stress σ

Strain ε

Slope E

so P=EAε

Strain energy continued

2

2

0

1 1( , ) ( ) ( , ) ( , )

2 2

1 [ ( ) ( , )] ( , )

2

1 ( ) ( , )

2

Integrating yields the strain energy for axial tension

in a bar element:

1 ( ) ( , )

2

dV P x t dx P x t x t dx

EA x x t x t dx

EA x x t dx

V E A x x t dx

2

0

1 ( , ) ( )

2

u x tE A x dx

x

r

Virtual Reality (actually: virtual displacement)

A virtual displacement

Based on variational math

Small or infinitesimal

changes compatible with

constraints

No time associated with

change

Variation or

Change in:

Consequence of satisfying the constraint:

1

Constraint: ( ) , a constant

( )

Taylor expansion:

( )

0

n

i

i i

f

f c

f c

ff x c

x

f

rr

r

r r

r

rr

Virtual work

th

1

Suppose the mass is acted on by with system in

static equilibrium

0,

the principle of virtual work:

0

which states that if a system is in equi

i

i i i

n

i i

i

i

W

f

f r

F r

librium, the

work done by externally applied forces through a

virtual displacement is zero: 0

has an critical value

V

V

Dynamic Equilibrium

D'Alembert's Principle move inertia force to left side and

treat dynamics as statics. From Newton's law in terms of

change in momentum:

0

This allows us to use virtual work i

i i

F p F p

n the dynamic case:

0

0

i

i m

F p r

F r r

Hamilton’s Principle

( )

1 ( )

2

1( ) ( ), multiply by

2

( )

( )

d

dt

dm

dt

dW m T

dt

dT W m

dt

r r r r r r

r r r r

r r r r r r

r r

r r

Integrate this last expression

22

1

1

2 2

11

2

1

2

1

path indepence

( )

0

0, for conservative forces

0, Hamilton's principle

tt

tt

t t

tt

t

t

t

t

dT W dt m dt

dt

T W dt m

T W dt W V

T V dt

r r

r r

Lagrange’s Equation

1 2 3Let ( , , ... , ), called generalized coordinates

Let a generalized force (or moment)

The Lagrange formulation, derived from Hamilton's principle

for determining the equations of moti

n i

i

i

q q q q t q

WQ

q

r r

on are

i

i i i

d T T VQ

dt q q q

(4.143)

(4.144)

The Lagrangian, L

0, 1,2, (1.146)i i

d L Li n

dt q q

Let L = (T - U), called the Lagrangian

Then (4.145) becomes:

For the (common) case that the potential

energy does not depend on the velocity:

0i

U

q

Advantages

• Equations contain only scalar quantities

• One equation for each degree of freedom

• Independent of the choice of coordinate system since the

energy does not depend on coordinates

• See examples in Section 4.7 pages 369-377

• Useful in situations where F = ma is not obvious

22

212

212

21

21

21 )()( yyxxyx

Example of Generalize Coordinates

2211 and qq

How many dof?

What are they?

Are there constraints?

m1

(x1,y

1)

m2

(x2,y

2)

x

y

1

2

q1

q2

There are only 2 DOF and one choice is:

Example 4.7.3 (also illustrates linear approximation method)

)( ),( 21 tqtxq

Here G is mass center and e is

the distance to the elastic axis.

Let m denote the mass of the

wing and J the rotational inertia

about G.

Take the generalized coordinates to be:

Called the pitch and plunge model

.

e

x(t)

G

k1

k2 q t( )

2 21 1

2 2GT mx J

Computing the Energies

( ) ( ) sin ( )

( ) ( ) cos ( ) ( ) cos ( )

G

G

x t x t e t

dx t x t e t x t e t

dt

2 21 1[ cos ]

2 2T m x e J

The Kinetic Energy is

The relationship between xG and x is

So the kinetic energy is

22

21

2

1

2

1kxkU

Potential Energy and the Lagrangian

22 2 2

1 2

1 1 1 1 cos

2 2 2 2

L T U

m x e J k x k

The potential energy is:

The Lagrangian is:

2

1cos sin 0mx me em k x

2 2 2 2

2cos cos sin cos 0J me x me me k

Computing Derivatives for Equation (1.146)

1

2

1

1

[ os ]

sin

L Lm x e c

q x

d Lmx me me

dt x

L Lk x

q x

Now use the Lagrange equation to get:

Likewise differentiation with respect to q2 = θ yields:

Next Linearize and write in matrix form

1cos sin

1

2

2

0( ) ( ) 0

0( ) ( ) 0

km me x t x t

kme me J t t

Using the small angle approximations:

In matrix form this becomes:

Note that this is a dynamically coupled system

Next consider the Single Spring-Mass System and

compute the equation of motion using the Largranian

approach

2 2

2 2

1 1,

2 2

1 1

2 2

,

0 0

T mx U kx

L T U mx kx

L Lmx kx

x x

d L Lmx kx

dt x x