vibration.pdf

VIBRATION SIMULATION USING MATLAB

By

Park, Jeong Gyu

DEPARTMENT OF PRECISION ENGINEERING

KYOTO UNIVERSITY

KYOTO, JAPAN

MAY 2003

c© Copyright by Park, Jeong Gyu, 2003

Table of Contents

Table of Contents ii

1 Basics of Matlab 1

1.1 Making matrix . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1

1.2 Matrix Manipulations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2

1.3 Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3

1.4 Plotting . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3

1.5 Programming in MATLAB . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3

1.5.1 The m-files . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3

1.5.2 Repeating with for loops . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

1.5.3 If statements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

1.5.4 Writing function subroutines . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

1.6 Saving and Loading . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

1.7 Help . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

2 Single Degree of Freedom System 7

2.1 Free Vibrations of Single-Degree-of-Freedom Systems . . . . . . . . . . . . . . . . . . 7

2.1.1 Viscous Damping . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

2.2 Forced Vibration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

2.2.1 Direct Force Excitation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

2.2.2 Base Excitation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11

2.3 Simulation with MATLAB . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12

2.3.1 Transfer Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12

ii

2.3.2 State Space Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13

3 Multiple Degree of Freedom Systems 17

3.1 Some Basics Concepts for Linear Vibrating System . . . . . . . . . . . . . . . . . . . 17

3.1.1 Eigenvalue Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17

3.1.2 Orthogonality of normal modes . . . . . . . . . . . . . . . . . . . . . . . . . . 20

3.1.3 Normalization of Mode Shapes . . . . . . . . . . . . . . . . . . . . . . . . . . 21

3.1.4 Modal Coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22

3.2 Proportional Damping . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24

3.3 Modal Analysis of the Force Response . . . . . . . . . . . . . . . . . . . . . . . . . . 25

3.4 State-Space Approach . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27

3.4.1 Free Vibration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27

3.4.2 Forced Vibration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29

4 Design for Vibration Suppression 31

4.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31

4.2 Vibration Absorber . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31

4.2.1 SDOF with Undamped DVA . . . . . . . . . . . . . . . . . . . . . . . . . . . 32

4.2.2 SDOF with damped DVA . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34

4.3 Isolation Design . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36

4.3.1 Passive Isolators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36

4.3.2 Skyhook and Active Isolators . . . . . . . . . . . . . . . . . . . . . . . . . . . 38

4.3.3 Semi-active Isolators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41

5 Vibration of strings and rods 42

5.1 Strings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42

5.2 Rods . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42

6 Bending of Beam 44

6.1 Equation of Motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44

6.2 Eigenvalue Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45

6.2.1 Boundary condition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45

6.3 Some Properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48

iii

6.4 Forced Vibration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48

6.4.1 Point force excitation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49

6.4.2 Moment excitation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51

7 Plate 54

7.1 Plate in Bending . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54

7.2 Equation of Motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55

8 Approximate Method 57

8.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57

8.2 Rayleigh Ritz Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57

9 Finite Element Analysis 69

9.1 Euler-Bernoulli Beam . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 69

9.1.1 Basic relation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 69

9.1.2 Finite Element Modeling . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 70

9.2 Thin Plate Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72

9.2.1 formulation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72

9.3 Finite Element Modeling . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73

9.4 Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 77

iv

Chapter 1

Basics of Matlab

Matlab R©(MathWorks, Inc., )1 is an interactive program for numerical computation and data visu-alization. It is used extensively by vibration and control engineers for analysis and design. Thereare many different toolboxes available which extend the basic functions of MATLAB into differentareas.

1.1 Making matrix

Matlab uses variables that are defined to be matrices. A matrix is a collection of numerical valuesthat are organized into a specific configuration of rows and columns. Here are examples of matricesthat could be defined in Matlab.

A = [1 2 3 4;5 6 7 8;9 10 11 12]

Transpose of a matrix using the apostrophe

B=A’

C=[2,2,3

4,4,6

5,5,8]

The colon operation ’:’ is understood by Matlab to perform special and useful operations. If twointeger numbers are separated by a colon, Matlab will generate all of the integers between these twointegers.a = 1:8

generates the row vector,a = [ 1 2 3 4 5 6 7 8 ]

1see, http://www.mathworks.com/

1

2

If three numbers, integer or non-integer, are separated by two colons, the middle number is inter-preted to be a ”range” and the first and third are interpreted to be ”limits”. Thus

b = 0.0 : .2 : 1.0

generates the row vector

b = [ 0.0 .2 .4 .6 .8 1.0 ]

C=linspace(0,10,21)

D=logspace(-1,1,10)

eye(3)

zeros(3,2)

1.2 Matrix Manipulations

Element of matrixA(2,3)

Sizesize(A)

length(a)

TransposeA’

Column or row componentsA(:,3)

Matrix addition, subtraction and multiplicationD = B * C

D = C * B

If you have a square matrix, like E, you can also multiply it by itself as many times as you like byraising it to a given power.

E ∧ 3

Element addition, subtraction and multiplicationAnother option for matrix manipulation is that you can multiply the corresponding elements of twomatrices using the .* operator (the matrices must be the same size to do this).

E = [1 2;3 4]

F = [2 3;4 5]

G = E.*F

If wanted to cube each element in the matrix, just use the element-by-element cubing.

E. ∧ 3

3

1.3 Functions

Matlab includes many standard functions. In Matlab sin and pi denotes the trigonometric functionsine and the constant π.fun=sin(pi/4)

To determine the usage of any function, typehelp function-name

[Example] Verify the variables i, j, cos, exp,log, log10 in MATLAB

1.4 Plotting

One of Matlab most powerful features is the ability to create graphic plots. Here we introduce theelementary ideas for simply presenting a graphic plot of two vectors. ExamplePlot the sin(x)/x in the interval [π/100, 10π]————————————–>>x=pi/100:pi/100:10*pi

>>y=sin(x)./x

>>plot(x,y)

>>grid

————————————-

1.5 Programming in MATLAB

1.5.1 The m-files

It is convenient to write a number of lines of Matlab code before executing the commands. Filesthat contain a Matlab code are called the m-files.

Table 1.1: Basic Matrix Functions

Symbol Explanationsinv Inverse of a matrixdet Determinant of a matrix

trace Summation of diagonal elements of a matrix

4

0 5 10 15 20 25 30 35-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

Figure 1.1: Sin(x)/x

Table 1.2: Basic Plotting Command

Command Explanationsplot(x,y) A Cartesian plot of the vectors x and ysubplotloglog A plot of log(x) vs log(y)

semilogx(x,y) A plot of log(x) vs ysemilogy(x,y) A plot of x vs log(y)

title placing a title at top of graphics plotxlabelylabelgrid Creating a grid on the graphics plot

5

1.5.2 Repeating with for loops

• the for loopsSyntax of the for loop is shown belowfor n=0:10

x(n+1)=sin(pi*n/10)

end

The for loops can be nestedH=zeros(5)

for k=1:5

for l=1:5

H(k,l)=1/(k+l-1)

end

end

1.5.3 If statements

If statements use relational or logical operations to determine what steps to perform in the solutionof a problem.

• the general form of the simple if statement is

if expression

commands

end

In the case of a simple if statement, if the logical expression is true, the commands is executed.However, if the logical expression is false, the command is bypassed and the program control jumpsto the statement that follows the end statement.

• The if-else statement

The if-else statement allows one to execute one set of statements if logical expression is true and adifferent set of statements if the logical statement is false. The general form is

if expression

commands(evaluated if expression is true)

else

commands(evaluated if expression is false)

end

6

1.5.4 Writing function subroutines

function [mean,stdev] = stat(x)

n = length(x);

mean = sum(x) / n;

1.6 Saving and Loading

All variables in the workspace can be viewed by command whos or who.To save all variables from the workspace in binary MAT-filesave FILENAME

An ASCII file is a file containing characters in ASCII format, a format that is independent of ’mat-lab’ or any other executable program. You can save variables from the workspace in ASCII formatwith optionsave filename.dat variable -ascii

To load variables you can use load command.load FILENAME

To clear variables you can use load command.clear

1.7 Help

To learn more about a function you can use help.>> help for

If you do not remember the exact name of a function you can use lookfor

>>lookfor sv

Chapter 2

Single Degree of Freedom System

In this chapter we will study the responses of systems with a single degree of freedom. It is impor-tant topic to master, since the complicated multiple-degree-of-freedom systems(MDOF) can oftentreated as if they are simple collections of several single-degree-of-freedom(SDOF) systems. Oncethe responses of SDOF are understood, the study of complicated MDOF becomes relatively easy.

2.1 Free Vibrations of Single-Degree-of-Freedom Systems

2.1.1 Viscous Damping

For the free vibration of a single-degree-of-freedom system with mass m, spring constant k, andviscous damping c, the system undergoes a dynamic displacement x(t) measured from the staticequilibrium position of the mass. Applying Newton’s law, the equation of motion of the system isrepresented by

m

c

k

F

x

Figure 2.1: Single degree of freedom system.

7

8

mx + cx + kx = 0 (2.1.1)

subject to the initial conditions x(0) = x0 and x(0) = v0. If we divide (2.1.1) by m we can reexpressit in terms as

x + 2ζωnx + ω2nx = 0 (2.1.2)

where ωn =√

k/m is natural angular frequency and ζ = c/(2√

km) is the damping ratio.

To solve the damped system of equation (2.1.2), assuming

x = Aest (2.1.3)

Substituting equation (2.1.3) into equation (2.1.2) yields an algebraic equation in the form

s2 + 2ζωns + ω2n = 0 (2.1.4)

The solutions of equation (2.1.4) are given by

s1,2 = −ζωn ± ωn

√(ζ2 − 1) (2.1.5)

There are three possible cases:

(a) Overdamped MotionIn this case, the damping ratio is greater than 1 (ζ > 1). The discriminant of equation (2.1.5) ispositive, resulting in a pair of distinct real roots. The solution of equation (2.1.1) then becomes

x(t) = e−ζωnt(Ae−ωn

√ζ2−1t + Beωn

√ζ2−1t) (2.1.6)

which represents that the vibration will not occur since the damping force is so large that therestoration force from the spring is not sufficient to overcome the damping force.

(b) Underdamped MotionIn this case the damping ratio is less than 1 (0 < ζ < 1) and the discriminant of equation (2.1.5) isnegative, resulting in a complex conjugate pair of roots. The solutions for this case can be expressedas

x(t) = e−ζωnt(Aejωn

√1−ζ2t + Be−jωn

√1−ζ2t)

= e−ζωnt(Aejωdt + Be−jωdt)

= e−ζωnt(C cos ωdt + D sinωdt)

= Xe−ζωnt sin(ωdt + φ)

(2.1.7)

where j =√−1, X and φ are constants. The the damped natural frequency is denoted by

ωd =√

1 − ζ2ωn

(c) Critically Damped MotionIn this last case, the damping ratio is exactly 1(0zeta = 1). The solution takes the form

x(t) = (A + Bt)e−ωnt (2.1.8)

where the constants A and B are determined by the initial conditions.

9

Homework 2.1.1. Evaluate the constants A and B in equation (2.1.7) using the initial conditions

x(0) = x0 and v(0) = v0.

Homework 2.1.2. Describe the definition of logarithmic decrement in free vibration.

2.2 Forced Vibration

2.2.1 Direct Force Excitation

For a single-degree-of-freedom system with viscous damping and subjected to a forcing function F (t)as shown in figure 2.1, the equation of motion can be written as

mx + cx + kx = F (t) (2.2.1)

The complete solution to equation (2.2.1) consists of two parts, the homogenous solution (the com-plementary solution) and the particular solution. The homogenous solution is the same as the freevibration which was described in last section. It is often common to ignore the transient part ofthe total solution and focus only on the steady-state response. Taking Laplace transformation of asecond order differential equation with zero initial conditions, the transfer function is

X(s)F (s)

=1/m

s2 + 2ζωns + ω2n

(2.2.2)

where ωn =√

k/m, ζ = c/2mωn.

Substituting jω for s to calculate the frequency response, where j is the imaginary operator:

X(jω)F (jω)

=1/mω2

[(ωn/ω)2 − 1] + 2jζ(ωn/ω)(2.2.3)

Example 2.2.1. Plot the amplitude and phase angle of the single degree of freedom system. �

Example MATLAB Code

————————————————————————————————–

clf; clear all;

m = 1;

zeta = 0.1:0.1:1;

k = 1;

wn = sqrt(k/m);

10

w = logspace(-1,1,400);

rad2deg = 180/pi;

s = j*w;

for cnt = 1:length(zeta)

xfer(cnt,:)=(1/m) ./ (s.∧2 + 2*zeta(cnt)*wn*s + wn∧2);

mag(cnt,:) = abs(xfer(cnt,:));

phs(cnt,:) = angle(xfer(cnt,:))*rad2deg;

end


figure(1)

loglog(w,mag(cnt,:),’k-’)

title(’SDOF frequency response magnitudes for zeta = 0.2 to 1.0 in steps of 0.2’)

xlabel(’Frequency(rad/sec)’)

ylabel(’Magnitude’)

grid

hold on

end

hold off


figure(2)

semilogx(w,phs(cnt,:),’k-’)

title(’SDOF frequency response phases for zeta = 0.2 to 1.0 in steps of 0.2’)

xlabel(’Frequency(rad/sec)’)

ylabel(’Phase’)

grid

hold on

end

hold off

——————————————————————————————–

11

10-1

100

101

10-3

10-2

10-1

100

101

Frequency(rad/sec)

Magnitude

10-1

100

101

-180

-160

-140

-120

-100

-80

-60

-40

-20

0

Frequency(rad/sec)

Phase

Figure 2.2: SDOF

2.2.2 Base Excitation

Often, machines are harmonically excited through elastic mounting, which may be modeled bysprings and dashpots. For example, an automobile suspension system is excited by road surface.

Consider the single degree of freedom system in Figure 2.3(a). The structure with mass m isconnected to the base by stiffness, k, and damping with viscous damping coefficient c. The equation

12

x

y

m

k

c

x

y

m

c k

(a) (b)

Figure 2.3: Free diagram of base excited single degree of freedom system.

of motion ismx + c(x − y) + k(x − y) = 0 (2.2.4)

(a)Derive the displacement transmissibility, X/Y and plot the magnitude and phase.(b) The transmitted force by the base excitation to the structure is FT = k(x − y) + c(x − y).The force transmissibility, FT /kY is defined as the dimensionless relation between maximum basedisplacement Y and the transmitted force magnitude FT .Derive the force transmissibility and plotas function of frequency ratio.

[Homework2]Sky hook damperConsider the single degree of freedom system in figure 2.3(b).The structure with mass m is connectedto the base by stiffness, k. Let us suppose that the viscous damping with viscous damping coefficientc is connected to the sky. (a)Derive the displacement transmissibility, X/Y and plot the magnitudeand phase.(b) Derive the force transmissibility, FT = k(x − y), and plot as function of frequency ratio.

2.3 Simulation with MATLAB

2.3.1 Transfer Function

The linear time invariant(LTI) systems can be specified by transfer functions. The correspondingcommand is :sys=tf(num,den)The output sys is a model-specific data structure.

Example 2.3.1. Sample Matlab code for Bode plot

13

m=1

zeta=0.1

k=1

wn=sqrt(k/m)

den=[1 2*zeta*wn wn∧2]

num=[1/m]

sys=tf(num,den)

bode(sys)

Example 2.3.2. The function lsim simulates the response to more general classes of inputs. For

example,

t=0:0.01:50;

u=sin(t);

lsim(sys,u,t)

simulates the response of the linear system sys to a sine wave for a duration of 50 seconds.

Table 2.1: Basic Commends for Time and Frequency Response

Command Explanationsbode(sys) Bode plot

nyquist(sys) Nyquist plotstep(sys) step response

impulse(sys) impulse responseinitial(sys, x0) undriven response to initial conditionlsim(sys,u,t,x0) response to input u

2.3.2 State Space Analysis

It is desirable to change the system equation for an n d.o.f system with n second order differentialequation to 2n first order differential equations. The first order form of equations for the system iscalled as state space form.

Start by solving equation second order differential equations.

mx + cx + kx = F (t) (2.3.1)

14

we define the state vector asx(t) =

[x(t) x(t)

]T

(2.3.2)

Then, adding the identity x = x, equation (2.3.1) can be written in the state form as

x(t) = Ax(t) + BF (2.3.3)

where the system matrix A and the input matrix B are :

A =

[0 1

−m−1k −m−1c

](2.3.4)

and

B =

[0

m−1

](2.3.5)

Schematically, a Single Input Single Output(SISO) state space system is represented as shown in

B + C +

A

D

f(t)Input

y(t)Output

System Matrix

x(t)

Direct Transmission Matrix

dx(t)/dt

Integrate

Figure 2.4: State space system block diagram

Figure (2.4). The scalar input u(t) is fed into both the input matrix B and the direct transmissionmatrix D. The output of the input matrix is a n × 1 vector, where n is the number of states. Theoutput is fed into a summing junction to be added to the output of the C matrix.

The output of the B matrix is added to the feedback term coming from the system matrix and isfed intro an integrator block. The output matrix has as many rows as outputs, and has as manycolumns as states, n.

15

To account for the case where the desired output is not just the states but is some linear combinationof the states, and output matrix C is defined to relate the outputs to the states. Also, a matrix D,know as the direct transmission matrix, is multiplied by the input F (t) to account for outputs thatare related to the inputs but that bypass the states.

y(t) = Cx(t) + DF (2.3.6)

The output matrix C has as many rows as outputs required and as many columns as states. Thedirect transmission matrix D has the same number of columns as the input matrix B and as manyrows as the output matrix C.

Example 2.3.3. Numerically compute the free vibration of mass-spring-damper system using ini-

tial function in MATLAB. � Example MATLAB Code

m=1;d=0.1;k =1;

A=[0 1;-k/m -c/m];

C=[1 0];

sys=ss(A,[],C,[]);

x0=[10,0];

initial(sys,x0)

Initial Condition Results

Time (sec)

Amplitude

0 20 40 60 80 100 120-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

Figure 2.5: Initial condition results

The result of free vibration of the one degree of freedom system is shown in figure 2.5.

16

Homework 2.3.4. One of the common excitation in vibration is a constant force that is applied for

a short period of time and then removed. Numerically calculate the response of mass-spring-dashpot

system to this excitation in MATLAB.

mx + cx + kx = Fo[1 − H(t − t1)] (2.3.7)

where H is Heaviside function. stepfun is useful command to solve this problem.

Chapter 3

Multiple Degree of FreedomSystems

3.1 Some Basics Concepts for Linear Vibrating System

3.1.1 Eigenvalue Problem

In the previous chapters a single degree of freedom system with a single mass, damper and springwas considered. Real systems have multiple degrees of freedom and their analysis is complicatedby the large number of equations involved. To deal with them, matrix are used. The equation ofmotion for n-degree of freedom equation can be written as

[m]{x} + [c]{x} + [k]{x} = [bf ]{f} (3.1.1)

where the mass [m], damping [c], and stiffness [k] matrices are symmetric.

First consider undamped vibration without excitation force. The system can be solved by assuminga harmonic solution of the form

x = uejωt (3.1.2)

Here,u is a vector of constants to be determined, ω is a constant to be determined.

Substitution of this assumed form of the solution into the equation of motion yields

(−ω2M + K)uejωt = 0 (3.1.3)

Note that the scalar ejωt �= for any value of t and hence equation (3.1.1) yields the fact that ω andu must satisfy the vector equation

(−ω2M + K)u = 0

17

18

Note that this represents two algebraic equations in the three unknowns; ω, u1, u2 where u =[ u1 u2 ]T .

This equation is satisfied for any u if the determinant of the above equation is zero.∣∣−ω2M + K∣∣ = 0 (3.1.4)

The simultaneous solution of equation (3.1.4) results in the values of parameter ω2. The ω is calledas eigenvalues of the problem.

Once the value of ω is established, the value of the constant vector u can be found by solvingequation (3.1.3).

Example 3.1.1. Consider the system with two masses represented in figure 3.1.

c1

k1

x1

k2

c3

k3

x2

c2

m 1 m 2

F1 F2

Figure 3.1: 2dof

The equations of motion become[m1 0

0 m1

]{x1

x2

}+

[c1 + c2 −c2

−c2 c2 + c3

]{x1

x2

}+

[k1 + k2 −k2

−k2 k2 + k3

]{x1

x2

}=

{F1(t)

F2(t)

}

(3.1.5)To determine the natural frequencies and natural mode shapes of the system, the undamped freevibration of the system is first considered. Thus the equations reduce to[

m1 0

0 m1

]{x1

x2

}+

[k1 + k2 −k2

−k2 k2 + k3

]{x1

x2

}=

{0

0

}(3.1.6)

Consider a numerical example for the system shown in figure (3.1). Let c1 = c2 = c3 = 0, m1 = 5kg,m2 = 10kg, k1 = 2N/m, k2 = 2N/m, k3 = 4N/m. Substituting in equation (3.1.6) yields[

5 0

0 10

]{x1

x2

}+

[4 −2

−2 6

]{x1

x2

}=

{0

0

}(3.1.7)

19

Assume harmonic responses of the of the form x1 = A1exp(iωt) and x2 = A2exp(iωt). Equation(3.1.6) becomes

ω2

[5 0

0 10

]{A1

A2

}=

[4 −2

−2 6

]{A1

A2

}(3.1.8)

Solving Eigenvalue Problem with MATLAB

The eigenvalue problem of a matrix is defined as

Au = λu (3.1.9)

and generalized eigenvalue problem isKu = λMu (3.1.10)

The eig is subroutine for computing the eigenvalues and the eigenvectors of the matrix A or>>[V,D]=eig(A)

>>[V,D]=eig(K,M)

The eigenvalues of system are stored as the diagonal entries of the diagonal matrix D and theassociated eigenvectors are stored in columns of the matrix V .

� Example MATLAB Code

m=[5 0 ;0 10]; k=[4 -2;-2 6]; [v,d]=eig(k,m)

The function eig in MATLAB gives unsorted eigenvalues, so it will be help to make sorting theeigenvalues of the system.

function [u,wn]=eigsort(k,m);Omega=sqrt(eig(k,m));[vtem,d]=eig(k,m);[wn,isort]=sort(Omega);il=length(wn);for i=1:ilv(:,i)=vtem(:,isort(i));enddisp(’The natural frequencies are (rad/sec)’)disp(’ ’)wndisp(’ ’)disp(’The eigenvectors of the system are’)v

20

——————————————————————The two natural frequencies are ω1=0.6325 rad/s, ω2=1 rad/s

The eigenvectors are u1 ={

1 1}T

, u2 ={

1 −0.5}T

m 1 m 2

m 1 m 2

1 1

1

-1/2

Figure 3.2: Mode shapes for the two degree of freedom system

3.1.2 Orthogonality of normal modes

The modes are orthogonal with respect to the mass matrix and stiffness matrix.

{u}T2 [m]{u}1 = 0

{u}T2 [k]{u}1 = 0

(3.1.11)

Mass normalizing equation (3.1.11), we can get the general relations as

{u}Ti [m]{u}i = mi, i = 1, 2

{u}Ti [k]{u}i = miω

2i = ki, i = 1, 2

(3.1.12)

where mi and ki is called modal mass and modal stiffness for the i-th modal vector of vibration.The numerical values of the mode shape will be used to determine the modal mass and modal

stiffness. The mode shapes were found to be u1 ={

1 1}T

for ω1 =√

2/5 rad/s, and u2 ={1 −0.5

}T

for ω1 = 1 rad/s

Verification with MATLAB

u1 = v(:, 1);u2 = v(:, 2);u1′ ∗ [m] ∗ u1 = 15

21

u1′ ∗ [m] ∗ u2 = 0u2′ ∗ [m] ∗ u2 = 7.5k1 = ω2

1 ∗ m1 = 6k2 = ω2

2 ∗ m2 = 15/2

3.1.3 Normalization of Mode Shapes

While above relations are related to the mass and stiffness of the modal space, it is important toremember that the magnitude of these quantities depends upon the normalization of the modalvectors. Therefore, only the combination of a modal vector together with the associated modalmass and stiffness represent a unique absolute characteristic concerning the system being described.When we scaled the eigenvector such that mi = 1, the equation (3.1.12) becomes

{u}Ti [m]{u}i = 1, i = 1, 2

{u}Ti [k]{u}i = ω2

i , i = 1, 2(3.1.13)

This meas that mi is not unique. There are several ways to normalize the mode shapes.

(1) The mode shapes can be normalized such that the modal mass mi is set to unity.(2) The largest element of the mode shape is set to unity.(3) A particular element of the mode shape is set to unity.(4) The norm of the mode vector is set to unity.

Example 3.1.2. Using the previous two degree of freedom example, normalize the modal vectors

such that {u}Ti [m]{u}i = 1, i = 1, 2 The mass normalization of the first and second natural modes

are

{u}1 =1√m1

1

1

=

1√15

1

1

{u}2 =1√m2

1

−1/2

=

1√15/2

1

−1/2

The orthogonality of modes permit us to transform the coupled equations of motion defined inphysical coordinate to uncoupled system in the modal coordinate.

22

3.1.4 Modal Coordinates

In solving the equations of motion for an undamped system (3.1.6), the major obstacle encounteredwhen trying to solve for the system response x for a particular set of exciting forces and initialconditions, is the coupling between the equations. The coupling is seen in terms of non-zero offdiagonal elements.

If the system of equations could be uncoupled, so that we obtained diagonal mass and stiffnessmatrices, then each equation would be similar to that of a single degree of freedom system, and couldbe solved independent of each other. The process of deriving the system response by transformingthe equations of motion into an independent set of equations is known as modal analysis

Thus the coordinated transformation we are seeking, is one that decouples the system. The newcoordinate system can be found referring to orthogonal properties of the mode shapes discussed inequation (3.1.12) and (3.1.13).

{x(t)} =n∑

i=1

{u}iqi(t) (3.1.14)

where the physical coordinate, {x(t)} are related with the normal modes, {u}i and the normaldecoupled coordinate, qi.

Equation (3.1.14) may be written in matrix form as

{x(t)} = [P ]{q(t)} (3.1.15)

where [P ] is called the modal matrix . Thus, the modal matrix for a 2-DOF system can appear as

[P ] = [ {u}1 {u}2 ] (3.1.16)

Substituting equation (3.1.16) into the general equation (3.1.1), we obtain as

[m][P ]{q} + [c][P ]{q} + [k][P ]{q} = {f} (3.1.17)

Multiplying on the left by [P ]T ,

[P ]T [m][P ]{q} + [P ]T [c][P ]{q} + [P ]T [k][P ]{q} = [P ]T {f} (3.1.18)

We know that orthogonality of the modes with respect to mass and stiffness matrices. Assumingthat the viscous damping can be decoupled by modal matrix, we obtain

qi(t) + 2ζiωiq(t) + ω2i qi(t) = Ni(t), i = 1, 2, · · · (3.1.19)

where Ni(t) is

Ni(t) ={u}T

i {f(t)}{u}T

i [m]{u}i=

{u}Ti {f(t)}mi

(3.1.20)

The ratio in equation (3.1.20){u}T

i

{u}Ti [m]{u}i

(3.1.21)

23

is called modal participation factor. The displacement can be expressed as

x =∞∑

i=1

{u}iqi =∞∑

i=1

{u}i{u}Ti {f(t)}

mi[(ω2i − ω2) + 2iζiωi]

(3.1.22)

where ωi is the natural frequency in the i-th mode. If eigenvector {u}i is mass normalized, {u}Ti [m]{u}i =

1.

Numerical Simulation with MATLAB

[P ]T ∗ [m] ∗ [P ] =

(1 0

0 1

)

[P ]T ∗ [k] ∗ [P ] =

(0.4 0

0 1

)

[P ]T =

(0.2582 0.2582

0.3651 −0.1826

)

Thus the equations of motion in modal coordinate are[1 0

0 1

]{q1

q2

}+

[0.4 0

0 1

]{q1

q2

}=

{0.2582f1 + 0.2582f2

0.3651f1 − 0.1826f2

}=

{f

′1

f′2

}(3.1.23)

The matrix equation of (3.1.23) can be written in terms of algebraic differential equations

q1 + 0.4q1 = f1′

q2 + q2 = f2′ (3.1.24)

Hence, the system equations have been uncoupled by using the modal matrix as a coordinate trans-formation.

Example 3.1.3. Calculate the response of the system illustrated in figure (3.3) to the initial dis-

placement x(0) =[

1 1]T

with x(0) =[

0 0]T

using modal analysis.

The initial conditions in modal space become

{q(0)} = [P ]T ∗ {x(0)} =[

0.5164 0.1826]T

{q(0)} = [P ]T ∗ {x(0)} =[

0 0]T

The modal solution of equation (3.1.24) is

q1(t) = q1(0) cos(ω1t) = 0.5164 cos(0.6325t)

24

5 10

1

2 2 4

2/5

11

x1 x2

q1 q2

f1 f2

f ’2f ’1

Figure 3.3: The undamped two degree of freedom system and broken down to two single degree offreedom systems

q2(t) = q2(0) cos(ω2t) = 0.1826 cos(t)

Using the transformation x(t) = Pq(t) yields that the solution in physical coordinates is

x(t) =

(0.1333 cos(0.6325t) + 0.0667 cos(t)

0.1333 cos(0.6325t) − 0.0333 cos(t)

)

3.2 Proportional Damping

Damping is present in all oscillatory systems. As there are several types of damping, viscous,hysteretic, coulomb etc., it is generally difficult to ascertain which type of damping is representedin a particular structure. In fact a structure may have damping characteristics resulting from acombination of all types. In many cases, however, the damping is small and certain simplifyingassumptions can be made. The most common model for damping is proportional damping definedas

[c] = α[m] + β[k] (3.2.1)

where [c] is damping matrix and α, β are constants. For the purposes of most practical problems,the simpler relationship will be sufficient.

Caughey1 showed that there exists a necessary and sufficient condition for system (3.1.1) to becompletely uncoupled is that [m]−1[c] commute with [m]−1[k].

([m]−1[c])([m]−1[k]) = ([m]−1[k])([m]−1[c]) (3.2.2)

1T.K. Caughey, ”Classical Normal Modes in Damped Linear Systems”, Journal of Applied Mechanics, Vol 27,Trans. ASME, pp.269-271, 1960

25

0 10 20 30 40 50 60 70 80 90 100-0.2

-0.1

0

0.1

0.2

0.3

0 10 20 30 40 50 60 70 80 90 100-0.2

-0.1

0

0.1

0.2

Figure 3.4: Time response of mass 1 and mass 2

or[c][m]−1[k] = [k][m]−1[c] (3.2.3)

3.3 Modal Analysis of the Force Response

The forced response of a multiple-degree-of-freedom system can also be calculated by use of modalanalysis.

Example 3.3.1 (See Example 4.6.1 in Inman, pp.296). For this example, let m1 = 9kg,

m2 = 1kg,k1 = 24N/m, and k2 = 3kg. Assume that the damping is proportional with α = 0 and

β = 0.1, so that c1 = 2.4Ns/m, and c1 = 0.3Ns/m. Also assume that F1 = 0, and F2 = 3cos2t.

Calculate the steady-state response.

9 0

0 1

x1

x2

+

2.7 −0.3

−0.3 0.3

x1

x2

+

27 −3

−3 3

x1

x2

=

0 0

0 1

F1

F2

(3.3.1)

26

Numerical Simulation with MATLAB

[P ] =

(−0.2357 −0.2357

−0.7071 0.7071

)

[P ]T ∗ [m] ∗ [P ] =

(1 0

0 1

)

[P ]T ∗ [c] ∗ [P ] =

(0.2 0

0 0.4

)

[P ]T ∗ [k] ∗ [P ] =

(2 0

0 4

)

[P ]T [B] =

(0 −0.7071

0 0.7071

)

Hence the decoupled modal equations become

q1 + 0.2q + 2q1 = −0.7071 ∗ 3 ∗ cos 2t

q2 + 0.4q + 4q2 = 0.7071 ∗ 3 ∗ cos 2t(3.3.2)

Comparing the coefficient of qi to 2ζiωi yields

ζ1 = 0.22√

2

ζ2 = 0.22∗2

Thus the damped natural frequencies becoms

ωd1 = ω1

√1 − ζ2

1 � 1.41

ωd2 = ω2

√1 − ζ2

2 � 1.99

Note that while the force F2 is applied only to mass m2, it becomes applied to both coordinate whentransformed to modal coordinates. Let the particular solutions of equations (3.3.2) be q1p and q2p.

c1

k1

x1

k2

x2

c2

m 1 m 2

F1 F2

Figure 3.5: Damped two-degree-of-freedom system

27

The steady state solution in the physical coordinate system is

xss(t) = [P ]qp(t) =

(−0.2357q1p(t) − 0.2357q2p(t)

−0.7071q1p(t) + 0.7071q2p(t)

)

3.4 State-Space Approach

Simulation by state-space method is a much easier way to obtain the systems response when com-pared to computing the response by modal analysis. However, the modal approach is needed toperform design and to gain insight into the dynamics of the system. In this section the simulationmethod for free vibration and forced vibration by state space formulation will be discussed.

3.4.1 Free Vibration

Consider the forced response of a damped linear system. The most general case can be written as

[m]{x} + [c]{x} + [k]{x} = 0 (3.4.1)

with initial conditionx(0) = x0 x(0) = x0

Again it is useful to write this expression in a state-space form by defining the two n × 1 vectorsy1 = x and y2 = x, then the equations (3.4.1) becomes

y(t) = Ay(t) (3.4.2)

where

A =

[0 I

−m−1k −m−1c

](3.4.3)

The eigenvalues λi will appear in complex conjugate pairs in the form

λi = −ζωi − jωi

√1 − ζ2

i

λi = −ζωi + jωi

√1 − ζ2

i

(3.4.4)

Example 3.4.1. Consider the system shown in figure 3.1. Calculate the response of the system

to the initial condition using state-space method. Let c1 = c3 = 0, c2 = 0.2N·s/m, m1 = 2kg,

m2 = 1kg, k1 = 0.2N/m, k2 = 0.05N/m, k3 = 0.05N/m. The initial condition of m1 is 0.1m and let

the other parameters be all zero.


28

—————-dof2ini.m————————-

m1=2;m2=1;

d1=0; d2=0.2; d3=0;

k1=0.2; k2=0.05;k3=0.05;

m=[m1 0;0 m2]; d=[d1+d2 -d2; -d2 d2+d3]; k=[k1+k2 -k2;-k2 k2+k3];

A=[zeros(2,2),eye(2);-inv(m)*k,-inv(m)*f]; C = [1 0 0 0];

x0=[0.1 0 0 0];

sys=ss(A,[],C,[])

initial(sys,x0)

The simulation result is shown in figure (3.6).

Response to Initial Conditions

Time (sec)

Am

plitu

de

0 50 100 150 200 250 300 350 400 450 500−0.1

−0.08

−0.06

−0.04

−0.02

0

0.02

0.04

0.06

0.08

0.1

To:

Y(1

)

Figure 3.6: Time response of mass 1

Homework 5 Consider the system shown in figure 3.5. Let m1 = 10kg, m2 = 1kg, k1 = 0.5N/m,k2 = 0.05N/m, and c1 = 0, c2 = 0.2N·s/m. The initial condition of m1 is 0.1m and let the otherparameters be all zero. Plot the transient response of mass 1.

29

3.4.2 Forced Vibration

[m]{x} + [c]{x} + [k]{x} = [bf ]{f} (3.4.5)

with initial conditionx(0) = x0 x(0) = x0

The state-space equations arey(t) = Ay(t) + Bf (3.4.6)

where

A =

[0 I

−m−1k −m−1c

](3.4.7)

and

B =

[0

m−1bf

](3.4.8)

Example 3.4.2. Compare the frequency response function of the two degree of freedom shown

in figure (3.1) between c2 = 0 and c2 = 0.2N·s/m. The other parameters are as follows. Let

m1 = 2kg,m2 = 1kg, k1 = 0.2N/m, k2 = 0.05N/m, k3 = 0.05N/m and c1 = c3 = 0 N·s/m and the

excitation force F2 be zero.


—————-dof2frf.m————————-

m1=2;m2=1; c1=0; c2=0.0; c3=0;k1=0.2; k2=0.05;k3=0.05;

Bf=[1; 0]; m=[m1 0; 0 m2]; damp=[c1+c2 -c2; -c2 c2+c3]; K=[k1+k2 -k2;-k2 k2+k3];

A=[zeros(2,2),eye(2);-inv(m)*k,-inv(m)*damp];

B=[zeros(2,1); inv(m)*bf];C = [1 0 0 0]; D=zeros(size(C,1), size(B,2))

sys=ss(A,B,C,D)

d1=0; d2=0.02; d3=0;

damp1=[c1+c2 -c2; -c2 c2+c3];

Adamp=[zeros(2,2),eye(2);-inv(m)*k,-inv(m)*damp1];

sysdamp=ss(Adamp,B,C,D)

w=linspace(0.1, 1, 800)

30

bode(sys,sysdamp,w)

—————————— the result of simulation is shown in figure (3.7).

Bode Diagram

Frequency (rad/sec)

Pha

se (

deg)

Mag

nitu

de (

dB)

−40

−20

0

20

40

60

80

10−1

100

−180

−90

0

90

180

270

360

Figure 3.7: Frequency response fucntion of mass 1

Chapter 4

Design for Vibration Suppression

4.1 Introduction

A Dynamic Vibration Absorber (DVA) is a device consisting of a reaction mass, a spring elementwith appropriate damping that is attached to a structure in order to reduce the dynamic responseof the structure. The frequency of dynamic absorber is tuned to a particular structural frequency sothat when that frequency is excited external force. The concept of DVA was first applied by Frahmin 1909 to reduce the rolling motion of ships as well as hull vibrations. A theroy for the DVA waspresented later by Ormondroyd and Den Hartog (1928)1. The detailed study of optimal tuning anddamping parameters was discussed in Den Hartog’s on Mechanical Vibration (1940) book 2 .

4.2 Vibration Absorber

Figure (4.1) shows a SDOF system having mass m and stiffness k, subjected to external forcing. Adynamic absorber with mass m2, stiffness k2, and dashpot c2 is attached to the primary mass. Whatwe now have is a 2DOF problem rather than the original SDOF. The m1, c1, k1 system is referred toas the primary system, and m2, c2, k2 system is known as the secondary system. The displacementof primary mass and absorbing mass are x1 and x2, respectively. With this notation, the governingequations take the form

1J.Ormondroyd, and J.P.Den Hartog, ”The theory of the dynamic vibration absorber”, Trans. ASME, 50, 1928,pp. 9-15

2J.P.Den Hartog, Mechanical vibration, Dover, 4th ed. Reprint, 1984

31

32

[m1 0

0 m2

]{x1

x2

}+

[c1 + c2 −c2

−c2 c2

]{x1

x2

}+

[k1 + k2 −k2

−k2 k2

]{x1

x2

}=

{F (t)

0

}

(4.2.1)

We set xi = Re[Xiejωt] for steady-state response, which leads to the following complex amplitude-

frequency equations,

c1

k1

x1

k2

x2

c2

m 1 m 2

F1

Figure 4.1: SDOF system coupled with a DVA

[[(k1 + k2) + jω(c1 + c2) − m1ω

2] −(k2 + jωc2)

−(k2 + jωc2) k2 + jωc2 − m2ω2

]{X1

X2

}=

{F

0

}(4.2.2)

4.2.1 SDOF with Undamped DVA

Let us consider the case where damping is negligible, c1 = c2 = 0. We then find from equation(4.2.2) that

X1 = F(k2 − m2ω

2)[(k1 + k2) − ω2m1](k2 − m2ω2) − k2

2

(4.2.3)

X2 = Fk2

[(k1 + k2) − ω2m1](k2 − m2ω2) − k22

(4.2.4)

where the determinant of the system of coefficients in equation is

∆(ω) = [(k1 + k2) − ω2m1](k2 − m2ω2) − k2

2. (4.2.5)

First, note from equation (4.2.3) that the magnitude of steady-state vibration, x1 becomes zerowhen the absorber parameters k2 and m2 is chosen to satisfy the tuning condition

ω2 = k2/m2 (4.2.6)

33

In this case the steady-state motion of the absorber mass is calculated from equation (4.2.4)

X2 = − F

k2(4.2.7)

With the main mass standing still and the secondary mass having a motion −F/k2exp(jωt) the forcein the damper spring varies as −Fexp(jωt), which is actually equal and opposite to the externalforce.

For simplifications we want to bring equation (4.2.3) and (4.2.4) into a dimensionless form and forthat purpose we introduce the following parameters:

xst = F/k1 ; static deflection of primary system.ω2

1 = k1/m1 ; natural frequency of primary systemω2

2 = k2/m2 ; natural frequency of secondary systemµ = m2/m1 ; mass ratio=secondary mass/primary mass

With this definitions, also note thatk2

k1= µ

ω22

ω21

= µf2 (4.2.8)

where frequency ratio f is f = ω2/ω1.

Then equationa (4.2.3)and (4.2.4) becomes

X1

xst=

1 − β2

[1 − β2][1 + µf2 − (fβ)2] − µf2(4.2.9)

X2

xst=

1[1 − β2][1 + µf2 − (fβ)2] − µf2

(4.2.10)

where frequency ratio β is β = ω/ω2.

The absolute value of this system is plotted in figure (4.2) for the case µ = 0.25.

In fact, if the driving frequency shifts such that |X1/xst| > 1, the force transmitted to the primarysystem is amplified and the absorber system is not an improvement over the original design of theprimary system.

Let us consider the case that the frequency ratio f = 1, (i.e., ω2 = ω1, or k2/m2 = k1/m1)

For this special case, equations (4.2.9) and (4.2.10) becomes

X1

xst=

1 − β2

(1 − β2)(1 + µ − β2) − µ(4.2.11)

X2

xst=

1(1 − β2)(1 + µ − β2) − µ

(4.2.12)

The natural frequencies are determined by setting the denominators equal to zero :

(1 − β2)(1 + µ − β2) − µ = 0

β4 − fβ2(2 + µ) + 1 = 0(4.2.13)

34

0 0.5 1 1.5 20

0.5

1

1.5

2

2.5

3

3.5

4

4.5

5

Normalized magnitude

frequency ratio β

Figure 4.2: Plot of normalized magnitude of the primary mass versus the normalized driving fre-quency

The solutions areβ2 = (1 + µ/2) ±

√µ + µ2/4 (4.2.14)

This relation is plotted graphically in figure (4.3). Note that as µ is increased, the natural frequenciessplit farther apart.

Homework 4.2.1. Inman’s Book Example 5.3.1

Homework 4.2.2. Inman’s Book Example 5.3.2

4.2.2 SDOF with damped DVA

The equations of motion are given in matrix form by equation (4.2.1). Note that these equations can-not necessarily be solved by using the modal analysis technique of Chapter 3 because the equationsdo not decouple (KM−1C �= CM−1K).

The steady-state response of equation (4.2.1) can be obtained by assuming a solution of the formby assuming a solution of the form

x(t) = X(t)ejωt =

[X1

X2

]ejωt (4.2.15)

where X1 is the amplitude of vibration of the primary mass and X2 is the amplitude of vibration ofthe absorber mass. From the (4.2.2), we obtain that

35

0.6 0.8 1 1.2 1.4 1.60

0.05

0.1

0.15

0.2

0.25

0.3

0.35

0.4

0.45

0.5

Frequency ratio β

mass ratio

µ

Figure 4.3: Plot of mass ratio versus system natural frequency(normalized to the frequency of thesecondary system

X1

F=

(k2 − m2ω2) + jc2ω

det([K] − ω2[M ] + jω[C])(4.2.16)

X2

F=

k2 + jc2ω

det([K] − ω2[M ] + jω[C])(4.2.17)

which expresses the magnitude of the response of the primary mass and secondary mass, respectively.Note that these values are complex numbers.

First, consider the case for which the internal damping of the primary system is neglected (c1 = 0).Using complex arithmetic, the amplitude of the motion of the primary mass can be written as thereal number

X21

F 2=

(k2 − m2ω2)2 + (c2ω)2

[(k1 − m1ω2)(k2 − m2ω2) − m2k2ω2]2 + [k1 − (m1 + m2)ω2]2c22ω

2(4.2.18)

It is instructive to examine this amplitude in terms of the dimensionless ratios introduced for theundamped vibration absorber. The amplitude x1 is written in terms of the static deflection xst = F/k

of the primary system. In addition, consider the mixed damping ratio defined by

ζ =c2

2m2ω1(4.2.19)

where ω =√

k1/m1. Using the standard frequency ratio r = ω/ω1, the ratio of natural frequencies

36

f = ω2/ω1, and the mass ratio µ = m2/m1, equation (4.2.18) can be written as

X1

xst=

√(2ζr)2 + (r2 − f2)2

(2ζr)2(r2 − 1 + µr2)2 + [µr2f2 − (r2 − 1)(r2 − f2)]2(4.2.20)

which expresses the dimensionless amplitude of the primary system. Note that the amplitude of theprimary system response is determined by four physical parameter values: mass ratio µ, the ratioof the decoupled natural frequencies f the ratio of the driving frequency to the primary naturalfrequency r and the damping ratio of absorber ζ.

These four parameters can be considered as design variables and are chosen to give the smallestpossible value of the primary mass’s response, x1 for a given application.

It is instructive to verify this result for several particular cases

Homework 4.2.3. Plot the compliance curves when the parameters of absorbers are: 1. ζ2 = ∞2. ζ2 = 0

3. ζ = 0.10

4. ζ = 0.10

for mass ratio µ = 1/20 and frequency ratio f = 1.

The dynamic vibration absorber said to be optimally tuned and damped when the two resonancepeaks are equal in magnitude. The optimal frequency ratio f and damping ratio are given as

f =1

1 + µ(4.2.21)

ζ =

√3µ

8(1 + µ)3(4.2.22)

Homework 4.2.4. Obtain the optimal frequency ratio f and optimal damping ratio ζ when mass

ratio is µ = 1/20.

4.3 Isolation Design

4.3.1 Passive Isolators

In figure (4.5), a single-degree-of-freedom vehicle model is shown with (a) passive, (b) skyhooksuspensions.

The passive system using linear elements has the equation of motion.

x + 2ζωn(x1 − x0) + ω2n(x1 − x0) = 0 (4.3.1)

37

0.6 0.7 0.8 0.9 1.0 1.1 1.2 1.30

2

4

6

8

10

12

14

16

=1/20f=1

zeta=0.10

zeta=0.32

zeta=inf.

zeta=0

x 1/xst

Excitation frequency ratio (r)

Figure 4.4: Amplitudes of the main mass for various values of absorber damping. All curves passthrough the fixed points

x1

x0

m

k c

Figure 4.5: Schematic of passive isolators

38

where ω2n = k/m, and ζ = c/2

√km. For the base excitation problem it is assumed that the base

moves harmonically such thatx0 = X0exp(jωbt) (4.3.2)

where X(0) denotes the amplitude of the base motion and ωb represents the frequency of the baseoscillation.

The displacement of mass divided by the amplitude of base excitation is obtained as

X1

X0=

[1 + (2ζr)2

(1 − r2)2 + (2ζr)2

]1/2

(4.3.3)

where the frequency ratio r = ωb/ωn.

The ratio is called the displacement transmissibility. Another quantity of interest in the base exci-tation problem is the force transmitted to the mass as the result of a harmonic displacement of thebase. Hence the force transmitted to the mass is the sum of the force in the spring and the force inthe damper

FT (t) = k(x1 − x0) + c(x1 − x0) (4.3.4)

The force transmissibility is defined as

FT

kX0= r2

[1 + (2ζr)2

(1 − r2)2 + (2ζr)2

]1/2

(4.3.5)

Figure (4.6) is a frequency response plot from equation (4.3.3).

4.3.2 Skyhook and Active Isolators

When active suspensions are used the suspension force can be generated based on control strategies.using optimal control theory and a commonly used quadratic performance criterion it was shown 3

that an optimum single-degree-of-freedom isolator must generate suspension force as

Fa/m = −2ζωnx1 − ω2n(x1 − x0) = 0 (4.3.6)

leading to a sprung mass equation of motion as

x + 2ζωnx1 + ω2n(x1 − x0) = 0 (4.3.7)

whici is the same as he governing equation of skyhook isolators shown in figure (4.7).

The displacement of mass divided by the amplitude of base excitation is obtained as

X1

X0=

[1

(1 − r2)2 + (2ζr)2

]1/2

(4.3.8)

where the frequency ratio r = ωb/ωn

3Bender, E.K., Optimum linear preview control with application to vehicle suspension, ASME, Journal of basicengineering, 90(2), June 1968, pp.213-221.

39

10-1

100

101

10-1

100

Freqeucney ratio

Displacement Transmissibility

ζ=1 ζ=0.8

ζ=0.4

ζ=0.6

ζ=0.2

Figure 4.6: Displacement transmissibility for passive isolator

m

k

c

x0

x1

x1

x0

m

Fa

Figure 4.7: Schematic of passive isolators

40

10-1

100

101

10-1

100

Freqeucney ratio

Displacement Transmissibility

ζ=0.2 ζ=0.4

ζ=0.6

ζ=0.8

ζ=1

Figure 4.8: Displacement transmissibility for skyhook isolator

41

4.3.3 Semi-active Isolators

Semi-active suspensions respresents a compromise between passive and active ones. The concept ofsemi-active suspension was first proposed by Crosby and Karnopp4 in 1973

m

k Fs

x1

x0

Figure 4.9: Schematic of semi-active isolators

F′s = Fs/m =

{2ζωn, x1(x1 − x0) > 0

0, x1(x1 − x0) < 0

}(4.3.9)

4Crosby, M.J., and Karnopp, D.C., The active damper - a new concept for shock and vibration control, The shockand vibration bulletin, 43(4), June 1973, pp.119-133

Chapter 5

Vibration of strings and rods

5.1 Strings

5.2 Rods

Consider the vibation of an elastic rod (or bar) of length L and of varying cross-sectional area shownin figure (5.1). The forces on the infinitesimal element summed in the x direction are

L

xx x+dx

F F+dF

u(x,t)

Figure 5.1: Cantilevered rod in longitudinal vibration along x

F + dF − F = ρA(x)dx∂u(x, t)

∂t2(5.2.1)

where u(x, t) is the deflection of the rod in the x direction. From the solid mechanics,

F = EA(x)∂u(x, t)

∂x(5.2.2)

42

43

where E is the Young’s modulus. The differential form of F becomes

dF =∂F

∂xdx (5.2.3)

from the chain rule for partial derivatives. Substitution of equation (5.2.2) and (5.2.3) into (5.2.1)and dividing by dx yields

ρA(x)∂2u(x, t)

∂t2=

∂

∂x

(EA(x)

∂u(x, t)∂x

)(5.2.4)

When A(x) is a constant this equation becomes

∂2u(x, t)∂t2

=(

E

ρ

)∂2u(x, t)

∂x2(5.2.5)

The quantity v =√

E/ρ defines the velocity of propagation of the displacement (or stress wave) inthe rod.

Chapter 6

Bending of Beam

6.1 Equation of Motion

x

z y

t

b

Figure 6.1: Beam

The equation of motion of Euler-Bernoulli Beam is

m(x)∂2w

∂t2+ c

∂w

∂t+ EI

∂4w

∂x4= f(x, t) (6.1.1)

where,m is mass per unit length of beam defined as m = ρA. If no damping and no external force isapplied so that c = 0, f(x, t) = 0, and EI(x) and m(x) are assumed to be constant, equation (6.1.1)simplifies

∂2w

∂t2+

EI

m

∂4w

∂x4= 0 (6.1.2)

Note that the free vibration equation (6.1.2) contains four spatial derivatives and hence requiresfour boundary conditions. The two time derivatives requires that two initial conditions, one for thedisplacement and one for the velocity.

44

45

6.2 Eigenvalue Problem

For the eigenvalue problem, assume the product solution as

w(x, t) = W (x)F (t) (6.2.1)

where W (x) depends on the spatial position alone and F (t) depends on time alone. Introducingequation (6.2.1) into equation (6.1.2), we can obtain the following equation as

d4W (x)dx4

− β4W (x) = 0 (6.2.2)

where β4 = ω2mEI , 0 < x < L.

L

x

t

Figure 6.2: Cramped-Free transverse beam

6.2.1 Boundary condition

Clamped-free

The boundary conditions for the clamped-free case are

W (0) = 0dW (x)

dx |x=0 = 0d2W (x)

dx2 |x=L = 0d3W (x)

dx3 |x=L = 0

(6.2.3)

The solution of equation (6.2.2) is

W (x) = C1 sin βx + C2 cos βx + C3 sinh βx + C4 cosh βx (6.2.4)

Applying the boundary conditions for x = 0, we find

C2 + C4 = 0

C1 + C3 = 0(6.2.5)

so that the eigenfunction is reduced to x = L, we get

C1(sin βL + sinhβL) + C2(cos βL + cosh βL) = 0

C1(cos βL + cosh βL) − C2(sin βL − sinh βL) = 0(6.2.6)

46

Equating the determinant of the coefficients to zero, we obtain the characteristic equation[(sin βL + sinh βL) (cos βL + cosh βL)

(cos βL + cosh βL) −(sin βL − sinh βL)

]{C1

C2

}=

{0

0

}(6.2.7)

The characteristic equation iscos βL cosh βL = −1 (6.2.8)

From the numerical analysis β1L = 1.875, β2L = 4.694, β3L = 7.855

ω1 = (1.875)2√

EImL4 rad/sec

ω2 = (4.694)2√

EImL4 rad/sec

ω3 = (7.855)2√

EImL4 rad/sec

(6.2.9)

We obtain the corresponding eigenfunctions

Wr(x) = Cr(cos βx − cosh βx) + Crsin βL−sinh βLcos βL+cosh βL (sin βx − sinh βx)

= Ar[(sin βrL − sinhβrL)(sin βrx − sinh βrx) + (cos βrL + cosh βrL)(cos βrx − cosh βrx)](6.2.10)

Example 6.2.1. The geometric and material properties are

ρ(Density) L(Length) b(Width) t(Thickness) E

2750 kg/m3 340 mm 22 mm 2 mm 7.00×1010N/m3

The natural frequencies are

ω1 = 88.6 rad/sec = 14.1 Hz

ω2 = 555.2 rad/sec = 88.4 Hz

ω3 = 1554.7 rad/sec = 247.4 Hz

(6.2.11)

Example 6.2.2. Plot the mode shapes of clamped-free beam with the same dimension specified

above example. Normalize the eignefunction as∫ L

0

W 2i dx = 1

A1 = 0.56461, W1(x) = A1[1.72[cos(5.51x) − cosh(5.51x)] − 1.26[sin(5.51x) − sinh(5.51x)]]



[Homework 7] Calculate the natural frequency and plot first four mode shapes of beam withfree-free boundary condition .

47

0.05 0.1 0.15 0.2 0.25 0.3

0.5

1

1.5

2

2.5

3

3.5

0.05 0.1 0.15 0.2 0.25 0.3

-2

-1

1

2

3

0.05 0.1 0.15 0.2 0.25 0.3

-3

-2

-1

1

2

Figure 6.3: First three eigenfunctions for clamped-free beam

48

6.3 Some Properties

The normal modes must satisfy the equation of motion and its boundary conditions. The normalmodes Wi are also orthogonal functions satisfying the relation

∫ L

0m(x)Wi(x)Wj(x)dx = 0 for j �= i

and Mi for j = i.

From the expansion theorem for self-adjoint distributed systems, the solution of equation (6.1.2) canbe expressed as

w(x, t) =∞∑

i=1

Wi(x)qi(t) (6.3.1)

The generalized coordinate qi(t) can be determined from Lagrange’s equation by establishing thekinetic and potential energies.

6.4 Forced Vibration

The forced response of a beam can be calculated using modal analysis just as in the lumped system.The approach again uses the orthogonality condition of the unforced system’s eigenfunctions toreduce the calculation of the response to a system of decoupled modal equations for the time response.

m∂2w

∂t2+ c

∂w

∂t+ EI

∂4w

∂x4= f(x, t) (6.4.1)

First, expand the applied force p(x, t) in terms of the modes

f(x, t) =∞∑

i=1

fi(t)Wi(x) (6.4.2)

Multiply both sides of this equation by Wj and then integrate over the beam span,

fi(t) =∫ L

0

f(x, t)Wi(x)dx (6.4.3)

Substituting equation (6.3.1) and (6.4.2) into equation (6.4.1) we obtain

∞∑i=1

[mWi(x)qi(t) + cWiq(t) + EIW′′′′i (x)qi(t)] =

∞∑i=1

fi(t)Wi(x) (6.4.4)

We know that the modes are satisfying

EIW′′′′i (x) = ω2

i mWi(x)

Substituting this relation into equation (6.4.4) leads to

∞∑i=1

[mWi(x)qi(t) + cWiq(t) + ω2i mWi(x))qi(t)] =

∞∑i=1

fi(t)Wi(x) (6.4.5)

49

It is convenient to normalize the eigenfunction as∫ L

0

W 2i (x)dx = 1, i = 1, 2, · · · (6.4.6)

Since the eigenfunction Wi(t) are not zero, equation (6.4.5) becomes infinite set of independentmodal equations:

qi(t) + 2ζiωiq(t) + ω2i qi(t) = fi(t)/m, i = 1, 2, · · · (6.4.7)

6.4.1 Point force excitation

The excitation force becomesf(x, t) = F (t)δ(x − x1) (6.4.8)

Equation (6.4.3) becomes

fi(t) =∫ L

0

F (t)δ(x − x1)Wi(x)dx = F (t)Wi(x1) (6.4.9)


qi(t) + 2ζiωiq(t) + ω2i qi(t) = F (t)Wi(x1)/m, i = 1, 2, · · · (6.4.10)

0 50 100 150 200 250 300 350 400 450 50010

−7

10−6

10−5

10−4

10−3

10−2

10−1

100

Frequency (Hz)

FR

F (

m/N

)

Figure 6.4: The frequency response function(FRF) obtained from the modal model.

Example 6.4.1. Obtain the frequency response function of the cantilever beam. The specification

of beam is the same with the Example 6.2.1. The excitation position, xa is 0.34mm and sensing

position, xs is 0.2m. The example MATLAB code is printed below and the frf was printed at figure

(6.4)

50


————————–cantifrf.m————————————-

clear

m = 3; n = 1; z = 0.001; rho = 2750; E = 70e9;

L = 0.34; b = 0.022; t = 0.002;

A = t*b; Is = t∧3*b/12; mass=rho*A;

xa = [0.34];

xs = [0.2];

global betaL beta Ar

betaL=[1.875104 4.694091 7.854757];

beta=betaL/L;

Ar=[0.56461 0.031393 0.00133];

wn=[88.6 555.2 1554.7];

M = eye(m,m);

K = diag(wn(1:m).∧2,0);

Damp=diag(2*z*wn(1:m),0);

Bf=zeros(m,n); ys=zeros(m,n);

for i = 1:n

for r = 1:m

Bf(r,i)=cantimode(r,xa)/mass;

end

end

for i = 1:n

for r = 1:m

ys(r,i)=cantimode(r,xs);

end

end

A = [zeros(m,m),eye(m);-inv(M)*K,-inv(M)*Damp];

B =[zeros(m,1);inv(M)*Bf(:,1)];

C = [ys’,zeros(n,m)];

D =zeros(size(Cc,1),size(B,2));

w = linspace(0,500*2*pi,800);

[mag,phs] = bode(A,B,C,D,1,w);

semilogy(w/2/pi,mag(:,1))

——————————————————————-

———-cantimode.m———————————————-

function y = cantimode(r,x)

51


y=Ar(r)*((sin(betaL(r))-sinh(betaL(r)))*(sin(beta(r)*x)-sinh(beta(r)*x))

+(cos(betaL(r))+cosh(betaL(r)))*(cos(beta(r)*x)-cosh(beta(r)*x)));

—————————————————————

Homework 8 Obtain the frequency response function of the beam with free-free boundary condition.The specification of the system is the same with above example.

6.4.2 Moment excitation

The excitation force becomes

f(x, t) = Mo∂

∂x[δ(x − x2) − δ(x − x1)] (6.4.11)


fi(t) =∫ L

0

Mo∂

∂x[δ(x − x2) − δ(x − x1)]Wi(x)dx = Mo[W

′i (x2) − W

′i (x1)] (6.4.12)


qi(t) + 2ζiωiq(t) + ω2i qi(t) = Mo[W

′i (x2) − W

′i (x1)]/m, i = 1, 2, · · · (6.4.13)

Example 6.4.2. Plot the frequency response function of the cantilever beam excited by coupled

moment.

ρ(Density) L(Length) b(Width) t(Thickness) E x1 x12

2750 kg/m3 340 mm 22 mm 2 mm 7.00×1010N/m3 48 mm 80 mm

W′r(x) = Ar[βr(sin βrL − sinhβrL)(cos βrx − cosh βrx) − βr(cos βrL + cosh βrL)(sin βrx + sinh βrx)]

(6.4.14)

� Example MATLAB Code ⇒ cantipzt.m

clear m = 3;

n = 1;

z = 0.001;

rho = 2750;

E = 70e9;

L = 0.34;

b = 0.022;

52

t = 0.002;

A = t*b;

Is = t∧3*b/12;

mass=rho*A;

xa1 = [0.048];

xa2 = [0.080];

xs = [0.340];


betaL=[1.875104 4.694091 7.854757];

beta=betaL/L; Ar=[0.56461 0.031393 0.00133];

wn=[88.6 555.2 1554.7];

M = eye(m,m);

K = diag(wn(1:m).∧2,0);

Damp=diag(2*z*wn(1:m),0);

Bf=zeros(m,n); ys=zeros(m,n);

for i = 1:n

for r = 1:m

Bf(r,i)=(dcantimode(r,xa2)-dcantimode(r,xa1))/mass;

end

end

for i = 1:n

for r = 1:m

ys(r,i)=cantimode(r,xs);

end

end

A = [zeros(m,m),eye(m);-inv(M)*K,-inv(M)*Damp];

B =[zeros(m,1);inv(M)*Bf(:,1)]; C = [ys’,zeros(n,m)];

D =zeros(size(C,1),size(B,2));

w = linspace(0,500*2*pi,800);

[mag,phs]=bode(A,B,C,D,1,w); semilogy(w/2/pi,mag(:,1))

—————-

function y = dcantimode(r,x)


y1=beta(r)*(sin(betaL(r))-sinh(betaL(r)))*(cos(beta(r)*x)-cosh(beta(r)*x));

y2=-beta(r)*(cos(betaL(r))+cosh(betaL(r)))*(sin(beta(r)*x)+sinh(beta(r)*x));

y=Ar(r)*(y1+y2);

53

0 50 100 150 200 250 300 350 400 450 50010

−6

10−5

10−4

10−3

10−2

10−1

100

Frequency (Hz)

FR

F (

m/T

)

Figure 6.5: The frequency response function(FRF) excited by PZT.

Chapter 7

Plate

7.1 Plate in Bending

The bending behavior of plates can be understood with a direct extension of what we have alreadylearned about the bending of beams.

1ρx

=∂2w

∂x2,

1ρy

=∂2w

∂y2(7.1.1)

let u and v be components of displacements at any point in the plate,

u(x, y, z) = −z∂w

∂x, v(x, y, z) = −z

∂w

∂y(7.1.2)

The strain are given as follows:

{εb} =

εx

εy

γxy

=

∂u∂x

∂v∂y

∂u∂y + ∂v

∂x

= −z

∂2w∂x2

∂2w∂y2

2 ∂2w∂x∂y

(7.1.3)

Hooke’s law for plane stress relates these strains to the stress resultants,

σx =E

1 − ν2(εx + νεy) = − Ez

1 − ν2

(∂2w

∂x2+ ν

∂2w

∂y2

)(7.1.4)

σy =E

1 − ν2(εy + νεx) = − Ez

1 − ν2

(∂2w

∂y2+ ν

∂2w

∂x2

)(7.1.5)

τxy = −2Gz∂2w

∂x∂y= − Ez

1 + ν

∂2w

∂x∂y(7.1.6)

The stress-strain relationships take the matrix form as

{σb} = [Db]{εb} (7.1.7)

54

55

where

[Db] =E

1 − ν2

1 ν 0

ν 1 0

0 0 1−ν2

(7.1.8)

Each stress resultant is multiplied by its respective moment arm, yielding the following moments

Mx =∫ h/2

−h/2

zσxdz = −D

(∂2w

∂x2+ ν

∂2w

∂y2

)(7.1.9)

My =∫ h/2

−h/2

zσydz = −D

(∂2w

∂y2+ ν

∂2w

∂x2

)(7.1.10)

Mxy =∫ h/2

−h/2

zτxydz = −D(1 − ν)∂2w

∂x∂y(7.1.11)

where D = Eh3/12(1−ν2) is called the flexural rigidity of the plate. Equation (7.1.9)-(7.1.11) relatemoments to deflection w.

D

(∂4w

∂x4+ 2

∂4w

∂x2∂y2+

∂4w

∂y4

)+ ρh

∂2w

∂t2= p(x, y, t) (7.1.12)

orD∇4w + ρhw = p(x, y, t) (7.1.13)

7.2 Equation of Motion

The bending energy expression for the thin plate are

V =12

∫v

{σb}T {εb}dυ (7.2.1)

The bending energy becomes

V =12

∫v

{εb}T [Db]{εb}dυ (7.2.2)

The kinetic energy of the plate is given by

T =12

∫A

ρhw2dA (7.2.3)

The response of the structure is defined in physical coordinates as a series expansion over thegeneralized coordinates:

w =N∑

r=1

φr(x, y)qr(t) (7.2.4)

56

Substituting equation (7.2.4) into equation (7.2.3), one obtains an expression for the entry of theith row and j th column of the matrix

Ms,ij = ρh

∫ b

0

∫ a

0

φi(x, y)φj(x, y)dxdy (7.2.5)

Substituting equation (7.2.4) into equation (7.2.2), one obtains an expression for the entry of theith row and j th column of the stiffness matrix

Ks,ij = Ds

∫ b

0

∫ a

0

{∂2φi

∂x2

∂2φj

∂x2+

∂2φi

∂y2

∂2φj

∂y2+ νs

(∂2φi

∂x2

∂2φj

∂y2+

∂2φi

∂y2

∂2φj

∂x2

)+ 2(1 − νs)

∂2φi

∂x∂y

∂2φj

∂x∂y

}dxdy

(7.2.6)where Ds = h3

12Es

1−ν2s

Chapter 8

Approximate Method

8.1 Introduction

It is too difficult to obtain closed form solution for many problems that are more complex than agroup of lumped spring-mass systems or a simple continuous system, such as a string. This sectionpresents methods to obtain approximate solutions. With the techniques to be introduced in thischapter, we can analyze quite general systems efficiently and accurately.

8.2 Rayleigh Ritz Method

The Rayleigh Ritz method obtains an approximate solution to a differential equation with givenboundary conditions using the functional of the equation. The procedure of this method can besummarized in two steps as given below:1. Assume an admissible solution which satisfies the geometric boundary condition and containsunknown coefficients.2. Substitute the assumed solution into the kinetic and potential energy and find the unknowncoefficients.

Example 8.2.1. A clamped-pinned beam with dynamic vibration absorber.

we must select basis functions satisfy the boundary conditions that φ(x) = dφ(x)/dx = 0 at x = 0

and φ(x) = 0 at x = L. Hence the following functions satisfy these conditions.

φr(x) =x

Lsin(

rπx

L) (8.2.1)

Three-term Ritz series are considered in this example. the transverse component of beam w is

57

58

expressed as summation of Ritz function as

w(x, t) =3∑

r=1

φr(x)qr(t) (8.2.2)

and let the general displacement of ma be denoted as q4.

The kinetic energy and potential energy are

T =12

∫ L

0

mbw2dx +

12maq2

4 (8.2.3)

V =12

∫ L

0

EbIb

(∂2w

∂x2

)2

dx +12k[w(L/2, t) − q4]2 (8.2.4)

where mb is the mass per unit length of beam.

Substituting equation (8.2.2) into equation (8.2.3) and (8.2.4), one obtains

T =12

3∑r=1

3∑s=1

(mb

∫ L

0

φrφsdx

)qr qs +

12maq2

4 (8.2.5)

V =12EbIb

3∑r=1

3∑s=1

∫ L

0

(∂2φr

∂x2

∂2φs

∂x2dx

)qrqs +

12k

[3∑

r=1

3∑s=1

φr(L/2)φs(L/2)

]qrqs

− k3∑

r=1

φr(L/2)qrq4 +12kq2

4

(8.2.6)

The mass and stiffness matrices are

[M ] = mbL

0.1413 −0.0901 0.019 0

−0.0901 0.1603 −0.0973 0

0.019 −0.0973 0.1639 0

0 0 0 µ

(8.2.7)

L

L/2

ma

k

Figure 8.1: A clamped-pinned beam with dynamic vibration absorber

59

and

[K] =EbIb

L3

43.376 + 0.25κ −74.570 75.873 − 0.25κ −κ/2

−74.570 368.323 −459.529 0

75.873 − 0.25κ −459.529 1559.3 + 0.25κ κ/2

−κ/2 0 κ/2 κ

(8.2.8)

where µ = ma/(mbL) and κ = kL3/EbIb.

The generalized force are

Q(t) =∫ L

0

f(x, t)φr(x)dx (8.2.9)

——-pinclamp.nb——-

Let the non-dimensional resonant angular frequency be wr =√

mbL4

EbIbwr. The resulting eigensolution

when κ = 0,and ma = 0, are

w1 w2 w3

Ritz Method(N=3) 15.7563 50.6438 109.9433

Vibration Table 15.4182 49.9648 104.2477

The mode shapes are

q1 q2 q3

Mode 1 1 0.1561 -0.0021

Mode 2 0.5047 1.0000 0.1723

Mode 3 0.3058 0.6543 1.000

The natural frequency when κ = 15.75,and µa = 0.1,are

file.pc.m w1 w2 w3 w4

Ritz Method(N=3) 11.1152 17.7539 50.7127 110.0512

The mode shapes are

q1 q2 q3 q4

Mode 1 0.4282 0.0742 -0.0029 1.0000

Mode 2 1.0000 0.1457 0.0003 -0.4992

Mode 3 0.5101 1.0000 0.1731 -0.0110

Mode 4 0.3043 0.6536 1.000 0.0046

60

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

0.2

0.4

0.6

0.8

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1−0.5

0

0.5

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1−0.5

0

0.5

1

Figure 8.2: Mode shapes of pin-clamped beam/ pcmode1.m

0 50 100 15010-7

10-6

10-5

10-4

10-3

10-2

10-1

100

101

Figure 8.3: Frequency response function of clamped-pinned beam/ pcmode1.m

61

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1−0.5

0

0.5

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1−0.5

0

0.5

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1−0.5

0

0.5

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1−0.5

0

0.5

Figure 8.4: Mode shapes of pin-clamped beam with dynamic absorber/ pcmode2.m

0 50 100 15010-6

10-5

10-4

10-3

10-2

10-1

100

101

Figure 8.5: Frequency response function of clamped-pinned beam with dynamic vibration/pcfrf2.m

62

The mode shapes of beam with absorber mass are shown in figure (8.4)

Example 8.2.2. Beam with free-free boundary condition.

Ten-term Ritz series are considered in this example. the transverse component of tennis racket w is

expressed as summation of Ritz function as

w(x, t) =10∑

r=1

φr(x)qr(t) (8.2.10)

A schematic diagram of tennis racket and ball model was shown in figure (8.6).

L L

x

z

Figure 8.6: A schematic diagram of beam with free-free boundary condition.

The translate rigid mode can be expressed by φ1(x) = 1. To represent rotational rigid mode we

select the basis function to be φ2(x) = x/L. Because there are no geometric boundary condition to

satisfy, the power series can be selected as basis function.

φr(x) =( x

L

)r−1

, r = 1, 2, · · · , 10 (8.2.11)


T =12

∫ L

−L

mbw2dx (8.2.12)

V =12

∫ L

−L

EbIb

(∂2w

∂x2

)2

dx (8.2.13)


T =12

10∑r=1

10∑s=1

(mb

∫ L

−L

φr(x)φs(x)dx

)qr qs (8.2.14)

V =12

10∑r=1

10∑s=1

(EbIb

∫ L

−L

∂2φr

∂x2

∂2φs

∂x2dx

)qrqs (8.2.15)

63

The elements of the matrices can be computed as follows :

Mrs = mbLr+s−1 [1 − (−1)r+s−1], r, s = 1, 2, · · · , 10

Krs = 0, r = 1, 2, and s = 1, 2 · · · , 10

Krs = 0, s = 1, 2, and r = 1, 2 · · · , 10

Krs = EbIb(r−1)(r−2)(s−1)(s−2)L3(r+s−5) [1 + (−1)r+s], r, s = 3, 4, · · · , 10

(8.2.16)

When the system parameters are EbIb = 121Nm, mb = 0.355kg/0.685m, 2L = 0.685m, the natural

frequencies of the free-free beam are

ω1 ω2 ω3 ω4 ω5 ω6

Ritz Method(N=6)(Hz) 0 0 116.9 329.3 1157.6 2361.7

Ritz Method(N=8)(Hz) 0 0 116.0 319.7 669.7 116.6

Ritz Method(N=10)(Hz) 0 0 116 320 628 104.2

Vibration Table 0 0 116 319.4 626.6 1035.9

Example 8.2.3. Tennis racket and ball

Ten-term Ritz series are considered in this example. A schematic diagram of tennis racket and ball

model was shown in figure (8.7). The displacements of mB1 and mB2 are denoted as q11 and q12.

L

Ls

m B

kS

L

m B

kB

x

z

Figure 8.7: A schematic diagram of tennis racket and ball model

64


T =12

∫ L

−L

mbw2dx +

12mB1q

211 +

12mB2q

212 (8.2.17)

V =12

∫ L

−L

EbIb

(∂2w

∂x2

)2

dx +12kS [w(LS , t) − q11]2 +

12kB [qN+1 − qN+2]2 (8.2.18)


T =12

N∑r=1

N∑s=1

(mb

∫ L

−L

φr(x)φs(x)dx

)qr qs +

12mB1q

2N+1 +

12mB2q

2N+2 (8.2.19)

V =12

N∑r=1

N∑s=1

(ERIR

∫ L

−L

∂2φr

∂x2

∂2φs

∂x2dx

)qrqs +

12ks

[N∑

r=1

N∑s=1

φr(Ls)φs(Ls)

]qrqs

− kS

N∑r=1

φr(Ls)qrqN+1 +12ksq

2N+1 +

kB

2[q2

N+1 − 2qN+1qN+2 + q2N+2]

(8.2.20)

The elements of the mass matrix can be computed as follows:

Mrs = mbLr+s−1 [1 − (−1)r+s−1], r, s = 1, 2, · · · , N

MN+1,N+1 = mB1

MN+2,N+2 = mB2

(8.2.21)

The elements of the stiffness matrix are

Krs,racket = EbIb(r−1)(r−2)(s−1)(s−2)L3(r+s−5) [1 + (−1)r+s] r, s = 3, 4, · · · , N

Krs,coupled = ksφr(Ls)φs(Ls), r, s = 1, 2 · · · , N

Kr,N+1 = KN+1,r = −kSφr(Ls), r = 1, 2 · · · , N

KN+1,N+1 = kB + kS

KN+1,N+2 = KN+2,N+1 = −kB

KN+2,N+2 = kB

(8.2.22)

EI = 121Nm,2L = 0.685m,mB = 0.355kg/0.685m, Ls = 0.1575, mB1 = mB2 = 0.028kg, kS =

4.15e4N/m, kB = 7.98e4N/m.

Calculate the natural frequencies of the tennis racket and ball system.

ω1 ω2 ω3 ω4 ω5 ω6

Ritz Method(N=8)(rad/s) 0 0 113.5 142.4 325.5 409.9

Ritz Method(N=10)(Hz) 0 0 114 142 325 410

Ritz Method(N=12)(rad/s) 0 0 114 142 325 410

The mode shapes are

65

Mode q1 q2 q3 q4 q5 q6 q7 q8 q9 q10 q11 q12

Mode 1 -0.2 1 0 0 0 0 0 0 0 0 0.26 0.26

Mode 2 1 -0.2 0 0 0 0 0 0 0 0 0.91 0.91

Mode 3 0.25 -0.01 -1. -0.07 0.3 0.03 -0.03 0.0 -0.01 0. 0.17 0.2

Mode 4 0.32 -0.04 -0.8 0.56 0.7 -0.28 -0.5 -0.02 0.15 0.05 -0.72 -1

Mode 5 0. 0.37 -0.02 -1. 0.05 0.73 -0.04 -0.27 0.01 0.04 0.02 -0.04

Mode 6 -0.05 -0.37 0.38 0.97 -0.89 -0.85 0.76 0.45 -0.24 -0.11 1 -0.76

-0.2 0 0.2

-1

0

1

mode1

-0.2 0 0.2

-1

0

1

mode2

-0.2 0 0.2-1

-0.5

0

0.5mode3

-0.2 0 0.2-0.5

0

0.5mode4

-0.2 0 0.2

-0.2

-0.1

0

0.1

0.2

mode5

-0.2 0 0.2

-0.2

-0.1

0

0.1

0.2

mode6

Figure 8.8: Mode shapes of tennis racket and ball model

66

Example 8.2.4. A cantilever beam with a surface mounted active material. Calculate the natural

frequencies and mode shapes by Ritz method.

φr =x

Lsin(

rπx

2L) (8.2.23)

L

x

z

x1

x2

EbIb m b

EpIp m p

Figure 8.9: A schematic diagram of beam with configured with a distributed piezoelectric device.


T =12

∫ L

0

mbw2dx +

12

∫ L

0

2mpw2(H[x − x1] − H[x − x2])dx (8.2.24)

V =12

∫ L

0

EbIb

(∂2w

∂x2

)2

dx +12

∫ L

0

EpIeq

(∂2w

∂x2

)2

(H[x − x1] − H[x − x2])dx (8.2.25)

where mb is the mass per unit length of beam, EIeq is the effective stiffness per unit length of the

combined piezoceramic elements.

Now, we expand the solution in terms of a finite set of comparison function:

w =N∑

r=1

φr(x)qr(t) (8.2.26)


T =12

N∑r=1

N∑s=1

[(mb

∫ L

0

φr(x)φs(x)dx

)+(

2mp

∫ x2

x1

φr(x)φs(x)dx

)]qr qs (8.2.27)

V =12

N∑r=1

N∑s=1

[(∫ L

0

EbIb∂2φr(x)

∂x2

∂2φs(x)∂x2

dx

)+(

2EpIeq

∫ x2

x1

∂2φr(x)∂x2

∂2φs(x)∂x2

dx

)]qrqs

(8.2.28)

67

Msrs = mb

∫ L

0

φr(x)φs(x)dx

Mprs = 2mp

∫ x2

x1

φr(x)φs(x)dx

Ksrs =∫ L

0

EbIb∂2φr(x)

∂x2

∂2φs(x)∂x2

dx

Kprs = 2EpIeq

∫ x2

x1

∂2φr(x)∂x2

∂2φs(x)∂x2

dx

(8.2.29)

where the equivalent area moment acting on a beam is written as

Ieq =bpt

3p

12+ bptp

(tb2

+tp2

)2

(8.2.30)

Let the non-dimensional resonant angular frequency be wr =√

mbL4

EbIbwr. The resulting eigensolu-

tions of cantilever beam are

w1 w2 w3 w4

Ritz Method(N=4) 3.5222 22.2704 67.41100 309.79199

Ritz Method(N=6) 3.5172 22.1179 62.3783 124.1380

Vibration Table 3.5160 22.0345 61.6972 120.9019

The specification of the cantilever beam are

ρ(Density) L(Length) b(Width) t(Thickness) Eb

2750 kg/m3 340 mm 25.4 mm 2 mm 7.00×1010N/m3

The natural frequencies of the cantilever beam are

f1 f2 f3 f4

Ritz Method(N=6) 14.11 Hz 88.70 Hz 250.16 Hz 497.84 Hz

The specification of the piezo element are given as

ρ(Density) Lp(Length) bp(Width) tp(Thickness) Ep x1 x2

8200 kg/m3 32.7 mm 22 mm 0.22 mm 1.45 ×1011N/m3 30 mm 62.7 mm

The natural frequencies of the cantilever beam with piezoelectric materials are

f1 f2 f3 f4

Ritz Method(N=6) 16.37 Hz 93.24 Hz 250.17 Hz 495.25 Hz

68

Homework 8.2.5. A cantilever beam with lumped mass M is shown in figure. Calculate the

natural frequencies and mode shapes by Ritz method.

L

x

z

EbIb m b

M

Figure 8.10: A schematic diagram of cantilever beam with lumped mass

Chapter 9

Finite Element Analysis

9.1 Euler-Bernoulli Beam

9.1.1 Basic relation

The beam with length 2a and constant cross-sectional area A is shown in figure (??).

x

z

a a

pz

Figure 9.1: Mode shapes of tennis racket and ball model

We assume that the stress component σy, σz, τxy, andτyz are zero. It also assumes that the planesections which are normal to the undeformed axis remain plane after bending. With this assumption,the axial displacement u at a distant z from the neutral axis is

u(x, z) = −z∂w

∂x(9.1.1)

The strain component is

εx =∂u

∂x= −z

∂w2

∂x2(9.1.2)

γxz =∂u

∂z+

∂w

∂x= 0 (9.1.3)

69

70

The strain energy stored in the elements is given by

U =12

∫v

σxεxdV (9.1.4)

The normal stress is given byσx = Eεx (9.1.5)

Substituting equation (9.1.5)and (9.1.2) into equation (9.1.4) gives , since dV = dA · dx

U =12

∫ a

a

EIy

(∂w2

∂x2

)dx (9.1.6)

whereIy =

∫A

z2dA (9.1.7)

The kinetic energy is given by

T =12

∫ a

a

ρAw2dx (9.1.8)

The virtual work done by external force for the element is

δWext =12

∫ a

a

pzδwdx (9.1.9)

9.1.2 Finite Element Modeling

The displacement function can be represented by a polynomial having four constants

w = α1 + α2ξ + α3ξ2 + α4ξ

3 (9.1.10)

The expression(9.1.10) can be written in the following matrix form

w = 1, ξ, ξ2, ξ3�

α1

α2

α3

α4

(9.1.11)

Differentiating equation (9.1.10) gives

aθy = a∂w

∂x=

∂w

∂ξ= α2ξ + 2α3ξ + 3α4ξ

2 (9.1.12)

Evaluating (9.1.10) and (9.1.12) at ξ = ∓1 gives

w1

aθy1

w2

aθy2

=

1 −1 1 −1

0 1 −2 3

1 1 1 1

0 1 2 3

α1

α2

α3

α4

(9.1.13)

71

Solving for {α} gives{α} = [C]e{w}e (9.1.14)

where{v}T

e = w1, θy1, w2, θy2� (9.1.15)

[m]e =ρAa

105

78 22a 27 −13a

22a 8a2 13a −6a2

27 13a 78 −22a

−13a −6a2 −22a 8a2

(9.1.16)

[k]e =EIy

2a3

3 3a −3 3a

3a 4a2 −3a 2a2

−3 −3a 3 −3a

3a 2a2 −3a 4a2

(9.1.17)

72

9.2 Thin Plate Theory

9.2.1 formulation

The bending energy expression for the thin plate are

U =12

∫v

{σb}T {εb}dV (9.2.1)

where bending stress and strain components are

{σb} = {σx σy τxy}T , {εb} = {εx εy γxy}T (9.2.2)

The stress-strain relationships take the form

{σb} = [Db]{εb} (9.2.3)

where

[Db] =E

1 − ν2

1 ν 0

ν 1 0

0 0 1−ν2

(9.2.4)

In deriving the energy functions for plate bending, the basic assumptions are that the direct stressin the transverse direction, σz, is zero. Also, a straight line normal to the middle surface of theundeformed plate remains normal after deformation. Therefore, the displacements u and v are givenby

u(x, y, z) = −z∂w

∂x, v(x, y, z) = −z

∂w

∂y(9.2.5)

where w(x, y) denotes the displacement of middle surface in the z-direction. The strain are given asfollows:

{εb} =

εx

εy

γxy

=

∂u∂x

∂v∂y

∂u∂y + ∂v

∂x

= −z

κx

κy

κxy

(9.2.6)

where the curvature vector κ is given as follows:

{κ} = {κx κy κxy}T = {∂2w

∂x2

∂2w

∂y22

∂2w

∂x∂y}T (9.2.7)

Using (9.2.6), the strain matrix can be written in the form

{εb} = −z{κ} (9.2.8)

Substituting (9.2.8) and (9.2.3) into (9.2.1)and integrating with respect to z gives

U =12

∫A

h3

12{κ}T [Db]{κ}dA (9.2.9)

73

The kinetic energy of the plate is given by

T =12

∫A

ρhw2dA (9.2.10)

The normal displacement of plate, w, and the two rotations have relation as follows:

θx =∂w

∂y, θy = −∂w

∂x(9.2.11)

9.3 Finite Element Modeling

In terms of the master coordinates (9.2.11) becomes

θx =1b

∂w

∂η, θy = −1

a

∂w

∂ξ(9.3.1)

Since the rectangular element has 12 degrees of freedom, the displacement function can be repre-sented by a polynomial having twelve terms, that is

w = α1 + α2ξ + α3η + α4ξ2 + α5ξη + α6η

2

+α7ξ3 + α8ξ

2η + α9ξη2 + α10η

3 + α11ξ3η + α12ξη

3(9.3.2)

The expression(9.3.2) can be written in the following matrix form

w = 1, ξ, η, ξ2, η ξ, η2, ξ3, η ξ2, η2 ξ, η3, η ξ3, η3 ξ�{α} (9.3.3)

where{α} = α1, α2, α3, · · · , α12� (9.3.4)

Evaluating w, bθx, and aθy at ξ = ∓1, η = ∓1gives

{w}e = [A]e{α} (9.3.5)

where{w}T

e = w1, bθx1, aθy1, · · · , w4, bθx4, aθy4� (9.3.6)

74

and

[A]e =

1 −1 −1 1 1 1 −1 −1 −1 −1 1 1

0 0 1 0 −1 −2 0 1 2 3 −1 −3

0 −1 0 2 1 0 −3 −2 −1 0 3 1

1 1 −1 1 −1 1 1 −1 1 −1 −1 −1

0 0 1 0 1 −2 0 1 −2 3 1 3

0 −1 0 −2 1 0 −3 2 −1 0 3 1

1 1 1 1 1 1 1 1 1 1 1 1

0 0 1 0 1 2 0 1 2 3 1 3

0 −1 0 −2 −1 0 −3 −2 −1 0 −3 −1

1 −1 1 1 −1 1 −1 1 −1 1 −1 −1

0 0 1 0 −1 2 0 1 −2 3 −1 −3

0 −1 0 2 −1 0 −3 2 −1 0 −3 −1

(9.3.7)

Solving (9.3.5) for {α}gives{α} = [A]−1

e {w}e (9.3.8)

where

A−1 =18

2 1 −1 2 1 1 2 −1 1 2 −1 −1

−3 −1 1 3 1 1 3 −1 1 −3 1 1

−3 −1 1 −3 −1 −1 3 −1 1 3 −1 −1

0 0 1 0 0 −1 0 0 −1 0 0 1

4 1 −1 −4 −1 −1 4 −1 1 −4 1 1

0 −1 0 0 −1 0 0 1 0 0 1 0

1 0 −1 −1 0 −1 −1 0 −1 1 0 −1

0 0 −1 0 0 1 0 0 −1 0 0 1

0 1 0 0 −1 0 0 1 0 0 −1 0

1 1 0 1 1 0 −1 1 0 −1 1 0

−1 0 1 1 0 1 −1 0 −1 1 0 −1

−1 −1 0 1 1 0 −1 1 0 1 −1 0

(9.3.9)

75

The shape function in master elements are given as follows:

φ1

φ2

φ3

φ4

φ5

φ6

φ7

φ8

φ9

φ10

φ11

φ12

=18

2 − 3 η + η3 − 3 ξ + 4 η ξ − η3 ξ + ξ3 − η ξ3

1 − η − η2 + η3 − ξ + η ξ + η2 ξ − η3 ξ

−1 + η + ξ − η ξ + ξ2 − η ξ2 − ξ3 + η ξ3

2 − 3 η + η3 + 3 ξ − 4 η ξ + η3 ξ − ξ3 + η ξ3

1 − η − η2 + η3 + ξ − η ξ − η2 ξ + η3 ξ

1 − η + ξ − η ξ − ξ2 + η ξ2 − ξ3 + η ξ3

2 + 3 η − η3 + 3 ξ + 4 η ξ − η3 ξ − ξ3 − η ξ3

−1 − η + η2 + η3 − ξ − η ξ + η2 ξ + η3 ξ

1 + η + ξ + η ξ − ξ2 − η ξ2 − ξ3 − η ξ3

2 + 3 η − η3 − 3 ξ − 4 η ξ + η3 ξ + ξ3 + η ξ3

−1 − η + η2 + η3 + ξ + η ξ − η2 ξ − η3 ξ

−1 − η + ξ + η ξ + ξ2 + η ξ2 − ξ3 − η ξ3

(9.3.10)

The finite element approximation of the displacement w over a given element with n nodes has theform

w = N1(ξ, η) N2(ξ, η) N3(ξ, η) N4(ξ, η)�{w}e

= N(ξ, η)�{w}e

(9.3.11)

where{w}T

e = w1 θx1 θy1 · · · w4 θx4 θy4� (9.3.12)

andN(ξ, η) = φ1 bφ2 aφ3 · · · φ10 bφ11 aφ12� (9.3.13)

Substituting (9.3.11)into (9.2.10) gives

Te =12{w}T

e [m]e{w}e (9.3.14)

where

[m]e =∫

A

ρhN(ξ, η)�T N(ξ, η)�dξdη

= ρhab

∫ 1

−1

∫ 1

−1

N(ξ, η)�T N(ξ, η)�dξdη

(9.3.15)

is the element inertia matrix.

Substitution (9.3.11) and (9.2.7) into (9.2.9)gives

Ue =12{w}T

e [k]e{w}e (9.3.16)

76

where the stiffness element matrix is

[k]e =∫

A

h3

12[B]T [Db][B]dA (9.3.17)

and the curvature vector in finite element [B] is

[B] =

∂2

∂x2

∂2

∂y2

2 ∂2

∂x∂y

N(ξ, η)� =

1a2

∂2

∂ξ2

1b2

∂2

∂η2

2ab

∂2

∂ξ∂η

N(ξ, η)� (9.3.18)

The expression of curvature vector with respect to the shape function is as follows

[B] =

1a2

∂2φ1∂ξ2

ba2

∂2φ2∂ξ2

1a

∂2φ3∂ξ2 · · · 1

a2∂2φ10∂ξ2

ba2

∂2φ11∂ξ2

1a

∂2φ12∂ξ2

1b2

∂2η1∂η2

1b

∂2φ2∂η2

ab2

∂2φ3∂η2 · · · 1

b2∂2φ10∂η2

1b

∂2φ11∂η2

ab2

∂2φ12∂η2

2ab

∂2φ1∂ξ∂η

2a

∂2φ2∂ξ∂η

2b

∂2φ3∂ξ∂η · · · 2

ab∂2φ10∂ξ∂η

2a

∂2φ11∂ξ∂η

2b

∂2φ12∂ξ∂η

(9.3.19)

where derivatives of shape functions are as follows :

∂2

∂ξ2

φ1

φ2

φ3

φ4

φ5

φ6

φ7

φ8

φ9

φ10

φ11

φ12

=14

3 ξ − 3 η ξ

0

1 − η − 3 ξ + 3 η ξ

−3 ξ + 3 η ξ

0

−1 + η − 3 ξ + 3 η ξ

−3 ξ − 3 η ξ

0

−1 − η − 3 ξ − 3 η ξ

3 ξ + 3 η ξ

0

1 + η − 3 ξ − 3 η ξ

(9.3.20)

77

∂2

∂η2

φ1

φ2

φ3

φ4

φ5

φ6

φ7

φ8

φ9

φ10

φ11

φ12

=14

3 η − 3 η ξ

−1 + 3 η + ξ − 3 η ξ

0

3 η + 3 η ξ

−1 + 3 η − ξ + 3 η ξ

0

−3 η − 3 η ξ

1 + 3 η + ξ + 3 η ξ

0

−3 η + 3 η ξ

1 + 3 η − ξ − 3 η ξ

0

(9.3.21)

∂2

∂ξ∂η

φ1

φ2

φ3

φ4

φ5

φ6

φ7

φ8

φ9

φ10

φ11

φ12

=18

4 − 3 η2 − 3 ξ2

1 + 2 η − 3 η2

−1 − 2 ξ + 3 ξ2

−4 + 3 η2 + 3 ξ2

−1 − 2 η + 3 η2

−1 + 2 ξ + 3 ξ2

4 − 3 η2 − 3 ξ2

−1 + 2 η + 3 η2

1 − 2 ξ − 3 ξ2

−4 + 3 η2 + 3 ξ2

1 − 2 η − 3 η2

1 + 2 ξ − 3 ξ2

(9.3.22)

9.4 Example

The natural frequency of simple supported thin plate is given by

ωmn = π2[(m

L1)2 + (

n

L2)2]

√DE

ρh(9.4.1)

where

DE =Eh3

12(1 − υ2)(9.4.2)

is the flexural rigidity of the plate.

78

The speculation of plate is as follows ρ = 1, h = 1, E = 48, L1 = 8, L2 = 8, υ = 0.5 and 4×4 elementare used in finite element model.

Mode Number Analytical solution(rad/s) Finite Element Solution(rad/s)

ω11 0.7123 0.6875

ω12 = ω21 1.7807 1.6977

ω22 2.8491 2.5488

ω13 = ω31 3.5414 4.0931

ω23 = ω32 4.6298 4.0931

ω33 6.4104 5.4184

vibration.pdf

Documents

term ritz series

ritz function asw

optimal frequency ratio

free boundary condition

dynamic vibration absorber

frequency response function

frequency response function

decoupled modal equations