Vibration

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-les . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 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 dierent toolboxes available which extend the basic functions of MATLAB into dierentareas.

1.1 Making matrix

Matlab uses variables that are dened to be matrices. A matrix is a collection of numerical valuesthat are organized into a specic conguration of rows and columns. Here are examples of matricesthat could be dened 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

2If three numbers, integer or non-integer, are separated by two colons, the middle number is inter-preted to be a range and the rst 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 3Element 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

31.3 Functions

Matlab includes many standard functions. In Matlab sin and pi denotes the trigonometric functionsine and the constant .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 [/100, 10]>>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-les.

Table 1.1: Basic Matrix Functions

Symbol Explanationsinv Inverse of a matrixdet Determinant of a matrixtrace Summation of diagonal elements of a matrix

40 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

51.5.2 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 isif 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 statementThe if-else statement allows one to execute one set of statements if logical expression is true and adierent 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

61.5.4 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-lesave FILENAME

An ASCII le is a le 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 Newtons law, the equation of motion of the system isrepresented by

m

c

k

F

x

Figure 2.1: Single degree of freedom system.

7

8mx + 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 + 2nx + 2nx = 0 (2.1.2)

where n =

k/m is natural angular frequency and = c/(2km) 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 + 2ns + 2n = 0 (2.1.4)

The solutions of equation (2.1.4) are given by

s1,2 = n n

(2 1) (2.1.5)There are three possible cases:

(a) Overdamped MotionIn this case, the damping ratio is greater than 1 ( > 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) = ent(Aen

21t + Ben

21t) (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 sucient to overcome the damping force.

(b) Underdamped MotionIn this case the damping ratio is less than 1 (0 < < 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) = ent(Aejn

12t + Bejn

12t)

= ent(Aejdt + Bejdt)

= ent(C cosdt + D sindt)

= Xent sin(dt + )

(2.1.7)

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

d =

1 2n

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

x(t) = (A + Bt)ent (2.1.8)

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

9Homework 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 denition 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 gure 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 dierential equation with zero initial conditions, the transfer function is

X(s)F (s)

=1/m

s2 + 2ns + 2n(2.2.2)

where n =

k/m, = c/2mn.

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

X(j)F (j)

=1/m2

[(n/)2 1] + 2j(n/) (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 + wn2);mag(cnt,:) = abs(xfer(cnt,:));

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

end


gure(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 o


gure(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 o

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 stiness, k, and damping with viscous damping coecient 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 dened 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 gure 2.3(b).The structure with mass m is connectedto the base by stiness, k. Let us suppose that the viscous damping with viscous damping coecientc 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 specied by transfer functions. The correspondingcommand is :sys=tf(num,den)The output sys is a model-specic 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 wn2]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 dierentialequation to 2n rst order dierential equations. The rst order form of equations for the system iscalled as state space form.

Start by solving equation second order dierential equations.

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

14

we dene 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

m1k m1c

](2.3.4)

and

B =

[0

m1

](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)/dtIntegrate

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 dened 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 gure 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[1H(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 stiness [k] matrices are symmetric.

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

x = uejt (3.1.2)

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

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

(2M + K)uejt = 0 (3.1.3)Note that the scalar ejt = for any value of t and hence equation (3.1.1) yields the fact that andu must satisfy the vector equation

(2M + K)u = 0

17

18

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

This equation is satised for any u if the determinant of the above equation is zero.2M + K = 0 (3.1.4)The simultaneous solution of equation (3.1.4) results in the values of parameter 2. The is calledas eigenvalues of the problem.

Once the value of 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 gure 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 c2c2 c2 + c3

]{x1

x2

}+

[k1 + k2 k2k2 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 rst considered. Thus the equations reduce to[

m1 0

0 m1

]{x1

x2

}+

[k1 + k2 k2k2 k2 + k3

]{x1

x2

}=

{0

0

}(3.1.6)

Consider a numerical example for the system shown in gure (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 22 6

]{x1

x2

}=

{0

0

}(3.1.7)

19

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

2

[5 0

0 10

]{A1

A2

}=

[4 22 6

]{A1

A2

}(3.1.8)

Solving Eigenvalue Problem with MATLAB

The eigenvalue problem of a matrix is dened as

Au = u (3.1.9)

and generalized eigenvalue problem isKu = Mu (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 1=0.6325 rad/s, 2=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 stiness 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 = mi2i = ki, i = 1, 2

(3.1.12)

where mi and ki is called modal mass and modal stiness 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

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

1 1}T

for 1 =

2/5 rad/s, and u2 ={1 0.5

}Tfor 1 = 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 = 21 m1 = 6k2 = 22 m2 = 15/2

3.1.3 Normalization of Mode Shapes

While above relations are related to the mass and stiness 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 stiness 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 = 2i , 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 rst and second natural modesare

{u}1 = 1m1

1

1

= 1

15

1

1

{u}2 = 1m2

1

1/2

= 1

15/2

1

1/2

The orthogonality of modes permit us to transform the coupled equations of motion dened 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 odiagonal elements.

If the system of equations could be uncoupled, so that we obtained diagonal mass and stinessmatrices, 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 stiness matrices. Assumingthat the viscous damping can be decoupled by modal matrix, we obtain

qi(t) + 2iiq(t) + 2i qi(t) = Ni(t), i = 1, 2, (3.1.19)

where Ni(t) is

Ni(t) ={u}Ti {f(t)}{u}Ti [m]{u}i

={u}Ti {f(t)}

mi(3.1.20)

The ratio in equation (3.1.20){u}Ti

{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[(2i 2) + 2iii]

(3.1.22)

where i 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 + 0.2582f20.3651f1 0.1826f2

}=

{f

1

f2

}(3.1.23)

The matrix equation of (3.1.23) can be written in terms of algebraic dierential 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 gure (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(1t) = 0.5164 cos(0.6325t)

24

5 10

1

2 2 4

2/51

1

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(2t) = 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 dicult 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 denedas

[c] = [m] + [k] (3.2.1)

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

Caughey1 showed that there exists a necessary and sucient 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 = 0 and

= 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

x1x2

+

2.7 0.30.3 0.3

x1x2

+

27 33 3

x1x2

=

0 0

0 1

F1F2

(3.3.1)

26

Numerical Simulation with MATLAB

[P ] =

(0.2357 0.23570.7071 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 0.70710 0.7071

)

Hence the decoupled modal equations become

q1 + 0.2q + 2q1 = 0.7071 3 cos 2tq2 + 0.4q + 4q2 = 0.7071 3 cos 2t

(3.3.2)

Comparing the coecient of qi to 2ii yields

1 = 0.2222 = 0.222

Thus the damped natural frequencies becoms

d1 = 1

1 21 1.41d2 = 2

1 22 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 dening 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

m1k m1c

](3.4.3)

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

i = i ji

1 2ii = i + ji

1 2i

(3.4.4)

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

to the initial condition using state-space method. Let c1 = c3 = 0, c2 = 0.2Ns/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.

Example MATLAB Code

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 gure (3.6).

Response to Initial Conditions

Time (sec)

Ampl

itude

0 50 100 150 200 250 300 350 400 450 5000.1

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 gure 3.5. Let m1 = 10kg, m2 = 1kg, k1 = 0.5N/m,k2 = 0.05N/m, and c1 = 0, c2 = 0.2Ns/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 condition

x(0) = x0 x(0) = x0

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

where

A =

[0 I

m1k m1c

](3.4.7)

and

B =

[0

m1bf

](3.4.8)

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

in gure (3.1) between c2 = 0 and c2 = 0.2Ns/m. The other parameters are as follows. Letm1 = 2kg,m2 = 1kg, k1 = 0.2N/m, k2 = 0.05N/m, k3 = 0.05N/m and c1 = c3 = 0 Ns/m and theexcitation force F2 be zero.

Example MATLAB Code

-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 gure (3.7).

Bode Diagram

Frequency (rad/sec)

Phas

e (de

g)M

agni

tude

(dB)

40

20

0

20

40

60

80

101 100180

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 rst 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 Hartogs on Mechanical Vibration (1940) book 2 .

4.2 Vibration Absorber

Figure (4.1) shows a SDOF system having mass m and stiness k, subjected to external forcing. Adynamic absorber with mass m2, stiness 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 c2c2 c2

]{x1

x2

}+

[k1 + k2 k2k2 k2

]{x1

x2

}=

{F (t)

0

}

(4.2.1)

We set xi = Re[Xiejt] 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) + j(c1 + c2)m12] (k2 + jc2)

(k2 + jc2) k2 + jc2 m22

]{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 nd from equation(4.2.2) that

X1 = F(k2 m22)

[(k1 + k2) 2m1](k2 m22) k22(4.2.3)

X2 = Fk2

[(k1 + k2) 2m1](k2 m22) k22(4.2.4)

where the determinant of the system of coecients in equation is

() = [(k1 + k2) 2m1](k2 m22) k22. (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

2 = 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 = Fk2

(4.2.7)

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

For simplications 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 deection of primary system.21 = k1/m1 ; natural frequency of primary system22 = k2/m2 ; natural frequency of secondary system = m2/m1 ; mass ratio=secondary mass/primary mass

With this denitions, also note thatk2k1

= 2221

= f2 (4.2.8)

where frequency ratio f is f = 2/1.

Then equationa (4.2.3)and (4.2.4) becomes

X1xst

=1 2

[1 2][1 + f2 (f)2] f2 (4.2.9)

X2xst

=1

[1 2][1 + f2 (f)2] f2 (4.2.10)

where frequency ratio is = /2.

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

In fact, if the driving frequency shifts such that |X1/xst| > 1, the force transmitted to the primarysystem is amplied 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., 2 = 1, or k2/m2 = k1/m1)

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

X1xst

=1 2

(1 2)(1 + 2) (4.2.11)

X2xst

=1

(1 2)(1 + 2) (4.2.12)

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

(1 2)(1 + 2) = 04 f2(2 + ) + 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

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

The solutions are2 = (1 + /2)

+ 2/4 (4.2.14)

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

Homework 4.2.1. Inmans Book Example 5.3.1

Homework 4.2.2. Inmans 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 (KM1C = CM1K).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)ejt =

[X1

X2

]ejt (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

mass ratio

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

X1F

=(k2 m22) + jc2

det([K] 2[M ] + j[C]) (4.2.16)

X2F

=k2 + jc2

det([K] 2[M ] + j[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

X21F 2

=(k2 m22)2 + (c2)2

[(k1 m12)(k2 m22)m2k22]2 + [k1 (m1 + m2)2]2c222(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 deection xst = F/kof the primary system. In addition, consider the mixed damping ratio dened by

=c2

2m21(4.2.19)

where =

k1/m1. Using the standard frequency ratio r = /1, the ratio of natural frequencies

36

f = 2/1, and the mass ratio = m2/m1, equation (4.2.18) can be written as

X1xst

=

(2r)2 + (r2 f2)2

(2r)2(r2 1 + r2)2 + [r2f2 (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 , the ratioof the decoupled natural frequencies f the ratio of the driving frequency to the primary naturalfrequency r and the damping ratio of absorber .

These four parameters can be considered as design variables and are chosen to give the smallestpossible value of the primary masss 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. 2 = 2. 2 = 0

3. = 0.10

4. = 0.10

for mass ratio = 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 + (4.2.21)

=

3

8(1 + )3(4.2.22)

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

ratio is = 1/20.

4.3 Isolation Design

4.3.1 Passive Isolators

In gure (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 + 2n(x1 x0) + 2n(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/x

st

Excitation frequency ratio (r)

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

x1

x0

m

k c

Figure 4.5: Schematic of passive isolators

38

where 2n = k/m, and = c/2km. For the base excitation problem it is assumed that the base

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

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

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

X1X0

=[

1 + (2r)2

(1 r2)2 + (2r)2]1/2

(4.3.3)

where the frequency ratio r = b/n.

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 dened as

FTkX0

= r2[

1 + (2r)2

(1 r2)2 + (2r)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 = 2nx1 2n(x1 x0) = 0 (4.3.6)

leading to a sprung mass equation of motion as

x + 2nx1 + 2n(x1 x0) = 0 (4.3.7)

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

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

X1X0

=[

1(1 r2)2 + (2r)2

]1/2(4.3.8)

where the frequency ratio r = b/n3Bender, E.K., Optimum linear preview control with application to vehicle suspension, ASME, Journal of basic

engineering, 90(2), June 1968, pp.213-221.

39

10-1

100

101

10-1

100

Freqeucney ratio

Displacement Transmissibility

=1 =0.8

=0.4 =0.6

=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

=0.2 =0.4 =0.6

=0.8 =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 rst proposed by Crosby and Karnopp4 in 1973

m

k Fs

x1

x0

Figure 4.9: Schematic of semi-active isolators

Fs = Fs/m =

{2n, x1(x1 x0) > 00, 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 gure (5.1). The forces on the innitesimal 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 = A(x)dxu(x, t)t2

(5.2.1)

where u(x, t) is the deection 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 Youngs modulus. The dierential 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

A(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

)2u(x, t)

x2(5.2.5)

The quantity v =

E/ denes 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 dened as m = A. 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)simplies

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

4W (x) = 0 (6.2.2)

where 4 = 2mEI , 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 sinx + C2 cosx + C3 sinhx + C4 coshx (6.2.4)

Applying the boundary conditions for x = 0, we nd

C2 + C4 = 0

C1 + C3 = 0(6.2.5)

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

C1(sinL + sinhL) + C2(cosL + coshL) = 0

C1(cosL + coshL) C2(sinL sinhL) = 0(6.2.6)

46

Equating the determinant of the coecients to zero, we obtain the characteristic equation[(sinL + sinhL) (cosL + coshL)

(cosL + coshL) (sinL sinhL)

]{C1

C2

}=

{0

0

}(6.2.7)

The characteristic equation iscosL coshL = 1 (6.2.8)

From the numerical analysis 1L = 1.875, 2L = 4.694, 3L = 7.855

1 = (1.875)2

EImL4 rad/sec

2 = (4.694)2

EImL4 rad/sec

3 = (7.855)2

EImL4 rad/sec

(6.2.9)

We obtain the corresponding eigenfunctions

Wr(x) = Cr(cosx coshx) + Cr sinLsinhLcos L+coshL (sinx sinhx)= Ar[(sinrL sinhrL)(sinrx sinhrx) + (cosrL + coshrL)(cosrx coshrx)]

(6.2.10)

Example 6.2.1. The geometric and material properties are

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

2750 kg/m3 340 mm 22 mm 2 mm 7.001010N/m3

The natural frequencies are

1 = 88.6 rad/sec = 14.1 Hz

2 = 555.2 rad/sec = 88.4 Hz

3 = 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 specied

above example. Normalize the eignefunction as L0

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)]]A2 = 0.03139, W2(x) = A2[1.72[cos(13.81x) cosh(13.81x)] 1.75[sin(13.81x) sinh(13.81x)]]A3 = 0.00133, W3(x) = A3[1.71[cos(23.10x) cosh(23.10x)] 1.71[sin(23.1x) sinh(23.1x)]]

[Homework 7] Calculate the natural frequency and plot rst 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

L0

m(x)Wi(x)Wj(x)dx = 0 for j = iand 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 Lagranges 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 systems eigenfunctions toreduce the calculation of the response to a system of decoupled modal equations for the time response.

m2w

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) = L0

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) + EIWi (x)qi(t)] =

i=1

fi(t)Wi(x) (6.4.4)

We know that the modes are satisfying

EIWi (x) =

2i mWi(x)

Substituting this relation into equation (6.4.4) leads to

i=1

[mWi(x)qi(t) + cWiq(t) + 2i mWi(x))qi(t)] =i=1

fi(t)Wi(x) (6.4.5)

49

It is convenient to normalize the eigenfunction as L0

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

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

qi(t) + 2iiq(t) + 2i 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)(x x1) (6.4.8)

Equation (6.4.3) becomes

fi(t) = L0

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


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

0 50 100 150 200 250 300 350 400 450 500107

106

105

104

103

102

101

100

Frequency (Hz)

FRF

(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 specication

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 gure

(6.4)

50

Example MATLAB Codecantifrf.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 = t3*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 8Obtain the frequency response function of the beam with free-free boundary condition.The specication of the system is the same with above example.

6.4.2 Moment excitation

The excitation force becomes

f(x, t) = Mo

x[(x x2) (x x1)] (6.4.11)


fi(t) = L0

Mo

x[(x x2) (x x1)]Wi(x)dx = Mo[W i (x2)W

i (x1)] (6.4.12)


qi(t) + 2iiq(t) + 2i qi(t) = Mo[Wi (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.

(Density) L(Length) b(Width) t(Thickness) E x1 x122750 kg/m3 340 mm 22 mm 2 mm 7.001010N/m3 48 mm 80 mm

Wr(x) = Ar[r(sinrL sinhrL)(cosrx coshrx) r(cosrL + coshrL)(sinrx + sinhrx)]

(6.4.14)

Example MATLAB Code cantipzt.mclear 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 = t3*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 500106

105

104

103

102

101

100

Frequency (Hz)

FRF

(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.

1x

=2w

x2,

1y

=2w

y2(7.1.1)

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

u(x, y, z) = z wx

, v(x, y, z) = z wy

(7.1.2)

The strain are given as follows:

{b} =

x

y

xy

=

ux

vy

uy +

vx

= z

2wx2

2wy2

2 2w

xy

(7.1.3)

Hookes law for plane stress relates these strains to the stress resultants,

x =E

1 2 (x + y) = Ez

1 2(2w

x2+

2w

y2

)(7.1.4)

y =E

1 2 (y + x) = Ez

1 2(2w

y2+

2w

x2

)(7.1.5)

xy = 2Gz 2w

xy= Ez

1 + 2w

xy(7.1.6)

The stress-strain relationships take the matrix form as

{b} = [Db]{b} (7.1.7)

54

55

where

[Db] =E

1 2

1 0

1 0

0 0 12

(7.1.8)

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

Mx = h/2h/2

zxdz = D(2w

x2+

2w

y2

)(7.1.9)

My = h/2h/2

zydz = D(2w

y2+

2w

x2

)(7.1.10)

Mxy = h/2h/2

zxydz = D(1 ) 2w

xy(7.1.11)

where D = Eh3/12(12) is called the exural rigidity of the plate. Equation (7.1.9)-(7.1.11) relatemoments to deection w.

D

(4w

x4+ 2

4w

x2y2+

4w

y4

)+ h

2w

t2= p(x, y, t) (7.1.12)

orD4w + hw = p(x, y, t) (7.1.13)

7.2 Equation of Motion

The bending energy expression for the thin plate are

V =12

v

{b}T {b}d (7.2.1)

The bending energy becomes

V =12

v

{b}T [Db]{b}d (7.2.2)

The kinetic energy of the plate is given by

T =12

A

hw2dA (7.2.3)

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

w =N

r=1

r(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 = h b0

a0

i(x, y)j(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 stiness matrix

Ks,ij = Ds b0

a0

{2ix2

2jx2

+2iy2

2jy2

+ s

(2ix2

2jy2

+2iy2

2jx2

)+ 2(1 s)

2ixy

2jxy

}dxdy

(7.2.6)where Ds = h

3

12Es

12s

Chapter 8

Approximate Method

8.1 Introduction

It is too dicult 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 eciently and accurately.

8.2 Rayleigh Ritz Method

The Rayleigh Ritz method obtains an approximate solution to a dierential 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 satises the geometric boundary condition and containsunknown coecients.2. Substitute the assumed solution into the kinetic and potential energy and nd the unknowncoecients.

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

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

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

r(x) =x

Lsin(

rx

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

r(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

L0

mbw2dx +

12maq

24 (8.2.3)

V =12

L0

EbIb

(2w

x2

)2dx +

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

3r=1

3s=1

(mb

L0

rsdx

)qr qs +

12maq

24 (8.2.5)

V =12EbIb

3r=1

3s=1

L0

(2rx2

2sx2

dx

)qrqs +

12k

[3

r=1

3s=1

r(L/2)s(L/2)

]qrqs

k3

r=1

r(L/2)qrq4 +12kq24

(8.2.6)

The mass and stiness matrices are

[M ] = mbL

0.1413 0.0901 0.019 00.0901 0.1603 0.0973 00.019 0.0973 0.1639 0

0 0 0

(8.2.7)

L

L/2

ma

k

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

59

and

[K] =EbIbL3

43.376 + 0.25 74.570 75.873 0.25 /274.570 368.323 459.529 0

75.873 0.25 459.529 1559.3 + 0.25 /2/2 0 /2

(8.2.8)

where = ma/(mbL) and = kL3/EbIb.

The generalized force are

Q(t) = L0

f(x, t)r(x)dx (8.2.9)

-pinclamp.nb-

Let the non-dimensional resonant angular frequency be wr =

mbL4

EbIbwr. The resulting eigensolution

when = 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 = 15.75,and a = 0.1,are

le.pc.m w1 w2 w3 w4Ritz 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 10.5

0

0.5

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10.5

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 10.5

0

0.5

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10.5

0

0.5

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10.5

0

0.5

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10.5

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 gure (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) =10r=1

r(x)qr(t) (8.2.10)

A schematic diagram of tennis racket and ball model was shown in gure (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 1(x) = 1. To represent rotational rigid mode we

select the basis function to be 2(x) = x/L. Because there are no geometric boundary condition to

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

r(x) =( xL

)r1, r = 1, 2, , 10 (8.2.11)


T =12

LL

mbw2dx (8.2.12)

V =12

LL

EbIb

(2w

x2

)2dx (8.2.13)


T =12

10r=1

10s=1

(mb

LL

r(x)s(x)dx

)qr qs (8.2.14)

V =12

10r=1

10s=1

(EbIb

LL

2rx2

2sx2

dx

)qrqs (8.2.15)

63

The elements of the matrices can be computed as follows :

Mrs = mbLr+s1 [1 (1)r+s1], r, s = 1, 2, , 10Krs = 0, r = 1, 2, and s = 1, 2 , 10Krs = 0, s = 1, 2, and r = 1, 2 , 10

Krs =EbIb(r1)(r2)(s1)(s2)

L3(r+s5) [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

1 2 3 4 5 6

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 gure (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

LL

mbw2dx +

12mB1q

211 +

12mB2q

212 (8.2.17)

V =12

LL

EbIb

(2w

x2

)2dx +

12kS [w(LS , t) q11]2 + 12kB [qN+1 qN+2]

2 (8.2.18)


T =12

Nr=1

Ns=1

(mb

LL

r(x)s(x)dx

)qr qs +

12mB1q

2N+1 +

12mB2q

2N+2 (8.2.19)

V =12

Nr=1

Ns=1

(ERIR

LL

2rx2

2sx2

dx

)qrqs +

12ks

[N

r=1

Ns=1

r(Ls)s(Ls)

]qrqs

kSN

r=1

r(Ls)qrqN+1 +12ksq

2N+1 +

kB2

[q2N+1 2qN+1qN+2 + q2N+2](8.2.20)

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

Mrs = mbLr+s1 [1 (1)r+s1], r, s = 1, 2, , NMN+1,N+1 = mB1MN+2,N+2 = mB2

(8.2.21)

The elements of the stiness matrix are

Krs,racket =EbIb(r1)(r2)(s1)(s2)

L3(r+s5) [1 + (1)r+s] r, s = 3, 4, , NKrs,coupled = ksr(Ls)s(Ls), r, s = 1, 2 , NKr,N+1 = KN+1,r = kSr(Ls), r = 1, 2 , N

KN+1,N+1 = kB + kSKN+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.

1 2 3 4 5 6

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 q12Mode 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.

r =x

Lsin(

rx

2L) (8.2.23)

L

x

z

x1

x2

EbIb m bEpIp m p

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


T =12

L0

mbw2dx +

12

L0

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

V =12

L0

EbIb

(2w

x2

)2dx +

12

L0

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 eective stiness per unit length of the

combined piezoceramic elements.

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

w =N

r=1

r(x)qr(t) (8.2.26)


T =12

Nr=1

Ns=1

[(mb

L0

r(x)s(x)dx

)+(2mp

x2x1

r(x)s(x)dx)]

qr qs (8.2.27)

V =12

Nr=1

Ns=1

[( L0

EbIb2r(x)

x22s(x)

x2dx

)+(2EpIeq

x2x1

2r(x)x2

2s(x)x2

dx

)]qrqs

(8.2.28)

67

Msrs = mb L0

r(x)s(x)dx

Mprs = 2mp x2x1

r(x)s(x)dx

Ksrs = L0

EbIb2r(x)

x22s(x)

x2dx

Kprs = 2EpIeq x2x1

2r(x)x2

2s(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 specication of the cantilever beam are

(Density) L(Length) b(Width) t(Thickness) Eb2750 kg/m3 340 mm 25.4 mm 2 mm 7.001010N/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 specication of the piezo element are given as

(Density) Lp(Length) bp(Width) tp(Thickness) Ep x1 x28200 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

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

natural frequencies and mode shapes by Ritz method.

L

x

z

EbIb m bM

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 gure (??).

x

z

a a

pz

Figure 9.1: Mode shapes of tennis racket and ball model

We assume that the stress component y, z, xy, andyz 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 wx

(9.1.1)

The strain component is

x =u

x= z w

2

x2(9.1.2)

xz =u

z+

w

x= 0 (9.1.3)

69

70

The strain energy stored in the elements is given by

U =12

v

xxdV (9.1.4)

The normal stress is given byx = Ex (9.1.5)

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

U =12

aa

EIy

(w2

x2

)dx (9.1.6)

whereIy =

A

z2dA (9.1.7)

The kinetic energy is given by

T =12

aa

Aw2dx (9.1.8)

The virtual work done by external force for the element is

Wext =12

aa

pzwdx (9.1.9)

9.1.2 Finite Element Modeling

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

w = 1 + 2 + 32 + 43 (9.1.10)

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

w = 1, , 2, 3

1

2

3

4

(9.1.11)

Dierentiating equation (9.1.10) gives

ay = aw

x=

w

= 2 + 23 + 342 (9.1.12)

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

w1

ay1

w2

ay2

=

1 1 1 10 1 2 31 1 1 1

0 1 2 3

1

2

3

4

(9.1.13)

71

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

where{v}Te = w1, y1, w2, y2 (9.1.15)

[m]e =Aa

105

78 22a 27 13a22a 8a2 13a 6a227 13a 78 22a13a 6a2 22a 8a2

(9.1.16)

[k]e =EIy2a3

3 3a 3 3a3a 4a2 3a 2a23 3a 3 3a3a 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

{b}T {b}dV (9.2.1)

where bending stress and strain components are

{b} = {x y xy}T , {b} = {x y xy}T (9.2.2)

The stress-strain relationships take the form

{b} = [Db]{b} (9.2.3)

where

[Db] =E

1 2

1 0

1 0

0 0 12

(9.2.4)

In deriving the energy functions for plate bending, the basic assumptions are that the direct stressin the transverse direction, z, 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 wx

, v(x, y, z) = z wy

(9.2.5)

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

{b} =

x

y

xy

=

ux

vy

uy +

vx

= z

x

y

xy

(9.2.6)

where the curvature vector is given as follows:

{} = {x y xy}T = {2w

x22w

y222w

xy}T (9.2.7)

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

{b} = z{} (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{}T [Db]{}dA (9.2.9)

73

The kinetic energy of the plate is given by

T =12

A

hw2dA (9.2.10)

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

x =w

y, y = w

x(9.2.11)

9.3 Finite Element Modeling

In terms of the master coordinates (9.2.11) becomes

x =1b

w

, y = 1

a

w

(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 = 1 + 2 + 3 + 42 + 5 + 62

+73 + 82 + 92 + 103 + 113 + 123(9.3.2)

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

w = 1, , , 2, , 2, 3, 2, 2 , 3, 3, 3 {} (9.3.3)

where{} = 1, 2, 3, , 12 (9.3.4)

Evaluating w, bx, and ay at = 1, = 1gives

{w}e = [A]e{} (9.3.5)

where{w}Te = w1, bx1, ay1, , w4, bx4, ay4 (9.3.6)

74

and

[A]e =

1 1 1 1 1 1 1 1 1 1 1 10 0 1 0 1 2 0 1 2 3 1 30 1 0 2 1 0 3 2 1 0 3 11 1 1 1 1 1 1 1 1 1 1 10 0 1 0 1 2 0 1 2 3 1 30 1 0 2 1 0 3 2 1 0 3 11 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 11 1 1 1 1 1 1 1 1 1 1 10 0 1 0 1 2 0 1 2 3 1 30 1 0 2 1 0 3 2 1 0 3 1

(9.3.7)

Solving (9.3.5) for {}gives{} = [A]1e {w}e (9.3.8)

where

A1 =18

2 1 1 2 1 1 2 1 1 2 1 13 1 1 3 1 1 3 1 1 3 1 13 1 1 3 1 1 3 1 1 3 1 10 0 1 0 0 1 0 0 1 0 0 14 1 1 4 1 1 4 1 1 4 1 10 1 0 0 1 0 0 1 0 0 1 01 0 1 1 0 1 1 0 1 1 0 10 0 1 0 0 1 0 0 1 0 0 10 1 0 0 1 0 0 1 0 0 1 01 1 0 1 1 0 1 1 0 1 1 01 0 1 1 0 1 1 0 1 1 0 11 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:

1

2

3

4

5

6

7

8

9

10

11

12

=18

2 3 + 3 3 + 4 3 + 3 31 2 + 3 + + 2 3 1 + + + 2 2 3 + 3

2 3 + 3 + 3 4 + 3 3 + 31 2 + 3 + 2 + 3 1 + 2 + 2 3 + 3

2 + 3 3 + 3 + 4 3 3 31 + 2 + 3 + 2 + 3 1 + + + 2 2 3 3

2 + 3 3 3 4 + 3 + 3 + 31 + 2 + 3 + + 2 3 1 + + + 2 + 2 3 3

(9.3.10)

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

w = N1(, ) N2(, ) N3(, ) N4(, ){w}e= N(, ){w}e

(9.3.11)

where{w}Te = w1 x1 y1 w4 x4 y4 (9.3.12)

andN(, ) = 1 b2 a3 10 b11 a12 (9.3.13)

Substituting (9.3.11)into (9.2.10) gives

Te =12{w}Te [m]e{w}e (9.3.14)

where

[m]e =A

hN(, )T N(, )dd

= hab 11

11

N(, )T N(, )dd

(9.3.15)

is the element inertia matrix.

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

Ue =12{w}Te [k]e{w}e (9.3.16)

76

where the stiness element matrix is

[k]e =A

h3

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

and the curvature vector in nite element [B] is

[B] =

2

x2

2

y2

2 2

xy

N(, ) =

1a2

2

2

1b2

2

2

2ab

2

N(, ) (9.3.18)

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

[B] =

1a2

212

ba2

222

1a

232 1a2

2102

ba2

2112

1a

2122

1b2

212

1b222

ab2

232 1b2

2102

1b2112

ab2

2122

2ab

21

2a

22

2b

23 2ab

210

2a

211

2b212

(9.3.19)

where derivatives of shape functions are as follows :

2

2

1

2

3

4

5

6

7

8

9

10

11

12

=14

3 3 0

1 3 + 3 3 + 3

0

1 + 3 + 3 3 3

0

1 3 3 3 + 3

0

1 + 3 3

(9.3.20)

77

2

2

1

2

3

4

5

6

7

8

9

10

11

12

=14

3 3 1 + 3 + 3

0

3 + 3

1 + 3 + 3 0

3 3 1 + 3 + + 3

0

3 + 3 1 + 3 3

0

(9.3.21)

2

1

2

3

4

5

6

7

8

9

10

11

12

=18

4 3 2 3 21 + 2 3 21 2 + 3 24 + 3 2 + 3 21 2 + 3 21 + 2 + 3 24 3 2 3 21 + 2 + 3 21 2 3 24 + 3 2 + 3 21 2 3 21 + 2 3 2

(9.3.22)

9.4 Example

The natural frequency of simple supported thin plate is given by

mn = 2[(m

L1)2 + (

n

L2)2]

DEh

(9.4.1)

where

DE =Eh3

12(1 2) (9.4.2)

is the flexural rigidity of the plate.

78

The speculation of plate is as follows = 1, h = 1, E = 48, L1 = 8, L2 = 8, = 0.5 and 44 elementare used in nite element model.

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

11 0.7123 0.6875

12 = 21 1.7807 1.6977

22 2.8491 2.5488

13 = 31 3.5414 4.0931

23 = 32 4.6298 4.0931

33 6.4104 5.4184

Vibration

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