Vibs1_review_notes.pdf

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Mechanical Vibrations I

Lecture Notes -1- 06/16/06 12:25 PM


m

k c

x(t) f(t)

Author: Allyn W. Phillips, PhD

University of Cincinnati

UC Course Nr.: 20-MECH-480Revision: 16-Jun-2006

Copyright 2001-2006


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Course Outline

Topics

Review Dynamicso Units

o Complex Numberso Free Body Diagramso Absolute vs. Relative Motiono Differential Equationso Matrices & Linear Algebrao Kinematics

Rolling Contact Coordinate Systems

SDOF / MDOF Conceptso Newtons Method

Frequency, Damping, Mode Shape & Scaling

o Lagranges Method Energy Method (SDOF only)

o Eigenvalue / Eigenvector Solutions Superposition

o Free Vibrationso Forced Vibrations

Steady-State Transient

SDOF / MDOF Applicationso SDOF

Estimating Frequency & Damping

Half Power Log Decrement

Transmissibility Isolation

o MDOF Frequency Response Function Concepts

Stiffness / Compliance Impedance / Mobility

Modal Parameter Estimation Impedance / FRF Methods


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Course Material Sequence

Introduction to Mechanical Vibrations [AWP #1] Newtons Method [AWP #2]

o Units [TD 0]

o Matrices & Linear Algebra [TMH A]o Free Body Diagrams [TD 2.2, 5.1]

Single and Multiple Bodyo Absolute vs. Relative Motion

Force Balance

Mass absolute acceleration Damping relative velocity Stiffness relative displacement

o Linear & Torsional Rolling Contact

o Equation of Motion [TD 5.1-3]

Single DOF MCK v. RLC Equivalence [TMH 2.6]

Homogeneous Solution of Equation of Motion [AWP #4]o Differential Equations [TMH D]

Linear, Constant Coefficient

Harmonic Functions (sin, cos) [TD 1] Complex Exponentials

o Free Vibrations [TD 2.1, 2.6] Oscillatory Motion [TD 1]

Harmonic Motion Periodic Motion

o Vibration Terminology [TD 1] Frequency Damping [TD 2.6, 2.8, 2.9]

Viscous Coulomb Structural

Mode Shape [TD 5.1] Modal Scaling

o Eigenvalue / Eigenvector Solutions [TD C] Superposition

o Effects of Changing Mass, Stiffness, Damping

o Modal Parameter Estimation Damping

Log Decrement [TD 2.7, AWP #5] Lagranges Method [TD 7.23, TMH B, AWP #6]

o Energy Terms Kinetic


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Potential Dissipative

o Contrast with Newtons Method Global vs. Local Control Volume

o Equation of Motion Show Equivalence to Newtons Method

o Single DOF reducesEnergy Method [TD 2.3] Conservative System - only

Steady-State Solution of Equation of Motion [AWP #7]o Forced Vibrations [TD 3.1]

Harmonic Excitationo Laplace Transform [TD B]

Converts Differential Equations to Algebraic Equations Transfer Function

o Fourier Transform [TD 1.2, 13.7, DLB] Frequency Response Function [AWP #8]

MCK Variations SDOF Concept Applications [TD 3]o Harmonic Unbalance Force [TD 3.2, AWP #9]o Transmissibility [TD 3.5, AWP #10]o Isolation [TD 3.6, AWP #10]o Dampers

Mass Tuned Absorber

o Modal Parameter Estimation [AWP #12] Frequency Damping

Half Power [TD 3.10, AWP #8] Mode Shape Quadrature

MDOF Systems [AWP #11]o Equations of Motion

Matrix Solutiono Natural Frequencieso Mode Shapes

Transient Solution of Equation of Motion [TD 4, AWP #13]o Forced Vibrations

Impulse Excitation [TD 4.1]

Transient Excitation [TD 4.2]o Convolution [TMH 2.7]

Properties: F[f(t)*g(t)] = F[f(t)] F[g(t)]o Numerical Solutions [AWP #14]

Runge-Kutta

MDOF Detailed Overview (provided for reference) [AWP #15]


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#1 Introduction to Mechanical Vibrations

All mechanical systems vibrate or undergo oscillatory motion. However, in order to

discuss the behavior of vibrating systems, it is first necessary to establish some

nomenclature. There are several terms used to describe vibrations, among which are:oscillatory motion, periodic motion, harmonic motion, period, and frequency.

Oscillatory motionis any pattern of motion where the system under observation moves

back and forth across some equilibrium position, but does not necessarily have anyparticular repeating pattern.

Periodic motionis a specific form of oscillatory motion where the motion pattern repeatsitself with a uniform time interval. This uniform time interval is referred to as theperiod

and has units of seconds per cycle. The reciprocal of the period is referred to as the

frequencyand has units of cycles per second. This unit combination has been given a

special unit symbol and is referred to asHertz(Hz).

Harmonic motionis a specific form of periodic motion where the motion pattern can be

describe by either a sine or cosine. This motion is also sometimes referred to as simpleharmonic motion. Because the sine or cosine technically uses angles in radians, the

frequency term expressed in the units radians per second ( secrad ). This is sometimes

referred to as the circular frequency. The relationship between the frequency in Hz and

the frequency in secrad is simply the relationship,2 rad

cycle .

Natural frequencyis the frequency at which an undamped system will tend to oscillate

due to initial conditions in the absence of any external excitation. Because there is no

damping, the system will oscillate indefinitely.

Damped natural frequencyis frequency that a damped system will tend to oscillate due to

initial conditions in the absence of any external excitation. Because there is damping in

the system, the system response will eventually decay to rest.

Resonanceis the condition of having an external excitation at the natural frequency of the

system. In general, this is undesirable, potentially producing extremely large systemresponse.

Degrees-of-freedomis the number of independent coordinates necessary to describe them

configuration (state) of a system. It can also be expressed as the number of coordinatesused to describe the configuration of a system minus the number of independent

constraint equations between those coordinates. For example, a point in space has three

degrees-of-freedom ( , , )x y z and a rigid body in space has six ( , , , , , )x y zx y z .


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A decibelis a log scale measure of relative power. Specifically, it is ten times the log

base ten of the power to a reference power, 1010logref

PdB

P

=

. If the quantity being

referenced is not power based, then the quantities are first squared, yielding2

10 210log

ref

VdBV

=

or 1020logref

VdBV

=

.

An octaveis a frequency range where the upper frequency is twice the lower frequency.

A decadeis a frequency range where the upper frequency is ten times the lowerfrequency.

Bandwidthandfrequency spanboth terms describe a range of frequencies from low to

high. Bandwidth is particularly associated with the energy of the response of a systemand in this course is used in reference to damping (see section eight.)

Having established a small set of basic nomenclature, it is now possible to provide a briefoverview of the direction of this course. In this course, Mechanical Vibrations I, much of

the effort will be spent upon bringing together the many concepts and techniques

developed in previous courses and applying them to the solution of vibrations problems.

Note that, it is possible to solve simple, but representative, problems utilizing only

information previously learned. To help clarify which courses and knowledge are used,the following introductory example has been developed.


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Introductory Example

For the figure given, assume that the object rollswithout slipping. The objective will be to develop

the equation of motion and determine the naturalfrequency of the system. (A more completeexample would involve a general multi-degree of

freedom system, where the objective would be to

determine; the equations of motion, the systempole, mode shapes, and steady-state response.)

Step 1: Draw the free-body diagram for each independent rigid body in the system. Be

sure to include allexternal forces and moments and their points of application. (Recallthat the internal force resulting from the spring is due to the actual relativemotion of the

ends of the spring. Therefore the coordinatexis measured from thefree lengthof the

spring.) [Mechanics II, Kinematics & Dynamics]

Step 2: Write down all the force and

moment balance equations. [MechanicsII, Kinematics & Dynamics]

) fxF f k x f mx =

) 0yF N mg =

) f cgcg f r f r J + =

Step 3: Identify all relevant coordinateconstraint equations. [Mechanics II, Kinematics & Dynamics]

r= and therefore x r= & x r=

Step 4: Choose the desired independent coordinates and eliminate all the dependent

coordinates by substituting the relevant constraint equations. [Mechanics II, Kinematics

& Dynamics]

) fxF f k x f mx =

) 0yF N mg =

) f cgcgM f r f r J r+ =

Step 5: Evaluate the set of simultaneous differential equations to yield the equations ofmotion. [Mechanics II, Kinematics & Dynamics]

( )f t

( )x t

( )t

k

, cgm J

r

Rolls w/o slipping

=

f

k x

ff

mxcgJ

N

mg


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)yF N mg=

) 2cgfcgJ

f f xr

=

) 2& cgx cgJ

F M f k x f x mxr

=

Step 5a: Identify the actual equation of motion.

22

cgJm x k x f

r

+ + =

Step 6: Solve the simultaneous set of differential equations of motion. The total solution

( ( )Tx t ) involves the sum of two parts: the particular solution ( ( )p t ) and the

homogeneous solution ( ( )hx t ). [Differential Equations]

Step 6a: The particular solution involves knowing the exact form of the forcing function

( ( )f t ). [Differential Equations]

Step 6b: A second order homogeneous constant coefficient differential equation hascomplex exponentials as its solution. [Differential Equations, Numerical Methods]

( ) th t Ce= and therefore ( ) thx t C e

= & 2( ) th t C e=

2

20cg tJm k Ce

r

+ + =

therefore the system poles (eigenvalues) are:

1,2

2

cg

kj

Jm

r

=

+

and the homogeneous solution is:

1 21 2( ) t th t C e C e = +

Step 6c: Formulate the complete solution ( ( ) ( ) ( )T p ht x t x t = + ) and evaluate the initial

conditions to eliminate the constants of integration. [Differential Equations]


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Note: if this example had involved more than one independent dynamic coordinate, the

solution would have additionally required the solution of an eigenvalue/eigenvectorproblem. [Differential Equations, Numerical Methods]


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Lecture Notes -10- 06/16/06 12:25 PM

#2 - Newtons Method ( -or- d'Alemberts Method )

Force Balance

GF M q= ( -or- 0GF M q = )

Moment Balance

/P P G P PJ r M q= + ( -or- / 0P P G P PM J r M q = )

Typical problems with Newtons ( -or- d'Alemberts ) formulation:

Be sure to establish the number of degrees of freedom first and formulate allterms in only those variables. Clearly identify which degrees of freedom are

relative coordinates versus absolute coordinates. Also, clearly identify what willbe the positive direction of motion for each coordinate. Watch out for

rotational/translational problems. State any constraint relationships that relate

independent and dependent coordinates.

Evaluate the static balance for the problem in order to determine whether theorientation of the system in the gravitational field will effect the equations of

motion (Are the weights of the objects balanced by an initial static deflection in

the springs?). When in doubt, perform a static force balance to determine theappropriate constraint equation.

For displacement, velocity and acceleration terms, be sure to develop absoluteorrelativedisplacement, velocity and accelerations of appropriate points as required.

Watch out for 2-D and 3-D vector motions. Be sure to draw the appropriate free body diagrams for each mass (or combination

of masses) in the system.

o Whenever the system is separated in order to draw a free body diagram,replace the separation with the appropriate internal forces/moments (equal

and opposite forces/moments on each side of the separation).

o Do not move forces/moments arbitrarily from one mass to another. Theinternal forces account for the effects of one mass on another.

Develop one equation of motion for each degree of freedom of the system usingNewton's ( -or- d'Alemberts ) method. Be sure to watch for moving reference

frame issues. Also, check that the units are the same for each term in an equation

(Forces + Moments: NOT!) If necessary, once the exact equations of motion have been determined, linearize

the equations of motion by neglecting nonlinear terms in the equations of motion.Note that the linear equations of motion may not adequately describe the original

equations of motion if some of the terms that have been neglected are not

insignificant.

P

G


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Lecture Notes -11- 06/16/06 12:25 PM

Newton Example A

In order to solve Newtons Equation,

iF mx= ,

for the given single degree-of-freedomsystem, it is first necessary to draw the

complete free-body diagram. This involves

identifying allinternal and external forcesacting upon the degree-of-freedom.

(Recall that the internal force resulting

from the spring is due to the actual relativemotion of the ends of the spring. Thereforethe coordinatexis measured from thefree lengthof the spring.)

All the internaland externalforcesare combined to yield the actual

equation of motion.

f kx mg cx mx =

collecting all the coordinate based

terms to one side of the equation

and all the externally applied forcesto the other yields

mx cx kx f mg + + =

Since the mgterm ends up on the right hand side with the external force, the coordinate

x can be redefined to be from thestaticequilibrium position. As a result, the mgterm

can be eliminated as shown.

Evaluating the static equilibrium equation (i.e. 0x= , 0x= & 0f = .) yields,

kx mg =

Therefore the static deflection ( ) can be evaluated as

mg

k =

The coordinate system reference can be shifted to the static equilibrium position bydefining

m

k c

x(t) f(t)

k x

m

cx

= m

mx

mg

f


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Lecture Notes -12- 06/16/06 12:25 PM

sx x=

Since the static deflection ( ) is constant, obviously the velocity and acceleration isunchanged by the coordinate shift.

sx x= and s x=

Substituting into the original solution,

( )s s smx cx k x f mg + + + =

But recall that k mg = , therefore

s s smx cx kx f + + =

Note: Often the solution solved about the static equilibrium position will eliminate the

mg term when there are springs to carry the load. When this occurs, the mg term will

end up on the right hand side of the equation with the applied external forces.


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Lecture Notes -13- 06/16/06 12:25 PM

Newton Example B

Again, in order to solve Newtons Equation,

0 cgJ L mL = + ,

for the given single degree-of-freedomsystem, it is first necessary to draw the

complete free-body diagram. This involves

identifying allinternal and external forcesacting upon the degree-of-freedom. In this

case, there is no specific external moment applied at point o. Also, recall the parallel

axis theorem, 20 cgJ J mL= + . (See above for the general expression.)

All the internaland externalmoments arecombined to form the actual equation of motion.

Notice that neither the support force T nor the

centripetal acceleration force 2mL contribute

to the moment about point o. (However Twould contribute to the moments about the cg.)

( )2sin cgmgL J mL = +

collecting all the coordinate based terms to one side of the equation and all the externally

applied moments to the other yields

( )2 sin 0cgJ mL mgL + + =

This time, since the mgterm ends up on the left hand side (multiplied by the coordinate),

it is not possible to eliminate the mgterm from the solution. So the exactequation of

motion remains,


However, because the sinmgL term is non-linear in the desired coordinate, it ispossible to reduce the solution to a linearapproximation by using the first order

McClaurin Series expansion, sin for 1 , as follows.

( )2 0cgJ mL mgL + + =

,cgm J

L

=

cgJ

mg

T

mL2mL


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Lecture Notes -14- 06/16/06 12:25 PM

#3 - The Eigenvalue Problem

In many areas of engineering, a mathematical concept referred to as the eigenvalueproblem arises. While the general application of eigenvalue methods has been used to

solve a wide variety of mathematical, scientific and engineering problems, in this course,we will focus upon its application to the area of mechanical vibrations. Because of this,

we will not examine the eigenvalue/eigenvector problem in detail, but only its most basiccharacteristics.

Mathematically, the eigenvalue problem can be expressed as:

[ ]{ } { }A x x=

In other words, given a matrix [ ]A , the object is to find a vector { } that when multiplied

times the matrix [ ]A does not change direction, but only magnitude (and possibly sense.)If this vector can be found, it is referred to as an eigenvector of [ ]A , and the scalar ,which describes the change in magnitude and sense, is referred to as an eigenvalue of

[ ]A . (Note that in general, there will be as many eigenvalue/eigenvector pairs as the sizeof the matrix. Also, it is only defined for square matrices.)

The solution is essentially a two-step process. First identify the eigenvalues of the matrix,

and then solve for their associated eigenvectors.

Solving for the eigenvalues involves manipulating the equation form as follows.

[ ]{ } { } { }0A x x =

[ ]{ } [ ]{ } { }0A x I x =

[ ] [ ] { } { }0A I x =

At this point, we recognize the form as a null space solution problem. There are three

ways in which the equation can be satisfied, two of them, [ ] [ ] [ ]0A I = and

{ } { }0x = , are trivial solutions and are rejected. The third is that the matrix [ ] [ ]A I

is singular (or rank deficient) for certain values of . These values of are theeigenvalues of the matrix [ ]A . To solve for the s, recognize that a singular (or rankdeficient) matrix has a determinant of zero.

[ ] [ ] 0A I =


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Lecture Notes -15- 06/16/06 12:25 PM

Expanding this determinant results in a polynomial, known as the characteristic

polynomial, whose roots are the eigenvalues .

After identifying the eigenvalues, go back and solve the linear equations,

[ ] [ ] { } { }0A I x = , for each eigenvalue . However there is a problem, since the

coefficient matrix is singular (or rank deficient) when evaluated at the eigenvalues, wecannot simply solve the equation set. In general, the matrix will be rank deficient by one,

indicating that one variable is free, so the most straight-forward solution approach is to

simply set one of the vector { } elements equal to one and solve as normal. (We will not

get into the issues involved in the repeated root problem. In other words, where two or

more of the eigenvalues are equal.)


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Lecture Notes -16- 06/16/06 12:25 PM

Eigenvalue/Eigenvector Example

Given the following matrix, find its associated eigenvalues and eigenvectors.

[ ] 4 12 5

A =

First identify the eigenvalues,

[ ] [ ]4 1 1 0 4 1

02 5 0 1 2 5

A I

= = =

4 1(4 )(5 ) ( 1)( 2) 0

2 5

= =

2 9 18 0 + =

Which has the roots, 1 3 = and 2 6 = .

Next identify the eigenvector associated with each eigenvalue.

[ ] [ ] { } { }0A I x =

1 1

2 2

4 1 1 0 4 1 0

2 5 0 1 2 5 0

x x

x x

= =

For the first eigenvalue, 1 3 = ,

11

21

4 1 0

2 5 0

x

x

=

1 1

2 2

4 3 1 1 1 0

2 5 3 2 2 0

x x

x x

= =

Notice the rank deficient nature of the resulting coefficient matrix.

Set one element of the vector { }x equal to one (unity) to solve. For this example, we will

use the first element. (Note that we could actually set it equal to any arbitrary valueexcept zero and still solve. Occasionally an eigenvector will contain a zero, but we cannot

arbitrarily pick that value. Finally, if after picking a value, the resulting equations cannot

be solved, simply try again picking a different element.)

Letting 1 1x = ,


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Lecture Notes -17- 06/16/06 12:25 PM

2

11 1 0

2 2 0x

=

We can solve either equation for2

. Lets pick the first for simplicity.

[ ] 22

11 1 1 0x

x

= =

Which yields, 2 1x = . Therefore, the first eigenvalue/eigenvector pair is:

1 3 = , { }11

1x

=

Repeating the process for the second eigenvalue, 2 6 = , yields,

12

22

4 1 0

2 5 0

x

x

=

1 1

2 2

4 6 1 2 1 0

2 5 6 2 1 0

x x

x x

= =

Again letting 1 1x = , and solving for 2x , yields,

2

12 1 0

2 1 0x =

[ ] 22

12 1 2 0x

x

= =

Which yields, 2 2x = . Therefore, the second eigenvalue/eigenvector pair is:

2 6 = , { }21

2x

=


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Lecture Notes -18- 06/16/06 12:25 PM

Final Notes

Looking at the form of the eigenvalue problem one more time, it can be shown that it canactually be used to solve certain first order differential equations directly, those of the

form 'y Ay= . Assuming that the vector { } is of the form { } { }( ) tx t X e= , then theeigenvalue problem can be written as.

[ ]{ } { }A x x=

[ ]{ } { }( ) ( )A x t x t=

[ ]{ } { }t tA X e X e =

[ ]{ } { }A X X=

Which is really no different than the example problem solved above.

There are a few issues, which will be covered later, applying this method to the second

order differential equations that arise in structural dynamics, specifically the need to use

state-space expansion to convert the set second order differential equations to a larger

system of first order differential equations.

Finally, when using MATLAB to solve eigenvalue problems, while the eigenvalues

should be the same, the eigenvector will in general be scaled differently. MATLAB usesunity vector length for its scaling.


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Lecture Notes -19- 06/16/06 12:25 PM

#4 - Homogeneous Solution of the Equation of Motion

For the single degree offreedom (SDOF) system

shown, the equation of motionis:

( ) ( ) ( ) ( )mx t cx t k x t f t + + =

This differential equation canbe solved by any of several

methods, including: Laplace Transforms, Fourier Transforms, or by an assumed solution.

In this case, assuming a solution appropriate for a second-order, constant coefficientdifferential equation.

The general multi-degree of freedom problem can be formulated in matrix form as:

[ ]{ } [ ]{ } [ ]{ } { }( ) ( ) ( ) ( )x t C x t K x t f t+ + =

Using a solution appropriate for a linear, constant matrix coefficient, differential

equation, (namely complex exponentials), we have:

{ } { }( ) stx t X e= & { } { }( ) stf t F e= NOTE: thatX,Fandsare complex scalars.

By focusing on the homogeneous solution (i.e. 0f = ), we first evaluate the derivatives of

the assumed solution,{ } { }( ) stt X e= :

{ } { } { }( ) ( ) stddt x t x t X se= =

{ } { } { }2

2

2( ) ( ) stddt

x t x t X s e= =

By substituting the assumed velocity and acceleration into the original matrix expression

and collecting terms, we have: (NOTE: ste is non zero for all time and so may be

eliminated.)

[ ] [ ] [ ] { } { }2

0M s C s K X + + =

We should recognize this as a form of eigenvalue problem (or a null space solution

involvings.) Also, recognize that the solution { } { }0X = is the trivial solution and does

not contribute any useful information. Therefore, since the system matrix becomes rank

deficient for certain values ofs, the problem reduces to a polynomial root solver. (Thedeterminant of the system matrix is zero.)

m

k

c

x(t)f(t)


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Lecture Notes -20- 06/16/06 12:25 PM

[ ] [ ] [ ]2 0M s C s K+ + =

This determinant results in a polynomial whose roots are the poles of the system. This isknown as the characteristic polynomial. Often the determinant notation is dropped (but

still written equal to thescalarzero implying a determinant) and the matrix characteristicpolynomial is expressed as,

[ ] [ ] [ ]2 0M s C s K+ + =

Returning to the scalar (1 DOF) problem as initially presented, we have:

2 0ms cs k + + =

Since the determinant of a scalar is simply the scalar itself, the characteristic polynomial

is simply:

2 0ms cs k + + =

Since this is a quadratic equation, there are two roots, namely:

2

1,2

4

2

c c mk

m

=

This results in the homogeneous (transient) solutions form of:

1 2

1 2( ) t t

x t Ae A e = +

The characteristic behavior of this system solution is determined by the relationship of m,

c, & k.

2 4c mk> Two distinct real roots.2 4c mk= Two equal real roots.

2 4c mk< One pair of complexconjugate roots.

For the condition of 2 4c mk> , no oscillation occurs. This is called over-damped. Sinceno oscillation occurs, this is not particularly interesting from a vibration point of view

and we will not investigate it any further.

For the condition of 2 4c mk= , this is the boundary between oscillatory and non-oscillatory behavior. This is called critically-damped. We will return to this later.


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Lecture Notes -21- 06/16/06 12:25 PM

The interesting oscillatory behavior occurs when 2 4c mk< . This is called under-damped.When a system is under-damped, the system will respond to initial conditions with

harmonic oscillatory behavior.

First, we recognize the roots of the characteristic equation as the system poles ( r ). For

the given SDOF system, the roots have the following form:

2

1,22 2

r

c k cj

m m m

= =

Examining the two parts (real and imaginary) of the poles, we can write:

r r rj =

Where,

2rc

m

= is the damping for mode r(for single degree-of-freedom systems, it isoften written as d ) and

2

2r

k c

m m

=

is the dampednatural frequency for mode r(for single

degree-of-freedom system, it is often written as d )

Since 1 and 2 are complex conjugates, the homogeneous solution for the SDOF

problem can be written as:

**( ) t tx t Xe X e = +

If we let 0c , then rk

m . This is called the undampednatural frequency, written

as r (for single degree-of-freedom system, it is often written as n .) This is the

frequency at which the system will oscillate if there is no damping in the system.

Returning to the critically damped condition, where 2 4c mk= , we can now define a fewmore vibration parameters.

Defining the critical damping as:

2cc mk=

Allows us to define the damping ratioas:

c

c

c =

We can now express the damping and damped natural frequency as functions of the

undamped natural frequency and damping ratio, as follows:


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Lecture Notes -22- 06/16/06 12:25 PM

2r r r

c

m = =

2

21

2

r r r

k c

m m

= =

The single degree of freedom equation can now be written as:

2 ( )( ) 2 ( ) ( )r r rf t

x t x t x tm

+ + =

While each of these parameters has been developed from a single degree of freedom (1

DOF) point of view, the concepts are equally applicable to the multi-degree of freedom

case, as well. For this, we need to use a nomenclature that reflects the fact that there is

more than one system pole (with its conjugate). We do this by using the subscript r(indicating resonance) with each of the modal parameters, as follows.

r r rj = + is the system pole for mode r, where r is the damping and r is the

damped natural frequency for mode r. From this, we define 2 2r r r = + as the

undampednatural frequency for mode r. Then, the damping ratio is simply defined as,

rr

r

=

. The general homogeneous solution then takes on the following form.

{ } { } { }*

*( ) trtr

r rr

t X e X e= +

The outgrowth of this approach is that it is no longer necessary to know the actual mass,stiffness and damping of the system in order to identify its modal characteristics. This is asignificant advantage when applied to experimental modal analysis because, as will be

shown later, the system poles ( r s) can be obtained directly from measured Frequency

Response Functions.


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Lecture Notes -23- 06/16/06 12:25 PM

Homogeneous Example

Given a single degree-of-freedom mass/spring/damper system (MCK) with 5kgmass,

20N s m damping, and 1000Nm stiffness: identify the pole of the system ( r ), the

undamped natural frequency ( r ), and the damping ratio ( r ).

To calculate the system pole, evaluate the characteristic equation,

2 0ms cs k + + = 25 20 1000 0s s+ + =

which has the following roots. Note that the roots of the characteristic equation are the

system poles.

220 1000 20

2 5 5 2 5

N s N N sm m m

r jkg kg kg

=

( )2

2

22 200 2rad rad rad s sr sj =

2 14 radsr j =

The pole gives the damping ( r ) and the damped natural frequency ( r ) directly. Recall,

that to express the answer in units of Hertz, it is necessary to divide by 2 .

2rad

sr = -or-

1

0.318Hz Hz =

14 radsr = -or-7

2.228Hz Hz

=

For this problem, the undamped natural frequency ( r ) may be calculated from either

the magnitude of the system pole, or by evaluating the characteristic equation for 0c= .Both approaches yield the same solution.

2

22 14 200 14.14rad rad rad s sr r sj = = = = -or- 2.251Hz

221000 200 14.14

5

N

m rad rad sr skm kg = = = =

-or- 2.251Hz

Again, for this problem, the damping ratio may be calculated by either of twoapproaches, either directly from the system pole or from the systems mechanical

properties (MCK).


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Lecture Notes -24- 06/16/06 12:25 PM

20.1414

14.14

radsr

rrad

sr

= = =

-or- 14.14%

200.1414

2 2 5 1000

N sm

rN

c m

c c

c mk kg

= = = =

-or- 14.14%

Recognize one important point in the solution method. The direct (MCK) solutions are

only valid for a single degree-of-freedom (SDOF) system while the solution methods

based upon the system pole are valid for a system with any number of degrees-of-freedom (MDOF).


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Lecture Notes -25- 06/16/06 12:25 PM

#5 Estimating Modal Parameters Time Domain

There are many techniques, both simple and advanced, for estimating modal parameters

(frequency, damping, mode shape, and scaling) from measured experimental data. This

section will focus upon the single degree-of-freedom time domain technique known aslog decrement.

Log Decrement

The log decrement method is based entirely upon the characteristics of a single degree-

of-freedom free decay response.

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15-4

-3

-2

-1

0

1

2

3

4

Time (sec)

Amplitude

Recalling the form of solution, ( ) rtx t Xe= , it is possible to estimate the system pole

( r r rj = + ) directly from the observed response.

Estimating the damped natural frequency ( r ) is easy. First estimate the period of one

oscillation ( ), the reciprocal of the period, 1

, is the damped natural frequency ( r ) in

Hz.

2 radsr

= -or- 1 Hz

Estimating the damping ( r ) requires a little more effort. The development proceeds as

follows.

Begin by evaluating the position, ( )t , for two time instants one cycle apart.


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Lecture Notes -26- 06/16/06 12:25 PM

1 1 1

1( ) r r rt t j t x t Xe Xe e

= = 2 2 2

2( ) r r rt t j t x t Xe Xe e = =

Next evaluate the log decrement ( ) as:

1 1

2 2

1

2

( )ln ln

( )

r r

r r

t j t

t j t

x t Xe e

x t Xe e

= =

However, since the points are defined one cycle apart, 2 1t t = + . Substituting into the

above expression yields,

( ) ( )

1 1

1 1ln

r r

r r

t j t

t j t

Xe e

Xe e

+ +

=

Next, recall that cos sinje j = + . Hence,

( ) ( ) ( )11 11 12 2 2rr rr r rj tj t j tj t j t je e e e e e

++ += = = =

So the expression for the log decrement can be reduced to

( )( )( ) ( )

1

11

1

ln ln lnr

rr r

r

ttt

rt

ee e e

e

+

+

= = = =

Recalling that r r r = ,21r r r = and 2r

= . Substituting into the above

expression yields,

2 2

2 22

1 1

r r rr r r r r

rr r r

= = = = =

Therefore the log decrement equals,

2

2

1

r

r

=

If the damping ratio is small (eg.

0.1< -or- 10%), then the above

expression can be simplified to:

2 r

(Error less than 1% when 0.1< )

Multiple Cycle Solution

If time points multiple (n) periods apart

are used, the expression for the log

decrement is modified as follows:0

2

1 ( ) 2ln

( ) 1

r

n r

x t

n x t

= =

where: nis the number of cycles.

Which simplifies to:

01 ( )ln 2( )

r

n

x t

n x t

=


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Lecture Notes -27- 06/16/06 12:25 PM

Log Decrement Example

Using the figure given above, estimate both the damped natural frequency and the

damping ratio.

First, select two convenient time points to estimate the period ( ). Between 0t= seconds and 9t= seconds, there are 11 cycles.

90.8182

11sec

cycle

sec

cycles= =

Therefore the damped natural frequency is

111.222

9r

cyclesHz

sec = = -or- 7.679 rads

Second, select two convenient time points to estimate the log decrement ( ). Between

1.8t seconds and 6.7t seconds, there are 6 cycles with ( 1.8) 3x t and

( 6.7) 2x t .

0 01 ( ) 1 ( 1.8) 1 3ln ln ln 0.0676( ) 6 ( 6.7) 6 2n n

x t x t

n x t x t

= = = =

Since the log decrement ( ) is so small, first try the simplified approximation for

damping ratio ( ).

0.0676 0.01082 2

= = -or- 1.08%

(Error check: 0.1 < so the approximation is adequate.)

Just for completeness, since 0.1< , 21 1 , therefore

27.679

1

r radsr r

r

= =

-or- 1.222Hz

0.0108 7.679 0.0829rad rad s sr r r = = = -or- 0.0132Hz


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Lecture Notes -28- 06/16/06 12:25 PM

#6 - Lagranges Method

Lagranges Equation

i

i i i i

d T T U DF

dt q q q q

+ + =

Where:

T= Kinetic Energy U= Potential Energy D = Dissipative Energy iF= Externally applied force/moment

iq = Generalized coordinate

Generalized Coordinates

, , , , ,x y zq x y z =

Kinetic Energy

2 2

1 1

1 1

2 2

jm NN

i i

i i

T M x J

= =

= +

Potential Energy

( )21 1

1

2

k mN N

i i

i i

U K x M g x= =

= +

Dissipative Energy

2

1

1

2

cN

i

i

D C x=

=


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Lecture Notes -29- 06/16/06 12:25 PM

Lagranges Method

Typical problems with Lagrange formulations:

Be sure to establish number of degrees of freedom first and formulate all energy

terms in only those variables. Clearly identify which degrees of freedom arerelative coordinates versus absolute coordinates. Watch out for

rotational/translational problems.

For kinetic energy terms, be sure to formulate absolutevelocities before takingderivatives. Watch out for 2-D and 3-D vector motions.

For potential energy terms, be sure that the actual deflection, described by relativeand/or absolute coordinates, in spring elements is described. Watch out for 2-D

and 3-D vector motions.

There should be only one total kinetic energy equation, one total potential energyand one total dissipative energy equation for the system. The kinetic, potential anddissipative energy equations should involve only the N generalized coordinates

and the constants (mass, damping, stiffness) of the system. Apply the Lagrange Equation once for each generalized coordinate. For N degrees

of freedom, N generalized coordinates will yield N equations of motion.

If necessary, linearize the equations of motion by neglecting nonlinear terms inthe equations of motion. Note that the linear equations of motion may not

adequately describe the original equations of motion.

NOTE: For the single degree-of-freedom (only one coordinate), conservative (no

damping), un-forced problem (no external forces), there is a simplified method called the

Energy Method.

( ) 0d

T Udt

+ =


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Lecture Notes -30- 06/16/06 12:25 PM

Lagrange Example A

In order to solve the Lagrange Equation,

i

i i i i

d T T U D Fdt q q q q

+ + =

,

for the given single degree-of-freedom

system, it is first necessary to formulatethe energy equations: kinetic, potential and

dissipative.

First, formulating the kinetic energy yields, 212T mx=

Second, formulating the potential energy yields, 212U kx mgx= +

Finally, formulating the dissipative energy yields, 212

D cx=

Next, it is necessary to evaluate each of the Lagrange terms,

( ) ( )212d T d d

mx mx mxdt x dt x dt

= = =

( )212 0T

mxx x

= =

( )212U

kx mgx kx mg x x

= + = +

( )212D

cx cxx x

= =

Which are combined to form the actual equation of motion.

0mx kx mg cx f + + + =

mx cx kx f mg + + =

Since the mgterm ends up on the right hand side with the external force, the coordinatex can be redefined to be from thestaticequilibrium position. As a result, the mgterm is

eliminated as shown.

mx cx kx f + + =

m

k c

x(t) f(t)


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Lecture Notes -31- 06/16/06 12:25 PM

Lagrange Example B

Again, in order to solve the LagrangeEquation,

i

i i i i

d T T U DF

dt q q q q

+ + =

,

for the given single degree-of-freedomsystem, it is necessary to formulate the energy

equations: kinetic, potential and dissipative.

First, formulating the kinetic energy yields, ( )2

21 12 2cgT J m L = +

Second, formulating the potential energy yields, ( )1 cosU mgL =

Finally, formulating the dissipative energy yields, 0D=

Next, it is necessary to evaluate each of the Lagrange terms,

( )( ) ( )22 2 21 12 2cg cg cg d T d d J m L J mL J mLdt dt dt

= + = + = +

( )( )221 12 2 0cgT J m L

= + =

( )( )1 cos sinU

mgL mgL

= =

( )0 0D

= =

Which are combined to form the actual equation of motion.

2 0 sin 0 0cgJ mL mgL + + + =

This time, since the mgterm ends up on the left hand side (multiplied by the coordinate),

it is not possible to eliminate the mgterm from the solution. So the exactequation of

motion remains,


, cgm JL


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Lecture Notes -32- 06/16/06 12:25 PM

However, because the sinmgL term is non-linear in the desired coordinate, it is

possible to reduce the solution to a linearapproximation by using sin as follows.

( )2 0cgJ mL mgL + + =


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Lecture Notes -33- 06/16/06 12:25 PM

Energy Example A

In order to solve the Energy Equation,

( ) 0d T Udt

+ = ,

for the given single degree-of-freedom system, it

is first necessary to formulate the energy

equations: kinetic and potential.

Formulating the kinetic energy yields, 212

T mx= ,

and formulating the potential energy yields, 212U kx mgx= + .

Evaluating the Energy equation yields,

( ) ( )2 21 12 2 0d d

T U mx kx mgx mxx kxx mgxdt dt

+ = + + = + + =

Observe the oddform of the equation. There is an extra velocity term, x , that must be

canceled out to yield the actual equation of motion.

0mx kx mg + + =

mx kx mg + =

Since the mgterm ends up on the right hand side with the external force, the coordinate

x can be redefined to be from thestaticequilibrium position. As a result, the mgterm is

eliminated as shown.

0mx kx+ =

NOTE: Comparing the solution generated by Lagrange Method to that of the Energy

Method*shows that the Energy Method is not simply a single degree-of-freedom

Lagrange Solution, but rather a related, by not equivalent technique. Recognize also thatthe application of the Energy Method is extremelylimited, i.e. single degree-of-freedom,

conservative and un-forced. The Lagrange Method has no such limitations and is in fact acompletely general solution method, comparable in usefulness to Newtons Method.

*Note that the Energy Method is presented for completeness. You may encounter it in your studies of other

reference material, but because it is not a general technique, it will not be used in this course.

m

k

x(t)


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Lecture Notes -34- 06/16/06 12:25 PM

#7 - Steady-State Solution of Equation of Motion

The general solution to the equation of motion involves both the homogeneous and the

particular solution. While the form of the homogeneous solution is dictated by the nature

of the governing second order constant matrix coefficient differential equation, the formof the particular solution is controlled by the form of the forcing function (excitation).

Mathematically, there are infinite possibilities, but for practical applications, the form can

be restricted to harmonic functions. By limiting the form of the excitation to harmonicfunctions, the concept of steady-state solution can be defined. For vibration purposes, the

steady-state solution is the resulting response after all the initial condition transients have

decayed. When solving the response for the steady-state, the solution is most easily

formulated in either the Laplace Domain (s) or the Frequency Domain (). (NOTE: it

will be shown later that the Frequency Domain solution can be developed from the

Laplace Domain by letting s j= .)

Working with a representative single

degree-of freedom system (SDOF), theequation of motion is:

( ) ( ) ( ) ( )mx t cx t k x t f t + + =

For a harmonic excitation, the forcing

function can be represented as:

( ) stf t Fe=

For the assumed solution, the response will

have the compatible form,

( ) stx t Xe= , ( ) stx t sXe= and 2( ) stt s Xe=

where bothF&Xare complex scalars.

Substituting, the above solution forms into the differential equation of motion produces a

solution that must be valid for every value ofs. This results in the following LaplaceDomain solution.

2 st st st st ms Xe csXe kXe Fe+ + =

By collecting common terms, the expression reduces to:

( )2 st st ms cs k Xe Fe+ + =

m

k c

x(t) f(t)


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Lecture Notes -35- 06/16/06 12:25 PM

At this point the response/excitation ratio

can be calculated. This ratio is called theTransfer Function. It is normally written

as ( )H s .

21( ) ( )XH s s

F ms cs k=

+ +

While the Transfer Function is a

convenient mathematical model for

defining the input-output relationship,well suited to controls applications, it is

less appealing for vibrations purposes

primarily because it a continuousmathematical expression. However,

because the complex surface defined by

the ( )H s expression is analytic, any slicethrough the surface contains all the

information necessary to reconstruct theentire surface. This recognition leads to

the use of the Fourier Transform, that while its mathematical background is different, it is

effectively the Laplace Surface evaluated at s j= . (Another argument for the

equivalence of the Laplace Transform and the Fourier Transform lies in the fact that both

transforms start with the same Time Domain function, i.e. 1 1( ) ( ( )) ( ( ))g t L G s F G = = .)

Evaluating the Transfer Function at s j= yields another input-output form known as

the Frequency Response Function (FRF).

2

1( ) ( )

XH

F m j c k

=

+ +

By contrast with the Transfer Function,the Frequency Response Function is well

suited for vibration applications,

particularly experimental applications.While there is no discrete equivalent to the

Laplace Transform, there is a discrete

Fourier Transform that yields essentiallythe same information as the continuousintegral transform. This is particularly

important when working with discrete,

sampled time data.

By plotting both the Transfer Function and the Fourier Transform (over a limited range),

the equivalence of both transforms can be shown.

Multi-Degree of FreedomTransfer Function

The same basic development is

applicable to the multi-degree of freedom

system, as well.

Starting from the equation

[ ]{ } [ ]{ } [ ]{ } { }( ) ( ) ( ) ( )x t C x t K x t f t+ + =

and combining with

{ } { }( ) stt X e= & { } { }( ) stf t F e=

yields the matrix equation

[ ] [ ] [ ] { } { }2 st st s C s K X e F e + + =

By recognizing the basic definition of the

Transfer Function, the expression

becomes,

[ ] [ ] [ ]1

2( )H s M s C s K

= + +

Multi-Degree of FreedomFrequency Response Function

Just as the Transfer Function has a multi-degree of freedom form, so does the

Frequency Response Function.

By evaluating the multi-degree of

freedom (matrix) Transfer Function at

s j= , the multi-degree of freedom

Frequency Response Function results.

[ ] [ ] [ ]1

2( )H M j C K

= + +


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Lecture Notes -36- 06/16/06 12:25 PM

One interesting aspect of the difference, however, is that while the Transfer Functionrequires the plotting of two surfaces, the Frequency Response Function requires only two

curves, clearly a much more convenient representation.

Note that in addition to displacement per unit force input-output relationships, other

steady-state input-output relationships can be developed. Some of these alternative

relationships will be explored later as steady-state applications.

-1

-0.5

0

0.5

1

-10

-5

0

5

10

0

0.2

0.4

0.6

0.8

1

Sigma [rad/sec]Omega [rad/sec]

Magnitude

-1

-0.5

0

0.5

1

-10

-5

0

5

10

-4

-2

0

2

4


Phase[rad]

-10 -8 -6 -4 -2 0 2 4 6 8 100

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

Omega [rad/sec]

M

agnitude

-10 -8 -6 -4 -2 0 2 4 6 8 10-4

-3

-2

-1

0

1

2

3

4

Omega [rad/sec]

Phase[rad]


37/86


Lecture Notes -37- 06/16/06 12:25 PM

#8 - The Frequency Response Function (FRF)

In general, for most vibrations problems, the input/output relationship known as the

Frequency Response Function (FRF) will be used. This function describes the

input/output relationship on a frequency-by-frequency basis.

Although not generally referred to as a Frequency Response Function, the effect of agraphic equalizer on the sound from a stereo system can be described by thefrequency

response function. In this case, the input is the raw audio and the output is the modifiedaudio. By adjusting the amplification or attenuation (and relative phasing) of the various

bands, the equalizer can give the resulting audio a variety of sound characteristics, from

flat responseto enhancing male or female voices to concert hall sound effects.

Recall that the Frequency Response Function can be expressed by evaluating the Transfer

Function at s j= .

2

1( ) ( )

XH

F m j c k

=

+ +

By contrast with the Transfer Function,

the Frequency Response Function is well

suited for vibration applications,particularly experimental applications.

While there is no discrete equivalent to the

Laplace Transform, there is a discrete

Fourier Transform that yields essentiallythe same information as the continuous

integral transform. This is particularlyimportant when working with discrete,

sampled time data.

Input-Output Model

For the frequency response function, the following pictorial model is helpful. In fact,

when used for determining the steady state

response of a system, its application isextremely simple. It is merely a

multiplication in the frequency domain.

( ) ( ) ( )X H F =

Multi-Degree of FreedomFrequency Response Function

Just as the Transfer Function has a multi-degree of freedom form, so does the

Frequency Response Function.

By evaluating the multi-degree of

freedom (matrix) Transfer Function at

s j= , the multi-degree of freedom

Frequency Response Function results.

[ ] [ ] [ ]1

2( )H M j C K

= + +

( )H ( )F ( )X


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Lecture Notes -38- 06/16/06 12:25 PM

Complex Exponential vs. Sine/Cosine Relationship

The relationship between the complex exponential form and the sine/cosine form of the

can be developed as follows:

Starting with the complex exponential form for the forcing function,

*( ) j t j t f t Fe F e = +

and recalling the Euler identity,

cos sinie i = +

the relationship is easily developed. (Remember that Fis complex.) Expanding the

expression for ( )f t ,

( ) ( )(cos sin ) ( )(cos sin )R I R If t F jF t j t F jF t j t = + + +

Collecting real and imaginary terms yields,

( ) ( cos sin ) ( cos sin ) ( cos sin ) ( cos sin )R I I R R I I Rf t F t F t j F t F t F t F t j F t F t = + + + +

Observing that the imaginary terms cancel and that the real terms are equivalent yields,

( ) 2 cos 2 sinR If t F t F t =

Comparing with a traditional differential equations solution of ( ) cos sint A t B t = + shows that 2 RA F= and 2 IB F= . Despite the use of complex exponentials in the

development, clearly the resulting time domain waveform is real.

FRF Characteristics

This figure presents a typical

single degree-of-freedomfrequency response function.

The FRF is present two sided

(both positive and negativefrequency) to clearly show that

the FRF for positive and

negative frequencies arecomplex conjugates.

*( ) ( )H H =

-20 -15 -10 -5 0 5 10 15 2010

-5

10-4

10-3

10-2

Frequency [Hz]

Magnitude[m/

N]

-90

0

90

180

m = 5 kg ; c = 20 Ns/m ; k = 1000 N/m

Phase[deg]


39/86


Lecture Notes -39- 06/16/06 12:25 PM

Because of this complex conjugate relationship, usually only the positive frequency part

is plotted.

Predicted Response*

It is important to remember that although the FRF is defined for allfrequencies, the

steady-state response (output) for a linear system will always be at thesamefrequency as

the excitation (input). This prediction comes by manipulating the definition of thefrequency response function, as shown.

( ) ( ) ( )X H F =

For a single degree-of-freedom system, the predicted steady-state response has thefollowing form.

2( )( ) FX

m j c k

=

+ +

Peak Response

Another important aspect of the frequency response function involves the frequency of

maximum response. It will be designated by p .

Starting with the single degree-of-freedom frequency response function,

2

1( )H

m j c k

=

+ +

it is possible to calculate the frequency of maximum (peak) response. Formulating the

magnitude of the frequency response yields,

( ) ( )2 22

1( )H

k m c

= +

In order to identify the frequency of maximum response, it is necessary to evaluate thederivative of the frequency response magnitude with respect to frequency as follows.

*TODO: Show example F(w), show resulting X(w). (FRF continuous, F&X discrete) Express F(w) =

CONST . Show units of FRF = 1/k, compliance. Work example. Show MDOF FRF for comparison.


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Lecture Notes -40- 06/16/06 12:25 PM

( )( ) ( )

2 22

1( )

d dH

d dk m c

= +

( ) ( ) ( )( )1

2 2 22( )d d

H k m cd d

= +

( ) ( ) ( )( ) ( ) ( )( )3

2 22 222 21( )2

d dH k m c k m c

d d

= + +

( ) ( ) ( )( ) ( ) ( ) ( ) ( )3

2 2 22 2 21( ) 2 22

d d dH k m c k m k m c c

d d d

= + +

( ) ( ) ( )( ) ( )( ) ( )( )3

2 222 21( ) 2 2 2

2d H k m c k m m c c

d

= + +

In order to satisfy the maximum (or minimum), the derivative of the magnitude of the

frequency response function, ( )( ) 0pd

Hd

= , must equal zero. Since only the second

term in the above relation can be equal to zero (except at = ), the solution can bereduced to solving the following equation.

( )( ) ( )( )22 2 2 0p p pk m m c c + =

( )( )2 2 22 0p pkm m c + =

By dividing by the mass, 2m , the expression can be simplified by recognizing that

2 k

m = , 2

c

m = and 0p = cannot be the peak response for all possible MCK

combinations.

2

22 0pk c

m m

+ =

( ) ( )22 22 2 0p + =

2 2 2 22 0p =

( )2 2 21 2p =


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Lecture Notes -41- 06/16/06 12:25 PM

Therefore, the frequency of maximum (peak) response is as follows.

21 2p =

For comparison, recognize that the damped natural frequencyis 21r r = .

Resonant Response

Starting with the definition of resonance, = , (i.e. the response of the system to anexcitation at the undampednatural frequency of the system.), the amplitude of the system

response can be calculated.

Starting with the single degree-of-freedom frequency response function,

2

1( )H

m j c k

=

+ +

from the definition of resonance, = , (and 2k m= ) the resonant response, ( )H ,

equals

1 1 1 1 1( )

2 22cH

kj c j c j kj km j

= = = = =

If the damping is small (i.e. 0.1< ), then the peak response can be approximated as

1( ) ( )

2pH H

j k

=

Half Power Method

After defining the resonant frequency, it is possible to estimate the damping, , from the

shape of the frequency response function in the vicinity of the resonance. Again,assuming that the damping is small (i.e. 0.1< ), the half-power bandwidth points can

be calculated and then damping can be estimated from the bandwidth and the resonance

location.

Starting with the definition of the half-

power bandwidth points,

L

H

p

( )2

pH ( )pH


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Lecture Notes -42- 06/16/06 12:25 PM

1

2( ) ( ) ( )p L HH H H = =

and recognizing that for a lightly damped system, p , then expanding the single

degree-of-freedom frequency response yields,

2

1 1 1

22 j k k m j c =

+

Manipulating the expression yields a form from which the half-power bandwidth points

may be calculated.

2 2

2 2

1 1 1 1

221 1 2

cj j

k

= =

+ +

2

22 2 1 2j

= +

Squaring both sides yields

22 2 2 4 2

2 2 2

2 2 2 4 28 1 4 1 2 4

= + = + +

( ) ( )4 2

2 24 2 1 8 0

+ + =

Solving for

2

yields,

( ) ( ) ( )2

2 2 22 4 2 4 2 4 1 8

2

=

2

2 21 2 2 1

= +

Assuming is small (i.e. 1 ), then 2 0 , and the half-power bandwidth points are

approximated by,

2

1 2


43/86


Lecture Notes -43- 06/16/06 12:25 PM

Taking the difference between the points yields,

2 2

24H L H L H L

+

Using the approximation, 2H L + , allows reducing the expression to,

2H L

Therefore the damping factor can be approximated as,

2

H L

This approximation is also often written as2 c

f

f

or

2 c

BW

f .

While this method appears particularly simple, in practice, there are several difficulties

utilizing it effectively. Specifically, when using real sampled data, the ability to

determine the actual magnitude and location of ( )pH is limited by the measured data.

The actual sampled data will not in general include the exact frequency ( p ) and so both

the location and magnitude will be in error. Since the magnitude of ( )pH is used to

estimate the half-power bandwidth points, the location of L and H will be in error. Ingeneral, the errors tend to be the following: under estimating the magnitude of ( )pH ,

under estimating the magnitude of the half-power points, estimating L too low, and

estimating H too high. The compound effect of these errors is that generally the

damping factor ( ) is estimated too high.


44/86


Lecture Notes -44- 06/16/06 12:25 PM

FRF Example A

Given a single degree-of-freedom (SDOF) system with 5m kg= , 20N s mc = &

1000Nmk= , assuming a harmonic excitation of magnitude 20N@ 10Hz, determine the

response of the system to the given force.

Solution

First, identify the actual steady-state operating frequency.

10Hz= -or- 20 62.832rad rad s s =

Then, evaluate the SDOF frequency response function.

2 1( )H m j c k

= + +

( )2

1(20 )

5 20 20 20 1000

rads

rad rad N s N s s m m

Hkg j

= + +

( )1

(20 )18, 739 1, 257

rads

Nm

Hj

= +

( ) 6(20 ) 53.125 3.563 10rad ms NH j =

Finally, determine the response in the frequency domain.

( ) ( ) ( )X H F =

( ) 6(20 ) 53.125 3.563 10 20rad ms NX j N =

( ) 3(20 ) 1.063 0.071 10radsX j m = -or- 1.06 179.16 mm -or-

1.06 3.075rad mm

Therefore, the time domain response of the system to the given forcing function, ( )f t , is

expressed by*:

( ) 20cos(20 )f t t N=

( ) 1.06cos(20 3.075)t t mm=

*TODO: Need to show equivalence of two sided FRF solution.


45/86


Lecture Notes -45- 06/16/06 12:25 PM

#9 - Rotating Unbalance

For many situations, the source of excitation is an unbalance in a rotating machine part.

This internal (self) excitation is a function of the operation of the mechanism. However,

the techniques developed are applicable to this condition, as well.

By defining the position of the unbalance mass ( em ) as ( ) ( )t e t+ and the mass of the

system moving in translation only as em m , we

can write the equation of motion for this

representative single degree-of-freedom systemas:

( ( ) ( )) ( ) ( ) ( ) ( ) 0e em x t e t m m x t cx t k x t + + + + =

Rearranging terms yields:

( ) ( ) ( ) ( )emx t cx t k x t m e t + + =

Comparing this solution with the previously

developed equation of motion shows that the expression ( )em e t is equivalent in form to

the original external forcing function ( )f t . As such we can use the same harmonic

function assumed solution form to convert this to a steady-state solution.

By using both,

( ) st

x t Xe= , ( ) st

x t sXe= &2

( ) st

t s Xe= and

( ) ste t Ee= , ( ) ste t sEe= & 2( ) ste t s Ee=

where bothX&Eare complex scalars.

Substituting, the above solution forms into the differential equation of motion produces asolution that must be valid for every value ofs. This results in the following Laplace

Domain solution.

2 2st st st st

ems Xe csXe kXe m s Ee+ + =

By collecting common terms, the expression reduces to:

( )2 2st stems cs k Xe m Es e+ + =

Again, comparison with the Laplace Domain steady-state solution shows that2

eF m Es . Calculation of the effective Transfer Function yields:

m

k c

x(t)mee


46/86


Lecture Notes -46- 06/16/06 12:25 PM

2

2( ) ( )

e

X sH s s

m E ms cs k

= =

+ +

Or expressed in the Frequency Domain as:

2

2( ) ( )

e

XH

m E m j c k

= =

+ +

In the Frequency Domain, the effective forcing function is 2eF m E . As can be seen

from the form, the magnitude of the force is linearly proportional to both the unbalance

mass and eccentricity, however, it is proportional to the square of the rotational frequency

. Because of this, the unbalance is normally expressed in mass times length units ( em e )

and hence the above frequency response form (displacement / unbalance).

-1

-0.5

0

0.5

1

-10

-5

0

5

10

0

2

4

6

8

10


Magnitude

-1

-0.5

0

0.5

1

-10

-5

0

5

10

-4

-2

0

2

4


Phase[rad]

-10 -8 -6 -4 -2 0 2 4 6 8 100

1

2

3

4

5

6

7

8

9

O me a r ad /s ec

M

agnitude

-10 -8 -6 -4 -2 0 2 4 6 8 10-4

-3

-2

-1

0

1

2

3

4

Omega [rad/sec]

Phase[rad]


47/86


Lecture Notes -47- 06/16/06 12:25 PM

Rotating Example A

A machine of 100 kg mass has a 20 kg rotor with 5 mm eccentricity. The mounting

springs have 85 kNmk= and the damping is negligible. The operating speed is 600 rpm

and the unit is constrained to move vertically. (a)Determine the dynamic amplitude ofthe machine. (b)Redesign the mounting so that the dynamic amplitude is reduced to one

half of the original value, but maintaining the same natural frequency.

Solution

Given the following parameters:

100m kg= , 20em kg= , 5e mm= , 85 kNmk= ,

0N sm

c & 600rpm=

(a) Starting with the frequency response function

previously developed,

2

2( ) ( )

e

XH

m e m j c k

= =

+ +

Solve for the dynamic amplitude X by rearranging the expression and evaluating its

magnitude,

2

2( ) e

m eX

m j c k

=

+ +

The additional parameters needed are the unbalance and the rotational frequency in rads .

320 5 10 0.1em e kg m kg m= =

min

6002 20 62.83

60rad rad

s ss

rpm = = =

Evaluating the dynamic amplitude with 0c= yields,

2 2

2 2

(0.1 )(20 )

85,000 100 (20 )

radse

N radm s

m e kg mX

k m kg

= =

2

2

394.780.001274

309,784

kg m

s

kg

s

X m

= =

or 1.274mm

m

k c

x(t)mee


48/86


Lecture Notes -48- 06/16/06 12:25 PM

(b)There are two options for reducing the dynamic amplitude by half and yet keeping thesame natural frequency, increasing mass and increasing damping.

12

0.6372reduced original

X X mm= =

85,00029.15

100

Nm rad

s

k

m kg = = =

Option 1: increasing mass.

Manipulating the expression for dynamic amplitude

2 2 2

2 2 2 2( ) ( )

e e e

km

m e m e m eX

k m m m

= = =

yields an expression for the increased mass.

2 2

2 2 2 2

(0.1 )(20 )200

( ) 0.6372 ((29.15 ) (20 ) )

radse

rad rad s s

m e kg mm kg

X mm

= = =

However, keeping the same implies a corresponding increase in k. Therefore,

2 2200 (29.15 ) 170,000rad N s mk m kg = = = or 170 kNm

Notice how the solution makes sense intuitively. The operating speed is above the natural

frequency of the system. In that region, the frequency response function magnitude

(e

X

m e) approaches

1

m, so it would be expected that to reduce the dynamic amplitude by

half would require approximately doubling the mass.

Option 2: increasing damping.

Again, manipulating the expression for dynamic amplitude

2

2

em e

X m j c k

= + +

2 2

eX m j c k m e + + =

2

2 em e

m j c k X

+ + =


49/86


Lecture Notes -49- 06/16/06 12:25 PM

( )2

22

2 2

2( )

em ek m c

X

+ =

( )2

22

2 2

2( )

em ec k m

X

=

yields an expression for the necessary damping

( )2

22

2

2

2

em ek m

Xc

=

( )( )

22

22

2

2

0.1 (20 )85,000 100 (20 )

(0.006372 )

(20 )

rads

N radm s

rads

kg mkg

m

c

=

( )2 2

42 2

24 4

2 2

2 2

32

9 99 2

155.85 10309,784 383.87 10 95.966 10406.0 10 8,540

3,947.8 3,947.8

kg m

kgskg kg s

kgs ss

rad rad s s

mc

= = =

or more familiarly, 8,540N s mc = . This can be expressed as a fraction of critical damping

(c

c

c = ) by recalling the expression for the critical damping value, 2cc km= .

8,5401.465

2 2 85,000 100

N sm

Nm

c

km kg

= = =

or 146.5%

Notice that to reduce the dynamic response of this system utilizing only damping requires

an over-damped condition of nearly 150%.

For this problem, the mass addition is most likely to be the easiest and cheapest practical

solution.


50/86


Lecture Notes -50- 06/16/06 12:25 PM

#10 - Steady-State Applications

In this section, two different but related steady-state applications will be developed. The

first is the force transmissibility relationship. For transmissibility, the interest is in

determining the ratio of the force transmitted to the base ( ( )TF ) to the applied force

( ( )F ) as a function of frequency. The second application is vibration isolation. For

isolation, the interest is in determining the ratio of the response of the system ( ( )X ) to

the motion of the base ( ( )Y ) as a function of frequency. (This is sometimes referred to

as response ratio.) Just as in the earlier development of the Frequency Response Function(FRF), for both applications, the focus will be upon harmonic excitation.

Transmissibility

For transmissibility, we want to determine the ( )TFF

relationship. We begin with the

equation of motion for our representative single degree-of -freedom system. (NOTE: we

assume negligible motion of the base.)

( ) ( ) ( ) ( )mx t cx t k x t f t + + =

In addition, we need the restraining force at

the other end of the spring/damper.

( ) ( ) ( )T

cx t k x t f t + =

Applying the same transformationsdeveloped earlier, namely that the forcing

functions and responses can be express as

complex exponentials, and collecting terms,we get:

( )2 st st ms cs k Xe Fe+ + = and

( ) st st Tcs k Xe F e+ =

Finally, by taking the ratio of the second equation to the first, we get the desiredtransmissibility relationship.

2( )T

F cs ks

F ms cs k

+=

+ +

m

k c

x(t) f(t)

fT(t)


51/86


Lecture Notes -51- 06/16/06 12:25 PM


2( )T

F j c k

F m j c k

+=

+ +

Isolation

For isolation, we want to determine the ( )X

Y relationship. We begin with the equation

of motion for our representative single degree-of freedom system. (NOTE: this time thebase motion is NOT negligible, however there is no explicitly applied external force.)

Because there is motion for both the structure

and the base, we need to use the difference inmotion across the spring/damper.

( ) ( )( ) ( ) ( ) ( ) ( ) 0mx t c x t y t k x t y t + + =

Transforming the equation as before, we get:

( ) ( )2 0st st st st stms Xe c sXe sYe k Xe Ye+ + =

Collecting terms, and moving the base motion to

the other side of the equation yields,

2

( ) ( )

st st

ms cs k Xe cs k Ye+ + = +

Finally, by cross dividing the relevant terms, we get the desired isolation relationship.

2( )

X cs ks

Y ms cs k

+=

+ +


2( )

X j c k

Y m j c k

+=

+ +

m

k c

x(t)

y(t)


52/86


Lecture Notes -52- 06/16/06 12:25 PM

Conclusion

Comparing the expression for transmissibility to the expression for isolation, we find thatthey are identical. This interesting result implies that the process of determining the force

transmitted to the base is equivalent to determining the base motion transmitted to the

system.

-1

-0.5

0

0.5

1

-10

-5

0

5

10

0

2

4

6

8

10


Magnitude

-1

-0.50

0.5

1

-10

-5

0

5

10

-4

-2

0

2

4


Phase[rad]

-10 -8 -6 -4 -2 0 2 4 6 8 100

1

2

3

4

5

6

7

8

9

10

Omega [rad/sec]

Magnitude

-10 -8 -6 -4 -2 0 2 4 6 8 10-4

-3

-2

-1

0

1

2

3

4

Omega [rad/sec]

Phase[rad]


53/86


Lecture Notes -53- 06/16/06 12:25 PM

Transmissibility Example A

A vertical single-cylinder diesel engine of 500 kg mass is mounted on springs with

200 kNmk= and dampers with 0.2 = . The rotating parts are well balanced. The mass of

the equivalent reciprocating parts is 10 kg and the stroke is 200 mm. Find the dynamicamplitude of the vertical motion, the transmissibility, and the force transmitted to the

foundation, if the engine is operated at (a)200 rpm; (b)600 rpm.

Solution

Given the following parameters:

500m kg= , 10Rm kg= , 200stroke mm= ,

200 kNmk= & 0.2 =

(a)200 rpm

Starting with the single degree-of-freedom

frequency response function previously developed,

2

1( ) ( )

XH

F m j c k

= =

+ +

Solve for the dynamic amplitude X by rearranging the expression and evaluating its

magnitude,

2

( )( )

FX

m j c k

=

+ +

The additional parameters needed are the equivalent force, the damping and the operating

frequency expressed in rads . From the mechanical unbalance development, the equivalent

force can be expressed as 2eq eF m e= . Recognizing that the reciprocating mass is the

effective mass, e Rm m= , and that the effective eccentricity is half the stroke,

2strokee= , yields,

min

200 202 20.94

60 3rad rad

s ss

rpm = = =

2 210 0.1 (20.94 ) 438.7radseq eF m e kg m N= = =

2 2 0.2 200,000 500 4000N N sm mc km kg = = =

m

k c

x(t)

stroke


54/86


Lecture Notes -54- 06/16/06 12:25 PM

Evaluating the dynamic amplitude yields,

2 2

438.7

200,000 500 (20.94 ) (20.94 )(4000 )

eq

N rad rad N sm s s m

F NX

k m j c kg j = =

+ +

( ) 2438.7

0.005119, 324 83, 776 kg

s

NX m

j= =

+or 5.1mm

Evaluating the transmissibility yields,

2 2

200,000 (20.94 )(4000 )

200,000 500 (20.94 ) (20.94 )(4000 )

N rad N sm s m


k j c jTR

k m j c kg j

+ += =

+ +

( )( )

200,000 83,7762.52

19, 324 83, 776

Nm

Nm

jTR

j

+= =

+

From the definition of transmissibility, namely TF

TRF

= , the force transmitted to the

foundation can be determined.

2.52 438.7 1106T eqF TR F TR F N N= = = =

Notice that due to the characteristics of the system, the force transmitted to the

foundation is approximately 2.5 times the force applied to the system.

(b)600 rpm

Again solving for the dynamic amplitude Xusing the rearranged expression andevaluating its magnitude,

2

( )( )

FX

m j c k

=

+ +

The additional parameters needed are the equivalent force, the damping and the operating

frequency expressed in rads . From the mechanical unbalance development, the equivalent

force can be expressed as 2eq eF m e= . Recognizing that the reciprocating mass is the

effective mass, e Rm m= , and that the effective eccentricity is half the stroke,

2strokee= , yields,

min

6002 20 62.83

60rad rad

s ss

rpm = = =


55/86


Lecture Notes -55- 06/16/06 12:25 PM

2 210 0.1 (62.83 ) 3,948radseq eF m e kg m N= = =

2 2 0.2 200,000 500 4000N N sm mc km kg = = =

Evaluating the dynamic amplitude yields,

2 2

3,948

200,000 500 (62.83 ) (62.83 )(4000 )

eq


F NX

k m j c kg j = =

+ +

( ) 26 33,948

0.00221.774 10 251.3 10 kg

s

NX m

j= =

+ or 2.2mm

Evaluating the transmissibility yields,

2 2

200,000 (20.94 )(4000 )

200,000 500 (62.83 ) (62.83 )(4000 )

N rad N sm s m

N rad rad N s

m s s m

k j c jTR

k m j c kg j

+ += =

+ +

( )( )

3

6 3

200,000 251.3 100.179

1.774 10 251.3 10

Nm

Nm

jTR

j

+ = =

+

From the definition of transmissibility, namely TF

TRF

= , the force transmitted to the

foundation can be determined.

0.179 3,948 708T eqF TR F TR F N N= = = =

Notice that although the equivalent applied force is 900% larger than for the first case,

the force actually transmitted to the foundation is about 30% smaller.


56/86


Lecture Notes -56- 06/16/06 12:25 PM

#11 - Multiple Degree of Freedom Systems

MDOF Example

Starting with the simple two degree-of-freedom

system shown, the objective will be to determine

the equations of motion and the naturalfrequency and mode shapes.

The equations of motion can be found by eitherof two techniques, Newtons Method or

Lagranges Method.

Equations of Motion NewtonsMethod

As has been shown, Newtons Method involves

identifying all forces, both internal and external,acting on each body and then expressing those forces in terms of the given coordinates.

Therefore, the first action must be the development of thefree body diagramsfor each

body.

The given two degree-of-freedom

system has the two free body

diagrams, show at right.

Evaluating Newtons equation,

i

i

f ma=

, for each body yields a set

of equations describing the motion ofthe system.

( )1 1 1 2 1 2 1 1 1f k x k x x m g m x + =

( )2 2 2 1 2 2 2f k x x m g m x + =

Collecting and rearranging the terms

into standard form yields,

( )1 1 1 2 1 2 2 1 1m x k k x k x f m g + + = +

2 2 2 2 2 1 2 2m x k x k x f m g + = +

1( )x t 1( )f t

2( )x t 2( )f t

1k

2k

1m

2m

( )2 2 1k x x

2m

1 1k x

1m

( )2 1 2k x x

=1m

1 1m x

=2m

2 2m x

body #1

body #2

2m g

1m g

2f

1f


57/86


Lecture Notes -57- 06/16/06 12:25 PM

Or expressed in matrix form as,

1 1 1 2 2 1 1 1

2 2 2 2 2 2 2

0

0

m x k k k x f m g

m x k k x f m g

+ + = +

Recall that when the gravitational term ( im g) ends up on the force side of the equation,

the associated coordinate ( ix ) can be redefined from the static equilibrium position

thereby eliminating the im gterm from the equation. In this case, both terms are

eliminated.

1 1 1 2 2 1 1

2 2 2 2 2 2

0

0

m x k k k x f

m x k k x f

+ + =

Equations of Motion Lagranges Method

As has been shown, Lagranges Method involves identifying the kinetic,potentialand

dissipativeenergies for the entire system, then evaluating the Lagrange Equation for each

degree of freedom.

i

i i i i

d T T U DF

dt q q q q

+ + =

First, identifying the kinetic energy term,

2 21 11 1 2 22 2

T m x m x= +

And the potential energy term,

( )221 1

1 1 2 2 1 1 1 2 22 2U k x k x x m gx m gx= +

Provides all the information necessary to determine the equations of motion. Note that

there are no dissipative terms for this problem, so 0D= .

Second, evaluating the various partial derivatives for each degree of freedom, yields,

(First degree of freedom)

( ) ( )2 21 11 1 2 2 1 1 1 12 21 1

d T d d m x m x m x m x

dt x dt x dt

= + = =


58/86


Lecture Notes -58- 06/16/06 12:25 PM

( )( ) ( )( )221 11 1 2 2 1 1 1 2 2 1 1 2 2 1 12 21 1

1U

k x k x x m gx m gx k x k x x m g x x

= + = +

(Second degree of freedom)

( ) ( )2 2

1 11 1 2 2 2 2 2 22 2

2 2

d T d d m x m x m x m xdt x dt x dt

= + = =

( )( ) ( )221 11 1 2 2 1 1 1 2 2 2 2 1 22 22 2

Uk x k x x m gx m gx k x x m g

x x

= + =

Finally, collecting the various terms for each degree of freedom yields the equations of

motion.

( )1 1 1 2 1 2 2 1 1m x k k x k x f m g + + = +

2 2 2 2 2 1 2 2m x k x k x f m g + = +

Or expressed in matrix form as,

1 1 1 2 2 1 1 1

2 2 2 2 2 2 2

0

0

m x k k k x f m g

m x k k x f m g

+ + = +

Again recall, when the gravitational term ( im g) ends up on the force side of the equation,

then the associated coordinate ( ix ) can be redefined from the static equilibrium position

thereby eliminating the im gterm from the equation. In this case, both terms are

eliminated.

1 1 1 2 2 1 1

2 2 2 2 2 2

0

0

m x k k k x f

m x k k x f

+ + =

Notice that the actual equations of motion identified are identical, regardless of whether

Newtons Method or Lagranges Method was used.

Natural Frequency and Mode Shape

In order to identify the natural frequency and mode shape for a set of simultaneous

differential equations of motion, it is necessary to solve an eigenvalue problem. By

transformi

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