vibrations chapter 2 2013 post

7/28/2019 Vibrations Chapter 2 2013 Post

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Chapter 2 Modeling of Vibratory Systems

1ENME 361, Spring 2013

ENME361

Vibrations, Control, and Optimization ISpring 2013

Chapter 2

Modeling of Vibratory Systems

Acknowledgement: Professors B. Balachandran and E. B. Magrab

Assignment 2 (due on 2/12/13): 2.3, 2.8, 2.11, 2.20, and 2.27



In Chapter 2, we shall show how to do the follow ing:

Compute the mass moment of inertia of rotationalsystems

Determine the stiffness of various elastic componentsin translation and torsion and the equivalent stiffness

when many individual linear components are combined

Determine the stiffness of f luid and pendulum elements

Determine the potential energy of stiffness elements

Determine the damping for systems that have differentsources of dissipation

Construct models of vibratory systems


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Two examples from mechanical engineering

A Microelect romechanical System

x

ElectrostaticforceFe

Base acceleration y

End massm

Equivalent structurestiffnessk

Modeled as a sing le

degree-of-freedom

system



The Human Body

Multi-degree-of-freedom system


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2.1: Introduction

There are, in general, three elements thatcomprise a vibrating sys tem

Inertia element

Stores and releases kinetic energy

Stiffness element

Stores and releases potential energy

Dissipation element

Used to express energy loss in a system



xek

ec

em

Inertia Element

Stiffness Element

Dissipation Element

GeneralizedCoordinate

Model Construction


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Each of these elements has differentexcitation-response characteristics.

The excitation is in the form of a force or amoment and the corresponding response ofthe element is in the form of a displacement,velocity, or acceleration.



Inertia Elements:

Inertia

Element

Kinetic

EnergyInertia elements are characterized by a relationshipbetween an applied force (or moment) and the

corresponding acceleration response

Inertia

Element

Inertia

Element Forceand/or

moment

Acceleration


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Stiffness Elements:

Stiffness

Element

Potential

EnergyStiffness elements are characterized by a relationship

between an applied force (or moment) and the

corresponding displacement (or rotation) response

Stiffness

Element

Stiffness

Element

Force

and/or

moment

Displacement



Dissipation Elements:

Dissipation

Element

Energy

Loss Dissipation elements are characterized by a relationship

between an applied force (or moment) and the

corresponding t ranslational or rotational velocity

response

Dissipation

Element

Dissipation

Element Force

and/or

moment

Velocity


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Table 2.1: Units of components comprising a vibrating

mechanical system and their customary symbols

Quantity Units

Translational motionMass,mStiffness,kDamping,cExternal force,F

RotationalmotionMassmomentof inertia,JStiffness,kt

Damping,ctExternalmoment,M

kgN/mNs/mN

kgm2Nm/radNms/radNm



k c

m

O

ab

l

?

??

e

e

e

m

kc

?x


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2.2: Inertia Elements

Translational motion o f a mass: Is described as motion along the path followed

by the center of mass

The associated inertia property depends onlyon the total mass of the system and is

independent of the geometry of the mass

distribution of the system



Consider a mass translating with a velocity of

magnitude in the plane.

k

j

Oi

Y

X

Z

F i

x i

m

mx -X Y

,d

mdt

F p p rPrincipleof Linear Momentum:

d

F mxdt

F xm

i i

1

2

T m r rKineticEnergy:

2

1

2

1

2

T m x x

mT x

i i


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Rotational motions of a mass:

The inertia property is a function of the mass

distribution as described by it s mass moment of inertiaabout its center of mass or a fixed point O.

2O GJ J md

where

m is the mass of the element

JG is the mass moment of inertiaabout the center of mass

d is the distance from the centerof gravity to the point O

When the mass oscillates about a fixed po int O or a

pivot point O, the rotary inertia JO is given by the

parallel-axes theorem

z

G

O

d

,GJ m



Table 2.2: Mass moments of inertia about axis z normal

to the x-y plane and passing through the center of mass

L

zL/2

21

12G mL

R

z

G

21

2GJ mR

R22

5G mR

Slenderbar

Circulardisk

Sphere


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Example 2.1: Determination of mass moments of inertia

Uniform disk

m, J G

R

O

G

2O GJ J mR

From Table 2.2

21

2GJ mR

Therefore,

2 2 21 3

2 2OJ mR mR mR

2O GJ md

kgm2

kgm2

kgm2

From the parallel-axes theorem



c.g.L

L/2m

O

G

Uniform bar

From Table 2.2, we have that

21

12GJ mL

From the parallel-axes theorem,

we find

2

2 2 21 1

2 12 4 3O G

L mJ J m mL L mL

2

O GJ md


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Example 2.2: Slider mechanism: system w ith varying inertia

property

r

l ams

b

emb

me

ml

O

O

2O GJ md We will use the parallel-axestheorem. Hence, we need thedistances from each moving

members c.g. to point O.

a

a

e/2

b/2

r

c.g.

c.g.

b/2

ab

O

2 2 2

22 2

22 2

( ) 2 cos

( ) 2 cos

( ) 2 cos( )

b

e

r a b ab

a b a ab

a e a ae

Law of cosines



The rotary inertia JO of this system is given by

( ) ( ) ( )l s b eO m m m m

J J J J J

r

l ams

b

emb

me

ml

O

O

2

2 2 2

2 22 2

2 22 2

1

3

( ) ( ) 2 cos

( ) cos12 3

( ) cos( )12 3

l

s

b

e

m l

m s s

m b b b b

m e e e e

J ml

J mr m a b ab

b b

J m ma m a ab

e eJ m ma m a ae

where

2O GJ md

kgm2


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Example: Equivalent Mass of a System

Rigid link 1: rotateswithpulleyaboutO

?eqm 21

2eqmT x

2l

1l

1m

cr

Rigid link 2Cylinder

No slip

2m2k

cmcJ

( )y t

Pulley

m

pJ

O

1k

pr

( )x t

2l

1l

1m

cr


No slip

2m2k

cmcJ

( )y t2l

1l

1m

cr


No slip

2m2k

cmcJ

( )y t

Pulley

m

pJ

O

1k

pr

( )x tPulley

m

pJ

O

1k

pr

( )x t

( ) is "small"x t



( ) is "small"x t

( )( )p

x ttr

2

1

1,

2T mx

2

2

2

1 1

,2 2p p p

x

T J J r

22

2 1 13 1

1 1

2 2 3 p

ml xT J

r

11( ) ( ) ( )

p

ly t t l x t

r

m1k

( )x t

1l

1m

pJ

O

pr

( )t

( )y t


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2l

cr

2m

2k

cmcJ

( )y t

( )y t

( )t

1 1( ) ( ) ( )c c p

l lt t x t

r r r

11( ) ( ) ( )

p

ly t t l x t

r

2

2 14 2 2

1 1,

2 2 p

l xT my m

r

2

2 15

1 1,

2 2c c

p

l xT my m

r

2

2 16 1 1

2 2c c

c p

l xT J Jr r



2 22

2 1 1

2 2 2

1 1 12

1 1 1

2 2 2 3

1 1 1

2 2 2

p

p p

c c

p p c p

x ml xT mx J

r r

l x l x l xm m J

r r r r

1 2 3 4 5 6T T T T T T T

212

eqmT x

2 2 22

1 1 1 1 122 23

p

eq c c

p p p p c p

J ml l l lm m m m J

r r r r r r


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Examples of sp rings

Spring

Motion

Highway base isolation

using cylindri cal rubber

bearings

Wire rope isolators

Motion

Wire

Wire



Steel coi l

springs

Addi tional Examples of springs

Air springs

Motion

Steel cable springs use in

a chimney tuned mass

damper

Motion

Cablesprings


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2.3: Stiffness Elements

Fs is the internal

force acting with in

the stiffness

element.

It is called the

restoring force.

As the sti ffness element is deformed, energy

is stored in this element, and as the stiffness

element is un-deformed, energy i s released.

This energy is called the potential energy.

F

FSF

O

O

-S F F

StiffnessElement



Potential energy in a spring

The potential energy V is defined as the work done to

take the stiffness element from the deformed position

to the un-deformed position; that is, the work needed to

un-deform the element to its original shape.

0 0

0

0

-

-

Sx

x

x

xd F x dx

F x dx

V x

F x dx

= F x i i

i i

F

xi

j

F

xi

j


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2.3.2 Linear Springs

Linear Translation Spring

x

F

k

Single

spring

( )F x kx

where

F is the force applied to a linear spring (N)

k is the spring constant (N/m)

0 0 0

2

( ) ( )

1

2

x x x

V x F x dx kxdx k xdx

kx

N

(Nm)

The potential energy in the spring is



Note:

The stiffness constant k is a static concept , and hence, a

static loading is sufficient to determine this parameter.

0.1 0.05 0 0.05 0.11500

1000

500

0

500

1000

1500

x (m)

F

(N)

k=10568 N/m

Fitted curve

Data

Experimentally obtained data used to determine the linear spring constant k


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Linear Torsion Spring

( ) tk

kt

where

is the moment applied to a linear spring(Nm)

k t is the torsion spring constant (Nm/rad)

The potential energy in the torsion spr ing is

0 0

2

( ) ( )

12

t

t

V d k d

k

(Nmrad)



Combinations of L inear Springs

Two springs in series

F

x

1k 2k

1k 2k F

x

1k 2k

1k 2k

1k 2k

?ek

1k 2k1k 2k

?ek


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1 1

1

1

F k x

Fx

k

2 2

2

2

F k x

Fx

k

F1k 2kFFF

The force on each spring is the same

1 2

1 2 1 2

1 1F Fx x x Fk k k k



F1k 2k

1 2

1 2

1 2

1 2

1

1 2

1 1

e

kk

k k

kkk

k k

F x xk k

x


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1 2

1 2

e

kkk

k k

1k 2k

ek



N Springs in Series

ek

1k 2k

1Nk Nk

1

1 2 1

1 1 1 1e

N N

kk k k k


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The potential energy fo r two springs in series is

where V1(x1) is the potential energy associated with

the spring of stiffness k1 and V2(x2) is the potential

energy associated with the spring of stif fness k2.

Then,

1 2 1 1 2 2( , ) ( ) ( )V x x V x V x

2 21 2 1 1 2 2

1 1( , )

2 2V x x k x k x



Two springs in parallel

The displacements of both springs are equal

F

x

1k 2k1k 2k

F

x

1k 2k1k 2k

1k 2k ?ek


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1 1k xF 2 2k xF

2k

F

1F 2F

1F 2F

1k 2k

F

1F 2F

1F 2F

1k

1 2

1 2

1 2

kx k x

k k x

F F F

1 2

1 2e

xk k

k k k

F



1 2ek k k 1k 2k ek1k 2k ekek

1k 2k ekek 1Nk Nk

1 2 1e N Nk k k k k


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Axially loadedrodorcable:

AEk

L

: area of cross section

: Young's modulus

A

E,F x

L,A E

,F x

L,A E

A Few Equivalent Spr ing Constants of Common Struc tural

Elements Used in Vibratory Models (from Table 2.3)



Axially loaded rod or cable:

1 2

4

Eddk

L

: Young's modulusE,F x

E L

1d

2d

,F x

E L

1d

2d


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Hollow circular rod in torsion:

,

t

GIk

L

L,G I 4 4out in32

d dI

: polar moment of inertia

: shear modulus

I

G



CantileverBeam:

3

3,

0

EIk

a

a L

,F x

La

: area moment of inertia

: Young's modulus

I

E


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Pinned-pinned Beam:

2 2

3EI a b

k a b

,F x

a b


: Young's modulus

I

E



Clamped-clamped Beam:

3

3 3

3EI a bk

a b

,F x

a b


: Young's modulus

I

E


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1 2

,it t t tii

IGk k k k

L

: polar moment of inertia; : shear modulusI G

,

1L 2L

1IG 2IG




1

1 2

1 1,it ti

i t t

IGk k

L k k

,

1L 2L

1

IG 2

IG

: polar moment of inertia; : shear modulusI G


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L

AEk

3

3EIk

a

t

GIk

L

Engineering unit check:

N/m [(m2)(N/m2)/m]

N/m [(N/m2)(m4)/m3]

Nm [(N/m2)(m4)/m]



F

2k1k

Example: Equivalent spring constant

F

2k1k

Case 1 Case 2

1 2ek k k

1

1 2

1 1ek

k k


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Example: Equivalent spring constant

F

L

1

EI

2

EI

k

F

LL

1

EI

2

EI

k

F

k

1k

2k

F

k

1k

2k

k1

k2



Example: (Continuation)

F

k

1k

2k

1 2

1 23 3

3 3,

EI EIk k

L L

1

2

1

1 1ek kk k


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Example 2:

k

k

F

1k

F

a b

EI



Example 2: (Continuation)

1ek k k

k

F

1k

1 2 2

3EI a bk

a b


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Linear Torsion Springs

kt1 kt kt1kt

Springs in

parallelSprings in

series

1 2

1 2 1 2

( ) ( ) ( )

( )t t t t

te

k k k k

k

1 2

1 2

1 2

1 1

t t

t t te

k k

k k k

Potential energy Potential energy

1 2

21( )2

t tV k k 1 22 2

1 2 1 21 1( , )2 2

t tV k k



Example 2.7 Equivalent stiffness of springs in parallel:

removal of a restriction

k1k2

ba

F

a

x2x1

x1 x2

2 1 2

1 2

bx x x x

a b

b ax xa b a b

b

aF

F2=k2x2 F1 =k1 1 1 2

2 1

F F F

bF aF


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1

2

bF

F a b

aFF

a b

From the force and moment balance equations, we obtain

)(

)(

22

22

11

11

bak

aF

k

Fx

bak

bF

k

Fx

Therefore

1 2

2 22 1

21 2

( ) ( ) ( ) ( )

( )

b bF a aFx

a b k a b a b k a b

F k b ka

a b kk

and we obtain

1 2

b ax x x

a b a b

1 2

2 1

F F F

bF aF



Comparing this result to the form

e

Fx

k

the equivalent spring constant ke is determined as

21

22

221 )(

akbk

bakkke


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2.3.4: Other Forms of Potent ial Energy Elements

Fluid - Manometer (Ships at sea)

Compressed gas piston (Jackhammer)

Pendulums (Cranes, hoists on mov ing platforms)

The source of the restoring force is a fluid element or a

gravitational loading



Fluid Element - Manometer

g

Ao

The total displacement

of the fluid is 2x.

The total force of the

displaced fluid acting on

the rest of the fluid is

( ) 2m oF x gA x

Engineering unit check:

3 2 2

2 2 3

(N) ~ (kg/m ) (m/s ) (m ) (m)

~(kg)(m/s )(mm/m )

~N

oF g A x

The equivalent spring

constant of the system is

2

me

o

dFk dx

gA


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Pendulum System 1

L

L/2

c.g.

L/2

mgx)cos1(

2cos

22

LLLx

Since F = mgj, the potential energy increase is0

0

( ) ( )initial position

deformed position x

x

V x x mg dx

mgdx mgx

F dx j j

The change in height is given

by

i

j



( ) (1 cos )2

mgLV

This can be writ ten as

From the Taylor series approximation

...2

1cos2

we obtain2 2

( ) (1 cos ) 1 12 2 2 2 2

mgL mgL mgLV

from which we identify the equivalent spring constant

as

2e

mgLk


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L

m1

L

m1gx

Pendulum System 2

Using the previous results by

replacing L/2 with L

(1 cos )x L

The potent ial energy can be

approximated as

21( )2

eV k

where we replace m with m1 and L/2

with L to obtain

1ek mgL



L

m1

L

m1gx

L/2mg

Pendulum System 3

The potential energy of the bar

and the mass are obtained from

the previous results as

2 21

1 1( )

4 2V mgL mgL

we find that the equivalent spring constant is

12

e

mk m gL

21( )2

eV k

Comparing with


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Pendulum System 4

L

m1

L

m1g

x

L

O

QP

e1e2

i

j

We see that

21 cos2

Lx L

The potential energy is

1 1

1 1

21

( ) ( ) ( )

( )

1( )2

L LL

L xL x L x

V x mg dx mgx

mg L L x mgx

V mgL

F dx j j

=



We see that when the pendulum is inverted there is a

decrease in potential energy, since V() is negative.

Therefore,

1ek mgL

L

m1

Lm1g

x


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2.4: Dissipation ElementsDamping elements are assumed to have neither

inertia nor the means to store or release potential

energy.

The mechanical motion imparted to these elements

is converted to heat or sound and, hence, they are

called non-conservative ordissipative because this

energy i s not recoverable by the mechanical system.



London Millennium Footbridge and Dampers from Taylor Devices

Source: Taylor Devices, Damper Retrofit o f The London Millennium

Footbridge: A case study in Biodynamic Design

Space Satellite Damper Adaptedfor London Milliennium

Footbridge


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London Millennium Footbridge and Dampers from Taylor Devices

Source: Taylor Devices, Damper Retrofit o f The London Millennium

Footbridge: A case study in Biodynamic Design



There are four common types o f damping mechanisms

used to model vibratory systems.

(i) Viscous damping

(ii) Coulomb or dry fri ction damping

(iii) Material or solid o r hysteretic damping

(iv) Fluid damping

In all these cases, the damping force is usually expressed

as a function of velocity.


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2.4.1 Viscous Damping

When a viscous fluid flows through a slot or around

a piston in a cylinder, the damping force generated is

proportional to the relative velocity between the two

boundaries confining the fluid.

Fx

Viscous Fluid

PistonCylinder

F x xc Damping Coefficient

(unit: N/(m/s )Linear Case:



Energy Dissipation

The energy dissipated by a linear viscous damper

is given by

2 2dE Fdx Fxdt cx dt c x dt

The symbol used to represent a viscous damper is

orc

c


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In the case of a nonlinear viscous damper described by

a function the equivalent linear viscous dampingaround an operating speed is determined asfollows:

Linear viscous damping elements can be combined inthe same way that linear springs are, except that theforces are proportional to velocity instead of displacement.

l

e

x x

dF xc

dx

F x

lx x



Two Linear Dampers in Series: The force on eachdamper is the same

F

1c 2c

1x

2x

1c 2c


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?ec

1c 2c

Two Linear Dampers in Series:



FFFF1 1

1

1

F c x

Fx

c

2 2 1

2 1

2

F c x x

Fx x

c

1c 2c1x 2x1x

Two Linear Dampers in Series:

2 1 2 1x x x x 21 2 1 2

1 1F Fx F

c c c c


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1 2

1

1

2

2

1 2

1

1 2

2

2

1 1

e

cc

cF x x

c c

ccc c c

c

1

c2

c

F2

xTwo Linear Dampers in Series:



ec

1

1 2 1

1 1 1 1e

N N

cc c c c

1c 2c

1Nc NcN Linear Dampers in Series


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Two Linear Dampers in Paral lel : The velocit ies of both dampers are equal

1c 2c

F

2c1c

x



1 2ec c c

1c 2c ec

Two Linear Dampers in Paral lel : The velocit ies of both dampers are equal


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N Linear Dampers in Parallel

1 2 1e N Nc c c c c

1c 2c

ec1Nc Nc



Example: Equivalent Damping Coefficient. Assume is small.

3c

O

ab

l

2c1c 3c

O

ab

l

2c1c

O

d

ec

O

d

ec


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is "small"

3c

, O

ab

l

2c1c 1x 2x 3x3c

, O

ab

l

2c1c 1x1x 2x 3x

Example: Equivalent Damping Coefficient

1 2 3

, ,x a x b x l



Example: Equivalent Damping Coefficient

,

3c

O

2c1c

1x 2x 3x

1F 2F 3F

,

3c

O

2c1c

1x 2x 3x

1F 2F 3F

3c

O

2c1c

1x1x 2x 3x

1F 2F 3F


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1 1 1 1

2 2 2 2

3 3 3 3

F cx ca

F c x c b

F c x c l

1 2 3OM F a F b F l

O

1F 2F 3F

,



O

1F 2F 3F

2

1 2 3

1

2

2

2

1

3

2 3

O F a F b F l

ca a c b

M

ca c b c

c

l

b l l

,


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O

d

ec x

is "small" x d

,



e eF c x c d

2

e

O

e

Fd

c d

d

d

M

c

ec

x

O

F

,

ec

x

O

F

,

xx

O

F

, O

F

,


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2 2 21 2 3 2eca cb l dc c

OOM M

2 2 21 22

3

ec d

ca c b c l



2.5 Model Construction

Discrete system or a lumped-parameter system

Only discrete elements are used to model a

physical system

mass, stiffness, and damping

These three types of elements appear as

parameters in the governing equations of the

system.

Distributed-parameter system orcontinuous systemNo discrete elements are used to model the

system. One or more partial differential equations

are used to model the system. Equivalent to an

infinite number of discrete elements.


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The Human Body Revis ited

Multi-degree-of-freedom system



Machine-tool Cutting Process

Work pieceMachine bed

Tool

TurretTool slide

Beam fixed at each end

(machine bed)

Cantilever beam(work piece)

Cutting force

Turret

J , Mm

Tool

Physical system Vibratory model

Cutting force

Turret andslidestiffness and

damping

Lb, mb, Eb, Ib

Lw, mw, Ew, Iw

c

Coupled discrete and distr ibuted parameter model


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?

?

?

e

e

e

m

k

c

k

c

1m

O

2m

G 3, Gm J

a

b

l

g

Example: r2

r1



Glossary Chapter 2

Coulomb damping a damping model for dry f ri ct ion in

which the damping force has a constant magnitude and a

direction opposite to that of the motion

Dissipation element a dev ice that provides res is tance to

motion in the form of an irrecoverable loss of energy

Mass a measure of the amount of mater ial a body

contains; its weight is the mass times the gravity force in a

gravitational field; for translational motions, the ratio of

force to acceleration

Mass moment of inertia a measure o f the resistance of a

body to angular acceleration about a given axis; the ratio

of moment to angular acceleration


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Potential energy a scalar quant ity that represents the

energy that is associated with elastic forces and

gravitational forces

Spring a dev ice that recovers its or ig inal shape when

released after being distorted

Stiffness element a dev ice that stores and releases

potential energy

Viscous damping a damping model in which the

damping force has a magnitude linearly proportional to

the system speed and a direction opposite to that of the

motion

vibrations chapter 2 2013 post

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