vibrations chapter 3 governing equations

7/26/2019 Vibrations Chapter 3 Governing Equations

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Chapter 3 Single Degree-of-Freedom Systems

1Copyright 2008 B. Balachandran and E. B. Magrab ENME361, Fall 2015

ENME 361

Vibrations, Control, and Optimization I

Fall 2014

Chapter 3Governing Equations

B. BalachandranProfessor

Department of Mechanical Engineering


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2Copyright 2008 B. Balachandran and E. B. Magrab

In Chapter 3, we shall show how to do the following:

Obtain the governing equation of motion for singledegree-of-freedom translating and rotating systemsby using force balance and moment balancemethods

Obtain the governing equation of motion for singledegree-of-freedom translating and rotating systems

by using Lagranges equations

Determine the equivalent mass, equivalent stiffness,

and equivalent damping of a single degree-of-

freedom system

Determine the natural frequency and damping factor

of a system

ENME361, Fall 2015


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3Copyright 2008 B. Balachandran and E. B. Magrab ENME361, Fall 2008

Quick revision of ODEs

Second order, linear, homogeneous, differential equation withconstant coefficients

ad2y

dx2 + b

dy

dx+ cy = 0

y = epx

!ap2 + bp + c = 0"p =#b b2 # 4ac

2a

!x = C1ep

1x

+C2ep

2x

(forp1 "p2 )

x =epx

C3+C

4x( ) (forp1 =p2 =p)


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3.2.1 Force-Balance MethodsVertical Vibrations of a Spring-Mass-Damper System: Interested in

getting the corresponding equation of motion

( )str L x!= = + +r j j

st

mg

k! =

Inertia force:!m!!rj

2

2

( )G

G

d md

dt dt

d dm m

dt dt

= =

= =

vpF

v rf(t) is an externally

applied force

k

k

c

c

m x

dst

mgm

mg f(t)

mr!!

k(x+dst)

k(x+dst)X

Yj

iO

cr!

cr!

Lxst

State 1 (S & D areunstretched)

State 3 (S & D respond to thedynamics of the mass, which is atan instantaneous location xfrom

the static location of the S & D )

Need the force balance only inj

direct ion since the massoscillates only in the vertical

direction

State 2 (S & D are stretchedby dstby the hanging masswhich is at rest)

S: SPRING,

D: DAMPER

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A force balance along thej direction results in

f(t)j+ mgj

External forces acting on system

! "# $# ! kx + k!st( )j

Spring force acting on mass

! "# $## ! c

dr

dtj

Damping forceacting on mass

!"#!m

d2r

dt2 j

Inertia force

! "# $#

= 0

m

mg f(t)

m!!r

k(x+!st) c!r

Static-Equilibrium Position

The static-equilibrium positionof a system is theposition that corresponds to the systems reststate; that is, a position when Also thereis no external force, i.e., f(t)=0

Thus,

!r =!!r = 0.

and x= 0 is the static-equilibrium position of the system.

Another approach:Mass x acceleration = Sum ofthe external forces

kx + k!st

=mg

!!st =

mg

k

we takex = 0, asxhas been measured

from the static equilibrium position

!"#

$%&

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Therefore, the equation of motion for oscillations about thestatic-equilibrium position (i.e., we drop mg) is

2

2 ( )d x dxm c kx f t

dt dt + + =

Punch Press

Wing Vibrations

Some examples of external forces:

Fluctuating air pressure loading such

as that on the wing of an aircraftFluctuating electromagnetic forces

such as in a loudspeaker coil

Electrostatic forces that appear in

some microelectromechanical

devicesForces caused by an unbalanced

mass in rotating machinery

Buoyancy forces on floating systems

Impacts

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Horizontal Vibrations of a Spring-Mass-Damper System

k

k

c

c

x

m

f(t)

m!!x

c!x

kx

m

kx

c!x

X

Y

i

j

g

O

f(t)i

External forcesacting on system

!!

kxi

Spring forceacting on

mass

!!

c

dr

dti

Dampingforce acting

on mass

!!m

d2r

dt2 i

Inertia force

! "# $#

= 0

which also results in2

2 ( )

d x dxm c kx f t

dt dt + + =

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Force Transmitted to Fixed Surface

k(x+!st)

k(x+!st) c!x

c!x

FR

FR = k!st

Staticcomponent

!+ kx + c

dx

dt

Dynamiccomponent

! "# $#

We consider only the dynamic part

of the reaction force, therefore

Rd

dxF c kx

dt= +

On the other hand, the static part

is FRS =k!st =mg

Do the force balancesseparately on the mass,S/D, and the support; FRis the force to be exerted

on the support to keep itin its position

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Example 3.1: Wind driven oscillations about a systemsstatic equilibrium position

Wind typically generates a force on the structure that

consists of a steady-state part and a fluctuating part. Insuch cases,

f(t) = fss

steady!

statepart

!

+ fd(t)

fluctuatingpart

!

2

2 ( )

( )ss d

d x dxm c kx f t

dt dt

f f t

+ + =

= +

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To solve this equation, we assume that the displacement xconsists ofstatic component xoand a dynamic component xd(t). In other words, we

assume that presence of a steady-state force must trigger a steady-state

displacement. Thus,

( ) ( )o d

x t x x t= +

Substituting this equation into the governing equation, we get2

2 ( )d d

d o d ss

d x dx

m c kx kx f t f dt dt + + + = +

Therefore,

o ssx f k=

2

2 ( )d d d d

d x dxm c kx f t

dt dt + + =

and xd

is a solution to

(static-equilibrium position)

ENME361, Fall 2015


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3.2.2 Moment-Balance Methods

M(t)

Disc with rotary inertiaJG

about rotation axis

Axis of rotation

Shaft with equivalenttorsional stiffness kt

Housing filled with oil

k

JG

!!!

kt!ct

!!

M(t)

G

H = JG

!!k

The angular momentum about the center of mass of thedisc is

Since the rotary inertia JGand the unit vector kdo notchange with time

M =dH

dt= J

G

!!!k

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M(t)k

External momentacting on disk

!"#! k

t!k

Restoring moment due to shaft stiffness

!

! ct

d!

dtk

Damping moment due to oil in housing

!"#

!JG

d2!

dt2 k

Inertial moment

! "# $#

= 0

JG

!!!

kt!ct

!!

M(t)

G

Summing the moments about the axis ofrotation G, we obtain

2

2 ( )

G t t

d dJ c k M t

dt dt

! !!+ + =

which can be written as the following scalar equation

Remark: All linear single degree-of-freedom systems aregoverned by a linear second-order ordinary differentialequation with an inertia term, a stiffness term, a damping

term, and a term related to the external forcing imposedon the system.

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3.3.1: Natural Frequency: It is frequency by which a systemtends to oscillate in the absence of any damping or driving force.

In other words, it is the in-builtfrequency of the system.

https://www.youtube.com/watch?v=3mclp9QmCGs(Tacoma Bridge Disaster)

md2x

dt2 + c

dx

dt+ kx = f(t)

No damping or driving force

md

2x

dt2 + kx = 0

!x = Asin k

mt

!

"#

$

%&+B cos

k

mt

!

"#

$

%& = Asin !nt( )+ Bcos !nt( )

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3.3.1: Natural Frequency

Translation Vibrations: Natural Frequency

For translation oscillations of a single-degree-of-freedom system, we define the natural frequency as

2 2

N/m N/m 1 12 rad/s rad/s

kg Ns /m s sn n

kf

m

! "

# $= = = = = %& '

& '

2

2

m/s 1 12 rad/s

mn n

st

gf

s s! "

#

$ %= = = =& '

& '( )

where fn= "n/(2#) is also the natural frequencyexpressed in Hertz (Hz = 1/s).

For system that exhibit oscillations along the vertical

direction, since k"st= mg, this definition can also bewritten as

Remember this asa formula now

Note the changes in units

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Rotational Vibrations: Natural Frequency

!n

= 2!fn = kt

J rad/s kt

J= Nmkg !m2

= NmN !s2 /m !m2

= 1s2

"#$ %&'

Design Guidelines

For single degree-of-freedom systems, we can state

that

An increase in the stiffness or a decrease in the massor mass moment of inertia increases the naturalfrequency

A decrease in the stiffness or an increase in the massor mass moment of inertia decreases the naturalfrequency

The greater the static displacement, the lower thenatural frequency (for vertical oscillations)

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Period of Undamped Free Oscillations

1 2

sn n

T f

!

"

= =

Thus, increasing the natural frequency decreases theperiod and vice versa.

T fn

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3.3.2 Damping Factor

Translation Vibrations: Damping Factor

md2x

dt2 + c

dx

dt+ kx = f(t)

No driving force

md

2

xdt

2 + c dx

dt+ kx = 0

!p =

"c c2 " 4km

2k, wherex = ept

!" =c2

4km=

1

2

c

km

=

c

2m

m

k=

c

2m#n

ENME361, Fall 2015


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3.3.2 Damping Factor

Translation Vibrations: Damping Factor

2

Ns N = = 1

2 2 m kg/s kg!m/s2

n

n

cc c

m kkm

!"

!

# $= = = % &'

We see that $is a non-dimensional quantity

Critical Damping , Under-damping, and Over-damping

We define the quantity cc, called the critical damping,

as( )22 2 kg/s = Ns / (ms) = Ns/mc nc m km!= =

Then, the damping factor can be written as

c

c

c

! =

Remember this asa formula now; we

shall do itsderivation later

and !( )c=c

c

=1

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We now define four regions

Undamped:$= 0

Underdamped: 0 < $< 1

Critically damped: $= 1

Overdamped: $> 1

Rotational Vibrations: Damping Factor

2 2

Nsm N = = 1

2 kg!m /s kg!m/s2

t t

n t

c c

J k J!

"

# $= = % &

' (

We see that $for rotational motions is also a non-dimensional quantity.

ENME361, Fall 2015


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md2x(t)

dt2

kg(m/s2 )

! "# $#

+cdx(t)

dt(Ns/m)(m/s)

!"$

+kx(t)

(N/m)m

%

= f(t)

Governing Equation of Motion in Terms of Natural Frequencyand Damping Factor

subject to the initial conditions

d2x

dt2 +

c

m

dx

dt+

k

mx = f(t)

We divide this equation by mto obtain

(0)(0) (m), (m/s)

o o

dxx X V

dt= =

Consider the SDOF governing equation:

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d2x

dt2 +c

m

dx

dt +!n2

x =f(t)

m !d2x

dt2 + 2!"ndx

dt +!n2

x =f(t)

m

which can be written as

Also considering a dimensionless time

!n

2=

k

m

! = c2m!

n

!c

m=2!"

n

! ="

nt

!"n

2d2x

d#2 + 2$"

n

2dx

dt+"

n

2x =f(t)

m

%d2x

d#2 + 2$

dx

dt+x =

f(t)

m"n

2%

d2x

d#2 + 2$

dx

dt+x =

f(t)

k.

ENME361, Fall 2015


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Example 3.7: Effects of system parameters on thedamping ratio

An engineer finds that a vertical single degree-of-

freedom system with mass m, damping c, and springconstant k has too much static deflection "st. The

engineer would like to decrease "stby a factor of 2,

while keeping the damping ratio constant.

We shall show that there three ways in which one canachieve this goal.

!st

*=

!st

2(by problem defintion)

We always have

!n =

g

"st

#!n

*=

g

"st

* =

g

" / 2= 2

g

"st

= 2!n

(follows from the problem definition)

! =c

2m"n

=!* =c

*

2m*"n

* =

1

2

c*

2m*"n

(by problem definition)

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First Choice

For the first choice, let cremain constant (c=c*).

!! =!* yields,c

2m!n

=

1

2

c

2m*!n

!m* =m

2

CHOICE 1:

c*= c, m

*=

m

2, k

*= 2k

From !n

*= 2!

n,

k*

m* = 2

k

m"

k*

m* = 2

k

m" 2

k*

m= 2

k

m using m* =

m

2

#

$%

&

'(

" k* = 2k

ENME361, Fall 2015


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Second Choice

For the second choice, let mremain constant, i.e,

m*=m.

!! =!* yields,c

2m"n

=

1

2

c*

2m"n

" c* = 2c

From !n

*= 2!

n,

k*

m* = 2

k

m!

k*

m* = 2

k

m!k* = 2k using m* =m"#

$%

CHOICE 2:

c*= 2c, m

*=m, k

*= 2k

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Third Choice

For the last choice, let kremain constant, i.e., k=k*.

From !n

*= 2!

n,

k*

m* = 2

k

m!

k*

m* = 2

k

m!m* =

m

2 using k* =k"#

$%

Please note thatwe start from the

second condition,since kis involved

!" ="* yields,c

2m#n

=

1

2

c*

2m*#n

$c

2m#n

= 2c*

2m#n

using m* =m

2

%

&'

(

)*

$ c* =c

2

CHOICE 3:

c*=

c

2, m

*=

m

2, k

*=k

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3.4: Governing Equations with Different Types of Damping

Replace the viscous damping term with the appropriate expression thatrelates the force to velocity

Coulomb or Dry Friction Damping

md2x

dt2 + mgsgn( !x)

Nonlinear dryfriction force

! "# $#+ kx = f(t)

Fluid Damping

md

2x

dt

2 + c

d|!x |!x

Nonlinear fluiddamping force

!"# $#

+ kx = f(t)

Structural Damping

md

2x

dt2 + k!"sgn( !x) | x |

Nonlinear structuraldamping force

! "## $##+ kx = f(t)

md

2x(t)

dt2

kg(m/s2 )

! "# $#

+cdx(t)

dt(Ns/m)(m/s)

!"$

+kx(t)

(N/m)m%

= f(t)

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3.4: Governing Equations with Different Types of Damping

3.5.1: System with Base excitation

k cy(t)

x(t)m m

k(x!y) c(!x ! !y)

m!!x

Package

"

X

Yj

Base

O

md2x

dt2 + c

dx

dt+ kx =c

dy

dt+ ky (here F

ext =F

base =c

dy

dt+ ky)

The governing equation of motion is

The displacements y(t) and x(t) are measured from afixed point O located in an inertial reference frame and afixed point located at the systems static-equilibrium

position, respectively.

Example: Disturbance impartedon a vehicle by the road

Equivalent response of the mass m

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If the relative displacement is desired, then we let

)()()( tytxtz !=

and we get

2

2

2

2

dt

ydmkz

dt

dzc

dt

zdm !=++

where is the acceleration of the base.)(ty!!

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3.5.2: System with Unbalanced Rotating Mass

t! "=

2

2 x

d x dxM c kx N

dt dt + + =

aP O

=

L

!!

!e1 + L!

!

2

e2

e2 =! sin!i+ cos!j

e1e2

!

mo

"t

"t

Nx =!

mo (!!

x! !"

2

sin"

t)N

y =m

o!"

2 cos"t

Fan

Clothesdryer

kx xc!

Mx!!

M Ny

Nx

O

M

mo%

k c

wtY

X

i

jO O

Systems (fans, dryers) have certain degree of unbalance. This unbalance is modeledas a mass mo[located as a fixed distance %from the center of rotation of the spring-

mass (M)-damper system] that rotates with an angular speed ". Calculate thisunbalanced force

Here Nxand Nyare the reaction forces exerted on the point O (which is a point on the bodyM; hence Nxand Nyserve as applied forces on mass M) caused by the rotation of mass mo.In order to obtain Nxand Nywe first need to obtain the acceleration of mass mo.

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Consider a fixed point O located in an inertial reference frame. We first need to findthe acceleration of mass mowith respect to this point.

e1e2

!

mo

"t

"t

O

P

rP/ !O

=rO/ !O

+ rP/O

=rO/ !O

+ !e1

"d

dtrP/ !O( ) =

d

dtrO/ !O( )+ !

d

dte

1( ) =d

dtrO/ !O( )+ ! "k#e1( ) =

d

dtrO/ !O( )+ !"e2

"d

2

dt2 r

P/ !O( ) =d

2

dt2 r

O/ !O( )+d

dt!"e

2( ) = !!x( )i+ !"d

dte

2( ) = !!x( )i+ !" "k#e2( )

d2

dt

2 r

O/ !O( ) = !!x( )iis the acceleration of mass M or acceleration of point O w.r.t. point !O$

%

&'

(

)

= !!x( )i* !"2e1 = !!x( )i* !"2 sin"ti+ cos!tj( ) = !!x* !"2 sin"t( ) i* !"2 cos"tj

!Force exerted by the link on mass mo

: mo

!!x" #$2 sin$t( )i" #$2 cos$tj%& '(!Force exerted by the mass m

oon the link: "m

o!!x" !"2 sin"t( )i" !"2 cos"tj#$ %&

!Force exerted by the support O on the link (in order to keep the link in equilibrium) :

mo

!!x" !"2 sin"t( )i" !"2 cos"tj#$ %&

!Force exerted by the link on the support O: "mo

!!x" #$2 sin$t( )i" #$2 cos$tj%& '(This is the FORCE exerted on the mass M due to the rotation of mass mo

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!M!!x + c!x + kx = Nx ="m

o !!x" !"2 sin"t( )

# M +mo( ) !!x + c!x + kx =mo!"2 sin"t

#!!x +c

m!x +

k

mx =

mo!"

2

msin"t m = M +m

o( )$% &'

!"n =

k

m

!#st

=

mg

k

Magnitude of the unbalanced force = mo$"

2

m

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3.6: Lagranges Equations

Let us consider a system with Ndegrees of freedom

that is described by a set of N generalized coordinatesqi, i= 1, 2, ..., N.

These coordinates are unconstrained, independent

coordinates; that is, they are not related to each other

by geometrical or kinematical conditions.

Recall

Generalized coordinatesthe minimum number ofindependent coordinates needed to describe a

system

ENME361, Fall 2015

C Si l D f F d S


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where

: generalized velocities

T: kinetic energy of the system

V :potential energy of the system

D: the Rayleigh dissipation function

Qj: generalized force that appears in the jthequation

Lagranges equations have the form

NjQq

V

q

D

q

T

q

T

dt

d

jjjjj ,...,2,1

==

!

!+

!

!+

!

!

"##%

&&(

!

!

!!

!qj

ENME361, Fall 2015

Ch 3 Si l D f F d Sjq!jq!


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The generalized forces Qjare given by (assuming the system tobe composed of lbodies)

Qj = F

l! "

rl

"qjl

# + Ml!" l

" !qjl

#where

Fl: vector representations of the externally

applied forces on thelth

body

Ml: vector representations of the externally

applied moments on the lthbody

rl: position vector to the location where the force

Fl is applied!

l: lthbodys angular velocity about the axis alongwhich the considered moment is applied

ENME361, Fall 2015

Ch t 3 Si l D f F d S t


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Linear Vibratory Systems

For vibratory systems with linear inertial characteristics,

stiffness characteristics, and viscous dampingcharacteristics, the quantities T, V, and Dtake thefollowing form

T =1

2m

jn!q

j!qn

n=1

N

!j=1

N

!

V =1

2k

jnq

jq

nn=1

N

!j=1

N

!

D =1

2cjn

!qj!qn

n=1

N

!j=1

N

!

Nis the number of degrees of freedomMjn ; inertia coefficients

kjn; stiffness coefficients

cjn; damping coefficients

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Single-Degree-of-Freedom Systems: N= 1

11111 Qq

V

q

D

q

T

q

T

dt

d=

!

!+

!

!+

!

!

"##$

%

&&'

(

!

!

!!

where

Q1 = F

l

l

! "#r

l

!q1

+ Ml

l

! "#

l

! !q1

Linear Single-Degree-of-Freedom Systems

T =1

2m

jn!q

j!qn

n=1

1

!j=1

1

! =1

2m

11!q1

2

V =

1

2 kjnqjqnn=1

1

!j=1

1

! =

1

2 k11q12

D =1

2cjn

!qj!qn

n=1

1

!j=1

1

! =1

2c11

!q1

2

T =1

2m

e

!q1

2

V =

1

2 keq12

D =1

2ce

!q1

2

11

11

11

e

e

e

m m

k k

c c

=

=

=

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Upon substituting these results into Lagrangesequations, we obtain

d

dt

!T

! !q1

!"#

$%&'(T

!q1

+!D

! !q1

+!V

!q1

=Q1

d

dt

!

! !q1

1

2m

e!q1

2!"#

$%&

!

"#$

%&'

((q

1

1

2m

e!q1

2!"#

$%&+

'' !q

1

1

2ce!q1

2!"#

$%&+

''q

1

1

2k

eq1

2!"#

$%& = Q

1

ddt

me!q1( )! 0+ ce !q1 + keq1 =Q1me!!q1+ c

e

!q1+ k

eq1 =Q

1

or in expanded form2

1 1

12 ( )e e ed q dq

m c k q f t dt dt + + =T =

1

2

me

!q1

2

V =1

2keq1

2

D =1

2ce

!q1

2

ENME361, Fall 2015

Chapter 3 Single Degree of Freedom Systems


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Example 3.9; Equation of motion for a linear singledegree-of-freedom system

1 , ( ) , = , and 0l l lq x f t x= = =F j r j M

k c

mj

f(t)

x

1 0 ( ) ( )lll j

xQ f t f t q x

! != " + = " =! !#

r jF j

We first note that

The generalized force is

The system kinetic energy and potential energy are,respectively,

T =

1

2m

!x

2

V =

1

2 kx

2

and the dissipation function is

2

2

1xcD !=

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Upon substituting these results into Lagrangesequations, we obtain

d

dt

!

!!x

1

2m!x

2!"#

$%&

!"#

$%&' (

(x1

2m!x

2!"#

$%&+

''!x

1

2c!x

2!"#

$%&+

''x

1

2kx

2!"#

$%& = f(t)

d

dtm!x( )! 0+ c!x + kx = f(t)

Thus, the governing equation of motion is2

2 ( )

d x dxm c kx f t

dt dt + + =

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Example 3.10: Equation of motion for a system that translatesand rotates

k cr

(t)

G

m, JG !

Y

Z

k

i

j

O

We can choose either xor &as thegeneralized coordinate. Note x= r#.

We choose q1as the generalized

coordinate and recognize that

q1 =!, F

l = 0, M

l = M(t)k, and

l = !!k

Then the generalized force is

Q1= 0+ M

l!"

l

" !q1l

$ = M(t)k!"!

"!k= M(t)

The potential energy of the linear spring is

2 2 2 21 1 1( )2 2 2

V kx k r kr ! != = =

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Then the equivalent stiffness of the system is2

ek kr=

The kinetic energy of the disc is the sum of the kineticenergy due to translation of the center of mass of the discand the kinetic energy due to rotation about the center of

mass. Thus,

T =1

2m!x

2

Translationkinetic energy

!

+1

2J

G

!!2

Rotationalkinetic energy

!"#

=

1

2mr

2 !!2+

1

2J

G

!!2

=

1

2mr

2+

mr2

2

!

"#

$

%&!!

2=

1

2

3mr2

2

!

"#

$

%&!!

2

since JG= mr2/2.

k cr

(t)

G

m, JG !

Y

Z

k

i

j

O

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Then the equivalent mass of the system is23

2e

m mr=

The dissipation function takes the form

D =1

2c!x

2=

1

2c r

!!( )2

=

1

2cr

2( )!!2

Then the equivalent damping coefficient is2

ec cr=

The governing equation of motion then becomes

2

2

22 2 2

2

( )

3( )

2

e e e

d dm c k M t

dt dt

d dmr cr kr M t

dt dt

! !!

! !!

+ + =

+ + =

k cr

(t)

G

m, JG !

Y

Z

k

i

j

O

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Natural Frequency and Damping Factor

( )

2

2

2

2

2

3 2 3

2 2 3 2 2 3 6

en

e

e

e n

k kr k

m mr m

c cr c

m mr k m km

!

"!

= = =

= = =

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Glossary Chapter 3

Base excitation an input applied to the base of a system

Critically damped damping factor is equal to one

Damping factor a non-dimensional quantity relating the

amount of viscous dissipation in a system to the stiffness

and mass of the systemLagranges equations equations of motions derived from

a formulation based on the system kinetic energy,potential energy, and work expressed in terms of

generalized coordinates

Natural frequency the frequency at which an undamped

system will vibrate in the absence of external forces

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Overdamped system damping factor is greater thanone

Period of oscillation the reciprocal of a systemsoscillation frequency given in Hertz

Static equilibrium position the position of a system at

restUndamped system a system without damping

Underdamped system damping factor is greater than

zero and less than one

vibrations chapter 3 governing equations

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