vibrations chapter 3 2013 1

7/28/2019 Vibrations Chapter 3 2013 1

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Chapter 3 Single Degree-of-Freedom Systems

1ENME 361, Spring 2013

ENME 361

Vibrations, Control, and Optimization I

Spring 2013

Chapter 3

Governing Equations

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



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

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

by us ing force balance and moment balance

methods

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

by us ing Lagranges equations

Determine the equivalent mass, equivalent stiffness,and equivalent damping of a single degree-of-

freedom system

Determine the natural frequency and damping factorof a system


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Consider the principle of l inear momentum

( )d mm

dt

vF p a

F: The net external force vector acting on the system

0mF a =

3.2.1 Force-Balance Methods

-ma : inertial force

The sum of the external forces and inertial forces actingon the system is zero




Vertical Vibrations of a Spring-Mass-Damper System

( )f t

m

ckDetermine the governingequation of motion


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Vertical Vibrations of a Spring-Mass-Damper System

k

k

c

c

m x

st

mgm

mg f(t)

mx

k(x+st)

k(x+st)X

Yj

iO

cx

cx

L

xst



A force balance along thej direction results in

2

Damping force Inertia forceExternal forces acting Springforceactingacting on masson system onmass

( ) 0st

dx d xf t mg kx k c m

dt dt

2

j j j j j

m

mg f(t)

mr

k(x+st) cr cx

mx


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Therefore, the equation o f motion for oscillations about

the static-equilibrium posit ion is

2

2( )

d x dxm c kx f t

dt dt

Static-Equilibrium Position

The static-equilibrium position of a system is theposition that corresponds to the systems rest state;

that is, a position when

Thus,

0

stkx k mg

kx mg mg

x

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

0.x x



Force Transmit ted to Fixed Surface

k(x+st)

k(x+st) cx

cx

FR

Static Dynamiccomponent component

R st

dxF k kx c

dt

We consider only the dynamic part

of the reaction force, therefore

RddxF c kxdt


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Horizontal Vibrations of a Spring-Mass-Damper System

( )f tm

c

k

Determine the governing equation of motion



Horizontal Vibrations of a Spring-Mass-Damper System

k

k

c

x

m

f(t)

mx

cx

kx

m

kx

cx

X

Y

i

j

g

O

2

External forces Springforce Damping Inertia forceacting on system acting on forceacting

mass on mass

( ) 0dr d r

f t kx c mdt dt

2

i i i i

which also results in

2

2( )

d x dxm c kx f t

dt dt


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Punch Press

Wing Vibrations

Some examples of external forces:

Fluctuating air pressure loadingsuch as that on the wing of an

aircraft

Fluctuating electromagnetic forcessuch as in a loudspeaker coil

Electrostatic forces that appear insome microelectromechanical

devices

Forces caused by an unbalancedmass in rotating machinery

Buoyancy forces on floating systems

Impacts



3.2.2 Moment-Balance Methods

M(t)

Disc with rotary inertiaJGabout rotation axis

Axis of rotation

Shaft with equivalenttorsional stiffnesskt

Housing fi lled with oil

k

Principle of angular momentum:

G

dJ

dt

HM k 0GJ

M k =

Inertialmoment


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3.2.2 Moment-Balance Methods

G

J

kt ct

M(t)

G

2

2

Restoring moment due Damping moment due Inertial momentExternal momentto shaft stiffness to oil in housingactingondisk

( ) 0t t Gd d

M t k c Jdt dt

k k k k

Summing the moments about the axis of rotation G, we obtain



2

2

Restoring moment due Damping moment due Inertial momentExternal momentto shaft stiffness to oil in housingactingondisk

( ) 0t t Gd d

M t k c Jdt dt

k k k k

2

2( )G t t

d dc k M t

dt dt

which can be written as the following scalar equation


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Remark: All linear single degree-of-freedom systems aregoverned by a linear second-order ordinary differential

equation with an inertia term, a stiffness term, a damping

term, and a term related to the external forcing imposed

on the system.

2

2( )G t t

d dJ c k M t

dt dt

2

2( )

d x dxm c kx f t

dt dt



Example: Governing equation of motion for a planar pendulum

Z

(xp,yp,0)

X

Y

O

L

e1e2

k

Q

i

j

h

P

We shall determine the velocity and acceleration o f

the planar pendulum P with respect to point O

when it rotates with an angular velocity of . Thus,

2

P O Q O P Q

h L

r r r

j e

k

The position vectors are


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Governing Equation of Motion

2

2

( )d x c dx k f t x

dt m dt m m

We divide this equation by m to obtain

Consider the SDOF governing equation:

2

2( )

d x dxm c kx f t

dt dt

If we define2n

k

m 2

n

c

m

22

2

12 ( )

n n

d x dxx f t

dt dt m



3.3.1: Natural Frequency

Translation Vibrations: Natural Frequency

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 frequency expressedin Hertz (Hz = 1/s).

For translation oscillations of a single-degree-of-freedom

system, we define the natural frequency as

For system that exhibit oscil lations along the vertical

direction, since kst = mg , this definition can also be writtenas


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Rotational Vibrations: Natural Frequency

2 2 2 2

Nm Nm 12 rad/s

kg m N s /m m st

n n

kf

J

Design Guidelines

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

that

An increase in the sti ffness or a decrease in the massor mass moment of inertia increases the natural

frequency

A decrease in the stif fness or an increase in the massor mass moment of inertia decreases the natural

frequency

The greater the static d isplacement, the lower thenatural frequency



Period o f Undamped Free Oscillations

1 2s

n n

Tf

Thus, increasing the natural frequency decreases the

period and vice versa.

T fn


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3.3.2 Damping Factor

Translation Vibrations: Damping Factor

2

Ns N= =1

2 2 m kg/s kgm/s2

n

n

cc c

m kkm

We see that is a non-dimensional quantityCritical Damping , Under-damping, and Over-damping

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

Then, the damping factor can be written as

c

c

c

We define the quanti ty cc, called the critical damping, as

2Ns/m kgm/s s/m= =1

kg1/s kg1/s



We now define four regions

Undamped: = 0Underdamped: 0 < < 1Critically damped: = 1Overdamped: > 1

2 2

Nsm N= =1

2 kgm /s kgm/s2

t t

n t

c c

J k J

We see that for rotational motions i s also a non-dimensional quantity.

Rotational Vibrations: Damping Factor


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We notice that the product nt is dimensionless since(1/s)(s) = 1. Therefore, we define = nt and obtain

2

2

12 ( )

d x dxx f

d d k

2

22

12 ( )n n

d x dxx f t

dt dt m

Notice that in the absence of forcing, the motion of avibratory system can be described by just one system

parameter---- .

2

22 0

d x dxx

d d

in the absence of forc ing

Governing Equation of Motion in Terms of Natural Frequency

and Damping Factor



0 5 10 15 20 25 30 350

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

(nt)

y()

(x()/

st

)

= 0.15If

n=100 rad/s and

st=2 mm, then

=10 t=10/100=0.1 sy()=1 x()=12=2 mm

st

n

xy

t


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3.5.1: System with Base Exci tation

2

2

d x dxm c kx

dt dt

dyc ky

dt

The governing equation of motion is

The displacements y(t) and x(t) are measured from a fixed

point O located in an inertial reference frame and a fixedpoint located at the systems static-equilibrium position ,

respectively.



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

Fan

Clothes

dryer



3.5.2: System with Unbalanced Rotating Mass

t

2

2 x

d x dxc kx N

dt dt

2 2( sin ) ( cos )P O t t a i j

2

2

( sin )

cos

x o

y o

N m x t

N m t


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Thus, the governing equation i s

2

22

sn( ) ioo

d x dxM m c kxdt dt

m t

which is rewritten as2

2

2

( )sn2 i

n n

d x dxx

dt dt

Ft

m

where

2( )

o

n

o

m M m

k

m

F m



Example: Governing equation of motion for an inverted

pendulum


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Example: Governing equation of motion for an inverted

pendulum

m

c

ka

g

O

b

r

tk

OI

m

c

ka

g

O

b

r

tk

OI



3.6: Lagranges Equations

Let us consider a system with N degrees of freedom

that is described by a set ofN generalized coordinates

q i, 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 coord inatesthe minimum number ofindependent coordinates needed to describe a

system


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where

: generalized veloci ties

T : kinetic energy of the system

V : potential energy of the system

D : the Rayleigh dissipation func tionQj : generalized force that acts on thejth mass

Lagranges equations have the form

NjQq

V

q

D

q

T

q

T

dt

dj

jjjj

,...,2,1

jq



jqjq

The generalized forces Qj are given by

l lj l l

l lj j

Qq q

r F M

where

Fl : vector representations of the externally

applied forces

Ml : vector representations of the externally

applied moments

rl : position vector to the location where the force

is applied

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


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Linear Vibratory Systems

The quantities T, V, and D take the following fo rm

1 1

1 1

1 1

1

2

1

2

1

2

N N

jn j n

j n

N N

n j n

j n

N N

jn j n

j n

T m q q

V k q q

D c q q

N is the number of degrees of f reedom

Mjn ; inertia coefficientskjn ; stiffness coefficients

cjn ; damping coefficients



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

1

1111

Qq

V

q

D

q

T

q

T

dt

d

where

1

1 1

l ll l

l l

Qq q

r F M

Linear Single-Degree-of-Freedom Systems1 1

211 1

1 1

1 12

11 11 1

1 12

11 11 1

1 1

2 2

1 1

2 2

1 1

2 2

jn j n

j n

jn j n

j n

jn j n

j n

T m q q m q

V k q q k q

D c q q c q

21

2

1

21

1

2

1

2

1

2

e

e

e

T m q

V k q

D c q

11

11

11

e

e

e

m m

k k

c c


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Upon substitu ting these results into Lagranges

equations, we obtain

1

1 1 1 1

2 2 2 2

1 1 1 1 1

1 1 1 1

1 1 1 1

1 1 1 1

2 2 2 2

0

e e e e

e e e

d T T D V Qdt q q q q

dm q m q c q k q Q

dt q q q q

dm q c q k q Q

dt

or in expanded form

2

1 11 12e e e

d q dqm c k q Q

dt dt

21

21

21

1

2

1

21

2

e

e

e

T m q

V k q

D c q

1 1 1 1e e em q c q k q Q



Example 3.9; Equation of motion for a linear single

degree-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 ( ) ( )l

l

l j

xQ f t f t

q x

r jF j

We firs t note that

The generalized force is

The system kinetic energy and potential energy are,

respectively,

2 21 12 2

T mx V kx

and the dissipation func tion is2

2

1xcD

1

1111

Qq

V

q

D

q

T

q

T

dt

d


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Upon substitu ting these results into Lagranges

equations, we obtain

2 2 2 21 1 1 1 ( )2 2 2 2

0 ( )

dmx mx cx kx f t

dt x x x x

dmx cx kx f t

dt

Thus, the governing equation of motion is

2

2( )

d x dxm c kx f t

dt dt



Example 3.11: Governing equation for an inverted pendulum

The total rotational inertia of the

system is

1 2O O OJ J J

where, from the parallel axis theorem

2 2 2 21 1 1 1 1 1

2

2 222 2 2 2 2 2

2 2

5 5

1 1

12 2 3

O

O

m r m L m r L

Lm L m m L

We choose as the generalizedcoordinate and the system kinetic energytakes the form

2 21 2

2 2 2 21 1 2 2

1 1

2 2

1 2 1

2 5 3

O O OT J J J

m r L m L


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For small rotations about the upright posit ion, we can

express the translation of mass m1

as

1 1x L

Then, the system potential energy can be expressed as

2 2 221 1 1 2

2 221 1 1 2

1 1 1

2 2 2 2

1

2 2

LV kx m gL m g

LkL m gL m g

The dissipation function takes the form

2 2 21 1

1 1

2 2D cx cL



From the form ofT, V, and D, we find that2 2 2

1 1 1 2 2

2 21 1 1 2

21

2 1

5 3

2

e

e

e

m m r m L m L

Lk kL m gL m g

c cL

Thus,

2 21 1 1 2

1 2

2en

e O O

LkL m gL m g

k

m J J

We note that in order forke to bepositive,

22

21121

LgmgLmkL


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Example 3.11: Governing equation for an inverted pendulum

Equation of Motion :

0e e em c k

Natural Frequency:

2 21 1 1 2

1 2

2en

e o o

LkL m gL m g

k

m J J



Natural frequency of pendulum sys tem

In this case, the pendulum is hanging down.

The only changes to the previous results occurs in

the expression for the potential energy; that is,

2 2 221 1 1 2

2 221 1 1 2

1 1 1

2 2 2 2

1

2 2

LV kx m gL m g

LkL m gL m g

and the natural frequency becomes

221 1 1 2

1 2

2en

e O O

LkL m gL m g k

m J J


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We rewrite this expression as

2 21 1

1 1

22

1 1 21

1

21

5

n

m Lk m gL

m L

rm L

L

Ifm2


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Then the equivalent s tiffness of the system is2

ek kr

The kinetic energy of the disc is the sum of the kinetic

energy due to translation of the center of mass of the disc

and the kinetic energy due to rotation about the center of

mass. Thus,

2 2 2 2 2

Translation Rotationalkinetic energy kinetic energy

2 22 2 2

1 1 1 1

2 2 2 2

1 1 3

2 2 2 2

G GT mx J mr J

mr mr mr

since JG = mr2/2.

k cr

(t)

G

x

m, JG

Y

Z

k

i

j

O



Then the equivalent mass of the system is23

2em mr

The dissipation function takes the form

2

2 2 21 1 1

2 2 2D cx c r cr

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 d

m c k M t dt dt

d dmr cr kr M t

dt dt

k cr

M(t)

G

x

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



Addi tional Examples fo r Use of Lagranges Equations

NjQq

V

q

D

q

T

q

T

dt

dj

jjjj

,...,2,1

where

: generalized veloci ties

T : kinetic energy of the system

V : potential energy of the systemD : the Rayleigh dissipation function

Qj : generalized force that acts on thejth mass

jq


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Example 3.13: Governing equation for a translating

system with a pre-tensioned or pre-compressed springThe horizontal spring is pre-

tensioned with a tension T1, which is

produced by an initial extension of

the spring by an amount o; that is,T1 = k1o.The kinetic energy of the system is

21

2T mx

The potential energy of the system is

2 2

1 2

1 1

( ) 2 2oV x k L k x



T1

Fs

kx

L

2 2x L 2 2 2

2 2

1 ( / )

11

2 2

L L x L L x L L

x L xL L

L L

where

Then, the potential energy becomes2

2

2

1 2

1 1( )

2 2 2o

L xV x k k x

L

We note that2

1 2

31 1

2 2

12

2

2

o

o

V L x xk k x

x L L

k k xk x

L L

Tk x

L


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Since D = 0, Q1 = 0, q1 = x, and me = m, the Lagrange

equation becomes

21

22

0

0

d T T V

dt x x x

Td xm k x

dt L

Consequently, the natural frequency is

m

LTkn

/12

If the spring of constant k1 is compressed instead of being

in tension, then we can replace T1 by T1 and we find that2 1 /

nk T L

m



It is seen that the natural frequency can be made very low

by adjusting the compression of the spring with sti ffness

k1.

At the same t ime, the spring with s ti ffness k2 can be made

stiff enough so that the static disp lacement of the system

is not excessive.

This type of system is the basis of at least one commercial

product [Minus K Technology (www.minusk.com)].

m

k2

k1, T1

L

x2 1 /n

k T Lm


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Glossary Chapter 3

Base excitation an input applied to the base of a system

Critically damped damping fac to r is equal to one

Damping factor a non-d imensional quant ity relating the

amount of viscous d issipation in a system to the stiffness

and mass of the sys tem

Lagranges equations equations of motions der ived from

a formulation based on the system k inetic energy,

potential energy, and work expressed in terms of

generalized coordinates

Natural frequency the frequency at which an undampedsystem will vib rate in the absence of external forces



Overdamped system damping fac to r is greater than

one

Period of oscillation the reciprocal of a systems

oscillation f requency given in Hertz

Static equilibrium posit ion the pos it ion of a system at

rest

Undamped system a system without damping

Underdamped system damping fac to r i s g reater than

zero and less than one

vibrations chapter 3 2013 1

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