ME_302-L-Forced_Vibrations_V3.pdf

-ME 302-Theory of Machines II

HARMONIC EXCITATION OF

ME-302 THEORY OF MACHINES II Page: 1/24Kvan Azgn

HARMONIC EXCITATION OFSINGLE DEGREE-OF-FREEDOM

SYSTEMS

eq eq eq 0 0Initial Conditions

m x c x k x F(t), x 0 x , x 0 x

Forced Vibrations

x t is the total forced vibration response (work is done on the system by external forces)


For >0, only xp(t) has contribution to the long term behavior of x(t). That is why xp(t) is also called the steady state solution.

h p

h eq eq eq

p

x t x t x t

x t is the homogenous solution, the free vibration response (soln. of m x c x k x 0)

x t is t

he particular solution and it depends on the form of F(t)

ss pt

x t = lim x t =x t which is independent of initial conditions

eq eq eq 0 0Initial Conditions

m x c x k x F(t), x 0 x , x 0 x

Forced Vibrations


hx t x t px t

eq eq eq 0m x c x k x F(t) F sin t

Amplitude ofexcitation (N)

Harmonic Excitation

Frequency ofexcitation (rad/sec)

Harmonic excitation (with single frequency) commonly occur in practical problems, thus deserves special attention


special attention

Sinusoidalwavy road

c k

c k

M

Suspension

Any periodic excitation can be approximated by a truncated Fourier Series (sinusoidal) Expansion

eq eq

eq

eq eq

2n

eq

m x k x F(t)

k F(t)x x

m m

F(t)x x

m

where x is the independent generalized coordinate, is the generalized mass (meq or Ieq)and is the generalized stiffness (keq or kt,eq)

eqm

k

Harmonic Excitation of Undamped SDOF Systems


and is the generalized stiffness (keq or kt,eq)

When F(t) is harmonic, i.e.,

We need to introduce the complex excitation and complex solution

0 02 20 0n n

eq eq

F(t) F sin t OR F(t) F cos t

F Fx x sin( t) OR x x cos( t)

m m

0( )

i t

i t

F t F e

x t Xe

eqk

0 2 0

( )

i t

i tni t

eq

F t F e Fx x e

mx t Xe

2 i tx t X eiWhere



2 02

i t i t i t

n

eq

FX e Xe e

mi

22

i i

n

eq

Fe e

mi

X

2

2 0

2 0

2

2

since cos sin

i in

eq

in

eq

i

Fe e

m X

Fe

m X

e i

F

i

i

Note that this eq. is the char. equation where s terms are replaced by i terms



0

2 0

0 0

2

2 2

0

2

2

Imag. part : 0 sin sin 0 0 or

0 if Real part : cos

if

1

eq

n

n

neq

eeq neq

n

Fi

m X

F

m X

F F F

mk

Xk

2

2

1 for

1

n

q

staticncase

0 2 0

w hen

( )

n

n

n

n

i t

i tni t

eq

F t F e Fx x e

mx t tXe

2

2 n nn

i i t

n

ti ix t X X eteWhere



2 2 02

n n n ni t i t i t i tn

e

n n

q

FXe X e X

mi e etit

02

in

eq

Fi e

m X

00

2

2

since cos sin

2

i

eq

i

eq

i

n

n

Fe

m X

Fe

m X

e i

m XF

i

i



0

0

0

0

0

2Imag. part : 2 sin sin

Real part : 0 cos , since sin 02

2sin 1

2

eq

eq

n

n

n

n

eq

eq

eq

m XFi

m X F

F

m X

m X FX

F m

i

0 0 0Input : ( ) sinc so

i t i FF t F e tF t

Substituting X and into the assumed solution

2

0 1

1

eq

FX

k



0

Re( ( )) Im( ( ))

Re

0

( ( )) Im( ( ))

0

2

Input : ( )

0 if For , where

i

f

Output :

For ,

sin

sin

c s

cos

o

F t F t

i t n

n

nx t x t

i t

n

i FF t F e

x t Xe

x t tX

t

i X t

e

F t

X t

Re( ( )) Im( ( ))

s2

in

o2

c s

x t x t

tX t i tX t

2

1

eqstatic

ncase

0

2

eqn

FX

m

eq eq eq

eq eq

eq eq eq

2n n

eq

m x c x k x F(t)

c k F(t)x x x

m m m

F(t)x 2 x x

m

where x is the independent generalized coordinateis the generalized mass (meq or Ieq)is the generalized damping (c or c )c

eqm

Harmonic Excitation of Damped SDOF Systems


is the generalized damping (ceq or ct,eq)is the generalized stiffness (keq or kt,eq)

When F(t) is harmonic, i.e.,

We need to introduce the complex excitation andcomplex solution

eqc

0 0F(t) F sin t OR F(t) F cos t

0( )

i t

i t

F t F e

x t Xe

eqk

0 2 0

( )2

i t

i tn ni t

eq

F t F e Fx x x e

mx t Xe

2

i t

i t

ix t X e

x t iX eWhere



2 2 02

i t i t i t i tn n

eq

FX e X e Xe e

mi i

2 2 02

i i i

n n

eq

Fei e

mi e

X

2

2 0

2 0

2

2

2

2

i i in n

eq

in n

eq

i i

i i

Fe e e

m X

Fe

m XNote that this eq. is the char. equation where s terms are



0

0

2

2 2 0

0

2

since cos sin

2Imag. part : 2 sin sin

Real part : cos cos

i

n eq

n

eq

n eq

n

eq

e i

m XFi

m X F

mF

X

i

X

m F

equation where s terms are replaced by i terms

2

0 0

22

2 0

1

2

2

2

2

2sin ,cos

arctan , 2

And,

2

n eqn eq

n n

in n

eq

m Xm X

F F

Fe

m X

F

i



2 22

2

2 0

0 02 2 222

2

22

2

1 1

21 2

n n

eq

eq eq nn n

n n

F

m

F

m

X

FX

m

2

02 2

2

1

1 2

eq

n n

FX

k

0

Re( ( )) Im( ( ))

0 0Input : ( )

Outp

cos

cos

si

ut :

n

sin

i t

F t F t

i t

i F t

i X

F t F e

x t tXe

F t

X t

Substituting X and into the assumed solution



Re( ( )) Im( ( ))

Outp cosut : sin

x t x t

i Xx t tXe X t

Re

Im

0( ) i tF t F e

( )

i t

x t Xe

0r F0

2

2

22

1

1 2

eq

n n

Fr X

k

Re

Im

2 0

2 2

2 2

( )2

( )2 ( )

( ) 1( )

( ) 2

( ) 1 1 1 1( )

i t

n n

eq eq

n n

eq

eq n n

F eF tx x x

m m

F ss s X s

m

X sG s

F s m s s

X iiG



2 22 2

2

2

2

2 2

( )( ) 22

21 1 1( )

2

eq eq n nn n

n n

eq n n eq n

ii ii i

ii

F m m

Gm mi

G

0

2 2 222 2

2

2

2

2

2

222 22

1 1

21 2

21( ) arctan , 2

2

eqn

n n

X

F

n n

n n

eq n n

k

Gm

ii

2 Amplitude of forced response:

n

nXm Xk X XM

Some Definitions:



0

0

00 0 0 0

max. spring force:excitation force amp.

22 22

:Static deflection due to

1 2, = arctan

11 2

s

n

F

F

MFF F Fk

M


2 22

2

2

When =0, M at =1

For other values of , M reaches a maximum when

11 2 is a minimum

2 1 2 2 2 2 0

RM

dR

d

2 22

1

1 2

M


2 2

2max

max 2

4 1 2 0

1 2 exists for 0.5 corresponding to

1

2 1

Note that

M

2 2res n d n n1 2 1


2

M=1 when 0 for all values (this is static case where 0)

M 0 when (Due to inertia, i.e., x= )

Increasing decreases M for a given .

REM RKSA :

x

2 22

1

1 2

M


n

Increasing decreases M for a given .

For =0, M at 1 (i.e., )

For 0.5 M re

2res

max 2

aches a maximum (resonance) at 1 2

1with a value

2 1

For 0.5, 0 at 0, for 0.5 M decreases monotonically

with increasing , there is no resonance peak.

M

dM

d

2 22

1

1 2

M


=0

M


Increasing

=0.5

=0.707


For =0, response and excitation are in-phase for 1

and out-of-phase for 1

For 0, response and excitation are in-phase only when 0

REMARKS:

2

2= arctan

1


For 1 phase is -90 regardless of damping and

excitat

ion is in-phase with velocity.

For 1, - - , response leads the excitation.2

For 1, - , response is opposite to excitation.

1


2

2= arctan

1

=0


Increasing

Increasing

Example: For an undamped spring-mass system m=10kg, k=1kN/m, F0=50N. Determine the amplitude of oscillations for steady state case for the excitation frequencies 1=2rad/s and 2=20rad/s


rad/s 1010

1000

sin50

n

tkxxm


mm 67.1

3

1005.0

21

1

1000

50

11210

20 ,For

mm 208.596.0

1005.0

2.01

1

1000

50

112.010

2 ,For

10

522

02

222

522

01

111

mm

n

mm

n

Mk

FX

M

Mk

FX

M

Example: A part with m=2kg vibrates in a viscous medium. Result of a frequency sweep shows that a harmonic exciting force of amplitude 2.8N results in a maximum steady state vibration amplitude of 12mm and of period 0.2s. c=?


res

res

2 2 2res n

2 231.42

0.2

1 2 1 2 31.42, "k" and " " are unknown (1) 1 2 987.212 2

T

k k

m


res n

resmax res 2 2

0

22

1 2 1 2 31.42, "k" and " " are unknown (1) 1 2 987.212 2

1 0.012 1, "k" and " " are unknown (2)

2.82 1 2 1

2.8 1 2 987.21= ,

0.012 1 22 1

m

XM M

F k k

k 2 2

2 2 2 2

1 1

2 2

let u= 561.35 1 1.96 1 4 4

561.35 561.35 1.96 7.87 7.87 569.22 569.22 1.96 290.40 290.40 1 0

0.996545 0.99827Which one? Note that there would be n

0.034554 0.05878

u u u u

u u u u u u u u

u u

u uo resonance if 0.707

0.05878 substitute in (1) or (2) 1988.169

2 2 0.05878 1988.169 2 7.4135 /

k

c km Ns m

ME_302-L-Forced_Vibrations_V3.pdf

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