vibrations of 1 dof systems -...

VIBRATIONS OF 1 DOF SYSTEMS

VIBRATION: a periodic motion about an equilibrium position, such as the regular displacement of guitar string playing.

Vibrations are in general dangerous: they are associated with noise, malfunctioning and possibly fatigue failure.

Vibrations analysis of continuous systems is usually performed by means of lumped parameter model, i.e by substituting a continuous system with a discrete one with a finite number of DOF. We will thus study only the simple 1 DOF case as representative analysis for the most general case.


FREE VIBRATIONS

0i d el+ + =F F F 0mx cx kx+ + =

D’Alembert equation

i mx= −F Inertia Force

d cx= −F Damper Force

el kx= −F Spring Force

x

k

c

m

lumped parameters

m

continuous


FREE VIBRATIONS

x

m

k0mx kx+ =

The solution of this second order differential equation will have the following form

1 21 2( ) t tx t X e X eλ λ= +

where λ1,2 are the roots of the characteristic equation

1,24

2 nmk ki i

m mλ ω± −

= = ± = ±2 0m kλ + =

Let us start considering a system with no dumping. The dynamic equilibrium equation will be:


FREE VIBRATIONS

x

m

k

1 2( ) n ni t i tx t X e X eω ω−= +

Applying the Euler rule: cos sinize z i z= +

1 2 1 2( ) ( ) cos( ) ( ) sin( )n nx t X X t i X X tω ω= + + −

cos( ) sin( )n nA t iB tω ω= +

Exploiting the rotating vector representation, we have

( ) cos( )nx t C tω ϕ= +

Im

Re

ωnt

AB

C

φ


FREE VIBRATIONS

x

m

k The value of C and φ can be determined by the initial conditions of the system

( ) cos( )nx t C tω ϕ= +

0

0

(0)(0)

x xx x

=⎧⎨ =⎩

x

t

x0

C

2 / n Tπ ω =

0 0

00

(0) cossin(0) n

x x x Cx Cx x

ϕω ϕ

= =⎧ ⎧⇒⎨ ⎨ = −= ⎩⎩

0 0 0 0

0 00

(0) cos /sin / cos(0)

n

n

x x x C tg x xx C C xx x

ϕ ϕ ωω ϕ ϕ

= = = −⎧ ⎧ ⎧⇒ ⇒⎨ ⎨ ⎨= − == ⎩ ⎩⎩


x

m

k

c

0mx cx kx+ + =

The solution of this second order differential equation will have the same form

1 21 2( ) t tx t X e X eλ λ= +

where λ1,2 are the roots of the characteristic equation

2

1,24

2c c mk

mλ − ± −

=2 0m c kλ λ+ + =

FREE VIBRATIONS WITH DAMPING

In this case, the dynamic equilibrium equation will be:


x

m

k

c

21) 4c mk≥

23) 4c mk<

According to the sign of the term , it is possible to distinguish three cases:

2 4c mk−


The root λ1,2 are real number, producing an aperiodic damped motion of the mass m.

22) 4c mk=The root λ1,2 are coincident and real, producing an aperiodic damped motion of the mass m. Rare limit case

The root λ1,2 are complex and conjugate, producing a periodic damped motion of the mass m. Vibrations.


2crc mk=

cr

cc

ξ =

21,2 1

s

n niω

λ ξω ω ξ= − ± −

nkm

ω =


2

1,24

2c c mk

mλ − ± −

=

It can be convenient to rearrange the characteristic equation, introducing some significant term:

the damping value distinguish between periodic and aperiodic response of the system

the natural pulse of the undamped system

damping factor: the ration between the actual value of damping and its critical value



x

m

k

( ) cos( )nsx t Ce tξω ω ϕ−= +

x

t

x0

ne ξω−

ne ξω−−

c


FORCED VIBRATIONS

x

m

k

0( ) cos( )F t F tω=

c

( )( ) cos( ) sin( )i i i iF t A t B tω ω= +∑

Let us consider the effect of a periodic force acting on the system. Every periodic force can be decomposed in Fourier series, i.e

Thus, we may exploit the superposition principle and consider the system as loaded by a single sinusoidal force, without loss of generality


( ) g px t x x= +

FORCED VIBRATIONS

x

m

k0 cos( )F tω

c

( )i d el t+ + =F F F F

D’Alembert equation

0 cos( )mx cx kx F tω+ + =

gx = solution of the free motion

sx = specific integral solution

We assume the specific integral to be sinusoidal, with the same pulse of the applied force with a delay of φ and we limit our study to steady state vibrations

( ) cos( )g px t x x X tω ϕ= + = −


FORCED VIBRATIONS

x

m

k0 cos( )F tω

c

20cos( ) sin( ) cos( ) cos( )m X t c X t kX t F tω ω ϕ ω ω ϕ ω ϕ ω− − − − + − =

( ) cos( )x t X tω ϕ= −

( ) sin( )x t X tω ω ϕ= − −

2( ) cos( )x t X tω ω ϕ= − −

0 cos( )mx cx kx F tω+ + =


FORCED VIBRATIONS

x

m

k0 cos( )F tω

c

20sin( cos( ) cos( ))cos( ) F tkm t c t X tX Xωω ω ωω ϕϕ ω ϕ− − =+ −− −

Im

Re

ωt

F0

φ

-F0

kX

c Xω2m Xω


FORCED VIBRATIONS( ) ( )2 22 2 2

0X k m c Fω ω⎡ ⎤− − =⎢ ⎥⎣ ⎦( ) ( )2 22 2 20

2

X k m c F

ctgk m

ω ω

ωϕω

⎧ ⎡ ⎤− − =⎢ ⎥⎪ ⎣ ⎦⎪⎨⎪ =⎪ −⎩

2crc mk=cr

cc

ξ = nkm

ω =

Im

Re

ωt

F0

φ

-F0

kX

c Xω2m Xω

( ) ( )0

2 22

FX

k m cω ω=

− +

2

ctgk m

ωϕω

=−

0

222

/

1 cr

cr

F k

cm ck c kω ω

=⎛ ⎞⎛ ⎞

− + ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠

2

1

cr

cr

ccc k

mk

ω

ω=

−


FORCED VIBRATIONS

0

2 22

2

/

1 2nn

F kX

ω ωξωω

=⎛ ⎞ ⎛ ⎞− +⎜ ⎟ ⎜ ⎟

⎝ ⎠⎝ ⎠

2

2

2

1

n

n

tg

ωξω

ϕωω

=−

2crc mk=cr

cc

ξ = nkm

ω =

Im

Re

ωt

F0

φ

-F0

kX

c Xω2m Xω


FORCED VIBRATIONS

0

2 22

2

/

1 2nn

F kX

ω ωξωω

=⎛ ⎞ ⎛ ⎞− +⎜ ⎟ ⎜ ⎟

⎝ ⎠⎝ ⎠

2

2

2

1

n

n

tg

ωξω

ϕωω

=−


VIBRATIONS: ISOLATION OF THE BASEx

m

k0 cos( )F tω

c

sin( ) cos( )c X t kX tT ω ωω= − +

( ) cos( )x t X tω=

( ) sin( )x t X tω ω= −

T cx kx= +

Let us see how to design the suspension in order to reduce the force transmitted to the base.

Im

Reωt

kX

c Xω

T

φ

0( ) cos( )T t T tω ϕ= +



m

k0 cos( )F tω

c

2 2 20T X k c ω= +

2crc mk=cr

cc

ξ =nkm

ω =

20 2

41 mkT kXk

ξ ω= +

22

2 21 cr

cr

cckXc k

ω= +

2

1 2n

kX ωξω

⎛ ⎞= + ⎜ ⎟

⎝ ⎠

2n

ctgkω ωϕ ξ

ω= =

TIm

Reωt

kX

c Xω

φ



m

k0 cos( )F tω

c

2

0 1 2n

T kX ωξω

⎛ ⎞= + ⎜ ⎟

⎝ ⎠

0

2 22

2

/

1 2nn

F kX

ω ωξωω

=⎛ ⎞ ⎛ ⎞− +⎜ ⎟ ⎜ ⎟

⎝ ⎠⎝ ⎠

2

0

2 220

2

1 2

1 2

n

nn

TF

ωξω

ω ωξωω

⎛ ⎞+ ⎜ ⎟⎝ ⎠=

⎛ ⎞ ⎛ ⎞− +⎜ ⎟ ⎜ ⎟

⎝ ⎠⎝ ⎠

TIm

Reωt

kX

c Xω

φ


2

2 2n

ωω

>>

VIBRATIONS: ISOLATION OF THE BASE

0

0

1TF

<< 2 2nω ω>> k

2

0

2 220

2

1 2

1 2

n

nn

TF

ωξω

ω ωξωω

⎛ ⎞+ ⎜ ⎟⎝ ⎠=

⎛ ⎞ ⎛ ⎞− +⎜ ⎟ ⎜ ⎟

⎝ ⎠⎝ ⎠

0

0

1TF

<


VIBRATIONS: ISOLATION OF THE MASS

x

m

k

c

( ) cos( )y t Y tω=

Let us see how to design the suspension in order to isolate the suspended mass from the vibration of the frame.

y

( )el k x y= − −F

( ) ( ) 0mx c x y k x y+ − + − =

( )d c x y= − −F

i mx= −F



x

m

k

c

y( ) ( ) 0mx c x y k x y+ − + − =

( )s x y= −

ms cs ks my+ + = −

( ) cos( )y t Y tω=

( ) sin( )y t Y tω ω= −2( ) cos( )y t Y tω ω= −

2 cos( )ms cs ks m Y tω ω+ + =


Im

Re

ωtφ kS

c Sω2m Sω


2 cos( )ms cs ks m Y tω ω+ + =

x

m

k

c

y

cos( )s S tω ϕ= −

2 2sin cococos( s( ) ()) s )(m S kS tS m Yc tt tω ω ϕ ϕω ω ω ω ωϕ −− −− + =−

2m Yω

2m Yω−



x

m

k

c

yIm

Re

ωtφ kS

c Sω2m Sω

2m Yω

2m Yω−

2

2

2 22

21 2

n

nn

Yωω

ω ωξωω

=⎛ ⎞ ⎛ ⎞− +⎜ ⎟ ⎜ ⎟

⎝ ⎠⎝ ⎠

2

2 22

2

/

1 2nn

m Y kS ω

ω ωξωω

=⎛ ⎞ ⎛ ⎞− +⎜ ⎟ ⎜ ⎟

⎝ ⎠⎝ ⎠


2

2

2

1

n

n

tg

ωξω

ϕωω

=−


2

2

2 22

21 2

n

nn

YS

ωω

ω ωξωω

=⎛ ⎞ ⎛ ⎞− +⎜ ⎟ ⎜ ⎟

⎝ ⎠⎝ ⎠



s y= −

0x =

S Y=

ϕ π=

S Y=2

2 1n

ωω

>> k

ϕ π= 1ξ << c

s x y= −

EQUIVALENT STIFFNESS PARAMETERS

EQUIVALENT STIFFNESS OF SPRINGS IN PARALLEL

1 1

2 2

1 2 1 2

F k xF k xF F F k x k x

= Δ= Δ= + = Δ + Δ

1 2eqk k k= +

1k

2kF

xΔ

1 1 2

1

( )

eq

F x k kF k x= Δ += Δ 1

n

eq ii

k k=

= ∑

EQUIVALENT STIFFNESS OF SPRINGS IN SERIES

1 1

2 2

1 2( )eq

F k xF k xF k x x

= Δ= Δ= Δ + Δ

1k 2k

1xΔ 2xΔ

1 2

( )eq

F F Fk k k

= + 1 2

1 2eq

k kk

k k=

+

F

1

1 1n

ieq ik k=

= ∑

EQUIVALENT STIFFNESS PARAMETERS

vibrations of 1 dof systems -...

Documents