Prof Bower`s forced vibration summary slides

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Forced Vibrations – concept checklist
You should be able to:
1. Be able to derive equations of motion for spring-mass systems subjected
to external forcing (several types) and solve EOM using complex vars,
or by comparing to solution tables
2. Understand (qualitatively) meaning of ‘transient’ and ‘steady-state’
response of a forced vibration system (see Java simulation on web)
3. Understand the meaning of ‘Amplitude’ and ‘phase’ of steady-state
response of a forced vibration system
4. Understand amplitude-v-frequency formulas (or graphs), resonance, high
and low frequency response for 3 systems
5. Determine the amplitude of steady-state vibration of forced spring-mass
systems.
6. Deduce damping coefficient and natural frequency from measured forced
response of a vibrating system
7. Use forced vibration concepts to design engineering systems
EOM for forced vibrating systems
L0
k, L0
x(t)
1 d 2x
F(t)=F0 sin t
n2 dt 2
m



2 km
,
K
1
k
x(t)
L0
k, L0
1 d 2x
m
n2 dt 2
Base Excitation

n 
y(t)=Y0 sint
L0 x(t)
k, L0

2 dx
 x  KF0 sin  t
n dt
k
,
m
n 
External forcing



2 dx
2 dy 
 x  K y

n dt

dt
n


k
,
m


,
K 1
2 km
y(t)=Y0sint
m

m0
Y0 2
2 dx
K d2y
Rotor Excitation  2 dt 2   dt  x    2 dt 2  K  2 sin t
n
n
n
n
1 d 2x
n 
k
M


2 kM
m
K  0 M  m  m0
M
Steady-state solution – external force
L0
k, L0
x(t)
F(t)=F0 sin t
1 d 2x
m
n2

dt
2

2 dx
 x  KF (t )
n dt
n 
k
,
m


2 km
x(t )  X 0 sin  t   
X0 
KF0
1/2

2
2
2 2
 1   / n   2 / n  




  tan 1
2 / n
1   2 / n2
System vibrates at same frequency as force
Amplitude depends on forcing frequency, nat frequency, and damping coeft.
,
K
1
k
Steady-state solution – Base excitation
L0
x(t)
k, L0
1 d 2x
m
n2

dt
y(t)=Y0 sint
2


2 dx
2 dy 
 x  K y

n dt
n dt 

n 
k
,
m
x(t )  X 0 sin  t   
X0 

KY0 1   2 / n 

2 1/2
1/2

2
2
2 2
 1   / n   2 / n  




  tan 1
2 3 / n3
1  (1  4 2 ) 2 / n2


2 km
,
K 1
Steady-state solution – Rotor excitation
L0 x(t)
k, L0

Y
2 dx

xK 0
sin t
2
2
2

dt
n dt
n
n
y(t)=Y0 sint

m
2
1 d2x
m0
n 
k
M


2 kM
m
K  0 M  m  m0
M
x(t )  X 0 sin  t   
X0 
KY0 2 / n2
1/2

2
2
2 2
 1   / n   2 / n  




  tan 1
2 / n
1   2 / n2
Using forced vibration measurements to find
natural frequency and damping coefficient
Amplitude
Xmax
Xmax/ 2
  2  1



 max
Frequency

2max
n  max
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