Single Degree of Freedom – Free Vibration
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Single Degree of Freedom – Free Vibration
fixed / ground
c k
m
FBD of single mass
x
damper FspringF
m staticx
0
kxxcxm
maFFF xscx
Solution A: Critically damped
occurs noscillatio no , thereforenote
)()(
ratio damping the, 2
c let
(undamped)frequency natural g/x /let
1damped Critically A. Case
00021
n
static
txxxeetAAtx
m
mk
)(ζ
ntt
n
nn
Case A – Critically damped
Case B - Overdamped
frequency natural damped1
2
1,
2
1
1exp(1exp()(
2)(
noscillatio no -1 Overdamped B.Case
2
00000
22
121
nd
d
n
n
n
nnt
tsts
xxB
xxxA
tBtAetx
eAeAtx
)(ζ
n
Case B - Overdamped
Underdamped
andfrequency natural damped theis ,)1(
sincos)(
nOscillatio -10 ed UnderdampC. Case
1/22
000
nd
dd
nd
t txx
txetx
)ζ(
n
Case C - Underdamped
Example D - less damping
SDOF – Forced Vibration
tutu sin)( 0
)( tx
c k
m What force does the mass “feel?”
road
Summing the forces in the x direction:
c
k
tFtckukxxcxm
tkutcukuuckxxcxm
u(t)
tuututu
kuuckxxcxm
uxkuxcxm
1
02/1222
0
00
00
tanfor
)sin()sin()(
obtain we RHS, thesimplfyingby and
sincos
find we,for ngsubstituti therefore,
cos then,sin)( sincehowever
obtain weg,rearrangin
0)()(
ratio damping the, 2
c
andfrequency natural undamped the,)/(
for
])/2()/1[(
)sin()/(F)(
n
2/1
2/122220
m
mk
wtktx
n
nnss
From D. Eqns we obtain steady state solution
What force does “Mass” feel?
2/1
2222
2
0
t
)/2()/1(
)/2(1
isibility transmiss theof magnitude thewhere
)sin( show that can weAnd,
)(F
find wedamper, and spring thefrom results
mass the toed transmitt force theSince
nn
n
t
ssss
t
T
T
tTF
F
kxxckxxct
F
Example ¼ car model
symbol value unitsweight W 1000 lbf mass m 2.588 lbf/(in/sec^2)
spring k 100 lbs/indamper c lb/(in/sec)
damping ratio zeta 0.500 n/anatural frequency 6.216 rad/sec
0.989 Hzdamped nat. freq. 0.857 Hz
n
d
http://www.me.metu.edu.tr/courses/ME432/soft07.html
Effect of damping on forced vibration
Transmissibility
0.00
0.50
1.00
1.50
2.00
2.50
3.00
0.0 5.0 10.0 15.0
Forcing Frequency (Hz)
Tra
nsm
issib
ilit
y