philadelphia university faculty of engineering mechanical engineering department dr. adnan dawood...
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Philadelphia University Faculty of Engineering
Mechanical Engineering Department
Dr. Adnan Dawood Mohammed(Professor of Mechanical Engineering)
Mechanical VibrationsForced Vibration of a Single Degree of Freedom System
Harmonically Excited Vibration
Physical system
force offrequency theis
force of amplitude theis cos
sin
:forms following theof one may takesIt force. harmonic a is
motion ofEquation ...
tjo
oo
o
eFtF
FtFtF
tFtF
tF
tFkxxcxm
Harmonically Excited Vibration
amplitude' response statesteadty ' thecalled is X , X(t)
as written becan solution The .frequency same with the
harmonic is (solution) response t theexpect thacan wefrequency with
harmaonic is n)(excitatio force theBecause equation. homogenuosnon
for thesolution theisIt :response state(steady integral Particular .2
sincos)(
:as written assolution w The
1for chapter). previousin (describedequation shomogeneou the
forsolution theisIt :response) (Transientfunction ary Complement 1.
parts two
hassolution Its .homogenuos-non isequation aldifferentiorder 2nd This
(1)motion ofEquation ...
:form heConsider t
21
tjp
ddt
c
tjo
ex
tCtCetx
eFkxxcxm
n
Harmonically Excited Vibration
amplitude. response statesteady
theof magnitude"" thedenotes X , X
as written be nowcan integral particular for the There
X re wheXX
Response. and Forcebetween angle"
phase" theis tan where,
as By writing
X
:Xfor solve and (1)equation into integral particular theSubstitute
222
21-222
2
2
tjp
oj
j
o
ex
cmk
Fe
mk
cwecmk
jcmk
jcmk
F
Harmonically Excited Vibration
motion. theof conditions inatial theknowing
from dermined be toconstants are C2 and C1 that Note
X sincos)(
)()()(
:is (1)motion ofequation for the (response)solution totalThe
1
2tan
ratio.frequency theis where, 21
X
:as written bemay and X
21
21
222
tjdd
t
pc
n
o
etCtCetx
txtxtx
r
r
rrr
k
F
n
Harmonically Excited Vibration
222
st
sto
21
1M
(M)".Factor ion Magnificat" thecalledusually is
ratio The ).( deflection static" thecalledusually is k
F termThe .5
.frequency and amplitude
constant h motion wit harmonic a is ))(( response stateSteady 4.
me.certain ti aafter neglected becan and time
with decayst motion tha a represents ))(( responseTransient .3
cos)(cos form force For the 2.
sin)(sin form force For the 1.
:NOTES
rr
X
tx
tx
tXtxtFF(t)
tXtxtFF(t)
p
c
po
po
Damped Forced Vibration System
Graphical representation for Magnification factor M and ϕ.
Damped Forced Vibration System
Notes on the graphical representation of X.
For ζ = 0 , the system is reduced and becomes un-
damped.
for any amount of ζ > 0 , the amplitude of
vibration decreases (i.e. reduction in the
magnification factor M). This is correct for any
value of r.
For the case of r = 0, the magnification factor
equals 1.
The amplitude of the forced vibration
approaches zero when the frequency ratio ‘r’
approaches the infinity (i.e. M→0 when r → ∞)
Damped Forced Vibration System
Notes on the graphical representation for ϕ.
For ζ = 0 , the phase angle is zero for 0<r<1 and
180o for r>1.
For any amount of ζ > 0 and 0<r<1 , 0o<ϕ<90o.
For ζ > 0 and r>1 , 90o<ϕ<180o.
For ζ > 0 and r=1 , ϕ= 90o.
For ζ > 0 and r>>1 , ϕ approaches 180o.
Harmonically Excited Vibration
) small(for is
ntdisplaceme themeans which 2
X then 1)(r resonanceAt 3.
is
ntdisplaceme themeans which X then 1)(rfrequency high At 2.
is
ntdisplaceme themeans which X then 0)(rfrequency lowAt 1.
amplitude response s.s theis 21
X
: have We
:frequencynt with displaceme ofVariation
2
222
control.Damping
control. Mass
control. Stiffness
k
F
m
F
k
F
rr
kF
o
o
o
o
Harmonically Excited Vibration
2res
2
1/2-222
21 i.e
a asknown is which 21 when maximum is X
:givescondition This
0)X( when maximum is X
21 X
:as written becan amplitude response s.s The
n
n
o
r
dr
d
rrk
F
Frequency,Resonance
:amplitude Maximum ofFrequency
Forced Vibration due to Rotating Unbalance
tmekxxcxM t sin...
2
Forced Vibration due to Rotating Unbalance
below figure in theshown are and X of plots The
1
2tan
and
21X
or ,21
X
21
222
2
222
2
r
r
rr
rMme
rr
kme
t
Transmissibility of Force
)r-(1
r2tan ,
)2()1(
)2(1TR
TR bility,Transmissi Force theas TR define
)(
)(F
i.e, motion harmonic Assume
F
is foundation the toed transmittforce then the
,neglugible is foundation of deflection theIf
21
22
2
F
2tr
tr
rr
r
F
F
cjmk
Fcjk
Xex
xckxe
F
o
tr
F
o
tj
tj
Transmissibility of displacement (support motion)
Physical system:
Mathematical model: 0....
yxkyxcxm
The forcing function for the base excitation
Transmissibility of displacement (support motion)
Substitute the forcing function into the math. Model:
motion) (Harmonict Xsin x(t)Assume
tan
:Where
sincossin...
1
22
k
c
ckYA
tAtYctkYkxxcxm
2
1
222
2
222
22
d
1
2tan
21
21)bility(TRTransmissint Displaceme
r
r
rr
r
cmk
ck
Y
X
Transmissibility of displacement (support motion)
Graphical representation of Force or Displacement Transmissibility ((TR) and the Phase angle (
Example 3.1: Plate Supporting a Pump: A reciprocating pump, weighing 68 kg, is mounted at
the middle of a steel plate of thickness 1 cm, width 50 cm, and length 250 cm. clamped along two edges as shown in Fig. During operation of the pump, the plate is subjected to a harmonic force, F(t) = 220 cos (62.832t) N. if E=200 Gpa, Find the amplitude of vibration of the plate.
Example 3.1: solution
The plate can be modeled as fixed – fixed beam has the following stiffness:
The maximum amplitude (X) is found as:
mNx
xxkSo
mxxxbhIBut
l
EIk
/ 82.400,10210250
10667.4110200192 ,
10667.41101105012
1
12
1
192
32
99
493223
3
mmmk
FX o 32487.1
832.626882.400,102
2202
-ve means that the
response is out of phase
with excitation
Example 3.2:Find the total response of a single-degree-of-
freedom system with m = 10 kg, c = 20 N-s/m, k=4000 N/m, xo = 0.01m and = 0 when an external force F(t) = Fo cos(ωt) acts on the system with Fo = 100 N and ω = 10 rad/sec .
Solution a. From the given data
ox.
sradm
kn /20
10
4000
05.020102
20
2
nm
c
sradd
d
nd
/975.19
2005.01
1
2
2
5.020
10
n
r
Example 3.2: Solution
Total solution:X(t) = X c (t) + X p(t)
mk
Fost 025.0
4000
100
mrr
X st 3326.021 222
o
r
r814.3
1
2tan
21
0.066)-0t20.3326cos()-cos(19.97t20*.050-
e (t)
0.066)-tcos(X)-tncos(-
e (t)
tAx
tnAx
0.066)-0t20.3326cos(- )066.0-cos(19.97t--6.64ex(t)
-0.3325A and rad 0.066
0.438,
0.066)-0t220sin(*0.3326
-(-t)e*)-(19.97tcos)-(19.97t-19.97sin *-e)(
0.066)-0t20.3326cos()-cos(19.97t-e (t)
(1) 33187.0cos
cos0.066*0.3326cos
(2) and (1) From
(2) 0219.0sinsin97.190
0 0,at t
0
01.0 0,at t
t
tAtAtx
tAx
A
A
AA
tx
mtx