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8/17/2019 Machine Dynamics Vibration
1/44
Dr.Tamer Kepçeler 1
MACHINE DYNAMICSVIBRATION
Dr. Tamer Kepçeler 1
Dr. Tamer Kepçeler 2
What is vibration?
•Mechanical vibration is the motion of a particle or body whichoscillates about a position of equilibrium. Most vibrations in machinesand structures are undesirable due to increased stresses and enerylosses.
!f a particle is displaced throuh a distance x m from its equilibriumposition and released with no velocity" the particle will undero simpleharmonic motion"
Simple harmonic motion with a
circular motion of a point ma
Dr. Tamer Kepçeler #
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Dr.Tamer Kepçeler 2
Vi!ration fre"uenc# an$ perio$
• Time interval required for a system to complete a full cycle of the motion is the
period of the vibration.
• Number of cycles per unit time defines the frequencyof the vibrations
• Maximum displacement of the system from the equilibrium position is the
amplitude of the vibration.
Dr. Tamer Kepçeler $
The element for %i!ratin& #tem
• Ma
• Sprin&
• Dampin&
• 'orce
Dr. Tamer Kepçeler %
&
&
De&ree of 'ree$om
Minimum number of independent coordinates required
to determine completely the positions of all parts of a
system at any instant of time
Dr. Tamer Kepçeler '
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Dr.Tamer Kepçeler 3
Sin&le $e&ree of free$om #tem
Dr. Tamer Kepçeler (
&
&
&
θ
θ
Two $e&ree of free$om #tem
Dr. Tamer Kepçeler )
&1
&2
1θ
2θ
1θ 2θ
θ
&
Three $e&ree of free$om #tem
Dr. Tamer Kepçeler *
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Dr.Tamer Kepçeler 4
Example of Infinite-number-of-degrees-of-freedom
system:
1. Infinite num!er of $e&ree of free$om #tem are
terme$ continuous or distributed #tem
2. Finite num!er of $e&ree of free$om are terme$
discrete or lumped parameter #tem
() More accurate reult o!taine$ !# increain&
num!er of $e&ree of free$om
Dr. Tamer Kepçeler 1+
SI Birim Sitemi
*im Birim Sem!ol+,unlu- Metre m.ütle .ilo&ram -&/aman Sani#e .u%%et Newton N 0-&)m1234erilme 5acal 5a 0N1m23*6 7oule 7 0N)m34üç 8att 8 0713
're-an Hert, H, 0913Moment M N)m.:tleel Atalet Momenti 7 -&)m2
.eit Atalet Momenti I m;
Dr. Tamer Kepçeler 11
Harmonic motion
Dr. Tamer Kepçeler 12
T
t2sinAx π=
x=displacement (m,rad)
A=amplitude (m,rad)
t=time (s)
T=period (s)
t
x
A
T
,eriodic Motion- motion repeated after equal intervals of time
armonic Motion- simplest type of periodic motion
8/17/2019 Machine Dynamics Vibration
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Dr.Tamer Kepçeler 5
Simple harmonic motion with acircular motion of a point ma
Dr. Tamer Kepçeler 1#
)tsin(AtsinA-x
/2)tsin(AtcosAx
tsinAx
f 2T
2
22 π+ωω=ωω=
π+ωω=ωω=
ω=
π=π
=ω
&&
&
/
,0 0
& π2
tω
tsinA ω
Diplacement< %elocit# an$acceleration
Dr. Tamer Kepçeler 1$
AωA
2Aω
t
x
x
x&
x&&
tω
o90
o180
Dr. Tamer Kepçeler 1%
Spring elements
A spring is a type of mechanical link, which in most applications is assumedto have negligile mass and damping! The most common type of spring isthe helical"coil spring used in retractale pens and pencils, staplers, and
suspensions of freight trucks and other vehicles!
#n fact, any elastic or deformale ody or memer such as cale , ar,eam, shaft, or plate, can e considered as a spring!
8/17/2019 Machine Dynamics Vibration
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Dr.Tamer Kepçeler 6
Dr. Tamer Kepçeler 1'
Spring elements
$elical springs
Dr. Tamer Kepçeler 1(
%eaf springs
Sprin& characteritic
Dr. Tamer Kepçeler 1)
34
5 m4
34
5 m4
6inear sprincharacteristics
3on76inear sprincharacteristics
α
8/17/2019 Machine Dynamics Vibration
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Dr.Tamer Kepçeler 7
Sprin& contant
Dr. Tamer Kepçeler 1*
(N/m) x
F tank =α=
force
displacement
Sprin& contant ta!le
Dr. Tamer Kepçeler 2+
L
IEk =
L
AEk =
3
4
Rn64
dGk =
3L
EI3k =
L
IGk
p=
Dr. Tamer Kepçeler 21
682
3L
IE48k =
682
3L
IE192k =
682
3L7
IE768k =
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Dr.Tamer Kepçeler 8
Dr. Tamer Kepçeler 22
a b
&
y ( )222
x22bxL
LIE6
xbPy
ba
LIE3k −−==
δ
9! 63L
IE12k =
Dr. Tamer Kepçeler 2#
( ) 2aaL
IE3k
+=
6 a
( )a8L3a
IE24k
2 +=
6 a
Sprin& in paralleel
Dr. Tamer Kepçeler 2$
:1 :2 ≡
m
:e;
m &&
∑=
=++++=n
1i
in321eş k k ......k k k k
8/17/2019 Machine Dynamics Vibration
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Dr.Tamer Kepçeler 9
Sprin& in erie
Dr. Tamer Kepçeler 2%
m &
:e;
m &
≡:1:2
∑=
=++++=
n
1iin321eş k
1
k
1.....
k
1
k
1
k
1
k
1
E=ample>
Dr. Tamer Kepçeler 2'
:1 :2
:#
:$
m &
Determine the equivalentsprin constant of the ivensystem
E=ample>
Dr. Tamer Kepçeler 2(
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Dr.Tamer Kepçeler 10
E=ample>
Dr. Tamer Kepçeler 2)
:1
:2
:ç1
:ç2
m &
Determine the equivalentsprin constant of the ivensystem.
Find the equivalent spring constant of the system.
Obtain the differential equation and find the natural frequency
Dr. Tamer Kepçeler 2*
61 62
6
m
:1
:20
>
Solution
Dr. Tamer Kepçeler #+
Let’s find the equivalent spring constant of the system
"&ecause of the m!g weight force k ' spring affects & point as
2
Bk
gm=δ
0>
BδB / A
δ:1
:2
11
Ak L
L.g.m=δ
2
11
2
B / ALk
L.g.m=δ
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Dr.Tamer Kepçeler 11
Dr. Tamer Kepçeler #1
1L
Lg.m
7on the same way m. causes :1 strechin.The effect of mand e&tension on 0 point can be found respectively by
0lso" the this force causes e&tension on > point as ivenbelow?
The total displacement on point > is?
11
Ak L
L.g.m=δ
2
11
2
B / ALk
L.g.m=δ
2
11
2
2
B / ABLk
L.g.m
k
g.m+=δ+δ=δ
Dr. Tamer Kepçeler #2
7the equivalent sprin constant of thesystem is?
δ=
g.mk eş
2
2
2
11
2
121
2
11
2
2
eşLk Lk
Lk k
Lk
L.g.m
k
g.m
g.mk
+=
+
=
Dr. Tamer Kepçeler ##
7The equivalent sprin7mass system is?
≡ :e;
& xm &&
xk eş
0xk xm x-k xm amF eşeş =+=⇒=∑ &&&&
( )22
2
11
2
121
nLk Lk m
Lk k
+=ω rad8s
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Dr.Tamer Kepçeler 12
S-ill aement>
Dr. Tamer Kepçeler #$
ind the equivalent sprincoefficient.
L,d,E φ
L,I,E
&
Ma an$ Inertia Element
Dr. Tamer Kepçeler #%
The definition moments of inertia of mass related to rotatin mass-
@otatin a&is
D
dm
r
∫=D
2dmrJ
Problem: Orta no-ta?n$an mafall? %e a!it -eitli !ir @u!uun -:tleel atalet
momentinin !ulunma?
Dr. Tamer Kepçeler #'
&0
d&
6
y
&
682
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Dr.Tamer Kepçeler 13
Çözüm:
Dr. Tamer Kepçeler #(
dxAdV =dVdm ρ=
∫=D
2dmrJ
∫ ∫−
ρ=ρ=
2
L
2
L
2
L
2
L-
22 dxxAdxxAJ
32
L
2
L
3
LA12
1
3
xAJ ρ=ρ=
−
2Lm12
1J LAm =⇒ρ=
Example: Dic fin$ the mpment of inertia of ma)
Dr. Tamer Kepçeler #)
r
dr
θ
θd
R
dA
L
Solution:
Dr. Tamer Kepçeler #*
9lementary square-
9lementary volume-
9lementary mass-
dr.dsin.rdA θ=
dr.dsinr.L.dA.LdV θ==
dr.dL.r.sin.dV.dm θρ=ρ=
dr.d.Lr.dm θρ=⇒θ≅θ ddsin
∫∫∫π
πρ=θρ==
2
0
R
0
43
D
2 R..L.2
1dr.d.r.L.dmrJ
22 R.m2
1J L.R..V.m =⇒πρ=ρ=
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Dr.Tamer Kepçeler 14
+n$ampe$ 'ree Vi!ration
Dr. Tamer Kepçeler $+
The motion of equation of mass-spring system(vertical ):
Atatic equilibrium
&
&
t
6+
stδ
::
:
Dr. Tamer Kepçeler $1
stk δ
g.mG =
ree body diaram
0Fy =Σ
st
n
2n
st
st
g
m
k
gmk
.k g.mG
δ==ω
ω=δ
=
δ==
wheren
ω is the natural frequency of the system
Atatic equilibrium
Dr. Tamer Kepçeler $2
>y applyin 3ewtonBs second rule-
( ) Gx-kxm a.mF st ++δ=⇒=Σ &&
st.k g.mG δ==
0xkxm =+&&
&
)x(k st +δ
g.mG =
xm &&
Dynamic equilibrium-
8/17/2019 Machine Dynamics Vibration
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Dr.Tamer Kepçeler 15
Dr. Tamer Kepçeler $#
3ewtonCs second rule-
a.mF =Σ
&
:m
&
xm &&
xk m
0xkxm x-kxm =+⇒= &&&&
The motion of equation of mass-spring system(horizontal):
O!tainin& Motion of E"uation!# Ener Metho$
Dr. Tamer Kepçeler $$
This method can be applied if the system is?
• ndamped
• irst order system
sabitCEE pk ==+
( ) 0EEdt
dpk =+
O!tainin& motion of e"uation for
un$ampe$ free %i!ration
Dr. Tamer Kepçeler $%
0xkxm =+&&The solution for this system is iven as
tseAx =
where" 0 and s are interation constants.
ts2
ts
eAsx
eAsx
=
=
&&
&
8/17/2019 Machine Dynamics Vibration
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Dr.Tamer Kepçeler 16
Dr. Tamer Kepçeler $'
( ) 0eAk smst2 =+
where" 0e ,A ts ≠ dEr.
0k sm2 =+
This equation is called characteristic equation and theroot are"
n2,1 im
k s ω=−= mm
Dr. Tamer Kepçeler $(
!n this case the motion of equation is-
ti
2
ti
1
ts
2
ts
1nn21 eAeAeAeA)t(x
ωω− +=+=
01 and 02 can be found by initial conditions.
t.sin.it.coseti θθ=θ mm
( ) ( )( ) ( ) tsinAAitcosAA)t(x
tsinitcosAtsinitcosA)t(x
n21n21
nn2nn1
ω−−ω+=
ω+ω+ω−ω=
( )211 AAB += ve ( )212 AAB −=
Dr. Tamer Kepçeler $)
tsinBtcosB)t(x n2n1 ω+ω=
!nitial conditionsare assumed
=
=
⇒=
0
0
x)0(x
xx(0)
0t
&&
, xB 01 =n
02
xB
ω= &
tsinx
tcosx)t(x nn
0n0 ω
ω+ω=
&
8/17/2019 Machine Dynamics Vibration
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Dr.Tamer Kepçeler 17
Problem> O!tain the $iferential e"uation of the &i%en pen$ulum) 'in$ the naturalfre"uenc# an$ perio$)
Dr. Tamer Kepçeler $*
The lenth of the robe is 6 and there is aweihtless mass on the tip of the robe.
Solution-
gm
ϕ&&Jϕ L
Dr. Tamer Kepçeler %+
3ewtonCs second law-
ϕ=Σ &&TopJM
ϕ=ϕ sing.L-mJ &&
2LmJ = ϕ≅ϕ⇒
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Dr.Tamer Kepçeler 18
Solution>
Dr. Tamer Kepçeler %2
Disturb the system...and e&amine the forces...
x.k
ϕ
g.M
g.m
ϕ&&mJ
ϕ&&MJ
Dr. Tamer Kepçeler %#
ϕ=Σ &&TopJM
3ewtonFs second law-
ϕϕϕ−=ϕ+ϕ cosx.Lk-sing.LM-sin2
Lg.mJJ Mm &&&&
1cos sin 0 ≅ϕϕ≅ϕ⇒
8/17/2019 Machine Dynamics Vibration
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Dr.Tamer Kepçeler 19
Problem> o!tain the motion af e"uation an$fin$ the natural fre"uenc# of the &i%en #tem
Dr. Tamer Kepçeler %%
m"6 M&
:
Solution>
Dr. Tamer Kepçeler %'
ϕ &ϕ&&
mJ
xk
xM &&
M
m"6
Dr. Tamer Kepçeler %(
ϕ=Σ &&TopJM
x.L-k.LxMJ m =+ϕ &&&&
1cos sin 0 ≅ϕϕ≅ϕ⇒
8/17/2019 Machine Dynamics Vibration
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Dr.Tamer Kepçeler 20
Dr. Tamer Kepçeler %)
0LkLMLm3
1 222 =ϕ+ϕ
+ &&
0kMm3
1=ϕ+ϕ
+ &&
M3m
k 3
Mm3
1
k
m
k n
+=
+==ω rad8s
Problem> A6a?$a $en&e -onumun$a %erilen itemin $iferani#el $en-lemini @?-ar?pta!ii fre-an?n? heapla#?n?,
Dr. Tamer Kepçeler %*
:M &
m"r
Çözüm>
Dr. Tamer Kepçeler '+
ϕ
xk
M&
ϕ&&mJ
xM &&
8/17/2019 Machine Dynamics Vibration
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Dr.Tamer Kepçeler 21
Dr. Tamer Kepçeler '1
3ewtonCun 2. :anunu uyulanErsa"
ϕ=Σ &&TopJM
x.r-k.rxMJ m =+ϕ &&&&
ϕ=
ϕ=
ϕ=ϕ=
&&&&
&&
rx
rx
rsin.rx
1cos sin 0 ≅ϕϕ≅ϕ⇒
8/17/2019 Machine Dynamics Vibration
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Dr.Tamer Kepçeler 22
Çözüm>
Dr. Tamer Kepçeler '$
xk
ϕ&&mJ
&
xm &&
ϕ
T
r
Dr. Tamer Kepçeler '%
3ewtonCun 2. :anunu uyulanErsa"
2
m rm
2
1J =amF , JM Top =Σϕ=Σ &&
( ) rxkxmJ
xkxmT Tx-kxm
rTJ
m
m
+−=ϕ
+=⇒+=
−=ϕ
&&&&
&&&&
&&
ϕ=
ϕ=
ϕ=ϕ=
&&&&
&&
rx
rx
rsin.rx
yaGElabilir.
Dr. Tamer Kepçeler ''
( )
m3
k 2
m
k 0km
2
3
0k rrmrm2
1
rk rrmJ
n
222
m
==ω⇒=ϕ+ϕ
=ϕ+ϕ
+
ϕ+ϕ−=ϕ
&&
&&
&&&&
rad8s
8/17/2019 Machine Dynamics Vibration
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Dr.Tamer Kepçeler 23
Problem> A6a?$a $en&e -onumun$a %erilen itemin $iferani#el $en-lemini @?-ar?pta!ii fre-an?n? heapla#?n?,
Dr. Tamer Kepçeler '(
: :
M
m"r m"r
&
y
Çözüm>
Dr. Tamer Kepçeler ')
M
&
y
ϕ
Dr. Tamer Kepçeler '*
( ) 0EEdt
dpk =+
y2x =
22
k
222222
k
222
k
rm2
3M2E
rm2
1
2
1 2rm
2
1 2r4M
2
1E
J2
1 2ym
2
1 2xM
2
1E
ϕ
+=
ϕ+
ϕ+ϕ=
ϕ+
+=
&
&&&
&&&
ϕ=
ϕ=
&& ry
ry
ϕ=
ϕ=
ϕ=
&&&&
&&
r2x
r2x
r2x
8/17/2019 Machine Dynamics Vibration
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Dr.Tamer Kepçeler 24
Dr. Tamer Kepçeler (+
2
k k k
1
k
1
k
1eş
eş
=⇒+=
22222
eşp rk r42
k
2
1xk
2
1E ϕ=ϕ==
( )
( )[ ]
0rk2rm3M4
rk rm2
3M2
dt
d
0EEdt
d
22
2222
pk
=ϕϕ+ϕ+
ϕ+ϕ
+
=+
&&&
&
0≠ϕ& olmalE"
Dr. Tamer Kepçeler (1
( )
m34M
k 2
m
k
0k2m3M4
n+
==ω
=ϕ+ϕ+ &&
Dr. Tamer Kepçeler (2
2h
&2
h
:1
:2 :#
&1
ϕ
m1
m2
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Dr.Tamer Kepçeler 25
Çözüm>
Dr. Tamer Kepçeler (#
&2
h
&1
ϕ
m1
m2
2h
23xk
11xk
( )122 xxk −
11xm &&
22xm &&
Dr. Tamer Kepçeler ($
3ewtonCun 2. :anunu uyulanErsa"
JM Topϕ=Σ &&
( )
( ) h.xxk h.xk 2
h2.xxk h.xk h2.xmh.xm
12223
12212211
−+−
−−−=+ &&&&
ϕ=
ϕ=
ϕ=
&&&&
&&
hx
hx
hx
1
1
1
ϕ=
ϕ=
ϕ=
&&&&
&&
h2x
h2x
h2x
2
2
2
:oordinatlarE1x ve 2x ϕ enelle;tirilmi; KoordinatE cinsindenyaGElErsa"
Dr. Tamer Kepçeler (%
( ) ( )hhh2k hk 4hhh2k 2
hk hm4hm
2
2
32
2
1
2
2
2
1
ϕ−ϕ+ϕ−ϕ−ϕ−
ϕ−=ϕ+ϕ &&&&
( ) ( ) 0k 4k k m4m 32121 =ϕ+++ϕ+ &&
21
321n
m4m
k 4k k
m
k
+
++==ω rad8s
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Dr.Tamer Kepçeler 26
Dr. Tamer Kepçeler ('
( ) 2232
122
2
11p
2
22
2
11k
xk 2
1xxk
2
1xk
2
1E
xm2
1xm
2
1E
+−+=
+= &&
( ) 0EEdtd
pk =+
ϕ=
ϕ=
ϕ=
&&&&
&&
hx
hx
hx
1
1
1
ϕ=
ϕ=
ϕ=
&&&&
&&
h2x
h2x
h2x
2
2
2
:oordinatlarE1x ve 2x ϕ enelle;tirilmi; :oordinatE cinsindenyaGElErsa"
Dr. Tamer Kepçeler ((
( )
22
3
2
2
2
1p
22
3
2
2
22
1p
22
2
2
1k
22
2
22
1k
hk 2hk 2
1hk
2
1E
h4k 2
1hh2k
2
1hk
2
1E
hm2hm2
1E
h4m2
1hm
2
1E
ϕ
++=
ϕ+ϕ−ϕ+ϕ=
ϕ
+=
ϕ+ϕ=
&
&&
Dr. Tamer Kepçeler ()
( ) ( ) 0k 4k k m4m 32121 =ϕ+++ϕ+ &&
( ) 0EEdt
dpk =+
0hk 2hk 2
1hk
2
1 hm2hm
2
1
dt
d 223
2
2
2
1
22
2
2
1 =
ϕ
+++ϕ
+ &
( ) ( )[ ] 0hk 4hk hk hm4hm 232
2
2
1
2
2
2
1 =ϕϕ+++ϕ+ &&&
0≠ϕ& olmalE" bu durumda paranteG içi sEfEra e;it olmalEdEr.
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Dr.Tamer Kepçeler 27
Problem> A6a?$a $en&e -onumun$a %erilen itemin $iferani#el $en-lemini @?-ar?pta!ii fre-an?n? heapla#?n?,
Dr. Tamer Kepçeler (*
y
& m
M
r
:
Çözüm>
Dr. Tamer Kepçeler )+
ϕ
y
& m
M
:
r
( )
2r
x
2r
x
2r
x
ryryx
2
xy
2
xy
2
xy
&&&&
&&
&&&&
&&
=ϕ=ϕ=ϕ
ϕ=⇒ϕ=−
===
Dr. Tamer Kepçeler )1
( )
M2
34m
k
m
k 0xkxM
2
34m
0x 0xxk 4
1xM
8
3m
0xk 8
1xM
16
3m
2
1
dt
d 0EE
dt
d
xk 8
1
4
x k
2
1 yk
2
1E
xM16
3m
2
1
r4
x rM
2
1
2
1
4
xM
2
1xm
2
1E
J2
1yM
2
1xm
2
1E
n
22
pk
22
2
p
2
2
22
22
k
222
k
+==ω=+
+
≠=
+
+
=
+
+=+
===
+=++=
ϕ++=
&&
&&&&
&
&&&
&
&&&
8/17/2019 Machine Dynamics Vibration
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Dr.Tamer Kepçeler 28
Dr. Tamer Kepçeler )2
3ewtonCun 2. :anunu uyulanErsa"
&
1T
xm &&
m 1-Txm amF =⇒=Σ && 1
yM
ϕ
yk
ϕ&&J
1T 2TyM &&
21 TT-k yyM amF ++=⇒=Σ && 2
1" 2 Cnin içine :onursa"
2Txmk yyM =++ &&&& 3
rTrTJ JM21T −=ϕ⇒α=Σ && 4
Dr. Tamer Kepçeler )#
1 ve 3" 4 IHn içine :onursa"
( )
M2
3m4
k
m
k 0xkxM
2
3m4
0xk 2
1xM4
3m2
xm-xk 2
1-xM
2
1-xmxM
4
1
rxm2
xk
2
xM-rxm
r2
x rM
2
1
rxmk yyM-rx-mJ
n
2
+==ω=+
+
=+
+
−=
++−=
++=ϕ
&&
&&
&&&&&&&&
&&&&
&&&&
&&&&&&&&
Problem:A6a?$a-i titre6im iteminin :,erine m -:tlei h #:-e-liin$en $:6:p
#ap?6?#or) M -:tleininhare-et$en-lemini #a,?n?,)
Dr. Tamer Kepçeler )$
h
m
M
:
&
8/17/2019 Machine Dynamics Vibration
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Dr.Tamer Kepçeler 29
Çözüm:
Dr. Tamer Kepçeler )%
h
m
M
:
&m
:
M &
gh2V =
k
mgx0 =
mM
gh2mVx 00
+==&
( ) xmM &&+
kx
Dr. Tamer Kepçeler )'
m :Htlesi ile MCnin çarpE;tEJE anda:i momentumunu yaGalEm.
( )mM
gh2mxVVmM2ghmVm)(MVm
0000+
==⇒+=⇒+= &
m :Htlesinden dolayE : yayE bir mi:tar sE:E;Er.
k
mgx0st −==δ
3ewtonCun 2. :anunu uyulanErsa"
( ) ( ) 0xkxmM x-kxmM amF =++⇒=+⇒=∑ &&&&
Dr. Tamer Kepçeler )(
tsinx
tcosx)t(x nn
0n0 ω
ω+ω=
&
Aistemin tabii fre:ansE.
mM
k
m
k n
+==ω
AnHmsHG serbest titre;im hare:etinin ba;lanEç ;artlarEna baJlEhare:et den:lemi a;aJEda:i ibiydi-
deJerler yerine :onulura?
( )( )
+++
+−= t
mM
k sin
mMk
gh2mt
mM
k cos
k
mgtx
8/17/2019 Machine Dynamics Vibration
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Dr.Tamer Kepçeler 30
Problem> A6a?$a $en&e -onumun$a %erilen itemin $iferani#el $en-lemini @?-ar?pta!ii fre-an?n? heapla#?n?,
Dr. Tamer Kepçeler ))
1k
2k
&y
m1
m2r
0
>
α
Çözüm>
Dr. Tamer Kepçeler )*
1k
2k
&
m1
ym2
r
0>
ϕ
0C α
Dr. Tamer Kepçeler *+
α=ϕ
α=ϕ
α=ϕ
α=ϕ⇒ϕ==
α=
α=
α=⇒=α
cosr
x
cosr
x
cosr
x
cos
x r rzAB
tgxy
tgxy
tgx y x
ytg
&&&&
&&
&&&&
&&
8/17/2019 Machine Dynamics Vibration
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Dr.Tamer Kepçeler 31
Dr. Tamer Kepçeler *1
22
21p
22
2
2
1p
2
2
2
1p
xtgk 2
1k
2
1E
tgxk 2
1xk
2
1E
yk 2
1xk
2
1E
α+=
α+=
+=
2
2
2
2
2
21k
22
22
2
22
2
2
1k
222
21k
xcos
xm
4
1tgm
2
1m
2
1E
cosr
xrm
2
1
2
1tgxm
2
1xm
2
1E
J21ym
21xm
21E
&&
&&&
&&&
α+α+=
α+α+=
ϕ++=
Dr. Tamer Kepçeler *2
( )
( )
( )
( )
α+α+
α+==ω
=α++
α+α+
≠=
α++
α+α+
=
α++
α+α+
=+
22
2
21
2
21n
2
2122
2
21
2
2122
2
21
22
21
2
22
2
21
pk
cos
1m
2
1tgmm
tgk k
m
k
0xtgk k xcos
1m21tgmm
0x 0xxtgk k xcos
1m
2
1tgmm
0xtgk 2
1k
2
1x
cos
1m
4
1tgm
2
1m
2
1
dt
d
0EEdt
d
&&
&&&&
&
rad8s
Dr. Tamer Kepçeler *#
xk 1
yk 2
&
m1
ym2r
0>ϕ
0C α
ym2 &&
xm1 &&
ϕ&&J
12N
21N
8/17/2019 Machine Dynamics Vibration
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Dr.Tamer Kepçeler 32
Dr. Tamer Kepçeler *$
yk 2
ym2r
ym2 &&
ϕ&&J
12N
T
xk 1
m1 0C
α
xm1 &&
21N
&
T
ϕ
3ewtonCun 2. :anunu uyulanErsa"
α+−=⇒=Σ cosNyk ymamF 1222y && 1
r
JTTrJJM
ϕ=⇒=ϕ⇒ϕ=Σ
&&&&&& 2
α
+==
cos
yk ymNN 22
2112
&&
α−α
−−=
⇒=Σ
sinNcos
Txk xm
amF
2111
x
&& 3
Dr. Tamer Kepçeler *%
1 ve 2 e;itli:leri 3 nolu den:lemde yerine :onulursa"
( ) 0xtgk k xcos
1m
2
1tgmm
sincos
tgxk tgxm
cosr
cosr
xrm
2
1
xk xm
sincos
yk ym
cosr
Jxk xm
2
2122
2
21
22
2
2
11
2211
=α++
α+α+
αα
α+α−
αα−−=
αα
+−
α
ϕ−−=
&&
&&
&&
&&
&&&&&&
Problem> A6a?$a $en&e -onumun$a %erilen itemin $iferani#el $en-lemini @?-ar?p
ta!ii fre-an?n? heapla#?n?,)
Dr. Tamer Kepçeler *'
&M
m
0
:
6 6
8/17/2019 Machine Dynamics Vibration
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Dr.Tamer Kepçeler 33
Çözüm>
Dr. Tamer Kepçeler *(
&M
0
:
ϕ
gxAρ
xM &&
kx
ϕ&&J
Dr. Tamer Kepçeler *)
ϕ=
ϕ=
ϕ=⇒
8/17/2019 Machine Dynamics Vibration
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Dr.Tamer Kepçeler 34
Çözüm>
Dr. Tamer Kepçeler 1++
3
33
3
n
3
3
eş
3
3
eş
3
3eş
Lm
bhE4
m
L
bhE4
m
k
0yL
bhE4ym
y-k ym amF
LbhE4 k
12 hbI
LIE48k
===ω
=+
=⇒=Σ
===
&&
&&
rad8s
ym
:e;
Problem> anal#,e the motion of the #tem!# coni$erin& initial con$ition for 2) econ$
Dr. Tamer Kepçeler 1+1
&
y
c1
:1
c2:2
m1
m2
α
Çözüm>
Dr. Tamer Kepçeler 1+2
&
y
m1
m2
α
xc1 &
xk 1
yk 2 yc2 &
ym 2 &&
xm1 &&
8/17/2019 Machine Dynamics Vibration
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Dr.Tamer Kepçeler 35
Dr. Tamer Kepçeler 1+#
2
2
2
1p
2
2
2
1d
2221k
yk 2
1xk
2
1E
yc2
1xc
2
1E
ym2
1xm
2
1E
+=
+=
+=
&&
&&
tanxy
tanxy
tanx y x
y tan
α=
α=
α=⇒=α
&&&&
&&
( )
( )
( ) 222122
2
2
1p
22
21
22
2
2
1d
22
21
22
2
2
1k
xtank k 2
1tanxk
2
1xk
2
1E
xtancc2
1tanxc
2
1xc
2
1E
xtanmm2
1tanxm
2
1xm
2
1E
α+=α+=
α+=α+=
α+=α+=
&&&
&&&
Dr. Tamer Kepçeler 1+$
agrange equation for the eneraliGed coordinates-
0x
E
x
E
x
E
x
E
dt
d pdk k =∂
∂+
∂
∂+
∂
∂−
∂
∂
&&
( ) ( )
( ) ( )
( ) ( ) ( ) 0xtank k xtanccxtanmm
xtank k x
E xtancc
x
E 0
x
E
xtanmmx
E
dt
d xtanmm
x
E
2
21
2
21
2
21
2
21
p2
21
dk
2
21k 2
21k
=α++α++α+
α+=∂
∂α+=
∂
∂=
∂
∂
α+=
∂
∂α+=
∂
∂
&&&
&&
&&&
&&
Dr. Tamer Kepçeler 1+%
o30 Nm/s30c Nm/s40c
N/m600 k N/m700 k kg3m kg8m
21
2121
=α==
====
=
=⇒=
m/s0.05x
m0.01x 0t&
01046.41xx32.57x9.732 =++ &&&
( ) ( ) ( )
ωξ−ω
ωξ++ω= ωξ− tsin
1
xxtcosxetx d2
n
0n0d0
tn&
8/17/2019 Machine Dynamics Vibration
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Dr.Tamer Kepçeler 36
Dr. Tamer Kepçeler 1+'
( ) ( ) ( )[ ]t164.3sin014.0t164.3cos01.0etx
rad/s164.3284.013.31
rad/s3.373.9
41.1046
m
k
284.09.73 41.10462
32.57
mk 2
c
c
c
t9443.2
22
nd
n
kr
π+π=
π=−π=ξ−ω=ω
π===ω
====ξ
−
Dr. Tamer Kepçeler 1+(
Dr. Tamer Kepçeler 1+)
8/17/2019 Machine Dynamics Vibration
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Dr.Tamer Kepçeler 37
Problem> A6a?$a $en&e -onumun$a-i itemin %erilen $eer %e !a6lan&?@ 6artlar?na!al? olara- i@in hare-etini incele#ini,)
Dr. Tamer Kepçeler 1+*
:1
c
m1
:2m2
M"6
Çözüm>
Dr. Tamer Kepçeler 11+
m1
m2
M"6
ϕ
x xm 2 && xk 2
xm1 &&
xk 1
xc&
ϕ&&J
2ML12
1J
2
Lx
2
Lx
2
Lx
=
ϕ=ϕ=ϕ= &&&&&&
Dr. Tamer Kepçeler 111
3ewtonCun 2. :anunu uyulanErsa"
( ) 0k k cmmM3
1
04
Lk
4
Lk
4
Lc
4
Lm
4
LmML
12
1
2
Lxk -
2
Lxk -
2
Lxc-
2
Lxm
2
LxmJ
JM
2121
2
2
2
1
22
2
2
1
2
2121
Top
=ϕ++ϕ+ϕ
++
=ϕ
++ϕ+ϕ
++
=++ϕ
ϕ=∑
&&&
&&&
&&&&&&&
&&
8/17/2019 Machine Dynamics Vibration
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Dr.Tamer Kepçeler 38
Dr. Tamer Kepçeler 112
π=ϕ=
π=ϕ=
=
==
==
===
s / rad180
2x
rad180
6x
0t
m2L N.s/m250c
N/m10 25 k N/m10 51k
kg15m kg05m kg30M
00
00
3
2
3
1
21
&&
Dr. Tamer Kepçeler 11#
( ) ( ) ( )
ωξ−ω
ϕωξ+ϕ+ωϕ=ϕ ωξ− tsin
1
tcoset d2
n
0n0d0
tn&
rad/s33.707216.0135.71
07216.075 400002
250
mk 2
c
c
c
rad/s35.775
40000
m
k
04000025075
22
nd
kr
n
π=−π=ξ−ω=ω
====ξ
π===ω
=ϕ+ϕ+ϕ &&&
( ) ( ) ( )[ ]t33.7sin009093.0t33.7cos1047.0et t666.1 π+π=ϕ −
Dr. Tamer Kepçeler 11$
8/17/2019 Machine Dynamics Vibration
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Dr.Tamer Kepçeler 39
Dr. Tamer Kepçeler 11%
m:
cM"6
Problem> anal#,e the motion of the #tem!# coni$erin& initial con$ition for 2) econ$
Çözüm>
Dr. Tamer Kepçeler 11'
ϕ
mg
x
y
kx
yc&2ML
3
1J
2
Ly
2
Ly
2
Ly
Lx Lx Lx
=
ϕ=ϕ=ϕ=
ϕ=ϕ=ϕ=
&&&&&&
&&&&&&
ϕ&&MJ
ϕ&&mJ
Mg
Dr. Tamer Kepçeler 11(
3ewtonCun 2. :anunu uyulanErsa"
012000512
02
LMgmgLkL
4
LcML
3
1mL
yMgxmgLkx2
LycJJ
JM
22
22
Mm
=ϕ+ϕ+ϕ
=ϕ
+++ϕ+ϕ
+
−−−−=ϕ+ϕ
ϕ=∑
&&&
&&&
&&&&&
&&
8/17/2019 Machine Dynamics Vibration
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Dr.Tamer Kepçeler 40
N/m1012k Ns/m5c m1L kg6M kg10m3=====
π=ϕ=
π=ϕ=
=
s / rad180
5x
rad180
10x
0t
00
00
&&
Dr. Tamer Kepçeler 11)
( ) ( ) ( )
ωξ−ω
ϕωξ+ϕ
+ωϕ=ϕ
ωξ−
tsin1
tcoset d2n
0n0
d0
tn&
( ) ( ) ( ){ }t065.10sin00389.0t065.10cos1745.0et t2055.0 π+π=ϕ −
Dr. Tamer Kepçeler 11*
π≅−π=ξ−ω=ω
π===ω
====ξ
065.100065.01065.101
rad/s065.1012
12000
J
k
0065.012 120002
5
kJ2
c
c
c
22
nd
n
kr
Dr. Tamer Kepçeler 12+
8/17/2019 Machine Dynamics Vibration
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Dr.Tamer Kepçeler 41
+n$ampe$ 'orce$ Vi!ration
The t#pe of the force
• Compoller e=ternal force)
• The force !# un!alance$ mae)
• The force which come from &roun$)
Dr. Tamer Kepçeler 121
Lonsider that the compoller force is harmonic?
( ) ( ) ( ) ( ) ( ) ( )φ+ω=φ+ω== φ+ω tsinFtF , tcosFtF ,eFtF 00ti
0
Dr. Tamer Kepçeler 122
The equation of motion:
3ewtonCs 2nd law"
Fxkxm Fxkxm amF =+⇒+−=⇒=Σ &&&&
m &
:
( )tcosFF 0 ω=
m
xm &&
xk
( )tcosFF 0 ω=
How can we o!tain the e"uation of motion
Dr. Tamer Kepçeler 12#
( )tcosFF0 ω=
Differential equation of the system-
( )tcosFkxxm 0 ω=+&&
The eneral solution-
( ) ( ) ( ) ( )txtxtxtx öhg +==
omoenous solution-
( ) ( ) ( )tsinAtcosAtx n2n1h ω+ω=
yarEcE :uvvet harmoni: olduJu için Gel çGHm Ideharmoni: ve aynE fre:ansEna sahip olaca:tEr.
( )tF ( )txöω
( ) ( )tcosXtxö ω=
1
2
3
4
8/17/2019 Machine Dynamics Vibration
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Dr.Tamer Kepçeler 42
Dr. Tamer Kepçeler 12$
( ) ( )( ) ( )
( ) ( )tcosXtx
tsinXtx
tcosXtx
2
ö
ö
ö
ωω−=
ωω−=ω=
&&
& !
! 1
"
( ) ( ) ( )
( ) ( ) ( )2
00
2
0
2
m-k
F X tcosFtcosXmk
tcosFtcosk tcosXm
ω=⇒ω=ωω−
ω=ω+ωω−
( ) ( ) ( ) ( )tcosmk
FtsinAtcosAtx
2
0n2n1 ω
ω−+ω+ω= #
Dr. Tamer Kepçeler 12%
!nitial conditions"
=
=
=
=
00t
00t
xx
xx
&&ise"
n
0
2
2
001
xA
mk
FxA
ω=
ω−−=
&
( ) ( ) ( ) ( )tcosmk
Ftsin
xtcos
mk
Fxtx
2
0n
n
0n2
00
ωω−
+ωω
+ω
ω−−=
&
Dr. Tamer Kepçeler 12'
n X ω
ωthe variation of by
{
2
n
0
2
n
0
2
0
2
0
2
0
1
1
F
k X
1
k
F
m
k 1
k
F
k
m-1
k
F
Xmk
FX
2n
ω
ω−
=⇒
ω
ω−
=ω
−
=ω
=⇒ω−
=
ω
0mplitude ratio and e&tension factor(Genlik oranı veya büyütme faktörü)0F
k XR =
8/17/2019 Machine Dynamics Vibration
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Dr.Tamer Kepçeler 43
Dr. Tamer Kepçeler 12(
L0A9 1 - ?10n
<ω
ω<
( ) ( )tcosFtF0
ω=
tω
tω
( ) ( )tcosXtxö ω=
Dr. Tamer Kepçeler 12)
( ) ( )tcosFtF0
ω=
tω
tω
( ) ( )tcosXtxö
ω=
L0A9 2- ?1n
>ω
ω
Dr. Tamer Kepçeler 12*
( ) ( )tcosFtF 0 ω=
t
t
( )tx
L0A9 #- ?1n =ω
ω
8/17/2019 Machine Dynamics Vibration
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Problem> Anal#,e the motion of the #tem !# coni$erin& initial con$ition for 2 econ$)
Dr. Tamer Kepçeler 1#+
L,mcubuk
2,e,m ω
x
k
M
Çözüm:
2
cubuk Lm3
1J
Lx Lx Lx
=
ϕ=ϕ=ϕ= &&&&&&
tcosmeF
0F
2 ωω= 321
Dr. Tamer Kepçeler 1#1
ϕ
ϕ&&J
kx
x xM &&