Machine Dynamics Vibration

8/17/2019 Machine Dynamics Vibration

1/44

Dr.Tamer Kepçeler 1

MACHINE DYNAMICSVIBRATION

Dr. Tamer Kepçeler 1


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 #


2/44


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 '


3/44


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 *


4/44


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;


Harmonic motion


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


5/44


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!


6/44


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


7/44


Sprin& contant

Dr. Tamer Kepçeler 1*

(N/m) x

F tank =α=

force

displacement

Sprin& contant ta!le


L

IEk =

L

AEk =

3

4

Rn64

dGk =

3L

EI3k =

L

IGk

p=


682

3L

IE48k =

682

3L

IE192k =

682

3L7

IE768k =


8/44



a b

&

y ( )222

x22bxL

LIE6

xbPy

ba

LIE3k −−==

δ

9! 63L

IE12k =


( ) 2aaL

IE3k

+=

6 a

( )a8L3a

IE24k

2 +=

6 a

Sprin& in paralleel


:1 :2 ≡

m

:e;

m &&

∑=

=++++=n

1i

in321eş k k ......k k k k


9/44


Sprin& in erie


m &

:e;

m &

≡:1:2

∑=

=++++=

n

1iin321eş k

1

k

1.....

k

1

k

1

k

1

k

1

E=ample>


:1 :2

:#

:$

m &

Determine the equivalentsprin constant of the ivensystem

E=ample>



10/44


E=ample>


: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


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


11/44


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


12/44


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


13/44


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=


14/44


+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-


15/44


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

=

=

&&

&


16/44


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=

&


17/44


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⇒


18/44


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⇒


19/44


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 &&&&



20/44


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 &&


21/44


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



22/44


Çözüm>

Dr. Tamer Kepçeler '$

xk

ϕ&&mJ

&

xm &&

ϕ

T

r

Dr. Tamer Kepçeler '%


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


23/44



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


24/44


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


25/44


Çözüm>

Dr. Tamer Kepçeler (#

&2

h

&1

ϕ

m1

m2

2h

23xk

11xk

( )122 xxk −

11xm &&

22xm &&

Dr. Tamer Kepçeler ($


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


26/44


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.


27/44



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++=

&&

&&&&

&

&&&

&

&&&


28/44


Dr. Tamer Kepçeler )2


&

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

:

&


29/44


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


( ) ( ) 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


30/44



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

&&&&

&&

&&&&

&&


31/44


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


32/44


Dr. Tamer Kepçeler *$

yk 2

ym2r

ym2 &&

ϕ&&J

12N

T

xk 1

m1 0C

α

xm1 &&

21N

&

T

ϕ


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


33/44


Çözüm>

Dr. Tamer Kepçeler *(

&M

0

:

ϕ

gxAρ

xM &&

kx

ϕ&&J

Dr. Tamer Kepçeler *)

ϕ=

ϕ=

ϕ=⇒


34/44


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 &&


35/44


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&


36/44


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+)


37/44


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>


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= &&&&&&



( ) 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=∑

&&&

&&&

&&&&&&&

&&


38/44



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

&&


( ) ( ) ( )

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 −



39/44



m:

cM"6

Problem> anal#,e the motion of the #tem!# coni$erin& initial con$ition for 2) econ$

Çözüm>


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



012000512

02

LMgmgLkL

4

LcML

3

1mL

yMgxmgLkx2

LycJJ

JM

22

22

Mm

=ϕ+ϕ+ϕ

=ϕ

+++ϕ+ϕ

+

−−−−=ϕ+ϕ

ϕ=∑

&&&

&&&

&&&&&

&&


40/44


N/m1012k Ns/m5c m1L kg6M kg10m3=====

π=ϕ=

π=ϕ=

=

s / rad180

5x

rad180

10x

0t

00

00

&&


( ) ( ) ( )

ωξ−ω

ϕωξ+ϕ

+ωϕ=ϕ

ωξ−

tsin1

tcoset d2n

0n0

d0

tn&

( ) ( ) ( ){ }t065.10sin00389.0t065.10cos1745.0et t2055.0 π+π=ϕ −


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



41/44


+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$)


Lonsider that the compoller force is harmonic?

( ) ( ) ( ) ( ) ( ) ( )φ+ω=φ+ω== φ+ω tsinFtF , tcosFtF ,eFtF 00ti

0


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


( )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


42/44



( ) ( )( ) ( )

( ) ( )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= #


!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−−=

&


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 =


43/44



L0A9 1 - ?10n

<ω

ω<

( ) ( )tcosFtF0

ω=

tω

tω

( ) ( )tcosXtxö ω=


( ) ( )tcosFtF0

ω=

tω

tω

( ) ( )tcosXtxö

ω=

L0A9 2- ?1n

>ω

ω


( ) ( )tcosFtF 0 ω=

t

t

( )tx

L0A9 #- ?1n =ω

ω


44/44

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 &&

Documents

Machine Dynamics Vibration