Chapter5 Vibration Ahmedawad

V. VIBRATION OF UNDAMFED TWO

DEGREE OF FREEDOM SYSTEMS

Mohamed T. Hedaya MECHANICAL VIBRATIONS First Edition, p. 5.1

5.1. Free Vibration

5.1 .1 . Equation of Motion

mrXr--krXr-kz(xr-xa)

mzXz=-kz(xz-xr)

fTlr Xr + (kr +kz)xr

mz xz + (-kz) Xr

[*, oll*,i*[k,*t,L o mrl LxrJ L *k2

Iml {x}+[k]{x}={0}[mJ = mass matrix,

{x} = displacement vector,

+(-k2)x2 = 0

+(k2) x2 = 0

-t rlJ*,1 = Ioi

k2 I l*rj LoJ

k, (xixJ

k, (x5x,)

(differential equation of motion) (D.E.O.M.),

IkJ=stiffnessrnatrix,

i0) = zero vector.

Mohamed T. Hedaya MECHANICAL VIBRATIONS First Edition, p.5.2

&Bt-' \' -

l-"i ilG { (

>*+I

SCI ,'?^{ Jr.-. Y _J-t' I ---t-ll-:

.&,r,,,

It"

Fo,lA-it" ;

i'l/I A''- I l^ tl

\-, I ',

ao,;

l.:--!-rt,

I

""\

{;7 { ---,xI'l)-/ L-.,.. I

, i -' fl. 7' ,:.,,t'.rI

\tlt

++

l') O ,r Ze*<> sol..li.rrr

Ao*

*r11\s'

r lli I

fir_UN ,.,4t)i:-4-(nl,,,;.ll.1.'t"ij, .-. if )

t,-LflA-*ItP?

. _-__.]

-+J -!

+i Ji I r "'t--.1 - I

7----l-#

,i II

'2j+',r ^;^\ _ (_)

Ik

Assume {*,i - rAl B) -} J*,}

lxr) = t*J

cos (utt - F) -) turJThis means that, rn1 and mz are moving in harn

same circular frequency (u) and phase angle betr

This is called a principal, or normal, mode of vibratic

-*'[T' J,]i3i cos (,r,t-F) + [-:;i' ;:']i:

L -k2 - ro2me + t<, _] [eJ [o

rrlr Inz uo- [m1 kz + mz(kr + kz)] toz+ kr kz = 0 (C E )

The characteristic equation is a iqr)-dratic equation in ta-1.

Therefore, the possible solutions of the D.E.O.M. are

{x;} = {il} "o,

(,r, r - F,) or {l} = il;} "o' ('lo't-F')

Therefore, the general solution (E.O.M") is expressed as

{X;} = {l;}

.,"" (ar, t- Fr) . i3;} "o, (wzt- Fz) (z),

'-L), *uuo r..--'-1!.,,-- --- ,^,-,, tCI, ( 002Ar, Az, Br, Bz, F,, F, ar6 determindb*nylriitirf conditions [xr(0), x2(0), xr(0),

xr(o)1.

Assume{;;} ={l} cos(urt;; -; {iT=;{31*,.**This means that, IIll and mz are moving in harmonic motions with the

same circular frequency (u) and phase angle between them 0o or 180o.

Thisiscalledaprincipal,ornormal,modeofvibration.Substitution>

-*'[T' J,]i3i cos ('lr':t'F) + [-:;i' ;:']il] *. (,t ,u,= {:}

-'7 L -k2 -r'*,*L,_lleJ-tol tI,,

For nontrivial solution,

(- ur' rnr * kr + kzX- ,' *, + kz) - (- kzX- kz) = 0 -)Irlr ITlz t^t*- [m, kz + mz(kr + ke)]ulz+ kt kz = 0 (characteristic equationXC.E.)

Mohamed T. Hedaya MECHAI{ICAL VIBRATIONS First Edition, p. 5.3


[* r'*, ,-,k1 + k2 -kz l{li = {oi (1 ) ->L - k2 - a'mr+ t, _J leJ |'oj

(-,^r'fftr *kr +kz) A-kzB =0 (3)

- kz A * (- tt'ITlz * k2) B = 0 (4)

(3)and(4) ->amplituderatio(R)= E --uJ2m,r+kl +k' =- =k'1--/ A kz -o2mr+k,

Substitution by rrtr snd tr:2 gives two values for R, denoted by R1 and R2.

This gives Br = Ar Rr and Bz = Az Rz.

Therefore, the general solution (E.O.M.) (2) may be expressed as

[] = o, {;,}

,". (ur, t - Fr) + Az [i cos (to2t - Fz) (s),

Ar, Az, Fr, Fz are determined by initial conditions [x1(0), xz(0), i,'(0), Xr(0)].


5.{.2. Modes of Vibration

[x,'] l1'l , , ^\.. t'1lt-;l = o,t*,i cos (t'r' t - F,) a Az

t*ri cos (trr2t - 92) Q)

I General solution | = l- First mode

--l + F Second mode --{

. ln general, each of m1 and m2 moves in periodic motion consisting of two

harmonic motions with circular frequencies tll1 and {ri2.

' For Az = 0, E.O.M. i, I*'l f 1 I

..xr.J = ot

t*,,J cos (t'rr t - Fl) (first mode)'

then, the ratio of amplitudes of m1 and m2 is 1 : Rr * first mode shape.

. For Ar = 0, E.O.M. is {*,| } = n, {i } cos (ur2 r - Fz} (second mode),Lx,J . LR,] L 1., \

then, the ratio of amplitudes of mr and mz is 1 : Re ---+ second mode shape.


Example

Find cr)1 ond u)2, a,nd plot the mode shapes.

Find the equation of motion for initial conditions,

x1(0) = 0, x2(0) = 4, ,xr(O) = xe(0) = 0.

2mx,=-lkxr-k(x1 -x2)

mXz=.k(Xp.Xr)

2 m X, * (2k + k)xr +(- k)x2 = 0

mXz +(-k) xr+(k)x, = 0

[2* olj*,i. [2k+k -klj*,1=JoiL o *jlx,l - [ -r, x ]l*,j-iojlml {x}+[k]{x}={0} (differential equation of motion) (D.E.O.M.)

Mohamed T. Hedaya MECHANICAL VIBRATIONS First Editian, p. 5.7

[2* 0l I*,1 * [2k+k -kl J*,1 = JolL o m_l LxrJ [ -r< k .] L*r.J [.oJ


Assume {x;i

= {li cos (ut-F) -> {;;} =-ur, {l} *- (t^rt-F)

Substitution -),'[T -l]il] cos(,r,t B) .[1 J]{l} cos(rrt u,={:}

For nontrivial solution,

(-2 w'm + 3 kX- u' m + k) - k2 = 0 -)2m2ut' + (- 2 m k- 3 m k) ur'+ (3 k' - k') = S -)2m2 ulo -5 m k w2+zk2= 0 (characteristicequationxC.E,)

ZnW-ffi (CE)

,tir'=fi(Zrn)fR-

h)r = l--il2 m

[- 2ur2m + gxIL -k

2klmpkx"

*k lJAi = i- o2m + t_l [n] t

(6) ->

- 2ul2m + 3k k

k -to2m-rk

W'=Wz2=Zkl[fl -]

Rt= 2

Rz=-1


Rt= 2Rz=-1

First rnode Second mode

r'i1f,:Mode shapes ,\^"^9l

Mohamed T. Hedaya MECHANICAL VI First Edition, p. 5.10

in\l io.t Con rJi(io^tIro,

I\I .:.s._ iy),\ -1,.1

\-1 I j

r: I 1.N\, 1

'-\ln k :i iP"- 1*r \tt

f,:

+ A,Ct-) ,' t r-

',

II)cr"(c't*l'-''L\ J

<V/\t i'/\,

I rz \\X, }\-J

{^ i ,l

i 'u." i'

\ rT r\*"i

" y(c) =X,(o) * c;or7:fu, Ai,inF, ilI\.oJ ,i,

11 I

"\ l.

-'rJ./\-i tL i

[*r

,)lr l- tP

i( / 'J,r-l'tl)

\r , ,,r {,:2Jirr,(d,r,t f r

ll"\{t(,t.), * o,l\ IJ''JL 'J- 5z

i.r',i

r_r.. I

a/ l^/ V\

,\r\l

l-,\

,1 * '-lI,-/ t

. i,i,

;r07

(\ 1)CIt- )

+ i$z Aa

A,+ o

A'*u

*(J

r\J

S'nFrr rc), i,'r f^ : rI-

:A)' \.',".d, t +[2 J

a\Ali\r j ''i - '' JJ, (

\ *i i

(7\.t(-r\'\]\\-r J

r]

2_

A

Arx

AA

:

A,

:")

l1

t

Xf ir) =n, Y,

,-/, li _-.',tl

t\/1\ A2 j(''. t) I

ln l":\/1 j

/\/ir t,A-

, l[

O *q;r)na /1 r

+4.t

i.

-LJ=A

/'1

/ilJ {r r"h

A,= +Ar =

Atrz

The equation of motion (general solution) may be expressed as

ffi =o'fi,] 'o"

(t*rr t-Fr). " {;i 'o'

(c*rzt- F,),

tI- Ekk)t = V2m' tu), = il;-'Rr=2, Re=-1

lnitial conditions [x1(0) = 0, x2(0) = A, Xr(Q) = *e(0) = 0J -]Ar=A/3, Az=-A/3, Fr=Fz=g-)

Mohamed T. Hedaya tUfCUnfrttCAt V

I') ,o* [12) t,

[*,'l1|lxz.J

x

_13

x1

Ae

_243

,ffi') t{i,} *'[,ff') -' r=znt^t*Tzm


5.1.3. Vibration of Unrestrained Tvuo Rotor System

5.1.3.1, Restrained and unrestrained 2 D.O.F. sretems

" r/- -o-. \*/"

Restrained systems Unrestrained systems


5.1.3.2. Free Vibration af Unrestrained 2 Rotor Svstry

lr0r=-Q(0r-02)

lz0z--q(02-8r)

l,6r +(q)01 +(-q)02 =

l, 0, + (-q) 0, + (q) 0z .0

0

il ll i;;i .[ ; l] fi;]t'0.l

-l (

L0j

Assume {3;}

= {l} cos (ot-p) -> {3;}= "{;}

cos(u':t'B)

Mohamed T. Hedaya MECHANICAL VIBRATION$ First Edition, p. 5.14

{:J* * c

u)' :

[?,

R.,

P. * u,f i, _i"nrr' llt

t- l(I, +Ir) fi r t1'- / \zr' l- \fl/

-/Jr'i + \l"v *r2 ,y

t/1t)I * 4-

t.--!,-

c

-{- *---*F

-rt-_tLIl1

Ilr--.L - I

-"-.@*#/r'.P'

J

lr*t-- L

Substitution -)-,'[

;]{3} cos(u,t u,.[-;

L -q -*'tr*qlLaJFor nontrivial solution,

fl{3} cos(rrrt-u,=

i;}

= {:i r)

lrlzua-(lr+12)qal2=0,'[f, lzwz-(lr+lr) qj =o

uo2=0

(C"E.)-) ,5N, t'-: \-s *==l

wf=t#Mohamed T. Hedaya MECHANICAL VIBRATIONS First Edition, p. 5.15

rtrJO=0i ,;.-1 r, rnuh,

(^: uitr^t,!. .ro{r'\' 1' ri''

[-r'tr +Q *q

L -o -w'1,

[rT- flr/o [,,

. u.]

oiio^ Jun l - 'ii: ( i';l'ai '

= {3} (z) ->

h)1 =

\};u

I lnll1' j

- -r.d2 lr +Q

q

+q

*a2lz +Q

(r)'=uJo2=0-)

' 11 12

rRo='l

aRr=_

+ no \ i7\a cki;'n''-.u'


Q'*",

T"*\--.** -'-*--t - - t_J\

sLcJl

It

qV

[" .jef 55.]etd""'---re-. Uflilorrt

{r)=tu)o=0

Ro=1

0r=02

Shaft and rotors move as a rigid body

Mode shapes

k)=0)r=ffiRr = -hllz=-LzlLt

0r=Acos(trt1t-B)

0z= - tltll;) A eos (urt - F)

Elastic mode

Mohamed T. Hedaya M ECHAN ICAL VIBRATIOT.IS

Note

Node divides the shaft into single D.O. F.

equal to the system frequency.

Examplg

Torsional stiffness of shaft = q

First Edition, p. 5.17

systems having frequencies

I o2lNode

3q


,'f) P i') i x Y,t'.U3rr\t'i(rl ;'ri; rr-*

rlTOfced'

5, Z

Itia

e

l

LWo

eol,-'* &\,

c.rrl X,U

X, l'.'^f,r, r ce

1

Y-'\ L ,(o,-I +v<.lp

't

I1",:r,: t',

I

:i -,t {i^",L)O \1AA,-+

l-.Lf \r,^.n ::''lt\,

nLjllo( fcr*t h

?o" (r-'t 1' ,'

[,,1 t

ij'l, ':.r'.1"1+,'1.,]

I

t\-.:re(n\.:r G

/ ., \+ ' - U'Diaf Po\-r tto"r

1, - /- I IrJA,

r/il\.t

fro*rf.,fco

2-

t',t ( <trtf.,.t *rL

5"1.3.3. Free Vibration of Rotatins-Unresttalfled 2 Rotor Svgtem

. When the system is vibrating without rotation,

0r=Acbs(ul1t-B) and 0z= - (l1llz) A cos (t^trt - B) ->0r = - ur Asin (u1t- p) and 0z= urr (lrllz)Asin (trl1t- p)

. When the system is vibrating while rotating at a uniform speed Q,

0r = Q-ur Asin (t*t-F) and 62= CI + t';r (lrllz)Asin (urrt-F)

01

o


5.2. Harr-ngni$altv Excited Yibrqtion

5.2.1. Steady State Response

rnr Xr = - kr Xr - kz (xr - xr) + Fro cos tltrfiz Xz = - kz (xz - X.,)

rflr Xr + (kr+kz)xr +(-k2)x2 = Fro cos tltrfiz xz + (-kz) X1 +(k2) x2 = 0

[*, o I {*,} . [*, 1k, n,l {*,} = {t1o}.o. r,I o ,rl IxrJ L -kz kz .j [*ri - i o j --- *'

[m] {x} + [ k] {x} = {Fo} (differential equation of motion) (D,E.O.M.),

{Fo} = force amplitude vector.


-----+ ()-s\ u^n1e_

*

\

A.

{u1- ,..^it' ,'i ,l_ _ )rt1{)-l f"''' \r'r':{

.L, * I .J \o, :)t I

( Un k ^o w4 u'ttt $z ?,')

tl-,f ;:tt c+. ;'t i-' ,-'1

i.: ,''l \

,:. lk f r' ' r'.

I \:-7 /-t/.{)I \v

y * y'* ** i1*

,,1

X, --\./r i I ''

i""* J

^t^Pr/)a:,,

-J "*J& ci*j-*

en3\'t ..j

V +v-(

X 'Je

, !I

--* o

ICl.r [no* n ( $".,,..

\

r/ l)YA/ \ , L_l-,''I \

Vr\2 q licn in 2 unk, ow4

Chl c {r}c{of

2 *qu

jr*^,

BI [AI .,.-: Ir\ I l(LJJ L

I

q l-'r)l

-1

\r -Bl

L-. AJAb - tt

{x;} = ix;} "o*

,r,t -}Substitution -)-*'[T'

.1,]iX;] ,o,,r,t + [-:;i' -:']{X;i cos,ot = {?} cos u,t

L -k, -^'*r*kr]LxrJ LoJ

Assume the steady state solution

o [Z(ur)] {X} = {Fo} (8),

tz(t*)l = impedanee matrix,

(8) ->{X} = [Z(ur)1-1 gol

r[Z(ur)] =-k)2[m]+[k]{X} = amplitude vector,

fi;) ='ur'{x;} cos,r,t

F,o

Mohamed T. Hedaya M ECHA}.I ICAL VI B RATIONS First Edition , p.5.21

(9) ->

(10) ->

iX) = [Z(r^r)]-' {Fo},

(*t^r'*, + k, + k, )(-ur2m , i kr) -GI*, I =LX,,J

tz(r.,r)] =[-r'2m'+k'+k, :k, l-tL -kz -tJ'tnz*kr_j

kz IJF,.I-ut'm, +k1 +trl L o J

I- *'*, * kr lirrl

[-'I; "*'i cos,t

Equation of motion is

I*,I =L*, J (*t^l'*., + k, + k, )(-ul2m z +kz) *kr,


det[Z(crt)]

(-tu'm, + k, + k, )(-u.rzm z +kz) *kr' (10) ->

Notes

r Positive X1 rresns that, the motion of m; has a phase angle 0o, while

negative X; mesns that, the motion of mr has a phase angle 180o.

fx,, 1=* and lxrl =oo

r At u.r = 4f;lmr, (- w'mz + kz) = Q -)Xr=0 and X2--F1glk2


J- *'*, * tr lI t, ]

Example

mr = 100 kg

kr = 1000 Nlmm

trtz = 25 kg

kz = 160 N/mm

Fr = 1000 cos trrt N, t is in seconds.

Plot the frequency response curves.

v - 1000(-25w2 +160x103)Xt= effi' X2=1000x160x103

det[Z(urfl '

det[Z(ul)] = 2500 uro - 45 x 106 u]2 + 1G0 x 10e,

Attrr=69.83 rad/s or 1146 rad/s, lX., I =oo and lXrl =oo

' Jkffi = 80 radls and Fn I kz = 6.25 ffiril -)At trt = 80 rad/s, Xr = 0 and Xz= - 6.25 mm


At trr = 69.83 rad/s

At ul = 80 radls,

or 114.6 rad/s,

Xr=0 and

lx, l=oo and

Xz--6.25 mm

lx2l=*

lxrl(rnm

15

10

120 160 {lJ 200

15

lx,I(mm

10

II

/

\J

\/

r\

Mohamed T. l-ledaya MECHANICAL VIBRATIONS First Edition, p. 5.25

5.2.2. Vibration Absorber

Examnle

mr = 100 kg Iilz = 25 kg

k1 = 1000 N/mm

kz = 160 N/mm

Fr = 1000 cos urt N,

t is in seconds.

For (a) and (b), plot J X1 | versus ur.

For the main system,

k)n = Jqln1

= Jt o67t oo = 1oo radls (a) Main system (b) 2 D.O.F. system


E = k, x.= 1..(=L)

l-1..,

a rca i3- /Y1 t i: ; \o.lio^*.,,{", = lo, r* & opp'l il< J,,* t{i(.,"

L d,.,

su!.,r.;r'i<,J

Notes

o The amplitude of the main spring- 15

mass dystem (kr, m,) can be I X, I

reduced, through a certain range of (mrn)

ur, by addition of the spring-mass 10

system (kz, mz).

o After addition of (k2, mz), when

u =S7m2, fft1 stays stationary.

Then, Xz = - Fro I k2, ofid the force of

the spring kz on rflr is equal and

opposite to the exciting force.

r The system (kz, mz) is called a vibration absorber, or dynamic absorber.


Tuned absorber

Let unl = .8 andVmr

For any absorber, X1 = 0 when o) = b)nz.

ln many cases, absorber is designed so

that, Xr = 0 when kJ = td,.,r.SUch an

absorber is called tuned absorber.

For tuned absorber,

kn k.,=._)ml m2

ffi, =k?m1 k1

&)1 (rJnt

Un2

(l)2


tr Y eiti*d\ r.,

\;,r*;-*-\ 6t{*'-"*

\ f^L

Ir<+urr"3C,lJ

CO^r

f.U ^,

{JJ \

LUr

*..-.,---.*.**-*-.ff!

*o(.4ion "??_ D,o,

a.. -,LbPI

Let oU=

bJ

0nt

k2l-rkl

tu),- r, o12=Unt

ffi2=ml

.[r @2

0ntlx, l

Characteristic equation is

Ifir l'nz ,o- [*, kz + tTtz(kr + kr)] u'+ktkz=0,

and its roots E[€ trlr2 and {r22.

Division by k, kz -)f-Q+p) f+1=0 (1 1),

and its roots o[€ r12 and r22.

Mohamed T. Hedaya MECHAN'CAL VIBRATIONS First Edition, p. 5.29

ro-(z+p) f+1=0 (11) 3

has the roots r12 and r22 -; r

srtzarzz=Z+V Z

.ftfz=1

Every value for p gives one positive

value for rr and one positive value

for 12.

Note

Small p gives

1- small frequency range for. amplitude reduction of m1,

2- large amplitud€ of rfie.

,?

ft

0 0.2 0.4 0.6 0.8 tJ 1

Mohamed T. Hedaya MECHANICAL VIBRATIONS First Edition, p. 5S0

5.2.3. Forced Vibration of Unrestrained Two Rotor System

5.2.3. X., Sferaoff Sfafe Response

l,6r =^g (0r -02)+Trocosrut Ilz0z=-q(02-0r)

1., 6r + ( q) 01 + (-q) 02 = Tro cos tot

lrdz +(-q)0,+ (q) 0z = 0

[', ol I6,i . I q -ql It,] = {r,o} "o.,rtL0 lzl [.e, j L* q q_] [erj L 0 j

I I I {6} + [ q ] {0} = {To} (differential equation of rnotion) (D.E.O.M.),

t ll = inertia matrix, [q ]=torsional stiffness matrix,

te) = angular displacement vector, {To} = torque amplitude vector.


Assume the steady state solution

it,I = {e,o } .o, tirr -} i6.,l = - u), {?,t } .o. ,tlerl

- lerof ''os url EP

t6rl - - t^'

lerof '-o *

Substitution -F,'

[x i]iil:] "o*,* . [-; l]{3;:i "".

krt = {x'i cos urt

-'7 L -o -t^t'lrnq-ltorol-ioJtz(uI {00} = tTo} (2),

tz(t t)l = impedance matrix, i0oi = angular amplitude vector.

(12) ->too) = tz(o)l-' [|.o]


.L

i f *,*J1 f',l n{"uC

r') o Phon:**?:r) {* '

iuI

'[p{e,"1t",,,

,1,',, , fo

sl J, * ('\

)(

O or \EO

9," ( O"o A&BnoI

Tn ((Ll(y V \

I!

*r*l

/r(lV

a\n

/;\tr-)

e

TanQ.u* 3n

:1 ,

: -{l

**1-' {

Jrsr tj-r(l

0, O.)

l'o^ 1''S, SJ

q, f".,^ I

,?].,,,..,-* |

-o21,, +q -q l_,-q -wzlr+q)

{00} = tz(ur)l-' ft-o}, tz(t t)j

1

det[Z(u)]

T,o

ur' [f, l, @'-- (r, + l, )qJ

.')(13)

[-ur2lr*q J l{r,o}_,L o -ur'l,*q]1 oi-ie,o Ilerol

-

f - w'1,JL)Lq

[e,o Ile,l -

J- t^l't, + q]

Iq )COS trtt

The steady state response is

Je,l = liofe, J *' [], t, @'- (t, + t, iq]


{-,,r'r, * o} (13)

l.q)T,o

ur' [], l, @'- (1, + r, )q]

[e,o ]

ie,o i -

5.2.3.2. D:_ynamb Tarque on Shaft

Dynamic torque exerted by l, on shaft (Tq) = Q (0r - 0z)

1.,.s./' ' - ^

r T6 = T66 Co$ tdt, t Too = Q (0ro - 0zo) (4)

(13), (14) -> I do - [, ur{r t +b)q --

r Ar resonance (, = hr1 = ,F F ,, ), I Too | = *.

Mohamed T, Hedaya MECHANICAL VIBRATIONS First Edition, p. 5.34

t-ldo - -QTrolz at(l)=uJ1 = lroolIr l, w2 - (ll + l, )q'

Too

.tuJ(k)r--+Too>0 'u>h)r->Too<0


'UJ<b)t--+Too>0 'h)>{dr--+Too<05"2.3.3. Additign of a StatiL-Tarque

ln many cares, the external torque on 11 is given by

Tr = T* + Too cog UJt, T* is a constant static torque.

Therefore, the total torque exerted by 1., on shaft is given by

rT1 =T"+TooCOS1111t

T,

r;

T,

1

Moharned T. Hedaya MECF{ANICAL VIBRATIONS First Edition, p. 5.36

Tt = T. + Too COS UJt

5.2.3.4. ..Torsional Stress in Shaft

Each torque produces torsional stress in the shaft given by . = "Ind'

Thisgives . r, = r* * rd' costrlt, . ru = ft a rdO = TP

[*. *,]


5.3. EqHlvalent Torsignal Svsterns

5.3.1, Equivalent Shaft Length

Mohamed T. Hedaya MECHANICAL VIBRATIONS

Gnd^a,v

32 L"d4 d-4

LLe

Gn(Da*d4) Grd-arv

32 L"D4 - d4 d^4o ___ = ____:_

LLe

MECHANICAL VIBRATIONS First Edition, p. 5.38

\ /n llJ, \J, L-

V-1

GB

. d..

. L{-*

. "iL

s c&/n(

sqY?\a

tl1rn(

Lt'rn0*s.,*\.

.-*-*L5

HO,d rfin& = {nb \.,

tt,---.'

a,a-'e,r 4 '3

'(

\c\v/\'< OG

\

i\,\,"L

.:-' *l'")l/,lt

',',/

,4 ; 7 cos{,t)l

. t i"-t i t4p---:1}0.1 "') **

Sp*nJ,-f J4\ , f . ]*-n,_?

|:-i"

rn tf?l-,_*;-)'

5.3.2. Equivalent Mass Moment of lnertia of Slider Crank Mechanism1",. = moment of inertia of crank

ms = mass of slider

rncon = mass of connecting rod

lTlR, tTtB are masses at A and B equivalent to fficon

fT'16 =fflconb/LlTlg =fficonalL

r Equivalent rotating masses (m,t1) = l+ + fflAr*

r Equivalent reciprocating masses (ffi,.*") = lns + rln

r Equivalent moment of inertia of mechanism ( l* ) = (rn,o, * 1 m,"") r,


5.3.3. Equivalent Geared SystemGear i1)l"t^t Tl

Refrenseshaft

Refrenceshaft

Let n=

Gear {2}

Gearedshaft

Geared system

speed of geared shaftspead of reference shaft

Equivalent system

"l4u=n2 lo-P'Qz* = n2 qz

r Qie = tr'q,,"l:*=n2 la

e11*:P2 1'



533\r>

o

--r^

5.4. $olved Examples

Example 5.1

a =20bmm

lrlr = 6 kg

kr = ke = 240 N/m

b = 100 mm

fflz=4kg

k2 = 96 N/m

F = 2 cos 4t N, t is in seconds,

Find the natural frequencies.

Plot the mode shapes.

Find the steady state response.


Example 5.2

Without absorber, resonance occurs at

u = 100 rad/s.

Trial absorber, with ma$s 8.1 kg and tuned

to 100 rad/s, gave

First Edition, p. 5.4'1

hlr = 80 rad/s,

lxrl= 5 mm.

Find mr and kr.

An absorber tuned

used to give,

trlr s 85 rad/s and

Find fiiz, kz and X2.

125 rad/s,

to 100 rad/s is to be

trlzI115radls.


Example 5.3

lr = 0.5 kg.m'

q = 60000 N.mlrad

Tr = 500 + 320 cos 200t

lz = 0.25 kg.m'

t is in seconds.

Find wr.

Find the steady state response

Plot the total torque on the shaft versus time.

Mohamed T" Hedaya MECHANICAL VIBRATIONS First Edition, p. 5.43

Chapter5 Vibration Ahmedawad

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