VI. VIBRATION OF UNDAMPED

MULTIDEGREE OF FREEDOM

SYSTEMS

Mohamed T. Hedaya

6.1. Free VjE!'ation,

Example

mr Xr - - kr Xr - kz (x,, - xr)

nlz *z = - kz (xz - xr) - ks (xz - xs)

meXg=-ka(xs-xz)

IIlr Xr

rns Xs

tml tx)+[k]{x}={0}[mJ = mass matrix,

{x} = displacernent vector,

MECHANICAL VIBRATIONS First Edition, p.6.1

I,

+ (k.+kz) xr+(-kz) x2

mz xz + (-kz) x1+(k2+ft 3)x2+{-fu)xs = 0

+(-k s) x2+(k3) Xs = 0

-0

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

Ik]=stiffnessmatrix,

{0} = zero vector'

Mohamed T. Hedaya MECHANICAL VIBRATIONS First Edition, p.6.2

tml {x} + [k] {x} = {0} (D.E.O.M.)

Assume{x}={A}cos(urt.F),{A}=amplitudevector>

- urz [mJ {A} cos (ut - F) * t k] tA} cos (urt - F) = {0}, -}

[ -,,r' [m] + t k I l{A} = {0},

For nontrivial solution,

det[- rll2 [m] + [k]]= O (characteristicequation) (C.E.)

. C.E" is a third degree polynornial equation in u:2.

Rule

For n D.O.F. systems, C.E. is an nth degree polynomial equation in urz

MECHANICAL VIBRATIONS First Edition, p. 6.3

Free Vibration of Unrestrained Three Rotor System

Assume that, (qr I lr) * (qz / ls)

l, 6r = - Q (0r - 02)

bAz= - q (02 - 0r) - g (02 - 0s)

lr6s=-q(0s-0r)

tll{0} + [q ]{e} = {0} (differential equation of motion) (D.E,O.M.),

I lJ= inertia matrix, IqJ= torsional stiffness matrix,

t0) = angular displacement vector, {0} = zero vector.

Assume {x} = {A} cos (t..tt - F),

- {r2 [ | ] {A} cos (tot - F) * tq I {A} cos (urt- F) = {0}->[-ur'p]+tq1HA]={0}

Mohamed T. Hedaya MECHANICAL VIBRATIONS First Edition, p. 6.4

T_lirs/ ,lr,.n[,* , ,.L

I

tr- ttt.

-i

^\ -f-(l\ *L_

f ,r.J)

-L3

-**,Vcr&(s-

J! l_, t t,t^tQ{,f"4h du( a

*c-.yt*",i:,t'

{"1 ,,.i.*J'

, N*I si.;t{ 1, L ,{It,rf').\, h*li r/i(14...,. q, ;4-i- .,)

-llt.l*L

il tl'.,t

*-. .2[;tttn'.*t (t0].J;q ;,r1)':r1,"l-i\,',n

--p erc,.n/14 ('*l) s*\.v-*- -tt f'o".-..

flr< B)1oyq

i.a,\u"e \t"

"f L"*-.

n-o-c\'.

()4(q

t-or*

f ..r c:-<

.r*rd

[-ur'1]*tqll{A}={0} t

For nontrivial solution,

. |r iz lsl tr6 - [q,1( 11+12) lr+ Qz( 12+13) 1",] oo + [qrqz( 11+lr+;r; ] trt2 = 0 (C.8.)

t^

Rigid body mode

(q,/1,) . t^t,' . (qzllr)

First elastic modet\\-,, ,!',n' ,tl''l)

t$22 ) (qz / le)

Second elastic mode

Mohamed T. Hedaya MECHANICAL VIBMTIONS First Edition, p. 6.5

6.2. Hjr.rmon icallv Excited Vibration

Exarnple

rflr Xr - - kr Xr - kz (xr - xz) + Fro cos tot

mz xz= - kz (xz - xr) - ks (x2 - &)

msXg--ks(xa-xz)

rTlr Xr + (kr +kz) xt+(-kz) xz= F rc cos trtt

rfiz xz + (-kz) x1+(kr+k3) x2+(-k3)xs = 0

rns Xs +(-ks) x2+(k3) Xe = 0

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

{Foi = force amplitude vector.

Mohamed T. Hedaya MECHANICAL VIBRATIONS First Edition, p.6.6

lml {x} * [kI {x} = {Fo} (D"E.O.M.)

Assume the steady state solution

{x} = {X} cos rrrt,

tz(t t)l {X} = {Fo} (1), tz(t t)] = im'edance matrix

. Equation (1) represents 3 simultaneous linear equations in Xr, Xz, Xa.

Solution gives the values of X1, Xz, Xe.The steady state response is

{x} = {X} cos urt

Rule

For n D.O.F. systems, substitution by {x} = {X}cos rrrt gives n simultaneous

llnear equations in the amplitudes Xr, X2, ... Xn.

Mohamed T. Hedaya MECHANICAL VIBRATIONS First Edition, p. 6.7

Forced Vibration of Unrestrained Three Rotor System

Sfeady Sfafe Response

1., 6r = - Qr (0r - 0z) + Tro cos t'ot

l, 0, = - q, (02 - 0r) - q, (02 - 0s)

lsBs=-Qz(0s-0:)

+ (q,) 0r+(-qr) 0, = Tro cos trtt

lr dz + (-q,) 0r+(qr+qz)02+(-qz)0s = 0

lr or

+(-qz) 02+(Q2)0s = Q

t ll{6} + [ q ]t0] = {To} (differential equation of motion) (D.E.O.M.),

tTo) = torque amplitude vector.

Mohamed T. Hedaya MECHANICAL VIBRATIONS First Edition, p. 6.8

ls 0r

I:-

ir l',tt (t,. inlntur--* -..-.--*i r---- --] I O[r \ t'L'

6 ,b\/.a klr It()r u?l

fo ll

l-.-I- I

-q{l

sha? t^*)I \--l l:L

,0l orAx< \ {cs'"4'

I

=Lr9, =

{o rgu'x tr *"'"

o (O'L=--rr^'

"*.*L t fo

A\}Vrl'-t-ln L

\

i ,r1

l.t\zlD

A

I

T

^lL)ta .+ 5s

,-t^tu a

:l,?

rD r- iu,^,

\,," 4;.", ,4

,rln,n., q 'T['

I61t.." [,.^

Jr, ,ri ^

' 5-sn ''itr

i -.{..,\'-,.

t I I {6} * [q] te] = fi-o] (D.E.o.M.),

Assume the steady state solution

{0} = {00} cos rrlt,

[ - ur' I I J + t q ] I {00} = {To}-}

tz(t^:)l {00} = fl-o} [Z(to)] = im'edance matrix

. Equation (2) represents 3 simultaneous linear equations in 0ro, 0zo, Oso,

Solution gives the values of 01s, 0zo, Oeo. The steady slate response is

{0} = {00} cos r,,rt

Mohamed T. Hedaya MECHANICAL VIBRATIONS First Edition, p. 6.9

Dynamfc looues on Shafts

Dynamic torque exerted by l, on shaft (1) (Tor) = q, (0, - 0r)

I Tor = Toro CO$ ut, r Toro = Qr (0ro - 0zo)

Dynamic torque exerted by lz on shaft (2) (Toz) = Qe (02 - 0g)

o Toz = Tozo COS tJt, r Tczo = 9e (0zo - Oso)

Ad d it i o n al a Stati cJo rq u e

Total torque on each shaft is obtained from the relation,

total torque = static torque + dynamic torque

Iorsrbnal Sfresses in Shafrs

Each torque produces torsional stress in the shaft given by , = J 6I .

lr d"

Mohamed T. Hedaya MECHANICAL VIBRATIONS First Edition, p.6.10

6.3. Solved Example

Example 6.1

lr = 1 kg.m' lz=Zkg.m'

ls = 3 kg.m'

et=2x 106N.m/rad

Qz=4x106N.miradTr = 400 sin1000t N.m,

t is in seconds.

Find the natural frequencies.

Plot the mode shapes.

Find the steady state response.

Plot the dynamic torque on each shaft versus tirne.

t^

le,

I

Mohamed T. Hedaya MECHANICAL VIBRAT|ONS First Edition, p. 6.11