Chapter3 Vibration Ahmedawad

III. FREE VIBRATION OF DAMPED SINGLE

DEGREE OF FREEDOM

SYSTEMS

Mohamed T. Hedaya MECHANICAL VIBRATIONS First Edition, p. 3.1

3.1. Tvpes of Danrqins

3.1.1. Viscous Damping

Viscous damper (dashpot)

Damping force on mass = - c X,

c is the coefficient of viscous damping I N/(m/s) = N.s/m ].

\1,

Mohamed T. Hedaya MECHANICAL VIBRAT]ONS First Edition, p. 3.2

{:

%

.t{

)

\/,t4

J[1,n.

'-)

\,I

i

It,' , \Ii, {r" , .\f\.i

N

iI

a 1i); i -t:";ir ll\ r t<Vlg

llc,-IJ "rti\/ [ \( o)

i,..1-,

-1-\i1

hlL l,/',^.,/ Vi Ii

lJ ilL-

t

I

e,oul c- b /r^PM\

DSTt:rA

.l r'

N ).r, ".:tl- .i ,,'.

I. . r__ .ttiI i I -1',I i I r*t

-.t._==

i1---irt.\

l'/

-is

,,-..:.,

3.1.2. Hysteresis Damping (Structural Damping)

When an elastic member

is loaded and unloaded in

its elastic region, the

(Stress - Strain) relation

is not the same during

loading and unloading.

This is due to the friction

between the internal

layers of the material,

which indicates that actual materials are not perfectly elastic. This gives a

loop on the (Stress - Strain) diagram called Hysteresis loop.


Strain(displace rnent)

' Stress (force)

eresIS

The area of the loop on the (Force - Displacement) diagram of the

member denotes the energy disslpated by Hysteresis damping during one

cycle of loading and unloading of this member.

ln order to study the vibrations of systems with Hysteresis damping,

Hysteresis damping is replaced by the viscous damping which dissipates

the same energy per cycle. The coefficient of .this viscous damping is

called the coefficient of equivalent viscous damping (c").

Original system Equivalent system

Mohamed T. Hedaya MECHANICAL VIBRATIONS First Edition , p.3.4

AIlr-

/.I

L) L'

Ic( -+ k t,,',,n-l -, ' ,i'-)

InLr*uJ -."rn

'!;1 -ti4

At-= Trq kA".t-.

z*d

,t

For a system damped by Hysteresis damping and performing harmonic

motion x = A cos @t, it was found, experimentally, that, the energy

dissipated by Hysteresis darhping per cycle (AE) is proportional to the

maximum strain energy in the system during the cycle t1 t n2;. Tfris givesa ,2

AE=nqkA2 (1) (J,..".-,,i.*)

11 is called the Hysteresis damping factor, the structural damping factor or

the loss factor of the material causing the damping.

It can be proven that, the energy dissipated by viscous damping per cycle,

for the same motion is given by

AE = n cu rr: A2 (2)

(1) and (2) -> j ce =

nku)


3.1.3. Coulomb Damping (Ory Fniction)

Magnitude of damping force on mass = Fo = P N,

p is the kinematical coefficient of friction.

Damping force on mass is opposite to X.

x


[n# + cf oK]c=o

ffir'

i'

ur[: Z.Kn ,

rK, r"UK

L,l Kryr_i \ i-Jr

')* s1 (jJn

Nll(,r) =

K,1 ls

: - k*:l5''al

t: N-1,lt

(,0,., . " t.:

=

lrl=

+ C-lt + Kr:e

t- ..lc'*4km

2- rn {}h

3.2. Viscous Dampinq

mX=-cx-kxmX +cX+kx=0This is the differential equation of motion (D.O.F.M.).

Solution of this equation depends on the value of c relative to (2Jk m),

which is equal to (2 p1 urn). This solution is explained in Appendix A.2.

Let(=cl(2mr,.rn).

This gives three Gases: (1) ( > 1, (2) 1= 1, (3) ( < 1

Mohamed T. Hedaya MECHANICAL VIBRATIONS First Edition, p.3.7

Case 1: Overdamped System (c > Zmrrl") (( > 1)

E.O.M. o x = Ar er,t + Az e^r' , oAr,Az=(-#rl(;'l *t J rt

Ar and Az ar€ constants determined by initial conditions.

Motion is nonoscillatory.


)*''r,r.C"*ryr

r i'\ll.V l;(- . -)U1 tr -=r-r )C*--:B\,1-ltnr.-tt,(: Lr

Lulr,ll[r,t ^1 Il') 5 / n4 j

t ,/\

-J71,*,'; t7

-

'-A\:r-----=' '--

-r-heItr'tJ\rf . t ,a

{)tll nf'"J|,)J

*1,;{*. t rsclls

Case 2: Critically Damped $ystem (c = 2muln) {( = 1)

D.O.F.M. mX +cx+kx=0E.O.M. I X = (Ar + Ae t) e-rnf ,

A1 and 42 or€ constants determined by initial conditions.

Motion is nonoscillatory.

Mohamed T. Hedaya MECHANICAL VIBMTIONS First Edition, p.3.9

Effect ef Dampinq o0 Settlins Tipe

. With zero irlitial velocily, the time required for the system to reach its

equilibrium position increases as damping increases.

. Critically damped system takes the shortest time. Therefore, the pointers

of measuring instruments are usually designed to be critically damped.

Mohamed T. Hedaya MECHANICAL VIBRATIONS First Edition, p.3.10 i

Case 3: Underdamped $ystem (c < 2mor) (( < i)D.O.F.M. mX +cx+kx=0E.O.M. ox=as-(r.,rnt cos(t*6t-p), j h)d =h)n J71 ,

j.A and B are constants determined by initial conditions.

- 4g*(t^t"t

Motion is oscillatory with deeaying (amplitude) and eireii{ar frcrquency uro.


har r >nic- r' i:' \

'l ..^rl',.-i

Loa a r i th m i c d e-c re m e Q!

Xr- = **(r^r^(tr-tz) - *(t,r"(tz-t,) -;,x2

X', =

"(r^rnTox2

_ _ 2-_!!"

sJ"r

Mohamed T. Hedaya MECHANICAL VIBMTIONS First Edition, p.3.12

il'.1

l'i,t

x2

x1 -^o t

x26=(urnTo=qwn2I

h)6 uJn frI E,

2n(

^ff426 (= 2nU,lt:(' I is catted the lggryIhglg-desremgnt"

Generally, I1 = X'

= xt -x2 x3 x4

x1 =

X, * *, * I, ....... x xn = en6

Xn+1 X2 X3 X4 Xn+1

This means that, when ( is known, 6 can be determined from the relation

(6 = ZnQ..fi1), and then, the ratio (x1lxn*1) can be found.

xn

The relation r j1-Xn+1

MECHANICAL VIBRATIONS First Edition, p. 3.13

(3)

(4)

1ln x1

n Xn+1

6

J+"U;62^lt

- q'

i A usual practical way to determine ( is to let the system vibrate freely, to

determine the ratio between two amplitudes (x1 / xp*r)and use equation (3)

to find 5. Then, equation (4) can be applied for the determination of (.

Gritical damping coefficient {c"}

c" is defined as the damping coefficient required to make the system

critically damped, or the minimum damping coefficient which produces

nonoscillatory motion. lt is obtained from e c. = 2 m on"

Darnping factor (damping ratio) {0( is defined asthe ratio between cand c*. -p c (= c/ c. = c/ (2 m ron).


Motion with Different Values of Damping Factor(fl

MECHANICAL VIBRATTONS First Edition, p. 3.15

3.3. Hvsteresis DamEin$ (Strugtural Darnpinsl

Original systemjce=r1 k/tl (5)

ln free vibration, tll is equal to tllo, which

can not applied for direct calculation of c".

Equivalent system

depends on ce. So, equation (5)

Substitution for c" and u gives

,1rQ 6 tdn) =qk

Ldn

.\=2luEquations

1- 1,' (6) -> .21"2-l-t[:F (7)

(6) and (7) can be used for calculation of q and (" respectively.

.l 7l- 5e


3.4. GoulorTb Dampinq (D_rv Friction)

Consider the following case

k= 2 N/cm N=mg=2A

Fo=[.lN=6

Six equilibrium positions

Frictionless eguilibrium position

N

N

1cm:1

2cm

3cm


The mass can stay in static equilibrium within an equilibrium zone of tength

6 crn, 3 cm to the left of the frictionless equilibrium position and 3 cm to the

riQht of the frictionless equilibrium position.

r The ma$s of a spring-ma$s system can stay in static equilibrium on arough plane within an equilibrium zone of length 2A, A to the left of the

frictionless equilibrium position and A to the right of the frictionless

equilibrium position,

oA=UN/k= Fd/k.

o During free vibration, the mass stops when its velocity becomes zero

within the equilibrium zone.

MECHANICAL VIBRATIONS First Edition, p. 3.18Mohamed T. Hedaya

a -- ( , _B) *-13-fi (u\ \-unL *y J , k

Equation of motion

Exampl?

Considera spring-mass system with x(0) - Xr: and x(0) = 0.

1- First half cycle (from right to left) (X< 0) L

mX=-kx+FomX+kx=Fo

x = A cos(u.rnt - F) + Fd / k = A cos(,ri, - F) + A

lnitial conditions -) x = (Xo - Ai cos(t*nt) + 6Motion is harmonic with circular frequency

u)n , amplitude (Xo - A) and center Or.

Xtn+A=Xo-A-;>Xtn=Xo-2A (8)


2- Secon d half cycle (from left to right) $ > A)

mX=-kx-FomX+kx=-Fo

x = A cos(tont - F) - Fd / k = A cos(or,,t - B) - A

lnitial conditions -) x = (Xlz - A) cos(ulnt) - A

Motion is harmonic with circular frequency A

kln , amplitude (Xtn - A) and center O2. * A

Xr +A=X,z-A-;>Xr = X1lz' 2 b, {9)

.X

Mohamed T. Fledaya MECHANICAL VIBRATIONS First Edition, p. 3.20

Reduction of (apparent a

Xtn=Xo-2A (B)

Equations (8) and (9) s

2A everv half cvcle.

Reduction of (apparent amplitude)

Xtn=Xo-2A (B) Xr=Xtn-ZA, (9)

Equations (8) and (9) show that, the (apparent amplitude) is reduced by

2A every half cycle.

(Apparent amplitude) and mass position

r(Apparentamplitude)afternhalfcycles=Xnlz=Xo-2nA>

o Position of mass after n half cycles = xn1 = (-1)n (Xo - 2 n A)

iriass Etop

Mass stops after n half cycle, if (k Xnrz < Fo) , i.e. if (Xnrz 5 A), i.e. if the

mass is inside the equilibrium zone.

ln order to find n, put Xnrz s A. This gives

X6-2 n A<A-), -2nAs -Xo+A-) r n 2 5 -O.U2A


Examole

m=2kg k='1974N/m U=0.3x(0) = g x(0) = 29 mm (to the right)

Find time and position at which mass stops.

fk- trry = 31.41T radls*)n = {* = l-,

a = L = lN = Fmg - 0.3x2x9.81

= o.oo3 mkkk1s74

2L, 2 x 0.003

tso=9*r=!* 2n =q,. j"-=o.s $2 2 u)n 2 31.417

Xs/z= (-1)u(0.029 - 2x5x 0.003)=0.00t m - 1 mm(tothe right)


\i\cou\f) v' +CX t

X, = €n'x,*

',l"*rnnpk{-e

Q_L

'sL*

(_tJ; Lun

zl\tti t,u*r/

-/) a

1A'.-.,,.,.ir-,

(

\ st-a

ear6*b (pr^^ -.a \ -Yn -: k"r^_'L -{- I \

R.Jr.{ io. Fr'l r,rn .*i I

t?,t',

)J^l I (q cl,.

c'

['.,,1

30

x(mm)

20

10

0

-10

-20

-30


3.5,_$olved Examples.

Example 3.1

L=500 mm b=300 mm

m=6 kg

c = 20 N.s/m '

k = 200 N/m

0(0; = 6.7' c.W. 610; = s

Find time and position after two cycles.

x-_b 0

Uniform rod

Mohamed T. Hedaya

t"o \'.. \,\aa\. ',,

T 1"rU)n

2T

.**lwffi

V ' =q.r-{ J a* pJ

Example 3.2

L = 1200 mrn

m=20 kg

q = 0.04

x(0) = 19 mm

d=30 mm

E = 20x1010 N/m2

x(0) = g


Example 3.3

L=500 mm

k = 200 Nlm

0(0) = 6,7o C.W.

m=6 kg

To = 0.5 N.m

6101 = s


Find time and position at which rod stop{


Uniform rod

Chapter3 Vibration Ahmedawad

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Transcript of Chapter3 Vibration Ahmedawad