Philadelphia University Faculty of Engineering Mechanical Engineering Department Dr. Adnan Dawood Mohammed (Professor of Mechanical Engineering) Mechanical

Philadelphia University Faculty of Engineering

Mechanical Engineering Department

Dr. Adnan Dawood Mohammed(Professor of Mechanical Engineering)

Mechanical VibrationsForced Vibration of a Single Degree of Freedom System

Harmonically Excited Vibration

Physical system

force offrequency theis

force of amplitude theis cos

sin

:forms following theof one may takesIt force. harmonic a is

motion ofEquation ...

tjo

oo

o

eFtF

FtFtF

tFtF

tF

tFkxxcxm


amplitude' response statesteadty ' thecalled is X , X(t)

as written becan solution The .frequency same with the

harmonic is (solution) response t theexpect thacan wefrequency with

harmaonic is n)(excitatio force theBecause equation. homogenuosnon

for thesolution theisIt :response state(steady integral Particular .2

sincos)(

:as written assolution w The

1for chapter). previousin (describedequation shomogeneou the

forsolution theisIt :response) (Transientfunction ary Complement 1.

parts two

hassolution Its .homogenuos-non isequation aldifferentiorder 2nd This

(1)motion ofEquation ...

:form heConsider t

21

tjp

ddt

c

tjo

ex

tCtCetx

eFkxxcxm

n


amplitude. response statesteady

theof magnitude"" thedenotes X , X

as written be nowcan integral particular for the There

X re wheXX

Response. and Forcebetween angle"

phase" theis tan where,

as By writing

X

:Xfor solve and (1)equation into integral particular theSubstitute

222

21-222

2

2

tjp

oj

j

o

ex

cmk

Fe

mk

cwecmk

jcmk

jcmk

F


motion. theof conditions inatial theknowing

from dermined be toconstants are C2 and C1 that Note

X sincos)(

)()()(

:is (1)motion ofequation for the (response)solution totalThe

1

2tan

ratio.frequency theis where, 21

X

:as written bemay and X

21

21

222

tjdd

t

pc

n

o

etCtCetx

txtxtx

r

r

rrr

k

F

n


222

st

sto

21

1M

(M)".Factor ion Magnificat" thecalledusually is

ratio The ).( deflection static" thecalledusually is k

F termThe .5

.frequency and amplitude

constant h motion wit harmonic a is ))(( response stateSteady 4.

me.certain ti aafter neglected becan and time

with decayst motion tha a represents ))(( responseTransient .3

cos)(cos form force For the 2.

sin)(sin form force For the 1.

:NOTES

rr

X

tx

tx

tXtxtFF(t)

tXtxtFF(t)

p

c

po

po

Damped Forced Vibration System

Graphical representation for Magnification factor M and ϕ.


Notes on the graphical representation of X.

For ζ = 0 , the system is reduced and becomes un-

damped.

for any amount of ζ > 0 , the amplitude of

vibration decreases (i.e. reduction in the

magnification factor M). This is correct for any

value of r.

For the case of r = 0, the magnification factor

equals 1.

The amplitude of the forced vibration

approaches zero when the frequency ratio ‘r’

approaches the infinity (i.e. M→0 when r → ∞)


Notes on the graphical representation for ϕ.

For ζ = 0 , the phase angle is zero for 0<r<1 and

180o for r>1.

For any amount of ζ > 0 and 0<r<1 , 0o<ϕ<90o.

For ζ > 0 and r>1 , 90o<ϕ<180o.

For ζ > 0 and r=1 , ϕ= 90o.

For ζ > 0 and r>>1 , ϕ approaches 180o.


) small(for is

ntdisplaceme themeans which 2

X then 1)(r resonanceAt 3.

is

ntdisplaceme themeans which X then 1)(rfrequency high At 2.

is

ntdisplaceme themeans which X then 0)(rfrequency lowAt 1.

amplitude response s.s theis 21

X

: have We

:frequencynt with displaceme ofVariation

2

222

control.Damping

control. Mass

control. Stiffness

k

F

m

F

k

F

rr

kF

o

o

o

o


2res

2

1/2-222

21 i.e

a asknown is which 21 when maximum is X

:givescondition This

0)X( when maximum is X

21 X

:as written becan amplitude response s.s The

n

n

o

r

dr

d

rrk

F

Frequency,Resonance

:amplitude Maximum ofFrequency

Forced Vibration due to Rotating Unbalance

tmekxxcxM t sin...

2

Forced Vibration due to Rotating Unbalance

below figure in theshown are and X of plots The

1

2tan

and

21X

or ,21

X

21

222

2

222

2

r

r

rr

rMme

rr

kme

t

Transmissibility of Force

)r-(1

r2tan ,

)2()1(

)2(1TR

TR bility,Transmissi Force theas TR define

)(

)(F

i.e, motion harmonic Assume

F

is foundation the toed transmittforce then the

,neglugible is foundation of deflection theIf

21

22

2

F

2tr

tr

rr

r

F

F

cjmk

Fcjk

Xex

xckxe

F

o

tr

F

o

tj

tj

Transmissibility of displacement (support motion)

Physical system:

Mathematical model: 0....

yxkyxcxm

The forcing function for the base excitation


Substitute the forcing function into the math. Model:

motion) (Harmonict Xsin x(t)Assume

tan

:Where

sincossin...

1

22

k

c

ckYA

tAtYctkYkxxcxm

2

1

222

2

222

22

d

1

2tan

21

21)bility(TRTransmissint Displaceme

r

r

rr

r

cmk

ck

Y

X


Graphical representation of Force or Displacement Transmissibility ((TR) and the Phase angle (

Example 3.1: Plate Supporting a Pump: A reciprocating pump, weighing 68 kg, is mounted at

the middle of a steel plate of thickness 1 cm, width 50 cm, and length 250 cm. clamped along two edges as shown in Fig. During operation of the pump, the plate is subjected to a harmonic force, F(t) = 220 cos (62.832t) N. if E=200 Gpa, Find the amplitude of vibration of the plate.

Example 3.1: solution

The plate can be modeled as fixed – fixed beam has the following stiffness:

The maximum amplitude (X) is found as:

mNx

xxkSo

mxxxbhIBut

l

EIk

/ 82.400,10210250

10667.4110200192 ,

10667.41101105012

1

12

1

192

32

99

493223

3

mmmk

FX o 32487.1

832.626882.400,102

2202

-ve means that the

response is out of phase

with excitation

Example 3.2:Find the total response of a single-degree-of-

freedom system with m = 10 kg, c = 20 N-s/m, k=4000 N/m, xo = 0.01m and = 0 when an external force F(t) = Fo cos(ωt) acts on the system with Fo = 100 N and ω = 10 rad/sec .

Solution a. From the given data

ox.

sradm

kn /20

10

4000

05.020102

20

2

nm

c

sradd

d

nd

/975.19

2005.01

1

2

2

5.020

10

n

r

Example 3.2: Solution

Total solution:X(t) = X c (t) + X p(t)

mk

Fost 025.0

4000

100

mrr

X st 3326.021 222

o

r

r814.3

1

2tan

21

0.066)-0t20.3326cos()-cos(19.97t20*.050-

e (t)

0.066)-tcos(X)-tncos(-

e (t)

tAx

tnAx

0.066)-0t20.3326cos(- )066.0-cos(19.97t--6.64ex(t)

-0.3325A and rad 0.066

0.438,

0.066)-0t220sin(*0.3326

-(-t)e*)-(19.97tcos)-(19.97t-19.97sin *-e)(

0.066)-0t20.3326cos()-cos(19.97t-e (t)

(1) 33187.0cos

cos0.066*0.3326cos

(2) and (1) From

(2) 0219.0sinsin97.190

0 0,at t

0

01.0 0,at t

t

tAtAtx

tAx

A

A

AA

tx

mtx

Documents

Philadelphia University Faculty of Engineering Mechanical Engineering Department Dr. Adnan Dawood Mohammed (Professor of Mechanical Engineering) Mechanical