Philadelphia University Faculty of Engineering Mechanical Engineering Department Dr. Adnan Dawood...

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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

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

motion ofEquation ...

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

tCtCetx

eFkxxcxm

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

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

21-222

cwecmk

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

ratio.frequency theis where, 21

:as written bemay and X

etCtCetx

txtxtx

(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

tXtxtFF(t)

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.

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

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

amplitude response s.s theis 21

: have We

:frequencynt with displaceme ofVariation

control.Damping

control. Mass

control. Stiffness

1/2-222

21 i.e

a asknown is which 21 when maximum is X

:givescondition This

0)X( when maximum is X

:as written becan amplitude response s.s The

Frequency,Resonance

:amplitude Maximum ofFrequency

Forced Vibration due to Rotating Unbalance

tmekxxcxM t sin...

Forced Vibration due to Rotating Unbalance

below figure in theshown are and X of plots The

or ,21

Transmissibility of Force

r2tan ,

)2()1(

)2(1TR

TR bility,Transmissi Force theas TR define

i.e, motion harmonic Assume

is foundation the toed transmittforce then the

,neglugible is foundation of deflection theIf

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

:Where

sincossin...

tAtYctkYkxxcxm

21)bility(TRTransmissint Displaceme

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:

mxxxbhIBut

/ 82.400,10210250

10667.4110200192 ,

10667.41101105012

493223

FX o 32487.1

832.626882.400,102

-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

kn /20

05.020102

/975.19

2005.01

Example 3.2: Solution

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

Fost 025.0

X st 3326.021 222

r814.3

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

0.066)-tcos(X)-tncos(-

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

01.0 0,at t

tAtAtx

Philadelphia University Faculty of Engineering Mechanical Engineering Department Dr. Adnan Dawood...

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