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Lecture 2c

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0% found this document useful (0 votes)
28 views6 pages

Lecture 2c

Uploaded by

Yusuf Gul
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
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Download as PDF, TXT or read online on Scribd
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Mechanical

Vibrations

Prof. Dr. Kenan Y. Şanlıtürk


sanliturk@itu.edu.tr

Content

1. Introduction to Vibration and Free response


2. Response to Harmonic Excitation
3. General Forced Response
4. Multi-Degree-of-Freedom systems
5. Design for Vibration Suppression

Page 1
2. Response to Harmonic Excitation

• Alternative methods of solutions of EOM

• Geometric method

• Use of Complex functions, Frequency


Response Method

• Laplace Transform method

Alternative methods of solutions of EOM

 Up to now, method of “Undetermined Coefficients” is used.

 Now, three alternatives:


– Geometric solution
– Use of complex functions, Frequency Response Method
– Via Laplace Transform

Page 2
Geometric Solution

mx  cx  kx  F0 cos t
In steady-state condition
x, x, x
 There is 90 degree phase difference between

- displacement and velocity , and

- velocity and acceleration

 Elastic, damping and inertia forces as well as external forces can be


considered as vectors.

mx  cx  kx  f (t ),
f (t )  Fe jt , x p (t )  Xe jt

( 2 m  jc  k ) Xe jt  Fe jt


( 2 m  jc  k ) X  F

kX  jc X   2 mX  F

Page 3
Geometric Solution kX  jc X   2 mX  F

F
F

F 2  (k  m 2 ) 2 X 2  (c ) 2 X 2
F
X
(k  m 2 ) 2  (c ) 2

At different frequencies!

FRF: Frequency Response Function using complex algebra

mx(t )  cx(t )  kx(t )  f (t )


f (t )  Fe jt
x p (t )  Xe jt
( 2 m  jc  k ) Xe jt  Fe jt
F
X
(k  m )  j (c )
2

X 1 Notation:
H ( )    ( )
F (k  m )  j (c )
2
is also used.

Page 4
Frequency Response Function using complex algebra

X 1  c 
 e  j ,   tan 1  2 
F (k  m )  (c )
2 2 2
 k  m 
F
X e  j
(k  m )  (c )
2 2 2

if F  F0
F0
x p (t )  Xe jt  e j (t  )
(k  m 2 ) 2  (c ) 2

The real part of this last equation corresponds to the particular solution
we obtained before when f(t) = Fo cos(t)

Frequency Response Function

Steady-State condition

(ms 2  cs  k ) X ( s )  F ( s )

X ( s) 1
 H (s)  2 Transfer function
F (s) ms  cs  k

X ( ) 1
H ( )   Frequency Response
F ( ) k  m 2  jc Function

Page 5
Solution via Laplace Transform

 Time domain variable is transformed to complex s variable

 Differential equation is turned into algebraic equation

 Time response is obtained by Inverse Laplace Transform

Page 6

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