Two  Degree  of  Freedom  Systems

Dr./  Ahmed  NagibDecember  15,  2015

Two  Degree  of  Freedom  Systems

Real  systems  can  not  be  modelled  as  one  degree  of  freedom  system,  and  are  modelled  by  using  multiple  degree  of  freedom  systems.We  will  extend  the  previous  chapters  for  two  degree  of  freedom  system.

Examples  Two  Degree  of  Freedom  SystemsSystems  that  require  two  independent  coordinates  to  describe  their  motion  are  called  two  degree  of  freedom  systems.  Some  examples  of  systems  having  two  degrees  of  freedom  are  shown.

Examples  Two  Degree  of  Freedom  Systems

Examples  Two  Degree  of  Freedom  Systems

Two  Degree  of  Freedom  – Undamped SystemExample  1

Two  Degree  of  Freedom  – Undamped SystemExample  1

Two  Degree  of  Freedom  – Undamped SystemExample  1

Two  Degree  of  Freedom  – Undamped SystemExample  1

Two  Degree  of  Freedom  – Undamped SystemExample  1

Two  Degree  of  Freedom  – Undamped SystemExample  1

Two  Degree  of  Freedom  – Undamped SystemExample  1

Two  Degree  of  Freedom  – Undamped SystemExample  1

Two  Degree  of  Freedom  – Undamped SystemExample  1

Two  Degree  of  Freedom  – Undamped SystemExample  1

Two  Degree  of  Freedom  – Undamped SystemExample  1

Two  Degree  of  Freedom  – Undamped SystemExample  2

Two  Degree  of  Freedom  – Undamped SystemExample  2

Two  Degree  of  Freedom  – Undamped SystemExample  2

Two  Degree  of  Freedom  – Undamped SystemExample  2

Two  Degree  of  Freedom  – Undamped SystemExample  2

Two  Degree  of  Freedom  – Undamped SystemExample  2

Two  Degree  of  Freedom  – Undamped SystemExample  2

Two  Degree  of  Freedom  – Undamped SystemExample  2

Two  Degree  of  Freedom  – Undamped SystemExample  2

Two  Degree  of  Freedom  – Undamped SystemExample  2

Two  Degree  of  Freedom  – Undamped SystemExample  2

Two  Degree  of  Freedom  – Undamped SystemExample  2

Two  Degree  of  Freedom  – Undamped SystemExample  2

Two  Degree  of  Freedom  – Undamped SystemExample  2

Two  Degree  of  Freedom  – Undamped SystemExample  2

Two  Degree  of  Freedom  – Undamped SystemExample  2

Two  Degree  of  Freedom  – Forced  Vibration

Two  Degree  of  Freedom  – Forced  Vibration

Two  Degree  of  Freedom  – Forced  Vibration

Two  Degree  of  Freedom  – Forced  Vibration  ExampleFind  The  Steady  state  response  of  the  following  System

Two  Degree  of  Freedom  – Forced  VibrationExample

Two  Degree  of  Freedom  – Dynamic  Absorber

Two  Degree  of  Freedom  – Dynamic  AbsorberA  machine  or  system  may  experience  excessive  vibration  if  it  is  acted  upon  by  a  force  whose  excitation  frequency  nearly  coincides  with  a  natural   frequency  of  the  machine  or  system.  In  such  cases,  the  vibration  can  be  reduced  by  using  a  vibration  neutralizer  or  dynamic  vibration  absorber,  which  is  simply  another  spring-­mass  system.  The  dynamic  vibration  absorber  is  designed  such  that  the  natural  frequencies  of  the  resulting  system  are  away  from  the  excitation  frequency.  The  analysis  of  the  dynamic  vibration  absorber  will  be  considered  by  idealizing   the  machine  as  a  single  degree  of  freedom  system.

Two  Degree  of  Freedom  – Dynamic  Absorber

Two  Degree  of  Freedom  – Dynamic  Absorber

Two  Degree  of  Freedom  – Dynamic  Absorber

Two  Degree  of  Freedom  – Dynamic  Absorber

We  are  primarily  interested  in  reducing  the  amplitude  of  the  machine  (X1)  In  order  to  make  the  amplitude  of  zero,  the  numerator  of  Eq.  (X1)  should  be  set  equal  to  zero.  

This  gives

If  the  machine,  before  the  addition  of  the  dynamic  vibration  absorber,  operates  near  its  resonance,   .Thus  if  the  absorber  is  designed  such  that

where  

Two  Degree  of  Freedom  – Dynamic  Absorber

The  equations  of  X1 and  X2 can  be  rewritten  as

Two  Degree  of  Freedom  – Dynamic  Absorber

The  equations  of  X1 and  X2 can  be  rewritten  as

Two  Degree  of  Freedom  – Dynamic  Absorber

As  seen  before,  X1 =  0  at  ω =ω2 =ωa.  At  this  frequency  :

𝑋" = −𝑘&𝑘"𝛿() = −

𝑘&𝑘"×𝐹,𝑘&= −

𝐹,𝑘"  

This  shows  that  the  force  exerted  by  the  auxiliary  spring  is  opposite  to  the  impressed  force  (k2 X2 =  -­ Fo)  and  neutralizes  it,  thus  reducing  X1 to  zero.  The  size  of  the  dynamic  vibration  absorber  can  be  found  from  

k2 X2 =  m2 ω2 X2 =  -­ Fo

Thus  the  values  of  k2 and  m2 depend  on  the  allowable  value  of  X2

Two  Degree  of  Freedom  – Dynamic  Absorber

The  two  peaks  correspond  to  the  two  natural  frequencies  of  the  composite  system.  The  values  of  the  two  natural  frequencies  can  be  found  by  equating  the  denominator  of  the  following  equation  to  zero

,  which  leads  to:

1 +𝑘"𝑘&−

𝜔𝜔(

"1 −

𝜔𝜔1

"−𝑘"𝑘&= 0        𝑜𝑟,    

𝜔6 − 𝜔(" +𝜔1" +𝑘"𝑚&

𝜔" + 𝜔(". 𝜔1" = 0      

Two  Degree  of  Freedom  – Dynamic  Absorber

The  roots  of  the  previous  equation  is  given  by

which  can  be  seen  to  be  functions  of                                      and

Two  Degree  of  Freedom  – Dynamic  Absorber

The  roots  of  the  previous  equation  is  given  by

which  can  be  seen  to  be  functions  of                                      and

Two  Degree  of  Freedom  – Dynamic  Absorber

Two  Degree  of  Freedom  – Dynamic  AbsorberNotes:

1. It can be seen, that ωn1 is less than andωn2 is greater than the operating speed(ω) of the machine. Thus the machinemust pass through ωn1 during start-upand stopping. This results in largeamplitudes.

2. Since the dynamic absorber is tuned toone excitation frequency (ω), the steady-state amplitude of the machine is zeroonly at that frequency. If the machineoperates at other frequencies or if theforce acting on the machine has severalfrequencies, then the amplitude ofvibration of the machine may becomelarge.

Two  Degree  of  Freedom  – Dynamic  AbsorberNotes:

3. It can be seen from equations (31) and(32) that the amplitude of the absorber’smass (X2) is always much greater than thatof the main mass (X1). Thus the designshould be able to accommodate the largeamplitudes of the absorber mass.4. Since the amplitudes of m2 are expected tobe large, the absorber spring (k2) needs to bedesigned from a fatigue point of view.

Two  Degree  of  Freedom  – Dynamic  AbsorberNotes:

5. The variation of ωn1/ωa and ωn2/ωa asfunctions of the mass ratio m2/m1 are shownfor three different values of the frequencyratio ωa/ωs. It can be seen that the differencebetween ωn1 and ωn2 increases withincreasing the ratio of m2/m1.

Two  Degree  of  Freedom  – Dynamic  Absorber

•  If  ω hits  ω:& or  ω:& resonance  occurs•  Using  ;<

;=><1,  defines  useful  operating  

range  of  absorber•  In  this  range  some  absorption  still  occurs

Absorber Zone

Two  Degree  of  Freedom  – Dynamic  AbsorberAbsorber Zone

This illustrate that the useful operating range of absorber design is

(0.908 𝜔? <  𝜔 < 1.118 𝜔?) .

Hence if the driving frequency drifts within this range, the absorber design still offers some protection to the primary system by reducing its steady state vibration amplitude.

Two  Degree  of  Freedom  – Dynamic  Absorber

Two  Degree  of  Freedom  – Dynamic  AbsorberExample

Two  Degree  of  Freedom  – Modal  Analysis  

Two  Degree  of  Freedom  – Modal  Analysis    

Two  Degree  of  Freedom  – Modal  Analysis  

Two  Degree  of  Freedom  – Modal  Analysis  

Two  Degree  of  Freedom  – Modal  Analysis  

a)  Static  Coupling

a)  Static  +Dynamic  Couplings

Two  Degree  of  Freedom  – Modal  Analysis  

a)  Static  Coupling

a)  Static  +Dynamic  Couplings

Two  Degree  of  Freedom  – Modal  Analysis  

Two  Degree  of  Freedom  – Modal  Analysis  

Two  Degree  of  Freedom  – Modal  Analysis  

Two  Degree  of  Freedom  – Modal  Analysis  

Two  Degree  of  Freedom  – Modal  Analysis  

Two  Degree  of  Freedom  – Modal  Analysis  

Two  Degree  of  Freedom  – Modal  Analysis  

Two  Degree  of  Freedom  – Modal  Analysis  

Example 4.2.1 (Inman)

Calculate the Matrix 𝐾A For M = 9 00 1 K= 27 −3

−3 3Example 4.2.2 (Inman)

Calculate the Eigen values and vectors of Matrix 𝐾A andnormalize the Eigen vectors and check if they areorthogonal

Two  Degree  of  Freedom  – Modal  Analysis  

Two  Degree  of  Freedom  – Modal  Analysis  

Two  Degree  of  Freedom  – Modal  Analysis  

Two  Degree  of  Freedom  – Modal  Analysis  

Two  Degree  of  Freedom  – Modal  Analysis  

Two  Degree  of  Freedom  – Modal  Analysis  

Two  Degree  of  Freedom  – Modal  Analysis  

Two  Degree  of  Freedom  – Modal  Analysis  

Two  Degree  of  Freedom  – Modal  Analysis  

Two  Degree  of  Freedom  – Modal  Analysis  

Two  Degree  of  Freedom  – Modal  Analysis  

Two  Degree  of  Freedom  – Modal  Analysis  

Two  Degree  of  Freedom  – Modal  Analysis  

Two  Degree  of  Freedom  – Modal  Analysis  

Two  Degree  of  Freedom  – Modal  Analysis  

Two  Degree  of  Freedom  – Modal  Analysis  

Two  Degree  of  Freedom  – Modal  Analysis  

Two  Degree  of  Freedom  – Modal  Analysis  

P = 𝑣" 𝑣" = &"1 11 −1

So  that  PT P becomes

P = &"

&"

1 11 −1

1 11 −1

= &"1 + 1 1 − 11 − 1 1 + 1 = 1 0

0 1 = I

Calculate the Matrix PT 𝐾AP for the previous two degree of freedom example

𝑃H𝐾A𝑃 =121 11 −1

3 −1−1 3

121 11 −1

= &"1 11 −1

2 42 −4

= &"4 00 8 = 2 0

0 4 =ΛNote that the diagonal elements of the spectral matrix Λ are the natural frequencies 𝜔&and 𝜔". That is, from previous calculations, 𝜔&"=2 and 𝜔"" = 4, so that Λ=diag(𝜔&" 𝜔"") =diag(2 4) = 2 0

0 4

Two  Degree  of  Freedom  – Modal  Analysis  

𝑆 = 𝑀N& "⁄ 𝑃 = &"

&P

00 1

1 11 −1

𝑆 = &"

&P

&P

1 −1and

𝑆N& = 𝑃H𝑀N& "⁄ = &"1 11 −1

&P

00 1

𝑆 = &"3 13 −1

Two  Degree  of  Freedom  – Modal  Analysis  

The  given  initial  conditions  are𝑥a =

10 and  �̇�a =

00

The  initial  conditions  in  the  generalizedcoordinate  𝑟 will  be

𝑟 0 = 𝑆N&𝑥a =&"3 13 −1

10 =

P"P"

�̇� 0 = 𝑆N&�̇�a =&"3 13 −1

00 = 0

0

Two  Degree  of  Freedom  – Modal  Analysis  

So  that  𝑟& 0 = 𝑟" 0 = 3 2⁄ and  �̇�& 0 = �̇�" 0 = 0𝑥a =

10 and  �̇�a =

00

The  response  in  the  physical  coordinate  r 𝑡is  calculated  from

𝑟& 𝑡 =32sin 2  𝑡 +

𝜋2 =

32cos 2  𝑡

𝑟" 𝑡 =32sin 2  𝑡 +

𝜋2 =

32cos 2  𝑡

Two  Degree  of  Freedom  – Modal  Analysis  

The  response  in  the  physical  coordinate  𝑥 𝑡is  calculated  from

𝑥 𝑡 = 𝑆  𝑟 𝑡 =12

13

13

1 −1

32cos 2  𝑡

32cos 2  𝑡

𝑥 𝑡 =0.5 cos 2  𝑡 + cos 2  𝑡1.5 cos 2  𝑡 + cos 2  𝑡

which  is  identical  to  the  solution  obtained  in  Example  4.1.7

Two  Degree  of  Freedom  – Modal  Analysis  The  response  in  the  generalized  coordinate  𝑟 𝑡and  the  physical  coordinate  𝑥 𝑡  are  plotted  as  follows

Two  Degree  of  Freedom  – Modal  Analysis  

Two  Degree  of  Freedom  – Modal  Analysis  

Two  Degree  of  Freedom  – Modal  Analysis  

Two  Degree  of  Freedom  – Modal  Analysis  

Two  Degree  of  Freedom  – Modal  Analysis  

Two  Degree  of  Freedom  – Modal  Analysis