Lecture 19 Two DOF Systems Forced

8/12/2019 Lecture 19 Two DOF Systems Forced

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Rose-Hulman Institute of TechnologyMechanical Engineering

Vibrations

Todays Objectives:

Students will be able to:

a) Find the steady stateresponse of 2DOF forced

systems

b) Use Simulink for MDOF

systems

Forced two degree of freedom systems


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Vibrations

Finish Example from last time

=

++

++

+

0

0

0

0

2

22

2

112211

221121

0

GG x

LkLkkLkL

kLkLkkx

J

m

&&

&&

21

12

in LL

xx

s +

=

2

21

11

21

2G

2

2

1

1 xso xLL

Lx

LL

L

L

xx

L

xx GG

++

+=

=

k1 k2

m, J0

L1 L2

Lets use the displacement at bothends as our coordinates

x1

x2

Note:

==

==

0G0 0 MJM

dt

dL

FxmFdt

dP

sys

G

sysx

&&

&&

So


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Vibrations

Finish Example from last time

0221221

1

121

2 =+++

++ xkxkxLL

mL

xLL

mL

&&&&

222111

21

12000 LxkLxk

LL

xxJJM

dt

dLGsys =

+

==

&&&&&&

x1

x2

k1x1

k2x2

0222111121

02

21

0=+++

LxkLxkxLL

JxLL

J &&&&

In matrix form we get

0

2

1

2211

21

1

2

21

0

21

0

21

1

21

2

=

+

++

++x

x

LkLk

kk

x

x

LL

J

LL

JLL

mL

LL

mL

&&

&&

Notes

2212

21

11

21

2 xkxkxLL

mLx

LL

mLxmF

dt

dPG

sysx =+

++

== &&&&&&


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Vibrations

Finish Example from last time - Matlab

evec1 =

-0.0315 0.0032

0.0059 0.0574

eval1 =4981.3 0

0 6785.3

freq =

70.5785

82.3732

evec2 =

-1.0000 -0.5187

-0.7780 1.0000

eval2 =

4981.3 0

0 6785.3

freq2 =

70.5785

82.3732


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Vibrations

Forced vibrations 2 DOF systems

tsinF

x

x

kk

kk

x

x

m

m

=

+

00

0 1

2

1

2212

1211

2

1

2

1

&&

&&

{ } { } { }pc

xxx +=

tsinX

X

x

x

=

2

1

2

1

Consider the 2-DOF system with the following EOM:

We know there will be a particular solution (satisfies right hand side) and a

homogeneous solution (satisfied RHS = 0)

Lets look at the particular solution. Since the RHS is a sin lets assume

Substituting into EOM we get

=

0

1

2

1

2

22212

12

2

111 F

X

X

mkk

kmk


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Vibrations

Steady-state response (cont.)

( )[ ]Z

( )[ ] ( )[ ]

( )[ ]

=

=

00

111

2

1 F

Zdet

ZintadjoFZ

X

X

From previous page

=

0

1

2

1

2

22212

12

2

111 F

X

X

mkk

kmk

So, we can solve for {X1 X2}Tby premultiplying by [Z()]-1

Quick matrix review of the inverse of a 2x2 matrix:

=

ac

bd

bcaddc

ba 11


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Vibrations

( )[ ]( ) ( )( ) 21222222111 `kmkmkwZdet +=

( )[ ]( ) ( )( )( )( )22222121

22

2

22

121

=

=

mm

mmwZdet

The determinant of [Z()] is

This can be factored and written in terms of the naturalfrequencies as

( )

=

0

1 1

2

12212

12

2

222

2

1 F

mkk

kmk

ZX

X

So we get

X1 = X2 =


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Vibrations

Sketch these

( )( )( )22222121

1

2

2221

=

mm

FmkX

( )( )22222121112

2

=mm

FkX

X1

1 2

X2

1 2

These are frequency response plots


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Vibrations

Summary

1. Find equations of motion:

2. Assume simple harmonic motion

3. This gives

4. Find natural frequencies

5. Substitute in natural frequencies

and find natural modes

6. Write homogeneous solution

1. Find equations of motion:

2. To find the particular solution

assume:

3. This gives

4. Solve for {X}

[ ]{ } [ ]{ } { }0=+ xKxM &&

[ ] [ ]( ){ } { }02 =+ XKM

[ ]{ } [ ] { }02 =+ KxM

[ ] [ ]( ){ } { }02 =+ ii XKM

Homogeneous Solution 2DOF Steady State Solution

(with harmonic forcing)

[ ]{ } [ ]{ } { } tsinFxKxM =+&&

{ } { } tsinXx =

[ ] [ ]( ){ } { }FXKM =+ 2

{ } [ ] [ ]( ) { }FKMX 12 +=

{ } { } tieXx =

Known!

Forcing frequency

Natural frequency

Naturalmode

Magnitude of sint


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Vibrations

There is a nice website in Australia

http://www.mech.uwa.edu.au/bjs/Vibration/TwoDOF/default.html


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Vibrations

What do we do if we have damping?

[ ]{ } [ ]{ } [ ]{ } { }FxKxCxM =++ &&&

( ){ } ( )[ ] ( ){ } ( )[ ] ( ){ }sFsHsFsBsX == 1

[ ] [ ] [ ]( )

( )

( ){ } ( ){ }sFsXKCsMs

sB

=++

444 3444 212

If we have

If we take the Laplace Transform

so

If we let s = j( )[ ] matrixfunctionresponsefrequency=jH

We can determine this from

experimental data and use it for

system ID


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Vibrations

Using SimulinkSometimes you have to add small

damping to get rid of transient solution

0 50 100 150 200 250 300-0.5

0

0.5

0 50 100 150 200 250 300-1

0

1

0 50 100 150 200 250 300-0.5

0

0.5

0 50 100 150 200 250 300-1

0

1

Lecture 19 Two DOF Systems Forced

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Transcript of Lecture 19 Two DOF Systems Forced