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8/12/2019 Lecture 19 Two DOF Systems Forced
1/12
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
8/12/2019 Lecture 19 Two DOF Systems Forced
2/12
Rose-Hulman Institute of TechnologyMechanical Engineering
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
8/12/2019 Lecture 19 Two DOF Systems Forced
3/12
Rose-Hulman Institute of TechnologyMechanical Engineering
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 =+
++
== &&&&&&
8/12/2019 Lecture 19 Two DOF Systems Forced
4/12
Rose-Hulman Institute of TechnologyMechanical Engineering
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
8/12/2019 Lecture 19 Two DOF Systems Forced
5/12
Rose-Hulman Institute of TechnologyMechanical Engineering
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
8/12/2019 Lecture 19 Two DOF Systems Forced
6/12
Rose-Hulman Institute of TechnologyMechanical Engineering
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
8/12/2019 Lecture 19 Two DOF Systems Forced
7/12
Rose-Hulman Institute of TechnologyMechanical Engineering
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 =
8/12/2019 Lecture 19 Two DOF Systems Forced
8/12
Rose-Hulman Institute of TechnologyMechanical Engineering
Vibrations
Sketch these
( )( )( )22222121
1
2
2221
=
mm
FmkX
( )( )22222121112
2
=mm
FkX
X1
1 2
X2
1 2
These are frequency response plots
8/12/2019 Lecture 19 Two DOF Systems Forced
9/12
Rose-Hulman Institute of TechnologyMechanical Engineering
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
8/12/2019 Lecture 19 Two DOF Systems Forced
10/12
Rose-Hulman Institute of TechnologyMechanical Engineering
Vibrations
There is a nice website in Australia
http://www.mech.uwa.edu.au/bjs/Vibration/TwoDOF/default.html
8/12/2019 Lecture 19 Two DOF Systems Forced
11/12
Rose-Hulman Institute of TechnologyMechanical Engineering
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
8/12/2019 Lecture 19 Two DOF Systems Forced
12/12
Rose-Hulman Institute of TechnologyMechanical Engineering
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