multi degree of freedom systems_1
TRANSCRIPT
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DR. MUHD. HAFEEZ BIN ZAINULABIDINFACULTY OF MECHANICAL & MANUFACTURING ENGINEERING
UTHM
NOISE & VIBRATION
MULTI DEGREE OF FREEDOM SYSTEMS
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CHAPTER 2
Multi Degree of Freedom Systems
Figure source: http://cite.iiit.ac.in/vlab/Expr6.htm
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Multi Degree of Freedom Systems
1. Mass Spring System
2. Generalized Coordinates and GeneralizedForces
3. Lagranges Equations
4. General Equations in Matrix Form
5. Eigenvalues and Eigenvectors
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Mass Spring System
a) Newtons 2nd Law Approach
Figure 2(a) and 2(b)
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The application of Newtons 2nd law of motion to
mass mi gives
Equations of motion of the masses m1 and mnwhere i = 1, xo = 0 and i = n, xn+1 = 0.
1...,,3,2;
11
111111
!!
niFxk
xkkxkxcxccxcxm
iii
iiiiiiiiiiiiii
1221212212111Fxkxkkxcxccxm !
nnnnnnnnnnnnn
Fxkkxkxccxcxm ! 1111
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Equations of motion can be expressed in matrix
form as
where [m],[c] and [k] are the mass, damping and
stiffness matrices, respectively.
? A ? A ? A FxkxcxmTTTT !
? A
-
!
nm
m
m
m
m
.
/1///
.
.
.
000
000
000
000
3
2
1
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? A
-
yy
!
1
4
433
3322
221
000
00
000
00
000
nnnccc
c
ccc
cccc
ccc
c
.
.///
//1
.
.
.
? A
-
yy
!
1
4
433
3322
221
000
00
000
00
000
nnnkkk
k
kkk
kkkk
kkk
k
.
.///
//1
.
.
.
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are the displacement, velocity,acceleration and force vectors.
FxxxT
TTT and,,
,
2
1
y
y!
tx
tx
tx
x
n
T
,
2
1
y
y!
tx
tx
tx
x
n
T
,
2
1
y
y
tx
tx
tx
x
n
T
.
2
1
y
y!
tF
tF
tF
F
n
T
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In general form, the mass, damping and stiffness
matrices are
? A
-
!
nnnnn
n
n
n
mmmm
mmmm
mmmm
mmmm
m
.////
.
.
.
321
3333213
2232212
1131211
? A
-
!
nnnnn
n
n
n
cccc
cccccccc
cccc
c
.
////
.
.
.
321
3333213
2232212
1131211
? A
-
!
nnnnn
n
n
n
kkkk
kkkkkkkk
kkkk
k
.
////
.
.
.
321
3333213
2232212
1131211
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b) Influence Coefficients Approach
The equations of motion of a multi degree offreedom system can also be written in terms ofinfluence coefficients.
There are 3 main influence coefficients:
i. Stiffness Influence Coefficients
ii. Flexibility Influence Coefficientsiii. Inertia Influence Coefficients
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Stiffness Influence Coefficients
Defined as the force at point idue to a unitdisplacement at point jwhen all the points otherthan the point jare fixed.
Total force at point i,Fi can be found by summingup the forces due to all displacementsxj j = 1, 2,, n).
or
nixkFn
j jiji
...,,2,1,1
!!!
? AxkF TT
!
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Steps in determining the stiffness influence
coefficients.1. Assume x1 =1 and x2,x3, ,xn = 0. The set of
forces will maintain the system in the assumedconfiguration.
2. Write static equilibrium equations for eachmass.
3. Solve the equations to find the influence
coefficients.4. Repeat Step 1to 3by assuming x2 =1 and so
on.
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Example:
Consider multi degree of freedom mass springsystem shown in Figure 2(c).
Figure 2(c)
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x1=1
Figure 2(d)
0:
:
:
313
2212
11211
!
!
!
km
kkm
kkkm
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x2=1
Figure 2(e)
3323
23222
2121
:
:
0:
kkm
kkkm
kkm
!
!
!
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x3=1
Figure 2(f)
3333
3232
131
:
0:
0:
kkm
kkm
km
!
!
!
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Solution forx1=1
Solution forx2=1
Solution forx3=1
Stiffness matrix
0,, 312212111 !!! kkkkkk
3323222212 ,, kkkkkkk !!!
33332313 ,,0 kkkkk !!!
? A
-
!
33
3322
221
0
0
kk
kkkk
kkk
k
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Flexibility Influence Coefficients
Defined as the deflection at point idue to a unitload at point jwhen all the loads at other pointsother than the point jare zero.
Total deflection at point i,xi can be found bysumming up the contributions of all forces Fj.
or
niFaxn
j
jiji ...,,2,1,1
!! !
? AFaxTT
!
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Steps in determining the flexibility influence
coefficients.1. Assume F1 =1 and F2, F3, , Fn = 0.
2. Write static equilibrium equations for each mass.
3. Solve the equations to find the flexibilitycoefficients.
4. Repeat Step 1to 3by assuming F2 =1 and so on.
OR5. Find inverse [k] if the stiffness matrix is available.
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F1=1
Figure 2(g)
0:
:
1:
213133
21313112122
112121111
!
!
!
aakm
aakaakm
aakakm
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F2=1
Figure 2(h)
0:
1:
:
223233
22323122222
122221211
!
!!
aakm
aakaakm
aakakm
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F3=1
Figure 2(i)
1:
:
:
233333
23333132322
132321311
!
!!
aakm
aakaakm
aakakm
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Solution for F1=1
Solution for F2=1
Solution forF
3=1
Flexibility matrix
131
121
111
1
,
1
,
1
ka
ka
ka !!!
21
32
21
22
1
12
11,
11,
1
kka
kka
ka !!!
321
33
21
23
1
13
111,
11,
1
kkka
kka
ka !!!
? A
-
!
321211
21211
111
111111
11111
111
kkkkkk
kkkkk
kkka
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Inertia Influence Coefficients
Defined as the set of impulses applied at point itoproduce a unit velocity at point jand zero atevery other points.
Total impulse at pointi,
can be found bysumming up the impulses causing the velocities
or
jx
nixmFn
j
jiji
...,,2,1,1
~!!
!
? AxmFT
T!
~
iF~
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Steps in determining the inertia influence
coefficients.1. Assume and .
2. Write equilibrium equations for each mass.
3. Solve the equations to find the inertiacoefficients.
4. Repeat Step 1to 3 by assuming and so
on.
11 !x 0...,,, 32 !nxxx
12 !x
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Generalised Coordinates &
Generalised Forces
Equations of motion of a vibrating system can beformulated in a number of different coordinate
systems. n independent coordinates are necessary to
describe the motion of a system having n d.o.f.
Any set ofn independent coordinates is calledgeneralised coordinates, designated by
q1,q2,,qn.
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Generalised coordinates may be lengths, angles,
or any other set of numbers that define theconfiguration of the system at any time uniquely.
Generalised coordinates also independent of the
conditions of constraint. When external forces act on the system, the
configuration of the system changes.
New configuration of the system can be obtainedby changing the generalised coordinates qjby qj,j =1, 2, , n.
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IfUj denotes the work done in changing the
generalised coordinate qjby the amount qj, thecorresponding generalised forceQj can bedefined as
where Qj will be a force (moment) when qj is alinear (angular) displacement.
njq
UQj
j
j ...,,2,1, !x
!
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Example: Triple Pendulum
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To illustrate the concept of generalized coordinates, consider thetriple pendulum system shown.
The system can be specified by the six coordinates (xj,yj),
j =1, 2, 3.
However the coordinates are not independent but are constrainedby the relations
Since the coordinates (xj,yj) are not independent, they cannot be
called generalized coordinates. The angular displacement are used to specify the locations of
the masses without constraint.
Thus, they form a set of generalized coordinates .
23
2
23
2
23
2
2
2
12
2
12
2
1
2
1
2
1
lyyxx
lyyxx
lyx
!
!
!
jU
jjq U!
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Example:
The arrangement of the compressor, turbine and generator in athermal power plant is shown in figure above. This arrangementcan be considered as a torsional system whereJi denote the massmoments of inertia of the three components,Mti indicate theexternal moments acting on the components and kti represent the
torsional spring constants of the shaft between the components.Derive the equations of motion of the system using Lagrangesequations by treating the angular displacements of the componentsas generalized coordinates.
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Substituting [1], [2] and [3] into Lagranges equations to obtain
Equations of motion expressed in matrix form
3233333
2331223222
12212111
ttt
ttttt
tttt
MkkJ
MkkkkJ
MkkkJ
!
!!
UUU
UUUUUUU
!
-
-
3
2
1
3
2
1
33
3322
221
3
2
1
3
2
1
0
0
0000
00
t
t
t
tt
tttt
ttt
MM
M
kkkkkk
kkk
J
J
J
UU
U
UU
U
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General Equations in Matrix Form
Equations of motion of a multi d.o.f system can be derived inmatrix form from Lagranges equations
The kinetic and potential energy in matrix form
Rayleigh dissipation function
niFx
V
x
R
x
T
x
T
dt
di
iiii
...,,2,1, !!x
x
x
x
x
x
x
x
? A
? Axkx
xmxT
T
T
TT
T
T
2
1
2
1
!
!
? AxcxTT
T
2
1!
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From the theory of matrices
Substituting into Lagranges equation to obtain
? A ? A ? A
nixm
xmmxxm
x
T
T
i
TTT
i
...,,2,1,2
1
2
1
!!
!!
x
x
T
T
T
TTT
T
T
HHH
nix
Txm
x
T
dt
d
i
T
i
i
...,,2,10,and !!x
x!
x
x T
T
? A ? A ? A
nixk
xkkxxkx
T
i
TTT
i
...,,2,1,
21
21
!!
!!xx
TT
TTT
TTT
HHH
? A ? A ? A
nixc
xccxxcx
R
T
i
TTT
i
...,,2,1,
2
1
2
1
!!
!!x
x
T
T
T
TTT
T
T
HHH
? A ? A ? A FxkxcxmTTT
T !
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Eigenvalues and Eigenvectors
Solution of Eigenvalue Problem
Consider equation of motion in the form of
The equation can also be expressed aswhere
Premultiplying equation by [k]-1to obtain
where [I] is the identity matrix and
For nontrivial solution of
? A ? A? A 02TT
! Xmk [
? A ? A? A 0TT
! XmkP 21
[P !
? A ? A? A 0TT
! XDIP
? A ? A ? AmkD1!
X
T
? A ? A 0!!( DIP
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Example:
Find the natural frequencies and mode shapes of the systemshown in figure above for k1 = k2 = k3 = kand m1 = m2 = m3= m.
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Solution:
The dynamical matrix is
Where the stiffness, flexibility and mass matrix can be obtainedusing influence coefficient approach or Lagranges equation
and
Thus
? A ? A ? A ? A? AmamkD !! 1
? A
-
!
321
221
1111
ka ? A
-
!
100
010
001
mm
? A
-
!
321
220
111
k
mD
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Setting the characteristics determinant equal to zero to obtainthe frequency equation
where
Dividing throughout by
where
? A ? A ? A 0321
221
111
00
00
00
!
-
-
!!(k
mDI
P
P
P
P
P
0165312221
123
!!
EEEEEE EEE
EEE
k
m
k
m2
[
PE !!
2
1
[P !
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The roots of cubic equation are
Once the natural frequencies are known, the mode shapes oreigenvectors can be calculated using equation
where
m
k
k
m
44504.0,19806.0 1
2
1
1!!! [
[
E
m
k
k
m
8025.1,2490.3 3
2
3
3 !!! [[
E
m
k
k
m2471.1,5553.1 2
2
22 !!! [
[E
? A ? A? A 3,2,1,0 !! iXDI iiTT
P
!i
i
i
i
X
X
X
X
3
2
1T
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First Mode: Substituting the value of
From the first two rows of equation
Solutions
and
The mode shape
km0489.51 !P
!
-
-
-
0
0
0
321
221
111
100
010
001
0489.51
3
1
2
1
1
X
X
X
km
km
!
-
!
0
0
0
0489.221
20489.31
110489.4
13
1
2
1
1
X
X
X
11
1
3
1
2
1
1
1
3
1
2
20489.3
0489.4
XXX
XXX
!
!
11
1
2 8019.1 XX ! 1
1
1
3 2470.2 XX !
!
!
2470.2
8019.1
11
1
1
3
1
2
1
1
1X
X
X
X
X
T
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Second Mode: Substituting the value of
From the first two rows of equation
Solutions
and
The mode shape
!
-
-
-
0
0
0
321
221
111
100
010
001
6430.02
3
2
2
2
1
X
X
X
km
km
!
-
0
0
0
3570.221
23570.11
113570.0
23
2
2
2
1
X
X
X
21
2
3
2
2
2
1
2
3
2
2
23570.1
3570.0
XXX
XXX
!
!
21
2
2 4450.0 XX ! 2
1
2
3 8020.0 XX !
!
!
8020.0
4450.0
12
1
2
3
2
2
2
1
2X
X
X
X
X
T
km6430.02 !P
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Mode Shapes
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ALL THE BEST!