G. Leng, ME Dept, NUS 3. Mathematical Properties of MDOF Systems 3.1 The Generalized Eigenvalue Problem Recall that the natural frequencies ω and modes a are found from [ - ω 2 M + K ] a = 0 or K a = ω 2 M a Where M and K are the mass and stiffness matrices of the MDOF system NB : M & K are symmetric matrices, M = M T and K = K T
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G. L
eng,
ME
Dep
t, N
US
3. M
athe
mat
ical
Pro
pert
ies o
f MD
OF
Syst
ems
3.1
The
Gen
eral
ized
Eig
enva
lue
Prob
lem
Rec
all t
hat t
he n
atur
al fr
eque
ncie
s ωan
d m
odes
aar
e fo
und
from
[ -ω
2M
+
K ]
a=
0
orK
a=
ω2
M a
Whe
re M
and
K a
re th
e m
ass a
nd st
iffne
ss m
atric
es o
f the
MD
OF
syst
em
NB
: M
& K
are
sym
met
ric
mat
rices
, M =
MT
and
K =
KT
G. L
eng,
ME
Dep
t, N
US
This
is a
ctua
lly a
mor
e ge
nera
l ver
sion
of t
he e
igen
valu
epr
oble
m
A x
= λ
x
whe
re A
is a
squa
re m
atrix
and
the
unkn
owns
x a
nd λ
are
calle
d th
e ei
genv
ecto
r and
eig
enva
lue.
The
eige
nval
uesa
re o
btai
ned
by so
lvin
g a
char
acte
rist
ic e
quat
ion
det[
A -λ
I ]
=
0
And
for e
ach
eige
nval
ueyo
u ca
n fin
d th
e ei
genv
ecto
rs b
y so
lvin
g
[ A -λ
I ] x
=
0
Que
stio
n : W
here
’s th
e an
alog
y ?
G. L
eng,
ME
Dep
t, N
US
3.2
Ort
hogo
nal p
rope
rty
of n
atur
al m
odes
(eig
enve
ctor
s)
Ort
hogo
nal P
rope
rty:
The
nat
ural
mod
es a
re “
orth
ogon
al”
with
re
spec
t to
both
the
mas
s and
stiff
ness
mat
rices
Proo
f :
Con
side
r tw
o m
odes
i &
j of
the
syst
em
Prem
ultip
lyea
ch e
quat
ion
with
a m
ode
vect
or
G. L
eng,
ME
Dep
t, N
US
Sinc
e M
and
K a
re sy
mm
etric
Hen
ce su
btra
ctin
g th
e tw
o eq
uatio
n yi
elds
Sinc
e th
e na
tura
l fre
quen
cies
are
dis
tinct
G. L
eng,
ME
Dep
t, N
US
Hen
ce th
e na
tura
l mod
es a
re o
rthog
onal
with
resp
ect t
o th
e m
ass m
atrix
. Sim
ilarly
for t
he st
iffne
sssm
atrix
.
0=
a jT
Ka i
Hom
ewor
k : P
rove
this
G. L
eng,
ME
Dep
t, N
US
Exa
mpl
e: V
erify
the
orth
ogon
al p
rope
rty fo
r the
2 D
OF
syst
em
m2m
kk
k
+ve
x 1x 2
Rec
all t
he m
ass a
nd st
iffne
ss m
atric
es a
re :
−
−
k
kk
km
m2
22
00
G. L
eng,
ME
Dep
t, N
US
and
the
natu
ral f
requ
enci
es a
nd n
orm
al m
odes
are
:
ω1
=
0.7
96 √
(k/m
)
ω
2=
1.
538 √(
k/m
)
=
1732
.01a
−=
1732
.22a
a 1T
M a
2=
G. L
eng,
ME
Dep
t, N
US
Sim
ilarly
a 1T
K a
2=
Hom
ewor
k : V
erify
the
orth
ogon
al p
rope
rty fo
r oth
er e
xam
ples
G. L
eng,
ME
Dep
t, N
US
Que
stio
n : C
an y
ou su
gges
t ano
ther
way
to n
orm
aliz
e m
odes
?
Ans
wer
:
Mod
es n
orm
aliz
ed th
is w
ay a
re c
alle
d or
thon
orm
alm
odes
Que
stio
n : S
o w
hat’s
the
big
deal
abo
ut o
rthog
onal
ity?
The
big
deal
:
G. L
eng,
ME
Dep
t, N
US
3.3
Dec
oupl
ing
a M
DO
F Sy
stem
Let
a 1, .
.., a
Nbe
the
mod
es o
f an
N D
OF
syst
em :
M x
’’+
K x
=F
with
initi
al c
ondi
tions
x(0
) = x
oan
d x’
(0) =
vo
The
mod
al m
atri
x P
is o
btai
ned
by p
laci
ng th
ese
mod
e ve
ctor
s to
geth
er c
olum
n w
ise
P=
[ a1
...
aN
]
G. L
eng,
ME
Dep
t, N
US
Def
ine
a ch
ange
of c
oord
inat
es x
= P
y
Subs
titut
e in
the
EOM
:
and
initi
al c
ondi
tions
:
G. L
eng,
ME
Dep
t, N
US
Pre
mul
tiply
EO
M b
y PT
Pre
mul
tiply
initi
al c
ondi
tions
by
PTM
G. L
eng,
ME
Dep
t, N
US
By
the
orth
ogon
ality
of th
e m
odes
, PT
M P
and
PT
K P
are
di
agon
alm
atric
es.
How
so ?
PTM
P=
a 1T
[ M
]
[ a 1
...
aN
] ... a N
T
G. L
eng,
ME
Dep
t, N
US
The
syst
em d
ecou
ples
into
N S
DO
F eq
uatio
ns !
miy
i’’
+ k i
y i=
a iT F
y i(0
) =
( aiT
M x
o) /
mi
y i’(
0)
= ( a
iTM
vo
) / m
ii =
1, .
.., N
whe
re m
i=
a iT
M a
ik i
=a i
TK
ai
Are
we
done
?
G. L
eng,
ME
Dep
t, N
US
Que
stio
n : H
ow d
o w
e ge
t the
act
ual
resp
onse
x ?
Ans
wer
:
G. L
eng,
ME
Dep
t, N
US
Exa
mpl
e : D
ecou
plin
g a
MD
OF
syst
em
mm
kx 1x 2 k
k
M=
m
0K
=2k
-k0
m
-k2k
ω1
= √
(k/m
)ω
2=
√(
3k/
m)
=
111a
−=
112a
G. L
eng,
ME
Dep
t, N
US
Find
the
resp
onse
for i
nitia
l con
ditio
ns x
(0) =
{1,
0}T
and
x’(0
) =
{0,0
}T
Form
the
mod
al m
atrix
P=
The
mod
al m
ass m
atrix
PT M
P
G. L
eng,
ME
Dep
t, N
US
The
mod
al st
iffne
ss m
atrix
PT K
P
= Hen
ce th
e de
coup
led
EOM
for t
he m
odal
coo
rdin
ates
are
:
Wha
t els
e do
we
need
?
G. L
eng,
ME
Dep
t, N
US
Get
the
initi
al c
ondi
tions
for y
1an
d y 2
.
( PT
M P
) y(
0)
= PT
M x
o
G. L
eng,
ME
Dep
t, N
US
So w
e ne
ed to
solv
e ?
SD
OF
equa
tions
:
2
y 1’’
+ k
/my 1
=
0y 1
(0)
= 1
/2y 1
’(0)
=
0
y 2’’
+ 3
k/m
y2
=0
y 2(0
) =
-1/2
y 2’(
0) =
0
The
solu
tion
is :
G. L
eng,
ME
Dep
t, N
US
Fina
lly tr
ansf
orm
bac
k to
real
coo
rdin
ates
x 1= =
x 2= =
G. L
eng,
ME
Dep
t, N
US
Not
es
1. D
ecou
plin
g m
ay n
ot w
ork
if a
dam
ping
mat
rix is
pre
sent
.
Eg:
M x
'' +
C x
' +
K x
= F
An
exce
ptio
nal c
ase
is R
ayle
igh
dam
ping
whe
re
C
= a
M
+ b
K
G. L
eng,
ME
Dep
t, N
US
2. T
he tr
ansf
orm
atio
n x
= P
yis
bas
ical
ly a
sum
mat
ion
of
natu
ral m
odes
x=
[ a1
...
aN
] y
1 ... y N
=
y 1a 1
+
...
+
y
Na N
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