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16
52103-433 Introduction to M
echanical Vibration
Ch
ula
lon
gko
rn U
nive
rsity, T
HA
ILA
ND
Mech
an
ical E
ng
ineerin
g D
ep
artm
en
t
CH
AP
TE
R 7
FIN
ITE
EL
EM
EN
T
ME
TH
OD
(FE
M)
FO
R V
IBR
AT
ION
16
62103-433 Introduction to M
echanical Vibration
Ch
ula
lon
gko
rn U
nive
rsity, T
HA
ILA
ND
Mech
an
ical E
ng
ineerin
g D
ep
artm
en
t
Fin
ite E
lem
en
t Meth
od
(F
EM
) for V
ibra
tion
Ou
tline
s:
•Introduction
• and
of bar elements
• and
of beam elem
ents
•Lum
ped mass m
atrices
•Trusses
•M
odel reduction
KM
KM
16
72103-433 Introduction to M
echanical Vibration
Ch
ula
lon
gko
rn U
nive
rsity, T
HA
ILA
ND
Mech
an
ical E
ng
ineerin
g D
ep
artm
en
t
Vib
ratio
n o
f Fle
xib
le
Stru
ctu
res
so called “Distributed-P
arameter S
ystems
Suspension heads
in hard disk drives
Airplane w
ingstructure
Flexible
robot arm
16
82103-433 Introduction to M
echanical Vibration
Ch
ula
lon
gko
rn U
nive
rsity, T
HA
ILA
ND
Mech
an
ical E
ng
ineerin
g D
ep
artm
en
t
Mo
del o
f Fle
xib
le
Stru
ctu
res
Flexible structures can be m
odeled as bar, beam
, plate, shell, and etc.
Two w
ays to model:
1.A
na
lytic
al m
od
el
(beyond scope of this class)
•use the know
ledge of strength of m
aterial and dynamics to derive E
OM
.
•E
OM
s are in form of P
artial Differential
Equations (P
DE
).
•U
se discretization technique to solve P
DE
.
2.F
inite
Ele
me
nt M
od
el (F
EM
)
16
92103-433 Introduction to M
echanical Vibration
Ch
ula
lon
gko
rn U
nive
rsity, T
HA
ILA
ND
Mech
an
ical E
ng
ineerin
g D
ep
artm
en
t
FE
M: A
pp
roxim
atio
n
meth
od
Procedure:
•D
ivide (discretize) the structure into a num
ber of small sim
ple elements. T
he elem
ents are connected to each others by nodes.
•A
ll node displacement are chosen as
DO
Fs, so called nodal D
OF.
•D
isplacement function w
ithin the element
are approximated by a linear com
bination of low
-order polynomials.
•F
or vibration application, we construct the
matrices
and w
.r.t. all nodal DO
F
using energy method.
KM
d1
d2
d4
d3
di is nodal degree of
freedom
d1
d2Mdn
Elem
ent
nodes
17
02103-433 Introduction to M
echanical Vibration
Ch
ula
lon
gko
rn U
nive
rsity, T
HA
ILA
ND
Mech
an
ical E
ng
ineerin
g D
ep
artm
en
t
Bar E
lem
en
ts
From
solid mechanics, static displacem
ent of the bar is governed by
Ste
p I: d
ivid
e th
e b
ar in
to
ele
me
nts
.
3 elements, 4 nodes
, ,
, and are nodal
displacements.
Ste
p II: a
dis
pla
ce
me
nt
fun
ctio
n
Consider any elem
ent, e.g., the 1st elem
ent. A
ssume a displacem
ent function within the
element as
EA
x2
2
d du
0=
u1
t()u
2t()
u3
t()u
4t()
ux
t,(
)c
1t()x
c2
t()+
=
u(x,t)x
E, ρ, L
u1 (t)
u2 (t)
u3 (t)
u4 (t)
12
31
23
411
2
17
12103-433 Introduction to M
echanical Vibration
Ch
ula
lon
gko
rn U
nive
rsity, T
HA
ILA
ND
Mech
an
ical E
ng
ineerin
g D
ep
artm
en
t
B.C
.s:
,,
Hence
and
Rew
rite a displacement function in m
atrix form
:
Differentiate u w
.r.t. x:
x0
=u
xt,
()
u1
t()=
xl
=u
xt,
()
u2
t()=
c1
t()u
2t()
u1
t()–l
------------------------------=
c2
t()u
1t()
=
ux
t,(
)1
xl --,–
xl --u
1t()
u2
t()N
ut()
==
x∂ ∂
ux
t,(
)x
d dNu
t()1l --- ,
–1l ---
u1
t()
u2
t()B
ut()
==
=
Shape functions
11-x/l
x/lxl
xl
u2
u1 u(x,t)
17
22103-433 Introduction to M
echanical Vibration
Ch
ula
lon
gko
rn U
nive
rsity, T
HA
ILA
ND
Mech
an
ical E
ng
ineerin
g D
ep
artm
en
t
Ste
p III: C
on
stru
ct lo
ca
l and
Stra
in e
ne
rgy:
Strain energy of the bar elem
ent V(t) is
orwhere
is a local stiffness matrix defined as
KM
Vt()
12 ---E
Ax
∂ ∂u
xt,
()
2
xd0 l∫
=
Vt()
12 ---E
Au
TBTB
u[
]xd
0 l∫=
Vt()
12 --- uT
t() EAl -------
11–
1–1
ut()
12 --- uT
t()ku
t()=
=
k
kE
Al -------1
1–
1–1
=
17
32103-433 Introduction to M
echanical Vibration
Ch
ula
lon
gko
rn U
nive
rsity, T
HA
ILA
ND
Mech
an
ical E
ng
ineerin
g D
ep
artm
en
t
Kin
etic
en
erg
y:
Kinetic energy of the bar elem
ent T(t) is
where
is the bar density.S
ince
Therefore
orwhere
is a local mass m
atrix defined as
Tt()
12 ---Aρ
x()t
∂ ∂u
xt,
()
2
xd0 l∫
=ρx()
t∂ ∂
ux
t,(
)N
t∂ ∂
ut()
Nu
t()=
=
Tt()
12 ---Aρ
x()u
TNTN
u[
]xd
0 l∫=
Tt()
12 --- uT
t() ρA
l6 ---------
21
12
ut()
12 --- uT
t()mu
t()=
=
m
mρ
Al
6 ---------2
1
12
=
17
42103-433 Introduction to M
echanical Vibration
Ch
ula
lon
gko
rn U
nive
rsity, T
HA
ILA
ND
Mech
an
ical E
ng
ineerin
g D
ep
artm
en
t
Ste
p IV
: As
se
mb
ly
Let’s consider a three-element bar below
s.
For any one elem
ent,
In this case ,
Stra
in e
ne
rgy:
1-st element:
2-nd element:
kE
Al -------1
1–
1–1
=
lL3 ---
=k
3E
AL
-----------1
1–
1–1
=
V1
t()12 --- u
Tku
3E
A2
L-----------
u1
u2
T
11–
1–1
u1
u2
==
V2
t()3
EA
2L
-----------u
2
u3
T
11–
1–1
u2
u3
=
ll
l
L
u1 (t)
u2 (t)
u3 (t)
u4 (t)
17
52103-433 Introduction to M
echanical Vibration
Ch
ula
lon
gko
rn U
nive
rsity, T
HA
ILA
ND
Mech
an
ical E
ng
ineerin
g D
ep
artm
en
t
3-rd element:
Total strain energy is
orWith boundary conditions;
, then V
3t()
3E
A2
L-----------
u3
u4
T
11–
1–1
u3
u4
=
VV
1V
2V
3+
+=
Vt()
3E
A2
L-----------
u1
u2
u3
u4
T
11–
00
1–2
1–0
01–
21–
00
1–1
u1
u2
u3
u4
=
u0
t,(
)u
1t()
0=
=
Vt()
3E
A2
L-----------
0u2
u3
u4
T
11–
00
1–2
1–0
01–
21–
00
1–1
0u2
u3
u4
=
17
62103-433 Introduction to M
echanical Vibration
Ch
ula
lon
gko
rn U
nive
rsity, T
HA
ILA
ND
Mech
an
ical E
ng
ineerin
g D
ep
artm
en
t
V(t) can be reduced to
where K
is a global stiffness matrix.
, for the 3-element bar
Vt()
3E
A2
L-----------
u2
u3
u4
T
21–
0
1–2
1–
01–
1
u2
u3
u4
12 --- uTK
u=
=
K3
EA
L-----------
21–
0
1–2
1–
01–
1
=
17
72103-433 Introduction to M
echanical Vibration
Ch
ula
lon
gko
rn U
nive
rsity, T
HA
ILA
ND
Mech
an
ical E
ng
ineerin
g D
ep
artm
en
t
Again, consider the three-elem
ent bar:F
or any one element,
Kin
etic
en
erg
y:
1-st element:
2-nd element:
3-rd element:
Total kinetic energy is
or
mρ
Al
6 ---------2
1
12
ρA
L18
-----------2
1
12
==
T1
t()12 --- u
Tmu
ρA
L36
-----------u
1
u2
T
21
12
u1
u2
==
T2
t()ρ
AL
36-----------
u2
u3
T
21
12
u2
u3
=
T3
t()ρ
AL
36-----------
u3
u4
T
21
12
u3
u4
=
Tt()
T1
T2
T3
++
=
17
82103-433 Introduction to M
echanical Vibration
Ch
ula
lon
gko
rn U
nive
rsity, T
HA
ILA
ND
Mech
an
ical E
ng
ineerin
g D
ep
artm
en
t
With boundary conditions;
, then
V(t) can be reduced to
where M
is a global mass m
atrix.
, for the 3-element bar
Tρ
AL
36-----------
u1
u2
u3
u4
T
21
00
14
10
01
41
00
12
u1
u2
u3
u4
=
u0
t,(
)u
1t()
0=
=
Tρ
AL
36-----------
0u2
u3
u4
T
21
00
14
10
01
41
00
12
0u2
u3
u4
=
Tρ
AL
36-----------
u2
u3
u4
T
41
0
14
1
01
2
u2
u3
u4
12 --- uTM
u=
=
Mρ
AL
18-----------
41
0
14
1
01
2
=
17
92103-433 Introduction to M
echanical Vibration
Ch
ula
lon
gko
rn U
nive
rsity, T
HA
ILA
ND
Mech
an
ical E
ng
ineerin
g D
ep
artm
en
t
Notes:
•To analyze vibration of the structure, w
e use the global m
ass matrix
and global stiffness m
atrix to construct the m
atrix equation
Questions:
•C
an we determ
ine and
, if the elem
ent size is not uniform?
•H
ow accurate is the F
EM
result?
MK
Mu
t()K
ut()
+0
=
KM
18
02103-433 Introduction to M
echanical Vibration
Ch
ula
lon
gko
rn U
nive
rsity, T
HA
ILA
ND
Mech
an
ical E
ng
ineerin
g D
ep
artm
en
t
Co
mp
aris
on
of N
atu
ral
Fre
qu
en
cie
s
For L =
1m, ρ =
2700 kg/m3, and
N/m
2.
E7
107
×=
Exact ω
nω
n from F
EM
%difference
1-element m
odel:
ωn1 =
8,819 rad/sω
n1 = 7,998 rad/s
3-element m
odel:
ωn1 =
8,092 rad/s
ωn2 =
26,458 rad/s
ωn3 =
47,997 rad/s
ωn1 =
7,998 rad/s
ωn2 =
23,994 rad/s
ωn3 =
39,900 rad/s
1.18%
10.3%
20.3%
Rule of thum
b: At least tw
ice as many
elements m
ust be used than number of
accurate frequencies required.
10.3%
18
12103-433 Introduction to M
echanical Vibration
Ch
ula
lon
gko
rn U
nive
rsity, T
HA
ILA
ND
Mech
an
ical E
ng
ineerin
g D
ep
artm
en
t
Beam
Ele
men
ts
From
solid mechanics, static displacem
ent of the beam
is governed by
Ste
p I: d
ivid
e th
e b
ea
m in
to
ele
me
nts
.
1 element, 2 nodes
and are nodal linear displacem
ent. and
are nodal angular displacement.
Ste
p II: a
dis
pla
ce
me
nt fu
nc
tion
Assum
e a displacement function w
ithin the elem
ent as
x2 2
d dE
Ix
2
2
d dw
0=
w1
t()w
2t()
φ1
t()φ
2t()
wx
t,(
)c
1t()x
3c
2t()x
2c
3t()x
c4
t()+
++
=
w(x,t)
x
EA
, ρ, I, L
w1 (t)
w2 (t)
φ1 (t)
φ2 (t)
l
18
22103-433 Introduction to M
echanical Vibration
Ch
ula
lon
gko
rn U
nive
rsity, T
HA
ILA
ND
Mech
an
ical E
ng
ineerin
g D
ep
artm
en
t
B.C
.s:
, and
, and
Hence
and
Rew
rite a displacement function in m
atrix form
:
where
x0
=w
0t,
()
w1
t()=
x∂ ∂
w0
t,(
)φ
1t()
=
xl
=w
lt,
()
w2
t()=
x∂ ∂
wl
t,(
)φ
2t()
=
c1
t()1l 3 ---
2w
1w
2–
()
lφ
1φ
2+
()
+[
]=
c2
t()1l 2 ---
3w
2w
1–
()
l2φ
1φ
2+
()
–[
]=
c3
t()φ
1t()
=c
4t()
w1
t()=
wx
t,(
)N
1N
2N
3N
4
w1
t()
φ1
t()
w2
t()
φ2
t()
Nd
t()=
=
Shape functions
18
32103-433 Introduction to M
echanical Vibration
Ch
ula
lon
gko
rn U
nive
rsity, T
HA
ILA
ND
Mech
an
ical E
ng
ineerin
g D
ep
artm
en
t
00.1
0.20.3
0.40.5
0.60.7
0.80.9
10
0.2
0.4
0.6
0.8 100.1
0.20.3
0.40.5
0.60.7
0.80.9
1-0.2
-0.15
-0.1
-0.05 0
0.05
0.1
0.15
N1
x()1
3x
2
l 2--------
–2
x3
l 3--------
+=
N2
x()x
2x
2
l --------–
x3
l 2----
+=
N3
x()3
x2
l 2--------
2x
3
l 3--------
–=
N4
x()x
2l ----–
x3
l 2----
+=
N1 (x)
N3 (x)
N2 (x)
N4 (x)
1
1
x/l
x/l
18
42103-433 Introduction to M
echanical Vibration
Ch
ula
lon
gko
rn U
nive
rsity, T
HA
ILA
ND
Mech
an
ical E
ng
ineerin
g D
ep
artm
en
t
Ste
p III: C
on
stru
ct lo
ca
l and
Stra
in e
ne
rgy
Strain energy of the beam
element V
(t) is
where
is the flexural rigidity of the beam.
Plug
into and
integrate , w
e get
where
is a local stiffness matrix defined as
KM
Vt()
12 --- εσ12 ---
EI
x2 2
∂ ∂w
xt,
()
2
xd0 l∫
==
EI
wx
t,(
)N
x()dt()
=V
t()V
t()Vt()
12 --- dT
t()kd
t()=
k
kE
Il 3------
126
l12
–6
l
6l
4l 2
6l
–2
l 2
12–
6l
–12
6l
–
6l
2l 2
6l
–4
l 2
=
18
52103-433 Introduction to M
echanical Vibration
Ch
ula
lon
gko
rn U
nive
rsity, T
HA
ILA
ND
Mech
an
ical E
ng
ineerin
g D
ep
artm
en
t
Kin
etic
en
erg
y
Kinetic energy of the beam
element T
(t) is
where
is the beam density.
Since
Therefore
or
where
is a local mass m
atrix defined as
Tt()
12 ---Aρ
x()t
∂ ∂w
xt,
()
2
xd0 l∫
=ρx()
t∂ ∂
wx
t,(
)N
t∂ ∂
dt()
Nd
t()=
=
Tt()
12 ---Aρ
x()d
TNTN
d[
]xd
0 l∫=
Tt()
12 --- dT
t()md
t()=mm
ρA
l420---------
15622
l54
13l
–
22l
4l 2
13l
3l 2
–
5413
l156
22–
l
13l
–3
l 2–
22l
–4
l 2
=
18
62103-433 Introduction to M
echanical Vibration
Ch
ula
lon
gko
rn U
nive
rsity, T
HA
ILA
ND
Mech
an
ical E
ng
ineerin
g D
ep
artm
en
t
Ste
p IV
: As
se
mb
ly
Let’s consider a two-elem
ent beam below
s.
For one elem
ent,
For 2-elem
ent , then
kE
Il 3------
126
l12
–6
l
6l
4l 2
6l
–2
l 2
12–
6l
–12
6l
–
6l
2l 2
6l
–4
l 2
=
mρ
Al
420---------
15622
l54
13l
–
22l
4l 2
13l
3l 2
–
5413
l156
22–
l
13l
–3
l 2–
22l
–4
l 2
=
lL2 ---
=
w1 (t)
φ1 (t)
φ2 (t)
φ3 (t)
w2 (t)
w3 (t)
ll
18
72103-433 Introduction to M
echanical Vibration
Ch
ula
lon
gko
rn U
nive
rsity, T
HA
ILA
ND
Mech
an
ical E
ng
ineerin
g D
ep
artm
en
t
Stra
in e
ne
rgy
1-st element:
k8
EI
L3
---------
123
L12
–3
L
3L
L2
3L
–0.5
L2
12–
3L
–12
3L
–
3L
0.5L
23
L–
L2
=
mρ
AL
840
-----------
15611
L54
6.5L
–
11L
L2
6.5L
0.75L
2–
546.5
L156
11L
–
6.5L
–0.75
L2
–11
L–
L2
=
V1
t()12 --- d
1Tk
d1
w1
φ1
w2
φ2
Tk
w1
φ1
w2
φ2
==
18
82103-433 Introduction to M
echanical Vibration
Ch
ula
lon
gko
rn U
nive
rsity, T
HA
ILA
ND
Mech
an
ical E
ng
ineerin
g D
ep
artm
en
t
2-nd element:
Total strain energy is
Extended m
atrix form of V
(t) is
With fixed-free B
.C.s,
an
d
,
V(t) is then
V2
t()12 --- d
2Tk
d2
12 ---
w2
φ2
w3
φ3
Tk
w2
φ2
w3
φ3
==
VV
1V
2+
=
Vt()
12 ---
w1
φ1
w2
φ2
w3
φ3
T
w1
φ1
w2
φ2
w3
φ3
=
k
k
w0
t,(
)w
1t()
0=
=
x∂ ∂
w0
t,(
)φ
1t()
0=
=
18
92103-433 Introduction to M
echanical Vibration
Ch
ula
lon
gko
rn U
nive
rsity, T
HA
ILA
ND
Mech
an
ical E
ng
ineerin
g D
ep
artm
en
t
V(t) can be reduced to
where K
is a global stiffness matrix.
,
for th
e 2
-ele
me
nt b
ea
m
Vt()
12 ---
00w2
φ2
w3
φ3
T
00w2
φ2
w3
φ3
=
k
k
Vt()
8E
I
2L
3---------
w2
φ2
w3
φ3
T12
12+
()
3L
–3
L+
()
12–
3L
3L
–3
L+
()
L2
L2
+(
)3
L–
0.5L
2
12–
3L
–12
3L
–
3L
0.5L
23
L–
L2
w2
φ2
w3
φ3
12 --- dTK
d=
=
K8
EI
L3
---------
240
12–
3L
02
L2
3L
–0.5
L2
12–
3L
–12
3L
–
3L
0.5L
23
L–
L2
=
19
02103-433 Introduction to M
echanical Vibration
Ch
ula
lon
gko
rn U
nive
rsity, T
HA
ILA
ND
Mech
an
ical E
ng
ineerin
g D
ep
artm
en
t
Kin
etic
en
erg
y
1-st element:
2-nd element:
Total kinetic energy is
T1
t()12 --- d
1 Tmd
1 ˙
w1
φ1
w2
φ2
Tm
w1
φ1
w2
φ2
==
T2
t()12 --- d
2 Tmd
2 ˙
w2
φ2
w3
φ3
Tm
w2
φ2
w3
φ3
==
TT
1T
2+
=
19
12103-433 Introduction to M
echanical Vibration
Ch
ula
lon
gko
rn U
nive
rsity, T
HA
ILA
ND
Mech
an
ical E
ng
ineerin
g D
ep
artm
en
t
Extened m
atrix form of T
(t) is
With fixed-free B
.C.s,
an
d
,
T(t) is then
Tt()
12 ---
w1
φ1
w2
φ2
w3
φ3
T
w1
φ1
w2
φ2
w3
φ3
=
m
m
w0
t,(
)w
1t()
0=
=
x∂ ∂
w0
t,(
)φ
1t()
0=
=
Tt()
12 ---
00w2
φ2
w3
φ3
T
00w2
φ2
w3
φ3
=
m
m
19
22103-433 Introduction to M
echanical Vibration
Ch
ula
lon
gko
rn U
nive
rsity, T
HA
ILA
ND
Mech
an
ical E
ng
ineerin
g D
ep
artm
en
t
T(t) can be reduced to
where M
is a global mass m
atrix.
,
for th
e 2
-ele
ment b
eam
Tt()
12 --- ρA
L8
40-----------
w2
φ2
w3
φ3
T156
156+
()
11L
–11
L+
()
546.5
L–
11L
–11
L+
()
L2
L2
+(
)6.5
L0.75
L2
–
546.5
L156
11L
–
6.5L
–0.75
L2
–11
L–
L2
w2
φ2
w3
φ3
12 --- dTM
d=
=
Mρ
AL
840
-----------
3120
546.5
L–
02
L2
6.5L
0.75L
2–
546.5
L156
11L
–
6.5L
–0.75
L2
–11
L–
L2
=
19
32103-433 Introduction to M
echanical Vibration
Ch
ula
lon
gko
rn U
nive
rsity, T
HA
ILA
ND
Mech
an
ical E
ng
ineerin
g D
ep
artm
en
t
Lu
mp
ed
-Mass M
atric
es
• constructed from
kinetic energy are called “consistent-m
ass matrix” and they
are a full matrix.
•It is difficult to calculate
of the full m
atrix.
•A
n alternative method is to construct
as a “lum
ped-mass m
atrix”.
Consider a bar elem
ent:
Consider a beam
element:
M
M1–
M
ρ, A, L
mρ
AL
=
m/2
m/2
Mρ
AL
2-----------
10
01
=
ρ, A, L
mρ
AL
=
I13 ---
m2 ----
L2 ---
2
= m/2
m/2
Mρ
AL
2-----------
10
00
0L
2
12 ------0
0
00
10
00
0L
2
12 ------
=
19
42103-433 Introduction to M
echanical Vibration
Ch
ula
lon
gko
rn U
nive
rsity, T
HA
ILA
ND
Mech
an
ical E
ng
ineerin
g D
ep
artm
en
t
Tru
ss S
tructu
re
Let’s consider element 2:
Rew
rite in a matrix form
or
u3
t()U
3θ
U4
θsin
+cos
=
u4
t()U
5θ
U6
θsin
+cos
=
u3
t()
u4
t()θ
cosθ
sin0
0
00
θcos
θsin
U3
U4
U5
U6
=
1
2
3
u1
U2
U1
U5
U6u
2
U5
U6
u4
u3
U3
U4
U3 θ
θ
l l
l12
1
2
u3
U4
19
52103-433 Introduction to M
echanical Vibration
Ch
ula
lon
gko
rn U
nive
rsity, T
HA
ILA
ND
Mech
an
ical E
ng
ineerin
g D
ep
artm
en
t
where:
is the local coordinates of
element 2,
is the global
coordinates of element 2.
in g
lob
al c
oo
rdin
ate
s
Elem
ent 2:
where
is the global stiffness matrix of
element 2.
or
u2
t()Γ
U2
t()=
u2
u3
u4
T=
U2
U3
U4
U5
U6
T=
KV2
t()12 --- u
2 TKe u
212 --- U
2 TΓTK
e ΓU
212 --- U
2 TK2(
) U2
==
=
K2(
)
K2(
)E
Al -------
θcos
0
θsin
0
0θ
cos
0θ
sin
11–
1–1
θcos
θsin
00
00
θcos
θsin
=
K2(
)E
Al -------
θcos
() 2
θθ
cossin
θcos
() 2
–θ
θcos
sin–
θθ
cossin
θsin
() 2
θθ
cossin
–θ
sin(
) 2–
θcos
() 2
–θ
θcos
sin–
θcos
() 2
θθ
cossin
θθ
cossin
–θ
sin(
) 2–
θθ
cossin
θsin
() 2
=
19
62103-433 Introduction to M
echanical Vibration
Ch
ula
lon
gko
rn U
nive
rsity, T
HA
ILA
ND
Mech
an
ical E
ng
ineerin
g D
ep
artm
en
t
Elem
ent 1:
where:
is the global
coordinates of element 1,
for this example.
As
se
mb
ly:
where:
is th
e fu
ll
glo
ba
l co
ord
ina
tes,
is th
e e
xp
an
de
d g
lob
al
stiffn
ess m
atrix
asso
cia
ted
w
ith
.
V1
t()12 --- U
1 TK1(
) U1
=
U1
U1
U2
U5
U6
T=
K1(
)K
2()
=
Vt()
V1
t()V
2t()
+=
Vt()
12 --- U1 TK
1() U
112 --- U
2 TK2(
) U2
+12 --- U
TKU
==
UU
1U
2U
3U
4U
5U
6
T=
K
U
19
72103-433 Introduction to M
echanical Vibration
Ch
ula
lon
gko
rn U
nive
rsity, T
HA
ILA
ND
Mech
an
ical E
ng
ineerin
g D
ep
artm
en
t
in g
lob
al c
oo
rdin
ate
s
The full m
ass matrix
associated with
can be determined the sam
e way as
.
M
MU
K
19
82103-433 Introduction to M
echanical Vibration
Ch
ula
lon
gko
rn U
nive
rsity, T
HA
ILA
ND
Mech
an
ical E
ng
ineerin
g D
ep
artm
en
t
CH
AP
TE
R 8
VIB
RA
TIO
N
TE
ST
ING
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