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