June 30, 2020

Silent Engineering(Lecture 2)

Examples of Analysis for Various Plates

-Fixing conditions of plates andthe results of modal analysis -

Tokyo Institute of TechnologyDept. of Mechanical EngineeringSchool of Engineering

Prof. Nobuyuki Iwatsuki

１. Example of Modal Analysis 1.1 Modal analysis of peripherally simply

supported thin rectangular plate

x

y

a

b

O

Thickness: hDensity: ρYoung’s modulus: EPoison’s ratio: ν

Explained in last week!

Equation of motion:

0)2( 24

4

22

4

4

4

=−∂∂

+∂∂

∂+

∂∂ Wh

yW

yxW

xWD ωρ (1)

Boundary conditions:

0,0;,0

0,0;,0

2

2

2

2

=∂∂

==

=∂∂

==

yWWby

xWWax

　

　

(2)

NnMmb

yna

xmCyxW mn

,...,2,1,,...,2,1

,sinsin),(

==

=

　　　　

　　ππ

Mode of vibration:

(3)

)( 2

2

2

22

bn

am

hD

+=ρ

πω

Natural angular frequency:

Explained in last week!

(4)

By giving m and n, we can easily calculate mode of vibration and natural frequences.

For a=1326.8mm, b=749.5mm, h=4.5mm, E=205GPa, ν=0.3, ρ=7860kg/m3

(m,n)=(1,1)

(2,1)

(3,1)

(1,2)

(2,2)

(4,1)

Calculated mode shapes and natural frequenciesof peripherally simply supported thin rectangular plate (1)

Nodal lines

For a=1326.8mm, b=749.5mm, h=4.5mm, E=205GPa, ν=0.3, ρ=7860kg/m3

(m,n)=(3,2)

(5,1)

(4,2)

(1,3)

(2,3)

(3,3)

Calculated mode shapes and natural frequenciesof peripherally simply supported thin rectangular plate (2)

Nodal lines

For a=1326.8mm, b=749.5mm, h=4.5mm, E=205GPa, ν=0.3, ρ=7860kg/m3

(m,n)=(5,2)

(6,1)

(4,3)

(6,2)

(7,1)

(5,3)

Calculated mode shapes and natural frequenciesof peripherally simply supported thin rectangular plate (3)

Nodal lines

For a=1326.8mm, b=749.5mm, h=4.5mm, E=205GPa, ν=0.3, ρ=7860kg/m3

(m,n)=(6,2)

(6,3)

(7,3)

Calculated mode shapes and natural frequenciesof peripherally simply supported thin rectangular plate (4)

Nodal lines

1.2 Modal analysis of peripherally clamped thin rectangular plate

Boundary conditions:

0,0;,0

0,0;,0

=∂∂

==

=∂∂

==

ywwby

xwwax

　

　(5)

Resultantly we obtain

0,0;,0

0,0;,0

=∂∂

==

=∂∂

==

yWWby

xWWax

　

　

(6)

xa

b

O

Thickness: hDensity: ρYoung’s modulus: EPoison’s ratio: ν

y

Let assume vibration shape W(x,y) as same as simply supported plate as

byn

axmCyxW mn

ππ sinsin),( = (7)

Then the boundary conditions can be satisfied as

0),(),()0,(),0( ==== bxWyaWxWyW (8)

0sin),(

0sin)0,(

0sin),(

0sin),0(

≠−=∂

∂

≠=∂

∂

≠−=∂

∂

≠=∂

∂

axm

bnC

ybxW

axm

bnC

yxW

byn

amC

xyaW

byn

amC

xyW

mn

mn

mn

mn

ππ

ππ

ππ

ππHowever

(9)It is confirmed that Eq.(7)cannot satisfy the boundaryconditions.

There exist no exact solution for peripherally clamped rectangular plate.

2. Approximate Modal Analysis with Rayleigh-Ritz Method

2.1 Relation between energy and natural frequency

tAtx ωcos)( =

k

xm

Let’s consider a single DOF vibration system and assume its vibration displacement as .

The kinetic energy, K, and potential energy, P, are calculatedas

tkAkxtP

tmAxmtK

ω

ωω

222

2222

cos21

21)(

sin21

21)(

==

==

(10)

The maximum values of the both energies are calculated as

*max

2max

*max

222max

2121

PkAP

KmAK

==

== ωω(11)

*max

*max2

KP

mk==ω

Because K+P=const., therefore Kmax=Pmax . From Eq.(10), we can derive the natural angular frequency as

(12)Natural angular frequency can be calculated with maximum values of kinetic and potential energies which are calculated with the exact deformation

If the energy is calculated with the approximated deformation,the approximated natural angular frequency becomes higher.

Therefore the approximated natural angular frequency can be obtainedby searching vibration deformation which minimizes Eq.(12).

Rayleigh-Ritz method

2.2 Rayleigh-Ritz methodMode shape is assumed as a linear combination of trial functions whichsatisfy boundary conditions as

),(),( yxWCyxW ii

i∑=where

::),(

i

i

CyxW Trial functions

Coefficients

(13)

Kinetic energy of thin plate is calculated as

∫ ∑

∫

∫

=

=

=

S iii

S

S

dsyxWCh

dsyxWh

dstyxwhK

22

22

2max

),(2

),(2

]),,(21max[

ωρ

ωρ

ρ

　　　

　　　

Therefore

∫ ∑

=

S iii dsyxWChK

2*max ),(

2ρ

(14)

Potential energy of thin plate is calculated as strain energy as

dsyx

WC

y

WC

x

WC

y

WC

x

WCD

dsyx

WyW

xW

yW

xWD

dsyx

wyw

xw

yw

xwD

dsyx

wMywM

xwM

dsdzP

ii

iii

iii

i

S

ii

iii

i

S

S

Sxyyx

S

h

h xyxyyyxx

∂∂

∂−

∂

∂⋅

∂

∂−−

∂

∂+

∂

∂=

∂∂

∂−

∂∂⋅

∂∂

−−

∂∂

+∂∂

=

∂∂

∂−−

∂∂⋅

∂∂

−−

∂∂

+∂∂

=

∂∂

∂−+

∂∂

−+

∂∂

−=

++=

∑∑∑

∫∑∑

∫

∫

∫

∫ ∫−

22

2

2

2

2

2

2

2

2

2

22

2

2

2

22

2

2

2

2

22

2

2

2

22

2

2

2

2

2

2

2

2

2

2/

2/max

)1(2

2

)1(22

)1(22

max

221max

])(21max[

ν

ν

ν

γτεσεσ

　　　　　

　　　

　　　

　　　

　　　

*max

22

2

2

2

2

2

2

2

2

2

)1(2

2

P

dsyx

WCyWC

xWC

yW

xWCD

i

ii

i

ii

i

ii

S

ii

ii

=

∂∂

∂−

∂∂

⋅∂∂

−−

∂∂

+∂∂

=

∑∑∑

∫ ∑

ν　

　　　

(15)

Then natural angular frequency should be minimized with respect to the coefficients, Ci, as

= *

max

*max2 min

KPω (16)

Therefore minimal value of Eq.(16) should be calculated as

NiC

KP

i

,...2,1,0*max

*max

==∂

∂　　 (17)

Eq.(17) is modified as

( ) 0*max

*max

*max

*max

*max

2*max

*max*

max*max

*max

*max

*max

=∂∂

−∂∂

=∂∂⋅−⋅

∂∂

=∂

∂

KC

KKP

CP

KC

KPKC

P

CKP

iiii

i

(18)

(19)NiC

KC

P

ii

1,2,..,,0*max2

*max ==

∂∂

−∂∂

∴ 　　　　 ω

dsyxWyxWCh

dsyxWCyxWh

dsyxWCyxWCC

h

dsyxWChCC

K

jj S

ij

S jjji

S j jjjjj

i

S jjj

ii

),(),(

),(),(

),(),(

),(2

2*max

⋅=

=

⋅∂∂

=

∂∂

=∂∂

∑ ∫

∫ ∑

∫ ∑ ∑

∫ ∑

ρ

ρ

ρ

ρ

　　　

　　　

　　　

From Eq.(14) we obtain

(20)

From Eq.(15) we obtain

dsyx

WC

yxWD

dsyW

xW

CyW

CxWD

dsyW

xW

CyW

xWD

dsyx

WC

yW

CxW

C

yW

xW

CDC

PC

j

jj

S

i

S j

ijj

j

jj

i

jj

jj

S

ii

j

jj

j

jj

j

jj

S

jj

jj

ii

∂∂

∂⋅

∂∂∂

−+

∂∂⋅

∂

∂+

∂

∂⋅

∂∂

−−

∂

∂+

∂

∂⋅

∂∂

+∂∂

=

∂∂

∂−

∂

∂⋅

∂

∂−−

∂

∂+

∂

∂

∂∂

=∂∂

∑∫

∫ ∑∑

∑∫

∑∑∑

∫ ∑

22

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

22

2

2

2

2

2

2

2

2

2*

max

)1(2

)1(

)1(2

2

ν

ν

ν

　

　

　　　　

　　　

∫∑

∑ ∫

∑ ∫

∑ ∫

∂∂

∂⋅

∂∂∂

−+

∂∂⋅

∂

∂−−

∂

∂⋅

∂∂

−−

∂

∂+

∂

∂⋅

∂∂

+∂∂

=

S

ji

jj

i

j S

jj

j

j S

ij

jj

j S

iij

dsyx

Wyx

WCD

dsyW

xW

CD

dsyW

xWCD

dsyW

xW

yW

xWCD

22

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

)1(2

)1(

)1(

ν

ν

ν

　　

　　

　　

(21)

dsyxWyxWCh

dsyx

Wyx

WCDdsyW

xW

CD

dsyW

xWCDds

yW

xW

yW

xWCD

CK

CP

jj S

ij

S

ji

jj

i

j S

jj

j

j S

ij

jj

j S

iij

ii

),(),(

)1(2)1(

)1(

2

22

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

*max2

*max

⋅−

∂∂

∂⋅

∂∂∂

−+∂∂⋅

∂

∂−−

∂

∂⋅

∂∂

−−

∂

∂+

∂

∂⋅

∂∂

+∂∂

=

∂∂

−∂∂

∑ ∫

∫∑∑ ∫

∑ ∫∑ ∫

ωρ

νν

ν

ω

　　

　　　

　

　　

By substituting Eqs.(20), (21) into Eq.(19), we obtain

N×1 vector

1×Nvector

(22)[ ]

0

2

1

2

=

−=

N

iii

C

C

CC

BA

ω

For i=1~N, we obtain

N×Nmatrix N×1 vector

(23)

When angular velocity ω is given, this is a system of linear equations with respect to coefficients, Ci, in which the right side constant vector is zero vector.

[ ] [ ]( ) [ ]0

2

1

2 =

−

N

i

C

C

CC

BA

ω

[ ] [ ]{ } 0det 2 =− BA ω

Therefore the following equation should hold.

(24)

Equation with respect to ω2

Approximated natural angular frequency equation

The approximated natural angular frequency ωi (i=1~N) can be solvedwith numerical method like Newton method or bisection method.

By substituting the obtained approximated natural angular frequencyinto Eq.(23) , the coefficients for trial functions, Cj, can be calculatedas a ratio for each trial function.

The approximated mode shape can be obtained.

Rayleigh-Ritz method

3. Examples of Approximated Modal Analysis with Rayleigh-Ritz Method

xa

b

O

Thickness: hDensity: ρYoung’s modulus: EPoison’s ratio: ν

3.1 Ex.(1) Peripherally clamped thin rectangular platey

O L

Trial function is given as multiplication of eigenfunctions of Euler-Bernouilli beam which is clamped at both ends to satisfy boundary conditions.

x

Length: LCross-sectional area: ADensity: ρYoung’s modulus: EMoment of inertia of area: I

y w(x,t)

Equation of motion of Euler-Bernouilli beam:

0),(),(2

2

4

4

=∂

∂+

∂∂

ttxwA

xtxwEI ρ (25)

Assumption of solution with separation of variables:

)sincos)((),( tBtAxWtxw ωω += (26)

Rewritten equation of motion of Euler-Bernouilli beam:

0)()( 24

4

=−∂

∂ xWAx

xWEI ωρ (27)

Solution can be represented as:

LxD

LxD

LxD

LxDxW λλλλ sincossinhcosh)( 4321 +++= (28)

(29)whereEI

AL 244 ωρλ =

Boundary conditions:

0,0;,0 =∂∂

==x

WWLx 　 (30)

By substituting Eq.(30) into Eq.(28), we obtain

[ ]

=

=

− 0000

)(

cossincoshsinhsincossinhcosh

000101

4

3

2

1

4

3

2

1

DDDD

P

DDDD

LLLL

LL λ

λλλλλλλλλλλλ

λλ

(31)

Therefore

( )[ ]( ) ( ) 0coscosh1)(2det =⋅−= λλλλL

P (32)

By solving Eq.(32) with a numerical method such as Bisection Method, we obtain

i λi

1 4.7302 7.8533 10.996: :

Coefficients can be represented as

iii

iii

ii

iii

iii

ii

DD

DD

DD

DD

,0,4

,0,3

,0,2

,0,1

coscoshsinsinh

coscoshsinsinh

λλλλ

λλλλ

−+

=

−=−+

−=

=

(33)

Mode shape of Euler-Bernouilli beam can be represented as

ii

iii

iii

iiii L

xLx

Lx

LxDxW

λλλλα

λλαλλ

coscoshsinsinh

sinsinhcoscosh)( ,0

−+

=

−−−=

　　　　where

(34)

Therefore mode shape function of rectangular plate is given as

−−−⋅

−−−=

⋅=

∑

∑

by

by

by

by

ax

ax

ax

ax

C

yWxWCyxW

jyjyjy

jyjy

ixixix

ixix

jiij

jyji

ixij

,,,

,,

,,,

,,

,

,,

,

sinsinhcoscosh

sinsinhcoscosh

)()(),(

λλα

λλ

λλα

λλ

　　　　　　　　

　　　　

(35)

By substituting Eq.(35) into (20)~(24) , mode of vibration of peripherally clamped thinrectangular plate can be calculated.

For a=1326.8mm, b=749.5mm, h=4.5mm, E=205GPa, ν=0.3, ρ=7860kg/m3

1st mode

Calculated mode shapes and natural frequenciesof peripherally clamped thin rectangular plate (1)

2nd mode

3rd mode

4th mode

5th mode

6th mode

Nodal lines

For a=1326.8mm, b=749.5mm, h=4.5mm, E=205GPa, ν=0.3, ρ=7860kg/m3

7th mode

Calculated mode shapes and natural frequenciesof peripherally clamped thin rectangular plate (2)

8th mode

9th mode

10th mode

11th mode

12th mode

Nodal lines

For a=1326.8mm, b=749.5mm, h=4.5mm, E=205GPa, ν=0.3, ρ=7860kg/m3

13th mode

Calculated mode shapes and natural frequenciesof peripherally clamped thin rectangular plate (3)

14th mode

15th mode

16th mode

17th mode

18th mode

Nodal lines

"MODE VECTORS C(I,J)"1 50.0167230113 1

-0.395925959633D+00 0.261469656398D-12 -0.196069532541D-020.148695887244D-09 0.400866631526D-15 0.190104200322D-11-0.145661530073D-01 -0.621618927908D-13 0.735311214401D-030.468864192148D-10 0.138965128501D-15 0.768442021939D-12-0.239758620027D-02 -0.285212819456D-13 0.466018870508D-030.411960556242D-10 0.566142383589D-16 0.815220335506D-12-0.627345017925D-03 -0.110809664028D-13 0.227983263418D-032 69.3765119732 1

-0.158023361577D-09 0.267801773626D-16 -0.284719748538D-11-0.420678396258D+00 0.893629064034D-12 -0.715828296484D-020.198610039323D-09 0.177078085499D-16 0.296559544418D-11-0.118802297498D-01 -0.555693535908D-13 0.938536968024D-030.419495670410D-10 0.645322429740D-17 -0.174846680311D-11-0.257326640226D-02 -0.267700445405D-13 0.576242662412D-030.173769934106D-08 0.249883609096D-16 0.135765812117D-09

::

4 127.5364383246 10.517111782955D-11 -0.435714521570D+00 0.102828199432D-120.449235182804D-16 0.221952789501D-09 0.868313535296D-170.611942853424D-12 -0.334056298652D-01 0.102486688267D-130.161924962144D-15 0.908542873312D-10 0.476017589705D-170.372464771601D-13 -0.765134638114D-02 -0.508632064686D-140.164395560331D-16 0.674906109258D-10 0.204951987707D-170.143866797544D-13 -0.241083479328D-02 -0.235825415697D-14

5 146.1307429482 1-0.251002441355D-16 0.252445502971D-09 -0.504900167608D-17-0.476324496893D-11 0.450396073293D+00 -0.167592443744D-12-0.310780307386D-16 -0.313532992558D-09 0.139918147139D-160.135067676375D-11 0.316150232441D-01 0.711769530751D-130.147309174552D-16 -0.102441353811D-09 0.987196384084D-17-0.148105441711D-13 0.849966986000D-02 0.116135743445D-14-0.101450945683D-16 -0.121236351434D-08 0.160711511004D-17

Calculated coefficients of trial functions:

j

i

Natural frequencyModeNo.

Mode of vibration of peripherally clamped thin rectangular plate can be calculated with Rayleigh-Ritz method.

Comparison between simply supported and clamped platesMode No. Simply supported Clamped

Natural frequencyfi[Hz]

Nodal linesM,N

Natural frequencyfi[Hz]

Nodal linesM,N

1 25.7 0,0 50.0 0,02 44.3 1,0 69.4 1,03 75.3 2,0 103.2 2,04 84.0 0,1 127.5 0,15 102.6 1,1 146.1 1,16 118.7 3,0 150.2 3,07 133.6 2,1 178.1 2,18 174.6 4,0 211.8 4,09 177.1 3,1 223.9 3,110 181.2 0,2 244.3 0,211 199.8 1,2 262.7 1,212 230.9 2,2 283.1 2,213 232.9 4,1 285.6 5,014 242.8 5,0 293.9 4,115 274.3 3,2 338.6 3,216 301.2 5,1 355.8 5,117 323.5 6,0 372.0 6,018 330.2 4,2 396.3 4,2

Natural frequencies of clamped plate are higherthan those of simply supported plate.

Experimental Validation:

FFT Analyzer

Modal Analysis SoftwareLX-cada

Accelerometer

ExciterForcesensor

Setup for Experimental Modal AnalysisBy changing the position ofaccelerometer, transfer function betweenexciting force and acceleration responses

Mode No. Nodal linesM,N

Calculated Naturalfrequency

fi[Hz]

Measured Naturalfrequency

fi[Hz]

Error

%1 0,0 50.0 49.0 +2.00 2 1,0 69.4 68.7 +1.01 3 2,0 103.2 102.8 +0.39 4 0,1 127.5 126.1 +1.10 5 1,1 146.1 147.2 -0.75 6 3,0 150.2 150.8 -0.40 7 2,1 178.1 178.9 -0.45 8 4,0 211.8 209.0 +1.32 9 3,1 223.9 225.7 -0.80 10 0,2 244.3 － －

11 1,2 262.7 260.9 +0.69 12 2,2 283.1 282.4 +0.25 13 4,1 285.6 284.4 +0.42 14 5,0 293.9 295.9 -0.68 15 3,2 338.6 342.1 -1.03 16 5,1 355.8 356.5 -0.20 17 6,0 372.0 358.1 3.74 18 4,2 396.3 396.3 0.00

Comparison between calculated and measured natural frequency

Calculated natural frequencies agree very well with the measured values.

xa

b

O

Thickness: hDensity: ρYoung’s modulus: EPoison’s ratio: ν

3.2 Ex.(2) Cantilever rectangular platey

Trial function is given as multiplication of eigenfunctions of Euler-Bernouilli beams. The one is cantilever beam in x-direction andthe other is free beam.

Boundary conditions:

0,0;,0

0,0;

0,0;0

2

2

3

3

2

2

3

3

=∂∂

=∂∂

=

=∂∂

=∂∂

=

=∂∂

==

xW

xWLx

xW

xWLx

xWWx

　

　

　(36)

(37)

Cantilever beam:

Free beam:

In the same way as clamped beam, we obtain:

LxD

LxD

LxD

LxDxW λλλλ sincossinhcosh)( 4321 +++=

EIAL 2

4 ωρλ =where

Cantilever beam:01coscosh =+⋅ λλ

i λi

1 1.8752 4.6943 7.855: :

Free beam:01coscosh =−⋅ λλ

i λi

1 -2 -3 4.7304 7.853: :

Rigid body modes

They can be solved with a numerical method such as Bisection Method.

For a=230.0mm, b=180.0mm, h=4.0mm, E=205GPa, ν=0.3, ρ=7860kg/m3

1st mode

Calculated mode shapes and natural frequenciesof cantilever thin rectangular plate

2nd mode

3rd mode

4th mode

5th mode

Mode of vibration of cantileverthin rectangular plate can becalculated with Rayleigh-Ritzmethod.

Nodal lines

MODE VECTORS C(I,J)"1 64.7850935567 1

-0.496549661881D+00 0.141531270934D-11 0.312927146781D-02 0.169763719937D-14 0.267781219889D-03 -0.125916249787D-140.149666917864D-02 0.311306131174D-12 0.227650566620D-02 -0.555968777847D-16 0.397730128265D-03 -0.241215907211D-14-0.320445084408D-04 0.144911967917D-13 0.840684505912D-03 0.323163213895D-15 0.263867288233D-03 -0.152540044867D-140.628757941599D-04 0.195359309408D-13 0.211350819772D-03 0.498583019399D-16 0.121839659983D-03 -0.792642171238D-15-0.192882791224D-04 0.616781916663D-14 0.118081570015D-03 0.192985530512D-15 0.794087704003D-04 -0.536843922506D-150.126612179284D-04 0.204231723697D-14 0.362326761121D-04 -0.224721107550D-16 0.391765888874D-04 -0.281542236926D-152 191.1597375131 1

-0.193820581618D-16 0.120982618674D+01 0.112965676384D-14 0.251156499061D-02 0.570569916754D-15 0.155662321848D-03-0.369132351105D-15 0.183775153175D+00 -0.114129823456D-14 -0.950541625959D-03 -0.403863951435D-15 -0.106700272307D-030.476908422342D-15 0.497536048329D-03 0.111746624664D-14 0.147741729945D-03 0.336501620421D-15 0.125450528648D-03-0.233077889182D-15 0.712468782981D-02 -0.286164131758D-15 -0.343998917632D-03 -0.205757676868D-15 -0.173452507407D-040.262360658274D-15 -0.927696970695D-04 0.453121136376D-15 0.462834107393D-04 0.204984095097D-15 0.549134851936D-04-0.937197425874D-16 0.100744287826D-02 -0.142216211223D-15 -0.110510470306D-03 -0.113846160541D-15 -0.102042973458D-043 401.9113685018 10.176869850720D-02 -0.355135265824D-11 0.458396227472D-01 -0.107216887459D-13 0.300405387848D-02 -0.175520499758D-130.462058121764D+00 0.684510713024D-12 0.178183660776D-01 0.209362938905D-13 0.234845506158D-02 -0.122799917004D-13-0.327534734163D-02 0.167310793262D-12 0.116105532691D-02 0.315779103947D-14 -0.207382058918D-03 0.144386190063D-140.734569201300D-03 0.385783566373D-13 -0.387794260511D-03 0.371175106463D-14 -0.439959932841D-03 0.438304906393D-14-0.464336466688D-03 0.590845433847D-13 0.193149962356D-03 0.109706201255D-15 -0.123240466673D-03 0.953577071255D-150.208219316398D-03 0.228178776559D-14 -0.114317283046D-03 0.934441956349D-15 -0.143682535445D-03 0.180477480014D-144 663.3721279463 10.511711459158D-14 0.165283605014D+00 0.223688107877D-13 0.104520934381D-01 0.263897269511D-14 0.137481697614D-020.354291483125D-14 -0.109055844611D+01 -0.170302268102D-14 -0.935953353638D-02 -0.234219393955D-14 -0.230611402477D-030.303149220026D-14 -0.126235401287D+00 0.521428182371D-14 0.199683607769D-02 0.168294157853D-14 0.474128806448D-03-0.187970280638D-14 0.112097602093D-01 -0.365186469717D-14 -0.824975050054D-03 -0.156063215160D-14 -0.408446028357D-030.101557772952D-14 -0.107469259552D-01 0.170757322934D-14 0.714916323357D-03 0.898426360705D-15 0.152468437343D-03-0.921187952501D-15 0.173746308586D-02 -0.159334863293D-14 -0.304908026891D-03 -0.853941659040D-15 -0.184391043499D-035 761.3700132323 1

-0.218288842698D-02 0.818825365275D-13 -0.315692054502D+00 0.223774612558D-13 0.125796228974D-02 0.545300308753D-150.339498372509D-01 -0.159281137174D-12 -0.613047332145D-01 0.240465661454D-14 -0.132192447039D-03 -0.204415432183D-150.335771900450D-01 -0.159732124828D-13 -0.600358958411D-02 0.226170717125D-14 0.568641423395D-03 -0.333852997767D-14-0.728718276183D-02 0.664072960580D-14 -0.361203448906D-02 -0.541900583205D-15 -0.708818881420D-03 0.242769396199D-140.335246417596D-02 -0.276048806260D-14 -0.152775369680D-02 0.630425920603D-15 0.118531462583D-03 -0.556042336487D-15-0.172227112449D-02 0.148945404852D-14 -0.300068774780D-03 -0.310304600200D-15 -0.317223587181D-03 0.103421862092D-14

j

i

4. Concluding remarksSeveral examples of modal analysis of plate wereshown. (1)Exact solution of natural angular frequency and

mode shape of peripherally simply supportedthin rectangular plates can be easily calculated.

(2)There exists no exact solution for peripherallyclamped thin rectangular plate or cantilever thin rectangular plates.

(3)By applying Rayleigh-Ritz method with trialfunctions of eigenfunction of Euler-Bernouillibeams, various thin rectangular plates can be analyzed.