To

proceed

with

the

Legrange

equations,

it

is

clear

we

need

toknow

how

to

calculate

different

forms

of

strain

energy

and

kinetic

energyin

solids.

We

therefore

look

at

energies

in

a

deformed

beam.從拉氏方程出發(fā)，計算梁的變形能，動能。We

look

at

something

called

the

“Rayleigh

Ritz

method”Calculate

the

mode

sh s

of

a

beam.

This

is

a

generalized‘prototype’method

for

the

finite

element

method.瑞利--里茲法計算梁的振型—有限元方法的“原型”Lecture3--伯努力梁Flexural

vibration

of

beams梁的彎曲振動Euler–Bernoulli

beams(‘simple

beams’)The

Newtonian

solution.

解y,

displacement

vx

0dxx

LxQMxM

M

dxdxTotal

vertical

force

acting

on

the

element

dxNewton’s

second

law

of

motion

for

dx

giveswhereρ

is

density

and

A

the

cross-sectionalareaTotal

moment

acting

on

element

dx

about

its

centre

point

x

xY

Q

dx

Q

dx,Q

2vx

dx

Adx

t

2

,x

t

2Q

2v

A

,dx

QdxQdx

dx

M

C

M

dx

MMM

dx

x

2

2

xThe

last

term

can

be

neglected

(second

order

smallthe

othersty)

in

comparison

with（1）QMMM

x

dxdxNewton’s

second

law

of

motion

then

gives（動量矩定理）where

I

is

the

second

moment

of

area

of

the

cross-section（橫截面的二階矩）θ

is

angle

of

rotation

of

element

dxFor

‘simple’

beam

theory

weusually

neglect

the

rotational

inertia

on

the

RHS（簡梁，略去轉動慣性）RHS—rectangle

hollow

section2MM

Q

0xSubstituting

(2)

into

(1)

to

eliminate

Q

givesThe

moment

curvature

（曲率）relationship,assuming

that

all

the

displacementv

is

due

bending

is(any

displacement

due

to

shear

is

neglected

for‘simple’beams).略去剪切效應

02

M

2vx2

A

t

2（2）（3）x

t

2

x

t

2Q

2vdx

Qdx

Idx

A

,（1）This

is

the

(familiar?)

simple

beam

equation梁的彎曲振動方程This

is

Hooke’s

law

for

bending.‘Moment

=

Bendingstiffness

·

curvature’（彎曲剛度

曲率）like

‘F

=

k

x’Substituting

(4)

into

(3)

to

eliminate

M2vM

EI

x2×（4）4v

2vEI

x4

A

t

2

0A

t

2

,Q

2vx（1）xM

Q

0（2）t

2A

02

M

2vx2（3）用于繞中心軸轉動用于橫向運動（2）代入（1）略去。。。Strain

Energy

in

a

beam

bending

element梁的彎曲單元中的變形能Simple

beam

theory

assumes

plane

sections

（橫截面的平面假設）which

arenormal

to

the

un-deformed

axis

of

the

beam

remain

plane

after

bending

and

are

stillnormal

to

the

deformed

axis.y,

displacement

vThe

axial

displacement

of

a

point

within

the

beam

ataheight

y

is

given

bydxxuxux,

y

y

vxv

slope斜率u(x,y)yThe

strain

components

of

this

are

therefore應變分量...we

are

neglecting

shear

in

this

treatment

略去剪切效應The

strain

energy

is

therefore

given

byThe

normal

stress

is

given

byyu(x,y)v2xx

y

2

x

u

u

v

0y

xxySoVx

x2

U

1

dV

x

E

xand

dV

dAdx

x

EI

2

dx

2v

2

2v

22

0

x

1

L2

0

A1

L2

V1U

x

x

dV

Ey

dAdx

222

Awhere

I

1

y2dANotation

for

a

beam

in

bending

Lx,uyThe

Rayleigh

RitzMethodRayleigh

Ritz

method

for

a

beamy,vz,

wLegrange

equationsStrain

energyKinetic

energyConsider

one

forceid

T

dt

q

（1）Qi

Fy

x,t

U

dxz

EI

0x2

2vx,

t

22

1

L（2）

dx0

vxt

,

21

LT

A2

t（3）The

Rayleigh-Ritz

method

approximates

the

exact

solution

witha

finite

expansionof

the

form:

近似解A

series

expansion

..

could

bea

polynomialseries,

or

a

Fourier

orsomething

else

depending

on

whatk? (x)

are.where

{?

k(x),

k=

1,

2,

..

m}

is

a

set

of

prescribed

functions

of

x

which

arelinearly

independent

基函數事先選定and

qk(t)

are

unknown

functions

of

time,

t,

to

be

determined未知函數待求

m

kx

qk

t

k

1

Rev

x,t

The

prescribed

functions?

k(x)

are

required

to

satisfy

the

following

conditions:to

be

linearly

independent

i.e.

one

cannot

be

described

as

a

linearcombination

of

the

others.基函數之間相互獨立to

be

p

times

differentiable,where

p

is

the

order

of

highest

derivativeappearing

in

the

expression

of

the

strain

energy3)

satisfy

the

geometric

boundary

conditionsp

階可導滿足幾何邊界條件4)

form

a

complete

series

so

that2構成完備的函數序列i.e.

the

expansion

tends

to

an

accurateapproximation

as

you

take

more

and

more

terms.0lim

Lmk

1mx

qk

t

v

xk

The

prescribed

functions

?

k(x)might

be

常用的基函數1)polynomialfunctions

多項式函數2)trigonometric

functions(a

Fourier

expansion)三角函數（付氏展開形式）wherez

is

a

complex

number z

為復數3)

Legendrepolynomialswherem

32

m

mexpz

,

x

expz

,

xm1

1

2

x

expz

,

x勒讓德多項式2m

1xd

m

x

md

m

x1

m1

x

m

10

4)Chebyshev

polynomials

切比雪夫多項式kindSecond

kindThe

set

of

functiontaken

is

called

a

set

of

basisfunctions.

The

series

expansion

is

therefore

said

to‘span’

a

‘function

space’

–

i.e.

the

set

of

all

functionsthat

can

be

made

out

of

linear

combinations

of

this

setof

functions.基函數

函數空間

m

x

cosmcos1

xsin

cos1

xm

x

1sin

m

1cos

x1)

Strainenergy

應變能So

ifis

written

in

the

matrix

form:Notation

means

a

row

matrix.行陣Here’s

a

trick.

The

differentiation

only

has

to

be

done

to

thebasis

functions,because

the

coefficients

are

just

numbers

now.只對基函數求導數x2EIU

z

dx

2vx,

t

201

L2mvx,

t

k

xqk

t

k

1

qt

qt

m

tq

211

2

m

vxq

tt

x

,

q

t

qt

xqm

t

x22

q1

t

m

x22

1

2

x2

x22

22

x2vx,

t

2x

x

xandso

thatwhereWe

can

call

this[K]because

we

saw

that

Ugave[K]at

ofthe

last

lecture應變能給出剛度陣This

is

a

clever

dodge.

Becausethe

coefficientsof

the

series

arejust

numbers

we

can

take

themout

of

the

integral.

Instead

ofintegrating

some

awkwardfunction

we

now

only

integratethe

tamebasis

functions.在此只對基函數積分是十分聰明的t

x2Tqt

qx2

x2

2vx,

t

2

2xT

2x

K

qt

TZU

1

qt

T

EI21

2

2

q

t2dx

q

tx

0

xL

2xT

2xZK

EIdxx

x0

2

2L

2xT

2x2)

Kinetic

energyand

putting

inthen?

mass

×

velocity2

,

summed

forall

slices

dx

thick.

dx

t0

vx,

t

21

LT

A2

,

tqtqm

tq2121

vxt

,

t

t

t

qt

qt

qtm

tm

21

x21

xq

xxtandso

thatwheredt

dvx,t

1

T

ttMT

qt

TqtxxALT120

dx

x2M

A0動能給出質量陣Now,with

no

applied

forces,the

terms

of

the

Legrange

equations

give用拉氏方程可得控制方程as

we

saw

before.Assuming

harmonic

motion

for

the

qk(t)parameters

in

the

form

q(t)=Re{reiωt

},the

natural

frequencies

ωj

and

corresponding

modes

can

be

calculated

from

theeigen-value

problem

解特征值問題，可得固有頻率，振型等K

i

M

rj

0MqKq

0where and

the

mode

shs

are

given

by振型

2j

j111Rer1

Rer

Rer

Tj

x

x

Rer

Example:Flexural

Vibration

of

Clamped

–

Simply

Supported

Beamusing

theRayleigh-Ritz

methodBoundary

conditions:LeftRightLx,uy,vv0,

t

0v0,

t

x

0vL,

t

02vL,

t

0MZ

0

i.ex2The

Rayleigh-Ritz

method

approximates

the

transverse

displacement

with

a

finiteexpansion

of

the

form:橫向位移的有限項展開式Let’s

choose

the

basis

functions

to

beso

thatmk

km

xA

cos

tk

1v

x,

t

kL

xk

1

x

xv

m321

2

mL

xcos

tx,t

A

xm1L

x

A

x

L

x

A

xthen

the

four

conditions

necessary

to

implement

theRayleigh-Ritz

method

are

verified.In

particular

we

have

managed

to

arrange

that

the

geometric

boundary

conditions

areautomatically

satisfied

by

the

basis

functions

we

have

chosen.Left

end

side(clamped)

左邊界條件whereRight

end

side(simply-supported)

右邊界條件'k

1xkkk

1

0

0

0

kv0,

t

0v0,

t

0

0

0

0'k

1

k

1k

1k

1xkL

Lk

1

L

k

2

L

kvL,

t

0vL,

t

0

L

L

L

L

0'kk

x

dxx

k

1Lxk

1

k

2xk

1

Lk

1

d

Then ut

the

approximation

into

the

terms

of

the

Lagrange

equations.將近似解代入拉氏方程and

put

inEIU

Lz

dx

02x22

vx,

t

2

1T

2

01

LAv2

x,

t

dxvm

x,t

x

A

cvm

x,

t

x2

L

x1)

Strain

energy

L

xAm

cost

A

cost

A

cost

21m1x3

L

x

xthent

txxT

2x2

1m

xv,

t22t

TZK

EI02U

1

whereWewantthe

elements

of

the

matrixThis

is

just

a

column

times

a

row

so

the

terms

in

the

matrix

are

easy

to

define,

theyareLet’s

say

we

will

only

take

2

terms

in

the

series.That

way

we

will

get

a

twoby

twomatrix.The

point

is

that

these

arecalculable.

They

are

easyintegrals

to

evaluate

withonly

known

terms

in

themsince:

k

x

x

L

xk

11k

1kx

2

xkk

x1kk

1

k12xkkdx2dx2

xdk

x

k

xd

1

TZK

EI0Li0i,

j

z

j

1

jLx

j

2

j

1x

x

dx

iLx

i

2i

1

i

1K

EI

j

1

ji1002201,13

LLL

EI

4L2

x

12Lx2

12

x

4EIL34L

24Lx

36

x2

dx

EI

K

EI

2L

6x

dxsimilarly2,2K

24

5

EIL5

4EIL4

and1,22,1K

K2)

Kinetic

energy02T

Av

x,

t

dx21

L

m

q

txtxvm

,

2

,

t

Lthen

1

2

2

1

T

M0wherethen

the

elementsof

the

matrix

are

given

by:Again

just

using

our

two

functions

x2

L

xkk

1Since:

x

x

L

x

0

ij,and

x3

L

x105etc.7202x

1,1L

x

dx

ALM

ALthe

following

system

of

equation

for

free

vibrations

is

obtainedUsing

Lagrange's

equations

we

can

now

derive

the

equation

of

motion

for

freevibrationd

T

U

0dt

qs

qswhere

Mq

d

T

m

dt

q

d

T

dt

q

1

Kq

m

q

U

q

U

1MqKq

0Assuming

a

harmonic

solution

so

that:the

following

matrix

eigen-value

problem

is

obtainedK

MA

0if

we

write

2This

is

a

set

of

m

linear

homogenous

equations

in

the

unknownsThe

system

of

equations

has

a

non-zero

solutions

ifdetK

i

M

0

A3

A1

cost

22

cost

A

A1

A3

cost

A

cost

qt

A1,

A2

,

Amexpanding

the

determinant

gives

a

polynomial

of

degree

m

in

λ.

The

polynomialwill

have

m

roots

λ1,

λ2

,…λ

m.

These

roots

are

called

eigen-values.Remembering

that

λ

=ω

2

the

natural

frequencies

are

given

by

λj

=ωj

2For

each

eigenvalue

λj

there

is

a

unique

solution

(to

withinan

arbitrary

constant)to

the

equationThese

solutions

areknownas

eigenvectors.K

MA

0j

jforAWhen

combined

with

the

prescribed

functions

?

k

(x)

they

define

thesh s

ofode

shΦ

j(x)

is

given

byBy ng

our

eigen-

ysisytically

we

can

getresults

that

are

still

in

termsof

E,

ρ,

L,

I,

A.

It

is

only

a2by

2

matrix

eigen-valueproblem

after

all.I

have

only

put

specificvalues

in

so

that

I

can

plotthe

eigenvectors

in

.the

modes

in

an

approximate

sense

since

ea

jx

xAjGoing

back

to

our

example

calculationIf

we

have

a

steel

beamE

207

109

N

m2

7860

kg

m3with

geometryL

9.144

mI

1.774

104m4A

1.006

102

m2Then

we

get

the

two

by

twomatrix

equation105

4

4L4L3

4L424L3

A1.021.12

1.23

e111.0269

e12frequencies20.04

Hz97.72

Hz158593.7698

e50110105In

fact

we

can

do

a

bit

more

algebra

and

work

out

what

teral

Ki,j

andMi,j

willbe.

可以給出剛度陣和質量陣中元素的一般表達式：ij

1

1

xjijjii

dxxLjjiixdjxjLLLLiZj

K

EI

00ij

100,dx

2

1

2

1

1

21

2

jxijjL

i2idx

1

2

11

xT

TZK

EI0

i

j

1i

1

Li

ji

1i

j

2

j

1

L i

2

j

2

j

1

i

j

1

i

ji

1i

j

1

j

Li

j

1

i

2i

1

j

1

jLi