Slide Chpt05

7/30/2019 Slide Chpt05

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1

Finite Element Method

FEM FOR BEAMS

for readers of all backgrounds

G. R. Liu and S. S. Quek

CHAPTER 5:


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Finite Element Method by G. R. Liu and S. S. Quek2

CONTENTS

INTRODUCTION

FEM EQUATIONS

Shape functions construction

Strain matrix

Element matrices

Remarks

EXAMPLE Remarks


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INTRODUCTION

The element developed is often known as abeam element.

A beam element is a straight bar of anarbitrary cross-section.

Beams are subjected to transverse forces

and moments. Deform only in the directions perpendicular

to its axis of the beam.


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INTRODUCTION

In beam structures, the beams are joined

together by welding (not by pins or hinges).

Uniform cross-section is assumed.

FE matrices for beams with varying cross-

sectional area can also be developed

without difficulty.


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FEM EQUATIONS


Strain matrix

Element matrices


6/26



Consider a beam element

d1= v1

d2 = 1

d3= v2

d4= 2

= - a

= 1

= a

= 1

,

2a

0

Natural coordinate system:

a

x


7/26Finite Element Method by G. R. Liu and S. S. Quek7


3

3

2

210)( vAssume that

In matrix form:

3

2

1

0

321)(

v or p )()( Tv

)32(

11 2321

a

v

ax

v

x

v




1

1

1

(1) ( 1)

d(2)d

v v

vx

To obtain constant coefficientsfour conditions

2

2

1

(3) (1)

d(4)

d

v v

vx

Atx= a or = 1

Atx= a or = 1




4

3

2

1

321

321

2

2

1

1

0

1111

0

1111

aaa

aaa

v

v

or Ad ee

aa

aa

aa

aa

e

11

00

33

22

4

11A or

ee dA1




Therefore,

ev dN )(

where

)()()()()(4321

1 NNNNe

PAN

)1()(

)32()(

)1()(

)32()(

32

44

3

41

3

3242

3

41

1

a

a

N

N

N

Nin which



Strain matrix

yLvx

vy

x

uxx

2

2

Therefore,

NNNNB

22

2

22

2

ay

ay

xyyL

4321

NNNN Nwhere

)31(

2

,

2

3

)31(2

,2

3

43

21

aNN

a

NN(Second derivative of

shape functions)



Element matrices

dd][][1

d)()(dd

1

132

2

2

2

4

1

1

2

2

2

2

2

NNNN

NNBcBk

TzT

z

Ta

aA

T

V

e

a

EIa

aEI

xxx

AyEV

x

NNNNNNNN

NNNNNNNN

NNNNNNNN

NNNNNNNN

a

EIze d

41342414

41332313

42322212

41312111

1

13

k

2

22

3

4.

33

234

3333

2

asy

a

aaa

aa

a

EIzek

Evaluate

integrals



Element matrices

x

NNNNNNNN

NNNNNNNN

NNNNNNNN

NNNNNNNN

Aa

aAxAVTT

a

aA

T

V

e

d

dddd

44342414

43332313

42322212

41312111

1

1

1

1

NNNNNNm

2

22

8.

2278

6138

13272278

105

asy

a

aaa

aa

Aae

m

Evaluate

integrals



Element matrices

13

2

13

1

1

2

1

1

1

1

4

3

2

1

2

2

ddd

s

af

sy

s

af

sy

s

s

s

s

yf

S

s

T

V

b

T

e

m

faf

m

faf

m

f

m

f

N

N

N

N

afSfVf

y

y

f

NNf

eeeee fdmdk



Remarks

Theoretically, coordinate transformation can also be used

to transform the beam element matrices from the local

coordinate system to the global coordinate system.

The transformation is necessary only if there is more than

one beam element in the beam structure, and of which

there are at least two beam elements of different

orientations.

A beam structure with at least two beam elements of

different orientations is termed aframe orframework.



EXAMPLE

Consider the cantilever beam as shown in the figure. The beam is fixed

at one end and it has a uniform cross-sectional area as shown. The beam

undergoes static deflection by a downward load ofP=1000N applied at

the free end. The dimensions and properties of the beam are shown inthe figure.

P=1000 N

0.5 m

0.06 m

0.1 m

E=69 GPa =0.33


17/26Finite Element Method by G. R. Liu and S. S. Quek

17

EXAMPLE

Step 1: Element matrices

33 6 41 1

0.1 0.06 1.8 10 m12 12

zI bh

9 63

6 -2

3 0.75 3 0.75

69 10 1.8 10 0.75 0.25 0.75 0.125

3 0.75 3 0.752 0.25

0.75 0.125 0.75 0.25

3 0.75 3 0.75

0.75 0.25 0.75 0.1253,974 10 Nm3 0.75 3 0.75

0.75 0.125 0.75 0.25

e

K k



18

EXAMPLE

Step 1 (Contd):

1 1

1 16

2 2

2 2

?3 0.75 3 0.75 unknown reaction shear force

?0.75 0.25 0.75 0.125 unknow3,974 10 3 0.75 3 0.75

00.75 0.125 0.75 0.25

v Q

M

v Q P

M

DK F

n reaction moment

Step 2: Boundary conditions 1 1 0v

1 1

1 16

2 2

2 2

03 0.75 3 0.75

00.75 0.25 0.75 0.1253.974 10

3 0.75 3 0.75

00.75 0.125 0.75 0.25

v Q

M

v Q P

M



19

EXAMPLE Step 2 (Contd):

6 -23 0.75

3.974 10 Nm0.75 0.25

K

Therefore, Kd=F where

dT= [ v2 2] ,1000

0

F



20

EXAMPLE

Step 3: Solving FE equation

Two simultaneous equations

v2

= -3.355 x 10-4 m

2 = -1.007 x 10-3 rad

6

1 2 2

6 4 3

3.974 10 ( 3 0.75 )

3.974 10 [ 3 ( 3.355 10 ) 0.75 ( 1.007 10 )]

998.47N

Q v

6

1 2 2

6 4 3

3.974 10 ( 0.75 0.125 )

3.974 10 [ 0.75 ( 3.355 10 ) 0.125 ( 1.007 10 )]

499.73Nm

M v

Substitute

back intofirst two

equations of

Kd=F



21

Remarks

FE solution is the same as analytical solution

Analytical solution to beam is third order

polynomial (same as shape functions used) Reproduction property



22

CASE STUDY

Resonant frequencies of micro resonant transducer

Membrane

Bridge


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CASE STUDYNumber of 2-

node beam

elements

Natural Frequency (Hz)

Mode 1 Mode 2 Mode 3

10 4.4058 x 105

1.2148 x 106

2.3832 x 106

20 4.4057 x 105 1.2145 x 106 2.3809 x 106

40 4.4056 x 105 1.2144 x 106 2.3808 x 106

60 4.4056 x 105 1.2144 x 106 2.3808 x 106

Analytical

Calculations4.4051 x 105 1.2143 x 106 2.3805 x 106


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CASE STUDY

Mode 1 (0.44 MHz)

0

0.2

0.4

0.6

0.8

1

1.2

0 20 40 60 80 100

x (um)

Dy(um)


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CASE STUDY

Mode 2 (1.21MHz)

-1.5

-1

-0.5

0

0.5

1

1.5

0 20 40 60 80 100

x (um)

Dy(um)


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26

CASE STUDY

Mode 3 (2.38 MHz)

-1.5

-1

-0.5

0

0.5

1

1.5

0 20 40 60 80 100

x (um)

Dy(um)

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