1 Finite Element Method FEM FOR FRAMES for readers of all backgrounds G. R. Liu and S. S. Quek...

1

FFinite Element Methodinite Element Method

FEM FOR FRAMES

for readers of all backgroundsfor readers of all backgrounds

G. R. Liu and S. S. Quek

CHAPTER 6:

2Finite Element Method by G. R. Liu and S. S. Quek

CONTENTSCONTENTS INTRODUCTION FEM EQUATIONS FOR PLANAR FRAMES

– Equations in local coordinate system– Equations in global coordinate system

FEM EQUATIONS FOR SPATIAL FRAMES– Equations in local coordinate system– Equations in global coordinate system

REMARKS


INTRODUCTIONINTRODUCTION

Deform axially and transversely. It is capable of carrying both axial and transverse

forces, as well as moments. Hence combination of truss and beam elements. Frame elements are applicable for the analysis of

skeletal type systems of both planar frames (2D frames) and space frames (3D frames).

Known generally as the beam element or general beam element in most commercial software.


FEM EQUATIONS FOR PLANAR FEM EQUATIONS FOR PLANAR FRAMESFRAMES

Consider a planar frame element

Y, V

X, U

node 1 (u1, v1, z1)

x, , u y, v

z

z

l=2a

node 2 (u2, v2, z2)

1 1

2 1

3 1

4 2

5 2

6 2

diplacement components at node 1

diplacement components at node 2

ze

z

d u

d v

d

d u

d v

d

d


Equations in local coordinate systemEquations in local coordinate system

Combination of the element matrices of truss and beam elements

2

2

2

1

1

1

z

ze

v

u

v

u

d

Truss

Beam

From the truss element,

1 1 4 2

1 12 2

4 22

0 0 0 0

0 0 0 0 0

0 0 0 0

0 0

. 0 0

0

AE AEa a

trusse AE

a

d u d u

d u

d u

sy

k

(Expand to 6x6)



From the beam element,

3 2 3 2

2

3 2

3 1 5 2 6 22 1

3 3 3 3

2 2 2 2 2 12 3

2 3 1

3 35 22 2

26 2

( ) ( ) ( )( )

0 0 0 0 0 0

0

0

0 0 0

.

z z z z

z z z

z z

z

z z

EI EI EI EI

a a a aEI EI EI

a a a zbeame

EI EI

a aEI

za

d d v dd v

d v

d

d vsy

d

k

(Expand to 6x6)



0

00.

00

0000

00000

0000

2

22

sya

AE

aAE

aAE

trussek

3 2 3 2

2

3 2

3 3 3 3

2 2 2 22 3

2

3 3

2 22

0 0 0 0 0 0

0

0

0 0 0

.

z z z z

z z z

z z

z

EI EI EI EI

a a a aEI EI EI

a a abeame

EI EI

a aEI

a

sy

k+

a

EIa

EI

a

EIa

AEa

EI

a

EI

a

EIa

EI

a

EI

a

EI

a

EIa

AEa

AE

e

z

zz

zzz

zzzz

sy2

2

3

2

32

2

322

3

2

3

2

3

2

322

23

2

2323

.

00

0

0

0000

k



Similarly so for the mass matrix and we get

2

22

8

2278.

0070

61308

132702278

00350070

105

a

asy

aaa

aa

Aae

m

And for the force vector,

13

2

2

13

1

1

2

2

s

af

syy

sxx

s

af

syy

sxx

e

m

faf

faf

m

faf

faf

y

y

f


Equations in global coordinate systemEquations in global coordinate system

Coordinate transformation

ee TDd

where

j

j

j

i

i

i

e

D

D

D

D

D

D

3

13

23

3

13

23

D ,

100000

0000

0000

000100

0000

0000

yy

xx

yy

xx

ml

ml

ml

ml

T

D3i -2

D3i-1

D3j -2

D2j

2a

x

u1

u2

fs1

global node j local node 2

global node i local node 1

0

X

Y

o

x

y

v1

v2

D3j

D3j - 1

z

2

z1

D3i



D3i -2

D3i-1

D3j -2

D2j

2a

x

u1

u2

fs1

global node j local node 2

global node i local node 1

0

X

Y

o

x

y

v1

v2

D3j

D3j - 1

z

2

z1

D3i

cos( , ) cos

cos( , ) sin

j ix

e

j ix

e

X Xl x X

l

Y Ym x Y

l

cos( , ) cos(90 ) sin

cos( , ) cos

j iy

e

j iy

e

Y Yl y X

l

X Xm y Y

l

2 2( ) ( )e j i j il X X Y Y

Direction cosines in T:

(Length of element)



Therefore,

TkTK eT

e

TmTM eT

e

eT

e fTF


FEM EQUATIONS FOR FEM EQUATIONS FOR SPATIAL FRAMESSPATIAL FRAMES

Consider a spatial frame element

v1

u1 w1

x

y

z

1

2

z2

w2

x2

y2

z1

x1

y1

u2

v2

1 1

2 1

3 1

4 1

5 1

6 1

27

28

29

210

211

212

x

y

ze

x

y

z

d u

d v

d w

d

d

d

ud

vd

wd

d

d

d

d

Displacement components at node 1

Displacement components at node 2


Equations in local coordinate systemEquations in local coordinate system Combination of the element matrices of truss and

beam elements

3 2 3 2

3 2 3 2

1 21 21 1 1 1 2 2 2 2

2 23 3 3 3

2 2 2 2

3 3 3 3

2 2 2 2

2

0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0

0 0 0 0 0 0

0 0 0 0 0

z z z z

y y y y

y yx xz z

AE AEa a

EI EI EI EI

a a a a

EI EI EI EI

a a a a

GJa

e

u v w u v w

k

2

2

3 2

3 2

2

2 3

22 3

2

23 3

2 2

3 3

2 2

2

2

2

.

0 0

0 0 0 0 0

0 0 0 0

0 0 0 0 0

0 0 0

0 0

0 0

0

y y y

z z z

z z

y y

y

z

GJa

EI EI EI

a a aEI EI EI

a a a

AEa

EI EI

a a

EI EI

a a

GJa

EI

aEI

a

sy

v1

u1 w1

x

y

z

1

2

z2

w2

x2

y2

z1

x1

y1

u2

v2



2

2

2

22

22

22

8080070.0220782200078000007060001308060130008003500000700130270002207813000270220007800000350000070

105

aa

rsya

a

aaaaaa

rraa

aa

Aa

x

xx

e

m

whereA

Ir x

x 2



Y

X

Z

y x

z

D6i-1

D6i-2

D6i-3

D6i-4

D6i-5

D6j-2

D6j-1

D6j-3

D6j-4

D6j

D6i

d6

d5

d4

d3

d2 d1

d12

d11

d10

d9

d8

d7

D6j-5

y x

z

1

2 3



Coordinate transformation

ee TDd where

j

j

j

j

j

j

i

i

i

i

i

i

e

D

D

D

D

D

D

D

D

D

D

D

D

6

16

26

36

46

56

6

16

26

36

46

56

D ,

3

3

3

3

T000

0T00

00T0

000T

T

zzz

yyy

xxx

nml

nml

nml

3T



cos( , ), cos( , ), cos( , )

cos( , ), cos( , ), cos( , )

cos( , ), cos( , ), cos( , )

x x x

y y y

z z z

l x X m x Y n x Z

l y X m y Y n y Z

l z X m z Y n z Z

Direction cosines in T3


Equations in global coordinate systemEquations in global coordinate system Vectors for defining location and orientation of

frame element in space

Y

X

Z

x

x

1

2 3

12 VV

13 VV

y

y

z

y

)()( 1312 VVVV

2V

1

1V 1V

3V

1

1V

ZZYYXXV

1111

ZZYYXXV

2222

ZZYYXXV

3333

lkkl

lkkl

lkkl

ZZZ

YYY

XXX

k, l = 1, 2, 3

221

221

221122 ZYXVVal

ZZYYXXVV

21212112

ZZYYXXVV

31313113



frame element in space (cont’d)

Y

X

Z

x

x

1

2 3

12 VV

13 VV

y

y

z

y

)()( 1312 VVVV

2V

1

1V 1V

3V

1

1V

a

ZZxZxn

a

YYxYxm

a

XXxXxl

x

x

x

2),cos(

2),cos(

2),cos(

21

21

21

)()(

)()(

1312

1312

VVVV

VVVVz

Za

ZY

a

YX

a

X

VV

VVx

222

)( 212121

12

12

})()(){(2

1213131212131312121313121

123

ZYXYXYXZXZXZYZYA

z

221313121

221313121

221313121123 )()()( YXYXXZXZZYZYA



frame element in space (cont’d)

Y

X

Z

x

x

1

2 3

12 VV

13 VV

y

y

z

y

)()( 1312 VVVV

2V

1

1V 1V

3V

1

1V )(

2

1

)(2

1

)(2

1

21313121123

21313121123

21313121123

YXYXA

Zzn

XZXZA

Yzm

ZYZYA

Xzl

z

z

z

xzy

xzxzy

xzxzy

xzxzy

lmmln

nllnm

mnnml



Therefore,

TkTK eT

e

TmTM eT

e

eT

e fTF


REMARKSREMARKS

In practical structures, it is very rare to have beam structure subjected only to transversal loading.

Most skeletal structures are either trusses or frames that carry both axial and transversal loads.

A beam element is actually a very special case of a frame element.

The frame element is often conveniently called the beam element.


CASE STUDYCASE STUDY

Finite element analysis of bicycle frame


CASE STUDYCASE STUDYYoung’s modulus,

E GPaPoisson’s ratio,

69.0 0.33

74 elements (71 nodes)

Ensure connectivity



Constraints in all directions

Horizontal load



M = 20X



-9.68 x 105 Pa

-1.214 x 106 Pa

-6.34 x 105 Pa

-6.657 x 105 Pa

9.354 x 105 Pa

-5.665 x 105 Pa

-6.264 x 105 Pa

Axial stress

1 Finite Element Method FEM FOR FRAMES for readers of all backgrounds G. R. Liu and S. S. Quek...

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