3.MASAT_2012-2013_Timoshenko_2D

8/11/2019 3.MASAT_2012-2013_Timoshenko_2D

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Fundaments of Finite Element Anal sis

U n i v e r s i t y o f M i n h o

Department of Civil EngineeringGuimares, Portugal

[email protected] www.civil.uminho.pt


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2

TIMOSHENKO ELEMENTS FOR 2D FRAMES

Introduction

The2D frames are structures formed by bars that can be modelled by beamelements with axial deformability. The Euler-Bernoulli and the Timoshenko theories

are currently used.

In the Euler-Bernoulli theory it is assumed that the cross sections, orthogonal

Joaquim Barros

Modeling and Advanced Structural Analysis Techniques

University of Minho Department of Civil Engineering

to the longitudinal axis of the beam before deformation, remain plain andorthogonal to this axis after deformation.

In the Timoshenko theory it is assumed that the cross sections, orthogonal to

the longitudinal axis of the beam before deformation, remain plain but not

necessarily orthogonal to this axis after deformation => the shear deformation istaken into account.


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3

l1

2l , u l 2

1ll , u1

1ldu

1dl

Normal superfcie mdia

aps a deformao

ll , u2

Joaquim Barros



Inclinao do eixo da viga

1l

1l

Inclinao do eixo da viga

Normal superfcie mdia

aps a deformao

1l , u

Deformao admitida

Deformao real


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FORMULATION FOR THE TIMOSHENKO 2D FE

Displacement field

0),,(

)0,0,(),,()0,0,()0,0,(),,(

321

321321

3212321321

3

22

311

=

========

u

uuuu

l , u2l2

Joaquim Barros



l , u1 l11lu(l , )

3l3

l3

3l

l 3l2

2l


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5

Strain field

+

=

=

21

1

12

1

21

1

ddu

ddu

d

du

1 d

ud

1

ud

Joaquim Barros



+

=

+

=

0

01

2

1

1

3

3

2

f

c

a

d

d

udd

1da

221

23

=== ffd

d

3

2

1

=

dud

c

1

2

33

d

ud=


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6

1

2

33

d

ud=

l , ulu

l , u2l2

(l , )3l3

l

Joaquim Barros



l3

3l

2ldu

dl1

1

2l

dl

du

Deformao admitida

Deformao real


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7

Stress field

[ ]Tcaf =

1=af

21 =c

a f

= +

fa

lN 1 Ml3

l2

1l

3(l )

Joaquim Barros



(l )3

l2

l1

l l1 2 l l1 2

Distribuio

aproximada

Distribuio

real


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8

Stress-strain relationships

=

=

c

af

c

af

G

E

0

0

=

=

21

1

21

1

0

0

G

E

Joaquim Barros



=

=

=

c

f

a

c

f

a

c

f

a

G

E

E

G

E

E

00

00

00'

=

==

=

3

2

1

1

22

1

d

udG

EE

d

udE

c

ff

a


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9

Resultant stresses

dbd

V

M

N

ha

b

h

h

=

=

2/

2

2/

2/2

'

21

1

1

2

3

1

=

a

h

h

b

h

h

c

f

Edbd

dbd

G

E

Ea

2/

2/2

2

2/

2/2

Joaquim Barros



b h

c

f

22/

2

( )

=

=

c

f

a

c

b

h

h

f

b

h

h

Gbh

Ehb

Ehb

Gdbd

Edbd

*

3

2/

2/2

22/

2/

22

12


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10

Resultant stresses

=

=

c

f

a

AG

EI

EA

V

M

N

*2

3

2

3

1

( ) AhbA == **2

(l )3

l2

l1 3l

M1

Nl Vl2 2lV

Distribuio

aproximada

Distribuio

real

Momento

flector

Esforo

axial

Joaquim Barros



C=b/a

=6/(7+20K )

K=C/(1+C )

2a

=5/6 =6/7

l32b ht

2

2 =0.69 =0.32

t

l2l3

l2Esforo

de corte


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11

Internal work

[ ]

++=

+

+

=

=

)(

)(

)(

0

0

)(

int

e

e

e

c

fa

V

T

c

T

f

T

a

V c

fa

T

c

fa

V

Te

dVG

EE

dVG

E

dVW

Joaquim Barros



( ) ++++= )(eVc

T

cf

T

fa

T

ff

T

aa

T

a dVGEEEE

( ) ++=)(

23 1

*)(

inteL

c

T

cf

T

fa

T

a

edAGIEAEW

=

=

=

=

)( )(

13

)( )(

31

0

0

1

2/

2/2

11

2

1

2/

2/2

11

2

e e

e e

V L

h

haf

V L

h

hf

T

a

ddbd

udE

d

ddVE

ddbd

dEd

uddVE


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12

=)()(

11

11

11 ee

L

a

T

a

L

dAEdd

udEA

d

ud

Internal work due to axial deformation:

Internal work due to bending:

Joaquim Barros



=

)( )(

3

3

3

3

11

11e eL L

f

T

f dIEdd

dEId

d

=

)(

2

)(

3

2

23

2

1

*

1

1

*

1 ee L

c

T

c

L

dAGdd

udAG

d

ud

Internal work due to shear deformation :


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13

Finite element of 2 nodes

Displacements

333

222

111

2121111

2121111

2121111

)()()(

)()()(

)()()(

sNsNsusNusNsu

usNusNsu

+=+=

+=

l , u2 2

( ) )(

32

22

12

31

21

11

21

21

21

1

0000

0000

0000

)(

3

2

1

eeUN

u

u

u

u

NN

NN

NN

su

u

=

=

Joaquim Barros



2 l 3

1 l 3

u 2 l 221 lu

ji

3( , u )

3 1l , u1

21

1 l1u

12 lu

21

21

N (s ) = 1/2(1-s )1 1 1

1 12N (s ) = 1/2(1+s )


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B matrix

1

11

1

1

1

11

2

ds

ud

L

ds

ud

d

ds

d

uda

=

==

( ) 31

21

11

211 00

200

2a

u

u

dNdNs

=

Axial

Joaquim Barros



( ) )(

32

22

1211

ee

a UB

u

uss

=

( )

=

=

001

001

002

002

1

2

1

1

LL

ds

dN

Lds

dN

LB

e

a


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15

1

11

1

1

3

33

2

ds

d

L

ds

d

d

ds

d

df

=

===

( ) 31

21

11

211

200

200f

u

uu

ds

dN

Lds

dN

Ls

=

Bending

Joaquim Barros



( ) )(

32

22

ee

f UB

u

=

( )

=

=

LL

dsdN

LdsdN

LB ef

100

100

2002001

2

1

1


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Shear2

3

2

3

2

3

1

1

1 1

1

2

c

du

d

duds

d ds

du

L ds

=

=

=

( )12

31

21

11

2

1

21

1

11

20

20c

u

u

u

Nds

dN

LN

ds

dN

Ls

=

Joaquim Barros



( ) )(

32

22

ee

c UB=

( )

( ) ( )

+=

=

11

2

1

21

1

1

12

1101

2

110

2020

sL

sL

Nds

dNL

NdsdN

LB ec


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Stiffness matrix

[ ] ( )[ ] ( ) ( )[ ] ( ) ( )[ ] ( ) ++=)(

23 1

)(*)()(

inte

L

ee

c

Te

c

e

f

Te

f

e

a

Te

a

TeedUBAGBBIEBBAEBUW

( )[ ] ( )=)(

1

)(

eL

e

a

Te

a

e

a dBAEBk

Joaquim Barros



( ) ( )

3

( )

( )

1e

Te e e

f ff

L

k B E I B d =

( )

[ ]

( )

=

)(2 1

*)(

eL

e

c

Te

c

e

c

dBAGBk

)()()()( e

c

e

f

eekkkk a ++=


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18

=

000000

000000

001001

000000

000000

001001

)(

)(

e

e

aL

EAk

=

100100

000000000000

100100

000000

000000

)(

)( 3

e

e

L

EIk f

Joaquim Barros



=

320

620

210

210

000000

620

320

210

210

000000

22

22

)(*

)( 2

LLLL

LL

LLLL

LL

L

GAk

e

e

c


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19

++

= 620

320

2020

0000

****

****

)(

232232

2222

LGA

L

EIGALGA

L

EIGA

GA

L

GAGA

L

GAL

EA

L

EA

k e

Joaquim Barros



++

320620

20

20

0000

****

****

232232

2222

LGA

L

EIGALGA

L

EIGA

GA

L

GAGA

L

GALL


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( ) ( ) ( ) ( )

Te e e e

a f cB B B B =

=

=

*

2

3

00

0000

00

0000

GA

EIEA

D

DD

D

c

f

a

Joaquim Barros



( )[ ] ( )

( )

[ ]

( )

+

=

=

1

1 1

1

)(

2

)(

dsL

BDB

dBDBk

eTe

L

eTee

e


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21

Shear Locking

+

+

=

=

1

11

1

11

2

21111

dsL

EA

dsL

BAEBk aaa

+1 L

=> 1 PI

Joaquim Barros



+

=1

11

1

1

2

23

33333

dsL

EI

fff

+

+

=

=

1

11

2

1

*

1

11

*

)1(4

1

2

2

2

32333

dssAGL

dsLBAGBk ccc

=> 1 PI

=> 2 PI


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22

n. of integration points

n

Polynomial degree

2n-1

Normalized coordinates of the

integration points

si

Weights

Wi

1 1 0.0 2.0

2 3-1/3

1/3

1

1

3 5

-3/5

0

3/5

5/9

8/9

5/9

Joaquim Barros



4 7

-0.8611363116

-0.3399810436

0.3399810436

0.8611363116

0.3478548451

0.6521451549

0.6521451549

0.3478548451

5 9

-0.9061798459

-0.5384693101

0.0

0.5384693101

0.9061798459

0.2369268851

0.4786286705

0.568888889

0.4786286705

0.2369268851


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Example of shear locking effect

S

S

h

1

S-S

l1

L

E, A

l2

3

(l )

F

Joaquim Barros



(1)

31

u 21 l

1 l 3

u 22 l

2 l 3


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24

( ) ( )[ ] )1()1(11 QUkk cf =

=

++

6232

22

2

1

2

1

1

****

****

****

2

3

2

2222

232232

2222

F

RR

u

u

GAGAGAGA

LGA

L

EIGALGA

L

EIGA

GA

L

GAGA

L

GA

Joaquim Barros



++3262

****3

232232 LGAL

EIGALGAL

EIGA

=

+

0

32

2

3

2

232

22

2

2

**

**

Fu

LGA

L

EIGA

GA

L

GA


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QF

F

EI

L

EI

L

EI

L

EI

L

GA

L

u

=

+

+=

0

3

1

33

332

3

2

2

23

*

2

2

( )FLL

u

+

=2

3

*2

Joaquim Barros



32

2

23

3

=

=

L

h

h

L

=


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26

If the cantilever is modelled by one Euler-Bernoulli beam element (withoutconsidering the shear deformation) the compliance matrix of the structurewill be:

=

33

33

2

232

23

EI

L

EI

LEI

L

EI

L

F

( ) FEI

Lu

EBsc

exacto3

2

3

3

2

=

Joaquim Barros



Using the conventional theory of structures, based on the Euler-Bernoulli

theory, and considering the shear deformation, the compliance matrix hasthe following configuration:

+

=

33

332

2

232

23

*

EI

L

EI

L

EI

L

EI

L

GA

L

F

( ) FEI

L

GA

Lu

EBcc

exacto

+=

32

2 3

3

*2


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27

( )

( )( )34

343

3

3

1

22

2

3

3

*

2

2

3

32

2

2

+

+=

+

+==

EI

L

EIL

GAL

u

u

exacto

E

Joaquim Barros




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Adopting the reduced integration in a proper balance is obtained due to theexcessive stiffness introduced by the shear cofficients in the stiffness matrix

( )e

e

e

c LL

LLLL

LL

L

GAk

=

11

4242

2

1

2

1

22

)(*

)( 2

)()()( eeeQUk =

+

=

0

24332

3

2

2

23

*

2

2 F

LL

EI

L

EI

L

GA

L

u

Joaquim Barros



LLLL

4242

22

=

+

0

42

2

3

2

232

22

2

2

**

**

Fu

LGA

L

EIGA

GA

L

GA

233

EIEI

)()()( eeeQFU =

( ) FEIL

GA

L

FFu

+==32

2 4

3

*112

( ) 22

2

2

4

33

2

2

+==

EBsc

exactou

u


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Evolution of the error with the mesh refinement of the cantilever

Values of =/h

Number of elements 1 2 4 8 16

( )EBscexacto

Eu

u

2

2

2

2

= 0.750 0.938 0.984 0.996 0.996

Joaquim Barros




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Example

100kN/m

The frame 2D of the figure should be modelled by two Timoshenko finite elements of 2 nodes each. Usingthe selective integration scheme for the evaluation of the stiffness matrix and the full integration schemefor the resultant stresses, determine:

1. The stiffness matrix of the structure;2. The load vector of the structure;3. The displacement and reactions;4. The resultant stresses.

Data:

Joaquim Barros



km

450.6m

0.3m

0.005m

4m 6m

3m

EC = 30 GPa = 0.0Km = 500000 KN/m

3.MASAT_2012-2013_Timoshenko_2D

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