TMHL63, ht2, 2013 - Linköping University · 2013. 10. 31. · TMHL63, ht 2, 2013 Lecture 7; Introd. 2-dim. elastostatics; cont. (modified 2013-10-31) 3 Review- the CST-element x

TMHL63, ht 2, 2013

Lecture 7; Introd. 2-dim. elastostatics; cont.

(modified 2013-10-31) 1

TMHL63, ht2, 2013

Lecture 7;

Introductory 2-dimensional elastostatics; cont.

TMHL63, ht 2, 2013


(modified 2013-10-31) 2

Introductory 2-dimensional elastostatics; cont.

We will now continue our study of 2-dim. elastostatics, and focus on a

somewhat more advanced element then the CST-element, namely the

4-noded bilinear (and isoparametric) element.

x

y

en1

en3

en2eu2

ev1

eu1

ev2

eu4

ev4

),( 11 ee yx

),( 22 ee yx

),( 33 ee yx

),( 44 ee yxeu3

ev3en4

However, let us first start

with a quick review of the

CST-element!

TMHL63, ht 2, 2013


(modified 2013-10-31) 3

Review- the CST-element

x

y

en1

en3

en2eu2

ev1

eu1

ev2

eu3

ev3

We do not use any

elementwise/local

coordinate system;

only the global

one!

Counter-

clockwise

numbering of

the nodes ),( 11 ee yx

),( 22 ee yx

),( 33 ee yx

TMHL63, ht 2, 2013


(modified 2013-10-31) 4

The CST-element; cont.

eu

ef

ed

eN

D

F

eC

e

e

eE

eBen1

en3

en2

eu2

ev1

eu1

ev2

eu3

ek K

eT

e VB e

T

eC

On the previous lecture we found the

following transformation diagram

ev3

TMHL63, ht 2, 2013


(modified 2013-10-31) 5


eu

ef

ed

eN

D

F

eC

e

e

eE

eB

ek K

eT

e VB e

T

eC

xy

y

x

xy

y

xE

2/)1(00

01

01

1 2

en1

en3

en2

eu2

ev1

eu1

ev2

eu3

ev3

TMHL63, ht 2, 2013


(modified 2013-10-31) 6


eu

ef

ed

eN

D

F

eC

e

e

eE

eB

ek K

eT

e VB e

T

eC

e

e

xye

ye

xe

v

u

xy

y

x

0

0

en1

en3

en2

eu2

ev1

eu1

ev2

eu3

ev3

TMHL63, ht 2, 2013


(modified 2013-10-31) 7


eu

ef

ed

eN

D

F

eC

e

e

eE

eB

ek K

eT

e VB e

T

eC

e

e

e

e

e

e

eee

eee

e

e

v

u

v

u

v

u

NNN

NNN

v

u

3

3

2

2

1

1

321

321

000

000

The rest of the

diagram as usual!

Now, what do we

get for our 4-

noded element?

en1

en3

en2

eu2

ev1

eu1

ev2

eu3

ev3

TMHL63, ht 2, 2013


(modified 2013-10-31) 8

The bilinear 4-noded element; cont.

eu

ef

ed

eN

D

F

eC

e

e

eE

eB

ek K

tdABT

e _ e

T

eC

en1

en3

en2

eu2

ev1

eu1

ev2

eu4

ev4

eu3

ev3

en4

As for CST-element!

TMHL63, ht 2, 2013


(modified 2013-10-31) 9

eu

ef

ed

eN

D

F

eC

e

e

eE

eB

ek K

tdABT

e _ e

T

eC

en1

en3

en2

eu2

ev1

eu1

ev2

eu4

ev4

eu3

ev3

en4

Since the stress/strain fields will not be constant in a 4-noded

element, the strain energy must be written as an integral.

From this it follows (by MPE) that we get an integral in the

transformation diagram!


TMHL63, ht 2, 2013


(modified 2013-10-31) 10

eu

ef

ed

eN

D

F

eC

e

e

eE

eB

ek K

e

T

eC

e

e

e

e

e

e

e

e

eeee

eeee

e

e

v

u

v

u

v

u

v

u

NNNN

NNNN

v

u

4

4

3

3

2

2

1

1

4321

4321

000

0000

en1

en3

en2

eu2

ev1

eu1

ev2

eu4

ev4

eu3

ev3

en4

Not easy, as

in the CST-

case, to find

the shape

functions N1e

to N4e!

dAtBT

e _


TMHL63, ht 2, 2013


(modified 2013-10-31) 11

eu

ef

ed

eN

D

F

eC

e

e

eE

eB

ek K

e

T

eC

en1

en3

en2

eu2

ev1

eu1

ev2

eu4

ev4

eu3

ev3

en4

The basic method to set up the shape functions for a 4-noded element is

to introduce an additional elementwise/local (so called natural)

coordinate system. However, before that, we will look a little bit at the

lower left part of the transformation diagram!

tdABT

e _


TMHL63, ht 2, 2013


(modified 2013-10-31) 12

eu ed

eN e

eB

More specifically, let us split the matrix [∂ε] in two parts

(which is not against the law :)

en1

en3

en2

eu2

ev1

eu1

ev2

eu4

ev4

eu3

ev3

en4


TMHL63, ht 2, 2013


(modified 2013-10-31) 13

eu ed

eN e

eB

We have

e

e

e

e

e

e

e

e

xye

ye

xe

v

u

y

x

y

x

y

vx

v

y

ux

u

v

u

xy

y

x

0

0

0

0

0110

1000

0001

0110

1000

0001

0

0

IB

eu

i.e.

eIeIee uBuBu


TMHL63, ht 2, 2013


(modified 2013-10-31) 14

eu ed

eN

e

eB

The lower left part of the transformation diagram may thus be

rewritten as shown below.

And why is that fine, one may ask?

Well, as we soon will see, it opens the way to make an

advantage of an additionally introduced natural coordinate

system!

IB eu

Previously we had [∂ε] here!

en1

en3

en2

eu2

ev1

eu1

ev2

eu4

ev4

eu3

ev3

en4


TMHL63, ht 2, 2013


(modified 2013-10-31) 15

eu ed eN

e

eB

Let us now introduce the natural

coordinate system according to the

illustration, where the symbol n in

the transformation diagram

indicates that an entity/quantity

depends on the natural

coordinates.

IB eu

en1

en3

en2

eu2

ev1

eu1

ev2

eu4

ev4

eu3

ev3

en4

x

y

11

1

1

en1 en2

en3en4

enu

enu n nN

enT

eu3

ev3

eu2

ev2ev1

eu1

eu4ev4


TMHL63, ht 2, 2013


(modified 2013-10-31) 16

eu ed eN

e

eB

Bilinear shape functions are easy to set up

in the natural coordinates!

IB eu

en1

en3

en2

eu2

ev1

eu1

ev2

eu4

ev4

eu3

ev3

en4

x

y

1

1

1

1

en1 en2

en3en4

enu

enu n nN

enT

1 11

4

11N

N1


TMHL63, ht 2, 2013


(modified 2013-10-31) 17

eu ed eN

e

eB

IB eu

en1

en3

en2

eu2

ev1

eu1

ev2

eu4

ev4

eu3

ev3

en4

x

y

1

1

1

1

en1 en2

en3en4

enu

enu n nN

enT

1 11

4

12N

N2


TMHL63, ht 2, 2013


(modified 2013-10-31) 18

eu ed eN

e

eB

IB eu

en1

en3

en2

eu2

ev1

eu1

ev2

eu4

ev4

eu3

ev3

en4

x

y

1

1

1

1

en1 en2

en3en4

enu

enu n nN

enT

1

114

13N

N3


TMHL63, ht 2, 2013


(modified 2013-10-31) 19

eu ed eN

e

eB

IB eu

en1

en3

en2

eu2

ev1

eu1

ev2

eu4

ev4

eu3

ev3

en4

x

y

1

1

1

1

en1 en2

en3en4

enu

enu n nN

enT

1

114

14N

N4


TMHL63, ht 2, 2013


(modified 2013-10-31) 20

en1

en3

en2

eu2

ev1

eu1

ev2

eu4

ev4

eu3

ev3

en4

x

y

1

1

1

1

en1 en2

en3en4

1 114

11N

N1

As can be seen, we will for simplicity

skip the index n when writing the

natural shape functions!

However, there is no risk of confusing

things, since we only will set up the

shape functions in the natural

coordinates.


TMHL63, ht 2, 2013


(modified 2013-10-31) 21

114

11N

114

12N

114

13N

114

14N

4

1)0,0(1 N

4

1)0,0(2 N

4

1)0,0(3 N

4

1)0,0(4 N

e

e

e

e

e

e

e

e

eeee

eeee

e

e

v

u

v

u

v

u

v

u

NNNN

NNNN

v

u

4

4

3

3

2

2

1

1

4321

4321

000

0000

eeeee

eeeee

vvvvv

uuuuu

4321

4321

4

1)0,0(

4

1)0,0(

We note


TMHL63, ht 2, 2013


(modified 2013-10-31) 22

eu ed eN

e

eB

IB eu

11

1

1

en1 en2

en3en4 enu

enu n nN

enT

eu3

ev3

eu2

ev2ev1

eu1

eu4ev4

e

e

e

e

e

e

v

u

v

v

u

u

0

0

0

0

Please observe! These displacement

components depend on the natural

coordinates!


TMHL63, ht 2, 2013


(modified 2013-10-31) 23

eu ed eN

e

eB

IB eu

enu

enu n nN

enT

y

vx

v

y

ux

u

yx

yx

yx

yx

v

v

u

u

e

e

e

e

e

e

e

e

00

00

00

00

The sub-matrix found in the

transformation matrix [Tn]e is

called the Jacobean matrix of

the coordinate change, and

will be denoted [J]!

The chain rule implies

[Tn]e The bilinear 4-noded element; cont.

TMHL63, ht 2, 2013


(modified 2013-10-31) 24

eu

ed eN

e

eB

IB eu

enu

enu n nN

enT

Thus, the strain-displacement matrix [B]e is given by the following expression

nnenIe NTBB 1

Since we already are in full control of [BI], [∂n]

and [Nn], it just remains to find the inverse of

the transformation matrix [Tn]e! We have

J

JT

en0

0

1

11

0

0

J

JT

en

1121

1222

21122211

1 1

JJ

JJ

JJJJJ

see Eqs. 8.72 - 8.74 in the book


TMHL63, ht 2, 2013


(modified 2013-10-31) 25

eu ed eN

e

eB

IB eu

enu

enu n nN

enT

Thus, we must fix [J]!

nnenIe NTBB 1

J

JT

en0

0

If we choose to interpolate the coordinates

in the same way as the displacements

(so called isoparametry), we have

node

e

x

e

e

e

en

x

e

e

y

y

x

Ny

x

4

1

1

:

and can find [J] by differentiating

the expression for {x}e in an

appropriate way- ALL DONE!


TMHL63, ht 2, 2013


(modified 2013-10-31) 26

In order to calculate the element stiffness, we must take care of the integral

tdABEBk ee

T

ee

This integration (in the natural coordinates) is done numerically by so

called Gauss-quadrature (see the book), which basically means that we

evaluate the integrand at a number of points, and multiply with an

associated area. Depending on element type, a certain number of

evaluation points are needed, which for our 4-noded element i 4.

It must finally be noted that also the above element sometimes behaves

poorly, and that a higher order element then must be used.


TMHL63, ht2, 2013 - Linköping University · 2013. 10. 31. · TMHL63, ht 2, 2013 Lecture 7; Introd. 2-dim. elastostatics; cont. (modified 2013-10-31) 3 Review- the CST-element x

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