Finite Element Methodology Planar Line Element Planar Line Element v1 1 2 v2 21 v(x) x.

Finite Element MethodologyFinite Element MethodologyPlanar Line ElementPlanar Line Element

i

j

k1

2

3

1 2

y

Planar Line ElementPlanar Line Element

v1

1

2

v2

21

v(x)x

1 2

y

Planar Line ElementPlanar Line Element

qq

2

2

d v M

dx EI

BMD

SFDdM

Vdx

dVq

dx

Planar Line ElementPlanar Line Element

(1)

where q = intensity of lateral loadingand EI = flexural rigidity.

04

4

EI

q

dx

vd

Integrating, we have:Integrating, we have:

32

21

212

2

13

3

2axa

xa

dx

dvaxa

dx

vda

dx

vd

and 43

22

31

26axa

xaxav

or3

42

321 xxxv

Planar Line ElementPlanar Line Element

In matrix form, this becomes:In matrix form, this becomes:

4

3

2

1

321

a

a

a

a

xxxv

Ld ~~~

Planar Line ElementPlanar Line Element

We now turn our attention to the nodal deflections. We now turn our attention to the nodal deflections. The nodal rotations The nodal rotations can be found as follows. can be found as follows.

The slope at any point along the length of the The slope at any point along the length of the element is: element is:

2432 32 xx

dx

dv

20

1

xdx

dv

24322 32

xdx

dv

Planar Line ElementPlanar Line ElementWe can therefore construct a matrix relationship We can therefore construct a matrix relationship between the nodal deflections and the undetermined between the nodal deflections and the undetermined coefficients.coefficients.

4

3

2

1

2

32

2

2

1

1

3210

1

0010

0001

v

v

C ~~~

Planar Line ElementPlanar Line Element

2323

22

1

~

/1/2/1/2

/1/3/2/3

0010

0001

C

* This equation can also be interpreted as an

"interpolation formula", whereby intermediate values of displacements are determined from those at specific points (in this case, the nodes).

NLCLd 1

~~~~~~~~ *

Planar Line ElementPlanar Line Element

State of Strain:State of Strain:

The state of strain for the line element can The state of strain for the line element can

be represented by how “curved” it is, ie:be represented by how “curved” it is, ie:

2

2

dx

vd

4

3

2

1

432

2

620062

a

a

a

a

xxdx

vd

~~~ H

Planar Line ElementPlanar Line Element

~

1

~~~ CH

2

2

1

1

2323

22~

/1/2/1/2

/1/3/2/3

0010

0001

6200

v

v

x

2

2

1

1

232232~

62;

126;

64;

126

v

v

xxxx

Planar Line ElementPlanar Line Element

State of StressState of StressIn this problem we consider the state of stress In this problem we consider the state of stress to be defined by the bending moment at any to be defined by the bending moment at any section. Therefore we have for the line element:section. Therefore we have for the line element:

)(;~~~2

2

Ddx

vdEIM

~

1

~~~~ CHD

Planar Line ElementPlanar Line Element

v1*

v2*

V1

1

2

V2

21

Real Actions

Virtual Virtual displacementsdisplacements

Equilibrium Equation in Element Equilibrium Equation in Element Co-ordinatesCo-ordinates

virtual nodal displacements virtual nodal displacements associated virtual internal state of strainassociated virtual internal state of strain

*

~

1

~~

*

~ CH

Planar Line ElementPlanar Line ElementNow the internal virtual work on a length dx of the element is equal to Now the internal virtual work on a length dx of the element is equal to the product of the virtual strains and the real stresses the product of the virtual strains and the real stresses

dxCHDHC

dxCHDCHdx

TTT

TT

.

..

~

1

~~~~

1

~

*

~

~

1

~~~

*

~

1

~~~

*

~

We now integrate over the length of the element, noting that matrix We now integrate over the length of the element, noting that matrix H alone is dependent on x. Therefore, the total internal work is:H alone is dependent on x. Therefore, the total internal work is:

~

1

~~~~

1

~

*

~ CdxHDHC TTT

Equating internal and external work, we have:Equating internal and external work, we have:

~

1

~~~~

1

~

*

~~

*

~ CdxHDHCq TTTT

Planar Line ElementPlanar Line Element

Alernatively:Alernatively:

~~~

1

~~~~

1

~~ kCdxHDHCq TT

The element stiffness matrix is given by: The element stiffness matrix is given by:

1 1

~ ~ ~~ ~ ~

TTk C H DH dx C

2

~~~

361200

12400

0000

0000

6200

6

2

0

0

xx

xEIxEI

x

HDH TNow: Now:

Planar Line ElementPlanar Line Element

2

0~~~

361200

12400

0000

0000

EIdxHDH T

So that: So that:

6606

2020

0000

0000

~~~~1 EICdxHDH T

o

6606

2020

0000

0000

/1/100

/2/300

/1/210

/2/301

~~~~~2

32

2

32

11 EICdxHDHC TT

o

Planar Line ElementPlanar Line Element

Hence: Hence:

22

22

3

11

~

4626

612612

2646

612612

~~~~~

EICdxHDHCk TT

o

This stiffness matrix could have been constructed This stiffness matrix could have been constructed directly by considering the response of the line directly by considering the response of the line element to individual nodal displacements via the element to individual nodal displacements via the slope-deflection equations, slope-deflection equations, (see 421-307 notes ).(see 421-307 notes ).

Planar Line ElementPlanar Line Element

2

2

1

1

22

22

3

2

2

1

1

4626

612612

2646

612612

v

v

EI

M

q

M

q

y

y

Hence: Hence:

For conformability (matrix manipulative purposes) For conformability (matrix manipulative purposes) it is desirable to include the third planar force and it is desirable to include the third planar force and displacement displacement

2

2

2

1

1

1

22

22

3

2

2

2

1

1

1

460260

61206120

000000

260460

61206120

000000

v

u

v

u

EI

M

q

q

M

q

q

y

x

y

x

Equilibrium Equation in System Co-ordinates Equilibrium Equation in System Co-ordinates

Planar Line ElementPlanar Line Element

~~~

_

~TkTK T

NULL

NULL

T = ~ wherewhere

(Note: We would continue this formulation with an (Note: We would continue this formulation with an enhancement for axial effects – these have been enhancement for axial effects – these have been neglected in the present approach)neglected in the present approach)

AssembleAssemble sK '_

~~~~KQ

~~~q

Solve:Solve:

We’ve only just begun ………… ‘The Carpenters’We’ve only just begun ………… ‘The Carpenters’