Finite Element Finite Element Methodology Methodology Planar Line Element Planar Line Element i j k 1 2 3
Apr 01, 2015
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
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’