4nodequad
TRANSCRIPT
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MANE 4240 & CIVL 4240
Introduct ion to Finite Elements
Four-noded
rectangular element
Prof. Suvranu De
Reading assignment:
Logan 10.2 + Lecture notes
Summary:
Computation of shape functions for 4-noded quad
Special case: rectangular element
Properties of shape functions
Computation of strain-displacement matrix
Example problem
Hint at how to generate shape functions of higher order
(Lagrange) elements
Finite element formulation for 2D:
Step 1: Divide the body into finite elements connected to each
other through special points (nodes)
x
y
Su
STu
v
x
px
py
Element e
3
2
1
4
y
xv
u
1
2
3
4
u1
u2
u3
u4
v4
v3
v2
v1
=
4
4
3
3
2
2
1
1
v
u
v
u
v
u
v
u
d
Displacement approximation in terms of shape functions
Strain approximation in terms of strain-displacement matrix
Stress approximation
Summary: For each element
Element stiffness matrix
Element nodal load vector
dNu =
dBD=
dB =
= eVk dVBDBT
S
eT
b
e
f
SS
T
f
V
TdSTdVXf += NN
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Constant Strain Triangle (CST) : Simplest 2D finite element
3 nodes per element
2 dofs per node (each node can move in x- and y- directions)
Hence 6 dofs per element
x
y
u3
v3v1
u1
u2
v2
2
3
1
(x,y)
vu
(x1,y1)
(x2,y2)
(x3,y3)
332211
332211
vy)(x,Nvy)(x,Nvy)(x,Ny)(x,v
uy)(x,Nuy)(x,Nuy)(x,Ny)(x,u
++
++
Formula for the shape functions are
A
ycxbaN
A
ycxbaN
A
ycxbaN
2
2
2
3333
2222
1111
++=
++=
++=
12321312213
31213231132
23132123321
33
22
11
x1
x1
x1
det2
1
xxcyybyxyxa
xxcyybyxyxa
xxcyybyxyxa
y
y
y
triangleofareaA
===
===
===
==
where
x
y
u3
v3v1
u1
u2
v2
2
3
1
(x,y)
vu
(x1,y1)
(x2,y2)
(x3,y3)
Approximation of the strains
x y
u
v
u v
x
y
x
B dy
y x
= =
+
=
=
332211
321
321
332211
321
321
000000
2
1
y)(x,Ny)(x,Ny)(x,Ny)(x,Ny)(x,Ny)(x,N
y)(x,N0
y)(x,N0
y)(x,N0
0y)(x,N
0y)(x,N
0y)(x,N
bcbcbc
cccbbb
A
xyxyxy
yyy
xxxB
=
=
3
3
2
2
1
1
321
321
v
u
v
u
v
u
N0N0N0
0N0N0N
y)(x,v
y)(x,uu
dNu =
Approximation ofdisplacements Element stiffness matrix
= eVk dVBDBT
Atke
VBDBdVBDB
TT== t=thickness of the elementA=surface area of the element
Since B is constant
t
A
Element nodal load vector
S
eT
b
e
f
SST
f
VT dSTdVXf += NN
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Class exercise
= eTS
S
T
SdSTf N
For the CST shown below, compute the vector of nodal loads due to surface traction
x
y
fS3x
fS3yfS2y fS2x
2 3
1
=e
lS
along
T
SdSTtf
31 32N
py=-1(0,0) (1,0)
Class exercise
=
el
Salong
T
SdSTtf
31 32N
x
y
fS3x
fS3yfS2y fS2x
2 3
1
py=-1
=
1
0ST
The only nonzero nodal loads are
= =1
0 3222 x
yalongSdxpNtf
y
= =1
0 3233 x
yalongSdxpNtf
y
( ) ( )
x
xxy
xy
y
xy
y
y
y
xyy
A
xyyyxyx
A
xba
A
ycxbaN
yalong
=
=
=
=
+=
+=
++=
=
1
)(
)1(
0x1
0x1
x1
det
)1(
x1
x1
x1
det
222
231
1
3
2
11
1
33
22
11
11
13311322
0
222
322
(can you derive this simpler?)
2
)1)(1(1
0
1
0 3222
t
dxxt
dxpNtf
x
xyalongS y
=
=
=
=
=
Now compute
= =1
0 3233 x
yalongSdxpNtf
y
4-noded rectangular element with edges parallel to the
coordinate axes:
x
y
(x,y)
vu
(x1,y1)(x2,y2)
(x3,y3)
1 2
34(x4,y4)
2b
2a
=
4
1
iy)u(x,y)(x,ui
iN
=
4
1
iy)v(x,y)(x,vi
iN
4 nodes per element 2 dofs per node (each node can move in x- and y- directions)
8 dofs per element
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Generation of N1:
x
y
1 2
34
2b
2a
N1
l1(y)
l1(x)
1
1
21
21 )(
xx
xxxl
=At node 1
has the property
0)(
1)(
21
11
=
=
xl
xl
Similarly
41
41 )(
yy
yyyl
=
has the property
0)(
1)(
41
11
=
=
yl
yl
Hence choose the shape function at node 1 as
( )( )4241
4
21
2111
4
1)()( yyxx
abyy
yy
xx
xxylxlN =
==
( )( )
( )( )
( )( )
( )( )134
243
312
421
4
1
4
1
4
1
4
1
yyxxab
N
yyxxab
N
yyxxab
N
yyxxab
N
=
=
=
=
Using similar arguments, choose
Properties of the shape functions:
1. The shape functions N1, N2 , N3 and N4 are bilinear functions
of x and y
=nodesotherat
inodeatyx
0
''1),(N
i
3. Completeness
yy
xx
i
i
=
=
=
=
=
=
4
1i
i
4
1i
i
4
1i
i
N
N
1N
2. Kronecker delta property
3. Along lines parallel to the x- or y-axes, the shape functions
are linear. But along any other line they are nonlinear.
4. An element shape function related to a specific nodal point is
zero along element boundaries not containing the nodal point.
5. The displacement field is continuous across elements
6. The strains and stresses are not constant within an element
nor are they continuous across element boundaries.
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=
=
4
4
3
3
2
2
1
1
B
44332211
4321
4321
xy
v
u
v
u
v
u
v
u
y)(x,Ny)(x,Ny)(x,Ny)(x,Ny)(x,Ny)(x,Ny)(x,Ny)(x,N
y)(x,N0
y)(x,N0
y)(x,N0
y)(x,N0
0y)(x,N
0y)(x,N
0y)(x,N
0y)(x,N
xyxyxyxy
yyyy
xxxx
y
x
The strain-displacement relationship
=
yyxxyyxxyyxxyyxx
xxxxxxxx
yyyyyyyy
abB
13243142
3412
1234
0000
0000
4
1
Notice that the strains (and hence the stresses) are NOT constant within an element
The B-matrix (strain-displacement) corresponding to this element is
We will denote the columns of the B-matrix as
Computation of the terms in the stiffness matrix of 2D elements (recap)
x
y
(x,y)
v
u
1 2
34v4
v3
v2v
1
u1u
2
u3u4
1 2 3 4
1 2 3 4
1 1 2 2 3 3 4 4
N (x,y) N (x,y) N (x,y) N (x,y)0 0 0 0
N (x,y) N (x,y) N (x,y) N (x,y)0 0 0 0
N (x,y) N (x,y) N (x,y) N (x,y) N (x,y) N (x,y) N (x,y) N (x,y)
x x x x
y y y y
y x y x y x y x
u1 v1 u2 v2 u3 u4
v3 v4
1 1
1
1
1
1
N (x,y)0
N (x,y )0 ; ; and so on...
N (x,y)N (x,y )
u v
x
B By
yx
= =
= eVk dVBDBT
1 1 12 1 3 14 1 5 1 6 1 7 18
2 1 2 2 2 3 2 4 2 5 26 2 7 2 8
31 3 2 33 3 4 35 36 37 3 8
4 1 4 2 4 3 4 4 4 5 46 4 7 4 8
51 52 53 54 55 5 6 5 7 5 8
61 6 2 63 6 4 65 66 67 6 8
71 7 2 73 7 4 75 76 77 7 8
8 1 8 2 8 3 8 4 8 5 86 8 7 8 8
k k k k k k k k
k k k k k k k k
k k k k k k k k
k k k k k k k k k
k k k k k k k k
k k k k k k k k
k k k k k k k k
k k k k k k k k
=
u1
v1
u2
v2
u3
u4
v3
v4
u1 v1u
2v
2 u3u4v3
v4
The stiffness matrix corresponding to this element is
which has the following form
1 1 1 1 1 2
1 1 1 1
T T T
11 12 13
T T
21 21
B D B dV; B D B dV; B D B dV,...
B D B dV; B D B dV;.....
e e e
e e
u u u v u uV V V
v u v vV V
k k k
k k
= = =
= =
The individual entries of the stiffness matrix may be computed as follows
Notice that these formulae are quite general (apply to all kinds
of finite elements, CST, quadrilateral, etc) since we have not
used any specific shape functions for their derivation.
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Example
x
y
2
34
1
300 psi
1000 lb
3 in
2 inThickness (t) = 0.5 in
E= 30106psi
=0.25
(a) Compute the unknown nodal displacements.
(b) Compute the stresses in the two elements.
This is exactly the same problem that we solved in last class, except
now we have to use a single 4-noded element
Realize that this is a plane stress problem and therefore we need to use
psiE
D7
210
2.100
02.38.0
08.02.3
2
100
01
01
1
=
=
Write down the shape functions
( )( )
( )( )
( )( )
( )( )6
)3(
4
1
64
1
6
)2(
4
1
6
)2)(3(
4
1
134
243
312
421
yxyyxx
abN
xyyyxx
abN
yxyyxx
abN
yxyyxx
abN
==
==
==
==
20
23
03
00
yx
We have 4 nodes with 2 dofs per node=8dofs. However, 5 of these are fixed.
The nonzero displacements are
u2 u3 v3
Hence we need to solve
=
yfv
u
u
kkk
kkk
kkk
33
3
2
333231
232221
131211
0
0
Need to compute only the relevant terms in the stiffness matrix
2 2 2 3 2 3
3 2 3 3 3 3
3 2 3 3 3 3
T T T
11 12 13
T T T
21 22 13
T T T
31 22 13
B D B dV; B D B dV; B D B dV
B D B dV; B D B dV; B D B dV
B D B dV; B D B dV; B D B dV
e e e
e e e
e e e
u u u u u vV V V
u u u u u vV V V
v u v u v vV V V
k k k
k k k
k k k
= = =
= = =
= = =
u2 u3 v3
=
=
6
06
)2(
0
2
2
2
x
y
y
N
x
N
B u
=
=
6
06
0
3
3
3
x
y
y
N
x
N
B u
=
=
6
6
00
3
3
3
y
x
x
N
y
NB v
Compute only the relevant columns of the B matrix
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2 2
T
11
3 2
8 7 5 2
0 0
7
B D B dV
20.5 (0.1067 10 0.533 10 )( ) 3.33 10
6
0 .656 10
e u uV
x y
k
yx dxdy
= =
=
= +
=
Similarly compute the other terms
How do we compute f3y
ySyff
310003 +=
lb
dxx
dxNtf
x
x edgealongS y
225
2
3150
3)300)(5.0(
)300(
3
0
3
0 4333
=
=
=
=
=
=
4 3
36 22334
3
xxyNN
yy
edge =
==
==
lbffySy
1225100033
=+=
x
y
a a12
3 4
b
b
Corner nodes
+=
=
+
=
+
+=
224223
222221
2
)(
2
)(
2
)(
2
)(
2
)(
2
)(
2
)(
2
)(
b
yby
a
xaxN
b
yby
a
xaxN
b
yby
a
xaxN
b
yby
a
xaxN
5
6
7
89
Midside nodes
+=
=
=
+
=
2
22
2822
22
7
2
22
2622
22
5
2
)(
2
)(
2
)(
2
)(
b
yb
a
xaxN
b
yby
a
xaN
b
yb
a
xaxN
b
yby
a
xaN
Center node
2 2 2 2
9 2 2
a x b yN
a b
=
How about a 9-noded rectangle?
Question: Can you generate the shape functions of a 16-noded rectangle?
Note: These elements, whose shape functions are generated by multiplying the
shape functions of 1D elements, are said to belong to the Lagrange family