2.094 — Finite Element Analysis of Solids and Fluids
Fall ‘08
Lecture 6 - Finite element formulation, example, convergence
Prof. K.J. Bathe
6.1
MIT OpenCourseWare
Example
Reading:
Ex. 4.6 in
the text
t = 0.1,
KU = R;
�
K=
K (m) ;
R = RB + R s + R c + R r
�
T
(m)
K
=
B (m) C (m) B (m) d V (m)
�
(m)
RB ;
m
6.1.1
�
�
K�
(6.1)
(6.2)
V (m)
m
RB =
E, ν plane stress
(m)
RB
�
T
H (m) f B(m) d V (m)
=
(6.3)
V (m)
F.E. model
el. (2)
u1 u2 u3 u4
↓
↓
↓
⎡↓
� � � ×
= ⎢ .
⎢ ..
⎢
⎣
v1 v2 v3 v4
↓ ↓ ↓ ↓⎤
× × × ×
⎥
⎥
⎥
⎦
23
← u1
..
.
(6.4)
MIT 2.094
6. Finite element formulation, example, convergence
In practice,
⎛
⎞
u1
⎜ .. ⎟
⎜ . ⎟
⎜
⎟
⎜ u4 ⎟
⎜
⎟
�=B⎜
⎟
⎜ v1 ⎟
⎜ . ⎟
⎝ .. ⎠
v4
�
�
�
K� =
B T CB dV ;
el
V
(6.5)
where K is 8x8 and B is 3x8.
Assume we have K (8x8) for el. (2)
U1 U2 U3 U4 U5 · · · U11
↓
↓
↓
↓
↓
↓
↓
⎡
× × × × × × ×
..
..
..
⎢
⎢
.
.
.
⎢
⎢
..
..
..
⎢
.
.
.
⎢
K
= ⎢
����
�
�
�
⎢
assemblage
⎢
..
..
..
⎢
.
.
.
⎢
⎢
.
.
..
.
.
⎢
.
.
.
⎢
⎣
×
×
×
×
×
×
×
· · · U18
↓
↓
⎤
× ×
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎦
×
×
← U1
..
.
..
.
← U11
..
.
..
.
← U18
(6.6)
Consider,
�
�
�
HS = H�
T
H S f S d S;
RS =
S
�
H=
⎛
⎜
⎜
⎜
⎜
⎜
u=⎜
⎜
⎜
⎜
⎜
⎝
h1
0
u1
u2
..
.
u4
v1
..
.
h2
0
h3
0
h4
0
0
h1
0
h2
(6.7)
on surface
0
h3
0
h4
�
← u(x, y)
← v(x, y)
(6.8)
⎞
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎠
(6.9)
v4
24
MIT 2.094
6. Finite element formulation, example, convergence
�
�
HS = H�
y=+1
⎡ 1
2 (1 + x)
=⎣
0
(6.10)
1
2 (1
− x) 0
0
0
0
0
0
1
2 (1
0
+ x)
1
2 (1
0
0
⎤
⎦
− x)
0
(6.11)
0
From (6.7);
⎡
�
+1
RS =
−1
⎢
⎢
⎣
1
2 (1
1
2 (1
+ x)
− x)
0
0
⎤
0
�
�
⎥
0
0
⎥
(0.1) dx
1
⎦ −p(x)
� �� �
2 (1 + x)
1
thickness
(1
−
x)
2
(6.12)
⎡
⎤
0
⎢
⎥
0
⎥
RS = ⎢
⎣ −p0 (0.1)
⎦
−p0 (0.1)
6.1.2
(6.13)
Higher-order elements
Want h1 , h2 , h3 , h4 , h5
u(x, y) =
�5
i=1
hi u i .
hi = 1 at node i and 0 at all other nodes.
h5 = 12 (1 − x2 )(1 + y)
25
MIT 2.094
6. Finite element formulation, example, convergence
1
h5
2
1
h5
2
1
(1 + x)(1 + y) −
4
1
h2 = (1 − x)(1 + y) −
4
1
h3 = (1 − x)(1 − y)
4
1
h1 = (1 + x)(1 − y)
4
h1 =
(6.14)
(6.15)
(6.16)
(6.17)
Note:
�
We must have
u(x, y) =
�
�
i
hi = 1
hi = 1 to satisfy the rigid body mode condition.
h i ui
(6.18)
i
Assume all nodal point displacements = u∗ . Then,
u(x, y) =
�
h i u ∗ = u∗
�
i
h i = u∗
(6.19)
i
From (6.1),
�
�
�
K
(m)
U =R
(6.20)
U =R
(6.21)
m
� ��
B
(m) T
C
(m)
B
(m)
dV
(m)
�
V (m)
m
where C (m) B (m) U = τ (m) . (Assume we calculated U .)
��
m
�
T
B (m) τ (m) d V (m) = R
V
F
(6.22)
(m)
(m)
= R;
m
F
(m)
�
T
B (m) τ (m) d V (m)
=
V
(m)
Two properties
I. The sum of the F (m) ’s at any node is equal to the applied external forces.
26
(6.23)
MIT 2.094
6. Finite element formulation, example, convergence
II. Every element is in equilibrium under its F (m)
ˆ T F (m) = U
ˆT
U
�
�
T
V
B (m) τ (m) d V (m)
��
�
=�(m)
�
=
(6.24)
(m)
T
T
�(m) τ (m) d V (m)
(6.25)
V (m)
=0
(6.26)
ˆ T = virtual nodal point displacement.
where U
Apply rigid body displacement.
If we move the element virtually in the rigid body modes, �(m) is zero. Therefore the virtual work
obtained due to virtual motion of the element is zero. Then the element is in equilibrium under its F (m) .
27
MIT OpenCourseWare
http://ocw.mit.edu
2.094 Finite Element Analysis of Solids and Fluids II
Spring 2011
For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.