2.092/2.093 — Finite Element Analysis of Solids & Fluids I
Fall ‘09
Lecture 6 - Finite Element Solution Process
Prof. K. J. Bathe
MIT OpenCourseWare
In the last lecture, we used the principle of virtual displacements to obtain the following equations:
KU = R
K = ΣK
(m)
m
; K
(m)
�
(1)
B (m)T C (m) B (m) dV (m)
=
V (m)
R = RB + R S
RB =
RS =
(m)
ΣRB
m
(m)
ΣRS
m
;
;
�
(m)
RB
=
(m)
RS
=Σ
H (m)T f B(m) dV (m)
V
(m)
�
i
i(m)
H Sf
i(m)
Sf
T
i(m)
f Sf
i(m)
dSf
u(m)
=
↓
H (m) U
(2)
ε(m)
=
B (m) U
(3)
Note that the dimension of u(m) is in general not the same as the dimension of ε(m) .
Example: Static Analysis
Reading assignment: Example 4.5
1
Lecture 6
Finite Element Solution Process
2.092/2.093, Fall ‘09
Assume:
i. Plane sections remain plane
ii. Static analysis → no vibrations/no transient response
iii. One-dimensional problem; hence, only one degree of freedom per node
Elements 1 and 2 are compatible because they use the same U2 . Next, use a linear interpolation function.
⎡
u(1) (x) =
� �
1−
x
100
�
x
100
�
��
H (1)
⎤
U1
0 ⎣ U2 ⎦
� U
3
⎡
ε(1) (x) =
�
1
− 100
�
⎡
�
K
100
= E·1·
0
⎢
⎢
⎣
1
��100
B (1)
1
− 100
1
100
⎡
; u(2) (x) =
�
; ε(2) (x) =
0
�
=
1
E ⎢
⎢ −1
100 ⎣
0
−1 0
1
0
⎡
0 0
⎥
13E ⎢
⎥
⎢
0 ⎦+
0 1
3 · 80 ⎣
0
0 −1
�
�
⎤
⎤
�
x
x
80
1 − 80
��
H (2)
⎤
U1
⎣ U2 ⎦
� U
3
�
⎡
�
dx + E
0
0
⎡
�
⎤
U1
0 ⎣ U2 ⎦
� U
3
1
100
0
�
�
⎥�
⎥ − 1
⎦
100
�
0
80
�
0
1
− 80
��
1
80
B (2)
⎡
0
�
x 2⎢
⎢
1+
−1
40 ⎣ 80
1
80
⎤
U1
⎣ U2 ⎦
� U
3
�
⎤
⎥�
⎥ 0
⎦
1
− 80
1
80
�
dx
⎤
⎥
−1 ⎥
⎦
1
The “equivalent cross-sectional area” of element 2 is A =
areas of the end faces A = 1 and A = 9.
2
13
3
cm2 . This equivalent area must lie between the
Lecture 6
Finite Element Solution Process
⎡
K=
2.4
E ⎢
⎢ −2.4
240 ⎣
0
−2.4
0
2.092/2.093, Fall ‘09
⎤
⎥
−13 ⎥
⎦
13
15.4
−13
We note:
• Diagonal terms must be positive. If the diagonal terms are zero or negative, then the system is unstable
physically. A positive diagonal implies that the degree of freedom has stiffness at that node.
• K is symmetric.
• K is singular if rigid body motions are possible. To be able to solve the problem, all rigid body modes
must be removed by adequately constraining the structure. i.e. K is reduced by applying boundary
conditions to the nodes.
The K used to solve for U is, then, positive definite (det K > 0). This ensures that the elastic strain energy
is positive and nonzero for any displacement field U . In the analysis, each element is in equilibrium under
its nodal forces, and each node is in equilibrium when summing element forces and external loads.
Homework Problem 2
⎡
⎣
εxx
εzz
⎤
⎡
⎦=⎣
3
∂u
∂x
u
x
⎤
⎦
Lecture 6
Finite Element Solution Process
εzz is frequently called the “hoop strain”, εθθ .
2π(u + x) − 2πx
u
=
x
2πx
⎡
⎤
E ⎣ 1 ν ⎦
C=
1 − ν2
ν 1
εzz =
�
�
f B = ρω 2 R N/cm3
4
; R = x
2.092/2.093, Fall ‘09
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2.092 / 2.093 Finite Element Analysis of Solids and Fluids I
Fall 2009
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