2.094 — Finite Element Analysis of Solids and Fluids
Fall ‘08
Lecture 3 - Finite element formulation for solids and structures
Prof. K.J. Bathe
MIT OpenCourseWare
Reading:
Sec. 6.1-6.2
We need to satisfy at time t:
• Equilibrium
∂ tτij t B
+ fi = 0
∂ txj
t
(i = 1, 2, 3) in t V
t Sf
τij tnj = fi
(3.1)
(i = 1, 2, 3) on t Sf
(3.2)
• Compatibility
• Stress-strain law(s)
Principle of virtual displacements
�
�
�
t
τij teij d tV =
ui tfiB d tV +
tV
1
teij =
2
tV
�
∂ ui
∂uj
+ t
∂ t xj
∂ xi
tS
f
ui | t S
f
t Sf
fi
d tSf
(3.3)
�
(3.4)
• If (3.3) holds for any continuous virtual displacement (zero on tSu ), then (3.1) and (3.2) hold and
vice versa.
• Refer to Ex. 4.2 in the textbook.
10
MIT 2.094
3. Finite element formulation for solids and structures
Major steps
I. Take (3.1) and weigh with ui :
�t
�
τij,j + t fiB ui = 0.
(3.5a)
II. Integrate (3.5a) over volume t V :
�
�t
�
τij,j + t fiB ui d t V = 0
(3.5b)
tV
III. Use divergence theorem. Obtain a boundary term of stresses times virtual displacements on t S =
t
Su ∪ t Sf .
IV. But, on t Su the ui = 0 and on t Sf we have (3.2) to satisfy.
Result: (3.3).
Example
�
t
tV
τ11 te11 d tV =
�
t S
tS
f
ui f1 f d tSf
(3.6)
One element solution:
11
MIT 2.094
3. Finite element formulation for solids and structures
1
1
(1 + r) u1 + (1 − r) u2
2
2
1
1
t
t
u(r) = (1 + r) u1 + (1 − r) t u2
2
2
1
1
u(r) = (1 + r) u1 + (1 − r) u2
2
2
u(r) =
(3.7)
(3.8)
(3.9)
Suppose we know t τ11 , t V , t Sf , t u ... use (3.6).
For element 1,
t e11
�
tV
∂u
= t = B (1)
∂ x
�
u1
u2
for el. (1)
T
t
−→
te11t τ11 d V
�
(3.10)
�
[u1
u2 ]
tV
�
for el. (1)
−→
⎡
= ⎣ U1
����
u2
B (1)
T
t
τ11 d t V
��
�
(3.11)
= t F (1)
u2 ] t F (1)
⎤
� t (1) �
F̂
U 2 U 3⎦
����
0
[u1
(3.12)
(3.13)
u1
where
t
(1)
F̂1
t (1)
F̂2
(1)
(3.14)
t (1)
F1
(3.15)
= t F2
=
For element 2, similarly,
⎡
⎤
= ⎣U 1
U2
����
u2
U3 ⎦
����
�
0
t ˆ
(2)
F
�
(3.16)
u1
R.H.S.
⎡
�
U1
�
U2
��
U
T
U3
⎤
(unknown reaction at left)
⎣
⎦
0
t Sf
�
t
Sf · f
1
�
(3.17)
Now apply,
U
T
=
�
1
0 0
�
(3.18)
=
�
0
1 0
�
(3.19)
=
�
0
0 1
�
(3.20)
then,
U
T
then,
U
T
12
MIT 2.094
3. Finite element formulation for solids and structures
This gives,
⎡
�
t
F̂ (1)
0
�
�
+
t
0
F̂ (2)
�
⎤
unknown reaction
⎢
⎥
0
=⎣
⎦
t tSf t
f1 · Sf
(3.21)
We write that as
t
F = tR
�
�
t
F = fn t U1 , t U2 , t U3
(3.22)
(3.23)
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2.094 Finite Element Analysis of Solids and Fluids II
Spring 2011
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