2.094 — Finite Element Analysis of Solids and Fluids
Fall ‘08
Lecture 14 - Total Lagrangian formulation, cont’d
Prof. K.J. Bathe
MIT OpenCourseWare
Truss element. 2D and 3D solids.
�
t+Δt
τij δ t+Δteij d t+ΔtV =
t+ΔtV
�
0V
�
0V
0 Cijrs 0ers δ 0eij δ
t+Δt
t+Δt
0
0 Sij δ
0�ij δ V
(14.1)
t+Δt
(14.2)
R
R
⏐
⏐
� linearization
�
t
0
t+Δt
R−
0 Sij δ 0 ηij δ V =
�
0
=
t+Δt
V +
0V
0V
t
0
0 Sij δ 0eij δ V
(14.3)
Note:
δ 0eij = δ 0t�ij
varying with respect to the configuration at time t.
F.E. discretization
�
0
xi =
hk 0xki
t
xi =
k
t
ui =
�
�
t+Δt
hk txki
xi =
k
hk tuki
t+Δt
ui =
k
(14.4) into (14.3) gives
�t
�
t
0 KL + 0 KN L U =
�
�
hk t+Δtxki
(14.4a)
hk uik
(14.4b)
k
hk t+Δtuik
k
ui =
�
k
t+Δt
R − 0tF
(14.5)
57
MIT 2.094
14. Total Lagrangian formulation, cont’d
Truss
ΔL
0L
� 1 small strain assumption:
t
0K
=
E 0A
0L
⎡
⎢
=⎢
⎣
+
cos2 θ
cos θ sin θ
− cos2 θ
2
cos θ sin θ
sin θ − sin θ cos θ
cos2 θ
− cos2 θ − cos θ sin θ
2
− cos θ sin θ
− sin θ
sin θ cos θ
⎡
⎤
1
0 −1
0
t
⎥
P ⎢
0
0
1
−1
⎢
⎥
0L ⎣ −1
0
1
0 ⎦
0
1
0 −1
⎤
− cos θ sin θ
− sin2 θ ⎥
⎥
sin θ cos θ ⎦
sin2 θ
(14.6)
(notice that the both matrices are symmetric)
�
u1
v1
�
�
=
cos θ
− sin θ
sin θ
cos θ
��
�
u1
v1
(14.7)
Corresponding to the u and v displacements we have:
t
0K
=
=
E 0A
0L
⎡
1
⎢ 0
⎢
⎣ −1
0
(14.8)
0
0
0
0
⎤
−1 0
0 0 ⎥
⎥+
1 0 ⎦
0 0
⎡
t
P
0L
1
⎢ 0
⎢
⎣ −1
0
⎤
0 −1
0
0 −1 ⎥
1
⎥
0
1
0 ⎦
−1
0
1
58
(14.9)
MIT 2.094
14. Total Lagrangian formulation, cont’d
t
Q 0L = tP · Δ
⇒
P
Q=
0L
· Δ
(14.10)
t
where the boxed term is the stiffness. In axial direction,
0
E A
0L
�
t
P
0L
t
. But, in vertical direction,
P
0L
P
0L
is not very important because usually
is important.
⎡
⎤
− cos θ
⎥
⎢
t
t ⎢ − sin θ ⎥
0 F = P ⎣ cos θ ⎦
sin θ
2D/3D (e.g. Table 6.5)
(14.11)
2D:
�2 �
�2 �
1 ��
t
t
�
u
+
u
=
u
+
u
u
+
u
u
+
0 2,1
0 11
0 1,1
0 1,1 0 1,1
0 2,1 0 2,1
0 1,1
�
��
� 2�
��
�
0e11
(14.12)
0 η11
0�22
=· · ·
(14.13)
0�12
=· · ·
(14.14)
= ?
(14.15)
(Axisymmetric)
0�33
�
1 �� t � 2
−I
0U
2⎡
⎤
× × 0
⎣ × × 0 ⎦
t 2
0 0 ×
0U =
↑
t 2
( λ)
t
0�
=
�
�
2π 0x1 + tu1
d ts
λ= 0 =
d s
2π 0x1
t
u
=1+ 0 1
x1
(14.16)
(14.17)
t
t
0�33
�
�2
u1
1+ 0
− 1
x1
�
�2
t
u
1 tu1
= 0 1 +
x1
2 0x1
1
=
2
��
(14.18)
t
(14.19)
59
MIT 2.094
14. Total Lagrangian formulation, cont’d
t+Δt
0�33
0�33 =
t
u + u1
1
= 10
+
x1
2
t+Δt
0�33
− 0t�33 =
�t
�2
u 1 + u1
0x
1
t
u 1 u1
1
u1
0x + 0x · 0x + 2
1
1
1
(14.20)
�
�2
u1
0x
1
(14.21)
How do we assess the accuracy of an analysis?
• Mathematical model ∼ u
• F.E. solution ∼ uh
Find �u − uh � and �τ − τh �.
References
[1] T. Sussman and K. J. Bathe. “Studies of Finite Element Procedures - on Mesh Selection.” Computers
& Structures, 21:257–264, 1985.
[2] T. Sussman and K. J. Bathe. “Studies of Finite Element Procedures - Stress Band Plots and the
Evaluation of Finite Element Meshes.” Journal of Engineering Computations, 3:178–191, 1986.
60
Reading:
Sec. 4.3.6
MIT OpenCourseWare
http://ocw.mit.edu
2.094 Finite Element Analysis of Solids and Fluids II
Spring 2011
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