2.094 — Finite Element Analysis of Solids and Fluids
Fall ‘08
Lecture 13 - Total Lagrangian formulation, cont’d
Prof. K.J. Bathe
MIT OpenCourseWare
Example truss element. Recall:
Principle of virtual displacements applied at some time t + Δt:
�
t+Δt
τij δ t+Δteij d t+ΔtV =
t+ΔtV
�
0V
t+Δt
0 Sij
t+Δt
0�ij
0�ij
t+Δt
t+Δt
0
0 Sij δ
0�ij δ V
=
t+Δt
(13.1)
t+Δt
(13.2)
R
R
= 0tSij + 0 Sij
=
t
0�ij
(13.3)
+ 0�ij
(13.4)
= 0eij + 0 ηij
(13.5)
where 0tSij and 0t�ij are known, but 0 Sij and 0�ij are not.
�
1�
t
t
0ui,j + 0uj,i + 0uk,i 0uk,j + 0uk,j 0uk,i
2
�
1�
0 ηij =
0uk,i 0uk,j
2
0eij
=
Substitute into (13.2) and linearize to obtain
�
�
t
0
δ 0eij 0 Cijrs 0ers d 0V +
0 Sij δ 0 ηij d V =
0V
F.E. discretization gives
�t
�
t
0 KL + 0 KN L ΔU =
(13.6)
(13.7)
t+Δt
0V
t+Δt
R − 0tF
�
R−
0V
δ 0eij 0tSij d 0V
(13.8)
(13.9)
53
MIT 2.094
13. Total Lagrangian formulation, cont’d
�
t
0 KL
=
t
0 KN L
=
0V
�
0V
t T
t
0
0 BL 0 C 0 BL d V
(13.10)
t T
0 BN L
(13.11)
t
t
0
0 S 0 BN L d V
����
matrix
�
t
0F
=
0V
t T
0 BL
t
0
0 Ŝ d V
����
(13.12)
vector
The iteration (full Newton-Raphson) is
�
�
t+Δt (i−1)
t+Δt (i−1)
+
ΔU (i) =
0 KL
0 KN L
t+Δt
U (i) =
t+Δt
R−
t+Δt
U (i−1) + ΔU (i)
Truss element example
t+Δt (i−1)
0F
(13.13)
(13.14)
(p. 545)
Here we have to only deal with 0tS11 , 0e11 , 0 η11
0e11
0 η11
∂u1
∂ tu
∂uk
+ 0 k · 0
0
∂ x1
∂ x1 ∂ x1
�
�
1 ∂uk ∂uk
=
·
2 ∂ 0x1 ∂ 0x1
=
(13.15)
(13.16)
We are after
⎞
u11
⎜ u12 ⎟ t
⎟
= 0tBL ⎜
⎝ u21 ⎠ = 0 BL û
u22
⎛
0e11
ui =
2
�
(13.17)
hk uki
(13.18)
k=1
t
ui =
2
�
hk tuki
(13.19)
k=1
54
MIT 2.094
13. Total Lagrangian formulation, cont’d
∂u1
∂ tu1 ∂u1
∂ tu2 ∂u2
+
+
∂ 0x
∂ 0x1 ∂ 0x1
∂ 0x1 ∂ 0x1
�0 1
�
t 2
0
u1 = L + ΔL cos θ − L
�
�
t 2
u2 = 0L + ΔL sin θ
0e11
0e11
=
1 �
−1
=0
L
⎛
0 1
�
0
(13.20a)
(13.20b)
(13.20c)
û
⎞
⎜
⎟
⎜0
⎟
⎜ L + ΔL
⎟ 1 �
⎟·
+⎜
cos
θ
−
1
⎜
⎟ 0L −1
0L
⎜�
��
�⎟
⎝
⎠
0
1
�
0
û
∂ tu1
∂ 0x1
⎛
⎞
⎜
⎟
⎜0
⎟
⎜ L + ΔL
⎟ 1 �
⎜
+⎜
sin θ⎟
⎟ · 0L 0
0L
⎜�
⎟
��
�
⎝
⎠
−1
0
1
�
û
(13.20d)
∂ tu2
∂ 0x1
=0tBL û
(13.20e)
Hence,
0
0e11
=
L + ΔL �
− cos θ
2
( 0L)
− sin θ
cos θ
sin θ
�
û
(13.20f)
where the boxed quantity above equals 0tBL . In small strain but large rotation analysis we assume
ΔL � 0L,
0e11
=
0 η11
δ 0 η11
1 �
− cos θ
0L
1
=
2
�
− sin θ
cos θ
∂u1 ∂u1
∂u2 ∂u2
+ 0
0
0
∂ x1 ∂ x1
∂ x1 ∂ 0x1
sin θ
�
û
(13.20g)
�
(13.21a)
�
�
1 ∂δu1 ∂u1
∂u1 ∂δu1
∂δu2 ∂u2
∂u2 ∂δu2
=
+ 0
+ 0
+ 0
2 ∂ 0x1 ∂ 0x1
∂ x1 ∂ 0x1
∂ x1 ∂ 0x1
∂ x1 ∂ 0x1
�
�
∂δu1 ∂u1
∂δu2 ∂u2
=
+ 0
∂ 0x1 ∂ 0x1
∂ x1 ∂ 0x1
t
0 S11 δ 0 η11
=
�
∂δu1
∂ 0x1
∂δu2
∂ 0x1
��
� � tS
0
0 11
t
0
0 S11
�
��
�
∂u1
∂ 0x1
∂u2
∂ 0x1
(13.21b)
(13.21c)
�
(13.21d)
t
0S
�
∂u1
∂ 0x1
∂u2
∂ 0x1
�
=
1
0L
�
�
−1
0
0
−1
��
BN L
1 0
0 1
�
û
(13.21e)
�
55
MIT 2.094
0C
t
0 Ŝ
13. Total Lagrangian formulation, cont’d
=E
(13.22)
= 0tS11
(13.23)
Assume small strains
t
0K
EA
0L
⎡
cos2 θ
⎢
=⎢
⎣
sym
�
=
⎡
cos θ sin θ
sin2 θ
1
P ⎢
0
⎢
+0 ⎣
−1
L
0
�
t
− cos2 θ
− sin θ cos θ
cos2 θ
��
t
0 KL
0 −1
0
1
0
1
−1 0
��
t
0 KN L
⎤
− cos θ sin θ
− sin2 θ ⎥
⎥
sin θ cos θ ⎦
sin2 θ
�
⎤
0
−1 ⎥
⎥
0 ⎦
1
�
When θ = 0, 0tKL doesn’t give stiffness corresponding to u22 , but 0tKN L does.
56
(13.24)
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.