2.094 — Finite Element Analysis of Solids and Fluids
Fall ‘08
Lecture 12 - Total Lagrangian formulation
Prof. K.J. Bathe
We discussed:
�
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�
t
0X
∂ txi
=
∂ 0xj
t
0C
= 0tX T 0tX
⇒
d tx = 0tX d 0x,
d 0x =
(12.1)
(12.2)
�
d 0x = 0t X d tx
� t �−1 t
dx
0X
where 0t X =
0
� t �−1
∂ xi
=
0X
∂ txj
The Green-Lagrange strain:
� 1 �t
�
1 �t T t
t
0� =
0X 0X − I =
0C − I
2
2
�
(12.3)
(12.4)
Polar decomposition:
t
0X
= 0tR0tU ⇒ 0t� =
�
1 �� t � 2
−I
0U
2
(12.5)
We see, physically that:
where d t+Δtx and d tx are the same lengths
⇒ the components of the G-L strain do not
change.
Note in FEA
�
⎫
0
xi =
hk 0xki ⎪
⎪
⎬
k
for an element
�
t
⎪
ui =
hk tuki ⎪
⎭
(12.6)
k
t
xi = 0xi + tui →
for any particle
(12.7)
Hence for the element
�
�
t
xi =
hk 0xki +
hk tuki
k
=
�
(12.8)
k
hk
�0 k t k�
xi + ui
(12.9)
k
=
�
hk txkk
(12.10)
k
49
MIT 2.094
12. Total Lagrangian formulation
E.g., k = 4
2nd Piola-Kirchhoff stress
0
t
0S
Then
�
0V
=
ρ0 t 0 T
X τ tX
tρ t
t
0 Sij
→
δ 0t�ij d 0V =
�
tV
components also independent of a rigid body rotation
(12.11)
t
(12.12)
τij δ t
eij d tV = tR
We can use an incremental decomposition of stress/strain.
t+Δt
0S
t+Δt
0 Sij
t+Δt
0�
t+Δt
0�ij
= 0tS + 0 S
(12.13)
= 0tSij + 0 Sij
(12.14)
=
=
t
0� + 0�
t
0�ij + 0�ij
(12.15)
(12.16)
Assume the solution is kown at time t, calculate the solution at time t + Δt. Hence, we apply (12.12)
at time t + Δt:
�
t+Δt
t+Δt
0
t+Δt
R
(12.17)
0 Sij δ
0�ij d V =
0V
Look at δ 0t�ij :
�
1 �t
t
t
t
0ui,j + 0uj,i + 0uk,i 0uk,j
2�
�
1
∂δui
∂δuj
∂δuk ∂ tuk
∂ tuk ∂δuk
t
δ 0�ij =
+ 0 + 0 · 0 + 0 · 0
2 ∂ 0xj
∂ xi
∂ xi ∂ xj
∂ xi ∂ xj
�
�
1
δ 0t�ij =
δ 0ui,j + δ 0uj,i + δ 0uk,i 0tuk,j + 0tuk,i δ 0uk,j
2
δ 0t�ij = δ
(12.18a)
(12.18b)
(12.18c)
We have
t+Δt
0�ij
− 0t�ij = 0�ij
0�ij
(12.19)
= 0eij + 0 ηij
(12.20)
50
MIT 2.094
12. Total Lagrangian formulation
where 0eij is the linear incremental strain, 0 ηij is the nonlinear incremental strain, and
⎛
0eij
=
⎞
1⎜
⎟
t
t
⎝ u + u + u u + u u ⎠
2 0 i,j 0 j,i �0 k,i 0 k,j �� 0 k,i 0 k,j�
(12.21)
initial displ. effect
1
u u
0 ηij =
2 0 k,i 0 k,j
(12.22)
where
0uk,j
=
∂uk
,
∂ 0xj
uk =
t+Δt
uk − tuk
(12.23)
Note
δ t+Δt�ij = δ 0�ij
�
� δ 0t�ij = 0 when changing the configuration at t + Δt
�
(12.24)
From (12.17):
�
�t
��
�
δ 0eij + δ 0 ηij d 0V
0 Sij + 0 Sij
0V
�
=
0V
=
�t
� 0
t
0 Sij δ 0eij + 0 Sij δ 0eij + 0 Sij δ 0 ηij + 0 Sij δ 0 ηij d V
(12.25)
t+Δt
R
Linearization
⎛
⎞
t
t
0 KN L U
0 KL U
�
⎜� �� � �t �� �⎟ 0
⎜0 Sij δ 0eij + 0 Sij δ 0 ηij ⎟ d V =
⎝
⎠
(12.26)
t+Δt
�
R−
0V
0V
�
t
0
0 Sij δ 0eij d V
��
t
0F
(12.27)
�
We use,
0 Sij
� 0 Cijrs 0ers
(12.28)
We arrive at, with the finite element interpolations,
�t
�
t
t+Δt
R − 0tF
0 KL + 0 KN L U =
where U is the nodal displacement increment.
51
(12.29)
MIT 2.094
12. Total Lagrangian formulation
Left hand side as before but using (k − 1) and right hand side is
�
t+Δt (k−1) 0
t+Δt
= t+ΔtR −
d V
0 Sij δ
0�ij
(12.30)
0V
gives
t+Δt
R−
t+Δt (k−1)
0F
In the full N-R iteration, we use
�
�
t+Δt (k−1)
t+Δt (k−1)
+
ΔU (k) =
0 KL
0 KN L
(12.31)
t+Δt
R−
52
t+Δt (k−1)
0F
(12.32)
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2.094 Finite Element Analysis of Solids and Fluids II
Spring 2011
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