2.094 — Finite Element Analysis of Solids and Fluids
Fall ‘08
Lecture 18 - Solution of F.E. equations
Prof. K.J. Bathe
MIT OpenCourseWare
In structures,
Reading:
Sec. 8.4
F (u, p) = R.
(18.1)
In heat transfer,
F (θ) = Q
(18.2)
In fluid flow,
F (v, p, θ) = R
(18.3)
In structures/solids
F =
�
F (m) =
m
��
m
0V (m)
t (m) T t ˆ (m) 0 (m)
d V
0S
0 BL
Elastic materials
Example
p. 590 textbook
76
(18.4)
MIT 2.094
18. Solution of F.E. equations
Material law
t
0 S11
˜
0t�11
=E
(18.5)
In isotropic elasticity:
Ẽ =
t
0�
E (1 − ν)
,
(1 + ν) (1 − 2ν)
(ν = 0.3)
�
1 �� t � 2
1
=
U
−
I
⇒ 0t�11 =
2 0
2
(18.6)
��
0
L + tu
0L
�2
�
1
−1 =
2
��
t
u
1+ 0
L
�
�2
−1
(18.7)
where 0tU is the stretch tensor.
0
t
0 S11
=
ρ0
X tτ 0X T
tρ t 11 11 t 11
(18.8)
with
0
0
t X11
=
0L
L
,
+ tu
t
⇒ 0tS11 =
L
0L
0
0 0
ρ L = tρ tL
� 0 �2
0
L
L
t
τ11 = t tτ11
tL
L
L
1
∴ t tτ11 = Ẽ ·
L
2
��
t
t
u
1+ 0
L
ẼA
⇒ τ11 A = P =
2
t
(18.9)
��
(18.10)
�
�2
−1
t
u
1+ 0
L
(18.11)
��
�2
−1
t
u
1 +
0
L
�
(18.12)
This is because of the material-law assumption (18.5) (okay for small strains . . . )
Hyperelasticity
t
0W
= f (Green-Lagrange strains, material constants)
�
�
1 ∂0tW
∂0tW
t
+ t
0 Sij =
2 ∂ 0t�ij
∂ 0�ji
� t
�
∂ 0tSij
1 ∂ 0 Sij
C
=
+
0 ijrs
2 ∂ 0t�rs
∂ 0t�sr
77
(18.13)
(18.14)
(18.15)
MIT 2.094
18. Solution of F.E. equations
Plasticity
• yield criterion
• flow rule
• hardening rule
t
�
t−Δt
τ =
t
τ +
dτ
(18.16)
t−Δt
Solution of (18.1) (similarly (18.2) and (18.3))
Newton-Raphson Find U ∗ as the zero of f (U ∗ )
f (U ∗ ) =
R − t+ΔtF
�
�
�
�
�
∂f ��
t+Δt (i−1)
· U ∗ − t+ΔtU (i−1) + H.O.T.
=f
U
+
�
∂U t+ΔtU (i−1)
where
t+Δt
(18.17)
(18.18)
U (i−1) is the value we just calculated and an approximation to U ∗ .
t+Δt
Assume t+ΔtR is independent of the displacements.
�
� ∂ t+ΔtF ��
t+Δt
t+Δt (i−1)
�
0=
R−
F
−
· ΔU (i)
∂U � t+ΔtU (i−1)
(18.19)
We obtain
t+Δt
K (i−1) ΔU (i) =
t+Δt
K
(i−1)
t+Δt
R − t+ΔtF (i−1)
(18.20)
�
�
��
∂ t+ΔtF ��
∂F ��
=
=
∂U � t+ΔtU (i−1)
∂U � t+ΔtU (i−1)
(18.21)
Physically
t+Δt
(i−1)
K11
Δ
=
�
t+Δt
(i−1)
F1
�
(18.22)
Δu
78
MIT 2.094
18. Solution of F.E. equations
Pictorially for a single degree of freedom system
t
i = 1;
t+Δt
i = 2;
Convergence
K
K Δu(1) =
t+Δt
(1)
t+Δt
Δu
(2)
=
R − tF
R−
(18.23)
t+Δt
F
(1)
(18.24)
Use
�ΔU (i) �2 < �
��
2
�a�2 =
(ai )
(18.25)
(18.26)
i
But, if incremental displacements are small in every iteration, need to also use
� t+ΔtR − t+ΔtF (i−1) �2 < �R
18.1
(18.27)
Slender structures
(beams, plates, shells)
t
�1
Li
(18.28)
79
MIT 2.094
18. Solution of F.E. equations
Beam
e.g.
t
L
=
1
100
(4-node el.)
The element does not have curvature →
we have
a spurious shear strain
(9-node el.)
→ We do not have a shear (better)
→ But, still for thin structures, it has problems
like ill-conditioning.
⇒ We need to use beam elements. For curved structures also spurious membrane strain can be
present.
80
MIT OpenCourseWare
http://ocw.mit.edu
2.094 Finite Element Analysis of Solids and Fluids II
Spring 2011
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