2.094— Finite Element Analysis of Solids and Fluids
— Fall ‘08 —
MIT OpenCourseWare
Contents
1 Large displacement analysis of solids/structures
3
1.1
Project Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3
1.2
Large Displacement analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4
1.2.1
Mathematical model/problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4
1.2.2
Requirements to be fulfilled by solution at time t . . . . . . . . . . . . . . . . . . .
5
1.2.3
Finite Element Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5
1.2.4
Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
6
2 Finite element formulation of solids and structures
7
2.1
Principle of Virtual Work . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
8
2.2
Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
8
3 Finite element formulation for solids and structures
10
4 Finite element formulation for solids and structures
14
5 F.E. displacement formulation, cont’d
19
6 Finite element formulation, example, convergence
23
6.1
Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23
6.1.1
F.E. model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23
6.1.2
Higher-order elements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25
7 Isoparametric elements
28
8 Convergence of displacement-based FEM
33
9 u/p formulation
37
10 F.E. large deformation/general nonlinear analysis
41
11 Deformation, strain and stress tensors
45
12 Total Lagrangian formulation
49
1
MIT 2.094
Contents
13 Total Lagrangian formulation, cont’d
53
14 Total Lagrangian formulation, cont’d
57
15 Field problems
61
15.1 Heat transfer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61
15.1.1 Differential formulation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61
15.1.2 Principle of virtual temperatures . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62
15.2 Inviscid, incompressible, irrotational flow . . . . . . . . . . . . . . . . . . . . . . . . . . . . 64
16 F.E. analysis of Navier-Stokes fluids
65
17 Incompressible fluid flow and heat transfer, cont’d
71
17.1 Abstract body . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71
17.2 Actual 2D problem (channel flow) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71
17.3 Basic equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72
17.4 Model problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72
17.5 FSI briefly . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 75
18 Solution of F.E. equations
76
18.1 Slender structures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 79
19 Slender structures
81
20 Beams, plates, and shells
85
21 Plates and shells
90
2
2.094 — Finite Element Analysis of Solids and Fluids
Fall ‘08
Lecture 1 - Large displacement analysis of solids/structures
Prof. K.J. Bathe
1.1
MIT OpenCourseWare
Project Example
Physical problem
Reading:
Ch. 1 in
the text
“Simple” mathematical model
• analytical solution
• F.E. solution(s)
More complex mathematical model
• holes included
• large disp./large strains
• F.E. solution(s) ⇒
• How many finite elements?
We need a good error measure (especially for FSI)
“Even more complex” mathematical model
The “complex mathematical model” includes Fluid
Structure Interaction (FSI).
You will use ADINA in your projects (and homework) for structures and fluid flow.
3
MIT 2.094
1.2
1. Large displacement analysis of solids/structures
Large Displacement analysis
Lagrangian formulations:
• Total Lagrangian formulation
• Updated Lagrangian formulation
1.2.1
Reading:
Ch. 6
Mathematical model/problem
Given
Calculate
the
the
the
the
the
original configuration of the body,
support conditions,
applied external loads,
assumed stress-strain law
deformations, strains, stresses of the body.
Question Is there a unique solution? Yes, for infinitesimal small displacement/strain. Not necessarily
for large displacement/strain.
For example:
Snap-through problem
The same load. Two different deformed configurations.
4
MIT 2.094
1. Large displacement analysis of solids/structures
Column problem, statics
Not physical
1.2.2
t
R is in “direction” of bending moment ⇒ Not in
equilibrium.
Requirements to be fulfilled by solution at time t
I. Equilibrium of stresses (Cauchy stresses, forces per unit area in t V and on t Sf ) with the applied
body forces t f B and surface tractions t f Sf
II. Compatibility
III. Stress-strain law
1.2.3
Finite Element Method
I. Equilibrium condition means now
• equilibrium at the nodes of the mesh
• equilibrium of each finite element
II. Compatibility satisfied exactly
III. Stress-strain law satisfied exactly
5
MIT 2.094
1.2.4
1. Large displacement analysis of solids/structures
Notation
Cauchy stresses (force per unit area at time t):
t
τij
i, j = 1, 2, 3
t
τij = t τji
(1.1)
6
2.094 — Finite Element Analysis of Solids and Fluids
Fall ‘08
Lecture 2 - Finite element formulation of solids and structures
Prof. K.J. Bathe
MIT OpenCourseWare
Assume that on t Su the displacements are zero (and t Su is constant). Need to satisfy at time t:
• Equilibrium of Cauchy stresses t τij with applied loads
�
�
t T
τ = t τ11 t τ22 t τ33 t τ12 t τ23 t τ31
(2.1)
(For i = 1, 2, 3)
t
τij,j + t fiB
t
=
t
τij nj
=
t Sf
fi
=
(e.g.
0 in t V (sum over j)
(2.2)
t Sf
fi on t Sf (sum over j)
t
τi1 t n1 + t τi2 t n2 + t τi3 t n3
(2.3)
)
(2.4)
S
And: t τ11 t n1 + t τ12 t n2 = t f1 f
• Compatibility The displacements t ui need to be continuous and zero on t Su .
• Stress-Strain law
t
� �
τij = function t uj
(2.5)
7
Reading:
Ch. 1, Sec.
6.1-6.2
MIT 2.094
2. Finite element formulation of solids and structures
Principle of Virtual Work∗
2.1
�
t
τij t eij d t V =
�
tV
tV
t B
fi
ui d t V +
�
tS
t Sf
fi
S
u i f d t Sf
(2.6)
f
where
t eij
�
�
with ui �
t
=
1
2
�
∂ui
∂uj
+ t
∂ t xj
∂ xi
�
(2.7)
= 0
(2.8)
Su
2.2
Example
Assume “plane sections remain plane”
Principle of Virtual Work
�
�
t
t
τ11 t e11 d V =
tV
tV
t B
f1
t
�
t
u1 d V +
tS
S
Pr u 1 f d t S f
(2.9)
f
Derivation of (2.9)
t
�t
∗ or
τ11,1 + t f1B = 0
�
τ11,1 + t f1B u1 = 0
by (2.2)
(2.10)
(2.11)
Principle of Virtual Displacements
8
MIT 2.094
Hence,
�
2. Finite element formulation of solids and structures
�t
tV
�
τ11,1 + t f1B u1 d t V = 0
� tS
τ11 u1 � t Sf −
� �� u�
t
�
t
tV
S
u1 f tτ11 tSf
t
(2.12)
�
u1,1 τ d V +
���� 11
tV
u1 t f1B
d t V = 0
(2.13)
te11
where tτ11 | tS = tPr .
f
Therefore we have
�
�
t
t
e
τ
d
V
=
t 11
11
tV
S
u1 t f1B d t V + u1 f t Pr t Sf
tV
(2.14)
From (2.12) to (2.14) we simply used mathematics. Hence, if (2.2) and (2.3) are satisfied, then (2.14)
must hold. If (2.14) holds, then also (2.2) and (2.3) hold!
Namely, from (2.14)
�
�
� tS
u1,1 tτ11 d tV = u1 tτ11 � tSf −
u
tV
tV
u1 tτ11,1 d tV =
�
tV
S
u1 tf1B d tV + u1 f
tPr tSf
(2.15)
or
�
u1
tV
�t
�
�
S �
τ11,1 + tf1B d tV + u1 f tPr − tτ11 tSf = 0
�
Now let u1 = x 1 −
x
tL
��
(2.16)
�
τ11,1 + tf1B , where tL = length of bar.
t
Hence we must have from (2.16)
t
τ11,1 + tf1B = 0
(2.17)
and then also
t
Pr = tτ11
(2.18)
9
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)
13
2.094 — Finite Element Analysis of Solids and Fluids
Fall ‘08
Lecture 4 - Finite element formulation for solids and structures
Prof. K.J. Bathe
MIT OpenCourseWare
We considered a general 3D body,
Reading:
Ch. 4
The exact solution of the mathematical model must satisfy the conditions:
• Equilibrium within t V and on t Sf ,
• Compatibility
• Stress-strain law(s)
I. Differential formulation
II. Variational formulation (Principle of virtual displacements) (or weak formulation)
We developed the governing F.E. equations for a sheet or bar
We obtained
t
F = tR
(4.1)
where t F is a function of displacements/stresses/material law; and t R is a function of time.
Assume for now linear analysis: Equilibrium within 0 V and on 0 Sf , linear stress-strain law and small
displacements yields
t
F = K · tU
(4.2)
We want to establish,
KU (t) = R(t)
(4.3)
14
MIT 2.094
4. Finite element formulation for solids and structures
Consider
Û T =
�
U1
V1
W1
U2
· · · WN
�
(N nodes)
(4.4)
where Û T is a distinct nodal point displacement vector.
Note: for the moment “remove Su ”
We also say
�
Û T = U1
U2
U3
· · · Un
�
(n = 3N )
(4.5)
We now assume
⎤(m)
u
= ⎣ v ⎦
w
⎡
u(m) = H (m) Û ,
u(m)
(4.6a)
where H (m) is 3 x n and Û is n x 1.
�(m) = B (m) Û
(4.6b)
where B (m) is 6 x n, and
T
�(m) =
e.g. γxy
�
�xx �yy
∂v
∂u
=
+
∂x ∂y
�zz
γxy
γyz
γzx
�
We also assume
u(m)
�
(m)
=
H (m) Û
(4.6c)
=
(m)
Û
(4.6d)
B
15
MIT 2.094
4. Finite element formulation for solids and structures
Principle of Virtual Work:
�
�
T
T
� τ dV =
Û f B dV
V
(4.7)
V
(4.7) can be rewritten as
��
m
T
�(m) τ (m) dV (m) =
(m) T
��
V (m)
m
Û
fB
(m)
dV (m)
(4.8)
V (m)
Substitute (4.6a) to (4.6d).
�
�
�
�
T
(m) T
(m)
(m)
Û
B
τ
dV
=
V (m)
m
T
Û
�
��
H
(m) T
f
B (m)
(4.9)
�
dV
(m)
V (m)
m
τ (m) = C (m) �(m) = C (m) B (m) Û
Finally,
�
��
T
�
Û
�
B
(m) T
(4.10)
�
C
(m)
B
(m)
dV
(m)
Û =
V (m)
m
�
��
T
�
Û
�
m
H
(m) T
(4.11)
�
fB
(m)
dV (m)
V (m)
with
T
T
ˆ B (m)
�(m) = U
T
(4.12)
KÛ = RB
(4.13)
where K is n x n, and RB is n x 1.
Direct stiffness method:
�
K=
K (m)
(4.14)
m
RB =
�
(m)
RB
(4.15)
m
K
(m)
(m)
RB
�
T
B (m) C (m) B (m) dV (m)
=
�V
(m)
V
(m)
T
H (m) f B
=
(m)
(4.16)
dV (m)
(4.17)
16
MIT 2.094
4. Finite element formulation for solids and structures
Example 4.5 textbook
E = Young’s Modulus
Mathematical model
Plane sections remain plane:
F.E. model
⎡
⎤
U1
U = ⎣ U2 ⎦
U3
(4.18)
Element 1
⎡
u(1) (x) =
�
�
1−
x
100
x
�� 100
H (1)
⎤
U1
0 ⎣ U2 ⎦
� U
3
�
(4.19)
17
MIT 2.094
4. Finite element formulation for solids and structures
⎤
U1
0 ⎣ U2 ⎦
� U
3
(4.20)
x
80
(4.21)
⎡
�(1)
xx (x) =
�
1
− 100
�
1
��100
B (1)
�
Element 2
u(2) (x) =
�
0
�
�(2)
xx (x)
=
�
�
0
x
1 − 80
��
H (2)
1
1
− 80
80
��
B (2)
�
U
�
�
U
(4.22)
�
Then,
⎡
⎤
1 −1 0
0
13E ⎣
E ⎣
−1 1 0 ⎦ +
0
K=
240
100
0
0 0
0
⎡
⎤
0
0
1 −1 ⎦
−1 1
(4.23)
where,
�
�
A�
<
η=0
1
�
�
AE
E(1)
≡
L
100
� �
13 E
E · 13
=
3 80
3 · 80
� �� �
A∗
�
�
A∗
< A�
(4.24)
(4.25)
(4.26)
η=80
< 4.333 <
9
18
2.094 — Finite Element Analysis of Solids and Fluids
Fall ‘08
Lecture 5 - F.E. displacement formulation, cont’d
Prof. K.J. Bathe
MIT OpenCourseWare
For the continuum
Reading:
Ch. 4
• Differential formulation
• Variational formulation (Principle of Virtual Displacements)
Next, we assumed infinitesimal small displacement, Hooke’s Law, linear analysis
KU = R
(5.1a)
u(m) = H (m) U
K=
�
(5.1b)
K (m)
(5.1c)
(m)
(5.1d)
m
R=
�
RB
m
(m)
�(m) = B U
�
�
U T = U1 U2 · · · Un , (n = all d.o.f. of element assemblage)
�
T
K (m) =
B (m) C (m) B (m) dV (m)
(m)
�V
(m)
T
(m)
RB =
H (m) f B
dV (m)
(5.1e)
(5.1f)
(5.1g)
(5.1h)
V (m)
Surface loads
Recall that in the principle of virtual displacements,
�
Sf T
f Sf dSf
“surface” loads =
U
(5.2)
Sf
uS
H
(m)
S (m)
= HS
(m)
�
�
= H (m) �
U
(5.3)
(5.4)
evaluated at the surface
19
MIT 2.094
5. F.E. displacement formulation, cont’d
Substitute into (5.2)
�
(m) T
(m)
T
U
HS
fS
dS (m)
(5.5)
S (m)
for element (m) and one surface of that element.
Rs(m) =
�
HS
(m) T
fS
(m)
dS (m)
(5.6)
S (m)
Need to add contributions from all surfaces of all loaded external elements.
KU = RB + RS + Rc
(5.7)
where Rc are concentrated nodal loads.
Assume
• (5.7) has been established without any displacement boundary conditions.
• We, however, know nodal displacements Ub (rewriting (5.7)).
�
KU = R ⇒
Kaa
Kba
Kab
Kbb
��
Ua
Ub
�
�
=
Ra
Rb
�
(5.8)
Solve for Ua :
Kaa Ua = Ra − Kab Ub
(5.9)
where Ub is known!
Then use
Kba Ua + Kbb Ub = Rb + Rr
(5.10)
where Rr are unknown reactions.
Example 4.6 textbook
⎛
⎞
⎡
1
τxx
E
⎝ τyy ⎠ =
⎣ ν
1 − ν2
τxy
0
ν
1
0
0
0
1−ν
2
⎤⎛
⎞
�xx
⎦ ⎝ �yy ⎠
γxy
20
(5.11)
MIT 2.094
5. F.E. displacement formulation, cont’d
⎛
⎜
⎜
⎜
⎜
�
�
⎜
u(x, y)
=H⎜
⎜
v(x, y)
⎜
⎜
⎜
⎝
u1
u2
u3
u4
v1
v2
v3
v4
⎞
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎠
(5.12)
If we can set this relation up, then clearly we can get H (1) , H (2) , H (3) , H (4) .
u(m) = H (m) U
(5.13)
Also want �(m) = B (m) U . We want H. We could proceed this way
u(x, y) = a1 + a2 x + a3 y + a4 xy
(5.14)
v(x, y) = b1 + b2 x + b3 y + b4 xy
(5.15)
Express a1 . . . a4 , b1 . . . b4 in terms of the nodal displacements u1 . . . u4 , v1 . . . v4 .
(e.g.) u(1, 1) = a1 + a2 + a3 + a4 = u1 .
h1 (x, y) = 14 (1 + x)(1 + y) interpolation function
for node 1.
h2 (x, y) = 14 (1 − x)(1 + y)
21
MIT 2.094
5. F.E. displacement formulation, cont’d
h3 (x, y) = 14 (1 − x)(1 − y)
h4 (x, y) = 14 (1 + x)(1 − y)
u(x, y) = h1 u1 + h2 u2 + h3 u3 + h4 u4
(5.16)
v(x, y) = h1 v1 + h2 v2 + h3 v3 + h4 v4
(5.17)
⎛
�
u(x, y)
v(x, y)
�
�
=
h1
0
h2
0
�
h3
0
h4
0
0
h1
0
h2
0
h3
��
H (2x8)
⎜
⎜
⎜
�⎜
0 ⎜
⎜
h4 ⎜
�⎜
⎜
⎜
⎝
u1
u2
u3
u4
v1
v2
v3
v4
⎞
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎠
(5.18)
We also want,
⎛
⎛
⎞
⎡
�xx
h1,x
⎝ �yy ⎠ = ⎣ 0
γxy
h1,y
�
h2,x
0
h2,y
h3,x
0
h3,y
h4,x
0
h4,y
0
h1,y
h1,x
0
h2,y
h2,x
��
B (3x8)
∂u
∂x
∂v
=
∂y
∂u ∂v
+
=
∂x
∂y
0
h3,y
h3,x
0
h4,y
h4,x
⎜
⎜
⎤⎜
⎜
⎜
⎦⎜
⎜
⎜
�⎜
⎜
⎝
u1
u2
u3
u4
v1
v2
v3
v4
⎞
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎠
(5.19)
�xx =
(5.20)
�yy
(5.21)
γxy
(5.22)
22
2.094 — Finite Element Analysis of Solids and Fluids
Fall ‘08
Lecture 6 - Finite element formulation, example, convergence
Prof. K.J. Bathe
6.1
MIT OpenCourseWare
Example
Reading:
Ex. 4.6 in
the text
t = 0.1,
KU = R;
�
K=
K (m) ;
R = RB + R s + R c + R r
�
T
(m)
K
=
B (m) C (m) B (m) d V (m)
�
(m)
RB ;
m
6.1.1
�
�
K�
(6.1)
(6.2)
V (m)
m
RB =
E, ν plane stress
(m)
RB
�
T
H (m) f B(m) d V (m)
=
(6.3)
V (m)
F.E. model
el. (2)
u1 u2 u3 u4
↓
↓
↓
⎡↓
� � � ×
= ⎢ .
⎢ ..
⎢
⎣
v1 v2 v3 v4
↓ ↓ ↓ ↓⎤
× × × ×
⎥
⎥
⎥
⎦
23
← u1
..
.
(6.4)
MIT 2.094
6. Finite element formulation, example, convergence
In practice,
⎛
⎞
u1
⎜ .. ⎟
⎜ . ⎟
⎜
⎟
⎜ u4 ⎟
⎜
⎟
�=B⎜
⎟
⎜ v1 ⎟
⎜ . ⎟
⎝ .. ⎠
v4
�
�
�
K� =
B T CB dV ;
el
V
(6.5)
where K is 8x8 and B is 3x8.
Assume we have K (8x8) for el. (2)
U1 U2 U3 U4 U5 · · · U11
↓
↓
↓
↓
↓
↓
↓
⎡
× × × × × × ×
..
..
..
⎢
⎢
.
.
.
⎢
⎢
..
..
..
⎢
.
.
.
⎢
K
= ⎢
����
�
�
�
⎢
assemblage
⎢
..
..
..
⎢
.
.
.
⎢
⎢
.
.
..
.
.
⎢
.
.
.
⎢
⎣
×
×
×
×
×
×
×
· · · U18
↓
↓
⎤
× ×
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎦
×
×
← U1
..
.
..
.
← U11
..
.
..
.
← U18
(6.6)
Consider,
�
�
�
HS = H�
T
H S f S d S;
RS =
S
�
H=
⎛
⎜
⎜
⎜
⎜
⎜
u=⎜
⎜
⎜
⎜
⎜
⎝
h1
0
u1
u2
..
.
u4
v1
..
.
h2
0
h3
0
h4
0
0
h1
0
h2
(6.7)
on surface
0
h3
0
h4
�
← u(x, y)
← v(x, y)
(6.8)
⎞
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎠
(6.9)
v4
24
MIT 2.094
6. Finite element formulation, example, convergence
�
�
HS = H�
y=+1
⎡ 1
2 (1 + x)
=⎣
0
(6.10)
1
2 (1
− x) 0
0
0
0
0
0
1
2 (1
0
+ x)
1
2 (1
0
0
⎤
⎦
− x)
0
(6.11)
0
From (6.7);
⎡
�
+1
RS =
−1
⎢
⎢
⎣
1
2 (1
1
2 (1
+ x)
− x)
0
0
⎤
0
�
�
⎥
0
0
⎥
(0.1) dx
1
⎦ −p(x)
� �� �
2 (1 + x)
1
thickness
(1
−
x)
2
(6.12)
⎡
⎤
0
⎢
⎥
0
⎥
RS = ⎢
⎣ −p0 (0.1)
⎦
−p0 (0.1)
6.1.2
(6.13)
Higher-order elements
Want h1 , h2 , h3 , h4 , h5
u(x, y) =
�5
i=1
hi u i .
hi = 1 at node i and 0 at all other nodes.
h5 = 12 (1 − x2 )(1 + y)
25
MIT 2.094
6. Finite element formulation, example, convergence
1
h5
2
1
h5
2
1
(1 + x)(1 + y) −
4
1
h2 = (1 − x)(1 + y) −
4
1
h3 = (1 − x)(1 − y)
4
1
h1 = (1 + x)(1 − y)
4
h1 =
(6.14)
(6.15)
(6.16)
(6.17)
Note:
�
We must have
u(x, y) =
�
�
i
hi = 1
hi = 1 to satisfy the rigid body mode condition.
h i ui
(6.18)
i
Assume all nodal point displacements = u∗ . Then,
u(x, y) =
�
h i u ∗ = u∗
�
i
h i = u∗
(6.19)
i
From (6.1),
�
�
�
K
(m)
U =R
(6.20)
U =R
(6.21)
m
� ��
B
(m) T
C
(m)
B
(m)
dV
(m)
�
V (m)
m
where C (m) B (m) U = τ (m) . (Assume we calculated U .)
��
m
�
T
B (m) τ (m) d V (m) = R
V
F
(6.22)
(m)
(m)
= R;
m
F
(m)
�
T
B (m) τ (m) d V (m)
=
V
(m)
Two properties
I. The sum of the F (m) ’s at any node is equal to the applied external forces.
26
(6.23)
MIT 2.094
6. Finite element formulation, example, convergence
II. Every element is in equilibrium under its F (m)
ˆ T F (m) = U
ˆT
U
�
�
T
V
B (m) τ (m) d V (m)
��
�
=�(m)
�
=
(6.24)
(m)
T
T
�(m) τ (m) d V (m)
(6.25)
V (m)
=0
(6.26)
ˆ T = virtual nodal point displacement.
where U
Apply rigid body displacement.
If we move the element virtually in the rigid body modes, �(m) is zero. Therefore the virtual work
obtained due to virtual motion of the element is zero. Then the element is in equilibrium under its F (m) .
27
2.094 — Finite Element Analysis of Solids and Fluids
Fall ‘08
Lecture 7 - Isoparametric elements
Prof. K.J. Bathe
We want K =
MIT OpenCourseWare
�
V
B T CB dV , RB =
�
V
Reading:
Sec. 5.1-5.3
H T f B dV . Unique correspondence (x, y) ⇔ (r, s)
(r, s) are natural coordinate system or isoparametric coordinate system.
x=
4
�
hi xi
(7.1)
hi yi
(7.2)
i=1
y=
4
�
i=1
where
1
(1 + r)(1 + s)
4
1
h2 = (1 − r)(1 + s)
4
...
h1 =
u(r, s) =
4
�
(7.3)
(7.4)
hi ui
(7.5)
hi vi
(7.6)
i=1
v(r, s) =
4
�
i=1
28
MIT 2.094
7. Isoparametric elements
� = Bû ûT =
∂u
∂x
⎡
⎢
⎢
⎢
⎢
⎣
⎛
⎝
+
∂
∂r
∂
∂s
⎝
∂
∂x
∂
∂y
· · · v4
�
(7.7)
⎥
⎥
⎥ = Bû
⎥
⎦
∂v
∂x
⎞
⎡
(7.8)
∂x
∂r
∂y
∂r
∂x
∂s
∂y
∂s
⎠=⎣
�
⎛
u2
u1
⎤
∂v
∂y
∂u
∂y
�
��
∂
∂r
⎛
⎠ = J −1 ⎝
⎦⎝
∂
∂x
∂
∂y
⎞
(7.9)
⎠
�
J
⎞
⎤⎛
∂
∂s
⎞
(7.10)
⎠
J must be non-singular which ensures that there is unique correspondence between (x, y) and (r, s).
Hence,
� 1� 1
K=
B T CB t det(J ) dr ds
(7.11)
�
��
�
−1 −1
dV
�
1
�
1
Also, RB =
−1
H T f B t det(J ) dr ds
(7.12)
−1
Numerical integration (Gauss formulae) (Ch. 5.5)
K∼
=t
��
i
T
Bij
CBij det(Jij ) × (weight i, j)
(7.13)
j
2x2 Gauss integration,
(i = 1, 2)
(j = 1, 2) (weight i, j = 1 in this case)
(7.14)
(7.15)
29
MIT 2.094
7. Isoparametric elements
9-node element
x=
9
�
hi xi
(7.16)
hi yi
(7.17)
hi u i
(7.18)
hi vi
(7.19)
i=1
y=
9
�
i=1
u=
9
�
i=1
v=
9
�
i=1
Use 3x3 Gauss integration
For rectangular elements, J = const
Consider the following element,
Note, here we could use hi (x, y) directly.
�
J=
�
�
3 = 26
0
0
�
(7.20)
3
2
Then, we can determine the number of appropriate integration points by investigating the maximum
order of B T CB.
For a rectangular element, 3x3 Gauss integration gives exact K matrix. If the element is distorted,
a K matrix which is still accurate enough will be obtained, (if high enough integration is used).
30
MIT 2.094
7. Isoparametric elements
Convergence Principle of virtual work:
�
�T C� dV = R(u)
(7.21)
V
Find u, solution, in V, vector space (any continuous function that satisfies boundary conditions),
satisfying
�
�T C� dV = a(u, v) = (f , v) for all v, an element of V.
(7.22)
� �� �
� �� �
V
bilinear form
R(v)
Example:
Finite Element problem
Find uh ∈ Vh , where Vh is F.E. vector space such that
a(uh , vh ) = (f , vh ) ∀vh ∈ Vh
(7.23)
Size of Vh ⇒ # of independent DOFs (here it’s 12).
Note:
a(w, w)
� �� �
> 0 for w ∈ V
� 0)
(w =
2x (strain energy when imposing w)
Also,
a(wh , wh ) > 0 for wh ∈ Vh
(Vh ⊂ V, wh �= 0)
31
Reading:
Sec. 5.5.5,
4.3
MIT 2.094
Property I
7. Isoparametric elements
Define: eh = u − uh .
From (7.22), a(u, vh ) = (f , vh )
(7.24)
From (7.23), a(uh , vh ) = (f , vh )
(7.25)
Hence,
a(u − uh , vh ) = 0
(7.26)
a(eh , vh ) = 0
(7.27)
(error is orthogonal in that sense to all vh in F.E. space).
Property II
a(uh , uh ) ≤ a(u, u)
(7.28)
Proof:
a(u, u) = a(uh + eh , uh + eh )
(7.29)
� 0 by Prop. I
��
2a(u
= a(uh , uh ) + �
��
h , eh )
+ a(eh , eh )
� �� �
(7.30)
≥0
∴ a(u, u) ≥ a(uh , uh )
(7.31)
32
2.094 — Finite Element Analysis of Solids and Fluids
Fall ‘08
Lecture 8 - Convergence of displacement-based FEM
Prof. K.J. Bathe
MIT OpenCourseWare
(A) Find
u ∈ V such that a(u, v) = (f , v) ∀v ∈ V (Mathematical model)
(8.1)
a(v, v) > 0 ∀v ∈ V,
(8.2)
v=
� 0.
where (8.2) implies that structures are supported properly. E.g.
(B) F.E. Problem
Find
uh ∈ Vh such that a(uh , vh ) = (f , vh ) ∀vh ∈ Vh
a(vh , vh ) > 0
∀vh ∈ Vh ,
vh �= 0
(8.3)
(8.4)
Properties eh = u − uh
(I) a(eh , vh ) = 0
∀vh ∈ Vh
(8.5)
(II) a(uh , uh ) ≤ a(u, u)
(8.6)
33
MIT 2.094
8. Convergence of displacement-based FEM
�
�
(C) Assume Mesh�
�
�
e.g. Mesh�
h1
h1
�
�
“is contained in” Mesh�
�
�
not contained in Mesh�
h2
h2
We assume (C), but need another property (independent of (C))
(III) a(eh , eh ) ≤ a(u − vh , u − vh ) ∀vh ∈ Vh
(8.7)
uh minimizes! (Recall eh = u − uh )
Proof: Pick wh ∈ Vh .
�0
��
a(eh + wh , eh + wh ) = a(eh , eh ) + �
2a(e
+ a(wh , wh )
��
h , wh )
� �� �
(8.8)
≥0
Equality holds for (wh = 0)
a(eh , eh ) ≤ a(eh + wh , eh + wh )
= a(u − uh + wh , u − uh + wh )
(8.9)
(8.10)
Take wh = uh − vh .
a(eh , eh ) ≤ a(u − vh , u − vh )
(8.11)
Using property (III) and (C), we can say that we will converge monotonically, from below, to a(u, u):
34
MIT 2.094
8. Convergence of displacement-based FEM
Pascal triangle (2D)
⇒ 4-node element is complete to k = 1.
⇒ 9-node element is complete to k = 2.
(Ch. 4.3)
error in displacement ∼ C · hk+1
(8.12)
(C is a constant determined by the exact solution, material property. . . )
error in stresses ∼ C · hk
error in strain energy ∼ C · h2k
(8.13)
(← these C are different)
(8.14)
Hence,
E − Eh = C · h2k
(roughly equal to)
(8.15)
By theory,
log (E − Eh ) = log C + 2k log h
(8.16)
35
MIT 2.094
8. Convergence of displacement-based FEM
By experiment, we can evaluate log(E − Eh ) for different meshes and plot log(E − Eh ) vs. log h
We need to use graded meshes if we have high stress gradients.
Example
Consider an almost incompressible material:
�V = vol. strain
(8.17)
� · v → very small or zero
(8.18)
or
We can “see” difficulties:
p = −κ�V
κ = bulk modulus
(8.19)
As the material becomes incompressible (ν = 0.3 → 0.4999)
�
κ→∞
p → finite number
�V → 0
(8.20)
(Small error in �V results in huge error on pressure as κ → ∞, the constant C in (8.15) can be very large
⇒ locking)
36
2.094 — Finite Element Analysis of Solids and Fluids
Fall ‘08
Lecture 9 - u/p formulation
Prof. K.J. Bathe
MIT OpenCourseWare
We want to solve
Reading:
Sec. 4.4.3
I. Equilibrium
�
τij,j + fiB = 0 in Volume
S
τij nj = fi f
on Sf
(9.1)
II. Compatibility
III. Stress-strain law
Use the principle of virtual displacements
�
�T C� dV = R
(9.2)
V
We recognize that if ν → 0.5
�V → 0
(�V = �xx + �yy + �zz )
E
→∞
κ=
3(1 − 2ν)
p = −κ�V must be accurately computed
(9.3)
(9.4)
(9.5)
Solution
τij = κ�V δij + 2G��ij
(9.6)
where
�
1
δij = Kronecker delta =
0
i=j
i=
� j
(9.7)
Deviatoric strains:
�V
��ij = �ij −
δij
3
τij = −pδij + 2G��ij
(9.8)
�
p=−
τkk �
3
(9.9)
(9.2) becomes
�
�
�T � �
� C � dV +
�V κ�V dV = R
V
V
�
�
T
�� C � �� dV −
�TV p dV = R
V
(9.10)
(9.11)
V
37
MIT 2.094
9. u/p formulation
We need another equation because we now have another unknown p.
p + κ�V = 0
(9.12)
p (p + κ�V ) dV = 0
�V �
p�
−
p �V +
dV = 0
κ
V
(9.13)
�
(9.14)
For an element,
u = Hû
(9.15)
�
� = BD û
�V = BV û
p = Hp p̂
(9.16)
(9.17)
(9.18)
Plane strain (�zz = 0)
4/1 element
Example
�V = �xx + �yy
⎡
�xx − 31 (�xx + �yy )
⎢ �yy − 1 (�xx + �yy )
3
�� = ⎢
⎣
γxy
− 13 (�xx + �yy )
Reading:
Ex. 4.32 in
the text
(9.19)
⎤
⎥
⎥
⎦
(9.20)
� 0!
Note: �zz = 0 but ��zz =
p = Hp p̂ = [1]{p0 }
(9.21)
p(x, y) = p0
(9.22)
We obtain from (9.11) and (9.14)
�
��
� �
�
Kuu Kup
û
R
=
Kpu Kpp
p̂
0
(9.23)
�
T
�
BD
C BD dV
V
�
=−
BVT Hp dV
V
�
=−
HpT BV dV
V
�
1
=−
HpT Hp dV
κ
V
Kuu =
(9.24a)
Kup
(9.24b)
Kpu
Kpp
(9.24c)
(9.24d)
38
MIT 2.094
9. u/p formulation
In practice, we use elements that use pressure interpolations per element, not continuous between
elements. For example:
Then, unless ν = 0.5 (where Kpp = 0), we can use static condensation on the pressure dof’s.
Use p̂ equations to eliminate p̂ from the û equations.
�
�
−1
Kuu − Kup Kpp
Kpu û = R
(9.25)
(In practice, ν can be 0.499999. . . )
The “best element” is the 9/3 element. (9 nodes for displacement and 3 pressure dof’s).
p(x, y) = p0 + p1 x + p2 y
(9.26)
The inf-sup condition
Reading:
Sec. 4.5
⎡
⎤
=�V
� �� �
�
⎢
⎥
⎢ Vol qh � · vh dVol ⎥
inf
sup
⎢
⎥≥β>0
���� ���� ⎣
�qh � �vh �
⎦
� �� �
qh ∈Qh vh ∈Vh
(9.27)
for normalization
Qh : pressure space.
If “this” holds, the element is optimal for the displacement assumption used (ellipticity must also be
satisfied).
Note:
infimum = largest lower bound
supremum = least upper bound
For example,
inf {1, 2, 4} = 1
sup {1, 2, 4} = 4
inf {x ∈ R; 0 < x < 2} = 0
sup {x ∈ R; 0 < x < 2} = 2
(9.23) rewritten (κ = ∞, full incompressibility). Diagonalize using eigenvalues/eigenvectors.
For a mesh of element size h we want βh > 0 as we refine the mesh, h → 0
39
MIT 2.094
9. u/p formulation
For
(entry [3,1] in matrix) assume the circled entry is the minimum (inf) of
.
Also, all entries in the matrix not shown are zero.
Case 1 βh = 0
⎧
0 · uh |i = 0 (from the bottom equation)
⎨
⇒
α
·
u
|
+
0
· ph |j = Rh |i (from the top equation)
⎩ ���� h i
�
=0
⇒ no equation for ph |j
⇒ spurious pressure! (any pressure satisfies equation)
Case 2 βh = small = �
� · uh |i = 0
∴�·
⇒ uh |i = 0
ph |j + uh |i · α = Rh |i
Rh |i
⇒ ph |j =
�
�
⇒
displ. =
pressure →
0
large
�
as � is small
The behavior of given mesh when bulk modulus increases: locking, large pressures. See Example
4.39 textbook.
40
2.094 — Finite Element Analysis of Solids and Fluids
Fall ‘08
Lecture 10 - F.E. large deformation/general nonlinear analysis
Prof. K.J. Bathe
MIT OpenCourseWare
We developed
�
t
τij teij d tV = tR
Reading:
Ch. 6
(10.1)
tV
1
teij =
2
�
�
∂ui
∂uj
+ t
∂ txj
∂ xi
�
(10.2)
τij δt eij d tV = tR
t
tV
1
δt eij =
2
�
∂(δui ) ∂(δuj )
+
∂ t xj
∂ t xi
(10.3)
�
(≡ t eij )
(10.4)
In FEA:
t
F = tR
(10.5)
In linear analysis
t
F = K t U ⇒ KU = R
(10.6)
In general nonlinear analysis, we need to iterate. Assume the solution is known “at time t”
t
x=
0
x + tu
(10.7)
Hence t F is known. Then we consider
t+Δt
F =
t+Δt
R
(10.8)
Consider the loads (applied external loads) to be deformation-independent, e.g.
41
MIT 2.094
10. F.E. large deformation/general nonlinear analysis
Then we can write
t+Δt
F = tF + F
(10.9)
t+Δt
U = tU + U
(10.10)
where only t F and t U are known.
F ∼
= tK ΔU ,
t
K = tangent stiffness matrix at time t
(10.11)
From (10.8),
t
K ΔU =
t+Δt
R − tF
(10.12)
We use this to obtain an approximation to U . We obtain a more accurate solution for U (i.e.
using
t+Δt
K (i−1) ΔU (i) =
t+Δt
U
(i)
=
t+Δt
R − t+ΔtF (i−1)
t+Δt
U
(i−1)
+ ΔU
t+Δt
F)
(10.13)
(i)
(10.14)
Also,
t+Δt
F (0) = tF
t+Δt
K
(0)
(10.15)
t
= K
(10.16)
U (0) = tU
(10.17)
t+Δt
Iterate for i = 1, 2, 3 . . . until convergence. Convergence is reached when
�
�
�
�
�ΔU (i) � < �D
�
�2
� t+Δt
t+Δt (i−1) �
R−
F
�
� < �F
2
Note:
�a�2 =
��
2
(ai )
i
�
ΔU
(i)
=U
i=1,2,3...
ΔU (1) in (10.13) is ΔU in (10.12).
(10.13) is the full Newton-Raphson iteration.
How we could (in principle) calculate t K
Process
• Increase the displacement t Ui by �, with no increment for all t Uj , j �= i
• calculate
t+�
F
• the i-th column in tK = ( t+�F − tF ) /� =
∂ tF
.
∂ tUi
42
(10.18)
(10.19)
MIT 2.094
10. F.E. large deformation/general nonlinear analysis
So, perform
Pictorially,
⎡
..
⎢ .
⎢
t
K = ⎢ ...
⎣
..
.
this process for i = 1, 2, 3, . . . , n, where n is the total number of degrees of freedom.
..
.
..
. ···
..
.
⎤
⎥
⎥
⎥
⎦
A general difficulty: we cannot “simply” increment Cauchy stresses.
•
t+Δt
τij referred to area at time t + Δt
• t τij referred to area at time t.
We define a new stress measure, 2nd Piola - Kirchhoff stress,
refers to original configuration. Then,
t+Δt
0 Sij
= 0tSij + 0 Sij
t+Δt
0 Sij ,
where 0 in the leading subscript
(10.20)
The strain measure energy-conjugate to the 2nd P-K stress 0tSij is the Green-Lagrange strain 0t�ij
Then,
�
0V
t
0 Sij
Also,
�
0V
δ 0t�ij d 0V = tR
t+Δt
0 Sij
δ t+Δt0�ij d 0V =
(10.21)
t+Δt
R
(10.22)
Example
43
MIT 2.094
10. F.E. large deformation/general nonlinear analysis
t
F = tR
t+Δt
F =
(10.23)
t+Δt
R
(10.24)
(every time it is in equilibrium)
(10.13) and (10.14) give:
i = 1,
t+Δt
K (0) ΔU (1) =
t+Δt
R − t+ΔtF (0) ≡ fn( tU )
=
t+Δt
K (1) ΔU (2) =
t+Δt
U
(1)
t+Δt
(10.25)
(1)
(10.26)
R − t+ΔtF (1)
(10.27)
U
(0)
+ ΔU
i = 2,
t+Δt
t+Δt
U
(2)
=
t+Δt
U
(1)
+ ΔU
(2)
(10.28)
44
2.094 — Finite Element Analysis of Solids and Fluids
Fall ‘08
Lecture 11 - Deformation, strain and stress tensors
Prof. K.J. Bathe
MIT OpenCourseWare
We stated that we use
�
�
t
t
τij δ teij d V =
tV
Reading:
Ch. 6
t
0 Sij
0V
δ 0t�ij d 0V = tR
(11.1)
The deformation gradient We use txi = 0xi + tui
⎤
⎡ ∂ tx
∂ tx
∂ tx
1
⎢
⎢
⎢
t
⎢
X
=
0
⎢
⎢
⎣
1
1
∂ 0x1
∂ 0x2
∂ 0x3
t
t
t
∂ x2
∂ 0x1
∂ x2
∂ 0x2
∂ x2
∂ 0x3
∂ tx3
∂ 0x1
∂ tx3
∂ 0x2
∂ tx3
∂ 0x3
⎥
⎥
⎥
⎥
⎥
⎥
⎦
(11.2)
⎤
d tx1
d tx = ⎣ d tx2 ⎦
d tx3
⎡ 0
⎤
d x1
d 0x = ⎣ d 0x2 ⎦
d 0x3
⎡
(11.3)
(11.4)
Implies that
d tx = 0tX d 0x
(11.5)
(0tX is frequently denoted by 0tF or simply F , but we use F for
force vector)
We will also use the right Cauchy-Green deformation tensor
t
0C
= 0tX T 0tX
Some applications
45
(11.6)
MIT 2.094
11. Deformation, strain and stress tensors
The stretch of a fiber ( tλ):
� t �2
� t �2
d txT d tx
ds
λ = 0 T 0 = 0
d x d x
d s
(11.7)
The length of a fiber is
�
�1
d 0s = d 0xT d 0x 2
(11.8)
� 0 T t T��t
�
0
� t �2
d x 0X
0X d x
λ =
,
d 0s · d 0s
from (11.5)
(11.9)
Express
� �
d 0x = d 0s 0n
(11.10)
0
0
n = unit vector into direction of d x
� t �2 0 T t 0
⇒ λ = n 0 C n
∴ tλ =
(11.11)
(11.12)
� 0 T t 0 � 12
n 0C n
(11.13)
Also,
�
d tx̂
�T � t � � t � � t �
· d x = d ŝ d s cos tθ,
From (11.5),
�
ˆT
d 0x̂T 0tX
cos tθ =
��
t
0X
d 0x
d tsˆ d ts
0 0 T t 0
d ŝ n̂ 0 C n d 0s
=
d tsˆ · d ts
(a · b = �a��b� cos θ)
(11.14)
�
�
t ˆ
0X
≡ 0tX
�
(11.15)
(11.16)
0
∴ cos tθ =
n̂T 0tC 0n
(11.17)
tλ̂ tλ
Also,
0
t
ρ=
ρ
det 0tX
(see Ex. 6.5)
(11.18)
Example
Reading:
Ex. 6.6 in
the text
46
MIT 2.094
11. Deformation, strain and stress tensors
1
(1 + 0x1 )(1 + 0x2 )
4
h1 =
(11.19)
..
.
t
xi = 0xi + tui
=
4
�
hk txki ,
(11.20)
(i = 1, 2)
(11.21)
k=1
where txki are the nodal point coordinates at time t ( tx11 = 2, tx12 = 1.5)
Then we obtain
�
1
5 + 0x2
t
1
0X =
4 2 (1 + 0x2 )
At 0x1 = 0, 0x2 = 0,
�
�
1 5
�
t
=
0 X �0
4 12
xi = 0x2 =0
1 + 0x1
1
0
2 (9 + x1 )
1
�
(11.22)
�
(11.23)
9
2
The Green-Lagrange Strain
t
0�
=
� 1 �t
�
1 �t T t
0X 0X − I =
0C − I
2
2
∂
∂ txi
=
∂ 0xj
(11.24)
�0
�
xi + tui
∂ tui
=
δ
+
ij
∂ 0xj
∂ 0xj
We find that
�
1 �t
t
t
t
t
0�ij =
0ui,j + 0uj,i + 0uk,i 0uk,j ,
2
(11.25)
sum over k = 1, 2, 3
(11.26)
where
t
0ui,j
=
∂ tui
∂ 0xj
(11.27)
47
MIT 2.094
11. Deformation, strain and stress tensors
Polar decomposition of 0tX
t
0X
= 0tR 0tU
(11.28)
where 0tR is a rotation matrix, such that
t T t
0R 0R
=I
(11.29)
and 0tU is a symmetric matrix (stretch)
Ex. 6.9 textbook
�
t
0X
=
√
3
2
1
2
1
−
√2
3
2
��
4
3
0
0
�
(11.30)
3
2
Then,
� �2
= 0tX T 0tX = 0tU
�
1 �� t �2
t
−I
0� =
0U
2
t
0C
(11.31)
(11.32)
This shows, by an example, that the components of the Green-Lagrange strain are independent of a
rigid-body rotation.
48
2.094 — Finite Element Analysis of Solids and Fluids
Fall ‘08
Lecture 12 - Total Lagrangian formulation
Prof. K.J. Bathe
We discussed:
�
MIT OpenCourseWare
�
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)
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)
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
2.094 — Finite Element Analysis of Solids and Fluids
Fall ‘08
Lecture 15 - Field problems
Prof. K.J. Bathe
MIT OpenCourseWare
Heat transfer, incompressible/inviscid/irrotational flow, seepage flow, etc.
Reading:
Sec. 7.2-7.3
• Differential formulation
• Variational formulation
• Incremental formulation
• F.E. discretization
15.1
Heat transfer
Assume V constant for now:
S = Sθ ∪ Sq
θ(x, y, z, t) is unkown except θ|Sθ = θpr . In addition, q s
|Sq is also prescribed.
15.1.1
Differential formulation
I. Heat flow equilibrium in V and on Sq .
∂θ
II. Constitutive laws qx = −k ∂x
.
∂θ
qy = −k
∂y
∂θ
qz = −k
∂z
(15.1)
(15.2)
III. Compatibility: temperatures need to be continuous and satisfy the boundary conditions.
61
MIT 2.094
15. Field problems
Heat flow equilibrium gives
�
�
�
�
�
�
∂
∂θ
∂
∂θ
∂
∂θ
k
+
k
+
k
= −q B
∂x
∂x
∂y
∂y
∂z
∂z
(15.3)
where q B is the heat generated per unit volume. Recall 1D case:
unit cross-section
dV = dx · (1)
(15.4)
q |x − q |x+dx + q B dx = 0
�
∂qx
q|x − q|x +
dx + q B dx = 0
∂x
(15.5)
�
−
�
∂
∂x
∂
∂x
−k
�
k
∂θ
∂x
∂θ
∂x
�
�
(15.6)
� + qB �
�=0
dx
dx
�
(15.7)
= −q B
(15.8)
We also need to satisfy
∂θ
= qS
∂n
k
(15.9)
on Sq .
15.1.2
�
θ
Principle of virtual temperatures
∂
∂x
�
∂θ
k
∂x
�
+ ··· + q
B
�
=0
(15.10)
�
( θ�S = 0 and θ to be continuous.)
θ
�
�
θ
V
∂
∂x
�
�
�
∂θ
k
+ · · · + q B dV = 0
∂x
(15.11)
62
MIT 2.094
15. Field problems
Transform using divergence theorem (see Ex 4.2, 7.1)
�
�
�
�T
Sq
�
B
θ
kθ dV =
θq dV +
θ q S dSq
����
V
V
heat flow
⎛
∂θ
∂x
⎞
⎜
⎜
θ =⎜
⎜
⎝
∂θ
∂y
⎟
⎟
⎟
⎟
⎠
�
(15.12)
Sq
(15.13)
∂θ
∂z
⎡
k
k=⎣ 0
0
0
k
0
⎤
0
0 ⎦
k
(15.14)
Convection boundary condition
�
�
qS = h θe − θS
(15.15)
where θe is the given environmental temperature.
Radiation
�
� �4 �
4
q S = κ∗ (θr ) − θS
�
� �2 � � r
��
�
2
= κ∗ (θr ) + θS
θ + θS θr − θS
�
�
= κ θr − θS
(15.16)
(15.17)
(15.18)
where κ = κ(θS ) and θr is given temperature of source. At time t + Δt
�
�
�
� T t+Δt t+Δt �
S
t+Δt B
θ
k
θ dV =
θ
q dV +
θ t+Δtq S dSq
V
V
t+Δt
θ = tθ + θ
Let
t+Δt (i)
or
θ
t+Δt (0)
with
θ
=
t+Δt (i−1)
θ
(15.19)
Sq
(15.20)
+ Δθ
(i)
(15.21)
t
= θ
(15.22)
From (15.19)
�
�T
(i)
θ t+Δtk(i−1) Δθ � dV
V
�
�
�T
t+Δt � (i−1)
=
θ t+Δtq B dV −
θ t+Δtk(i−1)
θ
dV
V
V
�
�
��
�
(i−1)
(i)
S
+
θ t+Δth(i−1) t+Δtθ e − t+Δtθ S
+ ΔθS
dSq
(15.23)
Sq
where the ΔθS
(i)
term would be moved to the left-hand side.
We considered the convection conditions
�
�
�
S
θ t+Δth t+Δtθ e − t+Δtθ S dSq
(15.24)
Sq
The radiation conditions would be included similarly.
63
MIT 2.094
15. Field problems
F.E. discretization
θ = H1x4 · t+Δtθ̂4x1
t+Δt
t+Δt �
θ2x1
t+Δt S
for 4-node 2D planar element
(15.25)
= B2x4 · t+Δtθ̂4x1
S
θ =H ·
(15.26)
t+Δt
θ̂
(15.27)
For (15.23)
�
⎛
⎞
�
gives
(i)
θ t+Δtk(i−1) Δθ � dV =⇒ ⎝ ����
B T t+Δtk(i−1) ����
B dV ⎠ Δ ����
θ̂ (i)
� �� �
V
�T
V
�
4x2
θ t+Δtq B dV ⇒
V
�
2x4
2x2
(15.28)
4x1
H T t+Δtq B dV
(15.29)
V
�
θ
� T t+Δt (i−1) t+Δt � (i−1)
k
θ
��
dV ⇒
V
B T t+Δtk(i−1) BdV
�
V
�
θ
S T t+Δt (i−1)
h
�
t+Δt e
θ −
�
t+Δt S (i−1)
θ
t+Δt (i−1)
�
+ ΔθS
(i)
��
dSq
θ̂
��
known
(15.30)
�
=⇒
Sq
⎛
�
⎞⎞
(15.31)
T
Sq
15.2
⎛
S t+Δt (i−1)
h
H S ⎝ t+Δtθ̂ e − ⎝ t+Δtθ̂ (i−1) +Δ ����
H
θ̂ (i) ⎠⎠ dSq
� �� �
����
� �� �
� �� �
4x1
1x4
4x1
4x1
4x1
Inviscid, incompressible, irrotational flow
2D case: vx , vy are velocities in x and y directions.
�·v =0
∂vx
∂vy
+
=0
∂x
∂y
∂vx
∂vy
−
=0
∂y
∂x
or
Reading:
Sec. 7.3.2
(15.32)
(incompressible)
(15.33)
(irrotational)
(15.34)
Use the potential φ(x, y),
vx =
⇒
∂φ
∂x
vy =
∂φ
∂y
(15.35)
∂2φ ∂2φ
+ 2 = 0 in V
∂x2
∂y
(15.36)
(Same as the heat transfer equation with k = 1, q B = 0)
64
2.094 — Finite Element Analysis of Solids and Fluids
Fall ‘08
Lecture 16 - F.E. analysis of Navier-Stokes fluids
Prof. K.J. Bathe
MIT OpenCourseWare
Incompressible flow with heat transfer
Reading:
Sec.
7.1-7.4,
Table 7.3
We recall heat transfer for a solid:
Governing differential equations
(kθ,i ),i + q B = 0
in V
(16.1)
�
�
θ�
�
∂θ ��
�
� = qS �
∂n Sq
Sq
(16.2)
is prescribed, k
Sθ
Sθ ∪ Sq = S
Sθ ∩ Sq = ∅
(16.3)
Principle of virtual temperatures
�
�
�
S
B
θ,i kθ,i dV =
θq dV +
θ q S dSq
V
V
(16.4)
Sq
for arbitrary continuous θ(x1 , x2 , x3 ) zero on Sθ
For a fluid, we use the Eulerian formulation.
65
MIT 2.094
16. F.E. analysis of Navier-Stokes fluids
�
�
∂
(ρcp vθ)dx + conduction + etc
ρcp v θ|x − ρcp v θ|x +
∂x
(16.5)
In general 3D, we have an additional term for the left hand side of (16.1):
��
·�
v )θ − ρcp (v · �) θ
−� · (ρcp vθ) = −ρcp � · (vθ) = −ρcp (�
�
��
�
(16.6)
term (A)
where � · v = 0 in the incompressible case.
� · v = vi,i = div(v) = 0
(16.7)
So (16.1) becomes
�
�
(kθ,i ),i + q B = ρcp θ,i vi ⇒ (kθ,i ),i + q B − ρcp θ,i vi = 0
Principle of virtual temperatures is now (use (16.4))
�
�
�
�
θ,i kθ,i dV +
θ (ρcp θ,i vi ) dV =
θq B dV +
V
V
V
(16.8)
S
θ q S dSq
(16.9)
Sq
Navier-Stokes equations
• Differential form
τij,j + fiB = ρvi,j vj
(16.10)
with ρvi,j vj like term (A) in (16.6) = ρ(v · �)v in V .
�
�
∂vj
1 ∂vi
τij = −pδij + 2µeij
+
eij =
2 ∂xj
∂xi
(16.11)
• Boundary conditions (need be modified for various flow conditions)
Sf
τij nj = fi
on Sf
(16.12)
Mostly used as fn = τnn = prescribed, ft = unknown with possibly
inflow conditions).
And vi prescribed on Sv , and Sv ∪ Sf = S and Sv ∩ Sf = ∅.
66
∂vn
∂n
=
∂vt
∂n
= 0 (outflow or
MIT 2.094
16. F.E. analysis of Navier-Stokes fluids
• Variational form
�
�
�
�
v i ρvi,j vj dV +
eij τij dV =
v i fiB dV +
V
V
V
S
S
v i f fi f dSf
(16.13)
Sf
�
p� · vdV = 0
(16.14)
V
• F.E. solution
We interpolate (x1 , x2 , x3 ), vi , v i , θ, θ, p, p. Good elements are
×: linear pressure
◦: biquadratic velocities
(Q2 , P1 ), 9/3 element
9/4c element
Both satisfy the inf-sup condition.
So in general,
Example:
For Sf e.g.
τnn = 0,
∂vt
= 0;
∂n
(16.15)
67
MIT 2.094
and
∂vn
∂t
16. F.E. analysis of Navier-Stokes fluids
is solved for. Actually, we frequently just set p = 0.
Frequently used is the 4-node element with constant pressure
It does not strictly satisfy the inf-sup condition. Or use
Reading:
Sec. 7.4
3-node element with a bubble node.
Satisfies inf-sup condition
1D case of heat transfer with fluid flow, v = constant
Re =
vL
ν
Pe =
vL
α
α=
k
ρcp
Reading:
Sec. 7.4.3
(16.16)
• Differential equations
θ|x=0 = θL
kθ�� = ρcp θ� v
(16.17)
θ|x=L = θR
(16.18)
In non-dimensional form
1 ��
θ = θ�
Pe
Reading:
p. 683
(now θ�� and θ� are non-dimensional)
�
�
exp Pe
θ − θL
L x −1
⇒
=
θR − θL
exp (Pe) − 1
(16.19)
(16.20)
68
MIT 2.094
16. F.E. analysis of Navier-Stokes fluids
• F.E. discretization
θ�� = Peθ�
1
�
(16.21)
�
� �
1
θ θ dx + Pe
0
θθ� dx = 0 + { effect of boundary conditions = 0 here}
(16.22)
0
Using 2-node elements gives
1
2
(h∗ )
Pe =
(θi+1 − 2θi + θi−1 ) =
Pe
(θi+1 − θi−1 )
2h∗
vL
α
(16.23)
(16.24)
Define
Pee = Pe ·
�
h
vh
=
L
α
Pee
−1 −
2
�
(16.25)
�
θi−1 + 2θi +
�
Pee
− 1 θi+1 = 0
2
(16.26)
what is happening when Pee is large? Assume two 2-node elements only.
θi−1 = 0
θi+1 = 1
1
θi =
2
(16.27)
(16.28)
�
�
Pee
1−
2
(16.29)
69
MIT 2.094
16. F.E. analysis of Navier-Stokes fluids
θi =
1
2
�
1−
Pee
2
�
(16.30)
For Pee > 2, we have negative θi (unreasonable).
70
2.094 — Finite Element Analysis of Solids and Fluids
Fall ‘08
Lecture 17 - Incompressible fluid flow and heat transfer, cont’d
Prof. K.J. Bathe
17.1
MIT OpenCourseWare
Abstract body
Reading:
Sec. 7.4
Fluid Flow
Sv , Sf
Sv ∪ Sf = S
Sv ∩ Sf = 0
17.2
Heat transfer
Sθ , Sq
Sθ ∪ Sq = S
Sθ ∩ Sq = 0
Actual 2D problem (channel flow)
71
MIT 2.094
17.3
17. Incompressible fluid flow and heat transfer, cont’d
Basic equations
P.V. velocities
�
�
�
�
v i ρvi,j vj dV +
τij eij dV =
v i fiB dV +
V
V
V
S
S
v i f fi f dSf
(17.1)
Sf
Continuity
�
pvi,i dV = 0
(17.2)
V
P.V. temperature
�
�
�
�
θρcp θ,i vi dV +
θ,i kθ,i dV =
θq B dV +
V
V
V
S
θ q S dS
(17.3)
Sq
F.E. solution
xi =
�
hk xki
(17.4)
vi =
�
hk vik
(17.5)
θ=
�
hk θk
(17.6)
p=
�
h̃k pk
(17.7)
⎛
⇒ F (u) = R
17.4
⎞
v
u = ⎝ p ⎠ nodal variables
θ
(17.8)
Model problem
1D equation,
ρcp v
dθ
d2 θ
=k 2
dx
dx
(17.9)
(v is given, unit cross section)
Non-dimensional form (Section 7.4)
Pe
dθ
d2 θ
= 2
dx
dx
(17.10)
72
MIT 2.094
Pe =
17. Incompressible fluid flow and heat transfer, cont’d
vL
,
α
α=
k
ρcp
(17.11)
θ∗ is non-dimensional
�
�
exp PeLx − 1
θ − θL
=
exp (Pe) − 1
θR − θ L
(17.10) in F.E. analysis becomes
�
�
dθ
dθ dθ
θPe dV +
dV = 0
dx
V
V dx dx
(17.12)
(17.13)
Discretized by linear elements:
h∗ =
θ(ξ) =
�
�
ξ
ξ
1 − ∗ θi−1 + ∗ θi
h
h
h
L
(17.14)
For node i:
−θi−1 −
Pee
Pee
θi−1 + 2θi − θi+1 +
θi+1 = 0
2
2
(17.15)
where
Pee =
vh
α
�
�
h
= Pe
L
(17.16)
This result is the same as obtained by finite differences
�
1
�
� � =
2 (θi+1 − 2θi + θi−1 )
i
(h∗ )
�
θi+1 − θi−1
�
θ� � =
2h∗
i
73
(17.17)
(17.18)
MIT 2.094
17. Incompressible fluid flow and heat transfer, cont’d
Considered θi+1 = 1, θi−1 = 0. Then
θi =
1 − (Pee /2)
2
(17.19)
Physically unrealistic solution when Pee > 2. For this not to happen, we should refine the mesh—a very
fine mesh would be required. We use “upwinding”
dθ ��
θi − θi−1
� =
dx i
h∗
(17.20)
The result is
(−1 − Pee ) θi−1 + (2 + Pee ) θi − θi+1 = 0
Very stable, e.g.
�
θi−1 = 0
θi+1 = 1
⇒ θi =
(17.21)
1
2 + Pee
(17.22)
Unfortunately it is not that accurate. To obtain better accuracy in the interpolation for θ, use the
function
� x�
exp Pe L
−1
(17.23)
exp (Pe) − 1
The result is Pee dependent:
This implies flow-condition based interpolation. We use such interpolation functions—see references.
References
[1] K.J. Bathe and H. Zhang. “A Flow-Condition-Based Interpolation Finite Element Procedure for
Incompressible Fluid Flows.” Computers & Structures, 80:1267–1277, 2002.
[2] H. Kohno and K.J. Bathe. “A Flow-Condition-Based Interpolation Finite Element Procedure for
Triangular Grids.” International Journal for Numerical Methods in Fluids, 51:673–699, 2006.
74
MIT 2.094
17.5
17. Incompressible fluid flow and heat transfer, cont’d
FSI briefly
Lagrangian formulation for the structure/solid
Arbitrary Lagrangian-Eulerian (ALE) formulation Let f be a variable of a particle (e.g. f = θ).
Consider 1D
�
∂f
∂f
�
f˙�
=
+
v
(17.24)
∂t
∂x
particle
where v is the particle velocity. For a mesh point,
�
∂f
∂f
�
f ∗�
=
+
vm
∂t
∂x
mesh point
(17.25)
where vm is the mesh point velocity. Hence,
�
�
∂f
�
�
f˙�
= f ∗�
+
(v − vm )
∂x
particle
mesh point
(17.26)
Use (17.26) in the momentum and energy equations and use force equilibrium and compatibility at
the FSI boundary to set up the governing F.E. equations.
References
[1] K.J. Bathe, H. Zhang and M.H. Wang. “Finite Element Analysis of Incompressible and Compressible
Fluid Flows with Free Surfaces and Structural Interactions.” Computers & Structures, 56:193–213,
1995.
[2] K.J. Bathe, H. Zhang and S. Ji. “Finite Element Analysis of Fluid Flows Fully Coupled with
Structural Interactions.” Computers & Structures, 72:1–16, 1999.
[3] K.J. Bathe and H. Zhang. “Finite Element Developments for General Fluid Flows with Structural
Interactions.” International Journal for Numerical Methods in Engineering, 60:213–232, 2004.
75
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
2.094 — Finite Element Analysis of Solids and Fluids
Fall ‘08
Lecture 19 - Slender structures
Prof. K.J. Bathe
Beam analysis,
t
L
MIT OpenCourseWare
� 1 (e.g.
t
L
=
1
1
100 , 1000 , · · · )
Reading:
Sec. 5.4,
6.5
(plane stress)
� L
�
0
2
J=
0 2t
1
h2 = (1 − r) (1 + s)
4
1
h3 = (1 − r) (1 − s)
4
(19.1)
(19.2)
(19.3)
Beam theory assumptions (Timoshenko beam theory):
v2∗ = v3∗ = v2
t
u3∗ = u2 + θ2
2
t
∗
u 2 = u2 − θ 2
2
⎡
B∗ = ⎢
⎢
⎢
⎢
⎣
(19.4)
(19.5)
(19.6)
u∗2
v2∗
u∗3
v3∗
− 41 (1 + s) L2
0
− 14 (1 − s) L2
0
0
1
4 (1
− r) 2t
1
4 (1
−
r) 2t
− 14 (1 + s) L2
− 14 (1
0
− 14 (1 − r) 2t
81
−
⎤
r) 2t
− 14 (1 − s) L2
⎥ etc
⎥
⎥
⎥
⎦
(19.7)
MIT 2.094
19. Slender structures
u2
⎡
Bbeam
⎢
= ⎢
⎢ ···
⎢
⎣
− L1
v2
θ2
t
2L s
0
0
�
0
�
0
�
0
− L1
− 21 (1 − r)
⎤
∂u
∂x
⎛
⎜
⎥ ⎜
⎥ ⎜
⎥∼⎜
⎥ ⎜
⎦ ⎜
⎝
⎞
0
∂v
�
∂y
�
∂u
∂y
+
∂v
∂x
⎟
⎟
⎟
⎟
⎟
⎟
⎠
(19.8)
1
(1 − r)v2
2
1
st
u(r) = (1 − r)u2 − (1 − r)θ2
2
4
v(r) =
(19.9)
(19.10)
at r = −1,
v(−1) = v2
(19.11)
st
u(−1) = − θ2 + u2
2
(19.12)
Kinematics is
u(r) =
1
(1 − r)u2
2
(19.13)
results into �xx
→ �xx =
∂u 2
1
· =−
∂r L
L
(19.14)
st
(1 − r)θ2
4
(19.15)
u(r, s) = −
results into �xx , γxy
→ �xx =
γxy =
v(r) =
st
2L
∂u 2
1
· = − (1 − r)
∂s t
2
1
(1 − r)v2
2
(19.16)
(19.17)
(19.18)
results into γxy
→ γxy = −
82
1
L
(19.19)
MIT 2.094
19. Slender structures
For a pure bending moment, we want
1
1
− v2 − (1 − r)θ2 = 0
L
2
(19.20)
for all r! ⇒ Impossible (except for v2 = θ2 = 0) ⇒ So, the element has a spurious shear strain!
Beam kinematics (Timoshenko, Reissner-Mindlin)
γ=
�
dw
−β
dx
1 3
I=
bt
12
(19.21)
�
(19.22)
Principle of virtual work
�
L
EI
0
dβ dβ
dx + AS G
dx dx
�
L
�
0
dw
−β
dx
��
�
� L
dw
− β dx =
pwdx
dx
0
As = kA = kbt
(19.23)
(19.24)
To calculate k
� �2
�
�
V
1
1
2
(τa ) dA =
dAs
A
2G
2G
s
A
AS
Reading:
p. 400
(19.25)
where τa is the actual shear stress:
� � �2
�
t
− y2
3 V
2
τa = ·
� t �2
2 A
(19.26)
2
and V is the shear force.
⇒k=
Reading:
Ex. 5.23
5
6
(19.27)
83
MIT 2.094
19. Slender structures
Now interpolate
w(r) = h1 w1 + h2 w2
(19.28)
β(r) = h1 θ1 + h2 θ2
(19.29)
Revisit the simple case:
1+r
w1
2
1+r
β=
θ1
2
w=
(19.30)
(19.31)
Shearing strain
γ=
1+r
w1
−
θ1
L
2
(19.32)
Shear strain is not zero all along the beam. But, at r = 0, we can have the shear strain = 0.
θ1
w1
−
can be zero
L
2
(19.33)
Namely,
θ1
w1
2
−
= 0 for θ1 = w1
L
L
2
(19.34)
84
2.094 — Finite Element Analysis of Solids and Fluids
Fall ‘08
Lecture 20 - Beams, plates, and shells
Prof. K.J. Bathe
MIT OpenCourseWare
Timoshenko beam theory
The fiber moves up and rotates and its length does not change.
Principle of virtual displacement
�
EI
L
�
� � �T
β
β � dx + (Ak)G
0
0
L
(Linear Analysis)
�
dw
−β
dx
�T �
�
� L
dw
− β dx =
wT pdx
dx
0
(20.1)
Two-node element:
Three-node element:
For a q-node element,
�T
�
û = w1 · · · wq θ1 · · · θq
w = Hw û
β = Hβ û
�
�
Hw = h1 · · · hq 0 · · · 0
�
�
Hβ = 0 · · · 0 h1 · · · hq
dx
J=
dr
(20.2)
(20.3)
(20.4)
(20.5)
(20.6)
(20.7)
85
MIT 2.094
20. Beams, plates, and shells
dw
= J −1 Hw,r û
dx
� �� �
(20.8)
dβ
ˆ
= J −1 Hβ,r u
dx � �� �
(20.9)
Bw
Bβ
Hence we obtain
� � 1
�
EI
BβT Bβ det(J )dr + (Ak)G
−1
1
�
T
(Bw − Hβ ) (Bw − Hβ ) det(J )dr û
−1
�
1
=
HwT p det(J )dr
(20.10)
−1
Kû = R
(20.11)
K is a result of the term inside the bracket in (20.10) and R is a result of the right hand side.
For the 2-node element,
w1 = θ1 = 0
(20.12)
w2 , θ2 = ?
(20.13)
γ=
1+r
w2
−
θ2
L
2
(20.14)
We cannot make γ equal to zero for every r (page 404, textbook). Because of this, we need to use
about 200 elements to get an error of 10%. (Not good!)
Recall almost or fully incompressible analysis: Principle of virtual displacements:
�
�
�T � �
� C � dV +
�v (κ�v )dV = R
V
(20.15)
V
u/p formulation
�
�
T
�� C � �� dV −
�v pdV = R
V
V
�
�p
�
p
+ �v dV = 0
κ
V
(20.16)
(20.17)
But now we needed to select wisely the interpolations of u and p. We needed to satisfy the inf-sup
condition
�
qh � · vh dVol
≥β>0
(20.18)
inf sup Vol
����
����
�qh ��vh �
qh ∈Qh vh ∈Vh
86
MIT 2.094
20. Beams, plates, and shells
4/1 element:
We can show mathematically that this element does not satisfy inf-sup condition. But, we can also
show it by giving an example of this element which violates the inf-sup condition.
v1 = �,
above
�
v2 = 0 ⇒ � · vh for both elements is positive and the same. Now, if I choose pressures as
qh �vh dVol = 0,
hence (20.18) is not satisfied!
(20.19)
Vol
9/3 element
satisfies inf-sup
9/4-c
satisfies inf-sup
Getting back to beams
� L
�
�
EI
β βdx + (AkG)
0
�
0
L
�
�
dw
− β γ AS dx = R
dx
(20.20)
L
�
�
γ AS γ − γ AS dx = 0
(20.21)
dw
− β,
dx
(20.22)
0
where
γ=
from displacement interpolation
87
MIT 2.094
20. Beams, plates, and shells
γ AS = Assumed shear strain interpolation
(20.23)
2-node element, constant shear assumption. From (20.21),
�
L
�
0
�
� L
dw
�dx =
�dx
−β �
γ AS
γ AS �
γ AS
dx
0
�
+1
�
⇒−
−1
⇒ γ AS =
1+r
θ2
2
�
·
Reading:
Sec. 4.5.7
(20.24)
L
dr + w2 = γ AS · L
2
(20.25)
w2 − L2 θ2
L
(20.26)
γ AS (shear strain) is equal to the displacement-based shear strain at the middle of the beam.
Use γ AS in (20.20) to obtain a powerful element. For “our problem”,
γ AS = 0
�
⇒EI
0
L
θ2
2
(20.27)
�
�
β β � dx = M β �x=L
(20.28)
hence
L
w2 =
�� �
�
2
1
· L θ2 = M
⇒ EI
L
⇒ θ2 =
ML
,
EI
w2 =
(20.29)
M L2
2EI
(20.30)
(exact solutions)
88
MIT 2.094
20. Beams, plates, and shells
Plates
Reading:
Fig. 5.25,
p. 421
⎧
⎪
⎨ w = w(x, y) is the transverse displacement of the mid-surface
v = −zβy (x, y)
⎪
⎩
u = −zβx (x, y)
(20.31)
For any particle in the plate with coordinates (x, y, z), the expressions in (20.31) hold!
We use
w=
q
�
hi wi
(20.32)
i=1
βx = −
βy = +
q
�
i=1
q
�
hi θyi
(20.33)
hi θxi
(20.34)
i=1
where q equals the number of nodes. Then the element locks in the same way as the displacement-based
beam element.
89
2.094 — Finite Element Analysis of Solids and Fluids
Fall ‘08
Lecture 21 - Plates and shells
Prof. K.J. Bathe
MIT OpenCourseWare
Timoshenko beam theory, and Reissner-Mindlin plate theory
For plates, and shells, w, βx , and βy as independent variables.
w = displacement of mid-surface, w(x, y)
A = area of mid-surface
p = load per unit area on mid-surface
w = w(x, y)
(21.1)
w(x, y, z) = w(x, y)
(21.2)
The material particles at “any z” move in the z-direction as the mid-surface.
u(x, y, z) = −βx z = −βx (x, y)z
v(x, y, z) = −βy z = −βy (x, y)z
(21.3)
(21.4)
∂βx
∂u
= −z
∂x
∂x
∂βy
∂v
= −z
=
∂y
∂y
�
�
∂u ∂v
∂βx
∂βy
=
+
= −z
+
∂y
∂x
∂y
∂x
�xx =
(21.5)
�yy
(21.6)
γxy
∂βx
∂x
⎛
⎞
⎜
�xx
⎜
⎝ �yy ⎠ = −z ⎜
⎜
⎜
γxy
⎝
⎞
⎛
�
∂βy
∂y
∂βx
∂y
+
��
κ
(21.7)
∂βy
∂x
⎟
⎟
⎟
⎟
⎟
⎠
(21.8)
�
90
MIT 2.094
21. Plates and shells
∂w ∂u
∂w
− βx
+
=
∂x
∂z
∂x
∂w ∂v
∂w
− βy
=
+
=
∂y
∂z
∂y
γxz =
γyz
⎛
⎞
⎡
1
τxx
E
⎝ τyy ⎠ =
⎣ ν
1 − ν2
τxy
0
ν
1
0
(21.9)
(21.10)
0
0
1−ν
2
⎤⎛
⎞
�xx
⎦ ⎝ �yy ⎠ = C · �
γxy
(21.11)
(plane stress)
�
τxz
τyz
�
E
=
2(1 + ν)
�
�
γxz
γyz
= Gγ
(21.12)
Principle of virtual work for the plate:
� �
A
+ 2t
⎡
�
�xx
�yy
γ xy
− 2t
� �
+ 2t
k
A
− 2t
�
�
1
E ⎣
ν
1 − ν2
0
γ xz
γ yz
⎞
�xx
⎦ ⎝ �yy ⎠dzdA+
1−ν
γxy
2
�
��
�
�
�
1 0
γxz
G
dzdA =
wpdA
0 1
γyz
A
ν
1
0
0
0
⎤⎛
(21.13)
Consider a flat element:
⎛
⎜
⎜
⇒ Kb ⎜
⎝
w1
θx1
θy1
.
..
⎞
⎛
⎟
⎟
⎟ , also Kpl.
⎠
where Kb is 12x12 and Kpl.
str.
⎞
u1
⎜ v ⎟
str. ⎝ 1 ⎠
...
(21.14)
is 8x8.
91
MIT 2.094
21. Plates and shells
For a flat element:
⎛
�
⇒
Kb
0
0
Kpl. str.
⎜
⎜
⎜
⎜
⎜
�⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎝
w1
θx1
θy1
..
.
⎞
⎟
⎟
⎟
⎟
⎟
⎟
⎟
4 ⎟
θy ⎟ = · · ·
u1 ⎟
⎟
v1 ⎟
⎟
.. ⎟
. ⎠
v4
(21.15)
u=
�
hi u i
(21.16)
v=
�
hi vi
(21.17)
w=
�
hi wi
�
βx = −
hi θyi
�
βy =
hi θxi
(21.18)
(21.19)
(21.20)
From (21.13)
⎡
1
3
Et
⎣ ν
κT
12 (1 − ν 2 )
A
0
�
ν
1
0
0
0
1−ν
2
⎤
�
⎦ κdA +
γ T Gt · k
A
�
1
0
0
1
�
γdA
(21.21)
where k is the shear correction factor.
Next, evaluate
∂w ∂w ∂βx
∂x , ∂y , ∂x ,
. . . etc. ⇒ This element, as it is, locks!
This displacement-based element “locks in shear”. We need to change the transverse shear interpola­
tions.
γyz =
1
1
A
C
(1 − r)γyz
+ (1 + r)γyz
2
2
(21.22)
where
A
γyz
=
�
��
�
∂w
− βy ��
∂y
evaluated at A
(21.23)
from the w, βy displacement interpolations.
1
1
B
D
(1 − s)γxz
+ (1 + s)γxz
(21.24)
2
2
with this mixed interpolation, the element works. Called MITC interpolation (for mixed interpolatedtensional components)
γxz =
92
MIT 2.094
21. Plates and shells
Aside: Why not just neglect transverse shears, as in Kirchhoff plate theory?
βx = 0 ⇒ βx = ∂w
∂x
� 2
�
• Therefore we have ∂∂xw2 , · · · in strains, so we need continuity also for
� ∂w
�
∂x
· · ·
• If we do, γxz =
∂w
∂x −
•
w=
�
�
1
1
r(1 + r)w1 − r(1 − r)w2 + 1 − r2 w3
2
2
w2 and w3 never affect w1 (∴ w|r=1 = w1 ).
But,
∂w
1
1
= (1 + 2r)w1 − (1 − 2r)w2 − 2rw3
∂r 2
2
�
�
w2 and w3 affect ∂w
∂r 1 . ⇒ This results in difficulties to develop a good
element based on Kirchhoff theory.
With Reissner-Mindlin theory, we independently interpolate rotations such that this problem does
not arise.
For flat structures, we can superimpose the plate bending and plane stress element stiffness. For
shells, curved structures, we need to develop/use curved elements, see references.
References
[1] E. Dvorkin and K.J. Bathe. “A Continuum Mechanics Based Four-Node Shell Element for General
Nonlinear Analysis.” Engineering Computations, 1:77–88, 1984.
[2] K.J. Bathe and E. Dvorkin. “A Four-Node Plate Bending Element Based on Mindlin/Reissner Plate
Theory and a Mixed Interpolation.” International Journal for Numerical Methods in Engineering,
21:367–383, 1985.
[3] K.J. Bathe, A. Iosilevich and D. Chapelle. “An Evaluation of the MITC Shell Elements.” Computers
& Structures, 75:1–30, 2000.
[4] D. Chapelle and K.J. Bathe. The Finite Element Analysis of Shells – Fundamentals. Springer,
2003.
93
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2.094 Finite Element Analysis of Solids and Fluids II
Spring 2011
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