topic 5: finite element method
TRANSCRIPT
Topic 5: Finite Element Method
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Finite Element Method (1)
Main problem of classical variational methods (Ritz method etc.)
– difficult (op impossible) definition of approximation function
ϕ for non-trivial problems.
The solution can be a division of structure to n small parts with
simple shapes. Approximational functions ϕj can be defined
on them.
Becasue Π is a scalar value:
Πapprox. =n∑
j=1Πe,j, (1)
where Πe,j . . . potentional energy of j-th part (”finite element”).2
Finite Element Method (2)
The rest of solution can be identical to the Ritz method:
∂Π = 0. (2)
Note: here we will utilize the Lagrange variational principle. It
implies that unknown variables will be deformations. This will
be a ”deformational version” of the Finite Element Method.
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Finite Element Method (3)
Finite Element Method (FEM) alternatives:
• deformational – unknowns are displacements and rotations
(most common more - than 90% software use it),
• force – unknows are forces/moments
• mixed.
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Deformational variant of FEM
Unknowns are related to theory of elasticity:
• plane problem (shear walls, . . . ): u, v
• slabs (bending problems): w, ϕx, ϕy
• volume.spatial problems: u, v, w
Approximational functions mostly have form of polynomials.5
Derivation of finite elementfor trusses (1)
1 2
u u1 2
y
x
Unknown deformational parameters: u in every node.
Totally 2 unknown parameters on element:
{u1, u2}T .
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Derivation of finite elementfor trusses (2)
Geometrical equations:
εx =∂u
∂x(3)
In matrix form (ε = ∂T u):
{
εx}
=[
∂∂x
] {
u}
(4)
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Derivation of finite elementfor trusses (3)
Equilibrium equations:
∂σx
∂x+X = 0 (5)
In matrix form: (∂σ +X = 0):
[
∂∂x
] {
σx}
+{
X}
={
0}
(6)
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Derivation of finite elementfor trusses (4)
Constitutive equations:
σx = E × εx (7)
In matrix form (σ = D ε):
{
σx}
=[
E] {
εx}
(8)
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Derivation of finite elementfor trusses (5)
Approximation of unknowns (node displacements):
u(x) = a1 + a2 x (9)
In matrix form (u = U a):
{
u}
=[
1 x]
a1a2
(10)
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Derivation of finite elementfor trusses (6)
Approximation of unknowns in nodes 1, 2
(r = S a):
u1u2
=
1 x11 x2
a1a2
(11)
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Derivation of finite elementfor trusses (7)
By combination of ε = ∂T u and u = U a we can get ε = B a,
where B = ∂T U a:
{
εx}
=[
∂∂x
] [
1 x]
a1a2
(12)
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Derivation of finite elementfor trusses (8)
By combination of ε = ∂T u and u = U a we can get ε = B a,
where B = ∂T U a:
{
εx}
=[
0 1]
a1a2
(13)
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Derivation of finite elementfor trusses (9)
From r = S a:
a = S−1 r, (14)
kde:
S =
1 x11 x2
⇒ S−1 =
x2x2−x1
−x1x2−x1
−1x2−x1
−1x2−x1
(15)
Then instead of ε = Ba we can write:ε = B S−1 r:
{
εx}
=[
0 1]
x2x2−x1
−x1x2−x1
−1x2−x1
−1x2−x1
u1u2
. (16)
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Derivation of finite elementfor trusses (11)
Potential energy of internal forces:
Πi =1
2
∫
VεTσ d V =
1
2
∫
VεT D ε d V (17)
Potential energy of external forces:
Πe = −∫
VXT r d V −
∫
SpT r d S. (18)
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Derivation of finite elementfor trusses (12)
Potential energy of system:
Π =1
2
∫
VεT D ε d V −
∫
VXT r d V −
∫
SpT r d S. (19)
After replacemetn of ε and extraction of r:
Π =1
2rT
∫
VS−1T BT DBS−1 d V rT−
∫
VXT d V r−
∫
SpT d S r.
(20)
In short form:
Π =1
2rT K r − FT r. (21)
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Derivation of finite elementfor trusses (13)
By use of Lagrange variational principle: (∂ Π = min.) na (21):
K r = F, (22)
where K . . . stiffness matrix of finite element:
K =∫
VS−1T BT D B S−1 d V, (23)
F . . . load vector of finite element:
F = −∫
VXT d V −
∫
SpT d S. (24)
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Derivation of finite elementfor trusses (14)
For this particular finite element:
F = X + p. (25)
K =∫
VS−1T BT DB S−1dV = A
∫ L
0,S−1T BT DB S−1dx,
(26)
with matrices members shown:
K = A∫ L
0
x2x2−x1
−1x2−x1
−x1x2−x1
−1x2−x1
01
[
E] [
0 1]
x2x2−x1
−x1x2−x1
−1x2−x1
−1x2−x1
dx
(27)
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Derivation of finite elementfor trusses (15)
Full formula (simplified):
K = A
x2x2−x1
−1x2−x1
−x1x2−x1
−1x2−x1
01
[
E] [
0 1]
x2x2−x1
−x1x2−x1
−1x2−x1
−1x2−x1
∫ L
0dx
(28)
After modification (integration∫L0 dx = L and multiplication):
K = EAL
1(x2−x1)2
−1(x2−x1)2
−1(x2−x1)2
1(x2−x1)2
, x2−x1 = L ⇒ K =
EAL
−EAL
−EAL
EAL
,
(29)
which is a sfiffness matrix and it is identical to one that can be
derived by slope-deflection method/stiffness method.19
Derivation of finite elementfor trusses (16)
The unknowns can be computed by solution of linear equation
system:
Kere = Fe,
In full form:
EAL
−EAL
−EAL
EAL
u1u2
=
F1F2
(30)
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Derivation of finite elementfor trusses (17)
Expansion for two unknowns u and v in every node:
1 2
xu1 u2
v1 v2
y
EAL0 −EA
L0
0 0 0 0−EAL 0 EA
L 00 0 0 0
u1v1u2v2
=
F10F20
(31)
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Interpolation polynomials (1)
• Best convergence can be reached for full polynomials of
n-th grade (Zenısek et al).
• Number of constants (a1, a2, ...) have to be equal to number
of unknowns on finite element
• For this reason it in not always possible fo use full polyno-
mials
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Interpolation polynomials (2)
For one unknown x:
1. a1 + a2 x
2. a1 + a2 x + a3 x2
3. a1 + a2 x + a3 x2 + a4 x
3
4. a1 + a2 x + a3 x2 + a4 x
3 + a5 x4
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Interpolation polynomials (3)
For two unknowns x and y:
1. a1 + a2 x + a3 y
2. a1 + a2 x + a3 y + a4 xy + a5 x2 + a6 x
2
3. a1 + a2 x + a3 y + a4 xy + a5 x2 + a6 x
2 + a7 x3 + a8 y
3 +
a9 x y2 + a10 x2 y
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