Topic 5: Finite Element Method

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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