Finite Elements (I) Model Equation

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© Numerical Factory

Mètodes Numèrics: A First Course on Finite Elements

Dept. Matemàtiques ETSEIB - UPC BarcelonaTech

Finite Elements (I)

Following: Curs d’Elements Finits amb Aplicacions (J. Masdemont)http://hdl.handle.net/2099.3/36166


Model Equation


Finite Elements

Differential Equations (physical problem):

1-dim:Let’s assume that 𝑢(𝑥) is a magnitude (temperature, displacement, etc.)

1D Model Equation


Finite Elements


1-dim:

2-dim:

Let’s assume that 𝑢(𝑥) is a magnitude (temperature, displacement, etc.)

1D Model Equation

Now 𝑢(𝑥, 𝑦) is a magnitude

2D Model Equation


Finite Elements


1-dim:

2-dim:

Let’s assume that 𝑢(𝑥) is a magnitude (temperature, displacement, etc.)

1D Model Equation

Now 𝑢(𝑥, 𝑦) is a magnitude (temperature, etc.)

2D Model Equation

2on Order Terms


Finite Elements Procedure :

• First Step: To set up and express the equation at each element (element linear eq. system)

• Second Step: To assemble the contribution of each element (global linear eq. system)

• Third Step: To solve the linear eq. system

( 2x2 or 3x3 )

( NxN )

http://hdl.handle.net/2099.3/36166

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( 2x2 or 3x3 )

( NxN )






( 2x2 or 3x3 )

( NxN )


Element Decomposition


Finite Elements

Element decomposition (meshing):

1-dim: (a line)

𝑥 = 0 𝑥 = 𝐿


Finite Elements


1-dim: (a line)

𝑛2 𝑛3 𝑛4𝑛1

Ω1 Ω2 Ω3 Ω4

𝑛5

𝑥 = 0 𝑥 = 𝐿


Finite Elements


1-dim: (a line)

2-dim: (a 2D region)


Ω1 Ω2 Ω3 Ω4

𝑛5

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1D Elements


• For 1D domains, generically, the elements are defined as segments Ω𝑒 = [𝑥𝑖 , 𝑥𝑖+1] that covers de complete domain 𝛀.

Finite Elements 1D

(local) Nodes1 2

Linear Element 𝛀

(local) Nodes

1 2

Quadratic Element 𝛀3

Ω𝑒

Ω𝑒


• For 1D domains, generically, the elements are defined as segments Ω𝑒 = [𝑥𝑖 , 𝑥𝑖+1] that covers de complete domain 𝛀.

Finite Elements 1D

(local) Nodes

1 2

Linear Element 𝛀

(local) Nodes1 2

Quadratic Element 𝛀3

Ω𝑒

Ω𝑒


Let’s assume that 𝑢(𝑥) is a magnitude (temperature, displacement, etc.) that we want to compute in the nodes 𝑛𝑖 of one element Ω𝑒 the usual FEM notation is:

𝑢 𝑛𝑖 = 𝑢𝑖𝑒

For the linear case:

Finite Elements 1D

𝛀Ω𝑒

𝑢1𝑒 𝑢2

𝑒


Let’s assume that 𝑢(𝑥) is a magnitude (temperature, displacement, etc.) that we want to compute in the nodes 𝑛𝑖 of one element Ω𝑒 the usual FEM notation is:

𝑢 𝑛𝑖 = 𝑢𝑖𝑒

For the quadratic case:

Finite Elements 1D

𝛀Ω𝑒

𝑢1𝑒 𝑢3

𝑒𝑢2𝑒


When we consider the total domain, nodes of consecutive elements must be identify in order to obtain continuous solutions: (last node equals the next first one)

𝑢𝑁𝑒−1 = 𝑢1

𝑒 , 𝑢𝑁𝑒= 𝑢1

𝑒+1

For the linear case (𝑁 =2):

Nodes enumeration

𝛀Ω𝑒

𝑢2𝑒−1 = 𝑢1

𝑒

Ω𝑒−1 Ω𝑒+1

𝑢2𝑒 = 𝑢1

𝑒+1

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𝑒+1

For the linear case (𝑁 =2):

Nodes enumeration

𝛀Ω𝑒

𝑢2𝑒−1 = 𝑢1

𝑒

Ω𝑒−1 Ω𝑒+1

𝑢2𝑒 = 𝑢1

𝑒+1

𝑢1𝑒−1 𝑢2

𝑒−1 𝑢1𝑒 𝑢2

𝑒 𝑢1𝑒+1 𝑢2

𝑒+1





𝑒+1

For the quadratic case (𝑁 =3):

Nodes enumeration

𝛀Ω𝑒

𝑢3𝑒−1 = 𝑢1

𝑒

Ω𝑒−1


Nodes enumeration

• Global enumeration: Once we have identify the connected nodes, we rename them using a global

enumeration.

Global Nodes


Ω1 Ω2 Ω3 Ω4

𝑛5


Connectivity Matrix

• Connectivity Matrix: Says the global nodes attached

to each element.• Example:

Global Nodes


Ω1 Ω2 Ω3 Ω4

𝑛5

𝑒𝑙𝑒𝑚 =

12

23

34

45

One row for each element

Node numbers


Connectivity Matrix

• Connectivity Matrix: Says the global nodes attached

to each element.• Example:

Global Nodes


Ω1 Ω2 Ω3 Ω4

𝑛5

𝑒𝑙𝑒𝑚 =

12

23

34

45

One row for each element

Node numbers


Stiff Matrix

Element Stiff Matrix (𝑲𝒆): Is the one related to the physical problem stated for each element (this is the thought part of the method).

Because it is associated to each element, its size agrees with the number of nodes in each element.

1-dim linear element (two nodes) 𝑲𝒆 is a 2x2 matrix

1-dim quadratic element (three nodes) 𝑲𝒆 is a 3x3 matrix

2-dim linear Triangular element (three nodes) 𝑲𝒆 is a 3x3 matrix

2-dim linear Quadrilateral element (four nodes) 𝑲𝒆 is a 4x4 mat

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Stiff Matrix






2-dim linear Quadrilateral element (four nodes) 𝑲𝒆 is a 4x4 mat

𝐾𝑒 =𝑘11𝑒 𝑘12

𝑒

𝑘21𝑒 𝑘22

𝑒


Stiff Matrix






2-dim linear Quadrilateral element (four nodes) 𝑲𝒆 is a 4x4 mat 𝐾𝑒 =

𝑘11𝑒 𝑘12

𝑒 𝑘13𝑒

𝑘21𝑒 𝑘22

𝑒 𝑘23𝑒

𝑘31𝑒 𝑘32

𝑒 𝑘33𝑒


Stiff Matrix





2-dim linear Triangular elem (three nodes) 𝑲𝒆 is a 3x3 matrix

2-dim linear Quadrilateral elem (four nodes) 𝑲𝒆 is a 4x4 mat


Stiff Matrix








Stiff Matrix







Usually 𝐾𝑒 are symmetric matrices


Global Stiff Matrix

• Global Stiff Matrix (𝑲) : Is the assembled matrix of all 𝑲𝒆 element stiff matrices.

In a generic way, for a 1dim problem, 𝑲 is squarematrix of dimension the number of global nodes.

Example:

𝑲 =

𝑘11𝑘21

𝑘12𝑘22

𝑘13 𝑘14 𝑘15𝑘23 𝑘24 𝑘25

𝑘31 𝑘32 𝑘33 𝑘34 𝑘35𝑘41𝑘51

𝑘42𝑘52

𝑘43 𝑘44 𝑘45𝑘53 𝑘54 𝑘55

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Global Stiff Matrix

• Global Stiff Matrix (𝑲) : Is the assembled matrix of all 𝑲𝒆 element stiff matrices.

In a generic way, for a 1dim problem, 𝑲 is square matrix of dimension the number of global nodes.

Example:

𝑲 =

𝑘11𝑘21

𝑘12𝑘22

𝑘13 𝑘14 𝑘15𝑘23 𝑘24 𝑘25

𝑘31 𝑘32 𝑘33 𝑘34 𝑘35𝑘41𝑘51

𝑘42𝑘52

𝑘43 𝑘44 𝑘45𝑘53 𝑘54 𝑘55


Ω1 Ω2 Ω3 Ω4

𝑛5


Element Assembly


1D Elements Assembly

• Assembly

𝑲 =

𝑘11𝑘21

𝑘12𝑘22

𝑘13 𝑘14 𝑘15𝑘23 𝑘24 𝑘25

𝑘31 𝑘32 𝑘33 𝑘34 𝑘35𝑘41𝑘51

𝑘42𝑘52

𝑘43 𝑘44 𝑘45𝑘53 𝑘54 𝑘55


Ω1 Ω2 Ω3 Ω4

𝑛5

𝑒𝑙𝑒𝑚 =1⋮4

2⋮5

𝐾1 =𝑘111 𝑘12

1

𝑘211 𝑘22

1



• Assembly

𝑲 =

𝑘111

𝑘211

𝑘121

𝑘221

𝑘13 𝑘14 𝑘15𝑘23 𝑘24 𝑘25

𝑘31 𝑘32 𝑘33 𝑘34 𝑘35𝑘41𝑘51

𝑘42𝑘52

𝑘43 𝑘44 𝑘45𝑘53 𝑘54 𝑘55


Ω1 Ω2 Ω3 Ω4

𝑛5


2⋮5

𝐾1 =𝑘111 𝑘12

1

𝑘211 𝑘22

1



• Assembly

𝑲 =

𝑘11𝑘21

𝑘12𝑘22

𝑘13 𝑘14 𝑘15𝑘23 𝑘24 𝑘25

𝑘31 𝑘32 𝑘33 𝑘34 𝑘35𝑘41𝑘51

𝑘42𝑘52

𝑘43 𝑘44 𝑘45𝑘53 𝑘54 𝑘55


Ω1 Ω2 Ω3 Ω4

𝑛5


2⋮5



• Assembly

𝑲 =

𝑘11𝑘21

𝑘12𝑘22

𝑘13 𝑘14 𝑘15𝑘23 𝑘24 𝑘25

𝑘31 𝑘32 𝑘33 𝑘34 𝑘35𝑘41𝑘51

𝑘42𝑘52

𝑘43 𝑘44 𝑘45𝑘53 𝑘54 𝑘55


Ω1 Ω2 Ω3 Ω4

𝑛5


2⋮5

𝑘22 = 𝑘221 + 𝑘11

2

Contribution of two elements

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• Assembly

𝑲 =

𝑘11𝑘21

𝑘12𝑘22

𝑘13 𝑘14 𝑘15𝑘23 𝑘24 𝑘25

𝑘31 𝑘32 𝑘33 𝑘34 𝑘35𝑘41𝑘51

𝑘42𝑘52

𝑘43 𝑘44 𝑘45𝑘53 𝑘54 𝑘55


Ω1 Ω2 Ω3 Ω4

𝑛5


2⋮5



• Assembly

𝑲 =

𝑘11𝑘21

𝑘12𝑘22

𝑘13 𝑘14 𝑘15𝑘23 𝑘24 𝑘25

𝑘31 𝑘32 𝑘33 𝑘34 𝑘35𝑘41𝑘51

𝑘42𝑘52

𝑘43 𝑘44 𝑘45𝑘53 𝑘54 𝑘55

The rest of the elements in 𝑲 are zero


Ω1 Ω2 Ω3 Ω4

𝑛5


2⋮5



• Example: Suppose that for each element its local Stiff matrix is constant (we’ll see later how to compute it)


𝑒

𝑘21𝑒 𝑘22

𝑒 = 𝐶 ·1 −1−1 1





𝑒

𝑘21𝑒 𝑘22

𝑒 = 𝐶 ·1 −1−1 1

For the bar:


Ω1 Ω2 Ω3 Ω4

𝑛5

𝑲 = 𝐶

1−1

−11

0 0 00 0 0

0 0 0 0 000

00

0 0 00 0 0





𝑒

𝑘21𝑒 𝑘22

𝑒 = 𝐶 ·1 −1−1 1

For the bar:


Ω1 Ω2 Ω3 Ω4

𝑛5

𝑲 = 𝐶

1−1

−12

0 0 0−1 0 0

0 −1 1 0 000

00

0 0 00 0 0





𝑒

𝑘21𝑒 𝑘22

𝑒 = 𝐶 ·1 −1−1 1

For the bar:


Ω1 Ω2 Ω3 Ω4

𝑛5

𝑲 = 𝐶

1−1

−12

0 0 0−1 0 0

0 −1 2 −1 000

00

−1 2 −10 −1 1

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• Exercise: Consider the problem of the previous bar, but using now quadratic elements.


Ω1 Ω2

𝑛5

𝐾𝑒 =4 2 −12 16 2−1 2 4

Modify the previous steps in order to obtain the assembly matrix when only two elements are taken



• Exercise: Consider the problem of the previous bar, but using now quadratic elements.


Ω1 Ω2

𝑛5

𝐾𝑒 =4 2 −12 16 2−1 2 4

𝑲 =

𝑘111

𝑘211

𝑘121

𝑘221

𝑘13 𝑘14 𝑘15𝑘23 𝑘24 𝑘25

𝑘31 𝑘32 𝑘33 𝑘34 𝑘35𝑘41𝑘51

𝑘42𝑘52

𝑘43 𝑘44 𝑘45𝑘53 𝑘54 𝑘55

Modify the previous steps in order to obtain the assembly matrix when only two elements are taken


2D Element Assembly



• Example of local and global 2D nodes enumeration for linear triangular elements

Hint: Notice that the local enumeration must becounter-clockwise in order to preserve orientation




Connectivity matrix

𝐵 =

1523

2345

3256




Connectivity matrix

𝐵 =

1523

2345

3256

Compute the global stiff matrix for this example: 𝐾𝑒 =𝑘11 ⋯ 𝑘16⋮ ⋱ ⋮𝑘61 ⋯ 𝑘66

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For linear triangular elements:

Let’s do the assembly process for this example.

If 𝑢 𝑥 is a 1D magnitude (temperature)For each element the Stiffness Matrix isa 3x3 matrix

𝐾𝑒 =

𝑘11𝑒 𝑘12

𝑒 𝑘13𝑒

𝑘21𝑒 𝑘22

𝑒 𝑘23𝑒

𝑘31𝑒 𝑘32

𝑒 𝑘33𝑒

, 𝑢 =

𝑢1𝑒

𝑢2𝑒

𝑢3𝑒



The stiff matrix is a 6x6 matrix

Each row is associated to the corresponding node: row 1 with node 1 We will fill the matrix row by row



First row:The 𝑘11 element is the relation of global node 1 with itself

this means the relationship between:

local node 1 of the element Ω1 and itself: 𝑘111

Row 1 :

𝐾 =

𝑘111



First row:The 𝑘12 element is the relation of global node 1 with node 2


local node 1 of Ω1 and local node 2 of Ω1: 𝑘121

Row 1 :

𝐾 =

𝑘111 𝑘12

1





local node 1 of Ω1 and local node 3 of Ω1: 𝑘131

Row 1 :

𝐾 =

𝑘111 𝑘12

1 𝑘131




No connection through an element exist !!

The same happens with nodes 5 and 6

Row 1 :

𝐾 =

𝑘111 𝑘12

1 𝑘131 0

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No connection through an element exist !!

The same happens with nodes 5 and 6

Row 1 :

𝐾 =

𝑘111 𝑘12

1 𝑘131 0 0 0



Symmetries:

Because of stiff matrix symmetries, we can fill the first column.

col 1 :

𝐾 =

𝑘111 𝑘12

1 𝑘131

𝑘211

𝑘311

0 0 0

000


2D Elements AssemblySecond row:

The 𝑘22 element is the relation of global node 2 with itself:




Row 2 :

𝐾 =

𝑘111 𝑘12

1 𝑘131

𝑘211 𝑘22𝑘311

0 0 0

000

𝑘22 = 𝑘221 + 𝑘33

2 + 𝑘113



The 𝑘23 element is the relation of global node 2 with node 2:

local node 2 of Ω1 and node 3 of Ω1: 𝑘231


Row 2 :

𝐾 =

𝑘111 𝑘12

1 𝑘131

𝑘211 𝑘22 𝑘23𝑘311

0 0 0

000

𝑘23 = 𝑘231 + 𝑘32

2





Row 2 :

𝐾 =

𝑘111 𝑘12

1 𝑘131

𝑘211 𝑘22 𝑘23𝑘311

0 0 0𝑘24

000

𝑘24 = 𝑘123






Row 2 :

𝐾 =

𝑘111 𝑘12

1 𝑘131

𝑘211 𝑘22 𝑘23𝑘311

0 0 0𝑘24 𝑘25

000

𝑘25 = 𝑘312 + 𝑘13

3

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No connection exist

Row 2 :

𝐾 =

𝑘111 𝑘12

1 𝑘131

𝑘211 𝑘22 𝑘23𝑘311

0 0 0𝑘24 𝑘25 0

000

𝑘26 = 0



Symmetries:

Because of stiff matrix symmetries, we can fill the first column.

col 2 :

We left as an exercise to fill all the matrix!!




Finite Elements (II)Meshing - Weak Formulation



Meshing


Meshing 2D domains

• Meshing a general domain is a difficult problem. We’ll not study it in depth in our course.

• Two main concerns when meshing a domain are:

– Good fitting of the domain

– Good Numerical properties (stability)


Meshing 2D domains

• Classification:

– Structured Mesh:

are identified by regular connectivity

– Unstructured Mesh


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Meshing 2D domains

• Classification:

– Structured Mesh:

are identified by regular connectivity

– Unstructured Mesh


Meshing 2D domains

Mesh quality:

– Aspect Ratio: It is the ratio of

longest to the shortest side in an element.

Best = 1

Acceptable < 5


Meshing 2D domains

Mesh quality:

– Skewness:


Meshing 2D domains

Mesh quality:

– Inscribed-Circumscribed ratio:


Meshing 2D domains• Mesh refinement: More elements where

physical features are changing


Model Equation

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2D-Model Equation

For 2D problems we will use the model equation. A 2on order PDE for 𝑢 = 𝑢 𝑥, 𝑦 (primary variable)

Defined on a 2-dim domain Ω, with 𝑎𝑖𝑗 𝑥, 𝑦 and 𝑓 𝑥, 𝑦

known functions.


2D-Model Equation

• Notation: In many books you can find the expressions

𝛻 · 𝑢 ≡𝜕𝑢

𝜕𝑥+𝜕𝑢

𝜕𝑦, if 𝑢 = 𝑢 𝑥, 𝑦

𝛻 · 𝑢1, 𝑢2 ≡ 𝑑𝑖𝑣 𝑢 ≡𝜕𝑢1𝜕𝑥

+𝜕𝑢2𝜕𝑦

, if 𝑢 =𝑢1(𝑥, 𝑦)

𝑢2 𝑥, 𝑦

𝛻𝑢 ≡ 𝑔𝑟𝑎𝑑 𝑢 ≡𝜕𝑢

𝜕𝑥,𝜕𝑢


• Example: Poisson equation−𝛻 · 𝑎𝛻𝑢 = 𝑓

If 𝑎 =const, −𝑎𝛻 · 𝛻𝑢 ≡ −𝑎 𝛻2𝑢 ≡ −𝑎∆𝑢 = 𝑓

Laplacian Operator


2D-Model Equation


𝛻 · 𝑢 ≡𝜕𝑢

𝜕𝑥+𝜕𝑢


𝛻 · 𝑢1, 𝑢2 ≡ 𝑑𝑖𝑣 𝑢 ≡𝜕𝑢1𝜕𝑥

+𝜕𝑢2𝜕𝑦

, if 𝑢 =𝑢1(𝑥, 𝑦)

𝑢2 𝑥, 𝑦


𝜕𝑥,𝜕𝑢




Laplacian Operator


2D-Model Equation


𝛻 · 𝑢 ≡𝜕𝑢

𝜕𝑥+𝜕𝑢


𝛻 · 𝑢1, 𝑢2 ≡ 𝑑𝑖𝑣 𝑢 ≡𝜕𝑢1𝜕𝑥

+𝜕𝑢2𝜕𝑦

, if 𝑢 =𝑢1(𝑥, 𝑦)

𝑢2 𝑥, 𝑦


𝜕𝑥,𝜕𝑢




Laplacian Operator


2D-Model Equation


𝛻 · 𝑢 ≡𝜕𝑢

𝜕𝑥+𝜕𝑢


𝛻 · 𝑢1, 𝑢2 ≡ 𝑑𝑖𝑣 𝑢 ≡𝜕𝑢1𝜕𝑥

+𝜕𝑢2𝜕𝑦

, if 𝑢 =𝑢1(𝑥, 𝑦)

𝑢2 𝑥, 𝑦


𝜕𝑥,𝜕𝑢




Laplacian Operator


• Poisson equation:

It corresponds to the 2D model equation

with

𝑎11 = 𝑎22 = 𝑎, 𝑎12 = 𝑎21 = 𝑎00 = 0

2D-Model Equation

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Weak Formulation: intuition


Idea of the Weak Formulation

• We need the solution of the model equation (a PDE equation)

• Instead we use an integral equation:

(1)

(2)




• Instead we use an integral equation:

(1)

(2)That’s hard!!!




• Instead, we use an integral equation:

(1)

(2)





If we know how to solve (2) then, what do we know about the solution of (1) ?

(1)

(2)





(1)

(2)

𝐹 𝑢, 𝑥, 𝑦 𝑑𝑥𝑑𝑦 = 0If we have

Question:

𝐹 𝑢, 𝑥, 𝑦 = 0

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(1)

(2)

𝐹 𝑢, 𝑥, 𝑦 𝑑𝑥𝑑𝑦 = 0If we have

Answer: NO!!!

𝐹 𝑢, 𝑥, 𝑦 = 0





(1)

(2)

What about if we have?

𝐹 𝑢, 𝑥, 𝑦 = 0

𝐹 𝑢, 𝑥, 𝑦 𝑑𝑥𝑑𝑦 = 0𝜔1 𝑥, 𝑦 ·

⋮

𝐹 𝑢, 𝑥, 𝑦 𝑑𝑥𝑑𝑦 = 0𝜔𝑛 𝑥, 𝑦 ·

⋮





(1)

(2)

Answer: ….almost!!!

𝐹 𝑢, 𝑥, 𝑦 = 0

𝐹 𝑢, 𝑥, 𝑦 𝑑𝑥𝑑𝑦 = 0𝜔1 𝑥, 𝑦 ·

⋮

𝐹 𝑢, 𝑥, 𝑦 𝑑𝑥𝑑𝑦 = 0𝜔𝑛 𝑥, 𝑦 ·

⋮





(1)

(2)

If the functions 𝜔𝑖 𝑥, 𝑦 are the shape functions (the base of the polynomials) the solution of (2) is a good approximation of the solution of (1).This solution is known as the weak solution of equation (1)


Weak Formulation


Weak Formulation

Using a general weight function 𝜔 = 𝜔(𝑥, 𝑦), we write the integral expression:

In 2D, we have the divergence theorem (from Gauss)

𝜕Ω𝑘 is the boundary of the domain Ω𝑘, and 𝑛 is the normalvector to 𝜕Ω (pointing external)

𝜔(𝑥, 𝑦)(2)

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Weak Formulation



𝜕Ω𝑘 is the boundary of the domain Ω𝑘, and 𝑛 is the normalvector to 𝜕Ω (pointing external)

(2) 𝜔(𝑥, 𝑦)


𝜔(𝑥, 𝑦)

Weak Formulation



𝜕Ω𝑘 is the boundary of the domain Ω𝑘, and 𝑛 is the normal vector to 𝜕Ω (pointing external)

How do we make the divergence appear in the equation (2)?

(2)


Weak Formulation

We need to introduce some notation, let’s say

Deriving now the products and we get:

Then, the first part of the weak form says

𝜔(𝑥, 𝑦)


Weak Formulation


Deriving now the products 𝜔𝐹1 and 𝜔𝐹2 we get:

𝜕

𝜕𝑥𝜔𝐹1 =

𝜕𝜔

𝜕𝑥𝐹1 +𝜔

𝜕𝐹1𝜕𝑥

𝜕

𝜕𝑥𝜔𝐹2 =

𝜕𝜔

𝜕𝑥𝐹2 +𝜔

𝜕𝐹2𝜕𝑥


Weak Formulation



𝜕

𝜕𝑥𝜔𝐹1 =

𝜕𝜔

𝜕𝑥𝐹1 +𝜔

𝜕𝐹1𝜕𝑥

𝜕

𝜕𝑥𝜔𝐹2 =

𝜕𝜔

𝜕𝑥𝐹2 +𝜔

𝜕𝐹2𝜕𝑥

−𝜔𝜕𝐹1𝜕𝑥

=𝜕𝜔

𝜕𝑥𝐹1 −

𝜕

𝜕𝑥𝜔𝐹1

that we can rewrite


=𝜕𝜔

𝜕𝑥𝐹2 −

𝜕

𝜕𝑥𝜔𝐹2


Weak Formulation




=𝜕𝜔

𝜕𝑥𝐹1 −

𝜕

𝜕𝑥𝜔𝐹1 −𝜔

𝜕𝐹2𝜕𝑥

=𝜕𝜔

𝜕𝑥𝐹2 −

𝜕

𝜕𝑥𝜔𝐹2

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Weak Formulation



Then, the first part of the weak form (2) says

𝜔(𝑥, 𝑦)

𝜔(𝑥, 𝑦)


=𝜕𝜔

𝜕𝑥𝐹1 −

𝜕

𝜕𝑥𝜔𝐹1 −𝜔

𝜕𝐹2𝜕𝑥

=𝜕𝜔

𝜕𝑥𝐹2 −

𝜕

𝜕𝑥𝜔𝐹2


Weak Formulation

The last two terms are the divergence terms. Therefore,

where .

For a triangle is:Γ𝑘 positive oriented


Weak Formulation


where .



Weak Formulation


.


Divergence Theorem


Weak Formulation


.


Divergence Theorem

where Γ𝑘 = 𝜕Ω𝑘 with positive orientation


Weak FormulationThese last terms, coming from the divergence theorem are related to the boundary and they are denoted by:

Finally using this definition we rewrite the weak form (2) as:

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2D integral on Ω𝑘




1D integral on Γ𝑘


Weak FormulationTo solve these integrals:

1. We first substitute 𝜔 = 𝜓𝑖𝑘 𝑥, 𝑦 for 𝑖 = 1…𝑛 (the shape

functions)

2. We interpolate the unknown function and their derivatives

3. Substitute all in equation (2)

𝑢 𝑥, 𝑦 =

𝑗=1

𝑛

𝑢𝑗𝑘𝜓𝑗

𝑘(𝑥, 𝑦)𝜕𝑢

𝜕𝑥=

𝑗=1

𝑛

𝑢𝑗𝑘𝜕𝜓𝑗

𝑘

𝜕𝑥




functions)


𝑢 𝑥, 𝑦 =

𝑗=1

𝑛



𝜕𝑥=

𝑗=1

𝑛


𝑘

𝜕𝑥




functions)


𝑢 𝑥, 𝑦 =

𝑗=1

𝑛



𝜕𝑥=

𝑗=1

𝑛


𝑘

𝜕𝑥

Fundamental trick!!!!Due to the interpolation, instead of a continuous unknown function

the only unknowns are the values at the nodes. Only few points!!!

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functions)


3. Substitute all in equation (2)

𝑢 𝑥, 𝑦 =

𝑗=1

𝑛



𝜕𝑥=

𝑗=1

𝑛


𝑘

𝜕𝑥


Weak FormulationGrouping the unknown terms 𝑢𝑗

𝑘

Which is in fact a linear system of equations



𝑘


𝑘11𝑘 ⋯ 𝑘1𝑛

𝑘

⋮ ⋱ ⋮𝑘𝑛1𝑘 ⋯ 𝑘𝑛𝑛

𝑘

𝑢1𝑘

⋮𝑢𝑛𝑘

=𝐹1𝑘

⋮𝐹𝑛𝑘

+𝑄1𝑘

⋮𝑄𝑛𝑘



𝑘



𝑘


𝑘

𝑢1𝑘

⋮𝑢𝑛𝑘

=𝐹1𝑘

⋮𝐹𝑛𝑘

+𝑄1𝑘

⋮𝑄𝑛𝑘



𝑘



𝑘


𝑘

𝑢1𝑘

⋮𝑢𝑛𝑘

=𝐹1𝑘

⋮𝐹𝑛𝑘

+𝑄1𝑘

⋮𝑄𝑛𝑘



𝑘



𝑘


𝑘

𝑢1𝑘

⋮𝑢𝑛𝑘

=𝐹1𝑘

⋮𝐹𝑛𝑘

+𝑄1𝑘

⋮𝑄𝑛𝑘

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𝑘



Weak FormulationThen, for each element you need to compute two terms to build the element linear system:

• The stiff matrix


𝑘


𝑘

𝑢1𝑘

⋮𝑢𝑛𝑘

=𝐹1𝑘

⋮𝐹𝑛𝑘

+𝑄1𝑘

⋮𝑄𝑛𝑘




• The 𝐹’s vector


𝑘


𝑘

𝑢1𝑘

⋮𝑢𝑛𝑘

=𝐹1𝑘

⋮𝐹𝑛𝑘

+𝑄1𝑘

⋮𝑄𝑛𝑘




• The 𝐹’s vector


𝑘


𝑘

𝑢1𝑘

⋮𝑢𝑛𝑘

=𝐹1𝑘

⋮𝐹𝑛𝑘

+𝑄1𝑘

⋮𝑄𝑛𝑘

The Q’s vector depends on the Boundary Conditions and is computed once you assemble the global system


Notation:

𝑢 = 𝑢(𝑥, 𝑦) is named primary variable

is named secondary variable

Boundary Conditions (BC):

𝑢𝐴 = 𝑢(𝑥𝐴) is an essential BC (fix the primary variable)

is a natural BC (fix the secondary variable)

Weak Formulation

Remember


Notation:

𝑢 = 𝑢(𝑥, 𝑦) is named primary variable

is named secondary variable

Boundary Conditions (BC):

𝑢𝐴 = 𝑢(𝑥𝐴) is an essential BC (fix the primary variable)

is a natural BC (fix the secondary variable)

Weak Formulation

Remember

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Weak FormulationNotation from the global system of equations:

We will denote by:



We will denote by:



We will denote the different terms by:










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Computing the integrals


Computing the Integrals

To build the linear system you need to compute terms like these ones:

In general you need to compute numerically these 2D integrals. For that we will use Gauss integration methods that will be introduced later.

For some easy cases there are some explicit formulas that we present next.


Computing the Integrals

To build the linear system you need to compute terms like these ones:

In general you need to compute numerically these 2D integrals. For that we will use Gauss integration methods that will be introduced later.

For some easy cases there are some explicit formulas that we present next.


Triangles

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Computing the Integrals: Triangles

• If we consider constant coefficients for the model equation

We have to compute

In the case of a general linear triangular element

𝑲𝑖𝑗𝑘,11 = 𝑎11 ඵ

Ω𝑘

𝜕𝜓𝑖𝑘

𝜕𝑥𝑥, 𝑦

𝜕𝜓𝑗𝑘

𝜕𝑥𝑥, 𝑦 𝑑𝑥𝑑𝑦 = 𝑎11

1

4𝐴𝑘𝛽𝑖𝛽𝑗

𝑣1

𝑣2

𝑣3 Ω𝑘

𝜓𝑖𝑘 𝑥, 𝑦 = 𝑎𝑖 + 𝛽𝑖 𝑥 + 𝛾𝑖 𝑦, 𝑖 = 1,2,3

𝑎𝑖 =𝑥𝑗𝑦𝑘 − 𝑥𝑘𝑦𝑗

2𝐴𝑘

𝛽𝑖 =𝑦𝑗 − 𝑦𝑘

2𝐴𝑘

𝛾𝑖 =𝑥𝑘 − 𝑥𝑗

2𝐴𝑘


Ω𝑘

𝜕𝜓𝑖𝑘

𝜕𝑥𝑥, 𝑦

𝜕𝜓𝑗𝑘

𝜕𝑥𝑥, 𝑦 𝑑𝑥𝑑𝑦

𝜕𝜓𝑖𝑘

𝜕𝑥𝑥, 𝑦 = 𝛽𝑖

(𝑖, 𝑗, 𝑘) cyclic permutations

𝜕𝜓𝑖𝑘

𝜕𝑦𝑥, 𝑦 = 𝛾𝑖




We have to compute

In the case of a general linear triangular element𝑣1

𝑣2

𝑣3 Ω𝑘

𝜓𝑖𝑘 𝑥, 𝑦 =

𝑎𝑖+𝛽𝑖 𝑥+𝛾𝑖 𝑦

2𝐴𝑘, 𝑖 = 1,2,3

𝑎𝑖 = 𝑥𝑗𝑦𝑘 − 𝑥𝑘𝑦𝑗𝛽𝑖 = 𝑦𝑗 − 𝑦𝑘𝛾𝑖 = 𝑥𝑘 − 𝑥𝑗


Ω𝑘

𝜕𝜓𝑖𝑘

𝜕𝑥𝑥, 𝑦

𝜕𝜓𝑗𝑘


𝜕𝜓𝑖𝑘

𝜕𝑥𝑥, 𝑦 = 𝛽𝑖


𝜕𝜓𝑖𝑘

𝜕𝑦𝑥, 𝑦 = 𝛾𝑖




We have to compute



Ω𝑘

𝜕𝜓𝑖𝑘

𝜕𝑥𝑥, 𝑦

𝜕𝜓𝑗𝑘


1


𝑣1

𝑣2

𝑣3 Ω𝑘𝑲𝑖𝑗𝑘,11 = 𝑎11 ඵ

Ω𝑘

𝜕𝜓𝑖𝑘

𝜕𝑥𝑥, 𝑦

𝜕𝜓𝑗𝑘


𝜕𝜓𝑖𝑘

𝜕𝑥𝑥, 𝑦 =

𝛽𝑖2𝐴𝑘

𝜕𝜓𝑖𝑘

𝜕𝑦𝑥, 𝑦 =

𝛾𝑖2𝐴𝑘



2𝐴𝑘, 𝑖 = 1,2,3






We have to compute



Ω𝑘

𝜕𝜓𝑖𝑘

𝜕𝑥𝑥, 𝑦

𝜕𝜓𝑗𝑘


1


𝑣1

𝑣2

𝑣3 Ω𝑘𝑲𝑖𝑗𝑘,11 = 𝑎11 ඵ

Ω𝑘

𝜕𝜓𝑖𝑘

𝜕𝑥𝑥, 𝑦

𝜕𝜓𝑗𝑘




2𝐴𝑘, 𝑖 = 1,2,3



𝜕𝜓𝑖𝑘

𝜕𝑥𝑥, 𝑦 =

𝛽𝑖2𝐴𝑘

𝜕𝜓𝑖𝑘

𝜕𝑦𝑥, 𝑦 =

𝛾𝑖2𝐴𝑘




Ω𝑘

𝜕𝜓𝑖𝑘

𝜕𝑥𝑥, 𝑦

𝜕𝜓𝑗𝑘

𝜕𝑥𝑥, 𝑦 𝑑𝑥𝑑𝑦 =

𝑎114𝐴𝑘

𝛽𝑖𝛽𝑗


Ω𝑘

𝜕𝜓𝑖𝑘

𝜕𝑥𝑥, 𝑦

𝜕𝜓𝑗𝑘

𝜕𝑦𝑥, 𝑦 𝑑𝑥𝑑𝑦 =

𝑎124𝐴𝑘

𝛽𝑖𝛾𝑗

𝑲𝑖𝑗𝑘,21 =

𝑎214𝐴𝑘

𝛾𝑖𝛽𝑗


Ω𝑘

𝜕𝜓𝑖𝑘

𝜕𝑦𝑥, 𝑦

𝜕𝜓𝑗𝑘

𝜕𝑦𝑥, 𝑦 𝑑𝑥𝑑𝑦 =

𝑎224𝐴𝑘

𝛾𝑖𝛾𝑗


Ω𝑘

𝜓𝑖𝑘 𝑥, 𝑦 𝜓𝑗

𝑘 𝑥, 𝑦 𝑑𝑥𝑑𝑦 =𝑎00𝐴𝑘12

2 1 11 2 11 1 2

𝐹𝑘 =𝑓𝑘𝐴𝑘3

111

If the vertices of the triangle are 𝑣𝑖 = 𝑥𝑖, 𝑦𝑖 we define:

𝛽𝑖 = 𝑦𝑗 − 𝑦𝑘𝛾𝑖 = − 𝑥𝑗 − 𝑥𝑘


𝐴𝑘 is triangle area

𝑣1

𝑣2

𝑣3

Ω𝑘

All together:



• In the case of a general linear triangular rectangle element for the Poisson’s Equation

(a11 = a22 = 𝑐, a12 = a21 = a00 = 0)

𝑣1

𝑣2

𝑣3

𝑎

𝑏Ω𝑘

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• In the case of a general linear triangular rectangle element for the Poisson’s Equation

(a11 = a22 = 𝑐, a12 = a21 = a00 = 0)

𝐾𝑘 =𝑐

2𝑎𝑏

𝑏2 −𝑏2 0−𝑏2 𝑎2 + 𝑏2 −𝑎2

0 −𝑎2 𝑎2

𝑣1

𝑣2

𝑣3

𝑎

𝑏Ω𝑘

𝐹𝑘 =𝑓𝑘𝐴𝑘3

111

=𝑓𝑘𝑎𝑏

6

111

The formula:

Simplifies to:


Rectangles


Computing the Integrals: Rectangles


In the case of a rectangular quadrilateral

𝑎

𝑏Ω𝑘





𝑎

𝑏Ω𝑘





𝑎

𝑏Ω𝑘




Finite Elements (III)1D Finite Elements



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1D Finite Elements


The 1D case: Weak Formulation

Similar to the 2D case, the model equation for the 1D case is:

If Ωk = [𝑥𝐴, 𝑥𝐵], the variational formulation gives:

Like in 2D, the term 𝑄 ≡ 𝑎1𝑑𝑢

𝑑𝑥, and because the outer normal orientation, we

have 𝑄𝑖𝑘 = −𝑎1

𝑑𝑢

𝑑𝑥ȁ𝑥=𝑋𝐴 and 𝑄𝑖

𝑘 = 𝑎1𝑑𝑢

𝑑𝑥ȁ𝑥=𝑋𝐵 Notice the minus sign on

the left node.




For a 1D element Ωk = [𝑥𝐴, 𝑥𝐵], after passing to the integral equation, using the shape 1D functions (Lagrange Polynomials) and interpolate, the weak or variational formulation gives:




If Ωk = [𝑥𝐴, 𝑥𝐵], the weak or variational formulation gives:


The 1D case: Weak Formulation• As a linear equations system it is written:



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Notice that now these are 1D integrals!!





Like in 2D, the term 𝑄 ≡ 𝑎1𝑑𝑢

𝑑𝑥· 𝑛 , and because the outer normal orientation,

we have 𝑄𝑖𝑘 = −𝑎1

𝑑𝑢

𝑑𝑥ȁ𝑥=𝑋𝐴 and 𝑄𝑖

𝑘 = 𝑎1𝑑𝑢

𝑑𝑥ȁ𝑥=𝑋𝐵 .

𝑥𝐴

Ω𝑘𝑛

− +

𝑥𝐵

Notice the minus sign on the left node.

𝑛


Computing the integrals


The 1D Linear element

Consider now the linear reference element Ω𝑅 = [−1,1]

The shape functions can be written for 𝜉 ∈ [−1,1]

𝜓1𝑅 𝜉 =

1

21 − 𝜉 𝜓2

𝑅 𝜉 =1

21 + 𝜉





The idea is to use integral properties to pass every other linear element to the reference one.

𝜓1𝑅 𝜉 =

1

21 − 𝜉 𝜓2

𝑅 𝜉 =1

21 + 𝜉

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The idea is to use integral properties to pass every other linear element to the reference one.

For a general element Ω𝑘 = [𝑥𝐴, 𝑥𝐵] with 𝑥 ∈ [𝑥𝐴, 𝑥𝐵]we have the change of variables with the interval [−1,1] is:

for ℎ𝑘 = 𝑥𝐵 − 𝑥𝐴𝑥 =ℎ𝑘2

𝜉 + 1 + 𝑋𝐴

𝜉 =2

ℎ𝑘𝑥 − 𝑥𝐴 − 1

Or the inverse:

𝜓1𝑅 𝜉 =

1

21 − 𝜉 𝜓2

𝑅 𝜉 =1

21 + 𝜉



• Therefore, the change of variables in the integrals gives:






• For the second integral



• For the second integral

with




with

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with

=1

2=1

2




with

= ±1

2= ±

1

2

1



• The constant case:

Consider now the case where 𝑎1 𝑥 = 𝑎1𝑘, 𝑎𝑜 𝑥 = 𝑎0

𝑘, =





𝑘, =





𝑘, =





𝑘, =

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𝑘, =





𝑘, =





𝑘, =




collecting all the terms we have


The 1D Quadratic element

• When we consider quadratic elements, the shape functions are

• In the constant coefficient case we obtain:


1D Finite ElementsExamples

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Linear Elasticity 1D

Linear Elasticity 1D equation:

with 𝐸 𝑥 the material elasticity function (Young modulus), 𝐴 𝑥the section area and 𝑢 𝑥 the displacement.


Constant Pillar



• Example 1: Constant loaded column

Let’s assume E·A constant (homogeneous column).

11



• This is a particular case of the model equation

with (if the column weight is not consider).

as we learned before, the problem can be stated as



• This is a particular case of the model equation

with (if the column weight is not consider).

as we learned before, the problem can be stated as


Linear Elasticity 1DAfter assembly the system we obtain

Finally, we impose

Solution:

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Finally, we impose

Solution:



Finally, we impose

Solution:



Finally, we impose

Solution:



Finally, we impose

Unknowns primary: 𝑈2, 𝑈3 , 𝑈4 , 𝑈5 and also secondary 𝑄11



Finally, we impose



Finally, we impose

Solution:

(Reaction force on the ground)

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Variable Pillar


• Example 2: Concrete Pyramidal Column with variable section area


𝑓(𝑥)


• Example 2: Concrete Pyramidal Column with variable

section area


𝑓(𝑥)

Now the inner weight of the column is also taken into account

𝑓 𝑥 = −𝜔𝑑𝑉

𝑑𝑥

𝜔 been the concrete specific weight


• Example 2: Concrete Pyramidal Column with variable

section area


𝑓(𝑥)

Model equation:𝑎1 𝑥 = 𝐸 𝑥 · 𝐴 𝑥 non constant𝑎0 𝑥 = 0


𝑑𝑥internal weight force


Example 2: Concrete Pyramidal Column

Consider


(concrete specific weight)


Linear Elasticity 1DNow, the section area is not constant

and therefore, the internal forces 𝑓 𝑥 = 𝜔𝑑𝑉

𝑑𝑥for unit

length can be expressed by the product

(Hint: This can be obtained by the derivative of the formula for the volume of column above a level x.

𝑉 𝑥 =1

2𝐴 3 + 𝐴 𝑥 · ℎ

)

Wide value as a function 𝑏(𝑥)𝑥 ȁ 0 3𝑏ȁ 1.5 0.5

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𝑑𝑥for unit



𝑉 𝑥 =1

2𝐴 3 + 𝐴 𝑥 · ℎ

)

𝑏(𝑥) =

𝑥 ȁ 0 3𝑏ȁ 1.5 0.5Wide value as a function 𝑏(𝑥)




𝑑𝑥for unit



𝑉 𝑥 =1

2𝐴 3 + 𝐴 𝑥 · ℎ

)

𝑏(𝑥) =

𝑥 ȁ 0 3𝑏ȁ 1.5 0.5

𝐴 𝑥 = 0.5 · 𝑏(𝑥)

Each section is a rectangle:

Wide value as a function 𝑏(𝑥)




𝑑𝑥for unit



𝑉 𝑥 =1

2𝐴 3 + 𝐴 𝑥 · ℎ

)

𝑏(𝑥) =

𝑥 ȁ 0 3𝑏ȁ 1.5 0.5

𝐴 𝑥 = 0.5 · 𝑏(𝑥)

Each section is a rectangle:

Wide value as a function 𝑏(𝑥)


Linear Elasticity 1DThe internal forces corresponding to the pillar weight can be computed from the specific weight


𝑑𝑥

for unit length can be expressed by the product


𝑉 𝑥 =1

2𝐴 3 + 𝐴 𝑥 · ℎ

)



• Here, to obtain the linear system

we need to compute

That gives different values for each element. Considering the first one, , we obtain:

𝜓1𝑘 𝑥 =

𝑥 − 𝑥𝐵𝑥𝐴 − 𝑥𝐵

=𝑥 − 𝑥𝐵−ℎ𝑘

𝜓2𝑘 𝑥 =

𝑥 − 𝑥𝐴𝑥𝐵 − 𝑥𝐴

=𝑥 − 𝑥𝐴ℎ𝑘




we need to compute


𝜓1𝑘 𝑥 =

𝑥 − 𝑥𝐵𝑥𝐴 − 𝑥𝐵

=𝑥 − 𝑥𝐵−ℎ𝑘

𝑑𝜓1𝑘

𝑑𝑥𝑥 =

−1

ℎ𝑘

𝜓2𝑘 𝑥 =

𝑥 − 𝑥𝐴𝑥𝐵 − 𝑥𝐴

=𝑥 − 𝑥𝐴ℎ𝑘

𝑑𝜓2𝑘

𝑑𝑥𝑥 =

1

ℎ𝑘

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we need to compute

That gives different values for each element. Considering the first one, , here ℎ𝑘 = 1,we obtain:




we need to compute





we need to compute





we need to compute




• For the three elements we have:

and next, assembling the system

With BC

Solution:© Numerical Factory


• For the three elements we have:

and next, assemble the system

With BC

Solution:

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With BC


essential condition


With BC

Solution:


essential BC:

natural BC:

𝑄1 =𝐸

64𝑈1 − 4𝑈2 +

25𝜔

72= −4.3586𝑒 + 04




Finite Elements (IV)Boundary Conditions



FEM 2D Natural Boundary Conditions


Boundary Conditions

1 2

𝐿𝑒

1D – Boundary Conditions (BC)

𝑄1𝑘 = −𝑎1

𝑑𝑢

𝑑𝑥𝑄2𝑘 = +𝑎1

𝑑𝑢

𝑑𝑥𝑄𝑖𝑘 = 𝑎1

𝑑𝑢

𝑑𝑥· 𝑛


Boundary Conditions

1 2

𝐿𝑒


𝑄1𝑘 = −𝑎1

𝑑𝑢


𝑑𝑢


𝑑𝑢

𝑑𝑥· 𝑛


𝑄𝑖𝑘 = න

Γ𝑘𝑞𝑛𝜓𝑖 𝑥, 𝑦 𝑑𝑠 where

and Γ𝑘 ≡ 𝜕Ωk is the boundary of the element Ω𝑘 .


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Boundary Conditions

1 2

𝐿𝑒


𝑄1𝑘 = −𝑎1

𝑑𝑢


𝑑𝑢


𝑑𝑢

𝑑𝑥· 𝑛


𝑄𝑖𝑘 = න

𝜕Ω

𝑞𝑛𝜓𝑖 𝑥, 𝑦 𝑑𝑠 where

Triangular element

and Γ𝑘 ≡ 𝜕Ωk is the boundary of the element Ω𝑘 .


Boundary Conditions

If we consider a triangular element

Γ1

Γ2

Γ3

1 2

3


Boundary Conditions


Γ1

Γ2

Γ3

1 2

3


Boundary Conditions


Γ1

Γ2

Γ3

1 2

3


Boundary Conditions


Γ1

Γ2

Γ3

1 2

3


Boundary Conditions


Γ1

Γ2

Γ3

1 2

3

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( 𝑄𝑖𝑗𝑘 means the flux on node 𝑖 corresponding to the contribution of edge 𝑗 )

Boundary Conditions


with

Γ1

Γ2

Γ3

1 2

3



Boundary Conditions


with

Γ1

Γ2

Γ3

1 2

3



Boundary Conditions


with

Γ1

Γ2

Γ3

1 2

3


BC on Triangles


Boundary Conditions

Triangular Shape Functions (a plane)

𝜓𝑖𝑘 𝑥, 𝑦 = 𝛼 + 𝛽𝑥 + 𝛾𝑦


Boundary Conditions

(0, 0)

(1, 0.5)

(0.5, 1)

1

2

3Triangular Shape Functions (a plane)

Triangular Element𝜓𝑖𝑘 𝑥, 𝑦 = 𝛼 + 𝛽𝑥 + 𝛾𝑦

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Boundary Conditions

(0, 0)

(1, 0.5)

(0.5, 1)

1

2

3Triangular Shape Functions

Triangular Element

𝜓1𝑘(𝑥, 𝑦)



Boundary Conditions

(0, 0)

(1, 0.5)

(0.5, 1)

1

2


Triangular Element

𝜓1𝑘(𝑥, 𝑦) 𝜓2

𝑘(𝑥, 𝑦)



Boundary Conditions

(0, 0)

(1, 0.5)

(0.5, 1)

1

2


Triangular Element

𝜓1𝑘(𝑥, 𝑦) 𝜓2

𝑘(𝑥, 𝑦) 𝜓3𝑘(𝑥, 𝑦)



Boundary Conditions

• Restriction of 𝜓𝑖𝑘(𝑥, 𝑦) to the boundaries

First function: 𝜓1𝑘(𝑥, 𝑦)

𝜓11𝑘 𝑥, 𝑦 ቚ

Γ1= 1 −

𝑠

ℎ1, 𝑠 ∈ [0, ℎ1]

Restriction to Γ1

1

0

ℎ1𝑠 = 0 𝑠 = ℎ1



Boundary Conditions



Γ2= 0

Restriction to Γ2

0

ℎ2𝑠 = 0 𝑠 = ℎ2


0𝜓1𝑘(𝑥, 𝑦)


Boundary Conditions




Γ3=

𝑠

ℎ3, 𝑠 ∈ [0, ℎ3]

Restriction to Γ3

1

0

ℎ3𝑠 = 0 𝑠 = ℎ3


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Boundary Conditions


Second function: 𝜓2𝑘(𝑥, 𝑦)


Γ1=

𝑠

ℎ1, 𝑠 ∈ [0, ℎ1]

Restriction to Γ1

1

0

ℎ1𝑠 = 0 𝑠 = ℎ1


Boundary Conditions




Γ2= 1 −

𝑠

ℎ2, 𝑠 ∈ [0, ℎ2]

Restriction to Γ2

1

0

ℎ2𝑠 = 0 𝑠 = ℎ2


Boundary Conditions




Γ3= 0

Restriction to Γ3

0

ℎ3𝑠 = 0 𝑠 = ℎ3

0


Boundary Conditions


Third function: 𝜓3𝑘(𝑥, 𝑦)


Γ1= 0

Restriction to Γ1


Γ2=

𝑠

ℎ2, 𝑠 ∈ [0, ℎ2]

Restriction to Γ2


Γ3= 1 −

𝑠

ℎ3, 𝑠 ∈ [0, ℎ3]

Restriction to Γ3


Boundary Conditions

• For the shape functions, they can be seen as the 1D Lagrange’s polynomial associated to the edge

ℎ𝑗𝑘 is the length of the j-th edge of the triangle

𝑣3

𝑣1𝑣2𝑒1

𝑒2𝑒3

Edge notation:


Boundary Conditions

• Balance, of course, applies to interior faces

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Boundary Conditions

• Balance, of course, applies to interior faces


Boundary Conditions

• Balance, of course, applies to interior faces and only the ones on the boundary have to be consider


Constant BC on Triangles


Constant Boundary Conditions

• Consider the case where a constant BC is applied to one edge of the triangle

2

3

𝑞𝑛𝑘 = ቐ

𝑞0 on Γ10 on Γ20 on Γ3




2

3

𝑞𝑛𝑘 = ቐ


𝑄1𝑘 = 𝑄11

𝑘 +𝑄13𝑘

Node 1




2

3

𝑞𝑛𝑘 = ቐ


𝑄1𝑘 = 𝑄11

𝑘 +𝑄13𝑘

Node 1

𝑄11𝑘 = න

Γ1

𝑞𝑛𝜓11𝑘 𝑠 𝑑𝑠 =

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2

3

𝑞𝑛𝑘 = ቐ


𝑄1𝑘 = 𝑄11

𝑘 +𝑄13𝑘

Node 1

𝑄11𝑘 = න

Γ1


= 𝑞0 0ℎ1 1 −

𝑠

ℎ1𝑑𝑠 =

1

2𝑞0ℎ1




2

3

𝑞𝑛𝑘 = ቐ


𝑄1𝑘 = 𝑄11

𝑘 +𝑄13𝑘

Node 1

𝑄11𝑘 = න

Γ1


= 𝑞0 0ℎ1 1 −

𝑠

ℎ1𝑑𝑠 =

1

2𝑞0ℎ1

𝑄13𝑘 = න

Γ3

𝑞𝑛𝜓13𝑘 𝑠 𝑑𝑠 = 0




2

3

𝑞𝑛𝑘 = ቐ


𝑄2𝑘 = 𝑄21

𝑘 +𝑄22𝑘

Node 2




2

3

𝑞𝑛𝑘 = ቐ


𝑄2𝑘 = 𝑄21

𝑘 +𝑄22𝑘

Node 2

𝑄21𝑘 = න

Γ1





2

3

𝑞𝑛𝑘 = ቐ


𝑄2𝑘 = 𝑄21

𝑘 +𝑄22𝑘

Node 2

𝑄21𝑘 = න

Γ1


= 𝑞0 0ℎ1 𝑠

ℎ1𝑑𝑠 =

1

2𝑞0ℎ1




2

3

𝑞𝑛𝑘 = ቐ


𝑄2𝑘 = 𝑄21

𝑘 +𝑄22𝑘

Node 2

𝑄21𝑘 = න

Γ1


= 𝑞0 0ℎ1 𝑠

ℎ1𝑑𝑠 =

1

2𝑞0ℎ1

𝑄22𝑘 = න

Γ2


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2

3

𝑞𝑛𝑘 = ቐ


𝑄3𝑘 = 𝑄32

𝑘 +𝑄33𝑘

Node 3




2

3

𝑞𝑛𝑘 = ቐ


𝑄3𝑘 = 𝑄32

𝑘 +𝑄33𝑘

Node 3

𝑄32𝑘 = න

Γ2





2

3

𝑞𝑛𝑘 = ቐ


𝑄3𝑘 = 𝑄32

𝑘 +𝑄33𝑘

Node 3

𝑄32𝑘 = න

Γ2


𝑄33𝑘 = න

Γ3





2

3

𝑞𝑛𝑘 = ቐ


𝑄1𝑘

𝑄2𝑘

𝑄3𝑘

=𝑞0ℎ12

110

The constant value is distributed between nodes 1 and 2


Linear BC on Triangles


2

3

Linear Boundary Conditions

• Consider now a linear function applied at edge 1

𝑞𝑛𝑘 =

𝑞0𝑠

ℎ1on Γ1

0 on Γ20 on Γ3

𝑄1𝑘 = 𝑄11

𝑘 +𝑄13𝑘

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2

3



𝑞𝑛𝑘 =

𝑞0𝑠

ℎ1on Γ1

0 on Γ20 on Γ3

𝑄1𝑘 = 𝑄11

𝑘 +𝑄13𝑘


2

3



𝑞𝑛𝑘 =

𝑞0𝑠

ℎ1on Γ1

0 on Γ20 on Γ3

𝑄1𝑘 = 𝑄11

𝑘 +𝑄13𝑘

𝑄13𝑘 = 0


2

3



𝑞𝑛𝑘 =

𝑞0𝑠

ℎ1on Γ1

0 on Γ20 on Γ3

𝑄1𝑘 = 𝑄11

𝑘 +𝑄13𝑘

𝑄2𝑘 = 𝑄21

𝑘 +𝑄22𝑘


2

3



𝑞𝑛𝑘 =

𝑞0𝑠

ℎ1on Γ1

0 on Γ20 on Γ3

𝑄1𝑘 = 𝑄11

𝑘 +𝑄13𝑘

𝑄2𝑘 = 𝑄21

𝑘 +𝑄22𝑘


2

3



𝑞𝑛𝑘 =

𝑞0𝑠

ℎ1on Γ1

0 on Γ20 on Γ3

𝑄1𝑘 = 𝑄11

𝑘 +𝑄13𝑘

𝑄2𝑘 = 𝑄21

𝑘 +𝑄22𝑘

𝑄22𝑘 = 0


2

3



𝑞𝑛𝑘 =

𝑞0𝑠

ℎ1on Γ1

0 on Γ20 on Γ3

𝑄1𝑘 = 𝑄11

𝑘 +𝑄13𝑘

𝑄2𝑘 = 𝑄21

𝑘 +𝑄22𝑘

𝑄3𝑘 = 𝑄32

𝑘 +𝑄33𝑘

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2

3



𝑞𝑛𝑘 =

𝑞0𝑠

ℎ1on Γ1

0 on Γ20 on Γ3

𝑄1𝑘 = 𝑄11

𝑘 +𝑄13𝑘

𝑄2𝑘 = 𝑄21

𝑘 +𝑄22𝑘

𝑄3𝑘 = 𝑄32

𝑘 +𝑄33𝑘

𝑄32𝑘 = 0, 𝑄33

𝑘 = 0


2

3



𝑞𝑛𝑘 =

𝑞0𝑠

ℎ1𝑘 on Γ1

0 on Γ20 on Γ3

𝑄1𝑘 = 𝑄11

𝑘 +𝑄13𝑘

𝑄2𝑘 = 𝑄21

𝑘 +𝑄22𝑘

𝑄3𝑘 = 𝑄32

𝑘 +𝑄33𝑘

𝑄32𝑘 = 0, 𝑄33

𝑘 = 0

𝑄1𝑘

𝑄2𝑘

𝑄3𝑘

=𝑞0ℎ1

𝑘

6

120


Different Boundary Conditions

• As a final case now we have de contribution

of the two previous cases. In principle,

2

3

𝑄1𝑘

𝑄2𝑘

𝑄3𝑘

=𝑞1ℎ3

𝑘

2

101

+𝑞0ℎ1

𝑘

6

120

=

𝑞1ℎ3𝑘

2+1

6𝑞0ℎ1

𝑘

1

3𝑞0ℎ1

𝑘

𝑞1ℎ3𝑘

2

𝑞𝑛𝑘 =

𝑞0𝑠

ℎ1𝑘 on Γ1

0 on Γ2𝑞1 on Γ3





2

3

𝑄1𝑘

𝑄2𝑘

𝑄3𝑘

=𝑞1ℎ3

𝑘

2

101

+𝑞0ℎ1

𝑘

6

120

𝑞𝑛𝑘 =

𝑞0𝑠

ℎ1𝑘 on Γ1






2

3

𝑄1𝑘

𝑄2𝑘

𝑄3𝑘

=𝑞1ℎ3

𝑘

2

101

+𝑞0ℎ1

𝑘

6

120

𝑞𝑛𝑘 =

𝑞0𝑠

ℎ1𝑘 on Γ1






2

3

𝑄1𝑘

𝑄2𝑘

𝑄3𝑘

=𝑞1ℎ3

𝑘

2

101

+𝑞0ℎ1

𝑘

6

120

=

𝑞1ℎ3𝑘

2+1

6𝑞0ℎ1

𝑘

1

3𝑞0ℎ1

𝑘

𝑞1ℎ3𝑘

2

𝑞𝑛𝑘 =

𝑞0𝑠

ℎ1𝑘 on Γ1


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• Example of local and global nodes enumeration for linear triangular elements

Connectivity matrix

𝐵 =

1523

2345

3256

Hint: Notice that the local enumeration must becounter-clockwise in order to preserve orientation



• For linear triangular elements

If 𝒖 𝒙 is a 1D magnitude (temperature)For each element the Stiffness Matrix isa 3x3 matrix

𝐾𝑒 =

𝑘11𝑒 𝑘12

𝑒 𝑘13𝑒

𝑘21𝑒 𝑘22

𝑒 𝑘23𝑒

𝑘31𝑒 𝑘32

𝑒 𝑘33𝑒

, 𝑢 =

𝑢1𝑢2𝑢3


𝐵 =

1523

2345

3256


𝐵 =

1523

2345

3256


𝐵 =

1523

2345

3256


𝐵 =

1523

2345

3256

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2D Assembly of mixed elements

• Let’s consider the example:

Although it is not usual,

we can mix different type of

elements:

Triangular + Rectangular


2D Assembly of element equations

• The connectivity matrix in this case is not uniform

𝐾𝑒 =

𝑘11𝑒 𝑘12

𝑒 𝑘13𝑒

𝑘21𝑒 𝑘22

𝑒 𝑘23𝑒

𝑘31𝑒 𝑘32

𝑒 𝑘33𝑒

, e=1,2,3,4𝐾𝑒 =

𝑘11𝑒 𝑘12

𝑒

𝑘21𝑒 𝑘22

𝑒

𝑘13𝑒 𝑘14

𝑒

𝑘23𝑒 𝑘24

𝑒

𝑘31𝑒 𝑘32

𝑒

𝑘41𝑒 𝑘42

𝑒

𝑘33𝑒 𝑘34

𝑒

𝑘43𝑒 𝑘44

𝑒

, e=5,6

TriangularRectangular



• The global stiffness matrix is

with



• Let’s consider a simple example to explain flux balance and BC for the assembled system

2 2



• Here the balance must be imposed on nodes 2 and 4

remember that 𝑄𝑖𝑗𝑘 means

the flux on node 𝑖 corresponding to

the contribution of edge 𝑗

Consider node 2:

by construction we also have therefore

, that have to be defined on the BC of the problem.







Consider node 2:



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Consider node 2:



Finite Elements (I) Model Equation

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Transcript of Finite Elements (I) Model Equation