MECH593 Introduction to Finite Element Methods

Finite Element Analysis of 2-D Problems

Dr. Wenjing Ye

2-D Discretization

Common 2-D elements:

2-D Model Problem with Scalar Function- Heat Conduction

• Governing Equation

0),(),(),(

yxQy

yxT

yx

yxT

x in W

• Boundary Conditions

Dirichlet BC:

Natural BC:

Mixed BC:

Weak Formulation of 2-D Model Problem

• Weighted - Integral of 2-D Problem -----

• Weak Form from Integration-by-Parts -----

0 ( , )w T w T

wQ x y dxdyx x y y

T Tw dxdy w dxdy

x x y y

( , ) ( , )( , ) 0

T x y T x yw Q x y dA

x x y y

Weak Formulation of 2-D Model Problem• Green-Gauss Theorem -----

where nx and ny are the components of a unit vector, which is normal to the boundary G.

sinjcosijninn yx

Ω

❑ 𝜕𝜕 𝑥 (𝜅𝑤 𝜕𝑇

𝜕𝑥 )𝑑𝑥𝑑𝑦=∮Γ

❑

(𝜅𝑤 𝜕𝑇𝜕𝑥 )𝑛𝑥𝑑𝑠

Ω

❑ 𝜕𝜕 𝑦 (𝜅𝑤 𝜕𝑇

𝜕 𝑦 )𝑑𝑥𝑑𝑦=∮Γ❑

(𝜅𝑤 𝜕𝑇𝜕 𝑦 )𝑛 𝑦𝑑𝑠

Weak Formulation of 2-D Model Problem

• Weak Form of 2-D Model Problem -----

EBC: Specify T(x,y) on G

NBC: Specify on Gx y

T Tn n

x y

where is the normal

outward flux on the boundary G at the segment ds.

( )n x y

T Tq s i j n i n j

x y

Ω

❑ [𝜕𝑤𝜕𝑥 (𝜅 𝜕𝑇𝜕 𝑥 )+ 𝜕𝑤𝜕𝑦 (𝜅 𝜕𝑇𝜕 𝑦 )−𝑤𝑄 (𝑥 , 𝑦 )]𝑑𝑥𝑑𝑦−∮Γ

❑

𝑤 [(𝜅 𝜕𝑇𝜕 𝑥 )𝑛𝑥+(𝜅 𝜕𝑇𝜕 𝑦 )𝑛𝑦 ]𝑑𝑠=0

FEM Implementation of 2-D Heat Conduction – Shape Functions

Step 1: Discretization – linear triangular element (T3)

T1

T3

T2

332211 TTTT

Derivation of linear triangular shape functions:

Let ycxcc 2101

Interpolation properties

1 at node

0 at other nodesi

i

ith

0 1 1 2 1

0 1 2 2 2

0 1 3 2 3

1

0

0

c c x c y

c c x c y

c c x c y

1

0 1 1

1 2 2

2 3 3

1 1

1 0

1 0

c x y

c x y

c x y

1

1 1 2 3 3 2

1 2 2 2 3

3 3 3 2

1 11

1 1 02

1 0 e

x y x y x yx y

x y x y y yA

x y x x

Same 3 1 1 3

2 3 1

1 3

1

2 e

x y x yx y

y yA

x x

1 2 2 1

3 1 2

2 1

1

2 e

x y x yx y

y yA

x x

FEM Implementation of 2-D Heat Conduction – Shape Functions

linear triangular element – local (area) coordinates

T1

T3

T2

A3

A1

A2

f1

f2

f3

𝜙1=(1 𝑥 𝑦 )

2 𝐴𝑒 {𝑥2𝑦 3−𝑥3 𝑦2

𝑦 2− 𝑦3

𝑥3−𝑥2}= 𝐴1

𝐴𝑒

=𝜉

𝜙2=(1 𝑥 𝑦 )

2 𝐴𝑒 {𝑥3 𝑦1−𝑥1 𝑦3

𝑦3− 𝑦1

𝑥1−𝑥3}= 𝐴2

𝐴𝑒

=𝜂

𝜙3=(1 𝑥 𝑦 )

2 𝐴𝑒 {𝑥1 𝑦2−𝑥2 𝑦1

𝑦1− 𝑦2

𝑥2−𝑥1}= 𝐴3

𝐴𝑒

=1−𝜉−𝜂=𝜁

T3

T2

T1

𝜉=1

𝜉=0

𝜂=1

𝜂=0

FEM Implementation of 2-D Heat Conduction – Shape Functions

quadratic triangular element (T6) – local (area) coordinates

T1

T3

T2

T3

T2

T1

𝜉=1

𝜉=0

𝜂=1

𝜂=0

T4

T6

T5

𝜉=0.5T4

T6

T5

Serendipity Family – nodes are placed on the boundary

for triangular elements, incomplete beyond quadratic

Interpolation Function - Requirements

• Interpolation condition

• Take a unit value at node i, and is zero at all other nodes

• Local support condition

• fi is zero at an edge that doesn’t contain node i.

• Interelement compatibility condition

• Satisfies continuity condition between adjacent elements over any element boundary that includes node i

• Completeness condition

• The interpolation is able to represent exactly any displacement field which is polynomial in x and y with the

order of the interpolation function

Formulation of 2-D 4-Node Rectangular Element – Bi-linear Element (Q4)

1 1 2 2 3 3 4 4( , )u u u u u

f1 f2

f3

f4

Note: The local node numbers should be arranged in a counter-clockwise sense. Otherwise, the area Of the element would be negative and the stiffness matrix can not be formed.

Let

x

h

1 2

34𝜙1=

14

(1−𝜉 ) (1−𝜂 )

𝜙2=14

(1+𝜉 ) (1−𝜂 )

𝜙3=14

(1+𝜉 ) (1+𝜂 )

𝜙4=14

(1−𝜉 ) (1+𝜂 )

Formulation of 2-D 4-Node Rectangular Element – Bi-linear Element (Q4)

Physical domain (physical element) Reference domain (master element)

x

h

x

h

x

y 12

34

1 2

34

• Weak Form of 2-D Model Problem -----

Assume approximation:

and let w(x,y)=fi(x,y) as before, then

e

j ji iijK dxdy

x x y y

where

FEM Implementation of 2-D Heat Conduction – Element Equation

𝑇=∑𝑗=1

𝑛

𝑇 𝑗𝜙 𝑗

Ω 𝑒

❑ [ 𝜕𝑤𝜕𝑥 (𝜅 𝜕𝑇𝜕𝑥 )+ 𝜕𝑤𝜕𝑦 (𝜅 𝜕𝑇𝜕𝑦 )−𝑤𝑄 (𝑥 , 𝑦 )]𝑑𝑥𝑑𝑦+∮Γ𝑒

❑

𝑤𝑞𝑛𝑑𝑠=0

Ω 𝑒

❑ [ 𝜕𝜙 𝑖

𝜕 𝑥𝜅

𝜕𝜕 𝑥 (∑𝑗=1

𝑛

𝑇 𝑗𝜙 𝑗)+ 𝜕𝜙 𝑖

𝜕 𝑦𝜅

𝜕𝜕 𝑦 (∑

𝑗=1

𝑛

𝑇 𝑗𝜙 𝑗)−𝜙𝑖𝑄 (𝑥 , 𝑦 )]𝑑𝑥𝑑𝑦+∮Γ 𝑒

❑

𝜙𝑖𝑞𝑛𝑑𝑠=0∑𝑗=1

𝑛

𝐾 𝑖 𝑗𝑇 𝑗=Ω𝑒

❑

𝜙𝑖𝑄 (𝑥 , 𝑦 )𝑑𝑥𝑑𝑦−∮Γ 𝑒

❑

𝜙𝑖𝑞𝑛𝑑𝑠

23 23 23 31 23 12

23 31 31 31 31 12

23 12 31 12 12 12

4 e

l l l l l l

K l l l l l lAl l l l l l

Linear Triangular Element Equation

1 1

2 2

3 3

Q q

F Q q

Q q

∑𝑗=1

𝑛

𝐾 𝑖 𝑗𝑇 𝑗=Ω𝑒

❑

𝜙𝑖𝑄 (𝑥 , 𝑦 )𝑑𝑥𝑑𝑦−∮Γ 𝑒

❑

𝜙𝑖𝑞𝑛𝑑𝑠

2 3 3 2

11 2 3

3 2

1

2 e e

x y x yx y A

y yA A

x x

3 1 1 3

22 3 1

1 3

1

2 e e

x y x yx y A

y yA A

x x

1 2 2 1

33 1 2

2 1

1

2 e e

x y x yx y A

y yA A

x x

where is the length vector from the ith node to the jth node.

Assembly of Stiffness Matrices

𝐹 𝑖❑(𝑒)=

Ω𝑒

❑

𝜙 𝑖𝑄 (𝑥 , 𝑦 )𝑑𝑥𝑑𝑦 −∮Γ 𝑒

❑

𝜙 𝑖𝑞𝑛𝑑𝑠=∑𝑗=1

𝑛𝑒

𝐾 𝑖𝑗❑(𝑒)𝑇 𝑗❑

(𝑒)

, , , , ,

Imposing Boundary Conditions

The meaning of qi:

(1) (1) (1)1 12 23 31

(1) (1)12 23

(1) (1) (1) (1) (1) (1) (1) (1) (1)2 2 2 2 2

(1) (1) (1) (1)2 2

n n n n

h h h

n n

h h

q q ds q ds q ds q ds

q ds q ds

(1) (1) (1)1 12 23 31

(1) (1)23 31

(1) (1) (1) (1) (1) (1) (1) (1) (1)3 3 3 3 3

(1) (1) (1) (1)3 3

n n n n

h h h

n n

h h

q q ds q ds q ds q ds

q ds q ds

(1) (1) (1)1 12 23 31

(1) (1)12 31

(1) (1) (1) (1) (1) (1) (1) (1) (1)1 1 1 1 1

(1) (1) (1) (1)1 1

n n n n

h h h

n n

h h

q q ds q ds q ds q ds

q ds q ds

1

2

3

11

2

3

1

2

3

11

2

3

Imposing Boundary Conditions

(1) ( 2)23 41

(1) (2)n nh hq qEquilibrium of flux:

(1) ( 2) (1) ( 2)23 41 23 41

(1) (1) (2) (2) (1) (1) (2) (2)2 1 3 4; n n n n

h h h h

q ds q ds q ds q ds

FEM implementation:

(1) (2)2 2 1q q q

(1) (1)12 23

(1) (1) (1) (1) (1)2 2 2n n

h h

q q ds q ds

Consider

( 2) ( 2)12 41

(2) (2) (2) (2) (2)1 1 1n n

h h

q q ds q ds

(1) ( 2)12 12

(1) (1) (2) (2)2 2 1n n

h h

q q ds q ds (1) ( 2)31 34

(1) (1) (2) (2)3 3 4n n

h h

q q ds q ds

(1) (2)3 3 4q q q

(1) (1)23 31

(1) (1) (1) (1) (1)3 3 3n n

h h

q q ds q ds ( 2) ( 2)34 41

(2) (2) (2) (2) (2)4 4 4n n

h h

q q ds q ds

Calculating the q Vector

Example:

293T K0nq

1nq

2-D Steady-State Heat Conduction - Example

0nq

0.6 m

0.4 m

A

BC

DAB:

CD: convectionCm

Wh o 250 CT o25

DA and BC: CT o180

x

y