enis-fem chapter 2

8/3/2019 Enis-fem Chapter 2

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Finite Element Method Modeling Dr. Slim Choura

2Finite elementanalysis of two-

dimensional

Problems


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Finite Element Method Modeling Dr. Slim ChouraPage 47

The objective of this chapter is to extend the basic steps discussed earlier for one-

dimensional problems to two-dimensional boundary-value problems involving a single

dependent variable.

2.1 Boundary Value Problems2.2 The Model EquationConsider the problem of finding the solution u of the second-order partial differential

equation

11 12 21 22 00 0u u u u

a a a a a u f x x y y x y

+ + + =

(2. 1)

where ( )2,1, =jiaij , 00a and f , and the specified boundary conditions are given. The

Poisson equation corresponds to aaa 2211 == and 12 21 00 0a a a= = = :

( )a u f = in (2. 2)

where is the gradient operator defined by

x y

= +

i j (2. 3)

2 2

2

2 2x y

= = +

(2. 4)

with 2 defined as the Laplacian operator, and i and j are the unit vectors along the x

andy axes, respectively. Equation (2.2) in the cartesian coordinate system takes the form:

u ua a f

x x y y =

(2. 5)

2.2.1 Finite Element DiscretizationIn two dimensions, there is more than one simple geometric shape that can be used in a

finite element (see figure 2.1). As will be discussed, a triangle is the easiest geometric

shape, followed by a rectangle.


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Figure 2.1. A typical triangular element

The general rule of mesh generation for finite element formulations includes the

following:

1. Select elements that characterize the governing equations of the problem.2. The number, shape, and type of elements should be such that the geometry of the

domain is represented as accurately as desired.

3. The density of elements should be such that regions of large gradients of thesolution are adequately modeled.

4. Mesh refinements should vary gradually from high density regions to low-densityregions. Iftransition elements are used, they should be used away from critical

regions (i.e., regions of large gradients). Transition elements are those that

connect lower-order elements to higher-order elements (e.g., linear to quadratic).

2.2.2 Weak FormIntegrate the resulting equation over the element domain e :

( ) ( )e

1 2 00 0w F F a u f dxdyx y

+ =

(2. 6)where

e

ds e

i

j x

y

n


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1 11 12 2 21 22u u u u

F a a F a a x y x y

= + = +

(2. 7)

In the second step, we distribute the differentiable equally between u and w . First we

note the following identity:

( ) ( )1 21 1 2 2F Fw w

w F wF w F wF x x x y y y

= =

(2. 8)

Using the gradient (or divergence) theorem:

( )e e

1 1 xwF dxdy wF n ds

x

=

(2. 9)

( )e e

2 2 ywF dxdy wF n ds

y

=

(2. 10)where

xn and yn are the direction cosines of the unit normal vector

cos sinx yn n = + = +i j i j (2. 11)

on the boundary e , and ds is the arc length of infinitesimal line element along the

boundary (see figure 2.1). Using (2.8), (2.9) and (2.10) in (2.6) we get

e

e

11 12 21 22 00

11 12 21 22

0

x y

w u u w u ua a a a a wu wf dxdy

x x y y x y

u u u uw n a a n a a ds

x y x y

= + + + +

+ + +

(2. 12)

From the above equation, u is the primary variable and

11 12 21 22n x y

u u u uq n a a n a a

x y x y

+ + +

(2. 13)

is the secondary variable of the formulation.


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The third and last step is to use the definition (2.13) in (2.13) and write the weak form of

(2.1) as:

e

e

11 12 21 22 000=

n

w u u w u ua a a a a wu wf dxdy

x x y y x y

w q ds

+ + + +

(2. 14)

or

( ) ( ),B w u l w= (2. 15)

where

( )e

11 12 21 22 00,w u u w u u

B w u a a a a a wu dxdy x x y y x y

= + + + +

(2. 16)

( )e e

nl w wf dxdy wq ds

= (2. 17)The quadratic functional can be obtained when the ( ) ,B is symmetric ( 12 21a a= )

( ) ( ) ( )1 ,2

I u B u u l u= (2. 18)

2.2.3 Finite Element ModelSuppose that u is approximated over a typical element e by the expression

( ) ( ) ( )

1

, , ,

n

e e e

j j

j

u x y U x y u x y

=

= (2. 19)where eju is the value of

eU at the jth

node ( ),j jx y of the element, ande

j are the

Lagrange interpolation functions, with the property

( ),ei j j ijx y = (2. 20)

Substituting (2.19) into (2.14), we obtain


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e

e

11 12

1 1

21 22

1 1

00

1

0

n ne e

j je e

j j

j j

n ne e

j je e

j j

j j

n

e e

j j n

j

wa u a u

x x y

wa u a u

y x y

a w u wf dxdy w q ds

= =

= =

=

= +

+ +

+

(2. 21)

Since we need n independent algebraic equations to solve for the n unknowns

1 2, , ... ,e e enu u u we choose n independent functions for w : 1 1, , ,e e enw = . The ith

algebraic equation is obtained by substituting ei

w = into (2.21):

e

11 12 21 22 00

1

0

( 1, 2, , )

e

e

ne e e ee e j j j j e ei i

i j j

j

e e

i i n

a a a a a dxdy u

x x y y x y

f dxdy q ds i n

=

= + + + +

=

(2. 22)

or

1

n

e e e e

ij j i i

j

K u f Q

=

= + (2. 23)where

11 12 21 22 00

e

e e

e e e ee e j j j je e ei i

ij i j

e e e e

i i i n i

K a a a a a dxdy

x x y y x y

f f dxdy Q q ds

= + + + +

= =

(2. 24)

In matrix notation, (2.24) takes the form

{ } { } { }e e e eK u f Q = + (2. 25)


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Note that e eij ji

K K= only when 12 21a a= . Equation (2.25) represents the finite element

model of (2.1).

2.2.4 Interpolation FunctionsAn examination of the variational form (2.14) and the finite element matrices in (2.24)

shows that ei should be at least linear functions ofx and y . We shall consider linear

triangular elements and linear rectangular elements.

2.2.4.1. Linear triangular elementConsider the linear approximation

( ) 1 2 3,eU x y c c x c y= + + (2. 26)

We must rewrite the approximation (2.26) such it satisfies the conditions

( ),e e e ei i iU x y u= (2. 27)

where ( ),e ei ix y (i = 1, 2, 3) are the global coordinates of the three vertices of the triangle

e

Thus,

( )

( )

( )

e

1 1 1 1 2 1 3 1

e

2 2 2 1 2 2 3 2

e

3 3 3 1 2 3 3 3

,

,

,

u U x y c c x c y

u U x y c c x c y

u U x y c c x c y

= + +

= + +

= + +

(2. 28)

In matrix form:

1 1 1 1

2 2 2 2

3 3 3 3

1

1

1

u x y c

u x y c

u x y c

=

(2. 29)

1

2

3


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Solving for the cs we obtain

( )( )

( )

1 1 1 2 2 3 3

2 1 1 2 2 3 3

3 1 1 2 2 3 3

1

21

21

2

e

e

e

c u u uA

c u u uA

c u u uA

= + +

= + +

= + +

(2. 30)

where

1 2 3

2e

A + +

= (2. 31)

is the area of the triangle and i , i and i are the geometric constants

( )

( ); and , and permute in a natural order

i j k k j

i j k

i j k

x y x y

y y i j k i j k

x x

=

=

=

(2. 32)

Substituting for ic from (2.30) into (2.26), we get

( ) ( ) ( ) ( )

( )

1 1 2 2 3 3 1 1 2 2 3 3 1 1 2 2 3 3

3

1

1,

2

,

e

e

e e

i i

i

U x y u u u u u u x u u u y

A

u x y

=

= + + + + + + + +

=

(2. 33)

where ei are the linear interpolation functions for the triangular element

( )1

( 1, 2, 3)2

e

i i i i

e

x y iA

= + + = (2. 34)

The linear interpolation functions ei are shown in figure (2.2). They have the properties

( ),e e ei j j ijx y = (i,j = 1, 2, 3) (2. 35)

3 3 3

1 1 1

1, 0, 0e e

e i i

i

i i i

x y

= = =

= = =

(2. 36)


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Figure 2.2. Linear interpolation functions for the three-node triangular element

Example 2.1

Consider the triangular element shown in figure 2.3.

Figure 2.3. A triangular element with element nodes and coordinates

It can be shown that

1 2 3

1 2 3

1 2 3

11 5 1

1 3 2

2 1 3

= = =

= = =

= = =

and

( ) ( ) ( )1 2 31 1 1

11 2 5 3 1 2 37 7 7

e e e x y x y x y = = + = +

2.2.4.2. Linear rectangular elementHere we consider an approximation of the form

( ) 1 2 3 4,eU x y c c x c y c xy= + + + (2. 37)

1

2

3

1

2

3

1

3

1

2

32

1

1

1

1

2

3

(2 , 1)

(5 , 3)

(3 , 4)

x

y


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and use a rectangular element of sides a and b (see figure 2.4).

(a)

(b)

Figure 2.2. Linear rectangular elements and its interpolation functions

For the sake of convenience we choose a local coordinate system ( )yx , to derive the

interpolation functions. We assume that

( ) 1 2 3 4,eU x y c c x c y c xy= + + + (2. 38)

and require

( )( )

( )

( )

1 1

2 1 2

3 1 2 3 4

4 1 3

0,0,0

,

0,

e

e

e

e

u U cu U a c c a

u U a b c c a c b c ab

u U b c c b

= =

= = +

= = + + +

= = +

(2. 39)

Solving for ic ( 1, 2, 3, 4i = ), we obtain

1 1c u= 2 12

u uc

a

= 4 13

u uc

b

= 3 4 1 24

u u u uc

ab

+ = (2. 40)

x

x

y y

a

b

1

4 3

2

1

2

3

4

1

1

1

2

3

4

1

2

1

2

3

4

1

3

1

2

3

4

14


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Substituting these into (2.38), we obtain

( ) ( )

4

1

, ,e e e

i i

i

U x y u x y=

=

(2. 41)

where

1 2 3 41 1 1 1e e e e x y x y x y x y

a b a b a b a b

= = = =

(2. 42)

2.2.5 Evaluation of Element Matrices, and VectorsThe exact solution of the element matrices eK and { }

ef in (2.24) is, in general, not

easy. Therefore, they are evaluated using numerical integration techniques. However,

when ija , 00a , and f are element-wise constant, it is possible to evaluate the integrals

exactly over the linear triangular and rectangular elements.

Let eK in (2.24) be rewritten as the sum of basic matrices S ( 2,1,0, = ):

00 11 12 12 22

00 11 12 21 22

TeK a S a S a S a S a S = + + + + (2. 43)

where

e

ij i, j,S dxdy

= (2. 44)

2.2.5.1Element matrices for a linear triangular element

For a triangle, the following exact integral formulae are available for evaluating the

integrals. Let

m n

mn I x y dxdy

= (2. 45)Then we have


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00 e

I A= (area of the triangle)

10

e I A x=

3

1

13

i

i

x x

=

=

01

e I A y=

3

1

13

i

i

y y

=

=

3

11

1

912

e

i i

i

A I x y xy

=

= +

3

2 2

20

1

912

e

i

i

A I x x

=

= +

3

2 2

02

1

912

e

i

i

A I y y

=

= +

(2. 46)

Using the linear interpolation functions (2.34) in (2.44), and noting that

2

i i

e

x A

=

2

i i

e

y A

=

(2. 47)

we obtain:

( ) ( ){

( ) }

11 12 22

00

20 11 02

1 1 1

4 4 4

1

4

1

ij i j ij i j ij i j

e e e

ij i j i j j i i j j i

e

i j i j j i i j

e

S S S A A A

S x yA

I I I A

= = =

= + + + +

+ + + +

(2. 48)

In view of the identity2

3

i i i e x y A + + = [which follows from (2.32) and (2.46)] for

an element-wise constant value ofef f= , we have

( )1 1

2 3

e

i e i i i e e f f x y f A = + + = (2. 49)

which implies an equal distribution among the loads. Once the coordinates of the element

nodes are known, one can compute iii and, from (2.32) and substitute into (2.48) to

obtain the element matrices, which is turn can be used in (2.43) to obtain the element

matrix eK . For example, when 002112 and, aaa are zero, and 2211 and aa are element-

wise constant, we have


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( )11 221

4

e e e e e e e

ij i j i j

e

K a aA

= + (2. 50)

2.2.5.2 Element matrices for a linear rectangular elementWhen ( ), 0, 1, 2ija i j = and f are constants, we can use the interpolation functions of

(2.42) expressed in the local coordinates, which are related to the local coordinates by

1

e x x x= + 1e y y y= + dx dx= dy dy= (2. 51)

where ( )1 1,e ex y are the global coordinates of node 1 of element e with respect to the

global coordinate system. For example, we have

1 1

1 1

00

0 0

e e

e e

x a y b a b

ij i j i j

x y

S dxdy dxdy

+ +

= = where a and b are the lengths along the x and y axes of the element. Consider the

coefficient

2 2

00

11 1 1

0 0 0 0

1 1

9

a b a b

x y abS dxdy dxdy

a b

= = =

Similarly, we can evaluate all the matrices S with the aid of the integral identities

2

0

13

a

x adx

a

=

,0

16

a

x x adx

a a

=

,0

12

a

x adx

a

=

,0

2

a

x adx

a= (2. 52)

We have

11 12

22 00

2 2 1 1 1 1 1 1

2 2 1 1 1 1 1 111 1 2 2 1 1 1 16 4

1 1 2 2 1 1 1 1

2 1 1 2 4 2 1 2

1 2 2 1 2 4 2 1

1 2 2 1 1 2 4 26 36

2 1 1 2 2 1 2 4

bS Sa

a abS S

b

= =

= =

(2. 53)


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{ } { }1

1 1 1 14

T f fab=

2.2.5.3 Evaluation of the boundary integralsHere we consider the evaluation of the boundary integrals of the type

( )e

e e ei n iQ q s ds

= (2. 54)where enq is a function of the distance s along the boundary

e .

Let us consider the linear triangular element shown in figure 2.5. The linear interpolation

functions associated with this element are given by (2.34).

Figure 2.5. The linear triangular element in the global and local coordinate systems

Now let us choose a coordinate system (s , t) with its origin at node 1 and the coordinate s

parallel to the side connecting nodes 1 and 2. The two coordinate system (x ,y) and (s , t)

are related by

1 1 1x a b s c t = + + 2 2 2y a b s c t = + +

The constants 1a , 1b , 1c , 2a , 2b and 2c can be determined from the following

conditions:

when 0s = , 0t= , 1x x= , 1y y=

when as = , 0=t , 2xx = , 2yy =

when cs = , bt = , 3xx = , 3yy =

1

i

j x

y

Side 1

Side 3

t

s

a

2

Side 2b c

3


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We obtain

( ) ( )( ) ( )

1 2 1 1 2 3

1 2 1 1 2 3

, 1

, 1

s c c t

x s t x x x x x xa a a b

s c c t y s t y y y y y y

a a a b

= + + +

= + + +

(2. 55)

These expressions allow us to express ( )yxi , as ( )tsi , , which can be evaluated on

the side connecting nodes 1 and 2 by setting 0=t in ( )tsi , :

( ) ( ) ( ) ( )( ),0 ,0 , ,0i i is s x s y s =

( ) ( )1 2 1s

x s x x xa

= + ( ) ( )1 2 1s

y s y y ya

= +

For instance,

( )

( )

1 1 1 1 2 1 1 2

1 2 3

11 1

2

11 1

2

s s s ss x x y y

A a a a a

s s

A a a

= + + + +

= + + =

where the definitions of 1 , 1 and 1 have been used to rewrite the entire expression.

Similarly,

( )2s

sa

= ( )3 0s =

where 12a h= is the length of side 1-2. When ( ),i x y are evaluated on side 3-1 of the

element, we obtain

( )113

ss

h = ( )2 0s = ( )3

13

1s

sh

=

where the s coordinate is taken along the side 3-1, with origin at node 3, and13h is the

length of side 3-1. Thus evaluation of ei

Q involves the use of appropriate 1-D

interpolation functions and the known variation of nq on the boundary:

( ) ( ) ( ) ( ) ( ) ( )1 2 2 3 3 1

1 2 3

e

i i n i n i n

e e e

i i i

Q s q s ds s q s ds s q s ds

Q Q Q

= + +

+ +

(2. 56)


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where ji denotes the integral over line connecting node i to nodej, the s coordinate is

taken from node i to node j, with the origin at node i, and eiJ

Q is defined as the

contribution to eiQ from enq on sideJ(see figure 2.5) of the elemente

:

( ) ( )side

e

iJ i n

J

Q s q s ds= (2. 57)For example,

( ) ( ) ( )1 1 1 11 2 3 11 2 3 10ee

n n n

Q q s ds q ds q ds

= = + +

The contribution from side 2-3 is zero, because 1 is zero on side 2-3 of a triangular

element. For a rectangular, element, eQ1 has contributions from sides 1-2 and 4-1,

because 1 is zero on sides 2-3 and 3-4.

Example

Figure 2.6. Evaluation of boundary integrals in the finite element analysis

eh

0q

12

3

s

Case 1

eh

0q

12

3

s

Case 2

eh

0q 23

1

s

Case 4

eh

0

q

123

Case 3

0q

4

65

1q


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Consider the evaluation of the boundary integral eiQ in (2.54) for the four cases of ( )sq

and finite elements shown in figure 2.6. For each case, we must use the ( )sq and the

interpolation functions associated with the type of boundary element (i.e., linear or

quadratic).

Case 1: ( ) constant0 == qsq ; linear element:

( )e

0 0

0

0 0 1, 2, 3

eh

e

i i iQ q ds q ds i

= = + + = where

( )1 1e

ss

h = ( )2

e

ss

h = ( )3 0s =

We have

( )1 0 111

2

e e

eQ q h Q= = ( )2 0 21

1

2

e e

eQ q h Q= = 3 0

eQ =

Case 2: ( ) 0 (linear variation)e

sq s q

h= ; linear element:

( )e

0

0

0

1, 2, 3eh

e

i i i

e e

qsQ q ds s ds i

h h

= = = where

( )1 1e

ss

h = ( )2

e

ss

h = ( )3 0s =

We have

( )1 0 111

6

e e

eQ q h Q= = ( )2 0 21

1

3

e e

eQ q h Q= = 3 0

eQ =

Case 3: ( ) constant0 == qsq ; quadratic element:

( )e

01, 2, ... , 6e

i iQ q ds i

= = where

( )12

1 1e e

s ss

h h

=

( )2

41

e e

s ss

h h

=

( )3

21

e e

s ss

h h

=


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and 4 , 5 and 6 are zero on side 1-2-3. We have

( )1 0 111

6

e e

eQ q h Q= = ( )2 0 214

6

e e

eQ q h Q= = ( )3 0 311

6

e e

eQ q h Q= =

Case 4: ( )sq as shown in figure 2.6; linear element:

( ) ( )e

0 1 1 2 3

121 2 2 3

0 0e e e ei i i i i i i

sQ q s ds q ds q ds Q Q Q

h

= = + + = + = we obtain:

1 0 0 12 11

12 121 2

2 0 1 0 12 1 23 21 22

12 12 231 2 2 3

3 1 1 23 32

232 3

11 0 0 ( )

6

1 11 0 ( )

3 2

10 0 ( )

2

e e

e e e

e e

s sQ q ds q h Q

h h

s s sQ q ds q ds q h q h Q Q

h h h

sQ q ds q h Q

h

= + + = =

= + + = + = =

= + + = =

2.2.6 Assembly of Element EquationsThe assembly of finite element equations is based on the continuity of primary variables

and the equilibrium of secondary variables. We illustrate the procedure by considering a

finite element mesh consisting of a triangular element and a quadrilateral element (see

figure 2.7).

(a)

1

2

3

5

4

1 2

Side 3

Side 2

Side 1

Side 3

Side 1

1

2

3

1 2

3

4

1

11 11

1

12 12

1 2

22 22 11

14

15

1 2

23 23 14

Global Local

0

0

K K

K K

K K K

K

K

K K K

+

+

1

4

35

2

1 41

3

4

1

2

23

32

11

1

3

32

2

7

6

1

14 13

1 2

34 43 31

17

1 2 3 4

4 33 11 11 11

2 3

45 12 13

56

3 4

47 12 13

Global Local

0

0

K K

K K K

K

K K K K K

K K K

K

K K K

+

+ + +

+

+


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(b)

Figure 2.7. Assembly of finite element coefficient matrices (a) assembly of two elements

(b) assembly of several elements

From the finite element mesh shown in figure 2.7a, we note the following connectivity

relations between the global and element nodes:

=

3542

321B (2. 58)

where indicates that there is no entry. The correspondence between the local and global

nodal values is (see figure 2.7a)

1

1 1u U= ,1 2

2 1 2u u U= = ,1 2

3 4 3u u U= = ,2

2 4u U= ,2

3 5u U= (2. 59)

Next we use the balance of secondary variables. At the interface between the two

elements, the fluxes from them should be equal in magnitude and opposite in sign. In

figure 2.7a, the interface is along the side connecting global nodes 2 and 3. Hence, the

internal flux 1nq on side 2-3 of element 1 should balance the flux

2

nq on side 4-1 of

element 2

( ) ( )1 22 3 4 1

n nq q

= or ( ) ( )1 22 3 1 4

n nq q

= (2. 60)

In the finite element method, we impose the above relation in a weighted-residual sense:

1 2

23 14

1 1 2 2

2 1n n

h h

q ds q ds = 1 223 14

1 1 2 2

3 4n n

h h

q ds q ds = (2. 61)where e

pqh denotes the length of the side connecting node p to node q of the elemente

.

The above equations can be written in the form

1 223 14

1 1 2 2

2 10

n n

h h

q ds q ds + = 1 223 14

1 1 2 2

3 40

n n

h h

q ds q ds + = (2. 62)or


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1 2

22 14 0Q Q+ = 1 2

32 44 0Q Q+ = (2. 63)

where eiJQ denotes the part of eiQ that comes from sideJof element e [see (2.57)]

( ) ( )side

e e e

iJ n i

J

Q q s s ds= (2. 64)

The element equations of the two elements are written first. For the model at hand there

is only one primary degree of freedom per node. For the triangular element, the element

equations are of the form

1 1 1 1 1 1 1 111 1 12 2 13 3 1 1

1 1 1 1 1 1 1 1

21 1 22 2 23 3 2 2

1 1 1 1 1 1 1 1

31 1 32 2 33 3 3 3

K u K u K u f Q

K u K u K u f Q

K u K u K u f Q

+ + = +

+ + = +

+ + = +

(2. 65)

For the rectangular element, the element equations are given by

2 2 2 2 2 2 2 2 2 2

11 1 12 2 13 3 14 4 1 1

2 2 2 2 2 2 2 2 2 2

21 1 22 2 23 3 24 4 2 2

2 2 2 2 2 2 2 2 2 2

31 1 32 2 33 3 34 4 3 3

2 2 2 2 2 2 2 2 2 2

41 1 42 2 43 3 44 4 4 4

K u K u K u K u f Q

K u K u K u K u f Q

K u K u K u K u f Q

K u K u K u K u f Q

+ + + = +

+ + + = +

+ + + = +

+ + + = +

(2.

66)

In order to impose the balance of secondary variables in (2.63), we must add the second

equation of element 1 to the first equation of element 2, and also add the third equation of

element 1 to the fourth equation of element 2 (using the global-variable notation (2.59)):

( ) ( ) ( )( ) ( ) ( )

1 1 2 1 2 2 2 1 2 1 2

21 1 22 11 2 23 14 3 12 4 13 5 2 1 2 1

1 1 2 1 2 2 2 1 2 1 2

31 1 32 41 2 33 44 3 42 4 43 5 3 4 3 4

K U K K U K K U K U K U f f Q Q

K U K K U K K U K U K U f f Q Q

+ + + + + + = + + +

+ + + + + + = + + +

Now we can impose the conditions in (8.52) by setting appropriate portions of the

expressions in parentheses on the right-hand sides of the above equations equal to zero:


21/41


( ) ( )

( )

1 2 1 1 1 2 2 2 2

2 1 21 22 23 11 12 13 14

1 1 1 2 2 2 2

21 23 22 14 11 12 13

Q Q Q Q Q Q Q Q Q

Q Q Q Q Q Q Q

+ = + + + + + +

= + + + + + +

( ) ( )

( )

1 2 1 1 1 2 2 2 2

3 4 31 32 33 41 42 43 44

1 1 1 2 2 2 2

31 33 32 44 41 42 43

Q Q Q Q Q Q Q Q Q

Q Q Q Q Q Q Q

+ = + + + + + +

= + + + + + +

2.2.7 PostprocessingThe finite element solution at any point (x ,y) in an element e is given by

( ) ( )1

, ,

n

e e e

j j

j

U x y u x y=

=

(2. 67)

and its derivatives are computed as:

1

neeje

j

j

Uu

x x

=

=

,1

neeje

j

j

Uu

y y

=

=

(2. 68)

Equations (2.67) and (2.68) can be used to compute the solution and its derivatives at any

point (x , y) in the element. It is useful to generate, by interpolation from (2.67),information needed to plot contours of eU and its gradient.

2.2.8 Axisymmetric Problems

Model equation

( )11 22 001 ,

u ura a a u f r z

r r r z z

+ =

(2. 69)

where 00a , 11a , 22a and f are given as functions ofrandz. For example, this equation

arises in the study of heat transfer in cylindrical geometries.

Weak form

Following the three-step procedure, we write the weak form of (2.69)

0

0


22/41


( i ) ( )11 22 001 0 ,

e

u uw ra a a u f r z rdrdz

r r r z z

= +

( ii ) ( )11 22 00 0 ,e

w u w ura ra wa ru wrf r z drdz

r r z z

= + +

11 22

e

r r

u uw ra n ra n ds

r z

+

( iii ) ( )11 22 00

0 ,e

w u w ura ra wa ru wrf r z drdz

r r z z

= + +

e nwq ds

(2. 70)

where w is the weight function andnq is the normal flux,

11 22

n r r

u uq r a n a n

r z

= +

(2. 71)

Finite element model

Let us assume that ( )zru , is approximated by

( ) ( )

1

, ,

n

e e e

j j

j

u U r z u r z

=

= (2. 72)The interpolation functions ( ),ej r z are the same as those developed in (2.34) and (2.42)

for triangular and rectangular elements, with x r= and y z= . Substitution of (2.72) for

u and ej for w into the weak form gives the ith

equation of the finite element model:

e e

11 22 00

1

0

e

n e ee ej j e e ei i

i j j

j

e e

i i n

a a a rdrdz u

r r z z

f rdrdz q ds

=

= + +

(2. 73)

or


23/41


1

0

n

e e e e

ij j i i

j

K u f Q

=

= (2. 74)where

11 22 00

e

e ee ej je e ei i

ij i j

K a a a rdrdz

r r z z

= + +

e

e ei i

f f rdrdz

= , ee ei i nQ q ds

= (2. 75)Exact evaluation of the integrals in e

ijK and e

if for polynomial forms of ija and f

is

possible. However, we evaluate them numerically using the numerical integration

methods.

2.2.9 Example: Poisson EquationConsider the Poisson equation

02 fu = or 02

2

2

2

f

y

u

x

u=

+

in (2. 76)

0u

y

=

in a square region (see figure 2.8). The boundary condition of the problem is

0u = on (2. 77)

We wish to solve the problem using the finite element method.

0=

nu

y

x

0=u

0=u

Domain used for the

triangular-elementmeshes

0=

y

u

0=

xu

Line of symmetry


24/41


(a)

(b) (c)

Figure 2.8. (a) Geometry and computational domain, and boundary conditions (b) coarsefinite element mesh of linear triangular elements (c) refined finite element mesh of linear

triangular elements

Note that the problem at hand has symmetry about x = 0 and y = 0 axes; it is also

symmetric about the diagonal linex = y (see figure 2.8a).

Solution by linear triangular elements

At a first choice, we use a uniform mesh of four linear triangular elements to represent

the domain shown in figure 2.8b. Note that elements 1, 3 and 4 are identical in orientation

as well as geometry. Element 2 is geometrically identical with element 1, except that it is

oriented differently. If we number the local nodes of element 2 to match those of element

1 then all four elements have the same element matrices, and it is necessary to compute

them only for element 1.

We consider element 1 as the typical element, with its local coordinate system ( )yx , .

Suppose that the element dimensions, i.e., length and height, are a and b, respectively.

The coordinates of the element nodes are

( ) ( )0,0, 11 =yx , ( ) ( )0,, 22 ayx = , ( ) ( )bayx ,, 33 =

0=

nu

y

x

0=u

0=

yu

1 2 4

3 5

6

1

2

3

4

1 2

3

1

1

1 2

2

2

3

3

3

1 2 4

35

15

11

23

45

67

6

7

8

9

10

11

12

13

14

1 1 1

1 1 1

1 1 1

11

11

1

1


25/41


Hence, the parameters i , i and i are given by

( ) ( ) ( )

1 2 3 3 2 2 3 1 1 3 3 1 2 2 1

1 2 3 2 3 1 3 1 2

1 2 3 2 3 1 3 1 2

0 0

0

0

x y x y ab x y x y x y x y

y y b y y b y y

x x x x a x x a

= = = = = =

= = = = = =

= = = = = =

The element coefficients eijK and

e

if are given by

2 2

1 2 2 2 2

2 2

01

2 0

b b

K b a b aab a a

= +

,

{ }

1 0

1

16 1

f abf

=

(2. 78)

The element matrix in (2.78) is valid for the Laplace operator 2 on any right-angled

triangle with sides a and b in which the right-angle is at node 2, and the diagonal line of

the triangle connects node 3 and node 1. Note that the off-diagonal coefficient associated

with the nodes on the diagonal line is zero for a right-angled triangle. These observations

can be used to write the element matrix associated with the Laplace operator or any right-

angled triangle. For example, if the right-angled corner is numbered as node 1, and the

diagonal-line nodes are numbered as 2 and 3 (following the counter-clockwise numbering

scheme), we have (note that a denotes the length of side connecting nodes 1 and 2)

2 2 2 2

1 2 2

2 2

1

02

0

a b b a

K b bab

a a

+

=

For the mesh shown in figure 2.8b, we have

1 2 3 4K K K K = = = , { } { } { } { }1 2 3 4 f f f f = = =

For a = b, the coefficient matrix in (2.78),

1

1 1 01

1 2 12

0 1 1

K

=

(2. 79)


26/41


The assembled coefficient matrix for the finite element mesh is 66 , because there are

six global nodes, with one unknown per node. The assembled matrix can be obtaineddirectly by using the correspondence between the global nodes and the local nodes,

expressed through the connectivity matrix

=

653

542

235

321

B (2. 80)

The assembled system of equations is

1

11

1 2 3

2 3 12

1 2 4

3 3 2 10

3

24

2 3 4

5 1 3 2

4

6 3

1 1 0 0 0 0 1

1 4 2 1 0 0 3

0 2 4 0 2 0 31

0 1 0 2 1 0 12 24

0 0 2 1 4 1 3

0 0 0 0 1 1 1

QU

Q Q QU

U Q Q Qf

QU

U Q Q Q

U Q

+ + + +

= +

+ +

(2. 81)

The sums of the secondary variables at global nodes 2, 3 and 5 are

1 2 3

2 3 1 2Q Q Q Q+ + = , 1 2 43 2 1 3

Q Q Q Q+ + = , 2 3 41 3 2 5Q Q Q Q+ + = (2. 82)

At nodes 1, 4 and 6 we have 11 1Q Q= , 32 4

Q Q= and 43 6Q Q= .

The specified boundary conditions on the primary degrees of freedom of the problem are

4 5 60U U U= = = (2. 83)

The specified secondary degrees of freedom are (all due to symmetry)

1 2 3 0Q Q Q= = = (2. 84)

Since4U , 5U and 6U are known, the secondary variables at these nodes, i.e., 4Q , 5Q

and6Q are unknown, and can be obtained in the post-computation.


27/41


Since the only unknown primary variables are (1U , 2U , and 3U ), and ( 4U , 5U , and 6U )

are specified to be zero, the condensed equations for the primary unknowns can be

obtained by deleting rows and columns 4, 5 and 6 from the system (2.81). In retrospect, it

would have been sufficient to assemble the element coefficients associated with the

global nodes 1, 2 and 3, i.e., writing out equations 1, 2 and 3:

1 1 1 1

11 12 13 1 1

1 1 2 3 1 2 1 2 3

21 22 33 11 23 32 2 2 3 1

1 1 2 1 2 4 1 2 4

31 32 23 33 22 11 3 3 2 1

0

0

0

K K K U f

K K K K K K U f f f

K K K K K K U f f f

+ + + = + + + + + + + +

(2. 85)

The unknown secondary variables 4Q , 5Q and 6Q can be computed either from the

equations (i.e., from equilibrium)

1 14 1 12 1

2 3 4 1 3 2 4

5 1 3 2 13 31 12 21 2

4 4

3 31 36

0 0

0

0 0

Q f K U

Q f f f K K K K U

f K U Q

= + + + + +

(2. 86)

or from their definitions (2.82) and (2.56). For example, we have

3 3 3 3 3 3 3

4 22 2 2 2

1 2 2 3 3 1

n n n

Q Q q dx q dy q ds

= = + +

(2. 87)

where

( )31 2

1 2

0 0, 0n x y xu u u

q n n n x y y

= + = = =

( ) ( )32 3

2 3

1, 0n x y x xu u u

q n n n n x y x

= + = = =

( ) ( )3 32 22 3 1 3

23

1 , 0y

h

= =

Thus,

23

3

4 22

230

1

h

u yQ Q dy

x h

= =

where

u

x

from the finite element interpolation is


28/41


33

3

3

1

2

j

j

j

uu

x A

=

=

We obtain ( 23h a= ,

3

1 a = ,2

32A a= , 4 5 0U U= = )

3

3 323

4 2

31

0.54

j j

j

hQ u U

A

=

= = (2. 88)Using the numerical values of the coefficients eijK and

e

if (with 10 =f ), we write the

condensed equations for 1U , 2U , and 3U as

1

2

3

0.5 0.5 0 11

0.5 2.0 1.0 324

0 1.0 2.0 3

U

U

U

=

(2. 89)

Solving (2.89) fori

U (i = 1, 2, 3), we obtain

1 0.31250U = 2 0.22917U = 3 0.17708U =

and from (2.86), we have

3

1223 4

32 22 2

4

32 3

1 0 0.5 0 0.19791713 0 0 1 0.302083

243 0 0 0 0.041667

UQQ Q U

Q U

+ = + =

(2. 90)

By interpolation, 322

Q , for example, is equal to25.0 U , and it differs from

3

22Q computed

from equilibrium by the amount 32

1

24f

=

.

Solution by linear rectangular elements

Note that we cannot exploit the symmetry along the diagonal yx= to our advantage when

we use a mesh of rectangular elements. Therefore, we use a 22 uniform mesh of

rectangular elements (see figure 2.9) to discretize a quadrant of the domain.


29/41


Four-element mesh Sixteen-element mesh

Figure 2.9. Finite element discretization by linear rectangular elements

Note that no discretization error is introduced in this case. Since all elements are

identical, we shall compute the element matrices for only one element, say element 1. We

have

( ) ( ) ( ) ( )1 2 3 41 2 1 2 2 1 2 4 1 2 2 x y x y x y x y = = = = (2. 91)

0.5 0.5

0 0

j je i i

ijK dxdy

x x y y

= +

(2. 92)0.5 0.5

0

0 0

ei i

f f dxdy= Evaluating these integrals, we obtain {see (2.53): 11 22eK S S = + }

eQ44

eQ14

eQ43

e

Q11

e

Q21

eQ33

eQ22

eQ32

0=xu

0=u

0=

y

u

1 2

4

3

56

1 2

3 4

1 2 43 5

1

6

987

0=u

21 3 4

65 7 8

9

16

121110

14 1513

10

15

20

25

11

16

21 22 23 24

0=xu

0=u

0=u

0=

y

u


30/41


4 1 2 1

1 4 1 21

2 1 4 16

1 2 1 4

eK

=

{ }

1

2

3

4

1

11

116

1

e

e

e

e

e

Q

QF

Q

Q

= +

(2. 93)

where

( ) ( )

( ) ( )

2 3

1 2

4 1

3 4

0

0

x y

e (e) (e)

i n i n i y x a

x yx y

(e) (e)

n i n i y b x

x y

Q q x , y dx q x , y dy

q x , y dy q x , y dy

= =

= =

= +

+ +

(2. 94)

and ( )ii y,x denote the local coordinates of the element nodes (and 2 1 3 4a x x x x= =

and 4 1 3 2b y y y y= = ).

The condensed equations are:

1 1 1 1

11 12 14 13 1

1 1 2 1 1 2

12 22 11 24 23 14 2

1 1 1 3 1 3

14 24 44 11 43 12 4

1 1 2 1 3 1 2 4 3

13 23 14 43 12 33 44 11 22 5

1 1

1 1

1 2 1

2 1 2

1 3

4 1

1 2 4 3

3 4 1 2

K K K K U

K K K K K K U

K K K K K K U

K K K K K K K K K Uf Q

f f Q

f f

f f f f

+ + + +

+ + + + +

+ + = +

+ + + +

2

1

1 3

4 1

1 2 4 3

3 4 1 2

Q

Q Q

Q Q Q Q

+ + + +

(2.95)

The boundary conditions on the secondary variables are:

1

1 0Q = 1 2

2 1 0Q Q+ = 1 3

4 1 0Q Q+ = (2. 96)

and the balance of secondary variables at global node 5 requires

1 2 4 3

3 4 1 2 0Q Q Q Q+ + + =

Thus, we have

1

2

4

5

4 1 1 2 1

1 8 2 2 21 1

1 2 8 2 26 16

2 2 2 16 4

U

U

U

U

=

(2. 97)


31/41


The solution of these equations is

31071.01=

U 24107.02=

U 24107.04=

U 19286.05=

U

The secondary variables3Q , 6Q and 9Q at nodes 3, 6 and 9, respectively, can be

computed from the equations ( 223 QQ = , 42

2

36 QQQ += , 439

QQ = )

123 2 31 32 34 35

22 4

6 3 2 61 62 64 65

44

3 91 92 94 9595

12 2 2

2 21 34

22 4 2 2 4

3 2 31 34 21

44 4

3 31

5

0 00 0

0 0 0

UQ f K K K K

UQ f f K K K K

U f K K K K Q

U

U

f K K U f f K K K

Uf K

U

= + +

= + + +

1

2

4

5

1 0 1 0 2 0.166971 1

2 0 2 0 2 0.2696416 6

1 0 0 0 2 0.12679

U

U

U

U

= + =

(2. 98)

It can be shown that the exact solution to Equation (2.76) is given by the following series

solution

( ) ( )( )

( )203

1

1 cos cosh 1, 1 4 2 1

2 2cosh

n

n n

n

n nn

y xfu x y y n

=

= + =

(2. 99)

The finite element solutions obtained using two different meshes of triangular elements

and two different meshes of rectangular elements are compared in Table 2.1 with the 50-

term series solution (atx = 0 for varyingy) in (2.99) ( 10 =f ).


32/41


Triangular elements Rectangular elements Series

solutiony 4 elements 16 elements 4 elements 16 elements

0.00 0.3125 0.3013 0.3107 0.2984 0.2947

0.25 0.2709* 0.2805 0.2759* 0.2824 0.2789

0.50 0.2292 0.2292 0.2411 0.2322 0.2293

0.75 0.1146*

0.1393 0.1205*

0.1414 0.1397

1.00 0.0000 0.0000 0.0000 0.0000 0.0000

*Interpolated values

Table 2.1. Comparison of the finite element solution u(0 ,y) with the series solution

The finite element solution obtained using 16 triangular elements is the most accurate one

when compared with the series solution. The accuracy of the triangular element mesh is

due to the large number it has compared with the number of elements in the rectangular

element mesh for the same size of domain.

The solution u and its gradient can be computed at any interior point of the domain. For a

point (x ,y) in the element e , we have

( ) ( )

1

, ,

n

e e

j j

j

U x y u x y

=

= (2. 100)

( )

1

,

ne

e i

y i

i

Uq x y u

y y

=

= =

, ( )1

,

ne

e i

x i

i

Uq x y u

x x

=

= =

(2. 101)

2.2.10 Example: Heat ConductionConsider steady-state heat conduction in an isotropic rectangular region of dimension 3a

by 2a (see figure 2.10a).


33/41


Figure 2.10. Finite element analysis of a heat conduction problem over a rectangular

domain

We wish to determine the temperature distribution using the finite element method in the

region and the heat required at the boundaryx = 3a to maintain it at zero temperature.

We note that the problem at hand is governed by (no internal heat generation ( )00 =f and

no convection boundary conditions)

2 0k u = (2. 102)

where kis conductivity of the medium. Hence the finite element model of the problem is

given by

{ } { } { } { }( )0e e e eK u Q f = =

insulated

insulated

ax

TT 6cos0

=

0=T

x

y

a3 a2

1 2 3 4

1 2 3

654

1 2 3 4

8765

9 10 11 12

12

36

54

9 10 11 12

8765 7

8

9

12

11

10

(a)

(b)

(c)


34/41


where eiu is the temperature at node i of the element

e , and

e

j je i iijK k dxdy

x x y y

= +

e

ei n iQ q ds

=

(2. 103)

Triangular element mesh (12 elements) (figure 2.10b)

By renumbering the element nodes, all elements can be made to have a common

geometric shape, and thus the elements need to be computed only for a single element.

Thus, for a typical element of the mesh of triangles in figure 2.10b, the element

coefficient matrix is: (see 2.79)

1 1 0

1 2 12

0 1 1

e kK

=

(2. 104)

The boundary conditions require that

4 8 12 0U U U= = = , 9 0U T= , 10 01

32

U T= , 11 01

2U T=

1 2 3 5 0F F F F = = = = (zero heat flow due to insulated boundary) (2. 105)

We first write the six finite elements equations for the six unknown primary variables.

These equations come from nodes 1, 2, 3, 5, 6, and 7:

1

2

3

05

06

7 0

02 1 0 1 0 0

01 4 1 0 2 0

00 1 4 0 0 2

1 0 0 4 2 02 2

0 2 0 2 8 2 3

0 0 2 0 2 8

U

U

Uk k

TU

TU

U T

=

(2. 106)

The solution of these equations is (in oC)

1 00.6362U T= , 2 00.5510U T= , 3 00.3181U T=

5 00.7214U T= , 6 00.6248U T= , 7 00.3607U T= (2. 107)

The exact solution of (2.102) for the boundary conditions shown in figure (2.10a)


35/41


( )( ) ( )

( )0

cosh / 6 cos / 6,

cosh / 3

y a x aT x y T

= (2. 108)

Evaluating the exact solution at the nodes, we have (inoC)

1 00.6249T T= , 2 00.5412T T= , 3 00.3124T T=

5 00.7125T T= , 6 00.6171T T= , 7 00.3563T T= (2. 109)

The heat at node 4, for example, can be compared from the fourth finite element

equation:

5

4 2 41 1 42 2 43 3 44 4 45 5 46 6 47 7 48 8F Q K U K U K U K U K U K U K U K U = = + + + + + + + +

(2. 110)

Noting that

41 42 43 44 45 46 47 48 49 4(10) 4(11) 4(12)0K K K K K K K K K K K K = = = = = = = = = = = =

and 084 ==UU , we obtain

5

4 2 3 0

10.1591

2F Q kU kT = = = (in W) (2. 111)

Rectangular element mesh (6 elements) (figure 2.10c)

The element coefficient matrix is given by (2.43) and (2.53) with 00 0a = , 11 22a a k= = ,

12 21 0a a= = , and a = b = 1:

4 1 2 1

1 4 1 2

2 1 4 16

1 2 1 4

e kK

=

, { } { }0ef = (2. 112)

The present mesh of rectangular elements is node-wise equivalent to the triangular

element mesh considered in figure 2.10b. Hence, the boundary conditions (2.105) are

valid for the present case. The six finite element equations for the unknowns1U , 2U , 3U ,

5U , 6U and 7U have the same form as before.


36/41


1

2

3

0 05

6 0 0 0

70 0

04 1 0 1 2 001 8 1 2 2 2

00 1 8 0 2 2

31 2 0 8 2 06 6

2 2 2 2 16 2 2 3

0 2 2 0 2 16 3

U

U

Uk k

T TU

U T T T

U T T

= +

+ +

+

(2. 113)

Their solution is

1 00.6128U T= , 2 00.5307U T= , 3 00.3064U T=

5 00.7030U T=

, 6 00.6088U T=

, 7 00.3515U T=

(2. 114)

The heat at node 4 is given by

2

3 43 3 47 7 3 7 0

20.1682 (in W)

6 6

k kQ K U K U U U T = + = =

Triangles Rectangles Analytical

solutionx y 23 46 23 46

0.0 0.0 0.6362 0.6278 0.6128 0.6219 0.6249

0.5 0.0 0.6064 0.6007 0.6036

1.0 0.0 0.5510 0.5437 0.5307 0.5386 0.54121.5 0.0 0.4439 0.4398 0.4419

2.0 0.0 0.3181 0.3139 0.3064 0.3110 0.3124

2.5 0.0 0.1625 0.1610 0.1617

0.0 1.0 0.7214 0.7148 0.7030 0.7102 0.71250.5 1.0 0.6904 0.6860 0.6882

1.0 1.0 0.6248 0.6190 0.6088 0.6150 0.6171

1.5 1.0 0.5054 0.5022 0.5038

2.0 1.0 0.3607 0.3574 0.3515 0.3551 0.3563

2.5 1.0 0.1850 0.1838 0.1844

Table 2.2. Comparison of the nodal temperatures ( ) 0/, TyxT using various finite element

meshes, with the analytical solution

2.2.11 Example: Heat ConvectionThe governing equation for steady-state heat transfer in plane systems is a special case of

(2.1), and is given by


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( ),x yT T

k k f x y x x y y

=

(2. 115)

where T is the temperature (inoC ), xk and yk are the thermal conductivities (in W m

-1

oC

-1) along the x andy directions, respectively, and f is the internal heat generation per

unit volume (in W m-3). For a convective boundary, the natural boundary condition is a

balance of energy transfer across the boundary due to conduction and/or convection (i.e.,

Newtons law of cooling):

( ) x x y y nT T

k n k n T T qx y

+ + =

(2. 116)

where is the convective conductance (in W m-2oC 1),

T is the ambient temperature

of the surrounding, and nq is the specified heat flow. It is the presence of the term

( )

TT that requires some modification of Equation (2.14)-(2.17). These become

( )

( ) ( )

0

,

e e

e e

x y x x y y

x y n

w T w T w wk k wf dxdy w k n k n ds

x x y y x y

w T w T

k k wf dxdy w q T T ds x x y y

B w T l w

= + +

= +

=

(2. 117)

where

( ),e e

x y

w T w T B w u k k dxdy wTds

x x y y

= + +

(2. 118)

( )e e e

nl w wf dxdy wT ds wq ds

= + +

(2. 119)

The finite element model of (2.115) is obtained by substituting the finite element

approximation of the form

( ) ( )

1

, ,

n

e e

j j

j

T x y T x y

=

= (2. 120)for T and ej for w into (2.117)


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( )1

n

e e e e e

ij ij j i i

j

K H T F P

=

+ = + (2. 121)

e

e e

e e

j je i i

ij x y

e e e e

i i n i i i

e e e e

ij i j i i

K k k dxdy x x y y

F f dxdy q ds f Q

H ds P T ds

= +

= + +

= =

(2. 122)

The coefficients and eie

ijPH (due to convection) for a linear triangular element (see

figure 2.11) are defined by

12 23 31

12 23 31

12 23 31

0 0 0

12 23 31

12 23 31

0 0 0

e e e

e e e

h h h

e e e e e e e e e e

ij i j i j i j

h h h

e e e e e e e

i i i i

H ds ds ds

P T ds T ds T ds

= + +

= + +

(2. 123)

where eij is the film coefficient (assumed to be constant) for the side connecting nodes i

andj of the element e , ijT

is the ambient temperature on that side, and eijh is the length

of the side. For a rectangular element, the expressions in (2.123) must be modified to

account for four line integrals on four sides of the element.

Figure 2.11. Triangular and quadrilateral elements with node numbers and local

coordinates for the evaluation of the boundary integrals

For a linear triangular element, the matrices eH and { }eP are given by

1

Side 1

Side 3

s

2

Side 2

3

s

s

1

Side 1

Side 4

s 2

Side 2

4ss

3

Side 3

s


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23 23 31 3112 12

2 1 0 0 0 0 2 0 1

1 2 0 0 2 1 0 0 06 6 60 0 0 0 1 2 1 0 2

e e e ee e

e h hhH

= + +

(2. 124)

{ }23 3112

23 23 31 3112 12

1 0 1

1 1 02 2 2

0 1 1

e e e ee e

e T h T hT hP

= + +

(2. 125)

For a linear rectangular element, they are given by

23 2312 12

34 34 41 41

2 1 0 0 0 0 0 0

1 2 0 0 0 2 1 00 0 0 0 0 1 2 06 6

0 0 0 0 0 0 0 0

0 0 0 0 2 0 0 1

0 0 0 0 0 0 0 0

0 0 2 1 0 0 0 06 6

0 0 1 2 1 0 0 2

e ee e

e

e e e e

hhH

h h

= +

+ +

(2. 126)

{ }23 3412 41

23 23 34 3412 12 41 41

1 0 0 1

1 1 0 00 1 1 02 2 2 2

0 0 1 1

e e e ee e e e

e T h T hT h T hP

= + + +

(2. 127)

As an example, consider the heat transfer in a rectangular region of dimensions a by b,

subject to the boundary conditions shown in figure 2.12.

Figure 2.12. Domain and boundary conditions for convective heat transfer

1 2 3

765

1 2 3 4

9876

11 12 13 14

insulated

insulated

( ) 0=+ TTxTk

0TT=

x

y a

b 8

4

15

10

5


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The heat transfer in the region is governed by

0x y

T T

k k x x y y

=

(2. 128)

The finite element model of the equation is given by

{ } { } { } { } { }( )0e e e e e eK H u Q P f + = + = (2. 129)

where eiu denotes the temperature at node i of the element

e . The element matrices are

2 2 1 1 2 1 1 2

2 2 1 1 1 2 2 1

1 1 2 2 1 2 2 16 61 1 2 2 2 1 1 2

ye xkk r

K

r

= +

( )23 23

0 0 0 0

0 2 1 04, 8

0 1 2 06

0 0 0 0

e e

e hH e

= =

( )

23

23 23

0

1

4, 812

0

e e

e T h

P e

= =

(2. 130)

where

1

2 21

4

bb

ra

a

= =

There are 10 nodal temperatures that are to be determined and heats at all nodes except

nodes 7, 8, 9, and 10 are to be computed. To illustrate the procedure, we write algebraic

equations for only representative temperatures and heats.

Node 10 (for temperatures)

( ) ( ) ( )

( ) ( ) ( )

4 4 4 4 8 4 4 8 8 8

31 4 32 32 5 43 21 9 33 33 22 22 10 24 14

8 8 4 4 8 8 4 8

23 23 15 3 3 2 2 3 2 (known)

K U K H U K K U K H K H U K U

K H U Q P Q P P P

+ + + + + + + + +

+ + = + + + = +


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Node 14 (for heat 14Q )

( ) ( )7 8 7 7 8 8 7 7 8 814 3 4 31 8 32 41 9 42 10 34 13 33 44 14 43 15Q Q Q K U K H U K U K U K K U K U + = + + + + + + +

From the boundary conditions, we know the temperatures at nodes 11-15 (i.e., 11U , 12U ,

13U , 14U , 15U are known). Substituting the values ofe

ijK , e

ijH and e

iP , we obtain explicit

form of the algebraic equations. For example, the algebraic equation corresponding to

node 10 is

4 5 9

10 14 15

21 1 1 12 26 6 12 6

2 1 1 12

3 2 6 6 2 2

y y y y

x x x x

y y y

x x x

k k k k

k r U k r b U k r k r U r r r r

k k kb bk r U k r U k r U bT

r r r

+ + + + + + +

+ + + + + + =

enis-fem chapter 2

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