Download - Enis-fem Chapter 2
-
8/3/2019 Enis-fem Chapter 2
1/41
Finite Element Method Modeling Dr. Slim Choura
2Finite elementanalysis of two-
dimensional
Problems
-
8/3/2019 Enis-fem Chapter 2
2/41
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.
-
8/3/2019 Enis-fem Chapter 2
3/41
Finite Element Method Modeling Dr. Slim ChouraPage 48
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
-
8/3/2019 Enis-fem Chapter 2
4/41
Finite Element Method Modeling Dr. Slim ChouraPage 49
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.
-
8/3/2019 Enis-fem Chapter 2
5/41
Finite Element Method Modeling Dr. Slim ChouraPage 50
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
-
8/3/2019 Enis-fem Chapter 2
6/41
Finite Element Method Modeling Dr. Slim ChouraPage 51
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)
-
8/3/2019 Enis-fem Chapter 2
7/41
Finite Element Method Modeling Dr. Slim ChouraPage 52
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
-
8/3/2019 Enis-fem Chapter 2
8/41
Finite Element Method Modeling Dr. Slim ChouraPage 53
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)
-
8/3/2019 Enis-fem Chapter 2
9/41
Finite Element Method Modeling Dr. Slim ChouraPage 54
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
-
8/3/2019 Enis-fem Chapter 2
10/41
Finite Element Method Modeling Dr. Slim ChouraPage 55
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
-
8/3/2019 Enis-fem Chapter 2
11/41
Finite Element Method Modeling Dr. Slim ChouraPage 56
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
-
8/3/2019 Enis-fem Chapter 2
12/41
Finite Element Method Modeling Dr. Slim ChouraPage 57
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
-
8/3/2019 Enis-fem Chapter 2
13/41
Finite Element Method Modeling Dr. Slim ChouraPage 58
( )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)
-
8/3/2019 Enis-fem Chapter 2
14/41
Finite Element Method Modeling Dr. Slim ChouraPage 59
{ } { }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
-
8/3/2019 Enis-fem Chapter 2
15/41
Finite Element Method Modeling Dr. Slim ChouraPage 60
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)
-
8/3/2019 Enis-fem Chapter 2
16/41
Finite Element Method Modeling Dr. Slim ChouraPage 61
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
-
8/3/2019 Enis-fem Chapter 2
17/41
Finite Element Method Modeling Dr. Slim ChouraPage 62
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
=
-
8/3/2019 Enis-fem Chapter 2
18/41
Finite Element Method Modeling Dr. Slim ChouraPage 63
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
+
+ + +
+
+
-
8/3/2019 Enis-fem Chapter 2
19/41
Finite Element Method Modeling Dr. Slim ChouraPage 64
(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
-
8/3/2019 Enis-fem Chapter 2
20/41
Finite Element Method Modeling Dr. Slim ChouraPage 65
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:
-
8/3/2019 Enis-fem Chapter 2
21/41
Finite Element Method Modeling Dr. Slim ChouraPage 66
( ) ( )
( )
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
-
8/3/2019 Enis-fem Chapter 2
22/41
Finite Element Method Modeling Dr. Slim ChouraPage 67
( 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
-
8/3/2019 Enis-fem Chapter 2
23/41
Finite Element Method Modeling Dr. Slim ChouraPage 68
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
-
8/3/2019 Enis-fem Chapter 2
24/41
Finite Element Method Modeling Dr. Slim ChouraPage 69
(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
-
8/3/2019 Enis-fem Chapter 2
25/41
Finite Element Method Modeling Dr. Slim ChouraPage 70
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)
-
8/3/2019 Enis-fem Chapter 2
26/41
Finite Element Method Modeling Dr. Slim ChouraPage 71
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.
-
8/3/2019 Enis-fem Chapter 2
27/41
Finite Element Method Modeling Dr. Slim ChouraPage 72
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
-
8/3/2019 Enis-fem Chapter 2
28/41
Finite Element Method Modeling Dr. Slim ChouraPage 73
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.
-
8/3/2019 Enis-fem Chapter 2
29/41
Finite Element Method Modeling Dr. Slim ChouraPage 74
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
-
8/3/2019 Enis-fem Chapter 2
30/41
Finite Element Method Modeling Dr. Slim ChouraPage 75
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)
-
8/3/2019 Enis-fem Chapter 2
31/41
Finite Element Method Modeling Dr. Slim ChouraPage 76
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 ).
-
8/3/2019 Enis-fem Chapter 2
32/41
Finite Element Method Modeling Dr. Slim ChouraPage 77
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).
-
8/3/2019 Enis-fem Chapter 2
33/41
Finite Element Method Modeling Dr. Slim ChouraPage 78
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)
-
8/3/2019 Enis-fem Chapter 2
34/41
Finite Element Method Modeling Dr. Slim ChouraPage 79
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)
-
8/3/2019 Enis-fem Chapter 2
35/41
Finite Element Method Modeling Dr. Slim ChouraPage 80
( )( ) ( )
( )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.
-
8/3/2019 Enis-fem Chapter 2
36/41
Finite Element Method Modeling Dr. Slim ChouraPage 81
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
-
8/3/2019 Enis-fem Chapter 2
37/41
Finite Element Method Modeling Dr. Slim ChouraPage 82
( ),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)
-
8/3/2019 Enis-fem Chapter 2
38/41
Finite Element Method Modeling Dr. Slim ChouraPage 83
( )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
-
8/3/2019 Enis-fem Chapter 2
39/41
Finite Element Method Modeling Dr. Slim ChouraPage 84
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
-
8/3/2019 Enis-fem Chapter 2
40/41
Finite Element Method Modeling Dr. Slim ChouraPage 85
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
+ + + + + + + + +
+ + = + + + = +
-
8/3/2019 Enis-fem Chapter 2
41/41
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
+ + + + + + +
+ + + + + + =