FEM 6 Wtd resid and Scalar field.ppt

1
Method of Weighted Residuals
Introduction
The method of weighted residuals (Galerkin’s method
of weighted residuals) is emphasized as a tool for finite
element formulation for essentially any field problem
governed by a differential equation.
Method of Weighted Residuals
The method of weighted residuals (MWR) is an
approximate technique for solving boundary value problems
that utilizes trial functions satisfying the prescribed boundary
conditions and an integral formulation to minimize error, in an
average sense, over the problem domain.

2
Given a differential equation of the general form
D[y(x), x] = 0 a < x < b (5.1)
subject to homogenous boundary conditions
y(a) = y(b) = 0 (5.2)
the method of weighted residuals seeks an approximate
solution in the form
   
1
*
n
i i
i
y x c N x


(5.3)
where y* is the approximate solution expressed as the
product of ci unknown constant parameters to be determined
and Ni(x) trial functions.

3
The major requirement placed on the trial functions is
that they be admissible functions; that is, the trial functions
are continuous over the domain of interest and satisfy the
specified boundary conditions exactly.
In addition, the trial functions should be selected to
satisfy the “physics” of the problem in a general sense. Given
these somewhat lax conditions, it is highly unlikely that the
solution represented by Equation 5.3 is exact.

4
Instead, on submission of the assumed solution into
the differential Equation 5.1, a residual error (hereafter
simply called residual) results such that
R(x) = D[y*(x), x] ≠ 0 (5.4)
where R(x) is the residual. Note that the residual is also a
function of the unknown parameters ci.
The method of weighted residuals requires that the
unknown parameters ci be evaluated such that
    0 1,
b
i
a
w x R x dx i n
 

where wi(x) represent n arbitrary weighting functions.
(5.5)

5
We observe that, on integration, Equation 5.5 results in
n algebraic equations, which can be solved for the n values of
ci. Equation 5.5 expresses that the sum (integral) of the
weighted residual error over the domain of the problem is zero.
Owing to the requirements placed on the trial functions,
the solution is exact at the end points (the boundary conditions
must be satisfied) but, in general, at any interior point the
residual error is nonzero.

6
Method of weighted residual techniques are
•Point collocation
•Subdomain collocation
•Least squares, and
•Galerkin’s method
In Galerkin’s weighted residual method, the weighting
functions are chosen to be identical to the trial functions; that
is,
wi(x) = Ni(x) i = 1,n (5.6)

7
Therefore, the unknown parameters are determined via
        0 1,
b b
i i
a a
w x R x dx N x R x dx i n
  
 
again resulting in n algebraic equations for evaluation of the
unknown parameters. The following examples details of the
procedure.
(5.7)

8
Example 5.1
Use Galerkin’s method of weighted residuals to obtain an
approximate solution of the differential equation
2
2
2
10 5 0 1
d y
x x
dx
   
with boundary conditions y(0) = y(1) = 0.
Solution
The presence of the quadratic term in the differential
equation suggests that trial functions in polynomial from are
suitable. For homogeneous boundary conditions at x = a and
x = b, the general form
N(x) = (x – xa)p (x – xb)q

9
with p and q being positive integers greater than zero,
automatically satisfies the boundary conditions and is
continuous in xa ≤ x ≤ xb. Using a single trial function, the
simplest such form that satisfies the stated boundary
conditions is
N1(x) = x(x-1)
Using this trial function, the approximate solution per
Equation 5.3 is
y*(x) = c1x(x – 1)
and the first and second derivates are

10
 
1
2
1
2
*
2 1
*
2
dy
c x
dx
d y
c
dx
 

respectively. (We see, at this point, that the selected trial
solution does not satisfy the physics of the problem, since
we have obtained a constant second derivative. The
differential equation is such that the second derivative must
be a quadratic function of x. Nevertheless, we continue the
example to illustrate the procedure.)

11
Substitution of the second derivative of y*(x) into the
differential equation yields the residual as
R(x;c1) = 2c1 – 10x2 – 5
which is clearly nonzero. Substitution into Equation 5.7 gives
  
1
2
1
0
1 2 10 5 0
x x c x dx
   

which after integration yields c1 = 4, so the approximate
solution is obtained as
y*(x) = 4x(x-1)
    0 1,
b
i
a
N x R x dx i n
 
 (5.7)

12
 
2 3
2
1
2
10
10 5 5
3
dy d y x
dx x dx x C
dx dx
     
 
 
3 4 2
1 1 2
10 5 5
5
3 6 2
dy x x x
y x dx x C dx C x C
dx
 
       
 
 
 
1
5 5
0
6 2
C
  
For this relatively simple example, we can compare the
approximate solution result with the exact solution, obtained
by integrating the differential equation twice as follows:
Applying the boundary condition y(0) = 0 gives C2 = 0, while
the condition y(1) = 0 becomes

13
from which C1 = -10/3. Hence, the exact solution is given by
  4 2
5 5 10
6 2 3
y x x x x
  

14
Example 5.2
Obtain a two-term Galerkin solution for the problem of
Example 5.1 using the trial functions
N1(x) = x(x-1) N2(x) = x2(x-1)
Solution
The two-term approximate solution is
y*(x) = c1x(x – 1) + c2x2(x-1)
and the second derivative is
 
2
1 2
2
*
2 2 3 1
d y
c c x
dx
  

15
Substituting into the differential equation, we obtain the
residual
R(x; c1, c2) = 2c1+ 2c2 (3x-1) – 10x2 – 5
Using the trial functions as the weighting functions per
Galerkin’s method, the residual equations become
   
   
1
2
1 2
0
1
2 2
1 2
0
1 2 2 3 1 10 5 0
1 2 2 3 1 10 5 0
x x c c x x dx
x x c c x x dx
 
     
 
 
     
 



16
1 2
1 2
4
0
3 6 3
2 3
0
6 15 4
c c
c c
   
   
1 2
19 5
6 3
c c
 
After integration and simplification, we obtain the algebraic
equations
Simultaneous solution results in
so the two-term approximate solution is
   
2 3 2
19 5 5 3 19
* 1 1
6 3 3 2 6
y x x x x x x x
      

17
For comparison, the exact, one-term and two-term solutions
are plotted in Figure 5.2. The differences in the exact and
two-term solutions are barely discernible.

18
Example 5.3
Use Galerkin’s method of weighted residuals to obtain a one-
term approximation to the solution of the differential equation.
2
2
4 0 1
d y
y x x
dx
   
with boundary conditions y(0) = 0, y(1) = 1.
Solution
Here the boundary conditions are not homogeneous, so a
modification is required. Unlike the case of homogeneous
boundary conditions, it is not possible to construct a trial
solution of the form c1N1(x) that satisfies both stated boundary
conditions.

19
Instead, we assume a trial solution as
y* = c1N1(x) + f(x)
where N1(x) satisfies the homogeneous boundary
conditions and f(x) is chosen to satisfy the non
homogeneous condition. (Note that, if both boundary
conditions were nonhomogeneous, two such functions
would be included.) One such solution is
y* = c1x(x – 1) + x
which satisfies y(0) and y(1) = 1 identically.
Substitution into the differential equation results in the
residual

20
 
2
2 2
1 1 1 1 1 1 1
2
*
; * 4 2 4 2 3
d y
R x c y x c c x c x x x c x c x c x
dx
           
      
1 1
2
1 1 1 1 1
0 0
; 1 2 3 0
N x R x c dx x x c x c x c x dx
     
 
    2
5 5 1
* 1
6 6 6
y x x x x x x
    
and the weighted residual integral becomes
While algebraically tedious, the integration is
straightforward and yields
c1 = 5/6
so the approximate solution is

21
As in the previous example, we have the luxury of
comparing the approximate solution to the exact solution,
which is
y(x) = 4x – 3.565 sinx
The approximate solution and the exact solution are
shown in Figure 5.3 for comparison. Again, the agreement is
observed to be reasonable but could be improved by adding a
second trial function.

23
Example 5.4
Solve the problem of Example 5.1 and 5.2 by assuming a
general polynomial form for the solution as
y*(x) = c0 + c1x + c2x2+…
Solution
For a first trial, we take only the quadratic form
y*(x) = c0 + c1x + c2x2
and apply the boundary conditions to obtain
y*(0) =0 = c0
y*(1) = 0 = c1 + c2

24
The second boundary conditions equations show that c1 and c2
are not independent if the homogeneous boundary condition is
to be satisfied exactly. Instead, we obtain the constraint
relation c2 = -c1. The trial solution becomes
y*(x) = c1x + c2x2 = c1x – c1x2 = c1 x(1-x)
and is the same as the solution obtained in Example 5.1.
Next we add the cubic term and write the trial solution as
y*(x) = c0 + c1x + c2x2+ c3x3

25
Application of the boundary conditions results in
y*(0) = 0 = c0
y*(1) = 0 = c1 +c2 + c3
so we have the constraint relation
c1 + c2 + c3 =0
Expressing the constraint as c3 = - (c1 + c2), the trial solution
becomes
y*(x) = c1x + c2x2 + c3x3 = c1x + c2x2 - (c1 + c2) x3
= c1 x(1-x2)+c2x2(1-x)

26
and we have obtained two trial functions, each identically
satisfying the boundary conditions. We add the quartic term
and examine the trial solution.
y*(x) = c0 + c1x + c2x2+ c3x3+ c4x4
and the boundary conditions give
c0 = 0
c1 + c2 + c3 + c4 = 0

27
We use the constraint relation to eliminate (arbitrarily) c4 to
obtain
y*(x) = c1x + c2x2+ c3x3 – (c1 + c2 + c3)x4
= c1x(1-x3) + c2x2(1-x2) + c3x3(1-x)
Substituting into the differential equation, the residual is
found to be
R(x; c1,c2,c3) = – 12c1x2 + c2(2 – 12x2) + c3(6x – 12x2)
– 10x2 -5

28
1
2
3
4
10
3
5
2
0
5
6
c
c
c
c
 



  4 2
5 5 10
*
6 2 3
y x x x x
  
If we set that the residual expression equal to zero and
equate coefficients of powers of x, we find that the residual
is exactly zero if
so that
and we have obtained the exact solution.

29
THE GALERKIN FINITE ELEMENT METHOD
 
2
2
0
d y
f x a x b
dx
   
The concept of minimizing the residual error is readily
adapted to the finite element context using the Galerkin
approach as follows
subject to boundary conditions
y(a) = ya y(b) = yb (5.9)
The problem domain is divided into M “elements” (Figure 5.4a)
bounded by M + 1 values xi of the independent variable, so
that x1 = xa and xM+1 = xb to ensure inclusion of the global
boundaries.
(5.8)

31
An approximate solution is assumed in the form
   
1
1
*
M
i i
i
y x y n x


 
where yi is the value of the solution function at x = xi and ni(x)
is a corresponding trial function.
(5.10)
Note that, in this approach, the unknown constant
parameters ci of the method of weighted residuals become
unknown discrete values of the solution function evaluated
at specific points in the domain.

32
There also exists a major difference in the trial functions. As
used in Equation 5.10, the trial functions ni(x) are nonzero
only a small portion of the global problem domain. Specifically,
a trial function ni(x) is nonzero only in the interval xi-1 < x <
xi+1, and for ease of illustration, we use linear functions
defined as follows.
 
 
 
1
1
1
1
1
1
1 1
0
i
i i i
i i
i
i i i
i i
i i i
x x
n x x x x
x x
x x
n x x x x
x x
n x x x x x






 

  


  

  
(5.11)

33
The trial functions are simply linear interpolation functions
such that the value of the solution y(x) in xi < x < xi+1 is a linear
combination of adjacent “nodal” values yi and yi+1. The first
four trial functions are as shown in Figure 5.4b, and we
observe that, in the interval x2 ≤ x ≤ x3, for example, the
approximate solution as given by Equation 5.10 is
      3 2
2 2 3 3 2 3
3 2 3 2
*
x x x x
y x y n x y n x y y
x x x x
 
   
 
(The trial functions used here are linear but higher – order
functions can also be used)
(5.12)

34
     
   
2 2
1 1
2 2
1 1
*
;
M M
i i i
i i
d y d
R x y f x y n x f x
dx dx
 
 
   
   
   
   
 
       
   
2
1
2
1
; 0
b b
a a
x x M
j i j i i
i
x x
d
n x R x y dx n x y n x f x dx
dx


 
  
 
 

 
Substitution of the assumed solution (5.10) into the
governing Equation 5.8 yields the residual
to which we apply Galerkin’s weighted residual method, using
each trial function as a weighting function, to obtain
(5.13)
j = 1, M+1 (5.14)

35
In light of Equation 5.11 and Figure 5.4b, we observe that, in
any interval xj ≤ x ≤ xj+1, only two of the trial functions are
nonzero. Taking this observation into account, Equation 5.14
can be expressed as
     
   
1 2
1 1
2
0 1, 1
j
j
x
j j j j j
x
d
n x y n x y n x f x dx j M
dx

 
 
    
 
 

Integration of Equation 5.15 yields M+1 algebraic equations in
the M+1 unknown nodal solution values yj, and these
equations can be written in the matrix form
[K] {y} = {F} (5.16)
where [K] is the system “stiffness” matrix, {y} is the vector of
nodal “displacements” and {F} is the vector of nodal “forces”.
(5.15)

36
Element Formulation
If the exact solution for Equation 5.8 is obtained, then that
solution satisfies the equation in any subdomain in (a, b) as
well. Consider the problem
 
2
1
2
0 j j
d y
f x x x x
dx

   
where xj and xj+1 are contained in (a, b) and define the nodes
of a finite element. The appropriate boundary conditions
applicable to Equation 5.17 are
y(xj) = yj y(xj+1)= yj+1 (5.18)
(5.17)

37
  1
1 1
1
j
j j
j j
x x
N x x x x
x x




  

 
2 1
1
j
j j
j j
x x
N x x x x
x x



  

and these are the unknown values of the solution at the end
points of the subdomain. Next we propose an approximate
solution of the form
y(e) (x) = yjN1(x) + yj+1N2(x) (5.19)
where superscript (e) indicates that the solution is for the finite
element and the interpolation functions are now defined as
(5.20a)
(5.20b)

38
Note the relation between the interpolation functions defined
in Equation 5.20 and the trial functions in Equation 5.11. The
interpolation functions correspond to the overlapping portions
of the trial functions applicable in a single element domain.
Also note that the interpolation functions satisfy the conditions
N1(x = xj) = 1 N1(x = xj+1) = 0
N2(x = xj) = 0 N2(x = xj+1) = 1 (5.21)
such that the element boundary (nodal) conditions, Equation
5.18, are identically satisfied. Substitution of the assumed
solution into Equation 5.19 gives the residual as

39
(5.22)
 
 
 
       
2 2
1 1 1 2
2 2
; , 0
e
e
i j j j
d y d
R x y y f x y N x y N x f x
dx dx
 
 
     
 
   
   
 
 
1 1 2
1 2
; , 0 1,2
j j
j j
x x e
e
i i j i
x x
d y
N x R x y y dx N x f x dx i
dx
 

 
   
 
 
 
 
 
   
1 1
2
2
0 1, 2
j j
j j
x x
e
i i
x x
d y
N x dx N x f x dx i
dx
 
  
 
where the superscript is again used to indicate that the
residual is for the element. Applying the Galerkin weighted
residual criterion results in
or
as the element residual equations.
(5.23)
(5.24)

40
 
   
   
1 1 1
0 1, 2
j j j
j j
j
x x x
e e
i
i i
x x
x
dN
dy dy
N x dx N x f x dx i
dx dx dx
  
   
 
 
   
 
1 1
1
1
j j
j j j
x x
e e
x x x
dN dy dy
dx N x f x dx
dx dx dx
 
 
 
 
   
 
1 1
1
2
2
j j
j j j
x x
e e
x x x
dN dy dy
dx N x f x dx
dx dx dx
 

 
 
Applying integration by parts to the first integral results in
which, after evaluation of the nonintegral term and
rearranging is equivalent to the two equations, is
(5.26a)
(5.26b)
(5.25)

41
Note that, in arriving at the form of Equation 5.26, explicit
use has been made of Equation 5.21 in evaluation of the
interpolation functions at the element nodes.
Integration of Equation 5.24 by parts results in three
benefits [2]:
1. The highest order of the derivatives appearing in the
element equations has been reduced by one.
2.As will be observed explicitly, the stiffness matrix was
made symmetric. If we did not integrate by parts, one of
the trial functions in each equation would be
differentiated twice and the other trial function not
differentiated at all.

42
3. Integration by parts introduces the gradient boundary
conditions at the element nodes. The physical significance
of the gradient boundary conditions becomes apparent in
subsequent physical applications.
   
 
2 2
1 1 1
1 1 2
1 2 1
x x e
x x x
dN dN dN dy
y y dx N x f x dx
dx dx dx dx
 
  
 
 
 
   
 
2 2
1 1 2
2 1 2
1 2 2
2
x x e
x x x
dN dN dN dy
y y dx N x f x dx
dx dx dx
dx
 
  
 
 
 
Setting j = 1 for notational simplicity and substituting Equation
5.19 into Equation 5.26 yields
(5.27a)
(5.27b)

43
11 12 1 1
21 22 2 2
k k y F
k k y F
    

   
 
    
2
1
, 1,2
x
j
i
ij
x
dN
dN
k dx i j
dx dx
 

which are of the form
The terms of the coefficient (element stiffness) matrix are
defined by
and the element nodal forces are given by the right – hand
sides of Equations 5.27.
(5.28)
(5.29)

44
Galerkin procedure for element formulation is followed
and the system equations are assembled in the usual manner
of the direct stiffness method, the resulting system equations
are identical in every respect to those obtained by the
procedure represented by Equation 5.13.
It is important to observe that, during the assembly
process, when two elements are joined at a common node as
in Figure 5.5, for example, the assembled system equation
for the node contains a term on the right-hand side of the
form.

45
   
4 4
3 4
x x
dy dy
dx dx
 
If the finite element solution were the exact solution, the first
derivatives for each element indicated in expression 5.30
would be equal and the value of the expression would be
zero.
(5.30)

46
Example 5.5
Use Galerkin’s method to formulate a linear finite element for
solving the differential equation
2
2
4 0 1 2
d y dy
x x x
dx
dx
    
subject to y(1) = y(2) = 0.
4 0
d dy
x x
dx dx
 
 
 
 
Solution
First, note that the differential equation is equivalent to
which after two direct integrations and application of
boundary conditions, has the exact solution

47
  2 3
1
2
y x x In x
In
  
      2 1
1 1 2 2 1 2
2 1 2 1
x x x x
y x N x y N x y y y
x x x x
 
   
 
2
1
4 0 1,2
x
i
x
d dy
N x x dx i
dx dx
 
 
  
 
 
 
 

For the finite element solution, the simplest approach is to
use a two-node element for which the element solution is
assumed as
where y1 and y2 are the nodal values. The residual equation
for the element is

48
2 2 2
1 1 1
4 0 1,2
x x x
i
i i
x x x
dN
dy dy
N x x dx xN dx i
dx dx dx
   
 
which becomes, after integration of the first term by parts,
2
2 2
1
1 1
1 2
1 2 4 1,2
x
x x
i
i i
x
x x
dN dN dN dy
x y y dx N x xN dx i
dx dx dx dx
 
   
 
 
 
 
 
2 2
1
1 1
2
1 2 1
2
2 1
2 1
1
4
x x
x
x x
x x
dy
x y y dx x x dx
dx x x
x x

  


 
Substituting the element solution form and rearranging, we
have
Expanding the two equations represented by the last result
after substitution for the interpolation functions and first
derivates yields

49
 
 
2 2
2
1 1
1
2 1 2
2
2 1
2 1
1
4
x x
x
x x
x x
dy
x y y dx x x dx
dx x x
x x

  


 
Integration of the terms on the left reveals the element
stiffness matrix as
 
 
2 2
2 1
2
2 1
1 1
1 1
2
e x x
k
x x

  
    
  
  
while the gradient boundary conditions and nodal forces are
evident on the right-hand side of the equations.

50
To illustrate, a two-element solution is formulated by taking
equally spaced nodes at x = 1,1.5,2 as follows.
 
1.5
1
1
1
1.5
4 1.166666...
1.5 1
x
F x dx

 


 
1.5
1
2
1
1
4 1.33333...
1.5 1
x
F x dx

 


Element 1
x1 = 1 x2 = 1.5 = 2.5

51
 
2
2
1
1.5
2
4 1.166666...
2 1.5
x
F x dx

 


 
2
2
1
1.5
1.5
4 1.83333...
2 1.5
x
F x dx

 


Element 2
x1 = 1.5 x2 = 2 k = 3.5

52
 
 
1
2
1
1
1
2
1.1667
2.5 2.5
2.5 2.5
1.3333 1.5
x
x
dy
dx
y
dy
y
dx
 
 
 
 

     

   
 

     
   
 
 
 
 
2
3
2
1
2
2
1.1667 1.5
3.5 3.5
3.5 3.5
1.8333 2
x
x
dy
dx
y
dy
y
dx
 
 
 
 

     

   
 

     
   
 
 
The element equations are then
Denoting the system nodal values as Y1, Y2, Y3 at x = 1,1.5,2,
respectively, the assembled system equations are
1
3
1
2
3
1.1667
2.5 2.5 0
2.5 6 3.5 3
0 3.5 3.5
1.8333 2
x
x
dy
dx
Y
Y
Y dy
dx
 
 
 
  
   
   
 
   
   
 
   
 

     
 
 
 

53
Applying the global boundary conditions Y1 = Y3 = 0, the
second of the indicated equations gives Y2 = -0.5 and
substitution of this value into the other two equations yields
the values of the gradients at the boundaries as
1 3
2.4167 1.7917
x x
dy dy
dx dx
 
For comparison, the exact solution gives
 
1 3
2
1.5 0.5049 2.3281 1.8360
x x
dy dy
y x Y
dx dx
    

54
1
5
1
2
3
4
5
0.5417
4.5 4.5 0 0 0
4.5 10 5.5 0 0 1.25
0 5.5 12 6.5 0 1.5
0 0 6.5 14 7.5 1.75
0 0 0 7.5 7.5
0.9583 2
x
x
dy
dx
Y
Y
Y
Y
Y dy
dx
 
 
 
  
   
 
   
  
 
   
 
 
   
   
     
  
     
 
    
     
 
 
While the details will be left as an end-of-chapter problem, a
four-element solution for this example (again, using equally
spaced nodes xi (1, 1.25, 1.5, 1.75, 2)) results in the global
equations

55
Applying the boundary conditions Y1 = Y5 = 0 and solving the
remaining 3 x 3 system gives the results
1
5
2
3
4
0.4026
0.5047
0.3603
2.350
1.831
x
x
Y
Y
Y
dy
dx
dy
dx






56
 
 
2
2
0
x
x
d u x
d d
E E
dx dx dx


  
     
1 1 2 2 1 2
1
x x
u x u N x u N x u u
L L
 
    
 
 
APPLICATION OF GALERKIN’S METHOD TO
STRUCTURAL ELEMENTS
Spar Element
The elastic bar or spar element, equilibrium equation is
obtained
where we assume constant elastic modulus. Denoting
element length by L, the displacement field is discretized by
(5.32)
(5.31)

57
 
2 2
2 2
0
0 1,2
L
i i
V
d u d u
N x E dV N E Adx i
dx dx
   
  
   
   
 
0
0
L
L
i
i
dN du du
AE dx N AE
dx dx dx
 
  
 

And, since the domain of interest is the volume of the element,
the Galerkin residual equations become
where dV = A dx and A is the constant cross-sectional area of the
element. Integrating by parts and rearranging, we obtain
which, utilizing Equation 5.32, becomes
(5.34)
(5.33)

58
 
1
1 1 2 2 0 0
0
0
L
x x
x
dN d du
AE u N u N dx AE AE A
dx dx dx
 
 

     

 
2
1 1 2 2
0
L
x L x L
x L
dN d du
AE u N u N dx AE AE A
dx dx dx
 
 

   

(5.35a)
(5.35b)
From the right sides of Equation 5.35, we observe that, for
the bar element, the gradient boundary condition simply
represents the applied nodal force since σA = F.
Equation 5.35 is readily combined into matrix form as

59
where the individual terms of the matrix are integrated
independently.
1 1 1 2
1 1
1 2 2 2 2 2
0
L
dN dN dN dN
u F
dx dx dx dx
AE dx
dN dN dN dN u F
dx dx dx dx
 
     

     
     
 
 
 (5.36)
Carrying out the indicated differentials and integrations, we
obtain
1 1
2 2
1 1
1 1
u F
AE
u F
L
    
 

   
 

     
which is the same result as obtained earlier for the bar element.
(5.37)

60
Beam Element
Application of the Galerkin method to the beam element
begins with consideration of the equilibrium conditions of a
differential section taken along the longitudinal axis of a
loaded beam as depicted in Figure 5.7 where q(x) represents
a distributed load expressed as force per unit length.

61
  0
dV
V V dx q x dx
dx
 
    
 
 
 
dV
q x
dx

  0
2
dM dV dx
M dx M V dx dx q x dx
dx dx
 
 
     
   
 
dM
V
dx
 
Whereas q may vary arbitrarily, it is assumed to be constant
over a differential length dx. The condition of force equilibrium
in the y direction is
from which
Moment equilibrium about a point on the left face is expressed
as
which (neglecting second-order differentials) gives
(5.41)
(5.40)
(5.39)
(5.38)

62
 
2
2
d M
q x
dx

2
2
z
d v
M EI
dx

Combining Equations 5.39 and 5.41, we obtain
Recalling, form the elementary strength of materials theory, the
flexure formula corresponding to the sign convections of Figure
5.7 is
v represents displacement in the y direction, which in
combination with Equation 5.42 provides the governing
equation for beam flexure as
(5.42)
(5.43)

63
 
2 2
2 2
z
d d v
EI q x
dx dx
 

 
 
(5.44)
         
4
1 1 2 1 3 2 4 2
1
( ) i i
i
v x N x v N x N x v N x N x
  

    
   
2
1
2 2
2 2
0 1,4
x
i z
x
d d v
N x EI q x dx i
dx dx
 
 
  
 
 
 
 

Galerkin’s finite element method is applied by taking the
displacement solution in the form
The element residual equations are
(5.46)
(5.45)

64
   
2 2 2
1 1
1
3 3
3 3
0 1,4
x x x
i
i z z i
x x
x
dN
d v d v
N x EI EI dx N q x dx i
dx
dx dx
   
 
2 3
2 3
z z
dM d d v d v
V EI EI
dx dx dx dx
 
   
 
 
Integrating the derivative term by parts and assuming a
constant EIz, we obtain
and since
(5.48)
(5.47)

65
we observe that the first term of Equation 5.47 represents the
shear force conditions at the element nodes. Integrating
again by parts and rearranging gives
 
2 2
2 2
1 1 1 1
2 2 3 2
2 2 3 2
1,4
x x
x x
i i
z i i z z
x x x x
d N dN
d v d v d v
EI dx N q x dx N EI EI i
dx
dx dx dx dx
   
 
and, per Equation 5.43, the last term on the right introduces
the moment conditions at the element boundaries.
(5.49)

66
    
k F
 
2
1
2
2
2 2
, 1,4
x
j
i
ij z
x
d N
d N
k EI dx i j
dx dx
 

 
2 2
2
1 1 1
3 2
3 2
1,4
x x
x
i
i i i z z
x x x
dN
d v d v
F N q x dx N EI EI i
dx
dx dx
   

Equation 5.49 can be written in the matrix form
where the terms of the stiffness matrix are defined by
which is identical to results previously obtained by other
methods. The terms of the element force vector are defined by
(5.51a)
(5.50)

67
     
2
2 2
1 1
1
1,4
x
x x
i
i i i x x
x
dN
F N q x dx N V x M x i
dx
   

or, using Equations 5.43 and 5.48,
where the integral term represents the equivalent nodal force
and moments produced by the distributed load. If q(x) = q =
constant (positive upward), substitution of the interpolation
functions into Equation 5.51 gives the element nodal forces
vector as
(5.51b)

68
(5.52)
 
1
2
1
2
2
2
2
12
2
12
qL
V
qL
M
F
qL
V
qL
M
 

 
 
 

 
 
 
 

 
 
 
 
 
 
Where two beam elements share a common node, one of
two possibilities occurs regarding the shear and moment
conditions

69
1. If no external force or moment is applied at the node, the
shear and moment values of Equation 5.52 for the
adjacent elements are equal and opposite, canceling in
the assembly step.
2. If a concentrated force is applied at the node, the sum of
the boundary shear forces for the adjacent elements must
equal the applied force.

70
One dimensional Heat Conduction
Application of the Galerkin finite element method to one-
dimensional, steady-state heat conduction is developed with
reference to Figure 5.8a. Figure 5.8b shows the control
volume of differential length dx of the body, which is
assumed to be of constant cross-sectional area and uniform
material properties.

72
The principle of conservation of energy is applied to obtain
the governing equation as follows:
Ein + Egenerated = Eincrease + Eout (5.53)
Equation 5.53 states that the energy entering the control
volume plus energy generated internally by any heat source
present must equal the increase in internal energy plus the
energy leaving the control volume. For the volume of Figure
5.8b, during a time interval dt, Equation 5.53 is expressed as
x
x x
q
q Adt QAdxdt U q dx Adt
x

 
   
 

 
(5.54)

73
where
qx = heat flux across boundary (W/m2)
Q = internal heat generation rate (W/m3)
U = internal energy (W).
The last term on the right side of Equation 5.54 is a
two-term Taylor series expansion of qx(x,t) evaluated at
x + dx. Note the use of partial differentiation, we assume that
the dependent variables vary with time as well as spatial
position.
The heat flux is expressed in terms of the temperature
gradient via Fourier’s law of heat conduction:

74
x x
T
q k
x



(5.55)
where kx = material thermal conductivity in the x direction
(W/m-oC) and T = T(x ,t) is temperature. The increase in
internal energy is
U c Adxdt

 
where
c = material specific heat (J/kg-oC)
ρ = material density (kg/m3)
(5.56)
Substituting Equations 5.55 and 5.56 into 5.54 gives
x
T
QAdxdt c Adxdt k Adxdt
x x

 
 
   
 
 
  (5.57)

75
2
2
x
T T
Q c k
t x

 
 
 
/ 0,
T t
  
2
2
0
x
T
k Q
x

 

Assuming that the thermal conductivity is constant, Equations
5.57 becomes
We for only in steady-state heat conduction and for the steady
state the governing equation for steady-state, one -
dimensional conduction is obtained as
(5.59)
(5.58)

76
Next, the Galerkin finite element method is applied to
Equation 5.59 to obtain the element equations. A two-node
element with linear interpolation functions is used and the
temperature distribution in an element expressed as
T(x) = N1(x)T1 + N2(x)T2 (5.60)
where T1 and T2 are the temperatures at nodes 1 and 2, which
define the element, and the interpolation functions N1 and N2
are given by Equation 5.20. As in previous examples,
substitution of the discretized solution (5.60) into the
governing differential Equation 5.55 results in the residual
integrals:

77
 
2
1
2
2
0 1,2
x
x i
x
T
k Q N x Adx i
x
 

  
 

 
 (5.61)
where we note that the integration is over the volume of the
element, (i.e) the domain of the problem, with dV = A dx.
Integrating the first term by parts yields
   
1 2 2
1 1 1
0 1,2
x x x
i
x i x i
x x x
dN
dT dT
k AN x k A dx A Q N x dx i
dx dx dx
   
  (5.62)

78
2 2
1
1 1
1 1 2
1 2 1
x x
x x
x
x x
dN dN dN dT
k A T T dx A Q N dx k A
dx dx dx dx
 
  
 
 
 
2 2
2
1 1
2 1 2
1 2 2
x x
x x
x
x x
dN dN dN dT
k A T T dx A Q N dx k A
dx dx dx dx
 
  
 
 
 
Evaluating the first term at the limits as indicated, substituting
Equation 5.60 into the second term, and rearranging,
Equation 5.58 results in the two equations
(5.63)
(5.64)
Equations 5.63 and 5.64 are of the form
[k]{T} = {fQ} + {fg} (5.65)

79
2
1
, 1,2
x
l m
lm x
x
dN dN
k k A dx l m
dx dx
 

2
1
2
1
1 1
2 2
x
Q
x
x
Q
x
f A QN dx
f A QN dx




where [k] is the element conductance (“stiffness”) matrix
having terms defined by
The first term on the right-hand side of Equation 5.65 is the
nodal “force” vector arising from internal heat generation with
values defined by
(5.67)
(5.66)

80
and vector {fg} represents the gradient boundary conditions
at the element nodes. Performing the integrations indicated
in Equation 5.66 gives the conductance matrix as
 
1 1
1 1
x
k A
k
L

 
  

  (5.68)
while for constant internal heat generation Q, Equation 5.67
results in the nodal vector
  2
2
g
QAL
f
QAL
 
 
 
 
 
 
 
(5.69)

81
The element gradient boundary conditions, using Equation
5.55, described by
  1 1
2
2
x x
g x
x
x
dT
q
dx
f k A A
dT q
dx
 

   
   
 
   

   
 
 
 
are such that, at internal nodes where elements are joined,
the values for the adjacent elements are equal and opposite,
cancelling mathematically.
(5.70)

82
Example 5.6
The circular rod depicted in Figure 5.9 has an outside diameter
of 60mm, length of 1 m, and is perfectly insulated on its
circumference. The left half of the cylinder is aluminum, for
which kx = 200W/m-oC and the right half is copper having
kx = 389 W/m-oC. The extreme right end of the cylinder is
maintained at a temperature of 80oC, while the left end is
subjected to a heat input rate 4000 W/m2. Using four equal-
length elements, determine the steady-state temperature
distribution in the cylinder.

84
 
  
2
200 / 4 0.06
1 1 1 1 1 1
2.26 /
1 1 1 1 1 1
0.25
o
x
al
k A
k W C
L

  
     
  
     
  
     
 
  
2
389 / 4 0.06
1 1 1 1 1 1
4.40 /
1 1 1 1 1 1
0.25
o
x
cu
k A
k W C
L

  
     
  
     
  
     
Solution
The elements and nodes are chosen as shown in the bottom
of Figure 5.9. For aluminum elements 1 and 2, the
conductance matrices are
while, for copper elements 3 and 4,

85
Applying the end conditions T5 = 80oC and q1 = 4000 W/m2,
the assembled system equations are
 
1
2
2
3
4
5
5
2.26 2.26 0 0 0 4000
2.26 4.52 2.26 0 0 0
0.06
0 2.26 6.66 4.40 0 0
4
0 0 4.40 8.80 4.40 0
0 0 0 4.40 4.40 80
11.31
0
0
0
0.0028
T
T
T
T
q
q

  
   
 
   
   
   
   
 
  
   
     
 
     
 
 
   
   
 
 
 
 
 
 
 
 

 
 

86
1
2
3
4
2.26 2.26 0 0 11.31
2.26 4.52 2.26 0 0
0 2.26 6.66 4.40 0
0 0 4.40 8.80 352.0
T
T
T
T
  
   
 
   
     
  
   
 
     
     

   
 
1
2
3
4
2.26 2.26 0 0 11.31
0 2.26 2.26 0 11.31
0 2.26 6.66 4.40 0
0 0 4.40 8.80 352.0
T
T
T
T
  
   
 
   
    
  
   
 
     
     

   
 
Accounting for the known temperature at node 5, the first four
equations can be written as
The system of equations is triangularized (used here simply to
illustrate another solution method) by the following steps.
Replace the second equation by the sum of the first and
second to obtain

87
1
2
3
4
2.26 2.26 0 0 11.31
0 2.26 2.26 0 11.31
0 0 4.40 4.40 11.31
0 0 4.40 8.80 352.0
T
T
T
T
  
   
 
   
    
  
   
 
    
     

   
 
1
2
3
4
2.26 2.26 0 0 11.31
0 2.26 2.26 0 11.31
0 0 4.40 4.40 11.31
0 0 0 4.40 363.31
T
T
T
T
  
   
 
   
    
  
   
 
    
     
   
 
Next, replace the third equation by the sum of the second
and third
Finally, replace the fourth with the sum of the third and fourth to
obtain

88
The triangularized system then gives the nodal temperatures
in succession as
4
3
2
1
82.57
85.15
90.14
95.15
o
o
o
o
T C
T C
T C
T C




The fifth equation of the system is
-4.40T4 + 4.40(80) = -0.0028q5
which, on substitution of the computed of T4, results in
q5 = 4038.6 W/m2

89
Scalar Field Problems
Introduction
Quantities such as temperature, pressure, and stream
potentials are scalar in nature. In two-dimensional steady –
state heat conduction, for example, the temperature field
T(x,y) is the unknown to be determined.
The general Helmholtz equation, given by
0
z y z
k k k Q
x x y y z z
  

 
     
   
    
 
   
     
   
 
(10.1)

90

 
, ,
x y z
 

Together with boundary conditions on and its derivatives.
In the Eq.10.1, is the field variable that is to be
be determined. Table 10.1 lists some of the engineering
problems described by Eq. 10.1.
Problem Equation Field Variable Parameter
Boundary
Conditions
Heat conduction Temperature, T Thermal
conductivity, k
Torsion Stress function,
θ
Potential flow Stream function,
ψ
2 2
2 2
0
T T
k Q
x y
 
 
  
 
 
 
 
0 0
,
T
T T k q
n
T
k h T T
n


  


  

2 2
2 2
2 0
x y
 
 
 
  
 
 
  0
 
2 2
2 2
0
x y
 
 
 
 
 
 
  0
 

Table 10.1 Examples of Scalar Field Problems in Engineering

91
Table 10.1 Examples of Scalar Field Problems in Engineering
Problem Equation Field Variable Parameter
Boundary
Conditions
Seepage and
groundwater flow
Hydraulic
potential,
Hydraulic
conductivity, k
Electric potential Electric
potential, u
Permittivity,
Fluid flow in
ducts
Nondimensional
velocity, W
Acoustics
Pressure, p
(complex)
Wave number,
2 2
2 2
0
k Q
x y
 
 
 
  
 
 
 

0
0
n
y
 







2 2
2 2
u u
x y

 
 
  
 
 
 
 0 , 0
u
u u
n

 

2 2
2 2
1 0
W W
X Y
 
 
  
 
 
 
0
W 
2 2
2
2 2
0
p p
k p
x y
 
 
  
 
 
 
2 2 2
/
k c


0
0
,
1
p p
p
v
ik c n






92
, , 0
x y
T k k k and
 
   
2 2 2 2
/ / 0,
T x T y Q
     
For example, if we set and consider
only x and y, we get which describes
the heat-conduction problem for temperature T, where k is
the thermal conductivity and Q is the heat source/sink.
STEADY-STATE HEAT TRANSFER
Heat is transferred in the form of conduction, convection,
and thermal radiation. Only conduction and convection
modes are treated here.

93
The heat flow through the wall of a heated room on a winter
day is an example of conduction. The conduction process
is quantified by Fourier’s law. In a thermally isotropic
medium. Fourier’s law for two-dimensional heat flow is
given by
x y
T T
q k q k
x y
 
 
 
(10.2)
where T = T(x,y) is a temperature field in the medium, qx
and qy are the components of the heat flux (W/m2), k is the
thermal conductivity (W/m. 0C), and are the
/ , /
T x T y
   
temperature gradients along x and y, respectively.

94
In convection heat transfer, there is transfer of energy
between a fluid and a solid surface as a result of a
temperature difference. There can be free or natural
convection, such as the circulation pattern set up while
boiling water in a kettle due to hot water rising and cooler
water moving down, or there can be forced convection,
such as when the fluid flow is caused by a fan. The
governing equation is of the form
 
s
q h T T
  (10.3)

95
where q is the convective heat flux (W/m2), h is the convection
heat-transfer coefficient or film coefficient (W/m2. 0C), and Ts and
are the surface and fluid temperatures respectively. The
film coefficient h is the property of the flow and depends on
T
various factors, such as whether convection is natural or forced,
whether the flow is laminar or turbulent, the type of fluid, and the
geometry of the body.

96
One-Dimensional Heat Conduction
Governing equation Consider heat conduction in a plane
wall with uniform heat generation (Fig. 10.2). Let A be the
area normal to the direction of heat flow and let Q(W/m3) be
the internal heat generated per unit volume. A control
volume is shown in Fig.10.2. Since the heat rate (heat flux X
area) that is entering the control volume plus the heat rate
generated equals the heat rate leaving the control volume,
we have
dq
qA QAdx q dx A
dx
 
  
 
 
(10.4)

97
Canceling qA from both sides yields
dq
Q
dx
 (10.5)
Substituting Fourier’s law
dT
q k
dx
  (10.6)

98
into Eq. 10.5 results in
0
d dT
k Q
dx dx
 
 
 
 
Usually, Q is called a source when positive (heat is generated)
and is called a sink when negative (heat is consumed).
(10.7)
Boundary conditions Consider the wall of a tank containing a
hot liquid at a temperature T0 with an airstream of temperature T
passed on the outside, maintaining a wall temperature of TL at
the boundary (Fig. 10.3a). The boundary conditions for this
problem are

99
0
0
x
T T


 
L
x L
q h T T

 
(10.8)
(10.9)
As another example, consider a wall, as shown in Fig.10.3b,
where the inside surface is insulated and the outside is a
convection surface. Then, the boundary conditions are
 
0
0 L
x x L
q q h T T
 
   (10.10)

100
  1 1 2 2
e
T N T N T
NT
  

The one-dimensional element To apply the finite element
method, the problem is discretized in the x dimension, as
shown in Fig.10.4a. The temperatures at the various nodal
points, denoted by T, are the unknowns (except at node 1,
where T1 = T0). Within a typical element e (Fig.10.4b), whose
local node numbers are 1 and 2, the temperature field is
approximated using shape functions N1 and N2 as
(10.11)

102
   
1 2
1 / 2, 1 / 2,
N N
  
   
 
1
2 1
2
1
x x
x x
   

2 1
2
d dx
x x
 

where varies from -1 to +1, N = [N1,N2],
and Te = [T1, T2]T. Noting the relations
(10.12)
we have
 
2 1
2 1
2
1
1, 1 e
dT dT d
dx d dx
dN
x x d
T
x x






 

(10.13a)
e
T
.

103
e
T
dT
B T
dx

 
2 1
1
1, 1
T
B
x x
 

or
where
(10.14)
(10.13b)
0 2.
800 25 / .
o
T C and h W m C
  
Example 10.1
A composite wall consists of three materials, as shown in
Fig.E10.1a. The outer temperature is T0 = 20oC.
Convection heat transfer takes place on the inner surface
of the wall with Determine the
temperature distribution in the wall.

105
Solution: A three-element finite element model of the wall is
shown in Fig.E10.1b. The element conductivity matrices are
   
 
1 2
3
1 1 1 1
20 30
1 1 1 1
0.3 0.15
1 1
50
1 1
0.15
T T
T
k k
k
 
   
 
   
 
   

 
  

 
The global K = ΣkT is obtained from these matrices as
1 1 0 0
1 4 3 0
66.7
0 3 8 5
0 0 5 5
K

 
 
 
 

 
 
 

 

106
Now, since convection occurs at node 1, the constant h =25 is
added to the (1,1) location of K. This results in
1.375 1 0 0
1 4 3 0
66.7
0 3 8 5
0 0 5 5
K

 
 
 
 

 
 
 

 
Since no heat generation Q occurs in this problem, the heat
rate vector R consists only of in the first row. That is,
hT
R = [25 X 800, 0, 0, 0]T

107
The specified temperature boundary condition T4 = 200C,
will now be handled by the penalty approach. We choose C
based on
4
4
max 10
66.7 8 10
ij
C K
 
  
Now, C gets added to (4,4) location at K, while CT4 is
added to the fourth row of R. The resulting equations are
1
2
3
4
4
25 800
1.375 1 0 0
0
1 4 3 0
66.7
0
0 3 8 5
0 0 5 80005 10672 10
T
T
T
T

  
 
 
 
 
 
     
  
   
 
     
     
 
     

108
The solution is
T = [304.6, 119.0, 57.1, 20.0]T0C
Comment. The boundary condition T4 = 200C can also be
handled by the elimination approach. The forth row and
column of K is deleted, and R is modified according to
Eq. 3.70. The resulting equations are
1
2
3
1.375 1 0 25 800
66.7 1 4 3 0
0 3 8 0 6670
T
T
T
 
 
   
   
 
  
   
 
   
 
 
   
 
which yields
[T1, T2, T3] = [304.6, 119.0, 57.1]0C

109
Heat flux boundary condition Certain physical situations
are modeled using the boundary condition
q = q0 at x =0 (10.28)
where qo is a specified heat flux on the boundary. If q = 0,
then the surface is perfectly insulated. A nonzero value of q0
occurs, for example, due to an electrical heater or pad
where one face is in contact with the wall and the other face
is insulated

110
Comment on forced and natural boundary conditions In
this problem, boundary conditions of the type T = T0, which is
on the field variable itself, are called forced boundary
conditions. On the other hand, the boundary condition 0 0 ,
x
q q
 
or equivalently, -kdT/dxx=0 = q0 is called a natural boundary
condition involving the derivative of the field variable.
Example 10.2
Heat is generated in a large plate (k = 0.8 W/m0C) at the rate
of 4000 W/m3. The plate is 25 cm thick. The outside surfaces
of the plate are exposed to ambient air at 300C with a
convective heat-transfer coefficient of 20 W/m2. 0C. Determine
the temperature distribution in the wall.

111
/ 0.8/.0625 12.8,
k  
 
12.8 12.8 0
12.8 25.6 12.8
0 12.8 12.8 20
K
 

 
  
 
 
 
 
Solution: The problem is symmetric about the centerline of
the plate. A two-element finite element model is shown in Fig.
E10.2. The left end is insulated (q = 0) because no heat can
flow across a line of symmetry. Noting that
we have
The heat rate vector is assembled from the heat source (Eq.
10.25) as well as due to convection as
R = [125 250 (125 + 20 x 30)]T

112
Solution of KT = R yields
[T1, T2, T3] = [94.0, 84.3, 55.0]0C

113
One dimensional Heat Transfer in Thin Fins
A fin is an extended surface that is added onto a structure to
increase the rate of heat removal.
Consider a thin rectangular fin as shown in Fig.10.6. The
problem can be treated as one dimensional. The governing
equation may be derived from the conduction equation with
heat source, given by
0
d dT
k Q
dx dx
 
 
 
 
The convection heat loss in the fin can be considered as a
negative heat sources

115
   
c
Pdx h T T
Q
A dx



 
c
Ph
T T
A

   (10.32)
where P = perimeter of fin and Ac = area of cross section.
Thus, the governing equation is
  0
c
d dT Ph
k T T
dx dx A

 
  
 
 
In case the base of the fin is held at T0 and the tip of the fin is
insulated (heat going out of the tip is negligible). The boundary
conditions are then given by
T = T0 at x =0 (10.33a)
q = 0 at x = L (10.33b)

116
Example 10.3
A metallic fin, with thermal conductivity k = 360 W/m.0C, 0.1 cm
thick, and 10 cm long, extends from a plane wall whose
temperature is 2350C. Determine the temperature distribution
and amount of heat transferred from the fin to the air at 200C
with h = 9 W/m2. 0C. Take the width of fin to be 1 m.
Solution Assume that the tip of the fin is insulated. Using a
three-element finite element model (Fig.E10.3) and
assembling as given previously, we find that Eq.10.40
, ,
T T
K H R
yields

117
2
2
3
2 3
4
2
3
2 1 0 4 1 0
9 3.33 10
360
1 2 1 1 4 1
3.33 10 3 10
0 1 1 0 1 2
2 10711 235
9 20 3.33 10
2 0
10
1 0
T
T
T

 


 
  
   
   
 
   
    
 
   
   
 
   

     
 
 
   
      
 
   
   
   
The solution is
[T2, T3, T4] = [209.8, 195.2, 190.5]0C
The total heat loss in the fin can now be computed as
e
e
H H


118
The loss He in each element is
 
e av s
H h T T A

 
where As = 2 x (1 x 0.0333)m2, and Tav is the average
temperature within the element.
We obtain
Hloss = 334.3 W/m
Two – Dimensional Steady – State Heat Conduction
Determine the temperature distribution T(x,y) in a long,
prismatic solid in which two-dimensional conduction effects are
important. Once the temperature distribution is known, the heat
flux can be determined from Fourier’s law.

120
Differential equation Consider a differential control volume in
the body, as shown in Fig. 10.8. The control volume has a
constant thickness in the z direction. The heat generation is
denoted by Q (W/m3). Since the heat rate (= heat flux x area)
entering the control volume plus the heat rate generated
equals the heat rate coming out, we have (Fig. 10.8)


121
y
x
x y x y
q
q
q dy q dx Qdx dy q dx dy q dy dx
x y
    

 

 
     
 
 
 
   
(10.41)
or, upon canceling terms,
0
y
x
q
q
Q
x y


  
 
(10.42)
/ /
x y
q k T x and q k T y
     
0
T T
k k Q
x x y y
 
   
 
  
 
 
   
   
Substituting for into Eq.10.42, we get
the heat diffusion equation
(10.43)
We note that this partial differential equation is a special case
of the Helmholtz equation given in Eq. 10.1.

Boundary Conditions The governing equation, Eq.10.43, has
to be solved together with certain boundary conditions. These
boundary conditions are of three types as shown in Fig. 10.9:
122
1)Specified temperature T = To on St,
2)Specified heat flux qn = qo on Sq, and
3)Convection on Sc.
 
n
q h T T
 
The interior of the body is denoted by A, and the boundary is
denoted as S = (ST + Sq + Sc). Further, qn is the heat flux
normal to the boundary. The sign convection adopted here for
specifying qo is that qo > 0 if heat is flowing out of the body,
while qo < 0 if heat is flowing into the body.

124
The triangular element The triangular element (Fig.10.10)
will be used to solve the heat-conduction problem. Extension
to quadrilateral or other isoparametric elements follows in a
similar manner as discussed earlier for stress analysis.

125
Consider a constant length of the body perpendicular to the
x,y plane. The temperature field within an element is given by
T = N1T1 + N2T2 + N3T3
or
T = NTe (10.44)
where  
, ,1
N    
   are the element-shape functions and
Te = [T1, T2, T3]T.
Using Isoparametric method
x = N1x1 + N2x2 +N3x3
y = N1y1 + N2y2 + N3y3 (10.45)

126
Further, the chain rule of differentiation yields
T T x T y
x y
T T x T y
x y
  
  
    
 
    
    
 
    
(10.46)
or
T T
x
J
T
T
y



  
 
   
 
   

   


   

   
  
 
(10.47)

127
In Eq. 10.47, J is the Jacobian matrix given by
13 13
23 23
x y
J
x y
 
 
 
(10.48)
where xij = xi – xj, yij = yi – yj, and = 2Ae, where Ae is the
det J
area of the triangle. Equation 10.47 yields
1
T
T
x
J
T T
y




 

 
 
  

   

   
 
   

   

   
(10.49a)
23 13
23 13
1 0 1
1
0 1 1
det
e
y y
T
x x
J
 
   
    
 
 
 
(10.49b)

128
which can be written as
e
T
T
x
B T
T
y

 
 

 

 

 

 
 
(10.50)
where
 
 
23 13 13 23
23 13 23 13
1
det
T
y y y y
B
J x x x x
 
 
  
 
 
 
(10.51a)
23 31 12
32 13 21
1
det
y y y
x x x
J
 
 
 
(10.51b)

129
Example 10.4
A long bar of rectangular cross section, having thermal
conductivity of 1.5W/m oC is subjected to the boundary
conditions shown in Fig.E10.4a. Two opposite sides are
maintained at a uniform temperature of 180oC; one side is
insulated, and the remaining side is subjected to a convection
process with Determine the
2
25 50 / . .
o o
T C and h W m C
  
temperature distribution in the bar.

130
Solution A five-node, three-element finite element model
of the problem is shown in Fig.E10.4b, where symmetry
about the horizontal axis is used. Note that the line of
symmetry is shown as insulated, since no heat can flow
across it.

131
The element matrices are developed as follows. The element
connectivity is defined as in the following table:
Element 1 2 3 local
1 1 2 3
2 5 1 3 global
3 5 4 3

132
23 31 12
32 13 21
1
det
T
y y y
B
x x x
J
 
  
 
 
 
 
1
2
3
0.15 0.15 0
1
0 0.4 0.4
0.06
0.15 0.15 0.3
1
0.4 0.4 0
0.12
0.15 0.15 0
1
0 0.4 0.4
0.06
T
T
T
B
B
B

 
  

 
 
 
  

 

 

  

 
We have
For each element,

133
T T
e T T
k kA B B yields

 
      
1 1 1
1.5 0.03 T
T T T
k B B

1 2 3
0.28125 0.28125 0
0.28125 2.28125 2.0
0 2.0 2.0

 
 
  
 
 

 
Then,
 
2
5 1 3
1.14 0.86 0.28125
0.86 1.14 0.28125
0.28125 0.28125 0.5625
T
k
 
 
 
  
 
 
 
 
 
3
5 4 3
0.28125 0.28125 0
0.28125 2.28125 2.0
0 2.0 2.0
T
k

 
 
  
 
 

 

134
Now the matrices hT for elements with convection edges are
developed. Since both elements 1 and 3 have edges 2-3 ( in
local node numbers) as convection edges, the formula
2 3
0 0 0
0 2 1
6
0 1 2
T
h
h 
 
 
  
 
 
can be used, resulting in
   
1 2
1 2 3 5 4 3
0 0 0 0 0 0
0 2.5 1.25 0 2.5 1.25
0 1.25 2.5 0 1.25 2.5
T T
h h
   
   
 
   
   
   

135
The matrix K = Σ(kT + hT) is now assembled. The elimination
approach for handling the boundary conditions T = 180oC at
nodes 4 and 5 results in striking out these rows and columns.
However, these fourth and fifth rows are used subsequently
for modifying the R vector. The result is
1 2 3
1.42125 0.28125 0.28125
0.28125 4.78125 0.75
0.28125 0.75 9.5625
K
 
 
 
  
 
 
 
 

136
 
2 3
0 1 1
2
hT
r  
 
     
 
1
1 2 3
50 25 0.15
0 1 1
2
r 
Now the heat-rate vector R is assembled from element
convection contributions. The formula
results in
and
     
 
3
5 4 3
50 25 0.15
0 1 1
2
r 

137
Thus,
 
1 2 3
93.75 0 1 2
T
R
In the elimination approach, R gets modified according to
Eq.7.30. Solution of KT = R then yields
[T1, T2, T3] = [124.5, 34.0, 45.5]oC

138
TORSION
Consider a prismatic rod of arbitrary cross-sectional shape,
which is subjected to a twisting moment M as shown in
Fig.10.13. The problem is to determine shearing stress τxz and
τyz (Fig. 10.14) and the angle of twist per unit length, α.
It can be shown that the solution of such problems, with
simply connected cross sections, reduces to solving the two-
dimensional equation

140
2 2
2 2
2 0 in A
x y
 
 
  
 
θ = 0 on S (10.74)
(10.73)
where A is interior and S is the boundary of the cross
section. Again, we note that Eq. 10.73 is a special case of
Helmholtz’s equations given in Eq.10.1.
0
z y z
k k k Q
x x y y z z
  

 
     
   
    
 
   
     
   
 
(10.1)

141
In Eq.10.74, θ is called the stress function, since once θ is
known, then shearing stresses are obtained as
xz yz
G G
y x
 
   
 
 
 
(10.75)
with α determined from
2
A
M G dA
 
  (10.76)
where G is the shear modulus of the material. The finite
element method for solving Eqs. 10.73 and 10.74.

142
where N = [ξ, η, 1- ξ - η] are the usual shape functions, and
θe = [θ1, θ2, θ3]T are the nodal values of θ. Furthermore, we
have the isoparametric relations.
1 1 2 2 3 3
1 1 2 2 3 3
x N x N x N x
y N y N y N y
  
  
x y
x
x y
y
 
  


  
  
    
 
     
   
   
 

   

  
 
   
  
   
    
   
(10.78)
Triangular Element
The stress function θ within a triangular element is
interpolated as
θ = Nθe (10.77)

143
or
T
T
J
x y
   
 
 
 
   
  
 
   
   
where the Jacobian matrix is given by
13 13
23 23
x y
J
x y
 
 
 
(10.79)
det J
T
e
B
x y
 

 
 

 
 
 
where xij = xi – xj, yij = yi – yj, and = 2Ae. The preceding
(10.80a)
equations yield

144
or
T e
yz xz G B
   
 
 
  (10.80b)
where
23 31 12
32 13 21
1
det
y y y
B
x x x
J
 
  
 
(10.81)
The fact that identical relations also apply to the heat-
conduction problem in the previous section shown the
similarity of treating all field problems by the finite element
method.

145
Example 10.5
Consider the shaft with a rectangular cross section shown in
Fig. E10.5a. Determine, in terms of M and G, the angle of
twist per unit length.

146
Element 1 2 3
1 1 3 2
2 3 4 2
3 4 5 2
4 5 1 2
Solution A finite model of a quadrant of this cross section
is shown in Fig. E10.5b. We define the element connectivity
as in the following table:
Using the relations
23 31 12
32 13 21
1
det
y y y
B
x x x
J
 
  
 

147
and
k = AeBTB
we get
   
1 1
1 2 3
1.042 0.292 1.333
1.5 1.5 0
1 1
1.042 1.333
2 2 4
6 2
2.667
B k
Symmetric

 

   
  
   
 
   
 
Similarly,
 
2
3 4 2
1.042 0.292 0.75
1
1.042 0.75
2
1.5
k
Symmetric
 
 
 
 
 
 
 

148
 
3
4 5 2
1.042 0.292 1.333
1
1.042 1.333
2
2.667
k
Symmetric

 
 
 
 
 
 
 
4
5 1 2
1.042 0.292 0.75
1
1.042 0.75
2
1.5
k
Symmetric
 
 
 
 
 
 
 
Similarly, the element load vector f = (2Ae/3) [1, 1, 1]T for
each element is
 
2
2 1, 2, 3, 4
2
i
f i
 
 
 
 
 
 

149
We can now assemble K and F. Since the boundary
conditions are
θ3 = θ4 = θ5 = 0
we are interested only in degrees of freedom 1 and 2. Thus,
the finite element equations are
1
2
2.084 2.083 4
1
2.083 8.334 8
2


  
   

   
 

   
 
The solution is
[θ1, θ2] = [7.676 , 3.838]
Consider the equation
2
A
M G dA
 
 

150
  
/3 1, 1, 1 ,
e
e
N dA A we get


 
1 2 3
2 4
3
e e e
e
e
A
M G   
 
   
 
 

0.004
M
G
 
Using θ = Nθe, and noting that
This multiplication by 4 is because the finite element model
represents only one-quarter of the rectangular cross section.
Thus, we get the angle of twist per unit length to be

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