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Lecture 2
FEM in 1D: assembly of system matrix, cooling fin
(Lecture notes taken by Jaeseok Heo and Jason Andrus)

Rayleigh-Ritz Discrete Model.
Using a generalized minimization we define:
min G (u ( x))
where u ( x)  S
Substitute u ( x) 
G : S 
and
N
 u  ( x)
j
j
1
where  j (x) is a shape function that could be a trig function or polynomial, ect.
Thus the new problem is:
 n

min G   u j j ( x) 
 1

where u n
and
F : n  
 n

F  u   G   u j j ( x )  .
 1

Now Set :
F (u )
 0.
u j
This yields n equations for n u j  thereby introducing many possible approximate
solutions to the differential equation.

Finite Element Discrete Model: Variational
Recall that we can represent the potential energy of the string by:
J (u ) 
T
2

L
0
L
u x2 ( x)   f u
0
where the set of functions is defined as:
S  {u :[0 L ]  
u (0) and u ( L) given
and

L
0
u x2 ( x)  }.
Let us define u as:
u
n
 u  ( x)
j 0
j
j
Let us recall that φj(x) is a shape function and can be represented as shown below:
For a domain with five nodes and four elements:
We can expect the space function to behave:
1
φ4
φ0
h
L-h
1
xi-h
xi
xi +h
We can now take the partial derivative of the potential energy:

ui
2
L
 T L n
 n


J   u j j ( x)  
    u j j ( x)   
ui  2 0  0
 0

x 0
 n

f   u j j ( x )  
 0
 
1
Let the integral be broken into intervals by element:

L
0

h
0

2h
 ... 
L
L h
h
Thus we get:

J
ui


ui

xi

xh  h

ui

xi  h
xi
Introducing some notation let:
ei  [ xi , xi  h]
where ui corresponds to system node i.
If we restrict u e ( x)  u ( x) ei then we get the following behavior
u(x)
1
2
where the circled numbers represent the corresponding element node numbers.
We can then represent the system by a combination of it’s nodes using the following
expression:
i  nod (e, i ) where
i  system number
e  element number
i  element node number
Note that in a 1-D system the element node # will be 1 or 2, in a 2-D system the element
node# will be 1,2, or 3, and in a 3-D system the element node # will be 1,2,3, or 4

Linear shape function properties.
N2 ei
N1 ei
N2
ei-1
ei-1
xi
ei
Given a shape function we need to know:   x2 ,  f  where φ is a combination of N1 and N2.

L
0
N1ex 
1
 xh

dx    h  1


0
h x
h


h
2
1
1
0  N    h  h  h
Also need to know :
L
e 2
1x
 N 
L
e 2
1
0
2h
 xh
  
 
0
h 
6

h
2
In general this can be represented by:
N
m
1
N 2n 
m! n !
h
(m  n  1)!

Assembly by nodes for string.
L
L
0
0
T  uxx   f 
Let us define f I as:
f I   f j j ( x)
T
L
0
  u  ( x)       f  ( x) 
L
j
 T 
L
0
j
xj
j
x

ix
j
0
 jx ( x)ix u j  
j
j
i
   (x)  f
L
0
j
i
j
xi
These terms are nonzero if:
j  i 1
j i
j  i 1
Thus the nonzero terms are:
1
2
1
h
4h
h
 ui 1  ui  ui 1  fi 1 
fi  fi 1
h
h
h
6
6
6
We define u as:
u   u j j ( x )
j : all nodes
j
The matrix A may be assembled by inspecting each row of the matrix, that is, fixing i and varying j = i-1,
i and i+1. The 5x5 matrix below can be assembled either by rows or four 2x2 element matrices.
A55

x x
 x xv v


v vw w

w wz


z





z
z 
Assembly by elements and element matrices for string.
u   u j ( x) j : all elements
e
j


e
J   u j ( x)     
e
 j
 j j

The integral can be broken into intervals:

 
e
J   u j ( x)     
ei1
ei
ui  j

ui-1
ui
ui+1
ei
ui  u1ei or ui  u2ei

J u e  k11u1e  k12u2e  d1e
e
u1
 

J u e  k21u1e  k22u2e  d 2e
u2e
 
k
Element Matrices k e   11
 k21
 1
 h
k12 

T

k22 
 1
 h
1
 
h

1 
h 

Heat in a cooling fin with derivative boundary condition.
 Kuxx  Cu  f
u (0) given
Kux ( L)  c us  u( L) 
T
hot
W
x=L
c  0 :insulated
c   u ( L)  us : given
0  c   : partial insulation
TW  Ku xx    2T  2W  c  us  u 
If c   , then the problem is
 Kuxx  Cu  f
u (0), u ( L) given .
The weak equation of formed by    Ku xx  Cu     f  ,  (0)   ( L)  0
  Ku 
x
x
 Cu    f 
New element matrix is
h
 K
C

h
3
ke  
 K  C h
6
 h
K
h
C 
h
6
.
K
h 
C
h
3 
