Page 97 - Fundamentals of The Finite Element Method for Heat and Fluid Flow

THE FINITE ELEMENT METHOD
obey certain continuity and compatibility conditions. These conditions will be discussed
later in the text.
n
e e 89
I(T ) = I (T ) (3.248)
e=1
Thus, instead of working with a functional defined over the whole solution region, our
attention is now focused on a functional defined for the individual elements. Hence,
n
e
δI = δI = 0 (3.249)
e=1
e
where the variation in I is taken only with respect to the r nodal values associated with
the element e,thatis,
∂I e ∂I e
= = 0 with j = 1, 2,... ,r (3.250)
∂T ∂T j
Equation 3.250 comprises a set of r equations that characterize the behavior of the
element e. The fact that we can represent the functional for the assembly of elements
as a sum of the functional for all individual elements provides the key to formulating
individual element equations from a variational principle. The complete set of assembled
finite element equations for the problem is obtained by adding all the derivatives of I,as
given by Equation 3.250, for all the elements. We can write the complete set of equations as
n e
∂I ∂I
= = 0 with i = 1, 2,... ,M (3.251)
∂T i ∂T i
e=1
The problem is complete when the M set of equations are solved simultaneously for
the M nodal values of T . We now give the details for formulating the individual finite
element equations from a variational principle.
2 2 2
1 ∂T e ∂T e ∂T e
e e
I = k x + k y + k z − 2GT d
2
∂x ∂y ∂z

e 1 e 2
+ qT ds + h(T − T a ) ds (3.252)
2
S 2e S 3e
with
 
 T 1
 
 
e T 2
T = [N]{T}= [N 1 ,N 2 ,... ,N r ] = N 1 T 1 + N 2 T 2 + ··· N r T r (3.253)
...
 
 
T r
and
∂T e
= N 1
∂T 1