Page 98 - Fundamentals of The Finite Element Method for Heat and Fluid Flow

90
∂T
∂T 2
∂T e e = N 2 THE FINITE ELEMENT METHOD
= N r (3.254)
∂T r
or
 
N 1


∂T e  N 2  T
= ={N}= [N] (3.255)
∂{T} ...
 
 
N r
The gradient matrix is written as
∂T ∂N 1 ∂N 2 ∂N r
 e   
...
 
   
∂x ∂x ∂x
 
   
 ∂x   T 1 
     
∂T  ∂N 1 ∂N 2 ∂N r  T 2
 e   
{g}= =  ...  = [B]{T} (3.256)
 ∂y ∂y 

 ∂y  ···
  ∂y   
   
∂N 1 ∂N 2 ∂N r T r
 e   
 ∂T 
...
 
 
∂z ∂z ∂z ∂z
Consider
e
 
 ∂T 
 
 
 
  ∂x 




e e e k x 00  ∂T e 

T ∂T ∂T ∂T
{g} [D]{g}=  0 k y 0 
∂x ∂y ∂z  ∂y 
 
00 k z  e 
 ∂T 
 
 
 
∂z
 
e 2 e 2 e 2
∂T ∂T ∂T
= k x + k y + k z (3.257)
∂x ∂y ∂z
substituting into Equation 3.252, we have
1 " # 1
e T e e e 2
I = {g} [D]{g}− 2GT d
+ qT ds + h(T − T a ) ds (3.258)
2
S 2e S 3e 2
T
T
T
From Equation 3.256 we can substitute {g} [D]{g}= {T} [B] [D][B]{T} and mini-
mizing the integral, we have (employing Equation 3.255)
∂I e 1 T 1 T
= 2[B] [D][B]{T}d
− 2G[N] {T}d
∂{T}
2
2

T T
+ q[N] {T}ds + h[N] {T}ds
S 2e S 3e

T
− h[N] T a ds = 0 (3.259)
S 3e