Page 58 - Fundamentals of The Finite Element Method for Heat and Fluid Flow

THE FINITE ELEMENT METHOD
50
where,
1
N i =
2A (a i + b i x + c i y)
1
N j = (a j + b j x + c j y)
2A
1
N k = (a k + b k x + c k y) (3.39)
2A
and
a i = x j y k − x k y j ; b i = y j − y k ; c i = x k − x j
a j = x k y i − x i y k ; b j = y k − y i ; c j = x i − x k
a k = x i y j − x j y i ; b k = y i − y j ; c k = x j − x i (3.40)
If we evaluate N i at node i, where the coordinates are (x i ,y i ), then we obtain
2A
1
(N i ) i = (x j y k − x k y j ) + (y j − y k )x i + (x k − x j )y i = = 1 (3.41)
2A 2A
Similarly, it can readily be verified that (N j ) i = (N k ) i = 0.
Thus, we see that the shape functions have a value of unity at the designated vertex
and zero at all other vertices. It is possible to show that
N i + N j + N k = 1 (3.42)

everywhere in the element, including the boundaries.
The gradients of the temperature T are given by
∂T ∂N i ∂N j ∂N k b i b j b k
= T i + T j + T k = T i + T j + T k
∂x ∂x ∂x ∂x 2A 2A 2A
∂T ∂N i ∂N j ∂N k c i c j c k
= T i + T j + T k = T i + T j + T k (3.43)
∂y ∂y ∂y ∂y 2A 2A 2A
or  
∂T
   
 
∂x b i b j b k
  1 T i 
{g}= = T j = [B]{T} (3.44)
 ∂T  2A c i c j c k  
T k
 
∂y
 
It should be noted that both ∂T /∂x and ∂T /∂y are constants within an element as
b i ,b j ,b k and c i ,c j ,c k are constants for a given triangle. Hence, the heat fluxes q x and q y
are also constants within a linear triangular element. Since the temperature varies linearly
within an element, it is possible to draw the isotherms within a linear triangle and this is
illustrated in the following example.
Example 3.2.2 As an illustration of the method of calculation, let us calculate the temper-
ature, T and heat fluxes q x and q y within an element for the data given in Table 3.2
Calculate the temperature T , and the heat flux components q x and q y at (2.0, 1.0) if the
◦
thermal conductivity of the material is 2 W/cm K. Draw the isothermal line for 60 Cinthe
triangle.