Page 142 - Fundamentals of The Finite Element Method for Heat and Fluid Flow

STEADY STATE HEAT CONDUCTION IN MULTI-DIMENSIONS
134
Therefore, for the first node
N 1 = 1 (a 1 + b 1 x 4 + c 1 y 4 ) (5.22)
2A
where
a 1 = x 2 y 3 − x 3 y 2 = 2500.00
b 1 = y 2 − y 3 = 0.0
c 1 = x 3 − x 2 =−50.00 (5.23)

At point 4, (x = 40, y = 40), from Equation 5.22, we get (2A = 2500)
1
N 1 = (5.24)
5
Similarly, it can be verified that N 2 = 1/5 and N 3 = 3/5. Note that N 1 + N 2 + N 3 = 1.
The substitution of these shape function values into Equation 5.21 results in a value of
◦
T 4 = 160 C.
The heat flux in the x and y directions are calculated as
∂T 10 2
q x =−k =− (b 1 T 1 + b 2 T 2 + b 3 T 3 ) =−20 W/cm (5.25)
∂x 2500
2
Similarly, it can be shown that q y =−20 W/cm . It should be noted that the flux is
constant over a linear triangular element.
From Examples 5.2.1 and 5.2.2, the demonstration of problems involving constant tem-
perature boundary conditions is clear. It is therefore essential to move on to an example
with more complicated boundary conditions. Thus, in the following example, a conduction
problem is considered, which has mixed boundary conditions.

Example 5.2.3 Determine the temperature distribution in a square plate of unit thickness
size 5 cm as shown in Figure 5.9. The upper triangular half has an internal heat generation
3
of 1.2 W/cm , while the lower half has a point source of 5 W/cm in the thickness direction
(point source on a two-dimensional plane) at the point (1, 1) cm. In addition to the above
heat sources, the bottom side of the plate is insulated, the right vertical side is subjected to
a temperature of 100 C, the top side is subjected to a convective heat transfer boundary
◦
2
◦
condition with a heat transfer coefficient of h = 1.2 W/cm K and T a = 30 C and the left
2
vertical side is subjected to a uniform heat flux of 2 W/cm . Assume a thermal conductivity
of 2 W/cm C.
◦
To make the solution procedure simple, the plate is divided into two triangular elements
as shown in Figure 5.10. The elemental equations of both elements can be set up separately
using the formulation discussed (Equations 5.8 and 5.9). For the first element, a 1 = 25.0,
b 1 =−5.0, c 1 =−5.0, a 2 = 0.0, b 2 = 5.0, c 2 = 0.0, a 3 = 0.0, b 3 = 0.0, c 3 = 5.0.
The stiffness matrix for element 1 is
 
 2   2
b b 1 b 2 b 1 b 3 c
t  1 1 c 1 c 2 c 1 c 3 
[K] 1 = k x  b 1 b 2 b 2 b 2 b 3  + k y  c 1 c 2 c 2 c 2 c 3  (5.26)
2 2
4A  2 2 
b 1 b 3 b 2 b 3 b c 1 c 3 c 2 c 3 c
3 3