Page 53 - Fundamentals of The Finite Element Method for Heat and Fluid Flow

THE FINITE ELEMENT METHOD
Example 3.2.1 Calculate the temperature of an 8-cm long bar at a distance of 5 cm from
◦
one end where the temperature is 120 C with the other end at a temperature of 200 C.
Assume the temperature variation between the two end points as being linear. ◦ 45
From Equation 3.8, the temperature distribution over an element can be written as
T = N i T i + N j T j (3.16)
where, at x = 5cm
x j − x 3
N i = =
x j − x i 8
x − x i 5
N j = = (3.17)
x j − x i 8
◦
Substituting into Equation 3.16, we get T = 170 C. Note that N i + N j = 1.

3.2.2 One-dimensional quadratic element

We can see from Figure 3.3(d) that a better approximation for the temperature profile could
be achieved if we use parabolic arcs over each element rather than linear segments. The
function T(x) would therefore be quadratic in x within each element and is of the form
T(x) = α 1 + α 2 x + α 3 x 2 (3.18)

We now have three parameters to determine and hence we need the temperature at
one more point in addition to two end points of an element. We choose the mid-point in
addition to the end values to get the following equations for the temperature at these three
locations,

T i = α 1
l 2
l
T j = α 1 + α 2 + α 3
2 2
T k = α 1 + α 2 l + α 3 l 2 (3.19)
From the above three equations, we obtain the following values for the three constants
α 1 , α 2 and α 3 .
α 1 = T i
1
α 2 = (−3T i + 4T j − T k )
l
2
α 3 = 2 (T i − 2T j + T k ) (3.20)
l
Substituting the values of α 1 , α 2 and α 3 , into Equation 3.18 and collating the coefficients
of T i ,T j and T k , we get
3x 2x x x x x
2 2 2
T = T i 1 − + + T j 4 − 4 + T k 2 − (3.21)
l l 2 l l 2 l 2 l