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EXAMPLE 31.2 oach yields equations that are identical to those derived above. Calculus (for example, see Allaire, 1985). For the Element Equation for a Heated Rod Problem Statement. Employ Eq. (31.26) to develop the, element equations for a 10-cm rod with boundary conditions of $T(0, t) = 40$ and $T(10, t) = 200$ and a uniform heat source of $f(x) = 10$. Employ four equal-size elements of length $= 2.5$ cm. Solution. The heat source term in the first row of Eq. (31.26) can be evaluated by substituting Eq. (31.3) and integrating to give $\int_0^{2.5} \frac{2.5 - x}{2.5} 10 \, dx = 12.5$ Similarly, Eq. (31.4) can be substituted into the heat source term of the second row of Eq. (31.26), which can also be integrated to yield $\int_0^{2.5} \frac{x - 0}{2.5} 10 \, dx = 12.5$ These results along with the other parameter values can be substituted into Eq. (31.26) to give $0.4T_1 - 0.4T_2 = \frac{dT}{dx}(x_1) + 12.5$ and $-0.4T_1 + 0.4T_2 = \frac{dT}{dx}(x_2) + 12.5$ -0.4 -0.4 0 -0.4 0 0.8 0 -0.4 0 0 0 -0.4 0.8 -0.4 0 0 1 -0.4 0.4 $\begin{bmatrix} T_1\\T_2\\T_3\\T_4\\T_5 \end{bmatrix} = \begin{Bmatrix} -dT_1/dx + 12.5\\25\\25\\25\\dT_5/dx + 12.5 \end{Bmatrix}$

          EXAMPLE 31.2
oach yields equations that are identical to those derived above.
Calculus (for example, see Allaire, 1985). For the
Element Equation for a Heated Rod
Problem Statement. Employ Eq. (31.26) to develop the, element equations for a 10-cm
rod with boundary conditions of $T(0, t) = 40$ and $T(10, t) = 200$ and a uniform heat
source of $f(x) = 10$. Employ four equal-size elements of length $= 2.5$ cm.
Solution. The heat source term in the first row of Eq. (31.26) can be evaluated by
substituting Eq. (31.3) and integrating to give
$\int_0^{2.5} \frac{2.5 - x}{2.5} 10 \, dx = 12.5$
Similarly, Eq. (31.4) can be substituted into the heat source term of the second row of
Eq. (31.26), which can also be integrated to yield
$\int_0^{2.5} \frac{x - 0}{2.5} 10 \, dx = 12.5$
These results along with the other parameter values can be substituted into Eq. (31.26)
to give
$0.4T_1 - 0.4T_2 = \frac{dT}{dx}(x_1) + 12.5$
and
$-0.4T_1 + 0.4T_2 = \frac{dT}{dx}(x_2) + 12.5$
-0.4
-0.4
0
-0.4
0
0.8
0
-0.4
0
0
0
-0.4
0.8
-0.4
0
0
1
-0.4
0.4
$\begin{bmatrix} T_1\\T_2\\T_3\\T_4\\T_5 \end{bmatrix} = \begin{Bmatrix} -dT_1/dx + 12.5\\25\\25\\25\\dT_5/dx + 12.5 \end{Bmatrix}$

EXAMPLE 31.2
oach yields equations that are identical to those derived above.
Calculus (for example, see Allaire, 1985). For the
Element Equation for a Heated Rod
Problem Statement. Employ Eq. (31.26) to develop the, element equations for a 10-cm
rod with boundary conditions of T(0, t) = 40 and T(10, t) = 200 and a uniform heat
source of f(x) = 10. Employ four equal-size elements of length = 2.5 cm.
Solution. The heat source term in the first row of Eq. (31.26) can be evaluated by
substituting Eq. (31.3) and integrating to give
∫0^2.5(2.5 - x)/(2.5) 10 dx = 12.5
Similarly, Eq. (31.4) can be substituted into the heat source term of the second row of
Eq. (31.26), which can also be integrated to yield
∫0^2.5(x - 0)/(2.5) 10 dx = 12.5
These results along with the other parameter values can be substituted into Eq. (31.26)
to give
0.4T1 - 0.4T2 = (dT)/(dx)(x1) + 12.5
and
-0.4T1 + 0.4T2 = (dT)/(dx)(x2) + 12.5
-0.4
-0.4
0
-0.4
0
0.8
0
-0.4
0
0
0
-0.4
0.8
-0.4
0
0
1
-0.4
0.4
< b m a t r i x >
= -dT1/dx + 12.5
25
25
25
dT5/dx + 12.5

Added by Brandy N.

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