Problem 7.9

Problem 7.9#

The local linear shape functions are defined in 7.56 as

\[\begin{equation*} N_a (\xi) = \frac{1}{2} \left[ 1 + (-1)^a \, \xi \right] \end{equation*}\]

and the local diffusivity matrix is defined in 7.63 as

\[\begin{equation*} k_{ab}^A = \int_{x_A}^{x_{A+1}} \left( \partial_x N_a \right) \left( \partial_x N_b \right) \: \mathrm{d}x \quad. \end{equation*}\]

To get the derivatives of the shape functions, we need to use the chain rule

\[\begin{equation*} \partial_x N_a = \frac{\mathrm{d}N_a}{\mathrm{d} \xi}\frac{\mathrm{d} \xi}{\mathrm{d} x}. \end{equation*}\]

yielding

\[\begin{equation*} k_{ab}^A = \int_{x_A}^{x_{A+1}} \frac{\mathrm{d}N_a}{\mathrm{d} \xi} \frac{\mathrm{d}N_b}{\mathrm{d} \xi} \left[ \frac{\mathrm{d} \xi}{\mathrm{d} x} \right]^2 \: \mathrm{d}x \quad. \end{equation*}\]

Next we can change the variable of integration as

\[\begin{equation*} \mathrm{d}x = \frac{\mathrm{d}x }{\mathrm{d}\xi}\: \mathrm{d}\xi \end{equation*}\]

and the limits

\[\begin{align*} x = x_A &\rightarrow \xi = -1 \\ x = x_{A+1} &\rightarrow \xi = +1 \end{align*}\]

to produce

\[\begin{equation*} k_{ab}^A = \int_{-1}^{1} \frac{\mathrm{d}N_a}{\mathrm{d} \xi} \frac{\mathrm{d}N_b}{\mathrm{d} \xi} \left[ \frac{\mathrm{d} \xi}{\mathrm{d} x} \right]^2 \: \frac{\mathrm{d}x }{\mathrm{d}\xi}\: \mathrm{d}\xi = \int_{-1}^{1} \frac{\mathrm{d}N_a}{\mathrm{d} \xi} \frac{\mathrm{d}N_b}{\mathrm{d} \xi} \frac{\mathrm{d} \xi}{\mathrm{d} x} \: \: \mathrm{d}\xi \end{equation*}\]

as in 7.64. Now let us evaluate this integral using

\[\begin{equation*} \frac{\mathrm{d}N_a}{\mathrm{d} \xi} = \frac{1}{2} (-1)^a \quad, \end{equation*}\]
\[\begin{equation*} \frac{\mathrm{d}N_b}{\mathrm{d} \xi} = \frac{1}{2} (-1)^b \quad, \end{equation*}\]
\[\begin{equation*} \frac{\mathrm{d}\xi}{\mathrm{d} x} = \frac{2}{\Delta x_A} \quad, \end{equation*}\]

which gives

\[\begin{align*} k_{ab}^A &= \frac{1}{4} \int_{-1}^{1} (-1)^a (-1)^b \frac{2}{\Delta x_A} \: \: \mathrm{d}\xi \\ &= \frac{1}{4} \int_{-1}^{1} (-1)^{a+b} \frac{2}{\Delta x_A} \: \: \mathrm{d}\xi \: = \frac{(-1)^{a+b}}{\Delta x_A} \end{align*}\]