Problem 7.12
Eqn 7.85 tells us that
\[\begin{equation*}
M_{AB} = \int_0^1 N_A\,N_B \: \mathrm{d}x
\end{equation*}\]
\(M_{11}\)
This is the first node in the global domain, meaning it will only have contributions from the first element.
Given that, from Eqns 7.56 and 7.57,
\[\begin{align*}
N_1 &= \frac{1}{2} \left[ 1 - \xi \right] \\
\frac{\mathrm{d} x}{\mathrm{d} \xi} &= \frac{1}{2} \Delta x_1
\end{align*}\]
we may write the integral as
\[\begin{align*}
M_{11} &= \int_{-1}^1 N_1\,N_1 \frac{\mathrm{d} x}{\mathrm{d} \xi} \: \mathrm{d}\xi \\
&= \int_{-1}^1 \frac{\Delta x_1 }{8} \left[ 1 - \xi \right]^2 \: \mathrm{d}\xi \\
&= \frac{\Delta x_1 }{8} \left[ \frac{\xi^3}{3} - \xi^2 + \xi \right]^1_{-1} \\
&= \frac{\Delta x_1 }{3} \quad.
\end{align*}\]
Note here the important change in the integration limits from \(x = [0, 1]\) to \(\xi = [-1, 1]\).
\(M_{A-1, \,A}\)
This part in the global matrix is only contributed to by a single element, \(A-1\). Given that, from Eqns 7.56 and 7.57,
\[\begin{align*}
N_1 &= \frac{1}{2} \left[ 1 - \xi \right] \\
N_2 &= \frac{1}{2} \left[ 1 + \xi \right] \\
\frac{\mathrm{d} x}{\mathrm{d} \xi} &= \frac{1}{2} \Delta x_{A-1}
\end{align*}\]
we may write the integral as
\[\begin{align*}
M_{A-1, \,A} &= \int_{-1}^1 N_1\,N_2 \frac{\mathrm{d} x}{\mathrm{d} \xi} \: \mathrm{d}\xi \\
&= \int_{-1}^1 \frac{\Delta x_{A-1} }{8} \left( 1 - \xi \right)\left( 1 + \xi \right) \: \mathrm{d}\xi \\
&= \frac{\Delta x_{A-1} }{8} \left[ \xi - \frac{\xi^3}{3} \right]^1_{-1} \\
&= \frac{\Delta x_{A-1} }{6} \quad.
\end{align*}\]
\(M_{A, \,A}\)
In this case a global node may have contributions from two different elements (the last node of the \(A-1\) element) and the first node of the \(A\)th element. Hence we consider the integral for both elements and assemble them. For the \(A-1\)’th element, \(a=2\) for
\[\begin{align*}
N_{a} &= \frac{1}{2} \left[ 1 + (-1)^{a} \, \xi \right] \\
\end{align*}\]
meaning the integral for this element is
\[\begin{align*}
M_{AA} &= \int_{-1}^1 N_2\,N_2 \frac{\mathrm{d} x}{\mathrm{d} \xi} \: \mathrm{d}\xi \\
M_{AA} &= \int_{-1}^1 \frac{1}{4} (1 + \xi)^2 \frac{\Delta x_{A-1}}{2} \: \mathrm{d}\xi \quad = \frac{1}{3}\Delta x_{A-1}
\end{align*}\]
Now let us consider the \(A\)’th element. In this case we are focused on the first shape function
\[\begin{align*}
M_{AA} &= \int_{-1}^1 N_2\,N_2 \frac{\mathrm{d} x}{\mathrm{d} \xi} \: \mathrm{d}\xi \\
M_{AA} &= \int_{-1}^1 \frac{1}{4} (1 - \xi)^2 \frac{\Delta x_{A}}{2} \: \mathrm{d}\xi \quad = \frac{1}{3}\Delta x_{A}
\end{align*}\]
hence the overall contribution to this global (assembled) matrix is
\[\begin{equation*}
M_{AA} = \frac{1}{3}\Delta x_{A-1} + \frac{1}{3}\Delta x_{A} \quad .
\end{equation*}\]
