Problem 7.4

Starting with Eqn. 7.28 and taking our answers from Problem 7.3:

\[ \Sigma_{B=1}^n \mathrm{K}_{AB} \, \mathrm{d}_b = \mathrm{F}_{A} \]

where the matrix is

\[\begin{split} \mathrm{K} = \begin{bmatrix} 2 & -2 \\ -2 & 4 \end{bmatrix} \quad. \end{split}\]

The force vector can be written explicitly from Eqns 7.42-7.43, with the assumption of \(h=0\), as

\[\begin{split} \mathrm{F} = \begin{bmatrix} H_0 \\ 2\theta_1 \end{bmatrix} \end{split}\]

such that the overall linear system of equations is

\[\begin{split} \begin{bmatrix} 2 & -2 \\ -2 & 4 \end{bmatrix} \begin{bmatrix} d_1 \\ d_2 \end{bmatrix} = \begin{bmatrix} H_0 \\ 2\theta_1 \end{bmatrix} \quad. \end{split}\]

To determine an expression for \(\theta(x)\), we seek the coefficients \(d_B\) such that we can write \(\theta(x)\) in terms of the shape functions in Eqn. 7.26:

\[ \theta(x) = \Sigma_{B=1}^n d_B N_B(x) + \theta_1 N_{n+1}(x) \]

where \(n=2\) for this case.

Solving for the coefficients \(d_B\)

In this case it is straightforward to solve this linear system of equations by inverting the matrix \(K\). The inverse of this 2 x 2 matrix, \(K^{-1}\), is

\[\begin{split} \begin{bmatrix} 1 & \frac{1}{2} \\ \frac{1}{2} & \frac{1}{2} \end{bmatrix} \end{split}\]

and so the coefficients may be determined by

\[\begin{split} \begin{bmatrix} d_1 \\ d_2 \end{bmatrix} = \begin{bmatrix} 1 & \frac{1}{2} \\ \frac{1}{2} & \frac{1}{2} \end{bmatrix}\begin{bmatrix} H_0 \\ 2\theta_1 \end{bmatrix} \quad. \end{split}\]

such that

\[\begin{split} d_1 = H_0 + \theta_1 \\ d_2 = \frac{1}{2} H_0 + \theta_1 \end{split}\]

Applying the coefficients \(d_B\)

Substituting the coefficients \(d_1\) and \(d_2\) into

\[ \theta(x) = d_1 N_1 + d_2 N_2 + \theta_1 N_{n+1}(x) \]

will provide the final solution for \(\theta\). Because the form of the shape functions is not the same at each part of the domain (there are two elements) we need to consider each element separately to be certain. We will see, however, that each element provides the same solution for \(\theta\).

Element 1: \(0 \leq x \leq \frac{1}{2}\)

Applying the linear shape functions in 7.39-7.41 as well as the coefficients for \(d_1\) and \(d_2\), while noting \(d_{n+1}\) = \(d_3\) = \(\theta_1\):

\[ \theta(x) = (H_0 + \theta_1)(1-2x) + 2x\,(\frac{1}{2} H_0 + \theta_1) \]

since \(N_3\) is 0 here. Hence, this reduces to

\[ \theta(x) = \theta_1 + (1-x)H_0 \]

Element 2: \(\frac{1}{2} \leq x \leq 1\)

In this case \(N_1\) = 0 and so

\[ \theta(x) = 2(\frac{1}{2} H_0 + \theta_1)(1-x) + \theta_1 (2x-1) \]

which also reduces to

\[ \theta(x) = \theta_1 + (1-x)H_0 \]