Question 3.5: Figure 3.27a shows a FEA model for a 2D heat transfer proble...

Figure 3.27a shows a FEA model for a 2D heat transfer problem.
The model is to find the temperature distribution, and the FEA model consists of 10 nodes for three triangular elements and three rectangle elements. Build a system model based on the given element models as follows:
The stiffness matrices [K_{i}] for elements are found as

Triangle elements (1), (4) and (6) :     Rectangle elements (2), (3) and (4) :

[K]^{(e)}= \begin{bmatrix}0.5 & -0.5 & 0 \\ -0.5 & 1 & -0.5 \\ 0 & -0.5 & 0.5 \end{bmatrix}      [K]^{(e)}= \begin{bmatrix}4 & -1 & -2 & -1 \\ -1& 4& -1 & -2\\ -2& -1& 4 &-1\\-1&-2&-1&4 \end{bmatrix}

3.5
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Each node in the FEA element has one DoF, and the model has 10 nodes.
Therefore, the system model has N = 10 DoF, and [K] for the system has a size of 10 × 10.
The system model is assembled from element models. The model as N_{e}=6 elements, the involved DoF in these elements are described in Fig. 3.27b.

To use Eq. (3.91) to assemble system model, [A_{i}] (i = 1, 2, …, 6) are determined based on Fig. 3.27b as

\left[K\right]_{N\times N} = \sum\limits_{i= 1}^{i= N_{e} }{\left[A_{i} \right] ^ {T} \left[K_{i} \right]\left[A_{i} \right] }         (3.91)

[A_{1} ]=\begin{bmatrix} 1 & 0& 0& 0& 0& 0& 0& 0& 0& 0 \\ 0 & 1& 0& 0& 0& 0& 0& 0& 0& 0 \\ 0& 0& 0& 0& 1& 0& 0& 0& 0& 0 \end{bmatrix}

 

[A_{2} ]=\begin{bmatrix}0 & 1& 0& 0& 0& 0& 0& 0& 0& 0 \\ 0 & 0& 1& 0& 0& 0& 0& 0& 0& 0 \\ 0& 0& 0& 0& 0& 1& 0& 0& 0& 0 \\ 0& 0& 0& 0& 1& 0& 0& 0& 0& 0 \end{bmatrix}

 

[A_{3} ]=\begin{bmatrix}0 & 0& 1& 0& 0& 0& 0& 0& 0& 0 \\ 0 & 0& 0& 1& 0& 0& 0& 0& 0& 0 \\ 0& 0& 0& 0& 0& 0& 1& 0& 0& 0 \\ 0& 0& 0& 0& 0& 1& 0& 0& 0& 0 \end{bmatrix}

 

[A_{4} ]=\begin{bmatrix} 0 & 0& 0& 0& 1& 0& 0& 0& 0& 0 \\ 0 & 0& 0& 0& 0& 1& 0& 0& 0& 0 \\ 0& 0& 0& 0& 0& 0& 0& 1& 0& 0 \end{bmatrix}

 

[A_{5} ]=\begin{bmatrix}0 & 0& 0& 0& 0& 1& 0& 0& 0& 0 \\ 0 & 0& 0& 0& 0& 0& 1& 0& 0& 0 \\ 0& 0& 0& 0& 0& 0& 0& 0& 1& 0 \\ 0& 0& 0& 0& 0& 1& 0& 1& 0& 0 \end{bmatrix}

 

[A_{6} ]=\begin{bmatrix} 0 & 0& 0& 0& 0& 0& 0& 1& 0& 0 \\ 0 & 0& 0& 0& 0& 0& 0& 0& 1& 0 \\ 0& 0& 0& 0& 0& 0& 0& 0& 0& 1 \end{bmatrix}

 

Substituting the above assistivematrices and given stiffnessmatrices in Eq. (3.91) yield the stiffness matrix of the system as

[\boldsymbol{K}]_{10 \times 10}=\sum_{i=1}^{i=6}\left[\boldsymbol{A}_i\right]^T\left[\boldsymbol{K}_i\right]\left[\boldsymbol{A}_i\right]=\left[\begin{array}{cccccccccc} 0.5 & -0.5 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ -0.5 & 0.5 & -1 & 0 & -1.5 & -2 & 0 & 0 & 0 & 0 \\ 0 & -1 & 8 & -1 & -2 & -2 & -2 & 0 & 0 & 0 \\ 0 & 0 & -1 & 4 & 0 & -2 & -1 & 0 & 0 & 0 \\ 0 & -1.5 & -2 & 0 & 5 & -1.5 & 0 & 0 & 0 & 0 \\ 0 & -2 & -2 & -2 & -1.5 & 13 & -2 & -1.5 & -2 & 0 \\ 0 & 0 & -2 & -1 & 0 & -2 & 8 & -2 & -1 & 0 \\ 0 & 0 & 0 & 0 & 0 & -1.5 & -2 & 5 & -2 & 0 \\ 0 & 0 & 0 & 0 & 0 & -2 & -1 & -1.5 & 5 & -0.5 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & -1 & 0.5 \end{array}\right]     (3.39)

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