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# 计算机代写|机器学习代写Machine Learning代考|COMP7703 Brick Elements

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## 计算机代写|机器学习代写Machine Learning代考|Lagrange type elements

The Lagrange type brick elements can be developed in precisely the same manner as the 2D rectangular elements described in Chapter 7. Consider a brick element with $n_d=$ $(n+1)(m+1)(p+1)$ nodes shown in Figure 9.14. The element is defined in the domain of $(-1 \leq \xi \geq 1,-1 \leq \eta \geq 1,-1 \leq \zeta \geq 1)$ in the natural coordinates $\xi, \eta$ and $\zeta$. Due to the regularity of the nodal distribution along the $\xi, \eta$ and $\zeta$ directions, the shape function of the element can be simply obtained by multiplying one-dimensional shape functions with respect to the $\xi, \eta$ and $\zeta$ directions using the Lagrange interpolants defined in Eq. (4.82) [Zienkiewicz et al., 2000]:
$$N_i=N_I^{1 D} N_J^{1 D} N_K^{1 D}=l_I^n(\xi) l_J^m(\eta) l_K^p(\varsigma)$$
Due to the delta function property of the 1D shape functions given in Eq. (4.83), it is easy to confirm that the $N_i$ given by Eq. (9.66) also has the delta function property.

## 计算机代写|机器学习代写Machine Learning代考|Serendipity type elements

The method used in constructing the Lagrange type of elements is very systematic. However, the Lagrange type of elements is not very widely used, due to the presence of the interior nodes. A serendipity type of brick elements without interior nodes is created by inspective construction methods as described in Chapter 7 for $2 \mathrm{D}$ rectangular elements.

Figure 9.15(a) shows a 20-nodal tri-quadratic element. The element has eight corner nodes and twelve mid-side nodes. The shape functions in the natural coordinates for the quadratic brick element are given as follows:
$$\begin{array}{ll} N_j=\frac{1}{8}\left(1+\xi_j \xi\right)\left(1+\eta_j \eta\right)\left(1+\zeta_j \varsigma\right)\left(\xi_j \xi+\eta_j \eta+\varsigma_i \varsigma-2\right) \ & \text { for corner nodes } j=1, \ldots, 8 \ N_j=\frac{1}{4}\left(1-\xi^2\right)\left(1+\eta_j \eta\right)\left(1+\zeta_j \varsigma\right) & \text { for mid-side nodes } j=10,12,14,16 \ N_j=\frac{1}{4}\left(1-\eta^2\right)\left(1+\xi_j \xi\right)\left(1+\zeta_j \varsigma\right) & \text { for mid-side nodes } j=9,11,13,15 \ N_j=\frac{1}{4}\left(1-\varsigma^2\right)\left(1+\xi_j \xi\right)\left(1+\eta_j \eta\right) & \text { for mid-side nodes } j=17,18,19,20 \end{array}$$
where $\left(\xi_j, \eta_j\right)$ are the natural coordinates of node $j$. It is very easy to observe that the shape functions possess the delta function property. The shape function is constructed by simple inspections, making use of the shape function properties. For example, for corner node 2 (where $\xi_2=1, \eta_2=-1, \zeta_2=-1$ ), the shape function $N_2$ has to pass the following four planes as shown in Figure 9.16 to ensure its vanishing at remote nodes:
\begin{aligned} 1+\xi=0 \Rightarrow \text { vanishes at nodes } 1,4,5,8,11,15,19,20 \ \eta-1=0 \Rightarrow \text { vanishes at nodes } 3,4,7,8,10,14,18,19 \ \varsigma-1=0 \Rightarrow \text { vanishes at nodes } 5,6,7,8,13,14,15,16 \ \xi-\eta-\varsigma-2=0 \Rightarrow \text { vanishes at nodes } 9,12,17 \end{aligned}

## 计算机代写|机器学习代写Machine Learning代考|Lagrange type elements

Lagrange 型砖单元可以按照与第 7 章中描述的二维矩形单元完全相同的方式开发。考虑一个砖单元 $n_d=$ $(n+1)(m+1)(p+1)$ 节点如图 9.14 所示。该元筙在以下域中定义
$(-1 \leq \xi \geq 1,-1 \leq \eta \geq 1,-1 \leq \zeta \geq 1)$ 在自然坐标 $\xi, \eta$ 和 $\zeta$. 由于节点分布沿 $\xi, \eta$ 和 $\zeta$ 方向，单元的形函数 可以简单地通过将一维形函数乘以相对于 $\xi, \eta$ 和 (使用等式中定义的拉格朗日揷值的方向。(4.82)
N_i=N_I^{1 D} N_J^{1 D} N_K^{1 D}=l_I^n(\xi) l_J^m(\eta) l_K^p(\varsigma)


## 计算机代写|机器学习代写Machine Learning代考|Serendipity type elements

$N_j=\frac{1}{8}\left(1+\xi_j \xi\right)\left(1+\eta_j \eta\right)\left(1+\zeta_j \varsigma\right)\left(\xi_j \xi+\eta_j \eta+\varsigma_i \varsigma-2\right) \quad$ for corner nodes $j=1, \ldots, 8 N_j=\frac{1}{4}\left(1-\xi^2\right)\left(1+\eta_j \eta\right)\left(1+\zeta_j \varsigma\right) \quad$ fol

$1+\xi=0 \Rightarrow$ vanishes at nodes $1,4,5,8,11,15,19,20 \eta-1=0 \Rightarrow$ vanishes at nodes $3,4,7,8,10,14,18,19 \varsigma-1=0 \Rightarrow$ vanish

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