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# 数学代写|有限元方法作业代写finite differences method代考|Higher-order elements: the isoparametric concept

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## 数学代写|有限元方法作业代写finite differences method代考|A two-point boundary-value problem

In Chapter 3 , the finite element philosophy was developed and illustrated by means of examples in which the interpolation functions were linear polynomials. There is no reason to choose linear functions only; indeed, higher-order polynomials are frequently used and in principle cause no more difficulty. We shall illustrate the approach using the Galerkin finite element method of Sections 3.2-3.5.

$$-u^{\prime \prime}=f(x), \quad 0<x<1, \quad u(0)=1, \quad u^{\prime}(1)+2 u(1)=3 .$$
Suppose that the region $0 \leq x \leq 1$ is divided into $E$ elements, each element having three nodes; a typical element is shown in Fig. 4.1. In terms of its three nodal values, the variable $u$ may be uniquely interpolated as a quadratic polynomial, the interpolation being given by
$$u^{e}=\mathbf{N}^{e} \mathbf{U}^{e},$$
where $\mathbf{U}^{e}=\left[\begin{array}{lll}U_{1} & U_{2} & U_{3}\end{array}\right]^{T}$ and the shape function matrix is
$$\mathbf{N}^{e}=\frac{1}{2}\left[\begin{array}{lll} \xi^{2}-\xi & 2\left(1-\xi^{2}\right) & \xi^{2}+\xi \end{array}\right] .$$

## 数学代写|有限元方法作业代写finite differences method代考|Higher-order rectangular elements

The shape functions for the bilinear rectangular element, discussed in Section 3.7, were obtained by taking products of the Lagrange linear interpolation polynomials. This idea may be extended to develop higher-order rectangular elements; for example, the shape functions for the elements in Fig. $4.3$ are obtained by taking products of Lagrange quadratic, cubic and quartic polynomials.

Although the Lagrange family is convenient to use and easy to set up, it does have two drawbacks:

1. The elements have many internal nodes and, in the overall system, these contribute to one element only. This may, however, be overcome by the ‘condensation’ procedure of Section 4.5.
2. The expressions for the shape functions contain relatively high-order terms, while some lower-order terms are missing.

## 数学代写|有限元方法作业代写FINITE DIFFERENCES METHOD代考|Higher-order triangular elements

The quadratic, cubic and quartic triangular elements are shown in Fig. 4.5. The shape functions are very easy to generate in terms of the area coordinates given by eqn (3.62).

Corner nodes:
$$N_{i}=L_{i}\left(2 L_{i}-1\right)$$
Mid-side nodes:
$$N_{4}=4 L_{1} L_{2}, \text { etc. }$$
(b) Cubic element.
Corner nodes:
$$N_{i}=\frac{1}{2}\left(3 L_{i}-1\right)\left(3 L_{i}-2\right) .$$
Side nodes:
\begin{aligned} &N_{4}=\frac{9}{2} L_{1} L_{2}\left(3 L_{1}-1\right), \ &N_{5}=\frac{9}{2} L_{1} L_{2}\left(3 L_{1}-2\right), \text { etc. } \end{aligned}
Node at centroid:
$$N_{10}=27 L_{1} L_{2} L_{3} .$$
(c) Quartic element. The shape functions are obtained in Exercise 4.5.
For all these triangular elements, the function values along a side are uniquely determined by the nodal values on that side. Consequently, these elements are conforming elements. The element matrices for both triangular and rectangular elements are obtained in the usual way; however, it is not difficult to see that the integrals involved make hand calculation impracticable for the cubic and higher-order elements. For this reason, it is usual to use numerical integration to obtain the matrices.

## 数学代写|有限元方法作业代写FINITE DIFFERENCES METHOD代考|A TWO-POINT BOUNDARY-VALUE PROBLEM

ñ和=12[X2−X2(1−X2)X2+X].

## 数学代写|有限元方法作业代写FINITE DIFFERENCES METHOD代考|HIGHER-ORDER RECTANGULAR ELEMENTS

1. 元素有许多内部节点，在整个系统中，这些节点仅对一个元素有贡献。然而，这可以通过第 4.5 节的“浓缩”程序来克服。
2. 形状函数的表达式包含相对高阶项，而缺少一些低阶项。

## 数学代写|有限元方法作业代写FINITE DIFFERENCES METHOD代考|HIGHER-ORDER TRIANGULAR ELEMENTS

ñ一世=大号一世(2大号一世−1)

ñ4=4大号1大号2, 等等
b立方元素。

ñ一世=12(3大号一世−1)(3大号一世−2).

ñ4=92大号1大号2(3大号1−1), ñ5=92大号1大号2(3大号1−2), 等等

ñ10=27大号1大号2大号3.
C四次元。形状函数在练习 4.5 中获得。