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# 数学代写|有限元方法代写finite differences method代考|ENGR7961 Field Variable Interpolation

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## 数学代写|有限元代写Finite Element Method代考|Field Variable Interpolation

Consider now a triangular element of thickness $h$. The nodes of the element are numbered 1 , 2 and 3 counter-clockwise, as shown in Figure 7.4. For 2D solid elements, the field variable is the displacement, which has two components $(u$ and $v)$, and hence each node has two Degrees Of Freedom (DOFs). Since a linear triangular element has three nodes, the total number of DOFs of a linear triangular element is six. For the triangular element, the local coordinate of each element can be taken as the same as the global coordinate, since there are no advantages in specifying a different local coordinate system for each element.

Now, let us examine how a triangular element can be formulated. The displacement $\mathbf{U}$ is generally a function of the coordinates $x$ and $y$, and we express the displacement at any point in the element using the displacements at the nodes and shape functions. It is therefore assumed that (see Section 3.4.2)
$$\mathbf{U}^h(x, y)=\mathbf{N}(x, y) \mathbf{d}_e$$
where the superscript $h$ indicates that the displacement is approximated, and $\mathbf{d}_e$ is a vector of the nodal displacements arranged in the order of
$$\mathbf{d}_e=\left{\begin{array}{l} u_1 \ v_1 \ u_2 \ v_2 \ u_3 \ v_3 \end{array}\right} \text { displacements at node } 1$$

and the matrix of shape functions $\mathbf{N}$ is arranged as
in which $N_i(i=1,2,3)$ are three shape functions corresponding to the three nodes of the triangular element. Equation (7.1) can be explicitly expressed as
\begin{aligned} & u^h(x, y)=N_1(x, y) u_1+N_2(x, y) u_2+N_3(x, y) u_3 \ & v^h(x, y)=N_1(x, y) v_1+N_2(x, y) v_2+N_3(x, y) v_3 \end{aligned}
which implies that each of the displacement components at any point in the element is approximated by an interpolation from the nodal displacements using the shape functions. This is because the two displacement components are basically independent from each other. The question now is how can we construct these shape functions for our triangular element that satisfies the sufficient requirements: delta function property; partitions of unity; and linear field reproduction.

## 数学代写|有限元代写Finite Element Method代考|Shape Function Construction

Development of the shape functions is normally the first, and most important, step in developing finite element equations for any type of element. In determining the shape functions $N_i$ ( $\left.i=1,2,3\right)$ for the triangular element, we can of course follow exactly the standard procedure described in Sections 3.4 .3 and 4.2.1, by starting with an assumption of the displacements using polynomial basis functions with unknown constants. These unknown constants are then determined using the nodal displacements at the nodes of the element. This standard procedure works in principle for the development of any type of element, but may not be the most convenient method. We demonstrate here another slightly different approach for constructing shape functions. We start with an assumption of shape functions directly using polynomial basis functions with unknown constants. These unknown constants are then determined using the property of the shape functions. The only difference here is that we assume directly the shape function instead of the displacements. For a linear triangular element, we assume that the shape functions are linear functions of $x$ and $y$. They should, therefore, have the form of
\begin{aligned} & N_1=a_1+b_1 x+c_1 y \ & N_2=a_2+b_2 x+c_2 y \ & N_3=a_3+b_3 x+c_3 y \end{aligned}
where $a_i, b_i$ and $c_i(i=1,2,3)$ are constants to be determined. Equation (7.5) can be written in a concise form,
$$N_i=a_i+b_i x+c_i y, \quad i=1,2,3$$

## 数学代写|有限元代写Finite Element Method代考|Field Variable Interpolation

$$\mathbf{U}^h(x, y)=\mathbf{N}(x, y) \mathbf{d}_e$$

\left 缺少或无法识别的分隔符

$$u^h(x, y)=N_1(x, y) u_1+N_2(x, y) u_2+N_3(x, y) u_3 \quad v^h(x, y)=N_1(x, y) v_1+N_2(x, y) v_2+N_3(x, y) v_3$$

## 数学代写|有限元代写Finite Element Method代考|Shape Function Construction

$$N_1=a_1+b_1 x+c_1 y \quad N_2=a_2+b_2 x+c_2 y N_3=a_3+b_3 x+c_3 y$$

$$N_i=a_i+b_i x+c_i y, \quad i=1,2,3$$

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