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# 数学代写|有限元代写Finite Element Method代考|EG55M1 Shape Function Construction

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

Consider a structure consisting of a number of trusses or bar members. Each of the members can be considered as a truss/bar element of uniform cross-section bounded by two nodes $\left(n_d=2\right)$. Consider a bar element with nodes 1 and 2 at each end of the element, as shown in Figure 4.1. The length of the element is $l_e$. The local $x$-axis is taken in the axial direction of the element with the origin at node 1 . In the local coordinate system, there is only one DOF at each node of the element, and that is the axial displacement. Therefore, there is a total of two DOFs for the element, i.e. $n_e=2$. In the FEM discussed in the previous chapter, the displacement in an element should be written in the form
$$u^h(x)=\mathbf{N}(x) \mathbf{d}_e$$
where $u^h$ is the approximation of the axial displacement within the element, $\mathbf{N}$ is a matrix of shape functions that possess the properties described in Chapter 3 , and $\mathbf{d}_e$ should be the vector of the displacements at the two nodes of the element:
$$\mathbf{d}_e=\left{\begin{array}{l} u_1 \ u_2 \end{array}\right}$$
The question now is, how can we determine the shape functions for the truss elements?
We follow the standard procedure described in Section 3.4.3 for constructing shape functions, and assume that the axial displacement in the truss element can be given in a general form
where $u^h$ is the approximation of the displacement, $\alpha$ is the vector of two unknown constants, $\alpha_0$ and $\alpha_1$, and $\mathbf{p}$ is the vector of polynomial basis functions (or monomials). For this particular problem, we use up to the first order of polynomial basis. Depending upon the problem, we could use a higher order of polynomial basis functions. The order of polynomial

## 数学代写|有限元代写Finite Element Method代考|Strain Matrix

As discussed in Chapter 2 , there is only one stress component $\sigma_x$ in a truss, and the corresponding strain can be obtained by
$$\varepsilon_x=\frac{\partial u}{\partial x}=\frac{u_2-u_1}{l_e}$$
which is a direct result from differentiating Eq. (4.11) with respect to $x$. Note that the strain in Eq. (4.12) is a constant value in the element.

It was mentioned in the previous chapter that we would need to obtain the strain matrix, B, after which we can obtain the stiffness and mass matrices. In this case, this can be easily done. Equation (4.12) can be re-written in a matrix form as
$$\varepsilon_x=\frac{\partial u}{\partial x}=L \mathbf{N} \mathbf{d}_e=\mathbf{B d}_e$$
where the strain matrix $\mathbf{B}$ has the following form for the truss element:
$$\mathbf{B}=L \mathbf{N}=\frac{\partial}{\partial x}\left[\begin{array}{ll} 1-x / l_e & x / l_e \end{array}\right]=\left[\begin{array}{cc} -1 / l_e & 1 / l_e \end{array}\right]$$

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

。在局部坐标系中，单元的每个节点只有一个自由度，即轴向位移。因此，该元青共有两个自由度，即 $n_e=2$. 在上一章讨论的 FEM 中，单元中的位移应写为
$$u^h(x)=\mathbf{N}(x) \mathbf{d}_e$$

〈left 的分隔符缺失或无法识别

## 数学代写|有限元代写Finite Element Method代考|Strain Matrix

$$\varepsilon_x=\frac{\partial u}{\partial x}=\frac{u_2-u_1}{l_e}$$

$$\varepsilon_x=\frac{\partial u}{\partial x}=L \mathbf{N} \mathbf{d}_e=\mathbf{B} \mathbf{d}_e$$

$$\mathbf{B}=L \mathbf{N}=\frac{\partial}{\partial x}\left[\begin{array}{ll} 1-x / l_e & x / l_e \end{array}\right]=\left[\begin{array}{ll} -1 / l_e & 1 / l_e \end{array}\right]$$

## MATLAB代写

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