# 数学代写|有限元方法代写Finite Element Method代考|Reduced Integration Element (RIE)

#### Doug I. Jones

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## 数学代写|有限元方法代写Finite Element Method代考|Reduced Integration Element (RIE)

When equal interpolation of $w_h^e(x)$ and $\phi_x^e(x)$ is used $(m=n)$, all sub-matrices in Eq. (5.3.15) are of the same order: $n \times n$, where $n$ is the number of terms in the polynomial (or $n-1$ is the degree of interpolation). The element coefficient matrices $K_{i j}^{11}, K_{i j}^{12}$ as well as the first part of $K_{i j}^{22}$ are evaluated exactly. The second part of $K_{i j}^{22}$ is to be evaluated using reduced integration. For the choice of linear interpolation functions, and for element-wise constant values of $G_e A_e K_s$ and $E_e I_e$, the matrices in Eq. (5.3.15) for this case have the following explicit values (when $k_f^e=0$ ):
$$\begin{gathered} \mathbf{K}^{11}=\frac{G_e A_e K_s}{h_e}\left[\begin{array}{rr} 1 & -1 \ -1 & 1 \end{array}\right], \quad \mathbf{K}^{12}=\frac{G_e A_e K_s}{2}\left[\begin{array}{rr} -1 & -1 \ 1 & 1 \end{array}\right] \ \mathbf{K}^{22}=\frac{E_e I_e}{h_e}\left[\begin{array}{rr} 1 & -1 \ -1 & 1 \end{array}\right]+\frac{G_e A_e K_s h_e}{4}\left[\begin{array}{lr} 1 & 1 \ 1 & 1 \end{array}\right] \end{gathered}$$
where one-point integration is used to evaluate the second part of $\mathbf{K}^{22}$. Note that $\mathbf{K}^{11}, \mathbf{K}^{12}$, and the first part of $\mathbf{K}^{22}$ can be evaluated exactly with onepoint quadrature (i.e., numerical integration) when $E_e I_e$ and $G_e A_e K_s$ are constant because the integrands of these coefficients are constant. Hence, one-point integration for $\mathbf{K}^{\alpha \beta}$ satisfies all requirements. The resulting beam element is termed the reduced integration element (RIE). Expressing Eq. (5.3.15), with $\mathbf{K}^{\alpha \beta}$ from Eq. (5.3.25a), and rearranging the nodal vector, we obtain
$$\frac{G_e A_e K_s}{4 h_e}\left[\begin{array}{cccc} 4 & -2 h_e & -4 & -2 h_e \ -2 h_e & h_e^2\left(1+4 \mu_e\right) & 2 h_e & h_e^2\left(1-4 \mu_e\right) \ -4 & 2 h_e & 4 & 2 h_e \ -2 h_e & h_e^2\left(1-4 \mu_e\right) & 2 h_e & h_e^2\left(1+4 \mu_e\right) \end{array}\right]\left{\begin{array}{c} w w_1^e \ S_1^e \ w_2^e \ s_2^e \end{array}\right}=\left{\begin{array}{c} q_1^e \ 0 \ q_2^e \ 0 \end{array}\right}+\left{\begin{array}{c} Q_1^e \ Q_2^{\prime} \ Q_3^{\prime} \ Q_4^{\prime} \end{array}\right}$$
where
$$\mu_e=E_e I_e / G_e A_e K_s h_e^2$$
To resemble Eq. (5.3.24), we can write the element equations of the RIE element as
$$\begin{gathered} \frac{E_e I_e}{6 \mu_e h_e^3}\left[\begin{array}{cccc} 6 & -3 h_e & -6 & -3 h_e \ -3 h_e & h_e^2\left(1.5+6 \mu_e\right) & 3 h_e & h_e^2\left(1.5-6 \mu_e\right) \ -6 & 3 h_e & 6 & 3 h_e \ -3 h_e & h_e^2\left(1.5-6 \mu_e\right) & 3 h_e & h_e^2\left(1.5+6 \mu_e\right) \end{array}\right]\left{\begin{array}{c} w_1^e \ s_l^e \ w_2 \ s_2^* \end{array}\right}=\left{\begin{array}{c} q_1^e \ 0 \ q_2^e \ 0 \end{array}\right}+\left{\begin{array}{c} Q_1^e \ Q_2^e \ Q_3^2 \ Q_4^3 \end{array}\right} \ q_i^e=\int_0^{h_e} \psi_i^e q_e d \bar{x},(i=1,2)\left(\psi_i^e\right. \text { are linear) } \ \mu_e=\frac{E_e I_e}{G_e A_e K_s h_e^2} \end{gathered}$$

## 数学代写|有限元方法代写Finite Element Method代考|Governing Equations

In this section, the finite element model of axisymmetric bending of circular plates using the classical plate theory (i.e., the theory in which transverse shear strain is assumed to be zero) is developed. We select the cylindrical coordinate system $(r, \theta, z)$ such that $r$ is the radial coordinate outward from the center of the plate $(0 \leq r \leq R), z$ denotes the transverse coordinate $(-H / 2 \leq$ $z \leq H / 2$ ), where $H$ is the total thickness of the plate, and $\theta$ is the angular coordinate ( $0 \leq \theta \leq 2 \pi)$, as shown in Fig. 5.4.1.

Governing equations of circular plates subjected to loads as well as boundary conditions that are independent of the angular coordinate $\theta$ can be formulated in terms of the radial coordinate $r$ alone. Thus, the axisymmetric bending of circular plates is a one-dimensional problem, as discussed next (see Fig. 5.4.2).

The total displacements $\left(u_r, u_\theta, u_z\right)$ along the three coordinate directions $(r, \theta, z)$, as implied by the Love-Kirchhoff hypothesis for plates, which is the same as the Euler-Bernoulli hypothesis for beams, are given by
$$u_r(r, z)=-z \frac{d w}{d r}, u_\theta(r, z)=0, u_z(r, z)=w(r)$$
where $w$ denotes the transverse displacement of a point on the midplane of the plate. The nonzero linear strain components referred to the cylindrical coordinate system are given by (see Reddy [1, 2] and Ugural [6])
$$\varepsilon_{r r}=\frac{d u_r}{d r}=-z \frac{d^2 w}{d r^2}, \varepsilon_{\theta \theta}=\frac{u_r}{r}=-\frac{z}{r} \frac{d w}{d r}$$

# 有限元方法代考

## 数学代写|有限元方法代写Finite Element Method代考|Reduced Integration Element (RIE)

$$\begin{gathered} \mathbf{K}^{11}=\frac{G_e A_e K_s}{h_e}\left[\begin{array}{rr} 1 & -1 \ -1 & 1 \end{array}\right], \quad \mathbf{K}^{12}=\frac{G_e A_e K_s}{2}\left[\begin{array}{rr} -1 & -1 \ 1 & 1 \end{array}\right] \ \mathbf{K}^{22}=\frac{E_e I_e}{h_e}\left[\begin{array}{rr} 1 & -1 \ -1 & 1 \end{array}\right]+\frac{G_e A_e K_s h_e}{4}\left[\begin{array}{lr} 1 & 1 \ 1 & 1 \end{array}\right] \end{gathered}$$

$$\frac{G_e A_e K_s}{4 h_e}\left[\begin{array}{cccc} 4 & -2 h_e & -4 & -2 h_e \ -2 h_e & h_e^2\left(1+4 \mu_e\right) & 2 h_e & h_e^2\left(1-4 \mu_e\right) \ -4 & 2 h_e & 4 & 2 h_e \ -2 h_e & h_e^2\left(1-4 \mu_e\right) & 2 h_e & h_e^2\left(1+4 \mu_e\right) \end{array}\right]\left{\begin{array}{c} w w_1^e \ S_1^e \ w_2^e \ s_2^e \end{array}\right}=\left{\begin{array}{c} q_1^e \ 0 \ q_2^e \ 0 \end{array}\right}+\left{\begin{array}{c} Q_1^e \ Q_2^{\prime} \ Q_3^{\prime} \ Q_4^{\prime} \end{array}\right}$$

$$\mu_e=E_e I_e / G_e A_e K_s h_e^2$$

$$\begin{gathered} \frac{E_e I_e}{6 \mu_e h_e^3}\left[\begin{array}{cccc} 6 & -3 h_e & -6 & -3 h_e \ -3 h_e & h_e^2\left(1.5+6 \mu_e\right) & 3 h_e & h_e^2\left(1.5-6 \mu_e\right) \ -6 & 3 h_e & 6 & 3 h_e \ -3 h_e & h_e^2\left(1.5-6 \mu_e\right) & 3 h_e & h_e^2\left(1.5+6 \mu_e\right) \end{array}\right]\left{\begin{array}{c} w_1^e \ s_l^e \ w_2 \ s_2^* \end{array}\right}=\left{\begin{array}{c} q_1^e \ 0 \ q_2^e \ 0 \end{array}\right}+\left{\begin{array}{c} Q_1^e \ Q_2^e \ Q_3^2 \ Q_4^3 \end{array}\right} \ q_i^e=\int_0^{h_e} \psi_i^e q_e d \bar{x},(i=1,2)\left(\psi_i^e\right. \text { are linear) } \ \mu_e=\frac{E_e I_e}{G_e A_e K_s h_e^2} \end{gathered}$$

## 有限元方法代写

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## MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。

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