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# 数学代写|有限元方法代写finite differences method代考|Timoshenko Frame Element Based on RIE

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## 数学代写|有限元代写Finite Element Method代考|Timoshenko Frame Element Based on RIE

Since the element stiffness matrix of the RIE is the same as the CIE, the element stiffness matrix in the global coordinates as given in Eq. (6.3.18) is valid for the frame element based on the RIE. The contributions of the axial and transverse distributed forces to the nodes are given by
$$\bar{f}_i^e=\int_0^{h_e} f(\bar{x}) \psi_i^e(\bar{x}) d \bar{x}, \quad \bar{q}_i^e=\int_0^{h_e} q(\bar{x}) \psi_i^e(\bar{x}) d \bar{x}$$
where $\psi_i^e$ are the linear interpolation functions. Then the element force vector for the RIE frame element is given by
$$\mathbf{F}^e=\left{\begin{array}{c} \bar{F}_1^c \cos \alpha_c-\bar{F}_2^c \sin \alpha_c \ \bar{F}_1^e \sin \alpha_c+\bar{F}_2^e \cos \alpha_c \ \bar{F}_3^c \ \bar{F}_4^e \cos \alpha_c-\bar{F}_5^e \sin \alpha_c \ \bar{F}_4^e \sin \alpha_e+\bar{F}_5^e \cos \alpha_c \ \bar{F}_6^c \end{array}\right}, \quad\left{\begin{array}{c} \bar{F}_1^c \ \bar{F}_2^c \ \bar{F}_3^c \ \bar{F}_4^e \ \bar{F}_5^e \ \bar{F}_6^e \end{array}\right}=\left{\begin{array}{c} \bar{f}_1^e \ \bar{q}_1^c \ 0 \ \bar{f}_2^e \ \bar{q}_2^e \ 0 \end{array}\right}+\left{\begin{array}{c} \bar{Q}_1^c \ \bar{Q}_2^e \ \bar{Q}_3^e \ \bar{Q}_4^e \ \bar{Q}_5^e \ \bar{Q}_6^e \end{array}\right}$$

## 数学代写|有限元代写Finite Element Method代考|Lagrange Multiplier Method

In the Lagrange multiplier method the problem is reformulated as one of
determining the stationary (or critical) points of the modified function $F_L$ ( $x$, $y)$
$$F_L(x, y)=f(x, y)+\lambda G(x, y)$$
subject to no constraints. Here $\lambda$ (a number) denotes the Lagrange multiplier. The solution to the problem is obtained by setting partial derivatives of $F_L$ with respect to $x, y$, and $\lambda$ to zero:
$$\begin{array}{r} \frac{\partial F_L}{\partial x}=\frac{\partial f}{\partial x}+\lambda \frac{\partial G}{\partial x}=0 \ \frac{\partial F_L}{\partial y}=\frac{\partial f}{\partial y}+\lambda \frac{\partial G}{\partial y}=0 \ \frac{\partial F_L}{\partial \lambda}=G(x, y)=0 \end{array}$$
Thus, there are three equations in three unknowns $(x, y, \lambda)$. In the Lagrange multiplier method a new variable, Lagrange multiplier, is introduced with each constraint equation.

## 数学代写|有限元代写Finite Element Method代考|Timoshenko Frame Element Based on RIE

$$\bar{f}_i^e=\int_0^{h_e} f(\bar{x}) \psi_i^e(\bar{x}) d \bar{x}, \quad \bar{q}_i^e=\int_0^{h_e} q(\bar{x}) \psi_i^e(\bar{x}) d \bar{x}$$

$$\mathbf{F}^e=\left{\begin{array}{c} \bar{F}_1^c \cos \alpha_c-\bar{F}_2^c \sin \alpha_c \ \bar{F}_1^e \sin \alpha_c+\bar{F}_2^e \cos \alpha_c \ \bar{F}_3^c \ \bar{F}_4^e \cos \alpha_c-\bar{F}_5^e \sin \alpha_c \ \bar{F}_4^e \sin \alpha_e+\bar{F}_5^e \cos \alpha_c \ \bar{F}_6^c \end{array}\right}, \quad\left{\begin{array}{c} \bar{F}_1^c \ \bar{F}_2^c \ \bar{F}_3^c \ \bar{F}_4^e \ \bar{F}_5^e \ \bar{F}_6^e \end{array}\right}=\left{\begin{array}{c} \bar{f}_1^e \ \bar{q}_1^c \ 0 \ \bar{f}_2^e \ \bar{q}_2^e \ 0 \end{array}\right}+\left{\begin{array}{c} \bar{Q}_1^c \ \bar{Q}_2^e \ \bar{Q}_3^e \ \bar{Q}_4^e \ \bar{Q}_5^e \ \bar{Q}_6^e \end{array}\right}$$

## 数学代写|有限元代写Finite Element Method代考|Lagrange Multiplier Method

$$F_L(x, y)=f(x, y)+\lambda G(x, y)$$

$$\begin{array}{r} \frac{\partial F_L}{\partial x}=\frac{\partial f}{\partial x}+\lambda \frac{\partial G}{\partial x}=0 \ \frac{\partial F_L}{\partial y}=\frac{\partial f}{\partial y}+\lambda \frac{\partial G}{\partial y}=0 \ \frac{\partial F_L}{\partial \lambda}=G(x, y)=0 \end{array}$$

## MATLAB代写

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