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# 数学代写|有限元方法代写finite differences method代考|Discretization of the Domain

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## 数学代写|有限元代写Finite Element Method代考|Discretization of the Domain

The domain $\Omega=(0, L)$ of the straight beam shown in Fig. 5.2.3(a) is divided into a set of, say, $N$ line elements, a typical element being $\Omega^e=\left(x_a^e, x_b^e\right)$, as indicated in Fig. 5.2.3(b). Although the element is geometrically the same as that used for bars, the number and form of the primary and secondary unknowns at each end point, which constitutes a node, are dictated by the weak formulation of the differential equation, (5.2.10). We isolate a typical element $\Omega^e=\left(x_a^e, x_b^e\right)$ and construct the weak form of Eq. (5.2.10) over the element. The weak form provides the form of the primary and secondary variables of the problem. The primary variables are kinematic quantities that are required to be continuous throughout the domain, while the secondary variables are kinetic entities that are required to satisfy equilibrium conditions. When the secondary variables are physically not meaningful, the integration-by-parts step that yields them should not be carried out.
The weak forms of problems in solid mechanics can be developed either from the principle of virtual work (i.e., the principle of virtual displacements or virtual forces) or from the governing differential equations. Here we start with the given differential equation, Eq. (5.2.10), and using the three-step procedure obtain the weak form. We shall also consider the principle of virtual work in the sequel.
Suppose that $w_h^e$ is the finite element approximation of $w$ and let $v_i^e$ be a weight function over the element $\Omega^e=\left(x_a^e, x_b^e\right)$. Following the three-step procedure illustrated in Example 2.4.2, we write
\begin{aligned} 0= & \int_{x_a^e}^{x_b^e} v_i^e\left[\frac{d^2}{d x^2}\left(E_e I_e \frac{d^2 w_h^e}{d x^2}\right)+k_f^e w_h^e-q_e\right] d x \ = & \int_{x_a^e}^{x_b^e}\left[-\frac{d v_i^e}{d x} \frac{d}{d x}\left(E_e I_e \frac{d^2 w_h^e}{d x^2}\right)+k_f^e v_i^e w_h^e-v_i^e q_e\right] d x+\left[v_i^e \frac{d}{d x}\left(E_e I_e \frac{d^2 w_h^e}{d x^2}\right)\right]{x_a^e}^{x_b^e} \ = & \int{x_a^e}^{x_b^e}\left(E_e I_e \frac{d^2 v_i^e}{d x^2} \frac{d^2 w_h^e}{d x^2}+k_f^e v_i^e w_h^e-v_i^e q_e\right) d x \ & +\left[v_i^e \frac{d}{d x}\left(E_e I_e \frac{d^2 w_h^e}{d x^2}\right)-\frac{d v_i^e}{d x}\left(E_e I_e \frac{d^2 w_h^e}{d x^2}\right)\right]_{x_a^e}^{x_b^e} \end{aligned}
where $\left{v_i^e(x)\right}$ is a set of weight functions that are twice differentiable with respect to $x$. Note that, in the present case, the first term of the equation is integrated twice by parts to trade two differentiations to the weight function $v_i^e$, while retaining two derivatives of the dependent variable, $w_l^e$; that is, the differentiation is distributed equally between the weight function $v_i^e$ and the transverse deflection $w_h^e$. Because of the two integrations by parts, there appear two boundary expressions, which are to be evaluated at the two boundary points $x=x_a^e$ and $x=x_b^e$. Examination of the boundary terms indicates that the bending moment $M_h^e=-E_e I_e d^2 w_h^e / d x^2$ and shear force $V_h^e=-(d / d x)\left(E_e I_e d^2 w_h^e / d x^2\right)$ are the secondary variables and $\left(v_i^e \sim\right) w_h^e$ and slope $\left(d v_i^e / d x \sim\right) d w_h^e / d x$ are the primary variables. Thus, the weak form indicates that the boundary conditions for the EBT involve specifying one element of each of the following two pairs:
$$\left(w, V=\frac{d M}{d x}\right), \quad\left(\theta_x \equiv-\frac{d w}{d x}, M\right)$$
Mixed boundary conditions involve specifying a relationship between the variables of each pair:
Vertical spring: $V+k_s w=0 ; \quad$ Torsional spring: $M+\mu_s \theta_x=0$
where $k_{\mathrm{s}}$ and $\mu_{\mathrm{s}}$ are the stiffness coefficients of linear and torsional springs, respectively.

## 数学代写|有限元代写Finite Element Method代考|Approximation Functions

The weak form in Eq. (5.2.13) requires that the approximation $w_h^e(x)$ of $w(x)$ over a finite element should be such that it is twice-differentiable and satisfies the interpolation properties; that is, satisfies the following geometric “boundary conditions” of the element, as illustrated in Fig. 5.2.4:
$$w_h^e\left(x_a\right) \equiv \Delta_1^e, \quad w_h^e\left(x_b^e\right) \equiv \Delta_3^e, \quad \theta_x^e\left(x_a^e\right) \equiv \Delta_2^e, \quad \theta_x^e\left(x_b^e\right) \equiv \Delta_4^e$$
Note that $x_a^e$ and $x_b^e$ are the global coordinates of nodes 1 and 2, respectively. In satisfying the essential (or geometric) boundary conditions in Eq. (5.2.16), the approximation automatically satisfies the continuity conditions. Hence, we pay attention to the satisfaction of the conditions in Eq. (5.2.16), which forms the basis for the derivation of the interpolation functions of the EulerBernoulli beam element.
Since the approximation functions to be derived are valid over the element domain, it is simpler to derive them in terms of the local coordinate $\bar{x}$ with origin at node $1, \bar{x}=x-x_a^e$ Since there are a total of four conditions in an element (two per node), a four-parameter polynomial must be selected for $w_h^e$ :
$$w(\bar{x}) \approx w_h^e(\bar{x})=c_1^e+c_2^e \bar{x}+c_3^e \bar{x}^2+c_4^e \bar{x}^3$$
Note that the minimum continuity requirement (i.e., the existence of a nonzero second derivative of $w_h^e$ in the element) is automatically met. In addition, the cubic approximation of $w_h$ allows computation of the shear force, which involves the third derivative of $w_h^e$. Next, we express $c_i^e$ in terms of the primary nodal variables
$$\Delta_1^e=w_h^e(0), \quad \Delta_2^e=-\left.\frac{d w_h^e}{d \bar{x}}\right|{\bar{x}=0}, \Delta_3^e=w_h^e\left(h_e\right), \Delta_4^e=-\left.\frac{d w_h^e}{d \bar{x}}\right|{\bar{x}=h_e}$$
such that the conditions (5.2.16) are satisfied:
\begin{aligned} & \Delta_1^e=w_h^e(0) \quad=c_1^e \ & \Delta_2^e=-\left.\frac{d w_h}{d x}\right|{x=x_a}=-\left.\frac{d w_h^e}{d \bar{x}}\right|{\dot{x}=0}=-c_2^e \ & \Delta_3^e=w_h^e\left(h_e\right)=c_1^e+c_2^e h_e+c_3^e h_e^2+c_4^e h_e^3 \ & \Delta_4^e=-\left.\frac{d w_h^e}{d x}\right|{x=x_b}=-\left.\frac{d w_h^e}{d \bar{x}}\right|{\bar{x}=h_e}=-c_2^e-2 c_3^e h_e-3 c_4^e h_e^2 \end{aligned}

## 数学代写|有限元代写Finite Element Method代考|Discretization of the Domain

\begin{aligned} 0= & \int_{x_a^e}^{x_b^e} v_i^e\left[\frac{d^2}{d x^2}\left(E_e I_e \frac{d^2 w_h^e}{d x^2}\right)+k_f^e w_h^e-q_e\right] d x \ = & \int_{x_a^e}^{x_b^e}\left[-\frac{d v_i^e}{d x} \frac{d}{d x}\left(E_e I_e \frac{d^2 w_h^e}{d x^2}\right)+k_f^e v_i^e w_h^e-v_i^e q_e\right] d x+\left[v_i^e \frac{d}{d x}\left(E_e I_e \frac{d^2 w_h^e}{d x^2}\right)\right]{x_a^e}^{x_b^e} \ = & \int{x_a^e}^{x_b^e}\left(E_e I_e \frac{d^2 v_i^e}{d x^2} \frac{d^2 w_h^e}{d x^2}+k_f^e v_i^e w_h^e-v_i^e q_e\right) d x \ & +\left[v_i^e \frac{d}{d x}\left(E_e I_e \frac{d^2 w_h^e}{d x^2}\right)-\frac{d v_i^e}{d x}\left(E_e I_e \frac{d^2 w_h^e}{d x^2}\right)\right]{x_a^e}^{x_b^e} \end{aligned} 其中$\left{v_i^e(x)\right}$是一组对$x$二阶可微的权函数。请注意，在本例中，方程的第一项通过分部积分两次，以对权重函数$v_i^e$进行两次微分，同时保留因变量$w_l^e$的两次导数;即微分在权函数$v_i^e$和横向挠度$w_h^e$之间均匀分布。由于采用了两次分部积分法，出现了两个边界表达式，分别在两个边界点$x=x_a^e$和$x=x_b^e$处求值。对边界项的检验表明，弯矩$M_h^e=-E_e I_e d^2 w_h^e / d x^2$和剪力$V_h^e=-(d / d x)\left(E_e I_e d^2 w_h^e / d x^2\right)$是次变量，$\left(v_i^e \sim\right) w_h^e$和斜率$\left(d v_i^e / d x \sim\right) d w_h^e / d x$是主变量。因此，弱形式表明EBT的边界条件涉及指定以下两对中的每一对的一个元素: $$\left(w, V=\frac{d M}{d x}\right), \quad\left(\theta_x \equiv-\frac{d w}{d x}, M\right)$$ 混合边界条件包括指定每对变量之间的关系: 垂直弹簧:$V+k_s w=0 ; \quad$扭转弹簧:$M+\mu_s \theta_x=0$ 其中$k{\mathrm{s}}$和$\mu_{\mathrm{s}}$分别为线性弹簧和扭转弹簧的刚度系数。

## 数学代写|有限元代写Finite Element Method代考|Approximation Functions

(5.2.13)式中的弱形式要求$w(x)$在有限元上的近似$w_h^e(x)$是二次可微的，并满足插值性质;即满足单元的几何“边界条件”，如图5.2.4所示:
$$w_h^e\left(x_a\right) \equiv \Delta_1^e, \quad w_h^e\left(x_b^e\right) \equiv \Delta_3^e, \quad \theta_x^e\left(x_a^e\right) \equiv \Delta_2^e, \quad \theta_x^e\left(x_b^e\right) \equiv \Delta_4^e$$

$$w(\bar{x}) \approx w_h^e(\bar{x})=c_1^e+c_2^e \bar{x}+c_3^e \bar{x}^2+c_4^e \bar{x}^3$$

$$\Delta_1^e=w_h^e(0), \quad \Delta_2^e=-\left.\frac{d w_h^e}{d \bar{x}}\right|{\bar{x}=0}, \Delta_3^e=w_h^e\left(h_e\right), \Delta_4^e=-\left.\frac{d w_h^e}{d \bar{x}}\right|{\bar{x}=h_e}$$

\begin{aligned} & \Delta_1^e=w_h^e(0) \quad=c_1^e \ & \Delta_2^e=-\left.\frac{d w_h}{d x}\right|{x=x_a}=-\left.\frac{d w_h^e}{d \bar{x}}\right|{\dot{x}=0}=-c_2^e \ & \Delta_3^e=w_h^e\left(h_e\right)=c_1^e+c_2^e h_e+c_3^e h_e^2+c_4^e h_e^3 \ & \Delta_4^e=-\left.\frac{d w_h^e}{d x}\right|{x=x_b}=-\left.\frac{d w_h^e}{d \bar{x}}\right|{\bar{x}=h_e}=-c_2^e-2 c_3^e h_e-3 c_4^e h_e^2 \end{aligned}

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