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• Statistical Inference 统计推断
• Statistical Computing 统计计算
• Advanced Probability Theory 高等概率论
• Advanced Mathematical Statistics 高等数理统计学
• (Generalized) Linear Models 广义线性模型
• Statistical Machine Learning 统计机器学习
• Longitudinal Data Analysis 纵向数据分析
• Foundations of Data Science 数据科学基础

数学代写|有限元方法代写Finite Element Method代考|Stresses in a Timoshenko beam

As in the case of the Euler-Bernoulli beam, by using the Hooke’s law, the moment curvature relationship is expressed as follows:
\begin{aligned} &M=\int_{-h / 2}^{h / 2} \sigma_x y b d y=-E \frac{d \theta}{d x} \int_{-h / 2}^{h / 2} y^2 b d y \ &M=-E I \frac{d \theta}{d x} \end{aligned}
Also as in the case of the Euler-Bernoulli beam an internal shear stress develops in the beam due to transverse loading. The shear stress is expressed by using the Hooke’s law, as $\tau_{x y}=G \gamma_{x y}$, where $G$ is the shear modulus of the material. The shear force resultant $V$ is obtained by integrating the shear stress $\tau_{x y}$ through the thickness $(-h / 2 \leq z \leq h / 2)$ of the beam, as follows:
$$V=\int_{-h / 2}^{h / 2} \tau_{x y} b d z=\int_{-c / 2}^{c / 2}\left(G \gamma_{x y}\right) b d z=k\left(G A \gamma_{x y}\right)$$
where $A$ is the cross-sectional area of the beam (Fig. 2.16). Note that, in this derivation it is assumed that the shear strain $\gamma_{x y}$ is constant thought the beam thickness. The factor $k$ is the shear correction factor which compensates for assuming that $\gamma_{x y}$ does not vary in $y$-direction.

Using the geometric relation given in Eq. (2.131) the shear force resultant $V$ can be expressed as follows:
$$V=k G A\left(\frac{d v}{d x}-\theta\right)$$
Thus the number of unknowns for the Timoshenko beam theory can be reduced to two $(v, \theta)$. Once the shear force resultant is known the shear stress $\tau_{x y}$ can be found as follows:
$$\tau_{x y}=\frac{V}{A}=k G\left(\frac{d v}{d x}-\theta\right)$$
The stress-displacement relationships for the Timoshenko beam are summarized in Table 2.1.

数学代写|有限元方法代写Finite Element Method代考|Heat transfer equation in a one-dimensional solid

Consider a general one-dimensional problem where the temperature inside the body varies with time and there is a heat source/sink within a body.

Fig. 2.20B shows a small volume of such a body with width $d x$ in the direction of conduction. Heat flows in the direction normal to the area $A$, as shown in the figure. The body is considered infinitely wide in all other directions, so that the heat transfer may be considered one dimensional.

The energy balance requires that the rate of change of heat energy in this volume to be equal to the sum of the heat energy flowing across the boundary of the volume per unit time and the heat energy generated inside the volume per unit time. Mathematically, this is stated as follows:
$$q_x^h-q_{x+d x}^h+Q A d x=\rho c A d x \frac{\partial T}{\partial t}$$
The conducted components of the heat flow are expressed by using the Fourier’s law,
Energy conducted in left face: $\quad q_x^h=-k_x A \frac{\partial T}{\partial x}$
Energy conducted out right face: $\quad q_{x+d x}^h=-\left[k_x A \frac{\partial T}{\partial x}+\frac{\partial}{\partial x}\left(k_x A \frac{\partial T}{\partial x}\right) d x\right]$
The energy is generated within the solid due to a heat source $Q$ which generates energy per unit volume. The unit of $Q$ is $\mathrm{W} / \mathrm{m}^3$. The total energy generated in the given volume then becomes,
Energy generated within unit volume: $Q A d x$
$(2.169)$
The balance of the conducted and generated energy should be zero, when the system is at steady state. But, it is possible for the energy to have a transient condition before a steady state is established. The excess energy is stored inside the body, and it is proportional to the transient gradient of temperature,
Energy stored internally: $\rho c(A d x) \frac{\partial T}{\partial t}$

有限元方法代考

数学代写|有限元方法代写有限元法代考| Timoshenko梁中的应力

\begin{aligned} &M=\int_{-h / 2}^{h / 2} \sigma_x y b d y=-E \frac{d \theta}{d x} \int_{-h / 2}^{h / 2} y^2 b d y \ &M=-E I \frac{d \theta}{d x} \end{aligned}

$$V=\int_{-h / 2}^{h / 2} \tau_{x y} b d z=\int_{-c / 2}^{c / 2}\left(G \gamma_{x y}\right) b d z=k\left(G A \gamma_{x y}\right)$$
，其中$A$为梁的截面积(图2.16)。注意，在此推导中，假定剪应变$\gamma_{x y}$是梁厚不变的。因子$k$是剪切修正因子，它补偿了假设$\gamma_{x y}$在$y$方向上不变化

$$V=k G A\left(\frac{d v}{d x}-\theta\right)$$

$$\tau_{x y}=\frac{V}{A}=k G\left(\frac{d v}{d x}-\theta\right)$$
Timoshenko梁的应力-位移关系总结在表2.1

数学代写|有限元方法代写Finite – Element Method代考|一维固体中的传热方程

2.20B显示这种体在传导方向上的体积很小，宽度为$d x$。热流的方向与区域$A$垂直，如图所示。物体在所有其他方向上都被认为是无限宽的，因此传热可以被认为是一维的 能量平衡要求该体积内的热能变化率等于单位时间内流经体积边界的热能与单位时间内在体积内产生的热能之和。从数学上讲，这表述如下:
$$q_x^h-q_{x+d x}^h+Q A d x=\rho c A d x \frac{\partial T}{\partial t}$$

$(2.169)$

有限元方法代写

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