数学代写|CE595 Finite Element Method

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这是普渡大學有限元方法的代写成功案例。

CE595课程简介

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CE595 Finite Element Method HELP（EXAM HELP， ONLINE TUTOR）

Solution: First, we know that the significant load taken by a pile is vertical. So the axial deformation and stress will be the significant quantities, and the lateral deformation due to the Poisson effect may be neglected. Second, we wish to consider the static case. The governing equations of this simplified problem can be obtained using Newton’s second law and a uniaxial stressstrain (constitutive) relation for the material of the bar. The reader should note the similarity in the derivation of the equations of this solid mechanics problem with the heat flow problem discussed earlier.

Figure 1.2.3 shows an element of length $\Delta x$ with axial forces acting at both ends of the element, where $\sigma_x$ denotes stress i.e., force per unit area; $\mathrm{N} / \mathrm{m}^2$ in the $x$ direction, which is taken positive upward; $f(x)$ denotes the body force measured per unit length $(\mathrm{N} / \mathrm{m})$. Hence, $\left[A \sigma_x\right]x$ is the net tensile force on the volume element at $x$ and $\left[A \sigma_x\right]{x+\Delta x}$ is the net tensile force at $x+$ $\Delta x$. The resistance of the surrounding medium is $m_j u P$ per unit length in the direction opposite to the force $Q_0$, where $P$ is the perimeter. Then setting the sum of the forces to zero i.e., applying Newton’s second law in the $x$ direction yields
$$
-\left[A \sigma_x\right]x+\left[A \sigma_x\right]{x+\Delta x}+f \Delta x-m_f u P \Delta x=0
$$
Dividing throughout by $\Delta x$ and taking the limit $\Delta x \rightarrow 0$, we obtain
$$
\frac{d}{d x}\left(A \sigma_x\right)-c u+f(x)=0
$$
where $c=m_j P$.
The stress $\sigma_x$ can be related to the axial displacement using Hooke’s law and straindisplacement relation
$$
\sigma_x=E \varepsilon_x, \quad \varepsilon_x=\frac{d u}{d x}
$$
where $E$ is Young’s modulus $\left.\mathrm{N} / \mathrm{m}^2\right), u(x)$ denotes the axial displacement $(\mathrm{m})$ and $\varepsilon_x$ is the axial strain $(\mathrm{m} / \mathrm{m}$. Again, Eq. (1.2.25) represents a constitutive equation. Note that a system can have several constitutive relations, each depending on the phenomenon being studied. The study of heat transfer in a bar required us to employ Fourier’s law to relate temperature gradient to heat flux, and the study of deformation of the same bar subjected to axial forces requires us to use Hooke’s law to connect stress to displacement gradient.

Solution: We begin with the following general first-order differential equation [of which Eq. (1.2.4) is a special case]
$$
\frac{d u}{d t}=f(t, u), \quad 00$ subject to the initial (i.e., time $t=0$ ) condition $u(0)=u_0$. We approximate the derivative at time $t=t_i$ by
$$
\left.\left(\frac{d u}{d t}\right)\right|{t=t_i} \approx \frac{u\left(t{i+1}\right)-u\left(t_i\right)}{t_{i+1}-t_i}
$$
We note that the derivative at $t=t_i$ is replaced by its definition except that we did not take the limit $\Delta t \equiv t_{i+1}-t_i \rightarrow 0$; that is why it is an approximation. We also note that the slope of $u$ at $t_i$ in Eq. (1.3.2) is based on values of $u$ at $t_i$ and $t_{i+1}$. We can also take it to be the slope at $t=t_{i+1}$
$$
\left.\left(\frac{d u}{d t}\right)\right|{t=t{i+1}} \approx \frac{u\left(t_{i+1}\right)-u\left(t_i\right)}{t_{i+1}-t_i}
$$
The formula in Eq. (1.3.2) is called, for obvious reasons, forward difference, while that in Eq. (1.3.3) is called backward difference. The forward difference scheme is also known as Euler’s explicit scheme or first-order Runge-Kutta method. For increasingly small values of $\Delta t$, we hope that both approximations have a decreasingly small error in the computation of the slope. However, the two schemes will have different numerical convergence and stability behavior, as will be discussed in more detail in Chapter 7.
Substituting Eq. (1.3.2) into Eq. (1.3.1) at $t=t_i$, we obtain the forward difference formula
$$
u_{i+1}=u_i+\Delta t f\left(u_i, t_i\right), \quad u_i=u\left(t_i\right), \Delta t=t_{i+1}-t_i
$$
Equation (1.3.4) can be solved, starting from the known value $u_0$ of $u(t)$ at $t$ $=0$, for $u_1=u\left(t_1\right)=u(\Delta t)$. This process can be repeated to determine the values of $u$ at times $t=\Delta t, 2 \Delta t, \ldots, n \Delta t$. Of course, there are higher-order finite difference schemes that are more accurate than the Euler’s scheme and we will not discuss them as they are outside the scope of this study. Note that we are able to convert the ordinary differential equation, Eq. (1.3.1), to an algebraic equation, Eq. (1.3.4), which needs to be evaluated at different times to construct the time history of $u(t)$.

Solution: First we divide the domain $(0, L)$ into a finite set of $N$ intervals of equal length $\Delta x$, as shown in Fig. 1.3.2(b). Then we approximate the second derivative of $\theta=T-T_{\infty}$ directly using the centered difference scheme [error is of order $O(\Delta x)^2$ ] $\left(\frac{d^2 \theta}{d x^2}\right){x=x_i} \approx\left(\frac{\theta{i-1}-2 \theta_i+\theta_{i+1}}{(\Delta x)^2}\right)$

This approximation involves three points, called mesh points, at which function evaluations are required. The mesh points are separated by a distance of $\Delta x$. Using the above approximation in Eq. (1.2.19), we obtain
$$
-\left(\theta_{i-1}-2 \theta_i+\theta_{i+1}\right)+(m \Delta x)^2 \theta_i=0 \text { or }-\theta_{i-1}+\left[2+(m \Delta x)^2\right] \theta_i-\theta_{i+1}=0
$$
Equation (1.3.8) is valid for any mesh point $x=x_i, i=1,2, \ldots, N$, at which the solution is not known. The formula contains values of $\theta$ at three mesh points $x=x_{i-1}, x_i$, and $x=x_{i+1}$ at a time. Note that Eq. (1.3.8) is not used at mesh point $x=x_0=0$ because the temperature is known there, as given in Eq. (1.2.20). However, use of Eq. (1.3.8) at mesh point $x=x_N=L$ requires the knowledge of the fictitious value $\theta_{N+1}$ (as we see in the sequel, we never have to deal with such fictitious values in the finite element method). The forward finite difference approximation of the second boundary condition in Eq. (1.2.20) at mesh point $x_N=L$ can be used to determine $\theta_{N+1}$ :
$$
\frac{\theta_{N+1}-\theta_N}{\Delta x}+\frac{\beta}{k} \theta_N=0 \Rightarrow \theta_{N+1}=\left(1-\frac{\beta \Delta x}{k}\right) \theta_N
$$
Application of the formula in Eq. (1.3.7) to mesh points at $x_1, x_2, \ldots, x_N$ yields
$$
\begin{array}{r}
-\theta_0+D \theta_1-\theta_2=0 \
-\theta_1+D \theta_2-\theta_3=0 \
-\theta_2+D \theta_3-\theta_4=0 \
\cdots \quad \cdots \quad \cdots \
-\theta_{N-1}+D \theta_N-\theta_{N+1}=0
\end{array}
$$
where $D=\left[2+(m \Delta x)^2\right]$. Equation (1.3.9) can be used to eliminate $\theta_{N+1}$ from the last equation in (1.3.10). Then Eq. (1.3.10) consists of $N$ equations in $N$ unknowns, $\theta_1, \theta_2, \ldots, \theta_N$.

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