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: The three basic features of the finite element method in the present case take the following form. First, the division of the perimeter of a circle into a collection of line segments. In theory, we need infinite number of such line elements to represent the perimeter; otherwise, the value computed will have some error. Second, writing an equation for the quantity of interest (perimeter) over an element (line segment) in this case is exact, because the approximated perimeter is a straight line. Third, the assembly of elements amounts to simply adding the element lengths to obtain the total value. Although this is a trivial example, it illustrates several (but not all) ideas and steps involved in the finite element analysis of a problem. We outline the steps involved in computing an approximate value of the perimeter of the circle. In doing so, we introduce certain terms that are used in the finite element analysis of any problem.

Finite element discretization. First, the domain (i.e., the perimeter of the circle) is represented as a collection of a finite number $n$ of subdomains, namely, line segments, as shown in Fig. 1.4.3(b). This is called discretization of the domain. Each subdomain (i.e., line segment) is called an element. The collection of elements is called the finite element mesh. The elements are connected to each other at points called nodes. In the present case, we discretize the perimeter into a mesh of five $(n=5)$ line segments, as shown in Fig. 1.4.3(c).

The line segments can be of different lengths. When all elements are of the same length, the mesh is said to be uniform; otherwise, it is called a nonuniform mesl.

Element equations. A typical element (i.e., line segment, $\Omega_e$ ) is isolated and its required properties, i.e., length, are computed by some appropriate means. Let $h_e$ be the length of element $\Omega_e$ in the mesh. For a typical element $\Omega_e, h_c$ is given, as can be seen from $\Gamma$ ig. 1.4.3(d), by
$$
h_e=2 R \sin \frac{1}{2} \theta_e
$$
where $R$ is the radius of the circle and $\theta_e<u$ is the angle subtended by the line segment. The above equations are called element equations. Ancient mathematicians most likely made measurements, rather than using Eq. (1.4.1), to find $h_e$ (they did not know $\pi$ ).

Assembly of element equations and solution. The approximate value of the perimeter of the circle is obtained by putting together the element properties in a meaning[ul way; this process is called the assembly of the element equations. It is based, in the present case, on the simple idea that the total perimeter of the polygon $\Omega_h$ (assembly of elements) is equal to the sum of the lengths of individual elements:
$$
P_n=\sum_{e=1}^n h_e
$$
Then $P_n$ represents an approximation to the actual perimeter, $P$. If the mesh is uniform, or $h_e$ is the same for each of the elements in the mesh, then $\theta_e=2 \pi / n$, and we have
$$
P_n=n\left(2 R \sin \frac{\pi}{n}\right)
$$

Convergence and error estimate. For this simple problem, we know the exact solution: $P=2 \pi R$. We can estimate the error in the approximation and show that the approximate solution $P_n$ converges to the exact value $P$ in the limil as $n \rightarrow \infty$. Consider a typical element $\Omega_e$. The error in the approximation is equal to the difference between the length of the sector and that of the line segment [see Fig. 1.4.3(d)]
$$
E_e=\left|S_e-h_e\right|
$$
where $S_e=R \theta_e$ is the length of the sector. Thus, the error estimate for an element in the mesh is given by
$$
E_e=R\left(\frac{2 \pi}{n}-2 \sin \frac{\pi}{n}\right)
$$
The total error (called global error) is given by multiplying $E_e$ by $n$ :
$$
E=n E_e=2 R\left(\pi-n \sin \frac{\pi}{n}\right)=2 \pi R-P_n=P-P_n
$$
We now show that $E$ goes to zero as $n \rightarrow \infty$. Letting $x=1 / n$, we have
$$
P_n=2 R n \sin \frac{\pi}{n}=2 R \frac{\sin \pi x}{x}
$$
and
$$
\lim {n \rightarrow \infty} P_n=\lim {x \rightarrow 0}\left(2 R \frac{\sin \pi x}{x}\right)=\lim _{x \rightarrow 0}\left(2 \pi R \frac{\cos \pi x}{1}\right)=2 \pi R
$$
Hence, $E_n$ goes to zero as $n \rightarrow \infty$. This completes the proof of convergence.

Solution: The minors are
$$
\begin{array}{lll}
M_{11}(\mathbf{A})=\left|\begin{array}{rr}
4 & 3 \
-3 & 5
\end{array}\right|, & M_{12}(\mathbf{A})=\left|\begin{array}{rr}
1 & 3 \
2 & 5
\end{array}\right|, & M_{13}(\mathbf{A})=\left|\begin{array}{rr}
1 & 4 \
2 & -3
\end{array}\right| \
M_{21}(\mathbf{A})=\left|\begin{array}{rr}
5 & -1 \
-3 & 5
\end{array}\right|, & M_{22}(\mathbf{A})=\left|\begin{array}{rr}
2 & -1 \
2 & 5
\end{array}\right|, & M_{23}(\mathbf{A})=\left|\begin{array}{rr}
2 & 5 \
2 & -3
\end{array}\right| \
M_{31}(\mathbf{A})=\left|\begin{array}{rr}
5 & -1 \
4 & 3
\end{array}\right|, & M_{32}(\mathbf{A})=\left|\begin{array}{rr}
2 & -1 \
1 & 3
\end{array}\right|, & M_{33}(\mathbf{A})=\left|\begin{array}{ll}
2 & 5 \
1 & 4
\end{array}\right|
\end{array}
$$
To compute the determinant, we use the definition in Eq. (2.2.48) and expand by the first column and obtain
$$
\begin{aligned}
|A| & =\sum_{i=1}^3(-1)^{i+1} a_{i 1}\left|A_{i 1}\right|=a_{11} M_{11}(\mathbf{A})-a_{21} M_{21}(\mathbf{A})+a_{31} M_{31}(\mathbf{A}) \
& =a_{11}\left|\begin{array}{rr}
4 & 3 \
-3 & 5
\end{array}\right|-a_{21}\left|\begin{array}{rr}
5 & -1 \
-3 & 5
\end{array}\right|+a_{31}\left|\begin{array}{rr}
5 & -1 \
4 & 3
\end{array}\right| \
& =2[(4)(5)-(3)(-3)]+(-1)[(5)(5)-(-1)(-3)]+2[(5)(3)-(-1)(4)] \
& =2(20+9)-(25-3)+2(15+4)=74
\end{aligned}
$$
We note that the determinant also can be computed by expanding by the first row:
$$
\begin{gathered}
|A|=a_{11} M_{11}(\mathbf{A})-a_{12} M_{12}(\mathbf{A})+a_{13} M_{13}(\mathbf{A})=2 \times 29-5 \times(-1)+(-1) \times \
(-11)=74
\end{gathered}
$$
To compute the inverse of the given matrix we first compute the cofactors:
$$
\begin{aligned}
& \operatorname{cof}{11}(\mathbf{A})=(-1)^2 M{11}(\mathbf{A})=4 \times 5-(-3) 3=29 \
& \operatorname{cof}{12}(\mathbf{A})=(-1)^3 M{12}(\mathbf{A})=-(1 \times 5-3 \times 2)=1 \
& \operatorname{cof}{13}(\mathbf{A})=(-1)^4 M{13}(\mathbf{A})=1 \times(-3)-2 \times 4=-11 \
& \operatorname{cof}{21}(\mathbf{A})=(-1)^3 M{21}(\mathbf{A})=-[5 \times 5-(-3)(-1)]=-22 \
& \operatorname{cof}{22}(\mathbf{A})=(-1)^4 M{22}(\mathbf{A})=[2 \times 5-2 \times(-1)]=12 \
& \operatorname{cof}{23}(\mathbf{A})=(-1)^5 M{23}(\mathbf{A})=-[2 \times(-3)-2 \times 5]=16
\end{aligned}
$$
and so on. Then the Adj (A) is given by cof11 (A) cof12 (A) cof13 (A)
$$
\operatorname{Adj}(\mathbf{A})=\left[\begin{array}{lll}
\operatorname{cof}{11}(\mathbf{A}) & \operatorname{cof}{12}(\mathbf{A}) & \operatorname{cof}{13}(\mathbf{A}) \ \operatorname{cof}{21}(\mathbf{A}) & \operatorname{cof}{22}(\mathbf{A}) & \operatorname{cof}{23}(\mathbf{A}) \
\operatorname{cof}{31}(\mathbf{A}) & \operatorname{cof}{32}(\mathbf{A}) & \operatorname{cof}_{33}(\mathbf{A})
\end{array}\right]^{\mathrm{T}}=\left[\begin{array}{rrr}
29 & -22 & 19 \
1 & 12 & -7 \
-11 & 16 & 3
\end{array}\right]
$$
The inverse of $\mathbf{A}$ can be now computed using Eq. (2.2.51):
$$
\mathbf{A}^{-1}=\frac{1}{74}\left[\begin{array}{rrr}
29 & -22 & 19 \
1 & 12 & -7 \
-11 & 16 & 3
\end{array}\right]
$$
It can be easily verified that $\mathbf{A A}^{-1}=\mathbf{A}^{-1} \mathbf{A}=\mathbf{I}$.

Solution: The only nonzero linear strain is
$$
\varepsilon_{x x}=\frac{d u}{d x}-z \frac{d^2 w}{d x^2}
$$
The strain energy density is
$$
\begin{aligned}
U_0 & =\int_0^{\varepsilon_{x x}} \sigma_{x x} d \varepsilon_{x x}=\int_0^{\varepsilon_{x x}} E \varepsilon_{x x} d \varepsilon_{x x} \
& =\frac{E}{2} \varepsilon_{x x}^2=\frac{E}{2}\left(\frac{d u}{d x}-z \frac{d^2 w}{d x^2}\right)^2
\end{aligned}
$$
The strain energy $U$ is given by
$$
\begin{aligned}
U & =\int_{\Omega} U_0 d \Omega=\int_0^L \int_A \frac{E}{2}\left(\frac{d u}{d x}-z \frac{d^2 w}{d x^2}\right)^2 d A d x \
& =\frac{1}{2} \int_0^L\left[A_{x x}\left(\frac{d u}{d x}\right)^2+D_{x x}\left(\frac{d^2 w}{d x^2}\right)^2\right] d x
\end{aligned}
$$
where $A_{x x}$ and $D_{x x}$ are defined as (for the case in which $E$ is at most a function of $x$ )
$$
A_{x x}=\int_A E d A=E A, \quad D_{x x}=\int_A z^2 E d A=E I
$$
we have
$$
U=\frac{1}{2} \int_0^L\left[E A\left(\frac{d u}{d x}\right)^2+E I\left(\frac{d^2 w}{d x^2}\right)^2\right] d x
$$
where we have used the fact (because $x$-axis is taken along the geometric centroid)
$$
\int_A z d A=0
$$

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