### 数学代写|计算方法代写computational method代考|Post-solution operations

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## 数学代写|计算方法代写computational method代考|Post-solution operations

Following assembly of the coefficient matrix and enforcement of the essential boundary conditions (when applicable) the resulting system of simultaneous equations is solved by one of several methods designed to exploit the symmetry and sparsity of the coefficient matrix. The solvers are classified into two broad categories; direct and iterative solvers. Optimal choice of a solver in a particular application is based on consideration of the size of the problem and the available computational resources.
At the end of the solution process the finite element solution is available in the form
$$u_{F E}=\sum_{j=1}^{N_{u}} a_{j} \varphi_{j}(x)$$
where the indices reference the global numbering and $N_{u}$ is the number of degrees of freedom plus the number of Dirichlet conditions.

The basis functions are decomposed into their constituent shape functions and the element-level solution records are created in the local numbering convention. Therefore the finite element solution on the $k$ th element is available in the following form:
$$u_{F E}^{(k)}=\sum_{j=1}^{p_{k}+1} a_{j}^{(k)} N_{j}(\xi)$$

## 数学代写|计算方法代写computational method代考|Computation of the quantities of interest

The computation of typical engineering quantities of interest (QoI) by direct and indirect methods is outlined in this section.
Computation of $u_{F E}\left(x_{6}\right)$
Direct computation of $u_{F E}$ in the point $x=x_{0}$ involves a search to identify the element $I_{k}$ in which point $x_{0}$ lies and, using the inverse of the mapping function defined by eq. (1.60), the standard

coordinate $\xi_{0} \in I_{\mathrm{st}}$ corresponding to $x_{0}$ is determined:
$$\xi_{0}=Q_{k}^{-1}\left(x_{0}\right)=\frac{2 x_{0}-x_{k}-x_{k+1}}{x_{k+1}-x_{k}}$$
and $u_{F E}\left(x_{0}\right)$ is computed from
$$u_{F E}\left(x_{0}\right)=\sum_{j=1}^{p_{k}+1} a_{j}^{(k)} N_{j}\left(\xi_{0}\right)$$
Direct computation of $u_{F E}^{\prime}\left(\boldsymbol{x}{\theta}\right)$ Direct computation of $u{F E}^{\prime}$ in the point $x_{0}$ involves the computation of the corresponding standard coordinate $\xi_{0} \in I_{\mathrm{st}}$ using eq. (1.83) and evaluating the following expression:
$$\left(\frac{d u_{F E}}{d x}\right){x=x{0}}=\frac{2}{\ell_{k}}\left(\frac{d u_{F E}}{d \xi}\right){\xi=\xi{0}}=\frac{2}{\ell_{k}} \sum_{j=1}^{P_{k}+1} a_{j}^{(k)}\left(\frac{d N_{j}}{d \xi}\right){\xi=\xi{0}}$$
where $\ell_{k} \stackrel{\text { def }}{=} x_{k+1}-x_{k}$. The computation of the higher derivatives is analogous.
Remark $1.8$ When plotting quantities of interest such as the functions $u_{F E}(x)$ and $u_{F E}^{\prime}(x)$, the data for the plotting routine are generated by subdividing the standard element into $n$ intervals of equal length, $n$ being the desired resolution. The QoIs corresponding to the grid-points are evaluated. This process does not involve inverse mapping. In node points information is provided from the two elements that share that node. If the computed QoI is discontinuous then the discontinuity will be visible at the nodes unless the plotting algorithm automatically averages the QoIs.
Indirect computation of $u_{F E}^{\prime}\left(x_{0}\right)$ in node points
The first derivative in node points can be determined indirectly from the generalized formulation. For example, to compute the first derivative at node $x_{k}$ from the finite element solution, we select $v=N_{1}\left(Q_{k}^{-1}(x)\right)$ and use
$$\int_{x_{k}}^{x_{k+1}}\left(\kappa u_{F E}^{\prime} v^{\prime}+c u_{F E} v\right) d x=\int_{x_{k}}^{x_{k+1}} f v d x+\left[\kappa u_{F E}^{\prime} v\right]{x=x{k+1}}-\left[\kappa u_{F E}^{\prime} v\right]{x=x{k}} .$$
Test functions used in post-solution operations for the computation of a functional are called extraction functions. Here $v=N_{1}\left(Q_{k}^{-1}(x)\right)$ is an extraction function for the functional $-\left[\kappa u_{F E}^{\prime}\right]{x=x{k}}$. This is because $v\left(x_{k}\right)=1$ and $v\left(x_{k+1}\right)=0$ and hence
\begin{aligned} -\left[\kappa u_{F E}^{\prime}\right]{x=x{k}} &=\int_{x_{k}}^{x_{k+1}}\left(\kappa u_{F E}^{\prime} v^{\prime}+c u_{F E} v\right) d x-\int_{x_{k}}^{x_{k+1}} f v d x \ &=\sum_{j=1}^{p_{k}+1} c_{1 j}^{(k)} a_{j}^{(k)}-r_{1}^{(k)} \end{aligned}
where, by definition; $c_{i j}^{(k)}=k_{i j}^{(k)}+m_{i j}^{(k)}$.

## 数学代写|计算方法代写computational method代考|Nodal forces

which is the exact solution. The choice $v=1-x$ was exceptionally fortuitous because it happens to be the Green’s function (also known as the influence function) for $u^{\prime}(0)$. Therefore the extracted value is independent of the solution $u \in E^{0}(I)$.
Let us choose $v=1-x^{2}$ for the extraction function. In this case
$$u^{\prime}(0)=v(\bar{x})-\int_{0}^{1} u^{\prime} v^{\prime} d x=\frac{15}{16}+2 \int_{0}^{1} u^{\prime} x d x$$
Substituting $u_{F E}^{\prime}$ for $u^{\prime}$ :
\begin{aligned} \int_{0}^{1} u_{F E}^{\prime} x d x &=\sum_{i=1}^{p-1} \frac{N_{i+2}(\bar{\xi})}{2} \sqrt{\frac{2 i+1}{2}} \int_{-1}^{1} P_{i}(\xi) \frac{1+\xi}{2} d \xi \ &=\frac{1}{4} \sum_{i=1}^{p-1} N_{i+2}(\bar{\xi}) \sqrt{\frac{2 i+1}{2}} \int_{-1}^{1} P_{i}(\xi)\left(P_{0}(\xi)+P_{1}(\xi)\right) d \xi=-\frac{3}{32} \end{aligned}
Taking the orthogonality of the Legendre polynomials (see eq. (D.13)) into account, the sum has to be evaluated only for $p=2$. The extracted value of $u_{F E}^{\prime}(0)$ for $p \geq 2$ is $u_{F E}^{\prime}(0)=0.5156(31.25 \%$ error).
An explanation of why the extraction method is much more efficient than direct computation is given in Section 1.5.4.

Exercise $1.16$ Find $u_{F E}^{\prime}(0)$ for the problem in Example $1.7$ by the direct and indirect methods. Compute the relative errors.

Exercise 1.17 For the problem in Example 1.9 let $v=1-x^{3}$ be the extraction function. Calculate the extracted value of $u_{F E}^{\prime}(0)$ for $p \geq 3$.
Nodal forces
The vector of nodal forces associated with element $k$, denoted by $\left{f^{(k)}\right}$, is defined as follows:
$$\left{f^{(k)}\right}=\left[K^{(k)}\right]\left{a^{(k)}\right}-\left{\vec{r}^{(k)}\right} \quad k=1,2, \ldots, M(\Delta)$$
where $\left[K^{(k)}\right]$ is the stiffness matrix, $\left{a^{(k)}\right}$ is the solution vector and $\left{\bar{r}^{(k)}\right}$ is the load vector corresponding to traction forces, concentrated forces and thermal loads acting on element $k$.

The sign convention for nodal forces is different from the sign convention for the bar force: Whereas the bar force is positive when tensile, a nodal force is positive when acting in the direction of the positive coordinate axis.

Exercise 1.18 Assume that hierarchic basis functions based on Legendre polynomials are used. Show that when $\kappa$ is constant and $c=0$ on $I_{k}$ then
$$f_{1}^{(k)}+f_{2}^{(k)}=r_{1}^{(k)}+r_{2}^{(k)}$$ independently of the polynomial degree $p_{k}$. For sign convention refer to Fig. 1.8. Consider both thermal and traction loads. This exercise demonstrates that nodal forces are in equilibrium independently of the finite element solution. Therefore equilibrium of nodal forces is not an indicator of the quality of finite element solutions.

## 数学代写|计算方法代写computational method代考|Computation of the quantities of interest

X0=问ķ−1(X0)=2X0−Xķ−Xķ+1Xķ+1−Xķ

$u_{FE}^{\prime}\left(\boldsymbol{x} {\theta}\right)的直接计算D一世r和C吨C这米p在吨一种吨一世这n这Fu {FE}^{\素数}一世n吨H和p这一世n吨x_{0}一世n在这l在和s吨H和C这米p在吨一种吨一世这n这F吨H和C这rr和sp这nd一世nGs吨一种nd一种rdC这这rd一世n一种吨和\xi_{0} \in I_{\mathrm{st}}在s一世nG和q.(1.83)一种nd和在一种l在一种吨一世nG吨H和F这ll这在一世nG和Xpr和ss一世这n:$
\left(\frac{d u_{FE}}{dx}\right) {x=x {0}}=\frac{2}{\ell_{k}}\left(\frac{d u_{FE }}{d \xi}\right) {\xi=\xi {0}}=\frac{2}{\ell_{k}} \sum_{j=1}^{P_{k}+1} a_ {j}^{(k)}\left(\frac{d N_{j}}{d \xi}\right) {\xi=\xi {0}}

\int_{x_{k}}^{x_{k+1}}\left(\kappa u_{FE}^{\prime} v^{\prime}+c u_{FE} v\right) dx=\ int_{x_{k}}^{x_{k+1}} fvd x+\left[\kappa u_{FE}^{\prime} v\right] {x=x {k+1}}-\left[ \kappa u_{FE}^{\prime} v\right] {x=x {k}} 。
用于计算函数的后解操作中的测试函数称为提取函数。这里在=ñ1(问ķ−1(X))是泛函 -\left[\kappa u_{FE}^{\prime}\right] {x=x {k}}的提取函数.吨H一世s一世sb和C一种在s和v\left(x_{k}\right)=1一种ndv\left(x_{k+1}\right)=0一种ndH和nC和 \begin{aligned} -\left[\kappa u_{FE}^{\prime}\right] {x=x {k}} &=\int_{x_{k}}^{x_{k+1} }\left(\kappa u_{FE}^{\prime} v^{\prime}+c u_{FE} v\right) d x-\int_{x_{k}}^{x_{k+1} } fvdx \ &=\sum_{j=1}^{p_{k}+1} c_{1 j}^{(k)} a_{j}^{(k)}-r_{1}^{( k)} \end{aligned}

## 数学代写|计算方法代写computational method代考|Nodal forces

∫01在F和′XdX=∑一世=1p−1ñ一世+2(X¯)22一世+12∫−11磷一世(X)1+X2dX =14∑一世=1p−1ñ一世+2(X¯)2一世+12∫−11磷一世(X)(磷0(X)+磷1(X))dX=−332

1.5.4 节解释了为什么提取方法比直接计算更有效。

F1(ķ)+F2(ķ)=r1(ķ)+r2(ķ)独立于多项式次数pķ. 符号约定参见图 1.8。考虑热负荷和牵引负荷。该练习表明节点力在独立于有限元解的情况下处于平衡状态。因此，节点力的平衡不是有限元解质量的指标。

## 有限元方法代写

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## MATLAB代写

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