User:Eml5526.s11.team01.roark/Mtg35

[[media: Fe1.s11.mtg35.djvu| Mtg 35]]: Fri, 25 Mar 11 [[media: Fe1.s11.mtg35.djvu| Page 35-1]]

Example

[[media: Fe1.s11.mtg34.djvu|(1) p. 34-5]]: Consider element e=5 figure [[media: Fe1.s11.mtg30.djvu| p. 30-1]]



$$\displaystyle x_{i}^{\left( 5 \right)}={{\varphi }^{\left( 5 \right)}}\left( \mathbf \xi \right)=\sum\limits_{I=1}^{{{N}_{I}}\left(\mathbf \xi  \right)}x_{i,I}^{\left( 5 \right)},\,\,i=1,2$$

$$\displaystyle \left\{ \begin{matrix} x_{1}^{\left( 5 \right)} \\ x_{2}^{\left( 5 \right)} \\ \end{matrix} \right\}=\left\{ \begin{matrix} {{x}^{\left( 5 \right)}} \\ {{y}^{\left( 5 \right)}} \\ \end{matrix} \right\}=\sum\limits_{I=1}^{{{N}_{I}}\left(\mathbf \xi \right)}\left\{ \begin{matrix} x_{I}^{\left( 5 \right)} \\ y_{I}^{\left( 5 \right)} \\ \end{matrix} \right\}$$ End Example

[[media: Fe1.s11.mtg35.djvu| Page 35-2]] HW 6.4:

FB, pp. 148-149, problems. 6.1, 6.2, 6.3 End HW 6.4

Comparison of $\displaystyle {{\mathbf k}^{e}}$ in parent coordinate $\displaystyle \left\{ {{\xi }_{i}} \right\}$: continued [[media: Fe1.s11.mtg34.djvu|(5) p. 34-3]]:


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$$\displaystyle {{\mathbf k}^{e}}=\left[ k_{IJ}^{e} \right]$$
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$$\displaystyle k_{IJ}^{e}=\int_^ – {\left( \nabla N_{I}^{e} \right)\cdot \mathbf K}\cdot \left( \nabla N_{J}^{e} \right)d{{w}^{e}}$$
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$$\displaystyle k_{IJ}^{e}=\underbrace{\int_^ – {\left[ {{\nabla }_{x}}N_{I}^{e}\left(\mathbf \xi \right) \right]\cdot\mathbf K\left( x \right)}\cdot \left[ {{\nabla }_{x}}N_{J}^{e}\left( \mathbf \xi  \right) \right]dw_{x}^{e}}_{Note!\,\,Convert\,to\,\left\{ {{\xi }_{i}} \right\}\,\,coord.}$$ [[media: Fe1.s11.mtg34.djvu|(1) p. 34-4]]:
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$$\displaystyle \underbrace{dw_{x}^{e}}_{d\Omega =d{{x}_{1}}d{{x}_{2}}}=\underbrace{\det \mathbf J\left(\mathbf \xi \right)}_{J\left(\mathbf \xi  \right)}\underbrace{dw_{\xi }^{e}}_{d{{\xi }_{1}}d{{\xi }_{2}}=d\mathbf \omega }$$
 *  (4)
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$$\displaystyle \mathbf K\left(\mathbf x \right)= \mathbf K\left( {\mathbf {x}^{e}}\left(\mathbf \xi \right) \right)= \mathbf K\underbrace{\left( {{\varphi }^{e}}\left(\mathbf \xi  \right) \right)}_{\left( 1 \right)\,p.\,34-5}$$
 *  (5)
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$$\displaystyle {{\nabla }_{x}}N_{I}^{e}\left(\mathbf \xi \right)=\left[ \begin{matrix} \frac{\partial }{\partial {{x}_{1}}} \\ \frac{\partial }{\partial {{x}_{2}}} \\ \end{matrix} \right]N_{I}^{e}\left(\mathbf \xi \right)$$ [[media: Fe1.s11.mtg35.djvu| Page 35-3]]
 *  (6)
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$$\displaystyle \frac{\partial }{\partial {{x}_{i}}}=\frac{\partial }{\partial {{\xi }_{j}}}\frac{\partial {{\xi }_{j}}}{\partial {{x}_{i}}}\Rightarrow \left\lfloor \frac{\partial }{\partial \underbrace_{col}} \right\rfloor =\left\lfloor \frac{\partial }{\partial \underbrace_{row}} \right\rfloor \left\lfloor \frac{\partial \overbrace^{row}}{\partial \underbrace_{col}} \right\rfloor $$
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$$\displaystyle \underbrace{\left\lfloor \frac{\partial }{\partial {{x}_{1}}}\frac{\partial }{\partial {{x}_{2}}} \right\rfloor }_{\nabla _{x}^{T}}=\underbrace{\left\lfloor \frac{\partial }{\partial {{\xi }_{1}}}\frac{\partial }{\partial {{\xi }_{2}}} \right\rfloor }_{\nabla _{\xi }^{T}}\underbrace{\left\lfloor \begin{matrix} \frac{\partial {{\xi }_{1}}}{\partial {{x}_{1}}} & \frac{\partial {{\xi }_{1}}}{\partial {{x}_{2}}} \\ \frac{\partial {{\xi }_{2}}}{\partial {{x}_{1}}} & \frac{\partial {{\xi }_{2}}}{\partial {{x}_{2}}} \\ \end{matrix} \right\rfloor }_{{{\mathbf J}^{-1}}\left(\mathbf \xi \right)}$$
 *  (2)
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$$\displaystyle {{\nabla }_{x}}={{\mathbf J}^{-T}}\left(\mathbf \xi \right){{\nabla }_{\xi }}$$
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[[media: Fe1.s11.mtg35.djvu|(6) p. 35-2]]


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$$\displaystyle {{\nabla }_{x}}N_{I}^{e}\left(\mathbf \xi \right)={{\left( {{\mathbf J}^{e}} \right)}^{-T}}\left(\mathbf \xi  \right){{\nabla }_{\xi }}N_{I}^{e}\left(\mathbf \xi  \right)$$
 *  (4)
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[[media: Fe1.s11.mtg35.djvu|(1) p. 34-4]]
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$$\displaystyle {{\mathbf J}^{e}}\left(\mathbf \xi \right)=\left[ \frac{\partial x_{i}^{e}\left(\mathbf \xi  \right)}{\partial {{\xi }_{j}}} \right]$$ [[media: Fe1.s11.mtg35.djvu| Page 35-4]]
 *  (5)
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[[media: Fe1.s11.mtg34.djvu|p. 34-1]]: G2DM1.0/D1(Dataset 1): $$\displaystyle \Omega =\mathbf \omega =\square $$ biunit square
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PDE: [[media: Fe1.s11.mtg33.djvu|(4) p. 33-2]]: $$\displaystyle \mathbf K=\mathbf I$$ (Identity Matrix) $$\displaystyle f=0$$, $$\displaystyle \frac{\partial u}{\partial t}=0$$

Essential boundary condition: g=2 on $$\displaystyle \partial \Omega $$

Natural boundary condition: none
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HW 6.5:

Use 2DLIBF [[media: Fe1.s11.mtg29.djvu|(4) p. 29-2]] with n=m=2,4,6,8,… to solve G2DM1.0/D1 for $$\displaystyle {{u}^{h}}$$ until accuracy $$\displaystyle {{10}^{-6}}$$ at center $$\displaystyle \left( x,y \right)=\left( 0,0 \right)$$

Note: Verify your results with any FE code, with detailed documentation.

End HW 6.5