User:Eml5526.s11.team5.JA/Mtg32

Mtg 32: Fri, 18 Mar 11

[[media: Fe1.s11.mtg32.djvu| Page 32-1 ]]

Summary:

Force/Heat Source


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$$ \tilde{\mathbf{F}}^{\color{Red}e} = \mathbf{L}^{{\color{Red}e}^{\color{Blue}T}} \mathbf{f}^{\color{Red}e}, \tilde{\mathbf{F}} = \sum_{\color{Red}e = 1}^{\color{Red}nel} \tilde{\mathbf{F}}^{\color{Red}e} \ $$

$$
 *  $$ \displaystyle {\color{Red}(3)}
 * }

$$
 * }
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 * }

End Summary

Derivation: Going from  local to global 

Equilibrium at element level:

Example: Bar element




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$$ k^{\color{Red}e}

\begin{bmatrix} 1    & -1   \\ -1    & 1 \end{bmatrix}

\begin{Bmatrix} d_1^{\color{Red}e}\\ d_2^{\color{Red}e} \end{Bmatrix} =

\begin{Bmatrix} f_1^{\color{Red}e}\\ f_2^{\color{Red}e} \end{Bmatrix}

$$

$$
 *  $$ \displaystyle
 * }

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$$ k^{\color{Red}e} = \frac{E^{\color{Red}e} A^{\color{Red}e}}{L^{\color{Red}e}} \ $$

$$
 *  $$ \displaystyle
 * }

, where

$$ E^{\color{Red}e} \ $$ = Young's modulus of element $$ {\color{Red}e} \ $$

$$ A^{\color{Red}e} \ $$ = Area of cross section of element $$ {\color{Red}e} \ $$

$$ L^{\color{Red}e} \ $$ = Length of element $$ {\color{Red}e} \ $$

For more details: EML 4500 Fall 08 : http://en.wikiversity.org/w/index.php?title=User:Eml4500.f08.bike.mcdonald/homework_report_3&oldid=347273

End Example

WRF


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$$ \mathbf{w}^{\color{Red}e} \cdot {\color{Blue}[} \mathbf{k}^{\color{Red}e} \mathbf{d}^{\color{Red}e} - \mathbf{f}^{\color{Red}e} {\color{Blue}]} = 0 \ \forall \ \mathbf{w}^{\color{Red}e} \ $$

$$
 *  $$ \displaystyle {\color{Red}(1)}
 * }

Local - to - global relationship:


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$$ \mathbf{d}^{\color{Red}e} = \mathbf{L}^{\color{Red}e} \tilde{\mathbf{d}} \ $$

$$
 *  $$ \displaystyle {\color{Red}(2)}
 * }


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$$ \mathbf{w}^{\color{Red}e} = \mathbf{L}^{\color{Red}e} \tilde{\mathbf{w}} \ $$

$$
 *  $$ \displaystyle {\color{Red}(3)}
 * }


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(1)-(3): $$ \left ( \mathbf{L}^{\color{Red}e} \tilde{\mathbf{w}} \right ) {\color{Red} \cdot}  \left [ \mathbf{k}^{\color{Red}e} \left ( \mathbf{L}^{\color{Red}e} \tilde{\mathbf{d}} \right ) - \mathbf{f}^{\color{Red}e} \right ] = 0 \ \forall \ \tilde{\mathbf{w}} \ $$

$$
 *  $$ \displaystyle {\color{Red}(4)}
 * }

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$$ \Rightarrow \tilde{\mathbf{w}} {\color{Red}\cdot}  \mathbf{L}^{{\color{Red}e}^{\color{Blue}T}}  \left [ \mathbf{k}^{\color{Red}e} \mathbf{L}^{\color{Red}e} \tilde{\mathbf{d}}  - \mathbf{f}^{\color{Red}e} \right ] = 0 \ \forall \ \tilde{\mathbf{w}} \ $$

$$
 * <p style="text-align:right"> $$ \displaystyle {\color{Red}(5)}
 * }

[[media: Fe1.s11.mtg32.djvu| Page 32-3 ]]

Same thought process applied to  mass/capacitance matrix .

Scalar PDE in 2D, 3D: Heat, membrane, Laplace

Transient heat:  Consider $$ \forall \ w \ \in \Omega \ $$



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$$ {\color{Blue}Q_1 } \ $$ = Heat flow into $$ \omega \ $$ though $$ \partial \omega \ $$

$$
 * <p style="text-align:right"> $$ \displaystyle
 * }

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$$ {\color{Blue}Q_2 } \ $$ = Heat generated in $$ \omega \ $$ by heat source $$ f( \mathbf{x}, t) \ $$

$$
 * <p style="text-align:right"> $$ \displaystyle
 * }

, where $$ \mathbf{x} = {\color{Blue} \left ( x_1, x_2, x_3 \right ) = (x, y, z ) }  \ $$

[[media: Fe1.s11.mtg32.djvu| Page 32-4 ]]

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$$ {\color{Blue}Q_3 } \ $$ = Heat in $$ \omega \ $$ due ot change in temperature $$ u \ $$

$$
 * <p style="text-align:right"> $$ \displaystyle
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Balance of heat: [[media: Fe1.s11.mtg6.djvu| Mtg. 6 -> 1D ]]

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$$ {\color{Blue}Q_1} = \underbrace_ \int_{ \partial \omega} \mathbf{q} \cdot \mathbf{n} \ d( \partial \omega) \ \underbrace{=}_{\color{Blue}Divergence \ Theorem} {\color{Red}-} \int_{ \omega} {\rm div}(\mathbf{q}) \ d \omega \ $$

$$
 * <p style="text-align:right"> $$ \displaystyle {\color{Red}(2)}
 * }

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$$ {\color{Blue}Q_2} = \int_{\omega} f( \mathbf{x}, t) \ d \omega \ $$

$$
 * <p style="text-align:right"> $$ \displaystyle {\color{Red}(3)}
 * }

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$$ {\color{Blue}Q_3} = \int_{\omega} \rho c \frac{ \partial u }{ \partial t} \ d \omega \ $$

$$
 * <p style="text-align:right"> $$ \displaystyle {\color{Red}(4)}
 * }

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$$ \Rightarrow {\color{Red}-} \int_{\omega} {\rm div}(\mathbf{q}) \ d \omega + \int_{\omega} f( \mathbf{x}, t) \ d \omega = \int_{\omega} \rho c \frac{ \partial u }{ \partial t} \ d \omega \ \forall \ \omega \ \in \ \Omega \ $$

$$
 * <p style="text-align:right"> $$ \displaystyle {\color{Red}(5)}
 * }

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$$ \Rightarrow \int_{ \omega} \left [ {\color{Red}-} {\rm div}(\mathbf{q}) + f  - \rho c \frac{ \partial u }{ \partial t} \right ] \ d \omega = 0 \ \forall \ \omega \ \in \ \Omega \ $$

$$
 * <p style="text-align:right"> $$ \displaystyle
 * }

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$$ \Rightarrow {\color{Red}-} {\rm div}(\mathbf{q}) + f  - \rho c \frac{ \partial u }{ \partial t} = 0 \ \forall \ \underbrace{\mathbf{x}}_{\color{Blue} point \  'x' = \left ( x_1, x_2, x_3  \right )} \ \in \ \Omega \ $$

$$
 * <p style="text-align:right"> $$ \displaystyle {\color{Red}(6)}
 * }