User:Eml5526.s11.team5/sub7

= 7.1  and  =

Given: 2D Strong Form and BCs

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$$\Omega=\bar{w}=\Box$$ biunit square $$
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 *  $$ \displaystyle                                                                          (Eq. 7.1.1)
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$$div(\underline{K} \cdot grad (u))+f=\rho c \frac{\partial u}{\partial t}$$ $$
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 *  $$ \displaystyle                                                                          (Eq. 7.1.2)
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$$ \underline{K}= \underline{I}$$ (identity matrix) $$
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 *  $$ \displaystyle                                                                          (Eq. 7.1.3)
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$$f=0 \quad$$,  $$\frac{\partial u}{\partial t}= 0$$ $$
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 *  $$ \displaystyle                                                                          (Eq. 7.1.4)
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Essential boundary condition:
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 * $$g=2 \quad$$ on  $$\partial\Omega \quad$$

$$
 *  $$ \displaystyle                                                                          (Eq. 7.1.5)
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Natural boundary condition:
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 * none

$$
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Find: Temperature inside and error at (0,0)

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Solve for $$u^h$$ until error is less then $$10^{-6}$$ at center (x,y)=(0,0) for: $$
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1) Static (steady state): $$f(x)=1 \quad$$ in $$\Omega=\Box$$
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 * 1a) 2d LIBF


 * 1b) 2d LLEBF


 * 1b1) Uniform mesh


 * 1b2) Non-uniform mesh

2) Dynamic(transient):$$\rho c=3 \quad$$


 * Initial condition $$u(x,t=0)=xy \qquad \forall x \in \Omega$$

$$
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2a)
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 * $$f(x,t)=0 \qquad \forall x \in \Omega,\ \forall t >0$$


 * $$g(x,t)=2 \mbox{ on }\Gamma_g=\partial \Omega$$


 * 2d LIBF

$$
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<span id="(1)">
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2b)
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 * $$f(x,t)=1 \qquad \forall x \in \Omega,\  \forall t >0$$


 * $$g(x,t)=2 \mbox{ on } \Gamma_g=\partial \Omega$$


 * 2b1) 2d LIBF


 * 2b2) 2d LLEBF


 * 2b2a) Uniform mesh


 * 2b2b) non-uniform mesh

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

In general by Prof. Vu Quoc's

$$ \tilde{m}(w,u) + \tilde{k}(w,u) = \tilde{f}(w,u) \ $$

$$ \tilde{m}(w,u) = \int_\Omega w \rho c \frac{\partial u}{\partial t} d \Omega \ $$

$$ \tilde{k}(w,u) = \int_{\Omega^e} \triangledown{w} \kappa(x, y) \triangledown{u} d \Omega \ $$

$$ \tilde{f}(w) = - \int_{\Gamma_h}w h d \Gamma_h + \int_\Omega w f d \Omega \ $$

$$ K_{ij}^e = \int_{\Omega^e} \triangledown{b^e_i}'(x, y) \kappa(x, y) \triangledown{b^e_j}'(x, y) d \omega_x \ $$

, where $$ x = x^e \ $$, where $$x^e \ $$ represents local node.

For linear 2D case with boundary being square

Bases, $$ b_i \ $$ are defined to be $$ L_{[I,J]} (x, y) = N^e_{[I,J]} (x, y) \ $$

where $$ N^e_{[I,J]}(x, y) = N_I^e(x)N_J^e(y) \ $$ and $$ I, J \ $$ corresponds to numbering of nodes on element scale.



also noting that it's a rectangular shape, such that the distance from x: 1-to-2 = 1-to-3 and node 1 = node 4 in 'x' coordiantes, y: 1-to-4 = 1-to-3 and node 1 = node 2 in 'x' coordinates, we substitute $$ y^e_4 $$ for $$ y^e_3 \ $$ and $$ y^e_1 $$ for $$ y^e_2 \ $$ and the same for other coordinates

$$ N^e_{[1,1]} (x, y) = \frac{x^e-x^e_2}{x^e_1 - x^e_2} \frac{y^e-y^e_4}{y^e_1 - y^e_4} = \frac{1}{4}(x-\xi)(y - \eta) = b_1^e (x,y) \ $$

$$ N^e_{[2,1]} (x, y) = \frac{x^e-x^e_1}{x^e_2 - x^e_1} \frac{y^e-y^e_4}{y^e_1 - y^e_4} = \frac{1}{4}(x-\xi)(y - \eta)  = b_2^e(x,y)\ $$

$$ N^e_{[2,2]} (x, y) = \frac{x^e-x^e_1}{x^e_2 - x^e_1} \frac{y^e-y^e_1}{y^e_4 - y^e_1} = \frac{1}{4}(x-\xi)(y - \eta)  = b_3^e (x,y) \ $$

$$ N^e_{[1,2]} (x, y) = \frac{x^e-x^e_2}{x^e_1 - x^e_2} \frac{y^e-y^e_1}{y^e_4 - y^e_1} = \frac{1}{4}(x-\xi)(y - \eta)  = b_4^e (x,y) \ $$

$$ \mathbf{BN} =

\begin{bmatrix} \frac{\partial N^e_{[1,1]} (x, y) }{\xi} & \frac {\partial N^e_{[2,1]} (x, y) }{\xi} & \frac {\partial N^e_{[2,2]} (x, y) }{\xi} & \frac{\partial N^e_{[1,2]} (x, y) }{\xi}   \\

\frac{\partial N^e_{[1,1]} (x, y) }{\eta} & \frac {\partial N^e_{[2,1]} (x, y) }{\eta} & \frac {\partial N^e_{[2,2]} (x, y) }{\eta} & \frac{\partial N^e_{[1,2]} (x, y) }{\eta}

\end{bmatrix}

\ $$

$$ [\mathbf{x} \ \mathbf{y}] \ $$

Jacobian Matrix = BN *[mathbf{x} \ mathbf{y}]

taking the $$ x^e \ $$ and $$ y^e \ $$ values from the definition of parent element and substituting into Jacobian matrix we get:

$$ \mathbf{J}^e =

\begin{bmatrix} \frac{\partial x}{\partial \xi} &  \frac{\partial y}{\partial \xi}    \\ \\ \frac{\partial x}{\partial \eta} &  \frac{\partial y}{\partial \eta} \end{bmatrix} =

\begin{bmatrix} \frac{1}{2} (x_2^e - x_1^e) & 0    \\ \\ 0 &  \frac{1}{2} (y_4^e - y_1^e) \end{bmatrix}

$$

From given

$$ d \omega_x = det(\mathbf{J}^e) d \omega_\xi = d \omega_x = \frac{1}{2} (x_2^e - x_1^e)*\frac{1}{2} (y_4^e - y_1^e) d \omega_\xi \ $$

or

$$ d \omega_\xi = \frac{ d \omega_x}{det(\mathbf{J}^e)} = d \omega_\xi = \frac{ d \omega_x}{\frac{1}{2} (x_2^e - x_1^e)*\frac{1}{2} (y_4^e - y_1^e)} $$

With this last information we have a completely defined stiffness element matrix $$ K^e \ $$. Also note that since we have, 4 degrees of freedom in 1 element node, we do expect the gram matrix $$ \Gamma \ $$ to be 4x4, therefore, $$ \kappa_{4x4} \ $$

$$ K_{ij}^e = \int_{\Omega^e} \triangledown{b^e_i}'(x, y) \kappa(x, y) \triangledown{b^e_j}'(x, y) d \omega_x \ $$

$$ K_{ij}^e = \int_{-1}^{+1}

\begin{bmatrix} \triangledown{b^e_1} \triangledown{b^e_1} & \triangledown{b^e_1} \triangledown{b^e_2} & \triangledown{b^e_1} \triangledown{b^e_3} & \triangledown{b^e_1} \triangledown{b^e_4}      \\ & \triangledown{b^e_2} \triangledown{b^e_2} & \triangledown{b^e_2} \triangledown{b^e_3} & \triangledown{b^e_2}\triangledown{b^e_4}      \\ & &  \triangledown{b^e_3} \triangledown{b^e_3} & \triangledown{b^e_3} \triangledown{b^e_4}      \\ symetry & &   & \triangledown{b^e_4} \triangledown{b^e_4} \end{bmatrix}

\begin{bmatrix} 1 & 0 & 0 & 0      \\ 0 & 1 &  0 & 0      \\ 0 & 0 &  1 & 0      \\ 0 & 0 &  0 & 1      \end{bmatrix}

\left ( d \frac{1}{2} (x_2^e - x_1^e)*\frac{1}{2} (y_4^e - y_1^e) d \omega_\xi \right ) \ $$

note, to save some space in typing the formula (else it stretched out of the page), the independent variables for $$ b_i^e (\xi, \eta) \ $$ were avoided

for the initial square of length of '1', we divided into 4 smaller squares each 0.5 length, the element nodes stiffness matrix came out to be:

K(:,:,1) = [ 1/24, -1/96, -1/48, -1/96] [ -1/96,  1/24, -1/96, -1/48] [ -1/48, -1/96,  1/24, -1/96] [ -1/96, -1/48, -1/96,  1/24] K(:,:,2) = [ 1/24, -1/96, -1/48, -1/96] [ -1/96,  1/24, -1/96, -1/48] [ -1/48, -1/96,  1/24, -1/96] [ -1/96, -1/48, -1/96,  1/24]

this answer shows 2 elements stiffness matrices located along 'x' axes. Noting that because of the symmetry of each elements, the equivalent spacing of each node in elements, and because the material matrix $$ \kappa \ $$ is identity matrix, all stiffness element matrices will be the same. The only differ in element stiffness matrix will occur when we will change the number of stiffness.

,where

$$ b^e_1(\xi, \eta) = \frac{1}{A^e}(x_1^e \frac{1-\xi}{2} + [ \frac{1+\xi}{2}- 1 ] x^e_2)(y_1^e \frac{1-\eta}{2} + [ \frac{1+\eta}{2} - 1 ] ) y^e_4) \ $$

$$ b_2^e(\xi, \eta) =  -\frac{1}{A^e}( [\frac{1-\xi}{2}- 1]x^e_1 + x_2^e \frac{1+\xi}{2})(y_1^e \frac{1-\eta}{2} +  [\frac{1+\eta}{2} - 1] y^e_4)  \ $$

$$ b_3^e (\xi, \eta)  =  \frac{1}{A^e}( [\frac{1-\xi}{2}- 1]x^e_1 + x_2^e \frac{1+\xi}{2})([\frac{1-\eta}{2} -1]y^e_1 + y_4^e \frac{1+\eta}{2} ) \ $$

$$  b_4^e (\xi, \eta) =  -\frac{1}{A^e}(x_1^e \frac{1-\xi}{2} + [\frac{1+\xi}{2}- 1]x^e_2)( [\frac{1-\eta}{2}-1]y^e_1 + y_4^e \frac{1+\eta}{2})  \ $$

force
From prof. Vu Quoc's notes for

$$ f^e(w) = - \int_{\Gamma_h} wh d {\Gamma_h} + \int_\omega wf d \omega \ $$

$$ f^e(w) = - \int_{\Gamma_h} wh d {\Gamma_h} + \int_\omega w1 d \omega \ $$

converting to local coordinates

$$ f^e(w) = - \int_{\omega_\xi}

\begin{bmatrix} \triangledown{b^e_1}      \\ \triangledown{b^e_2}    \\ \triangledown{b^e_3}     \\ \triangledown{b^e_4} \end{bmatrix}

2* \left ( \frac{1}{4} A_e \ d\omega_\xi \right ) + \int_{\omega_\xi}

\begin{bmatrix} \triangledown{b^e_1}      \\ \triangledown{b^e_2}    \\ \triangledown{b^e_3}     \\ \triangledown{b^e_4} \end{bmatrix}

\left( \frac{1}{4} A_e \ d\omega_\xi \right ) =

- \int_{\omega_\xi}

\begin{bmatrix} \triangledown{b^e_1}      \\ \triangledown{b^e_2}    \\ \triangledown{b^e_3}     \\ \triangledown{b^e_4} \end{bmatrix}

\left( \frac{1}{4} A_e \ d\omega_\xi \right ) \ $$

F(:,:,1) = [ 1/16,  1/16] [ -1/16,  1/16] [ -1/16, -1/16] [  1/16, -1/16] F(:,:,2) = [ 1/16,  1/16] [ -1/16,  1/16] [ -1/16, -1/16] [  1/16, -1/16]

Global Matrix creation
Then we constructed the Global stiffness matrix by the process of superimposing the local stiffness coefficients. Matlab output below shows the process of superimposing the elements to global stiffness matrix.

The matlab code for global matrix construction was written with respect to this element naming system, and then it was extended to encompass other numbers of elements



Original element matrices

K(:,:,1) = [ 1/24, -1/96, -1/48, -1/96] [ -1/96,  1/24, -1/96, -1/48] [ -1/48, -1/96,  1/24, -1/96] [ -1/96, -1/48, -1/96,  1/24]

K(:,:,2) = [ 1/24, -1/96, -1/48, -1/96] [ -1/96,  1/24, -1/96, -1/48] [ -1/48, -1/96,  1/24, -1/96] [ -1/96, -1/48, -1/96,  1/24] Superpositions to global matrix

KGlobal = 1/24         -1/96           0             -1/96          -1/48           0              0              0              0             -1/96           1/12          -1/96          -1/48          -1/48          -1/48           0              0              0              0             -1/96           1/24           0             -1/48          -1/96           0              0              0             -1/96          -1/48           0              1/12          -1/48           0             -1/96          -1/48           0             -1/48          -1/48          -1/48          -1/48           1/6           -1/48          -1/48          -1/48          -1/48           0             -1/48          -1/96           0             -1/48           1/12           0             -1/48          -1/96           0              0              0             -1/96          -1/48           0              1/24          -1/96           0              0              0              0             -1/48          -1/48          -1/48          -1/96           1/12          -1/96           0              0              0              0             -1/48          -1/96           0             -1/96           1/24

FGlobal = 1/8           0             -1/8            1/4            0             -1/4            1/8            0             -1/8

Boundary Conditions
Noting that essential BCs were specified to be on all sides of the 2x2 square. It would mean (viewing from the above global node figure) all nodes around node 5 will be tagged as essential BC.



Since we can either eliminate the nodes form Global Stiffness matrix to construct another one with Temperature ('u' function) as unknown or just '0' the node values in Global Stiffness matrix and then do inverse and multiply by Force matrix to find other non-constrained dofs.

We wrote a matlab code that automatically constrains the associate nodes and puts their contribution to force function.

Here how code applies BC's

KGlobal = 1/24         -1/96           0             -1/96          -1/48           0              0              0              0             -1/96           1/12          -1/96          -1/48          -1/48          -1/48           0              0              0              0             -1/96           1/24           0             -1/48          -1/96           0              0              0             -1/96          -1/48           0              1/12          -1/48           0             -1/96          -1/48           0             -1/48          -1/48          -1/48          -1/48           1/6           -1/48          -1/48          -1/48          -1/48           0             -1/48          -1/96           0             -1/48           1/12           0             -1/48          -1/96           0              0              0             -1/96          -1/48           0              1/24          -1/96           0              0              0              0             -1/48          -1/48          -1/48          -1/96           1/12          -1/96           0              0              0              0             -1/48          -1/96           0             -1/96           1/24

Kchange = 0             0              0              0              0              0              0              0              0              0              0              0              0              0              0              0              0              0              0              0              0              0              0              0              0              0              0              0              0              0              0              0              0              0              0              0              0              0              0              0              1/6            0              0              0              0              0              0              0              0              0              0              0              0              0              0              0              0              0              0              0              0              0              0              0              0              0              0              0              0              0              0              0              0              0              0              0              0              0              0              0              0

FGlobal = -1/16          0              1/16          -1/8            0              1/8           -1/16           0              1/16

Fchange = -7/48         -7/48           0             -1/6            1/3            1/12          -7/48          -7/48           0

Then the matrices were reduced to the ones of the importance ( Global K matrix after the BCs where applied was isolated so no 0s would be present and attributing F function value(s) where picked by code as well)

KchangeF = 1/6

FchangeF = 1/3

Finally Temperature at unknown nodes was found by using this formula

$$ u = K^{-1}F \ $$

u = 2

Also output for 4x4 element free node values came out to be:

dcalc = 2             2              2              2

as you may see, only 1 node, in this example was present, so only 1 numerical value was generated.



Discussion
Note 1 important aspect of this output. That is, with forcing function changed to 1, the output of free degrees of freedom is still '2' as if no forcing function was present to begin with. I guess it might be explainable that because of constrained boundary conditions at all 4 ends, in order to keep the temperature constant at all 4 ends while heat was generated, the generated heat had to be instantaneously dissipated though the boundaries of the area, so that boundary temperature would stay unchanged.

As in matrix/numeral explanation, it should be noted, that because essential boundary condition is applied over all 4 sides, it has integration limits (working boundary range) as all element boundary. Therefore, with intorduction with heat source, force instead of being modified, was only scaled (remember force equation):

for f = 0 (HW 6.6)

$$ f^e(w) = - \int_{\omega_\xi}

\begin{bmatrix} \triangledown{b^e_1}      \\ \triangledown{b^e_2}    \\ \triangledown{b^e_3}     \\ \triangledown{b^e_4} \end{bmatrix}

2* \left ( \frac{1}{4} A_e \ d\omega_\xi \right ) \ $$

for f= 1 (HW 7.1)

$$ f^e(w) = - \int_{\omega_\xi}

\begin{bmatrix} \triangledown{b^e_1}      \\ \triangledown{b^e_2}    \\ \triangledown{b^e_3}     \\ \triangledown{b^e_4} \end{bmatrix}

2* \left ( \frac{1}{4} A_e \ d\omega_\xi \right ) + \int_{\omega_\xi}

\begin{bmatrix} \triangledown{b^e_1}      \\ \triangledown{b^e_2}    \\ \triangledown{b^e_3}     \\ \triangledown{b^e_4} \end{bmatrix}

\left( \frac{1}{4} A_e \ d\omega_\xi \right ) =

- \int_{\omega_\xi}

\begin{bmatrix} \triangledown{b^e_1}      \\ \triangledown{b^e_2}    \\ \triangledown{b^e_3}     \\ \triangledown{b^e_4} \end{bmatrix}

\left( \frac{1}{4} A_e \ d\omega_\xi \right ) \ $$

you may see that force term is a scale only term. Moreover, notice the force global term

For 4 element at f = 0: FGlobal = 1/8           0             -1/8            1/4        0            -1/4            1/8            0             -1/8

Fchange = 5/24          7/48          -1/16           7/24         -1/3           -5/24           5/24           7/48          -1/16

For 4 element at f = 1:

FGlobal =

-1/16          0              1/16          -1/8            0             1/8           -1/16           0              1/16

Fchange =

-7/48         -7/48           0             -1/6           -1/3            1/12          -7/48          -7/48           0

For 9 element at f = 1

FGlobal =

-1/36          0              0              1/36          -1/18           0        0            1/18          -1/18           0        0            1/18          -1/36           0              0              1/36

Fchange =

-7/108        -7/108         -7/108          0            -11/108         5/54           5/54           1/108         -5/54         5/54           5/54           1/54          -7/108         -7/108         -7/108          0

, where boxed answer shows free nodes.

Note that force matrix DOES differ, but note that because of the similarity of force functions (scalar multiples of each other, namely, caused because forcing function is a constant and essential boundary condition acts on the same area as forcing function), the global force matrix, since it contained '0's on initial force function, will contain '0's on any other force function, as long as forcing function is scalar and essential B.C is the same as whole element B.C. Therefore, with same used stiffness matrix 'K' the adjusted force matrix with initial values of '0' on free nodes, always came out to be the same for specific number of elements with differing scalar value of forcing function.

As a result, for free node displacements, we observed the constant value of '2'.

--Eml5526.s11.team5.JA 20:51, 6 April 2011 (UTC)

Transient Term
$$ \tilde{m}(w,u) = \int_\Omega w \rho c \frac{\partial u}{\partial t} d \Omega = \int_\Omega w xy \frac{\partial u}{\partial t} d \Omega =

\int_{\omega_\xi}

\begin{bmatrix} \triangledown{b^e_1}      \\ \triangledown{b^e_2}    \\ \triangledown{b^e_3}     \\ \triangledown{b^e_4} \end{bmatrix}

\left( x_1^e \frac{1-\xi}{2} + x_2^e \frac{1+\xi}{2} \right ) \left( y_1^e \frac{1-\eta}{2} + y_4^e \frac{1+\eta}{2} \right ) \frac{\partial u}{\partial t} \left( \frac{1}{4} A_e \ d\omega_\xi \right ) \ $$

Overall Equation
$$ \int_{\omega_\xi}

\begin{bmatrix} \triangledown{b^e_1}      \\ \triangledown{b^e_2}    \\ \triangledown{b^e_3}     \\ \triangledown{b^e_4} \end{bmatrix}

\left( x_1^e \frac{1-\xi}{2} + x_2^e \frac{1+\xi}{2} \right ) \left( y_1^e \frac{1-\eta}{2} + y_4^e \frac{1+\eta}{2} \right ) \frac{\partial u}{\partial t} \left( \frac{1}{4} A_e \ d\omega_\xi \right ) + \int_{\omega_\xi}

\begin{bmatrix} \triangledown{b^e_1} \triangledown{b^e_1} & \triangledown{b^e_1} \triangledown{b^e_2} & \triangledown{b^e_1} \triangledown{b^e_3} & \triangledown{b^e_1} \triangledown{b^e_4}      \\ & \triangledown{b^e_2} \triangledown{b^e_2} & \triangledown{b^e_2} \triangledown{b^e_3} & \triangledown{b^e_2}\triangledown{b^e_4}      \\ & &  \triangledown{b^e_3} \triangledown{b^e_3} & \triangledown{b^e_3} \triangledown{b^e_4}      \\ symetry & &   & \triangledown{b^e_4} \triangledown{b^e_4} \end{bmatrix}

\begin{bmatrix} 1 & 0 & 0 & 0      \\ 0 & 1 &  0 & 0      \\ 0 & 0 &  1 & 0      \\ 0 & 0 &  0 & 1      \end{bmatrix}

\left ( d \frac{1}{2} (x_2^e - x_1^e)*\frac{1}{2} (y_4^e - y_1^e) d \omega_\xi \right ) = - \int_{\omega_\xi}

\begin{bmatrix} \triangledown{b^e_1}      \\ \triangledown{b^e_2}    \\ \triangledown{b^e_3}     \\ \triangledown{b^e_4} \end{bmatrix}

2* \left ( \frac{1}{4} A_e \ d\omega_\xi \right ) \ $$

Static using 2d LLEBF non-uniform mesh








Here the answer is also 2 at every node, which agrees with the uniform mesh LLEBF. There is a slight difference between nodes which can be attributed to matlab roundoff error.

Dynamic (case 2b) using 2d LLEBF uniform mesh








Dynamic (case 2b) using 2d LLEBF non-uniform mesh






Eml5526.s11.team5.savery

Eml5526.s11.team5.smith 20:37, 21 April 2011 (UTC)

Eml5526.s11.team5.smith 21:11, 21 April 2011 (UTC)

= 7.2 Finding the static and dynamic solution of a bi-unit square using 2D LIBF =

P.D.E
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$$T div (grad u) + f = \rho \ \frac{\partial^2u}{\partial t^2} $$ $$
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 * <p style="text-align:right"> $$ \displaystyle                                                                     (Eq. 7.2.1)
 * }

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$$ \quad T = 4 $$ $$
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 * <p style="text-align:right"> $$ \displaystyle                                                                     (Eq. 7.2.2)
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$$ f (\mathbf{x}) = 1 $$ $$
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 * <p style="text-align:right"> $$ \displaystyle                                                                     (Eq. 7.2.3)
 * }

<span id="(1)">
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$$ \quad \rho = 3 $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle                                                                     (Eq. 7.2.4)
 * }

<span id="(1)">
 * {| style="width:100%" border="0"

$$ \quad Basis \ Function \ : \ 2D \ Lagrange \ Interpolation \ Basis \ Function $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle\!
 * }

Essential Boundary Condition
<span id="(1)">
 * {| style="width:100%" border="0"

$$ \quad g = 0 $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle                                                                     (Eq. 7.2.5)
 * }

Static(steady state)
<span id="(1)">
 * {| style="width:100%" border="0"

1. Find the static solution $$ \quad U^h $$ to $$ \quad 10^{-6} $$ accuracy at the center. $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle\!
 * }

Dynamic(transient)
<span id="(1)">
 * {| style="width:100%" border="0"

1. Fundamental Frequency and Fundamental Period $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle\!
 * }

<span id="(1)">
 * {| style="width:100%" border="0"

2. Plot $$ u^h(0,t) \ vs \ t \quad  for \ t \ \epsilon \ [0,2T_1] $$ for symmetric and Non-Symmetric $$ \quad u^0 $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle\!
 * }

<span id="(1)">
 * {| style="width:100%" border="0"

3. Produce a movie of the vibrating membrane for symmetric and Non-Symmetric $$ \quad u^0 $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle\!
 * }

Solution
Substituting the given values in the P.D.E, we get, <span id="(1)">
 * {| style="width:100%" border="0"

$$\quad 4 \ div (grad u) \ + \ 1 \ = \ 0 $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle\!
 * }

<span id="(1)">
 * {| style="width:100%" border="0"

$$ 4 \left ( \frac{\partial^2u}{\partial x^2} + \frac{\partial^2u}{\partial y^2}\right ) + 1 = 0$$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle\!
 * }

<span id="(1)">
 * {| style="width:100%" border="0"

$$ \frac{\partial^2u}{\partial x^2} + \frac{\partial^2u}{\partial y^2} = \frac{-1}{4} $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle                                                                     (Eq. 7.2.6)
 * }

2D LIBF

<span id="(1)">
 * {| style="width:100%" border="0"

$$ N_I (x,y) = L_{i,m}(x).L_{j,n}(y) \quad Where \ I = i+(j-1)m $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle                                                                     (Eq. 7.2.7)
 * }

Where,

<span id="(1)">
 * {| style="width:100%" border="0"

$$ L_{i,m}(x) = \prod_{k=1\neq i}^{m}\frac{x-x_k}{x_i - x_k} $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle\!
 * }<span id="(1)">
 * {| style="width:100%" border="0"

$$ L_{j,n}(x) = \prod_{k=1\neq j}^{m}\frac{y-y_k}{y_j - y_k} $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle\!
 * }

Consider the Weak Form

<span id="(1)">
 * {| style="width:100%" border="0"

$$  \tilde{m}(w,u) + \tilde{k}(w,u) = \tilde{f}(w) $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle                                                                     (Eq. 7.2.8)
 * }

where,

<span id="(1)">
 * {| style="width:100%" border="0"

$$m(w,u) = \int\limits_\Omega {w\rho c\frac} d\Omega $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle\!
 * }

<span id="(1)">
 * {| style="width:100%" border="0"

$$ k(w,u) = \int\limits_\Omega {\nabla wK\nabla u} d\Omega  $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle\!
 * }

<span id="(1)">
 * {| style="width:100%" border="0"

$$ f(w,u) = - \int_{\Gamma h} {whd{\Gamma _h}}  + \int\limits_\Omega  {wf} d\Omega  $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle\!
 * }

Static case
For static case, $$ \frac{\partial u}{\partial t} = 0 $$ the above equation can be written as,

<span id="(1)">
 * {| style="width:100%" border="0"

$$ \quad K d = F $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle\!
 * }

where,

<span id="(1)">
 * {| style="width:100%" border="0"

$$ {K_{ij}} = \int_ {(\nabla N_I^e)I(\nabla N_J^e)d{w^e}} $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle                                                                     (Eq. 7.2.9)
 * }

Using the above equations in MATLAB, we get,

K =  33.3257  -13.5386  -13.5386    2.8639 -13.5386  33.3257    2.8639  -13.5386       -13.5386    2.8639   33.3257  -13.5386         2.8639  -13.5386  -13.5386   33.3257

F = [0.5625  0.5625  0.5625  0.5625]^T

d = [0.0617  0.0617  0.0617  0.0617]^T

Dynamic case
For this case, It is given that the value of mass density $$ \quad \rho \ = \ 3$$

Free Vibration: Solve Eigen Value problem
The equation given to solve the Eigen value problem is,

<span id="(1)">
 * {| style="width:100%" border="0"

$$ \mathbf{K} \mathbf{\phi} = \lambda \mathbf{M}\mathbf{\phi}  $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle                                                                     (Eq. 7.2.10)
 * }

Where,

<span id="(1)">
 * {| style="width:100%" border="0"

$$\quad \mathbf{\phi} - Eigen \ vectors $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle\!
 * }

<span id="(1)">
 * {| style="width:100%" border="0"

$$ \quad \lambda - Eigen \ values $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle\!
 * }

Using the above equation and the expression for K, F matrix, we can calculate the fundamental frequency and fundamental time period.

Fundamental Frequency = 0.23725

Fundamental Time Period = 4.2149

Transient Analysis with symmetric Uo
Initial conditions

<span id="(1)">
 * {| style="width:100%" border="0"

$$u(\mathbf{x},t=0) = u^h_s(x) \ \forall \  \mathbf{x} \ \epsilon  \ \Omega  $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle                                                                     (Eq. 7.2.11)
 * }

<span id="(1)">
 * {| style="width:100%" border="0"

$$ \dot{u} (\mathbf{x},t=0) = 0 \ \forall \  \mathbf{x} \ \epsilon  \ \Omega $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle                                                                     (Eq. 7.2.12)
 * }

<span id="(1)">
 * {| style="width:100%" border="0"

$$T = 4, \qquad f(\mathbf{x},t) = 0 \ in \  \Omega  $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle                                                                     (Eq. 7.2.13)
 * }

By using the above initial condition, we can produce the movie of the vibrating membrane.

Movie of the vibrating membrane
The Movie of dynamic membrane has been uploaded in the youtube and it can be seen from the link below

FE Analysis of a Dynamic Membrane - symmetric case

For T = 0,



Transient Analysis with Non-symmetric Uo
In this case, initial u value is given by,

<span id="(1)">
 * {| style="width:100%" border="0"

$$u(\mathbf{x},t=0) = (x+1)(y+\frac{1}{2}) cos (\frac{\pi}{2}x) cos (\frac{\pi}{2}y) \ \forall \  \mathbf{x} \ \epsilon  \ \Omega$$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle                                                                     (Eq. 7.2.14)
 * }

By using the above initial condition, we can produce the movie of the vibrating membrane.

Movie of the vibrating membrane
The Movie of dynamic membrane for the non-symmetric case, has been uploaded in the youtube and it can be seen from the link below

FE Analysis of a Dynamic membrane - Non Symmetric case



--Eml5526.s11.team5.srv 21:06, 21 April 2011 (UTC)

MATLAB code for producing the movie of the vibrating membrane
--Lokeshdahiya 21:09, 21 April 2011 (UTC)

= 7.3 =

Given/Find: Static Solution of the Circle
<span id="(1)">
 * {| style="width:100%" border="0"

Let Omega a circle with a one-unit radius. Find the static solution such that: $$T=4$$ $$f(\underline x)=1$$ in Omega $$g=0$$ on $$\Gamma _g=\partial \Omega $$
 * style="width:95%" |
 * style="width:95%" |

all to 10^-6 accuracy at the center. Quadrilateral and Triangular elements should be used but for only 1/4 of the circle. Plot the deformed shape in 3D. $$
 * <p style="text-align:right"> $$ \displaystyle
 * }

Solution: Static Solution of the Circle Using Quadrilateral and Triangular Elements to an Accuracy of 10^-6
<span id="(1)">
 * {| style="width:100%" border="0"

The Partial Differential Equation for a vibrating membrane is the following via transient analysis: $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle
 * }

<span id="(1)">
 * {| style="width:100%" border="0"

T(\textrm{div \ grad \ u})+\textrm{f}=\rho \frac{\partial ^2u}{\partial t^2} $$ $$
 * style="width:95%" |$$
 * style="width:95%" |$$
 * <p style="text-align:right"> $$ \displaystyle
 * }

<span id="(1)">
 * {| style="width:100%" border="0"

Where: $$ u(\underline x,t)=$$ Transverse Displacement $$T=\quad$$ Tension (Force/Length) = Constant $$\textrm{f}(\underline x,t)=$$ Distributed Transverse Force $$\rho (\underline x)=$$ Mass Density (Mass/Area) & $$\textrm{div \ grad\ u}=\frac{\partial }{\partial x_i}\frac{\partial u}{\partial x_j}$$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle
 * }

<span id="(1)">
 * {| style="width:100%" border="0"

The Weighted Residual Form is as follows: $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle
 * }

<span id="(1)">
 * {| style="width:100%" border="0"

\int_{\Omega } w\left [ T\frac{\partial ^2u}{\partial x_i\partial x_j}+\textrm{f}-\rho \frac{\partial ^2u}{\partial t^2} \right ]d\Omega =0 $$ $$
 * style="width:95%" |$$
 * style="width:95%" |$$
 * <p style="text-align:right"> $$ \displaystyle
 * }

<span id="(1)">
 * {| style="width:100%" border="0"

Which can thereby be broken into three separate terms: $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle
 * }

<span id="(1)">
 * {| style="width:100%" border="0"

\int_{\Omega } wT\frac{\partial ^2u}{\partial x_i\partial x_j}d\Omega =\mathbb{K} $$ $$
 * style="width:95%" |$$
 * style="width:95%" |$$
 * <p style="text-align:right"> $$ \displaystyle
 * }

<span id="(1)">
 * {| style="width:100%" border="0"

\int_{\Omega } w\textrm{f}\ d\Omega $$ $$
 * style="width:95%" |$$
 * style="width:95%" |$$
 * <p style="text-align:right"> $$ \displaystyle
 * }

<span id="(1)">
 * {| style="width:100%" border="0"

\int_{\Omega } w\rho \frac{\partial ^2u}{\partial t^2}\ d\Omega=\tilde{m}(w,u) $$ $$
 * style="width:95%" |$$
 * style="width:95%" |$$
 * <p style="text-align:right"> $$ \displaystyle
 * }

<span id="(1)">
 * {| style="width:100%" border="0"

This final equation will be used later in the Weak Form. For now, the first term above (set equal to $$\mathbb{K}$$), is equivalent to: $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle
 * }

<span id="(1)">
 * {| style="width:100%" border="0"

\mathbb{K}=\int_{d\Omega } \frac{\partial }{\partial x_i}(wT\frac{\partial u}{\partial x_j})\ d\Omega-\int_{\Omega } \frac{\partial w}{\partial x_i}T\frac{\partial u}{\partial x_j}\ d\Omega $$ $$
 * style="width:95%" |$$
 * style="width:95%" |$$
 * <p style="text-align:right"> $$ \displaystyle
 * }

<span id="(1)">
 * {| style="width:100%" border="0"

And by using divergence theorem, the first term of the equation above may be converted to yield: $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle
 * }

<span id="(1)">
 * {| style="width:100%" border="0"

\mathbb{K}=\int_{\partial \Omega } n_iwT\frac{\partial u}{\partial x_j}d(\partial \Omega)-\int_{\Omega } \frac{\partial w}{\partial x_i}T\frac{\partial u}{\partial x_j}\ d\Omega $$ $$
 * style="width:95%" |$$
 * style="width:95%" |$$
 * <p style="text-align:right"> $$ \displaystyle
 * }

<span id="(1)">
 * {| style="width:100%" border="0"

Now since $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle
 * }

<span id="(1)">
 * {| style="width:100%" border="0"

\partial \Omega =\Gamma _g\cup \Gamma _h\ \textrm{but }\ w=0\ \ \textrm{on}\ \ \Gamma _g $$ $$
 * style="width:95%" |$$
 * style="width:95%" |$$
 * <p style="text-align:right"> $$ \displaystyle
 * }

<span id="(1)">
 * {| style="width:100%" border="0"

and the following are true: $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle
 * }

<span id="(1)">
 * {| style="width:100%" border="0"

\frac{\partial w}{\partial x_i}=\textrm{grad}\ w=\bigtriangledown w $$ $$
 * style="width:95%" |$$
 * style="width:95%" |$$
 * <p style="text-align:right"> $$ \displaystyle
 * }

<span id="(1)">
 * {| style="width:100%" border="0"

\frac{\partial u}{\partial x_j}=\textrm{grad}\ u=\bigtriangledown u $$ $$
 * style="width:95%" |$$
 * style="width:95%" |$$
 * <p style="text-align:right"> $$ \displaystyle
 * }

<span id="(1)">
 * {| style="width:100%" border="0"

Then the K equation becomes: $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle
 * }

<span id="(1)">
 * {| style="width:100%" border="0"

\mathbb{K}=\int_{\Gamma _h}(n_i\ wT\frac{\partial u}{\partial x_j}\ \partial \Gamma _h)-\int_{\Omega }\bigtriangledown wT\bigtriangledown u\ d\Omega $$ $$
 * style="width:95%" |$$
 * style="width:95%" |$$
 * <p style="text-align:right"> $$ \displaystyle
 * }

<span id="(1)">
 * {| style="width:100%" border="0"

Otherwise written as: $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle
 * }

<span id="(1)">
 * {| style="width:100%" border="0"

\mathbb{K}=\int_{\Gamma _h}w\ \underline nT\cdot \textrm{grad \ u}\ \partial \Gamma _h-T\int_{\Omega }\bigtriangledown w\bigtriangledown u\ d\Omega $$ $$
 * style="width:95%" |$$
 * style="width:95%" |$$
 * <p style="text-align:right"> $$ \displaystyle
 * }

<span id="(1)">
 * {| style="width:100%" border="0"

where the last term is conveniently used to write ultimately the weak form of the weighted residual form which is the following: $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle
 * }

<span id="(1)">
 * {| style="width:100%" border="0"

\tilde{\textrm{f}}(w)=\tilde{m}(w,u)+\tilde{k}(w,u)\ \ \ \forall w \ \ \ s.t.\ \ \ w=0\ \ \ on\ \ \ \Gamma _g $$ $$
 * style="width:95%" |$$
 * style="width:95%" |$$
 * <p style="text-align:right"> $$ \displaystyle
 * }

<span id="(1)">
 * {| style="width:100%" border="0"

\tilde{m}(w,u)=\int_{\Omega } w\rho \frac{\partial ^2u}{\partial t^2}\ d\Omega $$ $$
 * style="width:95%" |$$
 * style="width:95%" |$$
 * <p style="text-align:right"> $$ \displaystyle
 * }

<span id="(1)">
 * {| style="width:100%" border="0"

\tilde{k}(w,u)=T\int_{\Omega }\bigtriangledown w\bigtriangledown u \ d\Omega $$ $$
 * style="width:95%" |$$
 * style="width:95%" |$$
 * <p style="text-align:right"> $$ \displaystyle
 * }

<span id="(1)">
 * {| style="width:100%" border="0"

\tilde{\textrm{f}}(w)=\int_{\Gamma _h}w \underline n \ T\ \textrm{grad u}\ \partial \Gamma _h+\int_{\Omega }w\textrm{f}\ d\Omega $$ $$
 * style="width:95%" |$$
 * style="width:95%" |$$
 * <p style="text-align:right"> $$ \displaystyle
 * }

<span id="(1)">
 * {| style="width:100%" border="0"

From the Discrete Weighted Form, $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle
 * }

<span id="(1)">
 * {| style="width:100%" border="0"

$$\underline M_{FF}\underline d_F+\underline K_{FF}\underline d_F=\underline F_F-(\underline M_{FE}g+\underline K_{FE}g)$$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle
 * }

<span id="(1)">
 * {| style="width:100%" border="0"

But because g=0 from the boundary conditions, as well as M due to the static requirement, the displacements may be derived: $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle
 * }

<span id="(1)">
 * {| style="width:100%" border="0"

$$\underline K \underline d=\underline F$$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle
 * }

<span id="(1)">
 * {| style="width:100%" border="0"

$$\therefore \underline d=\underline K^{-1}\underline F$$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle
 * }

<span id="(1)">
 * {| style="width:100%" border="0"

Now to solve for the displacements: $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle
 * }

<span id="(1)">
 * {| style="width:100%" border="0"

$$N_I(x,y)=L_{i,m}(x)\cdot L_{j,n}(y) $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle
 * }

<span id="(1)">
 * {| style="width:100%" border="0"

Where $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle
 * }

<span id="(1)">
 * {| style="width:100%" border="0"

$$I=i+(j-1)m $$ and $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle
 * }

<span id="(1)">
 * {| style="width:100%" border="0"

$$L_{i,m}(x)=\prod_{K=1, K\neq i}^{m}\left ( \frac{x-x_k}{x_i-x_k} \right ) $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle
 * }

<span id="(1)">
 * {| style="width:100%" border="0"

And setting the Shape Function to the following: $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle
 * }

<span id="(1)">
 * {| style="width:100%" border="0"

$$N=\begin{bmatrix} \frac{1}{4}(1-\xi _1)(1-\xi _2)\\ \frac{1}{4}(1+\xi _1)(1-\xi _2)\\ \frac{1}{4}(1-\xi _1)(1+\xi _2)\\ \frac{1}{4}(1-\xi _1)(1+\xi _2) \end{bmatrix} $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle
 * }

<span id="(1)">
 * {| style="width:100%" border="0"

The deformed shape coordinates may further be derived from these following two equations, where the summations are taken across all coordinate points taken from Abaqus output files: $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle
 * }

<span id="(1)">
 * {| style="width:100%" border="0"

$$X_{1}^{e}=\varphi_{1}^{e}(\xi )=\sum_{I=1}^{nel}N_I(\xi )x_{1,I}^{e} $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle
 * }

<span id="(1)">
 * {| style="width:100%" border="0"

$$(y=)X_{2}^{e}=\varphi_{2}^{e}(\xi )=\sum_{I=1}^{nel}N_I(\xi )x_{2,I}^{e} $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle
 * }

<span id="(1)">
 * {| style="width:100%" border="0"

Using the following Jacobian matrix: $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle
 * }

<span id="(1)">
 * {| style="width:100%" border="0"

$$J=\left [ \frac{\partial x_i}{\partial \xi _j} \right ]=\begin{bmatrix} \frac{\partial x_1}{\partial \xi _1} & \frac{\partial x_1}{\partial \xi _2}\\ \frac{\partial x_2}{\partial \xi _1} & \frac{\partial x_2}{\partial \xi _2}\\ \end{bmatrix} $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle
 * }

<span id="(1)">
 * {| style="width:100%" border="0"

The following equations may now be utilized: $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle
 * }

<span id="(1)">
 * {| style="width:100%" border="0"

$$\underline B_I(\xi )=\bigtriangledown _xN_{I}^{e}(\xi )=(\underline J^e)^{-T}(\xi )\bigtriangledown_\xi N_{I}^{e}(\xi ) $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle
 * }

<span id="(1)">
 * {| style="width:100%" border="0"

$$k_{I3}^{e}=\int_{\omega =el}N_I(\xi )\cdot T\cdot \underline B_J (\xi )J(\xi )d(el) $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle
 * }

<span id="(1)">
 * {| style="width:100%" border="0"

$$F_i=\int_{\omega =el}N_I(\xi )f\ det(J(\xi ))d(el) $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle
 * }

<span id="(1)">
 * {| style="width:100%" border="0"

Implementing MATLAB at this point, in conjunction with output data files from Abaqus which contain element coordinate points for three separate quadrilateral meshes and three separate triangular meshes (included), the final displacements may be derived, to an accuracy of 10^-6: $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle
 * }

<span id="(1)">
 * {| style="width:100%" border="0"

The above MATLAB script for the 6 different meshes generate the following results; the first three show quadrilateral elements in order of increasing elements. The last three show triangular elements in order of increasing elements. $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle
 * }

The figure to the left shows a unit-less displacement of: 0.0623258

The figure to the left shows a unit-less displacement of: 0.0624613

The figure to the left shows a unit-less displacement of: 0.0625011

The figure to the left shows a unit-less displacement of: 0.0624613

The figure to the left shows a unit-less displacement of: 0.0639197

The figure to the left shows a unit-less displacement of: 0.0619208

The figure to the left shows a unit-less displacement of: 0.0621967

Comparision With ABAQUS
The problem can be converted to an equivalent Heat problem with the following assumptions:

Conductance K=4,

Temperature T on the boundary = 0

Heat generation f=1

The equivalent model was analyzed in ABAQUS and the following result was obtained

The Temperature at the center turned out to be 6.25e-02 which is equal to what we got using MATLAB.



--Eml5526.s11.team5.vijay 21:17, 21 April 2011 (UTC)

--Eml5526.s11.team5.mcdaniel 21:23, 21 April 2011 (UTC)

= 7.4 : Newton's Law of Cooling http://en.wikiversity.org/w/index.php?title=File:Fe1.s11.mtg41.djvu&page=5 =

Given
Data Set G2DM2.0/D1

$$ \Omega = Domain \quad $$

$$ K=I, f(x,t)=0$$

$$ \frac {\partial u}{\partial t} = 0, ~g=2, ~h=3$$

$$ H=.5 ~u_\infty =.1$$

$$ \partial \Omega = \Gamma_g \bigcup \Gamma_h \bigcup \Gamma_H $$

and, The Partial Differential Equation is:

Find
1) The weak form for heat conduction in 2-D with boundary convection. 2) Develop semi-discrete equations (ODEs). Give detailed expression of all quantities. 3) Solve G2DM2.0/D1 using 2D LIBF till $$\displaystyle 10^{-6} $$ accuracy at the center. Plot solution in 3D with contour.

1.The Weak Form for heat conduction in 2D with boundary convection
From the PDE (Eqn 6.10.1) the weak form can be written as,

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Using integration by parts,

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Substituting into the integral we obtain,

<span id="(1)">
 * {| style="width:100%" border="0"


 * style="width:95%" |$$
 * style="width:95%" |$$

\int\limits_{\text{ }\!\!\Omega\!\!\text{ }}{\frac{\partial }{\partial {{x}_{i}}}\left( w{{K}_{ij}}\frac{\partial u}{\partial {{x}_{j}}} \right)}d\text{ }\!\!\Omega\!\!\text{ }-\int\limits_{\text{ }\!\!\Omega\!\!\text{ }}{\frac{\partial w}{\partial {{x}_{i}}}\left( {{K}_{ij}}\frac{\partial u}{\partial {{x}_{j}}} \right)}d\text{ }\!\!\Omega\!\!\text{ +}\int\limits_{\text{ }\!\!\Omega\!\!\text{ }}{wf\left( x,t \right)}d\text{ }\!\!\Omega\!\!\text{ }-\int\limits_{\text{ }\!\!\Omega\!\!\text{ }}{w\rho c\frac{\partial u}{\partial t}}d\text{ }\!\!\Omega\!\!\text{ }=0 $$ $$
 * <p style="text-align:right"> $$ \displaystyle (Eq. 7.10.3)
 * }

Applying the Gauss Theorem on the first term gives,

<span id="(1)">
 * {| style="width:100%" border="0"

\int\limits_{\text{ }\!\!\Omega\!\!\text{ }}w{{K}_{ij}}\frac{\partial u}{\partial{{x}_{j}}}{{n}_{i}}d\text{ }\!\!\Gamma\!\!\text{ }-\int\limits_{\text{ }\!\!\Omega\!\!\text{ }}{\frac{\partial w}{\partial {{x}_{i}}}\left( {{K}_{ij}}\frac{\partial u}{\partial {{x}_{j}}} \right)}d\text{ }\!\!\Omega\!\!\text{ +}\int\limits_{\text{ }\!\!\Omega\!\!\text{ }}{wf\left( x,t \right)}d\text{ }\!\!\Omega\!\!\text{ }-\int\limits_{\text{ }\!\!\Omega\!\!\text{ }}{w\rho c\frac{\partial u}{\partial t}}d\text{ }\!\!\Omega\!\!\text{ }=0 $$ $$
 * style="width:95%" |$$
 * style="width:95%" |$$
 * <p style="text-align:right"> $$ \displaystyle (Eq. 7.10.4)
 * }

The Heat Flux is given by,

<span id="(1)">
 * {| style="width:100%" border="0"

q_{i}=-K_{ij}\frac{\partial u}{\partial x_{j}} $$ $$
 * style="width:95%" |$$
 * style="width:95%" |$$
 * <p style="text-align:right"> $$ \displaystyle
 * }

Substituting the heat flux and rearranging the terms in Eq. 7.10.4 gives,

<span id="(1)">
 * {| style="width:100%" border="0"

\int\limits_{\text{ }\!\!\Omega\!\!\text{ }}{-w{{q}_{i}}{n}_{i}}d\text{ }\!\!\Gamma\!\!\text{ }-\int\limits_{\text{ }\!\!\Omega\!\!\text{ }}{\frac{\partial w}{\partial {{x}_{i}}}\left( {{K}_{ij}}\frac{\partial u}{\partial {{x}_{j}}} \right)}d\text{ }\!\!\Omega\!\!\text{ +}\int\limits_{\text{ }\!\!\Omega\!\!\text{ }}{wf\left( x,t \right)}d\text{ }\!\!\Omega\!\!\text{ }-\int\limits_{\text{ }\!\!\Omega\!\!\text{ }}{w\rho c\frac{\partial u}{\partial t}}d\text{ }\!\!\Omega\!\!\text{ }=0 $$ $$
 * style="width:95%" |$$
 * style="width:95%" |$$
 * <p style="text-align:right"> $$ \displaystyle (Eq. 7.10.5)
 * }

The first term can be rewritten as,

<span id="(1)">
 * {| style="width:100%" border="0"

\int\limits_{\Gamma }{-w{{q}_{i}}{{n}_{i}}d\Gamma =}\int\limits_{-w{{q}_{i}}{{n}_{i}}d{{\Gamma }_{g}}}-\int\limits_{-w{{q}_{i}}{{n}_{i}}d{{\Gamma }_{h}}}-\int\limits_{-w{{q}_{i}}{{n}_{i}}d{{\Gamma }_{H}}} $$ $$
 * style="width:95%" |$$
 * style="width:95%" |$$
 * <p style="text-align:right"> $$ \displaystyle\!
 * }

Select $$ \quad w $$ such that $$\quad w=0 $$ on $$\quad {{\Gamma }_{g}}$$

Organizing, Eq. 7.10.5 then becomes the following weak form

<span id="(1)">
 * {| style="width:100%" border="0"

\int\limits_{\text{ }\!\!\Omega\!\!\text{ }}{w\rho c\frac{\partial u}{\partial t}d\text{ }\!\!\Omega\!\!\text{ +}}\int\limits_{\text{ }\!\!\Omega\!\!\text{ }}{\frac{\partial w}{\partial {{x}_{i}}}\left( {{K}_{ij}}\frac{\partial u}{\partial {{x}_{j}}} \right)d\text{ }\!\!\Omega\!\!\text{ }}+\int\limits_{wHu}d{{\text{ }\!\!\Gamma\!\!\text{ }}_{H}}=-\int\limits_{whd{{\text{ }\!\!\Gamma\!\!\text{ }}_{h}}}+\int\limits_{wH{{u}_{\infty }}d{{\text{ }\!\!\Gamma\!\!\text{ }}_{H}}}+\int\limits_{\Omega }{wf\left( x,t \right)d\text{ }\!\!\Omega\!\!\text{ }},$$ $$ $$ \quad \forall w $$ such that $$\displaystyle w=0 \quad on \quad {{\Gamma }_{g}} $$
 * style="width:95%" |$$
 * style="width:95%" |$$
 * <p style="text-align:right"> $$ \displaystyle (Eq.7.10.6)
 * }

2) Develop Semidiscrete Equations (ODE's)
Let $$ u=[ N]{{u}_{e}}\quad $$ and $$ \quad w=[ N ]{{w}_{e}},$$ where $$\displaystyle [N] $$ is the Lagrange polynomial and $${u}_{e}~ and ~ {w}_{e} $$ are the nodal values of $$ u \quad $$ and $$w \quad $$ respectively. Now, take the derivative of basis functions with respect to x1 and x2

Substituting the Langrangian form into the weak form (Eq 7.10.6), yields the semi-dicrete form:

From Eq. 6.10.8, we can obtain the Heat Source vector, Conductivity matrix, and Capacitance Matrix.

The heat source vector is $$\tilde{f}(w)= -\int\limits_{h{{[N]}^{T}}d{{\Gamma }_{h}}}+\int\limits_{H{{[N]}^{T}}{{u}_{\infty }}d{{\Gamma }_{h}}}+\int\limits_{\Omega }{{{[N]}^{T}}f(x,t)d\Omega } $$

The Conducitivity matrix is $$\tilde{k}(w,u)= \int\limits_{\Omega }{{{[B]}^{T}}[B]d\Omega }+\int\limits_{H{{[N]}^{T}}[N]d{{\Gamma }_{h}}} $$

The Capacitance Matrix is $$\tilde{m}(w,u)= \int\limits_{\Omega }{\rho c{{[N]}^{T}}[N]\frac{\partial u}{\partial t}d\Omega } $$

3.Solve G2DM2.0/D1 Using 2D LIBF till $$10^{-6}$$ accuracy at center. Plot Solution in 3D w/Contour.
To solve the problem using LIBF, divide the x-axis and Y-axis into 2 parts 2D LIBF

<span id="(1)">
 * {| style="width:100%" border="0"

$$ N_I (x,y) = L_{i,m}(x).L_{j,n}(y) \quad Where \ I = i+(j-1)m $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle\!
 * }

Where,

<span id="(1)">
 * {| style="width:100%" border="0"

$$ L_{i,m}(x) = \prod_{k=1\neq i}^{m}\frac{x-x_k}{x_i - x_k} $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle\!
 * }<span id="(1)">
 * {| style="width:100%" border="0"

$$ L_{j,n}(x) = \prod_{k=1\neq j}^{m}\frac{y-y_k}{y_j - y_k} $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle\!
 * }

Consider a case of 9 nodes as shown below



The trial solution is

The trial solution must satisfy the essential boundary condition. Throughout boundary $$ \Gamma_g, \quad $$ the temperature should be a constant 2. Thus $$ d_1, d_2, d_3, d_6, d_9 =2. \quad $$ --Eml5526.s11.team5.smith 21:22, 21 April 2011 (UTC)

=Contributing Members =