text:
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

$$
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 $$ \displaystyle\!
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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

$$
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 $$ \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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$$T div (grad u) + f = \rho \ \frac{\partial^2u}{\partial t^2} $$ $$
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 $$ \displaystyle                                                                     (Eq. 7.2.1)
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$$ \quad T = 4 $$ $$
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 $$ \displaystyle                                                                     (Eq. 7.2.2)
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$$ f (\mathbf{x}) = 1 $$ $$
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 $$ \displaystyle                                                                     (Eq. 7.2.3)
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$$ \quad \rho = 3 $$ $$
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 $$ \displaystyle                                                                     (Eq. 7.2.4)
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$$ \quad Basis \ Function \ : \ 2D \ Lagrange \ Interpolation \ Basis \ Function $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle\!
 * }

Essential Boundary Condition

 * {| style="width:100%" border="0"

$$ \quad g = 0 $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle                                                                     (Eq. 7.2.5)
 * }

Static(steady state)

 * {| style="width:100%" border="0"

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

Dynamic(transient)

 * {| style="width:100%" border="0"

1. Fundamental Frequency and Fundamental Period $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle\!
 * }

 * {| 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 $$ \quad u^0 $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle\!
 * }

 * {| style="width:100%" border="0"

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

Solution
Substituting the given values in the P.D.E, we get,
 * {| style="width:100%" border="0"

$$\quad 4 \ div (grad u) \ + \ 1 \ = \ 0 $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle\!
 * }

 * {| style="width:100%" border="0"

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

 * {| style="width:100%" border="0"

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

2D LIBF

 * {| style="width:100%" border="0"

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

Where,

 * {| style="width:100%" border="0"

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

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

Consider the Weak Form

 * {| style="width:100%" border="0"

$$  \tilde{m}(w,u) + \tilde{k}(w,u) = \tilde{f}(w) $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle                                                                     (Eq. 7.2.8)
 * }

where,

 * {| style="width:100%" border="0"

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

 * {| style="width:100%" border="0"

$$ k(w,u) = \int\limits_\Omega {\nabla wK\nabla u} d\Omega  $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle\!
 * }

 * {| style="width:100%" border="0"

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

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

 * {| style="width:100%" border="0"

$$ \quad K d = F $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle\!
 * }

where,

 * {| style="width:100%" border="0"

$$ {K_{ij}} = \int_ {(\nabla N_I^e)I(\nabla N_J^e)d{w^e}} $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \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,

 * {| style="width:100%" border="0"

$$ \mathbf{K} \mathbf{\phi} = \lambda \mathbf{M}\mathbf{\phi}  $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle                                                                     (Eq. 7.2.10)
 * }

Where,

 * {| style="width:100%" border="0"

$$\quad \mathbf{\phi} - Eigen \ vectors $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle\!
 * }

 * {| style="width:100%" border="0"

$$ \quad \lambda - Eigen \ values $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \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

 * {| style="width:100%" border="0"

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

 * {| style="width:100%" border="0"

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

 * {| style="width:100%" border="0"

$$T = 4, \qquad f(\mathbf{x},t) = 0 \ in \  \Omega  $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \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,

 * {| style="width:100%" border="0"

$$u(\mathbf{x},t=0) = (x+1)(y+\frac{1}{2}) cos (\frac{\pi}{2}x) cos (\frac{\pi}{2}y) \ \forall \  \mathbf{x} \ \epsilon  \ \Omega$$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \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

 * {| 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. $$
 *
 $$ \displaystyle
 * }

Solution: Static Solution of the Circle Using Quadrilateral and Triangular Elements to an Accuracy of 10^-6

 * {| style="width:100%" border="0"

The Partial Differential Equation for a vibrating membrane is the following via transient analysis: $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

 * {| style="width:100%" border="0"

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

 * {| 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) & $$\textrm{div \ grad\ u}=\frac{\partial }{\partial x_i}\frac{\partial u}{\partial x_j}$$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

 * {| style="width:100%" border="0"

The Weighted Residual Form is as follows: $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

 * {| 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%" |$$
 * style="width:95%" |$$
 *
 $$ \displaystyle
 * }

 * {| style="width:100%" border="0"

Which can thereby be broken into three separate terms: $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

 * {| style="width:100%" border="0"

\int_{\Omega } wT\frac{\partial ^2u}{\partial x_i\partial x_j}d\Omega =\mathbb{K} $$ $$
 * style="width:95%" |$$
 * style="width:95%" |$$
 *
 $$ \displaystyle
 * }

 * {| style="width:100%" border="0"

\int_{\Omega } w\textrm{f}\ d\Omega $$ $$
 * style="width:95%" |$$
 * style="width:95%" |$$
 *
 $$ \displaystyle
 * }

 * {| style="width:100%" border="0"

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

 * {| 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: $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

 * {| 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%" |$$
 * style="width:95%" |$$
 *
 $$ \displaystyle
 * }

 * {| style="width:100%" border="0"

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

 * {| 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%" |$$
 * style="width:95%" |$$
 *
 $$ \displaystyle
 * }

 * {| style="width:100%" border="0"

Now since $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

 * {| style="width:100%" border="0"

\partial \Omega =\Gamma _g\cup \Gamma _h\ \textrm{but }\ w=0\ \ \textrm{on}\ \ \Gamma _g $$ $$
 * style="width:95%" |$$
 * style="width:95%" |$$
 *
 $$ \displaystyle
 * }

 * {| style="width:100%" border="0"

and the following are true: $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

 * {| style="width:100%" border="0"

\frac{\partial w}{\partial x_i}=\textrm{grad}\ w=\bigtriangledown w $$ $$
 * style="width:95%" |$$
 * style="width:95%" |$$
 *
 $$ \displaystyle
 * }

 * {| style="width:100%" border="0"

\frac{\partial u}{\partial x_j}=\textrm{grad}\ u=\bigtriangledown u $$ $$
 * style="width:95%" |$$
 * style="width:95%" |$$
 *
 $$ \displaystyle
 * }

 * {| style="width:100%" border="0"

Then the K equation becomes: $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

 * {| 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%" |$$
 * style="width:95%" |$$
 *
 $$ \displaystyle
 * }

 * {| style="width:100%" border="0"

Otherwise written as: $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

 * {| 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%" |$$
 * style="width:95%" |$$
 *
 $$ \displaystyle
 * }

 * {| 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: $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

 * {| 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%" |$$
 * style="width:95%" |$$
 *
 $$ \displaystyle
 * }

 * {| style="width:100%" border="0"

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

 * {| style="width:100%" border="0"

\tilde{k}(w,u)=T\int_{\Omega }\bigtriangledown w\bigtriangledown u \ d\Omega $$ $$
 * style="width:95%" |$$
 * style="width:95%" |$$
 *
 $$ \displaystyle
 * }

 * {| 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%" |$$
 * style="width:95%" |$$
 *
 $$ \displaystyle
 * }

 * {| style="width:100%" border="0"

From the Discrete Weighted Form, $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

 * {| style="width:100%" border="0"

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

 * {| 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: $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

 * {| style="width:100%" border="0"

$$\underline K \underline d=\underline F$$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

 * {| style="width:100%" border="0"

$$\therefore \underline d=\underline K^{-1}\underline F$$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

 * {| style="width:100%" border="0"

Now to solve for the displacements: $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

 * {| style="width:100%" border="0"

$$N_I(x,y)=L_{i,m}(x)\cdot L_{j,n}(y) $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

 * {| style="width:100%" border="0"

Where $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

 * {| style="width:100%" border="0"

$$I=i+(j-1)m $$ and $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

 * {| style="width:100%" border="0"

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

 * {| style="width:100%" border="0"

And setting the Shape Function to the following: $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

 * {| style="width:100%" border="0"

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

 * {| 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: $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

 * {| style="width:100%" border="0"

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

 * {| style="width:100%" border="0"

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

 * {| style="width:100%" border="0"

Using the following Jacobian matrix: $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

 * {| style="width:100%" border="0"

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

 * {| style="width:100%" border="0"

The following equations may now be utilized: $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

 * {| style="width:100%" border="0"

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

 * {| style="width:100%" border="0"

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

 * {| style="width:100%" border="0"

$$F_i=\int_{\omega =el}N_I(\xi )f\ det(J(\xi ))d(el) $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

 * {| 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: $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

 * {| 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. $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \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,

Using integration by parts,

Substituting into the integral we obtain,

 * {| 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 $$ $$
 *
 $$ \displaystyle (Eq. 7.10.3)
 * }

Applying the Gauss Theorem on the first term gives,

 * {| 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%" |$$
 * style="width:95%" |$$
 *
 $$ \displaystyle (Eq. 7.10.4)
 * }

The Heat Flux is given by,

 * {| style="width:100%" border="0"

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

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

 * {| 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%" |$$
 * style="width:95%" |$$
 *
 $$ \displaystyle (Eq. 7.10.5)
 * }

The first term can be rewritten as,

 * {| 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%" |$$
 * style="width:95%" |$$
 *
 $$ \displaystyle\!
 * }

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

Organizing, Eq. 7.10.5 then becomes the following weak form

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

 * {| style="width:100%" border="0"

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

Where,

 * {| style="width:100%" border="0"

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

$$ L_{j,n}(x) = \prod_{k=1\neq j}^{m}\frac{y-y_k}{y_j - y_k} $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \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 =

title: User:Eml5526.s11.team5/sub7


text: User talk:Cs4008

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/

title: User talk:Cs4008


text:
BoM Product Structure

Students will analyse BoMs, derive various types of BoMs (Single Level, Indented and Summarised) and learn about implosions and explosions. They will learn key phrases and definitions used throughout the BoM process.

What is the Bill of Materials (BoM)?
A Bill of Material is an inclusive definition of a final product that includes a list of the items, ingredients and/or materials and the quantities of each needed to manufacture, assemble or mix an end product.

A BoM for a house would include the following sub-assemblies: cement block, lumber, shingles, doors, windows, plumbing, electric, heating etc. Each subassembly also contains a BoM; the heating system is made up of the furnace, ducts etc.

There are several different forms of the BoM, which include:

 * Single Level Bill of Material (Figure 1)
 * Indented Bill of Material (Figure 2)
 * Summarised Bill of Material (Figure 3)

These are basic examples, a BoM generally contains more information, such as, an item's effective date, item number, description etc.

A summarised BoM is useful as it shows the total usage of each item needed for the product in a single list, but results in the duplication of the product's assemblies. For every product BoM that uses a duplicated assembly, it must be changed whenever there is a change in the assembly.

The indented BoM is useful as it helps one to understand the structure of the product, but all the parts belonging to an assembly are repeated(look at B, E and F). This can be a problem: lets imagine that the BoM is much bigger than the one provided in Figure 2, it will lead to a massive duplication of data.

As a solution to the duplication problem, each assembly is held once in a ‘single level’ BoM. In Figure 1, level one shows the single level BoM of Product 1, it then has pointers to each single level BoM of its assemblies. This allows changes to an assembly to occur in one place. A single level BoM (Level 1 of Figure 1) is sufficient when a product is assembled at one time from a set of purchased parts and raw materials, but it does not adequately describe a product that has sub assemblies. The BoM is also known as the Product Structure, and the Product Structure helps to describe a product with sub assemblies. Although, the Product Structure does not specify the quantity of items needed for a product, therefore, the Bill of Materials (BoM) and the BoM Product Structure are used together.

BoM Product Structure
To illustrate the Product Structure; a tree like structure can be drawn (Figure (4 to be provided)). The Product Structure provides a hierarchical classification of the items which form a product. Items can take many forms: raw materials; intermediate items; items; assemblies; parts; components.

Raw materials appear at the bottom of the structure. In the example opposite; E, F, D and C are raw materials. B is a component of A, Intermediate Items are those in between. There are 2 intermediate items in the example; B and A. The Product Structure helps to display and describe the relationships between one item and another, it does not specify the sequence of assembly. The assembly sequence is important to determine what operations should be performed by a worker, and what kind of work stations are needed. So, the assembly sequence is provided in an assembly chart.

BoM Implosion and Explosion
A BoM explosion (Figure 5) helps to break an assembly or sub-assembly into its components or raw materials. For example, an engineer needs to see what is required to construct Assembly A, using the explosion, they can see a list of all Assembly A's components.

A BoM implosion (Figure 6) helps to link a component to an assembly, sub-assembly or product. For example, an engineer needs to see where Raw Material E is used, using the implosion, they can see the range of assemblies, sub-assemblies or products where Raw Material E is used.

Where is the BoM used?

The BoM is stored in a database and it is used in Materials Requirement Planning systems (Figure 7) in conjunction with the MPS and Inventory file to determine exactly how many components are needed in order to produce the quantity of finished products stated in a MPS and to determine the items for which purchase requisitions and production orders must be released. MRP systems help to determine: what to order; how much to order; when to order; when to schedule delivery. They also help to plan and control inventory, support detailed capacity planning and support priority planning on the shop floor.

Accounting use the BoM to cost the product and it is also used for kitting production orders, order entry and engineering purposes.

Test yourself
Consider a plug.

The Product Structure for the plug is as follows:

1. Create a single level explosion for the 13 AMP Square Pin Plug

2. Pick the odd one out: Negative Pin Assembly, Earth Pin Assembly, Final Assembly Cover Screw or Fuse Holder Assembly

3. List the Components of 'Positive Lead Hold Assembly'

4. List all Raw Materials

5. True or False:
 * Assembly has many components
 * Assembly has many sub-assemblies
 * Part has many Assemblies
 * Part has many components

6. Where do raw materials tend to appear in the Product Structure?

7. What is meant by each of the following:
 * a) A single-level explosion?
 * b) A indented implosion?
 * c) A summarised explosion?

8. For the Product Structure below, derive the:
 * a) Summarised BoM
 * b) Indented BoM
 * c) Single-level BoM

Test yourself - Answers
Provide answers to the questions!

1. 13 AMP Square Pin Plug
 * Wire Retainer Screws
 * Negative Pin Assembly
 * 13 amp Top Cover
 * 13 amp Bottom Cover
 * Earth Pin Assembly
 * Final Assembly Cover Screw
 * Fuse Holder Assembly
 * Retainer

2. Final Assembly Cover Screw because it is a raw material. All the others as assemblies.

3. Positive Lead/Fuse Holder Assembly, Screw

4. There are 14 Raw Materials used; Positive Lead Holder, Fuse Holder, Screw (3 per), Positive Pin, 13 amp Fuse, Earth Pin, Negative Pin, Retainer, Final Assembly Cover Screw, 13 amp Bottom Cover, 13 amp Top Cover, Wire Retainer Screws.

5. a) True
 * b) True
 * c) True
 * d) True

6. They tend to be children

7. a) A single-level explosion: A list of what components are needed to build the Parent item.
 * b) An indented implosion: A listing of all components used in an assembly. There are not totals provided. When each component is used it is repeated..
 * c) Summarised explosion: The total quantity of each component required to build the Parent part. The main difference from the single-level explosion is the need to aggregate the quantities. The first example indicates the quantity of sub-parts required for the part whenever it is required. It does not indicate how many of the subparts are needed to build the Parent item.

8.
 * a) Summarised BoM

 * b) Indented BoM

 * c) Single Level BoM

Glossary
Assembly : A group of components which are put together in manufacturing (e.g. cooking).

Component : A raw material, ingredient, part or sub assembly, but not a product, that is used in the manufacture of a higher-level assembly, compound, or other item.

Cumulative lead time : the longest time it takes to traverse the Bill of Materials (BOM) from the leaf back to the root ; it is the time it takes to accomplish the activity the BOM represents.

Intermediate Item : Any material which is neither a product nor a raw material but is consumed in the making of products.

Item : Any unique manufactured or purchased part or assembly. May be a product, raw material, intermediate item, but not in process.

MPS : A Master Production Schedule is a plan for production, staffing, inventory, etc.

MRP : Material Requirements Planning, a type of application software system used to schedule and monitor the use of components and other materials in a manufacturing operation.

Part : Any one-piece item which may be manufactured or purchased. Typically manufactured at Fabrication.

Raw Material : The material which is bought from a supplier and used directly or indirectly in producing a product.

Resource : Any asset which is scarce or limited and therefore needs to be planned and managed(Materials, Human Resources, production facilities).

title: BoM Product Structure


text: File:Hw7 p2 team 1.png

title: File:Hw7 p2 team 1.png


text: File:Hw7 p2 team 1 1.png

title: File:Hw7 p2 team 1 1.png


text:
User:SimplyIrresistible

I am Simply / Irresistible

My other user pages:

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simple:User:SimplyIrresistible

title: User:SimplyIrresistible


text:
User talk:SimplyIrresistible

SimplyIrresistible 04:56, 22 April 2011 (UTC)

title: User talk:SimplyIrresistible


text:
User:Alkox

HÊĹĹö Wikiversity, I am an Undergrad and a learner. Learning is my hobby....... And I am really Thankful to the Wikiversity for Promoting Learning by providing the Contents free......

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text: Fluid mechanics for MAP/to do

title: Fluid mechanics for MAP/to do


text:
Template:Fluid mechanics for MAP/to do

To edit this page go to this template page

 * Rework the figures of the fluid statics
 * Rework the figures of the fluid_dynamics
 * Add an application section at the end of each chapter

title: Template:Fluid mechanics for MAP/to do


text: File:Le Moigne model.png

title: File:Le Moigne model.png


text:
User:Bobthesob

Hi! I'm a guy in the mid twenties from Sweden. Currently I'm studying social science and development studies at the university in Uppsala, Sweden. I have a big interest in open source and projects like that.

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title: User:Bobthesob


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text: File:University of Westminster logo.JPG

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text:
Hinduism/Bhagavad Gita

The Shrimad Bhagavad Gita ('The Song by God'), often referred to as the Gita, is a 700-verse Hindu scripture that is part of the epic Mahabharata (chapters 23–40 of Bhishma Parva), dated to the second half of the first millennium BCE and exemplary for the emerging Hindu synthesis. It is considered to be one of the holy scriptures for Hinduism.

title: Hinduism/Bhagavad Gita


text: User:Leutha/sandbox

title: User:Leutha/sandbox


text: File:University of Surrey Logo (RGB 281 - 457).jpg


text: Template:University of Surrey

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User:Bjung

Hello. I'm mainly active on Wikipedia.

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User talk:Bobthesob

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You don't need to be an educator to edit. You only need to be bold to contribute and to experiment with the sandbox or your userpage. See you around Wikiversity! Jtneill - Talk - c 21:57, 22 April 2011 (UTC)

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text: Category:Intercultural competence

title: Category:Intercultural competence


text: File:Left yin-yang.png

title: File:Left yin-yang.png


text:
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User talk:Marie.sustayta

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User talk:Pranay.suhas93

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User talk:Brmejia

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You don't need to be an educator to edit. You only need to be bold to contribute and to experiment with the sandbox or your userpage. See you around Wikiversity! --Leutha 08:52, 23 April 2011 (UTC)

title: User talk:Brmejia


text:
User talk:Jjvdl

 Hello Jjvdl, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. After you leave a comment on a talk page, remember to sign and date; it helps everyone follow the threads of the discussion. The signature icon in the edit window makes it simple. All users are expected to abide by our Privacy policy, Civility policy, and the Terms of Use while at Wikiversity.

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You don't need to be an educator to edit. You only need to be bold to contribute and to experiment with the sandbox or your userpage. See you around Wikiversity! --Leutha 08:53, 23 April 2011 (UTC)

title: User talk:Jjvdl


text:
User talk:Vmj

 Hello Vmj, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. After you leave a comment on a talk page, remember to sign and date; it helps everyone follow the threads of the discussion. The signature icon in the edit window makes it simple. All users are expected to abide by our Privacy policy, Civility policy, and the Terms of Use while at Wikiversity.

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You don't need to be an educator to edit. You only need to be bold to contribute and to experiment with the sandbox or your userpage. See you around Wikiversity! --Leutha 08:53, 23 April 2011 (UTC)

title: User talk:Vmj


text:
User talk:Ashishasitis

 Hello Ashishasitis, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. After you leave a comment on a talk page, remember to sign and date; it helps everyone follow the threads of the discussion. The signature icon in the edit window makes it simple. All users are expected to abide by our Privacy policy, Civility policy, and the Terms of Use while at Wikiversity.

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You don't need to be an educator to edit. You only need to be bold to contribute and to experiment with the sandbox or your userpage. See you around Wikiversity! --Leutha 08:53, 23 April 2011 (UTC)

title: User talk:Ashishasitis


text:
User talk:Katrina~enwikiversity

 Hello Katrina, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. After you leave a comment on a talk page, remember to sign and date; it helps everyone follow the threads of the discussion. The signature icon in the edit window makes it simple. All users are expected to abide by our Privacy policy, Civility policy, and the Terms of Use while at Wikiversity.

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You don't need to be an educator to edit. You only need to be bold to contribute and to experiment with the sandbox or your userpage. See you around Wikiversity! --Leutha 08:54, 23 April 2011 (UTC)

Your account will be renamed
Hello,

The developer team at Wikimedia is making some changes to how accounts work, as part of our on-going efforts to provide new and better tools for our users like cross-wiki notifications. These changes will mean you have the same account name everywhere. This will let us give you new features that will help you edit and discuss better, and allow more flexible user permissions for tools. One of the side-effects of this is that user accounts will now have to be unique across all 900 Wikimedia wikis. See the announcement for more information.

Unfortunately, your account clashes with another account also called Katrina. To make sure that both of you can use all Wikimedia projects in future, we have reserved the name Katrina~enwikiversity that only you will have. If you like it, you don't have to do anything. If you do not like it, you can pick out a different name.

Your account will still work as before, and you will be credited for all your edits made so far, but you will have to use the new account name when you log in.

Sorry for the inconvenience.

Yours, Keegan Peterzell Community Liaison, Wikimedia Foundation 23:41, 17 March 2015 (UTC)

Renamed
 This account has been renamed as part of single-user login finalisation. If you own this account you can |log in using your previous username and password for more information. If you do not like this account's new name, you can choose your own using this form after logging in: . -- Keegan (WMF) (talk) 06:13, 19 April 2015 (UTC)

title: User talk:Katrina~enwikiversity


text:
User talk:Spainxxiv

 Hello Spainxxiv, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. After you leave a comment on a talk page, remember to sign and date; it helps everyone follow the threads of the discussion. The signature icon in the edit window makes it simple. All users are expected to abide by our Privacy policy, Civility policy, and the Terms of Use while at Wikiversity.

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You don't need to be an educator to edit. You only need to be bold to contribute and to experiment with the sandbox or your userpage. See you around Wikiversity! --Leutha 08:54, 23 April 2011 (UTC)

title: User talk:Spainxxiv


text:
Wikiversity talk:Community Review/Custodianship process

Ideas for resolutions
Just a few thoughts that have been bouncing around my brain a while:
 * 1. make it a separate usergroup that's just like custodian but can be removed locally if there are any concerns, if the user goes on break during the mentorship period, etc. This would increase the "no big deal" factor, as well as give us more freedom to experiment with giving the tools temporarily to course instructors, etc.

 * 2. Require consent of at least 2 crats before starting a mentorship.

 * 3. If someone is unable to find a willing mentor or there are not 2 crats agreeing to promote, allow the person to use an "rfa" to request self-mentoring.

 * 4. formalize mentorship a bit, perhaps requiring a log with comments on a public page where the mentored custodian can give us their thinking behind each tool usage.

It would be nice to get this squared away. --SB_Johnny talk 10:09, 23 April 2011 (UTC)
 * While the separate usergroup idea has merit, i.e., locally removable privileges, presumably removable by any 'crat, it's not clear to me that the incidence of any necessity for this is great enough to justify the change, given that it is easy to get a steward to lift a bit if there is good cause. I've proposed a Candidates for Custodianship/Standard stop agreement to make it simple to stop any problematic actions by a probationary custodian, and to make it easy to have a steward be shown some diffs, and remove the tools. Even absent that, there is no example of a rogue probationary custodian where it was difficult to stop the problem. While I'm aware that some might claim I was so, I'd given permission for SB_Johnny to yank the tools at any time, and I'd also stated I'd stop what I was doing if asked by any custodian, and, in fact, I was begging for additional custodial attention.

 * What's the problem that the fixes above are intended to fix? What went wrong? --Abd 02:58, 8 June 2011 (UTC)

 * The problem arises when there is a mentorship that members of the community might be concerned about. There should be some checks and balances to prevent abuse. For example, we've had a few cases over the years where custodians have been demoted by community processes: should they just be able to become custodians again if one particular 'crat decides to mentor and/or promote? It's a bad loophole, period. --SB_Johnny talk 12:54, 8 June 2011 (UTC)
 * No example has been shown of damage from such a "loophole." Rather, one disgruntled desysopped permanent custodian complained loudly when SB_Johnny followed the policy, which that custodian had previously relied upon, and brought in SPAs to support this axe-grinding. "Concern" does not equal "abuse" or "loophole." This discussion could spin out into a dirty-laundry fest, do we need to look back at what happened, and the role that this or that editor or bureaucrat played in it? Or can we move ahead, for the welfare of Wikiversity? I'm hoping for the latter possibility.
 * The existing policy allows users to readily obtain and work with the tools, and there is no doubt but that security can be increased (I proposed and agreed to the stop agreement as an example of how to do this), but the policy, as-is, is not broken and change to it should be incremental. Making the process complicated and difficult is not a sound approach. --Abd 14:23, 8 June 2011 (UTC)


text:
User talk:Laughingforever

 Hello Laughingforever, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. After you leave a comment on a talk page, remember to sign and date; it helps everyone follow the threads of the discussion. The signature icon in the edit window makes it simple. All users are expected to abide by our Privacy policy, Civility policy, and the Terms of Use while at Wikiversity.

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You don't need to be an educator to edit. You only need to be bold to contribute and to experiment with the sandbox or your userpage. See you around Wikiversity! --Leutha 10:46, 23 April 2011 (UTC)

title: User talk:Laughingforever


text:
User talk:King mak

 Hello King mak, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. After you leave a comment on a talk page, remember to sign and date; it helps everyone follow the threads of the discussion. The signature icon in the edit window makes it simple. All users are expected to abide by our Privacy policy, Civility policy, and the Terms of Use while at Wikiversity.

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You don't need to be an educator to edit. You only need to be bold to contribute and to experiment with the sandbox or your userpage. See you around Wikiversity! --Leutha 10:46, 23 April 2011 (UTC)

title: User talk:King mak


text:
User talk:Somaiawael

 Hello Somaiawael, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. After you leave a comment on a talk page, remember to sign and date; it helps everyone follow the threads of the discussion. The signature icon in the edit window makes it simple. All users are expected to abide by our Privacy policy, Civility policy, and the Terms of Use while at Wikiversity.

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You don't need to be an educator to edit. You only need to be bold to contribute and to experiment with the sandbox or your userpage. See you around Wikiversity! --Leutha 10:46, 23 April 2011 (UTC)

title: User talk:Somaiawael


text:
User talk:Kalash~enwikiversity

 Hello Kalash, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. After you leave a comment on a talk page, remember to sign and date; it helps everyone follow the threads of the discussion. The signature icon in the edit window makes it simple. All users are expected to abide by our Privacy policy, Civility policy, and the Terms of Use while at Wikiversity.

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You don't need to be an educator to edit. You only need to be bold to contribute and to experiment with the sandbox or your userpage. See you around Wikiversity! --Leutha 10:47, 23 April 2011 (UTC)

Your account will be renamed
Hello,

The developer team at Wikimedia is making some changes to how accounts work, as part of our on-going efforts to provide new and better tools for our users like cross-wiki notifications. These changes will mean you have the same account name everywhere. This will let us give you new features that will help you edit and discuss better, and allow more flexible user permissions for tools. One of the side-effects of this is that user accounts will now have to be unique across all 900 Wikimedia wikis. See the announcement for more information.

Unfortunately, your account clashes with another account also called Kalash. To make sure that both of you can use all Wikimedia projects in future, we have reserved the name Kalash~enwikiversity that only you will have. If you like it, you don't have to do anything. If you do not like it, you can pick out a different name.

Your account will still work as before, and you will be credited for all your edits made so far, but you will have to use the new account name when you log in.

Sorry for the inconvenience.

Yours, Keegan Peterzell Community Liaison, Wikimedia Foundation 23:40, 17 March 2015 (UTC)

Renamed
 This account has been renamed as part of single-user login finalisation. If you own this account you can |log in using your previous username and password for more information. If you do not like this account's new name, you can choose your own using this form after logging in: . -- Keegan (WMF) (talk) 06:12, 19 April 2015 (UTC)

title: User talk:Kalash~enwikiversity


text:
User talk:Zeyathura~enwikiversity

 Hello Zeyathura, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. After you leave a comment on a talk page, remember to sign and date; it helps everyone follow the threads of the discussion. The signature icon in the edit window makes it simple. All users are expected to abide by our Privacy policy, Civility policy, and the Terms of Use while at Wikiversity.

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You don't need to be an educator to edit. You only need to be bold to contribute and to experiment with the sandbox or your userpage. See you around Wikiversity! --Leutha 10:47, 23 April 2011 (UTC)

Your account will be renamed
Hello,

The developer team at Wikimedia is making some changes to how accounts work, as part of our on-going efforts to provide new and better tools for our users like cross-wiki notifications. These changes will mean you have the same account name everywhere. This will let us give you new features that will help you edit and discuss better, and allow more flexible user permissions for tools. One of the side-effects of this is that user accounts will now have to be unique across all 900 Wikimedia wikis. See the announcement for more information.

Unfortunately, your account clashes with another account also called Zeyathura. To make sure that both of you can use all Wikimedia projects in future, we have reserved the name Zeyathura~enwikiversity that only you will have. If you like it, you don't have to do anything. If you do not like it, you can pick out a different name.

Your account will still work as before, and you will be credited for all your edits made so far, but you will have to use the new account name when you log in.

Sorry for the inconvenience.

Yours, Keegan Peterzell Community Liaison, Wikimedia Foundation 23:48, 17 March 2015 (UTC)

Renamed
 This account has been renamed as part of single-user login finalisation. If you own this account you can |log in using your previous username and password for more information. If you do not like this account's new name, you can choose your own using this form after logging in: . -- Keegan (WMF) (talk) 06:25, 19 April 2015 (UTC)

title: User talk:Zeyathura~enwikiversity


text: Template:Infobox ASGELE

title: Template:Infobox ASGELE


text: Template:Infobox ASGELE/doc

title: Template:Infobox ASGELE/doc


text: File:Red Herrings (Simulation).Introduction.pdf


text: File:Red Herrings (Simulation).Fishermen.pdf


text: File:Red Herrings (Simulation).Regulators.pdf


text: File:Red Herrings (Simulation).Scientists.pdf


text:
Talk:Forbidden Kingdom (strategic simulation)

As anyone played? Does anyone know how long it takes to play?Leutha 15:35, 23 April 2011 (UTC)

We played the game two or three times at Mercyhurst as part of my school project. We allocated an hour and a half per game. It takes about 20 minutes per round because you have to meet with the other players sort of like Diplomacy. As I recall the game lasted maybe 4 rounds because people tended to become competitive rather than cooperative or submissive, and that leads to failure in this game. Pnoble805 (talk) 17:44, 3 March 2012 (UTC)


text:
User talk:Acab852

 Hello Acab852, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. After you leave a comment on a talk page, remember to sign and date; it helps everyone follow the threads of the discussion. The signature icon in the edit window makes it simple. All users are expected to abide by our Privacy policy, Civility policy, and the Terms of Use while at Wikiversity.

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You don't need to be an educator to edit. You only need to be bold to contribute and to experiment with the sandbox or your userpage. See you around Wikiversity! --Leutha 16:12, 23 April 2011 (UTC)

title: User talk:Acab852