text:
User:Eml5526.s11.team3.ushnish12/Homework 4

For each of the family of basis functions, and for $$\displaystyle{{\Gamma }_{g}}=\{\beta \}$$, find the corresponding basis functions satisfying

{ $$\displaystyle {{b}_{j}}(x), j=0,1,2...n $$ } such that $$\displaystyle {{b}_{0}}(\beta)\ne 0$$   and   $$\displaystyle {{b}_{j}}(\beta)= 0, j = 1,2,3...n $$

$$\displaystyle i> {{F}_{p}}={{x}^{j}}, j=0,1,2...n $$

$$\displaystyle ii> {{F}_{p}}=\cos jx, j=0,1,2...n $$

$$\displaystyle iii> {{F}_{p}}=1,\sin jx, j=1,2...n $$

$$\displaystyle iv> {{F}_{p}}=\cos jx,\sin kx, j=0,1,2...n , k=1,2...n$$

$$\displaystyle v> {{F}_{p}}={{e}^{jx}}, j=0,1,2...n $$

soln
$$\displaystyle i> $$ For the Constraint Breaking Function to be satisfied, we take the basis function for the Polynomial case so that it satisfies the Essential Boundary Condition $$\Gamma _{g}=\left \{ \beta\right \}$$ is -

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$$\displaystyle
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{{F}_{p}}^{-}={{(x-\beta)}^{j}}, j=0,1,2...n

$$ (Eq )<1>
 *
 * }

For $$\displaystyle j=0 $$ at  $$\displaystyle x= \beta, {{F}_{p}}^{-} = {(x-\beta)^0} = 1$$

For $$\displaystyle j=1 $$ at  $$\displaystyle x= \beta, {{F}_{p}}^{-} = {(\beta-\beta)^1} = 0$$

For $$\displaystyle j=2 $$ at  $$\displaystyle x= \beta, {{F}_{p}}^{-} = {(\beta-\beta)^2} = 0$$

For $$\displaystyle j=n $$ at  $$\displaystyle x= \beta, {{F}_{p}}^{-} = {(\beta-\beta)^n} = 0$$

Hence the Constraint Breaking Function is satisfied by this polynomial basis function where

$$\displaystyle {{b}_{0}}(\beta)\ne 0$$  and   $$\displaystyle {{b}_{j}}(\beta)= 0, j = 1,2,3...n $$

Now for the given problem, the value of $$\displaystyle \beta =4 $$

The plot of $$\displaystyle {{F}_{p}}$$ and $$\displaystyle{{F}_{p}}^{-} $$ is generated from Matlab.

Matlab Code:

$$\displaystyle ii>$$ While considering the Cosine basis function $$\displaystyle {{F}_{c}}=\cos jx $$, we are unable to find such a transformation for which $$\displaystyle {{b}_{0}}(\beta)= 0$$.

So we make the transformation of
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$$\displaystyle {{b}_{0}}(\beta)= 1, {{F}_{c}}^{-}= \cos (2j-1)\pi x/2\beta, j=1,2...n
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 * style="width:92%; padding:10px; border:2px solid #ff0000" |

$$ (Eq )<2>
 *
 * }

For $$\displaystyle j=1 $$ at  $$\displaystyle x= \beta, {{F}_{c}}^{-}= \cos (2-1)\pi \beta/2\beta= \cos \pi/2 = 0$$

For $$\displaystyle j=2 $$ at  $$\displaystyle x= \beta, {{F}_{c}}^{-}= \cos (4-1)\pi \beta/2\beta= \cos 3\pi/2 = 0$$

For $$\displaystyle j=3 $$ at  $$\displaystyle x= \beta, {{F}_{c}}^{-}= \cos (6-1)\pi \beta/2\beta= \cos 5\pi/2 = 0$$

For $$\displaystyle j=n $$ at  $$\displaystyle x= \beta, {{F}_{c}}^{-}= \cos (2n-1)\pi \beta/2\beta= \cos (2n-1)\pi/2 = 0$$

Hence the Constraint Breaking Function is satisfied by this Cosine basis function where

$$\displaystyle {{b}_{0}}(\beta)\ne 0$$  and   $$\displaystyle {{b}_{j}}(\beta)= 0, j = 1,2,3...n $$

Now for the given problem, the value of $$\displaystyle \beta =4 $$

The plot of $$\displaystyle {{F}_{c}}$$ and $$\displaystyle{{F}_{c}}^{-} $$ is generated from Matlab.

Matlab Code:

$$\displaystyle iii>$$ While considering the sine basis function $$\displaystyle {{F}_{s}}=\sin jx $$, we have taken the  transformation  $$\displaystyle {{b}_{0}}(\beta)= 1$$.

So we make the transformation for the sin function as of
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$$\displaystyle {{b}_{0}}(\beta)= 1, {{F}_{s}}^{-}= \sin (j\pi x/\beta), j=1,2...n
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 * style="width:92%; padding:10px; border:2px solid #ff0000" |

$$ (Eq )<3>
 *
 * }

For $$\displaystyle j=1 $$ at  $$\displaystyle x= \beta, {{F}_{s}}^{-}= \sin 1\pi \beta/\beta= \sin \pi = 0$$

For $$\displaystyle j=2 $$ at  $$\displaystyle x= \beta, {{F}_{s}}^{-}= \sin 2\pi \beta/\beta= \sin 2\pi = 0$$

For $$\displaystyle j=3 $$ at  $$\displaystyle x= \beta, {{F}_{s}}^{-}= \sin 3\pi \beta/\beta= \sin 3\pi = 0$$

For $$\displaystyle j=n $$ at  $$\displaystyle x= \beta, {{F}_{s}}^{-}= \sin n\pi \beta/\beta= \sin n\pi = 0$$

Hence the Constraint Breaking Function is satisfied by this Sine basis function where

$$\displaystyle {{b}_{0}}(\beta)\ne 0$$  and   $$\displaystyle {{b}_{j}}(\beta)= 0, j = 1,2,3...n $$

Now for the given problem, the value of $$\displaystyle \beta =4 $$

The plot of $$\displaystyle {{F}_{s}}$$ and $$\displaystyle{{F}_{s}}^{-} $$ is generated from Matlab.

Matlab Code:

$$\displaystyle iv>$$ While considering the Fourier basis function $$\displaystyle {{F}_{f}}=\cos jx,\sin kx $$, we are unable to find such a transformation for which $$\displaystyle {{b}_{0}}(\beta)= 0$$.

So we make the transformation of

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$$\displaystyle
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{{b}_{0}}(\beta)= 1, {{F}_{f}}^{-}= \cos (2j-1)\pi x/2\beta, j=1,2...n , \sin (k\pi x/\beta), k=1,2...n

$$ (Eq )<4>
 *
 * }

For $$\displaystyle j=1 $$ at  $$\displaystyle x= \beta, {{F}_{f}}^{-}= \cos (2-1)\pi \beta/2\beta= \cos \pi/2 = 0$$

For $$\displaystyle k=1 $$ at  $$\displaystyle x= \beta, {{F}_{f}}^{-}= \sin 1\pi \beta/\beta= \sin \pi = 0$$

For $$\displaystyle j=2 $$ at  $$\displaystyle x= \beta, {{F}_{f}}^{-}= \cos (4-1)\pi \beta/2\beta= \cos 3\pi/2 = 0$$

For $$\displaystyle k=2 $$ at  $$\displaystyle x= \beta, {{F}_{f}}^{-}= \sin 2\pi \beta/\beta= \sin 2\pi = 0$$

For $$\displaystyle j=3 $$ at  $$\displaystyle x= \beta, {{F}_{f}}^{-}= \cos (6-1)\pi \beta/2\beta= \cos 5\pi/2 = 0$$

For $$\displaystyle k=3 $$ at  $$\displaystyle x= \beta, {{F}_{f}}^{-}= \sin 3\pi \beta/\beta= \sin 3\pi = 0$$

For $$\displaystyle j=n $$ at  $$\displaystyle x= \beta, {{F}_{f}}^{-}= \cos (2n-1)\pi \beta/2\beta= \cos (2n-1)\pi/2 = 0$$

For $$\displaystyle k=n $$ at  $$\displaystyle x= \beta, {{F}_{f}}^{-}= \sin n\pi \beta/\beta= \sin n\pi = 0$$

Hence the Constraint Breaking Function is satisfied by this Fourier basis function where

$$\displaystyle {{b}_{0}}(\beta)\ne 0$$  and   $$\displaystyle {{b}_{j}}(\beta)= 0, j = 1,2,3...n $$

Now for the given problem, the value of $$\displaystyle \beta =4 $$

The plot of $$\displaystyle {{F}_{f}}$$ and $$\displaystyle{{F}_{f}}^{-} $$ is generated from Matlab.

Matlab Code:

$$\displaystyle v>$$ While considering the Exponential basis function $$\displaystyle {{F}_{e}}={{e}^{jx}}$$, we are unable to find such a transformation for which $$\displaystyle {{b}_{0}}(\beta)= 0$$.

So we make the transformation of

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$$\displaystyle
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{{b}_{0}}(\beta)= 1, {{F}_{e}}^{-}= {{e}^{j(x-\beta)}}-1, j=1,2...n

$$ (Eq )<5>
 *
 * }

For $$\displaystyle j=1 $$ at  $$\displaystyle x= \beta, {{F}_{e}}^{-}= {{e}^{1(\beta-\beta)}}-1= {e}^0 - 1 = 1-1 = 0$$

For $$\displaystyle j=2 $$ at  $$\displaystyle x= \beta, {{F}_{e}}^{-}= {{e}^{2(\beta-\beta)}}-1= {e}^0 - 1 = 1-1 = 0$$

For $$\displaystyle j=3 $$ at  $$\displaystyle x= \beta,{{F}_{e}}^{-}= {{e}^{3(\beta-\beta)}}-1= {e}^0 - 1 = 1-1 = 0$$

For $$\displaystyle j=n $$ at  $$\displaystyle x= \beta,{{F}_{e}}^{-}= {{e}^{n(\beta-\beta)}}-1= {e}^0 - 1 = 1-1 = 0$$

Hence the Constraint Breaking Function is satisfied by this polynomial basis function where

$$\displaystyle {{b}_{0}}(\beta)\ne 0$$  and   $$\displaystyle {{b}_{j}}(\beta)= 0, j = 1,2,3...n $$

Now for the given problem, the value of $$\displaystyle \beta =4 $$

The plot of $$\displaystyle {{F}_{e}}$$ and $$\displaystyle{{F}_{e}}^{-} $$ is generated from Matlab.

Matlab Code:

2nd epart
F = {$$\displaystyle 1,e^x,e^{2x}$$} = {$$\displaystyle b_{1}(x),b_{2}(x),b_{3}(x)$$} $$\Gamma = \begin{bmatrix} \Gamma_{11}&\Gamma_{12}&\Gamma_{13} \\ \Gamma_{21}&\Gamma_{22}&\Gamma_{23} \\ \Gamma_{31}&\Gamma_{32}&\Gamma_{33} \\ \end{bmatrix}$$

$$\;\Gamma_{ij}=\int_{\Omega}b_{i}(x)b_{j}(x)dx$$ $$\Gamma_{11} =  = \int_{-2}^{4}1 dx = 6$$ $$\Gamma_{12} = \Gamma_{21}= = \int_{-2}^{4}e^x dx =54.4628 $$ $$\Gamma_{13} = \Gamma_{31}= = \int_{-2}^{4}e^{2x} dx =1490.47 $$ $$\Gamma_{22} =  = \int_{-2}^{4} e^{2x} dx = 1490.47$$ $$\Gamma_{23} = \Gamma_{32}= = \int_{-2}^{4}e^{3x} dx =54251.6 $$ $$\Gamma_{33} =  = \int_{-2}^{4}e^{4x} dx =2.2215*10^6 $$

$$\Gamma = \begin{bmatrix} 6 &54.4628&1490.47 \\ 54.4628&1490.47&54251.6  \\ 1490.47&54251.6&2.2215*10^6  \\ \end{bmatrix}$$

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$$\displaystyle
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\Gamma = \begin{bmatrix} 6 &54.4628&1490.47 \\ 54.4628&1490.47&54251.6  \\ 1490.47&54251.6&2.2215*10^6  \\ \end{bmatrix}

$$ (Eq )<6>
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 * }

The determinant of the matrix is given by:-

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$$\displaystyle
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Det \begin{bmatrix} 6 &54.4628&1490.47 \\ 54.4628&1490.47&54251.6  \\ 1490.47&54251.6&2.2215*10^6  \\ \end{bmatrix} =1.13603*10^9

$$ (Eq )<7>
 *
 * }

Hence we can say that the function $$\displaystyle e^{jx} $$ is independent on $$\displaystyle \Omega $$

b. Generating a cylinder

The CGX commands used for the generation of a disc are- PNT, LINE, GSUR, ELTY, MESH and plus.

The commands are the same used to generate a disc and are explained above.-

c. Generating a sphere

The CGX commands used for the generation of a sphere are- PNT, LINE, GSUR, ELTY, MESH.

The commands are the same used to generate a disc and are explained above.-

d. Generating a sphere-volume

The CGX commands used for the generation of a sphere are- PNT, LINE, GSUR, ELTY, MESH and GBOD.

The commands are the same used to generate a disc and are explained above except GBOD.-

i> GBOD- This command is used to define or redefine a body in the most basic way. Each body must have five to seven surfaces to be mesh-able. The syntax to print a point with co-ordinates is

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

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

'gbod' |'!' 'NORM' '+|-' '+|-' -> .. ( 5-7 surfaces )

$$
 * }

An example is

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$$\displaystyle
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 * style="width:92%; padding:10px; border:2px solid #ff0000" |

GBOD \quad B001 \quad NORM \quad - \quad S001 \quad + \quad S002 \quad - \quad S005 \quad - \quad S004 \quad - \quad S003\quad -\quad  S006

$$
 * }

This command will create a body with name B001. NORM is a necessary placeholder which may be used for functionality in the future. The surfaces are followed with a + or - sign in front which in an indiacation of the orientation of each surface.

vi> plus- This command is used to display the entities of an additional set on the Calculix Graphics page. The syntax of plus is

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

'plus' ['n'|'e'|'f'|'p'|'l'|'s'|'b'|'S'|'L']['a'|'d'|'p'|'q'|'v'] -> 'w'|'k'|'r'|'g'|'b'|'y'|'m'|'i'

$$
 * }

, where Nodes =n, Elements =e, Faces =f, Points =p, Lines =l, Surfaces =s, Bodies =b, Nurb Surfaces =S and Nurb Lines =L and White =w, Black =k, Red =r, Green =g, Blue =b, Yellow =y, Magenta =m

An example is

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$$\displaystyle
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 * style="width:92%; padding:10px; border:2px solid #ff0000" |

plus \quad ea\quad all

$$
 * }

This command will display the whole image on the Calculix graphics window.


text:
User:Eml5526.s11.team6/hwk4

=Problem 4.1 Solving problem 2.9 for $$ \phi = \frac{\pi}{2}$$ using specific basis function=

Given
Solving a differential equation using Weight Residual Form. The partial differential equation is:

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$$ \frac{\partial}{\partial x}\left(2\frac{\partial u}{\partial x}\right) + 3 = 0 $$
 * $$\displaystyle (Eq.9.1) $$
 * }
 * }

The boundary conditions are:
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$$ u\left(1\right) = 0$$

$$\frac{du}{dx}\left(0\right) = -4 $$ Weighting funtion to be used:
 * }
 * }

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$$ b_{i} = \left \{ 1, \sin\left ( x \right ),\sin\left (  2x \right )... \right \} $$
 * $$\displaystyle (Eq.9.2) $$
 * }
 * }

Which is correction function when $$\phi = \frac{\pi}{2}$$

Find

 * 1 Find an approximate solution of the form $$U^{h}=\sum_{i=0}^{n}d_{i}b_{i}$$ for n=2
 * 2 Find two equations that enforce the boundary conditions
 * 3 Project the weight residues
 * 4 Display the equations in matrix form
 * 5 Solve for $$\displaystyle d $$
 * 6 Construct $$U^{h}$$ and plot $$U^{h}$$ and $$u $$
 * 7 Repeat 9.1 to 9.6 for n = 4 and n = 6

Find an approximate solution with n=2
For n = 2:
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$$ d_{i}=\begin{bmatrix} d0 & d1 & d2 \end{bmatrix} $$
 * $$\displaystyle (Eq.9.3) $$
 * }
 * }

and

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$$ b_{i}=\begin{bmatrix} 1 & \sin  \left ( x\right ) & \sin  \left ( 2x\right ) \end{bmatrix} $$
 * $$\displaystyle (Eq.9.4) $$
 * }
 * }

Therefore:

Find two equations that enforce the boundary conditions
The boundary conditions are:

And

Project the weight residues
In this case changing the function u(x) by the approximated function $$U^{h}$$ in the PDE (Eq.9.1), we found that P(u) is:

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$$ P\left ( U^{h} \right )=2\frac{\partial^2 U^{h}}{\partial x^2}+3=-2\left ( d1\sin \left (x \right ) +4d2\sin \left (2x \right )\right )+3 $$
 * $$\displaystyle (Eq.9.8) $$
 * }
 * }

The equation 9.8 can be written as:

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$$ P\left ( U^{h} \right )= -2\begin{bmatrix}0 & \sin \left (x\right ) & 4\sin \left (2x \right ) \end{bmatrix}\begin{bmatrix} d0\\ d1\\

d2\end{bmatrix}+3 $$
 * $$\displaystyle (Eq.9.9) $$
 * }
 * }

Projecting the residue we have:

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$$ \int_{a}^{b}b(x)P\left ( U^{h} \right )= 0 $$ After substitution of (Eq.9.4) and (Eq.9.9) in (Eq.9.10) the following equation is obtained
 * $$\displaystyle (Eq. 9.10) $$
 * }
 * }

Display the equations in matrix form
Performing the product and then the integration indicated in (Eq.9.11) we have:

The latter equation is clearly of the form $$\displaystyle Kd=F$$. We can observe that K is not symmetric.

Solve for $$ d $$
In (Eq.9.12) we have three equations; but, as we need to enforce the boundary conditions we take the only one of them and solve it together with (Eq.9.6) and (Eq.9.7). From these equations we now have the system:

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$$ \begin{bmatrix} 1 &\sin\left (1 \right )   & \sin\left (2  \right ) \\ 0 &\cos \left (0 \right ) &2\cos \left (0  \right ) \\ 0 & 2\frac{\sin^3\left(1\right)}{3} & 2-\frac{\sin\left(4\right)}{2} \end{bmatrix}\begin{bmatrix} d0\\ d1\\

d2\end{bmatrix}=\begin{bmatrix} 0\\ 4\\

2.12\end{bmatrix}$$
 * $$\displaystyle (Eq.9.13) $$
 * }
 * }

After solving the system in (Eq.9.13) we obtain d:

Construct $$U^{h}$$ and plot $$U^{h}$$ and $$u$$
Replacing d in (Eq.9.5) we obtain the approximated solution:

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$$ U^{h}=-4.65-7.34\sin \left (x \right )+1.67\sin \left ( 2x \right ) $$
 * $$\displaystyle (Eq.9.14) $$
 * }
 * }

On the other hand, the exact solution of the PDE (Eq.9.1) with the given boundary condition is:

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$$ u\left ( x \right )=-\frac{3}{4}x^{2}-4x+\frac{19}{4} $$
 * $$\displaystyle (Eq.9.15) $$
 * }
 * }

In Fig.9.1 we show the exact and the approximated solution.

Plot the graph for $$ u\left(x\right)$$ and $$u^h\left(x\right)$$
Using following generalized code for any value of n, we get plots for n=4 and n=6

The error at x=0.5 is plotted for different values of n

Note:From the plot it is clear that error increases with increase in value of n for x=0.5

=Problem 4.2 "Obtain the weak form for the equations of heat conduction"=

Given
The equations of heat conduction

with the boundary conditions and The condition on the right is a convection condition.

Find
The weak form for the equations of heat conduction.

Solution
If we choose an arbitrary function w(x) and multiply the governing equation (4.2.1) and the convection boundary condition (4.2.3), we have

Do integrate by parts with equation (4.2.4), we have

$$ \left. {w\left( x \right)Ak\frac} \right|_0^{10} - \int_0^{10} {\fracAk\frac} dx + \int_0^{10} s dx = 0 $$

Then use the natural boundary condition (4.2.5), we have

where the weighting function satisfy

Therefore the weak form for the equations of heat conduction is to find T such that satisfy (4.2.2) and (4.2.6) with the constrain that satisfy (4.2.7).

=Problem 4.3 Finding weak form for heat conduction problem=

Given
Given the strong form for the heat conduction problem in a circular plane:

with the essential boundary conditions and natural boundary condition

where R is the total radius of the plate, s is the heat source per unit length along the plate radius. T is the temperature and k is the conductivity. Assume that k, s and R are given.

Find
a. Construct the weak form for the above strong form.

b. Use quadratic trial (candidate) solution of the form $$T= \alpha_0+\alpha_1 r+\alpha_2 r^2$$ and weight function of the same form to obtain a solution of the weak form.

c. Solve the differential equation with the boundary conditions and show that the temperature distribution along the radius is given by

Solution
a.

Introducing the arbitrary weighted function
Weak form can be found as follows, with weighted function

Now, multiply eq. 4.3.1 by $$ \displaystyle w $$ and integrate between 0 to R.

Use integration by parts for the first term in eq. 4.3.6

Using eq. (4.3.3) and eq. (4.3.5), we get

Put eq. (4.3.8) in eq. (4.3.7), we get the weak form as follows

b.

Finding solution of the weak form using quadratic trial solution
Now, consider the trial solution

By given essential boundary condition eq. (4.3.2) and by natural boundary condition eq. (4.3.3) Therefore, we get Put this eq. (4.3.14) in eq. (4.3.10), we get Hence, By eq. (4.3.2) and eq. (4.3.16) we get,

Taking arbitrary function $$ \displaystyle w $$ such that $$\displaystyle \beta_1=0$$, by comparing with eq. (4.3.15) Put this in weak form eq. (4.3.9), we get Integrating, factoring out $$\beta_0,\beta_2 $$ and rearranging the terms gives As the above must hold for arbitrary weight functions, it must hold for arbitrary $$\displaystyle\beta_0 $$ and $$\displaystyle\beta_2 $$. Therefore, the coefficients of $$\displaystyle\beta_0 $$ and $$\displaystyle\beta_2 $$ must vanish, which gives the following algebraic equation in $$\displaystyle\alpha_0 $$ and $$\displaystyle\alpha_2 $$: Solving above, second row gives Therefore, From eq. (4.3.18)

c.

Proof for the temperature distribution equation
Therefore, the solution is Hence, the solution for the given differential equation is

=Problem 4.4 Developing the weak form for the Torsion Bar=

GIVEN
Referring to figure 4.4.1 given below for a cylindrical circular bar in torsion -

Strong Form - $$

\displaystyle      \frac{d}\left( {JG\frac{d\phi}} \right) + m = 0 $$    With $$ \displaystyle 0 \leqslant x \leqslant l $$ $$\displaystyle Eq. 4.4.1 $$

Natural Boundary Condition at x = l :

M (x = l ) = $$ \displaystyle \left( {JG\frac{d\phi}} \right)_{}^{x = l} = \overline M =12 $$ $$\displaystyle Eq. 4.4.2 $$

Essential Boundary Condition at x = 0 :

$$ \displaystyle \phi \left( {x = 0} \right) = \overline \phi =4 $$ $$\displaystyle Eq. 4.4.3 $$

Where,

m : Distributed moment per unit length M : Torsion Moment $$ \displaystyle \phi $$ : Angle of rotation , G : Shear modulus J : Polar moment of inertia = $$ \frac{\pi C^2}{2} $$ $$\displaystyle Eq. 4.4.4 $$ C : Radius of the cylindrical bar

FIND
a) To construct the weak form for the strong form of the circular torsion bar as given in $$\displaystyle Eq. 4.4.1 $$

b) Taking the values given in data set Mtg 9 and to integrate the differential equation given in $$\displaystyle Eq. 4.4.1 $$

c) To find the integration constants using the boundary conditions given in $$ \displaystyle Eq. 4.4.2$$ and $$ \displaystyle Eq. 4.4.3$$,

Developing the weak form
Let us assume a weighting function w (x) such that $$ \displaystyle w (x=0) = 0 $$. The condition is applied on w (x) as w (x) is chosen arbitrarily and needs to vanish at x = 0 where the essential boundary condition is applied.

Hence, now the weak form becomes -

$$ \displaystyle \frac{d}\left( {JG\frac{d\phi}} \right) + m = 0  $$

$$ \displaystyle i.e.$$ $$ \displaystyle \frac{d}\left( {J(x)G(x)\frac{d\phi}} \right) + m(x) = 0 $$ $$\displaystyle Eq. 4.4.5 $$

Note - Here we assume that J, G and m are functions of x and are NOT constants.

Integrating by parts for the first term in $$ \displaystyle Eq. 4.4.5 $$ as per the integration rule -

$$ \displaystyle   \int\limits_a^b {udv = } \left[ {uv} \right]_a^b - \int\limits_a^b {vdu} $$ $$\displaystyle Eq. 4.4.6 $$

and assuming -

$$ \displaystyle \ u = w (x)$$    and  $$ \displaystyle \   v = \left[ {\left( {JG\frac} \right) } \right] $$  ,we get

$$ \displaystyle \left[ {w(x)\left( {JG\frac} \right)} \right]_0^l - \int\limits_0^l {\frac\left( {JG\frac} \right)} dx + \int\limits_0^l {w(x).m} dx = 0 $$

$$ \displaystyle \left[ {w(x)\left( {JG\frac} \right)} \right]_{}^{x = l} - \left[ {w(x)\left( {JG\frac} \right)} \right]_{x = 0}^{} - \int\limits_0^l {\frac\left( {JG\frac} \right)} dx + \int\limits_0^l {w(x).m} dx = 0 $$
$$\displaystyle Eq. 4.4.7 $$

Now as per the choice of the weight function, $$ \displaystyle /w (x= 0) = 0 $$. Hence the second term in the $$ \displaystyle Eq. 4.4.7 $$ vanishes.

Also using $$ \displaystyle Eq. 4.4.2 $$, the first term in $$ \displaystyle Eq. 4.4.7 $$ can be replaced by $$ \displaystyle {\text{w}}\left( l \right).\overline M $$ = $$12 \displaystyle {\text{w}}\left( l \right)$$ and JG=2+3x and m(x)=5x

Hence after applying the sign changes, the weak form of $$ \displaystyle Eq. 4.4.1 $$ becomes -

Integration of the strong form to get the exact solution
We use the general 1-Dimensional Model Data set 1 from Mtg 9 - Page 9-2 The genearal 1-Dimensional Model Data set 1 given above provides us to assume that $$ \displaystyle \ m (x) = 5x $$

Hence the strong form given in $$ \displaystyle Eq. 4.4.1 $$ converts to the following -

1.

$$ \displaystyle \frac{d}\left( {JG\frac} \right)+ 5x=0 $$

With  $$ \displaystyle\ 0 \leqslant x \leqslant l   $$
$$\displaystyle Eq. 4.4.9 $$

Integrating $$ \displaystyle Eq. 4.4.9 $$ with respect to x, substituting JG=2+3x,we get, -

$$ \displaystyle 1.\left[ {\left( {(2+3x)\frac} \right)} \right] =- \left[ { {\frac} } \right] + C $$

$$ \displaystyle \left[ {\left( {\frac} \right)} \right] =- \left[ { {\frac} } \right] +\frac{C}{2+3x} $$

From the boundary conditions,

$$\left[ {\left( {\frac} \right)} \right]_{x=1}^{}=\frac{12}{5}$$
 *
$$\displaystyle (Eq. 4.4.10) $$

$$\frac{12}{5}=\frac{-5}{2(5)}+\frac{C}{5} $$

Rearranging and integrating step 2 of above $$ \displaystyle Eq. 4.4.10 $$ again with respect to x using integration by parts as given in $$ \displaystyle Eq. 4.4.6 $$ with constant D involved, we get -

$$ \displaystyle \int \left( {\frac} \right)dx = \int \frac{\frac{29}{2}- \left[\frac {5x^2}{2} \right]}{2+3x} $$

$$ \displaystyle \left[ {\phi } \right] = \frac{29}{2}\int\frac{1}{2+3x} - \frac{5}{2}\int\frac{x^2}{2+3x} + D $$

$$ \displaystyle \left[ {\phi } \right]= \frac{29}{2} \frac{log(2+3x)}{3}-\frac{5}{2}\frac{8log(2+3x)+9x^2-12x}{54}+D $$

Now using $$ \displaystyle Eq. 4.4.3 $$ with $$ \displaystyle \phi \left( {x = 0} \right) = \overline \phi=4$$ ,we can update above equation as -

$$ \displaystyle D= 4 + \frac{5}{2}*\frac{8log(2)}{54}-\frac{29}{2}\frac{log(2)}{3}$$
$$\displaystyle Eq. 4.4.12 $$

Substituting the values

$$\phi=\frac{29}{6}log(2+3x)-\frac{5}{2}\left[\frac{4}{27}log(2+3x)-\frac{x^2}{6}-\frac{2x}{9}\right]+0.91 $$

=Problem 4.5 Modifying Basis Functions Using Constraint Breaking Solutions (CBS)=

Given
Refer to lecture slide [[media:Fe1.s11.mtg21.djvu|21-1]] for problem assignment. For $$\Gamma_g=\left [ \beta \right ]$$, and each of the following basis function families:

Find
1) Find the corresponding family $$ \overline{F}_{I}=\left [ \overline{b}_{j} \right ] $$ satisfying CBS, i.e.,

Choose $$ \left [ b_{i}(x), \; i=0,1,2,... \right ] $$ such that $$ b_{0}(\beta)\neq 0 $$ and $$ b_{i}(\beta)=0 \quad\text{for}\quad \; i=1,2,...,n $$

Let $$ \Omega = ] \; \alpha, \beta \; [ \; = \; ] -2,4 \; \; [ $$

2) Given that $$ \overline{b}_{0}(x)=constant; \quad\text{plot}\quad b_{j} \quad\text{and}\quad \overline{b}_{j} \quad\text{for}\quad j=1,2,3 $$

3) Show $$ \left [ e^{jx}, \; j=0,1,2,... \right ] $$ is linearly independent on $$ \Omega $$.

Find corresponding family of basis functions that satisfy the CBS
1) Find corresponding family of basis functions that satisfy the CBS Given that $$ \Gamma_{g} = \left [ x= \beta \right ] $$

Polynomial Basis
Shift the set of basis functions (Eq. 4.5.1) so that they cross the x-axis at the point $$ x=\beta \; $$

Therefore,

Cosine
Shift the set of basis functions (Eq. 4.5.2) so that they cross the x-axis at the point $$ x=\beta \; $$

Therefore,

Sine Basis
Shift the set of basis functions (Eq. 4.5.3) so that they cross the x-axis at the point $$ x=\beta \; $$

Therefore,

Fourier Basis
In order to shift the set of basis functions (Eq. 4.5.4) so that they cross the x-axis at the point $$ x=\beta \; $$, we must employ the same tactics used for the Sine and Cosine basis functions.

Therefore,

Exponential Basis
Shift the set of basis functions (Eq. 4.5.5) so that they cross the x-axis at the point $$ x=\beta \; $$. This involves not only shifting the functions across the x-axis, but the functions must also be shifted in the negative y-direction because ordinarily the function does not cross the x-axis.

Therefore,

Plot Original and CBS Basis Functions Over the Domain
2) (Eq. 4.5.6), (Eq. 4.5.6), (Eq. 4.5.6), (Eq. 4.5.6), and (Eq. 4.5.6) are plotted along the x-domain $$ \Omega = ] \; \alpha, \beta \; [ \; = \; ] -2,4 \; \; [ $$.  These plots are shown in the figures below.

Determine Linear Independence of Exponential Basis Functions Over the Given Domain
3) Show $$ \left [ e^{jx}, \; j=0,1,2,... \right ] $$ is linearly independent on $$ \Omega $$.

Given that

where

and

Given that i=1,3 and j=1,3 (Eq. 4.5.16) takes the following form

The order of $$ b_i \;$$ and $$ b_j \;$$ does not affect the caluclation of (Eq. 4.5.18), therefore (Eq. 4.5.19) is a symmetric matrix. (Eq. 4.5.19) is calculated as follows.

Substituting (Eq. 4.5.20-25) into (Eq. 4.5.19) yields the following Gram matrix.

The determinant of (Eq. 4.5.26) is

$$ det[ \Gamma ] $$ is equal to a non-zero number, therefore, $$ \left [ b_i \right ] $$ is linearly independent.

= Problem 4.6 Derivation of strong and weak form equations for an elastic bar with an applied variable distributed spring.=

Given
Consider the elastic bar with a variable distributed spring p(x) along its length shown below in Figure 4.6.1. The distributed distributed s[pring imposes an axial force on the bar in proportion to the displacement. The bar is of length l, cross-sectional area A(x), Young's modulus E(x) with body force b(x) and boundary conditions shown in the following figure.

The above problem statement is given in A First Course in Finite Elements FB p74 Problem 3.9.

Find
a. Construct the strong form. b. Construct the weak form.

Solution
The bar must satisfies the following conditions: 1. The bar must be in equilibrium. 2. The bar must satisfy Hooke's Law σ(x)=E(x)ε(x). 3. The displacement field must be compatible. 4. It must satisfy the strain-displacement equation.

The following Figure 4.6.2 shows the internal force balance of a segment of the bar.

a. Strong Form
Performing a force balance on the section in Figure 4.6.2 yields the following differential equation 4.6.1.

Dividing the above equation through by dx to simplify terms yielded the following equation 4.6.2.

Now take the limit as Δx goes to 0. The result is presented below in equation 4.6.3.

By Hooke's Law:

Substituting Eq.4.6.6 into Eq.4.6.5 and then plugging that result into Eq.4.6.4 yielded the following equation.

Substituting Eq.4.6.7 into Eq.4.6.3 yields the final strong for differential equation as follows:

b. Weak Form
For the weak form, we multiply the strong form solution by the weighting function w and integrate the result over the length of the bar 0 to l as shown in the following equation:

Integrating the first term of Eq.4.6.8 by using integration by parts yielded the following equation:

Knowing that: $$A(x)E(x)\frac{du}{dx}=\bar{t}$$ and accounting for the b(x)-p(x) term yielded the following:

The final weak form becomes the following:

=Problem 4.7: Introduction to Calculix=

Given
Introduction to Calculix - A non-linear Finite Element Analysis Code

Calculix is an open source Finite Element Analysis code with the authors acknowledging that naming conventions and input style formats for CalculiX are based on those used by ABAQUS, a proprietary and commercial, general purpose finite element code but the results obtained from CalculiX are in no way connected to ABAQUS.

Website for informational and software download (Linux executable) purposes - http://dhondt.de/ Website for informational and software download (Windows executable) purposes - www.bconverged.com/calculix

Calculix in all has 2 major modules, namely- 1. CGX ( Calculix Graphics Module ) - This program is designed to generate and display finite elements (FE) and results coming from CalculiX CrunchiX (ccx).The Concept and File Format sections give some background on functionality and mesher capabilities. The Getting Started section describes how to run the verification examples you should have obtained along with the code of the program.

2. CCX (Calculix CrunchiX (Solver) Module) - This is a theoretical section giving some background on the analysis types, elements, materials etc. Then, an overview is given of all the available keywords in alphabetical order, followed by detailed instructions on the format of the input deck. If CalculiX does not run because your input deck has problems, this is the section to look at. The last section contains a description of the verification examples you should have obtained along with the code of the program. If you try to solve a new kind of problem you haven't dealt with in the past, check this section for examples. You can also use this section to check whether you installed CalculiX correctly (if you do so with the compare script and if you experience problems with some of the examples, please check the comments at the start of the corresponding input deck). Finally, the User's Manual ends with some references used while writing the code.

Find
a) Installation of the Calculix Graphics Module and to explain the installation process for layman(CGX) b) To access the CGX manual c) To sign up with used group for asking the queries,if any and to gain access to the archives where we can access the email conversation from various users,a source of possible learning d) To reproduce the basic inbuilt examples from CGX data base installed on your machine, with a detailed report, about entities such as - 1. Dics 2. Cylinder 3. Sphere 4. Sphere Volume 5. Airofoil The report shall include explanation to all the CGX commands used to generate above example geometries along with their meshes using screen-shots of the useful explanatory steps.

For Linux
For a naiver Linux user now using Windows, we strongly recommend using Ubuntu as your linux operating system which can be easily downloaded and installed at Ubuntu

To install Calculix for Linux, start by downloading the source code and examples available at http://dhondt.de/cgx_2.2.all.tar.bz2. Right now you need open a terminal by selecting that from Applications as in the following picture.

Before installation, you should check openGL on your system by typing the following words in your terminal:

find /usr/* -name libGL* -print

And the following results show that you have the openGL-library which is needed.

/usr/lib/mesa/libGL.so.1 /usr/lib/mesa/libGL.so.1.2 /usr/lib/libGLU.so.1 /usr/lib/libGLU.so.1.3.070701

Another thing you need is c++ complier since we have to compile the source code. You can check where is your c++ complier by typing

which c++

If you don't have these essential things, please typing the following to download and install them

sudo apt-get install linux-kernel-headers sudo apt-get install build-essential

Then we start to install cgx.

1.Unpack the file cgx_2.2.all.tar.bz2 in "/usr/local". An easy way is first exacting the compressed file using your right mouse menu and then copy to the destination.

2.Then enter the folder where the source code located by typing

cd /usr/local/CalculiX/cgx_2.2/src/

3.Run "make" as a root and copy the binary "cgx" to "/usr/local/bin/cgx" with

cp cgx /usr/local/bin/cgx

4.Run "latex cgx" in "/usr/local/CalculiX/cgx_2.2/doc" for three times and then run "latex2html cgx".

5.Now your CalculiX has been installed and please enjoy

6.In order to check the installation, type

cgx -b dummy.fbd

and you can see a window appear on your screen as following.

Note:

1.If you are counting any problems for being unable to find any *.h, you can type the following to ensure that you have install all the essential libraries before building the source codes

sudo apt-get install build-essential libsdl1.2debian libsdl1.2-dev libgl1-mesa-dev libglu1-mesa-dev libsdl-image1.2 libsdl-image1.2-dev

For Windows
To install CalculiX for Windows, start by downloading the executable file available at http://www.bconverged.com/download.php.

Before installing CalculiX, we recommend that you create a folder that you will want to use as your workspace for CalculiX files.

Access the zip file, CalculiX_2_2_win_002.zip, and run the executable CalculiX_2_2_win_002.exe. Review the license agreement and select "I agree" when you are prompted to as shown below.

The next screen will ask you to select the directory to which Calculix will be installed. You may use the default directory or click "Browse" to select the directory of your choosing. Once you have the desired directory in the "destination folder" field (1) click the "next" button (2) to continue.

You will now be shown a prompt which allows you to choose which components and/or extensions of the program you wished to have installed. It is recommended to use the default settings unless you are an advanced user. Click "Install" to begin the installation.

When the installation begins you will be asked to specify a folder or directory to use as your workspace. Choose the folder you created prior to starting the installation on the file tree and click "OK"

CalculiX will now begin installing onto your computer. When completed you will be presented with the screen below. Specify if you'd like to view the help file by checking or un-checking the corresponding box and then click "Finish" to complete the installation process.

Congratulations !! You have completed installing CalculiX on your Windows based computer.

Access to CGX manual
To locate the user manual after installation of the software, begin by clicking on the windows start button. All programs will open and select the folder bConverge. This folder will open with all of the calculix programs listed, however to access the manual double-click on the help icon inside of the bConverged folder. This action will open a new window containing several folders. Expand the Calculix folder and then the CGX folder. This will bring you to the Calculix User's Manual -Calculix GraphiX, Version 2.2- as shown below.

To navigate through the manual any of the blue links in the initial index will take the user to that section of the the manual as shown in the above picture. Pressing ctrl+f will allow the user to find a particular word within the manual as well. At the top of the manual, there are four buttons Next, Up, Previous, and Contents as shown below. The function of these buttons is presented below.

Next: Flips to the next page of the manual. Up: Returns to the previously viewed screen before a link was clicked. Previous: Flips to the previous page of the manual. Contents: Takes the user to the index of links of all contents in the manual.

Additional help while Getting Started- The Calculix User Group
Student can join the yahoo user group dedicated to the forums related to knowledge sharing about Calculix. In the very friendly environment of these forums, the student can present their difficulties and queries and can get benefited by the knowledge of many other Calculix users. Student will be given a choice of joining this user group while in some final steps of the software installation. Even if the student misses to join it at the time of installation, he/she can join the group later using help on Calculix or from Calculix website online. With the help of this user group, the student will be able to have access to the email archives of the group members which at times can become useful to answer at least some of the queries or for informational purposes.

Reproducing the basic inbuilt examples from CGX- A report
Here we present the steps, codes, explanation to all the CGX commands used in the codes to generate the sample inbuilt example geometries in CGX database after installation, along with their meshes using screen-shots of the useful explanatory steps. Please refer to the CGX manual (as explained in the above section) for more details.

Figure
Following figure 4.7.1 shows the geometry and the mesh generated by CGX for a disc in four 90 degree segments-

Geometry generation code
Following figure 4.7.2 shows the code to generate the geometry and the mesh by CGX for the disc in Figure 4.7.1. Explanation of the commands used is given in the succeeding subsection.

Explanation of the commands used
This subsection provides the explanation of the CGX commands used in Figure 4.7.2

PNT
The format for entering a point in CGX is as follows -

'pnt' <name(char<9)>|'!' [<x> <y> <z>]|

[  ]|

[<P1> <P2> ]|

[<setname(containing nodes)>]

This keyword is used to define or redefine a point. There are four possibilities to define a point.

To define a point just with coordinates:

pnt p1 11 1.2 34

or,

pnt ! 11 1.2 34

where the name is chosen automatically.

It is also possible to create points on a line or in the direction from P1 to P2 by defining a spacing (ratio) and number-of-points:

pnt ! L1 0.25 3

or

pnt ! P1 P2 0.25 3

This will create 3 new points at the positions 0.25, 0.5 and 0.75 times the length of the line or the distance from P1 to P2.

And it is also possible to create points on the positions of existing nodes. The command

pnt ! set

will create new points on the positions of the nodes included in the specified set.

Usually when points are defined interactive the command qpnt is used. Please refer to the CGX manual for its details

LINE
The format for entering a Line is as follows-

'line' <name(char<9)>|'!' <p1> <p2> <cp|seq>

This keyword is used to define or redefine a line. A line depends on points.

A line can only be defined if the necessary points are already defined. There are three different types of lines available.

The straight line -

line l1 p1 p2 4

is defined by: its name l1 (the name could have up to 8 characters), by the points p1 and p2 and optionally by the division.

The arc

line ! p1 p2 cp 4

needs a center point cp. The radius changes linear from p1 to p2 if the center-point cp is excentric. The name is chosen automatically (triggered by the character !).

The spline

line l1 p1 p2 seq 4

needs a so called sequential-set, seq (use the command seqa to define such a set). This set seq stores the spline points between the end-points in the right order. The spline function is described in [14]. Usually, a line is defined interactively with qlin.

GSUR
The format for entering a surface is as follows-

'gsur' <name(char<9)>|'!' '+|-' 'BLEND| ' '+|-' <line|lcmb> '+|-' -> <line|lcmb> .. (3-5 times)

This keyword is used to define or redefine a surface in the most basic way (see also qsur). Each surface must have three to five lines or combined lines (see lcmb) to be mesh-able. However, the recommend amount of edges is four. For example,

gsur S004 + BLEND - L002 + L00E + L006 - L00C

will create the surface S004 with a mathematically positive orientation indicated by the + sign after the surface name.

The keyword BLEND indicates that the interior of the surface will be defined according to reference [12] given as- S. A. Coons, 'Surfaces for computer-aided design of space forms'. Project MAC, MIT (1964). Revised to MAC-TR-41 (1967) or a NURBS surface is referenced.

Use a + or - in front of the lines or lcmbs to indicate the orientation. These signs will be corrected automatically if necessary. If automatic name generation is desired, then use ! instead of a name.

ELTY
The format used to assign a specific element type to a set of entities and generating the mesh, is as follows-

'elty' [ ] ['be2'|'be3'|'tr3'|'tr3u'|'tr6'|'qu4'|-> 'qu8'|'he8'|'he8f'||'he8i'|'he8r'|'he20'|'he20r']

This keyword is used to assign a specific element type to a set of entities (see section Element Types in the appendix). The element name is composed of the following parts: The leading two letters define the shape (be: beam, tr: triangle, qu: quadrangle, he: hexahedra), then the number of nodes and at last an attribute describing the mathematical formulation or other features (u: unstructured mesh, r: reduced integration, i: incompatible modes, f: fluid element for ccx). If the element type is omitted, the assignment is deleted. If all parameters are omitted, the actual assignments are posted:

elty

will print only the sets with assigned elements. Multiple definitions are possible. For example,

elty all

deletes all element definitions. If the geometry was already meshed, the mesh will NOT be deleted. If the mesh command is executed again after new assignments has taken place, additional elements will be created.

elty all he20

assigns 20 node brick-elements to all bodies in the set all.

elty part1 he8 <p?

redefines that definition for all bodies in the set part1.

elty part2 qu8

assigns 8 node shell elements to all surfaces in set part2.

Figure
Following figure 4.7.3 shows the geometry and the mesh generated by CGX for a cylinder in four 90 degree segments-

Geometry generation code
Following figure 4.7.4 shows the code to generate the geometry and the mesh by CGX for the cylinder in Figure 4.7.3.

Explanation of the commands used is given in the succeeding subsection.

Explanation of the commands used
Please refer to the 'Disc' section above for explanation of all the following CGX commands used to build the geometry and mesh of the cylinder Following commands are used to generate the geometry and mesh of the cylinder in figure 4.7.3 -

'PNT For entering a point - (refer to the 'Disc' subsection above for explanation)

'LINE For entering a Line - (refer to the 'Disc' subsection above for explanation)

'GSUR For entering a surface - (refer to the 'Disc' subsection above for explanation)

'ELTY To assign a specific element type to a set of entities and generating the mesh - (refer to the 'Disc' subsection above for explanation)

Figure
Following figure 4.7.5 shows the geometry and the mesh generated by CGX for a sphere created from 32 blended surfaces.-

Geometry generation code
Following code was used to generate the geometry by the CGX for the sphere in Figure 4.7.5.

Explanation of the commands used is given in the succeeding subsection.

Explanation of the commands used
Please refer to the 'Disc' section above for explanation of all the following CGX commands used to build the geometry and mesh of the sphere Following commands are used to generate the geometry and mesh of the sphere in figure 4.7.5 -

'PNT For entering a point - (refer to the 'Disc' subsection above for explanation)

'LINE For entering a Line - (refer to the 'Disc' subsection above for explanation)

'GSUR For entering a surface - (refer to the 'Disc' subsection above for explanation)

'ELTY To assign a specific element type to a set of entities and generating the mesh - (refer to the 'Disc' subsection above for explanation)

Figure
Following figure 4.7.8 shows the geometry and the mesh generated by CGX for a sphere volume segment

Geometry generation code
Following figure 4.7.9 shows the code to generate the geometry and the mesh by CGX for the cylinder in Figure 4.7.8.

Explanation of the commands used is given in the succeeding subsection.

Explanation of the commands used
Please refer to the 'Disc' section above for explanation of all the following CGX commands used to build the geometry and mesh of the cylinder Following commands are used to generate the geometry and mesh of the cylinder in figure 4.7.3 -

'PNT For entering a point - (refer to the 'Disc' subsection above for explanation)

'LINE For entering a Line - (refer to the 'Disc' subsection above for explanation)

'GSUR For entering a surface - (refer to the 'Disc' subsection above for explanation)

'ELTY To assign a specific element type to a set of entities and generating the mesh - (refer to the 'Disc' subsection above for explanation)

Figure
Following figure 4.7.9 and figure 4.7.10 show the geometry and the mesh generated by CGX for an airfoil using many 4 nodes element-

Geometry generation code
Following codes show how to generate the geometry and the mesh by CGX for the cylinder in Figure 4.7.l3.

Explanation of the commands used is given in the succeeding subsection.

Explanation of the commands used
Please refer to the 'Disc' section above for explanation of all the following CGX commands used to build the geometry and mesh of the airfoil

Following commands are used to generate the geometry and mesh of the airfoil in figure 4.7.9

'PNT

For entering a point - (refer to the 'Disc' subsection above for explanation)

'LINE

For entering a Line - (refer to the 'Disc' subsection above for explanation)

'GSUR

For entering a surface - (refer to the 'Disc' subsection above for explanation)

'ELTY

To assign a specific element type to a set of entities and generating the mesh - (refer to the 'Disc' subsection above for explanation)

'SEQA

'seqa' ['pnt' .. <=>]|              ['afte'|'befo'  .. <=>]|              ['end' .. <=>]

This keyword is used to create or redefine a set marked as a sequential set. This set is used to define splines (see line, with the command qlin such a set is automatically created). To begin such a set type for example:

SEQA Q003 PNT P004 P005 P006 P00M P00N

Because of the parameter PNT the program will create or overwrite the set Q003. The command will continue in the next line if the sign = is found:

SEQA Q003 PNT P004 P005 P006 = P007 P008 P009

The parameter AFTE will insert additional points after the first specified point. The parameter BEFO will insert additional points before the first specified point and the parameter END will add additional points to a sequential set.

'LCMB

'lcmb' <name(char<9)>|'!' ['+|-' '+|-' '+|-' -> ..(up to 14 lines)]| ['ADD' '+|-' '+|-' -> '+|-' ..(up to 14 lines)]

This keyword is used to define, extend or redefine a combined line (lcmb). Combined lines are necessary if the edge of a surface should be made of more than one line. Usually the user does not create lcmb's directly. They are created automatically during the process of defining a surface with the command qsur. There is no limitation to the number of lines in a combined line. However with one command, not more than 14 lines can be specified at a time. To specify more than that or to extend an existing lcmb a modify command has to follow. For example,

lcmb U260 + U249 - U248 - U247 - U243 - U237 - U236 - U231 - U219

defines the lcmb U260 with 8 lines and their orientation in the lcmb. The following command

lcmb U260 ADD - U218 - U217

extends the lcmb U260 by two additional lines.

'SETA

'seta' '!'|'n'|'e'|'p'|'l'|'c'|'s'|'b'|'S'|'L'|'se' <name ..> | ['n'|'e' '-' ]

This keyword is used to create or redefine a set (see also qadd). All entities like points or bodies and so on must be stored at least in one set to be reachable. The set all is created automatically at startup and will be open (see seto) all the time unless explicitly closed (see setc). To add points to the set dummy type:

seta dummy p p1 p2

This will add the points p1 and p2 to the set. The following entities are known:

Nodes n, Elements e, Faces f, Points p, Lines l, Surfaces s, Bodies b, Nurb Surfaces S, Nurb Lines L and names of other sets se. If other sets are added then the contents are NOT automatically added. Please use also comp do to add the contents.

The program will automatically determine the type of the entities if not specified, but then the names must be unique. More than one name can be specified. A minus sign between two names specifies a range of names with steps of steps.

With the '!' key instead of a setname the program generates automatic sets with system defined setnames and stores entities in it. The only case were this can be used is to separate independent meshes stored in the basic set. The single independent meshes are then referenced by new setnames:

seta ! all

=Problem 4.8 Determining Mass Matrix [M] and Force Matrix [F] for a Dynamic Elastic PDE=

Given
Refer to lecture slide [[media:Fe1.s11.mtg23.djvu|23-4]] for problem assignment. Given the following Strong Form of a dynamic PDE:

where the weak form of (Eq. 4.8.1) is given as

and the following constants have been defined as:

Find
Consider a trial (candidate) solution and weight functions of the form

Find $$ \underline{ \tilde{m} } $$ and $$ F \left ( t \right ) $$

Determining Mass Matrix [M]
Given

and

and also, the order of index i and j does not affect the outcome of $$ \underline{\tilde m} $$ in (Eq. 4.8.6). Therefore, $$ \underline{\tilde m} $$ is a symmetric matrix.

Solving for $$ \underline{\tilde m} $$,

Placing (Eq. 4.8.8-17) into their respective places in the mass matrix, and applying its symmetric properties yields the following mass matrix,

Determining Force Matrix [F]
To determine $$ F \left ( t \right ) $$, we use the following equation for an elastic bar,

The first variable we will solve for is the matrix $$ \underline{K}_{FF} $$, which is found using the following equation,

As with the matrix $$ \underline{\tilde m} $$, the matrix $$ \underline{K}_{FF} $$ is also symmetric. This is because the order of index i and j does not affect the outcome of $$ \underline{K}_{FF} $$ in (Eq. 4.8.20).

Solving for $$ \underline{K}_{FF} $$,

Placing (Eq. 4.8.21-26) into their respective places in the stiffness matrix, and applying its symmetric properties yields the following stiffness matrix,

Next, we will determine the matrix $$ underline{d}_{F} $$. This is done by solving for the constants $$ [\alpha_0,\alpha_1,\alpha_2,\alpha_3] $$ in (Eq. 4.8.4).

First, the weight function must disappear at the essential boundary condition, i.e., $$w(3) = 0$$.

Using this requirement in (Eq. 4.8.5), we have the following equation,

Likewise, to satisfy the essential boundary condition of the strong form, $$u(3) = sin(2t)$$

Next we will evaluate the derivatives of the trial solution and the weight function.

Plugging $$\displaystyle (Eq. 4.8.30)$$, $$\displaystyle (Eq. 4.8.31)$$, and (Eq. 4.8.3) into $$\displaystyle (Eq. 4.8.2)$$,

where

and

(Eq. 4.8.33) and (Eq. 4.3.34) are substituted into (Eq. 4.8.32), and (Eq. 4.8.32) is reduced using Matlab. The code used is provided at the end of this problem. The terms $$ \beta_1, \beta_2, \quad\text{and}\quad \beta_3 $$ from the reduced form of (Eq. 4.8.32) are isolated in the following equation.

Equation $$\displaystyle (Eq. 4.8.35)$$ must hold for any arbitrary weight function $$w(x)$$, so long as it satisfies the homogeneous essential boundary condition. It then also follows that it must also hold for any arbitrary $$\beta_1$$, $$\beta_2$$, and $$\beta_3$$, leading to the following algebraic equation:

Pre-multiplying both sides of $$\displaystyle (Eq. 4.8.36)$$ by the inverse of the coefficient matrix on the left side yields:

We can obtain $$ \underline{d}_{F}'' $$ by taking the second time derivative of (Eq. 4.8.39), i.e.,

The last variable we need to obtain to solve for the force matrix in (Eq. 4.8.19) is $$ \underline{m}_{FF} $$. The subscript "FF" denotes the components that are not associated with the essential boundary condition. As a result, $$ \underline{m}_{FF} $$ is obtained by excluding row i=0 and column j=0 from (Eq. 4.8.18), i.e.,

Finally, (Eq. 4.8.27), (Eq. 4.8.39), (Eq. 4.8.40), and (Eq. 4.8.41) are substituted into (Eq. 4.8.19) in order to determine the force matrix.

Matlab Code
clc; clear all; close all; syms a0 a1 a2 a3 b0 b1 b2 b3 A E x C t real; a0 = sin(2*t); b0 = 0; u = a0 + a1*(x-3) + a2*(x-3)^2 + a3*(x-3)^3; w = b0 + b1*(x-3) + b2*(x-3)^2 + b3*(x-3)^3; dudx = a1 + 2*a2*(x-3) + 3*a3*(x-3)^2; dwdx = b1 + 2*b2*(x-3) + 3*b3*(x-3)^2; int(dwdx*A*E*dudx,x,1,3) + 0.1*A*(-b1*2+b2*4-b3*8) - int(2*x*w,x,1,3) - int(w*2*(-4*sin(2*t)),x,1,3) C = 2*[2,-4,8;-4,32/3,-24;8,-24,288/5] Cinv = C^-1 Cinv*C Cinv*[A/5-20/3+16*sin(2*t);-2*A/5+8-64/3*sin(2*t);4*A/5-56/5+32*sin(2*t)] F = [8,-12,19.2;-12,19.2,-32;19.2,-32,54.8571]*[-32*sin(2*t);-8*sin(2*t);0]+ 2*[2,-4,8;-4,32/3,-24;8,-24,285/5]*[-3.95+8*sin(2*t);-1.5+2*sin(2*t);-0.1667]

=Contributing Members=

Eml5526.s11.team6.gravois 15:51, 2 March 2011 (UTC)

Eml5526.s11.team6.deshpande 15:56, 2 March 2011 (UTC)

Eml5526.s11.team6.tupsakhare 16:06, 2 March 2011 (UTC)

eml5526.s11.team6.lee 16:23, 2 March 2011 (UTC)

Eml5526.s11.team6.vork 16:52, 2 March 2011 (UTC)

Eml5526.s11.team6.kurth 19:10, 2 March 2011 (UTC)

Eml5526.s11.team6.joglekar 01:58, 3 March 2011 (UTC)

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

title: User:Eml5526.s11.team6/hwk4


text:
User:Leighblackall/An ethical framework for ubiquitous learning

Read the current draft on Google Docs

This is a work in progress, perpetually it would seem... The first draft was completed 31 May 2011, and feedback is always sought openly from anyone with interest. Comments are recorded on the discussion page, and are being incorporated into the work below. For continuing developmental notes, feedback and review, please see the notes below and the the discussion page.

Leighblackall (talk) 00:09, 2 October 2012 (UTC)

Abstract
Ubiquitous learning has come to mean learning which is mediated by pervasive computing devices and digital media. This paper considers ubiquitous learning as a deliberate process with or without computing devices or digital media. An ethical framework is offered, for considering ubiquitous learning more holistically, and as a way for technology and other concerns to be determined by ethics. It follows David Holgrem’s format used to popularise Permaculture design, and draws from the theories and critiques of Ivan Illich, Neil Postman, Jean Lave and Etienne Wenger, Christopher Alexander and colleagues, Richard Stallman and others. A range of present day examples of projects and initiatives are used to show how such an ethical approach to ubiquitous learning is being realised.

Keywords: Ubiquitous learning, networked learning, ethics, peripheral participation, situationism, Illich

Developments

 * 9 August 2017 In preparation for a talk in Seoul, where I will try to continue this project into a paper called Humanist technology, I searched "Humanist Technology" and came across a valuable blog post by Michael Sacasas called Humanist technology criticsm compiled in 2015. In it he cites a good number of significant works across a 200 year period that have contributed to the question around technology and humanity. IN the end of the post he attempts a distillation of those works into a set of 5 principles that should be built into my project here:
 * "That said, I would suggest that a humanist critique of technology entails a preference for technology that:
 * (1) operates at a humane scale,
 * (2) works toward humane ends,
 * (3) allows for the fullest possible flourishing of a person’s capabilities,
 * (4) does not obfuscate moral responsibility, and
 * (5) acknowledges certain limitations to what we might quaintly call the human condition.

Mike Ananny, Kate Crawford] First Published December 13, 2016], notes to follow...
 * 20 July 2017 Connecting Data and Power, as a Learning Analytics CoP forms at RMIT and the possibility of using morphing this Ethical Framework into something relating to Data
 * 21 February 2017 On discovering a range of new features in the web service ResearchGate, I returned to this project to do a quick edit and upload it as a Working Paper and project.
 * 16 December 2016 After Joyce Seitzinger mentioned Paul Prinsloo's work interrogating there ethics of data, I had a brief exchange with Paul on his blog post Decolonising the collection, analyses and use of student data: A tentative exploration/proposal then through Twitter, Paul offered a link to [http://journals.sagepub.com/doi/full/10.1177/1461444816676645 Seeing Without Knowing: Limitations of the transparency ideal and its application to algorithmic accountability
 * October 2014: I cited this work, along with the Proposal for Intellectual Property Policy and Open Education Practices in a submission to La Trobe University's Future Ready restructure - where they were apparently seeking critical thinkers to help review and shape new operating principles. Because of this submission, I'm picking up the editing after a year or more away.
 * Rick Hayes has introduced me to a number of philosophers that I'll eventually get to look through, but Albert Borgmann is sticking, and his book Technology and the Character of Contemporary Life: A Philosophical Inquiry may need to be incorporated into this work.
 * La Trobe University sends out a communication to all staff referring to the new legislation on data privacy. In that is a reference to an "international benchmark" of 7 principles called Privacy by Design. I need to consider these in this project.

 * 29 June 2013: Work in progress presented to the 2013 IEEE International Symposium on Technology and Society
 * 31 May 2011: Reviewed edit of first draft completed
 * 25 May 2011: Peer review of first draft completed
 * 4 May 2011: (approximate date) First draft seeking peer review


text: File:Egm6341.s11.team3.HW4.6.example.jpg


text:
User:Eml5526.s11.team01/Hwk4

=Problem 4.1: Solving using WRF with a change in Basis Function= Solving a differential equation using Weight Residual Form.

Given
The partial differential equation is:

The boundary conditions are:

$$ u\left(1\right) = 0$$

$$\frac{du}{dx}\left(0\right) = -4 $$

As weighting function should be used:

Find

 * 1. Find an approximate solution of the form $$U^{h}=\sum_{i=0}^{n}d_{i}b_{i}$$ for $$\displaystyle n=2$$
 * 2. Find two equations that enforce the boundary conditions
 * 3. Project the weight residues
 * 4. Display the equations in matrix form
 * 5. Solve for $$\displaystyle d $$
 * 6. Construct $$\displaystyle U^{h}$$ and plot $$\displaystyle U^{h}$$ and $$\displaystyle u$$
 * 7. Repeat 1 to 6 for $$\displaystyle n=4$$ and $$\displaystyle n=6$$

Find an approximate solution with n=2
When n = 2 we have:
 * {| style="width:100%" border="0"

$$d_{i}=\left[ \begin{matrix} d_{0} & d_{1} & d_{2} \\ \end{matrix} \right]$$
 * $$\displaystyle (4.1.3) $$
 * }
 * }

and

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

$$b_{i}=\left[ \begin{matrix} 1 & \sin \left( x \right) & \sin \left( 2x \right) \\ \end{matrix} \right]$$
 * $$\displaystyle (4.1.4) $$
 * }
 * }

Therefore:

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

 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * style="width:2%; padding:1px; border:2px solid #FF0000" |

$$U^{h}=d_{0}+d_{1}\sin \left( x \right)+d_{2}\sin \left( 2x \right)$$
 * $$\displaystyle (4.1.5) $$
 * style= |
 * }

Find two equations that enforce the boundary conditions
The boundary conditions are:
 * {| style="width:100%" border="0"

 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * style="width:2%; padding:1px; border:2px solid #FF0000" |

$$u\left( 1 \right)=0=d_{0}+d_{1}\sin \left( 1 \right)+d_{2}\sin \left( 2 \right)$$ Differentiating
 * $$\displaystyle (4.1.6) $$
 * style= |
 * }
 * {| style="width:100%" border="0"

$$\frac{du}{dx}=0+d_{1}\cos \left( x \right)+2d_{2}\cos \left( 2x \right)$$
 * $$$$
 * }
 * }

And evaluating

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

$$\frac{du}{dx}(0)=-4=d_{1}+2d_{2}$$
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * $$\displaystyle (4.1.7) $$
 * style= |
 * }

Project the weight residues
In this case changing the function $$u\left( x \right)$$ by the approximated function $$\displaystyle U^{h}$$ in the PDE $$\left(4.1.1\right)$$, we found that $$P\left( u \right)$$ is:

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

$$P\left( U^{h} \right)=2\frac{\partial ^{2}U^{h}}{\partial x^{2}}+3=-2d_{1}\sin \left( x \right)-8d_{2}\sin \left( 2x \right)+3=0$$
 * $$\displaystyle (4.1.8) $$
 * }
 * }

$$\left(4.1.8\right)$$ can be written as:
 * {| style="width:100%" border="0"

$$P\left( U^{h} \right)=\left[ \begin{matrix} 0 & -2d_{1}\sin \left( x \right) & -8d_{2}\sin \left( 2x \right) \\ \end{matrix} \right]\left[ \begin{matrix} d_{0} \\ d_{1} \\ d_{2} \\ \end{matrix} \right]+3$$
 * $$\displaystyle (4.1.9) $$
 * }
 * }

Projecting the residue we have:

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

$$ \int_{a}^{b}b(x)P\left ( U^{h} \right )= 0 $$ After substitution of $$\left(4.1.4\right) and \left(4.1.9\right) into \left(4.1.10\right)$$, the following equation is obtained
 * $$\displaystyle (4.1.10) $$
 * }
 * }
 * {| style="width:100%" border="0"

 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * style="width:2%; padding:1px; border:2px solid #FF0000" |

$$\int_{0}^{1}{\left[ \begin{matrix} 1 \\   \sin \left( x \right)  \\ \sin \left( 2x \right) \\ \end{matrix} \right]}\left[ \begin{matrix} 0 & 2d_{1}\sin \left( x \right) & 8d_{2}\sin \left( 2x \right) \\ \end{matrix} \right]dx\left[ \begin{matrix} d_{0} \\ d_{1} \\ d_{2} \\ \end{matrix} \right]=3\int_{0}^{1}{\left[ \begin{matrix} 1 \\   \sin \left( x \right)  \\ \sin \left( 2x \right) \\ \end{matrix} \right]}dx$$
 * $$\left(4.1.11\right)$$
 * style= |
 * }

Display the equations in matrix form
Performing the product and then the integration indicated in $$\left(4.1.1\right)$$, we have
 * {| style="width:100%" border="0"

 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * style="width:2%; padding:1px; border:2px solid #FF0000" |

$$\left[ \begin{matrix} 0 & 2\sin \left( x \right) & 8\sin \left( 2x \right) \\ 0 & 2\sin ^{2}\left( x \right) & 8\sin \left( x \right)\sin \left( 2x \right) \\ 0 & 2\sin \left( x \right)\sin \left( 2x \right) & 8\sin ^{2}\left( x \right) \\ \end{matrix} \right]\left[ \begin{matrix} d_{0} \\ d_{1} \\ d_{2} \\ \end{matrix} \right]=\left[ \begin{matrix} 3 \\   1.3791  \\   2.1242  \\ \end{matrix} \right]$$
 * $$\left(4.1.12\right)$$
 * style= |
 * }

The latter equation is clearly of the form $$\displaystyle Kd=F$$. We can observe that matrix K is not symmetric.

Solve for $$\displaystyle d $$
In equation $$\left(4.1.12\right)$$ we have three equations; but, as we need to enforce the boundary conditions we take the only one of them and solve it together with equations $$\left(4.1.6\right)$$ and $$\left(4.1.7\right)$$ From these equations we now have the system:

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

$$\left[ \begin{matrix} 1 & .8415 & .9093 \\   0 & 1 & 2  \\   0 & .7944 & 4.7568  \\ \end{matrix} \right]\left[ \begin{matrix} d_{0} \\ d_{1} \\ d_{2} \\ \end{matrix} \right]=\left[ \begin{matrix} 1 \\   -4  \\   2.1242  \\ \end{matrix} \right]$$
 * $$\left(4.1.13\right) $$
 * }
 * }

After solving the system in $$\left(4.1.13\right)$$ we obtain d:
 * {| style="width:100%" border="0"

 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * style="width:2%; padding:1px; border:2px solid #FF0000" |

$$\left[ \begin{matrix} d_{0} \\ d_{1} \\ d_{2} \\ \end{matrix} \right]=\left[ \begin{matrix} 4.6608 \\   -7.3471  \\   1.6735  \\ \end{matrix} \right]$$
 * $$$$
 * style= |
 * }

Construct $$U^{h}$$ and plot $$U^{h}$$ and $$u$$
Replacing $$\displaystyle d_{i}$$ in $$\left(4.1.5\right)$$ we obtain the approximated solution:

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

$$ \displaystyle U^{h}=4.6608-7.3471x-1.6735x^{2}$$

 * $$\left(4.1.14\right)$$
 * }
 * }

On the other hand, the exact solution of the PDE $$\left(4.1.1\right)$$ with the given boundary condition is:

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

$$ u\left ( x \right )=-\frac{3}{4}x^{2}-4x+\frac{19}{4} $$
 * $$\left(4.1.15\right)$$
 * }
 * }

We observe that the approximated and the exact solution are not the same. Perhaps they would be closer if we increased the number of terms. Figure 1.1 we show the exact and the approximated solution.

Repeat 1.1 to 1.6) for i = 4 and i = 6
Now we repeat the procedure using n = 4 and n=6 in Eq.1.4

As the procedure is the same shown before, we develop a routine using Matlab to perform the products, the integration and to solve the system of equations. The routine is shown below in the appendix. Figures 1.2 and 1.3 show the results of n=4 and n=6, respectively. Figure 1.4 shows the error versus the number of terms.

We see that with increase in terms, the error increases as well.

Appendix
 Matlab Code:

Contributing Members
Solved and posted by James Roark 22:14, 28 February 2011 (UTC)

=Problem 4.2: Weak form for the equation of heat conduction= Fish and Belystschko page73, Problem 3.5.

Obtain the weak from for the equation of heat conduction with the boundary conditions $$ T\left(0\right) = 100$$ and $$ q\left(10\right) = h\bar{T} $$. The condition on the right is a convection condition.

Multiplying by the weight function
Multiplying the first two equations by the weight funtion we can write:

Applying integration by parts
Applying integration by parts to $$\left(4.2.2a\right)$$ we ca write:

Therefore we can write that:

Using the natural boundary condition we can write:

Which yields

Weak form for 1D heat conduction problems
Therefore the weak form for 1D heat conduction problems can be written as follows:

Contributing Members
Solved and posted by Cedric Adam, 1 March 2011

=Problem 4.3: Weak form for heat conduction problem in a circular plate=

Given
Fish and Belystschko page73, Problem 3.6.

Given the strong form for the heat conduction problem in a circular plate:

Natural boundary condition :

Essential boundary condition :

where $$ \displaystyle R $$ is the total radius of the plate, $$\displaystyle s $$ is the heat source per unit length along the plate radius, $$\displaystyle  T $$ is the temperature and $$\displaystyle  k $$ is the conductivity.

Find
Assume $$ k, ~s~ $$ and $$ ~R~ $$ are given:

1. Construct the weak form for the above strong form.

2. Use quadratic trial (candidate) solutions of the form $$\displaystyle T = \alpha_{0} + \alpha_{1}r + \alpha_{2}r^2 $$ and weight functions of the same form to obtain a solution of the weak form.

3. Solve the differential equation with the boundary conditions and show that the temperature distribution along the radius is given by

Construction of the weak form
First we multiply the equation in the strong form by the weight function and integrate over the domain.

The middle term cancels out when evaluated from $$\displaystyle 0 to  R $$ which yields

Quadratic trial (candidate) solutions
The trial solution $$ T\left ( r \right )$$ and the weight function $$ w\left ( r \right )$$ can be written as follows:

Using the boundary conditions, the coefficients of the trial solutions can be solved. First the you take the derivative of the trial solution and set it to the natural boundary condition.

Therefore $$\displaystyle \alpha_1 = 0 $$

Now use the essential boundary

Therefore

which solves for our two temperature functions

Now solve for the weight functions using $$ w \left( R \right) = 0 $$

Which yields

And the derivative of the weight function is

Before substitution, when the middle term is evaluated from $$ \left( 0 to R \right) $$ it goes to zero

Therefore we can cancel out the middle term and substitute in the rest

Integrating we get:

After evaluating the domain, the equation simplifies to:

Factoring out $$\displaystyle \beta_1 $$ and $$\displaystyle \beta_2 $$ we get:

Assuming that $$\displaystyle \beta_1 $$ and $$\displaystyle \beta_2 $$ equal zero, $$\displaystyle \alpha_2 $$ can be solved:

Now $$\displaystyle \alpha_2 $$ can be substituted into $$ T \left( r \right) $$ to obtain the solution of the weak form

Solve the differential equation
Using the strong form the differential equation can be solved by taking the integral:

Using the natural boundary condition $$ \frac{dT}{dr} \left(r = 0 \right) = 0 \rightarrowc_1 = 0 $$

Integrate once again we get:

Using the essential boundary condition $$ T \left( r = R \right) = 0 $$ solves for:

substituting in $$\displaystyle c_2 $$ yields the temperature distribution

Contributing Members
Solved and posted by Harrison Sheffield 12:19, 1 March 2011 (UTC)

=Problem 4.4: Weak form for circular bar in torsion=

Given
Fish and Belystschko page73, Problem 3.7.

The strong form for the circular bar in torsion (Figure 4.1) is the following:

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

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

$$\frac{d}{dx}\left( JG\frac{d\phi }{dx} \right)+m=0,\quad \quad \quad 0\le x\le 1,$$

 *  $$\displaystyle (4.4.1) $$
 * }
 * }

where $$\displaystyle JG=(2+3x),\quad m(x)=5x .$$

Natural boundary condition:

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

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

$$\frac{d\phi }{dx}(x=1)=-6,$$

 *
 $$\displaystyle (4.4.2) $$
 * }
 * }

Essential boundary condition:

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

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

$$\phi \left( x=0 \right)=\bar{\phi }=4,$$

 *
 $$\displaystyle (4.4.3) $$
 * }
 * }

where $$\displaystyle m(x)$$ is a distributed moment per unit length, $$\displaystyle M$$ is the torsion moment, $$\displaystyle \phi $$ is the angle of rotaion, $$\displaystyle G$$ is the shear modulus and $$\displaystyle J$$ is the polar moment of inertia given by $$\displaystyle J=\pi {{C}^{4}}/2$$ , where $$\displaystyle C$$ is the radius of the circular shaft.

Find
1. Construct the weak form for the circular bar in torsion.

2. Integrate the differential equation given above. Find the integration constants using boundary conditions.

3. Plot the result of $$\phi(x)$$.

Construct the weak form for the circular bar in torsion
As the weight function must vanish on the essential boundaries, we consider all smooth weight functions w(x) such that w(0)=0. The trial solutions must satisfy the essential boundary condition $$\phi \left( x=0 \right)=4,$$.

We start by multiplying the governing equation and the natural boundary condition over the domains where they hold by an arbitrary weight function:
 * {| style="width:100%" border="0"

$$ \int_{0}^{1} w [\frac{\mathrm{d} }{\mathrm{d} x}( (2+3x)\frac{\mathrm{d} \phi}{\mathrm{d} x})+5x]dx = 0 \forall w(x), $$

 *
$$\displaystyle (4.4.4) $$
 * }
 * }

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

$$ [w(\frac{\mathrm{d} \phi}{\mathrm{d} x}+6 )]_{x=1} = 0   \forall w(1). $$

 *
$$\displaystyle (4.4.5) $$
 * }
 * }

Next we integrate (4.4.4) by parts, then we get the following result:

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

$$ \int_{0}^{1} w [\frac{\mathrm{d} }{\mathrm{d} x}((2+3x)\frac{\mathrm{d} \phi}{\mathrm{d} x})]dx = \left. \left( w(2+3x)\frac{\text{d}\phi }{\text{d}x} \right) \right|_{x=0}^{x=1} - \int_{0}^{1} \frac{\mathrm{d} w }{\mathrm{d} x}(2+3x)\frac{\mathrm{d} \phi}{\mathrm{d} x}dx $$

 *
$$\displaystyle (4.4.6) $$
 * }
 * }

We have constructed the weight functions so that w(0)=0; therefore, the first term on the RHS of (4.4.6) vanishes at x=0. Substituting (4.4.6) into (4.4.4) gives:

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

$${{\left. \left( w(2+3x)\frac{\text{d}\phi }{\text{d}x} \right) \right|}_{x=1}}-\int_{0}^{1}{\frac{\text{d}w}{\text{d}x}}(2+3x)\frac{\text{d}\phi }{\text{d}x}dx+5\int_{0}^{1}{wxdx}=0 \forall w(x)\quad with\quad w(0)=0.$$

 *
$$\displaystyle (4.4.7) $$
 * }
 * }

Substituting (4.4.5) into the first term of (4.4.7) gives:

Integrate the differential equation and find the integration constants using boundary conditions
(Eq.4.1) yields:

Apply the boundary condition, $$\frac{d\phi }{dx}(x=1)=-6$$ gives:

So,

Apply the boundary condition $$\phi \left( 0 \right)=4  $$ gives:

Then we have the following result:

MATLAB Source for Plot
The following MATLAB code was used to plot the result.

Contributing Members
Solved and posted by Yu Hao 1 March 2011

=Problem 4.5: Various Basis Functions satisfying the Constraint Breaking Solution=

Given

 * a) Constraint Breaking Solution which needs to be satisfied is $$b_{0}(\beta)\neq 0$$ and $$b_{i}(\beta)=0$$, i=1,2,3,....,n.
 * b) Consider $$\Omega=[-2,4]$$

Find
For $$\Gamma _{g}=\left \{ \beta\right \}$$ and each of the following Families of Basis Functions Find the Corresponding Basis Functions Satisfying the Given Constraint Breaking Solution. Plot original and required basis function for $$j=1,2,3$$
 * 1. $$F_{p}:=\left \{ x^{j}, j=0,1,2,...\right \}$$.
 * 2. $$F_{c}:=\left \{ \cos(jx), j=0,1,2,... \right \}$$.
 * 3.$$F_{s}:=\left \{ 1,\sin(jx), j=1,2,... \right \}$$.
 * 4. $$F_{f}:= \left \{\cos(jx),j=0,1,2,...\sin(jx),j=1,2,...\right\}$$
 * 5. $$F_{e}:=\left \{ e^{jx},j=0,1,2,.. \right \}$$.
 * 6. Show that $$\left \{ e^{jx},j=0,1,2,.. \right \}$$ is Linear Independent on $$\Omega$$.

1) Find the Corresponding Polynomial Basis Function $$F_{p}:=\left \{ x^{j}, j=0,1,2,...\right \}$$ Satisfying the Given Constraint Breaking Solution
If we choose $$b_{i}\left ( x \right )=\left(x-\beta\right)^{j}$$, it satisfies the given Constraint Breaking Solution as below,

and

So the Polynomial Basis Function satisfying the given constraint breaking solution is,

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

$$b_{j}\left ( x \right )=\left \{\left(x-\beta\right)^{j}, j=0,1,2,...\right \}$$ Plot Between the original and the required basis function is plotted in the MATLAB using the following code.
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 *
$$\displaystyle (4.5.3) $$
 * style= |
 * }

Appendix
 Matlab Code:

2) Find the Corresponding Cosine Basis Function $$F_{c}:=\left \{ \cos(jx), j=0,1,2,... \right \}$$ Satisfying the Given Constraint Breaking Solution
If we select $$b_{j}\left ( x \right )=\cos(j(x-\beta)), j=0$$

It will yield, $$b_{0}\left ( \beta\right)=\cos(0)=1\neq0$$

and further

$$b_{j}\left ( x \right )=\cos(\frac{\pi}{2}(2j+1)\frac{x}{\beta}), j=1,2,...$$

it will yield,

$$b_{j}\left ( \beta \right )=\cos(n\frac{\pi}{2})=0, n = 1, 3, 5, ...$$

So the Cosine Basis Function satisfying the given constraint breaking solution is,

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

$$b_{j}\left ( x \right )=\left \{\cos(j(x-\beta)), j=0\right \}$$
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * style="width:2%; padding:1px; border:2px solid #FF0000" |

&

$$b_{j}\left ( x \right )=\left \{\cos(\frac{\pi}{2}(2j+1)\frac{x}{\beta}), j=1,2,...\right \}$$ Plot of the original and the required Basis Function is plotted in the MATLAB using the following code
 *
$$\displaystyle (4.5.4) $$
 * style= |
 * }

Appendix
 Matlab Code:

3) Find the Corresponding Sine Basis Function $$F_{s}:=\left \{1,\sin(jx), j=1,2,... \right \}$$ Satisfying the Given Constraint Breaking Solution
If we select,

$$b_{j}\left ( x \right )=\sin(j\pi\frac{x}{\beta}), i=1,2,...$$

It yields,

$$b_{j}\left ( \beta \right )=\sin(n\pi)=0, n = 1,2,3,...$$

So the Sine Basis Function satisfying the given constraint breaking solution is,

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

$$b_{j}\left ( x \right )=\left \{1,\sin(j\pi\frac{x}{\beta}), j=1,2,...\right \}$$
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 *
$$\displaystyle (4.5.5) $$
 * style= |
 * }

Appendix
 Matlab Code:

===4) Find the Corresponding Fourier Basis Function $$F_{f}:=\left \{\cos(jx),j=0,1,2,... \sin(jx),j=1,2,...\right\}$$ Satisfying the Given Constraint Breaking Solution===

If we select, According to $$\displaystyle (4.5.4)$$

$$b_{j}\left ( x \right )=\cos(j(x-\beta)), j=0$$

It will yield, $$b_{0}\left ( \beta\right)=\cos(0)=1\neq0$$

and further

$$b_{j}\left ( x \right )=\cos(\frac{\pi}{2}(2j+1)\frac{x}{\beta}), j=1,2,...$$

it will yield,

$$b_{j}\left ( \beta \right )=\cos(n\frac{\pi}{2})=0, n=1,3,5,...$$

Also,

If we select, According to $$\displaystyle (4.5.5)$$,

$$b_{j}\left ( x \right )=\sin(j\pi\frac{x}{\beta}), i=1,2,...$$

It yields,

$$b_{j}\left ( \beta \right )=\sin(n\pi)=0,n=1,2,3,...$$

Eventually,

The Fourier Basis Function Satisfying the given Constraint Breaking Solution is,

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

$$b_{j}\left ( x \right )=\left \{1,\cos(\frac{\pi}{2}(2j+1)\frac{x}{\beta}), j=1,2,...,
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * style="width:2%; padding:1px; border:2px solid #FF0000" |

\sin(j\pi\frac{x}{\beta}), j=1,2,...\right \}$$ Plot of the original and the required Basis Function is plotted in the MATLAB using the following code.
 *
$$\displaystyle (4.5.6) $$
 * style= |
 * }

Appendix
 Matlab Code:

===5) Find the Corresponding Exponential Basis Function $$F_{e}:=\left \{ e^{jx},j=0,1,2,... \right\}$$ Satisfying the Given Constraint Breaking Solution===

If we select,

$$b_{j}\left ( x \right)=\left \{e^{jx\beta}, i=0\right \}$$

and further,

$$b_{j}\left ( x \right)=\left \{e^{j(x-\beta)}-1, i=1,2,...\right \}$$

Eventually, The Fourier Basis Function Satisfying the Given Constraint Breaking Solution is,

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

$$b_{j}\left ( x \right )=\left \{1,e^{j(x-\beta)}-1, i=1,2,...\right \}$$ Plot of the original and the required Basis Function is plotted in the MATLAB using the following code.
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 *
$$\displaystyle (4.5.7) $$
 * style= |
 * }

Appendix
 Matlab Code:

6) Show that $$\left \{ e^{jx},j=0,1,2,.. \right \}$$ is Linear Independent on $$\Omega$$
We have,

Calculating $$\Gamma (F)$$ for the given interval $$\Omega [-2,4]$$,we get

Therefore,

To check the Linear Independent Behavior we need to check the Determinant of this matrix. So calculating the determinant of this matrix in the MATLAB Using the Following code,

Appendix
 Matlab Code:

The Determinant arrives as $$1.1145\times (10)^{9}$$ which is a Non-zero number.

So we can say that, $$\left \{ e^{jx},j=0,1,2,.. \right \}$$ is Linear Independent on $$\Omega$$.
=Problem 4.6: Strong and weak form for elastic bar with variable distributed spring=

Given
Fish and Belystschko page74, Problem 3.9.

Consider an elastic bar with a variable distributed spring $$\displaystyle p\left( x \right)$$ along its length as shown in Figure 6.1. The distributed spring imposes an axial force on the bar in proportion to the displacement. Consider a bar of length l, cross-sectional area $$\displaystyle A\left( x \right)$$, Young’s modulus $$\displaystyle  E\left( x \right)$$ with body force $$\displaystyle  b\left( x \right)$$ and boundary conditions as shown in Figure 6.1

Find

 * 1. Construct the strong form.
 * 2. Construct the weak form.

Construct the strong form
From figure 6.1, construct a free body diagram Balance the forces in figure 6.2 to obtain:

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

$$\left[ p\left( x+\frac{dx}{2} \right)+b\left( x+\frac{dx}{2} \right) \right]dx+F\left( x+dx \right)-F\left( x \right)=0$$
 *
$$\displaystyle (4.6.1) $$
 * }
 * }

Rearranging:

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

$$\left[ p\left( x+\frac{dx}{2} \right)+b\left( x+\frac{dx}{2} \right) \right]+\frac{F\left( x+dx \right)-F\left( x \right)}{dx}=0$$
 *
$$\displaystyle (4.6.2) $$
 * }
 * }

Let $$dx\to 0$$, therefore

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

$$\left[ p\left( x \right)+b\left( x \right) \right]+\frac{dF}{dx}=0$$
 *
$$\displaystyle (4.6.3) $$
 * }
 * }

Note that $$F=\sigma A\left( x \right)=E\left( x \right)\varepsilon A\left( x \right)=E\left( x \right)A\left( x \right)\frac{du}{dx}$$

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

$$\left[ p\left( x \right)+b\left( x \right) \right]+\frac{d}{dx}\left[ E\left( x \right)A\left( x \right)\frac{du}{dx} \right]=0$$
 *
$$\displaystyle (4.6.4) $$
 * }
 * }

Apply boundary conditions, let $$u\left( l \right)=\bar{u}$$and if $$\bar{t}>0$$ at $$x=l$$, then $$\bar{t}=\left. E\left( x \right)\frac{du}{dx} \right|_{x=0}=0$$. We arrive at the strong form

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

$$\left[ p\left( x \right)+b\left( x \right) \right]+\frac{d}{dx}\left[ A\left( x \right)E\left( x \right)\frac{du}{dx} \right]=0$$
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 *
$$\displaystyle (4.6.5a) $$
 * style= |
 * }
 * {| style="width:100%" border="0"

$$A\left( x \right)\left( E\left( x \right)\frac{du}{dx}+\bar{t} \right)=0$$
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 *
$$\displaystyle (4.6.5b) $$
 * style= |
 * }

Construct the weak form
Let $$p\left( x \right)+b\left( x \right)=\hat{b}\left( x \right)$$in equation 6.5a.

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

$$\frac{d}{dx}\left[ A\left( x \right)E\left( x \right)\frac{du}{dx} \right]+\hat{b}\left( x \right)=0$$
 *
$$\displaystyle (4.6.6) $$
 * }
 * }

Following the procedure outlined in Fish & Belytschko section 3.2, page 47, multiply equations 6.5a and 6.5b by an arbitrary function $$w\left( x \right)$$such that $$w\left( l \right)=0$$and integrate to obtain

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

$$\int\limits_{0}^{l}{w\left( x \right)\left[ \frac{d}{dx}\left[ A\left( x \right)E\left( x \right)\frac{du}{dx} \right]+\hat{b}\left( x \right) \right]}=0$$
 *
$$\displaystyle (4.6.7a) $$
 * }
 * }

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

$$\left. w\left( x \right)A\left( x \right)\left( E\left( x \right)\frac{du}{dx}+\bar{t} \right) \right|_{x=0}=0$$
 *
$$\displaystyle (4.6.7b) $$
 * }
 * }

Rewrite equation 4.6.7a

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

$$\int\limits_{0}^{l}{w\left( x \right)\frac{d}{dx}\left[ A\left( x \right)E\left( x \right)\frac{du}{dx} \right]}dx+\int\limits_{0}^{l}{w\left( x \right)\hat{b}\left( x \right)}dx=0$$
 *
$$\displaystyle (4.6.8) $$
 * }
 * }

Integrate the first term of equation 4.6.8 by parts

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

$$\left. \left[ w\left( x \right)\underbrace{A\left( x \right)E\left( x \right)}_{\sigma } \right] \right|_{0}^{l}-\int\limits_{0}^{l}{\frac{dw}{dx}A\left( x \right)E\left( x \right)\frac{du}{dx}}dx+\int\limits_{0}^{l}{w\left( x \right)\hat{b}\left( x \right)}dx=0$$
 *
$$\displaystyle (4.6.9) $$
 * }
 * }

Since $$w\left( l \right)=0$$, this term drops out as zero. Noting that $$\left. {\bar{t}} \right|_{x=0}=\left. \sigma \right|_{x=0}$$

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

$$\int\limits_{0}^{l}{\frac{dw}{dx}A\left( x \right)E\left( x \right)\frac{du}{dx}}dx=\left. \left[ w\left( x \right)A\left( x \right)\bar{t} \right] \right|_{0}+\int\limits_{0}^{l}{w\left( x \right)\hat{b}\left( x \right)}dx$$
 *
$$\displaystyle (4.6.10) $$
 * }
 * }

Replacing $$\hat{b}$$ we arrive at the weak form

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

$$\int\limits_{0}^{l}{\frac{dw}{dx}A\left( x \right)E\left( x \right)\frac{du}{dx}}dx=\left. \left[ w\left( x \right)A\left( x \right)\bar{t} \right] \right|_{0}+\int\limits_{0}^{l}{w\left( x \right)\left[ b\left( x \right)+p\left( x \right) \right]}dx$$
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 *
$$\displaystyle (4.6.11) $$
 * style= |
 * }

Contributing Members
Solved and posted by James Roark 02:51, 2 March 2011 (UTC)

=Problem 4.7: working with calculix. From lecture 23 =

Given
Calculix is a nonlinear Finite Element code, open source, Abaqus-like

Find

 * 1. From the page http://www.dhondt.de/ download and install Calculix
 * 2. Read the manual, sign up with user group to as questions if any. Also access archive.
 * 3. Reproduce basic examples: disk, cylinder, sphere, sphere-volume, airfril.
 * 4. write a report for dummies explaining how to install and all the commands used in basic examples.

Download and instal Calculix
The program was succesfully downloaded and instaled

Read the manual, sign pu with user group t ask questions if any
The manual was downloaded but it was not succesfully opened from the page given in the problem, but the manual was found in the page http://web.mit.edu/calculix_v2.0/CalculiX/cgx_2.0/doc/cgx/index.html where it could be easily opened. From there the manual was read.

Reproduce basic examples: disk, cylinder, sphere, sphere-volume, airfril
The basics examples were reproduced: disc, cylinder, sphere, sphere_seg, and sphere_vol they are shown in above in the attached document.

Write report for "dummies" explining how to install and run cgx and all the commands used in basic examples
A document with a simple explanation about the installation and the construction of the simple examples listed above is shown in the pdf document

http://upload.wikimedia.org/wikiversity/en/3/35/Fe1.s11.team1.hw4.Basic_Calculix_Manual.pdf

It was found difficult to work with this interface because not icons or graphic interfaces are available, instead you have to remember all the commands and its sintaxis. to counter the beforementioned problem there are some better options like using a program where the pre and post process is made in a very easy way by using a completely graphic interphase and you can write your own conde in the program that you prefer. One of those options is the program GID which can be download free going to the page:

http://www.gid-usa.com/index.html

With the free version of GiD you can work with a limited number of nodes, but it is enough to solve simple problems that are usefull for your training.

Contributing Members
Solved and posted by Fernando Casanova 18:50, 28 February 2011 (UTC)

=Problem 4.8: Based off Problem 3.4=

Given
Based off Problem 3.4

Given the strong form

and

with

Find
Find $$ \tilde{M} $$

Assuming

$$ A = 1, E = 2, \bar{m} = 3 $$

Find $$ F \left(t \right) $$

Assuming

$$ n = 3 $$

$$ u^{h} \left ( p,t \right ) = g\left ( t \right ) = sin2t $$

From the strong form
From the strong form we can easily obtain the weak form. Find $$\displaystyle u(x) $$ smooth enough and satisfy the essential boundary condition:

such that:

since

Thus

From discrete weak form
From discrete weak form we get:

=Contributing Members=

title: User:Eml5526.s11.team01/Hwk4


title: File:Fe1.s11.team1.hw4.fig1.1.jpg


text:
Template:Preschool Language and Skills

Guide | 1 | 2 | 3 | 4 | 5 | 6 |

title: Template:Preschool Language and Skills


title: File:Fe1.s11.team1.hw4.fig1.2.jpg


title: File:Fe1.s11.team1.hw4.fig1.3.jpg


title: File:Fe1.s11.team1.hw4.fig1.4.jpg


text:
Preschool Language and Skills/2

 * Created: 2011 02 28 | Percent completed: [[Image:100%.png]]

Reinforcer
Positive reinforcement, by definition, is an item or activity that when presented after a behavior increases the the behavior in the future. Simply, the item or activity is only a reinforcer if it increases an identified behavior.
 * Definition

Guidelines for use of reinforcers

 * 1) Sanitize the environment
 * 2) * Limit the availability and visibility of reinforcers, toys, and items. Put them up, onto a shelf in boxes that are labeled or in a locked room.
 * 3) * Every time a fun item appears, it is because you made it available, thereby making you more valuable.
 * 4) Approach the student with something fun
 * 5) * Approach the student with a reinforcer in an outstretched hand.
 * 6) * Make it obvious that you are approaching with something fun.
 * 7) Pair your voice and the environment with reinforcement
 * 8) * While pairing with the student’s reinforcers, say the student’s name and use short phrases to describe what you are doing.
 * 9) Make sure that what you are offering is more desirable then their current situation
 * 10) * Do not interrupt the student’s fun to do something less reinforcing.
 * 11) Make activities more fun
 * 12) * When a student is playing with a toy or engaged in a fun activity, do things that make it more enjoyable.
 * 13) Become a conditioned reinforcer
 * 14) * A conditioned reinforcer is a type of reinforcer that obtains its value by having been paired with other reinforcers.
 * 15) * Students with autism who have many conditioned reinforcers will learn important skills more easily.
 * 16) * The goal of pairing is for the student to like to be with the staff and to approach him or her without hesitation.
 * 17) * The student should be looking for ways to be with you; not escape from you.
 * 18) * Be sure this happens BEFORE you place instructional demands (ex. identifying body parts, calendar, imitation, etc.).
 * 19) * Only routine instructions should be given. For example, being asked to sit for lunch as part of the daily routine.

 * Goals
 * 1) Have at least 10 reinforcer items/activities.  If you don't, shorten teaching time and build up reinforcers.
 * 2) Parent/therapist as reinforcer (does child come to you willingly)
 * 3) Will take a food reinforcer from a table and eat it.
 * 4) Will take a food reinforcer from hand and eat it.  If not, start with popping food reinforcer into mouth at first.
 * 5) Will take a toy reinforcer and play with it safely (appropriately or inappropriately)
 * 6) No free access to reinforcer.  If the student has free access, he/she might satiate on the item and then you will no longer be able to use the reinforcer as a teaching tool.

Remember: you want the student to run TO you, not AWAY from you.

title: Preschool Language and Skills/2


text:
User:Egm5526.s11.team4.zou/HW4

= Problem 4.1 - Residue Projection for Modified Trigonometric Function Basis =

This problem refers to the previous Problem 2.9 and Problem 3.11

Problem Statement
Given the differential equation in strong form,

Consider the basis

Let, $$\displaystyle n=2,4$$ or $$\displaystyle 6$$ where $$\displaystyle ndof=n+1=3, 5$$ or $$\displaystyle 7$$ where ndof is the number of degrees of freedom and $$\displaystyle d=\left\{ {{d}_{j}};j=0,1,2.....n \right\}$$.

 * 1) Find two equations that enforce boundary conditions for

 * 1) Find $$\displaystyle 1,3$$ or $$\displaystyle 5$$ more equation(s) to solve for $$\displaystyle \mathbf{d}={{({{d}_{j}})}_{n\times 1}}

(j=0,1,2....n)$$by projecting the residue $$\displaystyle P({{u}^{h}}$$)on the basis function$$\displaystyle {{b}_{k}}(x)$$with $$\displaystyle k=1,2,3\cdots \cdots

$$ such that the additional equation is linearly independent from the above equation in part2.

 * 1) Display $$\displaystyle 3/5/7$$ equations in matrix form $$\displaystyle \mathbf{KD=F}$$ Observe the symmetry properities of$$\displaystyle \mathbf{K}$$

 * 1) Solve for $$\displaystyle \mathbf{d}$$

 * 1) Construct $$\displaystyle {{u}^{h}}(x)$$and plot $$\displaystyle {{u}^{h}}(x)$$ versus $$\displaystyle {{u}_{exact}}(x)$$

Exact Solution
With the general solution

After applying boundary conditions

For the case n=2
since $$\displaystyle {{b}_{j}}=\left\{ 1,\sin x,\sin 2x \right\}$$ with $$\displaystyle {{d}_{j}}=\left\{ {{d}_{0}},{{d}_{1}},{{d}_{2}} \right\}$$,

Boundary Conditions:

Additional equation:

For the third condition, we project the residue on the basis function.

Where:

hence,

and

thus the projection is

Putting equation 4.1.6, 4.1.7 and 4.1.11 into matrix form, we can have

Result:

For the case n=4
The approximating solution

Enforcing the boundary conditions:

There still three additional equations which been derived from enforcing the projection of $$\displaystyle P(u)$$ on basis to be zero:

putting equations 4.1.15 to 4.1.20 in matrix form yields:

For the case n=6
The approximating solution

Enforcing the boundary conditions:

There still three additional equations which been derived from enforcing the projection of $$\displaystyle P(u)$$on basis to be zero:

putting equations 4.1.25 to 4.1.31 in matrix form yields:

Thus

Comparison of error
The maximum error for each case is listed below

It is clearly seen that for the cases n=4 and n-6 the errors are negligible.

Problem Statement

 * 1) Download the windows-version of Calculix from its website, and then install it on computer.
 * 2) Run the CalculiX Command: [Start] [Programs] [bConverged][CalculiX] [CalculiX Command]
 * 3) Reproduce the basic examples
 * 4) Write a report

A. The “disc” problem
1. First switch to the directory of working ‘E:\Documents\ccx’.

2. Then input the command ‘cgx –b disc.fbd’, there will appear another command “CalculiX Graphix”.

3. Input the command” pnt [  ]” and then plot all the points with the command ‘plot p all b’. For example, ”PNT py  0.0  1.0 0.0” is used to define a point which is located at (0,1,0). The figure is showed below.

4. Input the command” line    ” and the “gsur '+|-'BLEND| ' '+|-'  '+|-' -> .. (3-5 times)and then plot all the lines and surfaces with command ‘plus’. For example, “LINE L001 P00I P001 p0 4 ” is used to define a arc line which pass through points P00I and P001 and its center point is p0 and radius is 4. The figure is showed below.

5. You can use mouse on the Graphix command to rotate and zoom in or out the figure.

6. Input the command ‘ELTY all QU4’ and ‘mesh all’, and then left-click on the marginal area on the Graphix, then on ’Viewing’ and continue to ‘show all elements with light’ and ‘Toggle Element edges’, as showed in following figures

B. The “cylinder” problem
1. First switch to the directory of working ‘E:\Documents\ccx’.

2. Then input the command ‘cgx –b cylinder.fbd’, there will appear another command “CalculiX Graphix”.

3. Input the command” pnt [  ]” and then plot all the points. Input the command” line    ” and the “gsur '+|-'BLEND| ' '+|-'  '+|-' -> .. (3-5 times)and then plot all the lines and surfaces. The figure is showed below

4. Use the command “seta |'n'|'e'|'p'|'l'|'c'|'s'|'b'|'S'|'L'|'se' <name ..> ['n'|'e' <name” to define some sets to be used in further loading operations. For example, “SETA p1 p p1” is to define a set of points which contains point p1.

5. You can use mouse on the Graphix command to rotate and zoom in or out the figure.

6. Input the command ‘ELTY all QU4’ and ‘mesh all’, and then left-click on the marginal area on the Graphix, then on ’Viewing’ and continue to ‘show all elements with light’ and ‘Toggle Element edges’, as showed in following figure.

C. The “sphere” problem
1. First switch to the directory of working ‘E:\Documents\ccx’

2. Then input the command ‘cgx –b sphere.fbd’, there will appear another command “CalculiX Graphix”.

3. Input the command” pnt [<x> <y> <z>]” and then plot all the points with the command ‘plot p all b’. For example, ”PNT py  0.0  1.0 0.0” is used to define a point which is located at (0,1,0). The figure is showed below.

4. Input the command” line <p1> <p2> <cp|seq> ” and the “gsur '+|-'BLEND| ' '+|-' <line|lcmb> '+|-' -><line|lcmb> .. (3-5 times)and then plot all the lines and surfaces with command ‘plus’. For example, “LINE L001 P00I P001 p0 104 ” is used to define a arc line which pass through points P00I and P001 and its center point is p0 and radius is 104. The figure is showed below.

5. Input the command ‘ELTY all QU4’ and ‘mesh all’, and then left-click on the marginal area on the Graphix, then on ’Viewing’ and continue to ‘show all elements with light’ and ‘Toggle Element edges’, as showed in following figure.

D. The “sphere-volume” problem
1. First switch to the directory of working ‘E:\Documents\ccx’

2. Then input the command ‘cgx –b sphere-volu.fbd’, there will appear another command “CalculiX Graphix”.

3. Input the command” pnt [<x> <y> <z>]” and then plot all the points. Input the command” line <p1> <p2> <cp|seq> ” and the “gsur '+|-'BLEND| ' '+|-' <line|lcmb> '+|-' -><line|lcmb> .. (3-5 times)and then plot all the lines and surfaces. The figure is showed below

4. Input the command” gbod 'NORM' '+|-' '+|-' ->( 5-7 surfaces )” to define a volume. For example, “GBOD B001 NORM + A006 - A003 - A004 + A002 + A001” is to define a volume which consists of these above areas.

5. Input the command ‘ELTY all he20’ and ‘mesh all’, and then left-click on the marginal area on the Graphix, then on ’Viewing’ and continue to ‘show all elements with light’ and ‘Toggle Element edges’, as showed in following figure.

E. The “airfoil” problem
1. First switch to the directory of working ‘E:\Documents\ccx’

2. Then input the command ‘cgx –b airfoil.fbd’, there will appear another command “CalculiX Graphix”.

3. Input the command” pnt [<x> <y> <z>]” and then plot all the points with the command ‘plot p all b’. For example, ” PNT P05D     0.00118       -5.37045        0.00000” is used to define a point which is located at (0.00118,-5.37045,0.00000). The figure is showed below.

4. Input the command ”seqa ['pnt' .. <=>]|['afte'|'befo' .. <=>]|['end' .. <=>]” to define a sequential set. For example, “SEQA S005 pnt  P010 P00C P00A P01C” is used to define a sequential point set which pass through points P010 P00C P00A P01C sequentially.

5. Input the command ”lcmb ['+|-' '+|-' '+|-' ->  ..(up to 14 lines)]| ['ADD' '+|-' '+|-' -> '+|-' ..(up to 14 lines)]” to combine lines. For example, “ LCMB C001 + L015 + L01X + L01Y + L016” is used to combine the lines of C001, L015 + L01X + L01Y, L016.

6. Input the command” line <p1> <p2> <cp|seq> ” and the “gsur '+|-'BLEND| ' '+|-' <line|lcmb> '+|-' -><line|lcmb> .. (3-5 times)and then plot all the lines and surfaces with command ‘plus’. For example, “LINE L001 P00I P001 p0 104 ” is used to define a arc line which pass through points P00I and P001 and its center point is p0 and radius is 104. The figure is showed below.

7. Input the command ‘ELTY all QU4’ and ‘mesh all’, and then left-click on the marginal area on the Graphix, then on ’Viewing’ and continue to ‘show all elements with light’ and ‘Toggle Element edges’, as showed in following figure.

=Problem 4.8 Finding Mass Matrix in weak form=

Given:
The strong form:

and

with:

Find
the mass matrix $$\displaystyle \tilde{M}$$

Solution:
Form the strong form, we can easily obtain the weak form: Find $$\displaystyle U(x)$$ smooth enough and satisfy the essential boundary condition:$$\displaystyle u(x=3)=0.001$$ such that:

since $$\displaystyle {{u}^{h}}(x=3,t)=\sin (2t)$$

Thus:

Consider

So

From the discrete weak form,

title: User:Egm5526.s11.team4.zou/HW4