text:
User:Eml5526.s11.team01/Hwk5

 HOMEWORK 5

=Problem 5.1: Weak form with appropriate functions such as Polynomial, Fourier & Exponential until convergence.=

Given
General 1-D Model 1 Data set 1. The Partial Differential equation for this data set is,

Find
Solve the Weak Form of the above general model 1 data set 1 for the Functions such as

1)Polynomial Basis Function,

2)Exponential Basis Function &

3)Fourier Basis Function

to find out the convergence point.

1) Polynomial Basis Function
Solve the Differential Equation

in the interval (0,1) with the boundary conditions

The Weak Form of the given Differential Equation will be obtained as follows

So we need to integrate the above equation using integration by parts, after which we will get the following equation,

We now have to select the appropriate Polynomial Basis Function which will follow the constraint breaking solution. so selecting

Clearly,

We need, So Choosing,

We now have, Using the essential Boundary condition we have, Therefore, In the identical way we can also write,

With the condition,

Therefore,

Now we need to calculate the derivative of $$U$$ & $$w$$, Therefore,

Substituting this into the weak form of $$\left(5.1.9\right)$$, we get,

Cancelling $$c_{i}$$ and Organizing, we get, Simplifying the last equation we get,

We can obtain the value of $$\displaystyle a_{j}$$ from the above equation.

The analytical exact solution on the other hand is obtained as,

therefore,

The Values of p & r are obtained by using the following boundary conditions,

To solve this system of equations we developed a MATLAB code. It was found out that with $$n=2$$ a good approximation is obtained. However with$$n=8$$ the error was smaller than $$10^{-6}$$. The MATLAB code is as follows.

Appendix
 Matlab Code:

The analytical and the numerical solution along with the Error between the two are plotted as below.

2) Fourier Basis Function
To follow the constraint Breaking Solution we choose,

We need to satisfy the essential boundary condition So we choose the basis Function such that So we can write, Now the trial solution will be as follows,

Now we need to solve for the remaining coefficients By constructing a Stiffness & Force Matrices as follows.

Here the first equation gets satisfied for even values of i & j,

Whereas 2nd equation is satisfied for odd values of i & j,

further 3rd equation is satisfied for remaining combinations of i & j.

Here the first equation gets satisfied for even values of i & j,

Whereas 2nd equation is satisfied for odd values of i & j,

We developed a MATLAB code to generate the Force and the stiffness Matrix and We get the following values.

We can now use the relationship which is, $$\displaystyle Kd=F$$ to solve for $$\displaystyle d$$.

Ultimately the trial solution of the equation is,

We have developed at MATLAB code to solve these system of equations & to deliver the Error between analytical and the numerical solution.

Appendix
 Matlab Code:

The analytical and the numerical solution along with the Error between the two are plotted as below. From the Plots we can explicitly say that the solution converges at n=9.

3) Exponential Basis Function
To follow the constraint Breaking solution we choose: Clearly, And as we need We choose, Eventually we have, From the essential boundary condition, we have, Therefore, In the identical way we have, Using the condition, Therefore Derivative of U & w will be,

Replacing in the weak form of $$\left(5.1.5\right)$$ we get,

Cancelling $$c_{i}$$ and then organizing we obtain, Performing the integration we obtain following equation,

The value of the $$a_{j}$$can be calculated from the above equation. we developed a MATLAB code to find the error between the Exact and the analytical solution. The MATLAB code is as follows.

Appendix
 Matlab Code:

The following two figures shows the basis Function and the Error between the Analytical and the numerical solution. It can be observed that the Exponential basis Function converges somewhat Slowly than the Polynomial basis Function. For exponential the convergence was achieved with 10 terms whereas with polynomial, it was achieved with only 8 functions. Moreover the stiffness matrix obtained with exponential functions was bad-conditioned for a number of functions greater than 8.

=Problem 5.2: Weak form with appropriate functions such as Polynomial, Fourier & Exponential until convergence.=

Given
General 1-D Model 1 Data set 1. The Partial Differential equation for this data set is,

Find
Solve the Weak Form of the above general model 1 data set 1 for the Functions such as

1)Polynomial Basis Function,

2)Fourier Basis Function &

3)Exponential Basis Function

to find out the convergence point.

Solution
Solve the differential equation $$ \frac{d}{dx} \left[ \left(2+3x \right)  \frac{du}{dx} \right]+5x = 0 $$ in the interval (0,1) with the boundary conditions

$$ u\left(1 \right)= 4 ; \frac{du}{dx} \left(0 \right)= -6 $$

The weak form was obtained in the latter section:

As we do not know $$ \frac{du}{dx} \left(1\right) $$ then we chose $$ w \left(1 \right) = 0 $$ ; moreover $$ u^h \left(1 \right)= 4 $$

1) Polynomial Basis Function
Polynomial basis functions To follows the constrain braking solution, we chose:

Clearly $$ b_1=1 $$, and as we need that $$ b_i \left(1 \right) = 0, i = 1,2,... $$;we chose $$ \beta=1 $$. Then we have $$ U^h \left(x\right)= a_0+a_1 \left(x-1 \right)+a_2 \left(x-1\right)^2+... $$

From the essential boundary condition we have: $$ U^h \left(1 \right)= 4= a_0 $$

Therefore:

In a similar way we have

With the condition: $$ w \left(1 \right)= 0= c_0 $$

Therefore:

The derivative of U and w are:

Replacing in the weak form we have:

Canceling $$ c_i $$ and Organizing we obtain:

Performing the integration we obtain:

From the latter equation the coefficients $$ a_j $$ can be obtained.

The analytical general solution is:

With the boundary conditions we obtain the constants:

 Matlab Code:

2) Fourier Basis Function
Now the differential equation is solved using Fourier basis function and the boundary conditions:

To follow the constrain braking solution we chose:

For Cosine Clearly $$ b_1=1 $$, and as we need that $$ b_j \left(1 \right) = 0,j=0,1,2,... ~$$ ; we chose $$ \beta = 1  $$.

For Sine Clearly $$ b_1=1 $$, and as we need that $$ b_k \left(1 \right) = 0,k=1,2,... ~$$ ; we chose $$ \beta = 1  $$.

Then we have $$ U^h \left(x \right)= a_0 cos \left(x-1 \right)-1 + a_1 cos2 \left( x -1 \right)-1 ... $$

$$ U^h \left(x \right)= a_1 sin \left(x-1 \right) + a_2 sin2 \left( x -1 \right) ... $$

From the essential boundary condition we have: $$ U^h \left(1 \right)= 4=a_0 $$

Therefore:

By constructing a Stiffness & Force Matrices as follows.

Here the first equation gets satisfied for even values of i & j,

Whereas 2nd equation is satisfied for odd values of i & j,

further 3rd equation is satisfied for remaining combinations of i & j.

Here the first equation gets satisfied for even values of i & j,

Whereas 2nd equation is satisfied for odd values of i & j,

We developed a MATLAB code to generate the Force and the stiffness Matrix and We get the following values.

We can now use the relationship which is, $$\displaystyle Kd=F$$ to solve for $$\displaystyle d$$.

Thus the trial solution is:

 Matlab Code:

3) Exponential Basis Function
Now the differential equation is solved using exponential basis function and the boundary conditions:

To follow the constrain braking solution we chose:

Clearly $$ b_1=1 $$, and as we need that $$ b_i \left(1 \right) = 0,i=1,2,... ~$$ ; we chose $$ \beta =1 $$.

Then we have $$ U^h \left(x \right)= a_0+a_1 \left(e^{\left(x-1 \right)}-1\right)+a_2 \left(e^{2\left(x-1\right)}-1\right)+... $$

From the essential boundary condition we have: $$ U^h \left(1 \right)= 4=a_0 $$

Therefore:

In similar way we have:

With the condition: $$ w(1) = 0 = c_1 $$

Therefore:

The derivative of U and w are:

Replacing in the weak form we have:

Canceling $$ c_i $$ and organizing we obtain:

Performing the integration we obtain:

The following figure shows the exact and numerical solutions. Although good approximation was obtained at $$ n = 3 $$, the convergence for an error of order $$10^{-6} $$ was only achieved with $$ n=29 $$ and to $$n $$ greater than $$ 8 $$, a bad-conditioned (close to singular) stiffness matrix was obtained. Figure 2.4 shows the error.

This results shows that polynomial is better basis function because the convergence is obtained faster; moreover for other problems, the integration may not be possible to perform analytically as in this case, making necessary that the program perform the integration numerically which will increase the time of numerical calculations and decrease the accuracy.

 Matlab Code:

=Problem 5.3: Solve the General 1-D Model 1.0/Data 1b using Weak Form with the Lagrange Interpolation Basis Functions (LIBF)=

Explain how the LIBF basis functions are used as the constraint breaking solution (CBS)
The basis functions are the following:

Solution
1.

So for the LIBF, at the essential boundary condition, there is only one term which does not equal to zero, the other terms are equal to zero. So the CBS is satisfied.

2.

Plot all the LIBF used:

For m=4:

m=6:

m=8：

3.

Solve for d, we have:

=Problem 5.4: Solve the General 1-D Model 1.0/Data 1 using Weak Form with the Lagrange Interpolation Basis Functions (LIBF)=

Explain how the LIBF basis functions are used as the constraint breaking solution (CBS)
The basis functions are the following:

Solution
1.

So for the LIBF, at the essential boundary condition, there is only one term which does not equal to zero, the other terms are equal to zero. So the CBS is satisfied.

2.

Plot all the LIBF used:

For m=4:

m=6:

m=8：

3.

Solve for d, we have:

So, we have:

4.

m=4:

m=6:

m=8：

=Problem 5.5: Calculix=

Given
Given the disk problem in homework four [| Working with Calculix]

Find

 * 5.1 Extract node information to include node numbers and coordinates, and element information to include element numbers and element nodes.
 * 5.2 Generate 3 meshes of same disk with triangular elements.
 * 5.3 Install ccx, run examples, write report for “dummies”

5.1 Node and Element information
A disk was created in Problem 4.7 in the previous homework. CGX was opened with this file, and the command “prnt se 1” was used to extract the information. This was copied to a file and is available at the following link http://en.wikiversity.org/wiki/File:Fe1.s11.team1.hw5.node.element.info.pdf

5.2 Three Meshes
Element type was changed using the command “elty all tr3”, “etly all tr3u” and “elty all tr6” and then using the “mesh all” command to mesh the disk. Refer to figures 5.1, 5.2 and 5.3

5.3 CCX
CCX was installed and opened. The manual spells out the program well; this was downloaded, but can be found at http://web.mit.edu/calculix_v2.0/CalculiX/ccx_2.0/doc/ccx/index.html. Examples were run, please see the following report for findings. http://en.wikiversity.org/wiki/File:CCX.pdf

Contributing Members
Posted by James Roark 01:59, 23 March 2011 (UTC)

=Problem 5.6: Quadratic Lagrangian Elemental Basis Function.=

Given
Consider any Quadratic Function.

Find
Plot of the Individual Lagrangian Quadratic Function and the combined plot of all the three Lagrangian Quadratic Functions.

Solution
For any Quadratic Function the Number of nodes are always 3. Therefore we can write, We will select the global node numbers as follows, We have according to the Lagrangian Interpolation Basis Function, Which is the Polynomial of degree (n-1)

So, Substituting the values from Equation $$\displaystyle (5.6.2)$$, we get,

In the Similar way,

&

We Developed a MATLAB code to plot all these Quadratic Lagrangian Basis Function. This very MATLAB code is as follows.

Appendix
 Matlab Code:

The plots of all the Individual Lagrangian Quadratic Function and the Combined plot of all the Three Functions are as follows.

=Problem 5.7: Linear Lagrangian Element Basis Function= Solving a ordinary differential equation using Linear Lagrangian Element Basis Function. From Mtg.30

Given
The partial differential equation is:

The boundary conditions are:

$$ u\left(0\right) = 4$$

$$\frac{du}{dx}\left(1\right) = -6 $$

Find

 * 1. Explain how Linear Lagrangian Element Basis Function (LLEBF) are used as Costrain Breaking Solution
 * 2. Plot all LLEBF
 * 3. Solve the system dividing the interval [0,1] in 3 divisions (m = 4) and continue dividing until convergence of 10^-6 of the error at x = 0.5
 * 4. Plot the exact and the aproximated solution
 * 5. Plot the error as a function of the number of points

Explain how LLEBF are used as CBS
A function $$b_{i}$$ is used as CBS if: $$b_{0}\left(\beta\right) = 1$$, and $$b_{i}\left(\beta\right) = 0; i = 1, 2, ...$$.

A funtion u(x) can be interpolated by using linear lagrange polynomial considering an element of two points (nodes): 1 and 2; the function will be given by:

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

$$u\left ( x \right )=u_{1}\frac{x-x_{2}}{x_{1} - x_{2}}+u_{2}\frac{x-x_{1}}{x_{2} - x_{1}} =u_{1}N_{1}+u_{2}N_{2} $$
 * $$\displaystyle (5.7.2) $$
 * }
 * }

Where $$u_{1}, u_{2}$$ are the nodal values of the funtion at the nodes 1 and 2 respectively.

The Lagrange Polynomial has the property of have the value of 1 at the respective node an 0 at the others; for example in equation (5.7.2), if an esential boundary condition is applied on node 1, whe have that $$N_{1} = b_{1} = 1$$ and $$N_{2} = b_{2} = 0$$ at the node 1 where the essential boundary condition is applied. In this case we only have two fuctions (b1 and b2); therefore this polynomial follows CBS. Symilar aproach can be used if the essential boundary condition is at node 2.

Plot all LLEBF used
Fig. 5.7.1 shows the two basis functions described in Eq. (5.7.2); in the figure, clearly can be seen that N1 is equall to 1 at node 1 and 0 at node 2, and N2 is equal to 0 at node 1 and equal to 1 at node 2. The plot are obviously straight lines because the functions are linear.

In the present problem we have to start working with 4 poins (3 elements) and increase the number of points until that the error at x = 0.5 is smaller than 10^-6. We are asked to plot all the basis functions used, but the basis funtion used are similar for all elements does not matter the size of the element. When the number of elements is increased, the lenght of the elements is decreased, but the basis fuction are the same as that shown in Fig. 5.7.1 for all the elements. Obviously a function is valid only inside its respective element

Solve the system dividing the interval [0,1] in 3 divisions (m = 4) and continue dividing until convergence of 10^-6 of the error at x = 0.5
Firs we obtain the weak form of the differential equation; to do that we multiply it for a weight funtion and integrate from 0 to 1:

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

$$\int_{0}^{1}w\left ( \frac{d}{dx} \left [ \left ( 2+3x \right ) \frac{du}{dx}\right ]+5x\right )dx=0 $$
 * $$\displaystyle (5.7.3) $$
 * }
 * }

Performing integration by parts in the first term we obtain:

As we do not know the derivative of u at x = 0, we chose w(0) = 0. The global stiffness matrix will be assembled from the stiffness matrix of each element. To get the element stiffness matrix we change the cordinate x to the cordinate $$\epsilon$$ which is equal to -1 at the node 1 and is equal to 1 at the node 2, as indicated in the Fig. 5.7.2

The new cordinate $$\epsilon$$ is given by:

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

$$\varepsilon = \frac{2\left ( x-x_{1} \right )}{x_{2}-x_{1}}-1=\frac{2\left ( x-x_{1} \right )}{l}-1 $$ In the last step, it was used the fact that $$x_{2}-x_{1} = l$$; the lengt of the element. Taking the derivative on both sides of equation (5.7.5) we have:
 * $$\displaystyle (5.7.5) $$
 * }
 * }

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

$$ d\varepsilon =\frac{2dx}{l}$$
 * $$\displaystyle (5.7.6) $$
 * }
 * }

Replacing the cordinate x in the lagrange polynomial we obtain:

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

$$u\left ( \varepsilon \right )=\begin{bmatrix} \frac{1-\varepsilon }{2} & \frac{1+\varepsilon }{2} \end{bmatrix}\begin{bmatrix} u_{1}\\

u_{2}\end{bmatrix}=\left [ N \right ]\begin{bmatrix} u_{1}\\

u_{2}\end{bmatrix} $$
 * $$\displaystyle (5.7.7) $$
 * }
 * }

In symilar way w can be written as:

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

$$\left [w\left(x\right)= N \right ]\begin{bmatrix} c_{1}\\

c_{2}\end{bmatrix}=\begin{bmatrix} c_{1} & c_{2} \end{bmatrix}\left [N \right ]^{T} $$ As in the weak form, we need the derivatives of u and w with respect to x, we calculathe those derivatives by using the chain rule:
 * $$\displaystyle (5.7.8) $$
 * }
 * }

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

$$\frac{du}{dx}=\frac{du}{d\varepsilon }\frac{d\varepsilon }{dx}=\frac{du}{d\varepsilon }\frac{2}{l} $$
 * $$\displaystyle (5.7.9) $$
 * }
 * }

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

$$\frac{dw}{dx}=\frac{dw}{d\varepsilon }\frac{d\varepsilon }{dx}=\frac{dw}{d\varepsilon }\frac{2}{l} $$ From Eq. (5.7.7) we can take the derivative with respect to \epsilon to obtain:
 * $$\displaystyle (5.7.10) $$
 * }
 * }
 * {| style="width:100%" border="0"

$$\frac{du}{d\varepsilon }=\begin{bmatrix} -\frac{1}{2} & \frac{1}{2} \end{bmatrix}\begin{bmatrix} u_{1}\\

u_{2}\end{bmatrix}=\left [ B \right ]\begin{bmatrix} u_{1}\\

u_{2}\end{bmatrix} $$ In similar way the derivative of w is:
 * $$\displaystyle (5.7.11) $$
 * }
 * }

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

$$\frac{dw}{d\varepsilon }=\begin{bmatrix} c_{1} & c_{2} \end{bmatrix}\left [ B \right ]^{T} $$
 * $$\displaystyle (5.7.12) $$
 * }
 * }

Making the change of variables on the stiffness matrix and on the force vector we get:

To assemble the system we build a Matlab code (shown below) wich basically follow the algoritm described on Fig. 5.7.3. Inside the code, the integration was done by using the Gauss quadrature method. In that method, the integral of a funtion is given by:

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

$$ \int_{-1}^{1}f\left ( \varepsilon \right )d\varepsilon =\sum p_{i}f\left ( \varepsilon _{i} \right ) $$
 * $$\displaystyle (5.7.15) $$
 * }
 * }

Where $$p_{i}$$ are weighting coefficients. As in Equation (5.7.14) we have a polynomial of order 2, we need at least 2 coefficients to exactly calculate the integral. In this case the integral is given by:

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

$$ \int_{-1}^{1}f\left ( \varepsilon \right )d\varepsilon =f\left ( \frac{1}{\sqrt{3}} \right )+f\left ( \frac{-1}{\sqrt{3}} \right ) $$
 * $$\displaystyle (5.7.16) $$
 * }
 * }

Plot the exact and the aproximated solution
Fig. 5.7.4 and Fig. 5.7.5 shows the exact and aproximated solution obtained with 4 and 6 nodes respectively. It can be observed that even with 4 nodes the numerical solution aproximates well to the analytical.

To calculate the error, at x = 0.5 is the mid point of the domain, we obtain the value at x = 0.5 interpolating the two neighboring values. The interpolation was done by using the same Lagrange polynomial described before. We found that even though a good aproximation was obtained even with low number of elements (n = 4, Fig. 7.5.4) the convergence to a tolerance of 10^-6 was not obtained with a number of elements lower than 100. Therefore we decrease the tolerance to 10^-3. With this this tolerance the convergence still was achieve with a large number of elements (19) Fig.7.5.6 shows the error as a funtion of the number of elements.

Appendix
 Matlab Code:

=Problem 5.8: Linear Lagrangian Element Basis Function= Solving a ordinary differential equation using Linear Lagrangian Element Basis Function. From Mtg.30

Given
The partial differential equation is:

The boundary conditions are:

$$ u\left(1\right) = 4$$

$$\frac{du}{dx}\left(0\right) = -6 $$

Find

 * 1. Explain how Linear Lagrangian Element Basis Function (LLEBF) are used as Costrain Breaking Solution
 * 2. Plot all LLEBF
 * 3. Solve the system dividing the interval [0,1] in 3 divisions (m = 4) and continue dividing until convergence of 10^-6 of the error at x = 0.5
 * 4. Plot the exact and the aproximated solution
 * 5. Plot the error as a function of the number of points

Solution
This problem is similar to 5.7 but only change the boundary conditions, therefore, the weak form is the same exept that now we chose w(1)=0 and impose the boundary condition u(1) = 4. As the problem is similar, we use the same Matlab routine developed for porblem 5.7. As all the develpment is the same that in problem 5.7 here we only present the results. Fig. 5.8.1 and Fig. 5.8.2 show the analytical and numerical solution obtained with 4 and 6 nodes respectively; again, although good aproximation was obtained even with 4 nodes, no convergence was achieved even with 100 nodes; therefore, the tolerance was decreased to 10^-3. With this new tolerance, convergence was achieved with 8 nodes. Fig. 5.8.3 shows the error as funtion of the number of nodes.

=Contributing Members=

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University of Florida/Egm6321/F10.TEAM1.WILKS/Mtg7

Mtg 7: Tues, 08Sept 09

[[media: Egm6321.f09.mtg7.djvu | Page 7-1]]
Mediawiki Hints: Alt+Shift+E = Edit Alt+Shift+P = Preview Alt+Shift+S = Save [[media: Egm6321.f09.mtg6.djvu | p.6-3]]: Case2: $$ h_xN=0 \ \Rightarrow \ -h_yM+h(N_x-M-y)=0 \Rightarrow \ \frac{h_y}{h}=\frac{1}{M} \left ( N_x-M_y \right )=-g(y) \ $$ HW: Complete details to obtain $$ h_y \ $$


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Engineering Experience 4: Design a Small Solar Vehicle/Nl/Team PM2

Het team
Ons team bestaat uit 7 studenten Industriele wetenschappen uit het 2de bachelor aan de Groep T hogeschool. Er werd ons gevraagd om een kleine zonnewagen te bouwen beginnende met een zonnepaneeltje en een DC-motor. Op deze blog zal regelmatig een update geschreven worden over de voortgang van het project.

Materiaal
- Balsahout - achteruitkijkspiegel - 4 mindiscs - 1 zonnepaneel - 1 stalen as van 33 cm lang met een diameter van 4mm - 1 motor - 1 suikerblok - 1 schakelaar

Site
Op onze site worden alle bestanden bijgehouden en wat meer informatie gegeven:

www.borasol.tk

Blog
Week 1

De naam is gekozen! Het wordt Borasol!!

We zijn deze week al goed beginnen werken.. Het moest wel want vandaag is er al een deadline! 4 documenten moeten er al afgegeven worden. Maar door goed in team te werken en goed door te werken zijn we er geraakt!

Nu gaan we het logo nog even ontwerpen --> Zie latere post!

Week 2

We hebben feedback gehad over onze gantt-chart en wbs.

Enkele dingen waren er niet zo goed of ontbraken. Dit zullen we tegen volgende week aanpassen.

Nu gaan we vrijdag nog eens afspreken om nog verder te werken en wat te brainstormen over ons concept. Verder hebben we ook een facebookgroep opgericht en zo kunnen we hier rap iets op posten en zal er ook rapper reactie zijn omdat tegenwoordig facebook meer bekeken wordt dan mails.

Week 3

Deze week hebben we het definitieve ontwerp gekozen. Het frame zal bestaan uit een T-profiel met een beweeglijke achteruitkijkspiegel in het midden waar het zonnepaneel op bevestigd zal worden. Minidiscs zullen dienen als wielen en een metalen staaf als aandrijfas. Het frame zal uit balsahout bestaan omdat dit een licht maar vrij stevige houtsoort is. De afzondelijke onderdelen van de wagen zullen vooral met epoxylijm aan elkaar bevestigd worden omdat deze lijm ook zeer veel gebruikt wordt bij modelbouw.

Week 4

We zijn beginnen bouwen aan het frame. Het hout is gesneden en aan elkaar geplakt. Later deze week hebben we ook de achteruitkijkspiegel bevestigd op het frame en zijn we op zoek gegaan naar de gepaste tandwielen. Deze hebben we echter nog niet gevonden.

Week 5

Maandag: De bouw van de zonnewagen is vandaag in zijn laatste fase gegaan voor het testen kan beginnen. De juiste tandwielen zijn eindelijk gevonden, nu nog op de assen zetten en we kunnen testen!

Donderdag: Normaal gingen de tandwielen er vandaag op gezet worden maar we hebben een probleem met de overbrenging tussen het kleine tandwiel van de motor en die dat de as zal aandrijven.

Week 6

Deze week was het deadline voor Case I en Case Simulink. Door technische problemen is de deadline voor de Simulink-case niet gehaald. De andere deadline werd wel gehaald. Simulink was een moeilijk deel voor ons team omdat er niemand er echt inzicht in kreeg ook al werd hier veel tijd in gestoken. Er werd dus een expert bij geroepen en vroegen we deze om hulp. Hij begreep er echter ook niet veel van en verstond niet goed wat we moesten doen waardoor hij niet veel hulp kon bieden. Simulink zou dus later afgewerkt worden zodat dit voor het eindverslag in orde zou zijn.

Week 7

Omdat onze tandwielen nog altijd niet goed waren, besloten we om nieuwe te gaan kopen. Hiervoor zochten we enkele modelbouw-winkels op. Nadat we enkele adressen gevonden hadden vertrokken er 3 teamleden naar de winkels. Echter bleek dit verloren moeite te zijn omdat die adressen verouderd waren en er dus maar op 1 adres van de 5 een winkel bestond die ook nog eens op die dag gesloten was. Het was echter een vermoeiende dag geweest waardoor het voor deze week gedaan was met de tandwielen omdat er ook nog voor andere vakken gewerkt moest worden omdat teamleden test hadden of een andere deadline

Week 8

Het wagentje was zo goed als af. We hadden de oude tandwielen er op gemonteerd en een extra as gemaakt zodat we de nieuwe en goede tandwielen( ratio van ongeveer 6 a 7) er onmiddellijk op konden vastmaken. Ondertussen was simulink ook volledig afgewerkt en werd hier een verslag van gemaakt wat later hier terug te vinden zal zijn.

Week 9

Dinsdag Deze week konden we het wagentje testen. Omdat de goede tandwielen nog niet waren aangekomen door een vertraging bij het opsturen gingen we het wagentje testen met de oude tandwielen. We wister echter wel al op voorhand dat het wagentje waarschijnlijk niet ging bewegen omdat de ratio te klein was. Onze voorspelling gold ook en het wagentje ging niet vooruit terwijl de as wel aangedreven werd als er geen weerstand op de wielen stond. Het enige wat we nu konden doen was wachten op de tandwielen.

Vrijdag Eindelijk kregen we onze tandwielen toegestuurd en konden we ze monteren op de as. Alles werd vastgemaakt en dit werd dan een weekend gerust gelaten zodat dit kon drogen.

Week 10

Maandag 02/05

Vandaag kregen we te horen dat het morgen race gaat zijn. Om de laatste stukjes vast te maken en en de look van de wagen nog te veranderen kwamen we nog een voorlaatste keer samen. Samen werd er gewerkt, zowel aan de laatste berekeningen en aan het wagentje zelf. Omdat het werk aan het wagentje zelf uiteindelijk enkel nog plakwerk was, was dit echter wel een vermoeiende vergadering. Er moest telkens gewacht worden totdat de lijm droog was zodat er iets anders op geplakt kon worden; dit duurde echter langer als verwacht waardoor de klok steeds verder tikte.

Nadat alles geplakt was, ging iedereen naar huis en werd er door een teamlid later op die avond nog een likje verf gegeven aan het autootje zelf en werd het getest. De test verliep feilloos, echter kon men niet zien of het wagentje echt op de grond reed omdat alles eerst opzij werd gezet om te drogen.

Dinsdag 03/05 was het de grote dag.. De race was daar! Omdat het bordje waar ons logo op stond kapot was gegaan in week 9 zijn twee teamleden nog rap naar Fablab gereden om een nieuwe te maken. Het resultaat was des te mooier omdat het deze keer geplakt werd op de zwarte ondergrond van het autootje.

Het autootje was klaar voor een eerste test. Met een lamp van 500W probeerden we het wagentje op een stenen ondergrond. Het wagentje reed en iedereen was al een beetje opgelucht. Nu nog op de race zelf.

Na een laatste check en laatste aanpassingen, waren we klaar voor de race. Om ons zonnepaneel des te efficiënter te maken hadden we deze in een doos gelegd tussen koelelementen. Dit zou er voor zorgen dat er een grotere temperatuurwisseling zou zijn en dus een grotere efficiëntie.

Als team vertrokken we naar het martelarenplein. Toen we aankwamen waren er al teams aan het testen. Omdat vorige test ook goed ging dachten we dit ook te doen.

We maakten het autootje klaar en lieten het rijden. Het vertrok echt snel maar net op de helling aangekomen stopte het opeens. Nadat we nagekeken hadden wat er mis was bleek dat het kleine tandwiel die de as aandreef er af was gevallen. Dit was een grote teleurstelling voor ons allemaal. We hadden echter nog lijm bij maar we hadden toch allemaal bang dat dit niet op tijd droog zou zijn.

De eerste race vertrokken we wel maar vielen we stil omdat een ander onderdeel los was gekomen. Echter was dit al een week geplakt geweest maar kwam opeens trug los.

Na het vastmaken van dit stuk was het tijd voor de tweede race. Hier vertrok ons wagentje en geraakte tot aan de helling maar bleef steken in het begin van de helling omdat het tandwieltje niet meer goed bevestigd was ten gevolge dat het nog niet goed droog was.

Week 11

Dinsdag: Deze week moeten alle documenten binnen voor vrijdag 18u De meeste documenten zijn al af of zo goed als af en zal er dus vandaag en morgen gewerkt worden aan het eindverslag zodat dit donderdag nagelezen kan worden.

Woensdag : Alle vergaderverslagen en andere afzonderlijke documenten zijn verzameld en worden morgen op de wiki gezet. Verder is er ook al een voorlopige versie geschreven van het eindverslag en zal dit morgen nagekeken worden en aangepast waar nodig.


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User talk:Csmayes

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

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You don't need to be an educator to edit. You only need to be bold to contribute and to experiment with the sandbox or your userpage. See you around Wikiversity! --mikeu talk 17:23, 21 March 2011 (UTC)

Many small edits
Hi, Csmayes, and thanks for all your edits on the collaborative learning topic. If I could make one suggestion, it would be to make one edit, rather than many small ones. One useful tool is the 'preview' button, which helps you see the updated version of the article. It helps you by removing the need to save your changes then click 'edit' on the article. It also keeps the number of items in the recent changes down, which, although a small point, can be useful. Thanks, and keep up the great work! Gaidheal1 20:05, 14 April 2011 (UTC)

title: User talk:Csmayes


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User:Eml5526.s11.team2.reiss/topsecret

Given
[]

Solution
For m=4

We have four nodes at 0, .25, .5, .75, 1.

The Lagrange interpolation functions for m=4 are

title: User:Eml5526.s11.team2.reiss/topsecret


text: Category:Radio Interferometer Telescope

title: Category:Radio Interferometer Telescope


text:
User:Yongtao-Li/NM.HW5.4

=Problem 5.4 a Mass-Spring-Damper with Gaussian noise and Cauchy noise=

GIVEN
An ideal mass-spring-damper system with mass m, spring constant k and viscous damper of damping coefficient c is subject to an oscillatory force u. This can be illustrated in the following image.

FIND
1)Derive equations of motion in terms of d, c, k, m, u.

2)Let $$ {\mathbf{x}} = \left\{ {\begin{array}{ccccccccccccccc} d \\  {\dot d} \end{array}} \right\} = \left\{ {\begin{array}{ccccccccccccccc}   \\ \end{array}} \right\}$$. Find $$\left( {{\mathbf{F}},{\mathbf{G}}} \right) $$

3)Find $${c_{cr}}$$ in terms of k, m st. this system is critically damped.

4)Let k=1, m=1/2, $$ {{\mathbf{x}}_0} = {\left[ {0.8, - 0.4} \right]^T} $$

a)For u=0, plot $${{\mathbf{x}}_k}$$ for $$c = \frac{1}{2}{c_{cr}},{c_{cr}},\frac{3}{2}{c_{cr}}$$.

b)For u=0.5 Gaussian noise and $$c = \frac{3}{2}{c_{cr}}$$, plot $${{\mathbf{x}}_k}$$.

c)For u=0.5 Cauchy noise and $$c = \frac{3}{2}{c_{cr}}$$, plot $${{\mathbf{x}}_k}$$.

SOLUTION
1) Since the system is subject to an oscillatory force generated from the spring, we have this oscillatory force

and also a damping force from the damper:

Since we can use [http://en.wikipedia.org/wiki/Newton%27s_laws_of_motion#Newton.27s_second_law Newton's second law

Therefore we can derive equations of motion as following:

2) From Eq. 5.4.4, we can solve for $${\ddot d}$$ as following:

Therefore we can rewrite $${\dot x}$$ in matrix form

If we make a discretization as

and use forward Euler method

We can have

3) Since we can rewrite Eq. 5.4.4 as

where $$\omega_0 = \sqrt{ k \over m } $$, $$\zeta = { c \over 2 \sqrt{m k} }$$

When $$\zeta = 1$$, the system is said to be critically damped. A critically damped system converges to zero faster than any other and without oscillating. Therefore we have $${c_cr}$$ indicated in Eq. 5.4.12 and can look for damping for more explanations.

4) a) The following matlab codes are used to solve the problem without any noise, and the damping coefficients can be chose to create different cases with under-damping, critical damping or over-damping.

Comment:

Different c values cause different damping cases. From the pic above, we can find out a critically damped system converges to zero faster than any other and without oscillating.

b) The following matlab codes are used to solve the problem with the Gaussian Random Noise and over-damping case.

Comment:

When the system has Gaussian noise, the convergence can be archived, however the final result is not exactly zero since there is still noise all the time and the final result oscillated around zero.

c) The following matlab codes are used to solve the problem with the Cauchy Noise and over-damping case.

Comment:

When the system has Cauchy noise, the same result was generated as Gaussian noise. And the oscillation around the zero is much larger than the case with Gaussian noise.

title: User:Yongtao-Li/NM.HW5.4


text:
File:Gear+ratio.pdf

Licensing
for school project

title: File:Gear+ratio.pdf


text:
File:Panel supporter.jpg

Licensing
for school project

title: File:Panel supporter.jpg


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text: Musical spatialism

title: Musical spatialism


text: File:Nm1.s11.team5.HW5.P2.matlabcode.pdf


text:
File:Componets.pdf

Licensing
for school project

title: File:Componets.pdf


title: File:Fe1.s11.mtg33.djvu


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text:
Data Visualization/Visualisation of quantitative information

Description
Overviews levels of measurement and graphical approaches to analysis of univariate data.

Slides

 * Visualisation of quantitative information (Slideshare)
 * Handout:
 * [[Media:Visualiation of quantitative information 3 slides per page.pdf|Visualisation of quantitative information]] (3 slides per page)
 * [[Media:Visualiation of quantitative information 6 slides per page.pdf|Visualisation of quantitative information]] (6 slides per page)

Reading
Denscombe, M. (2010). The good research guide for small-scale social research projects (4th ed.) NY: McGraw Hill. (Ch 13: Quantitative data)


text:
User:Eml5526.s11.team2/homework4homework4 Xfile

=Problem 5.1: Deriving the Weak Form From the Strong Form (Data 1b)=

Given: Strong Form, Boundary Conditions, and Basis Functions
The Strong Form from lecture 9-1 and Fish and Belytschko p. 73:

Find: The Weak Form in Order to Solve for $$ u^h(x) $$ with Following Basis Functions
Solve for $$ \displaystyle u(x) $$ and $$ \displaystyle u^h(x) $$ using the following basis functions until the error $$ \displaystyle e(n) =|u^h(0.5)-u(0.5)| \leq 10^{-6}$$.

 * 1) Polynomial Basis Function  $$ \displaystyle \mathcal{F}_P :=\{ x^j,j=0,1,2,...\} $$
 * 2) Exponential Basis Function $$ \displaystyle \mathcal{F}_e :=\{ e^{jx},j=0,1,2,...\} $$
 * 3) Fourier Basis Function     $$ \qquad \displaystyle \mathcal{F}_F :=\{cos(jx),j=0,1,2,...,sin(kx),k=1,2,...\} $$

On the domain defined by,

1. Approximating Solution Using Polynomial Basis Function
Here is the matlab code used to obtain the approximating solutions:

Multiplying Equ 2.20 by w(x) and integrating within the domain:

Next we will integrate by parts to obtain,

Since $$ \displaystyle w(x=0) = 0 $$ we can simplify (2.22) in the following way to obtain the weak form,

Now we can use the polynomial basis functions,

Which allows us to write the weak form (2.25) in terms of our matrices $$ \displaystyle {\color{red}\mathbf{K}_{FF}} $$, $$ \displaystyle {\color{red}\mathbf{d}_{F}} $$ and  $$ \displaystyle {\color{red}\mathbf{F}_{F}} $$.

Next we wish to obtain the solution to $$ \displaystyle \mathbf{d_F} $$ by solving the following system of equations,

Solving for the approximating solution, the d matrix can transposed and multiplied by the basis function matrix:

Using the provided matlab code to obtain the d coefficients and applying to the basis function:

n=8
We found that for $$ \displaystyle n=8 $$ the error was of the order $$ \displaystyle 10^{-6} $$. Using the Matlab code and rounding off to the nearest hundredths place gives the matrices $$ \displaystyle \left[ \mathbf{K_{FF}} \right]_ $$ and $$ \displaystyle \left[ \mathbf{F_{F}} \right]_ $$

Using our Matlab code we obtain the following solution for $$ \displaystyle \left[ \mathbf{d_{F}} \right]_ $$. We have chosen to report these values in the long format to demonstrate that rounding error has not occurred.

Plot: Comparing the exact solution $$ u(x) $$ to $$ u_h(x) $$ for $$ n=2,4,6,8 $$
The plots for solutions and error at x=.5 are as follows:

Plot: Showing the error $$ e(n) = |u(x)-u_h(x)| $$
For the polynomial approximate solution, with the increasing number of nodes, the error decreases. For this particular instance, the error XXXXXXX

2. Approximating Soultion Using Fourier Basis Function
Here is the matlab code used to obtain the approximating solutions:

Using the same process as the polynomial solution and using the general equation from equation 2.21, the fourier solution is obtained the same way. W(x) is now equation 5.35 from HW4

Using the provided matlab code to obtain the d coefficients and applying to the basis function:

n=8
Using the Matlab code and rounding off to the nearest hundredths place gives the matrices $$ \displaystyle \left[ \mathbf{K_{FF}} \right]_ $$ and $$ \displaystyle \left[ \mathbf{F_{F}} \right]_ $$

Using our Matlab code we obtain the following solution for $$ \displaystyle \left[ \mathbf{d_{F}} \right]_ $$. We have chosen to report these values in the long format to demonstrate that rounding error has not occurred.

Plot: Comparing the exact solution $$ u(x) $$ to $$ u_h(x) $$ for $$ n=2,4,6,8 $$
The plots for solutions and error at x=.5 are as follows:

3. Approximating Soultion Using Exponential Basis Function
Here is the matlab code used to obtain the approximating solutions:

Using the same process as the polynomial solution and using the general equation from equation 2.21, the exponential solution is obtained the same way. W(x) is now equation 5.41 from HW4

Using the provided matlab code to obtain the d coefficients and applying to the basis function:

n=8
Using the Matlab code and rounding off to the nearest hundredths place gives the matrices $$ \displaystyle \left[ \mathbf{K_{FF}} \right]_ $$ and $$ \displaystyle \left[ \mathbf{F_{F}} \right]_ $$

Using our Matlab code we obtain the following solution for $$ \displaystyle \left[ \mathbf{d_{F}} \right]_ $$. We have chosen to report these values in the long format to demonstrate that rounding error has not occurred.

Plots
The plots for solutions and error at x=.5 are as follows:

=Problem 5.2: Deriving the Weak Form From the Strong Form (Data 1)=

Given: Strong Form, Boundary Conditions, and Basis Functions
The Strong Form from lecture 9-1 and Fish and Belytschko p. 73:

Find: The Weak Form in Order to Solve for $$ u^h(x) $$ with Following Basis Functions
Solve for $$ \displaystyle u(x) $$ and $$ \displaystyle u^h(x) $$ using the following basis functions until the error $$ \displaystyle e(n) =|u^h(0.5)-u(0.5)| \leq 10^{-6}$$.

 * 1) Polynomial Basis Function  $$ \displaystyle \mathcal{F}_P :=\{ x^j,j=0,1,2,...\} $$
 * 2) Exponential Basis Function $$ \displaystyle \mathcal{F}_e :=\{ e^{jx},j=0,1,2,...\} $$
 * 3) Fourier Basis Function     $$ \qquad \displaystyle \mathcal{F}_F :=\{cos(jx),j=0,1,2,...,sin(kx),k=1,2,...\} $$

On the domain defined by,

1. Approximating Solution Using Polynomial Basis Function
Here is the matlab code used to obtain the approximating solutions:

Multiplying Equ 2.20 by w(x) and integrating within the domain:

Next we will integrate by parts to obtain,

Since $$ \displaystyle w(x=1) = 0 $$ we can simplify (2.22) in the following way to obtain the weak form,

Now we can use the polynomial basis functions,

Which allows us to write the weak form (2.25) in terms of our matrices $$ \displaystyle {\color{red}\mathbf{K}_{FF}} $$, $$ \displaystyle {\color{red}\mathbf{d}_{F}} $$ and  $$ \displaystyle {\color{red}\mathbf{F}_{F}} $$.

Next we wish to obtain the solution to $$ \displaystyle \mathbf{d_F} $$ by solving the following system of equations,

Solving for the approximating solution, the d matrix can transposed and multiplied by the basis function matrix:

Using the provided matlab code to obtain the d coefficients and applying to the basis function:

n=8
We found that for $$ \displaystyle n=8 $$ the error was of the order $$ \displaystyle 10^{-6} $$. Using the Matlab code and rounding off to the nearest hundredths place gives the matrices $$ \displaystyle \left[ \mathbf{K_{FF}} \right]_ $$ and $$ \displaystyle \left[ \mathbf{F_{F}} \right]_ $$

Using our Matlab code we obtain the following solution for $$ \displaystyle \left[ \mathbf{d_{F}} \right]_ $$. We have chosen to report these values in the long format to demonstrate that rounding error has not occurred.

Plot: Comparing the exact solution $$ u(x) $$ to $$ u_h(x) $$ for $$ n=2,4,6,8 $$
The plots for solutions and error at x=.5 are as follows:

Plot: Showing the error $$ e(n) = |u(x)-u_h(x)| $$
For the polynomial approximate solution, with the increasing number of nodes, the error decreases. For this particular instance, the error XXXXXXX

2. Approximating Soultion Using Fourier Basis Function
Here is the matlab code used to obtain the approximating solutions:

Using the same process as the polynomial solution and using the general equation from equation 2.21, the fourier solution is obtained the same way. W(x) is now equation 5.35 from HW4

Using the provided matlab code to obtain the d coefficients and applying to the basis function:

n=4
etc...

n=8
Using the Matlab code and rounding off to the nearest hundredths place gives the matrices $$ \displaystyle \left[ \mathbf{K_{FF}} \right]_ $$ and $$ \displaystyle \left[ \mathbf{F_{F}} \right]_ $$

Plot: Comparing the exact solution $$ u(x) $$ to $$ u_h(x) $$ for $$ n=2,4,6,8 $$
The plots for solutions and error at x=.5 are as follows:

3. Approximating Soultion Using Exponential Basis Function
Here is the matlab code used to obtain the approximating solutions:

Using the same process as the polynomial solution and using the general equation from equation 2.21, the exponential solution is obtained the same way. W(x) is now equation 5.41 from HW4

Using the provided matlab code to obtain the d coefficients and applying to the basis function:

n=8
Using the Matlab code and rounding off to the nearest hundredths place gives the matrices $$ \displaystyle \left[ \mathbf{K_{FF}} \right]_ $$ and $$ \displaystyle \left[ \mathbf{F_{F}} \right]_ $$

Using our Matlab code we obtain the following solution for $$ \displaystyle \left[ \mathbf{d_{F}} \right]_ $$. We have chosen to report these values in the long format to demonstrate that rounding error has not occurred.

The plots for solutions and error at x=.5 are as follows:

Matlab
=Problem 5.3: Torsion of a Bar = Modified version of HW 4.4 defined on page 2 of Lecture 21 from Fish & Belytschko Page 73 Problem 3.7

Given
Where the Weak Form for a bar in torsion can be defined as

Make use of the Lagrange Interpolation Basis Function (LIBF) with uniform discretization (equidistant nodes) using $$ \displaystyle m = 4,6,8   $$ where the LIBF can defined as follows

Find
 a) Explain how Langrange Interpolation Basis Functions (LIBF) are used as Constraint Breaking Solutions b) Plot all LIBF used c) Use matlab quad function or WolframAlpha.com to integrate d) Plot approximate vs exact solution and convergence error versus equidistant nodes (m)

(a)
Langrange Interpolation Basis Functions satisfy the condition of the constraint breaking solution by assuming a value of zero at every node except for the node where the LIBF is defined there it assumes a value of 1. It acts like the Kronecker Delta

(b)
The following is the LIBF for the $$ \displaystyle m = 4 $$

The following is the LIBF for the $$ \displaystyle m = 6 $$

The following is the LIBF for the $$ \displaystyle m = 8 $$

Plot: Comparing the exact solution $$ u(x) $$ to $$ u_h(x) $$ for $$ n=2,4,6,8 $$
The plots for solutions and error at x=.5 are as follows:

Matlab Code
=Problem 5.4: Torsion of a Bar = Modified version of HW 4.4 defined on page 2 of Lecture 21 from Fish & Belytschko Page 73 Problem 3.7

Given
Where the Weak Form for a bar in torsion can be defined as

Make use of the Lagrange Interpolation Basis Function (LIBF) with uniform discretization (equidistant nodes) using $$ \displaystyle m = 4,6,8   $$ where the LIBF can defined as follows

Find
 a) Explain how Langrange Interpolation Basis Functions (LIBF) are used as Constraint Breaking Solutions b) Plot all LIBF used c) Use matlab quad function or WolframAlpha.com to integrate d) Plot approximate vs exact solution and convergence error versus equidistant nodes (m)

(a)
Langrange Interpolation Basis Functions satisfy the condition of the constraint breaking solution by assuming a value of zero at every node except for the node where the LIBF is defined there it assumes a value of 1. It acts like the Kronecker Delta

(b)
The following is the LIBF for the $$ \displaystyle m = 4 $$

The following is the LIBF for the $$ \displaystyle m = 6 $$

The following is the LIBF for the $$ \displaystyle m = 8 $$

(c)
Our solution $$ u \left ( x \right ) $$ and our weight functions can be approximated using the following definitions,

To approximate our solution u, we divide our x domain into m nodal subdivisions. Additionally, we will use m LIBF to approximate our solution. These LIBF will have the following form,

Therefore, for m=4, we take the first 4 Lagrange Interpolation Basis Functions, or more specifically,

Here we want to develop equations for $$ \tilde{K_{KK}} $$ and $$ \tilde{F_K} $$ and solve for $$ \tilde{d_{K}} $$, similar to the previous problems.

Plot: Comparing the exact solution $$ u(x) $$ to $$ u_h(x) $$ for $$ n=2,4,6,8 $$
The plots for solutions and error at x=.5 are as follows:

Matlab Code
= Problem 5.5 : Calculix-Using CCX module=

Disk Problem
After building disc in cgx module then we might want to know that properties of nodes,lines, surfaces, and bodies so now lets look at how we can get more information what we have done for our model.

First we have to open our program. To do this run the following ;

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

$$  \displaystyle [All Programs]\Rightarrow [bConverged]\Rightarrow [Calculix]\Rightarrow [Calculix Command] $$
 * style="width:95%" |
 * style="width:95%" |
 * }

Open calculix and go to directory where the basic examples stays in your harddrive.
 * {| style="width:100%" border="0"

$$  \displaystyle ...\Rightarrow [Program Files]\Rightarrow [bConverged] \Rightarrow [Calculix] \Rightarrow [Cgx] \Rightarrow [Examples] \Rightarrow [Basic] $$
 * style="width:95%" |
 * style="width:95%" |
 * }

Then type to open disc example:

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

$$  \displaystyle [cgx]\Rightarrow [disc] $$
 * style="width:95%" |
 * style="width:95%" |

 * }

Now we want to learn about point properties. To see point properties in our graph secren type

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

$$  \displaystyle \begin{matrix} plot & pa   & all \end{matrix} $$
 * style="width:95%" |
 * style="width:95%" |
 * }

We need to mesh disc to see the mesh properties to mesh the disc please type following steps

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

$$  \displaystyle \begin{matrix} elty & all & qu8 \end{matrix} $$
 * style="width:95%" |
 * style="width:95%" |
 * }

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

$$  \displaystyle \begin{matrix} mesh & all \end{matrix} $$
 * style="width:95%" |
 * style="width:95%" |

 * }

What we have done so far is that we opened disc and meshed it with quadratic elements. So we are ready to see element properties on our graph screen. To see properties of elements type,
 * {| style="width:100%" border="0"

$$  \displaystyle \begin{matrix} plot & ma  & all \end{matrix} $$
 * style="width:95%" |
 * style="width:95%" |
 * }

Also we can get more specific properties form the special editor of calculix called 'SciTE.' We can go to examples directory and just basically right click on disc example and open with by editing. SciTE automatically will be opened.

From below we can easly see the properties of all components of disc. Especially point properties are boxed.

Generating Different Meshes
We will do the same procedures as we have done before. Our model is prepared. So we will mesh our model and take properties of different kinds of meshes for 2-D. Our meshes for 2-D are:

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

$$  \displaystyle \begin{matrix} tr3 & 3 node & 2D element \\ gu4 & 4 node & 2D element\\ qu8 & 8 node & 2D element \end{matrix} $$     (N)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }
 * {| style="width:100%" border="0"

$$  \displaystyle \begin{matrix} Mesh & with  & tr3 \end{matrix} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * {| style="width:100%" border="0"

$$  \displaystyle \begin{matrix} Mesh & with  & qu4 \end{matrix} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * {| style="width:100%" border="0"

$$  \displaystyle \begin{matrix} Mesh & with  & qu8 \end{matrix} $$
 * style="width:95%" |
 * style="width:95%" |
 * }

CCX module
In this module we will analyze for unknowns that we are interested in. This unknowns can be displacements or temperatures. Also we can find fluxes between these nodes.

For the first step go to directory where beam example stays in your harddrive.

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

$$  \displaystyle ....[Program Files]\Rightarrow [bConverged]\Rightarrow [Calculix]\Rightarrow [ccx]\Rightarrow [test] $$
 * style="width:95%" |
 * style="width:95%" |
 * }

Find beampl.inp and right click on it. Then edit it by selecting open with. SciTe automatically will be opened. Every command inside is ready so we have assumed the we have completed by the mesh and ready for solving.So click tools and click 'solve'.

You have to see that $$ \displaystyle Job Finished $$  Then click tools and click 'post prosess'.

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

$$  \displaystyle Result 1 $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * {| style="width:100%" border="0"

$$  \displaystyle Result2 $$
 * style="width:95%" |
 * style="width:95%" |
 * }

=Problem 5.6: The Quadratic Lagrangian Element Basis Function = Lecture 30-6

Given:
Given the Linear Lagrangian Element Basis Function,

from Lecture 30-6

Find:
The Quadratic Lagrangian Element Basis Function

Solution:
The above is the depiction of the basis function satisfying the CBS for element node i-1 The above is the depiction of the basis function satisfying the CBS for element node i The above is the depiction of the basis function satisfying the CBS for element node i+1 Next we overlay the three functions to show the shape function for the quadratic Lagrangian

=Problem 5.7 Weak form solution using LLEIF=

Given
Solve the following discrete weak form with Linear Lagrange Element Interpolation Basis.

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

$$  \displaystyle \sum {{c_i}\left[ {\sum {{{\tilde K}_{ij}}{d_j} - {{\tilde F}_i}} } \right]} = 0 $$     (3.7.1)
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Where

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$$  \displaystyle {{\tilde K}_{ij}} = \int\limits_0^1 {{{b'}_i}(2 + 3x){{b'}_j}dx} $$     (3.7.2)
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$$  \displaystyle {F_i} = {b_i}\left( 1 \right)\left( {12} \right) + \int_0^1 {{b_i}\left( x \right)5x} dx $$ (3.7.3)
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Such that

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$$  \displaystyle {u^h}\left( 0 \right) = \sum {{d_j}{b_j}\left( 0 \right) = 4} $$     (3.7.4)
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$$  \displaystyle {w^h}\left( 0 \right) = \sum {{c_i}{b_i}\left( 0 \right) = 0} $$     (3.7.5)
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d. Plot $${u^h}$$ vs $${u}$$
Plot $${u^h}\left( {0.5} \right) - u\left( {0.5} \right)$$ vs n (total no. of dof's)

a. Using LLEBF in constraint breaking solution(CBS)
We have following constraint to be satisfied to use weak form

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$$  \displaystyle {w^h}\left( 0 \right) = \sum {{c_i}{b_i}\left( 0 \right) = 0} $$     (3.7.6)
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At node 1 (i.e. at x=0) all the basis function are zero except 1.

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$$  \displaystyle {w^h} = {c_1}{b_1} + {c_2}{b_2} + {c_3}{b_3} + {c_4}{b_4} + \cdots $$     (3.7.7)
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$$  \displaystyle {w^h} = {c_1}(1) + {c_2}(0) + {c_3}(0) + {c_4}(0) + \cdots $$     (3.7.8)
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$$  \displaystyle {w^h} = {c_1} $$     (3.7.9)
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For
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$$  \displaystyle {w^h}\left( 0 \right) = 0 $$     (3.7.10)
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$$  \displaystyle {c_1} = 0 $$     (3.7.11)
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To satisfy homogeneous boundary condition while using Linear Lagrange Element Interpolation Functions, we just have to select our $$w^h$$ such that $${c_1} = 0$$.

c Formulation of K and F on element level
We can divide the total domain into elemental domain.

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$$  \displaystyle \Omega = \bigcup\limits_{e = 1}^{nel} $$     (3.7.12)
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Values of K and F will be summation of K and F over the elements
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$$  \displaystyle {{\tilde K}_{ij}} = \sum\limits_{e = 1}^{nel} {K_{ij}^e} $$     (3.7.13)
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$$  \displaystyle K_{ij}^e = \int\limits_ {{{b'}^e}_i(2 + 3x){{b'}^e}_jdx} $$     (3.7.14)
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$$  \displaystyle {{\tilde F}_i} = \sum\limits_{e = 1}^{nel} {F_i^e} $$     (3.7.15)
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$$  \displaystyle {F_i}^e = \int_ {b_i^e5xdx} $$     (3.7.16)
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There will be two basis function for each element. Generalised values of these basis function will be as follow

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$$  \displaystyle {b_e} = \frac $$     (3.7.17)
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$$  \displaystyle {b_{e + 1}} = \frac $$     (3.7.18)
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d. Solution using Matlab
Plot of exact solution and approximate solution with 2 elements

Plot of exact solution and approximate solution with 4 elements

Plot of exact solution and approximate solution with 6 elements

Plot of exact solution and approximate solution with 8 elements

Plot of error vs dof's

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$$  \displaystyle err = u\left( {0.5} \right) - {u^h}\left( {0.5} \right) $$     (3.7.19)
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$$  \displaystyle dof's = nel + 1 $$     (3.7.20)
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where nel= number of elements.

for 16 elements

dof's=16+1=17

comment
Solution is converging with increasing no. of elements. For n=16 error is less than 0.0001.

=Problem 5.8:Using LLEBF to solve G1DM1.0/D1=

Given
G1DM1.0/D1 : Model for elastic bar problem was given in meeting 9.

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$$  \displaystyle \frac{\mathrm{d} }{\mathrm{d} x}\left [ (2+3x)\frac{\mathrm{d} u}{\mathrm{d} x} \right ]+5x=0 $$     (8.1)
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$$  \displaystyle u(1)=4 $$     (8.2)
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$$  \displaystyle \frac{\mathrm{d} u(x=0)}{\mathrm{d} x}=-3 $$     (8.3)
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Weak Form
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$$  \displaystyle \int_{\Omega }\frac{\mathrm{d} w}{\mathrm{d} x}AE\frac{\mathrm{d} u}{\mathrm{d} x}dx-\left ( wAt \right )_{\Gamma _{h}}-\int_{\Omega }wbdx=0 \Rightarrow \Omega =]0,1[ $$     (8.4)
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We have to manipulate weak form in order to use for element matrices. As to element mesh we use weak form for each elements and then we can take summation for all elements to get global matrices.

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$$  \displaystyle \sum\limits_{e=1}^{nel}{{{w}^{eT}}\left\{ \int\limits_{{{x}_{1}}^{e}}^{{{x}_{2}}^{e}}{{{B}^{eT}}{{A}^{e}}{{E}^{e}}{{B}^{e}}dx{{d}^{e}}-\int\limits_{{{x}_{1}}^{e}}^{x_{2}^{e}}{{{N}^{eT}}bdx-{{({{N}^{eT}}{{A}^{e}}\overline{t})}_{x=0}}}} \right\}=0} $$     (8.5)
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Approximate functions gives as in F&B Book,

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$$  \displaystyle u^{h}=\sum_{e=1}^{nel}N^{e}d^{e}=\left ( \sum_{e=1}^{nel}N^{e}L^{e}\right )\mathbf{d} $$     (8.6)
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$$  \displaystyle w^{h}=\sum_{e=1}^{nel}N^{e}w^{e}=\left ( \sum_{e=1}^{nel}N^{e}L^{e}\right )\mathbf{w} $$     (8.7)
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Where N is called as 'shape functions' and L is called 'gather matrices'.

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$$  \displaystyle N^{e}=\frac{1}{l^{e}}[\begin{matrix} x_{2}^{e}-x & x-x_{1}^{e} \end{matrix}] $$     (8.8)
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$$  \displaystyle B^{e}=\frac{1}{l^{e}}[\begin{matrix} -1 & 1 \end{matrix}] $$     (8.9)
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Analytical Solution

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$$  \displaystyle \int \left ( (2+3x)\frac{\mathrm{d} u}{\mathrm{d} x} \right )dx=-\int (5x)dx+c_{1} $$     (8.10)
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$$  \displaystyle 2\frac{\mathrm{d} u}{\mathrm{d} x}(x=0)=0+c_{1} \Rightarrow c_{1}=-6 $$     (8.11)
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$$  \displaystyle u=-\frac{5}{2}\int \frac{x^2}{2+3x}dx-\int \frac{6}{2+3x}dx+c_{2} $$     (8.12)
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$$  \displaystyle -\frac{5}{2}\left [ \frac{1}{54}(9x^2-12x+8log(3x+2)-12) \right ]-\left [ \frac{1}{3}log(3x+2) \right ]+c_{2} $$     (8.13)
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Using other boundry condition ;

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$$  \displaystyle c_{2}=\frac{119}{36}+\frac{64}{27}log(5) $$     (8.14)
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$$  \displaystyle u(x)=-\frac{5}{12}x^2+\frac{5}{9}x-\frac{64}{27}log(2+3x)+\frac{139}{36}+\frac{64}{27}log(5) $$     (8.15)
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FEA solutions
When evaluating unknowns on distributed nodes our shape functions must meet some conditions.Our approximate functions must be complete in each element. As saying completeness we are saying that by using our nodal values we must be able to identify coefficients in our approximate solutions. After some manipulation we conclude with shape functions. For each nodes we have shape functions and for middle nodes shape functions come from adjacent elements.

For example two element mesh ;
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$$  \displaystyle N=N^{(1)}L^{(1)}+N^{(2)}L^{(2)}=[\begin{matrix} N_{1}^{(1)} & N_{2}^{(1)}+N_{1}^{(2)} & N_{2}^{(2)} \end{matrix}]
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$$     (8.16)
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For this example at x1 (first node) only N1 will be 1. Others will be zero. So we verify CBS.

Using 4 elements

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$$  \displaystyle \Omega =]0,1[ $$ [[image:Egm6321.f10.team2.oztekin.Elemtdiv.png|center]]
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Lets find shape functions for 4 elements.

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$$  \displaystyle\begin{align} &N^{(1)}=\frac{1}{0.25}[\begin{matrix} 0.25-x & x \end{matrix}]\\ &N^{(2)}=\frac{1}{0.25}[\begin{matrix} 0.5-x & x-0.25 \end{matrix}]\\ &N^{(3)}=\frac{1}{0.25}[\begin{matrix} 0.75-x & x-0.5 \end{matrix}]\\ &N^{(4)}=\frac{1}{0.25}[\begin{matrix} 1-x & x-0.75 \end{matrix}] \end{align} $$     (8.17)
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These shape functions are useful for us to make connection between nodes. After defining these tools now we can find stiffness matrices for all elements. Then after finding all of stiffness matrices for elements we can construct global stiffness matrix.

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$$  \displaystyle K^{e}=\int_{x_{1}^{e}}^{x_{2}^{e}}B^{eT}A^{e}E^{e}B^{e}dx $$     (8.18)
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$$  \displaystyle\begin{align} {{K}^{e}}&=\int\limits_{x_{1}^{e}}^{x_{2}^{e}}{\frac{1}\left[ \begin{matrix} -1 \\   1  \\ \end{matrix} \right]}(2+3x)\frac{1}[\begin{matrix} -1 & 1 \\ \end{matrix}]dx \\ &=\frac{1}\left[ \begin{matrix} 1 & -1 \\   -1 & 1  \\ \end{matrix} \right]\left[ 2(x_{2}^{e}-x_{1}^{e})+\frac{3}{2}(x_{2}^{e2}-x_{1}^{e2}) \right] \end{align} $$     (8.19)
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$$  \displaystyle {{K}^{e}}=\frac{1}\left[ \begin{matrix} 1 & -1 \\   -1 & 1  \\ \end{matrix} \right]\left[ 2+\frac{3}{2}(x_{2}^{e}+x_{1}^{e}) \right] $$     (8.20)
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$$  \displaystyle K^{(1)}=\begin{bmatrix} 9.5 & -9.5\\ -9.5& 9.5 \end{bmatrix} $$     (8.21)
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$$  \displaystyle K^{(2)}=\begin{bmatrix} 12.5 & -12.5\\ -12.5& 12.5 \end{bmatrix} $$     (8.22)
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$$  \displaystyle K^{(3)}=\begin{bmatrix} 15.5 & -15.5\\ -15.5& 15.5 \end{bmatrix} $$     (8.23)
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$$  \displaystyle K^{(4)}=\begin{bmatrix} 18.5 & -18.5\\ -18.5& 18.5 \end{bmatrix} $$     (8.24)
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We can construct global stiffness matrix now. This will be distribution of all four element stiffness matrices into appropriate places. For our convenience i will use direct method. But it is useful using below equation for computing purposes.

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$$  \displaystyle K=\sum_{e=1}^{nel}L^{eT}K^{e}L^{e}=L^{1T}K^{1}L^{1}+L^{2T}K^{2}L^{2}+L^{3T}K^{3}L^{3}+L^{4T}K^{4}L^{4} $$     (8.25)
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$$  \displaystyle K=\left[ \begin{matrix} \begin{matrix} \begin{matrix} \begin{matrix} 9.5 \\   -9.5  \\ \end{matrix}  \\ 0 \\   0  \\   0  \\ \end{matrix} & \begin{matrix} \begin{matrix} -9.5 \\   22  \\ \end{matrix}  \\ -12.5 \\   0  \\   0  \\ \end{matrix}  \\ \end{matrix} & \begin{matrix} \begin{matrix} 0 \\   -12.5  \\ \end{matrix}  \\ 28 \\   \begin{matrix} -15.5 \\   0  \\ \end{matrix}  \\ \end{matrix} & \begin{matrix} \begin{matrix} 0 \\   0  \\ \end{matrix}  \\ -15.5 \\   34  \\   -18.5  \\ \end{matrix} & \begin{matrix} \begin{matrix} 0 \\   0  \\ \end{matrix}  \\ 0 \\   -18.5  \\   18.5  \\ \end{matrix}  \\ \end{matrix} \right]
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$$     (8.26)
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The next step we have to find force matrices. Which ha two components. Lets first find force matrix at natural boundary condition and second distributed force matrix on our domain.

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$$  \displaystyle f_{\Gamma _{h}}=-(N^{eT}A^{e}b)_{\Gamma _{h}^{e}}=-N^{eT}(x_{1})(2+3x)_{x=0}(-3)=6N^{eT}(x_{1}) $$     (8.27)
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$$  \displaystyle f_{{\Gamma }^{(1)}}=6\left[ \begin{matrix} N_{1}^{(1)}({{x}_{1}}) \\ N_{2}^{(1)}({{x}_{2}}) \\ \end{matrix} \right]=\left[ \begin{matrix} 6 \\   0  \\ \end{matrix} \right]
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$$     (8.28)
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$$  \displaystyle\begin{align} &f_{{\Gamma }^{(2)}}=f_{{\Gamma }^{(3)}}=f_{{\Gamma }^{(4)}}=6\left[ \begin{matrix} N_{1}^{(e)}({{x}_{1}}) \\ N_{2}^{(e)}({{x}_{2}}) \\ \end{matrix} \right]=\left[ \begin{matrix} 0 \\   0  \\ \end{matrix} \right] &for & e=2,3,4 \end{align} $$     (8.29)
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$$  \displaystyle\begin{align} f_{\Gamma _{h}} &=\sum_{e=1}^{nel}L^{eT}f_{\Gamma _{h}}^{e} \\ &=\left[ \begin{matrix} \begin{matrix} 6 \\   0  \\ \end{matrix}  \\ 0 \\   0  \\   0  \\ \end{matrix} \right]
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\end{align} $$     (N)
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$$  \displaystyle\begin{align} f_{\Omega }^{e} & =\int_{x_{1}^{e}}^{x_{2}^{e}}N^{eT}b(x)dx\\ &=\int\limits_{x_{1}^{e}}^{x_{2}^{e}}{\frac{1}\left[ \begin{matrix} x_{2}^{e}-x \\ x-x_{1}^{e} \\ \end{matrix} \right]}5xdx \\ &=\frac{5}\left[ \begin{matrix} \frac{x_{2}^{e}(x_{2}^{e2}-x_{1}^{e2})}{2}-\frac{(x_{2}^{e3}-x_{1}^{e3})}{3} \\ \frac{(x_{2}^{e3}-x_{1}^{e3})}{3}-\frac{x_{1}^{e}(x_{2}^{e2}-x_{1}^{e2})}{2} \\ \end{matrix} \right]
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\end{align} $$     (8.30)
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$$  \displaystyle f_{\Omega }^{(1)}=\left[ \begin{matrix} 0.05208334 \\   0.10416666  \\ \end{matrix} \right] $$     (8.31)
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$$  \displaystyle f_{\Omega }^{(2)}=\left[ \begin{matrix} 0.2083334 \\   0.2604166  \\ \end{matrix} \right] $$     (8.32)
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$$  \displaystyle f_{\Omega }^{(3)}=\left[ \begin{matrix} 0.36458334 \\   0.41666666  \\ \end{matrix} \right] $$     (8.33)
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$$  \displaystyle f_{\Omega }^{(4)}=\left[ \begin{matrix} 0.52083334 \\   0.57291666  \\ \end{matrix} \right] $$     (8.34)
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$$  \displaystyle\begin{align} f_{\Omega }&=\sum_{e=1}^{4}L^{eT}f_{\Omega }^{e} &\left[ \begin{matrix} \begin{matrix} 0.05208334 \\   0.3125  \\ \end{matrix}  \\ 0.625 \\   0.9375  \\   0.57291666  \\ \end{matrix} \right]
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\end{align} $$     (8.35)
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$$  \displaystyle \left[ \begin{matrix} \begin{matrix} 9.5 \\   -9.5  \\ \end{matrix} & \begin{matrix} -9.5 \\   22  \\ \end{matrix} & \begin{matrix} 0 \\   -12.5  \\ \end{matrix} & \begin{matrix} \begin{matrix} 0 & 0 \\ \end{matrix}  \\ \begin{matrix} 0 & 0 \\ \end{matrix}  \\ \end{matrix} \\ 0 & -12.5 & 28 & \begin{matrix} -15.5 & 0 \\ \end{matrix}  \\ 0 & 0 & -15.5 & \begin{matrix} 34 & -18.5 \\ \end{matrix}  \\ 0 & 0 & 0 & \begin{matrix} -18.5 & 18.5 \\ \end{matrix}  \\ \end{matrix} \right]\left[ \begin{matrix} \begin{matrix} {{d}_{1}} \\ {{d}_{2}} \\ \end{matrix} \\ {{d}_{3}} \\ {{d}_{4}} \\ 4 \\ \end{matrix} \right]=\left[ \begin{matrix} \begin{matrix} 0.05208334 \\   0.3125  \\ \end{matrix}  \\ 0.625 \\   0.9375  \\   0.57291666  \\ \end{matrix} \right]+\left[ \begin{matrix} \begin{matrix} 6 \\   0  \\ \end{matrix}  \\ 0 \\   0  \\   0  \\ \end{matrix} \right]+\left[ \begin{matrix} \begin{matrix} 0 \\   0  \\ \end{matrix}  \\ 0 \\   0  \\   {{r}_{1}}  \\ \end{matrix} \right]
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$$     (8.36)
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$$  \displaystyle \left[ \begin{matrix} {{d}_{1}} \\ {{d}_{2}} \\ {{d}_{3}} \\ {{d}_{4}} \\ \end{matrix} \right]=\left[ \begin{matrix} 2.0257 \\   1.3886  \\   0.8794  \\   0.4285  \\ \end{matrix} \right]
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$$     (8.37)
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Comparing results with exact ones at the nodes;

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$$  \displaystyle\begin{align} &\left[ \begin{matrix} 2.0257 \\   1.3886  \\   0.8794  \\   0.4285  \\ \end{matrix} \right]+\left[ \begin{matrix} 4 \\   4  \\   4  \\   4  \\ \end{matrix} \right]=\left[ \begin{matrix} 6.0257 \\   5.3886  \\   4.8794  \\   4.4285  \\ \end{matrix} \right]
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&{{d}_{exact}}=\left[ \begin{matrix} 6.0331 \\   5.3911  \\   4.8802  \\   4.4286  \\ \end{matrix} \right]

\end{align} $$     (8.38)
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$$  \displaystyle 4 elements $$
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Using 6 elements
Shape functions for 6 elements
 * {| style="width:100%" border="0"

$$  \displaystyle\begin{align} & N^{(1)}=\frac{1}{0.1666}[\begin{matrix} 0.1666-x & x \end{matrix}]\\ &N^{(2)}=\frac{1}{0.1666}[\begin{matrix} 0.3333-x & x-0.16666 \end{matrix}]\\ & N^{(3)}=\frac{1}{0.1666}[\begin{matrix} 0.5-x & x-0.3333 \end{matrix}]\\ &N^{(4)}=\frac{1}{0.1666}[\begin{matrix} 0.6666-x & x-0.5 \end{matrix}]\\ & N^{(5)}=\frac{1}{0.1666}[\begin{matrix} 0.8332-x & x-0.6666 \end{matrix}]\\ & N^{(6)}=\frac{1}{0.1666}[\begin{matrix} 1-x & x-0.8332 \end{matrix}]\\
 * style="width:95%" |
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\end{align} $$     (8.39)
 * <p style="text-align:right">
 * }

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

$$  \displaystyle \left[ \begin{matrix} \begin{matrix} \begin{matrix} \begin{matrix} 13.5048 & -13.5048 & 0 & 0 \\ \end{matrix} & 0 & 0 & 0  \\ \end{matrix} \\ \begin{matrix} \begin{matrix} -13.5048 & 30.0105 & -16.5057 & 0 \\ \end{matrix} & 0 & 0 & 0  \\ \end{matrix} \\ \begin{matrix} \begin{matrix} 0 & -16.5057 & 36.0132 & -19.5075 \\ \end{matrix} & 0 & 0 & 0  \\ \end{matrix} \\ \begin{matrix} \begin{matrix} 0 & 0 & -19.5075 & 42.0157 \\ \end{matrix} & -22.5084 & 0 & 0  \\ \end{matrix} \\ \end{matrix} \\ \begin{matrix} \begin{matrix} 0 & 0 & 0 & -22.5084 \\ \end{matrix} & 48.0164 & -25.508 & 0  \\ \end{matrix} \\ \begin{matrix} \begin{matrix} 0 & 0 & 0 & 0 \\ \end{matrix} & -25.508 & 54.008 & -28.5  \\ \end{matrix} \\ \begin{matrix} \begin{matrix} 0 & 0 & 0 & 0 \\ \end{matrix} & 0 & -28.5 & 28.5  \\ \end{matrix} \\ \end{matrix} \right]\left[ \begin{matrix} \begin{matrix} {{d}_{1}} \\ {{d}_{2}} \\ {{d}_{3}} \\ {{d}_{4}} \\ \end{matrix} \\ {{d}_{5}} \\ {{d}_{6}} \\ 4 \\ \end{matrix} \right]=\left[ \begin{matrix} \begin{matrix} 6.0232199942 \\   0.13890282  \\   0.277972087  \\   0.416641681  \\ \end{matrix}  \\ 0.555277743 \\   0.694970317  \\   0.394287575  \\ \end{matrix} \right]+\left[ \begin{matrix} \begin{matrix} 0 \\   0  \\   0  \\   0  \\ \end{matrix}  \\ 0 \\   0  \\   {{r}_{1}}  \\ \end{matrix} \right]
 * style="width:95%" |
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$$     (8.40)
 * <p style="text-align:right">
 * }

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

$$  \displaystyle \left[ \begin{matrix} \begin{matrix} {{d}_{1}} \\ {{d}_{2}} \\ {{d}_{3}} \\ \end{matrix} \\ {{d}_{4}} \\ {{d}_{5}} \\ {{d}_{6}} \\ \end{matrix} \right]=\left[ \begin{matrix} \begin{matrix} 2.0292 \\   1.5831  \\   1.2098  \\ \end{matrix}  \\ 0.8797 \\   0.575  \\   0.2845  \\ \end{matrix} \right]
 * style="width:95%" |
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$$     (8.41)
 * <p style="text-align:right">
 * }

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

$$  \displaystyle\begin{align} & \left[ \begin{matrix} \begin{matrix} 2.0292 \\   1.5831  \\   1.2098  \\ \end{matrix}  \\ 0.8797 \\   0.575  \\   0.2845  \\ \end{matrix} \right]+\left[ \begin{matrix} \begin{matrix} 4 \\   4  \\   4  \\ \end{matrix}  \\ 4 \\   4  \\   4  \\ \end{matrix} \right]=\left[ \begin{matrix} \begin{matrix} 6.0292 \\   5.5831  \\   5.2098  \\ \end{matrix}  \\ 4.8797 \\   4.575  \\   4.2845  \\ \end{matrix} \right]
 * style="width:95%" |
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& {{d}_{exact}}=\left[ \begin{matrix} \begin{matrix} 6.0331 \\   5.5853  \\   5.2109  \\ \end{matrix}  \\ 4.8802 \\   4.5753  \\   4.2847  \\ \end{matrix} \right]

\end{align} $$     (8.42)
 * <p style="text-align:right">
 * }

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

$$  \displaystyle 6 elements $$
 * style="width:95%" |
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 * }

Using 8 elements
Shape functions for 8 elements
 * {| style="width:100%" border="0"

$$  \displaystyle\begin{align} & N^{(1)}=\frac{1}{0.125}[\begin{matrix} 0.125-x & x \end{matrix}]\\ &N^{(2)}=\frac{1}{0.125}[\begin{matrix} 0.25-x & x-0.125 \end{matrix}]\\ & N^{(3)}=\frac{1}{0.125}[\begin{matrix} 0.375-x & x-0.25 \end{matrix}]\\ &N^{(4)}=\frac{1}{0.125}[\begin{matrix} 0.5-x & x-0.375 \end{matrix}]\\ & N^{(5)}=\frac{1}{0.125}[\begin{matrix} 0.625-x & x-0.5 \end{matrix}]\\ & N^{(6)}=\frac{1}{0.125}[\begin{matrix} 0.75-x & x-0.625 \end{matrix}]\\ & N^{(7)}=\frac{1}{0.125}[\begin{matrix} 0.875-x & x-0.75 \end{matrix}]\\ & N^{(8)}=\frac{1}{0.125}[\begin{matrix} 1-x & x-0.875 \end{matrix}]\\ \end{align} $$     (8.43)
 * style="width:95%" |
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 * }

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

$$  \displaystyle \left[ \begin{matrix} 17.5 & -17.5 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\   -17.5 & 38 & -20.5 & 0 & 0 & 0 & 0 & 0 & 0  \\   0 & -20.5 & 44 & -23.5 & 0 & 0 & 0 & 0 & 0  \\   0 & 0 & -23.5 & 50 & -26.5 & 0 & 0 & 0 & 0  \\   0 & 0 & 0 & -26.5 & 56 & -29.5 & 0 & 0 & 0  \\   0 & 0 & 0 & 0 & -29.5 & 62 & -32.5 & 0 & 0  \\   0 & 0 & 0 & 0 & 0 & -32.5 & 68 & -35.5 & 0  \\   0 & 0 & 0 & 0 & 0 & 0 & -35.5 & 74 & -38.5  \\   0 & 0 & 0 & 0 & 0 & 0 & 0 & -38.5 & 38.5  \\ \end{matrix} \right]\left[ \begin{matrix} \begin{matrix} \begin{matrix} {{d}_{1}} \\ {{d}_{2}} \\ {{d}_{3}} \\ \end{matrix} \\ {{d}_{4}} \\ {{d}_{5}} \\ {{d}_{6}} \\ \end{matrix} \\ {{d}_{7}} \\ {{d}_{8}} \\ {{d}_{9}} \\ \end{matrix} \right]=\left[ \begin{matrix} \begin{matrix} \begin{matrix} \begin{matrix} 6.01302084 \\   0.078125  \\ \end{matrix}  \\ 0.15625 \\ \end{matrix}  \\ 0.234375 \\   0.31250004  \\   0.390625  \\ \end{matrix}  \\ 0.46874373 \\   0.546875  \\   0.29947916  \\ \end{matrix} \right]+\left[ \begin{matrix} \begin{matrix} \begin{matrix} 0 \\   0  \\   0  \\ \end{matrix}  \\ 0 \\   0  \\   0  \\ \end{matrix}  \\ 0 \\   0  \\   {{r}_{1}}  \\ \end{matrix} \right]
 * style="width:95%" |
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$$     (8.44)
 * <p style="text-align:right">
 * }

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

$$  \displaystyle \left[ \begin{matrix} \begin{matrix} \begin{matrix} \begin{matrix} {{d}_{1}} \\ {{d}_{2}} \\ \end{matrix} \\ {{d}_{3}} \\ \end{matrix} \\ {{d}_{4}} \\ {{d}_{5}} \\ \end{matrix} \\ {{d}_{6}} \\ {{d}_{7}} \\ {{d}_{8}} \\ \end{matrix} \right]=\left[ \begin{matrix} \begin{matrix} \begin{matrix} 2.0312 \\   1.6876  \\ \end{matrix}  \\ 1.3904 \\   1.1246  \\   0.8800  \\ \end{matrix}  \\ 0.6497 \\   0.4286  \\   0.213  \\ \end{matrix} \right]
 * style="width:95%" |
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$$     (8.45)
 * <p style="text-align:right">
 * }

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

$$  \displaystyle\begin{align} &\left[ \begin{matrix} \begin{matrix} \begin{matrix} 2.0312 \\   1.6876  \\ \end{matrix}  \\ 1.3904 \\   1.1246  \\   0.8800  \\ \end{matrix}  \\ 0.6497 \\   0.4286  \\   0.213  \\ \end{matrix} \right]+\left[ \begin{matrix} \begin{matrix} \begin{matrix} 4 \\   4  \\ \end{matrix}  \\ 4 \\   4  \\   4  \\ \end{matrix}  \\ 4 \\   4  \\   4  \\ \end{matrix} \right]=\left[ \begin{matrix} \begin{matrix} \begin{matrix} 6.0312 \\   5.6876  \\ \end{matrix}  \\ 5.3904 \\   5.1246  \\   4.88  \\ \end{matrix}  \\ 4.6497 \\   4.4286  \\   4.213  \\ \end{matrix} \right]
 * style="width:95%" |
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&{{d}_{exact}}=\left[ \begin{matrix} \begin{matrix} \begin{matrix} 6.0331 \\   5.6866  \\ \end{matrix}  \\ 5.3911 \\   5.1249  \\   4.8802  \\ \end{matrix}  \\ 4.6498 \\   4.4286  \\   4.213  \\ \end{matrix} \right]

\end{align} $$     (8.46)
 * <p style="text-align:right">
 * }

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

$$  \displaystyle 8 elements $$
 * style="width:95%" |
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 * }

=References=

=Contributing Members & Referenced Lecture=


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User:Eml5526.s11.team6.deshpande/hwk5

Homework 5 = Problem 5.1 Solution of the PDE using weak form.=

Find
Solve the following PDE using weak form with appropriate $$F_I$$ I= Polynomial function, Fourier function, Exponential function until convergence of $$u^h(0.5)$$
 * {| style="width:100%" border="0"

$$ \frac{\partial}{\partial x}\left [ (2+3x)\frac{\partial u}{\partial x} \right ]+5x=0 \qquad \forall  x   \epsilon \left ]0,1   \right [$$ (5.1.1)
 * }
 * }

Given

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$$ u(0)=4, \qquad -\frac{\partial u}{\partial x}_{(x=1)}=6 $$
 *  (5.1.2)
 * }
 * }

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

Now, multiply eq. 5.1.1 by $$ \displaystyle w $$ and integrate between 0 to 1.

Use integration by parts for the first term in eq. 5.1.4

Using eq. (5.1.2) and eq. (5.1.3), we get

Put eq. (5.1.6) in eq. (5.1.5), we get the weak form as follows

a.Using the polynomial basis function

For getting appropriate constraint breaking solution, we choose Also, Take $$ \displaystyle \beta=0$$ for the constraint breaking solution. Therefore, From eq. 5.1.2 and eq. 5.1.10, we can say that Therefore, eq. 5.1.10 becomes Similarly, the weighted function can be written as Eq. 5.1.3 gives, Therefore eq. 5.1.13 becomes Hence, from eq. 5.2.12 and 5.2.15 Put this in eq. 5.1.7, we get Using eq. 5.1.15 we can find $$ \displaystyle w(1) $$ and put in eq. 5.1.18, we get By integrating and verifying with| Wolfram Alpha, we get The following figure shows comparison of analytical and numerical solution for different values of n. As expected, the numerical solution approaches analytical solution as n increases.

Approximate solution converges to analytical solution at n=4. To minimize the error,higher values of n need to be considered.

= Problem 5.2 Solution of the PDE using weak form.=

Find
Solve the following PDE using weak form with appropriate $$F_I$$ I= Polynomial function, Fourier function, Exponential function until convergence of $$u^h(0.5)$$
 * {| style="width:100%" border="0"

$$ \frac{\partial}{\partial x}\left [ (2+3x)\frac{\partial u}{\partial x} \right ]+5x=0 \qquad \forall  x   \epsilon \left ]0,1   \right [$$ (5.2.1)
 * }
 * }

Given

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

$$ u(1)=4, \qquad -\frac{\partial u}{\partial x}_{(x=0)}=6 $$
 *  (5.2.2)
 * }
 * }

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

Now, multiply eq. 5.2.1 by $$ \displaystyle w $$ and integrate between 0 to 1.

Use integration by parts for the first term in eq. 5.2.4

Using eq. (5.2.2) and eq. (5.2.3), we get

Put eq. (5.2.6) in eq. (5.2.5), we get the weak form as follows

a.Using the polynomial basis function

For getting appropriate constraint breaking solution, we choose Also, Take $$ \displaystyle \beta=1$$ for the constraint breaking solution. Therefore, From eq. 5.2.2 and eq. 5.2.10, we can say that Therefore, eq. 5.2.10 becomes Similarly, the weighted function can be written as Eq. 5.2.3 gives, Therefore eq. 5.2.13 becomes Hence, from eq. 5.2.12 and 5.2.15 Put this in eq. 5.2.7, we get Using eq. 5.2.15 we can find $$ \displaystyle w(0) $$ and put in eq. 5.2.18, we get By integrating and verifying with| Wolfram Alpha, we get Following Matlab code is used to find coefficients a_j by changing values of i and j in the above equation The figure shows comparison of analytical and numerical solution for polynomial function. At n=5,numerical solution approaches analytical solution.Error decreases below $$ 10^{-6}$$ for n=10

title: User:Eml5526.s11.team6.deshpande/hwk5


text:
User:Eml5526.s11.t1.ab/HW5

=Problem 5.1: Weak form with appropriate functions such as Polynomial, Fourier & Exponential until convergence.=

Given
General 1-D Model 1 Data set 1. The Partial Differential equation for this data set is,

Find
Solve the Weak Form of the above general model 1 data set 1 for the Functions such as

1)Polynomial Basis Function,

2)Exponential Basis Function &

3)Fourier Basis Function

to find out the convergence point.

1) Polynomial Basis Function
Solve the Differential Equation

in the interval (0,1) with the boundary conditions

The Weak Form of the given Differential Equation will be obtained as follows

So we need to integrate the above equation using integration by parts, after which we will get the following equation,

We now have to select the appropriate Polynomial Basis Function which will follow the constraint breaking solution. so selecting

Clearly,

We need, So Choosing,

We now have, Using the essential Boundary condition we have, Therefore, In the identical way we can also write,

With the condition,

Therefore,

Now we need to calculate the derivative of $$U$$ & $$w$$, Therefore,

Substituting this into the weak form of $$\left(5.1.9\right)$$, we get,

Cancelling $$c_{i}$$ and Organizing, we get, Simplifying the last equation we get,

We can obtain the value of $$\displaystyle a_{j}$$ from the above equation.

The analytical exact solution on the other hand is obtained as,

therefore,

The Values of p & r are obtained by using the following boundary conditions,

To solve this system of equations we developed a MATLAB code. It was found out that with $$n=2$$ a good approximation is obtained. However with$$n=8$$ the error was smaller than $$10^{-6}$$. The MATLAB code is as follows.

Appendix
 Matlab Code:

The analytical and the numerical solution along with the Error between the two are plotted as below.

2) Fourier Basis Function
To follow the constraint Breaking Solution we choose,

We need to satisfy the essential boundary condition So we choose the basis Function such that So we can write, Now the trial solution will be as follows,

Now we need to solve for the remaining coefficients By constructing a Stiffness & Force Matrices as follows.

Here the first equation gets satisfied for even values of i & j,

Whereas 2nd equation is satisfied for odd values of i & j,

further 3rd equation is satisfied for remaining combinations of i & j.

Here the first equation gets satisfied for even values of i & j,

Whereas 2nd equation is satisfied for odd values of i & j,

We developed a MATLAB code to generate the Force and the stiffness Matrix and We get the following values.

We can now use the relationship which is, $$\displaystyle Kd=F$$ to solve for $$\displaystyle d$$.

Ultimately the trial solution of the equation is,

We have developed at MATLAB code to solve these system of equations & to deliver the Error between analytical and the numerical solution.

Appendix
 Matlab Code:

The analytical and the numerical solution along with the Error between the two are plotted as below. From the Plots we can explicitly say that the solution converges at n=9.

3) Exponential Basis Function
To follow the constraint Breaking solution we choose: Clearly, And as we need We choose, Eventually we have, From the essential boundary condition, we have, Therefore, In the identical way we have, Using the condition, Therefore Derivative of U & w will be,

Replacing in the weak form of $$\left(5.1.5\right)$$ we get,

Cancelling $$c_{i}$$ and then organizing we obtain, Performing the integration we obtain following equation,

The value of the $$a_{j}$$can be calculated from the above equation. we developed a MATLAB code to find the error between the Exact and the analytical solution. The MATLAB code is as follows.

Appendix
 Matlab Code:

The following two figures shows the basis Function and the Error between the Analytical and the numerical solution. It can be observed that the Exponential basis Function converges somewhat Slowly than the Polynomial basis Function. For exponential the convergence was achieved with 10 terms whereas with polynomial, it was achieved with only 8 functions. Moreover the stiffness matrix obtained with exponential functions was bad-conditioned for a number of functions greater than 8.

=Problem 5.6: Quadratic Lagrangian Elemental Basis Function.=

Given
Consider any Quadratic Function.

Find
Plot of the Individual Lagrangian Quadratic Function and the combined plot of all the three Lagrangian Quadratic Functions.

Solution
For any Quadratic Function the Number of nodes are always 3. Therefore we can write, We will select the global node numbers as follows, We have according to the Lagrangian Interpolation Basis Function, Which is the Polynomial of degree (n-1)

So, Substituting the values from Equation $$\displaystyle (5.6.2)$$, we get,

In the Similar way,

&

We Developed a MATLAB code to plot all these Quadratic Lagrangian Basis Function. This very MATLAB code is as follows.

Appendix
 Matlab Code:

The plots of all the Individual Lagrangian Quadratic Function and the Combined plot of all the Three Functions are as follows.

title: User:Eml5526.s11.t1.ab/HW5


text:
User:Eml5526.s11.team2/homework4homework4 Xfile/cleveland

=Problem 5.5: =

Guide to Using the ccx Solver
=Problem 5.1=

Concise Formulation of Problem
Specifically we want to determine $$ \displaystyle u^h(x) $$ by solving to following second order differential equation subject to the essential and natural boundary conditions. On the domain defined by,

Find: The Weak Form in Order to Solve for $$ u^h(x) $$ with Following Basis Functions
Solve for $$ \displaystyle u(x) $$ and $$ \displaystyle u^h(x) $$ using the following basis functions until the error $$ \displaystyle e(m) =|u^h(0.5)-u(0.5)| \leq 10^{-6}$$.

 * 1) Polynomial Basis Function  $$ \displaystyle \mathcal{F}_P :=\{ x^j,j=0,1,2,...\} $$
 * 2) Exponential Basis Function $$ \displaystyle \mathcal{F}_e :=\{ e^{jx},j=0,1,2,...\} $$
 * 3) Fourier Basis Function     $$ \qquad \displaystyle \mathcal{F}_F :=\{cos(jx),j=0,1,2,...,sin(kx),k=1,2,...\} $$


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User:Matt~enwikiversity

Hello
I am an educator and learner in the areas of IT, Business and Sustainability i.e sustainable practices (eg: agriculture) and community (eg: processes). Within these areas I am very interested in how people learn practices and processes and how these may change from unsustainable to sustainable.

title: User:Matt~enwikiversity


text:
User talk:Matt~enwikiversity

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title: User talk:Matt~enwikiversity


text:
User:Eml5526.s11.team7.jb/work2

For Polynomial Basis
The corresponding family $${{\bar{F}}_{P}}=\left\{ {{{\bar{b}}}_{j}} \right\}$$ of polynomial basis function satisfying the Constraint Breaking Solution (CBS) for $$\Gamma_g = \left \{ 1 \right \}$$ is:

$${{F}_{P}}=\left\{ {{\left( x-1 \right)}^{j}},\text{ }j=0,1,2,\ldots \right\}$$

Now that the family of basis functions are determined, we recall that the discrete weak form is defined by

$$ \mathbf{\tilde{d}} \cdot \mathbf{b}( \beta ) = g \qquad \text{and} \qquad \mathbf{\tilde{c}} \cdot \left[ \mathbf{\tilde{M}} \mathbf{\tilde{d}}^{(s)} + \mathbf{\tilde{K}} \mathbf{\tilde{d}} - \mathbf{\tilde{F}} \right] = 0 $$

where the expression $$\displaystyle \mathbf{\tilde{M}} \mathbf{\tilde{d}}^{(s)} = 0 $$ for the static case. Then the discrete weak form is simplified to:

\mathbf{K} \mathbf{d} = \mathbf{F} \qquad \Rightarrow \qquad \mathbf{K} \mathbf{d} = \mathbf{F}_F - \mathbf{K}_{FE} g. $$

The first constant, $$\displaystyle d_0$$, is determined from the formulation of the CBS. It is known that:

 * $$\displaystyle u^h(\beta) = d_0 b_0(\beta)=g$$

 * $$\displaystyle d_0= \frac{g}{b_0(\beta)}=4/1=4$$

Now the matrices used to determine the rest of the coefficients are determined as follows. Note that the number of terms needed in the approximate solution, $$\displaystyle n$$, is determined by the convergence interval given. The first few terms of each matrix computation is shown. A Matlab script was written to iterate these computation for increasing values of $$\displaystyle n$$ until the desired accuracy was reached.

\displaystyle

{F_F} = \left\{ {{F_i},i = 1,2,...,n} \right\} = \left\{ {\tilde f\left( \right),i = 1,2,...,n} \right\}

$$

\displaystyle

\tilde f({b_i}) = {b_i}(\alpha )h + \int\limits_\alpha ^\beta {{b_i}fdx}

$$

\displaystyle

h = {a_2}(\alpha )\frac = (2+3x)(\alpha )\frac = 2*6=12

$$

\displaystyle

\tilde f({b_1}) = {b_1}(0)h + \int\limits_0^1 {{b_1}(5x)dx}=1*12+\int\limits_0^1  {{(x-1)}(5x)dx}=-77/6

$$

\displaystyle

\tilde f({b_2}) = {b_2}(0)h + \int\limits_0^1 {{b_2}(5x)dx}=-1*12 + \int\limits_0^1  {{(x-1)^2}(5x)dx}=149/12

$$

$$ \displaystyle

\therefore {F_F} = \left[ {\begin{array}{ccccccccccccccc} {-77/6} \\    {149/12} \\   {.} \\   {.} \\  {.} \end{array}} \right]

$$

\displaystyle

{K_{FE}} = \left\{ {{K_{0j}};j = 1,...,n} \right\} = \left\{ {\tilde k\left( {{b_0},{b_j}} \right),j = 1,...,n} \right\}

$$

\displaystyle

\tilde k({b_i},{b_j}) = \int\limits_0^1 {b{'_i}{a_2}b{'_j}dx}

$$

\displaystyle

\tilde k({b_0},{b_1}) = \int\limits_0^1 {b{'_0}{a_2}b{'_1}dx} = \int\limits_0^1 {b{'_0}(2+3x)b{'_1}dx}  = 0 = \tilde k({b_0},{b_2}) = \tilde k({b_0},{b_3}) = 0

$$

\displaystyle

\therefore {K_{EF}} = \left[ {\begin{array}{ccccccccccccccc} 0 \\  0 \\   . \\ .\\ . \end{array}} \right]

$$

And：

\displaystyle

\because {K_{FE}} = {K_{EF}}^T = \left[ {\begin{array}{ccccccccccccccc} 0 \\  0 \\   . \\ .\\ . \end{array}} \right]

$$

\displaystyle

{K_{FF}} = \left[ {{K_{ij}};i,j = 1,2,...} \right] = \tilde k({b_i},{b_j})

$$

\displaystyle

\tilde k({b_1},{b_1}) = \int\limits_0^1 {b{'_1}{2+3x}b{'_1}dx} = \int\limits_0^1 {1*(2+3x)*1dx}=7/2

$$

\displaystyle

\tilde k({b_1},{b_2}) = \int\limits_0^1 {b{'_1}{2+3x}b{'_2}dx} = \int\limits_0^1 {1*(2+3x)*2(x-1)dx}=-3

$$

\displaystyle

\tilde k({b_2},{b_1}) = \int\limits_0^1 {b{'_2}{2+3x}b{'_1}dx} = \int\limits_0^1 {2(x-1)*(2+3x)*1dx}=-3

$$

\displaystyle

\tilde k({b_2},{b_2}) = \int\limits_0^1 {b{'_2}{2+3x}b{'_2}dx} = \int\limits_0^1 {2(x-1)*(2+3x)*2(x-1)dx}=11/3

$$

\displaystyle \therefore {K_{FF}} = \begin{bmatrix} 7/2&-3 &.&. \\  -3&11/3  & .&.\\ . &.  & .&.\\ . &.  & .&. \end{bmatrix} $$

The matrices are then solved for the coefficients:

\displaystyle

Kd = F $$

\displaystyle

F = {F_F} - K_{FE}g = {F_F} $$

\displaystyle

K_{FF}d_F=F_F $$

\displaystyle

d_F=K_{FF}^{-1} F_F $$

The approximate solution, $$\displaystyle u^h(x)$$ is defined as,

 * $$\displaystyle u^h(x)= d_0b_0+\sum_{j=1}^{n}d_jb_j(x) \qquad \text{where} \quad d_0=4 \ \text{and} \ b_0=1$$.

As stated above a Matlab script was written to iterate $$\displaystyle n$$ until the desired accuracy was reached. At $$\displaystyle n=6$$ the error was $$\displaystyle  O(3.4091*10^{-6})$$. The following plots were generated:

title: User:Eml5526.s11.team7.jb/work2


text: File:Egm6342.s11.team3.hw5 3a.sskim.png

title: File:Egm6342.s11.team3.hw5 3a.sskim.png


text: File:Egm6342.s11.team3.hw5 3b.sskim.png

title: File:Egm6342.s11.team3.hw5 3b.sskim.png


text: File:Egm6342.s11.team3.hw5 3c.sskim.png

title: File:Egm6342.s11.team3.hw5 3c.sskim.png


text: File:Hw5 6.jpg

title: File:Hw5 6.jpg


text: File:Egm6342.s11.team3.hw5 4.sskim.png

title: File:Egm6342.s11.team3.hw5 4.sskim.png