text:
User:Eml5526.s11.team3.hylon/Homework 6

=Problem 6.6 Solve G2DM1.0/D1 using 2D LLEBF until accuracy 10^{-6} at center(0,0)=

Problem Statement
Refer to lecture notes [[media:fe1.s11.mtg35.djvu|35-4]], [[media:fe1.s11.mtg36.djvu|36-6]] and [[media:fe1.s11.mtg37.djvu|37-1]] for more detailed description.

Given
Strong form for the heat problem is given in Lecture 33-2:

where $$\boldsymbol{\kappa}\left( {x,t} \right) = \mathbf{I} $$ $$\qquad\text{and}\quad $$   $$f\left( {x,t} \right) = 0 $$   and $$\frac{\partial u}{\partial t}\left( {x,t} \right) = 0 $$

so we get

Essential boundary conditions, Natural boundary conditions,

2). Integrate element stiffness matrix in parent coordinate
Find appropriate u to integrate stiffness matrix exactly with Gauss-Legendre Quadrature. Given detailed argument.

3). Use distorted quadrilateral meshes, compare results with uniform meshes
Meshes can be generated from CGX, and analysis can be done using Fish and Belytschko's code and CCX.

Find Exact solution
The solution for the PDE $$\frac{\partial^2 u}{\partial x^2}+\frac{\partial^2 u}{\partial y^2}=0$$ will be of the form

$$u(x,y)=c_{1}xy+c_{2}x+c_{3}y+c_{4}$$ where,  $$c_{1},c_{2},c_{3} and c_{4}$$  are constants.

Details of the solution of PDE can be find at EqWorld

The essential BC is specified through out the boundary of the square hence u=2 for the points on the 4 vertices of the square, i.e for (-1,-1);(-1,1);(1,-1) and (1,1) we have u=2.

Substituting for these four points into the equation for u above and solving them simultaneously we get

$$c_{1}=0$$ ,

$$c_{2}=0$$ ,

$$c_{3}=0$$

and $$c_{4}=2$$

hence the exact solution of the PDE is given by

Results and plots
Below is the analysis results:

Comparison with using distorted quadrilateral meshes
=Problem 6.9 Reproduce Exercise8.3, 8.4 and Problem 8.6 in Textbook = Refer to P200, P205 and P212 of Fish & Belytschko's A First Course in Finite Elements, 2007, John Wiely & Sons, Ltd, for more detailed description.

Problem Statement
The conductivity is isotropic and K=5. The temperature T=0 is prescribed along bottom edge and left edge.

The heat flux q=0 and q=20 are prescribed on the right edge and top edge, respectively.

A constant heat source s=6 is applied over the plate.

Analysis Results
Using provided Matlab codes, we can obtain following results for using different number of element.

Instruction on using the provided Matlab codes:
heat2d.m is the main program;

When running heat2d.m, it will call functions of include_flags, [K,f,d] = preprocessor, [ke, fe] = heat2Delem(e), [K,f] = assembly(K,f,e,ke,fe), src_and_flux(f), [d,f_E] = solvedr(K,f,d) and postprocessor(d) step by step.

In the include_flags function, we define global variables which may be used in following functions and subprograms.

In the [K,f,d] = preprocessor function, we define how many elements we gonna use in this program, and there are three types provided, i.e. 1element, 16elements and 64elements. While doing analysis, we can change element number here. In each element type input file, we define material property, essential BCs, natural BCs, mesh generation and also plot the meshed figure.

In the [ke, fe] = heat2Delem(e) function, we compute the elemental conductance matrix and nodal flux vector. In this function, the Gauss function for integration is also called.

In the [K,f] = assembly(K,f,e,ke,fe) function, we assemble the elemental conductance matrix and the heat source vector.

In the src_and_flux(f) function, we Compute and assemble nodal boundary flux vector and point sources.

In the [d,f_E] = solvedr(K,f,d) function, we partition and solve the system of equations.

In the postprocessor(d) function, we plot temperature and flux.

Matlab Codes

From below website, we can download all the required Matlab Codes. All Rights Reserved.

The specific codes for this case are in the package of Chapter 8.

Matlab Codes donwload

Problem Statement
Since the exercise haven’t specify the parameters a and b in the textbook, from the plot provided, we take a=0.25 and b=1 for following analysis. For heat equation with isotropic conductivity and K=1, the corresponding source term is given by
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$$\displaystyle
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s=-{{\nabla }^{2}}T=2a\frac{1}{\sqrt{{{x}^{2}}+{{y}^{2}}}}-4

$$ The essential boundary conditions on $${{\Gamma }_{T}}$$ are
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$$\displaystyle
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T(r=a)=0,T(x=\pm b,y)={{a}^{2}}+{{b}^{2}}+{{y}^{2}}-2a\sqrt{{{y}^{2}}+{{b}^{2}}}

$$ The natural boundary conditions on $${{\Gamma }_{q}}$$ are ($$\bar{q}=-k{{n}^{T}}\nabla T$$)
 * }.
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$$\displaystyle
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\begin{align} & \bar{q}(x,y=b)=-\frac{\partial T}{\partial y}(x,y=b)=2b(\frac{2a}{\sqrt{{{x}^{2}}+{{y}^{2}}}}-1) \\ & \bar{q}(x,y=-b)=-\frac{\partial T}{\partial y}(x,y=-b)=2b(1-\frac{2a}{\sqrt{{{x}^{2}}+{{y}^{2}}}}) \\ \end{align}

$$
 * }

Analysis Results
Based on provided Matlab codes for exercise 8.3, page 200, we modified the code to match the problem 8.4, page 205.

Below is the analysis results:

Problem Statement
Consider a chimney constructed of two isotropic materials: dense concrete ($$k=2.0W{}^{\circ }{{C}^{-1}}$$) and bricks ($$k=0.9W{}^{\circ }{{C}^{-1}}$$).

The temperature of the hot gases on the inside surface of the chimney is $$140{}^{\circ }C$$, whereas the outside is exposed to the surrounding air, which is at $$T=10{}^{\circ }C$$. The dimensions of the chimney (in meters) are shown below.

For the analysis, exploit the symmetry and consider 1/8 of the chimney cross sectional area. Consider a mesh of eight elements as shown below. Determine the temperature and flux in the two materials.

Analyze the problem with 2×2, 4×4 and 8×8 quadrilateral elements for 1/8 of the problem domain. A2×2 finite element mesh is shown in Figure below. Symmetry implies insulated boundary conditions on edges AD and BC. Note that element boundaries have to coincide with the interface between the concrete and bricks.

Analysis Results
Based on provided Matlab codes for exercise 8.3, page 200, we modified the code to match the problem 8.6, page 212.

Below is the analysis results:

title: User:Eml5526.s11.team3.hylon/Homework 6


text:
User:Egm6341.s11.team2.no one needs to knowHW6

=Problem 6.1: Higher Order Trapezoid Rule Error (HOTRE) = Reference: Problems 2, and 3

''' This problem was solved without referring to S10. '''

A. Finding the Polynomials $$ {p_6}(t) $$ and $$ p_7(t) $$ and $$ E $$
Referring to Lecture 30,$$ \displaystyle P_6(t) $$ and $$ \displaystyle p_7(t) $$ can be written as follows:

$$ \displaystyle {p_6}(t) = \int\limits_{ - 1}^1 {{p_5}(t).dt = } {c_1}\frac + {c_3}\frac + {c_5}\frac + {c_7} $$

$$ \displaystyle  {p_7}(t) = \int\limits_{ - 1}^1 {{p_6}(t).dt = } {c_1}\frac + {c_3}\frac + {c_5}\frac + {c_7}t + {c_8}  $$

Knowing that $$ \displaystyle p_7(t) $$ is an off function and $$ \displaystyle p_7(0)= 0 $$ and $$ \displaystyle p_7(-1)=p_7(-1)=0 $$, the coefficients $$ \displaystyle c_7 $$ and $$ \displaystyle c_8 $$ can be calculated.

$$ \displaystyle p_7(0)=0 \Rightarrow c_8=0 $$

$$ \displaystyle p_7(1)=0 \Rightarrow c_7= 31/15120 $$

where

In order to fine the error, referring to Lecture 30, page 6, Equation (2) :

$$\displaystyle E = \left[ {{p_2}{g^{(1)}} + {p_4}{g^{(3)}}} \right]_{ - 1}^{ + 1} - \int\limits_{ - 1}^1 {{p_5}{g^{(5)}}.dt = } \left[ {{p_2}{g^{(1)}} + {p_4}{g^{(3)}}} \right]_{ - 1}^{ + 1} + \left[ \right]_{ - 1}^{ + 1} - \int\limits_{ - 1}^1 {{p_6}{g^{(6)}}.dt}  $$

$$\displaystyle E = \left[ {{p_2}{g^{(1)}} + {p_4}{g^{(3)}} + {p_6}{g^{(5)}}} \right]_{ - 1}^{ + 1} + \underbrace {\left[ \right]_{ - 1}^{ + 1}}_{ = 0} - \int\limits_{ - 1}^1 {{p_7}{g^{(7)}}.dt} $$

B. Finding $$ t_k(x) $$
Referring to Lecture 30, page 2:

$$ x(t) = t\frac{h}{2} + \frac{2} $$

Therefore:

note that :

$$\displaystyle  {t_k} = \frac{2}{h}[{x_k} - (\frac{2})] = \frac{2}{h}[\frac{2}] = \frac{2}{h}[\frac{2}] =  - 1  $$

and

$$\displaystyle {t_{k + 1}} = \frac{2}{h}[{x_{k + 1}} - (\frac{2})] = \frac{2}{h}[\frac{2}] = \frac{2}{h}[\frac{h}{2}] = 1   $$

C. Finding $$ d_1$$, $$ d_2 $$, and $$ d_3 $$
Referring to Lecture 31, page 2, Equation (2):

$$ {d_r} = \overline  = \frac $$

therefore:

$$ {d_1} = \overline  = \frac $$

and

$$ {d_2} = \overline  = \frac $$

and

$$ {d_3} = \overline  = \frac $$

=Problem 6.2: Proof of Higher Order Trapezoid Rule Error (HOTRE) =

Reference: Problem 5

''' This problem was solved without referring to S10. '''

Given: The Error for the Trapezoid Rule
The error of the trapezoid rule (Referring to Lecture 30, Page 2) is:

$$\displaystyle E^T_n= \sum_{k=0}^{n-1} \left[  \int_{x_{k}}^{x_{k+1}}f(x)dx-\frac{h}{2} \left\{ f(x_k)+f(x_{k+1}) \right\}  \right] $$

Find: The Error in the Higher Order Trapezoid Rule

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E=\sum_{r=1}^{\ell} 2 \, \overline{d}_{2r} \, h^{2r-1}\Bigg[f^{(2r-1)}(b)-f^{(2r-1)}(a)\Bigg]-\frac{h^{2\ell}}{2^{2\ell}} \sum_{k=0}^{n-1} \int_{x_k}^{x_{k+1}} \, p_{2\ell}(t_{k}(x)) \, f^{(2\ell)}(x)dx $$
 * $$\displaystyle
 * $$\displaystyle
 * }.
 * }.

where,

 * $$\displaystyle \overline{d}_{2r} = \frac{p_{2r}(1)}{2^{2r}} $$
 * $$\displaystyle x \in [a,b] $$

Solution
Referring to Referring to Lecture 30, Page 2, the error can be written as:

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$$\displaystyle E^T_n= \frac{h}{2}\sum_{k=0}^{n-1} \left[  \int_{-1}^{+1}g_k(t)dt- \left\{ g_k(-1)+g_k(+1) \right\}  \right] $$     (6.2.1)
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 *
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where

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$$\displaystyle g_k(t) = f(x(t)) $$     (6.2.2)
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 *
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Please see Lecture 30, Page 2 for the more information on the transformation.

Using successive steps of integration by part method, the error can be written as:

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$$\displaystyle \begin{align} E^T_n & = \frac{h}{2} \left[ \sum_{k=0}^{n-1}\Big[p_2(t)g_k^{(1)}(t)+p_4(t)g_k^{(3)}(t_k)+p_6(t)g_k^{(5)}(t)+......+p_{2\ell}(t)g_k^{(2\ell-1)}(t)\Big]_{-1}^{+1}- \sum_{k=0}^{n-1} \left[ \int_{-1}^{1}p_{2\ell}(t)g_k^{(2\ell)}(t) dt \right]\right]\\ &= \frac{h}{2} \left[ \sum_{r=1}^{\ell}\sum_{k=0}^{n-1}\Big[p_{2r}(t)g_k^{(2r-1)}(t)\Big]_{-1}^{+1} - \sum_{k=0}^{n-1} \left[    \int_{-1}^{1}p_{2\ell}(t)g_k^{(2\ell)}(t) dt \right]\right] \end{align} $$     (6.2.3)
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 *
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note that $$\displaystyle p_{2r}(1)=p_{2r}(-1) $$, therefore:

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$$\displaystyle \begin{align} E^T_n &= \frac{h}{2} \left[ \sum_{r=1}^{\ell}\sum_{k=0}^{n-1}\Big[p_{2r}(+1)g_k^{(2r-1)}(+1)-p_{2r}(-1)g_k^{(2r-1)}(-1)\Big]- \sum_{k=0}^{n-1} \left[   \int_{-1}^{1}p_{2l}(t)g_k^{(2\ell)}(t) dt \right]\right]\\ & \\ &= \frac{h}{2} \left[ \sum_{r=1}^{\ell}\sum_{k=0}^{n-1} p_{2r}(+1) \Big[g_k^{(2r-1)}(+1)-g_k^{(2r-1)}(-1)\Big]- \sum_{k=0}^{n-1} \left[   \int_{-1}^{1}p_{2l}(t)g_k^{(2\ell)}(t) dt\right]\right] \end{align} $$     (6.2.4)
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 *
 * }

Referring to Lecture 30, Page 3, Equation 2:

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$$\displaystyle g_k^{(i)}(t)=\left(\frac{h}{2}\right)^if^{(i)}(x(t)),\,\,\,\,\ x \in [x_k,x_{k+1}] $$     (6.2.5)
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 *
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which results in:

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$$\displaystyle \begin{align} &g_k^{(2r-1)}(+1)=\left(\frac{h}{2}\right)^{2r-1}f^{(2r-1)}(x_{k+1}),\\ &g_k^{(2r-1)}(-1)=\left(\frac{h}{2}\right)^{2r-1}f^{(2r-1)}(x_{k}),\\ &g_k^{(2\ell)}(t)=\left(\frac{h}{2}\right)^{2\ell}f^{(2\ell)}(x_{t}) \end{align} $$     (6.2.6)
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 *
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Plug equation 2.6 in equation 2.4:

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$$\displaystyle \begin{align} E^T_n & = \left(\frac{h}{2}\right) \left[ \sum_{r=1}^{\ell} p_{2r}(+1) \left(\frac{h}{2}\right)^{2r-1} \sum_{k=0}^{n-1} \Big[f^{(2r-1)}(x_{k+1})-f^{(2r-1)}(x_k)\Big]- \sum_{k=0}^{n-1} \left[   \int_{x_k}^{x_{k+1}}p_{2\ell}(t_k(x))\left(\frac{h}{2}\right)^{2\ell}f^{(2\ell)}(x) dx \right] \right]\\ &= \sum_{r=1}^{\ell} h^{2r} \frac{p_{2r}(+1)}{2^{2r}} \Big[f^{(2r-1)}(x_{n})-f^{(2r-1)}(x_0)\Big]-  \left(\frac{h}{2}\right)^{2\ell+1} \sum_{k=0}^{n-1} \left[ \int_{x_k}^{x_{k+1}}p_{2\ell}(t_k(x))f^{(2\ell)}(x) dx \right] \end{align} $$     (6.2.7)
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where we may define:

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$$\displaystyle \overline{d}_{2r} = \frac{p_{2r}(+1)}{2^{2r}} $$ (6.2.8)
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$$\displaystyle x_n=b, x_0=a $$ (6.2.7)
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 Please check the final answer since the factor for the second summation has a different power than the given equation ( in the problem statement, I could not figure where is the problem

=Problem 6.3 Verify and compare coefficients in HOTRE(Higher order trapezoidal rule error) from Bernoulli numbers=

''' This problem was solved without referring to S10. '''

Given: The Bernoulli Number in the Following Series
$$\displaystyle {B_{2r}}$$ are the Bernoulli numbers with even index, which can be determined from the Taylor series expansion

Where the coefficients can be determined using the following equation:

1. The Bernoulli Coefficients for 2n = 2,4,6
$$\displaystyle \bar{d_{2}}, \; \bar{d_{4}}, \; \bar{d_{6}}$$

1. Determine the Bernoulli Coefficients for 2n = 8,10
$$\displaystyle \bar{d_{8}}, \; \bar{d_{10}}$$ using Eq.6.3.1 and compare the results with Eq.6.3.2

Solution
1) We can obtain the Taylor series of $$x\,coth(x)$$ using wolframalpha:

Compare Eq.6.3.3 and Eq.6.3.1, we can obtain that

The above results are identical with what we have using Eq.6.3.2 in Problem 6.1.

2) From Eq.6.3.3 and compare it to Eq.6.3.1, we have

Comment: The Eq.6.3.1 is slightly different from the equation given in Mtg31-3, however this equation should be correct after our comparison.

=Problem 6.4 Redo Proof of HOTRE=

From the lecture slide Mtg 31-4

''' This problem was solved without referring to S10 homework. '''

Problem Statement:Using a Different Approach
Redo the proof of the HOTRE by another approach: cancel all the odd order derivative of $$\displaystyle g$$, and try to find where this approach collapse.

Solution: Proof of HOTRE Using a Different Approach
We have known form the class that the error at the panel $$\displaystyle k$$ can be expressed as:

If we keep on this integration by part and using $$\displaystyle p(t)$$ to cancel all the odd order derivative of $$\displaystyle {{g}_{k}}(t)$$, we have,

Similar to the process in step 2ab and step 3ab on the lecture, when we use $$\displaystyle p(t)$$ to cancel all the odd order derivative of $$\displaystyle {{g}_{k}}(t)$$, i.e. using $$\displaystyle {{p}_{2r}}$$ to cancel all $$\displaystyle {{g}_{2r-1}}(t)$$ by letting $$\displaystyle {{p}_{2r}}(1)={{p}_{2r}}(-1)=0$$, we are actually ensuring that all $$\displaystyle {{p}_{2r+1}}$$ are odd functions.

Hence,

Then Eq (6.4.2) becomes,

And the total error is a summation of $$\displaystyle {{E}_{k}}$$,

Reason why this approach fail
Then we can see the critical weakness of the approach of cancelling odd order derivative of $$\displaystyle {{g}_{k}}(t)$$ lies in the summation $$\displaystyle \sum\limits_{k=0}^{n-1}{\left[ g_{k}^{(2r)}(1){\color{red}+}g_{k}^{(2r)}(-1) \right]}$$.

Because the sign between $$\displaystyle g_{k}^{(2r)}(1)$$ and $$\displaystyle g_{k}^{(2r)}(-1)$$ is "plus" rather than "minus" then it can ‘ t cancel terms when sum according to $$\displaystyle k$$ making the final expression too complex compared to the approach stated in the lecture and also making the computation time enlongated.

Note that the corresponding summation in the preferred approach is with a "minus" sign reducing the summation symbol from 2 to 1.

=Problem 6.5 Calculation using Recurrence Formula=

''' This problem was solved by referencing Team 2, S10 homework 5, Problem8. '''

From the lecture slide Mtg 32-3

Given: The Recurrence Relation and Coefficience
For $$\displaystyle {{p}_{2i}}$$ and $$\displaystyle {{p}_{2i+1}}$$:

And the relation for the coefficient $$\displaystyle {{c}_{2i+1}}$$:

Solution:
From the S10 work we have known,

Then by Eq(6.5.3) we have,

Then by the recurrence formula Eq(6.5.1) and (6.5.2) we have:

and also,

=Problem 6.6 - Understanding Kessler's Code=

''' This problem was solved without referring to S10. '''

Given: Kessler's Code
Kessler's code, given in his paper :

Find: The Results Given by Kessler's Code and Develop a Better Understanding of HORTE
In order to better understand Kessler's code and the HOTRE, improve upon the work of s10 and explain the code line by line. Then, run Kessler's code for $$i=1,2,3$$ to reproduce his results. The collective work of s10 is referenced here under HW5, Problem 10.

Solution: A Line-by-Line Synopsis
The following box presents Kessler's code commented line-by-line to describe the algorithm for using the HOTRE. Running Kessler's code with n = 3 results in the following output:

These with the additional last line written into the code, the output matches the tables presented in Kessler's paper:

=Problem 6.7: Determination of Ellipse and Bifolium Arc Lengths =

''' This problem was solved with reference to S10 Matlab code. '''

Given: Definition of an Ellipse and Bifolium
The equation defining an ellipse,
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$$\displaystyle \left( \frac{x}{a} \right)^2 + \left( \frac{y}{b} \right)^2 = 1 $$     (6.7.1)
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Where the eccentricity is given by,
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$$\displaystyle e=\left(1- \frac{b^2}{a^2} \right)^{1/2} $$     (6.7.2)
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 *
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The parameterized form of the ellipse can be rewritten in terms of $$ r(\theta) $$ and the eccentricity as follows,
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$$\displaystyle \begin{align} r(\theta ) &= \frac\\ a&=10\\ b&=3 \end{align} $$     (6.7.3)
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 *
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Furthermore the circumference of the Ellipse, often referred to as the Elliptic Integral of the 2nd Kind, is shown here,
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$$\displaystyle \begin{align} & i) \quad C = \int_0^{2\pi } {dl} = 4aE(e)\\ & ii) \quad dl = \left[1-e^2cos^2 t \right]^{1/2} dt \\ \implies \quad &iii) \quad E(e) = {\int_0^{\frac{\pi }{2}} {\left[ {1 - {e^2}\sin (\theta )} \right]} ^{\frac{1}{2}}}d\theta \end{align} $$     (6.7.4)
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 *
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The equation for the bifolium is,
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$$\displaystyle r(\theta ) = 2\sin (\theta ){\cos ^2}(\theta ) $$     (6.7.5)
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 *
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1. The Parameterized Form of the Ellipse
Using the trigonometric relationships
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$$\displaystyle \begin{align} x &= acos(t)\\ y &= bsin(t) \end{align} $$     (6.7.6) Determine $$ \displaystyle dl $$ and show equation (6.7.4.iii) holds.
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 *
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A) For an Ellipse

 * Elliptical Integral of the 1st Kind Integrating (6.7.3)
 * Then with the following parametrized definitions determine the arc length $$ \displaystyle \overset{\frown}{PQ}$$ using:
 * Composite Trapezoid
 * Romberg Quadrature Rule
 * Clenshaw-Curtis Rule
 * chebfun/sum command

 * Elliptical Integral of the 2nd Kind Using (6.7.4iii)
 * Using the error for the composite trapezoid rule find $$ \displaystyle n $$ such that
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$$\displaystyle $$     (6.7.7)
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 * E_{n}| \leq h^2\frac{(b-a)}{12}M_{2}
 *
 * }

 * Then with the following parametrized definitions determine the arc length $$ \displaystyle \overset{\frown}{PQ}$$ using:
 * Composite Trapezoid
 * Romberg Quadrature Rule
 * Clenshaw-Curtis Rule
 * chebfun/sum command

B) For the Bifolium

 * Determine the folium length using the following numerical integration methods:
 * Composite Trapezoid
 * Romberg Quadrature Rule
 * Clenshaw-Curtis Rule
 * chebfun/sum command

1. The Parameterized Form of the Ellipse
Using the trigonometric relationships
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$$\displaystyle \begin{align} x &= acos(t)\quad \implies \quad dx=-acos(t)\\ y &= bsin(t)\quad \implies \quad dy=bcos(t) \end{align} $$     (6.7.8)
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 *

 * }

An infinitesimal change in $$ \displaystyle l $$ written as $$ \displaystyle dl $$ can be described in terms of the infinitesimal change in $$ \displaystyle x $$ and $$ \displaystyle y $$ as,
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$$\displaystyle dl = {\left[ {d{x^2} + d{y^2}} \right]^{\frac{1}{2}}} $$     (6.7.9)
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 * }

Substituting the relations shown in (6.7.8) into (6.7.9) we get,
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$$\displaystyle dl = {\left[ {{a^2}{{\sin }^2}(\theta ) + {b^2}{{\cos }^2}(\theta )} \right]^{\frac{1}{2}}}d\theta $$    (6.7.10) Next we will do some algebra and arrive a the desired result,
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 *

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$$\displaystyle \begin{align} dl &= a{\left[ {{{\sin }^2}(\theta ) + \frac{{\cos }^2}(\theta )} \right]^{\frac{1}{2}}}d\theta \\ &= a{\left[ {1 - {{\cos }^2}(\theta ) + \frac{{\cos }^2}(\theta )} \right]^{\frac{1}{2}}}d\theta\\ &= a{\left[ {1 - \underbrace {\left( {1 - \frac} \right)}_{{\cos }^2}(\theta )} \right]^{\frac{1}{2}}}d\theta\\ &= a{\left[ {1 - {e^2}{{\cos }^2}(\theta )} \right]^{\frac{1}{2}}}d\theta \end{align} $$     (6.7.11)
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 * }

Determination of $$ n $$
To determine $$\displaystyle n $$ let us examine the error for the composite trapezoid rule.
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$$\displaystyle $$
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 * E_{n}| \leq \frac{(b-a)h^2}{12}M_{2}
 *

 * }

The above error for can be rewritten using
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$$\displaystyle \begin{align} h &= \frac{b-a}{n} \\ M_{2} &= \max_{\zeta \in [a,b]}|f^{(2)}(\zeta)| \\ \end{align} $$     (6.7.13)
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 * }

And after substitution we get,
 * {| style="width:100%" border="0"

$$\displaystyle n^2 \leq \frac{(b-a)^3}{12}\frac{1}{|E_{n}|} \max_{\zeta \in [a,b]}|f^{(2)}(\zeta)| $$     (6.7.14)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

A) For an Ellipse
Where $$ \displaystyle a=10 $$ and $$ \displaystyle b=3 $$ using,
 * Composite Trapezoid
 * Romberg Quadrature Rule
 * Clenshaw-Curtis Rule
 * chebfun/sum command

Elliptical Integral of the 1st Kind
The elliptical integral of the 1st kind is described by the following equations,
 * {| style="width:100%" border="0"

$$\displaystyle \begin{align} I &= \int_{0}^{2\pi }{\sqrt{{{r}^{2}}+{{\left( \frac{dr}{d\theta } \right)}^{2}}}}d\theta \\ r(\theta) &= \frac{a(1-e^2)}{1-e \, cos(\theta)} \\ e &= \sqrt{1-{{\left( \frac{b}{a} \right)}^{2}}} \end{align} $$     (6.7.15)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Elliptical Integral of the 2nd Kind
The elliptical integral of the 2nd kind is described by the following equation,
 * {| style="width:100%" border="0"

$$\displaystyle I=4a\int_{0}^{\pi /2}{\sqrt{1-{{e}^{2}}{{\sin }^{2}}\theta }}\ d\theta =4aE(e) $$     (6.7.16)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

B) For the Bifolium

 * Determine the folium length using the following numerical integration methods:
 * Composite Trapezoid
 * Romberg Quadrature Rule
 * Clenshaw-Curtis Rule
 * chebfun/sum command

Numerical Results
The table below summarizes the numerical results for four methods of numerical integration discussed above:
 * The elliptical integral of the 1st kind
 * We obtained the fastest convergence using the Romberg Tables with an elapsed time of 0.701108 seconds.
 * The composite trapezoid method performed the worst taking 2.886114 seconds with the number of iterations/nodes $$ n=385 $$.
 * Both the fast and standard Clenshaw Curtis methods took just over 1 second, however the fast Clenshaw Curtis method required 22 more iterations until convergence was achieved.
 * The elliptical integral of the 2nd kind
 * All four methods performed well for the elliptical integral of the 2nd kind.
 * Convergence was best for the Romberg and Composite Trapezoid methods requiring an elapsed time of about 0.60 seconds.
 * Not far behind were both the Clenshaw Curtis methods which took about 0.65-0.66 seconds to achieve convergence.
 * The Bifolium
 * For the bifolium numerical integration the results among the four methods varied the most.
 * Romberg significantly out performed all other methods only taking 0.73 seconds for convergence.
 * The second fastest convergence was for the trapezoid method and took 7.01 seconds and n=1097 to converge.
 * Unlike the trend observed for the elliptical integrations, the bifolium had faster convergence using the fast Clenshaw Curtis method than the standard, taking 13.53 and 40.79 seconds respectively.

=Problem 6.8: Derive Arc Length Equation=

''' This problem was solved without referring to S10. '''

From the lecture slide Mtg 33-3

Given: The Definition of an Arc Length
The formula for arc length from lecture 32-4:

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

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

\widehat{PQ} = \int_{\theta_{P}}^{\theta_{Q}}d\theta\left[r^2+\left(\frac{dr}{d\theta}\right)^{2}\right]^{\frac{1}{2}}. $$     (6.8.1)
 *

 * }

The law of cosines:

Triangle OAB:

 * HWK6 P8.jpg

where

Find: The Formula for the Arc Length of an Ellipse
Derive the formula for arc length of ellipse using triangle OAB and the law of cosines.

Solution
Applying the law of cosines formula to the OAB triangle case gives:

Substituting in Eqs 8.3-5 into Eq 8.6 gives

Taylor series expansion of $$\displaystyle \cos (d\theta) $$ is

Because theta is assumed to be small, we only need to take the first two terms of the Taylor series expansion of $$\displaystyle \cos (d\theta) $$ and sub into Eq 8.8.

The last term in Eq 8.11 $$\displaystyle rdrd\theta^2$$ is zero because the three derivatives shrink faster than the two derivative elements $$\displaystyle dr^2$$ and $$\displaystyle d\theta^2$$.

The remainder is

Taking the square root of both sides of Eq 8.12 then integrating from $$\displaystyle\theta_P$$ to $$\displaystyle\theta_Q$$ gives,

= Problem 6.9: Identification of Basis Functions $$\displaystyle \overline{N_i}$$ =

''' This problem was solved without referring to S10. '''

Given: $$ Z(s) $$ Expressed as a Linear Combination with its Basis
The relationship between $$\displaystyle Z(s)$$ and $$\displaystyle d_i$$(degree of freedom):

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

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

\begin{align} Z(s)&=\sum_{i=0}^{3}c_i s^i \\ &= \sum_{i=1}^{4} \overline{N_i}(s) \, d_i\\ \end{align}

$$     (6.9.1-2)
 *

 * }

where,

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

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

\begin{align} \overline{N_i}(s) &= basis\,\, function\\ d_i(s) &= degrees\,\, of\,\, freedom\\ d_1&=Z_i \\ d_2&=\dot{Z_i} \\ d_3&=Z_{i+1} \\ d_4&=\dot{Z}_{i+1}\\ \end{align}

$$
 *

 * }

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

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

\dot{Z}=\frac{dZ}{dt} = \frac{dZ}{ds} \frac{ds}{dt}

$$    (6.9.3)
 *

 * }

Find: The Basis Function for $$ i=1,2,3,4$$
Identify and plot the the basis function $$\displaystyle \left\{\overline{N_i}(s) \right\}\,\, ,i=1,2,3,4 $$.

Solution
From the Eq 1 of 35-4(lecture slide)

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

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

\begin{align} \begin{Bmatrix} c_0 \\ c _1 \\ c _2 \\ c_3 \end{Bmatrix} &= \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ -3 & -2 & 3 & -1 \\ 2 & 1 & -2 & 1 \end{bmatrix} \begin{Bmatrix} Z_i \\ Z_i^' \\ Z_{i+1} \\ Z_{i+1}^' \end{Bmatrix} \\ &=\begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ -3 & -2 & 3 & -1 \\ 2 & 1 & -2 & 1 \end{bmatrix} \begin{Bmatrix} d_1 \\ hd_2 \\ d_3 \\ hd_4 \\ \end{Bmatrix}. \\ &=\begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ -3 & -2 & 3 & -1 \\ 2 & 1 & -2 & 1 \end{bmatrix} \begin{Bmatrix} \overline{d}_1 \\ \overline{d}_2 \\ \overline{d}_3 \\ \overline{d}_4 \\ \end{Bmatrix}. \\

\end{align}

$$     (6.9.3-5)
 *

 * }

where,

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

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

\begin{align} \overline{d}_1&=d_1 \\ \overline{d}_2&= h \, d_2 \\ \overline{d}_3&=d_3 \\ \overline{d}_4&= h \, d_4 \\ \end{align}

$$
 *

 * }

Expand the matrix equation.

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

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

\begin{align} &c_0= \overline{d}_1 \\ &c_1= \overline{d}_2 \\ &c_2= -3\overline{d}_1 -2\overline{d}_2+3\overline{d}_3-\overline{d}_4 \\ &c_3= 2\overline{d}_1 +1\overline{d}_2-2\overline{d}_3+\overline{d}_4 \\

\end{align}

$$     (6.9.6-9)
 *

 * }

Substitute $$\displaystyle c_i$$ to Eq 9.1

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

$$\displaystyle \begin{align} Z(s)&=\sum_{i=0}^{3}c_i s^i \\ &= c_0 + c_1 s^1 + c_2 s^2 + c_3 s^3 \\ &= \overline{d}_1 + \overline{d}_2 s^1 + (-3\overline{d}_1 -2\overline{d}_2+3\overline{d}_3-\overline{d}_4) s^2 + (2\overline{d}_1 +1\overline{d}_2-2\overline{d}_3+\overline{d}_4) s^3 \end{align}
 * style="width:95%" |
 * style="width:95%" |

$$     (6.9.10-12)
 *

 * }

Rearrange in terms of $$\displaystyle \overline{d}_i $$

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

$$\displaystyle \begin{align} Z(s)& = (2s^3-3s^2+1)\overline{d}_1+(s^3-2s^2+s)\overline{d}_2+(-2s^3+3s^2)\overline{d}_3+(s^3-s^2)\overline{d}_4 \\ &=\overline{N}_1(s) \, \overline{d}_1+\overline{N}_2(s) \, \overline{d}_2+\overline{N}_3(s) \, \overline{d}_3+\overline{N}_4(s) \, \overline{d}_4 \end{align} $$     (6.9.13-14)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Comparing the LHS and RHS of the equation gives
 * {| style="width:100%" border="0" cellpadding="5" cellspacing="0"

(6.9.15-18)
 * style="width:10%" |
 * style="width:80%" |
 * style="width:80%" |

 * style="width:80%" |

 * }

Plot of $$\displaystyle \overline{N_i} $$.

=References=

=Contributing Members & Referenced Lecture=

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


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$$\left( \begin{matrix}  6 & y  \\   {{a}_{21}} & {{a}_{22}}  \\ \end{matrix} \right)\left( \begin{matrix}   {{a}_{11}} & {{a}_{12}}  \\   {{a}_{21}} & {{a}_{22}}  \\ \end{matrix} \right) $$

title: User:Eml5526.s11.team6.tupsakhare/hwk7


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User:Eml5526.s11.team7.jin/HW6

=Problem 6.2 Solving the General 1D problem with Quadratic Lagrangian element basis function (QLEBF)=

Problem Description
Similar to HW5.{2,4,8}, solve G1DM1.0/D1b using Quadratic Lagrangian Element Basis Function (QLEBF) with uniform discretization (equidistant element nodes) until convergence of $${{u}^{h}}(0.5)$$ to $$O({{10}^{-6}})$$. ( $$nel=2,4,6,8,...$$).

 * 1. For $$nel=2$$, compute $$\mathbf{\tilde{K}}=\sum\limits_{e=1}^{2}$$, with $${{\mathbf{\tilde{K}}}^{e}}$$ by (1)p.30-5, display $${{\mathbf{\tilde{K}}}^{e}},e=1,2$$.

 * 2. Compute $${{\mathbf{k}}^{e}}$$, $${{\mathbf{L}}^{e}}$$, for $$e=1,2$$.

 * 3. Compute $${{\tilde{\Kappa }}^{e}}={{\mathbf{L}}^{e}}^{T}{{\mathbf{k}}^{e}}{{\mathbf{L}}^{e}}$$, for $$e=1,2$$, compare to the result by (1).

 * 4. Plot all QLEBF for $$nel=3$$.

 * 5. Plot $$u_^{h}$$ vs. $$u$$, $$\left[ u_^{h}(0.5)-u(0.5) \right]$$ vs. $$\tilde{n}$$

Part 1
For element 1, the 3 basis functions are:
 * {| style="width:100%" border="0"

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

b_1^1 = \frac \cdot \frac = 8{x^2} - 6x + 1

$$     (6.2.1)
 *
 * }
 * {| style="width:100%" border="0"

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

b_2^1 = \frac \cdot \frac = - 16{x^2} + 8x

$$     (6.2.2)
 *
 * }
 * {| style="width:100%" border="0"

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

b_3^1 = \frac \cdot \frac = 8{x^2} - 2x

$$     (6.2.3) By (1)p.30-5,
 *
 * }
 * {| style="width:100%" border="0"

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

K_{ij}^e = \int\limits_ {b{{_i^e}^\prime }(x){a_2}(x)b{{_j^e}^\prime }(x)dx}

$$     (6.2.4) Thus, we have
 *
 * }
 * {| style="width:100%" border="0"

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

K_{11}^1 = \int\limits_0^{0.5} {\left( {16x - 6} \right)\left( {2 + 3x} \right)\left( {16x - 6} \right)dx} = \frac{6}

$$     (6.2.5)
 *
 * }
 * {| style="width:100%" border="0"

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

K_{12}^1 = K_{21}^1 = \int\limits_0^{0.5} {\left( {16x - 6} \right)\left( {2 + 3x} \right)\left( { - 32x + 8} \right)dx} =  - \frac{3}

$$     (6.2.6)
 *
 * }
 * {| style="width:100%" border="0"

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

K_{13}^1 = K_{31}^1 = \int\limits_0^{0.5} {\left( {16x - 6} \right)\left( {2 + 3x} \right)\left( {16x - 2} \right)dx} = \frac{6}

$$     (6.2.7)
 *
 * }
 * {| style="width:100%" border="0"

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

K_{22}^1 = \int\limits_0^{0.5} {\left( { - 32x + 8} \right)\left( {2 + 3x} \right)\left( { - 32x + 8} \right)dx} = \frac{3}

$$     (6.2.8)
 *
 * }
 * {| style="width:100%" border="0"

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

K_{23}^1 = K_{32}^1 = \int\limits_0^{0.5} {\left( { - 32x + 8} \right)\left( {2 + 3x} \right)\left( {16x - 2} \right)dx} =  - \frac{3}

$$     (6.2.9)
 *
 * }
 * {| style="width:100%" border="0"

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

K_{33}^1 = \int\limits_0^{0.5} {\left( {16x - 2} \right)\left( {2 + 3x} \right)\left( {16x - 2} \right)dx} = \frac{6}

$$     (6.2.10) Then
 *
 * }
 * {| style="width:100%" border="0"

$$ \displaystyle
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |

{{\mathbf{\tilde K}}^1} = \left[ {\begin{array}{ccccccccccccccc} {65/6}&{ - 38/3}&{11/6}&0&0 \\  { - 38/3}&{88/3}&{ - 50/3}&0&0 \\   {11/6}&{ - 50/3}&{89/6}&0&0 \\   0&0&0&0&0 \\   0&0&0&0&0 \end{array}} \right]

$$     (6.2.11) Similar to element 1, when  e=2
 *
 * }
 * {| style="width:100%" border="0"

$$ \displaystyle
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |

{{\mathbf{\tilde K}}^2} = \left[ {\begin{array}{ccccccccccccccc} 0&0&0&0&0 \\  0&0&0&0&0 \\   0&0&{107/6}&{ - 62/3}&{17/6} \\   0&0&{ - 62/3}&{136/3}&{ - 74/3} \\   0&0&{17/6}&{ - 74/3}&{131/6} \end{array}} \right]

$$     (6.2.12) Thus,
 *
 * }
 * {| style="width:100%" border="0"

$$ \displaystyle
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |

{\mathbf{\tilde K}} = \sum\limits_{e = 1}^2  = \left[ {\begin{array}{ccccccccccccccc} {65/6}&{ - 38/3}&{11/6}&0&0 \\  { - 38/3}&{88/3}&{ - 50/3}&0&0 \\   {11/6}&{ - 50/3}&{98/3}&{ - 62/3}&{17/6} \\   0&0&{ - 62/3}&{136/3}&{ - 74/3} \\   0&0&{17/6}&{ - 74/3}&{131/6} \end{array}} \right]

$$     (6.2.13)
 *
 * }

Part 2
By (1)p.31-3,
 * {| style="width:100%" border="0"

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

k_{ij}^e = \int\limits_ {b{{_i^e}^\prime }(x){a_2}(x)b{{_j^e}^\prime }(x)dx}

$$     (6.2.14) For element 1
 *
 * }
 * {| style="width:100%" border="0"

$$ \displaystyle
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |

{{\mathbf{k}}^1} = \left[ {\begin{array}{ccccccccccccccc} {65/6}&{ - 38/3}&{11/6} \\  { - 38/3}&{88/3}&{ - 50/3} \\   {11/6}&{ - 50/3}&{89/6} \end{array}} \right]

$$     (6.2.15)
 *
 * }
 * {| style="width:100%" border="0"

$$ \displaystyle
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |

{{\mathbf{L}}^1} = \left[ {\begin{array}{ccccccccccccccc} 1&0&0&0&0 \\  0&1&0&0&0 \\   0&0&1&0&0 \end{array}} \right]

$$     (6.2.16) For element 2
 *
 * }
 * {| style="width:100%" border="0"

$$ \displaystyle
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |

{{\mathbf{k}}^2} = \left[ {\begin{array}{ccccccccccccccc} {107/6}&{ - 62/3}&{17/6} \\  { - 62/3}&{136/3}&{ - 74/3} \\   {17/6}&{ - 74/3}&{131/6} \end{array}} \right]

$$     (6.2.17)
 *

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

$$ \displaystyle
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |

{{\mathbf{L}}^2} = \left[ {\begin{array}{ccccccccccccccc} 0&0&1&0&0 \\  0&0&0&1&0 \\   0&0&0&0&1 \end{array}} \right]

$$     (6.2.18)
 *

 * }

Part 3
For element 1
 * {| style="width:100%" border="0"

$$ \displaystyle
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |

{{\mathbf{\tilde {\rm K}}}^1} = {{\mathbf{L}}^1}^T{{\mathbf{k}}^1}{{\mathbf{L}}^1} = \left[ {\begin{array}{ccccccccccccccc} 1&0&0 \\  0&1&0 \\   0&0&1 \\   0&0&0 \\   0&0&0 \end{array}} \right]\left[ {\begin{array}{ccccccccccccccc} {65/6}&{ - 38/3}&{11/6} \\  { - 38/3}&{88/3}&{ - 50/3} \\   {11/6}&{ - 50/3}&{89/6} \end{array}} \right]\left[ {\begin{array}{ccccccccccccccc} 1&0&0&0&0 \\  0&1&0&0&0 \\   0&0&1&0&0 \end{array}} \right] = \left[ {\begin{array}{ccccccccccccccc} {65/6}&{ - 38/3}&{11/6}&0&0 \\  { - 38/3}&{88/3}&{ - 50/3}&0&0 \\   {11/6}&{ - 50/3}&{89/6}&0&0 \\   0&0&0&0&0 \\   0&0&0&0&0 \end{array}} \right]

$$     (6.2.19) For element 2
 *

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

$$ \displaystyle
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |

{{\mathbf{\tilde {\rm K}}}^2} = {{\mathbf{L}}^1}^T{{\mathbf{k}}^1}{{\mathbf{L}}^1} = \left[ {\begin{array}{ccccccccccccccc} 0&0&0 \\  0&0&0 \\   1&0&0 \\   0&1&0 \\   0&0&1 \end{array}} \right]\left[ {\begin{array}{ccccccccccccccc} {107/6}&{ - 62/3}&{17/6} \\  { - 62/3}&{136/3}&{ - 74/3} \\   {17/6}&{ - 74/3}&{131/6} \end{array}} \right]\left[ {\begin{array}{ccccccccccccccc} 0&0&1&0&0 \\  0&0&0&1&0 \\   0&0&0&0&1 \end{array}} \right] = \left[ {\begin{array}{ccccccccccccccc} 0&0&0&0&0 \\  0&0&0&0&0 \\   0&0&{107/6}&{ - 62/3}&{17/6} \\   0&0&{ - 62/3}&{136/3}&{ - 74/3} \\   0&0&{17/6}&{ - 74/3}&{131/6} \end{array}} \right]

$$     (6.2.20) The result is exactly same as achieved in part 1.
 *

 * }

Part 4
The plot of QLEBF for nel = 3:

Part 5
The plots of $$u_^{h}$$ vs. $$u$$,

The plots of $$\left[ u_^{h}(0.5)-u(0.5) \right]$$ vs. $$\tilde{n}$$

=Problem 6.9 Finite Element Formulation for Multidimensional Scalar Field Problems=

Problem Description
Consider the heat conduction problem depicted in Figure 6.9.1. The coordinates are given in meters. The conductivity is isotropic, with $$\mathbf{D}=k\left[ \begin{matrix} 1 & 0 \\   0 & 1  \\ \end{matrix} \right]$$, and $$k=5\text{W}{}^{\text{o}}{{\text{C}}^{\text{-1}}}$$. The temperature $$T=0$$ is prescribed along edges AB and AD. The heat fluxes $$\bar{q}=0$$ and $$\bar{q}=20\text{W}{{\text{m}}^{\text{-1}}}$$ are prescribed on edges BC and CD, respectively. A constant heat source $$s=6\text{W}{{\text{m}}^{\text{-2}}}$$ is applied over the plate. Consider this problem modeled with 16 quadrilateral finite elements as shown in Figure 6.9.2.

Solve this problem using the finite element code given in Section 12.5.

Solution
By using the matlab codes downloaded from download, The post-processing results for temperature and flux are shown in Figure 6.9.3 and Figure 6.9.4.

All the matlab codes for this problem can be found in the file folder 'ch08'.

The main program is given in heat2d.m file. It calls the following functions: include_flags, [K,f,d] = preprocessor, [ke, fe] = heat2Delem(e), [K,f] = assembly(K,f,e,ke,fe), f = src_and_flux(f), [d,f_E] = solvedr(K,f,d) and postprocessor(d).

In the function of [K,f,d] = preprocessor, input file, which defines material properties, mesh data, vector and matrices initializations, essential and natural conditions and mesh generation, is read.

In the function of [ke, fe] = heat2Delem(e), the element conductance matrix and nodal source vector are integrated using Gauss quadrature. And then assembled by the function of [K,f] = assembly(K,f,e,ke,fe).

The f = src_and_flux(f) function calculates the nodal boundary flux vector.

In the postprocessor(d) function, the results are plotted in Figure 6.9.3 and Figure 6.9.4.

Problem Description
Refer to p205-207 of F&B for detailed problem description.

Problem Description
Consider a chimney constructed of two isotropic materials: dense concrete ($$k=2.0W{}^{\circ }{{C}^{-1}}$$) and bricks ($$k=0.9W{}^{\circ }{{C}^{-1}}$$). The temperature of the hot gases on the inside surface of the chimney is $$140{}^{\circ }C$$, whereas the outside is exposed to the surrounding air, which is at $$T=10{}^{\circ }C$$. The dimensions of the chimney (in meters) are shown below. For the analysis, exploit the symmetry and consider 1/8 of the chimney cross sectional area. Consider a mesh of eight elements as shown below. Determine the temperature and flux in the two materials. Analyze the problem with 2×2, 4×4 and 8×8 quadrilateral elements for 1/8 of the problem domain. A2×2 finite element mesh is shown in Figure below. Symmetry implies insulated boundary conditions on edges AD and BC. Note that element boundaries have to coincide with the interface between the concrete and bricks.

title: User:Eml5526.s11.team7.jin/HW6


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User talk:Merlissimo

 Hello Merlissimo, 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 15:35, 30 March 2011 (UTC)

title: User talk:Merlissimo


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text: Category:Collaborative play writing

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Loyola University Chicago - Project Eden

Project Eden is an edible landscaping initiative at Loyola University Chicago, to increase the campuses current edibility

Initiative
As modern urbanites our knowledge of agriculture and food production may be limited to what we see in the grocery store. In spite of our higher educations and our lucrative careers, we often fail to realize how our fundamental survival is inherently dependent on the experience and quality of our agriculture. However food production has been removed from the majority of the population, leaving little opportunity to learn let alone develop any intimacy with our food. These problems of distance and ignorance are recurring themes in the faults of the modern food system. Food is grown in one location than shipped a great distance to the rest of society, locking in a spiritual, physical and mental distance from our food. This distance is inadvertently causing us to make blind choices about our food in daily life, resulting in negative downstream effects in fields such as nutrition, mental health, food security etc. In looking to provide opportunity and again immerse people to again interact with their food production, contemporary urban agriculture resources will have to be established.
 * Fresh Produce travels an average of 1,500 miles
 * Fewer than 2% of people in the US farm
 * Community gardeners and their children eat healthier, nutrient rich diets than non gardeners

Vision
Imagine a campus where fruit trees line the sidewalks, grape vines scale vertical walls, and annual vegetables fill landscaping garden beds. The Garden of Eden? No, it’s Loyola University’s Lake Shore Campus in downtown Chicago, where students, faculty ,and community members come together to assume responsibility for stewardship and maintenance of this edible landscape. The university encourages this contemporary urban agriculture through curricula and community-wide events. Individuals commune with each other and their environment through food and work.
 * Havanas able to produce over 50% of it’s produce needs
 * 40% of Toronto residents grow some of their own food
 * Victory gardens supplied up to 40% of the produce in America during WWII

Methods and Products
Map Link to map Soil
 * Current edible plant species on campus were surveyed
 * Recording longitudinal location with a Garmen GPSMAP® 60Cx device, then interfaced longitudinal coordinates with Google Maps
 * The edible map has been published online, where clickable points of interest for the plants have been placed forwarding the user to more information.

Soil were sent out for analysis from four locations across campus: Winthrop and Loyola ave, Halas Field, Cudahy Science, and Mundelein Center. No hazardous materials were found in soil test results, however soil conditions on the lakeshore campus were found to be strongly alkaline
 * Tests were performed for: Soil Density, pH, cation exchange capacity
 * Nutrient levels of: P,K, Ca, Mg, NO3
 * Micronutrient capacity of: B, Mn, Zn, Cu, Fe, S
 * Metals of concern: Pb, Cd, Ni, Cr

Responsible Stewardship Documents in the Cloud section, in effort to share and further urban agriculture from an individual level to an institutional level.
 * Plants were attempted to be maintained and cared for using organic and ecologically friendly methods first, Organic methods used may not have fallen into criteria for qualification of USDA Organic.
 * Data, documents and observation have been published on the free-source software of Wikipedia in the Wikiversity

Events and scholastic workdays were created to promote and engage the student body in activities of the project. Three main events were created:
 * 1) 1.Welcome Week Victory Garden workshop.
 * 2) 2.Fall planting
 * 3) 3.Spring Planting

New Plant Installations

 * Based on campus survey of map and soil plants which were native, and suitable to soil and local lakeside climate conditions were selected.
 * Over 200 new native perennial edible plants were installed on campus including plants such as black cherry, concord grape and beach plum.

Featured areas of interest

 * Winthrop farms ( Loyola and Winthrop Ave )

 * LUREC

Unlike the other edible landscaping initiatives on occasion the planting of edible plants has occurred around campus. Some locations with edible plants planted by these dareing students include:
 * Guerrilla spaces
 * Blueberry bushes left of the moon sculpture next the the IC
 * Raspberry Bush next at Flanner Hall entrance.

Thoughts and Future
The goal of this project has been working to create the essential resources to enable urban agriculture to take place at a university level. Urban agriculture is a commitment to education; an education that is progressive, holistic, interdisciplinary and encourages participation at all levels. In doing so people can then organically experience the intimacy of food production as a component of their daily lives. However in a place like a university the time a student spends at the university for their career is around 4 years. Institutionalization of an urban agriculture program as component of the university experience will enable students (and community ) to reap the benefits of a true agri-culture.

Project History

 * Arbor Day
 * Winthrop

Involvement
Growers' Guild Loyola, the campus gardening club, have been much of the backbone in working to introduce edible plants to the Loyola University Chicago campus. The club has organized events, workshops and landscape tours to promote the campus edible landscape.

Center for Urban Environmental Research Policy (CUERP) has been the promotional administrative force encouraging the institutionalization of Project Eden. Instructor Adam Schubel of CUERP has built the project as a part of the course STEP:Food Systems, offered at Loyola University Chicago


text:
User:Eml5526.s11.team4/HW6

= Problem 6.1 - QLEBF with G1DM1.0/D1b =

Problem Statement
Solve data set G1DM1.0/D1b using Quadrilateral Lagrangian Element Basis Functions(QLEBF)with uniform discretization (equidistant element nodes) nel (number of elements) = 2,4,6,8,... This problem is similar to Homework problems 5.1, 5.3, and 5.7. Recall: The data set G1DM1.0/D1b is defined by an elastic bar with variable cross-section: (Image from Charles Cook, HW1.1) $$\Omega=\left\{0,1\right\}$$ $$\Gamma_{g}=\left\{0\right\}, g = 4$$ $$\Gamma _{h}=\left\{1\right\}, h = 12$$ $$a_{2}\left\{x\right\}=2+3x$$ $$f\left\{x,t\right\}=5x$$ $$\frac{\partial^{2} u}{\partial t^{2}}=0$$ 1.) For nel = 2, compute $$\mathbf{\tilde{K}}=\sum_{e=1}^{2}\mathbf{\tilde{K}^{e}}$$ with $$\mathbf{\tilde{K}^{e}}$$ from Equation (1) from lecture notes page 30-5. Display $$\mathbf{\tilde{K}^{e}}$$, e = 1,2. 2.) Compute $$\mathbf{k}^{e}, \mathbf{L}^{e}$$ for e = 1,2. 3.) Compute $$\mathbf{\tilde{K}^{e}}=(\mathbf{L}^{e})^{T}\mathbf{k}^{e}\mathbf{L}^{e}$$ for e = 1,2. Compare to the results from sections 1. 4.) Plot all QLEBF for nel = 3. 5.) Plot $$u_{\tilde{n}}^{h}$$ versus u,$$ \left [u_{\tilde{n}}^{h}(.5)-u(.5) \right ]$$ versus $$\tilde{n}$$.

solution
1.For nel=2, comp $$\tilde{K}=\sum\limits_{e=1}^{2}$$ ,with by (1)p30.5,display $${{\tilde{K}}^{e}}$$, e=1,2 e=1: $${{{\tilde{K}}}^{1}}=\left[ \begin{matrix} {10.833333333333334} & {-12.666666666666666} & {1.833333333333333} & {0} & {0} \\  {-12.666666666666666} & {29.333333333333332} & {-16.666666666666668} & {0} & {0} \\   { 1.833333333333333} & {-16.666666666666668} & {14.833333333333334}  & {0} & {0}\\   { 0} & {0} & {0}  & {0} & {0}\\   { 0} & {0} & {0}  & {0} & {0}\\ \end{matrix} \right]$$ e=2: $${{{\tilde{K}}}^{2}}=\left[ \begin{matrix} { 0} & {0} & {0} & {0} & {0}\\   { 0} & {0} & {0}  & {0} & {0}\\   { 0} & {0} &   {17.833333333333332} & {-20.666666666666668} & {2.833333333333334}  \\   { 0} & {0} &   {-20.666666666666668} & { 45.333333333333336} & {-24.666666666666668}  \\   { 0} & {0} &   { 2.833333333333334} & {-24.666666666666668} & {21.833333333333332}  \\ \end{matrix} \right]$$ Then,sum on e=1,2:

$$\tilde{K}=\left[ \begin{matrix} {10.833333333333334} & {-12.666666666666666} & {1.833333333333333} & {0} & {0} \\  {-12.666666666666666} & {29.333333333333332} & {-16.666666666666668} & {0} & {0} \\   { 1.833333333333333} & {-16.666666666666668} & {32.666666666666664}  &{-20.666666666666668} & {2.833333333333334}\\   { 0} & {0} &   {-20.666666666666668} & { 45.333333333333336} & {-24.666666666666668}  \\   { 0} & {0} &   { 2.833333333333334} & {-24.666666666666668} & {21.833333333333332}  \\ \end{matrix} \right]$$

2.comp $${{k}^{e}}$$ , $${{L}^{e}}$$, for e=1,2 e=1:

$${{k}^{1}}=\left[ \begin{matrix} {10.833333333333334} & {-12.666666666666666} & {1.833333333333333} \\   {-12.666666666666666} & {29.333333333333332} & {-16.666666666666668}  \\   { 1.833333333333333} & {-16.666666666666668} & {14.833333333333334}  \\ \end{matrix} \right]$$

$${{L}^{1}}=\left[ \begin{matrix} 1 & 0 & 0 & 0 & 0 \\   0 & 1 & 0 & 0 & 0  \\   0 & 0 & 1 & 0 & 0  \\ \end{matrix} \right]$$

e=2:

$${{k}^{2}}=\left[ \begin{matrix} {17.833333333333332} & {-20.666666666666668} & {2.833333333333334} \\   {-20.666666666666668} & { 45.333333333333336} & {-24.666666666666668}  \\   { 2.833333333333334} & {-24.666666666666668} & {21.833333333333332}  \\ \end{matrix} \right]$$

$${{L}^{2}}=\left[ \begin{matrix} 0 & 0 & 1 & 0 & 0 \\   0 & 0 & 0 & 1 & 0  \\   0 & 0 & 0 & 0 & 1  \\ \end{matrix} \right]$$

3.comp $${{{\tilde{K}}}^{e}}={{L}^{e}}^{T}{{k}^{e}}{{L}^{e}}$$, for e=1,2. compare to 1). e=1: $${{{\tilde{K}}}^{1}}=\left[ \begin{matrix}  {10.833333333333334} & {-12.666666666666666} & {1.833333333333333} & {0} & {0} \\   {-12.666666666666666} & {29.333333333333332} & {-16.666666666666668} & {0} & {0} \\   { 1.833333333333333} & {-16.666666666666668} & {14.833333333333334}  & {0} & {0}\\   { 0} & {0} & {0}  & {0} & {0}\\   { 0} & {0} & {0}  & {0} & {0}\\ \end{matrix} \right]$$ e=2: $${{{\tilde{K}}}^{2}}=\left[ \begin{matrix}   { 0} & {0} & {0}  & {0} & {0}\\   { 0} & {0} & {0}  & {0} & {0}\\   { 0} & {0} &   {17.833333333333332} & {-20.666666666666668} & {2.833333333333334}  \\   { 0} & {0} &   {-20.666666666666668} & { 45.333333333333336} & {-24.666666666666668}  \\   { 0} & {0} &   { 2.833333333333334} & {-24.666666666666668} & {21.833333333333332}  \\ \end{matrix} \right]$$ Compare with result on 1), we can see that the result is the same by using these two methods. 4.plot all QLEBF for nel=3 5.plot $$\displaystyle {{u}^{h}}$$ vs $$\displaystyle u$$,when number of element is 2,3,4,6,respectively. plot $$\displaystyle \left[ u_{nel}^{h}(0.5)-u(0.5) \right]$$ vs number of element

The error at x=0.5 can be listed in the following table: From the table, it can be found that the error will be lower than 10^(-6) when the number of elements is greater than n=14.

Comments:
It is observed that the QLEBF method converges to the accuracy of 10^(-6) with much less element number than LLEBF method in homework 5.7. This problem was solved by shengfeng yang

= Problem 6.2 - QLEBF with G1DM1.0/D1 =

Problem Statement
Solve data set G1DM1.0/D1 using Quadrilateral Lagrangian Element Basis Functions(QLEBF)with uniform discretization (equidistant element nodes) nel (number of elements) = 2,4,6,8,... This problem is similar to Homework problems 5.2, 5.4, and 5.6. Recall: The data set G1DM1.0/D1 is defined by an elastic bar with variable cross-section: (Image from Charles Cook, HW1.1) $$\Omega=\left\{0,1\right\}$$ $$\Gamma_{g}=\left\{1\right\}, g = 4$$ $$\Gamma _{h}=\left\{0\right\}, h = 12$$ $$a_{2}\left\{x\right\}=2+3x$$ $$f\left\{x,t\right\}=5x$$ $$\frac{\partial^{2} u}{\partial t^{2}}=0$$ 1.) For nel = 2, compute $$\mathbf{\tilde{K}}=\sum_{e=1}^{2}\mathbf{\tilde{K}^{e}}$$ with $$\mathbf{\tilde{K}^{e}}$$ from Equation (1) from lecture notes page 30-5. Display $$\mathbf{\tilde{K}^{e}}$$, e = 1,2. 2.) Compute $$\mathbf{k}^{e}, \mathbf{L}^{e}$$ for e = 1,2. 3.) Compute $$\mathbf{\tilde{K}^{e}}=(\mathbf{L}^{e})^{T}\mathbf{k}^{e}\mathbf{L}^{e}$$ for e = 1,2. Compare to the results from sections 1. 4.) Plot all QLEBF for nel = 3. 5.) Plot $$u_{\tilde{n}}^{h}$$ versus u,$$ \left [u_{\tilde{n}}^{h}(.5)-u(.5) \right ]$$ versus $$\tilde{n}$$.

solution
1.For nel=2, comp $$\tilde{K}=\sum\limits_{e=1}^{2}$$ ,with by (1)p30.5,display $${{\tilde{K}}^{e}}$$, e=1,2 e=1: $${{{\tilde{K}}}^{1}}=\left[ \begin{matrix} {10.833333333333334} & {-12.666666666666666} & {1.833333333333333} & {0} & {0} \\  {-12.666666666666666} & {29.333333333333332} & {-16.666666666666668} & {0} & {0} \\   { 1.833333333333333} & {-16.666666666666668} & {14.833333333333334}  & {0} & {0}\\   { 0} & {0} & {0}  & {0} & {0}\\   { 0} & {0} & {0}  & {0} & {0}\\ \end{matrix} \right]$$ e=2: $${{{\tilde{K}}}^{2}}=\left[ \begin{matrix} { 0} & {0} & {0} & {0} & {0}\\   { 0} & {0} & {0}  & {0} & {0}\\   { 0} & {0} &   {17.833333333333332} & {-20.666666666666668} & {2.833333333333334}  \\   { 0} & {0} &   {-20.666666666666668} & { 45.333333333333336} & {-24.666666666666668}  \\   { 0} & {0} &   { 2.833333333333334} & {-24.666666666666668} & {21.833333333333332}  \\ \end{matrix} \right]$$ Then,sum on e=1,2:

$$\tilde{K}=\left[ \begin{matrix} {10.833333333333334} & {-12.666666666666666} & {1.833333333333333} & {0} & {0} \\  {-12.666666666666666} & {29.333333333333332} & {-16.666666666666668} & {0} & {0} \\   { 1.833333333333333} & {-16.666666666666668} & {32.666666666666664}  &{-20.666666666666668} & {2.833333333333334}\\   { 0} & {0} &   {-20.666666666666668} & { 45.333333333333336} & {-24.666666666666668}  \\   { 0} & {0} &   { 2.833333333333334} & {-24.666666666666668} & {21.833333333333332}  \\ \end{matrix} \right]$$

2.comp $${{k}^{e}}$$ , $${{L}^{e}}$$, for e=1,2 e=1:

$${{k}^{1}}=\left[ \begin{matrix} {10.833333333333334} & {-12.666666666666666} & {1.833333333333333} \\   {-12.666666666666666} & {29.333333333333332} & {-16.666666666666668}  \\   { 1.833333333333333} & {-16.666666666666668} & {14.833333333333334}  \\ \end{matrix} \right]$$

$${{L}^{1}}=\left[ \begin{matrix} 1 & 0 & 0 & 0 & 0 \\   0 & 1 & 0 & 0 & 0  \\   0 & 0 & 1 & 0 & 0  \\ \end{matrix} \right]$$

e=2:

$${{k}^{2}}=\left[ \begin{matrix} {17.833333333333332} & {-20.666666666666668} & {2.833333333333334} \\   {-20.666666666666668} & { 45.333333333333336} & {-24.666666666666668}  \\   { 2.833333333333334} & {-24.666666666666668} & {21.833333333333332}  \\ \end{matrix} \right]$$

$${{L}^{2}}=\left[ \begin{matrix} 0 & 0 & 1 & 0 & 0 \\   0 & 0 & 0 & 1 & 0  \\   0 & 0 & 0 & 0 & 1  \\ \end{matrix} \right]$$

3.comp $${{{\tilde{K}}}^{e}}={{L}^{e}}^{T}{{k}^{e}}{{L}^{e}}$$, for e=1,2. compare to 1). e=1: $${{{\tilde{K}}}^{1}}=\left[ \begin{matrix}  {10.833333333333334} & {-12.666666666666666} & {1.833333333333333} & {0} & {0} \\   {-12.666666666666666} & {29.333333333333332} & {-16.666666666666668} & {0} & {0} \\   { 1.833333333333333} & {-16.666666666666668} & {14.833333333333334}  & {0} & {0}\\   { 0} & {0} & {0}  & {0} & {0}\\   { 0} & {0} & {0}  & {0} & {0}\\ \end{matrix} \right]$$ e=2: $${{{\tilde{K}}}^{2}}=\left[ \begin{matrix}   { 0} & {0} & {0}  & {0} & {0}\\   { 0} & {0} & {0}  & {0} & {0}\\   { 0} & {0} &   {17.833333333333332} & {-20.666666666666668} & {2.833333333333334}  \\   { 0} & {0} &   {-20.666666666666668} & { 45.333333333333336} & {-24.666666666666668}  \\   { 0} & {0} &   { 2.833333333333334} & {-24.666666666666668} & {21.833333333333332}  \\ \end{matrix} \right]$$ Compare with result on 1), we can see that the result is the same by using these two methods. 4.plot all QLEBF for nel=3 5.plot $$\displaystyle {{u}^{h}}$$ vs $$\displaystyle u$$,when number of element is 2,3,4,6,10,respectively.

plot $$\displaystyle \left[ u_{nel}^{h}(0.5)-u(0.5) \right]$$ vs number of element

The error at x=0.5 can be listed in the following table: From the table, it can be found that the error will be lower than 10^(-6) when the number of elements is greater than n=8.

Comments:
It is observed that the QLEBF method converges to the accuracy of 10^(-6) with much less element number than LLEBF method in homework 5.8. This problem was solved by shengfeng yang

=Problem 6.3 Reproduce a tutorial=

Assigned in the lecture slide Mtg33

Problem statement
Reproduce all steps in the Beede’s tutorial

Part1: Buliding Geometry and Meshing
Opening the SciTE text editor inplanted in the bConverged Calculix pacage, and paste the following content into it then save as an .fbd file, like beam.fbd

pnt p1 0 0 0 pnt p2 25 0 0 pnt p3 50 0 0 pnt p4 75 0 0 pnt p5 100 0 0 line l1 p1 p2 25 line l2 p2 p3 25 line l3 p3 p4 25 line l4 p4 p5 25 seta lines l l1 l2 l3 l4 swep lines sweplines tra 0 10 0 10 seta surfaces s A001 A002 A003 A004 swep surfaces swepsurface tra 0 0 1 1 elty all he8 mesh all plot m all

The beam created:

This beam.fbd will reproduce the same geometry property as stated in the tutorial. For detailed explanation of the lines, please refer to the the Calculix cgx online documentary or previous wikiversity tutorials by our team.

Part2: Exporting Mesh, Loads, and Boundary Conditions
Open the SciTE editor again and paste such content, then overwrite the previous beam.fbd file. The new file pretty much the same geometry and mesh as the previous but use different name for the mesh and exports the nodal information to a .msh file.

seto beam pnt p1 0 0 0 pnt p2 25 0 0 pnt p3 50 0 0 pnt p4 75 0 0 pnt p5 100 0 0 line l1 p1 p2 25 line l2 p2 p3 25 line l3 p3 p4 25 line l4 p4 p5 25 seta lines l l1 l2 l3 l4 swep lines sweplines tra 0 10 0 10 seta surfaces s A001 A002 A003 A004 swep surfaces swepsurface tra 0 0 1 1 setc beam elty beam he8 mesh beam send beam abq rot -z frame

Then followed the steps of the tutorial as it use the “interactive” style of operating Calculix.

The nodal information in beam.msh file:

Setting nodes for the fixed end:

Boundary information:

The nodal information for the load:

Part3: Writing an Input File for the CCX Solver
Use SciTE to create a new file named beam.inp and write down the following content,

*HEADING Model: Calculix Beam Input File *INCLUDE,INPUT=beam.msh *BOUNDARY *INCLUDE,INPUT=fixed_123.bou *MATERIAL,NAME=EL *ELASTIC 30000000, 0.3 *SOLID SECTION,ELSET=Ebeam,MATERIAL=EL *STEP *STATIC *DLOAD *INCLUDE,INPUT=load.dlo *NODE FILE U *EL FILE S *END STEP

Then click on the SciTE menu “Tools->Solve”, when the “Job done!” is prompt we can click on the menu Tools->Post Process thus we open the result file. Choose “Datasets->STRESS” and then “Entities->Mises” than we can see the result just same as the tutorial.

For detailed explanation on the .inp file, please refer to the Calculix ccx online documentary or our team’s previous homework. .

Part4: Post Processing Results in CGX
By manipulating the cgx module for post processing we have the following results:

The displacement distribution when the bending is seen:

The displacement distribution at a certain plain:

The Graph function of post processing:

Problem Statement
Recall matrix algebra: (1) $$(\mathbf{A}\mathbf{B})^{T}=\boldsymbol{B}^{T}\boldsymbol{A}^{T}$$ (2) $$(\boldsymbol{A}^{-1})^{T}=(\boldsymbol{A}^{T})^{-1}=\boldsymbol{A}^{-T}$$ Given the matrices: $$ \boldsymbol{A}=\begin{bmatrix} 1 & 1 & 1\\ 2 & -1 & 3\\ 3 & 2 & 6 \end{bmatrix} $$ $$ \boldsymbol{B}=\begin{bmatrix} 1 & 3 & 5\\ 1 & -4 & 1\\ 2 & 5 & 8 \end{bmatrix} $$ Verify the matrix algebra properties shown above (properties 1 and 2). Find and explain the syntax of Wolframalpha

Solution
Condition 1

(1a) is equal to (1b), this proves that $$(\mathbf{A}\mathbf{B})^{T}=\boldsymbol{B}^{T}\boldsymbol{A}^{T}$$

The wolfram alpha calculation for $$(\mathbf{A}\mathbf{B})^{T}$$ is given in the following pdf file

Click on the pdf file to see the second page.

The wolfram alpha calculation for $$\boldsymbol{B}^{T}\boldsymbol{A}^{T}$$ is given in the following pdf file

Click on the pdf file to see the second page.

The Wolfram alpha results match with the manually calculated results of condition 1 of problem statement.This validates the proof.

Condition 2

(2a) is equal to (2b).This proves that $$(\boldsymbol{A}^{-1})^{T}=(\boldsymbol{A}^{T})^{-1}=\boldsymbol{A}^{-T}$$

The wolfram alpha calculation for $$(\boldsymbol{A}^{-1})^{T}$$ is given in the following pdf file

The wolfram alpha calculation for $$(\boldsymbol{A}^{T})^{-1}$$ is given in the following pdf file

Click on the pdf file to see the second page.

The Wolfram alpha results match with the manually calculated results of condition 2 of problem statement.This validates the proof.

More details on the appropriate syntax can be found at the following link: http://reference.wolfram.com/mathematica/tutorial/BasicMatrixOperations.html

Problem Statement
Table of Gauss points and weights (used in Gauss Quadrature integration), from wikipedia

Verify this table against Fish and Belytschko's table on p. 89, and the NIST Handbook. Perform problems 4.6, 4.7, and 4.9 on p. 91.

Verification of Gauss points and weights
Evaluating the wikipedia table into decimal values:

From Fish and Belytschko, the numerical values are listed below:

From the NIST Digital Library of Mathematical Functions, the 5-point values are:

Comparing the three tables above it is clear they are in agreement. Further, the values are confirmed with the MATLAB script at the end of this problem which can calculate the points and weights for $$n$$ points.

FB Problem 4.6
Use Gauss quadrature to obtain exact values for the following integrals. Verify by analytical integration.

For each integration the number of points needed is given by

part a
Solved with Wolfram Alpha here

For the integration of this second order polynomial we need 2 points. Using the matlab script at the end of this problem's solution to get the quadrature points and weights we integrate with the following script.

The MATLAB scripts calculates the integration to be 25.333333333333329, which is within round-off error of the analytic result found (thus exact for a numerical result).

part b
$$\int_{-1}^{1}\left(\xi^4+2\xi^2\right)d\xi = 26/15$$

Solved with Wolfram Alpha here

For a fourth-order polynomial we need three integration points. The following code finds the integration result exactly.

The numerical integration provided a result of 1.733333333333333, a numerically exact result.

part c
The MATLAB function has been created and used in the above parts. The function can be seen at the end of this problem.

FB Problem 4.7
Use three-point Gauss quadrature to evaluate the following integrals. Compare to the analytical integral.

part a
NOTE: There is a typo in the text here. The kernel uses $$\xi$$ and $$x$$ inconsistently.

Evaluated with Wolfram Alpha here.

Evaluating numerical also finds a value of zero (0.0).

part b
NOTE: There is a typo in the text here. The kernel uses $$\zeta$$ and $$\xi$$ inconsistently.

Evaluated with Wolfram Alpha here.

Here, numerical integration with three points does not match the analytical result. The numerical result with three points is 1.529961646516273. This is expected as we are not integrated a polynomial but a trig function (Gauss quadrature is based upon interpolating polynomials of increasing order).

FB Problem 4.8
The integral $$\int_{-1}^{1}\left(3\xi^3+2\right)d\xi$$ can be integrated exactly using two-point Gauss quadrature. How is the accuracy affected if

part a
one-point quadrature is employed

part b
three-point quadrature

Solution
First note that the integration with two-points results in a value of 4.0. With one-point the value is also 4.0. With three-points the value is also 4.0 as expected.

This result is somewhat unexpected and is the result of coincidence and the integral mean value theorem. The integral mean value theorem states that there is a point within a domain that is valued the same as the integration value. For this integral, this point so happens to be at zero, the location where the sample point is taken for one quadrature point. Thus, for this particular equation, with this particular domain of integration, the single point predicts the exact value correctly.

Problem Statement
Problem 1: The heat conduction problem in Example 8.1 ( Fish and Belytschko ) is modeled with 16 quadrilateral finite elements.The problem is solved manually using the finite element code given in Section 12.5

Solution
1.) The problem is solved using the MATLAB code given in section 12.5 of companion site of the text book Fish and Belytschko.On running the code we receive the following post process result plots for temperature and flux figs (a),(b). Fluxes are calculated by looping over the number of elements.For the four node quadrilateral there are four Gauss points as shown in fig (c).

2.) Reproduce Example 8.4 (page 205) from Fish and Belytschko. 3.) Complete problem 8.6 (page 212) from Fish Belytschko.

title: User:Eml5526.s11.team4/HW6


text:
User talk:Emajid

 Hello Emajid, 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 23:13, 30 March 2011 (UTC)

title: User talk:Emajid


text:
User talk:Kottivijay

 Hello Kottivijay, 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 23:13, 30 March 2011 (UTC)

title: User talk:Kottivijay


text:
User talk:TransControl

 Hello TransControl, 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 23:13, 30 March 2011 (UTC)

title: User talk:TransControl


text:
User talk:Nithinlaxman

 Hello Nithinlaxman, 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! --Abd 23:52, 30 March 2011 (UTC)

i feel its going good while learning kannada in this way but can u please continue updating this regularly with regards --Vidya4u.46 06:07, 20 February 2012 (UTC)vidya sagar 20-feb-2012 11:35 A.M.

Your page Open Educational Resources for School Teachers from Developing Nations
I saw your page and did a little work on it. Some comments:

 * I used a subpage for participants, instead of the separate top level page you used. In general, if you create pages which are dependent upon another page, being part of a class or some data for it, or the like, it's better to use subpages. This will more closely match the Wikibooks practice, and it keeps things organized better.

 * You use the  tag a lot. That's normally not necessary. Double-return paragraphing is common, and that always separates the lines. If you use * to bullet a list or # to number it, you don't need to break the preceding lines, those tags will automatically break. So too will :, used for indentation.

 * Local links should generally be presented as such rather than using URLs. It's a lot easier! You can also link to wikipedia with a local link preceded by "w:", i.e., you can read a lousy article on my favorite topic at Cold fusion, for which I wrote Cold fusion.

 * Asking people to publish their email address on-wiki is asking them to expose themselves to massive spam. I suggest letting people sign up by user name only, and they can email you through the Wikiversity interface, which will show you their Wikiversity user name and the email they have registered with Wikiversity, assuming they have email enabled (which is always a good idea, in fact. No reason not to do it.)

 * When you make a personal comment such as "I'd like to thank...," I suggest you sign it with --~ which will substitute your signature (normally your user name as a link to your user page) and a date stamp. Three tildes does just the user name....

 * Thanks for working on educational resources here.

One more thing: that page title is extremely long. I do wonder if it's been over-specified. What about Open Educational Resources for Teachers? Are there other kinds of teachers who wouldn't benefit from this, and does a teacher in a developing nation need something different from a perhaps-isolated teacher in a developed nation? Or the whole class you are developing could be a seminar or class under the main Open educational resources page, which might be better. And give you a much shorter title! Notice that the main resource on OER has no capital letters except on the first word, which is more common Wikiversity style for page names, unless the name is truly a proper noun. Is it? (If so, we should rename the main OER page! -- I think we should.) --Abd 23:52, 30 March 2011 (UTC)

 * Thank you for the suggestions, They are indeed helpful in designing a better wikiversity page. Regarding the Course title page, I agree it is long but I dont have enough time to send the change notices now, so will stick with it as of now. I do agree with the e mail id of the participants, and I plan to change them to the wiki format soon.--Nithin Lakshmana 05:20, 31 March 2011 (UTC)

title: User talk:Nithinlaxman


text:
User:Jonmsmith

Jon Michael Smith is a retired NASA Senior Executive. Mr. Smith was heavily involved in initial Space Shuttle Program's marketing project. He sold flights on the Space Shuttle to countries and firms that required space launch services. In his mid career years he led the award winning Advanced Communications Satellite program and served as the Commercialization Manager for the Space Operations Management Office, that led to the commercialization of NASA's polar communications network. In his later career years he returned to the Space Shuttle program and participated in the closeout planning of the Shuttle program.

After the Columbia accident, Mr. Smith was asked to lead a small team of managers to study of the Management Root Cause of the Space Shuttle Columbia accident and re3commend corrective actions. This was a challenging assignment because there were no "Management" Root Cause (MRC) techniques to guide the study team. The team developed unique approaches to analyzing managements role in disasters and identifying compelling corrective actions.

Mr. Smith retired from the NASA Johnson Space Center in Houston, Texas in January 2007. Now, he is the proprietor of Jon M. Smith and Associates (JMSA), a consulting firm. He is a subject matter expert on flight guidance and control systems, systems engineering and program management and control.

Mr. Smith is also known for his developments in Mathematics. His most notable is the development of the T-integrator. The T-Integrator is an accurate numerical integrator that can be tuned to the problem it is being used to solve. The T-Integrator is a numerical integrator based on information theory, not approximation theory. It is known for containing classical approximation theory integrators as special cases of the more modern information theory integrators. He published his work in The AIAA Journal of Dynamic Systems, Measurement and Control and presented this engineering applications in his book, published by John Wiley and Sons, entitled Mathematical Modeling and Digital Simulation for Engineers and Scientists. The T-Integrator is also published in the CRC Concise Encyclopedia of Mathematics. He continues to develop this numerical integration concept in his retirement years.

title: User:Jonmsmith


text:
User talk:Jonmsmith

 Hello Jonmsmith, 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.

To get started, you may

 * Take a guided tour and learn to edit.
 * Visit a (kind of) random project.
 * Browse Wikiversity, or visit a portal corresponding to your educational level: pre-school, primary, secondary, tertiary, non-formal education.
 * Find out about research activities on Wikiversity.
 * Explore Wikiversity with the links to your left.

 * Read an introduction for teachers and find out how to write an educational resource for Wikiversity.
 * Give feedback about your initial observations
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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 00:05, 31 March 2011 (UTC)

title: User talk:Jonmsmith


text: Category:Bash programming

title: Category:Bash programming


text: Category:Digital Libraries

title: Category:Digital Libraries


text:
User:Egm6321.f10.team2.oztekin/Newwwwwww

=Problem 6.3:Using Calculix to reproduce Beede's Models=

Calculix is an open source program that we can solve our FEA problems. Bconverged is a really good provider of calculix that would offer engineers work between different platforms. After downloading calculix software from bconverged we will have two modules of calculix and text editor called SciTE. These two modules that belongs to calculix are $$ \displaystyle\mathit{cgx} $$ and $$\displaystyle\mathit{ccx} $$. In the first module of calculix we can design our solid models and define boundary conditions on it. The solver,crunchix (ccx), is solver and post processor.Lets start building models that are presented by Beede's.Beede's examples. Also this cite and more information is provided by Dr. Vu-Quoc in our EML5526 FEA class page. Class page

FEA BEAM PART 1:Getting started
As a first step we have to open SciTE text editor that is a place where we type our commands and view our models. To run it

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

$$  \displaystyle [Start] \Rightarrow [All Programs]\Rightarrow [bConverged]\Rightarrow [SciTE] $$
 * style="width:95%" |
 * style="width:95%" |

 * }

Here are some information about SciTE editor.

This is very important to understand that we will use two typing methods to type our commands. This i a very beautiful property of calculix. We can type in text editor and that 'run ' all of the codes in it or we can type directly by clicking on the graph screen and execute the code by hitting enter.

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

$$  \displaystyle [open SciTE]\Rightarrow [File]\Rightarrow [Save As]\Rightarrow [beam.fbd] $$
 * style="width:95%" |
 * style="width:95%" |

 * }

The step presented above is a essential step. Because you can not run the codes without saving!!!

 * Type these codes into your SciTE editor. Then hit F10 key or click tools--post processor to run.
 * pnt p1 0 0 0
 * pnt p2 100 0 0
 * line l1 p1 p2 25

Let's get familiar with our codes.

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

$$  \displaystyle \begin{matrix} pnt \Rightarrow name & x coord. & y coord. & z coord. & \end{matrix} $$
 * style="width:95%" |
 * style="width:95%" |

 * }

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

$$  \displaystyle \begin{matrix} line \Rightarrow name & start point & end point & division & \end{matrix} $$
 * style="width:95%" |
 * style="width:95%" |

 * }

 * Proceed with expanding commands in text editor and creating four lines and five nodes. We are partially creating this big line because if we had constructed the line in one step the mesh would have been arbitrary. So to avoid arbitrariness we are doing more steps. We want equal divisions among mesh elements.

 * Here how we can view our model in different aspects. In the area surrounded by thin black surface use this properties of mouse.

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

$$  \displaystyle \begin{matrix} LMB & rotate \\ MMB & zoom\\ RMB & pan \end{matrix} $$
 * style="width:95%" |
 * style="width:95%" |
 * }

 * While cgx window is active it means that we are working in 'interactive mode.' Type these commands while cgx window is active.
 * plot pa all
 * plus la all

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

$$  \displaystyle \begin{matrix} plot & n(nodes) & e(elements) &f(faces) & p(points) & l(lines) & s(surfaces) & b(bodies) & all(set name) \end{matrix} $$
 * style="width:95%" |
 * style="width:95%" |

 * }

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

$$  \displaystyle \begin{matrix} plus & n(nodes) & e(elements) &f(faces) & p(points) & l(lines) & s(surfaces) & b(bodies) & all(set name) \end{matrix} $$
 * style="width:95%" |
 * style="width:95%" |

 * }

Commands are same for plot and plus. But there is a slight difference between them. Plot command plots only one type of entity removes other entities from graph screen but plus command adds entities to the other entities that are already visual in graph screen.

 * Type seta lines l l1 l2 l3 l4
 * swep lines sweplines tra 0 10 0 10
 * seta surfaces s A001 A002 A003 A004
 * swep surfaces swepsurfaces tra 0 0 1 1

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

$$  \displaystyle \begin{matrix} seta & type & elements. in.set & \end{matrix} $$
 * style="width:95%" |
 * style="width:95%" |

 * }

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

$$  \displaystyle \begin{matrix} swep & set & fx &fy & fz &  tra& dx & dy & dz \end{matrix} $$
 * style="width:95%" |
 * style="width:95%" |

 * }

As we see above we have created body from swepping surfaces, surfaces from swepping lines. When we create a high order entity lower order entities will be kept in order to define the high order one. For example if we create surface lines and points will be kept.

After constructing the body now we can mesh our body.We have different kinds of elements for 1D,2D,and 3D meshes. Type these commands in interactive mode.
 * elty all he8
 * mesh all
 * plot m all

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

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

 * }

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

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

 * }

By using different types of views we can get different aspects of our mesh and other entities.

FEA BEAM PART 2:Exporting mesh,loads,and boundary conditions
In this section again we start with our text editor and open beam.fbd. Now we are updating written codes in it. New codes
 * seto beam
 * pnt commands
 * line commands
 * sets and swep codes
 * setc beam
 * defining and setting up mesh

Between seto and setc commands we have defined set named beam and we put all of the entities in this general set. Also we have defined other sets just before we did in the last section.If we hit the F10 key and run our new codes beam.msh will be created automatically.Opening beam.msh yields following

As we see above two sets defined in this document. Node and element sets. Looking at element sets we can see that elements are 3D elements with 8 nodes. The style of presenting data and storing is calculix convenience.

Here we have defined another set named fixed to assign fixed nodes in it. View the appropriate section of the beam and type quadd fixed while in interactive mode. Then press r and move the cursor down and press r to create select box. Be careful to create select box that can fit all tha nodes that we are interested in. Press a to active mode a then press n to select nodes in the box. You will see the node properties in the output section and by pressing q we will have finished our job.

As mentioned in the article we have two different boundary conditions. We have defined fixed dots where we will apply restriction on their displacement. Also turn the view to see faces and viewing elements,we found element '751'.And we have applied pressure force its upward surface. And we applied two force vectors on its two nodes.

We can show element correspondence with their families.

FEA BEAM PART 3
In this section as you see in the figure we created new text file but with an extension .inp. This means that now we are using ccx commands. Generally start with '*'. First we are taking input parameter from *.msh and *.123 documents. Between STEP and ENDSTEP we are also defining material properties and selensting Solve and Post process tabs from the tools bar we can find stresses along the beam.

FEA BEAM PART 4

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

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

 * }

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

$$  \displaystyle \begin{matrix} Part & Of  & Animation \end{matrix} $$
 * style="width:95%" |
 * style="width:95%" |

 * }

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

$$  \displaystyle \begin{matrix} Y stress & At  & Portion \end{matrix} $$
 * style="width:95%" |
 * style="width:95%" |

 * }


text: File:Egm6321.f10.team2.oztekin.SciTE view.png


text: File:Egm6321.f10.team2.oztekin.Linewithtwopoints.png


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text:
User:Eml5526.s11.team2.brenner/hwk6

=Problem 6.1 Using Quadratic Lagrange Element Basis Functions (QLEBF) obtain the trial solution from Weak Form=

Problem Statement
Use QLEBF with uniform discretization (equidistant element nodes) for $$ \displaystyle nel = 2,4,6,8 $$

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

Essential Boundary Function  $$ u \left(0\right)=4 $$, and

Natural Boundary Function  $$ n(1)a_2(1)\frac{\partial u(x=1)}{\partial x}=12$$

$$\implies (1)(2+3x)\frac{\partial u(x=1)}{\partial x}=12$$

$$\implies \frac{\partial u(x=1)}{\partial x}=\frac{12}{5}

$$

$$ \Gamma_g = \left[0\right] \qquad \Gamma_h = \left[1\right] $$

Find
1) For number of elements $$ \displaystyle nel = 2 $$, compute $$ \displaystyle \tilde{K} = \sum_{e=1}^2 \tilde{K}^e $$ with $$ \displaystyle \tilde{K}^e $$ given by (6.1.1) display $$ \displaystyle \tilde{K}^e, e = 1,2 $$

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

$$ \displaystyle \tilde{K_{ij}}^e = \int_{\omega^e}{{b_i}^e}^\prime(x^e)a_2(x^e){{b_j}^e}^\prime(x^e)dx^e $$ (6.1.1)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

2) Compute $$ \displaystyle {k}^e, {L}^e for e = 1,2 $$

3) Compute $$ \displaystyle \tilde{K}^e = {{L}^e}^T {k}^e {L}^e for e =1,2 $$ compare to item 1) above

4) Plot all QLEBF for $$ \displaystyle nel = 2 $$

5) Plot $$ \displaystyle {u^h}_\tilde{n} $$ vs. $$ \displaystyle u $$ and $$ \displaystyle {u^h}_\tilde{n}(0.5) - u(0.5) $$ vs $$ \displaystyle \tilde{n} $$

title: User:Eml5526.s11.team2.brenner/hwk6


text:
User:Chrismo/Photography

=Photographic Composition= My photo's for completing this course on Photographic_Composition

I will try to collect 20 images in the following categories.

Lesson 6: Blur the Background
Working towards 20 images for the course.

title: User:Chrismo/Photography


text:
University Edible Landscaping

Edible Landscaping
Edible landscaping is the conversion of land spaces, especially marginal land spaces such as flowerbeds along a sidewalk, into working spaces growing food producing plant otherwise known as edible plants. This can be achieved through exercising techniques of permaculture where plants are selected to grow with complementary plants while respecting the immediate local environmental conditions. Using topiary and ornamental design with these edible plants also offers the opportunity to convert the otherwise ornamental botanic gardens into productive working spaces. That is, an edible landscape can use techniques of permaculture, which selects plants to grow with complementary plants while respecting the immediate local environmental conditions. Topiary and ornamental designs convert edible plants into ornamental botanic shapes that combine the concept of a botanical garden and food-productive working spaces.

Edible landscaping is not a new idea; edible landscaping dates back well over two thousand years to the hanging gardens of Babylon. The Garden of Eden is also a biblical example. Edible landscaping was also an underlying principle to medieval monastic gardens.

Australian naturist, Bill Mollison created a renewed interest in edible landscaping in the 1970's, through his concept of permaculture. Permaculture is a landscape design technique designed to form harmonious balances between input and outputs.

University Involvement
Universities as a center of higher learning have an obligatory role in the community to research and introduce edible landscapes.

These edible landscapes are a matter of social justice The majority of food in this country travels 1500 miles to reach a consumer’s plate. The fossil fuels necessary to transport produce this absurd distance creates a massive carbon footprint that contributes to global warming. Additionally, this food is oftentimes grown with hazardous pesticides and herbicides, affecting the health of the farm workers. Edible landscapes can demonstrates that there are alternatives to industrial agriculture. The landscape will spur discussion of current farming practices and sustainable agricultural practices while providing students with food that is grown for the nourishment of body and spirit.

Key benefits of an edible landscape in university

 * Edible landscapes can serve as a living landscape for academic exercises
 * Edible landscaping is a mater of social justice to allow access of nutritious food to all people.
 * Positive mental health.

List of Universities participating in edible landscaping projects.

 * Loyola University Chicago - Project Eden

 * University Of Minnesota - UM Edible Landscape
 * Scripps College - Food and Sustainability Community Project

title: University Edible Landscaping


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text:
Talk:Happiness/Happiness and Life

Freedom of science
Why is this here? It's a personal essay, it should be on a blog or something. Does this meet our criteria for inclusion in some way? Mglovesfun 12:29, 31 March 2011 (UTC)
 * Essays are generally allowed in the Wikiversity. Here is the freedom of science. "Wikiversity is a Wikimedia Foundation project devoted to learning resources, learning projects, and research for use in all levels, types, and styles of education from pre-school to university, including professional training and informal learning. We invite teachers, students, and researchers to join us in creating open educational resources and collaborative learning communities."--Nobody60 (talk) 13:46, 27 May 2012 (UTC)

title: Talk:Happiness/Happiness and Life


text:
Talk:Happiness/Philosophy of Happiness

Why is this here? It's a personal essay, it should be on a blog or something. Does this meet our criteria for inclusion in some way? Mglovesfun 12:30, 31 March 2011 (UTC)
 * Be tolerant. Be happy. The philosophy of happiness is an important part of the philosophy. Maybe you have another opinion than I, but that´s normal in science. That leads to discussion and discussion is a motor of progress.--Nobody60 13:41, 4 August 2011 (UTC)
 * I believe it should stay. A University should create discussion, as Nobody60 said. The joy about open content allows others to share ideas and learn from new perspectives, and I believe that's what Wikiversity is about. Also this was a very insightful read, my professor actually got me into Philosophy. Probably one of the few classes that showed me that learning and understanding was a joy and celebration of life. It's something that is beyond the scope of the finite, an adventurous step into the unknown. It's awesome. &  Heartwarming (discuss • contribs) 03:03, 27 December 2015 (UTC)

title: Talk:Happiness/Philosophy of Happiness


text:
User:Mglovesfun

Wikimedia contributor, mainly Wiktionary but sometimes follow links to other projects. See wikt:User:Mglovesfun.

title: User:Mglovesfun


text: File:Egm6321.f10.team2.oztekin.Step7.png