text:
User:Eml5526.s11.team2/template

=Template For Homework Submission= On HW 1 I was able to go back and change all of the formatting so it was consistent throughout the entire report. For HW2 I was not able to do this because I had less time this week then I had last week. All team member should make an effort to write their reports using the same format and do so the first time through. I hope that these

1. Label Equations (Problem"#"."1")
Problem 1 would have equations that look like this

And so on...............

Not Like these!

For n=1
Text Here

For n=2
Text Here

For n=1
Text Here

For n=2
Text Here

For n=1
Text Here

For n=2
Text Here

3. Center Aligned even for Multiple Equations
The only EXPECTION would be for the problem statement which can be left aligned. To be consistent if the problem statement has a theorem such IMVT the equation should be numbered and center align if you are to reference it later on.

Notice how the following equations are center aligned and still match up nicely. Also notice how a very long equation can span several lines with the simple use of "\begin{align}...\end{align}", where an ampersand "&" aligns them and all lines end with a "\\" except the last.

4. Box the final solution in RED and CENTERED
Go into EDIT on this part and copy and paste the format. It both centers and boxes.

4.Problems that have a star *
As Vu-Quoc mentioned in class a statement such as this should be included at the beginning of the problem.

 This problem was solved referring to Team 3, Hwk 1, S10 homework

=Some Cool Stuff to Make Our Work Stick Out=

title: User:Eml5526.s11.team2/template


text:
User:Eml5526.s11.team01/Hwk3

=Problem 1: Solving using WRF=

Solving a differential equation using Weight Residual Form. From lecture 13

Given
The partial differential equation is:

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

$$ \frac{\partial}{\partial x}\left(2\frac{\partial u}{\partial x}\right) + 3 = 0 $$
 * $$\displaystyle (Eq.1.1) $$
 * }
 * }

The boundary conditions are:

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

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

A weighting funtion should be used:

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

$$ b_{i} = b_{i}=\left ( x+1 \right )^{i} $$
 * $$\displaystyle (Eq.1.2) $$
 * }
 * }

Find

 * 1.1 Find an approximate solution of the form $$U^{h}=\sum_{i=0}^{n}d_{i}b_{i}$$ for n=2
 * 1.2 Find two equations that enforce the boundary conditions
 * 1.3 Project the weight residues
 * 1.4 Display the equations in matrix form
 * 1.5 Solve for $$\displaystyle d $$
 * 1.6 Construct $$U^{h}$$ and plot $$U^{h}$$ and $$u$$
 * 1.7 Repeat 1.1 to 1.6 for n = 4 and n = 6

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

$$ d_{i}=\begin{bmatrix} d0 & d1 & d2 \end{bmatrix} $$
 * $$\displaystyle (Eq.1.3) $$
 * }
 * }

and

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

$$ b_{i}=\begin{bmatrix} 1 & \left ( x+ 1\right) & \left ( x+ 1\right )^{2} \end{bmatrix} $$
 * $$\displaystyle (Eq.1.4) $$
 * }
 * }

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

$$ U^{h}=d0+d1\left ( x+1 \right )+d2\left ( x+1\right )^{2} $$
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * style="width:2%; padding:1px; border:2px solid #FF0000" |

 * $$\displaystyle (Eq. 1.5) $$
 * style= |
 * }

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

$$ u\left(1\right)=0=d0+2d1+4d2$$
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * style="width:2%; padding:1px; border:2px solid #FF0000" |

 * $$\displaystyle (Eq. 1.6) $$
 * style= |
 * }

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

$$ \frac{du}{dx}(0)=-4 = d1+2d2$$
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * style="width:2%; padding:1px; border:2px solid #FF0000" |

 * $$\displaystyle (Eq. 1.7) $$
 * style= |
 * }

1.3 Project the weight residues
In this case changing the function u(x) by the approximated function $$U^{h}$$ in the PDE (Eq.1.1), we found that P(u) is:

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

$$ P\left ( U^{h} \right )=2\frac{\partial^2 U^{h}}{\partial x^2}+3=2d2+3 $$
 * $$\displaystyle (Eq.1.8) $$
 * }
 * }

The (Eq.1.8) can be written as:

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

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

d2\end{bmatrix}+3 $$
 * $$\displaystyle (Eq.1.9) $$
 * }
 * }

Projecting the residue we have:

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

$$ \int_{a}^{b}b(x)P\left ( U^{h} \right )= 0 $$ After substitution of (Eq.1.4) and (Eq.1.9) in (Eq.1.10) the following equation is obtained
 * $$\displaystyle (Eq. 1.10) $$
 * }
 * }
 * {| style="width:100%" border="0"

$$ 2\int_{0}^{1}\begin{bmatrix} 1\\ \left (x+ 1\right ) \\ \left ( x+1\right )^{2} \end{bmatrix}\begin{bmatrix} 0 &0 & 2\end{bmatrix}dx\begin{bmatrix} d0\\ d1\\
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * style="width:2%; padding:1px; border:2px solid #FF0000" |

d2\end{bmatrix}=-3\int_{0}^{1} \begin{bmatrix} 1\\ \left ( x+1\right ) \\

\left ( x+1 \right )^{2}\end{bmatrix}dx $$

 * $$\displaystyle (Eq. 1.11) $$
 * style= |
 * }

1.4 Display the equations in matrix form
Performing the product and then the integration indicated in (Eq.1.11) we have:
 * {| style="width:100%" border="0"

$$ \begin{bmatrix} 0 &0 & 4 \\ 0 & 0& 6 \\ 0 & 0& 3\frac{28}{3}\end{bmatrix}\begin{bmatrix} d0\\ d1\\
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * style="width:2%; padding:1px; border:2px solid #FF0000" |

d2\end{bmatrix}=\begin{bmatrix} -3\\ -\frac{9}{2}\\

-21\end{bmatrix} $$

 * $$\displaystyle (Eq. 1.12) $$
 * style= |
 * }

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

1.5 Solve for $$ d $$
In (Eq.1.12) we have three equations; but, as we need to enforce the boundary conditions we take the only one of them and solve it together with (Eq.1.6) and (Eq.1.7). From these equations we now have the system:

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

$$ \begin{bmatrix} 1 &2   & 4 \\ 0 &1 &2 \\ 0 & 0& 4 \end{bmatrix}\begin{bmatrix} d0\\ d1\\

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

-3\end{bmatrix}$$
 * $$\displaystyle (Eq.1.13) $$
 * }
 * }

After solving the system in (Eq.1.13) by using a calculator, we obtain d:
 * {| style="width:100%" border="0"

$$\begin{bmatrix} d0\\ d1\\
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * style="width:2%; padding:1px; border:2px solid #FF0000" |

d2\end{bmatrix}=\begin{bmatrix} 8\\ -2.5\\ -0.75\end{bmatrix}$$

 * $$$$
 * style= |
 * }

1.6 Construct $$U^{h}$$ and plot $$U^{h}$$ and $$u$$
Replacing d in (Eq.1.5) we obtain the approximated solution:

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

$$ U^{h}=8-2.5\left ( x+1\right )-0.75\left ( x+1\right )^2 $$
 * $$\displaystyle (Eq.1.14) $$
 * }
 * }

On the other hand, the exact solution of the PDE (Eq.1.1) with the given boundary condition is:

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

$$ u\left ( x \right )=-\frac{3}{4}x^{2}-4x+\frac{19}{4} $$
 * $$\displaystyle (Eq.1.15) $$
 * }
 * }

We observe that the aproximated and the exact solution are the same; therefore, increasing the number of terms in b(x) will not improve the solution

In Fig.1.1 we show the exact and the approximated solution.

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

As the procedure is the same shown before, we develop a routine using Matlab to perform the products, the integration and to solve the system of equations. The routine is show in the appendix. The error was also Zero for i = 4, and i = 6.

Appendix
 Matlab Code:

=Problem 2: Spring System =

Given
Spring System with closed ends.

Find

 * 2.1 Number the Element and Nodes.
 * 2.2 Assemble the Global Stiffness and Force Matrix.
 * 2.3 Partition the system and solve for the Nodal Displacements.
 * 2.4 Compute the Reaction Forces.

2.1 Number the Element and Nodes.
The Spring Systems is shown in the figure.

Superscripts to each of the stiffness values denotes the element number. Furthermore, Node Numbers are also shown in the figure. As shown in the figure, there are 4 nodes.

2.2 Assemble the Global Stiffness and Force Matrix.
For Element Number 1,$$I=1 $$ and $$J=4$$,

$$K^{(1)}=k\begin{bmatrix} 1& -1\\ -1 & 1 \end{bmatrix}\Rightarrow \tilde{K}=\begin{bmatrix} k& 0 & 0 & -k\\ 0 & 0 & 0 & 0\\ 0& 0 & 0 & 0\\ -k& 0 & 0 & k \end{bmatrix}$$

For Element Number 2,$$I=4 $$ and $$J=3$$,

$$K^{(2)}=2k\begin{bmatrix} 1& -1\\ -1 & 1 \end{bmatrix}\Rightarrow \tilde{K}=\begin{bmatrix} 0& 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0& 0 & 2k& -2k\\ 0& 0 & -2k & 2k \end{bmatrix}$$

For Element Number 3,$$I=3 $$ and $$J=2$$,

$$K^{(3)}=k\begin{bmatrix} 1& -1\\ -1 & 1 \end{bmatrix}\Rightarrow \tilde{K}=\begin{bmatrix} 0& 0 & 0 & 0\\ 0 & k & -k & 0\\ 0& -k & k& 0\\ 0& 0 & 0& 0 \end{bmatrix}$$

For Element Number 4,$$I=1 $$ and $$J=3$$,

$$K^{(4)}=3k\begin{bmatrix} 1& -1\\ -1 & 1 \end{bmatrix}\Rightarrow \tilde{K}=\begin{bmatrix} 3k& 0 & -3k & 0\\ 0 & 0& 0& 0\\ -3k& 0 & 3k& 0\\ 0& 0 & 0& 0 \end{bmatrix}$$

Assembled Global Systems Matrix will be given by,
 * {| style="width:100%" border="0"

$$K=\sum_{e=1}^{4}\tilde{K}^{e}=\begin{bmatrix} k+3k & 0 & -3k & -k\\ 0 & k & -k & 0\\ -3k & -k &6k &-2k \\ -k & 0 & -2k & 3k \end{bmatrix}$$
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * style= |
 * }

Where Superscript $$'e'$$ denotes the Element Number.

The Displacement and Global Force Matrices for this system are given by,

2.3 Partition the system and solve for the Nodal Displacements.
Eventually, The Global System of Equations is Given by,

$$\begin{bmatrix} k+3k & 0 & -3k & -k\\ 0 & k & -k & 0\\ -3k & -k &6k &-2k \\ -k & 0 & -2k & 3k \end{bmatrix}\begin{bmatrix} 0\\ 0\\ u_{3}\\ u_{4} \end{bmatrix}=\begin{bmatrix} r_{1}\\ r_{2}\\ 50\\ 0 \end{bmatrix}$$
$$\displaystyle (Eq.2.1) $$

As the first two displacements are prescribed, We can Partition the system after 2 Rows and 2 Columns,

And Can form the Matrix System of Equivalent to,

$$\begin{bmatrix} K_{E} & K_{EF}\\ K_{EF}^{T} & K_{F} \end{bmatrix}\begin{bmatrix} \bar{d}_{E}\\ d_{F} \end{bmatrix}=\begin{bmatrix} r_{E}\\ f_{F} \end{bmatrix}$$

Where,

$$K_{E}=\begin{bmatrix} 4k & 0\\ 0 & K \end{bmatrix}, K_{EF}=\begin{bmatrix} -3k & -k\\ -k& 0 \end{bmatrix},K_{EF} ^T=\begin{bmatrix} -3k & -k\\ -k & 0 \end{bmatrix}, K_{F}=\begin{bmatrix} 6k & -2k\\ -2k & 3K \end{bmatrix}, \bar{d}_{E}=\begin{bmatrix} 0\\ 0\\ \end{bmatrix}, d_{F}=\begin{bmatrix} u_{3}\\ u_{4} \end{bmatrix},r_{E}=\begin{bmatrix} r_{1}\\ r_{2} \end{bmatrix}, f_{F}=\begin{bmatrix} 50\\ 0 \end{bmatrix}$$

Solving the Following Reduced System of Equations,

$$K_{E}\bar{d}_{E}+K_{EF}d_{F}=r_{E}, K_{EF}^T\bar{d}_{E}+K_{F}d_{F}=f_{F}$$

Using $$\displaystyle (Eq.2.1) $$ we can form the Following Simultaneous Equations,

$$ 6k.u_{3}-2k.u_{4}=50$$
$$\displaystyle (Eq.2.2) $$

$$ -2k.u_{3}-3k.u_{4}=0$$
$$\displaystyle (Eq.2.3) $$

Solving These 2 Simultaneous Equations, We get,

2.4 Compute the Reaction Forces.
Now, Utilizing the First row of the math>\displaystyle (Eq.2.1) we can get

And Further, Utilizing the Second row of the math>\displaystyle (Eq.2.1) we can get,

=Problem 3: Truss Structure=

Given
The truss Structure Shown in the Figure. Nodes A & B are fixed. A Force of 10N Acts in the Positive x-Direction at node C. Young's Modulus is $$E=10^{11}Pa$$ and the Cross Sectional Area for all bars are $$ A= 2.10^{-2}$$

Find

 * 3.1 Number the Element and Nodes
 * 3.2 Assemble the Global Stiffness and Force Matrix
 * 3.3 Partition the system and solve for the Nodal Displacements
 * 3.4 Compute the Stresses & Reactions

3.1 Number the Element and Nodes
The Truss System Along with the Element Numbers and the Node Numbers Written on it, is Shown in the figure.

3.2 Assemble the Global Stiffness and Force Matrix
For Element Number 1,$$I=1 $$ and $$J=3$$, Also Element 1 is inclined at 90 deg to the Positive X-direction,$$\phi^{(1)} =90^{0}$$, $$cos(90)=0,sin(90)=1$$, $$l=1$$,

Using the Following formula for obtaining the Elemental Stiffness Matrix for all the elements,

$$K^{e}=\frac{AE}{l}\begin{bmatrix} (cos\phi ^{e})^{2} & cos\phi ^{e}.sin\phi ^{e} & -(cos\phi ^{e})^{2} & -cos\phi ^{e}.sin\phi ^{e} \\ cos\phi ^{e}.sin\phi ^{e} & (sin\phi ^{e})^{2} & -cos\phi ^{e}.sin\phi ^{e} &-(sin\phi ^{e})^{2} \\ -(cos\phi ^{e})^{2} & -cos\phi ^{e}.sin\phi ^{e} & (cos\phi ^{e})^{2} & cos\phi ^{e}.sin\phi ^{e}\\ -cos\phi ^{e}.sin\phi ^{e} & -(sin\phi ^{e})^{2} & cos\phi ^{e}.sin\phi ^{e} & (sin\phi ^{e})^{2} \end{bmatrix}$$
$$\displaystyle (Eq.3.1) $$

Utilizing above equation for Element 1,

$$K^{(1)}=\frac{AE}{l}\begin{bmatrix} 0 & 0 & 0 & 0\\ 0 &1 &  0&-1 \\ 0 & 0 & 0 & 0\\ 0 & -1 & 0 & 1 \end{bmatrix}$$

For Element Number 2,$$I=2$$ and $$J=4$$, Also Element 2 is inclined at 90 deg to the Positive X-direction,$$\phi^{(2)} =90^{0}$$, $$cos(90)=0,sin(90)=1$$, $$l=1$$, Therefore,

$$K^{(2)}=\frac{AE}{l}\begin{bmatrix} 0 & 0 & 0 & 0\\ 0 &1 &  0&-1 \\ 0 & 0 & 0 & 0\\ 0 & -1 & 0 & 1 \end{bmatrix}$$

For Element Number 3,$$I=2$$ and $$J=3$$, Also Element 3 is inclined at 135 deg to the Positive X-direction,$$\phi^{(3)} =135^{0}$$, $$cos(135)=\frac{-1}{\sqrt{2}},sin(135)=\frac{1}{\sqrt{2}}$$, $$l=\sqrt{2}$$, Therefore,

$$K^{(3)}=\frac{AE}{l}\begin{bmatrix} \frac{1}{2\sqrt{2}} & -\frac{1}{2\sqrt{2}} & -\frac{1}{2\sqrt{2}} & \frac{1}{2\sqrt{2}}\\ -\frac{1}{2\sqrt{2}} &\frac{1}{2\sqrt{2}} &  \frac{1}{2\sqrt{2}}&-\frac{1}{2\sqrt{2}} \\ -\frac{1}{2\sqrt{2}} & \frac{1}{2\sqrt{2}} & \frac{1}{2\sqrt{2}} & -\frac{1}{2\sqrt{2}}\\ \frac{1}{2\sqrt{2}} & -\frac{1}{2\sqrt{2}} & -\frac{1}{2\sqrt{2}} & \frac{1}{2\sqrt{2}} \end{bmatrix}$$

For Element Number 4,$$I=3$$ and $$J=4$$, Also Element 4 is inclined at 90 deg to the Positive X-direction,$$\phi^{(2)} =0^{0}$$, $$cos(0)=1,sin(0)=0$$, $$l=1$$, Therefore,

$$K^{(4)}=\frac{AE}{l}\begin{bmatrix} 1 & 0 & -1 & 0\\ 0 &0 &  0&0\\ -1 & 0 & 1 & 0\\ 0 & 0 & 0 & 0 \end{bmatrix}$$

Making the Assembly of all these 4 Elemental Matrices, we get the following Global Stiffness Mtria for the given Truss System,

$$\displaystyle (Eq.3.2) $$ Now, Only force is acting in positive X-Direction at C, and there will be Reactions at A & B.

Therefore, The Force Matrix Can be Assembled as,

$$\displaystyle (Eq.3.3) $$

3.3 Partition the system and solve for the Nodal Displacements
From $$\displaystyle (Eq.3.2) $$ & $$\displaystyle (Eq.3.3)  $$ The Global System os Equation will be,

$$AE \begin{bmatrix} 0 & 0 &0 & 0 & 0 & 0 &  0& 0\\ 0 & 1 & 0 & 0 & 0 & -1 & 0 &0 \\ 0 & 0 & \frac{1}{2\sqrt{2}} & \frac{-1}{2\sqrt{2}} & -\frac{1}{2\sqrt{2}} & \frac{1}{2\sqrt{2}} &0  &0 \\ 0 & 0 & -\frac{1}{2\sqrt{2}} &1+\frac{1}{2\sqrt{2}} & \frac{1}{2\sqrt{2}} & -\frac{1}{2\sqrt{2}} & 0 & -1\\ 0 & 0 & -\frac{1}{2\sqrt{2}} & \frac{1}{2\sqrt{2}} & 1+\frac{1}{2\sqrt{2}} &-\frac{1}{2\sqrt{2}} & -1 &0 \\ 0 &-1 &\frac{1}{2\sqrt{2}}  & -\frac{1}{2\sqrt{2}} & -\frac{1}{2\sqrt{2}} &1+\frac{1}{2\sqrt{2}}  & 0 & 0\\ 0 & 0 & 0 & 0 & -1 & 0 & 1 & 0\\ 0 & 0 & 0 & -1 & 0 & 0 & 0 & 1 \end{bmatrix}.\begin{bmatrix} 0\\ 0\\ 0\\ 0\\ u_{3x}\\ u_{3y}\\ u_{4x}\\ u_{4y} \end{bmatrix}=\begin{bmatrix} r_{1x}\\ r_{1y}\\ r_{2x}\\ r_{2y}\\ 0\\ 0\\ 10\\ 0 \end{bmatrix}$$

As the Displacements are prescribed untill second node, Partitioning the systems after second row and column and further comparing with the following equation,

$$\begin{bmatrix} K_{E} & K_{EF}\\ K_{EF}^{T} & K_{F} \end{bmatrix}\begin{bmatrix} \bar{d}_{E}\\ d_{F} \end{bmatrix}=\begin{bmatrix} r_{E}\\ f_{F} \end{bmatrix}$$

we get, $$\bar{d_{E}}=\begin{bmatrix} 0\\ 0\\ 0\\ 0 \end{bmatrix},\ d_{F}=\begin{bmatrix} u_{3x}\\ u_{3y}\\ u_{4x}\\ u_{4y} \end{bmatrix},\ r_{E}=\begin{bmatrix} r_{1x}\\ r_{1y}\\ r_{2x}\\ r_{2y} \end{bmatrix},\ f_{F}=\begin{bmatrix} 0\\ 0\\ 10\\ 0 \end{bmatrix},\ K_{F}=\begin{bmatrix} 1+\frac{1}{2\sqrt{2}} & -\frac{1}{2\sqrt{2}} & -1 &0 \\ -\frac{1}{2\sqrt{2}} &1+\frac{1}{2\sqrt{2}} &0  & 0\\ -1 & 0 & 1 & 0\\ 0 & 0 & 0 &1 \end{bmatrix},\ K_{EF}=\begin{bmatrix} 0 & 0 & 0 & 0\\ 0 &-1 &0  & 0\\ -\frac{1}{2\sqrt{2}} & \frac{1}{2\sqrt{2}} &  0& 0\\ \frac{1}{2\sqrt{2}} & -\frac{1}{2\sqrt{2}} &0 &-1 \end{bmatrix}$$

Eventually, we can obtain,

$$d_{F}=K_{F}^{-1}(f_{F}-K_{EF}^T.\bar{d}_{E})$$

But $$\bar{d}_{E}$$ is a Zero Matrix, which will yield, $$d_{F}=K_{F}^{-1}.f_{F}$$ Solving this system in MATLAB, Using the following code, we get,

 Matlab Code:

$$\displaystyle (Eq.3.4) $$

These are the Unknown Nodal Displacements.

3.4 Compute the Stresses & Reactions
Now the Reactions can be calculated as,

$$r_{E}=K_{E}.\bar{d}_{E}+K_{EF}.d_{F}$$

But again $$\bar{d}_{E}$$ is a Zero Matrix, Which will give, $$r_{E}=K_{EF}.d_{F}$$.

Using the Following MATLAB Code to obtain the Reaction Matrix,

 Matlab Code:

$$\displaystyle (Eq.3.5) $$

These are the Unknown Reactions. Now We can calculate the Stress in each element using the following formula,

$$\sigma ^e=\frac{E^e}{l^e}\begin{bmatrix} -cos\phi ^{e} &-sin\phi ^{e} & cos\phi ^{e} & sin\phi ^{e} \end{bmatrix}.d^e$$
$$\displaystyle (Eq.3.6) $$

Now we can utilize the above equation to calculate the Stresses for all the elements,

For element 1, $$\phi =90^0, l=1$$ &

$$d^1=\begin{bmatrix} u_{1x}\\ u_{1y}\\ u_{3x}\\ u_{3y} \end{bmatrix}=\frac{1}{2\times 10^{9}}\begin{bmatrix} 0\\ 0\\ 38.2843\\ 10 \end{bmatrix}$$

Using the Following MATLAB code,  Matlab Code:

Therefore,

In the similar fashion, For Element 2, Utilizing $$\displaystyle (Eq.3.6) $$ we get,

For Element 3, Utilizing $$\displaystyle (Eq.3.6) $$ we get,

Finally, For Element 4, Utilizing $$\displaystyle (Eq.3.6) $$ we get,

=Problem 4 : Show that $$\displaystyle \alpha$$ does not equal $$\displaystyle   \beta$$ =

Given
Given

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

$$\alpha =b_{i}\left[ a_{2}^{\prime }b_{j}^{\prime }+a_{2}b_{j}^{\prime \prime } \right]$$ And
 *
$$\displaystyle (Eq. 3.1) $$
 * }
 * }

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

$$\beta =b_{j}\left[ a_{2}^{\prime }b_{i}^{\prime }+a_{2}b_{i}^{\prime \prime } \right]$$
 *
$$\displaystyle (Eq. 3.2) $$
 * }
 * }

Find
Show that $$\alpha \ne \beta $$

Solution
Let

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

$$b_{i}\left( x \right)=\cos \left( ix \right)$$ And
 *
$$\displaystyle (Eq. 3.3) $$
 * }
 * }
 * {| style="width:100%" border="0"

$$b_{j}\left( x \right)=\cos \left( jx \right)$$ Assume that $$i=1$$ and $$j=2$$. Therefore, equations 3.3 and 3.4 become:
 *
$$\displaystyle (Eq. 3.4) $$
 * }
 * }
 * {| style="width:100%" border="0"

$$b_{i}\left( x \right)=\cos \left( x \right)$$
 *
$$$$
 * }
 * }

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

$$b_{j}\left( x \right)=\cos \left( 2x \right)$$ Differentiating:
 *
$$$$
 * }
 * }
 * {| style="width:100%" border="0"

$$b_{i}^{\prime }\left( x \right)=-\sin \left( x \right)$$
 *
$$$$
 * }
 * }

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

$$b_{j}^{\prime }\left( x \right)=-2\sin \left( 2x \right)$$
 *
$$$$
 * }
 * }

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

$$b_{i}^{\prime \prime }\left( x \right)=-\cos \left( x \right)$$
 *
$$$$
 * }
 * }

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

$$b_{j}^{\prime \prime }\left( x \right)=-4\cos \left( 2x \right)$$ Plugging in values and re-arranging
 *
$$$$
 * }
 * }
 * {| style="width:100%" border="0"

$$\alpha =-2a_{2}^{\prime }\sin \left( 2x \right)\cos \left( x \right)-4a_{2}^{\prime \prime }\cos \left( 2x \right)\cos \left( x \right)$$
 *
$$\displaystyle (Eq. 3.5) $$
 * }
 * }

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

$$\beta =-a_{2}^{\prime }\sin \left( x \right)\cos \left( 2x \right)-a_{2}^{\prime \prime }\cos \left( x \right)\cos \left( 2x \right)$$ By inspection of terms $$\displaystyle a_{2}^{\prime }$$ and $$\displaystyle a_{2}^{\prime \prime }$$ it can be shown that
 *
$$\displaystyle (Eq. 3.6) $$
 * }
 * }
 * {| style="width:100%" border="0"

$$\alpha \ne \beta $$
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 *
$$\displaystyle ( Eq. 3.7) $$
 * style= |
 * }

=Problem 5: Equivalent Stiffness of a Spring =

Given
A spring is given in Figure 5.1 as follows:

Find
Show that the equivalent stiffness of a spring aligned in the x direction for the bar of thickness t with a centered square hole shown in Figure 5.1 is can be described by

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

$$ k=\frac{5Etab}{(a+b)l} $$

 *
$$\displaystyle (Eq. 5.1) $$
 * }
 * }

Solution
The left end is constrained, i.e., prescribed displacement is zero at the left end.

Assume there is a force F acting on the right end, and the nodes are numbered in (Fig.5.2).

The Element stiffness matrices are the following:

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

$$ \mathbf{K}^{\left ( 1\right )}  =\begin{bmatrix} k^{\left ( 1 \right )} & -k^{\left ( 1 \right )} \\ -k^{\left ( 1 \right )} & k^{\left ( 1 \right )} \end{bmatrix},

\mathbf{K}^{\left ( 2\right )}  =\begin{bmatrix} k^{\left ( 2 \right )} & -k^{\left ( 2 \right )} \\ -k^{\left ( 2 \right )} & k^{\left ( 2 \right )} \end{bmatrix},

\mathbf{K}^{\left ( 3\right )}  =\begin{bmatrix} k^{\left ( 3 \right )} & -k^{\left ( 3 \right )} \\ -k^{\left ( 3 \right )} & k^{\left ( 3 \right )} \end{bmatrix}$$

 *
$$\displaystyle (Eq. 5.2)$$
 * }
 * }

By direct assembly, the global stiffness matrix is:

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

$$ \mathbf{K}=\begin{bmatrix} k^{\left ( 1 \right )} & -k^{\left ( 1 \right )} & 0 & 0\\ -k^{\left ( 1 \right )} & k^{\left ( 1 \right )} + k^{\left ( 2 \right )} & -k^{\left ( 2 \right )} & 0\\ 0 & -k^{\left ( 2 \right )}  & k^{\left ( 2 \right )} + k^{\left ( 3 \right )}  & -k^{\left ( 3 \right )}   \\ 0 & 0 & -k^{\left ( 3 \right )}  & k^{\left ( 3 \right )} \end{bmatrix} $$

 *
$$\displaystyle (Eq. 5.3) $$
 * }
 * }

The displacements and force matrices for the system are:

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

$$ \mathbf{d}=\begin{bmatrix} 0\\ u_{2}\\ u_{3}\\ u_{4} \end{bmatrix},

\mathbf{f}=\begin{bmatrix} 0\\ 0\\ 0\\ F \end{bmatrix},

\mathbf{r}=\begin{bmatrix} r_{1}\\ 0\\ 0\\ 0\end{bmatrix} $$

 *
$$\displaystyle (Eq. 5.4) $$
 * }
 * }

The global system of equations is given by:

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

$$ \begin{bmatrix} k^{\left ( 1 \right )} & -k^{\left ( 1 \right )} & 0 & 0\\ -k^{\left ( 1 \right )} & k^{\left ( 1 \right )} + k^{\left ( 2 \right )} & -k^{\left ( 2 \right )} & 0\\ 0 & -k^{\left ( 2 \right )}  & k^{\left ( 2 \right )} + k^{\left ( 3 \right )}  & -k^{\left ( 3 \right )}   \\ 0 & 0 & -k^{\left ( 3 \right )}  & k^{\left ( 3 \right )} \end{bmatrix}

\begin{bmatrix} 0\\ u_{2}\\ u_{3}\\ u_{4} \end{bmatrix}=

\begin{bmatrix} r_{1}\\ 0\\ 0\\ F \end{bmatrix} $$

 *
$$\displaystyle (Eq. 5.5) $$
 * }
 * }

Solve the matrix equation above; we can have the following four equations

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

$$ \left\{\begin{matrix} -k^{\left ( 1 \right )}u_{2}=r_{1} \\ u_{2}\left ( k^{\left (1 \right )} + k^{\left (2 \right)} \right )- u_{3}k^{\left (2 \right)}=0 \\ -u_{2}k^{\left (2 \right )} + u_{3}\left ( k^{\left (2 \right )} + k^{\left (3 \right)} \right ) - u_{4}k^{\left (3 \right)}=0  \\ -u_{3}k^{\left (3 \right)}+u_{4}k^{\left (3 \right)}=F \end{matrix}\right. $$

 *
$$\displaystyle (Eq. 5.6) $$
 * }
 * }

Because we have the following relationship:

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

$$ k=\frac{AE}{l} $$

 *
$$\displaystyle (Eq. 5.7) $$
 * }
 * }

Then we can get the following results:
 * {| style="width:100%" border="0"

$$ k^{\left ( 1 \right )}=\frac{10atE}{l}, k^{\left ( 2 \right )}=\frac{5btE}{l}, k^{\left ( 3 \right )}=\frac{10atE}{l} $$

 *
$$\displaystyle (Eq. 5.8) $$
 * }
 * }

Substitute (Eq.5.8) into (Eq.5.6), we can have the following result:

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

$$ u_{4}=\frac{Fl\left ( a+b \right )}{5abtE} $$

 *
$$\displaystyle (Eq. 5.9) $$
 * }
 * }

where $$u_{4}$$ is the displacement of node 4.

And then,

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

$$ \frac{F}{u_{4}}=\frac{5Etab}{\left ( a+b \right )l} $$

 *
$$\displaystyle (Eq. 5.10) $$
 * }
 * }

which yields:

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

$$ k=\frac{5Etab}{\left ( a+b \right )l} $$
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * style="width:2%; padding:1px; border:2px solid #FF0000" |

 *
$$\displaystyle ( Eq. 5.11) $$
 * style= |
 * }

=Problem 6: Three Bar Structure =

Given
Given a three bar structure subjected to the prescribed load at point C equal to $$ 10^3 ~N $$ as shown in Figure 2.19 of the book. The Young's modulus is $$ E = 10^{11} ~Pa $$, the cross-sectional area of the bar BC is $$ 2 \times 10^{-2} ~m^2 $$ and that of BD and BF is $$ 10^{-2} ~m^2 $$. Note that point D is free to move in the x-direction. Coordinates of joints are given in meters.

Find

 * 6.1 Construct the global stiffness matrix and load matrix.

 * 6.2 Partition the matrices and solve for the unknown displacements at Point B and displacement in the x-direction at point D.

 * 6.3 Find stresses in the three bars.

 * 6.4 Find the reactions at nodes C,D,F.

6.1 Construct the global stiffness matrix and load matrix
For Element 3, $$ \phi = 45^o,I = 3,J =4 $$

$$ K^{\left(3\right)} = \frac{10^{-2} \times 10^{11}}{\sqrt{2}} \begin{bmatrix} \frac{1}{2} &\frac{1}{2} &-\frac{1}{2} &-\frac{1}{2}\\ \frac{1}{2} &\frac{1}{2} &-\frac{1}{2} &-\frac{1}{2}\\ -\frac{1}{2} &-\frac{1}{2} &\frac{1}{2} &\frac{1}{2}\\ -\frac{1}{2} &-\frac{1}{2} &\frac{1}{2} &\frac{1}{2} \end{bmatrix} $$

For Element 2, $$ \phi = 90^o,I = 2,J = 4 $$

$$ K^{\left(2\right)} = \frac{2 \times 10^{-2} \times 10^{11}}{1} \begin{bmatrix} 0 &0  &0  &0\\  0  &1  &0 &-1\\  0  &0  &0 &0 \\  0  &-1  &0 &1 \end{bmatrix} $$

For Element 1, $$ \phi = 135^o,I = 1,J =4 $$

$$ K^{\left(1\right)} = \frac{10^{-2} \times 10^{11}}{\sqrt{2}} \begin{bmatrix} \frac{1}{2} &-\frac{1}{2} &-\frac{1}{2} &\frac{1}{2}\\ -\frac{1}{2} &\frac{1}{2} &\frac{1}{2} &-\frac{1}{2}\\ -\frac{1}{2} &\frac{1}{2} &\frac{1}{2} &-\frac{1}{2}\\ \frac{1}{2} &-\frac{1}{2} &-\frac{1}{2} &\frac{1}{2} \end{bmatrix} $$

Global Stiffness Matrix

$$ K = 10^{9} \begin{bmatrix} \frac{1}{2 \sqrt{2}} &-\frac{1}{2 \sqrt{2}} &0  &0  &0  &0  &-\frac{1}{2 \sqrt{2}}  &\frac{1}{2 \sqrt{2}} \\ -\frac{1}{2 \sqrt{2}} &\frac{1}{2 \sqrt{2}} &0  &0  &0  &0  &\frac{1}{2 \sqrt{2}}  &-\frac{1}{2 \sqrt{2}} \\ 0 &0 &0  &0  &0  &0  &0  &0 \\ 0 &0  &0  &2  &0  &0  &0  &-2 \\ 0 &0  &0  &0  &\frac{1}{2 \sqrt{2}}  &\frac{1}{2 \sqrt{2}}  &-\frac{1}{2 \sqrt{2}}  &-\frac{1}{2 \sqrt{2}} \\ 0 &0 &0  &0  &\frac{1}{2 \sqrt{2}}  &\frac{1}{2 \sqrt{2}}  &-\frac{1}{2 \sqrt{2}}  &-\frac{1}{2 \sqrt{2}} \\ -\frac{1}{2 \sqrt{2}} &\frac{1}{2 \sqrt{2}} &0  &0  &-\frac{1}{2 \sqrt{2}}  &-\frac{1}{2 \sqrt{2}}  &\frac{1}{\sqrt{2}}  &0 \\ \frac{1}{2 \sqrt{2}} &-\frac{1}{2 \sqrt{2}} &0  &-2  &-\frac{1}{2 \sqrt{2}}  &-\frac{1}{2 \sqrt{2}}  &0  &\frac{1}{\sqrt{2}}+2 \end{bmatrix} $$

$$ d = \begin{bmatrix} 0\\ 0\\ 0\\ 0\\ u_{3x}\\ 0\\ u_{4x}\\ u_{4y} \end{bmatrix} f = \begin{bmatrix} 0\\ 0\\ 0\\ 0\\ 0\\ 0\\ 10^3\\ 0 \end{bmatrix} r = \begin{bmatrix} r_{1x}\\ r_{1y}\\ r_{2x}\\ r_{2y}\\ 0\\ r_{3y}\\ 0\\ 0 \end{bmatrix} $$

Therefore

$$ load~matrix = \begin{bmatrix} r_{1x}\\ r_{1y}\\ r_{2x}\\ r_{2y}\\ 0\\ r_{3y}\\ 10^3\\ 0 \end{bmatrix} $$

6.2 Partition the matrices and solve
Global system of equations:

$$ 10^9 \begin{bmatrix} \frac{1}{2 \sqrt{2}} &-\frac{1}{2 \sqrt{2}} &0  &0 &\mid &0  &0  &-\frac{1}{2 \sqrt{2}}  &\frac{1}{2 \sqrt{2}} \\ -\frac{1}{2 \sqrt{2}} &\frac{1}{2 \sqrt{2}} &0  &0 &\mid  &0  &0  &\frac{1}{2 \sqrt{2}}  &-\frac{1}{2 \sqrt{2}} \\ 0 &0 &0  &0 &\mid &0  &0  &0  &0 \\ 0 &0 &0  &2 &\mid &0  &0  &0  &-2 \\ - &- &- &- &\mid &- &- &- &- \\ 0 &0 &0  &0 &\mid &\frac{1}{2 \sqrt{2}}  &\frac{1}{2 \sqrt{2}}  &-\frac{1}{2 \sqrt{2}}  &-\frac{1}{2 \sqrt{2}} \\ 0 &0 &0  &0 &\mid &\frac{1}{2 \sqrt{2}}  &\frac{1}{2 \sqrt{2}}  &-\frac{1}{2 \sqrt{2}}  &-\frac{1}{2 \sqrt{2}} \\ -\frac{1}{2 \sqrt{2}} &\frac{1}{2 \sqrt{2}} &0  &0 &\mid &-\frac{1}{2 \sqrt{2}}  &-\frac{1}{2 \sqrt{2}}  &\frac{1}{\sqrt{2}}  &0 \\ \frac{1}{2 \sqrt{2}} &-\frac{1}{2 \sqrt{2}} &0  &-2 &\mid &-\frac{1}{2 \sqrt{2}}  &-\frac{1}{2 \sqrt{2}}  &0  &\frac{1}{\sqrt{2}}+2 \end{bmatrix} \begin{bmatrix} 0\\ 0\\ 0\\ 0\\ --\\ u_{3x}\\ 0\\ u_{4x}\\ u_{4y} \end{bmatrix} = \begin{bmatrix} r_{1x}\\ r_{1y}\\ r_{2x}\\ r_{2y}\\ --\\ 0\\ r_{3y}\\ 10^3\\ 0 \end{bmatrix} $$

Partitions

$$ \bar{d}_{E} = \begin{bmatrix} u_{1x}\\ u_{1y}\\ u_{2x}\\ u_{2y} \end{bmatrix} = \begin{bmatrix} 0\\ 0\\ 0\\ 0 \end{bmatrix},

d_{F}= \begin{bmatrix} u_{3x}\\ 0\\ u_{4x}\\ u_{4y} \end{bmatrix},

r_{E} = \begin{bmatrix} r_{1x}\\ r_{1y}\\ r_{2x}\\ r_{2y} \end{bmatrix} ,

f_{F} = \begin{bmatrix} 0\\ r_{3y}\\ 10^3\\ 0 \end{bmatrix} $$

$$ K_{EF} = \begin{bmatrix} 0 &0 &-\frac{1}{2 \sqrt{2}}  &\frac{1}{2 \sqrt{2}} \\ 0 &0 &\frac{1}{2 \sqrt{2}}  &-\frac{1}{2 \sqrt{2}} \\ 0 &0 &0  &0 \\ 0 &0  &0  &-2\ \end{bmatrix},

K_{EF}^{T} = \begin{bmatrix} 0 &0 &0  &0 \\ 0 &0  &0  &0 \\ -\frac{1}{2 \sqrt{2}} &\frac{1}{2 \sqrt{2}}  &0  &0 \\ \frac{1}{2 \sqrt{2}} &-\frac{1}{2 \sqrt{2}} &0  &-2 \end{bmatrix},

K_{F} = \begin{bmatrix} \frac{1}{2 \sqrt{2}} &\frac{1}{2 \sqrt{2}} &-\frac{1}{2 \sqrt{2}}  &-\frac{1}{2 \sqrt{2}} \\ \frac{1}{2 \sqrt{2}} &\frac{1}{2 \sqrt{2}} &-\frac{1}{2 \sqrt{2}}  &-\frac{1}{2 \sqrt{2}} \\ -\frac{1}{2 \sqrt{2}} &-\frac{1}{2 \sqrt{2}} &\frac{1}{\sqrt{2}}  &0 \\ -\frac{1}{2 \sqrt{2}} &-\frac{1}{2 \sqrt{2}} &0  &\frac{1}{\sqrt{2}}+2 \end{bmatrix},

K_{E} = \begin{bmatrix} \frac{1}{2 \sqrt{2}} &-\frac{1}{2 \sqrt{2}} &0  &0 \\ -\frac{1}{2 \sqrt{2}} &\frac{1}{2 \sqrt{2}} &0  &0 \\ 0 &0 &0  &0 \\ 0 &0  &0  &2 \end{bmatrix} $$

Unknown displacement at B is solved by

$$ K_{EF}^{T}\bar{d}_{E} + K_{F}d_{F} = f_{F} $$

Subtracting the first term from both sides of the above equation and premultiply by $$ K_F^{-1} $$, we obtain

$$ d_F = K_F^{-1} \left( f_F - K_{EF}^T \bar{d}_E \right) $$

Solving for the displacements we get: $$ d_{F}= 10^{-6} \begin{bmatrix} 1+2\sqrt{2}\\ 0\\ \frac{1}{2}+2\sqrt{2}\\ \frac{1}{2} \end{bmatrix} ~m $$

6.3 Find stresses in the three bars
Element 3 Stress

$$ \sigma = 0 ~Pa $$

Element 2 Stress

$$ \sigma = \frac{E}{l} \begin{bmatrix} 0 &-1 &0  &1 \end{bmatrix} 10^{-6} \begin{bmatrix} \frac{1}{2} + 2\sqrt{2}\\ \frac{1}{2}\\ 0\\ 0 \end{bmatrix} = -5 \times 10^4 ~Pa $$

Element 1 Stress

$$ \sigma = \frac{E}{l} \begin{bmatrix} \frac{\sqrt{2}}{2} &-\frac{\sqrt{2}}{2} &\frac{\sqrt{2}}{2}  &\frac{\sqrt{2}}{2} \end{bmatrix} 10^{-6} \begin{bmatrix} \frac{1}{2} + 2\sqrt{2}\\ \frac{1}{2}\\ 0\\ 0 \end{bmatrix} = \sqrt{2} \times 10^5 ~Pa $$

6.4 Find the reactions at nodes C,D,F.
Reactions can be solved using this equation and the global matrix

$$ r_E = K_E \bar{d}_E + K_{EF} d_F $$

$$ load~matrix = \begin{bmatrix} r_{1x}\\ r_{1y}\\ r_{2x}\\ r_{2y}\\ 0\\ r_{3y}\\ 10^3\\ 0 \end{bmatrix}

= \begin{bmatrix} -10^3\\ 10^3\\ 0\\ -10^3\\ 0\\ 0\\ 10^3\\ 0 \end{bmatrix} ~ N $$

=Problem 7: Solving using WRF= Solving a differential equation using Weight Residual Form. From lecture 17

Given
The partial differential equation is:

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

$$ \frac{\partial}{\partial x}\left(2\frac{\partial u}{\partial x}\right) + 3 = 0 $$
 *
$$\displaystyle (Eq. 7.1) $$
 * }
 * }

The boundary conditions are:

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

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

As weighting function should be used:

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

$$ b_{i} = b_{i}=\left ( x \right )^{i} $$
 *
$$\displaystyle (Eq. 7.2) $$
 * }
 * }

Find

 * 7.1 Find an approximate solution of the form $$U^{h}=\sum_{i=0}^{n}d_{i}b_{i}$$ for n=2
 * 7.2 Find two equations that enforce the boundary conditions
 * 7.3 Project the weight residues
 * 7.4 Display the equations in matrix form
 * 7.5 Solve for $$\displaystyle d $$
 * 7.6 Construct $$U^{h}$$ and plot $$U^{h}$$ and $$u$$
 * 7.7 Repeat 7.1 to 7.6 for n = 4 and n = 6

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

$$d_{i}=\left[ \begin{matrix} d_{0} & d_{1} & d_{2} \\ \end{matrix} \right]$$
 *
$$\displaystyle (Eq.7.3) $$
 * }
 * }

and

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

$$b_{i}=\left[ \begin{matrix} 1 & \left( x \right) & \left( x \right)^{2} \\ \end{matrix} \right]$$
 *
$$\displaystyle (Eq.7.4) $$
 * }
 * }

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

$$U^{h}=d_{0}+d_{1}\left( x \right)+d_{2}\left( x \right)^{2}$$
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 *
$$\displaystyle (Eq. 1.5) $$
 * style= |
 * }

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

$$u\left( 1 \right)=0=d_{0}+d_{1}+d_{2}$$ Differentiating
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 *
$$\displaystyle (Eq. 7.6) $$
 * style= |
 * }
 * {| style="width:100%" border="0"

$$\frac{du}{dx}=0+d_{1}+2d_{2}x$$
 *
$$$$
 * }
 * }

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

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

$$\frac{du}{dx}(0)=-4=d_{1}$$

 *
$$\displaystyle (Eq. 7.7) $$
 * style= |
 * }

7.3 Project the weight residues
In this case changing the function $$u\left( x \right)$$ by the approximated function $$U^{h}$$ in the PDE (Eq.7.1), we found that $$P\left( u \right)$$ is:

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

$$P\left( U^{h} \right)=2\frac{\partial ^{2}U^{h}}{\partial x^{2}}+3=4d_{2}+3$$
 *
$$\displaystyle (Eq.7.8) $$
 * }
 * }

The (Eq.7.8) can be written as:

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

$$P\left( U^{h} \right)=\left[ \begin{matrix} 0 & 0 & 4 \\ \end{matrix} \right]\left[ \begin{matrix} d_{0} \\ d_{1} \\ d_{2} \\ \end{matrix} \right]+3$$
 *
$$\displaystyle (Eq.7.9) $$
 * }
 * }

Projecting the residue we have:

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

$$ \int_{a}^{b}b(x)P\left ( U^{h} \right )= 0 $$ After substitution of equations 7.4 and 7.9) in 7.10, the following equation is obtained
 *
$$\displaystyle (Eq. 7.10) $$
 * }
 * }
 * {| style="width:100%" border="0"

$$\int_{0}^{1}{\left[ \begin{matrix} 1 \\   x  \\ x^{2} \\ \end{matrix} \right]}\left[ \begin{matrix} 0 & 0 & 4 \\ \end{matrix} \right]dx\left[ \begin{matrix} d_{0} \\ d_{1} \\ d_{2} \\ \end{matrix} \right]=-3\int_{0}^{1}{\left[ \begin{matrix} 1 \\   x  \\ x^{2} \\ \end{matrix} \right]}dx$$
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 *
$$\displaystyle (Eq. 7.11) $$
 * style= |
 * }

7.4 Display the equations in matrix form
Performing the product and then the integration indicated in equation 7.11, we have
 * {| style="width:100%" border="0"

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

$$\left[ \begin{matrix} 0 & 0 & 4 \\   0 & 0 & 2  \\   0 & 0 & \frac{4}{3}  \\ \end{matrix} \right]\left[ \begin{matrix} d_{0} \\ d_{1} \\ d_{2} \\ \end{matrix} \right]=\left[ \begin{matrix} -3 \\   -\frac{3}{2}  \\ -1 \\ \end{matrix} \right]$$
 *
$$\displaystyle (Eq. 7.12) $$
 * style= |
 * }

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

7.5 Solve for $$\displaystyle d $$
In equation 7.12 we have three equations; but, as we need to enforce the boundary conditions we take the only one of them and solve it together with equations 7.6 and 7.7. From these equations we now have the system:

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

$$\left[ \begin{matrix} 1 & 1 & 1 \\   0 & 1 & 2  \\   0 & 0 & 2  \\ \end{matrix} \right]\left[ \begin{matrix} d_{0} \\ d_{1} \\ d_{2} \\ \end{matrix} \right]=\left[ \begin{matrix} 0 \\   -4  \\   -3  \\ \end{matrix} \right]$$
 *
$$\displaystyle (Eq.7.13) $$
 * }
 * }

The latter is a linear system and can be solved by using any calculator. After solving the system in (Eq.7.13) we obtain d:
 * {| style="width:100%" border="0"

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

$$\left[ \begin{matrix} d_{0} \\ d_{1} \\ d_{2} \\ \end{matrix} \right]=\left[ \begin{matrix} \frac{5}{2} \\ -1 \\   -\frac{4}{3}  \\ \end{matrix} \right]$$
 *
$$$$
 * style= |
 * }

7.6 Construct $$U^{h}$$ and plot $$U^{h}$$ and $$u$$
Replacing $$d_{i}$$ in equation 7.5 we obtain the approximated solution:

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

$$U^{h}=\frac{5}{2}-x-\frac{4}{3}x^{2}$$
 *
$$\displaystyle (Eq.7.14) $$
 * }
 * }

On the other hand, the exact solution of the PDE (Eq.7.1) with the given boundary condition is:

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

$$ u\left ( x \right )=-\frac{3}{4}x^{2}-4x+\frac{19}{4} $$
 *
$$\displaystyle (Eq.7.15) $$
 * }
 * }

We observe that the approximated and the exact solution are not the same. Perhaps they would be closer if we increased the number of terms. Figure 7.1 we show the exact and the approximated solution.

7.7 Repeat 7.1 to 7.6) for i = 4 and i = 6
Now we repeat the procedure using n = 4 and n=6 in Eq.7.4

As the procedure is the same shown before, we develop a routine using Matlab to perform the products, the integration and to solve the system of equations. The routine is shown below in the appendix. Figures 7.2 and 7.3 show the results of n=4 and n=6, respectively. Figure 7.4 shows the error versus the number of terms.

Appendix
 Matlab Code:

=Problem 8: Weak Form Proof=

Given
The Strong form is:
 * {| style="width:100%" border="0"

$$\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} \left ( AE\frac{\mathrm{d} u}{\mathrm{d} x} \right ) + 2x=0$$ on      1<x<3,
 *
$$\displaystyle (Eq. 8.1a)  $$
 * }
 * }

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

$$ \displaystyle \sigma (1)=\left ( E\frac{\mathrm{d} u}{\mathrm{d} x} \right )_{x=1}=0.1, $$

 *
$$\displaystyle (Eq. 8.1b)  $$
 * }
 * }

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

$$ \displaystyle u(3)=0.001 $$

 *
$$\displaystyle (Eq. 8.1c)  $$
 * }
 * }

Find
Show that the weak form is:
 * {| style="width:100%" border="0"

$$ \int_{1}^{3}\frac{\mathrm{d} w}{\mathrm{d} x}AE\frac{\mathrm{d} u}{\mathrm{d} x}dx = -0.1\left ( wA \right )_{x=1}+\int_{1}^{3}2xwdx,\forall w$$ with $$w(3)=0$$

 *
$$\displaystyle (Eq. 8.2) $$
 * }
 * }

Solution
Equation (8.1b) is a condition on the derivative of u(x), so it is a natural boundary conditon; (8.1c) is a condition on u(x), so it is an essential boundary condition. Therefore, as the weight function must vanish on the essential boundaries, we consider all smooth weight functions w(x) such that w(3)=0. The trial solutions must satisfy the essential boundary condition u(3) = 0.001.

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

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

 *
$$\displaystyle (Eq. 8.3) $$
 * }
 * }

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

$$ [wA(E\frac{\mathrm{d} u}{\mathrm{d} x}-0.1)]_{x=1} = 0, \forall w(1). $$

 *
$$\displaystyle (Eq. 8.4) $$
 * }
 * }

Next we integrate (Eq.8.3) by parts, then we get the following result:

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

$$ \int_{1}^{3} [w (\frac{\mathrm{d} }{\mathrm{d} x}AE\frac{\mathrm{d} u}{\mathrm{d} x})]dx = \left (wAE\frac{\mathrm{d} u}{\mathrm{d} x} \right )\mid _{x=1}^{x=3} - \int_{1}^{3}  \frac{\mathrm{d} w }{\mathrm{d} x}AE\frac{\mathrm{d} u}{\mathrm{d} x}dx $$

 *
$$\displaystyle (Eq. 8.5) $$
 * }
 * }

We have constructed the weight functions so that w(3)=0; therefore, the first term on the RHS of (Eq.8.5) vanishes at x=3. Substituting (Eq.8.5) into (Eq.8.3) gives:

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

$$ -\left (wAE\frac{\mathrm{d} u}{\mathrm{d} x} \right )\mid _{x=1} - \int_{1}^{3}  \frac{\mathrm{d} w }{\mathrm{d} x}AE\frac{\mathrm{d} u}{\mathrm{d} x}dx + \int_{1}^{3} 2xwdx=0, \forall w(x) $$ with w(3)=0. Substituting (Eq.8.4) into the first term of (Eq.8.6) gives:
 *
$$\displaystyle (Eq. 8.6) $$
 * }
 * }

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

$$ \int_{1}^{3} \frac{\mathrm{d} w }{\mathrm{d} x}AE\frac{\mathrm{d} u}{\mathrm{d} x}dx = -0.1(wA)_{x=1} + \int_{1}^{3} 2xwdx, \forall w(x) $$ with w(3)=0.
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 *
$$\displaystyle ( Eq. 8.7) $$
 * style= |
 * }

=Problem 9: A solution to the weak form in Problem 3.1=

Consider a trial (candidate) solution of the form $$ u\left ( x \right ) = \alpha_{0} + \alpha_{1}\left ( x-3 \right ) $$ and a weight function of the same form. Obtain a solution to the weak form in Problem 3.1. Check the equilibrium equation in the strong form in Problem 3.1; is it satisfied?

Check the natural boundary condition; is it satisfied?

Solution
$$ u\left ( x \right ) = \alpha_{0} + \alpha_{1}\left ( x-3 \right ) $$

$$ w\left ( x \right ) = \beta_{0} + \beta_{1}\left ( x-3 \right ) $$

Using boundary conditions from Problem 3.9

$$ w \left(3 \right) = 0 \therefore  \beta_{0} = 0 $$

$$ u \left(3 \right) = \alpha_{0} = 0.001 $$

Only one unknown parameter and one arbitrary parameter remain

$$ u\left ( x \right ) = 0.001 + \alpha_{1}\left ( x-3 \right ) \frac{du}{dx} = \alpha_{1} $$

$$ w\left( x \right) = \beta_{1} \left(x - 3 \right) \frac{dw}{dx} = \beta_{1} $$

Now substitute into Problem 3.1 weak form

$$ \int_{1}^{3} \frac{\mathrm{d} w }{\mathrm{d} x}AE\frac{\mathrm{d} u}{\mathrm{d} x}dx =  -0.1(wA)_{x=1} + \int_{1}^{3} 2xwdx$$

$$ \int_{1}^{3} \beta_{1}\alpha_{1}EAdx + 0.1(\beta_{1} \left(x-3 \right)A)_{x=1} - \int_{1}^{3} 2\beta_{1}x \left(x-3 \right)dx = 0 $$

Integrating and factoring out $$ \beta_1 $$ we get:

$$ \beta_{1} \left( 2 \alpha_{1} EA - 0.2A + \frac{20}{3} \right) = 0 $$

The $$ \beta_1 $$ goes to zero and $$ \alpha_1 $$ can be solved:

$$ \alpha_{1} = \frac{0.1A - \frac{10}{3}}{EA} $$

Therefore the solution to the weak form is

$$ u \left( x \right) = 0.001 + \frac{0.1A - \frac{10}{3}}{EA} \left( x-3 \right) $$

Checking for equilibrium in the strong form, values are substituted into

$$ \frac{d}{dx} \left( AE \frac{du}{dx} \right) + 2x = 0 $$

This yields $$ 2x = 0 $$ which does not satisfy.

Checking the natural boundary condition, this equation is used

$$ \sigma \left(1 \right) = \left(E \frac{du}{dx} \right)_{x=1} = 0.1 $$

Substituting in $$ \alpha_1 $$ we get

$$ 0.1 - \frac{\frac{10}{3}}{A} = 0.1 $$

Thus, this boundary condition will satisfy only for large values of $$ A $$

=Problem 10: Fish and Belytschko Problem 3.4=

Repeat Problem 3.3 with the trial solution $$ u\left ( x \right ) = \alpha_{0} + \alpha_{1}\left ( x-3 \right ) + \alpha_{2}\left ( x-3 \right )^{2}$$.

Find
Consider a trial (candidate) solution of the form $$ u\left ( x \right ) = \alpha_{0} + \alpha_{1}\left ( x-3 \right ) + \alpha_{2}\left ( x-3 \right )^{2}$$ and a weight function of the same form. Obtain a solution to the weak form in Problem 3.1. Check the equilibrium equation in the strong form in Problem 3.1; is it satisfied?

Check the natural boundary condition; is it satisfied?

Given
The weak from is given by $$ \int_{1}^{3} \frac{\mathrm{d} w }{\mathrm{d} x}AE\frac{\mathrm{d} u}{\mathrm{d} x}dx = -0.1(wA)_{x=1} + \int_{1}^{3} 2xwdx, \forall w $$ with $$  w \left(3 \right) = 0 $$

Solution
The trial solution $$ u\left ( x \right )$$ and the weight function $$ w\left ( x \right )$$ can be written as follows:

$$ u\left ( x \right ) = \alpha_{0} + \alpha_{1}\left ( x-3 \right ) + \alpha_{2}\left ( x-3 \right )^{2}$$

$$ w\left ( x \right ) = \beta_{0} + \beta_{1}\left ( x-3 \right ) + \beta_{2}\left ( x-3 \right )^{2}$$

When:

$$ w \left(3 \right) = 0; \Rightarrow \beta_{0} = 0 $$

and for:

$$ u \left(3 \right) \Rightarrow \alpha_{0} = 0.001 $$

$$ u\left ( x \right )$$ and $$ w\left ( x \right )$$ can therefore be re-written as follows:

$$ u\left ( x \right ) = 0.001 + \alpha_{1}\left ( x-3 \right )+\alpha_{2}\left ( x-3 \right )^{2}$$

$$ w\left( x \right) = \beta_{1} \left(x - 3 \right)+\beta_{2}\left ( x-3 \right )^{2}$$

Whith their respective derivatives:

$$\frac{du}{dx} = \alpha_{1}+2\alpha_{2}\left(x-3\right)$$

$$\frac{dw}{dx} = \beta_{1}+2\beta_{2}\left(x-3\right)$$

Now substituting into the weak form yields:

$$ \int_{1}^{3} AE[\beta_{1}+2\beta_{2}\left(x-3\right)][\alpha_{1}+2\alpha_{2}\left(x-3\right)]dx = -0.1[A(\beta_{1}\left(x-3\right)+\beta_{2}\left(x-3\right)^{2})]_{x=1} + \int_{1}^{3} 2x[\beta_{1}\left(x-3\right)+\beta_{2}\left(x-3\right)^{2}]dx$$

which can be re-arranged as:

$$ \int_{1}^{3} AE[\beta_{1}+2\beta_{2}\left(x-3\right)][\alpha_{1}+2\alpha_{2}\left(x-3\right)]dx - 0.1[A(\beta_{1}\left(x-3\right)+\beta_{2}\left(x-3\right)^{2})]_{x=1} - \int_{1}^{3} 2x[\beta_{1}\left(x-3\right)+\beta_{2}\left(x-3\right)^{2}]dx = 0$$

Evaluating the integral we get:

$$ AE \left(2\alpha_{1}\beta_{1}-4\alpha_{1}\beta_{2}-4\alpha_{2}\beta_{1}+\frac{32\alpha_{2}\beta_{2}}{3}\right)- 0.1[A(4\beta_{2}-2\beta_{1})] - \left(8\beta_{2}-\frac {20\beta_{1}}{3}\right)=0 $$

Factoring out $$\left( \beta_{1}\right)$$ and $$\left( \beta_{2}\right)$$ we get:

$$\beta_{1}\left[2AE\left(\alpha_{1}-2\alpha_{2}\right)-0.2A+\frac{20}{3}\right]+\beta_{2}\left[4AE\left(\frac{8}{3}\alpha_{2}-\alpha_{1}\right)+0.4A-8\right]=0$$

$$\displaystyle {{\beta }_{1}}$$ and $$\displaystyle {{\beta }_{2}}$$ must be zero since they were arbitrarely chosen and because the above equation must hold. Therefore we can write both terms of the above equation equal to zeor as folows:

$$ \left[2AE\left(\alpha_{1}-2\alpha_{2}\right)-0.2A+\frac{20}{3}\right]=0$$

and

$$ \left[4AE\left(\frac{8}{3}\alpha_{2}-\alpha_{1}\right)+0.4A-8\right]=0 $$

Solving for $$\displaystyle {{\alpha }_{1}}$$ and $$\displaystyle {{\alpha }_{2}}$$ we find that:

$$\displaystyle {{\alpha }_{1}}=\frac{\frac{116}{3}+0.2A}{2AE} $$

and

$$\displaystyle {{\alpha }_{2}}=\frac{8}{AE} $$

The solution to the weak form can be written as

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

$$ u(x)= 0.001+\frac{\frac{116}{3}+0.2A}{2AE}(x-3)-\frac{8}{AE}(x-3)^{2} $$
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 *
$$\displaystyle (Eq. 10.1) $$
 * style= |
 * }

Checking if equilibrium in the strong form is satisfied, substituted into the equations:

$$ \frac{d}{dx} \left( AE \frac{du}{dx} \right) + 2x = 0 $$

Yields the following:

$$\displaystyle {2x-16=0}$$

which does not satisfy.

Checking if the natural boundary condition is satisfied, the following equation can be used:

$$ \sigma \left(1 \right) = \left(E \frac{du}{dx} \right)_{x=1} = 0.1 $$

=Problem 11: Solving using WRF=

Solving a differential equation using Weight Residual Form. From lecture 17

Given
The partial differential equation is:

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

$$ \frac{\partial}{\partial x}\left(2\frac{\partial u}{\partial x}\right) + 3 = 0 $$
 *
$$\displaystyle (Eq.11.1) $$
 * }
 * }

The boundary conditions are:

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

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

As weighting funtion should be used:

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

$$ b_{i} =1+a_{i}\cos \left ( ix \right )+b_{i}\sin \left ( ix \right ) $$
 *
$$\displaystyle (Eq.11.2) $$
 * }
 * }

Find

 * 11.1 Find an approximate solution of the form $$U^{h}=\sum_{i=0}^{n}d_{i}b_{i}$$ for n=2
 * 11.2 Find two equations that enforce the boundary conditions
 * 11.3 Project the weight residues
 * 11.4 Display the equations in matrix form
 * 11.5 Solve for $$\displaystyle d $$
 * 11.6 Construct $$U^{h}$$ and plot $$U^{h}$$ and $$u$$
 * 11.7 Repeat 1.1 to 1.6 for n = 4 and n = 6

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

$$ d_{i}=\begin{bmatrix} d0 & d1 & d2 & d3 & d4

\end{bmatrix} $$
 *
$$\displaystyle (Eq.11.3) $$
 * }
 * }

and

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

$$ b_{i}=\begin{bmatrix} 1 & \cos\left ( x\right) & \sin\left ( x\right) & \cos\left ( 2x\right) & \sin\left ( 2x\right)\end{bmatrix} $$
 *
$$\displaystyle (Eq.11.4) $$
 * }
 * }

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

$$ U^{h}=d0+d1\cos\left ( x \right )+d2\sin\left ( x\right )+ d3\cos\left ( 2x \right )+d4\sin\left ( 2x\right ) $$
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * style="width:2%; padding:1px; border:2px solid #FF0000" |

 *
$$\displaystyle (Eq. 11.5) $$
 * style= |
 * }

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

$$ u\left(1\right)=0=d0+d1\cos\left(1\right)+d2\sin\left(1\right)+d3\cos\left(2\right)+d4\sin\left(2\right)$$
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * style="width:2%; padding:1px; border:2px solid #FF0000" |

 *
$$\displaystyle (Eq. 11.6) $$
 * style= |
 * }

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

$$ \frac{du}{dx}(0)=-4 = d2+2d4$$
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * style="width:2%; padding:1px; border:2px solid #FF0000" |

 *
$$\displaystyle (Eq. 11.7) $$
 * style= |
 * }

11.3 Project the weight residues
In this case changing the function u(x) by the approximated function $$U^{h}$$ in the PDE (Eq.11.1), we found that P(u) is:

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

$$ P\left ( U^{h} \right )=2\frac{\partial^2 U^{h}}{\partial x^2}+3=2d2+3 $$
 *
$$\displaystyle (Eq.11.8) $$
 * }
 * }

The (Eq.11.8) can be written as:

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

$$ P\left ( U^{h} \right )= -2\begin{bmatrix}0 & \cos\left(x\right) & \sin\left(x\right) & 4\cos\left(2x\right) &4\sin\left(2x\right)\end{bmatrix}\begin{bmatrix} d0\\ d1\\ d2\\ d3\\ d4\end{bmatrix}+3 $$
 *
$$\displaystyle (Eq.11.9) $$
 * }
 * }

Projecting the residue we have:

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

$$ \int_{a}^{b}b(x)P\left ( U^{h} \right )= 0 $$ After substitution of (Eq.11.4) and (Eq.11.9) in (Eq.11.10) the following equation is obtained
 *
$$\displaystyle (Eq. 11.10) $$
 * }
 * }
 * {| style="width:100%" border="0"

$$ 2\int_{0}^{1}\begin{bmatrix} 1\\ \cos\left (x\right ) \\ \sin\left ( x\right )\\ \cos\left (2x\right ) \\ \sin\left ( 2x\right ) \end{bmatrix}\begin{bmatrix} 0 &\cos\left(x\right) & \sin\left(x\right)&\cos\left(2x\right) & \sin\left(2x\right)\end{bmatrix}dx\begin{bmatrix} d0\\ d1\\ d2\\ d3\\ d4\end{bmatrix}=3\int_{0}^{1} \begin{bmatrix} 1\\ \cos\left ( x\right ) \\
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * style="width:2%; padding:1px; border:2px solid #FF0000" |

\sin\left ( x \right ) \\ \cos\left ( 2x\right ) \\

\sin\left ( 2x \right ) \end{bmatrix}dx $$

 *
$$\displaystyle (Eq. 11.11) $$
 * style= |
 * }

11.4 Display the equations in matrix form
Performing the product and then the integration indicated in (Eq.1.11) we have:
 * {| style="width:100%" border="0"

$$ \begin{bmatrix} 0 &1.68 & 0.91 &3.63& 5.66\\ 0 & 1.45& 0.70 & 3.55&4.49\\ 0 &0.70 & 0.54 &0.81 & 3.17\\ 0 & 0.88& 0.20 & 3.24& 1.65\\ 0 & 1.12& 0.79& 1.65& 4.75\end{bmatrix}\begin{bmatrix} d0\\ d1\\ d2\\ d3\\ d4\end{bmatrix}=\begin{bmatrix} 3\\ 2.52\\ 1.37\\ 1.36\\ 2.12\end{bmatrix} $$
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * style="width:2%; padding:1px; border:2px solid #FF0000" |

 *
$$\displaystyle (Eq. 11.12) $$
 * style= |
 * }

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

11.5 Solve for $$ d $$
In (Eq.11.12) we have three equations; but, as we need to enforce the boundary conditions we take the only one of them and solve it together with (Eq.11.6) and (Eq.11.7). From these equations we now have the system:

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

$$ \begin{bmatrix} 0 &\cos\left(1\right)   & \sin\left(1\right) &\cos\left(2\right)   & \sin\left(2\right)\\ 0 &0 &1 &0  & 2\\ 0  & -1.68  & -0.91 & -3.63  & -5.66\\ 0 &-1.45 & -0.70 & -3.55   & -4.49\\ 0 & -0.70& -0.54 & -0.81   & -3.17 \end{bmatrix}\begin{bmatrix} d0\\ d1\\ d2\\ d3\\

d4\end{bmatrix}=\begin{bmatrix} 0\\ -4\\ -3\\ -2.52\\ -1.37\end{bmatrix}$$
 *
$$\displaystyle (Eq.11.13) $$
 * }
 * }

After solving the system in (Eq.11.13), by using the code shown in apedix, we obtain d:
 * {| style="width:100%" border="0"

$$\begin{bmatrix} d0\\ d1\\ d2\\ d3\\ d4\end{bmatrix}=\begin{bmatrix} 0.67\\ 4.88\\ -4.72\\ -0.80\\ -0.36\end{bmatrix}$$
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * style="width:2%; padding:1px; border:2px solid #FF0000" |

 *
$$$$
 * style= |
 * }

11.6 Construct $$U^{h}$$ and plot $$U^{h}$$ and $$u$$
Replacing d in (Eq.11.5) we obtain the approximated solution:

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

$$ U^{h}=0.67+4\cos\left(x\right)-4.72\sin\left(x\right)-0.8\cos\left(2x\right)-0.36\sin\left(2x\right) $$
 *
$$\displaystyle (Eq.11.14) $$
 * }
 * }

On the other hand, the exact solution of the PDE (Eq.11.1) with the given boundary condition is:

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

$$ u\left ( x \right )=-\frac{3}{4}x^{2}-4x+\frac{19}{4} $$
 *
$$\displaystyle (Eq.11.15) $$
 * }
 * }

11.7 Repeat 11.1 to 11.6) for i = 4 and i = 6
Now we repeat the procedure using i = 4 and i=6 in Eq.11.4

As the procedure is the same shown before, we develop a routine using Matlab to perform the products, the integration and to solve the system of equations. The routine is show in the appendix.

In Fig.11.1 we show the exact and the approximated solution for i = 2,4, and 6. In that figure, it can be observed that all the numerical solutions fits very well the analytical solution.

We found a small error for i = 2 (0.0003); however the error lightly increases for i = 4 and decreases again for i = 6. As all the solutions are close to the analytical, we believe that this error is mainly due to round-off error which is increased with i = 4 and i = 6 due to the higher number of operations that the program has to perform. Fig. 11.2 shows the error variation as a function of i

Appendix
 Matlab Code:

=Contributing Members=
 * James Roark
 * Abhijeet Bhalerao
 * Fernando Casanova
 * Yu Hao
 * Harrison Sheffield
 * Cedric Adam

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

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Classical Chinese/Exercise 1: Structure

{Which of these has correct word order? - 吾食不一果 + 吾食一果 - 吾一果食 - 一果食吾
 * type=""}

{Which of these has correct word order? - 老人不跑能 + 老人不能跑 - 老人跑不能 - 不能老人跑
 * type=""}

{Which of these has correct word order? - 吾施(give)台於書 - 吾台施(give)書 - 施(give)台吾書 + 吾施(give)書於台
 * type=""}

{Which of these has the correct word order? - 走兔市 - 走市兔 - 兔市走 + 兔走市
 * type=""}

{Which of these has the correct word order? + 鳥高飛 - 鳥飛高 - 高飛鳥 - 飛高鳥
 * type=""}

{Where should the character 治 be placed if you wanted to write the sentence the king rules the country? - 治王國 + 王治國 - 國治王 - 王國治
 * type=""}

{Where should the character 大 be placed if you wanted to write the sentence I see a large/big mountain? - 見吾大山 - 吾見山大 - 大山吾見 + 吾見大山
 * type=""}

{Where should the character 鳥 be placed if you wanted to write the sentence the black bird eats? + 黑鳥食 - 食鳥黑 - 黑食鳥 - 鳥黑食
 * type=""}

{Which sentence would be considered past tense? - 今老走 - 君說臣 - 吾施金 + 昔人不能飛
 * type=""}

{Which sentence would be considered past tense? - 來年兵攻國 + 昨兵攻國 - 國兵攻城 - 兵攻國
 * type=""}

{Which sentence would be considered past tense? - 今國無君 + 前代有君 - 此日無須食 - 子將載國史
 * type=""}

{Which sentence would be considered future tense? - 兵將攻京 - 昨兵攻京 - 兵攻京 + 兵將攻京
 * type=""}

title: Classical Chinese/Exercise 1: Structure


text:
Teaching EFL Listening via FUN WITH ENGLISH Books

This is a collection of lesson plans and related materials designed for 7th and 8th grade teachers in China who are using the FUN WITH ENGLISH [FWE] book series to teach listening English as a Foreign Language [EFL]. Hopefully this will also be valuable for other EFL teachers in China as well as any teacher of EFL (and likely ESL) around the world. Finally, this collection is not perfect and will always be open to improvements. Please make changes as you see fit.

To-do

 * Add a nutshell up top.


text:
Talk:Peace studies/Peace in Islam

no topic hahaha zahid amin pakistan

[[Media:Example.ogg]]
islam is a true religion zahid amin frm pakistan

IDEA OF SHERSHAH ZARGARZADA PEACE IS ALSO MAKE STABILITY IN EVERY SOCIETY

title: Talk:Peace studies/Peace in Islam


text:
User:Supasin suic

SCM Software

I am working in manufacturing (Factory)

Intelex's web-based Supplier Management Software System takes a complete lifecycle approach to supplier relationship management (SRM). The software considers each stage of the process from the initial vendor registration and setup, to ongoing supplier performance assessment, to implementing corrective/preventive actions and measuring the success of those actions. In addition, it can be categorize suppliers by company name, supplier number, performance rating, etc.

The benefit of this packaging

1.We can save time and it is very convenience due to the web- based type ‘s package, every staffs are able to instantly access, change, or update information, sustaining an up to date and accurate supplier list anywhere there is an internet connection.

2.we can evaluate the supplier rating in order to know performance to identify top performing/underachieving suppliers.

3.We can do streamline workflow by tracking all supplier/vendor data in a central portal that's accessible organization-wide in order to improve productivity in your business while cutting down on costs --

hi im Charyanaan and Supasin Group

What is SugarCRM?

SugarCRM Inc. is the 800-pound gorilla in the open-source CRM category. Founded in 2004 by John Roberts, Clint Oram and Jacob Taylor, the Sugar open-source code has been downloaded more than 3 million times. The company has received $26 million in venture financing and employs more than 100 people. More than 12,000 companies use SugarCRM including Honeywell International, Starbucks Corp., First Federal Bank and BDO Seidman LLP. SugarCRM is written in PHP and is compatible with the MySQL database. You can easily import data from another CRM solution, or even just a standard spreadsheet using the standard Comma Separated Value (CSV) format. You can import directly into the database, or use the import wizard found in the shortcut menus through out the application. If you need additional help, SugarCRM offers data migration packages to assist you with importing your data SugarCRM is an open-source CRM platform which covers the basic needs of contact, account, campaign and sales management.

It's benefits are with the selling premium cable TV subscription package on UBC shop around:

• It is very easy to customize

• It is easy to expand

• It can be integrated with Microsoft Outlook

It's drawbacks are:

• It is not the best platform for more complex integration projects

• The user interface is not optimal for recurring tasks

What can we offer besides SugarCRM?

JBOSS Process Engine provides a very functional platform for integration, task and process management. A single web application is created to provide a unified interface, while all the information and actions are pulled together via web services.

title: User:Supasin suic


text:
User:Methinee.P

The implementation of a software-as-a-service (SaaS) billing solution, hotel chain Guoman & Thistle has reduced the number of invoices that were subject to error. The complex nature of the old software, only one employee was able to access supply chain data and deal with the operation and management of this function. Inaccurate data also often led to delays, or incorrect payments being sent to suppliers. For the hotel industry, suppliers can range from food and drinks manufacturers to laundry services, personnel suppliers, and building maintenance services – all with different needs. The system has simplified data exchange for these organizations and accommodated retrospective invoicing.

Benefit
 * A properly run supply-chain management program can save a hotel 8 percent to 15.
 * An old system or an ineffective system and re-engineer it to a software-based.
 * Time and money often provide the motivation needed to implement a program that is becoming more essential as operating margins grow tighter.

--

Compiere Inc. offers a complete ERP (Enterprise Resource Planning) and CRM application with support for marketing and sales, human resources, inventory control and more. There is even a full business-accounting package, something often seen in open-source software. Compiere received a $6 million injection of venture capital in June 2006. Compiere is written in Java, JavaScript and PL/SQL, and it is compatible with JDBC and Oracle databases.

Compiere Inc. is a computer systems software, CRM software with ERP systems included in the same compiere Inc. computer software, open source software has functions supported. Corporate and enterprise computer systems consisting Spears 6 system is the key.

1. Partner Relationship Management System (Partner Management).

2. The successor price to pay (Requisition to Pay).

3. Systems Customer Relationship Management (Customer Management).

4. System bid to keep cash (Quote to Cash).

5. Supply Chain Management (Supply Chain Management).

6. Analytics system performance (Performance Analysis).

In addition, compiere Inc. also includes a computer system are many facilities such as the system stores on the Web (Web Store). Audio All Formats systems are working together successfully. Homogeneous computer systems compiere Inc. software open-source software was designed to weaken the refinement features to adapt to particular individual users without compromising any computer capable. Compiere Inc. able to work in all languages and can be customized to work business in many ways. Characteristics that do well in various countries around the world for computer compiere Inc. suitable for small Business - to - medium (SME).

title: User:Methinee.P


text:
User:Apawan.L

'''You as selling premium cable TV Subscription packages on UBC. Shop around for a open source customer relations packages that will help market upgrade to current customer.'''

Hipergate

Hipergate is an open source web base application suite with intranet CRM, sales automation, email marketing, content management, a bug tracker, a project manager, groupware, a calendar, forums, file sharing, and directory services. It is based on Java /JSP for PostgreSQL, Oracle, and SQL Server.

It's mission is to cover a full range of technical requirements in any organization. All applications are addresses from Internet Explorer, without needing any other additional software in the client computer. This suite is multi-company capable, and can be used in a single company, a corporate group or working as an ASP solution capable of serving an unlimited quantity of single customers.

What makes hipergate unique? •	It is completely free. •	It is the only Open Source application with such a wide functional coverage. •	Runs on multiple databases and operating systems •	Has a unitary conceptual design and implementation. •	Is adequate for being used by Application Service Providers. •	Supports multiple languages. •	Has a professional services support network. •	Is well documented.

In this case, we can use Hipergate for selling premium cable TV Subscription packages on UBC by selling via E-shopping in the Website with give customer the Information, Catalogs and the payment processing.

title: User:Apawan.L


text:
User:Usanee.J

'''You as selling premium cable TV Subscription packages on UBC. Shop around for a open source customer relations packages that will help market upgrade to current customer.'''

Hipergate

Hipergate is an open source web base application suite with intranet CRM, sales automation, email marketing, content management, a bug tracker, a project manager, groupware, a calendar, forums, file sharing, and directory services. It is based on Java /JSP for PostgreSQL, Oracle, and SQL Server. It's mission is to cover a full range of technical requirements in any organization. All applications are addresses from Internet Explorer, without needing any other additional software in the client computer. This suite is multi-company capable, and can be used in a single company, a corporate group or working as an ASP solution capable of serving an unlimited quantity of single customers.

What makes hipergate unique?

•	It is completely free. •	It is the only Open Source application with such a wide functional coverage. •	Runs on multiple databases and operating systems •	Has a unitary conceptual design and implementation. •	Is adequate for being used by Application Service Providers. •	Supports multiple languages. •	Has a professional services support network. •	Is well documented.

In this case, we can use Hipergate for selling premium cable TV Subscription packages on UBC by selling via E-shopping in the Website with give customer the Information, Catalogs and the payment processing.

title: User:Usanee.J


text:
User talk:Charles Jeffrey Danoff/CELTA log

What does it mean to be a 'dancing bear'? --Brylie 07:30, 6 February 2011 (UTC)


text:
User talk:Sarkar~enwikiversity

 Hello Sarkar, 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
 * Discuss Wikiversity issues or ask questions at the colloquium.
 * Chat with other Wikiversitans on #wikiversity-en.
 * Follow Wikiversity on twitter (http://twitter.com/Wikiversity) and identi.ca (http://identi.ca/group/wikiversity).

You don't need to be an educator to edit. You only need to be bold to contribute and to experiment with the sandbox or your userpage. See you around Wikiversity! --Leutha 08:49, 6 February 2011 (UTC)

Your account will be renamed
Hello,

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

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

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

Sorry for the inconvenience.

Yours, Keegan Peterzell Community Liaison, Wikimedia Foundation 23:42, 17 March 2015 (UTC)

Renamed
 This account has been renamed as part of single-user login finalisation. If you own this account you can |log in using your previous username and password for more information. If you do not like this account's new name, you can choose your own using this form after logging in: . -- Keegan (WMF) (talk) 06:21, 19 April 2015 (UTC)

title: User talk:Sarkar~enwikiversity


text:
User talk:Chef.ishwarjagrit

 Hello Chef.ishwarjagrit, 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
 * Discuss Wikiversity issues or ask questions at the colloquium.
 * Chat with other Wikiversitans on #wikiversity-en.
 * Follow Wikiversity on twitter (http://twitter.com/Wikiversity) and identi.ca (http://identi.ca/group/wikiversity).

You don't need to be an educator to edit. You only need to be bold to contribute and to experiment with the sandbox or your userpage. See you around Wikiversity! --Leutha 08:50, 6 February 2011 (UTC)

title: User talk:Chef.ishwarjagrit


text:
User talk:Tzafra

 Hello Tzafra, 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
 * Discuss Wikiversity issues or ask questions at the colloquium.
 * Chat with other Wikiversitans on #wikiversity-en.
 * Follow Wikiversity on twitter (http://twitter.com/Wikiversity) and identi.ca (http://identi.ca/group/wikiversity).

You don't need to be an educator to edit. You only need to be bold to contribute and to experiment with the sandbox or your userpage. See you around Wikiversity! --Leutha 08:50, 6 February 2011 (UTC)

title: User talk:Tzafra


text:
User talk:Muneendra~enwikiversity

 Hello Muneendra, 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
 * Discuss Wikiversity issues or ask questions at the colloquium.
 * Chat with other Wikiversitans on #wikiversity-en.
 * Follow Wikiversity on twitter (http://twitter.com/Wikiversity) and identi.ca (http://identi.ca/group/wikiversity).

You don't need to be an educator to edit. You only need to be bold to contribute and to experiment with the sandbox or your userpage. See you around Wikiversity! --Leutha 08:51, 6 February 2011 (UTC)

Your account will be renamed
Hello,

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

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

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

Sorry for the inconvenience.

Yours, Keegan Peterzell Community Liaison, Wikimedia Foundation 23:39, 17 March 2015 (UTC)

Renamed
 This account has been renamed as part of single-user login finalisation. If you own this account you can |log in using your previous username and password for more information. If you do not like this account's new name, you can choose your own using this form after logging in: . -- Keegan (WMF) (talk) 06:16, 19 April 2015 (UTC)

title: User talk:Muneendra~enwikiversity


text:
User talk:Beth21~enwikiversity

 Hello Beth21, 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
 * Discuss Wikiversity issues or ask questions at the colloquium.
 * Chat with other Wikiversitans on #wikiversity-en.
 * Follow Wikiversity on twitter (http://twitter.com/Wikiversity) and identi.ca (http://identi.ca/group/wikiversity).

You don't need to be an educator to edit. You only need to be bold to contribute and to experiment with the sandbox or your userpage. See you around Wikiversity! --Leutha 08:51, 6 February 2011 (UTC)

Your account will be renamed
Hello,

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

Unfortunately, your account clashes with another account also called Beth21. To make sure that both of you can use all Wikimedia projects in future, we have reserved the name Beth21~enwikiversity that only you will have. If you like it, you don't have to do anything. If you do not like it, you can pick out a different name.

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

Sorry for the inconvenience.

Yours, Keegan Peterzell Community Liaison, Wikimedia Foundation 23:28, 17 March 2015 (UTC)

Renamed
 This account has been renamed as part of single-user login finalisation. If you own this account you can |log in using your previous username and password for more information. If you do not like this account's new name, you can choose your own using this form after logging in: . -- Keegan (WMF) (talk) 06:04, 19 April 2015 (UTC)

title: User talk:Beth21~enwikiversity


text:
User talk:Moxit1045

 Hello Moxit1045, 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
 * Discuss Wikiversity issues or ask questions at the colloquium.
 * Chat with other Wikiversitans on #wikiversity-en.
 * Follow Wikiversity on twitter (http://twitter.com/Wikiversity) and identi.ca (http://identi.ca/group/wikiversity).

You don't need to be an educator to edit. You only need to be bold to contribute and to experiment with the sandbox or your userpage. See you around Wikiversity! --Leutha 08:51, 6 February 2011 (UTC)

title: User talk:Moxit1045


text:
User talk:Philosokitteh

 Hello Philosokitteh, 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
 * Discuss Wikiversity issues or ask questions at the colloquium.
 * Chat with other Wikiversitans on #wikiversity-en.
 * Follow Wikiversity on twitter (http://twitter.com/Wikiversity) and identi.ca (http://identi.ca/group/wikiversity).

You don't need to be an educator to edit. You only need to be bold to contribute and to experiment with the sandbox or your userpage. See you around Wikiversity! --Leutha 08:52, 6 February 2011 (UTC)

title: User talk:Philosokitteh


title: SPIR608/Games


text:
User talk:Kate moncrieff

 Hello Kate moncrieff, 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
 * Discuss Wikiversity issues or ask questions at the colloquium.
 * Chat with other Wikiversitans on #wikiversity-en.
 * Follow Wikiversity on twitter (http://twitter.com/Wikiversity) and identi.ca (http://identi.ca/group/wikiversity).

You don't need to be an educator to edit. You only need to be bold to contribute and to experiment with the sandbox or your userpage. See you around Wikiversity! Jtneill - Talk - c 11:44, 6 February 2011 (UTC)

 Woof! My name is Jack Russell. I am a dog and a Wikiversity mascot. I am pretty new around Wikiversity. Perhaps we can learn together! Tail wag...

Images
Hi Kate - Just a heads up that you'll need to add a license to your image uploads. Sincerely, James. -- Jtneill - Talk - c 11:44, 6 February 2011 (UTC)

File:DSCF0267.jpg
Hi! Thanks for uploading the file File:DSCF0267.jpg. Unfortunately, it isn't currently in use and doesn't have any licensing information, so as part of a cleanup I've tagged it to be removed. If you would like to keep the file here, though, that would be great - all you need to do is add an appropriate license if the copyright belongs to you, or if it is otherwise available under a compatible free license. If you need any help, just ask - I'm very happy to assist. - Bilby 02:21, 5 October 2011 (UTC)

What is the copyright of File:Ibar4-09.jpg?
Hi! Thanks for uploading the file File:Ibar4-09.jpg. Unfortunately, when you did so you didn't include information about the license, and the file will need a license or other copyright information in order to stay at Wikiversity.

You can easily fix this by adding on the page, and filling out who the owner is and the copying terms. In particular, we need to know who owns the rights to the file and what the terms of use are. In regard to the latter, Wikiversity accepts open content, public domain, and fair use works (see Wikiversity's Copyright policy). You can place a on the file page to signify the copyright terms. If you need any help, feel free to leave me a message, and I'll do what I can to assist. -Bilby 00:39, 4 November 2011 (UTC)

Files Missing License Information
Thank you for uploading files to Wikiversity. See Media for copyright and license requirements for Wikiversity files. All files must have copyright and/or license information added to the file.

Instructions for adding copyright and/or license information are available at License tags. Files must be updated within seven days or they may be removed without further notice. See Requests For Deletion for more information.

The following files are missing copyright and/or license information:
 * File:Ibar4-09.jpg
 * File:Spkate.jpg

MaintenanceBot (discuss • contribs) 03:48, 11 November 2013 (UTC)

title: User talk:Kate moncrieff


text:
User talk:ABDULLAH AHSAN

 Hello ABDULLAH AHSAN, 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
 * Discuss Wikiversity issues or ask questions at the colloquium.
 * Chat with other Wikiversitans on #wikiversity-en.
 * Follow Wikiversity on twitter (http://twitter.com/Wikiversity) and identi.ca (http://identi.ca/group/wikiversity).

You don't need to be an educator to edit. You only need to be bold to contribute and to experiment with the sandbox or your userpage. See you around Wikiversity! --Leutha 13:01, 6 February 2011 (UTC)

title: User talk:ABDULLAH AHSAN


text:
User:KBlott/templates

Hello world!

title: User:KBlott/templates


text:
User:KBlott/echo

""

title: User:KBlott/echo


text: User:KBlott/sandbox

title: User:KBlott/sandbox


text: User:KBlott/templates/sum

title: User:KBlott/templates/sum


text:
User talk:Naval Thakral

 Hello Naval Thakral, 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
 * Discuss Wikiversity issues or ask questions at the colloquium.
 * Chat with other Wikiversitans on #wikiversity-en.
 * Follow Wikiversity on twitter (http://twitter.com/Wikiversity) and identi.ca (http://identi.ca/group/wikiversity).

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! --Ottava Rima (talk) 14:46, 6 February 2011 (UTC)

File:450px-Naval.jpg
Hi! Thanks for uploading the file File:450px-Naval.jpg. It is a great photo, but unfortunately it isn't currently in use and doesn't have any licensing information, so as part of a cleanup I've tagged it to be removed. If you would like to keep the file here, though, that would be great - all you need to do is add an appropriate license if the copyright belongs to you, or if it is otherwise available under a compatible free license. If you need any help, just ask - I'm very happy to assist. - Bilby 02:08, 5 October 2011 (UTC)

title: User talk:Naval Thakral


text:
Template:Sum

This template adds two numbers. For example "21.509999999999998" returns "21.51". Please ignore the expression error.

title: Template:Sum


text:
Portal:Game design/Intro

Welcome to the Game Design Portal. This portal gives access to a range of game design resources and projects available on Wikiversity.

title: Portal:Game design/Intro