text:
User:Eml5526.s11.team2.sandhu

[HW7]

=My Page= =Homework 3=

=Problem 3.3 Finite element analysis of a four bar plane structure=

Statement
Consider the truss structure given in Figure. Nodes A and B are fixed.A force equal to 10N acts in the positive x-direction at node C. Coordinates of joints are given in meters. Young’s modulus is $$E={{10}^{11}}$$pascals and the cross-sectional area for all bars are $$A=2*{{10}^{-2}}{{m}^{2}}$$.

Given
Force at node $$ C = 10N$$, $$ E={{10}^{11}}$$pascal and $${{A}^{1}}={{A}^{2}}={{A}^{3}}={{A}^{4}}=2*{{10}^{-2}}{{m}^{2}}$$

To Find
a. Number the elements and nodes.

b. Assemble the global stiffness and force matrix.

c. Partition the system and solve for the nodal displacements.

d. Compute the stresses and reactions.

Solution
a. There are four elements and four nodes. We start numbering nodes from the fixed nodes and then arbitrarily assign numbers to other nodes.

b. Stiffness for a element in two dimensional coordinate system inclined to x axis at angle $$\phi $$ is given by

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

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

\displaystyle {K^e} = {k^e}\left[ {\begin{array}{ccccccccccccccc} &{\cos {\phi ^e}\sin {\phi ^e}}&{ - {{\cos }^2}{\phi ^e}}&{ - \cos {\phi ^e}\sin {\phi ^e}} \\ {\cos {\phi ^e}\sin {\phi ^e}}&&{ - \cos {\phi ^e}\sin {\phi ^e}}&{ - {{\sin }^2}{\phi ^e}} \\ { - {{\cos }^2}{\phi ^e}}&{ - \cos {\phi ^e}\sin {\phi ^e}}&&{\cos {\phi ^e}\sin {\phi ^e}} \\ { - \cos {\phi ^e}\sin {\phi ^e}}&{ - {{\sin }^2}{\phi ^e}}&{\cos {\phi ^e}\sin {\phi ^e}}& \end{array}} \right] $$      (3.3.1)
 *
 * }

For element 1 $$ {\phi ^1} = {90^o} $$, Global node numbers 1 & 3
 * {| style="width:100%" border="0"

$$  \displaystyle {K^1} = AE\left[ {\begin{array}{ccccccccccccccc} 0&0&0&0 \\  0&1&0&{ - 1} \\   0&0&0&0 \\   0&{ - 1}&0&1 \end{array}} \right] $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

For element 2 $$ {\phi ^2} = {135^o} $$,Global node numbers 2 & 3

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

$$  \displaystyle {K^2} = AE\left[ {\begin{array}{ccccccccccccccc} {\frac{1}}&{ - \frac{1}}&{ - \frac{1}}&{\frac{1}} \\ { - \frac{1}}&{\frac{1}}&{\frac{1}}&{ - \frac{1}} \\ { - \frac{1}}&{\frac{1}}&{\frac{1}}&{ - \frac{1}} \\ {\frac{1}}&{ - \frac{1}}&{ - \frac{1}}&{\frac{1}} \end{array}} \right] $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

For element 3 $$ {\phi ^3} = {90^o} $$,Global node numbers 2 & 4
 * {| style="width:100%" border="0"

$$  \displaystyle {K^3} = AE\left[ {\begin{array}{ccccccccccccccc} 0&0&0&0 \\  0&1&0&{ - 1} \\   0&0&0&0 \\   0&{ - 1}&0&1 \end{array}} \right] $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

For element 4 $$ {\phi ^4} = {0^o} $$,Global node numbers 3 & 4
 * {| style="width:100%" border="0"

$$  \displaystyle {K^4} = AE\left[ {\begin{array}{ccccccccccccccc} 1&0&{ - 1}&0 \\  0&0&0&0 \\   { - 1}&0&{ - 1}&0 \\   0&0&0&0 \end{array}} \right] $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Global matrix can be found by adding corresponding elements of each matrix.
 * {| style="width:100%" border="0"

$$  \displaystyle K = AE\left[ {\begin{array}{ccccccccccccccc} 0&0&0&0&0&0&0&0 \\  0&1&0&0&0&{ - 1}&0&0 \\   0&0&{\frac{1}}&{ - \frac{1}}&{ - \frac{1}}&{\frac{1}}&0&0 \\ 0&0&{ - \frac{1}}&{1 + \frac{1}}&{\frac{1}}&{ - \frac{1}}&0&{ - 1} \\ 0&0&{ - \frac{1}}&{\frac{1}}&{1 + \frac{1}}&{ - \frac{1}}&{ - 1}&0 \\ 0&{ - 1}&{\frac{1}}&{ - \frac{1}}&{ - \frac{1}}&{1 + \frac{1}}&0&0 \\ 0&0&0&0&{ - 1}&0&1&0 \\  0&0&0&{ - 1}&0&0&0&1 \end{array}} \right] $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Matrix $$d$$ and $$ f+r$$ will be as follows

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

$$  \displaystyle d = \left[ {\begin{array}{ccccccccccccccc} 0 \\  0 \\   0 \\   0 \\    \\    \\    \\ \end{array}} \right]f + r = \left[ {\begin{array}{ccccccccccccccc} \\   \\    \\    \\   0 \\   0 \\   {10} \\   0 \end{array}} \right] $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Now writing global system of equations
 * {| style="width:100%" border="0"

$$  \displaystyle Kd = f + r $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

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

$$  \displaystyle AE\left[ {\begin{array}{ccccccccccccccc} 0&0&0&0&\vline & 0&0&0&0 \\ 0&1&0&0&\vline & 0&{ - 1}&0&0 \\ 0&0&{\frac{1}}&{ - \frac{1}}&\vline & { - \frac{1}}&{\frac{1}}&0&0 \\ 0&0&{ - \frac{1}}&{1 + \frac{1}}&\vline & {\frac{1}}&{ - \frac{1}}&0&{ - 1} \\ \hline 0&0&{ - \frac{1}}&{\frac{1}}&\vline & {1 + \frac{1}}&{ - \frac{1}}&{ - 1}&0 \\ 0&{ - 1}&{\frac{1}}&{ - \frac{1}}&\vline & { - \frac{1}}&{1 + \frac{1}}&0&0 \\ 0&0&0&0&\vline & { - 1}&0&1&0 \\ 0&0&0&{ - 1}&\vline & 0&0&0&1 \end{array}} \right]\left[ {\begin{array}{ccccccccccccccc} 0 \\  0 \\   0 \\   0 \\ \hline \\   \\    \\ \end{array}} \right] = \left[ {\begin{array}{ccccccccccccccc} \\   \\    \\    \\ \hline 0 \\  0 \\   {10} \\   0 \end{array}} \right] $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

c. The partitions are done according to the fixed number of nodes. There are two fixed node so matrices are partitioned after fourth row and fourth column.
 * {| style="width:100%" border="0"

$$  \displaystyle {K_E} = AE\left[ {\begin{array}{ccccccccccccccc} 0&0&0&0 \\  0&1&0&0 \\   0&0&{\frac{1}}&{ - \frac{1}} \\ 0&0&{ - \frac{1}}&{1 + \frac{1}} \end{array}} \right] {K_{EF}} = AE\left[ {\begin{array}{ccccccccccccccc} 0&0&0&0 \\  0&{ - 1}&0&0 \\   { - \frac{1}}&{\frac{1}}&0&0 \\ {\frac{1}}&{ - \frac{1}}&0&{ - 1} \end{array}} \right] $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

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

$$  \displaystyle {K_F} = AE\left[ {\begin{array}{ccccccccccccccc} {1 + \frac{1}}&{ - \frac{1}}&{ - 1}&0 \\ { - \frac{1}}&{1 + \frac{1}}&0&0 \\ { - 1}&0&1&0 \\  0&0&0&1 \end{array}} \right] {{\tilde d}_E} = \left[ {\begin{array}{ccccccccccccccc} 0 \\  0 \\   0 \\   0 \end{array}} \right] $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

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

$$  \displaystyle {d_F} = \left[ {\begin{array}{ccccccccccccccc} \\   \\    \\ \end{array}} \right] {r_E} = \left[ {\begin{array}{ccccccccccccccc} \\   \\    \\ \end{array}} \right] {f_F} = \left[ {\begin{array}{ccccccccccccccc} 0 \\  0 \\   {10} \\   0 \end{array}} \right] $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

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

$$  \displaystyle {K_{EF}}^T{\cancelto{0}{{\tilde d}_E}} + {K_F}{d_F} = {f_F}
 * style="width:95%" |
 * style="width:95%" |

$$      (3.3.2)
 *
 * }

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

$$  \displaystyle {K_F}{d_F} = {f_F} $$      (3.3.3)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

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

$$  \displaystyle {d_F} = K_F^{ - 1}{f_F} $$      (3.3.4)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

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

$$  \displaystyle {d_F} = \left[ {\begin{array}{ccccccccccccccc} {0.1914} \\  {0.05} \\   {0.2414} \\   0 \end{array}} \right] \times {10^{ - 7}} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

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

$$  \displaystyle {r_E} = {K_E}{\cancelto{0}{{\tilde d}_E}} + {K_{EF}}{d_F} $$      (3.3.5)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

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

$$  \displaystyle {r_E} = {K_{EF}}{d_F} $$      (3.3.6)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

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

$$  \displaystyle {r_E} = \left[ {\begin{array}{ccccccccccccccc} 0 \\  { - 10} \\   { - 10} \\   { 10} \end{array}} \right] $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

d. Stresses in the elements.

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

$$  \displaystyle {\sigma ^e} = \frac\left[ {\begin{array}{ccccccccccccccc} { - \cos {\varphi ^e}}&{ - \sin {\varphi ^e}}&{\cos {\varphi ^e}}&{\sin {\varphi ^e}} \end{array}} \right]{d^e} $$      (3.3.7)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

For element 1 $$\varphi = {90^o}$$
 * {| style="width:100%" border="0"

$$  \displaystyle {d^1} = \left[ {\begin{array}{ccccccccccccccc} \\   \\    \\ \end{array}} \right] = \left[ {\begin{array}{ccccccccccccccc} 0 \\  0 \\   {0.1914} \\   {0.05} \end{array}} \right] \times {10^{ - 7}} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

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

$$  \displaystyle {\sigma ^1} = \frac{1}\left[ {\begin{array}{ccccccccccccccc} 0&{ - 1}&0&1 \end{array}} \right]\left[ {\begin{array}{ccccccccccccccc} 0 \\  0 \\   {0.1914} \\   {0.05} \end{array}} \right] \times {10^{ - 7}} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

For element 2 $$\varphi = {135^o}$$

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

$$  \displaystyle {d^2} = \left[ {\begin{array}{ccccccccccccccc} \\   \\    \\ \end{array}} \right] = \left[ {\begin{array}{ccccccccccccccc} 0 \\  0 \\   {0.1914} \\   {0.05} \end{array}} \right] \times {10^{ - 7}} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

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

$$  \displaystyle {\sigma ^2} = \frac{1}\left[ {\begin{array}{ccccccccccccccc} {\frac{1}}&{ - \frac{1}}&{ - \frac{1}}&{\frac{1}} \end{array}} \right]\left[ {\begin{array}{ccccccccccccccc} 0 \\  0 \\   {0.1914} \\   {0.05} \end{array}} \right] \times {10^{ - 7}} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

For element 3 $$\varphi = {90^o}$$

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

$$  \displaystyle {d^3} = \left[ {\begin{array}{ccccccccccccccc} \\   \\    \\ \end{array}} \right] = \left[ {\begin{array}{ccccccccccccccc} 0 \\  0 \\   {0.2414} \\   0 \end{array}} \right] \times {10^{ - 7}} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

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

$$  \displaystyle {\sigma ^3} = \frac{1}\left[ {\begin{array}{ccccccccccccccc} 0&{ - 1}&0&1 \end{array}} \right]\left[ {\begin{array}{ccccccccccccccc} 0 \\  0 \\   {0.2414} \\   0 \end{array}} \right] \times {10^{ - 7}} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

For element 4 $$\varphi = {0^o}$$

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

$$  \displaystyle {d^4} = \left[ {\begin{array}{ccccccccccccccc} \\   \\    \\ \end{array}} \right] = \left[ {\begin{array}{ccccccccccccccc} {0.1914} \\  {0.05} \\   {0.2414} \\   0 \end{array}} \right] \times {10^{ - 7}} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

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

$$  \displaystyle {\sigma ^4} = \frac{1}\left[ {\begin{array}{ccccccccccccccc} { - 1}&0&1&{} \end{array}} \right]\left[ {\begin{array}{ccccccccccccccc} {0.1914} \\  {0.05} \\   {0.2414} \\   0 \end{array}} \right] \times {10^{ - 7}} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

=Problem 3.6 Finite element analysis of three bar hanging structure=

Statement
Given the three-bar structure subjected to the prescribed load at point C equal to $$ 10^3 $$ N as shown in Figure 2.19.The Young’s modulus is         $$ E = {10^{11}}$$ Pa, the  cross-sectional area of the bar  BC is          $$2*{10^{ - 2}}{m^2}$$ 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.

Given
Force at node$$ c = {10}^{3}N$$. Young's Modulus $$E = {10^{11}}$$ Pa. $${{A}^{1}}={{A}^{3}}=2*{{10}^{-2}}{{m}^{2}}$$ and $${{A}^{2}}={{10}^{-2}}{{m}^{2}}$$ Point D is free to move in x direction only.

To Find
a. Construct the global stiffness matrix and load matrix. b. Partition the matrices and solve for the unknown displacements at point B and displacement in the x-direction at point D. c. Find the stresses in the three bars. d. Find the reactions at nodes C, D and F.

Solution
a. There are three elements and four nodes. Stiffness for a element in two dimensional coordinate system inclined to x axis at angle $$\phi $$ is given by

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

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

\displaystyle {K^e} = {k^e}\left[ {\begin{array}{ccccccccccccccc} &{\cos {\phi ^e}\sin {\phi ^e}}&{ - {{\cos }^2}{\phi ^e}}&{ - \cos {\phi ^e}\sin {\phi ^e}} \\ {\cos {\phi ^e}\sin {\phi ^e}}&&{ - \cos {\phi ^e}\sin {\phi ^e}}&{ - {{\sin }^2}{\phi ^e}} \\ { - {{\cos }^2}{\phi ^e}}&{ - \cos {\phi ^e}\sin {\phi ^e}}&&{\cos {\phi ^e}\sin {\phi ^e}} \\ { - \cos {\phi ^e}\sin {\phi ^e}}&{ - {{\sin }^2}{\phi ^e}}&{\cos {\phi ^e}\sin {\phi ^e}}& \end{array}} \right] $$      (3.6.1)
 *
 * }

For element 1 $$ {\phi ^1} = {{-45}^o} $$, Global node numbers 1 & 4
 * {| style="width:100%" border="0"

$$  \displaystyle {K^1} = AE\left[ {\begin{array}{ccccccccccccccc} {1/2}&{-1/2}&{-1/2}&{1/2} \\  {-1/2}&{1/2}&{1/2}&{-1/2} \\   {-1/2}&{1/2}&{1/2}&{-1/2} \\   {1/2}&{-1/2}&{-1/2}&{1/2} \end{array}} \right] $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

For element 2 $$ {\phi ^2} = {{-90}^o} $$,Global node numbers 2 & 4

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

$$ \displaystyle {K^2} = A_{BC}E\left[ {\begin{array}{ccccccccccccccc} 0&0&0&0 \\  0&1&0&{ - 1} \\   0&0&0&0 \\   0&{ - 1}&0&1 \end{array}} \right] $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

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

$$  \displaystyle {K^2} = AE\left[ {\begin{array}{ccccccccccccccc} 0&0&0&0 \\  0&2&0&{ - 2} \\   0&0&0&0 \\   0&{ - 2}&0&2 \end{array}} \right] $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

For element 3 $$ {\phi ^4} = {{-135}^o} $$,Global node numbers 3 & 4
 * {| style="width:100%" border="0"

$$ \displaystyle {K^3} = AE\left[ {\begin{array}{ccccccccccccccc} {1/2}&{1/2}&{-1/2}&{-1/2} \\  {1/2}&{1/2}&{-1/2}&{-1/2} \\   {-1/2}&{-1/2}&{1/2}&{1/2} \\   {-1/2}&{-1/2}&{1/2}&{1/2} \end{array}} \right] $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Global matrix can be found by adding corresponding elements of each matrix.

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

$$  \displaystyle K = AE\left[ {\begin{array}{ccccccccccccccc} {\frac{1}{2}}&{ - \frac{1}{2}}&0&0&0&0&{ - \frac{1}{2}}&{\frac{1}{2}} \\ { - \frac{1}{2}}&{\frac{1}{2}}&0&0&0&0&{\frac{1}{2}}&{ - \frac{1}{2}} \\ 0&0&0&0&0&0&0&0 \\  0&0&0&2&0&0&0&{ - 2} \\   0&0&0&0&{\frac{1}{2}}&{\frac{1}{2}}&{ - \frac{1}{2}}&{ - \frac{1}{2}} \\ 0&0&0&0&{\frac{1}{2}}&{\frac{1}{2}}&{ - \frac{1}{2}}&{ - \frac{1}{2}} \\ { - \frac{1}{2}}&{\frac{1}{2}}&0&0&{ - \frac{1}{2}}&{ - \frac{1}{2}}&1&0 \\ {\frac{1}{2}}&{ - \frac{1}{2}}&0&{ - 2}&{ - \frac{1}{2}}&{ - \frac{1}{2}}&0&3 \end{array}} \right] $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

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

$$  \displaystyle d = \left[ {\begin{array}{ccccccccccccccc} 0 \\  0 \\   0 \\   0 \\    \\   0 \\    \\ \end{array}} \right] f + r = \left[ {\begin{array}{ccccccccccccccc} \\   \\    \\    \\   0 \\    \\    \\   0 \end{array}} \right] $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Now writing global system of equations
 * {| style="width:100%" border="0"

$$  \displaystyle Kd = f + r $$ Swapping the row 5&6 of $$K$$,$$d$$ and $$f+r$$ matrices, and column 5&6 of$$K$$ matrix.
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }
 * {| style="width:100%" border="0"

$$  \displaystyle \left[ {\begin{array}{ccccccccccccccc} {\frac{1}{2}}&{ - \frac{1}{2}}&0&0&0&\vline & 0&{ - \frac{1}{2}}&{\frac{1}{2}} \\ { - \frac{1}{2}}&{\frac{1}{2}}&0&0&0&\vline & 0&{\frac{1}{2}}&{ - \frac{1}{2}} \\ 0&0&0&0&0&\vline & 0&0&0 \\ 0&0&0&2&0&\vline & 0&0&{ - 2} \\ 0&0&0&0&{\frac{1}{2}}&\vline & {\frac{1}{2}}&{ - \frac{1}{2}}&{ - \frac{1}{2}} \\ \hline 0&0&0&0&{\frac{1}{2}}&\vline & {\frac{1}{2}}&{ - \frac{1}{2}}&{ - \frac{1}{2}} \\ { - \frac{1}{2}}&{\frac{1}{2}}&0&0&{ - \frac{1}{2}}&\vline & { - \frac{1}{2}}&1&0 \\ {\frac{1}{2}}&{ - \frac{1}{2}}&0&{ - 2}&{ - \frac{1}{2}}&\vline & { - \frac{1}{2}}&0&3 \end{array}} \right]\left[ {\begin{array}{ccccccccccccccc} 0 \\  0 \\   0 \\   0 \\   0 \\ \hline \\   \\ \end{array}} \right] = \left[ {\begin{array}{ccccccccccccccc} \\   \\    \\    \\    \\ \hline 0 \\   \\   0 \end{array}} \right] $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

b. The partitions are done according to the fixed number of nodes. There are two fixed node so matrices are partitioned after fourth row and fourth column.

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

$$  \displaystyle {K_E} = \left[ {\begin{array}{ccccccccccccccc} {\frac{1}{2}}&{ - \frac{1}{2}}&0&0&0 \\ { - \frac{1}{2}}&{\frac{1}{2}}&0&0&0 \\ 0&0&0&0&0 \\  0&0&0&2&0 \\   0&0&0&0&{\frac{1}{2}} \end{array}} \right] {K_{EF}} = \left[ {\begin{array}{ccccccccccccccc} 0&{ - \frac{1}{2}}&{\frac{1}{2}} \\ 0&{\frac{1}{2}}&{ - \frac{1}{2}} \\ 0&0&0 \\  0&0&{ - 2} \\   {\frac{1}{2}}&{ - \frac{1}{2}}&{ - \frac{1}{2}} \end{array}} \right] $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

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

$$  \displaystyle {K_F} = \left[ {\begin{array}{ccccccccccccccc} {\frac{1}{2}}&{ - \frac{1}{2}}&{ - \frac{1}{2}} \\ { - \frac{1}{2}}&1&0 \\ { - \frac{1}{2}}&0&3 \end{array}} \right] {d_F} = \left[ {\begin{array}{ccccccccccccccc} \\   \\ \end{array}} \right] {f_F} = \left[ {\begin{array}{ccccccccccccccc} 0 \\   \\   0 \end{array}} \right] $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

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

$$  \displaystyle {{\tilde d}_E} = \left[ {\begin{array}{ccccccccccccccc} 0 \\  0 \\   0 \\   0 \\   0 \end{array}} \right] {r_E} = \left[ {\begin{array}{ccccccccccccccc} \\   \\    \\    \\ \end{array}} \right] $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

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

$$  \displaystyle {K_{EF}}^T{\cancelto{0}{{\tilde d}_E}} + {K_F}{d_F} = {f_F}
 * style="width:95%" |
 * style="width:95%" |

$$      (3.6.2)
 *
 * }

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

$$  \displaystyle {K_F}{d_F} = {f_F} $$      (3.6.3)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

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

$$  \displaystyle {d_F} = K_F^{ - 1}{f_F} $$      (3.6.4)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

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

$$  \displaystyle {d_F} = \left[ {\begin{array}{ccccccccccccccc} {0.30} \\  {0.25} \\   {0.05} \end{array}} \right] \times {10^{ - 5}}m $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

c. Stresses in the elements.

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

$$  \displaystyle {\sigma ^e} = \frac\left[ {\begin{array}{ccccccccccccccc} { - \cos {\varphi ^e}}&{ - \sin {\varphi ^e}}&{\cos {\varphi ^e}}&{\sin {\varphi ^e}} \end{array}} \right]{d^e} $$      (3.3.7)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

For element 1 $$\varphi = {{-45}^o}$$
 * {| style="width:100%" border="0"

$$  \displaystyle {d^1} = \left[ {\begin{array}{ccccccccccccccc} \\   \\    \\ \end{array}} \right] = \left[ {\begin{array}{ccccccccccccccc} 0 \\  0 \\   {0.25} \\   {0.05} \end{array}} \right] \times {10^{ - 5}} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

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

$$  \displaystyle {\sigma ^1} = \frac\left[ {\begin{array}{ccccccccccccccc} { - \frac{1}}&{\frac{1}}&{\frac{1}}&{ - \frac{1}} \end{array}} \right]\left[ {\begin{array}{ccccccccccccccc} 0 \\  0 \\   {0.25} \\   {0.05} \end{array}} \right] \times {10^{ - 5}} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

For element 2 $$\varphi = {{-90}^o}$$
 * {| style="width:100%" border="0"

$$  \displaystyle {d^1} = \left[ {\begin{array}{ccccccccccccccc} \\   \\    \\ \end{array}} \right] = \left[ {\begin{array}{ccccccccccccccc} 0 \\  0 \\   {0.25} \\   {0.05} \end{array}} \right] \times {10^{ - 5}} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

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

$$  \displaystyle {\sigma ^2} = \frac{1}\left[ {\begin{array}{ccccccccccccccc} 0&1&0&{ - 1} \end{array}} \right]\left[ {\begin{array}{ccccccccccccccc} 0 \\  0 \\   {0.25} \\   {0.05} \end{array}} \right] \times {10^{ - 5}} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

For element 3 $$\varphi = {{-135}^o}$$
 * {| style="width:100%" border="0"

$$  \displaystyle {d^1} = \left[ {\begin{array}{ccccccccccccccc} \\   \\    \\ \end{array}} \right] = \left[ {\begin{array}{ccccccccccccccc} {0.30} \\  0 \\   {0.25} \\   {0.05} \end{array}} \right] \times {10^{ - 5}} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

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

$$  \displaystyle {\sigma ^3} = \frac\left[ {\begin{array}{ccccccccccccccc} {\frac{1}}&{\frac{1}}&{ - \frac{1}}&{ - \frac{1}} \end{array}} \right]\left[ {\begin{array}{ccccccccccccccc} {0.30} \\  0 \\   {0.25} \\   {0.05} \end{array}} \right] \times {10^{ - 5}} $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

d. Reactions at node C(node 2),D(node 3) & F(node 1)

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

$$  \displaystyle {r_E} = {K_E}{\cancelto{0}{{\tilde d}_E}} + {K_{EF}}{d_F} $$      (3.6.5)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

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

$$  \displaystyle {r_E} = {K_{EF}}{d_F} $$      (3.6.6)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

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

$$  \displaystyle ${r_E} = \left[ {\begin{array}{*{20}{c}} { - {{10}^3}} \\   \\   0 \\   { - {{10}^3}} \\   0 \end{array}} \right]Newton $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Now as element '3' is free to move in x direction so reaction $$$$ will be zero. Adding this element to $${r_E}$$ matrix.

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

$$  \displaystyle $$      (N)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

Given
$$\left\{ {{\underline{a}}_{i}},i=1,2,......n \right\}$$ is a orthonormal basis.

i.e. $${{\underline{a}}_{i}}.{{\underline{a}}_{j}}={{\delta }_{ij}}$$

$$\left[ {{b}_{jk}} \right]=\left( \begin{matrix}  1 & 1 & 1  \\   2 & -1 & 3  \\   3 & 2 & 6  \\ \end{matrix} \right)$$

$${{\underline{b}}_{j}}={{\underline{b}}_{jk}}{{\underline{a}}_{k}}$$

i.e.

$$\begin{align} & {{\underline{b}}_{1}}={{\underline{a}}_{1}}+{{\underline{a}}_{2}}+{{\underline{a}}_{3}} \\ & {{\underline{b}}_{2}}=2{{\underline{a}}_{1}}-{{\underline{a}}_{2}}+3{{\underline{a}}_{3}} \\ & {{\underline{b}}_{3}}=3{{\underline{a}}_{1}}+2{{\underline{a}}_{2}}+6{{\underline{a}}_{3}} \\ \end{align}$$

also

$$\underline{v}=5{{\underline{a}}_{1}}-7{{\underline{a}}_{2}}-4{{\underline{a}}_{3}}$$

To Find
1) Find $$\left[ {{b}_{jk}} \right]$$

2) Find $$\underline{\Gamma }\left( {{\underline{b}}_{1}},{{\underline{b}}_{2}},{{\underline{b}}_{3}} \right)$$, which is also equal to $$\underline{K}$$  also find $$\text{det}\left[ \underline{\Gamma } \right]$$

3) Solve $$\underline{F}=\left\{ {{F}_{i}} \right\}=\left\{ {{\underline{b}}_{i}}.\underline{v} \right\}$$

4) Solve $$\underline K \underline d = \underline F $$ for $$d = \left[  \right]$$

5) Use $${\underline w _i}.\underline{\underline P} \left( {\underline v } \right) = 0$$ $$\forall i = 1,2,.......n$$ to find $$ \underline {\overline K } \underline d = \underline {\overline F } $$. What will be $$\underline {\overline K } $$ and $$\underline {\overline F } $$ for this case.

6) Solve for $$\underline d $$ for this case and compare to the value calculated in 4).

7) Observe the symmetric properties of $$\underline K $$ and $$\underline {\overline K } $$. Discuss the pros and cons of two methods.

Background Theory
$$\left\{ {{\underline{b}}_{i}},i=1,2.....n \right\}$$ is a basis for $${{\mathbb{R}}^{n}}$$ not rectilinear orthonormal.

$$\underline{v}$$ is any vector,such that

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

$$  \displaystyle \underline v = \sum\limits_{j = 1}^n $$      (2.1)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

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

$$  \displaystyle \sum\limits_{j = 1}^n {{{\underline b }_j}{v_j} - \underline v = 0} $$      (2.2)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

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

$$  \displaystyle \underline{\underline P} \left( {\underline v } \right) = 0 $$      (2.3)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

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

$$  \displaystyle \underline{\underline P} \left( v \right) = \sum\limits_{j = 1}^n {{{\underline b }_j}{v_j} - \underline v } $$      (2.4)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Successively multiply by $${\underline b _i}\left( {i = 1,2,......n} \right)$$ equations with n unknowns are formed

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

$$  \displaystyle {\underline b _i}.\underline{\underline P} \left( {\underline v } \right) = 0 $$      (2.5)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

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

$$  \displaystyle {\underline b _i}\sum\limits_{j = 1}^n  = {\underline b _i}.\underline v $$ (2.6)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

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

$$  \displaystyle \sum\limits_{j = 1}^n {{{\underline b }_i}.{{\underline b }_j}{v_j}} = {\underline b _i}.\underline v $$ (2.7)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

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

$$  \displaystyle {\left[ \right]_{n \times n}}{\left[  \right]_{n \times 1}} = {\left[  \right]_{n \times 1}} $$      (2.8)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

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

$$  \displaystyle \underline K \underline d = \underline F $$ (2.9)
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * <p style="text-align:right">
 * }

Solution
1) Using MATLAB to find determinant. $$\det \left[ \right] = -8$$

2) Elements of Stiffness matrix $$\underline K $$ are given by equation number (7),(8) & (9) and shown in following matrix.
 * {| style="width:100%" border="0"

$$  \displaystyle \underline K = \left( {\begin{array}{ccccccccccccccc}  {{{\underline b }_1}.{{\underline b }_1}}&{{{\underline b }_1}.{{\underline b }_2}}&{{{\underline b }_1}.{{\underline b }_3}} \\   {{{\underline b }_2}.{{\underline b }_1}}&{{{\underline b }_2}.\underline  }&{{{\underline b }_2}.{{\underline b }_3}} \\   {{{\underline b }_3}.{{\underline b }_1}}&{{{\underline b }_3}.{{\underline b }_2}}&{{{\underline b }_3}.{{\underline b }_3}} \end{array}} \right) $$      (2.10)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

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

$$  \displaystyle \underline K = \left( {\begin{array}{ccccccccccccccc}  3&4&{11} \\   4&{14}&{22} \\   {11}&{22}&{49} \end{array}} \right) $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

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

$$  \displaystyle \det \left( {\underline K } \right) = 64 $$      (2.11)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

3) As shown in equation numbers (7) & (8).

$$\underline F = \left\{  \right\} = \left\{ {{{\underline b }_i}.\underline v } \right\}$$

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

$$  \displaystyle \left( {\begin{array}{ccccccccccccccc}  \\    \\ \end{array}} \right) = \left( {\begin{array}{ccccccccccccccc}  {{{\underline b }_1}.\underline v } \\   {{{\underline b }_2}.\underline v } \\   {{{\underline b }_3}.\underline v } \end{array}} \right) $$      (2.12)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

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

$$  \displaystyle \left( {\begin{array}{ccccccccccccccc}  \\    \\ \end{array}} \right) = \left( {\begin{array}{ccccccccccccccc}  { - 6} \\   5 \\   { - 23} \end{array}} \right) $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

4) To find the value of $$\underline d$$

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

$$  {\underline K ^{ - 1}} = \left( {\begin{array}{ccccccccccccccc}  {3.1562}&{0.7187}&{ - 1.0312} \\   {0.7187}&{0.4062}&{ - 0.3437} \\   { - 1.0312}&{ - 0.3437}&{0.4062} \end{array}} \right) $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

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

$$  \displaystyle \underline d = {\underline K ^{ - 1}}\underline F $$ (2.13)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

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

$$  \displaystyle \underline d = \left( {\begin{array}{ccccccccccccccc}  {8.375} \\   {5.625} \\   { - 4.875} \end{array}} \right) $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

5) Derivation for $$\underline {\overline K } \underline d = \underline {\overline F } $$ where $$\left\{ {{{\underline w }_i},i = 1,2.....n} \right\}$$ is a line independent family of vectors. Consider $${w_i} = {a_i}$$$$\left\{ {{{\underline a }_i},i = 1,2,......n} \right\}$$ is an orthonormal basis.

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

$$  \displaystyle {\underline w _i}.\underline{\underline P} \left( {\underline v } \right) = 0 $$      (2.14)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Putting value of $$\underline{\underline{P}}\left( v \right)$$ from equation number (4).

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

$$  \displaystyle {\underline a _i}\sum\limits_{j = 1}^n  = {\underline a _i}.\underline v $$ (2.15)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

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

$$  \displaystyle \sum\limits_{j = 1}^n {{{\underline a }_i}.{{\underline b }_j}{v_j}} = {\underline a _i}.\underline v $$ (2.16)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

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

$$  \displaystyle {\left[ \right]_{n \times n}}{\left[  \right]_{n \times 1}} = {\left[  \right]_{n \times 1}} $$      (2.17)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

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

$$  \displaystyle \underline {\overline K } \underline d = \underline {\overline F } $$      (2.18)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Where $$\underline{\overline{K}}$$ and $$\underline{\overline{F}}$$ are matrices having following elements.

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

$$  \displaystyle \underline {\overline K } = \left( {\begin{array}{ccccccccccccccc}  {{{\underline a }_1}.{{\underline b }_1}}&{{{\underline a }_1}.{{\underline b }_2}}&{{a_1}.{{\underline b }_3}} \\   {{{\underline a }_2}.{{\underline b }_1}}&{{{\underline a }_2}.\underline  }&{{a_2}.{{\underline b }_3}} \\   {{{\underline a }_3}.{{\underline b }_1}}&{{{\underline a }_3}.{{\underline b }_2}}&{{{\underline a }_3}.{{\underline b }_3}} \end{array}} \right) $$      (2.19)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

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

$$  \displaystyle \overline {\underline F } = \left( {\begin{array}{ccccccccccccccc}  {{{\underline a }_1}.\underline v } \\   {{{\underline a }_2}.\underline v } \\   {{{\underline a }_3}.\underline v } \end{array}} \right) $$      (2.20)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

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

$$  \displaystyle \underline {\overline K } = \left( {\begin{array}{ccccccccccccccc}  1&2&3 \\   1&{ - 1}&2 \\   1&3&6 \end{array}} \right) $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

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

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

$$\overline{\underline{F}}=\left( \begin{matrix}  5  \\   -7  \\   -4  \\ \end{matrix} \right)$$

 * }
 * }

6) To find $$\underline d $$

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

$$  \displaystyle \underline d = {\underline {\overline K } ^{ - 1}}\underline {\overline F } $$      (2.21)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

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

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

$${{\overline{\underline{K}}}^{-1}}=\left( \begin{matrix}  1.500 & 0.375 & -0.875  \\   0.500 & -0.375 & -0.125  \\   -0.500 & -0.125 & 0.375  \\ \end{matrix} \right)$$

 * }
 * }

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

$$  \displaystyle \underline d = \left( {\begin{array}{ccccccccccccccc}  {8.375} \\   {5.625} \\   { - 4.875} \end{array}} \right) $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

The value of $$\underline d$$ is same irrespective of method adopted to calculate it.

7) The matrix $$\underline{K}$$ is a symmetric matrix whereas matrix $$\underline{\overline{K}}$$ is not symmetric. Pros & Cons of first approach:- Pros:-
 * 1) The matrix $$\underline{K}$$ is a symmetric matrix.

Cons:-
 * 1) Calculations become more involved.

Pros & Cons of first approach:- Pros:-
 * 1) The matrix $$\underline{\overline{K}}$$ is not a symmetric matrix.

Cons:-

solution

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

$$  \displaystyle m\left( {x + \frac{2}} \right) = \rho \left( {x + \frac{2}} \right)\frac{1}{2}\left[ {h\left( x \right) + h\left( {x + dx} \right)} \right]b $$     (1)
 * style="width:95%" |
 * style="width:95%" |
 * }

multiply both sides with $$dx$$

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

$$  \displaystyle m\left( {x + \frac{2}} \right)dx = \left[ {\rho \left( {x + \frac{2}} \right)\frac{1}{2}\left[ {h\left( x \right) + h\left( {x + dx} \right)} \right]b} \right]dx $$     (2) expanding $$ h(x+dx)$$ with taylor series
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }
 * {| style="width:100%" border="0"

$$  \displaystyle m\left( {x + \frac{2}} \right)dx = \left[ {\rho \left( {x + \frac{2}} \right)\frac{1}{2}\left[ {h\left( x \right) + h\left( x \right) + \fracdx \ldots } \right]b} \right]dx $$     (3)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

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

$$  \displaystyle m\left( {x + \frac{2}} \right)dx = \left[ {\rho \left( {x + \frac{2}} \right)\left[ {h\left( x \right) + \frac\frac{2} \ldots } \right]b} \right]dx $$     (4)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

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

$$  \displaystyle m\left( {x + \frac{2}} \right)dx = \left[ {\rho \left( {x + \frac{2}} \right)\left[ {h\left( {x + \frac{2}} \right)} \right]b} \right]dx $$     (5) now expanding $$ \rho \left( {x + \frac{2}} \right)$$ and $$h\left( {x + \frac{2}} \right)$$ by taylor series
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }
 * {| style="width:100%" border="0"

$$  \displaystyle m\left( {x + \frac{2}} \right)dx = \left[ {\left[ {\rho \left( x \right) + \frac\frac{2} \ldots } \right]\left[ {h\left( x \right) + \frac\frac{2} \ldots } \right]b} \right]dx $$     (6)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

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

$$  \displaystyle m\left( {x + \frac{2}} \right)dx = \left[ {\rho \left( x \right)h\left( x \right) + \rho \left( x \right)\frac\frac{2} + h\left( x \right)\frac\frac{2} + \frac\frac{2}\frac\frac{2} \ldots } \right]bdx $$     (7)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Neglecting higer order terms and considering first three terms of expansion

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

$$  \displaystyle m\left( {x + \frac{2}} \right)dx = \left[ {\rho \left( x \right)h\left( x \right) + \frac\frac{2}} \right]bdx $$     (8)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

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

$$  \displaystyle m\left( {x + \frac{2}} \right)dx = \left[ {\rho \left( x \right)h\left( x \right)dx + \frac\frac{2}} \right]b $$     (9)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Neglecting second term due to presence of $$d{x^2}$$

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

$$  \displaystyle m\left( {x + \frac{2}} \right)dx = \left[ {\rho \left( x \right)h\left( x \right)dx} \right]b $$     (10)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

as $$bh\left( x \right) = A\left( x \right)$$

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

$$  \displaystyle m\left( {x + \frac{2}} \right)dx = \rho \left( x \right)A\left( x \right)dx $$     (11)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

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

$$  \displaystyle m\left( {x + \frac{2}} \right) = \rho \left( x \right)A\left( x \right) $$     (12)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

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

$$  \displaystyle m\left( {x + \frac{2}} \right) = m\left( x \right) $$     (13)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Now PDE for the general cross section as proved in 1.1.1

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

$$  \displaystyle \frac{\partial }\left( {E\left( x \right)A\left( x \right)\frac} \right) + f\left( {x,t} \right) = m\frac $$     (14)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

For given rectangular cross section put $$m = m\left( {x + \frac{2}} \right)$$
 * {| style="width:100%" border="0"

$$  \displaystyle \frac{\partial }\left( {E\left( x \right)A\left( x \right)\frac} \right) + f\left( {x,t} \right) = m\left( {x + \frac{2}} \right)\frac $$     (15)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

From euation no.(13) $$m\left( {x + \frac{2}} \right) = \rho \left( x \right)A\left( x \right)$$

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

$$  \displaystyle \frac{\partial }\left( {E\left( x \right)A\left( x \right)\frac} \right) + f\left( {x,t} \right) = \rho \left( x \right)A\left( x \right)\frac $$     (16)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Now put $$A\left( x \right) = bh\left( x \right)$$

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

$$  \displaystyle \frac{\partial }\left( {E\left( x \right)bh\left( x \right)\frac} \right) + f\left( {x,t} \right) = \rho \left( x \right)bh\left( x \right)\frac $$     (17)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

As $$b$$ is a constant, taking it out of partial differential

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

$$  \displaystyle b\frac{\partial }\left( {E\left( x \right)h\left( x \right)\frac} \right) + f\left( {x,t} \right) = b\rho \left( x \right)h\left( x \right)\frac $$     (18)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Dividing both sides of equation with $$b$$

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

$$  \displaystyle \frac{\partial }\left( {E\left( x \right)h\left( x \right)\frac} \right) + \frac{b} = \rho \left( x \right)h\left( x \right)\frac $$     (19) This partial differential equation represent a bar with varying rectangular cross section.
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

title: User:Eml5526.s11.team2.sandhu


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

title: File:Eml5526.s11.team2.FEAHW2pr9.png


text: File:Egm6321.f10.team2.oztekin.FEAHW2pr6pi2.png


text:
User:Ruvald

Mark Ruvald Pedersen, s072095

One of the most interesting courses at DTU: kurser.dtu.dk

title: User:Ruvald


text:
User:Metadawson

Leave a message for Metadawson

Metadawson's Free the Knowledge Blog

Alexia Dawson
Campbell, California

Software Engineering Author at Apple, Inc.

Post-Master's certificate student in San Jose State University's College of Information, studying Archiving
 * Subjects of interest: open data, free knowledge, information access, semantics, and data governance
 * Career goal: Manage digitization at a company like Internet Archive or Wikimedia Foundation

title: User:Metadawson


text:
User:Eguerrero739

Edward Guerrero

MD Anderson Cancer Center Houston, Tx

title: User:Eguerrero739


text:
User talk:Eguerrero739

 Hello Eguerrero739, 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) 17:57, 2 February 2011 (UTC)

title: User talk:Eguerrero739


text:
User talk:Metadawson

 Hello Metadawson, 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) 18:20, 2 February 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:Wow serafin head.jpg

MaintenanceBot (discuss • contribs) 03:49, 11 November 2013 (UTC)

title: User talk:Metadawson


text:
SPIR608 Political Simulation and Gaming/2011/Vietnam 1955 Scenario

Operation Pale Charlie At the end of the Second World War states were confronting the new reality of resistance to empire. Operation Pale Charlie takes 13-25 players on a multi-power politics simulation of 1950s Vietnam, giving them key roles in France, China, Russia, USA, Britain, French Indo-China and the Vietminh, otherwise known as the Indo-Chinese Communist Party and associates. Alliances made and broken, policies forged in the heat of battles and events, and nuclear brinkmanship met head on. Featuring Operation Castor, the Battle of Dien Bien Phu, Algeria, the Suez Canal question, and the Geneva Peace Conference. Individual and confidential state objectives, politicians and certainty, and the boundaries to leadership tested. Full situation briefings provided.


text:
User:Metadawson/Blog

=My Wikiversity Blog - Free the knowledge= Copy this link and paste it into your aggregator. Add entry

Linked Data
Yesterday I got my copy of Linked Data: Evolving the Web into a Global Data Space. The book was release in February. I'm looking forward to getting my understanding of Web data up-to-date. This is the broad area of study I am looking into to find a niche for my PhD. I want to get started in my research within the next 5 years. By then the industry may have changed drastically, so I need to stay abreast of the technology in use. Metadawson 16:31, 23 March 2011 (UTC)

Freeing Knowledge
“We can make knowledge available online more easily, more widely and more cheaply,” Willinsky said. There is an excellent article found at The Stanford Daily. The efforts are focused on making research articles and journals more available to everyone. The Public Knowledge Project itself will be one of my topics of research in the coming months. Metadawson 16:58, 3 March 2011 (UTC)

Berkman Center
I just found the Berkman Center for Internet & Society which is a organization at Harvard University interested in many aspects of Internet knowledge and law. I never knew where Larry Lessig works. I don't think there are any opportunities there for me to do research, but they are doing some interesting work. 16:43, 3 March 2011 (UTC)

Gaming inspiration
Last night, in three hours of concentrated effort my online guild took down five raid bosses. These are big bad guys that take total coordination of 10 or 25 people. I had already killed and gotten treasure from the first three bosses on a previous date, so I chose to sit out and let other guild members play. I did minor chores in the game while listening to my guild (in VOIP) work together, fail and try again. I celebrated their wins with them and shared in our group success. After the first three bosses were dead one of my guild-mates offered to leave the group of 10 to make room for me to play. I joined in the group and helped to take down two more bosses. I did not get any treasure, but I was extremely satisfied that my friends and I conquered our foes and some of them got some nice rewards. I tell this story because this was the happiest I have been in several days. I felt accomplished, accepted and part of something bigger. Today I watched Jane McGonigal's Ted talk on gaming to save the world. With the recent extreme highs and lows in my life I realize I have to do something different. I need to try to integrate my online and real lives to make me feel successful and valuable in life. My challenge now is to determine how. I've accepted the quest, but there are no directions and no wiki to help. Right now, I am excited and scared and a bit confused. Hopefully I can maintain the hope. Metadawson 16:05, 2 March 2011 (UTC)

In Control 2011 conference
Monday-Tuesday, Feb. 21-22, I attended the second annual In Control conference organized by Environments for Humans. The topics included HTML5, CSS3, microformats, jQuery, design, user experience and testing. It was a fabulously energizing two days. The presenters were all friendly and had more exciting information than they could share in the limited time. It is so nice to be with fellow nerds that are into intelligent (Web) design. I've given myself a huge design project to put all I learned into practice. Metadawson 19:07, 24 February 2011 (UTC)

An Open Web, open to edits
A new ebook An Open Web begins with a dire warning that the Web is now closed. Reading further in the introduction I found two sentences that make me want to read this collaborative work.
 * "This book will take the view that the Open Web is an essential technology and cultural practice for the future of the Internet and human society. The Web as we know it has had a positive and even revolutionary impact on key areas of science, technology, politics and culture."

Perhaps this is not a doom and gloom warning about the rise of proprietary information. I will have to give it a read and report back. Metadawson 17:56, 7 February 2011 (UTC)

Mozilla Drumbeat to make CC easy
Today "Learning freedom and the web" posted an article "Announcing Open Attribute". The new tools strive to make posting open content with the proper Creative Commons licensing a super easy task. So far the tools are available as Firefox and Chrome addons. When I have some content to post, I'll try it out. Metadawson 17:45, 7 February 2011 (UTC)

More training needed for a cool job
I'm getting more and more interested in research every day. I have thought about looking for a PhD program in the next five years. I found this job listing today. It gives me some skills to shoot for in order to be ready for my next career. Wikimedia Foundation Job Opening --Metadawson 20:14, 2 February 2011 (UTC)

Joining Wikiversity
February 2nd, I added my name to a list for the Composing free and open online educational resources class on Wikiversity. It looks like the course should have began in January but has been postponed. We will see if there is a new start date announced.

Signing up to support Wikipedia
January 31st I completed the "Wikipedia Campus Ambassador" application. It looks like the second ever training session took place in January so I was late in applying. I hope the Wikimedia group hosts another training session this year. I am too excited to wait until 2012. Metadawson 18:25, 2 February 2011 (UTC)

Getting Started
January 30th I stumbled on Wikimedia's ambassador program. After reading up on the movement to promote free knowledge I excitedly threw my hat into the ring. I am so thrilled to have a new venture to pursue. Extracurricular and intellectual pursuits are important to me. Without such challenges I feel I am not living up to my potential. Check in with this blog for the story of my adventure. Metadawson 18:25, 2 February 2011 (UTC)

title: User:Metadawson/Blog


text:
User:Metadawson/Blog/rss

   Metadawson's Wikiversity Blog http://en.wikiversity.org/wiki/User:Metadawson/Blog My Wikiversity Blog - Free the knowledge! en

Blogging on Wikiversity http://en.wikiversity.org/wiki/User:Metadawson/Blog#Blogging_on_Wikiversity http://en.wikiversity.org/wiki/User:Metadawson/Blog#Blogging_on_Wikiversity Documenting my journey from meta dawson to wiki dawson. I'll record my progress as well as my mistakes in hopes of smoothing the process if possible. 18:44, 2 February 2011 (UTC) Metadawson

title: User:Metadawson/Blog/rss


text: File:Plotux.jpg

title: File:Plotux.jpg


text: File:Hw2 3 chart1.jpg

title: File:Hw2 3 chart1.jpg


text: File:Hw2 3 chart2.png

title: File:Hw2 3 chart2.png


title: File:Figure2.2.png


text: Wikiversity:Social contract

title: Wikiversity:Social contract


text:
Wikiversity talk:Social contract

This page was started to explore an idea that was discussed at colloqium. --mikeu talk 20:56, 2 February 2011 (UTC)

Re: Conflict Resolution
I presume the type of conflicts that arise are about editing articles (from what i've seen of Wikipedia) or procedural matters to do with how Wikiversity is run. If the problem is that the structure of some part of Wikiversity is constantly changed back and forth to suit the views of two conflicting editors then the resultant instability could be nullified by using an algorithm (either automated or enacted by people) that detects the identity of the editors causing the oscillation and removes the offending series of edits by both  the editors. Presumably if the offending edits are not vandalism both view points may have some validity. Much conflict can be avoided if people remember that dialog is only useful if done in the spirit of educating and learning at the same time.

 * What's needed in these situations is to boil the "he said" / "she said" into a succinct presentation of point and counter-point that doesn't simply replay endlessly, but becomes as concise as possible so that the conflict can be analyzed by independent parties so that a resolution may be crafted. —Moulton 01:03, 15 February 2011 (UTC)

Issues to address
A preliminary shortlist.

 * Collaboration
 * Commitment
 * Dedication
 * Due process
 * Focus
 * Honor
 * Privacy (that includes outing elswehere)
 * Quality
 * Respect (that includes behaviour elsewhere)
 * Support

Regards, Guido den Broeder 00:53, 15 February 2011 (UTC)

title: Wikiversity talk:Social contract