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


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$$
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\displaystyle {K^e} = {k^e}\left[ {\begin{array}{ccccccccccccccc} &{\cos {\phi ^e}\sin {\phi ^e}}&{ - {{\cos }^2}{\phi ^e}}&{ - \cos {\phi ^e}\sin {\phi ^e}} \\ {\cos {\phi ^e}\sin {\phi ^e}}&&{ - \cos {\phi ^e}\sin {\phi ^e}}&{ - {{\sin }^2}{\phi ^e}} \\ { - {{\cos }^2}{\phi ^e}}&{ - \cos {\phi ^e}\sin {\phi ^e}}&&{\cos {\phi ^e}\sin {\phi ^e}} \\ { - \cos {\phi ^e}\sin {\phi ^e}}&{ - {{\sin }^2}{\phi ^e}}&{\cos {\phi ^e}\sin {\phi ^e}}& \end{array}} \right] $$      (3.3.1)
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For element 1 $$ {\phi ^1} = {90^o} $$, Global node numbers 1 & 3
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$$  \displaystyle {K^1} = AE\left[ {\begin{array}{ccccccccccccccc} 0&0&0&0 \\  0&1&0&{ - 1} \\   0&0&0&0 \\   0&{ - 1}&0&1 \end{array}} \right] $$
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For element 2 $$ {\phi ^2} = {135^o} $$,Global node numbers 2 & 3


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$$  \displaystyle {K^2} = AE\left[ {\begin{array}{ccccccccccccccc} {\frac{1}}&{ - \frac{1}}&{ - \frac{1}}&{\frac{1}} \\ { - \frac{1}}&{\frac{1}}&{\frac{1}}&{ - \frac{1}} \\ { - \frac{1}}&{\frac{1}}&{\frac{1}}&{ - \frac{1}} \\ {\frac{1}}&{ - \frac{1}}&{ - \frac{1}}&{\frac{1}} \end{array}} \right] $$
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For element 3 $$ {\phi ^3} = {90^o} $$,Global node numbers 2 & 4
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$$  \displaystyle {K^3} = AE\left[ {\begin{array}{ccccccccccccccc} 0&0&0&0 \\  0&1&0&{ - 1} \\   0&0&0&0 \\   0&{ - 1}&0&1 \end{array}} \right] $$
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For element 4 $$ {\phi ^4} = {0^o} $$,Global node numbers 3 & 4
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$$  \displaystyle {K^4} = AE\left[ {\begin{array}{ccccccccccccccc} 1&0&{ - 1}&0 \\  0&0&0&0 \\   { - 1}&0&{ - 1}&0 \\   0&0&0&0 \end{array}} \right] $$
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Global matrix can be found by adding corresponding elements of each matrix.
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$$  \displaystyle K = AE\left[ {\begin{array}{ccccccccccccccc} 0&0&0&0&0&0&0&0 \\  0&1&0&0&0&{ - 1}&0&0 \\   0&0&{\frac{1}}&{ - \frac{1}}&{ - \frac{1}}&{\frac{1}}&0&0 \\ 0&0&{ - \frac{1}}&{1 + \frac{1}}&{\frac{1}}&{ - \frac{1}}&0&{ - 1} \\ 0&0&{ - \frac{1}}&{\frac{1}}&{1 + \frac{1}}&{ - \frac{1}}&{ - 1}&0 \\ 0&{ - 1}&{\frac{1}}&{ - \frac{1}}&{ - \frac{1}}&{1 + \frac{1}}&0&0 \\ 0&0&0&0&{ - 1}&0&1&0 \\  0&0&0&{ - 1}&0&0&0&1 \end{array}} \right] $$
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Matrix $$d$$ and $$ f+r$$ will be as follows


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$$  \displaystyle d = \left[ {\begin{array}{ccccccccccccccc} 0 \\  0 \\   0 \\   0 \\    \\    \\    \\ \end{array}} \right]f + r = \left[ {\begin{array}{ccccccccccccccc} \\   \\    \\    \\   0 \\   0 \\   {10} \\   0 \end{array}} \right] $$
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Now writing global system of equations
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$$  \displaystyle Kd = f + r $$
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$$  \displaystyle AE\left[ {\begin{array}{ccccccccccccccc} 0&0&0&0&\vline & 0&0&0&0 \\ 0&1&0&0&\vline & 0&{ - 1}&0&0 \\ 0&0&{\frac{1}}&{ - \frac{1}}&\vline & { - \frac{1}}&{\frac{1}}&0&0 \\ 0&0&{ - \frac{1}}&{1 + \frac{1}}&\vline & {\frac{1}}&{ - \frac{1}}&0&{ - 1} \\ \hline 0&0&{ - \frac{1}}&{\frac{1}}&\vline & {1 + \frac{1}}&{ - \frac{1}}&{ - 1}&0 \\ 0&{ - 1}&{\frac{1}}&{ - \frac{1}}&\vline & { - \frac{1}}&{1 + \frac{1}}&0&0 \\ 0&0&0&0&\vline & { - 1}&0&1&0 \\ 0&0&0&{ - 1}&\vline & 0&0&0&1 \end{array}} \right]\left[ {\begin{array}{ccccccccccccccc} 0 \\  0 \\   0 \\   0 \\ \hline \\   \\    \\ \end{array}} \right] = \left[ {\begin{array}{ccccccccccccccc} \\   \\    \\    \\ \hline 0 \\  0 \\   {10} \\   0 \end{array}} \right] $$
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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.
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$$  \displaystyle {K_E} = AE\left[ {\begin{array}{ccccccccccccccc} 0&0&0&0 \\  0&1&0&0 \\   0&0&{\frac{1}}&{ - \frac{1}} \\ 0&0&{ - \frac{1}}&{1 + \frac{1}} \end{array}} \right] {K_{EF}} = AE\left[ {\begin{array}{ccccccccccccccc} 0&0&0&0 \\  0&{ - 1}&0&0 \\   { - \frac{1}}&{\frac{1}}&0&0 \\ {\frac{1}}&{ - \frac{1}}&0&{ - 1} \end{array}} \right] $$
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$$  \displaystyle {K_F} = AE\left[ {\begin{array}{ccccccccccccccc} {1 + \frac{1}}&{ - \frac{1}}&{ - 1}&0 \\ { - \frac{1}}&{1 + \frac{1}}&0&0 \\ { - 1}&0&1&0 \\  0&0&0&1 \end{array}} \right] {{\tilde d}_E} = \left[ {\begin{array}{ccccccccccccccc} 0 \\  0 \\   0 \\   0 \end{array}} \right] $$
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$$  \displaystyle {d_F} = \left[ {\begin{array}{ccccccccccccccc} \\   \\    \\ \end{array}} \right] {r_E} = \left[ {\begin{array}{ccccccccccccccc} \\   \\    \\ \end{array}} \right] {f_F} = \left[ {\begin{array}{ccccccccccccccc} 0 \\  0 \\   {10} \\   0 \end{array}} \right] $$
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$$  \displaystyle {K_{EF}}^T{\cancelto{0}{{\tilde d}_E}} + {K_F}{d_F} = {f_F}
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$$      (3.3.2)
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$$  \displaystyle {K_F}{d_F} = {f_F} $$      (3.3.3)
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$$  \displaystyle {d_F} = K_F^{ - 1}{f_F} $$      (3.3.4)
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$$  \displaystyle {d_F} = \left[ {\begin{array}{ccccccccccccccc} {0.1914} \\  {0.05} \\   {0.2414} \\   0 \end{array}} \right] \times {10^{ - 7}} $$
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$$  \displaystyle {r_E} = {K_E}{\cancelto{0}{{\tilde d}_E}} + {K_{EF}}{d_F} $$      (3.3.5)
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$$  \displaystyle {r_E} = {K_{EF}}{d_F} $$      (3.3.6)
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$$  \displaystyle {r_E} = \left[ {\begin{array}{ccccccccccccccc} 0 \\  { - 10} \\   { - 10} \\   { 10} \end{array}} \right] $$
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d. Stresses in the elements.


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$$  \displaystyle {\sigma ^e} = \frac\left[ {\begin{array}{ccccccccccccccc} { - \cos {\varphi ^e}}&{ - \sin {\varphi ^e}}&{\cos {\varphi ^e}}&{\sin {\varphi ^e}} \end{array}} \right]{d^e} $$      (3.3.7)
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For element 1 $$\varphi = {90^o}$$
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$$  \displaystyle {d^1} = \left[ {\begin{array}{ccccccccccccccc} \\   \\    \\ \end{array}} \right] = \left[ {\begin{array}{ccccccccccccccc} 0 \\  0 \\   {0.1914} \\   {0.05} \end{array}} \right] \times {10^{ - 7}} $$
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$$  \displaystyle {\sigma ^1} = \frac{1}\left[ {\begin{array}{ccccccccccccccc} 0&{ - 1}&0&1 \end{array}} \right]\left[ {\begin{array}{ccccccccccccccc} 0 \\  0 \\   {0.1914} \\   {0.05} \end{array}} \right] \times {10^{ - 7}} $$
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For element 2 $$\varphi = {135^o}$$


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$$  \displaystyle {d^2} = \left[ {\begin{array}{ccccccccccccccc} \\   \\    \\ \end{array}} \right] = \left[ {\begin{array}{ccccccccccccccc} 0 \\  0 \\   {0.1914} \\   {0.05} \end{array}} \right] \times {10^{ - 7}} $$
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$$  \displaystyle {\sigma ^2} = \frac{1}\left[ {\begin{array}{ccccccccccccccc} {\frac{1}}&{ - \frac{1}}&{ - \frac{1}}&{\frac{1}} \end{array}} \right]\left[ {\begin{array}{ccccccccccccccc} 0 \\  0 \\   {0.1914} \\   {0.05} \end{array}} \right] \times {10^{ - 7}} $$
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For element 3 $$\varphi = {90^o}$$


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$$  \displaystyle {d^3} = \left[ {\begin{array}{ccccccccccccccc} \\   \\    \\ \end{array}} \right] = \left[ {\begin{array}{ccccccccccccccc} 0 \\  0 \\   {0.2414} \\   0 \end{array}} \right] \times {10^{ - 7}} $$
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$$  \displaystyle {\sigma ^3} = \frac{1}\left[ {\begin{array}{ccccccccccccccc} 0&{ - 1}&0&1 \end{array}} \right]\left[ {\begin{array}{ccccccccccccccc} 0 \\  0 \\   {0.2414} \\   0 \end{array}} \right] \times {10^{ - 7}} $$
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For element 4 $$\varphi = {0^o}$$


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$$  \displaystyle {d^4} = \left[ {\begin{array}{ccccccccccccccc} \\   \\    \\ \end{array}} \right] = \left[ {\begin{array}{ccccccccccccccc} {0.1914} \\  {0.05} \\   {0.2414} \\   0 \end{array}} \right] \times {10^{ - 7}} $$
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$$  \displaystyle {\sigma ^4} = \frac{1}\left[ {\begin{array}{ccccccccccccccc} { - 1}&0&1&{} \end{array}} \right]\left[ {\begin{array}{ccccccccccccccc} {0.1914} \\  {0.05} \\   {0.2414} \\   0 \end{array}} \right] \times {10^{ - 7}} $$
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=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


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$$
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\displaystyle {K^e} = {k^e}\left[ {\begin{array}{ccccccccccccccc} &{\cos {\phi ^e}\sin {\phi ^e}}&{ - {{\cos }^2}{\phi ^e}}&{ - \cos {\phi ^e}\sin {\phi ^e}} \\ {\cos {\phi ^e}\sin {\phi ^e}}&&{ - \cos {\phi ^e}\sin {\phi ^e}}&{ - {{\sin }^2}{\phi ^e}} \\ { - {{\cos }^2}{\phi ^e}}&{ - \cos {\phi ^e}\sin {\phi ^e}}&&{\cos {\phi ^e}\sin {\phi ^e}} \\ { - \cos {\phi ^e}\sin {\phi ^e}}&{ - {{\sin }^2}{\phi ^e}}&{\cos {\phi ^e}\sin {\phi ^e}}& \end{array}} \right] $$      (3.6.1)
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For element 1 $$ {\phi ^1} = {{-45}^o} $$, Global node numbers 1 & 4
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$$  \displaystyle {K^1} = AE\left[ {\begin{array}{ccccccccccccccc} {1/2}&{-1/2}&{-1/2}&{1/2} \\  {-1/2}&{1/2}&{1/2}&{-1/2} \\   {-1/2}&{1/2}&{1/2}&{-1/2} \\   {1/2}&{-1/2}&{-1/2}&{1/2} \end{array}} \right] $$
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For element 2 $$ {\phi ^2} = {{-90}^o} $$,Global node numbers 2 & 4


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$$ \displaystyle {K^2} = A_{BC}E\left[ {\begin{array}{ccccccccccccccc} 0&0&0&0 \\  0&1&0&{ - 1} \\   0&0&0&0 \\   0&{ - 1}&0&1 \end{array}} \right] $$
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$$  \displaystyle {K^2} = AE\left[ {\begin{array}{ccccccccccccccc} 0&0&0&0 \\  0&2&0&{ - 2} \\   0&0&0&0 \\   0&{ - 2}&0&2 \end{array}} \right] $$
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For element 3 $$ {\phi ^4} = {{-135}^o} $$,Global node numbers 3 & 4
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$$ \displaystyle {K^3} = AE\left[ {\begin{array}{ccccccccccccccc} {1/2}&{1/2}&{-1/2}&{-1/2} \\  {1/2}&{1/2}&{-1/2}&{-1/2} \\   {-1/2}&{-1/2}&{1/2}&{1/2} \\   {-1/2}&{-1/2}&{1/2}&{1/2} \end{array}} \right] $$
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Global matrix can be found by adding corresponding elements of each matrix.


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$$  \displaystyle K = AE\left[ {\begin{array}{ccccccccccccccc} {\frac{1}{2}}&{ - \frac{1}{2}}&0&0&0&0&{ - \frac{1}{2}}&{\frac{1}{2}} \\ { - \frac{1}{2}}&{\frac{1}{2}}&0&0&0&0&{\frac{1}{2}}&{ - \frac{1}{2}} \\ 0&0&0&0&0&0&0&0 \\  0&0&0&2&0&0&0&{ - 2} \\   0&0&0&0&{\frac{1}{2}}&{\frac{1}{2}}&{ - \frac{1}{2}}&{ - \frac{1}{2}} \\ 0&0&0&0&{\frac{1}{2}}&{\frac{1}{2}}&{ - \frac{1}{2}}&{ - \frac{1}{2}} \\ { - \frac{1}{2}}&{\frac{1}{2}}&0&0&{ - \frac{1}{2}}&{ - \frac{1}{2}}&1&0 \\ {\frac{1}{2}}&{ - \frac{1}{2}}&0&{ - 2}&{ - \frac{1}{2}}&{ - \frac{1}{2}}&0&3 \end{array}} \right] $$
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$$  \displaystyle d = \left[ {\begin{array}{ccccccccccccccc} 0 \\  0 \\   0 \\   0 \\    \\   0 \\    \\ \end{array}} \right] f + r = \left[ {\begin{array}{ccccccccccccccc} \\   \\    \\    \\   0 \\    \\    \\   0 \end{array}} \right] $$
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Now writing global system of equations
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$$  \displaystyle Kd = f + r $$ Swapping the row 5&6 of $$K$$,$$d$$ and $$f+r$$ matrices, and column 5&6 of$$K$$ matrix.
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$$  \displaystyle \left[ {\begin{array}{ccccccccccccccc} {\frac{1}{2}}&{ - \frac{1}{2}}&0&0&0&\vline & 0&{ - \frac{1}{2}}&{\frac{1}{2}} \\ { - \frac{1}{2}}&{\frac{1}{2}}&0&0&0&\vline & 0&{\frac{1}{2}}&{ - \frac{1}{2}} \\ 0&0&0&0&0&\vline & 0&0&0 \\ 0&0&0&2&0&\vline & 0&0&{ - 2} \\ 0&0&0&0&{\frac{1}{2}}&\vline & {\frac{1}{2}}&{ - \frac{1}{2}}&{ - \frac{1}{2}} \\ \hline 0&0&0&0&{\frac{1}{2}}&\vline & {\frac{1}{2}}&{ - \frac{1}{2}}&{ - \frac{1}{2}} \\ { - \frac{1}{2}}&{\frac{1}{2}}&0&0&{ - \frac{1}{2}}&\vline & { - \frac{1}{2}}&1&0 \\ {\frac{1}{2}}&{ - \frac{1}{2}}&0&{ - 2}&{ - \frac{1}{2}}&\vline & { - \frac{1}{2}}&0&3 \end{array}} \right]\left[ {\begin{array}{ccccccccccccccc} 0 \\  0 \\   0 \\   0 \\   0 \\ \hline \\   \\ \end{array}} \right] = \left[ {\begin{array}{ccccccccccccccc} \\   \\    \\    \\    \\ \hline 0 \\   \\   0 \end{array}} \right] $$
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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.


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$$  \displaystyle {K_E} = \left[ {\begin{array}{ccccccccccccccc} {\frac{1}{2}}&{ - \frac{1}{2}}&0&0&0 \\ { - \frac{1}{2}}&{\frac{1}{2}}&0&0&0 \\ 0&0&0&0&0 \\  0&0&0&2&0 \\   0&0&0&0&{\frac{1}{2}} \end{array}} \right] {K_{EF}} = \left[ {\begin{array}{ccccccccccccccc} 0&{ - \frac{1}{2}}&{\frac{1}{2}} \\ 0&{\frac{1}{2}}&{ - \frac{1}{2}} \\ 0&0&0 \\  0&0&{ - 2} \\   {\frac{1}{2}}&{ - \frac{1}{2}}&{ - \frac{1}{2}} \end{array}} \right] $$
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$$  \displaystyle {K_F} = \left[ {\begin{array}{ccccccccccccccc} {\frac{1}{2}}&{ - \frac{1}{2}}&{ - \frac{1}{2}} \\ { - \frac{1}{2}}&1&0 \\ { - \frac{1}{2}}&0&3 \end{array}} \right] {d_F} = \left[ {\begin{array}{ccccccccccccccc} \\   \\ \end{array}} \right] {f_F} = \left[ {\begin{array}{ccccccccccccccc} 0 \\   \\   0 \end{array}} \right] $$
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$$  \displaystyle {{\tilde d}_E} = \left[ {\begin{array}{ccccccccccccccc} 0 \\  0 \\   0 \\   0 \\   0 \end{array}} \right] {r_E} = \left[ {\begin{array}{ccccccccccccccc} \\   \\    \\    \\ \end{array}} \right] $$
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$$  \displaystyle {K_{EF}}^T{\cancelto{0}{{\tilde d}_E}} + {K_F}{d_F} = {f_F}
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$$      (3.6.2)
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$$  \displaystyle {K_F}{d_F} = {f_F} $$      (3.6.3)
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$$  \displaystyle {d_F} = K_F^{ - 1}{f_F} $$      (3.6.4)
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$$  \displaystyle {d_F} = \left[ {\begin{array}{ccccccccccccccc} {0.30} \\  {0.25} \\   {0.05} \end{array}} \right] \times {10^{ - 5}}m $$
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c. Stresses in the elements.


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$$  \displaystyle {\sigma ^e} = \frac\left[ {\begin{array}{ccccccccccccccc} { - \cos {\varphi ^e}}&{ - \sin {\varphi ^e}}&{\cos {\varphi ^e}}&{\sin {\varphi ^e}} \end{array}} \right]{d^e} $$      (3.3.7)
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For element 1 $$\varphi = {{-45}^o}$$
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$$  \displaystyle {d^1} = \left[ {\begin{array}{ccccccccccccccc} \\   \\    \\ \end{array}} \right] = \left[ {\begin{array}{ccccccccccccccc} 0 \\  0 \\   {0.25} \\   {0.05} \end{array}} \right] \times {10^{ - 5}} $$
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$$  \displaystyle {\sigma ^1} = \frac\left[ {\begin{array}{ccccccccccccccc} { - \frac{1}}&{\frac{1}}&{\frac{1}}&{ - \frac{1}} \end{array}} \right]\left[ {\begin{array}{ccccccccccccccc} 0 \\  0 \\   {0.25} \\   {0.05} \end{array}} \right] \times {10^{ - 5}} $$
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For element 2 $$\varphi = {{-90}^o}$$
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$$  \displaystyle {d^1} = \left[ {\begin{array}{ccccccccccccccc} \\   \\    \\ \end{array}} \right] = \left[ {\begin{array}{ccccccccccccccc} 0 \\  0 \\   {0.25} \\   {0.05} \end{array}} \right] \times {10^{ - 5}} $$
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$$  \displaystyle {\sigma ^2} = \frac{1}\left[ {\begin{array}{ccccccccccccccc} 0&1&0&{ - 1} \end{array}} \right]\left[ {\begin{array}{ccccccccccccccc} 0 \\  0 \\   {0.25} \\   {0.05} \end{array}} \right] \times {10^{ - 5}} $$
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For element 3 $$\varphi = {{-135}^o}$$
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$$  \displaystyle {d^1} = \left[ {\begin{array}{ccccccccccccccc} \\   \\    \\ \end{array}} \right] = \left[ {\begin{array}{ccccccccccccccc} {0.30} \\  0 \\   {0.25} \\   {0.05} \end{array}} \right] \times {10^{ - 5}} $$
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$$  \displaystyle {\sigma ^3} = \frac\left[ {\begin{array}{ccccccccccccccc} {\frac{1}}&{\frac{1}}&{ - \frac{1}}&{ - \frac{1}} \end{array}} \right]\left[ {\begin{array}{ccccccccccccccc} {0.30} \\  0 \\   {0.25} \\   {0.05} \end{array}} \right] \times {10^{ - 5}} $$
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d. Reactions at node C(node 2),D(node 3) & F(node 1)


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$$  \displaystyle {r_E} = {K_E}{\cancelto{0}{{\tilde d}_E}} + {K_{EF}}{d_F} $$      (3.6.5)
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$$  \displaystyle {r_E} = {K_{EF}}{d_F} $$      (3.6.6)
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$$  \displaystyle ${r_E} = \left[ {\begin{array}{*{20}{c}} { - {{10}^3}} \\   \\   0 \\   { - {{10}^3}} \\   0 \end{array}} \right]Newton $$
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Now as element '3' is free to move in x direction so reaction $$$$ will be zero. Adding this element to $${r_E}$$ matrix.


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$$  \displaystyle $$      (N)
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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


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$$  \displaystyle \underline v = \sum\limits_{j = 1}^n $$      (2.1)
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$$  \displaystyle \sum\limits_{j = 1}^n {{{\underline b }_j}{v_j} - \underline v = 0} $$      (2.2)
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$$  \displaystyle \underline{\underline P} \left( {\underline v } \right) = 0 $$      (2.3)
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$$  \displaystyle \underline{\underline P} \left( v \right) = \sum\limits_{j = 1}^n {{{\underline b }_j}{v_j} - \underline v } $$      (2.4)
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Successively multiply by $${\underline b _i}\left( {i = 1,2,......n} \right)$$ equations with n unknowns are formed


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$$  \displaystyle {\underline b _i}.\underline{\underline P} \left( {\underline v } \right) = 0 $$      (2.5)
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$$  \displaystyle {\underline b _i}\sum\limits_{j = 1}^n  = {\underline b _i}.\underline v $$ (2.6)
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$$  \displaystyle \sum\limits_{j = 1}^n {{{\underline b }_i}.{{\underline b }_j}{v_j}} = {\underline b _i}.\underline v $$ (2.7)
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$$  \displaystyle {\left[ \right]_{n \times n}}{\left[  \right]_{n \times 1}} = {\left[  \right]_{n \times 1}} $$      (2.8)
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$$  \displaystyle \underline K \underline d = \underline F $$ (2.9)
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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.
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$$  \displaystyle \underline K = \left( {\begin{array}{ccccccccccccccc}  {{{\underline b }_1}.{{\underline b }_1}}&{{{\underline b }_1}.{{\underline b }_2}}&{{{\underline b }_1}.{{\underline b }_3}} \\   {{{\underline b }_2}.{{\underline b }_1}}&{{{\underline b }_2}.\underline  }&{{{\underline b }_2}.{{\underline b }_3}} \\   {{{\underline b }_3}.{{\underline b }_1}}&{{{\underline b }_3}.{{\underline b }_2}}&{{{\underline b }_3}.{{\underline b }_3}} \end{array}} \right) $$      (2.10)
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$$  \displaystyle \underline K = \left( {\begin{array}{ccccccccccccccc}  3&4&{11} \\   4&{14}&{22} \\   {11}&{22}&{49} \end{array}} \right) $$
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$$  \displaystyle \det \left( {\underline K } \right) = 64 $$      (2.11)
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3) As shown in equation numbers (7) & (8).

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


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$$  \displaystyle \left( {\begin{array}{ccccccccccccccc}  \\    \\ \end{array}} \right) = \left( {\begin{array}{ccccccccccccccc}  {{{\underline b }_1}.\underline v } \\   {{{\underline b }_2}.\underline v } \\   {{{\underline b }_3}.\underline v } \end{array}} \right) $$      (2.12)
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$$  \displaystyle \left( {\begin{array}{ccccccccccccccc}  \\    \\ \end{array}} \right) = \left( {\begin{array}{ccccccccccccccc}  { - 6} \\   5 \\   { - 23} \end{array}} \right) $$
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4) To find the value of $$\underline d$$


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$$  {\underline K ^{ - 1}} = \left( {\begin{array}{ccccccccccccccc}  {3.1562}&{0.7187}&{ - 1.0312} \\   {0.7187}&{0.4062}&{ - 0.3437} \\   { - 1.0312}&{ - 0.3437}&{0.4062} \end{array}} \right) $$
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$$  \displaystyle \underline d = {\underline K ^{ - 1}}\underline F $$ (2.13)
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$$  \displaystyle \underline d = \left( {\begin{array}{ccccccccccccccc}  {8.375} \\   {5.625} \\   { - 4.875} \end{array}} \right) $$
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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.


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$$  \displaystyle {\underline w _i}.\underline{\underline P} \left( {\underline v } \right) = 0 $$      (2.14)
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Putting value of $$\underline{\underline{P}}\left( v \right)$$ from equation number (4).


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$$  \displaystyle {\underline a _i}\sum\limits_{j = 1}^n  = {\underline a _i}.\underline v $$ (2.15)
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$$  \displaystyle \sum\limits_{j = 1}^n {{{\underline a }_i}.{{\underline b }_j}{v_j}} = {\underline a _i}.\underline v $$ (2.16)
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$$  \displaystyle {\left[ \right]_{n \times n}}{\left[  \right]_{n \times 1}} = {\left[  \right]_{n \times 1}} $$      (2.17)
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$$  \displaystyle \underline {\overline K } \underline d = \underline {\overline F } $$      (2.18)
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Where $$\underline{\overline{K}}$$ and $$\underline{\overline{F}}$$ are matrices having following elements.


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$$  \displaystyle \underline {\overline K } = \left( {\begin{array}{ccccccccccccccc}  {{{\underline a }_1}.{{\underline b }_1}}&{{{\underline a }_1}.{{\underline b }_2}}&{{a_1}.{{\underline b }_3}} \\   {{{\underline a }_2}.{{\underline b }_1}}&{{{\underline a }_2}.\underline  }&{{a_2}.{{\underline b }_3}} \\   {{{\underline a }_3}.{{\underline b }_1}}&{{{\underline a }_3}.{{\underline b }_2}}&{{{\underline a }_3}.{{\underline b }_3}} \end{array}} \right) $$      (2.19)
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$$  \displaystyle \overline {\underline F } = \left( {\begin{array}{ccccccccccccccc}  {{{\underline a }_1}.\underline v } \\   {{{\underline a }_2}.\underline v } \\   {{{\underline a }_3}.\underline v } \end{array}} \right) $$      (2.20)
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$$  \displaystyle \underline {\overline K } = \left( {\begin{array}{ccccccccccccccc}  1&2&3 \\   1&{ - 1}&2 \\   1&3&6 \end{array}} \right) $$
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$$\overline{\underline{F}}=\left( \begin{matrix}  5  \\   -7  \\   -4  \\ \end{matrix} \right)$$


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6) To find $$\underline d $$


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$$  \displaystyle \underline d = {\underline {\overline K } ^{ - 1}}\underline {\overline F } $$      (2.21)
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$${{\overline{\underline{K}}}^{-1}}=\left( \begin{matrix}  1.500 & 0.375 & -0.875  \\   0.500 & -0.375 & -0.125  \\   -0.500 & -0.125 & 0.375  \\ \end{matrix} \right)$$


 * }
 * }


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$$  \displaystyle \underline d = \left( {\begin{array}{ccccccccccccccc}  {8.375} \\   {5.625} \\   { - 4.875} \end{array}} \right) $$
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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

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$$  \displaystyle m\left( {x + \frac{2}} \right) = \rho \left( {x + \frac{2}} \right)\frac{1}{2}\left[ {h\left( x \right) + h\left( {x + dx} \right)} \right]b $$     (1)
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multiply both sides with $$dx$$


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$$  \displaystyle m\left( {x + \frac{2}} \right)dx = \left[ {\rho \left( {x + \frac{2}} \right)\frac{1}{2}\left[ {h\left( x \right) + h\left( {x + dx} \right)} \right]b} \right]dx $$     (2) expanding $$ h(x+dx)$$ with taylor series
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$$  \displaystyle m\left( {x + \frac{2}} \right)dx = \left[ {\rho \left( {x + \frac{2}} \right)\frac{1}{2}\left[ {h\left( x \right) + h\left( x \right) + \fracdx \ldots } \right]b} \right]dx $$     (3)
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$$  \displaystyle m\left( {x + \frac{2}} \right)dx = \left[ {\rho \left( {x + \frac{2}} \right)\left[ {h\left( x \right) + \frac\frac{2} \ldots } \right]b} \right]dx $$     (4)
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$$  \displaystyle m\left( {x + \frac{2}} \right)dx = \left[ {\rho \left( {x + \frac{2}} \right)\left[ {h\left( {x + \frac{2}} \right)} \right]b} \right]dx $$     (5) now expanding $$ \rho \left( {x + \frac{2}} \right)$$ and $$h\left( {x + \frac{2}} \right)$$ by taylor series
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$$  \displaystyle m\left( {x + \frac{2}} \right)dx = \left[ {\left[ {\rho \left( x \right) + \frac\frac{2} \ldots } \right]\left[ {h\left( x \right) + \frac\frac{2} \ldots } \right]b} \right]dx $$     (6)
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$$  \displaystyle m\left( {x + \frac{2}} \right)dx = \left[ {\rho \left( x \right)h\left( x \right) + \rho \left( x \right)\frac\frac{2} + h\left( x \right)\frac\frac{2} + \frac\frac{2}\frac\frac{2} \ldots } \right]bdx $$     (7)
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Neglecting higer order terms and considering first three terms of expansion


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$$  \displaystyle m\left( {x + \frac{2}} \right)dx = \left[ {\rho \left( x \right)h\left( x \right) + \frac\frac{2}} \right]bdx $$     (8)
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$$  \displaystyle m\left( {x + \frac{2}} \right)dx = \left[ {\rho \left( x \right)h\left( x \right)dx + \frac\frac{2}} \right]b $$     (9)
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Neglecting second term due to presence of $$d{x^2}$$


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$$  \displaystyle m\left( {x + \frac{2}} \right)dx = \left[ {\rho \left( x \right)h\left( x \right)dx} \right]b $$     (10)
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as $$bh\left( x \right) = A\left( x \right)$$


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$$  \displaystyle m\left( {x + \frac{2}} \right)dx = \rho \left( x \right)A\left( x \right)dx $$     (11)
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$$  \displaystyle m\left( {x + \frac{2}} \right) = \rho \left( x \right)A\left( x \right) $$     (12)
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$$  \displaystyle m\left( {x + \frac{2}} \right) = m\left( x \right) $$     (13)
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Now PDE for the general cross section as proved in 1.1.1


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$$  \displaystyle \frac{\partial }\left( {E\left( x \right)A\left( x \right)\frac} \right) + f\left( {x,t} \right) = m\frac $$     (14)
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For given rectangular cross section put $$m = m\left( {x + \frac{2}} \right)$$
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$$  \displaystyle \frac{\partial }\left( {E\left( x \right)A\left( x \right)\frac} \right) + f\left( {x,t} \right) = m\left( {x + \frac{2}} \right)\frac $$     (15)
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From euation no.(13) $$m\left( {x + \frac{2}} \right) = \rho \left( x \right)A\left( x \right)$$


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$$  \displaystyle \frac{\partial }\left( {E\left( x \right)A\left( x \right)\frac} \right) + f\left( {x,t} \right) = \rho \left( x \right)A\left( x \right)\frac $$     (16)
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Now put $$A\left( x \right) = bh\left( x \right)$$


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$$  \displaystyle \frac{\partial }\left( {E\left( x \right)bh\left( x \right)\frac} \right) + f\left( {x,t} \right) = \rho \left( x \right)bh\left( x \right)\frac $$     (17)
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As $$b$$ is a constant, taking it out of partial differential


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$$  \displaystyle b\frac{\partial }\left( {E\left( x \right)h\left( x \right)\frac} \right) + f\left( {x,t} \right) = b\rho \left( x \right)h\left( x \right)\frac $$     (18)
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Dividing both sides of equation with $$b$$


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$$  \displaystyle \frac{\partial }\left( {E\left( x \right)h\left( x \right)\frac} \right) + \frac{b} = \rho \left( x \right)h\left( x \right)\frac $$     (19) This partial differential equation represent a bar with varying rectangular cross section.
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