User:Egm6321.f10.team2.oztekin/Newwwww

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

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


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


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


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


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

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


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

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

Using 4 elements

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$$  \displaystyle \Omega =]0,1[ $$
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Lets find shape functions for 4 elements.


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


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


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


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


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

\end{align} $$     (8.38)
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$$  \displaystyle 4 elements $$
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Using 6 elements
Shape functions for 6 elements
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$$  \displaystyle\begin{align} & N^{(1)}=\frac{1}{0.1666}[\begin{matrix} 0.1666-x & x \end{matrix}]\\ &N^{(2)}=\frac{1}{0.1666}[\begin{matrix} 0.3333-x & x-0.16666 \end{matrix}]\\ & N^{(3)}=\frac{1}{0.1666}[\begin{matrix} 0.5-x & x-0.3333 \end{matrix}]\\ &N^{(4)}=\frac{1}{0.1666}[\begin{matrix} 0.6666-x & x-0.5 \end{matrix}]\\ & N^{(5)}=\frac{1}{0.1666}[\begin{matrix} 0.8332-x & x-0.6666 \end{matrix}]\\ & N^{(6)}=\frac{1}{0.1666}[\begin{matrix} 1-x & x-0.8332 \end{matrix}]\\
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\end{align} $$     (8.39)
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$$  \displaystyle \left[ \begin{matrix} \begin{matrix} \begin{matrix} \begin{matrix} 13.5048 & -13.5048 & 0 & 0 \\ \end{matrix} & 0 & 0 & 0  \\ \end{matrix} \\ \begin{matrix} \begin{matrix} -13.5048 & 30.0105 & -16.5057 & 0 \\ \end{matrix} & 0 & 0 & 0  \\ \end{matrix} \\ \begin{matrix} \begin{matrix} 0 & -16.5057 & 36.0132 & -19.5075 \\ \end{matrix} & 0 & 0 & 0  \\ \end{matrix} \\ \begin{matrix} \begin{matrix} 0 & 0 & -19.5075 & 42.0157 \\ \end{matrix} & -22.5084 & 0 & 0  \\ \end{matrix} \\ \end{matrix} \\ \begin{matrix} \begin{matrix} 0 & 0 & 0 & -22.5084 \\ \end{matrix} & 48.0164 & -25.508 & 0  \\ \end{matrix} \\ \begin{matrix} \begin{matrix} 0 & 0 & 0 & 0 \\ \end{matrix} & -25.508 & 54.008 & -28.5  \\ \end{matrix} \\ \begin{matrix} \begin{matrix} 0 & 0 & 0 & 0 \\ \end{matrix} & 0 & -28.5 & 28.5  \\ \end{matrix} \\ \end{matrix} \right]\left[ \begin{matrix} \begin{matrix} {{d}_{1}} \\ {{d}_{2}} \\ {{d}_{3}} \\ {{d}_{4}} \\ \end{matrix} \\ {{d}_{5}} \\ {{d}_{6}} \\ 4 \\ \end{matrix} \right]=\left[ \begin{matrix} \begin{matrix} 6.0232199942 \\   0.13890282  \\   0.277972087  \\   0.416641681  \\ \end{matrix}  \\ 0.555277743 \\   0.694970317  \\   0.394287575  \\ \end{matrix} \right]+\left[ \begin{matrix} \begin{matrix} 0 \\   0  \\   0  \\   0  \\ \end{matrix}  \\ 0 \\   0  \\   {{r}_{1}}  \\ \end{matrix} \right]
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$$     (8.40)
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$$  \displaystyle \left[ \begin{matrix} \begin{matrix} {{d}_{1}} \\ {{d}_{2}} \\ {{d}_{3}} \\ \end{matrix} \\ {{d}_{4}} \\ {{d}_{5}} \\ {{d}_{6}} \\ \end{matrix} \right]=\left[ \begin{matrix} \begin{matrix} 2.0292 \\   1.5831  \\   1.2098  \\ \end{matrix}  \\ 0.8797 \\   0.575  \\   0.2845  \\ \end{matrix} \right]
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$$     (8.41)
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$$  \displaystyle\begin{align} & \left[ \begin{matrix} \begin{matrix} 2.0292 \\   1.5831  \\   1.2098  \\ \end{matrix}  \\ 0.8797 \\   0.575  \\   0.2845  \\ \end{matrix} \right]+\left[ \begin{matrix} \begin{matrix} 4 \\   4  \\   4  \\ \end{matrix}  \\ 4 \\   4  \\   4  \\ \end{matrix} \right]=\left[ \begin{matrix} \begin{matrix} 6.0292 \\   5.5831  \\   5.2098  \\ \end{matrix}  \\ 4.8797 \\   4.575  \\   4.2845  \\ \end{matrix} \right]
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& {{d}_{exact}}=\left[ \begin{matrix} \begin{matrix} 6.0331 \\   5.5853  \\   5.2109  \\ \end{matrix}  \\ 4.8802 \\   4.5753  \\   4.2847  \\ \end{matrix} \right]

\end{align} $$     (8.42)
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$$  \displaystyle 6 elements $$
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Using 8 elements
Shape functions for 8 elements
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$$  \displaystyle\begin{align} & N^{(1)}=\frac{1}{0.125}[\begin{matrix} 0.125-x & x \end{matrix}]\\ &N^{(2)}=\frac{1}{0.125}[\begin{matrix} 0.25-x & x-0.125 \end{matrix}]\\ & N^{(3)}=\frac{1}{0.125}[\begin{matrix} 0.375-x & x-0.25 \end{matrix}]\\ &N^{(4)}=\frac{1}{0.125}[\begin{matrix} 0.5-x & x-0.375 \end{matrix}]\\ & N^{(5)}=\frac{1}{0.125}[\begin{matrix} 0.625-x & x-0.5 \end{matrix}]\\ & N^{(6)}=\frac{1}{0.125}[\begin{matrix} 0.75-x & x-0.625 \end{matrix}]\\ & N^{(7)}=\frac{1}{0.125}[\begin{matrix} 0.875-x & x-0.75 \end{matrix}]\\ & N^{(8)}=\frac{1}{0.125}[\begin{matrix} 1-x & x-0.875 \end{matrix}]\\ \end{align} $$     (8.43)
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$$  \displaystyle \left[ \begin{matrix} 17.5 & -17.5 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\   -17.5 & 38 & -20.5 & 0 & 0 & 0 & 0 & 0 & 0  \\   0 & -20.5 & 44 & -23.5 & 0 & 0 & 0 & 0 & 0  \\   0 & 0 & -23.5 & 50 & -26.5 & 0 & 0 & 0 & 0  \\   0 & 0 & 0 & -26.5 & 56 & -29.5 & 0 & 0 & 0  \\   0 & 0 & 0 & 0 & -29.5 & 62 & -32.5 & 0 & 0  \\   0 & 0 & 0 & 0 & 0 & -32.5 & 68 & -35.5 & 0  \\   0 & 0 & 0 & 0 & 0 & 0 & -35.5 & 74 & -38.5  \\   0 & 0 & 0 & 0 & 0 & 0 & 0 & -38.5 & 38.5  \\ \end{matrix} \right]\left[ \begin{matrix} \begin{matrix} \begin{matrix} {{d}_{1}} \\ {{d}_{2}} \\ {{d}_{3}} \\ \end{matrix} \\ {{d}_{4}} \\ {{d}_{5}} \\ {{d}_{6}} \\ \end{matrix} \\ {{d}_{7}} \\ {{d}_{8}} \\ {{d}_{9}} \\ \end{matrix} \right]=\left[ \begin{matrix} \begin{matrix} \begin{matrix} \begin{matrix} 6.01302084 \\   0.078125  \\ \end{matrix}  \\ 0.15625 \\ \end{matrix}  \\ 0.234375 \\   0.31250004  \\   0.390625  \\ \end{matrix}  \\ 0.46874373 \\   0.546875  \\   0.29947916  \\ \end{matrix} \right]+\left[ \begin{matrix} \begin{matrix} \begin{matrix} 0 \\   0  \\   0  \\ \end{matrix}  \\ 0 \\   0  \\   0  \\ \end{matrix}  \\ 0 \\   0  \\   {{r}_{1}}  \\ \end{matrix} \right]
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$$     (8.44)
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$$  \displaystyle \left[ \begin{matrix} \begin{matrix} \begin{matrix} \begin{matrix} {{d}_{1}} \\ {{d}_{2}} \\ \end{matrix} \\ {{d}_{3}} \\ \end{matrix} \\ {{d}_{4}} \\ {{d}_{5}} \\ \end{matrix} \\ {{d}_{6}} \\ {{d}_{7}} \\ {{d}_{8}} \\ \end{matrix} \right]=\left[ \begin{matrix} \begin{matrix} \begin{matrix} 2.0312 \\   1.6876  \\ \end{matrix}  \\ 1.3904 \\   1.1246  \\   0.8800  \\ \end{matrix}  \\ 0.6497 \\   0.4286  \\   0.213  \\ \end{matrix} \right]
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$$     (8.45)
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$$  \displaystyle\begin{align} &\left[ \begin{matrix} \begin{matrix} \begin{matrix} 2.0312 \\   1.6876  \\ \end{matrix}  \\ 1.3904 \\   1.1246  \\   0.8800  \\ \end{matrix}  \\ 0.6497 \\   0.4286  \\   0.213  \\ \end{matrix} \right]+\left[ \begin{matrix} \begin{matrix} \begin{matrix} 4 \\   4  \\ \end{matrix}  \\ 4 \\   4  \\   4  \\ \end{matrix}  \\ 4 \\   4  \\   4  \\ \end{matrix} \right]=\left[ \begin{matrix} \begin{matrix} \begin{matrix} 6.0312 \\   5.6876  \\ \end{matrix}  \\ 5.3904 \\   5.1246  \\   4.88  \\ \end{matrix}  \\ 4.6497 \\   4.4286  \\   4.213  \\ \end{matrix} \right]
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&{{d}_{exact}}=\left[ \begin{matrix} \begin{matrix} \begin{matrix} 6.0331 \\   5.6866  \\ \end{matrix}  \\ 5.3911 \\   5.1249  \\   4.8802  \\ \end{matrix}  \\ 4.6498 \\   4.4286  \\   4.213  \\ \end{matrix} \right]

\end{align} $$     (8.46)
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$$  \displaystyle 8 elements $$
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