User:Eml5526.s11.team7.jb/57work

Problem Description
Solve the general 1-D model for an elastic bar with data set 1b (G1DM1.0/D1b) using the Lagrangian Interpolation Basis Function until convergence of $$\displaystyle u^h (0.5) \rightarrow O(10^{-6}) $$.
 * 1. Explain how the basis functions satisfy the constraint breaking solution (CBS).
 * 2.Plot these basis functions.
 * 3.Use matlab to integrate.
 * 4.Make Plots of u^h vs u and u^h(.5)-u(.5) vs m

Solution
The Lagrangian Interpolation Basis Function (LIBF) is given by the following equation and generates a polynomial of degree $$\displaystyle n-1$$:


 * $$L_{i,n}\left (x\right )=\prod_{k=1}^{n}\frac{x-x_{k}}{x_{i}-x_{k}},k\neq i$$

For a linear function the number of nodes is 2 ($$\displaystyle n=2$$). For uniform discretization with $$\displaystyle nel=4$$ the domain $$\displaystyle \Omega$$ will be subdivided into 4 equal length elements each having two nodes. The LIBFs for the first element $$\displaystyle n=1$$ over the elemental domain $$\displaystyle \omega^{(1)}=[0,.25]$$ can be written as follows:


 * $$L_{1,2}^{(1)}\left (x\right)=b_{1}^{(1)}=\frac{(x-x_2)}{x_1-x_2}=\frac{(x-.25)}{0-.25}=-4(x-.25)$$


 * $$L_{2,2}^{(1)}\left (x\right)=b_{2}^{(1)}=\frac{(x-x_1)}{x_2-x_1}=\frac{(x-0)}{.25-0}=4(x)$$

Recall the constraint breaking solution (CBS) states the basis functions must satisfy $$\displaystyle \left\{ b_i(x), i=0,1,2,...\right\}$$ such that $$\displaystyle b_0(\beta) \neq 0$$ and $$\displaystyle b_i(\beta) = 0$$ for $$\displaystyle i=1,2,...,n$$.

It can be seen that these basis functions for the first element satisfy the CBS at each node. For the first node, the basis function has a value of one at that node and a value of zero at every other node (there is only one other node for the linear LIBF). In other words the basis function is equal to the kronicker delta at each node.


 * $$b_{1}^{(1)}(\beta^{(1)}_1)=L_{1,2}^{(1)}\left (0\right)=b_{1}^{(1)}(0)=-4(0-.25)=1$$


 * $$L_{1,2}^{(1)}\left (.25\right)=b_{1}^{(1)}(.25)=-4(.25-.25)=0$$

Similarly for the second node, the basis function has a value of one at that second node and a value of zero at the first node.

A Matlab script was written to plot the LIBFs at each node for each element. The Matlab script is given followed by the plots. The first plot shows each of the LIBFs plotted across the entire domain. The second plot is more informative as the plot is divided into the elemental domains and lines corresponding to the values zero and one are present. The legend states the node number, $$n$$, and element number, $$e$$, of each LIBF.



The global matrices are formed by summing the matrices pertaining to each element:


 * $$\displaystyle \mathbf{K}_{ij} = \sum_{e=1}^{nel}\mathbf{K}^{e}_{ij}$$

The element matrices are formed as in problems 5.1 and 5.2 as follows:


 * $$\displaystyle \mathbf{F}^{e}_{F}=b_{i}^{(1)}(\alpha_{i}^{e} )h + \int_{\omega^{e}}^{.} {{b_{i}^{(1)}}fdx}$$



\displaystyle

h = {a_2}(\alpha )\frac = (2+3\alpha)\frac = 5*12/5=12

$$


 * $$\displaystyle \mathbf{F}^{(1)}_{1}=b_{1}^{(1)}(0)h + \int_{0}^{.25} {{b_{1}^{(1)}}fdx}=1*12+\int_{0}^{.25} -4(x-.25)(5x)=12.0520833$$


 * $$\displaystyle \mathbf{F}^{(1)}_{2}=b_{2}^{(1)}(.25)h + \int_{0}^{.25} {{b_{2}^{(1)}}fdx}=1*12+\int_{0}^{.25} 4x(5x)=12.104167$$

$$ \displaystyle

\therefore {F_F} = \left[ {\begin{array}{ccccccccccccccc} {12.0520833} \\    {12.104167} \end{array}} \right]

$$


 * $$\displaystyle \mathbf{K}^{e}_{ij}=\int_{\omega^{e}}^{.} b_{i}^{e'}(x^e)a_2(x^e)b_{j}^{e'}(x^e)dx$$



\displaystyle

\mathbf{K^{(1)}_{FF}} = \left\{ {\mathbf{K^{(1)}_{ij}};i,j = 1,...,n} \right\}

$$



\displaystyle

\mathbf{K^{(1)}_{11}}= \int\limits_0^{.25} b_{1}^{(1)'}a_2b_{1}^{(1)'}=\int\limits_0^{.25}-4*(2+3x)*-4=32(.25)+24(.25)^2=9.5

$$

\displaystyle

\mathbf{K^{(1)}_{12}}= \int\limits_0^{.25} b_{1}^{(1)'}a_2b_{2}^{(1)'}=\int\limits_0^{.25}-4*(2+3x)*4=-32(.25)-24(.25)^2=-9.5

$$



\displaystyle

\mathbf{K^{(1)}_{21}}=\mathbf{K^{(1)}_{12}} =-9.5

$$

\displaystyle

\mathbf{K^{(1)}_{22}}= \int\limits_0^{.25} b_{1}^{(1)'}a_2b_{1}^{(1)'}=\int\limits_0^{.25}4*(2+3x)*4=32(.25)+24(.25)^2=9.5

$$



\displaystyle \therefore \mathbf{K^{(1)}_{FF}} = \begin{bmatrix} 9.5&-9.5\\ -9.5&9.5 \end{bmatrix} $$



\mathbf{d_{F}^{(1)}} = \mathbf{K}_{FF}^{(1)-1} \mathbf{F_{F}^{(1)}} $$