User:Eml4500.f08.qwiki.bishop/Lecture 33

=Meeting 33 Wednesday November 12, 2008=

FEM via PVW (continued)
As we saw in Lecture ___, the assumed solution u(x) for a two-node element is a linear interpolation between the nodal unknowns, and is expressed as $$ u(x) = N_i(x) d_i + N_{i+1}(x) d_{i+1}$$. To obtain a solution for the linear displacement at $$x_{i+1} $$, we must evaluate $$ u(x) $$ for $$ u(x_{i+1})$$. Doing so yields:
 * $$ u(x_{i+1}) = N_i(x_{i+1}) d_i + N_{i+1}(x_{i+1}) d_{i+1} = d_{i+1} $$
 * $$ u(x_{i+1}) = \Bigl(\frac{x - x_{i+1}}{x_i - x_{i+1}}*d_i + \frac{x - x_i}{x_{i+1} - x_i} * d_{i+1}\Bigr) \Bigl\vert_{x = x_{i+1}} $$
 * $$ u(x_{i+1}) = \Bigl(\frac{x_{i+1} - x_{i+1}}{x_i - x_{i+1}}*d_i + \frac{x_{i+1} - x_i}{x_{i+1} - x_i} * d_{i+1}\Bigr) = \Bigl(\frac{0}{x_i - x_{i+1}}*d_i + \frac{x_{i+1} - x_i}{x_{i+1} - x_i} * d_{i+1}\Bigr)

= 0 * d_i + 1 * d_{i+1} $$ Therefore, one can see that $$ u(x_{i+1}) = d_{i+1} $$.

Since the weighting function w(x) is a linear function similar to $$ u(x) $$, one can linearly interpolate between $$ x_i \text{ and } x_{i+1} $$ to solve for $$w(x) \text{ : } $$
 * $$ w(x) = \Bigl(\frac{x - x_{i+1}}{x_i - x_{i+1}}*w_i + \frac{x - x_i}{x_{i+1} - x_i} * w_{i+1}\Bigr)

= N_i(x)w_i + N_{i+1}(x)w_{i+1} $$.

Element Stiffness Matrix for Element i
In practice, most finite element solutions are written in matrix notation to ease understanding of the system. For simple elements such as the one pictured above from Lecture ___, matrix notation is not necessary. However, for complicated elements it is nearly impossible to express element solutions without using matrix notation. In this section, we will show how to transform from the continuous stiffness equation $$ \beta = \int_{0}^{L} \frac{\delta w}{\delta x}(EA)\frac{\delta u}{\delta x}dx $$ to a discrete stiffness matrix $$ \mathbf{w}*(\mathbf{K} \mathbf{d}) $$. To begin, we must first note that
 * $$ \frac{\delta w}{\delta x} = [N_i^{'}w_i + N_{i+1}^{'}w_{i+1}] \text { and }

\frac{\delta u}{\delta x} = [N_i^{'}d_i + N_{i+1}^{'}d_{i+1}] \text{.}$$

$$\text{Therefore, } \beta = \int_{0}^{L} \frac{\delta w}{\delta x}(EA)\frac{\delta u}{\delta x}dx  \text{ can be expressed as } $$
 * $$ \beta = \int_{x_i}^{x_{i+1}} [N_i^{'}w_i + N_{i+1}^{'}w_{i+1}](EA)[N_i^{'}d_i + N_{i+1}^{'}d_{i+1}]dx \text{, where } N_i^{'} = \frac {\delta N_i(x)}{\delta x} \text{ and } N_{i+1}^{'} = \frac {\delta N_{i+1}(x)}{\delta x}.$$

Note that u(x) and $$ \frac{\delta u(x)}{\delta x} \quad $$can also be expressed in matrix form, where
 * $$ u(x) =

\lfloor N_i(x) \quad N_{i+1}(x)\rfloor \begin{bmatrix} d_i \\ d_{i+1} \\ \end{bmatrix} = \mathbf{N(x)}_{1x2} \begin{bmatrix} d_i \\ d_{i+1} \\ \end{bmatrix} \text{, and }$$


 * $$\frac{\delta u(x)}{\delta x} =

\lfloor N_i^{'}(x) \quad N_{i+1}^{'}(x)\rfloor \begin{bmatrix} d_i \\ d_{i+1} \\ \end{bmatrix} = \mathbf{B(x)}_{1x2} \begin{bmatrix} d_i \\ d_{i+1} \\ \end{bmatrix} $$

Similarly, w(x) and $$ \frac{\delta w(x)}{\delta x} \quad $$can also be expressed in matrix form, where
 * $$ w(x) = \mathbf{N(x)}_{1x2}

\begin{bmatrix} w_i \\ w_{i+1} \\ \end{bmatrix} \text{, and } $$


 * $$ \frac{\delta w(x)}{\delta x} = \mathbf{B(x)}_{1x2}

\begin{bmatrix} w_i \\ w_{i+1} \\ \end{bmatrix} $$

The element DOFs for element i are shown in the picture below:

The picture above shows that: Therefore, we may express equivalent DOFs in matrix format:
 * Node 1 of element i is equivalent to node i
 * Node 2 of element i is equivalent to node i+1
 * $$ \begin{bmatrix}

d_i \\ d_{i+1} \\ \end{bmatrix} = \begin{bmatrix} d_1^{(i)} \\ d_2^{(i)} \\ \end{bmatrix} = \mathbf{d}^{(i)} $$


 * $$ \begin{bmatrix}

w_i \\ w_{i+1} \\ \end{bmatrix} = \begin{bmatrix} w_1^{(i)} \\ w_2^{(i)} \\ \end{bmatrix} = \mathbf{w}^{(i)} $$

By substituting equivalent values for $$\frac{\delta u(x)}{\delta x} \quad$$ and $$ \frac{\delta w(x)}{\delta x} \quad $$ into $$ \beta = \int_{x_i}^{x_{i+1}} \frac{\delta w}{\delta x}(EA) \frac{\delta u}{\delta x}dx \quad $$, the continuous stiffness equation reduces to
 * $$ \beta = \int_{x_i}^{x_{i+1}} (\mathbf{B} \mathbf{w}^{(i)})(EA)(\mathbf{B} \mathbf{d}^{(i)})dx $$

(Move this and/or expain) Goal: $$ \beta = \mathbf{w}^{(i)} * (\mathbf{k}^{(i)} \mathbf{d}^{(i)}) $$

Since $$(EA) $$, $$(\mathbf{B} \mathbf{w}^{(i)})  $$ , and $$(\mathbf{B} \mathbf{d}^{(i)}) $$ are scalars, we may rearrange the terms in any fashion. Doing so allows us to express the continuous stiffness equation as


 * $$ \beta = \int_{x_i}^{x_{i+1}} (EA)(\mathbf{B} \mathbf{w}^{(i)})(\mathbf{B} \mathbf{d}^{(i)})dx $$

Recall that by the definition of a transpose:
 * $$ (\mathbf{B} \mathbf{w}^{(i)})^T = (\mathbf{w}^{(i)T} \mathbf{B}^T) = (\mathbf{w}^{(i)} \mathbf{B}^T)$$
 * $$ (\mathbf{B} \mathbf{d}^{(i)})^T = (\mathbf{d}^{(i)T} \mathbf{B}^T) = (\mathbf{d}^{(i)} \mathbf{B}^T)$$

Due to the fact that $$\mathbf{w}^{(i)} $$ and $$ \mathbf{d}^{(i)} $$ are independent from x, we may remove them from the integral:
 * $$ \beta = \mathbf{w}^{(i)} * \Bigl (\int_{x_i}^{x_{i+1}} \mathbf{B}^T(EA) \mathbf{B} dx \Bigr) \mathbf{d}^{(i)} $$

From the above equation, one may easily see that the stiffness equation $$ \beta = \mathbf{w}^{(i)} * \Bigl(\int_{x_i}^{x_{i+1}} \mathbf{B}^T(EA) \mathbf{B} dx \Bigr) \mathbf{d}^{(i)} $$ reduces to $$ \beta = \mathbf{w}^{(i)} * (\mathbf{k}^{(i)} \mathbf{d}^{(i)})$$. Therefore, we have solved for the stiffness matrix $$ \mathbf{k}^{(i)} $$:
 * $$ \mathbf{k}^{(i)} = \int_{x_i}^{x_{i+1}} \mathbf{B}^T(EA) \mathbf{B} dx $$

It is also known that
 * $$ B(x) = \lfloor N_i^{'}(x) \quad N_{i+1}^{'}(x)\rfloor = \Bigl \lfloor \frac{1}{x_i - x_{i+1}} \quad \frac{1}{x_{i+1} - x_i} \Bigr \rfloor = \Bigl \lfloor \frac{1}{-L^{(i)}} \quad \frac{1}{L^{(i)}} \Bigr \rfloor \quad$$, where the length of element i equals $$\quad L^{(i)} = x_{i+1} - x_i $$