User:Eml4500.f08.qwiki.bishop/HW7

=Lecture 35 - Monday, November 17, 2008=

Frame Elements
A frame element consists of a truss (axial deformation) element plus the beam (transverse deformation) element.

A model frame with 2 elements is shown below:



Because the connection between the two elements is rigid, the angle between them will stay constant after deformation.

The free body diagrams of each element is shown below:



\underbrace{f_i^{(e)}}_{\text{general forces}}$$
 * In general, $$\underbrace{d_i^{(e)}}_{\text{general displacement}}\rightarrow


 * e=1,2
 * i=1,...,6

$$\underbrace{d_3^{(e)},d_6^{(e)}}_{\text{rotational dofs}}\rightarrow \underbrace{f_3^{(e)},f_6^{(e)}}_{\text{bending moment}}$$

2-D Frame Global DoF's
The labeling system for the global degrees of freedom is shown below.



There are now 9 total degrees of freedom. Therefore, the following statements are true:


 * 2 element stiffness matrix: $$\underline{k}_{(6x6)}^{(e)}$$, e=1,2
 * Global stiffness matrix: $$\underline{k}_{(9x9)}=A\underline{k}_{(6x6)}^{(e)}$$

The 9x9 global stiffness matrix will now look like this.



=Lecture 36 - Wednesday, November 19, 2008= \frac{EA}{L} & 0 & 0 & \frac{-EA}{L} & 0 & 0\\ & \frac{12EI}{L^{3}} & \frac{6EI}{L^2} & 0 & \frac{-12EI}{L^{3}} & \frac{6EI}{L^2}\\ & & \frac{4EI}{L} & 0 & \frac{-6EI}{L^2} & \frac{2EI}{L}\\ & &  & \frac{EA}{L} & 0 & 0\\ & sym & &  & \frac{12EI}{L^{3}} & \frac{-6EI}{L^2}\\ & &  &  &  & \frac{4EI}{L} \end{bmatrix}$$ \tilde{\mathbf{d}}^{(e)}_1\\ \tilde{\mathbf{d}}^{(e)}_2\\ \tilde{\mathbf{d}}^{(e)}_3\\ \tilde{\mathbf{d}}^{(e)}_4\\ \tilde{\mathbf{d}}^{(e)}_5\\ \tilde{\mathbf{d}}^{(e)}_6 \end{Bmatrix},\tilde{\mathbf{f}}^{(e)}=\begin{Bmatrix} \tilde{\mathbf{f}}^{(e)}_1\\ \tilde{\mathbf{f}}^{(e)}_2\\ \tilde{\mathbf{f}}^{(e)}_3\\ \tilde{\mathbf{f}}^{(e)}_4\\ \tilde{\mathbf{f}}^{(e)}_5\\ \tilde{\mathbf{f}}^{(e)}_6 \end{Bmatrix}$$
 * $$\tilde{\mathbf{k}}=\begin{bmatrix}
 * $$\tilde{\mathbf{k}}^{(e)}_{6x6}\tilde{\mathbf{d}}^{(e)}_{6x1}=\tilde{\mathbf{f}}^{(e)}_{6x1}$$
 * $$\tilde{\mathbf{d}}^{(e)}=\begin{Bmatrix}
 * Where $$\tilde{\mathbf{d}}^{(e)}_3=\mathbf{d}^{(e)}_3, \tilde{\mathbf{d}}^{(e)}_6=\mathbf{d}^{(e)}_6, \tilde{\mathbf{f}}^{(e)}_3=\mathbf{f}^{(e)}_3, \tilde{\mathbf{f}}^{(e)}_6=\mathbf{f}^{(e)}_6$$

Analyzing Dimensions
\tilde{d}_1 \end{bmatrix}=L=\begin{bmatrix} \tilde{d}_i \end{bmatrix}\rightarrow i=1,2,4,5$$ \tilde{d}_1 \end{bmatrix}=1$$
 * $$\begin{bmatrix}
 * $$\begin{bmatrix}
 * $$\sigma =E\varepsilon\rightarrow[\sigma] =[E][\varepsilon] $$

=Lecture 37 - Friday, November 21, 2008= Dimensional analysis can be performed on the element stiffness matrix using the following steps.

$$\left[\varepsilon \right]=\frac{\left[d u \right]}{\left[dx \right]}=\frac{L}{L}=1$$

$$\left[\sigma \right]=\left[E  \right]=\frac{F}{L^{2}}$$

$$\left[A \right]={L^{2}}$$

$$\left[I \right]={L^{4}}$$

Using the above relations, each element's dimension in the elemental stiffness matrix can be discovered. When each element is multiplied by its corresponding displacement, it can be seen that each element returns a force. The following are two examples from the element stiffness matrix.

$$\left[ \frac{EA}{L}\right]=\left[\tilde{k}_{11} \right]=\frac{(F/L^2)(L^2)}{L}=\frac{F}{L}$$

$$\left[\tilde{k}_{11} \tilde{d}_1 \right]=\left[\tilde{k}_{11} \right]\left[\tilde{d}_1 \right]=F$$

$$\left[\tilde{k}_{23}\tilde{d}_3 \right]=\left[\tilde{k}_{23} \right]\left[\tilde{d}_3 \right]=\frac{6\left[ E\right]\left[I \right]}{\left[ L^2\right]}=\frac{1\cdot (F/L^2)(L^4)}{L^2}=F$$

Using the above techniques, dimensional analysis can be performed on the entire element stiffness matrix.

$$\tilde{\mathbf{k^e}}=\begin{bmatrix} \frac{EA}{L} & 0 & 0 & \frac{-EA}{L} & 0 & 0\\ & \frac{12EI}{L^{3}} & \frac{6EI}{L^2} & 0 & \frac{-12EI}{L^{3}} & \frac{6EI}{L^2}\\ & & \frac{4EI}{L} & 0 & \frac{-6EI}{L^2} & \frac{2EI}{L}\\ & &  & \frac{EA}{L} & 0 & 0\\ & sym & &  & \frac{12EI}{L^{3}} & \frac{-6EI}{L^2}\\ & &  &  &  & \frac{4EI}{L} \end{bmatrix}$$

$$\tilde{\mathbf{k^e}}=\begin{bmatrix} \frac{F}{L} & 0 & 0 & \frac{-F}{L} & 0 & 0\\ & \frac{F}{L} & F & 0 & \frac{-F}{L} & F\\ & & FL & 0 & -F & FL\\ & &  & \frac{F}{L} & 0 & 0\\ & sym & &  & \frac{F}{L} & -F\\ & &  &  &  & FL \end{bmatrix}$$

The element force displacement relationship in global coordinates can be found from the element force displacement relationship in local coordinates as follows:

$$\mathbf{k^{(e)}_{6x6}d^{(e)}_{6x1}=f^{(e)}_{6x1}}$$

$$\mathbf{k^{(e)}_{6x6}=\tilde{T}^{(e)T}_{6x6}\tilde{k}^{(e)}_{6x6}\tilde{T}^{(e)}_{6x6}}$$

$$\begin{Bmatrix} \tilde{d}_1\\ \tilde{d}_2\\ \tilde{d}_3\\ \tilde{d}_4\\ \tilde{d}_5\\ \tilde{d}_6 \end{Bmatrix}=\begin{bmatrix} \mathbf{R_{2x2}}& &0  & 0 & 0 & 0\\ & & 0 & 0 & 0 &0 \\ 0& 0 & 1 & 0 & 0 &0 \\ 0& 0 & 0 & \mathbf{R_{2x2}} &  & 0\\ 0& 0 & 0 & &  & 0\\ 0& 0 &0  & 0 & 0 & 1 \end{bmatrix}\begin{Bmatrix} d_1\\ d_2\\ d_3\\ d_4\\ d_5\\ d_6 \end{Bmatrix}$$

Here the 6x6 matrix including the $$\mathbf{R_{2x2}}$$ matrices is the $$\tilde{T}^{(e)}_{6x6}$$.

The derivation of $$\tilde{k}^{(e)}$$ from the principal of virtual work, focusing only on bending effects, provides the following:

$$\frac{d}{dx^2}\left( (EI)\frac{d^2v}{dx^2}\right)-f_t(x)=m(x)\ddot{v}$$

=Lecture 38 - Monday, November 24, 2008=

Derivation of $$\tilde{k}^{(e)}$$
The motivation for the following is to deform the shape of the truss, and interpret the transverse displacement of v(λ) element. Continuing from aboves lecture 37 the derivation of $$\tilde{k}^{(e)}$$ from the Principle of Virtual Work, focusing only on the bending effect, is shown below.

$$-\frac{\partial^{2}}{\partial x^{2}}\left((EI)\frac{\partial ^{2}v}{\partial x^{2}} \right) + f_{t}(t)=m(x)\ddot{v}$$

$$+\frac{\partial}{\partial x}\left((EA)\frac{\partial u}{\partial x} \right)+f_{a}(x,t)=m(x)\ddot{u}$$

The two above equations can be described in the table below:

PVW for beams: $$\int_{0}^{L}{W(x)\left[ -\frac{\partial^{2}}{\partial x^{2}}\left((EI)\frac{\partial ^{2}v}{\partial x^{2}} \right) + f_{t}(t)-m(x)\ddot{v}\right]}$$ for all possible W(x), dx=0

Integration by Parts
To perform integration by parts, various new assigned equations are introduced to make it simpler to follow.

$$\mathbf{\alpha} := \int_{0}^{L}{W(x)\left[ -\frac{\partial^{2}}{\partial x^{2}}\left((EI)\frac{\partial ^{2}v}{\partial x^{2}} \right)\right]dx} $$

In equation α, it can be broken up to separate parts which will then be used to perform integration by parts. The separate parts are as follows:

$$\mathbf{s(x)}= W(x) $$

$$ \mathbf{s^{'}(x)}= \frac{dW(x)}{dx}$$

$$\mathbf{r(x)} = \frac{\partial}{\partial x}\left\{ (EI)\frac{\partial ^{2}v}{\partial x^{2}} \right\} $$

$$\mathbf{r^{'}(x)} = \frac{\partial}{\partial x}\left(\frac{\partial}{\partial x}\left\{ (EI)\frac{\partial ^{2}v}{\partial x^{2}} \right\} \right)$$

Now that the parts of the equation are identified, the theory behind the integration is as follows:

$$\int sdr=sr-\int rds$$

Inputing the values to the above equation yeilds:

$$ \mathbf{\alpha} := \left[W(x)\frac{\partial}{\partial x}\left\{ (EI)\frac{\partial ^{2}v}{\partial x^{2}} \right\}\right]_{0}^{L}-\int_{0}^{L}{\frac{dW(x)}{dx}\frac{\partial}{\partial x}\left\{ (EI)\frac{\partial ^{2}v}{\partial x^{2}} \right\}dx}$$

Looking at α, it is evident that a second integration by parts is required. Below defines the new parts to the equation.

$$\mathbf{ \beta_{1}} :=\left[W(x)\frac{\partial}{\partial x}\left\{ (EI)\frac{\partial ^{2}v}{\partial x^{2}} \right\}\right]_{0}^{L}$$

$$\mathbf{s(x)}= \frac{dW(x)}{dx}$$

$$ \mathbf{s^{'}(x)}= \frac{d^{2}W(x)}{dx^{2}}$$

$$\mathbf{r(x)} = (EI)\frac{\partial ^{2}v}{\partial x^{2}} $$

$$\mathbf{r^{'}(x)} = \frac{\partial}{\partial x}\left\{ (EI)\frac{\partial ^{2}v}{\partial x^{2}} \right\}$$

Inputing these variables into the theoretical equation yields:

$$\mathbf{\alpha} := \beta_{1}-\left[\frac{dW(x)}{dx}(EI)\frac{\partial ^{2}v}{\partial x^{2}}\right]_{0}^{L}+\int_{0}^{L}{\frac{d^{2}W(x)}{dx^{2}}(EI)\frac{\partial ^{2}v}{\partial x^{2}}dx}$$

Therefore, α becomes the following completed integration by parts.

$$\mathbf{\alpha} := \left[W(x)\frac{\partial}{\partial x}\left\{ (EI)\frac{\partial ^{2}v}{\partial x^{2}} \right\}\right]_{0}^{L}-\left[\frac{dW(x)}{dx}(EI)\frac{\partial ^{2}v}{\partial x^{2}}\right]_{0}^{L}+\int_{0}^{L}{\frac{d^{2}W(x)}{dx^{2}}(EI)\frac{\partial ^{2}v}{\partial x^{2}}dx}+ \int_{0}^{L}{Wf_{t}dx}-\int_{0}^{L}{Wm\ddot{v}dx} $$ for all possible W(x)

Stiffness Term
Here the stiffness term is defined as δ = $$\int_{0}^{L}{\frac{d^{2}W(x)}{dx^{2}}(EI)\frac{\partial ^{2}v}{\partial x^{2}}dx}$$. Now to focus on this stiffness term by deriving the beam stiffness matrix and identifying the beam shape functions.

‎

The transverse displacement (V) is defined below:

$$V(\tilde{x})=N_{2}(\tilde{x})d_{2}(\tilde{x})+N_{3}(\tilde{x})d_{3}(\tilde{x})+N_{5}(\tilde{x})d_{5}(\tilde{x})+N_{6}(\tilde{x})d_{6}(\tilde{x})$$

Recall that $$u(\tilde{x})=N_{1}(\tilde{x})d_{1}(\tilde{x})+N_{4}(\tilde{x})d_{4}(\tilde{x})$$ is that axial displacement (u) equation. The plots of the axial displacements are shown below.

‎

The plots of the transverse diplacements are shown below.



The equations for each of the displacements is defined below.

$$N_{2}(\tilde{x}) = 1 - \frac{3\tilde{x}^{2}}{L^{2}}+\frac{2\tilde{x}^{3}}{L^{3}}$$

$$N_{3}(\tilde{x}) = \tilde{x} - \frac{2\tilde{x}^{2}}{L}+\frac{\tilde{x}^{3}}{L^{2}}$$

$$N_{5}(\tilde{x}) = \frac{3\tilde{x}^{2}}{L^{2}}-\frac{2\tilde{x}^{3}}{L^{3}}$$

$$N_{6}(\tilde{x}) = - \frac{\tilde{x}^{2}}{L}+\frac{\tilde{x}^{3}}{L^{2}}$$

=Lecture 39 - Monday, December 1, 2008= Continuing from the last lecture, the plots for $$N_{5}(\tilde{x})$$ and $$N_{6}(\tilde{x})$$ are shown below.



The following equation will be useful in the computation of $$u(\tilde{x})$$ and $$v(\tilde{x})$$.

$$\underline{\tilde{d}}_{6x1}^{(e)}=\underline{\tilde{T}}_{6x6}^{(e)}\underline{d}_{6x1}^{(e)}$$


 * Note that $$\underline{d}_{6x1}^{(e)}$$ is known after solving the Finite Element system.

Computation of $$u(\tilde{x})$$ and $$v(\tilde{x})$$


$$\underline{u}(\tilde{x})$$ can be expressed as follows.


 * $$\underline{u}(\tilde{x})=u(\tilde{x})\vec{\tilde{i}}+v(\tilde{x})\vec{\tilde{j}}$$


 * $$\underline{u}(\tilde{x})=u_{x}(\tilde{x})\vec{i}+u_{y}(\tilde{x})\vec{j}$$

Now, $$u(\tilde{x})$$ and $$v(\tilde{x})$$ can be computed using the following equations.


 * $$u(\tilde{x})=N_{1}(\tilde{x})d_{1}(\tilde{x})+N_{4}(\tilde{x})d_{4}(\tilde{x})$$


 * $$v(\tilde{x})=N_{2}(\tilde{x})d_{2}(\tilde{x})+N_{3}(\tilde{x})d_{3}(\tilde{x})+N_{5}(\tilde{x})d_{5}(\tilde{x})+N_{6}(\tilde{x})d_{6}(\tilde{x})$$

Then, $$u_{x}(\tilde{x})$$ and $$u_{y}(\tilde{x})$$ can be computed from $$u(\tilde{x})$$ and $$v(\tilde{x})$$

u_{y}(\tilde{x}) \end{Bmatrix}=\underline{R}^{T}\begin{Bmatrix}u(\tilde{x})\\ v(\tilde{x})\end{Bmatrix}$$
 * $$\begin{Bmatrix}u_{x}(\tilde{x})\\

u(\tilde{x})\\ v(\tilde{x})\\ \end{Bmatrix}=\begin{bmatrix} N_1 & 0 & 0 & N_4 & 0 & 0\\ 0 & N_2 & N_3 & 0 & N_5 & N_6\\ \end{bmatrix}\begin{Bmatrix} \tilde{d}_1^{(e)}\\ \tilde{d}_2^{(e)}\\ \tilde{d}_3^{(e)}\\ \tilde{d}_4^{(e)}\\ \tilde{d}_5^{(e)}\\ \tilde{d}_6^{(e)}\\ \end{Bmatrix}$$
 * $$\begin{Bmatrix}

N_1 & 0 & 0 & N_4 & 0 & 0\\ 0 & N_2 & N_3 & 0 & N_5 & N_6\\ \end{bmatrix}$$
 * $$\mathbb{N}(\tilde{x})=\begin{bmatrix}

Therefore, the final statement can be made.

$$\begin{Bmatrix}u_{x}(\tilde{x})\\ u_{y}(\tilde{x}) \end{Bmatrix}=\underline{R}^{T}\mathbb{N}(\tilde{x})\underline{\tilde{T}}^{(e)}\underline{d}^{(e)}$$

=Meeting 40, Friday, December 5, 2008=

Dimensional Analysis
As was shown in Meeting 39, the transfer function in Eqn. 2 transforms a 6x1 matrix consisting of transverse displacements and rotations to a 2x1 matrix of real linear displacements. To understand how the transfer function operates, it is wise to ensure that the resultant is dimensionally correct. To begin, recall Eqn. 2 follows the form:


 * $$ \begin{bmatrix}

u(\tilde{x}) \\ v(\tilde{x}) \\ \end{bmatrix} = \begin{bmatrix} N_1 \quad 0 \quad 0 \quad  N_4 \quad 0 \quad 0 \\ \quad 0 \quad N_2 \quad N_3 \quad 0 \quad N_5 \quad N_6 \\ \end{bmatrix} \begin{bmatrix} \tilde{x}^{(e)}_1 \\ \tilde{x}^{(e)}_2 \\ \tilde{x}^{(e)}_3 \\ \tilde{x}^{(e)}_4 \\ \tilde{x}^{(e)}_5 \\ \tilde{x}^{(e)}_6 \\ \end{bmatrix} $$

\begin{bmatrix} u(\tilde{x}) \\ v(\tilde{x}) \\ \end{bmatrix} = \mathbb{N} \mathbf{\tilde{d}}^{(e)} $$

The resultant matrix of Eqn. 2 from Meeting 39 is a 2x1, consisting of $$ u(\tilde{x}) \text{ and  } v(\tilde{x}) $$, which are real axial coordinate system displacements. The dimensions on $$ u(\tilde{x}) \text{ and  } v(\tilde{x}) $$ are length. Put simply $$ \begin{bmatrix} u(\tilde{x}) \\ \end{bmatrix} = L \text{ and } \begin{bmatrix} u(\tilde{x}) \\ \end{bmatrix} = L \text{ where } [i] \text{ is used to express the dimension of a given variable i }$$.

From Meeting 31, we found that Ni and Ni+1 exhibit the following relationship:
 * $$N_{i}(x)=\frac{x-x_{i+1}}{x_{i}-x_{i+1}}$$
 * $$N_{i+1}(x)=\frac{x-x_{i}}{x_{i+1}-x_{i}}$$

From the above equations, it is evident that both Ni and Ni+1 are unitless because a length in the numerator is divided by a length term in the denominator. Knowing this, we can write
 * $$ [N_i] = 1 \text{ and } [N_{i+1}] =1 $$.

In Meeting 32, we learned that axial displacement is calculated as a linear combination of nodal displacement multiplied by the shape function at node i and the respective value at node i+1
 * $$u(x)=N_{i}(x)d_{i}+N_{i+1}(x)d_{i+1}$$

Similarly, by examining the $$ \mathbb{N} $$ matrix, one can see that for the axial displacement $$ u(\tilde{x}) \quad $$,
 * $$u(\tilde{x}) = N_1 \tilde(d)_1 + N_4 \tilde{d}_4 $$
 * $$\quad N_1(\tilde{x}) = N_i(x) \text{ and } N_4(\tilde{x}) = N_{i+1} $$

Knowing this, one can infer that N1 and N4 are also unitless.

For the $$\mathbb{N} \quad $$ matrix,

\begin{bmatrix} N_1 \\ \end{bmatrix} = \begin{bmatrix} N_4 \\ \end{bmatrix} = 1 $$

\begin{bmatrix} u(\tilde{X}) \\ \end{bmatrix} = \begin{bmatrix} N_1 \\ \end{bmatrix} \begin{bmatrix} \tilde{d}^{(e)}_1 \\ \end{bmatrix} + \begin{bmatrix} N_4 \\ \end{bmatrix} \begin{bmatrix} \tilde{d}^{(e)}_4 \\ \end{bmatrix} = 1 $$
 * $$\begin{bmatrix}

u(\tilde{X}) \\ \end{bmatrix} = 1 * L + 1 * L $$
 * $$\therefore

\begin{bmatrix} u(\tilde{X}) \\ \end{bmatrix} = L $$

Similarly, it is wise to prove that $$ \begin{bmatrix} v(\tilde{X}) \\ \end{bmatrix} \quad $$ is dimensionally correct.
 * $$ v(\tilde{x}) = N_2 \tilde{d}_2 + N_3 \tilde{d}_3 + N_5 \tilde{d}_5 + N_6 \tilde{d}_6 $$

The table below shows the dimensions of each term in the above equation

From the above table, one can see that
 * $$[v(\tilde{x})] = [N_2] [\tilde{d}_2] + [N_3] [\tilde{d}_3] + [N_5] [\tilde{d}_5] + [N_6] [\tilde{d}_6] $$
 * $$[v(\tilde{x})] = 1*L + L*1 + 1*L + L*1 = L $$

From the above work, it is easy to see that Eqn. 2 from Meeting 39 is dimensionally correct, in that it transforms a matrix of displacement (unit length) and rotational (dimensionless) values into a matrix of displacements (length only).

Derivation of Shape Functions
To gain a thorough understanding of the shape functions for a beam, we will derive each on individually. The specified beam shape functions are listed on p. 34_4 (Meeting 34), and include $$N_2, N_3 , N_5, N_6 $$. Their respective plots are shown on p.38_3 (insert link)

Recall that the governing PDE for beams is given by:
 * $$ -\frac{d^2}{d \tilde{x}}((EX) \frac{d^2 V}{d \tilde{x}^2}) + f(\tilde{x}) = m(\tilde{x})\ddot{v} = 0$$

To derive the shape functions, we will examine the PDE for a static case and without a distributed load. In other words, $$f(\tilde{x}) = 0 \text{ and  } m \ddot{v} = 0 $$. Therefore, the PDE reduces to
 * $$\frac{d^2}{d \tilde{x}^2}((EX)\frac{d^2v}{d \tilde{x}^2}) = 0 $$

Furthermore, we will consider the case where EI = constant. As a results, we may express the governing PDE as
 * $$\frac{d^4}{d \tilde{x}^4}v = 0 $$

We integrated the PDE four times and obtain a general solution with four constants of integration:
 * $$v(\tilde{x}) = C_0 + C_1 \tilde{x}^1 + C_2 \tilde{x}^2 + C_3 \tilde{x}^3 $$

Deriving N_2(x)
To obtain $$ N_2 (\tilde{x}) \quad $$, we must apply the following boundary conditions (BCs):
 * $$ v(0) = 1 \text{, } v(L) = 0 $$
 * $$ v'(0) = 0 \text{, } v'(L) = 0 $$

Since there are four BCs and four unknowns, we are able to solve for C_0, C_1, C_2, and C_3: By solving the system of equations, we find that
 * $$v(\tilde{x}) = C_0 + C_1 \tilde{x}^1 + C_2 \tilde{x}^2 + C_3 \tilde{x}^3 $$
 * $$ v(0) = 1 = C_0 $$
 * $$ v(L) = 1 + C_1 L + C_2 L^2 + C_3 L^3 $$
 * $$ v'(\tilde{x}) = C_1 + 2C_2 \tilde{x} + 3C_3 \tilde{x}^2 $$
 * $$ v'(0) = C_1 = 0$$
 * $$ v'(L) = 2C_2 L + 3C_3 L^2 = 0 $$
 * $$C_0 = 1 $$
 * $$C_1 = 0 $$
 * $$C_2 = \frac{-3}{L^2} $$
 * $$C_3 = \frac{2}{L^3} $$

Therefore, we've found that
 * $$N_2(\tilde{x}) = 1 - \frac{3\tilde{x}^2}{L^2} + \frac{2\tilde{x}^3}{L^3} $$

Deriving N_3(x)
To obtain $$ N_3 (\tilde{x}) \quad $$, we must apply the following boundary conditions (BCs):
 * $$ v(0) = 1 \text{, } v(L) = 0 $$
 * $$ v'(0) = 1 \text{, } v'(L) = 0 $$

Since there are four BCs and four unknowns, we are able to solve for C_0, C_1, C_2, and C_3: By solving the system of equations, we find that
 * $$v(\tilde{x}) = C_0 + C_1 \tilde{x}^1 + C_2 \tilde{x}^2 + C_3 \tilde{x}^3 $$
 * $$ v(0) = 1 = C_0 $$
 * $$ v(L) = 1 + C_1 L + C_2 L^2 + C_3 L^3 $$
 * $$ v'(\tilde{x}) = C_1 + 2C_2 \tilde{x} + 3C_3 \tilde{x}^2 $$
 * $$ v'(0) = C_1 = 0$$
 * $$ v'(L) = 2C_2 L + 3C_3 L^2 = 0 $$
 * $$C_0 = 0 $$
 * $$C_1 = 1 $$
 * $$C_2 = \frac{-2}{L} $$
 * $$C_3 = \frac{2}{L^2} $$

Therefore, we've found that
 * $$N_3(\tilde{x}) = \tilde{x} - \frac{2\tilde{x}^{2}}{L}+\frac{\tilde{x}^{3}}{L^{2}}$$

Deriving N_5(x)
To obtain $$ N_3 (\tilde{x}) \quad $$, we must apply the following boundary conditions (BCs):
 * $$ v(0) = 0 \text{, } v(L) = 1 $$
 * $$ v'(0) = 0 \text{, } v'(L) = 0 $$

Since there are four BCs and four unknowns, we are able to solve for C_0, C_1, C_2, and C_3: By solving the system of equations, we find that
 * $$v(\tilde{x}) = C_0 + C_1 \tilde{x}^1 + C_2 \tilde{x}^2 + C_3 \tilde{x}^3 $$
 * $$ v(0) = 1 = C_0 $$
 * $$ v(L) = 1 + C_1 L + C_2 L^2 + C_3 L^3 $$
 * $$ v'(\tilde{x}) = C_1 + 2C_2 \tilde{x} + 3C_3 \tilde{x}^2 $$
 * $$ v'(0) = C_1 = 0$$
 * $$ v'(L) = 2C_2 L + 3C_3 L^2 = 0 $$
 * $$C_0 = 0 $$
 * $$C_1 = 0 $$
 * $$C_2 = \frac{3}{L^2} $$
 * $$C_3 = \frac{-2}{L^3} $$

Therefore, we've found that
 * $$N_5(\tilde{x}) = \frac{3\tilde{x}^{2}}{L^{2}}-\frac{2\tilde{x}^{3}}{L^{3}} $$

Deriving N_6(x)
To obtain $$ N_3 (\tilde{x}) \quad $$, we must apply the following boundary conditions (BCs):
 * $$ v(0) = 0 \text{, } v(L) = 0 $$
 * $$ v'(0) = 0 \text{, } v'(L) = 1 $$

Since there are four BCs and four unknowns, we are able to solve for C_0, C_1, C_2, and C_3: By solving the system of equations, we find that
 * $$v(\tilde{x}) = C_0 + C_1 \tilde{x}^1 + C_2 \tilde{x}^2 + C_3 \tilde{x}^3 $$
 * $$ v(0) = 1 = C_0 $$
 * $$ v(L) = 1 + C_1 L + C_2 L^2 + C_3 L^3 $$
 * $$ v'(\tilde{x}) = C_1 + 2C_2 \tilde{x} + 3C_3 \tilde{x}^2 $$
 * $$ v'(0) = C_1 = 0$$
 * $$ v'(L) = 2C_2 L + 3C_3 L^2 = 0 $$
 * $$C_0 = 0 $$
 * $$C_1 = 0 $$
 * $$C_2 = \frac{-2}{L} $$
 * $$C_3 = \frac{1}{L^2} $$

Therefore, we've found that
 * $$N_6(\tilde{x}) = - \frac{\tilde{x}^{2}}{L}+\frac{\tilde{x}^{3}}{L^{2}} $$

=Lecture 41 - Monday, December 8, 2008=

Finding the new k values
So in general,


 * $$\tilde{k}_{ij}=\int_{0}^{L}{\frac{d^{2}N_{i}}{dx^{2}}(EI)\frac{d^{2}N_{j}}{dx^{2}}dx}$$ where i,j = 2,3,5,6

Elastodynamics (trusses, frames, 2-D and 3-D elasticity)
From the model problem on P. 31.1 of the lecture notes

Discrete PVW:


 * $$\mathbf\bar{w}[ \mathbf\bar{M} \mathbf\ddot\bar{d} + \mathbf\bar{K} \mathbf\bar{d} - \mathbf\bar{F}] = 0$$ for all $$\mathbf\bar{w}$$

for $$\mathbf\bar{w}$$ the boundary conditions are already applied

this gives us Equation 1


 * $$\mathbf\bar{M} \mathbf\ddot\bar{d} + \mathbf\bar{K} \mathbf\bar{d} = \mathbf\bar{F} $$

with Boundary Conditions
 * $$\mathbf\bar{d(0)} = \mathbf\bar{d_{0}} $$
 * $$\mathbf\dot\bar{d(0)} = \mathbf\bar{v_{0}}$$

Standard Eigenvalue Problem


 * $$\mathbf{A} \mathbf{x} = \lambda \mathbf{x}$$

this is because B is an identity matrix and therefore doesn't effect the equation

$$\lambda = w^{2}$$. . . eigenvalue

$$ (\lambda_{i}, \phi_{i}) $$. . . eigenpairs, for i = 1,. . ., n

=Mediawiki vs. WebCT=


 * Number of team members familiar with E-Learning: 6
 * Number of team members not familiar with E-Learning: 0

After learning how to use MediaWiki this semester, team qwiki has learned many advantages MediaWiki has over WebCT. One of the obvious advantages is the ability for anyone to publicly access anyone's work or information. This is useful in our FEA class because we can look at other team members pages to help us understand the lecture material in preparation for a test.

Another advantage is that work done by one team member can be edited by another member. Therefore, errors can easily be corrected by anyone without having to go through the hassle of contacting the team member that made the error.

It is also very easy to add files into a MediaWiki page. All that must be done is to upload the file into WikiMedia Commons. The Commons page can support a wide array of file types, making it convenient to share files on your page. These files can also be licensed so that an individuals is credited with his or her work.

For this class, entering in equations into our pages was a necessity. WebCT does not support the Latex language that MediaWiki does. Entering in equations was simple and could be easily edited using Latex.

The team feels the biggest advantage MediaWiki has over WebCT is the accessibility of MediaWiki. WebCT slows down the more users are accessing it at a time, making it hard to use during times of high use, such as the end of a semester and during exams. The service can sometimes shut down all together if the server is down. We have heard stories of students not being able to finish their work in time because WebCT was down the night before an assignment was due. MediaWiki is worldwide, and it is unlikely that it would be inaccessible for an extended period of time.

The group consensus is that for this class, MediaWiki is the preferred medium over WebCT. The only flaw we can complain about is the ability for anyone to edit our work because it is a public website. However, any vandalism can be fixed because the website history logs every version ever made of a page. Therefore, it is only an inconvenience to change pages back to how they were. Team qwiki did not experience any such occurrence.

=Contributing Members=
 * Andrea Booher - EML4500.f08.qwiki.booher 23:06, 7 December 2008 (UTC)
 * Michael Berry - Eml4500.f08.qwiki.berry 05:42, 9 December 2008 (UTC)
 * David Nobles - Eml4500.f08.qwiki.nobles 16:03, 9 December 2008 (UTC)
 * Chris Bishop - Eml4500.f08.qwiki.bishop 18:44, 9 December 2008 (UTC)
 * Lee Harris - Eml4500.f08.qwiki.harris 20:09, 9 December 2008 (UTC)
 * Eric Gatch - Eml4500.f08.qwiki.gatch 20:31, 9 December 2008 (UTC)