text:
User:Eas4200c.f08.vqcrew.c/Homework 7

 See my comments below. Eml4500.f08 21:54, 10 December 2008 (UTC)

Non-Symmetric Thin Walled Cross-Section

Now that the formulas have been laid out for non-symmetric case, as seen in HW6, it is now time to look at a problem utilizing these equations. The non-symmetric case can be seen in Fig 1. This case is vary similar to the symmetrical cross-section in example problem 5.2 in the book, however the areas will vary by stringer for the non-symmetrical case. The figure also shows the location of the centroid which was calculated in the HW section below, and the shear flow across each skin panel.

 * $$A_1 = A \;$$

 * $$A_2 = 2A \;$$

 * $$A_3 = 3A \;$$

 * $$A_4 = 4A \;$$

In evaluating Fig 1., the mean value theorem will be used in analyzing the first moment of the area, Q_y and Q_z, so the equations can be modified to the following:

 * $$ Q_z = \int_A \bar{y} \,dA = \bar{y} A $$

 * $$ Q_y = \int_A \bar{z} \,dA = \bar{z} A $$

The Area to be used for the overall moments will have to be the sum of the area of the components. Neglecting the skin and spar webs the equation for the overall area the sum of the area of $$A_i\;$$

 * $$A = \sum_{i=1}^{4}{A_i}\;$$

From the first moment of area equations and the known area it is now possible to examine the shear flow along the path s. Looking at Fig 1., V_z is zero for this case so the shear flow equations becomes the following:

 * $$ q(s) = (k_{yz}Q_z^{(s)} - k_z Q_y^{(s)}) V_z $$

Now if we break down the shear flow equation and examine each variable it can be determined the each variable is actually independent of the past s.

 * $$\left.\begin{matrix}

1) & V_z & \\ 2) & k_{yz} & k_z\\ 3) & Q_z & Q_y \end{matrix}\right\}$$ independent of s

From the two first moment of inertia equations it can also be seen the they are constant between two stringers. This is due to all areas are concentrated of stringers, thus constant and independent of s between the stringers.

These relations yield the final assumption that the shear flow q(s) is constant between two stringers, but q(s) would increment (jump) when crossing over a stringer.

Now that we have the following relation with the stringers, it is now time to lay out a game plan to solve for the shear flows along the path s.

Step 1: Find $$ (\bar{y}, \bar{z})$$

Step 2: Find $$ I_y, I_z, I_{yz}\;$$

Step 3: Find $$ k_y, k_z, and k_yz \;$$

Step 4: Follow path "s" to find $$q_{1-2}, q_{2-3}, q_{3-4}\;$$

$$ q_{1-2}\;$$ represents the shear flow in skin panel 1-2. This can be represented in the same form as the shear flow along any path.

 * $$ q_{12} = (k_{yz}Q_z^{12} - k_z Q_y^{12}) V_z $$

Where the first moment area of inertia is local around each panel. Where $$y_1$$ is the y component of the string with respect to the origin at the centroid, and similarly for $$z_1$$

 * $$ Q_z^{12} = y_1 A_1 $$

 * $$ Q_y^{12} = z_1 A_1 $$

A quick look for reference of the following skin panel 2-3 shows what the first moment of inertia would resemble in that case.

 * $$ Q_z^{23} = y_1 A_1 + y_2 A_2 $$

A look at a different type of non-symmetrical case might help better explain different approaches that could be taken when looking for the shear flow. Fig 2. shows different paths and areas for each path that could be used for solving such a shape.

The area for this case that was split into two section is the sum of the two sections:

 * $$ A = A_{\hat{s}} + A_{\hat{s}, L} $$

Mini Plan
Single-Cell Sections
 * - without stringers
 * - with stringers

Multi-Cell Sections
 * - without stringers
 * - with stringers

Without Stringers
Question: Can the setup shown in Figure 3 resist $$\displaystyle V_{z}$$?

Answer: No.

 * $$R^{z} = R_{AB}^{z} + R_{BA}^{z}$$

$$\displaystyle R^{z}$$ is the total resultant in the z-direction

$$R_{AB}^{z}$$ is the reslutant of q in AB

$$R_{BA}^{z}$$ is the resultant of q in BA

 * $$R_{AB}^{z} = -q\bar{A^{'}B^{'}} = R_{BA}^{z}$$

 * so, $$\displaystyle R^{z} = 0$$

Therefore, the configuration can not resist any $$\displaystyle V_{z}$$.

With Stringers

Figure 4 shows that q(s) is piecewise constant with respect to the curvilinear coordinate, s. (i.e. $$\displaystyle q_{ij}$$ = constant within each panel)

However, the value for q is not the same throughout all of the panels.

 * $$q_{12} \neq q_{23} \neq q_{31}$$

The presence of the stringers causes a non-constant q.

 * $$R^{z} = V_{z} \neq 0$$

Superposition can be used to solve for the values of $$\displaystyle q_{ij}$$ in each section by solving for values of q in the case without stringers, Figure 5, and for the case including an imaginary cut, Figure 6, in one of the sections. This is a valid use of superposition due to the linearity of the relationship.

 * $$q_{12} = q + \tilde{q}_{12}$$
 * $$q_{23} = q + \tilde{q}_{23}$$
 * $$q_{31} = q + \tilde{q}_{31}$$

Analysis Algorithm
Objectives: One unknown q ($$\displaystyle \tilde{q_{ij}}$$ are known after solving P1), need to solve one equation for one unknown.

Method: Knowing the values for $$\displaystyle V_{y}$$ and $$\displaystyle V_{z}$$

 * 1) Solve P2 for $$\displaystyle \tilde{q_{12}}$$ and $$\tilde{q_{23}}$$

 * 2) Moment Equation - take the moment about any point in the plane (y,z)

 * a) Superposition: $$\displaystyle q_{ij} = q + \tilde{q_{ij}}$$

 * b) Select point $$\displaystyle \bar{O}$$ in the plane (y,z)

Without Stringers

The same type of analysis can be performed for multi-cell sections as for single cell sections. Figure 7 shows the multi-cell case without stringers. The resultant of all of the shear flows in each cell can be calculated by the sum of the resultants in every cell.

 * $$\displaystyle R^{z} = \sum_{i = 1}^{N_{cells}}R_{i}^{z}$$

Similar to the single case, the resultant for each cell can be found, therefore yielding the result for the entire multi-cell section. Once again, this value is equal to zero. Without the support of the stringers, this setup can not resist $$\displaystyle V_{z}$$.

 * $$\displaystyle R_{i}^{z} = 0 \Rightarrow R^{z} = 0 $$

With Stringers

The true shear flow, $$\displaystyle q_{ij}$$, is equal to the sum of the closed cell constant shear flow, q, and the open cell piecewise constant shear flow, $$\displaystyle \tilde{q_{ij}}$$.

 * $$\displaystyle q_{ij} = q + \tilde{q_{ij}}$$

In order to solve for all of the true shear flows, imaginary cuts must be made into some of the walls. This must be done carefully to ensure that no stringers are left unattached to the rest of the assembly. Figure 8 shows the overall problem including no cuts and all of the stringers. Figure 9 is displaying the case in which the stringers are removed so that the closed cell shear flow can be calculated. Figure 10 represents the case in which the cuts have been made and the open cell piecewise constant shear flows can be found. The total resultant of the multi-cell case with stringers can be calculated by the sum of the resultants of P1 and P2.

 * $$\displaystyle R^{z} = R^{z_1} + R^{z_2}$$

Recall: $$\displaystyle R^{z_1} = 0$$

Therefore: $$\displaystyle R^{z} = R^{z_2} = V_{Z} \neq 0$$

Solving P2 - Euler Cut Principle
The Euler Cut Principle involves finding the equilibrium of each stringer. This is basically a statics problem where the sum of the forces, which are given through the shear flows, are equal to zero. Stringer 3 is shown in Figure 11 in two dimensions and again in three dimensions in Figure 12, with the stringer expanded into the x-direction. The sum of the forces in the x-direction can be expressed as follows:

 * $$\sum{F_x}=0=\int_{A_3}^{}{\left[\sigma _{xx}(x+dx) -\sigma _{xx}(x)\right]}dA_3=\int_{A_3}^{}{\left[\frac{d\sigma _{xx}}{dx}dx + HOT\right]}dA_3=\left[ \tilde{q_{31}}-\tilde{q_{23}}-\tilde{q_{43}}\right]dx$$

Thus we solve for $$\tilde{q_{31}}$$ as:

 * $$\tilde{q_{31}}=\tilde{q_{23}}+\tilde{q_{43}}+q^{(3)}$$

Where $$q^{(3)} \;$$ is the contribution to shear flow by Stringer 3:

 * $$q^{(3)}=-\int_{A_3}^{}{\frac{d\sigma _{xx}}{dx}}dA_3$$

The integral can be solved for $$q^{(3)} \;$$ as:

 * $$q^{(3)}=-\left(k_yV_y-k_{yz}V_z \right)Q_z^{(3)}-\left(k_zV_z-k_{yz}V_y \right)Q_y^{(3)}$$

where:

 * $$V_y=\frac{dM_z}{dx}$$
 * $$V_z=\frac{dM_y}{dx}$$
 * $$Q_z^{(3)}=\int_{A_3}^{}{ydA_3}$$
 * $$Q_y^{(3)}=\int_{A_3}^{}{zdA_3}$$

The same procedure can be used to solve for Stringer 2:

 * $$\tilde{q_{23}}= \underbrace{\tilde{q_{12}}}_{flow in}-\underbrace{\tilde{q_{24}}}_{flow out}+q^{(2)}$$

The value of $$q^{(2)} \;$$ can be computed as done with Stringer 3 with the values of $$Q_z^{(2)}$$ and $$Q_y^{(2)}$$ as:

 * $$Q_z^{(2)}=y_2A_2$$
 * $$Q_y^{(2)}=z_2A_2$$

where $$y_2 \;$$ and $$z_2 \;$$ are the y and z coordinates of Stringer 2, and $$A_2 \;$$ is the area of Stringer 2.

The same procedure can be followed on Stringer 4 to find $$q^{(4)} \;$$ and $$\tilde{q_{24}}$$. Hence, we can also deduce $$\tilde{q_{31}}$$.

Using the principles of superposition, the values of q can be determined.

 * $$q_{ij}=\tilde {q}_{ij} + q_{k}$$

On the right side of the equations, the tilde terms are the knowns and the non-tilde terms are unknowns.

 * $$q_{12}=\tilde {q}_{12} + q_{1}$$
 * $$q_{23}=\tilde {q}_{23} + q_{1}$$
 * $$q_{31}=\tilde {q}_{31} + q_{1}$$
 * $$q_{24}=\tilde {q}_{24} + q_{2}$$
 * $$q_{43}=\tilde {q}_{43} + q_{2} - q_{3}$$
 * $$q_{41}=\tilde {q}_{41} + q_{3}$$

These equations reveal that there are three unknowns, $$q_1, q_2, q_3 \;$$, therefore three equations are needed to solve for these values to complete the problem. The three equations to be use are: 1) Moment equation: Take the moment of $$V_y \;$$, $$V_z \;$$, and $$\left\{q_{12},...,q_{41} \right\}$$ about any convenient point (usually where the lines of action of $$V_y \;$$ and $$V_z \;$$ intersect). 2) $$\theta _1=\theta _2 \;$$ 3) $$\theta _2=\theta _3 \;$$ Equations 2 and 3 are the compatibility equations.

To recap, in order to solve P2 for each cell you must follow path "$$s_i \;$$" and then determine the equilibrium of each stringer on the path. There are two ways to accomplish this: 1) Complete method using FBD as shown above 2) A consequence of the first method, as exemplified in Figure 13:

 * $$\tilde{q_{j6}} = \tilde{q_{j2}} - \tilde{q_{j5}} - \tilde{q_{j8}} + q^{(j)}$$

Also, the left side of the equation simply becomes negative if the direction of $$\tilde{q_{j6}} \;$$ changes (and becomes $$\tilde{q_{6j}}$$), as shown in Figure 14.

 * $$\tilde{q_{6j}} = \tilde{q_{j2}} - \tilde{q_{j5}} - \tilde{q_{j8}} + q^{(j)}$$

Using the Euler Cut Principle brings forth a question, what happens if the object is cut such that one stringer is isolated as shown in Figure 15? Due to the cuts made, $$\tilde{q_{23}}=\tilde{q_{31}}=\tilde{q_{34}}=0$$. We then sum the forces acting on the FBD shown in Figure 16.

 * $$\underbrace{\tilde{q_{31}}}_{0} = \underbrace{\tilde{q_{23}}}_{0} - \underbrace{\tilde{q_{34}}}_{0} + q^{(3)}$$

This shows that $$q^{(3)}=0 \;$$, which is not possible. Therefore, when using the Euler Cut method a stringer must not be isolated from the rest of the body.

Shear Buckling
Expressing $$\begin{Bmatrix} C_{22}, C_{13}, C_{31}, C_{33}\end{Bmatrix}$$ in terms of $$C_{11}$$ and for $$\vartheta = 1.5$$ we can accomplish:

 * 1) Finding $$\lambda$$ for $$\vartheta = 1.5$$
 * 2) Evaluate numerically for $$k_{sys}$$ in equation 30
 * 3) Find $$[k_{ij}]$$

 * $$\begin{bmatrix} k_{12} & k_{23} & \cdots & k_{25} \\ \vdots & \ddots & & \vdots \\ \vdots & & \ddots & \vdots \\k_{32} & \cdots & \cdots & k_{55}\end{bmatrix} \begin{bmatrix} C_{22} \\ \vdots \\ \vdots \\ C_{33}\end{bmatrix} = \begin{bmatrix} -\frac{u_z}{a}C_{11} \\ 0 \\ 0 \\ 0\end{bmatrix}$$

Then solve $$\begin{Bmatrix} C_{22}, C_{13}, C_{31}, C_{33}\end{Bmatrix}$$ intersect with $$C_{11}$$. If the $$K$$ matrix is expressed as $$\underline{\overline{K}}$$ then the inverse of said matrix is $$\underline{\overline{K}}^{-1}$$. The values for $$\begin{Bmatrix} C_{22}, C_{13}, C_{31}, C_{33}\end{Bmatrix}$$ can be related as

 * $$\begin{Bmatrix} C_{22}, C_{13}, C_{31}, C_{33}\end{Bmatrix} = \underline{\overline{K}}^{-1}\ \begin{bmatrix} -\frac{u_z}{a}C_{11} \\ 0 \\ 0 \\ 0\end{bmatrix}$$

Resulting in

 * $$u_z = C_{11}sin(\frac{\pi x}{a})sin(\frac{\pi y}{b}) + C_{22}sin(\frac{2\pi x}{a})sin(\frac{2\pi y}{b}) + C_{13}sin(\frac{\pi x}{a}) sin(\frac{3\pi y}{b}) + C_{31}sin(\frac{3\pi x}{a})sin(\frac{\pi y}{b}) + C_{33}sin(\frac{3\pi x}{a})sin(\frac{3\pi y}{b})$$

Answer to Jared's Question of Airfoil Cell Lacking Exposed Surface Area - Part 2
In the case of an airfoil lacking exposed surface area, the resulting equation of shear flow would be as follows:

 * $$ 0=q^{(1)}+q^{(2)}+q^{(4)}$$

Which cannot be true or even possible, since:

 * $$ 0=\sum_{e=1}^{4}q^{(e)}$$

This equation can also be expressed as:

 * $$ 0=q^{(1)}+q^{(2)}+q^{(3)}+q^{(4)}\; \therefore \; q^{(1)}+q^{(2)}+q^{(4)}=-q^{(3)}\neq 0$$

To prove that the sum of all the shear flows is zero (as shown above) the following equation, derived earlier, can be utilized:

 * $$ q^{(e)}=n_z Q_{z}^{(e)}+n_{y} Q_{y}^{(e)} $$

It should be noted that n in the above equation does not represent a unit normal directional component, as was the case previously. In the above equation:

 * $$ n_z = -(k_y V_y -k_{yz}V_z) $$
 * $$ n_y = -(k_z V_z -k_{yz}V_z) $$

Now, the above expression for shear flow, q, can be used to reinforce the fact that the sum of all the individual shear flows is zero.

 * $$ \sum_{e=1}^4 q^{(e)}=n_z \sum_{e=1}^4 Q_z ^{(e)} + n_y \sum_{e=1}^4 Q_y^{(e)} $$

Using the definition of $$ Q_z $$ and $$ Q_y $$ along with the Mean Value Theorem, it can be shown that:

 * $$ \sum_{e=1}^4 Q_z ^{(e)}=0 \; and \; \sum_{e=1}^4 Q_y^{(e)}=0 \; \therefore \; \sum_{e=1}^4 q^{(e)}=0 $$

Read and Report: Section 4.2 Bidirectional Bending
When considering a beam characterized by bidirectional bending, the approximate displacement expansions, u, v and w, are as follows:

 * $$ u = u_0 (x) + z \psi_y (x) + y \psi_z (x) $$
 * $$ v = v_0 (x) $$
 * $$ w = w_0 (x) $$

In the above equations, $$ \psi_y $$ and $$ \psi_z $$ represent rotations about the y and z axes, respectively, using the right-hand rule about the respective axes to determine the sign of each. Consequently, the strains are as follows:

 * $$ \epsilon_{xx} = \frac{\delta u}{\delta x} = \frac{du_0}{dx}+z\frac{d \psi_y}{dx}+y\frac{d \psi_z}{dx} $$
 * $$ \gamma_{xy} = \frac{\delta v}{\delta x} + \frac{\delta u}{\delta y} = \frac{dv_0}{dx}+\psi_z $$
 * $$ \gamma_{xz} = \frac{\delta w}{\delta x} + \frac{\delta u}{\delta z} = \frac{dw_0}{dx}+\psi_y $$

Using the assumption that $$ \gamma_{xy} = 0 $$ and $$ \gamma_{xz} = 0 $$ simplifies the last two expressions to:

 * $$ \psi_z = -\frac{dv_0}{dx} $$
 * $$ \psi_y = -\frac{dw_0}{dx} $$

which when subsituted into the equation for strain in the x direction yield:

 * $$ \epsilon_{xx} = \frac{du_0}{dx} - y\frac{d^2v_0}{dx^2} - z\frac{d^2w_0}{dx^2} $$

If $$ N_x = 0 $$, then $$ \frac{du_0}{dx}=0 $$, further simplifying the equation of x-axis strain to:

 * $$ \epsilon_{xx} = - y\frac{d^2v_0}{dx^2} - z\frac{d^2w_0}{dx^2} $$

It is next desirable to apply this result to the y and z bending moment equations. The definition of these moments is as follows:

 * $$ M_y = \int_A \int z \sigma_{xx}dA = -E \int_A \int (yz\frac{d^2 v_0}{dx^2}+z^2\frac{d^2 w_0}{dx^2})dA = -EI_{yz}\frac{d^2 v_0}{dx^2} - EI_y\frac{d^2 w_0}{dx^2} $$
 * $$ M_z = \int_A \int y \sigma_{xx}dA = -EI_{yz}\frac{d^2 w_0}{dx^2} - EI_z\frac{d^2 v_0}{dx^2} $$

where

 * $$ I_y = \int_A \int z^2 dA $$
 * $$ I_z = \int_A \int y^2 dA $$
 * $$ I_{yz} = \int_A \int yz dA $$

In the definition of bending moment equations, the quadratic terms can be expressed as the curvature, i.e.:

 * $$ \chi_y = \frac{d^2 v_0}{dx^2} \; and \; \chi_z = \frac{d^2 w_0}{dx^2} $$

Now, the equations can be rewritten in matrix form:

 * $$ \begin{bmatrix} M_y \\ M_z \end{bmatrix} = \begin{bmatrix} I_y & I_{yz} \\ I_{yz} & I_z \end{bmatrix} \begin{bmatrix} -E\chi_z \\ -E\chi_y \end{bmatrix} = \mathbf{I} \begin{bmatrix} -E\chi_z \\ -E\chi_y \end{bmatrix} $$

 * $$ \epsilon_{xx} = -y\chi_y - z\chi_z = \begin{bmatrix} z & y \end{bmatrix} \begin{bmatrix} -\chi_z \\ -\chi_y \end{bmatrix} $$

 * $$ \sigma_{xx} = E\epsilon_{xx} = E \begin{bmatrix} z & y \end{bmatrix} \begin{bmatrix} -\chi_z \\ -\chi_y \end{bmatrix} = E \begin{bmatrix} z & y \end{bmatrix} \frac{1}{E} \mathbf{I^{-1}} \begin{bmatrix} M_y \\ M_z \end{bmatrix} = \begin{bmatrix} z & y \end{bmatrix} \mathbf{I^{-1}} \begin{bmatrix} M_y \\ M_z \end{bmatrix} $$

where

 * $$ \mathbf{I^{-1}} = \frac{1}{D} \begin{bmatrix} I_z & -I_{yz} \\ -I_{yz} & I_y \end{bmatrix} $$

and

 * $$ D = I_y I_z - (I_{yz})^2 $$

Similarly, the shear flow equation can be converted to matrix form:

 * $$ q = -\int_A \int \frac{d\sigma_{xx}}{dx}dA $$

where

 * $$ \frac{d\sigma_{xx}}{dx} = \begin{bmatrix} z & y \end{bmatrix} \mathbf{I^{-1}} \begin{bmatrix} \frac{dM_y}{dx} \\ \frac{dM_z}{dx} \end{bmatrix} $$

Recall that:

 * $$ V_y = \frac{dM_y}{dx} $$
 * $$ V_z = \frac{dM_z}{dx} $$

and

 * $$ Q_y = \int_A \int z dA $$
 * $$ Q_z = \int_A \int y dA $$

MATLAB Coding Problem Continued

 * {| class="collapsible collapsed"

!Full MATLAB Code - Single Celled Airfoil, Shear Analysis
 *
 *
 * }

 * {| class="collapsible collapsed"

!Full MATLAB Code - Multi-Celled Airfoil, Shear Analysis
 *
 *
 * }

 * {| class="collapsible collapsed"

!Full MATLAB Code - Critical Buckling Shear Stresses and Buckling Modes
 *
 *
 * }
 * {| class="collapsible collapsed"

!Full MATLAB Code - Verification of Lambda Consider the 2x2 $$\mathbf{K}\;$$ matrix:
 *
 *

\displaystyle \left[ \begin{array}{ll} \frac{\lambda (1 + \vartheta^2)^2}{\vartheta^2} &	 \frac{4}{9} \\	 \frac{4}{9} &	 \frac{16 \lambda (1 + \vartheta^2)^2}{\vartheta^2} \end{array} \right] $$

The determinant of this matrix yields the following:

 * $$\left(\frac{\lambda (1 + \vartheta^2)^2}{\vartheta^2}\right)\left(\frac{16 \lambda (1 + \vartheta^2)^2}{\vartheta^2}\right) - \frac{16}{81} = 0$$

Reducing,

 * $$\lambda^2 = \frac{\vartheta^4}{9(1+\vartheta^2)^2}$$

Thus,

 * $$\lambda = \pm \frac{1}{9} \frac{\vartheta^2}{(1+\vartheta^2)}$$

 * }

 * {| class="collapsible collapsed"

!Code Output - Single Celled Airfoil, Shear Analysis
 *
 *
 * }
 * {| class="collapsible collapsed"

!Code Output - Multi-Celled Airfoil, Shear Analysis
 *
 *
 * }

Figure 17 shows the minimum critical buckling stresses versus aspect ratio. The blue line is for the 2 equation method, the red line for the 5 equation method, and the green line for clamped conditions. As expected, the clamped conditions stresses are higher. Figure 18 shows the buckling mode shapes perspective view.

The calculated minimum shear stress in the skin panels is approximately 2 orders of magnitude lower than the theoretical buckling stress. While this may be evidence of a coding error, assuming there are no coding issues this would indicate the skin panels are not particularly effective at handling shear loads.

LaTeX Code Editor
The provided LaTeX equation editor was initially quite useful, since no members of the group had any prior experience using this coding language. The interface is very straightforward and easy to use, and since the code is auto-generated, it can be cut and pasted directly into the MediaWiki editing field. The editor also included a preview feature, so it was readily apparent if any mistakes were made during the coding process. However, as the semester progressed, the experience gained from manipulating these codes in the HW reports began to render the editor less efficient. Once one has a basic working knowledge of LaTeX coding, it is far easier to just code a tilde above a bold symbol than it is to individually search out the appropriate buttons to generate the code with the editor. While it is still useful for generating larger codes, such as blank matrices and vectors, most of the required coding for the homework assignments is much more easily done by hand than it is using the editor. In short, the editor is a wonderful learning tool for LaTeX coding, but it less useful for experienced users.

InkScape
Initially, the group generated figures and charts for the homework reports using various different softwares. While there was nothing wrong with the individual submissions, compiling a report using figures generated from multiple sources is less than ideal. It is important for these sorts of assignments to maintain a consistent look and feel throughout, as you would with any sort of paper or technical report. In order to better achieve this state, the group elected to start using only InkScape to generate non-MATLAB figures. Employing InkScape allowed the group to generate consistently uniform figures for the assignments, and the difference is very noticeable in a positive way. Furthermore, InkScape is a vector graphics program, which means that a given image will appear the same regardless of of the size it is scaled to, which is very convenient for thumbnailing on MediaWiki. The end result was that the overall quality of the group's submissions was greatly improved after the adoption of InkScape as the preferred figure generating tool.

WikiEd
At the start of the semester, some of the group members elected to employ WikiEd to assist with the generation of MediaWiki content. The experience with this extension to MediaWiki closely paralleled that of the LaTeX code editor. At first, the extension was very helpful. As with the LaTeX editor, the interface for WikiEd is very straightforward and easy to use. Anytime any non-text code is required, one can simply push a button and the code is generated in the editing field. This was very helpful at the beginning of the semester when the group was still getting familiar with MediaWiki. However, as the semester went on, the extensions became more and more cumbersome. In many cases, it is much faster to simply type in the required code than it is to go hunt for the appropriate button to generate it using WikiEd. Also, it became apparent later on that if one is typing code into a WikiEd-modified editing field, and one accidentally clicks the back button on the browser (or hits the browser back key on the keyboard, if applicable), one loses all of the work entrered into the editing field since the last save. Needless to say, this is an extremely irritating bug. As with the LaTeX editor, the WikiEd extension was initially quite useful for learning purposes but became more of a burden as the semester continued.

MediaWiki vs. E-Learning
It is the general opinion of this group that E-Learning is the slightly superior option for this kind of assignment. While MediaWiki is far more powerful in terms of content handling; that is, ability to handle equations, videos, audio files, etc., it is slightly less user-friendly and has less organizational tools. Although E-Learning is notorious for server issues, generally speaking this can be avoided by approaching assignments in a timely manner. Perhaps the biggest advantage over MediaWiki when using E-Learning is that E-Learning has the built-in capability for transmitting grades to students in a confidential manner. MediaWiki has no method for handling this, and as such requires the use of email for grade distribution. All in all, while MediaWiki is a very powerful tool and a useful tool to be familiar with, E-Learning is still a superior option for class use.

Contributing Team Members
Steven Hepsworth Eas4200c.f08.vqcrew.f 21:46, 7 December 2008 (UTC)

Keith Javier Stober Eas4200c.f08.vqcrew.b 21:56, 8 December 2008 (UTC)

Kevin Klauk Eas4200c.f08.vqcrew.E 01:43, 9 December 2008 (UTC)

Casey Barnard Eas4200c.f08.vqcrew.g 04:29, 9 December 2008 (UTC)

Darin Toscano Eas4200c.f08.vqcrew.d 12:13, 9 December 2008 (UTC)

Adam Edstrand Eas4200c.f08.vqcrew.a 18:29, 9 December 2008 (UTC)

John Saxon Eas4200c.f08.vqcrew.c 18:59, 9 December 2008 (UTC)

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/

title: User:Eas4200c.f08.vqcrew.c/Homework 7


text:
User:Eml4500.f08.ateam.boggs.t/week 15

Lecture 39

title: User:Eml4500.f08.ateam.boggs.t/week 15


text:
User:Eml4500.f08.ramrod.c/12-1-08

Meeting 39 December 1, 2008

$$\mathbf{\tilde{d}^{(e)}}_{6x1} = \mathbf{\tilde{T}^{(e)}}_{6x6}\underbrace{\mathbf{d^{(e)}}_{6x1}}$$
 * $$\mbox{known after solving FE system}$$

Compute $$U(\tilde{x})\; \mbox{and} \; V(\tilde{x})$$

$$\mathbf{U(\tilde{x})} = U(\tilde{x})\vec{\tilde{i}} + V(\tilde{x})\vec{\tilde{j}}$$

$$\mathbf{U(\tilde{x})} = U_x(\tilde{x})\vec{i} + U_y({x})\vec{j}$$

Compute $$U(\tilde{x})\; \mbox{and}\; V(\tilde{x})$$ using Equations (1) and (2) from above

Compute $$U_x(\tilde{x}) \; \mbox{and} U_y(\tilde{x}) \; \mbox{from} \; U(\tilde{x}) \; \mbox{and} \; V(\tilde{x})$$
$$\begin{Bmatrix} U_x(\tilde{x})\\ V(\tilde{x}) \end{Bmatrix} = \mathbf{R}^T\begin{Bmatrix}U(\tilde{x})\\V(\tilde{x})\end{Bmatrix}$$

 * Note: The RT matrix can be found above.

$$\begin{Bmatrix}U(\tilde{x})\\V(\tilde{x})\end{Bmatrix} = \underbrace{\begin{bmatrix} N_1 & 0 & 0 & N_4 & 0 & 0\\ 0 & N_2 & N_3 & 0 & N_5 & N_6 \end{bmatrix}}_{\mathbb{N}(\tilde{x})}\begin{Bmatrix} \tilde{d}_1^{(e)}\\ \tilde{d}_2^{(e)}\\ \tilde{d}_3^{(e)}\\ \tilde{d}_4^{(e)}\\ \tilde{d}_5^{(e)}\\ \tilde{d}_6^{(e)} \end{Bmatrix}$$

$${\color{Blue}\begin{Bmatrix} U_x(\tilde{x})\\ U_y(\tilde{x}) \end{Bmatrix}= \mathbf{R}^T\mathbf{\mathbb{N}}(\tilde{x})\mathbf{\tilde{T}}^{(e)}\mathbf{d}^{(e)}}$$

Pros

 * Can find information on basically anything.

Cons

 * Anyone can vandalize your contributions.
 * Have to learn Wiki code to make contributions.

title: User:Eml4500.f08.ramrod.c/12-1-08


text: User:Erakura

title: User:Erakura


text: User talk:Erakura

title: User talk:Erakura


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User:Eas4200c.f08.gator.reger/Week13,14

Meeting 36, Weds 11/19
Upcoming plans:

 * S.) Single-cell Sections
 * S.1) Without Stringers
 * S.2) With Stringers
 * M.) Multi-cell Sections
 * M.1) Without Stringers
 * M.2) With Stingers

S.1) Without Stringers
Can the section from Figure 1 resist transverse shear $$V_x\,$$

$$R^z = R^z_{AB} + R^z_{BA}\,$$ Where $$ R^z_{AB} = q\bar{A'B'} = -R^z_{AB}\,$$ Therefore: $$R^z = 0\,$$

S.2) With Stringers
Looking at Figure 2 (Neglecting contribution of web to bending)

$$ R^z = V_z \,$$ Principal of superposition due to linearity, non-constant shear flow. $$q_{12} = q + \tilde{q_{12}} ,..., q_{ij} = q + \tilde{q_{ij}}\,$$ Where $$ q_{12}\,$$ is not equal to $$ q_{23}\,$$ is not equal to $$ q_{31}\,$$ But $$ q_{ij}(s)\,$$ is constant.

Analytic Algebra
Observation: One unknown $$ q\,$$ (since $$\tilde{q_{ij}}\,$$ is known after solving P2): Need one equation. Method: 1.) Solve problem P2 (Given $$ V_z, V_y\,$$) for all \tilde{q_{ij}}\, 2.) Moment Equation: Take moment about any point in the plane ($$y,z\,$$) 2.1) Superposition: $$ q_{ij} = q + \tilde{q_{ij}}\,$$ 2.2) Select point $$\bar{\theta}\,$$ in plane ($$y,z\,$$) 2.3) $$\Sigma{\theta}\,$$ Moment of ($$ V_z, V_y\,$$)

title: User:Eas4200c.f08.gator.reger/Week13,14


text:
User:Sobreira

User:
 * wikipedia:gl:User:Sobreira
 * wikipedia:en:User:Sobreira
 * wiktionary:gl:User:Sobreira
 * wiktionary:en:User:Sobreira

See at w:Galician language

title: User:Sobreira


title: File:Figu1.PNG


title: File:Figu2.PNG


text:
User:Eas4200c.f08.gator.edwards/Week Seven Class Notes

 Class Notes | (archived)

 Matlab Program Continued | (archived)

Contributing Team Members:

The following members contributed to this report:

Robert Reger Eas4200c.f08.gator.reger 22:09, 5 December 2008 (UTC)

Greg Edwards Eas4200c.f08.gator.edwards 19:04, 7 December 2008 (UTC)

Eas4200c.f08.spars.prey 07:48, 9 December 2008 (UTC)

Justin Dickinson Eas4200c.f08.gator.Dickinson 09:33, 9 December 2008 (UTC)

William Mueller Eas4200c.f08.gator.mueller 19:58, 9 December 2008 (UTC)

Taylor Lusk Eas4200c.f08.gator.lusk 21:10, 9 December 2008 (UTC)


text:
User:Eas4200c.f08.gator.edwards/Wk7 Class Notes

Meeting 36, Wed 11/19
Upcoming plans:

 * S.) Single-cell Sections
 * S.1) Without Stringers
 * S.2) With Stringers
 * M.) Multi-cell Sections
 * M.1) Without Stringers
 * M.2) With Stingers

S.1) Without Stringers
Can the section from Figure 1 resist transverse shear $$V_x\,$$

Resultant force on single cell section, Rz $$R^z = R^z_{AB} + R^z_{BA}\,$$ Where $$R^z_{AB} = q\bar{A'B'} = -R^z_{AB}\,$$ and $$R^z_{BA} = q\bar{B'A'} = -R^z_{AB}\,$$ Therefore: $$R^z = 0\,$$

S.2) With Stringers
Looking at Figure 2 The contribution of bending due to the web is assumed to be zero, therefore each section between stringers has a constant shear flow, q0.

$$ R^z = V_z \,$$ Principal of superposition due to linearity, non-constant shear flow. $$q_{12} = q + \tilde{q}_{12} ,..., q_{ij} = q + \tilde{q}_{ij}\,$$ Where $$ q_{12}\,$$ is not equal to $$ q_{23}\,$$ is not equal to $$ q_{31}\,$$ But $$ q_{ij}(s)\,$$ is constant.

Analytic Algebra
Observation: One unknown $$ q\,$$ (since $$\tilde{q}_{ij}\,$$ is known after solving P2): Need one equation. Method: 1.) Solve problem P2 (Given $$ V_z, V_y\,$$) for all $$ \tilde{q}_{ij}\,$$ 2.) Moment Equation: Take moment about any point in the plane ($$y,z\,$$) 2.1) Superposition: $$ q_{ij} = q + \tilde{q}_{ij}\,$$ 2.2) Select point $$\bar{\theta}\,$$ in plane ($$y,z\,$$) 2.3) $$\Sigma{\theta}\,$$ Moment of ($$ V_z, V_y\,$$)

Analytic Algebra cont.
Method cont: 3.) Back to superposition:  $$q_{ij}=q+\tilde{q}_{ij}$$   where:  $$q_{ij}\ =\ $$ true shear flow  $$q\ =\ $$closed cell constant shear flow  $$\tilde{q}_{ij}\ = \ $$ open cell piecewise constant shear flow

M.1) Multicell without stringers
$$R^{z}=\sum_{i=1}^{n_{cells}}R_{i}^{z}$$ $$R^z_i=0$$ $$R^z=0$$

M.2) Multicell with stringers
Solve by using the fact that P = P1 + P2, or $$R^z=R^{z1}+R^{z2}$$

and by recalling: $$R^{z1}=\sum_iR_i^{z1}=0$$ $$R^z=R^{z2}=V_z$$

Solving P2
Referring to Figure 5 for stringer 3, and using the fact that:

$$\sum{F_x=0}$$,

$$\int_{A_3}\left[\sigma_{xx}\left(x+dx \right)- \sigma_{xx}\left(x \right)\right]dA_3$$,

$$\left[ -\tilde{q}_{23}-\tilde{q}_{43}-\tilde{q}_{31}\right]dx$$

we have:

$$\tilde{q}^{(3)}=\tilde{q}_{23}+\tilde{q}_{43}+q^{(3)}$$

$$q^{(3)}=-\int_{A_3}\frac{d\sigma_{xx}}{dx}dA_3$$

$$\ $$

Recall: $$\qquad\ V_y=\frac{dM_z}{dx}\ \ \ \ V_z=\frac{dM_y}{dx}$$

$$q^{(3)}=-\left(k_yV_y-k_{yz}V_z\right) Q_z^{(3)}-\left(k_zV_z-k_{yz}V_y\right) Q_y^{(3)}$$ $$Q_z^{(3)}=\int_{A_3}y\ dA_3 \ \ \ Q_y^{(3)}=\int_{A_3}z\ dA_3$$

Now for Stringer 2:

$$\tilde{q}_{23}=\tilde{q}_{12}-\tilde{q}_{24}+q^{(2)}$$

Where $$\tilde{q}_{12}=\tilde{q}_{24}=0\ \ \ \ $$ because of the cut

$$q^{(2)}$$ is computed as in (1)

denoting $$y_2$$and $$z_2$$ as the y and z coordinates of stringer 2 and $$A_2$$ as the area of stringer 2:

$$Q_z^{(2)}=y_2A_2$$ $$Q_y^{(2)}=z_2A_2$$

Now for Stringer 4:

$$\tilde{q}_{43}=\tilde{q}_{24}-\tilde{q}_{41}+q^{(4)}$$

Where $$\tilde{q}_{24}=\tilde{q}_{41}=0\ \ \ \ $$ because of the cut

$$q^{(4)}$$ is computed as above for $$q^{(2)}$$

Hence deduce $$\tilde{q}_{31}\ \rightarrow$$finished P2

Superposition (again)
$$q_{ij}=\tilde{q}_{ij}+q_k$$ $$q_{12}=\tilde{q}_{12}+q_1$$ $$q_{23}=\tilde{q}_{23}+q_1$$ $$q_{31}=\tilde{q}_{31}+q_1$$ $$q_{24}=\tilde{q}_{24}+q_2$$ $$q_{43}=\tilde{q}_{43}+q_3$$ $$q_{41}=\tilde{q}_{41}+q_3$$

3 unknowns: $$q_1$$,$$q_2$$,$$q_3$$

Therefore, we need 3 equations:
 * 1) Moment Equation: Take mom. of $$V_y$$,$$V_z$$,and$$\left\{q_{12},...q_{41} \right\}$$ about any convenient point (usually where lines of action of $$V_y$$and $$V_z$$ intersect)
 * 2) $$\theta_1=\theta_2$$
 * 3) $$\theta_2=\theta_3$$ where $$\theta$$ is computed with true shear flow

Meeting 39, Mon 12/1
Solving P2: For each cell
 * Follow path $$S_i\,$$
 * Write equilibrium for each stringer path:

We can go about this in 2 ways:

1.) Complete method: Free Body Diagram.

2.) Consequence of first method.

$$\bar{q_{j6}} = \bar{q_{j2}} - \bar{q_{j5}} - \bar{q_{j8}} + q^{(j)}\,$$

Q: What if we cut cell walls such that one stringer was isolated? A: It cannot be solved in this manner. After curring the three cell walls surrounding "3" we see that this allows no shear flow in "3" therefore making the equation equal to zero and invalid.

Meeting 40, Fri 12/5
 Plate Buckling

Plate buckling shape under shear. Express $$\left[C_{22},C_{13},C_{31},C_{33} \right]$$ in terms of $$\displaystyle C_{11}$$. Use $$\vartheta =1.5$$.

The first step is to find $$\displaystyle \lambda$$ for $$\vartheta =1.5$$.

$$  \displaystyle \lambda =  \left[ \frac{\vartheta^4}{81 (1 + \vartheta^2)^4} \left\{ 1 	 + 	 \frac{81}{625} +	 \frac{81}{25} \left(	   \frac{1 + \vartheta^2}{1 + 9 \vartheta^2}	 \right)^2 +	 \frac{81}{25} \left(	   \frac{1 + \vartheta^2}{9 + \vartheta^2}	 \right)^2 \right\} \right]^{1/2} $$

$$\displaystyle \lambda=0.02875$$

Once the $$\displaystyle \lambda$$ is calculated, the next step is to evaluate $$\textbf{K}$$

$$  \displaystyle \textbf{K}= \left[ \begin{array}{lllll} \frac{\lambda (1 + \vartheta^2)^2}{\vartheta^2} &	 \frac{4}{9} &	 0	 &	 0	 &	 0	 \\	 \frac{4}{9} &	 \frac{16 \lambda (1 + \vartheta^2)^2}{\vartheta^2} &	 - \frac{4}{5} &	 - \frac{4}{5} &	 \frac{36}{25} \\	 0	 &	 - \frac{4}{5} &	 \frac{\lambda (1 + 9 \vartheta^2)^2}{\vartheta^2} &	 0	 &	 0	 \\	 0	 &	 - \frac{4}{5} &	 0	 &	 \frac{\lambda (9 + \vartheta^2)^2}{\vartheta^2} &	 0	 \\	 0	 &	 \frac{36}{25} &	 0	 &	 0	 &	 \frac{\lambda (9 + 9 \vartheta^2)^2}{\vartheta^2} \end{array} \right] \left\{ \begin{array}{l} C_{11} \\	 C_{22} \\	 C_{13} \\	 C_{31} \\	 C_{33} \end{array} \right\} =  \left\{ \begin{array}{l} 0	 \\	 0	 \\	 0	 \\	 0	 \\	 0     \end{array} \right\} $$

Using $$\vartheta =1.5$$ and $$\lambda=0.02875$$

$$  \left[ \begin{array}{lllll} 0.13497 	 &	 0.44444	 &	 0	 &	 0	 &	 0	 \\	 0.44444	 &	 2.15944  	 &	 - 0.8	 &	 - 0.8	 &	 1.44	 \\	 0	 &	 - 0.8	 &	 5.76997  	 &	 0	 &	 0	 \\	 0	 &	 - 0.8	 &	 0	 &	 1.61719  	 &	 0	 \\	 0	 &	 1.44	 &	 0	 &	 0	 &	 10.93219        \end{array} \right] \left\{ \begin{array}{l} C_{11} \\	 C_{22} \\	 C_{13} \\	 C_{31} \\	 C_{33} \end{array} \right\} =  \left\{ \begin{array}{l} 0	 \\	 0	 \\	 0	 \\	 0	 \\	 0     \end{array} \right\} $$

The matrices above can be reduced to

$$  \left[ \begin{array}{lllll} 2.15944 	 &	 - 0.8	 &	 - 0.8	 &	 1.44	 \\	 - 0.8	 &	 5.76997  	 &	 0	 &	 0	 \\	 - 0.8	 &	 0	 &	 1.61719  	 &	 0	 \\	 1.44	 &	 0	 &	 0	 &	 10.93219        \end{array} \right] \left\{ \begin{array}{l} C_{22} \\	 C_{13} \\	 C_{31} \\	 C_{33} \end{array} \right\} =  \left\{ \begin{array}{l} \frac {-4}{9}C_{11} \\	 0	 \\	 0	 \\	 0     \end{array} \right\} $$

Then by setting arbitrarily $$\displaystyle c_{11}=0$$, $$\left[C_{22},C_{13},C_{31},C_{33} \right]$$ can be solved for in terms of $$\displaystyle C_{11}$$. The results from MATLAB are as follows:

$$  \left\{ \begin{array}{l} C_{22} \\	 C_{13} \\	 C_{31} \\	 C_{33} \end{array} \right\} =  \left\{ \begin{array}{l} -0.3038C_{11} \\	 -0.0421C_{11} \\	 -0.1503C_{11} \\	 0.0400C_{11} \end{array} \right\} $$

$$U_{z}=C_{11}\sin (\frac{\pi x}{a})\sin (\frac{\pi y}{b})+C_{22}\sin (\frac{2\pi x}{a})\sin (\frac{2\pi y}{b})+C_{13}\sin (\frac{\pi x}{a})\sin (\frac{3\pi y}{b})+C_{31}\sin (\frac{3\pi x}{a})\sin (\frac{\pi y}{b})+C_{33}\sin (\frac{3\pi x}{a})\sin (\frac{3\pi y}{b})$$

Plot $$\displaystyle U_{z}$$ as a function of (x,y)

Meeting 41, 8 Dec 2008
Timoshenko Beam Theory

This theory accounts for the transverse shear deformation and rotational inertia effects in short beams that are made of sandwiched composites or those that are prone to high-frequency agititaion when the wavelength of the frequency matches the thinckness of the beam. The equation that is developed from this theory is a 4th order with a 2nd order spatial derivative term as well. The additional mechanisms of the deformation lowers the stiffness of a beam, leading to a larger deflection under a static load.

This theory allows for vibrations, and may be described with the coupled linear partial differential equations

\rho A\frac{\partial^{2}u}{\partial t^{2}} = \frac{\partial}{\partial x}\left( A\kappa G \left(\frac{\partial u}{\partial x}-\theta\right)\right) + w $$

\rho I\frac{\partial^{2}\theta}{\partial t^{2}} = \frac{\partial}{\partial x}\left(EI\frac{\partial \theta}{\partial x}\right)+A\kappa G\left(\frac{\partial u}{\partial x}-\theta\right) $$

The dependent variables are:

$$u$$, the translational displacement of the beam

$$\theta$$, the angular displacement.

Note: unlike the Euler-Bernoulli theory, the angular deflection is another variable and not approximated by the slope of the deflection.

Also:

 * $$\rho$$ is the density of the beam material (but not the linear density).
 * $$A$$ is the cross section area.
 * $$E$$ is the elastic modulus.
 * $$G$$ is the shear modulus.
 * $$I$$ is the second moment of area.
 * $$\kappa$$, called the Timoshenko shear coefficient, depends on the geometry. Normally, $$\kappa = 5/6$$ for a rectangular section.
 * $$w$$ is a distributed load (force per length).

These parameters are not necessarily constants.

In determining the shear coefficient it must satisfy:

 * $$\int_A \tau dA = \kappa G A \theta\,$$

For a static beam, the equations can be decoupled:

\frac{\partial^2}{\partial x^2}\left(EI\frac{\partial \theta}{\partial x}\right) = -w $$

\frac{\partial u}{\partial x} = \theta - \frac{1}{A\kappa G} \frac{\partial}{\partial x}\left(EI\frac{\partial \theta}{\partial x}\right) $$

it is seen that the Timeoshenko beam theory for this static case is equivalent to the Euler-Bernoulli theory when the last term above is neglected.

An approximation that is valid when:

\frac{EI}{L A \kappa G} \ll 1 $$

where $$L$$ is the length of the beam and $$H$$ is the maximum deflection.

Wikiversity vs. e-learning
In our opinion, Wikiversity provides a much better learning environment than e-Learning does. Some of the reasoning for this conclusion includes:
 * Wikiversity allows anybody with internet access to collaborate on a variety of topics, while e-Learning is limited to only the students in the class given access
 * The speed of e-Learning depends upon however many students need to be accessing the site at one time, so if it is evening time when many students are doing homework, it is noticeably slower. This is compared to Wikiversity, whose speed is independent upon how many UF students are doing homework.
 * Wikiversity is a tool that we can use on our resume compared to the fact that e-Learning is not
 * Wikiversity expertise translates into being able to contribute to other sister software using MediaWiki whether it is Wikiversity or some system our employer has set up.
 * Articles created on Wikiversity are accessible long after the course they were created for is over. On e-Learning, whenever the professor decides to close the class after it is over, the notes and everything else becomes permanently inaccessible.
 * On the other hand though, students must spend time learning and familiarizing themselves with the MediaWiki format, while all students are already used to using e-Learning in their classes.

Recommended Software to Improve Productivity
Throughout the course, different types of software were used to create the homework reports. Writing the equations was made much easier by using the codecogs webpage. Another equation editing webpage that is helpful is also the sitmo software. For the figure drawing, creating a drawing in Microsoft Word is usually the easiest, but you can also use this project draw software. When it came to just writing the articles, it was easiest to type it directly into the editing box.

The following members contributed to this report:
Robert Reger Eas4200c.f08.gator.reger 22:10, 5 December 2008 (UTC)

Greg Edwards Eas4200c.f08.gator.edwards 19:20, 7 December 2008 (UTC)

Jacob Papp Eas4200c.f08.team12.papp 01:17, 9 December 2008 (UTC)

Eas4200c.f08.spars.prey 07:47, 9 December 2008 (UTC)

Justin Dickinson Eas4200c.f08.gator.Dickinson 09:36, 9 December 2008 (UTC)

William Mueller Eas4200c.f08.gator.mueller 19:59, 9 December 2008 (UTC)

Taylor Lusk Eas4200c.f08.gator.lusk 21:11, 9 December 2008 (UTC)


text:
Talk:La Littérature Française

Specific Literary Periods
I have started a section on the page where French Literature is broken down by century and each century has its own page on Wikiversity, to (possibly) be used as the main pages for individual literature classes.

I will create the page for the 18th Century myself and use it in conjunction with the e-mail list I created not long ago. The idea is that the Wikiversity page can be the place where assignments and lectures are posted, but we will handle discussion on the list. Thoughts? KirbyPuckettFan 02:39, 6 December 2008 (UTC) (talk)

title: Talk:La Littérature Française


text: File:TwobartrussmultiEA.png

title: File:TwobartrussmultiEA.png


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User:Eml4500.f08.team.foskey.ckf/hw7c

Electric Pylon Frame System

Problem Statement
This problem uses the electric pylon code electric_pylon.mprovided earlier, and solves the system as a frame system. This problem is identical in concept to the earlier electric pylon problem, except that the earlier problem was a truss system and did not account for bending moments in the bars.

The assignment again involved scaling the height of the truss to a max height of 60m, applying a 1000N load, and calculating the stresses in the elements. The elements under the highest tensile and compressive stresses are identified,as well as the elements under the highest shear forces. The setup of the problem is seen below.

MATLAB Code (Loading Code)
The following MATLAB code is used to apply a 1000N downward load to the far right tip of the frame system. The unloaded and loaded frames are plotted, and the stresses in the elements are evaluated. The code is commented throughout, and explained further below.

Additional Functions
This loading code uses three functions: PlaneFrameElement, PlaneFrameResults, and NodalSoln.

PlaneFrameElement
This function works by receiving an element's modulus of elasticity, cross sectional area, moment of inertia, and the coordinates of the elements endpoints. The length of the element is calculated based on the endpoint coordinates, and the director cosines are calculated using the length and coordinates.

Then the local element stiffness matrix k_ is found. This matrix is converted into its global coordinate system counterpart k by using the transformation matrix T. All three of these matrices are the matrices used for a frame system, which accounts for the bending moment degrees of freedom.

PlaneFrameResults
This function works by receiving an element's modulus of elasticity, cross sectional area, moment of inertia, the coordinates of the elements endpoints, and the displacements of the element. The length of the element is calculated based on the endpoint coordinates, and the director cosines are calculated using the length and coordinates. The displacement matrix d is created in local coordinates using the transformation matrix T on the global displacements. The strain, eps, is found by dividing the total element elongation by the element's original length. The stress, sigma, is found by multiplying the strain by the modulus of elasticity e. The axial force in the element is then found by multiplying the stress by the cross sectional area A of the element.

NodalSoln
The function NodalSoln reads in the global stiffness matrix, the applied loads, a matrix ,debc, containing the degrees of freedom with specified values (in this case the constrained degrees of freedom), and a matrix, ebcVals ,with the values of the nodes (in this case zeros). The function first calculates the number of global degrees of freedom. Through several steps, the function ultimately calculates the global displacements, and the reactions (in a global coordinate system) of the bars. The displacement matrix, and reaction matrix is returned to the original call function. This function is the same as the function used for the truss system because it is not dependent on the number of degrees of freedom at each node, only the overall total number of degrees of freedom.

Explanation Of The Loading Code
The pylon generated by the originally provided code is to be loaded with a 1000N downward force on the right tip. The original, unloaded frame is to be plotted with the deformed frame, and the elements undergoing the highest stresses (tensile and compressive) are to be marked.

The code for loading the pylon can be broken down into three sub-parts: 1) converting the provided code to a usable form for analyzing the frame system, 2) plotting the frames, and 3) determining the maximum stresses.

In the first part of the loading code (given the file name "electricframe.m") takes the data from the provided code and converts it into variables that will work in a format of solving the frame similar to earlier MATLAB projects involving truss systems (see the original Electric Pylon Truss Problem) The loading code begins by running the provided code to get all of the data that that code generates. This eliminates the need to retype all of the original code material into a new code. Next, the loading code determines the highest point of the originally created truss and determines a scaling factor. The highest point is determined by setting a dummy variable "keep" equal to zero initially, and then checking the y coordinate value of each node to see if it is greater than the "keep" value. If it is, then that value becomes the new "keep" value, until all nodes have been checked. The scaling factor was found using the form

$$ScalingFactor = \frac{60}{keep}$$

Every node coordinate is then multiplied by the scaling factor, increasing the size of the pylon to a maximum height of 60m.

Next, the loading code is given the parameters of the elements. Each element has a Young's Modulus of 200GPa, and a cross sectional area of 4cm2. Since the problem is a frame system, a moment of inertia value is also needed. The bars were assumed to have square cross sections. This leads to calculating the moment of inertia to be

$$I = \frac{b*h^3}{12} = \frac{A^2}{12}$$

Where I is the moment of inertia, b is the base of the cross section, and h is the height. Since the cross section is a square, b and h are equal. The end result is one twelfth the square of the cross sectional area.

The load P is also entered as -1000, because the load is downward (in the negative y direction). The code then creates a 91x2 matrix of the nodal coordinates, with each node represented by a row, the x coordinates in the first column, and the y coordinates in the second row. The total number of displacements is also determined by multiplying the number of nodes by 3, which is the number of degrees of freedom for each node( horizontal, vertical, and a rotation).

The loading program then creates the connectivity array by looping through each element and filling the connectivity array conn with the associated values from the node_connect matrix of the provided code. The location matrix master array lmm is then created. This is accomplished by using the node_connect matrix form the provided code. To fill lmm, one has to realize that each node has three associated degrees of freedom. Using this, the matrix is filled in for elements 1 and 2 like so:

(Note: Only the bold values are actually put into the lmm matrix)

This is accomplished by taking the global node number, decreasing it by one, multiplying it by three, and adding 1 for the horizontal dof, 2 for the vertical, or 3 for the rotation. For example, in element 1, the global nodes associated with it are 1 and 3, and the global dofs are 1, 2, 3, 7, 8 and 9. Looking at the code:

%create location matrix master array using data from original code lmm=zeros(n_elem,6); for i=1:n_elem d1= 3*(node_connect(1,i)-1)+1; d2= 3*(node_connect(1,i)-1)+2; d3 = 3*(node_connect(1,i)-1)+3; d4= 3*(node_connect(2,i)-1)+1; d5= 3*(node_connect(2,i)-1)+2; d6= 3*(node_connect(2,i)-1)+3; lmm(i,1)=d1; lmm(i,2)=d2; lmm(i,3)=d3; lmm(i,4)=d4; lmm(i,5)=d5; lmm(i,6)=d6; end

and realizing that node_connect(1,i) will give the global node 1, and that node_connect(2,i) will give 3 (the other global node), it follows that d1, d2, d3, d4, d5, and d6 will equal 1, 2, 3, 7, 8 and 9 respectively. This makes the first row of lmm to be

$$ lmm=\begin{bmatrix} 1 & 2 & 3 & 7 & 8 & 9 \\ \end{bmatrix} $$

From there, the code uses parts of older codes to constrain nodes 1 and 2, and to set up the force matrix R. The code then calculates the elemental stiffness matrices and the global stiffness matrix by using the PlaneFrameElement function. The code then calculates the strain, stress and axial force in each element using the NodalSoln and PlaneFrameResults functions.

The second part of the loading code plots the frames. The code begins by closing the plot created by the provided code. This is why, when running the code, a plot will appear and quickly disappear. The code then creates a matrix of the displaced nodal coordinates defnodes by adding the displacements found in part 1 to the original nodal coordinates. The code then plots the original frame by creating vectors for each element that go from the node at one end, to the node at the other end of an element. This process is repeated for the deformed frame, this time using the coordinates of the dislocated nodes.

In the third part of the loading code, the results matrix from part 1 is examined. Since the second column of the results matrix represents the stress in an element, that value is examined. The stress value of the first element is examined first, and it is checked to be a compressive force or a tensile force (a compressive stress will be negative, and a tensile stress will be positive). The value is then stored the highest stress. The next element's stress is examined, and determined if it is compressive or tensile. The value is then compared to the highest stress. If it is of greater magnitude, it becomes the highest stress, and its element number is remembered. This process continues until all elements have been checked.

To find the highest nodal rotation, a loop check every third displacement from d, until the highest rotation and its correspond in node is found. Then the conn matrix is used to find the elements that touch that node. The displacements for those elements are found and converted to their axial equivalents. The displacement equations are differentiated twice to obtain the shear equations, which are then used to find the maximum shear in the elements touching the node with the highest rotation.

The code ends by plotting the elements under the highest stresses in different colors so that they might be easily identified, using the same process explained earlier. The plot is titled, axis are labeled and a legend is created.

Results Output
Running the loading code "electricframe.m" generates the following output in MATLAB

Comments On Output
Due to the fact that there are 91 elements in the frame system, the elemental stiffness matrices and the global stiffness matrix are not shown in the output. Displaying these matrices would have made the output far too long, and the global stiffness matrix would be too large to display on a computer screen in a logical manner. The matrix would be broken up and displayed in segments.

The output does display three matrices, the displacement matrix d, the reactions matrix reactions and the results matrix results.

In the displacement matrix, every third value represents a rotational displacement. The first six values of d are zero because nodes 1 and 2 were constrained.

The reactions matrix shows the forces acting at the constrained nodes. This information can be understood more easily in table format

The results matrix shows the strain, stress and axial forces in each element, and is again easier to understand in table form

The final few lines of output identify the elements under the highest stresses and roations. Based upon the output, the element under the highest tensile stress is element 77, which is under 2123 Pa of tensile stress. The element under the highest compressive stress is element 32, which is under 1.632 Pa of compressive stress.

"rotate_node" is the node with the highest rotational displacement (node 33), which is 0.14978 radians counter clockwise. The two elements touching that node are element 55 and 77. The shear force in element 55 is 1.5257e+009 N/m2 and the shear force in element 77 is 1.6221e+009 N/m2.

Resulting Figure
Running the above MATLAB code generates the following figure

In the figure, the red dotted lines represent the undeformed frame and the blue solid lines represent the deformed frame. The light blue element is the element under the highest tensile stress, and the green element is the element under the highest compressive stress. Node 33 is identified, as it is the node under the highest rotational displacement. Element 55 in pink is identified as the element with one of the highest shear stresses. Element 77 is identified by the arrow, as it is element under the highest tensile stress, as well as the highest shear stress.

Notes: The arrow and node 33 identifier in the figure were added manually, and element 77 will be pink when running the MATLAB code above. These things were changed manually to better convey the message of the problem.

It should also be noted that this figure is an "incomplete" figure. A complete figure would show the elements as curved lines, showing the bending the elements occur. This would have been cumbersome to show in the figure as there are so many elements, and the coding for this would have been considerably more difficult.

Frame System vs. Truss System
Compared to the same load in a frame system, as seen in the Original Electric Pylon Problem one notices from the figures that the frame system as a whole deforms less than the truss system. This is because the frame system accounts for bending of the elements and rotations at the nodes. Also it is interesting to note that different elements are now under the highest stresses. In the frame system, the element with the highest compressive stress is now located farther from the actual applied load.

Also it should be noted that the stresses and forces in the elements is lower in the frame system. This is because the forces are now distributed as axial loads and as bending moments. There are also no shear forces in the truss problem, as there are no bending moments in the truss.

title: User:Eml4500.f08.team.foskey.ckf/hw7c


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User:NPThomas

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Nepal/Newspapers

There are not many daily newspapers in Nepal. The case is same in the weekly papers. There are many which are registered but not many of them are under continuous publication. Some of the newspapers which are popular in Nepal are 1)Kantipur 2)The Himalayan Times(english) 3)Nepal Samacharpatra 4)Kathmandu Post(english)

title: Nepal/Newspapers


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Category:Translators fr-eng

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User:Carlmisullivan

My name is Carl Sullivan, I am a full time student. I just discovered Wikiversity and am intrigued by its possibilities.

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User:KirbyPuckettFan/Sandbox

This is my Sandbox.

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Radio Interferometer Telescope

Introduction
This project deals with making a astronomical interferometer (a type of a radio telescope) based on the techniques of aperture synthesis. This telescope is designed to work at a radio frequency of 74 MHz.

The basic idea is to measure complex cross-correlation coefficients between voltage signals from two antennas separated by a distance, $$b$$. This technique, called interferometry, is used to obtain higher angular resolution, which may not be attainable due to modest sizes of individual antenna (or dishes) involved in observations.

Methodology
We plan to use half-wave dipole antennas to sense electromagnetic waves from astronomical sources at $$74 $$ MHz. This signal will be down-converted to $$1.9$$ MHz using a mixer, after necessary amplification. A small band around 73.9 MHz, of about $$ 0.25$$ MHz will be recorded using a direct-voltage recorder system. This raw, 2-bit quantized data will be stored on a PC for a later processing by customized computer programs.

These programs have been termed as `software correlator'. Our code for cross-correlation is based on lookup-table approach, details of which will be available on a separate page titled BITS Correlator.

title: Radio Interferometer Telescope


text:
User:Eas4200c.f08.aeris.krammer/HW7

Minor edit. Formatting of picture sizing edited. Content remains the same. Eas4200c.f08.aeris.mcrae 23:30, 12 December 2008 (UTC)

Unsymmetric Thin-Walled Cross Sections

We know that for a symmetric thin-walled cross section that $$\ q_{s} = -\int\int_{A_{s}}^{}{\frac{d\sigma_{xx}}{dx}dA} $$

For unsymmetric thin-walled cross sections under bidirectional bending, the bending stress $$\left( \sigma_{xx}\right)$$ must be taken into account. Doing so, we arrive at the following equation for sheer flow.

 * $$\ q_{s} = -\left(k_{y}V_{y} - k_{yz}V_{z} \right)Q_{z} + \left(k_{z}V_{z} - k_{yz}V_{y} \right)Q_{y} $$

We can begin our analysis of the figures to the right using the Mean Value Theorem to find the first moments of area around the z and y axes.

 * $$\ Q_{z} = \int \int_{A_{s}}^{}{ydA} = \bar{y}A $$

 * $$\ Q_{y} = \int \int_{A_{s}}^{}{zdA} = \bar{z}A $$

For the case of stringers of different areas, as shown in Figure 1, the area to be used in this equation is $$\ A = \sum_{i=1}^{4}{A_{i}} $$. For the case shown in Figure 2, the area used in this equation is $$\ A = A_{\hat{s}} + A_{\hat{s},L} $$.

Using this, as well as equations from the previous homework, we can break down analysis of unsymmetric thin-walled cross sections, such as the two shown on the right in Figure 1 and Figure 2, into the following four steps.

 * 1. Find $$ \left(\bar{y}, \bar{z} \right) $$.
 * 2. Find $$ I_{y} $$, $$ I_{z} $$, and $$ I_{yz} $$.
 * 3. Find $$ k_{y} $$, $$ k_{z} $$, and $$ k_{yz} $$.
 * 4. Follow path $$ s $$ to find $$ q_{12} $$, $$ q_{23} $$, and $$ q_{34} $$.

Shear Flow in a Single and Multi-Cell Section
There are different scenarios that can be analyzed to obtain a better understanding of shear flow in sections. The scenarios are divided into Single-Cell sections and Multi-Cell sections. In a Single-Cell section, two more scenarios can be analyzed. One without stringers and another with stringers. This can also be done for Multi-Cell sections.

The following plan was used to abbreviate and guide the analysis of shear flow through each different scenario:

S) Single-Cell sections S1) Single-Cell section without stringers (most basic analysis) S2) Single-Cell section with stringers

M) Multi-Cell sections

M1) Multi-Cell section without stringers M2) Multi-Cell section with stringers

To analyze S1, the cell is looked at as a thin-walled cross-section. The shear flow through the section must be found.

To analyze the shear flow in S2, the problem is divided into two parts using the superposition principal. The first part is to obtain the shear flow in a cell without stringers. The second part is to analyze a cell with stringers and applying a cut in one of its shear paths. The shear flows from each part is then added together to obtain the shear flow in S2.

The shear flow within each section, between stringers, are not equal to each other but are constant.

Therefore shear flow of S2 is also constant.

To conclude the analysis, solve for the shear flow in the second part and take the moment about a point selected on the plane of the section.

Multi-Cell Sections
Using superposition, we are able to use the following equation when solving for shear flow: qij = q + qijN where,qij, q, and qijN are true shear flow, closed cell constant shear/flow, and open cell piecewise constant shear flow, respectively. This property can be applied to a multi-cell section with or without stringers.

Case 1: Without Stringers

Rz = ΣRiz where Rz is the resultant of the z-axis. This equation explains a very useful property for this case: Riz = 0 so, Rz = 0 This resultant can be expanded to any number of cells.

Case 2: With Stringers

Using the equation from the previous case, we can apply it to this case by                         Rz1 = ΣRiz1 = 0 This is true for this case because when the summation is expanded, we get: Rz = Rz1 + Rz2. Each term is the resultant of P, resultant of P1 and resultant of P2 respectively. For this case, the resultants of P and P2 are equal to each other which causes P1 = 0.

To reinforce the use of this property, let us do an example and calculate the equilibrium for each stringer in P2.

Equilibrium in Stringers
Equilibrium needs to be obtained in all the stingers when calculating shear flow. There are 2 ways to compute the shear flow: one method has been shown earlier, and the second way to compute the shear flow is the consequence of this first method.

We can analyze what happens of we cut the walls and isolate one stringer.

Shear flow in a cell can be analyzed by making a cut so it can be treated as an open section with an existing constant shear flow. We can deduce the following relation from the figure with an isolated stringer due to the cuts:

$$\displaystyle \tilde{q}_{23}=\tilde{q}_{31}=\tilde{q}_{34}=0$$

We can also see from the equilibrium equation that

$$\displaystyle \tilde{q}_{23}=\tilde{q}_{31}=\tilde{q}_{34}+q^{(3)}$$

From the above equations we can see that $$\displaystyle q^{(3)}$$ must be 0, which cannot be true. Therefore we can say that in making cuts no part of the section should be isolated.

Solving P2
The resulting equation for shear flow would be $$\displaystyle 0=q^{(1)}+q^{(2)}+q^{(4)}$$ However, this relationship is not true, and it is not even possible. The following explains why.

$$\displaystyle 0=q^{(1)}+q^{(2)}+q^{(3)}+q^{(4)}$$ Or this relationship can be rewritten as $$\displaystyle 0=\sum_{4}^{e=1}{}q^{(e)}$$

Therefore, $$\displaystyle q^{(1)}+q^{(2)}+q^{(4)}=-q^{(3)}\neq 0$$

We can prove that the sum of all shear flows is zero $$\displaystyle q^{(e)}=n_{z}Q_{z}^{(e)}+n_{y}Q_{y}^{(e)}$$ Where, $$\displaystyle n_{z}:=-(k_{y}V_{y}-k_{yz}V_{z})$$ $$\displaystyle n_{y}:=-(k_{z}V_{z}-k_{yz}V_{y}) $$

Therefore the sum of all shear flows is $$\displaystyle \sum_{4}^{e=1}{}q^{(e)}=n_{z}\sum_{4}^{e=1}{}Q_{z}^{(e)}+n_{y}\sum_{4}^{e=1}{}Q_{y}^{(e)}$$ However, $$\displaystyle \sum_{4}^{e=1}{}Q_{y}^{(e)}=0$$ and $$\displaystyle \sum_{4}^{e=1}{}Q_{z}^{(e)}=0$$

Since Q is the first area moment of inertia, computed respect to the coordinate system which has its origin centered at y-z.

Therefore, $$\displaystyle Q(\hat{z})=\int_{A(\hat{z})}^{}{}zdA$$ And $$\displaystyle Q(\hat{y})=0=z_{c}\int_{A}^{}{}dA$$ The relationship above is due to the mean value theorem.

Shear Flow
In order to plot the buckling shape, we express the coefficients $$\left\{\begin{matrix} C_{22} & C_{13} & C_{31} & C_{33} \end{matrix} \right\}$$ in terms of $$ C_{11} $$. Using an aspect ratio $$\left( \vartheta \right)$$ equal to 1.5 we can break our analysis down to the following four steps.

 * 1. Find $$ \lambda $$.

 * $$ \displaystyle \lambda = \left[ \frac{\vartheta^4}{81 (1 + \vartheta^2)^4} \left\{1 + \frac{81}{625} + \frac{81}{25} \left( \frac{1 + \vartheta^2}{1 + 9 \vartheta^2} \right)^2 + \frac{81}{25} \left( \frac{1 + \vartheta^2}{9 + \vartheta^2} \right)^2 \right\} \right]^{1/2} = 0.0288 $$

 * 2. Evaluate $$ K_{5x5} $$ numerically.

\left[ \begin{array}{lllll} \frac{\lambda (1 + \vartheta^2)^2}{\vartheta^2} &	 \frac{4}{9} &	 0	 &	 0	 &	 0	 \\	 \frac{4}{9} &	 \frac{16 \lambda (1 + \vartheta^2)^2}{\vartheta^2} &	 - \frac{4}{5} &	 - \frac{4}{5} &	 \frac{36}{25} \\	 0	 &	 - \frac{4}{5} &	 \frac{\lambda (1 + 9 \vartheta^2)^2}{\vartheta^2} &	 0	 &	 0	 \\	 0	 &	 - \frac{4}{5} &	 0	 &	 \frac{\lambda (9 + \vartheta^2)^2}{\vartheta^2} &	 0	 \\	 0	 &	 \frac{36}{25} &	 0	 &	 0	 &	 \frac{\lambda (9 + 9 \vartheta^2)^2}{\vartheta^2} \end{array} \right] \left\{ \begin{array}{l} C_{11} \\	 C_{22} \\	 C_{13} \\	 C_{31} \\	 C_{33} \end{array} \right\} =  \left\{ \begin{array}{l} 0	 \\	 0	 \\	 0	 \\	 0	 \\	 0     \end{array} \right\} $$

 * 3. Truncate $$ K_{5x5} $$ into $$ K_{4x4} $$.

\mathbf \bar K _{4 \times 4} = \left[ \begin{array}{llll} \frac{16 \lambda (1 + \vartheta^2)^2}{\vartheta^2} &	 - \frac{4}{5} &	 - \frac{4}{5} &	 \frac{36}{25} \\	 - \frac{4}{5} &	 \frac{\lambda (1 + 9 \vartheta^2)^2}{\vartheta^2} &	 0	 &	 0	 \\	 - \frac{4}{5} &	 0	 &	 \frac{\lambda (9 + \vartheta^2)^2}{\vartheta^2} &	 0	 \\	 \frac{36}{25} &	 0	 &	 0	 &	 \frac{\lambda (9 + 9 \vartheta^2)^2}{\vartheta^2} \end{array} \right] \left\{ \begin{array}{l} C_{22} \\	 C_{13} \\	 C_{31} \\	 C_{33} \end{array} \right\} =  \left\{ \begin{array}{l} \frac{-4 C_{11}}{9} \\	 0	 \\	 0	 \\	 0     \end{array} \right\} $$

 * 4. Solve for $$\left\{\begin{matrix} C_{22} & C_{13} & C_{31} & C_{33} \end{matrix} \right\}$$ in terms of $$ C_{11} $$, noting that $$ K^{-1} $$ is the inverse of $$ K $$.

 * $$\begin{Bmatrix}

C_{22} \\ C_{13} \\ C_{31} \\ C_{33} \end{Bmatrix} = K^{-1} \begin{Bmatrix} -\frac{4}{9} C_{11} \\ 0 \\ 0 \\ 0 \end{Bmatrix}$$

 * $$u_{z} = C_{11} sin(\frac{\pi x}{a}) * sin (\frac{\pi y}{b}) + C_{22} sin(\frac{2 \pi x}{a}) * sin (\frac{2 \pi y}{b}) + C_{31} sin(\frac{\pi x}{a}) * sin (\frac{3 \pi y}{b}) + C_{31} sin(\frac{3\pi x}{a}) * sin (\frac{\pi y}{b})+ C_{33} sin(\frac{3\pi x}{a}) * sin (\frac{3\pi y}{b})$$

Bidirectional Bending Continued
Let us continue to examine the beam with an arbitrarily-shaped cross section. As shown in the figure, we assume that the x-axis coincides with the centroidal axis. We also assume that no torsion is present since the loads (P) pass through the shear center.

Under this bidirectional bending the displacement is a function of x, y and z variables. The displacement equations can be defined as:  $$\displaystyle u=u_{0}(x)+z\psi _{y}(x)+y\psi _{y}(x)$$

 $$\displaystyle v=v_{0}(x)$$

$$\displaystyle w=w_{0}(x)$$

Where $$\displaystyle u_{0}$$ denotes the displacement in the x direction, $$\displaystyle v_{0}$$ denotes the displacement in the y direction and $$\displaystyle w_{0}$$ denotes the displacement in the z direction. The subscript 0 refers to the neutral axis. $$\displaystyle \psi _{y}$$ and $$\displaystyle \psi _{z}$$ are rotations about the cross section y and z axes respectively.

The corresponding strains are
$$\varepsilon _{xx}=\frac{\partial u}{\partial x}=\frac{du_{0}}{dx}+z\frac{d\psi _{y}}{dx}+y\frac{d\psi _{z}}{dx}$$

$$\gamma _{xy}=\frac{\partial v}{\partial x}+\frac{\partial u}{\partial y} =\frac{dv_{0}}{dx}+\psi _{z}$$
 $$\gamma _{xz}=\frac{\partial w}{\partial x}+\frac{\partial u}{\partial z} =\frac{dw_{0}}{dx}+\psi _{y}$$

We can assume that $$\displaystyle \gamma _{xy}=\gamma _{xz}=0$$. This yields to $$\displaystyle \gamma _{z}=-\frac{dv_{0}}{dx}$$ $$\displaystyle \gamma _{y}=-\frac{dw_{0}}{dx}$$

Substituting these results back to the bending strain equation we get

$$\varepsilon _{xx}=\frac{du_{0}}{dx}-y\frac{d^{2}v_{0}}{dx^{2}}-z\frac{d^{2}w_{0}}{dx^{2}}$$

Since, $$\displaystyle \frac{du_{0}}{dx}=0$$ and $$\displaystyle N_{x}=0$$, we can simplify the equation above:

$$\varepsilon _{xx}=-y\frac{d^{2}v_{0}}{dx^{2}}-z\frac{d^{2}w_{0}}{dx^{2}}$$

Therefore, we can define the bending moments about the y and z axes
 $$M_{y}=\int_{A}^{}{}\int z\sigma _{xx}dA= -E\int_{A}^{}{}\int (yz\frac{dv_{0}}{dx^{2}}+z^{2}\frac{dw_{0}}{dx^{2}})dA=-EI_{yz}\frac{d^{2}v_{0}}{dx^{2}}-EI_{y}\frac{d^{2}w_{0}}{dx^{2}}$$

$$M_{z}=\int_{A}^{}{}\int y\sigma _{xx}dA= =-EI_{z}\frac{d^{2}v_{0}}{dx^{2}}-EI_{yz}\frac{d^{2}w_{0}}{dx^{2}}$$

Where the inertia moments are defined such The moment of inertia about the y-axis $$\displaystyle I_{y}=\int_{A}^{}{}\int z^{2}dA$$ The moment of inertia about the z-axis $$ \displaystyle I_{z}=\int_{A}^{}{}\int y^{2}dA$$ And the product of inertia is $$\displaystyle I_{yz}=\int_{A}^{}{}\int yzdA$$

The moment equations above can be expressed in matrix form using the following notations for the curvatures: $$ \displaystyle X_{y}={d^{2}v_{0}}{dx^{2}}$$ and $$\displaystyle X_{z}={d^{2}w_{0}}{dx^{2}}$$

$$\begin{Bmatrix} M_{y}\\ M_{z} \end{Bmatrix} =\begin{vmatrix} I_{y} & I_{yz}\\ I_{yz}& I_{z} \end{vmatrix}\begin{Bmatrix} -EX_{z}\\ -EX_{y} \end{Bmatrix}$$

Also, strain can be expressed in a similar manner using the curvature relations above.

$$\varepsilon _{xx}=-yX_{y_{}}-zX_{z}=\begin{bmatrix} z & y \end{bmatrix}\begin{Bmatrix} -X_{z}\\ -X_{y}

\end{Bmatrix}$$

 $$\sigma _{xx}=E\varepsilon _{xx}=E\begin{bmatrix} z & y \end{bmatrix}\begin{Bmatrix} -X_{z}\\ -X_{y} \end{Bmatrix}=E\begin{bmatrix} z & y \end{bmatrix}\frac{1}{E}I^{-1}\begin{Bmatrix} M_{y}\\ M_{z} \end{Bmatrix}$$

The inverse of the 2 by 2 matrix is

$$I^{-1}=\frac{1}{D}\begin{vmatrix} I_{z} & -I_{yz}\\ -I_{yz} & I_{y} \end{vmatrix}

$$

And the determinant D is defined as $$\displaystyle D=I_{y}I_{z}-(I_{yz})^{2}$$

We can directly derive the martix form of the shear flow. We know that the general expression for shear flow is
$$q=-\int_{A}^{}{}\frac{d\sigma _{xx}}{dx}dA$$ We also need a relationship between the sear and the moment
$$\begin{matrix} d\sigma_{xx}\\ dx \end{matrix} = \begin{bmatrix} z & y \end{bmatrix} I^{-1}\begin{Bmatrix}

\frac{dM_{y}}{dx}\\ \frac{dM_{z}}{dx} \end{Bmatrix}$$

From here we can conclude that

$$V_{y}=\frac{dM_{y}}{dx}, V_{z}=\frac{dM_{z}}{dx}$$ and

$$Q_{y}=\int_{A}^{}{}zdA, Q_{z}=\int_{A}^{}{}ydA$$

Matlab HW7: Determining the True Shear Flow in Different-Celled NACA Airfoils
The problem statement in this homework was to determine the true shear flows of two different cases of the NACA airfoil subject to shear forces. Part 1 was to determine these shear flows in a single-cell airfoil and part 2 was to determine these shear flows in a double-cell airfoil. The airfoils in both cases have four stringers: 2 positioned at a quarter of the chord length and 2 positioned at three quarters of the chord length. In each case, the problem was solved using superposition of the shear flows in a case analyzed without stringers and a case with stringers. The equation of moment and the compatibility equation were also used to find the true shear flows. The code in which was developed to solve this problem is shown in the collapsible boxes below.

For part 1, the code starts out with finding the values ky, kz, and kyz by using the moment of inertias obtained by previous homework assignments. Then, it goes about solving the P2 problem as posed in lecture. The trailing edge is the portion of the skin that is designated to be cut. Here, Qy and Qz are calculated at each stringer node and the equilibrium equations of the shear flows at each stringer are written. This solves for all the q tilda's in P2. After these shear flows are obtained, it is necessary to find a the shear flow of the two cells (q_11 and q_21) solving for the P1 problem as posed in lecture. The indices in the notation are described as follows: the first subscript is the cell number, and the second subscript is the part of the homework the shear flow corresponds to. To do this, the moment equation is used and the moment was taken about point D. To find the moments contributed by the shear flows found in P2, the code was written to sweep from stringer to stringer that had a non-zero shear flow. At each sweep, an infinitesimal length of the skin was calculated and multiplied by the shear flow through that skin portion to obtain a force. Once that force was determined, the perpendicular distance from point D to the line of action of the force was obtained and then multiplied to the force to get an infinitesimally small moment. All the moments were added together to get the moment produced by each shear flow from P2. Along with the compatibility equation, q_11 and q_21 are obtained. Back to superposition, the shear flows of P1 and P2 are added together to get the true shear flows.

For part 2, the code uses the same moments as part 1 because nothing has changed. However, the change is in the moment equation. The moment equation now has three unknowns (q_12 and q_22 and q_32) so it alone is not sufficient in solving for the P1 shear flows. This is where the compatibility equation comes in. We know that the twist angle of the first cell is equal to the twist angle of the second and third cells. The moment and compatibility equations are arranged to that it has coefficients multiplied to q_12 and q_22 and q_32. Then, the equations were arranged in matrix form and solved for in Matlab. Then, superposition was used to find the true shear flow in the three-cell case.

Next we analyzed buckling from shear flow. The critical buckling stress is analyzed two different ways. First we approximate the critical buckling by truncating the equation below to two variables, $$C_{11}$$ and $$C_{22}$$.

$$  \displaystyle \left[ \begin{array}{lllll} \frac{\lambda (1 + \vartheta^2)^2}{\vartheta^2} &	 \frac{4}{9} &	 0	 &	 0	 &	 0	 \\	 \frac{4}{9} &	 \frac{16 \lambda (1 + \vartheta^2)^2}{\vartheta^2} &	 - \frac{4}{5} &	 - \frac{4}{5} &	 \frac{36}{25} \\	 0	 &	 - \frac{4}{5} &	 \frac{\lambda (1 + 9 \vartheta^2)^2}{\vartheta^2} &	 0	 &	 0	 \\	 0	 &	 - \frac{4}{5} &	 0	 &	 \frac{\lambda (9 + \vartheta^2)^2}{\vartheta^2} &	 0	 \\	 0	 &	 \frac{36}{25} &	 0	 &	 0	 &	 \frac{\lambda (9 + 9 \vartheta^2)^2}{\vartheta^2} \end{array} \right] \left\{ \begin{array}{l} C_{11} \\	 C_{22} \\	 C_{13} \\	 C_{31} \\	 C_{33} \end{array} \right\} =  \left\{ \begin{array}{l} 0	 \\	 0	 \\	 0	 \\	 0	 \\	 0     \end{array} \right\} $$

becomes,

$$  \displaystyle \left[ \begin{array}{ll} \frac{\lambda (1 + \vartheta^2)^2}{\vartheta^2} &	 \frac{4}{9} \\	 \frac{4}{9} &	 \frac{16 \lambda (1 + \vartheta^2)^2}{\vartheta^2} \end{array} \right] \left\{ \begin{array}{l} C_{11} \\	 C_{22} \end{array} \right\} =  \left\{ \begin{array}{l} 0	 \\	 0     \end{array} \right\} $$

Since we are truncating the original equation, we lose some accuracy. By keeping the original 5x5 matrix and solving for the eigenvalues, we obtain the second equation for critical buckling stress. Therefore, the two equations for critical buckling stress become:

$$  \displaystyle (\sigma_{xy})_{cr}^{\star,2} =  \pm k_c^{\star,2} \frac{\pi^2 D}{b^2 h}  \, \ k_c^{\star,2} =  \frac{\pi^2}{32} \frac{9 (1 + \vartheta^2)^2}{\vartheta^3} $$

And

$$  \displaystyle (\sigma_{xy})_{cr}^{\star,5} =  \pm k_c^{\star,5} \frac{\pi^2 D}{b^2 h}  \, \ k_c^{\star,5} =  \frac{\pi^2}{32} \frac{1}{\lambda \vartheta} $$

By plotting the two equations with the parameters from the airfoil panel FB, we can see the error that occurs by truncating. It also becomes evident that the error increases at larger values of $$a/b$$.

If we add the plot for a clamped boundary condition we see that the critical buckling stress increases as expected. From the plot it is evident that a value of $$K=4.1$$ is not possible; hence, this value was an error in the experiment. A larger plot is shown below.

Relating the work above to the NACA airfoil, we can calculate the shear stress through the panel BF with the calculated shear flow.

$$ N_{xy}^{} = \sigma_{xy} h $$

$$ \sigma_{xy} =N_{xy}^{} / h $$

Since the thickness, $$h$$, of the panel is given as 2 mm, the shear stress is calculated as follows:

$$ \sigma_{xy} = \frac{5 623.7 (N/m)} {0.002 (m)} = 2,811,850 (Pa)$$

For large values of a/b, this value approaches the critical buckling predicted above. Since this is an airfoil, it is safe to assume $$a/b$$ will be much smaller than one. For this assumption, the stresses on the airfoil will not induce buckling of the panels.

E-learning versus MediaWiki
From previous experience using E-learning, once the server gets busy, the entire system is slow. Using Wikiversity and MediaWiki, the traffic of the server never seems to overwhelm the system and cause slow loading and sluggish performance. This increases the accessibility and reliability of MediaWiki. E-learning's information is only available for a semester and then whatever is posted is taken off. On Wikiversity, the information available on it is essentially ever-lasting so long as the user and Wikiversity continues to exist. Using E-learning, the user interface of the system was more cluttered and unorganized than Wikiversity. I believe that Wikiversity has a more user-friendly system. Learning to use MediaWiki was not difficult although typing in the mathematical formulas were very tedious. As for listing MediaWiki experience on a resume, this is a very minor thing. The MediaWiki software is not difficult to use. MATLAB experience, C++ programming experience or web page design experience are far better things to have and list on a resume. Also, a positive aspect of the wiki system is that anyone can edit the information provided by the wiki pages. This allows for universal editing, and accessing. MediaWiki allows for text, images, equations, video, and audio to be displayed on one page all at once, available for immediate use. E-Learning presents the media in file or link form. Text, equations and some images are commonly presented as Word documents on E-Learning. Video, audio and some images are presented as separate files, or as links to those files. Both E-Learning and MediaWiki can present any media type; they just differ on how it is presented. E-Learning has its advantages by presenting the material as files. Printing documents is simpler when they are already in document form, as opposed to trying to print a webpage. Also, if any of the video or audio files were wanted to be on an mp3 player or some other portable device, they can be moved quickly and simply with E-Learning’s file format.

Recommended Software
One helpful Firefox plugin used to edit Wiki articles is wikEd. It features a rich-text WYSIWYG editor with many powerful tools.

Other helpful links included an HTML-to-Wiki converter and an Excel-to-Wiki converter.

For many of the images the open source software, Inkscape, and the vector-based Adobe Illustrator (can be accessed in most UF Computer Labs) were used. And for the equations the latex equation editor proved very helpful.

Contributing Members
Jin Yu Guan Eas4200c.f08.aeris.guan 23:28, 8 December 2008 (UTC)

Delvis Stoute Eas4200c.f08.spars.stoute 06:25, 9 December 2008 (UTC)

Nelson Caceres Eas4200c.f08.aeris.caceres 07:59, 9 December 2008 (UTC)

Josh Holladay Eas4200c.f08.aeris.holladay 16:24, 9 December 2008 (UTC)

Anett Krammer Eas4200c.f08.aeris.krammer 18:27, 9 December 2008 (UTC)

Jesse McRae Eas4200c.f08.aeris.mcrae 02:50, 10 December 2008 (UTC)

title: User:Eas4200c.f08.aeris.krammer/HW7


text:
User:Carabera

For my userpage, see User:Carabera on Wikipedia.

title: User:Carabera


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User:Jmasaki

User: jmasaki
I am a current medical student with a background in biology, microbiology, and immunology. Masters degree: Microbiology Bachelors degree: Biology Other academic specialties: Cancer research, computer science, and Japanese language & culture.

title: User:Jmasaki


text: File:TA 1 and 4.jpg

title: File:TA 1 and 4.jpg


text: File:TA 2 and 5.JPG

title: File:TA 2 and 5.JPG


text: File:TA 3 and 6.JPG

title: File:TA 3 and 6.JPG


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User:Eas4200c.f08.aero.e/HW7

=Week 13 - 11/24 through 11/28=

Solving Problem P2:
Equation of each stringer (Euler cut principle)

Stringer 3

$$ \sum F_x = 0 = \int_A [ \sigma_{xx}(x + dx) - \sigma_xx(x)]dA_3 + [-\tilde{q}_{23} - \tilde{q}_{43} + \tilde{q}_{31}] dx $$

$$ q^{3} = [ \sigma_{xx}(x + dx) - \sigma_xx(x)] $$

So

$$ \tilde{q}_{31} = \tilde{q}_{23} + \tilde{q}_{43} - q^{(3)} $$

The contribution to shear flow by stringer 3 is

$$ q^{(3)} = -\int_{A3} \frac{d\sigma_{xx}}{dx} dA_3 $$

Recall:

$$ V_y = \frac{dM_z}{dx} \;\; V_z = \frac{dM_x}{dx} $$

$$ q^{(3)} = -(K_yV_y - K_{yz}V_z)Q_z^{(3)} - (K_zV_z - K_{yz}V_y)Q_y^{(3)} $$

The equation above will be called Equation 1.

Stringer 2

$$ \tilde{q}_{23} = \tilde{q}_{12} - \tilde{q}_{24} + q^{(2)} $$

Where $$ \tilde{q}_{12} = \tilde{q}_{24} = 0 $$ because of the cuts.

Use Equation 1 to solve for $$\tilde{q}_{23} = q^{(2)} $$.

Stringer 4

$$ \tilde{q}_{43} = \tilde{q}_{24} - \tilde{q}_{41} + q^{(4)} $$

Where $$ \tilde{q}_{24} = \tilde{q}_{41} = 0 $$ because of the cuts.

Use Equation 1 to solve for $$\tilde{q}_{43} = q^{(4)} $$.

Superposition (again)

$$ q_{ij} = \tilde{q}_{ij} + q_k $$

This leads to

$$ q_{12} = \tilde{q}_{12} + q_1 $$

$$ q_{23} = \tilde{q}_{23} + q_1 - q_2 $$

$$ q_{31} = \tilde{q}_{31} + q_1 - q_3 $$

$$ q_{24} = \tilde{q}_{24} + q_2 $$

$$ q_{43} = \tilde{q}_{43} + q_3 - q_2 $$

$$ q_{41} = \tilde{q}_{41} + q_4 $$

The $$\tilde{q}$$s are known, and the qs are not. Witch leaves 3 unknowns; q1, q2, andq3. So 3 equations are needed. These are

1) Moment equation: Take the moment of Vy, Vz, and {q12,...., q41} about any convenient point (usually where lines of action Vy and Vz intersect).

2) Compatibility equation: $$ \theta_1 = \theta_2 $$

3) Compatibility equation: $$ \theta_2 = \theta_3 $$

--EAS4200C.F08.AERO.CLK04D 17:28, 9 December 2008 (UTC)

Wednesday
Thanksgiving Break

Friday
Thanksgiving Break

=Week 14 - 12/1 through 12/5=

Monday
As a continuation to problem (P) from last lecture

*Follow path "$$S_{i}$$"

*Equilibrium of each stringer on path $$S_{i}$$

There are two methods to solve the problem.

1) Complete method. (Free body diagram) see lecture 38-1

$$q_{j6}=q_{2j}-q_{j5}-q_{j8}+q^{(3)}$$

2) Method is a consequence of 1st method.

$$-q_{6j}=q_{2j}-q_{j5}-q_{j8}+q^{(3)}$$

--Eas4200c.f08.aero.arena 20:48, 9 December 2008 (UTC)

Wednesday
Exam 3

Friday
The following concepts will be applied to the NACA Airfoil and can be found in the Matlab section of this homework report.

Plotting buckling shape under shear expresion $$\left\{C_{22}, C_{13}, C_{31}, C_{33} \right\}$$ in terms of:

$$C_{11}$$ and for $$V=1.5$$

The following are the steps to follow:

 * 1) 1 Find λ for $$V=1.5$$

 * 1) 2 Evaluate num $$K\ddot{}_{5x5}$$

 * 1) 3 Remember that $$K\ddot{} = \left[K_{ij} \right]$$

Then we have the following matrix:

$$\begin{vmatrix} K_{22} & K_{23}&  K_{24}&  K_{25}\\ K_{32} & K_{33}&  K_{34}&  K_{35}\\ K_{42} & K_{43}&  K_{44}&  K_{45}\\ K_{52} & K_{53}&  K_{54}&  K_{55}\\ \end{vmatrix}\begin{Bmatrix} C_{22}\\ C_{13}\\ C_{31}\\ C_{33} \end{Bmatrix} = \begin{Bmatrix} -\frac{4}{9}C_{11}\\ 0\\ 0\\ 0\\

\end{Bmatrix} $$

So, we solve for: $$\left\{C_{22}, C_{13}, C_{31}, C_{33} \right\}$$ in terms of $$C_{11}$$

Now, we know that: $$K\ddot{}^{-1}$$ is the inverse of $$K\ddot{}$$

Then;

$$\begin{Bmatrix} C_{22}\\ C_{13}\\ C_{31}\\ C_{33} \end{Bmatrix} = K\ddot{}^{-1}\begin{Bmatrix} \frac{-4}{9}C_{11}\\ 0\\ 0\\ 0 \end{Bmatrix}$$

Now, we see that:

$$u_{z}=C_{11}\sin \left(\frac{\pi x}{a} \right)\sin \left(\frac{\pi y}{b} \right)+C_{22}\sin \left(\frac{2\pi x}{a} \right)\sin \left(\frac{2\pi y}{b} \right)+C_{13}\sin \left(\frac{\pi x}{a} \right)\sin \left(\frac{3\pi y}{b} \right)+C_{31}\sin \left(\frac{3\pi x}{a} \right)\sin \left(\frac{\pi y}{b} \right)+C_{33}\sin \left(\frac{3\pi x}{a} \right)\sin \left(\frac{3\pi y}{b} \right)$$

Finally, we set $$C_{11}=1$$ and plot $$u_{z}\left(x, y \right)$$

-- Now, we proceed to continue the lecture of 12/01/08.

So we answer the problem on meeting 39-2 (end of 12/01/08) in two parts:

Part1: Equilibrium of isolated stringer (Already done)

Part2: Closed cell, equilibrium of stringer.

Stringer 1:

$$q\tilde{}_{12} = q\tilde{}_{31}+q\tilde{}_{41}+q^{(1)}$$

Where $$q\tilde{}_{12} $$ and $$ q\tilde{}_{41} $$ are unknowns and $$ q\tilde{}_{31} = 0$$

Stringer 2:

$$q\tilde{}_{24} = q\tilde{}_{12}-q\tilde{}_{23}+q^{(2)}$$

Where $$ q\tilde{}_{23} = 0$$

Stringer 4:

$$q\tilde{}_{41} = q\tilde{}_{24}+q\tilde{}_{34}+q^{(4)}$$

Where $$q\tilde{}_{34}$$ and $$q\tilde{}_{24}= 0$$

Then, we have that:

$$q\tilde{}_{41} = \left(q\tilde{}_{41}+q^{(1)} +q^{(2)}+q^{(4)}\right)$$

The $$q\tilde{}_{41}$$'s cancel out and we finally obtain:

$$0=q^{(1)}+q^{(2)}+q^{4}=-q^{(3)}$$

(This problem will be continued on 12/08/08)

--Eas4200c.f08.aero.gabo28 07:31, 9 December 2008 (UTC)

=Week 15 - 12/8=

Monday
12/5 continued (page 40-3)

$$ 0 = q^{(1)} + q^{(2)} + q^{(4)} $$

"Not Possible"

What we really have is,

$$ 0 = \sum_{e=1}^4 q^{(e)} \;\;\; \Rightarrow \;\;\; q^{(1)} + q^{(2)} + q^{(4)} = -q^{(3)} $$

Where,

$$ q^{(e)} = n_zQ^{(e)}_z + n_yQ^{(e)}_y $$

$$ n_z = -(K_yV_y - K_{yz}V_z) $$

$$ n_y = -(K_zV_z - K_{yz}V_y) $$

$$ \sum_{e=1}^4q^{(e)} = n_z \sum_{e=1}^4Q^{(e)}_z + n_y \sum_{e=1}^4Q^{(e)}_y $$

$$ \sum_{e=1}^4Q^{(e)}_z = 0 $$

$$ \sum_{e=1}^4Q^{(e)}_y = 0 $$

If we have,

$$ Q_z(\hat{z}) = \int_{A(\hat{z})} z dA $$

If $$A(\hat{z}) = A$$

$$ Q_y = \int_A z dA = 0 = z_c \int_A dA $$

This is true because, for A $$ \quad z_c = 0$$.

From Sect. 4.2, we have the approximate displacement expansions given by

$$ u = u_0(x) + z\psi_y(x) + y\psi_z(x) $$

$$ v = v_0(x) $$

$$ w = w_0(x) $$

Where $$\psi_y$$ and $$\psi_z$$ are rotations of the cross-section about the y and z axes.

The corresponding strains are

$$ \varepsilon_{xx} = \frac{\delta u}{\delta x} = \frac{du_0}{dx} + z \frac{d\psi_y}{dx} + y \frac{d\psi_z}{dx} $$

$$ \gamma_{xy} = \frac{\delta v}{\delta x} + \frac{\delta u}{\delta y} = \frac{dv_0}{dx} + \psi_z $$

$$ \gamma_{xz} = \frac{\delta w}{\delta x} + \frac{\delta u}{\delta z} = \frac{dw_0}{dx} + \psi_y $$

Simplifying with the assumption $$ \psi_{xy} = \psi_{xz} = 0 $$ yields the relations

$$ \psi_z = -\frac{dv_0}{dx} $$

$$ \psi_y = -\frac{dw_0}{dx} $$

Subbing this in to the bending strain and using the argument that $$ \frac{du_0}{dx} = 0 $$ if $$N_x = 0$$ yields

$$ \varepsilon_{xx} = -y \frac{d^2v_0}{dx^2} - z \frac{d^2w_0}{dx^2} $$

The bending moments about the y and z axes, respectively, are defined as

$$ M_y = \iint_{A} z \sigma_{xx} dA = -E \iint_A \left( yz \frac{d^2v_0}{dx^2} + z^2 \frac{d^2w_0}{dx^2} \right) $$

$$ = -EI_{yz} \frac{d^2v_0}{dx^2} - EI_y \frac{d^2w_0}{dx^2} $$

$$ M_z = \iint_{A} y \sigma_{xx} dA= -EI_{z} \frac{d^2v_0}{dx^2} - EI_{yz} \frac{d^2w_0}{dx^2} $$

Where

$$ I_y = \iint_A z^2 dA $$ moment of inertia about y-axis

$$ I_z = \iint_A y^2 dA $$ moment of inertia about z-axis

$$ I_{yz} = \iint_A yz dA $$ product of inertia

And

$$ \frac{d^2v_0}{dx^2} = X_y $$

$$ \frac{d^2w_0}{dx^2} = X_z $$

Where Xy and Xz are the curvatures.

In matrix form this can be written

$$ \begin{bmatrix} M_y \\ M_z \end{bmatrix} = -E \begin{bmatrix} I_y & I_{yz} \\ I_{yz} & I_z \end{bmatrix} \begin{bmatrix} X_z \\ X_y \end{bmatrix} $$

And

$$ \varepsilon_{xx} = -yX_y - zX_z = \begin{bmatrix} z & y \end{bmatrix} \begin{bmatrix} -X_z \\ -X_y \end{bmatrix} $$

So plugging into the stress equation and solving for the matrix $$ \begin{bmatrix} -X_z \\ -X_y \end{bmatrix} $$ we get

$$ \sigma_{xx} = E \varepsilon_{xx} = E \begin{bmatrix} z & y \end{bmatrix} \begin{bmatrix} -X_z \\ -X_y \end{bmatrix} = E \begin{bmatrix} z & y \end{bmatrix} \left( \frac{1}{E} \underline{I}^{-1} \begin{bmatrix} M_y \\ M_z \end{bmatrix} \right ) $$

Where

$$ \underline{I}^{-1} = \frac{1}{D} \begin{bmatrix} I_z & -I_{yz} \\ -I_{yz} & I_y \end{bmatrix} $$

$$ D = I_yI_z + (I_{yz})^2 $$

Shear flow

$$ q = -\int_A \frac{d \sigma_{xx}}{dx}dA $$

$$ \frac{d \sigma_{xx}}{dx} = \begin{bmatrix} z & y \end{bmatrix} \underline{I}^{-1} \begin{bmatrix} \frac{dM_y}{dx} \\ \frac{dM_z}{dx} \end{bmatrix} $$

Where

$$ V_y = \frac{dM_y}{dx} $$

$$ V_z = \frac{dM_z}{dx} $$

And

$$ Q_y = \int_A zdA $$

$$ Q_z = \int_A ydA $$

This leads directly to the shear flow in matrix form.

--Eas4200c.f08.aero.E 01:38, 9 December 2008 (UTC)

=Matlab= Results from code: (using 2000 segments) q_along_trailing_edge = -9.9964e+005 q_from_F_to_B = -2.8788e+006 q_from_B_to_E = -2.3337e+006 q_from_E_to_H = -1.8031e+006

--Eas4200c.f08.aero.r 19:45, 9 December 2008 (UTC)

=Mediawiki vs. E-leaning=

EAS4200C.Fall08.AERO.Watlington.VG 20:29, 9 December 2008 (UTC)

title: User:Eas4200c.f08.aero.e/HW7


text: File:TA Vector u.JPG

title: File:TA Vector u.JPG


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User:Eml4500.f08.Ateam.philbin/HW7

Beam Shape Functions
The beam shape functions for the six forces is as follows.

Note the similarities between the corresponding horizontal and vertical forces as well as the moments. The vertical sides of the first four shapes have a magnitude of 1. The initial slope of mode 3 and the final slope of mode 6 both have a magnitude of 1. The final displacement vector can be written in the following form.

$$v(\tilde{x}) = N_2(\tilde{x})\hat{d}_2 + N_3(\tilde{x})\hat{d}_3 + N_5(\tilde{x})\hat{d}_5 + N_6(\tilde{x})\hat{d}_6$$

$$u(\tilde{x}) = N_2(\tilde{x})\hat{d}_1 + N_2(\tilde{x})\hat{d}_4 $$

In order to compute these components, the relationship developed in a previous lecture will be used.

$$ \tilde{d}^{(e)}=\tilde{T}^{(e)}\ \tilde{d}^{(e)}$$

This relationship is used to reorient the axial coordinate system to the global coordinate system.

Computing the displacement vector
The displacement vector can be written in terms of its components, both in the axial and global coordinate systems.

$$\mathbf{u}(\tilde{x})=u(\tilde{x})\vec{\tilde{i}}+v(\tilde{x})\vec{\tilde{j}}\quad (axial\ coordinates)$$

$$\mathbf{u}(\tilde{x})=u_x(\tilde{x})\vec{i}+v_y(\tilde{x})\vec{j}\quad (global\ coordinates) $$

In order to find $$u_x(\tilde{x})\ and\ v_y(\tilde{x})$$ a transformation matrix $$R^T$$ will be used.

$$\begin{Bmatrix} u_x(\tilde{x})\\v_y(\tilde{x})\end{Bmatrix} = R^T \begin{Bmatrix} u(\tilde{x}) \\ v(\tilde{x})  \end{Bmatrix} $$

Now the equation for $$\begin{Bmatrix} u(\tilde{x}) \\ v(\tilde{x})  \end{Bmatrix}$$ must be found. It is related to the displacements and the equation derived from the mode shapes.

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

Using these equations, they can be rewritten to give the final form of the global displacement vector.

$$\begin{Bmatrix} u_x(\tilde{x})\\v_y(\tilde{x})\end{Bmatrix} = R^T \begin{bmatrix} N_1 &0&0&N_4&0&0\\0&N_2&N_3&0&N_5&N_6\end{bmatrix} \tilde{T}^{(e)} d^{(e)}$$

Team View of Mediawiki
Mediawiki is an interesting medium for our class to use to complete our homework. Its learning curve is hard at first but once the proper syntax is learned, very professional work can be created. The math function in particular can create very precise and systematic functions that are easy to understand. Since all members of the group are using the same editors, a cohesive presentation can be achieved throughout each homework assignment. Since there are multiple groups working on each assignment, once turned in, each student has access to many different views and approaches to each problem.

Suggested software for Mediawiki
Most of the aspects of each homework assignment are similar and flow throughout the assignment. However, it was noticed that the programs that each student used for creating images were not the same. Therefore, some images were more plain that the others. In order to create a more uniform look, each group should choose an image software program at the beginning of the class and stick with it throughout the course.

Besides the Paint program that comes on every Windows PC, there are several programs available online that have enhanced features. Some such as Paint.NET and e-Paint are available for free and offers more advanced tools that the basic paint program. Another program that can easily be found online is GIMP (GNU Image Manipulation Program) which is more designed to modify images. Inkscape is a vector based graphics editor which can create Scalable Vector Graphics (SVG) file format. This is different that the usual scalar approach and allows the size of images to be easily changed with little loss of quality.

Anyone of these programs would be effective at creating the images that were needed for this class. In future courses, if a group were to decide on one, it would increase the overall continuity between individual contributions.

12/08/08
$$\tilde{k}_{23}=\frac{6EI}{L^2}= \int_{0}^{L} \frac{d^2N_2}{dx^2}(EI)\frac{d^2N_3}{dx^2}\, dx$$

Or to write this in a more generalized form:

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

Discrete Principal of Virtual Work
$$\bar{w}\cdot[\bar{M}\ddot{\bar{d}}+\bar{k}\bar{d}-\bar{F}]=0 \quad\ for\ all\ \bar{w}$$

The bar notation denotes that the boundary conditions have already been applied. The ordinary differential equations (second order in time) and the initial conditions governing the elastodynamics of the discrete continuity problem are as follows.

$$\bar{M}\ddot{\bar{d}}+\bar{k}\bar{d}=\bar{F}(t)\quad\ (1)$$

$$\bar{d}(0)=\bar{d}_0\quad \quad (initial\ position)$$

$$\dot{\bar{d}}(o)=\bar{v}_0 \quad \quad (initial\ velocity)$$

Eigenvalue Solution
1) Consider the unforced vibration problem:

$$\bar{M}\ddot{v}+ \bar{k} v = 0 \quad \ (2)$$

Assume $$v(t)=(sin {wt})\mathbf{\phi}$$. Here $$\phi$$ is a column matrix of size (nx1) that is not time dependent.

$$\ddot{v}=-w^2sin {wt}\phi$$

Plug the previous equation back into equation (1):

$$-w^2 sin{wt} \bar{M}\phi + sin {wt} \bar{k} \phi = 0 $$

This can be simplified to the generalized eigenvalue problem:

$$\bar{k}\phi = w^2 \bar{M} \phi$$

An even more generalized version can be written in terms of matrices A and B.

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

Once evaluated, the following relations are made.

$$\mathbf{\lambda} = w^2$$

$$Mode\ i=\begin{cases} V_i(t)=(sin {w_it})\phi_i \\ i=1,...,n\end{cases}$$

2) Model superposition method:

Orthogonal property of eigenvalues is shown in the following equation:

$$\phi_{i} \bar{M} \phi_{j} = \delta _{i j} = \begin{cases} 1, & \mbox{if }\ {i=j} \\ 0, & \mbox{if }\ {i \ne j} \end{cases}$$

Here $$ \mathbf{\delta}_{ij}$$ is the Kronecker delta which has a variety of uses including digital processing.

$$\bar{M}\phi_j=\lambda^{-1}_j\phi^T_i\bar{k}\phi_j$$

$$\phi_i^T\bar{M}\phi_j=\delta_{ij}\lambda^{-1}_j\phi^T_i\bar{k}\phi_j$$

$$\phi^T_i\bar{k}\phi_j=\lambda_j \delta_{ij}$$

Next look at:

$$\bar{M}\ddot{\bar{d}}+\bar{k}\bar{d}=\bar{F}\quad \ (3)$$

$$\bar{d}(0)=\bar{d}_0$$

$$\dot{\bar{d}}(0)=\bar{v}_0$$

then

$$\bar{d}(t)=\sum_{i=1}^n \xi_i(t)\phi_i $$

Now plug the previous equation into equation (3):

$$\bar{M}(\sum_j \xi_j\phi_j)+\bar{k}(\sum_j\xi_j\phi_j)=\bar{F}$$

where

$$\ddot{\bar{d}}=\sum_j \xi_j\mathbf{\phi_j}\quad and\ \bar{d}=sum_j\xi_j\mathbf{\phi_j}$$

Finally,

$$\sum_j \ddot{\xi}_j(\mathbf{\phi_i^T}\mathbf{\bar{M}}\mathbf{\phi_j})+ \sum_j \ddot{\xi}_j(\mathbf{\phi_i^T}\mathbf{\bar{k}}\mathbf{\phi_j})=\mathbf{\phi^T_i} \mathbf{F}$$

$$\ddot{\xi}_i+\lambda_i\xi_i=\phi^T_i\mathbf{F}$$

title: User:Eml4500.f08.Ateam.philbin/HW7


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User talk:Lurkingfate

Welcome
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