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)