User:Eas4200c.f08.blue.a/Homework7

Mean Value Theorem and Stringers
Previously, we were discussing the problem with a set of 4 stringers with equal sizes versus a set of 4 stringers with varying sizes.



This leads to the discussion on the Mean Value Theorem. The Mean Value Theorem states that:



$$\int zdA = \bar z A$$

(neglect skin and sparwebs)



Since:
 * V2
 * KY2, K2
 * Q2, Qy

are all independent of 's' (the path), then shear flow will then be independent of q(s)

When crossing a stringer:


 * 1) Find (Yc, Xc)
 * 2) Find Kx, Kz, Kyz
 * 3) Find Iz, Iy, Izy
 * 4) Follow Path (s) to find:

q12, q23, q34

Where q12 is the shear flow in panel 12



where: (y1 is the y coordinate of stringer 1)

where:

Notice:

Single and Multi Cell Shear Flow
A plan for attacking the problem of solving for the shear flow in a multi-cell section with stringers was introduced.

This miniPLAN is shown below. S) Singe-cell sections                   S.1)  without stringers S.2) with stringers               M)  Multi-cell sections M.1) without stringers                   M.2)  with stringers

First, a single-cell section with no stringers(as in Figure 1) is analyzed to find the shear flow q through the entire section (S.1 from the miniPLAN). The section can be viewed as a thin walled cross section. As is shown below, a thin walled cross section cannot resist transverse shear (Vz).



Figure 1: Shear flow (q) in a single-cell section

Next, a single-cell section with stringers is analyzed as is shown in. The solution to this problem (P) can be analyzed by decomposing the problem into two parts: A section with stringers must be considered without stringers (as in problem P1 in Figure 3) and an open cross section with stringers (as in problem P2 in Figure 4).



Figure 2: Single-cell section with stringers



Figure 3: Single-cell section without stringers



Figure 4: Open single-cell section with stringers

P=P1+P2

It is known that the sum of shear forces in P is not equal to zero and that the shear flows between stringers in P are not equal but the shear flow is constant in the within each parcel. Therefore, the presence of stringers results in a non-constant shear flow.

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

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

$$q_{ij}(s)=const$$

Problem P1 is solved just as in S.1 of the miniPLAN and the resulting shear flows between stringers can be solved with the superposition of P1 and P2.

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

The method for solving the problem consists of the following steps:

1) Solve P2 for $$\tilde{q}_{12},\tilde{q}_{23},\tilde{q}_{31} $$

2) Take the moment about any point in the plane of the section

2.1) Superposition $$q_{ij}=q+\tilde{q}_{ij} $$

2.2) Selection point in plane and find moment about that point

3) Back to superposition

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

Multi Cell Shear Flow
M. Multicell

First, consider a case without stringers



For n cells we have:

$$R^{\bar{z}}+\sum_{i=1}^{n cells}{R^{\bar{z}}_{i}}$$

and

$$R^{\bar{z}}=0$$

The probelm can be broken down using superposition. the first part of the problem consists of computing the shear flow of the web. The second problem consists of computing the shear in the stringers. The cuts to the walls have been made in read int eh figure below.



The shear flow across all of the cut sections can be assumed to be zero. Once the solutions to the two parts of the problem have been solved, they can simply be added together like so:

$$R^{\bar{z}}=R^{z_{1}}+R^{z_{2}}$$

Note that $$R^{z_{1}}=0$$

and thus,

$$R^{\bar{z}}=R^{z_{2}}=V_{z}$$

Continuation of Mulit Cell Shear Flow and Euler Cut Principle
Continuation of problem using Euler cut principle Stringer 3: Refer to the image below



$$\sum{F_{x}}=0$$

$$\int_{A_{3}}^{}{\left[\sigma_{xx}(x+dx)-\sigma_{xx}(x) \right]dA_{3}}+\left[\tilde{-q_{23}}-\tilde{q_{43}}+\tilde{q_{31}} \right]dx$$

Taylor series: $$\frac{d\sigma_{xx}dx}{dx} + h.o.t.$$

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

$$q^{3}\Rightarrow$$ stringer 3 contribution $$-\int_{A_{3}}^{}{\frac{d\sigma_{xx}}{dx}}dA_{3}$$

Recall $$v_{y}=\frac{\partial Mz}{\partial x}$$, $$v_{z}=\frac{\partial My}{\partial x}$$

$$q^{(3)}=-\left(k_{y}v_{y}-k_{yz}v_{z}\right)Q_{z}^{(3)}-\left(k_{z}v_{z}-k_{yz}v_{y} \right)Q_{y}^{(3)}$$

$$Q_{z}^{(3)}=\int_{A_{3}}^{}{y}dA_{3}$$ and  $$Q_{y}^{(3)}=\int_{A_{3}}^{}{z}dA_{3}$$

String 2

$$\tilde{q_{23}}=\tilde{q_{12}}+\tilde{q_{24}}+\tilde{q}^{(2)}$$ but because of the way you cut the cells $$\tilde{q_{12}}$$ and $$\tilde{q_{24}}$$ equal zero.

$$q^{(2)}$$ is computed the same as $$q^{(3)}$$ except $$A_{3}$$ becomes $$A_{2}$$

$$Q_{z}^{(2)}=y_{2}A_{2}$$ and $$Q_{y}^{(2)}=z_{2}A_{2}$$

Stringer 4 is $$\tilde{q}_{43}=\tilde{q}_{24}-\tilde{q}_{41}+q^{(4)}$$ but $$\tilde{q}_{24}$$ and $$\tilde{q}_{41}$$ become zero because of the cut of the cells.$$q^{(4)}$$ can be computed as above for $$q^{(2)}$$

This marks the end of problem 2

Now returning to the superposition principle

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

$$q_{12}=\tilde{q}_{12}+q_{1}$$

$$q_{23}=\tilde{q}_{23}+q_{1}$$

All of the $$\tilde{q}$$ are known

Continuing this process for $$q_{24}$$, $$q_{43}$$, and $$q_{41}$$

$$q_{24}=\tilde{q}_{24}+q_{1}$$

$$q_{43}=\tilde{q}_{43}+q_{1}$$

$$q_{41}=\tilde{q}_{41}+q_{1}$$

To solve problems, such as problem 2, you'll have a certain number of unknown shear stresses and you'll get the same amount of equations to solve for the unknowns from,

1)The moment equation: Take the moment of $$v_{y}$$, $$v_{z}$$, and $$\left\{q_{12}, ..., q_{41} \right\}$$ about any convenient point (usually where lines of action of $$v_{y}$$, $$v_{z}$$ cross)

2)$$\theta _{1}=\theta _{2}$$

3)$$\theta _{2}=\theta _{3}$$

Continuation of Stringer Shear Force
Resulting for the consequences of the first method,

$$\tilde q_{j6}=\tilde q_{2j}-\tilde q_{j5}- \tilde q_{j8}+q(j)$$

then if the $$\tilde q_{6j}$$ is reversed

then

$$-\tilde q_{j6}=\tilde q_{2j}-\tilde q_{j5}- \tilde q_{j8}+q(j)$$

The question was raised that if one stringer was isolated if the system could still be solved.

this is not true because when one stringer is isolated it would yield that the q of the actual stringer is zero which is known not to be true.

$$\tilde q_{j6}=\tilde q_{2j}-\tilde q_{j5}- \tilde q_{j8}+q(j)$$ which would yeild $$0=0+0+0+q(j)$$ and q(j) is a number that is non zero $$q(j) \neq 0$$ so this yields that the system is unsolvable whenever a stringer is isolated.

Plate Buckling Problems
Refer to the course page for more information with regards to this lecture. Equation numbers referenced in this lecture come from the aforementioned page.

Plotting buckling shapes under shear stress:

Express $$\displaystyle\{C_{22},C_{13},C_{31},C_{33} \}$$ in terms of $$\displaystyle C_{11}$$ for $$\displaystyle \vartheta = 1.5$$ 1) Find $$\displaystyle \lambda$$ for $$\displaystyle \vartheta = 1.5$$ using equation (30).
 * $$\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}$$ 2) Evaluate $$\displaystyle \mathbf K_{5 \times 5}$$ numerically in equation (26).
 * $$\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\}$$ 3) $$\displaystyle \mathbf K = [K_{ij}]$$
 * $$\displaystyle \begin{bmatrix} 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{bmatrix} \begin{Bmatrix}C_{22}\\C_{13}\\C_{31}\\C_{33} \end{Bmatrix} = \begin{Bmatrix} -\frac{4}{9}C_{11}\\0\\0\\0 \end{Bmatrix}$$

Solve for $$\displaystyle\{C_{22},C_{13},C_{31},C_{33} \}$$ in terms of $$\displaystyle C_{11}$$
 * $$\displaystyle \{C\}= [K]^{-1}\begin{Bmatrix} -\frac{4}{9}C_{11}\\0\\0\\0 \end{Bmatrix}$$
 * $$\displaystyle u_z = C_{11}\sin\left(\frac{\pi x}{a}\right)\sin\left(\frac{\pi y}{b}\right) + C_{22}\sin\left(\frac{2\pi x}{a}\right)\sin\left(\frac{2\pi y}{b}\right) + C_{13}\sin\left(\frac{\pi x}{a}\right)\sin\left(\frac{3\pi y}{b}\right) + C_{31}\sin\left(\frac{3\pi x}{a}\right)\sin\left(\frac{\pi y}{b}\right) + C_{33}\sin\left(\frac{3\pi x}{a}\right)\sin\left(\frac{3\pi y}{b}\right)$$

Set $$\displaystyle C_{11} = 1$$, plot $$\displaystyle u_z$$.

Results:
 * $$\displaystyle \lambda = 0.0288$$
 * $$\displaystyle \mathbf K = \begin{bmatrix}   2.1597  & -0.8000 &  -0.8000   & 1.4400\\

-0.8000 &  5.7706     &    0    &     0\\   -0.8000    &     0  &  1.6174    &     0\\    1.4400     &    0    &     0  & 10.9334\end{bmatrix}$$
 * $$\displaystyle \begin{Bmatrix}C_{22}\\C_{13}\\C_{31}\\C_{33} \end{Bmatrix}= \mathbf K^{-1}\begin{Bmatrix} -\frac{4}{9}\\0\\0\\0 \end{Bmatrix} = \begin{Bmatrix}  -0.3037\\

-0.0421\\  -0.1502\\    0.0400\end{Bmatrix}$$ Setting a = 3, b=2 (to keep the plate aspect ratio = 1.5).
 * $$\displaystyle u_z = \sin\left(\frac{\pi x}{3}\right)\sin\left(\frac{\pi y}{2}\right) -0.3037\sin\left(\frac{2\pi x}{3}\right)\sin\left(\frac{2\pi y}{2}\right) -0.0421\sin\left(\frac{\pi x}{3}\right)\sin\left(\frac{3\pi y}{2}\right) -0.1502\sin\left(\frac{3\pi x}{3}\right)\sin\left(\frac{\pi y}{2}\right) + 0.0400\sin\left(\frac{3\pi x}{3}\right)\sin\left(\frac{3\pi y}{2}\right)$$

Note: (continued from lecture on 12/1/08) Answer: in 2 parts Part 1: Equilibrium of isolated stringer Part 2: Closed cell, equilibrium of stringers Stringer 1:
 * $$\displaystyle \tilde{q_{12}} = \tilde{q_{31}} + \tilde{q_{41}} + q^{(1)}$$
 * $$\displaystyle \tilde{q_{31}} = 0$$
 * $$\displaystyle \Rightarrow \tilde{q_{12}} = \tilde{q_{41}} + q^{(1)}$$

Stringer 2:
 * $$\displaystyle \tilde{q_{24}} = \tilde{q_{12}} - \tilde{q_{23}}+q^{(2)}$$
 * $$\displaystyle \tilde{q_{23}} = 0$$
 * $$\displaystyle \Rightarrow \tilde{q_{24}} = \tilde{q_{12}} +q^{(2)}$$

Stringer 4:
 * $$\displaystyle \tilde{q_{41}} = \tilde{q_{24}} + \tilde{q_{34}} + q^{(4)} $$
 * $$\displaystyle \tilde{q_{34}} = 0$$
 * $$\displaystyle \tilde{q_{24}} = \tilde{q_{12}} +q^{(2)} = \tilde{q_{41}} + q^{(1)}+q^{(2)}$$
 * $$\displaystyle \Rightarrow \tilde{q_{41}} = \tilde{q_{41}}+q^{(1)}+q^{(2)}+q^{(4)}$$

Matlab Analysis
This Matlab assignment again analyzes the NACA 2415 airfoil. This time, it looks at the shear flow in the panels (shear flow and spar webs) when it is subjected to a transverse load. The load forces are a result of the lift and drag on the NACA 2415 airfoil.

Matlab Code
Now, here is the m-file used to calculate the Plate Buckling Characteristics:

Review of Wikiversity and Comparison to WebCT
This course relied heavily on the MediaWiki software for homework reports. As a group, we felt that this was not a very effective method of learning aerospace structures. Throughout the year, we spent more time formatting the Wiki pages and making them look good as opposed to working practice problems and working on Matlab projects. Writing equations and uploading pictures was a somewhat time consuming process that we feel took away from the learning experience. (By now we are all experts at using it, but even now it is still very time consuming)

When compared to the WebCT software that UF provides, MediaWiki is by far superior to WebCT for what this class uses it for. For online collaboration of teammates and making notes availiable to the entire class, MediaWiki is the way to go. It is definately the way of the future for online/teamwork collaboration, but we feel that it is still not user friendly enough to warrant a switch to it right now.

It was very interesting learning this software and using these new teaching methods this semester. This review is simply a collection of our thoughts about the MediaWiki software and the influence on the class.

Team Members Who Contributed to this Report
Mark Barry Eas4200c.f08.blue.a 14:13, 8 December 2008 (UTC)

Thomas Sexton EAS4200c.f08.blue.c 23:37, 9 December 2008 (UTC)

Marc Pelletier EAS4200c.f08.blue.b 17:10, 8 December 2008 (UTC)

Jeffrey Shea EAS4200c.f08.Blue.E 19:10, 9 December 2008 (UTC)

Clay Robertson eas4200c.f08.blue.f 21:48, 9 December 2008 (UTC)

Christian Garabaya

Wei-Teck Lee Eas4200c.f08.blue.g 02:10, 9 December 2008 (UTC)