User:Eas4200c.f08.aero6.ortega/HW7

Vu-Quoc example of open cross-section with varying stringer areas


mean value theorem:

$$Q_{z} = \int_{A} ydA = \bar{y}\int_{A}da = \bar{y}A$$

$$Q_{y} = \int_{z}dA = \bar{z}A$$

$$A = \sum^{4}_{i=1}A_{i}$$(neglecting the skin and the spar webs)

Recall from chapter 5.5 of C.T. Sun's Mechanics of Aircraft Structures $$q(s) = (k_{yz}Q_{z} - k_{z}Q_{y})V_{z}\!$$

Notice the shear flow is constant along the webs because the area is restricted to the stringers. Also:

1)$$V_{z}\!$$

2)$$k_{yz}, k_{z}\!$$

3)$$Q_{z}, Q_{y}\!$$

Are all independent of the surface contour. However, $$q(s)$$ would increment or "jump" when crossing a stringer.

Step 1. find ($$\bar{y}_{c}, \bar{z}_{c}$$)

Step 2. find $$I_{y}, I_{z}, I_{yz}\!$$

Step 3. find $$k_{y}, k_{z}, k_{yz}\!$$

Step 4. follow the contour path (s) to find $$q_{12}, q_{23}, q_{34}$$

where

$$q_{12} = (k_{yz}Q_{z}^{12} - k_{z}Q_{y}^{12})V_{z}$$

$$Q_{z}^{12} = y_{1}A_{1}$$ where $$y_{1}$$ is the y-coordinate for stringer 1.

$$Q_{y}^{12} = z_{1}A_{1}$$

note$$Q_{z}^{23} = y_{1}A_{1} + y_{2}A_{2}$$

Note that the contour path is not fixed. If beginning the contour from the opposite direction, the terms are as follows:



to be continued...

Mini-plan
S) Single-cell section
 * S.1) without stringers
 * S.2) with stringers

M) multi-cell sections
 * M.1) without stringers
 * M.2) with stringers

S.1) without stringers
$$q = c = $$constant shear flow

Question: can this set up resist $$V_{z}$$ (transverse shear flow)?-No.

$$R^{z} = R_{AB}^{z} + R_{BA}^{z}$$ where R is the resultant of q

$$R_{AB}^{z} = -q\bar{A'B'} = -R_{BA}^{z} ==R^{z} = 0$$

S.2) with stringers
Neglecting the contribution to web bending:

When you cross a stringer, you have a jump. This is a somewhat step-wise function.$$q_{12} \ne q_{23} \ne q_{31}$$. However, $$q_{ij}(s) = c = \!$$constant within each panel. Therefore q(s) is piece-wise constant throughout the contour.

$$R^{z} = V_{z} \ne  0\!$$. Thus this is a non-constant shear flow.

The principle of Super Position is a powerful tool for solving these problems and is readily available due to the linearity of the components.

$$q_{12} = q + \bar{q_{12}}$$

$$q_{23} = q + \bar{q_{23}}$$

$$q_{31} = q + \bar{q_{31}}$$

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

One "q" ($$\bar{q_{ij}}$$ are known after solving the case without stringers) and one equation for each unknown.

Method
Data:$$V_{y}, V_{z}$$

1) Solve "2" for $$\bar{q_{12}}, \bar{q_{23}} (\bar{q_{31}} = 0)$$

2) Moment equation: take the moment about any point in the y-z plane
 * 2.1) Super-position:$$q_{ij} = q + \bar{q_{ij}}$$
 * 2.2) Select a point ( denoted $$\bar{o}$$) in plane y-z
 * $$\sum_{\bar{o}}Moments = \sum_{\bar{o}} of q_{ij}$$

3) Back to Super position

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

where $$q_{ij}$$ is the true shear flow, $$q$$ is the closed-cell constant shear flow and $$\bar{q_{ij}}$$ is the open-cell piece-wise constant shear flow.

M.1) Multi-cell without stringers
q = 0 at the cuts and is constant through the contour of the skin, thus q = 0 as shown.

$$R^{z} = R^{z1} + R^{z2} = V_{z}\!$$ where $$R^{z1} = 0\!$$

Looking at stringer 3


$$\sum F_{x} = 0 = \int_{A_{3}} [\theta_{xx}(x + dx) - \theta_{xx}(x)]dA_{3} = [-\bar{q_{23}} -\bar{q_{43}} + \bar{q_{31}}]dx$$. Notice the integrand is simply a Taylor series expansion.

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

$$q^{(3)} = -\int_{A_{3}}\frac{d\theta_{xx}}{dx}dA_{3}$$ This is the contribution to shear flow by stringer 3.

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


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$$q^{(3)} = -(k_{y}V_{y} - k_{yz}V_{z})Q_{z}^{(3)} - (k_{z}V_{z} - k_{yz}V_{y})Q_{y}^{(3)}$$ where $$Q_{z}^{(3)} = \int_{A_{3}}ydA_{3}$$
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and $$Q_{y}^{(3)} = \int_{A_{3}}zdA_{3}$$

Now the process is repeated for the other stringers.

Stringer 2
$$\bar{q_{23}} = \bar{q_{12}} - \bar{q_{24}} + q^{(2)}$$ where $$\bar{q_{12}} = \bar{q_{24}} =0$$ and $$q^{(2)}$$ is the shear flow due to bending at stringer 2. Thus

$$\bar{q_{23}} = q^{(2)}$$ which is solved by the same structure to that of stringer 3.

$$Q_{z}^{(2)} = y_{2}A_{2}$$ where $$y_{2}$$ is the y-coordinate of stringer 2

$$Q_{y}^{2} = z_{2}A_{2}$$

Stringer 4
$$\bar{q_{43}} = \bar{q_{24}} - \bar{q_{41}} + q^{(4)}$$ again $$\bar{q_{24}} =\bar{q_{41}} =0$$ and $$q^{(4)}$$ is solved just like that of stringer 3. Due to $$\bar{q_{31}}$$:

Super-position (again)

$$q_{ij} = \bar{q_{ij}} + q_{k}$$

$$q_{12} = \bar{q_{12}} + q_{1}$$

$$q_{23} = \bar{q_{23}} + q_{1}$$

$$q_{31} = \bar{q_{31}} + q_{1}$$

$$q_{24} = \bar{q_{24}} + q_{2}$$

$$q_{43} = \bar{q_{43}} + q_{2}$$

$$q_{41} = \bar{q_{41}} + q_{2}$$

Hence there are 3 unknowns ($$q_{1}, q_{2}, q_{3}$$) and 3 equations:

1) Moment equation (usually take at a convenient point such as to cancel the contribution of $$V_{y}, V_{z}$$

2) $$\theta_{1} = \theta_{2}\!$$ where $$\theta\!$$ must be computed with the true shear flow!

3)$$\theta_{2} = \theta_{3}\!$$

Stringers continued (lecture on 12/08/08)
$$ 0 = q^{(1)} + q^{(2)} + q^{(4)} \!$$

This not true or possible so...

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

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

$$ n_{z} := -(k_{y}V_{y} - k_{yz}V_{z}) \!$$

$$n_{y} := -(k_{z}V_{z} - k_{yz}V_{y}) \!$$

Therefore

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



$$Q_{y}(\hat{z})=\int zdA = 0 = z_{c}dA \!$$

NACA airfoil back of envelope verification

Given a rectangle with points $$\bar{B}, \bar{E}, \bar{F}, \bar{H} \!$$

$$ \bar{B}\bar{E} = \bar{F}\bar{H} = .5(BE + FH) $$

Bidirectional bending Continuation (from Lecture 12/8/2008)
$$M_y=-EY_{yz}\frac{d^2v_o}{dx^2}-EI_y\frac{d^2w_o}{dx^2}$$ $$M_y=-EY_{z}\frac{d^2v_o}{dx^2}-EI_{yz}\frac{d^2w_o}{dx^2}$$

$$ \begin{Bmatrix} M_y\\ M_z \end{Bmatrix} = \begin{vmatrix} I_y & I_{yz}\\ I_{yz} & Iz \end{vmatrix}

\begin{Bmatrix} -Ex_z\\ -Ex_y \end{Bmatrix}

$$

$$\epsilon_{xx} = -yx_y - zx_z = \begin{bmatrix} z & y \end{bmatrix} \begin{Bmatrix} -x_z\\ -x_y \end{Bmatrix}$$

$$ \sigma_{xx} = E\epsilon_{xx} = E \begin{bmatrix} z & y \end{bmatrix} \begin{Bmatrix} -x_z \\ -x_y \end{Bmatrix} = E \begin{bmatrix} z & y \end{bmatrix} \frac{1}{E} I^{-1} \begin{Bmatrix} M_y \\ M_z \end{Bmatrix} = \begin{bmatrix} z & y \end{bmatrix} I^{-1} \begin{Bmatrix} M_y \\ M_z \end{Bmatrix} $$

where we know

$$ I^{-1} = \frac{1}{D} \begin{bmatrix} I_z & -I_{yz} \\ -I_{yz} & I_y \end{bmatrix} $$ and $$ D = I_y I_z - \left(I_{yz}\right)^2 \,$$

For the shear flow equation in matrix notation:

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

where we know

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

and we also know

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

and

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

and additionally, we know

$$ Q_y = \int_A z \quad dA \,$$

and

$$ Q_z = \int_A y \quad dA \,$$

Back to the NACA Airfoil
The following MATLab Code calculates the shear flow in each segment of a single-cell and three-celled NACA 4-digit series airfoil. The results for the NACA 2415 (with spar webs at quarter and three-quarter chord in the multicell airfoil) are as follows:

For the single cell,

The $$q^{(1)}\!$$, $$q^{(2)}\!$$, $$q^{(3)}\!$$ for the three-celled airfoil are,

And the $$\sigma_{BF}\!$$ and $$\sigma_{EH}\!$$ values are,

Plate Buckling
Using the following equation from Dr. Vu-Quoc's plate buckling website, [Plate Buckling], we calculate $$\displaystyle\lambda$$ using an aspect ratio, $$\vartheta$$, of 1.5

$$  \displaystyle \lambda =  \left[ \frac{\vartheta^4}{81 (1 + \vartheta^2)^4} \left\{ 1 	 + 	 \frac{81}{625} +	 \frac{81}{25} \left(	   \frac{1 + \vartheta^2}{1 + 9 \vartheta^2}	 \right)^2 +	 \frac{81}{25} \left(	   \frac{1 + \vartheta^2}{9 + \vartheta^2}	 \right)^2 \right\} \right]^{1/2} = 0.0739 $$

Using these two values we can complete the following matrix which is also found in the Plate Buckling website:

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

$$  \displaystyle \left[ \begin{array}{lllll} 0.3471 	 &	 0.4444	 &	 0	 &	 0	 &	 0	 \\	 0.4444	 &	 5.5533  	 &	 -0.8000	 &	 -0.8000	 &	 1.4400	 \\	 0	 &	 -0.8000	 &	 14.8382  	 &	 0	 &	 0	 \\	 0	 &	 -0.8000	 &	 0	 &	 4.1588  	 &	 0	 \\	 0	 &	 1.4400	 &	 0	 &	 0	 &	 28.1135        \end{array} \right] \left\{ \begin{array}{l} C_{11} \\	 C_{22} \\	 C_{13} \\	 C_{31} \\	 C_{33} \end{array} \right\} =  \left\{ \begin{array}{l} 0	 \\	 0	 \\	 0	 \\	 0	 \\	 0     \end{array} \right\} $$

We can rearrange this matrix as follows:

$$  \displaystyle \left[ \begin{array}{lllll} 5.5533 	 &	 -0.8000	 &	 -0.8000	 &	 1.4400	 \\	 -0.8000	 &	 14.8382  	 &	 0	 &	 0	 \\	 -0.8000	 &	 0	 &	 4.1588  	 &	 0	 \\	 1.4400	 &	 0	 &	 0	 &	 28.1135        \end{array} \right] \left\{ \begin{array}{l} C_{22} \\	 C_{13} \\	 C_{31} \\	 C_{33} \end{array} \right\} =  \left\{ \begin{array}{l} 0.4444C_{11} \\	 0	 \\	 0	 \\	 0     \end{array} \right\} $$

Using Matlab we can then express $$C_{22}, C_{13}, C_{31}, C_{33}$$ in terms of $$C_{11}$$:

$$\left\{ \begin{array}{l} C_{22} = 0.07630C_{11} \\	 C_{13} = -0.00411C_{11} \\	 C_{31} = -0.01468C_{11} \\	 C_{33} = 0.00391C_{11} \end{array} \right\}$$

We can also derive the transverse displacement of the beam using the following formulas:

$$  \displaystyle \psi (x,y) =   c_{mn} \sin \left(     \frac      {m \pi x}      {a}   \right) \sin \left(     \frac      {n \pi y}      {b}   \right) \, \ {\rm for} \ m,n = 1, 2, 3, \ldots $$

$$  \displaystyle u_z (x,y) =  \sum_{m=1}^{\infty} \sum_{n=1}^{\infty} \psi_{mn} (x,y) $$

These equations and the above 'C' terms yield the following specific bending:

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

$$         = C_{11}sin(\frac{\pi x}{a})sin(\frac{\pi y}{b})+$$ $$           0.07630C_{11}sin(\frac{2\pi x}{a})sin(\frac{2 \pi y}{b})+$$ $$           -0.00411C_{11}sin(\frac{\pi x}{a})sin(\frac{3 \pi y}{b})+$$ $$           -0.01468C_{11}sin(\frac{3 \pi x}{a})sin(\frac{\pi y}{b})+$$ $$           0.00391C_{11}sin(\frac{3 \pi x}{a})sin(\frac{3 \pi y}{b})$$

If we set $$C_{11} = 1, a = 3, abd b = 2$$ we are left with:

$$u_z (x,y) = sin(\frac{\pi x}{3})sin(\frac{\pi y}{2})+$$ $$           0.07630sin(\frac{2\pi x}{3})sin(\frac{\pi y}{1})+$$ $$           -0.00411sin(\frac{\pi x}{3})sin(\frac{3 \pi y}{2})+$$ $$           -0.01468sin(\frac{\pi x}{1})sin(\frac{\pi y}{2})+$$ $$           0.00391sin(\frac{\pi x}{1})sin(\frac{3 \pi y}{2})$$

The following figure shows the amount of deflection the beam undergoes at specific (x,y) locations. Where the x location or the y location is zero the bar undergoes zero deflection. It undergoes its greatest deflection near the center of the beam which makes sense if we assume that the beam is fixed on all four edges.



Recommended Software
Writing Wiki Articles:


 * MediaWiki is similar in many ways to HTML, so having some knowledge in that language is very helpful. Now that so many web companies use this language for their websites, having the basic knowhow and ability to design in MediaWiki is a real advantage in the workplace.

Writing LaTex Equations:


 * Knowing how to use the LaTex Equation Editor is vital for programming in MediaWiki, and is useful for displaying equations in MATLab, however Microsoft Word 2007 has an equation editor which is much easier and far more user friendly. The learning curve is virtually non-existent and the equations can be saved as mpegs or jpegs and then uploaded.  For novices, this is probably not a bad alternative.

Drawing Figures:


 * Again, Microsoft Word 2007 provided our group with the best figure drawing capabilities, whether it was drawing simple boxes and circles or complex airfoils or curved surfaces. MS Paint and Adobe Photoshop were also a convenient tools.

Contributing Team Members
Kris Loper Eas4200c.f08.aero6.loper 17:52, 9 December 2008 (UTC)

Matt Inman Eas4200c.f08.aero6.inman 17:56, 9 December 2008 (UTC)

Eduardo Villalba Eas4200c.f08.aero6.villalba 20:30, 9 December 2008 (UTC)

Scott Chastain Eas4200c.f08.aero6.chastain 20:44, 9 December 2008 (UTC)

Felipe Ortega Eas4200c.f08.aero6.ortega 21:33, 9 December 2008 (UTC)

Marlana Behnke Eas4200c.f08.aero6.behnke 21:46, 9 December 2008 (UTC)