User:Uf.team.aero/HW7

= Shear Loading on thin walled open section =

Consider the following open section consisting of 4 stringers:

The steps involved to solve for the shear flow:
 * 1) Find the location of the centroid $$\left(y_c,z_c \right)$$
 * 2) Find $$I_y,I_z,I_{yz}$$
 * 3) Find $$k_y,k_z,k_{yz}$$
 * 4) Follow the path "s" to find the shear flow

= Shear Loading Resistance of Thin-walled Sections =

It is necessary for the analysis of aircraft structures to understand the response of an arbitrary thin-walled section to an applied shear load. As such, the following outline will be used in analysis.

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

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

The analysis is as follows:

S.1) Single-cell Sections Without Stringers
It is known from previous work that shear flow in a thin-wall can be assumed constant. The resultant of the shear flow can be found as the line integral of the shear flow projected onto the axis of interest. The figure below shows an arbitrary shape containing some shear flow.

To calculate the resultant of the shear flow along the z-axis, consider the shear flows from A to B and B to A.


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

As a result, the single-cell section without stringers is shown to be incapable of resisting a shear loading in the z-direction, VZ. Since the Z-axis is chosen arbitrarily, any direction can chosen. Thus, in general, a single-cell section without stringers is incapable of resisting any shear loading whatsoever.

S.2) Single-cell Sections With Stringers
There exist single-cell sections with stringers, such that the shear flow is no longer constant about the perimeter of the section. It is assumed that the thin walls play very little role in shear loading resistance, such that the shear flow can be assumed piece-wise constant between stringers. This is shown in the figure below.



In the above figure,


 * $$ q_{ij} = constant $$

where subscript ij represents the shear flow in the wall along the path from stinger i to stringer j. Furthermore,


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

It is proposed that for a balance of forces, the resultant from all shear flows must balance the applied shear loading condition. For example, if some shear load in the z-direction is placed on the section, then the resultant of shear flow is given as


 * $$ R_Z = V_Z \neq 0 $$

To analyze a section with stringers, the previous method and equations used for an open section must be modified to account for a closed section. Specifically, the available equations are applicable to an open section, such that a path can be drawn beginning at a free surface (where the shear flow is known to be zero. As such, the method of superposition is used to calculate the total shear flow in a closed section with stringers.

First a constant shear flow, q0 is assumed to be present throughout the perimeter of the section as if no stringers are present as shown in the figure below.



Next, the original section is assumed to be cut at a given point in the wall between two stringers. Thus, the resulting section is an open section with a free edge. Since the shear flow at a free edge is zero, and since the shear flow is assumed to be constant between two stringers, the wall section containing the cut is defined to have zero shear flow. Then, the method and equations for analyzing an open section can be implemented to find the piece-wise constant shear flows between the stringers, namely q12', q23', q31'. For the given cut location, it is seen that $$ q_{31}' = 0 $$. This is shown in the figure below.



The result of the superposition is given in equation form as


 * $$ q_{ij} = q_0 + q_{ij}' $$

-Notes from 8 Dec-


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

or


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

Note, from 38_2


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

where


 * $$ n_{z}=-(k_y V_y-k_{yz}V_z)$$

and (for HW7 answer)


 * $$ n_{y}=-(k_z V_z-k_{yz}V_y)$$

so


 * $$ \Sigma q^{(e)}=n_z \sum_{1 }^{4 } Q_z^{(e)}+n_y \sum_1^4Q_y^{(e)}$$

on 41_2,we spoke of ways to determine Q for an area. Refer to the following figure:



Where,


 * $$Q_y ( \hat{z})= \int zdA$$


 * $$M_y=-EI_{yz} \frac{ \partial ^2 v_o}{ \partial z^2} - EI_y \frac{\partial ^2 w_o}{\partial x^2 } $$

and


 * $$M_z=-EI_{z} \frac{ \partial ^2 v_o}{ \partial x^2} - EI_{yz} \frac{\partial ^2 w_o}{\partial x^2 } $$

Where


 * $$\frac{ \partial ^2 v_o}{ \partial x^2}$$

can be noted as $$\chi_y$$

and


 * $$\frac{\partial ^2 w_o}{\partial x^2 }$$

can be noted as $$ \chi_z$$

The equations are rearranged in the following manner to find q for the cell:


 * $$ \left [ \begin{array}{rr} M_y\\ M_z \end{array} \right ] = \left [ \begin{array}{cc} I_y & I_{yz} \\ I_{yz} & I_z \end{array} \right ] \left [ \begin{array}{cc} -E X_z \\ -E X_y \end{array} \right ] $$


 * $$\epsilon _{xx} = -y \chi_y - z \chi_z=  \left [ \begin{array}{cc} z & y \end{array} \right ] \begin{Bmatrix}

- \chi _z \\ - \chi _y \end{Bmatrix}$$


 * $$\sigma _{xx} = E \epsilon _{xx} = E \left [ \begin{array}{cc} z & y \end{array} \right ] \begin{Bmatrix}

- \chi _z \\ - \chi _y \end{Bmatrix} =E \left [ \begin{array}{cc} z & y \end{array} \right ]\frac{1}{E} \check{I}^{-1}\begin{Bmatrix} - M _y \\ - M _z \end{Bmatrix}$$

where


 * $$\check{I} = \begin{bmatrix}

I_y & I_{yz}\\ I_{yz} & I_z \end{bmatrix}$$

and


 * $$\check{I}^{-1} = \frac{1}{D}\begin{bmatrix}

I_z & -I_{yz}\\ -I_{yz} & I_y \end{bmatrix}$$

and to find q:


 * $$q=-\int_{a}^{}{\frac{\partial \sigma _{xx}}{\partial x}}dA$$

where


 * $${\frac{\partial \sigma _{xx}}{\partial x}}=\begin{bmatrix}

z & y \end{bmatrix}I^{-1}\begin{bmatrix} \frac{dM_y}{dx}\\ \\ \frac{dM_z}{dx} \end{bmatrix}$$

M.1) Multicell Sections Without Stringers
NOTES FOR NOV. 21

The figure below depicts an example of a multicell section without stringers



The resultant of the shear flow for a figure of this case can be found by


 * $$R^z= \sum_{i=1}^n R_i^z$$

where n is the number of cells and $$R_i^z$$ is the resultant of the shear flow in the i-th cell.

It has been previously shown that any cell without strings is incapable of resisting shear in the Z direction and therefore


 * $$R_i^z = 0$$

for all i. It then directly follows that


 * $$R^z = 0$$

for multicell sections without stringers.

M.2) Multicell Sections With Stringers
Below is an example of a multicell section with stringers.



To solve for the resultant shear flow in such a system two states need to be analyzed then combined with the principle of superposition. State P1 below shows the closed cell constant shear flow, whereas state P2 below shows the open cell piecewise constant shear flow.





States P1 and P2 shown above are components of the total resultant shear flow, which can be formulated as


 * $$R^z = R^{z1}+R^{z2}$$

where $$R^z$$ is the resultant force of the section depicted at the beginning of the section, $$R^{z1}$$ is the resultant of the state P2, and $$R^{z2}$$ is the resultant of the state P2. Furthermore, from the analysis of single celled section we can expect that


 * $$R^{z1} = \sum{i=1}^n R_i^{z1} = 0 \qquad$$ and $$\qquad R^{z2} = R^{z2} V_2 \ne 0$$

Stringer 3
$$\displaystyle q^{(3)} = -\int_{A_3} \frac{d \sigma_{XX}}{dx} dA_3$$ The 3 designates that this is the shear flow contributed by stringer 3.



Recall:

$$\displaystyle V_y = \frac{M_z}{dx}$$

$$\displaystyle V_z = \frac{M_y}{dx}$$

p.34-3: $$ \displaystyle q_{(3)} = -(k_y V_y - k_{yz} V_z) Q_z^{(3)} - (k_z V_z - k_{yz} V_y) Q_y^{(3)}$$ → Eq. 1 $$ \displaystyle => Q_z^{(3)} = \int_{A_s} y dA$$ and $$ \displaystyle Q_y^{(3)} = \int_{A_s} z dA $$

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

Because of the way the cell was cut: $$\displaystyle \tilde q_{12} = 0$$ $$\displaystyle \tilde q_{24} = 0$$

Therefore $$\displaystyle \tilde q_{23} = q^{(2)}$$

$$ \displaystyle q_{(2)} = -(k_y V_y - k_{yz} V_z) Q_z^{(2)} - (k_z V_z - k_{yz} V_y) Q_y^{(2)}$$ → Eq. 1 $$ \displaystyle => Q_z^{(2)} = \int_{A_s} y dA = y_2 A_2$$ and $$ \displaystyle Q_y^{(2)} = \int_{A_s} z dA = z_2 A_2$$

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

Because of the way the cell was cut: $$\displaystyle \tilde q_{24} = 0$$ $$\displaystyle \tilde q_{41} = 0$$

Therefore $$\displaystyle \tilde q_{43} = q^{(4)}$$

$$ \displaystyle q_{(4)} = -(k_y V_y - k_{yz} V_z) Q_z^{(4)} - (k_z V_z - k_{yz} V_y) Q_y^{(4)}$$ → Eq. 1 $$ \displaystyle => Q_z^{(4)} = \int_{A_s} y dA = y_4 A_4$$ and $$ \displaystyle Q_y^{(4)} = \int_{A_s} z dA = z_4 A_4$$

Superposition (again)
$$\displaystyle q_{ij} = \tilde q_{ij} + q_{k}$$

$$\displaystyle q_{12} = \tilde q_{12} + q_{1}$$

$$\displaystyle q_{31} = \tilde q_{31} + q_{1}$$

$$\displaystyle q_{24} = \tilde q_{24} + q_{2}$$

$$\displaystyle q_{43} = \tilde q_{43} + q_{4}$$

$$\displaystyle q_{41} = \tilde q_{41} + q_{1}$$

=> 3 Unknowns $$\displaystyle q_1, q_2, and q_3$$: need 3 equations
 * 1) Moment Equation: Take moment of $$\displaystyle V_y, V_z, and {q_{12} ... q_{41}}$$ about any convenient point. (Usually where lines of action of $$\displaystyle V_y and V_z$$ intersect)
 * 2) $$\displaystyle \theta_1 = \theta_2$$
 * 3) $$\displaystyle \theta_2 = \theta_3$$

Solving Problem P2 continued
There are two methods to solve the shear flows of the section with hypothetical cuts (Problem P2):


 * 1) Complete method: free body diagram
 * 2) Consequence of complete method: conservation of shear flows

To expedite the solution of simple problems, we will further examine the conservation of shear flow approach. It is seen that the flow into or out of a stringer, both from the wall sections and the stringer itself, must be conserved. That is, the shear flow arriving at the stringer location must equal that leaving the stringer location. For example, consider the figure below.



To solve for $$ q_{j6}' $$, consider the summation of inflows and outflows.


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

Thus, the outward shear flow from stringer j to stringer 6 is equal to the inward shear flows from 2 to j and the contribution from stringer j itself less the outward shear flow towards stringers 5 and 8. In this manner, shear flow about stringer j is conserved.

Likewise,


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

It is seen that the shear flow in from stringer 2 to j and the shear flow from stringer j itself are supplying the shear outflows to stringers 5, 6 and 8. Again, this makes sense from a conservation perspective.

Isolating a Stringer
What if a closed, multicell section is cut in such a manner that a single stringer is isolated? What are the results? Is this allowed?

To answer this question, consider a cut of the interior walls such that stringer 3 is completely isolated. In this case, $$ q_{23}' = q_{31}' = q_{34}' = 0 $$. The resulting shear flow conservation of stringer 3 would produce


 * $$ q_{31}' = q_{23}' - q_{34}' + q^{(3)} $$

Based upon the previous statement that the shear flows in the cut walls are identically zero, the above statement would yield


 * $$ q^{(3)} = -\int_{A_3}^{}{\frac{d \sigma_{xx}}{dx}dA_3} = 0 $$.

Since the area A3 is nonzero and the normal stress is nonzero, this cannot be true. As such, the method of cutting a multicell section cannot be performed in such a way as to isolate any one stringer.

A continuation from stringer isolation
When looking at a closed cell, stringer that is in equilibrium, may help in simplification for the following stringers:

For stringer 1, we may write the following equation for equilibrium:


 * $$\tilde{q_{12}} = \tilde{q}_{31} + \tilde{q}_{41} + q_1$$

where $$\tilde{q}_{31}=0$$

After simplification, the equation becomes:


 * $$\tilde{q_{12}} = \tilde{q}_{41} + q_1$$

For stringer 2, accordingly we may write the following equation for equilibrium:


 * $$\tilde{q_{24}} = \tilde{q}_{12} + \tilde{q}_{23} + q_2$$

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

After simplification, the equation becomes:


 * $$\tilde{q_{24}} = \tilde{q}_{12} + q_2$$

= Plotting Buckling Shape under Shear Stress =

First, we can express $$\begin{Bmatrix} C_{22}, C_{13}, C_{31}, C_{33}\end{Bmatrix}$$ in terms of $$C_{11}$$.

For $$\vartheta = 1.5$$ we can accomplish the following:


 * Find $$\lambda$$ for $$\vartheta = 1.5$$ using equation (30)


 * Evaluate for $$k_{sys}$$ using equation (26)


 * Calculate $$[k_{ij}]$$


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

Then we will need to solve $$\begin{Bmatrix} C_{22}, C_{13}, C_{31}, C_{33}\end{Bmatrix}$$ intersect with $$C_{11}$$. If $$K$$ matrix is expressed as $$\underline{\overline{K}}$$ then its inverse is $$\underline{\overline{K}}^{-1}$$. Therefore, the calculated values for $$\begin{Bmatrix} C_{22}, C_{13}, C_{31}, C_{33}\end{Bmatrix}$$ can be related as:


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

Looking at this result,


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

= NACA Airfoil Continued =

This weeks homework incorporated finding the stresses in the airfoil due to shear loads imparted on the wing by drag and lift forces. This would be calculated for both single cell and multicell sections. For the first single cell section, we started with finding just the shear flow and determined from our code the following data:


 * {| style="width:100%" border="0"

Single Cell Airfoil
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |

Length of FB = 0.25343 m

q_HF = 626.4307 N/m

q_FB = -16006.5727 N/m

q_BE = 626.4307 N/m

q_EH = 32480.5985 N/m


 * }



By dividing our obtained value of shear flow in each member by the thickness of the panel, 2 mm, would could compute the value of shear stress in the panel. Doing this for the panel EH, we determined $$\tau = 1.624e7\ Pa$$. This value is close to the experimental clamped critical stress for large aspect ratios. This is something we would expect as an aircraft wing has a relatively large aspect ratio.

The image to the right shows the buckling shape when adding the five mode; m = 1,2,1,3,3 and n = 1,2,3,1,3.

= Comparison of WebCT & Mediawiki =

As a team, we are in agreement that Mediawiki and WebCT each have benefits and drawbacks. The overriding comparisons are in the areas of collaboration, layout and privacy.

As a means of collaboration, Mediawiki excels. It is established for the expressed purpose of collaboration. It benefits from common internet accessibility, the history of edits and a common document. When all group members are able to see the same document real time and edit as necessary, it makes things much smoother. The ability to track which user made which edits is very useful for gauging contributions and correcting mistakes. WebCT uses discussion boards for collaboration, which severely limits the scope of projects capable. Since no common document is established, there is no means to understand edits rather than constantly exchanging a master document via email or messaging. Ultimately, for this class, the scope of collaboration did not extend as far as was perhaps intended. In general, the assignment was divided based on lecture days and each person made their own posts with little editing. In particular, collaboration across groups was very limited if it even existed at all.

The layout of WebCT lends itself to a more logical repository of common files, such as syllabus, homework assignments, solutions, grades, etc. The file structure is very straightforward for WebCT. However, Mediawiki doesn't have as good a hierarchical structure. This can be confusing and circular at times.

Because WebCT requires an individual login and provides the ability to privately upload individual assignments, it outperforms Mediawiki in terms of privacy. WebCT is a much more individualized platform as compared to Mediawiki. Also, the universal accessibility of Mediawiki reduces the problems of limited bandwidth often encountered during high traffic periods by WebCT.

Regardless of the platform, we feel that submitting homework and notes as was done in this class is not the most effective use of time. It would have been far easier to write out homework assignments and solutions by hand rather than in an electronic format. It was often frustrating to type out the Latex code for equations, taking far longer than just writing it by hand. Furthermore, drawing figures by means of Visio or Inkscape was also troublesome and frustrating. The return on investment for this time was very small. It served more to frustrate and waste time than to reinforce learning. Mediawiki certainly represents a great resource, but may be better used for presenting a final report or project rather than day-to-day homework assignments.

= Contributing Team Members =


 * Jared Lee --Eas4200c.f08.aero.lee 21:57, 21 November 2008 (UTC)
 * Ray Strods --Eas4200c.f08.aero.strods 20:32, 24 November 2008 (UTC)
 * Oliver Oyama --Eas4200c.fo8.aero.oyama 20:29, 8 December 2008 (UTC)
 * Thomas McGilvray--Eas4200c.f08.aero.mcgilvray 23:24, 8 December 2008 (UTC)
 * William Kurth --Eas4200c.f08.aero.kurth 15:44, 9 December 2008 (UTC)
 * Gonzalo Barcia --Eas4200c.f08.aero.barcia 16:44, 9 December 2008 (UTC)
 * Melisa Gaar -- Eas4200c.f08.aero.gaar 21:02, 9 December 2008 (UTC)