User:Eas4200c.f08.vqcrew.f

$$\frac{W_f}{W_o} = \frac{T}{W_o}\frac{t}{Isp}$$

User:Eas4200c.f08.vqcrew.f/hw3

User:Eas4200c.f08.vqcrew.f/hw4

User:Eas4200c.f08.vqcrew.f/hw5

User:Eas4200c.f08.vqcrew.f/hw6

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 #. 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 #., 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 #., 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 #. 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:


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