User:EAS4200C.F08.WIKI.A/Homework7

=Shear Flow-Open Section=



Mean (Average) Value Theorem (MVT)

$$\int_{A}^{}{\bar{y}dA} = \bar{y}_{c}\int_{A}^{}{dA}$$

$$=\bar{y}A$$

$$\int_{A}^{}{\bar{z}dA} = \bar{z}_{c}A$$

With $$\bar{z}_{c}$$ denoting the average z value.

$$A = \sum_{i=1}^{4}{A_{i}}$$

(Neglect skin and spar webs)

This is a similar problem as was done in homework 5.

Equation 5.5 in the book is given as:

$$q(s) = (k_{yz}Q_{z}(s) - k_{z}Q_{y}(s))V_{z}$$

Since: $$1) V_{z} 2) k_{yz},k_{z} 3) Q_{z}, Q_{y}$$ are all concentrated in the stringers and are all independent of s, this implies that the shear flow q(s) is constant between 2 stringers.

However, q(s) would increment (jump) when crossing a stringer.

Now, to find q(s), the following recipe is given:

Step 1:

Find $$(\bar{y}_{c}, \bar{z}_{c})$$ by the following formulas:

$$\bar{y} = \sum{\frac{yA}{A}}$$

$$\bar{z} = \sum{\frac{zA}{A}}$$

Step 2:

Find $$I_y, I_z, I_{yz}$$

Step 3:

Find $$k_y, k_z, k_{yz}$$

Step 4:

Follow the path of 's' to find $$q_{12}, q_{23}, q_{34}$$

Where the subscripts denote the shear flow from one stringer to the next.

It should be noted that:

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

and so on. In other words, each Q is the sum of the Q's before it.

Superposition revisited
3) Back to Superposition :

$$q_{ij} = q + q_i$$ where $$q_{ij}$$ is true shear flow $$q$$ is the closed cell constant shear flow $$q_i$$ is the open cell piecewise constant shear flow

M.1)W/o stringers


$$R^z=\sum_{i=1}^{n cells}{R^z_i}$$ where $$n=$$ the number of cells

$$R^z_1=0$$

$$R^z=0\;$$

M.2)With Stringers


$$R^z=R^{z1}+R^{z2}\;$$ where $$R^z \;$$ is the resultant of P

$$R^{z1}\;$$ is the resultant of P1 and $$R^{z2}\;$$ is the resultant of P2.

Recalling $$R^z=\sum_{i=1}^{n cells}{R^z_i}=0\;$$ provides us with the formula:

$$R^z=R^{z2}=V_z\neq 0\;$$

Equilibrium of Stringer Three
We will now look at the equilibrium of stringer three. The two and three dimensional drawings of this stringer and its spar webs are shown below.



We will first take the sum of the forces in the x-direction.

 $$\displaystyle \sum{F_x} = 0 = \int_{A_3}^{}{\left[ \sigma_{xx}(x+dx) - \sigma_{xx}(x)\right]dA_3\left[-\tilde{q}_{23} - \tilde{q}_{43} +\tilde{q}_{31}\right]dx}$$

The first term of the integral is the taylor series expansion $$\displaystyle \frac{d\sigma_{xx}}{dx}dx $$ plus the higher order terms. the sumation of the force in the x-direction can therefore be writen as the following;

$$\displaystyle \tilde{q}_{31} - \tilde{q}_{23} +\tilde{q}_{43}+q^{(3)}= 0$$

The term $$\displaystyle q^{(3)}$$ is the contribution to the shear flow by stringer three and equals the following.

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

If you recall that the shear flow in the y- and z-directions are $$\displaystyle V_y = \frac{dM_z}{dx}, V_z = \frac{dM_y}{dx}$$ respectively, the contribution of the shear flow by stringer three can be rewriten as the following;

$$\displaystyle q^{(3)} = -(K_yV_y - K_zV_z)Q_z^{(3)} - (K_zV_z - K_{yz}V_y)Q_y^{(3)}$$

where  $$\displaystyle Q_z^{(3)} = \int_{A_3}^{}{ydA_3}$$   and   $$\displaystyle Q_y^{(3)} = \int_{A_3}^{}{zdA_3}$$

Equilibrium of Stringer 2
The process is the same as before. The results of the equilibrium of stringer two are shown below.

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

$$\displaystyle \tilde{q}_{12} = \tilde{q}_{24} = 0$$ because of cuts

$$\displaystyle Q_z^{(2)} = y_2A_2 $$ and $$\displaystyle Q_z^{(2)} = z_2A_2$$ $$(y_2,z_2)$$ are coordinates of stringer 2

$$\displaystyle q^{(2)}$$ is computed as before.

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

$$\displaystyle \tilde{q}_{24} = \tilde{q}_{41} = 0$$ because of cuts

$$\displaystyle Q_z^{(4)} = y_4A_4 $$ and $$\displaystyle Q_z^{(4)} = y_4A_4$$ $$(y_2,z_2)$$ are coordinates of stringer 4

Equilibrium of Stringer 3 Revisited
Now we can deduce the equation of $$\displaystyle \tilde{q}_{31}$$ because we now know the components of the equation.

$$\displaystyle \tilde{q}_{31} - \tilde{q}_{23} + \tilde{q}_{43}+q^{(3)}= 0$$

$$\displaystyle \tilde{q}_{31} = \tilde{q}_{23} - \tilde{q}_{43} + q^{(3)} = q^{(2)} - q^{(4)} + q^{(3)}$$

Total Shear in the Potato
With the information provided above we can no go back and use superposition in and add all the components to find the shear flow in the original potato.

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

$$\displaystyle q_{12} = \tilde{q}_{12} + q_1 = 0 + q_1$$

$$\displaystyle q_{24} = \tilde{q}_{24} + q_2 = 0 + q_2$$

$$\displaystyle q_{43} = \tilde{q}_{43} + q_4 = q^{(4)} + q_4$$

$$\displaystyle q_{31} = \tilde{q}_{31} + q_3 = q^{(2)} - q^{(4)} + q^{(3)} + q_3$$

$$\displaystyle q_{41} = \tilde{q}_{41} + q_4 = 0 + q_4$$

We now have three unknowns $$\displaystyle q_1,q_2,q_3$$ so we need three equations. the first equation is the moment equation. You need to take the moment of $$\displaystyle V_y, V_z $$ and $$\left\{q_{12},...........q_{41} \right\}$$ about any point. this point is usually where the lines of action of $$\displaystyle V_y, V_z $$ intersect. The second and third equations are the compatibility equations which are shown below.

<p style="text-align:center;">$$\displaystyle \theta_1 = \theta_2$$

<p style="text-align:center;">$$\displaystyle \theta_2 = \theta_3$$

Another Method
Solving P2: For each cell

-Follow path $$S_i$$

-Find the equilibrium of each stringer on this path: There are two ways of solving

1) Complete method FBD as seen before

2) Consequence of the first method:

<p style="text-align:center;">$$\tilde{q}_{j6}=\tilde{q}_{2j}-\tilde{q}_{j5}-\tilde{q}_{j8}+\tilde{q}^{j}$$



In this example, $$\tilde{q}_{6j}$$ is now going in the opposite direction as previous but the answer will remain the same as long as the conventions stay consistent.

<p style="text-align:center;">$$-\tilde{q}_{6j}=\tilde{q}_{2j}-\tilde{q}_{j5}-\tilde{q}_{j8}+\tilde{q}^{j}$$



Note: What if we cut cell walls such that one stringer was isolated?





<p style="text-align:center;">$$\tilde{q}_{31}=\tilde{q}_{23}-\tilde{q}_{34}+\tilde{q}^{3}$$

This equation says that $$\tilde{q}^{3}= 0$$ which is not true. Therefore you cannot isolate the stinger in this way because it results in an answer that is not valid.

Trying to make different cuts to isolate stringer 2 will result in the same false answer.

<p style="text-align:center;"> $$\tilde{q}_{31}=\tilde{q}_{12}-\tilde{q}_{24}+\tilde{q}^{2}$$

<p style="text-align:center;">$$\tilde{q}^{2}= 0$$

Cont'd from Mtg 39
Similarly, developing an equilibrium equation for stringer one yields

<p style="text-align:center;"> $$\displaystyle \tilde q_{12} = \tilde q_{31} + \tilde q_{41} + q^{(1)} $$

where $$\displaystyle \tilde q_{12} $$ and $$\displaystyle \tilde q_{41} $$ are unknown. $$\displaystyle \tilde q_{31} $$ becomes zero due to the cut between stringers three and one. Doing the same for stringer two yields

<p style="text-align:center;"> $$\displaystyle \tilde q_{24} = \tilde q_{12} - \tilde q_{23} + q^{(2)} $$

In this case, $$\displaystyle \tilde q_{23} $$ becomes zero due to the cut between stringers two and three. Again, developing the equation for stringer four yields

<p style="text-align:center;"> $$\displaystyle \tilde q_{41} = \tilde q_{24} + \tilde q_{34} + q^{(4)} $$

Substituting the equations for stringers one and two into this equation gives us

<p style="text-align:center;"> $$\displaystyle \tilde q_{41} = ( \tilde q_{41} + q^{(1)} ) + q^{(2)} + \tilde q_{34} + q^{(4)} $$

Due to the cut between stringers three and four, $$\displaystyle \tilde q_{34} $$ becomes zero. In addition, $$\displaystyle \tilde q_{41} $$ cancels out, leaving

<p style="text-align:center;"> $$\displaystyle 0 = q^{(1)} + q^{(2)} + q^{(4)} = -q^{(3)} $$

This, of course is not true, or even possible. The following steps explain why this is not true.

<p style="text-align:center;"> $$\displaystyle 0 = q^{(1)} + q^{(2)} + q^{(3)} + q^{(4)} $$

<p style="text-align:center;"> $$\displaystyle 0 = \sum_{e=1}^4 q^{(e)} $$

Recall,

<p style="text-align:center;"> $$\displaystyle q^{(e)} = n_z Q_z^{(e)} + n_y Q_y^{(e)} $$

<p style="text-align:center;"> $$\displaystyle n_z := -(K_y V_y - K_yz V_z) $$

<p style="text-align:center;"> $$\displaystyle n_y := -(K_z V_z - K_yz V_y) $$

Thus,

<p style="text-align:center;"> $$\displaystyle \sum_{e=1}^4 q^{(e)} = n_z \sum_{e=1}^4 Q_z^{(e)} + n_y \sum_{e=1}^4 Q_y^{(e)} $$

Where $$\displaystyle \sum_{e=1}^4 Q_z^{(e)} $$ and $$\displaystyle \sum_{e=1}^4 Q_y^{(e)} $$ must both be equal to zero.



$$\displaystyle Q_y $$ can be expressed as

<p style="text-align:center;"> $$\displaystyle Q_y (\hat z) = \int_{A(\hat z)} z dA $$

<p style="text-align:center;"> $$\displaystyle Q_y = \int_A z dA = 0 = z_c \int_A dA $$

Notice that the last part of the above equation is from the mean value theorem. $$\displaystyle z_c $$ would be zero in this case, which is not possible.

Steps for Shear Buckling
The steps for plotting the buckling shape and expressing the coefficients $$\displaystyle \{ C_{11}, C_{22}, C_{13}, C_{31}, C_{33} \} $$ in terms of $$\displaystyle C_{11} $$ for the aspect ratio $$\displaystyle \vartheta = 1.5 $$ are as follows.

1) Find $$\displaystyle \lambda $$ for $$\displaystyle \vartheta = 1.5 $$ using equation 30 from Professor Vu-Quoc's page on plate buckling.

<p style="text-align:center;"> $$  \displaystyle \lambda =  \left[ \frac{\vartheta^4}{81 (1 + \vartheta^2)^4} \left\{ 1 	 + 	 \frac{81}{625} +	 \frac{81}{25} \left(	   \frac{1 + \vartheta^2}{1 + 9 \vartheta^2}	 \right)^2 +	 \frac{81}{25} \left(	   \frac{1 + \vartheta^2}{9 + \vartheta^2}	 \right)^2 \right\} \right]^{1/2} $$

2) Evaluate $$\displaystyle \mathbf K _{5 \times 5} $$ numerically using equation 26 from Professor Vu-Quoc's page on plate buckling.

<p style="text-align:center;"> $$  \displaystyle \mathbf \bar K _{5 \times 5} = \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]

$$

3) Solve $$\displaystyle \{ C_{11}, C_{22}, C_{13}, C_{31}, C_{33} \} $$ in terms of $$\displaystyle C_{11} $$.

<p style="text-align:center;"> $$  \displaystyle \left[ \begin{array}{lllll} K_{22} &	 K_{23} &	 K_{24} &	 K_{25} \\	 K_{32} &	 K_{33} &	 K_{34} &	 K_{35} \\	 K_{42} &	 K_{43} &	 K_{44} &	 K_{45} \\	 K_{52} &	 K_{53} &	 K_{54} &	 K_{55} \end{array} \right] \left\{ \begin{array}{l} C_{22} \\	 C_{13} \\	 C_{31} \\	 C_{33} \end{array} \right\} =  \left\{ \begin{array}{l} -{4 \over 9} C_{11} \\	 0	 \\	 0	 \\	 0     \end{array} \right\} $$

<p style="text-align:center;"> $$ \displaystyle u_z = C_{11} \sin{\pi x \over a} \sin{\pi y \over b} + C_{22} \sin{2 \pi x \over a} \sin{2 \pi y \over b} + C_{13} \sin{\pi x \over a} \sin{3 \pi y \over b} + C_{31} \sin{3 \pi x \over a} \sin{\pi y \over b} + C_{33} \sin{3 \pi x \over a} \sin{3 \pi y \over b} $$

Setting $$\displaystyle C_{11} = 1 $$, $$\displaystyle u_z(x,y) $$ can be plotted.

Proof of Stress Equation
[READ AND REPORT SEC. 4.2 (eqn 4.22a to eqn 4.28c)]

The following is a proof for the stress equation:

<p style="text-align:center;"> $$\sigma _{xx}=\begin{bmatrix} z & y \end{bmatrix} \begin{bmatrix} K_y & -K_{yz}\\ -K_{yz} & K_z \end{bmatrix}\begin{Bmatrix} M_y\\M_z \end{Bmatrix}$$

First, the moments can be written as

<p style="text-align:center;"> $$\displaystyle M_y = - EI_{yz} {d^2 V_o \over dx^2} - EI_y {d^2 W_o \over dx^2} $$

<p style="text-align:center;"> $$\displaystyle M_z = - EI_{z} {d^2 V_o \over dx^2} - EI_{yz} {d^2 W_o \over dx^2} $$

Let the curvatures $$\displaystyle {d^2 V_o \over dx^2} = \chi_y $$ and $$\displaystyle {d^2 W_o \over dx^2} = \chi_z $$

In matrix form, whe have

<p style="text-align:center;"> $$\displaystyle \begin{Bmatrix} M_y\\M_z \end{Bmatrix} = \begin{bmatrix} I_y & I_{yz}\\ I_{yz} & I_z \end{bmatrix}\begin{Bmatrix} -E \chi_z \\ -E \chi_y \end{Bmatrix} $$

<p style="text-align:center;"> $$\displaystyle \varepsilon_{xx} = -y \chi_y - z \chi_z = \begin{bmatrix} z & y \end{bmatrix} \begin{Bmatrix} \chi_z \\ \chi_y \end{Bmatrix} $$

Thus,

<p style="text-align:center;"> $$\displaystyle \sigma_{xx} = E \varepsilon_{xx} = E \begin{bmatrix} z & y \end{bmatrix} \begin{Bmatrix} -\chi_z \\ -\chi_y \end{Bmatrix} = E \begin{bmatrix} z & y \end{bmatrix} {1 \over E} \textbf{I}^{-1} \begin{Bmatrix} M_y \\ M_z \end{Bmatrix} $$

Notice that $$\displaystyle E $$ cancels out. Also note that

<p style="text-align:center;"> $$\displaystyle \textbf{I}^{-1} = {1 \over D} \begin{bmatrix} I_z & -I_{yz}\\ -I_{yz} & I_y \end{bmatrix} $$

<p style="text-align:center;"> $$\displaystyle D= I_y I_z - (I_{yz})^2 $$

Recall the expression for the shear flow

<p style="text-align:center;"> $$\displaystyle q = -\int_A {d \sigma_{xx} \over dx} dA $$

Thus,

<p style="text-align:center;"> $$\displaystyle {d \sigma_{xx} \over dx} = \begin{bmatrix} z & y \end{bmatrix} \textbf{I}^{-1} \begin{Bmatrix} {dM_y \over dx} \\ {dM_z \over dx} \end{Bmatrix} $$

<p style="text-align:center;"> $$\displaystyle V_y = {dM_y \over dx} $$

<p style="text-align:center;"> $$\displaystyle V_z = {dM_z \over dx} $$

<p style="text-align:center;"> $$\displaystyle Q_y = \int_A zdA $$

<p style="text-align:center;"> $$\displaystyle Q_z = \int_A ydA $$

= MATLAB Airfoil Coding =

Single Cell Airfoil
EDU>> NewAirfoil(2415,.5,[0 0 0])

Single Cell

q_0 =

2.9437e+004

q_hf =

2.9437e+004

q_fb =

-2.2512e+005

q_be =

3.2195e+004

q_eh =

2.8555e+005

Note: All values in N/m

Multi Cell Airfoil
EDU>> NewAirfoil(2415,.5,[0 0 0])

Multi Cell

q_1 =

3.2721e+004

q_2 =

2.6482e+004

q_3 =

3.7939e+004

q_hf =

3.7939e+004

q_fb =

-2.2807e+005

q_be =

3.5480e+004

q_eh =

2.8260e+005

q_w1 =

-6.2391e+003

q_w2 =

1.1456e+004

Note: All values in N/m

Also, it can be seen here that the shear flow values in the skin panels are similar between the single and multi cell cases. This verifies the validity of the two cases.

= Buckling Analysis =

<p style="text-align:center;">$$ \displaystyle K_{2\times 2} = \left[ \begin{array}{ll} \frac{\lambda (1 + \vartheta^2)^2}{\vartheta^2} &	 \frac{4}{9} \\	 \frac{4}{9} &	 \frac{16 \lambda (1 + \vartheta^2)^2}{\vartheta^2} \end{array} \right] $$

<p style="text-align:center;"> $$ \displaystyle \det K_{2\times 2} = \frac{16\lambda ^2(1+\vartheta ^2)^4}{\vartheta ^4}-\frac{16}{81} = 0 $$

<p style="text-align:center;"> $$ \displaystyle \lambda ^2 = \frac{\vartheta ^4}{81(1+\vartheta ^2)^4} $$

<p style="text-align:center;"> $$ \displaystyle \lambda = \pm \frac{1}{9}\frac{\vartheta ^2}{(1+\vartheta ^2)^2}$$

<p style="text-align:center;">

=Mediawiki vs. E-learning=

During this course, we used wikimedia to turn in our homework. Our homework consisted of writing up the class notes and doing some matlab assignments. We were asked to form teams to work on this Wikimedia homework. The format of these homework assignments forced us to attend class once every two weeks in order to get the notes for our assigned day. When doing this type of work, Wikimedia is a much more convenient and usable program to upload and format than E-learning. E-learning was not designed to do what was require for this course. Wikimedia's ability to save all archived versions of pages as well as the ability to easily interface with equation editors made it user friendly for students. It also allowed for easier grade formatting and uploading. I see E-learning as an older, less usable and less versatile version of Wikimedia.

We do believe that Wikimedia is a more versatile program than E-learning and was a much better decision for us to accomplish our homework as assigned. However, We do not believe that Wikimedia allowed me to learn Aerospace structures effectively. We would prefer to have traditional homework problems that would allow us to practice the method rather than regurgitating theory that we may or may not understand. Also, assigning Matlab to try to patch this significant problem only compounded it because most of us have never used Matlab, adding another program we needed to learn to succeed in this class, without learning any actual Aerospace Structures. We believe assigning homework in this way has hindered our learning. An argument could be made that we can always look back at these pages to find information about structures. We would respond to this argument by saying that none of our homework was graded carefully enough to catch errors or that it was written with any understanding or authority of the subject it was discussing. We would much rather turn to a validated source like a textbook than look on Wikiveristy. We feel that this style hindered our learning experience. Because we have learned near nothing in the class, we would venture to say that the homework's effectiveness on our learning experience was an utter failure.

Eas4200c.f08.WIKI.E 21:10, 9 December 2008 (UTC)

=Individual Contributions=

Christopher Pergola EAS4200C.F08.WIKI.A 17:53, 9 December 2008 (UTC)

Braden Barnes Eas4200c.f08.wiki.b 22:59, 8 December 2008 (UTC)

Eric Viale Eas4200c.f08.wiki.c 01:11, 9 December 2008 (UTC)

Michael Rolle Eas4200c.f08.wiki.d 05:26, 9 December 2008 (UTC)

Chris Fontana Eas4200c.f08.WIKI.E 20:25, 9 December 2008 (UTC)

Jeff Huenink Eas4200c.f08.wiki.f 21:32, 9 December 2008 (UTC)