User:Eas4200c.f08.aeris.mcrae

Problem 1.1
Problem Statement: We have a hollow rectangular beam, of side lengths a and b, and wall thickness t. We wish to find optimal b that will maximize the load-bearing capacity of the beam.

Assumptions:
 * $$\ t << a $$ (given)
 * $$\ t << b $$ (given)
 * $$\ L = 2 (a+b) $$ (given)
 * $$\ M = T $$ (given)
 * $$\ \sigma_{allowable} = 2 \tau_{allowable} $$ (given)

Governing Equations:
 * $$\ \sigma = \frac{T}{2 a b t}\,$$

This problem must be solved for 2 separate cases to arrive at a conclusion. The first case is where $$ \sigma_{max} = \sigma_{allowable}$$, and the second case is where $$\tau_{max} = \tau_{allowable}$$.

Case 1: Assume $$\sigma_{max} = \sigma_{allowable}$$.

Recall that $$\sigma = \frac{M z}{I}\,$$ with $$z = \frac{b}{a}\,$$.

We can now solve for M, which as can be seen below is a function of the ratio of I to b.
 * $$M = \frac{2 I \sigma_{max}}{b}\,$$
 * $$M = \frac{2 I \sigma_{allowable}}{b}\,$$
 * $$M = 2 \sigma_{allowable}* \frac{I}{b}\,$$

In order to maximize M we need to either maximize I, minimize b, or do a combination of both. Before we look at this further, we need to get an expression for I in terms of the given variables.


 * $$ I = \sum_{i=1}^4 \frac{b_i h_i^3}{12}\ + A_i d_i^2$$
 * $$ I = 2[\frac{a t^3}{12} + a t (\frac{b}{2})^2] + 2(\frac{t b^2}{12})$$
 * $$ I = \frac{a t}{12} (t^2 + 3 b^2) + 2(\frac{t b^2}{12})$$

Going back to the given assumptions, we recall that
 * $$\ t << b $$ --> $$\ t^2 << b^2 $$ --> $$\ t^2 << 3 b^2 $$

We can now assume that the wall thickness is negligible, and neglecting the t term we can now substitute our I back into the term we are trying to maximize, $$ \frac{I}{b}$$.


 * $$ \frac{I}{b} = \frac{t b}{6} (3 a + b)$$

Recalling that $$ a = \frac{L}{2} - b$$, we can simplify to just one variable, b.


 * $$ \frac{I}{b} = \frac{t b (3 L - 4 b)}{12}$$

We note that this is a downward opening parabola, and using some simple calculus we solve for the vertex at $$ b = \frac{3 L}{8}$$.

Substituting back into the original assumption $$\ L = 2 (a+b) $$, we find that $$ a = \frac{L}{8}$$.

Thus, in Case 1 ($$ \sigma_{max} = \sigma_{allowable} $$) we find that $$ a = \frac{L}{8}$$ and $$ b = \frac{3 L}{8}$$.

Case 2: Assume $$\tau_{max} = \tau_{allowable}$$ ($$\tau_{max}$$ reaches $$\tau_{allowable}$$ first).


 * $$ \tau = \frac{T}{2abt} = \tau_{max} = \tau_{allow} $$

Solving for T, we find that $$\ T = T_{max} = (2t \tau_{allow})(ab)_{max} $$.

Note that here $$ ab_{max} $$ is a variable, and we may treat $$ 2t /tau_{allow} $$ as a constant.

STILL NEED ---> show a = b = L/4 <---

STILL NEED ---> check to see if Case 2 (and maybe Case 1) is acceptable (see notes) <---

Homework Statement from 9/10
Find the moment of inertia, with respect to y, of a circular cross section and a u-shaped cross section, as shown below.

CASE 1: Beam with circular cross section.

We begin with the definition of moment of inertia, I. Specifically here we are calculating moment of inertia with respece to the y-axis, Iy.


 * $$\ I_{y}=\int r^{2} dm $$

Solving for this integral gives us the moment of inertia of the circle in Case 1.


 * $$\ I_{y}=\int r^{2} dm$$


 * $$\ I_{y}=\int r^{2} \pi r dr$$


 * $$\ I_{y}=\int \pi r^{3} dr$$


 * $$\ I_{y}=\frac{\pi}{4} r^{2}$$

CASE 2: Beam with u-shaped cross section.

As can be seen in the diagram, we can divide the "u-shaped" beam into 3 separate parts, with the symmetrical side piece being equal. We can calculate the moments of inertia of these three parts separately, and then combine them using the Parallel Axis Theorem.

We know that the moment of inertia of a rectangle is equal to $$\frac{b^{3} h}{12}$$.

From here we can calculate the moment of inertia for each of the 3 pieces, centered about each centroid.

For the 2 symmetrical pieces.


 * $$\ I_{y}=\frac{b^{3} h}{12}$$


 * $$\ I_{y}=\frac{c^{3} b}{12}$$

Using the Parallel Axis Theorem, we can move the centroid of these moments to the centroid about the x-axis.


 * $$\ I_{y}= \frac{c^{3} b}{12} + \frac{a + c^{2} b c}{4}$$

Noting we have two of these pieces we can multiply this moment by 2.


 * $$\ I_{y}=2 \frac{c^{3} b}{12} + 2 \frac{a + c^{2} b c}{4}$$

For the bottom piece.


 * $$\ I_{y}=\frac{b^{3} h}{12}$$


 * $$\ I_{y}=\frac{a^{3} c}{12}$$

Because the centroid of this piece is already on the x-axis, we do not need to use the Parallel Axis Theorem here.

Combining the pieces

We can combine the moments of the calculated pieces to obtain the total moment of inertia with respect to the y-axis, Iy.


 * $$\ I_{y,total}=\frac{c^{3} b}{6} + \frac{a^{3} c}{12} + \frac{(a + c)^{2} b c}{2}$$

HW3 - 9/22
Open thin-walled sections are very common members in aircraft structures. Sheer flow travels through these members, as can be seen by the directional arrows in the figure below.

A sheer force is induced through this member due to the sheer flow through the member.


 * $$\ dF=q*dl=q(dl_{x}\hat{i}+dl_{z}\hat{k}) $$


 * $$\ dF=q*dl=q(dl*cos(\theta)*\hat{i}+dl*sin(\theta)*\hat{k}) $$

We now integrate, from A to B as given in the figure above, to find the resultant force.


 * $$\ F=\int_{A}^{B}{dF}=q\left[\left( \int_{A}^{B}{dx}*\hat{i}\right) + \left( \int_{A}^{B}{dz}*\hat{k}\right) \right]$$


 * $$\ F=q\left(a*\hat{i}+b*\hat{k} \right)$$

NOTE: q is a constant with respect to x & z.

Now that we have the resulting sheer force vector we can calculate the resultant magnitude of this vector. Noting that this sheer force is broken down into an x and z component, we use Pythagorean Theorem to find the resulting magnitude.


 * $$\ \left| \left| F\right|\right|=\sqrt{F_{x}^2+F_{z}^2} $$


 * $$\ \left| \left| F\right|\right|=\sqrt{(q*a*\hat{i})^2+(q*b*\hat{k})^2}$$


 * $$\ \left| \left| F\right|\right|=q*\sqrt{a^{2}+b^{2}} $$

Strain Matrix (October 15th)
In a 3-D body under strain, this strain can be broken down into a component matrix, as seen below.

$$\varepsilon =\begin{bmatrix} \varepsilon _{xx} & \varepsilon _{xy} & \varepsilon _{xz}\\ \varepsilon _{yx} & \varepsilon _{yy} & \varepsilon _{yz}\\ \varepsilon _{zx} & \varepsilon _{zy} & \varepsilon _{zz} \end{bmatrix} =\begin{bmatrix} \varepsilon _{11} & \varepsilon _{12} & \varepsilon _{13}\\ \varepsilon _{21} & \varepsilon _{22} & \varepsilon _{23}\\ \varepsilon _{31} & \varepsilon _{32} & \varepsilon _{33} \end{bmatrix} =\left[\varepsilon _{ij} \right]$$

Due to symmetry, there are a several equalities we can make.

$$\varepsilon _{xy}=\varepsilon _{yx}$$

$$\varepsilon _{yz}=\varepsilon _{zy}$$

$$\varepsilon _{xz}=\varepsilon _{zx}$$

We can now see that this 3x3 strain matrix has only 6 components, $$\varepsilon _{xx}, \varepsilon _{yy},, \varepsilon _{zz}, \varepsilon _{xy}, \varepsilon _{yz}, \varepsilon _{xz}$$.

The equation for each component can be expressed as $$\varepsilon _{ij}=\frac{1}{2}\left(\frac{\partial u_{i}}{\partial x_{j}}+\frac{\partial u_{j}}{\partial x_{i}}\right)$$

Stress Tensor Matrix

Similarly, we can make some equalities to the stress tensor matrix, which is seen below.

$$\sigma =\begin{bmatrix} \sigma _{xx} & \sigma _{xy} & \sigma _{xz}\\ \sigma _{yx} & \sigma _{yy} & \sigma _{yz}\\ \sigma _{zx} & \sigma _{zy} & \sigma _{zz} \end{bmatrix} =\begin{bmatrix} \sigma _{11} & \sigma _{12} & \sigma _{13}\\ \sigma _{21} & \sigma _{22} & \sigma _{23}\\ \sigma _{31} & \sigma _{32} & \sigma _{33} \end{bmatrix} =\left[\sigma _{ij} \right]$$

$$\sigma _{xy}=\sigma _{yx}$$

$$\sigma _{yz}=\sigma _{zy}$$

$$\sigma _{xz}=\sigma _{zx}$$

We see that the stress tensor matrix, like the strain matrix, also has 6 components.

HW5 MatLAB Problem Statement
In this assignment we will modify the MatLAB code from the previous homework to include stringers attaching the spars to the flange, as can be seen in the figure below.



Given an input NACA airfoil (NACA 2415 shown in the above figure), we are to compute the centroid of the stringers (as well as the airfoil), the components of the moment of inertia tensor, the highest normal bending stress and location of this stress, and the ultimate bending moments of the airfoil.

We first compute this data neglecting the bending effects of the skin and the partition walls (spar webs). We are then to recompute the data, only this time NOT neglecting the bending effects of the skin or partition walls (or the stringers themselves). A comparison of the difference will then be made so we can access the contribution of these bending effects.

In order to make these calculations, we need to know some initial data on the stringers to be used. We first make an assumption on the bending moments, shown below.

$$\ M_{y}=-125,000 N*cm $$

$$\ M_{x}=50,000 N*cm $$

We also need to know the cross sectional area of these stringers, which is given below.

$$\ A_{B}=2 cm^{2} $$

$$\ A_{E}=2 cm^{2} $$

$$\ A_{H}=1 cm^{2} $$

$$\ A_{F}=1 cm^{2} $$

For the second part of our calculations, we are to take into account the bending affects of the skin, partition walls, and stringers themselves. We can note that the stringers are essentially stamped out sheet metal which is then bent into the L-shape as seen in the first figure. Because of this, thickness of the stringers are uniform, and we let this thickness equal 5 mm, or 0.005 m. Knowing this, we can compute the width of the metal sheet used to manufacture the stringers, as seen below.

$$\ \frac{\frac{0.0002}{2}}{0.5}=0.02m=2cm $$

Because the thin skin would buckle under high normal bending stress acting on the plane of the skin, we will next find the highest normal bending stress in any of the stringers, and take a closer look at that stringer. Assuming the ratio of My and Mz are constant, we will then set this maximum normal bending stress to the ultimate stress for our material, steel 300M. Upon doing this, we are now able to compute the ultimate bending moments My and Mz for our full blown airfoil (which we will be imputing as the NACA 2415 airfoil).

11/14 - Meeting 34


General equation for asymmetric cross sections:
 * $$\ \int_{A_{s}}^{}{ \frac{d \sigma_{xx}}{dx}dA} = -q_{s}$$

Symmetric Case:


 * $$\ \sigma_{xx} = \frac{M_{y}z}{I_{y}} $$


 * $$\ q_{s} = - \frac{V_{z}}{I_{y}} \int_{A_{s}}^{}{zdA} = \frac{-V_{z}Q_{y}}{I_{y}} $$        (eq. 5.2)


 * $$\ Q_{y} = \int_{A_{s}}^{}{zdA} = z_{c}A_{s} $$

Asymmetric Case:


 * $$\ \sigma _{xx} = \left(k_{y}M_{z} - k_{yz}M_{y} \right)y + \left(k_{z}M_{y} - k_{yz}M_{z} \right)z $$         (eq. 5.3)

with


 * $$\ k_{y} = \frac{I_{y}}{D} $$

and


 * $$\ k_{yz} = \frac{I_{yz}}{D} $$

and


 * $$\ k_{z} = \frac{I_{z}}{D} $$

This can be also written in matrix form:


 * $$\ \sigma _{xx} = \begin{bmatrix} y & z \end{bmatrix}\begin{bmatrix} k_{y} & -k_{yz}\\ -k_{yz} & k_{z} \end{bmatrix} \begin{Bmatrix} M_{z} \\ M_{y} \end{Bmatrix} $$


 * $$\ \sigma _{xx} = \begin{bmatrix} y & z \end{bmatrix}\begin{bmatrix} k_{y}M_{z} & -k_{yz}M_{y}\\ -k_{yz}M_{z} & k_{z}M_{y} \end{bmatrix} $$

We can also flop the order of variables, and come up with the preferred form of this equation:


 * $$\ \sigma _{xx} = \begin{bmatrix} z & y \end{bmatrix}\begin{bmatrix} k_{z} & -k_{yz}\\ -k_{yz} & k_{y} \end{bmatrix} \begin{Bmatrix} M_{y} \\ M_{z} \end{Bmatrix} $$

We particularize to symmetric cross section, and find that $$\ I_{yz} = 0 $$. Proof of this is shown in the tab below.

Consider $$\ M_{z} = 0 $$

We know the following relations.


 * $$\ I_{yz} = 0 $$
 * $$\ D = I_{y}I_{z} $$
 * $$\ k_{yz} = 0 $$
 * $$\ k_{z} = \frac{1}{I_{y}} $$

Combining these equations we find that $$\ \sigma_{xx} = \frac{M_{y}z}{I_{y}} $$.

Asymmetric Case


 * $$\ q_{s} = -\left( k_{y}V_{y} - k_{yz}V_{z} \right) Q_{z} - \left( k_{z}V_{z} - k_{yz}V_{y} \right) Q_{y} $$          (eq. 5.5)


 * $$\ Q_{z} = \int_{A_{s}}^{}{y*dA} $$


 * $$\ Q_{y} = \int_{A_{s}}^{}{z*dA} $$

For stringer web sections, if the thickness (t) of the skin and spar webs is very small we can neglect the computations of Iy, Iz, Iyz, Qy, and Qz. We use only the areas of the stringers.

Unsymmetric Thin-Walled Cross Sections


We know that for a symmetric thin-walled cross section that $$\ q_{s} = -\int\int_{A_{s}}^{}{\frac{d\sigma_{xx}}{dx}dA} $$

For unsymmetric thin-walled cross sections under bidirectional bending, the bending stress $$\left( \sigma_{xx}\right)$$ must be taken into account. Doing so, we arrive at the following equation for sheer flow.


 * $$\ q_{s} = -\left(k_{y}V_{y} - k_{yz}V_{z} \right)Q_{z} + \left(k_{z}V_{z} - k_{yz}V_{y} \right)Q_{y} $$

We can begin our analysis of the figures to the right using the Mean Value Theorem to find the first moments of area around the z and y axes.


 * $$\ Q_{z} = \int \int_{A_{s}}^{}{ydA} = \bar{y}A $$


 * $$\ Q_{y} = \int \int_{A_{s}}^{}{zdA} = \bar{z}A $$

For the case of stringers of different areas, as shown in Figure 1, the area to be used in this equation is $$\ A = \sum_{i=1}^{4}{A_{i}} $$. For the case shown in Figure 2, the area used in this equation is $$\ A = A_{\hat{s}} + A_{\hat{s},L} $$.

Using this, as well as equations from the previous homework, we can break down analysis of unsymmetric thin-walled cross sections, such as the two shown on the right in Figure 1 and Figure 2, into the following four steps.


 * 1. Find $$ \left(\bar{y}, \bar{z} \right) $$.
 * 2. Find $$ I_{y} $$, $$ I_{z} $$, and $$ I_{yz} $$.
 * 3. Find $$ k_{y} $$, $$ k_{z} $$, and $$ k_{yz} $$.
 * 4. Follow path $$ s $$ to find $$ q_{12} $$, $$ q_{23} $$, and $$ q_{34} $$.

Shear Flow
In order to plot the buckling shape, we express the coefficients $$\left\{\begin{matrix} C_{22} & C_{13} & C_{31} & C_{33} \end{matrix} \right\}$$ in terms of $$ C_{11} $$. Using an aspect ratio $$\left( \vartheta \right)$$ equal to 1.5 we can break our analysis down to the following four steps.


 * 1. Find $$ \lambda $$.


 * $$ \displaystyle \lambda = \left[ \frac{\vartheta^4}{81 (1 + \vartheta^2)^4} \left\{1 + \frac{81}{625} + \frac{81}{25} \left( \frac{1 + \vartheta^2}{1 + 9 \vartheta^2} \right)^2 + \frac{81}{25} \left( \frac{1 + \vartheta^2}{9 + \vartheta^2} \right)^2 \right\} \right]^{1/2} = 0.0288 $$


 * 2. Evaluate $$ K_{5x5} $$ numerically.



\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. Truncate $$ K_{5x5} $$ into $$ K_{4x4} $$.



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


 * 4. Solve for $$\left\{\begin{matrix} C_{22} & C_{13} & C_{31} & C_{33} \end{matrix} \right\}$$ in terms of $$ C_{11} $$, noting that $$ K^{-1} $$ is the inverse of $$ K $$.


 * $$\begin{Bmatrix}

C_{22} \\ C_{13} \\ C_{31} \\ C_{33} \end{Bmatrix} = K^{-1} \begin{Bmatrix} -\frac{4}{9} C_{11} \\ 0 \\ 0 \\ 0 \end{Bmatrix}$$


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

Stability Equations

 * $$\ \dot{x} = Ax + B \eta $$


 * $$\ \begin{bmatrix}

\Delta \dot{u}\\ \Delta \dot{\beta}\\ \Delta \dot{\alpha}\\ \Delta \dot{q}\\ \Delta \dot{r}\\ \Delta \dot{\theta}\\ \Delta \dot{\psi}\\ \end{bmatrix} = A \begin{bmatrix} u\\ \beta\\ \alpha\\ q\\ r\\ \theta\\ \psi\\ \end{bmatrix} + B \begin{bmatrix} pitch flap\\ yaw flap\\ \end{bmatrix} $$


 * $$\ A = \begin{bmatrix}

-0.40617261082 & 0 & 0 & 0 & 0 & -0.00807575185 & 0\\ 0 & -7.34602885379 & 18.84954000000 & 0 & -1 & 0 & -0.00807575185\\ 0 & 18.84954000000 & -9.38398809293 & 1 & 0 & -0.00807574113 & 0\\ 0 & 0 & -11351.24919089994 & -16.41759631431 & 17.95643978307 & 0 & 0\\ 0 & 31751.40272182580 & 0 & -17.95643978307 & -15.42011712577 & 0 & 0\\ 0 & 0 & 0 & 1 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 1 & 0 & 0 \end{bmatrix} $$


 * $$\ B = \begin{bmatrix}

0 & 0\\ 0 & -0.000167397975141\\ -0.000164984586141 & 0\\ -2.899278907682155 & 0\\ 0 & 2.198365483499154\\ 0 & 0\\ 0 & 0 \end{bmatrix} $$


 * $$\ C = \begin{bmatrix}

1 & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 1 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 1 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 1 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 1 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 0 & 0 & 0 & 1 \end{bmatrix} $$


 * $$\ D = \begin{bmatrix}

0 & 0\\ 0 & 0\\ 0 & 0\\ 0 & 0\\ 0 & 0\\ 0 & 0\\ 0 & 0 \end{bmatrix} $$