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=EAS4200c: Aerospace Structures - Class Notes=

HW 1
See Homework 1 Report

Lecture 6 - 9/8/08
Problem 1.1 Continued

$$\displaystyle \sigma = \frac{M z}{I}$$ ,

$$\displaystyle =>\sigma_{max}= \frac{M b}{2 I}$$

$$\displaystyle => M = \frac{2 I \sigma_{max}}{b} = 2\sigma_{allowable} (\frac{I}{b})$$

$$M_{max} = 2 \sigma_{allowable} (\frac{I}{b})_{max}$$

$$a = \frac{L}{2} - b$$

$$ (\frac{I}{b})_{max} = \frac{I}{b} => M_{max} = 2\sigma_{all}(\frac{I}{b}) = T_{max}$$

$$\tau_{max} = \frac{T_{max}}{2abt} = \frac{M_{max}}{2abt}$$

$$\tau_{allowable} = \frac{\sigma_{allowable}}{2}$$

Verify:

Case 1: $$\tau_{max} < \tau_{allowable}$$ - this case is acceptable (b/a) ratio

Case 2: $$\tau_{max} > \tau_{allowable}$$ - this case is not acceptable (b/a) ratio

Assume that the torsional shear stress $$\tau$$ is constant is along the wall.

Lecture 7 - 9/10/08
Problem 1.1 Continued

Case 1: Assume σ max = σ allowable

Recall: $$\displaystyle \sigma = \frac{M z}{I}$$ ,       $$z = \frac{b}{2}$$

Solving for Moment: $$M = \frac{2 I \displaystyle \sigma}{b}$$

$$M = 2\displaystyle \sigma ( \frac{I}{b} )$$   :(I/b) is a function of the length dimensions for the cross section (a,b)

Recall Assumptions:
 * 1) L = 2(a+b) = Constant
 * 2) M = T
 * 3) σ = 2τ

Solving for dimension a : a = (L / 2) - b. Solving the parameter for a will give us the opportunity to solve the moment equation as a function of only b.

Now maximize the moment equation:  $$M = 2\displaystyle \sigma ( \frac{I}{b} )$$ max. This equation shows that in order to maximize the Moment (M), the ratio of (I/b) must be maximized.

I = Second Moment of Inertia. I can be found by using the Parallel Axis Theorem.

I = I y + Ad 2

Iundefined is the area moment of inertia through the parallel axis to the center of mass

Iy is the area moment of inertia through the center of mass

A is the surface area

d is the distance the y axis to the cg axis

Calculation for beam in Problem 1.1: $$I= 2( \frac{b^3 t}{12} ) + 2( \frac{3a b^2 t}{12} )$$

Simplify: $$I= ( \frac{t b^2}{6} )(3a + b)$$

Maximizing $$( \frac{I}{b} ) = ( \frac{t b}{6} )(3a + b)$$, Remember $$a= ( \frac{L}{2} ) - 2$$

Now: $$( \frac{I}{b} ) = ( \frac{t b}{12} )(3L + 4b)$$

Solving the equation for b yields two distinct solutions: $$b = 0$$   and   $$b = \frac{3L}{4}$$

These two points are the zero crossings on the graph for a parabola. To find the desired solcution (b max) you need to find the maximum of that parabola. This can be done easily by realizing that the maximum must be half way between the two zero crossings. This gives the solution to the Case 1 as the following.

b 1 = $$ ( \frac{3L}{8} )$$

Now that we have solved for b, we can now evaluate the entire beam structure.

Plugging b into the parameter equation $$a = L - ( \frac{b}{2} )$$    yields a 1 = $$ ( \frac{L}{4} )$$

Thus comparing the Length of the two sides gives: $$ ( \frac{b^1}{a^1} ) = 3$$

The relationship between the second moment of inertia to the side be can also help solve for the Moment M.

$$M = \frac{2 I \displaystyle \sigma}{b}$$

Case 1 Momentum: $$M = ( \frac{3t L^2}{32} )\displaystyle \sigma$$

Lecture 8 - 9/12/08
Motivation: Study of a box beam (Problem 1.1) with a rectangular cross-section is a model for aerospace structures (fuselage, wings).

Problem 1.1 continued shear stress due to T max : $$\displaystyle \tau = ( \frac{T^1}{2a^1 b^1 t} )$$ This derivation is from slide 5-2 in the lecture notes.

Due to Assumption 2 T max = M max, and plugging that relationship and the calculated values for a and b into the shear stress yields:

$$\displaystyle \tau_{max} = ( \frac{M^1}{2 (\frac{L}{8}) (\frac{3L}{8}) t} )$$

Thus: $$\displaystyle \sigma = ( \frac{32}{3t L^2} ) M^1$$

This solution looks shows that: $$\tau_{max} = \displaystyle \sigma_{allow}$$

Recall Assumption 3) $$\displaystyle \sigma_{allow} = 2\tau_{allow}$$

This gives the result: $$2\displaystyle \tau_{allow} > \tau_{allow}$$ This relationship is not acceptable. The relationship would also mean that: $$\displaystyle \tau_{max} > \tau_{allow}$$. The relationship means failure for the beam.

Case 2: $$\tau_{max} = \displaystyle \tau_{allow}$$    and verify    $$\displaystyle \sigma_{max} < \sigma_{allow}$$

$$\displaystyle \tau = \frac{T}{2a b t} =  \tau_{max}  =  \tau_allow$$

Solving for the Torque (T) : $$T = 2t\displaystyle \tau_{allow}(ab)$$

$$2t\displaystyle \tau_{allow}$$ is a constant. So if you were trying to maximize the Torque that would mean that you would have to maximize the product of $$ab$$.

Lecture 9 - 9/15/08
Problem 1.1 Case 2 Continued: $$\tau_{max} = \displaystyle \tau_{allow}$$

$$\tau_{max} = \frac{T}{2 a b t}$$

replace τmax with τallow and move the denominator to the left hand side.

Now, we have $$T = ( 2 t \tau_{allow} ) * ( a b )$$ where we want to maximize the last factor...

Proof:

$$T_{max} = ( 2 t \tau_{allow} ) * ( a b )_{max}$$

Remember that $$L = 2 * ( a + b )$$ from which we get $$a = \frac{L}{2} - b$$

Multiply through by a factor of b to get $$( a b ) = \frac{L b}{2} - b^2$$

Taking the derivative of both sides of this equation gives: $$\frac{d( a b )}{db} = \frac{L}{2} - 2b$$

Setting this derivative equal to zero and rearranging the equation reveals: $$\frac{L}{2} = 2b$$

from which we get $$b = \frac{L}{4}$$

Plugging this answer back into $$L = 2 * ( a + b )$$ gives the answer $$a = \frac{L}{4}$$

From Lecture 8: $$a^{(2)} = b^{(2)} = \frac{L}{4}$$

Solving to the maximum Torque. $$T_{max} = (2t\displaystyle \tau_{allow})( \frac{L}{4} )^2$$

$$T_{max} = ( \frac{1}{8} ) t L^2 \displaystyle \tau_{allow} =  M_{max}^{(2)}$$     By Assumption 2.

$$M_{max}^{(2)} = \frac{t L^{2}}{16} * \sigma_{allow}$$     By Assumption 3.

$$\sigma_{allow} = \frac{16 M_{max}^{(2)}}{t L^{2}}$$

$$b = b^{(2)} = \frac{L}{4}$$

$$\frac{I}{b} = \frac{t b^{(2)}}{6}*(3a+b^{(2)})$$

Replacing $$b^{(2)}$$ with $$\frac{L}{4}$$ and rearranging we get

$$\frac{I}{b} = \frac{t L^2}{24}$$

Recall $$\sigma_{max}^{(2)} = \frac{M_{max}^{(2)} b^{(2)}}{I^{(2)} 2}$$

Replacing $$\frac{b}{I}$$ in the above equation and rearranging yields:

$$\sigma_{max} = \frac{3}{4}*\sigma_{allow}$$ which is less than the maximum allowable stress! This case is a success! Thus, $$b^{(2)}=\frac{L}{4}$$

Lectures 11 - 9/19/08
Q: Why are the lass of a stringer not completely "vertical"?

A:One of the reasons is a very simple one, storage. It is easier to store the stringers if the walls are not completely vertical.

Shear Panel:

Hooke's Law

$$\displaystyle \tau = G \displaystyle \gamma$$

G = shear modulus

$$\displaystyle \gamma$$ = shear strain, counterpart to normal strain $$\displaystyle \epsilon$$ , but describes the change in the angle between two lines in the material due to deformation.

$$\displaystyle \gamma = \frac{dv}{dy} + \frac{du}{dx} = \frac{du_x}{dy} + \frac{du_y}{dx}$$

u = u x displacement along x direction v = u y displacement along y direction

Assigned Problems
Problem 1.7 Compare the load-carrying capabilities of two beams having the respective cross-sections shown in Fig. 1.19. Use bending rigidity as the criterion for comparison. It is given that R o =10cm, t=1cm, and the two cross-sections have the same area.

Bending rigidity is defined as the... and is determined by multiplying the Young's modulus by the moment of inertia of the cross section. In order to compare the two cross-sections, we must first determine the unknown dimension (R 1 ) of the second beam.



$$A^{(1)} = A^{(2)}$$

$$A^{(1)}=\displaystyle \pi R^2$$

Plugging in the value of R=10cm gives the Area for Case 1:

$$A^{(1)}=\displaystyle \pi (10)^2 = 100\pi    =   314.2 cm^2$$

$$A^{(2)}= 2\displaystyle \pi R_1^2 + 2R_o(t)   = A^{(1)}$$

$$A^{(2)}= 2\displaystyle \pi R_1^2 + 2(10cm)(1cm)   = 314.2 cm^2$$

Solving for R 1: $$R_1 = 6.84 cm^2$$

Next, we compute the moment of inertia for both beams:

$$ I_y = \int R_o^2\,dA                                    dA = \pi R dr $$

=$$\int R^2(\displaystyle \pi R)\,dr $$

= $$ \displaystyle \pi \int R_o^3\,dr $$

$$ I_y = \frac{\displaystyle \pi }{4} R_o^4 $$

$$ I_y = \frac{\displaystyle \pi }{4} (10cm)^4 $$

$$ I_y = 2500\displaystyle \pi  = 7854 cm^4 $$

Now Calculate I 2 :

$$I_2 = \frac{1}{12}t(2R_o)^3 + 2( \frac{1}{4} \displaystyle \pi R_1^4 + (\displaystyle \pi R_o^2 (R_o +R_1)^2))$$

$$I_2 = \frac{1}{12}(1cm)(2(10cm))^3 +2[ \frac{1}{4} \displaystyle \pi (6.84cm)^4 + (\displaystyle \pi (6.84cm)^2 (10cm + 6.84 cm)^2)]     =      I_2 = 87468.5 cm^4$$

Thus I 2 > I 1. Assuming that the sections are made of the same material, we can conclude that beam 2 has a higher bending rigidity due to the larger second moment of inertia.

Extra Student Contributions
Works Cited

Sun, C.T. Aerospace Stuctures, 2006

www.Merriam-Webster.com

2007-02-17 20:31 Keenan Pepper 525×315×1 (1262 bytes) Parallel axes rule for area moment of inertia. Image made by MH 21:51, July 9, 2005 (UTC)

Contributing Team Members HW 2
Steven Hepsworth Eas4200c.f08.team12.hepsworth 21:51, 23 September 2008 (UTC)

Brian Taylor Eas4200c.f08.team12.taylor 20:53, 23 September 2008 (UTC)

Victoria WatlingtonEAS4200C.Fall08.Team12.Watlington.VG 03:51, 24 September 2008 (UTC)

= Homework 3 =

Curved Panels
Looking at the shear flow in a curved panel: $$d\vec{F} = qd\vec{l} = q(dl_{y}\hat{j} + dl_{z}\hat{k})$$
 * $$= q(dlcos\theta\ \hat{j} \ +\ dlsin\theta \ \hat{k})$$

Therefore the resultant shear force vector is:


 * $$\vec{F} = \int_{A}^{B}{d\vec{F}} = q \left( \int_{A}^{B}dy\ \hat{j}+ \int_{A}^{B}dz\ \hat{k}\right)$$
 * $$\vec{F} = q\left(a\ \hat{j} + b\ \hat{k} \right)$$

Now, the magnitude of the resultant shear force vector is:
 * $$\parallel \vec{F} \parallel = \sqrt{F_{y}^{2}+F_{z}^{2}}$$
 * $$=q\sqrt{a^{2}+b^{2}}$$


 * $$=qd$$


 * $$\parallel \vec{F} \parallel =qd$$

Now we will relate the magnitude of the resultant shear force vector (R) to the torque $$\ T=2q\bar{A}$$
 * $$\vec{T}= T\hat{i}$$


 * $$dF = qdl$$


 * $$d\vec{T}=\vec{r}$$ x $$d\vec{F}$$


 * $$ dT = rdFsin\theta= \rho dF$$


 * $$ = \rho q dl$$


 * $$dA = \frac{1}{2}\ \rho dl$$


 * $$T=\oint dT = q\oint \rho dl = 2q\int_{\bar{A}}dA$$


 * $$\Rightarrow T=2q\bar{A}$$

Circular Cross-Section
Uniform bar with circular cross section non warping.

$$T = \iint_A{r\tau \; \mathrm{d}A} = \int_A{rG(r\theta) \; \mathrm{d}A} = G\theta \int_A{r^2 \; \mathrm{d}A} = G\theta J = \frac{\pi G\theta}{2}r^4$$

Hollow thin walled cross section.

$$r_i = a \; r_o = b$$

$$J = \frac{\pi}{2} \left( b^4 - a^4 \right) = \frac{\pi}{2}(b-a)(b+a)(b^2+a^2) = \frac{\pi}{2}(t)(2\bar{r})(2\bar{r}^2) = 2\pi t\bar{r}^3$$

So the torque equation becomes

$$T=2 \pi G \theta t {\bar r}^3$$

NACA Airfoils
For the Matlab portion of the homework assignment, a portion is dedicated to finding the area enclosed by the airfoil. To do this, the airfoil will be broken up into many small segments, each of area dA. All of these segment areas will then be added up to come up with a total area. The second image on the right shows an enlarged region of the first image of the airfoil.

Looking at the image of the enlarged section, to find the area:
 * $$ d\vec{A} = \frac{1}{2} \ \vec{r}\ $$ x $$\ \vec{PQ}$$
 * $$=||d\vec{A}|| \ \ \hat{i}$$
 * $$ d\vec{A} = \frac{1}{2} \ \vec{r}\ $$ x $$\ \vec{PQ}$$
 * $$=||d\vec{A}|| \ \ \hat{i}$$
 * $$=||d\vec{A}|| \ \ \hat{i}$$

Torsion of Non-Circular Bars
warping = axial displacement along x-axis (i.e. along bar length) of a point on the deformed (rotated) cross-section.

Note: $$\vec{PP'}\quad$$ is the displacement vector
 * $$u_{z}\quad$$ is the z-component of the displacement vector
 * $$u_{y}\quad$$ is the y-component of the displacement vector

From the figure, we can see that the displacement is perpendicular to OP, but why? We can see that $$OP=OP'=R\qquad$$ and that $$u_{z} = Rsin(\alpha )\quad$$ and with $$\alpha\quad$$ being small, $$u_{z} = R\alpha\quad$$ By the same logic: $$u_{y}=R(1-cos(\alpha ))=R(1-1) = 0\quad$$. Therefore we can approximate the displacement as being perpendicular to OP.

 Kinematic Assumptions 
 * Starting with the definition of the rate of twist $$\theta = \frac{\alpha}{x}\quad \rightarrow \alpha = \theta x$$

$$u_{y}=-\theta x z \qquad \qquad (3.11)$$ $$u_{z}=PP'cos(\beta ) =OP\alpha cos(\beta ) = \alpha y $$ $$u_{z}=\theta x y \qquad \qquad (3.12)$$ $$u_{z}\quad$$ is the z displacement of a point on the cross-section. $$u_{x}=\theta \psi (y,z) \qquad \qquad (3.13)$$ The three equations above are the Kinematic Assumptions
 * Warping displacement along x-axis

A. Kinematic Assumptions
(Section 3.2 in textbook) See above three equations.

B. Strain-Displacement Relationship
(Section 3.2 in textbook) Strain is a non-dimensional value representing the ratio of the amount the material has deformed in one direction versus the original length of the material in that direction. $$\epsilon = \frac{\delta\ell}{\ell_{0}}$$

C. Equilibrium Equation for Stresses
(Chapter 2, Section 3.6) If a body is in equilibrium the stress field must satisfy these equations everywhere:

$$\frac{\partial\sigma_{xx}}{\partial x} + \frac{\partial\tau_{yx}}{\partial y} + \frac{\partial\tau_{zx}}{\partial z} = 0$$

$$\frac{\partial\tau_{xy}}{\partial x} + \frac{\partial\sigma_{yy}}{\partial y} + \frac{\partial\tau_{zy}}{\partial z} = 0$$

$$\frac{\partial\tau_{xz}}{\partial x} + \frac{\partial\tau_{yz}}{\partial y} + \frac{\partial\sigma_{zz}}{\partial z} = 0$$

$$\tau_{xy}=\tau_{yx}\quad\tau_{yz}=\tau_{zy}\quad\tau_{xz}=\tau_{zx}$$

D. Prandtl Stress Function $$\phi$$
(3.2, Eq. 3.15) The Prandtl stress function is defined such that its derivative in the y-direction equals the shear stress in the plane perpendicular to the y-direction (the x-z plane), and that its derivative in the x-direction equals the negative shear stress in the plane perpendicular to the x-direction (the y-z plane). $$\tau_{xy} = \frac{\partial \phi}{\partial y}$$ $$\tau_{yz} = -\frac{\partial \phi}{\partial x}$$

E. Strain Compatibility Equation
(Eq. 3.17) $$\frac{\partial\gamma_{yz}}{\partial x} - \frac{\partial\gamma_{xz}}{\partial y} = 2\Theta $$

F. Equation for $$\phi$$
(Eq. 3.19) Combining the compatibility equation for torsion and Prandtl's stress function we get: $$\frac{\partial^{2}\phi}{\partial x^{2}} + \frac{\partial^{2}\phi}{\partial y^{2}} = -2G\Theta$$

G. Boundary Conditions for $$\phi$$
(Eq. 3.24) The traction free boundary condition tz = 0 is now: $$\frac{\partial\phi}{\partial s} = 0$$ Meaning that the stress function is constant at the surface. For a solid regular body the constant is arbitrary and is often made to be 0 for simplicity. $$\phi = 0$$

H. Torque
$$T=2\int\int_{A}^{}{}\phi dA$$ (Eq. 3.25)

$$T=GJ\theta $$

$$J=\frac{-4}{\nabla^{2}\phi }\int \int _{A}\phi dA$$

(pg. 74)

I. Thin-walled Cross-section
Ad-hoc assumption on shear flow. Formal derivative (Section 3.5). We previously did the end of this section, which is easier than the beginning.

$$T = 2q\bar{A}$$

J. Twist Angle $$\theta$$: Method 1
$$\theta = \frac{1}{2G\bar{A}}\oint{\frac{q}{t}ds}$$

K. Section 3.6 on multicell thin-walled cross-section
Looking at the picture of the multicell wing we find that the number of cells, $$n_{c}$$ is equal to 3.

K1.
$$T=2\sum_{i = 1}^{n_{c}}{q_{i}\bar{A}_{i}}$$


 * Where $$q_{i}$$ = shear flow in cell i
 * and $$\bar{A}_{i}$$ = the "average" area of cell i

Define: $$T_{i}=2q_{i}\bar{A}_{i}\qquad$$ torque generated by one cell
 * $$\Rightarrow T=\sum_{i=1}^{n_{c}}{T_{i}}$$

K2.
Shape of airfoil is "rigid" in the yz plane (can warp out of plane). $$\theta = \theta _{1} = \ldots = \theta _{n_{c}}$$

$$\theta _{i}=\frac{1}{2G_{i}\bar{A}_{i}}\oint\frac{q_{i}}{t_{i}}ds$$


 * $$G_{i}$$= shear modulus of cell i.
 * $$t_{i}$$= thickness of cell i.
 * $$t_{i}(s)$$= curvilinear coordinate along cell wall.

Shear Flow: $$T = 2q \bar{A}$$
 * $$q=\frac{T}{2\bar{A}} \qquad$$ express q as a function of T

Twist Angle:
 * $$\theta = \frac{1}{2G\bar{A}}\sum_{j=1}^{3}{\frac{q_{j}l_{j}}{t_{j}}}\qquad$$ with l as the length of the segment and j as the index number.

NOTE:


 * $$\int \qquad$$: elongated S, standing for summation (continuous)
 * $$\sum \qquad$$: (discrete) sum

Area of a Triangle
Prove $$A = \frac{1}{2}bh$$

First begin by considering a rectangle with base of BD and height of AD. The area of this rectangle would be (BD)(AD). Now, if we just consider half of this rectangle (triangle ABD) we would have an area of $$\frac{1}{2}$$(BD)(AD). This is the area of the triangle pictured including the area enclosed by the dotted lines. If we subtract the area enclosed by the dotted lines from the area we just found, we will get the area of the triangle we are interested in. The area of the dotted line would be $$\frac{1}{2}$$(AD)(CD) by the same logic as before. Performing this, we find that $$A = \frac{1}{2}\left(AD \right)\left(BD \right)- \frac{1}{2}\left(AD \right)\left(CD \right) = \frac{1}{2}\left(AD \right)\left[ (BD)-(CD)\right]$$

Looking at the picture, we can see that by definition BD - CD = b and AD = h. Plugging these relationships into the above equation yields $$A = \frac{1}{2}bh$$

Polar Moment of Inertia of Circular Cross-Section
For a circle of radius a, the polar moment of inertia ($$J_{0}$$) can be calculated as follows:

$$J_{0} = \int_{A}\rho^{2}dA = \int_{0}^{a}\int_{0}^{2\pi}\rho^{2}(\rho d\theta)d\rho $$
 * $$= \int_{0}^{a} \rho^{2}\left(2\pi\rho d\rho \right) $$
 * $$= 2\pi\int_{0}^{a}\rho^{3}d\rho = \frac{1}{2}\pi a^{4}$$
 * $$\Rightarrow J_{0} = \frac{1}{2}\pi a^{4}$$

Solid vs. Hollow Cross-Section
Compare a solid circular cross section to a hollow thin walled circular cross section.

$$r_{o}^{(a)}=1cm$$

$$r_{i}^{(b)}=5cm$$

$$t^{(b)}=0.1cm$$

$$A^{(a)}=\pi r^{2}=\pi (1cm)^{2}=3.1415cm^{2}$$

$$A^{(b)}=\pi (r_{o}^{2}-(r_{o}-t_{b})^{2})=\pi (5cm^{2}-(5cm-0.1cm)^{2})=3.1102cm^{2}$$

$$J^{(a)}=\frac{\pi }{2}r^{4}=\frac{\pi }{2}(1cm)^{4}=1.5708cm^{4}$$

$$J^{(b)}=2\pi t\bar{r}^{3}=2\pi t(\frac{2r_{o}-t^{(b)}}{2})^{3}=2\pi (0.1cm)(\frac{2(5cm)-(0.1cm)}{2})^{3}=76.207cm^{4}$$

$$\frac{J^{(a)}}{J^{(b)}}=\frac{1.5708cm^{4}}{76.207cm^{4}}=0.0206$$

Find $$r_{i}^{(c)}$$ if $$t^{(c)}=0.02r_{o}^{(c)}$$ and $$J^{(b)}=J^{(c)}$$.

$$J^{(b)}=76.207cm^{4}=2\pi (0.02r_{i}^{(c)})(\frac{2(r_{i}^{(c)}+(0.02r_{i}^{(c)}))}{2})^{3}=1.06684r_{o}^{(c)}=J^{(c)}$$

$$r_{i}^{(c)}=2.907$$

Single Cell Section Area


To calculate the area of this figure, we break it down into two smaller parts: a semicircle and a triangle. The area of a semicircle is $$\frac{1}{2}\pi R^{2}\quad$$ with R being $$\frac{b}{2}$$ = 1 m. The area of a triangle is $$\frac{1}{2}bh\quad$$ with the base being a (4 m) and the height being b (2 m).

Therefore the overall area is:

$$\bar{A} = \frac{1}{2}\pi \left( \frac{b}{2}\right)^{2} + \frac{1}{2}ab $$
 * $$= \frac{1}{2}\pi \left( \frac{1}{2}\right)^{2} + \frac{1}{2}\left(4 \right)\left(2 \right)$$
 * $$=\frac{\pi}{8} + 4 \quad m.$$
 * $$= \frac{1}{2}\pi \left( \frac{1}{2}\right)^{2} + \frac{1}{2}\left(4 \right)\left(2 \right)$$
 * $$=\frac{\pi}{8} + 4 \quad m.$$
 * $$=\frac{\pi}{8} + 4 \quad m.$$
 * $$=\frac{\pi}{8} + 4 \quad m.$$

MATLAB Airfoil code
The following MATLAB function was created to plot the shape of a NACA 4-digit series airfoil. It also calculates the area and centroid of the cross section. Following the function is the implementation and output, copied out of the command window.

EDU>> airfoil(2415);

Cross Sectional Area A = 0.1013 Centroid Location C = (0.4203, 0.0157)

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