User:Eas4200c.fo8.aero.lee/HW3

= Curved Panels =





The expression for the finite force applied to a finite unit length of a curved panel is $$d\vec{F}=q\,d\vec{l}$$. However, to analyze the curved panel with respects to an x and y component, we can make use of the fact that $$d\vec{l}=dl_{y}\,\hat{j}+dl_{z}\,\hat{k}$$ to render the equation:


 * $$d\vec{F}=q\left(dl_{y}\,\hat{j}+dl_{z}\,\hat{k} \right)$$

For $$dl_{y}$$ and $$dl_{z}$$ we can implement the polar coordinate transformations:
 * $$dl_{y}=dl\,cos(\theta)$$
 * $$dl_{z}=dl\,sin(\theta)$$

(remembering that we are in the yz plane)



and hence:
 * $$dl\,cos(\theta)=dy$$
 * $$dl\,sin(\theta)=dz$$

Integrate to find the total force
 * $$\vec{F}=\int d\vec{F}=q\int dy\,\hat{j}+dz\,\hat{k}$$
 * $$\vec{F}=q\left(\Delta y\hat{j}+\Delta z\hat{k} \right)=F_{y}\,\hat{j}+F_{z}\,\hat{k}$$

For our application we can solve for the magnitude of the force finally arriving to the equation:
 * $$\parallel\vec{F}\parallel = \sqrt{{F_y}^2+{F_z}^2}=q\sqrt{{\Delta y}^2+{\Delta z}^2}$$

= Closed Thin Walled Cross Sections =





We can relate our resultant force, $$R= \parallel\vec{F}\parallel$$, with the torque, $$T=\,2\,q\,\bar{A}$$ by:
 * $$d\vec{T}={\vec{r}}\times{d\vec{F}}$$

by letting $$\rho$$ be the perpendicular distance from the point of reference to the corresponding finite force, and by remembering that $$d\vec{F}=q\,d\vec{l}$$, we can conclude:
 * $$dT={\rho}\,q\,dl$$
 * $$T=\int dT=q\int \rho\,dl=q\int 2\,dA$$
 * $$T=2\,q\,\bar{A}$$

= Torsion of Uniform, circular Bars =





It is first important to define the rate of twist $$\theta$$ as:
 * $$\theta=\frac{d\alpha}{dx}=rate\,\,of\,\, twist$$

AND
 * $$\gamma=r\frac{d\alpha}{dx}=r\,\theta$$

Also, the second polar area moment of Inertia is defined as:
 * $$J=\int \int _Ar^2\,dA$$

To start, our equation for torque is:
 * $$T=\int \int _Ar\,\tau\,dA$$

By substituting in that $$\tau=G\,\gamma$$ and $$\gamma=r\,\theta$$ we get:
 * $$T=\int \int _A r\,G\,(r\,\theta)=G\,\theta\int \int _A r^2dA$$
 * $$T=G\,\theta\,J$$

= Torsion of Uniform, Non-circular Bars =

Torsion of uniform, non-circular bars typically leads to an occurrence of warping of the cross section. Warping is when a point in the cross section under torsion, subject to deformation, will undergo a axial displacement along the length of bar. A simple example of warping is rolling up a sheet of paper then applying a torque - the rolled sheet will noticeably elongate along its axis.

Figure 4 below shows a typical depiction of displacement in a cross section view of a non-circular bar in torsion.



As can be seen above, the displacement vector, $$\overrightarrow{P P'}$$ is perpindicular to the original position vector, $$\overrightarrow{O P}$$. This method of describing the displacement vector is chosen because with the assumption that $$\alpha$$ is small the y and z components of the displacement vector can easily be calculated. Figure 4a below shows the y and z components of displacement for the case $$\beta = 0$$.



By applying basic trigonometry to the figure above, $$u_y$$ and $$u_z$$ can be solved for in terms of $$\alpha$$:
 * $$ u_y = R(1-\cos{\alpha}) \qquad u_z = R\sin{\alpha}$$

If the assumption that $$\alpha\approx0$$ is made, then $$\cos{\alpha}\approx1$$ and $$\sin{\alpha}\approx0$$. These equations then become:
 * $$u_y\approx0 \qquad u_z\approx{R\alpha}$$

For the general case as shown in figure 4, R becomes the initial distance OP and the y-component of the displacement of point P can be expressed as
 * $$ u_{y} = -(PP')sin\beta = -(OP\alpha)sin\beta = -\alpha(OP\sin\beta) = -\alpha z_{p} $$

Similarly the z-component of the displacement of point P is


 * $$ u_{z} = (PP')cos\beta = (OP\alpha)cos\beta = \alpha(OP\cos\beta)=\alpha y_{p} $$

Recalling the definition of $$ \theta $$


 * $$ u_{y} = -\theta xz_{p} $$


 * $$ u_{z} = \theta xy_{p} $$

Generalizing for all points:


 * $$ u_{y} = -\theta xz $$ (Eq. 1)


 * $$ u_{z} = \theta xy $$ (Eq. 2)

For warping displacement in the axial direction (i.e. x-direction) it can be shown that


 * $$ u_{x} = \theta \psi (y,z) $$ (Eq. 3)

= Road Map for Analysis of Wing Torsion of a Multicell Section =


 * 1) Kinematic Assumptions (Equations (1), (2) and (3) above) [Sun, Sec. 3.2]
 * 2) Strain Displacement Relationship [Sun, Sec. 3.2]
 * 3) Equilibrium Equation for Stresses [Sun, Ch. 2, Sec. 3.2]
 * 4) Prandtl Stress Function $$ \phi $$ [Sun, Sec. 3.2, Eqn. 3.15]
 * 5) Strain Compatibility Equation [Sun, Sec. 3.2, Eqn. 3.17]
 * 6) Equation for $$ \phi $$ [Sun, Sec. 3.2, Eqn. 3.19]
 * 7) Boundary Conditions for $$ \phi $$ [Sun, Sec. 3.2, Eqn. 3.24]
 * 8) Calculation of Torque


 * $$ T = 2\int\int_{A}^{}{\phi dA} = GJ\theta $$ [Sun, Sec. 3.2, Eqn. 3.25]


 * $$ J = \frac{-4}{\bigtriangledown^2 \phi } \int \int_{A}^{}{\phi dA} $$


 * 9. Formal Derivation of Torque and Shear Flow Relationship


 * $$ T = 2q\bar{A} $$ [Sun, Sec. 3.5, Eqn. 3.48]


 * 10. Calculation of Twist Angle $$ \theta $$


 * $$ \theta = \frac{1}{2G\bar{A}}\oint_{}^{}{\frac{q}{t}ds} $$ [Sun, Sec. 3.5, Eqn. 3.56]


 * 11. Solution of Multicell Thin-walled Sections  [Sun, Sec. 3.6]

= Use of "RoadMap" to Solve a Single Cell Section Airfoil =

We can simplify the analysis of a complex, multi-cell airfoil by looking at a semi-circle attached to a triangle.



Average Area ($$\bar{A}$$)
To find $$\bar{A}$$ we simply look at the area of each shape sepratly.
 * $$\bar{A}=\bar{A}_{Semi-Circle}+\bar{A}_{Triangle}$$
 * $$\bar{A}=\frac{1}{2}\left[\pi\left(\frac{b}{2} \right)^2\right]+\frac{1}{2}\,b\,a$$
 * $$\bar{A}=5.57$$

Shear flow ($$q$$)
Remembering our derived equation for Torque, $$T=2\,q\,\bar{A}$$, we can solve for the shear flow:
 * $$q=\frac{T}{2\,\bar{A}}=\frac{T}{11.14}$$

Twist angle ($$\theta$$)
Our twist angle equation is:
 * $$\theta=\frac{1}{2\,G\,\bar{A}}\oint{\frac{q}{t}ds}$$

However when looking at each segment of the wall, we have 3 separate sections with constant thickness, so the integral portion becomes a sum:
 * $$\theta=\frac{1}{2\,G\,\bar{A}}\sum_{i=1}^{3}{\frac{q_i\,l_i}{t_i}}$$
 * $$\theta=\frac{q}{11.14\,G}\left (\frac{2\,\pi}{2\,t_1}+\frac{4}{t_2}+\frac{4.47}{t_3} \right)$$
 * $$\theta=10\frac{T}{G}$$

= Proof of Triangle Area Formula =

The area of a triangle is given as:


 * $$ A = \frac{1}{2}bh $$

To prove this equation, the area of triangle CDE is subtracted from the area of triangle BDE. Notice that area of a right triangle is exactly half the area of a square whose sides are equal to the legs of the triangle. Thus:


 * $$ A_{BDE} = \frac{1}{2}(BD)(DE) = \frac{1}{2}(BD)h $$

and


 * $$ A_{CDE} = \frac{1}{2}(CD)(DE) = \frac{1}{2}(CD)h $$

Subtracting the areas to solve for the triangle of interest:


 * $$ A_{BCE} = A_{BDE} - A_{CDE} = \frac{1}{2}(BC-CD)h = \frac{1}{2}bh $$

= Second Polar Area Moment of Inertia for a Circle = The definition of the Second Polar Area Moment of Inertia is given as:


 * $$ J = \int r^2 dA $$

For a circle of radius a, the second polar moment of inertia can be solved via a double integral as


 * $$ J = \int r^2 dA = \int_{0}^{2\pi}\int_{0}^{a}{r^2}rdrd\theta = \frac{1}{4}a^4\left[\int_{0}^{2\pi}d\theta \right] = \left[\frac{1}{4}a^4 \right]\left[2\pi \right] = \frac{\pi }{2}a^4 $$

= Second Polar Moment of Inertia of a Thin-walled Cylinder =



For a thin-walled cylinder of inner radius r_{i} = a </math and outer radius $$ r_{o} = b $$ the average radius can be given as:


 * $$ \bar{r} = \frac{b+a}{2} $$

The square of the average radius is:


 * $$\bar{r}^2 = \left( \frac{b+a}{2} \right)^2 = \frac{1}{4}\left(b^2 + a^2 + 2ab \right)$$

If the assumption is made that the wall is thin such that thickness, $$ t = b-a $$, is much less than a or b, then


 * $$ \bar{r}^2 = \frac{1}{4}\left(2b^2 + 2a^2 \right) = \frac{1}{2}\left(b^2 + a^2 \right) $$

Then,the second polar area moment of inertia for the thin-walled section can be given as


 * $$ J = \int r^2 dA = \int_{0}^{2\pi}\int_{a}^{b}{r^2}rdrd\theta = \frac{1}{4}(b^4 - a^4)\left[\int_{0}^{2\pi}d\theta \right] = \left[\frac{1}{4}(b^4 - a^4) \right]\left[2\pi \right] = \frac{\pi }{2}(b^4 - a^4) $$

Expanding the last term provides


 * $$ J = \frac{\pi }{2}(b-a)(b+a)(b^2 - a^2) $$

Recalling the definitions for $$ t $$, $$ \bar{r} $$ and $$ \bar{r}^2 $$ from above


 * $$ J = \frac{\pi }{2}t(2\bar{r})(2\bar{r}^2) = 2\pi t\bar{r}^3 $$

= Comparison of a solid, circular cross-section to a hollow, thin-walled, circular cross-section =

Given


A solid, circular cross-section has radius $$ r_{o}^{(a)} = 1 cm $$. A hollow, thin-walled, circular cross-section has inner radius $$ r_{i}^{(b)} = 5 cm $$ and thickness $$ t = 0.1 cm $$.

Find
a) compute $$ A^{(a)} $$ and $$ A^{(b)} $$

b) compute $$ J^{(a)} $$ and $$ J^{(b)} $$

c) compare $$ \frac{J^{(a)}} {J^{(b)}} $$

d) find radius $$ r_{i}^{(c)} $$ of thin-walled section with thickness $$ t = 0.02r_{i}^{(c)} $$ such that $$ J^{(c)} = J^{(a)} $$ and compare $$ \frac{A^{(a)}} {A^{(c)}} $$

a)

 * $$ A^{(a)} = \pi (r_{o}^{(a)})^2 = \pi(1 cm^2) = \pi cm^2 $$


 * $$ A^{(b)} = \pi ((r_{i}^{(b)}+t)^2 - (r_{i}^{(b)})^2) = \pi((5.1 cm)^2 -(5cm)^2) = 1.01\pi cm^2 $$

b)

 * $$ J^{(a)} = \frac{\pi }{2}(r_{o}^{(a)})^4 = \frac{\pi }{2}(1cm)^4 = \frac{\pi }{2}cm^4 $$


 * $$ J^{(b)} = \frac{\pi }{2}((r_{i}^{(b)} t) - r_{i}^{(b)})^4 = \frac{\pi }{2}((5.1cm)^4 - (5cm)^4 = 25.76\pi cm^4 $$

c)

 * $$ \frac{J^{(a)}}{J^{(b)}} = \frac{\frac{\pi}{2}cm^4}{25.76\pi cm^4} = 0.0194 $$

d)
To determine the proper $$ r_{i}^{(c)} $$, set the second polar area moment of inertia equations for a thin-walled section and a solid section equal to one another and solve for the unkonwn:


 * $$ \frac{\pi }{2}\left[\left(1.02r_{i}^{(c)} \right)^4 - \left(r_{i}^{(c)}\right)^4 \right] = \frac{\pi}{2}\left(r^{(a)}\right)^4 = \frac{\pi}{2} cm^4 $$

thus,


 * $$ r_{i}^{(c)} = 12.13 cm $$

To compare areas, calculate the area $$ A^{(c)} $$ and compare to $$ A^{(a)} $$ determined above.


 * $$ A^{(c)} = \pi\left[\left(r_{o}^{(c)}\right)^{2} - \left(r_{i}^{(c)}\right)^{2}\right] = 5.945\pi cm^2 $$

Comparing the two areas,


 * $$ \frac{A^{(a)}}{A^{(c)}} = 0.168 $$

= Analysis of Multi-Cell Sections =

It is important to understand how to analyze a multi-cell section because of their widespread use in common structures such as wings. We can apply what we already know about a single cells to our analysis of multi-cells by simply adding or integrating all of the single cell sections.

If we look at the torque generated by the $$i^{th}$$ single cell, we get the formula:
 * $$T_i=2\,q_i\,\bar{A}_i$$

To apply this equation to a multi-cell structure, we simply add all of the cells. Given that $$n_c$$ is the number of cells:
 * $$T=\sum_{i=1}^{n_c}{T_i}=2\sum_{i=1}^{n_c}{q_i\,\bar{A}_i}$$

As for the rate of twist, $$\theta$$, since the structure is a rigid body, $$\theta={\theta}_1=...={\theta}_{n_c}$$ such that the rate of twist attributed to the $$i^{th}$$ cell is:
 * $${\theta}_i=\frac{1}{2\,G_i\,\bar{A}_i}\oint{\frac{q_i}{t_i}ds}$$

= NACA Airfoil Matlab Project =

The purpose of this project is to form the basis for future structural analysis of thin walled bodies, with emphasis on airfoils. The code should be able to plot the body, determine the average area (Ā), and find the location of the centroid. These values are found through splitting the perimeter of the bodies into small discrete segments and then using properties of the cross product to determine the area from a point in space and a quadrature to determine the centroid.

= Contributing Members = The following Team Aero members contributed to this report.

Jared Lee --Eas4200c.f08.aero.lee 03:06, 1 October 2008 (UTC)

Ray Strods --Eas4200c.f08.aero.strods 03:53, 7 October 2008 (UTC)

Oliver Oyama --Eas4200c.fo8.aero.oyama 17:14, 7 October 2008 (UTC)

William Kurth --Eas4200c.f08.aero.kurth 11:53, 8 October 2008 (UTC)

Gonzalo Barcia --Eas4200c.f08.aero.barcia 12:30, 8 October 2008 (UTC)