User:Eas4200c.f08.vqcrew.E/Homework4

Quadrature of a NACA Airfoil


The area of any shape can be determined by using a collection of many small triangles. The triangles are defined by an arbitrary origin, O, and an infinitesimal length on the shape. Since the infinitesimal length and perpendicular distance from the origin to the surface are known, the infinitesimal area, dA, can be calculated. Figure ## shows this process for an airfoil.


 * $$ \bar{A}_{airfoil} = \bar{A}_{1} + \bar{A}_{2} = \bar{A}_{1} - \left|\bar{A}_{2}\right|$$

In this case, the upper surface yields a positive area and the lower surface yields a negative area due to the location of the origin and the nature of the right-hand rule for vectors. The total area of the airfoil is computed as a sum of all of the small areas for each length along the airfoil contour. Figure ## provides a more detailed picture of how each area is calculated. The example shown is for the lower surface, therefore the cross-product of the vectors $$\textstyle \vec{r}$$  and $$\textstyle \vec{PQ}$$ is negative. The negative cross-product leads to a negative area for the lower surface in this configuration.


 * $$\displaystyle d\vec{A} = \frac{1}{2}(\vec{r}\times\vec{PQ})$$



Problems with Trapezoidal Method of Quadrature
The process of dividing the area of a shape in to a series of small trapezoids may be an accurate way to determine the total area for some shapes in some configurations, but it is not a general solution. For example, Figure ## shows the area of an airfoil being divided into trapezoids horizontally. For this shape in this orientation, the result will not be very accurate due to the errors near the trailing edge. The thin trailing edge is not conducive to being split up in this manner. It is true that if the trapezoids were aligned vertically the result would more accurately represent the true area, but a more elegant and beautiful solution can be reached. Using the triangle method described above, a more accurate result for the area can be found regardless of the location of the origin or orientation of the airfoil.

Single Cell Airfoil


For a single cell airfoil the shear flow is constant.


 * $$\displaystyle q = q_{1} = q_{2} = q_{3}$$

The rate of twist, $$\theta$$, can be determined using the known lengths and thickness of each segment as well as the average area, $$\bar{A}$$, computed in Homework 3.


 * $$\displaystyle \theta = \frac{1}{2G\bar{A}}q\sum_{j = 1}^{3}{\frac{l_{j}}{t_{j}}}$$


 * $$\displaystyle \theta= \frac{1}{2G\bar{A}}q[\frac{\frac{\pi}{2}(b)}{t_{1}} + \frac{a}{t_{2}}+ \frac{\sqrt{a^{2}+b^{2}}}{t_{3}}]$$

Substituting in the known values each segment length and thickness and the average area, the result for the twist angle can be found.


 * $$\displaystyle \theta= \frac{111.29 q}{G}$$

Now, the maximum shear stress, $$\displaystyle \tau_{max}$$, can be determined using the constant shear flow and minimum thickness of a segment.


 * $$\displaystyle \tau_{max} = \frac{q}{min[t_{1},t_{2},t_{3}]}$$

If $$\textstyle \tau_{max} = \tau_{allow}$$ (given), and since $$\textstyle q = \frac{T}{2\bar{A}}$$, the allowable torque can be written as


 * $$\displaystyle T_{allow} = 2\bar{A}\tau_{allow}min[t_{1},t_{2},t_{3}]$$

For a known value of $$\textstyle \tau_{allow}$$ equal to 100 GPa, $$\textstyle T_{allow}$$ can be determined.


 * $$\displaystyle T_{allow} = 2\bar{A}\tau_{allow}min[t_{1},t_{2},t_{3}] = 2(5.5708 m^{2})(10^{11} Pa)(0.008 m) = 8.913\times10^{9} N/m$$