User:Eas4200c.f08.blue.a/Lecture 18

Quadrature allows an observer outside of a NACA airfoil to find the cross sectional area. An example of this is shown below in Figure 1.

Figure 1

$$\bar{A}=\bar{A_{1}}+\bar{A_{2}} =\bar{A_{1}}\left|\bar{A_{2}} \right|$$



In this close up, $$d\vec{A}$$ is defined as:

$$d\vec{A}=\frac{1}{2}\vec{r}+\vec{PQ}=-\left|dA \right|\vec{i}$$

Dividing the area into trapezoids and adding them is another method used to find the area under a curve called the (Trapezium rule), but it is not as elegant or effective as triangular quadrature in more complicated applications such as finding the area of the oddly shaped airfoil in Figure 2 below.

Figure 2

Returning to the idealized single cell airfoil example from above, where the shear flow is constant, the twist angle can be found using the formula found in step J of the road-map for torsional analysis of an aircraft wing.



$$\Theta =\frac{1}{2G\bar{A}}\oint_{}^{}{\frac{q}{t}ds}$$

because $$\left(q \right)$$ is constant,

$$\left(q \right)=\left( q_{1}\right)=\left( q_{2}\right)=\left( q_{3}\right)$$

$$\Theta =\frac{1}{2G\bar{A}}q\sum_{j=1}^{3}{\frac{l_{j}}{t_{j}}}$$ $$=\frac{1}{2G\bar{A}}q\left[\frac{2\pi \left(\frac{b}{2} \right)}{t_{1}}+\frac{a}{t_{2}}+\frac{\sqrt{a^{2}+b^{2}}}{t_{3}} \right]=q\left(\frac{1}{2\left(5.57m^{2} \right)G} \right)\left(\frac{\pi \left( 1m\right)}{0.008m}+\frac{4m}{0.01m} +\frac{\sqrt{16m^{2}+4m^{2}}}{0.01m} \right)=\frac{q}{\left( 11.14m^{2}\right)G}\left(1239.91 \right)=111.24m^{-2}\frac{q}{G}$$

The Max Shear Stress $$\left( \tau _{max}\right)$$ can also be found now using the formula:

$$ \tau _{max}=\frac{q}{min\left\{t_{1},t_{2},t_{3} \right\}}$$

If the maximum shear stress is equal to the allowable shear stress and from the formula to find shear flow, the allowable torque $$\left[ T_{allow}\right]$$ can be found.

$$\tau _{max}=\left( \tau _{allow}\right)$$

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

$$T_{allow}=2\bar{A}\tau _{allow}\left(min\left\{ t_{1},t_{2},t_{3}\right\} \right)$$

Therefore, if $$\tau _{allow}=100GPa$$, and because $$t_{1}=0.008m$$, $$t_{2}=t_{3}=0.01m$$

$$T_{allow}=2\bar{A}\tau _{allow}t_{1}=2\left(5.57m^{2} \right)\left(100GPa \right)\left(0.008m \right)=8.91\left(10^{9} \right)Nm$$