User:Eas4200c.f08.gator.edwards/Wk4 Class Notes

Single Cell Airfoil


Shear flow is constant $$q = q_1 = q_2 = q_3$$

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

$$\qquad= \frac{1}{2 G \bar{A}}\ q\left[ \frac{\pi\left( \frac{b}{2}\right)}{t_{1}} + \frac{a}{t_{2}}+\frac{\sqrt{a^2+b^2}}{t_3}\right]$$

$$\ $$

$$\qquad = (\frac{1}{qG})141.13$$

Maximum Shear Stress $$\qquad \tau_{max}$$

$$\tau _{max}=\frac{q}{min \ of\ ( t_1 ,t_2,t_3) }$$

If $$\tau _{max} = \tau _{allow}\ \ \ $$ (given)

and since $$\ \ \ q = \frac{t}{2\bar{A}}\ \ $$

then $$\ \ T_{allow}=2\bar{A}\tau_{allow}min\left\{ t_1,t_2,t_3 \right\}$$

Let $$\ \ \tau_{allow} = \ 100GPa.\ \ $$ Find $$T_{allow}$$

$$T_{allow} = $$7028 MN•m

Multicell Airfoil


Put 2 partitions in airfoil and find rate of twist by dividing c into 3 parts

We will go over a specific example first and a generalization later. The first constraint equation can be considered the torque of the multicell airfoil

$$T=T_{1}+T_{2}=2q_{1}\bar{A}_{1}+2q_{2}\bar{A}_{2}$$

The second being that the airfoil is a rigid body and must twist equally throughout the airfoil

$$\theta _{1} = \theta _{2}$$

These areas for the simplified airfoil are easily calculated with

$$\bar{A}_{1}=(a)(c)$$ $$\bar{A}_{2}=(b)(c)$$

To find the angle of twist as previously shown in the road map of Homework 3 is

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

For each cell the angle is represented as

$$\theta _{1}=\frac{1}{2G\bar{A}_{1}} \left[\frac{q_{1}}{t_{1}}(a)+\frac{q_{1}}{t_{1}}(c)+\frac{q_{1}}{t_{1}}(a)+\frac{q_{1}}{t_{12}}(c)-\frac{q_{2}}{t_{12}}(c) \right]$$

$$\theta _{2}=\frac{1}{2G\bar{A}_{2}} \left[\frac{q_{2}}{t_{2}}(c)+\frac{q_{2}}{t_{2}}(b)+\frac{q_{2}}{t_{2}}(b)+\frac{q_{2}}{t_{12}}(c)-\frac{q_{1}}{t_{12}}(c) \right]$$

q1 and q2 can be considered unkowns and since we now have to constraint equations we can solve for both q's. This is shown in detail in the homework section representing q in terms of T.

Next, using the two expresions for $$\theta_{}$$ an expresion between $$\theta_{}$$ and T based on the notion that

$$\theta _{1} = \theta _{2}=...= \theta _{c_{n}}$$

Which is in turn equal to

$$=\frac{T}{2GJ}$$

From this we can now easily solve for J

Strain Matrix
There are 6 components to the 3D strain matrix because of symmetry $$$$.

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

$$\varepsilon _{12}=\varepsilon _{21}$$

$$\varepsilon _{23}=\varepsilon _{32}$$

$$\varepsilon _{13}=\varepsilon _{31}$$

The equation for each element of the 3D strain matrix

$$\varepsilon _{ij}=\frac{1}{2}\left(\frac{\partial u_{i}}{\partial x_{j}}+\frac{\partial u_{j}}{\partial x_{i}}\right)$$

And because of this equation there are only 6 independent component of the 3D strain matrix. Also the stress tensor matrix only has 6 independent components as well.

$$\sigma =\left[\sigma _{ij} \right]_{3x3}$$