User:Eas4200c.f08.gator.edwards/Wk4 Homework Problems

Ideal and Rectangular airfoil
The skin thickness is 2mm, spar thickness 3mm, the c value is 0.5 m. First the ideal airfoil will be calculated, a rectangular two cell section, a single cell airfoil, and then compared to a three cell airfoil with spars at .25c and .75c from the leading edge.

Ideal Airfoil



$$\bar{A}=\frac{\pi }{2}(h^{2})+\frac{1}{2}(h)(l)$$

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

$$\theta =\frac{q}{2\bar{A}G}\oint{\frac{ds}{t}}$$

$$\theta =\frac{q}{2\bar{A}G}\left[\frac{h\pi }{2t_{skin}} +\frac{l}{t_{skin}}+\frac{\sqrt{h^{2}+l^{2}}}{t_{skin}} \right]$$

$$J=\frac{T}{\theta G}$$

Rectanglular Two Cell



$$T=2\sum_{i=1}^{2}{q_{i}\bar{A}_{i}}$$ gives the next three equations

$$q_{1}=\frac{T}{2\bar{A}_{1}}$$, $$q_{2}=\frac{T}{2\bar{A}_{2}}$$

$$\theta _{i}=\frac{1}{2\bar{A}_{i}G}\oint{\frac{q.ds}{t}}$$

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

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

$$J=\frac{T}{\theta G}$$

Strain Matrix Symmetry
Since the equation for each element of the strain matrix is given by

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

So the term $$\varepsilon _{ji}$$ is equal to the term $$\varepsilon _{ij}$$

$$\varepsilon _{ji}=\frac{1}{2}\left[\frac{\partial u_{j}}{\partial x_{i}}+\frac{\partial u_{i}}{\partial x_{j}} \right]$$

This proves that the strain matrix is symmetric.

Hooke's Law
Normal Strains:

$$\epsilon _{xx}=\frac{\sigma _{xx}}{E} - \frac{\nu \ \sigma_{yy}}{E} - \frac{\nu \  \sigma_{zz}}{E}$$

$$\epsilon _{yy}=-\frac{\nu \ \sigma _{xx}}{E} + \frac{\sigma_{yy}}{E} - \frac{\nu \  \sigma_{zz}}{E}$$

$$\epsilon _{zz}=-\frac{\nu \ \sigma _{xx}}{E} - \frac{\nu \  \sigma_{yy}}{E} + \frac{\sigma_{zz}}{E}$$

$$\ $$

Shear Strains:

$$\gamma _{xy} = 2\epsilon _{xy} = \frac{\tau _{xy}}{G}$$

$$\gamma _{yz} = 2\epsilon _{yz} = \frac{\tau _{yz}}{G}$$

$$\gamma _{zx} = 2\epsilon _{zx} = \frac{\tau _{zx}}{G}$$

Strains in a Uniform bar
From the Kinematic Assumptions derived earlier:

$$u_{y} = -\Theta x z$$

$$u_{z} = \Theta x y$$

$$u_{x} = \Theta \Psi(y,z)$$

(In this coordinate system the x-axis lies along the axial direction and the cross-section lies in the y-z plane)

The equation for normal strains is the partial derivative of the displacement taken with respect to the direction of the displacement.

$$\epsilon_{xx} = \frac{\partial u_{x}}{\partial x} $$

$$\epsilon_{yy} = \frac{\partial u_{y}}{\partial y} $$

$$\epsilon_{zz} = \frac{\partial u_{z}}{\partial z} $$

Substituting the displacements into these formulae renders:

$$\epsilon_{xx} = \frac{\partial u_{x}}{\partial x} = \frac{\partial(\Theta\Psi(y,z))}{\partial x} = 0$$

$$\epsilon_{yy} = \frac{\partial u_{y}}{\partial y} = \frac{\partial(-\Theta x z)}{\partial y} = 0$$

$$\epsilon_{zz} = \frac{\partial u_{z}}{\partial z} = \frac{\partial(\Theta x y)}{\partial z} = 0$$

The equation for shear strains is the partial derivative of the displacement in direction a taken with respect to b, plus the partial derivative of the displacement in direction b with respect to a.

$$\gamma_{ab} = \frac{\partial u_{a}}{\partial b} + \frac{\partial u_{b}}{\partial a}$$

Hooke's Law for Isotropic Elasticity
$$\begin{Bmatrix}\epsilon _{11}\\\epsilon _{22}\\\epsilon _{33}\\\gamma _{23}\\\gamma _{31}\\ \gamma _{12}\end{Bmatrix}= \begin{Bmatrix} \frac{1}{E} & \frac{-\nu}{E} & \frac{-\nu}{E} & 0 & 0 & 0\\ \frac{-\nu}{E}& \frac{1}{E}& \frac{-\nu}{E} & 0 &  0&0 \\ \frac{-\nu}{E}& \frac{-\nu}{E} & \frac{1}{E} & 0 & 0 & 0\\ 0 & 0 & 0& \frac{1}{G} & 0 &0 \\ 0&0 & 0 & 0 & \frac{1}{G} & 0\\ 0& 0 & 0&0  & 0 & \frac{1}{G} \end{Bmatrix}\begin{Bmatrix} \sigma _{11}\\ \sigma _{22}\\ \sigma _{33}\\ \sigma _{23}\\ \sigma _{31}\\ \sigma _{12} \end{Bmatrix}$$

$$\ $$

Which can also be written as:

$$\ $$

$$\begin{Bmatrix}\epsilon _{11}\\\epsilon _{22}\\\epsilon _{33}\\\epsilon _{23}\\\epsilon _{31}\\ \epsilon _{12}\end{Bmatrix}= \begin{Bmatrix} \frac{1}{E} & \frac{-\nu}{E} & \frac{-\nu}{E} & 0 & 0 & 0\\ \frac{-\nu}{E}& \frac{1}{E}& \frac{-\nu}{E} & 0 &  0&0 \\ \frac{-\nu}{E}& \frac{-\nu}{E} & \frac{1}{E} & 0 & 0 & 0\\ 0 & 0 & 0& \frac{1}{2G} & 0 &0 \\ 0&0 & 0 & 0 & \frac{1}{2G} & 0\\ 0& 0 & 0&0  & 0 & \frac{1}{2G} \end{Bmatrix}\begin{Bmatrix} \sigma _{11}\\ \sigma _{22}\\ \sigma _{33}\\ \sigma _{23}\\ \sigma _{31}\\ \sigma _{12} \end{Bmatrix}$$

Notes on the Poisson's Ratio
For steel: $$\ \nu = 0.3$$

For cork: $$\ \nu = 0$$

For rubber: $$\ \nu = 0.5$$

$$\ $$

For most materials, $$\ 0\leq \nu \leq 0.5 \qquad \ $$ but some materials have been developed that have a negative poisson ratio.

$$\ $$

$$\ $$

$$\ $$

$$\ $$

$$\ $$

$$\ $$

$$\ $$

$$\ $$

$$\ $$

$$\ $$

$$\ $$

$$\ $$

$$\ $$

$$\ $$

$$\ $$

$$\ $$

$$\ $$

$$\ $$

$$\ $$

$$\ $$

$$\ $$

$$\ $$

$$\ $$

$$\ $$

$$\ $$

$$\ $$

$$\ $$