User:Eas4200c.f08.aero.lee/HW4

= The Multicell Airfoil = First we will look at a specific example of how to analyze a multicell airfoil; the generalization will be provided later. Below is a schematic of a simple approximation of a two-celled airfoil.



For this example, $$t_1 = 0.3$$ cm, $$t_2 = 0.5$$ cm, $$t_{12} = 0.4$$ cm, $$a = 30$$ cm, $$b = 60$$ cm, and $$c = 40$$ cm.

To solve for the total effect of torsion on such an airfoil, contributions from each cell must be taken into account. Each cell will undergo its own shear flow and torque, and the overall torque can be equated as the sum of the torque in each cell.


 * $$T = T_1 + T_2 \qquad \qquad (1)$$

where $$T_1$$ and $$T_2$$ are the torque in cells 1 and 2, respectively. Using the equation $$T = 2q\bar A$$, where $$q$$ is shear flow and $$\bar A$$ are shear flow and average area. Applying this and equation 1 above becomes:


 * $$T = 2q_1\bar {A_1}+2q_2\bar {A_2}$$

where


 * $$\bar {A_1} = ac$$
 * and
 * $$\bar {A_2} = bc$$

It should be noted that both cells have the same rate of twists, $$\theta_1$$ and $$\theta_2$$, respectively.


 * $$\theta_1 = \theta_2 \qquad \qquad (2)$$

Also, an expression for $$\theta$$ has been previously derived and for each cell it is as follows


 * $$\theta_1 = \frac {1}{2G\bar {A_1}}\oint \frac {q_1}{t_1(s)}\, ds$$


 * and


 * $$\theta_2 = \frac {1}{2G\bar {A_2}}\oint \frac {q_2}{t_2(s)}\, ds$$

For this simplified model these equations become:


 * $$ \theta_1 = \frac{1}{2G\bar{A_1}}\oint_{}^{}{\frac{q_1}{t_1}ds} = \frac{1}{2G\bar{A_1}}\left(2\frac{q_1}{t_1}a + \frac{q_1}{t_1}c + (q_1 - q_2)\frac{c}{t_{12}} \right) $$


 * $$ \theta_2 = \frac{1}{2G\bar{A_2}}\oint_{}^{}{\frac{q_2}{t_2}ds} = \frac{1}{2G\bar{A_2}}\left(2\frac{q_2}{t_2}b+ \frac{q_2}{t_2}c + (q_2 - q_1)\frac{c}{t_{12}} \right) $$

= Different Methods for Determining Ā =





Two methods that were discussed in class was approximating area by using the triangular quadrature method shown in figure 2, and the trapezoidal approximation shown in figure 3. It has been deemed that using the triangular quadrature method is more elegant.

= Ad Hoc Engineering Method =

The equation $$ \theta = \frac{1}{2G\bar{A}}\oint_{}^{}{\frac{q}{t}ds} $$ can be developed in an ad hoc manner. Consider a uniform bar of non-circular cross-section subject to twist.



The displacement of point P to P' can be given as $$ \frac{PP'}{OP} = tan(\alpha) \approx \alpha $$ using a small angle approximation. Projecting the displacement onto a path perpendicular to OP' provides $$ PP'' = PP'cos(\alpha) $$. It is also seen that $$ OPcos\alpha = OP'' $$ via a project of OP across angle $$ \alpha $$ Thus,


 * $$ PP = (OPtan\alpha)cos\alpha = (OPcos\alpha)tan\alpha = (OP)tan\alpha $$

Recall from a previous lecture that $$ OP = r $$ and $$ OP'' = \rho $$. Then, the displacement of P in the direction perpendicular to the surface of the bar, using a small angle approximation, is given as


 * $$ PP'' = (rcos\alpha)tan\alpha = \rho tan\alpha \approx \rho \alpha $$

Also recall the definition of the rate of twist, $$ \theta $$, along a section of length dx is defined as $$ \theta = \frac{\alpha}{dx} $$. Examining the shear strain that results along an incremental section of the bar, substituting the above equations, it is seen that


 * $$ \gamma = \frac{PP''}{dx} = \frac{\rho \alpha}{dx} = \rho \left(\frac{\alpha}{dx}\right) = \rho \theta $$

Remembering that stress is calculated in the curvilinear coordinate system


 * $$\tau=G\,\gamma=G\rho\,\theta$$


 * $$\tau(s)=G\rho(s)\,\theta(x)$$

We can Integrate along the contour to arrive at our targeted equation


 * $$\oint{\tau(s)ds}=\oint{G\rho(s)\,\theta(x)ds}=G\,\theta(x)\oint{\rho(s)ds}$$


 * $$\oint{\frac{q(s)}{t(s)}ds}=G\,\theta(x)\,2\,\bar{A}$$


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

= Normal Strains as Defined by the Theory of Elasticity =

The Theory of Elasticity gives us the equations of displacement within a bar subjected to a twist rate, $$ \theta $$, as:


 * $$ u_x = \theta\psi (y,z) $$


 * $$ u_y = -\theta xz $$


 * $$ u_z = \theta xy $$

where $$ \psi (y,z) $$ is a function of only y and z.

= Shear Strains as Defined by the Theory of Elasticity =

From above, remember that the Theory of Elasticity gives us the following kinematic assumptions:




 * $$ u_x = \theta\psi (y,z) $$


 * $$ u_y = -\theta xz $$


 * $$ u_z = \theta xy $$

Looking at the following figure, Figure 5:

Looking at the equations they vary slightly from those used in "Aircraft Structures" by C.T. Sun primarily due to the fact that different coordinate systems are used. Looking at the figure, the simple transformation in notation can be made between each of the axes that are shown.

Using the Kinematic Assumptions and that the relationship between normal strains and shear strains the following results are obtained:


 * $$\displaystyle \epsilon_{xx} = \displaystyle \epsilon_{yy} = \displaystyle \epsilon_{zz} = \displaystyle \gamma_{yz} = 0$$

=Looking at Strain Compenents in 3D=

When looking at the strain coefficients in 3D, we know that there are 9 coefficients, arranged in a 3X3 matrix, given by the matrix :

$$\mathbf{\displaystyle \epsilon} =  \begin{bmatrix}

\epsilon_{xx} & \epsilon_{xy} & \epsilon_{xz} \\ \epsilon_{yx} & \epsilon_{yy} & \epsilon_{yz} \\ \epsilon_{zx} & \epsilon_{zy} & \epsilon_{zz} \end{bmatrix} $$

Using this matrix, we now need to convert this matrix into indicial notation. Indicial Notation is used here and allows us to substitute the following values:


 * $$ x \Leftrightarrow 1, y \Leftrightarrow 2 , z \Leftrightarrow 3 $$

Using indicial notation, the previous matrix, $$\mathbf{\displaystyle \epsilon}$$ can be rewritten as:


 * $$\mathbf{\displaystyle \epsilon} = \begin{bmatrix}

\epsilon_{11} & \epsilon_{12} & \epsilon_{13} \\ \epsilon_{21} & \epsilon_{22} & \epsilon_{23} \\ \epsilon_{31} & \epsilon_{32} & \epsilon_{33} \end{bmatrix}$$

This matrix may also be rewritten once again by using the Tensor Calculus Theory of Relativity, and be written in Tensorial Notation. Using this notation, the matrix is rewritten as follows:


 * $$\mathbf{\displaystyle \epsilon} = \begin{bmatrix}

\epsilon_{ij}\end{bmatrix}$$ $$$$

where, ij range from 1 to 3 and the following hold true:


 * i represents the row index
 * j represents the column index

Using the calculated matrix of $$\displaystyle \epsilon_{[ij]}$$ is formed in tensorial notation, the symmetry of the strain tensor can be given as follows:


 * $$\displaystyle \epsilon_{ij} = \frac{1}{2} (\frac{\partial u_i}{\partial x_j} + \frac{\partial u_j}{\partial x_i})$$

By using this symmetrical formula, there are many advantages that include the need to only remember one formula and not have to constantly keep track of the x, y, and z variables.

Taking the coordinates of x,y,z to be $$ x_1, x_2 ,  x_3 $$ respectively the symmetrical strain tensor can now be rewritten making the following substitutions: $$ x \Leftrightarrow x_1 , y \Leftrightarrow x_2 , z \Leftrightarrow x_3 $$

and written as:


 * $$\epsilon_{11} = \displaystyle \epsilon_{xx} = \frac{1}{2} (\frac{\partial u_1}{\partial x_1} + \frac{\partial u_1}{\partial x_1})=  \frac{\partial u_1}{\partial x_1} \Rightarrow \frac{\partial u_x}{\partial x}$$

=Stress Component in 3D= Similarly from the proof of $$\displaystyle \epsilon$$ we know that for the stress tensor $$\displaystyle \sigma$$ there will only be 6 independent solutions to solve for the 9 coefficients of the matrix. This 3x3 tensor $$\displaystyle \sigma$$ can be described much like the other tensor $$\displaystyle \epsilon$$ was.


 * $$\mathbf{\displaystyle \sigma} = \begin{bmatrix}

\sigma_{ij}\end{bmatrix}$$ $$$$

where similarly, here, ij range from 1 to 3 and the following conditions hold true:


 * i represents the row index
 * j represents the column index

 Q:  Does the symmetry of $$\displaystyle \epsilon$$ related to the isotropy of the material?

Referring to the following equation (1) from the notes dealing with the shear strains defined by the theory of elasticity, we obtain the following equation:

$$\displaystyle \epsilon _{xx}=\epsilon _{yy}=\epsilon _{zz}=\gamma _{yz}=0$$

Due to $$\displaystyle \epsilon$$ - $$\displaystyle \sigma$$ relationship this equation can also be rewritten as following:

$$\displaystyle \sigma _{xx}=\sigma _{yy}=\sigma _{zz}=\tau _{yz}=0$$

=Relation of Strain to Elastic Modulus and Poisson's Ratio= We wish to relate normal strain, εxx, to elastic modulus, E, and Poisson's Ratio, ν. Poisson's ratio, named after Simeon Denis Poisson, is defined as:
 * $$\nu =-\frac{lateral strain}{axial strain} =-\frac{\epsilon_{y}}{\epsilon_{x}}=-\frac{\epsilon_{z}}{\epsilon_{x}}$$

(for an axial load in the x direction, for example.)

When taking a simple rigid object and compressing it in one direction of the material will tend to deform in the opposite direction, a direction that is perpendicular to the force. As shown in the illustration above in Figure 7, the compression in the x direction causes the rigid structure to deform (positively) in the y direction.

Some common values for Poisson's Ratio are given as:

For Rubber,$$\nu_{rubber} = 0.5 $$

For steel,$$\nu_{steel} = 0.3 $$

For a simple cork,$$\nu_{cork} = 0 $$

Using Hooke's Law:
 * $$\epsilon _{xx}=\frac{\sigma _{xx}}{E}-\nu \frac{\sigma _{yy}}{E}-\nu \frac{\sigma _{zz}}{E}$$

We now wish to find $$\epsilon _{yy}$$ and $$\epsilon _{zz}$$:
 * $$\epsilon _{yy}=-\nu\frac{\sigma _{xx}}{E}+\frac{\sigma _{yy}}{E}-\nu \frac{\sigma _{zz}}{E}$$
 * $$\epsilon _{zz}=-\nu\frac{\sigma _{xx}}{E}-\nu\frac{\sigma _{yy}}{E}+ \frac{\sigma _{zz}}{E}$$

For shear strain:
 * $$\gamma _{xy}=2\epsilon _{xy}=\frac{\tau _{xy}}{G}$$

We also wish to find $$\gamma _{yz}$$ and $$\gamma _{zx}$$. In the same manner,


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

and


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

=Hooke's Law for Isotropic Elasticity= Hooke's Law for Isotropic Elasticity can be defined in matrix form as:



\begin{Bmatrix} \epsilon _{xx}\\ \epsilon _{yy}\\ \epsilon _{zz}\\ \gamma _{yz}\\ \gamma _{zx}\\ \gamma _{xy}\\ \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 _{xx}\\ \sigma _{yy}\\ \sigma _{zz}\\ \sigma _{yz}\\ \sigma _{zx}\\ \sigma _{xy}\\ \end{Bmatrix} $$ and since, for example,
 * $$\gamma _{xy}=2\epsilon _{xy}=\frac{\sigma _{xy}}{G}

$$ then
 * $$\epsilon _{xy}=\frac{\sigma _{xy}}{2G}$$

the matrix becomes:
 * $$\begin{Bmatrix}

\epsilon _{xx}\\ \epsilon _{yy}\\ \epsilon _{zz}\\ \epsilon _{yz}\\ \epsilon _{zx}\\ \epsilon _{xy} \end{Bmatrix} = \begin{vmatrix} \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{vmatrix}

\begin{Bmatrix} \sigma _{xx}\\ \sigma _{yy}\\ \sigma _{zz}\\ \sigma _{yz}\\ \sigma _{zx}\\ \sigma _{xy} \end{Bmatrix}$$

= A note on Poisson's ratio = For most materials, Poisson's ratio is positive, and is between 0 and 0.5. For example, steel has a ratio of 0.3, meaning an axial load in the x direction will result in some deformation in the y and z directions. It is interesting to note, however, that cork (used in wine bottles) has a ratio of zero. Also, some materials have a negative Poisson’s ratio. When subjected to strain in a longitudinal axis, the transverse strain in the material will actually be positive, and the material would therefore increase in that direction as well.

= Simplified Airfoil Example = A simplified airfoil is given in the figure below. It has a semicircular leading edge of diameter b = 2 m and a triangular trailing edge with one side of length a = 4 m. Thicknesses are $$ t_{1} = 0.008\ m,\ t_{2} = t_{3} = 0.01\ m $$. It is desired to determine the average area $$ \bar{A} $$, maximum torque, and the torsional constant J of the airfoil.



=NACA Airfoil Sample Roadmap= Using the following example we will outline the procedures to begin to setup the matlab project of this homework assignment by breaking up the multicell airfoil into 4 distinct areas, defined as: $$\bar {A_1}, \bar {A_{21}}, \bar {A_{22}},$$ and $$\bar {A_{3}}$$. The breakdown is shown below in Figure 9.



Similarly to what will be done in the HW 4 Matlab assignment, each of the areas will be calculated and added to develop the following sum:


 * $$\bar {A} = \bar {A_1}+\bar {A_{21}}+\bar {A_{22}}+\bar {A_3}$$

Here $$\bar {A_2}$$ represents the area of the 2nd airfoil section that is defined likewise as: $$\bar {A_2} =\bar {A_{21}}+\bar {A_{22}}$$. For simplicity this area section is divided into two sections ($$\bar {A_{21}}$$ and $$\bar {A_{22}}$$), by the dividing bisecting line, from E to F,($$\bar {EF}$$).

In Matlab, the total area will be taken by summing the individual sections. For these sections, the initial point will be set and the area calculated by sweeping areas for specific points from approximate semi circles and triangles. After the sum is taken from all of the components, careful consideration will need to be taken to compare and ensure that the results obtained in this homework 4 Matlab approximation are within 1% of the previous results obtained in the Matlab assignment in homework 3.

First to calculate area $$\bar {A_{1}}$$ we need to set the center point at $${P_{0}=D}$$ and sweep the area of the circle from the following points: $${B}\rightarrow{L}\rightarrow{E}$$

To first calculate the area of the subsection, $$\bar {A_{21}}$$, the center point will be set at $${P_{0}=E}$$ and sweep the area of the sectional triangle from points: $${F}\rightarrow{B}$$

To calculate the area of the second subsection, $$\bar {A_{22}}$$, the center point will be set at $${P_{0}=F}$$ and sweep the area of the sectional triangle from points: $${E}\rightarrow{H}$$

Finally, to calculate the area of $$\bar {A_{3}}$$ the center point will be set at $${P_{0}=G}$$ and sweep the area of the final triangle from the points: $${H}\rightarrow{T}\rightarrow{F}$$

= NACA Airfoil Matlab Code Continued =

This weeks matlab code involved adding functions to determine the torsional constant, J, for both single cell airfoils as well as multi-celled bodies.

= Matlab-code Certification =

I, the undersigned, certify that I can read, understand, and write matlab codes, and can thus contribute effectively to my team.

--Eas4200c.f08.aero.mcgilvray 16:16, 21 October 2008 (UTC)

--Eas4200c.f08.aero.lee 00:21, 21 October 2008 (UTC)

--Eas4200c.f08.aero.kurth 08:59, 21 October 2008 (UTC)

--Eas4200c.fo8.aero.oyama 16:30, 23 October 2008 (UTC)

--Eas4200c.f08.aero.strods 03:38, 24 October 2008 (UTC)

--Eas4200c.f08.aero.barcia 24:28, 24 October 2008 (UTC)

= Contributing team members =

--Eas4200c.f08.aero.mcgilvray 16:16, 21 October 2008 (UTC)

--Eas4200c.f08.aero.lee 00:21, 21 October 2008 (UTC)

--Eas4200c.f08.aero.kurth 08:59, 21 October 2008 (UTC)

--Eas4200c.fo8.aero.oyama 16:30, 23 October 2008 (UTC)

--Eas4200c.f08.aero.strods 03:38, 24 October 2008 (UTC)

--Eas4200c.f08.aero.barcia 14:29, 24 October 2008 (UTC)