User:Eas4200c.f08.aero6.ortega/HW4

 Comment: The deadline for HW4 was 21:00 UTC, 24 Oct 08, and the official version submitted before the deadline was 00:02, 24 October 2008 and can be found here. The above official version did not contain the list of all contributing members, who added their signature to the version 12:06, 31 October 2008 here. However, this link was not originally updated in the HW table. This updated page is identical to the version that was submitted by the deadline, except that the signatures of the Contributing Team Members have been updated, and this comment box was added. Here is a comparison between these two versions. This is a link to the updated final submission Eas4200c.f08.aero6.behnke 15:18, 12 January 2009 (UTC) (UTC)

In a single-cell airfoil, shear flow is constant.

$$q = q_{1} = q_{2} = q_{3}\! $$

$$\Theta = \frac{1}{2G \bar{A}}\, q \sum_{j=1}^{3} \frac{l_{j}}{t_{j}} = \frac{1}{2G \bar{A}}\, q \left [\, \frac{\pi\frac{b}{2}}{t_{1}} + \frac{a}{t_{2}} + \frac{\sqrt{a^{2} + b^{2}}}{t_{3}}\, \right ]$$

Knowing the dimensions of the airfoil, where $$t_{1} = 0.08\ m\!$$ and $$t_{2} = t_{3} = 0.01\ m\!$$, also $$a = 4\ m\!$$ and $$b = 2\ m\!$$, the rate of twist angle may be computed.

Also,

$$\bar{A} = \frac{1}{2} \,a\,b + \frac{1}{2} \pi \frac{b}{2}^{2} = \frac{1}{2} 4\cdot 2 + \frac{1}{2} \pi \frac{2}{2}^{2} = 5.571\ m^{2}$$

Thus the rate of twist may be computed as

$$\Theta = \frac{1}{\bar{A}}\, q \sum_{j=1}^{3} \frac{l_{j}}{t_{j}} = \frac{1}{2\, G\, \left (5.571\right )}\, q \left [\, \frac{\pi\frac{2}{2}}{0.08} + \frac{4}{0.01} + \frac{\sqrt{4^{2} + 2^{2}}}{0.01}\, \right ] = \frac{79.56\, q}{G}$$

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

If $$\tau_{max} = \tau_{allow}\!$$ (given) and since $$q = \frac{T}{2\bar{A}}$$ then $$\ T_{allow} = 2\, \bar{A}\, \tau_{allow}\cdot min[t_{1},t_{2},t_{3}]$$

Letting $$\tau_{allow} = 100\, GPa,\ T_{allow} = 2\left (5.571\right )\left (100\right )\left (0.01\right ) = 11.14\, Nm\!$$

Specific Example
First Generalization

Find $$\Theta\!$$ as a function of $$T\!$$ and $$J\!$$ (torsional constant)

(1) $$T = T_{1} + T_{2} = 2Q_{1}\bar{A_{1}} + 2q_{2}\bar{A_{2}}$$

$$\bar{A_{1}} = ac\!$$ and $$\bar{A_{2}} = bc\!$$

rate of twist $$\Theta_{1} = \frac{1}{2G \bar{A_{1}}} \oint \frac{q_{i}}{t_{1}s}ds$$

Where $$\oint \frac{q_{i}}{t_{1}s}ds = \frac{q_{1}}{t_{1}}(2a + c) + \frac{(q_{1} - q_{2})}{t_{12}}c$$

$$\Theta_{2} = \frac{1}{2G\bar{A_{2}}}\left [\frac{2bq_{2}}{t_{2}} + \frac{cq_{2}}{t_{2}} + \frac{\left (q_{2}- q_{1}\right )}{t_{12}}c\right ]$$

$$\Theta = \Theta_{1} = \Theta_{2}\!$$

HW
The entire airfoil must twist at the same rate if it is assumed to be a rigid body. Solving for $$q_{1}\!$$ and $$q_{2}\!$$ in terms of $$T\!$$, the $$\Theta = \frac{T}{GJ}\!$$ and deduce $$J\!$$.

NACA 2415 HW
Graph a three-cell airfoil with the first partition at 0.25 chord and the second at 0.75 chord. This problem contains three unknowns...

(1) $$T = 2\sum_{i=1}^{3} q_{i}\bar{A_{i}}$$ in terms of T

(2) $$\Theta_{1} = \Theta_{2}\!$$

(3) $$\Theta_{2} = \Theta_{3}\!$$

Note that all three cells are made of the same material so they have the same shear modulus which will cancel in the following mathematics.

Find $$J\!$$

Theory Derivation
So far, we have already derived the following equations.

$$T = GJ\Theta\!$$

$$T = 2Q\bar{A}\!$$

Also the ad-hoc rate of twist equation.

$$\Theta = \frac{1}{2G\bar{A}} \oint \frac{q}{t}ds$$ Now we will derive the rate of twist equation for a non-uniform cross-section subject to twist.

The displacement $$PP\, '\!$$ due to $$\alpha\!$$:

$$\frac{PP\, '}{OP} = \tan \alpha$$

For small angles, the $$\tan \alpha = \alpha\!$$

Projecting the displacement $$PP\, '\!$$ in the direction perpendicular to $$OP\, '\!$$:

$$P\, '' = PP\, '\cos \alpha = \left (OP\tan \alpha\right )\cos \alpha = \left (OP\cos \alpha\right )\tan \alpha$$

where $$OP\cos \alpha=OP\, ''\!$$

Recall $$OP = r\!$$ (radial chord) and $$OP\, '' = \rho\!$$

$$PP\, ' = \left ( r\cos \alpha \right )\tan \alpha\!$$

where $$r\cos \alpha = \rho\!$$

and $$\tan \alpha = \alpha\!$$

So $$PP\, '\!$$ is the displacement of $$P\!$$ in the direction tangent to the lateral surface of the bar.

Strain $$\gamma = \frac{PP\, ''}{dx} = \frac{\rho\alpha}{dx} = \rho\Theta \ $$, where $$\ \Theta = \frac{\alpha}{dx}$$

Rafael made the comment in class that $$\alpha\!$$ is very small and may be denoted as $$d\alpha\!$$.

Recall Hooke's Law $$\tau = G\gamma = G\rho\Theta\!$$

$$\tau\!$$ is a variable dependent of the contour of the cross-section as is $$\rho\!$$, while $$\Theta\!$$ is relative to the x-axis. Integrating $$\tau\!$$ along the contour of the wall denoted by $$C\!$$ yields:

$$\oint_{C}\tau(s)ds = G\Theta\oint_{C}\rho(s)ds$$

where

$$\tau(s)\!$$ is $$\frac{q(s)}{t(s)}$$

$$\oint_{C}\rho(s)ds = 2\bar{A}$$

In general $$q\!$$ is constant and hence the expression from p.20.2 is derived.

Ad-hoc Rate of Twist Explanation
What is ad-hoc about the above derivation of $$\Theta\!$$ expressed on p.20.2 and the derivation of $$T = 2q\bar{A}$$?

1) strain $$\gamma\!$$ must be obtained using the displacement of $$P\!$$ in the direction tangent to $$C\!$$ at $$P\!$$, but $$PP\, ''$$ on p.20.2 is not necessarily tangent to $$C\!$$ (but very close).

2) $$\tau = \frac{q}{t}$$ is obtained from the ad-hoc assumption that $$\tau\!$$ was uniform across the wall thickness.

3) Inconsistency in assumption on size of $$ \alpha \!$$: to get line $$PP\, '$$, we assumed $$ \alpha \!$$ small; to get $$PP\, ''$$,we assumed $$ \alpha \!$$ is finite $$ (cos(\alpha) \!$$ and $$ P = r(cos(\alpha) \!$$; then we reintroduce small $$ \alpha \!$$ following that.

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Formal Justification by Elasticity Theory
A. Kinematic Assumption

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

Where $$\Theta\!$$ is a constant with respect to $$x\!$$ for a uniform bar.

$$U_{y}(x,z) = -\Theta xz\!$$

$$U_{z}(x,y) = \Theta xy\!$$

To transfer equations in Sun[2006] to those using a unified notation, all that is needed is a cyclic transformation of variables. Thus the following relationships may be expressed:

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

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

$$\epsilon_{yy} = \frac{\delta u_{y}(x,z)}{\delta y} = 0$$

$$\epsilon_{zz} = \frac{\delta u_{z}(x,y)}{\delta z} = 0$$

$$\gamma_{yz} = \frac{\delta u_{y}(x,z)}{\delta z} + \frac{\delta u_{z}(x,y)}{\delta y} = 0$$

since $$\frac{\delta u_{y}(x,z)}{\delta z} = -\Theta x$$

and $$\frac{\delta u_{z}(x,y)}{\delta y} = \Theta x$$

There are six strain components in a 3D derivative, even though the strain matrix is represented as shown below

$$ \epsilon_{ij} \Big|^{i=x,y,z}_{j=x,y,z} = \begin{bmatrix} \epsilon_{xx} & \epsilon_{xy} & \epsilon_{xz} \\ \epsilon_{yx} & \epsilon_{yy} & \epsilon_{yz} \\ \epsilon_{zx} & \epsilon_{zy} & \epsilon_{zz} \end{bmatrix}$$

The matrix can also be represented as

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

$$\epsilon_{12} = \epsilon_{21} \!$$  and  $$\epsilon_{xy} = \epsilon_{yx}\!$$

Therefore, there are six independent components of $$\epsilon$$

There is a similar situation for the stress tensor.

The symmetry of $$\epsilon$$ is not related to isotropy. Isotropic elasticity is related to the material behavior

Due to the stress-strain relation... $$\sigma _{xx} = \sigma _{yy} = \sigma _{zz} = \tau _{yz} = 0 \!$$

Tensorial Notation
The strain state can be represented as a tensor in multiple ways. The first uses the coordinate axis as subscripts to indicated directions. A second more universal method uses numbered indexes.

$$ \underline{\epsilon} = \begin{bmatrix} \epsilon_{xx}     & \epsilon_{xy} & \epsilon_{xz}      \\ \epsilon_{yx} & \epsilon_{yy} & \epsilon_{yz} \\ \epsilon_{zx}     & \epsilon_{zy} & \epsilon_{zz} \end{bmatrix} = \begin{bmatrix} \epsilon_{11}     & \epsilon_{12} & \epsilon_{13}      \\ \epsilon_{21} & \epsilon_{22} & \epsilon_{23} \\ \epsilon_{31}     & \epsilon_{32} & \epsilon_{33} \end{bmatrix} $$

The numbered indexes allow for the use of more generalized formulas in terms of indexes i and j. For instance, the strain equation can be written as:

$$\epsilon_{ij} = \frac{1}{2}(\frac{\partial{u_{i}}}{\partial{x_{j}}} + \frac{\partial{u_{j}}}{\partial{x_{i}}}) $$

Where in $$\epsilon_{ij}$$, $$i$$ corresponds to the row and $$j$$ corresponds to the column and:

$$x \longrightarrow x_{1}$$

$$y \longrightarrow x_{2}$$

$$z \longrightarrow x_{3} $$

Relating strains and stresses to Young's Modulus, E, Poisson's Ratio, $$\nu$$, and the Shear's Modulus, G
Most materials have a Poisson's Ratio between 0 and 0.5
 * Steel for example has a $$\nu = 0.3$$
 * Rubber has a $$\nu = 0.5$$
 * Cork has a $$\nu = 0$$

Normal Strains:
 * $$ \epsilon_{xx} = \frac{\sigma_{xx}}{E} - \frac{ \nu \sigma_{yy}}{E} - \frac{ \nu \sigma_{zz}}{E}$$


 * $$ \epsilon_{yy} = \frac{\sigma_{yy}}{E} - \frac{ \nu \sigma_{xx}}{E} - \frac{ \nu \sigma_{zz}}{E}$$


 * $$ \epsilon_{zz} = \frac{\sigma_{zz}}{E} - \frac{ \nu \sigma_{xx}}{E} - \frac{ \nu \sigma_{yy}}{E}$$

Shear Strains:
 * $$ \gamma_{xy} = 2 \epsilon_{xy} = \frac{ \tau_{xy}}{G}$$


 * $$ \gamma_{xz} = 2 \epsilon_{xz} = \frac{ \tau_{xz}}{G}$$


 * $$ \gamma_{yz} = 2 \epsilon_{yz} = \frac{ \tau_{yz}}{G}$$

Due to symmetry of the $$ \epsilon_{ij}$$ matrix, $$\epsilon$$ and $$\sigma$$ can be represented in column form:
 * $$\begin{Bmatrix}

\epsilon_{ij} \end{Bmatrix}$$ = $$\begin{Bmatrix} \epsilon_{11} \\ \epsilon_{22} \\ \epsilon_{33} \\ \epsilon_{23} \\ \epsilon_{31} \\ \epsilon_{12} \\ \end{Bmatrix}$$    and      $$\begin{Bmatrix} \sigma_{ij} \end{Bmatrix}$$ = $$\begin{Bmatrix} \sigma_{11} \\ \sigma_{22} \\ \sigma_{33} \\ \sigma_{23} \\ \sigma_{31} \\ \sigma_{12} \\ \end{Bmatrix}$$ Using Hooke's Law for isotropic elasticity:
 * $$\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}$$
 * or
 * $$\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}$$

Airfoil Study Using Matlab
The airfoil was then evaluated by using the program from multiple points within the airfoil. The point of origin, $$P_{o}$$, was changed multiple times while sweeping through smaller arcs. The combination of these arcs encompassed the entire airfoil. The procedure was as follows.

1. Set $$P_{o} = D$$, then sweep from point B $$ \rightarrow $$ L $$\rightarrow $$ E

2. Set $$P_{o} = E$$, then sweep from point B $$ \rightarrow $$ F

3. Set $$P_{o} = F$$, then sweep from point E $$ \rightarrow $$ H

4. Set $$P_{o} = G$$, then sweep from point H $$ \rightarrow $$ T $$\rightarrow $$ F

Matlab Certification

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

Kris Loper Eas4200c.f08.aero6.loper 19:51, 22 October 2008 (UTC)

Scott Chastain Eas4200c.f08.aero6.chastain 22:09, 23 October 2008 (UTC)

Matt Inman Eas4200c.f08.aero6.inman 23:58, 23 October 2008 (UTC)

70.171.31.121 03:58, 26 October 2008 (UTC)

Eduardo Villalba Eas4200c.f08.aero6.villalba 18:19, 27 October 2008 (UTC)

Contributing Team Members
Kris Loper Eas4200c.f08.aero6.loper 19:51, 22 October 2008 (UTC)

Scott Chastain Eas4200c.f08.aero6.chastain 22:11, 23 October 2008 (UTC)

Matt Inman Eas4200c.f08.aero6.inman 23:58, 23 October 2008 (UTC)

MarlanaBehnke Eas4200c.f08.aero6.behnke 12:06, 31 October 2008 (UTC) (previously forgot to log in)

Felipe OrtegaEas4200c.f08.aero6.ortega 18:15, 27 October 2008 (UTC)

Eduardo Villalba Eas4200c.f08.aero6.villalba 18:19, 27 October 2008 (UTC)