User:Eas4200c.f08.aero.e/Week 6

=Monday=

Homework #3
NACA 4 digit airfoil series

For homework number four teams will develop Matlab code to plot and analyze a NACA airfoil. This homework assignment will be an ongoing project with the ultimate going of developing Matlab code to do structural analysis of a wing. For the first assignment teams should:


 * Plot the airfoil.
 * Compute location of the airfoil.
 * Plot the centroid of the airfoil with cross hair.
 * Compute A_bar (from any point in the plane).

The user will input the number of segments and the coordinate of the upper and lower surfaces of the airfoil.

Matlab code is provided to plot a straight line. Teams should modify the code to plot the entire airfoil. Given some wing thickness the code must then find the centroid of the airfoil segment. A summation of the individual segment centroids will yield the overall airfoil centroid.

For each small airfoil the perpendicular distance from the origin (or other point) will need to be calculated.

This distance is assigned the variable $$\rho$$ as shown in the illustration below for vector $$\overrightarrow{PQ}$$



Note,

$$ \overrightarrow{dT} = \overrightarrow{r} $$ $$ \mathbf{x}$$ $$\overrightarrow{dF} = q\overrightarrow{r}$$ $$ \mathbf{x}$$ $$\overrightarrow{PQ} = (2dA)\overrightarrow{i}$$

$$ T = \oint_C \,dT $$

$$= q \oint_C \rho \, dl $$

$$=2q \int\limits_A \, dA$$

END OF HOMEWORK 3

Torsion of Uniform, non-circular bars
When a torque is applied to a uniform bar it leads to warping along the longitudinal axis of the bar. The amount of warping depends on the applied torque.

Warping - axial displacement along longitudinal axis of a point on the deformed (rotated) cross section.



Virtual Displacement:

Uy = y - component of displacement vector $$\overrightarrow{PP'}$$

Uz = z - component of $$\overrightarrow{PP'}$$

Note: Horizontal Bar

Matlab: NACA 4 Digit Airfoil Series
=Wednesday=

 Torsion Analysis 

PP' is a vector that describes displacement.

$$u_{y}= $$ y-Displacement (Horizontal)

$$u_{y}= $$ Z-Displacement (Vertical)


 * $$\theta =\frac{\alpha }{x},$$

$$\theta = $$ rate of twist

$$u_{y}= -\theta xz,$$

$$u_{z}= PP^'cos\beta ,$$

$$u_{z}= OP\alpha cos\beta ,$$

$$u_{z}= \alpha y,$$

$$OP\alpha cos\beta =\alpha y,$$

$$OP cos\beta = y,$$

$$u_{z}=\theta +xy,$$

$$u_{x}=\theta \psi(x,y) ,$$

source: http://www.3dscientifics.com/airfoil_cfd.htm

Roadmap to torsional analysis of aircraft wing

A. Kinematic assumption,  (sec. 3.2)

B. Strain - Displacement relations,    (sec. 3.2)

C. Equilibrium equation - stresses,    (sec. 3.6)

D. Prandtl stress $$\phi$$,    (sec. 3.15)

E. Strain compatibility,   (sec. 3.17)

F. Equation for $$\phi$$,    (sec. 3.19)

G. Boundary consitions for $$\phi$$,    (sec. 3.24)

H.

$$T=2\int\int_{A}^{}{}\phi dA$$

$$J=\frac{-4}{\nabla^{2}\phi }\int \int _{A}\phi dA$$

I. Thin-walled cross-section,  (Pg. 5-2) Ad Hoc assumption on shear flow $$T = 2q\bar{A}$$ Formal derivation can be found in section 3.5

J. Twist angle $$\theta$$,  Method 1  (sec.3.56)

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

K. Section 3.6 on multicell thin-walled cross-sections.

=Friday= 10/3/08 Lecture 17

Roadmap continued...

K. Multi-cell section (cell i=1,...$$n_c$$):

K1. The total torque T for a multi-cell section is defined as:

$$T=2\sum_{i=1}^{n_c}{q_i\bar{A}_i}$$

where $$q_i$$ is the shear flow in cell i, and $$\bar{A}_i$$ is the average area in cell i.

Justification for Torque equation

Define $$T_i=2q_i\bar{A}_i$$

as the torque generated by one cell. It then follows that T is the sum of all individual cell torques.

K2. The shape of air foil is rigid in the (y,z) plane (however, it can warp in and out of the plan in the x-direction.)

This rigidity leads to the following relation about the angles of rotation for each cell:

$$\theta =\theta _1=\theta _n$$

and $$\theta =\frac{1}{2G_i\bar{A}_i}\oint_{}^{}{\frac{q_i}{t_i}ds}$$

where $$G_i$$ is the shear modulus of the cell and $$t_i$$ is the thickness of each segment. $$t_i$$ can also be a function of $$s$$, the curvilinear coordinate along the cell wall.

Example

Now, let us refer back to Problem 1.1 to demonstrate the above equations in a single cell section.

First, calculate the area: $$\bar{A}=\frac{1}{2}\pi(\frac{b}{2})^2 + \frac{1}{2}(b*a)$$ = $$\bar{A}=\frac{1}{2}\pi(\frac{2}{2})^2 + \frac{1}{2}(2*4) = 5.57$$

Then, Shear flow: $$T= 2q\bar{A}$$ => $$q=\frac{T}{2\bar{A}}$$

Finally Twist Angle: $$\theta =\frac{1}{2}G\bar{A}\sum_{j=1}^{n_c} q_j\frac{l_j}{t_j}$$, where $$j$$ is the index for segment number within the cell.

NOTE:

integral sign: $$\int$$, standing for summation (continuous)

summation: $$\sum{}$$, (discrete) sum

This method is derived from the Riemann sum, which is computed by dividing a body into smaller rectangles and adding their areas together to arrive at a total area for the body.