User:EAS4200C.F08.AERO.CLK04D

11/7

Example of $$\overrightarrow{t}$$ in aircraft

$$\displaystyle \overrightarrow{n} = n_y\overrightarrow{j} + n_z\overrightarrow{k}$$

$$\displaystyle ||n||= 1 $$

$$\overrightarrow{n} =$$ unit normal vector



$$ \displaystyle dz=ds cos \theta | n_y=cos\theta$$

$$\displaystyle dy=ds sin \theta | n_y=cos\theta$$

$$\displaystyle \sum F_y=0= -\sigma_{yy}*(dz*1)$$ (Where 1 is the unit depth along the x-dir)


 * $$\displaystyle -\sigma_{yz}*(dz*1) + t_y(ds*1)$$

$$\displaystyle 0 = -\sigma_{yy}*ds*n_y - \sigma_{yz}*ds*n_y + t_y*ds$$

Canceling ds we arrive at the following equation:

$$\displaystyle \sigma_{yy} n_y + \sigma_{yz}n_z = t_y $$ (1)

Note: $$[t_y] = \frac{F}{L^3}$$ => Units of force/area

$$\overrightarrow{t} =$$ traction vector (distributed surface force)

$$\displaystyle [t_y] = [\sigma] $$

$$\displaystyle t_z = \sigma_{yz}*n_y + \sigma_{zz}*n_z $$ (2)

Combining (1) and (2)

$$\displaystyle \begin{bmatrix} t_y \\ t_z \end{bmatrix}\, = \begin{bmatrix} \sigma_{yy} & \sigma_{yz} \\ \sigma_{zy} & \sigma_{zz} \end{bmatrix}\, \begin{bmatrix} n_y \\ n_z \end{bmatrix}\,$$ (3)

(3) can be generalized to the 3 dimensional case:

$$\begin{bmatrix} t_1 \\ t_2 \\ t_3 \end{bmatrix}\, = \begin{bmatrix} \sigma_{11} & \sigma_{12} & \sigma_{13} \\ \sigma_{21} & \sigma_{22} & \sigma_{23} \\ \sigma_{31} & \sigma_{32} & \sigma_{33} \end{bmatrix}\, \begin{bmatrix} n_1 \\ n_2 \\ n_3 \end{bmatrix}\,$$ (4)

$$\displaystyle t_i = \sum_{j=1}^{3} \sigma_{ij}n_{j} $$

11/10

For Homework Watch Ruth Goldberg Video Section 7.7 (Plate buckling)

Torsional Analysis Continued:

Case 1: Thin walled cross section (closed) e.g. NACA airfoil

Compare to fig 3.11



Case 2: Solid cross section



Now we consider the case of uniform bar with a solid, circular cross section. Refer to road map, expression for T and J in terms of $$\displaystyle \phi$$

$$ T = 2\int\int\limits_A\phi dA$$

$$J = \frac{-4}{\nabla^2\phi}\int\int\limits_A \phi dl $$

$$\phi(y,z) = C(\frac{y^2}{a^2} + \frac{z^2}{b^2} - 1)$$

(Note: b = 0 for a circle)

Revisiting the road map,

Kinematic assumption (Compare to pg. 70)

4 - Stain components (Compare to pg. 70)

4 zero stress components (Compare to pg. 70)

$$ \epsilon - \sigma $$ relationship (2 forms: tensorial or engineering)

Rewrite $$ \epsilon - \sigma $$ relation

$$ \varepsilon_{ij} = \begin{bmatrix} {\underline A}_{3x3} & {0}_{3x3}\\ {0}_{3x3} & {\underline B}_{3x3}\\ \end{bmatrix} \sigma_{ij} $$

Let $$ \begin{bmatrix} {\underline A}_{3x3} & {0}_{3x3}\\ {0}_{3x3} & {\underline B}_{3x3}\\ \end{bmatrix} = \underline C $$

An advantage of using this form is that $$\underline C$$ can be easily inverted.

$$ \sigma_{ij} = \begin{bmatrix} {\underline A}^{-1} & \underline 0\\ \underline 0 & {\underline B}^{-1}\\ \end{bmatrix} \varepsilon_{ij} $$

$$ \begin{bmatrix} {\underline A}^{-1} & \underline 0\\ \underline 0 & {\underline B}^{-1}\\ \end{bmatrix} = \underline {C}^{-1} $$

When the two matrices are dotted into each other we get the identity matrix.

$$\underline C \underline {C}^{-1} = \underline I$$

Goal: prove 0 stress components.

$$\sigma _{xx}, \sigma _{yy} , \sigma _{zz}, \tau_{yz} = 0$$

$$\epsilon _{xx}, \varepsilon_{yy} , \varepsilon_{zz}, \gamma_{yz} = 0$$

This implies,

$$ \sigma_{ij} = \begin{bmatrix} {\underline A}^{-1} & \underline 0\\ \underline 0 & {\underline B}^{-1}\\ \end{bmatrix} \begin{bmatrix} 0\\ 0\\ 0\\ 0\\ \varepsilon_{31}\\ \varepsilon_{12}\\ \end{bmatrix} $$

(Note, using numerical notation as introduced in MIT OCW assigned reading)

$$\sigma _{11}, \sigma _{22} , \sigma _{33} = 0$$

$$\sigma _{23} = 2G\varepsilon_{23} = \frac{1}{2}\gamma_{23} = \frac{1}{2}\gamma_{yz}$$

$$\sigma _{31} = 2G\varepsilon_{31}$$

Roadmap step C:

Implicit reading: Section 2.4 in book (Non-Uniform stress field)

Consider 1-D case as a model:



Multi Cell Airfoil (Sec. 3.6)

Recall:

$$ T = 2q\bar A $$

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

Specific example first (generalization later)

Compare this example with page 94 in textbook.



$$ T = 2\sum_{i=1}^1 {q}_{i} \bar {A}_{i}  $$

Find theta as a function of T and J(torsional constant)

$$T = {T}_{1} + {T}_{2} = 2{q}_{1}\bar {A}_{1} + 2{q}_{2} \bar {A}_{2} $$   (1)

$$\bar {A}_{1} = ac, \bar {A}_{1} = bc$$

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

$$\oint\frac{{q}_{1}}{{t}_{1}}ds$$ q and t can be thought of as functions of (s), their curvilinear coordinates

therefore we can replace by discrete summation,

$${\theta}_{1} = \frac{1}{G\bar {A}_{1}} 2\frac{{q}_{1}a}{{t}_{1}} + \frac{{q}_{1}c}{{t}_{1}} + \frac{({q}_{1} - {q}_{2})c}{{t}_{12}}$$  (3)

$${\theta}_{2} = \frac{1}{G\bar {A}_{2}} 2\frac{{q}_{2}a}{{t}_{2}} + \frac{{q}_{2}c}{{t}_{2}} + \frac{({q}_{2} - {q}_{1})c}{{t}_{12}}$$  (4)

Since they are rigidly connect cells 1 and 2 must have the same rate of twist angle.

Therefore,

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

Think of q1 and q2 as 2 unknowns. Equations (1) and (2) are the 2 equations for the 2 unknowns (q1 and q2) with T being a variable. Use (1) and (2) to find an expression for q1 and q2 in terms of T.

$${q}_{1} = {\beta}_{1}T {q}_{2} = {\beta}_{2}T$$

Next use the experssion (3) or (4) to find the expression between theta and T

$$\theta = {\theta}_{1} = {\theta}_{2} = \frac{T}{GJ} $$

Finally from this equation the torsional constant J can be deduced.

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

Class Notes - 9/3/08
$$T_{max}= (2t\tau_{allow})(\frac{L}{4})^2 T_{max}$$

$$ = \frac{1}{8}tL^2\tau_{allow} = M_{max} $$

$$M_{max} = \frac{tL^2}{16} \sigma_{allow}$$

$$ => \sigma_{allow} = \frac{16{M}_{max}}{tL^2} $$

Recall: $$f(b) = \frac{I(b)}{b}$$

$$f(b) = \frac{I}{b} = \frac{3L - 4b}{12} $$

$$= \frac{tL^2}{24} $$ (Recall that b = $$\frac{L}{4}$$)

$$\sigma_{max} = \frac{{M}_{max}b}{2I} = {M}_{max} \frac{12}{tL^2}$$ = $$\frac{tL^2}{16} \sigma$$allow

Since, $$\sigma_{max} = \frac{12}{16} \sigma_{allowable} < \sigma_{allowable}$$

Case 2 is acceptible

Therefore, the optimal cross section for case two is a square.

=Wednesday=

Weight is a critical component in the design of aircraft structures. In order to minimize weight proper materials must be selected. Ideally materials used in aerospace applications have high stiffness and strength while still remaining light weight.

Stiffness is described by Young's Modulus ($$ \displaystyle E = \displaystyle \epsilon/ \displaystyle\sigma $$) the ratio of stress over strain. A higher Young's modulus corresponds with a higher material stiffness.

NOTE: Young's modulus is valid only in the elastic range (shown below).

The strength of a material is measured by the materials yield stress ($$\displaystyle \sigma$$y). Obviously a higher yield stress corresponds with a stronger material.

A materials toughness represents its ability to resist fracture. This is also know as fracture toughness.



Shown above is a graph of stress versus strain for a typical metal. Note that the graph is not exact and only for conceptual purposes. From the graph it is apparent that the ratio of stress and strain is linear inside the elastic range. Outside of this range the material deforms plasticly until the point of failure. Inside the plastic range a positive slope represents hardening of the material while a negative slope represents softening.

The Figure 1.1 to the right lists various categories of material properties and the materials which possess those properties.

The structure of an aircraft has two primary components:


 * 1) Geometry - Common aircraft geometries include monocoque shell like constructions and semimonocoque stiffened shell like structures.
 * 2) Material - Common aircraft materials include aluminum, titanium, and composite materials.

Below is a picture of the Boeing 787 Dreamliner which illustrates the aircraft's material composition. Composite materials make up fifty percent of the 787's total weight.

Boeing 787 $$Insert formula here$$