User:Eas4200c.f08.radsam/HW4

 Comment: This page is identical to the previous one, except that the "October 15th: Meeting 22" was added, which was previously completed but accidentally omitted. Here is a comparison between these two versions. Eas4200c.f08.radsam.a. 23:40, 31 October 2008 (UTC) =Homework #4=

=Class Notes=

October 6th: Meeting 18
 Discussion on Quadrature of an Airfoil and Continuation of Single-Cell Airfoil Analysis  Note: Quadrature of NACA airfoil (Figure 1).

$$\bar{A}=\bar{A_1}+\bar{A_2}=\bar{A_1}-|\bar{A_2}|\,$$ where $$\bar{A_1}>0\,$$ and $$\bar{A_2}<0\,$$.

How about using trapezoidals to integrate (Figure 2)? You could, but there are a couple of disadvantages:
 * There is change in curvature, and
 * It is just not as elegant as the other quadrature (triangles, as in Figure 3).



Back to single-cell airfoil (Pg. 17-3 in notes):

Shear flow is constant:

$$q=q_1=q_2=q_3\,$$

$$\theta=\frac{q}{2G\bar{A}}\sum_{j=1}^3\frac{l_j}{t_j}=\frac{q}{2G\bar{A}}\left [ \frac{2t_1}+\frac{a}{t_2}+\frac{\sqrt{a^2+b^2}}{t_3}\right ]=(HW)q\,$$ where $$\theta\,$$ is the rate of twist and $$HW\,$$ is to compute the numerical factor.

If $$\tau_{max}=\tau_{all}\,$$ (which is given), then the max shear stress, $$\tau_{max}\,$$, is given by: $$\tau_{max}=\frac{q}{min\{t_1,t_2,t_3\}}\,$$

And since $$q=\frac{T}{2\bar{A}}\,$$, $$T_{all}=2\bar{A}\tau_{all}*min\{t_1,t_2,t_3\}\,$$

Now, let $$\tau_{all}=100~GPa\,$$, find $$T_{all}\,$$ (HW).

Contributed by radsam.a.

October 8th: Meeting 19
Given: $$t_1 = 0.3cm\,$$

$$t_2 = 0.5cm\,$$

$$t_{12} = 0.4cm\,$$

$$a = 30cm\,$$

$$b = 60cm\,$$

$$c = 40cm\,$$

Find $$\theta\,$$ as a function of torque $$T\,$$ and torsional constant $$J\,$$ Due to superposition of shear flows, we know the total torque is the summation of $$T_1\,$$ and $$T_2\,$$

 $$T = T_1 + T_2 = 2 \bar A_1\ q_1 + 2 \bar A_2\ q_2\,$$ $$Eqn. 1\,$$  Computing the areas for each one of the cells:

 $$\bar A_1\ = ac\,$$ and  $$\bar A_2\ = bc$$ The expression for the twist angle $$\theta\,$$ was previously computed. Refer to 'Eqn. 3.56' in "Mechanics of Aircraft Structures":

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

The computed area is substituted into twist angle equation:

 $$\theta_1\ = \frac{1}{2 G \bar A_1} \left [ \frac{2 q_1 a}{t_1} + \frac{q_1 c}{t_1} + \frac{(q_1-q_2) c}{t_{12}} \right ]$$ $$Eqn. 2\,$$  The same procedure followed above is used to obtain the twist angle for cell 2:

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

 $$\theta_2\ = \frac{1}{2 G \bar A_2} \left [ \frac{2 q_2 b}{t_2} + \frac{q_2 c}{t_2} + \frac{(q_2-q_1) c}{t_{12}} \right ]$$ $$Eqn. 3\,$$

$$c_1\,$$ and $$c_2\,$$ have the same rate of twist angle, therefore: $$\theta_1\ = \theta_2\,$$ $$Eqn. 4\,$$

With the four equations above, expressions for $$q_1\,$$ and $$q_2\,$$ in terms of $$T\,$$ may be obtained. For instance:  $$q_1 = \beta_1\ T\,$$

$$q_2 = \beta_2\ T\,$$

Where: $$\beta_1\ = 0.01\,$$ and $$\beta_2 = 0.016\,$$ in N/m. "Computation"

In addition, a general expression for twist angle $$\theta\,$$ may be obtained: $$\theta\ = \theta_1\ = \theta_2\ = \frac{T}{G J}$$ Contributed by radsam.d

October 10th: Meeting 20
Recalling...

Torsion: $$T = GJ\theta\,$$


 * T is the torque (N·m or ft·lbf)
 * G is the shear modulus or more commonly the modulus of rigidity and is usually given in gigapascals (GPa), lbf/in2 (psi), or lbf/ft2
 * J is the torsion constant for the section . It is identical to the polar moment of inertia for a round shaft or concentric tube only. For other shapes J must be determined by other means. For solid shafts the membrane analogy is useful, and for thin walled tubes of arbitrary shape the shear flow approximation is fairly good, if the section is not re-entrant. For thick walled tubes of arbitrary shape there is no simple solution, and FEA may be the best method.
 * $$\theta\,$$ is the angle of twist in radians.

Once $$\theta\,$$ is found as a function of T, use...

$$\theta = \frac{T}{GJ}\,$$

or

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

...to find J

HW #4:


 * 3-cell NACA 2415 airfoil
 * 2 partition walls
 * 1st wall at quarter chord from leading edge
 * 2nd wall at three quarter chord from leading edge

3 unknowns... $$q_{1}, q_{2}, q_{3}\,$$:


 * 1) $$T=2\sum_{i=1}^3 q_{i} \bar{A}\,$$
 * 2) $$\theta_{1}=\theta_{2}\,$$
 * 3) $$\theta_{2}=\theta_{3}\,$$

Finding J...

Note: need thickness t, but G cancels out...


 * Now Theory derivation

$$T=GJ\theta\,$$ (done)

$$T=2q\bar{A}$$  (done)

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

Uniform bar with non-circular cross section subject to twist





Displacement $$\underline{PP'}$$ due to $$\alpha\,$$

$$\frac{PP'}{O\underline{P}} = \alpha\,$$ (for small alpha)

Projection Displacement $$\underline{PP'}$$ on the direction perpendicular to $$O\underline{P'}:$$

$$\underline{PP''}=\underline{PP'} cos{\alpha}\,$$

$$\underline{PP''}= (O \underline{P} tan{\alpha}) cos{\alpha}\,$$

$$=(O\underline{P} cos{\alpha}) tan{\alpha}$$

''where... $$O\underline{P} cos{\alpha} = O\underline{P}\,$$

Recall:

$$O\underline{P} = r \,$$ (radial coord)

$$O\underline{P''} = \rho \,$$

$$PP''= (r cos{d}) tan{\alpha}\,$$


 * where: $$PP''$$ is displacement of $$\underline{P}$$ in direction tangent to lateral surface of bar


 * $$r( cos{d})\,$$ is approximately $$\rho\,$$


 * $$tan{\alpha}\,$$ is approximately $$\alpha\,$$

Strain:

$$\gamma=\frac{\underline{PP''}}{dx} = \frac{\rho \alpha}{dx} = \rho \theta\,$$

with... $$\theta = \frac{\alpha}{dx}$$ rate of twist angle (alpha very small!)

Contribution by radsam.z

October 13th: Meeting 21
$$\theta\ = \frac{d \alpha}{dx} $$

Hooke's Law:

$$\tau\ = G\gamma = G\rho\theta = G\rho \frac{d \alpha}{dx}$$

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

$$\oint \tau(s) ds\ = G\theta\oint \rho(s) ds $$

but

$$\tau = \frac{q}{t}$$

and

$$\oint \rho(s) ds = 2 \bar A $$

Hence expression for theta p.21.2

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

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 Ca t P but $$PP'$$ on p.20.2 is not necessarily tangent to $$C'$$ (but actually close)

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

Formal justification (derivative) by elasticity theory: Roadmap p.16.2
A. Kinematic Assumption p.16.1

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

considered constant w,r,t, x: uniform bar

$$\displaystyle u_y(x,z) = -\theta_xz$$

$$\displaystyle u_z(x,y) = +\theta_xy$$

To transform the equation in SUN[2006] to those using our unified notation, use cyclic permutation.

$$\displaystyle \epsilon_{xx} = \epsilon_{yy} = \epsilon_{zz}$$

$$\epsilon_{xx} = \frac{dU_x}{dx}(y,z) = 0$$

$$\gamma_{yz} = \frac{du_y}{dz} + \frac{du_z}{dx} = 0$$

October 15th: Meeting 22
 Multicell Airfoil 



Po is defined as the observation point


 * 1) Set Po = D, sweep area is B-L-E
 * 2) Set Po = E, sweep area is F-B
 * 3) Area A22: Set Po = F, sweep area is E-H
 * 4) Area A3: Set Po = G, sweep area is H-T-F

Total Area = A1 + A21 + A22 + A3

Note: 3rd "Ad-hoc" point in engineering derivative: Inconsistency in assumption on size of alpha. To get line PP', assume alpha is small. To get PP", assume alpha is finite and p=rcos(alpha). Reintroduce small angle approximation after that.

Back to formal derivation:

Question: How many strain components exist in three dimensions?

Answer: SIX: $$\varepsilon _{xx}, \varepsilon _{yy}, \varepsilon _{zz}, \varepsilon _{xy}, \varepsilon _{xz}, \varepsilon _{yz}$$

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

The representative equation is: $$\varepsilon = \frac{1}{2}\left(\frac{du_{i}}{dx_{j}} + \frac{du_{j}}{dx_{i}}\right)$$

Question:  Is symmetry of epsilon related to isotropy of material?

Answer:  NO. Isotropic elasticity is related to material behavior (sigma-epsilon relationship or Young's Modulus).

Contribution by radsam.s

October 17th: Meeting 23
 Three-Dimensional Strain 

Slide 23.1a
Some say thank God it's Friday. I say too bad it's Friday. - Loc Vu-Quoc.

Question: How do we relate strains to Young's Modulus ($$E\,$$) and Poisson's Ratio ($$\nu\,$$)? Answer: (Steve) We use the relationship presented in Hooke's Law.
 * Note: Robert Hooke didn't just study the field of Elasticity; he was also a pioneer in the field of Microscopy, publishing the extremely influential Micrographia on the subject, at the age of 28.

$$\epsilon_{xx}=\sigma_{xx}/E\,$$ But as you pull a piece of rubber, it gets thinner. This lateral construction is due to the Poisson's Ration, $$\nu\,$$.

To correct for this, we re-write Hooke's law as: $$\epsilon_{xx}=\sigma_{xx}/E-\nu \sigma_{yy}/E -\nu \sigma_{zz}/E\,$$

And for shear strains: $$\gamma_{xy}=2\epsilon_{xy}=\tau_{xy}/G\,$$

From lecture 22-3, we have the following matrix:

$$\mathbf{\epsilon} = \begin{bmatrix} \epsilon_{11} & \epsilon_{12}  & \epsilon_{13} \\ \epsilon_{21} & \epsilon_{22} & \epsilon_{23} \\ \epsilon_{31} & \epsilon_{32} & \epsilon_{33} \end{bmatrix}= \epsilon_{ij}\,$$

and, using Voigt Notation (due to the symmetry of the 3x3 matrices $$\mathbf{\epsilon} = \begin{bmatrix}  \epsilon_{ij} \end{bmatrix} \,$$, $$\mathbf{\sigma} = \begin{bmatrix}   \sigma_{ij} \end{bmatrix} \,$$) we can arrange $$\epsilon_{ij}\,$$ and $$\epsilon_{ij}\,$$ into column matrices.

Slide 23.1b
The resultant 6x1 matrices are the following:
 * $$\epsilon_{ij} = \begin{bmatrix}

\epsilon_{11} \\ \epsilon_{22} \\ \epsilon_{33} \\ \epsilon_{23} \\ \epsilon_{31} \\ \epsilon_{12} \end{bmatrix}\,$$ and :$$\epsilon_{ij} = \begin{bmatrix} \sigma_{11} \\ \sigma_{22} \\ \sigma_{33} \\ \sigma_{23} \\ \sigma_{31} \\ \sigma_{12} \end{bmatrix}\,$$
 * Note: The original 3x3 column was symmetric, so $$\epsilon_{13}=\epsilon_{31}\,$$. We choose this notation in our 6x1 matrix rather than the $$\epsilon_{13}\,$$ that should be used to produce a beautiful, cyclic permutation (23 followed by 31, followed by 12.)

Hooke's Law for Isotropic Elasticity:
 * $$\begin{bmatrix}

\epsilon_{11} \\ ... \\ ... \\ \gamma_{23} \\ ... \\ ... \end{bmatrix} = \begin{bmatrix} 1/E & -\nu/E & -\nu/E & 0 & 0 & 0 \\ ... & ... & ... & ... & ... & ... \\ ... & ... & ... & ... & ... & ... \\   0 & 0 & 0 & 1/G & 0 & 0 \\ ... & ... & ... & ... & ... & ... \\ ... & ... & ... & ... & ... & ... \end{bmatrix} \begin{bmatrix} \sigma_{11} \\ \sigma_{22} \\ \sigma_{33} \\ \sigma_{23} \\ \sigma_{31} \\ \sigma_{12} \end{bmatrix}\,$$
 * Note: $$\sigma_{xy}=\tau_{xy}\,$$

For row four, $$\gamma_{23}=\gamma_{yz}\,$$ (Jared, Gary) $$=\tau_{yz}/G\,$$.

On Poisson's Ratio: Three Poisson's Ratios to note are listed in the table below. The first, for Steel, is noted because it is a value similar to all the materials we will deal with as Aerospace Engineers. The significance of the second and third, for cork and rubber respectively, is noted in homework 23.5. Nearly all materials are within the range of $$ 0 < \nu < 0.5 \,$$. Recently though, in the scientific journal Science (one of the two Premiere scientific journals, the opposite being Nature) Material Scientists reported making materials with Poisson's Ratios of $\nu < 0\,$. In other words, through Material Science and Engineering, materials which compress under tension, and expand under compression, have now been created.

Significant Poisson's Ratios:

Reading Assignments
MIT OpenCourseWare (OCW) Professor Lagace unit 4 (PDF) MIT OpenCourseWare (OCW) Professor Lagace unit 12 (PDF)

=Homework Problems and Solutions=

=References=

=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.radsam.m 19:01, 18 October 2008 (UTC) Eas4200c.f08.radsam.a. 18:55, 19 October 2008 (UTC) "Eas4200c.f08.radsam.z 16:02, 2 November 2008 (UTC)" Eas4200c.f08.radsam.r 03:59, 20 October 2008 (UTC) Eas4200c.f08.radsam.e 21:36, 21 October 2008 (UTC) Eas4200c.f08.radsam.s 12:57, 22 October 2008 (UTC) "Eas4200c.f08.radsam.a. 19:24, 23 October 2008 (UTC)"

=Homework 4 Contributors= Eas4200c.f08.radsam.e 19:31, 17 October 2008 (UTC) Eas4200c.f08.radsam.m 20:16, 18 October 2008 (UTC) Eas4200c.f08.radsam.a 19:07, 19 October 2008 (UTC) Eas4200c.f08.radsam.d 22:12, 19 October 2008 (UTC) Eas4200c.f08.radsam.z 23:50, 19 October 2008 (UTC) Eas4200c.f08.radsam.r 03:58, 20 October 2008 (UTC) Eas4200c.f08.radsam.s 13:00, 22 October 2008 (UTC)