User:Eas4200c.f08.aero.e/HW6

=Week 11 - 11/3 through 11/7=

Plate Buckling
From Plate Buckling by Prof. Loc Vu-Quoc

This page talks about 4 different subjects. They are the compressive load and shear load, with simply-supported boundary conditions, and with Clamped boundary conditions.

Compressive Load
This image is of a rectangle of a and b under a compressive load. The thickness is represented by h.

#1 Simply-supported boundary conditions

The Buckling mode shapes is defined as $$\displaystyle \psi_{mn} (x,y)$$

$$ \displaystyle \psi (x,y) = c_{mn}\sin\left(\frac{m \pi x}{a}\right)\sin\left(\frac{n \pi y}{b}\right)\, \ {\rm for} \ m,n = 1, 2, 3, \ldots $$

where cmn is a constan, and (m,n) are the number of halfwave-lengths along (x,y), respectively.

The critical buckling load is defined as $$\displaystyle (P_x)_{cr}$$

$$ \displaystyle (P_x)_{cr} = k_c \frac{\pi ^2 D}{b} $$

Where kc and D, the plate bending stiffness, are

$$ \displaystyle k_c(m, a/b) = \left(\frac{m b}{a} + \frac{a}{m b}\right)^2 $$

$$ \displaystyle D = \frac{E h^3}{12 (1 - \nu^2)} $$

Then stated is the dimensions of;

[I] as length to the 4th [σ] as force over length squared, where є is 1 [EI] as force times length squared [D] as force time length, where $$\nu$$ is 1 [(Px)cr] as force, with kc is 1

Critical buckling stress is denoted below

$$ \displaystyle (\sigma_{xx})_{cr} = \frac{(P_x)_{cr}}{bh} = k_c\frac{\pi ^2 D}{b^2 h} $$

With the minimum, denoted with a star, critical load is

$$ \displaystyle (\sigma_{xx})_{cr}^\star = k_c^\star\frac{\pi ^2 D}{b^2 h} $$

where

$$  \displaystyle k_c^\star (a/b) =  \min_{m = 1, 2, \cdots} \ k_c (m, a/b) =  \min_{m = 1, 2, \cdots} \left(     \frac      {m b}      {a}      +      \frac      {a}      {m b}   \right)^2 =  \left(      \frac      {m^\star b}      {a}      +      \frac      {a}      {m^\star b}   \right)^2 $$

$$  \displaystyle m^\star (a/b) =  \underset{m=1,2, \cdots}{\rm argmin} \ k_c (m, a/b) =  \underset{m=1,2, \cdots}{\rm argmin} \left(     \frac      {m b}      {a}      +      \frac      {a}      {m b}   \right)^2 $$

And then for a simply-supported column, the critical load is

$$  \displaystyle (P_{cr})_{column} =  \frac {\pi^2 EI} {L^2} $$

#2 Clamped boundary conditions

An Excel file is given to find the critical stress given by the equation

$$  \displaystyle \sigma^\prime =  K   \frac{E}{1 - \nu^2} \left(     \frac{t}{b}   \right)^2 $$

from the book by Young & Budynas, Roark's formula for stress adn strain, McGraw-Hill, 7th ed., 2002, p.730.

In this equation t = h, and

$$  \displaystyle K  = k_c \frac{\pi^2}{12} $$

The estimated buckling load for a clamped plate close to being square from Timoshenko & Gere [1961], Theory of elastic stability, McGraw-Hill, p.386.

$$  \displaystyle \psi (x,y) =  \frac {c} {4}  \left(      1 - \cos \frac{2 \pi x}{a}   \right) \left(     1 - \cos \frac{2 \pi y}{b}   \right) $$

Shear Load
This image is the same rectangle under a shearing load.

#1 Simply-supported boundary conditions

The shear force per unit length is represented as

$$  \displaystyle N_{xy} =  \sigma_{xy} h $$

The critical buckling shear load is found by equating the strain energy and the work done by external shear load, and solving for the shear force.

The strain energy, U, and the work, W, are denoted as

$$  \displaystyle U  = \frac{D}{2} \frac{\pi^4 a b}{4} \sum_{m=1}^{\infty} \sum_{n=1}^{\infty} C_{mn}^2 \left(     \frac{m^2}{a^2}      +      \frac{n^2}{b^2}   \right)^2 $$

$$  \displaystyle W  = - 4 N_{xy} \sum_{m=1}^{\infty} \sum_{n=1}^{\infty} \sum_{p=1}^{\infty} \sum_{q=1}^{\infty} C_{mn} C_{pq} \frac {mnpq} {(m^2 - p^2)(n^2 - q^2)} $$

So

$$  \displaystyle (N_{xy})_{cr} =  -   \frac{ab D}{32} \frac {     \displaystyle \sum_{m=1}^{\infty} \sum_{n=1}^{\infty} C_{mn}^2 \left(        \frac{m^2 \pi^2}{a^2}	 +         \frac{n^2 \pi^2}{b^2}      \right)^2 }  {      \displaystyle \sum_{m=1}^{\infty} \sum_{n=1}^{\infty} \sum_{p=1}^{\infty} \sum_{q=1}^{\infty} C_{mn} C_{pq} \frac{mnpq}{(m^2 - p^2)(n^2 - q^2)} } $$

To find Cmn and Cpq to minimize the critical load, take the derivative with respect to Cmn and set it equal to zero.

Letting the aspect ratio of the plated be $$ \displaystyle \vartheta :=  \frac{a}{b} $$

$$  \displaystyle \lambda :=  -   \frac{\pi^2}{32 \vartheta} \frac{\pi^2 D}{b^2 h (\sigma_{xy})_{cr}} $$

The first 5 of these equations in matrix form are

$$  \displaystyle \left[ \begin{array}{lllll} \frac{\lambda (1 + \vartheta^2)^2}{\vartheta^2} &	 \frac{4}{9} &	 0	 &	 0	 &	 0	 \\	 \frac{4}{9} &	 \frac{16 \lambda (1 + \vartheta^2)^2}{\vartheta^2} &	 - \frac{4}{5} &	 - \frac{4}{5} &	 \frac{36}{25} \\	 0	 &	 - \frac{4}{5} &	 \frac{\lambda (1 + 9 \vartheta^2)^2}{\vartheta^2} &	 0	 &	 0	 \\	 0	 &	 - \frac{4}{5} &	 0	 &	 \frac{\lambda (9 + \vartheta^2)^2}{\vartheta^2} &	 0	 \\	 0	 &	 \frac{36}{25} &	 0	 &	 0	 &	 \frac{\lambda (9 + 9 \vartheta^2)^2}{\vartheta^2} \end{array} \right] \left\{ \begin{array}{l} C_{11} \\	 C_{22} \\	 C_{13} \\	 C_{31} \\	 C_{33} \end{array} \right\} =  \left\{ \begin{array}{l} 0	 \\	 0	 \\	 0	 \\	 0	 \\	 0     \end{array} \right\} $$

Witch is a symmetry diagonal matrix.

Case 1

To find the minimum critical buckling shear stress use the top left 2x2 matrix and set the determinant equal to zero. The eigenvalues are

$$  \displaystyle \lambda =  \pm \frac{1}{9} \frac{\vartheta^2}{(1 + \vartheta^2)^2} $$

Then the minimum critical buckling shear stress is

$$  \displaystyle (\sigma_{xy})_{cr}^{\star,2} =  \pm k_c^{\star,2} \frac{\pi^2 D}{b^2 h}  \, \ k_c^{\star,2} =  \frac{\pi^2}{32} \frac{9 (1 + \vartheta^2)^2}{\vartheta^3} $$

The 2 next to the star means 2x2 matrix and it is not dependent on the direction.

this will give a 15% error witch will increase with the increas in the aspect ratio.

Case 2

Using all 5 equations and setting the determinant to zero yields

$$  \displaystyle \lambda =  \left[ \frac{\vartheta^4}{81 (1 + \vartheta^2)^4} \left\{ 1	 +	 \frac{81}{625} +	 \frac{81}{25} \left(	   \frac{1 + \vartheta^2}{1 + 9 \vartheta^2}	 \right)^2 +	 \frac{81}{25} \left(	   \frac{1 + \vartheta^2}{9 + \vartheta^2}	 \right)^2 \right\} \right]^{1/2} $$

And

$$  \displaystyle (\sigma_{xy})_{cr}^{\star,5} =  \pm k_c^{\star,5} \frac{\pi^2 D}{b^2 h}  \, \ k_c^{\star,5} =  \frac{\pi^2}{32} \frac{1}{\lambda \vartheta} $$

#2 Clamped boundary conditions

The critical shear load is simply give by

$$  \displaystyle (\sigma_{xy})_{cr} =  K   \frac {E} {1 - \nu^2} \left(     \frac{h}{b}   \right)^2 $$

Where in Young & Budynas, Roark's formula for stress and strain, McGraw-Hill, 7th edition, 2002, p.732, a table of kc are given for aspect ratios from 1 to infinity. So K can be found.

It is also seen in experimental results a K = 4.1 for large aspect ratios. This may be due to imperfections in the experimental plate specimens.

--Eas4200c.f08.aero.E 06:38, 20 November 2008 (UTC)

Wednesday
$$\sigma_{xx}=\frac{d\phi}{dz}$$

$$\sigma_{zx} = -\frac{d\phi}{dy}$$

The above equations automatically satisfy the following equation: $$\frac{d\sigma_{yx}}{dy} + \frac{dz\sigma_{x}}{dz} = 0$$

In continuation from the previous lecture and lecture 27-2.

$$\sum_{3}^{i=1}{}\frac{2\sigma_{ij}}{dx_{i}} = 0$$ for j = 1, 2, 3

$$\begin{bmatrix} d\sigma_{ij}\\ dx_{i} \end{bmatrix} = \frac{F}{L^{3}} = \frac{Force}{Volume}$$

Recall roadmap on lecture 16-2. Prandtl stress funtion $$\phi$$

$$\sigma_{yx} = \frac{d\phi}{dz}$$

$$\sigma_{zx} = -\frac{d\phi}{dy}$$

$$\phi \Rightarrow $$Plays a role of a potential function.

$$(\sigma_{yx},\sigma_{zx})$$ comp. of "gradient" of $$\phi$$

Recall: Scalar funtion (x,y,z)

Gradient vector $$f(x,y,z) = \frac{df}{dx}i+\frac{df}{dy}j+\frac{df}{dz}k$$

f = potential

$$\frac{d}{dy}(\frac{d\phi}{dz})+\frac{d}{dz}(-\frac{d\phi}{dy})=\frac{d^{2}\phi}{dydz}-\frac{d^{2}\phi}{dzdy}=0$$

Since $$\phi$$ is continuous and smooth then the second dericative is interchangable

$$i.e. \Rightarrow\frac{d^{2}\phi}{dydz}=\frac{d^{2}\phi}{dzdy}$$

$$\frac{d^{2}\phi}{dy^{2}}+\frac{d^{2}\phi}{dz^{2}}=-2G\theta$$

Derive

$$[t]_{3x1} = [\sigma]_{3x3}{n}_{3x1}$$, equation 3.19 Chapter 2

{t} → Component of traction force

[σ] → Component of stress

{n} → Component of normal vector Ň

2-d Case on Page 29

Homework and reading assignments

1.) Read and report equation 3.14

It is known from a previous lecture that the stress-strain relations give the following:

$$\sigma_{xx}=\sigma_{yy}=\sigma_{zz}=\tau_{xy}=0$$

So, the only nonvanishing stress components are $$\tau_{yz}, and, \tau_{xz}$$ By thinking of the displacement field then it can be seen that these two stress components are independent of Z. If no body forces are considered then the equilibirum equations reduce to the following:

$$\frac{d\sigma_{yx}}{dy} + \frac{dz\sigma_{x}}{dz} = 0$$

2.) Read and report equation 3.15 to 3.19

Prandtl introduced the stress function Φ(X,Y), also τ_{zx} and τ_{yz} derived from Φ automatically satisfy the equations of equilibium. Now using the strain-displacements relations the following equation can be derived:

$$\frac{d\gamma_{yz}}{dx} - \frac{d\gamma_{xz}}{dy} = 2\theta$$

This is the compatibility equation for torsion, now thinking in terms of the prandtl stress-strain relations and the equation is derived

$$\frac{d^{2}\phi}{dx^{2}} + \frac{d^{2}\phi}{dy^{2}} = -2G\theta$$

2.) Read and report until the end of section 3.2

Section 3.2 mainly deals with torsion of uniform bars. Much like a bending moment a torque is a familiar load in          aircraft structures. Two assumptions are needed. One being that plane sections of a shaft remain plane and circular and the other that diameters remain straight after deformation are needed in deriving the deformation and stress fields.

There are two methods of sloving torsion of solid shafts, the Prandtl stress fuction method and the Saint-Venant warping function method. Both methods lead to the same results.

The torsion problem reduces the following two requirements; obtaining the stress fuction and satisfying the boundary conditions. Then the total torque is obtained by integrating dT over the entire cross-section.

--Eas4200c.f08.aero.arena 07:24, 20 November 2008 (UTC)

Friday
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)

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

--EAS4200C.F08.AERO.CLK04D 16:12, 21 November 2008 (UTC)

=Week 12 - 11/10 through 11/14=

Monday
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)

--EAS4200C.F08.AERO.CLK04D 16:10, 21 November 2008 (UTC)

Wednesday
The following graph will help us understand the derivation.

We know that:

$$r = \sqrt{x^{2}+y^{2}}$$

And, we also must recall that:

$$T = 2\oint_{A}^{}{}\phi dA=2C\left(\frac{J}{a^{2}}-A \right)$$ and $$J=\int_{A}^{}{}r^{2}dA = \frac{1}{2}\pi a^{4}$$

where $$A = \pi*a^{2}$$

so we must remember that: $$\begin{cases} \sigma _{yx}=\frac{\partial \phi }{\partial z}=-G\theta z \\ \sigma _{zx}=-\frac{\partial \phi }{\partial y}=G\theta y \end{cases}$$

-- Homework

Read and report through the end of section 3.3

 Report 

Bars with circular cross-section

We consider a uniform bar with circular cross-section

We see that: $$x^{2}+y^{2}= a^{2}$$

And we assume the stress function of the following way: $$\phi = C\left(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{a^{2}} -1\right)$$

Substituting the above expression into: $$\frac{\partial ^{2}\phi }{\partial x^{2}}+\frac{\partial ^{2}\phi }{\partial y^{2}}=-2G\theta $$

we get: $$C = -\frac{1}{2}a^{2}G\theta $$

Which solve the torsion problem and tells us that the center axis is the center of twist.

Now, we know that torque is: $$T=2\int_{}^{}{}\int_{A}^{}{}\phi dxdy$$

The using the above values and plug them in the Torque equation we get:

$$T = 2C\int_{}^{}{}\int_{A}^{}{}\left(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{a^{2}}-1 \right)dxdy$$

That equals:

$$T = 2C\left(\frac{I_{p}}{a^{2}}-A \right)$$

where $$I_{p} = \int_{}^{}{}\int_{A}^{}{}r^{2}dA = \frac{1}{2}\pi a^{4}$$ and is the polar moment of inertia and A is the area of the cross-section.

And because $$A = \pi a^{2} $$ then $$t T = -\frac{2CI_{p}}{a^{2}} = \theta GI_{p}$$

So, the shear stresses are:

$$\begin{cases} \tau _{xz}=\frac{\partial \phi }{\partial y}= 2C\frac{y}{a^{2}} = -G\theta y \\ \tau _{yx}=-\frac{\partial \phi }{\partial x}=-2C\frac{x}{a^{2}}=G\theta x \end{cases}$$

Now, let's consider the figures below and concentrate on figure (a)

The stress vector are: $$t_{x}=t_{y}=0 $$ and $$  t_{z} = \tau _{xz}n_{x} + \tau _{yz}n_{y}$$

And since:

$$\begin{cases} n_{x}= cos\beta = \frac{x}{r}  \\ n_{y}=sin\beta = \frac{y}{r} \end{cases}$$

Then we get:

$$t_{z}=-G\theta \frac{xy}{r}+G\theta \frac{xy}{r}=0$$

We notice that the radial shear stress $$\tau _{rz}$$ vanishes because it is equal to $$t_{z}$$

Now, let's look at figure (b) of the above graph.

The unit vector on the surface is

$$n_{z}=0$$, $$n_{x}=sin\beta =\frac{y}{r}$$, $$ n_{y}=-cos\beta =-\frac{x}{r}$$

So, substituting the above equations as well as: $$\begin{cases} n_{x}= cos\beta = \frac{x}{r}  \\ n_{y}=sin\beta = \frac{y}{r} \end{cases}$$

into $$t_{z}=\tau _{xz}n_{x}+\tau _{yz}n_{y}=\frac{\partial \phi }{\partial y}n_{x}-\frac{\partial \phi }{\partial x}n_{y}$$

we get:

t_{z}=-G\theta r

We also see that on the z face the tangential shear stress is τz. And we end up getting:

$$\tau =-t_{z}=G\theta r$$

Where we use $$T=\theta =\theta GI_{p}$$ to get rid of θ

And finally, after algebra manipulation we get that $$w=u_{x}=0$$

Which tells us that for circular cross-sections bars there's no warping.

-- Homework

Using the equations: $$\begin{cases} \gamma _{yx} = \frac{\partial u_{x}}{\partial y}-\theta z \\ \gamma _{zx} = \frac{\partial u_{x}}{\partial z} + \theta y \end{cases}$$

and

$$\begin{cases} \tau _{yx} = \frac{\partial \theta }{\partial z} = \frac{2Cz}{a^{2}} = -G\theta z  \\ \tau _{zx} = -\frac{\partial \theta }{\partial y} =- \frac{2Cy}{a^{2}} = -G\theta y \end{cases}$$

And the δ-ε relations to show that

$$u_{x}\left(y,z \right) = 0$$

Solution

We know that:

$$\begin{cases} \tau _{yx} = \frac{\partial \theta }{\partial z} = \frac{2Cz}{a^{2}} = -G\theta z  \\ \tau _{zx} = -\frac{\partial \theta }{\partial y} =- \frac{2Cy}{a^{2}} = -G\theta y \end{cases}$$

Which can be rewritten as:

$$\begin{cases} \sigma_{yx} = -G\theta z&  \\ \sigma_{zx} = G\theta y& \end{cases} $$

Where we plug the σ values into the gamma values and we get:

$$\begin{cases} \gamma _{yx} = \frac{\partial u_{x}}{\partial y} - \theta z = \frac{\sigma _{yx}}{G} \Rightarrow -\theta z  \\ \gamma _{zx} = \frac{\partial u_{x}}{\partial z} + \theta y = \frac{\sigma _{zx}}{G} \Rightarrow \theta y \end{cases}$$

In a more simple way:

$$\begin{cases} \frac{\partial u_{x}}{\partial y} - \theta z =  -\theta z  \\ \frac{\partial u_{x}}{\partial z} + \theta y =  z\theta y \end{cases}$$

Solving for the partial part of the equation, we get:

$$\begin{cases} \frac{\partial u_{x}}{\partial y} = 0\\ \frac{\partial u_{x}}{\partial z} = 0 \end{cases} $$

Integrating the partial derivatives we get a constant and we realize that U is independent of x and y. -- Now, we proceed to start chapter 5

Motivation:

Now, if we draw a free body diagram of the above image in the following manner:

And the idea is to apply the principles developed in future clases for the NACA airfoil. Concepts and derivations for this graphs and chapter will be developed in the upcoming classes.

--Eas4200c.f08.aero.gabo28

Friday
$$\int_{A_s}\frac{d\sigma_{xx}}{dx}dA=-q_s$$(general for unsymmetrical cross section)

Symmetric

$$\sigma_{xx}=\frac{M_yz}{I_y}$$

$$q(s)=\frac{-V_zQ}{I_y}$$ $$Q=\int_{A_s}z dA=z_c A_s$$

Unsymmetrical Cross Section

$$\sigma_{xx}=(K_yM_-K_{yz}M_y)y+(K_zM_y-K_{yz}M_z)z$$

$$K_y=\frac{I_y}{D}$$ $$K_{yz}=\frac{I_{yz}}{D}$$ $$K_z=\frac{I_z}{D}$$

Where D is the determinant of I
 * $$\mathbf{\sigma_{xx}} = \begin{bmatrix}

y & z\end{bmatrix}$$ $$\begin{bmatrix} K_y & -K_{yz} \\ -K_{yz} & K_z\end{bmatrix}$$ $$\begin{bmatrix}

M_z\\ M_y\end{bmatrix}$$ Particularize to symmetric cross section, I_{yz}=0 consider $$M_z=0$$ $$I_{yz}=0$$ so $$D=I_yI_z$$ $$K_y=\frac{I_y}{D}=\frac{1}{I_z}$$, $$K_{yz}=0$$, $$K_z=\frac{I_c}{D}=\frac{1}{I_y}$$

so $$\sigma_{xx}=\frac{zM_y}{I_y}$$

Non-Symmetric $$q(s)= -(K_yV_y-K_{yz}V_z)Q_z -(K_zV_z-K_{yz}V_y)Q_y$$

stringer-web sections: thickness t of skin and spar webs are very small, so neglect in comp. $$I_y I_z I_{yz} Q_y Q_z$$ (use only areas of stringers)

EAS4200C.Fall08.AERO.Watlington.VG 21:57, 21 November 2008 (UTC)

=Matlab=

--Eas4200c.f08.aero.r 16:55, 21 November 2008 (UTC)--Eas4200c.f08.aero.r 16:55, 21 November 2008 (UTC)--Eas4200c.f08.aero.r 16:55, 21 November 2008 (UTC)--Eas4200c.f08.aero.r 16:55, 21 November 2008 (UTC)--Eas4200c.f08.aero.r 16:55, 21 November 2008 (UTC) --20:15, 21 November 2008 (UTC)20:15, 21 November 2008 (UTC)20:15, 21 November 2008 (UTC)~--Eas4200c.f08.aero.r 20:15, 21 November 2008 (UTC)