text:
User:Eas4200c.f08.carbon.orear/HW6/Lecture Notes

 See my comments below. Please don't remove these comment boxes; just add your comment to these comment boxes in case you fix the problems. Eml4500.f08 13:27, 23 November 2008 (UTC)

=Lecture Notes=

Equations of Equilibrium (Continued)
Using dimensional analysis, we can find the dimension, which is not the same as the unit. All measures have units, but some are dimensionless.

To find the dimension of $$\begin{bmatrix} \frac{f}{A} \end{bmatrix}$$, let's analyze each individual component:

$$[f] = \frac{F}{L}$$ and $$[A] = L^2$$

→ $$\begin{bmatrix} \frac{f}{A} \end{bmatrix} = \frac{F}{L^3}$$

This same methodology can be applied to any measurement such as stress and strain. The ad-hoc proof below show that you can approach the same dimension no matter what form of a measure you use:

$$[\sigma ] = \frac{F}{L^{2}}$$ and $$[dx] = L \Rightarrow \begin{bmatrix} \frac{d\sigma }{dx} \end{bmatrix} = \frac{[d\sigma ]}{[dx]} = \frac{F/L^{2}}{L} = \frac{F}{L^{3}} = [f]$$

Below is the dimensional analysis for strain, and shows that strain is non-dimensional (but still has a unit).

Unit: $$\varepsilon = \frac{du}{dx} = \frac{\Delta L}{L}$$

Dimension: $$[\varepsilon ] = \frac{[du]}{[dx]} = \frac{L}{L} = 1$$ (non-dimensional)

Also, $$\nu = -\frac{\varepsilon _{yy}}{\varepsilon _{xx}} \Rightarrow [\nu ] = \frac{[\varepsilon _{yy}]}{[\varepsilon _{xx}]} = 1$$

 * {| class="toccolours collapsible collapsed" width="60%" style="text-align:left"

!Contribution Eigenvalues and Buckling, from team Carbon
 * Eigenvalues are very important in determining material capabilities and making judgments for better designs. A study at Rice University using eigenvalue analysis showed that a column would be strongest if it were widest at the top, middle, and bottom of the column and skinny 1/4 distance from the top and 1/4 from the bottom.
 * Eigenvalues are very important in determining material capabilities and making judgments for better designs. A study at Rice University using eigenvalue analysis showed that a column would be strongest if it were widest at the top, middle, and bottom of the column and skinny 1/4 distance from the top and 1/4 from the bottom.

This study showed that this is the most material-conscious and strength-conscious way to build a column. Eigenvalues showed that naturally a column, regardless of its shape buckles at the top and lower quarter location. To conserve material, the thinning of these sections would not make a contribution to the buckling.

In addition, during testing of the Boeing 747 crash simulations, when falling flat on the underside the bottom of the fuselage would collapse at varying angles. The goal was to allow it collaps and keep the cabin level with the ground. To achieve this, engineers designed using eigenvalues the support plates as shown below with an indentation in the middle so that during a crash the plates would buckle uniformly and keep the cabin level.

Source: http://ceee.rice.edu/Books/LA/eigen/

Contribution by carbon.w
 * }

Torsional Analysis (Continued)
From lectures 27 and 29 (specifically pages 27-2 and 29-1) we recall that:

$$\left[\frac{\partial\sigma_{ij}}{\partial x_{i}}\right] = \frac{F}{l^3}$$ = force/volume

Recall that we are working our way through the "roadmap." Specifically, lets look at part D. The Prandtl Stress Funtion, $$\phi$$

(Eq. 1) $$\sigma_{yx} = \frac{\partial\phi}{\partial z}, \sigma_{zx} = - \frac{\partial\phi}{\partial y} $$

$$\phi$$ plays the role of a potential function, and $$\sigma_{yx}, \sigma_{zx}$$ are the components of the "gradient" of $$\phi$$ with respect to y and z.

Recall there is a scalar function f(x,y,z), where f is a potential function. The gradient of f is then a vector given by: $$\triangledown f(x,y,z) = \frac{\partial f}{\partial x}\hat i + \frac{\partial f}{\partial y}\hat j + \frac{\partial f}{\partial z}\hat k $$

Equation 1 (above) automatically satisfies Equation 1 from lecture 27-1:

$$\frac{\partial}{\partial y}\left(\frac{\partial \phi}{\partial z}\right) + \frac{\partial}{\partial z}\left(\frac{- \partial \phi}{\partial y}\right) = \frac{\partial^2 \phi}{\partial y \partial z} - \frac{\partial^2 \phi}{\partial z \partial y}$$

 * Note: If $$\phi$$ is constant and smooth, then the 2nd mixed derivative is interchangeable, i.e.: $$\frac{\partial^2 \phi}{\partial y \partial z} = \frac{\partial^2 \phi}{\partial z \partial y}$$

$$=> \frac{\partial^2 \phi}{\partial y^2} + \frac{\partial^2 \phi}{\partial z^2} = -2G\theta$$, where the left hand side of this equation is the Laplacian ($$ \triangledown^2 \phi$$) of $$\phi$$

Next, let's derive $$t_{3x1}=\sigma_{3x3}n_{3x1}$$, where t is the components of the traction force, sigma is the components of the stress tensor, and n is the components of the normal to the surface.

Let's look at the 2-D case first:

Below is an example of $$\hat t $$ in an aircraft structure:

Distributed Surface Forces

$$\displaystyle \boldsymbol n=n_y \boldsymbol j+n_z \boldsymbol k$$ $$\displaystyle ||\boldsymbol n||=1$$

Where n is the unit normal vector

$$\displaystyle dz=ds cos \theta      <=>       n_y=cos\theta$$ $$\displaystyle dy=ds sin \theta      <=>       n_y=sin\theta$$

$$\displaystyle \sum F_y=0= -\sigma_{yy} (dz*1)$$

Note: (1) is the unit vector along the x-axis

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

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

If we divide both sides by ds, and move t to the other side, we get the equation:

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

 * Note:
 * $$[t_y]=\frac {F}{L^2} = \frac {Force}{Area}$$
 * t- tranction vector (distributed surface force)
 * $$\displaystyle [t_y]=[r]$$

$$\sum F_y = 0$$ $$\displaystyle t_z=\sigma_{yz} n_y + \sigma_{zz} n_z$$

Adding both equations for $$\displaystyle t_y$$ and $$\displaystyle t_z$$

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

Where we observe $$\displaystyle \sigma_{yz}=\sigma_{zy}$$, making the matrix symmeteric.

$$\displaystyle t_1 = \sum_{j=1}^{3} \sigma_{ij}n_{j} $$
 * where i is the row and j is the column.

Rube-Goldberg and How we Learn
In this Lecture we were introduced to "How to run an airplane: Rube-Goldberg Device." A video was shown which can be found externally here.

 * {| class="toccolours collapsible collapsed" width="60%" style="text-align:left"

!Wiki-Homework: Our own Rube-Goldberg Device, from team Carbon We can relate the Rube-Goldberg device to the experience of this class. In some sense, it is much like our team Wiki-Homework reports. An assortment of working parts comes together to perform a common task. It is as though sections of our homework are like steps along the Rube-Goldberg device itself in that any individual section cannot stand alone, nor can the whole system be complete without the contribution of every distinct section.

The whole system is visible to an outside observer, and looking in, one can see all of the constructive, creative processes that produce the overall result. Where the Rube-Goldberg device is in fact a miniaturization of a marvelous engineering feat, teamwork on these assignments fully exercises the collective knowledge and efforts of multiple individuals.

Contribution by carbon.orear
 * }

Boundary Conditions for a Closed Thin-Wall
We looked at equations 3.40 and 3.41 given below.

Φ = C0 on S0

Φ = C1 on S1

and noted the boundary condition that

Φ = 0 on the lateral surface of the bar.

Torsional Analysis Continued
We begin with the figure:

The torque can be equated as

$$T = 2 \int_{ A}^{ } \phi dA = 2C( \frac{J }{ {a }^{2 }  } - A)$$

The area can simply be calculated using the radius $$a$$ as

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

The polar moment of inertia is also calculated using the radius, $$a$$

$$J = \int_{ A}^{ }  {r }^{ 2} dA =  \frac{1 }{2 } \pi  { a}^{ 4} $$

With

$$T=GJ\theta$$

According to Part D of the road map, the Prandtl Stress Function states

$$ \sigma_{zx} = \frac{d \phi }{dz } = -G\theta z $$  Normal to the y-axis in the x-axis direction

$$\sigma_{zx}= \frac{-d \phi }{ dy}  = G\theta y$$  Normal to the z-axis in the x-axis direction

 * {| class="toccolours collapsible collapsed" width="60%" style="text-align:left"

!Proof of $$\sigma_{yx}, \sigma_{zx}$$, from team Carbon
 * Taking the partial derivatives, with respect to y and z yields, of
 * Taking the partial derivatives, with respect to y and z yields, of

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

Yields

$$ \Rightarrow \sigma_{yx} = = \frac{-d \phi }{ dy}  = 2 \frac{  { y}^{ 2} }{ { a}^{2 }  }, \sigma_{yx} = 2 \frac{  { z}^{ 2} }{ { a}^{2 }  } $$

Substituting $$\phi = = \frac{-d \phi }{ dz}  = C( \frac{  {y }^{ 2} }{  {a }^{2 } } + \frac{ { z}^{2 }  }{  {a }^{2 } } - 1 )$$ into

$$ \frac{ { d}^{2 } \phi }{ d { y}^{ 2} } + \frac{ { d}^{2 } \phi }{ d { z}^{ 2} } = -2G\theta$$

Yields

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

Substituting into

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

Therefore proving that

$$ \sigma_{zx} = \frac{d \phi }{dz } = -G\theta z $$

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

Contribution by carbon.guidry
 * }

Also to be noted $$\tau = \frac{ Tr}{ J}$$

Proof of No Warping
In the case of no warping, the displacement, U, along the axial, x-axis, direction of the wing will be zero. Therefore when the wing is twisted, there is no displacement along the x-axis of the wing, and no warping.

 * {| class="toccolours collapsible collapsed" width="60%" style="text-align:left"

!HW Proof of No Warping, from team Carbon
 * Using the equations already proven above and the relation between stress and strain, this zero displacement or
 * Using the equations already proven above and the relation between stress and strain, this zero displacement or

$$U_{x}(y,z) = 0$$

Relating strain to stress gives

$$ \gamma_{yx} = \frac{\sigma_{yx} }{G } =  \frac{ dU_{x}}{ dy} - \theta y $$

and

$$ \gamma_{zx} = \frac{\sigma_{zx} }{G } =  \frac{ dU_{x}}{ dz} + \theta z $$

Remembering that

$$\sigma_{yx} = -G\theta z$$

$$\sigma_{zx} = -G\theta y$$

Substitution then of the above equation gives

$$ \frac{-G\theta z }{G } = \frac{ dU_{x}}{ dy} - \theta z $$

$$ \frac{-G\theta y }{G } = \frac{ dU_{x}}{ d} - \theta y $$

Therefore proving that

$$\frac{ dU_{x}}{ dy} = 0$$

$$\frac{ dU_{x}}{ dz} = 0$$

and therefore

$$U_{x}(y,z) = 0$$

Contribution by carbon.guidry
 * }

 Error: See my comments in Team VQCrew (with annotation). Eml4500.f08 14:04, 23 November 2008 (UTC)

Introduction to Shear Flow for Thin-Walled Cross section
Before the next section we remind ourselves of shear and bending moment along a cantilever beam with non-constant loading.

Flexural Shear Flow in Open Thin-Walled Sections
Taking the shaded area in account from the above object the flexural shear flow as $$\delta x$$ approaches 0. This is a general equation and can be used for either symmeterical or unsymmeterical cross sections. $$\displaystyle \int_{A_s}^{} \frac{d\sigma_{xx}}{dx} dA = -q_A$$

If a cross section is symmeterical about one of the axis (y or z), then $$I_{yz}=0$$.Using the above equations for general cross sections along with what we know about the properties of symmeteric cross sections we can simplify and come up with the following equations.

Symmeteric Cross Section Equations $$\sigma_{xx}= \frac{M_y z}{I_y}$$ $$q(s)=-\frac{V_z Q_y}{I_y}$$ $$Q_y=\int_{A_s}^{}z dA = z_c A_s$$

Unsymmeteric Cross Section Equations This equation is a modifided from equation (4.29) from the Sun 2006 textbook. $$\displaystyle \sigma_{xx} = (k_y M_z - k_{yz} M_y)y + (k_z M_y - k_{yz} M_z)z$$ $$k_y= \frac {I_y}{D}$$ $$k_{yz}= \frac {I_{yz}}{D}$$ $$k_z = \frac {I_z}{D}$$ where $$D=I_yI_z-I_{yz}^2$$

$$\sigma_{xx} = \begin{bmatrix}z & y\end{bmatrix}_{1x2}\begin{bmatrix}k_z & -k_{yz}\\ -k_{yz} & k_y\end{bmatrix}\begin{Bmatrix}M_y \\ M_z\end{Bmatrix}$$

We want to particularize to a symmetric cross-section so $$I_{yz}=0$$. Consider $$M_z=0$$:

$$I_{yz}=0 => D = I_yI_z$$

$$k_y = \frac{1}{I_z}, k_z=\frac{1}{I_y}, k_{yz}=0$$

$$\sigma_{xx} = \frac{ zM_{y}}{ I_{y}} $$

Non Symmetrical Case

$$\displaystyle q(s) = -(k_{y}V_{y} - k_{yz}V_{z})Q_{z} - ((k_{z}V_{z} - k_{yz}V_{y})Q_{y}$$

$$ Q_{z} = \int_{ A_{s}}^{ } y dA$$, $$Q_{y} =  \int_{ A_{s}}^{ } z dA $$

 * {| class="toccolours collapsible collapsed" width="60%" style="text-align:left"

!HW Particularize symmetrical cross section, from team Carbon
 * For the second moment of inertia about the y-axis then the z-axis, the symmetry would be at the centroid if the cross section is symmetrical. Because there is symmetry about the centroid, integration on either side would cancel out and equal zero.
 * For the second moment of inertia about the y-axis then the z-axis, the symmetry would be at the centroid if the cross section is symmetrical. Because there is symmetry about the centroid, integration on either side would cancel out and equal zero.

Considering if $$M_{z} = 0$$ and that $$I_{yz} = 0$$. From previous lectures, it was shown that by definition

$$D:= I_{y}I_{z} - (I_{yz})^{2}$$. Therefore, $$D = I_{y}I_{z}$$

$$\Rightarrow k_{y} = \frac{I_{y}}{D} = \frac{I_{y}}{I_{y}I_{z}} = \frac{1}{I_{z}}$$

$$k_{yz} = 0$$

$$k_{z} = \frac{I_{z}}{I_{y}I_{z}} = \frac{1}{I_{y}}$$

$$\sigma _{xx} = \begin{bmatrix} z & 0 \end{bmatrix}\begin{bmatrix} \frac{1}{I_{y}} & 0\\ 0 & \frac{1}{I_{z}} \end{bmatrix}\begin{Bmatrix} M_{y}\\ 0 \end{Bmatrix}$$

$$\sigma _{xx} = \begin{bmatrix} z & 0 \end{bmatrix} \begin{Bmatrix} \frac{M_{y}}{I_{y}}\\ 0 \end{Bmatrix}$$

$$\Rightarrow \sigma _{xx} = \frac{zM_{y}}{I_{y}}$$

Contribution by carbon.w
 * }

 * {| class="toccolours collapsible collapsed" width="60%" style="text-align:left"

!HW Matrix Form of q(s), from team Carbon
 * The following equation can can be written in matrix form $$q(s) = -(k_{y}V_{y} - k_{yz}V_{z})Q_{z} - (k_{z}V_{z} - k_{yz}V_{y})Q_{y}$$
 * The following equation can can be written in matrix form $$q(s) = -(k_{y}V_{y} - k_{yz}V_{z})Q_{z} - (k_{z}V_{z} - k_{yz}V_{y})Q_{y}$$

$$q(s) = \begin{bmatrix} -Q_{z} & -Q_{y} \end{bmatrix}\begin{bmatrix} k_{y} & -k_{yz}\\ -k_{yz} & k_{z} \end{bmatrix}\begin{Bmatrix} V_{y}\\ V_{z} \end{Bmatrix}$$

To arrive at the particular case where $$q(s) = -\frac{V_{z}Q_{y}}{I_{y}}$$, make the assumption that $$Q_{z} = 0$$ and $$I_{yz} = 0$$

Therefore $$D = I_{y}I_{z}$$ and this can be plugged into the k-values:

$$\Rightarrow k_{y} = \frac{I_{y}}{D} = \frac{I_{y}}{I_{y}I_{z}} = \frac{1}{I_{z}}$$

$$k_{yz} = 0$$

$$k_{z} = \frac{I_{z}}{I_{y}I_{z}} = \frac{1}{I_{y}}$$

The matrix then becomes

$$q(s) = \begin{bmatrix} 0 & -Q_{y} \end{bmatrix}\begin{bmatrix} \frac{1}{I_{z}} & 0\\ 0 & \frac{1}{I_{y}} \end{bmatrix}\begin{Bmatrix} V_{y}\\ V_{z} \end{Bmatrix} = q(s) = -\frac{V_{z}Q_{y}}{I_{y}}$$

Contribution by carbon.w
 * }

... 34-4 ...

Plate Buckling
The full plate buckling notes can be found here and were not created by Team Carbon. The key points of the HW 6 problem have been noted below.

Consider the factor with x as variable in the expression for u_z, i.e.,

$$\sin ( \frac{m \pi x}{a} )$$

Find the period T of this function by considering the following equation:

$$\sin ( \frac{m \pi (x + T)}{a} ) = \sin ( \frac{m \pi x}{a} )$$

We see that for this equation to hold (and T to not equal 0 in all cases), T must be a multiple of $$2 \pi$$ because the sin function is $$2 \pi$$ periodic.

Show that m is indeed the number of half wave-lengths.

We can take the second derivative of this function and plot it (though not shown here). The calculus shows that m is the distance between inflection points of the curve of the buckling plate. Inflection points designate changes in the sign of the curvature, thus each wave has two inflection points. The distance between inflection points, m, is therefore the number of half-wavelengths.

 Incomplete plate buckling section; missing perspective view of buckling mode shapes; see Team Aero. Eml4500.f08 11:35, 25 November 2008 (UTC)

These Views are located in the MATLAB section of our homework, Here User:Eas4200c.f08.carbon.orear/HW6/MATLAB Eas4200c.f08.carbon.clausen 15:09, 3 December 2008 (UTC)

Examination of e-learning and Wiki-Media
It was asked of each group that the members come to a consensus on the benefits of e-learning or MediaWiki, deciding which is more advantageous as a learning tool for students and why. As our group begins to weigh these tools, we will examine the following features of each. In HW 7 we will give testimony to our experiences with each and make a group decision on the better learning system.

 Note: There are many points that need to be included in the comparison, as per my e-mail on this matter, e.g., availability and accessibility of the work over the years (you may not have access to your work after the end of the semester with E-learning, but you can access your work and the work of all other teams and my feedback in pointing out the best features many many years to come with Mediawiki on Wikiversity), skills for job searching (it is unlikely that you can list E-learning on your resume, but you can list Mediawiki as skills that could be useful for your future employer), incorporation of other media (sound and video recordings) in the work (you can with Mediawiki, you cannot with E-learning), develop new technical text-processing skills such as latex (Mediawiki yes, E-learning no), etc. Reread my e-mail on this matter. Regarding privacy: you can create (structured) anonymous wiki usernames as per my instruction at the beginning of the semester; only me and my TAs would know your identity for grading purpose, just like with E-learning; see for example Team Radsam. Eml4500.f08 14:15, 23 November 2008 (UTC)


text:
User:Eas4200c.f08.carbon.orear/HW6/Contributors

Brian Taylor Eas4200c.f08.carbon.taylor 19:27, 16 November 2008 (UTC)

Jeff O'Rear Eas4200c.f08.carbon.orear 13:46, 19 November 2008 (UTC)

Jeff Wright Eml4500.f08.group.w 02:58, 21 November 2008 (UTC)

Beau Guidry Eas4200c.f08.carbon.guidry 03:53, 21 November 2008 (UTC)

Abby Botoh Eas4200c.f08.carbon.booth 03:55, 21 November 2008 (UTC)

Aaron Clausen Eas4200c.f08.carbon.clausen 13:47, 21 November 2008 (UTC)

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/