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User:Eas4200c.f08.wiki.c/homework report 6

=Meeting 30=

The torsional analysis previously started is continued by looking at the dimensions of the derivative of stress:

$$[\frac{\partial \sigma _{ij}}{\partial x_{i}}] = \frac{F}{L^3} = \frac{force}{volume}$$

Recalling - D. Prandtl stress function - from the road map previously constructed in class we see that:

$$\sigma _{yx} = \frac{\partial \phi }{\partial z}$$

and

 $$ \sigma _{zx} = -\frac{\partial \phi }{\partial y}$$

where $$\phi$$ plays the role of a potential function and $$(\sigma _{yx}, \sigma _{zx}$$ are composed of the "gradient" of $$\phi$$ with respect to (y,z).

As a review of the gradient operators, recall that for scalar functions f(x,y,z):  $$\bigtriangledown f(x,y,z) = \frac{\partial f}{\partial x}\vec{i} +\frac{\partial f}{\partial y}\vec{j} +\frac{\partial f}{\partial z}\vec{k}$$

Where f is the potential function.

Now, substituting the values for $$\sigma _{yx}$$ and $$\sigma _{zx}$$ into the previously given Laplace's Equation we get:

$$\frac{\partial }{\partial y}(\frac{\partial \phi}{\partial z}) + \frac{\partial }{\partial z}(-\frac{\partial \phi}{\partial y}) = 0$$

Which can be rearranged to show that:  $$\frac{\partial ^2 \phi}{\partial y \partial x} = \frac{\partial ^2 \phi}{\partial z \partial y}$$

Due to $$\phi$$ being a continuous and smooth function, the second derivatives are interchangeable in the order they are performed.

Likewise:

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

The lefthand side can also be written as $$\bigtriangledown ^2 \phi$$ which is the Laplacian of $$\phi$$.

$$[t]_{3x1} = [\sigma]_{3x3}[n]_{3x1}$$

Where t is the component of the traction force t, $$\sigma$$ is the component of the stress tensor, and n is the component of the normal vector $$\vec{n}$$

A graphical depiction of the 2-D case is shown below:

=MATLAB=

Above is a graph of $$\sigma_{xx}$$ at all y positions in the middle wing section on the top skin. Due to the graph being concave up, it can be seen that the load is compressive on the top skin of the wing section.

Above is a graph of $$\sigma_{xx}$$ at all y positions in the middle wing section on the bottom skin. Due to the graph being concave down, it can be seen that the load is tensile on the top skin of the wing section.

The average compressive stress was found to be 1.4423e+008.


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User:Castrophicus

I'm using wikiversity to aid my learning Japanese, Spanish, and Turkish.

title: User:Castrophicus


text:
User:EML4500.F08.RAMROD.B/Homework 6

=Continuous vs Discrete=

The following chart compares the continuous and discrete cases. Both cases are formulated using the principal of virtual work.

For the continuous case, the equations are valid for all $$\displaystyle w(x)$$ at $$\displaystyle w(0)=0$$ which is the boundary equation for the elastic bar. For the discrete case, the equations are valid for all $$\displaystyle \mathbf{w}$$. Here the boundary conditions are eliminated here.

Stiffness Term
the below picture depicts an elastic bar with all nodes.

This photo depicts the element displacement more closely.

Here we will assume the displacement $$\displaystyle u(x) $$ for $$\displaystyle x_i\leq x\leq x_{i+1}$$. This means that $$\displaystyle x \epsilon [x_i,x_{i+1}]$$ where the character $$\displaystyle \epsilon $$ means 'belongs to'.

Motivation For Linear Interpolation of $$\displaystyle u(x)$$ For The Two Bar Truss
The function $$\displaystyle u(x) $$ must be interpolated because the deformed shape is a straight line which means there must have been an assumption of displacement between two nodes. This allows a connection between the two cases where there are only axial displacements, or zero transverse displacements. The below picture shows this.

title: User:EML4500.F08.RAMROD.B/Homework 6


text:
User:Eas4200c.f08.blue.a/Homework6

Lecture Notes
Below are the lecture notes for HW #6. They are divided by date with a summary of each section at the top.

Intro To Torsional Analysis
This lecture covered deriving the shear stress matrices for a given load F.

$$ F = \frac{P}{A} $$

The shear stress can then be computed in terms of x,y,and z coordinates as well as the force F.

These results will be used in our discussion of torsional analysis.

Torsional Analysis 1
This lecture continued the discussion of torsional analysis that was done in previous lectures.

Continuing Torsional Analysis:

Dimensional analysis, from lecture 29, is applied to the differential stress formula from lecture 27 to show that the differential of stress over distance x has the dimensions of force over volume.

$$\left[\frac{\delta \sigma _{ij}}{\delta x_{i}} \right]=\frac{F}{L^{3}}=\frac{force}{volume}$$

In equation 3.14 in Mechanics of aircraft structures by C.T. Sun, it is shown that the of partial derivatives of shear force in the x and y directions is equal to zero.

$$\frac{\partial \tau _{xz}}{\partial x}+\frac{\partial \tau _{yz}}{\partial y}=\frac{\partial \sigma _{xz}}{\partial x}+\frac{\partial \sigma _{yz}}{\partial y}=0$$

From the roadmap described in lecture 16, the Prandtl stress function $$\phi $$ can be found using the following formulas (Sun 3.15):

$$\sigma _{yx}=\frac{\delta \phi }{\delta z}$$ $$\sigma _{zx}=\frac{-\delta \phi }{\delta y}$$

As can be seen, the Prandtl stress function plays the role of a potential function. $$\left(\sigma _{yx},\sigma _{zx} \right)$$ is a component of the "gradient" of the Prandtl function with respect to y and z.

Recall that the gradient of a scalar function f(x,y,z) is simply the partial differential of f in each direction (x, y, and z) or (i, j, and k).

$$\vec{\bigtriangledown} f(x,y,z)=\frac{\partial f}{\partial x}\hat{i}+\frac{\partial f}{\partial y}\hat{j}+\frac{\partial f}{\partial z}\hat{k} $$

In lecture 27, the class found that the sum of the partial derivative with respect to y of stress in the yx direction and the partial derivative with respect to z of the stress in the zx direction results in zero. Now we can replace the stresses with the Prandtl stress function equivalents.

$$\frac{\partial \sigma _{yx}}{\partial y}+\frac{\partial \sigma _{xz}}{\partial z}=0$$

$$\frac{\partial }{\partial y}\left(\frac{\partial \phi }{\partial z} \right)+\frac{\partial }{\partial z}\left(\frac{-\partial \phi }{\partial y} \right)=0$$

$$\frac{\partial ^{2}\phi }{\partial y\partial z}-\frac{\partial ^{2}\phi }{\partial z\partial y}=0$$

Since $$\phi$$ is continuous and smooth then the second mixed derivative is interchangable.

Now the Prandtl stress function $$\phi (x,y)$$ can be introduced in terms of the shear stress as in equation 3.15 (Sun, 70).

$$\tau _{xz}=\frac{\partial \phi }{\partial y}$$ $$\tau _{yz}=-\frac{\partial \phi }{\partial x}$$

It follows that:

$$\gamma _{xz}=\frac{\partial w}{\partial x}+\frac{\partial u}{\partial z}=\frac{\partial w}{\partial x}-\theta y$$

$$\gamma _{yz}=\frac{\partial w}{\partial y}+\frac{\partial v}{\partial z}=\frac{\partial w}{\partial y}+\theta x$$

$$\frac{\partial \gamma _{yz}}{\partial x}-\frac{\partial \gamma _{xz}}{\partial y}=2\theta $$

$$\gamma _{yz}=\frac{1}{G}\tau _{yz}$$

$$\gamma _{xz}=\frac{1}{G}\tau _{xz}$$

$$\frac{\partial \tau _{yz}}{\partial x}-\frac{\partial \tau _{xz}}{\partial y}=2G\theta $$

$$\frac{\partial ^{2}\phi }{\partial x^{2}}+\frac{\partial ^{2}\phi }{\partial y^{2}}=-2G\theta $$

$$\frac{\partial ^{2}\phi }{\partial y^{2}}-\frac{\partial ^{2}\phi }{\partial z^{2}}=-2G\theta=\bigtriangledown ^{2}\phi $$

This equation for the Prandtl Equation is equivalent to step F of the Torsional analysis Roadmap.

To find a function for the Prandtl Function, a two dimensional case is first looked at:

In the above image, a cross section of a non-uniform bar is shown with a force (t) pu.

This image represents the stress vector (t) on an infinitesimal surface of a non-uniform beam.

Torsional Analysis 2
This lecture covered how to obtain the stress matrices

$$\vec{t}=t_{y}\vec{j}+t_{z}\vec{k}$$

It can be infered that:

$$dz=dscos\theta $$  $$n_{y}=cos\theta $$

$$dy=dssin\theta $$  $$n_{z}=sin\theta $$

Summing the forces in the y directon:

$$\sum{F_{y}}=0=-\sigma _{yy}(dz)-\sigma _{yz}(dy)+t_{y}(ds)$$

and using the above relations:

$$t_{y}=\sigma _{yy}n_{y}+\sigma _{yz}n_{z}$$

Similarly,

$$\sum{F_{z}}=0=-\sigma _{zz}(dy)-\sigma _{yz}(dz)+t_{z}(ds)$$

and again, substituting, we find:

$$t_{z}=\sigma _{zz}n_{z}+\sigma _{yz}n_{y}$$

In matrix form this becomes:

$$\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}$$

Note: $$\sigma _{zy}= \sigma _{yz}$$ (because $$\sum{M_{x}}=\sum{M_{y}}=\sum{M_{z}}=0$$), but in order to keep with the unified notation, and to be compatible with the indicial notation, the difference is retained.

The generalized form is as follows:

$$\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}$$

ShorthandSymbollically it would written as

$$\begin{Bmatrix} t_{i} \end{Bmatrix}_{3x1}

=\begin{bmatrix} \sigma _{ij} \end{bmatrix}_{3x3}

\begin{Bmatrix} n_{j} \end{Bmatrix}_{3x1}$$

Plate Buckling 1
This lecture covered two cases of buckling for different cross sections.

Torsional Analogy: Continued from last lecture, there are two types of buckling

Plate Buckling 2
This lecture covered finding the shear stress in a thin walled cross section.

$$r=(y^2+z^2)^{1/2}$$

$$T=2\int \phi dA=2c(\frac{J}{a^2}-\pi a^2)$$

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

$$T= GJ\theta$$

$$\sigma_{yz}=\frac{\delta \phi}{\delta z}=-G \theta_z$$

$$\sigma_{zy}=\frac{\delta \phi}{\delta y}= -G\theta_y$$

$$\tau=\frac{T r}{J}$$

$$u_x(x,y)=0$$

i,e no warping

since

$$\gamma_{yx}=\frac{\delta_{yx}}{G}=\frac{\delta_u}{\delta_y}-\theta_z$$

$$\gamma_{zx}=\frac{\delta_{zx}}{G}=\frac{\delta_u}{\delta_z}-\theta_y$$

$$\sigma_{yx}=-G\theta_z$$

$$\sigma_{zx}=G\theta_y$$

therefore

$$\delta u=(G+1)\theta_z\ \delta y=(G+1)\theta_y \delta z$$

since there is no warping

$$\delta y= /delta z = 0$$

therefore

$$u=o$$

Torsional Analysis and Plate Buckling Combined
This lecture finalizes the discussion of torsional analysis on a non uniform object.

The figure above shows an unsymmetric thin-walled section. $$\displaystyle y_c$$ and $$\displaystyle z_c$$ are measured from the centroid of the shaded area. As labeled in the figure, the thickness $$\displaystyle t_s$$ can be a function of s, the curvilinear coordinate.

A general equation for thin-walled sections is the following (5.1):
 * $$\iint_{A_s} \frac{d\sigma_{xx}}{dx} dA = -q_s$$

A comparison of equations for symmetric and unsymmetric sections: (Note: in the following symmetric cases, the section is symmetric about the y-axis)

Symmetric: $$\displaystyle \sigma_{xx} = \frac{M_yz}{I_y}$$

Unsymmetric: $$\displaystyle \sigma_{xx} = (k_yM_z - k_{yz}M_y)y + (k_zM_y - k_{yz}M_z)z$$

where
 * $$\displaystyle k_y = \frac {I_y}{D}$$
 * $$\displaystyle k_{yz} = \frac {I_{yz}}{D}$$
 * $$\displaystyle k_z = \frac {I_z}{D}$$
 * $$\displaystyle D = I_yI_z - I_{yz}^2$$

The above equation for an unsymmetric section can be rewritten in matrix form to become
 * $$\displaystyle \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} = \begin{bmatrix} z & y \end{bmatrix} \begin{bmatrix} k_z & -k_{yz} \\ -k_{yz} & k_y \end{bmatrix} \begin{Bmatrix} M_y \\ M_z \end{Bmatrix} $$

The equation for an unsymmetric section can also be particularized to become the equation for a symmetric section (considering $$\displaystyle M_z$$ to be equal to zero). With a symmetric section, $$\displaystyle I_{yz}$$ is equal to zero.

 * $$\displaystyle M_z = 0 $$
 * $$\displaystyle I_{yz} = 0 $$
 * $$\displaystyle\therefore D = I_yI_z$$
 * $$\displaystyle k_y = \frac {I_y}{I_yI_z} = \frac {1}{I_z}$$
 * $$\displaystyle k_{yz} = \frac {I_{yz}}{I_yI_z} = 0$$
 * $$\displaystyle k_z = \frac {I_z}{I_yI_z} = \frac {1}{I_y}$$
 * $$\displaystyle\therefore \sigma_{xx} = (k_yM_z - k_{yz}M_y)y + (k_zM_y - k_{yz}M_z)z = (0-0)y + (k_zM_y - 0)z = \frac{M_yz}{I_y}$$

Continuing with our comparisons of equations between symmetric and unsymmetric sections:

Symmetric: $$\displaystyle q(s) = -\frac {V_zQ_y}{I_y}$$
 * $$Q_y = \int_{A_s} z\,dA = z_cA_s$$

Unsymmetric: $$\displaystyle q(s) = -(k_yV_y - k_{yz}V_z)Q_z - (k_zV_z - k_{yz}V_y)Q_y$$
 * $$Q_y = \int_{A_s} z\,dA$$
 * $$Q_z = \int_{A_s} y\,dA$$

The above equation for an unsymmetric section can be rewritten in matrix form to become
 * $$\displaystyle 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} = -\begin{bmatrix} Q_y & Q_z \end{bmatrix} \begin{bmatrix} k_z & -k_{yz} \\ -k_{yz} & k_y \end{bmatrix} \begin{Bmatrix} V_z \\ V_y \end{Bmatrix} $$

As with the equation for $$\displaystyle \sigma_{xx}$$, the equation for $$\displaystyle q(s)$$ with an unsymmetric section can be particularized to become the equation with a symmetric section. Similar to the case for $$\displaystyle I_{yz}$$ proven above, $$\displaystyle Q_z$$ is equal to zero due to symmetry about the y-axis. Again, $$\displaystyle I_{yz}$$ is also equal to zero.
 * $$\displaystyle Q_z = \int_{A_s} y\,dA = 0$$
 * $$\displaystyle k_y = \frac {I_y}{I_yI_z} = \frac {1}{I_z}$$
 * $$\displaystyle k_{yz} = \frac {I_{yz}}{I_yI_z} = 0$$
 * $$\displaystyle k_z = \frac {I_z}{I_yI_z} = \frac {1}{I_y}$$
 * $$\displaystyle \therefore q(s) = -(k_yV_y - k_{yz}V_z)Q_z - (k_zV_z - k_{yz}V_y)Q_y = -(k_yV_y-0)0 - (k_zV_z - 0)Q_y = -\frac {V_zQ_y}{I_y}$$

When dealing with stringer-web sections, it is assumed that the thickness $$\displaystyle t$$ of the skin and spar web is very small, therefore their areas are neglected in the computation of $$\displaystyle I_y, I_z, I_{yz}, Q_y, Q_z$$. The only areas used are the areas of the stringers (as shown in Fig. 2 to the left).

With a section that is symmetric about the y-axis, the areas will be set such that $$\displaystyle A_3 = A_2$$ and $$\displaystyle A_4 = A_1$$. However, for the unsymmetric case, the areas will be set such that $$\displaystyle A_1 = A,\,\, A_2 = 2A,\,\, A_3 = 3A,\,\, A_4 = 4A$$.

Matlab Code and Results
The Matlab code for this homework covers a plate buckling analysis for the NACA Airfoil. First, the compressive stresses are computed in the X direction so they can be compared to the buckling stress. An analysis is done comparing the stress in each direction for a variety of aspect ratios. This leads to the final comparison of the compressive stresses to buckling stress.

Matlab Code
This code is always being modified to optimize its performance and meet the requirements of the class. As revisions are made, they will be posted.

The m-file used in this HW Report is called NACA_Airfoil_Buckling_Stress.

Note: The NACA_Airfoil_Generator m-file is needed when running the Bucking Stress Program.

Now, here is the m-file used to calculate the Plate Buckling Characteristics:

Team Members that Contributed to This Report
Mark Barry Eas4200c.f08.blue.a 20:20, 19 November 2008 (UTC)

Thomas Sexton EAS4200c.f08.blue.c 00:15, 21 November 2008 (UTC)

Christian Garabaya Eas4200c.f08.blue.d 01:32, 20 November 2008 (UTC)

Clay Robertson eas4200c.f08.blue.f 00:21, 21 November 2008 (UTC)

Jeffrey Shea Eas4200c.f08.blue.e 23:10, 19 November 2008 (UTC)

Marc Pelletier EAS4200c.f08.blue.b 15:27, 21 November 2008 (UTC)

Wei-Teck Lee Eas4200c.f08.blue.g 05:40, 21 November 2008 (UTC)

title: User:Eas4200c.f08.blue.a/Homework6


text:
Draft talk:Archive/2024/WikiService/115

I suggest to insert a link at the bottom of the page where to go next in the course. --tomaschwutz 20:30, 19 November 2008 (UTC)

title: Draft talk:Archive/2024/WikiService/115


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

Composite Laminate Analysis: MATLAB Code Study
The code below, developed by carbon.orear and carbon.clausen, is used to analyze stress and strain in a composite laminate. This material is relevant to this course and is presented in Chapter 8 of the Sun book. It has been added to this Homework as an additional contribution and so that students may get an understanding of this material beyond the scope of this course. The material covered in Chapter 8 is extensive and it is uncertain at this time whether much of this course will focus on it.

For additional information, students can enroll in Dr. Ifju's Aerospace Structural Composites Course offered in Spring 2009.

This code is extensive, but has been slightly commented for ease of understanding. Most of the outputs are not cohesive without reading the comments in the code. Plots generated must be connected by hand as seen in the second example plot.

The Test Case is Example 8.4. from pg. 288 in the Sun book.

Case Test
EXAMPLE 8.4 in the Sun book has been used as a test model. A known strain input was given instead of a known Nx.

EDU>> layers

How many layers are there? 8

What is the thickness of each layer? .127/1000

What is the value of E (in GPa) in the (1) direction? 140

What is the value of E (in GPa) in the (2) direction? 10

What is the value (in GPa) of G12? 7

What is the value of the Poisson ratio in the (12) direction? .3

For the given properties the number of layers and Q matrix is given: 8

Q =

1.4091e+002 3.0194e+000            0 3.0194e+000 1.0065e+001            0 0           0  7.0000e+000

-5.0800e-004 -3.8100e-004 -2.5400e-004 -1.2700e-004 0 1.2700e-004 2.5400e-004 3.8100e-004 5.0800e-004

layer 1

What is the fiber orientation for this layer in radians? pi/4

layer 2

What is the fiber orientation for this layer in radians? -pi/4

layer 3

What is the fiber orientation for this layer in radians? 0

layer 4

What is the fiber orientation for this layer in radians? pi/2

layer 5

What is the fiber orientation for this layer in radians? pi/2

layer 6

What is the fiber orientation for this layer in radians? 0

layer 7

What is the fiber orientation for this layer in radians? -pi/4

layer 8

What is the fiber orientation for this layer in radians? pi/4

7.8540e-001 -7.8540e-001 0 1.5708e+000 1.5708e+000 0 -7.8540e-001 7.8540e-001

(:,:,1) =

4.6252e+001 3.2252e+001  3.2710e+001 3.2252e+001 4.6252e+001  3.2710e+001 3.2710e+001 3.2710e+001  3.6233e+001

(:,:,2) =

4.6252e+001 3.2252e+001 -3.2710e+001 3.2252e+001 4.6252e+001 -3.2710e+001 -3.2710e+001 -3.2710e+001 3.6233e+001

(:,:,3) =

1.4091e+002 3.0194e+000            0 3.0194e+000 1.0065e+001            0 0           0  7.0000e+000

(:,:,4) =

1.0065e+001 3.0194e+000  4.2585e-016 3.0194e+000 1.4091e+002  7.5859e-015 4.2585e-016 7.5859e-015  7.0000e+000

(:,:,5) =

1.0065e+001 3.0194e+000  4.2585e-016 3.0194e+000 1.4091e+002  7.5859e-015 4.2585e-016 7.5859e-015  7.0000e+000

(:,:,6) =

1.4091e+002 3.0194e+000            0 3.0194e+000 1.0065e+001            0 0           0  7.0000e+000

(:,:,7) =

4.6252e+001 3.2252e+001 -3.2710e+001 3.2252e+001 4.6252e+001 -3.2710e+001 -3.2710e+001 -3.2710e+001 3.6233e+001

(:,:,8) =

4.6252e+001 3.2252e+001  3.2710e+001 3.2252e+001 4.6252e+001  3.2710e+001 3.2710e+001 3.2710e+001  3.6233e+001

ABD info 6.1843e-002 1.7918e-002            0 1.7918e-002 6.1843e-002  1.7347e-018 0 1.7347e-018  2.1962e-002

0           0            0            0            0            0            0            0 -4.2352e-022

4.8977e-009 2.4994e-009  8.0404e-010 2.4994e-009 3.8257e-009  8.0404e-010 8.0404e-010 8.0404e-010  2.8473e-009

ans =

1.7652e+001 -5.1144e+000 4.0397e-016 -5.5026e-030 -9.9505e-030  6.4451e-029 -5.1144e+000 1.7652e+001 -1.3943e-015  1.8992e-029  3.4343e-029 -2.2245e-028 4.0397e-016 -1.3943e-015 4.5533e+001 -6.2022e-013 -1.1216e-012  7.2645e-012 -5.5026e-030 1.8992e-029 -6.2022e-013  3.0904e+008 -1.9515e+008 -3.2163e+007 -9.9505e-030 3.4343e-029 -1.1216e-012 -1.9515e+008  4.0111e+008 -5.8161e+007 6.4451e-029 -2.2245e-028 7.2645e-012 -3.2163e+007 -5.8161e+007  3.7671e+008

Input the midplane strains in a vector of the form [x y xy]. [1000e-6 0 0]

midstrains =

1.0000e-003           0            0

Input the midplane curvatures (in m^-1) in a vector of the form [x y xy]. [0 0 0]

midcurvs =

0    0     0

1.0000e-003 1.0000e-003  1.0000e-003  1.0000e-003  1.0000e-003  1.0000e-003  1.0000e-003  1.0000e-003 0           0            0            0            0            0            0            0            0            0            0            0            0            0            0            0

1.0000e-003 1.0000e-003  1.0000e-003  1.0000e-003  1.0000e-003  1.0000e-003  1.0000e-003  1.0000e-003 0           0            0            0            0            0            0            0            0            0            0            0            0            0            0            0

4.6252e-002 4.6252e-002  1.4091e-001  1.0065e-002  1.0065e-002  1.4091e-001  4.6252e-002  4.6252e-002 3.2252e-002 3.2252e-002  3.0194e-003  3.0194e-003  3.0194e-003  3.0194e-003  3.2252e-002  3.2252e-002 3.2710e-002 -3.2710e-002           0  4.2585e-019  4.2585e-019            0 -3.2710e-002  3.2710e-002

4.6252e-002 4.6252e-002  1.4091e-001  1.0065e-002  1.0065e-002  1.4091e-001  4.6252e-002  4.6252e-002 3.2252e-002 3.2252e-002  3.0194e-003  3.0194e-003  3.0194e-003  3.0194e-003  3.2252e-002  3.2252e-002 3.2710e-002 -3.2710e-002           0  4.2585e-019  4.2585e-019            0 -3.2710e-002  3.2710e-002

1.5000e-003 5.0000e-004  1.0000e-003  1.0000e-003  1.0000e-003  1.0000e-003  5.0000e-004  1.5000e-003 5.0000e-004 1.5000e-003  1.0000e-003  1.0000e-003  1.0000e-003  1.0000e-003  1.5000e-003  5.0000e-004 1.1102e-019 1.1102e-019  5.0000e-004 -5.0000e-004 -5.0000e-004  5.0000e-004  1.1102e-019  1.1102e-019

1.5000e-003 5.0000e-004  1.0000e-003  1.0000e-003  1.0000e-003  1.0000e-003  5.0000e-004  1.5000e-003 5.0000e-004 1.5000e-003  1.0000e-003  1.0000e-003  1.0000e-003  1.0000e-003  1.5000e-003  5.0000e-004 1.1102e-019 1.1102e-019  5.0000e-004 -5.0000e-004 -5.0000e-004  5.0000e-004  1.1102e-019  1.1102e-019

7.1963e-002 7.1963e-002  1.4091e-001  3.0194e-003  3.0194e-003  1.4091e-001  7.1963e-002  7.1963e-002 6.5421e-003 6.5421e-003  3.0194e-003  1.0065e-002  1.0065e-002  3.0194e-003  6.5421e-003  6.5421e-003 -7.0000e-003 7.0000e-003            0 -8.5725e-019 -8.5725e-019            0  7.0000e-003 -7.0000e-003

7.1963e-002 7.1963e-002  1.4091e-001  3.0194e-003  3.0194e-003  1.4091e-001  7.1963e-002  7.1963e-002 6.5421e-003 6.5421e-003  3.0194e-003  1.0065e-002  1.0065e-002  3.0194e-003  6.5421e-003  6.5421e-003 -7.0000e-003 7.0000e-003            0 -8.5725e-019 -8.5725e-019            0  7.0000e-003 -7.0000e-003


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text: User:Eas4200c.f08.carbon.orear/HW6/Composites2


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Wikiversity talk:FAQ/Editing/Edit summary

I did not understand Help:Edit_summary, could someone, please explain or demonstrate it? --tomaschwutz 21:46, 19 November 2008 (UTC)
 * When you want to create a new subject in the talk page, you need to used the '+' tab to create it. If you want to add a response to an existing text, just use the 'Modify' link of the thread.Crochet.david 16:27, 20 November 2008 (UTC)


text:
Learn SignWriting for American Sign Language/Handshapes

Basic Handshapes
In ASL, there are about 150 distinct handshapes. In SignWriting, the handshape symbols are all derived from 3 basic handshapes:
 * /Closed fist/
 * The S and index finger handshapes are both derived from the closed fist, since the fingers touch the palm in both cases.
 * /Open fist/
 * The O and D handshapes are derived from the open fist, since the fingers contact each other, not the palm.
 * /Flat palm/
 * The claw or 5-hand handshapes are derived from the flat palm, since the fingers do not contact each other or the palm.

Orientation

 * The palm of the hand is represented in white; the back of the hand in black.
 * Signs in the wall plane have no space at the knuckle joint; signs in the floor plane have a small space present.

Combined, these provide easy identifiers for the hand's position in space.


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User talk:Mike Peel

 Hello Mike Peel, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. After you leave a comment on a talk page, remember to sign and date; it helps everyone follow the threads of the discussion. The signature icon in the edit window makes it simple. To get started, you may

 * Take a guided tour and learn to edit.
 * Visit a (kind of) random random project.
 * Browse Wikiversity, or visit a portal corresponding to your educational level: pre-school, primary, secondary, tertiary, non-formal education.
 * Find out about research activities on Wikiversity.

 * Read an introduction for teachers and find out how to write an educational resource for Wikiversity.
 * Give feedback about your initial observations
 * Discuss Wikiversity issues or ask questions at the colloquium.
 * Chat with other Wikiversitans on #wikiversity-en.

And don't forget to explore Wikiversity with the links to your left. Be bold to contribute and to experiment with the sandbox or your userpage, and see you around Wikiversity! --Cormaggio talk 22:07, 19 November 2008 (UTC)

title: User talk:Mike Peel


text:
Learn SignWriting for American Sign Language/Introduction

SignWriting is a writing system for visuospatial languages, such as American Sign Language. Previous methods of writing such languages have relied on knowledge of another language (as in the case of English glosses) or have been found insufficient for everyday use (as in the case of Stokoe notation).

SignWriting uses iconic symbols to represent the body, it's movement, dynamics, facial expression, etc. It is written in columns going down the page, contrary to English which is written in lines from top to bottom, left to right.


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Wikipedia service-learning courses/101

Creating content in Wikimedia projects (articles, textbooks, learning projects) involves writing wiki hypertext.

Therefore, it is a good manner to get aquainted with the rules and to develop your skills.

Wiki Editing
Wikipedias Wikipedia:Tutorial gives an nice overview.

Skim through the following pages. You will probably coming back to these pages when writing content.

 * Help:Editing and its subpages.
 * Wikipedia:Edit summary legend

Concept questions
 * Why is it good to give ft in the summary field?

Assignment
Congratulations! You have an overview of the basics of creating wiki content. So you can start to create your first page.

If you do not have done so already, please [ create an account], log in, and come back to this page.

A click on the following link lets you create a subpage named My first page on your user page (you create it by clicking then on "Start the … page"). Open this new page in a new browser window, so that you can read this learning project and edit your new page at the same time. On this new page, create content about anything you like. You could describe how you found out about Wikiversity and what you expect to learn here. Use headings, lists and other formattings. Link back to this article.

Style of Articles
There is a wealth of information, especially in Wikipedia. Do not be afraid. Do not spent too much time with reading. Try to pick up the most important things at the beginning and the end of the pages and try to remember what else is there. So you can come back to it, when you are actually writing and policies are covered in more detail in a later course (Wikipedia_service-learning_courses/110)

 * Wikiversitie's Manual of Style
 * Wikipedias's Five Pillars
 * How to copy-edit wikipedia articles
 * Wikipedia's Manual of Style
 * Wikibook's Manual of Style

Concept questions
 * What are the common things in style between the Wikimedia projects? You may extend the content of the following page.
 * Wikipedia_service-learning_courses/Styles in common in wikimedia projects
 * List also some differences. Why do these differences make sense? You may extend the content of the following page.
 * Wikipedia_service-learning_courses/Differences in styles in wikimedia projects

English Language
For improving your English you can explore the Department of English as a second language

You can post questions about English grammar and usage at Wikipedia's language and grammar desk.

The following sections introduce you to several more advanced concepts that you should been aware of.

Images
How do I link to images? How are floating images positioned with the text? How do I upload an image?
 * Help:Images
 * Specifying placement
 * Create the link to the (not yet uploaded) image with specifying the intended image name.
 * e.g. [[Image:Tk_train.jpg|right]]
 * The link appears in red, the same as an non-existent page
 * When you click on this link, you are guided to upload the image.

More details of preparing images and using Wikimedia Commons are delt with in a later course.

 * Assignment
 * 1) Edit your My first page to display the Wikiversity logo (Wikiversity-logo.svg), scaled to 200 pixels.
 * 2) Add a new picture, that you upload to Wikiversity. Experiment with different positionings of text and the image.

Tables
How can I create tables in wiki syntax?
 * Help:Tables

 * Assignment
 * 1) Take notice of tables and their formatting when you are browsing mediawiki pages. Inspect the source code (via edit page) to learn how others have formatted tables.
 * 2) Edit your My first page to include a simple table. You make take some time to experiment with different formattings.

Namespaces

 * What is an namespace?
 * Help:Namespace (read first section)
 * What namespace are common to Wikimedia projects?
 * Namespace
 * What special namespaces are there in different projects?
 * Namespaces
 * Wikipedia:Namespaces
 * Wikibooks:Namespaces

Subpages
What is a subpage? Do you have an example? Where are subpages used most often?
 * Subpages explained on Meta
 * yes: Link/a/b.
 * on userpages: w:Wikipedia:Subpages
 * on portals
 * on archives of user talk pages.

Interlanguage and interwiki links
How can I link to pages the same wikimedia project in another language? How can I link to other wikimedia projects? What are the prefixes and shortcuts for the Wikimedia sister projects? How can I avoid that the prefix is displayed in the link? How can someone running an own Mediawiki setup the prefixes?
 * Help:Interlanguage links
 * basic syntax: language code:Title
 * example: Plankton
 * InterWikimedia links
 * example: InterWikimedia links : InterWikimedia links
 * Shortcuts and Templates
 * This is called the pipe trick: add a pipe "|" at the end of the link
 * example: Main Page : Main Page
 * Setup in Mediawiki

 * Assignment:
 * 1) Edit your My first page to add some links to articles in the Wikimedia sister projects
 * 2) Edit your My first page to add some links to articles of wikiversity in another language.

Final assignment

 * create your first wikipedia article
 * w:Wikipedia:Your_first_article

Where to go from here

 * check out the other courses from the Wikipedia service-learning courses
 * Wikipedia, a list of learning resources on Wikipedia

title: Wikipedia service-learning courses/101


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User:Eas4200c.f08.blue.a/Lecture 30

'''Torsional Analogy: Continued from last lecture, there are two types of buckling '''

title: User:Eas4200c.f08.blue.a/Lecture 30


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User:EML4500.f08.FEABBQ.Hamdan/HW-7

notes form November-17-08

$${\tilde{k}}^{(e)}_{6x6} {\tilde{d}}^{(e)}_{6x1}={\tilde{f}}^{(e)}_{6x1}$$

$${\tilde{d}}^{(e)}_{6x1}=\begin{Bmatrix} {\tilde{d}}^{(e)}_{1} \\ {\tilde{d}}^{(e)}_{2}\\ {\tilde{d}}^{(e)}_{3}\\ {\tilde{d}}^{(e)}_{4}\\ {\tilde{d}}^{(e)}_{5}\\ {\tilde{d}}^{(e)}_{6} \end{Bmatrix} $$   ,    $${\tilde{f}}^{(e)}_{6x1}=\begin{Bmatrix} {\tilde{f}}^{(e)}_{1} \\ {\tilde{f}}^{(e)}_{2}\\ {\tilde{f}}^{(e)}_{3}\\ {\tilde{f}}^{(e)}_{4}\\ {\tilde{f}}^{(e)}_{5}\\ {\tilde{f}}^{(e)}_{6} \end{Bmatrix}$$

$$\tilde{z}=z $$ ( Rotation about Z-axis)

Moments about the Z-axis are equal in both coordinates: $$\tilde{f}^{(e)}_{3}=f^{(e)}_{3}, \tilde{f}^{(e)}_{6}=f^{(e)}_{6}$$

$$

\tilde{k}=\begin{bmatrix} \frac{EA}{L} &0 &0  &\frac{-EA}{L}  &0  &0 \\ 0 &\frac{12EI}{L^{2}} &\frac{6EI}{L^2} &0 &\frac{-12EI}{L^3}  &\frac{6EI}{L^2} \\ 0&0 &\frac{4EI}{L}  &0  & \frac{-6EI}{L^2} &\frac{2EI}{L} \\ \frac{-EA}{L}& 0 & 0 & \frac{EI}{L}&  0& 0\\ 0&\frac{-12EI}{L^3} &\frac{-6EI}{L^2}  & 0 &  \frac{12EI}{L^3}& \frac{-6EI}{L^2}\\ 0&\frac{6EI}{L^2} & \frac{2EI}{L} &  0& \frac{-6EI}{L^2} &\frac{4EI}{L} \end{bmatrix}_{(6X6)}$$

 * Dimension Analysis:

For Displacement the dimension $$\begin{bmatrix} \tilde{d_1} \end{bmatrix}=L= \begin{bmatrix} \tilde{d_i} \end{bmatrix} \rightarrow   i= 1,2,4,5 $$

$$L\rightarrow$$ Lenght

$$\begin{bmatrix} \tilde{d_i} \end{bmatrix} \rightarrow$$ Dimension of $$\tilde{d}_i$$

The rotation has a dimension of 1 $$\begin{bmatrix} \tilde{d_3} \end{bmatrix}=1= \begin{bmatrix} \tilde{d_6} \end{bmatrix} $$

$$AB=R\cdot \theta$$

$$AB\rightarrow$$ Arc Length

$$\theta \rightarrow$$ Angle

$$ \theta =\frac{AB}{R}$$

$$\left[\theta \right]=\frac{\left[AB \right]}{\left[R \right]}=\frac{L}{L}=1$$

Notes from December 01-08
Computing $$U(\tilde{x}), V(\tilde{x)}$$ :

Local system: $$U(\tilde{x})=U(\tilde{x})\vec{\tilde{i}}+V(\tilde{x})\vec{\tilde{j}}$$

Global System:$$ U(\tilde{x})=U_{x}(\tilde{x})\vec+V_{y}(\tilde{x})\vec $$

Computing $$U(\tilde{x}), V(\tilde{x)}$$ using
 * $$V(\tilde{x)}=N_{2}(\tilde{x})\tilde{d_2}+N_{3}(\tilde{x})\tilde{d_3} +N_{5}(\tilde{x})\tilde{d_5}+N_{6}(\tilde{x})\tilde{d_6}$$
 * $$U(\tilde{x)}=N_{1}(\tilde{x})\tilde{d_1}+N_{4}(\tilde{x})\tilde{d_4}$$

Compute $$U_x(\tilde{x}), V_y(\tilde{x)}$$ from $$U(\tilde{x}), V(\tilde{x)}$$

$$\begin{Bmatrix} U_x(\tilde{x)}\\ U_y(\tilde{x)} \end{Bmatrix}= R^T \begin{Bmatrix} U(\tilde{x})\\V(\tilde{x})

\end{Bmatrix}$$

$$\begin{Bmatrix} U(\tilde{x})\\V(\tilde{x})

\end{Bmatrix}=\begin{bmatrix} N_1 &0 & 0 &N_4  &0  & 0\\ 0 &N_2 &N_3  & 0 & N_5 & N_6 \end{bmatrix}\begin{Bmatrix} \tilde{d_1^{(e)}}\\ \tilde{d_2^{(e)}}\\ \tilde{d_3^{(e)}}\\ \tilde{d_4^{(e)}}\\ \tilde{d_5^{(e)}}\\ \tilde{d_6^{(e)}} \end{Bmatrix}$$

let: $$ \begin{bmatrix} N_1 &0 & 0 &N_4  &0  & 0\\ 0 &N_2 &N_3  & 0 & N_5 & N_6 \end{bmatrix} =\mathbb{N(\tilde{\mathrm{x})}}$$

$$\begin{Bmatrix} U_x(\tilde{x)}\\ U_y(\tilde{x)} \end{Bmatrix}=R^T \mathbb{N(\tilde{\mathrm{x})}}\tilde{T}^{(e)}d^{(e)}$$

title: User:EML4500.f08.FEABBQ.Hamdan/HW-7


text: File:Drawing 11-17-08.jpg

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User:Eml4500.f08.ateam.mcnally/Week 8

Continueing the FEM via PVW
From a previous lecture on meeting 31 about the linear interpolation of U(x), W(x) can be shown by applying the same principle.

i.e.

$$ W(x)=N_{i}(x)W_{i}+N_{i+1}(x)W_{i+1} $$

Now the element stiffness matrix for element I can be written as:

$$ \beta = \int_{x_{i}}^{x_{i+1}}{[N_{i}'W_{i}+N_{i+1}'W_{i+1}](EA)} $$

with

$$[N_{i}'W_{i}+N_{i+1}'W_{i+1}]=W'(x)$$

$$[N_i'd_i+N_{i+1}'d_{i+1}]=U'(x)$$

$$ N_i':=\frac{dN_i(x)}{dx} $$

Likewise for $$N_{i+1}'$$

NOTE

$$ u(x)=\begin{bmatrix} N_i(x) & N_{i+1}(x) \end{bmatrix} \begin{Bmatrix} d_i\\ d_{i+1} \end{Bmatrix} =\mathbf{N(x)_{(1x1)}} \begin{Bmatrix} d_i\\ d_{i+1} \end{Bmatrix} $$

$$ \frac{du(x)}{dx}=\begin{bmatrix} N_i'(x) & N_{i+1}'(x) \end{bmatrix} \begin{Bmatrix} d_i\\ d_{i+1} \end{Bmatrix} =\mathbf{B(x)} \begin{Bmatrix} d_i\\ d_{i+1} \end{Bmatrix} $$

Similarly W(x) can be shown as:

$$ W(x)=\mathbf{N(x)}\begin{Bmatrix} W_i\\ W_{i+1} \end{Bmatrix} $$

$$ \frac{dW(x)}{dx}=\mathbf{B(x)}\begin{Bmatrix} W_i\\ W_{i+1} \end{Bmatrix} $$

Now, let us recall the element degrees of freedom as discussed in a previous lecture

therefore

$$ \begin{Bmatrix} d_i\\ d_{i+1}

\end{Bmatrix} = \begin{Bmatrix} d_1^{(i)}\\ d_2^{(i)} \end{Bmatrix} = \mathbf{d^{(i)}} $$

$$ \begin{Bmatrix} W_i\\ W_{i+1}

\end{Bmatrix} = \begin{Bmatrix} W_1^{(i)}\\ W_2^{(i)} \end{Bmatrix} = \mathbf{W^{(i)}} $$

The element stiffness matrix can now be written as

$$ \beta =\int_{x_{i}}^{x_{i+1}}{{BW}^{(i)}_{(1x1)}(EA)_{(1x1)}{Bd}^{(i)}_{(1x1)}}dx = {W}^{(i)}\cdot({k}^{(i)}{d}^{(i)}) $$

$$ \beta =\int_{x_i}^{X_{i+1}}(EA)({BW^{(i)}})_{(1x1)}\cdot({Bd^{(i)}}){(1x1)}dx

$$

$$ ({BW^{(i)}})^T({Bd^{(i)}}) $$

where

$$ {BW^{(i)T}} = {W^{(i)T}B^T} = {W^{(i)} \cdot B^T} $$

$$ \beta= \mathbf{W^{(i)}}\cdot\int \mathbf{B^T}EA\mathbf{B}dx\mathbf{d^{(i)}} $$

$$ \mathbf{k^{(i)}}_{(2x2)}=\int_{x_i}^{x^{i+1}}{\mathbf{B(x)^T}}_{(2x1)}(EA)_{1x1}\mathbf{B(x)}_{(1x2)}dx $$

If we consider $$ EA $$ to constant the matrix $$ {k^{(i)}} $$ can be written as

$$\textbf{k}^{(i)} = \frac{EA}{L^{(i)}}\begin{bmatrix} 1 & -1 \\ -1 & 1 \end{bmatrix}$$

The next step is the transformation of variable coordinate system from:

$${x}$$ to $$ \tilde{x} $$

Given

$$ d\tilde{x}=dx $$

$$ \tilde{x}:=x-x_i $$

Now $$ {k^{(i)}} $$ can be written as

$$ \mathbf{k^{(i)}}=\int_{\tilde{x}=0}^{\tilde{x}=L^{(i)}}{\mathbf{B^T(\tilde{x})}}EA(\tilde{x})\mathbf{B(\tilde{x}})d(\tilde{x}) $$

The following problem outlined by the image below can be solved using the transformed coordinate system.

Given:

With the following functions for the modulus elasticity and cross sectional area $$ {k^{(i)}} $$ matrix can be written as

$$A(\tilde{x}) = N_1^{(i)}(\tilde{x})A_1 + N_2^{(i)}(\tilde{x})A_2$$

$$E(\tilde{x}) = N_1^{(i)}(\tilde{x})E_1 + N_2^{(i)}(\tilde{x})E_2$$

$$ \mathbf{k^{(i)}}=\int_{\tilde{x}=0}^{\tilde{x}=L^{(i)}}{\mathbf{B^T(\tilde{x})}}{[N_1^{(i)}(\tilde{x})A_1 + N_2^{(i)}(\tilde{x})A_2]}{[N_1^{(i)}(\tilde{x})E_1 + N_2^{(i)}(\tilde{x})E_2]}(\tilde{x})\mathbf{B(\tilde{x}})d(\tilde{x}) $$

title: User:Eml4500.f08.ateam.mcnally/Week 8


text:
User:EML4500.f08.Ateam.Miller/homework6

11/14/08
Our goal is to use linear interpolation to determine $$N_i(\tilde{x})$$. If $$\tilde{x}= x-x_i$$, then

$$N_i(\tilde{x})=\frac{\tilde{x}-\tilde{x}_{i+1}}{\tilde{x}_{i}-\tilde{x}_{i+1}}$$

Now, the shape function can be written as follows:

$$ N_2^{(i)}(\tilde{x})=\frac{\tilde{x}}{L^{(i)}}= \begin{cases} & \text{0 at } \tilde{x}= 0\\ & \text{1 at } \tilde{x}= L \end{cases} $$

Our next goal is to compare the general $$\mathbf{k}^{(i)}$$ to the stiffness matrix obtained by using

$$\frac{1}{2}(A_1+A_2)$$ and $$\frac{1}{2}(E_1+E_2)$$, where $$E_1 \neq E_2$$. The resulting stiffness matrix is

$$ \frac{(E_1+E_2)(A_1+A_2)}{4L^{(i)}} \begin{bmatrix} 1 & -1\\ 1 & -1\\ \end{bmatrix} =\mathbf{k}_{avg}^{(i)} $$

This matrix is also known as the average stiffness matrix.

Now, we want to determine $$\mathbf{k}^{(i)}-\mathbf{k}_{avg}^{(i)}$$

$$ \mathbf{k}^{(i)}-\mathbf{k}_{avg}^{(i)}= \frac{(A_1+A_2)}{4L^{(i)}}(2E-E_1-E_2) $$

These results show that the average value of the stiffness matrix is not equal to the value of the stiffness matrix evaluated at the average value of x. This can be verified with the Mean Value Theorem.

Mean Value Theorem
Consider the 2D mass located in the xy plane. Its centroid is located at $$(\bar{x},\bar{y})$$.

The Mean Value Theorem (MVT) states

$$\int_{x=a}^{x=b}{f(x)dx}=f(\bar{x})[b-a]$$

for $$\bar{x}\epsilon[a,b]$$ or $$a \leq \bar{x} \leq b$$.

Similarly,

$$\int_{x=a}^{x=b}{f(x)g(x)dx}=f(\bar{x})g(\bar{x})[b-a]$$

for $$a \leq \bar{x} \leq b$$.

However,

$$f(\bar{x})\neq \frac{1}{b-a}\int_{a}^{b}{f(x)dx}$$

where $$\frac{1}{b-a}\int_{a}^{b}{f(x)dx}$$ is the average value of f.

Similarly,

$$g(\bar{x})\neq \frac{1}{b-a}\int_{a}^{b}{g(x)dx}$$

where $$\frac{1}{b-a}\int_{a}^{b}{f(x)dx}$$ is the average value of g.

title: User:EML4500.f08.Ateam.Miller/homework6


text:
User:Mopelia

Seitenaufruf um Uhr am Monday, den 29. July 2024.

title: User:Mopelia


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Template:Mortuary Science TOC

 * This module is

 * This topic is
 * —

 * Module tutor
 * {| cellpadding="4" cellspacing="0"

 * User: — Wikiversity talk page | E-mail tutor
 * }
 * Please contact the module tutor, the person who maintains this page, to request help or to give feedback.

title: Template:Mortuary Science TOC


text: Template talk:Mortuary Science TOC

title: Template talk:Mortuary Science TOC


text:
Social psychology (psychology)/Assessment/Essay/Cover page

Download, insert, and complete this electronic cover sheet.


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User:Eas4200c.f08.aero.e/HW5

 * [|Lectures: Week 9 --- 10/20 through 10/24]

 * [|Lectures: Week 10 --- 10/27 through 10/31]

title: User:Eas4200c.f08.aero.e/HW5


text:
User:Eas4200c.f08.aero.e/HW4

 * [|Lectures: Week 7 --- 10/6 through 10/10]

 * [|Lectures: Week 8 --- 10/13 through 10/17]

title: User:Eas4200c.f08.aero.e/HW4


text:
User:Eas4200c.f08.aero.e/HW1

 * [|Lectures: Week 1 --- 8/25 through 8/29]

 * [|Lectures: Week 2 --- 9/3 through 9/5]

title: User:Eas4200c.f08.aero.e/HW1


text:
User:Eas4200c.f08.aero.e/HW2

 * [|Lectures: Week 3 --- 9/8 through 9/12]

 * [|Lectures: Week 4 --- 9/15 through 9/19]

title: User:Eas4200c.f08.aero.e/HW2


text:
User:Eas4200c.f08.aero.e/HW3

 * [|Lectures: Week 5 --- 9/22 through 9/26]

 * [|Lectures: Week 6 --- 9/29 through 10/3]

title: User:Eas4200c.f08.aero.e/HW3


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text: Portal talk:North American History/Archive


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User:Eml4500.f08.gravy.jad/Notes6.g

Continuous PVW Case
Continuing Case: PVW of Dynamics of Elastic Bars

PDE: $$\ \frac{d}{dx} \left \lbrack (EA) \frac{du}{dx} \right \rbrack + f = m \ddot{u} \,$$ Equation(1)

$$\ \Rightarrow \mathbf{m} \mathbf{\ddot{d}} + \mathbf{k} \mathbf{d} = \mathbf{F} \,$$ Equation(2)

Derive (2) from (1): $$\ \int_{x=0}^{x=L} W(x) \left \lbrace \frac{d}{dx} \left \lbrack EA \frac{du}{dx} \right \rbrack + f - m \ddot{u} \right \rbrace \, dx = 0 \,$$ for all possible W(x). Where w(x) is the weighting function.

(1) $$\ \Rightarrow \,$$(3) trivial (3) $$\ \Rightarrow \,$$(1) not trivial (3) rewritten as: $$\ \int W(x)g(x)dx =0 \,$$for all W(x) Since (3) holds for all W(x), select W(x)=g(x), then (3) becomes $$\ \int g^{2}(x)dx = 0 \,$$

Integration by Parts
Integration by Parts: r(x), s(x)

$$\ (rs)' = r's + rs' \,$$ where $$\ r'=frac{dr}{dx} \,$$ and $$\ s'=frac{ds}{dx} \,$$

$$\ \int (rs)' = rs - \int rs' \,$$ $$\ \Rightarrow \int r's = rs - \int rs' \,$$

Recall cont. PVW (equation 3)

1st term: $$\ r(x)=(E,A) \frac{du}{dx}, s(x)=w(x) \,$$

By integrating by parts: $$\ \int_{x=0}^{x=L} \underbrace{w(x)}_{s} \frac{d}{dx} \underbrace{\left \lbrack (EA) \frac{du}{dx} \right \rbrack}_{r} dx = \left \lbrack W(EA) \frac{du}{dx} \right \rbrack |_{x=0}^{x=L} \,$$

$$\ = W(L) \left \lbrack (EA)(L) \right \rbrack \frac{du}{dx}(L,t) - W(0) \left \lbrack (EA)(O) \right \rbrack \frac{du}{dx} (0,t) \,$$

Now consider model prob:

At x=0, select W(x) such that W(0)=0 (ie. kinetically dismiss-able)

Motivation: Discrete PVW applied to Eq $$\ \mathbf{W}_{6x1} \cdot \underbrace{ \mathbf{k}_{6x1} \mathbf{d}_{2x1}-\mathbf{F}_{6x1} }_{6x1} = 0_{1x1} \,$$

$$\ F^{T} = \left \lfloor F_{1} F_{2} F_{3} F_{4} F_{5} F_{6} \right \rfloor \,$$

Where $$\ F_{1}, F_{2}, F_{5}, F_{6} \,$$ are unknown reactions. Since $$\ \mathbf{W} \,$$ can be selected arbitraily, select $$\ \mathbf{W} \,$$ such that $$\ W_{1}=W_{2}=W_{5}=W_{6}=0 \,$$, so to eliminate equations involving unknown reactions, thus eliminating 1,2,5,6; $$\ \mathbf{\overline{k}} \mathbf{\overline{d}}=\mathbf{\overline{F}} \,$$

Please note: $$\ \overline{\mathbf{W}} \cdot (\mathbf{\overline{kd}} - \mathbf{\overline{F}})=0 \,$$ for all $$\ \mathbf{W} \,$$

This concludes the motivation.

Unknown reaction term $$\ N(0,t) = \left \lbrack (EA)(0) \right \rbrack \frac{du}{dx} (0,t) \,$$

Continuing PVW: $$\ W(L)F(L) - \int_{0}^{L} \frac{dW}{dx}(EA) \frac{du}{dx} dx + \int_{0}^{L} W(x) \left \lbrack f-m \ddot{u} \right \rbrack dx=0 \,$$ for all W(x) such that W(0)=0.

$$\ \Rightarrow \int_{0}^{L} W(m \ddot{u}) dx + \int_{0}^{L} \frac{dW}{dx} (EA) \frac{du}{dx} dx = W(L)F(t) + \int_{0}^{L} Wf dx \,$$ for all W(x) such that W(0)=0.

Comparison between Constant and Discrete Setting
Constant Setting (PVW)

Inertia : $$\ \int_{0}^{L} Wm \ddot{u} dx = \alpha \,$$

Stiffness : $$\ \int_{0}^{L} \frac{dW}{dx} (EA) \frac{du}{dx} dx = \beta \,$$

Applied force : $$\ W(L)F(t) + \int_{0}^{L} Wf dx = \gamma \,$$

For all W(x) such that: $$\ \underbrace{W(0)=0}_{b.c.} \,$$

Discrete setting (PVW)

Inertia : $$\ \overline{\mathbf{W}} \cdot ( \overline{\mathbf{md}}) (1) \,$$

Stiffness : $$\ \overline{\mathbf{W}} \cdot ( \overline{\mathbf{kd}}) (2) \,$$

Applied force : $$\ \overline{\mathbf{W}} \cdot \overline{\mathbf{F}} (3) \,$$

For all W(x) such that: b.c.'s eliminated

Stiffness term:

Assume displacement u(x) for $$\ x_{i} \le x \le x_{i+1} \,$$

Motivation for linear interpolation of u(x)

Deformed shape is a straight line. ie. there was an implicit assumption of linear interpolation of displacement between 2 nodes

consider the case where there are only axial-displacement (ie. zero transverse displacement)

Question: Express V(x) in terms of $$\ d_{i} = u(x_{i}) \,$$ and $$\ d_{i+1} = u_(x_{i+1}) \,$$ as a linear function in x (ie. linear interpolation

$$\ u(x) = N_{i}(x) d_{i} + N_{i+1}(x)d_{i+1} \,$$ $$\ where N_{i}(x) and N_{i+1}(x) are linear functions in x \,$$

Solving Ni(x)
Rube-Goldberg device by (Walter Benjamin which is a German name so the W sounds like a V and the J sounds like a Y much like the Volkswagen where the V sounds like a F and the W sounds like a V) is a device which performs a simple task in a complicated manner. More information can be obtain here

"Honesty, imagination,and ethics" are important when it comes to writing these homework reports. Because everyone post their homework online it is important to remain honest and abide by your ethics and do your own work. When writing up the reports it takes imagination to use narrative in the reports.

From Continuous Principal of Virtual Work to Discrete Principal of Virtual Work
Discrete Principal of Virtual Work can be achieved using Lagrangian interpolation which is the motivation for the formation of Ni(x) and Ni+1(x).

1) Ni(x) and Ni+1(x) are linear straight lines, thus any linear combination of Ni and Ni+1 is also linear and in particular the expression for u(x) shown previously in this report.

$$N_i(x)= \alpha_i + \beta _ix$$, with ($$\alpha_i + \beta _i$$) are numbers

Linear combination of Ni and Ni+1
The Linear combination of Ni and Ni+1 is calculated as the following:

$$\ N_id_i + N_{i+1}d_{i+1}= (\alpha_i + \beta _ix)d_i +(\alpha_{i+1} + \beta _{i+1}x)d_{i+1} = (\alpha_{i}d_{i} +\alpha_{i+1}d_{i+1}) + (\beta _{i}d_{i}+\beta _{i+1}d_{i+1}) \,$$

which is clearly a linear function of x.

2)Recall equation for u(x)(the interpolation of u(x)) from previous in this report

$$\ u(x_i)= \underbrace{N_i(x_i)}_1 d_i+ \underbrace{N_{i+1}(x_i)}_0 d_{i+1}= d_i \,$$

Prove of equation $$u(x_{i+1})= d_{i+1}$$
The same interpolation can be used for w(x) so     $$\ w(x)= N_i(x)w_i+ N_{i+1}(x)w_{i+1} \,$$

The Element Stiffness matrix for element i
The Element Stiffness matrix for element i can be developed using the steps and equations which follow.

$$ \beta= \int_{x_i}^{x_{i+1}} [ \underbrace{N'_iW_i + N'_{i+1}W_{i+1}}_{W'(x)} (EA) \underbrace{N'_id_i + N'_{i+1}d_{i+1}}_{u'(x)}]dx $$

Where $$ N^{\prime}_i:= \frac{d N_i(x)}{dx}$$ Likewise for $$N^{\prime}_{i+1}$$ and written in matrix form for simplification.

$$ u(x)= \underbrace{\begin{bmatrix} N_{i}(x) && N_{i+1}(x) \end{bmatrix}}_{\vec{N}_{1x2}(x)} \begin{Bmatrix} d_i \\ d_{i+1}  \end{Bmatrix}_{2x1}  $$

Recall the element degrees of freedom

$$ \begin{Bmatrix} d_{i} \\ d_{i+1} \end{Bmatrix}= \begin{Bmatrix} d^{(i)}_1 \\ d^{(i)}_{2}  \end{Bmatrix}= \mathbf{d}^{(i)}$$

$$ \begin{Bmatrix} W_i \\ W_{i+1} \end{Bmatrix}= \begin{Bmatrix} W^{(i)}_1 \\ W^{(i)}_{2}  \end{Bmatrix}= \mathbf{W}^{(i)}$$

$$ \beta= \int_{x_i}^{x_{i+1}} \underbrace{(\mathbf{BW}^{(i)}) }_{1x1} \underbrace{(EA)}_{1x1} \underbrace{(\mathbf{Bd}^{(i)}}_{1x1}dx = \mathbf{W}^{(i)} \cdot (\mathbf{k}^{(i)} \mathbf{d}^{(i)}) $$

$$ \beta= \int_{x_i}^{x_{i+1}}(EA) \underbrace{(\mathbf{BW}^{(i)}) }_{1x1}\cdot  \underbrace{(\mathbf{Bd}^{(i)})}_{1x1}dx = \int_{x_i}^{x_{i+1}}(EA)(\mathbf{BW}^{(i)})^T(\mathbf{Bd}^{(i)})dx $$

$$= \int_{x_i}^{x_{i+1}}(EA)(\mathbf{W}^{(i)T}\mathbf{B}^T)(\mathbf{Bd}^{(i)})dx= \int_{x_i}^{x_{i+1}}(EA)(\mathbf{W}^{(i)} \cdot \mathbf{B}^T)(\mathbf{Bd}^{(i)})dx$$

$$ \beta=\mathbf{W}^{(i)} \cdot (\int_{x_i}^{x_{i+1}}\mathbf{B}^T(EA)\mathbf{B}dx) d^{(i)}$$

$$ k^{(i)}_{2x2}= \int_{x_i}^{x_{i+1}} \underbrace{\mathbf{B}^T(x)}_{2x1} \underbrace{(EA)(x)}_{1x1} \underbrace{\mathbf{B}  (x)}_{1x2} dx$$

$$B(x)= \begin{bmatrix} 1^{st} Term && \frac{1}{L^i} \end{bmatrix}$$ where $$L^{(i)}= x_{i+1} -x_i $$ (which is the length of element i).

Considering EA is a constant
Consider EA = constant, then the equation for k simplifies to:

$$ k^{(i)} {=} \frac{EA}{L^{(i)}} \begin{vmatrix} 1 & -1 \\ -1 & 1 \end{vmatrix} $$

Transformation of variable coordinates from x to $$ \tilde{x}$$
$$\tilde{x}:= x-x_i$$ and $$d\tilde{x}=dx$$

$$ k^{(i)}= \int_{\tilde{x}=0}^{\tilde{x}=L^{(i)}} \mathbf{B}^T(\tilde{x}) (EA)(\tilde{x}) \mathbf{B}(\tilde{x}) d\tilde{x}$$

The tapered bar can be drawn and solved using the tilde axis

$$ A(\tilde{x}) = N_1^{(i)}(\tilde{x})A_1 + N_2^{(i)}(\tilde{x})A_2$$

$$ E(\tilde{x}) = N_1^{(i)}(\tilde{x})E_1 + N_2^{(i)}(\tilde{x})E_2$$

$$k^{(i)}_{2x2}$$ derived
$$N_1^{(i)}(\tilde{x}) = N_1^{(i)}(x)$$ which was solve eariler in this report.

$$N_2^{(i)}(\tilde{x})= \frac{\tilde{x}}{L^{(i)}}= \bigg\{ \begin{matrix} 0 && at, \tilde{x}=0 \\ 1 && at, \tilde{x}=L^{(i)} \end{matrix}$$

k(1)
Another way to solve the tapered bar is using the average area and modulus of elasticity.

$$\underbrace{\frac{E(A_1+A_2)}{L}}_{Average Area} \begin{bmatrix}1&&-1 \\1&& 1 \end{bmatrix}= \mathbf{k}^{(i)}$$

Remark
Recall the mean value theorem (MVT) and its relation to the centroid:

MVT : $$\int_{x=a}^{x=b} f(x) dx= f(\bar{x})[b-a]$$

for $$\bar{x}\in[a,b]$$

$$\in$$ symbol means in so $$ a \le \bar{x} \ge b$$

centroid$$= \int_A xdA= \bar{x}\int_A = \bar{x}A$$

$$\int_{a}^{b}f(x)g(x)dx= f(\bar{x})g(\bar{x})[b-a]$$

where $$a \le \bar{x} \ge b$$

but $$f(\bar{x}) \ne \underbrace{\frac{1}{b-a} \int_{a}^{b}f(x)dx}_{Average value of f}$$

and $$g(\bar{x}) \ne \underbrace{\frac{1}{b-a} \int_{a}^{b}g(x)dx}_{Average value of g}$$

Modification of 2-bar truss code to accommodate general k(i)

The first modification to the 2-bar truss code was adding the values for E1(i), E2(i), A1(i), and A2(i) to the matrices at the beginning of the code. This was done with a 2x2 matrix for both E and A. The columns represent the elements, and the rows represent the nodes. Therefore, e(1,2) is E1(2). These values were then used to caculate k(i) using the following general formula:

$$k_{2x2}(\tilde{x})^{(i)} = \frac{\begin{bmatrix} 2 \left ( A_1 * E_1 \right ) +  \left ( A_2 * E_1 \right ) + \left ( A_1 * E_2 \right ) + 2\left ( A_2 * E_2 \right ) \end{bmatrix}}{6*L^{(i)}}\begin{bmatrix} 1 & -1 \\ -1 & 1 \end{bmatrix}$$

Then, the values of E1(i), E2(i), A1(i), and A2(i) were fed into the modified versions of PlaneTrussElement.m and PlaneTrussResults.m that are shown below.

The results were then plotted on the same graph as the deformed two bar truss without tapering. The blue dashed line represents the initial conditions, the green line is the non-tapered deformation, and the red line is the deformation of the tapered bars.

Electric pylon with the FE mesh
The following matlab code was used to solve the electric pylon. The functions: PlaneTrussElement, PlaneTrussResults,and NodalSoln were obtained from the Fundamental Finite Element Analysis and Applications book website. Here

The element number for the element with the highest tensile stress is represented by Tbn and equals 81. The highest tensile stress is represented by Ts and equals 9.0511e+005

The element number for the element with the highest compressive stress is represented by Cbn and equals 55. The highest compressive stress is represented by Cs and equals -8.6957e+005

This problem is statically indeterminate. One reaction was given but there are too many reactions remaining to solve using the static equations.

Contributing Team Members
The following students contributed to this report:

Jessica Dyer Eml4500.f08.gravy.jad 18:09, 21 November 2008 (UTC)

Brian Eggeman Eml4500.f08.gravy.bje 23:46, 20 November 2008 (UTC)

Sean Sullivan Eml4500.f08.gravy.sms 03:42, 21 November 2008 (UTC)

Mikal McFarlane Eml4500.f08.gravy.mmm 15:01, 21 November 2008 (UTC)

Patrick Clayton Eml4500.f08.gravy.psc 12:14, 21 November 2008 (UTC)

Justin Tennant Eml4500.f08.gravy.JMT 19:45, 21 November 2008 (UTC)

title: User:Eml4500.f08.gravy.jad/Notes6.g


title: File:Plate3.jpg


text: Template:Psychology/Start page

title: Template:Psychology/Start page


text: Teamwork

title: Teamwork


text:
Egyptology/Resources

Time Estimation
Time estimation for the completion of the required material in this section.
 * Lecture = 30 min.
 * Required Reading & Review Questions = 40-50 min.
 * Test Yourself Questions = 30-40 min.

Lecture
Coming Soon

Outline
 * The Old Kingdom (2700-2134 B.C.)
 * The Middle Kingdon (2134-1640 B.C.)
 * The New Kingdom (1640-1070 B.C.)
 * Egypt in Decline (1070-525 B.C.)
 * Society and Economy
 * Language, Script, Writing and Literature
 * The Egyptian Religion
 * The Sciences in Ancient Egypt
 * Art and Architecture

Required Reading

 * Ancient Egypt. Read the whole text.
 * From the Egyptian pantheon essay read the first part only.

Review Questions

 * Outline the poltical situation during the Old Kingdom.
 * What was life like in the New Kingdom period.
 * Describe the administrative structure.
 * Describe the social structure.
 * Describe agriculture.
 * What were the hyreroglyphics?
 * Who were Osiris, Isis, Horus, and Ra
 * What is mummification?

Additional Information

 * Pharaoh Khufu/Cheops
 * Pharaoh Tutankhamun
 * Pharaoh Ramesses II
 * Ancient Egyptian religion
 * Ancient Egyptian medicine
 * Giza pyramid complex

title: Egyptology/Resources


text:
Portal talk:Christian studies

Mormonism
"Views the Mormons within the context of Christianity" —There are several problems with this inclusion. 1. The Mormons are viewed many different ways within Christianity. As a Restorationist movement, the ways they fit into the greater context of Christianity is complicated. Furthermore, elements of the Latter Day Saint movement have merged with branches from without it, creating odd fusions of LDS branches. 2. The Latter Day Saint movement includes adherents which are not properly "Mormons", and "Mormons" as a descriptor can actually refer to individuals who are not adherents of any branch within the Latter Day Saint movement. "Mormonism" as a term should be avoided entirely and replaced with "Latter Day Saint Movement" unless there are compelling reasons to do otherwise. 3. What is "the context of Christianity"? The Jade Knight (d'viser) 09:11, 20 November 2008 (UTC)

title: Portal talk:Christian studies


text:
Talk:What is History

To be honest, I think every single course in the School of History could use this as an introductory element. I would recommend placing this at large for the entire school, instead of tailoring or including it in a specific course. You're welcome, of course, to list it within European History to 1648, but this seems too valuable to let it be claimed by that course alone. Does that make sense? The Jade Knight (d'viser) 09:20, 20 November 2008 (UTC)
 * Be my guest! Feel free to use it in any other part of the school or course etc, as long as it will be usefull to others.Dilos1 17:29, 20 November 2008 (UTC)
 * Is it okay if I remove the course-specific banner at the top? The Jade Knight (d'viser) 06:07, 21 November 2008 (UTC)
 * No, but there must be a link back to the History of Europe to 1648 main project page for those who will be attending the specific course. I suggest we either move it to the bottom of the page or make it smaller, more unoticable or add one more link back to the School of History page. I do not mind removing the banner but as the links on the left part of the Wiki pages under the Wikiversity logo are very general, new and inexpienced users may be lost. What do you think? Dilos1 06:51, 21 November 2008 (UTC)
 * Certainly, it would be appropriate to link to the School of History—I'm just thinking if this gets used for multiple courses, it might be odd having the reference to one specific course at the top. The Jade Knight (d'viser) 08:59, 21 November 2008 (UTC)
 * OK. let's make this a general school of history page. We can discuss it again if s'thing comes up in the future.Dilos1 12:01, 21 November 2008 (UTC)

Categories
You're welcome to put these where you like, but as an FYI, they're traditionally placed at the bottom of the page, just above inter-wiki links. The Jade Knight (d'viser) 07:47, 22 November 2008 (UTC)

title: Talk:What is History


text:
User talk:Princeof500

 Hello Princeof500, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. After you leave a comment on a talk page, remember to sign and date; it helps everyone follow the threads of the discussion. The signature icon in the edit window makes it simple. To get started, you may

 * Take a guided tour and learn to edit.
 * Visit a (kind of) random random project.
 * Browse Wikiversity, or visit a portal corresponding to your educational level: pre-school, primary, secondary, tertiary, non-formal education.
 * Find out about research activities on Wikiversity.

 * Read an introduction for teachers and find out how to write an educational resource for Wikiversity.
 * Give feedback about your initial observations
 * Discuss Wikiversity issues or ask questions at the colloquium.
 * Chat with other Wikiversitans on #wikiversity-en.

And don't forget to explore Wikiversity with the links to your left. Be bold to contribute and to experiment with the sandbox or your userpage, and see you around Wikiversity! --The Jade Knight (d'viser) 09:44, 20 November 2008 (UTC)

French
Hey there. Are you in PDT/PST? If so, you're in my time zone. At any rate, I'd be glad to Mentor you. Do you have any previous exposure to French? The Jade Knight (d'viser) 09:46, 20 November 2008 (UTC)

 Actually, I'm in UTC+7 not -7, not quite sure if compatible...
 * Oh, goodness! That is a bit of a difference.  What country are you in?  Australia?  The Jade Knight (d'viser) 05:41, 21 November 2008 (UTC)

 Will be moving to Australia next year, but this year I am in Malaysia.
 * Okay. FYI:  When responding at Wikiversity, it's customary to respond on the other user's talk page, and to sign your responses (using four tildes).  Well, I'm far enough West that if you'll be in Australia I may be as good as a resource as any, and if you're not interested in audiochat, we don't need to be on at the same time, at any rate.
 * Do you have any prior exposure to French? What is your interest in French, and what are your goals with the language (short or long term)?  The Jade Knight (d'viser) 07:58, 22 November 2008 (UTC)
 * French is useful. For now, just contact me via my talk page if you need anything.  FYI:  I started learning French when I was about your age, in middle school.  I continued working on it through high school, took a little bit in college, and visited French Polynesia to study it more.  I've also been to Montréal.  I'm not fluent in French, but I'm quite proficient in it.
 * What have you done to start learning French? The Jade Knight (d'viser) 12:27, 23 November 2008 (UTC)

 Sorry for replying at my talk page again... but will you please check your talkpage, I have already tryed to contact you via your talkpage. Look for the topic:"FRENCH"( without quotes)... Princeof500 13:05, 23 November 2008 (UTC)
 * I already said, just use my talk page to communicate with me. How is BBC Steps working out for you?  The Jade Knight (d'viser) 13:10, 23 November 2008 (UTC)
 * BBC Steps appears to include the written forms. What about it is unsatisfactory for you?  The Jade Knight (d'viser) 13:18, 23 November 2008 (UTC)
 * Yes, do you notice that when you listen to the voices, it shows text in the box to the left? By default that is the French text, though you can change it to show English or have sound only.  The Jade Knight (d'viser) 13:41, 23 November 2008 (UTC)
 * I am on the Wikiversity IRC Channel. I also have AIM.  The Jade Knight (d'viser) 13:53, 23 November 2008 (UTC)

Hello, Prince! I'm really not on Wikiversity all that much these days, but I'd still be happy to help you out. Can you get a chat client, like AIM, MSN (WLM), Yahoo!, Skype, or GMail? The Jade Knight (d'viser) 16:24, 25 August 2009 (UTC)

Possibly Malay
Can you read this? If so, can you tell me what it is? The Jade Knight (d'viser) 11:22, 25 November 2008 (UTC)

title: User talk:Princeof500


text:
Talk:The Early Humans and The Neolithic Revolution

I'd like to note somewhere that this section covers prehistory, and is thus somewhat speculative and less important to learn. While a familiarity of the topics covered in this section are helpful to give a greater perspective, I'd consider this section "optional". Your thoughts? The Jade Knight (d'viser) 09:57, 20 November 2008 (UTC)
 * Well, I disagree. To begin with, as this is not a course that one would be tested on, any part of it may be skipped by the potential student if he/she thinks that is boring or irrelevant. On the other hand, this section gives the student an idea of what is roughly the current state of our scientific knowledge on the origins of the human race and why there were humans in the Near East and Egypt before they arrived in Europe. Speculation exists only in the specifics of prehistory. The general picture is clear and can be read by the student in the selections.

 * The reason I have included paragraphs from the books by Darwin as well as the key phrases “natural selection” and “the survival of the fittest” is that this will help students later on to understand the conflict beginning in England in the 18th c., spread to Europe and later to the States between evolutionaries and creationists. Also, they will understand, the pretext of the "survival of the fittest" used as an excuse for European imperialism in the 19th c. This part is extremely important and valuable for later on in the History of Europe.

 * Every topic, reading, map, artifact etc in this course has a reason of existence here. I do not just put here whatever I find available. The structure and the content are very carefully selected and used for the purpose of giving the student the full picture of European History and civilization, its origins and cultural exchanges. Dilos1 17:56, 20 November 2008 (UTC)

 * I am aware that you make your selections deliberately (and that you are clearly knowledgeable about the subject matter). In this case, I happen to find this section, while interesting and useful in general, unnecessary to any study of European history.  It may certainly be appropriate to bring Darwin and his writings in at a later point when discussing Social Darwinism or the conflict between creationists and evolutionists.  However, neither of those are elements of ancient European History.  The fact that remains that the specifics of this sort of history (which, as being unrecorded, is not properly history in the more specific technical sense) are still highly speculative, even if decent idea about the generalities are known.  However, I think we should start learners in stuff better well-known than this, and in history; this chapter could easily belong to an Anthropology course.  The Jade Knight (d'viser) 06:13, 21 November 2008 (UTC)
 * OK. We have different conceptions of the background that one needs in order to study proper ancient history. The truth is that you do not need the pre-history of proper Europe to study the Middle Ages but you may need it in order to study the Celts, for example.
 * However, as you will come up with a condensed version of the whole “Emergence of Civilization” section and we place links together on the main page, this resolves the issue. Dilos1 06:57, 21 November 2008 (UTC)

I edited this page to simplify it. Dilos1 07:21, 24 November 2008 (UTC)
 * And here is the record of what you had previously added, so it may be easily accessed in case someone wishes to expand on it in the future. The Jade Knight (d'viser) 07:37, 24 November 2008 (UTC)


text:
User:Tomaschwutz/microbiology

Resources

 * Microbiology - Wikipedia, the free encyclopedia
 * Textbook of Bacteriology
 * Microbiology/Virology/Immunology/Bacteriology/Parasitology Text Book On-line
 * Microbiology/Virology/Immunology/Bacteriology/Parasitology Text Book On-line
 * Did God make pathogenic viruses? (copyright protected, only link)
 * http://henge.bio.miami.edu/mallery/150/index.htm
 * http://henge.bio.miami.edu/mallery/150/index.htm

Images

 * Picture of Adeno-Virus
 * Diagrammatical representation of human immunodeficiency virus (HIV)
 * bacteriophage injection
 * DNA is inserted into the bacterial chromosome
 * Replication cycle of a bacteriophage

title: User:Tomaschwutz/microbiology


text:
User talk:Nabsta2k7

 Hello Nabsta2k7, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. After you leave a comment on a talk page, remember to sign and date; it helps everyone follow the threads of the discussion. The signature icon in the edit window makes it simple. To get started, you may

 * Take a guided tour and learn to edit.
 * Visit a (kind of) random random project.
 * Browse Wikiversity, or visit a portal corresponding to your educational level: pre-school, primary, secondary, tertiary, non-formal education.
 * Find out about research activities on Wikiversity.

 * Read an introduction for teachers and find out how to write an educational resource for Wikiversity.
 * Give feedback about your initial observations
 * Discuss Wikiversity issues or ask questions at the colloquium.
 * Chat with other Wikiversitans on #wikiversity-en.

And don't forget to explore Wikiversity with the links to your left. Be bold to contribute and to experiment with the sandbox or your userpage, and see you around Wikiversity! Jtneill - Talk - c 12:05, 20 November 2008 (UTC)

 Hi! Our name is Vicky-Verity. We are twins and are Wikiversity mascots. We think Wikiversity is a great place to learn and make friends (when we aren't sleeping). Leave us a message!

12 Inch Production
Could you tell us some more about the educational purpose of this page? -- Jtneill - Talk - c 12:07, 20 November 2008 (UTC)

title: User talk:Nabsta2k7


text:
Systems theory

 * Systems theory
 * Systems Theory

title: Systems theory


text:
User:Princeof500

Princeof500 here! I am an eleven year old boy. Just signed up so please give me some tips!

title: User:Princeof500


text:
User:Eas4200c.f08.aeris.krammer/HW6

 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.

Matlab code section updated Eas4200c.f08.aeris.holladay 02:58, 9 December 2008 (UTC)

Copyright Violation/Plagiarism updated Contributing Members section updated Eas4200c.f08.aeris.caceres 06:12, 9 December 2008 (UTC)

Proof that $$ I_{yz} = 0 $$ expanded, Figure 2 added. Eas4200c.f08.aeris.mcrae 06:18, 9 December 2008 (UTC)

Dimensional Analysis
Using dimensional analysis, we are able to provide a strategy for choosing relevant data and forming a relationship between them. Let us take the example of normal stress. The standard equation for normal stress is  σ = F/A  where F is force and A is area. Breaking this equation down into components gives us the following:  [f] = F/L  and [A] = L2 where F is the force, L is the length, and [ ] is "The Dimension Of..." When organized together, we get  [f/A] = (F/L)/L2   =  F/L3 also known as force per volume or a body force.

Another example can be used for strain  (Є) . Strain is calculated as  Є = ΔL/L  where ΔL is the total deformation of the original length and L is the original length. This analysis shows that strain is a unitless parameter. This can be verified through dimensional analysis:  Є = du/dx → ΔL/L  therefore,  [Є] = [du]/[dx] = L/L → 1  which confirms that strain is indeed unitless.

This idea of dimensional analysis has many practical uses. Take for instance Poisson's Ratio  (υ)

 A = Єxx  B = Єyy   C = σy   D = σx Poisson's Ratio is the ratio of transverse contraction strain to longitudinal extension strain in the direction of stretching force. [Rod Lakes, University of Wisconsin]

Taking the y-component of normal strain at point y and relating it to the x-component of normal strain at point x gives us:  υ = -Єyy/Єxx  therefore,  [υ] = [Єyy]/[Єxx] → 1  Poisson's Ratio is a unitless parameter and this analysis does validate that reasoning.

Torsional Analysis Continued
We can recall that $$[\frac{\partial \sigma _{ij}}{\partial x_{i}}]=\frac{F}{l^{3}}$$, which is a ratio of the force and volume.

As we have determined earlier, the torque is the moment couple that acts about the longitudinal axis of a shaft. Working on the D part of the Roadmap, we can recall the Prandtl stress function $$\displaystyle \phi$$, which plays the role of a potential function. $$\sigma _{yx}=\frac{\partial \phi }{\partial z}$$ $$\sigma _{zx}= -\frac{\partial \phi }{\partial y}$$ $$\displaystyle \sigma _{yx}$$ and $$\displaystyle \sigma _{zx}$$ are components of the gradient of the Prandtl function with respect to $$\displaystyle y$$ and $$\displaystyle z$$. We can also determine that there is a scalar function $$\displaystyle f(x, z, y)$$ and the gradient of this  function is a vector and it can be expressed by  the following equation. $$\triangledown f(x,y,z) = \frac{\partial f}{\partial x}\hat{i}+\frac{\partial f}{\partial y} \hat{j}+\frac{\partial f}{\partial z} \hat{k} $$ We know that  $$\displaystyle \sigma _{yx}$$ and  $$\displaystyle \sigma _{zx}$$   have nonzero values. We can assume uniform stress through the length of the bar, and having static equilibrium, we can write
$$\frac{\partial \sigma _{xz} }{\partial z}+\frac{\partial \sigma _{yx} }{\partial y} =0$$

Now, we can now substitute the previously determined  $$\displaystyle \sigma _{yx}$$ and  $$\displaystyle \sigma _{zx}$$   in terms of  $$\displaystyle \phi$$  into this equation:
$$\frac{\partial }{\partial y}(\frac{\partial \phi }{\partial z}) + \frac{\partial }{\partial z} (-\frac{\partial \phi }{\partial y}) =0$$ This equation can be rearranged and expressed in the following form:
$$\frac{\partial ^{2}\phi }{\partial y \partial x}=\frac{\partial ^{2}\phi }{\partial z \partial y}$$ Since the function  $$\displaystyle \phi$$  is continuous and smooth, the second derivative is interchangeable. As an example, let's look at the equation below:
$$\frac{\partial ^{2} \phi}{\partial y \partial z}= \frac{\partial ^{2} \phi}{\partial z \partial y}$$ With the help of the earlier determined strain-displacement relations and stress - strain relations we can write the following relation in terms of the Prandtl function:
$$\frac{\partial ^{2} \phi}{\partial y ^{2}} + \frac{\partial ^{2} \phi}{\partial z ^{2}} = -2G \theta$$ The left side of this equation is the Laplacian $$\triangledown  ^2 \phi $$

Now, we can derive the stress vector or traction
 $$\displaystyle t_{3x1}=\sigma _{3x3}n_{3x1}$$ Where $$\displaystyle t$$ is the traction force component, $$\sigma$$ is the stress tensor component, and $$\displaystyle n$$ is the component normal to the surface. Examining the general 2D case first as shown in the figure, let's take a look at the uniform bar with an arbitrarily shaped cross section. Or we can also take a look at a more specific example as shown in the second image.

The stress vector can be determined  on the surface with the help of the normal vector  $$\bar{n}$$. Since we are working in the 2D zy-plane, we can see that $$\displaystyle n_{x}=0$$. Therefore, we can write the following relations
$$t_{x}=\sigma _{yx}n_{y}+\sigma _{zx}n_{z}=\frac{\partial \phi}{\partial y}n_{y}-\frac{\partial \phi}{\partial z}n_{z}$$
$$\displaystyle t_{y}=0 $$
$$\displaystyle t_{z}=0$$

Also, $$\displaystyle n_{y}$$ and $$\displaystyle n_{z}$$ can be expressed with the help of the angle the horizontal and vertical axis makes the tangent to the contour line at the examined point. Let us call this angle $$\displaystyle \eta$$. Then
$$n_{y}= cos \eta =\frac{\partial y}{\partial s}$$ and
$$n_{z}= sin \eta =-\frac{\partial z}{\partial s}$$ Therefore ,
$$t_{x}=\frac{\partial \phi}{\partial z}\frac{dz}{ds}+ \frac{\partial \phi}{\partial y}\frac{dy}{ds}=\frac{d \phi}{ds}$$

The traction boundary condition is $$\displaystyle t_{x}=0$$ and this can be rewritten as $$t_{x}=0=\frac{d \phi}{ds}$$ From this equation we can conclude that $$\displaystyle \phi$$ is a constant on the surface. Thus, the boundary condition can be expressed as  $$\displaystyle \phi =0$$

Also, working on the cross-section shear stresses, considering a differential area, we can determine the total resultant torque
$$\displaystyle 2\int \int_{A}^{}{} \phi dxdy$$ There is a relationship between the applied torque T and the rate of twist $$\displaystyle \theta$$ that results from this torque, namely
$$\displaystyle J=\frac{T}{G \theta}$$ In this equation $$\displaystyle J$$ is called the torsion constant.

Deriving Traction Force Equation cont...
Recall that the equation we are trying to derive is the following.
$$\displaystyle [t]_{3x1}=[\sigma]_{3x3}[n]_{3x1}$$ Traction force is the stress applied to a lateral surface. For example, stresses due to aerodynamic forces on a wing such as drag or even friction can be a traction force. The goal is to essentially derive the three components of the traction force matrix from the stress tensor and the surface's normal vector. This expression will be obtained by geometrical manipulation. Let's consider the following figure to illustrate an infinitesimally small surface element.
 In the diagram, ds is the length of the lateral surface, n is the unit normal vector of the lateral surface, and the base and height of this triangular element is dy and dz respectively. Now, in order to relate the traction force to the stress tensor components, the angle of the lateral surface with respect to the coordinate system must be utilized. First, the base and height of the triangle will be written in terms of the length of the lateral surface. Thus, it is simple to see the following by simple geometry.
 $$\displaystyle dz=dscos(\theta)$$

$$\displaystyle dy=dssin(\theta)$$ Similarly, the normal vector can be written in terms of the angle $$\theta$$. Realize that the magnitude of the normal vector is one, the following can be written.
 $$||\vec{n}||=1$$ $$\displaystyle n_y=cos(\theta)$$ $$\displaystyle n_z=sin(\theta)$$ To continue with the derivation, the next logical step is to write the equilibrium equations of the forces in the z and y directions. To simplify things, the depth in and out of the page will be assumed to be one (the area of the base and the height will be the base and the height respectively). Although this is the derivation for the 2D case, the process here can be extrapolated with another axis and equations for the 3D case can also be derived. Let's equate forces in the z and y directions.
 $$\sum {F}_y=0=-\sigma_{yy}dz-\sigma_{yz}dy +t_yd_s$$ Substituting the base and height will the lateral surface length and its respect direction cosine, we get the following equation. $$\displaystyle =-\sigma_{yy}ds n_y-\sigma_{yz}ds n_z +t_yd_s$$ It is possible to divide out the ds terms resulting in... $$\displaystyle t_y= \sigma_{yy}n_y+\sigma_{yz}n_z ... (1)$$ The sum of the forces in the z direction can be found in the below collapsible box. Once the equations 1 and 2 are obtained, this can be arranged into a matrix notation. In book the subscripts are different in that it uses the (x,y) coordinate system. Using cyclic permutation, we have the following equation.
 $$\begin{bmatrix} t_y\\ t_z\\ \end{bmatrix} = \begin{bmatrix} \sigma_{yy} & \sigma_{yz}\\ \sigma_{zy} & \sigma_{zz}\\ \end{bmatrix} \begin{bmatrix} n_y\\ n_z\\ \end{bmatrix}...(3) $$

The above equation is true because $$\sigma_{yz}=\sigma_{zy}$$. Equation 3 is written this way so that the first subscript indicates the row of the equation. In indicial notation, the equation is rewritten as the following.
 $$\begin{bmatrix} t_1\\ t_2\\ \end{bmatrix} = \begin{bmatrix} \sigma_{11} & \sigma_{12}\\ \sigma_{21} & \sigma_{22}\\ \end{bmatrix} \begin{bmatrix} n_1\\ n_2\\ \end{bmatrix}...(3') $$

This can be projected into the third dimension. Generalizing this equation into the 3D case in indicial notation yields the following.
$$ \begin{bmatrix} t_1\\ t_2\\ t_3\\ \end{bmatrix} = \begin{bmatrix} \sigma_{11} & \sigma_{12}& \sigma_{13}\\ \sigma_{21} & \sigma_{22}& \sigma_{23}\\ \sigma_{31} & \sigma_{32}& \sigma_{33}\\ \end{bmatrix} \begin{bmatrix} n_1\\ n_2\\ n_3\\ \end{bmatrix}...(4) $$ The advantage of using the indicial notation is that it can be written in a compact way that still provides all the information of the matrix notation.
 $$t_i=\sum_{j=1}^3{\sigma_{ij}n_j},i=1,2,3$$ Now, the original equation is derived.
$$\displaystyle [t_i]_{3x1}=[\sigma_{ij}]_{3x3}[n_i]_{3x1} $$     From the road map, we have the following three equations.
$$\displaystyle T=2\int\int_{A}^{}{}\phi dA$$ $$\displaystyle T = GJ\theta $$ $$\displaystyle J=\frac{-4}{\nabla^{2}\phi }\int \int _{A}\phi dA$$ Combining these three equations and solving for the Lagrangian of phi, we have the following equation.
$$ \frac{\partial^2{\phi}}{\partial{y}^2}+\frac{\partial^2{\phi}}{\partial{z}^2}=-2G\theta $$ where $$\phi$$ is constant on the lateral surface of the bar.

Bending and Torsional Analysis Continued
To continue torsional analysis, let us consider two cases of different cross sections. Case 1 is a thin-walled cross section that is closed. An example of this kind of cross section would be a NACA airfoil. It is two lateral surfaces denoted by S with corresponding subscripts as shown in the figure below. Case 2 is a solid cross section with only one lateral surface. Both cross sections are uniform and non circular.
 In case 2, solid sections with a single contour boundary have phi as an arbitrary constant and thus can be chosen to be zero. This is due to a traction free boundary condition meaning that the traction force t_z is equal to zero. In case 1, using the Prandtl stress function phi, the stress-free boundary condition says that the derivative of phi with respect to the contour length is zero on lateral surfaces o and i. This means that phi on S_o and S_i are different constants. Differing from case 2, these two constants cannot be set to equal zero. Now consider a uniform bar with circular, solid cross section. The Prandtl stress function of this geometry is as follows.

$$ \phi(y,z)=c\left(\frac{y^2}{a^2}+\frac{z^2}{b^2}-1\right) $$ where a=b=radius and thus constituting a circle. If a did not equal b, the shape of the cross section would be an ellipse. Applying equation 3.19 in the book, it is possible to solve for the factor C.
$$
 * }

\nabla^2{\phi}=\frac{\partial^2{\phi}}{\partial{y}^2}+\frac{\partial^2{\phi}}{\partial{z}^2}=-2G\theta$$ $$\nabla^2{\left[C(\frac{y^2}{a^2}+\frac{z^2}{a^2}-1)\right]}={C}\nabla^2{\left[(\frac{y^2}{a^2}+\frac{z^2}{a^2}-1)\right]}$$ $$=C\nabla{(\frac{2y}{a^2}+\frac{2z}{a^2})}=\frac{2C}{a^2}\nabla{(y+z)}=\frac{4C}{a^2}$$ Applying the compatibility equation yields the following $$C=-\frac{a^2}{2}G\theta$$

Bars with Circular Cross-Sections
Proving that a bar with a circular cross-section does not warp along its longitudinal axis or z-axis, one must first consider the contour shape of the boundary.

In this case, a circle

 represented by the equation... $$x^{2}+y^{2}=r^{2}$$

...is used to express the stress function for the circular bar, with 'C' being defined as...

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

...from the compatibility equation of torsion.

The stress function is expressed as...

 $$\phi =C\left(\frac{x^{2}}{r^{2}}+\frac{y^{2}}{r^{2}}-1 \right)$$

Using this formula...

 $$T=2\int _{A}\int \phi dx dy$$

...one can calculate the torque in terms of 'C' and the circular contour. The torque can then be expressed as...

 $$T=\theta \cdot J \cdot G$$

Acknowledging that shear stress is simply...

 $$\tau =\frac{Tr}{J}$$

...the torque is then divided by the polar moment of inertia and multiplied by the radius, or 'X' or 'Y'. It can be seen that the shear stress is equivalent to...

 $$\tau_{xz} =\sigma _{xz}=-G\theta y=2C\frac{y}{a^{2}}=\frac{\partial \phi }{\partial y}$$

$$\tau_{yz} =\sigma _{yz}=G\theta x=-2C\frac{x}{a^{2}}=-\frac{\partial \phi }{\partial x}$$

It is seen that...

 $$\sigma _{xz}=-G \cdot \theta \cdot y$$

$$\sigma _{yz}=G \cdot \theta \cdot x$$

can be substituted into the strain-displacement relations...

 $$\gamma _{xz}=\frac{\sigma_{xz}}{G}=\frac{\partial u_{z}}{\partial x}-\theta y$$

$$\gamma _{yz}=\frac{\sigma_{yz}}{G}=\frac{\partial u_{z}}{\partial y}+\theta x$$

...simplifying to...

 $$\frac{\partial u_{z}}{\partial x}=0$$

$$\frac{\partial u_{z}}{\partial y}=0$$

...which results in...

 $$u_{z}=0$$ .

 * }

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

Shear Flow in Open, Thin-Walled Sections

General equation for asymmetric cross sections:
 * $$\ \int_{A_{s}}^{}{ \frac{d \sigma_{xx}}{dx}dA} = -q_{s}$$

Symmetric Case:

 * $$\ \sigma_{xx} = \frac{M_{y}z}{I_{y}} $$

 * $$\ q_{s} = - \frac{V_{z}}{I_{y}} \int_{A_{s}}^{}{zdA} = \frac{-V_{z}Q_{y}}{I_{y}} $$        (eq. 5.2)

 * $$\ Q_{y} = \int_{A_{s}}^{}{zdA} = z_{c}A_{s} $$

Asymmetric Case:

 * $$\ \sigma _{xx} = \left(k_{y}M_{z} - k_{yz}M_{y} \right)y + \left(k_{z}M_{y} - k_{yz}M_{z} \right)z $$         (eq. 5.3)

with

 * $$\ k_{y} = \frac{I_{y}}{D} $$

and

 * $$\ k_{yz} = \frac{I_{yz}}{D} $$

and

 * $$\ k_{z} = \frac{I_{z}}{D} $$

This can be also written in matrix form:

 * $$\ \sigma _{xx} = \begin{bmatrix} y & z \end{bmatrix}\begin{bmatrix} k_{y} & -k_{yz}\\ -k_{yz} & k_{z} \end{bmatrix} \begin{Bmatrix} M_{z} \\ M_{y} \end{Bmatrix} $$

 * $$\ \sigma _{xx} = \begin{bmatrix} y & z \end{bmatrix}\begin{bmatrix} k_{y}M_{z} & -k_{yz}M_{y}\\ -k_{yz}M_{z} & k_{z}M_{y} \end{bmatrix} $$

We can also flop the order of variables, and come up with the preferred form of this equation:

 * $$\ \sigma _{xx} = \begin{bmatrix} z & y \end{bmatrix}\begin{bmatrix} k_{z} & -k_{yz}\\ -k_{yz} & k_{y} \end{bmatrix} \begin{Bmatrix} M_{y} \\ M_{z} \end{Bmatrix} $$

We particularize to symmetric cross section, and find that $$\ I_{yz} = 0 $$. Proof of this is shown in the tab below.

Consider $$\ M_{z} = 0 $$

We know the following relations.

 * $$\ I_{yz} = 0 $$
 * $$\ D = I_{y}I_{z} $$
 * $$\ k_{yz} = 0 $$
 * $$\ k_{z} = \frac{1}{I_{y}} $$

Combining these equations we find that $$\ \sigma_{xx} = \frac{M_{y}z}{I_{y}} $$.

Asymmetric Case

 * $$\ q_{s} = -\left( k_{y}V_{y} - k_{yz}V_{z} \right) Q_{z} - \left( k_{z}V_{z} - k_{yz}V_{y} \right) Q_{y} $$          (eq. 5.5)

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

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

For stringer web sections, if the thickness (t) of the skin and spar webs is very small we can neglect the computations of Iy, Iz, Iyz, Qy, and Qz. We use only the areas of the stringers.

HW6 MatLAB Problem
For this homework assignment we take a closer look at the effects of plate buckling. We begin by modifying the NACA airfoil code to calculate the stresses along the top and bottom skin panels. Using the equation for bending stress we calculate the stresses from point B to F and then from point E to H. The results are diplayed below. From the graph we can see that the top panel is under compression while the bottom panel is under tension. Also, the maximum absolute stress occurs near point B as determined last time.

 The figures you plotted were not the perspective view of the buckling mode shapes; see Team Aero. Eml4500.f08 15:42, 23 November 2008 (UTC)

The graph above is a view along the edge of a rectangular panel of dimensions 1 m by 1.5 m. From the graph we can can see the effect of the m in the number of half waves. Another way to determine m is using the period of the function. For the second graph (m=2), one complete cycle is shown so the period is $$a$$ whereas the first graph has period $$2a$$. Thus respectively, $$m=1$$ and $$m=2$$. Plotting the coefficient k_c versus the aspect ratio a/b for different values of m it becomes evident that the number of half waves affects the width of k_c at minimum values. When plotting the critical stress values for our airfoil panel, we see that the minimum stress value occurs when $$a/b=1$$. By looking at the equation for critical stress we can see why this is true. $$\sigma_{cr}=k_{c}*\frac{\pi^2*D}{b^2*h}$$ Since $$b^2$$ is in the denominator it will greatly increase the critical stress at lower values, hence the steep increase to the left of $$a/b=1$$. From the previous graph for $$m=1$$ we see that k_c is minimized at $$a/b=1$$, hence the minumum in critical stress at $$a/b=1$$.

Next we look at the clamped boundary condition for plate buckling.

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

Looking at the edge (y = b) of the panel we do not see any displacement in the z-direction for the clamped boundary condition. Examining the equation for Ψ we see that at y = b, the second term becomes $$cos(2\pi)$$. Therefore, the displacment along the y = 0 and y = b edge of the plate will be zero. However, if we look at the profile at y/2 we can see the maximum displacement as displayed below.

Contributions
Jesse W. McRae Eas4200c.f08.aeris.mcrae 19:43, 21 November 2008 (UTC) Josh Holladay Eas4200c.f08.aeris.holladay 19:57, 21 November 2008 (UTC) Anett Krammer Eas4200c.f08.aeris.krammer 20:34, 21 November 2008 (UTC) Jin Yu Guan Eas4200c.f08.aeris.guan 21:34, 21 November 2008 (UTC) Nelson Caceres Eas4200c.f08.aeris.caceres 21:45, 21 November 2008 (UTC)

title: User:Eas4200c.f08.aeris.krammer/HW6


text:
User:Eml4500.f08.jamama.justin/HW6

Rube Goldberg Device: This is a device that accomplishes a simple task through many unnecessary steps, and in a very indirect fashion.

An example of this type of device can be seen in the following youtube link: []

NOTE: Honesty, imagination, ethics. It is always very important to do your own work, and not plagiarize other people's work. Honesty is the best policy.

Cont PVW to discrete PVW.

Lagrangian Interpolation.

 * Motivation for form of Ni(x) and Ni+1(x)

1) Mi(x) and Mi+1(x) are linear (straight lines), thus any linear combination of Ni and Ni+1 are also linear and in the particular expression for U(x) on Page 31-3.

$$N_{i}(x)=\alpha _{i}+\beta _{i}x $$ with $$(\alpha _{i},\beta _{i})$$ being numbers.

$$N_{i+1}(x)=\alpha _{i+1}+\beta _{i+1}x$$ Linear combo of Ni and $$(\alpha _{i},\beta _{i})$$ numbers.

$$N_{i+1}$$: $$N_{i}d_{i}+N_{i+1}d_{i+1}=(\alpha _{i}+\beta _{i}x)d_{i}+(\alpha _{i+1}+\beta _{i+1}x)d_{i+1}$$ $$=(\alpha _{i}d_{i}+\alpha _{i+1}d_{i+1})+(\beta _{i}d_{i}+\beta _{i+1}d_{i+1})x$$ is clearly a function of x.

2) Recall equation for U(x) (interp of U(x)) on Page 31-3:

$$U(x_{i})=N_{i}(x_{i})d_{i}+N_{i+1}(x_{i})d_{i+1}=d_{i}$$

=1 =0

title: User:Eml4500.f08.jamama.justin/HW6


text: File:Hw6 1.jpg

title: File:Hw6 1.jpg


text:
User:Eas4200c.f08.nine.F/Homework 6

Group nine - Homework 6

Prandtl Stress Function
The Prandtl stress function is given by the following equation:

$$\sigma_{yx} = \frac{\delta \phi}{\delta z}, \, \, \sigma_{zx} = - \frac{\delta \phi}{\delta y}$$

where $$\phi$$ plays the role of a potential function.

This equation automatically satisfies $$\frac{\delta}{\delta y}\left(\frac{\delta\phi}{\delta z}\right) + \frac{\delta}{\delta z}\left(-\frac{\delta\phi}{\delta y}\right)$$

$$\,\, = \frac{\delta^{2}\phi}{\delta y\delta z} - \frac{\delta^{2}\phi}{\delta z\delta y} = 0$$

$$\phi$$ is constant and smooth so the second mixed derivative is interchangeable.

Now, let us take a look at the strain-displacement relations,

$$\gamma_{yx} = \frac{\delta u}{\delta y} + \frac{\delta v}{\delta x}, \,\,\,\,\, \gamma_{zx} = \frac{\delta u}{\delta z} + \frac{\delta w}{\delta x}$$

$$ \Rightarrow \,\,\,\,\begin{matrix}(1a)\\(1b)\end{matrix}\,\,\,\,\,\left\{\begin{matrix}\gamma_{yx} = \frac{\delta u}{\delta y} - \theta_{z}\\ \gamma_{zx} = \frac{\delta u}{\delta z} + \theta_{y} \end{matrix}\right.$$

We can now obtain the compatibility equation for torsion shown below:

$$\Rightarrow \,\,\,\,\frac{\delta\gamma_{zx}}{\delta y} - \frac{\delta\gamma_{yx}}{\delta z} = 2\theta$$

And now by using the stress-strain relations, we obtain the following

$$\gamma_{zx} = \frac{1}{G}\sigma_{zx}, \,\,\,\,\,\gamma_{yx} = \frac{1}{G}\sigma_{yx}$$

And from the compatibility equation for torsion, we get

$$\frac{\delta\sigma_{zx}}{\delta y} - \frac{\delta\sigma_{yx}}{\delta z} = 2G\theta$$

And now taking the Laplacian of $$\phi$$ and putting back in terms of the Prandtl stress function, we obtain the following,

$$(2)\,\,\,\,\,\frac{\delta^{2}\phi}{\delta x^{2}} + \frac{\delta^{2}\phi}{\delta y^{2}} = -2G\theta$$

Now, the torsion problem is reduced to finding the stress function $$\phi$$ while using the given boundary conditions. Also, their is no applied loads on the lateral surface of the bar, and therefore, the stress vector $$t$$ (traction) must vanish.

$$\left\{t\right\}_{3x1} = \left[\sigma\right]_{3x3}\left\{n\right\}_{3x1}$$

By using the unit normal $$n$$, the stress vector can be evaluated on the lateral surface where $$n_{x} = 0$$. Therefore,

$$\begin{Bmatrix}t_{x}\\t_{y}\\t_{z}\end{Bmatrix} = \begin{bmatrix}0&\sigma_{xy}&\sigma_{xz}\\\sigma_{xy}&0&0\\\sigma_{zx}&0&0\end{bmatrix} \begin{Bmatrix}0\\n_{y}\\n_{z}\end{Bmatrix}$$

$$\Rightarrow \,\,\,\,t_{y} = 0 \,\,\,\,\,t_{z} = 0$$

$$\Rightarrow \,\,\,\,t_{x} = \sigma_{yx}n_{y} + \sigma_{zx}n_{z} = -\frac{\delta\phi}{\delta z}n_{y} + \frac{\delta\phi}{\delta y}n_{z}$$

By using the Figure to the right, we can see that,

$$ \, dy = sin\theta ds$$

$$ \, dz = cos\theta ds$$

$$\,\,\,\,\,\Rightarrow \,\,\,\,n_{y} = cos\theta$$

$$\,\,\,\,\,\Rightarrow \,\,\,\,n_{z} = sin\theta$$

Using the above relations we obtain the following for $$t_{x}$$

$$t_{x} = \frac{\delta\phi}{\delta z}\frac{dz}{ds} + \frac{\delta\phi}{\delta y}\frac{dy}{ds} = \frac{d\phi}{ds}$$

And now the traction free boundary condition $$\left(t_{x} = 0\right)$$ is given by the following,

$$\frac{d\phi}{ds} = 0\,\,\,\,\,\Rightarrow \,\,\,\,\phi = $$ constant on the lateral surface of the bar.

Bars with Circular Cross-Sections

We are now going to take at look a uniform bar with a circular cross-section and show that there is no warping under torsion. The origin is chosen to be at the centroid of the bar, so that the boundary contour is given by the following equation.

$$y^{2} + z^{2} = a^{2}\,\,\,$$ where a is the radius of the circular boundary.

The stress function satisfying the boundary condition of the previous section is given below:

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

Substituting this equation into Eqn.(2), we obtain the following,

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

We can now solve for the torque as shown,

$$ \begin{align} T & = 2C\int\int_A\left(\frac{y^{2}}{a^{2}} + \frac{z^{2}}{a^{2}} - 1\right)dydz\\ & = 2C\int\int_A\left(\frac{r^{2}}{a^{2}} - 1\right)dA\\ & = 2C\left(\frac{I_{p}}{a^{2}} - A\right) \end{align} $$

where $$I_{p} = \int\int_A r^{2}dA = \frac{1}{2}\pi a^{4}$$  and  $$\,\,A = \pi a^{2}$$

Because $$a^{2}A = 2I_{p}$$  we have

$$T = -\left(\frac{2CI_{p}}{a^{2}}\right) = \theta GI_{p} = \theta GJ$$

The shear stresses become

$$\begin{matrix}(3a)\\(3b)\end{matrix}\,\,\,\,\, \left\{\begin{matrix}\sigma_{yz} = \frac{\delta\phi}{\delta z} = 2C\frac{z}{a^{2}} = -G\theta z\\ \,\,\,\,\,\sigma_{xz} = -\frac{\delta\phi}{\delta y} = -2C\frac{y}{a^{2}} = G\theta y \end{matrix}\right.$$

By taking a look stress vector t on the lateral surface of the bar and having $$n_{y} = \frac{y}{r}$$ and $$n_{z} = \frac{z}{r}$$ and by also using Eqns(3a, 3b), we obtain the following,

$$t_{x} = -G\theta\frac{yz}{r} + G\theta\frac{yz}{r} = 0$$

Now let us take a wedge out of the bar as shown in the Figure. So the unit vector on the wedged surface is given by

$$(4)\,\,\,\,\, n_x = 0 \,\,\,\,\,\,\,\,\,\, n_y = -cos\beta = -\frac{z}{r} \,\,\,\,\,\,\,\,\,\, n_z = sin\beta = \frac{y}{r}$$

Substituting Eqns(3a, 3b) with Eqn(4) into $$\,t_z = \sigma_{yx}n_y + \sigma_{zx}n_z$$ we obtain the following

$$\,t_z -G\theta r\,\,\,\,\,\Rightarrow\,\,\,\,\,\tau = -t_x = G\theta r$$

Using $$\,T = \theta GJ$$, we can eliminate $$\,\theta$$ to obtain the above expression in terms of the torque.

$$\tau = \frac{Tr}{J}$$

Finally, using Eqn(1) and Eqn(3) and the stress-strain relations, we can see that the warping is zero.

$$\,w = 0$$

Therefore, a uniform bar with a circular cross-section has no warping under torsion.

General Case for 2-D

$$\sum{F_y}=0=-\sigma_{yy}(dz(1))-\sigma_{yz}(dy(1))+t_y(ds(1))$$

Were $$\sigma_{yy}$$ is the unit depth along x-axis.

$$\sum{F_y}=0=t_y=\sigma_{yy}n_y+\sigma_{yz}n_z$$

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

$$t_y=\frac{F}{L^2}$$

Were F equals force and $$L^2$$ equals the area.

t is the traction vector (distributed surface force)

$$t_y=\sigma$$

$$\left[\begin{array}{ccc}t_y\\t_z\end{array}\right]$$=$$\left[\begin{array}{ccc}\sigma_{yy}&\sigma_{yz}\\\sigma_{zy}&\sigma_{zz}\end{array}\right]$$$$\left[\begin{array}{ccc}n_y\\n_z\end{array}\right]$$

$$\sigma_{yz}=\left[\begin{array}{ccc}\sigma_{yy}&\sigma_{yz}\\\sigma_{zy}&\sigma_{zz}\end{array}\right]$$

$$\left[\begin{array}{ccc}t_1\\t_2\\t_3\end{array}\right]$$=$$\left[\begin{array}{ccc}\sigma_{11}&\sigma_{12}&\sigma_{13}\\\sigma_{21}&\sigma_{22}&\sigma_{23}\\\sigma_{31}&\sigma_{32}&\sigma_{33}\end{array}\right]$$$$\left[\begin{array}{ccc}n_1\\n_2\\n_3\end{array}\right]$$

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

$$i=1,2,3$$

Were i is the row and j is the collumn in the matrix

Chapter 5: Flexural Shear Flow in Thin-Walled Sections

General equation for an unsymmetric cross-section $$\int_{A_s}^{} \frac{d \sigma_{xx}}{dx} dA=-q_s=-t\tau_{xs} $$

 General Unsymmetric Case $$\sigma_{xx}=(k_yM_z-k_{yz}M_y)y +(k_zM_y-k_{yz}M_z)z$$ $$k_y=\frac{I_y}{detI}$$ $$k_{yz}=\frac{I_{yz}}{detI}$$ $$k_z=\frac{I_z}{detI}$$

The symmetric cross section is derived from these assumptions: $$I_{yz} = 0$$ $$M_z=0$$ Therefore: $$detI=I_yI_z$$ $$k_y=\frac{1}{I_z}$$ $$k_{yz}=0$$ $$k_z=\frac{1}{I_y}$$  General Symmetric Case (along y-axis) $$\sigma_{xx}=\frac{M_y z}{I_y}$$ $$q(s)=\frac{-V_zQ_y}{I_y}$$ $$Q_y=\int_{A_s}^{}zdA=Z_cA_s$$

Stringers Stringer-Web Sections: thickness t of skin and spar webs are very small. With this fact the following components can be neglected:$$I_y,I_z,I_{yz},Q_y,Q_z$$. In the figure to the right would be symmetrical about the y-axis if $$A_2=A_3$$ and $$A_1=A_4$$. A non symmetrical case would have $$A_1=A$$, $$A_2=2A$$, $$A_3=3A$$ and $$A_4=4A$$.

Homework 34-3
Put equation (5.5) in matrix form

$$ q_s =-(k_yV_y-K_{yx}V_z)Q_z-(k_zV_z-k_{yz}V_y)Q_y$$  eq(5.5)

Matrix form: $$\displaystyle q(s) = -\begin{bmatrix} Q_y & Q_z \end{bmatrix} \begin{bmatrix} k_z & -k_{yz} \\ -k_{yz} & k_y \end{bmatrix} \begin{Bmatrix} V_z \\ V_y \end{Bmatrix} $$ Noting that: $$Q_z=\int_{A_s}^{}ydA$$ and $$Q_y=\int_{A_s}^{}zdA$$

MATLAB Code
{| class="toccolours collapsible collapsed" style="width:100%" ! MATLAB CODE

Sample Run of Code (NACA Plot)
NACA Airfoil calculation program Enter first digit of airfoil: 2 Enter second digit: 4 Enter the third and fourth digits: 15 Enter Py: 0 Enter Pz: 0 Enter number of segments: 60

The average area is: 0.103 The minumum number segments required to have the average area accurate within 1 percent is: 24.000

Figure 1 shows the cross-section of the NACA airfoil and the centroid line

 * }

Contributing Team Members
The following students contributed to this report:

Felix Izquierdo Eas4200c.f08.nine.F 17:34, 20 November 2008 (UTC) Ricardo Albuquerque Eas4200c.f08.nine.r 18:39, 20 November 2008 (UTC) Dave Phillips Eas4200C.f08.nine.d 19:46, 19 November 2008 (UTC) Stephen Featherman Eas4200c.f08.nine.s 16:49, 20 November 2008 (UTC)

title: User:Eas4200c.f08.nine.F/Homework 6


text:
User:Michael J Swassing

See also:

 * Michael J Swassing, (user page on Wikipedia).
 * Michael J Swassing, (Facebook).
 * Michael J Swassing, (LinkedIn).
 * Michael J Swassing - Knol (beta)

title: User:Michael J Swassing


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User:Mackenzie112195

Hello everyone! My name is Mackenzie and I am in the 8th grade. I do not know a lot yet, but what I do know, I am willing to share. When I grow up, I am going to live in South Carolina and have 2 majors from Ohio State University: one in international relations and one in veterinary medicine. If you want to talk, just leave a message on my page and I will get back to you as soon as I can. Thanks!

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