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User:Eml4500.f08.jamama.adam/homework5

Second Part of the matlab
From the Book Web Site, the following Matlab code for the analysis of the space truss system:

Running the code, the following results were collected:

With the results from the Matlab code, different plots of the deformation of the Space struss system were created. To accomplish this task the following Matlab code was written:

Once executed, the first graph that was created presented the 3D deformation of the space truss in solid red lines againts the initial position in broken blue lines.

Figure 1. 3D Deformation of the Space Truss.

Then a projection view of the deformation along the x-axix was created.

Figure 2. Projection along the x-axis.

Following the same fashion, several more graphs were created illustrating a projection along the z-axis (Figure 3), a projection along the y-axis (Figure 4) and a perspective view from an observer's point of view located at (-2, -2, 3) (Figure 5).

Figure 3. Projection along the z-axis.

Figure 4. Projection along the y-axis.

Figure 5. Prespective View from an Observer's point ov view: (-2, -2, 3).

title: User:Eml4500.f08.jamama.adam/homework5


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User:Eas4200c.f08.gator.edwards/Wk5 Class Notes

Kinematic Assumptions
4 zero strain components

4 zero stress components

Rewrite $$\ \epsilon -\sigma\ $$ relationship

$$\left\{\epsilon_{ij} \right\}=\begin{bmatrix} \textbf A &0\\ 0& \textbf{B} \end{bmatrix}\left\{\sigma_{ij} \right\}$$

$$\left\{\sigma_{ij} \right\}=\begin{bmatrix} \textbf A^{-1} &0 \\ 0&\textbf B^{-1} \end{bmatrix}\left\{\epsilon_{ij} \right\}$$

To verify they are inverses, multiply together to get identity matrix

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

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

$$\left\{\sigma_{ij} \right\}=\begin{bmatrix} \textbf A^{-1} &0 \\ 0&\textbf B^{-1} \end{bmatrix}\begin{bmatrix} 0\\ 0\\ 0\\ 0\\ \epsilon_{31}\\ \epsilon_{12} \end{bmatrix}$$

$$\sigma_{11}=0$$

$$\sigma_{22}=0$$

$$\sigma_{33}=0$$

$$\sigma_{23}=2G\epsilon_{23}=0$$

$$\sigma_{31}=2G\epsilon_{31}$$

$$\sigma_{12}=2G\epsilon_{12}$$

Now we move on to part C of the roadmap: Equilibrium Equation for stresses

Bidirectional Bending

$$M_{y}=\int_{A}z\sigma_{xx}dA$$

Similarly for x.

Moment of Inertia: $$I_{y},\ I_{z},\ I_{yz}$$

$$I_{y}=\int_{A}z^2dA$$

$$I_{z}=\int_{A}y^2dA$$

$$I_{yz}=\int_{A}yzdA$$

$$\sigma_{xx}=E\epsilon_{xx}$$

$$\sigma_{xx}=\frac{I_{y}M_z-I_{yz}M_y}{I_yI_z-(I_{yz})^2}\ y\ +\frac{I_{z}M_y-I_{yz}M_z}{I_yI_z-(I_{yz})^2}\ z$$

Recall:

$$I=\begin{bmatrix} I_{11} &I_{12} &I_{13} \\ & \begin{vmatrix} I_{22} & I_{23}\\ & I_{33} \end{vmatrix} \\ & & \end{bmatrix}$$

Note: the denominator of the above equation is the determinant of the I matrix.

$$D:=I_yI_z-(I_{yz})^2

=I_{22}I_{33}-(I_{23})^2$$

is the determinant of $$\begin{bmatrix} I_{22} &I_{23} \\ I_{32}&I_{33} \end{bmatrix}$$

Neutral Axis$$\rightarrow$$Bending Moment Stress = 0 $$\ (\sigma_{xx}=0)$$

Goal
Our goal is to prove:

$$\frac{\partial\sigma_{yx}}{\partial y}+\frac{\partial\sigma_{zx}}{\partial z}=0\ \ \ \ (1)$$

Now let's use indicial notation for the above equation, where x=1, y=2, and z=3:

$$\frac{\partial\sigma_{21}}{\partial x_2}+\frac{\partial\sigma_{31}}{\partial x_3}=0\ \ \ \ (2)$$

Recall:

$${\partial \sigma_{xx} \over \partial x} + {\partial \sigma_{yx} \over \partial y} + {\partial \sigma_{zx} \over \partial z}=0\ \ \ \ (3)$$

In indicial notation equation 3 becomes:

$$\sum_{i=1}^{3}\frac{\partial\sigma_{ij}}{\partial x_i}=0\ \ \ \ (4)$$

Derivation of $$\qquad \qquad \frac{\partial \sigma }{\partial x}+\frac{f(x)}{A}=0$$
First take the sum of all of the forces in the x direction of our 1-D model.

$$\sum{F_{x}}=0=-\sigma (x)A+\sigma (x+dx)A+f(x)dx$$

$$\displaystyle 0=A(-\sigma (x)+\sigma (x+dx))+f(x)dx$$

The term multiplied by A is expanded using a Taylor series expansion. The higher order terms are ignored so it is replaced with the equivalent below.

$$-\sigma (x)+\sigma (x+dx)\approx \frac{\partial \sigma }{\partial x}dx$$

This is then substituted back into the equation.

$$0=\frac{\partial \sigma }{\partial x}dx+f(x)dx$$

The dx can be factored out and the final equation is shown below.

$$\frac{\partial \sigma }{\partial x}+\frac{f(x)}{A}=0$$

Derivation of Equation 4
Now, let's investigate the non-uniform stress field in 3-D, but without applied load and focusing on the x-direction only (i.e. w/o the other stress components to avoid cluttering the figure)

$$\sum{F_{x}}=0=dydz\left[-\sigma _{xx}(x,y,z) + \sigma _{xx}(x+dx,y,z) \right]+dzdx\left[-\sigma _{yx}(x,y,z) + \sigma _{yx}(x,y+dy,z) \right]+dxdz\left[-\sigma _{zx}(x,y,z) + \sigma _{zx}(x,y,z+dz) \right]$$

$$0=dxdydz\left[\frac{\partial \sigma _{xx}}{\partial x} + \frac{\partial \sigma _{yx}}{\partial y} + \frac{\partial \sigma _{zx}}{\partial z}\right]$$

$$0=\frac{\partial \sigma _{xx}}{\partial x} + \frac{\partial \sigma _{yx}}{\partial y} + \frac{\partial \sigma _{zx}}{\partial z}$$

Note: $$ \sigma_{ij}$$ is the normal stress normal to the i axis facet in the j axis direction.

$$\Rightarrow \ Equation\ 3\ (and\ therefore\ 4)\   has\  been\  proved$$

Prandtl Stress function Φ
The Prandtl stress function is a special case of the Morera stress functions, in which it is assumed that A=B=0 and C is a function of x and y only.

$$[\sigma_{yx}]=(\frac{\partial\Phi}{\partial z}) $$

$$[\sigma_{zx}]=(\frac{\partial\Phi}{\partial y})$$

Φ acts as a potential function. ($$\sigma_{yx}$$, $$\sigma_{zx}$$) are components of Φ with respect to (y,z)

A brief mention should be made of the Maxwell stress functions. Maxwell stress functions are defined by assuming that the Beltrami stress tensor $$\Phi_{mn}$$ tensor is restricted to be of the form :
 * $$\Phi_{ij}=

\begin{bmatrix} A&0&0\\ 0&B&0\\ 0&0&C \end{bmatrix} $$

Substituting the expressions for the stress into the Beltrami-Michell equations yields the expression of the elastostatic problem in terms of the stress functions :

 * $$\nabla^2 A=\nabla^2 B=\nabla^2 C=\left(

\frac{\partial^2 A}{\partial x^2}+ \frac{\partial^2 B}{\partial y^2}+ \frac{\partial^2 C}{\partial z^2}\right)/(2-\nu),$$

These must also yield a stress tensor which obeys the specified boundary conditions.

The stress tensor which automatically obeys the equilibrium equation may now be written as :

$$\sigma_x= \frac{\partial^2\Phi_{yy}}{\partial z \partial z} + \frac{\partial^2\Phi_{zz}}{\partial y \partial y} -2\frac{\partial^2\Phi_{yz}}{\partial y \partial z}$$

$$\sigma_y= \frac{\partial^2\Phi_{xx}}{\partial z \partial z} +\frac{\partial^2\Phi_{zz}}{\partial x \partial x} -2\frac{\partial^2\Phi_{zx}}{\partial z \partial x}$$

$$\sigma_z= \frac{\partial^2\Phi_{yy}}{\partial x \partial x} +\frac{\partial^2\Phi_{xx}}{\partial y \partial y} -2\frac{\partial^2\Phi_{xy}}{\partial x \partial y}$$

$$\sigma_{xy}=-\frac{\partial^2\Phi_{xy}}{\partial z \partial z} -\frac{\partial^2\Phi_{zz}}{\partial x \partial y} +\frac{\partial^2\Phi_{yz}}{\partial x \partial y} +\frac{\partial^2\Phi_{zx}}{\partial y \partial z}$$

$$\sigma_{yz} =-\frac{\partial^2\Phi_{yz}}{\partial x \partial x} -\frac{\partial^2\Phi_{xx}}{\partial y \partial z} +\frac{\partial^2\Phi_{zx}}{\partial y \partial x} +\frac{\partial^2\Phi_{xy}}{\partial z \partial x}$$

$$\sigma_{zx}=-\frac{\partial^2\Phi_{zx}}{\partial y \partial y} -\frac{\partial^2\Phi_{yy}}{\partial z \partial x} +\frac{\partial^2\Phi_{xy}}{\partial z \partial y} +\frac{\partial^2\Phi_{yz}}{\partial x \partial y}$$

References


text:
User:Eml4500.f08.gravy.jad/Notes5.g

 See my comments below. Please don't remove these comments in case you amend your work; just add a comment in these comment boxes to say what you did. Eml4500.f08 12:56, 10 November 2008 (UTC) The 2-bar truss code was corrected and the error was explained. This is a link to the orginal page.Orginal and here is the differences between the pages Differences Eml4500.f08.gravy.mmm 00:26, 12 November 2008 (UTC)

FEA Notes and Homework 5

Principal of virtual work
Justification of eliminating rows 1,2,5,6 to obtain K2x2 in 2-bar truss:

as described in an earlier report Elimination of dofs

FD relation $$\mathbf{K}_{2X2} \mathbf{d}_{6x1}= \mathbf{F}_{6x1} $$

becomes

$$\mathbf{Kd-F}= \mathbf{0}_{6x1}$$  (Equation 1)

The Principal of virtual work gives the equation :$$ \mathbf{W}_{6x1} \cdot \overbrace{( \mathbf{Kd-F}))}^{6x1} = 0_{1x1}$$(Equation 2)

Is true for all W values Where $$\mathbf{W}_{6x1}$$ is the weighting matrix.

So (Equation 1)$$\iff$$ (Equation 2)

Proof:

A)(Equation 1)$$\Rightarrow$$ (Equation 2) is trivial

B)To prove (Equation 1)$$\Rightarrow$$ (Equation 2) we must choose different values for W

Choice 1: Select 'W' such that W1=1, W2=W3=W4=W5=W6=0 so

$$\mathbf{W^T}= \begin{bmatrix} 1 & 0 & 0 & 0 & 0 & 0 \end{bmatrix}$$

$$ \mathbf{W} \cdot ( \mathbf{Kd-F} ) = 1[ \sum^6_{j=1} \mathbf{K}_{1j} \mathbf{d}_j- \mathbf{F}_1] + 0[ \sum^6_{j=1} \mathbf{K}_{2j} \mathbf{d}_j- \mathbf{F}_2] + 0 [ \sum^6_{j=1} \mathbf{K}_{3j} \mathbf{d}_j- \mathbf{F}_3] + 0 [ \sum^6_{j=1} \mathbf{K}_{4j} \mathbf{d}_j- \mathbf{F}_4] + 0 [ \sum^6_{j=1} \mathbf{K}_{5j} \mathbf{d}_j- \mathbf{F}_5] + 0 [ \sum^6_{j=1} \mathbf{K}_{6j} \mathbf{d}_j- \mathbf{F}_6]  $$

so the result becomes  $$\sum^6_{j=1} \mathbf{K}_{1j} \mathbf{d}_j= \mathbf{F}_1 $$ (First Equation)

Choice 2: Select 'W' such that W2=1, W1=W3=W4=W5=W6=0 so

$$\mathbf{W^T}= \begin{bmatrix} 0 & 1 & 0 & 0 & 0 & 0 \end{bmatrix}$$

$$ \mathbf{W} \cdot ( \mathbf{Kd-F} ) = 0[ \sum^6_{j=1} \mathbf{K}_{1j} \mathbf{d}_j- \mathbf{F}_1] + 1[ \sum^6_{j=1} \mathbf{K}_{2j} \mathbf{d}_j- \mathbf{F}_2] + 0 [ \sum^6_{j=1} \mathbf{K}_{3j} \mathbf{d}_j- \mathbf{F}_3] + 0 [ \sum^6_{j=1} \mathbf{K}_{4j} \mathbf{d}_j- \mathbf{F}_4] + 0 [ \sum^6_{j=1} \mathbf{K}_{5j} \mathbf{d}_j- \mathbf{F}_5] + 0 [ \sum^6_{j=1} \mathbf{K}_{6j} \mathbf{d}_j- \mathbf{F}_6]  $$

so the result becomes  $$\sum^6_{j=1} \mathbf{K}_{2j} \mathbf{d}_j= \mathbf{F}_2 $$ (Second Equation)

Choice 3: Select 'W' such that W3=1, W1=W2=W4=W5=W6=0 so

$$\mathbf{W^T}= \begin{bmatrix} 0 & 0 & 1 & 0 & 0 & 0 \end{bmatrix}$$

$$ \mathbf{W} \cdot ( \mathbf{Kd-F} ) = 0[ \sum^6_{j=1} \mathbf{K}_{1j} \mathbf{d}_j- \mathbf{F}_1] + 0[ \sum^6_{j=1} \mathbf{K}_{2j} \mathbf{d}_j- \mathbf{F}_2] + 1 [ \sum^6_{j=1} \mathbf{K}_{3j} \mathbf{d}_j- \mathbf{F}_3] + 0 [ \sum^6_{j=1} \mathbf{K}_{4j} \mathbf{d}_j- \mathbf{F}_4] + 0 [ \sum^6_{j=1} \mathbf{K}_{5j} \mathbf{d}_j- \mathbf{F}_5] + 0 [ \sum^6_{j=1} \mathbf{K}_{6j} \mathbf{d}_j- \mathbf{F}_6]  $$

so the result becomes  $$\sum^6_{j=1} \mathbf{K}_{3j} \mathbf{d}_j= \mathbf{F}_3 $$ (Third Equation)

Choice 4: Select 'W' such that W4=1, W1=W2=W3=W5=W6=0 so

$$\mathbf{W^T}= \begin{bmatrix} 0 & 0 & 0 & 1 & 0 & 0 \end{bmatrix}$$

$$ \mathbf{W} \cdot ( \mathbf{Kd-F} ) = 0[ \sum^6_{j=1} \mathbf{K}_{1j} \mathbf{d}_j- \mathbf{F}_1] + 0[ \sum^6_{j=1} \mathbf{K}_{2j} \mathbf{d}_j- \mathbf{F}_2] + 0 [ \sum^6_{j=1} \mathbf{K}_{3j} \mathbf{d}_j- \mathbf{F}_3] + 1 [ \sum^6_{j=1} \mathbf{K}_{4j} \mathbf{d}_j- \mathbf{F}_4] + 0 [ \sum^6_{j=1} \mathbf{K}_{5j} \mathbf{d}_j- \mathbf{F}_5] + 0 [ \sum^6_{j=1} \mathbf{K}_{6j} \mathbf{d}_j- \mathbf{F}_6]  $$

so the result becomes  $$\sum^6_{j=1} \mathbf{K}_{4j} \mathbf{d}_j= \mathbf{F}_4 $$ (Fourth Equation)

Choice 5: Select 'W' such that W5=1, W1=W2=W3=W4=W6=0 so

$$\mathbf{W^T}= \begin{bmatrix} 0 & 0 & 0 & 0 & 1 & 0 \end{bmatrix}$$

$$ \mathbf{W} \cdot ( \mathbf{Kd-F} ) = 0[ \sum^6_{j=1} \mathbf{K}_{1j} \mathbf{d}_j- \mathbf{F}_1] + 0[ \sum^6_{j=1} \mathbf{K}_{2j} \mathbf{d}_j- \mathbf{F}_2] + 0 [ \sum^6_{j=1} \mathbf{K}_{3j} \mathbf{d}_j- \mathbf{F}_3] + 0 [ \sum^6_{j=1} \mathbf{K}_{4j} \mathbf{d}_j- \mathbf{F}_4] + 1 [ \sum^6_{j=1} \mathbf{K}_{5j} \mathbf{d}_j- \mathbf{F}_5] + 0 [ \sum^6_{j=1} \mathbf{K}_{6j} \mathbf{d}_j- \mathbf{F}_6]  $$

so the result becomes  $$\sum^6_{j=1} \mathbf{K}_{5j} \mathbf{d}_j= \mathbf{F}_5 $$ (Fifth Equation)

Choice 6: Select 'W' such that W6=1, W1=W2=W3=W4=W5=0 so

$$\mathbf{W^T}= \begin{bmatrix} 0 & 0 & 0 & 0 & 0 & 1 \end{bmatrix}$$

$$ \mathbf{W} \cdot ( \mathbf{Kd-F} ) = 0[ \sum^6_{j=1} \mathbf{K}_{1j} \mathbf{d}_j- \mathbf{F}_1] + 0[ \sum^6_{j=1} \mathbf{K}_{2j} \mathbf{d}_j- \mathbf{F}_2] + 0 [ \sum^6_{j=1} \mathbf{K}_{3j} \mathbf{d}_j- \mathbf{F}_3] + 0 [ \sum^6_{j=1} \mathbf{K}_{4j} \mathbf{d}_j- \mathbf{F}_4] + 0 [ \sum^6_{j=1} \mathbf{K}_{5j} \mathbf{d}_j- \mathbf{F}_5] + 1 [ \sum^6_{j=1} \mathbf{K}_{6j} \mathbf{d}_j- \mathbf{F}_6]  $$

so the result becomes  $$\sum^6_{j=1} \mathbf{K}_{6j} \mathbf{d}_j= \mathbf{F}_6 $$ (Sixth Equation)

Hence the equation becomes $$\mathbf{Kd}= \mathbf{F} $$ which is Equation 1 = For the Principal of virtual work when you account for the boundary conditions d1 =  d2 =  d5=  d6=0 The $$\mathbf{K}$$  matrix becomes the  $$\mathbf{\overline{K}}$$  matrix.

The weighing coefficients must be "kinetically admissible" which means they cannot violate the boundary conditions so W1 = W2 =  W5=  W6=0  where the weighing coefficients represent the virtual displacement.

The $$\overline{ \mathbf{K}}$$ is created as described in an eariler report Elimination of dofs

$$\mathbf{W}_{6x1} \cdot (

\underbrace{\mathbf{Kd}}_{\Downarrow}-\mathbf{F}) $$

$$\begin{bmatrix} && \\&& \end{bmatrix}_{6x2} -\begin{Bmatrix} \\ \\ \end{Bmatrix}_{2x1}

$$

$$= \begin{Bmatrix} W_3 \\ W_4 \end{Bmatrix} (\mathbf{\overline{K}_{2x2} \overline{d}_{2x1}- \overline{F}_{2x1} })$$ (Equation 3)

for all values of $$\begin{Bmatrix} W_3 \\ W_4 \end{Bmatrix} $$

with $$\mathbf{\overline{K}}=\begin{bmatrix} K_{33} && K_{34} \\ K_{43} && K_{44}\end{bmatrix} $$

$$\mathbf{\overline{d}}= \begin{Bmatrix}d_{3} \\ d_{4} \end{Bmatrix} $$

$$ \overline{F } = \begin{Bmatrix}F_{3} \\ F_{4} \end{Bmatrix} $$

Fixing the bug in the 2-bar truss code
There was a bug in the code solving the 2-bar truss solved in and earlier report 2-bar truss. The code was originally written for a 6-bar truss and was modified for the 2-bar truss. Unfortunately an error was created. The 2-bar truss had different values for the E's and A's for the elements,but the 6-bar truss had the same E's and A's for all of the elements. This was fixed by substituting the e and A in the following for loop with e(i) and A(i):

 The above array "results" was incomplete; it showed only the first column of this array. The reason was because this team used the incomplete code PlaneTrussResults from Team Delta_6; see my comments in "The best of HW5". Eml4500.f08 12:59, 10 November 2008 (UTC)

The code was actually not incomplete, but the way it was written would not work in the code used for the two-bar truss. The following modifactions were made to PlaneTrussResults.m to make it work in the two-bar truss code:

This line was added to the end of the code: results = [eps, sigma, force];

And the "[eps, sigma, force]" in the first line of code was changed to "results". This gave the desired values for the results variable in the two-bar truss code, and the fixed results were changed in the code above. Eml4500.f08.gravy.sms 9:52, 10 November 2008 (UTC)

Principal of virtual work with axial dofs
The Principal of virtual work can also be expressed with the equation $$q^{(1)}_{2x1}= \underline{T}_{2x4} d^{(i)}_{4x1}$$

This is much like how FEA course grades are calculated, $$Grade= \alpha_o \cdot HW grade + \sum^3_{i=1} Exam_i $$

Deriving $$\mathbf{K}^{(e)}_{4x4}= \mathbf{T}^{(e)T}_{4x2} \mathbf{\hat{K}}^{(e)}_{2x2} \mathbf{T}^{(e)}_{2x4} $$

Remember from previous reports [http://en.wikiversity.org/w/index.php?title=User:Eml4500.f08.gravy.jad/Notes3.g&oldid=348085#Similarly_for_node_2_:_HW. Axial dofs equation]and Force Displacement

That:

$$\mathbf{\hat{K}}^{(e)}_{2x2} \mathbf{q}^{(e)}_{2x1}= \mathbf{p}^{(e)}_{2x1} \Longrightarrow \mathbf{\hat{K}}^{(e)} \mathbf{q}^{(e)}- \mathbf{p}^{(e)}= \mathbf{0}^{(1)}_{2x2}$$ (Equation 1)

$$ \mathbf{\hat{W}}_{2x1} \cdot \underbrace{(\mathbf{\hat{K}}^{(e)} \mathbf{q}^{(e)}- \mathbf{p}^{(e)}}_{2x1)}= \mathbf{0}_{1x1} $$ (Equation 2)

We showed (Equation 1)$$\iff$$ (Equation 2) earlier in this report

Recall: $$\mathbf{q}^{(e)}_{2x1}= \mathbf{T}^{(e)}_{2x4} \mathbf{d}^{(e)}_{4x1} $$

Similarly

$$ \mathbf{\hat{W}}^{(e)}_{2x1}= \mathbf{T}^{(e)}_{2x4} \mathbf{W}^{(e)}_{4x1} $$

$$\ \mathbf{\hat{W}}_{2x1} = \,$$virtual axial displacement, corresponding $$\ \mathbf{q}^{e}_{2x1} \,$$

$$\ \mathbf{W}_{4x1} = \,$$virtual displacement in global coordinate system, with corresponding $$\ \mathbf{d}^{e}_{4x1} \,$$

Now if you replace (Equation 3) and (Equation 4) into (Equation 2), the end result is:

$$\ \left ( \mathbf{T}^{(e)} \mathbf{W} \right ) \cdot \left\lbrack \hat{\mathbf{k}}^{(e)} \left ( \mathbf{T}^{(e)} \mathbf{d}^{(e)} \right ) - \mathbf{P}^{(e)} \right\rbrack = 0 \,$$for all $$\ \mathbf{W}_{4x1} \,$$ (Equation 5)

Recall: $$\ \left ( \mathbf{A}_{pxq} \mathbf{B}_{qxr} \right )^{T} = \mathbf{B}^{T} \mathbf{A}^{T} \,$$ (Equation 6)

As an Ex:

$$\ \mathbf{A}_{2x3} = \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \end{bmatrix}_{2x3} \,$$ $$\ \mathbf{B}_{3x3} = \begin{bmatrix} 7 & 8 & 9 \\ 1 & 2 & 3 \\ 4 & 5 & 6 \end{bmatrix}_{3x3} \,$$

Proof of equation $$( \mathbf{T}^{(e)}W)\cdot [\mathbf{\hat{k}^{(e)}q^{(e)} - p^{(e)}}] = 0 $$
Show that the following equation is true:

$$( \mathbf{T}^{(e)}W)\cdot [\mathbf{\hat{k}^{(e)}q^{(e)} - p^{(e)}}] = 0 $$

Continuous case with Partial Differential Equations(PDE)
So far, we are dealing with the discrete case [not continuous][matrices], up until now...

NOW, the continuous case: (PDEs)

$$\ \Rightarrow \,$$Motivational Model Problem: Elastic bar with varying A(x), E(x), subjected to varying axial load (distributed) + concentrated load + inirtia force(dynamics) [the distributed and concentrated loads are both time dependent]

$$\ \sum F_{x} = 0 = -N(x,t) + N(x + dx, t) + f(x,t)dx - m(x)dx \ddot{u} \,$$

$$\ \sum F_{x} = \frac{dN}{dt} (x,t)dx + H.O.T + f(x,t)dx - m(x)dx \ddot{u} \,$$ (Equation 1)

Recall the Taylor series exp: $$\ f(x+dx) = f(x) + \frac{df}{dx}dx+ \frac{1}{2} \frac{d^{2}f}{dx^{2}}(x)dx+... \,$$from the third term onwards are are H.O.Ts = Higher Order Terms

(Equation 1) $$\ \Rightarrow \frac{dN}{dx} + f = m \ddot{u} \,$$ Equation of motion (EOM) (Equation 2)

$$\ N(x,t) = A(x) \sigma(x,t) \,$$where $$\ \sigma(x,t) = E(x) \epsilon(x,t) \,$$where $$\ \epsilon(x,t) = \frac{du}{dx}(x,t) \,$$ (Equation 3)

Combining (Equation 2) and (Equation 3) yields: $$\ \frac{d}{dx} \left \lbrack A(x)E(x) \frac{du}{dx} + f(x,t) = m(x) \ddot{u} \right \rbrack \,$$where $$\ \ddot{u} = \frac{d^{2}u}{dt^{2}} \,$$ (Equation 4)

(Equation 4) describes the PDE of motion.

Now, we need two boundary conditions (2nd order derivative with respect to x), AND two initial conditions (2nd derivative with respect to t) [initial displacement and initial velocity]

$$\ u(0,t) = 0 = u(L,t) \,$$

1)$$\ u(0,t)=0 \,$$, 2)$$\ N(L,t) = F(t) \,$$; $$\ A(L) \sigma (L,t) = A(L)E(L) \epsilon (L,t) \,$$where $$\ \epsilon (L,t) = \frac{du}{dx} (x,t) \,$$

$$\ \Rightarrow \frac{du}{dx} (L,t) = \frac{F(t)}{A(L)E(L)} \,$$@ x=l

Initial conditions: @$$\ t=0, u(x,0) = u(x) \,$$known function displacement, and $$\ \frac{du}{dx}(x,0) = \dot{u}(x,0)= \overline{V}(x) \,$$known function velocity

FEM Analysis 3-Bar Space Truss Using MATLAB
The following Matlab code was taken from the 3-bar space truss system found in the textbook on page 230. The code uses three other functions: SpaceTrussElement.m, NodalSoln.m, and SpaceTrussResults.m. These functions were obtained from the online resources located at the book's website here.

This space truss is found to be statically determinate. So taking the sum of the forces in the x,y and z direction yields the same reactions as as the FEA method. Which are

R1= 20375

R2= 13214

R3= -23148

Contributing Team Members
The following students contributed to this report:

Jessica Dyer Eml4500.f08.gravy.jad 09:04, 7 November 2008 (UTC)

Mikal McFarlane Eml4500.f08.gravy.mmm 15:21, 5 November 2008 (UTC)

Justin Tennant Eml4500.f08.gravy.JMT 22:28, 6 November 2008 (UTC)

Patrick Clayton Eml4500.f08.gravy.psc 06:48, 7 November 2008 (UTC)

Sean Sullivan Eml4500.f08.gravy.sms 10:45, 7 November 2008 (UTC)

Brian Eggeman Eml4500.f08.gravy.bje 15:59, 7 November 2008 (UTC)

title: User:Eml4500.f08.gravy.jad/Notes5.g


text:
User:Eml4500.f08.bottle.loschak/HW5

Homework 5
The next section of HW 5 will be discussing truss systems in three dimensional space, which had previously not been considered. The axial force displacement relation, commonly referred to as the FD relation, is very similar when compared to the 2-D truss element. Note that the equation of the 2-D truss element is: $$\mathbf{kd}=\mathbf{F}$$ We also recall that the FD relation for axial displacement of each element in the notation used in class is as follows: $$\frac{EA}{L}\begin{pmatrix} 1 & -1\\ -1 & 1 \end{pmatrix} \begin{pmatrix} d_{1}\\d_{2} \end{pmatrix} = \begin{pmatrix} P_{1}\\P_{2} \end{pmatrix} \Rightarrow \mathbf{k^{(e)}q^{(e)}}=\mathbf{p^{(e)}}$$ The main difference for the 3-D truss is that each matrix is a bit larger to account for the movement in the z-direction. The local degrees of freedom become the 6x1 matrix $$\mathbf{d^{(e)}}=\begin{Bmatrix} d_{1}^{(e)} \\ d_{2}^{(e)} \\ d_{3}^{(e)} \\ d_{4}^{(e)} \\ d_{5}^{(e)} \\ d_{6}^{(e)} \end{Bmatrix}$$ and the local forces become the 6x1 matrix below. $$\mathbf{f^{(e)}}=\begin{Bmatrix} f_{1}^{(e)} \\ f_{2}^{(e)} \\ f_{3}^{(e)} \\ f_{4}^{(e)} \\ f_{5}^{(e)} \\ f_{6}^{(e)} \end{Bmatrix} $$ Now the transform matrix T(e) must contain two extra columns, but the axial displacement q(e) and axial load p(e) will remain as 2x1 matrices. The transform matrix will become: $$\mathbf{T^{(e)}}=\begin{pmatrix} l_{s} & m_{s} & n_{s} & 0 & 0 & 0\\ 0 & 0 & 0 & l_{s} & m_{s} & n_{s} \end{pmatrix}$$ The transform matrix for the 3-D truss system was derived similarly to the 2-D transform matrix in lecture 12. The figure below displays the unit vectors in the normal x-y coordinate axis along with the rotated coordinate axis. The displacement vector for a local node is defined as follows. $$ \bar{d}_{1}^{(e)}=d_{1}^{(e)}\hat{i}+d_{2}^{(e)}\hat{j}+d_{3}^{(e)}\hat{k} $$ It is possible to take the dot product of the displacement vector and the rotated unit vector and carry out the steps listed below. $$ q_{1}^{(e)}=(d_{1}^{(e)}\hat{i}+d_{2}^{(e)}\hat{j}+d_{3}^{(e)}\hat{k})\hat{\tilde{i}} $$ $$ q_{1}^{(e)}=d_{1}^{(e)}(\hat{i}\cdot\hat{\tilde{i}})+d_{2}^{(e)}(\hat{j}\cdot\hat{\tilde{i}})+d_{3}^{(e)}(\hat{k}\cdot\hat{\tilde{i}}) $$ $$ q_{1}^{(e)}=l^{(e)}d_{1}^{(e)}+m^{(e)}d_{2}^{(e)}+n^{(e)}d_{3}^{(e)} $$ $$ q_{1}^{(e)}=\begin{bmatrix} l^{(e)} & m^{(e)} & n^{(e)} \end{bmatrix} \begin{Bmatrix} d_{1}^{(e)}\\ d_{2}^{(e)}\\ d_{3}^{(e)} \end{Bmatrix} $$ And Similarly, for local node 2, the following can also be said. $$ q_{2}^{(e)}=\begin{bmatrix} l^{(e)} & m^{(e)} & n^{(e)} \end{bmatrix} \begin{Bmatrix} d_{4}^{(e)}\\ d_{5}^{(e)}\\ d_{6}^{(e)} \end{Bmatrix} $$ Combining the axial displacements into one expression results in the following expression: $$ \begin{Bmatrix} q_{1}^{(e)}\\ q_{2}^{(e)} \end{Bmatrix}= \begin{bmatrix} l^{(e)} & m^{(e)} & n^{(e)} & 0 & 0 & 0\\ 0 & 0 & 0 & l^{(e)} & m^{(e)} & n^{(e)} \end{bmatrix} \begin{Bmatrix} d_{1}^{(e)}\\ d_{2}^{(e)}\\ d_{3}^{(e)}\\ d_{4}^{(e)}\\ d_{5}^{(e)}\\ d_{6}^{(e)} \end{Bmatrix} $$ This expression can be written in its most simple form as the following: $$\mathbf{q}^{(e)}=\mathbf{T}^{(e)}\mathbf{d}^{(e)}$$ Similarly, the local forces can be multiplied by the same transform matrix to equal the axial forces. $$\mathbf{p}^{(e)}=\mathbf{T}^{(e)}\mathbf{f}^{(e)}$$

title: User:Eml4500.f08.bottle.loschak/HW5


title: File:Hw5transform.png


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Talk:Introduction to Computers/Computer types

what is a computer?

Missing mini-computers - an older term for a computer midsized in power and functionality between micro and mainframe. Mini's morphed into the more modern server. Most modern mainframes act as both a server to PC's as well as doing their own processing jobs.


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User talk:Dvorap

 Hello Dvorap, 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 everything on 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.
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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 09:52, 5 November 2008 (UTC)

 Hi! My name is WikiversityJack. I am a pumpkin and a Wikiversity mascot. My aunt told me that fun + learning = Wikiversity. Let me know if I can help you out.

Language and tests
Hi Dvorap! What language are you contributing in? We may be able to help direct you to a better project for your contributions. This is the English language Wikiversity. For testing, please use the Sandbox or your user page (I've undone/delete some of the content you created). Please let us know if we can help. Sincerely, James. -- Jtneill - Talk - c 09:52, 5 November 2008 (UTC)
 * It appears to be Hebrew. Do we have a Hebrew Wikiversity?  The Jade Knight (d'viser) 10:00, 5 November 2008 (UTC)
 * Thanks, JK. Dvorap also emailed me (thanks) and (I hope you don't mind) is from an Israeli university and is exploring teaching and learning options for lecturers. So, welcome Dvorap, I hope we can help. First, the bad news - unfortunately, http://he.wikiversity.org does not exist yet :(. But there is a Hebrew page on beta Wikiversity, which is the beginnings, and this would probably be the best place for learning project materials in Hebrew and for related experiments: http://beta.wikiversity.org/wiki/עמוד ראשי. Dvorap also asked (in email) about ways of protecting pages so they can only be edited by staff/students in a course. This is not really possible (in this way). A learning management system would probably be best for that way. However, on Wikiversity we found that it is rare that other participants interfere unproductively in course content and pages can be protected if needed. Another page which might be of interest: School and university projects. An example, I've been running a university unit on social psychology. Feel free to ask any other questions here or at the Colloquium. Sincerely, James. -- Jtneill - Talk - c 12:06, 5 November 2008 (UTC)

title: User talk:Dvorap


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Introduction to Biomembranes

Biomembrane Transport
A biomembrane is a structure present in all cells, being either the external plasma membrane or internal membranes (e.g. endoplasmic reticulum, Golgi apparatus). Membranes exist in order to allow the cell to control and change its internal environment, and to protect important organelles from environmental damage or attack from pathogens. All Membranes consist of at least three substances: The ratio between lipid and proteins depends on the function of the cell, some examples of different cell membrane compositions are shown in the table below.
 * Lipids
 * Proteins
 * Carbohydrates (Glyco-Proteins, -lipids, sterols for structure)

Membranes as Lipid Bilayers
Membranes are made of a lipid by layer of phospholipids which are made up of a hydrophilic "polar" heads(water) and hydrophobic fatty acid tails (oil). Phospholipids will spontaneously form lipid bilayers in aqueous solution; tails aggregate and exclude water from a core, the structure is stabilised by tail-tail/ head-head interactions. The polar head region's charge is variable and fatty acid (R) chains vary in length from C12 to C22 as well as in their degree of unsaturation (number of double bonds). Each C=C bond produces a rigid kink in the the phospholipid tail. In 1972 Singer and Nicholson proposed the "Fluid Mosaic" model suggesting that the membrane itself is a two dimensional sea of mobile lipid in which proteins diffuse or 'float'.

Mechanisms by which substances cross the membrane
__INDEX__ __NEWSECTIONLINK__
 * Diffusion
 * For hydrophobic molecules, occurs through the phospholipid bilayer down a concentration gradient (i.e. from a region of high concentration of the substance to a region of low concentration)
 * For hydrophillic molecule diffusion, see Transport
 * Osmosis (net water flow)
 * As water is hydrophillic, it cannot pass through the hydrophobic cell membrane, and so requires special hydrophillic proteins called aquaporins in order to assist their transport.
 * Transport (Passive or Active)
 * Transport via proteins. This can be passive (i.e. down a concentration gradient, no energy input) or active (against a concentration gradient, requiring energy input)

title: Introduction to Biomembranes


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Diffusion and osmosis

Diffusion

Simple diffusion is known as the random movement of solute molecules in a solvent tending towards an equilibrium (evenly-distributed) state, a better way to understand this is that diffusion is a type of passive transport where the net movement of molecules of a solute from a region of greater concentration move to a region of lower concentration.

Rate of Diffusion
The rate of Diffusion (J) is proportional to the concentration gradient over distance x (membrane thickness) $$J\propto\frac{dc}{dx}$$

Rate of diffusion is independent of diffusion of other substances although it is related to
 * Physical properties of solute and solvent molecules (e.g. size, electric charge)
 * temperature
 * electric field

Diffusion of a solute is rapid over short distances, although the membrane is a selective barrier to diffusion and the concentration gradient lies within the membrane barrier. To truly understand the mechanism of diffusion it must be described quantitatively as Fick's Law of Diffusion

Fick's Law of Diffusion
From a biological perspective Fick's First law $$J=-D\frac{\partial\phi}{\partial x}$$ gave rise to the Formula $$J={P\cdot A\cdot (c_2 - c_1)}\,\!$$

 * J = the rate of diffusion (mol/sec)
 * P = the permeability coefficient of solute across membrane (m/sec)
 * A = membrane area (m2)
 * C2-C1 = The concentration difference often termed $$\Delta c $$

J is also related to the diffusion coefficient in the membrane (D),$$\Delta c $$ within the membrane and the membrane thickness (x) and can be calculated by means of this equation
 * $$J={D\cdot A\cdot \left(\frac {\Delta c_m}{x}\right)}\,\!$$

The Diffusion coefficient has various factors contributing to it and can be calculated using the formula $$D=\frac{R\cdot T}{(6\pi \cdot \eta \cdot r)}$$

 * R = Gas constant (8.3 J/K.mol)
 * T = Absolute temperature (K)
 * η = Viscosity of barrier
 * r = Radius of diffusing molecule

The hydrophobic lattice structure of the bilayer acts like a viscous molecular "sieve", also values for P and D are specific for membrane AND diffusing molecules. The solute concentration within a membrane depends on Kp. Kp is the lipid-water coefficient and is described as
 * $$K_p=\frac {\Delta C_m}{\Delta C}$$.

For a hydrophobic molecule Kp < 1 and for a hydrophilic molecule Kp < 1, permeability can then be calculated including Kp in the equation and thus gives us $$P=K_p\cdot\frac{D}{x}$$

Important factors for predicting passive permeability
frictional effects
 * Molecular size - Small,P↑ ; large . P↓
 * Molecular shapre - Straight, P↑ ; globular P↓
 * Membrane Viscostiy - Short R chains, -C=C-, Tο

Lipid solubility (Kp)
 * Kp high such as O2, CO2, anaesthetics, lipophilic group then P↑
 * Kp low such as sugars, amino acids, ions, polar/charged groups then P↓

Osmosis
Osmosis is the net solvent flow, water moves from regions of higher to lower (more negative) water potentials, showing bulk flow. In animals water potential = Osmotive potential (zero for pure water; increasingly negative as solute concentration increases)

Osmolarity is proportional to the concentration of dissolved solutes and inversely proportional to osmotic potential. A common example to describe osmolarity is red blood cells in different salt concentrations.

If a solution contains a higher concentration of salt, than the solution's water flows into the cell. This causes the cell to swell and possibly bursts. This solution is known as Hyperosmotic.

Solutions that are at equilibrium with the cell salt concentration are known Isosmotic.

If a cell is placed in water, in which the salt concentration in the water is smaller than inside the cell, the water will flow out of the cell, causing it to shrink. This type of solution is know as Hyposmotic.

title: Diffusion and osmosis


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User:Owlbuster

I am a student of Physics at the University of Modena and Reggio Emilia.

title: User:Owlbuster


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The Resting Membrane Potential

The Resting potential
The resting potential of a cell is the membrane potential that would be maintained if there were no action potentials, post-synaptic potentials, or other active changes in the membrane potential. In most cells the resting potential has a negative value, which by convention means that there is excess negative charge inside compared to outside. The resting potential is mostly determined by the concentrations of the ions in the fluids on both sides of the cell membrane and the ion transport proteins that are in the cell membrane. How the concentrations of ions and the membrane transport proteins influence the value of the resting potential is outlined below.

How is the resting membrane potential produced and maintained?
 * The Donnan Effect
 * Membrane Selectivity (Difference in permeabilities of ions)
 * Active Transport (Na/K ATPase)

The Donnan Effect - Simplified the donnan effect can be described as large impermeant negatively charged intracellular molecules attracting positively charged ions such as Na+, K+ and repelling negative ones such Cl-

Membrane selectivity - Membrane selectivity is the difference in permeabilities between different ions for example the plasma membrane is far more permeable to potassium than sodium.

Active transport - Active transport is the mediated process of moving particles across biological membrane against a concentration gradient. If the process uses chemical energy, such as from ATP, it is termed primary active transport. Secondary active transport involves the use of an electrochemical gradient.

The three factors above all responsible for the resting membrane potential and also maintaining it at a steady voltage.

The Nernst Equation
The Nernst equation is a mathematical equation which can be applied physiologically to cells to produce Equilibrium potentials (The transmembrane potential difference required to exactly balance a given concentration gradient) for certain ions, the Nerst equation is named after the German physical chemist who first formulated it, Walther Nernst.

The equation is as follows
 * $$W=RTln \frac{[X] compartment 1}{[X] compartment 2}$$

Where-
 * W = Chemical work
 * R = Gas Constant
 * T = Absolute temperature(oK)

Electrical working in the opposite direction $$ EFz $$

 * E = The potential deifference across the membrane
 * F = Faradays Constant (96500 coulombs/mole)
 * z = Vaelncy of ion

At equlibrium the two above equations can be expressed together as
 * $$ EFz=RTln \frac{[X]_1}{[X]_2}$$

If the equation is then solved for $$E$$
 * $$E_i=\frac{RT}{Fz}ln \frac{[X]_1}{[X]_2}$$

This is the Nernst equation as it is used in physiology to calcualte an equilibrium potential generated by a given ion gradient. The term $$\frac{RT}{Fz}ln$$ can be replaced by a temperature-dependent conversion factor:

At 20oC $$E_i=58\log_{10} \left ( \frac{[X]_1}{[X]_2} \right )$$

At 37oC (body temperature) $$E_i=61\log_{10} \left ( \frac{[X]_1}{[X]_2} \right )$$

The Nernst equation doesn't precisely calcualte the membrane potential because it is rarely at one equlibrium potential. Another equation was formulated allowing us to produce more accurate values for the membrane potential, The Goldman-Hodgkin-Katz equation.

The Goldman-Hodgkin-Katz Equation
The Goldman-Hodgkin-Katz (GHK) voltage equation is used in cell membrane physiology to determine the potential across a cell's membrane (Em) taking into account all of the ions that are permeant through that membrane. The discoverers of this are David E. Goldman of Columbia University, and the English Nobel laureates Alan Lloyd Hodgkin and Bernard Katz.

GHK equation
 * $$E_m= \frac{RT}{F}\ln{ \left( \frac{ P_{Na^{+}}[Na^{+}]_\mathrm{out} + P_{K^{+}}[K^{+}]_\mathrm{out} + P_{Cl^{-}}[Cl^{-}]_\mathrm{in} }{ P_{Na^{+}}[Na^{+}]_\mathrm{in} + P_{K^{+}}[K^{+}]_{\mathrm{in}} + P_{Cl^{-}}[Cl^{-}]_\mathrm{out} } \right) }$$

Again just like the Nersnt equation the first term can be simplified by a temperature dependent conversion factor producing this at 20oC
 * $$E_m=58\log_{10}{ \left( \frac{ P_{Na^{+}}[Na^{+}]_\mathrm{out} + P_{K^{+}}[K^{+}]_\mathrm{out} + P_{Cl^{-}}[Cl^{-}]_\mathrm{in} }{ P_{Na^{+}}[Na^{+}]_\mathrm{in} + P_{K^{+}}[K^{+}]_{\mathrm{in}} + P_{Cl^{-}}[Cl^{-}]_\mathrm{out} } \right) }$$

The Cl term of the equation is inverted so that z (valency) can be omitted from the first term. In many physiological applications the Cl term can be omitted altogether to yield the equation

 * $$E_m=58\log_{10}{ \left( \frac{ P_{Na^{+}}[Na^{+}]_\mathrm{out} + P_{K^{+}}[K^{+}]_\mathrm{out} }{ P_{Na^{+}}[Na^{+}]_\mathrm{in} + P_{K^{+}}[K^{+}]_{\mathrm{in}} } \right) }$$

Cl can be left out of the equation because in the steady state Cl does not contribute to the setting of E_m, Cl is said to be passively permeable and rather than the Cl gradient setting the Em, Em sets the Cl gradient. In other words Ecl- is in equilibrium with Em, $$E_{Cl}=E_m$$ this confers membrane stability.

The single most imporatnt factor in setting Em is the potassium gradient

/Quiz/

title: The Resting Membrane Potential


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Coordinate systems

This resource focuses on an introduction suitable for an introductory college physics course in electromagnetism.

We describe three different coordinate systems, known as Cartesian, cylindrical and spherical. The cylindrical system is closely related to polar coordinates.

See also /Derivation of formulas/

Cartesian Coordinate

 * Cartesian-coordinate-system.svg

Mathematical expression of a point on Cartesian coordinate
 * A (x,y)

Polar Coordinate

 * Examples_of_Polar_Coordinates.svg

Mathematical expression of a point on Cartesian coordinate
 * A (r,$$\theta$$)

Conversion Coordinate

 * Polar_to_cartesian.svg
 * $$r = \sqrt{x^2 + y^2}$$
 * $$\theta = Tan^{-} \frac{y}{x}$$
 * $$x(\theta) = r Cos \theta$$
 * $$y(\theta) = r Sin \theta$$

Cartesian Coordinates
To the left: Illustration of a Cartesian coordinate plane. Four points are marked and labeled with their coordinates: (2, 3) in green, (−3, 1) in red, (−1.5, −2.5) in blue, and the origin (0, 0) in purple.

To the right: A three dimensional Cartesian coordinate system, with origin O and axis lines X, Y and Z, oriented as shown by the arrows. The tick marks on the axes are one length unit apart. The black dot shows the point with coordinates x = 2, y = 3, and z = 4, or (2,3,4).

Line element in Cartesian coordinates
The distance between points $$(x_1,y_1,z_1)$$ and $$(x_2,y_2,z_2)$$ is
 * $$l = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2+ (z_2-z_1)^2} .$$

In other coordinate systems, it is best to focus on the distance between two points that are close together. For our purposes, it is best to replace the above formula with:

 * $$d\ell = \sqrt{(dx)^2 + (dy)^2+ (dz)^2} =\sqrt{\vec d\ell\cdot\vec d\ell}$$

where,

 * $$\vec d\ell = \hat x dx + \hat y dy + \hat z dz $$.

is known as a line element.

Line integral of a vector field in Cartesian coordinates
So far we have discussed the integral of a scalar field, $$f=f(\vec r)$$. An example of a scalar field is the temperature as a function of position (at any given time). Wind velocity as a function of position would be a vector field. In Cartesian coordinates it is easy to analyze a vector field as being equivalent to three scalar fields, one for each direction: Let $$\vec F(\vec r) = f(\vec r)\hat i + g(\vec r)\hat j + h(\vec r)\hat k \,$$ (so that Fx=f, Fy=g, and Fz=h). The line integral of the vector field from $$\vec r = \vec a$$ to $$\vec r = \vec b$$ can now be defined as the limit of a Riemann sum:

 * $$\int_\vec{a}^\vec{b}\vec F\cdot d\vec\ell\approx\sum_j \vec{F_j}\cdot\Delta\vec\ell \equiv \sum_j F_j \ \Delta\ell \cos\theta$$

Line integral of a uniform field over a straight path
Line integrals are trivial if the vector field is uniform, meaning that the field does not depend on position. A well known example is the acceleration of gravity at Earth's surface, under the assumption that we may approximate the Earth as 'flat': $$\vec F_0 = -g\hat y$$ where g=9.8m/s2, and the 0 subscript reminds us that the field is a constant. Constant terms may be factored out of either a Riemann sum or an integral: If $$\vec F$$ does not depend on position, then:

 * $$\sum_j \vec F_j\cdot\Delta\vec\ell_j\ =\vec F\cdot\sum_j \Delta\vec\ell

\leftrightarrow \int_\vec a^\vec b \vec F\cdot d \vec\ell\,$$ $$=\vec F\cdot\int_\vec a^\vec b d\vec\ell$$

Note how the integral notation permits us to state the endpoint, $$\vec b$$ and starting point, $$\vec a$$, of the path. This can be evaluated to assume:

$$ =\int_\vec a^\vec b \vec F\cdot d \vec\ell=\vec F\cdot (\vec b -\vec a)\,$$$$=F\;|\vec b-\vec a|\cos\theta$$

 * See the Escher Waterfall to appreciate the technology that would be made possible if weight were not a conservative force.

Differential volume element in Cartesian coordinates
The Cartesian volume element is,

 * $$dV=dxdydz,$$

which easily permits a volume integral over a rectangular box. The integral of a function f=f(x,y,z) over the box is the triple integral, which is often informally written using a single integral, sign $$\int$$.

 * $$\int f\,dV=\int_{x1}^{x2}dx \overbrace{\int_{y1}^{y2}dy \underbrace{\int_{z1}^{z2}dz\,f(x,y,z)}_{\mathrm{function\ of\ x\ and\ y}}}^{\mathrm{function\ of\  x\ only}}$$

Note how the variable of integration (x or y or z) will disappear each time the integral is evaluated between the endpoints. This is a scalar integral if the function, $$f=f(x,y,z)$$. In general, the limits may be functions, for example (z1,z2) may be functions of x and y, while (y1,y2) may be functions of x. Fortunately such complexity is not required in the volume integrals needed to understand electromagnetism.

Volume integral over a rectangular box
If none of the limits, (z1,z2) and (y1,y2), in the above integral are functions, then the integral is over a rectangular box. The most elementary example of such an integral occurs when $$f(x,y,z)=1$$:

 * $$\int 1 dV=(x_2-x_1)\times(y_2-y_1)\times(z_2-z_1)\equiv \iiint 1 dV=volume$$

This is just the volume of the box.

Differential surface element in Cartesian coordinates
Surface integrals can also be viewed as the limit of a Riemann sum, but can be more difficult to construct than line or volume integrals unless the geometry is very simple. The differential element of surface area can either be a scalar, $$dA$$, or the vector element, $$d\vec A = \hat n dA$$. Here, $$\hat n$$ is normal to the surface.

 * $$\int\vec F\cdot d\vec A\approx\sum_j \vec{F_j}\cdot\Delta\vec A \equiv \sum_j (F_j)\ (\Delta A_j )\cos\theta$$

At the moment of this writing a | Wikipedia page states that a surface element in Cartesian coordiantes is,

 * $$d\vec A=dxdy\hat z + dydz\hat x + dzdx\hat y\,,$$

which is at best misleading. In practice, one learns electromagnetism by doing very simple surface integrals where only one term is present.

A surface integral over a square with its normal parallel to a Cartesian axis
The vector surface area element,

 * $$d\vec A = dA \hat n = dxdy\hat z$$

occupies rectangle parallel to the xy-plane. If the integral is of the form $$\int \vec F\cdot d\vec A$$, and the area is a square of length L centered at the origin, and occupying the plane, z=H, then:

$$\int \vec F\cdot d\vec A = \int F_zdA=\int_{-L/2}^{+L/2}dx\int_{-L/2}^{+L/2}dy F_z(x,y,H) $$

In contrast to the one-dimensional integrals of a first-year calculus course, limits are not always used to define the region over which the field is integrated. Instead it is customary to state the integral, $$\int f\,dA $$, and then use words to describe the area (e.g. "where the integral is over a square with sides of length, L, centered at the origin and occupying the xy plane"). Very simply defined surface integrals can also be defined using limits. For example,

 * $$\int \vec F\cdot\,dA =\int_{-L/2}^{L/2}dx \int_{-L/2}^{L/2}dy F_z(x,y,0)\, $$

actually does describe a surface integral over a square with sides of length, L, that occupies the xy plane, and is centered at the origin. For example, if L=4, and

 * $$F_z(x,y,z)=(1 +z)y^2\Rightarrow F_z(x,y,0)=y^2$$,

we have:

 * $$\int F_z\,dA =

\underbrace{ \int_{-4/2}^{4/2}dx  }_4 \underbrace{ \int_{-4/2}^{4/2}y^2dy  }_{(2)   \left(\frac 1 3 \right)  \left(2^3\right)   } = 4 + 16/3 = \frac{28}{3}$$

Polar coordinates
In the two-dimensional polar coordinate system, the displacement vector, $$\vec r$$ is specified by the distance to the origin, $$r$$, and the angle, $$\theta$$, measured with respect to the x-axis. The unit vectors (shown in red), $$\hat r$$ and $$\hat\theta$$ are orthonormal, but change direction as the angle changes. It is evident from the two figures that:

 * $$d\vec\ell = \hat r dr + \hat\theta rd\theta$$

 * $$d\ell = \sqrt{ (dr)^2+r^2(d\theta)^2 }=

\left[ \left(\frac{dr}{d\tau}\right)^2  + r^2 \left(\frac{d\theta}{d\tau}\right)^2    \right]^{1/2}$$

 * $$dA = rdrd\theta$$

Cylindrical coordinates

The cylindrical coordinate system specifies point positions by the symbols $$(\rho, \varphi, z)$$, as shown in the figure. The coordinate $$z$$ can be positive or negative, while $$\rho$$ is always positive. Sometimes the Greek r,$$\rho$$ (called rho), is written in Latin form, $$r$$, in order to avoid confusion with rho as charge density, and also in order to emphasize the close relationship between polar and cylindrical coordinates.

The line element is
 * $$\vec d\mathbf{\ell} = \mathrm d\rho\,\boldsymbol{\hat \rho} + \rho\,\mathrm d\varphi\,\boldsymbol{\hat\varphi} + \mathrm dz\,\mathbf{\hat z}.$$

The volume element is
 * $$\mathrm{d}V = \rho\,\mathrm{d}\rho\,\mathrm{d}\varphi\,\mathrm{d}z.$$

The surface element in a surface of constant radius $$\rho$$ (a vertical cylinder) is

 * $$\mathrm{d}S_\rho= \rho\,\mathrm{d}\varphi\,\mathrm{d}z.$$

The surface element in a surface of constant azimuth $$\varphi$$ (a vertical half-plane) is

 * $$\mathrm{d}S_\varphi= \mathrm{d}\rho\,\mathrm{d}z.$$

The surface element in a surface of constant height $$z$$ (a horizontal plane) is

 * $$\mathrm{d}S_z= \rho\,\mathrm{d}\rho\,\mathrm{d}\varphi.$$

Spherical coordinates

 * $$\mathrm{d}\vec\ell = \mathrm{d}r\,\boldsymbol{\hat r} + r\,\mathrm{d}\theta \,\boldsymbol{\hat\theta } + r \sin{\theta} \, \mathrm{d}\varphi\,\mathbf{\boldsymbol{\hat \varphi}}.$$

While three surface elements exist (one for each direction), the only one commonly used to introduce electrodynamics is in the radial direction:

 * $$\hat r dA = \hat r r^2\sin\theta\,\mathrm{d}\theta\,\mathrm{d}\varphi.$$

Recall that the radian can be defined in terms of an arclength on a circle: $$d\theta\equiv ds/r$$. By analogy, the solid angle can be defined through an area on a sphere:

 * $$\mathrm{d}\Omega\equiv\frac{dA}{r^2}=\sin\theta\,\mathrm{d}\theta\,\mathrm{d}\varphi,$$

where dA is an area element taken on the surface of a sphere of radius, r, centered at the origin.

The volume element is spherical coordinates is:

 * $$\mathrm{d}V=r^2 \sin \theta \,\mathrm{d}r\,\mathrm{d}\theta\,\mathrm{d}\varphi.$$

Introductory discussions of electromagnetism often involve spherical symmetry, in which fields do not depend on the two directional coordinates (φ and θ). This permits integration over both variables:

$$\int d\Omega = \int_0^{2\pi}d\varphi \int_0^\pi sin\theta d\theta =\int_0^{2\pi}d\varphi \int_{-1}^1 d(\cos\theta)=4\pi$$

We have just shown that the solid angle associated with a sphere is 4π steradians (just as the circle is associated with 2π radians).

Divergence, gradient and Laplacian (differential operators)
from https://en.wikipedia.org/w/index.php?title=Divergence&oldid=605214437

Cartesian coordinates
The del operator acting on a scalar field $$f=f(\vec r)$$ is the gradient of that field:

 * $$\vec\nabla f= \boldsymbol{\hat x}\frac{\partial f}{\partial x} + \boldsymbol{\hat y}\frac{\partial f}{\partial y} + \mathbf{\hat z}\frac{\partial f}{\partial z},$$

The divergence of a the vector field $$\vec F=\vec F(\vec r)$$ is
 * $$\operatorname{div}\, \vec F

= \vec\nabla\cdot\vec F = \frac{\partial F_x}{\partial x}+\frac{\partial F_y}{\partial y} + \frac{\partial F_z}{\partial z}\,. $$

The Laplacian $$\nabla^2$$ of a scalar field is

 * $$ \nabla^2 f =\vec\nabla\cdot\vec \nabla f

= + {\partial^2 f \over \partial x^2 } + {\partial^2 f \over \partial y^2 } + {\partial^2 f \over \partial z^2 }. $$

Cylindrical coordinates
The del operator acting on a scalar field $$f=f(\vec r)$$ is the gradient of that field:

 * $$\vec\nabla f= \boldsymbol{\hat \rho}\frac{\partial f}{\partial \rho} + \boldsymbol{\hat \varphi}\frac{1}{\rho}\frac{\partial f}{\partial \varphi} + \mathbf{\hat z}\frac{\partial f}{\partial z},$$

The divergence of a the vector field $$\vec F=\vec F(\vec r)$$ is
 * $$\operatorname{div}\, \vec F

= \vec\nabla\cdot\vec F = \frac 1\rho \frac{\partial}{\partial\rho} (\rho F_\rho) + \frac 1\rho \frac{\partial F_\varphi}{\partial\varphi} + \frac{\partial F_z}{\partial z}\,. $$

The Laplacian $$\nabla^2$$ of a scalar field is

 * $$ \nabla^2 f =\vec\nabla\cdot\vec \nabla f

= {1 \over \rho} {\partial \over \partial \rho} \left( \rho {\partial f \over \partial \rho} \right) + {1 \over \rho^2} {\partial^2 f \over \partial \varphi^2} + {\partial^2 f \over \partial z^2 }. $$

Advanced topics in vector integrals
The most well-known integrals of vector calculus are deliberately chosen to be simple because the intent is usually to depict important concepts in physics. This section describes moderately difficult geometries.

Line integral of scalar field
The most elementary line integral is along an axis. For example,

 * $$\int_{x1}^{x2} f(x, y_0, z_0)\,dx$$

is the line integral of $$f(\vec r)$$ from the point $$(x_1,y_0,z_0)$$ to $$(x_2,y_0,z_0)$$. A line an an arbitrary direction may be defined parametrically, e.g., by:

 * $$\vec r = \vec r_0 + \vec\beta \tau$$

To define scalar line integral, we need to convert the vector path differential into a scalar by taking the magnitude. It is customary to label a differential in the vector $$\vec r$$ using the symbol $$d\vec\ell$$ instead of $$d\vec r$$ because the latter falsely hints at a change in radius, as we shall soon see:

 * $$d\vec\ell \equiv d\vec r = \vec\beta d\tau\Rightarrow d\ell=\beta d\tau\;,$$

where $$\beta\equiv |\vec\beta|$$ is the vector magnitude. Writing the path differential $$\vec d\ell$$ as $$\vec dr$$ would lead to great confusion when the (scalar) magnitude is taken because $$dr\equiv |d\vec r|$$ creates a conflict with the differential in radius if spherical coordinates are used: $$dr= d|\vec r| \ne |d\vec r|$$. The scalar line integral over a scalar field, $$f=f(\vec r)$$ is therefore:

 * $$\int_{\tau 1}^{\tau 2} f(x_0+\beta_x\tau, y_0+\beta_y\tau, z_0+\beta_z\tau)\,\sqrt{\beta_x^2+\beta_y^2+\beta_z^2}\,d\tau$$

is the line integral of $$f(\vec r)$$ from the point $$(x_1,y_0,z_0)$$ to $$(x_2,y_0,z_0)$$.

If $$\beta = 1$$, the three components of $$\vec\beta$$ are the cosines that each coordinate axis (x-y-z) makes with $$\vec\beta$$. The most convenient way to express the differential line vector and scalar are:

 * $$d\vec\ell = \hat xdx + \hat ydy$$

 * $$ d\ell =\sqrt{dx^2+dy^2}=

\left[ \left(\frac{dx}{d\tau}\right)^2 + \left(\frac{dx}{d\tau}\right)^2     \right]^{1/2}d\tau$$

The vector line integral in polar coordinates
$$\vec F \cdot d\vec\ell = F_xdx+F_ydy = F_r dr + F_\theta rd\theta$$

$$\vec F \cdot d\vec\ell = \left(F_x\hat i + F_y\hat j\right)\cdot \left( \hat i dx + \hat j dy\right)$$ $$= F_x dx (\hat i\cdot \hat i) + F_y dy (\hat j\cdot \hat j) + (junk)(\hat j \cdot \hat i)$$

where $$\hat i\cdot \hat i =\hat j\cdot \hat j = 1$$ and $$\hat i\cdot\hat j = 0$$.

For polar coordinates, use $$d\vec\ell = \hat r dr + rd\theta \hat \theta$$

Appendix: Teaching the vector
The "vector" is a nontrivial concept that arises even in Riemannian calculus where an n-dimensional space of variables is defined, and the concept of metric is used to generate vector and tensor fields, all without that fundamental sense of "direction" that students use to define "vector".

A students' introduction to coordinates starts with points on a two-dimensional x-y plane. There are a number of more or less equivalent way to label these points. We begin with the displacement vector:

 * $$\vec r = x\hat i + y\hat j \rightarrow x\hat x + y\hat y\rightarrow x_1\hat e_1 + x_2\hat e_2$$

This is essentially an ordered pair of variables that represent the real numbers x and y, and the formalism is easily extended to an arbitrary number of dimensions: :$$z\hat k \rightarrow  z\hat z\rightarrow x_3\hat e_3$$

One can define the displacement vector as an instruction to move a certain distance in a certain direction. In a Euclidean space, there is no difference between specifying a location (x,y) and an instruction to step from the origin in a certain direction to the point (x,y).

One advantage that vector notation offers is the ability to use symbols to represent a vector, and subscripts to represent the components:

 * a=ax i + ay j and b=bx i + by j

This notation uses boldface instead of arrows and hats to distinguish between vectors and scalars. (In other words, a, and $$\vec a$$ are different way to express the same thing.) The scalar (physics) or scalar (mathematics) is just a fancy word for "ordinary number". For example, ax and ay are scalars that form the two components of vector a. A famous use of different symbols to represent different vectors is F=ma, which says that the vector acceleration, a, multiplied the scalar, m, equals the net force, F (the latter being a vector because it has magnitude and direction).

Wikipedia links

 * Vector calculus identities
 * Surface integral
 * Cartesian coordinates
 * Polar coordinates
 * Cylindrical cordinates
 * Cylindrical coordinate system
 * Spherical coordinate system
 * Unit vector  (Wikipedia)
 * standard basis (Wikipedia)
 * Vector Calculus

other links

 * Math.com Homepage

title: Coordinate systems


text:
Georgian language

Georgian is the official language of the Republic of Georgia. There are approximately 6 million native speakers of Georgian worldwide. The vast majority of Georgian speakers are located in Georgia but significant size communities can also be found in Iran, Azerbaijan, Turkey and Russia. Georgian belongs to the Southern branch of the Caucasian family of languages. Its closest relations are Svan and Megrelian, which are spoken primarily in Northwest Georgia. Georgian possesses a number of dialects but is most commonly grouped into either Eastern or Western dialects. Standard Georgian is based on Eastern dialects. Georgian employs it own unique form for writing. Called Mkhedruli (the military/secular writing), the Georgian alphabet consists of thirty-three letters: 5 vowels and 28 consonants.

Georgian Language Course For Everybody is specially organized for international people, who are interested in learning Georgian Language in Tbilisi and be familiarized with Georgian culture. Staying in our country is an exciting and unique way to learn about Georgian history, culture and traditions. It's a wonderful opportunity to study and practice Georgian conversation.

Georgian Language Course contains: speaking, listening, grammar, vocabulary, reading, writing and special course about Georgian history and culture.

Course is designed both for beginners and peoples with skill of basics in Georgian Language.

Our teachers are all highly qualified, experienced and committed, who take a personal interest in each of our student. All teaching materials and textbooks are provided FREE for every student.

More information, please visit:
 * Georgian Language Courses in Tbilisi, Georgia

title: Georgian language


text: File:1-D Case.JPG

title: File:1-D Case.JPG


text:
User:Eas4200c.f08.blue.a/Homework5

October 20, 2008
Reading page 70 in textbook

$$\left\{\epsilon_{ij} \right\}_{6x1} = \begin{bmatrix} \bar{A}_{3x3} & o\\ 0 & \bar{B}_{3x3} \end{bmatrix}_{6x6}\left\{\sigma_{ij} \right\}_{6x1}$$

The $$\sigma$$ - $$\epsilon$$ relationship: Assuming diagonal coefficient is not zero

$$\left\{\sigma_{ij} \right\}_{6x1} = \begin{bmatrix} \bar{A}^{-1}_{3x3} & 0\\ 0 & \bar{B}^{-1}_{3x3} \end{bmatrix}_{6x6}\left\{\epsilon_{ij} \right\}_{6x1}$$

If $$\begin{bmatrix} \bar{A}^{-1}_{3x3} & 0\\ 0 & \bar{B}^{-1}_{3x3} \end{bmatrix}_{6x6}$$ we can varify that the relationship by $$\bar{c}^{-1}\bar{c} = \bar{I}$$ then $$\bar{c}^{-1}\bar{c} = \begin{bmatrix} \bar{A}^{-1}\bar{A} & 0 \\ 0 & \bar{B}^{-1}\bar{B} \end{bmatrix} = \bar{I}$$ therefore,

$$\left\{\sigma_{ij} \right\} = \begin{bmatrix} \bar{A}^{-1} & 0 \\ 0 & \bar{B}^{-1} \end{bmatrix}\begin{Bmatrix} 0\\ 0\\ 0\\ 0\\ \epsilon_{31}\\ \epsilon_{12} \end{Bmatrix}$$

Road map

Equilibrium Equation for Stress in a nonuniform stress field (section 2.4 in the book)

Consider a one dimensional case as a model.For a nonuniform field, the distribution is not constant, therefore $$f\left(x \right)\neq constant$$

and showing the shear forces acting on this

October 27, 2008
Bidirectional Bending

$$M_{y}=\int_{aA}^{}{z\sigma _{xx}dA}$$

$$M_{z}=\int_{aA}^{}{y\sigma _{xx}dA}$$

Moment of Inertia Tensors:

$$I_{yy}=\int_{A}^{}{z^{2}dA}$$

$$I_{zz}=\int_{A}^{}{y^{2}dA}$$

$$I_{yz}=\int_{A}^{}{zydA}$$

Recall that, by Hooke's Law :

$$\sigma _{xx}=E\varepsilon _{xx}$$

With the above equations this becomes:

$$\sigma _{xx}=(\frac{I_{yy}M_{z}-I_{yz}M_{y}}{I_{yy}I_{zz}-(I_{yz})^{2}})y$$+$$(\frac{I_{zz}M_{y}-I_{yz}M_{z}}{I_{yy}I_{zz}-(I_{yz})^{2}})z$$

Where the denominator of both terms is the determinant of bottom right corner of the Moment of inertia matrix (shown below):

$$I=\begin{bmatrix} I_{11} &I_{22} &I_{13} \\ I_{21}&I_{22} &I_{23} \\ I_{31}& I_{32} & I_{33} \end{bmatrix}$$

In addition, the location of the neutral axis can be found by setting $$\sigma _{xx}=0$$.

$$\sigma _{xx}=(\frac{I_{yy}M_{z}-I_{yz}M_{y}}{I_{yy}I_{zz}-(I_{yz})^{2}})y$$+$$(\frac{I_{zz}M_{y}-I_{yz}M_{z}}{I_{yy}I_{zz}-(I_{yz})^{2}})z=M_{yy}+M_{zz}=0$$

Therefore, $$z=(-\frac{M_{y}}{M_{z}})y=\tan (\beta) y$$

October 29, 2009
Continuing from the equation of equilibrium in terms of $$\bar{\sigma}$$

The goal is to find $$\frac{\partial \sigma_{yx}}{\partial y}+\frac{\partial \sigma_{zx}}{\partial z}=0$$ (equation 3.14 in textbook) Lets use indicial notation $$\frac{\partial \sigma_{21}}{\partial x_{2}}+\frac{\partial \sigma_{31}}{\partial x_{3}}=0$$

Recall from section 2.4 that

$$\frac{\partial \sigma_{xx}}{\partial x}+\frac{\partial \tau_{yx}}{\partial y}+\frac{\partial \tau_{zx}}{\partial z}=0$$ (equation 2.21 in textbook)

It follows, using the indicial notation that the three equilibrium equations become

$$\frac{\partial \sigma_{11}}{\partial x_{1}}+\frac{\partial \tau_{21}}{\partial x_{2}}+\frac{\partial \tau_{31}}{\partial x_{3}}=0$$ (equation 2.21 in textbook)

$$\frac{\partial \tau_{12}}{\partial x_{1}}+\frac{\partial \sigma_{22}}{\partial x_{2}}+\frac{\partial \tau_{32}}{\partial x_{3}}=0$$ (equation 2.22 in textbook)

$$\frac{\partial \tau_{13}}{\partial x_{1}}+\frac{\partial \tau_{23}}{\partial x_{2}}+\frac{\partial \sigma_{33}}{\partial x_{3}}=0$$ (equation 2.23 in textbook)

Therefore the summation notation of the equilibrium equations is as follows

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

October 31, 2008
Goal to determine the four zero stress components.

Obtain the equations resulting from

$$\sum_{i=1,j=1}^3\frac{\delta \sigma_{ij}}{\delta x_i}=0$$

which would result in

$$\frac{\delta \sigma_{11}}{\delta x_{1}}+\frac{\delta \sigma_{12}}{\delta x_{2}}+\frac{\delta \sigma_{13}}{\delta x_{3}}=0$$

$$\frac{\delta \sigma_{21}}{\delta x_{1}}+\frac{\delta \sigma_{22}}{\delta x_{2}}+\frac{\delta \sigma_{23}}{\delta x_{3}}=0$$

$$\frac{\delta \sigma_{31}}{\delta x_{1}}+\frac{\delta \sigma_{32}}{\delta x_{2}}+\frac{\delta \sigma_{33}}{\delta x_{3}}=0$$

One Dimensional model problem:

$$\sum F_2=0=-\sigma_w A+\sigma_{x+dx}A+f(x)dx$$

$$0=A[\sigma_{x+dx}-\sigma_x]+f(x)dx$$

Taylor series expansion $$=\frac{d \sigma}{dx}+$$ higher order terms

Now neglecting the higher order terms

$$\frac{\delta \sigma}{dx}+\frac{f(x)}{A}=0$$

So for all the forces in the X direction the sum is

$$\sum Fx =0= dy dz [-\sigma_{xx}(x,y,z)+\sigma_{xx}(x+dx),y,z)]+dzdx[-\sigma_{yx}(x,y,z)+\sigma_{yx}(x,y+dy,z)]+dxdy[\sigma_{zx}(x,y,z)+\sigma_{zx}(x,y,z+dz)]=0$$

which will yield

$$0=dx dy dz (\frac{\delta \sigma_{xx}}{\delta_x}+\frac{\delta \sigma_{xy}}{\delta_y}+\frac{\delta \sigma_{xz}}{\delta_z})$$

So for all the forces in the Y direction the sum is

$$\sum Fy =0= dy dz [-\sigma_{yx}(x,y,z)+\sigma_{yx}(x+dx),y,z)]+dzdx[-\sigma_{yy}(x,y,z)+\sigma_{yy}(x,y+dy,z)]+dxdy[\sigma_{yz}(x,y,z)+\sigma_{yz}(x,y,z+dz)]=0$$

which will yield

$$0=dx dy dz (\frac{\delta \sigma_{yx}}{\delta_x}+\frac{\delta \sigma_{yy}}{\delta_y}+\frac{\delta \sigma_{yz}}{\delta_z})$$

So for all the forces in the Z direction the sum is

$$\sum Fy =0= dy dz [-\sigma_{zx}(x,y,z)+\sigma_{zx}(x+dx),y,z)]+dzdx[-\sigma_{zy}(x,y,z)+\sigma_{zy}(x,y+dy,z)]+dxdy[\sigma_{zz}(x,y,z)+\sigma_{zz}(x,y,z+dz)]=0$$

which will yield

$$0=dx dy dz (\frac{\delta \sigma_{zx}}{\delta_x}+\frac{\delta \sigma_{zy}}{\delta_y}+\frac{\delta \sigma_{zz}}{\delta_z})$$

NACA Airfoil Matlab Code
Note 1: The code here uses x and y as the reference axes( as used in the textbook as opposed to y and z).

Note 2: The bending moment code is still being worked on at this time. What has been developed to this point is included and will be fully functional in a few days.

Note 3. Initially, i had planned on only having 1 m-file, but after adding more functionality to this project, i have split everything up into individual m-files ( for easier understanding).

Team Members That Contributed To This Report
Mark Barry Eas4200c.f08.blue.a 14:16, 5 November 2008 (UTC)

Clay Robertson eas4200c.f08.blue.f 13:42, 6 November 2008 (UTC)

Marc Pelletier EAS4200c.f08.blue.b 05:18, 7 November 2008 (UTC)

Jeffrey Shea EAS4200c.f08.Blue.E 19:24, 6 November 2008 (UTC)

Christian Garabaya Eas4200c.f08.blue.d 08:25, 7 November 2008 (UTC)

Tom Sexton EAS4200c.f08.blue.c 22:27, 6 November 2008 (UTC)

title: User:Eas4200c.f08.blue.a/Homework5


text: Category:Amateur astronomy

title: Category:Amateur astronomy


text:
User talk:Jayrawrrr

 Hello Jayrawrrr, 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 everything on 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.
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 * 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! SB_Johnny talk 15:01, 5 November 2008 (UTC)

title: User talk:Jayrawrrr


text:
Consensus in a wiki environment/Practical consensus

if we refer to consensus we could describe what ought to be (normative) and what is (empirical)

normative consensus
The agreement of all participants due to an succinct and effective persuasive statement in a dialogue or argument between opposing parties.

empirical consensus
The end of one of the parties ability or desire to argue and resist a well accepted statement from the opposing party.

Futility of normative consensus
I philosophically stand by the assertion that normative consensus or rare to none existent in the discussions of mankind given a wide enough audience. Further more if there are opposing parties, it is again rare to none existant that both parties would end a argument or discussion in normative consensus.

It would seem difficult to me, than to assume that in wikiversity, we seek to achieve normative consensus on any one thing.

consequence
thus although a exchange of opposing statements might be useful to the community, it will not be enough to completely end any sincere argument.

not everything is an argument
This dour consequence should be balanced by the fact that not everything is an argument. Indeed it must not be. The whole wikimedia model is based on Collaboration between knowledgable participants, and they must minimize arguments or the resulting document will be so unified and disorganized that noone will be able to read it.

in fact I think most wiki messes are the result of arguments

Are all statements arguments?
I say that Not everything is an argument but to some people it is. Any editor has the right to move a discussion into a argument (creating the need for empirical consensus). but this is a heady responseability. One should do it only by the thought that you represent others that would strongly feel this way. education must not be brainwashing, and alternate perspectives need welcomed for the educational effect we all wish to experience.

if one user invokes an argument, it is an argument and we ought to seek Empirical consensus. Explaining our position, but clear to the practical reality that the user probably will not agree. We hope for more polite forms of empirical consensus.

minimize arguments by
carefully stating ones expertise amending all arguments with an appeal from higher expertise
 * ;) unfortunately these experts sometimes argue as much as anybody

Keeping content specific and within material you know well Being modest Being polite stating things as clearly as possible
 * We all do as well as we can. I know. but we need to just keep this in mind.

Lastly seeking Empirical consensus, immediately. rather than delude ourselves that we are going to convince everyone --- but philosophically we have no right to be restrain opposing viewpoints (outside of ethical and moral policy).

we can achieve Empirical consensus
if:
 * one side agrees to end disruptive arguments
 * the two sides agree to a grudging compromise
 * one side agrees to state their position in one document and the other side agrees to state their postion in another document
 * * nicest would be for each side to link the opposing side, thus giving visitors the chance to freely understand each side's point of view

 * or LASTLY (and worst) if the other side (or individuals in the other side), are prevented from further alterations

Conclusion
Normative (ideal) consensus is a unreasonable goal for any sincere argument. Participants must do what they can to minimize arguing. We need to look for empirical consensus, immediately. we also need to realize that in extreme cases, we are going to have to revoke peoples ability to edit if they won't agree to disagree. The opposing party, the Minority should have the right, then to state his position (provided it doesn't break ethical or moral policy).--Jolie 17:14, 5 November 2008 (UTC)

Points to ponder? Under what circumstances does a user have the right to keep opposing viewpoints off his edited document, project, etc. What if one side won't coorperate/ be polite/ or is inconsistant in his argument?--Jolie 17:30, 5 November 2008 (UTC)


text:
User:Eml4500.f08.RAMROD.A/MyContributionHW5

Virtual Displacement Terms
ŵ2x1 = virtual axial displacement, with corresponding q(e)2x1

w4x1 = virtual axial displacement in global coordinates system, with corresponding d(e)4x1

Equation (5)
$$ (T^{(e)}w)\cdot [\hat{K}^{(e)}(T^{(e)}d^{(e)})-P^{(e)}] = 0 $$   for all w4x1

We want to prove equation (5) using matrix identity equation (6):

Equation 6: $$ (\mathbf{AB})^T=\mathbf{B}^T\mathbf{A}^T $$

Homework Example to verify equation (6)
$$ A=\begin{bmatrix}1 & 2 & 3\\4 & 5 & 6 \end{bmatrix}, B=\begin{bmatrix}7 & 8 & 9\\1 & 2 & 3\\4 & 5 & 6\end{bmatrix} , B^T = \begin{bmatrix}7 & 1 & 4\\ 8 & 2 & 5\\ 9 & 3 & 6\end{bmatrix} , A^T=\begin{bmatrix}1 & 4\\ 2 & 5\\ 3 & 6\end{bmatrix} $$

$$ AB=\begin{bmatrix}1\cdot7+2\cdot1+3\cdot4 & 1\cdot8+2\cdot2+3\cdot5 & 1\cdot9+2\cdot3+3\cdot6 \\4\cdot7+5\cdot1+6\cdot4 & 4\cdot8+5\cdot2+6\cdot5 & 4\cdot9+5\cdot3+6\cdot6 \end{bmatrix} =\begin{bmatrix}21 & 27 & 33\\57 & 72 & 87\end{bmatrix} $$

$$(AB)^T = \begin{bmatrix}21 & 57\\27 & 72\\ 33 & 87\end{bmatrix} $$

$$ B^TA^T = \begin{bmatrix}21 & 57\\27 & 72\\ 33 & 87\end{bmatrix} $$

This proves Equation (6): $$ (\mathbf{AB})^T=\mathbf{B}^T\mathbf{A}^T $$

We can recall that $$ \underline{a}\cdot\underline{b} = \underline{a}^T\underline{b} $$, where a and b are nx1 and aT is 1xn

It follows that from Eq(1):

$$ (T^{(e)}w)^T[\hat{k}^{(e)}(T^{(e)}d^{(e)})-P^{(e)}] = 0 $$, for all w,

and using Eq (6) above, that:

$$ w^TT^{(e)T}[\hat{k}^{(e)}(T^{(e)}d^{(e)})-P^{(e)}] = 0 $$

and that:

$$ w\cdot[(T^{(e)T}\hat{k}^{(e)}T^{(e)})d^{(e)}-(T^{(e)T}P^{(e)})] = 0 $$, for all w4x1

Also:

$$ w\cdot[\hat{k}^{(e)}d^{(e)}-f^{(e)}] = 0 $$, for all w.

All the above proves:

$$ k^{(e)}_{4x4} = T^{(e)T}_{4x2}\hat{k}^{(e)}_{4x2}T^{(e)}_{2x4} $$

Principal of Virtual Work
Equation 2:

$$ \hat{w}\cdot(\hat{k}^{(e)}q^{(e)}-P^{(e)}) = 0 $$, for all w

Equation 3:

$$ q^{(e)}_{2x1}=T^{(e)}_{2x4}d^{(e)}_{4x1} $$

and finally equation 4:

$$ \hat{w} = T^{(e)}w $$

Problems of the Continuous Case (PDE's)
The new problems will contain varying parameters

Example of problem:

Elastic bar with varying cross section A(x), varying modulus E(x), subjective to varying axial load (distributed), a concentrated load and inertia force (dynamic).

note: load will be time dependent


text:
User:Eml4500.f08.ateam.mcnally/Week 7

Meeting 27 Wednesday October 29th 2008
Today’s lecture was a continuation of previous lecture and further understanding and proof of the Principle of Virtual Work. The following variables must be defined for the proof and will be used for rest of this section. The real axial displacement is $$q_{2x1}^{(e)}$$

$$\hat{W}_{2x1}$$ is the virtual axial displacement that corresponds to the axial displacement $$q_{2x1}^{(e)}$$.

$$W_{4x1}$$ is the virtual axial displacement in the global coordinate system that corresponds to the real displacements in global coordinates $$d^{(e)}_4x1$$.

The following equations correspond to the real axial and virtual axial displacements. Equation 3 $$ q_{2x1}^{(e)}=T_{2x4}^{(e)}d_{4x1}^{(e)} $$

Equation 4 $$ \hat{W}_{2x1}=T_{2x4}^{(e)}W_{4x1} $$

The following equation can then be obtained by replacing equations 3 and 4 into equation 2 given in lecture meeting 26

$$ \left( T^{(e)}W\right)\cdot \left[\hat{k}_{2x2}^{(e)}\left(T^{(e)}d^{(e)} \right)-P_{2x1}^{(e)} \right]=0 $$ for all $$ W_{4x1}$$

The Equation above can then be simplified and solved using principles of Matrix multiplication and Matrix transposes. See the following proof and example for further explanation.

The transpose of two matrices that are multiplied together is equal to the transpose of each individual matrix multiplied together.

Given: $$ \left(A_{pxq}B_{qxr} \right)^{T}=B_{rxq}^{T}A_{qxp}^{T} $$

A and B are given as the following matices

$$ A_{2x3} $$ and $$ B_{3x3} $$

$$ A=\begin{bmatrix} 1 & 2 & 3\\ 4 & 5 & 6 \end{bmatrix} $$

$$ B=\begin{bmatrix} 7 & 8 & 9\\ 1 & 2 & 3\\ 4 & 5 & 6 \end{bmatrix} $$

$$ AB=\begin{bmatrix} 1 & 2 & 3\\ 4 & 5 & 6 \end{bmatrix} \begin{bmatrix} 7 & 8 & 9\\ 1 & 2 & 3\\ 4 & 5 & 6 \end{bmatrix}=\begin{bmatrix} 21 & 27 & 33\\ 57 & 72 & 87 \end{bmatrix} $$

$$ \left( AB\right)^{T}=\begin{bmatrix} 21 & 27 & 33\\ 57 & 72 & 87 \end{bmatrix}^{T}=\begin{bmatrix} 21 & 57\\ 27 & 72\\ 33 & 87 \end{bmatrix} $$

$$ B^{T}A^{T}=\begin{bmatrix} 7 & 1 & 4\\ 8 & 2 & 5\\ 9 & 3 & 6 \end{bmatrix}\begin{bmatrix} 1 & 4\\ 2 & 5\\ 3 & 6 \end{bmatrix}=\begin{bmatrix} 21 & 57\\ 27 & 72\\ 33 & 87 \end{bmatrix} $$

Elimination of the dot product in equation 2 can conducted using the following priciple:

$$ a_{nx1}\cdot b_{nx1}=a_{1xn}^{T}b_{nx1} $$

This simplifies equation 2 into:

$$ \left( T^{(e)}W\right)^{T}\left[\hat{k}^{(e)}\left(T^{(e)}d^{(e)} \right)-P^{(e)} \right]=0_{1x1} $$

Which can be further simplified by using the fact that transpose of two matrices that are multiplied together is equal to the transpose of each individual matrix multiplied together resulting in:

$$ W^{T}T^{(e)T}\left[\hat{k}{(e)}\left(T^{(e)}d^{(e)} \right)-P^{(e)} \right]=0_{1x1} $$

By applying the reverse of the principle above the final manipulation of equation 2 becomes:

$$ W \cdot \left[\left(T^{(e)T}\hat{k}^{(e)}T^{(e)} \right)d^{(e)} - \left(T^{(e)T}P^{(e)} \right) \right]= 0_{1x1}$$ for all $$ W_{4x1} $$

By realizing that $$T^{(e)T}\hat{k}^{(e)}T^{(e)} $$ equals element stiffness matrix, $$k^{(e)}$$, and that $$T^{(e)T}p^{(e)} $$ equals to the element force matrix, $$f^{(e)}$$. This relation in equation 2 simplifies

$$ W \cdot \left[k^{(e)}d^{(e)}-f^{(e)}\right]= 0$$

Since this has been proven for all $$ W_{4x1}$$ in previous lectures the equation above reduces to the original force-displacement relation discussed in previous lectures about the Finite Element Method FEM

$$ k^{(e)}d^{(e)}= f^{(e)} $$

Proving the principle of virtual work.

This is final section for FEM method using the discrete case with matrices.

The next section will use partial differential equations for the non continuous case. The model motivation for this problem is an elastic bar with varying cross section area A(x), Young’s Modulus E(X) (such as composite materials) which are subjected to varying axial load distributions, concentrated load distributions, and inertia forces. See the image below and following lecture notes for further explanation.

title: User:Eml4500.f08.ateam.mcnally/Week 7


text: File:Fivebarprobsetup.png

title: File:Fivebarprobsetup.png


text: File:Fivebartrussplot.png

title: File:Fivebartrussplot.png


text:
The Resting Membrane Potential/Quiz

{What are the 3 main factors that produce and maintain the Membrane potential? - Donnan Equilibrium, Membrane selectivity, concentration of ions. + Membrane selectivity, Active transport, Donnan Equilibrium. - Membrane selectivity, Donnan Equilibrium, Excess negative charge. - Active transport, Excess negative charge, concentration of ions.
 * type=""}

{The Nernst equation is used to obtain an accurate value for Em? - TRUE. + FALSE.
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{What is the most imporatnt factor in setting the resting membrane potential? + K+ concentration gradient - Concentration of ion - Na+ permeability - Active transport
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{Using the Nernst equation calculate the Ek+ if the intracellular concentration is 410 mmol/l and the extracellular is 22mmol/l assuming that the temperature is 20oC. What is your answer? - +69 - -79 + -74 - +54
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{What would the units be for the product from the Nernst equation? + mV-millivolts - A-Amps - C-Coulombs - Ion/Sec
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title: The Resting Membrane Potential/Quiz


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User:Eml4500.f08.RAMROD.F/HW5

Mtg 26 Monday October 27
We are going to try to derive element stiffness matrix by Principle of Virtual Work.

$$K^{(e)}=\underline{T}^{(e)^\tau} \hat{\underline{K}}^{(e)}\underline{T}^{(e)}$$

Recalling the force displacement with axial degree of freedom $$\underline{q}^{(e)}$$

$$\hat{\underline{K}}^{(e)}_{2x2}\cdot \underline{q}^{(e)}_{2x1} = \underline{P}^{(e)}_{2x1}$$

$$\hat{\underline{K}}^{(e)}_{2x2}\cdot \underline{q}^{(e)}_{2x1} - \underline{P}^{(e)}_{2x1}= \mathbf{0}_{2x1}$$ Equation $$(1)$$

Principle of Virtual Work applied here: $$\hat{\mathbf{W}}\bullet \left(\hat{\underline{K}}^{(e)}_{2x2}\cdot \underline{q}^{(e)}_{2x1} - \underline{P}^{(e)}_{2x1}\right)= \mathbf{0}_{2x1}$$ Equation $$(2)$$

We showed equation $$(1) \Leftrightarrow (2)$$

Recall $$\mathbf{\underline{q}}^{(e)}= T^{(e)}\mathbf{\underline{d}}^{(e)}$$

Similarly $$\hat{\underline{W}}^{(e)}_{2x1}= \bar{T}^{(e)}_{2x4}\mathbf{\underline{W}}^{(e)}_{4x1}$$

FOR KEVIN

$$K^{(e)}=\underline{T}^{(e)^\tau} \hat{\underline{K}}^{(e)}\underline{T}^{(e)}= k^{(e)}\begin{bmatrix} l^{2^{(e)}} & lm^{(e)} & lmn^{(e)} & -l^{2^{(e)}} & -lm^{(e)} & -ln^{(e)}\\ ln^{(e)} & n^{2^{(e)}} & mn^{(e)} & -lm^{(e)} & -m^{2^{(e)}} & -mn^{(e)}\\ lmn^{(e)} & mn^{(e)} & n^{2^{(e)}} & -lmn^{(e)} & -mn^{(e)} & -n^{2^{(e)}}\\ -l^2^ & -lm^{(e)} & -lmn^{(e)} & l^{2^{(e)}} & lm^{(e)} & lm^{2^{(e)}}\\ -lm^{(e)} & -m^{2^{(e)}} & -mn^{(e)} & lm^{(e)} & m^{2^{(e)}} &mn^{(e)} \\ -ln^{(e)} & -mn^{(e)} & -n^{2^{(e)}} & lm^{2^{(e)}} & mn^{(e)} & n^{2^{(e)}} \end{bmatrix}$$

title: User:Eml4500.f08.RAMROD.F/HW5


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User:Eas4200c.f08.vqcrew.E/Homework5

Linear Stress-Strain Relations
The expressions giving the strain induced by a stress can be rewitten to allow for the deriviation of a second expression relating the stress induced for a given strain.
 * $$ \begin{bmatrix} \varepsilon_{11} \\ \varepsilon_{22} \\ \varepsilon_{33} \\ \varepsilon_{23} \\ \varepsilon_{13} \\ \varepsilon_{12} \end{bmatrix} = \begin{bmatrix} \frac{1}{E} & \frac{-\nu}{E} & \frac{-\nu}{E} & 0 & 0 & 0 \\ \frac{-\nu}{E} & \frac{1}{E} & \frac{-\nu}{E} & 0 & 0 & 0 \\ \frac{-\nu}{E} & \frac{-\nu}{E} & \frac{1}{E} & 0 & 0 & 0 \\ 0 & 0 & 0 & \frac{1}{2G} & 0 & 0 \\ 0 & 0 & 0 & 0 & \frac{1}{2G} & 0 \\ 0 & 0 & 0 & 0 & 0 & \frac{1}{2G} \end{bmatrix} \begin{bmatrix} \sigma_{11} \\ \sigma_{22} \\ \sigma_{33} \\ \sigma_{23} \\ \sigma_{13} \\ \sigma_{12} \end{bmatrix} $$

Becomes,

 * $$\left\{\varepsilon_{ij} \right\}_{6x1} = \begin{bmatrix}A_{3x3} & 0_{3x3} \\ 0_{3x3} & B_{3x3} \end{bmatrix}\left\{\sigma_{ij} \right\}_{6x1}$$

Where $$A_{3x3} = \begin{bmatrix} \frac{1}{E} & \frac{-\nu}{E} & \frac{-\nu}{E} \\ \frac{-\nu}{E} & \frac{1}{E} & \frac{-\nu}{E} \\ \frac{-\nu}{E} & \frac{-\nu}{E} & \frac{1}{E}\end{bmatrix}$$ and $$B_{3x3} = \begin{bmatrix}\frac{1}{2G} & 0 & 0 \\ 0 & \frac{1}{2G} & 0 \\ 0 & 0 & \frac{1}{2G}\end{bmatrix}$$

Solving for the stress tensor,

 * $$\left\{\sigma_{ij} \right\}_{6x1} = \begin{bmatrix}A^{-1}_{3x3} & 0_{3x3} \\ 0_{3x3} & B^{-1}_{3x3} \end{bmatrix}\left\{\varepsilon_{ij} \right\}_{6x1}$$

In order to verify this result, a new matrix is defined as

 * $$\begin{bmatrix}c\end{bmatrix}_{6x6} = \begin{bmatrix}A_{3x3} & 0_{3x3} \\ 0_{3x3} & B_{3x3} \end{bmatrix}$$ with inverse equal to $$\begin{bmatrix}c\end{bmatrix}^{-1}_{6x6} = \begin{bmatrix}A^{-1}_{3x3} & 0_{3x3} \\ 0_{3x3} & B^{-1}_{3x3} \end{bmatrix}$$

Due to the definition of the inverse, it is known that the multiplication of matrix c with its inverse will yield the identity matrix, I.

 * $$\begin{bmatrix}c\end{bmatrix}_{6x6}\begin{bmatrix}c\end{bmatrix}^{-1}_{6x6} = \begin{bmatrix}I\end{bmatrix}_{6x6}$$

 * $$\begin{bmatrix}c\end{bmatrix}\begin{bmatrix}c\end{bmatrix}^{-1} = \begin{bmatrix}AA^{-1} & 0 \\ 0 & BB^{-1} \end{bmatrix} = \begin{bmatrix}I\end{bmatrix}$$

Now, the stress-strain relation can be found using the inverted matrices, A and B.

 * $$\left\{\sigma_{ij} \right\} = \begin{bmatrix}A^{-1} & 0_{3x3} \\ 0 & B^{-1} \end{bmatrix}\left\{\varepsilon_{ij} \right\} = \begin{bmatrix} \frac{E(v-1)}{(2v^2+v-1)} & \frac{-Ev}{(2v^2+v-1)} & \frac{-Ev}{(2v^2+v-1)} & 0 & 0 & 0\\ \frac{-Ev}{(2v^2+v-1)} & \frac{E(v-1)}{(2v^2+v-1)}  & \frac{-Ev}{(2v^2+v-1)} & 0 & 0 & 0\\ \frac{-Ev}{(2v^2+v-1)} & \frac{-Ev}{(2v^2+v-1)} & \frac{E(v-1)}{(2v^2+v-1)}  & 0 & 0 & 0\\ 0 & 0 & 0 & \frac{1}{2G} & 0 & 0\\ 0 & 0 & 0 & 0 & \frac{1}{2G} & 0\\ 0 & 0 & 0 & 0 & 0 & \frac{1}{2G} \end{bmatrix}\begin{Bmatrix}0 \\ 0 \\ 0 \\ 0 \\ \varepsilon_{31} \\ \varepsilon_{12} \end{Bmatrix} \;$$

title: User:Eas4200c.f08.vqcrew.E/Homework5


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User:EML4500.F08.RAMROD.B/Homework 5.1

=3-D Truss Relationships=

This process is very similar to that of the 2-D case. Therefore we will be taking the same steps to derive the Force-Displacement relationships. The elemental axial force displacement relationship is identical to the 2-D case and is shown below.

$$\displaystyle (k^e)(q^e)=p^e$$

Since the 3-D case will have three displacements at each node, the element displacement matrix will be a six by one (6x1) as will the element force matrix. These matrices are shown below.

 $$\displaystyle d^e = \begin{Bmatrix}d_1^e\\ d_2^e\\ d_3^e\\ d_4^e\\ d_5^e\\ d_6^e \end{Bmatrix}$$ $$\displaystyle f^e = \begin{Bmatrix}f_1^e\\ f_2^e\\ f_3^e\\ f_4^e\\ f_5^e\\ f_6^e \end{Bmatrix}$$

For the 3-D case the two by six (2x6) transformation matrix $$T^e$$ that relates the axial degrees of freedom to the global degrees of freedom is shown below.

 $$\displaystyle T^e = \begin{bmatrix} l^e & m^e & n^e & 0 & 0 & 0 \\ 0 & 0 & 0 & l^e & m^e & n^e \end{bmatrix}$$

This transformation matrix can be used to relate the axial and global degrees of freedom and forces as shown below.

$$\displaystyle \begin{bmatrix}l^e & m^e & n^e & 0 & 0 & 0 \\ 0 & 0 & 0 & l^e & m^e & n^e\end{bmatrix} \begin{Bmatrix}d_1\\ d_2\\ d_3\\ d_4\\ d_5\\ d_6\end{Bmatrix}=\begin{Bmatrix}q_1\\ q_2\end{Bmatrix}$$

It can also be used to relate the global forces to the local axial forces. This is shown below.

$$\displaystyle \begin{bmatrix}l^e & m^e & n^e & 0 & 0 & 0 \\ 0 & 0 & 0 & l^e & m^e & n^e\end{bmatrix} \begin{Bmatrix}f_1\\ f_2\\ f_3\\ f_4\\ f_5\\ f_6\end{Bmatrix}=\begin{Bmatrix}p_1\\ p_2\end{Bmatrix}$$

title: User:EML4500.F08.RAMROD.B/Homework 5.1


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Amateur astronomy

Amateur Astronomy is astronomy done by non-professionals who either recreationally observe or hope to contribute to current scientific research.

Amateur astronomy is applied astronomy and involves observing celestial objects. Amateur astronomy can be further divided by the complexity of the tools that are utilized. It is a misconception that serious observers must use expensive, complex tools and that only such observers can contribute to scientific research. Amateurs however must use the appropriate tool, however if they hope to advance professional research. With some areas of research, amateurs are now entirely unable to assist professionals as they cannot afford the appropriate tool. (With the exception of an amateur observing some kind of very unusual rare event, like the amateur that recently found a new nebula. )

What is the Appeal?
To look up in the sky on a dark night and see the immensity of space and the intricate glow of stars and clouds in the nearly incomprehensibly distant reaches of space is a special and wonderful experience.

As an amateur astronomer I still find much joy in simply looking at the stars. As I have had the opportunity to show people the sky and its beautiful sights, I would say that nearly 1/2 of all people have little interest in the night sky and probably wouldn't want to observe through a telescope.

Of the half left
 * 2/10 are distinctly let down by any observing experience.
 * 5/10 (of half) are people that find the view interesting, but do not find the sight mesmerizing
 * 2/10 are intrigued and/or mesmerized. they welcome/ask for other things to observe.
 * 1/10 people are so interested in what they see that you have problem getting them away from the telescope.

Some advice is to learn about what your looking at and carefully observe for subtle features. Are there voids, dark spots, whats its shape? Bring your imagination into play a little. Remember looking for shapes in the clouds when you were a kid? whats the shape look like to you. Sometimes these blobs are a little like art; it takes some whimsy to see the mastery of their design. Do NOT expect showy, eye catching images such as the photos taken from the Hubble or even masterful images taken from someone with a 2,000$ CCD camera. Images with appear generally in black and white and might appear blurry, smoky, or turbulent (although I caution the user that the stars should be small bright pinpricks. Blurry, smoky or distorted stars means either the scope is unfocused or has optical flaws.)

Divisions of Amateur Astronomy

 * Stargazing. Observations of objects in space done with the unaided eye.  Few of these observations are of assistance to professional astronomers.

 * Backyard Astronomy. Visual observations of objects using telescopic instrumentation.  Despite the wild popularity and definite increases in performance with imaging, there is still solid scientific contribution in ordinary visual backyard astronomy.

 * Astronomical imaging. Utilizing telescoping instrumentation to create an image.   The most serious observers have automated telescopes,and images nearly more expansive than their telescope.

 * Internet astronomy. Innovations of imaging, internet and automated instrument have created an oddity, an astronomical observer whom never observes (that is goes out sets up equipment,and looks at the stars). Observers can now record astronomical information with no knowledge of the stars and with little aesthetic appreciation of the night sky.

What can you observe?
Although you can observe many different things you need to bear in mind that some objects require special equipment to observe them well. I will link some popular targets and recommended equipment.

Stargazing; These objects are often can be seen with the unaided eyes
 * aurorae. Ionized gas that collect near the earth magnetic pole.  Can be seen with the unaided eye, Wide-Angle  astrophotography is useful
 * Conjunction_(astronomy_and_astrology). Alignments of the planets can be very pleasant to watch and notice, again many people enjoy photographing them Landscape_photography
 * satellite. are man-made objects that orbit the earth. They can appear very bright. (as an example one kind of satellite have giant solar panels that can reflect the sunlight shortly after sunset creating an  satellite_flare)
 * meteor or meteor showers. Small asteroids; rocks, and dust that enter the atmosphere create bright fiery streaks in the skies (popularly called falling stars). Meteors and meteor showers are best seen by the unaided eye or captured using wide angle  astrophotography

Backyard astronomy; These objects are best seen using a telescope or other special equipment
 * comets A icy object that that orbits the sun, while occasionally comets can be seen by the unaided eye.  Normally there are several comets in the sky that can be seen with a telescope.  Often these comets do not have an observable tail.
 * galaxy A large group of stars and gas.  Galaxies are some of the distant sights easily seen in a telescope.
 * nebula A cloud of dust and gas.  If the cloud is thick enough to reflect starlight we see a reflection nebula.  If the cloud has stars near its core that ionize the cloud, then the cloud becomes  fluorescent.  This is called a emission nebula.
 * open_cluster A gravitational bound group of stars.  Stars are normally created in small groups that condense from a parent nebula.  Open clusters often showing whimsical designs and great diversity.
 * globular_cluster A tight massive groups of many stars. these groups are normally tight accumulations of thousands to millions of stars.
 * double_star are apparently single stars that when observed in a telescope consist of two stars. They have interesting color contrast.
 * planets of the solar system of the 7 planets of the solar system; three will show detail in a modest telescope; ( Jupiter, Saturn, Mars) although mars will show details for a short while every 2 years (near its opposition).
 * moon shows intricate in any telescope and is a very good target for telescopic observation.
 * sun Using special adapters you can alter a telescope to be able to view the sun.
 * disclaimer it is potentially very dangerous to observe the sun in any telescope, any observer must carefully ensure that his equipment will shield his eyes from blinding amounts of light.

Imaging; Some objects are only well observed when imaged, either photographically or electronically
 * variable_stars Stars can at times in their life cycle become variable. It is useful for science to monitor and describe long term trends on any variable stars.  Even Amateurs without Imagining equipment can contribute to scientific knowledge by measuring the star brightness
 * variable_stars Stars can at times in their life cycle become variable. It is useful to science to monitor and describe long term trends on any variable stars.  Of course the most precise work is done using a CCD.
 * supernova Supernova are large explosions of a star that occurs to some stars as then reach the end of the life cycle.
 * asteroids Asteroids are large rocks in space. Imaging the sky can detect new asteroids and refine orbital information.  Scientists carefully check the skies for  potentially_hazardous_asteroid that might strike the earth.
 * See also Earth-impact_events

title: Amateur astronomy


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User talk:ShakespeareFan00/Sfan00 IMG

See also User talk:ShakespeareFan00

 Hello Sfan00 IMG, 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 everything on 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! --mikeu talk 20:55, 5 November 2008 (UTC)

Well, I found the midi file at http://www.digitalhymnal.org/dhymn.cfm?hymnNumber=85. I don't know if you meant playing that on the piano for wikiversity; it could be good. Or maybe you mean creating a midi for playing at Wiki Campus Radio. Maybe I score it using Noteworthy composer evaluation program because I guess that midi is not royalty free. --Davichito 20:09, 6 December 2008 (UTC)

bees
good background article --JWSchmidt 13:37, 20 April 2009 (UTC)

 * Thanks :) - I am not sure what the cite for the French paper is as it was only mentioned as such on the TV Broadcast item,

not in the BBC websites notes.. Sfan00 IMG 14:14, 20 April 2009 (UTC)

tracking virus source
I looked at the Wikinews article. I'm not sure that there is really much of a story there. It sounds like people will be investigating and looking for old samples that are positive for this new virus. Until those results are obtained, all the speculation does not mean much. --JWSchmidt 16:06, 5 May 2009 (UTC)

Note from a yank
hi, there ;) I've been thinking of starting something on the more general topic of direct democracy.  The Great Repeal Bill would make an interesting case study of using wiki participation for public policy.  Although I don't have a vested interest in the outcome of the Bill, I really think there is much that can be learned from this project.  I'll be heading out of town tomorrow, but I'll give some thought to this when I get back.  --mikeu talk 21:03, 30 July 2009 (UTC)

title: User talk:ShakespeareFan00/Sfan00 IMG


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User:Justinwhitaker~enwikiversity

Hi there.

I am Justin Whitaker, a Ph.D. student at Goldsmiths University of London with Damien Keown focusing on Buddhist Ethics.

I am from Montana, USA, and currently work in Missoula for The University of Montana at the Center for Ethics. I have a BA in Philosophy from UM-Missoula, an MA in Buddhist Ethics from Bristol University, UK, and plan to teach Buddhist Studies "when I grow up." Wish me luck. Along the way I hope to contribute to the Buddhism/Buddhist Ethics area here. I have also studied and taught Tibetan Buddhism, so I might lurk around there as well.

title: User:Justinwhitaker~enwikiversity