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User:Eml4500.f08.delta 6.krueger/Notes Monday October 27th

Principle of Virtual Work (PVW)

First the weighting coefficient needs to be defined, or reinforced, and an example needs to be given to emphasize its use and importance. We will let w equal the weighting coefficient.

w = weighting coefficient

Now an Example:

Lets look at the calculation of a final course grade. Where: Exam 1 = 82 and is worth 15 % of final grade Exam 2 = 74 and is worth 20 % of final grade Exam 3 = 92 and is worth 25 % of final grade Hw grade = 87 and is worth 40 % of final grade

The formula to calculate the final grade is given as:

$$ finalgrade = w_0(Hw grade) + \sum_{i=1}^{3}w_iExam_i$$

Where $$ w_i$$ and $$ w_0$$ are the exam weighting coefficients and the homework(Hw) weighting coefficient. Now plugging into the above formula:

$$ finalgrade = .4(87) + .15(82) + .2(74) + .25(92) = 84.9$$

Now to emphasize how the weighting coefficients will change the final value, we'll let the Hw grade and all the exam grades remain the same but change the worth to the following: Exam 1 worth 30 % Exam 2 worth 30 % Exam 3 worth 30 % Hw grade worth 10 %

The final grade now becomes:

$$ finalgrade = .1(87) + .3(82) + .3(74) + .3(92) = 83.1$$

This shows the importance of the weighting coefficients. The importance of a value, within an equation, can be changed by changing the weighting coefficient of that value. The final value will also be changed, as shown above, by changing the weighting coefficients.

Now we will return to the Principle of Virtual Work. Recall: $$K^{(e)}= T^{(e)^(T)} * \hat{K}^{(e)} * T^{(e)}$$

Where $$K^{(e)}$$ is a 4x4 matrix, $$T^{(e)^(T)}$$ is a 4x2 matrix, $$\hat{K}^{(e)}$$ is a 2x2 matrix and $$T^{(e)}$$ is a 2x4 matrix.

Question: How do we use the Principle of Virtual Work?

First Recall the Force Displacement Relationship (FD Rel) with respect to the axial degrees of freedom ($$q^{(e)}$$):

Eqn (1)$$\hat{K}^{(e)}q^{(e)} = p^{(e)}$$

Take note that the matrix p is lower case. This implies it is in reference to the element not the global coordinate system.

We then move the p matrix to the left hand side of the equation. Equation (1) is now in the same form as the FD Rel $$Kd-F=0$$.

$$\hat{K}^{(e)}q^{(e)} - p^{(e)}= 0$$

The equation is now multiplied by the weighting coefficient, which is a 2x2 matrix, to give the Principle of Virtual Work:

Eqn (2)$$w\cdot(\hat{K}^{(e)}q^{(e)} - p^{(e)})= 0$$

This holds true for all weighting coefficient matrices. In previous notes and home work we have shown that $$ Eqn(1) \Leftrightarrow Eqn(2)$$. We can ow exploit the Principle of Virtual Work in order to get expressions in the global coordinate system. Recall: Eqn (3)$$q^{(e)}=T^{(e)}d^{(e)}$$ Where $$q^{(e)}$$ is the real displacement and is a 2x1 matrix. $$T^{(e)}$$ is a 2x4 matrix and $$d^{(e)}$$ is a 4x1 matrix. $$w$$ corresponds to the axial degrees of freedom. Since we are using virtual work the axial weighting coefficient needs to be converted into the virtual weighting coefficient. This is done using Eqn (4) below: Eqn (4)$$\hat{w}=T^{(e)}w$$ Where $$\hat{w}$$ is a 2x1 matrix, $$T^{(e)}$$ is a 2x4 matrix and $$w$$ is a 4x1 matrix.


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User:Eml4500.f08.gravy.mmm/HW5

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 5: Select 'W' such that W5=1, W1=W2=W3=W4=W6=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 Sean
There was a bug in the code solving the 2-bar truss solved in and earier report inserst link. The code was orgnialy written for a 6-bar truss and was modified for the 2-bar truss. Unfornualy 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.

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)

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

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

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

title: User:Eml4500.f08.gravy.mmm/HW5


text:
Biomembrane Transport

 * Introduction to Biomembranes
 * Diffusion and Osmosis
 * The Resting Membrane Potential
 * Action potentials
 * Transporter Kinetics

title: Biomembrane Transport


text:
User:Crux~enwikiversity

Add

= My sandbox pages=
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title: User:Crux~enwikiversity


text: File:HW 5 mtg 28.JPG

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User:EAS4200C.F08.WIKI.A/Homework5

=Completing the Stress Strain Relation =

kinematic assumption drop down

4 zero strain components

5 zero stress components

$$\varepsilon -\sigma $$ relation in two forms (Tensorial 23-2 or Engineering 23-1b)

Rewrite $$\varepsilon -\sigma $$ relation: Full (not diagonal)

$$\begin{Bmatrix} \varepsilon _{i,j} \end{Bmatrix}_{6x1}= \begin{vmatrix} \mathbf{A}_{3x3} & \mathbf{0}_{3x3}\\ \mathbf{0}_{3x3} & \mathbf{B}_{3x3} \end{vmatrix} \begin{Bmatrix} \sigma _{i,j} \end{Bmatrix}_{6x1}$$

$$\varepsilon -\sigma $$ relation

$$\begin{Bmatrix} \sigma _{i,j} \end{Bmatrix}= \begin{vmatrix} \mathbf{A}_{3x3}^{-1} & \mathbf{0}_{3x3}\\ \mathbf{0}_{3x3} & \mathbf{B}_{3x3}^{-1} \end{vmatrix} \begin{Bmatrix} \varepsilon _{i,j} \end{Bmatrix}$$

$$\mathbf{C}_{6x6}= \begin{vmatrix} \mathbf{A}_{3x3} & \mathbf{0}_{3x3}\\ \mathbf{0}_{3x3} & \mathbf{B}_{3x3} \end{vmatrix}$$

$$ \mathbf{C}_{6x6}\cdot \mathbf{C}^{-1}_{6x6}=\mathbf{I}_{6x6}$$

$$\mathbf{C}_{6x6}\cdot \mathbf{C}^{-1}_{6x6}=\begin{vmatrix} \mathbf{A_{3x3}A}_{3x3}^{-1} & \mathbf{0}_{3x3}\\ \mathbf{0}_{3x3} & \mathbf{B_{3x3}B}_{3x3}^{-1} \end{vmatrix}=\mathbf{I_{6x6}} $$

We are trying to prove that $$ \sigma _{xx}=\sigma _{yy}=\sigma _{zz}=\gamma _{yz}=0$$

$$\begin{Bmatrix} \sigma _{i,j} \end{Bmatrix}=

\begin{vmatrix} \mathbf{A}_{3x3}^{-1} & \mathbf{0}_{3x3}\\ \mathbf{0}_{3x3} & \mathbf{B}_{3x3}^{-1} \end{vmatrix}\begin{Bmatrix} 0\\ 0\\ 0\\ 0\\ \varepsilon _{31}\\ \varepsilon _{12} \end{Bmatrix}$$ $$\begin{matrix} \sigma _{11}=0\\ \sigma _{22}=0\\ \sigma _{33}=0\\ \sigma _{23}=2G\varepsilon _{23}\\ \sigma _{31}=2G\varepsilon _{31}\\ \sigma _{21}=2G\varepsilon _{12} \end{matrix}$$

Consider the 1D stress model

=Bidirectional Bending Recipe=

The following is a general recipe of equations used for solving a bidirectional bending problem.

The moment in the y-direction is

 $$\displaystyle M_y = \int_A z \sigma_{xx} dA $$

Similarly, the moment in the z-direction is

 $$\displaystyle M_z = \int_A y \sigma_{xx} dA $$

Where $$\displaystyle dA = dydz $$

The moment of intertia in the y-direction is

 $$\displaystyle I_y = \int_A z^2 dA $$

Similarly, the moment of intertia in the z-direction is

 $$\displaystyle I_z = \int_A y^2 dA $$

 $$\displaystyle I_{yz} = \int_A yz dA $$

The stress in the x-direction is

 $$\displaystyle \sigma_{xx} = E \epsilon_{xx} = {I_y M_z - I_{yz} M_y \over I_y I_z - (I_{yz})^2} y + {I_z M_y - I_{yz} M_z \over I_y I_z - (I_{yz})^2} z $$

Recall that

 $$\displaystyle I = \begin{bmatrix} I_{11} & I_{12} & I_{13} \\ I_{21} & I_{22} & I_{23} \\ I_{31} & I_{32} & I_{33} \end{bmatrix}$$

Note that the determinat is defined as

 $$\displaystyle D := I_y I_z - (I_{yz})^2 = I_{22} I_{33} - (I_{23})^2 $$

for $$\displaystyle \begin{bmatrix} I_{22} & I_{23} \\ I_{32} & I_{33} \end{bmatrix}$$

 $$\displaystyle I_{32} = I_{zy} = \int_A zy dA = I_{yz} = I_{23} $$

At the neutral axis, or where $$\displaystyle \sigma_{xx} = 0 $$

 $$\displaystyle \sigma_{xx} = m_y y + m_z z $$

Thus,

 $$\displaystyle z = (- {m_y \over m_z})y = (\tan \beta) y $$

=Elastic Bar Analysis Continued=

We will now continue the equation of equilibrium in terms of $$\sigma$$. The goal is to obtain the following formula.

 $$\displaystyle \frac{\partial \sigma_{yx}}{\partial y} + \frac{\partial \sigma _{zx}}{\partial z} = 0$$

In indicial notation the equation can be represented as follows;

 $$\displaystyle \frac{\partial \sigma_{21}}{\partial x_2} + \frac{\partial \sigma _{31}}{\partial x_3} = 0$$

Now we must recall equation 2.21 from the text which is the above equation but with the term $$\displaystyle \frac{\partial \sigma_{xx}}{\partial x}$$ added in.

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

This equation represented in indicial notation is represented as the following

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

All three cases ($$j= 1,2,3$$) are shown below.

$$\displaystyle j=1, \frac{\partial \sigma_{11}}{\partial x_1} +  \frac{\partial \sigma_{21}}{\partial x_2} + \frac{\partial \sigma_{31}}{\partial x_3}$$

$$\displaystyle j=2, \frac{\partial \sigma_{12}}{\partial x_1}  + \frac{\partial \sigma_{22}}{\partial x_2}  +  \frac{\partial \sigma_{32}}{\partial x_3}$$

$$\displaystyle j=3, \frac{\partial \sigma_{13}}{\partial x_1}  +  \frac{\partial \sigma_{23}}{\partial x_2}  +  \frac{\partial \sigma_{33}}{\partial x_3}$$

Noting that there are four zer stress components that come from the stress-strain relation, these equations reduce to the following.

$$\displaystyle 0=\sigma_{11}=\sigma_{22}=\sigma_{33}=\sigma_{23}$$

$$\displaystyle j=1, \frac{\partial \sigma_{21}}{\partial x_2} + \frac{\partial \sigma_{31}}{\partial x_3}$$

$$\displaystyle j=2, \frac{\partial \sigma_{12}}{\partial x_1} +  \frac{\partial \sigma_{32}}{\partial x_3}$$

$$\displaystyle j=3, \frac{\partial \sigma_{13}}{\partial x_1} + \frac{\partial \sigma_{23}}{\partial x_2}$$

We will now derive the equation of equilibrium in terms of $$\sigma$$. To do so we will call upon the one dimentional model problem that we previously looked at. For convenience it is depicted below.

[\Image:]

First we must sum the forces in the x direction. doing so results in the following equation.

$$\sum{F_x} = 0 = -\sigma_{x}A + \sigma(x+dx)A + f(x)dx = A\left[\sigma(x+dx) - \sigma(x) \right] + f(x)dx$$

Taking the Taylor Series expansion of the part of the equation in the brackets (mutliplied by A) and ignoring the higher order terms yields the following.

 $$\displaystyle f(x+dx) = f(x) + \frac{df(x)}{dx}dx + \frac{1}{2}\frac{d^2f}{dx^2}xdx^2....$$    Taylor Series Expansion

$$\displaystyle \frac{d\sigma}{dx} + \frac{f(x)}{A} = 0....$$   Result Ignoring Higher Order Terms

 * A is the applied load - Body force = force/volume
 * f(x) = force/length

Now, non-uniform stress field is in 3-D, but without an applied load, focusing on the x-direction only (i.e. without the other stress component to avoid cluttering the figure)

Below is an image of shear on a 3-D cube :

Note: the first part of the superscript refers to the normal direction of the cube face (facet), and the second part refers to the direction of shear.

$$\sum{F_x} = {\color{red}dydz[-\sigma_{xx}(x,y,z) + \sigma_{xx}(x+dx,y,z)]} + {\color{blue}dzdx[-\sigma_{yx}(x,y,z) + \sigma_{yx}(x,y+dy,z)]} + {\color{green}dxdy[-\sigma_{zx}(x,y,z) + \sigma{zx}(x,y,z+dz)]}$$
 * The red term has facets normal to the x-axis, the blue has facets normal to the y-axis, and the green has facets normal to the z-axis.

$$0 = $$( dxdydz )$$[\frac{d\sigma_{xx}}{dx} + \frac{d\sigma_{yx}}{dy} + \frac{d\sigma_{zx}}{dz}]$$

This ends Step C of the Road Map.

=Matlab Analysis=

This picture represents the NACA airfoil with stringers attached. The analysis is done below.

Matlab Code
The Matlab Code used in the following analysis is provided below.

'Part 1' is commented out when doing the analysis for 'Part 2' and vice versa.

Stringer Analysis
The above shows the diagram of the NACA2415 airoil with only the stringers considered in the analysis and ignoring the contributions of the skin. The important results are given below.

I_22 =

3.5173e-005

I_23 =

9.5347e-007

I_33 =

6.6954e-007

beta =

1.5010

maxsigmaxx_ylocation =

0.4999

maxsigmaxx_zlocation =

-7.8576e-004

maxsigma_xx =

4.1620e+008

M_zult =

2.1625e+005

M_yult =

-8.6498e+004

Full NACA2415 Airfoil Analysis
The diagram above shows the centroid location and neutral axis of the airfoil when the skin is considered along with the stringers. The important results are as follows:

I_22combined =

3.5628e-005

I_23combined =

7.9850e-006

I_33combined =

1.2106e-004

beta =

0.1771

maxsigmaxx_ylocation =

0.4999

maxsigmaxx_zlocation =

-7.8576e-004

maxsigma_xx =

4.1620e+008

A close up of the difference in the centroids of the stringers and the total airfoil is provided below for easier analysis.

As seen, the location and magnitude of the max normal stress barely changes, which is expected due to the centroid barely changing with the addition of the skin to the analysis.

The highest bending stress is in a stringer occurs in stringer H due to its distance from the neutral axis. It's value is

maxsigma_xx1 =

3.3823e+008

This causes the ultimate bending moments experienced by the stringers to be:

M_zult =

3.1349e+003

M_yult =

-1.2539e+003

=Individual Contributions=

Christopher Pergola EAS4200C.F08.WIKI.A 20:06, 7 November 2008 (UTC)

Braden Barnes Eas4200c.f08.wiki.b 17:53, 4 November 2008 (UTC)

Eric Viale Eas4200c.f08.wiki.c 17:19, 7 November 2008 (UTC)

Jeff Huenink Eas4200c.f08.wiki.f 19:30, 7 November 2008 (UTC)

Chris Fontana Eas4200c.f08.WIKI.E 20:33, 7 November 2008 (UTC)

Michael Rolle Eas4200c.f08.wiki.d 21:36, 7 November 2008 (UTC)

title: User:EAS4200C.F08.WIKI.A/Homework5


text:
User:Eas4200c.f08.ZYX/HW Week 9-10

 Note: No matlab code problem. Team coordinator and another team member missing (dropped?); HW5 report submitted late. Team to be disbanded?

Please don't remove this comment in case you amend your work; just add a comment box to say what you did. Eml4500.f08 15:02, 11 November 2008 (UTC)

title: User:Eas4200c.f08.ZYX/HW Week 9-10


text:
User:EML4500.F08.RAMROD.B/Homework 5

The following is a FBD of an infinitesimal section area of the elastic bar with varying properties.

The bottom force that is not labeled in the picture can be represented as the following equation;

$$\displaystyle A(x) \partial x \rho(x) \frac{\partial^2 u}{\partial t^2}$$

Within this equation $$ A(x) \partial x \rho(x)$$ is equal to $$\displaystyle m(x)$$ which has the units of $$ \frac{mass}{length}$$. Now we proceed to sum the forces in the x-direction and the results are as follows;

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

where $$\ddot{u}$$ is equal to the acceleration as given in Newton's notation. Now we perform a Taylor Series expansion on this equation but we will be ignoring the higher order terms. This is performed below.

$$\displaystyle \sum{F_x}= \frac{\partial N}{\partial x}(x,t) + (Higher Order Terms ) + f(x,t) - m(x)\ddot{u}$$  $$\displaystyle f(x+dx) = f(x) + \frac{df}{dx}(x)dx + \frac{1}{2}\frac{d^2f}{dx^2} +.....$$  $$\displaystyle \frac{\partial N}{\partial x} + f = m\ddot{u}$$

The second equation is part of the expansion while the last equation is known as the equation of motion for the elastic bar which ignores the higher order terms of the expansion. The equation of motion is otherwise abbreviated as $$\displaystyle EOM$$. An expression for the force $$N(x,t)$$ can be obtained and the results are found below.

$$\displaystyle N(x,t) = A(x)\sigma(x) = A(x) E(x) \varepsilon (x,t) = A(x) E(x) \frac{\partial u}{\partial x}(x,t)$$

Putting this expression into the $$EOM$$ yields the partial differential equation of motion (PDE of Motion) for the elastic bar.

$$\displaystyle \frac{\partial }{\partial x}\left[ A(x) E(x) \frac{\partial u}{\partial x}\right] + f(x,t) = m(x) \ddot{u}$$

In order to solve this equation two boundary conditions along with two initial conditions are needed. The two initial conditions are initial displacement and initial velocity. For the boundary conditions, two cases need to be considered. They are depicted below.

For case a, the two boundary conditions are as follows;

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

$$\displaystyle \frac{\partial u}{\partial x}(L,t) = \frac{F(t)}{A(L)E(L)}$$

For case two the boundary conditions are as follows;

$$\displaystyle u(0,t) = o$$

$$\displaystyle N(L,t) = F(t)$$

The initial conditions are specific to different problems

title: User:EML4500.F08.RAMROD.B/Homework 5


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Skygazing

Skygazing, or more literally gazing at the sky, is an activity for leisure or with an interest in amateur astronomy. Astronomical observations are generally made with the naked eye or with basic optical aids. Simple naked-eye observations of the sky can reveal a great deal about the basics of astronomy and give a better understanding of the cosmos, while instruments, such as telescopes, are used to study deep space. Many different celestial objects can be viewed while skygazing during both night and daytime.

Diurnal (daytime) observation
The extreme brightness of the Sun saturates the sky during the day and prevents naked eye observation of less luminous objects, with the exception of the Moon and occasionally Venus.

Looking directly at the Sun can be very damaging for your eyes. Not even sunglasses will prevent eye damage, so special filtered glasses fitted with materials such as metalized PET film are essential. These can also be used to observe solar eclipses. Projecting an image of the Sun on to a surface using a piece of card with a pinhole can also be very effective, and safe. NEVER look directly at the Sun through a telescope or binoculars as this will cause instant and permanent blindness. Beware that some solar filters supplied with cheaper telescopes are not safe enough. Only filters clearly identified as complying with current safety standards should be used. and

Eclipses

 * /Solar eclipse lab on a sunny day/

Sunspots
Sunspots are difficult to see with the naked eye but can be safely viewed using projection as described above. They are effectively solar disturbances on the Sun's surface and can change over time as conditions within the Sun change, somewhat like storms on the Sun. By observing the Sun over time the appearance of sunspots changes due to both the Sun's rotation and each storm's evolution.

Aurora Viewing
Viewing Aurorae is a typical sky watching activity especially at higher magnetic latitudes.

Cometary Viewing
Comets can be easy to locate and photograph and an excellent reason to sky watch.

Atmospheric Optical Phenomena
Phenomena linked to the Sun and Moon: Other interesting observations in relation to the Sun and Moon (even Venus at times) can be made, although their observation depends on the particular atmospheric conditions. These can be easier to observe since they do not always require any eye protection.

 * Solar halo
 * A phenomenon that can be seen mainly in winter and at altitude, it takes the form of a big luminous club, a little extended in its width and centered on the Sun. It is produced by the refraction of the solar rays through a fine and uniform layer of high altitude clouds, the cirrostratus. Often requires eye protection, as the Sun is within the field of view.

 * Rainbow
 * Visible during or after a rainy period with a partially clear sky, it is a bow of white light refracted into all the visible colures of the spectrum by solar rays passing through raindrops. In optimal conditions, a second less luminous bow, with reversed colures, can be observed (see picture at right), the space between the two being slightly darker than the rest of the sky.

 * Parhelions (sun dogs)
 * Produced like the solar halo but by different clouds and occurring more frequently, these are two luminous spots, often in diffracted colors like in a rainbow, situated on both sides of the Sun at similar distances very close to the club of the associated halo.

Nocturnal (nighttime) observation
With good viewing conditions, during a dark and moonless night, it is possible to distinguish about 3000 stars in the celestial sphere.

Moon

The queen of the night - it is our only natural satellite. Naked eye observation allows us to track its changes over the monthly cycle.

 * Phases of the Moon
 * The following diagram explains how the Moon's phases are produced by the relative positions of the Moon, the Earth and the Sun in space. [[Image: TerreOrbiteLuneEtPhases.png|thumb|center|500px|Orbit of the Moon and phases seen from the Earth]]
 * Its light coming only from the reflection of the Sun's light on its surface, the Moon will present the appearance a thin crescent at twilight or at dawn when it will be located between the Earth and the Sun; a half disc for half of the night when it will be the same distance from the Sun as our planet and finally a complete disc all night long when it will be opposite the Sun in relation to the Earth. The sight of the setting crescent Moon against blue sky at twilight is well worth staying up late for.
 * The effect of the path of its illuminating solar rays can also be seen: in its first rising phase or its final setting phase, when it is only a crescent, one can often see the dark part of its disk weakly illuminated, permitting us to distinguish the form of the complete disc. This is due to the solar rays falling on the Earth reflecting toward the Moon and illuminating the side that is in darkness. This long route makes for a weak light reaching us, but it is sufficient to distinguish it.

 * Lunar craters
 * These are pits or depressions in the surface of the Moon, produced by great impacts of gigantic meteoroids which mostly took place billions of years ago. They range in size from huge walled plains more than a hundred miles across to microscopic pits. The smallest craters which can be glimpsed through ordinary binoculars are about twenty miles across. These craters are most common in the light-colored Lunar highlands. They are named after historical figures, mostly scientists.

 * Lunar Maria (seas)

 * Of different and darker composition than the rest of the surface, the Maria (singular "mare", Latin for sea) are composed of basalt. These flat areas of ancient frozen lava form the familiar pattern of dark spots which can be seen with the naked eye. They are given the names of fanciful bodies of water since early observers believed them to be literal seas. Charts of the Moon are available to aid the observer in identifying lunar features.

 * Lunar Eclipses
 * Following the same principle as the solar eclipses, the lunar eclipses only take place when the Moon is full and that the Earth is positioned between the Moon and the Sun. The diameter of the shadow of our planet being a lot bigger than the one of our satellite, these take place more frequently and have the same appearance wherever the viewer is located on Earth. At the time of the total phase, the Moon remains visible and has an orange color that is due to refracted solar rays tinted by the terrestrial atmosphere.

 * Lunar halo
 * Invoked by the same meteorological phenomenon as the solar halo, this presents itself however as the appearance of a luminous disc of a more vivid diffuse form with a reduced diameter than its solar equivalent.

Planets
Over a period of days a viewer will notice a few stars move relative to the others. They are actually planets. To distinguish a planet from a star, it is helpful to know that stars twinkle while planets have a steadier light, the stars being much further away and so presenting as a point rather than the disc of a planet. Once you have found a planet, it is interesting to know more about it and even to the naked eye, much can be discovered. Indeed, all visible planets have some features unique to them:

 * Mercury is rarely visible from Earth since its orbit keeps it very close to the Sun. [[Image: Lune Venus.jpg|thumb|right|250px|Crescent Moon and Venus]]

 * Venus, also known as the "Shepherd's star", is white in color, and appears brighter than any object in the sky after the Sun and Moon. Visible near twilight and dawn, like Mercury, Venus is an interior planet (one whose orbit lies between the Sun and the Earth) and appears to follow the Sun relative to our perspective (its maximum elongation is 47°). Its apparent magnitude varies depending on its phase (very much like the Moon) as well as its distance from Earth.

 * Mars is not unusually bright, but it can be distinguished from other celestial bodies by its reddish light. A casual observer (over the period of several days) may notice that it seems to make a U-turn (stops and then moves in retrograde): this is a property Mars shares with all the outer planets, and is owed to Earth's relative movement around the Sun. In the case of Mars, this takes place roughly every two years, with the period of greatest change in its apparent motion taking place over two months.

 * Jupiter  is pale in color. Although bright enough to be confused with Venus, a viewer can recognize it instantly: if one observes the equivalent of Venus in the middle of the night, it is Jupiter.

 * Saturn is considerably less bright than Jupiter (Uranus and Neptune are dimmer still).

Besides the planets, there are many other celestial curiosities:

Milky Way

The Milky Way is a denser region of stars in comparison to the rest of the sky. The image above represents the plane of our galaxy as viewed from within.

Stay one night in a place far from the bright lights of big cities so that your eyes get used to darkness, relax and wait. You will see a gigantic irregular milky band crossing the sky. This appearance gives its name since Greek antiquity. Scanning the myriad of stars that constitute the Milky Way is one of the greatest spectacles of sky gazing.

Constellations

These are not strictly speaking celestial objects since they constitute a grouping of stars making the shape of a figure, in general that of an animal or a mythological being, this nomenclature dating back to the Ancient Greek times for the stars seen in the Northern Hemisphere. Charts are available that show a complete view of what stars are visible at any given time on Earth. Astronomy 101 is based on these star charts, enabling the reader to navigate amongst the stars, using the Polaris star as a celestial north-pointing compass to be able to find the brightest of the stars: the Andromeda galaxy or the most luminous star of the sky (Sirius of Canis Major) for example. Punch cards with holes for finding and aligning the stars of the constellations were once popular for star gazing. This nineteenth century punched star chart by British engraver Sidney Hall (right photo) is an example of one used for star gazing in the northern hemisphere.

Meteors
During extended sky gazing you will occasionally notice streaks of light crossing the sky very quickly: "shooting stars". These are meteoroids that often weigh less than a gram but ignite when heated up by friction as they penetrate into the denser terrestrial atmosphere. One can see several dozen in one night. Some nights are especially favorable for their observation because the Earth, in its orbit, regularly crosses clouds of meteoroids well known to astronomers (see "meteor shower" for the dates).

Other phenomena are also visible to the naked eye, such as the comets, interesting and sometimes magnificent like Halley's comet as seen in 1910. Also galaxies, star clusters and nebulae are visible, but only appear as small milky patches, save for the remarkable Pleiades in the constellation of Taurus where one can distinguish the different stars.

Binocular gazing

Binoculars are very useful when you wish to observe bright, large astronomical objects. Thanks to them it is possible to see lunar craters. In spite of the distance between us and the Moon, one can observe the relief of these craters along the terminator, the separation line between the illuminated and darkened parts of the Moon. Lunar features are emphasized in this zone where sunlight strikes at a low angle and casts long shadows. This spectacle is a good start to sky gazing due to its ease.

Binoculars are good for the observation of large nebulas and the occasional bright comet. The reason is due to their very nature: the binocular enlarges images and adds brightness compared to naked eye views. Very extended objects can be seen in their entirety due to the wide field of view (which may not be the case with a telescope and its greater magnification) and with improved clarity and contrast compared to the naked eye. The Orion nebula is one of the most luminous and one of the easiest to locate. It is situated in the constellation of Orion, a constellation visible in winter, big and easily recognizable with its rectangular form and a short row of three bright stars forming the belt of Orion. One can also observe the cluster called the Pleiades, a stellar group composed of over fifteen stars which can be found by extending one of the diagonals of the rectangle of Orion toward the Northwest. Also visible in the late summer, fall and winter is another striking spectacle which lies beyond our own Milky way galaxy, the Andromeda galaxy. Though faintly visible to the naked eye, locating this object requires one to know how to identify the main constellations (see Locations of the constellations). The constellation Andromeda is situated under Cassiopeia in relation to the pole star. While viewing the Beta star of Andromeda in the binoculars, one ascends very slightly toward Cassiopeia until one sights a first small star, then one ascends again very slightly until one sees a fuzzy patch of light which is the heart of the Andromeda galaxy. If sky conditions are good, you may also see a surrounding very diffuse oval that represents the arms of the galaxy. This vast collection of stars is located 2.5 million light years away! It is one of the more distant objects that can be seen with common binoculars.

With experience, so long as the binoculars are held steadily and with decent atmospheric conditions, viewers endowed with good vision will be able to discern the four Galilean moons of Jupiter, even with simple 8x35.

Choice of binoculars
Their features are determined by two numbers: the first number indicates the magnification, the second the diameter of the lenses in the front, or aperture. Large apertures are recommended because they will collect more light and so reveal fainter objects, and give (for the same magnification) a larger field of view. Thus, while a birdwatcher might prefer a compact 8x35 binocular, a sky gazer will do better with a larger 10x50 glass.

Another number to consider is the exit pupil, which is the ratio of the objective lens diameter to the magnification. For optimal night viewing, the exit pupil should be greater than or equal to the diameter of the viewer's pupil under viewing conditions. For instance, assume the pupil enlarges to 5mm under given lighting conditions. 10x50 optics with an exit pupil of 5.0 would be preferable to 8x35's with an exit pupil of 4.375.

Observations with a refracting telescope
A refracting telescope is an instrument containing at least two lenses which focus light into an image at the focal plane. An eyepiece situated at the focus functions as a magnifying glass which works by permitting the eye of the observer to focus on this image at very close range, causing it to appear magnified. A good refracting telescope can be an instrument that one retains all his or her life, even after the acquisition of a bigger telescope.

The usually small aperture and general precision of refracting telescopes makes them well suited to the observation of objects which are bright and detailed, such as the Moon and planets. Even a small model of 60 mm diameter will reveal some detail on planets, and much detail on the Moon.

Jupiter is an ideal target for the first-timer equipped with a refracting telescope. Its observation lets us see the four main companions of the planet which are the Galilean moons as well as some details of the surface of the planet. It shows how much astronomical observation is a discipline of patience. A better view will require a more powerful telescope whose operation requires a finer mastery of the basics of astronomy. However using a telescope less effective than all those sold nowadays Galileo discovered the moons of Jupiter and confirmed Copernicus's theory: the Earth moves around the Sun, and not the reverse which was commonly believed at the time! With a refracting telescope, it is also possible to follow the phases of Venus and the change of its visible diameter with the passing of the months. Mars appears like an orange disc, but often without much detail. One can also follow the fluctuation of its visible diameter over the year. When Mars is nearest to the Earth, it is possible to distinguish its polar cap.

The most distant bright planet that one can observe with the refracting telescope is Saturn. If the conditions of observation are good, it unveils the very beautiful spectacle of its rings. One can follow the change of their appearance. In 2002, they were seen at their widest angle and best presentation, and will be seen in profile in 2010. They will then completely invisible for a few days, after which they will again appear as a thin, bright line as the angle gradually increases once more. Meanwhile, their appearance changes from year to year. With experience it is also possible to distinguish the large moon Titan.

The refracting telescope is a suitable instrument for surveying the Sun, but extreme precautions must be taken to avoid burning the retina and permanent blindness. NEVER look directly at the sun through any telescope or binoculars. The only safe method is projection onto a screen, using an appropriate Sun filter to avoid burning or damage to the telescope optics. When these precautions have been taken, the Sun is revealed in all its glory. It can be seen to evolve from day to day and its rotation can be clearly observed.

A refracting telescope also allows us to distinguish binary stars, nebulae (M42) and globular clusters (M13). Finally, let us not forget the Moon, which offers a multitude of details: craters, mountains, etc. As with binoculars, observation of the terminator reveals the best detail, notably the relief of the Moon.

Choice of the refracting telescope
The main optical problem of refracting telescopes is chromatic aberration (color fringing). When one observes a planet, the Moon, or a bright star at high magnification, it will be surrounded by a diffuse glow of unfocussed color, usually blue or violet. This effect can be minimized by the use of a lens with a long focal length, but this can result in an unwieldy instrument. Refractors can be made essentially free of false color using various panchromatic designs, many of which use three lenses (a triplet) as opposed to the two lenses (a doublet) found in the more common achromatic instruments. This system is costly. Refractors of this type can be less awkward because lenses of a shorter focal length can be used, resulting in a shorter telescope. It is difficult to construct refractors of more than 150 mm aperture because of the expense of the raw glass and the possibility of breakage in manufacturing. Let's add that a refractor is expensive in relation to a telescope of any other design of the same size. 60 mm diameter refractors are cheap, but from 100 mm and up they can be three times as expensive (or more) than a reflecting telescope of the same aperture. On the other hand a refracting telescope can be transported easily because it doesn't readily go out of alignment. Also in a refracting telescope, the objective is not obstructed in part by the secondary mirror that one finds in reflecting telescopes, which enhances the quality of the picture, the full surface of objective being used to collect light. The best choice (but also the costliest) is an apochromatic telescope that corrects all aberrations (chromatic and spherical).

Observation with a reflecting telescope
A reflecting telescope is made not of lenses but of mirrors. Being less costly to manufacture, one can, for the same price as a refracting telescope, acquire an instrument larger diameter that gives access to deeper space. Nevertheless, to take advantage of the power of a reflecting telescope, it is necessary to have a good observation site away from the lights of the city, otherwise the use of a good refracting telescope is preferable.

The major advantage of the reflecting telescope over the refracting telescope is its lower manufacturing cost, which allows the purchase, at a reasonable price, of an instrument of bigger diameter, giving a brighter image for observing distant and weakly luminous objects. Also chromatic aberration does not exist with this type of instrument, but the secondary mirror partially obscures the primary mirror, resulting in a loss of brightness of the order of 5 to 10%.

A reflecting telescope, unlike a refractor, needs some maintenance: the primary mirror has a certain degree of freedom in the tube and can in some cases (a physical shock for example) go out of alignment, requiring readjustment which can be done by the owner. This same mirror having a very fine coat of aluminium on its surface, deteriorates in contact with air and has a life of 8 to 10 years. Specialist firms can refurbish the mirror.

With a 150mm reflecting telescope, the viewer is able to distinguish the spiral arms of some galaxies and details in many star clusters. With such an instrument, most of the Messier objects can be appreciated in good detail. These instruments are also very useful when they are used for planet-gazing which they reveal, thanks to their better resolution, a multitude of details as the Great Red Spot of Jupiter, visible with a telescope of 200 mm or the Cassini Division in the rings of Saturn. It becomes possible to follow the changes in appearance of the main planets of the solar system with the passing of the months, and the craters of the Moon appear with all their details on the terminator.

A sufficiently powerful reflecting telescope (300mm) opens the way to hunt for the comets, the holy grail of the amateur astronomer who dreams of being the first to discover a new object to which they will give their name. Comet hunters constitute a world apart from the average astronomy hobbyist. Besides needing expensive instruments, comet research requires great rigor because it demands systematic observations, but some hobbyists count close to ten of these objects on their scorecard.

Whatever the type of observation, it is while getting involved in astrophotography that one gets the best out of your instrument. Lengthening the time of exposure, the improved brightness and contrast of the picture reveals the finest details. The best results are obtained using a CCD sensor connected to a computer. These sensors are contained in all electronic devices capable of taking images (webcam, digital camera, cell phone, etc.). The sensors in these devices can be used in CCD astrophotography, but the best pictures are taken with specialist monochromatic sensors. In any case, a little tinkering is necessary. The astronomy hobbyist who wants to become an amateur astronomer must start learning the fundamental principles of optics, as instruments of optimal performance cannot be bought in a store.

Types of reflecting telescope
There are two popular types of reflecting telescopes : the Newtonian and the Catadioptric.

Newtonian telescopes
Newtonian telescopes are characterized by longish tubes, slightly shorter in their focal length and are composed of a parabolic main mirrors fixed to the bottom of the tubes with flat secondary mirrors close to the opening, oriented at 45°, that directs the light outwards through the eyepiece. Observation is therefore from the side of the tube. Unfortunately the open tube allows dust to enter which deposits on the mirror. Another disadvantage is that the temperature inside the tube is slightly higher than the temperature of the surrounding environment (at least at the beginning of the night); the hotter air, while escaping, creates turbulence that reduces the quality of the image.

Catadioptric telescopes
Catadioptric telescopes combine specifically shaped mirrors and lenses to form an image. This is usually done so that the telescope can have an overall greater degree of error correction than their all-lens or all-mirror counterparts, with a consequently wider aberration-free field of view.

Schmidt
A Schmidt uses all spherical surfaces, making it an instrument that is easy to manufacture. It employs a cassegrain design with a compound curve corrector plate at the front of the tube that "corrects" all the aberration a spherical system would have (and also gives the telescope the advantage of having a sealed tube). The secondary mirror in a Schmidt is mounted on the corrector plate.

Maksutov
A Maksutov is similar to the Schmidt except it uses a thick concave corrector plate and the secondary is usually a silvered spot on the backside of the corrector (a "spot-mak"). This design has the added advantage over a Schmidt in that it never needs alignment, all optical elements are fixed in alignment.

Gallery
 Image: AuroraAustralisDisplay.jpg|Aurora Australis Display Over Wellington 2001

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Talk:Skygazing

Transwiki import
This page was imported from Skygazing due to w:Wikipedia:Articles for deletion/Skygazing. --mikeu talk 19:01, 4 November 2008 (UTC)

 * Technical note: I was only able to import the most recent version due a failure in the Import retrieving all of the revision history. --mikeu talk 19:22, 4 November 2008 (UTC)

Copyedit
This article is in dire need of editing to improve its English. I had a look at its previous history at wikipedia (linked above), but that's no better so I'll have a go myself. Kram 10:45, 27 January 2010 (UTC)

OK, I cleaned it up a bit, but as the article is a bit of an orphan, I don't want to spend too much time on it if its not linked from anywhere. Also I don't agree with some of its claims ... eg the bit about distinguishing stars from planets by their twinkling seems a bit dubious to me. Though not being an expert I don't want to change anything too much. I may check back sometime later. Kram 19:35, 27 January 2010 (UTC)

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User:EAS4200C.F08.WIKI.A/My Contribution: 5

We will no continue the equation of equilibrium in terms of $$\sigma$$. The goal is to obtain the following formula.

 $$\displaystyle \frac{\partial \sigma_{yx}}{\partial y} + \frac{\partial \sigma _{zx}}{\partial z} = 0$$

In indicial notation the equation can be represented as follows;

 $$\displaystyle \frac{\partial \sigma_{21}}{\partial x_2} + \frac{\partial \sigma _{31}}{\partial x_3} = 0$$

Now we must recall equation 2.21 from the text which is the above equation but with the term $$\displaystyle \frac{\partial \sigma_{xx}}{\partial x}$$ added in.

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

SEE EQUATIONS 2.22 AND 2.23 TO COMPARE.

This equation represented in indicial notation is represented as the following  $$\sum_{i=1}^{3}{\frac{\partial \sigma_{ij}}{\partial x_i}} = 0$$ for $$j=1,2,3$$

All three cases ($$j= 1,2,3$$) are shown below.

$$\displaystyle j=1, \frac{\partial \sigma_{11}}{\partial x_1} +  \frac{\partial \sigma_{21}}{\partial x_2} + \frac{\partial \sigma_{31}}{\partial x_3}$$  $$\displaystyle j=2, \frac{\partial \sigma_{12}}{\partial x_1}  + \frac{\partial \sigma_{22}}{\partial x_2}  +  \frac{\partial \sigma_{32}}{\partial x_3}$$  $$\displaystyle j=3, \frac{\partial \sigma_{13}}{\partial x_1}  +  \frac{\partial \sigma_{23}}{\partial x_2}  +  \frac{\partial \sigma_{33}}{\partial x_3}$$

Noting that there are four zer stress components that come from the stress-strain relation, these equations reduce to the following.

$$\displaystyle 0=\sigma_{11}=\sigma_{22}=\sigma_{33}=\sigma_{23}$$  $$\displaystyle j=1, \frac{\partial \sigma_{21}}{\partial x_2} + \frac{\partial \sigma_{31}}{\partial x_3}$$  $$\displaystyle j=2, \frac{\partial \sigma_{12}}{\partial x_1} +  \frac{\partial \sigma_{32}}{\partial x_3}$$
$$\displaystyle j=3, \frac{\partial \sigma_{13}}{\partial x_1} + \frac{\partial \sigma_{23}}{\partial x_2}$$

We will now derive the equation of equilibrium in terms of $$\sigma$$. To do so we will call upon the one dimentional model problem that we previously looked at. For conveienence it is depicted below.

[\Image:]

First we must sum the forces in the x direction. doing so results in the following equation.

$$\sum{F_x} = 0 = -\sigma_{x}A + \sigma(x+dx)A + f(x)dx = A\left[\sigma(x+dx) - \sigma(x) \right] + f(x)dx$$

Taking the Taylor Series expansion of the part of the equation in the brackets (mutliplied by A) and ignoring the higher order terms yields the following.

 $$\displaystyle f(x+dx) = f(x) + \frac{df(x)}{dx}dx + \frac{1}{2}\frac{d^2f}{dx^2}xdx^2....$$    Taylor Series Expansion

$$\displaystyle \frac{d\sigma}{dx} + \frac{f(x)}{A} = 0....$$   Result Ignoring Higher Order Terms


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User:Eml4500.f08.gravy.bje/Notes5

3-D Space Trusses
The axial FD relation for 3-D space trusses can be expressed the same as for the 2-D truss element

$$\ \ \frac{EA}{L} \begin{Bmatrix} 1 & -1 \\ -1 & 1 \end{Bmatrix} \begin{Bmatrix} d_{1} \\ d_{2} \end{Bmatrix}  =  \begin{Bmatrix} P_{1} \\ P_{2} \end{Bmatrix} \,$$

where E = elastic modulus of the material, A = area of the cross section of the element, L = length of the element, d1 and d2 are the displacements along the axis of the element, and P1 and P2 are possible applied axial loads at the end.

This can also be expressed as:

$$k_{1} d_{1} = r_{1}\,$$

The degrees of freedom for this system is represented as:

$$\ \ d = \begin{Bmatrix} u_{1} \\ v_{1} \\ w_{1} \\ u_{2} \\ v_{2} \\ w_{2} \end{Bmatrix}_{(6x1)} \,$$

where u, v, and w are your degrees of freedom in the x, y, and z directions respectively

The element forces for this system is represented as:

$$\ \ r = \begin{Bmatrix} F_{x1} \\ F_{y1} \\ F_{z1} \\ F_{x2} \\ F_{y2} \\ F_{z2} \end{Bmatrix}_{(6x1)} \,$$

For 3-D space trusses the transformation matrix is given in terms of three director cosines:

$$l^{(e)} = \frac{x_{2} - x_{1}}{L}$$

$$m^{(e)} = \frac{y_{2} - y_{1}}{L}$$

$$n^{(e)} = \frac{z_{2} - z_{1}}{L}$$

With these director cosines, the transformation between the local and global degrees of freedom can be expressed as:

$$ \begin{Bmatrix}d_{1} \\ d_{2} \end{Bmatrix} = \begin{bmatrix} l^{(e)} & m^{(e)} & n^{(e)} & 0 & 0 & 0 \\ 0 & 0 & 0 & l^{(e)} & m^{(e)} & n^{(e)} \end{bmatrix} \begin{Bmatrix} u_{1} \\ v_{1} \\ w_{1} \\ u_{2} \\ v_{2} \\ w_{2} \end{Bmatrix} $$

This expression is also used to show the relation between axial dofs and element dofs in global coordinates

Similarly, the relationn between axial forces and element forces in global coordinates is:

$$ \begin{Bmatrix}F_{1} \\ F_{2} \end{Bmatrix} = \begin{bmatrix} l^{(e)} & m^{(e)} & n^{(e)} & 0 & 0 & 0 \\ 0 & 0 & 0 & l^{(e)} & m^{(e)} & n^{(e)} \end{bmatrix} \begin{Bmatrix} F_{x1} \\ F_{y1} \\ F_{z1} \\ F_{x2} \\ F_{y2} \\ F_{z2} \end{Bmatrix} $$

The derivation for the transformation matrix is as follewed:

$$\ d_1^{(e)}=l^{(e)}u_1^{(e)} + m^{(e)}v_1^{(e)} + n^{(e)}w_1^{(e)} = \begin{bmatrix} l^{(e)} & m^{(e)} & n^{(e)} \end{bmatrix} \begin{Bmatrix} u_1 \\ v_1 \\ w_1 \end{Bmatrix} \,$$

$$ \begin{Bmatrix} d_1^{(e)} \\ d_2^{(e)} \end{Bmatrix}_{(2x1)} = \begin{bmatrix} l^{(e)} & m^{(e)} & n^{(e)} & 0 & 0 & 0 \\ 0 & 0 & 0 & l^{(e)} & m^{(e)} & n^{(e)} \end{bmatrix}_{(2x6)} \begin{Bmatrix} u_{1} \\ v_{1} \\ w_{1} \\ u_{2} \\ v_{2} \\ w_{2} \end{Bmatrix}_{(6x1)} $$

where the transformation matrix is equal to:

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

The global stiffness matrix can be calculated as follows:

$$\ \underline{k}^{(e)} = \underline{T}^{(e)T} \underline{k}^{(e)} \underline{T}^{(e)}$$

Where: $$\ \underline{T}^{(e)} = \begin{bmatrix} l^{(e)} & m^{(e)} & n^{(e)} & 0 & 0 & 0 \\ 0 & 0 & 0 & l^{(e)} & m^{(e)} & n^{(e)} \end{bmatrix}$$ and $$\ \underline{k}^{e)} = k^{(e)} * \begin{bmatrix} 1 & -1 \\ -1 & 1 \end{bmatrix}$$

Since $$k^{(e)}$$ is a scaler, we can move it within the equation without changing the problem. So after moving the scaler and inserting in the matrices, the following equation is found:

$$\ \underline{k}^{(e)} = k^{(e)} \begin{bmatrix} l^{(e)} & 0\\ m^{(e)} & 0 \\ n^{(e)} & 0 \\ 0 & l^{(e)} \\ 0 & m^{(e)} \\ 0 & n^{(e)} \end{bmatrix} \begin{bmatrix} 1 & -1 \\ -1 & 1 \end{bmatrix} \begin{bmatrix} l^{(e)} & m^{(e)} & n^{(e)} & 0 & 0 & 0 \\ 0 & 0 & 0 & l^{(e)} & m^{(e)} & n^{(e)}  \end{bmatrix}$$

Solving the first set of matrices gives:

$$\ \underline{k}^{(e)} = k^{(e)} \begin{bmatrix} l^{(e)} & -l^{(e)} \\ m^{(e)} & -m^{(e)} \\ n^{(e)} & -n^{(e)} \\ -l^{(e)} & l^{(e)}\\ -m^{(e)} & m^{(e)} \\ -n^{(e)} & n^{(e)} \end{bmatrix}  \begin{bmatrix} l^{(e)} & m^{(e)} & n^{(e)} & 0 & 0 & 0 \\ 0 & 0 & 0 & l^{(e)} & m^{(e)} & n^{(e)} \end{bmatrix}$$

Solving the second set of matrices gives:

$$\ \underline{k}^{(e)} = k^{(e)}\begin{bmatrix} (l^{(e)})^2 & l^{(e)}m^{(e)} & n^{(e)}l^{(e)} & -(l^{(e)})^2 & -l^{(e)}m^{(e)} & -n^{(e)}l^{(e)}  \\  l^{(e)}m^{(e)} & (m^{(e)})^2 & m^{(e)}n^{(e)} & -l^{(e)}m^{(e)} & -(m^{(e)})^2 & -m^{(e)}n^{(e)} \\ n^{(e)}l^{(e)} & m^{(e)}n^{(e)} & (n^{(e)})^2 & -n^{(e)}l^{(e)} & -m^{(e)}n^{(e)} & -(n^{(e)})^2 \\ -(l^{(e)})^2 & -l^{(e)}m^{(e)} & -n^{(e)}l^{(e)} & (l^{(e)})^2 & l^{(e)}m^{(e)} & n^{(e)}l^{(e)} \\ -l^{(e)}m^{(e)} & -(m^{(e)})^2 & -m^{(e)}n^{(e)} & l^{(e)}m^{(e)} & (m^{(e)})^2 & m^{(e)}n^{(e)} \\ -n^{(e)}l^{(e)} & -m^{(e)}n^{(e)} & -(n^{(e)})^2 & n^{(e)}l^{(e)} & m^{(e)}n^{(e)} & (n^{(e)})^2 \end{bmatrix} $$

Which is the definition of the stiffness matrix, therefore the equation is true.

Combining all of our equations we get our FD relation for the global stiffness matrix, the dofs, and the forces:

$$\ \ \frac{EA}{L} \begin{bmatrix} (l^{(e)})^2 & l^{(e)}m^{(e)} & n^{(e)}l^{(e)} & -(l^{(e)})^2 & -l^{(e)}m^{(e)} & -n^{(e)}l^{(e)} \\  l^{(e)}m^{(e)} & (m^{(e)})^2 & m^{(e)}n^{(e)} & -l^{(e)}m^{(e)} & -(m^{(e)})^2 & -m^{(e)}n^{(e)} \\ n^{(e)}l^{(e)} & m^{(e)}n^{(e)} & (n^{(e)})^2 & -n^{(e)}l^{(e)} & -m^{(e)}n^{(e)} & -(n^{(e)})^2 \\ -(l^{(e)})^2 & -l^{(e)}m^{(e)} & -n^{(e)}l^{(e)} & (l^{(e)})^2 & l^{(e)}m^{(e)} & n^{(e)}l^{(e)} \\ -l^{(e)}m^{(e)} & -(m^{(e)})^2 & -m^{(e)}n^{(e)} & l^{(e)}m^{(e)} & (m^{(e)})^2 & m^{(e)}n^{(e)} \\ -n^{(e)}l^{(e)} & -m^{(e)}n^{(e)} & -(n^{(e)})^2 & n^{(e)}l^{(e)} & m^{(e)}n^{(e)} & (n^{(e)})^2 \end{bmatrix} \begin{bmatrix} u_(e) \\ v_(e) \\ w_(e) \\ u_(e+1) \\ v_(e+1) \\ w_(e+1) \end{bmatrix} = \begin{bmatrix} F_ex \\ F_ey \\ F_ez \\ F_(e+1)x \\ F_(e+1)y \\ F_(e+1)z \end{bmatrix} $$

title: User:Eml4500.f08.gravy.bje/Notes5


text:
French/Ballade des Pendus

Author Biography (English)

Ce poème est de temps en temps intitulé "L'Épitaphe de Villon."

Texte
Frères humains qui après nous vivez, N'ayez les cœurs contre nous endurcis, Car, se pitié de nous pauvres avez, Dieu en aura plus tôt de vous mercis. Vous nous voyez ci attachés cinq, six: Quant de la chair que trop avons nourrie, Elle est piéça devorée et pourrie, Et nous, les os, devenons cendre et poudre De notre mal personne s'en rie; Mais priez Dieu que tous nous veuille absoudre!

Se frères vous clamons, pas n'en devez Avoid dédain, quoique fûmes occis Par justice. Toutefois, vous savez Que tous hommes n'ont pas bon sens rassis; Excusez-nous, puisque sommes transis, Envers le fils de la Vièrge Marie, Que sa grâce ne soit pour nous tarie, Nous préservant de l'infernale foudre. Nous sommes morts, âme ne nous harie, Mais priez Dieu que tous nous veuille absoudre!

La pluie nous a débués et lavés, Et le soleil desséchés et noircis, Pies, corbeaux, nous ont les yeux cavés, Et arraché la barbe et les sourcils. Jamais nul temps nous ne sommes assis; Puis çà, puis là, comme le vent varie, A son plaisir sans cesser nous charrie, Plus becquetés d'oiseaux que dés à coudre. Ne soyez donc de notre confrérie; Mais priez Dieu que tous nous veuille absoudre!

Prince Jésus, qui sur tous a maîtrie, Garde qu'Enfer n'ait de nous seigneurie: A lui n'ayons que faire ne que soudre. Hommes, ici n'a point de moquerie; Mais priez Dieu que tous nous veuille absoudre!

title: French/Ballade des Pendus


text:
User:Eml4500.f08.team.foskey.ckf/hw5c

MATLAB Assignment The Three Bar Space Truss

The Three Bar Problem
The Three Bar Truss problem is from Example 4.2 on page 230 of the text Fundamental Finite Element Analysis and Applications by M. Asghar Bhatti. The problem setup is shown in the figure below.

All of the bars have a Young's Modulus of 200GPa. The cross sectional area of elements 1 and 2 is 200mm2 and 600mm2 for element 3. The applied load P is 20kN. The nodal coordinates are as follows: Node 1 : (0.96, 1.92, 0) m Node 2 : (-1.44, 1.44, 0) m Node 3 : (0, 0, 0) m

Comments On The Main Code
The code used to solve the three bar space truss problem has comments throughout that help identify variables, and functions of parts of the code.

The code above systematically calculates the needed values to completely solve the problem. The code receives the element properties, and the nodal positions. Next it calculates the lengths of each bar, then creates the connectivity array and location matrix master array. The applied force matrix is then created. The global stiffness matrix is then calculated by looping through all of the elements. With this information, the output results shows the elemental stiffness matrices, the global stiffness matrix, the applied loads (in a global coordinate system), the displacements and the reactions. The results matrix displays the strain, stress, and axial forces in columns, for each element ( each row).

With all the important values calculated and displayed, the code next focuses on plotting the deformed and undeformed trusses. Each node is given a number, and each dislocated node is given a new number. Next, the code assigns each end of an element a node number (either and original node or a new dislocated node). The code then plots each element based on the node data. Line styles and color are changed after plotting the undeformed truss so that the deformed truss can be easily identified.

Part of the requirements of this assignment was to create plots of the truss in several views. To accomplish this, the plotting section of the code is embedded within a loop. Each time the loop is plotted, it is plotted in a different view perspective, adjusting the axis dimensions if necessary.

Also, for the benefit of the user, the two perspective view plots make use of MATLAB's "rotate3d" function. This allows the user to click on the figure and manually rotate the figure.

Functions used
SpaceTrussElement NodalSoln

SpaceTrussResults

Comments On The Functions
The functions above are modified versions of the functions used for plane trusses, as seen and explained here.

The function SpaceTrussElement behaves like the function PlaneTrussElement in that it calculates the length of an element, the director cosines, and the elemental stiffness matrix for an element from the values it is given (Young's Modulus, cross sectional area, and element end coordinates). The differences between the two functions is seen in the length calculation, where z coordinates are taken into consideration, the calculation of an additional director cosine for the third dimension, and a stiffness matrix that accounts for the third dimension.

SpaceTrussResults behaves like PlaneTrussResults, except it accounts for the z direction in the calculations of element length, director cosines, and in the transform matrix.

NodalSoln remains unchanged and is the same as the function used in the two bar truss problem.

Click here for the plane truss functions and their explanations.

Results Output
Running the above main code for the three bar space truss gives the following output:

Comments On the Results Output
The output results shows the elemental stiffness matrices( in local and global coordinate systems), the global stiffness matrix, the applied loads (in a global coordinate system), the displacements and the reactions(both in the global coordinate system). The results matrix displays in three columns the strain, stress, and axial forces (respectively) for each element. Each element is represented by a row, with row 1 being element 1 and so forth.

Note on Units: The units of the output values are as follows:

Stiffness Matrices, k_local, k, and K : Newtons per meter (N/m) Applied Loads, R : Newtons (N) Global Displacements, d : millimeters (mm) Reaction Forces, reactions : Newtons (N) Results, results : Dimensionless (m/m) for strain, Pascals (pa) for stress, and Newtons (N) for axial force

Figures

Figure 1 above shows the undeformed truss in red dotted lines, and the deformed truss in blue solid lines. It is shown as a view down the x axis, resulting in a plot in the yz plane.

Figure 2 again shows the undeformed truss in red, and the deformed truss in blue. The view for this plot is down the y axis, in the xz plane.

Figure 3 shows the deformed and undeformed trusses in the xy plane, down the z axis.

Figure 4 is a perspective view generated as a view of the trusses from the point (-2, -2, 3).

Figure 5 is also a perspective view, but from the point (-2, -1.5, 1). It was felt that this viewpoint offered a better view of the deformed truss system than the suggested viewpoint used in Figure 4.

In all five figures, the undeformed truss is shown in red dotted lines, and the deformed truss is shown in blue solid lines. Nodes and Elements were not labeled as it would severely clutter the plots, and it was not part of the assignment.

title: User:Eml4500.f08.team.foskey.ckf/hw5c


text: File:2 Bar Truss Qwiki.jpg

title: File:2 Bar Truss Qwiki.jpg


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title: File:3dfig2.png


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text: File:2 Bar Truss DOFs Qwiki.jpg

title: File:2 Bar Truss DOFs Qwiki.jpg


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text:
User:EAS4200c.f08.Blue.E/Oct31st

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 D 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

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

title: User:EAS4200c.f08.Blue.E/Oct31st


text:
User:EML4500.f08.RAMROD.E/HW5

Six Bar Truss Code for Elements with Varying E This Part goes below Dan's Plot of 6 bar truss
Next, the six bar truss Matlab code used above, was modified to allow for different values of E for each element. This code uses all the same functions as above (i.e. PlaneTrussElement.m, PlaneTrussResults.m and NodalSoln.m). The modified code is shown below. The results from this situation, as well as the constant E case and the undeformed truss system are shown in the plot below. The varying E case is shown by the dotted nodes while the undeformed case is shown dashed. The element and node numbers are suppressed to avoid cluttering the diagram. The code used to create this plot is very similar to that shown above for the six bar truss with constant E, but repeated for the two deformation cases.

Continuation of Pergola's HW 5, Goes below Pergola's part
Using the Principle of Virtual Work, the global FD relationship is given by:

$$\displaystyle K*(d)=F$$

Also, we can represent the elemental matrix $$k^e$$ in terms of the axial matrix $$\hat{k}$$ and the transformation matrix, $$\displaystyle T$$. This is shown below:

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

Lastly, the relationship between the element forces in global coordinates, $$\displaystyle f^e$$, and the axial forces, $$\displaystyle p^e$$, can be found using the transformation matrix, $$\displaystyle T$$.

$$\displaystyle f^e(T)=p^e$$

Or more detailed:

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

Using the above derived equation for $$\displaystyle k^e$$, we can give the actual values for its elements. Multiplying out the right side of the equation gives us the following:

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

$$k^e = \begin{bmatrix} 1 &-1 & 1\\ -1 & 1 & -1\\  1&  -1&1 \end{bmatrix}\begin{bmatrix} l^e & m^e & n^e & 0 & 0 & 0\\ 0 & 0 & 0 & l^e & m^e & n^e \end{bmatrix}\begin{bmatrix} l^e & 0\\ m^e & 0\\ n^e & 0\\ 0 & l^e\\ 0 & m^e\\ 0 & n^e \end{bmatrix}$$

This is the elemental stiffness matrix, $$\displaystyle k^e$$ for element e.

title: User:EML4500.f08.RAMROD.E/HW5


text:
User:Eas4200c.f08.blue.a/Lecture 28

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 D 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

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

title: User:Eas4200c.f08.blue.a/Lecture 28


text: File:6 truss varying e.jpg

title: File:6 truss varying e.jpg


text: File:Sixbarcode.jpg

title: File:Sixbarcode.jpg


text: File:3-bar space truss system setup.jpg

title: File:3-bar space truss system setup.jpg


text: File:Sixbarsetup.png

title: File:Sixbarsetup.png


title: File:Threebar3dsetup.png


text:
User:Eml4500.f08.team.foskey.ckf/hw4matlab

 Comment: This MATLAB portion of HW4 was not included in the original submission. The email stating that this problem was required was misread. Eml4500.f08.team.foskey.ckf 03:32, 5 November 2008 (UTC)

MATLAB Assignment The five bar truss

The Five Bar Truss Problem
The Five Bar Truss problem is from Example 1.4 on page 25 of the text Fundamental Finite Element Analysis and Applications by M. Asghar Bhatti. The code for analyzing the truss was provided, with the intent being to study the code, plot the trusses, and learn how to write code for future problems.

The setup of the Five Bar Truss can be seen below.

The given values for the problem are as follows:

Cross Sectional Areas of Elements
 * Element 1: 40 cm2
 * Element 2: 40 cm2
 * Element 3: 40 cm2
 * Element 4: 40 cm2
 * Element 5: 40 cm2

Young's Modulus
 * Element 1: 200 GPa
 * Element 2: 200 GPa
 * Element 3: 200 GPa
 * Element 4: 200 GPa
 * Element 5: 70 GPa

Applied Load P = 150 kN

Provided Code
The following code was obtained from the textbook's supplemental material website

Comments On The Provided MATLAB Code
The provided code begins by declaring all of the initial property values of the elements, and the applied load. The code also creates the connectivity and location matrix master arrays. The code then loops through the elements (based upon how the elemental properties were declared) to create elemental stiffness matrices. The applied force matrix is then created. Next, the code runs the NodalSoln function to calculate the displacements and reactions. To finish the code, the PlaneTrussResults function is run in a series of loops (again, dictated by the way the element properties were defined) to calculate the strain, stress, and axial forces in the elements.

The functions NodalSoln, PlaneTrussResults, and PlaneTrussElement are needed in addition to the above code in order for the code to run. Those functions and their explanations can be found here.

Plotting Code
The following code uses the provided code above to calculate needed values, then plots the undeformed and deformed trusses.

Comments On The Plotting Code
The above plotting code is used with the provided code to plot the deformed and undeformed trusses. The plotting code begins by running the provided code, called "fivebartruss_originalcode". ( A copy of the original provided code found near the top of this page was saved as a file called "fivebartruss_originalcode.m" so that any changes made to the original code that caused errors would not cause any errors in the plotting.) By running the provided code, all of the needed values are calculated and kept in memory by MATLAB. Once the original code has run, the plotting code rescales all of the nodal coordinates from millimeters to meters, and then calculates the lengths of the elements. The plotting code then assigns each node a position, and then assigns each element end a position. The code then plots the undeformed truss and the deformed truss in a connect-the-dots style by looping through the position numbers. To end the plotting code, the element and node numbers labeled on the figure, and a legend is created.

Results Output
The following results output can be obtained from either the original provided code, or from the plotting code.

Comments On The Results Output
The results output shows the global stiffness matrix K, the force matrix R, the global displacement matrix d, the reactions matrix reactions, and a results matrix. The matrix results shows the axial strain, axial stress, and axial forces of each element in columns 1, 2 and 3 (respectively). Each row represents an element, with row 1 corresponding to element 1, and row 5 corresponding to element 5.

The units of the output values are as follows:

Global Stiffness Matrix K : Newtons per meter (N/m) Applied Loads R : Newtons (N) Global Displacements d : millimeters (mm) Reaction Forces reactions : Newtons (N) Results results : Dimensionless (m/m) for strain, Pascals (pa) for stress, and Newtons (N) for axial force

Figure
The figure generated by the plotting code can be seen below. The undeformed truss (before loading) is seen in red dotted lines, and the deformed truss (after loading) is shown in blue solid lines. The elements are labeled near the midpoints of the undeformed elements, but moved slightly so that they were not ambiguous. The undisplaced nodes are labeled in red, and the displaced nodes (nodes 2 and 3) are labeled in blue. Nodes 1 and 4 are fixed points, and are only labeled once.


text:
User:Eas4200c.f08.radsam/HW5

 Comment: The Matlab code has been corrected and updated. Here is the HW5 original version. Eas4200c.f08.radsam.m 01:21, 11 November 2008 (UTC) =Homework #5=

=Class Notes=

October 20th: Meeting 24
 Exam Review

The important topics that were discussed in class and will be covered in the second exam are:
 * Kinematic assumption
 * 4 Zero-strain components
 * 4 Zero-stress components
 * Expressing the $$\varepsilon = \sigma$$ relationship
 * Rewriting the $$\varepsilon = \sigma$$ relationship

$$\varepsilon = \begin{bmatrix} A & 0\\ 0 & B \end{bmatrix} \left\{\sigma \right\}$$

and $$ \sigma = \begin{bmatrix} A^{-1} & 0\\ 0 & B^{-1} \end{bmatrix} \left\{\varepsilon \right\}$$

There is a way to check the validity of this method. The product of the matrix and its inverse should yield the identity matrix.

$$\sigma _{11}=\sigma _{22}=\sigma _{33}=0\,$$

$$\sigma _{23}=2G\varepsilon _{23}$$    $$\sigma _{31}=2G\varepsilon _{31}  $$     $$\sigma _{12}=2G\varepsilon _{12}$$

 Equilibrium Equation for Stresses in a Nonuniform Stress Field  Referring back to the roadmap from several class periods ago: Part C invovles finding the equilibrium equation for stresses in a nonuniform stress field.

Here, we will consider the 1-D model as well as two different loading conditions. The first is a uniform applied stress to a beam element and the second is a nonuniform stress field.

 * Figure 1 shows a beam element under uniform stress. Here, the axial load is evenly distributed and f(x) is constant.

 * A section dx of the beam element can be analyzed in terms of the forces and stresses applied to it. Since the stress is uniform, the shear stresses are zero.  Notice also that the normal stresses are equal on both sides.

 * For the non-uniform case, Figure 3 shows that the distributed load contributes some shear stress to the differential element. This causes the normal stress on the right side of the element to be different than the axial stress on the left side.  Because of the distributed load, the normal stresses are not equal, which means that f(x) is non-constant.

 * These observations are important for our analysis of the equilibrium equations, and will be referenced in later lectures.

Contribution by radsam.s

October 22nd: Meeting 25
Exam 2 was taken during this meeting.

October 27th: Meeting 26
 Bidirectional Bending
 * Section 4.2 in the textbook. As assigned reading read section 4.1 in the textbook to catch up, paying special attention to example 4.1 on page 124.

For the NACA Airfoil assignment for homework 5, we will have stringers at points B, E, H, F (see class notes 22-1). Note that skin and spar web do not participate in bending; only the stringers do. We need the areas of stringers $$A_{B},~A_{E},~A_{H},~A_{F}\,$$.

Slide 1:
 Bidirectional Bending Recipe
 * Note: We refer to this as a recipe because we're simply presenting the formula without a derivation.
 * Note: This is material is covered in the applied reading of section 4.2.

$$M_y=\int_A z \sigma_{xx} dA   \,$$ (equation 4.26) Similarly, for $$M_z\,$$, see equation 4.27 in the book.
 * Note: $$dA=dydz\,$$, so $$\int_A dA = \iint dydz\,$$

Slide 2:
Moment of intertia tensor: $$I_y,~I_z,~I_{yz}\,$$
 * Note: $$I_y=I_{yy}=I_{22}\,$$ ; $$I_z=I_{zz}=I_{33}\,$$ ; $$I_{yz}=I_{23}\,$$

$$Iy= \int_A z^2 dA\,$$ (equation 4.28a)

$$Iz= \int_A y^2 dA\,$$ (equation 4.28b)

$$Iyz = \int_A yz dA\,$$(equation 4.28c)

$$\sigma_{xx}=\Epsilon \varepsilon_{xx} = \frac{I_{y}M_{z} - I_{yz}M_{y}}{I_{y}I_{z}-(I_{yz})^{2}}y + \frac{I_{z}M_{y}-I_{yz}M_{z}}{I_{y}I_{z}-I_{yz}^{2}}z\,$$ (Equation 4.29)

But surely everyone can see such a recipe is mindless learning. Our class transcends mere formulas. This means a derivation is required.

Recall: $$\mathbf{I} = \begin{bmatrix} I_{11} & I_{12} & I_{13} \\ I_{21} & I_{22} & I_{23} \\ I_{31} & I_{32} & I_{33} \\\end{bmatrix}$$ The determinant of the bottom right 4x4 matrix ($$I_{22}\,$$,$$ I_{23}\,$$,$$I_{32}\,$$,$$ I_{33}\,$$) is the denominator of the $$\sigma_{xx}\,$$ formulas seen above.

Slide 3:

I_{22} & I_{23} \\ I_{32} & I_{33} \end{bmatrix}\,$$.
 * Note: $$D=I_y I_z - (I_{yz})^2\,$$ and can be written as $$D=I_{22} I_{33} - (I_{23})^2\,$$ which is in turn the determinant of matrix $$ \begin{bmatrix}

 * Note: $$I_{32}=I_{zy}:= \int_A zy dA\,$$, and $$zy=yz\,$$ therefore $$I_{yz}=I_{23}\,$$

Neutral Axis, where: $$\sigma_{xx}=0\,$$ From equation 4.29: $$\sigma_{xx}=m_{yy} + m_{zz} = 0\,$$ $$z=(\frac{-m_{y}}{m_{z}})y = (\tan{\beta})y\,$$ $$tan(\beta)=-tan(\alpha)\,$$

Contributed by radsam.z and radsam.e

October 29th: Meeting 27
 Roadmap Continued: Equation of Equilibrium for Stresses

In order to continue our journey into the land of exciting airfoil computation, we must now look at the equation of equilibrium in terms of $$\sigma\,$$, which is the stress tensor with 6 components.

Our goal is:

$$(1)\frac{\partial\sigma_{yx}}{\partial y}+\frac{\partial\sigma_{zx}}{\partial z}=0\,$$ (Eqn. 3.14)

or, using (absolutely awesome) indicial notation (sadly, cyclic permutation need not be applied):

$$(2)\frac{\partial\sigma_{21}}{\partial x_{2}}+\frac{\partial\sigma_{31}}{\partial x_{3}}=0\,$$

Also, recall (as in Section 2.4 in your book):

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

Similarly, see (2.22) and (2.23) --> HW5.

In indicial notation (2.21), (2.22), (2.23) can be expressed as:

$$(4)\sum_{i=1}^3 \frac{\partial \sigma_{ij}}{\partial x_i}=0\mbox{for}j=1,~2,~3\,$$

Contributed by radsam.a

October 31th: Meeting 28
For simplification purposes, we look at a a 1-dimensional model first. Equating all forces on the x-direction, we obtain:

$$\sum\ F_x = 0 = -\sigma_{(x)}\ A + \sigma_{(x + dx)}\ A + f_{(x)} dx\,$$

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

Applying Taylor series to the term inside the brackets, it becomes:

$$\frac{\partial\sigma_{x}}{\partial\ x} dx + higher~order~terms\,$$

Recall:

$$F_{(x + dx)} = F_{(x)} + \frac{\partial\ F_{(x)}}{\partial\ dx} dx + \frac{1}{2} \frac{\partial ^2 F_{(x)}}{\partial x^2} (dx)^2 + .~.~.\,$$

Neglecting higher order terms:

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

Where $$\frac{f_{(x)}}{A}\,$$ is the applied load.

Now, we may look at a non-uniform 3-D field without applied loads, and focusing on the x-direction.

Figure 1 shows an infinitesimal element in which the stress is not uniform. However, the element must remain in equilibrium, therefore the six equations of equilibrium must be satisfied. For example, forces along the x-direction are:

$$\sum\ F_x = 0 = dy dz [ -\sigma_{xx}\ (x,y,z) + \sigma_{xx}\ (x+dx,y,z)]\,$$ Facets with normal X

$$ + dz dx [ -\sigma_{yx}\ (x,y,z) + \sigma_{yx}\ (x,y+dy,z)]\,$$ Facets with normal Y

$$ + dx dy [ -\sigma_{zx}\ (x,y,z) + \sigma_{zx}\ (x,y,z+dz)]\,$$ Facets with normal Z

$$0 = (dx dy dz) \left [ \frac{\partial\sigma_{xx}}{\partial\ x} + \frac{\partial\sigma_{yx}}{\partial\ y} + \frac{\partial\sigma_{zx}}{\partial\ z} \right ]\,$$

Contributed by radsam.d and picture by radsam.a

=Homework Problems and Solutions=

=Homework #5 Contributors= Eas4200c.f08.radsam.a. 00:22, 5 November 2008 (UTC) Eas4200c.f08.radsam.d 22:52, 6 November 2008 (UTC) Eas4200c.f08.radsam.r 00:04, 6 November 2008 (UTC) Eas4200c.f08.radsam.s 03:37, 7 November 2008 (UTC) Eas4200c.f08.radsam.e 17:55, 7 November 2008 (UTC) Eas4200c.f08.radsam.z 19:32, 7 November 2008 (UTC) Eas4200c.f08.radsam.m 22:00, 7 November 2008 (UTC)

title: User:Eas4200c.f08.radsam/HW5


text:
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User:Eas4200c.f08.wiki.f/10-31-08

Neglect h.o.t.

 * → $$\frac{d\sigma}{dx} + \frac{f(x)}{A} = 0$$

 * Notes:
 * A is the applied load
 * Body force = force/volume
 * f(x) = force/length

Now, non-uniform stress field is in 3-D, but without an applied load, focusing on the x-direction only (i.e. without the other stress component to avoid cluttering the figure)

Below is an image of shear on a 3-D cube :

Note: the first part of the superscript refers to the normal direction of the cube face (facet), and the second part refers to the direction of shear.

$$\sum{F_x} = {\color{red}dydz[-\sigma_{xx}(x,y,z) + \sigma_{xx}(x+dx,y,z)]} + {\color{blue}dzdx[-\sigma_{yx}(x,y,z) + \sigma_{yx}(x,y+dy,z)]} + {\color{green}dxdy[-\sigma_{zx}(x,y,z) + \sigma{zx}(x,y,z+dz)]}$$
 * The red term has facets normal to the x-axis, the blue has facets normal to the y-axis, and the green has facets normal to the z-axis.

$$0 = $$( dxdydz )$$[\frac{d\sigma_{xx}}{dx} + \frac{d\sigma_{yx}}{dy} + \frac{d\sigma_{zx}}{dz}]$$

This ends Step C of the Road Map.

title: User:Eas4200c.f08.wiki.f/10-31-08


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User:Eas4200c.f08.gator.edwards/Week Five Class Notes

 Class Notes (current)

 Matlab Program Continued (current)

Contributing Team Members:

The following members contributed to this report:

Greg Edwards Eas4200c.f08.gator.edwards 01:02, 5 November 2008 (UTC)

William Mueller Eas4200c.f08.gator.mueller 01:48, 5 November 2008 (UTC)

Bryan Jackson Eas4200c.f08.spars.prey 08:33, 7 November 2008 (UTC)

Justin Dickinson Eas4200c.f08.gator.Dickinson 11:27, 7 November 2008  (UTC)

Taylor Lusk Eas4200c.f08.gator.lusk 21:17, 7 November 2008 (UTC)


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text:
Electricity/Introduction

Let us begin with a few fundamental ideas about electricity to anchor all of your future lessons...

Fundamental Background
First of all, you might ask yourself, 'what exactly is electricity?'

At the level of atomic physics, electricity is one of the fundamental forces of nature, arising from an "electric charge" possessed by subatomic particles (notably protons and electrons).

Electricity in Practice
The preceding description in terms of fundamental particles and forces is perhaps too theoretical for practical use. What is electricity in practice?

Electric charge
The tiny imbalance of positive and negative charges discussed above is caused by electrons moving (in enormous numbers at the atomic scale but very tiny numbers percentage-wise) away from the atomic nuclei to which they belong.

The unit of quantity of electric charge is the coulomb, abbreviated "C", and named after Charles-Augustin Coulomb. It is 6.241 x 1018 elementary charges. Or equivalently, the elementary charge (charge on a single electron or proton) is 1.602 x 10-19 coulombs.

Electric current
Flow of charge—what we call current—could be thought of as flow of positive charge in one direction, or of negative charge in the other direction. By convention, current flow is always considered to be the flow of positive charge, even though that is contrary to the actual flow of (negative) electrons. Hence the current is considered to flow out of the positive terminal of a battery, through whatever is being powered, and back into the negative terminal, even though electron flow through the wires is in the opposite direction.

The unit of current flow is the ampere, abbreviated "A", which is a flow rate of one coulomb per second. It is named after André-Marie Ampère. In the hydraulic analogy, it is analogous to the amount of flow through a pipe, or over a dam, etc., which might be measured in liters per second with the hydraulic analogy.

Kirchoff's Laws
There are two famous "laws" of electrical engineering, known as Kirchhoff's Voltage Law and Kirchhoff's Current Law. They are really just straightforward applications of what has been discussed above.

Kirchhoff's Voltage Law (sometimes abbreviated KVL) The sum of the voltage drops around any closed path in a circuit is zero

This is just conservation of energy—a quantity of electric charge could pick up energy for free by going around a closed loop with a nonzero total voltage drop. Remember that "voltage drop" between two points can be either positive or negative.

Kirchhoff's Current Law (sometimes abbreviated KCL) The sum of the currents flowing into a point in a circuit is zero

This is just conservation of charge. Remember that current flow "into" a point is negative if the current is flowing out. now

Ohm's Law
There is a similar principle for electricity: Ohm's law, discovered by Georg Simon Ohm in the 1820's Ohm's law states that The current through a resistive medium is proportional to the applied voltage
 * $$V = IR\,$$  Voltage equals current times resistance.
 * Also stated, of course, as:
 * $$I = \frac{V}{R}$$
 * $$R = \frac{V}{I}$$

Here are a few approximate conductivities, in siemens per meter:
 * Superconductors: infinity (because of complicated quantum-mechanical phenomena)
 * Silver: 6 * 107
 * Silicon: 2 * 10-3 (but, in semiconductor materials, it is "doped" with impurities, giving it a much higher conductivity)
 * Glass: about 10-12
 * Air: about 10-14
 * Paraffin: about 10-17
 * Teflon: about 10-23

Watt's Law
Watt's law, discovered by Watt in the 1820's, states that the power through a resistive medium is proportional to the applied voltage

 * $$P = IV = $$  Power equals current times voltage.
 * Also stated, of course, as:
 * $$I = \frac{P}{V}$$
 * $$V = \frac{P}{I}$$

"I-squared-R" and Joule's law
One sometimes encounters phrases like "I-squared-R losses" in a wire. This is a combination of the power formula and Ohm's law.
 * $$P = VI = (IR) I = I^2 R\,$$

Notice that the amount of power lost in a wire is proportional to the square of the current.

The phenomenon of heat being evolved when electricity passes through a conductor is sometimes called Joule's law. It's really just a consequence of conservation of energy and the equivalence of heat to other forms of energy.

Symbols and Abbreviations
The symbols used for quantities in electrical engineering can be confusing, since the symbol for a quantity may be different from the symbol for the units in which it is measured. We list the common symbols here, even though we have not yet defined all of the concepts involved.

Electrical charge

 * is typically denoted $$Q$$, measured in coulombs, abbreviated $$C$$.

Current

 * is typically denoted $$I$$, measured in amperes, abbreviated $$A$$.

An ampere is, among other things, a coulomb per second.

 * Example: $$V = IR\,$$  Ohm's law; voltage equals current times resistance.
 * Example: "$$I = 28mA\,$$"  "The current is 28 milliamperes."

Voltage

 * is typically denoted $$V$$ (or sometimes $$E$$), and measured in volts, abbreviated $$V$$. The use of "$$E$$" stands for "emf" (electro-motive force.)

A volt is, among other things, a joule per coulomb.

 * Example: "$$V = 28mV\,$$"  "The voltage is 28 millivolts."

Resistance

 * is typically denoted $$R$$, measured in ohms, abbreviated with the capital Greek omega: $$\Omega$$.

An ohm is, among other things, a volt per ampere.

 * Example: "$$R = 2.7K\Omega\,$$"  "The resistance is 2.7 kilohms."

Conductance

 * It is sometimes useful to speak of the reciprocal of resistance. This is called conductance, and is typically denoted $$G$$, traditionally measured in "mhos" ("mho" is "ohm" spelled backwards), abbreviated with an upside-down omega: $$\mho$$.  A less flippant term than "mho" has been adopted: the siemens, abbreviated $$S$$.

A mho/siemens is, among other things, an ampere per volt.

 * Example: $$I = VG\,$$  Ohm's law rewritten in terms of conductance.
 * Example: "$$G = 65m\mho\,$$"  "The conductance is 65 millimhos."
 * Example: "$$G = 65mS\,$$"  "The conductance is 65 millisiemens."

Capacitance

 * is typically denoted $$C$$, measured in farads, abbreviated $$f$$.

A farad is, among other things, a second per ohm, or a coulomb per volt.

 * Example: $$t = RC\,$$  The time constant is the resistance times the capacitance.
 * Example: "$$C = 75pf\,$$"  "The capacitance is 75 picofarads."

Inductance

 * is typically denoted $$L$$, measured in henries, abbreviated $$h$$.

A henry is, among other things, an ohm-second.

 * Example: $$f = \frac{1}{2\ \pi\ \sqrt{L\ C}}$$  is the formula for the frequency of a resonant circuit.
 * Example: "$$L = 120nh\,$$"  "The inductance is 120 nanohenries."

Power

 * is typically denoted $$P$$, measured in watts, abbreviated $$W$$.

A watt is, among other things, a joule per second, or a volt-ampere.

 * Example: $$P = VI\,$$  The power is the voltage times the current.
 * Example: "$$P = 75W\,$$"  "The power is 75 watts."

The frequency

 * of an oscillation or signal is typically denoted $$f$$, measured in hertz, abbreviated $$Hz$$.

A hertz is really just a reciprocal second. In fact, the unit of frequency used to be just "cycles per second" or simply "cycles".

 * Example: "$$f = 102.5 Mc\,$$"  "102.5 megacycles on the FM dial" (old way.)
 * Example: "$$f = 102.5 MHz\,$$"  "102.5 megahertz on the FM dial" (new way.)

It happens that, in a lot of the mathematical formulas the unit of radians per second is superior. A frequency in radians per second is $$2\ \pi$$ times the frequency in hertz. When measured this way, the symbol $$f$$ is replaced with the lower-case Greek omega: $$\omega$$. Many occurrences of $$2\ \pi$$ disappear from various formulas when radians are used.

 * Example: $$\omega = \frac{1}{\sqrt{L\ C}}$$  is the formula for the frequency of a resonant circuit, in radians per second.
 * Example: $$X_C = \frac{1}{2\ \pi\ f\ C}$$  is the formula for capacitive reactance, calibrated in hertz.
 * Example: $$X_C = \frac{1}{\omega\ C}$$  is the formula for capacitive reactance, calibrated in radians per second.
 * Example: "$$\omega = 644.0265mrad/s\,$$"  "644.0265 megaradians per second on the FM dial" (not known to have ever been announced.)

Static Electricity
Electricity manifests itself in two seemingly different ways. They are different manifestations of the same thing. "Static" electricity was known to the ancients. It involves very high voltages and very low currents—the currents are so low that one doesn't always realize how high the voltage is. In order for such high voltages to persist somewhere, the insulation must be extremely good, that is, the resistances must be extremely high. Fortunately for the history of science, materials like glass, amber, and some types of rubber and other materials, have the necessary high resistance. (The word "electricity" comes from the Greek word for amber.)

The voltages involved with static electricity are high enough to make the leaves of an electroscope move, but the currents are so low that we would normally not notice any effect of the current.

"Current" Electricity
"Current" electricity involves lower voltages and currents large enough to power light bulbs, motors, and such. Rubbing a glass rod with a piece of silk can't come anywhere near to providing the required level of sustained current. To get the required sustained levels of current requires either ongoing chemical reactions (as in a battery) or electromechanical devices (as in a generator.) The ability to do this was discovered in the 18th and 19th centuries by Luigi Galvani, Alessandro Volta and Michael Faraday.

The study of electronics and electrical engineering involve "current" electricity almost exclusively.

Series and Parallel Connections
The terms "series" and "parallel" are actually used quite loosely by electrical engineers, to describe aspects of circuit topology.

 * Series connection

 * Parallel connection
 * [[Image:Seriesparallel.jpg|right|thumb|500 px|Schematic diagram showing series and parallel connections.]]

 * 2 port network

How Are the Units Defined?
As one can tell from the connections among the electrical units and the physical units such as newtons, joules, and watts, a lot of care went into the design of the system of units. But how is the coulomb defined? Why is it equal to 6.241 x 1018 elementary charges?

The coulomb and volt are defined in terms of the ampere, so that a volt times an ampere equals a watt, that is, a newton-meter per second. And a coulomb is an ampere-second.

The ampere was determined experimentally, as follows: it is the amount of current flowing in each of two infinitely long and infinitely thin parallel wires separated by a distance of one meter that causes an attractive magnetic force of $$2 \times 10^{-7}$$ newtons per meter along the length of the wires. Why the factor of $$2 \times 10^{-7}$$? This was chosen to make the volt a reasonable quantity relative to the voltage coming out of batteries.

This definition of the units in terms of the magnetic force has the effect of defining the fundamental constant of the magnetic force, labeled $$\mu\,$$, to be $$4 \pi \times 10^{-7}\,$$. The fundamental constant of the electric force, labeled $$\epsilon\,$$, is related to this, according to the formula
 * $$c = \frac{1}{\sqrt{\epsilon \mu}}\,$$

where $$c\,$$, the speed of light, was traditionally determined experimentally.

The Next Lecture
For the next lecture, see Introduction to Electricity II.

title: Electricity/Introduction


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