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User talk:Prasadpepakayala

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

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title: User talk:Prasadpepakayala


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Czech Language/Body Parts

This vocab resource includes a detailed list of parts of the human body and other anatomical features.

General Anatomical Vocab
kloub = joint končetina = limb kost = bone krev = blood kůže = skin orgán = organ prst = digit (a finger or toe) sval = muscle

Top-down (en-cs)

 * head = hlava
 * hair = vlasy
 * ears = uši
 * face = tvář
 * brow = čelo/obočí
 * cheeks = líce
 * eyes = oči
 * jaw = čelist
 * mouth = ústa
 * lips = rty
 * teeth = zuby
 * tongue = jazyk
 * nose = nos
 * neck = krk
 * throat = hrdlo
 * shoulders = ramena
 * arms = paže/ruce
 * elbows = lokty
 * forearms = předloktí
 * wrists = zápěstí
 * hands = ruce
 * fingers = prsty
 * thumb = palec
 * chest = hruď
 * breasts = prsa
 * abdomen = břicho
 * navel = pupek
 * pelvis = pánev
 * hip = bok
 * groin = slabina
 * penis = pyj/penis
 * vagina = pochva/vagína
 * back = záda
 * backside = zadek
 * buttocks = hýždě
 * legs = nohy
 * thighs = stehna
 * knees = kolena
 * ankles = kotníky
 * feet = nohy [same as legs]
 * heels = paty
 * toes = prsty [same as fingers]

Alphabetical (cs-en)

 * bok = hip
 * břicho = abdomen
 * čelist = jaw
 * čelo = brow
 * hlava = head
 * hrdlo = throat
 * hruď = chest
 * hýždě = bottom, bum
 * jazyk = tongue
 * kolena = knees
 * kotníky = ankles
 * krk = neck
 * líce = cheeks
 * lokty = elbows
 * nohy = legs, feet
 * nos = nose
 * obočí = brow, eye-brows
 * oči = eyes
 * palec = thumb
 * paty = heels
 * pánev = pelvis
 * paže = arms
 * penis = penis
 * pochva = vagina, vulva [also means "sheath" or "scabbard"]
 * prsa = breasts
 * prsty = fingers, toes, digits
 * pupek = navel
 * pyj = penis
 * rty = lips
 * ruce = hands, arms
 * slabina = groin
 * stehna = thighs
 * tvář = face
 * uši = ears
 * ústa = mouth
 * vagína = vagina
 * vlasy = hair(pl)
 * záda = back
 * zadek = backside
 * zuby = teeth

cs-en

 * mozek = brain
 * plíce = lung
 * srdce = heart
 * žaludek = stomach
 * dvanáctníku = duodenum
 * lačník = jejunum
 * ileum = ileum
 * tenké střevo = small intestine
 * tlusté střevo = large intestine
 * tračník = colon
 * Sigmoideum = sigmoid colon
 * konečník = rectum
 * řiť = anus
 * játra = liver
 * slezina = spleen

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

title: Czech Language/Body Parts


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User:Iglika kazanaklieva

INTRODUCTION
My name is Iglika Kazanaklieva and I'm 3rd year student of Primary School Pedagogics in SU "St. Kliment Ohridski" in Sofia, Bulgaria. I'm the tutor of this years "Internet- and Project Competence”.

Last year i took part in the project and my theme was about "Problem Based Learning Through Games". It was very fun and helpful to work on a project like this. My goal was to see if this method will give good results based on the Bulgarian educational system.

CONTACT
iggy_pop@abv.bg

title: User:Iglika kazanaklieva


text:
User:Eml4500.f08.FEABBQ/HW4

Notes for Oct. 6, 2008
Connectivity Array "conn"

Consider the two bar truss system:

In this diagram the relationship between the local and global nodes can be seen. These relations are stored in the "connectivity array" in our Matlab Code:

$$ conn = \begin{bmatrix} 1 & 2 \\ 2 & 3 \\ \end{bmatrix} $$

$$conn(e,j)=$$ global node number of local node $$j$$ at element $$e$$

Location Matrix Master Array "lmm"

The Location Matrix Master Array stores the equation number for the element stiffness coefficient at each local degree of freedom

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

$$lmm(i,j)=$$ equation number (global degree of freedom #) for element stiffness coefficient to the $$j$$th local degree of freedom number $$i$$

Notes for Oct. 8, 2008
GOAL: To find $$\underline{\tilde{T}}^{(e)}$$ that transforms the set of local DOFs, $$\underline{d}^{(e)}$$, to another set of local DOFs, $$\underline{\tilde{d}}^{(e)}$$ , such that $$\underline{\tilde{T}}^{(e)}$$ is invertible.

Figure 2 below shows an element with both the original and transformed local DOFs.

Let:

$$\underline{\tilde{d}}^{(e)}=\underline{\tilde{T}}^{(e)}\underline{d}^{(e)}$$ (1)

Since,

$$\tilde{d_1}^{(e)}=\left[ \begin{matrix} l^{(e)}m^{(e)} \end{matrix} \right] \left[ \begin{matrix}{d_1}^{(e)}\\{d_2}^{(e)}\end{matrix} \right]$$ and,

$$\tilde{d_2}^{(e)}=\left[ \begin{matrix} -m^{(e)}l^{(e)} \end{matrix} \right] \left[ \begin{matrix}{d_1}^{(e)}\\{d_2}^{(e)}\end{matrix} \right]$$

we see that,

$$\left[ \begin{matrix}\tilde{d_1}^{(e)}\\\tilde{d_2}^{(e)}\end{matrix} \right]=\left[ \begin{matrix}l^{(e)} & m^{(e)}\\-m^{(e)} & l^{(e)}\end{matrix} \right]\left[ \begin{matrix}{d_1}^{(e)}\\{d_2}^{(e)}\end{matrix} \right]$$

Defining the 2x2 matrix above as $$\underline{R}^{(e)}$$, equation (1) becomes:

$$\left[ \begin{matrix}\tilde{d_1}^{(e)}\\\tilde{d_2}^{(e)}\\\tilde{d_3}^{(e)}\\\tilde{d_4}^{(e)}\end{matrix} \right]=\left[ \begin{matrix} \underline{R}^{(e)} & \underline{0}\\\underline{0} & \underline{R}^{(e)}\end{matrix} \right]\left[ \begin{matrix}{d_1}^{(e)}\\{d_2}^{(e)}\\{d_3}^{(e)}\\{d_4}^{(e)}\end{matrix}\right]$$

Figure 3 below shows an element with its corresponding transformed forces and DOFs.

Here the forces are defined by:

$$\underline{\tilde{f}}^{(e)}=k^{(e)}\left[\begin{matrix}1&0&-1&0\\0&0&0&0\\-1&0&1&0\\0&0&0&0\end{matrix}\right]\underline{\tilde{d}}^{(e)}$$

or

$$\underline{\tilde{f}}^{(e)}=\underline{\tilde{k}}^{(e)}\underline{\tilde{d}}^{(e)}$$

Notes for October 10
considering the case where $$ \mathbf{\tilde{d}_{4}^{(e)}} \neq 0     $$,     and       $$  \mathbf{\tilde{d}_{1}^{(e)}}=\mathbf{\tilde{d}_{2}^{(e)}}=\mathbf{\tilde{d}_{3}^{(e)}} = 0   $$

$$ \mathbf{\tilde{f}_{4x1}^{(e)}}= \mathbf{\tilde{k}_{4x4}^{(e)}}\mathbf{\tilde{d}_{4x1}^{(e)}}= \mathbf{0_{4x1}}  $$

The zero matrix $$\mathbf{0_{4x1}} $$ is the fourth col. of matrix$$\mathbf{\tilde{k}_{4x4}^{(e)}}$$

Interpretation of transverse dof's:

$$ \mathbf{\tilde{d}_{4x1}^{(e)}}= \mathbf{\tilde{T}_{4x4}^{(e)}}\mathbf {d_{4x1}^{(e)}}  $$

$$ \begin{Bmatrix} \tilde{d}_1^{(e)} \\ \tilde{d}_2^{(e)}\\ \tilde{d}_3^{(e)} \\\tilde{d}_4^{(e)} \end{Bmatrix} = \left[ \begin{matrix} R_{2x2}^{(e)} & 0_{2x2} \\  0_{2x2}& R_{2x2}^{(e)} \end{matrix} \right] \begin{Bmatrix} d_1^{(e)} \\ d_2^{(e)} \\ d_3^{(e)} \\ d_4^{(e)} \end{Bmatrix} $$

Similarly for: $$  \mathbf{\tilde{f}^{(e)}}= \mathbf{\tilde{T}^{(e)}}\mathbf{f^{(e)}}  $$

$$ \begin{Bmatrix} \tilde{f}_1^{(e)} \\ \tilde{f}_2^{(e)}\\ \tilde{f}_3^{(e)} \\\tilde{f}_4^{(e)} \end{Bmatrix} = \left[ \begin{matrix} R_{2x2}^{(e)} & 0_{2x2} \\  0_{2x2}& R_{2x2}^{(e)} \end{matrix} \right] \begin{Bmatrix} f_1^{(e)} \\ f_2^{(e)} \\ f_3^{(e)} \\ f_4^{(e)} \end{Bmatrix} $$

Also: $$\mathbf{\tilde{k}^{(e)}}\mathbf{\tilde{d}^{(e)}}=\mathbf{\tilde{f}^{(e)}}$$

$$\mathbf{\tilde{k}^{(e)}}\mathbf{\tilde{T}^{(e)}}\mathbf{\tilde{d}^{(e)}}=\mathbf{\tilde{T}^{(e)}}\mathbf{f^{(e)}}$$

if $$\mathbf{\tilde{T}^{(e)}}$$ is invertible then:

$$\left[\mathbf{\tilde{T}^{(e)^{-1}}}\mathbf{\tilde{k}^{(e)}}\mathbf{\tilde{T}^{(e)}}\right]\mathbf{{d}^{(e)}}=\mathbf{f^{(e)}}$$

Where $$\mathbf{\tilde{T}^{(e)}}$$ is block diagonal matrix, consider a general block diagonal matrix A

$$\mathbf{A} = \left[ \begin{matrix} D_{1} &...& 0  \\  ...& D_{2}&...\\ 0&...&D_{n} \end{matrix} \right]$$

$$\mathbf{D}_{1}, \mathbf{D}_{2}, .... , \mathbf{D}_{s}   $$     are matrices

$$\mathbf{A}^{-1}= ?$$

$$\mathbf{B}^{-1}=\left[ \begin{matrix} d_{11} &...& 0 \\  ...& d_22&...\\ 0&...&d_(nn) \end{matrix} \right] $$

$$\mathbf{B}=Diag\left[ \mathbf{d}_{11},\mathbf{d}_{22}, ...... , \mathbf{d}_{nn}  \right] $$

$$\mathbf{B}^{-1}=Diag\left[ \dfrac{1}{d}_{11},\dfrac{1}{d}_{22}, ...... , \dfrac{1}{d}_{nn}  \right] $$

Assuming $$\mathbf{d}_{ii}\neq 0 $$ for i = 1,2, ....,n for a block diagonal matrix A

$$\mathbf{A}=Diag\left[ \mathbf{D}_{1},\mathbf{D}_{2}, ...... , \mathbf{D}_{s}  \right] $$

$$\mathbf{A}^{-1}=Diag\left[ \mathbf{D}_{1}^{-1},\mathbf{D}_{2}^{-1}, ...... , \mathbf{D}_{s}^{-1}  \right] $$

$$\mathbf{\tilde{T}^{(e)^{-1}}} =Diag\left[\mathbf{R}^{(e)^{-1}}, \mathbf{R}^{(e)^{-1}}  \right] $$

$$\mathbf{R}^{(e)^{T}}=\left[ \begin{matrix} l^{(e)} & -m^{(e)} \\  m^{(e)} & l^{(e)} \end{matrix} \right]$$

$$\mathbf{R}_{2x2}^{(e)^{T}}\mathbf{R}_{2x2}^{(e)}=\left[ \begin{matrix} l^{(2)}+m^{(2)} & 0 \\  0 & l^{(2)}+m^{(2)} \end{matrix} \right] = \left[ \begin{matrix} l^{(2)}+m^{(2)} =1 & 0  \\  0 & l^{(2)}+m^{(2) =1} \end{matrix} \right] =\left[ \begin{matrix} 1 & 0  \\  0 & 1 \end{matrix} \right] = \mathbf{I}_{2x2}$$

From this we can conclude that $$\mathbf{R}_{2x2}^{(e)^{(-1)}}=\mathbf{R}_{2x2}^{(e)^{T}}$$

Also $$\mathbf{\tilde{T}^{(e)^{-1}}} =Diag\left[\mathbf{R}^{(e)^{T}}, \mathbf{R}^{(e)^{T}}  \right] = (Diag\left[\mathbf{R}^{(e)^{T}} , \mathbf{R}^{(e)^{T}}   \right])^{T}= \mathbf{\tilde{T}^{(e)^{T}}}$$ $$\mathbf{\tilde{T}^{(e)^{-1}}}=\mathbf{\tilde{T}^{(e)^{T}}}$$

$$\left[\mathbf{\tilde{T}^{(e)^{T}}}\mathbf{\tilde{k}^{(e)}}\mathbf{\tilde{T}^{(e)}}\right]\mathbf{d^{(e)}}=\mathbf{f^{(e)}}$$

The above relation agrees with the fundamental relation $$\mathbf{k^{(e)}}\mathbf{d^{(e)}}=\mathbf{f^{(e)}}$$

Where : $$ \mathbf{k^{(e)}}=\mathbf{\tilde{T}^{(e)^{T}}}\mathbf{\tilde{k}^{(e)}}\mathbf{\tilde{T}^{(e)}}$$

Notes for October 13, 2008
The four zero-eigenvalues of the $$ \underline{K}$$ matrix correspond to
 * 3 Rigid Body Modes
 * 1 Mechanism

HW 1
Plot the eigenvectors corresponding to the zero eigenvalues of the 2-bar truss system; interpret the results.

Before we do this assignment, it is important to understand the nature of these mode shapes. These mode shapes (that correspond to the zero eigenvalues) may come out as a linear combination of the pure mode shapes (pure rigid body motions and pure mechanism(s)). For example, take a body that has only two rigid body motions: horizontal and directly vertical. The linear combination of these two pure modes would be a diagonal motion. ''In Structural Mechanics, the eigenvalues represent the frequencies while the eigenvectors represent the mode shapes during free vibration. The lower frequencies (eigenvalues) tend to have more prominent vibration.''

Eigenvalue Problem: $$ \underline{K}\underline{v}=\lambda\underline{v} $$

Let {$$ \underline{u}_{1}, \underline{u}_{2}, \underline{u}_{3}, \underline{u}_{4} $$} be the pure eigenvectors corresponding to the four zero eigenvalues.

$$ \underline{K}_{_{6x6}}\underline{u}_{i_{6x1}}=0\cdot\underline{u}_{i_{6x1}}=\underline{0}_{_{6x1}} $$, where i=1,...,4.

A linear combination of {$$ \underline{u}_{i}, i=1,...,4$$}

$$ \sum_{i=1}^4 \alpha_{i_{1x1}} \underline{u}_{i_{6x1}} =: \underline{W}_{_{6x1}} $$, where $$ \alpha_{i} $$ are real numbers.

Note: the $$=:$$ means "equality definition."

$$ \underline{W} $$ is also an eigenvector corresponding to a zero eigenvalue: $$ \underline{K}\underline{W}=\underline{K}(\sum_{i=1}^4\alpha_{i} \underline{u}_{i})=\sum_{i=1}^4 \alpha_{i}(\underline{K}\underline{u}_{i})=\underline{0}=0\cdot\underline{W} $$

From the two-bar truss, we have the following output of $$ \underline{K} $$

And the following is the are the eigenvectors and eigenvalues, using the notation [V,D], where V are the eigenvectors and D are the eigenvalues.

The first four column vectors of V, which correspond to the four zero eigenvalues, are the eigenvectors for those zero eigenvalues. The matrix V is called the modal matrix.

The following are the four column vectors (eigenvectors) plotted as mode shapes. Note: they are shown in the order from the first column to the fourth column of the modal matrix.

HW2
Solve for $$ \underline{\overline{K}}\underline{v}=\lambda\underline{v} $$, the stiffness matrix for the constrained system below. Plot the eigenvectors corresponding to the zero eigenvalues for case (a).

The following is the Matlab code which solves for the global stiffness matrix, $$ \underline{K} $$, for the three bar truss in case (a) above.

The following is the output for the stiffness matrix of the three bar truss in case (a)

Using the method to solve for the eigenvectors of case(a), we get the following. Note: the first five column vectors correspond to zero eigenvalues.

The first five eigenvectors above correspond to zero eigenvalues, therefore their mode shapes are plotted below in order.

The Matlab code which solves for case (a) can be modified to solve for the stiffness matrix, $$ \underline{K} $$ for the four bar truss, case (b)

The following is the output for the stiffness matrix, $$ \underline{K} $$, for the four bar truss system.

Notes for Oct. 15, 2008
Justification of assembly of element stiffness matrices, $$ \mathbf{k^{(e)}} $$ into global stiffness matrices, $$ \mathbf{k} $$.

Consider the example of a 2-bar truss. Recall the element freebody diagram relation of,

$$ \mathbf{k_{4x4}^{(e)}} \mathbf{d_{4x1}^{(e)}} = \mathbf{f_{4x1}^{(e)}} $$

The equilibrium of global node 2 will be determined using:

1. The Euler cut principle

2. The Free Body Diagrams of element 1 and element 2

3. And for node 2, the identity of global dof's to element dof's for both elements 1 & 2

From these the equilibrium of global node 2 is:

$$ F_{x} = 0 = -f_{3}^{(1)} - f_{1}^{(2)} = 0 \qquad (1) $$

$$ F_{y} = 0 = P -f_{4}^{(1)} - f_{2}^{(2)} = 0 \qquad (2)$$

Notes for Oct. 17, 2008
Rearranging equations 1 and 2 give us the following two conditions:

 * $$ \mbox{Eq (1)}\quad \Rightarrow \quad f_3^{(10} + f_1^{(2)} = 0 \qquad \qquad (1')$$

 * $$ \mbox{Eq (2)}\quad \Rightarrow \quad f_4^{(1)} + f_2^{(2)} = P \qquad \qquad (2')$$

Further expanding each of these equations will prove the correct assembly of the third and fourth rows of the global stiffness matrix. The above equations in terms of local degrees of freedom are

 * $$ (1'):\qquad [k_{31}^{(1)}d_1^{(1)} + k_{32}^{(1)}d_2^{(1)} + k_{33}^{(1)}d_3^{(1)} + k_{34}^{(1)}d_4^{(1)}]\; +\; [k_{11}^{(2)}d_1^{(2)} + k_{12}^{(2)}d_2^{(2)} + k_{13}^{(2)}d_3^{(2)} + k_{14}^{(2)}d_4^{(2)}] = 0$$

 * $$ (2'):\qquad [k_{41}^{(1)}d_1^{(1)} + k_{42}^{(1)}d_2^{(1)} + k_{43}^{(1)}d_3^{(1)} + k_{44}^{(1)}d_4^{(1)}]\; +\; [k_{21}^{(2)}d_1^{(2)} + k_{22}^{(2)}d_2^{(2)} + k_{23}^{(2)}d_3^{(2)} + k_{24}^{(2)}d_4^{(2)}] = P$$

Now replacing the local displacement degrees of freedom with their global equivalent:

 * $$ (1'):\qquad [k_{31}^{(1)}d_1 + k_{32}^{(1)}d_2 + k_{33}^{(1)}d_3 + k_{34}^{(1)}d_4]\; +\; [k_{11}^{(2)}d_3 + k_{12}^{(2)}d_4 + k_{13}^{(2)}d_5 + k_{14}^{(2)}d_6] = 0$$

 * $$ (2'):\qquad [k_{41}^{(1)}d_1 + k_{42}^{(1)}d_2 + k_{43}^{(1)}d_3 + k_{44}^{(1)}d_4]\; +\; [k_{21}^{(2)}d_3 + k_{22}^{(2)}d_4 + k_{23}^{(2)}d_5 + k_{24}^{(2)}d_6] = P$$

Once again, further expanding and collecting likes terms for each:

 * $$ (1'):\qquad k_{31}^{(1)}d_1 + k_{32}^{(1)}d_2 + (k_{33}^{(1)}+k_{11}^{(2)})d_3 + (k_{34}^{(1)}+k_{12}^{(2)})d_4 + k_{13}^{(2)}d_5 + k_{14}^{(2)}d_6 = 0$$

 * $$ (2'):\qquad k_{41}^{(1)}d_1 + k_{42}^{(1)}d_2 + (k_{43}^{(1)}+k_{21}^{(2)})d_3 + (k_{44}^{(1)}+k_{22}^{(2)})d_4 + k_{23}^{(2)}d_5 + k_{24}^{(2)}d_6 = 0$$

As one can see these two expressions are exactly what would be found in the matrix multiplication of the 3rd and 4th rows of the global force-displacement relation. Therefore, this sufficiently proves the accuracy of the previous assembly of the global stiffness matrix.

In general, the assembly of element stiffness matrices for any number of elements into the global stiffness matrix can be expressed as follows:

 * $$\underline{K}_{\ n\, x\, n} = A_{e=1}^{nel}\ \ \underline{k}_{\ ned\ x\ ned}^{(e)}$$

where
 * $$\quad \underline{K}$$ - Global stiffness matrix
 * n - total number of global DoFs before eliminating b.c.'s
 * $$\underline{k}^{(e)}$$ - element stiffness matrix of element e
 * ned - number of element DoFs    *note*- ned is much smaller than n
 * $$A_{e=1}^{nel}$$ - assembly operator for e = 1,2,...,nel
 * nel - number of elements

One should note that the assembly operator is necessary rather than a simple summation because each element stiffness matrix is smaller in dimension and only a sub-matrix of the final global stiffness matrix. Therefore the assembly operator denotes implementing element stiffness matrices into the proper corresponding locations within the global stiffness matrix.

Principal of Virtual Work (PVW)
PVW is a concept that was previously used to reduce the global stiffness matrix, $$\underline{K}_{\, 6\, x\, 6}$$, to the modified stiffness matrix, $$\underline{\overline{K}}_{\, 2\, x\, 2}$$.

This concept is key in the derivation of finite element methods for solutions to partial differential equations. For an example, consider the force-displacement relation for a simple bar or spring:

 * $$Kd = F$$
 * $$ \Rightarrow Kd - F = 0 \qquad \qquad (3)$$

Implementing the PVW, equation 3 above is equivalent to:

 * $$ W \cdot (Kd - F) = 0 \qquad \mbox{for all W} \qquad \qquad (4)$$

It should be noted that

 * $$ (3) \Rightarrow (4) \qquad \mbox{trivial}$$

 * $$ (4) \Rightarrow (3) \qquad \mbox{NOT trivial}$$

Since equation 4 is assumed to be true for all W, then careful selection of values of W can infer things about the relation that is equation 3. For example, in the case that W = 1, equation 4 becomes

 * $$ 1 \cdot (Kd - F) = 0 \qquad \Rightarrow \qquad Kd - F = 0$$

thus, proving equation 3.

Matlab Homework
HW: Modify the Matlab code for the two bar truss to solve the 3 bar truss system:

--

Back to pg 12: Method 2 to derive k(e)

-transform a system with 4 degrees of freedom into a system also with 4 degrees of freedom (as opposed to 2) so that the transform matrix is now 4x4 and hopefully invertible.

People who contributed to HW4
Andy Koby 14:44, 17 October 2008 (UTC)

Nik Vitt 02:29, 20 October 2008 (UTC)

Esmail Hadman 11:51, 20 October 2008 (UTC)

William Kurth 12:37, 21 October 2008 (UTC)

Stark 19:09, 24 October 2008 (UTC)

title: User:Eml4500.f08.FEABBQ/HW4


text:
Earth-impact events/Discussion

Background
What have you heard about objects from space striking the Earth? Are you concerned about this? Has the media done a good job of informing the public? --mikeu talk 13:45, 17 October 2008 (UTC)

Questions

 * Can comets strike the Earth in the next 200 years? If so, which is the probability? Davichito 18:42, 17 October 2008 (UTC)

 * Yes, they could. But, most comets are in orbits that do not bring them close to the Earth.  So it is much rarer.  The closest that one will come to our planet (so far as we currently know) will be P/2000 G1 (LINEAR) on 2016-Mar-22.  It will be 12.7 LD at closest, which is much farther than the other "close approaches" talked about in the page.  See the list at the NASA/JPL Near Earth Object Program for more detailed and up to date information.  The probability of any of the comets in that list colliding with the earth in the next 200 years is extremely small.  --mikeu talk 19:01, 17 October 2008 (UTC)

 * In case an asteroid is approaching the Earth and there is a risk of collision, can actual technology be used to deviate the orbit of such object? And how can it be done? Davichito 18:42, 17 October 2008 (UTC)

 * There are a number of asteroid deflection strategies that have been proposed for planetary defense. The first step is to identify the objects early enough to allow the many years it would take to implement a plan to modify the orbit.  This is often referred to as the Spaceguard program.  There has been recent research on the use of a gravitational tractor to deflect asteroids.  There is a new Asteroid Deflection Research Center that has been created to develop these technologies.  --mikeu talk 19:37, 17 October 2008 (UTC)

 * About comets: There is a novel by Jules Verne: Off on a comet which is about a group of people taken to a comet, by accident. Also, I heard that Europe plans to make a trip to a comet in the next 10 years or so. Is that possible with our current technology? What advantages (scientific/technological/resources) can we get by going to a comet?

 * The Rosetta spacecraft was launched in 2004 and will study comet 67P/Churyumov-Gerasimenko when the spacecraft arrives there in 2014. It will release a probe which will land on the comet.  The main purpose of the mission is to understand the early history of the solar system, and how it formed.  The spacecraft also studied asteroid 2867 Steins on September 5, 2008.  --mikeu talk 20:11, 17 October 2008 (UTC)

 * I have recently read a relatively new hypothesis: The Impact And Exit Event, which proposes that the outer planets of our Solar System were created as a direct consequence of a collision between Mars and 'another planet'. Other results of the suggested collisions were that the remnant core of the obliterated planet is now orbiting close to the Sun (Mercury), the resultant accumulation of gases around large gravity wells (Jupiter, Saturn, Uranus) - and the ring systems of each, the creation of the asteriod belt (the rocky remnants of the obliterated planet), the creation of most of the geology of Earth, and the creation of the Moon. The latter two, the hypothesis suggests were caused by another remnant of the Mars collision - an impacting body that collided with Earth. There is no mention of whether the author believes that Earth is at risk of another impact anytime soon. Comets do get a mention - the impact and exit event hypothesis also suggests that comets are the icy remnants of the oceans and atmospheres of the colliding planets that were jettisoned into space following the event, and that they are a natural consequence. An extract from the hypothesis is available: http://www.theimpactandexitevent.com/takla_makan_screenshot.html . I am receptive to the hypothesis, because of what I have subsequently researched. I could do with others taking a look?

 * Thanks for the heads up on this hypothesis. Having read the entire content of the impact and exit event (obtained from Amazon) I have a few comments to make. 1) The collation of what is claimed within the book as 'a timeline of interconnected events' is in fact quite plausible. The context within which this 'timeline of events' is placed provides a sense of awareness rarely found in other (dare I say it?) pseudo-scientific publications. 2) The visual evidence used is sometimes difficult to grasp in terms of an overall 'single event' narrative. However, upon further research of my own I have found that imagery is an extremely important element of the hypothesis, so much so that I understand the difficulties involved in expressing the central thread of the hypothesis. 3) Much of the derogatory comments in some areas of the Internet are clearly based upon ignorance of the details stated within the hypothesis, which is unfortunate. Summary: It took a while for the breadth and scope of this hypothesis to formulate in my head, but now that it has I find myself searching - and finding - my own supporting evidence. If nothing else the hypothesis has a strong element of interactivity, which adds to its appeal.

 * I have just finished reading the Impact And Exit Event and wonder if this hypothesis has been peer reviewed? It is quite a piece of work but I cannot locate any papers referring to it on the internet (apart from the authors' own website from where I purchased my copy and discussions on a few obscure forums) Regards and thanks.

Do we have any defense at all against meteors, black holes, etc.?
Note: this section started at

Like, big ones! In all seriousness, if astronomers detected a continent sized meteor headed our way with say, a year's warning, -couldn't we do something? Or if a black hole was heading our way couldn't we use some kind of antimatter type force to push it towards the moon? Or for that matter, if the moon ever broke free (from a black hole?) and started slowly drifting towards us, could we use some greater force to fastly push it away? When our sun expands during it's death throes, how will life forms of that era deal with it?--Dr. Carefree (talk) 09:49, 3 January 2009 (UTC)

 * No problem. We have loads of asteroid deflection strategies.--Shantavira|feed me 10:15, 3 January 2009 (UTC)

 * Um what? Antimatter type force? Our current ability to produce antimatter is very limited and our ability to use it for practical purposes almost non-existant. And I'm not quite certain what the use of pushing a blackhole to the moon is. Plus the concept of us moving a black hole anytime soon, if ever, seem absurd to me. As to what humans will do, billions of years from now if they still exist and live on earth, I don't think many people have given it that serious thought although I'm sure there's something in science fiction Nil Einne (talk) 10:51, 3 January 2009 (UTC)

 * In practical terms, not only do we have no defense - we don't truly know what form that defense should take (because we don't know enough about the meteors themselves) and we don't have the ability to detect meteors early enough to make a difference.

 * Let's look at those problems individually:

 * Detection: With a large mass on a trajectory that's pointing it at us, the earlier we do something, the easier it is. To take an extreme example: Suppose we only have a few minutes warning...we might have to try to deflect a meteor when it's one earth diameter away from hitting us square-on at the equator...we'd have to bend it's path by 30 degrees to make it miss us instead.  To deflect a mountain moving at a thousand miles a second through an angle of 30 degrees in just a few seconds would probably take more energy than we have on the entire planet - it truly can't be done.  However, if we know that the meteor is a threat (say) 20 years before it hits us - then the amount of deflection we need is a microscopic fraction of a degree and a really gentle nudge would suffice to save our planet.  So early detection - and (most important) accurate orbital parameter determination - is a massive priority both because it gives us time (it might take 5 years to put together a strategy for deflecting this particular rock, building the spacecraft and getting it launched towards it's target) and it reduces the magnitude of our response.
 * Analysis: There are many possible categories of threat. Comets are mostly ice.  Meteors come in several varieties - some are essentially solid chunks of metal, others are solid chunks of rock, still others may be loose collections of small bolders, pebbles or even dust.  Right now, we don't know which is which - which ones are the most common - whether large, dangerous objects are predominantly of one kind or another.  We know (for example) that Comet Shoemaker-Levy broke up into a dozen pieces as it descended towards Jupiter - if we'd had to deflect that comet and we'd sent (say) a single large nuclear bomb then a whole range of disasterous possibilities come to mind: (a) The comet might break up before our rocket gets there and we can now only deflect one out of a dozen large, dangerous chunks. (b) Our bomb might actually do nothing more than break up the comet prematurely without deflecting it's course at all.  So until we "know our enemy" - we're kinda screwed.   We need to send lots of probes out to look in detail at a statistically reasonable collection of comets and meteors - and do lots of science to figure out what's out there.
 * Deflection/Destruction: The problem is that breaking up a large meteor or comet without deflecting it's path doesn't help us. The total damage to the earth from a single rock that's the size of a mountain that weighs a million tonnes is precisely the same as for a million car-sized rocks weighing one tonne each or for a trillion rocks the size of basket balls weighing a few kilograms each.  Simply smashing the meteor into pieces doesn't help at all!  (The number of movies that get this fact wrong is truly astounding!)  So we have to think in terms of deflection - not destruction.  If we have enough time (see "Detection" above) then something as simple as a heavy spacecraft that flies along parallel to the course of the meteor for a few years and provides a REALLY subtle gravitational shift - might be enough.  That's great because it works just as well with a flying rubble heap as it does for a mountain of nickle-iron or a million tonne dirty snowball.  However, getting big, heavy things out of earth orbit and flying as fast as a meteor requires a heck of a lot of fuel and a huge amount of up-front planning.  We certainly don't know about these threats early enough to do that reliably.   So then we're left with hitting the thing hard with a big bomb, hitting it hard with an 'impactor' or nudging it more gently with rocket motors.   None of those things will work for a flying rubble pile.  For solid bodies - that'll work.  We can build a probe with a rocket motor on it.  Make a soft-landing onto the object and start firing our rocket.  So long as the object is strong enough to take that pressure without breaking up - or without our rocket sinking into the surface or tilting sideways and deflecting the rock in the wrong direction - that could work.  But it's more complicated than that if the object is spinning (as many of them are) - because now the rocket has to fire intermittantly when the rock is at the correct orientation or else the miracle of 'spin stabilisation' (which is what makes bullets fly in a nice straight line) will frustrate our efforts.

 * So it's safe to say that right now, we're defenseless on all three levels. Our detection ability is getting slowly better - we have surveyed some of the very largest rocks - and we're tracking their orbits.  Perhaps we can now see mountain-sized rocks soon enough - but something a lot smaller than a mountain (like maybe a school-bus-sized rock) can take out a city - and we're nowhere even close to being able to track those soon enough or accurately enough.  NASA have sent out probes to several meteors and comets to take a close look at them - and we've even tried firing an 'impactor' at one them...but we have a long way to go.  A lot of people are thinking about deflection/destruction strategies...but no governments are building rockets and putting them into storage ready for the day when we'll need them - and funding for the entire process is distinctly lacking.

 * At some point we (as a species) need to seriously consider having a colony somewhere away from the Earth. There is always the possibility of the ultimate planet killer coming along that's too fast, too large, too unstable and/or too close to do anything about.  Having a colony of humans living on (say) Mars with a self-sufficient life-style and a large enough gene-pool is the ultimate way to ensure the survival of the species come-what-may.

 * SteveBaker (talk) 12:43, 3 January 2009 (UTC)

 * If you'd like to learn more about it and help create "coursework", there's some activity on Wikiversity at Earth-impact events/Discussion, as well as an opportunity to use your imagination and research skills for colonizing off-planet at Lunar Boom Town. (And SteveBaker: mind if I copy over what you just wrote above? Good detail there!) --SB_Johnny | talk 12:56, 3 January 2009 (UTC)
 * All contributions to the ref desk fall under the GFDL - so long as you are using them under those same terms, you are welcome to take whatever you need. SteveBaker (talk) 13:45, 3 January 2009 (UTC)

 * Surely it would help to break it up as small objects can be burned up in the earths atmosphere, the amount of mass burned off an object will be proportional to the surface area of the object, which dramatically increases if the object is broken up. To take my point to its logical conclusion, an asteroid of a large given mass would cause significant damage, whereas the same mass of dust colliding with the earth would most likely just flouresce in the atmosphere as it "burned". —Preceding unsigned comment added by 84.92.32.38 (talk) 14:12, 3 January 2009 (UTC)
 * The net kinetic energy that has to be absorbed by the Earth system (atmo included) remains exactly the same. While a single large rock would probably cause more damage (at least more localized damage) based on impact, even the distributed vaporization of a massive asteroid would be a catastrophe.  A school bus sized rock, sure -- vaporize it.  A dino killer?  That won't work. — Lomn 14:52, 3 January 2009 (UTC)
 * Our main capability in asteroid defense is early warning. Here is an incomplete list of early warning systems that I found:
 * Pan-STARRS a (nearly complete) telescope which will continuously survey the sky for asteroid threats
 * Near Earth Asteroid Tracking (NEAT) An organization which has several dedicated telescopes for asteroid detection and tracking.
 * Spaceguard, a loose affiliation of asteroid detection programs/enthusiasts (including a formal program at NASA).
 * Lincoln Near-Earth Asteroid Research (LINEAR) a NASA/USAF/MIT joint project for asteroid detection/tracking
 * Lowell Observatory Near-Earth-Object Search (LONEOS) a completed 15-year project which detected hundreds of asteroids.
 * Minor Planet Center an organization which is the recognized official center for collecting and publishing orbit data about all known asteroids and comets.
 * Catalina Sky Survey another sky survey designed to identify asteroids.
 * Spacewatch an University of Arizona program to discover near earth objects.
 * Near Earth Object Program (NEOP) a NASA effort to study the list of known asteroids and determine the impact risk.
 * We currently know of around 5500 (see the NEOP page) Near Earth Objects, and hundreds more are being discovered every year. So as you can see, we humans have actually put a fair amount of effort into detecting impact threats from space.  No one has ever tested any potential asteroid deflection systems, but as Steve says, early detection is key.  --Bmk (talk) 15:44, 3 January 2009 (UTC)

 * That doesn't seem quite right... if gazillions of little pieces vaporize in the atmosphere or even hit the earth (somewhat more slowed down by atmospheric friction), they wouldn't also vaporize large amounts of actual earth materials as well (which would cause snowstorms in Havana, etc.). Am I missing something? --SB_Johnny | talk 15:33, 3 January 2009 (UTC)

 * This is little as in much smaller than a mountain, not little as in the size of a pebble. The pieces will still be much too large to burn up in the atmosphere. — DanielLC 19:11, 5 January 2009 (UTC)

(edit conflict)A meteor the size of a continent would be larger than the Moon. The largest asteroid we know of, Ceres, is only about 1/4 the diameter of the moon, and it's big enough to be called a dwarf planet. There are plenty of other potential risks, for example a rouge star passing through the inner solar system would disrupt the trajectory of many asteroids and comets, flinging some toward Earth. The chances for a large black hole to pass through the solar system are very small, since it would be more massive than the sun but smaller than New York. Small black holes could evaporate to Hawking radiation. Remember that the asteroid/comet that probably caused the Cretaceous-Tertiary extinction (the one that killed the dinosaurs) was only about 15 km (9 miles) in diameter. As for the sun expanding, we could maybe move to Mars, but chances are we won't even survive that long. Most estimates predict there is a "high probability" for us to become extinct within the next three million years or so. Anyway, an asteroid bigger than, say, Lake of the Woods, would probably crash through the Earth's crust, exposing its mantle, causing further problems. Anyway, there are plenty of other potential doomsday events that could affect us in the near future (try exitmundi, [warning, popups]), and many of those pose a larger threat to us than the likelihood of an asteroid hitting Earth (which will, with 100% probability, eventually happen). In fact, one potentially catastrophic scenario is already unfolding, and could affect us in our lifetime, yet many are refusing to do anything about it. It's called global warming. ~AH1(TCU) 15:35, 3 January 2009 (UTC)

 * The sun won't just slowly expand to engulf the earth - it'll put out huge pulses of hard radiation and do all sorts of other ill-behaved things along the way. There is probably no place in the solar system where we could survive that event.  However, the sun blowing up is a fairly predictable event.  We can give a fairly accurate prediction as to when that'll happen - and doubtless if our ancestors survive that long - they'll know to a very accurate degree when this is going to happen.   So they'd have time to do something to escape to another solar system.  The problem with meteors and comets is that they are more random - and the best response is to try to deflect them somehow.  We'll certainly have a few thousand or even millions of years notice that the sun is going to give up on us - but we'd be lucky to get 20 years notice of an earth-smasher en route.  Black holes are not worth bothering about - there are no big-but-slow ones nearby - and we don't care too much about small ones.  Fast-but-big ones are impossible to deal with - if one of those comes by, there is nothing we can do.  Meteors and comets are in the middle ground - they DO wipe out entire ecologies (the Dinosaurs - and there was a report out a couple of days ago suggesting that the Clovis people of North America were wiped out by a comet/meteor impact) - and we could do something about it with present-day technology if we put our minds to it.  The odds of you or I personally being wiped out by one of these things is tiny - but it's one of the top risks for humanity as a species - so I think we should spend commensurate effort on solving the problem.  Global warming should be higher on our list - but comet/meteor protection ought to be up there in the top ten goals of humanity over the next 100 years. SteveBaker (talk) 02:31, 4 January 2009 (UTC)

We have one rock-solid defense against asteroids: Bruce Willis. —Preceding unsigned comment added by 79.122.10.173 (talk) 14:12, 4 January 2009 (UTC)

Activities
Have you seen a movie or a read a book that depicts a fictional impact event? Did you think that it was realistic? See, and add to, the examples here and describe your thoughts below:

What have you read in the news about asteroid close approaches or impacts? Did the news article provide enough information for you to understand the risks? Recent news coverage of "near misses" are listed here.

title: Earth-impact events/Discussion


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Wikiversity talk:Importers

I changed the link to the main page in Meta, cause it says in the Wikipedia page that it is a copy from meta. If someone thinks I did wrong, please revert. ---- (profile | chit chat | email) 14:11, 17 October 2008 (UTC)

title: Wikiversity talk:Importers


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Wikiversity talk:Think tank

"The Think Tank is intended to serve a similar function to the "WikiProjects" on Wikipedia" - I thought the schools and topics are for the exact same purpose. How about topic:current affairs?Hillgentleman | //\\ |Talk 14:40, 17 October 2008 (UTC)

 * The use of schools and topics has long been confusing. This is multidiscipline in any case, so probably better in the WV namespace. --SB_Johnny talk 15:02, 17 October 2008 (UTC)

 * "Wikiversity:Think tank" sounds like a page for reflections and visions on wikiversity. Usually a project: page is about the project itself.  WikiProject in wikipedia lies in the wikipedia:wikiproject psedo-namespace simply because they have no other choice - but we do.  The title "wikivesity:think tank" is very much as confusing.  But of course it is all up to the participants in this topic.  (However, I picture someone like Cormaggio or Thuvack thinking: "Why did't I take the name first.." :-)). Best, --Hillgentleman | //\\ |Talk 18:00, 18 October 2008 (UTC)

 *  Actually I was going for something with "squad" in the name... Cormaggio get's the credit for the name :-). --SB_Johnny talk 18:35, 18 October 2008 (UTC)

 * Hillgentleman makes a good point. The Jade Knight (d'viser) 22:41, 18 October 2008 (UTC)

 * I am interested in being involved and would like to have a better idea on how a think tank works, do we all think about the proposed topic, write something and come to a consensus? at any rate all these administrative details need defined before anything comes of this project.  I will be following the idea as it unfolds.--Jolie 13:07, 20 October 2008 (UTC)

 * Well, we could do it that way. I am also unclear on what kind of process SB_Johnny has in mind for creating these new projects.  So, I was just bold and started Earth-impact events quickly.  If there are events in the news it might just be better to start a stub, and then refine the administrative details and learning project structure as we go along.  Feel free to make suggestions on how you think this might work, or propose ideas for topics to include.  --mikeu talk 14:17, 20 October 2008 (UTC)

integration of help desk and content creation
Every time someone has a question and is motivated enough to post it on the help desk, it would be a good chance to create a learning resource around the question, even if it is only answered partially. Sometimes we would have some simple practical problems in our work; we may ask people or we may solve them by ourselves (maybe with google); and it is a waste not to share it.

--Hillgentleman | //\\ |Talk 15:46, 23 January 2009 (UTC)

 * I completely agree. I think the new "forums" being developed are probably an improvement over the help desk for this, since the individual topics will have distinct subpages... perhaps the help desk threads could be organized in a similar manner as well. --SB_Johnny talk 16:13, 23 January 2009 (UTC)

 * By new "forums" do you mean new projects or something like this which is currently being developed, I'm slightly confused by the term. DarkObsidian @en.Meta-Wiki 19:17, 23 January 2009 (UTC)

Interactive learning
I'm not sure if this is the correct location to ask and suggest or to go to the Help desk page. My question is about an interactive learning in this wikivirsity through supporting additional media such as animated flash pages, and scripting features as in JScript. The reason I suggest this is to facilitate the learner with all possible ways to browse and interact with some self pre-programmed study. an example of this is identifying locations and related points on an image or a map.--Email4mobile 16:36, 22 June 2009 (UTC)

title: Wikiversity talk:Think tank


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2008 global economic crisis

The global economic crisis of 2008 was triggered initially by problems in the US credit and housing markets, but from there has spread to the rest of the world's economies. This initiative will try to explain how this happened, what measures have been taken to reduce the bad effects, and so forth.

Describing Global Economic Crisis
It is interesting to understand the how an economic or financial crisis emanates before going into the details of its effects, consequences and remedies. The modern world has witnessed several financial crisis during the last 100 years. These crisis have largely been associated with classical booms and busts in credit markets that results in a boom and bust in banking and securities markets. A typical starting point of crisis is a shock in macro economic environment resulting in a changed profit outlook and enhanced bank lendings to fuel the economic wheel. The optimistic sentiments about future outlook lead to an abnormal hike in prices of specific assets and herding henceforth. Then suddenly this buble implodes and cut to size the investors expectation in such an unmanageable way that the widening gap between profits and speculative prices becomes hard to fil. This further moves up the interest rates and results in liquidity dry ups in market. Businesses start to contract in the wake of increasing real values of debt under disinflationary pressures. This all process put that banks in a situation of distress which lead to an ultimate failure if become persistent. The government intervention becomes imminent either in form of bailout, liquidation or forced closure etc.

Methods

 * Survey wikimedia participants on how it has affected them?
 * Survey wikimedia participants on their understanding?
 * Come up with timelines and connections showing how the problem spread from one sector to another, and from one national economy to another.
 * Discursive materials for floating ideas and solutions, discussion preferably including both economists and non-economists.

title: 2008 global economic crisis


text: Template:User Wikiversity

title: Template:User Wikiversity


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Managing a private wiki

MediaWiki software is freely licensed, so anyone with access to a server can start their own wikis. Some hosting companies specialize in it as well.

What you need

 * /MySQL/ or /PostgreSQL/
 * /Apache/ or /Internet Information Services/
 * /PHP/
 * /MediaWiki/

On the server

 * /LocalSettings.php/
 * /AdminSettings.php/
 * /Extensions/
 * /Upgrading/

title: Managing a private wiki


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Talk:Managing a private wiki

This should probably not duplicate documentation already available on MediaWiki.org. We should think more about what information is needed that's not already covered there. – Mike.lifeguard | @en.wb 16:15, 17 October 2008 (UTC)

 * Well, I had in mind something a little more "dummy-friendly", with a much better index of what to do and how, examples of php files that give you a wiki setup without having to add things one at a time, etc. I can say from personal experience that MediaWiki can be hard to navigate and understand, so a more "step-by-step" guide would be helpful (at least for me!). --SB_Johnny talk 16:25, 17 October 2008 (UTC)
 * Yep - that's not duplicating documentation already available at mediawiki.org :D – Mike.lifeguard | @en.wb 17:32, 17 October 2008 (UTC)

 * Huh? (Sorry, not sure what you mean) :-). --SB_Johnny talk 19:40, 17 October 2008 (UTC)
 * What you've suggested isn't covered on MW.org - it's mostly extension-specific information and other development-oriented information. That sort of thing shouldn't be duplicated (but what you've suggested isn't covered there, so that's where this project should likely focus). – Mike.lifeguard | @en.wb 22:32, 17 October 2008 (UTC)

Mike, I want to allow template looping during transclusion; do you know how to hack the parser to do it? Hillgentleman | //\\ |Talk 20:12, 17 October 2008 (UTC)
 * Well, I'm not clear what you mean by "template looping during transclusion" but I'm almost certain the answer is "no" - sorry. – Mike.lifeguard | @en.wb 22:32, 17 October 2008 (UTC)

title: Talk:Managing a private wiki


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

 Class Notes (current) archived

 Homework Problems (current) archived

 Matlab Program Continued (current) archived

 Matlab Certification

I, the undersigned, certify that I can read, understand, and write matlab codes, and can thus contribute effectively to my team.

Greg Edwards Eas4200c.f08.gator.edwards 01:41, 24 October 2008 (UTC)

Taylor Lusk Eas4200c.f08.gator.lusk 06:35, 24 October 2008 (UTC)

William Mueller Eas4200c.f08.gator.mueller 16:21, 24 October 2008 (UTC)

Jacob Papp Eas4200c.f08.team12.papp 13:22, 24 October 2008 (UTC)

Justin Dickinson Eas4200c.f08.gator.dickinson 13:20, 24 October 2008 (UTC)

Contributing Team Members:

The following members contributed to this report:

Greg Edwards Eas4200c.f08.gator.edwards 03:04, 18 October 2008 (UTC)

William Mueller Eas4200c.f08.gator.mueller 21:20, 23 October 2008 (UTC)

Justin Dickinson Eas4200c.f08.gator.dickinson 23:12, 23 October 2008 (UTC)

Taylor Lusk Eas4200c.f08.gator.lusk 08:15, 24 October 2008 (UTC)

Jacob Papp Eas4200c.f08.team12.papp 13:22, 24 October 2008 (UTC)


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User:Eas4200c.f08.gator.edwards/Wk4 Class Notes

Single Cell Airfoil

Shear flow is constant $$q = q_1 = q_2 = q_3$$

$$\theta = \frac{1}{2 G \bar{A}}\ q \sum_{j=1}^{3}\frac{l_{j}}{t_{j}}$$

$$\qquad= \frac{1}{2 G \bar{A}}\ q\left[ \frac{\pi\left( \frac{b}{2}\right)}{t_{1}} + \frac{a}{t_{2}}+\frac{\sqrt{a^2+b^2}}{t_3}\right]$$

$$\ $$

$$\qquad = (\frac{1}{qG})141.13$$

Maximum Shear Stress $$\qquad \tau_{max}$$

$$\tau _{max}=\frac{q}{min \ of\ ( t_1 ,t_2,t_3) }$$

If $$\tau _{max} = \tau _{allow}\ \ \ $$ (given)

and since $$\ \ \ q = \frac{t}{2\bar{A}}\ \ $$

then $$\ \ T_{allow}=2\bar{A}\tau_{allow}min\left\{ t_1,t_2,t_3 \right\}$$

Let $$\ \ \tau_{allow} = \ 100GPa.\ \ $$ Find $$T_{allow}$$

$$T_{allow} = $$7028 MN•m

Multicell Airfoil

Put 2 partitions in airfoil and find rate of twist by dividing c into 3 parts

We will go over a specific example first and a generalization later. The first constraint equation can be considered the torque of the multicell airfoil

$$T=T_{1}+T_{2}=2q_{1}\bar{A}_{1}+2q_{2}\bar{A}_{2}$$

The second being that the airfoil is a rigid body and must twist equally throughout the airfoil

$$\theta _{1} = \theta _{2}$$

These areas for the simplified airfoil are easily calculated with

$$\bar{A}_{1}=(a)(c)$$ $$\bar{A}_{2}=(b)(c)$$

To find the angle of twist as previously shown in the road map of Homework 3 is

$$\theta _{i}=\frac{1}{2G\bar{A}_{i}}\int \frac{q_{i}}{t_{i}}ds$$

For each cell the angle is represented as

$$\theta _{1}=\frac{1}{2G\bar{A}_{1}} \left[\frac{q_{1}}{t_{1}}(a)+\frac{q_{1}}{t_{1}}(c)+\frac{q_{1}}{t_{1}}(a)+\frac{q_{1}}{t_{12}}(c)-\frac{q_{2}}{t_{12}}(c) \right]$$

$$\theta _{2}=\frac{1}{2G\bar{A}_{2}} \left[\frac{q_{2}}{t_{2}}(c)+\frac{q_{2}}{t_{2}}(b)+\frac{q_{2}}{t_{2}}(b)+\frac{q_{2}}{t_{12}}(c)-\frac{q_{1}}{t_{12}}(c) \right]$$

q1 and q2 can be considered unkowns and since we now have to constraint equations we can solve for both q's. This is shown in detail in the homework section representing q in terms of T.

Next, using the two expresions for $$\theta_{}$$ an expresion between $$\theta_{}$$ and T based on the notion that

$$\theta _{1} = \theta _{2}=...= \theta _{c_{n}}$$

Which is in turn equal to

$$=\frac{T}{2GJ}$$

From this we can now easily solve for J

Strain Matrix
There are 6 components to the 3D strain matrix because of symmetry $$$$.

$$\varepsilon =\begin{bmatrix} \varepsilon _{xx} & \varepsilon _{xy} & \varepsilon _{xz}\\ \varepsilon _{yx} & \varepsilon _{yy} & \varepsilon _{yz}\\ \varepsilon _{zx} & \varepsilon _{zy} & \varepsilon _{zz} \end{bmatrix} =\begin{bmatrix} \varepsilon _{11} & \varepsilon _{12} & \varepsilon _{13}\\ \varepsilon _{21} & \varepsilon _{22} & \varepsilon _{23}\\ \varepsilon _{31} & \varepsilon _{32} & \varepsilon _{33} \end{bmatrix} =\left[\varepsilon _{ij} \right]$$

$$\varepsilon _{12}=\varepsilon _{21}$$

$$\varepsilon _{23}=\varepsilon _{32}$$

$$\varepsilon _{13}=\varepsilon _{31}$$

The equation for each element of the 3D strain matrix

$$\varepsilon _{ij}=\frac{1}{2}\left(\frac{\partial u_{i}}{\partial x_{j}}+\frac{\partial u_{j}}{\partial x_{i}}\right)$$

And because of this equation there are only 6 independent component of the 3D strain matrix. Also the stress tensor matrix only has 6 independent components as well.

$$\sigma =\left[\sigma _{ij} \right]_{3x3}$$


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User:Prashanthns

Easier to find me on my wikipedia user page

title: User:Prashanthns


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User:Eas4200c.f08.gator.edwards/Wk4 Homework Problems

Ideal and Rectangular airfoil
The skin thickness is 2mm, spar thickness 3mm, the c value is 0.5 m. First the ideal airfoil will be calculated, a rectangular two cell section, a single cell airfoil, and then compared to a three cell airfoil with spars at .25c and .75c from the leading edge.

Ideal Airfoil

$$\bar{A}=\frac{\pi }{2}(h^{2})+\frac{1}{2}(h)(l)$$

$$T=2q\bar{A}$$

$$\theta =\frac{q}{2\bar{A}G}\oint{\frac{ds}{t}}$$

$$\theta =\frac{q}{2\bar{A}G}\left[\frac{h\pi }{2t_{skin}} +\frac{l}{t_{skin}}+\frac{\sqrt{h^{2}+l^{2}}}{t_{skin}} \right]$$

$$J=\frac{T}{\theta G}$$

Rectanglular Two Cell

$$T=2\sum_{i=1}^{2}{q_{i}\bar{A}_{i}}$$ gives the next three equations

$$q_{1}=\frac{T}{2\bar{A}_{1}}$$, $$q_{2}=\frac{T}{2\bar{A}_{2}}$$

$$\theta _{i}=\frac{1}{2\bar{A}_{i}G}\oint{\frac{q.ds}{t}}$$

$$\theta _{1}=\frac{1}{2\bar{A}_{1}G}\left[\frac{q_{1}(2a+h)}{t_{skin}}+\frac{q_{12}h}{t_{spar}} \right]$$

$$\theta _{1}=\frac{1}{2\bar{A}_{2}G}\left[\frac{q_{2}(2b+h)}{t_{skin}}+\frac{q_{12}h}{t_{spar}} \right]$$

$$J=\frac{T}{\theta G}$$

Strain Matrix Symmetry
Since the equation for each element of the strain matrix is given by

$$\varepsilon _{ij}=\frac{1}{2}\left[\frac{\partial u_{i}}{\partial x_{j}}+\frac{\partial u_{j}}{\partial x_{i}} \right]$$

So the term $$\varepsilon _{ji}$$ is equal to the term $$\varepsilon _{ij}$$

$$\varepsilon _{ji}=\frac{1}{2}\left[\frac{\partial u_{j}}{\partial x_{i}}+\frac{\partial u_{i}}{\partial x_{j}} \right]$$

This proves that the strain matrix is symmetric.

Hooke's Law
Normal Strains:

$$\epsilon _{xx}=\frac{\sigma _{xx}}{E} - \frac{\nu \ \sigma_{yy}}{E} - \frac{\nu \  \sigma_{zz}}{E}$$

$$\epsilon _{yy}=-\frac{\nu \ \sigma _{xx}}{E} + \frac{\sigma_{yy}}{E} - \frac{\nu \  \sigma_{zz}}{E}$$

$$\epsilon _{zz}=-\frac{\nu \ \sigma _{xx}}{E} - \frac{\nu \  \sigma_{yy}}{E} + \frac{\sigma_{zz}}{E}$$

$$\ $$

Shear Strains:

$$\gamma _{xy} = 2\epsilon _{xy} = \frac{\tau _{xy}}{G}$$

$$\gamma _{yz} = 2\epsilon _{yz} = \frac{\tau _{yz}}{G}$$

$$\gamma _{zx} = 2\epsilon _{zx} = \frac{\tau _{zx}}{G}$$

Strains in a Uniform bar
From the Kinematic Assumptions derived earlier:

$$u_{y} = -\Theta x z$$

$$u_{z} = \Theta x y$$

$$u_{x} = \Theta \Psi(y,z)$$

(In this coordinate system the x-axis lies along the axial direction and the cross-section lies in the y-z plane)

The equation for normal strains is the partial derivative of the displacement taken with respect to the direction of the displacement.

$$\epsilon_{xx} = \frac{\partial u_{x}}{\partial x} $$

$$\epsilon_{yy} = \frac{\partial u_{y}}{\partial y} $$

$$\epsilon_{zz} = \frac{\partial u_{z}}{\partial z} $$

Substituting the displacements into these formulae renders:

$$\epsilon_{xx} = \frac{\partial u_{x}}{\partial x} = \frac{\partial(\Theta\Psi(y,z))}{\partial x} = 0$$

$$\epsilon_{yy} = \frac{\partial u_{y}}{\partial y} = \frac{\partial(-\Theta x z)}{\partial y} = 0$$

$$\epsilon_{zz} = \frac{\partial u_{z}}{\partial z} = \frac{\partial(\Theta x y)}{\partial z} = 0$$

The equation for shear strains is the partial derivative of the displacement in direction a taken with respect to b, plus the partial derivative of the displacement in direction b with respect to a.

$$\gamma_{ab} = \frac{\partial u_{a}}{\partial b} + \frac{\partial u_{b}}{\partial a}$$

Hooke's Law for Isotropic Elasticity
$$\begin{Bmatrix}\epsilon _{11}\\\epsilon _{22}\\\epsilon _{33}\\\gamma _{23}\\\gamma _{31}\\ \gamma _{12}\end{Bmatrix}= \begin{Bmatrix} \frac{1}{E} & \frac{-\nu}{E} & \frac{-\nu}{E} & 0 & 0 & 0\\ \frac{-\nu}{E}& \frac{1}{E}& \frac{-\nu}{E} & 0 &  0&0 \\ \frac{-\nu}{E}& \frac{-\nu}{E} & \frac{1}{E} & 0 & 0 & 0\\ 0 & 0 & 0& \frac{1}{G} & 0 &0 \\ 0&0 & 0 & 0 & \frac{1}{G} & 0\\ 0& 0 & 0&0  & 0 & \frac{1}{G} \end{Bmatrix}\begin{Bmatrix} \sigma _{11}\\ \sigma _{22}\\ \sigma _{33}\\ \sigma _{23}\\ \sigma _{31}\\ \sigma _{12} \end{Bmatrix}$$

$$\ $$

Which can also be written as:

$$\ $$

$$\begin{Bmatrix}\epsilon _{11}\\\epsilon _{22}\\\epsilon _{33}\\\epsilon _{23}\\\epsilon _{31}\\ \epsilon _{12}\end{Bmatrix}= \begin{Bmatrix} \frac{1}{E} & \frac{-\nu}{E} & \frac{-\nu}{E} & 0 & 0 & 0\\ \frac{-\nu}{E}& \frac{1}{E}& \frac{-\nu}{E} & 0 &  0&0 \\ \frac{-\nu}{E}& \frac{-\nu}{E} & \frac{1}{E} & 0 & 0 & 0\\ 0 & 0 & 0& \frac{1}{2G} & 0 &0 \\ 0&0 & 0 & 0 & \frac{1}{2G} & 0\\ 0& 0 & 0&0  & 0 & \frac{1}{2G} \end{Bmatrix}\begin{Bmatrix} \sigma _{11}\\ \sigma _{22}\\ \sigma _{33}\\ \sigma _{23}\\ \sigma _{31}\\ \sigma _{12} \end{Bmatrix}$$

Notes on the Poisson's Ratio
For steel: $$\ \nu = 0.3$$

For cork: $$\ \nu = 0$$

For rubber: $$\ \nu = 0.5$$

$$\ $$

For most materials, $$\ 0\leq \nu \leq 0.5 \qquad \ $$ but some materials have been developed that have a negative poisson ratio.

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text: File:5-bar truss system.jpg

title: File:5-bar truss system.jpg


text:
Managing a private wiki/LocalSettings.php

This is the file that holds all the configuration settings for your wiki. When making changes, make them here and never in DefaultSettings.php, which is for default settings. LocalSettings.php overrides the defaults, so that's where config settings for your wiki go.

Settings

 * /Default user options/ (setting user preferences for new accounts)
 * /Access/ (who can read, edit, or create accounts)
 * ../Adding extensions/

Example
Here's a sampling of some common settings (not a complete file):


text:
User:BrianDC2DO

This is me trying to learn more about medicine.

title: User:BrianDC2DO


text:
User talk:BrianDC2DO

Hit me with your best shot...fire away.

title: User talk:BrianDC2DO


text:
Managing a private wiki/Extensions

Vanilla MediaWiki is quite limited in comparison to the wikis run by the Wikimedia Foundation. We use many extensions to get enhanced functionality; here are some you may wish to consider using:
 * Extension:CategoryTree
 * Dynamically navigate the category structure

 * Extension:CrossNamespaceLinks
 * Lists links across namespaces that should not exist on Wikiversity

 * Extension:ExpandTemplates
 * Expands templates, parser functions and variables to show expanded wikitext and preview rendered page

 * Extension:Renameuser
 * Adds a special page to rename a user (need renameuser right)

 * DynamicPageList
 * Outputs a bulleted list of the most recent items residing in a category, or an intersection of several categories
 * This one is really useful!

 * Extension:ImageMap
 * Allows client-side clickable image maps using tag

 * Extension:ParserFunctions
 * Enhance parser with logical functions
 * This one is really useful!

 * SyntaxHighlight
 * Provides syntax highlighting using GeSHi Highlighter

 * mw:Extension:OggHandler
 * Handler for Ogg Theora and Vorbis files, with JavaScript player

 * Extension:AntiSpoof
 * Blocks the creation of accounts with mixed-script, confusing and similar usernames

 * Extension:ConfirmEdit & FancyCaptcha
 * Simple captcha implementation

 * Extension:Gadgets
 * Lets users select custom CSS and JavaScript gadgets in their preferences

 * Extension:MWSearch
 * MWSearch extension is a MediaWiki backend to fetch search results from MediaWiki Lucene-based search engine.

 * Extension:SpamBlacklist
 * Regex-based anti-spam tool
 * Can be quite difficult to use effectively; requires knowledge of regex

 * Extension:Title Blacklist
 * Allows administrators to forbid creation of pages (including uploads) and user accounts

title: Managing a private wiki/Extensions


text:
User:MeganSurette

Hi! My name is Megan Surette and I am a junior at University of North Carolina at Charlotte. I am a double major in Elementary Education and Sociology with a minor is Child and family development and want to teach somewhere in the K-2 range.

title: User:MeganSurette


text:
User:Ofey~enwikiversity

Also known as Hevosen

title: User:Ofey~enwikiversity


text:
User:Eas4200c.f08.radsam/HW4

 Comment: This page is identical to the previous one, except that the "October 15th: Meeting 22" was added, which was previously completed but accidentally omitted. Here is a comparison between these two versions. Eas4200c.f08.radsam.a. 23:40, 31 October 2008 (UTC) =Homework #4=

=Class Notes=

October 6th: Meeting 18
 Discussion on Quadrature of an Airfoil and Continuation of Single-Cell Airfoil Analysis  Note: Quadrature of NACA airfoil (Figure 1).

$$\bar{A}=\bar{A_1}+\bar{A_2}=\bar{A_1}-|\bar{A_2}|\,$$ where $$\bar{A_1}>0\,$$ and $$\bar{A_2}<0\,$$.

How about using trapezoidals to integrate (Figure 2)? You could, but there are a couple of disadvantages:
 * There is change in curvature, and
 * It is just not as elegant as the other quadrature (triangles, as in Figure 3).

Back to single-cell airfoil (Pg. 17-3 in notes):

Shear flow is constant:

$$q=q_1=q_2=q_3\,$$

$$\theta=\frac{q}{2G\bar{A}}\sum_{j=1}^3\frac{l_j}{t_j}=\frac{q}{2G\bar{A}}\left [ \frac{2t_1}+\frac{a}{t_2}+\frac{\sqrt{a^2+b^2}}{t_3}\right ]=(HW)q\,$$ where $$\theta\,$$ is the rate of twist and $$HW\,$$ is to compute the numerical factor.

If $$\tau_{max}=\tau_{all}\,$$ (which is given), then the max shear stress, $$\tau_{max}\,$$, is given by: $$\tau_{max}=\frac{q}{min\{t_1,t_2,t_3\}}\,$$

And since $$q=\frac{T}{2\bar{A}}\,$$, $$T_{all}=2\bar{A}\tau_{all}*min\{t_1,t_2,t_3\}\,$$

Now, let $$\tau_{all}=100~GPa\,$$, find $$T_{all}\,$$ (HW).

Contributed by radsam.a.

October 8th: Meeting 19
Given: $$t_1 = 0.3cm\,$$

$$t_2 = 0.5cm\,$$

$$t_{12} = 0.4cm\,$$

$$a = 30cm\,$$

$$b = 60cm\,$$

$$c = 40cm\,$$

Find $$\theta\,$$ as a function of torque $$T\,$$ and torsional constant $$J\,$$ Due to superposition of shear flows, we know the total torque is the summation of $$T_1\,$$ and $$T_2\,$$

 $$T = T_1 + T_2 = 2 \bar A_1\ q_1 + 2 \bar A_2\ q_2\,$$ $$Eqn. 1\,$$  Computing the areas for each one of the cells:

 $$\bar A_1\ = ac\,$$ and  $$\bar A_2\ = bc$$ The expression for the twist angle $$\theta\,$$ was previously computed. Refer to 'Eqn. 3.56' in "Mechanics of Aircraft Structures":

 $$\theta_1\ = \frac{1}{2 G \bar A_1} \oint \frac{q_1}{t_1} ds\,$$

The computed area is substituted into twist angle equation:

 $$\theta_1\ = \frac{1}{2 G \bar A_1} \left [ \frac{2 q_1 a}{t_1} + \frac{q_1 c}{t_1} + \frac{(q_1-q_2) c}{t_{12}} \right ]$$ $$Eqn. 2\,$$  The same procedure followed above is used to obtain the twist angle for cell 2:

 $$\theta_2\ = \frac{1}{2 G \bar A_2} \oint \frac{q_2}{t_2} ds\,$$

 $$\theta_2\ = \frac{1}{2 G \bar A_2} \left [ \frac{2 q_2 b}{t_2} + \frac{q_2 c}{t_2} + \frac{(q_2-q_1) c}{t_{12}} \right ]$$ $$Eqn. 3\,$$

$$c_1\,$$ and $$c_2\,$$ have the same rate of twist angle, therefore: $$\theta_1\ = \theta_2\,$$ $$Eqn. 4\,$$

With the four equations above, expressions for $$q_1\,$$ and $$q_2\,$$ in terms of $$T\,$$ may be obtained. For instance:  $$q_1 = \beta_1\ T\,$$

$$q_2 = \beta_2\ T\,$$

Where: $$\beta_1\ = 0.01\,$$ and $$\beta_2 = 0.016\,$$ in N/m. "Computation"

In addition, a general expression for twist angle $$\theta\,$$ may be obtained: $$\theta\ = \theta_1\ = \theta_2\ = \frac{T}{G J}$$ Contributed by radsam.d

October 10th: Meeting 20
Recalling...

Torsion: $$T = GJ\theta\,$$

 * T is the torque (N·m or ft·lbf)
 * G is the shear modulus or more commonly the modulus of rigidity and is usually given in gigapascals (GPa), lbf/in2 (psi), or lbf/ft2
 * J is the torsion constant for the section . It is identical to the polar moment of inertia for a round shaft or concentric tube only. For other shapes J must be determined by other means. For solid shafts the membrane analogy is useful, and for thin walled tubes of arbitrary shape the shear flow approximation is fairly good, if the section is not re-entrant. For thick walled tubes of arbitrary shape there is no simple solution, and FEA may be the best method.
 * $$\theta\,$$ is the angle of twist in radians.

Once $$\theta\,$$ is found as a function of T, use...

$$\theta = \frac{T}{GJ}\,$$

or

$$J=\frac{T}{G\theta\,}$$

...to find J

HW #4:

 * 3-cell NACA 2415 airfoil
 * 2 partition walls
 * 1st wall at quarter chord from leading edge
 * 2nd wall at three quarter chord from leading edge

3 unknowns... $$q_{1}, q_{2}, q_{3}\,$$:

 * 1) $$T=2\sum_{i=1}^3 q_{i} \bar{A}\,$$
 * 2) $$\theta_{1}=\theta_{2}\,$$
 * 3) $$\theta_{2}=\theta_{3}\,$$

Finding J...

Note: need thickness t, but G cancels out...

 * Now Theory derivation

$$T=GJ\theta\,$$ (done)

$$T=2q\bar{A}$$  (done)

$$\theta=\frac{1}{2G\bar{A}}\oint \frac{q}{t} ds\,$$

Uniform bar with non-circular cross section subject to twist

Displacement $$\underline{PP'}$$ due to $$\alpha\,$$

$$\frac{PP'}{O\underline{P}} = \alpha\,$$ (for small alpha)

Projection Displacement $$\underline{PP'}$$ on the direction perpendicular to $$O\underline{P'}:$$

$$\underline{PP''}=\underline{PP'} cos{\alpha}\,$$

$$\underline{PP''}= (O \underline{P} tan{\alpha}) cos{\alpha}\,$$

$$=(O\underline{P} cos{\alpha}) tan{\alpha}$$

''where... $$O\underline{P} cos{\alpha} = O\underline{P}\,$$

Recall:

$$O\underline{P} = r \,$$ (radial coord)

$$O\underline{P''} = \rho \,$$

$$PP''= (r cos{d}) tan{\alpha}\,$$

 * where: $$PP''$$ is displacement of $$\underline{P}$$ in direction tangent to lateral surface of bar

 * $$r( cos{d})\,$$ is approximately $$\rho\,$$

 * $$tan{\alpha}\,$$ is approximately $$\alpha\,$$

Strain:

$$\gamma=\frac{\underline{PP''}}{dx} = \frac{\rho \alpha}{dx} = \rho \theta\,$$

with... $$\theta = \frac{\alpha}{dx}$$ rate of twist angle (alpha very small!)

Contribution by radsam.z

October 13th: Meeting 21
$$\theta\ = \frac{d \alpha}{dx} $$

Hooke's Law:

$$\tau\ = G\gamma = G\rho\theta = G\rho \frac{d \alpha}{dx}$$

$$\tau(s)\ = G\rho(s)\theta(x)$$

$$\oint \tau(s) ds\ = G\theta\oint \rho(s) ds $$

but

$$\tau = \frac{q}{t}$$

and

$$\oint \rho(s) ds = 2 \bar A $$

Hence expression for theta p.21.2

$$\theta\ = \frac{1}{2 G \bar A} \oint \frac{q}{t} ds\,$$

What is ad-hoc about the above derivation of theta expressed on p.20.2 and the derivation of $$ T = 2q \bar A$$ ?

1) Strain gamma must be obtained using the displacement of $$P$$ in the direction tangent to Ca t P but $$PP'$$ on p.20.2 is not necessarily tangent to $$C'$$ (but actually close)

2) $$\tau = \frac{q}{t}$$ obtained from ad-hoc assumption that $$\tau$$ was uniform across wall thickness.

Formal justification (derivative) by elasticity theory: Roadmap p.16.2
A. Kinematic Assumption p.16.1

$$\displaystyle u_x(y,z) = \theta \psi(y,z)$$

considered constant w,r,t, x: uniform bar

$$\displaystyle u_y(x,z) = -\theta_xz$$

$$\displaystyle u_z(x,y) = +\theta_xy$$

To transform the equation in SUN[2006] to those using our unified notation, use cyclic permutation.

$$\displaystyle \epsilon_{xx} = \epsilon_{yy} = \epsilon_{zz}$$

$$\epsilon_{xx} = \frac{dU_x}{dx}(y,z) = 0$$

$$\gamma_{yz} = \frac{du_y}{dz} + \frac{du_z}{dx} = 0$$

October 15th: Meeting 22
 Multicell Airfoil

Po is defined as the observation point

 * 1) Set Po = D, sweep area is B-L-E
 * 2) Set Po = E, sweep area is F-B
 * 3) Area A22: Set Po = F, sweep area is E-H
 * 4) Area A3: Set Po = G, sweep area is H-T-F

Total Area = A1 + A21 + A22 + A3

Note: 3rd "Ad-hoc" point in engineering derivative: Inconsistency in assumption on size of alpha. To get line PP', assume alpha is small. To get PP", assume alpha is finite and p=rcos(alpha). Reintroduce small angle approximation after that.

Back to formal derivation:

Question: How many strain components exist in three dimensions?

Answer: SIX: $$\varepsilon _{xx}, \varepsilon _{yy}, \varepsilon _{zz}, \varepsilon _{xy}, \varepsilon _{xz}, \varepsilon _{yz}$$

$$\begin{bmatrix} \varepsilon _{xx} & \varepsilon _{xy} & \varepsilon _{xz} \\ \varepsilon _{yx} & \varepsilon _{yy} & \varepsilon _{yz} \\ \varepsilon _{zx} & \varepsilon _{zy} & \varepsilon _{zz} \end{bmatrix}$$

The representative equation is: $$\varepsilon = \frac{1}{2}\left(\frac{du_{i}}{dx_{j}} + \frac{du_{j}}{dx_{i}}\right)$$

Question:  Is symmetry of epsilon related to isotropy of material?

Answer:  NO. Isotropic elasticity is related to material behavior (sigma-epsilon relationship or Young's Modulus).

Contribution by radsam.s

October 17th: Meeting 23
 Three-Dimensional Strain

Slide 23.1a
Some say thank God it's Friday. I say too bad it's Friday. - Loc Vu-Quoc.

Question: How do we relate strains to Young's Modulus ($$E\,$$) and Poisson's Ratio ($$\nu\,$$)? Answer: (Steve) We use the relationship presented in Hooke's Law.
 * Note: Robert Hooke didn't just study the field of Elasticity; he was also a pioneer in the field of Microscopy, publishing the extremely influential Micrographia on the subject, at the age of 28.

$$\epsilon_{xx}=\sigma_{xx}/E\,$$ But as you pull a piece of rubber, it gets thinner. This lateral construction is due to the Poisson's Ration, $$\nu\,$$.

To correct for this, we re-write Hooke's law as: $$\epsilon_{xx}=\sigma_{xx}/E-\nu \sigma_{yy}/E -\nu \sigma_{zz}/E\,$$

And for shear strains: $$\gamma_{xy}=2\epsilon_{xy}=\tau_{xy}/G\,$$

From lecture 22-3, we have the following matrix:

$$\mathbf{\epsilon} = \begin{bmatrix} \epsilon_{11} & \epsilon_{12}  & \epsilon_{13} \\ \epsilon_{21} & \epsilon_{22} & \epsilon_{23} \\ \epsilon_{31} & \epsilon_{32} & \epsilon_{33} \end{bmatrix}= \epsilon_{ij}\,$$

and, using Voigt Notation (due to the symmetry of the 3x3 matrices $$\mathbf{\epsilon} = \begin{bmatrix}  \epsilon_{ij} \end{bmatrix} \,$$, $$\mathbf{\sigma} = \begin{bmatrix}   \sigma_{ij} \end{bmatrix} \,$$) we can arrange $$\epsilon_{ij}\,$$ and $$\epsilon_{ij}\,$$ into column matrices.

Slide 23.1b
The resultant 6x1 matrices are the following:
 * $$\epsilon_{ij} = \begin{bmatrix}

\epsilon_{11} \\ \epsilon_{22} \\ \epsilon_{33} \\ \epsilon_{23} \\ \epsilon_{31} \\ \epsilon_{12} \end{bmatrix}\,$$ and :$$\epsilon_{ij} = \begin{bmatrix} \sigma_{11} \\ \sigma_{22} \\ \sigma_{33} \\ \sigma_{23} \\ \sigma_{31} \\ \sigma_{12} \end{bmatrix}\,$$
 * Note: The original 3x3 column was symmetric, so $$\epsilon_{13}=\epsilon_{31}\,$$. We choose this notation in our 6x1 matrix rather than the $$\epsilon_{13}\,$$ that should be used to produce a beautiful, cyclic permutation (23 followed by 31, followed by 12.)

Hooke's Law for Isotropic Elasticity:
 * $$\begin{bmatrix}

\epsilon_{11} \\ ... \\ ... \\ \gamma_{23} \\ ... \\ ... \end{bmatrix} = \begin{bmatrix} 1/E & -\nu/E & -\nu/E & 0 & 0 & 0 \\ ... & ... & ... & ... & ... & ... \\ ... & ... & ... & ... & ... & ... \\   0 & 0 & 0 & 1/G & 0 & 0 \\ ... & ... & ... & ... & ... & ... \\ ... & ... & ... & ... & ... & ... \end{bmatrix} \begin{bmatrix} \sigma_{11} \\ \sigma_{22} \\ \sigma_{33} \\ \sigma_{23} \\ \sigma_{31} \\ \sigma_{12} \end{bmatrix}\,$$
 * Note: $$\sigma_{xy}=\tau_{xy}\,$$

For row four, $$\gamma_{23}=\gamma_{yz}\,$$ (Jared, Gary) $$=\tau_{yz}/G\,$$.

On Poisson's Ratio: Three Poisson's Ratios to note are listed in the table below. The first, for Steel, is noted because it is a value similar to all the materials we will deal with as Aerospace Engineers. The significance of the second and third, for cork and rubber respectively, is noted in homework 23.5. Nearly all materials are within the range of $$ 0 < \nu < 0.5 \,$$. Recently though, in the scientific journal Science (one of the two Premiere scientific journals, the opposite being Nature) Material Scientists reported making materials with Poisson's Ratios of $\nu < 0\,$. In other words, through Material Science and Engineering, materials which compress under tension, and expand under compression, have now been created.

Significant Poisson's Ratios:

Reading Assignments
MIT OpenCourseWare (OCW) Professor Lagace unit 4 (PDF) MIT OpenCourseWare (OCW) Professor Lagace unit 12 (PDF)

=Homework Problems and Solutions=

=References=

=Matlab Code Certification= I, the undersigned, certify that I can read, understand, and write matlab codes, and can thus contribute effectively to my team.

Eas4200c.f08.radsam.m 19:01, 18 October 2008 (UTC) Eas4200c.f08.radsam.a. 18:55, 19 October 2008 (UTC) "Eas4200c.f08.radsam.z 16:02, 2 November 2008 (UTC)" Eas4200c.f08.radsam.r 03:59, 20 October 2008 (UTC) Eas4200c.f08.radsam.e 21:36, 21 October 2008 (UTC) Eas4200c.f08.radsam.s 12:57, 22 October 2008 (UTC) "Eas4200c.f08.radsam.a. 19:24, 23 October 2008 (UTC)"

=Homework 4 Contributors= Eas4200c.f08.radsam.e 19:31, 17 October 2008 (UTC) Eas4200c.f08.radsam.m 20:16, 18 October 2008 (UTC) Eas4200c.f08.radsam.a 19:07, 19 October 2008 (UTC) Eas4200c.f08.radsam.d 22:12, 19 October 2008 (UTC) Eas4200c.f08.radsam.z 23:50, 19 October 2008 (UTC) Eas4200c.f08.radsam.r 03:58, 20 October 2008 (UTC) Eas4200c.f08.radsam.s 13:00, 22 October 2008 (UTC)

title: User:Eas4200c.f08.radsam/HW4


text:
Talk:Managing a private wiki/LocalSettings.php

Probably needs a whole separate page, but what's been confusing to me is the order in which things need to be on the file (includeonce this, extensions on the bottom, etc.). Could someone clarify that a bit? --SB_Johnny talk 19:42, 17 October 2008 (UTC)


title: User:Eas4200c.f08.ZYX.reger/Week7,8


text:
User:Hillgentleman/no final decision on irc

IRC is great and you can talk day and night there. It is a great place for collaborations and problem solving. But whatever is agreed on irc need not be Wikiversity consensus. Therefore there can be no final decision about Wikiversity on irc.


text:
User:Alistairjh

AlistairJH is a Wikipedia user who likes to write and edit articles about sport.

I write articles on 606 about sport, where people are free to comment, under the nickname 2o-DeMoN-o8. I also play Yahoo! Soccer Manager.

I also take lots of photos, there are three on Wikipedia and they are of sport.

Blogs
I used to blog online, but then I got a bit bored of it. I also played games online, but I got bored of them too. I get bored very easily.

My Stuff
My talk on Wikiversity My Contributions to Wikiversity

Wiki Notice
Although I have sites on many Wiki things, the main one is here: Wikipedia

title: User:Alistairjh


text: File:Merita.jpg

title: File:Merita.jpg


text:
User:Ryanmack

I'am a student at the Art Institute of Seattle trying to Figure out how to use this site for my general education class

title: User:Ryanmack


text:
User talk:Ryanmack

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

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And don't forget to explore Wikiversity with the links to your left. Be bold to contribute and to experiment with the sandbox or your userpage, and see you around Wikiversity! Jtneill - Talk - c 03:17, 18 October 2008 (UTC)

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title: User talk:Ryanmack


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User:EML4500.f08.A-team.VandenBerg/HW1

 Comment: This is HW1 located in the user namespace of Lukas Vandenberg --EML4500.f08.A-team.VandenBerg 20:12, 19 September 2008 (UTC)

Today is Wednesday, September 3, 2008

Things to be covered in class:
 * Trusses
 * Matrix Method (Located in Chapter 4 of the Book)
 * MIT’s Open Course Ware (Located on the wiki page)

Truss with elastic (deformable) bars:
Figure 1

 * There are four unknowns in this example with only three equations. This is then defined as a statistically indeterminate example.

Global (whole structure) FBD:
FBD= Free Body Diagram
 * Everything needs to be in equilibrium. Supports are removed and replaced with reaction forces that are not known.

Figure 2

 * Global refers to the whole structure.
 * Local refers to each element separately.
 * A number with a circle around is known as a global node, while a number with a triangle around it is a bar element. The number one with a triangle around it is bar element 1 from global node one to two, while the number 2 with a triangle around it is bar element two from global node two to three.
 * “P” is the known force, and all the “R” terms are the four unknown reaction forces.

Two Free Body Diagrams of Two Bar Elements:
Free Body Diagram of Body One:

Figure 3

 * The subscript on the "f" also stands for the degree of freedom (d.o.f.).
 * So, instead of always writing "degree of freedom" in long-hand every time, it can just be written as the d.o.f.
 * The subscript on the "f" is for the internal force on the element, which is represented by the number in parenthesis in the superscript. For this example the superscript would take on the values of one and two, while the subscript would take on the values of one, two, three, or four.
 * In this example the numbering could either be done the way it is, or done such that the local node numbers would be reversed.

Free Body Diagram of Body Two:

Figure 4

 * When labeling the forces with their degrees of freedom (also known as d.0.f.) you should always start with the horizontal direction and then proceeding on to the vertical direction.

You can also label bar element two in the following manner, with both approaches being correct

Figure 5

 * Always make sure to include parenthesis around the element number in superscript over the force (f) and displacement.

Force Displacement (also known as the F.D.) Relation Recall:
Figure 6

 * K=Stiffness
 * d=Distance Stretched
 * f=Force

The Force Displacement (F.D.) Relation of a one-dimensional spring element with one end fixed: f=kd The Force Displacement (F.D.) Relation of a one-dimensional spring element with two ends free:

Figure 7

 * fH, dH -internal displacement force

Figure 8

The spring in the picture has been stretched displacing the nodes.

For the following matrix, statics is used to fill it in:

$$ \begin{array}{cccc} \begin{Bmatrix} f_1 \\ f_2 \end{Bmatrix} & = & \begin{bmatrix} k & -k \\ -k & k \end{bmatrix} & \begin{Bmatrix} d_1 \\ d_2 \end{Bmatrix} \\ \text{2 x 1}~\text{Matrix} & & \text{2 x 2} ~ \text{Matrix} & \text{2 x 1} ~ \text{Matrix} \\ \text{row x column} \end{array} $$ Figure 9

5-1 Meeting 5: Friday, 5 September 08, EML4500

Reading assignments:

 * Chapter 4: Trusses, Beams, and Frames
 * Chapter 1: By Picture
 * Section 1.1: Discretization
 * 1.1  Plane Truss Elem.
 * 1.2  Assembly of Elem. Eqs. (Ex. 4.1: Five-Bar truss)
 * 1.3  Elem. Solu. and Model Validity
 * 1.4  Plane Truss Elem.

Steps to solve simple truss system:
1. Global picture (description) -At structure level: *global d.o.f.’s (displ. d.o.f’s) --> unknowns in general *global forces -Actually --> displ. d.o.f.’s are partitioned into: *a known part: e.g. fixed d.o.f.’s, constraints *an unknown part: solved using FEM -Similarly for the global forces: *a known part: applied forces *an unknown part: reactions

2. Element Picture -Element d.o.f.’s and element forces:
 * either in global or local coordinate system

3. Global FB relation -Element stiffness matrices in global coordinates -Element force matrices in global coordinates -Assembly of element stiffness matrix and element force matrix into global FB    relation:  K * d = F  “free-free system” (unconstrained)  K=singular
 * K --> N x N
 * d --> N x 1
 * F --> N x 1

4. Elimination of known d.o.f.’s to reduce the global FB relation -Stiffness matrix non-singular --> invertible -K * d = F     *K --> M x M      *d --> M x 1 M < N      *F --> M x 1 -M = number of unknown displ. d.o.f.’s -N = number of both known & unknown d.o.f.’s -K is non-singular --> K(-1) exists *d = K(-1) * F

5. Compute element forces from now known d --> elem. stresses

6. Compute reactions (unknown forces)

Specific example to see how method work.

Data:

-L(1) = 1 -L(2) = 2

Cross-sectional Area:
-A(1) = 1 -A(2) = 2

Young’s Modulus:
-E(1) = 3 -E(2) = 5

-Global picture: Global d.o.f.’s:

Numbering the displ. d.o.f.’s:
-Follow order of global node number -For each node, follow the order of global coord. axes, number displ d.o.f.’s for that node -Global forces --> same thing

Getting Started
Quick example of how text is input and shown in Matlab

% Matlab

EDU>> a = [ 1 2; 2 1 ]

a =

1    2     2     1

EDU>> a*a

ans =

5    4     4     5

EDU>> quit

Matrices
This is a matrices tutorial showing a user how to input different matrices, computing different mathematical equations. Some of these equations are matrix multiplication, matrix addition, matrix division, matrix inverses, and determinates.

EDU>> %Matlab EDU>> a = [ 1 2; 3 4 ]

a =

1    2     3     4

EDU>> a

a =

1    2     3     4

EDU>> a * a

ans =

7   10    15    22

EDU>> b = [ 1 2; 0 1 ]

b =

1    2     0     1

EDU>> a*b

ans =

1    4     3    10

EDU>> b*a

ans =

7   10     3     4

EDU>> a + b

ans =

2    4     3     5

EDU>> s = a + b

s =

2    4     3     5

EDU>> inv (s)

ans =

-2.5000   2.0000    1.5000   -1.0000

EDU>> s * inv (s)

ans =

1.0000  -0.0000         0    1.0000

EDU>> s/s

ans =

1    0     0     1

EDU>> s\s

ans =

1    0     0     1

EDU>> inv (s) * s

ans =

1.0000        0         0    1.0000

EDU>> a/b

ans =

1    0     3    -2

EDU>> a\b

ans =

-2.0000  -3.0000    1.5000    2.5000

EDU>> c = [ 1 1; 1 1 ]

c =

1    1     1     1

EDU>> inv(c); Warning: Matrix is singular to working precision. EDU>> det(a)

ans =

-2

EDU>> det(c)

ans =

0

EDU>> quit

Vectors in MatLab
Quick way to show vector algebra in Matlab with matrices:

EDU>> a = [ 1 2; 3 4 ]

a =

1    2     3     4

EDU>> b = [1;0]

b =

1    0

EDU>> a\b

ans =

-2.0000   1.5000

EDU>> quit

Editing Systems of Equations
Linear equations can be easily manipulatedwith use of matlab and matrices.

If you have to equations: ax + by = p    cx + dy = q

These two equations can be better understood, written, and compacted using matlab. AX = B

Where the coeffient matrix is the A

A = [ a b; c d ]

with the unknown matrix with the x and y coefficients

Loops in MatLab
A loop is a programming tool in Matlab that is used to do repetitive work. It saves the user time. In the tutorial, matrix a represents the probabilities that two island populations will move or stay put on their side of the island. x is a column vector representing the initial populations.

a=    0.8  0.1 0.2 0.9

x=    1 0

Successive population states can be predicted by multiplying the two matrices. Using a loop, the operation can be repeated a defined number of times using the following form:

>> a = [ 0.8 0.1; 0.2 0.9 ] >> x = [ 1; 0 ] >> for i = 1:20, x = a*x, end

This operation will be repeated 20 times, and the populations of the island can be predicted 20 time units away.

Graphing in MatLab
Here are some images of the types of graphing that MatLab can perform, with the description of the code used to draw the graph directly beneath its corresponding image. Some of the nuances of Matlab graphing can be seen here, such as changing the color (function call "colormap" in images 5 and 6), removing mesh lines (function call "shading faceted" in image 5), and using the function name "sphere" to create a sphere with default dimensions.

IMAGE 1

EDU>> x = 0:.01:6; y = cos(x); plot(x,y)

IMAGE 2

EDU>> t = -3:.1:3; x = cos(1.5*t); y = sin(0.5*t); z = t.^2; plot3(x,y,z)

IMAGE 3

EDU>> surf(magic(10))

IMAGE 4

EDU>> [x,y] = meshgrid(-3:.01:3, -3:.01:3); z = (x.^3 - y.^3); mesh(z)

IMAGE 5

EDU>> surf(eye(10)/hilb(10)); colormap(cool); shading faceted

IMAGE 6

EDU>> sphere; colormap(flag); shading flat

 Mtg. 6; Monday 8, September 2008  EML 4500

2) Elem. Picture

3) Global Force Displacement (F.D.)

Contributions for HW1
Contributors:

Lukas Vandenberg --EML4500.f08.A-team.VandenBerg 20:12, 19 September 2008 (UTC)

Brian Morford --EML4500.f08.A-team.morford 20:12, 19 September 2008 (UTC)

Joseph Rieth --EML4500.f08.A-team.rieth 20:13, 19 September 2008 (UTC)

Joseph Melvin --EML4500.f08.A-team.melvin 20:23, 19 September 2008 (UTC)

Brian Kirley --EML4500.f08.A-team.kirley 20:32, 19 September 2008 (UTC)

Daniel Robinson --Eml4500.f08.a-team.robinson 20:32, 19 September 2008 (UTC)

title: User:EML4500.f08.A-team.VandenBerg/HW1


text:
User:Eml4500.f08.group/Homework4/Lecture Notes

Lecture 18
connectivity array

Consider the two bar truss system below

The connectivity array:

$$conn=\begin{bmatrix} 1 & 2\\ 2& 3 \end{bmatrix} \begin{matrix} element 1\\ element  2 \end{matrix}$$

Where the numbers 1,2,3 inside the matrix represent the global node number and the columns represent the local node number.

In matlab: conn(e,j)= the global node number of local node j of element e.

Location matrix master array:

$$lmm=\begin{bmatrix} 1 & 2&3&4\\ 3& 4&5&6 \end{bmatrix} \begin{matrix} element 1\\ element  2 \end{matrix}$$

Where the numbers in the matrix represent the global degree of freedom number, or the equation number in the global stiffness matrix K

In matlab: lmm(e,j)= the equation number(global degree of freedom number) for the element stiffness coefficient corresponding to the jth local degree of freedom number.

=Derivation of Element Stiffness Matrix=

Lecture 19: Split, please contribute at least a fifth of the notes
Goal: WE want to find a transformation matrix ~T(e) that transforms one set of local element degrees of freedom, d(e), into another set of local degrees of freedoms, ~d(e), such that the transformation matrix can be inverted.

$$ \tilde{d^{(e)}}_{4x1} = \tilde{T^{(e)}}_{4x4} d^{(e)}_{4x1}$$

Equation (1) $$ \tilde{d_{1}}^{(e)}=\begin{bmatrix} l^{(e)}& m^{(e)} \end{bmatrix} \begin{Bmatrix} d_{1}^{(e)}\\ d_{2}^{(e)}

\end{Bmatrix} $$

Derivation: $$\tilde{d_{1}^{(e)}}=\vec{d}_{1}^{(e)}\cdot \vec{\tilde{i}}$$ $$=(d_{1}^{(e)}\vec{i}+d_{2}^{(e)}\vec{j})\cdot \vec{\tilde{i}}$$ $$=d_{1}^{(e)}(\vec{i}\cdot \vec{\tilde{i}})+d_{2}^{(e)}(\vec{j}\cdot \vec{\tilde{i}})$$ $$ =cos\theta ^{(e)}d_{1}^{(e)} + sin\theta ^{(e)}d_{2}^{(e)} $$ From the definition of l(e) equal to cosθ(e) and m(e) equal to sinθ(e), the above equation reduces to: $$\tilde{d_{1}^{(e)}}=l^{e}d_{1}^{e}+m^{e}d_{2}^{e}$$ In matrix form reduces to: $$\tilde{d_{1}^{(e)}}=\begin{bmatrix} l^{(e)} & m^{(e)} \end{bmatrix} \begin{Bmatrix} d_{1}^{(e)} \\ d_{2}^{(e)} \end{Bmatrix}$$

Equation (2) $$ \tilde{d_{2}}^{(e)}=\begin{bmatrix} -m^{(e)}& l^{(e)} \end{bmatrix} \begin{Bmatrix} d_{1}^{(e)}\\ d_{2}^{(e)}

\end{Bmatrix} $$

Derivation: $$\tilde{d_{2}^{(e)}}=\vec{d}_{1}^{(e)}\cdot \vec{\tilde{j}}$$ $$ =-sin\theta ^{(e)}d_{1}^{(e)} + cos\theta ^{(e)}d_{2}^{(e)} $$ From the definition of l(e) equal to cosθ(e) and m(e) equal to sinθ(e), the above equation reduces to: $$\tilde{d_{2}^{(e)}}=-m^{e}d_{1}^{e}+l^{e}d_{2}^{e}$$ In matrix form reduces to: $$\tilde{d_{2}^{(e)}}=\begin{bmatrix} -m^{(e)} & l^{(e)} \end{bmatrix} \begin{Bmatrix} d_{1}^{(e)} \\ d_{2}^{(e)} \end{Bmatrix}$$

Put equation (1) and equation (2) together in matrix form:

$$ \begin{Bmatrix} \tilde{d_{1}}^{(e)}\\ \tilde{d_{2}}^{(e)}

\end{Bmatrix}=\begin{bmatrix} l^{(e)}& m^{(e)}\\-m^{(e)}& l^{(e)} \end{bmatrix} \begin{Bmatrix} d_{1}^{(e)}\\ d_{2}^{(e)}

\end{Bmatrix} $$

Where $$R^{(e)}= \begin{bmatrix} l^{(e)} & m^{(e)} \\ -m^{(e)} & l^{(e)} \end{bmatrix}$$

$$\mathbf{\tilde{d}}^{(e)}_{4x1} = \mathbf{\tilde{T}}^{(e)}_{4x4}\textbf{d}^{(e)}_{4x1} \Rightarrow \begin{Bmatrix} \tilde{d^{(e)}_{1}}\\ \tilde{d^{(e)}_{2}}\\ \tilde{d^{(e)}_{3}}\\ \tilde{d^{(e)}_{4}} \end{Bmatrix} = \begin{bmatrix} \textbf{R}^{(e)}_{2x2} & \textbf{0}_{2x2}\\ \textbf{0}_{2x2} & \textbf{R}^{(e)}_{2x2} \end{bmatrix}\begin{Bmatrix} d^{(e)}_{1}\\ d^{(e)}_{2}\\ d^{(e)}_{3}\\ d^{(e)}_{4} \end{Bmatrix}$$

$$\mathbf{\tilde{f}}^{(e)} = k^{(e)}\begin{bmatrix} 1 & 0 & -1 & 0\\ 0 & 0 & 0 & 0\\ -1 & 0 & 1 & 0\\ 0 & 0 & 0 & 0 \end{bmatrix}\mathbf{\tilde{d}}^{(e)}$$

$$\mathbf{\tilde{f}}^{(e)}_{4x1} = \mathbf{\tilde{k}}^{(e)}_{4x4}\mathbf{\tilde{d}}^{(e)}_{4x1}$$

Considering the case where $$\mathbf{\tilde{d}}^{(e)}_{4} \neq 0 \Rightarrow \mathbf{\tilde{d}}^{(e)}_{1} = \mathbf{\tilde{d}}^{(e)}_{2} = \mathbf{\tilde{d}}^{(e)}_{3} = 0$$

$$\mathbf{\tilde{f}}^{(e)}_{4x1} = \mathbf{\tilde{k}}^{(e)}_{4x4}\mathbf{\tilde{d}}^{(e)}_{4x1} = \textbf{0}_{4x1}$$, the 4th column of the stiffness matrix.

Interpretation of Transverse Degrees of Freedom

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

!HW Force relationship, from team group
 * Using the relationship between distance and T, we can find the relationship between force and T: $$\mathbf{\tilde{d}}^{(e)}_{4x1} = \mathbf{\tilde{T}}^{(e)}_{4x4}\textbf{d}^{(e)}_{4x1}$$:
 * Using the relationship between distance and T, we can find the relationship between force and T: $$\mathbf{\tilde{d}}^{(e)}_{4x1} = \mathbf{\tilde{T}}^{(e)}_{4x4}\textbf{d}^{(e)}_{4x1}$$:

$$\mathbf{\tilde{f}}^{(e)}_{4x1} = \mathbf{\tilde{T}}^{(e)}_{4x4}\textbf{f}^{(e)}_{4x1} \Rightarrow \begin{Bmatrix} \tilde{f^{(e)}_{1}}\\ \tilde{f^{(e)}_{2}}\\ \tilde{f^{(e)}_{3}}\\ \tilde{f^{(e)}_{4}} \end{Bmatrix} = \begin{bmatrix} \textbf{R}^{(e)}_{2x2} & \textbf{0}_{2x2}\\ \textbf{0}_{2x2} & \textbf{R}^{(e)}_{2x2} \end{bmatrix}\begin{Bmatrix} f^{(e)}_{1}\\ f^{(e)}_{2}\\ f^{(e)}_{3}\\ f^{(e)}_{4} \end{Bmatrix}$$

contribution by carbon.w
 * }

In addition, using the force-distance relationship:

$$\mathbf{\tilde{k}}^{(e)}\mathbf{\tilde{d}}^{(e)} = \mathbf{\tilde{f}}^{(e)} \Rightarrow \mathbf{\tilde{k}}^{(e)}\mathbf{\tilde{T}}_{(e)}\mathbf^{(e)} = \mathbf{\tilde{T}}_{(e)}\mathbf^{(e)}$$

If $$\mathbf{\tilde{T}}_{(e)}$$ is invertible, then the relationship becomes:

$$[\mathbf{\tilde{T}}^{(e)-1}\mathbf{\tilde{k}}^{(e)}\mathbf{\tilde{T}}^{(e)}]\mathbf^{(e)} = \mathbf^{(e)}$$

$$\mathbf{\tilde{T}}_{(e)}$$ is a block diagonal matrix. The general form of a block diagonal matrix is as follows:

$$\textbf{A} = \begin{bmatrix} \textbf{D}_{1} & & \textbf{0}\\ & ... & \\ \textbf{0} & & \mathbf{D_{s}} \end{bmatrix}$$

Using a simpler example of the diagonal matrix, to display more clearly the format, we can determine A-1:

$$\textbf{B} = \begin{bmatrix} d_{11} & & & \textbf{0}\\ & d_{22} & & \\ & & ... & \\ \textbf{0} & & & d_{nn} \end{bmatrix} = Diag[d_{11}, d_{22},...,d_{nn}]$$

$$\mathbf{B^{-1}} = Diag[\frac{1}{d_{11}}, \frac{1}{d_{22}},...,\frac{1}{d_{nn}}]$$

The above is true if we assume that dii does not equal zero for i=1,...,n and a block diagonal matrix would be:

$$\mathbf{A} = Diag[\mathbf{D_{1}},...,\mathbf{D_{s}}]$$ $$\mathbf{A^{-1}} = Diag[\mathbf{D^{-1}_{1}},...,\mathbf{D^{-1}_{s}}]$$ Therefore $$\mathbf\tilde{T}^{(e)-1} = Diag[\mathbf{R^{(e)-1}},\mathbf{R^{(e)-1}}]$$

$$\mathbf{R^{(e)T}} = \begin{bmatrix} l^{(e)} & -m^{(e)}\\ m^{(e)} & l^{(e)} \end{bmatrix}$$

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

!In-class Team Exercise, from team group l^{(e)} & -m^{(e)}\\ m^{(e)} & l^{(e)} \end{bmatrix} * \begin{bmatrix} l^{(e)} & m(e)\\ -m^{(e)} & l^{(e)} \end{bmatrix} = [(l^{(e)})^{2}  (m^{(e)})^{2}][(-l^{(e)}m^{(e)})   (m^{(e)}l^{(e)}]$$
 * $$\mathbf{R^{(e)T}}*\mathbf{R^{(e)}} = \begin{bmatrix}
 * $$\mathbf{R^{(e)T}}*\mathbf{R^{(e)}} = \begin{bmatrix}

$$\mathbf{R^{(e)T}}*\mathbf{R^{(e)}} = \begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix}_{2x2} = \mathbf{\bar{I}_{2x2}}$$ (the identity matrix)

contribution by carbon.w
 * }

$$\Rightarrow \mathbf{R^{(e)-1}} = \mathbf{R^{(e)T}}$$

$$\mathbf{\tilde{T}^{(e)-1}} = Diag[\mathbf{R^{(e)T}}, \mathbf{R^{(e)T}}] = (Diag[\mathbf{R^{(e)}}, \mathbf{R^{(e)}}])^{T}$$

$$\Rightarrow \mathbf{\tilde{T}^{(e)-1}} = \mathbf{\tilde{T}^{(e)T}}$$

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

!HW Force relationship, from team group
 * Using the relationship shown above,
 * Using the relationship shown above,

$$[\mathbf{\tilde{T}}^{(e)-1}\mathbf{\tilde{k}}^{(e)}\mathbf{\tilde{T}}^{(e)}]\mathbf{\tilde{d}}^{(e)} = \mathbf{\tilde{T}}_{(e)}\mathbf{\tilde{f}}^{(e)}$$,

and plugging in $$\mathbf{\tilde{T}^{(e)-1}} = \mathbf{\tilde{T}^{(e)T}}$$, we arrive at:

$$[\mathbf{\tilde{T}}^{(e)T}\mathbf{\tilde{k}}^{(e)}\mathbf{\tilde{T}}^{(e)}]\mathbf{\tilde{d}}^{(e)} = \mathbf{\tilde{f}}^{(e)}$$

$$\mathbf{k}^{(e)} = \mathbf{\tilde{T}}^{(e)-1}\mathbf{\tilde{k}}^{(e)}\mathbf{\tilde{T}}^{(e)} $$

contribution by carbon.w
 * }

Lecture 22
Justification of assembling local element stiffness matrices into a global stiffness matrix. Consider the example of the 2-bar truss system discussed during the 9th lecture.

First we note the global displacements and forces that correspond to the element displacements and forces as shown below.

$$\begin{bmatrix} k^{(1)}_{11} & k^{(1)}_{12} & k^{(1)}_{13} & k^{(1)}_{14}\\ k^{(1)}_{21} & k^{(1)}_{22} & k^{(1)}_{23} & k^{(1)}_{24}\\ k^{(1)}_{31} & k^{(1)}_{32} & k^{(1)}_{33} & k^{(1)}_{34}\\ k^{(1)}_{41} & k^{(1)}_{42} & k^{(1)}_{43} & k^{(1)}_{44} \end{bmatrix}*\begin{Bmatrix} D1\\ D2\\ D3\\ D4 \end{Bmatrix}=\begin{Bmatrix} F1\\ F2\\ F3\\ F4 \end{Bmatrix}$$ $$\begin{bmatrix} k^{(2)}_{11} & k^{(2)}_{12} & k^{(2)}_{13} & k^{(2)}_{14}\\ k^{(2)}_{21} & k^{(2)}_{22} & k^{(2)}_{23} & k^{(2)}_{24}\\ k^{(2)}_{31} & k^{(2)}_{32} & k^{(2)}_{33} & k^{(2)}_{34}\\ k^{(2)}_{41} & k^{(2)}_{42} & k^{(2)}_{43} & k^{(2)}_{44} \end{bmatrix}*\begin{Bmatrix} D3\\ D4\\ D5\\ D6 \end{Bmatrix}=\begin{Bmatrix} F3\\ F4\\ F5\\ F6 \end{Bmatrix}$$

Next we can fill in the global stiffness matrix by observing that the columns of this stiffness matrix correspond to global displacements and the rows correspond to the global forces as shown below.

$$\begin{bmatrix} & (D1) & (D2) & (D3) & (D4) & (D5) & (D6)\\ (F1) & K11 & K12 & K13 & K14 & K15 & K16\\ (F2) & K21 & K22 & K23 & K24 & K25 & K26\\ (F3) & K31 & K32 & K33 & K34 & K35 & K36\\ (F4) & K41 & K42 & K43 & K44 & K45 & K46\\ (F5) & K51 & K52 & K53 & K54 & K55 & K56\\ (F6) & K61 & K62 & K63 & K64 & K65 & K66 \end{bmatrix}$$

My plugging the local stiffness values into the above matrix, we achieve the global stiffness matrix as shown below. This matrix exactly matches the global stiffness matrix for this 2-bar truss system which was given in meeting 9. $$\begin{bmatrix} k_{11}^{(1)}&k_{12}^{(1)}&k_{13}^{(1)}&k_{14}^{(1)} &0  &0 \\ k_{21}^{(1)}&k_{22}^{(1)} &k_{23}^{(1)}  &k_{24}^{(1)}  &0  &0 \\ k_{31}^{(1)}&k_{32}^{(1)} &k_{33}^{(1)}+k_{11}^{(2)}  &k_{34}^{(1)}+k_{12}^{(2)}  &k_{13}^{(2)}  &k_{14}^{(2)} \\ k_{41}^{(1)}&k_{42}^{(1)} &k_{43}^{(1)}+k_{21}^{(2)}  &k_{44}^{(1)}+k_{22}^{(2)}  &k_{23}^{(2)}  &k_{24}^{(2)} \\ 0& 0&  k_{31}^{(2)}&k_{32}^{(2)}  &k_{33}^{(2)}  &k_{34}^{(2)} \\ 0& 0&  k_{41}^{(2)}&k_{42}^{(2)}  &k_{43}^{(2)}  &k_{44}^{(2)} \end{bmatrix}$$

Equilibrium of Node 2

Based on the above element diagrams, the equilibrium of node 2 can be shown as follows.

$$\sum{F_{x}}=0=-f^{(1)}_{3}-f^{(2)}_{1}=0$$

$$\sum{F_{y}}=0=P-f^{(1)}_{4}-f^{(2)}_{2}=0$$

Lecture 23
From equations (1) and (2), we find:

Eq. (1) $$=> f_3^{(1)} + f_1^{(2)} = 0$$

Eq. (2) $$=> f_4^{(1)} + f_2^{(2)} = P$$

Futhermore using the relation f=kd, we find:

(1) $$[k_{31}^{(1)}d_1^{(1)} + k_{32}^{(1)}d_2^{(1)} + k_{33}^{(1)}d_3^{(1)} + k_{34}^{(1)}d_4^{(1)}] + [k_{11}^{(2)}d_1^{(2)} + k_{12}^{(2)}d_2^{(2)} + k_{13}^{(2)}d_3^{(2)} + k_{14}^{(2)}d_4^{(2)}] = 0 $$

(2) $$[k_{41}^{(1)}d_1^{(1)} + k_{42}^{(1)}d_2^{(1)} + k_{43}^{(1)}d_3^{(1)} + k_{44}^{(1)}d_4^{(1)}] + [k_{21}^{(2)}d_1^{(2)} + k_{22}^{(2)}d_2^{(2)} + k_{23}^{(2)}d_3^{(2)} + k_{24}^{(2)}d_4^{(2)}] = P $$

Next, we want to transform from local to global dofs:

(1) $$[k_{31}^{(1)}d_1 + k_{32}^{(1)}d_2 + k_{33}^{(1)}d_3 + k_{34}^{(1)}d_4] + [k_{11}^{(2)}d_3 + k_{12}^{(2)}d_4 + k_{13}^{(2)}d_5 + k_{14}^{(2)}d_6] = 0 $$

(2) $$[k_{41}^{(1)}d_1 + k_{42}^{(1)}d_2 + k_{43}^{(1)}d_3 + k_{44}^{(1)}d_4] + [k_{21}^{(2)}d_3 + k_{22}^{(2)}d_4 + k_{23}^{(2)}d_5 + k_{24}^{(2)}d_6] = P $$

Thus, we have obtained the 3rd and 4th rows of the global stiffness matrix, K, as seen on P. 9-1.

Next, let's look at the Assembly of $$\boldsymbol k^{(e)}$$, where e=1,...,nel (the number of elements), into the global stiffness matrix K:

$$\boldsymbol K = \overset{e=nel}{\underset{e=1}{A}} \boldsymbol k^{(e)}$$

where,

n: is the total number of global dofs before eliminating boundary conditions.

ned: is the number of elemental dofs (ned< kd-F=0$$

(4) $$ <=> w(kd-F)=0$$ for all w.

Proof:

(3) => (4) is trivial.

(4) => (3) is not trivial. Since (4) is valid for all w, you can select and arbitrary w, say w=1, then (4) becomes:

1. $$(kd-F)=0 =>$$ (3)