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History of Germany

Welcome to the History of Germany''' Course at Wikiversity. In this course we will study German History since Antiquity to the Modern Day. Good luck... ''' --EarlGrey2005

Rules of this course

 * Students must do homework.
 * Students must ask questions about Germany.
 * Students must take quizzes.
 * Successful students will become assistants.

If you wish to take this course, please add this template to your user page:

Schedule
The aim of this course is that to give some knowledge about German History on a student level.. Students have to do their homework, follow the lessons, solve the quizzes and if they need, they should ask their questions to their instructor. If the students aren't interested in their homework, they don't spend any effort for them, don't follow the lessons and don't solve the quizzes, they will add in to inactive students list. Before being added to inactive students list, they will be warned. When the 1st quarter is completed, the most succesful students who obey the rules of this lesson will be chosen as assistant students. In the 2nd quarter, the most succesful students(that were already assistant students) will be chosen as an assistant professor. In the 3rd and 4th quarter, the most successful student(that was a assistant professor) will be chosen as professional student. When the course is over, successful students will get an award from the instructor.

General Quizzes

 * Guiz over Week I

Quarter 1: Germanic Tribes, 750 BC - 768 AD

 * Week I:  Germanic Peoples
 * Week II: Roman Expansion and Teutoburg Forest
 * Week III: Stem Duchies
 * Week IV: Frankish Empire
 * Week V:  Who were the Frankish Kings?
 * Week VI: Frankish Expansion

Quarter 2: Early Holy Roman Empire

 * Week VII: Christmas Day, 800 AD
 * Week VIII: East Francia
 * Week IX:  Otto the Great
 * Week X: Church and State
 * Week XI: Change and Reform
 * Week XII: The Protestant Reformation

Quarter 3: Late Holy Roman Empire to the German Empire

 * Week XIII: Thirty Years War
 * Week XIV:  War of the Austrian Succession
 * Week XV: French Revolution
 * Week XVI: German Confederation
 * Week XVII: Franco-Prussian War
 * Week XVIII: Road to World War I

Quarter 4: World War I to Modern Day

 *  Week XIX:  World War One
 *  Week XX:  German Revolution of 1918-19
 *  Week XXI:  Weimar Republic
 *  Week XXII:  Nazi Germany and World War Two
 *  Week XXIII:  Cold War in Germany
 *  Week XXIV:  Modern Day Germany

Homework this week
No Homework Yet

Questions
is even a non-german allowed to participate to the course? I wait impatiently the answer

Students
If you want to be a student of this course, please add your user name below: FargoPiperStone pythonliving DrMonymona Stuart Halley Victoria.Engelherz

TyroneB1

 * 1) EarlGrey2005

Resources

 * en.wiki History of Germany
 * Brittanica History of Germany
 * HistoryWorld History of Germany

title: History of Germany


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User:Zakman93

Hi I am user Zakman 93 and I Instruct the History Of Germany Course In the European History Department and Grade 8 Honors Literature and American Literature in the English Department.

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

title: User:Zakman93


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Talk:History of Germany

Anyone interested in joining this course please email me --Zakman93 03:14, 6 November 2008 (UTC). This course needs several students to run. The first lecture will be posted in January. If by January several students do not sign up for this course, it will not run until the following January.--Zakman93 03:14, 6 November 2008 (UTC)

Its been over a year since I was on this sight. Since no one has expressed any interest, I will make this course non interactive, instead of a the course that I had planned where I developed activities based on student imput. --Zakman93 23:29, 11 November 2009 (UTC)--Zakman93 23:29, 11 November 2009 (UTC)

title: Talk:History of Germany


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Spanish 2/Chapter 3 (Daily Activities)

Getting Ready

 * acostarse - to go to bed
 * afeitarse - to shave
 * arreglarse - to fix
 * bañarse - to take a bath
 * cepillarse - to brush
 * cortarse el pelo - to cut one's hair
 * despertarse - to wake up
 * ducharse - to take a shower
 * levantarse - to get up
 * lavarse - to wash
 * pedir prestado(a) - to borrow
 * pintarse - to paint, to polish
 * ponerse - to put on
 * prepararse - to get ready
 * secarse - to dry
 * vestirse - to get dressed

Note: Acostarse has a stem change when conjugated (o to ue). Despertarse has a stemchange of (e to ie). Vestirse conjugates as (e to i).

Things needed to get ready

 * agua de colonia - cologne
 * anillo - ring
 * cepillo - brush
 * cinturón - belt
 * desodorante - deodorant
 * ducha - shower
 * gel - gel
 * joyas - jewelry
 * oro - gold
 * plata - silver
 * labios - lips
 * maquillaje - make-up
 * peine - comb
 * pelo - hair
 * salón de belleza - beauty salon
 * secador - blow dryer
 * toalla - towel
 * uñas - nails

Special events

 * audición - audition
 * boda - wedding
 * cita - date
 * concurso - contest
 * evento especial - special event

Feelings

 * entusiasmado(a) - excited
 * nervioso(a) - nervous
 * tranquilo(a) - calm
 * feliz  - happy

Other useful words

 * antes de - before
 * cómodo(a) - comfortable
 * depende - it depends
 * elegante - elegant
 * lentamente - slowly
 * luego - then
 * por ejemplo - for example
 * rápidamente - quickly
 * te ves - you look

Reflexive verbs
To say that people do something to or for themselves, reflexive verbs are used. You know a verb is reflexive if its infinite form ends in se. The reflexive pronouns in Spanish are me, te, se, nos, and os. Each pronoun corresponds to a different subject. Here is the conjugation of the present-tense of the reflexive verb acostarse:

 * me acuesto - I go to bed
 * te acuestas - you go to bed (singular)
 * se acuesta - he/she goes to bed
 * nos acostamos - we go to bed
 * os acostáis - you go to bed (plural)
 * se acuestan - they go to bed

Ser (to be)
Ser is one form of "to be" in Spanish. Use ser to talk about:
 * what a person or thing is
 * what a person or thing is like
 * where a person or thing is from
 * what a thing is made of
 * to whom something belongs

Here is the conjugation of ser:
 * soy - I am
 * eres - you are (singular)
 * es - he/she is
 * somos - we are
 * sois - you are (plural)
 * son - they are

Here is the conjugation of ser in the preterit tense:
 * fui - I was
 * fuiste - you were (singular)
 * fue - he/she was
 * fuimos - we were
 * fuisteis - you were (plural)
 * fueron - they were

Estar (to be)
Estar is the other form of "to be" in Spanish. Use estar to talk about:
 * how a person or thing is at the moment
 * how someone feels
 * where a person or thing is located

Here is the conjugation of estar:
 * estoy - I am
 * estás - you are (singular)
 * está - he/she is
 * estamos - we are
 * estáis - you are (plural)
 * están - they are

Here is the conjugation of estar in the preterit tense:
 * estuve - I am
 * estuviste - you are (singular)
 * estuvo - he/she is
 * estuvimos - we are
 * estuvisteis - you are (plural)
 * estuvieron - they are

Possessive adjectives
Spanish possessive adjectives have a long form that follows the noun. These forms are often used for emphasis. Sometimes, possessive adjectives may be used without the noun. To clarify or emphasize possession, you can use de + a noun or pronoun instead of suyo.

 * mío(a, -os, -as) - mine
 * el coche mío - my car / this car of mine
 * ¿De quién es la mochila? Es mía. No, es de ella. - Whose backpack is this? It's mine.  No, it's hers.
 * tuyo (a, -os, -as) - yours (singular/familiar)
 * suyo (a, -os, -as) - yours (singular/formal), his, hers, its
 * nuestro (a, -os, -as) - ours
 * vuestro (a, -os, -as) - yours (plural/familiar)
 * suyo (a, -os, -as) - yours (plural/formal), theirs

Vocabulario adicional

 * dientes - teeth
 * esmalte de uñas - nail polish
 *  espuma de afeitar - shaving foam
 * fijador - hair spray
 * lápiz de labios - lipstick
 * loción - lotion
 * loción astringente - astringent
 * loción humectante - moisturizing lotion
 * loción para después de afeitarse - aftershave lotion
 * maquinilla de afeitar - razor
 * sombra de ojos - eye shadow

Cultural Insight (Ponchos)
Ponchos, also known in other places as: (Huipil y sarape) are the typical clothing worn in the altiplano, a plateau in the Andes Mountains between Bolivia and Peru. The poncho is also the traditional clothing of Argentina, Chile, Colombia, Ecuador, Guatemala, and Mexico, especially in the mountainous regions. Ponchos are usually made of the wool from llama, alpacas, or other animals.

title: Spanish 2/Chapter 3 (Daily Activities)


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User:Eml4500.f08.team.foskey.ckf/hw5a

MATLAB Assignment  Debugging the Two Bar Truss problem

Two Bar Truss Code
The original two bar truss code was a modification of an earlier code that evaluated a truss composed of bars with eqaul material and physical properties. However, the two bar truss problem was composed of bars with unequal properties. This new, "debugged" code corrects that issue. The debugged code is a s follows:

Additional Functions
The following functions are used in the main code of TwoBarTrussDebugged.m, and are needed for the main code to run.

What The Function PlaneTrussElement Does
The function PlaneTrussElement reads in the Young's Modulus, the cross sectional area of a bar, and the coordinates of the endpoints of a two dimensional bar. The function then uses the coordinates to calculate the length of the bar. Next, the function calculates the director cosines. Finally, the function calculates the elemental stiffness matrix of the bar. This elemental stiffness matrix is returned to the original call function.

What NodalSoln Does
The function NodalSoln reads in the global stiffness matrix, the applied loads, a matrix ,debc, containing the degrees of freedom with specified values (in this case the constrained degrees of freedom), and a matrix, ebcVals ,with the values of the nodes (in this case zeros). The function first calculates the number of global degrees of freedom. Through several steps, the function ultimately calculates the global displacements, and the reactions (in a global coordinate system) of the bars. The displacement matrix, and reaction matrix is returned to the original call function.

What PlaneTrussResults Does
The function PlaneTrussResults reads in the Young's Modulus, the cross sectional area, the coordinates of the endpoints of a bar, and the displacements of a two dimensional (plane) bar. The function first calculates the length of the bar, and then calculates the director cosines. Next, the function creates a transform matrix T that transforms the global displacements into axial displacements. From there, the function calculates the strain, stress, and axial forces acting on the bar. These values are then returned to the calling function.

Comments on the changes to the code
As stated above, the original two bar truss code was based on a problem for which the bars making up the truss all had the same Young's Modulus and cross sectional area values. In that first problem these values were constants and were coded as such. In the two bar truss problem, each truss has a different value for its Young's Modulus (e) and cross sectional area (A). In the two bar truss problem, those values are coded as 2x1 matrices. However, when the PlaneTrussResults function is called, the original call is made, in which the values for "e" and "A" were constants. Since in the two bar truss problem these values are matrices, the entire matrix is passes, instead of just the values for one of the bars.

To correct the "bug" in the code, the values in the function call were changed to reflect that they needed to be elements of a matrix.

As a result, the last nine lines of code goes from the incorrect

to the corrected

with the only true change being the addition of "(i)" after the variables "e" and "A" in the function call of PlaneTrussResults.

To see the complete Incorrect code for the two bar truss, click here

Results
Running the above TwoBarTrussDebugged.m file generates the following results output:

k_local = 0.75       -0.75        -0.75         0.75  k = 0.5625     0.32476      -0.5625     -0.32476      0.32476       0.1875     -0.32476      -0.1875      -0.5625     -0.32476       0.5625      0.32476     -0.32476      -0.1875      0.32476       0.1875  k_local = 5   -5    -5     5  k = 2.5        -2.5         -2.5          2.5         -2.5          2.5          2.5         -2.5         -2.5          2.5          2.5         -2.5          2.5         -2.5         -2.5          2.5  K = 0.5625     0.32476      -0.5625     -0.32476            0            0      0.32476       0.1875     -0.32476      -0.1875            0            0      -0.5625     -0.32476       3.0625      -2.1752         -2.5          2.5     -0.32476      -0.1875      -2.1752       2.6875          2.5         -2.5            0            0         -2.5          2.5          2.5         -2.5            0            0          2.5         -2.5         -2.5          2.5  R = 0    0     0     7     0     0  d = 0           0        4.352       6.1271            0            0  reactions = -4.4378     -2.5622       4.4378      -4.4378  results = 1.7081      5.1244       5.1244       0.6276        3.138        6.276

Comments on results
Running the TwoBarTrussDebugged.m file provided the above results. The results show the local stiffness matrix, and elemental stiffness matrix (in the global coordinate system) for element one, and two. The results next show the global stiffness matrix, K, the applied loads R, and global displacements d. The reactions matrix gives the reaction forces acting at each node. The results matrix gives the strain (in column 1), stress (column 2), and axial forces (column 3) acting in element 1 ( in row 1), and element 2 ( in row 2).

By debugging the code, the results output is corrected, specifically the actual results matrix. The original results matrix is a 2x4 matrix of essentially garbage values generated by a function call that passes an entire matrix, as opposed to a single value from that matrix.

The original, incorrect results output can be seen at HW2 MATLAB portion

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User:Eml4500.f08.team.foskey.ckf/hw5b

MATLAB Assignment The Six Bar Truss Problem

The Six Bar Truss Problem
The following codes address the six bar truss problem as seen in Example 4.1 on page 226 of the text Fundamental Finite Element Analysis and Applications by M. Asghar Bhatti. This is a three part assignment, and each part is commented after its code. The figure below shows the problem setup

In the problem, P is equal to 20kN and $$ \alpha $$ is 30o. The Young's Modulus and cross sectional areas of the bars varies between Case 1 and Case 2 as explained following the MATLAB code for each case.

Comments And Explanation Of Case 1 Code
The code for the six bar truss above is commented throughout as best as possible to explain the function of key sections and variables.

In case 1, the Young's Modulus for all six of the bars is 200GPa, and the cross sectional area is 0.001 m2 for all six bars

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

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

The final part of the code labels the nodes and elements, as well as creates a legend

Comments And Explanation Of Case 2 Code
Again, the code for the six bar truss (Case 2) is commented throughout as best as possible.

In Case 2, each bar (element) of the truss has its own values for its Young's Modulus.

These values were:

E (1) = 150 GPa E (2) = 180 GPa E (3) = 200 GPa E (4) = 200 GPa E (5) = 220 GPa E (6) = 250 GPa

The values for the cross sectional area remains constant at 0.001 m2.

The Case 2 code works exactly as the code for Case 1 (see Comments And Explanation Of Case 1 Code above), with the exception that the values for the Young's Moduli are stored as a matrix rather than a constant variable. As a result, each time a value for the modulus is used, it is used as the value of an element in that matrix.

Comments On The Comparison
The comparison code basically runs the code for Case 1 and Case 2 back to back. The output results is the results for Case 2 first, followed by the results for Case 1. The results are preceded by text indicating the case for which they correspond.

The key feature in the comparison code is the figure generated. The comparison code plots the undeformed truss, the deformed truss in Case 1 and the deformed truss in Case 2 in one figure. The nodes and elements are not labeled in this figure to avoid cluttering the figure.

Note: The comparison code is considerable less commented as it is essentially a combination of the individual case codes.

For code comments, see the code for either Case 1 or Case 2 (above).

A Note On Functions Used
The codes for Case 1, Case 2, and the comparison code all use the functions PlaneTrussElement, PlaneTrussResults, and NodalSoln. These functions and explanations of what they do can be seen here

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

For the comparison results, the same results is generated twice, first for Case 2, and then for Case 1. At the beginning of each case's output, text states for which case the results are for.

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

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

Case 1 Figure
The figure below is generated by running the Case 1 code. The undeformed truss is shown in red dotted lines, and the deformed truss is shown in blue solid lines. Global nodes and elements are labeled on the deformed truss.

Since nodes 1, 3, and 4 were pinned, those nodes remain the same for both trusses. Node 5 moves slightly, and node 2 moves considerably, as this was the point at which the load was applied.

Note: To better show the deformed truss, the displacements were multiplied by a scaling factor of 5. This results in exaggerated displacements when the truss is plotted.

Case 2 Figure
The figure below is generated by running the Case 2 code. The undeformed truss is shown in red dotted lines, and the deformed truss is shown in blue solid lines. Global nodes and elements are labeled on the deformed truss.

Since nodes 1, 3, and 4 were pinned, those nodes remain the same for both trusses. Node 5 moves very slightly, and node 2 moves in the direction of the applied load. Since the bars of the truss had different material properties, the truss deforms slightly differently than in Case 1, but overall the deformation is similar.

Note: To better show the deformed truss, the displacements were multiplied by a scaling factor of 5. This results in exaggerated displacements when the truss is plotted.

Comparison of Cases 1 and 2 Figure
The figure below is created when the comparison code is run. It shows the undeformed truss in red dotted lines, the deformed truss for Case 1 in green solid lines, and the deformed truss for Case 2 in blue solid lines. Element and node numbers are neglected to avoid cluttering the figure.

By putting all three trusses in one figure, one can clearly see the difference that changing the Young's moduli makes on the deformations. It is interesting to note that node 2 moves more in Case 2, but node 5 almost not at all. Node 5 moves more in Case 1.

Note: To better show the displacements, the displacements of the deformed trusses are both multiplied by a scaling factor of 5.

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User:Eml4500.f08.ateam/hw 6

 See my comments below. Please don't remove these comment boxes; just add your comment to these comment boxes in case you fix the problems. Eml4500.f08 19:52, 26 November 2008 (UTC)

A-Team Teampage

Homework Report #6

This page is intended for Homework Report #6 of the A-Team.

Due date: Friday, 21 November 08, 5 pm EST = 22:00 UTC.

The Equation of Motion
The equation of motion can be derived by prescribing initial conditions to any system. The initial conditions for the system at t = 0 can be defined below.

$$ u\left(x,t=0\right) = \overline{u}(x) $$ for the given function of x, a known displacement function,

and

$$\frac{\partial u}{\partial t}(x,t=0)={\dot{u}}(x,t=0)=\overline{v}(x) $$ for a known velocity function.

The principle of virtual work (continuous) of dynamics of the elastic bar can be shown as a partial differential equation:

$$\frac{\partial}{\partial x}\left[(EA)\frac {\partial u}{\partial x}\right]+ f = m{\ddot{u}}\ \ \ \ (1) $$

The Discrete Equation of Motion can then be represented as

$$-\mathbf{K}\mathbf{d} + \mathbf{F} = \mathbf{M}\mathbf{\ddot{d}}$$

or $$\mathbf{M}\mathbf{\ddot{d}} + \mathbf{K}\mathbf{d} = \mathbf{F}\ \ \ \ (2) $$

Where $$\mathbf{M}$$ is a mass matrix.

$$\frac{\partial }{\partial x}\left[\left(EA \right)\frac{\partial }{\partial x} \right] = -Kd$$

$$f\, =\, F$$

$$m\ddot{u} = M\ddot{d} $$

Note that equation (2) is the equation of motion for a system with multiple degrees of freedom. A system with a single degree of freedom will be shown in equation 3.

For a single degree of freedom

$$\mathbf{m}\mathbf{\ddot{d}} + \mathbf{k}\mathbf{d} = \mathbf{F}$$

All of these equations have been derived with the help of the following integral:

$$\int_{x=0}^{x=L} W(x)\left({\frac{\partial}{\partial x}\left[(EA)\frac{\partial u}{\partial x}\right] + f - m\ddot{u}}\right)dx = 0 \ \ \ \ (3)$$

for all possible $$\mathbf {W(x)}$$, where $$\mathbf {W(x)}$$ is the weighting coefficient. As seen before, going from equation 1 to equation 3 is trivial but it is important to know how to get from equation 3 to equation 1. Equation 3 can be rewritten as:

$$\int_{x=0}^{x=L} w(x)g(x)dx = 0$$ for all $$\mathbf {w(x)}$$

Since equation 3 holds true for all values of $$\mathbf{w(x)}$$ by the principle of virtual work, we can choose $$ \mathbf {w(x) = g(x)}$$. This changes equation 3 to

$$\int_{x=0}^{x=L} g^2(x)dx = 0$$

Integrating by Parts
Integrating by parts is needed when trying to integrate two functions multiplied together. The formula for integrating by parts can be seen below. The symbols 'r' and 's' will represent the two different functions of x in this example.

$$ \left(rs\right)^'=r^'s+rs^'$$

Where: $$ r^' = \frac{dr}{dx} $$ and $$ s^'=\frac{ds}{dx} $$

Integrating both sides forms the equation:

$$\int \left(rs\right)^' = \int r^'s + \int rs^'$$

This leads to the final equation of:

$$ \int r^'s = rs - \int rs^'$$

Now to continue with the Principle of Virtual Work, recall equation 3:

$$\int_{x=0}^{x=L} W(x)\left({\frac{\partial}{\partial x}\left[(EA)\frac{\partial u}{\partial x}\right] + f - m\ddot{u}}\right)dx = 0 $$

For the first term, let $$r(x) = (EA){\frac{\partial u}{\partial x}}$$ and $${s\left(x\right) = W\left(x\right)}$$

After integrating by parts, the equation yields:

$$ \int_{x=0}^{x=L} W(x)\frac{\partial}{\partial x}\left[(EA)\frac{\partial u}{\partial x}\right]dx = \left[W(EA)\frac{\partial u}{\partial x}\right]_{x=0}^{x=L} - \int_{x=0}^{x=L}\frac{\partial W}{\partial x}(EA)\frac{\partial u}{\partial x}dx$$

Where the minus sign on the right side of the equation relates to the - Kd in the step between equations 1 and 2 above.

Replacing the x's with either L or 0, sets the right side equal to:

$$ = W(L)(EA)(L)\frac{\partial u}{\partial x}(L,t) - W(0)(EA)(0)\frac{\partial u}{\partial x}(0,t) - \int_{0}^{L}\frac{\partial w}{\partial x}(EA)\frac{\partial u}{\partial x}dx $$

Now consider the model equation with boundary conditions:

At x = 0, select W(x) such that $$W\left(0\right) = 0$$ or so that it is kinematically admissible.

Motivation: Discrete PVW applied to equation below:

$$ \mathbf{{W}_{6x1}\left(\left[K \right]_{6x2}\begin{Bmatrix} d_{3} \\ d_{4}\end{Bmatrix}_{2x1} - {F}_{6x1}\right)=0_{1x1}}$$

Where $$\left(\left[K \right]_{6x2}\begin{Bmatrix} d_{3} \\ d_{4}\end{Bmatrix}_{2x1} - {F}_{6x1}\right)$$ becomes a 6x1 matrix and $$ \mathbf{F}^{T}=\left[F_{1}F_{2}F_{3}F_{4}F_{5}F_{6}\right] $$ where F1, F2, F5, and F6 are unknown reactions.

Note that W can be selected arbitrarily, but for simpliciity select W such that W1 = W2 = W5 = W6 = 0. This is so all equations involving unknown reactions can be eliminated, or rows 1,2,5 and 6 are eliminated. Equation 1 shown below can then be formed.

$$ \mathbf{\overline{K}_{2x2}\overline{d}_{2x1}=\overline{F}_{2x1}} $$

Where $$\mathbf\overline{d} = \begin{Bmatrix} d_{3} \\ d_{4}\end{Bmatrix}$$ and $$\mathbf\overline{F} = \begin{bmatrix}F_{3} \\ F_{4}\end{bmatrix}$$.

Note that $$ \mathbf{\overline{W}\left(\overline{K}\overline{d}-\overline{F}\right)}=0 $$ (2) is actually the step before equation 1.

$$ W(L)F(t) - \int_{0}^{L}\frac{\partial w}{\partial x}(EA)\frac{\partial u}{\partial x}dx+\int w(x)[f - m\ddot{u}]dx=0 $$

This becomes the final equation:

$$ \int_{0}^{L}w(m\ddot{u})dx+\int_{0}^{L} \frac{\partial w}{\partial x}(EA)\frac{\partial u}{\partial x}dx

= W(L)F(t) + \int_{0}^{L}wfdx $$

These equations are for all $$ W\left(x\right)$$ such that $$ W\left(0\right) = 0 $$

Two cases of PVW

For the rest of this section, we will be focusing on element i in the above figures.

Assume the displacement for u(x) for $$x_i \le x \le x_{i+1}$$. (i.e., x $$\epsilon [x_i, x_{i+1}]$$)

The motivation for the linear interpolation of u(x) is shown below using a 2-bar truss.

The deformed shape is a straight line. This means there was an implicit assumption of the linear interpolation of the displacement between two nodes. Consider the case where there is only axial displacement, or zero transverse displacements. The question that needs to be solved is:

Express u(x) in terms of di = u(xi) and di+1 = u(xi+1) as a linear function in x.

The answer to this question is defined by linear interpolation, or:

$$u(x) = N_i\left(x\right)d_i + N_{i+1}\left(x\right)d_{i+1}$$

where $$N_i\left(x\right)$$ and $$N_{i+1}\left(x\right)$$ are linear functions of x.

Continuous PVW to Discrete PVW
Langrangian Interpolation

Motivation for form of $$N_i\left(x\right)$$ and $$N_{i+1}\left(x\right)$$

1) $$N_i\left(x\right)$$ and $$N_{i+1}\left(x\right)$$ are linear (straight lines), thus any linear combination of $$N_i\left(x\right)$$ and $$N_{i+1}\left(x\right)$$ is also linear, and in particular the expression for $$u\left(x\right)$$.

$$N_i\left(x\right) = \alpha_i + \beta_i x \qquad \quad \quad \ \ (with \ \alpha_i\, +\, \beta_i\, being\ real\ numbers) $$

$$N_{i+1}\left(x\right) = \alpha_{i+1} + \beta_{i+1} x \qquad (with \ \alpha_{i+1}\ +\, \beta_{i+1}\,  being\ real\ numbers) $$

Linear combination of $$N_i\left(x\right)$$ and $$N_{i+1}\left(x\right)$$:

$$N_i\left(x\right)d_i + N_{i+1}\left(x\right)d_{i+1} = (\alpha_i + \beta_i x)d_i + (\alpha_{i+1} + \beta_{i+1} x)d_{i+1}$$

$$=(\alpha_{i}\ d_i + \alpha_{i+1}\ d_{i+1})+ (\beta_{i}\ d_{i} + \beta_{i+1}\ d_{i+1}) x$$

Since the linear combination of this function is also a function of x (as shown in the last line), the initial assertion has been confirmed.

2) Recall the equation for $$ u(x)$$ (the interpolation of $$ u(x)$$) from the previous meeting:

$$N_i\left(x\right) = \frac {x - x_{i+1}}{x_{i} - x_{i+1}}$$

$$N_{i+1}\left(x\right) = \frac {x - x_{i}}{x_{i+1} - x_{i}}$$

Using this relationship and plugging in $$x\, =\, x_i$$

$$N_i\left(x\right) = \frac {x_{i} - x_{i+1}}{x_{i} - x_{i+1}}\ =\ 1 $$

$$N_{i+1}\left(x\right) = \frac {x_{i} - x_{i}}{x_{i+1} - x_{i}}\ = \ 0$$

Plugging these values into the equation for u(x):

$$u(x_i) = N_i\left(x_i\right)d_i + N_{i+1}\left(x_i\right)d_{i+1}$$

$$u(x_i) = (1)\, d_i + (0)\, d_{i+1}$$

FEM Via Principle of Virtual Work
The finite element method uses a weighting matrix W to solve for the solution. The first set of solutions is for a bar element i. The PVW solution followes a similar pattern with the displacement matrix being replaced by the weighting matrix W.

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

Apply the same interpretation for the W(x) matrix: For example,

$$ W(x) = N_i\, W_i + N_{i+1}\, W_{i+1} $$

Element Stiffness Matrix for element i:

$$ \int_{x_i}^{x_{i+1}}{} [N_iW_i + N_{i+1}W_{i+1}](EA)[N_id_i + N_{i+1}d_{i+1}]dx $$

The Definition of $$ N_i\, and\, N_{i+1} $$ can be seen below:

$$ N_i = \frac{dN_i(x)}{dx} $$       $$ N_{i+1} = \frac{dN_{i+1}(x)}{dx} $$

Note:

$$ u(x) = \N_i(x) \N_{i+1}(x)| \begin{Bmatrix} d_i\\ d_{i+1} \end{Bmatrix} $$

The N(x) matrix is a 1x2 matrix and the d matrix is 2x1.

Taking the derivative yeilds the following:

$$ \frac{du(x)}{dx} = \N_i'(x) \N'_{i+1}(x)| \begin{Bmatrix} d_i\\ d_{i+1} \end{Bmatrix} $$

The N'(x) matrix is a 1x2 matrix and is called the B(x) matrix.

Similarly for the W matrix from the Principle of Virtual Work (PVW)
$$ W(x) = N(x) \begin{Bmatrix} W_i\\ W_{i+1} \end{Bmatrix} $$

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

Recall the Element DOFs:

$$ \begin{Bmatrix} d_i\\ d_{i+1} \end{Bmatrix} = \begin{Bmatrix} d_1^i\\ d_2^i \end{Bmatrix} d^{(i)} $$

$$ \begin{Bmatrix} W_i\\ W_{i+1} \end{Bmatrix} = \begin{Bmatrix} W_1^i\\ W_2^i \end{Bmatrix} W^{(i)} $$

From these relations, we can find the integral that contains both W and d.

$$ \int_{x_i}^{x_{i+1}}{} (BW^i)(EA)(Bd^i)dx $$

The B matrix relates d to W. (Principle of Virtual Work)

Our Goal now is to derive $$ W^i\, (k_id_i) $$ from the previous integral.

Re-write the integral with the EA in the front:

$$ \int_{x_i}^{x_{i+1}}{} (EA)(BW^i)(Bd^i)dx $$

Take the Transpose of the BW factor to eliminate the B matrix:

$$ \int_{x_i}^{x_{i+1}}{} (EA)(BW^i)^T(Bd^i)dx $$

Distribute the Transpose inside of the Parentheses:

$$ \int_{x_i}^{x_{i+1}}{} (EA)(B^TW^{iT})(Bd^i)dx $$

The W matrix is equal to its transpose so it can be removed:

$$ \int_{x_i}^{x_{i+1}}{} (EA)(B^TW^{i})(Bd^i)dx $$

What we are left with is ( With the W matrix pulled outside the integrand because it does not vary with x):

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

The B matrix is a function of x and is equal too:

$$ B(x) = \|\frac{-1}{L^i} \|\frac{-1}{L^i}| $$

Note: $$ L^{i}\, =\, x_{i+1}\, -\, x_i $$

Transform of variable from $$x$$ to $$ \tilde{x}$$:

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

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

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

 Error: See my comments in Team Bike (with annotation). Perhaps because your two teams were working together on this part, so you get the same error; it's OK to collaborate across the teams, but be careful of getting the same mistakes. Eml4500.f08 19:52, 26 November 2008 (UTC)

Determining Average Stiffness Matrix
Our goal is to use linear interpolation to determine $$N_i(\tilde{x})$$. If $$\tilde{x}= x-x_i$$, then

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

Now, the shape function can be written as follows:

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

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

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

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

This matrix is also known as the average stiffness matrix.

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

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

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

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

The Mean Value Theorem (MVT) states

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

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

Similarly,

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

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

However,

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

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

Similarly,

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

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

MATLAB: Eigenvalue Analysis of Rectangular Three-Bar and Four-Bar Truss Systems
This problem is a second attempt at plotting the eigenvectors for the truss problem described in Homework Report #4. The unstable three-bar truss system is depicted as the truss at the left of the of the following figure while the stable four-bar truss is shown at the right.

Unstable Three-Bar Truss System
The global stiffness matrix for the problem is assembled as follows.

Using the global stiffness matrix, the following MATLAB command can be used to determine the eigenvalues and vectors for the truss. These are displayed as follows with V corresponding to the eigenvectors and D corresponding to the eigenvalues.

Plotting Eigenvectors
By plotting the columns of the eigenvector matrix that correspond to the zero eigenvalues, a mode of the truss can be depicted. In this case, the zero eigenvalues fall in columns 1, 2, 3, 4, and 5 of the eigenvalue matrix, D. Therefore, the first five columns of the eigenvector matrix, V, are plotted similarly to plotting displacements. These values are simply added to the original nodal positions to give the mode shape.

The following code was appended to the original truss problem MATLAB file to plot the eigenvectors.

First Column Eigenvectors
This plot represents the eigenvectors corresponding to the first column of the eigenvector matrix, V.

Second Column Eigenvectors
This plot represents the eigenvectors corresponding to the second column of the eigenvector matrix, V.

Third Column Eigenvectors
This plot represents the eigenvectors corresponding to the third column of the eigenvector matrix, V.

Fourth Column Eigenvectors
This plot represents the eigenvectors corresponding to the fourth column of the eigenvector matrix, V.

Fifth Column Eigenvectors
This plot represents the eigenvectors corresponding to the fifth column of the eigenvector matrix, V.

Stable Four-Bar Truss System
The global stiffness matrix for the problem is assembled as follows.

Using the global stiffness matrix, the following MATLAB command can be used to determine the eigenvalues and vectors for the truss. These are displayed as follows with V corresponding to the eigenvectors and D corresponding to the eigenvalues.

Plotting Eigenvectors
By plotting the columns of the eigenvector matrix that correspond to the zero eigenvalues, a mode of the truss can be depicted. In this case, the zero eigenvalues fall in columns 1, 2, 3, and 4 of the eigenvalue matrix, D. Therefore, the first five columns of the eigenvector matrix, V, are plotted similarly to plotting displacements. These values are simply added to the original nodal positions to give the mode shape, just as in the previous truss system.

The following code was appended to the original truss problem MATLAB file to plot the eigenvectors.

First Column Eigenvectors
This plot represents the eigenvectors corresponding to the first column of the eigenvector matrix, V.

Second Column Eigenvectors
This plot represents the eigenvectors corresponding to the second column of the eigenvector matrix, V.

Third Column Eigenvectors
This plot represents the eigenvectors corresponding to the third column of the eigenvector matrix, V.

Fourth Column Eigenvectors
This plot represents the eigenvectors corresponding to the fourth column of the eigenvector matrix, V.

MATLAB: Electric Pylon Structure Analysis
The truss system in this problem is a electric-line pylon consisting of 91 elements. The MATLAB code for accurately assembling this truss system was obtained from Dr. Loc Vu-Quoc and can be found here or seen below.

The objective of this problem is to perform additional MATLAB programming to find specific properties of the truss system based on given specifications and an applied load P. P is applied vertically downward at the far right tip of the arm of the pylon. All elements are constructed of 300M steel and have identical cross-sectional areas. Properties for this material were given by Dr. Vu-Quoc and verified online here. They are displayed in the table below.

Running the provided code produces the following plot showing the structure of the electric pylon truss system.

Analysis Requirements

 * Scale the electric pylon so that its height is 60 m
 * Plot the undeformed shape (dotted line) and the deformed shape (solid line)
 * Compute the axial stress in each bar
 * Display the highest tensile stress and the highest compressive stress and the corresponding elements in which they occur
 * Indicate these elements with an arrow on the figure of undeformed and deformed shapes
 * Determine if the problem is statically determinant and provide justification through argument or MATLAB programming
 * Construct the lumped mass matrix of the electric pylon
 * Solve the generalized eigenvalue problem
 * Find the lowest three eigenpairs and, in 3 separate plots, plot the eigenvectors as deformed shapes (solid line) superposed onto the undedeformed shape (dotted line)
 * Find the 3 lowest vibrational periods of the electric pylon

MATLAB Analysis Code
The code below is the  file written to solve the problem as assigned. This program requires additional MATLAB functions to run. These functions are also detailed below.

Additional Functions
Similar to the five-bar truss system, the electric pylon truss system analysis MATLAB program requires additional functions to run. Some of hese functions are found at the course textbook website, Fundamental Finite Element Analysis and Applications. They include,  , and. Other essential functions include  and. These functions were independently written and cannot be found with the preceding functions. The details of these functions are outlined below.

This function generates the element stiffness matrix for each element. When the function is called it calculates the X and Y coordinate positions of the element ends from the coordinates passed in. It produces the element length from this information. Next it calculates the direction cosines of the element. Using the Young's modulus and cross-sectional area of the element, the element stiffness matrix, k, is assembled. The stiffness matrix is passed back to the.

This function solves the force-displacement relationship for each node, resulting in the displacements and reactions at each node. This information is passed back to the.

This function outputs the axial strain, axial stress, and axial force of each element.

This function returns the length of an element based on the element coordinates it receives. It was independently written and cannot be found with the preceding functions.

This function was written to output the reduced matrices, $$\bar{M}$$ and $$\bar{K}$$. It was independently written and cannot be found with the preceding functions.

Running Analysis Code
The code below is the complete results from running the MATLAB program. Each part is detailed in later sections of this homework report.

Plot of Undeformed and Deformed Truss System
The extended MATLAB code plots the deformation of the truss system as follows. To scale the electric pylon truss to be 60m tall, the nodal positions were multiplied by a factor of 8.0753701211306. It is important to note that the deformation has been magnified by a factor of 100 for viewing purposes.

Axial Stress of Elements
The axial stresses for each element are simply the second column of the results array. They are shown above as the result  when the MATLAB code is run. They are also shown here.

Highest Tensile Stress
As shown on the plot and results from running, the greatest tensile stress occurs in Element 81. The value of the stress is 9.0511x106 Pa.

Highest Compressive Stress
As shown on the plot and results from running, the greatest compressive stress occurs in Element 55. The value of the stress is 8.6957x106 Pa. It is important to note that the negative sign has been removed from this stress because it has been denoted as compressive stress. In the stress table above, negative stresses denote compressive stress. This aligns with common stress notation.

Lumped Mass Matrix
The lumped mass matrix was assembled as a diagonal matrix of the masses of both degrees of freedom for each node. To determine the masses, each elements length was multiplied by its cross-sectional area and its density. Once the global nodes of each end of the element were known, the half of the element's mass was assigned to both nodes. Continuing this for each element created the lumped mass matrix.

Eigenpair Analysis
The three smallest eigenvalues for the truss system were determined using the generalized eigenvalue problem shown below. In this relation $$\bar{K}$$ is the reduced stiffness matrix and $$\bar{M}$$ is the reduced lumped mass matrix.

$$\bar{K}\nu =\lambda \bar{M}\nu$$

By plotting the columns of the eigenvector matrix that correspond to the smallest eigenvalues, modes of the truss can be depicted. In this case, the smallest eigenvalues were 132.16, 2468.4 and 2940.3. Therefore, the columns of the eigenvector matrix, V, that correspond to these values are plotted similarly to plotting displacements. These values are simply added to the original nodal positions to give the mode shape.

The following code was appended to the original truss problem MATLAB file to plot the eigenvectors. It is important to note that the eigenvectors were multiplied by a factor of 10 for viewing purposes.

Smallest Eigenpair Plot
This plot represents the eigenvectors corresponding to the smallest eigenvalue.

Second Smallest Eigenpair Plot
This plot represents the eigenvectors corresponding to the second smallest eigenvalue.

Third Smallest Eigenpair Plot
This plot represents the eigenvectors corresponding to the third smallest eigenvalue.

Lowest Vibrational Periods of the Electric Pylon
To find the three lowest vibration periods of the electric pylon, the eigenvalues are set equal to the angular frequencies as shown in the equation below.

$$\lambda = \omega^2$$

The angular frequencies are converted to vibrational frequencies as follows.

$$f = \frac{\omega}{2\cdot \pi}$$

From the vibrational frequencies, the vibrational periods can be determined using the following equation.

$$T = \frac{1}{f} = \frac{2\cdot \pi}{\omega}$$

Following these relations reveals the three lowest vibrational periods, shown in the table below.

Statically Determinant?
The electric pylon structure is not statically determinate. While two nodes of the system are fixed, these two nodes act as the termination point of four elements. Thus, there are a total of four unknown reaction forces in a system with only three equations with which to attempt to solve them. This cannot be solved using the principles of statics.

It is important to note that, while similar in number of fixed nodes to previous systems analyzed, this system differs in that the fixed nodes contain more than one element.

 Note: See my comments in Team Bike (with annotation). Eml4500.f08 19:52, 26 November 2008 (UTC)

Contributing Team Members
The following students contributed to this report:

Mark Barry Eml4500.f08.Ateam.barry 04:24, 20 November 2008 (UTC)

Daniel DiDomenico Eml4500.f08.ateam.didomenico 16:01, 17 November 2008 (UTC)

Sean Miller EML4500.f08.Ateam.Miller 18:17, 21 November 2008 (UTC)

Jil Paquette Eml4500.f08.Ateam.paquette 15:00, 19 November 2008 (UTC)

Matthew Philbin Eml4500.f08.Ateam.philbin 19:53, 20 November 2008 (UTC)

Ian Slevinski Eml.f08.Ateam.Slevinski 23:03, 20 November 2008 (UTC)

title: User:Eml4500.f08.ateam/hw 6


text:
User:Eml4500.f08.ateam/hw 7

A-Team Teampage

Homework Report #7

This page is intended for Homework Report #7 of the A-Team.

Due date: Tuesday, 09 December 08, 5 pm EST = 22:00 UTC.

Two-Bar Frame System
In this section frame structure is introduced. Analysis of frame structures is very similar to analysis of truss structures but there are a few key differences outline below. The properties of each element are shown below.

Element 1: $$E_1^{(1)} = 2.0$$ $$ E_2^{(1)} = 4.0$$ $$A_1^{(1)} = 0.5$$ $$ A_2^{(1)} = 1.5$$

Element 2: $$E_1^{(2)} = 3.0$$ $$ E_2^{(2)} = 7.0$$ $$A_1^{(2)} = 1.0$$ $$ A_2^{(1)} = 3.0$$

Model Frame With Two Elements:

Element 1 is the on the left while Element 2 is on the right.

In the previous picture, the frame has a rigid connection between the two elements. This means that when the frame is deformed, there is no change in the angle between the elements. This leads to bending moments at either end of each element. We will now consider the free body diagrams of each element of the frame:

It must be noted that in the diagrams, $$d_i^{(e)}$$ and $$f_i^{(e)}$$ are generalized forces:    $$d_3^{(e)}$$ and $$d_6^{(e)}$$ are the rotational degrees of freedom and $$f_3^{(e)}$$ and $$f_6^{(e)}$$ are the bending moments.

Now we can consider the global degrees of freedom for the frame.

With the two $$6x6$$ element stiffness matrices, we can develop the $$9x9$$ global stiffness matrix $$\underline{K}_{9x9} = A\underline{k}_{6x6}^{(e)}$$

The element force displacement equation in local coordinates is represented by:

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

Where the local element stiffness matrix is:

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

It is important to note that $$\tilde{f_3}^{(e)} = f_3^{(e)}$$ and $$\tilde{f_6}^{(e)} = f_6^{(e)}$$ because they are moments about the z axis and are unaffected by the differences in local and global coordinate systems.

Dimensional Analysis
When performing dimensional analysis, the following is the notation that will be used to represent the dimension of:

$$\left[\tilde{d_i}\right]$$ where i = 1,2,4,5

Where $$\left[\tilde{d_1}\right] = L$$ and $$\left[\tilde{d_3}\right] = 1 = \left[\tilde{d_6}\right]$$

From the previous figure, R can be found to help calculate the arc length, AB : $$\widehat{AB} = R\theta$$

Note that $$\theta$$ should be in radians for proper use in the following equation.

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

Now stress and strain can be found which will help verify the values of the stiffness matrix.

$$[\sigma] = [E][\epsilon]$$

$$[\varepsilon ] = \frac{[du]}{[dx]} = \frac{L}{L} = 1$$

$$[\sigma] = [E] = \frac{F}{L^2}$$

$$[A] = L^2$$

$$[I] = L^4$$

$$[\frac{EA}{L}] = [\tilde{k}_{11}] = \frac{(\frac{F}{L^2})L^2}{L} = \frac{F}{L}$$

$$[\tilde{k}_{11}\tilde{d}_1] = [\tilde{k}_{11}][\tilde{d}_1] = F$$

$$[\tilde{k}_{23}\tilde{d}_3] = \frac{6[E][I]}{L^2} = \frac{(\frac{F}{L^2})L^4}{L^2} = F$$

Using the principle of virtual work and focusing only on the bending effect, the expression for the matrix $$\tilde{k}$$ becomes:

$$\frac{\partial^2 }{\partial x^2}[(EI)\frac{\partial^2v }{\partial x^2}] - f_t(x) = m(x)\ddot{v}$$

Principle of Virtual Work for Beams
Motivation: The deformed shape of the tress element and the interpolation of the transverse displacements v(s), where $$s = \tilde{x}$$.

The PVW for beams can be defined as:

$$ \int_{0}^{L}{w(\tilde{x})}\left[-\frac{\partial ^2}{\partial x^2} \left((EI)\frac{\partial ^2v}{\partial x^2}\right) + f_t - m\ddot{v}\right]dx = 0$$ (1) for all possible w(x)

Begin the integration by parts for the first term:

$$\alpha = \int_{0}^{L}{w(\tilde{x})}\frac{\partial^2}{\partial x^2}\left\{(EI)\frac{\partial^2v}{\partial x^2}\right\}dx$$

where $$r(x) = \left\{(EI)\frac{\partial^2v}{\partial x^2}\right\}$$

and where $$r'(x) = \frac{\partial}{\partial x}\left(\frac{\partial}{\partial x}\left\{(EI)\frac{\partial^2v}{\partial x^2}\right\}\right)$$

This equation becomes: $$\alpha = \left[w\frac{\partial}{\partial x}\left\{(EI)\frac{\partial^2v}{\partial x^2}\right\}\right]^L_0 - \int_{0}^{L}\frac{dw}{dx}\frac{\partial}{\partial x}\left\{(EI)\frac{\partial^2v}{\partial x^2}\right\}dx$$

where $$s'(x) = \frac{\partial w}{\partial x}$$

and $$\beta_1 = \left[w\frac{\partial}{\partial x}\left\{(EI)\frac{\partial^2v}{\partial x^2}\right\}\right]^L_0$$.

After integrating by parts again, $$\alpha$$ becomes:

$$ = \beta_1 - \left[\frac{dw}{dx}\left(EI\right)\frac{\partial ^2v}{\partial x^2}\right]^L_0 + \int_{0}^{L}{\frac{\partial^2w}{\partial x^2}\left(EI\right)\frac{\partial ^2v}{\partial x^2}}dx$$

where $$\beta_2 = \left[\frac{dw}{dx}(EI)\frac{\partial ^2v}{\partial x^2}\right]^L_0$$

and the last integral can be replaced with the symbol $$\gamma$$. Note that this integral is symmetric. Equation (1) can now be rewritten and shortened to:

$$ - \beta_1 + \beta_2 - \gamma + \int_{0}^{L}{wf_tdx} - \int_{0}^{L}{wm\ddot{v}dx} = 0$$ for all possible w(x).

Stiffness Term $$\gamma$$
The focus is now on the stiffness term $$\gamma$$ to derive the beam stiffness matrix and to identify the beam shape function.

For this beam, the equation for $$v(\tilde{x})$$ can be seen below.

$$ v(\tilde{x}) = N_2(\tilde{x})\tilde{d}_2 + N_3(\tilde{x})\tilde{d}_3 + N_5(\tilde{x})\tilde{d}_5 + N_6(\tilde{x})\tilde{d}_6$$

Recall that $$u(\tilde{x}) = N_1(\tilde{x})\tilde{d}_1 + N_4(\tilde{x})\tilde{d}_4$$

$$N_2(\tilde{x}) = 1 - \frac{3\tilde{x}^2}{L^2} + \frac{2\tilde{x}^3}{L^3}\,\,\,\,\,\,\,\tilde{d}_2 $$

$$N_3(\tilde{x}) = \tilde{x} - \frac{2\tilde{x}^2}{L} + \frac{\tilde{x}^3}{L^2}\,\,\,\,\,\,\,\tilde{d}_3 $$

$$N_5(\tilde{x}) = \frac{3\tilde{x}^2}{L^2} - \frac{2\tilde{x}^3}{L^3}\,\,\,\,\,\,\,\tilde{d}_5$$

$$N_6(\tilde{x}) = -\frac{\tilde{x}^2}{L} + \frac{\tilde{x}^3}{L^2}\,\,\,\,\,\,\,\tilde{d}_6$$

Beam Shape Functions
The beam shape functions for the six forces are as follows:

Note the similarities between the corresponding horizontal and vertical forces as well as the moments. The vertical sides of the first four shapes have a magnitude of 1. The initial slope of mode 3 and the final slope of mode 6 both have a magnitude of 1. The final displacement vector can be written in the following form.

$$v(\tilde{x}) = N_2(\tilde{x})\hat{d}_2 + N_3(\tilde{x})\hat{d}_3 + N_5(\tilde{x})\hat{d}_5 + N_6(\tilde{x})\hat{d}_6$$

$$u(\tilde{x}) = N_2(\tilde{x})\hat{d}_1 + N_2(\tilde{x})\hat{d}_4 $$

In order to compute these components, the relationship developed in a previous lecture will be used.

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

This relationship is used to reorient the axial coordinate system to the global coordinate system.

Computing the Displacement Vector
The displacement vector can be written in terms of its components, both in the axial and global coordinate systems.

$$\mathbf{u}(\tilde{x})=u(\tilde{x})\vec{\tilde{i}}+v(\tilde{x})\vec{\tilde{j}}\quad (axial\ coordinates)$$

$$\mathbf{u}(\tilde{x})=u_x(\tilde{x})\vec{i}+v_y(\tilde{x})\vec{j}\quad (global\ coordinates) $$

In order to find $$u_x(\tilde{x})\ and\ v_y(\tilde{x})$$ a transformation matrix $$R^T$$ will be used.

$$\begin{Bmatrix} u_x(\tilde{x})\\v_y(\tilde{x})\end{Bmatrix} = R^T \begin{Bmatrix} u(\tilde{x}) \\ v(\tilde{x})  \end{Bmatrix} $$

Now the equation for $$\begin{Bmatrix} u(\tilde{x}) \\ v(\tilde{x})  \end{Bmatrix}$$ must be found. It is related to the displacements and the equation derived from the mode shapes.

$$\begin{Bmatrix} u(\tilde{x})\\v(\tilde{x})\end{Bmatrix} = \begin{bmatrix} N_1 &0&0&N_4&0&0\\0&N_2&N_3&0&N_5&N_6\end{bmatrix} \begin{Bmatrix} \tilde{d}_1^{(e)} \\ \tilde{d}_2^{(e)}\\\tilde{d}_3^{(e)}\\\tilde{d}_4^{(e)}\\\tilde{d}_5^{(e)}\\\tilde{d}_6^{(e)}  \end{Bmatrix} $$

Using these equations, they can be rewritten to give the final form of the global displacement vector.

$$\begin{Bmatrix} u_x(\tilde{x})\\v_y(\tilde{x})\end{Bmatrix} = R^T \begin{bmatrix} N_1 &0&0&N_4&0&0\\0&N_2&N_3&0&N_5&N_6\end{bmatrix} \tilde{T}^{(e)} d^{(e)}$$

Dimensional Analysis of Displacements
Axial Displacements

$$\left[u \right]=L$$

$$\left[N_{1} \right]= \left[N_{4} \right]=1$$

$$\left[N_{1} \right]\left[\tilde{d_{1}} \right]+\left[N_{4} \right]\left[\tilde{d_{4}} \right]$$

Transverse Displacements

$$\left[v \right]=L$$

$$\left[N_{2} \right]\left[\tilde{d_{2}} \right]=L$$

$$\left[N_{3} \right]\left[\tilde{d_{3}} \right]=L$$

Derivation of Beam Shape Functions
$$N_2,\ N_3,\ N_5,\ N_6$$

Refer to plots shown above to derive the beam shape functions.

The governing partial differential equation (PDE) for beams is as follows.

$$\frac{\delta ^{2}}{\delta x^{2}}\left\{\left(EI \right) \frac{\delta^{2}v }{\delta x^{2}}\right\}=0$$

Considering EI to be constant reveals the following.

$$\frac{\delta ^{4}}{\delta x^{4}}v=0$$

By integrating this four times we obtain four constants in a relation for V(x).

$$V\,(x) = c_0+c_1x^1+c_2x^2+c_3x^3$$

In order to obtain $$N_2(x)$$ where $$\tilde{x}\equiv x$$ the following boundary conditions must be considered to solve for the four constants determined above.

$$V\,(0)=1$$

$$V\,(L)=0$$

$$V\,'(0)=V\,'(L)=0$$

Evaluating V(x) at the boundary conditions yields

$$V(0)=1=c_0+c_1(0)+c_2(0)^2+c_3(0)^3=c_0\Rightarrow c_0=1$$

$$V'(0)=0=c_1+2c_2(0)+3c_3(0)^2=c_1\Rightarrow c_1=0$$

$$V(L)=0=1+0+c_2(L)^2+c_3(L)^3=1+c_2L^2+c_3L^3\Rightarrow (1)$$

$$V'(L)=0=0+2c_2(L)+3c_3(L)^2=2c_2L+3c_3L^2\Rightarrow (2)$$

From equation (2),

$$c_3=-\frac{2}{3}\frac{c_2}{L}$$

Now, inserting c3 from above into equation (1) yields

$$0=1+c_2L^2+\left( -\frac{2}{3}\frac{c_2}{L}\right)L^3\Rightarrow c_2=-\frac{3}{L^2}$$

and $$c_3=-\frac{2}{3}\frac{1}{L}\left({-\frac{3}{L^2}} \right)=\frac{2}{L^3}$$

Now, inserting c0, c1, c2, and c3 into V(x),

$$V(x)=1-\frac{3}{L^2}x^2+\frac{2}{L^3}x^3$$

at the N2 boundary conditions. This expression is the same as the N2(x) found previously.

Likewise for $$N_3$$ the following boundary conditions are considered.

$$V\,(0)=0$$

$$V\,(L)=0$$

$$V\,'(0)=1$$

$$V\,'(L)=0$$

Evaluating V(x) at the boundary conditions yields

$$V(0)=0=c_0+c_1(0)+c_2(0)^2+c_3(0)^3=c_0\Rightarrow c_0=0$$

$$V'(0)=1=c_1+2c_2(0)+3c_3(0)^2=c_1\Rightarrow c_1=1$$

$$V(L)=0=1+0+c_2(L)^2+c_3(L)^3=1+c_2L^2+c_3L^3\Rightarrow (1)$$

$$V'(L)=0=1+2c_2(L)+3c_3(L)^2=2c_2L+3c_3L^2\Rightarrow (2)$$

Using MATLAB to solve the system of 2 equations and 2 unknowns for equations (1) and (2),

$$c_2=\frac{-2}{L}$$

$$c_3=\frac{1}{L^2}$$

Now, inserting c0, c1, c2, and c3 into V(x),

$$V(x)=1x-\frac{2}{L}x^2+\frac{1}{L^2}x^3$$

at the N3 boundary conditions. This expression is the same as the N3(x) found previously.

Likewise for $$N_5$$ the following boundary conditions are considered.

$$V\,(0)=0$$

$$V\,(L)=1$$

$$V\,'(0)=0$$

$$V\,'(L)=0$$

Evaluating V(x) at the boundary conditions yields

$$V(0)=0=c_0+c_1(0)+c_2(0)^2+c_3(0)^3=c_0\Rightarrow c_0=0$$

$$V'(0)=0=c_1+2c_2(0)+3c_3(0)^2=c_1\Rightarrow c_1=0$$

$$V(L)=1=0+0+c_2(L)^2+c_3(L)^3=c_2L^2+c_3L^3=1\Rightarrow (1)$$

$$V'(L)=0=0+2c_2(L)+3c_3(L)^2=2c_2L+3c_3L^2=0\Rightarrow (2)$$

Using MATLAB to solve the system of 2 equations and 2 unknowns for equations (1) and (2),

$$c_2=\frac{3}{L^2}$$

$$c_3=\frac{-2}{L^3}$$

Now, inserting c0, c1, c2, and c3 into V(x),

$$V(x)=\frac{3}{L^2}x^2-\frac{2}{L^3}x^3$$

at the N5 boundary conditions. This expression is the same as the N5(x) found previously.

Likewise for $$N_6$$ the following boundary conditions are considered.

$$V\,(0)=0$$

$$V\,(L)=0$$

$$V\,'(0)=0$$

$$V\,'(L)=1$$

Evaluating V(x) at the boundary conditions yields

$$V(0)=0=c_0+c_1(0)+c_2(0)^2+c_3(0)^3=c_0\Rightarrow c_0=0$$

$$V'(0)=0=c_1+2c_2(0)+3c_3(0)^2=c_1\Rightarrow c_1=0$$

$$V(L)=0=0+0+c_2(L)^2+c_3(L)^3=c_2L^2+c_3L^3=0\Rightarrow (1)$$

$$V'(L)=0=0+2c_2(L)+3c_3(L)^2=2c_2L+3c_3L^2=0\Rightarrow (2)$$

Using MATLAB to solve the system of 2 equations and 2 unknowns for equations (1) and (2),

$$c_2=\frac{-1}{L}$$

$$c_3=\frac{1}{L^2}$$

Now, inserting c0, c1, c2, and c3 into V(x),

$$V(x)=\frac{-1}{L}x^2+\frac{1}{L^2}x^3$$

at the N6 boundary conditions. This expression is the same as the N6(x) found previously.

Element Stiffness Matrix Coefficient Derivation
The coefficients containing EA have already been derived. This leaves the terms containing an EI component. These coefficients are derived here.

$$\tilde{k_{22}}=\frac{12EI}{L^{3}}=\int_{0}^{L}{\frac{d^{2}N_{2}}{dx^{2}}(EI)\frac{d^{2}N_{2}}{dx^{2}}dx}$$

$$\tilde{k}_{23}=\frac{6EI}{L^2}= \int_{0}^{L} \frac{d^2N_2}{dx^2}(EI)\frac{d^2N_3}{dx^2}\, dx$$

Or to write this in a more generalized form:

$$\tilde{k}_{ij}= \int_{0}^{L} \frac{d^2N_i}{dx^2}(EI)\frac{d^2N_j}{dx^2}\, dx\quad i,j=2,3,5,6$$

Discrete Principal of Virtual Work
$$\bar{w}\cdot[\bar{M}\ddot{\bar{d}}+\bar{k}\bar{d}-\bar{F}]=0 \quad\ for\ all\ \bar{w}$$

The bar notation denotes that the boundary conditions have already been applied. The ordinary differential equations (second order in time) and the initial conditions governing the elastodynamics of the discrete continuity problem are as follows.

$$\bar{M}\ddot{\bar{d}}+\bar{k}\bar{d}=\bar{F}(t)\quad\ (1)$$

$$\bar{d}(0)=\bar{d}_0\quad \quad (initial\ position)$$

$$\dot{\bar{d}}(o)=\bar{v}_0 \quad \quad (initial\ velocity)$$

Eigenvalue Solution
1) Consider the unforced vibration problem:

$$\bar{M}\ddot{v}+ \bar{k} v = 0 \quad \ (2)$$

Assume $$v(t)=(sin {wt})\mathbf{\phi}$$. Here $$\phi$$ is a column matrix of size (nx1) that is not time dependent.

$$\ddot{v}=-w^2sin {wt}\phi$$

Plug the previous equation back into equation (1):

$$-w^2 sin{wt} \bar{M}\phi + sin {wt} \bar{k} \phi = 0 $$

This can be simplified to the generalized eigenvalue problem:

$$\bar{k}\phi = w^2 \bar{M} \phi$$

An even more generalized version can be written in terms of matrices A and B.

$$\mathbf{A} x=\lambda \mathbf{B} x$$

Once evaluated, the following relations are made.

$$\mathbf{\lambda} = w^2$$

$$Mode\ i=\begin{cases} V_i(t)=(sin {w_it})\phi_i \\ i=1,...,n\end{cases}$$

2) Model superposition method:

Orthogonal property of eigenvalues is shown in the following equation:

$$\phi_{i} \bar{M} \phi_{j} = \delta _{i j} = \begin{cases} 1, & \mbox{if }\ {i=j} \\ 0, & \mbox{if }\ {i \ne j} \end{cases}$$

Here $$ \mathbf{\delta}_{ij}$$ is the Kronecker delta which has a variety of uses including digital processing.

$$\bar{M}\phi_j=\lambda^{-1}_j\phi^T_i\bar{k}\phi_j$$

$$\phi_i^T\bar{M}\phi_j=\delta_{ij}\lambda^{-1}_j\phi^T_i\bar{k}\phi_j$$

$$\phi^T_i\bar{k}\phi_j=\lambda_j \delta_{ij}$$

Next look at:

$$\bar{M}\ddot{\bar{d}}+\bar{k}\bar{d}=\bar{F}\quad \ (3)$$

$$\bar{d}(0)=\bar{d}_0$$

$$\dot{\bar{d}}(0)=\bar{v}_0$$

then

$$\bar{d}(t)=\sum_{i=1}^n \xi_i(t)\phi_i $$

Now plug the previous equation into equation (3):

$$\bar{M}(\sum_j \xi_j\phi_j)+\bar{k}(\sum_j\xi_j\phi_j)=\bar{F}$$

where

$$\ddot{\bar{d}}=\sum_j \xi_j\mathbf{\phi_j}\quad and\ \bar{d}=sum_j\xi_j\mathbf{\phi_j}$$

Finally,

$$\sum_j \ddot{\xi}_j(\mathbf{\phi_i^T}\mathbf{\bar{M}}\mathbf{\phi_j})+ \sum_j \ddot{\xi}_j(\mathbf{\phi_i^T}\mathbf{\bar{k}}\mathbf{\phi_j})=\mathbf{\phi^T_i} \mathbf{F}$$

$$\ddot{\xi}_i+\lambda_i\xi_i=\phi^T_i\mathbf{F}$$

Team View of Mediawiki
Mediawiki is an interesting medium for our class to use to complete our homework. Its learning curve is hard at first but once the proper syntax is learned, very professional work can be created. Mediawiki has one main advantage, as well as disadvantage, in the fact that anybody can access it. A password is not needed to access the information, unlike in the e-Learning system where only members are able to view it. This is a disadvantage because now anyone can edit the information to something that could be false. The math function in particular can create very precise and systematic functions that are easy to understand. Since all members of the group are using the same editors, a cohesive presentation can be achieved throughout each homework assignment. Since there are multiple groups working on each assignment, once turned in, each student has access to many different views and approaches to each problem.

With the new method of teaching with MediaWiki, there is a dissenting opinion about using it for coursework. In using the MediaWiki software, there are several issues that make it not appealing for classes. The first reason is the difficulty and time it takes to create the Wiki entries. The Wiki pages contain figures and equations, each of which must be entered using a special Wiki format. This takes an enormous amount of time to do when first exposed to it. Even after mastering the MediaWiki software, it still takes time to enter all the equations. In doing so, the focus of the course shifts from learning the Finite Element Method to creating a nice-looking Wiki homework report. FEM is an extremely powerful method and we, as students at UF, have a unique opportunity to take a class entirely devoted to it. Not many universities require this as an undergraduate course. Unfortunately, the implementation of Wiki homework reports greatly hampers the learning experience for the course.

The time spent formatting the Wiki pages takes away from the time that could be spent learning and working practice problems. For students that are taking many classes, time spent making wiki pages is time that could be better used on coursework.

The final reason the MediaWiki software is not acceptable for coursework is the concern for cheating. Because the entries are stored on the public domain, it is accessible to anyone, including students in the same class. This allows for students to copy other students work and submit it as their own.

Comparison of MediaWiki and E-Learning
MediaWiki has many advantages over E-Learning (ELS). MediaWiki allows for collaboration to produce professional-looking reports and projects. There is currently no way in ELS to collaborate in a group project. The editing ability of MediaWiki is far greater than that of ELS. Pages can be organized with many parameters and many types of media can be uploaded. A table of contents can be generated and linked to sections within a page.

MediaWiki also has the ability to store a greater amount of information for an indefinite time. Anything posted using MediaWiki is permanent and can always be accessed unless deletion is requested. An ELS account can be closed by the professor when the course has been complete. Additionally, the resources of MediaWiki are superior to those of ELS. During peak usage times, ELS is rendered useless while Wikiversity has the capability to handle a large server load.

The advantage ELS has over MediaWiki is privacy. MediaWiki is public domain, and anything posted or uploaded can be seen and sometimes used by others. This creates a large opportunity for plagiarism and cheating. In addition, grades and evaluations can be posted on ELS so that only the student can view them. MediaWiki does not have this capability.

Suggested Software to Improve Productivity
Most of the aspects of each homework assignment are similar and flow throughout the assignment. However, it was noticed that the programs that each student used for creating images were not the same. Therefore, some images were more plain that the others. In order to create a more uniform look, each group should choose an image software program at the beginning of the class and stick with it throughout the course.

Besides the Paint program that comes on every Windows PC, there are several programs available online that have enhanced features. Some such as Paint.NET and e-Paint are available for free and offers more advanced tools that the basic paint program. Another program that can easily be found online is GIMP (GNU Image Manipulation Program) which is more designed to modify images. Inkscape is a vector based graphics editor which can create Scalable Vector Graphics (SVG) file format. This is different that the usual scalar approach and allows the size of images to be easily changed with little loss of quality.

Other common programs useful for creating neat looking images include those in the Microsoft Office Suite. If one has the latest version, MS Word is excellent for the easy production of images. If the 2003 version must be used, MS PowerPoint is also good for generating clean, higher resolution images.

Anyone of these programs would be effective at creating the images that were needed for this class. In future courses, if a group were to decide on one, it would increase the overall continuity between individual contributions.

Additionally, if it is decided that the course must have an online component, it may be advisable to explore Google Documents. While Google Docs does not have some of the editing capability of MediaWiki, it may be a more viable alternative. The learning curve is not nearly as high as that of WikiMedia. Users can assemble documents, spreadsheets, presentations, and forms to be shared with others and even published. These forms are similar to MS Suite applications which are familiar to almost everyone. Like Wikiversity, Google Docs also allows collaboration for projects. An advantage of Google Docs in this area is it has the capability for multiple users to edit the same file at the same exact moment. Google's software saves the file almost constantly, storing edits as users make them.

Much like Wikiversity, each revision of a file is saved and the file can be reverted to a selected revision at anytime. Google Docs also displays the revision history in a more informative script, displaying exactly what changes were made, and their author. A document behaves much like MS Word in that images can be added and a table of contents can be generated. Users could still use the Latex Equation Editor and paste in the images of equations instead of the code. Also, tables can be generated within a document or spreadsheet.

A great advantage of Google Documents is privacy. A team member can create a file and choose to share it with members of the group, the TA's, and the professor. No one else can access or see the team's work. This would prevent a large amount of copied work. There would also be no need to link the homework report to a homework table before a deadline. The grader simply has to access the file that has been shared with them.

Using this method simply requires the creation of a Google Account much like creating a Wikiversity account. Creating a Wikiversity account had no other use outside classwork as it was strictly recommended not to edit other articles using the Wiki account. A Google account can have a Gmail account and calendar in addition to other benefits. The Gmail account can be linked to access UF Webmail - like WebCT, a somewhat outdated interface - and provide the student an email address upon graduation when their  account is closed. Google's services have proven to be highly useful and reliable and would possibly be of benefit to this course.

MATLAB: Two-Bar Truss System - Varying Cross-sectional Areas
The set-up and data for this two-bar truss system are seen below.

In this problem, two different methods of assembling the element stiffness matrices were explored. In one method, the varying cross-sectional areas are accounted for using the following relation general stiffness matrix formula.

$$\hat{k}^{(i)}_{gen}=\frac{1}{6\cdot L}\left(2E_{1}A_{1}+(E_{1}A_{2} + E_{2}A_1)+2E_2A_2\right)\begin{bmatrix} 1 & -1\\ -1 & 1 \end{bmatrix} $$

Additionally, the areas were averaged and assembled into the average stiffness matrix. This relation is as follows.

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

This function generates the element stiffness matrix for each element. When the function is called it calculates the X and Y coordinate positions of the element ends from the coordinates passed in. It produces the element length from this information. Next it calculates the direction cosines of the element. Using the Young's modulus and cross-sectional area of the element, the element stiffness matrix, k, is assembled. The stiffness matrix is passed back to the.

This function solves the force-displacement relationship for each node, resulting in the displacements and reactions at each node. This information is passed back to the.

This function outputs the axial strain, axial stress, and axial force of each element.

Plot of Deformed Shape and Deformed Shape of Two-Bar Truss System Using Average Areas
This plot depicts the deformation of the two-bar truss system using the average stiffness matrix in red and the deformation of the system using the general stiffness matrix in green. The difference is clearly shown in the plot and the table below.

MATLAB: Two-Bar Frame System
This two-bar frame system is very similar to the one analyzed in | Homework Report #2 and further in | Homework Report #5. It uses the same data as before but, for this case, Element 1 is changed to a frame element, creating a frame system. The cross-section for this system is assumed to be square. The applied load, P, has a value of 7.

This functions generates the element stiffness matrices an the element R matrix. It acts similarly to the  seen in previous problems.

This function solves the force-displacement relationship for each node, resulting in the displacements and reactions at each node. This information is passed back to the.

Plot of Undeformed and Deformed Shapes
This plot depicts the undeformed and deformed shapes of the two-bar frame system. Additionally, the deformed shape of the two-bar truss system is plotted in green for comparison. This deformation is scaled by a factor of 0.5 for viewing purposes.

From the plot, the difference in frame elements and truss elements is apparent. The frame element was divided into 20 segments to show its complete deformation as a curve. The fixed boundary conditions cause an additional degree of freedom at each node, resulting in rotation or bending of the bar.

MATLAB: Electric Pylon Frame System
The structure in this problem is a electric-line pylon consisting of 91 elements. The MATLAB code for accurately assembling this system was obtained from Dr. Loc Vu-Quoc and can be found here or seen below.

The objective of this problem is to expand upon analysis performed in the previous homework report, | Homework Report #6. For this problem, the electric pylon truss system was converted to a frame system with each element having a square cross-section. The applied load P is applied vertically downward at the far right tip of the arm of the pylon. Clamped boundary conditions are applied at the base of the frame model, replacing the simply supported conditions of the previous problem. All elements are constructed of 300M steel and have identical cross-sectional areas. Properties for this material were given by Dr. Vu-Quoc and verified online here. They are displayed in the table below.

Analysis Requirements

 * Convert the electric pylon truss system to a frame system
 * Redo the static analysis using the same geometric, material and loading data as for Homework Report #6
 * Compare the nodal displacements of the node under the applied force for the frame model to those of the truss model
 * Identify the frame element or elements with the highest nodal bending moment and transverse shear force
 * Plot the undeformed shape, deformed shape of the truss model and "incomplete" deformed shape of the frame model using the same magnification factor
 * Determine if the problem is statically determinant and provide justification through argument or MATLAB programming
 * Compare the reactions obtained from the truss model to those from the frame model
 * Construct the lumped mass matrix of the electric pylon
 * Solve the generalized eigenvalue problem
 * Find the lowest three eigenpairs and, in 3 separate plots, plot the eigenvectors as deformed shapes superposed onto the undeformed shape
 * Find the three lowest vibrational periods of the electric pylon
 * Generate an animation of the three lowest vibrational periods of the electric pylon

MATLAB Analysis Code
The code below is the  file written to solve the problem as assigned. This program requires additional MATLAB functions to run. These functions are also detailed below.

This functions generates the element stiffness matrices an the element R matrix. It acts similarly to the PlaneTrussElement.m seen in previous problems.

This function returns three matrices consisting of the Axial Force Matrix,, the Bending Moment Matrix,  , and the Shear Force Matrix,. Each of these matrices lists the X and Y coordinates and the respective value for each element.

This function returns the length of an element based on the element coordinates it receives. It was independently written and cannot be found with the preceding functions.

This function solves the force-displacement relationship for each node, resulting in the displacements and reactions at each node. This information is passed back to the.

This function was written to output the reduced matrices, $$\bar{M}$$ and $$\bar{K}$$. It was independently written and cannot be found with the preceding functions.

Running Analysis Code
The code below is the complete results from running the MATLAB program. Each part is detailed in later sections of this homework report.

Plot of Deformed and Undeformed Shapes
The extended MATLAB code produces the following plot of the deformation of the truss system. The frame system is detailed in red while the truss system is shown in green. Since the systems overlap, it is difficult to view. The red frame system is plotted on top of the green truss system. To scale the electric pylon truss to be 60m tall, the nodal positions were multiplied by a factor of 8.0753701211306. It is important to note that the deformation has been magnified by a factor of 20 for viewing purposes.

Statically Determinant?
The electric pylon structure is not statically determinate. While two nodes of the system are fixed, these two nodes act as the termination point of four elements. Thus, there are a total of four unknown reaction forces in a system with only three equations with which to attempt to solve them. This cannot be solved using the principles of statics.

It is important to note that, while similar in number of fixed nodes to previous systems analyzed, this system differs in that the fixed nodes contain more than one element.

Comparison of Reactions
The reactions for the truss system and the frame system are almost identical. They are displayed in the following table.

Comparison of Nodal Displacements of Truss and Frame Systems
The nodal displacements of the frame and truss systems vary by minuscule amounts. The following picture magnifies the displacements for viewing. To better understand the scale, the distance between the two nodes is approximately one ten-thousandth of a meter.

The exact displacements are presented in the table below. The values for the truss and frame systems are almost identical.

Highest Nodal Bending Moment
As shown on the plot and results from running, the greatest nodal bending moment occurs in Element 81. The value of the moment is 6.2874 N-m.

Highest Transverse Shear Force
As shown on the plot and results from running, the highest transverse shear force occurs in Element 81. The value of the force is -5.4827 N.

Lumped Mass Matrix
The lumped mass matrix was assembled as a diagonal matrix of the masses of both degrees of freedom for each node. To determine the masses, each elements length was multiplied by its cross-sectional area and its density. Once the global nodes of each end of the element were known, the half of the element's mass was assigned to both nodes. Continuing this for each element created the lumped mass matrix.

Eigenpair Analysis
The three smallest eigenvalues for the truss system were determined using the generalized eigenvalue problem shown below. In this relation $$\bar{K}$$ is the reduced stiffness matrix and $$\bar{M}$$ is the reduced lumped mass matrix.

$$\bar{K}\nu =\lambda \bar{M}\nu$$

By plotting the columns of the eigenvector matrix that correspond to the smallest eigenvalues, modes of the truss can be depicted. In this case, the smallest eigenvalues were 132.86, 2471.2 and 2942.6. Therefore, the columns of the eigenvector matrix, V, that correspond to these values are plotted similarly to plotting displacements. These values are simply added to the original nodal positions to give the mode shape.

The following code was appended to the original truss problem MATLAB file to plot the eigenvectors. It is important to note that the eigenvectors were multiplied by a factor of 4 for viewing purposes.

Smallest Eigenpair Plot
This plot represents the eigenvectors corresponding to the smallest eigenvalue.

Second Smallest Eigenpair Plot
This plot represents the eigenvectors corresponding to the second smallest eigenvalue.

Third Smallest Eigenpair Plot
This plot represents the eigenvectors corresponding to the third smallest eigenvalue.

Vibrational Period Analysis
To find the three lowest vibration periods of the electric pylon, the eigenvalues are set equal to the angular frequencies as shown in the equation below.

$$\lambda = \omega^2$$

The angular frequencies are converted to vibrational frequencies as follows.

$$f = \frac{\omega}{2\cdot \pi}$$

From the vibrational frequencies, the vibrational periods can be determined using the following equation.

$$T = \frac{1}{f} = \frac{2\cdot \pi}{\omega}$$

Following these relations reveals the three lowest vibrational periods, shown in the table below. In comparison to the values obtained from the electric pylon truss system, the vibrational periods are similar but shorter. This is due to the added rigidity of the frame system. The values do not seem to vary much at all but it is important to remember that for both the frame and truss systems, the actual deformation was extremely small, so this difference in vibrational period is significant.

Eigenmode Analysis of the Electric Pylon Truss System
movies were created by adding the following MATLAB code segment to the code from Homework Report #6. These files were converted to  format for compatibility with Wikimedia.

Creating Videos for Upload to Wikimedia Commons
A more detailed explanation of how to generate and convert video files for upload to Wikimedia Commons can be found here or seen in the collapsible table below.

Animation of Lowest Eigenmode
This animation represents the lowest eigenmode of the electric pylon system.

Animation of Second Lowest Eigenmode
This animation represents the second lowest eigenmode of the electric pylon system.

Animation of Third Lowest Eigenmode
This animation represents the third lowest eigenmode of the electric pylon system.

Contributing Team Members
The following students contributed to this report:

Mark Barry Eml4500.f08.Ateam.barry 03:46, 9 December 2008 (UTC)

Daniel DiDomenico Eml4500.f08.ateam.didomenico 14:31, 25 November 2008 (UTC)

Sean Miller EML4500.f08.Ateam.Miller 18:31, 9 December 2008 (UTC)

Jil Paquette Eml4500.f08.Ateam.paquette 01:35, 9 December 2008 (UTC)

Matthew Philbin Eml4500.f08.Ateam.philbin 12:30, 07 December 2008 (UTC)

Ian Slevinski Eml4500.f08.Ateam.Slevinski 05:34, 09 December 2008 (UTC)

title: User:Eml4500.f08.ateam/hw 7


text:
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 * Chat with other Wikiversitans on #wikiversity-en.

And don't forget to explore Wikiversity with the links to your left. Be bold to contribute and to experiment with the sandbox or your userpage, and see you around Wikiversity! --mikeu talk 04:53, 4 November 2008 (UTC)

title: User talk:Stacym87


title: File:Sixbartrusscase1.png


title: File:Sixbartrusscase2.png


title: File:Sixbar1v2.png


text:
User talk:Soelo

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

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

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

And don't forget to explore Wikiversity with the links to your left. Be bold to contribute and to experiment with the sandbox or your userpage, and see you around Wikiversity! --The Jade Knight (d'viser) 08:34, 4 November 2008 (UTC)

Thanks for your contribution to my Irish project. The Jade Knight (d'viser) 08:34, 4 November 2008 (UTC)

title: User talk:Soelo


text:
User talk:Zakman93

 Hello Zakman93, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. After you leave a comment on a talk page, remember to sign and date; it helps everyone follow the threads of the discussion. The signature icon in the edit window makes it simple. To get started, you may

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

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

And don't forget to explore Wikiversity with the links to your left. Be bold to contribute and to experiment with the sandbox or your userpage, and see you around Wikiversity! --The Jade Knight (d'viser) 08:56, 4 November 2008 (UTC)

Welcome to the School of History. Let me know if I can help out in any way. The Jade Knight (d'viser) 08:56, 4 November 2008 (UTC)

American Lit
It may be helpful if you listed planned authors/texts that you would work with so that those who come by the page may find out what your course will be about. Ottava Rima (talk) 21:24, 4 November 2008 (UTC)

title: User talk:Zakman93


text:
User talk:Jeanluc~enwikiversity

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

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

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

And don't forget to explore Wikiversity with the links to your left. Be bold to contribute and to experiment with the sandbox or your userpage, and see you around Wikiversity! --The Jade Knight (d'viser) 09:55, 4 November 2008 (UTC)

French Mentoring
Salut, Jeanluc. You are listed at the French stream as being a fluent French speaker interested in teaching French. The French Mentoring project has recently been launched, and if you would be willing to add your name to that list, I will make sure you are put in contact with appropriate students who would like to learn from you. The Jade Knight (d'viser) 09:57, 4 November 2008 (UTC)

Your account will be renamed
Hello,

The developer team at Wikimedia is making some changes to how accounts work, as part of our on-going efforts to provide new and better tools for our users like cross-wiki notifications. These changes will mean you have the same account name everywhere. This will let us give you new features that will help you edit and discuss better, and allow more flexible user permissions for tools. One of the side-effects of this is that user accounts will now have to be unique across all 900 Wikimedia wikis. See the announcement for more information.

Unfortunately, your account clashes with another account also called Jeanluc. To make sure that both of you can use all Wikimedia projects in future, we have reserved the name Jeanluc~enwikiversity that only you will have. If you like it, you don't have to do anything. If you do not like it, you can pick out a different name.

Your account will still work as before, and you will be credited for all your edits made so far, but you will have to use the new account name when you log in.

Sorry for the inconvenience.

Yours, Keegan Peterzell Community Liaison, Wikimedia Foundation 23:39, 17 March 2015 (UTC)

Renamed
 This account has been renamed as part of single-user login finalisation. If you own this account you can |log in using your previous username and password for more information. If you do not like this account's new name, you can choose your own using this form after logging in: . -- Keegan (WMF) (talk) 06:12, 19 April 2015 (UTC)

title: User talk:Jeanluc~enwikiversity


text:
French Mentoring/Templates

Native Speakers
Salut,. You are listed at the French stream as being a native French speaker interested in teaching French. The French Mentoring project has recently been launched, and if you would be willing to add your name to that list, I will make sure you are put in contact with appropriate students who would like to learn from you. ~

Fluent Speakers
Salut,. You are listed at the French stream as being a fluent French speaker interested in teaching French. The French Mentoring project has recently been launched, and if you would be willing to add your name to that list, I will make sure you are put in contact with appropriate students who would like to learn from you. ~

Proficient Speakers
Salut,. You are listed at the French stream as being a proficient French speaker interested in teaching French. The French Mentoring project has recently been launched, and if you would be willing to add your name to that list, I will make sure you are put in contact with appropriate students who would like to learn from you. ~

Advanced Learners
Salut,. You are listed at the French stream as being an advanced learner of French who is interested in teaching French. The French Mentoring project has recently been launched, and if you would be willing to add your name to that list, I will make sure you are put in contact with appropriate students who would like to learn from you. ~

Proficient Students
Salut,. You are listed at the French stream as being a proficient student of French who is interested in learning French. The French Mentoring project has recently been launched, and if you add your name to that list, I will do my best to put you in contact with an appropriate Mentor who would be glad to help you. ~

Intermediate Students
Salut,. You are listed at the French stream as being an intermediate student of French who is interested in learning French. The French Mentoring project has recently been launched, and if you add your name to that list, I will do my best to put you in contact with an appropriate Mentor who would be glad to help you. ~

Advanced Beginning Students
Salut,. You are listed at the French stream as being a beginning student of French with prior French-language experience, who is interested in learning French at Wikiversity. The French Mentoring project has recently been launched, and if you add your name to that list, I will do my best to put you in contact with an appropriate Mentor who would be glad to help you. ~

Beginning Students
Salut,. You are listed at the French stream as being a beginning student of French who is interested in learning French. The French Mentoring project has recently been launched, and if you add your name to that list, I will do my best to put you in contact with an appropriate Mentor who would be glad to help you. ~

title: French Mentoring/Templates


text:
User talk:Xequ

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

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

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

And don't forget to explore Wikiversity with the links to your left. Be bold to contribute and to experiment with the sandbox or your userpage, and see you around Wikiversity! --The Jade Knight (d'viser) 10:18, 4 November 2008 (UTC)

French Mentoring
Salut, Xequ. You are listed at the French stream as being an advanced learner of French who is interested in teaching French. The French Mentoring project has recently been launched, and if you would be willing to add your name to that list, I will make sure you are put in contact with appropriate students who would like to learn from you. The Jade Knight (d'viser) 10:19, 4 November 2008 (UTC)

title: User talk:Xequ


text:
User:Peacemaker69

 Hello Peacemaker69, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. After you leave a comment on a talk page, remember to sign and date; it helps everyone follow the threads of the discussion. The signature icon in the edit window makes it simple. To get started, you may

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

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

And don't forget to explore Wikiversity with the links to your left. Be bold to contribute and to experiment with the sandbox or your userpage, and see you around Wikiversity! --Computer tom 11:45, 4 November 2008 (UTC)

 Hi , I am Percy. Nice to meet you. I love the beach, swimming, fishing, and I am currently doing some research at Wikiversity on oceanography. I am also learning to play piano. What do you like? Can I be your mascot?

title: User:Peacemaker69


text:
User:Ctwyd

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

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

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

And don't forget to explore Wikiversity with the links to your left. Be bold to contribute and to experiment with the sandbox or your userpage, and see you around Wikiversity! --Computer tom 11:46, 4 November 2008 (UTC)

 Hello , we are the Baah and Bahh twins and we'd both like to welcome you to Wikiversity. While you sit back and explore Wikiversity, enjoy some of our fresh goat milk. You can help us learn by creating new learning resources or help us by improving the guided tours and other exploration tools. We look forward to seeing you around. Baaaahbye for now.

title: User:Ctwyd


text:
User:Computer tom

Hello all!

title: User:Computer tom


text:
User talk:MdMcAlister

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

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

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

And don't forget to explore Wikiversity with the links to your left. Be bold to contribute and to experiment with the sandbox or your userpage, and see you around Wikiversity! --Computer tom 11:54, 4 November 2008 (UTC)

title: User talk:MdMcAlister


text:
User talk:Computer tom

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

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

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

And don't forget to explore Wikiversity with the links to your left. Be bold to contribute and to experiment with the sandbox or your userpage, and see you around Wikiversity! --mikeu talk 12:00, 4 November 2008 (UTC)

title: User talk:Computer tom


text: File:Example bar.png

title: File:Example bar.png