text:
User:Jenny O/Further Reading

Based on further reading this page relates to or expands on some of the issues raised in my e-portfolio.

Students as research participants
These results generally support the claim that there is an overreliance on students as participants in social psychology research. More recent data could not be found. Bodner's study has quite a small sample size.

In a sample of psychological research studies (N = 169) from 1999, Bodner (2006) found a 25:75 student to non-student participant ratio. However, in the sub-set of social psychology studies (n = 32) this ratio was 50:50.

Winter, North and Sugar (2001) also examined the use of students as research participants. Their sample of psychological studies (N = 1719) was taken at three ten year intervals: 1975, 1985 and 1995, across a number of psychology journals. Overall, 68.31% of these studies used undergraduate student participants exclusively. When considered by year and sub-discipline, studies in the Journal of Experimental Psychology had the highest student participant rate at 95.8% in 1995. The Journal of Personality and Social Psychology had fairly consistent rates over three decades: 70.1% (1975), 65.6% (1985), and 70.6% (1995).

References

Bodner, T.E. (2006). Designs, participants, and measurement methods in psychological research. Canadian Psychology, 47 (4), 263-272.

Wintre, M. G., North, C., & Sugar, L.A. (2001). Psychologists’ response to criticisms about research based on undergraduate participants: A
 * developmental perspective. Canadian Psychology, 42 (3), 216-225.

What is culture?
"There are no shortage of ways in which culture can be conceptualised" (Singelis, 2000)

Culture is the man-made part of the environment (Herskovits, 1955)

Culture is “the set of artifacts and mentifacts (ie ideas, beliefs, conventions) more common among its members than among outsiders” (Poortinga & van Hemert, 2001, p.1034)

Groups
Fiske, S. T. (2004). Small groups: Ongoing interactions. In S. T. Fiske (2004). Social beings: A core motives approach to social psychology. (Ch 12, pp.459 - 506). Hoboken, NJ: John Wiley.

title: User:Jenny O/Further Reading


text:
Women and Technology What Do They DO?

Free virtual online educational and motivational tools for school children and women in technology industries or considering a career in technology available at Doing IT Around the World

The tools provide an interview profile with 36 women across the globe (including Antarctica), photographs and a diary of what they did on August 11 2008. Additional links are within each role models profile for further learning.

The role models represent a wide diversity of women in technology industries. Their ages span from early 20’s through to near retirement. Their areas of technology passions span traditional information technology (e.g. CTO, software, hardware, support, consulting, sales and marketing services) to communications, knowledge management, governance, data mining, open source advocacy, web design, graphic art, instrumentation design, configuration engineering, computational biology, microelectronics, molecular biology, research, geo spatial, educational, medical and legal technologists.

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

title: Women and Technology What Do They DO?


text:
Portal talk:History

 * Could someone take a look at the code here and tell me why in the world this page is being categorized as European and Chinese history? The Jade Knight 05:38, 19 September 2008 (UTC)
 * Figured it out. The Jade Knight 05:40, 19 September 2008 (UTC)

 * Cleaned up History Portal categories, more left to do. William Hamblin 18:40, 15 May 2010 (UTC)

 * I've done a little work on History_Of_Scotland. Could someone take a look and give me some feedback? Berek 10:05, 16 August 2011 (UTC)

 * Is anyone currently active in this space? User:SecretlyHistoric SecretlyHistoric (discuss • contribs) 04:01, 27 December 2021 (UTC)
 * I kinda am... but I'm focused on Introduction to US History at the moment and don't have time for other projects. More help around this portal would be greatly appreciated. I also see that you're new to Wikiversity - welcome! Please ask if you have any questions and have fun on our project!!! —Atcovi (Talk - Contribs) 04:48, 27 December 2021 (UTC)
 * @Atcovi Thank you! I'm probably going to take a few days and get acquainted with the site, content guidelines, and all that good stuff. :) I'll definitely reach out, especially once I get my feet under me. SecretlyHistoric (discuss • contribs) 02:46, 28 December 2021 (UTC)

title: Portal talk:History


text:
User talk:JWSchmidt/Blog/15 September 2008

Training: namespace
A handy solution is in the subpages of the sandbox. --Hillgentleman|Talk 05:43, 19 September 2008 (UTC)


text:
How to test the difference between two correlations

An inferential test can be performed to determine whether the correlations for two samples are likely to represent real difference in a population.

Further information and resources for conducting such comparisons include:
 * How to Determine Whether Two Correlation Coefficients are Significant
 * Two Correlation coefficients (Calculator)
 * Significance of the Difference Between Two Correlation Coefficients (Calculator)


text:
Portal:Mineralogy

Topic space on Wikiversity is comparable to a university department. For example, minerals is a resource offered by the department of mineralogy of the school of geology.

This topic page is for organizing the development of Mineralogy content on Wikiversity. If you are knowledgeable in any area Mineralogy, feel free to improve upon what you see, we would greatly appreciate your contributions.

Learning projects and resources
Learning materials and learning projects can be used by multiple departments. Please cooperate with other departments that use the same learning resource.

Remember, Wikiversity has adopted the "learning by doing" model for education. Lessons should center on learning activities for Wikiversity participants. We learn by doing.

Also, select a descriptive name for each learning project. Learning projects can be listed in alphanumeric order by code. Code numbers are optional, however.

 * Mineralogy
 * Search for Mineralogy at Wikiversity

Wikimedia

 * Mineralogy on wikipedia

title: Portal:Mineralogy


text:
User:Eml4500.f08.lulz.strack/hw2

Element stiffness matrix
$$ \bar{k}^{(e)}= \begin{bmatrix} k_{11}^{(e)} & k_{12}^{(e)} & k_{13}^{(e)} & k_{14}^{(e)} \\ k_{21}^{(e)} & k_{22}^{(e)} & k_{23}^{(e)} & k_{24}^{(e)} \\ k_{31}^{(e)} & k_{32}^{(e)} & k_{33}^{(e)} & k_{34}^{(e)} \\ k_{41}^{(e)} & k_{42}^{(e)} & k_{43}^{(e)} & k_{44}^{(e)} \end{bmatrix}; e = 1,2 $$

For an elastic bar element:

$$ \bar{k}^{(e)}= K^{(e)} \begin{bmatrix} (l^{(e)})^2 & l^{(e)}m^{(e)} & -(l^{(e)})^2 & -l^{(e)}m^{(e)} \\ l^{(e)}m^{(e)} & (m^{(e)})^2 & -l^{(e)}m^{(e)} & -(m^{(e)})^2 \\ -(l^{(e)})^2 & -l^{(e)}m^{(e)} & (l^{(e)})^2 & l^{(e)}m^{(e)} \\ -l^{(e)}m^{(e)} & -(m^{(e)})^2 & l^{(e)}m^{(e)} & (m^{(e)})^2 \end{bmatrix}; K^{(e)} = \frac{E^{(e)}A^{(e)}}{L^{(e)}} $$

$$ l^{(e)} = \frac{(x_j-x_i)}{L^{(e)}}; m^{(e)} = \frac{(y_j-y_i)}{L^{(e)}} $$

Where x and y are the global coordinates of the nodes i and j at either end of the bar element.

Using MATLAB
Global coordinates of nodes:

nodes = [x1, y1; x2, y2; x3, y3];

>> nodes = [0, 0; 4*cos(pi/6) 4*sin(pi/6); 4*cos(pi/6)+2*cos(pi/4) 4*sin(pi/6)-2*sin(pi/4)]

nodes =

0           0       3.4641            2       4.8783      0.58579

Directional forces l and m: >> l_1 = (3.4641-0)/4

l_1 =

0.86603

>> m_1 = (2-0)/4

m_1 =

0.5

>> l_2 = (4.8783-3.4641)/2

l_2 =

0.7071

>> m_2 = (0.5858-2)/2

m_2 =

-0.7071

Elemental Stiffness Matrices: >> k_1 = 3*1/4*[l_1^2 l_1*m_1 -l_1^2 -l_1*m_1; l_1*m_1 m_1^2 -l_1*m_1 -m_1^2; -l_1^2 -l_1*m_1 l_1^2 l_1*m_1; -l_1*m_1 -m_1^2 l_1*m_1 m_1^2]

k_1 =

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_2 = 5*2/2*[l_2^2 l_2*m_2 -l_2^2 -l_2*m_2; l_2*m_2 m_2^2 -l_2*m_2 -m_2^2; -l_2^2 -l_2*m_2 l_2^2 l_2*m_2; -l_2*m_2 -m_2^2 l_2*m_2 m_2^2]

k_2 =

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

title: User:Eml4500.f08.lulz.strack/hw2


text:
User:Eml4500.f08.Lulz.Layton.Eric/Hw1/HW1 contributing team members

Contributing Team Members
Aaron Fisher - Eml4500.f08.Lulz.fisher 11:46, 12 September 2008 (UTC)

Eric Layton - Eml4500.f08.Lulz.Layton.Eric 14:46, 19 September 2008 (UTC)

Sam Miorelli - Eml4500.f08.lulz.abcd 08:03, 19 September 2008 (UTC)

Ben Mitchell - Eml4500.f08.Lulz.mitchell.bm 20:24, 12 September 2008 (UTC)

John Saxon - Eml4500.f08.Lulz.js 03:36, 19 September 2008 (UTC)

Andrew Strack - Eml4500.f08.lulz.strack 07:57, 19 September 2008 (UTC)


text:
Category talk:Physical fitness

any one interested in physical therapy or physiotherapy topics invited.

title: Category talk:Physical fitness


text:
User:Jenny O/Relationships

The need to belong
Humans have a strong desire to connect with others and form close, lasting relationships. This ‘need to belong’ is a fundamental part of our role as social and cultural beings. Social animals need others to survive and reproduce. This can be achieved by forming relationships (e.g, group membership, marriage) and gaining social acceptance (e.g., yielding to peer pressure, modeling acceptable behaviours). However, engaging in socially unacceptable behaviour or attitudes can lead to prejudice (e.g., racism, homophobia, sexism) and ostracism. Culturally, humans connect with others and seek validation and recognition of their relationships. For example in Western societies open marriage and all forms of  polygamy are considered as being socially unacceptable.

However, this may not be so in other social or cultural contexts (see Polygamy World Map).

The human need to belong has two components: Regular social contact with others, and close, stable, mutually intimate contact. Partially meeting or failing to meet these needs can result in significant dissatisfaction or ill health.

Those unable to satisfy their need to belong may seek bonds with other things (e.g., an animal or a television show). An interesting example of this comes from the BBC documentary Love me, love my doll about men who purchase made-to-order life-like silicon dolls as companions (See Love me, love my doll). The issue here is not the doll as a sexual object but the doll as a companion - being treated as human being. The doll is given a name, dressed and seated for meals or to watch television, taken out for a drive, to a cafe, on picnics, photographed, taken on holidays etc. What I find even more disturbing than the need to purchase these companions, is that someone actually identified (intentionally or unintentionally) that there was a market for such a commodity. (Buy a companion - a sad indictment of our isolating, valueless, consumer-driven Western society?)

Our textbook provides the Internet as an example of how many people create and maintain social connections, and in some cases a social life. The pervasive social use of mobile phones, text messaging and email are similar. But what is the quality of most of this ‘connectedness’? (See Tutorial Two: Communication). In some ways this is another indictment of our society as it demonstrates how impersonal and often superficial some of our human connections have become. (But are they less valuable? I suspect these connections are of great value to many people). Indeed, the human need for connection is so strong that people will go to extreme lengths to meet that need. The textbook describes isolated prisoners yelling down toilet bowls in the hope of being heard by others through the drain pipes.

Attraction
As the textbook concludes we tend to like good-looking, friendly, similar people or those who are ‘nice’ to us (p. 333). Apparently, people intuitively know what fosters attraction and use this to ingratiate others. I wonder how aware people are of this. Is this intuition stronger in more attractive people?

The matching hypothesis is an interesting concept. This hypothesis suggests that humans form relationships or friendships with others who are equally attractive. Baumeister & Bushman (2008) suggest that this is deeply embedded in our psyche and arises from our social (rather than cultural) needs (e.g., helping our kin: See The influences of culture & nature on social psychology). Overall people will still select partners and friends most similar to themselves.

 Overall people will still select partners and friends most similar to themselves

Promoting liking
Liking between individuals can be promoted through:
 * Rewards: Reinforcement theory; doing favours & praising others
 * e.g., A man buys a woman flowers, dinner or gifts

 * Reciprocity: Complimenting others, reciprocal liking, mimicry. (Balance Theory, Social Exchange Theory)
 * e.g., Street gang members dressing and behaving in a similar syle

 * Mere exposure: familiarity, shared experiences (good or bad)
 * e.g., Military reunions better patronised by those who fought together than those groups who did not experience active combat.
 * However, research has identitifed an important facet of propinquity (the close physical or psychological proximity of  people); familiarity or regular contact will  amplify your liking or disliking for an individual or group. Thus familiarity does not breed contempt or liking but intensifies the prevailing state of the relationship.

Beautiful is good effect
According to the textbook:
 * People prefer attractive over unattractive others (When all else is equal)
 * Attractive people are seen as being superior according to the "beautiful is good effect"
 * Attractive children fare better than other children
 * Being young and healthy equates to being a good mate and potential partner ( Evolutionary perspective)
 * An average face is more attractive than a distinctive face
 * Women are more attracted to rich or successful looking men

While I accept that there are some universal human features that make a person "good-looking" (e.g., facial symmetry - You could ask a plastic surgeon for the formula), I believe the textbook provides a very narrow view of how we conceptualise beauty or how we judge people as being asthetically pleasing. This online PowerPoint presentation offers a different perspective: Beauty & Culture

Beauty & Culture
The following summary is taken directy from the Beauty & Culture presentation:
 * Beauty is in the eye of the beholder [[Image:MeganColeman.jpg|thumb|right|200px|In modern Western society "thin is in"… In the United States, most people hold negative attitudes toward body fat]]

 * …a body is both a physical and a symbolic artifact, forged by nature and by culture at a particular moment in history (Sullivan, 2000)

 * The cultural and social context influences individual body choices and gives them social meaning (Sullivan 2000)

 * In Nigerian culture, a woman’s rotundity is a sign of good health, prosperity and charm. “Beauty is in the weight”, says a defender of the practice, “To be called a slim princess is an abuse” (Angeloni, 2001)

 * In modern Western society "thin is in"… In the United States, most people hold negative attitudes toward body fat

 * …in rich countries the correlation (between body weight and socio-economic status) is negative: People who are richer tend to be thinner. In undeveloped countries the correlation is positive: thinner people, in general, are usually poorer than the ones who weight more (Levy and Shiraev, 2001)

 * The Kayapo people of the Amazon, use scarification, ornaments for pierced lips, ears, and noses, body paint and different hair styles to distinguish social and age classes, as well as to adorn themselves for potential mates

 * Mass culture and the media have served to project the body ideal and turn gentle concern about appearance into obsessions

 * When powerful cultural ideologies and institutions change, body standards and customs change

 * In the early nineteenth century, European travelers and explorers expressed their disbelief at the "savage" and unusually decorated bodies of the natives, while asserting their own wildly interesting fashion statement consisting of wigs, large hats, painted faces, and body deformities caused by the wearing of narrow, pointed, and tight-fitting shoes

Rejection
Humans attach a significant power and importance to being socially accepted. Rejection equates to social exclusion, which occurs when a person is prevented from forming or maintaining a social bond with another person or group. Being excluded, rejected or ignored by others is referred to as ostracism, and can result in a range of psychological reactions.

I was quite astounded by the historical account of ostracism given in the textbook reporting the need to be named 6,000 times before being rejected by the community for 10 years (p. 342). Times have certainly changed. Based on those figures few people in our society would ever be ostracised (Except politicians at the polls). These days most people would be overjoyed to know that 6,000 have made the effort to comment on them!

Consequences of rejection
Reactions to rejection can be grouped using the the ABC Triad:

Affective:
 * Initially – emotional numbness (rather than sadness or anxiety)

 * Feelings are based on value of the relationship and clarity of rejection signal
 * Loneliness

Behavioural:
 * Undermined self-regulation resulting in impulsiveness, selfishness or aggressive tendencies
 * Conflicting social conscious and selfish impulses:
 * Less pro-social behaviours: decreased generosity, cooperation, and helpfulness
 * Increased deceitfulness: cheating or rule breaking
 * Skepticism of new interactions
 * Increased attention to interpersonal events

Cognitive:
 * A range of negative inner states
 * Rejection sensitivity – rejection expectancy and hypersensitivity
 * Diminished intellectual thought – e.g., decreased reasoning

(These reactions can be reversed with the prospect of social re-integration)

Considering the strong human need to belong it is understandable that prolonged social rejection can result in significant consequences. For example rejection in childhood (e.g., through prolonged bullying or due to individual differences) can led the development of stereotypes and self-fulfilling prophecy, loss of self-esteem, depression and social anxiety. Indeed, these issues can snow-ball and social exclusion has been linked to acts such as school shootings perpetrated by students: the ultimate anti-social retaliatory act.

Love
What I found most interesting about this chapter is my need to apply the content to all the relationships I know. Relationships are ignited with passionate love (intense longing or desire), while companionate love (mutual, caring and committed affection) enables a relationship to succeed and survive. The notion of love as either passionate or companionate  makes perfect sense to me. As do the disparate trajectories of passion and intimacy throughout a relationship. I also thought it was interesting that these constructs of love exist across cultures and are managed according to cultural norms. Not knowing a lot about arranged marriages, I have often wondered how love is dealt with in such circumstances. I could not imagine what it would be like to be married to someone you did not love (Although I expect there are many people who exist in those relationships and their marriages were not arranged!).

There are various theories of love. Sternberg’s Triangular Theory of Love (1988) is one such theory. This theory proposes that there are three basic elements of love: passion (motivational), commitment (cognitive) and intimacy (emotional). Different combinations of these elements result in differnt forms of love. Ideally love should contain equal measures of all three elements (Consummate Love).

Exchange Vs communal relationship
(Social Cognitive Perspective) Each type of relationship can be used to serve a purpose in different contexts.

Exchange relationships
 * Based on reciprocity & fairness
 * Common in broader society
 * Promote progress and wealth in society
 * e.g., most work or professional relationships

Communal relationships
 * Based on love and concern without reciprocity
 * Common in intimate interpersonal relationships
 * Desirable, healthier, mature
 * e.g., relationships with others at home or in families

Attachment and relationships
I have always been interested in the concept of attachment, particularly as I have watched my children grow. Initially it was thought that a childhood attachment style carried through to adult relationships. However, I tend to agree with the more recent opinion suggesting that attachment style is not static as environmental influences can change the way in which a person relates to others.

Neo-Freudians maintain that the type of relationship we develop with others stems from our childhood relationship with our primary caregiver. Bowlby and Ainsworth are well known for the early work in Attachment theory. These theorists maintain that the type of infant attachment to primary caregivers forms the basis of future relationships. (Also see Harlow and Harlow’s monkey experiment). Three attachment styles were identified: secure, anxious/ambivalent or avoidant. Shaver and colleagues extended these concepts to explain the romantic relationships of adults. For example, those with a secure attachment style become close and intimate without fear of abandonment, avoidant individuals maintain a distance from others while anxious/ambivalent individuals are dependent and insecure in their relationships.

More recently attachment has been conceptualised as a two-dimensional construct (rather than one-dimensional continuum). In this approach the interaction between anxiety (attitude toward self) and avoidance (attitude toward others) result in four styles of attachment (avoidant separates into fearful and dismissing avoidant). For example, a person with a fearful-avoidant attachment style would have high anxiety and avoidance, with a low opinion of his/herself, and avoid closeness. Regardless of the type of attachment style the basic human need to belong still exists. As a result those with a secure attachment style fare well, while those with an avoidant style experience significant tension between their need for connection and companionship and their need to maintain distance in their relationships.

Maintaining relationships
People tend to believe their relationships improve (positive illusions), while in reality relationships remain stable or decline. Once a reciprocal pattern of negative behaviours is established this downward spiral results in the relationships demise. Attributions also play a role in relationships: they are enhanced when positive acts are internally attributed and negative acts are externally attributed and break-down when the reverse applies. The Investment Model of Commitment suggests that satisfaction, alternatives and investments determine whether a person stays in a relationship.

A psychologist and mathematician Gottman (see e-reserve Gladwell, 2005) developed a method of predicting whether a couple will stay married by identifying emotional patterns in their interactions. Apparently, a ratio of five positive interactions to one negative interaction is required to maintain a relationship. According to Gottman analysing just three minutes of a couple’s interaction (slices of interactions) can assess this. An interesting part of this article described how individual personality patterns (fists) were derived from Morse code transmissions during the war. Even without the content of the message (what), revealing aspects of who and where could still be determined from the unique transmission of message. Indeed, Gottman was applying this pattern recognition (decoding) to marital relationships. As an aside, similar patterns of a person’s habits (and possibly personality) can be tracked and identified, particularly in this electronic age (e.g., telephone records, EFTPOS transactions, Internet usage, or even e-portfolio contributions!).

Sexuality
According to Diamond (2003) human relationships are based on two separate drives: attachment (love – gender independent) and sex (mating). These can be reinforced or in conflict.

Theories of sexuality include:
 * Social constructionist theories – sexual attitudes & behaviour shaped by society/culture
 * Evolutionary theory – sex drive innate (shaped by natural selection)
 * Social exchange theory – costs and benefits of interaction – sex is a female resource wanted by males

Sex and gender

The textbook addresses four stereotypes about sex and gender.

The following stereotypes are supported: 1. Men want more sex than women

2. Women are gatekeepers, controlling whether,when and how much sex occurs

The following stereotypes are not supported: 3. Men separate love and sex more than women

In fact...
 * Men accept sex without love
 * Women accept love without sex
 * Both genders find combined love and sex most beneficial

4. Women’s sexuality is attuned to natural forces, men's to culture

In fact...
 * Women's sexuality is more attunded to cultural forces
 * i.e., Women have higher erotic plasticity (sex drive shaped by contextual factors) than men

title: User:Jenny O/Relationships


text:
Talk:Power Generation/Introduction/Intro 1

Good edits
A big thank you to anonymous user(74.218.183.18). Very good edit (Please remember to log in when doing your edits). --Thuvack | talk 12:07, 19 September 2008 (UTC)


text:
User:Eas4200c.f08.vqcrew.E

 * Homework 1

 * Homework 2

 * Homework 3

 * Homework 4

 * Homework 5

 * Homework 6

 * Homework 7

title: User:Eas4200c.f08.vqcrew.E


text:
MediaWiki:Group-sysop-member

custodian

title: MediaWiki:Group-sysop-member


text:
User:Eml4500.f08.FEABBQ

Homework Pages:

 * Homework 1 -  Discussion
 * Homework 2 -  Discussion
 * Homework 3 -  Discussion
 * Homework 4 -  Discussion
 * Homework 5 -  Discussion
 * Homework 6 -  Discussion
 * Homework 7 -  Discussion

EML 4500 Class Homepage:

 * User:Eml4500.f08

FEABBQ Team Members:
--
 * Esmail Hamdan
 * J.Jayma
 * Grant Stark (Team Coordinator)
 * Nik Vitt
 * William Kurth
 * Andy Kobyljanec

title: User:Eml4500.f08.FEABBQ


text:
User:Eml4500.f08.FEABBQ/HW1

Homework 1
BACK TO FEABBQ

Due 09/12/08 at 17:00 EDT (i.e., 21:00 UTC)

 * Matlab Assignment (current version). Archived version for submission.

 * Lecture Notes (current version). Archived version for submission.

Homework 1 Contributing Members

 * Nik Vitt Eml4500.f08.FEABBQ.vitt 22:42, 23 September 2008 (UTC)

 * J.Jayma EML4500.f08.FEABBQ.Jayma 22:02, 16 September 2008 (UTC)

 * Esmail Hamdan EML4500.f08.FEABBQ.Hamdan16:35,17 September 2008 (UTC)

 * Andy Koby 16:31, 17 September 2008 (UTC)
 * William Kurth EML4500.f08.FEABBQ.Kurth 20:15, 19 September 2008 (UTC)

title: User:Eml4500.f08.FEABBQ/HW1


text:
User:Eml4500.f08.FEABBQ/HW1/Matlab

 Comment: This has been moved to the user namespace. Jose Jayma 16:48, 31 October 2008 (UTC).

Homework 1: Matlab Section
BACK TO FEABBQ/HW 1

Entering matrices
Matlab works completely with matrices.

Matrices are created through 3 different methods. 1. By an entered explicit list of elements 2. By generating a built-in statement or function 3. Loaded from an external file

A = [ 1 2 3; 4 5 6; 7 8 9] A = 1    2     3        4     5     6        7     8     9

Matlab also allows for complex numbers. You may use i or j.

a = [1 2;3 4] + i*[5 6;7 8] a = 1.0000 + 5.0000i  2.0000 + 6.0000i 3.0000 + 7.0000i  4.0000 + 8.0000i

If i or j are used as variables. A replacement imaginary unit may be created.

ii = sqrt(-1) ii = 0 + 1.0000i

Many built in functions are available in MATLAB. The function rand may be used to created a matrix with randomly generated entries.

rand (3) ans = 0.9501   0.4860    0.4565        0.2311    0.8913    0.0185        0.6068    0.7621    0.8214

rand (2,3) ans = 0.4447   0.7919    0.7382        0.6154    0.9218    0.1763

Function magic will create a martix with a magic square.

Function hilb will create a Hilbert matrix.

Individual numbers from a matrix may also be accessed by using a similar command example below.

Matrix operations, array operations
There are multiple matrix operations available. They can be used for scalars and matrices. They are

Operations can be made to operate entry wise by preceding them with a period.

Statements, expressions, variables; saving a session
MATLAB statements are usually of the form variable = expression To suppress printing of a statement you must follow the statement with a semicolon. MATLAB IS case sensitive. Therefore when creating variables keep in mind the case of the letter. The statement who will list the variables currently created and available.

who Your variables are: A   a    ans  ii

To clear a variable simply write clear variable. clear ii   who Your variables are: A   a    ans

A runaway display can be stopped by pressing CTRL-C.

Lastly to save a session simply press the same button or go to file->save. If a session is not saved it will be lost, thus losing all variables created.

Matrix building functions
Here are few examples

B = zeros (size(A)) B = 0    0     0         0     0     0         0     0     0

diag(A) ans = 1        5         9

diag(diag(A)) ans = 1    0     0         0     5     0         0     0     9

B = [A,zeros(3,2); zeros(2,3), eye(2)] B = 1    2     3     0     0         4     5     6     0     0         7     8     9     0     0         0     0     0     1     0         0     0     0     0     1

For, while, if and relations
These are all commands that will generally be used in M-files.

FOR will set a limit to the number of iterations that will run in a file. For instance

for x = 1:n q = x^2 end

This code will make q = x^2 as long x is between 1 and n.

WHILE will set an end to a loop by stating the values needed to exit the loop.

while 2^n < a         n = n + 1 end

This code will raise n by one as long as 2^n is less then a.

IF will not run a code unless the starting conditions mach the relation given with if.

if n<5 x = 3 + x         n = n + 1 end

RELATIONS Here are a list of the relational operators in MATLAB <   less than

>   greater than

<=  less than or equal

>=  greater than or equal

==  equal

~=  not equal

Relations may be connected by

&   and

 * or

~   not

Scalar functions
built in functions that operate on scalars or individual elements of a matrix. The most commonly used of these functions are as follows:

 * sin
 * asin (arcsin)
 * exp (exponential)
 * abs
 * round
 * cos
 * acos (arccos)
 * log (natural log)
 * sqrt
 * floor
 * tan
 * atan (arctan)
 * rem (remainder from division)
 * sign
 * ceil

Vector functions
functions perform actions on vectors in a "column by column" fashion and produce a row vector. The transpose can be used to operate on a row by row basis. Some of those functions are:

 * max
 * min
 * sort
 * sum
 * prod
 * median
 * mean
 * std
 * any
 * all

Matrix functions
The following is a list of matrix functions for Matlab. These functions can be accessed using different commands.
 * eig                (eigenvalues and eigenvectors)
 * chol               (cholesky factorization)
 * svd                (singular value decomposition)
 * inv                (inverse)
 * lu                 (LU factorization)
 * qr                 (QR factorization)
 * hess               (hessenberg form)
 * schur              (schur decomposition)
 * rref               (reduced row echelon form)
 * expm               (matrix exponential)
 * sqrtm              (matrix square root)
 * poly               (characteristic polynomial)
 * det                (determinant)
 * size               (size)
 * norm               (1-norm, 2-norm F-norm)
 * cond               (condition number in the 2-form)
 * rank               (rank)

Command line editing and recall
Matlab Editing and function keys:

 * left/right arrows scrolls the cursor left and right

 * up/down arrows scrolls up and down from line to line

 * backspace deletes the character left of the cursor

 * delete deletes the character left of the cursor

 * home moves the cursor to the beginning of the line

 * end moves the cursor to the end of the line

 * There is a graph for this section I just have to figure out how to add it.***
 * I guess you use the 'upload file' in the up-left corner, -andy

Submatrices and colon notation
Matlab allows for the simple creation of vectors with it's "colon notation", whose main operator is the ":", which means "everything." In the following example, a row vector is generated with a starting value of 0.2 and includes "everything" in between 0.2 and its final value, 1.2. The 0.2 in between the colons represents the step-size, meaning that the next value from the starting value is 0.2 over, and the next value after that is 0.2 over, and so on. Other examples of the use of the colon follow.

The following are more advanced uses of the colon operator. A(1:4,3)  first four entries of the third column A(:,3)    third column of A     A(:,[2 4]) columns 2 and 4 of A

M-files: script files, function files
Matlab has a built-in programming language, wherein commands that are usually typed at the prompt can be saved to a file (called an M-file because of its .m filename) and run all at once. There are two types of M-files: script files and function files.

A script is akin to a recipe, where commands are read top-down and executed in that order. Common commands (as well as more advanced commands) can be placed in a script so that they can be eaily executed. Inside of a script file, functions can also be called. Note that the definition of a function must exist in its own M-file, hence the second type of M-file, the function file. An example of function files follow.

Learning to write scripts and functions in M-files is essential to gaining proficiency in Matlab use.

Text strings, error messages, input
Like other programming environments/languages, Matlab has many ways of handling text. To assign a string of text to a variable, enclose the text in single quotes. This method is useful if the text is to be operated on or changed later on. To directly display text, use the  function, which is useful for simple messages. The following is a lighthearted example When programming, it is often desirable to handle errors and use text to display a readable message. The previous example shows the error message if the code was expecting a symmetric matrix and received otherwise. Another common use of text is to take in user input interactively. The previous examples takes in user input and assigns the data to the variable iter. Normal execution of the program then follows.

Managing M-files
The command  opens the M-file editor while simultaneously closing down the Matlab prompt. While this frees up RAM, it may be inconvenient to have to reopen Matlab and possibly redefine variables and other workspace entities. In order to keep Matlab open whenever running a command that would normally close Matlab, simply precede the command with an exclamation,. The following code opens the M-file editor; upon exiting the editor, Matlab is opened up in the same state as it was before closing. To sum up the introduction to M-file programming, we present common tools to use: debugging, setting directories and working with files. help dbtype   Opens the debugging help information pwd   Prints the current working directory. cd    Changes the directory. dir   Prints the contents of a directory. ls    Same as dir. The following is the output of the  command.

Comparing efficiency of algorithms: flops, tic, toe
The term flops is used in computer science to designate the number of floating point operations performed (hence the name flops). In Matlab, the command  can be used to gauge how computationally intensive a section of code (or an entire M-file or script) is numerically. The usage is detailed as follows: A special note, with version 7 of Matlab, flop counts are no longer available, though another method of gauging performance of a section of code is to use the  and   commands to time the length of execution. The usage is similar to : An example of this would be    tic; 123.54*200; toc Elapsed time is 0.000024 seconds.
 * 1) set the number of flops to 0 with the command .  This should happen at the beginning of your code.
 * 2) At the terminating point (the point after the code whose flops are desired), simply place the command.
 * 1) At the starting point of the code to test, enter the command.
 * 2) At the point where the desired execution time is located, enter the command.

Output format
Any standard computations performed in Matlab are done in double precision. It is possible to control the appearance of the output with the following commands. The following example used is  and the format option is followed by the output, format short ans = 2.3383e+004 format long ans = 2.338288893418174e+004

Hard Copy
Sometimes it is desirable to obtain a hardcopy of your Matlab work, i.e. a printed version. With the exception of graphics, it is possible to save everything on screen to a file with the following command, The  function can be controlled using the commands   and , which determine when to stop and restart the recording of on-screen contents, respectively.

Graphics
Matlab is capable of producing a variety of mathematical graphs and figures in 2 and 3 dimensions. The basic commands to use these functions are:

Planar 2d Plots

The plot(x,y) function will create a linear graph if x and y are both 1d vectors of the same length. The first variable in the plot function is the horizontal axis, and the second is graphed on the vertical. The easiest way of producing a 1d vector is using colon notation:

The following is a graph of a sin wave from -4 to 4. Adding shown in the code above is the syntax for adding titles and axis labels using the title and xlabel/ylabel commands respectively.

Streamlined code exists for graphing functions as well. First, an M-file must be created such as the following:

Using this code would now create a 2d graph from -1.5 to 1.5:

2d parametric graphs can also be created in Matlab. The following code and Figure 3 demonstrate:

Annotating Graphs

Graphs are easily annotated in Matlab as well. Typing in the following commands will label the active figure. The format is like:

Other visual features are available as well:

All of these visual modifications should be performed after the plot command.

Multiple Graphs on the Same Figure

There are different ways to place multiple graphs onto the same Matlab figure. The first is to overload the  function. The following code and Figure 4 show how this can be done:

The  command can also be used. When hold is on, every plot graphed will go onto the active figure. Turning hold off will create a new figure for each new graph, which is the Matlab default.

Using Different Colors and Line Types on Graphs

Sometimes the default blue solid line in Matlab graphs can get a little boring. To add a little variety to your graphs, there are several commands that can be incorporated into the  function:

These commands are used inside single quotation marks inside the  function. would plot a dashed red line and  would plot a green dotted line.

Other Types of Graphs

There are several other types of graphs which Matlab can produce. Although this guide does not have the details for them, they can all be easily found using the  command.

 * Polar
 * Bar
 * Hist
 * Quiver
 * Compass
 * Feather
 * Rose
 * Stairs
 * Fill
 * Subplot

Graphics Hardcopy

Oftentimes it is necessary to print a graph from Matlab. This is most easily done with the convenient 'Print' menu option on the Figure. It is normally easier though to use 'Save as' and save the graph as a .jpg or other type of graphics file. For the more adventurous, Matlab has a  command which will print the active graphics item. The  option can also be used to append a different saved file to the current graphics item for printing. Just use the format.

3-D Line Plots

There are few more impressive ways to show off Matlab skills than producing colorful 3-D plots. The  command can easily plot in 3 dimensions, just like   plots in 2 dimensions. The following example shows how a 3-D plot (Figure 6) can be parametrically created:

All of the settings applicable to 2-D plots (including axis labels and commands, titles, colors, etc) can be used on a 3-D plot as well.

3-D Mesh and Surface Plots

The mesh and surf functions can create 3 dimensional representations of matrices, either with a wire mesh or solid surface. If a single matrix can be inputed into either function, the z value is the matrix value at point (x,y) (row, col). Figures 7 and 8 show the mesh and surf functions being used to graph the 'eye' (identity) matrix:

3-D functions can also be graphed using mesh and surf in conjunction with the  function. Combined with colon notation,  creates an x-y rectangle in which a function is graphed. The following code and Figure 9 show how a 3-D function can be defined and graphed:

There are several ways to change the appearance of a 3-D graph in Matlab. First of all, the  command will affect the superimposed mesh lines on a surface graph. Various color profiles can also be experimented with. The following commands can be used like this:

Any of the functions to change the appearance should be used after the graph has been made.

Graphs can also be produced parametrically with M-files. Check the guide for further information on this, and Matlab's 'Handle Graphics' which can produce advanced graphs and plots.

Sparse Matrix Computations
Matlab typically handles matrices of which the contents are mostly nonzero. These matrices are considered "dense". A matrix with mostly zero entries is considered "sparse" and is worth defining for the possibility of ignoring the zero values and operating on the nonzero values only. The benefit of this is to reduce the memory footprint of the matrix and the computation time. Suppose a matrix is created by the following command, and that the matrix is filled with mostly zeros by the following command, the output of which looks like F = 9    4     0     0     0     0     2     0     7     0     0     0     0     8     1     0     0     0     0     0     4     3     6     0     0     0     0     8     2     4     0     0     0     0     1     4 The nonzero values of the matrix can be easily found with the command, followed by the output, S=sparse(F) S = (1,1)       9     (2,1)        2     (1,2)        4     (3,2)        8     (2,3)        7     (3,3)        1     (4,3)        4     (4,4)        3     (5,4)        8     (4,5)        6     (5,5)        2     (6,5)        1     (5,6)        4     (6,6)        4 The following commands will return the matrix to it's previous form and check it's storage mode, respectively F=full(S) issparse(F) Sparse matrices can also be generated directly as opposed to calling the sparse function on a full ("dense") matrix. A sparse diagonal matrix can be generated with the  command. The output is similar to calling the. m=6;n=6;e=ones(n,1);d=-2*e; T=spdiags([e,d,e],[-1,0,1],m,n) T = (1,1)      -2     (2,1)        1     (1,2)        1     (2,2)       -2     (3,2)        1     (2,3)        1     (3,3)       -2     (4,3)        1     (3,4)        1     (4,4)       -2     (5,4)        1     (4,5)        1     (5,5)       -2     (6,5)        1     (5,6)        1     (6,6)       -2  are the sparse analogs of , respectively. A more specific approach would be to define the nonzero values of the sparse matrix individually. For example (followed by output) i=[1 2 3 4 4 4];j=[1 2 3 1 2 3]; s=[5 6 7 8 9 10]; full(S),S = sparse(i,j,s,4,3) ans = 5    0     0     0     6     0     0     0     7     8     9    10     S = (1,1)       5     (4,1)        8     (2,2)        6     (4,2)        9     (3,3)        7     (4,3)       10 The nonzero entities were stored in the row vector. Another example of directly defining the nonzero values of a sparse matrix is    n=6;e=floor(10*rand(n-1,1));E=sparse(2:n,1:n-1,e,n,n) E = (2,1)       8     (3,2)        5     (4,3)        2     (5,4)        6     (6,5)        8 The output of Matlab arithmetic and functions will be in one of the two storage modes (either Full(F) or Sparse(S)). The following defines the output results of the manipulation done to either full or sparse matrices Sparse: S+S, S*S, S.*S, S.*F, S^n, S.^n, S\S, inv(S), chol(S), lu(S), diag(S) max(S), sum(S) Full: S+F, S*F, S\F, F\S

title: User:Eml4500.f08.FEABBQ/HW1/Matlab


text:
User:Eml4500.f08.FEABBQ/HW1/Notes

 Comment: This has been moved to the user namespace. Jose Jayma 16:48, 31 October 2008 (UTC).

Lecture Notes - Weeks 1,2
BACK TO FEABBQ/HW 1

The Following is a summary of the topics and information discussed during the first two weeks of EML4500 - Finite Element Analysis (FEA) at the University of Florida in the fall of 2008. A summary of all administrative information can be found  here .

Wikiversity
Wikiversity is just one of many projects formed under the parent organization of Wikimedia. Since Wikiversity is the platform in which the Professor, Dr. Loc Vu Quoc, plans on using for group collaboration and homework submission, most of the first week was dedicated to familiarizing the students with its structure.

Free Body Diagrams
Although a review from statics, the free body diagram (FBD) is an essential tool in FEA. In addition, because the systems for which FEA is used can be complex, it is necessary to form a rigorous convention for the labeling of all FBDs. Figures 1-3 below are the examples used in class to illustrate this convention.

Labeling Convention
In breaking a structure apart, it is important to differentiate between global FBDs (Fig. 2) and local FBDs (Fig. 3). For this purpose a circle is used when describing global nodes, while a square is used for local nodes. In addition, it is necessary to distinguish each piece, or element. A triangle is used for this purpose.

It is also important to assign a convention for the naming of forces and displacements. In global FBDs, as shown in Fig. 2, each force is titled R where the subscript denotes which node the force is being applied to, and the superscript shows the direction of the force. For local FBDs, as shown in Fig. 3, f denotes a force while d indicates a displacement. Here the subscript defines the particular local force or displacement, while the superscript indicates the element it corresponds to.

Force-Displacement Relation
The relationship between forces and displacements can be modeled by a combination of springs. The most simple example is a 1-D spring with one fixed end. In this case, the relationship between the force applied to the free end and the displacement of that end of the spring is given by:

$$f = k * d$$

To illustrate how matrices can be used for systems with more degrees of freedom (dof), a simple example is a 1-D spring with both ends being free. Here the force-displacement relation is given by:

$$ \begin{bmatrix} f_1 \\ f_2 \end{bmatrix} = \begin{bmatrix} k & -k \\ -k & k \end{bmatrix} \begin{bmatrix} d_1 \\ d_2 \end{bmatrix} $$

Recipe: Steps to Solving Simple Truss System Problems
$$\underline{K} \cdot \underline{d} = \underline{F}$$
 * 1) The first step of analyzing a truss system is to look at the entire system as a whole, also known as the global picture. The preliminary procedures in this analysis are first creating the global coordinate system, numbering the nodes in the system, as well as noting the global forces and displacement degrees of freedom (dofs) in the system.  As a convention, both the global forces and displacement dofs should be named starting with the first global node in the directions of the global coordinate system (x and y directions) and continued in this fashion for each successive node.  The aforementioned FBDs are good tools for the visualization of systems and the forces and dofs acting upon them.  For both the global dofs and the global forces, each contain some unknown and some known quantities.  The known values are usually given in the problem as contraints or parameters, such as a fixed node (zero displacement) or an applied force.  The unknown values are the values that will be solved for using finite element method (FEM).
 * 2) After the system is correctly identified, the next step is to analyze each element in the system separately. A FBD should be created for each element of the system, identifying the dofs and forces acting upon each particular element.  The nodes, dofs, and forces should be identified by the same convention mentioned in step 1, however either global or local element coordinates may be used.
 * 3) Once all the dofs and forces are identified for each element in the system, the next step is to create the element stiffness and force matrices. These matrices are assembled into the global force-displacement (FD) relation:

where K is a nxn stiffness matrix, d is a nx1 displacement dof matrix, F is a nx1 force matrix, and n is the number of both known and unknown dofs.
 * 1) Next, the known dofs are eliminated to reduce the global force displacement relations. The K,d, and F matrices above - that were originally nxn, nx1, and nx1 matrices, respectively - become mxm, mx1, and mx1, respectively, where m is the number of unknown only dofs and is less than n.  Assuming the new stiffness, K, matrix is an invertible matrix, the above equation can be rearranged to solve for the displacement matrix.
 * 2) Compute the element forces from the now known dofs and from this also calculate the element stress.
 * 3) Use these to compute the remaining unknown (reaction) forces.

title: User:Eml4500.f08.FEABBQ/HW1/Notes


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User:Eml4500.f08.FEABBQ/HW2

Homework 2

 * Lecture Notes (current version).

People who contributed to HW 2

 * J.Jayma EML4500.f08.FEABBQ.Jayma 01:57, 18 September 2008 (UTC)
 * Andy Koby 13:52, 19 September 2008 (UTC)
 * Esmail Hadman 01:44, 26 September 2008 (UTC)
 * Nikolas Vitt --Eml4500.f08.FEABBQ.vitt 04:55, 26 September 2008 (UTC)
 * William Kurth --Eml4500.f08.FEABBQ.Kurth14:35, 26 September 2008 (UTC)
 * Grant Stark -- Eml4500.f08.FEABBQ.StarkStark 16:53, 26 September 2008 (UTC)

title: User:Eml4500.f08.FEABBQ/HW2


text:
User:Eml4500.f08.FEABBQ/HW2/Notes

Class Notes: September 8, 2008

To show that the "vertical" components in the single element pictures from above you must take the moments about node 2 for each element.

Element 1: $$ \sum \mathbf{M_2} = F_1^{x} + (F_1^{y}*L_1) + F_2^{x} + F_2^{y} = 0 + (F_1^{y}*L_1) + 0 + 0 = 0$$

$$ \Rightarrow F_1^{y} = 0 $$

Element 2:

$$ \sum \mathbf{M_2} = F_3^{x} + (F_3^{y}*L_2) + F_2^{x} + F_2^{y} = 0 + (F_3^{y}*L_2) + 0 + 0 = 0$$

$$ \Rightarrow F_3^{y} = 0 $$

3) Global FD

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

$$\underline{k}^{(e)}$$ - Element Stiffness Matrix for Element e

$$\underline{d}^{(e)}$$ - Element Displacement Matrix of Element e

$$\underline{f}^{(e)}$$ - Element Force Matrix of Element e

The equation for $$\underline{k}^{(e)}$$ is,

$$\underline{k}^{(e)} = {k}^{(e)}$$ $$\left[ \begin{matrix} (l^{(e)})^{2} & l^{(e)}m^{(e)} & -(l^{(e)})^{2} & -l^{(e)}m^{(e)}\\ l^{(e)}m^{(e)} & (m^{(e)})^{2} & -l^{(e)}m^{(e)} & -(m^{(e)})^{2} \\ -(l^{(e)})^{2} & -l^{(e)}m^{(e)} & (l^{(e)})^{2} & l^{(e)}m^{(e)} \\ -l^{(e)}m^{(e)} & -(m^{(e)})^{2} & l^{(e)}m^{(e)} & (m^{(e)})^{2} \end{matrix} \right]$$

Axial stiffness of the bar element,

$${k}^{(e)}=\dfrac{E^{e}A^{e}}{(L^{e})}$$ $$ l^{(e)} = cos(\theta) ; m^{(e)} = sin(\theta) $$

$$l^{(e)}, m^{(e)}$$ are found through the director cosines method

$$ l^{(e)} = \mathbf{\tilde{i}} \cdot \mathbf{i} = cos(\theta^{(e)}) $$

$$ m^{(e)} = \mathbf{\tilde{i}} \cdot \mathbf{j} = sin(\theta^{(e)}) $$

$$ \mathbf{\tilde{i}} = cos(\theta^{(e)}) \mathbf{i} + sin(\theta^{(e)}) \mathbf{j} $$

$$ \mathbf{\tilde{i}} \cdot \mathbf{i} = (cos(\theta^{(e)}) \mathbf{i} + sin(\theta^{(e)}) \mathbf{j} ) \cdot \mathbf{i} $$

$$ \mathbf{\tilde{i}} \cdot \mathbf{i} = cos(\theta^{(e)}) \cdot \mathbf{i} \cdot \mathbf{i} + sin(\theta^{(e)}) \cdot \mathbf{i} \cdot \mathbf{j} = cos(\theta^{(e)}) \cdot (1) + sin(\theta^{(e)}) \cdot (0) = cos(\theta^{(e)})$$

$$ \mathbf{\tilde{i}} \cdot \mathbf{j} = (cos(\theta^{(e)}) \mathbf{i} + sin(\theta^{(e)}) \mathbf{j} ) \cdot \mathbf{j} $$

$$ \mathbf{\tilde{i}} \cdot \mathbf{j} = cos(\theta^{(e)}) \cdot \mathbf{i} \cdot \mathbf{j} + sin(\theta^{(e)}) \cdot \mathbf{j} \cdot \mathbf{j} = cos(\theta^{(e)}) \cdot (0) + sin(\theta^{(e)}) \cdot (1) = sin(\theta^{(e)})$$

Director Cosines
The director cosines are the components of the vector $$\mathbf{\tilde{i}}$$ with respect to the global coordinate axes, x and y:

$$\mathbf{\tilde{i}}=\cos \theta^{(e)}\mathbf{i}+\sin \theta^{(e)}\mathbf{j}$$

$$l^{(e)}$$ and $$m^{(e)}$$ are used to denote the magnitude of the i and j components, respectfully:

$$l^{(e)}=\cos \theta^{(e)}$$

$$m^{(e)}=\sin \theta^{(e)}$$

2-Bar Truss System, Continued

 * Element 1:
 * $$\theta^{(1)}=30^\circ, \qquad l^{(1)}=\cos 30^\circ = \frac{\sqrt{3}}{2}, \qquad m^{(1)}=\sin 30^\circ = \frac{1}{2}$$

 * $$k^{(1)}=\ \frac{E^{(1})A^{(1)}}{L^{(1)}}\ =\ \frac{(3)(1)}{(4)}\ =\ \frac{3}{4}$$

 * $$\underline{k}^{(1)}\ =\ \begin{bmatrix} k_{11}^{(1)} & \cdots & k_{14}^{(1)} \\ \vdots & \ddots & \vdots \\ k_{41}^{(1)} & \cdots & k_{44}^{(1)}\end{bmatrix}\ =\ \begin{bmatrix} k_{ij}^{(1)}\end{bmatrix}_{[4x4]}$$

 * $$k_{11}^{(1)}\ =\ k^{(1)}(l^{(1)})^2\ =\ \left (\frac{3}{4}\right )\left (\frac{\sqrt{3}}{2}\right )^2 = \frac{9}{16}$$

 * $$k_{12}^{(1)}\ =\ k^{(1)}(l^{(1)}m^{(1)})\ =\ \left (\frac{3}{4}\right )\left (\frac{\sqrt{3}}{2} \cdot \frac{1}{2}\right ) = \frac{3\sqrt{3}}{16}$$

 * $$k_{42}^{(1)}\ =\ -k^{(1)}(m^{(1)})^2\ =\ -\left (\frac{3}{4}\right )\left (\frac{1}{2}\right )^2 = -\frac{3}{16}$$

 * Observations:
 * Looking back to the equation for the element stiffness matrix, k (1), one may notice that only 3 elements need be calculated to determine the matrix. All elements are of one of three magnitudes, with varying signs.
 * Matrix k (1) is symmetric, meaning that kij(1) = kji(1).
 * In general,kij(e)=kji(e), which infers that the transpose of an element stiffness matrix is equal to itself.

 * This quality of the stiffness matrix allows for the following shorthand triangular form of the matrix:

 * $$\underline{k}^{(e)}\ =\ \begin{bmatrix} k_{11}^{(e)} & \cdots & k_{14}^{(e)} \\ & \ddots & \vdots \\ & & k_{44}^{(e)}\end{bmatrix}$$

 * Only the upper triangle of the matrix need be expressed; the empty lower triangle of the matrix is just the reflection of the upper part about the diagonal.

 * Element 2:
 * $$\theta^{(2)}=-45^\circ, \qquad l^{(2)}=\cos -45^\circ = \frac{\sqrt{2}}{2}, \qquad m^{(2)}=\sin -45^\circ = -\frac{\sqrt{2}}{2}$$

 * $$k^{(2)}=\ \frac{E^{(2})A^{(2)}}{L^{(2)}}\ =\ \frac{(5)(2)}{(2)}\ =\ 5$$

 * $$\underline{k}^{(2)}\ =\ \begin{bmatrix} k_{ij}^{(2)}\end{bmatrix}_{[4x4]}$$

 * $$k_{11}^{(2)}\ =\ k^{(2)}(l^{(2)})^2\ =\ 5\left (\frac{\sqrt{2}}{2}\right )^2 = 2.5$$

 * Observations:
 * Since both $$l^{(2)}$$ and $$m^{(2)}$$ have the same magnitude, all elements in k (2) will have the same magnitude just with appropriate sign.
 * Matrix k (2) is symmetric, meaning that its transpose is equal to itself.

 * Element Force-Displacement Relationship:
 * Now that the element stiffness matrices are known (see MATLAB solution at bottom), they can be applied to the element FD relationship:
 * $$\underline{k}_{\ 4x4}^{(e)} \cdot \underline{d}_{\ 4x1}^{(e)}\ =\ \underline{f}_{\ 4x1}^{(e)}$$

 * $$\underline{d}^{(e)}\ =\ \begin{Bmatrix} d_1^{(e)} \\ \vdots \\ d_4^{(e)} \end{Bmatrix} \qquad \qquad \underline{f}^{(e)}\ =\ \begin{Bmatrix} f_1^{(e)} \\ \vdots \\ f_4^{(e)} \end{Bmatrix}$$

Class Notes: September 12, 2008
Expressing the global force-matrix function can be simplified by using the compact notation:

$$\left [ K_{ij} \right ] \left \{ d_j \right \} = \left \{ F_i \right \}$$

$$\sum_{j=1}^6 K_{ij} d_j = F_i,\qquad i=1...6$$

How do you go from several local element matrices to one global matrix? An assembly plan! First you must identify the correspondence between the various elements and their degree's of freedom:

Class Notes: September 15, 2008
To assemble the global stiffness matrix, $$\underline{\overline{K}}$$, the local stiffness matrices,$$\underline{k}^{(1)}$$ and $$\underline{k}^{(2)}$$, are superimposed as shown in the figure below.

The elements which overlap are summed. The result is the global stiffness matrix shown below.

$$ \underline{\overline{K}} = \left[ \begin{array}{cccccc} k_{11}^{(1)} & k_{12}^{(1)} & k_{13}^{(1)} & k_{14}^{(1)} & 0 & 0 \\ k_{21}^{(1)} & k_{22}^{(1)} & k_{23}^{(1)} & k_{24}^{(1)} & 0 & 0 \\ k_{31}^{(1)} & k_{32}^{(1)} & (k_{33}^{(1)} + k_{11}^{(2)}) & (k_{34}^{(1)} + k_{12}^{(2)}) & k_{13}^{(2)} & k_{14}^{(2)} \\ k_{41}^{(1)} & k_{42}^{(1)} & (k_{43}^{(1)} + k_{21}^{(2)}) & (k_{44}^{(1)} + k_{22}^{(2)}) & k_{23}^{(2)} & k_{24}^{(2)} \\ 0 & 0 & k_{31}^{(2)} & k_{32}^{(2)} & k_{33}^{(2)} & k_{34}^{(2)} \\ 0 & 0 & k_{41}^{(2)} & k_{42}^{(2)} & k_{43}^{(2)} & k_{44}^{(2)} \\ \end{array} \right] $$

Therefore the global FD relationship, $$\underline{\overline{K}}$$ $$ \underline{\overline{d}} = \underline{\overline{F}}$$, is given by:

$$ \left[ \begin{array}{cccccc} k_{11}^{(1)} & k_{12}^{(1)} & k_{13}^{(1)} & k_{14}^{(1)} & 0 & 0 \\ k_{21}^{(1)} & k_{22}^{(1)} & k_{23}^{(1)} & k_{24}^{(1)} & 0 & 0 \\ k_{31}^{(1)} & k_{32}^{(1)} & (k_{33}^{(1)} + k_{11}^{(2)}) & (k_{34}^{(1)} + k_{12}^{(2)}) & k_{13}^{(2)} & k_{14}^{(2)} \\ k_{41}^{(1)} & k_{42}^{(1)} & (k_{43}^{(1)} + k_{21}^{(2)}) & (k_{44}^{(1)} + k_{22}^{(2)}) & k_{23}^{(2)} & k_{24}^{(2)} \\ 0 & 0 & k_{31}^{(2)} & k_{32}^{(2)} & k_{33}^{(2)} & k_{34}^{(2)} \\ 0 & 0 & k_{41}^{(2)} & k_{42}^{(2)} & k_{43}^{(2)} & k_{44}^{(2)} \\ \end{array} \right] $$ $$\left[ \begin{array}{l} d_1 \\ d_2 \\ d_3  \\ d_4 \\ d_5 \\ d_6 \end{array} \right] = \left[ \begin{array}{l} F_1 \\ F_2  \\ F_3  \\ F_4 \\ F_5 \\ F_6 \end{array} \right]$$

Class Notes: September 17, 2008
Certain degrees of freedom in our example problem are already known; the boundary conditions set these displacements to equal zero, while the displacements of global node 2 are to be determined. Setting $$ d_1 = d_2 = d_5 = d_6 =0 $$ is the same as deleting columns 1, 2, 5 and 6 in the stiffness matrix. Shown below are columns 3 and 4 and the simplified displacement vector..

$$\underline{K}\left[ \begin{array}{l} d_1=0 \\ d_2=0 \\ d_3=? \\ d_4=? \\ d_5=0 \\ d_6=0 \end{array} \right]= \left[ \begin{array}{ll} K_{13} & K_{14} \\ K_{23} & K_{24} \\ K_{33} & K_{34} \\ K_{43} & K_{44} \\ K_{53} & K_{54} \\ K_{63} & K_{64} \end{array} \right] \left[ \begin{array}{ll} d_3 \\ d_4 \end{array} \right] = \underline{F} $$

Since the stiffness matrix reduces down to a 6-by-2 size and the displacement reduces to a 2-by-1 matrix, the resulting force matrix will have dimensions 6-by-1.

Further reduction of this system of matrices can be performed by the Principal of Virtual Work, which allows us to delete the same rows from the stiffness matrix as we did the columns (1, 2, 5 and 6) as shown below.

$$\left[ \begin{array}{ll} K_{33} & K_{34} \\ K_{43} & K_{44} \end{array} \right] \left[ \begin{array}{l} d_3 \\ d_4 \end{array} \right] = \left[ \begin{array}{c} F_3=0 \\ F_4=P=7 \end{array} \right] $$

Since the goal now is to determine the displacements $$d_3$$ and $$d_4$$, we can rearrange the equation above to solve for these displacements.

$$\left[ \begin{array}{l} d_3 \\ d_4 \end{array} \right] = \underline{\overline{K}}^{-1} \left[ \begin{array}{c} 0 \\ P \end{array} \right] $$

We now have an inverse matrix. As a review...

$$\underline{\overline{K}}^{(-1)} = \dfrac{1}{det(\underline{\overline{K}})} \left[ \begin{array}{cc} K_{44} & -K_{34} \\ -K_{43} & K_{33} \end{array} \right]; det(\underline{\overline{K}}) = K_{33}K_{44}-K_{34}K_{43}$$

With the values $$K_{34} = K_{43} = -2.175$$, $$K_{33} = 3.0625$$ and $$ K_{44} = 2.6875 $$, we have

$$det(\underline{\overline{K}}) = 3.0625(2.6875)-(-2.175)(-2.175) = 3.4998$$

$$\underline{\overline{K}}^{(-1)} = \left[ \begin{array}{cc} 2.6875/3.4998 & 2.175/3.4998 \\ 2.175/3.4998 & 3.0625/3.4998 \end{array} \right] = \left[ \begin{array}{ll} 0.76790 & 0.62146 \\ 0.62146 & 0.87505 \end{array} \right] $$

$$\left[ \begin{array}{l} d_3 \\ d_4 \end{array} \right] = \left[ \begin{array}{ll} 0.76790 & 0.62146 \\ 0.62146 & 0.87505 \end{array} \right] \left[ \begin{array}{c} 0 \\ 7 \end{array} \right] = \left[ \begin{array}{cc} 4.35022 \\ 6.12535 \end{array} \right]$$

Since finding the inverse stiffness matrix was the prominent calculation, it is worthwhile to check the solution, especially given such a simple check as shown below.

$$\underline{\overline{K}}(\underline{\overline{K}}^{-1})=\underline{\overline{K}}^{-1}(\underline{\overline{K}})=\underline{I}=\left[ \begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array} \right]$$

The $$\underline{I}$$ above is called the Identity Matrix and results when a matrix is multiplied by it's own inverse.

 * A special note about the transpose of a matrix...

$$\underline{\overline{K}}^{T} = \left[ \begin{array}{cc} K_{33} & K_{43} \\ K_{34} & K_{44} \end{array} \right] $$

The transpose of the matrix $$\underline{\overline{K}}$$ is not the same as the matrix appearing in the $$\underline{\overline{K}}^{-1}$$ equation.

In order to compare to Statics, we compute the reaction forces.

5) Compute Reactions
We use two methods:

5.1) Use the element Force-Displacement relationships; $$\underline{k}^{(e)}\underline{d}^{(e)}=\underline{f}^{(e)}$$, where $$e=1,2$$.

Element 1 (e=1)

$$\underline{k}^{(1)}\underline{d}^{(1)}=\underline{f}^{(1)}$$. The contents of the stiffness matrix were previously calculated, and the displacement matrix is complete from our calculations above for the global displacements, while the force (reaction) matrix is unknown.

$$\underline{d}^{(1)}=\left[ \begin{array}{c} 0\\0\\4.352\\6.1271 \end{array} \right]$$

Element 2 (e=2)

Similarly, for $$\underline{d}^{(2)}$$, the stiffness matrix is already known and the following equation can be used to solve for the reactions of element 2.

$$\underline{k}^{(2)}\underline{d}^{(2)}=\underline{f}^{(2)}$$.

$$\underline{d}^{(2)}=\left[ \begin{array}{c} 4.352\\6.1271\\0\\0 \end{array} \right]$$

Class Notes: September 19, 2008
Revisiting element 1 and the known $$\underline{k}^{(1)}$$ and $$\underline{d}^{(1)}$$ matrices, we can solve for the reactions.

$$\left[ \begin{array}{cccc} 0.5625 & 0.32476 & ... & ...\\ ...\\...\\...\end{array} \right] \left[ \begin{array}{c} 0\\ 0\\ 4.352\\ 6.1271 \end{array} \right] = \left[ \begin{array}{c} \underline{f}^{(1)}_{1}\\ \underline{f}^{(1)}_{2}\\ \underline{f}^{(1)}_{3}\\ \underline{f}^{(1)}_{4} \end{array} \right]$$

Matrix multiplication results in the following values for the reactions of element 1...

$$ \left[ \begin{array}{c} -4.4378\\-2.5622\\4.4378\\2.5622 \end{array} \right] $$ where $$\underline{f}^{(1)}_{1}$$ and $$\underline{f}^{(1)}_{2}$$ are the reactions at the constrained node and $$\underline{f}^{(1)}_{3}$$ and $$\underline{f}^{(1)}_{4}$$ are the internal forces.

Similarly, for the stiffness and displacement matrices of element 2

$$\left[ \begin{array}{cccc} 2.5 & -2.5 & ... & ...\\ ...\\...\\...\end{array} \right] \left[ \begin{array}{c} 4.352\\ 6.1271\\ 0\\ 0 \end{array} \right] = \left[ \begin{array}{c} \underline{f}^{(2)}_{1}\\ \underline{f}^{(2)}_{2}\\ \underline{f}^{(2)}_{3}\\ \underline{f}^{(2)}_{4} \end{array} \right] = \left[ \begin{array}{c} -4.4377\\4.4377\\4.4377\\-4.4377 \end{array} \right]$$

Through the sum of the forces we can see that $$\Sigma F_{x} = \underline{f}^{(1)}_{1}+\underline{f}^{(1)}_{3}=-4.4378+4.4378=0$$ and $$\Sigma F_{y} = \underline{f}^{(2)}_{1}+\underline{f}^{(1)}_{4}=-2.5622+2.5622=0$$. We also note that $$ \Sigma M_{any point}=0$$. These three conditions satisfy the criteria for self-equilibrium.

HW for September 19, 2008
For element 1, if we call the reactions at local node 1, $$ -P^{(1)}_{1} $$, that is

$$ -P^{(1)}_{1}=\underline{f}^{(1)}_{1}+\underline{f}^{(1)}_{2}=-4.4378x-2.5622y$$

and if we also call the internal forces at node 2, $$ P^{(1)}_{2} $$ and $$ P^{(1)}_{2}=\underline{f}^{(3)}_{1}+\underline{f}^{(4)}_{2}=4.4378x+2.5622y$$

then $$ P^{(1)}_{2}=-(-P^{(1)}_{1})=P^{(1)}_{1} $$, further demonstrating equilibrium of element 1.

Similarly, for element 2, if we call local node 2, $$P^{(2)}_{2}$$, that is

$$ P^{(2)}_{2}=\underline{f}^{(2)}_{3}+\underline{f}^{(2)}_{4}=4.4377x-4.4377y$$

then $$ P^{(2)}_{1}= -P^{(2)}_{2} = \underline{f}^{(2)}_{3}+\underline{f}^{(2)}_{4}=-4.4377x+4.4377y$$

5.2) The second method for solving the two-bar truss problem involves using Statics. Here we use the Cut Principal by Euler.

To bring the Finite Element Solution into the picture, we verify the equilibrium of global node 2.

HW for September 19, 2008

For Node 2, $$ \Sigma F_{x}=-4.437+4.437=0$$ for the sum of the forces in the x-direction from both element 1 and element 2. The sum of the forces in the y-direction around global node 2 are also in equilibrium...

$$\Sigma F_{y}=7-4.4377-2.5622=0$$

where the forces are from the external force and the y-components of the forces from element 1 and element 2.

Matlab: Two-Bar Truss Example
The following is a matlab script and output for the two-bar truss example above.

title: User:Eml4500.f08.FEABBQ/HW2/Notes


text:
User:Ybit/Spanish

This is a personal learning page for user Ybit and other who help him to understand and talk Spanish as a second language.

1st lesson
--> than you see habl is a root and you just change terminations --> for a good declension there is also a need to remember personal pronouns: yo; tú; él, ella, Vd.; nosotros; vosotros; ellos, ellas, Vds.
 * What we did?
 * we get acquainted
 * have realised that you allready know a lot
 * realised that printed book is not equal to on-line book --> so we will continue with on-line version
 * studied about verbs and their declension:
 * all verbs ends on -r; reflexive verb ends on -rse = thats infinitive
 * there are 3 paradigms for declension according the verb ending: -ar (e.g. hablar), -er (e.g. comer) and -ir (e.g. vivir):
 * HABLAR - present tense
 * hablo
 * hablas
 * habla
 * hablamos
 * habláis
 * hablan
 * COMER - present tense
 * como
 * comes
 * come
 * comemos
 * coméis
 * comen
 * VIVIR - present tense
 * vivo
 * vives
 * vive
 * vivimos
 * vivís
 * viven
 * And how about reflexive verbs? Well, they are the same declension according paradigms, you just have to know how to change reflexive particle
 * LLAMARSE = LLAM-AR-SE - in the present tense is regular, so ti goes according HABLAR
 * me llamo
 * te llamas
 * se llama
 * nos llamamos
 * os llamáis
 * se llaman
 * we also spoke about the difference between r and rr, when rr has a tractor pronunciation
 * about the alphabet
 * started the first lesson on wikibooks

 * Homework
 * study 3 examples of regular verbs and reflexive verbs
 * find here and listen to CNN Plus: http://webteleradio.com/regiontv/spain.htm
 * memorise Spanish alphabet
 * write down a few sounds wich are simillar to rr pronunciation

 * Next lesson?
 * we will start with the test and review your homework
 * we will talk
 * we will finish the first lesson according and on-line textbook

Dictate

 * ahora
 * trabajo
 * yo
 * umedo - húmedo
 * note: humidity
 * un doras - Honduras
 * hola
 * yamarce - llamarse
 * remember:
 * me llamo
 * te llamas
 * se llama
 * nos llamamos
 * os llamáis
 * se llaman
 * ?karrar - querer
 * ?ike - Quique
 * pero
 * perro --> más sobre la pronunciación de la r
 * ?seboya - cebolla
 * in Czech language: cibule
 * colombia - Colombia
 * caracas - Caracas
 * ?kisana - gitana --> más sobre el flamenco --> y más sobre Andalucía
 * ?ikon - Gijón
 * madrid - Madrid
 * barcelona - Barcelona
 * ?caspeyano - castellano --> más sobre el castellano
 * ?saudi - Cádiz --> más sobre Andalucía
 * czecko - checo
 * ?supar - chupar
 * ?bida - vida
 * hache - ache
 * remember:
 * Spanish alphabet recording needed!
 * ocho
 * ?chupabida - chupavida
 * emane - Giménez
 * darcia - García
 * ahora
 * juan - Juan
 * jose - José
 * ajul - azul
 * no se - no sé
 * sabia - Sevilla --> más sobre Andalucía
 * ?bilbao - Bilbao
 * coledo - Toledo
 * ninyo - niño
 * anyo - año
 * españa - España
 * ?callego - gallego --> más sobre Galicia
 * ?gallego
 * ?itzel - Quetzal
 * is: Quetzacoatl
 * ?thickal - Tikal
 * ke - que
 * ?nalles - Yañéz
 * ?cardes - Cádiz
 * supapria - zapatería

 * mexíco - México
 * ? - Oaxaca
 * ishta wokstu? - Ixtacíhuatl --> why I wrote this word bad
 * exacto

--ybit 18:22, 21 September 2008 (UTC) corrected by--Juan 18:32, 21 September 2008 (UTC) --extended byJuan 07:47, 22 September 2008 (UTC)

Decline
hablar, comer, vivir, llamarse, escribir = 4 min

 * 1) hablar (tú) - hablas[[Image:Vista-clean.png|15px| ]]
 * 2) comer (nosotros) - comemos[[Image:Vista-clean.png|15px| ]]
 * 3) llamarse (ellos, ellas) - ? - se llama
 * 4) hablar (yo) - hablo[[Image:Vista-clean.png|15px| ]]
 * 5) escribir (Pedro) - escriba - escribe
 * 6) escribir (Vds.) - ? - escriben
 * 7) hablar (vosostros) - hablais - habláis
 * 8) llamarse (vosotros) - llama - os llamáis
 * 9) llamar (vosotros) - ? - llamáis
 * 10) llorar (yo) - ? - lloro
 * 11) comer (nosotros) - comemos[[Image:Vista-clean.png|15px| ]]
 * 12) vivir (tú) - vives[[Image:Vista-clean.png|15px| ]]
 * 13) hablar (él, ella) - habla[[Image:Vista-clean.png|15px| ]]
 * 14) trabajar (nosotros) - trabajamos[[Image:Vista-clean.png|15px| ]]
 * 15) vivir (vosotros) - vivimos - vivís
 * 16) trabajar (él, ella) - trabaja[[Image:Vista-clean.png|15px| ]]
 * 17) llorar (nosotros) - lloramos[[Image:Vista-clean.png|15px| ]]
 * 18) trabajar (yo) - trabajo[[Image:Vista-clean.png|15px| ]]
 * 19) comer (yo) - como[[Image:Vista-clean.png|15px| ]]
 * 20) vivir (yo) - vivo[[Image:Vista-clean.png|15px| ]]
 * 21) vivir (tú) - vives[[Image:Vista-clean.png|15px| ]]
 * 22) trabajar (ellos, ellas) - trabajan[[Image:Vista-clean.png|15px| ]]
 * 23) llamarse (tu) - - te llamas
 * 24) llamar (tu) - llamas[[Image:Vista-clean.png|15px| ]]
 * 25) presentar (él, ella, Vd.) - presenta[[Image:Vista-clean.png|15px| ]]
 * 26) presentar (yo) - presento[[Image:Vista-clean.png|15px| ]]
 * 27) comer (yo) - como[[Image:Vista-clean.png|15px| ]]
 * 28) presentar (Vds.) - ? - presentan
 * 29) trabajar (vosotros) - trabajamos - trabajáis
 * 30) hablar (Vds.) - ? - hablan

Correction: 12/30 entries = you should study more--Juan 08:48, 22 September 2008 (UTC)

What we did?

 * dictate
 * discovered the hispanic culture
 * conjugated
 * discovered raw food

Homework

 * go to Spanish (Ybit) and add information missing
 * have a look at: Spanish (Ybit) and study present tense within regular verbs for the next lesson

Next lesson?

 * we will repate this lesson
 * we will do lessons 1-3(4) from wikitextbokk

title: User:Ybit/Spanish


text:
User talk:Eml4500.f08.FEABBQ/HW2

Grade: 85.3

 * Who's submitting this?

 * Notes for the 10th are done, please browse over them to see if there are any typos (and to oogle at my hot skillz).  As for the comment at the top of the 8th, is the matlab solution at the bottom sufficient for that?  In the 10th's notes a lot of what was done was solving for the values of $$\underline{k}^{(1)}$$ and $$\underline{k}^{(2)}$$ which is what I believe that comment is referring too.  In my notes I said to refer to the matlab solution for the values for the matrices.  Anyways, sorry for it being last minute!  If you guys need to get a hold of me please call at (239)410-6903.  --EML4500.f08.FEABBQ.Kurth 18:43, 26 September 2008 (UTC)

 * Also, I went through and cleaned up some things around the lecture notes. There is still one comment on the very top for Sept 8 Andy Koby 14:27, 26 September 2008 (UTC)

 * Hey guys don't forget to sign your name with the ~ tag on the main HW page before we submit it. Andy Koby 14:14, 26 September 2008 (UTC)

 * I'll be working on the notes for the 10th today during/between classes...I have a good 6 hours or so to work on them so don't worry, they'll be done. - Billy

 * Ok guys I will not be available until afternoon today, so I will not be able to do anything before that. if something needs to be done on my behalf before noon please go ahead and help me out ,I think there is some more editing to be done on the homework, also we need to check what is wrong with all theses warnings " I though we fixed it"  I will see you all in class    ..          Esmail 09/26   ..  7 am

 * J: I had a couple of issues with the HW for the 19th. One small thing is I am wondering if you accidentally labeled the components for P2 of element 1 incorrectly. Also, I don't think it's correct to say that P1 = f1 + f2, etc. I mean, I know what you're getting at, but if anything you may want to say the magnitude of P = the RSS of those, or sum of forces... Lastly, I kind of feel what you did for the first HW problem is very similar to how I would approach the second HW for that day (the equilibrium of node 2) whereas the first homework is best answered using the two force member concept. I guess what I'm trying to say is I'm finding it hard to finish what you started without doing it differently. Let me know how you would like this to get done. Thanks. --Eml4500.f08.FEABBQ.vitt 04:53, 26 September 2008 (UTC)

 * Alright, the notes for the 15th are complete. I know he reviewed how to calculate some of the K values at the end of class that day, but wasn't sure if it was necessary to put there since it was just review. Also, I am aware that he began speaking about the elimination of known DOFs, but J's notes for the 17th begin with that and cover it thoroughly. Let me know if you think anything should be added. --Eml4500.f08.FEABBQ.vitt 04:40, 26 September 2008 (UTC)

Hey guys I have done all the diagrams the last two were posted on J's part, pleas everyone go and check his work adn remove any extra comments. Esmail Hadman 04:25, 26 September 2008 (UTC)

 * For future reference I am totally gonna start making these things up in class. What a pain in the ass. - Grant

Directional Cosines Diagram
 * I have fixed the mistake on the assembly drawing "thanks" and uploaded some more for the other parts . Esmail Hadman 03:00, 26 September 2008 (UTC)

2 force member concept

 * Alright thats what I got for the 8th for now. I'm gonna wrap them up tomorrow. - Grant

 * I uploaded the Matlab stuff that I think we were supposed to include in this HW. I also fixed on minor thing on the september 8 notes. JJ 02:42, 26 September 2008 (UTC)

 * Esmail, just one little thing I noticed in you assembly matrix diagram. The sub matrices in the d and F matrices are 2x1 rather than 2x2, just a minor fix.  Otherwise looks really good!  I'm doing the notes for the 10th now too.

 * I've got the 8th now. - Grant

 * Nik I have posted the assembly matrix diagram on my page use it where ever it needs to be...Esmail Hadman 01:42, 26 September 2008 (UTC)

 * Esmail, I have reduced the number of images required for Sept. 19 down to two and they are listed there. Let me know if you don't understand them, but you should have them in your notes. My sections are pretty much done, all I have to do is finish the equilibrium of node 2 which I'll do tomorrow. I have to study for Stats, I won't be able to do FEA anymore til then. JJ 19:50, 25 September 2008 (UTC)

 * Andy, Sept. 22 is due on HW 3... JJ 19:47, 25 September 2008 (UTC)

 * Maybe put which dates are done/need to be done on this AWESOME table? Andy Koby 19:45, 25 September 2008 (UTC)

 * I'll start doing some day when I get a chance, I have to leave the computer lab again but when I get back I'll pick a section... Also what's going on with all of our warnings? I tried changing the deletion tag, maybe I should change it back? Andy Koby 18:56, 25 September 2008 (UTC)

 * Esmail is doing September 8 and 10, I'm doing September 17 and 19. JJ 18:19, 25 September 2008 (UTC)

 * Sooooo, do we have any plan on how to finish this. Anyone want to claim stuff or assign stuff. I do some of the homework parts or make graphs, I just got to know what to do. -Grant

 * Meeting tomorrow at 10am in Marston 3rd Floor. Grant

 * What time are you guys meeting tomorrow? Esmail Hadman 00:20, 23 September 2008 (UTC)

 * The homework is all in your notes. JJ 03:11, 22 September 2008 (UTC)

 * Does anyone know exactly what homework things we need to do. If so could please list them.  Then we should start to split them up. Oh and personally I don't care if we do this stuff before the test. I'm ok with trying to finish it up whenever this week.

 * We don't have to split up the hw and notes anymore. It should all be one document, as per Vu-Quoc's instructions. JJ EML4500.f08.FEABBQ 03:21, 21 September 2008 (UTC)

 * I have posted the Element-1 diagram on the notes J [[Image:Drawing_09-17.jpg|center|thumb|100px]] Esmail Hadman 14:36, 19 September 2008 (UTC)

title: User talk:Eml4500.f08.FEABBQ/HW2


text: File:Drawing 09-17.jpg

title: File:Drawing 09-17.jpg


text:
User:Eml4500.f08.echo.lacroix/HW2

=Mon, Sept 15=

$$\mathbf{K}= \begin{bmatrix} k_{11}^{(1)} & k_{12}^{(1)} & k_{13}^{(1)} & k_{14}^{(1)} & 0 & 0\\k_{21}^{(1)} & k_{22}^{(1)} & k_{23}^{(1)} & k_{24}^{(1)} & 0 & 0\\k_{31}^{(1)} & k_{32}^{(1)} & k_{33}^{(1)}+k_{11}^{(2)} & k_{34}^{(1)}+k_{12}^{(2)} & k_{13}^{(2)} & k_{14}^{(2)}\\k_{41}^{(1)} & k_{42}^{(1)} & k_{43}^{(1)}+k_{21}^{(2)} & k_{44}^{(1)}+k_{22}^{(2)} & k_{23}^{(2)} & k_{24}^{(2)}\\0 & 0 & k_{31}^{(2)} & k_{32}^{(2)} & k_{33}^{(2)} & k_{34}^{(2)}\\0 & 0 & k_{41}^{(2)} & k_{42}^{(2)} & k_{43}^{(2)} & k_{44}^{(2)}\end{bmatrix}= \begin{bmatrix} K_{11} & K_{12} & K_{13} & K_{14} & K_{15} & K_{16}\\K_{21} & K_{22} & K_{23} & K_{24} & K_{25} & K_{26}\\K_{31} & K_{32} & K_{33} & K_{34} & K_{35} & K_{36}\\K_{41} & K_{42} & K_{43} & K_{44} & K_{45} & K_{46}\\K_{51} & K_{52} & K_{53} & K_{54} & K_{55} & K_{56}\\K_{61} & K_{62} & K_{63} & K_{64} & K_{65} & K_{66}\end{bmatrix}$$

$$K_{11}=k_{11}^{(1)}=\frac{9}{16}=0.5625$$

$$K_{12}=K_{21}=k_{12}^{(1)}=\frac{3\sqrt{3}}{16}=0.3248$$

$$K_{13}=K_{31}=k_{13}^{(1)}=-k_{11}^{(1)}=-0.5625$$

$$K_{14}=K_{23}=K_{32}=K_{41}=k_{14}^{(1)}=-k_{12}^{(1)}=-0.3248$$

$$K_{22}=k_{22}^{(1)}=\frac{3}{16}=0.1875$$

$$K_{24}=K_{42}=k_{24}^{(1)}=-k_{22}^{(1)}=-0.1875$$

$$K_{33}=k_{33}^{(1)}+k_{11}^{(2)}=	\begin{matrix} \frac{9}{16} \end{matrix}+\begin{matrix} \frac{5}{2} \end{matrix}=3.0625$$

$$K_{34}=k_{34}^{(1)}+k_{12}^{(2)}=	\begin{matrix} \frac{3\sqrt{3}}{16} \end{matrix}+\begin{matrix} \frac{-5}{2} \end{matrix}=-2.1752$$

$$K_{35}=K_{53}=k_{13}^{(2)}=-k_{11}^{(2)}=\frac{-5}{2}=-2.5$$

$$K_{36}=K_{45}=K_{54}=K_{63}=k_{14}^{(2)}=-k_{12}^{2}=2.5$$

$$K_{43}=K_{34}=k_{43}^{(1)}+k_{21}^{(2)}=k_{34}^{(1)}+k_{12}^{(2)}=-2.1752$$

$$K_{44}=k_{44}^{(1)}+k_{22}^{(2)}=\frac{3}{16}+\frac{5}{2}=2.6875$$

$$K_{46}=K_{64}=k_{24}^{(2)}=\frac{-5}{2}=-2.5$$

$$K_{55}=k_{33}^{(2)}=\frac{5}{2}=2.5$$

$$K_{56}=K_{65}=k_{34}^{(2)}=\frac{-5}{2}=-2.5$$

$$K_{66}=k_{44}^{(2)}=\frac{5}{2}=2.5$$

$$K_{15}=K_{16}=K_{25}=K_{26}=K_{51}=K_{52}=K_{61}=K_{62}=0$$

$$\mathbf{K}=\begin{bmatrix} 0.5625 & 0.3248 & -0.5625 & -0.3248 & 0 & 0\\0.3248 & 0.1875 & -0.3248 & -0.1875 & 0 & 0\\-0.5625 & -0.3248 & 3.0625 & -2.1752 & -2.5 & 2.5\\-0.3248 & -0.1875 & -2.1752 & 2.6875 & 2.5 & -2.5\\0 & 0 & -2.5 & 2.5 & 2.5 & -2.5\\0 & 0 & 2.5 & -2.5 & -2.5 & 2.5\end{bmatrix}$$

title: User:Eml4500.f08.echo.lacroix/HW2


text:
User talk:Eml4500.f08.FEABBQ

 Hello Eml4500.f08.FEABBQ, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. After you leave a comment on a talk page, remember to sign and date; it helps everyone follow the threads of the discussion. The signature icon in the edit window makes it simple. To get started, you may

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title: User talk:Eml4500.f08.FEABBQ


text:
User talk:Eas4200c.f08.gator.lusk

 Hello Eas4200c.f08.gator.lusk, 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.
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 * Browse everything on Wikiversity, or visit a portal corresponding to your educational level: pre-school, primary, secondary, tertiary, non-formal education.
 * Find out about research activities on Wikiversity.

 * Read an introduction for teachers and find out how to write an educational resource for Wikiversity.
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title: User talk:Eas4200c.f08.gator.lusk


text:
User:Eas4200c.f08.vqcrew.c/Homework 1/Contributing Team Members

HW1 Contributing Team Members
The following students contributed to this report: John Saxon Eas4200c.f08.vqcrew.c 03:26, 19 September 2008 (UTC) Adam Edstrand Eas4200c.f08.vqcrew.a 03:43, 19 September 2008 (UTC) Javier Stober Eas4200c.f08.vqcrew.b 13:22, 19 September 2008 (UTC) Darin Toscano Eas4200c.f08.vqcrew.d 04:02, 19 September 2008 (UTC) Kevin Klauk Eas4200c.f08.vqcrew.E 12:21, 19 September 2008 (UTC)


text:
User:Eas4200c.f08.vqcrew.c/Homework 2

= Class Notes =

Problem 1.1
Given Figure 1, find the optimum ratio $$\frac{b}{a}$$  to maximize the load bearing capability of the thin-walled beam.

The assumptions for this problem are as follows:
 * 1) The cross-sectional perimeter is constant, i.e. $$L=2(a+b)=Constant$$
 * 2) The torque $$T$$ is equal to the bending moment $$M$$, i.e. $$T = M$$
 * 3) The maximum allowable bending normal stress is twice the maximum allowable shear stress, i.e. $$\sigma_{allow} = 2\tau_{allow}$$

The assumption that $$t$$ is very small compared to $$a$$ and $$b$$ is also valid.

Since the cross-section walls are very thin, the shear stress can be assumed to be uniform along the walls, as shown in the shear flow diagram in Figure 2. The shear distribution and resultant shear vector $$V$$ for a small piece of the wall is also shown. This distribution is also mechanically equivalent to the linear distribution shown in Figure 3, where $$\tau =\frac{V}{t}$$.

The total torque ($$T$$) can be described as the sum of the torques on each wall of the box beam as follows: $$T = T_{AB}+T_{BC}+T_{CD}+T_{DA}$$

$$T_{AB}=(\frac{b}{2})Va=\frac{1}{2}\tau tab$$

$$T_{BC}=(\frac{a}{2})Vb=\frac{1}{2}\tau tab$$

$$T_{CD}=(\frac{b}{2})Va=\frac{1}{2}\tau tab$$

$$T_{DA}=(\frac{a}{2})Vb=\frac{1}{2}\tau tab$$

Therefore, $$T=2t\tau ab$$, and $$\tau =\frac{T}{2abt}$$.

There are two cases that must be analyzed in order to solve the problem:
 * 1) Assume the bending normal stress (σ) reaches $$\sigma_{allow}$$ first, then verify that $$\tau \le \tau_{allow}$$.
 * 2) Assume the shear stress (τ) reaches $$\tau_{allow}$$ first, then verify that $$\sigma \le \sigma_{allow}$$.

Case 1
Case 1 begins by recalling the formula for bending normal stress: $$\sigma =\frac{Mz}{I}$$, where $$M$$ is the bending moment, $$z$$ is the ordinate of a point on the axis perpendicular to the bending neutral axis, and I is the second area moment of inertia defined by $$I=\iint_A z^2 \ dy\, dz$$. In this case, $$z=\frac{b}{2}$$. Figure 4 shows each item pictorially.

Rewriting this equation yields $$M=\frac{2I\sigma _{max}}{b}$$, where $$\sigma _{max}=\sigma _{allow}$$ from the Case 1 assumption. We proceed to find $$M_{max}$$ by maximizing the ratio of $$\frac{I}{b}$$.

The expression for finding the moment of inertia $$I$$ of the box beam is: $$I=\sum_{i=1}^{4}{\left[\frac{b_{i}h_{i}^3}{12} + A_{i}d_i^2\right]}$$, where $$i$$ represents each wall of the box beam, varying from one to four. This simplifies to $$I=2 \frac{tb^3}{12}+2\left[\frac{at^3}{12}+\frac{atb^2}{4} \right]$$.

The first sum is the result of the vertical bars, and the second sum is the result of the horizontal bars. Because $$t\ll b$$, the $$t^3$$ term will be very small compared to $$b$$ and can be neglected. After some algebra, $$I$$ can be simplified to: $$I=\frac{tb^2}{6}\left(3a+b \right) $$.

Let us now define a function $$f(b)\equiv \frac{I}{b}$$ and maximize it to find the maximum bending moment, $$M_{max}$$. By using Assumption 1, we can substitute $$a$$ with $$a=\frac{L}{2}-b$$. Replacing this in the expression for $$I$$ results in the concave-downward quadratic shown in Figure 5: $$f(b)=\frac{tb}{12}\left(3L-4b \right)$$. By setting $$f(b)=0$$, the roots are shown to be $$b=0$$ and $$b=\frac{3L}{4}$$. Since the function is a quadratic, the maximum value of $$f(b)$$ is located halfway between the two roots at $$b^{(1)}=\frac{3L}{8}$$. By Assumption 1, we find $$a^{(1)}=\frac{L}{8}$$.

The maximum bending moment $$M_{max}^{(1)}$$ can then be written as:

$$M_{max}^{(1)} = \left(2\sigma _{allow} \right)\left(\frac{I^{(1)}}{b^{(1)}} \right) = \left(2\sigma _{allow} \right)\left(\frac{tb^{(1)}}{6} \right)\left(3a^{(1)}+b^{(1)} \right)$$

Substituting in for $$a^{(1)}$$ and $$b^{(1)}$$ yields: $$M_{max}^{(1)} = \frac{3tL^2}{32}\sigma _{allow}$$

This is also equal to $$T_{max}^{(1)}$$ based on Assumption 2. We will now use this assumption and the shear stress equation to determine the maximum shear stress for Case 1.

$$\tau _{max}=\frac{T_{max}^{(1)}}{2a^{(1)}b^{(1)}t} = \frac{\frac{3tL^2}{32}\sigma _{allow}}{2\frac{L}{8}\frac{3L}{8}t}$$

This reduces to:

$$\tau _{max}= \sigma _{allow}$$

Substituting Assumption 3 for $$\tau _{max}$$ gives:

$$\tau _{max}= 2 \tau _{allow}$$.

Therefore, Case 1 is an unacceptable solution because $$\tau _{max} > \tau _{allow}$$.

Case 2
For this case, the assumption is made that $$ \tau_{max} = \tau_{allow}$$  (i.e.  $$ \tau_{max}$$  reaches  $$ \tau_{allow}$$  first).

$$ \tau = \frac{T}{2abt} = \tau_{max} = \tau_{allow} $$

Rearrange to solve for the torque, T.

$$ T = (2t\tau_{allow})(ab) $$

$$ T_{max} = (2t\tau_{allow})(ab)_{max} $$

From Assumption 1, the perimeter of the beam is constant, L = 2(a + b). This allows for a to be solved in terms of only L and b.

$$a = \frac{L}{2} - b$$

Now, substituting in for a the maximum torque becomes

$$ T_{max} = (2t\tau_{allow})(\frac{bL}{2} - b^{2})_{max} $$

The first term in the maximum torque equation $$(2t\tau_{allow}) $$  is a constant. In order to maximize the torque, the second term $$(\frac{bL}{2} - b^{2})$$ must be maximized. A function f(b) is defined as the second term to allow for the maximum to be found.

$$ f(b) = (\frac{bL}{2} - b^{2}) $$

The derivative of f(b) is taken and set to zero to find the maximum of the function.

$$ f'(b) = (\frac{L}{2} - 2b) = 0 \Rightarrow b^{(2)} = \frac{L}{4} $$

Next, $$ a^{(2)}$$ is found through the perimeter relation.

$$ a^{(2)} = (\frac{L}{2} - \frac{L}{4}) = \frac{L}{4}$$

The value of $$ T^{(2)}$$ can now be determined and equated to the maximum moment by Assumption 2, T = M.

$$ T^{(2)}_{max} = (2t\tau_{allow})(\frac{L}{4})^{2} = \frac{1}{8}tL^{2}\tau_{allow} = M_{max}^{(2)}$$

The allowable normal stress can now be substituted in to the equation per Assumption 3, $$\sigma_{allow} = 2\tau_{allow}$$

$$M_{max}^{(2)} = \frac{1}{8}tL^{2}\tau_{allow} = \frac{1}{16}tL^{2}\sigma_{allow}$$

Solving for $$\sigma_{allow}$$  ,

$$\sigma_{allow} = \frac{16M_{max}^{(2)}}{tL^{2}}$$ Recall: $$I=\frac{tb^2}{6}(3a+b)$$

For $$ a=b=\frac{L}{4} $$

$$I^{(2)} = \frac{t(\frac{L}{4})^2}{6}(3(\frac{L}{4})+(\frac{L}{4})) = \frac{tL^3}{96}$$

Solve for $$\sigma_{max}^{(2)}$$  and substitute in values for $$I^{(2)}$$ and $$b^{(2)}$$ ,

$$\sigma_{max}^{(2)} = \frac{M_{max}^{(2)}b^{(2)}}{2I^{(2)}} = \frac{M_{max}^{(2)}\frac{L}{4}}{2\frac{tL^3}{96}} = \frac{12M_{max}^{(2)}}{tL^2}$$

Substituting in the previously found expression for $$M_{max}^{(2)}$$ ,

$$\sigma_{max}^{(2)} = \frac{12(\frac{1}{16}tL^{2}\sigma_{allow})}{tL^2} = \frac{12}{16}\sigma_{allow}$$

Since $$ \frac{12}{16}\sigma_{allow}$$ is less than $$\sigma_{allow}$$ , this configuration of dimensions for the thin-walled beam are acceptable.

Stringers, Shear Panels, and Shear Strain
= Homework Problems =

2nd Area Moment of Inertia of a Circle
Given a circle of radius R, the second area moment of inertia can be derived from the definition:
 * $$I_{y} = \iint_A z^2 \,dA$$

With circles it is often convenient to work with polar coordinates rather than with the traditional Cartesian coordinate system. To convert the above equation, a transformation must be done. Noting that

 * $$ \begin{align}

\\y &= r\ cos(\theta) \\z &= r\ sin(\theta) \\dA &= r\ dr\ d\theta \end{align}$$

Where the $$r$$ in the $$dA$$ term is the Jacobian of the transformation which is found by

 * $$J = \det\frac{\partial(x,y)}{\partial(r,\theta)}

=\begin{vmatrix} \frac{\partial x}{\partial r} & \frac{\partial x}{\partial \theta} \\ \frac{\partial y}{\partial r} & \frac{\partial y}{\partial \theta} \end{vmatrix} =\begin{vmatrix} \cos\theta & -r\sin\theta \\ \sin\theta & r\cos\theta \end{vmatrix} =r\cos^2\theta + r\sin^2\theta = r.$$

When substituting into the original definition, the following relation is obtained
 * $$I_{y} = \int_0^R \int_{0}^{2\pi} r^3\ sin^2(\theta)\ d\theta\ dr$$

and integrating
 * $$I_y = \frac{R^4}{4}\ \frac{2\pi}{2}$$

which yields
 * $$I_y = \frac{\pi R^4}{4}$$

Comparing a Circle to a Thin-Walled Three-Sided Box Beam of Equal Area
Given a circle of radius $$R$$ and a three-sided box beam of equal areas, the moments of inertia can be computed and compared. The moment of inertia for a circle was proven above and the moment of inertia of a rectangle is trivial and given as
 * $$I_y = \frac{b h^3}{12}$$

and applying the parallel axis theorem to the case of the three-sided beam would cause the resulting equation to be
 * $$I_y = \sum_{i=1}^3 \frac{b_i h_i^3}{12}+A_i d_i^2$$

the area of the circle will be
 * $$A_{circle} = \pi R^2$$

while the area of the three-sided box beam is
 * $$A_{box} = \sum_{i=1}^3 b_i h_i = 3 a t = \frac{3 a^2}{10}$$

Note: the relation $$t = \frac{a}{10}$$ was used in simplifying the area

Equating the two area terms
 * $$\pi R^2 = \frac{3 a^2}{10} \therefore a = R \sqrt{\frac{10 \pi}{3}}$$

Determining the $$I_y$$ for each case in terms of R yields for the circular case (case 1)
 * $$I_y^{(1)} = \frac{\pi R^4}{4} = .7854 R^4 $$

and for the three-sided box beam
 * $$I_y^{(2)} = 2 \frac{t a^3}{12} + (\frac{a t^3}{12} + (a t)(\frac{a}{2} + \frac{t}{2})^2)$$

plugging in $$t = \frac{a}{10}$$

 * $$I_y^{(2)} = \frac{2 a^4}{120} + (\frac{a^4}{12(10)^3} + \frac{a^2}{10}(\frac{a}{2} + \frac{a}{20})^2)$$

 * $$I_y^{(2)} = \frac{a^4}{60} + (\frac{a^4}{12000} + \frac{a^2}{10}(\frac{11 a}{20})^2)$$

 * $$I_y^{(2)} = \frac{a^4}{60} + \frac{a^4}{12000} + \frac{a^4(121)}{4000}$$

 * $$I_y^{(2)} = a^4(\frac{1}{60} +\frac{1}{12000} + \frac{121}{4000}) = .047 a^4 = .047(R\sqrt{\frac{10 \pi}{3}})^4 = 5.1541 R^4$$

 * $$\frac{I_y^{(2)}}{I_y^{(1)}} = \frac{5.1541 R^4}{.7854 R^4} = 6.5624$$

meaning that altering the cross section of a beam from circular to three-sided box while keeping the area the same will increase the moment of inertia by a factor of roughly 6.56.

Comparing a Circular Cross-Section to a Dumbbell Cross-Section of Equal Area
The area for the circular cross-section is
 * $$A_{circle} = \pi R_o^2$$

while the area for the dumbbell is
 * $$A_{dumbbell} = 2 \pi R_1^2 + 2 R_o t$$

equating the areas of each case and setting $$t = \frac{R_o}{10}$$ yields
 * $$\pi R_o^2 = 2 \pi R_1^2 + 2 R_o t \therefore R_1 = \sqrt{\frac{R_o^2}{2} - \frac{R_o^2}{10 \pi}} = .4682 R_o$$

The moment of inertia of a circle was derived above as
 * $$I_y^{(1)} = \frac{\pi R_o^4}{4} = .7854 R_o^4$$

while the moment of inertia of the dumbbell is
 * $$I_y^{(2)} = 2 (\frac{\pi R_o^4}{4} + \pi R_1^2(R_o + R_1)^2) + \frac{.1 R_o(2 R_0^3)}{12}$$

 * $$I_y^{(2)} = \frac{\pi}{2}(.04804) R_o^4 + \pi (.2192 R_o^2) (.2156 R_o^2) + .0666 R_o^4$$

 * $$I_y^{(2)} = 1.560 R_o^4 + .0666 R_o^4$$

 * $$I_y^{(2)} = 1.627 R_o^4$$

comparing the two moments of inertia
 * $$\frac{I_y^{(2)}}{I_y^{(1)}} = \frac{1.627 R_o^4}{.7854 R_o^4} = 2.0716$$

thus, altering the cross-section of a beam from circular to dumbbell-shaped while keeping the area the same serves to increase the moment of inertia by a factor of roughly 2.07

Proof that $$(a\ b)_{Max}$$ is a Square with Constant Perimeter
Given a rectangle of constant width, to maximize area it is intuitive that the optimal ratio of $$\frac{b}{a}$$ would be one, resulting in a square. Noting that L is constant and defining and substituting

F(b) = a\ b$$ $$ a = \frac{L}{2} - b$$ $$ F(b) = b\ (\frac{L}{2} - b) = \frac{L}{2}\ b - b^2$$

Recalling in calculus that setting the first derivative to zero and solving for the variable results in the critical number, which is a relative minimum or maximum of the function. Also noting that if the second derivative is negative at the critical number, that point is a relative maximum and alternatively if second derivative is positive at the critical number, that point is a relative minimum. Taking the first derivative and solving

\frac{dF(b)}{db} = \frac{L}{2} - 2b = 0$$ $$ b_{crit} = \frac{L}{4}$$

Meaning that at the point $$L/4$$, the function $$F(b)$$ is either a relative minimum or maximum. Taking the derivative again and plugging in the critical point

\frac{d^2F(b)}{db^2} = -2$$

Showing that when $$b = \tfrac{L}{4}$$ the function $$F(b)$$ is a maximum, showing that the area at that point is at its maximum. When $$b = \tfrac{L}{4}$$ and substituting to solve for a
 * $$a = \frac{L}{2} - \frac{L}{4} = \frac{L}{4}$$

This shows that $$a = b$$ meaning that the sides have equal length, resulting in the geometry of a square.

= Contributing Team Members = The following students contributed to this report:

John Saxon Eas4200c.f08.vqcrew.c 02:22, 29 September 2008 (UTC)

Adam Edstrand Eas4200c.f08.vqcrew.a 20:25, 26 September 2008 (UTC)

Javier Stober Eas4200c.f08.vqcrew.b 20:26, 26 September 2008 (UTC)

Darin Toscano Eas4200c.f08.vqcrew.d 20:08, 26 September 2008 (UTC)

Kevin Klauk Eas4200c.f08.vqcrew.E 16:22, 26 September 2008 (UTC)

=References=

title: User:Eas4200c.f08.vqcrew.c/Homework 2


text:
User:Eas4200c.f08.wiki.b/HW2

Case 1: Assume:
 * $$\displaystyle \sigma={Mz \over I}$$

Recall:
 * $$\displaystyle \sigma_{max}=\sigma_{allow}, z= {b \over 2} $$

Thus,
 * $$\displaystyle M= {2I\sigma_{max} \over b} $$
 * or $$\displaystyle M= (2\sigma_{allow})({I \over b})$$

Where $$\displaystyle (I/b)$$ is a function of $$\displaystyle a$$ and $$\displaystyle b$$. Recall from the first assumption:
 * $$\displaystyle L=2(a+b)=constant$$

Thus,
 * $$\displaystyle a={L \over 2}-b$$

Maximize $$\displaystyle M$$:
 * $$\displaystyle M_{max} = (2\sigma_{allow})({I \over b})$$
 * $$\displaystyle I = \sum_{i=1}^4 [{b_i(h_i)^3 \over 12} + A_i(d_i)^2] $$

 * $$\displaystyle = 2{tb^3 \over 12} + 2[{at^3 \over 12} + (at)({b \over 2})^2] $$

Let
 * $$\displaystyle \alpha = {at^3 \over 12} + (at)({b \over 2})^2 = \alpha ={at \over 12}[t^2+3b^2]$$

Since $$\displaystyle t<<b, t^2<<b^2, t^2<<3b^2 $$ Hence,
 * $$\displaystyle \alpha \approx {3ab^2t \over 12}$$

And therefore,
 * $$\displaystyle I \approx {2tb^3 \over 12} + {ab^2t \over 2} = {tb^2 \over 6}(3a+b)$$
 * $$\displaystyle f(b) := {I \over b} = {tb \over 6}(3a+b) $$
 * $$\displaystyle = {tb(3L-4b) \over 12} = \beta_0 + \beta_1 b^1 + \beta_2 b^2 $$

Where $$\displaystyle \beta_0 = 0, \beta_1 = {3Lt \over 12}, \beta_2 = -t/3 $$
 * $$\displaystyle {d^2f(b) \over db^2} = 2 \beta_2 = {-2t \over 3} < 0 $$
 * $$\displaystyle b = \begin{cases} 0 & \\ {3L \over 4} \end{cases} $$ [[Image:HW_2_9-10_2.JPG|thumb|400px|]]
 * $$\displaystyle {df(b) \over db} = \beta_1 + 2 \beta_2 b = 0 \Rightarrow b^{(1)} = {-\beta_1 \over 2 \beta_2} = {3L \over 8} $$
 * $$\displaystyle a^{(1)} = {L \over 2} - b^{(1)} = {L \over 8} $$

Thus,
 * $$\displaystyle {b^{(1)} \over a^{(1)}} = 3 $$
 * $$\displaystyle ({I \over b})_max = ({I^{(1)} \over b^{(1)} })$$
 * $$\displaystyle M_{max}^{(1)} = 2 \sigma_{allow}({I^{(1)} \over b^{(1)} }) = {3+L^2 \over 32} \sigma_{allow}$$

title: User:Eas4200c.f08.wiki.b/HW2


text: File:2 force body diagram.jpg

title: File:2 force body diagram.jpg


text: File:Element1 hw2.jpg

title: File:Element1 hw2.jpg