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User:Lizbennett/SOCIAL THINKING

Cognitive processes
Social thinking is about using your cognitive powers of thinking and perceiving when developing the self’s attitudes and behaviours. The human brain is complex and around the size of a large grapefruit weighing about 3 lbs or 1.5 kgs. It has two main systems: the automatic system - which makes quick, fairly accurate judgements and decisions; and the conscious system which works more slowly and thoroughly but can make more precise judgements and decisions. As humans have evolved they have adjusted their thinking patterns and introduced discriminatory behaviour and shortcuts to conserve the time spent on thinking. This is known as being a cognitive miser describes the individual’s reluctance to do much extra thinking. Connected to this, is our ability to use shortcuts or heuristics especially where we are already preoccupied and to counteract this individuals have learned automatic thinking behaviours.

Three main types of goals guide how people think – to find the right answer to the problem or question, to confirm the desired answer to a problem and to read the good answer of decision quickly. This introduces the question of automatic and controlled thinking – is thought automatic or is it controlled? The four elements that distinguish automatic from controlled thinking are intention, control, effort and efficiency. Automatic thinking is not guided by intention, are not subject to deliberate control, do not involve effort and are highly efficient. It involves very little effort because it involves knowledge structures – organised packets of information that are stored in memory, usually known as schemas or scripts. These are developed from early childhood through experience and guide the way we process information.

Schemas and scripts influence the way individuals perceive, interpret, judge and respond to events. Priming - the planting or activating of an idea stimulates further processing of a schema – is often used as a technique to trigger automatic processes and behaviours. Framing of messages is also used as a technique to influence – positively and negatively. Thought suppression uses two processes. One an automatic process that keeps a lookout for a reminder of the unwanted thought and the second is a controlled process that redirects attention away from the unpleasant thought. When the controlled process is relaxed the automatic process still watches for cues and may flood the mind with the unwanted thoughts.

Attributions give meaning to why individuals demonstrate different attitudes and behaviour. Attributions are the causal explanations individuals give for their own and others’ behaviours and for events in general. Heider (1958) put forward an explanation that explanations fall into two major categories: (a) internal factors such as ability, attitudes, personality, mood and effort and : (b) external factors such as task, other people or luck. Weiner (1971) proposed a two dimensional theory of attributions for success and failure. The first was internal versus external and the second was stable versus unstable. The four possible combinations of internal-external and stable-unstable yield the four main types of attributions that people make when they see themselves or someone else perform. Internal, stable attributions involve ability and this is very important because it involves permanent aspects of the self with people motivated to conclude that they have high ability. Internal, unstable attributions involve effort and this is unstable because it could change. Interestingly collectivist cultures emphasise effort whereas individualistic cultures emphasise ability. External, stable attributions point to task difficulty – success indicates task was easy, failure indicates it was hard. External unstable attributions involve luck – which indicates that there is very little credit or blame due to the person.

Self serving bias the tendency to take credit for success but deny blame for failure occurs because individuals want to interpret events that make them feel good. That means, they can discount their failures and maximise their successes. This links with the concept that individuals learn to think in ways that will help them get along better with others and thus accepted by others. Individuals also look and notice information that confirms their beliefs and discounts or ignores information that disconfirms their beliefs. This is known as the confirmation bias. Individuals love observing others and at times use their biases to make assumptions. The actor/observer bias is the tendency for actors to make external attributions and observers make internal attributions, put simply the actors attribute their behaviour to the situation and the observers attribute the actor’s behaviour to the actor. Within this is fundamental attribution error which is the tendency for observers to attribute other people’s behaviour to internal or dispositional causes and to downplay situational causes. The explanation for this is that behaviour is more noticeable that situation factors, individuals assign insufficient weight to the situation, the use of cognitive miser’s concept and language is richer in trait like terms than in situational terms.

Attribution uses the covariation principle which indicates that for something to be the cause of behaviour it must be present when the behaviour occurs and absent when the behaviour does not occur. Kelley proposes the attribution cube theory that people use three types of covariation information – consensus – whether other people would do the same thing in the same situation; consistency – whether the person typically behaves this way in this situation and distinctiveness – whether the person would behave differently in a different situation. Heuristics are used during the cognitive process to make judgements and inferences about the uncertain outcomes including the likelihood and frequency of an event. The four most common heuristics used are representative, availability, simulation, anchoring and adjustment.

As the information age has progressed, information is more freely and readily available. One of the issues raised is information overload or having too much information to comprehend or integrate to make a decision or remain informed. Common cognitive errors by individuals include the following heuristics – confirmation bias, conjunction fallacy, illusory correlation, base rate fallacy, gamblers fallacy, false consensus effect, false uniqueness effect, and statistical regression, illusion of control, magical thinking and counterfactual thinking. The following strategies are used to reduce cognitive errors – debiasing by providing information to improve decision-making, encouraging individuals to rely less on memory, to use explicit decision rules, to search for disconfirmatory information and to use meta-cognition or reflective approach to examine and reflect on the thinking process.

Attitudes, beliefs and consistency
Attitudes exist in substantial part to help guide behaviour. They are ideas that often determine how people will act. They differ from beliefs which are pieces of information about something, facts or opinions. Attitudes are for choosing, and beliefs are for explaining. Beliefs and attitudes both serve interpersonal functions. People have attitudes so they can differentiate between what they like and what they don’t like. They help us to adjust to new situations, seek out things that reward us and avoid those things that punish us. Attitudes influence what individuals do and won’t do. Individuals often have different evaluation of the same attitudes which can be unrelated and can serve different functions. Attitudes can be private not shared with others and at times individuals may not be aware of all of their own attitudes. Having an attitude often increases the speed, ease and quality of decision making. A meta-analysis of 88 studies found that attitudes are certain, stable, consistent, accessible, and based on direct experience are especially effective in predicting behaviour. Different explanations are given to how our attitudes are formed. Mere exposure effect, classical conditioning, operant condition and social learning all help form our attitudes. As individuals reflect on their attitudes they can become more extreme or adopt them more strongly. Studies have should that individuals are more accepting of evidence represented by in group members than by out- group members.

Belief and doubt are separate from understanding, but believing immediately automatically accompanies understanding. When given information, the automatic system automatically believes the information it is given. The conscious system can override this belief by deciding that it is false. Belief perseverance is the finding that once beliefs form, they are resistant to change, even if the information on which they are based is discredited. To validate and understand things correctly, individuals should cultivate the habit of trying out the opposite theory to whatever theory encountered. Assumptive worlds are the view that individuals live in social worlds based on certain beliefs or assumptions about reality. When problems or trauma occur individuals usually cope with the trauma and then go back to functioning normally. Coping can take many forms from self blame or blaming others and usually is part of a grieving process. Broadly, cognitive coping is the idea that beliefs play a central role in helping people cope with and recover from misfortunes. Within this, there are different elements, like downward comparison where individuals compare themselves to other individuals who are worse off. Other beliefs in cognitive coping relate to self esteem and control, the belief that all things have some useful or higher purpose. Beliefs help individuals cope with trauma or make the trauma seem compatible or even consistent with their beliefs. Religious beliefs appeals to some individuals who look to it to explain grand or small issues, to provide social support, a sense of meaning, purpose and direction for one’s life and an environment that fosters the development of virtues such as honesty or integrity. Religious beliefs can help people cope with stress and traumatic events. Individuals with irrational beliefs are usually more anxious, cope less well with terminal illnesses, become depressed, may be paralysed when needed to perform important behaviours and have lower levels of self esteem.

Consistency theories have three elements – specify the conditions required for consistency and inconsistency of cognitions; assume inconsistency is unpleasant and therefore motivates people to restore consistency; and specify the conditions required needed to restore consistency. In general people choose the path of least resistance to restore consistency. Heider’s balance theory proposes the idea that relationships among one person, the other person and an attitude object may be either balanced or unbalanced. These relationships can either involve attitudes or evaluations (sentiment) or belongingness (unit). Cognitive dissonance theory proposes that inconsistencies produce psychological discomfort, leading people to rationalise their behaviour or change their attitudes. Within this theory is effort justification or the finding that when people suffer or work hard or make sacrifices they will try to convince themselves that it is worthwhile, even being demeaned or embarrassed. Inconsistency between attitudes and behaviours is shown in the gap between general attitudes and specific behaviours, behaviour aggregation combining across many different behaviours on different occasions; the prominence of the attitude in the persons conscious mind and influence and the accessibility of the attitude to the mind.

Social influence
Social influence is seen as two major categories – normative influence the going along with the crowd in order to be liked and accepted and informational influence which refers to going along with the crowd because you think the crowd knows more than you do. Usually with informational influence is produced by ambiguous situations, so that individuals do not know how to behave and crisis situation where individuals don’t have time to think for themselves. Humans have a fundamental need to be accepted and evolutionary theory proposes that humans survive better if they belong to a group. To live together individuals usually need to agree on common beliefs, values, attitudes and behaviours that reduce in-group conflicts and act for the common good, thus conforming and behaving in a certain way. When individuals deviate from group norms they are usually rejected and studies found that people would rather agree with the group even though they knew it was wrong rather than suffer social rejection. Groups are quick to vote out the deviate or non conformist with rejection more likely when there are one or two non conformists that when there is a large number of nonconformists. Public compliance – where an individual outwardly goes along with the group but maintains a private, inner belief that the group is wrong is often practiced by individuals in order to comply.

Commitment by individuals to beliefs, attitudes or even to the group puts them in a position where they then find themselves acting consistently with that commitment. Not only do they feel internal pressure but also external pressure to behave consistently with that commitment. There are many techniques that individuals and groups use to influence individuals, usually based on the principle of commitment, consistency, reciprocation, scarcity and capturing and disrupting attention. These techniques are based on the individual doing something they would not usually do and the techniques are based on neutralising the conscious rational mind in order to bypass it.

The foot in the door technique is based on commitment and consistency, in which one starts with a small request in order to gain eventual compliance with a larger request. The low ball technique is also based on commitment in which one first gets the person to comply with a seemingly low cost request and only later reveals hidden additional costs. The bait and switch technique is based on commitment, in which one draws people in with an attractive offer that is unavailable and then switches them to a less attractive offer that is available. The labelling technique is another way of inducing compliance where one assigns a label to an individual and then requests a favour that is consistent with the label. This technique is also based on commitment and consistency and uses the individuals self concepts. Legitimization of paltry favours technique similarly, is based on the requester making a small amount of aid available.

The appreciation of reciprocity is deeply rooted in human nature where an individual feels guilty if someone does them a favour and they cannot repay it in some way. This loads up the threat of guilt feelings and increases compliance rates. Two techniques – door-in-the-face technique starts with an inflated request and then retreats to a smaller request that appears to be a concession. The request must not be too extreme so that it is seen as unreasonable and does not work if the requests made are done by separate individuals. That’s not all technique is similar, this is also based on personal obligation where one first makes an inflated request but before the person can answer yes or no, sweetens the deal by offering a discount or bonus. The techniques used with the scarcity principle are based on the product or service diminishing and so the individual’s chances to buy or be included are limited.

Individuals often try to catch the attention of others using the pique technique which uses a novel request to catch their attention. Similarly the disrupt-then-reframe technique is based on one disrupting critical thinking by introducing an unexpected element, then reframing the message in a positive light. Both these techniques try to capture or distract the individual with the pique technique placing emphasis on the individual thinking about the request more deeply and in the disrupt-then-reframe technique the individual is distracted and don’t pay too much attention to the original request or conversation.

Persuasion
Persuasion is an attempt to change an individual’s attitude. Aristotle proposed three components of persuasion being the speaker, the subject of the speech and the hearer to whom the speech is addressed. He suggests that the three elements necessary are emotional appeal, intellectual appeal and charisma. The key source of the message must usually be credible and likable. Credibility is based on expertise and trustworthiness of the source with likability based on similarity and physical attractiveness. Often convert communicators or individuals are perceived as credible sources because they are arguing against their own previously held attitudes and behaviours. The message given is usually based on facts or appeals to the emotions of the listener and often uses humour and/or fear as a slant.

Research has indicated that people who are in a good mood are more receptive to persuasive messages and fear appeals can change attitudes if they don’t induce too much fear and if the audience is told how to avoid the fearful outcome. The message must resonate with the audience’s intellect, need for thinking and concerns for now and in the future. The easiest people to persuade are young children followed by adolescents and young adults. By this stage attitudes are set in place and resistance to persuasion is evident. Cultural difference also plays a role with people from individualistic cultures placing more emphasis on the individual where the collectivist culture place more emphasis on the group. Alpha and Omega strategies are also used to persuade – when people are willing and able to listen to a message, strong arguments are more persuasive than weak arguments and vice versa.

Resisting persuasion is often much more difficult to put into practice. There are a number of techniques which can be used as resistance. One of these is the negative attitude change where we do exactly the opposite of what one is being persuaded to do. Others deal with delaying decisions in order to rationalise, reword the request in the individuals favour and pay attention to what other credible individual’s behaviours.

My Personal Reflections
Communication takes many forms and levels from verbal to non verbal and small talk to in depth complex discussions. During these engagements a mixture of automatic and controlled thinking with their various schemas and scripts are used. Within this, a feedback loop checks whether our audience understand the message and respond and this helps to adjust language, tone, speed and type of message with the ultimate desire to keep the interest of the listener. Deborah Tannen’s “9 to 5” describes the differences between conversation styles of men and women and how these can either help or limit promotional opportunities. The male rituals of bantering or avoiding the one-down position are common with women tending to downplay their authority and avoid boasting. The way in which men and women use their styles are used to judge their confidence, leadership and authority.

The use of body language often discriminates or confirms the message being given with eye avoidance, body position and stance and gestures being the key. In some cultures, eye avoidance is normal and acceptable. Malcolm Gladwell’s book “Blink” describes the concept of thin-slicing where we recognise patterns and make snap judgements and we do this process of editing unconsciously. He suggests that by trusting our instincts a snap judgement can be far more effective than a cautious decision and gives the example of a simulated war to bring his point home.

The messages we use can captivate or turn off the audience. Social influence is used neatly in these messages, using reciprocity and interest to draw attention and then persuasion to influence the decision or outcome. Tied in with this is our need to comply, be part of the group and reciprocate when it is our turn. A good salesman can push all these buttons and usually the average Joe will comply even though in a rational moment a different decision may have been made.

References

Baumeister, R.F. & Bushman, B.J. (2008). Social Psychology and Human Nature. 1st Ed. Thomson Learning Inc.

Tannen, D. (1994). Talking from 9 to 5. Virago

Gladwell, M. (2005). Blink The Power of Thinking without Thinking. Penguin Books

title: User:Lizbennett/SOCIAL THINKING


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User:Geoff Plourde/College

Restoring the Benefits of Free Knowledge: A Proposal for a Free University

Background
Knowledge has become more accessible over time. At first, it was restricted to those who could afford to study it. Now there is a free culture movement. This movement aims to bring the sum of all human knowledge to the world for free.Through volunteer projects that create educational resources, people no longer have to buy educational materials.

However, there is a facet of knowledge that is still restricted. While free culture has made a plethora of free educational resources, free instruction is virtually nonexistent. Formal education still has a price as students must pay to take courses for credit. There is no free way for the majority of the world's population to acquire a degree and better their lives, without cost. Therefore, in accordance with the principles of free culture, it is proposed that a tuition free online university be established. This free university would make for a better world by allowing anyone with internet access to study for and receive their degree, No longer would cost be a barrier to education. With these skills, people would be able to bring themselves out of poverty. The world would also gradually become more educated. A better educated populace would lead to smarter solutions and a better world.

Name and Governance
The tuition free online university would be called Frontier University. Like any other university, Frontier University would be a degree granting academic institution. The name was selected because this college will be a new frontier in education.

Frontier would be divided into schools and colleges that focus on specific areas. Each of these divisions would choose one of the University Regents. These Regents would be the main supervisory authority when sitting as a Board. In addition to providing University leadership, these divisions would make all recommendations for degrees. Despite the recommendation coming from a college, the main work and exact criteria would be developed by departments. Department chairs would serve as leaders of their departments and would collectively have the power to remove the dean. Deans would be division leaders and be elected by the faculty. The majority of schools would be exempt from departmentalization, as they would focus on a specific discipline. The deans of these schools would be elected and removed by their faculty, unlike colleges. In accordance with free culture principles of democracy, leadership would be expected to consult with colleagues prior to taking action, would be removable, and would continue as faculty.

Curriculum
All courses would be free of charge, making Frontier unique among academic institutions. Studies have shown that using information is the most effective way to retain it. Therefore it is anticpated that as many courses as possible will be designed around a learning by doing model.. As part of a course, students will work on content utilizable by society. This will provide both the world with educational resources and the student with a sense of accomplishment and pride. Memorization and practice (by rote) will also be used where necessary, as those techniques have been proven over history. Despite the model, every course will be taught at the same level or higher as those taught at a traditional university. Faculty would establish connections with other universities and be expected to compare curriculums.

As with a traditional university, when students have acquired enough course credits, they will be eligible to receive a degree. While tuition will be free, a small fee might need to charged to cover the cost of the degree. It is hoped that a grant would be secured to offset this cost though. All degrees would be the same as any other academic degree with all the powers thereof and would range from Associate to Doctoral level.

60 credits would be required for an Associate of Arts or Associate of Sciences degree. Half of these credits would be in general education to provide a more rounded graduate. The AA in general education would be entirely composed of general education and could be applied to the undergraduate general education requirements.

120 credits would be required for a Bachelor of Arts or a Bachelor of Science degree. As with AA and AS degrees, half of the credits would be in general education.

Graduate level degrees would be granted upon completion of a graduate course of study. Examples of such degrees are the Master of Arts and Master of Sciences degree. The M.A. would require a thesis and defense. The MS would require a research project. Graduate students might also be required to work on a undergraduate course in their discipline

Doctoral degrees would be granted to those who complete a doctoral course of study. For Ph.D.s a dissertation and research project would be required. For J.Ds, a rigorous curriculum would be required.

Many online universities sell honorary degrees. To maintain the reputability and value of degrees, honorary degrees shall only be issued by the Board of Regents, and only in exceptional cases.

Faculty
Faculty would be organized by department and consist of four levels of Professors. All faculty would be part time volunteers. Adjunct Professors would be volunteers that would teach under the supervision of another Professor. They would not be required to have any prerequisites other than knowledge of the subject. Associate Professors would be junior Professors and would be restricted to teaching lower division undergraduate courses. An Associate Professor must hold a graduate degree in the discipline. Assistant Professors are instructors that teach at the graduate level and hold doctoral degrees. Professors are the most senior faculty members. They have the most experience, a doctoral degree, and work with doctoral students. The Department Chair will always be the most senior Professor. All faculty will be members of the Academic Senate, regardless of their level.

College Preparatory Academy
Many individuals that this proposal is designed to serve may not have the skills necessary to succeed in college. To rectify this potential issue, a free college preparatory academy would be set up as a separate school of the university. The Academy would offer secondary level courses designed to fill gaps in students' knowledge, allowing them to succeed at a college level. Once a student has completed the courses necessary to succeed in college, they would be eligible to receive a college preparatory certificate. High school students would also be welcome and be eligible to receive a high school diploma and college preparatory certificate. As with the University program, there would be no charge to receive

These secondary courses would provide a project for Education students as they would be able to acquire real world experience. By teaching secondary courses, future teachers would strengthen their skills, acquire experience, and gain an advantage over others in the field.

Accreditation
Accreditation is crucial for alumni to receive recognition. The university as a whole would begin to seek accreditation as soon as possible. Individual programs would also seek specialized accreditation. A grant would be sought to cover the costs that this process would entail.

COURSE QUESTIONS

 * Q: How much are courses?
 * A: All courses offered by this University would be free of charge.

 * Q: How much are the books going to be?
 * A: Courses should be designed to allow any textbook to be used. Wherever possible, instructors will develop free online textbooks.

 * Q: How will courses compare to a regular university course?
 * A: Courses are designed to be at the same level or higher than that which is offered at a university.

 * Q:How do I get a degree for life experience?
 * A:Frontier does not grant life experience degrees. If you feel that you already know the necessary material, please contact departments about testing out of courses.

 * Q:Can I test out of a course?
 * A:Absolutely! While we do not grant degrees for life experience, we do recognize that some people may already be competent in the material offered. Each department would develop tests that would be administered to people who wish to test out of a course and will accept CLEP results.

 * Q: What would the AP credit policy be?
 * A: Any score of 3 or higher would allow credit for an introductory course. Each department will have the relevant information.

 * Q:I was in a degree program years ago, can I get credit?
 * A:Maybe, but you would need to demonstrate competency to receive credit. Contact relevant departments.

 * Q:What is the minimum age to take classes and work towards a degree?
 * A:Anyone who demonstrated the maturity to do college level work is welcome to be a student. There is no set age.

 * Q:What do I need to get a Bachelor degree?
 * A:You must complete 120 units of undergraduate coursework. 60 of those units must be in general education and 60 must be in the subject.

ACADEMY AND REFERRALS

 * Q: Do I need a high school diploma to receive a degree?
 * A: Not necessarily. If you pass the classes, you will receive a degree. If you have concerns about readiness or demonstrate inability, we have a College Preparatory Academy to fill gaps and prepare you for college coursework.

 * Q: What is a Academy Referral?
 * A: Students are referred to the Academy when they are lacking skills necessary to succeed. There are two types of referrals, general and subject. Subject referrals are issued when a department feels that you are unable to succeed in that subject. Each department chair has the authority to issue a subject referral upon consultation with faculty. General referrals are issued when a majority of a college's department chairs feel that you don't have the skills to succeed in college level coursework.

 * Q: Why was I referred to the Academy?
 * A: It was felt that you have knowledge gaps in an area that you need to address. Once those gaps are addressed, the referral will be cancelled.

 * Q: Can I take college courses while under referral to the Academy?
 * A: This depends on the referral. General referral students may not take college courses while the referral is outstanding. Subject referral students may take courses, provided they are in areas not mentioned in the referral.

 * Q:How do I get rid of the referral?
 * A:Referrals are corrective and not punitive. A college preparatory certificate will cancel any referral immediately. Once you demonstrate ability to succeed in a subject, departments will drop subject referrals. For general referrals, when you are able to succeed in a subject, the referral will be downgraded to a subject referral. Once you are able to succeed in all subjects, the referral will be cancelled.

FACULTY

 * Q: How do I receive a teaching appointment?
 * A: Appointments are made by the Department Chair in consultation with Department Faculty.

 * Q: Can I teach without a degree?
 * A: We welcome anyone who is willing and capable of teaching. Department Chairs can appoint anyone who is capable as a Adjunct Professor. Adjuncts teach under the supervision of a Professor and are restricted to the areas they are competent in.

 * Q: Are all faculty members qualified?
 * A: All faculty members are competent in the subject they teach.

 * Q: How much do you pay faculty?
 * A: We cannot afford to pay faculty members, as they are volunteers, and it would be impossible to have no tuition and salaried faculty. Faculty are paid in experience which increases their academic standing.

title: User:Geoff Plourde/College


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User:Eml4500.f08.ateam.nobrega/Homework Report Five

Proving Hookes Law through the Principle of Virtual Work
The equation $$ F = kd $$ can also be written as $$ kd - F = 0 $$ (Equation 3) or which is true for all values of W if $$ W(kd - F) = 0 $$ (Equation 4). This is a weak form of the Principle of Virtual Work. It is important to note that deriving Equation 4 from Equation 3 is a trivial solution, but deriving Equation 3 from Equation 4 is in fact Not trivial.

From the Force-Displacement relationship,

 * $$ F = kd $$

where F is a 6x1 matrix, k is a 6x6 matrix, and d is a 6x1 matrix, the Principle of Virtual Work can be derived as,

 * $$ W \cdot (kd - F) = 0 $$ (Equation 4)

This is true for all W. For this relationship to work W must be multiplied as a dot product, as 0 is not represented as a matrix. It is desired to show that Equation 3 can be derived from Equation 4 for all W.

To do this the first choice is to select W such that W1 = 1 and W2 through W6 = 0. W is originally defined as a 6x1 matrix but needs to be a 1x6 matrix so,

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

Equation 4 then becomes,

W \cdot (kd - F) = 1[\sum_{j=1}^6 k_{ij}d_{j} - F_{1}] + 0[\sum_{j=1}^6 k_{ij}d_{j} - F_{2}] + 0[\sum_{j=1}^6 k_{ij}d_{j} - F_{3}] + 0[\sum_{j=1}^6 k_{ij}d_{j} - F_{4}] + 0[\sum_{j=1}^6 k_{ij}d_{j} - F_{5}] + 0[\sum_{j=1}^6 k_{ij}d_{j} - F_{6}] = 0 $$

This becomes,

\sum_{j=1}^6 k_{ij}d_{j} - F_{1} = 0 $$

The second choice is to select 'W such that W1 equals 0, W2 equals 1, and W3 through W6 also equal 0. So,

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

Equation 4 then becomes,

W \cdot (kd - F) = 0[\sum_{j=1}^6 k_{ij}d_{j} - F_{1}] + 1[\sum_{j=1}^6 k_{ij}d_{j} - F_{2}] + 0[\sum_{j=1}^6 k_{ij}d_{j} - F_{3}] + 0[\sum_{j=1}^6 k_{ij}d_{j} - F_{4}] + 0[\sum_{j=1}^6 k_{ij}d_{j} - F_{5}] + 0[\sum_{j=1}^6 k_{ij}d_{j} - F_{6}] = 0 $$

This becomes,

\sum_{j=1}^6 k_{ij}d_{j} - F_{2} = 0 $$

This repeats as follows.

Choice 3,

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

Equation 4 then becomes,

W \cdot (kd - F) = 0[\sum_{j=1}^6 k_{ij}d_{j} - F_{1}] + 0[\sum_{j=1}^6 k_{ij}d_{j} - F_{2}] + 1[\sum_{j=1}^6 k_{ij}d_{j} - F_{3}] + 0[\sum_{j=1}^6 k_{ij}d_{j} - F_{4}] + 0[\sum_{j=1}^6 k_{ij}d_{j} - F_{5}] + 0[\sum_{j=1}^6 k_{ij}d_{j} - F_{6}] = 0 $$

This becomes,

\sum_{j=1}^6 k_{ij}d_{j} - F_{3} = 0 $$

Choice 4,

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

Equation 4 then becomes,

W \cdot (kd - F) = 0[\sum_{j=1}^6 k_{ij}d_{j} - F_{1}] + 0[\sum_{j=1}^6 k_{ij}d_{j} - F_{2}] + 0[\sum_{j=1}^6 k_{ij}d_{j} - F_{3}] + 1[\sum_{j=1}^6 k_{ij}d_{j} - F_{4}] + 0[\sum_{j=1}^6 k_{ij}d_{j} - F_{5}] + 0[\sum_{j=1}^6 k_{ij}d_{j} - F_{6}] = 0 $$

This becomes,

\sum_{j=1}^6 k_{ij}d_{j} - F_{4} = 0 $$

Choice 5,

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

Equation 4 then becomes,

W \cdot (kd - F) = 0[\sum_{j=1}^6 k_{ij}d_{j} - F_{1}] + 0[\sum_{j=1}^6 k_{ij}d_{j} - F_{2}] + 0[\sum_{j=1}^6 k_{ij}d_{j} - F_{3}] + 0[\sum_{j=1}^6 k_{ij}d_{j} - F_{4}] + 1[\sum_{j=1}^6 k_{ij}d_{j} - F_{5}] + 0[\sum_{j=1}^6 k_{ij}d_{j} - F_{6}] = 0 $$

This becomes,

\sum_{j=1}^6 k_{ij}d_{j} - F_{5} = 0 $$

Choice 6,

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

Equation 4 then becomes,

W \cdot (kd - F) = 0[\sum_{j=1}^6 k_{ij}d_{j} - F_{1}] + 0[\sum_{j=1}^6 k_{ij}d_{j} - F_{2}] + 0[\sum_{j=1}^6 k_{ij}d_{j} - F_{3}] + 0[\sum_{j=1}^6 k_{ij}d_{j} - F_{4}] + 0[\sum_{j=1}^6 k_{ij}d_{j} - F_{5}] + 1[\sum_{j=1}^6 k_{ij}d_{j} - F_{6}] = 0 $$

This becomes,

\sum_{j=1}^6 k_{ij}d_{j} - F_{6} = 0 $$

Hence $$ F = kd $$ (Equation 1).

To account for the boundary conditions of the two-bar truss where,

 * $$ d_{1} = d_{2} = d_{5} = d_{6} = 0 $$

the weighting coefficient must be "kinematically admissible" or cannot violate the boundary conditions. The weighting coefficients are the virtual displacements where,

 * $$ W_1 = W_2 = W_5 = W_6 = 0 $$

For all $$ W_3 $$ and $$ W_4 $$,

\begin{bmatrix} W_3 \\ W_4 \end{bmatrix} \cdot ( \begin{bmatrix}     k_{33} & k_{34} \\      k_{43} & k_{44} \end{bmatrix} \begin{bmatrix}      d_3 \\      d_4 \end{bmatrix} - \begin{bmatrix}      F_3 \\      F_4 \end{bmatrix} ) = 0 $$

Equation 1 can also be found through Equation 3 by adding F to both sides of the equation as shown below,

 * $$ kd - F = 0 $$

Adding F to both sides,

 * $$ kd - F + F = 0 + F $$

Here F cancels out on the left-hand side of the equation leaving,

 * $$ kd = F $$

The same derivation as shown above can be applied to the axial force displacement relationship. For example...

Deriving :$$ \mathbf k^{(e)} = \mathbf T^{(e)T} * \mathbf \hat k^{(e)} * \mathbf T^{(e)} $$

Recall Force Displacement relation with axial displacement degrees of freedom $$ \mathbf q^{(e)}$$

$$ \mathbf \hat k_{(2x2)}^{(e)} * \mathbf q_{(2x1)}^{(e)} = \mathbf p_{(2x1)}^{(e)} $$

$$ \rightarrow \mathbf \hat k^{(e)} * \mathbf q^{(e)} - \mathbf p^{(e)} = \mathbf 0 $$ : $$ (Eq.1) $$

$$ \mathbf \hat w_{(2x1)} * (\mathbf \hat k^{(e)} * \mathbf q^{(e)} - \mathbf p^{(e)})_{(2x1)} = 0 $$ for all $$ \mathbf \hat w_{(2x1)} $$ : $$ (Eq.2) $$

Recall: $$ \mathbf q_{(2x1)}^{(e)} = \mathbf T_{(2x4)}^{(e)} * \mathbf d_{(4x1)}^{(e)} $$ : $$ (Eq.3) $$

Similarly: $$ \mathbf \hat w_{(2x1)} = \mathbf T_{(2x4)}^{(e)} * \mathbf w_{(4x1)} $$ : $$ (Eq.4) $$

Note: $$ \mathbf \hat w_{(2x1)} $$ is the equivalent of the axial virtual displacement

If one were to apply all the possible $$ \mathbf \hat w $$ one derive the axial force displacement relationship as shown below...

$$ \mathbf \hat k_{(2x2)}^{(e)} * \mathbf q_{(2x1)}^{(e)} = \mathbf p_{(2x1)}^{(e)} $$

^Everything should go above this. This was the last day of notes

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

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

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

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

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

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

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

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

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

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

A and B are given as the following matices

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

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

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

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

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

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

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

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

This simplifies equation 2 into:

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

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

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

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

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

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

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

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

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

Proving the principle of virtual work.

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

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

Composite Materials
Composite materials consist of a strengthening fiber and matrix substrate. The addition of these fibers to a substrate is termed 'doping'. Due to the mixed material properties, composite materials often contain a varying elastic modulus. In other words, the Young's modulus is a function of position in the material.

Some examples of composite materials with this characteristic:

 * Carbon fiber
 * Fiber Glass

The Elastic Modulus is much higher in these materials in the fiber direction (longitudinal direction). However, perpendicular to the fibers (transverse direction) the Elastic Modulus is smaller as the fibers lose their strength in shear.

Free body diagram of a rigid body cross section

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

$$ \sum{F_x}=0=\frac{\delta N}{\delta x}(x,t)dx+H.O.T.+f(x,t)dx-m(x)\ddot{u}dx $$

Cancel dx and neglect the higher order terms (H.O.T.)

(Eqn. 1): $$ \sum{F_x}=0=\frac{\delta N}{\delta x}(x,t)+f(x,t)-m(x)\ddot{u} $$

Recall Taylor Series Expansion:

$$f(x+dx)=f(x)+\frac{df(x)}{dx}dx+[\frac{d^2f(x)}{2dx^2}+...] $$

Where the content inside the brackets represent higher order terms.

(Eqn. 2):  $$\frac{\delta N}{\delta x}+ f =m\ddot{u} $$

Where Equation 2 represents the equation of motion(EOM) of the elastic bar.

$$ N(x,t)=A(x)\sigma(x,t) $$ where: $$ \sigma(x,t)=E(x)\epsilon (x,t)$$ and: $$ \epsilon (x,t)=\frac{\delta u}{\delta x }(x,t)$$ (Eqn. 3)

Substituting Eqn. 3 into Eqn. 2:

$$ \frac{\delta }{\delta x}(A(x)E(x)\frac{\delta u}{\delta x})+f(x,t)=m(x)\ddot{u} $$, where: $$ \ddot{u}=\frac{\delta ^2u}{\delta t^2}$$

Partial Differential Equation of Motion
In order to integrate differential equations of motion, boundary conditions are needed to evaluate the constants of integration. For the problem above, this translates to two initial conditions (due to the 2nd order nature of the equations).

Boundary Conditions:

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

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

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

TWO-BAR Truss De-Bugged
Below is the debugged version of the two bar truss example. The problem was found at the end of the code for the results matrix. Initially the code did not take into account that the elements had different cross section areas (A) and Young's modulus of elasticy (e). This was corrected in the for statement and we were able to obtain the correct results.

SIX BAR Truss Plot

Derivation of Matrix Relations and Matrices
Section 2.4 in the Fundamental Finite Element Analysis and Applications textbook is concerned with space trusses. We can use our FEA knowledge and use the 2D plane truss concepts to derive expressions for a 3D space truss. It is important to note that the FD relations remain the same for each element.

 * $$\begin{bmatrix}

F_{1}\\ F_{2} \end{bmatrix} = \begin{bmatrix} k & -k\\ -k & k \end{bmatrix} \begin{bmatrix} d_{1}\\ d_{2} \end{bmatrix} = \begin{bmatrix} P_{1}\\ P_{2} \end{bmatrix} $$

The figure above depictsthe local and global coordinates for an axially loaded bar in three dimensions. The nodal degrees of freedom are u1, v1, w1, u2, v2 and w2.

The transofrmation betwwen local and global degrees of freedom:

d=Td_{l} $$

 * $$\begin{bmatrix}

d_{1}\\ d_{2} \end{bmatrix} = \begin{bmatrix} l_{s} & m_{s} & n_{s} & 0 & 0 & 0\\ 0 & 0 & 0 & l_{s} & m_{s} & n_{s} \end{bmatrix} \begin{bmatrix} u_{1}\\ v_{1}\\ w_{1}\\ u_{2}\\ v_{2}\\ w_{2} \end{bmatrix} $$

$$ l_{s}=\frac{x_{2}-x_{1}}{L}::: m_{s}=\frac{y_{2}-y_{1}}{L}::: l_{s}=\frac{z_{2}-z_{1}}{L} $$

$$ L=((x_{2}-x_{1})^{2} + (y_{2}-y_{1})^{2} + (z_{2}-z_{1})^{2}))^{0.5} $$

Now we must transform the elemental equations in local coordinates to the global coordinates relation. $$ k_{l}d_{l}=r_{l} $$ or $$ k_{l}Td=r_{l} $$ Mutliply by the transpose of T:

$$ T^{T}k_{l}Td=T^{T}r_{l} $$ Which simplifies to: kd=r

\frac{EA}{L} \begin{bmatrix} l_{s}^{2} & m_{s}l_{s} & n_{s}l_{s} & -l_{s}^{2} & -m_{s}l_{s} & -n_{s}l_{s}\\ m_{s}l_{s} & m_{s}^{2} & m_{s}n_{s} & -m_{s}l_{s} & -m_{s}^{2} & -m_{s}n_{s}\\ n_{s}l_{s} & m_{s}n_{s} & n_{s}^{2} & -n_{s}l_{s} & -m_{s}n_{s} & -n_{s}^{2}\\ -l_{s}^{2} & -m_{s}l_{s} & -n_{s}l_{s} & l_{s}^{2} & m_{s}l_{s} & n_{s}l_{s}\\ -m_{s}l_{s} & -m_{s}^{2} & -m_{s}n_{s} & m_{s}l_{s} & m_{s}^{2} & m_{s}n_{s}\\ -n_{s}l_{s} & -m_{s}n_{s} & -n_{s}^{2} & n_{s}l_{s} & m_{s}n_{s} & n_{s}^{2}\\ \end{bmatrix} \begin{bmatrix} u_{1}\\ v_{1}\\ w_{1}\\ u_{2}\\ v_{2}\\ w_{2}\\ \end{bmatrix} = \begin{bmatrix} F_{1x}\\ F_{1y}\\ F_{1z}\\ F_{2x}\\ F_{2y}\\ F_{2z}\\ \end{bmatrix} $$

If we assume that the concentrated loads are added directly to the global equations at the beginning of the assembly, then the F matrix is simply a 1x6 matrix of zeros. Using this knowledge we may now compute the nodal displacemnets. The axial displacments can be computed by transforming the global displacemnts into local coordinates. For example:
 * $$\begin{bmatrix}

d_{1}\\ d_{2} \end{bmatrix} = \begin{bmatrix} l_{s} & m_{s} & n_{s} & 0 & 0 & 0\\ 0 & 0 & 0 & l_{s} & m_{s} & n_{s} \end{bmatrix} \begin{bmatrix} u_{1}\\ v_{1}\\ w_{1}\\ u_{2}\\ v_{2}\\ w_{2} \end{bmatrix} $$ Therefore the axial displacement over the element is simply: $$ u(s)=(1-\frac{s}{L}\frac{s}{L})\begin{bmatrix} d_{1}\\ d_{2} \end{bmatrix} $$ This relation is true for s values greater than or equal to zero and less than or equal to L. Note that the axial strain can easily be computed by taking the derivative of this function with respect to time. Yielding: $$ strain=(-d_{1}+ d_{2})\frac{1}{L} $$

Contributing Team Members
Eml4500.f08.ateam.nobrega 01:01, 7 November 2008 (UTC)

Eml4500.f08.ateam.shah 04:41, 7 November 2008 (UTC)

Eml4500.f08.ateam.mcnally 04:45, 7 November 2008 (UTC)

Eml4500.f08.ateam.boggs.t 05:18, 7 November 2008 (UTC)

Eml4500.f08.ateam.carr 17:35, 7 November 2008 (UTC)

Eml4500.f08.ateam.rivero 17:49, 7 November 2008 (UTC)


text: File:Sixbartruss.jpg

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User:EML4500.f08.qwiki.booher

Wiki Website:

http://en.wikiversity.org/wiki/User:Eml4500.f08.qwiki.bishop

title: User:EML4500.f08.qwiki.booher


text: File:N EQ.JPG

title: File:N EQ.JPG


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title: File:Fixed bar.JPG


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text:
Brezhoneg Daoù/Lesson 7

{| cellpadding="10" cellspacing="5" style="width: 99%; background-color: inherit; margin-left: auto; margin-right: auto"
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Mutations
'''This lesson gives a full and rather complex view of mutations as exceptions are described together with the regular cases. For a more straightforward explanation please refer to the lesson in Brezhoneg Unan/Lesson 4'''

Definition
A mutation is the change of the first letter of a word according to different contexts.

Exemples : Mamm (mother) --> Ar vamm (the mother) Kozh (old)   --> Mamm gozh (Grand mother)

Table of mutations
Mutations in Breton obey to regular phonetics laws.

They are summarized in the table below (details to be found in the rest of this page) :

 * }

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Mutations to soften ("dre vlotaat")
Soft mutation happens in the following cases :

 * after a, da, daou, div, dindan, diwar, dre, e (possessive), eme, en em, en ur, hanter, holl, na, ne, pa, pe, re, war...
 * after the article for the feminine singular words and the related adjective
 * after the article for the masculine plural person names (except when the plural is in -où) and the related adjective

Please note :

 * D will not mute after names
 * K becomes C'H for names after the article when there is no other mutation
 * P, T, K will not mute for adjectives when the related names ends by another letter than L, M, N, R or a vowel.

Soft Mutation will be noted (S!) Following letters are affected: B    becomes	V D    becomes	Z K    becomes	G G    becomes	C'H GW   becomes	W P    becomes	B M    becomes	V T    becomes	D

Examples :

 * Breur (brother)      gives     Da vreur
 * Dant  (tooth)        gives     Da zant
 * Glin  (kneel)        gives     Da c'hlin

Mutations after the article
Here is a summary for mutations after the article :

Here is a presentation of the rules regarding the mutations after the aticle :

Go to the Quiz

Mutations of the adjective epithet
Here is a summary for mutations of the adjective epithet :

Here is a presentation of the rules regarding the mutations of the adjective epithet :

Go to the Quiz

 * }

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Mutations to harden ("dre galetaat")
Hard mutation happens in the following cases :

 * Possessive "ho" (yours)
 * After "az, "ez", "'z" (you - sg.)
 * After "da'z ..." (to yours ...)
 * After "ez" (in yours ...)

Hard Mutation will be noted (H!) Following letters are affected: B    becomes	P D    becomes	T G    becomes	K

Examples :

 * Breur (brother)      gives     Ho preur
 * Dant  (tooth)        gives     Ho tant
 * Glin  (kneel)        gives     Ho klin

 * }

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Mixed mutations
Mixed mutation happens in the following cases :

 * Present participe
 * Formed by "o" + Verb.

 * After verb particle "e"
 * After "ma" (That)

Mixed Mutation will be noted (M!) Following letters are affected: B    becomes	V D    becomes	T G    becomes	C'H GW   becomes	W M    becomes	V

Examples :

 * Deskiñ (to learn)    gives     O teskiñ
 * Gortoz (to wait)     gives     O c'hortoz

 * }

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Spirant mutations
Spirant mutation happens in the following cases :

 * After "ma" or "va" (mine)
 * After "he" (Hers)
 * After "o" (Theirs)
 * After numbers : tri, teir, pevar, peder, nav

Spirant Mutation will be noted (Sp!) Following letters are affected: P    becomes	F T    becomes	Z K    becomes	C'H

Examples :

 * Penn  (head)         gives     Va fenn
 * Kador (chair)        gives     Va c'hador

 * }

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Bloaz
Bloaz mutes into Vloaz after :

 * Pet ?
 * The cardinal numbers except after tri, pevar, pemp, nav

Exemple :

 * Pet vloaz eo bremañ ? Daouzek vloaz.

 * Pet vloaz eo bremañ ? Pemp bloaz.

Mintin
Mintin mutes into Vintin after :

 * Weekday names
 * Dec'h
 * Warc'hoazh

Exemple :

Dilun vintin, warc'hoazh vintin.

 * }

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Exercises
Go to the Quiz

 * }

title: Brezhoneg Daoù/Lesson 7


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

=6-Bar Truss: Case 1=

For the first case, all of the values for the modulus of elasticity and cross sectional area are the same.

The following code was compiled to produce the plot for the deformed and undeformed six bar truss.

=6-Bar Truss: Case 2=

For the second case, all of the values for the modulus of elasticity are different, while the values for the cross sectional area remain the same. The values for the different moduli can be found in the code below.

The following code was compiled to produce the plot for the deformed and undeformed six bar truss, including the deformed truss with the varying moduli of elasticity.

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


title: File:SP1.JPG


text:
User:Matthew M/Week 3 - Social Self

Week 3 - 05/07/08 – Social Self
A common question when asked to me which leaves me stunned and speechless at times would be ‘who are you really’? How do we as people define who we are, what makes us what we are? This generally confuses me and I utter words which describe what I like, what I do, how I am, what my personality is like, but I was never sure as to whether this actually answered the question at all. However, this weeks topic of the Social Self interested me greatly and helped me understand a little bit more. The question of ‘What is self?’ according to the textbook states that ‘self’ is generally combined of three separate parts. These three parts are; 1) Self-Knowledge or self-concept – this is the beliefs that people hold about themselves, 2) The interpersonal self of public self – this second part of self is related to people being social animals. This part helps people connect socially to other people and is the image of the person which is viewed by others, & 3) The agent or executive function – the last part of self is directed towards the actions which people take to do things. This part of the self helps a person to make choices in their life as well as hold control, not only over themselves, but others as well. After that the question then became, ‘what is the social self?”. According to the lecture notes the social self can be defined as the interpersonal or social self that engages face-to-face in social relations to other people. It appears that us as humans are all social animals and are group-based creatures. At times it even seems that the aim of our self is merely to become socially accepted and act ourselves out in social roles. We seem to do this in everyday life when changing who we are to become more accepted in a group or when we do things merely to please other people. It seems that throughout our everyday lives people underestimate how much our own behaviour will affect others and also how much we are affected by other peoples behaviour whether we like to believe it or not!!!

title: User:Matthew M/Week 3 - Social Self


text:
User:Matthew M/Week 4 - Social Thinking

Week 4 - 12/08/08 – Social Thinking
The subject of social psychology was initially influenced through behaviourism in the 1039’s to the 1950’s; however it was in the 1970’s when cognitive psychology increased the discussion and study of social thinking. What seemed to interest me this week was the topic under social cognition about the cognitive miser. Social Cognition is the study of thoughts about people and the social relationships which occur. This study seemed inevitable as it seems that humans tend to think about how other humans think a lot of the time. This is once again due to the fact that people are social animals and tend to always look for a way to be socially accepted. In order to become socially accepted we need to know what other people think of us and without that knowledge, it would seem like an impossible task. Human beings have the ability to think better than most other creatures on this earth and you would think that people would want to do something with this ability. However it could not be further from the truth. This is the part I found interesting, the fact that people always tend to be lazy or careless about their thinking. This is where the term ‘cognitive miser’ comes into play, which defines the human’s defiance to doing any unnecessary extra thinking. The belief is that we as people will automatically attempt to use as little cognitive effort at all so that we can save that effort for the things that we choose to think about. We generally avoid thinking too much or too hard. The human brain uses certain processing tools in order to help us think and some of these include stereotypes, schemas and scripts. Stereotypes are the beliefs on actions and traits which people automatically have when associating them with a particular group. Schemas are defined in the textbook as “knowledge structures that represent substantial information about a concept, its attributes and its relations to other concepts”. Lastly Scripts are a process in the human mind which consists of knowledge structures stored in the brain which will define certain situations as well as guide the behaviour of people.

title: User:Matthew M/Week 4 - Social Thinking


text:
User:Matthew M/Week 5 - Aggression

Week 5 - 19/08/08 – Aggression
This topic was particularly interesting to me and so far has stood out the most. With people I know, I have always thought questions as to why they seem to get so angry so easily towards other people when out, and at times even display signs of hostile aggression when not necessary. Aggression itself means the intent to harm another person when that person themselves in motivated to avoid that harm. However the behaviour which is classified as aggressive may vary depending upon cultures. Generally actions considered as antisocial behaviour in society can be either one which damages relationships, or against the norms of the culture. There are various beliefs about ‘aggressions’ some which believe that aggression is something which is classified as a learned behaviour where other believe that it is something that we are born with. Freud was one which believed that people are born with an innate instinct to make them behave in an aggressive manner, whereas on the other hand, the social learning theory seems to suggest that it is not an innate instinct but it is something we learn through the course of our lives. There seem to be different reasoning behind why aggression occurs and one which seemed to stand out to me was the frustration-aggression hypothesis. This hypothesis believes that people, when they are frustrated will always lead to them becoming aggressive and on the same line that when people are aggressive, there must always be some source of frustration involved. Even so according to the textbook studies have shown that frustration is no the whole cause of aggression. It was proposed that all states of negative effect can have an influence on aggression, factors such as the person being in an aversive emotional state or merely anger. An interesting thing which I found out was the correlation between age and aggression. Although younger children don’t seem to commit as many violent crimes as older people information in the textbook found that aggressive human beings are generally younger children. A study found that with interactions amongst children at a day-care setting 25% seemed to involve some form of physical aggression. Even adults or violent criminals don’t tend to resort to violence with 25% of their interactions!!!! Another interesting fact was that although media tends to show violence from men to women in relationships it is actually women which attack their relationship partner more often however, they don’t tend to cause as much harm. A chemical factor also seems to have a great influence on aggression and that is if increased levels of testosterone are involved along with alcohol there seems to be higher levels of aggressions, however if serotonin is decreased in the body along with an increase in vitamins it will help to reduce aggression. Luckily although it seems that mankind is doomed for a violent end, the cultures within society are what help people to restrain from violence and aggression.

title: User:Matthew M/Week 5 - Aggression


text:
User:Matthew M/Week 6 - Prejudice

Week 6 - 26/08/08 – Prejudice
This journal is getting slightly confusing since I seem to be jumping from place to place and being a week behind due to there being no tutorial in the first week. Anyway just a short bit on prejudice. This can be defined as a negative feeling toward an individual based on his or her membership in a particular group. This can range from things such as stereotyping people, to discrimination, social categorisation and racism. With things such as these there generally exists in the persons mind a difference between people who are like them and those which are not. This can be seen as the person’s in-group and out-group members. People’s in-group members generally consist of people who belong to the same group or categorisation as they do and in the case of out-group members it consists of people who are in different categorisations and groups. Common prejudices which occur are between races and cultures or even with people being intolerant of homosexuality within their own gender. It seems that one of the ways in prejudice can exist is due to in-group favouritism where people tend to look more favourably on others which are in their own group, and not so favourably on those which are not. Another way it exists is because people like to blame their misfortunes or problems on other people in the out-group. Even so changing prejudices does seem to be possible as long as people make conscious efforts to do so. It seems that people to hold prejudices are at time unaware of the impact it may have on the other people. People if stereotyped may conform to that stereotype in the belief that, that is the way in which they are meant to act, and on the other hand people may purposely act differently in order not to conform to stereotypes.

title: User:Matthew M/Week 6 - Prejudice


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User:Matthew M/Week 10 - Relationships

Week 10 - 23/09/08 – Relationships
The topics which were explored in this week were interesting and spoke of topics such as the need to belong, attraction, relationships and love. It seemed to be about things which every person will have to deal with at some point in their life. People are always wishing to belong and the term social acceptance seems to perfectly fit this wish. This term mean getting other people to like, respect, approve and want to have some form of relationship with you. An interesting theory which I found week was the social exchange theory. The belief that weighs social behaviour through costs and benefits of interaction such as the assumption that sex is a resource that women have and men want. In this theory it states that men try to provide resources such as money, attention, and respect or love in order to gain sex. This theory merely places an economic perspective on sex with a women’s sexuality as the supply and the men’s sexuality being the demand. The main idea here is that relationships are like an investment where we generally put the most effort into social situations where there is the probability of maximal benefits and minimal costs.

title: User:Matthew M/Week 10 - Relationships


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User:Matthew M/Week 11 - Groups

Week 11 - 30/09/08 – Groups
This week’s reading was on groups. This is a topic which personally I did not find as interesting as any of the others. Culture is usually a large factor in the forming of groups. This is topic which is greatly studied and in the evolution theory, the benefits of forming groups included; safety in numbers, group members can help each other to find food, and help to accomplish tasks in groups which would be rather difficult for a lone person to do. Groups are also good as when in cultural factions, they help to pass along information as well as experience which is learnt amongst the generations in order to make life easier for the younger ones. It is common in society for larger groups to have a hierarchy of power with leaders who have more power to hold control over the group. This can be seen if looking at a larger perspective like the president of the USA.

title: User:Matthew M/Week 11 - Groups


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User:Matthew M/Week 12 - Pro-Social

Week 12 - 07/10/08 – Pro-Social
The textbook defines pro-social behaviour as behaviours which involve doing good for others or society and build on relationships and allows society to function. The definition can also involve obeying rules, conforming to societal norms and cooperation and help which are all forms of pro-social behaviour. The definition of pro-social behaviour is not only behaviour which affects individuals and those around them but the whole of society. Pro-social behaviours also include things such as forgiveness, obedience, conformity, helping others and more. An interesting theory which popped up however was the kin selection theory. This theory is about the evolutionary tendency to help those which have the same genes. It is said that people are more likely to help a sibling over a nephew or a cousin, both of which have a lower level of the same genes. According to the textbook it has been shown that whether it is in life or death situations or everyday life, people tend to help family members and their closest relatives over other people. There are various reasons as to why people help others and one reason is altruistic helping which is motivated by empathy an emotional response which is selfless and wants only to reduce others distress. Another completely opposite reason is egoistic helping which is motivated only to reduce personal distress without consideration of others.

title: User:Matthew M/Week 12 - Pro-Social


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User:Matthew M/Week 13 - Environmental

Week 13 - 15/10/08 – Environmental
This was an interesting week as I had never previously even heard of environmental psychology. This topic ‘environmental psychology’ is the study of interactions and relations between people and their environments. The study of this topic seem to have close ties to environmental social sciences, ecological psychology, socio-architecture, environmental design research, behavioural geography and many more. This research started as the investigation of how the environment manages to affect a humans thought feelings and behaviours, however now it has completely changed with investigating how the actions of people manage to affect their environment. The way in which the environment can affect people I have never thought of before. It appears that nature based experiences with animals, nature scenes, and adventure therapy can all have a positive healthy effect on people. A fact I found rather interesting which I read in the lecture notes was that environmental problems are not only the concern of environmental scientists but of psychologists as well because environmental problems are primarily cause by human’s behaviours, feelings and attitudes. It seems that environment has such a great effect on people that with a nature scene for hospital rooms, it can benefit a patient’s recovery. This seems to be an interesting branch of psychology as well as a vital one since the fact that people can so greatly effect the environment, this study must continue in order to ensure that the environment will carry on.

title: User:Matthew M/Week 13 - Environmental


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Talk:Solar System, interactive/Earth/Solar System overview

why is uraus blue then in some photos it is green


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User talk:Intransient

Well done
Good job on setting up your account,. Let me know if I can do anything as you go along. 'Official' Wikiversity welcome below. -- Jtneill - Talk - c 22:08, 2 November 2008 (UTC)

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And don't forget to explore Wikiversity with the links to your left. Be bold to contribute and to experiment with the sandbox or your userpage, and see you around Wikiversity! Jtneill - Talk - c 22:08, 2 November 2008 (UTC)

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title: User talk:Intransient


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User:Eml 4500.f08.delta 6.specht/HW5

 Note: Derivation of 3-D bar element missing.

Cheating: Modified the "results" array for the 2-bar truss after I pointed out the problem in my lecture, without any change in the code and without putting a comment box to alert the readers; here is the comparison between the official version before the deadline and the cheated version. See my comments below.

Please don't remove this comment in case you amend your work; just add a comment box to say what you did. Eml4500.f08 12:20, 11 November 2008 (UTC)

=Homework 5=

Justification
We can justify the elimination of rows 1,2,5,6 to obtain $$\bar{\mathbf{K}}$$ in a 2-bar truss.

 * We begin with the FD relationship, writing it as $$K$$$$d$$ = $$F$$. Based on the model 2-bar system, we can generalize this method for any kind of application.

 * $$K_{6x6}$$$$d_{6x1}$$ = $$F_{6x1}$$

Therefore,
 * $$K$$$$d$$-$$F$$=$$0_{6x1}$$

 * Principle of Virtual Work:

 * $$w$$$$_{6x1}$$•($$K$$$$d$$-$$F$$)$$_{6x1}$$ = $$0_{1x1}$$; for all $$w$$$$_{6x1}$$

Where w is the weighting matrix, and the equation is equalized to zero as a sacalar (or 1x1 matrix). In other words, the weighting matix is dotted with the FD relationship matrix to obtain a scalar. Also, according to the wise Prof. Vu-Quoc, "$$w$$ was not selected out of frivolity," as it is called the weighting coefficient or matrix.We say that the FD relationship is equivalent to the PVW.

Proof of FD relationship and PVW equality

 * 1) FD relationship to PVW is trivial because if FD relationship is true, then $$w$$(0)=0, this being obvious
 * 2) We now want to show how to go from PVW to FD relationship.

 * We have the PVW: $$w$$$$_{6x1}$$•($$K$$$$d$$-$$F$$)$$_{6x1}$$ = $$0_{1x1}$$; for all $$w$$$$_{6x1}$$

 * Choice 1: Select $$w$$ such that $$w_{1}$$=1, $$w_{2}$$=...=$$w_{6}$$=0

 * Therefore,
 * $$w^{T}_{6x1}$$=$$\begin{bmatrix}

1 & 0 & 0 & 0 & 0 & 0 \end{bmatrix}$$
 * $$w$$•($$K$$$$d$$-$$F$$) = 1•$$\left[{\sum_{j=1}^{6}K_{1j}d_{j}-F_1} \right]$$+0•$$\left[{\sum_{j=1}^{6}K_{2j}d_{j}-F_2} \right]$$+...+0•$$\left[{\sum_{j=1}^{6}K_{6j}d_{j}-F_6} \right]$$=0
 * $$\sum_{j=1}^{6}{K_{1j}d_j}=F_1$$ (1$$^{st}$$equation)

 * Choice 2: Select $$w$$ such that $$w_{1}$$=0, $$w_{2}$$=1, $$w_{3}$$=...=$$w_{6}$$=0

 * Therefore,
 * $$w^{T}_{6x1}$$=$$\begin{bmatrix}

0 & 1 & 0 & 0 & 0 & 0 \end{bmatrix}$$
 * $$w$$•($$K$$$$d$$-$$F$$) = 0•$$\left[{\sum_{j=1}^{6}K_{1j}d_{j}-F_1} \right]$$+1•$$\left[{\sum_{j=1}^{6}K_{2j}d_{j}-F_2} \right]$$+0•$$\left[{\sum_{j=1}^{6}K_{3j}d_{j}-F_3} \right]$$+...=0
 * $$\sum_{j=1}^{6}{K_{2j}d_j}=F_2$$ (2$$^{nd}$$equation)

 * Choices 3 through 6 repeat this pattern.

Ultimately, we will have achieved the FD relationship.

PVW: Furthermore
According to the boundary conditions (BCs) of the 2-bar truss, $$d_1=d_2=d_5=d_6=0$$.

The weighting cofficient must be kinematically admissible. It cannot violate the BCs: $$w_1=w_2=w_5=w_6=0$$. Therefore, weighting coefficients go hand-in-hand with virtual displacements, as explained by calculus of variations.

For the 2-bar truss at hand, it is evident that $$w$$•($$K$$$$d$$-$$F$$) can be reduced to a rectangular matrix multiplied by a 2x1 matrix with a 6x1 matrix as a result.

 * $$w$$•($$K$$$$d$$-$$F$$)=$$\begin{bmatrix}

w_3\\ w_2 \end{bmatrix}$$•($$\bar{K}_{2x2}$$$$\bar{d}_{2x1}$$-$$\bar{F}_{2x1}$$)=0

Where,
 * $$\bar{K}$$=$$\begin{bmatrix}

K_{33} & K_{34}\\ K_{43} & K_{44} \end{bmatrix}$$
 * $$\bar{d}$$=$$\begin{Bmatrix}

d_3\\ d_4 \end{Bmatrix}$$
 * $$\bar{F}$$=$$\begin{Bmatrix}

F_3\\ F_4 \end{Bmatrix}$$

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

w = weighting coefficient

Now an Example:

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

The formula to calculate the final grade is given as:

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

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

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

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

The final grade now becomes:

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

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

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

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

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

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

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

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

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

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

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

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

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

Proving the Coordinate Systems Relationship
For this proof the following variables should be defined. $$q_{2x1}^{(e)}$$ is given as the real axial displacement. The variable $$\hat{W}_{2x1}$$ is the virtual axial displacement that corresponds to the axial displacement $$q_{2x1}^{(e)}$$. $$W_{4x1}d^{(e)}$$ is the virtual axial displacement in the global coordinate system that corresponds to the real displacement in global coordinates ($$d^{(e)}$$).

The equations for the real axial displacement and the virtual axial displacement are as follows:

$$ q_{2x1}^{(e)}=T_{2x4}^{(e)}d_{4x1}^{(e)} $$

$$ \hat{W}_{2x1}=T_{2x4}^{(e)}W_{4x1} $$ The equation for the axial force can be rearranged and multiplied by the weighting factor, which is the virtual axial displacement vector, in order to obtain:

$$ \hat{W}_{2x1}\cdot \left(\hat{k}_{2x2}^{(e)}q_{2x1}^{(e)}-p_{2x1}^{(e)} \right)=0_{1x1} $$

Once the equations for the real axial displacement and the virtual axial displacement are substituted into the equation given above, the following relation is obtained.

$$ \left( T^{(e)}W\right)\cdot \left[\hat{k}_{2x2}^{(e)}\left(T^{(e)}d^{(e)} \right)-p_{2x1}^{(e)} \right]=0 $$ This relation is true for all $$W_{4x1}$$. This equation can be further manipulated in order to eliminate the dot product that is present. The form of the equation obtained is below.

$$ \left( T^{(e)}W\right)^{T}\left[\hat{k}_{2x2}^{(e)}\left(T^{(e)}d^{(e)} \right)-p_{2x1}^{(e)} \right]=0_{1x1} W_{4x1} $$

By applying the relation below to the equation that is listed above, the equation above can be simplified to:

$$ W^{T}T^{(e)T}\left[\hat{k}_{2x2}^{(e)}\left(T^{(e)}d^{(e)} \right)-p_{2x1}^{(e)} \right]=0_{1x1} $$

This relation is obtained by using the following substitution for the dot product.

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

This same relation can be applied again in reverse once the transpose of the T matrix is distributed through the matrix containing the axial force-displacement relationship.

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

From this recognize that the value $$T^{(e)T}\hat{k}^{(e)}T^{(e)} $$ is equal to the element stiffness matrix, $$k^{(e)}$$, and that the value $$T^{(e)T}p^{(e)} $$ is equal to the element force vector, $$f^{(e)}$$. This relation can then be simplified to the force-displacement relation in global coordinates.

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

Since this is true for all $$ W_{4x1}$$, the original force-displacement,$$ k^{(e)}d^{(e)}= f^{(e)} $$, relation can be obtained. This is the second and more powerful method to prove the following relation that is used during this proof.

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

FEA and the Mechanics of Materials

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

$$ = \frac{dN}{dx}(x, t)dx + h.o.t. + f(x, t)dx + m(x)\ddot{u} $$

In the above equation, we can neglect the h.o.t. (higher order terms). If we recall Taylor Series Expansion:

$$ f(x+dx) = f(x) + \frac{df(x)}{dx}dx + \frac{1}{2}\frac{d^{2}f(x)}{dx^{2}} + ... $$

The term with the 1/2 coefficient and everything after that term are know as higher order terms. Equation (1) now turns into:

$$ \frac{dN}{dx} + f = m\ddot{u}\; \; \; \; \; \;\; \; \;\; \; \;Eq. (2) $$

This equation is also know as the equation of motion.

$$ N(x, t) = A(x)\sigma (x, t)\; \; \; \; \; \; \; \; \; \; \; \; Eq. (3) $$

$$ \sigma (x, t) = E(x)\varepsilon (x, t) $$

$$ \varepsilon (x, t) = \frac{du}{dx}(x, t) $$

The equation above is known as the Constitutive Relation. Now, if we substitute Equation (3) into Equation (2):

$$ \frac{d}{dx}[N(x, t) = A(x)\sigma (x, t)] + f(x, t) = m(x)\ddot{u} $$

This is know as the Partial Differential Equation of Motion. We need two boundary conditions, since this is a second order derivative with respect to x. These two conditions are the initial displacements and initial velocities.

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

$$ N(L, t) = A(L)\sigma (L, t) $$

$$ \sigma (L, t) = E(L)\varepsilon (L, t) $$

$$ \varepsilon (L, t) = \frac{du}{dx}(L, t) $$

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

Chapter 2: Elastic Bar Subject to Variable Properties and Loads
Up to this point the case of a discrete, non-continuous partial differential equation (PDE) has been discussed. The new material is concerning continuous PDEs. The problem that will be used to examine this type of analysis is an elastic bar that has a varying cross sectional area,$$A(x)$$, and modulus of elasticity, $$E(x)$$. This bar will be subject to varying axial loads (distributed loads), time dependent concentrated loads ($$f(x,t)$$), and dynamic inertial forces. An image of this case is shown below with the distributed load shown on the axis of the elastic bar.

Axial FD Relation
$$ k^{(e)}_{6x6}d^{(e)}_{6x1}=f^{(e)}_{6x1} $$

Element Displacements
$$ d^{(e)}=\begin{bmatrix} d_{1}^{(e)}\\ d_{2}^{(e)}\\ d_{3}^{(e)}\\ d_{4}^{(e)}\\ d_{5}^{(e)}\\ d_{6}^{(e)} \end{bmatrix} $$

Element Forces
$$ f^{(e)}=\begin{bmatrix} f_{1}^{(e)}\\ f_{2}^{(e)}\\ f_{3}^{(e)}\\ f_{4}^{(e)}\\ f_{5}^{(e)}\\ f_{6}^{(e)} \end{bmatrix} $$

Relation Between Axial DOF's and Element DOF's in Global Coordinates
$$ q^{(e)}=\begin{pmatrix} q_{1}^{(e)}\\ q_{2}^{(e)} \end{pmatrix}= \begin{bmatrix} l^{(e)} & m^{(e)} & n^{(e)} & 0 & 0 & 0\\ 0 & 0 & 0 & l^{(e)} & m^{(e)}  & n^{(e)} \end{bmatrix}\begin{pmatrix} d_{1}^{(e)}\\ d_{2}^{(e)}\\ d_{3}^{(e)}\\ d_{4}^{(e)}\\ d_{5}^{(e)}\\ d_{6}^{(e)} \end{pmatrix} =T_{2x6}^{(e)} d_{6x1}^{(e)} $$

Relation Between Axial Forces and Element Forces in Global Coordinates
$$ P^{(e)}=\begin{pmatrix} P_{1}^{(e)}\\ P_{2}^{(e)} \end{pmatrix}= \begin{bmatrix} l^{(e)} & m^{(e)} & n^{(e)} & 0 & 0 & 0\\ 0 & 0 & 0 & l^{(e)} & m^{(e)}  & n^{(e)} \end{bmatrix}\begin{pmatrix} f_{1}^{(e)}\\ f_{2}^{(e)}\\ f_{3}^{(e)}\\ f_{4}^{(e)}\\ f_{5}^{(e)}\\ f_{6}^{(e)} \end{pmatrix} =T_{2x6}^{(e)} f_{6x1}^{(e)} $$

Element Stiffness Matrix in Global Coordinates in Terms of Axial DOF's
$$ k^{(e)}= \frac{EA}{L} $$

$$ \hat{k}^{(e)}_{2x2}=k^{(e)}\begin{bmatrix} 1& -1 \\ -1 & 1 \end{bmatrix}=\frac{EA}{L}\begin{bmatrix} 1 & -1 \\ -1 & 1 \end{bmatrix} $$

$$ k^{(e)}_{6x6}=T^{(e)T}_{6x2}\hat{k}^{(e)}_{2x2}T^{(e)}_{2x6}=\frac{EA}{L}\begin{bmatrix} l^{(e)} & 0\\ m^{(e)} & 0\\ n^{(e)} & 0\\ 0 & l^{(e)}\\ 0& m^{(e)}\\ 0 & n^{(e)} \end{bmatrix}\begin{bmatrix} 1 & -1 \\ -1 & 1 \end{bmatrix}\begin{bmatrix} l^{(e)} &m^{(e)} & n^{(e)} & 0 & 0 & 0\\ 0& 0 & 0 & l^{(e)} & m^{(e)} & n^{(e)} \end{bmatrix} $$

$$ k^{(e)}_{6x6}=\frac{EA}{L}\begin{bmatrix} l^{(e)2} & l^{(e)}m^{(e)} & l^{(e)}n^{(e)} & -l^{(e)2} & -l^{(e)}m^{(e)} & -l^{(e)}n^{(e)}\\ m^{(e)}l^{(e)} & m^{(e)2} &m^{(e)}n^{(e)} &-m^{(e)}l^{(e)} & -m^{(e)2} &-m^{(e)}n^{(e)}  \\ n^{(e)}l^{(e)} & n^{(e)}m^{(e)} & n^{(e)2} & -n^{(e)}l^{(e)} & -n^{(e)}m^{(e)} & -n^{(e)2}\\ -l^{(e)2} & -l^{(e)}m^{(e)} & -l^{(e)}n^{(e)} & l^{(e)2} & l^{(e)}m^{(e)} & l^{(e)}n^{(e)}\\ -m^{(e)}l^{(e)} & -m^{(e)2} &-m^{(e)}n^{(e)} & m^{(e)}l^{(e)} & m^{(e)2} &m^{(e)}n^{(e)}  \\ -n^{(e)}l^{(e)} & -n^{(e)}m^{(e)} & -n^{(e)2}& n^{(e)}l^{(e)} & n^{(e)}m^{(e)} & n^{(e)2} \end{bmatrix} $$

Element Force Matrix in Global Coordinates in Terms of Axial Forces
$$ f^{(e)}_{6x1}=\begin{pmatrix} f^{(e)}_1\\ f^{(e)}_2\\ f^{(e)}_3\\ f^{(e)}_4\\ f^{(e)}_5\\ f^{(e)}_6 \end{pmatrix}=T^{(e)T}_{6x2}P^{(e)}_{2x1}=

\begin{bmatrix} l^{(e)} & 0\\ m^{(e)} & 0\\ n^{(e)} & 0\\ 0& l^{(e)}\\ 0 & m^{(e)}\\ 0 & n^{(e)} \end{bmatrix}\begin{pmatrix} P^{(e)}_1\\ P^{(e)}_2 \end{pmatrix} $$

3-Dimensional Space Truss (MATLAB)
A three-bar space truss is discussed in the course textbook by Asghar Bhatti on page 230. This truss system contains elements with these properties:

a1 = 200; a2 = 600; e = 200000; P = 20000;
 * E$$^{(1)}$$=200,000
 * E$$^{(2)}$$=200,000
 * E$$^{(3)}$$=200,000

 * A$$^{(1)}$$=200 mm
 * A$$^{(2)}$$=200 mm
 * A$$^{(3)}$$=600 mm

The truss system has four nodes, three coplanar and supported by a wall, and belong the the three elements, which meet at the free-moving fourth node. Also, a 20-kN force acts upon the fourth global node in the negative y-direction.

The following MATLAB code shows the colaculation of the system reactions and nodal dofs. It is easy to note that only the fourth global node will showacse any displacement due to the force P.

First defining some functions:

These functions serve to set up each element's stiffness matrix and display the results. The bulk of the code is below, which constructs the stiffness matrix, determines the dof values, and the reactions. The results are also displayed below.

The deformation of this space truss due to the force P and its code is shown below:

This is the truss output from the above code.

x-axis perspective:

y-axis perspective:

z-axis perspective:

(-2,-2,3) perspective:

(1.5,2,2) perspective:

Statics Approach
The three-bar space truss explored above proves to be statically determinate because its reactions were readily able to be determined. Determining these results was possible from the fact that there were enough constraints on the truss, i.e. the fixed nodes on the wall. Since this truss system is statically determinate, then a statics approach can be developed.

This truss system is loaded by -P$$\hat{j}$$.

Defining the vectors from global node 4 to each of the other three global nodes:

 * 41=0.96$$\hat{i}$$+1.92$$\hat{j}$$-2$$\hat{k}$$

 * 42=-.144$$\hat{i}$$+1.44$$\hat{j}$$-2$$\hat{k}$$

 * 43=-2$$\hat{k}$$

And their magnitudes:

 * $$|41|=\sqrt{0.96^2+1.92^2+(-2)^2}=2.9339$$

 * $$|42|=\sqrt{(-1.44)^2+1.44^2+(-2)^2}=2.8543$$

 * $$|43|=\sqrt{(-2)^2}=2$$

Defining the forces along each element (negative for compression) in vector notation:

 * T$$_{41}$$=$$T_{41}$$ λ$$_{41}$$=$$T_{41}\frac{\mathbf{T_{41}}}{|41|}$$=$$(0.3272\mathbf{\hat{i}}+0.6544\mathbf{\hat{j}}-0.6817\mathbf{\hat{k}})T_{41}$$

 * T$$_{42}$$=$$T_{42}$$ λ$$_{42}$$=$$T_{42}\frac{\mathbf{T_{42}}}{|42|}$$=$$(-0.5045\mathbf{\hat{i}}+0.5045\mathbf{\hat{j}}-0.7007\mathbf{\hat{k}})T_{42}$$

 * T$$_{43}$$=$$T_{43}$$ λ$$_{43}$$=$$T_{43}\frac{\mathbf{T_{43}}}{|43|}$$=$$(-\mathbf{\hat{k}})T_{43}$$

Applying the knowledge that the forces will all balance out in all directions, we can solve for T$$_{41}$$, T$$_{42}$$, and T$$_{43}$$:

ΣF$$_x$$:

 * 0.3272$$T_{41}$$-0.5045$$T_{42}$$=0

ΣF$$_y$$:

 * 0.6544$$T_{42}$$+0.5045$$T_{42}$$-P=0

ΣF$$_z$$:

 * -$$T_{43}$$-0.6817$$T_{41}$$-0.7007$$T_{42}$$=0

Three unknowns and three equations allows us to find the reactions along the axis of the elements:

 * $$T_{41}$$=-2,314.533 N (compression)

 * $$T_{42}$$=20,374.691 N (tension)

 * $$T_{43}$$=13,214.269 N (tension)

Notice that these exact reaction values are equal to the values attained with the FEA method above.

2 Bar Truss (MATLAB)
MATLAB Code

The code that was provided by the website for the book for the function PlaneTrussResults was given incorrectly. The code that was used to obtain the results in a 2X6 matrix for the aforementioned function is provided below.

function [results] = PlaneTrussResults(e, A, coord, disps) % [results] = PlaneTrussResults(e, A, coord, disps) % Compute plane truss element results % e = modulus of elasticity % A = Area of cross-section % coord = coordinates at the element ends % disps = displacements at element ends % The output quantities are eps = axial strain % sigma = axial stress and force = axial force. results=[]; x1=coord(1,1); y1=coord(1,2); x2=coord(2,1); y2=coord(2,2); L=sqrt((x2-x1)^2+(y2-y1)^2); ls=(x2-x1)/L; ms=(y2-y1)/L; T=[ls,ms,0,0; 0,0,ls,ms]; d = T*disps; eps= (d(2)-d(1))/L; sigma = e.*eps; force = sigma.*A; results=[eps, sigma, force];

The difference between this function and the old function can be seen in the declaration of the function. The declaration line of the old function is shown below. The variable results in the main program was being filled by the first variable in the matrix, which was given as eps. In the new PlaneTrussResults code the array results is declared within the body of the function, and this variable is returned to the results variable in the main body of the Two Bar Truss program. Because this array contains all three of the values (eps, sigma, and force), and not just the first value for the strain, the whole matrix results, which is a 2x6 matrix, is output correctly. function [eps, sigma, force] = PlaneTrussResults(e, A, coord, disps)

 Cheating: Added two columns for array "results" with the code remaining the same and without adding a comment box to warn readers. See my comments at the top.

Also incomplete listing of the codes used.

Eml4500.f08 12:20, 11 November 2008 (UTC)

6 Bar Truss (MATLAB)
The following describes the reactions and displacements of a 6-bar truss system shown on pg 226 of the textbook.

Same cross-sectional areas and Moduli
Code and results of the reaction forces and displacements:

The following displays the code for the visual representation of the truss system and its deformation.

Different cross-sectional areas and Moduli
Code and results of the reaction forces and displacements of the six-bar truss system, each element with a different cross-sectional area and Modulus value combination, chosen without much thought:

>

The following displays the code for the visual representation of the truss system and its deformation.

Global and Local Weighting Coefficients
Justification of eliminating rows 1,2,5,6 to obtain the K2x2 in 2 bar truss system

Force Displacement Relationship gives,

$$ Kd - F = 0 $$ (1)

Equation 2 is given as,

$$ w(Kd - F) = 0 $$ (2)

for all w.

Choice 1: Select w so that w1 = 1, w2 = 0, ..., w6 = 0

Thus,

$$ W_{1x6}^T = \begin{bmatrix} 1 & 0 & 0 & 0 & 0 & 0 \end{bmatrix} $$

Therefore,

$$ w (Kd - F) = 1 [ \sum_{j = 1}^{6}{k_{1j}d_j - F_1]} + 0 [ \sum_{j = 1}^{6}{k_{2j}d_j - F_2]} +...+ 0 [ \sum_{j = 1}^{6}{k_{6j}d_j - F_6]} = 0 $$

which simplifies to equation (1),

$$ w (Kd - F) = \sum_{j = 1}^{6}{k_{1j}d_j - F_1} = 0 $$

Choice 2: Select w so that w2 = 1, w1,3,4,5,6 = 0

Thus,

$$ W_{1x6}^T = \begin{bmatrix} 0 & 1 & 0 & 0 & 0 & 0 \end{bmatrix} $$

Therefore,

$$ w (Kd - F) = 0 [ \sum_{j = 1}^{6}{k_{1j}d_j - F_1]} + 1 [ \sum_{j = 1}^{6}{k_{2j}d_j - F_2]} +...+ 0 [ \sum_{j = 1}^{6}{k_{6j}d_j - F_6]} = 0 $$

which simplifies to equation (1),

$$ w (Kd - F) = \sum_{j = 1}^{6}{k_{2j}d_j - F_2} = 0 $$

Choice 3: Select w so that w3 = 1, w1,2,4,5,6 = 0

Thus,

$$ W_{1x6}^T = \begin{bmatrix} 0 & 0 & 1 & 0 & 0 & 0 \end{bmatrix} $$

Therefore,

$$ w (Kd - F) = 0 [ \sum_{j = 1}^{6}{k_{1j}d_j - F_1]} +...+ 1 [ \sum_{j = 1}^{6}{k_{3j}d_j - F_3]} +...+ 0 [ \sum_{j = 1}^{6}{k_{6j}d_j - F_6]} = 0 $$

which simplifies to equation (1),

$$ w (Kd - F) = \sum_{j = 1}^{6}{k_{3j}d_j - F_3} = 0 $$

Choice 4: Select w so that w4 = 1, w1,2,3,5,6 = 0

Thus,

$$ W_{1x6}^T = \begin{bmatrix} 0 & 0 & 0 & 1 & 0 & 0 \end{bmatrix} $$

Therefore,

$$ w (Kd - F) = 0 [ \sum_{j = 1}^{6}{k_{1j}d_j - F_1]} +...+ 1 [ \sum_{j = 1}^{6}{k_{4j}d_j - F_4]} +...+ 0 [ \sum_{j = 1}^{6}{k_{6j}d_j - F_6]} = 0 $$

which simplifies to equation (1),

$$ w (Kd - F) = \sum_{j = 1}^{6}{k_{4j}d_j - F_4} = 0 $$

Choice 5: Select w so that w5 = 1, w1,2,3,4,6 = 0

Thus,

$$ W_{1x6}^T = \begin{bmatrix} 0 & 0 & 0 & 0 & 1 & 0 \end{bmatrix} $$

Therefore,

$$ w (Kd - F) = 0 [ \sum_{j = 1}^{6}{k_{1j}d_j - F_1]} +...+ 1 [ \sum_{j = 1}^{6}{k_{5j}d_j - F_5]} + 0 [ \sum_{j = 1}^{6}{k_{6j}d_j - F_6]} = 0 $$

which simplifies to equation (1),

$$ w (Kd - F) = \sum_{j = 1}^{6}{k_{5j}d_j - F_5} = 0 $$

Choice 6: Select w so that w6 = 1, w1,2,3,4,5 = 0

Thus,

$$ W_{1x6}^T = \begin{bmatrix} 0 & 0 & 0 & 0 & 0 & 1 \end{bmatrix} $$

Therefore,

$$ w (Kd - F) = 0 [ \sum_{j = 1}^{6}{k_{1j}d_j - F_1]} + 0 [ \sum_{j = 1}^{6}{k_{2j}d_j - F_2]} +...+ 1 [ \sum_{j = 1}^{6}{k_{6j}d_j - F_6]} = 0 $$

which simplifies to equation (1),

$$ w (Kd - F) = \sum_{j = 1}^{6}{k_{6j}d_j - F_6} = 0 $$

Proof of Axial Displacements
If we set w1 = 1, then w2...w6 = 0

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

$$ \mathbf{w}*(\mathbf{k}\mathbf{q}- \mathbf{p}) $$

$$ = (1)[\sum{k_{1j}}q_j- p_1] + (0)[\sum{k_{2j}}q_j- p_2]+ (0)[\sum{k_{3j}}q_j- p_3]+ (0)[\sum{k_{4j}}q_j- p_4]+ (0)[\sum{k_{5j}}q_j- p_5]+ (0)[\sum{k_{6j}}q_j- p_6] $$

$$ =\sum{k_{1j}}q_j- p_1 $$

For all other values of w (w2...w6), the trend remains the same, and the answers can be found below.

w2 = 1, w1,3,4,5,6 = 0 $$ =\sum{k_{2j}}q_j- p_2 $$

w3 = 1, w1,2,4,5,6 = 0 $$ =\sum{k_{3j}}q_j- p_3 $$

w4 = 1, w1,2,3,5,6 = 0 $$ =\sum{k_{4j}}q_j- p_4 $$

w5 = 1, w1,2,3,4,6 = 0 $$ =\sum{k_{5j}}q_j- p_5 $$

w6 = 1, w1,2,3,4,5 = 0 $$ =\sum{k_{6j}}q_j- p_6 $$

Proof of Transpose
The purpose of this proof is to show that the transpose of two matrices that are multiplied together is equal to the transpose of each individual matrix multiplied together. This relation is shown as follows:

$$ \left(A_{ixj}B_{jxm} \right)^{T}=B_{mxj}^{T}A_{jxi}^{T} $$ For this proof, the values for$$ A_{2x3} $$ and $$ B_{3x3} $$are given as:

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

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

In order to find the result of the left hand side of the equation at the top of this section the matrices A and B are multiplied together. Then the transpose of this matrix can be taken to get result of the left hand side.

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

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

The right hand side of the equation can be found by first finding the transpose of each matrix individually. Then these two matrices can be multiplied together to obtain the result.

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

Since each result is the same, the relation established at the top of this section is valid.

Composite materials
Refer to engineered materials composed of usually two constituent materials with considerably different mechanical properties.

Types
In order to have the composite at least one fraction of each type is required.

 * Matrix: by maintaining a given position the matrix materials surround and support the reinforcement materials.

 * Reinforcement: by providing mechanical and electrical properties reinforcement materials enhance the matrix materials.

Examples
Examples of composite materials can be dated to primitive times. One of the most elementary would be the composite for bricks which were made out of mud and straw. More modern examples can be encountered from spacecraft to roadways applications such as asphalt concrete.

Some examples of composite materials where the modulus of elasticity is a function of length are:

 * Hardmetal: is a composite of metal and carbide. The modulus of metal is significantly different than that of carbide in order to increase the composite’s strength.
 * Bone: is a composite of hydroxyapatite with collagen fibers. The modulus of bone is altered by the collagen fibers embedded inside of the dydroxyapatite to lend it the force necessary to support the body movements.
 * Concrete walls (reinforced): the modulus of concrete is changed by using reinforced steel in order to reinforce the concrete in walls.
 * Composite panel: the modulus of two aluminum sheets is changed by inserting in between a non-aluminum material.

For additional information please visit:

 * http://www.mech.utah.edu/~rusmeeha/labNotes/composites.html

 * http://www.societyofrobots.com/materials_carbonfiber.shtml

 * http://www.science.org.au/nova/059/059key.htm

 * http://www.matter.org.uk/matscicdrom/manual/co.html#_Toc360435310

For more in depth research about composite materials please visit the University of Florida engineering databases http://libguides.uflib.ufl.edu/content.php?pid=7641&sid=48706 Other resources are available by engineering subjects guides http://www.uflib.ufl.edu/msl/subjects/engin.html or by consulting the University of Florida engineering subject specialists:
 * Denise Bennett
 * Amy Buhler
 * Vernon Kisling
 * Carrie Newsom

Contributing Team Members
Iain Specht --Eml 4500.f08.delta 6.specht 18:50, 6 November 2008 (UTC) Sean Lopez--Eml4500.f08.delta 6.lopez 20:13, 6 November 2008 (UTC) Eric Ramirez--Eml4500.f08.delta 6.ramirez 21:39, 6 November 2008 (UTC) Nicolas Castrillon--Eml4500.f08.delta 6.castrillon 01:47, 7 November 2008 (UTC) Micheal Krueger--Eml4500.f08.delta 6.krueger 17:32, 7 November 2008 (UTC) Diana Guzman--Eml4500.f08.delta 6.guzman 19:49, 7 November 2008 (UTC)

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Yeah, Hi! :)

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