text:
User:Eas4200c.f08.aero6.villalba/shear1-4

(1.4)
Problem statement: Use AS4/3501-6 carbon/epoxy composite to make the I-beam as stated in Problem 1.3. Compare its weight with that of the aluminum beam.

So, repeating the same process as with problem 1.3, we now use the AS4/3501-6 carbon/epoxy composite. Now we use the Young's modulus of this material:

 * $$E_{composite}=140 \ GPa$$

From problem 1.3, we know the expression for the moment of inertia of this I-beam (EQN1 ). This gave us the moment of inertia of the steel beam:

 * $$I_{steel} = 7.612 \times 10^{-6} \ m^4\!$$

From the problem statement $$E_{steel}\cdot I_{steel} = E_{composite}\cdot I_{composite}\!$$, we find $$I_{composite}\!$$ to be
 * $$I_{composite} = \frac{E_{steel} \cdot I_{steel}}{E_{composite}} = \frac{(200 \ GPa)(7.612 \times 10^{-6} \ m^4)}{140 \ GPa} = 1.087 \times 10^{-5} \ m^4$$

Now, rearranging EQN1 to solve for $$b\!$$ > we obtain the following:

 * $$b_{composite} = \frac{1}{2}\left(I_{composite}-\frac{t(h-2t)^3}{12}\right)\Bigg/\left(\frac{t^3}{12}+t\left(\frac{h-2t}{2}+\frac{t}{2}\right)^2\right) = 0.0843 \ m = 84.3 \ mm$$

This will yield the following cross-sectional area:

 * $$A_{composite} = 2 \cdot b_{composite} \cdot t+(h-2t) \cdot t = 1793 \ mm^2\!$$

Now, knowing $$\rho_{composite} = 1.55 \times 10^{-3} \ \frac{g}{mm^3}$$, we can obtain the mass-per-unit length and our weight-per-unit length

 * $$m_{composite}=\rho_{composite} \times A_{composite}\!$$
 * $$m_{composite}=\left(1.55 \times 10^{-3} \ \frac{g}{mm^3}\right) \left(1793 \ mm^2\right)\!$$
 * $$m_{composite}=2.78 \ \frac{g}{mm}\!$$

The weight-per-unit length of the composite beam is
 * $$W_{composite}=27.2 \ \frac {N}{m}$$

Therefore, the composite beam is more efficient weightwise than both the aluminum and the steel beams.


text:
User:Eas4200c.f08.aero6.villalba/shear1-5

(1.5)
Problem statement: Derive the relations given by equations [1.4] and [1.5] of the textbook.

If we have constant shear $$\tau\!$$ on a curved panel, we can break down the shear force into two components:

 * $$V_x=\tau\,t\,a \qquad \qquad [1.4] \!$$
 * $$V_y=\tau\,t\,b \qquad \qquad \, [1.5] \!$$

From Figure 2, we can find an expression for the differential resulting force $$d\vec V$$:

 * $$d\vec V=q\,\vec {dl}$$

Where $$q\!$$ is the shear flow and $$\vec {dl}$$ > is the differential length that is being studied.

Knowing $$\vec {dl}$$ > has two components (one in $$x\!$$ and another in $$y\!$$), we are able to work out our answer.

 * $$d\vec V=q\,\left( dl_x \, \hat i + dl_y \, \hat j\right)$$
 * $$=q\,\left( dl\, \cos \theta \, \hat i + dl\, \sin \theta \, \hat j\right)$$
 * $$=q\,\left( dx\, \hat i + dy\, \hat j\right)$$

Now, integrating both sides,

 * $$\int {d \vec V}=q\, \left[ \left( \int_{x_a}^{x_b} {dx} \right)\, \hat i + \left( \int_{y_a}^{y_b} {dy} \right)\, \hat j\right]$$

and we also know $$\int_{x_a}^{x_b} {dx}=a$$ and $$\int_{y_a}^{y_b} {dy}=b$$

therefore,

 * $$\vec V=q\, \left( a\, \hat i + b\, \hat j\right)=qa\, \hat i + qb\, \hat j$$

and knowing $$q = \tau \, t \!$$ we now have

 * $$\vec V=\tau t  a\ \hat i + \tau  t  b\ \hat j$$

So this gives us the two components previously stated,

 * $$V_x=\tau\,t\,a \!$$
 * $$V_y=\tau\,t\,b \!$$

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


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User:Eas4200c.f08.carbon.orear/HW3/Matlab

= Code =

NACAairfoil.m
Note: The code is currently unable to calculate the centroid. This issue is being debugged and resolved presently. Check back to the live page at a later time for this calculation and the appropriate debug section.

AirfoilArea.m
= Debugging =

Circle
With a circular airfoil, the Number of Segments such that the computed Area of the airfoil is accurate to within 1% of the exact value is 23 segments at which the computed area was 3.1115 m3. This is 0.95% away from the actual area of a unit circle, 3.14195 m3.

Code used to find ns_max

The area for a various number of segments is tabulated below.

Plotting the Area vs Number of segments up to the point of 1% accuracy we get the graph below.

At P0 = (0,0), Area computed equaled 3.1115

At P0 = (1,0), Area computed equaled 3.1115

At P0 = (1,-4), Area computed still equaled 3.1115

NACA Airfoil
With a circular airfoil, the Number of Segments such that the computed Area of the airfoil is accurate to within 1% of the previous value is 8 segments at which the computed area was 0.024616 m3. This is 0.77% away from the previous area of a unit circle, 0.024373 m3. At 10,000 segments, the computed area was 0.02569 m3.

Code used to find ns_max

The area for a various number of segments is tabulated below.

Plotting the Area vs Number of segments up to the point of 1% difference accuracy we get the graph below.

At P0 = (0,0), Area computed equaled 0.024616

At P0 = (.25, 0), Area computed equaled 0.024616

At P0 = (.25,-1), Area computed still equaled 0.024616


text: File:Hw3 fig4.jpg

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text:
Enterprise Resource Planning/Entrepreneurship

Programme
This is a module part of the Open Source ERP/Executive Masters course conducted by a private university.

Subject
ERP Entrepreneurship - Making Money Out Of Free Software

Objectives
The objectives of this subject are:
 * To study the various business models in the market
 * To align to the right business models for Open Source
 * To see avenues or gaps in the Open Source model to commercialise on

Learning Outcomes
On completion of this subject the students shall be able to:
 * Understand the commercial forces behind Open Source
 * Explore a new niche for the market that is viable and sustainable
 * Strategise collectively and leverage with market players
 * Foresee any risks and work out contigencies

Sypnosis
This subject examine how players make money with Open Source today. It explores the opportunities for the students to partake in.

Different Models
(20 hrs)
 * Fun Model - Story of Craiglist
 * Hobby Model - Story of Wikipedia
 * Brand Model - Story of Linux
 * Money Model - Story of RedHat
 * Services Model - Reinventing IBM

Exit Strategy
(20 hrs)
 * How Venture Capital works
 * How to protect a nudist colony
 * Branding Human Capital
 * Captive Market

Micro ISV Strategy
(20 hrs)
 * Doing everything yourself
 * MBA Crash Course
 * Marketing
 * Image
 * Clustering and Cloud Market

Charging Model
(20 hrs)
 * Per Hour or Per Day
 * Flat Rate or Piecemeal Package
 * Long term and micro transaction
 * Negotiation Strategy
 * Calculating the Sales Funnel

Business As Usual
(20 hrs) (to be cont'd)
 * Market Normalcy
 * Packaging
 * Service Contract
 * Warranty Clauses
 * Help Desk
 * Phone Support
 * On Site Service
 * Repair and parts


text:
User:Jenny O/Groups

I found Fisk's (2004) chapter Small Groups: Ongoing interactions far more interesting and informative than our textbook for this section (see Further Reading under Groups).

What is a group?
A group can be defined as two or more people who perceive themselves, or are perceived by others, as belonging together (Fiske, 2004). Aspects of belonging include: a shared identity, a common goal, interdependence and being distinct from an out-group. Interestingly, two people can constitute a group, albeit a special case. (However, I am not absolutely convinced of this: In my mind the combination of interactions between a couple - or dyad - are very different from the combination of interactions of three people. Adding just one person to a dyad increases the interactions exponentially). The degree to which a group makes a whole (in-group homogeneity) depends on similarity, proximity, interdependence and interaction (Gestalt principals). One's social identity (part of self-concept) is contextually based and derived from group membership. Self categorisation and social identity increase in-group favouritism and out-group disadvantage.

Human groups have social and cultural purposes. Groups offer animals advantages such as, safety, vigilance, shared resources and increased power (Evolutionary perspective). In addition humans can benefit from groups as groups facilitate complex communication and economic exchange, accumulation of knowledge, and role differentiation.

Concepts, theories and studies
There are a number of interesting concepts, theories and studies relating to groups: Deindividuation refers to the loss of self-awareness and individual accountability one can experience by becoming part of a group or mob. The individual actions of many Hutus during the Rwandan genocide may be attributed to this (see Ghosts of Rwanda). However, Fiske (2004) reports on more recent evidence suggesting that situation specific norms rather than deindivuation predict collective behaviour. According to the Social Identity Model of Deindividuation, a contextually salient group identity develops and group members express that identity in their behaviour. Fiske illustrates this with an example of the police action making a disparate group of protesters more homogeneous and thus making moderates more radical.

Social facilitation refers to the tendency to perform well while others are watching. This was identified in Triplett’s (1897) classic study of bicycle riders who performed better against other riders than the clock. However, further work in this area resulted in the Zyjonc’s Social Faciliation Theory which linked the presence of others with arousal to show that familiar tasks (practised or dominant response) improve in the presence of others, while unfamiliar tasks are impaired (see Yerkes-Dodson Law). For example, I have noticed, that people generally speak well in public if they are passionate about or well versed on the topic.

The Distraction Conflict Theory (Baron,1986) suggests that levels of attention or distraction affect performance. The presence of an audience results in an attentional conflict (attending to audience or task). This increases arousal and drive, leading to increased social facilitation (increased performance). This includes the “Somebody upstairs cares” syndrome (someone is actually intrested in what I am doing) and the Hawthorne Effect (conscious and unconscious modification of performance when being watched). Coupled with this is the cognitive concept of evaluation apprehension or one’s concern about being judged while performing a task – obviously not a problem experienced by  narcissists who crave attention and admiration.

Social loafing is an interesting phenomenon describing decreased individual effort on a collective task. According to Latane (1981) this occurs as individual responsibilty can be diffued in a group. Other contributing factors include: evaluation apprehension, the bad apple effect, or even not wanting to be a sucker (a uniquely human trait). So it seems reasonable that social loafing can be decreased by identifying group members and making them accountable.

Social Dilemmas describe a conflict of self and collective interest – an individual can gain more by defecting than cooperating. The commons dilemma (Originally named from grazing private flocks of sheep on the village green - the common) is one in which groups members all take from a shared resource, but if individuals take too much the communal resource is depleted (Fiske, 2004). That is, there are gains if all cooperate or losses if all compete. James provided the problem of global warming as a good example of this. Communication, personality, mood and the behaviour of others influence the use of common resources.

Group decision-making
Group decisions are not necessarily superior to or more balanced than individual decisions. They are different. It was originally thought that groups tended to make more risky decisions; however, current thinking suggests groups tend to make more polarised (extreme) decisions than individuals. This occurs due to individual attitudes, learning from others and conforming to a group stereotype. Groupthink refers to the tendency of group members to think alike as they are motivated by the need for consensus and minimum conflict. An illusion of invulnerability, a sense of moral superiority and underestimation of opponents also contribute. Groupthink is more likely to occur when the group is homogeneous, cohesive and isolated and the leader is strong or biased (Fiske, 2004). I imagine this is the type of thinking that occurs in cults.

Leadership
Leadership is very interesting and quite complex. A leader is a prototypical individual to whom others look for guidance. He or she essentially has more power than others and the ability to use social influence. Theories relating to leadership seem to have evolved over the past 50 years from a trait-based approach to behavioural and situational theories to the current transformational or transactional leadership theories. Transformational leaders seek to inspire and excite followers to improve performance. Whereas, transactional leaders use reward and punishment in the exchange of valued resources between leader and follower (Fiske, 2004). This approach emphasizes resource control, self-enhancement, shared understanding (roles and goals), and norm compliance.

We were shown a video during the lecture of leadership training. This is based on the premise that leadership skills can be taught. I actually question the validity of that process. The video really focused on confidence training for a woman who was unable to confidently lead a meeting. It might be a semantic argument but the notion of confidently leading a meeting and leadership are very different constructs in my mind. I believe there are very few people who are “true” and effective leaders (I suppose I adhere to the Great Man Theories – true leaders are born not made). The modern day leader or what is construed as leadership is usually more like facilitation or bullying than leading others.

title: User:Jenny O/Groups


text:
User talk:Sunnygirl~enwikiversity

Welcome
Hello Sunnygirl, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. After you leave a comment on a talk page, remember to sign and date; it helps everyone follow the threads of the discussion. The signature icon in the edit window makes it simple. To get started, you may
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And don't forget to explore Wikiversity with the links to your left. Be bold, and see you around Wikiversity! --Hillgentleman | //\\ |Talk 09:35, 8 October 2008 (UTC)

Your account will be renamed
Hello,

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

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

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User talk:Autodidact~enwikiversity

Welcome
Hello Autodidact, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. After you leave a comment on a talk page, remember to sign and date; it helps everyone follow the threads of the discussion. The signature icon in the edit window makes it simple. To get started, you may
 * Take a guided tour and learn to edit;
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 * Read and help develop our community policies；or
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And don't forget to explore Wikiversity with the links to your left. Be bold, and see you around Wikiversity! --Hillgentleman | //\\ |Talk 09:35, 8 October 2008 (UTC)

Your account will be renamed
Hello,

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

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

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

Sorry for the inconvenience.

Yours, Keegan Peterzell Community Liaison, Wikimedia Foundation 23:29, 17 March 2015 (UTC)

Renamed
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title: User talk:Autodidact~enwikiversity


text:
User:Jenny O/The Environment

Environmental psychology
Before embarking on this topic I had a very narrow view of environmental psychology. I found the issues regarding built environments and environmental design intensely interesting.

Oskamp and Schultz provide a simple definition of environmental psychology as the study of:  "the interactions and relations between people and their environments"

This includes built and natural environments.

Some important aspects of these environments include:

 * Density – the number of people per unit area

 * Crowding – subjective negative experiences relating to density. This is a form of environmental stress. Perceived control decreases the feeling of crowding.

 * Environmental stress - daily hassles or frustrations, noise, heat.

 * Environmental risk - natural disasters, disease, pollution, food contamination, and terrorism.

 * Environmental design - assessing and planning artificial designs and encompasses architectural psychology, consumer psychology, permaculture (permanent culture) and wayfinding (cognitive mapping of environments)

The built environment
The built environment plays a significant role in the affect, behaviour and cognitions of human beings (e.g., the effect of lighting on concentration; sounds and levels of stress; colours and emotions; space and privacy, and social behaviour in public places etc.) However, I also expect there are individual differences in the significance for some. For example, I am strongly attached to, and influenced by, my built and natural environments while others I know do not necessarily share the same feelings about their space or environment.

Sundstrom, Bell, Busby and Asmus (1996) (see Lecture 9 Readings) examine a range of psychological factors relating to built environments including: residential preference and satisfaction, housing design features, residential crowding, and place attachment. They also consider communal spaces such as workplaces, hospitals, schools, and prisons. It is well know in health that patients experience ICU-Syndrome, a delirium-like state or psychological dysfunction resulting (in part) from environmental aspects of time spent in the Intensive Care Unit (e.g., noise, lighting, movement, equipment, and being disconnected from the outside world). Sundstrom et al. also consider extreme living conditions and how people adapt in these situations (e.g., Outer space, Inuit settlements in North Alaska, and other polar regions). I thought it was interesting that environments such as detention centres and refugee camps have not been considered in the readings – maybe this is a more recent phenomenon, or are they equivalent to prisons?

Understandably windows and views are a very important part of our dwellings and workplaces. However, my husband told me about a recent experience of employing new graduates (in conditions of job oversupply). One of the most important criteria in considering the job was whether they would be given an office with a window rather than an internal workspace (A question the graduate would all ask very 'diplomatically' at interview). I was astonished to think that a new graduate had the audacity to ask such a question in a job interview, but maybe it has become a valid concern in maintaining psychological well-being when managing the stressful transition to corporate life!

Natural environments
Generally, people have a great affinity with natural environments. Some explain this in evolutionary terms as a genetic predisposition toward nature (e.g., the Bipohilia Hypothesis). The influence of nature on the human psyche is supported by observations such as our preference for natural rather than abstract artwork, our desire for dwellings with views of nature, and children’s affinity with animals in stories and characterisations. It is interesting that many social and moral lessons have been depicted through anthropomorphism (e.g., Aesop’s Fables, Alice in Wonderland,  The Wind in the Willows, and  Winnie-the-Pooh).

Nature and experiences with nature provide many positive effects. Research has been conducted in the effects of animals in human therapy (e.g., animal assisted therapy); natural scenes (e.g., in-patient hospital recovery – also see ICU-Syndrome); adventure therapy (e.g., outdoor education), and green exercise. The green prescription, or encouraging people to engage with nature to prevent and ameliorate physical and psychological dyfunction is being encouraged.

Environmental problems
Human behaviour is having a devastating effect on the natural environment, and by association on the human psyche. According to the Australian Bureau of Statistics (ABS, 2007), Australia’s environmental concerns relate to:

 * Population and urban trends (population growth & distribution, households, economic growth & consumption)
 * Human activities (energy & waste)
 * Atmosphere (greenhouse gases, climate change, ozone & air quality)
 * Water (consumption, conservation & management, and marine)
 * Landscape (land, forests & biodiversity)

Only yesterday, against the backdrop of the current world financial frenzy, | The Guardian newspaper reports: World is facing a natural resources crisis worse than the financial crunch. The article claims that humans are using 30% more resources than is sustainable, and two planets will be needed by 2030 to sustain our life-styles if we continue our current rate of consumption. Australia is apparently ranked as the fifth most destructive in the world.

Pro-environmental behaviours
Behavioural change is necessary to preserve resources and the natural environment. Attitudes do not manifest as behaviours. Although environmental awareness or a positive attitude toward the environment is helpful, strong pro-conservational norms (personal and societal), understanding, and commitment are needed to support behavioural changes. Implementing antecedent and consequent strategies can assist in effecting behavioural change:

Antecedent strategies:
 * Information campaigns e.g., government and other organizations: Clean-up Australia Day, Recycle Michael ad campaign. Best when used in the long term (cumulative effect)
 * Prompts. These are most effective when: practical and easy to use, and specific prompts are made (polite, trouble-free, close to point of response) e.g., Please place your rubbish in the bin provided.
 * Modelling or demonstrating desired behaviour.

Consequent strategies:
 * Reinforcement (can be short-lived) e.g., monetary incentives: tokens or coupons for use of public transport, rewards for the use of energy saving devices – solar hot water system or water tank rebates.
 * Punishment e.g., on-the-spot fines for littering, fines for industrial environmental pollution or waste, monetary disincentives (increasing the cost of petrol).
 * Feedback or providing information regarding the environmental effects or consequences of altered behaviours e.g electrical star ratings or water usage ratings on whitegoods, energy efficiency ratings for houses, mechanised feedback on households energy usage.

Other strategies include: Self-monitoring, fostering personal connections with nature and the environment, lobbying governments, supporting environmental organizations or groups.

A sustainable future
Developing a sustainable future for the world is one step toward protecting the environment (Oskamp & Schultz, 1998).

The IPAT formula can be used to assess the impacts of several variables influencing sustainability:

I = P x A x T

Where: I = Impact, P = Population, A = Amount of consumption per person (affluence), T = Technologies supporting affluence

The impact of future growth and economic development can be measured by setting the IPAT to a value of one as a baseline measure, and recalculating impact levels at regular intervals. Of several sustainability models assessed against this formula, a slow growth model was the most feasible. Slow growth emphasises population control and slowed economic growth, and would require more efficient technology and energy consumption. However, such changes are only possible with increased awareness, education, strong pro-conservational norms, commitment, and most saliently practising pro-environmental behaviours.

title: User:Jenny O/The Environment


text:
User:Eml4500.f08.RAMROD.A/MyContributionHW3

The 3 Bar Truss and the Moment Equation
A look at the 3 Bar Truss Setup

From Statics, We know that there are typically three useful equations:

$$\displaystyle \sum{Fx}=0 $$

$$\displaystyle \sum{Fy}=0 $$

$$\displaystyle \sum{M_A}=0 $$

Statically Indeterminate
For this truss setup, even the moment equation renders no useful information because all reactions pass through the same point, A, and the setup is statically indeterminate

Now, we ask the question: what if we tried to sum the moments about some arbitrary point, B?

We will need to look at another picture and vector math to find out answers:

Can we get Information Summing Moments about B?

A look at the vector algebra:

$$\displaystyle \sum{\overline{M_B}}= \overline{BA} \times \overline{F} $$ Thus, $$\displaystyle \sum{\overline{M_B}}= \overline{BA'} \times \overline{F} $$ for all A' on line of action of F (see image right) From Vector Algebra: $$\displaystyle \overline{BA'} = \overline{BA}+ \overline{AA'} $$ $$\displaystyle \overline{M_B} = (\overline{BA}+ \overline{AA'}) \times \overline{F} $$ $$\displaystyle \overline{M_B} = \overline{BA} \times \overline{F} + \overline{AA'} \times \overline{F} $$, With The latter cross product in the previous equation equal to zero, we are left with: $$\displaystyle \overline{M_B} = \overline{BA} \times \overline{F} + 0 $$ More Generally:

$$\displaystyle \sum_{i}^{}{B_i}=\sum_{i}^{}{\overline{BA_i} \times \overline}$$ Where Ai is any point on the line of action of F. Which Indicates: $$\displaystyle \sum{\overline{M_B}} = \overline{BA} \times \sum{\overline{F}} $$ We are already aware that: $$\displaystyle \sum{\overline{F}} = 0 $$ We can conclude that: $$\displaystyle \sum{\overline{M_B}} = \overline{BA} \times 0 = 0 $$ '''So, even if we take the moment from any arbitrary point, it will ultimately not provide any helpful information to solve the problem. The problem truly is statically indeterminate.

Three Truss Stiffness Matrix
Now we can solve for overall K by combining K(1), K(2) and K(3)

Matrix K $$\displaystyle K = \begin{bmatrix} k_{33}^{(1)}+k_{11}^{(2)}+k_{11}^{(3)} & k_{34}^{(1)}+k_{12}^{(2)}+k_{12}^{(3)}\\k_{43}^{(1)}+k_{21}^{(2)}+k_{21}^{(3)} & k_{44}^{(1)}+k_{22}^{(2)}+k_{22}^{(3)}\\k_{31}^{(2)}+k_{31}^{(3)} & k_{32}^{(2)}+k_{32}^{(3)}\\k_{41}^{(2)}+k_{41}^{(3)} & k_{42}^{(2)}+k_{42}^{(3)}\\k_{51}^{(3)} & k_{52}^{(3)}\\k_{61}^{(3)} & k_{62}^{(3)} \end{bmatrix} $$  =   $$\displaystyle K = \begin{bmatrix} k_{33} & k_{34}\\k_{43} & k_{44}\\k_{53} & k_{54}\\k_{63} & k_{64}\\k_{73} & k_{74}\\k_{83} & k_{84} \end{bmatrix} $$


title: File:3TrussB.jpg


text: File:AA'.jpg

title: File:AA'.jpg


text:
User:Eml4500.f08.echo.sell.cm/HW3

Solving for Unknown Displacements
Next, the elimination of known degree's of freedom can create the following matrices.

$$ d_1^{} = d_2^{} = d_5^{} = d_6^{}=0 $$

$$	\mathbf{K}\begin{Bmatrix} d_1=0\\d_2=0\\d_3\\d_4\\d_5=0\\d_6=0 \end{Bmatrix}=\begin{vmatrix} k_{13} & k_{14} \\ k_{23} & k_{24}\\k_{33} & k_{34}\\k_{43} & k_{44}\\k_{53} & k_{54}\\k_{63} & k_{64}\end{vmatrix}\begin{Bmatrix} d_3\\d_4\end{Bmatrix}=\mathbf{F}$$

Applying fixed boundary conditions allows us to delete the corresponding columns in the global stiffness matrix K

By the Principal of Virtual Work (PVW) we're able to delete the corresponding rows (in this case - 1, 2, 5, 6). Corresponding rows of F are also deleted.

The resulting FD relation is: $$ \begin{bmatrix} k_{33} & k_{34} \\ k_{43} & k_{44} \end{bmatrix} \begin{Bmatrix} d_3\\d_4\end{Bmatrix}=\begin{Bmatrix} F_3\\F_4\end{Bmatrix}$$

$$\mathbf{F}=\begin{Bmatrix} F_3\\F_4\end{Bmatrix}=\begin{Bmatrix} 0\\P\end{Bmatrix}$$

And P =>

What is the inverse of K without using a calculator?

Taking the determinant of K:

$$det \mathbf{K}=k_{33}k_{44}-k_{k34}k_{43}$$

$$\mathbf{K}^{-1}=\frac{1}{det\mathbf{K}}\begin{bmatrix} k_{44} & -k_{34} \\ -k_{43} & k_{33} \end{bmatrix}$$

Verifying:

$$\mathbf{K}\mathbf{K}^{-1}=\mathbf{K}^{-1}\mathbf{K}=\mathbf{I}=\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$$

This is essentially:

$$\frac{1}{det(\mathbf{K})}\begin{bmatrix} det(\mathbf{K}) & 0 \\ 0 & det(\mathbf{K}) \end{bmatrix}=\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$$

Note: $$\mathbf{K}^T=\begin{bmatrix} k_{33} & k_{43} \\ k_{34} & k_{44} \end{bmatrix}$$

Therefore: $$\mathbf{K}^{-1}\ne \ \frac{1}{det(\mathbf{K})}\mathbf{K}^T$$

Going back to the 2 bar truss problem...

$$\begin{Bmatrix} d_3 \\ d_4 \end{Bmatrix}=\mathbf{K}^{-1}\begin{Bmatrix} 0 \\ P \end{Bmatrix}=\begin{Bmatrix} 4.352 \\ 6.1271 \end{Bmatrix}$$

$$ \begin{Bmatrix} P_1^{(e)} \\ P_2^{(e)} \end{Bmatrix}_{2x1} = \mathbf{T}_{2x4}^{(e)} \begin{Bmatrix} f_1^{(e)} \\ f_2^{(e)} \\ f_3^{(e)} \\ f_4^{(e)} \end{Bmatrix}_{4x1}$$

$$\mathbf{P}_{}^{(e)} = \mathbf{T}_{}^{(e)} \mathbf{f}_{}^{(e)}$$

Recall the element axial Force-Displacement relationship

$$\mathbf{\hat{k}}_{2x2}^{(e)} = \mathbf{q}_{2x1}^{(e)} \mathbf{P}_{2x1}^{(e)}$$

where:

$$\mathbf{q}_{2x2}^{(e)} = \mathbf{T}_{2x1}^{(e)} \mathbf{d}_{2x1}^{(e)}$$

$$\mathbf{P}_{2x2}^{(e)} = \mathbf{T}_{2x1}^{(e)} \mathbf{f}_{2x1}^{(e)}$$

therefore:

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

 Goal:  We want to have $$\mathbf{k}_{}^{(e)} \mathbf{d}_{}^{(e)} = \mathbf{f}_{}^{(e)}$$.

So we need to move $$\mathbf{T}_{}^{(e)}$$ from the right hand side to the left hand side by "pre-multiplying" the equation by $$\mathbf{T}_{}^{(e) -1}$$.

Unfortunately since $$\mathbf{T}_{}^{(e)}$$ is a rectangular matrix it can not be inverted.

However through usage of the Principle of Virtual Work we find $$\mathbf{k}_{}^{(e)} = [\mathbf{T}_{}^{(e)T} \mathbf{\hat{k}}_{}^{(e)}\mathbf{T}_{}^{(e)}]$$

Which yields answer of $$[\mathbf{T}_{4x2}^{(e)T} \mathbf{\hat{k}}_{2x2}^{(e)}\mathbf{T}_{2x4}^{(e)}]_{4x4} \mathbf{d}_{4x1}^{(e)} = \mathbf{f}_{4x1}^{(e)}$$

and gives us $$\mathbf{k}_{}^{(e)} \mathbf{d}_{}^{(e)} = \mathbf{f}_{}^{(e)}$$

 HW:  Verify $$\mathbf{k}_{}^{(e)} = [\mathbf{T}_{}^{(e)T} \mathbf{\hat{k}}_{}^{(e)}\mathbf{T}_{}^{(e)}]$$

Using the Principle of Virtual Work $$\mathbf{\hat{k}}_{}^{(e)} = \mathbf{d}_{}^{(e)}) = \mathbf{f}_{}^{(e)}$$ because the transformation matrix $$\mathbf{T}_{}^{(e)}$$ is an arbitrary matrix. However through further evaluation of the equation $$\mathbf{\hat{k}}_{}^{(e)} = \mathbf{d}_{}^{(e)}) = \mathbf{f}_{}^{(e)}$$ we recognize that the sizes of the matrices prevent this operation from being carried out. We can re-size $$\mathbf{\hat{k}}_{}^{(e)}$$ by multiplying it by and identity matrix where the matrix transpose is multiplied by the matrix. So, $$\mathbf{T}_{}^{(e)T}\mathbf{T}_{}^{(e)}$$ is our identity matrix and by definition when multiply the identity matrix by the unit matrix we are left with a matrix the re-sized matrix. i.e. $$\mathbf{k}_{}^{(e)} = [\mathbf{T}_{}^{(e)T} \mathbf{\hat{k}}_{}^{(e)}\mathbf{T}_{}^{(e)}]$$

We could not solve as $$\mathbf{d}_{}^{} = \mathbf{K}_{}^{-1} = \mathbf{F}_{}^{}$$ because of the singulrity associated with matrix $$\mathbf{K}_{}^{}$$. ($$det\mathbf{K}_{}^{} = 0$$)

For an unconstrained strut system there are 3 possible rigid body motions in 2-D.(2trans. and 1 rat.)

These 3 motions are Axial Deformation, Rotation at Nodes and Translation at Nodes.

 HW:  Find the Eigenvalues of $$\mathbf{K}_{6x6}^{}$$ and make observations about the number of zero eigenvalues

dynamic evaluation problem $$\mathbf{K}_{}^{} \mathbf{v}_{}^{} = \lambda_{}^{} \mathbf{M}_{}^{} \mathbf{v}_{}^{}$$

where:

K = Stiffness Matrix

$$\mathbf{\lambda}$$ = eigenvalues(~vibration frequency)

M = Mass Matrix

$$(\mathbf{K}_{}^{} - \lambda_{}^{} \mathbf{M}_{}^{}) \mathbf{v}_{}^{} = 0$$

$$\begin{bmatrix} .5625 & .3248 & -.5625 & -.3248 & 0 & 0\\ .3248 & .1875 & -.3248 & -.1895 & 0& 0\\ -.5625 & -.3248 & 3.0625 & -2.1752 & -2.5 & 2.5\\ -.3248 & -.1875 & -2.1752 & 2.6875 & 2.5 & -2.5\\ 0 & 0 & -2.5 & 2.5 & 2.5 & -2.5\\ 0 & 0 & 2.5 & -2.5 & -2.5 & 2.5 \end{bmatrix}v = \mathbf{\lambda_{}^{}M_{}^{}}v$$

Solving yields the following eigenvalues: $$\mathbf{\lambda_{1}}$$ = 10.03, $$\mathbf{\lambda_{2}}$$ = 1.471, $$\mathbf{\lambda_{3}}$$ = 0, $$\mathbf{\lambda_{4}}$$ = 0, $$\mathbf{\lambda_{5}}$$ = .2983, $$\mathbf{\lambda_{6}}$$ = 0

Zeros occur in rows 3, 4 and 6 of lambda.

The eigenvector Matrix is: $$\begin{bmatrix} 0.0139 & 0.6174 & -0.6455 & 0.4494 & -0.6131 & 0.5997\\ 0.0080 & 0.3565 & -0.1291 & -0.6754 & 0.0754 & 0.1199\\ -0.5123 & -0.5409 & -0.6454 & -0.1682 & -0.6131 & 0.5997\\ 0.4904 & -0.4330 & -0.1291 & 0.3942 & 0.0754 & 0.1199\\ 0.4984 & -0.0765 & -0.2582 & -0.2812 & -0.3442 & 0.5014\\ -0.4984 & 0.0765 & 0.2582 & 0.2812 & 0.3442 & 0.0216 \end{bmatrix}$$

The zero eigenvalues correspond with zero stored elastic energy yields rigid body nodes

--C. Sell 23:30, 7 October 2008 (UTC)

title: User:Eml4500.f08.echo.sell.cm/HW3


title: File:K3Bar2.jpg


title: File:K3Bar3.jpg


text:
Internet and Project Competence

INTERNATIONAL PROJECT

{| id="Todo" cellspacing="0" cellpadding="0" style="width:100%; border:2px #A3D9BE solid;"  Internet- and Project Competence
 * colspan="2" style="background:#cef2e0; text-align:center; padding:2px; border-bottom:1px #A3D9BE solid;" |
 * colspan="2" style="background:#cef2e0; text-align:center; padding:2px; border-bottom:1px #A3D9BE solid;" |
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{| style="width:100%; background-color:transparent;"
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Background
We think that the basic objective of schools and universities is to prepare students for the knowledge society, i.e. to train their knowledge networking skills. Students should get used to search for information and to transform it into knowledge through individual and collaborative, participative techniques. In this context, we would like to stress the perspectives offered by various web 2.0 tools such as weblogs, wikis, video conferences and discussion boards. Thanks to all these different tools, people (students) from different countries and cultures can take part in the collective construction of knowledge by carrying out international projects. This implies, of course, a high degree of communicative competence and - besides technological skills - social skills such as

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Course description
Follow link for a brief course description: course description

current international project
We will carry out projects about primary school teaching, open learning and the ideas of formal and informal instruction together with students from Bulgaria, the USA and Germany. The professors involved are Leigh Ausband (USA), Iliana Mirtschewa (Bulgaria) and Klaudia Schultheis (Germany). Our students will work together on some of the topics and prepare presentations and thus create content which they will present online.

specified period
13th Oct – 1st Dec (6 weeks)

project schedule
Project Schedule

first instructions
Follow link for first instructions: Instructions
 * }

Working groups
working groups

participants
In order to sign up, please click on edit (here above) and replace "XY" by Your personal information (cf. Michael Kratky)

KWL CHART (template)
Please click on "edit" (here above), copy the code, paste it on your personal user site and insert Your text in the columns.

 * }

Resources
Common reading assingment for students: How Can ICT Literacy Skills Support Global Awareness at the K-5 Level? ''' questions to answer: 1. The reading on ICT/Global Awareness describes two examples of how ICT (Information and Communications Technology)skills can support global awareness. What do you think are the benefits to students who participate in these types of projects? What similar types of projects might you do with students when you begin teaching?''' Framework for the 21st Century '''questions to answer: 2. The Framework for 21st Century Learning list 4 areas that are important for a 21st education: Core Subjects and 21st Century Themes; Learning and Innovation Skills; Information, Media, and Technology Skills; and Life and Career Skills. Which of these, if any, are the schools in your area addressing? What does a teacher in training like yourself need to know and be able to do in order to teach 21st century skills to children?'''
 *
 *

Reading recommendation for the German students:
 * Martin Wellenreuther (2008): Lehren und Lernen - aber wie? Empirisch-experimentelle Forschungen zum Lehren und Lernen im Unterricht. 4. Aufl. Baltmannsweiler: Schneider Verlag Hohengehren
 * Falko Peschel (2006): Offener Unterricht. Idee - Realität - Perspektive und ein praxiserprobtes Konzept zur Diskussion. Teil I: Allgemeindidaktische Überlegungen. 4. Aufl. Baltmannsweiler: Schneider Verlag Hohengehren
 * Werner Jank & Hilbert Meyer (2008): Didaktische Modelle. 5. akt. Aufl. Berlin: Cornelsen Scriptor Verlag

Reading recommendation for the Bulgarian students:
 * Мирчева, Илиана. (2004): Проблеми на дидактиката на родинознанието и природознанието. София: ИК "Веда Словена - ЖГ". (виж разделите, посветени на отворените форми на обучение)
 * Мирчева, Илиана. (2006): Докосване до природата. София: ИК "Веда Словена - ЖГ". (виж разделите, посветени на отворените форми на обучение)
 * Петрова, Ваня. (2003): Работата по проект при запознаване с природната и обществената среда в І-ІV клас. Стара Загора: ИК “Кота”.
 * Boo, Max de. (2004): Using Science to Develop Thinking Skills at Key Stage 1. London: David Fulton Publishers. (мога да предоставя книгата на всеки от Вас, тъй като я няма в библиотеката)
 * Brubacher, M., R. Payne, K. Rickett. (1990): Perspectives on Small Group Learning. Oakville: Rubicon Publishing Inc. (има я в библиотеката на факултета)
 * Fried-Booth, Diana. Project Work. Oxford: Oxford University Press. (мога да предоставя книгата на всеки от Вас, тъй като я няма в библиотеката)

Reading recommendations for American students for some groups:

Group 1 – open education Humanism and Open Education http://chiron.valdosta.edu/whuitt/col/affsys/humed.html

Check for the next two in Atkins Library: A Learner's Bill of Rights (EJ709893) Author(s):	Rathbone, Charles H.

Source:	Phi Delta Kappan, v86 n6 p471 Feb 2005 Pub Date:	2005-02-01 Pub Type(s):	Journal Articles; Reports - Descriptive

Abstract: Celebrating its 35th year, the Fayerweather Street School of Cambridge, Massachusetts, is one of the few extant schools associated with the "open education" movement of the late Sixties. Not only has the school flourished, but it has managed to stick to its original progressive principles.

The Open Classroom: Were Schools Without Walls Just Another Fad? (EJ763250) Author(s):	Cuban, Larry

Source:	Education Next, v4 n2 p68-71 Spr 2004 Pub Date:	2004-00-00 Pub Type(s):	Journal Articles; Reports - Descriptive

Abstract: The open-classroom movement originated in British public elementary schools after World War II. American educators who adopted the trend viewed informal education--or, as they came to call it, open classrooms or open education

Group 3 - Types of instruction and learning theories Instructional Methods http://itc.utk.edu/~bobannon/in_strategies.html Direct Instruction http://olc.spsd.sk.ca/DE/PD/instr/direct.html Indirect Instruction http://olc.spsd.sk.ca/DE/PD/instr/indirect.html Learning Theories http://tip.psychology.org/theories.html

Group 5 – Interaction between children and nature Children and Nature http://www.childrenandnature.org/uploads/CNMovement.pdf

title: Internet and Project Competence


text: Internet and Project Competence/schedule


text:
User:Eml4500.f08.team.allen.oca/petition

EML 4500 Finite Element Analysis Class Petition


text:
Internet and Project Competence/Description

{| id="Todo" cellspacing="0" cellpadding="0" style="width:100%; border:2px #A3D9BE solid;"  Course Description
 * colspan="2" style="background:#cef2e0; text-align:center; padding:2px; border-bottom:1px #A3D9BE solid;" |
 * colspan="2" style="background:#cef2e0; text-align:center; padding:2px; border-bottom:1px #A3D9BE solid;" |
 * style="background:#f5fffa; padding:1.2em;" valign="top" |
 * style="background:#f5fffa; padding:1.2em;" valign="top" |

{| style="width:100%; background-color:transparent;"
 * style="width:50%; padding-right:1.2em;" valign="top" |

to put it into a nutshell...
The course „Internet- and Project-Competence“ has been designed to bring students to carry out real projects and to develop their communication-skills in order to enhance the flow of information between young people world wide. Within one semester, the students will

 * create an online community with international partners and search for information
 * carry out their projects (often combined with an excursion)
 * present the results of their projects on a website

Objectives
The main objective of IPC is to make students carry out projects collectively and to communicate with each other in order to produce knowledge. This meta-objective can be split up into the following items:

 * Capacity for teamwork
 * Capacity to resolve conflicts
 * Patience and perseverance
 * Planning skills
 * Communicative competence
 * Initiative
 * Web 2.0 literacy

 * style="width:50%; background:#f5fffa; border-left:1px solid #cef2e0; padding-left:1.2em;" valign="top" |

Course outline
Phase 1

 * Choice of a topic/country and formation of an online community
 * Drafting a thesis, developing a research design (questionnaires, interviews)
 * Preparation of the excursion/investigation

Phase 2

 * Students carry out projects, prove/disprove the validity of their thesis,
 * upload of the results in a self-created website,
 * and present their website to the public.

Certificate

Having passed the module, students may obtain the “IPC Certificate“, which assesses students‘ internet- and project competences (such as capacity for teamwork, communicative competence etc.)
 * }
 * }


text:
User:Eml4500.f08.team.dwyer.jd/HW3 (9-29 & 10-1)

Class Notes: September 29, 2008.
Using the Global Force-Displacement Relation

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

Only need to compute rows 1,2,5,6 to get F1, F2, F3, & F4.

Closing the loop between FEM and statics:

Virtual Displacement

 Two-Bar truss system:

By statics, reactions known and therefore the member forces; $$P_1^{(1)}$$ and $$P_2^{(1)}$$

Compute axial displacement degree of freedoms (amount of extension of bars):

$$q_2^{(1)} = \frac{P_1^{(1)}}{k^{(1)}} = \frac{P_2^{(1)}}{k^{(1)}}$$

$$ q_1^{(1)} = 0$$, (fixed global node 1)

$$q_1^{(2)} = \frac{-P_2^{(2)}}{k^{(2)}} = -AB$$

$$q_2^{(2)} = 0 $$, (fixed global node 3)


text:
Internet and Project Competence/description

{| id="Todo" cellspacing="0" cellpadding="0" style="width:100%; border:2px #A3D9BE solid;"  Course Description
 * colspan="2" style="background:#cef2e0; text-align:center; padding:2px; border-bottom:1px #A3D9BE solid;" |
 * colspan="2" style="background:#cef2e0; text-align:center; padding:2px; border-bottom:1px #A3D9BE solid;" |
 * style="background:#f5fffa; padding:1.2em;" valign="top" |
 * style="background:#f5fffa; padding:1.2em;" valign="top" |

{| style="width:100%; background-color:transparent;"
 * style="width:50%; padding-right:1.2em;" valign="top" |

to put it into a nutshell...
The course „Internet- and Project-Competence“ has been designed to bring students to carry out real projects and to develop their communication-skills in order to enhance the flow of information between young people world wide. Within one semester, the students will

 * create an online community with international partners and search for information
 * carry out their projects (often combined with an excursion)
 * present the results of their projects on a website

Objectives
The main objective of IPC is to make students carry out projects collectively and to communicate with each other in order to produce knowledge. This meta-objective can be split up into the following items:

 * Capacity for teamwork
 * Capacity to resolve conflicts
 * Patience and perseverance
 * Planning skills
 * Communicative competence
 * Initiative
 * Web 2.0 literacy

 * style="width:50%; background:#f5fffa; border-left:1px solid #cef2e0; padding-left:1.2em;" valign="top" |

Course outline
Phase 1

 * Choice of a topic/country and formation of an online community
 * Drafting a thesis, developing a research design (questionnaires, interviews)
 * Preparation of the excursion/investigation

Phase 2

 * Students carry out projects, prove/disprove the validity of their thesis,
 * upload of the results in a self-created website,
 * and present their website to the public.

Certificate

Having passed the module, students may obtain the “IPC Certificate“, which assesses students‘ internet- and project competences (such as capacity for teamwork, communicative competence etc.)
 * }
 * }


text:
User:Eml4500.f08.team.allen.oca/hw3

Solving the two bar truss by Infinitesimal Displacement
In order to complete the analysis of the two bar truss problem, the final configuration of the truss after the load has been applied must be found. To accomplish this, analysis must be done on the individual elements.

The length of segment AC in the figure can be found by treating the element as a spring, where the elongation length is equal to the force applied divided by the spring constant. In this manner, AC can be found by: $$ AC= \frac{P^{(1)}_2}{k^{(1)}}$$ $$ AC=\frac{5.1243}{\frac{3}{4}}=6.8324$$

The length of segment AB, as well as the coordinates of points B and C are solved as homework problems.HW from Oct 1

With the locations of points B and C known, the location of point D can be found by using the equations for the lines AB and BC.

To find the equation for a line containing a generic point P, we need to know the slope of the line, and the coordinates of P. If the angle of the line is known, the slope of the line is given as the "rise over run", or the y/x. Since the angle is $$\theta$$, the y portion of the line is $$sin\theta$$ and the x portion is $$cos\theta$$. y/x gives us $$ \frac{sin\theta}{cos\theta} = tan\theta$$. Using the point-slope formula, we arrive at $$ y-y_p = tan\theta(x-x_p)$$ where x_p and y_p are the x and y coordinates of point P. Since the lines BD and CD are perpendicular to AB and AC respectively, the slopes of BD and CD are the negative inverses of the lines they are perpendicular to. Thus the equations for lines BD and CD become: BD: $$ y-y_B = (tan\theta^{(2)})^{-1}(x-x_B)$$ CD:$$ y-y_C = (tan\theta^{(1)})^{-1}(x-x_C)$$

The location of point D is found in the homework problem. HW from Oct 1

The three bar truss
In the three bar truss diagram, the truss has the following parameters: $$     E^{(1)} = 2 $$ $$      A^{(1)} = 3 $$ $$      L^{(1)} = 5 $$ $$      E^{(2)} = 4 $$ $$      A^{(2)} = 1 $$ $$      L^{(2)} = L^{(1)} = 5 $$ $$      E^{(3)} = 3 $$ $$      A^{(3)} = 2 $$ $$      L^{(3)} = 10 $$ $$      P = 30 $$  $$\theta^{(1)} = 30 $$ $$   \theta^{(2)} = -30 $$ $$   \theta^{(3)} = 45 $$

In order to make solving the three bar truss system the most convenient, the local nodes are labeled as follows:

title: User:Eml4500.f08.team.allen.oca/hw3


text:
Internet and Project Competence/instructions

{| id="Todo" cellspacing="0" cellpadding="0" style="width:85%; border:2px #A3D9BE solid;"  International Project Competence - first instructions
 * colspan="2" style="background:#cef2e0; text-align:center; padding:2px; border-bottom:1px #A3D9BE solid;" |
 * colspan="2" style="background:#cef2e0; text-align:center; padding:2px; border-bottom:1px #A3D9BE solid;" |
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 * style="background:#f5fffa; padding:1.2em;" valign="top" |

{| style="width:100%; background-color:transparent;"
 * style="width:90%; padding-right:1.2em;" valign="top" |

Register with wikiversity

 * Create a user account at http://en.wikiversity.org/wiki/Wikiversity:Main_Page
 * Do the newcomers tour and get used to wikis: http://en.wikiversity.org/wiki/Help:The_original_tour_for_newcomers
 * Edit Your personal user page: upload a photo, present yourself (cf. http://en.wikiversity.org/wiki/User:Mk )

Sign up on the inscription list

 * Here: http://en.wikiversity.org/wiki/Wikiversity:IPC#participants

KWL Chart

 * Insert the KWL chart in Your user page: http://en.wikiversity.org/wiki/Wikiversity:IPC#KWL_CHART_.28template.29

Posting on discussion board

 * Please register with zum.de via the following link: http://www.zum.de/Foren/internet/auth/userform.php

 * You will be sent a confirmation email from "foren-admin@zum.de". Use the answer function of your email account and do not change the subject. Thus your account will be activated

 * Then log in on http://www.zum.de/Foren/internet/threads/thread414.php, enter Your user name and password and write a posting in which you briefly present yourself including the link to your wikiversity site. Write Your text directly into the box, then click on "Weiter zur Vorschau" which means "Show Preview" and then click on "publizieren und Nutzungsbedingungen anerkennen" which means "submit".

 * all threads of IPC: http://www.zum.de/Foren/internet/cgi/forum.cgi

Survey I

 * Answer the questions in Survey I: http://www.surveyshare.com/survey/take/?sid=76724
 * You will need the numerical identifier Your Professors will assign you before.

Come together

 * get in touch with international partners on our main discussion board: http://www.zum.de/Foren/internet/threads/thread414.php
 * Think about a topic you would like to work on or...
 * ... sign up for (a) topic(s) on the doodle-site: Doodle poll: IPC working groups

new thread for each group

 * Open a new thread as soon as you have found international partners:
 * Go to http://www.zum.de/Foren/internet/cgi/forum.cgi and...

 * ...click on "Neues Thema hinzufügen" (You find this link up left) and insert a name for your working group, i.e. the topic you are working on, for instance open education.

discussion on common reading
In your group discussion forum, please discuss the following questions on the common readings:

1. The reading on ICT/Global Awareness describes two examples of how ICT (Information and Communications Technology)skills can support global awareness. What do you think are the benefits to students who participate in these types of projects? What similar types of projects might you do with students when you begin teaching?
 *

2. The Framework for 21st Century Learning list 4 areas that are important for a 21st education: Core Subjects and 21st Century Themes; Learning and Innovation Skills; Information, Media, and Technology Skills; and Life and Career Skills. Which of these, if any, are the schools in your area addressing? What does a teacher in training like yourself need to know and be able to do in order to teach 21st century skills to children?
 *

week 4-7
WORK ON PROJECTS

week 8

 * upload the final version of your group powerpoint to your group wiki!

 * Please do the second survey at http://www.surveyshare.com/survey/take/?sid=76955

 * Remember the numerical identifier your instructors assigned you at the beginnig of the project.
 * }

 * style="width:10%; background:#f5fffa; border-left:1px solid #cef2e0; padding-left:1.2em;" valign="top" |

 * }


text:
User:Salmon of Doubt/monobook.css

/* Turn the "Save page" button green if I'm logged in */ INPUT#wpSave { background-color:#88ff88; }

title: User:Salmon of Doubt/monobook.css


text:
User:Eml4500.f08.team.foskey.ckf/HW3c

Solving the two bar truss by Infinitesimal Displacement
In order to complete the analysis of the two bar truss problem, the final configuration of the truss after the load has been applied must be found. To accomplish this, analysis must be done on the individual elements.

The length of segment AC in the figure can be found by treating the element as a spring, where the elongation length is equal to the force applied divided by the spring constant. In this manner, AC can be found by: $$ AC= \frac{P^{(1)}_2}{k^{(1)}}$$ $$ AC=\frac{5.1243}{\frac{3}{4}}=6.8324$$

The length of segment AB, as well as the coordinates of points B and C are solved as homework problems.HW from Oct 1

With the locations of points B and C known, the location of point D can be found by using the equations for the lines AB and BC.

To find the equation for a line containing a generic point P, we need to know the slope of the line, and the coordinates of P. If the angle of the line is known, the slope of the line is given as the "rise over run", or the y/x. Since the angle is $$\theta$$, the y portion of the line is $$sin\theta$$ and the x portion is $$cos\theta$$. y/x gives us $$ \frac{sin\theta}{cos\theta} = tan\theta$$. Using the point-slope formula, we arrive at $$ y-y_p = tan\theta(x-x_p)$$ where x_p and y_p are the x and y coordinates of point P. Since the lines BD and CD are perpendicular to AB and AC respectively, the slopes of BD and CD are the negative inverses of the lines they are perpendicular to. Thus the equations for lines BD and CD become: BD: $$ y-y_B = (tan\theta^{(2)})^{-1}(x-x_B)$$ CD:$$ y-y_C = (tan\theta^{(1)})^{-1}(x-x_C)$$

The location of point D is found in the homework problem. HW from Oct 1

The three bar truss
In the three bar truss diagram, the truss has the following parameters: $$     E^{(1)} = 2 $$ $$      A^{(1)} = 3 $$ $$      L^{(1)} = 5 $$ $$      E^{(2)} = 4 $$ $$      A^{(2)} = 1 $$ $$      L^{(2)} = L^{(1)} = 5 $$ $$      E^{(3)} = 3 $$ $$      A^{(3)} = 2 $$ $$      L^{(3)} = 10 $$ $$      P = 30 $$  $$\theta^{(1)} = 30 $$ $$   \theta^{(2)} = -30 $$ $$   \theta^{(3)} = 45 $$

In order to make solving the three bar truss system the most convenient, the local nodes are labeled as follows:

title: User:Eml4500.f08.team.foskey.ckf/HW3c


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Talk:Worldpolitics

Format
how do we distinguish who's talking and about what?

I intend to begin a thread of source concerning your itulip.com reseaches. but if we're not careful, it will get very unclear who's talking about what.--Jolie 15:49, 8 October 2008 (UTC)

 * We can have chat meetings on Wikiversity chat. This could atract other users. The front page should be shorter and more to the point. New pages can be created which can relate to this page. One addendum could be to get topics on separate articles.--Daanschr 14:30, 10 October 2008 (UTC)

 * Talking can be done on the talk page :-). There are several other options, it depends on what you and me want to do with this reading group.--Daanschr 14:32, 10 October 2008 (UTC)

 * Thank you, I confess I have looked and don't yet understand how to IRC (I installed Bersirc, but it isn't very intuitive). we can certainly talk here. perhaps we need to do periodic 'purge the chat'...  I'm still collecting my thoughts about the future that itulip.com predicts, there are some intricate economic factors that I am struggling to understand.

 * I will probably not be as active on the weekend. I have periodic time to think during the week, but it's not as easy during the weekend.--Jolie 22:49, 10 October 2008 (UTC)

Economics
The problem with economists is that they all predict something else. The economics on Itulip is pretty hardcore, i only understand half of it sometimes ;-).--Daanschr 10:07, 11 October 2008 (UTC)

 * The general idea that it is good for long-term growth that an economy produces products sounds very logical to me :-).--Daanschr 10:08, 11 October 2008 (UTC)

Here's something to ponder. If governments operate in deficits they create new dollars for no value. Inflation. Speculation ends up destroying dollars as money that is invested dissapears. Deflation.

Neither one is good, BUT there are opposite. Since money is simply a means of exchange, isn't two opposite foces that keep it in balance,sustainable?

Maybe these speculative 'bubbles' are keeping us from suffering from the incredibly large US federal deficit?--Jolie 12:13, 13 October 2008 (UTC)

 * It depends on who lends the money. Take Iceland for instance. Icelandic consumers and institutions have borrowed enormous amounts of money all over the world. Now, the country is bankrupt. The money was used to support a high standard of living, they didn't earn. The USA has borrowed enormous amounts of money worldwide and is now checked for it. A bubble can't help to solve this problem, because the dollar is a regional coin.

 * Perhaps a bubble can be good for making transactions of money, goods and services quickly available. A problem is that the owners of money in a bubble are using this money to invest in economically inefficient areas. The high prices in the housing market in the USA has let to the building of many houses which are now uninhabited. If the working hours and resources to build these houses were invested in roads, airports, education, than it would have been better for the economy in the long run. But, i am not an economist, so i don't know too much about it.--Daanschr 13:07, 13 October 2008 (UTC)

 * Itulip is about sustainability but it is a different kind of sustainability than only using the right resources. Also, transactions, the ownership of capital are part of the economy and can be of vital importance when it comes to the distribution of wealth.--Daanschr 13:09, 13 October 2008 (UTC)

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User:Eml4500.f08.Lulz.js/HW3

Axial displacement calculation for local node 2
During meeting 12, a calculation for the axial displacement of local node 1 for an arbitrary element e was considered. What follows is a similar calculation for local node 2 of the same element.

First, we define the displacement vector at local node 2 (Please note the [#] subscript was used in place of the "box" subscript used in class):

$$ \vec{\mathbf{d}^{(e)}_{[2]}} = d_{3}^{(e)} \vec{i} + d_{4}^{(e)} \vec{j} $$

Next, the axial displacement at node 2 is defined: $$ q_{2}^{(e)} = \vec{\mathbf{d}^{(e)}_{[2]}}\cdot\vec{\tilde{i}} $$

$$ q_{2}^{(e)} = (d_{3}^{(e)}\vec{i} + d_{4}^{(e)}\vec{j})\vec{\tilde{i}} $$

$$ q_{2}^{(e)} = d_{3}^{(e)}(\vec{i}\cdot\vec{\tilde{i}}) + d_{4}^{(e)}(\vec{j}\cdot\vec{\tilde{i}}) $$

From a previous homework question and the class notes, the following relations are valid:

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

Substituting,

$$ q_{2}^{(e)} = d_{3}^{(e)}\cos \theta^{(e)} + d_{4}^{(e)}\sin\theta^{(e)} $$

From the notes, the following relations are valid:

$$\cos\theta^{\left(e\right)} = l^{(e)}$$ $$\sin\theta^{\left(e\right)} = m^{(e)}$$

Substituting,

$$ q_{2}^{(e)} = d_{3}^{(e)}l^{(e)} + d_{4}^{(e)}m^{(e)} $$

Putting the above equation yields the following result, consistent with that of the axial displacement of local node 1:

$$ q_{2}^{(e)} = \begin{bmatrix} l^{(e)} & m^{(e)} \end{bmatrix}\begin{Bmatrix} d_{3}^{(e)}\\ d_{4}^{(e)} \end{Bmatrix} $$

Proof of $$\mathbf{k}^{(e)} = \mathbf{T}^{(e)^{T}} \mathbf{\hat{k}}^{(e)}\mathbf{T}^{(e)}$$
In this question we are asked to prove that the above matrix product generates the generic element stiffness matrix $$\mathbf{k}^{(e)}$$.

First, the three matrices used to generate the product:

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

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

$$\mathbf{\hat{k}}^{(e)} = \begin{bmatrix} 1&-1 \\ -1&1 \\ \end{bmatrix}$$

The product of the first two matrices:

$$\mathbf{\hat{k}}^{(e)}\mathbf{T}^{(e)} = \begin{bmatrix} 1&-1 \\ -1&1 \\ \end{bmatrix} \begin{bmatrix} l^{(e)}&m^{(e)}&0&0 \\ 0&0&l^{(e)}&m^{(e)} \end{bmatrix} = \begin{bmatrix} l^{(e)}&m^{(e)}&-l^{(e)}&-m^{(e)} \\ -l^{(e)}&-m^{(e)}&l^{(e)}&m^{(e)} \\ \end{bmatrix}$$

The entire product multiplied out:

$$\mathbf{T}^{(e)^{T}}\mathbf{\hat{k}}^{(e)}\mathbf{T}^{(e)} = \begin{bmatrix} l^{(e)}&0 \\ m^{(e)}&0 \\ 0&l^{(e)} \\ 0&m^{(e)} \\ \end{bmatrix} \begin{bmatrix} l^{(e)}&m^{(e)}&-l^{(e)}&-m^{(e)} \\ -l^{(e)}&-m^{(e)}&l^{(e)}&m^{(e)} \\ \end{bmatrix} $$ $$ = \begin{bmatrix} l^{(e)^{2}} & m^{(e)}l^{(e)} & -l^{(e)^{2}} & -m^{(e)}l^{(e)} \\ m^{(e)}l^{(e)} & m^{(e)^{2}} & -m^{(e)}l^{(e)} & -m^{(e)^{2}}\\ -l^{(e)^{2}} & -m^{(e)}l^{(e)} & l^{(e)^{2}} & m^{(e)}l^{(e)}\\ -m^{(e)}l^{(e)} & -m^{(e)^{2}} & m^{(e)}l^{(e)} & m^{(e)^{2}} \\ \end{bmatrix} $$

As shown in previous homework question, the element stiffness matrix:

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

Thus, it is easily shown that the following relation is indeed true:

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

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User:Eml4500.f08.team.dwyer.jd/HW3 (9-29)

Class Notes: September 29, 2008.
Using the Global Force-Displacement Relation

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

Only need to compute rows 1,2,5,6 to get F1, F2, F3, & F4.

Closing the loop between FEM and statics:

Virtual Displacement

 Two-Bar truss system: |center

By statics, reactions known and therefore the member forces; $$P_1^{(1)}$$ and $$P_2^{(1)}$$

Compute axial displacement degree of freedoms (amount of extension of bars):

$$q_2^{(1)} = \frac{P_1^{(1)}}{k^{(1)}} = \frac{P_2^{(1)}}{k^{(1)}}$$

$$ q_1^{(1)} = 0$$, (fixed global node 1)

$$q_1^{(2)} = \frac{-P_2^{(2)}}{k^{(2)}} = -AB$$

$$q_2^{(2)} = 0 $$, (fixed global node 3)


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User:Eas4200c.f08.aeris.krammer/HW4

 Comment: This page is identical to the previous version, except that notes on the Strain Matrix have been added (Item 7 in the Contents tab). Here you may find a comparison between this and the original version. Eas4200c.f08.aeris.mcrae. 15:44, 29 October 2008 (UTC)

Multicell Airfoil
Let's take a specific example first to find $$\displaystyle \theta$$ as a function of $$\displaystyle T$$ and $$\displaystyle J$$ (torsional constant) First, $$T=T_{1}+T_{2}=2q_{1}\bar{A_{1}}+2q_{2}\bar{A_{2}}  (1)$$ Where, $$\bar{A_{1}}=ac$$ $$\bar{A_{2}}=bc$$ Then, $$\theta_{1}=\frac{1}{2G\bar{A_{1}}}\oint_{}^{}{}\frac{q_{1}}{t_{1}(s)}ds=\frac{1}{2G\bar{A_{1}}}\left[\frac{2q_{1}b}{t_{1}} +\frac{q_{1}c}{t_{1}}+ \frac{(q_{1}-q_{2})c}{t_{1,2}}\right]   (2)$$ $$\theta_{2}=\frac{1}{2G\bar{A_{2}}}\oint_{}^{}{}\frac{q_{2}}{t_{2}(s)}ds=\frac{1}{2G\bar{A_{2}}}\left[\frac{2q_{2}b}{t_{2}} +\frac{q_{2}c}{t_{2}}+ \frac{(q_{2}-q_{1})c}{t_{1,2}}\right]   (3)$$ $$\displaystyle C_{1}$$ and $$\displaystyle C_{2}$$ have the same rate of twist angle, therefore $$\displaystyle \theta _{1}=\theta _{2}$$ Using equation (1), we can find an expression for $$\displaystyle q_{1}$$ and $$\displaystyle q_{2}$$ in terms of $$\displaystyle T$$: $$\displaystyle q_{1}=\beta _{1}T$$ $$\displaystyle q_{2}=\beta _{2}T$$ Where $$\displaystyle \beta _{1}$$ and $$\displaystyle \beta _{2}$$ is some number. Next, we can use equations (2) and (3) to find the to find the expression for $$\displaystyle \theta$$ and $$\displaystyle T$$, $$\displaystyle \theta =\theta _{1}=\theta _{2}=\frac{T}{2GJ}$$, and deduce $$\displaystyle J$$. Calculating with numerical values for the above equations: $$\displaystyle A_{1}=0.3m * 0.4 m =0.12m^2$$ $$\displaystyle A_{2}=0.6m * 0.4 cm =0.24m^2$$ Therefore,  $$\displaystyle T=0.24q_{1}+0.48q_{2}$$ Assuming $$\displaystyle G_{1}=G_{2}=G$$, we can solve for $$\displaystyle \theta _{1}$$ and $$\displaystyle \theta _{2}$$ $$ \theta _{1}=\frac{1}{0.24G}[\frac{1.2q_{1}}{0.003}+\frac{0.4q_{1}}{0.003}+\frac{0.4(q_{1}-q_{2})}{0.004}]$$ $$ \theta _{2}=\frac{1}{0.48G}[\frac{1.2q_{2}}{0.005}+\frac{0.4q_{2}}{0.005}+\frac{0.4(q_{2}-q_{1})}{0.004}]$$ Since $$\displaystyle \theta _{1}=\theta _{2}$$ equating the two equations above yields to $$\displaystyle 1366.6q_{1}=620q_{2}$$ Using equation $$\displaystyle (1)$$, we can express $$\displaystyle q_{1}$$ and $$\displaystyle q_{2}$$ as a function of $$\displaystyle T$$ $$\displaystyle q_{1}=0.77T$$  $$ \displaystyle q_{2}=1.698T$$ Where $$\displaystyle 0.77=\beta _{1}$$ and $$\displaystyle 1.698=\beta _{2}$$

Derivation of the Twist Angle in the Case of a Uniform, Non-Circular Cross Sectional Bar with Varying Shear Flow
Regarding the previous lecture, once $$\theta$$ is found in terms of T, use $$T=GJ\theta$$ or $$J=\frac{T}{G\theta}$$ to find the torsional constant J.  For this homework assignment, the NACA 2415 airfoil, referring to lecture 16.2, is partitioned into three cells. The cells are separated by two walls of assigned thickness. To perform the Matlab analysis similar to the homework 3, more information must be provided in order to properly calculate the "average" areas $$\bar{A}_1, \bar{A}_2, and \bar{A}_3$$ corresponding to the three cells of the airfoil and along with the torque (not indicated but will be treated as a constant), thickness of the walls and shell, contour length and shear modulus will allow us to calculate the shear flow on the airfoil. First, one wall is .25 times the chord length away from the leading edge, and a second wall is .75 times the chord length away from the leading edge. The shear modulus is not needed because the equations used will cancel out these terms. There will be three unknowns $$q_1,q_2, and q_3$$ and using the three following equations, these unknowns will be solved for. $$T=2\sum_{i=1}^3{q_i\bar{A}_i}$$ $$\displaystyle \theta_1=\theta_2$$ $$ \displaystyle \theta_2=\theta_3$$ Then, the torsional constant J will be found by the equation mentioned above. As we have already finished deriving the follow equations, $$T=GJ\theta, T=2q\bar{A}$$ Now we will derive the twist angle equation for a uniform, non-circular cross sectional with an engineering (ad hoc) method. $$\theta=\frac{1}{2G\bar{A}}\oint\frac{q}{t}ds$$ Let's consider this uniform, non-circular cross sectional bar subject to a torque. The bottom two pictures describe this type of bar with infinitesimally small axial length dx and the cross section is shown in the (y,z) plane.
 Displacement PP' is due to alpha, and approximating alpha as very small, we have the following expression for PP' and OP in a simple geometric triangular relation.
$$\frac{PP'}{OP}=tan(\alpha)$$ Now, project the displacement PP' in the direction orthogonal to OP'. In other words,
$$\displaystyle PP' = PP'' cos( \alpha)$$ $$\displaystyle PP=(OPtan(\alpha))cos(\alpha)=(OPcos(\alpha))tan(\alpha)=OPtan(\alpha)$$ Recall, that OP=R, the radial coordinate of the point P and thus $$\rho$$ must be the perpendicular distance from the origin to P which equals, in this case, OP. Knowing that, the following expression can be written.
$$\displaystyle PP''=(Rcos(\alpha))tan(\alpha)=\rho \alpha$$ Here, PP" is the displacement of P in the direction "tangent" to the lateral surface of the bar. Strain is $$\gamma=\frac{PP}{dx}$$ where dx is the distance between very close planes on the bar.
$$\gamma=\frac{PP}{dx}=\frac{\rho \alpha}{dx}=\rho \theta$$ Here theta is the rate of twist angle.

Formal Justification by Elasticity Theory

$$\displaystyle \tau =G\gamma=G\rho \theta$$ Where the rate of twist $$\displaystyle \theta =\frac{d\alpha }{dx}$$ Integrating along the contour wall
$$\oint_{C}^{}{}\tau (s) ds = G\theta (x)\oint_{C}^{}{}\varrho (s) ds$$ Where $$\displaystyle \tau (s) ds$$ is $$\displaystyle \frac{q(s)}{t(s)}$$ and $$\oint_{C}^{}{}\varrho (s) ds$$ is  $$\displaystyle 2\bar{A}$$ Hence we get the following expression for $$\displaystyle \theta$$
$$\theta =\frac{1}{2GA}\oint_{}^{}{}\frac{q}{t}ds$$ What is ad hoc about this derivation and the derivation of $$\displaystyle T=2q\bar{A}$$? 1) Strain must be obtained using the displacement of $$\displaystyle \vec{P}$$ in the direction tangent to $$\displaystyle C$$ at $$\displaystyle \vec{P}$$ but $$\displaystyle \vec{PP'}$$ is not necessarily tangent to $$\displaystyle C$$, although it is really close. 2) $$\displaystyle \tau =\frac{q}{s}$$ obtained from the ad-hoc assumption that $$\displaystyle \tau$$ was uniform across the wall thickness. Thus, we have the formal derivation by elasticity theory.

Roadmap: Kinematic Assumption
The kinematic assumptions are:
$$\displaystyle u_{x}(y,z)=\theta \psi (y,z)$$ (Where $$\displaystyle \theta$$ is considered to be constant with respect to $$\displaystyle x$$)
$$\displaystyle u_{y}(x,z)=-\theta xz$$
$$\displaystyle u_{z}(x,y)=\theta xy$$ The following relationship can be established:
$$\displaystyle \epsilon _{xx}=\epsilon _{yy}=\epsilon _{zz}=\gamma _{yz}=0$$ This relation can be proved by taking partial derivatives:
$$\epsilon _{xx}=\frac{\partial u_{x}(y,z)}{\partial x}=0$$
$$\epsilon _{yy}=\frac{\partial u_{y}(x,z)}{\partial y}=0$$
$$\epsilon _{zz}=\frac{\partial u_{z}(x,y)}{\partial z}=0$$
$$\gamma _{yz}=\frac{\partial u_{y}(x,z)}{\partial z}+\frac{\partial u_{z}(x,y)}{\partial y}=0$$ (Where $$\frac{\partial u_{y}(x,z)}{\partial z}=-\theta x$$ and $$\frac{\partial u_{z}(x,y)}{\partial y}=+\theta x$$)

NACA Airfoil: Triangle Rule
The triangle rule can be used to find the area of any 4-digit NACA Airfoil from an observer at any point. The total area would be found by:

$$A = A_{1} + A_{2}$$ $$ = A_{1} - |A_{2}|$$

To find the areas of $$A_{1}$$ and $$A_{2}$$, the triangle rule is used by the following:

This method works effectively because the observer is noted as the origin and each triangle stems from that origin. This allows for infinitely many triangles to be measured from a single point and an area to be calculated correctly and accurately. Another advantage to the triangle rule is that it can be used in any case.

 EXAMPLE 1

Although it may be easier to integrate a rigid body using other methods (i.e. Trapezoidal Rule), curvature changes, like the one shown above, may not be very easy to compute. The curve causes for tricky calculations which can be avoided by using triangles.

Single Cell Airfoil

Problem Statement: If shear flow is assumed to be constant (i.e. $$q = q_{1} = q_{2} = q_{3}$$), what is the angle of twist and the allowable torsion, $$T_{allow}$$.

First, we must calculate the rate of twist, θ,  θ = (q/2GA)Σ lj/tj . The variables are: 1. q = shear flow 2. lj = length 3. G = shear modulus of cell 4. tj = thickness in cell (curves along cell wall) 5. A = cell area

Applied to this single cell, the equation is:  θ = (q/2GA)[π*(b/2)/t1 + a/t2 + sqrt(a2 + b2)/t3]

Once the rate of twist is calculated we can then find the maximum shear stress, τmax. The formula for calculating τmax is:  τmax = q / min{t1, t2, t3}  where min{t1, t2, t3} refers to the smallest thickness value of the the three used.

Because this is a single cell,  τmax = τallow  and since $$q = T/2A$$, then  Tallow = 2A*τallow*min{t1, t2, t3}

Strain Matrix
In a 3-D body under strain, this strain can be broken down into a component matrix, as seen below.

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

Due to symmetry, there are a several equalities we can make.

$$\varepsilon _{xy}=\varepsilon _{yx}$$

$$\varepsilon _{yz}=\varepsilon _{zy}$$

$$\varepsilon _{xz}=\varepsilon _{zx}$$

We can now see that this 3x3 strain matrix has only 6 components, $$\varepsilon _{xx}, \varepsilon _{yy},, \varepsilon _{zz}, \varepsilon _{xy}, \varepsilon _{yz}, \varepsilon _{xz}$$.

The equation for each component can be expressed as $$\varepsilon _{ij}=\frac{1}{2}\left(\frac{\partial u_{i}}{\partial x_{j}}+\frac{\partial u_{j}}{\partial x_{i}}\right)$$

Stress Tensor Matrix

Similarly, we can make some equalities to the stress tensor matrix, which is seen below.

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

$$\sigma _{xy}=\sigma _{yx}$$

$$\sigma _{yz}=\sigma _{zy}$$

$$\sigma _{xz}=\sigma _{zx}$$

We see that the stress tensor matrix, like the strain matrix, also has 6 components.

Stress-Strain Relation
Normal Strains:
 $$\epsilon _{xx}=\frac{\sigma _{xx}}{E}-\frac{\nu \sigma _{yy}}{E}-\frac{\nu \sigma _{zz}}{E}$$
 $$\epsilon _{yy}=\frac{\sigma _{yy}}{E}-\frac{\nu \sigma _{xx}}{E}-\frac{\nu \sigma _{zz}}{E}$$
 $$\epsilon _{zz}=\frac{\sigma _{zz}}{E}-\frac{\nu \sigma _{xx}}{E}-\frac{\nu \sigma _{yy}}{E}$$ Shear Strains:
 $$\gamma _{xy}=2\epsilon _{xy}=\frac{\tau _{xy}}{G}$$
 $$\gamma _{yz}=2\epsilon _{yz}=\frac{\tau _{yz}}{G}$$
 $$\gamma _{zx}=2\epsilon _{zx}=\frac{\tau _{zx}}{G}$$ Another form, voigt notation, is used to describe normal strain and shear strain in a column matrix. This allows $$\epsilon=\begin{bmatrix} \epsilon_{ij} \end{bmatrix}_{3\times 3}$$ and $$\sigma=\begin{bmatrix} \sigma_{ij} \end{bmatrix}_{3\times 3}$$ to be written as:
 $$\begin{Bmatrix} \epsilon _{ij} \end{Bmatrix}_{6\times 1}=\begin{Bmatrix} \epsilon _{11}\\ \epsilon _{22}\\ \epsilon _{33}\\ \epsilon _{23}\\ \epsilon _{31}\\ \epsilon _{12}\\ \end{Bmatrix}_{6\times 1} \begin{Bmatrix} \sigma _{ij} \end{Bmatrix}_{6\times 1}=\begin{Bmatrix} \sigma _{11}\\ \sigma _{22}\\ \sigma _{33}\\ \sigma _{23}\\ \sigma _{31}\\ \sigma _{12}\\ \end{Bmatrix}_{6\times 1}$$

Hooke's Law for Isotropic Elasticity:
 $$\begin{Bmatrix} \epsilon _{11}\\ \epsilon _{22}\\ \epsilon _{33}\\ \epsilon _{23}\\ \epsilon _{31}\\ \epsilon _{12}\\ \end{Bmatrix}_{6\times 1}=\begin{vmatrix} \frac{1}{E} &\frac{-\nu }{E} &\frac{-\nu}{E}  &0  &0  &0 \\ \frac{-\nu }{E} &\frac{1}{E} &\frac{-\nu }{E}  &0  &0  &0 \\ \frac{-\nu }{E} &\frac{-\nu }{E} &\frac{1}{E}  &0  &0  &0 \\ 0 &0 &0  &\frac{1}{2G}  &0  &0 \\ 0 &0 &0  &0  &\frac{1}{2G}  &0 \\ 0 &0 &0  &0  &0  &\frac{1}{2G} \end{vmatrix}_{6\times 6}\begin{Bmatrix} \sigma _{11}\\ \sigma _{22}\\ \sigma _{33}\\ \sigma _{23}\\ \sigma _{31}\\ \sigma _{12}\\ \end{Bmatrix}_{6\times 1} $$ which can also be written in terms of gamma and tau:
 $$\begin{Bmatrix} \epsilon _{11}\\ \epsilon _{22}\\ \epsilon _{33}\\ \gamma _{23}\\ \gamma _{31}\\ \gamma _{12}\\ \end{Bmatrix}_{6\times 1}=\begin{vmatrix} \frac{1}{E} &\frac{-\nu }{E} &\frac{-\nu}{E}  &0  &0  &0 \\ \frac{-\nu }{E} &\frac{1}{E} &\frac{-\nu }{E}  &0  &0  &0 \\ \frac{-\nu }{E} &\frac{-\nu }{E} &\frac{1}{E}  &0  &0  &0 \\ 0 &0 &0  &\frac{1}{G}  &0  &0 \\ 0 &0 &0  &0  &\frac{1}{G}  &0 \\ 0 &0 &0  &0  &0  &\frac{1}{G} \end{vmatrix}_{6\times 6}\begin{Bmatrix} \sigma _{11}\\ \sigma _{22}\\ \sigma _{33}\\ \tau _{23}\\ \tau _{31}\\ \tau _{12}\\ \end{Bmatrix}_{6\times 1}$$

HW: Airfoil Matlab Code
For this homework assignment, our task was to modify the airfoil code from the last homework to make it work for multi-cell airfoils. The airfoil was to be divided at a quarter chord ($$c/4$$) and three-quarter chord ($$3c/4$$). The profile of the airfoil from last homework is shown below.

$$\bar{A}=0.0256 m^2$$ For this week we split the airfoil into three cells. The new profile for the airfoil is shown below.

$$\bar{A_{1}}=0.0072 m^2$$ $$\bar{A_{2}}=0.0159 m^2$$ $$\bar{A_{3}}=0.0026 m^2$$ The code had to be modified to calculate the area for each cell. To calculate the area for cell 1, we first found the closest location of $$y=c/4$$. Using the cross-product of two position vectors we were able to calculate the area to the leading edge of the airfoil. Next we calculated the area for cell 3 using a similar procedure. After finding the point of $$y=3c/4$$ we were able to calculate the area to the trailing edge. For the single-cell structure we obtain a torsional constant of 5.1118e-005. By adding two spars at c/4 and 3c/4 the torsional constant increases to 5.6071e-005. The additional spars did not increase the torsional constant by much.

Matlab Code Certification
I, the undersigned, certify that I can read, understand, and write matlab codes, and can thus contribute effectively to my team. Anett Krammer Eas4200c.f08.aeris.krammer 20:25, 23 October 2008 (UTC) Jin Yu Guan Eas4200c.f08.aeris.guan 20:29, 23 October 2008 (UTC) 02:45, 24 October 2008 (UTC) Nelson B Caceres Eas4200c.f08.aeris.caceres 14:13, 24 October 2008 (UTC) Josh Holladay Eas4200c.f08.aeris.holladay 19:36, 24 October 2008 (UTC) Jesse W. McRae Eas4200c.f08.aeris.mcrae 20:36, 28 October 2008 (UTC)

Contributions
Jin Yu Guan Eas4200c.f08.aeris.guan 19:01, 23 October 2008 (UTC) Anett Krammer Eas4200c.f08.aeris.krammer 19:55, 23 October 2008 (UTC) 02:44, 24 October 2008 (UTC) Nelson B Caceres Eas4200c.f08.aeris.caceres 14:13, 24 October 2008 (UTC) Josh Holladay Eas4200c.f08.aeris.holladay 19:36, 24 October 2008 (UTC) Jesse W. McRae Eas4200c.f08.aeris.mcrae 15:44, 29 October 2008 (UTC)

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User:Eml4500.f08.gravy.jad/NotesRedo2

 Comment: The differences between this page and the original page are new pictures, and added Matlab code for 2-bar truss. Here is a link to the original page HW2 report original And here is a link to a new page's history. comparison This page is a consolidation between the original page and the new page created so that the differences between the versions could be shown. Eml4500.f08.gravy.mmm 23:06, 5 November 2008 (UTC)

FEA Notes and Homework 2

Truss Problem description
Take a specific example to see how the recipe works.

Data:

Element Length

$$\ L^{(1)}=4 \,$$

$$\ L^{(2)}=2 \,$$

Young's Modules

$$\ E^{(1)}=3 \,$$

$$\ E^{(2)}=5 \,$$

Cross-sectional Area

$$\ A^{(1)}=1 \,$$

$$\ A^{(2)}=2 \,$$

Inclination angle

 * $$\ \theta^{(1)} = 30^o \,$$

 * $$\ \theta^{(2)} = -45^o \,$$

Solving a Truss Problem with Statics

First take the sum of the moments about global node 3 and solve for $$R_3$$ which is the reaction at global node 3 in the y-direction

$$ \sum {M_1}= P(4cos(30)) +R_3(4cos(30) + 2sin(45))$$

$$R_3= \frac{-P(4cos(30))} {4cos(30) + 2cos(45)} = -4.4378$$

Then use this information to take the sum of the forces in the y-direction and solve for $$R_1$$ which is the reaction at global node 1 in the y-direction.

$$\sum{F_y}= -4.4378 + P+R_1$$

$$\ R_1= -2.5622 \,$$

By observation it can be determined that there are no forces in the x-direction so $$R_1$$ and $$R_3$$ are the only reaction forces

Solving a Truss Problem with Finite Element Analysis
1)Global Picture

$$\begin{Bmatrix} f_1 \\\vdots & \\ f_6 \end{Bmatrix}_{1X6}=\begin{bmatrix} & &  \\  & k & \\ & &   \end{bmatrix}_{6X6}*    \begin{Bmatrix} d_1 \\\vdots & \\ d_6 \end{Bmatrix}_{1X6}$$

$$\begin{matrix} global& &global& &global \\ force & &stiffness& &displacement \\  column &   &matrix  & &column \\ matrix & & & &matrix \end{matrix}$$

2) Element Picture

3) Global FD at element level:

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

$$\ k^{(e)}\,$$ = the element stiffness matrix

$$\ d^{(e)}\,$$= the element displacement matrix

$$\ f^{(e)}\,$$= the element force matrix

$$k^{(e)}\begin{bmatrix} (l^{(e)})^2 & l^{(e)}m^{(e)} & -(l^{(e)})^2 & -(l^{(e)}m^{(e)})  \\  l^{(e)}m^{(e)} & (m^{(e)})^2 & -(l^{(e)}m^{(e)}) & -(m^{(e)})^2  \\ -(l^{(e)})^2 & -(l^{(e)}m^{(e)}) & (l^{(e)})^2 & l^{(e)}m^{(e)}\\ -(l^{(e)}m^{(e)}) & -(m^{(e)})^2 & l^{(e)}m^{(e)} & (m^{(e)})^2    \end{bmatrix}_{4X4} = \underline{k^{(e)}}$$

$$\ k^{(e)}= \,$$ $$\frac{E^{(e)}A^{(e)}}{L^{(e)}}$$ axial stiffness of bar element "e" where e=1,2

$$\ l^{(e)},m^{(e)} = \,$$ director cosines of x axis (goes from 1 to 2) with respect to global (x,y coordinates)

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

$$\ m^{(e)}= \vec i \cdot \vec j= cos\left (\frac{\pi}{2}- \theta^{(e)} \right )= sin\left (\theta^{(e)} \right )\,$$

$$\ \vec \tilde{i}=cos\left(\theta^{(e)} \right)\vec i + sin\left( \theta^{(e)}  \right )\vec j \,$$

$$\ \vec \tilde{i} \cdot \vec i = \vec \tilde{i} \left( cos\left(\theta^{(e)} \right)\vec i + sin\left( \theta^{(e)}  \right )\vec j \right) i= cos\left(\theta^{(e)}  \right) \vec i \cdot \vec i + sin\left( \theta^{(e)}  \right ) \vec i  \,$$

$$\ \vec \tilde{i} \cdot \vec j= HW \,$$

Element 1:
$$\ \theta^{(1)} = 30^o \,$$

$$\ l^{(1)}=cos\theta^{(1)}= cos 30^o= \frac{\sqrt{3}}{2} \,$$

$$\ m^{(1)}=sin\theta^{(1)}= sin 30^o= \frac{1}{2} \,$$

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

$$\ \underline{k^{(1)}}=\begin{bmatrix} k_{11}^{(1)} & k_{12}^{(1)} & k_{13}^{(1)} & k_{14}^{(1)}  \\  k_{21}^{(1)} & k_{22}^{(1)} & k_{23}^{(1)} & k_{24}^{(1)}  \\ k_{31}^{(1)} & k_{32}^{(1)} & k_{33}^{(1)} & k_{34}^{(1)}\\ k_{41}^{(1)} & k_{42}^{(1)} & k_{43}^{(1)} & k_{44}^{(1)}    \end{bmatrix}_{4X4}= \left [k_{ij}^1 \right ]_{4X4} \,$$

i=row number =1,2....4

j=column number =1,2....4

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

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

$$\ k_{42}^{(1)}=k_{24}^{(1)}= -k^{(1)}\left( m^{(1)} \right)^2=\frac{-3}{4}*\frac{1}{2}= \frac{-3}{16} \,$$

Observation:

1) Only need to compute 3 numbers. Other coefficients have same absolute value, just differ by (+ -)

2) Matrix $$ \underline{k^{(1)}}$$ is synmetrix, $$k_{ij}^{(1)}=k_{ji}^{(1)}$$

$$k_{13}^{(e)}=k_{31}^{(e)}$$ (just interchange the row and column index)

$$\ \underline{k^{(1)}}=\begin{bmatrix} \frac{3}{4}(\frac{\sqrt{3}}{2})^2 & \frac{3}{4}(\frac{\sqrt{3}}{2})(\frac{1}{2}) & -\frac{3}{4}(\frac{\sqrt{3}}{2})^2 & -\frac{3}{4}(\frac{\sqrt{3}}{2})(\frac{1}{2})  \\  \frac{3}{4}(\frac{\sqrt{3}}{2})(\frac{1}{2}) & \frac{3}{4}(\frac{1}{2})^2 & -\frac{3}{4}(\frac{\sqrt{3}}{2})(\frac{1}{2}) & -\frac{3}{4}(\frac{1}{2})^2  \\ -\frac{3}{4}(\frac{\sqrt{3}}{2})^2 & -\frac{3}{4}(\frac{\sqrt{3}}{2})(\frac{1}{2}) & \frac{3}{4}(\frac{\sqrt{3}}{2})^2 & \frac{3}{4}(\frac{\sqrt{3}}{2})(\frac{1}{2})\\-\frac{3}{4}(\frac{\sqrt{3}}{2})(\frac{1}{2}) & -\frac{3}{4}(\frac{1}{2})^2 & \frac{3}{4}(\frac{\sqrt{3}}{2})(\frac{1}{2}) & \frac{3}{4}(\frac{1}{2})^2    \end{bmatrix}_{4X4}=  \underline{k^{(1)}} _{4X4} \,$$

$$\ \underline{k^{(1)}}=\begin{bmatrix} \frac{9}{16}& \frac{3\sqrt{3}}{16} & -\frac{9}{16} & -\frac{3\sqrt{3}}{16}  \\  \frac{3\sqrt{3}}{16} & \frac{3}{16} & -\frac{3\sqrt{3}}{16} & -\frac{3}{16}  \\ -\frac{9}{16} & -\frac{3\sqrt{3}}{16} & \frac{9}{16} & \frac{3\sqrt{3}}{16}\\ -\frac{3\sqrt{3}}{16} & -\frac{3}{16} & \frac{3\sqrt{3}}{16} & \frac{3}{16}    \end{bmatrix}_{4X4}=  \underline{k^{(1)}} _{4X4} \,$$

In general, $$k_{ij}^{(e)}=k_{ji}^{(e)} or k^{(e) T} =k^{(e)}$$

The transpose of $$ k^{(e)}$$ is equal to $$ k^{(e)}$$

$$\ \underline{k^{1}}=\begin{bmatrix} k_{11}^{(e)} & k_{12}^{(e)} & k_{13}^{(e)} & k_{14}^{(e)}  \\   & k_{22}^{(e)} & k_{23}^{(e)} & k_{24}^{(e)}  \\  &  & k_{33}^{(e)} & k_{34}^{(e)}\\ sym. & &  & k_{44}^{(e)}    \end{bmatrix}_{4X4}= \left [k_{ij}^1 \right ]_{4X4} \,$$

Element 2:
$$\ k^{(2)}= \frac{E^{(2)}A^{(2)}}{L^{(2)}} \frac{\left(5\right ) \left(2\right )}{2}= 5\,$$

$$\ \theta^{(2)} = \frac{-\pi}{4}= -45^o \,$$

$$\ l^{(2)}=cos\theta^{(2)}= cos (\frac{-\pi}{4})= \frac{\sqrt{2}}{2} \,$$

$$\ m^{(2)}=sin\theta^{(2)}= sin (\frac{-\pi}{4})= \frac{-\sqrt{2}}{2} \,$$

$$\ k^{(2)} = \left [k_{ij}^{(2)} \right ]_{4X4} \,$$

$$\ k_{11}^{(2)} = k^{(2)}(l^{(2)})^2 = 5\frac{2}{4} = 2.5 \,$$

Observation:

1) Absolute value of all coefficients $$\ k_{ij}^{(e)} \,$$, $$\ e=2 \,$$ , $$\ (i,j)=1, ..., 4 \,$$ are the same => comp 1 coefficient. For other coefficients, and (+) or (-)

2) $$\ k ^{(2)T} = k^{(2)} \,$$ ie. $$\ k^{(2)} \,$$ sym

$$\ \underline{k^{(2)}}=\begin{bmatrix} \frac{5}{2}& \frac{5}{2} & -\frac{5}{2} & -\frac{5}{2}  \\  \frac{5}{2} & \frac{5}{2} & -\frac{5}{2} & -\frac{5}{2}  \\ -\frac{5}{2} & -\frac{5}{2} & \frac{5}{2} & \frac{5}{2}\\ -\frac{5}{2} & -\frac{5}{2} & \frac{5}{2} & \frac{5}{2}    \end{bmatrix}_{4X4}=  \underline{k^{(2)}} _{4X4} \,$$

Elem FD rel: $$\ k^{(e)}d^{(e)}= f^{(e)} \,$$
 * e=1,2

$$\begin{Bmatrix} d_1 \\\vdots & \\ d_6 \end{Bmatrix}_{1X6}, \begin{Bmatrix} f_1 \\\vdots & \\ f_6 \end{Bmatrix}_{1X6} $$

Global FD relationship:

$$\ \underline{K}_{nxn} \underline{d}_{nx1} = \underline{F}_{nx1} \,$$

there, n=6

$$\ \begin{bmatrix} k_{11} & k_{12} & k_{13} & k_{14} & k_{15} & k_{16}  \\ k_{21}& k_{22} & k_{23} & k_{24} & k_{25} & k_{26} \\ k_{31}& k_{32} & k_{33} & k_{34} & k_{35} & k_{36} \\ k_{41}& k_{42} & k_{43} & k_{44} & k_{45} & k_{46} \\ k_{51}& k_{52} & k_{53} & k_{54} & k_{55} & k_{56} \\ k_{61}& k_{62} & k_{63} & k_{64} & k_{65} & k_{66}  \end{bmatrix}  \begin{Bmatrix}  d_{1} \\ d_{2} \\ d_{3} \\ d_{4} \\ d_{5} \\ d_{6} \end{Bmatrix} = \begin{Bmatrix}  F_{1} \\ F_{2} \\ F_{3} \\ F_{4} \\ F_{5} \\ F_{6} \end{Bmatrix}\,$$

Global F-D Relationship
Noting the fact that the equations for the elemental $$\underline{d^{(e)}}$$ and global $$\underline{d}$$ overlap, as seen in the below picture, the following equations can be derived.

$$d_1 = d_1^{(1)}$$ $$d_2 = d_2^{(1)}$$ $$d_3 = d_3^{(1)} = d_1^{(2)}$$ $$d_4 = d_4^{(1)} = d_2^{(2)}$$ $$d_5 = d_3^{(2)}$$ $$d_6 = d_4^{(2)}$$

Below is a representation of the global $$\underline{K}$$ matrix, which is the stiffness matrix for the system. The two colored boxes represent the contributions of the elemental $$\underline{k^{(e)}}$$. The $$\underline{k^{(1)}}$$ matrix is shown in blue, while the red box represents $$\underline{k^{(2)}}$$.

$$\ \begin{bmatrix} K_{11}^{(1)} & K_{12}^{(1)} & K_{13}^{(1)} & K_{14}^{(1)} & 0 & 0  \\ K_{21}^{(1)}& K_{22}^{(1)} & K_{23}^{(1)} & K_{24}^{(1)} & 0 & 0 \\ K_{31}^{(1)} & K_{32}^{(1)} & (K_{33}^{(1)} + K_{11}^{(2)}) & (K_{34}^{(1)} + K_{12}^{(2)}) & K_{13}^{(2)} & K_{14}^{(2)} \\ K_{41}^{(1)}& K_{42}^{(1)} & (K_{43}^{(1)} + K_{21}^{(2)}) & (K_{44}^{(1)} + K_{22}^{(2)}) & K_{23}^{(2)} & K_{24}^{(2)} \\ 0 & 0 & K_{31}^{(2)} & K_{32}^{(2)} & K_{33}^{(2)} & K_{34}^{(2)} \\ 0 & 0 & K_{41}^{(2)} & K_{42}^{(2)} & K_{43}^{(2)} & K_{44}^{(2)}  \end{bmatrix}\,$$

EX:

$$\ K_{11} = k_{11}^{(1)} \,$$

$$\ K_{33} = k_{33}^{(1)} + k_{11}^{(2)} \,$$

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

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

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

$$ K_{33} = \frac{9}{16} + \frac{5}{2} = 3.0625 $$

$$ K_{43} = \frac{3\sqrt{3}}{16} + \frac{-5}{2} = 2.1752 $$

$$ K_{44} = k_{44}^{(1)} + k_{22}^{(2)} = \frac{3}{16} + \frac{5}{2} = 2.6875 $$ Mbr> $$\ \ = \begin{bmatrix} \frac{9}{16}& \frac{3\sqrt{3}}{16} & -\frac{9}{16} & -\frac{3\sqrt{3}}{16} & 0 & 0 \\  \frac{3\sqrt{3}}{16} & \frac{3}{16} & -\frac{3\sqrt{3}}{16} & -\frac{3}{16} & 0 & 0 \\ -\frac{9}{16} & -\frac{3\sqrt{3}}{16} & (\frac{9}{16} + \frac{5}{2}) & (\frac{3\sqrt{3}}{16} + \frac{5}{2}) & -\frac{5}{2} & -\frac{5}{2} \\ -\frac{3\sqrt{3}}{16} & -\frac{3}{16} & (\frac{3\sqrt{3}}{16} + \frac{5}{2}) & (\frac{3}{16} + \frac{5}{2}) & -\frac{5}{2} & -\frac{5}{2} \\  0 & 0 & -\frac{5}{2} & -\frac{5}{2} & \frac{5}{2} & \frac{5}{2} \\ 0 & 0 & -\frac{5}{2} & -\frac{5}{2} & \frac{5}{2} & \frac{5}{2}  \end{bmatrix}_{6X6}\,$$ $$\ \ = \begin{bmatrix} 0.5625 & 0.3248 & -0.5625 & -0.3428 & 0 & 0 \\  0.3248 & 0.1875 & -0.3248 & -0.1875 & 0 & 0 \\ -0.5625 & -0.3248 & (0.5625 + 2.5) & (0.3248 + 2.5) & -2.5 & -2.5 \\ -0.3248 & -0.1875 & (0.3248 + 2.5) & (0.1875 + 2.5) & -2.5 & -2.5 \\  0 & 0 & -2.5 & -2.5 & 2.5 & 2.5 \\ 0 & 0 & -2.5 & -2.5 & 2.5 & 2.5  \end{bmatrix}_{6X6}\,$$

Elimination of Known Degrees of Freedom
Knowing that global nodes 1 and 3 are actually fixed points, it is seen that there is no possibility of them moving. Therefore, the displacement at those points is zero.

$$\ d_1 = d_2 = d_5 = d_5 = d_6 = 0 \,$$

$$\ \underline{d}=\begin{Bmatrix} d_1 \\ d_2  \\  d_3 \\ d_4 \\ d_5 \\ d_6 \end{Bmatrix}= \begin{Bmatrix}  0 \\ 0 \\  d_3 \\ d_4 \\ 0 \\ 0 \end{Bmatrix}$$

Since 4 of the values for $$\underline{d}$$ are zero, we can reduce the system. Turning $$\underline{d}$$ into a 2x2 matrix, would mean that you would be able to reduce $$\underline{K}$$ to a 6x2 matrix. This is down by removing the columns that are associated with the rows within $$\underline{d}$$ that are zero. The new $$\underline{K}$$ is shown below.

$$\ \underline{K}=\begin{bmatrix} K_{13} & K_{14} \\ K_{23} & K_{24} \\  K_{33} & K_{34} \\ K_{43} & K_{44} \\ K_{53} & K_{54} \\ K_{63} & K_{64} \end{bmatrix}$$

This makes the F-D equation: $$\ \underline{K}_{6x2} \underline{d}_{2x1} = \underline{F}_{6x1} \,$$

The next step is to see that since global nodes 1 and 3 are fixed, we can able the concept of "Virtual Work" which will allow us to ignore their effects for the moment and let us turn $$\underline{F}$$ into a 2x1 by removing the components from those fixed nodes. However, by doing this, we need to adjust $$\underline{K}$$ to keep the equation true. We can further reduce $$\underline{K}$$ down to a 2x2, and the following equation results:

$$\ \underline{K}_{2x2} * \underline{d}_{2x1} = \underline{F}_{2x1} \,$$

Where $$\underline{K}$$ is equal to:

$$\ \underline{K}=\begin{bmatrix} K_{33} & K_{34} \\ K_{43} & K_{44} \end{bmatrix}$$ Since $$\underline{F}$$ is now only made up of the forces from global node 2, we now know the value of $$\underline{F}$$. $$\ \underline{F} = \begin{Bmatrix} F_3 \\ F_4 \end{Bmatrix}= \begin{Bmatrix}  0 \\ P \end{Bmatrix}$$ We also can solve for $$\underline{K}$$: $$\ \underline{K}=\begin{bmatrix} 3.0625 & 2.1752 \\ 2.1752 & 2.6875 \end{bmatrix}$$

At this point, the only term not known in the global F-D equation is $$\underline{d}$$. In order to solve the system, we must first solve for $$\underline{d}$$, which gives the following equation: $$\ \underline{d}_{2x1} = \underline{K}^{-1}_{2x2} * \underline{F}_{2x1} \,$$

Where $$\underline{K^{-1}}$$ is equal to: $$\ \underline{K^{-1}} = \frac{1}{det(\underline{K})}*\begin{bmatrix} K_{44} & -K_{43} \\ -K_{34} & K_{33} \end{bmatrix}$$ $$\ det(\underline{K}) = K_{33} * K_{34} - K_{34} * K_{43} $$

$$\ \underline{K^{-1}} = \begin{bmatrix} 0.7681 & -0.6217 \\ -0.6217 & 0.8753 \end{bmatrix}$$ Plugging everything in and solving for $$\underline{d}$$ gives:

$$\begin{Bmatrix} d_3 \\ d_4 \end{Bmatrix} = \underline{K^{-1}} \begin{Bmatrix} 0 \\ P \end{Bmatrix}$$

$$\begin{Bmatrix} d_3 \\ d_4 \end{Bmatrix} = \begin{bmatrix} 0.7681 & -0.6217 \\ -0.6217 & 0.8753 \end{bmatrix} * \begin{Bmatrix} 0 \\ P \end{Bmatrix} = \begin{Bmatrix} 4.352 \\ 6.1271 \end{Bmatrix} $$

$$\ \underline{d} = \begin{Bmatrix} 4.352 \\ 6.1271 \end{Bmatrix}$$

Computing Reactions
$$ Elem 1: \underline{k}^{(1)} \underline{d}^{(1)} = \underline{f}^{(1)} $$

$$\ d^{(1)} = \begin{bmatrix} 0 \\ 0 \\ 4.352 \\ 6.1271 \end{bmatrix}_{4X1}\,$$

$$\ = \begin{bmatrix} 0.5625 & 0.32476 & ... & ... \\ & & & \\ & & & \\ & & & \end{bmatrix}_{4X4} \begin{Bmatrix} 0 \\ 0 \\ 4.352 \\ 6.1271 \end{Bmatrix}_{4X1} \,$$

$$\ = \begin{bmatrix} -0.5625 & -0.32476 \\ ... & ... \\ & \\ & \end{bmatrix}_{4X2} \begin{Bmatrix} 4.352 \\ 6.1271 \end{Bmatrix}_{4X1} \,$$

$$\ = \begin{bmatrix} -4.4378 \\ -2.5622 \\ +4.4378 \\ +2.5622 \end{bmatrix}_{4X1}\,$$

$$\ = \begin{bmatrix} f_{1}^{(1)} \\ f_{2}^{(1)} \\ f_{3}^{(1)} \\ f_{4}^{(1)} \end{bmatrix}_{4X1}\,$$

 * f1(1) and f2(1) = Reactions

 * f3(1) and f4(1) = Internal Forces

 * Observation: Element 1 is in equilibrium

 * ΣFx = f1(1) + f3(1)

 * ΣFy = f2(1) and f4(1)

HW Redo New Pictures
Element 1:

Element 2:

ΣMat any point = 0

$$P_{1}^{(1)} = [(f_{1}^{(1)})^2 + (f_{2}^{(1)})^2]^ \frac{1}{2}$$

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

 * The same method to solve for element one can also be used to solve for element 2

HW: Verify equilibrium of element 1
Let $$P_{1}^{(1)} = f_{1}^{(1)} + f_{2}^{(1)} = -4.4378 i\hat{} -2.5622j\hat{}$$ and let $$P_{2}^{(1)} = f_{3}^{(1)} + f_{4}^{(1)} = 4.4378 i\hat{} + 2.5622j\hat{}$$

Note that $$P_{1}^{(1)} + P_{2}^{(1)} = 0$$ which illustrates that $$\sum{}F=0$$.

Also, note that $$P_{1}^{(1)} \times P_{2}^{(1)} = \left( -4.4378 i\hat{} -2.5622j\hat{} \right) \times \left( 4.4378 i\hat{} +2.5622j\hat{} \right) = \left( \left( -4.4378 \times 2.5622 \right) - \left( 4.4378 \times -2.5622 \right) \right) k\hat{} = 0$$.

This shows that $$P_{1}^{(1)}$$ and $$P_{2}^{(1)}$$ are in line with each other. Since all forces are in line with each other, $$\sum{}M_{anypoint} = 0$$.

HW: Verify equilibrium of node 2
Note that $$\sum{}F = P_{1}^{(1)} + P_{2}^{(2)} + P = \left( -4.4378 i\hat{} -2.5622j\hat{} \right) + \left( 4.4378 i\hat{} -4.4378j\hat{} \right) + 7j\hat{} = 0$$.

Also, since all forces are acting on the same point, $$\sum{}M_{2} = 0$$.

HW2 MATLab Redo HW problem
To solve the system above using an automated FEM process, we've turned to MATLab. Below is the code that was used:

In order to run this code, there are a few functions that are needed. All these functions were downloaded from the book's Website. They are listed below.

After running the preceding code, this was the output results: (Note: code was changed from MATLab laungue to a generic matrix in order to be easier to read.)

Contributing Team Members
The following students contributed to this report:

Jessica Dyer Eml4500.f08.gravy.jad 22:11, 25 September 2008 (UTC)

Mikal McFarlane Eml4500.f08.gravy.mmm 00:52, 26 September 2008 (UTC)

Patrick Clayton Eml4500.f08.gravy.psc 16:33, 26 September 2008 (UTC)

Brian Eggeman Eml4500.f08.gravy.bje 17:32, 26 September 2008 (UTC)

Justin Tennant Eml4500.f08.gravy.JMT 16:31, 26 September 2008 (UTC)

Sean Sullivan Eml4500.f08.gravy.sms 17:20, 26 September 2008 (UTC)

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