text:
User:Eas4200c.f08.spars.rondon/Homework 2

=Professor's Notes=

Problem 1.1

"1.1 The beam of a rectangular thin-walled section (i.e. t is very small) is designed to carry both bending moment M and torque T. If the total wall contour length L = 2(a + b) (see Figure 1) is fixed, find the optimum b/a ratio to achieve the most efficient section if M = T and $\sigma$allowable = 2$\tau$allowable. (Sun p. 17)"

Assumptions
There are four assumptions made in the problem statement.

Assumption 1:   $$\mathbf{L} = 2\mathbf{(a + b)} $$

Assumption 2:      $$\mathbf{M} = \mathbf{T} $$

Assumption 3:     $$\ \sigma_{allowable} = 2\tau_{allowable} $$

Assumption 4:     $$\ \mathbf{t<<a},\mathbf{t<<b}$$

Shear stress

 * The first step is to determine the relationship between the shear stress, $$\tau$$, and the torque, T. In actuality the shear stress takes a parabolic shape as shown in figure (2), but in this case since the thickness is very thin (assumption 4) the shear force can be approximated by a uniformly distributed shear flow q.

 * q = $$\tau$$t (3.47)

 * Where the shear flow is related to torque by,

 * dT = ρq ds

 * where ρ is equal to the lever arm and ds is the differential length of the very thin bar. Integrating, this equation becomes

 * T = ρqS

 * The torque in this problem can be broken up into the sum of the torques in each side, T = TAB + TBC + TCD + TDA. By symmetry TAB = TCD and TBC = TDA which simplifies the equation.

 * TAB = $$\frac{b}{2} (q\cdot a) = \frac{\tau abt}{2} $$

 * TAB = $$\frac{a}{2} (q\cdot b) = \frac{\tau abt}{2} $$

 * Therefore

 * T = $$ \mathbf{2\tau a b t} $$

 * $$ \tau = \frac{T}{2abt} $$

 * This is identical to to equation provided in the problem statement with no explanation.

Solution

 * In order to solve this problem it is necessary to consider two cases.

Case One

 * Assume the bending normal stress, $$ \sigma $$, reaches $$ \sigma_{allowable} $$ before shear stress, $$ \tau $$, reaches $$ \tau_{allowable} $$.
 * Recalling that the bending normal stress is related to the bending moment by ordinate of a point on an axis perpendicular to the bending neutral axis over the second moment of inertia according to the equation

 * $$ \sigma = \frac{Mz}{I} $$

 * From which the following relationship is derived

 * $$ \sigma_{allowable} = \frac{M_{max}b}{2I} $$ → $$ M_{max} = \frac{2I\sigma_{allowable}}{b} = 2\sigma_{allowable} (\frac{I}{b})_{max} $$

 * The second moment of inertia is represented by

 * $$ I = \sum_{i=1}^{4}{[\frac{b_i(h_i)^3}{12} + A_id_i^2]} = 2[\frac{tb^3}{12} + \frac{at^3}{12} + \frac{atb^2}{4}] = 2[\frac{tb^3}{12} + \frac{at}{12}(t^2 + 3b^2)]$$

 * Assuming that t is very small compared to b this equation can be approximated by

 * $$ I = 2[\frac{tb^3}{12} + \frac{3atb^2}{12}] = \frac{tb^2}{6}[b + 3a] $$

 * Recalling assumption 1 ($$ L = 2(a + b) $$) the variable a can be stated as

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

 * Which can then be used to express the second moment of inertia in terms of a single unknown variable b

 * $$ \frac{tb^2}{6}[3(\frac{L}{2} - b) + b] = \frac{tb^2}{6}[\frac{3L}{2} - 2b] = \frac{tb^2(3L - 4b)}{12} = \frac{1}{12}[-4tb^3 + 3Ltb^2]$$

 * Returning to the equation for Mmax it is evident that $$ 2\sigma_{allowable}$$ is a constant and therefore Mmax is maximized when $$ (\frac{I}{b}) $$ is maximized.

 * $$ f(b) := \frac{I}{b} = \frac{1}{12}[-4tb^2 + 3Ltb]$$
 * $$ f'(b) = -\frac{8tb}{12} + \frac{3Lt}{12} $$
 * $$ f''(b) = -\frac{8t}{12} $$

 * The negative second derivative indicates the function $$ f(b) $$ represents a concave parabola which reaches a maximum when it's slope, or first derivative, is equal to zero.

 * $$ f'(b^{(1)}) = 0 = -\frac{8tb^{(1)}}{12} + \frac{3Lt}{12}$$ → $$\frac{8tb^{(1)}}{12} = \frac{3Lt}{12} $$ → 8b(1) = 3L  → $$ b^{(1)} = \frac{3L}{8} $$

 * $$ b^{(1)} = \frac{3Lt}{12}$$ → $$ a^{(1)} = \frac{L}{2} - b^{(1)} = \frac{L}{2} - {3L}{8} = {L}{8} $$

 * $$ f(b^{(1)}) = f(\frac{3L}{8}) = \frac{1}{12}[-4t(\frac{3L}{8})^2 + 3Lt(\frac{3L}{8})] = \frac{1}{12} [\frac{9tL^2}{8} - \frac{36tL^2}{64}] = \frac{1}{12}[\frac{36tL^2}{64}] = \frac{3tL^2}{64}$$

 * Therefore

 * $$ M_{max} = 2\sigma_{allowable} \frac{3tL^2}{64}$$

 * Recalling assumption 2 ($$ M = T $$) and the equation relating the shear force to the torque ($$ \tau = \frac{T}{2abt} $$)

 * $$ \tau^{(1)}_{max} = \frac{T^{(1)}_{max}}{2a^{(1)}b^{(1)}t} = \frac{M^{(1)}_{max}}{2a^{(1)}b^{(1)}t} = \frac{2\sigma_{allowable}{\frac{3tL^2}{64}}} {2\frac{L}{8}\frac{3L}{8}t} = \sigma_{allowable}$$

 * Recall from assumption 3 ($$ \sigma_{allowable} = 2\tau_{allowable}) $$

 * $$ \tau^{(1)}_{max} = \sigma_{allowable} = 2\tau_{allowable} $$

 * $$ \tau^{(1)}_{max} > \tau_{allowable} $$

 * The maximum shear stress exceeds the allowable shear stress and therefore this solution is unacceptable.

Case Two

 * Assume the shear stress, $$ \tau $$, reaches $$ \tau_{allowable} $$ before the bending normal stress, $$ \sigma $$, reaches $$ \sigma_{allowable} $$.

 * Recall the equation determined earlier relating shear stress to torque ($$ \tau = \frac{T}{2abt} $$)

 * $$ \tau^{(2)}_{max} = \frac{T}{2abt} $$ → T(2)max = $$ 2\tau_{allowable} a^{(2)} b^{(2)} t $$

 * It's clear that maximizing the torque is equivalent to maximizing the product a(2) b(2) which recalling assumption 1 ($$ L = 2(a + b) $$) that product becomes

 * $$ \frac{L}{2}b^{(2)} + (b^{(2)})^2 = maximum$$

 * Taking the derivative

 * $$ \frac{L}{2} - 2b^{(2)} = 0 $$ → $$ 2b^{(2)} = \frac{L}{2} $$ → $$ b^{(2)} = \frac{L}{4} $$

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

 * Therefore the torque is maximized when a(2) = b(2) = $$ \frac{L}{4}$$, therefore

 * T(2)max = $$ 2\tau_{allowable} (\frac{L}{4})^2 t = \frac{1}{8}tL^2\tau_{allowable}$$

 * Recalling assumption 2 ($$\mathbf{M} = \mathbf{T} $$)

 * $$M^{(2)}_{max} = \frac{1}{8}tL^2\tau_{allowable}$$

 * Recalling assumption 3 ($$\ \sigma_{allowable} = 2\tau_{allowable} $$)

 * $$M^{(2)}_{max} = \frac{1}{16}tL^2\sigma_{allowable}$$ → $$\sigma_{allowable} = \frac{16M^{(2)}_{max}}{tL^2}$$

 * Recall from case one

 * $$ f(b) := \frac{I}{b} = \frac{1}{12}[-4tb^2 + 3Ltb]$$ → $$ f(b^{(2)}) = \frac{1}{12}[\frac{tL}{4}[3L - L] = \frac{tL^2}{24} $$

 * Recall also from case one

 * $$ \sigma_{max} = \frac{M_{max}b}{2I} $$ → $$ \sigma^{(2)}_{max} = \frac{1}{16}tL^2\sigma_{allowable}(\frac{12}{tL^2}) = \frac{12}{16}\sigma_{allowable} $$

 * $$\sigma_{max} < \sigma_{allowable}$$

 * The maximum normal stress does not exceed the allowable normal stress and therefore this solution is acceptable. Furthermore, it proves the optimal cross section is a square such that

 * $$ \frac{b}{a} = 1 $$

=Book Summary= In Chapter 2, the course begins introducing the concepts related to elasticity. Topics include displacement, stress (normal, uniform, principal, shear), strain, equilibrium, reviewing stress transformations, stress strain relations, elastic strain energy to name a few. These topics are covered are related to composites as well as non-composites to determine the failure charateristics of the materials.

2.1
2.1 is entitled "Concept of Displacement" which simply describes the movement of a point in a material from one position to another upon deformation of the material. This displacement is described by a vector, u. This vector has components of u, v, and w in the x, y, and z directions. Therefore, when a material is deformed, the new position of any point, denoted with a "prime" symbol, e.g. x', can be described by the following: x' = x + u, where x is the original position of the point in the material. The following figure illustrates a sample cross section which undergoes a deformation:

We seek to find an equation for displacement u based on x and uniform axial strain in a member, for example, a long rod. We assume some initial displacement u0 = u (x0). Then u1 = u(x0 + Δx) for the displacement of another point x0 + Δx. The elongation of the material between these two points is Δu = u1 - u0. Then the axial strain becomes ε = du/dx, and the equation for u is u = ε0 (x - x0) + u0. Again, this holds for any situation in which the strain is uniform in the member, if not, the displacement in one section will be different from that in another, and one must seek a more accurate solution or find an acceptable value for average displacement.

=Homework Problems=

Moments of Inertia of Circular Cross Sections vs Channels

What are the moments of inertia, Iy, for both case 1 and case 2 using polar coordinates. Compare the results to see which case is more efficient. Know that t = a/10 and that a = b. For case 1: Iy of a solid circular cross section is                                   $$Iy = \int\int_A Z^2 \,dy dz$$

with respect to the y and z axes. Z2 is also considered dA which, when the integral is taken, is the area of the circle $$A = \pi r^2$$. The formula then becomes $$Iy = \pi r^4/4 \,$$

where r is the perpendicular distance from the center to the edge of the circle.

For case 2: We know, since the mass is redistributed into a channel, that A1=A2. Since the area of a circle is πr2, then we can equate that to the area of the channel, 2(bt)+t(a-2t):

$$\pi r^2=2 (b t)+t (a-2 t)\,$$

Knowing that t=a/10, and b=a, then we can rewrite this as:

$$\pi r^2=2*(2 a/10)+(a/10)*(a-2(a/10))\,$$ This in turn simplifies to:

$$7 a^2/25=\pi r^2\,$$

so we can say:

$$a=(25 \pi/7)^{(1/2)} r\,$$

With this knowledge, we can use the parallel axis theorem for rectangles, shown to the right, to solve for the moment of inertia for the channel along the axis that goes through y=a/2 if y is the vertical coordinate, and the channel were to be standing on y=0.

$$I=((1/12) t a^3) + ((1/12) (a-2 t) (t^3) + (a t) (a/2)^2)\,$$

Once again using the relationship t=a/10 we can rewrite the moment of inertia equation as:

$$I=a^4 /120 + ((1/12) (4 a/5) (a^3 /1000) + (a^2 /10) (a^2 /4))\,$$ $$I=a^4 /120 + a^4/1500 + a^4 /40\,$$ $$I=17 a^4 /500\,$$

Recalling: $$a=(25 \pi/7)^{(1/2)} r\,$$ We can say: $$I=17 ((25 \pi/7)^{(1/2)} r)^4/500 = 85 \pi^2 r^4 /196\,$$

Dividing the I from case 1, and the I from case 2, by πr4 to get both I terms in simple numerical terms with which we can compare their values, we find: For I1: $$I=1/4=.25\,$$ And for I2: $$I=85 \pi /196=1.36\,$$

Dividing I2 by I1 we can see that the moment of inertia about the center of the channel is almost 5.5 times larger than the moment of inertia about the center of the circular cross section.

Problem 1.7

Compare the load carrying capabilities of two beams having the respective cross-sections shown in the figure. Use bending rigidity as the criterion for comparison. It is given that a = 4cm, t = 0.2cm, and the two cross-sections have the same area. (Sun, 18)

Case 1 To solve this, we first need the area of the square. The area of a square is $$A = a^2$$. Since a = 4cm, we get that A is 16cm2.

Now that we have the area, we need the moment of inertia. The formula used in this case is $$Iy = 1/12bh^3$$, where b is the length and h is the height. Since both b and h are 4cm, Iy = 21.33cm4.

Case 2 To solve the area, we need to break this cross section into parts; a rectangle and two squares. The area of a rectangle is $$A = at$$ where a is the length and t is the width. Using this data, A = 0.8cm2. The area of the squares on top and bottom are $$A = b^2$$. Since there are two squares, the area of both combined is 2b2. The total area of this cross section is the area of the rectangle + the area of both squares, which gives us 0.8cm2 + 2b2. Since both areas are equal to one another, our new equation is $$0.8cm^2 + 2b^2 = 16cm^2$$. Using algebra, we solve for b and find that it is 2.76cm.

The moment of inertia for this case will also be calculated in parts. For the rectangle, Iy = 1/12ta3 where t is the base and a is the height. When the values are plugged in, we get Iy = 1/12(0.2cm)(4cm)3. This gives a total of 1.067cm4. For the squares, the Iy formula is the same; Iy = 1/12bh3 where b is the base and h is the height. Since both b and h are the same value, Iy = 1/12(2.76cm)(2.76cm)4. This gives a total of 4.84cm4 for each square. The total Iy of this cross section is Itotal = 1.067cm4 + 4.84cm4 + 4.84cm4 which is 10.74cm4.

Comparison Although both cross sections have the same area, their moments of inertia are quite different. Iy for case 2 is much less and allows for less of a bending moment when used in practical applications. This allows for maximum potential of the materials used with less weight. A key ingredient in aerospace structures.

Problem 1.7 Modified

Find R1 such that the area of case 2 is equal to the area of case 1. Also, find the moment of inertia, Iy, for both cases and compare. R0 = 10cm and t = (1/10)*R0.

Case 1

The area of acircle is $$A = \pi r^2$$. Plugging in 10cm for R0 gives us $$A = \pi (10cm)^2$$ which is 314.159cm2.

The moment of inertia of a circle is $$I = \pi r^4/4$$. When we plug in R0 for r, the moment of inertia for this circle is 7853.98cm4.

Case 2

The area of case 2 will be broken down into three parts; a rectangle and two circles. For the rectangle, A=lw, where l is the length and w is the width. This will give the result of $$A = 20cm * 1cm$$ which is 20cm2. The area of both circles combined will be $$A = 2 \pi R^2$$. The total area will be the sum of both the rectangular and two circular areas. Since the areas of both cases are to be equivalent, we get $$314.159cm^2 = 20cm^2 + 2\pi R^2$$. by subtracting 20cm2 from both sides and dividing both sides by $$2\pi$$, we get R2 = 46.8169. When we square root both sides, the value of R1 is found to be 6.84cm

The total moment of inertia will also be broken down into parts; the rectangle and the two circles. The moment of inertia for a rectangle can be expressed as $$I = 1/12bh^3$$. The moment of inertia for each circle will use $$I = \pi r^4/4$$. Since b = 1cm and h = 20cm, plugging those values in gives the total of 666.667cm4 for the rectangular moment of inertia. For each circle, plug in 6.84cm for r and the result is 1719.15cm4. When totaled, I for case 2 is 1719.15cm4 + 1719.15cm4 + 666.667cm4= 4104.97cm4.

Comparison

Although both cases had the same area, it is shown that their moments of inertia are quite different. The moment of inertia for case 2 is much less which allows for much better resistance to bending.

Do stringers always have a rectangular cross section?

Stringers do not always have a rectangular cross section. They are made of bars with open, thin walled cross sections. (See Figure 9) This method is used to resist the bending moments imposed from external pressures on      an aircraft structure. It increases the amount of moment of inertia, while keeping the material fixed.

Why use angled beam-walls for stringers?

Angled beam-walls are used for stringers because this design allows for better stockpiling, manufacturing, and construction. Looking at Figure 10, it is shown that by the stringers being angled, more can fit when stacked on top of each other. This allows for maximizing a large quantity of stringers in a small storage area. When it comes to manufacturing, stringers are usually fabricated from flat sheets of metal by a stamper. When the stamper releases, it is easier for it not to get caught on an angled wall rather than a vertical wall (Dashed line of Figure 10). Lastly, when it comes time to rivet the stringers into place, it is easier with angled walls than with vertical walls. The rivet gun will not have to squeeze into place and every rivet will be accessible rather than closed.

=Contributing Team Members= Eduardo Rondon Eas4200c.f08.spars.rondon 15:22, 25 September 2008 (UTC)

Michael Lee   Eas4200c.f08.spars.lee 16:23, 25 September 2008 (UTC)

Eas4200c.f08.spars. 18:40, 25 September 2008 (UTC)

--Eas4200c.f08.spars.prey 07:28, 26 September 2008 (UTC)

Thomas McGilvray Eas4200C.f08.spars.mcgilvray 19:02, 26 September 2008 (UTC)


text:
User talk:Eas4200c.f08.spars.rondon/Homework 2

.

Homework 2
For Homework 2 I'll take over covering problem 1.1 as far as writing down the equations and explanations, it would be a big help if someone could add diagrams like he had in class though. Hopefully I'll post the finished problem 1.1 tonight. -Michael Lee

All right. Sounds good. This afternoon before I begin studying for fluids I'll try to scan up those diagrams and upload them. I'll throw them up on the wiki just to have them up, and as you write them you can copy them into whatever section you need. - Eddie

Eddie again, Putting up two messages I got. One from Bryan, one from Thomas:

"i'll take a look into Chp 2. mind you, it will more than likely be the smallest contribution of the two sections--Stability has homework due tomorrow as well....." - Bryan

"I'll do the Ch 2 review since I've already read a lot of it. --Eas4200C.f08.spars.mcgilvray 15:27, 25 September 2008 (UTC)" - Thomas

So I guess Delvis, you mind tackling the homeworks? I'll try helping you out as well. I'm almost done drawing out all the figures by the way, I'll have it scanned within an hour.

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


text:
User:Kozuch

See my Wikipedia user page!

title: User:Kozuch


text: User talk:NullofWest

title: User talk:NullofWest


text:
Wikiversity:Candidates for Custodianship/Salmon of Doubt

I am requesting custodianship on this project for a number of reasons.

Primarily, I intend to help with day-to-day maintenance. I have watched powerlessly as a pow-wow of three custodians were unable to clear a cross-namespace redirect for move. There are numerous edit histories on this project that are not GFDL compliant and require mergers. Image policy here requires a strong hand to fix - specifically, there is no robotic tagging and notification system for obvious copyvios. Users with bot-proficiency are necessary for a project in a growth stage, and I have that proficiency. Also, there are a lot of widely-used templates that require protecting, and the main page is regularly left open to vandalism - problems that I know how to solve.

Additionally, I would be extra vigilant in recent change patrolling - I have already been instrumental in dealing with one extremely vulgar vandal, and would be able to do more with the ability to stop vandalism in the project space without having to constantly revert while discussing vandal's intentions on their talk pages.

Finally, and perhaps more controversially, I believe that there is a serious problem with divulged personal information that cannot be solved without tools. In the event that otherwise private personal information about a user is disclosed, I will delete specific edits to protect the privacy of our contributors. If contributors to this project are driven away with threats to their health and well-being, there will be nothing left but people who make such threats. I do not intend to deal with this issue untill I have gained the trust of the full community, so consider it a longer-term project.

I will not protect pages to “win” edit wars, nor will I block established contributors without seeking broad consensus of other users, nor will I delete pages without wide consensus (unless those pages are blatant copyvios, only consist of invasions of privacy or where the sole author requests deletion). Three recent examples - the student portal and the albanian port history page and a case-study page I marked for deletion - where I supported deletion as an editor, I would determine as a custodian that there was no consensus to delete the pages and keep them.

Thank you for your consideration. I look forward to answering any questions you may have. Salmon of Doubt 19:40, 12 September 2008 (UTC)

Questions for the Candidate
Are you sure you can handle this, if I'm right users on this site must be registered here for three or four months or more and have the communities trust, but with you though is a bit of a mixture - I used to be known by a different username on this site and went into a probationary period but all it received was opposes due to me being too new and slightly went into problems with category creations - my advice for you is to leave this for now and try it in the next few months maybe then other users may permit this through, or they might allow it to go through now - but the community decides (including) the custodians. DarkMage 19:48, 12 September 2008 (UTC)

 * I am singularly positive that I can handle this. I do not see a time-registered requirement for custodianship, but if it is the will of the community that I wait longer, I am happy to do so. There are things I can fix right now, however, that have nothing to do with the conflicts that I am regetably involved in. At the very least, let me show my stuff. Limit me, as an example, to dealing only with users and issues I have no past history with and I'll prove myself. Salmon of Doubt 19:51, 12 September 2008 (UTC)

 * That maybe so, but other's may or may not agree to this - though if it does go through and someone does mentor you in using the tools then - you'll have to understand that you will be faced with many, many questions in the request for full custodianship, some of which maybe impossible to answer, though lets see if someone will mentor you or agree that you should wait a bit longer. DarkMage  19:55, 12 September 2008 (UTC)

 * No question is impossible to answer, thought it may be difficult to convince others you are right. I merely want the chance. Salmon of Doubt 19:56, 12 September 2008 (UTC)

 * Your first reply to me was to accuse me of being a Joan of Arc vandal, which is quite an offensive judgment to make especially for someone that seeks this candidacy. You also haven't answered questions related to the accusation of 'combatant' you made. Somehow you believe it is ok to make such hurtful accusations. Given the findings of controversy that I haven't brought here to Wikiversity until compelled like this, your accusations are not justified at all. Given a certain consensus that was stated on IRC about surrounding politics related to the past events that really had nothing to do about me, I have been very patient and hurt by such wrongful accusations then and now. 1) How are you going to solve this? 2) Why would you be happy to do it? 3) Are you going to update the revision link on the 'authority' you gave SB Johnny on his talk page after you answer these questions? Dzonatas 20:54, 12 September 2008 (UTC)

 * I regret not deleting that page when you removed your comments and will do so now. Salmon of Doubt 20:57, 12 September 2008 (UTC)
 * re the authority - What? I don't understand. SBJ has the authority to desysop me without cause. What would you propose I do? Salmon of Doubt 21:03, 12 September 2008 (UTC)
 * Ahh yes. In the event I substantively change my statement or make any additional commitments in response to questions I will update my diff. Salmon of Doubt 21:24, 12 September 2008 (UTC)

 * Deleting the page does not substantially change the accusation on that page and also on other pages that have taken place, which may contain the way you still feel even after deletion. I don't see an easy way to justify someone trying to be contrite by the act of evidence being hidden of past actions. It doesn't provide a solution to your conduct I have noticed as stated above. In fact, you stated when you would update your diff, but you haven't changed that to update it to include answer you make to my questions. I'm not SB Johnny, and I can't force you to make further commitments than what you conditionally set. It does, however, reveal your level of willingness to do it, and that you did not answer my question of 'why you would be happy to do it' . I don't think you are ready for this. Dzonatas 22:32, 12 September 2008 (UTC)

 * If I make any additional commitments, I'll certainly update the diff. There's little more I can do about my past mistakes than apologize and pledge to move on, so that's what I did. I found your statement confusing - what do you mean by "why you would be happy to do it?" To what do you refer? Salmon of Doubt 23:21, 12 September 2008 (UTC)

 * It is in reference to question 1: to solve. Dzonatas 16:03, 13 September 2008 (UTC)

 * Sorry, I still don't understand. Could you start over with your request? Salmon of Doubt 16:27, 13 September 2008 (UTC)

 * Yes, I could. I don't know what you don't understand about it, however. Dzonatas 17:21, 13 September 2008 (UTC)

Questions from JWSchmidt
Q1. "protect the privacy of our contributors" <-- can you suggest any improvements/changes to Privacy policy? Should we make that an official policy? --JWSchmidt 22:35, 12 September 2008 (UTC)

 * I did not see anything missing from that policy. I'll re review over the next few days. It should be official, yes. Salmon of Doubt

Q2. You archived active discussions on a talk page then obstinately refused to clean up the mess you had made. Why should the Wikiversity community trust you to clean up messed rather than make them? --JWSchmidt 03:01, 13 September 2008 (UTC)

 * I don't feel that I made a mess in that instance as it was trivial to revert my archiving of the page. I believe archiving the talk page would have solved the ongoing reversions on the page. That the alternative solution of my just allowing other users to pretend to be multiple individuals and deceptively sign their contributions was also successful is a demonstration of my improving acceptance of community norms here. Salmon of Doubt 11:33, 13 September 2008 (UTC)

 * Are you retaining for yourself the right to improperly archive pages in the future? When you do so, do you expect someone else to repair the damage you do? --JWSchmidt 16:16, 13 September 2008 (UTC)

 * No. I don't intend to improperly archive any pages. Salmon of Doubt 16:24, 13 September 2008 (UTC)

 * Based on my experience with you as a Wikiversity editor, I must entertain the hypothesis that when you say, "improperly", you might not be thinking about the standards of this community. Would you be willing to help author a Wikiversity guideline for how and when to archive talk pages and agree to abide by that policy? --JWSchmidt 17:37, 13 September 2008 (UTC)

 * We should work together to make a guideline, and after such time as it is approved, I will abide by it, and until such time I will archive no pages but my talk page or pages with over 50kbs of text, and then only sections that have not been edited in over 30 days. Salmon of Doubt 01:19, 14 September 2008 (UTC)

Q3. In my view there are two paths to custodianship, 1) demonstrate an understanding and commitment to the Wikiversity Mission or 2) have documented sysop experience at another WMF wiki and an interest in helping with technical matters at Wikiversity. As far as I can see, you meet neither of these requirements. Why should the Wikiversity community allow you to be a custodian? --JWSchmidt 03:14, 13 September 2008 (UTC)

 * I am unwilling to prove I fit on your second path, so based on your statement, I should not. Salmon of Doubt 11:33, 13 September 2008 (UTC)

 * Since you will not reveal all your alternate accounts (which I believe is the strongly suggested course of action at Wikipedia), would you be willing to authorize a checkuser investigation that would confirm that you have no more than two alternate accounts on WMF wikis? --JWSchmidt 16:16, 13 September 2008 (UTC)

 * I have only two active accounts on this Wiki, this account and my redirected bot account. I do not authorize anyone to go fishing for what my Wikipedia user name is, as that has no relevance on this totally independent project. Salmon of Doubt 16:24, 13 September 2008 (UTC)

 * I respect your right to create as many alternative accounts as you find useful. However, there is a recent case of a person who created multiple accounts on a WMF wiki project and apparently used those accounts in tandem to improperly influence community decisions. There have been recent calls for diligence at all WMF projects. You have attracted special attention to yourself because of your editing history at Wikiversity. Are you willing to say how many accounts you have created and/or have password-access to at en-Wikipedia? Are all of those accounts cross-identified by means of links that connect the user pages of the accounts? "I have only two active accounts on this Wiki" <-- do you have password access to any additional accounts at Wikiversity? Do you have any inactive accounts at Wikiversity? Do you coordinate off-wiki with other Wikimedians and act as a meat-puppet for them when you edit at Wikiversity? --JWSchmidt 17:37, 13 September 2008 (UTC)

 * I made accounts that were the same as Moulton's character names to prevent those names from being usurped by another user, but I scrambled their passwords. I have access to three (?) account at en-wikipedia - one that is my real name, one that is a bot linked to my real name and a secret account that I use when I want to do things and not be bothered - that final account is linked to my main account through knowledgeable individuals. Oh - I signed Salmon of Doubt up to SUL and accidentally made an edit on it on en, I think.

Q4. When you talked about using an unauthorized bot to control the ability of another Wikiversity participant to edit, some people were upset and indicated they did not feel safe at Wikiversity if you were free to make what they felt were terror threats. You indicated that you intended to use the bot to "remove links to malicious sites" At that time, I tried to get you to explain what you mean by "malicious sites". You replied, "that's none of your concern". Is it your intention as a custodian to have no concern when editors threaten to use unauthorized bots at Wikiversity? What is the correct way to deal with unwanted links to external sites? Why should the Wikiversity community trust you to follow proper procedures after you made a threat to violate the Wikiversity bot policy? "The operation of a bot requires approval" <-- have you used an unauthorized bot at Wikiversity? --JWSchmidt 03:44, 13 September 2008 (UTC)

 * Malicious sites are sites which either attempt to install malicious software on your computer (ency...dramatica), or sites where the accessing of the site tends to reveal personally identifying information to an individual with a history of disclosing that information. I do not intend to deal with user-conduct issues during my time as a probational custodian, including users who threaten to use unauthorized bots. Beyond that, how are you sure that such threats are not just special learning? The "correct" way to deal with unwanted links to external sites is to remove them. As I've said, I have used an unauthorized bot twice at Wikiversity - once to roll-back the contributions of a vandal who was later blocked indef for vandalizing and impersonation of a living person and once because test code escaped the corral. I reverted the one change made by the test code immediately. I will defend my use of automated tools to revert vandals, even if it is technically in violation of the bot policy if you so insist. Salmon of Doubt 11:33, 13 September 2008 (UTC)

 * At first I found it strange that a Wikiversity custodian would claim the right to violate policy...but now that I think about it, your mentor has done the same thing. Is there a legitimate way to prevent Wikiversity from having links to external sites? I'm not sure that your plans ("I will defend my use of automated tools") mesh with Wikiversity policy. Are you sure you cannot suggest additions to the proposed privacy policy? Maybe something like, "User:Salmon of Doubt has the unique right to, on his own, decide what constitutes a 'malicious site' and use his unauthorized bot to remove links to those sites." Then again, maybe part of your mentoring is about how to impose new policy on a whim without having to ever edit a policy page. --JWSchmidt 16:16, 13 September 2008 (UTC)

 * Yes. I could add those sites to the blacklist. I think you'll need to remain civil. Salmon of Doubt 16:24, 13 September 2008 (UTC)

 * I find it strange that only when prompted did you mention the legitimate way to prevent Wikiversity from having links to external sites. Do you intend to work with the Wikiversity community to make decisions about blacklist/whitelist or do you intend to use other means to impose your views of what constitutes a 'malicious site' at this community? --JWSchmidt 17:37, 13 September 2008 (UTC)

 * I don't intend to be involved in decision to blacklist/whitelist any sites at this project in the immediate future except to the extent that the sites are indisputably and currently malicious (nimp.org, for a timley example). Salmon of Doubt 01:19, 14 September 2008 (UTC)

Q5. When I was developing a page by constructing a learning project at the Student Union page you kept reverting my edits and falsely claimed that there was consensus to support your desire to keep the old undeveloped version. Why should the Wikiversity community trust you to be a custodian when you feel comfortable making a false claim of consensus while repeatedly reverting a fellow Wikiversity participant who is in the middle of building a learning resource? --JWSchmidt 04:35, 13 September 2008 (UTC)

 * There was no false claim of consensus made in that case - there was an actual consensus that your actions were counterproductive. I am thankful to Jade who eventually found a third-solution to the problem - moving your problematic and disruptive contributions to a third page. Salmon of Doubt 11:33, 13 September 2008 (UTC)

 * As a custodian, how do you intend to judge consensus for matters such as page deletion decisions? For the particular case of your edit waring at the student union page, can you count the number of editors who spoke for and against your claim of consensus? --JWSchmidt 16:16, 13 September 2008 (UTC)

 * I will evaluate the arguments of all of the parties and see if there are any valid objections raised. In the event of any valid objection. I would not use custodian tools. Consensus is not a vote, so I would not "count" editors. Salmon of Doubt 16:26, 13 September 2008 (UTC)

 * What is Wikiversity policy with respect to the edit reverts? Do you also agree to follow Wikiversity policy with respect to the edit reverts you perform in the future? "any valid objection" <-- What do you view as constituting a "valid objection"? --JWSchmidt 17:37, 13 September 2008 (UTC)

 * There is not currently a policy on reverts to the best of my knowledge. I no longer intend to revert the contributions of other editors outside of blatant copyvios and obvious vandalism. There is a policy on the use of rollback in edit wars (don't), but that seems to be all there is. Salmon of Doubt 01:19, 14 September 2008 (UTC)

 * When you were repeatedly reverting my good faith efforts to create a learning resource I pointed you towards this page: Rollback. Are you willing to accept the idea that the only time it is correct to revert an edit at Wikiversity without providing a specific reason in the edit summary is when reverting cases of obvious vandalism? --JWSchmidt 02:23, 14 September 2008 (UTC)

 * In principle, yes. There are times where it might be appropriate to rollback without edit summary but I can't currently put my finger on it. Salmon of Doubt 04:27, 14 September 2008 (UTC)

 * That learning project was highly controversial. Consensus (3-1) was to remove the information which you posted there.  Or do you believe, JWSchmidt, that you had consensus (1-3) in consistently re-adding the material?  The Jade Knight 04:17, 14 September 2008 (UTC)
 * Wikiveristy uses reasoned discourse and not numbers to determine consensus (or consensus). Hillgentleman|Talk 04:24, 14 September 2008 (UTC)
 * "Consensus (3-1)" <-- thanks for providing your (Jade Knight) count...I'm not sure how you arrived at that count. The reason for deleting the learning exercise seems to have been: "I do not want to learn about deletionism, so delete this learning resource about deletionism". I do not accept that as a valid justification for the removal of Wikiversity content from a page. Maybe there was another argument for deleting the learning exercise? --JWSchmidt 16:52, 14 September 2008 (UTC)

Q6. As a custodian, how many of your fellow Wikiversity participants do you expect to call a troll? --JWSchmidt 08:21, 13 September 2008 (UTC)

 * None. Like I've said, I made mistakes in the past and do not intend to repeat them. Salmon of Doubt 11:33, 13 September 2008 (UTC)

 * Have you apologized to all Wikiversity participants who you have called "troll" or accused of trolling? Will you apologize to them all? --JWSchmidt 16:16, 13 September 2008 (UTC)

 * Sure. Who needs apologizing to? I'm sorry I called any and all of you trolls. Salmon of Doubt 16:24, 13 September 2008 (UTC)

 * Thanks for making that apology. Since learning is a two-way street, can you ask your mentor to think about making the same apology? I seem to recall that he once referred to you as a troll (in #wikiversity-en). --JWSchmidt 17:37, 13 September 2008 (UTC)

 * I don't feel the need to be apologized to, but that you for thinking of me. Salmon of Doubt 01:19, 14 September 2008 (UTC)

Custodians willing to mentor

 * I will mentor on two conditions: (1.) you will act in exactly the manner you described above, and (2.) you will leave an edit on my talk giving me clear authority to ask a steward to revoke your tools without discussion, reason, or explanation. --SB_Johnny talk 20:04, 12 September 2008 (UTC)

 * I accept those conditions. Salmon of Doubt 20:04, 12 September 2008 (UTC)

 * Please make the edit. --SB_Johnny talk 20:06, 12 September 2008 (UTC)

 * Was crafting. Salmon of Doubt 20:08, 12 September 2008 (UTC)

✅ Good luck with the mop. --mikeu talk 02:24, 13 September 2008 (UTC)


text:
User:ΚΕΚΡΩΨ

 καὶ γὰρ ἐβούλοντο μιᾶναι τὴν γῆν ταύτην, μὴ γιγνώσκοντες τὴν ὕβριν ἣν ἐποίουν...

title: User:ΚΕΚΡΩΨ


text:
User talk:ΚΕΚΡΩΨ

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

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

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

And don't forget to explore Wikiversity with the links to your left. Be bold to contribute and to experiment with the sandbox or your userpage, and see you around Wikiversity! Jtneill - Talk - c 10:59, 20 September 2008 (UTC)

title: User talk:ΚΕΚΡΩΨ


text: Social psychology of climate change

title: Social psychology of climate change


text:
User:U3006567

=E-portfolio= Social psychology helps to make sense of the world me and assists me to understand different behaviours of other people. The interaction with other people becomes easier. It teaches how individual effort decreases as group size increases, often observed in real life large groups how people behave. Behaviourism seeks to explain all of psychology in terms of learning priniples of reward and punishment. This is supported by daily life events that every activity has consequences, if good you get a reward or bad be punished. ≈…––…12:54, 29 September 2008 --gladys U3006567 03:52, 14 October 2008 (UTC)gladys

Part 1
Social Self A man's social self is the recognition which the person gets from his freinds. We are not only gregarious animals, liking to be in sight of our fellow humans. We have innate propensity to get ourselves noticed and noticed favourably by our own kind. This we demonstrate by the the way dress taking extra care to impress others around us, making us feel good. Whatever, we do is always intentionally conducted to attract the attention of our humans. It is very important for us to be noticed by others; as was the comment that; that "No more fiendish punishment could be devised, were such a thing physically possible, than that one should be turned loose in society and remain absolutely unoticed by all the members therof". If no one turned round when we entered into a room, answered when we spoke or minded what we did, express approval. If none of these small things happened in our lives, we become angry, disapointed, feel bad and sad.--U3006567 11:11, 16 October 2008 (UTC) 16 October 2008

Part 2
The Self's Main Jobs Every person has and individual self. Everyone of us has a separate body and self. Self has three main parts. The first part consists of self-knowledge is to have self-awareness that enables the individual to develop elaborate sets of beliefs about themselves. Self-knowledge stores information about self-awreness, self-esteem and self-deception they are all important for self-maintenance and protection. The second part is the interpersonal self or public self is the part that hepls the person to interact with others socially.This makes people to adopt a particular image to convey to others.This image is often well gaurded to maintain the good-self, as bad-self would cause embarrassment to one's friends. The third part is the agent self or executive, is the working part thaty gets things done enables self to make choices and exert control. Control includes self control and control of other people. Agent self enables the person to make decisions on essntial matters such going to work or going out socialising with friends. --U3006567 12:33, 16 October 2008 (UTC)

Part 3
The Self's Main Jobs Human beings have a deep desire for self-knowledge, although, some people are more eager than others to learn about themselves. It is believed that no one is indifferent to self-knowledge. Reasons for wanting to understand self-knowledge, include three motives for self-knowledge, appraisal, self-enhancement and consistency motive. The simple desire to learn the truth about oneself is recognised as the first motive for wanting self-knowledge. The appraisal motive consists of a broad open-minded curiosity for information that is both important and reliable. Appraisal motive enables the person to make sensible choices of doing easy tasks first before doing difficult one after.Self-enhancement motive is the desire to learn favourable things about the self. Self-enhancement motive can exert bias, makes people to dismiss criticism while exaggerating praise for good qualities promoting vanity. The third motive is the consistency motive, the desire to get feedback that confirms what the person already believes about himself or herself. Once personal opinion is formed about self it makes it difficult to take other people's opinions. Thus, the person can be resistant to change.--U3006567 14:08, 16 October 2008 (UTC) Baumeister & Bushman, 2008</

Aggression
Here are some of the topics on aggression: Aggression is the act of extreme violence, a behaviour intended to harm another person. hostile aggression is a hot impulsive angry behaviour that is motivated by a desire to another person. The other form of aggression is known as the instrumental aggression, a cold premeditated, calculated behaviour that is motivated by some other goal of obtaining money or restoring justice. Examples of aggressive behaviours are school murders committed by disgruntled student for repeated teasing and bulling by their fellow students.--U3006567 14:37, 16 October 2008 (UTC)
 * Topic 1'Different types of aggression'

Aggression is an innate physiological reaction necessary for survival purposes used for protection and other needs. In other cases aggression is used to show power. Leaders of organisations often apply aggression to impose their powers, group leaders do the same to make to members, for them conform to commands. There other extremes of aggressions of violent acts, such as bashing another person. These sort of aggressive behaviours are usually motivated by frustration or anger for some reason or other. In social psychology explanations of aggression are divided into two classes, the biological as briefly explained above and the social which may be quite broad since it usually under free will. According to Vaughan & Hogg (2005) social psychology focus of social factors of aggression. Thus, social psychology addresses the classes of theories that incorporates the learning part of aggression. --U3006567 20:06, 27 October 2008 (UTC)
 * Topic 2 Is Aggression Innate or Learned?

 * Topic 3 Aggression can be learned

In social psychology, social learning is a broad range of behavioural approaches featuring processes such as acquisition of behaviour. The instigation of overt acts, including the maintenance of certain behaviours. Bandura (1973,1977, in Vaughan & Hogg 2005)recognised that if antisocial and pro-social behaviours as they are both influenced by biological factors connected to aggressiveness. Bandura conducted experiments to demonstrate that children can readily adopt mimicking (modelling) aggressive acts of others. Modelling is well recognised behaviour in studies of children,adopted by government organisations in parebtal teaching. Television advertisements show parents drinking alcohol, smoking cigarettes in the presence of children. These adverts are aimed at making parents aware that children mimic adult behaviour. If an adult acts aggressively toward another in the presence of the child, that child will go and do similar aggressive behaviou to others. Hence, Bandura's classic study in social psychology of "Sock it to the Bobo doll" is still current as demonstrated by the TV adverts of bad parenting. Aggression was recognised as a natural instinct,by Freud(1930)and termed it the' death instinct', that aggression is an innate process to healthily released. Studies of ethology, Lorenz (1966, Vaughan & Hogg 2005) recognized the positive value of survival it remains a parental responsibility to teach children to control their aggressive actions to avoid trouble or hurting another person.--203.173.12.219 04:30, 1 November 2008 (UTC)

Environmental psychology
Global warming 3
 * 1 iterm text
 * 2 item text
 * 3 item text

title: User:U3006567


text:
User talk:U3006567

Well done
Good job on setting up your account,. Let me know if I can do anything as you go along. 'Official' Wikiversity welcome below. -- Jtneill - Talk - c 13:23, 20 September 2008 (UTC)

 Hello U3006567, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. After you leave a comment on a talk page, remember to sign and date; it helps everyone follow the threads of the discussion. The signature icon in the edit window makes it simple. To get started, you may

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

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

And don't forget to explore Wikiversity with the links to your left. Be bold to contribute and to experiment with the sandbox or your userpage, and see you around Wikiversity! Jtneill - Talk - c 13:23, 20 September 2008 (UTC)

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

Hi
We are getting closer! I think. Did you post this reflection? Prosocial behavior. You can reply right here - this is your "talk page". -- Jtneill - Talk - c 11:58, 29 September 2008 (UTC)
 * Hi Gladys, I've moved your most recent contribution onto your user page (which are also calling your e-portfolio page). Here is that page: User:u3006567. Please edit it to include your other reflections. You can also find a link to it anytime you are logged in in the very top bar. -- Jtneill - Talk - c 13:08, 29 September 2008 (UTC)

Hi James Yes I did post my text reading some time ago, heading to my e-portfolio homepage --U3006567 19:06, 9 October 2008 (UTC)gladys

Oskar Schindler
Did you write about Oskar Schindler? - there is an "unclaimed" reflection here: Social psychology (psychology)/Lost contributions. If it is yours, then copy it to your page. If not, then reply to this message below. -- Jtneill - Talk - c 10:27, 1 October 2008 (UTC)

Finding your e-portfolio page
Hi Gladys, I will send you an email with a screenshot, but if you are logged in, you should be able to see a little person icon in the top center of the screen, with U3006567 in blue. Clicking on that will take you to your user page / e-portfolio page. -- Jtneill - Talk - c 21:57, 9 October 2008 (UTC)

Essay submission
You need to email me your essay as an attached file (do not post it on Wikiversity). See the instructions for how to do this here: Social psychology (psychology)/Assessment/Essay/Submission. -- Jtneill - Talk - c 22:30, 26 October 2008 (UTC)
 * I have successfully received your essay via email. -- Jtneill - Talk - c 11:07, 2 November 2008 (UTC)

Images
Info about adding images can be found here: Help:Images. I suggest you start with using some images that have already been used on some Wikiversity pages. That way you can look at the code on that page (e.g., by editing the page and copying the image syntax, then pasting on your page, and saving). As you get more confident, you can find images from Wiki commons e.g., at Social psychology on Wiki commons and use them on your page. Happy to help, but be aware that this takes time and at this stage, you may be better off concentrating on your text. -- Jtneill - Talk - c 11:27, 2 November 2008 (UTC)

title: User talk:U3006567


text:
Climate change

Aim of the project:
This learning project aims to support participants in their response to climate change through the collective identification, remixing, creation and use of learning resources on this topic.

Outcomes for the Learner-based Activiities
Please add any outcomes you would like this project to achieve. Also use the talk page to discuss these outcomes.
 * Find participants for the project
 * Discuss and select approaches to organise and implement the project
 * Work collectively to set the aims and objectives of the project
 * Negotiate a provisional outline of how these objective might be achieved
 * Collectively identify, locate, remix, edit and create learning activities and resources that will fulfil participants individual objectives.
 * Participate in the learning activities
 * Reflect on and evaluate these learning experiences
 * Refine the process and repeat

Learners Activity Page

 * /Learners Activity/ - Comment of Learner - response to performed role plays in the context of Climate Change. Suggestion of learner, how the learning environments about climate change can be improved.
 * (Data Driven Analysis) Explain the requirement of data driven analysis for climate change decision making. The artic ice produced layers of ice in the past. Are small bubbles or air encapsuled in the ice and how does that give information about greenhouse gas concentration in the past? How can the corresponding climate on earth be identified with other samples or fossiles on earth to clarify what climate could be expected? Look at the speed of increase of greenhouse emission in the atmosphere, how does that relate to mean temperatures and the application of the Moving Average on weather data to identify a long term trend. What are available the resources of peer-reviewed journals that you can find to support evidence-based decision making.
 * (Biodiversity) consider the climate change in the context of biodiversity. Try to describe possible links between Climate Change and Biodiversity as hypothesis. Then look for scientific evidence in peer-reviewed journals that support your previously mentioned links/hypothesis. What is the scientific evidence you find and what are the conclusions you draw from the scientific results?

Learning Activities

 * (Sustainable Development Goals) Analyse the Sustainable Development Goals and identify the contributions to Risk Management in the area of climate change.
 * (Risk Management) Analyse the generic approach describe in the Risk Management learning resource. Map risks to risk mitigation resource and analyse possible obstacles and challenges of the application risk mitigation strategies.
 * What is the difference between weather and climate?
 * (Polar Ice Shield - Atmospheric Diary) Snow encapsules in bubbles of the atmosphere in layers per year. The layers can be identified like layers in a tree trunc (dendrochronology). In the polar ice shield layers of ice aggegrate over years. This creates a climate diary of the atmosphere. If you combine this information with plants and animals living in a specific area at a specific time (e.g. proven by fossiles, paleogentics, ...) it is possible to estimate what climate is matching with the current Greenhouse gas concentration in the atmosphere?
 * (Risk Literacy) Why is it difficult to understand the consequences of climate change? "I am expose to 40 degrees celsius between Winter and Summer. I do not care about 2 degree temperature difference mentioned in global warming scenarios!"
 * (Responsibility) Discuss the responsibility of the current generations for the future generations. Create a role play and assign a role to different stakeholders. Discuss the responsibilty of the stakeholders and the requirements and constraints in which they perform decisions. What are the drivers for the decision to what extend is the climate change responsibility part of decision making? Try to identify conflicts of interest and discuss possible solutions for climate change risk mitigation.

Processes used in this project
This section will contain details of the various processes or methods used in this project in producing the stated outcomes. At the moment this is just a random list of some possible approaches. Please add your own suggestions or links to established approaches.

Participation and ownership
The nature of wikis encourages joint ownership of all the participants. The experience of other wiki projects can be used as a guide to encouraging participation.

Webquests
A webquest is an enquiry based learning activity were participants use the Internet to locate the resources they need to resolve the problem posed by the webquest. webquest.org

Reflective diary
This project encourages people to take a personal approach to the subject of climate change. Peoples feelings and emotions play a key part in any learning. One approach to paying attention to your feelings is to keep a reflective learning diary. In this you can record your immediate feelings about some aspect of your learning. Later you can reflect on these feelings and see how they effected your learning. If you wanted to share your reflective learning diary you could set up a page on your Wikiversity home page and add entries there. Alternatively you could setup a reflective learning blog and which can be either private or public.

Learning resources
Please add links to suitable learning resources in this section. Ideally these should be licensed under a creative commons or equivalent licence.
 * United Nations - Climate Change e-Learning Portal
 * Explore the Wikipedia content about Climate Change
 * C-Roads Climate Interactive: Create a role play for learners/students in which they pick up the role of different stakeholders, that negotate. The task is to represent the stakeholders interests and negotiate the actions for climate change risk mitigation. After negotiations define the emission peak year first and then calculate the total greenhouse gas reduction rate, reduction of deforrestation, ... for your activities. Discuss the results on the reduction of temperature increase until 2100.
 * Explain the need for /Risk Literacy/ about climate change and derive the purpose of the impact in the future.
 * Try to identify how many people will be affected by the rise of sea level
 * loss of agricultural areas,
 * urban area (try to quantify the affected population)
 * (AR Conference) Analyze the concept of an AR Conference in the context of climate change. Learner can meet like in a video conference in an AR Conference at different locations on earth to understand the impact of climate change on the specific location on earth. Augmented Reality can be used to explore the affected location without creating a huge carbon footprint by flying to that location. Furthermore it might be expensive and very time consuming to explore those place in the context of school, colleges and universities. How can Real World Labs be used and shared e.g. in Wikiversity so that those learning environments are created by the schools, colleges and universities at specific geolocation who shared with the other educational institutions all over world to be used for climate change Risk Literacy.
 * Due to climate change related risks like
 * rise of sea level of 3mm per year (e.g. need to relocate villages on the Fiji Islands)
 * increasing number of extreme events (see Disaster Management and UN-SPIDER)
 * alteration of spatial pattern of communicable disease according to spread of vectors like mosquitos (see Expert Focus Group for Space and Global Health)
 * loss of value of fossile fuel investments (see Global Challenges)

Essentials
This section contains links and references to key information about climate change. Please only add items to this list that you feel are essential to understanding climate change.

Reading

 * Summary of the IPCC Synthesis Report (PDF 2Mb) contains an overview of the latest IPCC report giving a state of the art overview of the science, consequences and mitigation of climate change and global warming
 * The Copenhagen Diagnosis report(PDF 3.3Mb) is an update of the science in the proceeding report and includes new information published in the three years since the IPCC report

Websites

 * The IPCC (Intergovernmental Panel on Climate Change) is the lead organisation for the international investigation into global warming and climate change.
 * The Copenhagen Diagnosis website synthesizes the most policy-relevant climate science published since the close-off of material for the last IPCC report
 * Real Climate is a blog by scientists working in climate science and an invaluable source for those who are scientifically inclined.

Introduction to the science

 * Climate Literacy Handbook

Short courses

 * Openlearn's global warming course provides a structured introduction to the science behind the debate from the people who brought you the Open University.
 * Global warming is a series of videos lectures by David Archer covering a 10-week course for non-science majors focuses on a single problem: assessing the risk of human-caused climate change.

Wikiversity projects
These are a list of existing projects on Wikiversity that relate to this project. If you find any other projects please add them. A description or summary of each project would also be useful.
 * /Personal views/ on climate change
 * Northern Arizona University/Environmental Ethics/Climate Change
 * /Scientific views/ on climate change
 * Juridical national measures on climate change
 * Visual thinking strategies
 * Peak oil, energy, and society

International Activities

 * 2017 UN Climate Change Conference (COP23), Bonn, Germany
 * 2016 UN Climate Change Conference (COP22), Marrakech

External Resources

 * Google Map of Climate Change: Zoom in on examples of the assessments of the National Climate Assessment and Development Advisory Committee.
 * Google Map of the Rise, Fall, and Migration of Civilization Due to Climate Change
 * United Nations Climate Change e-Learning Portal

title: Climate change


text: User:Kozuch/Survey responses

title: User:Kozuch/Survey responses


text:
User:Eml4500.f08.jamama.jan

Some of the notations

meta:Help:Displaying_a_formula

Submission 1
1st week & Mtg 5 (code is transfered from Wikipedia page)

Submission 2
Mtg 7

Submission 3
Mtg 12

Submission 4
Mtg 18

Submission 5
Submission 5 Matlab

Submission 6
Mtg 29 + Electric_Pylon code (extension of Justin Code)

Submission 7
Mtg 35

title: User:Eml4500.f08.jamama.jan


text:
User:Eml4500.f08.ateam.boggs.t/Homework Two

The Definition of a Moment
A force moment is a force applied to a system at a radial distance from the origin of that system. The SI units for a moment is the Newton-meter(N·m), the US Customary Units is the foot-pound(ft·lbf). The equation for a moment is:

$$ \overrightarrow{M_o} = \overrightarrow{F}\times\overrightarrow{r} $$

This is the definition of a moment about point O. A moment can also be referred to as the torque of a system.

A 2-Beam Truss System Solved Statically

$$\sum F_{x} = 0 = R_{1}cos(30) - R_{2}cos(-45)$$

$$\sum F_{y} = 0 = R_{1}sin(30) + R_{2}sin(-45) + P $$

$$ R_{1} = \frac{R_{2}cos(-45)}{cos(30)} = \frac{\sqrt{2}}{\sqrt{3}}R_{2} $$

$$ 0 = \frac{\sqrt{2}}{2\sqrt{3}}R_{2} - \frac{\sqrt{2}}{2}R_{2} + P $$

$$ 0 =\frac{\sqrt{2}(1 - \sqrt{3})}{2\sqrt{3}}R_{2} + P $$

$$ R_{2} = -\frac{2\sqrt{3}}{\sqrt{2}(1 - \sqrt{3})}P $$

$$ R_{1} = -\frac{2}{1 - \sqrt{3}}P $$

Element 1:
L(1) = 4 E(1) = 3 Ac(1) = 1 θ(1) = 30° - l(1) = cosθ(1) l(1) = cos30° = √ 3 /2 m(1) = sinθ(1) m(1) = sin30° = 1/2

Element 2:
L(2) = 2 E(2) = 5 Ac(2) = 2 θ(2) = -45° - l(2) = cosθ(1) l(2) = cos(-45°) = √ 2 /2 m(2) = sinθ(1) m(2) = sin(-45°) = √ 2 /2

The Stiffness Matrix:

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

k^{(e)} = \frac{E^{(e)}A^{(e)}}{L^{(e)}} $$

k^{(1)} = \frac{(3)\cdot(1)}{4} = \frac{3}{4} $$

k^{(2)} = \frac{(5)\cdot(2)}{2} = 5 $$

k_{11}^{(1)} = (l^{(1)})^2 $$

k_{11}^{(1)} = (\frac{\sqrt{3}}{2})^2 $$

k_{11}^{(1)} = \frac{3}{4} $$

k_{12}^{(1)} = l^{(1)}m^{(1)} $$

k_{12}^{(1)} = (\frac{\sqrt{3}}{2})\cdot(\frac{1}{2}) = \frac{\sqrt{3}}{4} $$

k_{22}^{(1)} = (m^{(1)})^2 $$

k_{22}^{(1)} = (\frac{1}{2})^2 = \frac{1}{4} $$

Due to symmetry, where k12(1) = k21(1),

K^{(1)} = \frac{3}{4}\cdot \begin{bmatrix} \frac{3}{4} & \frac{\sqrt{3}}{4} & -\frac{3}{4} & -\frac{\sqrt{3}}{4}\\ \frac{\sqrt{3}}{4} & \frac{1}{4} & -\frac{\sqrt{3}}{4} & -\frac{1}{4}\\ -\frac{3}{4} & -\frac{\sqrt{3}}{4} & \frac{3}{4} & \frac{\sqrt{3}}{4}\\ -\frac{\sqrt{3}}{4} & -\frac{1}{4} & \frac{\sqrt{3}}{4} & \frac{1}{4} \end{bmatrix} $$

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} $$

For Element 2,

k_{11}^{(2)} = (\frac{\sqrt{2}}{2})^2 $$

k_{11}^{(2)} = \frac{2}{4} = \frac{1}{2} $$

k_{12}^{(2)} = \frac{1}{2} $$

k_{22}^{(2)} = \frac{1}{2} $$

K^{(2)} = 5\cdot \begin{bmatrix} \frac{1}{2} & \frac{1}{2} & -\frac{1}{2} & -\frac{1}{2}\\ \frac{1}{2} & \frac{1}{2} & -\frac{1}{2} & -\frac{1}{2}\\ -\frac{1}{2} & -\frac{1}{2} & \frac{1}{2} & \frac{1}{2}\\ -\frac{1}{2} & -\frac{1}{2} & \frac{1}{2} & \frac{1}{2} \end{bmatrix} $$

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} $$


text: User talk:Eml4500.f08.jamama.jan

title: User talk:Eml4500.f08.jamama.jan


text:
User:Eml4500.f08.jamama.jan/Meeting7

Model of the bar turss system (continued)

Elem. 1
θ(1) = 30°

l(1) = cos(1)(θ) = cos(1)(30°) = $$ \frac{\sqrt{3}}{2} $$

m(1) = sin(1)(θ) = sin(1)(30°) = $$ \frac{1}{2} $$

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

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}$$ = [k$$_{ij}^{(e)}]$$4 × 4, where $$_{j = 1,2,3,4\  \Rightarrow \ column\ index}^{i = 1,2,3,4\ \Rightarrow \ row\ index}$$  and $$e $$ = number of element

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

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

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

Note: this matrix is symetircal with respect to the diagnol $$k_{xx}^{(1)}, where\; x= 1,2,3... \to k_{13}^{(1)}=k_{31}^{(1)}. $$

In genral, $$ k_{ij}^{(e)}= k_{ji}^{(e)}\ or\ [{\mathbf{k}^{(e)}}]^T = \mathbf{k}^{(e)}$$

Elem. 2

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

θ(2) $$= -45^\circ \Rightarrow _{m^{(2)} =\ sin(-45^\circ)\ =-\frac{\sqrt{2}}{2}}^{l^{(2)} =\ cos(-45^\circ)\ =\frac{\sqrt{2}}{2}} $$

$$\mathbf{k}^{(e)}$$ = [k$$_{ij}^{(e)}]$$4 × 4$$,\qquad \begin{matrix} k_{11}^{(2)} = k^{(2)} [l^{(2)}]^2 = 5 * [\frac{\sqrt{2}}{2}]^2 = \frac {5}{2}\\ k_{12}^{(2)} = k^{(2)} l^{(2)}m^{(2)} = 5 * \frac{\sqrt{2}}{2}* (-\frac{\sqrt{2}}{2}) = -\frac {5}{2}\\ k_{42}^{(2)} = k^{(2)} (-[m^{(2)}]^2) = 5 * (-[\frac{\sqrt{2}}{2}]^2) = -\frac {5}{2} \end{matrix} $$

Observe: Absolute values for all the coeficients are the same [k$$_{ij}^{(e)}]$$4 × 4, where $$ e =\ 2,and\ _{j = 1,2,3,4}^{i = 1,2,3,4}$$. All that differs are the positive/negative signs in front of the magnitudes of the coeficients.

Element F-D relation: $$ \mathbf{k}_{4 \times 4}^{e}*\mathbf{d}_{4\times1}^{e} = \mathbf{f}_{4\times1}^{e} $$

$$ \mathbf{d}^{e}\ ={ \begin{Bmatrix} d_1^{(e)}\\ \vdots\\ d_4^{(e)} \end{Bmatrix}}_{4 \times 1};\quad

\mathbf{f}^{e}\ = {\begin{Bmatrix} f_1^{(e)}\\ \vdots\\ f_4^{(e)} \end{Bmatrix}}_{4 \times 1} $$

Global F-D Relation("free-free" structure)

$$ \mathbf{K}_{n \times n}*\mathbf{d}_{n\times1} = \mathbf{F}_{n\times1} \Longrightarrow $$ there are $$\mathbf{6} $$ degrees of freedom (6 forces are acting upon the sturcture) $$\quad \Rightarrow \mathbf{n=6} $$

title: User:Eml4500.f08.jamama.jan/Meeting7


text:
User:Eml4500.f08.gravy.jad/Notes2

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

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

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

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

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

Obs:

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

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

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

title: User:Eml4500.f08.gravy.jad/Notes2


text:
Template:Edit this template

This template adds the edit link to the bottom of templates. [ edit]

title: Template:Edit this template


text:
User:Eml4500.f08.ateam.didomenico/hw 2

A-Team Teampage

Solving for element forces of Element 1
$$k^{(1)}d^{(1)} = f^{(1)}$$

$$\begin{Bmatrix} f^{(1)}_1 \\ f^{(1)}_2\\ f^{(1)}_3\\ f^{(1)}_4 \end{Bmatrix}$$ = $$\begin{bmatrix} 0.5625 & 0.32776 & -0.5625 & -0.32476 \\ 0.32476 & 0.1875 & -0.32476 & -0.1875 \\ -0.5625 & -0.32476 & 0.5625 & 0.32476 \\ -0.32476 & -0.1875 & 0.32476 & 0.1875 \end{bmatrix} $$ $$\begin{Bmatrix} 0 \\ 0 \\ 4.352 \\ 6.1271 \end{Bmatrix}$$

This relation can be reduced to the following by the observation that the first two terms in the element displacement matrix are zero. This is due to matrix algebra causing the first two columns of the element stiffness matrix will completely drop out.

 $$\begin{Bmatrix} f^{(1)}_1 \\ f^{(1)}_2\\ f^{(1)}_3\\ f^{(1)}_4 \end{Bmatrix}$$ = $$\begin{bmatrix} -0.5625 & -0.32476 \\ -0.32476 & -0.1875 \\ 0.5625 & 0.32476 \\ 0.32476 & 0.1875 \end{bmatrix} $$ $$\begin{Bmatrix} 4.352 \\ 6.1271 \end{Bmatrix}$$

Carrying out this relation reveals the element force matrix.

 $$\begin{Bmatrix} f^{(1)}_1 \\ f^{(1)}_2\\ f^{(1)}_3\\ f^{(1)}_4 \end{Bmatrix}$$ = $$\begin{Bmatrix} -4.4378 \\ -2.5622 \\ 4.4378 \\ 2.5622 \\ \end{Bmatrix}$$

In which, $$f^{(1)}_1$$ and $$f^{(1)}_2$$ are reaction forces while $$f^{(1)}_3$$ and $$f^{(1)}_4$$ are internal forces.

Element 1 is in equilibrium because the reaction forces and internal forces offset each other and the sum of the moments about any point on Element 1 is zero.

$$\sum F_x = f^{(1)}_1 + f^{(1)}_3 = 0$$

$$\sum F_x = f^{(1)}_2 + f^{(1)}_4 = 0$$

$$\sum M_{any} = 0$$

Solving for element forces of Element 2

$$P^{(1)}_1 = \sqrt{f^{(1)}_1 + f^{(1)}_2}$$

Accounting for the applied force
In order to proceed, the applied force P must be accounted for in the free-body diagram and the equilibrium of all nodes must be verified. The equilibrium of global nodes 1 and 3 were previously verified.

Statics method
There is a second method to solve for the reactions and internal forces of the 2-bar truss depicted here. This method uses the principles of statics, but also must employ the Euler Cut Principle. The Euler Cut Principle must be utilized because the truss is considered statically indeterminate.

Method 1

Method 2

There are multiple ways of applying the Euler Cut Principle to the truss. Each way is determined simply by where the cut is made in the truss system. Depending on this choice, there are two or three resulting free-body diagrams from which the forces are derived.

title: User:Eml4500.f08.ateam.didomenico/hw 2


text:
User:Eas4200c.f08.carbon.orear/HW2

=Homework 2=

Lecture Notes (current version) Lecture Notes (archived version)

Problem 1.1 (current version) Problem 1.1 (archived version)

Problem 1.7 (current version) Problem 1.7 (archived version)

HW2 Contributing Team Members (current version) HW Contributing Team Members (archived version)

title: User:Eas4200c.f08.carbon.orear/HW2


text:
User talk:Caswick

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

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

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

And don't forget to explore Wikiversity with the links to your left. Be bold to contribute and to experiment with the sandbox or your userpage, and see you around Wikiversity! --Emesee 20:19, 20 September 2008 (UTC)

title: User talk:Caswick


text:
User:Eas4200c.f08.carbon.orear/HW2/Lecture Notes

 Comment: This page is identical to the previous one, except that the "Relating Resultant Force to Torque" was added and minor corrections fixed. Also, this comment box was not made correctly before. Here is a comparison between these two versions. Eas4200c.f08.carbon.orear 04:18, 7 November 2008 (UTC)

Motivation
We are motivated to study the box beam with a rectangular cross-section (as in Homework Problem 1.1) as a model for aerospace structures such as the fuselage or wings. As we explore more complex geometries we will decide the helpful and problematic aspects of certain thin-walled cross-sections.

Uniaxial Tension and Mohr's Circle
Taking a closer look at uniaxial tension, we recall the mechanics approach using the Mohr's Circle. This approach begins with reporting the principal stresses. There is no shear in pure uniaxial tension. We plot normal stress on the horizontal axis (remembering that shear is zero at the principal stresses) and enscribe a circle with a diameter of σ.

The image below (also seen under HW Problem 1.1) shows an example of how we get the Mohr's circle for the case of uniaxial tension.

σxx and σyy are the axial and transverse principal stresses, respectively. σxx is denoted on the plot as simply σ and σyy has a value of 0 in this case (which is plotted at the origin. The length between these two points gives us the diameter of our Mohr's circle.

With the vertical axis denoting shear stress, τ, we see that the maximum positive allowable shear stress (τallow) occurs at the top of the circle, and the maximum negative allowable shear stress (-τallow) occurs at the bottom of the circle. In the circle shown these values are denoted 'allowable.'

The relationship σallow = 2 τallow is derived from the Mohr's circle visually once we understand these values. The radius in the vertical direction is multiplied by 2 to equal the diameter in the horizontl direction. Since it is a circle, this calculation makes sense.

Stringers (axial members)
The first question we want to answer is: Should we use rectangular cross-sectional bars?

The answer is No, not always. We should consider the example of a thin-walled open beam. Stringers are structural reinforcement components for the fuselage, and are subjected to a distributed load. It is important to have a lightweight beam that is resistant to bending and buckling. We can create cross-sections both open and closed that have the same properties (or very similar) in resistance of buckling. Why then use open cross-sections? Manufacturing is one important consideration. Stamping flat sheets of metal into beam elements yields a simple, lower cost alternative to extrusion of a closed thin-walled rectangular beam. Also, riveting and hot-dimpling require access that may only be obtained with an open cross-section. Artisans working on the aircraft must be able to easily and quickly work on the stringer component and a closed-wall cross-section becomes a hindrance. We can further examine stringer geometry to make it easy to stack and store many components without making significant changes to the structural properties that are advantageous to our design intentions.

Case 1

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

To convert the second moment of inertia equation in to polar coordinates you must substitute r and \theta for y and z.

 * $$ \begin{align}

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

Which gives us

 * $$I_{y}^{(1)}= \int_0^R \int_{0}^{2\pi} r^3\ sin^2(\theta)\ d\theta\ dr$$

and integrating
 * $$I_y^{(1)}= \frac{R^4}{4}\ \frac{2\pi}{2}$$

 * $$I_y^{(1)}= \frac{\pi R^4}{4}$$

Case 2

To find the second moment of inertia of the channel shown above we start with the original equation for I.

 * $$I_{y} = \iint_A z^2 \,dA$$

 * $$I_{y}^{(2)}= \int_{-a/2}^{a/2} \int_0^b z^2 \,dy dz - \int_{\frac{-(a-t)}{2}}^{\frac{a-t}{2}} \int_0^{b-t} z^2 \,dy dz $$

 * $$I_y^{(2)}= \frac{1}{3} b(\frac{a^3}{8} + \frac{a^3}{8}) - \frac{1}{3} (b-t)(2(\frac{(a-t)}{8}^3)$$
 * $$I_y^{(2)}= \frac{1}{12} ba^3 - \frac{1}{12} (b-t)(a-t)^3$$

 * $$A^{(2)}= 3at = A^{(1)} = \displaystyle\pi R^2$$
 * $$t=\frac{a}{10} $$
 * $$I_y^{(2)}= \frac{1}{12} ba^3 - \frac{1}{12} (b-\frac{a}{10})(a-\frac{a}{10})^3$$

Solving for R
 * $$R =\sqrt{ \frac{3a^2}{10\displaystyle \pi}}$$
 * $$I_y^{(1)}=\frac{9a^4}{400\displaystyle \pi}$$

When you compare the two cross sections you find that Case 2 has a much higher moment of inertia than Case 1

Hooke's Law for Shear and Simple Shear
Shear Stress is the product of the shear modulus and shear strain.

 * $$\displaystyle \tau = G \gamma$$

The shear modulus (G) is a property of the material in question. The shear strain (γ) is based on the deformation displacement in the transverse (or shear) direction. This direction is parallel to the face of the material acted upon. This means that we can think of shear stress as tangential, or parallel, stress.

 * $$\displaystyle \gamma = \frac{\partial u}{\partial y} + \frac{\partial v}{\partial x}$$

Note that v = uy and u = ux.

Simple shear is shown below.

We also want to define engineering shear strain as the total shear strain in the plane (here, x-y).

This value is exactly twice the tensorial shear strain, which only measures strain of x vs. y or y vs. x.

Shear Flow
Shear flow, (q) is force over contoured length.

 * $$\displaystyle q = \tau t$$

This means that shear flow is constant along the wall cross section despite wall thickness. This allows shear stress (τ) along the wall to be represented by shear flow (q) along the wall centerline. Shear flow forms a closed loop; this means that force resultants in the x and y directions due to shear flow are zero although a resultant torque is created as defined below.

 * $$\displaystyle dT = \rho q ds$$

The torque is dependent on the distance (ρ) from the origin to the wall segment (ds) and the shear flow (q) at that segment. Integrated, this becomes the following general equation for relating torque and shear flow.

 * $$\displaystyle T = \oint \rho q ds = \iint_\bar{A} 2 q dA = 2 q \bar{A}$$

Knowing that $$\bar{A} = ab$$ in the ad-hoc method used in problem 1.1 that shear flow is defined as $$q = \tau t$$, we find for problem 1.1, torque is defined as:

 * $$\displaystyle T = 2 (\tau t) (a b)$$

which can be rearranged as:

 * $$\displaystyle \tau = \frac{T}{2 a b t}$$

This proves that the ad-hoc method of elasticity seen in Problem 1.1 fits into our general case of elasticity as derived from shear flow.

Curved Panels
The concepts in this section, while discussed and continued in week three, are applicable and thus included in week two. The will be repeated for emphasis and completion in week three.

A shear panel (figure on right) is a thin sheet of material designed to take on an in-plane shear load. If the panel is of uniform thickness “t”, then the total shear forces in the x-direction and y-direction provided by the panel is given by

$$\displaystyle Vx = \tau*t*a = G*\gamma*t*a$$

$$\displaystyle Vy = \tau*t*b = G*\gamma*t*b$$

G is the shear modulus and $$\gamma$$ is the shear strain. Therefore the shear forces are proportional to the panel’s thickness and lateral dimensions.

The shear forces on a thin-walled curved panel (figure on lower right) under constant shear stress $$\tau$$ can be found in a similar way. Breaking down the shear force in horizontal and vertical components yields

$$\displaystyle Vx = \tau*t*a$$

$$\displaystyle Vy = \tau*t*a $$

Therefore the resulting components of the shear force can be related by

$$\displaystyle \frac{Vx}{Vy} = \frac{a}{b} $$

This proves that to handle a constant shear stress, a flat panel is most efficient because it saves unnecessary material from being used. This reduces weight which is one of the main concerns in the construction of airplanes and other aircraft structures.

The shear force on a thin curved panel can be found using shear flow. Referring to the diagram, shear flow, q, is constant with respect to the y and z axis and is defined by the following

$$\displaystyle q = \tau*t$$

With L being the length of the panel,

$$\displaystyle F = q*L$$

Therefore the force can be calculated by looking at a infinitesimal length of the panel and integrating over the entire length of the panel.

$$\displaystyle dF = q*dL = q*(dLy j + dLz k)$$

$$\displaystyle dF = q*(dL*cos(\Theta) j + dL*sin(\Theta) k)$$

The resultant shear force vector can then be calculated as

$$\displaystyle F = \int_{a}^{b}{dF} = q*[(\int_{a}^{b}{dy}) j + (\int_{a}^{b}{dz}) k]$$

$$\displaystyle F = q*(a j + b k)$$

Breaking it down into components

$$\displaystyle F = Fy j +Fz k$$

$$\displaystyle Fy= q*a$$

$$\displaystyle Fx = q*b$$

$$\displaystyle \frac{Fy}{Fz} = \frac{a}{b}$$

The magnitude of the resultant shear force is

$$\displaystyle \parallel F \parallel = [(Fy)$$2$$\displaystyle + (Fz)$$2$$\displaystyle ]$$1/2$$\displaystyle  = q*[a$$2$$\displaystyle  +b$$2$$\displaystyle ]$$1/2$$$$

Relating Resultant Force to Torque
The resultant force $$R = \displaystyle \parallel F \parallel$$ can be related to the torque by looking at the figure on the left with $$\displaystyle \vec{r}$$ being the position vector and $$\displaystyle \rho$$ being the perpendicular distance from the origin to dF. The torque can be derived by

$$\displaystyle d\vec{T}=\vec{r}Xd\vec{F} \Rightarrow dT  = p*dF $$

$$\displaystyle T = \oint_{}^{}{}dT = q*\oint_{}^{}{}p*dL$$ with p*dL = 2*dA

$$\displaystyle T = 2*q*\int dA$$

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

And therefore the relation can be visualized in the following equation:

$$\displaystyle T = \oint_{}^{}{} \rho *q*ds =\int \int_{\bar{A}}^{}{} 2*q*dA = 2*q*\bar{A}$$

=References=


text:
User:Eas4200c.f08.carbon.orear/HW2/Contributors

=HW 2 Contributing Team Members=

Jeffrey Wright Eas4200c.f08.carbon.w 20:14, 25 September 2008 (UTC)

Jeff O'Rear Eas4200c.f08.carbon.orear 19:07, 25 September 2008 (UTC)

Beau Guidry Eas4200c.f08.carbon.guidry 00:32, 26 September 2008 (UTC)

Aaron Clausen Eas4200c.f08.carbon.clausen 14:37, 26 September 2008 (UTC)

Abby Booth Eas4200c.f08.carbon.booth 19:11, 26 September 2008 (UTC)


title: File:Elem-2-jan.JPG


text:
Template:By school subject/Concept

 You can browse Wikiversity by school subject. The school subject portals aim to give a friendly and useful access route for users interested in pre-tertiary (pre-school, primary and secondary) education. The school subject portals aim to offer users an alternative non-university route to resources in their subject area which is free of the extreme specialisation which occurs at tertiary level and which is friendly and comprehensible to schoolchildren.

title: Template:By school subject/Concept


text:
Template:By education level/Concept

 You can browse Wikiversity by educational level. There is also a guided tour to these portals.

title: Template:By education level/Concept


text:
Template:By user type/Concept

 You can browse Wikiversity by user type. There are only a few of these, and the user type portals offer access routes based around the role you wish to play at Wikiversity.

title: Template:By user type/Concept


text:
Template:By completion status/Concept

 You can browse Wikiversity by the completion status of resources, which can help you find complete resources reading for using, or freshly started ones which you can help develop. Work also needs to be done on categorising resources by completion status, which is why some of the categories here are rather empty.

title: Template:By completion status/Concept


text:
Template:By educational level/Concept

 You can browse Wikiversity by educational level. There is also a guided tour to these portals.

title: Template:By educational level/Concept


text:
Template:By learning project/Concept

 During the course of its history, Wikiversity was instructed to organize itself more closely around "learning projects", so a structure appeared which offers an alternative way to explore content.

title: Template:By learning project/Concept


text:
Template:By department/Concept

 You can browse Wikiversity by department. Historically Wikiversity had something called "departments" which were put into the "Topic" namespace, and you can browse through them here.

title: Template:By department/Concept


text:
Template:By resource type/Concept

 You can browse Wikiversity by resource type. This route is rather rough and ready for the time being, as people's consistency and willingness to categorise their resources by type has been somewhat lacking. It is hoped that by spotlighting the "by resource type" exploration path, people will slowly improve the tagging of resources by resource type.

title: Template:By resource type/Concept


text:
Template:Departments by subject/Concept

 Here are departments.

title: Template:Departments by subject/Concept


text:
Template:By university school/Concept

 You can browse Wikiversity by "school", where "school" is understood in the US American sense of a major division of a university. Wikiversity is not a real university, but you can explore Wikiversity using the metaphor of a university. If you prefer the word "faculty" to "school", please skip to the box above.

title: Template:By university school/Concept


text:
Template:By university faculty/Concept

 You can browse Wikiversity by "faculty", where "faculty" is understood in the international (non-US American) sense of a major division of a university. Wikiversity is not a real university, but you can explore Wikiversity using the metaphor of a university. If you prefer the word "school" to "faculty", please skip to the box below.

title: Template:By university faculty/Concept


text:
Reactions of ionic halides

Detecting ionic halides

 * Most halides are soluble.

Aqueous halides with silver nitrate in the nucleus

 * We can detect them in solution by adding silver nitrate solution,
 * We get the corresponding silver halide forming (AgCl, AgBr, AgI) which are insoluble (except for AgF) so they form a precipitate:
 * AgF is soluble; no precipitate formed
 * AgCl forms a white precipitate
 * AgBr forms a cream precipitate
 * AgI forms a yellow precipitate

''An easy way to remember the results is that as we move down Group VII of the periodic table, the colours get "darker". So AgF is colourless because it is completely soluble, AgCl forms a white ppt, AgBr forms a cream ppt, and AgI forms a yellow ppt.''

''It is important to note that before this test is carried out the solution must be acidified so as to remove any carbonate ions or hydroxide ions that may be present in the solution as impurities. If they were allowed to remain these impurities would also form precipitates on reaction with silver nitrate, and so confuse the test.''

Aqueous halides with ammonia solutions
Distinguishing between white, cream and yellow can be difficult, so a second test can be used to confirm the results: The white precipitate formed with chloride ions will dissolve in dilute ammonia, the cream precipitate formed with bromide ions will dissolve only in concentrated ammonia, and the yellow precipitate formed with iodide ions will not dissolve in any concentration of ammonia. But the concentration changes as well.

Solid ionic halides and concentrated sulphuric acid
Solid ionic halides can be detected by their reactions with conc. sulphuric acid: So conc. sulphuric acid:
 * First, the corresponding hydrogen halide is formed
 * Then, the differences in the reactions depend on the reducing ability of the hydrogen halide and the moderately strong oxidising ability of concentrated sulphuric acid
 * HF is not a reducing agent but the other hydrogen halides are increasing so
 * does not react with HF or HCl
 * but does oxidise HBr and HI freeing the halogens
 * HBr reduces sulphuric acid to SO2
 * HI (being an even stronger reducing agent) reduces the acid to hydrogen sulphide

Sodium fluoride
NaF(s) + H2SO4(l) → HF(g) + NaHSO4(s) No further reaction occurs. Thats what you think!!!

Sodium chloride
NaCl(s) + H2SO4(l) → HCl(g) + NaHSO4(s) No further reaction occurs.

Sodium bromide

 * NaBr(s) + H2SO4(l) → HBr(g) + NaHSO4(s)
 * As conc. sulphuric acid is a moderately strong oxidising agent, it oxidises HBr to Br:
 * 2HBr(aq) + H2SO4(l) → Br2(g) + SO2(g) + 2H2O(l)
 * The aqueous bromide ion from HBr has been oxidised to form bromine:
 * 2Br- → Br2 + 2e-
 * The HBr and acid each provide 2H+ ions and the electrons from bromine are transferred to the sulphate ion:
 * 4H+ + SO42- + 2e- → SO2 + 2H2O
 * Adding these two equations together we get:
 * 2HBr(aq) + 2e- + H2SO4(l) → Br2(g) + 2e- + SO2(g) + 2H2O(l)
 * The electrons on either side of the equation cancel to give the overall equation for the reaction which is shown above

Sodium iodide
Here several reactions occur:
 * NaI(s) + H2SO4(l) → HI(g) + NaHSO4(s)
 * As usual the hydrogen halide is produced (hydrogen iodide in this case)

 * Next, two different reactions can occur:
 * The HI is oxidised to I2, and the sulphuric acid is reduced to sulphur dioxide and water
 * 2HI(aq) + H2SO4(l) → I2(g) + SO2(g) + 2H2O(l)
 * The HI is oxidised to I2, but this time the sulphuric acid is reduced to hydrogen sulphide and water
 * 8HI(aq) + H2SO4(l) → 4I2(s) + H2S(g) + 4H2O(l)

title: Reactions of ionic halides


text:
Orbital geometry

Figure 1 shows the geometry and characteristics of an elliptical orbit. F1 and F2 are called the focal points of the ellipse, a is called the semi-major axis, b is the semi-minor axis, and e is the eccentricity. For every orbit, the body being orbited is at one of the focal points of that orbit. For example, the sun is at a focal point of the Earth's orbit, and the Earth is at a focal point of the International Space Station's orbit. The distance between the center of the ellipse and either focal point is equal to the product of the semi-major axis and the eccentricity. For an ellipse, the eccentricity is between 0 and 1. For the special case where e is zero, the orbit geometry is a circle. A circle is an ellipse with a = b and both focal points in the same location.

For every ellipse, the sum of the distance between every point on the ellipse and each focal point is a constant value. This is illustrated in Figure 2 for a particular point. The following is true for an ellipse: c1+c2=2a

Figure 3 introduces many important parameters for elliptical orbits. The perifocal frame is defined by ie, ip, and with the third axis, ih, pointing out of the plane of the figure. Periapsis, designated by rp, is defined as the point in the orbit closest to the orbited body. Apoapsis, designated by ra, is defined as the point in the orbit farthest from from orbited body. There are specific terms that may be used in place of periapsis and apoapsis that are dependent on the body being orbited. For example, for orbits about the sun periapsis is called perihelion and apoapsis is called aphelion; for orbits about the earth periapsis is called perigee and apoapsis is called apogee.

title: Orbital geometry


title: File:1.1a.JPG


title: File:L5Fig1.JPG


text:
Chemistry (A-Level)

A brief overview of AS and A2 level Chemistry based around the Chemistry A specification

Unit 1 - Atoms, bonds and groups

 * Mole concept.

Unit 3 - Practical skills in chemistry 1

 * Practical skills in chemistry

Unit 4 - Rings, polymers and analysis

 * Rings, acids and amines
 * Polymers and synthesis
 * Analysis

Unit 5 - Equilibria, energetics and elements

 * Rates, equilibrium and pH
 * Energy
 * Transition elements

Unit 6 - Practical skills in chemistry 2

 * Practical skills in chemistry

title: Chemistry (A-Level)


title: File:L5Fig2.JPG


text:
The periodic table/Halogens

Halogens:

 * Flour ine
 * Hlorine
 * Chlorine
 * Chlorine's Brother
 * Bromine
 * Iodine
 * Astatine
 * Tennessine
 * Iodine
 * Astatine
 * Tennessine

Other resources

 * Reactions of ionic halides

title: The periodic table/Halogens


title: File:L5Fig3.JPG


text:
File:L5Fig4.JPG

Summary
Normal  0               false   false   false      EN-US   X-NONE   X-NONE                                                     MicrosoftInternetExplorer4

title: File:L5Fig4.JPG


title: File:L5Fig5.JPG


title: File:L5Fig6.JPG


title: File:L5Fig7.JPG


text:
User:W3asal

About Me
Math lover.

title: User:W3asal


text:
User talk:W3asal

Leave me a message here if you wish.

 Hello W3asal, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. After you leave a comment on a talk page, remember to sign and date; it helps everyone follow the threads of the discussion. The signature icon in the edit window makes it simple. To get started, you may

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

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

And don't forget to explore Wikiversity with the links to your left. Be bold to contribute and to experiment with the sandbox or your userpage, and see you around Wikiversity! Jtneill - Talk - c 00:14, 21 September 2008 (UTC)

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

title: User talk:W3asal


title: File:1.1b.JPG


title: File:1.1c.JPG


title: File:1.1d.JPG


text:
User:Eas4200c.f08.carbon.orear/HW2/Pb1.7

Problem
Make a comparison of the ability to bear loads of the two beams shown below using bending rigidity.

Solution

To find the value of R1, equate the areas of the two beams.

A1 = πRo2 = π m

A2 = 2tRo + 2πR12 = (.002 + 2πR12)m

→ R1 ≈ .06842m

Moments of Interia

Iy,1 = πRo^4/4 = (2.5*10-5)

Iy,2 = bh3/12 + 2[πR14/4 + πR12*d], where d = Ro + R1/2 ≈ .003983m

The moment of inertia of case 2 about the y-axis is much larger than that of case 2 indicating that the second beam has a larger load-bearing capability.


title: File:Pb1.7.JPG


text:
User:Mietchen/Citizendium

This page is meant for people who do not wish to register for Citizendium (my user page there). In case of problems with this page, please leave me an email or a tweet.

Orchid
I plan to nominate the article orchid for approval within the next two weeks and would welcome external feedback on it. To edit the article, you will need to register with your real name, and if you do not wish that, you can leave feedback here. Thank you! --Daniel Mietchen (talk) 20:29, 17 June 2009 (UTC)

Hawaiian alphabet
I have no special expertise in this but Hawaiian alphabet is also nearing approval, and so I would appreciate comments. Thanks a lot! --Daniel Mietchen (talk) 08:16, 21 June 2009 (UTC)

title: User:Mietchen/Citizendium


text:
Extraction of metals/Aluminium

Definitions

 * Cathode — Negative electrode
 * Positively charged ions (cations) collect at the cathode
 * Anode — Positive electrode
 * Negatively charged ions (anions) collect at the anode

The extraction process

 * Aluminium is extracted from its ore by electrolysis
 * This takes place in an iron bath lined with graphite called a cell
 * The graphite lining acts as the cathode
 * A series of graphite blocks act as the anode(s)
 * Pintu Aluminium

title: Extraction of metals/Aluminium