Physics problems - where to find similar exercices - Page 1
Forum section: electronics / general

author: gildasd
author_rank: Frequent Contributor
timestamp: March 09, 2015, 04:48:41 pm
content: Hello;Within my studies we have a small course of "resistance of material":The kind of problems we study are very different from the normal French or Belgian syllabus, so I'm struggling to find references:Here are 3 typical exercises that are annoying me (my methods end up filling 2 to 3 screen of my Ti Nspire transversally - so there must be a better way!):Top:3 Pillars in Al 70MPa, of surface 50mmm² receive a force of 800kN per meter. What is the force on each pillar.MiddleAt what temperature will a ring of AL with a bore of 24.993mm be released from a cylinder of diameter 25mm.Bottom:A a threaded bolt is heated to 80°, fitted by hand on a bronze spacer, and let to cool down to 25°. How big is the resulting force?If any of you know of a website or of a PDF that treats similar small problems, i'd be most grateful for a pointer.

author: Vgkid
author_rank: Super Contributor
timestamp: March 09, 2015, 05:01:02 pm
content: First - could you solve it by calculating the Force(80KN/M)/area,,,then divide that by 3...not working those out, just a guessMiddle - Coefficent of thermal expansion.http://physics.bu.edu/~duffy/py105/Temperature.htmlorhttp://www.webassign.net/question_assets/buelemphys1/chapter13/section13dash2.pdflast - coefficient of thermal expansion w/ youngs modulushttp://rockpile.phys.virginia.edu/arch21.pdf^^^ That might help, I can't recall doing a problem similiar to that last year.

author: IanB
author_rank: Super Contributor
timestamp: March 09, 2015, 05:34:23 pm
content: Quote from: gildasd on March 09, 2015, 04:48:41 pmHmm, I remember being taught how to solve those problems in an introductory mech. eng. class many years ago. I can't remember how to solve them now and I would have to work it out from scratch.As noted, they involve Young's modulus and the coefficient of thermal expansion. They are exercises in static analysis.You are basically going to write equations that must be satisfied and then solve them. At least one equation will involve dimensions (where items are in contact their dimensions are equal), and at least another equation will involve forces (where contact surfaces are in equilibrium then opposing forces are equal).I think you also will most readily write the equations in terms of deltas from an initial state. For example in the first problem, let the horizontal beam move downwards from the starting point in the diagram until equilibrium is reached. What changes has the system undergone at that point?

author: gildasd
author_rank: Frequent Contributor
timestamp: March 09, 2015, 05:50:43 pm
content: Thanks,Looking into your examples, at first glance it seems there is stuff to inspire me.We are using the formulae below:Lo = Original length.Delta L th = length variation due to heat.Alpha = Thermal expansion coef: Copper is 18x10^-6 for example.F = Force in NewtonsE = Young's Modulus.Sigma = compression in MPa

author: gildasd
author_rank: Frequent Contributor
timestamp: March 09, 2015, 09:41:12 pm
content: Thanks for the answers.I found great simplifications with your inputs.Only the 3 pillar one still stays overly complicated, I need two full pages yet I'm sure it can be solved in half of one...Cheers!

author: IanB
author_rank: Super Contributor
timestamp: March 09, 2015, 09:54:23 pm
content: Quote from: gildasd on March 09, 2015, 09:41:12 pmI will work on it this evening after work and see if I can figure it out. I'm sure I need a reason to brush up on my mechanical engineering

author: IanB
author_rank: Super Contributor
timestamp: March 09, 2015, 10:22:19 pm
content: Quote from: gildasd on March 09, 2015, 04:48:41 pmActually I don't understand the per meter part. Can you clarify what that means?

author: gildasd
author_rank: Frequent Contributor
timestamp: March 09, 2015, 10:42:41 pm
content: It means:800kNm x 0.2m = 160kN.So, in the "initial state" each pillar that "touches" has a force of 80kN on it.I forgot a 0...Here is the pdf with the materials, should be handy.

author: miguelvp
author_rank: Super Contributor
timestamp: March 10, 2015, 02:05:32 am
content: https://www.khanacademy.org/science/physicsNot just problems but free courses.They also have Algebra, Calculus, Matrices, Diff Equations, Vectors, etc...Edit: and I like the way Sal teaches.

author: Someone
author_rank: Super Contributor
timestamp: March 10, 2015, 02:16:14 am
content: Quote from: IanB on March 09, 2015, 10:22:19 pmIts a distributed load, so the displacement of the beam is parabolic until it hits that middle post and then it gets interesting with each side then being a parabola summed with the first parabola (at a rough approximation). With all the details these look suspiciously like assessable questions for coursework.

author: IanB
author_rank: Super Contributor
timestamp: March 10, 2015, 02:42:11 am
content: I understand the distributed load part now. But since no dimensions are given for the horizontal beam I think it can be assumed that the beam doesn't flex. Bendy beams would be the subject of a more advanced program of study than this one.

author: IanB
author_rank: Super Contributor
timestamp: March 10, 2015, 02:44:50 am
content: I worked the top problem for fun, and I noticed there is a "deliberate mistake" in the data given. The mistake is not obvious unless you do a sanity check on the results.

author: gildasd
author_rank: Frequent Contributor
timestamp: March 10, 2015, 07:36:19 am
content: Quote from: IanB on March 10, 2015, 02:44:50 amYou should find:189MPa for A and C21.4 MPa for B

author: gildasd
author_rank: Frequent Contributor
timestamp: March 10, 2015, 07:39:00 am
content: Quote from: miguelvp on March 10, 2015, 02:05:32 amThat was my first stop, but they don't have anything on material science.I have 1.25x10^9 points mostly from the videos and maths

author: John Coloccia
author_rank: Super Contributor
timestamp: March 10, 2015, 07:40:21 am
content: I still don't understand the 800kN/m business. What an odd way of specifying the force. You can't just multiply it by .2m because that only takes you to the center of the pillars. What about the rest of the pillars? You have no idea what the dimension is...you only know the area.Anyhow, the top one is straightforward. You figure out how much force it takes until all three are touching, and then from whatever is left, you work it out like a 3 pillar problem. Then you add it up for pillars A and C.

author: gildasd
author_rank: Frequent Contributor
timestamp: March 10, 2015, 07:41:43 am
content: Quote from: Someone on March 10, 2015, 02:16:14 amFor this exercice, consider the top bar to be infinitely rigid, no deformation except vertically on the posts.

author: IanB
author_rank: Super Contributor
timestamp: March 10, 2015, 12:10:19 pm
content: Quote from: gildasd on March 10, 2015, 07:36:19 amThe question asks for the force on each of the pillars. Force is measured in kN and MPa is a unit of pressure, so the answer cannot be given in MPa. Furthermore the three forces must add up to 160 kN, so this is one basic check on the correctness of the answer.I find that the forces on A, B and C are quite similar in magnitude.

author: gildasd
author_rank: Frequent Contributor
timestamp: March 11, 2015, 08:40:12 pm
content: Complete answer:

author: IanB
author_rank: Super Contributor
timestamp: March 11, 2015, 10:21:28 pm
content: Quote from: gildasd on March 11, 2015, 08:40:12 pmYes, I agree with that. But where did the 400 mm² cross-section of each pillar come from? In the original question you posted it seemed to be 50 mm². Without the same input data it was not possible to get the same numerical answers.

author: gildasd
author_rank: Frequent Contributor
timestamp: March 11, 2015, 10:30:26 pm
content: Quote from: IanB on March 11, 2015, 10:21:28 pmI blame my keyboardHowever this would have modified your result in Newtons but the MPa's would have been the same.Actually that's a good idea to check repeatability, I'll do that, thanks!

author: IanB
author_rank: Super Contributor
timestamp: March 11, 2015, 10:52:09 pm
content: Quote from: gildasd on March 11, 2015, 10:30:26 pmI do not think so. Stress is force over area. Since the area is reduced and the total force is the same, the stresses on each pillar must go up.

author: gildasd
author_rank: Frequent Contributor
timestamp: March 11, 2015, 11:28:04 pm
content: Scribble scribble scribble - AHA, a result!You are right.All three pillars go WAY above the 414x10^6 limit and get crushed!Is there a way to check this efficiently, without doing the whole process?

author: IanB
author_rank: Super Contributor
timestamp: March 12, 2015, 01:33:48 am
content: This is how I approached the problem. I don't think there is an obvious way of simplifying it. You basically have four equations in four unknowns.However, solving the problem first algebraically allows you to program the solution (programmable calculator?) and then plug in different numbers for the variables to try out different cases.

author: gildasd
author_rank: Frequent Contributor
timestamp: March 12, 2015, 03:24:06 pm
content: Very nice, using the E as a "pivot". Great stuff!

author: CatalinaWOW
author_rank: Super Contributor
timestamp: March 12, 2015, 03:55:27 pm
content: This thread demonstrates that in class as well as in real life, clarity in stating and understanding the problem is at least as important as the engineering and physics used in solving it.

author: gildasd
author_rank: Frequent Contributor
timestamp: March 12, 2015, 04:51:50 pm
content: Quote from: CatalinaWOW on March 12, 2015, 03:55:27 pmMoving goalpost are the standard fare of any project

author: gildasd
author_rank: Frequent Contributor
timestamp: March 14, 2015, 04:25:16 pm
content: Here is a follow-up exercise:The 3 supports are of different materials, we need to find the pressure on each one in MPa's.(The weight is horizontal and can't tilt).In the second half, the 3 supports are heated by 50°C and we need to check if the weight can touch all 3.(Sorry about the French, but it's pretty self explanatory).More seriously:The 3rd image is a simplification of a problem that I often encounter, but the methods and formulas that we have been given can only get you a rough approximation at best.And they take a couple of pages to resolve...

author: IanB
author_rank: Super Contributor
timestamp: March 14, 2015, 05:09:46 pm
content: Quote from: gildasd on March 14, 2015, 04:25:16 pmThat third image is a matter of resolving forces. The snag is that when the trusses AB and AC stretch the angles will alter. So for an exact solution it is necessary to write equations that apply to the final state with the final angle unknown. I think it will be necessary to use some triangle formulas like the cosine rule or the sine rule.

author: gildasd
author_rank: Frequent Contributor
timestamp: March 14, 2015, 05:42:42 pm
content: Quote from: IanB on March 14, 2015, 05:09:46 pmI'm going in that general direction... But it's HARD!This will probably need a small integral at one stage to cover the zone between the two angles...It's a bit frustrating that for such a common problem there is not a formula (digging through textbooks right now).

author: gildasd
author_rank: Frequent Contributor
timestamp: March 15, 2015, 12:40:02 am
content: I've tried to solve a problem nearly all day...With the tools at my disposition (only a 12 course on material science), I finally took an indirect route:I calculated the truss with a "wet finger in the breeze" force that was within the magnitude.With the resulting lengths I made a coefficient.With that coefficient, I obtained the proportion of deformation on the simplest part of the truss.Did a simple multiplication to get an "about there" result...Really trying to get this out of the way to do some electronics for my final project (concept validated today - yay!).

author: IanB
author_rank: Super Contributor
timestamp: March 15, 2015, 01:56:03 am
content: You can exploit the symmetry of the problem and consider only one half of the system. See the diagram below. Does this give you a hint? (I haven't solved it myself, but having gone thus far I am confident the solution will be attainable.)

author: gildasd
author_rank: Frequent Contributor
timestamp: March 15, 2015, 07:52:27 pm
content: Quote from: IanB on March 15, 2015, 01:56:03 amThanks!I used the symmetry too but used Pythagoras at the end to find EA'.I like what you did, but what does "k" represent?http://skyciv.com/free-truss-calculator/This is cool little tool to check final truss results.

author: IanB
author_rank: Super Contributor
timestamp: March 15, 2015, 09:21:05 pm
content: kis the "spring constant" of the truss.If we assume that the truss is a bar of cross-sectionA, lengthLand Young's modulusE, then we can write:F=EA/LdLorF=kdLwherek=EA/L

author: F4CPY
author_rank: Contributor
timestamp: March 16, 2015, 02:06:55 pm
content: HelloI think there's a better way to solve this.Try to apply PFS (principe fondamental de la statique) fundamental principle of staticat node AEvaluate each force in the bars (parametric function of F)Evaluate the sum of the traction energy of the three bars (N²*L)/(2*S*E)Apply Castigliano theorem : dérivative of the sum of the energy by the external force F equal displacement. (little displacements)Knowing displacement you will find F.Didier

author: gildasd
author_rank: Frequent Contributor
timestamp: March 16, 2015, 05:50:16 pm
content: Quote from: F4CPY on March 16, 2015, 02:06:55 pmThat's either brilliant or rather rude...

author: mrflibble
author_rank: Super Contributor
timestamp: March 16, 2015, 06:38:46 pm
content: Quote from: F4CPY on March 16, 2015, 02:06:55 pmHey, that sounds suspiciously much like D'Alembert's principle.Quote from: gildasd on March 16, 2015, 05:50:16 pmWell, if it's like the D'Alembert method like I think it is then it just might be brilliant.The concept of virtual work makes solving this type of problem totally easy.

author: IanB
author_rank: Super Contributor
timestamp: March 16, 2015, 06:57:28 pm
content: The theory is easy, but the algebra can become messy.The system here is non-linear for large displacements because the angle of the truss changes. The non-linearity introduces complexity. I am not sure that applying the principle of virtual work will in and of itself simplify this.However, if you assume that displacements are small relative to the dimensions of the system then you can linearize the system equations around the initial point and get a much simpler solution which is correct to the first order. This would probably be acceptable if the trusses were steel girders, but likely not be acceptable if the trusses were rubber bands.

author: gildasd
author_rank: Frequent Contributor
timestamp: March 21, 2015, 03:59:44 pm
content: Had the exam today and it went ok.I don't know how much an internet thanks is worth, but this is one.Thank-you to all that contributed, I might have not used your methods to the letter, but they certainly do open "mind doors".Thanks again!G.

author: Vgkid
author_rank: Super Contributor
timestamp: March 22, 2015, 05:08:34 am
content: What class are you taking.

author: gildasd
author_rank: Frequent Contributor
timestamp: March 22, 2015, 06:54:22 am
content: Engineering Watch Officer - Unlimited.That means ships of a minimum of 500Gt and 7.5kW to be relevant.