Physics problems - where to find similar exercices

[gildasd]

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.

Middle
At 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.

[Vgkid]

First - could you solve it by calculating the Force(80KN/M)/area,,,then divide that by 3...not working those out, just a guess
Middle - Coefficent of thermal expansion.
http://physics.bu.edu/~duffy/py105/Temperature.html
or
http://www.webassign.net/question_assets/buelemphys1/chapter13/section13dash2.pdf
last - coefficient of thermal expansion w/ youngs modulus
http://rockpile.phys.virginia.edu/arch21.pdf
^^^ That might help, I can't recall doing a problem similiar to that last year.

[IanB]

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!):

Hmm, 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?

[gildasd]

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 Newtons
E = Young's Modulus.
Sigma = compression in MPa

[gildasd]

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!

[IanB]

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...

I 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 

[IanB]

Top:
3 Pillars in Al 70MPa, of surface 50mmm² receive a force of 80kN per meter. What is the force on each pillar.

Actually I don't understand the per meter part. Can you clarify what that means?

[gildasd]

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.

[miguelvp]

https://www.khanacademy.org/science/physics

Not just problems but free courses.
They also have Algebra, Calculus, Matrices, Diff Equations, Vectors, etc...

Edit: and I like the way Sal teaches.

[Someone]

Top:
3 Pillars in Al 70MPa, of surface 50mmm² receive a force of 80kN per meter. What is the force on each pillar.

Actually I don't understand the per meter part. Can you clarify what that means?

Its 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.

[IanB]

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.

[IanB]

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.

[gildasd]

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.

You should find:
189MPa for A and C
21.4 MPa for B

[gildasd]

https://www.khanacademy.org/science/physics

Not just problems but free courses.
They also have Algebra, Calculus, Matrices, Diff Equations, Vectors, etc...

Edit: and I like the way Sal teaches.

That 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

[John Coloccia]

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.

[gildasd]

Top:
3 Pillars in Al 70MPa, of surface 50mmm² receive a force of 80kN per meter. What is the force on each pillar.

Actually I don't understand the per meter part. Can you clarify what that means?

Its 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.

For this exercice, consider the top bar to be infinitely rigid, no deformation except vertically on the posts.

[IanB]

You should find:
189MPa for A and C
21.4 MPa for B

The 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.

[gildasd]

Complete answer:

[IanB]

Complete answer: ...

Yes, 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.

[gildasd]

Complete answer: ...

Yes, 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.

I blame my keyboard 

However 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!

[IanB]

However this would have modified your result in Newtons but the MPa's would have been the same.

I 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.

[gildasd]

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?

[IanB]

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.

[gildasd]

Very nice, using the E as a "pivot". Great stuff!

[CatalinaWOW]

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.