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Mechanics of Materials basic exercise

  1. Nov 5, 2011 #1

    Femme_physics

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    1. The problem statement, all variables and given/known data

    A graded steel cylinder AISI 1030 is connected to a wall and loaded with axial forces.

    http://img542.imageshack.us/img542/3959/fbd1.jpg [Broken]


    Given:

    F = 30 [kN]
    d1 = 35mm
    d2 = 20mm
    Elastic Model: E = 2x105 MPa
    Safely coeffecient: 2
    Yield strength limit: 440 MPa


    Required:

    1) Describe the course of the internal forces along the beam and the course of the efforts.
    2) Check beam for strength
    3) Calculate the total distortion of the beam


    2. Relevant equations

    Presumably
    http://img831.imageshack.us/img831/533/neededformulas.jpg [Broken]

    3. The attempt at a solution

    Since I don't see a 3D drawing or references should I treat this problem like a 2D problem?


    1) I presume they're asking for a free body diagram here:

    http://img694.imageshack.us/img694/9629/freebodydiag.jpg [Broken]


    2)
    If I treat it like 2D:
    Before I can use the formula: Sigma = F/A

    I need to find A. So,
    A = 1 x 0.020 + 2 x 0.035 = 0.09m2

    sigma = F/A
    sigma = 30+30(2.5)/0.09
    sigma = 1166 N/mm2

    3) http://img511.imageshack.us/img511/3645/deltal.jpg [Broken]


    Is that correct?
     
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Nov 5, 2011 #2

    SteamKing

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    A = cross sectional area, not the area of the diagram

    When you are calculating axial stress, the applied force is spread over the area which is normal (at right angles to) the direction of the force.
     
  4. Nov 5, 2011 #3

    Femme_physics

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    Duly noted. Will correct it.

    You're saying it should be like this?:
    http://img10.imageshack.us/img10/7915/dongn.jpg [Broken]


    I'm not sure how this makes sense.
     
    Last edited by a moderator: May 5, 2017
  5. Nov 5, 2011 #4

    I like Serena

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    Hi Fp! :smile:


    It is a 3D problem.
    It's a side view of 2 cylinders.


    Yes, I think so too.
    But it puzzles me that they have an external force of F, and what appears to be an internal force of 2.5F.
    If it is really an internal force it couldn't be that big, so I guess we'll have to assume it's an external force that compresses the leftmost cylinder.

    Can you clarify that?


    As SteamKing said, you're trying to calculate the tensile stress, which is the force per area of cross section.
    We're talking about 2 cylinders here, with each their own cross section and tensile stress.
    If a tensile stress is greater than the yield strength limit, the beam would break.

    Do you know how to calculate the cross section of a cylinder?

    I think you need to calculate 2 distortions, one for each cylinder.
    And then you need to calculate the total distortion.
     
    Last edited by a moderator: May 5, 2017
  6. Nov 5, 2011 #5

    Femme_physics

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    Hi ILS. Was really hoping you'd still be here. If legends are true at your help-giving I'm good hands :smile: :wink:

    I see.


    Hmm. Well, it's saturday and I have no contact to my teacher. What do you think it is likely to be? It sounds like what ur surmising it is.

    Is my second Free Body Diagram (based on SteamKing's reply) correct?

    Yes, if I have internal diameter. I'm gonna ignore external diameter since it was not given to me.

    http://img638.imageshack.us/img638/9237/crosssectarea.jpg [Broken]

    So
    sigma = F/A
    sigma = 30+30(2.5)/0.03239
    sigma = 3241.74 N/mm2


    Well, you're the master/chief/sensei, I trust what you think.

    So I used the correct formula? I wasn't sure.
     
    Last edited by a moderator: May 5, 2017
  7. Nov 5, 2011 #6

    I like Serena

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    Well I was wondering when or if you'd be back. :tongue:
    Say, when the voting starts for special awards (soonish), will you vote for me? :wink:


    Since it's a basic exercise, I surmise that the left cylinder is compressed by 2.5F, and the right cylinder is expanded by F.
    Otherwise it would really be too complex for a basic exercise in tensile stress.


    No. Your first FBD was correct.
    I'm afraid your 2nd FBD makes no sense to you.
    (Did it make sense to you? :rolleyes:)

    Note that the forces you drew there extend over the entire cross section of the cylinder.


    You have combined both surfaces into one total surface.
    I'm afraid you're not supposed to do that.

    Btw, there's no such thing as an internal and external diameter.
    The cylinders are massive of uniform material.
    (It's not a pipe - that's in fluid hydraulics! :wink:)

    You need to calculate the tensile stress in the left cylinder, using its compressing force and its cross section.

    And separately, you need to calculate the tensile stress in the right cylinder, using the force that expands it and its cross section.



    The formula for cross sectional area, which is the surface of a circle, is correct.
    But you need to calculate and use only 1 surface at a time.
     
    Last edited by a moderator: May 5, 2017
  8. Nov 5, 2011 #7

    Femme_physics

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    Of course! What a question. Just link me in because I don't keep in touch of community activity.

    Makes sense. We're just starting, and this is as far as I know non-calculus based.


    I thought so! Glad you're here to set the record straight! :smile:

    http://img195.imageshack.us/img195/4654/penmark.jpg [Broken]

    Is that it?

    Do I keep the separate results for each one? Or do I combine them eventually?




    Oh! Thanks for helping me with the separation.

    OK :smile:

    http://img848.imageshack.us/img848/4015/penmark2.jpg [Broken]
     
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  9. Nov 5, 2011 #8

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    I'll get back to you on that! :smile:


    Hmm, what happened to the square-symbol in the circle-surface formula?

    Did you check your units?
    I think there should be a couple more or less zeroes in a couple of places.

    And I see you combined the forces for cylinder 2 by adding them, which looks good.
    But then for cylinder 1 you should subtract them to get the resulting force.



    No, you do not combine the tensile stresses.
    You're supposed to check each against the maximum stress the material can take.



    I have a couple of suggestions for improvement.
    Let me list them:

    1. Check your units. I recommend including them when you substitute the numbers.

    2. There should be a couple of parentheses in there.

    3. The surfaces-areas need to be updated with the new numbers.

    4. Cylinder 1 compresses, so it has a negative delta.

    5. The problem asks for the total distortion.
     
    Last edited by a moderator: May 5, 2017
  10. Nov 6, 2011 #9

    Femme_physics

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    Damn, I keep neglecting that. Even last year.

    Should I keep everything in mm? Since the final result should have the unit "mm". That's at least what the formulas say.

    Subtract what from what? F1 - F2? Or F2 - F1? Hmm, I fail to see the logic why.

    Ok, duly noted :smile:

    1) same question as I asked above

    2) Hmm, but the result turns out the same anyway, no?

    3) Oh, yea.

    4) Ah, will check!

    5) Delta l1 + Delta l2?
     
  11. Nov 6, 2011 #10

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    Yes, that would be fine.



    Can you draw an FBD for cylinder 1?



    Well, does cylinder 1 break? Or cylinder 2?
    That's what the problem asks.



    Same answer as above. :wink:


    Dunno, didn't check your calculations.


    Yea, baby! :tongue:


    Yep! :smile:
     
  12. Nov 6, 2011 #11

    Femme_physics

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    Got it. Force of habit I guess to convert mm to m, I guess.

    http://img411.imageshack.us/img411/9569/fizor.jpg [Broken]



    Was thinking to just do that for the FBD isolation:

    http://img513.imageshack.us/img513/4356/f123.jpg [Broken]

    Whereas F1 and F2 are the forces in the original diagram and F3 is the force applied by cylinder 2.

    Only calculations can tell me the answer, I'm still stuck at how to subtract the forces you talked about.

    Did you just Austin Powered me? :wink:
     
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  13. Nov 6, 2011 #12

    I like Serena

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    Much better!



    Hmm, I can't quite make it out.
    What are the magnitudes supposed to be?
    You used to be pretty good at drawing FBD's! :smile:

    Is F1 the force 2.5F?
    Shouldn't it be in the other direction then?

    Is F2 the force F?
    If so I suggest you draw it where it actually pushes against the cylinder.
    And also that you draw it a bit shorter (by about a factor 2.5 ;)).

    If F2 is the force from the original diagram, then that is also the force applied by cylinder 2.
    So how can F3 be that force?

    And what happened to the force that the wall applies?



    Yeah Baby.
     
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  14. Nov 6, 2011 #13

    Femme_physics

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    :biggrin:
    That's because I was drawing the other cylinder. Not the walled one.

    This is the walled one
    http://img442.imageshack.us/img442/6890/fixinginter.jpg [Broken]
     
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  15. Nov 6, 2011 #14

    I like Serena

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    Now that you did your dance of joy, could you switch around A1 and A2 please?
    (Just noticed that. :blushing:)



    Don't worry about the internal forces - they're not supposed to be in an FBD anyway. :wink:

    But what happened to the force that cylinder 2 exerts on cylinder 1?
    They are glued together or something, aren't they?
     
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  16. Nov 6, 2011 #15

    Femme_physics

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    You're right, I don't know why my logic said not to include it. But, perfectly legit and correct t include it.

    http://img573.imageshack.us/img573/8638/added.jpg [Broken]
     
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  17. Nov 6, 2011 #16

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    What about the direction of F?
    Shouldn't it go the other way?
     
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  18. Nov 6, 2011 #17

    Femme_physics

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    Wait, something is quite odd to me. When I isolate an object, aren't I supposed to draw all the forces acting ON THE OBJECT, as opposed to the forces the object is enacting on other objects that are not in the drawing?

    That's why I didn't include include a 3rd vector.
     
  19. Nov 6, 2011 #18

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    Yes. That is right.

    But I do not understand your conclusion.

    In total there are 3 external forces: F, 2.5F, and Nx.
    F acts on cylinder 2 and as a reaction, cylinder 1 pulls cylinder 2 back to the left.
    The reactive force between cylinder 1 and 2, is that cylinder 2 pulls cylinder 1 to the right.
     
  20. Nov 7, 2011 #19

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    Let's try again. :smile:

    Let's look *only* at F.

    In the total picture, the wall exerts an equal but opposite power.

    In the 2 separate FBD pictures, these forces have to balance out as well.

    attachment.php?attachmentid=40742&stc=1&d=1320662843.gif
     

    Attached Files:

  21. Nov 7, 2011 #20

    Femme_physics

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    You seemed to have ignored 2.5F, though, am I right?

    Sorry it takes me a while to reply sometimes. Appreciate your help. :smile:
     
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