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Thickness of a cylinder in a compound thin cylinder

  1. Apr 26, 2017 #1
    1. The problem statement, all variables and given/known data

    upload_2017-4-26_8-18-55.png

    Note: inner cylinder thickness = 2.5mm
    difference between common diameters before shrinking = 4.305x10^-3mm

    2. Relevant equations

    upload_2017-4-26_8-33-14.png
    upload_2017-4-26_8-33-26.png

    3. The attempt at a solution

    a) I am not sure where to start without either the inner or outer cylinder diameter but I think I need to use the following:

    The sum of the changes in diameters of the two cylinders is equal to the interference between the diameters before shrinkage:

    Δd1+Δd2=4.305x10^-3mm
    Pd1/2t1 + Pd2/2t2=4.305x10^-3mm

    Please help me get started
     
  2. jcsd
  3. Apr 26, 2017 #2
    edit the answer give for part a) is 2mm
     
  4. Apr 26, 2017 #3

    haruspex

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    Although the info E=200GPa is mentioned in part b, you will need this in part a.
    Δd1 and Δd2 represent a stretch in the outer cylinder and a compression of the inner one. In terms of those, what will the tension and compression be (per unit length)? What must the relationship be between the two forces?
     
  5. Apr 26, 2017 #4
    Will both cylinders not be compressed/shrunk?
    Sorry I do not follow
     
  6. Apr 26, 2017 #5

    haruspex

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    Perhaps I misunderstand the arrangement. Any diagram?
    I read it that two cylinders have been fabricated, one with an external diameter of slightly over 100mm, the other with an internal diameter slightly under. The larger will not slide over the smaller because of a 4.305x10^-3mm overlap. Through heating, the larger cylinder expands and is slid on, then allowed to cool. This will produce tension in the outer cylinder, increasing its internal diameter to 100mm, and compression in the inner cylinder, contracting its external diameter to 100mm.

    Do you read it differently?
     
  7. Apr 26, 2017 #6
    Unfortunately no diagraM.
    I imagine something like this:
    upload_2017-4-26_13-13-43.png

    I could be (most likely) wrong so I really don't know.
    All I do know is that the thickness of the outer cylinder must equal 2mm, I just don't know how to get there.
     
  8. Apr 26, 2017 #7

    haruspex

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    With set-up as you draw it, there would be no pressure between the two surfaces and no basis for calculating the other thickness.
    Try my version.
     
  9. Apr 26, 2017 #8
    Ah yes that makes sense.

    However I am still not sure how to start.
    I have P=200kPa
    t2=2.5mm
    D=100mm
    interference = 4.305x10^-3mm

    I can find the circumferential stress using upload_2017-4-26_8-33-14-png.197107.png

    = 200x10^3x100x10^-3/2x2.5x10^-3
    = 4MPa

    I don't see how I could use this to find t1.
    As far as the relationship between tension and compression forces I assume they are equal to each other once the common diameter of 100mm is established?
     
  10. Apr 26, 2017 #9

    haruspex

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    Yes.
    Good. You are given E. What radial compression would that lead to?
     
  11. Apr 27, 2017 #10
    Sorry for the delay I am in South Africa and your reply only came through at around 11pm last night.

    Okay so using E=200GPa

    Circumferential stress = PD/2tE
    =200x10^3x100x10^-3/2x2.5x10^-3x200x10^9
    =2x10^-5 Pa

    This doesn't seem right?

    Edit: in my textbook/study guide there is no formula for radial stress given "because it is very small and thus disregarded". I am only given formulas for circumferential (hoop) stress and longitudinal stress.
     
    Last edited: Apr 27, 2017
  12. Apr 27, 2017 #11
    If I use

    radial stress x radius = force
    radial stress =force/radius
    =200x10^3/50x10^-3
    =4MPa

    This is the same answer as I got earlier without using E.
    What would the next step be provided this is correct?
     
  13. Apr 27, 2017 #12

    haruspex

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    I did not mention radial stress. I asked about the radial compression, i.e. reduction in radius. The next question is the corresponding change in circumference.
     
  14. Apr 27, 2017 #13
    okay how about this:
    the sum of the changes in the diameters of the two cylinders
    is equal to the interference between the diameters before shrinkage

    Δdi+Δdo=4.305x10^-3mm

    Pd/2tE + Pd/2tE = 4.305x10^-3mm
    Pd/2tE + 2x10^-5 = 4.305x10^-3mm
    200x10^3xd/2x2.5x10^-3x200x10^9 + 2x10^-5 = 4.305x10^-3mm
    d = 21.425

    Ah I don't know :H
     
  15. Apr 27, 2017 #14

    haruspex

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    It would help me to follow your working if you would avoid plugging in numbers. Just keep everything algebraic. It's a very good style to adopt, having many advantages.
    ri = original outer radius of inner cylinder
    ro= original inner radius of outer cylinder
    ri-ro=δ (=4.305x10^-3mm)
    R= 100mm
    To= circumferential tension in outer cylinder, etc.
    wo = thickness of outer cylinder, etc.
    P= pressure between cylinders = 200kPa.

    Using those variables, what is the change in radius of the outer cylinder?
    What is the change in circumferential length of the outer cylinder?
    What equation does that give for To?
    What is the relationship between To and P?
    Same as above for inner cylinder.
     
  16. Apr 28, 2017 #15
    Sorry about that I am honestly just grasping at straws.

    I have tried to figure out where you want me to go with the above but I just cannot get it.
    R=radius? It should then be 100mm/2=50mm?

    Is there a specific equation that I am supposed to be using to find the solution?
     
  17. Apr 28, 2017 #16

    haruspex

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    Yes, sorry, I'm not used to diameters being given instead of radii.

    Forcing the cylinders to other than their relaxed radii results in a change in their circumferential lengths. That causes a compression in the one and a tension in the other. These forces must balance.

    Please try to answer my questions in turn. Where do you get stuck?
     
  18. Apr 28, 2017 #17
    No problem.
    Okay so the inner cylinder is under compression and the outer cylinder is under tension if I imagine the situation.
    So tension = compression but I don't know how to put this into an equation with the other variables.
     
  19. Apr 28, 2017 #18

    haruspex

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    Please try to answer my questions inpost #34 in sequence. The first one was:
     
  20. Apr 28, 2017 #19
    Using those variables, what is the change in radius of the inner cylinder? ri=4.305x10^-3mm +ro
    What is the change in circumferential length of the inner cylinder? Li=Lo-4.305x10^-3mm
    What equation does that give for Ti? Ti=P-To
    What is the relationship between Ti and P? The pressure between the cylinders increases the compression in the inner cylinder because the outer cylinder is pushing inwards onto it
     
  21. Apr 28, 2017 #20

    haruspex

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    As I asked, please do not plug in numbers yet, just use algebraic symbols. Much easier for me to follow what you do.
    Anyway, that does not give the change in radius of the inner cylinder. Its outer radius starts at ri. What is its final outer radius?
     
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