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Axial compressive load

  1. Nov 1, 2009 #1
    1. The problem statement, all variables and given/known data
    A steel pipe..has a diameter of 116mm...length of 1.8m...and wall thickness of 8mm...gets pounded down by an axial compressive load of 180kN... Lets say that E (the elasticity) = 200GPa...and V(the poisson's ratio) is 0.3... i need to work out the CHANGE IN WALL THICKNESS :confused: ...how?? anyone??:smile:


    2. Relevant equations
    change in diameter = -0.3 x (change in length / initial length) x initial diameter
    change in length = (bob x initial length) x E
    bob = Force / Area (CSA)
    CSA = Pi x radius squared

    3. The attempt at a solution
    I dont know where to go from here..HELP!!! :bugeye:
     
  2. jcsd
  3. Nov 1, 2009 #2

    Mapes

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    Hi Biru, welcome to PF. What is the physical interpretation of the Poisson's ratio?
     
  4. Nov 1, 2009 #3
    physical interpretation of poisson's ratio? how do you mean?
     
  5. Nov 1, 2009 #4

    Mapes

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    What is Poisson's ratio?
     
  6. Nov 1, 2009 #5
    0.3 haha sorry!
     
  7. Nov 1, 2009 #6

    Mapes

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    Yes, for this problem. But what does Poisson's ratio mean?
     
  8. Nov 1, 2009 #7
    its basically lateral strain / direct strain...the ratio between the two strains in other words...its mostly always 0.28 - 0.38... helps??
     
  9. Nov 1, 2009 #8

    Mapes

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    - lateral strain / axial strain to be precise. So can you apply this information to the change in wall thickness?
     
  10. Nov 1, 2009 #9
    rite..will i have to do it for both..the outside diameter and the inside diameter???
     
  11. Nov 1, 2009 #10
    so change in diameter = -0.3 x -(change in length/initial length) x initial diameter???
     
  12. Nov 1, 2009 #11

    Mapes

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    Looks like you've got the right idea. What's your final answer?
     
  13. Nov 1, 2009 #12
    umm...the wall thickness changes by 204 nano meters??
     
  14. Nov 1, 2009 #13

    Mapes

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    I'm getting a slightly different answer; how did you calculate yours?
     
  15. Nov 1, 2009 #14
    well first I worked out bob = Force / Area (CSA)..soo -180kn / pi x (116mm/2)squared!! got sumthing like -17.032 mega pascals...

    then i worked out the change in length..using..

    (bob x initial length) / Elasticity so (-17.032..MPa x 1.8) / 200GPa ..got me -153.289 nano meters...so bascially when the pipe got pounded by that 180kn the length shook by -153.29 nano meters..rite??

    then I used that to work out the change in diameter (of the outer circle)...so delta d got me..2.964 nano meters..so the outside diameter increased by 2.96 nano meters...

    then using the SAME change in length value (-153.29 nano meters)... I did the change in diameter again but this time using the inner circle so diameter of 100mm rather than 116mm... that got me 2.55 nano meters...

    so the inner circle increased from 100mm to 100.00000255 mm...and the outer circle increased from 116mm to 116.00000296mm ..the difference between them two numbers was 204 nano meter higher than 8mm..

    SO the wall thickness changed by 204 nano meters...NO??
     
  16. Nov 1, 2009 #15

    Mapes

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    Check that cross-sectional area again; it's a hollow pipe!
     
  17. Nov 1, 2009 #16
    umm...the only formula i kno is that.. pi x radius squared...

    so wat wud it be for hollow pipe??
    Pi x (bigger radius - smaller raidus) squared??
     
  18. Nov 1, 2009 #17

    Mapes

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    Not quite. Keep thinking about it.
     
  19. Nov 1, 2009 #18
    is it Pi/4 x ((d1^2) - (d2^2))??
     
  20. Nov 1, 2009 #19
    Or is it...Pi x r1 squared - Pi x r2 squared??
     
  21. Nov 1, 2009 #20
    i got -61.316 KPa for bob...is that rite??
    bob = force / area..
     
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