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Angle of twist / stress of thin wall having closed section

  1. Jan 3, 2017 #1
    1. The problem statement, all variables and given/known data
    why the mean area enclosed within the boundary of the centerline of tube thickness is (0.035)(0.057) ? Is it wrong ?

    Second question , why for the ds it's 2(57) and 2(35) respctively ? what is ds actually ?

    2. Relevant equations


    3. The attempt at a solution

    for the first question , Shouldnt it be (0.040-0.005-0.005) x ( 0.060 -0.003 -0.0030 = 0.054 ?
     

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  2. jcsd
  3. Jan 3, 2017 #2

    BvU

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    No, it's right. Mean height is 54+3/2+3/2 and mean width is 30+5/2+5/2.
    He's following the circumference of the mean area: the dashed line in figure 5-30 (d)
    Nothing again ? :smile:
    That's not an attempt, that's a question !
    Anyway, that would be following the inside of the tube, which is not the intention here.
     
  4. Jan 3, 2017 #3
    Do you mean the area enclosed by the hollow position + area of half thickness of the frame ?
     
  5. Jan 3, 2017 #4
    ok , i understand why it's 57mm and 35 mm , because it's the mean area enclosed within the boundary of the centerline of the tube thickness


    why the thickness shouldnt be 3/2 and 5/2 mm now ? Because for the area of 57x 35 mm , the thickness of the 'frame' has been halved , right ?
     
  6. Jan 3, 2017 #5

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    I think the book is perfectly clear. He's following
    which has a certain length and encloses a certain area. The tube thickness itself is of course not halved. (however, the expressions developed -- such as for 5-20 -- are valid only for thin walls)
     
  7. Jan 4, 2017 #6
    why The tube thickness itself is not halved ??? We can see that the figure in 7.30d , the thickness has been halved , right ?
    for the original frame , we have the inner length of 60-3-3 = 54m and 40-5-5 = 30m , for the mean area it's 54+(3/2) + (3/2) = 57 , and 30+(5/2)+(5/2) = 35 , so the thickness for the mean area is 3/2 and 5/2 respectively , right ? Which the thickness has been halved ?
     
  8. Jan 5, 2017 #7

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    The dashed line is simply an average circumference. If you cut the pipe lengthwise you get a long sheet with a certain width and a certain thickness (that varies with lateral position, so it's kept inside the ##\int ds##). No halving.

    See post #5 and check out the derivation of 5-20.
     
  9. Jan 5, 2017 #8
    CIRCUMFERENCE ? I didnt see any circle here
     
  10. Jan 5, 2017 #9

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  11. Jan 5, 2017 #10
    why the thickness shouldnt be the same through out the length of the pipe ?
     
  12. Jan 5, 2017 #11

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    If you cut it lengthwise and fold it flat, thickness is the same throughout the length of the pipe. That's in the problem statement. You get a long slab of approximately 174 mm wide with two sections of 3 mm thickness and two of 5 mm thickness.

    My estimate was/is that you could have figured that out for yourself. Idem circumference/perimeter :rolleyes: .
     
  13. Jan 5, 2017 #12
    Do you mean cut the beam at 2 locations( cut off a portion of the beam) and hold it flat ???
    How to fold it flat ? I couldnt imagine it at all
     
  14. Jan 5, 2017 #13

    BvU

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    You don't have to actually do it ! :smile:

    My advice: check out the derivation of 5-20
     
  15. Jan 5, 2017 #14
    it's not given in the book
     
  16. Jan 5, 2017 #15
    Can you explain about it ?
     
  17. Jan 5, 2017 #16

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  18. Jan 5, 2017 #17
    it's not the thickness of the wall bounded by the centerline ? it's still the same thickness before the centerline is drawn ?
     
  19. Jan 5, 2017 #18

    BvU

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    Yes to the latter. The shear stress is distributed over the entire thickness (5-18).
    I drew the centerline so you could see where the 174 came from; not to confuse you.
     
  20. Jan 5, 2017 #19
    174 ? where is it ?
     
  21. Jan 5, 2017 #20

    BvU

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    With the browser find function CTRL-F you can see that it is in post #11.

    My estimate was/is that you could have figured that out for yourself
     
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