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Shear stress in bolts help

  1. Jan 10, 2015 #1
    1. The problem statement, all variables and given/known data
    jan 2012 q3.jpg
    I've solved everything up to part d). I generally struggle with questions on bolts and I'm really not sure how to tackle this problem.

    2. Relevant equations
    τ = Tr/J, τ = 200MPa, T=1kNm

    3. The attempt at a solution
    I worked out the separation of the bolts (I think).
    Circumference around the inside: 2π(25mm) = 0.157m ∴ Separation: 0.157/2 = 0.0785m

    Not sure where to go next, if I'm honest. I tried this:

    T = 1kNm ⇒ F = 1kNm/0.0785m = 12.7kN
    so σ = F/2A as double shear so A = F/2σ = 12.7kN/2(200MPa) = 3.19x10-5m2

    so πr2 = 3.19-5 ⇒ r = 3.18mm ⇒ d = 6.37mm

    so I went wrong somewhere as the answer is 11.28mm

    any help appreciated. Thanks
     

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    Last edited: Jan 10, 2015
  2. jcsd
  3. Jan 10, 2015 #2

    SteamKing

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    It's not clear how the circumference of the solid shaft is useful here. You were instructed by the problem statement that friction between the shaft surface and the coupling was to be neglected. Therefore, all of the torque transmitted to the solid shaft must be resisted by the two bolts. What force on the bolts would produce an equal torque?

    This approach is not correct for the reason given above.
     
  4. Jan 10, 2015 #3
    Ah okay. So the for the force on each bolt:
    T = Fm ⇒ F = 1kNm/50mm = 20kN
    So 10kN on each bolt?
     
  5. Jan 10, 2015 #4

    SteamKing

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    You might want to check that figure of 10 kN/bolt.
     
  6. Jan 10, 2015 #5
    Oh okay. So would it be 20kN?
     
  7. Jan 10, 2015 #6

    SteamKing

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    Yes.
     
  8. Jan 10, 2015 #7
    And then use σ = F/A?
     
  9. Jan 10, 2015 #8
    so A = 20kN/200Mpa = 1x10-4
    then r = √(1x10-4/π) = 5.64mm
    so d = 11.3mm
    Thanks :D
     
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