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Number of bolts required for a flanged coupling?

  1. Jun 23, 2011 #1
    Hi everyone.

    This question is giving me a real headache (literally). It should be simple, but I’ve spend probably the last 4-5 hours trying to figure it out. The left side of my brain is redundant.

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

    The following specifications are used for the design of a flanged coupling between two coaxial shafts:

    Speed: 650rpm
    Power transmitted: 550 kW
    Bolt diameter: 12mm
    Pitch-circle diameter: 200mm
    Material: Mild steel
    Factor of safety: 4

    Determine the number of bolts required, assuming the bolts are equally loaded.

    2. Relevant equations

    Ultimate Shear Strength of mild steel = 360 N/mm^2

    Shear stress (MPa) = Force (Newtons) / Area (mm^2)

    Factor of safety = Ultimate Shear Strength / Shear Stress

    3. The attempt at a solution

    I cannot get anything down at all. And I also cannot understand why they have given me Speed and Power. Is there maybe some way to work out the force for the shear stress equation?

    Also for your information, this is question 27.11 from Val Ivanoff’s Engineering Mechanics.

    Any help would be much appreciated!
     
  2. jcsd
  3. Jun 23, 2011 #2

    tiny-tim

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    welcome to pf!

    hi jacoja06! welcome to pf! :smile:
    go back to basic definitions …

    power = energy/time = work/time (same thing)

    = force x distance/time

    = force x distance x angle/time :wink:
     
  4. Jun 23, 2011 #3
    Thanks for the reply and welcome! :)

    So using that formula;

    P = F x distance/time

    550kw = F x (650rpm x ((2 x pi)/60))

    F = 484.81N?
     
  5. Jun 23, 2011 #4

    tiny-tim

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    you've missed out the radius :redface:
     
  6. Jun 23, 2011 #5
    Oh, sorry... I didn't put the time in seconds in my last post. I was supposed to put;

    550kw = F x (650rpm x ((2 x pi)/60)/60)

    So;

    550kw = F x (68.0678rad/60)

    F = 484.81N

    Where would the radius come into the equation?
     
  7. Jun 23, 2011 #6

    SteamKing

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    Hint: Torque = Force * Distance
     
  8. Jun 23, 2011 #7

    SteamKing

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    You've also got an extra factor of 60 in your RPM conversion. Remember 1 min = 60 sec
     
  9. Jun 23, 2011 #8

    SteamKing

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    To solve the larger problem, how many bolts, you are trying to transmit a certain torque from one shaft to another at the coupling. You are given a standard size bolt to use, along with the material characteristics. How is the torque going to be transmitted? What type of stress will be set up in the bolts?
     
  10. Jun 23, 2011 #9
    Thanks, steamking.

    I think he distance part is confusing me the most at the moment. There isn't any specified distance I can use to find the force.
     
  11. Jun 23, 2011 #10

    gneill

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    The pitch-circle is the circle concentric with the shaft where the bolt holes are located.
     
  12. Jul 10, 2012 #11
    total torque to be transmitted => T=9550P/N (N-m)
    => 9550*550/650 = 8080N-m
    force on the coupling= 8080/0.250 =40400 N
    assuming its a mild steel of yield strength 240MPa and shear strength of 120 MPa,with a factor of safety of 4 the design shear strength is 30Mpa

    we have area of each bolt = Pi*12*12/4 =113.04mm2

    total load that a single bolt can take = 113*30 = 3390N

    so you will need 40400/3390 = 12 bolts.

    I hope this clarifies your question and i stand to be corrected if I'm wrong.
     
  13. Jul 10, 2012 #12
    speed and power helps you to find the torque..... you can use the equation I specified above, or you can also use this

    Power =(2*Pi*N*T)/60

    also, we design machines with factor of safety as a function of yield strength rather than UTS. :)
     
  14. Jul 10, 2012 #13
    Sorry!! the force on the coupling would be 8080/0.125 (consider the radius... not diameter. silly error :) )
    follow the same methodology
     
  15. Oct 22, 2012 #14
    My view on this problem:

    Bolt material: SANS 1431 GR 300WA (Mild steel). It is assumed that the minimum specified ultimate strength is 400 MPa and a proof strength of σ4.8 = 320 MPa which gives a shear yield strength of: 0.577x320 MPa = 185 MPa

    Shaft speed: N = 650 RPM
    Power transmitted: P = 550 kW
    Bolt diameter: D = 12 mm
    PCD = 200 mm
    Factor of safety: SF = 1.2

    The bolts will have to withstand two types of loading:
    Pre-tensile loading from fastening torque
    Shear loading from transferring required power

    Recommended bolt torque: 40 Nm
    This gives internal bolt force of:
    F=T⁄0.2D
    F=40⁄(0.2×(0.012))
    F=16.67 kN
    Tensile axial stress per bolt:
    σ_1=F⁄A
    σ_1=16670⁄((π×〖(9.85)〗^2)⁄4)
    σ_1=219 MPa

    Predict number of bolts required will be 8.
    Shear stress:
    P=F×v
    with
    v=ωr
    v=N(2π⁄60)r
    v=650(2π⁄60)(0.1)
    v=6.806 m/s
    this gives
    550000=F×6.806
    F=80.811 kN
    Hence
    F_(per bolt)=80811⁄8
    F_(per bolt)=10.1 kN

    Shear per bolt
    τ_(per bolt)=F⁄A
    τ_(per bolt)=10101⁄((π×(9.85)^2)⁄4)
    τ_(per bolt)=132.56 MPa

    If friction is neglected these stresses work together to give a Von Mises equivalent stress (In plane stress) of:
    σ_vm=√(〖σ_1〗^2+〖3(τ)〗^2 )
    σ_vm=√(〖219〗^2+〖3(132.56)〗^2 )
    σ_vm=317.3 MPa
    This gives a Safety factor of:

    SF=σ_4.8⁄σ_vm
    SF=320⁄317.3
    SF=1.008

    This indicates that when 8 bolts are to be used, it will be able to transfer the loads with a safety factor of 0.8% which is not adequate for the operation.








    Newly predict that the number of bolts required will be 12:
    Hence
    F_(per bolt)=80811⁄12
    F_(per bolt)=6734 N

    Shear per bolt
    τ_(per bolt)=F⁄A
    τ_(per bolt)=6734⁄((π×(9.85)^2)⁄4)
    τ_(per bolt)=88.37 MPa

    If friction is neglected these stresses work together to give a Von Mises equivalent stress (In plane stress) of:
    σ_vm=√(〖σ_1〗^2+〖3(τ)〗^2 )
    σ_vm=√(〖219〗^2+〖3(88.37)〗^2 )
    σ_vm=267 MPa
    This gives a Safety factor of:

    SF=σ_4.8⁄σ_vm
    SF=320⁄267
    SF=1.2

    This indicates that when 12 bolts are to be used, it will be able to transfer the loads with a safety factor of 20% which will be adequate for the operation

    Ideally the torque should be transmitted through friction alone
    With
    F=T⁄r
    And
    N=F⁄μ÷4

    With friction coefficient of 0.3 the amount of internal bolt force needed to transmit the power through friction alone is then:
    F=80811 N
    And
    N=80811⁄0.3÷12
    N=22.45 kN per bolt

    SF=16.67⁄22.45
    SF=0.75

    Provided that the bolts are torqued to the recommended value of 40 Nm the connection would not be able to transmit the torque through friction alone.

    For the ideal working, increase the number of bolts so that the power would be transmitted through friction alone, say 16 bolts...

    I see that this is an old thread, just thought my feedback would do some good...
     
  16. Oct 22, 2012 #15
    I have received nasty Warnings in the past from PF about revealing the entire solution to the OP without giving them an adequate chance to solve it themselves, and have had my replies removed from the thread. How come no one is receiving Notifications in this thread?
     
  17. Oct 22, 2012 #16

    tiny-tim

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    Hi Chestermiller! :smile:

    I'm not a mentor, but I'll guess that it's because the OP asked the question over a year ago, and …
    … probably has solved it himself by now! :wink:
     
  18. Oct 23, 2012 #17
    Yes I guessed that this is very old so the solution would have been found, seeing that you say that I gave the entire solution I believe my equations is correct?

    Just came upon the thread and saw no solution has been found that seems correct for me, and i tried giving a solution.

    Chestermiller, did not even know about the NASTY WARNINGS, but will keep it in mind for the future, thanks...
     
  19. Oct 23, 2012 #18
    You already have much good help in finding the simple solution, but it should be noted that in my experience shaft imbalance often becomes the dominant factor in real rotating flange design. Either because of a manufacturing error or because of a potential mechanical failure.
     
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