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Bolted assembly, friction force, clamping force

  1. Nov 15, 2015 #1
    I think I have solved my problem for a friction force between a bolted assembly, or at least I think did. If a torque was given of 267.6 in-lbs and a constant of k=0.2 and the diameter(d) =0.5,
    T=k*P*d should equal 267.6=0.2*P*0.5, resulting in P=2676lbs.
    Then using the equation:
    Tc=F*fc*dc/2
    Where Tc=Torque
    F= force =2676lbs
    fc= collar friction coefficient =0.15
    dc=mean collar diameter= dm+1.5dm/2=2.5dm/2=1.25dm: dc=dm*diameter=1.25*0.5=0.625
    Tc=2676*0.15*0.625/2
    Tc=125.438in-lbs
    Then
    Tc=ff*n
    Tc=Torque=125.438in-lbs
    ff=friction force
    n=center of bolt hole to outer edge of part =0.46inches
    125.438in-lbs=ff*0.46in
    ff=272.69lbs?
    Seems high for a frictional force for a 1/2 inch diameter bolt.
    Any help is greatly appreciated. Thank you for any help you can lend.
    IMG_0001.jpg
     
  2. jcsd
  3. Nov 20, 2015 #2
    Thanks for the post! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?
     
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