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Heavy duty door hinge design

  1. Jun 11, 2009 #1
    I working on a project to manufacture a door of size (W)1.5m and (H)2.5m that uses 7 butt hinges attached to the frame. The weight of the door is approx 300kg. Currently I have decided to position the hinges equally to each other but my production supervisor is advising against it. He suggest that about 3 hinges be placed closer at the top and the rest to be located equally at towards the bottom. How do I calculate the distance between hinges. Please help!
     
  2. jcsd
  3. Jun 11, 2009 #2

    Mech_Engineer

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    Why not just use 3 properly sized hinges?

    I think what your supervisor is getting at is that when the door is open and hanging from the hinges, the greatest shear force will be applied at the top-most and bottom-most hinges, so he wants you to goup the hinges such that there are more hinges where the shear force will be greatest.

    However, I don't think you need to use so many hinges (unless that is some sort of requirement). I have seen some pretty heavy doors hanging from 2 or 3 very big hinges.
     
  4. Jun 11, 2009 #3
    The loading on the top hinges by 300 Kg (660 pounds) is a lot of lateral force to put on 3 or 6 #10 screws in a door jam.
     
  5. Jun 11, 2009 #4

    Mech_Engineer

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    I seriously doubt this 300Kg door is being mounted on a standard residential or commercial door jam, but I could be wrong. I was imagining a big door for a secure room or safe, or walk-in freezer or something...
     
  6. Jun 11, 2009 #5
    thanx for your replies. The door will be mounted on specially made door jams and yes 7 hinges is the requirement from the customer. I referred to some notes & did the following calculations to justify my theories but am not sure if I missed anything!


    The radial load acting on top hinge (Fradial) = [Weight of door (W) x dist to ctr of gravity(d1) + External force (Fext) x door width (d2)] / dist between hinges (dhinge)
    = [(300*9.81*0.75) + (1800*1.5)]/dhinge
    = 4.9x10^3/dhinge

    Presuming the top hinge holds the weight of the door when opened;
    Fradial should be > weight of the door
    4.9x10^3/dhinge > 2.9 x 10^3
    dhinge < 1.69m

    So, if the distance between the hinges should be less than 1690mm, distance between all 7 hinges provided they are equally spaced = 1690/6 = 282mm.

    Is this correct?
     
  7. Jun 11, 2009 #6

    nvn

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    chrisya: Your calculations in post 5 don't look correct to me, so far.

    If you have seven equally-spaced hinges plus a vertical, external applied load of 1800 N at 1500 mm, then the horizontal force on the top and bottom hinge would be F = M*c/summation(y^2) = [(300*9.81)(750 mm) + (1800 N)(1500 mm)](1071.4 mm)/(3 571 430 mm^2) = 1472 N. And the vertical shear force on each hinge would be V = (300*9.81 + 1800)/7 = 677.6 N.
     
    Last edited: Jun 12, 2009
  8. Jun 11, 2009 #7
    How did you derive to that formulation. I referred to the guideline provided at this site
    http://www.allegiscorp.com/docs/hingeDesignGuide.pdf [Broken]
     
    Last edited by a moderator: May 4, 2017
  9. Jun 12, 2009 #8

    nvn

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    M = applied moment, c = extreme fiber distance from neutral axis, y = distance of each hinge from neutral axis.
     
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