What is the best way to position hinges on a heavy duty door design?

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The discussion focuses on the optimal positioning of seven butt hinges for a heavy-duty door measuring 1.5m by 2.5m and weighing approximately 300kg. The production supervisor recommends clustering three hinges at the top and distributing the others towards the bottom to better handle shear forces, while the original poster contemplates using all hinges equally spaced. Calculations are presented to determine the appropriate distance between hinges, suggesting that spacing should be less than 1690mm, resulting in approximately 282mm between each hinge if evenly distributed. However, another participant challenges the calculations, proposing a different method for determining the forces on the hinges based on applied loads. The conversation highlights the complexities involved in hinge placement for heavy doors and the importance of accurate calculations in ensuring structural integrity.
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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!
 
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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.
 
Mech_Engineer said:
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.
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.
 
Bob S said:
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.

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...
 
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?
 
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.
 
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How did you derive to that formulation. I referred to the guideline provided at this site
http://www.allegiscorp.com/docs/hingeDesignGuide.pdf
 
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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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