Heavy duty door hinge design

1. Jun 11, 2009

chrisya

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. Jun 11, 2009

Mech_Engineer

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.

3. Jun 11, 2009

Bob S

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.

4. Jun 11, 2009

Mech_Engineer

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...

5. Jun 11, 2009

chrisya

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?

6. Jun 11, 2009

nvn

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
7. Jun 11, 2009

chrisya

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
8. Jun 12, 2009

nvn

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