# Force on the top hinge of an open door

1. May 25, 2012

### LeeH

Please can one of you bright people help me, I am not very good at maths and would appreciate a line by line answer so I can copy it onto a spread sheet and use it in the future.

What I am trying to work out is what force will be excerted on a partition when a door is open at 90 degrees. If by calculating what the force on the top hinge is gives me this then that would be great, it doesnt have to be too accurate, just so I can get some rough idea of what the force will be.

Door leaf: 1.2M wide x 2.4M high = total 2.88M2
Weight: 20kg / M2, so 57.6kg in total

I have been trying to upload a photo of a typical door, but it keeps saing its too big so will try to add something after this first post.

Many thanks

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2. May 25, 2012

### Pkruse

How many hinges does the door have?

If more than two, then you have a statically indeterminate problem, which is much more difficult to solve. Then you would never be completely sure of the answer because of uncertainties in your input data.

If I were designing the door, I’d make a simplifying assumption that would be a bit conservative—but reasonable in the worst case not knowing how the hinges were installed and accounting for wear over time. I’d model the hinges as a pin connection on the top, and a sliding connection on the bottom. The vertical component of the load would be simply the weight of the door. To find the horizontal component, I’d sum the moments about the top pin and solve.

Then I’d be concerned about dynamic loads if someone slams the door or hangs something from it. So I’d put a safety factor of maybe three onto the result, and select hinges from a catalog of the next larger size.

3. May 25, 2012

### LeeH

Thanks for the response, to try to keep this simple, if the hinge was a continuous one then what be the force at the top acting on the partition, it's just like a sign sticking out of the wall then.

4. May 26, 2012

### haruspex

Treating the hinge as uniform, straight and elastic, the door will hang at some small angle to the vertical. Let the tangent of this be T. The horizontal force per unit length at height X will be proportional to T.X. The moment of this about the bottom of the hinge is K.T.X2. The total moment integrates this to get K.T.H3/3, where H is the height of the door.
If the door has width 2W, weight Mg, we have:
W.Mg = K.T.H3/3.
K.T = 3W.Mg/H3
The horizontal force per unit length near the top of the door is K.T.H = 3W.Mg/H2
There will also be a vertical force to balance the weight of the door. This will be a shear force on the hinge.

It's rather simpler with just one hinge at the top and one at the bottom. The horizontal force at the top is then W.Mg/H. The vertical force is hard to determine because it will depend on the precise positioning of the fittings. If the screw holes are slightly too close together on the doorway, most of the load will be on the lower hinge. If too far apart then on the upper.