I am trying to calculate the horizontal force created by a raised hangar door, but for the purpose of this question, we can consider the horizontal force of a 2x4 stud that is connected as shown in the illustration below.

The 2x4 is cut in half and connected with a hinge at the top and a hinge at the middle, and then is supported vertically with a cable (which in the door will pull the door up and down). Not pictured is a track which the 2x4 or door will be connected to that keeps the "bottom" of the door in alignment with the opening.

What I am trying to determine is the method to find the lateral force that is created, or in other words, how much force will the door put on the track when it is open?

I have intentionally not included weights or lengths or angles because I'm looking for a general solution.

enter image description here

  • Compute the centroid of the door's members, with the weight of the door acting at the centroid, you can compute the torque exerted at the beam's hinge. – whatsisname Sep 29 '20 at 15:47
  • That's an excellent question, however, I have to ask: If the bottom of the door (where the blue arrow starts) is in a track (against the jack stud where the blue arrow points, presumably), won't that hold the bottom of the door in place, eliminating the concern? Or are you trying to determine how strong that track and its fasteners need to be? – FreeMan Sep 29 '20 at 15:47
  • @FreeMan Exactly--the track (and wheel) will have to be strong enough to withstand the horizontal force. I would like to use standard garage-door hardware, but I don't want to risk the metal or the wheel breaking and having the entire door come crashing down. – brentonstrine Sep 29 '20 at 15:50
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    Once I learn it then I will know it. Right? That's how knowledge and experience work. Or do you never learn anything while working on a project? – brentonstrine Sep 29 '20 at 16:27
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    @isherwood: the OP isn't asking for help with a specific problem, he's trying to learn static force analysis from a Q&A site, a subject which is often a semester course in college. If he wants to know the loads for a specific door, that would be one thing, but he explicitly asked for "general solution" which means: take the college course. – whatsisname Sep 29 '20 at 16:34

Assume: The two halves of the door are of equal length and weight. (L and W) Let Theta be the angle between vertical, and the upper portion, measured from the opening. Thus, when the door is closed, theta is 0, and when open all the way (impossible in real life) theta is 90

Constraint: The door will form an isosceles triangle at all times when closing. (two equal length sides)

The X component of the length will be L sin( theta). H will be 2L cos(Theta)

Let R be the reaction of interest.

enter image description here

Take the sum of the torques about P (hinge pin at top) The forces P and T disappear as the moment arm is zero.

Sum Torque = 0 -> (L/2 sin(theta) * 2 W)(clockwise+) - R 2L cos(theta)(counterclockwise, -ve)
            = LW sin(theta) - 2RL cos(theta)
            = W sin(theta) - 2R cos(theta)
            2R cos(theta) = W sin(theta)
            R = W tan(theta)/2

Consider the results - if the door is closed, theta is zero and tan(theta) is zero.
If the door is fully open, tan(90) -> infinity and the universe ends.

If the door is open 89 degrees, then the force at R is 57/2 W. At 85 degrees, it's 11/2 W.

So you'll need to restrict the door opening with a stop in the track somewhere to keep things within reason. The greatest forces will occur at the top of the track.

  • My basic assumption is that this is a bifold door mounted sideways. – Chris Cudmore Sep 29 '20 at 18:10
  • Does it really approach infinity? If the bottom half of the door is slightly longer it's possible to open the door greater than 90°. You could made the part that sticks out on the bottom extremely lightweight so as to not affect the equation too much. It wouldn't approach infinity in reality. I think it would only approach infinity as the length of the door approaches infinity. For a finite length, there would be some reasonable finite max, no? – brentonstrine Sep 29 '20 at 21:00
  • @brentonstrine: it doesn't approach infinity in real life because real materials bend and deform, and the above analysis treats the beams as absolutely rigid. Which is just fine for a situation like this. – whatsisname Sep 29 '20 at 21:29
  • I mean, what values does it approach? Would it get to billions of newtons if it was made out of solid diamond? That doesn't pass the gut test for me. I would think force would have to stay within an order of magnitude of the weight of the whole door. – brentonstrine Sep 29 '20 at 23:57
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    @brentonstrine: not if the members are perfectly rigid and infinitely thin. We have a system here rotating about the hinge pin. As the door approaches 90deg open, we have a lever of basically constant length, torquing against a lever of ever decreasing length. As torque = force * distance, as that distance decreases, the force goes up and up. In the real world this doesn't happen, as things deform and changes the math. Give me a lever long enough and a fulcrum on which to place it, and I shall move the world. – whatsisname Sep 30 '20 at 1:31

The force is ZERO because in your picture the 2x4 is not attached.

If you want a theoretical answer then have an accurate picture. Seeing the end product might help as there are more than just at rest forces.

  • This would be better as a comment – whatsisname Nov 11 '20 at 7:04
  • @whatsisname - it is the right answer right now. Will delete it when he posts appropriate pictures for the fantasy calculations. – DMoore Nov 11 '20 at 7:13
  • Do you mean that in the picture the whole contraption will swing in the direction of the arrow? I have an idea. Imagine that the blue arrow is actually a physical arrow made out of steel and painted blue. There, now the picture is accurate and the 2x4 will hit a physical object and stop. Further, imagine that the arrow has a scale on it. How many pounds (or newtons or whatever) will it read? – brentonstrine Nov 11 '20 at 15:29

First, simplify the problem by assuming:

  • The door is at a 90° angle (perfectly horizontal). This creates the maximum load.
  • The center of mass is also 90° out from the top hinge (in reality it will be a little lower).
  • The center of mass is halfway between the two hinges. (1/4 the length of the fully extended door)

We know that torque = force * distance. We can use weight instead of force because the only force is gravity acting upon the mass--a.k.a. weight. Distance is the distance to the center of mass, which is half of the total distance. And torque in this situation manifests as a force pushing horizontally.

If you choose pounds (weight), feet (length), and foot-pounds (torque force) as your units, everything works out nicely.

So with that, we can find the answer easily with this:

horizontal torque = weight * distance

This gives us the horizontal torque, a.k.a. how much the wheel bearing pushes on the track.

  • At a length of 1 foot to the center of mass (i.e. a 4 foot tall door that folds up in half to become 2 foot long, putting the center of mass 1 foot out), the horizontal torque is equal to the weight of the door in pounds.
  • At a length of 2 feet, the horizontal torque is double the weight.
  • At 3 feet, it's triple the weight.
  • And so on.

I find this strategy very intuitive. I used this torque calculator.

  • When you say (or half the total length), are you saying “L” is half the length of the header? – Lee Sam Sep 29 '20 at 17:45
  • Not the header. What matters is the length of what's hanging from the header--so measuring from that hinge on the header down to the first hinge in the door would be the total length. The center of mass would be somewhere in the middle, approximately half way. That's why half the total length--it's the measure to the center of mass. – brentonstrine Sep 29 '20 at 20:53
  • What do you mean by horizontal torque? – Chris Cudmore Sep 30 '20 at 13:52
  • By horizontal, I mean perpendicular to vertical. Another word for this is "level". By horizontal torque, I am referring to the force created by the torque of the door hanging on the hinge which is directed horizontally. – brentonstrine Sep 30 '20 at 23:01

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