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I have an old bend/twisted wooden door. It has some character to it so I want to keep it and add square metal bars to straighten it. I have provided a sample drawing of the door and where it is twisted. I have drilled a hole in the upper most open part and if I pull with a scale I am able to straighten it with about 20kg (I haven't used a scale yet, but I will)

Twisted door, arrows show twists

I intend to put square metal bars on the bent sides (maybe even inside and out) and straighten the wood.

straightening bars

My question is what is the formula that I can use to calculate the size of the metal bar? For example I see there is a square profile bar 40x40x3mm that I will be able to buy and attach. So, is this going to be enough for the long side 2m and for the short side 1m? I can also order this square profile bar to be slightly bent and attach it so the two bends - the one of the wood and the one of the metal counter each other. But I am not really willing to go that way. I think its way too complicated.

Thank you in advance.

Edit:

  1. I think a visible rectangle bar on one or both faces will be stronger and more simple to attach, so I will go with that.
  2. No bend is acceptable - I want it straight.
  3. So the next question following the comment is how much of a bend do I need for what type of bar for each of the sides - short(1m) and long (2m)?
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  • $\begingroup$ If you don't go with the counter-bend, the reinforced door will still bend outwards, just less so. In which case you need to decide how much of a bend is acceptable and edit that into your question. Also, is your idea to attach this bar to one of the faces of the door (so that it's sticking out and visible on one side)? Or do you intend to actually get a U-shape and "wrap" the end of the door with the bar (which will probably require filing down the door a bit so that it still closes. $\endgroup$
    – Wasabi
    Commented Sep 26, 2018 at 19:35
  • $\begingroup$ Please check the edit I placed in the original question $\endgroup$
    – kdobrev
    Commented Sep 26, 2018 at 20:09

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If you estimate force needed to straighten the door applied at the corner is 20kg, it would be equivalent to the reaction of a uniformly loaded beam (your square bar) at one end.

So lets design the bar and for simplicity only the top bar, we can do the side later if we need to.

Lets say the door stays out of door jamb 2cm at the corner. So now we have all the data to design a bar or strap that has a deflection, built in bend, of 1/2 times 2cm =1cm and the 2 end reactions of 20kg, which for a 1m long strap mean omega is 40kg/m.

Deflection in a simply supported strap:

$$\delta = \frac {5 }{384 } \frac {\omega L^3}{EI} $$ This means $$0.01 = 5/384 \times40\times 1^3/(200GPa\times I)$$ $$ 2GPa\times I = 5/384\times 40 $$

This Gives the second moment, "I" of your strap and from there you can select the section of the bar that gives you optimal looks and required strength. Again this bar need to have a camber of 1cm built into it, so that after you attach it the door will be pressed into straight line.

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  • $\begingroup$ Thank you for this answer, although I did not understand it. I was trying to make sense reading en.m.wikipedia.org/wiki/Deflection_(engineering) but there I see your formula with L^4. Also I did not see the elasticity of my chosen bar 40x40x3mm. Also should I take into account to put just half of the force because I will be using two bars - vertical and horizontal. $\endgroup$
    – kdobrev
    Commented Sep 27, 2018 at 11:01
  • $\begingroup$ Thank you for this answer, although I did not understand it. I was trying to make sense reading en.m.wikipedia.org/wiki/Deflection_(engineering) but there I see your formula with L^4. Also I do not sundestand what is I. I suppose 2GPa is the elasticity of my chosen bar 40x40x3mm. Also should I take into account to put just half of the force because I will be using two bars - vertical and horizontal. $\endgroup$
    – kdobrev
    Commented Sep 27, 2018 at 11:12
  • $\begingroup$ @kdobrev no, you need to install a strap on top 1 meter long and calculate its dimension from the I. The power is 3 when you apply the total omega and 4 when you use omega/m.. if you install another one on the side vertically because of its length, it will contribute very little. $\endgroup$
    – kamran
    Commented Sep 27, 2018 at 15:05
  • $\begingroup$ I am honestly lost. I don't understand how only one strap will fix 2 curvatures. And since the I is some integral that I cannot compute I will provide a link with the possible materials I could use and please somebody pick one up for me, translate.google.com/… $\endgroup$
    – kdobrev
    Commented Sep 27, 2018 at 18:42
  • $\begingroup$ @Kamran is right. The Modulus of elasticity is going to be negligible for the size of angle iron chosen by virtue of its shape. Steel is incredibly strong so you really need to think about applied deformations to wood as "creep" due to moisture content. Think about adding a moisturiser to the door so that restoration is achieved faster with less damage. $\endgroup$
    – Rhodie
    Commented Jan 20, 2019 at 3:37

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