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I am designing a metal plate that will be laser-cut (or machine-cut) and then folded. I want to know how to size the pre-folded plate in order to get the right dimensions after folding.

Diagram of a folded 2mm aluminium plate with holes and window

My actual part is not exactly like this (I simplified it for ease of drawing) but it shows what I want to achieve. In this case it's a 2mm aluminium plate, the red arrows show the inner dimensions after folding that I want to specify and achieve. The holes must also line up and the window should be correctly placed.

Intuitively I would expect some compression along the inner part of the folds and stretching on the outer parts - ideally along the centre of the plate - but I don't know if this is what will happen.

Assuming the red arrows are 100mm each, should the plate be 300mm? I'm guessing not, so how do I calculate the radius of curvature that will be achieved and if I need to add (or remove) material at the folds in order to achieve my required dimensions?

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  • $\begingroup$ Some mechanical CAD systems will do this for you, asking questions about the material type and bending tool. Bend radius also matters - sharp folds in Al are not always recommended! $\endgroup$ – Brian Drummond Apr 23 '15 at 9:59
  • $\begingroup$ Alloy 5052 is a commonly available, relatively affordable aluminum that's decent for bending. If you try to put that bend in 6061 aluminum (the most common in America at least) it will probably crack before you get that bend made. $\endgroup$ – Ethan48 Apr 23 '15 at 13:15
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Your assumption is right! A 300mm long plate with two foldings won't do! This is because you need to take into account the bend allowance and the bend compensation!

But why is so?

Here is a diagram of what's going on: Bend allowance, bend deduction diagram


When you bend a material, part of it will extend (the external part of the bend), while another part will retract (the internal part).

The line (in the thickness of the plate) where the dimension doesn't change is called the Neutral line.

The neutral line is usually located between a third and a half of the material thickness (from the interior to the exterior of the bend). Meaning this line will hold its dimension, while the top surface (inner bend surface) will contract a little bit, and the bottom surface (outer bend surface) will expand a little bit.

Wikipedia has a nice little part on the calculation required, taking into account the angle of the bend, and the thickness of the material.

Here is the formula for Bend Allowance: $BA = A \left( \frac{\pi}{180} \right) \left( R + K \times T \right)$

Where, BA is the Bend Allowance, A is the bend angle in degrees, R is the inside bend radius, K is the ratio between the distance from the inside face to the Neutral line to the thickness of the material, usually $\frac{1}{3}$, and T is the thickness of the material.

And here is the formula for the Bend Deduction: $BD = 2 \left(R + T \right) \tan{ \frac{A}{2}} - BA$

Where, BD is the Bend Deduction, R is the inside bend radius, A is the bend angle in degrees, R is the inside bend radius, T is the thickness of the material and BA is the Bend Allowance.


In your case, you want to calculate the distance from the internal faces, not just for the straight part of the metal plate.

To do this, you just have to add the thickness of the material to your dimensions, i.e, your A length (from the picture above) is $(New Length) = (Old Length) + t * 2 $.

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  • $\begingroup$ Minor point: Every K-factor that I've seen (including on the wiki page linked) is $\leq 0.5$. So $2/3$ doesn't jibe with that. $\endgroup$ – Dan Apr 23 '15 at 13:55
  • $\begingroup$ I think the diagram is probably wrong/confusing - I added some comments on the Wikipedia talk page to this effect. $\endgroup$ – jhabbott Apr 23 '15 at 14:23
  • $\begingroup$ You're right! I confused myself! Correcting now! The diagram is effectively misleading or confusing in regards to the position of the neutral line. I think usually it's located more on the inside of the bend. $\endgroup$ – gromain Apr 23 '15 at 14:55
  • $\begingroup$ And I realise myself that my answer does not quite answer the original question, since it really needs more the bending compensation. I'll edit once again to fill in the details on this. $\endgroup$ – gromain Apr 23 '15 at 15:03
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Yes use Aluminum 5052 H32 for sheet metal parts with bends. r =or> T. I do it like this, get the lengths of the straight lines and set aside. The t/T is a little more than .50, say .53 until you get the real number. K= t/T, radius of the neutral line is inside r + t = r + .53T For a 90 degree bend, length of the bend is 2*pi*(r+t)/4 = pi*(r+t)/2 = pi*(r+.53T)/2, for any angle of bend, length of the bend is 2*pi*(r+t)*angle/360 length = staight lengths + length of bend, continue to add more bends if needed eg: angle 1-1/4in x 2-1/4in outside dimensions, 90deg bend, 1/8in thk aluminum 1/8in inside radius, remove t and r to get straights 1in and 2in add 2*pi*(.125+.53(.125))in/4 = 3in + .3in = 3.3in

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    $\begingroup$ This site can utilize mathjax. This helps to make formulae more readable. Please consider an edit to add this to make your answer easier to understand. $\endgroup$ – hazzey Aug 17 '16 at 1:10

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