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Sheet metal bending

Sheet bending force: what it depends on and how to estimate it, using a multi-sided pole section

The bending force has to be known in advance: if it exceeds the press capacity, the blank has to be split or bent another way. Below is what the force depends on, how it is estimated with one simple formula and two extended ones, and what comes out on a worked example of a plate for a multi-sided pole section. This is an estimate: for a real order we refine the force and the radius for the material and the tooling.

What bending force is and why it matters

In V-bending the plate lies on two edges of the die and the punch presses in the middle. The press has to bend the plate along the whole bend line in one stroke, so the calculated force is compared with the rated force of the press. Ours is 2000 t, with a bend length of 12 000 mm, and plate up to 25 mm thick and up to 12 m long (bending on the 2000 t press).

A multi-sided pole section is a typical case. The trapezoidal blank is bent along its whole length, and each face is a separate bend line as long as the section, up to 12 m. The length in the calculation is the longest possible, so the force is considerable.

What the force depends on

Several quantities appear in the calculation. Some work in every formula, others only in the extended ones. The table shows what each one means and how it affects the result.

QuantityEffectUsed in
Tensile strength σв (or yield strength σт) of the steelThe force grows in direct proportion: stronger steel needs a bigger forceσв in the simple formula, σт in the extended ones
Plate thickness sThe strongest factor: it enters squared, and a correction that depends on the ratio of s to A raises the force furtherall formulas
Bend line length LDirectly proportional: a line twice as long needs twice the forceall formulas
Die opening (V-width) AThe wider it is, the smaller the force; but a wider opening increases springbackall formulas
Die edge radius R and punch tip radius rThey change the lever arm on which the force actsextended formulas
Friction coefficient f of the plate on the dieFriction adds resistance as the plate slides over the die edgesextended formula no. 2
Plate angle αThe force changes during bending: it first rises, then fallsextended formulas
Reserve factor k1 and strain hardening factor k2They raise the result: k2 accounts for the metal hardening as it deformsextended formulas

The simple estimate

The simplest formula links the force to four quantities. Written out in text it reads:

F = (1 + 4·s/A) · σв · L · s² / A / 1000

Here F is the force in tonnes-force, σв is the tensile strength in kgf/mm², and s, A and L are the plate thickness, the die opening and the bend line length in millimetres. The bracketed factor is a thickness correction: the larger s is relative to A, the more it counts.

Take nominal values: σв = 42 kgf/mm² (about 412 MPa), s = 10 mm, A = 120 mm, L = 1000 mm. No steel grade is specified, so this is only a nominal plate. The result is F = (1 + 40/120) · 42 · 1000 · 100 / 120 / 1000 = 46.7 t, that is 46.7 t for each metre of bend line.

The extended estimate: two calculations with reserve

There are two more extended calculations, which take the tooling geometry and friction into account. Both use the plastic moment of the plate Mp = 1.5 · σт · L · s² / 6: the factor 1.5 converts the elastic section modulus of a rectangular section into the plastic one. After that the formulas differ.

Calculation no. 1 is based on work: the force times a small punch displacement Δh equals the work of the moment over a small rotation Δα. Δα is taken as 0.001°, converted to radians in the formula. F = k1 · k2 · 2 · Mp · Δα / (Δh · 9807), where Δh = (r + R + s) · (1/cos(α − Δα) − 1/cos α) + (A/2) · (tan α − tan(α − Δα)).

Calculation no. 2 is based on moment equilibrium with friction: F = k1 · k2 · 2 · Mp · (cos α + f · sin α) / ( ((A/2 − (r + R + s) · sin α) / cos α + f · s) · 9807 ). In both formulas the number 9807 converts newtons to tonnes-force, σт is in MPa, dimensions are in mm, and α is the plate angle measured from the initial flat position.

Input data of the example: σт = 335 MPa, L = 1000 mm, s = 10 mm, f = 0.15, A = 120 mm, R = 5 mm, r = 10 mm, k1 = 1.25 (reserve), k2 = 1.8 (hardening). The table shows the result for several angles.

Angle α, °F, calculation no. 1, tF, calculation no. 2, t
164.563.1
566.065.1
1067.067.0
1567.067.8
2066.067.7
3060.764.2
4545.450.9

How to read the difference between the estimates

The two extended estimates are close: 63 to 68 t over angles from 1° to 20°, with a maximum near 15°, after which the force falls. But they include the factor k1·k2 = 2.25. Without it the same formulas would give about 30 t. The simple formula gives 46.7 t for the same plate, which lies between these values.

Three estimates for one plate differ by almost a factor of two. This shows how approximate such a calculation is and why its figure cannot be taken as a rated value. The reserve and the hardening factor in this method are a way of not underestimating the load, not an exact physical value.

Worked example: a plate for a pole section

Take the same nominal steel (σв = 42 kgf/mm²) and the simple formula. A 12 m section means a bend line L = 12 000 mm. For a 10 mm plate with a die opening of 120 mm: F = (1 + 40/120) · 42 · 12 000 · 100 / 120 / 1000 = 560 t. The same by the metre: 46.7 t/m × 12 m = 560 t. Both routes gave the same result.

The table shows how the force changes when the thickness or the die opening changes. The opening widths here are only an illustration, not a list of our tooling: which die to use is decided when the order is prepared.

Plate thickness, mmDie opening, mmForce per metre, t/mForce over 12 m, t
812028.4340.5
1012046.7560.0
1212070.6846.7
108078.8945.0
1016032.8393.8
25200196.92362.5
25250147.01764.0
25300116.71400.0

What these numbers show

Raising the thickness from 10 to 12 mm (by 20 %) raises the force by 51 %. Widening the die opening from 120 to 160 mm lowers it by about 30 %. That is why a wider opening is chosen for thick plates, at the price of more springback.

As a guide: 2000 t over 12 m is about 167 t per metre of length, if the load is spread evenly. A 25 mm plate at a 200 mm opening needs 197 t/m by this estimate, which is more, while at 250 mm it needs 147 t/m, which fits within the guide. Whether a plate will actually go is decided from the press data sheet and the material, not from this table.

What this calculation does not give

It does not say by what angle to bend the plate past the target so that the right angle remains after springback: that depends on the steel strength, the radius-to-thickness ratio and the die opening. It does not give the minimum radius below which cracks appear: that radius grows with strength and thickness and is larger when the bend line runs along the rolling direction. Both questions are covered in the article on bending plate on the 2000 t press.

The calculation takes a nominal steel and an ideal plate. In reality the tensile and yield strength come from the certificate for the batch, and the state of the edges in the bend zone, with burrs removed and the hardened zone after thermal cutting ground off, also affects the result.

What to send for a refined estimate

For us to calculate the force and select tooling, the following is enough:

  • steel grade and plate thickness;
  • bend line length and blank dimensions;
  • bend angle and inside radius, or «the smallest possible»;
  • number of parts and tolerances.

If there is no drawing yet, send a sketch: KM and KMD design can bring it to a production drawing, and we do the bending, cutting and welding as part of steel fabrication. The bending service itself is described on the page sheet metal bending.

Standards and validity

The formulas in this article are an engineering estimate, not a requirement of the standards: we name no standard that the formulas themselves could be referred to. Plate properties are taken from the certificate for the batch and the standard to which the plate was supplied. Requirements for the execution of steel structures, including forming of parts, are in DSTU EN 1090-2:2019 «Execution of steel and aluminium structures. Part 2. Technical requirements for steel structures» (in force since 01.07.2020); structural calculation is under DBN V.2.6-198:2014 «Steel structures. Design standards» (in force; edition with Amendment No. 1). Current editions are checked in the design.

Updated:

Frequently asked questions about sheet bending force

How many tonnes are needed to bend a 10 mm plate 12 m long?

By the simple estimate, for a nominal steel with a tensile strength of 42 kgf/mm² and a 120 mm die opening, about 560 t. This is an illustration: for your plate the force depends on the steel grade and the opening width.

Is 2000 t enough for a 25 mm plate?

It depends on the die opening and the steel strength. By the same estimate over 12 m, an opening of 250 mm gave 1764 t and 200 mm gave 2362.5 t, which is more than the rating. For your plate we determine the opening and whether it can be bent when the order is prepared.

Why does thickness count squared?

Because the bending resistance of a plate grows as the square of its thickness. All else equal, a plate twice as thick needs more than four times the force: a correction that depends on the ratio of thickness to opening width is added to the square.

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