Press Brake Bend Deduction: 120 Brackets Coming Out Oversize and Overbent
A fabricator is producing 120 identical brackets from 3 mm S275JR mild steel sheet, 90° bends with outside flange dimensions of 50 mm and 50 mm, an internal bend radius of 6 mm and a K-factor of 0.42 for the R/T ratio in use. Blanks are sheared to length and formed on a 100 tonne press brake using a 24 mm V-die; the back gauge is set from the outside face of the tooling. The first 20 parts measure 53.2 mm on each outside flange instead of 50 mm, and the included angle comes out at 92° rather than 90°. On checking, the programmer simply set the blank length at 100 mm (50 + 50) because the drawing shows flange dimensions. Some parts also show a slight flange height variation of about 0.4 mm from one end of the bend to the other. Delivery is in three days and 80 blanks are already cut to 100 mm.
1. Calculate the correct flat blank length for these brackets, showing the bend allowance and bend deduction step by step, and explain why the 100 mm blank gave 53.2 mm flanges.
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Step 1 - Bend allowance (arc length of the neutral axis). BA = (π/180) × A × (R + K × t), with A = 90°. R + K·t = 6 + (0.42 × 3) = 6 + 1.26 = 7.26 mm. BA = 1.5708 × 7.26 = 11.40 mm.
Step 2 - Outside setback. For 90°, SB = (R + t) × tan(A/2) = (6 + 3) × 1 = 9.00 mm per side.
Step 3 - Bend deduction. BD = 2 × SB − BA = 18.00 − 11.40 = 6.60 mm.
Step 4 - Flat length. Flat = (outside flange 1) + (outside flange 2) − BD = 50 + 50 − 6.60 = 93.4 mm.
Why 100 mm gave 53.2 mm flanges. Cutting at the sum of the outside dimensions adds back the whole bend deduction twice over per part. Formed outside flange = (100 + 6.60) / 2 = 53.3 mm, which matches the 53.2 mm measured (the 0.1 mm difference is normal thickness and radius tolerance).
Practical consequences. The 80 blanks already cut at 100 mm are 6.6 mm too long. They cannot simply be re-sheared to 93.4 mm after forming because the bend line is already in the wrong place - the flanges would come out unequal. The realistic options are to shear the unbent blanks down to 93.4 mm before forming (workable if they have not been bent yet), or form them and use them on a non-critical job with the customer's written agreement. Get written sign-off; do not ship oversize parts against a drawing dimension.
2. The parts are springing back to 92°. Explain the cause and state what you would change, including whether the correction affects the bend deduction you just calculated.
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Cause. Springback is elastic recovery when the ram is released. The outer fibres have been stretched past yield but the inner fibres retain stored elastic energy, so the part opens up after unloading. On 3 mm S275 with R/t = 2, about 2° recovery on a 90° bend is typical and entirely predictable.
Correction. Set the ram to bottom (or coin) at 88° so that after recovery the part settles at 90°. Verify by trial on a scrap piece, measuring the angle with a digital protractor across the outside faces, and record the corrected ram depth on the set-up sheet so the next operator does not repeat the error. Do not chase the angle by adjusting the back gauge - that changes the flange dimensions and masks the problem.
Effect on the bend deduction. Strictly, reducing the bend angle changes both the arc length and the setback, so the theoretical bend deduction for 88° is marginally different (a little over 0.1 mm smaller). At this part size the difference is well inside the ±0.5 mm general tolerance and is swamped by material thickness variation, so the calculated 93.4 mm blank stands. What matters is that you recognise the effect exists and that the operator confirms the first-off part dimensionally before running the batch - the flange should be checked at the outside face, with the back gauge reference consistent between setter and inspector.
Documentation. Note the overbend value, tooling, die opening and the 93.4 mm blank on the job sheet so the next run starts from the proven set-up rather than from 100 mm again.
3. The foreman wants to run the remaining 40 blanks at 2.5 m length on the 100 tonne brake. Using a rule-of-thumb tonnage estimate of about 650 × t² / V kN per metre of bend for mild steel, check the machine capacity, and list the other reasons the flange height might vary by 0.4 mm along a bend.
Read the worked answer
Tonnage check. P ≈ 650 × t² / V = 650 × 3² / 24 = 650 × 9 / 24 = 244 kN per metre (about 25 tonnes per metre). Over a 2.5 m bend: 244 × 2.5 = 610 kN, roughly 62 tonnes. Against a 100 tonne machine this is within capacity - but only just over 60% utilisation, and the rule of thumb must be confirmed against the machine's own tonnage chart for the actual material tensile strength and die. Two further checks are needed: the punch tip load limit (narrow punches are rated per metre and can fail well below machine capacity), and whether the machine is rated for full-tonnage bending over the full 2.5 m bed length, not just at the centre. Note that if someone fits a 16 mm V-die to get a tighter radius, tonnage rises to about 366 kN/m, or roughly 91 tonnes over 2.5 m - close enough to the machine limit to be a real risk.
Causes of the 0.4 mm variation along the bend.
- Sheet thickness variation across the blank (hot-rolled mild steel can be 0.2–0.3 mm over nominal in patches), which changes the bend deduction locally.
- The blank not being square, or the bend line not parallel to the back gauge, so one end is formed against a different gauge reference.
- Die wear, a nicked die shoulder or swarf sitting in the V.
- The ram not being level or the machine's crowning not set for a 2.5 m length, giving less penetration at the ends.
- Grain direction: bending across the rolling direction behaves differently from bending along it.
- Operator holding the part rather than letting it follow the die, tilting it slightly.