Tools › Mechanical Design › Machining and Manufacturing › Sheet Metal Bend Allowance and Press Tonnage › Example

16-gauge steel channel with a return lip: setback, bend allowance, flat pattern, springback and press-brake tonnage by the FAA handbook method

Tool	Sheet Metal Bend Allowance and Press Tonnage
Source	Federal Aviation Administration (2023). Aviation Maintenance Technician Handbook: Airframe. FAA-H-8083-31B, chapter 4, pp. 4-58 to 4-73 (setback SB = K(R + T), bend allowance formula 2, Figure 4-128, relief holes); Federal Aviation Administration (1998). Acceptable Methods, Techniques, and Practices: Aircraft Inspection and Repair. AC 43.13-1B, par. 4-55 / 4-56; Gardiner, F. J. (1957). "The Spring Back of Metals." Transactions of the ASME, Vol. 79, pp. 1–9.
Level	Undergraduate

Problem

A 0.0598-in (16-gauge) low-carbon steel sheet (S_u 60 ksi, S_y 40 ksi, E 29 Msi) is bent into a channel with a return lip: flanges 1.0 / 2.0 / 1.0 / 0.5 in measured to the outside mold lines, three 90° bends of 0.06-in inside radius, 24 in long, on a press brake with a V-die of 8 t. Find the setback, bend allowance (FAA empirical formula) and bend deduction of each bend, the flats and the total developed width, the springback factor and die angle, the air-bend force per unit length and total (k = 1.33), the required capacity against a 50-kip press with a 1.2 factor, the minimum flange, the air-bend radius, the hole and relief-hole distances and the force to blank a 10-in perimeter at 0.75 S_u.

Inputs to Enter

t 0.0598 in, S_u 60 ksi, S_y 40 ksi, E 29 Msi, K-factor by the FAA empirical formula, outside mold-line dimensions, minimum radius 1 t (entered)
Flanges 1.0 / 2.0 / 1.0 / 0.5 in, bends 90° / 90° / 90° at R 0.06 in; bend length 24 in, V = 8 t, k 1.33, bottoming ×3, coining ×8, air radius 0.15 V, capacity factor 1.2, press 50 kip
Hole rule 2 t, slot rule 2.5 t, relief width 1 t; cut length 10 in, shear strength blank (0.75 S_u), stripping 0.10

Open the Sheet Metal Bend Allowance and Press Tonnage tool (loaded with this example)

Answers

Setback SB = tan 45° (0.06 + 0.0598) = 0.1198 in; BA = (0.01743 × 0.06 + 0.0078 × 0.0598) × 90 = 0.1361 in (K = 0.4475); BD = 2 × 0.1198 − 0.1361 = 0.1035 in
Flats 0.8802 / 1.7604 / 0.7604 / 0.3802 in; TDW = 3.7812 + 3 × 0.1361 = 4.1895 in (sum of flange dimensions 4.5 in less 3 × 0.1035)
Springback: x = 0.06 × 40 / (29,000 × 0.0598) = 0.001384, R_i/R_f = 0.99585, R_f = 0.06025 in, K_s = 0.99723, angle after springback 89.75°, die angle 90.25°
V = 0.4784 in; F/L = 1.33 × 60,000 × 0.0598² / 0.4784 = 596.5 lb/in (plastic hinge 448.5, elastic limit 199.3); F = 14.32 kip = 7.16 US tons; required 1.2 F = 17.18 kip, M.S. +1.91 against 50 kip; bottoming 42.95 kip, coining 114.5 kip
Minimum flange 0.4784 / (2 cos 45°) = 0.3383 in (every flange longer); air-bend radius 0.15 V = 0.0718 in; hole edge 2 t = 0.1196 in from the tangent line; relief hole 1/8 in (minimum governs); blanking 10 × 0.0598 × 45 ksi = 26.91 kip, with stripping and the factor 35.52 kip, M.S. +0.41

Notes

The tool's default inputs are this part; press Submit. The part and material are illustrative (stated on the page); the method is checked by test_sheet_metal_bending.php (212 checks) against the handbook's K chart (180 entries), its bend-allowance chart (229 cells), its U-channel, L-bracket, 45° and 135° examples, the AC 43.13-1B setback examples and Table 4-6, and the numpy script in References/Sheet-Metal/checks/ (fibre-integration springback, explicit part geometry, plastic hinge).