Sheet-Metal CAD/CAM Unfolding, Press-Brake Flat Patterns & Precision Fabrication
Sheet-Metal Bend Allowance, K-Factor & Bend Deduction Formula
Calculate Neutral-Axis Arc Bend Allowance BA (mm), Outside Setback OSSB (mm), Bend Deduction BD = 2*OSSB - BA (mm), and Laser-Cut Flat Blank Length (mm).
Why adding two outside flange dimensions L1 + L2 cuts your laser blank 5.74 mm TOO LONG on a 90° bend!
During sheet-metal press-brake bending of thickness T (mm) to complementary bend angle theta (deg, the angle swept by the bent flange: 90° for a right-angle bracket) around inside bend radius R (mm), inner fibers compress while outer fibers stretch—shifting the zero-strain neutral axis inward to t_na = K * T from the inside surface (where K-Factor is typically ~0.33 for tight R < T air bends and ~0.40–0.44 for standard R >= T air bends)! Neutral-axis arc length is the Bend Allowance BA = (PI() / 180) * theta * (R + K * T). Because mechanical drawings dimension brackets to the virtual sharp apex via Outside Flange Lengths L_1 and L_2, each flange includes Outside Setback OSSB = TAN((theta / 2) * (PI() / 180)) * (R + T)—so the CAD flat-pattern laser programmer MUST subtract Bend Deduction BD = 2 * OSSB - BA to cut exact Flat Blank Length L_flat = L_1 + L_2 - BD!
Try this 3.0 mm Mild-Steel Right-Angle Bracket first—Thickness T = 3.00 mm (B2), Inside_Radius R = 4.00 mm (B3), Bend_Sweep_Angle theta = 90.0° (B4), K_Factor = 0.42 (B5), Outside_Flange_1 = 80.00 mm (B6), and Outside_Flange_2 = 60.00 mm (B7): Neutral-Axis Bend Allowance BA equals 8.262 mm (E2), Outside Setback OSSB equals 7.000 mm (E3), Bend Deduction BD hits 2 * 7.000 - 8.262 = 5.738 mm (E4), and Exact Laser Flat Blank Length equals 80 + 60 - 5.738 = 134.262 mm (E5)!
| Sheet-Metal Unfolding KPI (T=3.0 mm, R=4.0 mm, 90° Sweep, K=0.42, 80x60 mm) | Formula | Verified Output |
|---|---|---|
| Bend Allowance BA (mm) | (PI()/180) * Angle_deg * (R + K * T) | 8.262 mm (11.017 mm @ 120° sweep) |
| Outside Setback OSSB (mm) | TAN((Angle_deg/2)*PI()/180) * (R + T) | 7.000 mm (12.124 mm @ 120° sweep) |
| Bend Deduction BD (mm) | 2 * OSSB - BA | 5.738 mm (13.232 mm @ 120° sweep!) |
| Exact Flat Blank Length (mm) | Flange_L1 + Flange_L2 - BD | 134.262 mm (126.768 @ 120° sweep) |
DIN 6935 / ASME Sheet-Metal Bend Deduction & Flat Blank Formula
Validate Thickness T > 0, Inside_Radius R >= 0, Bend_Sweep_Angle in (0°, 170°), K_Factor in [0.20, 0.50], and Outside Flanges larger than OSSB!
=LET(th_k, B2, rd_i, B3, an_g, B4, kf_c, B5, fl_a, B6, fl_b, B7, IF(OR(th_k <= 0, rd_i < 0, an_g <= 0, an_g >= 170, kf_c < 0.2, kf_c > 0.5), "Invalid Bend Input", LET(ba_m, (PI() / 180) * an_g * (rd_i + kf_c * th_k), os_b, TAN((an_g / 2) * (PI() / 180)) * (rd_i + th_k), IF(OR(fl_a <= os_b, fl_b <= os_b), "FLANGE SHORTER THAN SETBACK", ROUND(fl_a + fl_b - (2 * os_b - ba_m), 3)))))120° Acute Included Bracket Sweep Check (Set Bend Sweep Angle = 120.0°)
Bend deeper to a 120.0° sweep angle (60° acute included corner, set B4 = 120): TAN(60°) drives Outside Setback E3 up to 12.124 mm, more than doubling Bend Deduction E4 from 5.738 mm to 13.232 mm and shortening Flat Blank Length E5 to 126.768 mm!
=IF(ISTEXT(E5), E5, LET(ba_m, (PI() / 180) * B4 * (B3 + B5 * B2), os_b, TAN((B4 / 2) * (PI() / 180)) * (B3 + B2), ROUND(2 * os_b - ba_m, 3)))Try the example
Use the original values or change them to test your understanding.