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Bend allowance and blank length

Short answer

Bend allowance is the neutral axis length in the bend region, BA = (π/180)θ(R+Kt). For t=2, R=5, θ=90°, K=0.5, BA = (π/2)×(5+0.5×2) = 9.43 mm. Developed length = sum of flats (60+30+30) + 2×9.43 = 138.86 mm.

Define bend allowance. A sheet metal channel is made from 2 mm thick sheet with a flat base of 60 mm, two vertical flats (legs) of 30 mm each and two 90° bends with inside bend radius 5 mm. Find the developed (blank) length. Take the neutral axis factor K = 0.5 for the bend radius to thickness ratio used here. The given flat lengths are the straight portions between bend tangent points.

Bend allowance (BA) is the length of the neutral axis within the bent region. It is added to the flat lengths to get the developed length (blank length) of the sheet.

BA=π180 θ (R+K t)BA = \frac{\pi}{180}\,\theta\,(R + K\,t)

where θ\theta = bend angle in degrees, R = inside bend radius, t = thickness, K = location of the neutral axis from the inside surface (0.33 for R < 2t, 0.5 for R > 2t).

Data

t = 2 mm, R = 5 mm, θ\theta = 90°, K = 0.5, flats 60 mm, 30 mm and 30 mm.

R/t = 2.5 > 2, so K = 0.5 is acceptable.

Bend allowance for one bend

BA=π180×90×(5+0.5×2)=π2×6=9.425 mm\begin{aligned} BA &= \frac{\pi}{180}\times 90\times(5 + 0.5\times 2) \\ &= \frac{\pi}{2}\times 6 = 9.425\ \text{mm} \end{aligned}

Developed length

L=60+30+30+2×BA=120+18.85=138.85 mm\begin{aligned} L &= 60 + 30 + 30 + 2\times BA \\ &= 120 + 18.85 = 138.85\ \text{mm} \end{aligned}
SegmentLength (mm)
Base flat60
Two leg flats60
Two bends18.85
Total138.85

Answer: bend allowance = 9.43 mm per bend; blank length = 138.85 mm (about 139 mm).

✓ Answer checked by Claude Opus 5.5 (high effort)

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