---
title: "Bend allowance and blank length"
subject: "Workshop Technology"
chapter: 8
key: "semester_2/workshop_technology:bend-allowance-blank-length"
source: https://www.bctengineer.com/q/semester_2/workshop_technology/bend-allowance-blank-length
site: BCT Engineering
---

# Bend allowance and blank length

Workshop Technology · Chapter 8: Sheet Metal Works

- Asked in: year not given, 6 marks
- Video: [Understanding Bend Allowance and Bend Deduction | why it is important and how to calculate it](https://www.youtube.com/watch?v=LYzXBklYe_A) — Mechanical Design Adda
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_2/workshop_technology/bend-allowance-blank-length
- Whole chapter: https://www.bctengineer.com/md/semester_2/workshop_technology/chapter-8

## Question

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.

## 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.

## Answer

**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.

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

where $\theta$ = 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, $\theta$ = 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
$$
\begin{aligned} BA &= \frac{\pi}{180}\times 90\times(5 + 0.5\times 2) \\ &= \frac{\pi}{2}\times 6 = 9.425\ \text{mm} \end{aligned}
$$

#### Developed length
$$
\begin{aligned} L &= 60 + 30 + 30 + 2\times BA \\ &= 120 + 18.85 = 138.85\ \text{mm} \end{aligned}
$$

| Segment | Length (mm) |
|---|---|
| Base flat | 60 |
| Two leg flats | 60 |
| Two bends | 18.85 |
| Total | 138.85 |

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

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. What does bend allowance represent?
   A. The length of the neutral axis within the bent region, added to the flats to get the blank length
   B. The extra material removed from the inside of the bend
   C. The length of the inside surface of the bend
   D. The springback angle recovered after bending
   Answer: A. Bend allowance is the neutral-axis arc length in the bend; it is added to the flat lengths to get the developed length. (Key point: Meaning of bend allowance)

2. Which expression gives the bend allowance for a bend angle θ (degrees)?
   A. (π/180) θ R
   B. (π/180) θ (R + K t)
   C. (π/180) θ (R + t)
   D. θ (R + K t)
   Answer: B. The neutral axis lies K t from the inside surface, so its radius is R + K t. (Key point: Bend allowance formula)

3. For R = 5 mm, t = 2 mm, K = 0.5, θ = 90°, what is the bend allowance of one bend?
   A. 7.85 mm
   B. 11.0 mm
   C. 9.43 mm
   D. 18.85 mm
   Answer: C. BA = (π/2)(5 + 1) = 9.425 mm. 7.85 ignores the K t term; 18.85 is for two bends. (Key point: Bend allowance value)

4. What is the blank length of the channel (flats 60, 30, 30 mm and two bends)?
   A. 120 mm
   B. 138.85 mm
   C. 129.43 mm
   D. 157.7 mm
   Answer: B. L = 60 + 30 + 30 + 2 x 9.425 = 138.85 mm; both bends must be added. (Key point: Developed length)

## Related in Workshop Technology

- [Sheet metal cutting and forming operations](https://www.bctengineer.com/question.md?key=semester_2%2Fworkshop_technology%3Asheet-metal-operations-explained) — Workshop Technology, chapter 8
- [Turning time and metal removal rate](https://www.bctengineer.com/question.md?key=semester_2%2Fworkshop_technology%3Alathe-turning-time-mrr-numerical) — Workshop Technology, chapter 6
- [Drilling time and metal removal rate](https://www.bctengineer.com/question.md?key=semester_2%2Fworkshop_technology%3Adrilling-time-numerical) — Workshop Technology, chapter 5

## Related in other subjects

- [Sheet bending allowance, force, springback](https://www.bctengineer.com/question.md?key=semester_4%2Fmanufacturing_and_production_processes%3Abending-allowance-force-springback-numerical) — Manufacturing and Production Processes, chapter 6
- [Splice of 18 mm flats with M18 10.9 bolts](https://www.bctengineer.com/question.md?key=semester_6%2Fdesign_of_steel_and_timber_structure%3Atension-splice-m18-hsfg-109) — Design of Steel and Timber Structure, chapter 5
- [Siphon with 90 mm pipe, number and length](https://www.bctengineer.com/question.md?key=semester_4%2Fhydraulics%3Asiphon-90mm-pipe-total-length) — Hydraulics, chapter 2
- [Deep drawing blank size, hydroforming, spinning](https://www.bctengineer.com/question.md?key=semester_4%2Fmanufacturing_and_production_processes%3Adeep-drawing-hydroforming-spinning-formability) — Manufacturing and Production Processes, chapter 6