---
title: "Sheet bending allowance, force, springback"
subject: "Manufacturing and Production Processes"
chapter: 6
key: "semester_4/manufacturing_and_production_processes:bending-allowance-force-springback-numerical"
source: https://www.bctengineer.com/q/semester_4/manufacturing_and_production_processes/bending-allowance-force-springback-numerical
site: BCT Engineering
---

# Sheet bending allowance, force, springback

Manufacturing and Production Processes · Chapter 6: Sheet Metal Product Manufacturing Process

- Asked in: year not given, 3+2+3 marks
- Video: [sheet Metal 04 Bend Allowance Spring Back](https://www.youtube.com/watch?v=nPbU9FhvHdQ) — Sunil Mohod
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/manufacturing_and_production_processes/bending-allowance-force-springback-numerical
- Whole chapter: https://www.bctengineer.com/md/semester_4/manufacturing_and_production_processes/chapter-6

## Question

(a) A 3 mm thick L-shaped bracket is bent through 90 degree with an inside bend radius of 6 mm. The outside lengths of the two flanges are 60 mm and 80 mm. Find the flat blank length, taking the bend factor $K_{ba} = 0.33$. (b) Find the force for V-bending a 100 mm long, 3 mm thick strip of tensile strength 450 MPa, with a die opening of 16 times the thickness and $K_{bf} = 1.33$. (c) Explain springback and calculate the springback ratio and the angle to which a 2 mm sheet of yield strength 300 MPa and E = 200 GPa must be bent to give a final bend of 90 degree with an inside radius of 25 mm.

## Short answer

Blank length = 51 + 71 + (π/2)(6+0.33×3) = 133.0 mm. Bending force = 1.33×450×100×3²/48 = 11.2 kN. Springback is elastic recovery; ratio Ks = 0.946, thus overbend to 95.2° to obtain final 90° bend.

## Answer

#### (a) Bend allowance and blank length

The bend allowance is the length of the neutral axis within the bend:

$$
BA = \alpha\,(R + K_{ba}\,t) = \frac{\pi}{2}(6 + 0.33\times 3) = 1.5708\times 6.99 = 10.98\ \text{mm}
$$

Straight lengths (outside dimension minus radius and thickness for each flange):

$$
L_1 = 60 - (6+3) = 51\ \text{mm},\qquad L_2 = 80 - (6+3) = 71\ \text{mm}
$$

$$
L_{blank} = 51 + 71 + 10.98 = 133.0\ \text{mm}
$$

#### (b) Bending force

For V-die bending:

$$
F = \frac{K_{bf}\,(UTS)\,w\,t^2}{D}
$$

where $D = 16\times3 = 48$ mm is the die opening:

$$
F = \frac{1.33\times 450\times 100\times 3^2}{48} = 11\,222\ \text{N} \approx 11.2\ \text{kN}
$$

#### (c) Springback

**Springback** is the partial elastic recovery of a bent sheet when the load is removed. The bend radius and the included angle both increase. It rises with yield strength and bend radius, and falls with modulus and thickness. It is compensated by overbending, bottoming, or stretching the sheet during the bend.

Springback ratio from the standard relation:

$$
\frac{R_i}{R_f} = 4\left(\frac{R_i Y}{E t}\right)^3 - 3\left(\frac{R_i Y}{E t}\right) + 1
$$

$$
\frac{R_i Y}{E t} = \frac{25\times300}{200\,000\times 2} = 0.01875
$$

$$
\frac{R_i}{R_f} = 4(0.01875)^3 - 3(0.01875) + 1 = 0.9438 \;\Rightarrow\; R_f = 26.49\ \text{mm}
$$

The bend angle ratio follows $\dfrac{\alpha'_f}{\alpha'_i} = \dfrac{2R_i/t + 1}{2R_f/t+1}$:

$$
K_s = \frac{2(25)/2 + 1}{2(26.49)/2+1} = \frac{26}{27.49} = 0.9458
$$

Angle to which the sheet must be bent so that it relaxes to 90 degree:

$$
\alpha_i = \frac{90}{0.9458} = 95.2^\circ
$$

**Answer:** blank length $= 133.0$ mm; bending force $= 11.2$ kN; springback ratio $K_s = 0.946$, overbend to $95.2^\circ$ (springback about $5.2^\circ$).

## Check yourself

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

1. What does the bend allowance BA = alpha(R + K_ba t) represent?
   A. The length of the inside surface of the bend
   B. The length of the outside surface of the bend
   C. The length of the neutral axis within the bend
   D. The extra material added to both flanges for clamping
   Answer: C. Bend allowance is the arc length of the neutral axis in the bend, located at R + K t from the centre. (Key point: Bend allowance)

2. For the bracket with outside flange lengths 60 mm and 80 mm, R = 6 mm, t = 3 mm, how is each straight length found?
   A. Outside dimension minus R only
   B. Outside dimension minus t only
   C. Outside dimension plus (R + t)
   D. Outside dimension minus (R + t)
   Answer: D. The outside dimension includes the bend radius and thickness, so R + t is subtracted: 60 - 9 = 51 mm and 80 - 9 = 71 mm. (Key point: Straight flange length)

3. What is the flat blank length for the bracket?
   A. About 133 mm (51 + 71 + 10.98)
   B. About 150 mm (60 + 80 + 10.98)
   C. About 122 mm (51 + 71)
   D. About 144 mm (60 + 80 + 3 + 1)
   Answer: A. Blank length = two straight lengths plus the bend allowance = 51 + 71 + 10.98 = 133.0 mm. (Key point: Blank length)

4. Which expression gives V-die bending force, and what is D here?
   A. F = K_bf (UTS) w t / D, with D = 48 mm
   B. F = K_bf (UTS) w t^2 / D, with D = 16 x 3 = 48 mm
   C. F = K_bf (UTS) w t^2 / D, with D = 3 mm
   D. F = K_bf (UTS) w^2 t / D, with D = 16 mm
   Answer: B. Force rises with the square of thickness and falls with die opening; D = 16 t = 48 mm giving about 11.2 kN. (Key point: V-bending force)

## Related in Manufacturing and Production Processes

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## Related in other subjects

- [Bend allowance and blank length](https://www.bctengineer.com/question.md?key=semester_2%2Fworkshop_technology%3Abend-allowance-blank-length) — Workshop Technology, chapter 8
- [I-section: bending stress 80 mm above base](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Ai-section-bending-stress-80mm-above) — Strength of Materials, chapter 7
- [Bending stress diagram; moment in flanges](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Al-section-bending-stress-diagram-flange) — Strength of Materials, chapter 7