Sheet bending allowance, force, springback

(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 $5.2^\circ$).