Blanking force, clearance and die sizes

(a) Blanking and punching

Effect of clearance: the clearance is the gap between punch and die, given per side as a percentage of the sheet thickness (5-10 percent for steel). With the correct clearance, the cracks from the punch and die edges meet and give a clean edge, with a burnished zone about one third of the thickness and a small burr. Too small a clearance gives secondary shearing, a higher force and rapid tool wear. Too large a clearance gives a large rollover and burr, and a rough, tilted edge.

(b) Numerical

Blanking force

The shear area is the cut perimeter times the thickness:

$$F = \pi D\, t\, \tau = \pi\times 40\times 2\times 300 = 75\,398\ \text{N} = 75.4\ \text{kN}$$

Clearance and die sizes

$$c = 0.06\times 2 = 0.12\ \text{mm per side}$$

For blanking, the die decides the blank size:

$$D_{die} = 40.00\ \text{mm},\qquad D_{punch} = 40 - 2c = 40 - 0.24 = 39.76\ \text{mm}$$

Stripping force

$$F_s = 0.10\times 75.4 = 7.54\ \text{kN}$$

Total force the press must supply: $75.4 + 7.54 = 82.9$ kN, so a press of 100 kN (about 10 tonne) is selected.

Work done

$$W = F\times (p\,t) = 75\,398\times (0.4\times 2) = 60\,318\ \text{N\,mm} = 60.3\ \text{J}$$

The force can be reduced by grinding shear (about equal to the sheet thickness) on the cutting face, which spreads the cut over the stroke. In blanking the shear is ground on the die, so that the blank stays flat; in punching it is ground on the punch.

Answer: blanking force $= 75.4$ kN; punch $= 39.76$ mm, die $= 40.00$ mm; stripping force $= 7.54$ kN; work $= 60.3$ J.