Sheet Metal Product Manufacturing Process

(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 punch and die meet and give a clean edge, with a burnish zone about one third of thickness and a small burr. Too small a clearance gives secondary shear, 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 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; 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 on the punch face (equal to about the sheet thickness), which spreads the cutting over the stroke.

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

(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 $5.2^\circ$).

(a) Deep drawing calculation

Blank diameter from equal surface area of blank and cup (thickness constant, sharp corner):

$$D = \sqrt{d^2 + 4dh} = \sqrt{50^2 + 4\times50\times70} = \sqrt{16\,500} = 128.5\ \text{mm}$$

Draw ratio:

$$DR = \frac{D}{d} = \frac{128.5}{50} = 2.57$$

This is greater than the limiting draw ratio (LDR) of 2.0, so a single draw is not possible. First draw cup diameter $d_1 = D/2.0 = 64.2$ mm. The second draw (redraw) from 64.2 mm to 50 mm is a reduction of $1 - 50/64.2 = 22\%$, which is within the usual redraw limit of 25-30 percent. So two draws are needed.

Force of the first draw (approximate formula):

$$F = \pi d_1\, t\,(UTS)\left(\frac{D}{d_1} - 0.7\right) = \pi\times64.2\times1\times350\times(2.0-0.7) = 91.8\ \text{kN}$$

Blank-holder force is about one third of the drawing force, about 30.6 kN, to prevent wrinkling.

(b) Short notes

Hydroforming: a sheet or tube is pressed into a die by pressurised fluid (oil or water) instead of a rigid punch. In sheet hydroforming, a rubber diaphragm or fluid pressure forces the blank around the punch. In tube hydroforming, high internal pressure plus axial feeding expands the tube into the die (automobile frames, exhaust parts). It gives uniform thickness, complex shapes, less springback, and good finish, but the cycle time is longer and the equipment is costly.

Spinning: a flat disc or tube is clamped against a rotating mandrel and a roller or tool forces the metal over it. In conventional spinning the thickness stays nearly constant (cones, hemispheres, cookware, reflectors); in shear spinning the wall thins according to the sine law $t_f = t_0\sin\alpha$. Tooling is cheap and suits small batches of axisymmetric shapes.

(c) Formability assessment

• Tensile test: n-value (stretchability) and anisotropy ratio $r$ (deep drawability; larger $r$ is better).
• Limiting dome height / Erichsen cupping test: a punch stretches a clamped sheet until fracture; the depth indicates stretchability.
• Limiting draw ratio test (Swift cup test) for drawability.
• Forming limit diagram (FLD): plot of major strain against minor strain at which necking begins, obtained from grid-marked specimens; strains in the part must lie below the curve.

Answer: blank diameter $= 128.5$ mm, draw ratio $= 2.57$, two draws, first-draw force $= 91.8$ kN.