Sheet Metal Works

Sheet metal is generally 0.4 to 6 mm thick; operations are of cutting and forming types.

Blanking and piercing

Shearing

Cutting a sheet along a straight line between two blades. A guillotine shear has a fixed lower blade and a moving upper blade, inclined by a small rake to lower the force. Hand shears (snips) cut thin sheets.

   upper blade \
   ============ \___ sheet
   lower blade ______|

Bending

Forming a sheet along a straight axis by stressing it beyond its yield point. Air bending, V-bending and wiping/edge bending are done in press brakes. Springback occurs, so a slight over-bend is given. The minimum bend radius depends on the material ductility (about 1 to 2 times thickness).

Rolling

A sheet is passed through three rolls (pyramid or initial pinch type) to form cylinders and curved shapes. Moving the top roll changes the curvature.

Punching

A punch and die pair removes a slug and forms a hole; used for holes, slots and notches; turret presses punch many shapes using a CNC.

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

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

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

Given: outer D = 50 mm, hole d = 20 mm, t = 1.5 mm, $\tau$ = 320 MPa.

(a) Punch force

The force equals the shear area times the shear strength. The shear area is the perimeter cut times thickness.

$$\begin{aligned} \text{Cut perimeter} &= \pi(D + d) = \pi(50+20) = 219.9\ \text{mm} \\ F &= \text{perimeter}\times t\times\tau \\ &= 219.9\times 1.5\times 320 = 105{,}558\ \text{N} \end{aligned}$$

(b) Press capacity

$$F_{press} = 1.2\times 105.56 = 126.7\ \text{kN}$$

A 150 kN standard press would be selected.

(c) Work done

Approximate work per stroke: $W = F \times p\,t$, where p = 0.4 is the fraction of thickness penetrated before fracture.

$$W = 105{,}558\times 0.4\times 0.0015 = 63.3\ \text{J}$$

Note on shear: a shear angle on the punch lowers the peak force, but the work stays about the same.

Answer: (a) F = 105.6 kN; (b) press capacity = 126.7 kN (use 150 kN press); (c) W = 63.3 J.