Sheet Metal Nesting and Material Order
A cadet in the fabrication shop receives a work order for 400 identical spacer plates, each 150 mm long, 90 mm wide and 8 mm thick, cut from mild steel sheet stock of 2400 mm by 1200 mm by 8 mm. The plasma table needs a 6 mm gap between adjacent parts and a 6 mm margin at every sheet edge; parts are nested square to the sheet edges. The stores clerk ordered material by mass alone: he took the finished mass of the 400 plates, added 10 percent for offcuts, converted that to kilograms and ordered the nearest whole number of sheets. Mild steel density is 7850 kg per cubic metre, the price is $2.85 per kilogram, and sheet is sold only in whole sheets. The operator now proposes cutting the inter-part gap to 3 mm to save a sheet. Before any cutting starts, the cadet must confirm the number of sheets, the real material cost, and whether the operator's change saves anything at all.
1. How many sheets are actually required, and what mass of steel do the finished parts represent? Show the nesting and mass arithmetic, and state why the clerk's order is short.
Read the worked answer
Nesting. Part pitch along the sheet is 150 + 6 = 156 mm, and across it 90 + 6 = 96 mm.
- Columns: floor(2400 / 156) = 15 (15 x 156 = 2340 mm, leaving 60 mm edge strip, which satisfies the 6 mm margin).
- Rows: floor(1200 / 96) = 12 (12 x 96 = 1152 mm, leaving 48 mm).
- Yield: 15 x 12 = 180 parts per sheet, with 6 mm margins already covered.
Sheets. 400 / 180 = 2.22, so 3 sheets are needed (540 blanks, 60 spare). Two sheets give only 360 parts.
Mass check.
- Volume per plate = 0.150 x 0.090 x 0.008 = 1.08 x 10^-4 m^3, mass = 1.08 x 10^-4 x 7850 = 0.848 kg.
- 400 plates = 339.1 kg finished.
- One sheet: 2.4 x 1.2 x 0.008 = 0.02304 m^3, mass = 180.9 kg.
- Clerk's figure: 339.1 x 1.1 = 373.0 kg, which is 2.06 sheets, so he rounds to 2 sheets = 361.7 kg, enough metal by mass but not enough area for 400 nested parts.
Cost. 3 x 180.9 x $2.85 = $1,546. The mass method under-orders because scrap in nesting is a shape problem, not a 10% weight allowance.
2. The operator wants to reduce the inter-part gap from 6 mm to 3 mm to save a sheet. Does this work? Justify with the new pitch figures, and state what the real constraint is.
Read the worked answer
New pitch: 153 mm along and 93 mm across.
- Columns: floor(2400 / 153) = 15 (15 x 153 = 2295 mm; 16 would need 2448 mm).
- Rows: floor(1200 / 93) = 12 (12 x 93 = 1116 mm; 13 would need 1209 mm, which is 9 mm over the sheet).
- Yield: 15 x 12 = 180 parts per sheet - identical to the 6 mm case.
So the change saves nothing: 400 parts still need 3 sheets. The constraint is the short edge of the sheet, where the 12-row layout is already close to the limit; the long edge has spare strip that cannot be used because a partial column does not fit the 1200 mm direction.
Judgement. There is no benefit to be traded against the risks of a 3 mm gap: less tolerance for plasma kerf wander and sheet camber, a higher chance of parts fusing or tipping into the cut, and near-zero gain. Reject the change, or investigate a rotated/mixed nesting (for example turning alternate rows) or a larger sheet format, which are the options that can actually change the sheet count.