Stamping Math

How to Calculate Press Tonnage, Strokes Per Minute, and Coil Yield in Sheet Metal Stamping

The core stamping formulas worked in full: tonnage from cut perimeter, strokes per minute to output, and coil yield with real units and numbers.

Cutting tonnage is the first number to nail. The formula is Force (tons) = Perimeter (in) x Material Thickness (in) x Shear Strength (psi) / 2000. For a blank with a 24 in cut perimeter in 0.060 in cold-rolled steel at 50,000 psi shear strength, that is 24 x 0.060 x 50,000 / 2000 = 36 tons. Add roughly 20 to 30 percent for a stripper spring load and safety margin, so size the press near 47 tons. Run the same inputs through the Press Tonnage calculator, but always confirm shear strength from the mill cert, not a generic table, since HSLA grades push 60,000 to 70,000 psi.

Forming and drawing loads stack on top of cutting. A simple 90 degree air bend uses Tonnage per foot = (575 x t^2) / die opening, where t is thickness in inches. For 0.060 in steel over a 0.472 in die (8x thickness), that is (575 x 0.0036) / 0.472 = 4.4 tons per foot. A 10 in bend length needs 3.7 tons. When a progressive die punches, coins, and forms in one hit, sum every station's peak load, because the press must survive the worst simultaneous stroke, not the average.

Throughput starts with Press Strokes Per Minute. Parts per hour = SPM x 60 x cavities x efficiency. A press at 120 SPM running a single-out die at 85 percent uptime yields 120 x 60 x 1 x 0.85 = 6,120 good parts per hour. Feed length caps real SPM: with a 3.5 in progression and a servo feed limited to 1,400 in/min, max feed rate is 1,400 / 3.5 = 400 SPM, so the press, not the feed, is the constraint here. The Progressive Die Output tool chains these to a shift total.

Coil yield decides how much of the steel you actually sell. Yield percent = (Part net area x parts per coil) / total coil area x 100. A 0.060 in x 12 in wide coil weighing 5,000 lb at 0.283 lb/in^3 holds 5,000 / (0.283 x 0.060 x 12) = 24,528 running inches. At a 3.5 in progression that is 7,008 blanks per coil. If each blank nets 28 in^2 out of a 42 in^2 strip footprint, yield is 66.7 percent. The Coil Yield and Strip Layout Yield calculators expose exactly where the other 33 percent leaves as skeleton scrap.

Blank layout is where yield is won or lost. Compare one-up straight, two-up, and angled nesting on the Blank Size Optimization tool. Rotating an irregular blank 15 to 30 degrees often recovers 4 to 8 points of material. Keep web and edge margins realistic: minimum web between blanks is about 1x to 1.5x thickness, and edge margin about 1.5x to 2x thickness. For 0.060 in stock that means webs near 0.075 in and edges near 0.100 in. Shrinking margins below tool strength limits causes slug pulling and burr, so treat the geometric optimum as a ceiling, not a target.

Strip layout ties progression to width. Strip width = part width + 2 x edge margin + 2 x carrier width. A 2.0 in part with 0.10 in edges and 0.25 in carriers on each side needs 2.0 + 0.20 + 0.50 = 2.70 in wide coil. Progression = part length along strip + web. Multiply width x progression to get the strip footprint per part, then divide net part area by that footprint for utilization. Two-out layouts double parts per stroke but widen the coil, so recompute Press Tonnage since force scales with total cut perimeter across both parts.

Verify with a mass balance before committing. Input coil weight should equal finished part weight plus skeleton scrap plus slug scrap, within 1 to 2 percent. From the earlier coil, 5,000 lb in at 66.7 percent yield gives 3,335 lb of parts and 1,665 lb of scrap. Divide 3,335 lb by part weight to confirm the piece count matches the stroke count from your SPM math. When the two counts disagree by more than a couple percent, the error is usually an overstated yield or an unaccounted lead and tail scrap on each coil, typically 6 to 15 ft per coil.

Pull it together with one example line. A 150 SPM press, 3.5 in progression, single-out, 88 percent uptime runs 150 x 60 x 0.88 = 7,920 parts/hr. One 5,000 lb coil at 7,008 blanks lasts 7,008 / 7,920 = 0.88 hr, so you burn better than one coil per hour and need a coil every 53 minutes. That single figure drives your Coil Change Downtime planning and staffing. Every number above traces to a mill cert, a feed spec, or a coil weight, so keep those three inputs current and the rest of the stamping math stays honest.

Published 2026-07-01.