Blank Diameter for a Cylindrical Cup

A flat disc goes into the press and a cup comes out. How big does the disc have to be? The simplest possible answer sets the area of the blank equal to the area of the finished cup — the disc of the bottom plus the rectangle that the wall unrolls into — and solves for the diameter. That gives $\pi D^2/4 = \pi d^2/4 + \pi d h$, and rearranged, $D = \sqrt{d^2 + 4dh}$.

The first thing to do with the answer is not to cut a blank but to divide: $D/d$ is the drawing ratio, and it decides whether the part can be made at all. Below about 1.6 almost any formable sheet will manage it. Approaching 2.0 you need a good deep-drawing grade and a well-set die. Past about 2.2 no sheet material draws in a single operation, whatever the area arithmetic says, and the part becomes a redraw job — the diameter reduced in stages, sometimes with an anneal between them.

Now the honesty, because this equation is a starting point and not an answer. It assumes the sheet keeps its thickness everywhere, and it does not. Drawing thins the wall, hardest at the punch nose radius where the metal is stretched most, and it thickens the flange, where the outer material is squeezed circumferentially as it is pulled in. Volume is conserved through all of this; area is not, and the whole derivation rests on area. The formula also ignores the punch nose radius and the die profile radius completely, treating the corner between bottom and wall as a perfectly sharp line, and a real radiused corner consumes a different amount of material than a sharp one does.

How much does this matter? Less than it sounds, for a reason that has nothing to do with the arithmetic being good. A drawn cup almost never comes out with a level rim. Rolled sheet is anisotropic — its properties vary with direction relative to the rolling direction — and that variation makes the flange draw in faster in some directions than others, so the rim ends up with ears, usually four of them. The rim gets trimmed square afterwards regardless of how carefully the blank was calculated. So the practical procedure is: take this figure, add trim allowance, cut a trial blank, draw one, measure what actually happened, and adjust. Serious work uses finite-element simulation to develop the blank, which handles thinning, the die radii and the anisotropy together; this equation is what you use to decide roughly how much material a part will need and whether the process plan is plausible.

One structural note on the equation itself. Solved for the cup diameter it becomes a quadratic, $d^2 + 4hd - D^2 = 0$, whose positive root $d = \sqrt{4h^2 + D^2} - 2h$ is the only one that is a diameter. And the same approach — set the flat area equal to the formed area — is how blanks are developed for flanged and stepped shapes too, with the shape broken into elements whose areas are summed. The assumptions, and the trial-and-adjust that follows from them, are exactly the same.