Ask the Stamping Expert: How do we fix our slug pulling problem?

A: A very interesting phenomenon is going on here, and you have to see it to believe it. So try this:

Set the punch up on the bench, with the blanking face pointing straight up, and plug in the air. Now take a part you blanked and hold it about 1 in. away from the face of the punch. Orient the part to the punch face. Support the back of the part with the palm of your hand, allowing the air to hold the part in position directly over the face of the punch. Now start to move the part closer to the punch face. As you get closer there will be a point where the part will jump from your hand to the face of the punch. The part will remain stuck in position a few thousandths of an inch from the face of the punch. It is being held in place by the air current rushing across the face of the part between the part and punch face.

The force at work here is the same as that which makes a curve ball curve. When air runs faster on one side of a surface relative to the opposite side the surface, the object will move toward the faster air. When it happens to an airplane, it is called force lift. It’s very interesting that in this case, at about ¼ in. away, the phenomenon disappears and the air pushes the part away.

To solve this problem, add ejector pins to all five holes. I assume the blanking punch is 2 in. long. Counterbore the back side of the blanking punch air holes about 0.375 in. deep to accept the head of a standard, off-the-shelf steel round punch blank to be used as ejectors) at 0.095-in. body diameter (based on your 0.100-in. inside diameter through-holes). I assume the ejector punch head is about 0.1 in. thick. Grind the ejector punch blanks (again, to be your ejector pins) to 1.750 in. long. I believe the head on a standard 0.095-in.-OD punch blank is 0.156 in. dia. Make the counterbore diameter 0.010 in. bigger than the head diameter. This gives a nice slip-fit for the ejector punch body and the punch head. Grind two slots down the entire length of the ejector punch using a 0.025-in.-wide wheel with a hand-dressed full radius, 180 degrees from each other on the ejector punch body, about 0.025 in. deep. Assuming a 0.156-in.-OD head on a 0.095-in.-OD ejector punch, use this equation—0.156 - 0.095 = 0.061/2 = 0.0305—to determine how much bigger the ejector punch head is on the side of the ejector punch body. Again assuming a 0.156-in.-OD head, the slots will be 0.0305 + 0.025 = 0.0555 in. deep into the head OD in two places, 180 degrees from each other.

Insert the ejectors into the blank punch and assemble the tool. When activated by air, the pins will be driven down and stick out of the punch by 0.25 in., in much the same way that the shaft of an air cylinder is driven. Air also escapes through the grooves in the punch. When the tool is running, just after you blank the part at dead bottom of the stroke, you will turn on the air. As the tool begins the upstroke, the pins will push the part mechanically away from the face of the punch outside the zone of the phenomenon. Also, once this happens, the air escaping through the grooves will blow the part down through the tool.

Do some tests on the bench to get the ejectors as short as possible. The shorter they are, the faster you can run the tool. Don’t forget to turn the air off on the downstroke. That way, as the ejectors contact the material before blanking, they will simply back up and not damage the delicate 0.004-in.-thick material.