Plastics Technology

SEP 2018

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Let's start with the basics and build on that. Ejector pins are mounted in the ejector retainer plate by means of a through hole for their body and a counter-bore for their head. The typical head-diameter tolerance of a standard ejector pin is +0.000 to -0.010 in. from nominal. The counter-bore diameter for the head of a pin in the retainer plate is typically 1/32-in. larger than nominal. The typical head-thickness toler- ance for a standard ejector pin is +0.000 to -0.002 in. from nominal. The counter-bore depth for the head of a pin in the retainer plate is typically +0.001 to +0.003 in. deeper than nominal. Most pins have a 1/32-in. radius where the shaft of the pin meets the head, so clearance must be provided. Ejector-pin counter-bore sets, available through most molding supply companies, auto- matically add a chamfer to account for the radius on the pins. The tolerance for the diameter of an ejector pin varies depending on the size, type and manufacturer. It can be as much as -0.0010 in. below nominal to as little as -0.0000 in. below nominal. Keep in mind, that's a full thousandth. Most mold designers will make the through hole for the pin in the ejector retainer plate 1/64-in. larger than nominal. This allows a standard- size drill to be used and provides roughly no more than 0.008 in. per side of "float" in the event the through hole in the plate does not precisely align with the through hole in the core. The ejector retainer plate can have holes for the ejector pins, ejector sleeves, core pins, guide bushings, return pins, support pillars, lifters, screws and various other component parts. Basically, these plates look like Swiss cheese. Last month I discussed the problems associated with uneven loads on the ejector system. Cracking a fragile ejector retainer plate in half is one of those problems. Why Ejector Pins Break And How to Prevent It An ejector pin's unsupported length is a critical variable and a common cause of its bending, buckling and breaking. If you have ever watched an Olympic pole-vaulting event, you have seen how much the pole bends after the vaulter plants it in the box and leaps up to clear the bar. These poles can be over 16-ft long. If the vaulter holds the pole in the middle, instead of at the end, how much do you think that pole will bend? Obviously, not nearly as much. The material, hardness, diameter and even the length of the pole did not change. Only its unsup - ported length changed. The unsupported length is equal to the distance between two fixed or pivoting points of an object. This same premise also applies to ejector pins. A Swiss mathemati- cian and physicist by the name of Leonard Euler developed two formulas back in the mid-1700s that moldmakers can still use to predict whether an ejector pin will buckle and collapse. One formula is rather simple. It is called the Slenderness Ratio. This ratio is equal to the unsupported length of a slender column, such as an ejector pin, divided by the radius of gyration. Don't get excited. The radius of gyration for an ejector pin is simply its diameter divided by four. So the formula is: Slenderness Ratio = Unsupported Length ÷ Diameter ÷ 4 If the slenderness ratio is greater than 100, you can expect the pin will fail due to buckling. Here's the when and how to reduce the unsupported length of pins. By Jim Fattori PART 2 An ejector retainer plate can crack when subjected to an uneven load. Pin Diameter, in. Max. Unsupported Length, in. 1/32 or 0.0313 0.8 1/16 or 0.0625 1.6 5/64 or 0.0781 2.0 3/32 or 0.0938 2.3 7/64 or 0.1094 2.7 1/8 or 0.1250 3.1 9/64 or 0.1406 3.5 5/32 or 0.1563 3.9 11/64 or 0.1719 4.3 3/16 or 0.1875 4.7 13/64 or 0.2031 5.1 7/32 or 0.2188 5.5 15/64 or 0.2344 5.9 1/4 or 0.2500 6.3 TABLE 1 Pin Diam. Vs. Unsupported Length (For a Slenderness Ratio of 100) 46 SEPTEMBER 2018 Plastics Technology PTonline.com K now How TOOLING

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