I was asked the question…
Dan
“How one will decide the ejector pin diameter to eject the part safely
without any stress mark observed on the part?”
My Comments
To avoid permanent distortion of the plastic moldings, the number, location, and design of the ejector components must be developed to apply a low and uniform state of stress across the moldings. If the ejector force is uniformly distributed across many points in the mold cavity, then the molding will be uniformly ejected from the mold without any permanent distortion.
My Response
You must first have an estimate of the ejection force required to remove a molding from the core. The ejection force required to remove a molding from a mold core is a function of the normal force between the surface of the molding and the surface of the mold, together with the associated draft angle, the coefficient of friction, and the vacuum force. The coefficient of friction will vary from approximately .5 for highly polished surfaces to more than 1.0 for rough or textured surfaces.
The Ejection Force will then be:
Cos(Draft Angle) x Coefficient of Friction x Force(normal) + Vacuum force
There is a minimum push area that is required to avoid excessive shear stress on the plastic molding. More ejector pins will increase the pushing area and reduce the shear stress at each ejector pin location. The mold should be designed such that the perimeter around all the ejector pins provides a shear stress about one half the yield stress of the plastic material. The following equation relates the total perimeter of the ejector pins with the ejection force, plastic yield stress, and wall thickness.
Sum(Perimeter of Ejector Pins) > [2 x Ejection Force]
Plastic Yield Stress x Wall Thickness
If you cannot add additional ejector pins you will have to increase the ejector pin diameters until the sum of their perimeters satisfies the equation above. If more pins were added the diameters could be smaller as long as the equation above was satisfied.
-Andy