Welcome!

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.

I was recently asked the question…            

Kevin

“Do you have any way of easily calculating the residence time in screw / barrels?”

My Response

Most general purpose screws contain approximately 1-2 shots within the flights depending on the geometry. To calculate the residence time you will have to know your part volume, the volume of material in the barrel, and your cycle time. You can use a basic calculation for the barrel volume minus the screw volume (ignoring the flights because they are relatively small) to calculate the material your screw holds.

To calculate the volume of the barrel:

3.14 x (Barrel radius)^2 x (Barrel length)

To calculate the volume of the screw:

3.14 x (Average root radius of the screw)^2 x (Barrel length) 

Volume of material in the barrel/screw: Barrel volume-screw volume

As for the residence time:

If you calculate the volume of material in the barrel as 10 ounces and your part is 5 ounces your screw contains two shots of material. Multiply the number of shots in the barrel by your cycle time and you will get a residence time for the material.

For example: If you have a 25 second cycle time and your screw holds 2 shots the residence time is roughly 50 seconds (two cycle times).

The residence time will be a multiple of your cycle time plus or minus one cycle (depending on the ratio of part volume to barrel volume).

-Andy