## More Accurate Clamp Tonnage Calculations

I got this intriguing question the other day…

Jim
I am working with a part – family mould 1 + 1.  Here’s the scoop:

Projected Area:              89.75 “^2
Hold Pressure:               950 psi hyd.
Intensification ratio:        13.36
Plastic Pressure:           12,692 psi
Force:                           89.75 X 12,692 = 1,139,107 lbs.
Tonnage:                       1,139,107 lbs. / 2000 lbs./ton = 569T.

The tech guide says use 3.2T/in^2 = 3.2 X 89.75 = 287.2T.

The job was quoted in a 300T, and we run in 300T.

I would say, by theory, the plastic pressure based on hold pressure would be best calculation, but apparently not.  I don’t understand why.  It should be more accurate than the general rules of thumb that I’ve seen, based on part wall thickness etc. Any thoughts?
The design guides are actually great references, since they generalize the pressure losses in the runner and sprue.
This would be a great time to perform a pressure loss study to really get detailed numbers.
In such a study, you perform a series of short shots through the air, sprue, gates, and then short shot the cavity.
The peak pressures at each stage in the process should be recorded.
So, for example…
air: 2000psi
sprue: 3000psi
gates: 4000psi
short shot: 12000psi
(12000psi) – (4000psi) = 8000psi pressure applied to mold cavity
In this simplified case, only 2/3 of the applied pressure, (8000psi) / (12000psi) = 2/3 reaches the mold cavity. In this case, I would apply this 2/3 factor to the overall packing pressure. So if the packing pressure was 6000psi, I would calculate the tonnage using (6000psi) x (2/3) = 4000psi. This would actually provide a fudge factor since some pressure is lost within the mold cavity due to a pressure differential across the mold cavity.