Clamping Force Due to Mold Cavity Pressure

As shown in the exercise:
- as the diameter of the disk shaped part is varied the clamping force and part volume change in the same way,
- the change in volume of the spherical part and clamping force do not change in the same way as diameter is varied,
- most striking is that for the same diameter disk and sphere the clamping force is the same.

Interpreting these results leads to:
- since the disk volume depends only on cross sectional area for constant disk height, the clamping force depends on cross section area,
- sphere volume depends on diameter cubed and since clamping force varies differently with diameter than volume there is more evidence of the dependence of clamping force on area,
- this line of thought can be checked by calculating disk cross section area and clamping force as the product of the area and melt pressure,
- this idea should be made more specific and detailed.
A procedure for developing a quantitative result, that is a process model, is outlined below.

Fluid pressure acts normal to the mold surface. For a horizontal injection molding machine the clamping force is in the horizontal direction or the x direction. Force equilibrium requires that the force in the x direction due to the melt pressure is balanced by the clamping force so

F = Fx melt
Fx melt can be found by integrating the x direction component of the melt pressure p over the spherical surface area, A.

This procedure can be outlined by considering only a single mold cross section plane as in this sketch.

One half of the mold is shown with the clamping force and the melt pressure normal to the mold surface.
In the lower sketch a section of the mold representing a small area is shown. In this 2-dimensional section the area is seen as a line called dA.
The force in the x direction due to pressure p is

dF = ( p cosØ ) dA
Integration of this force over the semicircular area gives the total force due to p acting over the semi-circular line (area).

The grouping of the terms can be changed to give

dF = p ( cosØ dA )
In this form the cosØ dA term is a component of the area and is shown in the lower sketch. This area is called the projected area for the line (area) dA. The projected area of the sphere is a circular area as is shown in the upper right sketch.

A superficial summary is,
to calculate the clamping force we can use the horizontal component of p and integrate over the mold surface area
or use p acting over the projected area of the mold cavity

The formula for the clamping force needed to hold the two halves of a mold together is

F = p Aprojected
in which F is the clamping force, p is the melt pressure in the mold cavity and Aprojected is the projected area defined above.

This result shows that for a disk and a sphere of the same diameter, with mold cavities situated as shown in this exercise, the projected areas are equal and so the required clamping forces are equal.

A check: Using these general concepts it is possible to explain that,

- for the manufacture of a number, n, checkers per cycle,
- using a mold with the mold cavities layed out symmetrically around a central sprue,
- with the mold cavities connected to the sprue with equal length runners,
- and for full round (circular cross section) runners of constant diameter,
the clamping force component required due to melt pressure acting in runners, pr, is
Fr = n [ pr ( runner length )( runner diameter ) ]

© 2002 by Barney E. Klamecki all rights reserved.