Injection Molding - Clamp force calc - Sliding blocks

[Martin Spayne]

I would like to know if anyone has a calculation or formula for working out the theoretical clamp force that a standard Injection Moulding machine would need to use when "Sliding Dies" or "Sliding Blocks" are used. I use a proven formula for straight forward, two plate tooling but when it comes to moving cores on angle pins I have a problem, I think that theoretically there should be little difference and the reason why I need high clamp force is because the heel block or the core block is not hard on. Any suggestions?

[MrMold]

I believe the difference to be so slight that it's not worth woring about.

[lgreeves]

Your locking blocks will transfer the pressure of the machine to the slide. The pins will not have any pressure on them. A designed interference will give sealoff.

[Martin Spayne]

Surely the gate position would have an effect? If the component was gated in a position where the melt pressure could open the blocks then this should be avoided. Are there any more thoughts on this?

[spider]

The only thing that really should matter is the locking around the slide (sliding blocks?) It doesn't matter how much clamping pressure you put on the mold, if the heel flexes, it's useless. I usually like to have 3+ inches thick heel behind the slide to prevent that...(that is along with the locking angles on each side and a strait wall in front.)

Lot of times Mold shop will leave couple of thao preload on the slide heel when spoting the seal off, just to make sure it is pushing forward HARD.

[lgreeves]

The shot pressure should not excede the clamping force of the machine.

[jetski]

Can you send me a sketch. Angles of the drivers and cams can change things. We normaly go 20 on the drivers and 15 on the angle pins. I figure 1000 psi on the square in. of area pushing on the pins. It is actually less than this but it is a good save number to use. 1.00 dia pin on a cam will see less than 750 lbs of push on its face. So a 20 degree driver has a roughly 4:1 mechanical advantage. So from only the pin the mold would see 187 lbs of force trying to push it open. A cylinder would see 750 lbs of push (not 750 psi). A cylinder with .5 square inches of cross-section would need 1500 psi to hold the pin closed. I have used air cylinders to hold .25 dia blind core pins in location. Send me a drawing I'll look it over. I have been doing this since 1981. Christ I'm grey when the hell did that happen...Oh well it was the last hair on my head anyway.

[SORCHEROR]

what formula do you use?wall stock and material type change the factor to use,thinner walls and tougher flowing material need higher clamp pressure,molding surface area parralel to the platen is all that matter,vertical walls dont add tonage,but you do need to account for them on mold design for safe opperation with out failures,on slides with larger areas i usely design
and angle on the lock oppisite the slide to counter lock the back against the ejector(or movable)side of the mold,a ledge,or L shaped lock give more strenght to the lock

[MauricioB]

Martin
I realize this is quite a late answer but is going to be very helpful.

There is a web page: http://www.injecneering.com that we have created to give free access to applications for injection molding engineering. One of the five applications offered (Machine Selector) calculates the clamping force a tool needs. It does consider the effect of slides in the clamp force by calculating the force component in the platens' opening direction (see first picture below). It also considers the frictional force (reduces the clamp force). I agree with MrMold's comment that most of the times the effect of the slides in the clamp force is negligible. However, I have ran into tools where that is not the case as the slides' areas are bigger than the projected area (See second picture). I also like to always include it anyways just for the sake of accuracy.

Hope it helps
Mauricio Benavides