Calculating pressure from trapped silicone rubber

[ROLEX2018]

Hello forum first time posting in a long while.

I am trying to calculate pressure created by trapped silicone rubber in an enclosed cavity. To simply the condition imagine a silicone cylinder 1” in dia x 1” tall enclosed in a sealed steel tool, ultimately I would like to know pressure in psi excerpted on a laminate between silicone and tool.

I tried calculating pressure using bulk modulus but my number are too high from what I remember running a quick experiment with a press pressure was around 250 to 300 psi.

Thanks

 

[3DDave]

Rubbers are nearly incompressible, so any pressure calculation should be similar to that for liquids under the same condition.

 

[Compositepro]

Calculations are not often meaningful. The calculation you did is probably telling you how much pressure is possible (a lot). No pressure is generated at all until the silicone has expanded enough to fill the cavity. Then the voids in the composite part and the bagging must be compressed. Pressure is only generated by containment of the expanding silicone. Everything depends on the mold volume, part volume, silicone volume, and temperature change. These are rarely known accurately enough to yield useful results through calculation. Ultimately it usually does not matter what the exact pressure is, as long as it goes over 50-100 psi. The process is capable of generating over 10,000 psi.

 

[ROLEX2018]

Thanks for the response, my interest for the calculation is to be able to make a tool that is self contained meaning no need for a press to contain the pressure exceted by the expanding silicone. I have made tools with silicone matching the IML of the part but I would like to try and estimate gap I need to have between silicone and laminate to develop sufficient pressure to consolidate but not so much that we deform the tool.

Pressure I am trying to maintain is 80-120 psi cure is 250deg f, silicon is similar to aircast.

The calculations are made to guide the test and try process.

The other calculation approach I was going to attempt is assuming the interaction is made up of a series of springs were silicon expands as temperature increase and at the point of contact pressure ramps up,laminate has a lower spring rate and would take up most of the pressure while the tool surface is assumed fixed reviving all the pressure. Assuming a 1 sq inch area I would convert the force to pressure and use that to design the tool to avoid deflection.

Thanks again for your input.

 

[MikeHalloran]

I used to work in a factory where we made tiny 2-pc. ball valves for refrigeration tools.
We had a run of cracked (brass) bodies, which was eventually traced to a tolerance condition where the male plug end that compressed the ball seals was going too far into the female body thread, overcompressing an 0-ring, which, as noted, behaved as a liquid and expanded the body enough to crack it.
We had to change the screw machine cams to provide e.g. 2-3/4 turns of full thread in the one part, with a crazy tight tolerance of +/- 1/8 of a turn, or something like that. ... and just for fun, convince the new, small girl who was assembling the valves to use a shorter wrench.

My point is that using an elastomer as a spring by trapping it in a closed cavity and relying on its bulk modulus to provide a controlled force will probably cause you a lot of anguish, and give irreproducible results.

Can you instead enclose a die spring in your assembly, and use a metal piston and a silicone pad to spread its force over your laminate face?



Mike Halloran
Pembroke Pines, FL, USA