what is this Physics formula?

[smokehouse]

We have a plastic injection mold that has a horizontal slide that is hydraulicaly activated. Physical limitations mandate the use of only a small hydraulic cylinder. The hydraulic pressure however, is overcome by the molding pressure when plastic is injected into the cavity, causing the slide to move back ..

We elected to overcome this problem by installing a small vertical pan cylinder that is mounted to and is c'bored into the support plate. When activated vertically, the business end of this pan cylinder enters the horizontal slide carrier and thereby prevents the horizontal cylinder from retracting under molding pressure.

The vertical slide end that enters and interlocks to the horizontal slide has a 5 degree conical shape to it so that the end would measure about .500 dia + 5 deg/s.
Needless to say, the horizontal slide has the "female" counterpart to this.

Now to the formula:
Using a plastic inject pressure of 30,000 PSI, and with the frontal area of the horizontal slide being about 1" dia, how much UPWARD pressure is required of the vertical slide before it fails ( backs down) due to the 5 degree taper?

 

[insideman]

Force on primary slide= 30000pi/4=23562#.
Force on latch= 23562Sin5=2054#.

 

[smokehouse]

Insideman,
Thanks for the answer..If I follow your calculations, you determined the actual molding pressure on the slide area to be 23,562 lbs..
You then multiplied that by the sine of 5 degrees to get 2,054 lbs?
We are looking for the hydraulic pressure needed to keep the vertical slide in place..
Is 2,054 PSI the answer?
Thanks again..

 

[insideman]

No.

2054 is not the pressure needed, it is the force needed. You must size the cylinder and its pressure to provide that force.

For example, if you have 1,000 psi available, the piston area must be 2054/1000 = 2.054 sq in.

 

[insideman]

An afterthought: It would be better to have the locking cylinder move a lock block into place in such a way as to transmit NO force to it. I think this can be easily done.

 

[GregLocock]

Your calculation with the sin of 5 degrees in it is OK, but it ignores friction. The coeeficient of friction is likely to be enough to lock the pin in place, although it is a bit marginal. It varies a lot, depending on surface finish and contaminants. I'd make it 3 degrees just to be safe! Cheers

Greg Locock

 

[smokehouse]

We designed the vertical lock with a 5 degree angle so it would have some "lead-in". The lock end itself is 1/2" dia +5Deg/s and it enters to a depth of 1/2" into the slide carrier.
Because of the mold base size, we are restricted to a small pan cylinder for this vertical cylinder that will only take 500 PSI input.
I took an educated guess on the 5 degrees.
As mentioned, 3 degrees would hold better with less pressure required. However we went with 5 degrees to give it a little more "forgiveness" in case of slight misalignment.