# Thread: Basic pneumatic press compressive strength question

1. ## Basic pneumatic press compressive strength question

Hey guys,

So I'm helping the lady expand her soap making hobby. She currently uses a 1.5" cylinder press at 150 psi. So that means her press exerts a force of ~ 240 lbs.

Her press is attached to a 5 in^2 square block. So I believe that her square block exerts a compressive stress of.....

P = F/A = (240 lbs) / (5 in^2) = 48 psi. Correct?

Well, now I want to increase it from 1 soap, to say 40 soaps.......Well, with a multiple soap design I need space in-between the soap forms for wall thickness, etc. So say I need 8 in^2 for each soap. Okay, then for 40 soaps I need 8*40 = 320 in^2. Therefore, the force I need is F = P * A = (48 psi) * (320 in^2) = 15,360 lbs!!!!!

This confuses me. 320 in^2 is just a 18" by 18" square. Not very big. 50 psi also is not a lot. But if I'm understanding correctly to put a relatively low pressure (50 psi) across a relatively small area (18" x 18") I need to create basically 8 tons of force???

Am I following this correctly? Is 50 psi actually a pretty substantial compressive stress?

Thank you in advance for any help and guidance. I would like to design this press, but I didn't imagine I'd be making a 10 ton press. lol.  Reply With Quote

2. area of circle = pi * r^2 = 3.14157 * 0.75^2 = 1.77 in^2

force = pressure lbs/in^2 * area = 150 * 1.77 = 265.5 lbs axial force.

area of 5 in. square block = 5 * 5 = 25 in^2

265.5 lbs / 25 in^2 = 10.62 psi  Reply With Quote

3. Hey. Thanks so much for the reply! Unfortunately, the square block that presses the soap mold has an area of ~5 in^2 (~2.25" x 2.25"). It is not a 5" square.

I just couldn't belive that I really needed 8 tons of force to apply 50 psi over an area of 320 in^2. But I guess that's how the math works out.  Reply With Quote

compressive stress, force, pneumatic cylinder press, stamping press, strength 