What is the formula to figure out the PSI a pneumatics piston can hold?

hold? it sounds like you are asking how much pressure can you put in the piston? if so, that is spec’d out for the pistons by the manufacture

if you mean, how much force can the piston deliver when actuated, multiply the air pressure you are putting into the piston (PSI) times the area of the cylinder inside diameter (in sq inches)

for example, a piston with a 2" inside diameter is PI * R^2 = 3. 14 sq inches

so if you apply 60 psi it will push with 3.14 * 60 = 180 lbs of force.

Is that what you are asking?

The PSI a pneumatics piston can hold is usually written on the piston. You don’t have to figure it out.

if you were asking about pushing/pulling power, don’t forget, you must subtract the area of the piston rod from the area of the ID of the piston, when figuring out pulling power.

If your question relates to previous FIRST-legal pneumatic actuators, this PDF might be of use to you.

Thanks, I need to clarify my question. If the PSI isn’t on the piston, what formula do you use to figure it out? It’s for a project in my robotics class.

the max pressure a pneumatic piston can safely hold is a complicated thing to calculate. It would depend on how thick the walls of the piston are, the type of metal used, the threads on the connections, the seals inside the piston, the strenght of the rod

not to mention how strong the ends are so you dont blow the end of the cylinder out when it slams to one end under pressure.

there is no single formula - it really comes down to a design analysis to establish the spec for max pressure. and even that has variables: temp, humidity…

and finally , they aways include a safety factor- if the piston will fail at 300 psi they wont rate it for 299, they would rate it for 150 or 100 or 60 - depending one how much of a safety factor they want.

If you would like to treat it as a “thin-walled pressure vessel” I could get you the formula for that… it is:

Max Tangential (hoop) Stress = p * (di + t) / (2t)

p = pressure inside the vessel

di = inner diameter

t = wall thickness

Max stress for steel varies widely… from maybe 30,000 to 150,000 psi. Assuming you know this, you can re-arrange the above equation to solve for the max pressure inside. (then divide this by maybe 1.5 or 2 for a factor of safety, probably more since failure would be catastrophic!)

A longitudinal stress also exists because of the pressure on the end of the vessel.

this stress is equal to: (p*di) /(4*t)

If you want to think of these as pressurized cylinders that are NOT thin-walled, ie. the thickness of the wall is greater than 1/20th the radius, the formulas become more complicated.

Let me know what you need, I can come up with them pretty quick.

Matt

Hey, can you tell me how to build shoe-sized jet engines?

You bet, I just took an ME dual-level (ungrad / graduate) course at Purdue

**ME 535 :**

**Small Scale Jet Engines, Design and Small Volume Production for Non-Violent Competitve RC robots**.

It was taugh by Prof. Iam Liontuya

It’s the perfect course for FIRST. Unfortunately, I left my book and notes at school… I’ll have to get back to on the details. We’ll definitely be going jet engine this year. Gotta love highly skilled high schoolers on the CNC’s.

Matt

The maximum working pressure you can use is 60 psi. In order to determine the force of the cylinder multiply the pressure you are using times the area of the piston. See the pneumatics manual for a chart on this.