Pressure calculation, what am I doing wrong ?

[MechEng2]

I have a 110 PSI pneumatic line.
I have a 700 lbs linear force output.
I have 1.230" outside diameter spring using 0.250" wire.

My 110 PSI pneumatic line with cylinder is able to compress my 1.230" OD spring with 0.250" wire to 700 lbs linear force. My calculations reflect I should require about 916 PSI to achieve 700 lbs of force on a 1.230" diameter spring.
What am I doing wrong here, as I have a cylinder that can handle the 700 lbs of linear force output ? Based on my pneumatic calculations, I was thinking the max force output on 110 PSI should be about 330 lbs on my 1.230" OD.

 

[Andrew Mason]

Hi MechEng2. Welcome to PF!

I am not sure how your spring is connected to the pneumatic line. Is the spring connected to a piston? What is the area of the contact surface between the pneumatic line and the spring? (ie. if the spring compressed by a piston, what is the area of the piston?). How is that area related to the force on the spring and the pressure in the line?

You also have to tell us whether the 110 PSI is gauge pressure or absolute pressure.

AM

 

[MechEng2]

Thanks for the welcome and the help! The spring is located over a rod. It rests between to locating plates, essentially washers. the piston applies pressure to the spring and compresses the spring.
The piston is applying a load using a locating washer. I don't have that washer size. ~2.0"
Contact surface area of spring on piston locator is 0.7696 in^2.
The cylinder is a 4" cylinder, but I don't know if that matters.

110 psi is gauge pressure

 

[sophiecentaur]

What's the cross sectional area of the piston? The force will be Pressure times that Area. Are you assuming that the spring is the same diameter as the piston?
If the pressure is 110 PSI and the diameter of the piston is 1.23" then I think you should expect about 520 pounds force. That sort of implies the piston is in fact about 1.4 times the diameter of the spring if you are getting 700 lbs.
You seem to be using the calculated force from the spring to get 700lbs. You should be able to calibrate the spring with a known load on it, if you want to check the 700 figure. But the most likely thing is that you haven't used the right piston size, I think.
PS you need to check my arithmetic but I think it's ok and the factor 1.4 would fit in with the typical shape of a piston with a spring inside it??

 

[MechEng2]

What's the cross sectional area of the piston? The force will be Pressure times that Area. Are you assuming that the spring is the same diameter as the piston?
If the pressure is 110 PSI and the diameter of the piston is 1.23" then I think you should expect about 520 pounds force. That sort of implies the piston is in fact about 1.4 times the diameter of the spring if you are getting 700 lbs.
You seem to be using the calculated force from the spring to get 700lbs. You should be able to calibrate the spring with a known load on it, if you want to check the 700 figure. But the most likely thing is that you haven't used the right piston size, I think.
PS you need to check my arithmetic but I think it's ok and the factor 1.4 would fit in with the typical shape of a piston with a spring inside it??

I want to make sure we don't get mixed up. The cylinder is a 4" cylinder. The spring is on the outside. I use the cylinder to apply a force on the spring. Actually, the force on the spring is likely to be more than 700 lbs, but at 700 lbs the spring is at solid. I actually don't care what the final force output is, however I seem to be incorrectly calculating the max linear force capacity I can achieve out of my cylinder.

 

[MechEng2]

So then, if my pressure equals the cross section of my cylinder, and my cylinder is 4.0" with 110 PSI, then I am looking at max force output of 1381 lbs ?
4.0" cylinder, 2.0" R, (2 inch^2)x3.14x110 PSI= 1381.6 lbs
Assuming no load loss, is 1381.6 lbs a reasonable max force from my cylinder ?
At that point, do I care about my spring OD? The cylinder piston at 4.0" is dictating the force output, right ?

 

[lordvon]

When you are trying to compute the force from the pneumatic line, you have to use the cross-sectional area of the fluid chamber. Assuming it is 4":

r = 4 in / 2 = 2 in
A = pi * r ^ 2 = 12.566 in ^ 2
p = 110 psi
F = p * A = 1382.26 lbf

When you say "at 700 lbf the spring is solid", you mean it is fully compressed? The calculation above is certainly consistent with that.

 

[lordvon]

So then, if my pressure equals the cross section of my cylinder, and my cylinder is 4.0" with 110 PSI, then I am looking at max force output of 1381 lbs ?
4.0" cylinder, 2.0" R, (2 inch^2)x3.14x110 PSI= 1381.6 lbs
Assuming no load loss, is 1381.6 lbs a reasonable max force from my cylinder ?
At that point, do I care about my spring OD? The cylinder piston at 4.0" is dictating the force output, right ?

"Yes" to all of your questions, except "No" to caring about spring OD. It is separate from the fluid.

One caveat on the last question: you should be using the internal diameter of the piston, where all of the fluid is. The fluid is providing the pressure over that area.

 

[AZFIREBALL]

Can you provide an image of the setup?

 

[sophiecentaur]

you should be using the internal diameter of the piston,

To be clear, it's the outer diameter (virtually the same as the internal diameter of the cylinder) . The 'inner diameter ' could imply the size of the hole that the spring sits in.

 

[sophiecentaur]

The spring is on the outside.

I just read this and it has confused me. Do you mean the spring is not in the fluid or somehow wound round the outside of the cylinder? As @AZFIREBALL says, you need to give us a diagram.

 

[MechEng2]

see quick sketch of application.

 

[sophiecentaur]

A square cylinder or artistic licence?

 

[MechEng2]

A square cylinder or artistic licence?

external is square.

 

[sophiecentaur]

external is square.

OK.
This seems to be at an experimental stage. What is the final aim of the system? Springs have a bigger range of force when they are used in tension', rather than in compression. If you let it expand down into a hole in the table, the coils will not bunch up together. If you fix the top end of the spring and push through at the bottom end, you could get past 700lbs. Just an idea.

 

[MechEng2]

Based on calculations, I believe the 4" diameter cylinder cross section was the issue. I needed to account for that 4.0" diameter in my calculations as my driving force. Therefore, based on 110 PSI, my max force output will always be 1382 lbs regardless of spring size.

 

[sophiecentaur]

regardless of spring size.

Exactly. The spring has nothing to do with the force that the pneumatics can produce.

 

[sophiecentaur]

For accuracy, you must use the internal diameter of the cylinder for the area calculation.