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Can someone sense check my hydraulic force calculations?

Can someone sense check my hydraulic force calculations?

(OP)
Hi,

I've never specced any sort of hydraulic lifting system before and this has just come up on a project - just wondering if the collective minds here can sense check this for me to make sure I'm at least going about it the rightway? I feel like some of the values I'm getting for force etc are way out and I've never done this sort of application before so I don't know if my intuitive response is correct!

The application is a hatch which will open if struck by a force in excess of 5000N, and then a hydraulic cylinder will close the hatch again. This is shown in the attached image (roughly).

The approach that seems most obvious to me is a simple lever arm torque analysis to work out what size and pressure cylinder is required. In the attached image, the cylinder is indicated by the black lines. So for a torque calculation to get the force that will be applied to the line of action of the cylinder, I've taken 131.82 as lever arm, 1090 as radial arm and desired_force as the torque, using F = torque / (lever arm/radial arm) to get a target force, then using F = [pi/4(r2^2-r1^2)] * P to get a cylinder force which matches the target force (on the pull cycle of the cylinder) at a fixed pressure (in this case 2000psi if that's important). None of these values are restricted by design requirements at this stage, EXCEPT the desired force and the size of the hatch. The cylinder will be selected from a stock range from our hydraulics supplier, so I'm not too fussed about that.

I made a spreadsheet to rapidly iterate the changes to critical values - I've attached it, as well, in case I've just stuffed up a formula somewhere there.

Am I on the right track? I'm solo on this project and we're very busy at the moment so there's no-one else available to quiz sadly :( I think it's right from an mathematical perspective, but I'd rather not proceed unless I'm sure!

Thanks in advance!

-La.

RE: Can someone sense check my hydraulic force calculations?

F=(pi/4)*(D2^2-D1^2) not (pi/4)*(r2^2-r1^2)

Ted

RE: Can someone sense check my hydraulic force calculations?

I forgot to multply by p, pressure. F=area*pressure
Sorry.

Ted

RE: Can someone sense check my hydraulic force calculations?

(OP)
Right, so is d2^2 - d1^2 analogous to (radius of cylinder bore)^2 - (radius of cylinder rod)^2? And on the push cycle, it would just be d2^2?

RE: Can someone sense check my hydraulic force calculations?

D is diameter, r is radius. r=D/2 Area of a circle is (pi/4)*D^2 or pi*r^2.

Area of the annulus is (pi/4)*(D2^2-D1^2) where D2 is cylinder bore diameter and D1 is rod diameter. Or pi*(r2^2-r1^2) where r2 is cylinder bore radius and r1 is rod radius.

Push area would be (pi/4)*D2^2 or pi*r2^2.

Ted

RE: Can someone sense check my hydraulic force calculations?

The calculations above do not take into account that there will be some pressure on the cap side and some force will be needed ( lost ) to push the oil out the cap side when moving. What kind of valve are you going to use? There will be pressure drops across the valve. A simple bang-bang valve will not permit speed control.

Peter Nachtwey
Delta Computer Systems
http://www.deltamotion.com

RE: Can someone sense check my hydraulic force calculations?

Clarification needed:

The 5000N force is removed before the hydraulic cylinder tries to close the hatch? Or the hydraulics have to counter this force?
You have a 2000psi system to pressurize the hydraulics? This would mean that you need a cylinder with a bore area of 2.5 sq inches or a radius of 0.9 inches. larger is you want margin.
How are the hydraulics connected to the attach point? Rotational or linearly. You could have angular inefficiencies if it is linearly attached.

Why are you concerned with the cylinder rod in the force calculations? A hydraulic cylinder force is determined from Pressure divided by the area of the bore. This force is directly transferred to the hydraulic rod that is used for attachment.

Simply put hydraulic pump force is equal to the pressure times the bore/area, of the pump face.

You could add valves and or other restrictors to control the speed of the hydraulics.

Oh and Force is not a torque.

Force times the ratio of the arms will give to the force required to make your system static. (Be aware that the pivot point will experience the sum of the forces.) Also you mentioned forces in excess of 5000N so you will need to determine your upper limit too.

Good Luck

RE: Can someone sense check my hydraulic force calculations?

A good practice, we have adopted in doing spreadsheets, is to have a theory tab to explain where you got your reference equations from. Just FYI

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