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Force needed to open door 1

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waskillywabbit

Mechanical
Feb 26, 2004
302
Given:

174" ID Autoclave Side Hinged Door w/ accessories
Weight: 30,000 lbs (just picture it like opening a door in your house, only larger and heavier!)
5" bore hydraulic cylinder to open door
Thrust bearing holding weight of door
Pump pressure available 1500 psi

The cylinder is capable of moving the entire weight of the door w/ the appropriate pressure supplied (max 3000 psi) but how do I account for the fact that the cylinder is just pulling the door open, not lifting the door?

The door hinge bracketry sits on a thrust bearing to take the 30,000 load. How do I go about calculating the force needed to open the door? Obviously the cylinder doesn't have to move the full weight as it is resting on the thrust bearing. What percentage of the weight does the cylinder have to overcome to open the door?

Probably something simple for many of you fresh out of school but this old brain needs some help. Thanks.

Brian
WabFab
 
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Calculate the friction moment of the thrust bearing under the door weight.

M_friction = Bearing mean radius X door weight X bearing friction coefficient

Now:

F_cylinder = M_friction / vertical distance of cylinder pull point from the door hinges

The bearing dimensions and coefficient of friction is given in the bearing manufacturer catalog.
 
Hi wasskillywabbit,

If the door is hung on vertical hinge pins, (vertical axis of rotation) and swings horizontal then
israelkk is correct but probably meant:
F_cylinder = M_friction/HORIZONTAL distance of cylinder point from the door hinges.

Maybe the door is opened and closed by hand pumped hydraulics, but if you are using powered hydraulics please read on. The problem here is what happens when you stop the swinging door either on the closure seal or the swing open stop. Be sure and use a cushion cylinder, cushion stops and/or bi-directional flow control elements in your hydraulic circuit. Otherwise, you will be swinging a 15 ton hammer.
 
I meant vertical to the line connecting the hinges
 
Thanks for the replies! <applause>

I was on the right track in my thinking, but I couldn't find any relevant formulas in the books I have.

Everything is set up for the door to open nice and smooth and it works...just have to "prove" it on paper, you know how it goes.

Thanks again.

Brian
 
From where I come, first you have to show on paper that it works and then to actally build it and prove it.

This is the correct way otherwise you may end up with a too strong (expensive etc.) actuator or a too weak actuator where you will have to redesign the whole system (expensive etc.)

This is why engineers go to engineering school where they actually teach you the basics from where you can do all the needed calculations.
 
There are a lot more circumstances to the situation of which you have no idea and I have no control, such as experiential factors, previous working designs, etc...so thanks again for the help, but you can keep the lecture and condescending attitude...as where I come from, when someone says "thank you", you say "you're welcome" and move on.

Brian
 
You don't mention how long of stroke the cylinder has. OR mention anything about the linkage points where the cylinder attaches. If the cylinder is attached .25" from the hinge, and at a 170 degree angle from the door position you will need much more force than a cylinder mounted at a 90 degree angle from the face of the door and 15 inches from the hinge. It's all a matter of leverage, swing distance, and overall cylinder stroke. Depending on the mounting angle, the "usuable" force to rotate the door will change. A 1 in cylinder would be capable of opening a huge hanger door if it is attached to the door properly. It may need 10 feet of stroke, but it is possible....and how fast does the door need to open? If your pump only does 1/2 gpm quick calcs say your cylinder will only move about 6" in a minute....you don't want it too fast or too slow.

It's not as simple as will it work. The real question is; how well will it work, and is it the most affordable efficient method.
 
What velocity profile do you want the door to open with?

With basic open/close valves, a hydraulic cylinder is essentially a constant force machine. The geometry and kinematics of the connection of the cylinder to the door and a rigid mount may alter the force vs. degree of opening profile that the door sees.

Constant force means constant acceleration. The inertial forces associated with a 30,000 pound door are liekly not negligible, as ccw pointed out earlier.
 
You will have plenty of force to move the door with the area and pressure you have available you could actually pick up the door with the cylinder. I believe your bigger concern will be stopping the door movement. The buckling strength of the cylinder should be determined. The higher the speed of opening and closing the door will create more inertia to stop. The hydraulic system will also be stressed by the external load applied by decelerating or accelerating the mass. You should have a pressure relief valves in the cylinder circuits to avoid overloading the hydraulc system and effectively giving you a deceleration ramp on the door stopping. Flow controls will also be needed to make the door opening and door closing the same speed.

You will have to watch your mounting configuration for the cylinder the multiplication of forces could impact the system performance.
 
Your main concern will probably be when the cylinder gets near indeterminate. If the door opens 90 degrees in 15 seconds it will take about .33 seconds to accel/decell the door with a 12” lever(28,000 lb-ft), with a 6” lever(14,000 lb-ft) .66 seconds.

I guess my point is that you need to have something to dissipate the inertia. Built in cushions in cylinder, a bumper, separate hydraulic shocks or slow down ramp.

I would suggest separately mounted bumpers or shocks if the cylinder piston can bottom out at full extension since the angle usually gets ugly there. The further you can limit travel under 90 degrees the better.

Barry1961
 
Several people have brought out some valid points. You didn't mention whether your hydraulic system is dedicated to this function alone. Cylinder speed and load induced pressure will vary with oil temp and pressure fluctuations. You could make this simple or as complicated and more costly as you want. The simple route would be with a bang bang Directional Control Valve and Flow Control valves metering the flow or by using a feedback for door position and a proportional Directional Control Valve you could eliminate the need for the Flow Contols. Regardless an A and B to Tank relief valve would be a good idea to protect against load induced pressure. The cylider cushions at the end of stroke will certainly be a help. Maytag
 
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