Free body diagram and cylinder force help 2

[grunt58]

This is not homework. I've taken some engineering classes so I'm familiar with the theory and terms. I've just never been good at applying it.
I'm building a little enclosure with a pneumatic cylinder-actuated door. Need some help making the free-body diagram and force needed to open the door. If someone wouldn't mind sketching the free body diagram and writing the formula I'd appreciate it.

 

[3DDave]

It is simpler to make a graph of lifting the weight vs retracting the cylinder as an energy balance. Your CAD is already doing the trigonometry for you.

So, if 10 pounds of door is lifted 1 inch and the cylinder retracted 1/2 inch, the cylinder needs to supply 20 pounds of force to produce 10 in-lbs of lift.

 

[rb1957]

remove one element at a time and replace with forces ...
1) the actuator can only have loads along it's line of action, as it closes it pulls on the L brkt and an opposite force it applied to box at the other end (equal and opposite forces)
2) now you know the force applied to the L brkt, in general terms (at some angle, don't worry about that ... yet). The actuator will apply some force axially long the L-brkt, and some perpendicular ... against don't look at the details, just the concept. Now the other arm of the L brkt (attached to the door is supporting the weight of the door)
3) so lets remove the door. I'd have the upper L brkt react the weight of the door, and have the lower as just a guide.
4) now we can look at the L-brkt. This is a "three force member" ... look it up. Knowing two of the forces on the link, we know the direction of the 3rd force, generally.
5) now consider different positions, as drawn to start ... beginning, middle, end of door movement.

clear as mud ?

"Hoffen wir mal, dass alles gut geht !"
General Paulus, Nov 1942, outside Stalingrad after the launch of Operation Uranus.

 

[MintJulep]

Just by inspection, cylinder force needs to be about four times the door weight.

 

[Snickster]

The way to look at it is that the sum of forces and the sum of moments of the resulting forces about the pivot point of rotation has to be always equal to zero for static equilibrium to exists which basically it does since although there is movement there is no acceleration. The moment is the force times the perpendicular distance to the pivot point. In your case the resulting forces would be the weight of the door down and the pulling force of the cylinder.

Also if you break any resulting force into components then the sum of the forces and sum of the moments produced by the components also has to be equal to zero. Where now the distances are different than the perpendicualr distances when looking at the resulting forces because the components of the resulting forces are pointing in different direction than the resulting forces.

Also note that the lower support arm of the door does not support any force because it does not have an opposing moment at the pivot point which must be present to support any force.

If you provide all dimenstions of support arms/cylinder arm and angles I will make a free body diagram.

 

[grunt58]

If I missed any dimensions fudge it and note it on the free body diagram.

https://imgur.com/a/xqzaFfT

 

[hydtools]

The lower link is also a two-force member, the forces, equal and opposite, acting along its centerline.

Ted

 

[rb1957]

@grunt58 ... we've given you plenty of clues as to what to do.

if my post isn't clear, what are you missing ?

in your 3rd position, the actuator force is pretty much horizontal, the door weight is down (doh), so the line of action of the pivot point is along a line between the pivot point and the intersection of the horizontal actuator load and the vertical door weight (this is the three force body). Draw a force triangle, you know the weight, and the your scale the other forces (from the triangle).

Repeat for other positions ... the line of action of the door weight moves laterally, the actuator rotates (and extends/contracts). This is one of the tricky parts of designing this door opener ... ideally you need the actuator moving consistently as the actuator load is a function of it's extension

"Hoffen wir mal, dass alles gut geht !"
General Paulus, Nov 1942, outside Stalingrad after the launch of Operation Uranus.

 

[grunt58]

@rb My last reply was to @snick. They said they would draw up a free body diagram.

 

[rb1957]

ok I guess, but what is unclear about what we've told you, how to make the FBD ? If you need some further understanding, surely YT would help ?

You're a studant (and I'm a crabby olde fart) ... what will you learn if we do your homework ?

"Hoffen wir mal, dass alles gut geht !"
General Paulus, Nov 1942, outside Stalingrad after the launch of Operation Uranus.

 

[3DDave]

A free body diagram is a map of the path forces will take based on the map of the connections - how can you use a map you cannot make when the map of connections is your diagram?

That's why I suggested using energy instead and letting CAD do the math. Move any point some small amount and see how far the weight moves. Force times the distance is work (delta energy) in and equals force times distance for work out for rigid items and neglecting friction, no free body needed, just the CAD model.

 

[grunt58]

@Dave Just trying to educate myself. I'm interested in better understanding what you are saying. If I move the cylinder 1" the door will move in an arc x distance. Do you mean how far the center of mass moves or the door arm connection or? This is where I get tripped up. Reading and understanding is not how I learn. I do better by seeing and doing.

 

[rb1957]

COF again ... doing stuff on your own is IMHO the same as homework (ie not a professional problem). You're trying to learn FBDs, ok good. We've given you plenty of hints. If you're getting tripped up of the dynamic inertial forces (as things move), the common answer is these are neglected and we look at the systems as quasi-static snap shots. This is why we say things like the loads the actuator put on the structure are along it's line of action (at any particular moment it time, as per your three pictures). Now sure, you can add the dynamic forces, very difficult (what is the mass moment of inertia of the actuator ?).

If you're staring out learning FBDs, start with a static structure.

If you want to learn about dynamic mechanisms, that's a subtly different thing.

"Hoffen wir mal, dass alles gut geht !"
General Paulus, Nov 1942, outside Stalingrad after the launch of Operation Uranus.

 

[rb1957]

someone above posted that the lower link is a two force member. But if you include body forces, it becomes a three force member. the line of action of the two ends is the same, as the CG is along this line. The transverse forces at the CG will balance, and there'll be a small change in axial force.

The approach suggested is to replace elements with the forces they apply to the structure. The actuator and the lower link are two force members, the line of action is along the member. the angle link is a three force member. If this doesn't help, look up these terms,

"Hoffen wir mal, dass alles gut geht !"
General Paulus, Nov 1942, outside Stalingrad after the launch of Operation Uranus.

 

[3DDave]

Lift a weight up - force times vertical distance against gravity is the work done. You have a center of gravity mark on the door. Horizontal movement of the CG doesn't matter because gravity doesn't oppose that movement.

 

[rb1957]

yes, but that isn't a FBD. and ok, I know you know. The "difficult" part of the FBD is the load at the pivot of the L brkt.

but ... hummm ... the FBD of the door is not so simple. Humm, maybe you do need to include the dynamic/inertial loads ?
Part of me thinks the load in the lower link is zero, but then it's the thing keeping the door vertical.
The upper link is adding a load at an angle to the weight vector, so some transverse load, reacted by transverse inertial load ?
The CG of the door is moving along an arc ... it's not rotating, just translating.

So one thing you'll need is the spring curve of the actuator (to get a full picture of the kinematics).

In picture 2, with the links horizontal, weight is down, the inertial load is up, the upper link is carrying the weight (as transverse load), the lower link is ... doing nothing ? I think this shows that the load in the lower link is small, and changes sign as the door goes through the 2nd pic.

"Hoffen wir mal, dass alles gut geht !"
General Paulus, Nov 1942, outside Stalingrad after the launch of Operation Uranus.

 

[grunt58]

I got what I needed from a friend. My cylinder is well within the safety factor. I now have a Free Body to reference in the future.

 

[Snickster]

grunt58

Sorry but by the time you responded it got to be the weekend and I had things to do. I see you got someone you know to help you. If not I have time today to work on it. It no problem except it does take some time to draw up a sketch with details that you can understand.

 

[grunt58]

@snick I'd be curious as to what you come up with, if you don't mind. I don't want to trouble you though.

 

[Snickster]

OK I will work on it.