Can smaller bars be used in redesigning this lever mechanism?

[omarmorocci]

Hello everyone,
I am currently doing an internship and I am working on a part that opens and closes using a pneumatic cylinder as you can see bellow

Here is a force diagram

I would like to redesign this in order to use a cylinder with a shorter stroke and perhaps a slightly bigger diameter .My current idea is to use a four bar linkage

I am open to new Ideas

 

[Merlin3189]

I'm not sure I can see what you are doing. Is the load 1600N F₂ and is it always acting vertically down? Or is it perpendicular to FG? Do you need a force or do you need a torque?

Assuming ; F is the large cap head bolt below the hazard sign,;FGH is a rigid piece pivoted at F to the base and at H to the piston; G is the pivoted attachment of the load (1600N F₂ acting vertically down); I is the other end of the piston attached by pivot to the base;
then can you move the load attachment point G further from the pivot F? So that for a given movement of G, the piston point H moves less.
You appear to do something like that in your solution by making BA longer than FG. But I can't follow the logic of your solution, since placing P partway along CD means P will move less than D. You also have changed the direction of the piston, which looked in the photo to be constrained in its orientation by being fixed under the base.

Perhaps you caould annotate your photograph to indicate which bits move and in which direction. Also mark on the label points used in your diagram and use labels consistently throughout?

 

[Nidum]

Make cylinder(s) direct acting - no lever at all ?

 

[omarmorocci]

I'm not sure I can see what you are doing. Is the load 1600N F₂ and is it always acting vertically down? Or is it perpendicular to FG? Do you need a force or do you need a torque?

Assuming ; F is the large cap head bolt below the hazard sign,;FGH is a rigid piece pivoted at F to the base and at H to the piston; G is the pivoted attachment of the load (1600N F₂ acting vertically down); I is the other end of the piston attached by pivot to the base;
then can you move the load attachment point G further from the pivot F? So that for a given movement of G, the piston point H moves less.
You appear to do something like that in your solution by making BA longer than FG. But I can't follow the logic of your solution, since placing P partway along CD means P will move less than D. You also have changed the direction of the piston, which looked in the photo to be constrained in its orientation by being fixed under the base.

Perhaps you caould annotate your photograph to indicate which bits move and in which direction. Also mark on the label points used in your diagram and use labels consistently throughout?

The 1600 N F is the weight acting vertically down. , it can also bee seen as a torque of 160N/mm at F ( the big bolt)

Yes, you are right , I have moved the G away from A in order to reduce the force required in order to turn the flap.
If I redesign it then I can choose the new mounting position of the piston, I put the pneumatic cylinder like that as it resulted in less force requirement.
( but the pneumatic cylinder does indeed have to be bellow the base)

Alright,
So the the big bolt bellow the hazard sign is fixed . FGH is a fixed lever. It is pushed at H by the pneumatic cylinder.

In the new (very basic) design I am sort of using the four bar linkage in order to reduce the stroke and increase the force required , hence I can use a shorter cylinder with a slightly bigger diameter ( cheaper)

 

[omarmorocci]

Make cylinder(s) direct acting - no lever at all ?

How to connect it ?
Keeping in mind that the pneumatic cylinder has to be bellow the base of the structure as shown in the pic

 

[Merlin3189]

The 1600 N F is the weight acting vertically down. , it can also bee seen as a torque of 160N/mm at F ( the big bolt)

It sounds as if you really need a torque rather than 1200N or 1600N vertically down. So extending FG will not help.

Yes, you are right , I have moved the G away from A in order to reduce the force required in order to turn the flap.

You reduce the force but not the torque. You need only consider the stroke available from your new cylinder and how the linkage will convert that to the required angle of rotation. Then the cylinder will simply have to be made big enough to provide whatever force is needed.

If I redesign it then I can choose the new mounting position of the piston, I put the pneumatic cylinder like that as it resulted in less force requirement.
( but the pneumatic cylinder does indeed have to be bellow the base)

Seems contradictory. If you have to keep it below the base, you can't move it! (At least, not closer to the main pivot point.)
Below the base it looks as if it must be angled down away from the base, or at best parallel to it.

In the new (very basic) design I am sort of using the four bar linkage in order to reduce the stroke and increase the force required , hence I can use a shorter cylinder with a slightly bigger diameter ( cheaper)

But linkage CPD increases the cylinder stroke - D moves further than P. You would need to attach the cylinder at P and the lever BAP would have to become BAD.
But that linkage can't move in any case! If B and C are fixed pivots and BAP is rigid, then you have a triangle with two fixed corners and 3 rigid sides. The third point P cannot move. You would need to add a fourth member between CP and BAP with a floating pivot at each end.

You need to look at your device at both ends of its operation and see how your mechanism fits both extremes. (You will need to check the whole path, but the end points are most likely to show up any constraints on the movement of members.)

 

[omarmorocci]

It sounds as if you really need a torque rather than 1200N or 1600N vertically down. So extending FG will not help.You reduce the force but not the torque. You need only consider the stroke available from your new cylinder and how the linkage will convert that to the required angle of rotation. Then the cylinder will simply have to be made big enough to provide whatever force is needed.

I don't have any stroke available hence I'm free to choose any . I would like to design a mechanism that uses less storke and a little bit more force .
this will result in a cylinder with slightly bigger diameter and way shorter stroke hence making it cheaper

Seems contradictory. If you have to keep it below the base, you can't move it! (At least, not closer to the main pivot point.)
Below the base it looks as if it must be angled down away from the base, or at best parallel to it.

The best option is to make it parallel but when I played around with the program , I got much bigger forces required

But linkage CPD increases the cylinder stroke - D moves further than P. You would need to attach the cylinder at P and the lever BAP would have to become BAD.
But that linkage can't move in any case! If B and C are fixed pivots and BAP is rigid, then you have a triangle with two fixed corners and 3 rigid sides. The third point P cannot move. You would need to add a fourth member between CP and BAP with a floating pivot at each end.

B is fixed and can't be changed.
C is not yet fixed but will be , depending on the new design (hence we have the freedom to move this point)
BAP is not a rigid body , as A and P and D are all floating pivots

You need to look at your device at both ends of its operation and see how your mechanism fits both extremes. (You will need to check the whole path, but the end points are most likely to show up any constraints on the movement of members.)

The problem is that the system is only fixed in one end . There are many possibilities to mount the pneumatic cylinder

 

[Merlin3189]

... The best option is to make it parallel but when I played around with the program , I got much bigger forces required

Well if you manage to reduce the stroke, the force will need to increase won't it?
Work out ≤ work in, so force_inxdistance_in≥force_outxdistance_out
So force_in ≥ force_out x distance_out / distance_in
So if the output force and distance are a fixed requirement, then the input force is inversely proportional to the input distance (stroke)

... BAP is not a rigid body , as A and P and D are all floating pivots ...

Aah! Yes, I'd not noticed that. I was still thinking of the FGH rigid lever. So you are ok with that linkage.

The problem is that the system is only fixed in one end . There are many possibilities to mount the pneumatic cylinder

Not sure I follow this. I was talking about the start and end positions of the device when it opens / closes. What is the range of movement of each part?
The diagrams are all limited in showing a single position. Angles and forces will change as the device operates.

I thought the cylinder had to be fixed to the base of the "thing" and the lever GH was necessary to connect it to the fixed pivot at F. If you can move the cylinder to the side of the "thing" closer to F, then your problem is trivial (as Nidum said) - just move the cylinder as close to F as you need to reduce the stroke to what you want.
But if it has to be fixed to the bottom of the "thing" , I would have the lever FGH to transmit the movement from the base to the area round FG and another lever attached to the cylinder, like CD, which converts the available movement of the cylinder (stroke) to the required movement of H, and link the two levers with a floating link (PA) as you show. The calculation is, what is the required movement of H (or A), what is the available movement by the piston of D (or H), then that tells you the required ratio of the second lever attachments.
I've done a quick sketch of a 3:1 lever based on your idea and rotated the cylinder with a bell crank.

 

[omarmorocci]

Well if you manage to reduce the stroke, the force will need to increase won't it?
Work out ≤ work in, so forceinxdistancein≥forceoutxdistanceout
So forcein ≥ forceout x distanceout / distancein
So if the output force and distance are a fixed requirement, then the input force is inversely proportional to the input distance (stroke)

Yes that is true, but I am hoping to find a mechanical solution that will reduce the stroke significantly in comparision to the increase in force.
( For example in other system , pulleys are a mechanical solution that reduce the required force to lift something )

Not sure I follow this. I was talking about the start and end positions of the device when it opens / closes. What is the range of movement of each part?
The diagrams are all limited in showing a single position. Angles and forces will change as the device operates.

Here are the 2 concepts starting positions

here are their final position. The angle is approx 20 degrees

But if it has to be fixed to the bottom of the "thing" , I would have the lever FGH to transmit the movement from the base to the area round FG and another lever attached to the cylinder, like CD, which converts the available movement of the cylinder (stroke) to the required movement of H, and link the two levers with a floating link (PA) as you show. The calculation is, what is the required movement of H (or A), what is the available movement by the piston of D (or H), then that tells you the required ratio of the second lever attachments.
I've done a quick sketch of a 3:1 lever based on your idea and rotated the cylinder with a bell crank.

Yes , it has to be fixed bellow the Chute as there is no place on the sides.
My question is why is the one you drew with the bell crank better than the current one I have ?
Cause the goal is not to just make the stroke as short as possible but the buy the cheapest pneumatic cylinder , hence a huge force (3600 N) means double the diameter and 1/4 of the stroke .

 

[Merlin3189]

Yes that is true, but I am hoping to find a mechanical solution that will reduce the stroke significantly in comparision to the increase in force.

But as I said, you can't get more work out than you put in. So if you halve the stroke, you MUST double the force, if the load output does not change. Maybe the load can be changed?

( For example in other system , pulleys are a mechanical solution that reduce the required force to lift something )

but they increase the distance over which you have to pull the input.
(Strange that PF will immediately censor your post if you suggest an electrical machine with over unity efficiency, but people regularly suggest over unity mechanical systems and get away with it!)

Here are the 2 concepts starting positions & here are their final position. The angle is approx 20 degrees

Thanks, that's useful. Also your pdf report.

Yes , <the cylinder> has to be fixed bellow the Chute as there is no place on the sides.
My question is why is the one you drew with the bell crank better than the current one I have ?
Cause the goal is not to just make the stroke as short as possible but the buy the cheapest pneumatic cylinder , hence a huge force (3600 N) means double the diameter and 1/4 of the stroke .

Because it has to be underneath, I didn't want it to be mounted vertically, so I put it horizontallyand used the bell crank to convert to vertical motion, just like the original design. That was simply the next step in the development of your idea - first tidy all the inks to horizontal and vertical, so that I can see what is happening, then get the cylinder back in roughly the necessary orientation.

There are all sorts of ways of looking at it. In redrawing my final picture to clarify my intention, I realized that I was simply taking the original single lever and shortening it to reduce the piston stroke. But then the cylinder is to far from the original pivot. So I move the original lever (which is a bell crank) down to the cylinder and simply transmit the force vertically between two equal cranks.

The bellcrank here has 2.5:1 ratio, converting 90mm horizontal stroke into 36mm vertical movement. Then the near vertical link simply transmits that 1:1 to a matching arm on the operating mechanism.

 

[omarmorocci]

But as I said, you can't get more work out than you put in. So if you halve the stroke, you MUST double the force, if the load output does not change. Maybe the load can be changed?

Now I understood what you mean. Very naive of me not to see it from the beginning

Because it has to be underneath, I didn't want it to be mounted vertically, so I put it horizontallyand used the bell crank to convert to vertical motion, just like the original design. That was simply the next step in the development of your idea - first tidy all the inks to horizontal and vertical, so that I can see what is happening, then get the cylinder back in roughly the necessary orientation.

There are all sorts of ways of looking at it. In redrawing my final picture to clarify my intention, I realized that I was simply taking the original single lever and shortening it to reduce the piston stroke. But then the cylinder is to far from the original pivot. So I move the original lever (which is a bell crank) down to the cylinder and simply transmit the force vertically between two equal cranks.

Let me first clarify the available possible space
As you can see in broadpic1 , the chute has two stands (feet) those are parallel to the ground.
As you can see in broadpic3 , we have 700 mm bellow the curved part as it is empty.

I will simulate the drawing you drew and see what results I get.
I have another questions, with the new information, is it possible to come up with another mechanical system where only 1 bearing is used like the original system rather than two bearings like the second. As that might prove even more economical.

 

[Merlin3189]

is it possible to come up with another mechanical system where only 1 bearing is used like the original system rather than two bearings like the second. As that might prove even more economical.

I wondered about that! If you save something on a cheaper cylinder, but have to spend twice as much on extra fixings, bits and assembly costs, it's not much use.

I think it may be possible.
After my last post, I realized my sketch was not quite what my words said, so i did this, keeping the cylinder in the same place, but adding another pivot:

Now you ask about keeping to one pivot, I see that instead of reducing the vertical arm, maybe we increase the horizontal arm and keep a single pivot, if we can move that pivot a bit. (Move D to D') That still requires an extra pivot, if D had to be kept for another part of the mechanism behind the scenes (which is what I suspect.) It also has problems if E pivots about D behind the scenes, because the arc of E about D is not the same as the arc of E about D' and you'd have to allow a bit of slide at E.

All in all I can see why the mechanism was designed as it was and the cylinder constrained to be what it was. If DE has to rotate through 20^o then the arm down to the cylinder has to rotate 20^o and the stroke is determined by the length(*) of that arm x tan(20). If neither length can be altered significantly, you can only change things by adding extra mechanism with extra pivot(s). (* the length of the arm is really from the pivot to the end of the piston - FH in your original diagram.)

 

[omarmorocci]

I wondered about that! If you save something on a cheaper cylinder, but have to spend twice as much on extra fixings, bits and assembly costs, it's not much use.

Yes that is why I prefer to keep 1 fixed pivot if possible. If not , then I will do the financials to see if it's feasible.

I think it may be possible.
After my last post, I realized my sketch was not quite what my words said, so i did this, keeping the cylinder in the same place, but adding another pivot:

I'm now confused on what you mean, as that drawing isn't clear.( at the E' where 2 rods are connected)
Here's my drawing of your idea from your previous post

Now you ask about keeping to one pivot, I see that instead of reducing the vertical arm, maybe we increase the horizontal arm and keep a single pivot, if we can move that pivot a bit. (Move D to D') That still requires an extra pivot, if D had to be kept for another part of the mechanism behind the scenes (which is what I suspect.) It also has problems if E pivots about D behind the scenes, because the arc of E about D is not the same as the arc of E about D' and you'd have to allow a bit of slide at E.

Moving D is not an option because it's already at it's max. only E can be moved away from D

All in all I can see why the mechanism was designed as it was and the cylinder constrained to be what it was. If DE has to rotate through 20o then the arm down to the cylinder has to rotate 20o and the stroke is determined by the length(*) of that arm x tan(20). If neither length can be altered significantly, you can only change things by adding extra mechanism with extra pivot(s). (* the length of the arm is really from the pivot to the end of the piston - FH in your original diagram.)

So I suppose adding a bearing and another arm is the only possibility to reduce the stroke .
I will check the difference in price of cylinder and bearing cost and then see.

Another possibility is to reduce the cost is by using a smaller rod diameter that goes through D and E .

 

[Merlin3189]

I'm getting confused with all the different labels we have both used!
So going back to your original drawing with FGH, I'm assuming D is F and E is G.
D or F is the pivot which we can't move.
E or G is the point of attachment to the hidden device and can be moved away from D/F. (This operates over a fixed distance rather than a fixed angle?)
So if you increase the distance FG then G moves further for any given rotation about F, or equally, it moves the same distance for a smaller rotation.
The long arm GH moves through the the same angle as FG. So if the rotation is reduced, the stroke of the piston is reduced,
since stroke ∝ (angle of rotation) approximately.

Using your first diagram I just lengthen FG. It doesn't much matter where you rigidly attach GH to FG, because the distance from the pivot F to the piston attachment H is the important measurement.

I don't understand the bit about smaller diameter rod at D(F) and E(G)? The size of side arm FGH and the diameter of any axle or pivot rods through D/F and E/G would seem to be determined by strength requirements and so not open to negotiation - unless you can also find a cheaper way of providing the same strength.

Moving D is not an option because it's already at it's max. only E can be moved away from D

I can well understand that F (D) is fixed, but I don't understand "its max"? I assume there is a fixed point here because it is the hinge or pivot for whatever opens internally. I assume that is also used as the pivot for the external lever because it is convenient and avoids providing another strong anchor point.

The big question for me is still, how G operates the internal bit. If the internal bit is hinged at F and G simply pushes up on it to make it rotate, then moving G will not help. It will simply have to move further to cause the same opening angle. Effectively the load requirement would not be a simple linear motion, but a rotation of G about F, by so many degrees. Changing the FG arm length will not alter this angle, nor therefore the GH angle and the piston stroke. If this is the case then I think you would have to add another pivot somewhere in your solution.

I'm not sure my suggestions from a position of very incomplete knowledge are helping. Much as I'm enjoying juggling my ideas around here, I think I'll leave you to it for a while. I had thought by now some of the clever guys would have arrived. Perhaps if I shut up for a bit they might come in. I'll keep watching the thread and make any comments I can about whatever you come up with. Good luck with your project.

 

[omarmorocci]

I'm getting confused with all the different labels we have both used!
So going back to your original drawing with FGH, I'm assuming D is F and E is G.
D or F is the pivot which we can't move.
E or G is the point of attachment to the hidden device and can be moved away from D/F. (This operates over a fixed distance rather than a fixed angle?)
So if you increase the distance FG then G moves further for any given rotation about F, or equally, it moves the same distance for a smaller rotation.
The long arm GH moves through the the same angle as FG. So if the rotation is reduced, the stroke of the piston is reduced,
since stroke ∝ (angle of rotation) approximately.

Using your first diagram I just lengthen FG. It doesn't much matter where you rigidly attach GH to FG, because the distance from the pivot F to the piston attachment H is the important measurement. View attachment 104215

I don't understand the bit about smaller diameter rod at D(F) and E(G)? The size of side arm FGH and the diameter of any axle or pivot rods through D/F and E/G would seem to be determined by strength requirements and so not open to negotiation - unless you can also find a cheaper way of providing the same strength.


I can well understand that F (D) is fixed, but I don't understand "its max"? I assume there is a fixed point here because it is the hinge or pivot for whatever opens internally. I assume that is also used as the pivot for the external lever because it is convenient and avoids providing another strong anchor point.

The big question for me is still, how G operates the internal bit. If the internal bit is hinged at F and G simply pushes up on it to make it rotate, then moving G will not help. It will simply have to move further to cause the same opening angle. Effectively the load requirement would not be a simple linear motion, but a rotation of G about F, by so many degrees. Changing the FG arm length will not alter this angle, nor therefore the GH angle and the piston stroke. If this is the case then I think you would have to add another pivot somewhere in your solution.

I'm not sure my suggestions from a position of very incomplete knowledge are helping. Much as I'm enjoying juggling my ideas around here, I think I'll leave you to it for a while. I had thought by now some of the clever guys would have arrived. Perhaps if I shut up for a bit they might come in. I'll keep watching the thread and make any comments I can about whatever you come up with. Good luck with your project.

The rods I was talking about ( going through D and E ) were not chosen based on calculations but based on avaibility at the time , hence I say that maybe smaller bars can be used.

You cleared up so much .
Thank you for all your help, it was much appreciated.