Finding an unknown force

Simulation
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Hi all

I am in the process of sizing a lift that should be able to lift a load of 1.8 tons (~17.8kN).  One constraint I have is ergonomics, because the load will be placed on a surface of  850x200mm. I have to use one or more jacks to do this,  and I would need to know how to do that in order to find the force to lift that load at a certain speed. I will have to use several configurations of my system in order to find the most economical solution. Below is an image of the system I thought of using.

Thank you in advance for your help 

2

Hello

For the unknown force, I believe that our distinguished colleague @OBI WAN is quite aware.

More seriously  !

You must indicate what force you are talking about (cylinder, resistance of the axes, resistance of the scissor parts, etc...)

The height must also be indicated in the retracted position and in the top position at the top. It is important to know that the maximum effort is encountered when the table is practically flat (totally retracted with the minimum height) if you never go down with your load below 60 cm, then the efforts are much less.
You should also indicate if you have any overall dimension constraints, because the contact surface of  the 800 x 200 load does not presuppose the size of the table top.
Ergonomics have nothing to do with the problem.
Why do you make the assumption that you need several jacks? Especially since you don't say if it's a manual or motorized screw jack or if it's hydraulic jacks.
You don't specify whether the load will always stay in the center of the bed or not.
You don't say if the CG of the load  is in the middle of it or especially where the CG is located at the top.
You do not  say by indicating "a certain desired speed. ", nor the speed of elevation of the plate (and not that of the cylinder), nor whether the acceleration is instantaneous or progressive and over how many cm. You don't say why you need a precise speed or especially why (this is what will structure the choice of the Screw or Cylinder system)
You don't give any information about the scissor profiles or the resistance of the pins and yet this calculation is almost more structuring than the knowledge of the force of the cylinder itself.

I understand better the title "finding an unknown force"

Kind regards

 

 

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I admit that I have not been clear on the subject. Therefore

I'm talking about the strength of the cylinder, what effort is needed to overcome the 1.8 tons. 

The chainring must have a stroke of 510mm, after some research, I would need a cylinder stroke of 190-200mm.

I wasn't clear, the system should not exceed 800x200mm. On the height, I have no constraints.

The CG of the load will be in the middle and the load will be evenly  distributed across the tray. If you like, it's like putting a sledgehammer on the board that has the dimensions 800x200mm. 

Change of plans, they will be screw jacks, I need a gradual acceleration. I don't have a fixed speed. It is the operator who will choose how fast the elevator should rise. I set myself 10mm/s as the maximum speed of climb. 

Thank you for your answer, I'm learning, that's why I come here to improve myself and learn from other engineers. So we leave Obi Wan out of my problems. 

Tell me if I wasn't clear

 

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The form of your question suggested your status.

As we are not here to provide you with ready-made solutions, especially for a "simple" static problem, we will just give you some leads.
In this case, as I have just said, it is a problem of statics of forces. So I refer you to the course you must have had on the subject.

Show us your schematics, sketches and other calculations that you rely on to determine your cylinder stroke e.g. From there, we will be happy to guide you.
Is it a project of the company where you are learning or an exercise?
You are not talking about the minimum height your table should be (unless it is recessed into the floor).

By the way, ObiWan is one of the prominent members of this community ;-)

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Hello @davit27srg 

Obviously, the 3D model of the lift table does not yet exist, and that's good...
In my opinion, the approach consists of first simulating on a kinematic and static diagram (with Meca3D as far as I am concerned), in order to find the optimal arrangement with regard to the constraints imposed: stroke, speed, weight of the load... And pre-size the cylinder, once its stroke and the effort it must exert are known.
The work on a 3D model comes next, to possibly take into account the masses and inertia of the different parts, and to deal with the problem dynamically. Then comes the FE simulation to size the different elements...

Illustration of a quick study with "a bit random" dimensions, which gives the results below.
Data :
- Length of the lifting arms: 760 mm
- Table width: 800 mm
- Lifting stroke: 520 mm, at 13 mm/s speed, with soft start and stop
- Maximum load on the table:   18700 N
Results :
- Visible movement on the animation attachment. It is clear that the cylinder I have defined is not suitable in terms of travel. After modifying its attachment points with the frame and the lifting arm, the restarted simulation will give its verdict. The goal is to search for "the right geometric and kinematic layout" by iterations.

- Displacement (in red) and speed of the cylinder (in green)
The cylinder stroke is 210 mm.
The speed imposed on the table is 13 mm/s, with progressive start and stop phases over 10 s.
The maximum speed of the cylinder is 6mm/s. The simulation gives the speed law of the cylinder to obtain the expected law of the table.

- Cylinder force (in green), for a load of 18700 N on the table (in red):
The maximum value is 108 kN, in the lower position of the table, i.e. more than 6 times the load to be lifted.
The position of the load on the plate has no effect on the action of the cylinder.
It is clear that the provision I have adopted is not appropriate. This is a first draft, only the designer of the table is able to make the appropriate choices.

Good luck, at your disposal if you want to dig deeper into the question

M.Blt


tableelevatrice.mp4
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I would know how to solve this problem on paper, I want to know how to do it on SolidWorks because I think it's something that will come up often and even for later it can always be useful for me to know how to do it.

I made a very simplified 3D model, just to have an order of magnitude of the race to see what we find on the market. 

In fact, the elevator will be placed on a workbench, and on the elevator there will be a special support for a part that weighs 1.8 tons and requires precise height adjustment. This is a project of the company. 

I'm not clear in my explanations since I'm in the middle of it, I can't take a step back to be clearer for those who don't know this project. 

Thank you for your answer

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Hello @m.blt ,

That's exactly what I was looking to know! Thank you very much for taking the time to answer me!

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Hello @davit27srg

Happy to see you got the best answer.

As the subject is closed and quite fazit between us and serious  people I can tell you that I am not quite in line with the calculations.

What for!

1°) The overall dimensions of the table are 800 mm (we assume that this is the length) which means that if we take articulated bars right in the middle they can only be 800 mm. Less since the axles are inevitably inside the 800 mm zone not to be exceeded.

2°) we don't know if it is a hydraulic cylinder that will be used. You have to add a certain distance in addition to the useful run
3°) we don't know the low position of the table, I started 300 mm from the ground (to avoid too much effort at the start)

4°) we do not know the position of the rear cage of the cylinder in relation to the lower axis of one of the arms

5°) we do not know the position of the rod at the top of the arm. (This is decisive  for the race and the force to be exerted to lift the load.

6°) we know the height between the low and high position 0 to 560 mm

BRIEF!

a simple sketch under SW shows that it is not possible to have 560 mm of travel with a single scissor table but that it is necessary to switch to a double scissors at least.
Note that the simplest would be to have a horizontal cylinder located at the height of the low axes and even there the stroke must be  at least 425 mm but at least the overall length has a chance to pass (stroke + minimum space)
The cylinder for a simple scissors is not long enough for 560 mm because the size of the cylinder = Stroke + 100 to 200 mm of space, i.e. a minimum of 800 mm which does not fit in the table in the low position etc... etc....

With a double chisel and the cylinder placed correctly it is possible with a fairly low stroke but with a significant force of the cylinder.

For the necessary force it is even worse at the beginning it is at least 119 kN and after 300 mm it is ) 290 kN for 19kN of load to be lifted.

Moral : the stroke of the cylinder indicated by the first applicant is apparently wrong (it is indicated this:  after some research, I would need a cylinder stroke of 190-200mm.) You need a minimum of 345 but in reality more.

Between us , a table that lifts 1.8 tons and is only 200 mm wide, there will be deaths among the operators, especially since the table is itself on a trolley.

Making a double scissor table with an authorized width of only 200 mm is going to be sporty.

My point of view is that doing a static calculation without having the dimensions of a functional kinematics is only useful for sure from bad calculations, especially with single or double scissor tables.

Note that Solidworks motion allows you to know the forces and that Solidworks Simulation allows you to know the max stresses in the members and axes and that Solidworks CAD allows you to make a kinematics in a few minutes that holds up. A simple excel sheet does practically the same thing (except for precise RDM)

Kind regards

PS: The calculations I make are based on fully configurable excel sheets! In fact, single scissors or double scissors changes a lot of things ;:-)

1 Like
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Hello @Zozo_mp,

Thank you for your answer, I take note, I will look for an effective solution today. 

Kind regards

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Let's not confuse the pianist and the piano...

Hello @Zozo_mp 

I would like to respond to your message to our colleague @davit27srg , following his request for help.
I read this message very carefully, where you indicate in the introduction that you are "not quite in line with the calculations". I imagine that you are referring to the mechanical simulation that I proposed in response to this colleague's request.

Davit21srg's question concerns the determination of an effort in a system that he describes briefly by limiting himself to a few basic data: running, speed, load... My answer consists in a possibility of satisfying this request to calculate an effort (and the associated movement). The simulation uses a diagram whose principle is similar to that of the illustration that accompanies the question. In the absence of dimensions and precise specifications, I am only proposing avenues of analysis, and in no way a validation of the envisaged design.
Perhaps I am wrong to illustrate in detail the approaches I propose, on the basis of diagrams and other curves, imagining that it is an aid to understanding. But I have the idea that simply indicating "do a mechanical simulation" will not be of much help.

I have also taken care on several occasions to express reservations: "the cylinder [] is not adapted in terms of travel", "look for [] the right geometric and kinematic layout", or "It is clear that the layout I have chosen is not suitable".
The simulation that I propose is one tool among others whose purpose is to evaluate certain physical quantities of a system, quantities on which we can rely to validate - or reject - a preliminary design project. It is up to the person in charge of this project, in this case Davit27srg, to make this decision before proceeding with the design of this system.

On the other hand, I fully agree with your remarks about the lack of data to design the mechanism. It is clear that installing a screw jack capable of an effort of nearly 100 kN will not be an easy task in the space allocated. However, it is necessary to be able to know the value of this effort in order to become aware of the problem...

Later in your message, you yourself make a critical analysis of the system envisaged, then proposals for evolution (from single scissors to double scissors) based on results from a SolidWorks sketch or an Excel spreadsheet spreadsheet. Do you see a difference in nature between these tools and kinematic/static analysis with Meca3D? In all cases, these are simulations, to arrive at numerical values that determine the decision. The calculations themselves are not in question...

Davit27srg indicates that he is learning, and that he is looking to progress. He can discover two aspects of mechanical design through our respective answers: the "computational" dimension in the first place, regardless of the simulation tool used, and the ability to analyze and criticize the solutions, and to decide on the basis of, among other elements, the results of these simulations.

And to conclude, completely in line with the quality you attribute to us of "serious people"...
Kind regards,

M. Blt

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Hello @m.blt

I think we agree and don't think I'm criticizing meca3D, nor your approach ;-)

I especially wanted to insist, but as you have understood, on the fact that, given the very limited volume imposed, we could not start looking for a cylinder without having a functional kinematics.
A mini kinematics with a simple sketch shows that the simple scissors cannot work and this without calculations, hence the proposal of a double scissor table (and not a single scissor as in the image of the requester).

In the attached file I have put two possible positions for the cylinder and for me only the cylinder at the bottom horizontally presents the required conditions (stroke and size). However I didn't look if it went wide (I have doubts)
Personally I always proceed in order: fully functional kinematics and after calculations with sometimes with iterations if the calculations show a need for resizing.

If you were kind enough to do the calculations in mecha3D from this mini model attached, it would be nice and instructive.

I put the image because the PART file is in V 2019 that not everyone has

Kind regards

 

 


cinematique_table_elevatrice_encombrement_restreint.sldprt
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Hello @m.blt ,

I have Mecha 3D, the last time I used it was 4 years ago, could you tell me how you made a 2D simulation please? I only manage to get the 3D mecha bar when I'm in assembly so I have to have 3d parts right? 

Thanks in advance 

Kind regards

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@Zozo_mp , @davit27srg ...

Back to lifting with scissor table...
I just did a quick simulation on the same basis as my first answer from the point of view of the load and its motion, but with the kinematics you propose, with double scissors.
Some key findings:
From the point of view of the load, its speed and its displacement:
The speed (in green) follows a law imposed in trapezoid with a maximum value of 13 mm/s. The stroke (in pink) is 510 mm.

From the point of view of cylinder kinematics : speed (in green), maximum value of 3.6 mm/s. Displacement (in pink), total stroke 103 mm.

From the point of view of the actuator force (in green): maximum force in the lower position, 175 kN, i.e. about 10 times the value of the load (negative forces given the reference used for the actuator). The charge appears in pink for comparison.
The x-axis is at the top of the graph.

I took advantage of the diagram to look at the solution with an "oblique" cylinder. The stroke of the cylinder is 274 mm, its maximum speed is 7.5 mm/s.
From the point of view of effort, the pace is more regular, and the maximum value is about 60 kN (here, it is positive!).
Not really surprising, the race is 2.5 times longer...
The ratio of cylinder force to load is more favorable in this configuration. All that remains is to install a cylinder with this capacity.

All this in an ideal world without friction, disturbances, or deformations...
Have a good day everyone,

M.Blt

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