Scissor Lift Analysis

[BikeDaily]

Greetings,

I am analyzing a "stack table" with a scissor lifting mechanism. I am having difficulties when trying to analyze the structure with the hydraulic cylinder. The cylinder is fixed on the bottom and is pinned on the top. I have seen many examples in text books of scissor lifts and similar apparatus, and follow them quite well. But many don't have a cylinder, or if they do it is in a more convenient place such as on the bottom sliding joint. My case has me stumped.

Do I need to analyze the structure without the cylinder first?

I looked at just doing a FBD of the top beam that supports the external loads. I can solve this pretty easily. But I start having trouble when I take that through the scissor arms.

I have attached a sketch of an undimensioned model of the stack table.

Any and all thoughts are greatly appreciated.

Thanks,
CycleDaily

 

[Cockroach]

There are several issues with your design. Most obvious, the scissors do nothing for you since the cylinder does all the work. You need to put that cylinder horizontal between the Rollers and fixed anchor points.

But you would analyze this as a truss. Move the lift an incremental amount and hold; then find the forces on each member and the bearing loads at the pin in the middle.

Note the cylinder has two nodes, one at each end, so you need to look at forces in those connections also.

If you pick five positions between start and stop, equi-spaced, you can find the maximum point of stress to which you design the lift. All of the analysis is similar to that of a truss or structural elements with a dead load shown. Such a distributed load can be reduced to a force acting through the centroid of area.

Good luck with it.

Regards,
Cockroach

 

[fegenbush]

This is a good example of a statically indeterminate structure. Take the cylinder load as redundant and use one of the many methods for solving indeterminate systems. If doing this by hand, I would use the virtual work method or the Hardy Cross method. In reality, I would model it in my structural program and check the results.

 

[rb1957]

i see the scissors as doing two jobs ... keeping the bed flat, and reacting the imbalance load (as the load applied to the bed is not directly above the actuator).

the load in the two diagonal scissor legs will be the same (no? to keep sumFx =0)
so now you have two unknowns (actuator load and scissor leg load) reacting the applied load (two equations of equilibirium)

QED ...

 

[BikeDaily]

I understand there is no mechanical advantage when using the cylinder. (Which is quite often the reason for using a cylinder) The scissors are just to keep the table level while moving up and down through the stroke of the table. The scissors are especially important because the cylinder cannot be placed in the middle of the table and the load on the table can vary in location and magnitude.
I believe the example I have shown to be the most extreme, worst case scenario.

In this example I am concerned with the possibility of a side load (X direction) and moment being transferred through the cylinder. This is an especially large concern if the cylinder is fixed on one end, not pinned but actually fixed like I show.

Are my thoughts correct in thinking; in doing the design of the hydraulic circuit, it would be wise to have as low as pressure as possible. If not, unnecessary stress could be delivered through the scissors and in the horizontal direction of the cylinder. I also understand that when actually extending the cylinder, the pressure will not exceed the minimum amount required for it to move.

When solving this, am I able to use the hydraulic force for F(1) or do I need to solve for this force to determine cylinder size and pressure?

 

[rb1957]

why fix the cyclinder base ? why not pin both ends ?

 

[hydroman247]

^ +1 This is the standard for production scissor lifts.

 

[rb1957]

as i've posted, i think the bed is statically determinate (if you assume no horizontal force on the cyclinder and assuming the legs are axially loaded).
P1 = vertical load in scissor leg
P2 = load in actuator

2*P1+P2 = w1*D+w2*(F-D)
P1*(E-(A+B))-P1*(B-A) = w2*(F-D)*((F+D)/2-(A+B)) -w1*D*((A+B)-D/2)
P1 = [w2*(F-D)*((F+D)/2-(A+B)) -w1*D*((A+B)-D/2)]/[(E-(A+B))-(B-A)]

or something like ...

 

[Walterke]

^this
don't fix the base. Preferably, use some kind of guiding system to make sure the table can only move up/down.

I'm assuming w1 and w2 are given, so you should use these to calculate the minimum required F1 (lift force). From this, you can calculate what kind of cylinder (size+pressure) you need. Be sure to calculate the shaft for buckling as well.

You say:

the pressure will not exceed the minimum amount required for it to move.

This is false. Your pressure difference will determine the flow rate of your oil and thus the speed at which the table will go up/down.
Kinda forgot exactly how to calculate this, so you'll have to look it up (or someone here will help/correct me)

NX 7.5.5.4 with Teamcenter 8 on win7 64
Intel Xeon @3.2GHz
8GB RAM
Nvidia Quadro 2000

 

[rb1957]

thinking about it (for a minute), i guess the loads in the two scissor legs aren't the same, so that the legs are bending ... their vertical load is determined by sumFy and sum Mz and their horizontal loads would oppose each other, presumably something like the average of the the horizontal components of the two leg loads.

without horizontal load reacted by the cyclinder i guess it's a singly redundant structure (as above, use the cyclinder load as the redundancy, and solve by any of the regular methods).

but then ... looking at the free body of the bed and the two legs above the pivot point, what takes out the off-set moment between the load and the reaction (at the scissor pivot) ? bending of the legs ??

 

[BikeDaily]

One end of the cylinder is fixed because of the purchased cylinder and the size of the cylinder. The cylinder is fixed, not on the cap end, but on the "head" side of the cylinder. The actual cylinder is 60 in long and hangs below the table in a pit about 30 in. It is far easier to attach to the "head" end of the cylinder because it is geometrically within the table. This is a design that is currently in operation, therefore changing the design and orientation on the cylinder is not cost effective. Newer assemblies have a different scissor system. Long story short, this is the hand I have been dealt.

 

[Cockroach]

Your analysis would go something like the enclosed attachment. You need to make changes to allow for your specific design geometry, the mathematics would change for allowance of the hinge pin, as I noted in the model. But basically the same type of thing.

Regards,
Cockroach

 

[desertfox]

Hi BikeDaily

If this is the design currently in operation where are the analysis for them?
If you get yor hands on a previous calculation for this set up it should help you in your quest.

Regards

desertfox

 

[BikeDaily]

This was designed many years/decades ago, not engineered. So calculations don't exist. I'm sure the design process consisted of a few blokes flipping a coin on if the scissors are strong enough. Now the customer wants to increase their load carrying capability and I have to decide if it is strong enough.

 

[rb1957]

draw a free body of the bed ...
if two of the legs of the scissor legs are on rollers, then these can only react vertical (Y-) load.
the other leg can react X- and Y- loads, so can the cyclinder (being fixed at the base.

5 reactions, doubly redundant; but X- at the leg attmt is reacted by X- at the cyclinder, so singly redundant,
three Y- reactions for the applied loads;
but these X- loads are indeterminate from the bed FBD. i'd look at the scissor legs, applying and X- load at the LH attmt and seeing how the legs want to react this;
the cyclinder being unattached to the rest of the structure takes it's X- reaction down to the ground (together with it's Y- load).

 

[zekeman]

The first equation is incorrect. M1 is a moment at the base and has nothing to do with the equilibrium of the structure. F1 has to be a vertical force at the pin.
Finally, F1 is NOT known. If you make it an unknown you now have perfectly determinate problem.
Just remove M1 from the first equation and you have it.

 

[zekeman]

My bad,
Disregard my post. I saw the 1st equation was incorrect but the rest of my comments were wrong.
So, if M1 is not in eq 1, then I see 2 eq and 3 unknowns so far
Unknowns F1,Fe0 and Fa0
Need one more eq,energy
F1dy=w1(D)dy+w2(F-D)dy
Therefore
F1=W1D+w2(F-D)
Is this OK?

 

[rb1957]

isn't that just sumFy ? (the RHS looks like the applied load)
the RHS looks like PE of the load, but the LHS doesn't look like SE of the structure ?

 

[zekeman]

"isn't that just sumFy ?"
No, look at it again; sumFy would have a few other entries

"the RHS looks like PE of the load, but the LHS doesn't look like SE of the structure ? "

If a small man pushes up with a force F1 through a distance dy on the structure he puts in F1dy and this manifests itself into an increase of PE = to the RHS. Simple conservation of energy. Not SE.

 

[rb1957]

would not a change in bed height change the internal strain energy of the scissor legs ?
the RHS is PE of the applied loads, ok. The LHS is only the work done by the cyclinder force, and so IMHO is incomplete.