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Scissor Lift Analysis

Scissor Lift Analysis

Scissor Lift Analysis

(OP)
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

RE: Scissor Lift Analysis

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

RE: Scissor Lift Analysis

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.

RE: Scissor Lift Analysis

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 ...

RE: Scissor Lift Analysis

(OP)
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?

RE: Scissor Lift Analysis

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

RE: Scissor Lift Analysis

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

RE: Scissor Lift Analysis

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 ...

RE: Scissor Lift Analysis

^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:

Quote:

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

RE: Scissor Lift Analysis

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 ??

RE: Scissor Lift Analysis

(OP)
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.

RE: Scissor Lift Analysis

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

RE: Scissor Lift Analysis

(OP)
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.

RE: Scissor Lift Analysis

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).

RE: Scissor Lift Analysis

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.

RE: Scissor Lift Analysis

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?

RE: Scissor Lift Analysis

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 ?

RE: Scissor Lift Analysis

"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.

RE: Scissor Lift Analysis

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.

RE: Scissor Lift Analysis

"would not a change in bed height change the internal strain energy of the scissor legs ?"

Only second order


OK, you don't like this approach then, lets look at the legs. It is not difficult to see that each leg must have only vertical forces at the ends and they must be equal. Further,each puts out double that force on the scissor pin and since the pin is in equilibrium, the forces on both legs are equal.

Now, look at the top plate.
The two equal forces on the plate are opposite so the statememt
F1=sum of the weight loads is as I suggested.

This then solves the entire problem.
At each position of the lift F1 remains constant= to w1*l1+w2*l2 and the couple
M=F1*Lcm (Lcm= distance between cylinder and CM of weight load)
must be equal to the restraining couple from the legs or
Fq*(E-A)=F1*Lcm
Fq=F1*Lcm/(E-A)
E is variable
Fq is the solution for both legs at both ends. There are no horizontal forces at the leg ends




RE: Scissor Lift Analysis

there are no horizontal forces on the legs with rollers, sure. i don't know that you can say that for the pinned ends, given that the cyclinder is a cantilever.

the vertical forces (three of them) are easy enough to calc (bed as a free-body), not so sure about the scissor mechanism (i can see X- load at the bed being reacted at the scissor central pin, and so to the LH leg ground. the problem i see with the X- loads is that you can't determine them from the FBD of the bed (they're equal and opposite) and in the rest of the structure the legs react one X- force and it's reaction is carried by the cyclinder (so at ground level you'll have the cyclinder reactions (X- force and corresponding moment) being reacted by loads from the scissor legs (X- force and Y- couple).

looking quickly at the scissor legs, i see the Y- couple being reacted by and X- couple (applied at the pinned end and at the central pin), each scissor leg is a three force member (so the three forces have to intersect at a point).

RE: Scissor Lift Analysis

Technically you are right about the moment on the cylinder, but as a practical matter, you don't design the cylindrical mount to transmit a moment through its piston rod.

So, if we can agree to solve it as a pinned connection at the base, maybe this problem has the solution I suggested.This is not perfect, but not bad.

Otherwise, you will attempt to solve an indeterminate problem with assumptions that will yield no useful answer. The main assumption would be that that at limited angular movement the piston can exert a moment on the rod-- very problematical.

Basically I am modeling it as a piston moving vertically with a pinned connection between the rod and piston ( representing the inherent angular slop over limited angular motion the rod). This the same as pinning the base.




RE: Scissor Lift Analysis

if i look at a scissor leg as a free body, there are three forces (at the ends and at the central pivot). since the roller allows only vertical force, the other two pints must both have horizontal forces. This also follows from consideration of the couple of Y- forces at the two ends (ie there needs to be a couple in googly eyes.

so there is an X- force onto the bed (at the pinned connection), agreed?

then in consideration of the bed as a free body, there needs to be an X- reaction onto the cyclinder (whether we design for it or not ... if there's slop, then the bed will move slightly untill the slop is taken up, IMHO).

if you have the cyclinder horizontal, not attaching to the bed (as suggested above, and as we can't 'cause this thing is already made/designed) then there'd be only vertical forces onto the bed and the lower triangle of the scissor legs (below the pivot) would have to balance out the actuator load.

and that's the way i see it.

RE: Scissor Lift Analysis

I made my case, Rb made his.
Let the jury decide.
I'm done.

RE: Scissor Lift Analysis

Oh, I forgot to ask.
OK, so where is your solution?

RE: Scissor Lift Analysis

You can look at the friction of the rollers as being resisted by the end pins of the scissor member and piston end pin connection. Make an educated judgement by assigning half of the friction being resisted by the piston pinned connection to calculate the resisting moment at the base of the piston.

RE: Scissor Lift Analysis

I think Zekeman is correct in that the scissor mechanism ensures that the table and load always remain parallel with the ground when being lifted, therefore the hydraulic cylinder can be assumed to be pinned at both ends because the scissor prevents the whole thing from falling over.

There are no horizontally applied forces (only vertical), therefore all pin and slider reaction forces are vertical.

To calculate the lifting table pin and slider forces it is simply a matter of taking moments about the top LH scissor pin to calculate the top RH slider reaction, then by the sum of the vertical forces you can calculate the top LH pin reaction. The ground reactions are of equal value but opposite sign to these.

The cylinder force required is equal to the total load to be lifted (W1 + W2), plus the lifting table weight, plus 1/2 the weight of the scissor mechanism. The other 1/2 of the scissor mechanism is supported on the ground.

Regards, Ned

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