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
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
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
RE: Scissor Lift Analysis
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
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
RE: Scissor Lift Analysis
RE: Scissor Lift Analysis
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
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:
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)
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RE: Scissor Lift Analysis
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
RE: Scissor Lift Analysis
Regards,
Cockroach
RE: Scissor Lift Analysis
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
RE: Scissor Lift Analysis
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
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
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
the RHS looks like PE of the load, but the LHS doesn't look like SE of the structure ?
RE: Scissor Lift Analysis
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
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
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
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
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
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
Let the jury decide.
I'm done.
RE: Scissor Lift Analysis
OK, so where is your solution?
RE: Scissor Lift Analysis
RE: Scissor Lift Analysis
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