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

 

[zekeman]

"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

 

[rb1957]

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).

 

[zekeman]

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.

 

[rb1957]

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 X-).

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.

 

[zekeman]

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

 

[zekeman]

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

 

[chicopee]

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.

 

[Neilmo]

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