hi hope you are well in this current climate,
I am doing a project on Catia where i need to design a scissor jack like the one attached. however i wanted to know how i can calculate the dimensions and angles the criteria i have been given is it needs to hold 1.5 ton and have a minimum 30mm lifting capability
Let's see. Breaking this down into its separate components, start with the simplest.
The maximum delta lift distance, from max down to max lifted, is going to be 2L. At its lowest point, the distance from the floor to the vehicle would consist of only the height of the stand and that top block.
The maximum Fs, which would be the maximum tension experienced by the long screw which lifts and lowers the jack, would be F*sin(α), when α = just barely less than 90 degrees. (I suppose, at α = 90 degrees, the weight is held up entirely by the stand and the block, as they touch one another. None of the weight held up by the screw or by the L components.)
As the jack goes from full down to full up, that weight goes from being entirely held up by the screw (sin(90) = 1), to none of it being held up by the screw (sin(0) = 0).
At max lifted position, F1 = F2 = F3 = F4 = 0.5*F. That's because when fully lifted, the weight is held up equally by F1 and F3, on the left, and by F2 and F4, on the right, while at the same time, at that fully elevated position, the long screw holds up none of the weight. (No horizontal component of F, having to be countered by screw tension, when the jack is fully extended upward.)
At just barely above the fully lowered position, the tension of the screw F is countered by F1, F2, F3, and F4, each one of which has to be 0.5*F. In short, except perhaps at the fully lowered position, when the weight is held up only by the stand touching the block, F1 = F2 = F3 = F4 = 0.5*F. Those four members are always being compressed, each by 1/2 of the total weight F. The only difference being that they go from being entirely compressed just be the tension on the screw (at lowest elevation), to being entirely compressed just because they are directly holding up the entire weight F (fully lifted position). And all positions between these extremes.
So, to design the jack, you have to accommodate the maximum conditions each component encounters, plus of course, a healthy safety margin.
F=W+(WA/2) Tangent angle between the scissor arms and the horizontal
F= scissor lift
Kumar, I agree with your nicely brief reply, that the fraction of the weight F held up by the screw, compared with the weight held up vertically by the four brackets (L), is tan(α). So for example, when α = 45 degrees, tan(45) = opposite/adjacent = 1, an equal amount of weight is acting vertically on those four brackets, and horizontally on the screw.
What this misses, though, is the notion that the four brackets L are always being compressed by the same weight F. Because whether they are carrying the vertical component of the weight, or whether they are being compressed by the tension of the screw Fs, ultimately those four brackets are always carrying the weight F.
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