Scissor Lift Design
Scissor Lift Design
(OP)
Hi I'm just getting started on a scissor lift design and I'm trying to figure the reactions via free body diagram. Let me say that I'm no engineer so this isn't second nature to me. If anyone is kind enough to help a novice here is a link to a FBD that I drew up.
http://hosting.picgoo.com/uploads7/fbd_copy2.gif
(you may have to paste that in your browser navigation bar for it to work)
I feel like I'm missing a lot but I'm having a hard time picturing where the other reactions would be (if any).
All the connections are pinned, hinged at the left, rollers at the right side. I'm really interested in the load on member b-c which (in this case is a lead screw) I'm just treating like a pinned member. If anyone can offer advice assistance on the FBD that would be great, if you have any further input it would be much appreciated.
GNE
http://hosting.picgoo.com/uploads7/fbd_copy2.gif
(you may have to paste that in your browser navigation bar for it to work)
I feel like I'm missing a lot but I'm having a hard time picturing where the other reactions would be (if any).
All the connections are pinned, hinged at the left, rollers at the right side. I'm really interested in the load on member b-c which (in this case is a lead screw) I'm just treating like a pinned member. If anyone can offer advice assistance on the FBD that would be great, if you have any further input it would be much appreciated.
GNE





RE: Scissor Lift Design
Work=distance*force
Work.out = work.in ;(Assume 100% efficiency)
In upper/middle/lower position determine how much
the load moves (in mm) if the leadscrew(b-c) moves 1 mm.
Call the largest of these Qmax. ( you can calculate
it with trig, or make a model out of sticks )
LOAD*Qmax[mm] = FORCE.bc*1 [mm] ; LOAD,FORCE in same units
e.g. pound or kg. So at B-C the force is Qmax times larger
and the distance moved is Qmax times LESS.
<nbucska@pcperipherals DOT com> subj: eng-tips
RE: Scissor Lift Design
F=W/al*sin(2*theta1)
where l=length of the screw, F= screw compressive force and W is the vertical load placed anywhere on the platform.
Note that the solution is independent of the position of the load, the length of the scissor links, and theta2.
I used theta1 and theta2 since I haven't figured out how I can put in Greek letters.
If you are interested in the derivation (it involves some calculus) I can furnish it on request.
SK, consulting engineer
RE: Scissor Lift Design
RE: Scissor Lift Design
F=WL/2a(sin(thet1))
where I used the cap L for clarity instead of l which looks like a one and the argument in the sine function is in fact thet1, not 2thet1.