I have inherited a DIY build and I have some doubts about its design. Have a look at the following diagram: enter image description here

1 is the linear actuator. It's quite beefy 2000 lbs 12" stroke. It is installed in an aluminium channel. 4- Is a carriage, essentially a box section with little rollers on all sides, so it has no play within the channel and moves quite smoothly. This is connected to the actuator's rod. 3. Is a flat steel bar that links to an identical channel (2) and carriage assembly, though this doesn't have an actuator in it.

The system is supposed to lift a max 220 lbs load that is evenly distributed along the bar.

My questions are:

  1. While there don't seem to be any bending forces to the actuator itself, is it a safe design, as far as longevity of the components goes?
  2. is there anything I can do in channel 2 to support the system, e.g. a spring at the bottom, etc. to address any design problems?
  • 2
    $\begingroup$ I think a lot depends on the interface between 1 and 4 and on 4 generally. Can you post fotos? How long is 3? I think it will boil down to how much 4 deforms under load. $\endgroup$
    – mart
    Oct 27, 2020 at 16:49
  • $\begingroup$ The devil is in the details. Please post some photos if possible of the joint between 1 and 4 and also the rail system in 2 and 4. $\endgroup$
    – NMech
    Oct 27, 2020 at 17:28
  • $\begingroup$ As people have said it depends on the details, one thing you could do is if you can power both directions you could put a continuous Force spring on beam number two to offset a portion of the weight. $\endgroup$
    – MadHatter
    Oct 31, 2020 at 17:12

1 Answer 1


As a quick basic estimation approach we annotate the following:

  • The length of bar 3 = L

  • Top and bottom bearing force on each box $F_b$

  • The height and width of boxes 4 and its counterpart on channel 2 H and W. $$ \Sigma M=0 ,\quad 220*L/4 = H/2*4F_b \rightarrow \quad F_b=220*L/8H $$

Now you have to figure if the bearings and their connections are okay to take this load multiplied by a safety factor of say 3.


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