I have a basic question. Imagine a plate, with a hole drilled in each corner. Imagine 4 shafts in these 4 holes,attached to another plate.

Intuition tells me that it would take considerable effort to have the design as such that the plate slides without getting stuck.

Let's now replace one of those shafts by a leadscrew. Designs exist where the leads crew would allow to smoothly upper and lower said plate.

My intuition, however, indicates that I could never get such a conceptually simple design to work. Even with plain bearings.

What are the conditions to get this right?

  • $\begingroup$ Does your question refer to an actuation problem, as in how many actuators to use to control a mechanism, in your case a plate and how to do that with redundant actuation or does it refer to manufacturing quality and tolerancing? $\endgroup$ – 50k4 Jan 15 '18 at 20:49
  • $\begingroup$ that is lead screw not leads crew $\endgroup$ – jsotola Jan 16 '18 at 0:15
  • 1
    $\begingroup$ use cables and pulleys to move the other corners of the plate $\endgroup$ – jsotola Jan 16 '18 at 0:16

This is a great problem to hone mechanism design skills on.

There is a set of dimensions for which this mechanism works fine.

The condition for this is that the contact of each rod & hole, including the drive screw, remain on the correct side of the friction cone, for all free motion and bending of the elements.

You can start thinking about this by thinking of a single hole and rod. A thin plate with a large hole will jam. A thick plate with a small hole will slide.

A starting point for a basic model(equation) of this is to treat the contact between the rods and holes as flat surface contacts when estimating the friction, as well as assuming the hole edge loads do not start deforming material. You can resolve the external loads into normal contact forces and then compare the driving force (from your screw) to the friction force.

Below is a simple representation of the forces at one hole. Single Holes 2D Forces

Force $D$ is the driving, actuation force at some distance away from the hole. Force $C$ is the gravitational (and if significant, inertial) force. Forces $A$ and $B$ are the reaction forces of the edge of the hole on the rod.

For our simplified model, the basic rigid body assumptions are used. Capitals are vectors and $\times$ is the cross product. $F$ are forces, $P$ are positions.

$\sum {forces} = m a$ or $F_A + F_B + F_C + F_D = m a$

$\sum {moments} = I \alpha$ or $P_A\times F_A + P_B\times F_B + P_C\times F_C + P_D\times F_D = I \alpha$

where $\alpha = 0$ since we are sliding and not rotating.

Forces $A$ and $B$ have two components. The normal forces needed to balance the moments, and the axial forces that are calculated using a basic friction model.

$F_\mu = \mu F_{normal}$ where $\mu$ is the friction coefficient.

Keep in mind the friction model is discontinuous, and that the friction force component will equal $F_{D_z}+F_{C_z}$ until $F_{D_z}+F_{C_z}$ exceeds $\sum F_\mu$

Solving the above equations for $a$ is, I believe, the most basic equation that will let you estimate things.

If you need to be more accurate, or aggressive in your design, then you need to worry about the bearing loads and where plastic deformation will start destroying your mechanism.


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