4
$\begingroup$

I'm an EE student by trade, but I've started a project that requires some mechanical analysis, and I've run up against a bit of a wall in my calculations.

The system in question that I'm looking at is an actuator, connected via ball joint to a rod, which is then connected with a ball joint to a platform. During operation, the platform will be constrained to movement in either only the x direction, or only the y direction. I'm trying to find out, based on the force of the actuator and the geometry, what the X and Y components of theses forces will be.

enter image description here

(Apologies for the cartoonish drawing). The actuator is on the bottom left. For the purposes of calculation, I'm assuming that the rod is rigid, has zero mass, and that the actuator does not need to move to apply any amount of force.

I'm using static equilibrium to calculate the X and Y components of the force on the upper right hand side, which will be the X and Y force that my actuator applies to the moving platform.

The actuator will apply an initial force Fa to the rod. Since both supported ends are ball joints, I know that there can be no moments on either end of the rod, only forces acting in-line with it.

In the case of $\theta_M$ being equal to $\theta_R$, the solution is straightforward: all of the force is being applied in-line with the rod, so we set the sum of forces in X and Y equal to zero, and solve.

However I'm having trouble wrapping my head around what happens if $\theta_M \neq \theta_R$.

My naive approach to solving this would be as shown below:

enter image description here

Break down the initial applied force into components that are parallel with the rod, and components that are perpendicular to the rod.

Here's the part I'm unsure about. Since the actuator can't move, the perpendicular force should be cancelled by an equal and opposite normal force of the actuator pushing back against the rod, correct?

However something in the back of my head is telling me that I'm missing something.

Intuitively, if we were to apply the actuator force not in-line with the axis of the rod, I'd think that there would be some kind of lever action present. An example would explain what I'm trying to say best I think. Can I not analyze a system like this statically? Or am I just completely off base?

EDIT: to make the point I'm trying to get at clear, I was able to find a mechanism that outlines what I'm talking about:

enter image description here

If the wheel is driving the slider, won't the output force of the slider be at its peak value when members 2 and 3 are near parallel to one another? I feel like the same effect should be at play in my system above, but I don't know how to mathematically articulate that. Or my intuition about when the force is at peak value could be totally wrong.

$\endgroup$
  • $\begingroup$ If $\theta_m \neq \theta_r$ the rod is no longer in equilibrium, and that's why the static analysis fails. $\endgroup$ – Suba Thomas Jan 17 '17 at 19:00
  • $\begingroup$ Right, so I can't use static force analysis for something like this then? What techniques could I use to potentially determine the output force if I constrain my movement to either only X or only Y? Is my guess about leverage playing a factor in the crank and slider mechanism correct? $\endgroup$ – Platytude Jan 17 '17 at 21:06
  • $\begingroup$ 'Constrain my movement' is ambiguous. Do you mean just one end of the rod or both and what is the orientation of the rod? For example, if the rod is vertical and both ends are constrained to move only in the Y direction then the output force is the same as the input. $\endgroup$ – Suba Thomas Jan 17 '17 at 22:12
  • $\begingroup$ For the sake of the question, if the rod is at θr and the actuator is at θm (arbitrary angles), and the end of the rod is only allowed to move in the Y direction. I'm clear on what happens if the angles all line up, it's when they don't that I'm confused about $\endgroup$ – Platytude Jan 18 '17 at 0:08
1
$\begingroup$

Your equations are mostly correct, but you're missing the tangential load on the actuator:

$$\begin{array}{l} {F_a}_p & = & \text{Force parallel to the actuator arm at the joint} \\ {F_a}_t & = & \text{Force tangent to the actuator arm movement at the joint} \\ F_p & = & \text{Force parallel to the rod} \\ F_t & = & \text{Force perpendicular to the rod} \end{array} $$

$$F_p={F_a}_{p} \cos(\theta_m - \theta_r) - {F_a}_{t}sin(\theta_m - \theta_r)$$ $$0=F_t={F_a}_{p} \sin(\theta_m - \theta_r) + {F_a}_{t}cos(\theta_m - \theta_r)$$

These equations can be rearranged to give: $${F_a}_p=F_p \cos(\theta_m - \theta_r)$$ $${F_a}_t=F_p \sin(\theta_m - \theta_r)$$

Or rearranged again:

$$F_p = {F_a}_t \csc(\theta_m - \theta_r)$$

This does indeed give you a multiplying effect as the difference in angles approaches 0.

$\endgroup$
  • $\begingroup$ Thanks for the answer, especially given it's such an old question :P. I don't think our variable names quite match up, could you specify which forces Fap and Fat are referring to? Something else to note, the "actuators" in question are actually just large DC motors with a lever arm welded to one end of the shaft, so they should be able to take tons and tons of side loading with no problems. $\endgroup$ – Platytude May 31 '17 at 14:54
  • $\begingroup$ @Platytude Ok, I've changed the answer to be for an actuator with a rotating arm rather than a linear actuator. $\endgroup$ – Rick May 31 '17 at 16:19
  • $\begingroup$ Ahhhhhh that makes perfect sense now, I was missing the loading tangent to the actuator direction. Thanks again! It's not quite a rotary actuator, but it's similar to the arm on a servomotor. I'm only rotating it by a couple of degrees so it's "approximately linear" in the range that I care about. (something like this link $\endgroup$ – Platytude May 31 '17 at 19:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.