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Context: I'm designing a linkage mechanism as part of a university robotics design project. From previous analysis, I have determined the forces and torques applied to each member at all points in time throughout its cycle of operation.

For instance, here's the Magnitude of the forces acting on one member throughout a 60-second cycle (note: I'm not including the equations that describe these forces because they're very very long, and hopefully irrelevant to my question):

Magnitude of Forces

Magnitude of Forces acting on an example member, at each point in time.

Problem Since I know the forces at each pivot connection, I can resolve them (and the weight) parallel and perpendicular to each beam they act on. Designing for the total compressive/tensile force is therefore relatively simple.

Now I want to be able to design the members such that they can withstand the bending forces experienced during operation.

Current Attempt: I have resolved each force parallel and perpendicular to the beams it acts on, and was hoping to input the perpendicular forces into a shear force and bending moment analysis along the length of each member, as if it were a normal beam. However, as the linkage members are experiencing constant rotational and translational acceleration, they clearly aren't in static equilibrium like a typical structural beam! As such, when I plot the forces from one end to another, I end up with a big unbalanced discontinuity right at the end (rather than returning to zero as you'd usually expect):

Shear Force & Bending Moment Animation

Shear Force (Red) and Bending Moment (Blue) for an example member (same one as the force diagram above, where x is the length from A to C) in the linkage mechanism, animated through its operation cycle.

If I understand correctly, this implies that the internal forces due to applied forces at one end of the linkage don't propagate at all back down its length, which can't be accurate. So clearly I'm missing something...

Question: Is there any merit in this approach at all, or does the Shear Force and Bending Moment analysis I learned in 1st Year Structures fail as soon as something isn't in equilibrium? I expect this is the case, so what would the correct method be?

To be clear, I'm not asking anyone to do this analysis for me! I've just so far been unable to find any useful information on this aspect of linkage design, which really surprises me, so I can only assume that I'm missing something like the correct search term...

Thanks in advance for your help/advice!

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  • $\begingroup$ Is the robot moving fast. I mean if the robot is moving slowish then you can assme its static at each instance, since the Innertia part of the equation is small. Once it starts to dominate then i would do a flexible multibidy simulation because the analysis gets complicated really fast because the direction of the innertial forces may vary from timestep to timestep. Time to bust out your flexible multibody simulator. $\endgroup$
    – joojaa
    Apr 5 at 2:40
  • $\begingroup$ @joojaa thanks for your response. We do actually plan to eventually run some FEA our final CAD model, however the academics running this project have said they expect some bespoke maths and physics to back up any computer simulations we present. The linkage movement isn't exactly rapid, but as it'll be quite a substantial piece of machinery a lot of the forces involved are due to accelerating it between positions... but a static analysis will be our fall-back option failing anything else! $\endgroup$ Apr 5 at 8:56

1 Answer 1

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Yes, you are right. When there is an imbalance in the forces, the link will start to accelerate, or rotate.

Say we have a simply supported link, or beam on a pin-roller support. then the equilibrium equation becomes,

$$\Sigma M_{left}=0 \quad \Sigma P_n*x_n\downarrow+ R_{left}*L\uparrow = Torque $$

  • p = Force
  • R vertical reaction of support

$$T=\frac{dL}{dt}$$

  • T = torque
  • L= angular momentum = I$\omega$
  • I =moment of inertia of the link..
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  • $\begingroup$ Thanks for your answer - I've got some clarification questions if you don't mind. Do I understand correctly that "left" there means to the left of a cut made arbitrarily along the link/"beam"? Similarly, are P_n the forces applied to the link before that cut? I assume the moment of inertia must also be taken about the leftmost end of the beam? Secondly, does Torque refer to what's applied to the link, or is that the Bending Moment experienced by the link at the cut (and so the function plotted on a Bending Moment Diagram)? Is there a similar relationship for Shear Force? Cheers! $\endgroup$ Apr 5 at 8:43
  • $\begingroup$ @RobertGregson, no in this case there's no cut, it is the entire length of the member. the link is initially horizontal, with no axial load considered, Pn is any vertical load or shear applied at distance x. that may be even the right-hand joint shear imparted by the next member. This moment summation in static cases is zero, but if it does not add up to zero it will become a torque applied counterclockwise to the beam . that will cause the beam to rotate. however, we have isolated this link and ignored the change in forces from the adjacent links after the start of the rotation. $\endgroup$
    – kamran
    Apr 5 at 15:54
  • $\begingroup$ Ah ok, thank you for the clarification! If I understand correctly, this is actually the inverse of the analysis I've already completed, then - as I derived a force and moment balance for all the links, including the rate of change of translational and rotational momentum respectively, and solved these simultaneously to find the forces acting on each member such that they perform the desired motion (as illustrated in my first graph). What I was hoping to determine now is the internal stresses along the length of a member when it's subjected to these known, but unbalanced, forces? $\endgroup$ Apr 6 at 10:34
  • $\begingroup$ @RobertGregson you can do that by calculating the momentary balance of forces and reactions as if it were in equilibrium. You may draw the shear diagram, area of which is the moment, and calculate reactions. then compare these reactions to what you have the differents divided by L is the torque! $\endgroup$
    – kamran
    Apr 6 at 19:39

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