Watched an Abom79 video I’d seen before tonight which featured a parting blade with replaceable inserts. A parting blade doesn’t have a lot of space for holding down an insert; you can’t put a big screw in there, for example. So the insert was wedged into a sort of two-tine fork, which flexes elastically to admit it.
This means the force clamping the insert into the blade, perpendicular to the friction surface, is the same force you need to apply to force the tines apart to insert or remove the insert. To supply this force, there are round holes in the two tines, and an opening tool is supplied, consisting of a spacer between two parallel dowel pins that fit into the holes. One of the dowel pins is machined on an eccentric with a handle to rotate it relative to the spacer, moving it nearer to and farther from the other dowel pin, thus forcing the tines open. It’s sort of similar to circlip pliers in how it engages with the fork, although not in how it works.
This is a really appealing concept for a couple of reasons.
First, if we disregard friction and the compliance of the tool, the mechanical advantage is unlimited; the ratio between the length of the handle and the center-to-center distance of the eccentric provides the mechanical advantage, and the center-to-center distance (eccentric axis offset distance) can be any value down to zero. With an ordinary lever, to get a very short lever arm, you also have to make the lever arm thin, which limits the load it can take; no such limit exists with this eccentric lever, because the eccentric pin can be as large as it needs to be to resist the shear load (and, in the case where the hole is deep, bending loads).
Friction is still an issue: the surface of the dowel pin rotating half a turn in the hole creates a frictional moment opposing the action of the tool, whose lever arm is the radius of the pin, which must grow according to the square root of the fork-opening force in order for the pin not to shear off. So you have an opposing moment of F3/2, which in the usual case will be about as big as the work you’re actually doing, since the radius of the pin is typically a bit larger than the eccentric axis offset distance, and the coefficient of friction is typically a few times smaller than 1. Friction inside the tool may or may not be reduced with ball bearings or similar, but if not, it’s an additional comparable loss. Such losses may be desirable in this context to prevent back-drivability, but they limit its applicability.
Second, when the fork opens and closes, assuming parallel planar clamping surfaces, it doesn’t have any tendency to screw with the position of the insert it’s clamping down on, except in the direction of clamping and the two directions of rotation whose axes aren’t parallel to the direction of clamping. So it restrains the insert in six degrees of freedom, but purely with friction in three of those degrees, permitting any position in those three degrees. This is a nice improvement over things like jam nuts, which have a tendency to put slightly askew the position you were intending to hold steady.
Third, the mechanical advantage becomes infinite as the eccentric rotates to the position where the dowel pins are farthest apart, in the usual toggle-mechanism way, because the lever arm becomes zero.
We can ask, what is the maximum clamping force we can apply with this mechanism? There’s no inherent limit coming from the applied force (we can make the lever you rotate arbitrarily long, so given a fixed point, you can move the Earth with an arbitrarily small force) but there might be a limit coming from the energy, once friction is taken into account.
Also, the compliance of the thing you are clamping may itself provide a minimal energy cost: if under 10kN it elastically compresses 0.1 mm, you need to open the jaws by more than 0.1 mm in order to insert it in its uncompressed state, applying something more than 10kN in the process, probably only a little bit more if the spring clamp (which you are expanding) is more compliant than the thing being clamped. So that would require about a joule, plus frictional losses.
The compliance of the elastic part of the setup (the workpiece, in this case the clamp) can be reduced almost arbitrarily, as long as the compliance of the tool is smaller or at least not too much greater. Given a nominal Young’s modulus of 200 GPa for steel, and considering a 10 mm distance between the points where force is being applied, we can get a compliance of 5 microns per newton stretching a 100-micron-square steel wire, or 50 nm/N stretching (or compressing) a 1-mm-square steel rod, or 500 pm/N stretching (or compressing) something that averages out to a 10 mm block of steel. At this last compliance, 10kN would result in a 5 micron elongation, which is still plenty to switch between contact and non-contact regimes for things like electrical conductivity and so might be enough to clamp and unclamp something. As with brakes and clutches, the smaller the compliance, the smaller the energy that is needed to reach a given clamping force and thus a given stiction force. In this example, reaching 10kN of clamping force would only require 50 mJ plus frictional losses.
(Of course in the geometry described earlier the tension path between the prongs was not straight, increasing compliance, but there are rivet-like clamping geometries where the tension path is straight.)
Larger compliance may be desirable for resistance to shock loads: if the energy barrier to unclamping is 100 joules, shocks are much less likely to result in slippage than if it is 0.05 joules.
In cases where the elastic piece doesn’t have to be planar, the tool can be simplified to a single rigid body consisting of a handle perpendicular to a shaft consisting of two cylindrical sections with parallel axes. The shaft is inserted through a slot in the elastic piece into a round hole in another part of the elastic piece, and rotating the shaft with the handle then moves the round hole perpendicular to the length of the slot and to the shaft’s axis of rotation. This could reasonably be used for things like fasteners, though in the case of one tool to operate many fasteners, it might be cheaper to make the fasteners as simple as possible (like circlips) and put any extra complexity in the tool.
To reduce frictional losses, the actual holes can be mounted in rotating flexures, or the eccentric pin in the tool can be held in a pair of bearings to reduce friction. These bearings are mounted eccentrically inside a larger shaft mounted on its own bearings which is rotated by the handle. It’s possible but maybe not practical to avoid the use of large bearings in this case: the larger shaft can neck down to fit into two small bearings at the non-workpiece end, which are spaced far enough apart to handle the resulting moment.
Another way to reduce frictional losses is to provide some leverage within the flexing workpiece itself. In the case of a fork, you might clamp the workpiece half as far from the bifurcation as the distance from the bifurcation to the holes for the tool. Then the tool only has to apply half as much force as is applied to the thing being clamped, and the tool rotation experiences about half as much friction.
How big are those frictional losses?
Suppose that we are applying, again, 10kN of force with the tool, over a whole half turn, using a handle which cannot be more than 1 m long (the example in the video was about 150 mm long) and to which we can only apply 200 N of force. Perhaps the eccentric pin is made of a bearing bronze such as SAE 660 tin bronze. Its yield tensile strength is given as 125 MPa; its shear strength might be 0.6 of that, 75 MPa. To fail under 10kN of load, then, it needs to be 13 mm in diameter, giving a cross-sectional area of 132 mm2; 20 mm diameter would give a good safety factor. The eccentric axis distance could be as large as 20 mm, or actually even greater since when the lever arm is at its largest, the clip isn’t fully extended yet, so the force is not at its largest. If it’s 20 mm, we can use it to deform the clip by 40 mm, for a total useful work of 200 J.
That bronze is rated as having a frictional coefficient of 0.10, presumably on steel (though in 02001 Purcek et al. measured 0.68 when dry, so maybe 0.10 is with an oil film), so the pin rotating in the workpiece hole ramps up to 1 kN of friction. Half a turn is 62 mm, so we have 31 J of frictional losses rotating in the workpiece, and probably another 31 J of frictional losses where the shaft rotates inside the tool, for a total of 62 J losses, 76% efficiency. Ball bearings or similar could reduce these losses by about an order of magnitude, so they are more like 2% instead of 24%.
Note that this means you have to apply more force to the handle than the 200 N I was calculating with. More like 262 N. Except that the force is going up as the lever arm goes down; at 45 degrees you have 70.7% of the force, 7 kN, and also 70.7% of the lever arm, 14 mm, so only half that 200 N, 100 N, plus 70.7% of the frictional force, another 44 N at the handle, for a total of 144 N. A little lower, at 30 degrees, you have half the force, 5 kN, and 87% of the lever arm, so 43% of the 200 N plus 50% of the frictional force. I should plot this I guess.
This bronze is also rated as having 315 MPa compressive strength, so the hole in which that pin is turning would need to be at least 1.6 mm deep to avoid damaging the surface, maybe more like 3 mm deep.
This is a fairly terrifying tool configuration, though, sort of like garage-door-spring winding but at lower energy and much higher force. If you lose your grip on the handle, it is going to acquire 138 J of kinetic energy.
Suppose we are stretching a less compliant workpiece, so we only need to deform it 4 mm. If we use the same tool, we get about the same efficiency (a little better, actually, since much of the frictional loss is in parts of the circle that do very little real work) but we could alternatively redesign the tool to have an eccentric axis offset of only, say, 2 mm. This would reduce the moment from the workpiece from 200 N m to only 20 N m, but we still have 20 N m of frictional loads because the pin is still 20 mm in diameter. This enables us to reduce the handle length to 200 mm, to which we apply the same 200 N, but now at only 50% efficiency. This tool is no longer backdrivable by the workpiece’s elasticity, since the moment from the workpiece is equal to the moment from friction. (Even the larger configuration wasn’t backdrivable when fully toggled, because the frictional forces are at maximum and the lever arm is zero.)
We could maybe reduce the frictional losses further by using a mostly hardened steel pin with just a surface of bronze on it, enabling us to reduce the pin diameter by a factor of 2 or more without losing shear strength. Tool steels normally have (tensile yield) strengths of 1 GPa or higher.