Manipulator Torque (A Bit Scary for Us)

[Psychotechnoid]

After doing some calculations for my team's manipulator, I realized that a particular shaft needs to be able to take about 190 ft-lbs. of torque. While we have a motor and sprocket ratios that can handle it, we can't quite figure out how to set up the shaft itself and the hubs attached to it. We intend to use the Taigene van door motor mounted to a 22-tooth IFI sprocket linked to a 48-tooth IFI sprocket on the same shaft as a 9-tooth McMaster sprocket going to a 60-tooth IFI sprocket on a third shaft attached to a lever mechanism.

As stated, the shaft attached to the 60-tooth sprocket needs to be capable of exerting a maximum of about 190 ft-lbs. of torque. My problem is that I have absolutely no idea how to calculate the thickness of the shaft required, what material it should be comprised of, or how to couple the 60-tooth sprocket to it if the shaft must be thicker than .5 inches (since IFI sprockets have a 1.875" bolt circle, AndyMark hubs with a 1.875" bolt circle only go up to .5" bore, and I can't find any other hubs for this sort of application on the internet). Further, the only 9-tooth ANSI #35 roller chain sprockets that I can find with keyed bores use a 3/8" bore (McMaster 2500T11). Maybe this middle shaft really can be just a steel 3/8" diameter one like McMaster 1497K952 (since 190*9/60 is only 28.5 ft-lbs.), but I must admit that I really have no idea.

Perhaps this wouldn't be such a big deal if my school had a machine shop (to simply make really big hubs and widen the bore on the 9-tooth sprocket) or engineering classes (so that we could do these calculations) or my team had any mechanical engineers as mentors (so that they could tell us how to do these things), but as the situation is right now, we're pretty much stuck.

Can someone tell me how big the shafts to take 190 ft-lbs. should be and what material? Does anyone know whether a 3/8" keyed steel shaft would be sufficient for 28.5 ft-lbs. of torque? Is it possible to get something other than an AndyMark hub to translate these forces to the axles? Or would it be better to do a dead-axle design and put bearings in the sprockets instead (somehow linking the tiny 9-tooth one to the huge 48-tooth one)? Can regular ball bearings like the KoP ones even take these sort of loads? My team and I are almost completely out of options, and I hope that this design doesn't turn out to be impossible, too (we first tried a scissor lift... that will never happen again).

[Brad Voracek]

Well, I'm somewhat confused about the set up you are using, but is it something along the lines of this? http://i42.photobucket.com/albums/e3...0112081353.jpg,
at least the front part of that. I think we need a better understanding of what you are doing, but if you see the sprocket attached to the "arm", we ran that six inch sprocket to around a 2 inch one on the keyang motor, and were able to pick up the ball. We just used the plastic caps they give you with those motors, and hard mounted to a keyed half inch axle all the way across...


a good way to mount a lever is maybe attach it straight to the output sprocket, and have an axle to take some load but definitely not all of it, and if you can mounting straight onto the sprocket that helps, and is how we have done arms/levers in the past...


now that I think of it, I don't know how much help I was =x

[Betty_Krocker]

You may want to beef up the steel that you wish to use as an axle if i understand you correctly is what you are asking. Last year our team (2028) had a claw at the end of our design of arm tha only weighed 6 pounds, and we used the denso window motor with a 1 to 4 ratio (i think, at least it was for every 4 turns of the motor there was 1 turn for the pulley at the top) with pulleys and timing belts. Our axels were made out of .5 in solid steel. After about 3 matches, we noticed that we had BENT ALL of our axels! If you can find a better material than steel, i think you would be better off...

just throwing that out there...

[Psychotechnoid]

Actually, we were thinking something like this, but with a claw attached via another shaft to the arm segment so as to be able to "dunk" the ball over the overpass. But that claw's linkage isn't my concern at the moment - I'm more worried about the primary joint (in the upper left corner of the picture).

That is to say, our planned arm is a two-stage articulated system with a total reach of about 8-9 vertical feet (for hurdling purposes), and the linkage in question is the primary joint of the arm (hence the reason for the large torque requirement).

I do believe that mounting the arm directly to the 60-tooth output sprocket is probably the best means of attachment, but I still do not know how big to make the shaft to which said sprocket is attached, or how to attach the sprocket to the shaft (we intend to use a 60-tooth IFI sprocket to attach to this high-torque arm, but the only keyed hubs that I can find are these AndyMark ones, which seem grossly insufficient for this application).

The idea that a 6-pound claw at the end of maybe a 5-foot arm (30 ft-lbs.) bent a .5" solid steel shaft is a bit disturbing - it almost certainly rules out the 3/8" shaft in the middle of our assembly (which must take 28.5 ft-lbs.) unless we can find some funky alloys (or, God forbid, titanium, although if that is the only feasible avenue then we may well consider it).

I guess that I'm just looking for hard numbers here, or some way of calculating them.

[Brad Voracek]

In all actuality, if you attach the arm to the sprocket then you -don't- have to attach the sprocket to the axle, as it can just be free spinning and still work...


As you might be able to see in that picture, that's how they did it, and their sprocket looks free spinning on that axle although it is hard to tell.


If ultimately you do want to attach the axle to the sprocket, then something like that hub is really all I can think of... Sorry I'm not of more help to what you are asking. (and sometimes testing outweighs hard numbers. there can always be something you are forgetting, overcompensating for, etc...)

[RogerR]

i would strongly recomend using the pre-drilled bolt hole circle(s) to transmit the torque. these bolts will be able to handle farily high torque, leaving the shaft to deal with the relatively tame shear loading. also, i'd check the loading specs on the bearing, but you're probably not gonna overload them.

aditionally, i'd also double check your torque calculations; if i recall, the tiagene motor has a torque of 34 Nm, or roughly 25 ft*lbs; with a 60:9 reduction, you'll be stalling (or close to stalling, depending on the rotation direction) the motor. a stalled motor is never a happy motor. a good rule of thumb is to not load the motor more than 25% of its stall torque. also, is 190 ft*lbs correct? i have no idea what your mechanism is, but to generate a 190 ft*lb torque at the end of an 80" lever arm, you'd need almost 30 lbs of load, seems pretty hefty.

[Psychotechnoid]

Wow, I can't believe that I never thought of that! Of course I don't have to attach the sprocket to the axle if the mechanism itself is attached to the sprocket!

Now I just need to find a shaft material that can take the weight (or some way of taking the weight mostly off of the shaft), which will probably be quite high (since the chain on the back side of the 3.5825" radius sprocket will have to exert about 600 lbs. of force to balance the weight at the end of the lever on the other side - we should probably look into doubling up that 60-tooth sprocket and using double-strand chain, since 600 lbs. is almost twice as high as the working load that I recall seeing for ANSI #35 chain, but it should be doable at least). According to the IFI spec sheet, the sprocket can accept a .5" bearing, and if the shaft is really well-supported that should be possible (since there won't be any torque exerted on the shaft itself).

I guess I still need to work out how whether the middle shaft (with the 48-tooth and 9-tooth sprockets) can be attached to said sprockets or whether it will be necessary to somehow rig them together on a dead axle as well, but that seems to be a somewhat easier problem to overcome.

[MrForbes]

bolting the arm to the sprocket should work ok. The Igus 5/8" (from last year) or the new KOP 1" shaft and bushings is nice for making stuff like this pivot.

A 3/8" bolt is usually torqued to about 35 ft lbs when we use them on cars, so it should handle 28 ft lbs just fine, although the key may or may not be up to the task. you can calculate the shearing stress in it by figuring the torque is all transmitted to the key at the outer diameter of the shaft, and the key has a known cross section that must withstand that force.


and ditto on the motor torque, you want to load the motor much less than the stall torque, so it will last. 20% would be nice.

[Psychotechnoid]

Agreed on the bolt holes - mounting the arm directly to the holes in the sprocket seems to be a much better idea that attempting to mount the arm directly to the shaft itself.

I'm sorry if my explanation of the reduction was poorly done: what I meant was this: Taigene coupled to 22-tooth, linked to 48-tooth, coupled to 9-tooth, linked to 60-tooth - total reduction: ~14.5:1, and according to JVN's design calculator (my gracious thanks go to him), perfectly ok (close to the peak of the power curve). Stalled motor = hot motor = dead motor = bad situation.

Also, yes it is 190 ft-lbs. in a worst-case scenario (ball in claw, arm completely extended) - the arm, when fully extended with the claw, measures 108" long. While we cannot fully extend it and stay within the 80" bounding cylinder (the robot will have to do a "dunking" motion), it is important that the physical mechanisms be able to handle the arm being in this position for testing purposes and in case of programming glitches (we might get a penalty, but at least the arm won't break off).

I will definitely check the loading specs on the bearings as well - another point of good advice for which I am very grateful.

P.S. Are master links available for double-strand chain? Would we need a different kind of chain breaker for it? Or should I just use two loops of single-strand (on two 9-tooth sprockets on a keyed 3/8" shaft) and couple the double 60-tooth sprockets really tightly together to ensure even loading? Also, is 28.5 ft-lbs. an acceptable torque on a short 3/8" steel shaft, or will I need to find some sort of exotic material or larger-bored sprockets? Current opinions on the 3/8" shaft appear somewhat discouraging, and I'm wondering if there is any way to obtain perhaps 3/8" keyed titanium or something uber-strong. That shaft, of course, will also have to be supported against the large load from the 60-tooth sprocket linkage, although it will be pulled downward by the linkage to the van door motor as well (that will have to be supported really well, too).

[Psychotechnoid]

Ok, so assuming that I can find a nice alloy and that the key holds (would the key just shear in half if something went wrong?), a well-supported 3/8" shaft should not cause problems. How could we get the IFI sprocket to fit on a 1" shaft, though? The hole in the middle is 1.125" and meant for a .5" bearing, and we don't have access to any machining facilities. I guess that we could use something like McMaster's 5905K68 on the inside of the sprockets, but we would still have to increase the diameter of the hole in the middle by 1/8" (and I'm not sure that would even be safe). Or we could use McMaster bearing 7929K21 on a 5/8 shaft - actually, that seems pretty feasible. Thanks for the advice!

[MrForbes]

there's a formula for this somewhere....

J = pi d^4 /32

max shear stress = T r /J

i figure about 33,000 psi shear stress on a 3/8" shaft with 30 ft lbs torque

[MrForbes]

what's the OD of the 1" igus flanged bushing?

[Psychotechnoid]

Hmm... actually, that Igus bushing might work - it looks like the OD is 1.125 from the picture in the manual (I can't tell right now since I'm not at the team's work area to measure it), in which case no sprocket modification would be necessary. I'm somewhat concerned that the radial load on it (something over 600 lbs.) might be too high, though.

In the formula, what do J, d, T, and r represent (or is that T subscript r)? This appears to be the same formula that I found earlier and was confused about because I didn't know what the variables stood for.

[MrForbes]

we do that just to confuse you! and making the proper greek symbols is beyond my patience level

T = torque on shaft

r = radius of shaft

d = diameter of shaft

J = polar moment of inertia of shaft

And there are some assumptions made when the forumula is derived, so having a mechanics of materials book or a mechanical engineering handbook that explains it would be nice...but that's really 3rd year engineering school stuff....nice to see you're getting a head start

[MrForbes]

look up the igus specs on their online catalog, they list the allowable load, and you might be surprised how strong those little plastic bushings are!