My belt broke - motor's timing pulley isn't secure
- Thread starter swbluto
- Start date
12p3phPMDC
1 kW
- Joined
- Mar 16, 2009
- Messages
- 462
you can look up the shear strength of a roll pin. I can't quote the values off the top of my head...
but they are quite large. If you shear one with that motor, I'll dance like a chicken in the rain.
Yes, wollered is defined as the pulley wobulated on the shaft because the hole is worn out.
I doubt the pulley is hardened....you'll know when you try to drill it. If you can't drill it, it's hardened.
Is it aluminum or steel? (magnetic?)
but they are quite large. If you shear one with that motor, I'll dance like a chicken in the rain.
Yes, wollered is defined as the pulley wobulated on the shaft because the hole is worn out.
I doubt the pulley is hardened....you'll know when you try to drill it. If you can't drill it, it's hardened.
Is it aluminum or steel? (magnetic?)
12p3phPMDC said:
Ahhhh, good to learn new terms. Anyways, it was originally meant for a "flat" motor shaft as the middle of the pulley was D shaped, but I took a round file to to the pulley to make it circled as I didn't want to make drastic alterations to the motor's round shaft. It doesn't surprise me that it's not a perfect circle but I thought there'd be someway to secure it to the shaft so that it doesn't click. So, loctite 638 sounds like a winner? Or just a new pulley? I guess I might find something with a 8mm inner diameter but I don't really want to increase the tooth count too much as the current gear ratio is pretty ideal.
And it is magnetic, so I guess steel?
dontsendbubbamail
10 kW
- Joined
- May 28, 2008
- Messages
- 718
12p3phPMDC
1 kW
- Joined
- Mar 16, 2009
- Messages
- 462
As far a circlip groove:
you could spin up the motor and take slotting disc on a dremel and proceed carefully.
nice thing about pins, is that you need no flat spots, no threads, no glue, no circlips.
ok, I'll shut up about pins now.
you could spin up the motor and take slotting disc on a dremel and proceed carefully.
nice thing about pins, is that you need no flat spots, no threads, no glue, no circlips.
ok, I'll shut up about pins now.
dontsendbubbamail said:
Oh, that's cool. I was checking out the master index, http://www.wttool.com/category-exec/category_id/15714?utm_medium=cpc&utm_source=SiteChampion, and it appears they have one for every mm. So, I could get an 8mm rod if needed (This smaller motor has an 8mm shaft). That's kind of cool.
Anyways, the motor could sustain upto 4 bursting horsepower, but I'm only running at two. I take it small 1/16" roll pins are used for 50+ hp motors?
And you mention no glue. I was thinking the glue would be used to secure it to the shaft so that the pulley would no longer oscillate and click against the shaft - would a straighforward roll pin do that, too? It doesn't seem readily visualizable to me.
12p3phPMDC
1 kW
- Joined
- Mar 16, 2009
- Messages
- 462
I've never seen one on a 50 hp device...but that doesn't mean it's out of the realm of
possibility. I'd say that it's a matter of the material, it's diameter, and the application.
My experience is usually with cars and tools....
Cars are usually splines
Tools use set screws, pins, lots of stuff..
They are also used as mechanical fuses to protect expensive drivetrains
http://www.bulk-online.com/Forum/showthread.php?threadid=12379
You have to know the shear strength of the pin.
When you drill through the hub and the shaft you are creating
two shear zones on both sides of the shaft.
This is known as double shear.
http://www.engineersedge.com/roll_pin.htm
Lets say you had a 8mm moment..(back of envelope calcs here, please don't knock me to hard ME's).
with a 1/16 shear pin with 250 lbs of double shear force required to break it...
your motor is putting out 3 to 4 ft lbs of torque or so.
umm...with the pin 8mm away from the center of the shaft, you have 4 foot lbs/8mm ~ 150 lbs
So, it seems like a single 1/16 pin would do it. If you go to a 3/32 pin then your looking at much higher values.
You'll have shock loads and what not as well from the road transmitted back to the pin..
3/32 ~2.3 mm. just over a 1/4 of your shaft.
I'm dumb with ME stuff.......Miles?
I would just build it and see if broke.
The Loctite 638 doesn't fill big gaps which it sounds like you have, that's why Miles recommended a new pulley.
At some point, the fit of the machined parts just has to be better for whatever you are going to do.
If you oil the shaft and them jamb some JB weld in there to fill the gap, you might make it work.
A pin constrains the pulley rotationally and axially eliminating the need for a circlip.
possibility. I'd say that it's a matter of the material, it's diameter, and the application.
My experience is usually with cars and tools....
Cars are usually splines
Tools use set screws, pins, lots of stuff..
They are also used as mechanical fuses to protect expensive drivetrains
http://www.bulk-online.com/Forum/showthread.php?threadid=12379
You have to know the shear strength of the pin.
When you drill through the hub and the shaft you are creating
two shear zones on both sides of the shaft.
This is known as double shear.
http://www.engineersedge.com/roll_pin.htm
Lets say you had a 8mm moment..(back of envelope calcs here, please don't knock me to hard ME's).
with a 1/16 shear pin with 250 lbs of double shear force required to break it...
your motor is putting out 3 to 4 ft lbs of torque or so.
umm...with the pin 8mm away from the center of the shaft, you have 4 foot lbs/8mm ~ 150 lbs
So, it seems like a single 1/16 pin would do it. If you go to a 3/32 pin then your looking at much higher values.
You'll have shock loads and what not as well from the road transmitted back to the pin..
3/32 ~2.3 mm. just over a 1/4 of your shaft.
I'm dumb with ME stuff.......Miles?
I would just build it and see if broke.
The Loctite 638 doesn't fill big gaps which it sounds like you have, that's why Miles recommended a new pulley.
At some point, the fit of the machined parts just has to be better for whatever you are going to do.
If you oil the shaft and them jamb some JB weld in there to fill the gap, you might make it work.
A pin constrains the pulley rotationally and axially eliminating the need for a circlip.
It seems that something like 3/32 or 1/8 inch could work and it seems like it's well within the physical limits as you've posted. Thanks for the insight.
Now, I wonder about the glues. JB weld would probably work but how the heck do you remove it? It doesn't seem affected by "normal" high heat. It seems the loctite can be removed by "loctite solvent" or what is probably some normal household chemical and failing that, normal high heat seems possible... it shows 50% of normal strength at 150 degrees celsius, so quite possibly.
And, I noticed that it said bonding gaps could reach .25mm. The largest bonding gap is probably at max 1/4 a mm, but failing that, the majority of the shaft is under .25mm from the pulley's shaftway so it should still hold, shouldn't it? I think the pin would take off a lot of forces on the glue, so it wouldn't be too problematic. If that seems impractical, what about adding thin brass shims like Fechter had as a possibility and then using the loctite 638?
Now, I wonder about the glues. JB weld would probably work but how the heck do you remove it? It doesn't seem affected by "normal" high heat. It seems the loctite can be removed by "loctite solvent" or what is probably some normal household chemical and failing that, normal high heat seems possible... it shows 50% of normal strength at 150 degrees celsius, so quite possibly.
And, I noticed that it said bonding gaps could reach .25mm. The largest bonding gap is probably at max 1/4 a mm, but failing that, the majority of the shaft is under .25mm from the pulley's shaftway so it should still hold, shouldn't it? I think the pin would take off a lot of forces on the glue, so it wouldn't be too problematic. If that seems impractical, what about adding thin brass shims like Fechter had as a possibility and then using the loctite 638?
12p3phPMDC
1 kW
- Joined
- Mar 16, 2009
- Messages
- 462
If you did both pins and loctite on the shaft....I bet you'd be set.
Pinning it twice at opposing angles at different points on the shaft might also
help to tighten things up. And Loctite the roll pins in there just for good measure because of RPM.
Yea, if you can fit some thin sheet shims in there, that'll help tighten it up.
If it's on the order of 0.25 mm that's 10 thousanths of an inch...
I/8" is pushing it for me on a 8mm shaft.
1/8" is ~3.2mm. That's over 1/3 of the diameter.
That's alot of shaft to drill out.
I can't do the calcs, but
I don't think it's worth weakening the shaft
anymore than you have to by drilling it.
Pinning it twice at opposing angles at different points on the shaft might also
help to tighten things up. And Loctite the roll pins in there just for good measure because of RPM.
Yea, if you can fit some thin sheet shims in there, that'll help tighten it up.
If it's on the order of 0.25 mm that's 10 thousanths of an inch...
I/8" is pushing it for me on a 8mm shaft.
1/8" is ~3.2mm. That's over 1/3 of the diameter.
That's alot of shaft to drill out.
I can't do the calcs, but
I don't think it's worth weakening the shaft
anymore than you have to by drilling it.
12p3phPMDC said:
It sounds like two roll pins set perpendicularly to each other sounds best. I think it might not even need glue (Although I might add some just to be on the safer side, but I'm not sure what type I would use. It sounds like 638 might not be ideal and it's kind of expensive to get online, it seems.) as the pin would constrain the motion perpendicular to it and so two pins would constrain two perpendicular direction or essentially a plane of motion - since the oscillation occurs along this plane, they should disappear, it seems, and so would the clicking.
However, as elegant at that reasoning seems, I'm still kind of wary of it as it seems too simple.
Don't forget that when you compare 150 lb resistance to 250 lb limit, you are not taking into account a safety factor. Also, the loading can be dynamic, not static, so the actual load might be double that 150 lb estimate - now you're at 300 compared to 250. Still without a safety factor, BTW. Safety factors on something like a shear pin will be relatively high - factors of 6 come to mind on situations I've dealt with in the past. Even if you only used 2, you are comparing 250 to 600. Bring good walking shoes.
12p3phPMDC
1 kW
- Joined
- Mar 16, 2009
- Messages
- 462
250 lbs is for the 1/16" pin in a 300 series ss.
1150 lbs for a 3/32" pin in a 420 ss. 2250lbs for two.
see the link.
Now we are talking!
15:1 safety factor
In reference to the drill size chart # system:
Its not that weird if you think about it..
Its just a tap drill system. These are the bits
you need to undersize the hole just right for the tap, it's engineered that way.
I don't care what system it is as long as it works.
1150 lbs for a 3/32" pin in a 420 ss. 2250lbs for two.
see the link.
Now we are talking!
15:1 safety factor
In reference to the drill size chart # system:
Its not that weird if you think about it..
Its just a tap drill system. These are the bits
you need to undersize the hole just right for the tap, it's engineered that way.
I don't care what system it is as long as it works.
JS Tyro said:
I'm curious about that too. However, using my handy dandy simulator, I can check if it's close enough.
My simulator predicts 6 N.M. at 0 RPM which corresponds to ... *googles*...4.42 ft.lb. Whatever he did, it seems really close.
Anyways, I think I'm going to try 2 3/32" pins set perpendicularly, and I might use some of the blue "Loc-tite" stuff meant for bolts that I already have on the pins - I might also try some on the shaft, but I really want this thing to be removable so I can service the motor when needed so that's more of a last resort measure. I think I'll also look out for the 420 SS or something greater than 300 SS although I have my doubts that ACE's hardware and the like mentions the steel type.
Grinhill
10 kW
swbluto,
Sounds like you have plenty of good advice about the pulley.
With the new shaft option, and trying to cut a circlip groove, I like the dremel idea, combined with rotating the shaft in a battery drill while dremeling. The poor man's lathe. This would make the groove more accurate than slowly rotating the shaft by hand.
Here's my comment on belt tension: I would not attempt removing a belt unless you had a means of removing tension first. So, if you can remove the belt, then you aren't using enough belt tension.
As an example, I can't remove the belt on my bike unless I loosen my drive mounting bolts, and that's with a skinny 9mm belt which is 1.4 metres long! They are very difficult to stretch.
Sounds like you have plenty of good advice about the pulley.
With the new shaft option, and trying to cut a circlip groove, I like the dremel idea, combined with rotating the shaft in a battery drill while dremeling. The poor man's lathe. This would make the groove more accurate than slowly rotating the shaft by hand.
Here's my comment on belt tension: I would not attempt removing a belt unless you had a means of removing tension first. So, if you can remove the belt, then you aren't using enough belt tension.
As an example, I can't remove the belt on my bike unless I loosen my drive mounting bolts, and that's with a skinny 9mm belt which is 1.4 metres long! They are very difficult to stretch.
12p3phPMDC
1 kW
- Joined
- Mar 16, 2009
- Messages
- 462
Torque = 5250 x HP / rpm
Assumed 2000W power @ 4000 rpm
yields
~4 ft/lbs Torque
~4ft/lbs @ 8mm moment ~ 150 lbs.
(The reaction on the shear pins will be at 8mm right?)
Safety factor with one 2250 lb roll pin (3/32 made of 420 SS)
2250lbs in double shear/150 lb load = 15
Add two roll pins and it goes up....
Please correct me if you see something wrong here! I'm just an EE sticking my toes in ME waters...
Assumed 2000W power @ 4000 rpm
yields
~4 ft/lbs Torque
~4ft/lbs @ 8mm moment ~ 150 lbs.
(The reaction on the shear pins will be at 8mm right?)
Safety factor with one 2250 lb roll pin (3/32 made of 420 SS)
2250lbs in double shear/150 lb load = 15
Add two roll pins and it goes up....
Please correct me if you see something wrong here! I'm just an EE sticking my toes in ME waters...
12p3phPMDC said:
It shouldn't be 150lb. That assumes static loading. With dynamic loading, the exact load will be dependent on the nature of the loading characteristics (sinusoid, sawtooth, square etc). A factor of 2 is a back-of-the-envelope guesstimate that is often used. Hence that would halve the ratio of capacity to load. However, at 7.5, that would seem pretty good.
nieles
10 kW
the force will be at 1/5 D so with an 8mm diameter shaft the force will be at 4mm
so not 150 lbs but only 75
so not 150 lbs but only 75
nieles said:
As the point of force decreases in distance (Closer to the axis), the force increases. So it'd actually be 300 lbs.\
By the way, Wolfram Alpha is a pretty neat physics calculator. http://www78.wolframalpha.com/input/?i=force+of+4+foot+lbs+at+4+mm
Oh, but wait... Is the "double shear" force calculated differently? Since the torque would be exerted at the two points, perhaps the force at one point is half of the total force exerted?
12p3phPMDC
1 kW
- Joined
- Mar 16, 2009
- Messages
- 462
Ok, roll pins are rated for double shear based on the chart...
But is that based on the force at one shear location or the sum of two...
Since the forces are in rotation, i.e a force one way on top and the other way on bottom...
seems like you may need to account for both, i.e 600 lbs...
Well...regardless of where the numbers will fall, 3/32 seems sufficient.....1/16 definitely seems
too small, and 1/8" starts to take too much material out of the shaft. At some point, you gotta pick
something and run with it.
As far as the circlip idea, just fire up the motor using the controller you already have hooked up to spin it...
no need a power drill.
But is that based on the force at one shear location or the sum of two...
Since the forces are in rotation, i.e a force one way on top and the other way on bottom...
seems like you may need to account for both, i.e 600 lbs...
Well...regardless of where the numbers will fall, 3/32 seems sufficient.....1/16 definitely seems
too small, and 1/8" starts to take too much material out of the shaft. At some point, you gotta pick
something and run with it.
As far as the circlip idea, just fire up the motor using the controller you already have hooked up to spin it...
no need a power drill.
12p3phPMDC said:
I thought about this some more, and I came to the conclusion that the load is equally divided among the contact points.
Reasoning:
Imagine taking an infinite number of infinitely strong infinitesimally thin pins through the motor shaft and to the pulley (you might consider these pins as being joined together, like a "pin star" of some sort, if the multiple poles on one "pin" seems more similar than separate pins). If the force transmitted was summed on the basis of the calculation of one contact point, then the force on the pulley would be infinite. Reductio ad absurdum. Therefore, the forces are likely divided - So a 300 pound force on one contact point would probably translate to 150 pounds of force on each side of the roll pin, since there are two contact points. Having two roll pins would suggest a loading of 75 pounds at each contact point of a given roll pin (Although, this likely would vary upto somewhere under 150 lbs due to.. uhh... dynamics.).
So, I think the "double shear force" is, at maximum, 300 lbs. It might be 150, though, for a single pin - I just don't understand what "double shear lbs" is supposed to mean, exactly.
So, I think the highest quality 3/32" roll pin sounds pretty optimal.
Dudes - the torque is due to the two forces (shear at the pin where it exits the shaft's hole at two diametrically opposite points) separated by 8mm. Hence the calculation is either twice 4mm times one of the shear forces or once times 8mm times one force. 8mm is correct.
12p3phPMDC said:
When I saw this, two things stuck in my mind:
1 - How come after all these years, I've never seen this equation?
2 - Why is my gut feeling saying "This is wrong"?
I'm a CE from way back (specialty is structures, hence forces, stresses, dynamics etc). Not being a ME means I may have missed this as a short-cut equation. But it bugged me so I Googled "Torque = 5250 x HP / rpm" and found lots of web sites which all seem to have copied off the same original source. The gist of them is "Those guys that say 'Torque not power gives acceleration' are full of BS." Guess what?... they're wrong. I've recently posted "Crudely stated torque gives acceleration; power gives speed." So I'm one of those guys they don't like.
The equation is for a steady state situation where rpm = constant (i.e. no acceleration). So what does this tell you about real world acceleration? What does it tell you if "torque x 0 = Power"? - that's the condition where an electric motor generates maximum torque. That equation suggests that regardless of torque, power is zero if the motor is not turning. However, if you stall a motor, the power, ultimately - if nothing burns out before you reach steady state - is V^2/R, where R is the resistance of the coils in the motor. V is not zero just because rpm is zero. The equation can't take into account the fact that real world power is more than just mechanical power. Nor does it give us the maximum torque.
The previous calculation used an arbitrary rpm and power setting to determine the torque at that state. But what is the torque when the power is only generating heat and not moving the shaft? I know it's the sum of the moments due to the magnetic fields of the coils attracted to the magnets. But that is not dependent on total power. That peak torque, when accelerating from a standing start, is what is going to shear off the pins. Since torque increases with decreasing rpm (for a given power), then that torque is going to be greater than 4 lb-ft and the load on the shear pin will be higher than calculated.
I retract any statement I made about the safety factor being closer to 7.5. That was based on incorrect input.
So - how do you determine the torque at zero RPM without doing the magnetic field calcs?
12p3phPMDC
1 kW
- Joined
- Mar 16, 2009
- Messages
- 462
real world acceleration comes down to the torque curve.
you know the terms used, "flat", "fat", etc.
Generally, you want large torque available at all RPMs...
I can generate 10000 ft lbs with big old lever...But, I can't do it fast over a large distance.
But, I agree that you need to find the max torque and design around that...
but, what is it? is it the stall torque for a brushless DC with a sensorless controller?
What would be the best way to find the maximum torque this motor can crank out?
you know the terms used, "flat", "fat", etc.
Generally, you want large torque available at all RPMs...
I can generate 10000 ft lbs with big old lever...But, I can't do it fast over a large distance.
But, I agree that you need to find the max torque and design around that...
but, what is it? is it the stall torque for a brushless DC with a sensorless controller?
What would be the best way to find the maximum torque this motor can crank out?
