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Calculating load force applied to the end of a shaft | |||
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Posted by: WallyLux ® 01/19/2011, 21:15:19 Author Profile eMail author Edit |
I'm new to the forum and hope someone can help me out. I've attached a picture so hope it works. I have a ladle for pouring metal, the mass of the ladle itself is 60 kg and the max. mass of material that can fit in the ladle is 90 kg. Total mass 150 kg. The ladle has a tapered profile, so the top diameter is 340 mm, bottom diameter is 270 mm. The overall height is 460 mm. The ladle sits in a ring and this ring is attached to the end of a shaft. As the shaft rotates it pours the ladle in a teapot fashion in the direction of rotation of the shaft. The point of pivot on the ladle is centred in the x-axis and 150 mm from the top of the ladle in the y-axis. To find the torque applied by the ladle on the shaft can I treat the ladle like a beam, ie. treat the total mass as a distributed load, and therefore being able to calculate a point load and position. Then use this position as my distance from the shaft and apply the point load. I was going to treat the change in diameter as negligable as it is very small compared to the height of the ladle. The purpose of this is to select an appropriate size shaft and motor/gearbox to move it. I hope i have been clear and thanks in advance for any help.
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Posted by: RWOLFEJR ® 01/20/2011, 08:24:21 Author Profile eMail author Edit |
This sounds like a homework problem... (side bet buck?) Some things to think about...
Once you've decided if the arm will handle the bending... or bend an amount you feel is acceptable... then you'll need to look at the torqe or maybe think of it like twist... that will be required to tip this. If you're going to discount the changes in diameter of the arm then I'm hoping you'll be calculating using the smaller of the two diameters? Better to error to the safe side right? Material selection will also be important. All materials behave differently. A safety factor will need to be applied as well. I mean after all... would you want to be standing near this little ladle of molten metal that has been designed to meet "just enough to get by?" Hope this helps you out some...
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Posted by: WallyLux ® 01/20/2011, 21:05:27 Author Profile eMail author Edit |
Thanks Bob, No it's not homework, i'm at uni but i'm helping my dad to build a machine to make life easier. Sorry if i was unclear, it is the torque the ladle applies to the shaft that i'm interested in, and from that i can calculate twist. I need to know the torque necessary so i can choose a motor and gearbox. Bending moment and shear force is easy enough. Yes the load is on one end of the shaft, the other end of the shaft will be supported by two bearings approx 300 mm apart. Will just be using 1030 steel unless dimensions get ridiculous and as it is classed as lifting equipment and because of the environment it will be in i'm using a SF of 3-4. What i need confirmation on is whether i am correct when i say that the entire mass of the ladle and material acts through the CG of the assembly. And because the shaft that the ladle rotates around is offset to the CG of the ladle, the displacement between the CG and the shaft becomes my length that the force acts upon to determine torque? Can i simply assume that the ladle is full and that the material is solid or does the torque force actually increase as material is poured out and goes below the level of the pivot point? Thanks for the feedback, Mark |
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Posted by: Pinkerton ® 01/21/2011, 09:36:56 Author Profile eMail author Edit |
Hi Wally, I would not try to get too complicated with this. The difference in shaft size and HP will be tiny. It is not going to swing from 1/2hp to 20hp. I would assume the ladle to be filled to the center line of the shaft. i.e. Half full by your sketch. Calculate using crucible and contents as the load at the base point of the crucible. I would ignore any benefits that will be obtained by counter balancing as the crucible is tilted and the material flows past the center line and towards the pouring lip. You could get very technical and break it all down to balanced and un-balanced approaching the 45-deg rotation point but the difference will probably be in thousandths of an inch difference in shaft diameter and fractions of HP. Dave |
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Posted by: dalecyr ® 01/21/2011, 13:12:02 Author Profile eMail author Edit |
Dave: Just so *I* understand...
When you say "Calculate... ...at the base point of the crucible.",
I *think* that's what you mean,
Besides, 230mm is the more conservative number. Right? dale |
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Posted by: Pinkerton ® 01/21/2011, 14:21:31 Author Profile eMail author Edit |
Not quite Dale, Use the full mass of the crucible, plus the mass of material when filled to the shaft center-line. Because the crucible is tapered, the material will be a little less than half the 90Kg, but half would be fine. That means 60Kg crucible plus 45Kg material. You could get much finer results if this was going on the space shuttle, but for a gearbox and motor, rough enough. Dave |
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Posted by: rwolfejr ® 01/22/2011, 12:45:30 Author Profile eMail author Edit |
Yeah...
What Dave and Dale said... Also a consideration... You might want to think about a set-up similar to an old concrete mixer instead of a motor. A large diameter hand wheel and a gear reduction. I envision better control of your pour with the hand wheel rig over a motor. You can gear it down to make it easy to turn and not have the material sloshing with direction changes? Also cheaper than a motor with a brake. Good luck,
P.S.
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Posted by: WallyLux ® 01/23/2011, 17:03:29 Author Profile eMail author Edit |
Sounds good to me. Thanks for pointing out that i should be calculating the load when it is about half full. I overlooked the effect of the extra material and top part of the ladle balancing the load. Even though there is less material in the ladle at this point because of the shift in CG it makes a significant difference to the torque it applies. You're right that its not rocket science but I think I'll use the CG instead of the base point of the ladle. Because of the environment its in and its part of a machine that's classed as lifting equipment I was going to apply a SF of 3-4 anyway. The down side to that is that we need the machine to be as light as possible but applying this sort of SF means a big jump in motor and gearbox size. I've been told Cycloidal reduction boxes are good for high ratios and light weight. Has anyone dealt with these before. |
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Posted by: WallyLux ® 01/23/2011, 17:17:13 Author Profile eMail author Edit |
Bob: Thats the next challenge. We want to use a servo or encoded motor because... what for it... later on we want to semi-automate it. How likely it is, i don't know, but the goal is to be able to enter the parameters of a pour into a PLC, position it and press the big green 'GO' button. So because of the risks of power failure or any other unforseen issues, we'll need a manual over-ride as well. Which means effectively having 2 input shafts that have not physical attachment or dependence to each other, going to 1 ouput shaft. |
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Posted by: Pinkerton ® 01/23/2011, 20:29:36 Author Profile eMail author Edit |
Wally, For what you are intending I would stick with a worm-drive reduction gearbox. The great benefit of the worm drive is that is almost impossible to make it run backwards in a power failure situation. There is no telling how fast a cycloidal may snap back and slop molten material if the power cuts off. As for manual over-ride use a motor with a double ended shaft. One shaft end attaches to the gearbox, the other free end has a hand wheel. Providing you are not using a stepper motor for the future automation, there will be little difference in the required hand-torque to turn the motor. Rocket Science:
Also, I would not be applying such a high SF unless there was a very good reason to do so. From my guess of this final application, this seems like a predominantly fail-safe process, so I would opt for a 2:1 SF max. Just a thought-out-loud though. Dave Modified by Pinkerton at Mon, Jan 24, 2011, 10:01:47 |
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Posted by: WallyLux ® 01/25/2011, 00:07:53 Author Profile eMail author Edit |
Dave, What do you mean when you say the cycloidal may snap back? Do they tend to build up energy? I like the idea of their design but maybe a planetary gearhead would be better suited? The motor with the through shaft is a good idea, and one i am contemplating, but i think we will end up using a braked servo motor. But saying that if i had a 24V power supply they can be wired to release. As for the level of technicality, you're right, i just wanted to use the Rocket Science comment because you mentioned the space shuttle. Sort of like when you're playing golf and you actually get to yell 4!!! :D The high SF i am using is causing some issues when it comes to selecting components, so i'll probably have to re-assess it. The worm drive has its pros and cons but is looking like a good solution, as i wouldn't have to worry about having a disc brake on the shaft or having so many fail safes.
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Posted by: Pinkerton ® 01/25/2011, 09:51:56 Author Profile eMail author Edit |
Wally, I didn't meant to suggest a build up of energy, it would depend on the ratio involved as to how fast the crucible could roll back under weight in a power failure. The worm drive is all but fail-safe in a power failure. Dave |
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Posted by: WallyLux ® 01/26/2011, 16:25:01 Author Profile eMail author Edit |
Ok, i understand now. I was going to reply on the motor brake but not sure if i should be taking that risk.
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