Replies to this message

: Calculating load force applied to the end of a shaft
: Calculating load force applied to the end of a shaft -- WallyLux	Post Reply	Top of thread	Engineering Forum

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... What you've described is calculating for the bending of the arm/beam... not the torque load it will need to withstand. But yes you would want to also look at the bending. If the ladle is as shown in your picture, then you better look at that again before you select your beam bending equation. Looks to me like all the load is on one end of one arm? 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... Good luck, Bob

Post Reply Tell a Friend (must be logged in) Alert Admin About Post

Where am I? Original Top of thread

| |

: : Calculating load force applied to the end of a shaft
: : Calculating load force applied to the end of a shaft -- RWOLFEJR	Post Reply	Top of thread	Engineering Forum

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

Post Reply Tell a Friend (must be logged in) Alert Admin About Post

Where am I? Original Top of thread

| |

: : : Calculating load force applied to the end of a shaft
: : : Calculating load force applied to the end of a shaft -- WallyLux	Post Reply	Top of thread	Engineering Forum

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

Post Reply Tell a Friend (must be logged in) Alert Admin About Post

Where am I? Original Top of thread

| |

: : : : Calculating load force applied to the end of a shaft
: : : : Calculating load force applied to the end of a shaft -- Pinkerton	Post Reply	Top of thread	Engineering Forum

Posted by: dalecyr ® 01/21/2011, 13:12:02 Author Profile eMail author Edit

Dave: Just so *I* understand... "Total mass 150 kg." and... "The overall height is 460 mm." When you say "Calculate... ...at the base point of the crucible.", using a half full crucible, would that mean using 150kg/2 = 75kg (half the total load) and using 460mm/2 = 230mm (half the total height), putting ALL of the 75kg at the end (base) of the crucible, so the torque would be 75kg * 230mm? I *think* that's what you mean, because altho the CG will never be *quite* that far from the shaft, the differance between 230mm and the actual CG is not worth calculating for. Besides, 230mm is the more conservative number. Right? dale

Post Reply Tell a Friend (must be logged in) Alert Admin About Post

Where am I? Original Top of thread

| |

: : : : : Calculating load force applied to the end of a shaft
: : : : : Calculating load force applied to the end of a shaft -- dalecyr	Post Reply	Top of thread	Engineering Forum

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

Post Reply Tell a Friend (must be logged in) Alert Admin About Post

Where am I? Original Top of thread

| |

: : : : : : Calculating load force applied to the end of a shaft
: : : : : : Calculating load force applied to the end of a shaft -- Pinkerton	Post Reply	Top of thread	Engineering Forum

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, Bob P.S. Guess I owe you a buck...! Do you accept pay-pal?

Post Reply Tell a Friend (must be logged in) Alert Admin About Post

Where am I? Original Top of thread

| |

: : : : : : : Calculating load force applied to the end of a shaft
: : : : : : : Calculating load force applied to the end of a shaft -- rwolfejr	Post Reply	Top of thread	Engineering Forum

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.

Post Reply Tell a Friend (must be logged in) Alert Admin About Post

Where am I? Original Top of thread

| |

: : : : : : : : Calculating load force applied to the end of a shaft
: : : : : : : : Calculating load force applied to the end of a shaft -- WallyLux	Post Reply	Top of thread	Engineering Forum

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.

Post Reply Tell a Friend (must be logged in) Alert Admin About Post

Where am I? Original Top of thread

| |

: : : : : : : : : Calculating load force applied to the end of a shaft
: : : : : : : : : Calculating load force applied to the end of a shaft -- WallyLux	Post Reply	Top of thread	Engineering Forum

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: I didn't mean to imply that the design was getting elaborate, I was commenting to Dale, our resident Engineering-Technology Sponge, that the approach I was suggesting was a brute force approach. It could be done in a much more refined manner if weight and performance curves were critical such as in a space shuttle. Given that we are emptying a crucible of molten material, it does not need to be calculated to the last tiny drop. 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

Post Reply Tell a Friend (must be logged in) Alert Admin About Post

Where am I? Original Top of thread

| |

: : : : : : : : : : Calculating load force applied to the end of a shaft
: : : : : : : : : : Calculating load force applied to the end of a shaft -- Pinkerton	Post Reply	Top of thread	Engineering Forum

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.

Post Reply Tell a Friend (must be logged in) Alert Admin About Post

Where am I? Original Top of thread

| |

: : : : : : : : : : : Calculating load force applied to the end of a shaft
: : : : : : : : : : : Calculating load force applied to the end of a shaft -- WallyLux	Post Reply	Top of thread	Engineering Forum

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

Post Reply Tell a Friend (must be logged in) Alert Admin About Post

Where am I? Original Top of thread

| |

: : : : : : : : : : : : Calculating load force applied to the end of a shaft
: : : : : : : : : : : : Calculating load force applied to the end of a shaft -- Pinkerton	Post Reply	Top of thread	Engineering Forum

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.

Post Reply Tell a Friend (must be logged in) Alert Admin About Post

Where am I? Original Top of thread

|   |