I know this has nothing to do with diesels but can someone help me..

[XcumminsX]

i am really confused and i need to know how to do this problem. Why i cant understand it i dont know. My teacher wont help me out and my dad looked like this when i asked him . Yeah i am a machinist but we dont deal with this stuff, i guess thats for big factories or for engineers... .





1. what is the shaft size? (need to calcutlae for simple bending load)



2) determine a pulley configuration to turn that BLACK BOX at 725 rpm +/- 10 percent.



3) determine a standard motor size and rpm



4) determine a bearing configuration for the system.



If you guys do know this how do you get the answer i am sitting here making myself sick over this. Please help. For some motivation if someone does help me ill send you a cummins power badge p. s. i only have 2 power badges. PLEASE HELP ME IM A MORON :{

 

[dieselpilot1969]

what the heck is it?

 

[rrausch]

I dunno but here are some hints.



(1) Use standard shaft sizes. That is, use 1/2" 5/8" 3/4" etc. The larger the shaft diameter, and the larger the bearing the more load it can carry, and the longer it will last--better too big than too little, but you don't want to design a monster when something smaller and lighter would be better.



(2) What RPM is the motor turning? Take the anticipated motor RPM and divide it by 725. That will give you a number for you to use as a Multiplier. Then size your pulleys, again using standard sizes so that the diameter of the smaller pulley multiplied by your Multiplier equals the diameter of the larger pulley. Lets say your motor is turning at 1725 RPM. 1725 divided by 725 is about 2. 3. Then start looking at pulleys, and determine how much room you have for the pulley to swing and again, using standard sizes (not too big nor less than 2. 5") come up with a couple of pulleys where the diameter of the larger one is 2. 3 times the diameter of the smaller one.



(3) This depends on the system. What is the system being designed for?



(4) Bearings depend on load and RPM. These will be relitively low-rpm bearings, but I don't know the load. In some book somewhere there will be a table showing bearings and the load they can take. Sealed ball bearings are my favorite--I'm prejudiced--but they are cheap, dependable and easily replaced. Now then, YOU ARE NOT A MORON! You simply haven't learned this stuff yet. If you were a Moron, you'd be flipping burgers and blowing your wages on... (you get the picture!)

 

[KLockliear]

48" total black box on center... what does that mean? 48" diameter, radius or circumference?

 

[rrausch]

Also, what is the weight of the Black Box, and is this for a real application or is it a school problem. What does this machine actually do?

 

[Torque King Jay]

Questions

Does the black box move linearly along the 48" axis as if it were a ball screw on a lathe or a machining center and the load could be the turret or a column? If it does then I take it the weight if of the object is 1350 pounds and needs to move at speed of 200ft/min am I correct?



As far as the motor goes does it have to be electric or ISBe powered? (Had to add that)



Does the style of bearing matter??



Jay

 

[Phrogger57]

Problem posed

Try "Machinery's Handbook" for some ways to solve this (albeit in pieces). TKJay has a point- because of the impulses in a IC engine the HP requirement is higher than for an Electric Motor driven shaft. Also, consider is this a driving shaft or is the "black box" the driven load. Either way you need some idea of the load to properly size the motor. 1 giveaway: shafting(and motors) transmit more HP at higher RPM than lower. eg: 1 1/2" shaft will carry 4. 8 hp @ 100 rpm and 29hp @600rpm. Solve what you can, just make sure you keep the same assumptions throughout. -Eric

 

[Phrogger57]

Problem posed

Try "Machinery's Handbook" for some ways to solve this (albeit in pieces). TKJay has a point- because of the impulses in a IC engine the HP requirement is higher than for an Electric Motor driven shaft. Also, consider is this a driving shaft or is the "black box" the driven load. Either way you need some idea of the load to properly size the motor. 1 giveaway: shafting(and motors) transmit more HP at higher RPM than lower. eg: 1 1/2" shaft will carry 4. 8 hp @ 100 rpm and 29hp @600rpm. Solve what you can, just make sure you keep the same assumptions throughout. -Eric

 

[Phrogger57]

Problem posed

Try "Machinery's Handbook" for some ways to solve this (albeit in pieces). TKJay has a point- because of the impulses in a IC engine the HP requirement is higher than for an Electric Motor driven shaft. Also, consider is this a driving shaft or is the "black box" the driven load. Either way you need some idea of the load to properly size the motor. 1 giveaway: shafting(and motors) transmit more HP at higher RPM than lower. eg: 1 1/2" shaft will carry 4. 8 hp @ 100 rpm and 29hp @600rpm. Solve what you can, just make sure you keep the same assumptions throughout. -Eric



By the way, if you have access to a Grainger's catalog, that can help you with "standard" motor sizing.

 

[XcumminsX]

im pretty sure the black box stays right where it is. its staionary. for the bearings we get to make up which bearings all we did with bearing is lets say they are 90% efficient we just took whatever number and multiplied it by . 90 to get the right output. the black box needs to turn around 725 rpm. a i going to assume the "box" is actual not a box and has a diameter, what the diameter is i have no clue (didnt specify) but the load is 1350 lbs 200 ft per minute. .





On a note i took i wrote... ... . we are given force and velocity to give us a basic power relation.



Thats about all i can think of. .





THANK YOU FOR ALL HELPING I REALLY AM SHOCKED SO MANY PEOPLE ARE TRYING THIS! I need to know this stuff tomorrow for a final! the teacher will not help because if he does it will give away the answer so half the class is sitting there looking at him like this . Some teachers are just wrong. .

 

[XcumminsX]

This day is getting worse and worse, Today i had a sore throat all day and now i figure out its strep throat. I feel like ****. Guys if you know this please help.

Thanks

Nick

 

[XcumminsX]

guys a guy just asked me what material ... the teacher used bronze as an example. It just has to be a basic material. Thanks

 

[radar doctor]

Space Engine

Beleive you folks might watch a lot of Space shows on TV, Could this be the first stage of a Electro Magnetic Degaussing Agent?????

 

[rrausch]

Where's Illflem when we need him!

 

[Fireman]

XcumminsX,

I am assuming that this is a class assignment. If it is, your instructor isn't going to give you the answer. He has given YOU a problem to solve. You might ask him if the answer is your text book, if not ask him what other books he might suggest to find the solution. Then get in those books and dig, dig, dig. Part of your education is to learn how to solve problems, not just how to cut metal. I know this is not the answer you want, but the hard work and headaches will pay off one day. Don't short change yourself trying to take short cuts.

Fireman

 

[rbattelle]

X,



You still working on it, or have you already left for the exam? If not, I'll take a crack at it. Except it's definetly an application for Machinery's Handbook, and of course my copy is at work (day off today).



-Ryan

 

[Mathew Clausen]

I'll give you a hint on power ... . 1hp = 33,000 ft*lb/min.



Also - you can figure the effective diameter of the "black box" because you know the shaft speed is 725 rpm and that the weight is moving 200 ft/min.



200 ft/min = (725 rev/min) * (pi*dia ft/rev)



or rearrange to solve for the diameter (ft):



Effective dia (ft) = 200/(725*pi)



From that - the torque on the shaft can be found



Tq (ft-lb) = weight*eff radius. ]



Another useful formula - HP = Torque (ft*lb)*RPM / 5252.



Mathew

 

[XcumminsX]

Yeah guys i am still here. I didnt go to work today because of strep and my sinuses are so screwed up i cant breathe right and my vision is slightly blurred :{ . The one time i cant get sick i do...

 

[Torque King Jay]

Any results XcumminsX ?

I was sitting in a 2. 5 hour meeting today at work (boring) and thought about you problem. The whole Meyer's black box thing makes no sense if it were a rotating mass that weighed 1350 and is obviously less than 48" long. Lets solve that one.



Matt Clausen gave some great info that will help. I'll lay it out a little different for you.



200ft/min x 12= 2400in/min

2400in/min /(725rpm)= 3. 3103 inches per rev. That gives us the O. D. of the shaft.

You should be able to solve this.

3. 3103 O. D. / pie 3. 1415 = 1. 0537 inch shaft diameter.



Now for a shaft with a Meyer's whatever attached to at 1. 0537 inches in diameter to achieve 200 surface feet/min at 725rpm and have it less that 48" long with a weight of 1350lbs makes no sense in the real world.



Even if it were made of Iridium at density of 1413lbs/cubic foot (a 2 liter Coke bottle of it would weight about 99 lbs)



At 48" the shaft would be about 41. 86 cubic inches of Iridum and would weigh about 34lbs.



Sooo This has to be one of two things.

1) a constant driving load 1350ft/lbs that the motor is exerting on the shaft

2) A mass that is moving along the length of the shaft at 200ft/min and weight 1350lbs.



In the world I work in (machine tools) the weight of a column on a milling machine with a rapid traverse rate of 200ft/min (2400in/min) along an axis that is about 40 or so inches long makes alot more sense. You would just need to be given a pitch for the shaft (ball screw) to determine motor size/rpm.



If you want to keep figuring out the rest of the problem gear ratio for the pulleys etc. . Most common 3 phase a/c electric motors run at 3600 or 1800rpm. The rest could be a little challege but I think you can handle some of it.



XcumminsX find out just what exactly this thing is. Sorry for being long winded

Oo. Oo.

 

[Mathew Clausen]

I don't think there is any traversing in this problem. I interpret that there is just a 1350lb suspended load that is being lifted at 200 ft/min. There is 48" between bearing supports and the load is applied at the center (you need that info to calculate the bending stresses - which it says to assume as simple supported). This is a very typical design problem in mech engineering classes.



If you haven't figured it already - the power to lift 1350lb @200 ft/min comes to 8. 18hp. So - with some bearing loss and such I'd say the standard motor size would 10hp.



Mathew