Please help me do this Problem..

[XcumminsX]

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





notes = was given force and velocity to give a basic power relation.

 

[AKAMAC]

I'll get my old texts out tomorrow and help with HP and pulley ratio, but you will have to get some help on the shaft/load/bearing configuration. You don't give specs on metallurgy for the equation, big diff between 316 and aluminum.

Mike

 

[XcumminsX]

awesome if you can try to get it on here earlier in the day i have class at 5 :{

Thanks

Nick

 

[XcumminsX]

its not specified what material. He used Bronze as an example... . Any metal is good, he said keep it basic tho.

Thanks again

Nick

 

[AKAMAC]

http://www.massengineers.com/Documents/PowerEngineeringDictionary.htm

This site may answer alot of your questions. Use formulas.

Mike

 

[AKAMAC]

OK here we go.



HP=3. 0303(-5 exp)xFt lbs/min



lets go with a drive pully dia. of 5" and a standard motor RPM of 900.



driven pully dia. =drive pully x drive RPM/driven RPM



Since this is a test question I'll let you do the math, but will give you the hint that standard HP motors are 5,7. 5,10,15. These are in your spec. range.

I'll look back later, hope this helps. Mike

 

[AKAMAC]

I think I found a process shaft size of 1" would provide 196 ft/sec. That leaves you with bearing load configuration. Good luck Mike

 

[XcumminsX]

do you know what material the shaft was? did you find this info online? THANKS

Nick

 

[Doc Tinker]

I hope that problem is for extra credit - I have no clue and I don't want to hurt my head thinking about it.



Doc

 

[XcumminsX]

actually if i dont know how to do this i will fail my final and if i do that i fail the class :{ thats why i was hoping someone would help me understand how to do this

 

[AKAMAC]

I'll post as simple an explaination as I can when I get home. ~1hr

Mike

 

[XcumminsX]

ok thanks

 

[AKAMAC]

I'm going by 20 year old physics class memories so I hope that some P. E. will step in here and correct my calc's.



1) HP is equal to load at given rate. ie: ft pounds force per minute



This is the conversion that your question is giving values for.

Multiply 1350 (lbs) and the rate (200 ft/min. )

270000 ft pnd/ min (see 1 above)

I won't go into the math involved in the conversion factor but think of it as a constant. PLEASE

To convert (ft pnd/ min) to HP you multiply by this conversion factor (ref. from ulgys electrical hand book) 3. 0303 (exp minus 5)

This gives you 8. 18~ HP.

This is larger than 7. 5 and smaller than 10 as STD HP rating.

So lets not do the Yugo thing and go small. Better to over size the system.



2) Pulley size is determined by the following formula.

Speed of driven (shaft) is equal to driving pulley dia. (motor) multiplied by speed of motor (RPM) divided by the (shaft) driven pulley dia. Extrapolate the conversion and you end up finding the correct size of pulleys.

I used 5" because this will give you an even sized pulley for the driven shaft at standard motor rating of 900 rpm.



I gota break here and open a beer be right back

 

[AKAMAC]

OK



you have driving pulley, driving rpm and driven rpm.

5" times 900 divided by 750 = 6" (the driven pulley dia. )



3) The shaft size is determined by a load of integrated sequenced math that means nothing to me but will to you if you are going into fluid hydraulic engineering Sorry I digress.



I used the afore mentioned site to apply the diameter of several shaft sizes to get the 200 ft/ min required for this mess.



I'd like to know how the instructor in your class expected you to get this right if the knowledge he gave you did'nt cover applied physics. Not a flame, just curious.

 

[AKAMAC]

http://www.massengineers.com/Formulas/shaftspeed.htm



Punch in the numbers and find the answers... ... ... ... .



Not rocket science but it's fun to look smart.



I'm sorry if there is so much BS in these posts but I just had my son read this and he said I'll never make a teacher. Too much assumption and not enough explanation. Knowing and showing is where I fail, but you have to admit it LOOKS good.



Good luck on the exam's I'll look for your posts on results.



Mike

 

[XcumminsX]

hey thanks alot man, i am going to use your formulas on the test for saturday, since i have strep throat and a bunch of other problems i stayed home and didnt take the final today i also stayed home from work. Send me your address and ill send you the cummins badges

 

[AKAMAC]

We don't need no stink'in badges... ... .

We do this because we can.



If you look at the last site I posted you will see that there is a constant for the shaft size. Have fun and remember that the only wrong answer you give is the one the instructor sees is wrong.

 

[XcumminsX]

THANK YOU!

Thank you very much. I appreciate it alot!