design of shafts

Replies

• 

  Rifaa

  I cannot answer to a partial question. If you want us to help you then you should tell us more about what ever that you are doing.
• 

  vish4me

  the shaft carries about 15000 kgs distributed load. it has to rotate 180 degrees in 25 minutes. it is powered by an electric motor of 7 kW.
  i followed the design procedure given in the data hand book.
  the formulae used was d={16/pi*allowable((k*Mb)^2+(k*Mt)^2)^.5}^.333}.
  i gave it so that u can knw the method i followed.
  Mt=9550*1000*power/speed.i found out Mb by calculating the maximum bending moment from sfd and bmd diagram.
  bcos the speed was 0.02 rpm, Mt was in the order of 10^9 and Mb in the order of 10^6. hence the diameter came absurdly high.
  hence i wanted to know is there any other method to design shafts which rotate at slow speeds.<1 rpm.
  is the data i have given enough?
• 

  Rifaa

  Hmm. I got a better solution. Use gears to reduce the speed, it's more effective that way. and it will put less load on the motor too
• 

  Rohan_sK

  I think you just recheck the calculation of POWER and TORQUE, whether the formula used is correct( including the powers used ie ^x) and let us know. The use of gears would only reduce the speed but the load and power remains same, so Mt will not change.
• 

  vish4me

  i checked the formulae ......
  Mt=95550*1000*N/n where N is power in kW.
  n=speed of shaft in rpm.
  the formulae for dia is also the same i mentioned.
  Mt is coming high because the shaft is rotating very slowly . can u help me out.
• 

  Tkelly

  First question, how you decided the motor power without any calculations?
  I suggest you to speak to speed reducer manufacturer(worm gear box manufacturer) with available data. He will definitely guide you in selecting the gear box as well as motor. In your case I would suggest to go for double reduction gear box.
  
  Tom
  CAD Outsourcing Services India | Indovance Inc