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Power transmitted by a rotating circular section shaft

ÀÛ¼ºÀÚ Uploader : orion ÀÛ¼ºÀÏ Upload Date: 2019-11-09º¯°æÀÏ Update Date: 2020-01-06Á¶È¸¼ö View : 295

When the circular section shaft with diameter D (m) rotates at the speed N (rpm), find the maximum power that the shaft can deliver.

The allowable torsional stress of the shaft material is ¥ó (MPa).

The allowable torque of the shaft is as follows.

Ta = ¥ó*Jz/r = ¥ó*(¥ð/16)*D^3

Jz : polar moment of inertia of the circular section = (¥ð/2)*r^4

Since the power P = Ta*¥ø and ¥ø= 2*¥ð*N/60 (rad/s),

P = ¥ó*(¥ð/16)*D^3 * 2*¥ð*N/60

= (¥ð^2/8)*¥ó*D^3*N/60

Results are in MW, then expressing in kW is :

P = (125/60)*¥ð^2*¥ó*D^3*N

If ¥ó is used as the reference strength and the safety factor fs is applied, the equation is as follows.

P = (125/60)*¥ð^2*(¥ó/fs)*D^3*N

*** Âü°í¹®Çå[References] ***

P = (125/60)*¥ð^2*(¥ó/fs)*D^3*N
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