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Diameter of circular section shaft that transmits power

ÀÛ¼ºÀÚ Uploader : orion ÀÛ¼ºÀÏ Upload Date: 2019-11-05º¯°æÀÏ Update Date: 2020-01-07Á¶È¸¼ö View : 351

Find the required diameter of the circular section shaft that delivers the power P (MW).
The axis rotates at f (Hz) and the yield stress of the material is ¥ó (MN / m^2).

When the angular velocity of the axis is ¥ø (rad / s) and the transmitted power P (W) is the product of the torque T (Nm) and the angular velocity, then,

¥ø = 2*¥ð*f

P = T*¥ø = T*2*¥ð*f

T = P/(2*¥ð*f)

Since The maximum torque  Tmax = ¥ó*Jz/r and
the polar moment of inertia for the circular sectionJz = (¥ð/2)*r^4,

Tmax = ¥ó*(¥ð/2)*r^3

r = (2*Tmax/(¥ð*¥ó))^(1/3)

r = (P/(¥ð^2*f*¥ó))^(1/3)

This, the required diameter D is :

D = 2*(P/(¥ð^2*f*¥ó))^(1/3)

Applying the safety factor fs,

D = 2*(fs*P/(¥ð^2*f*¥ó))^(1/3)


*** Âü°í¹®Çå[References] ***

https://en.wikipedia.org/wiki/Polar_moment_of_inertia
D = 2*(fs*P/(¥ð^2*f*¥ó))^(1/3)
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