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현재 위치

Diameter of a circular cross-section shaft that transmits power

작성자 Uploader : 208st 작성일 Upload Date: 2023-06-05변경일 Update Date: 2023-06-05조회수 View : 116

Find the required diameter of the circular cross-section shaft carrying P(MW).
The shaft rotates at f (Hz), and the yield stress of the material is τ (MN/m^2).

When the angular velocity of the shaft is ω (rad/s), the transmitted power P (W) is the product of the torque T (N·m) and the angular velocity, so

ω = 2*π*f

P = T*ω = T*2*π*f

T = P/(2*π*f)

Since the maximum torque Tmax = τ*Jz/r and the polar moment of inertia of the circular section Jz = (π/2)*r^4,

Tmax = τ*(π/2)*r^3

r = (2*Tmax/(π*τ))^(1/3)

r = (P/(π^2*f*τ))^(1/3)

Therefore, the diameter D is:

D = 2*(P/(π^2*f*τ))^(1/3)

If we apply the factor of safety fs,

D = 2*(fs*P/(π^2*f*τ))^(1/3)

*** 참고문헌[References] ***

D = 2*(fs*P/(π^2*f*τ))^(1/3)

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현재 위치

Diameter of a circular cross-section shaft that transmits power

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