The 60-mm-diameter shaft rotates at 300 rev>min. This motion is caused by the unequal belt tensions on the pulley of 800 N and 450 N. Determine the power transmitted and the maximum shear stress developed in the shaft.

Power transmitted in the shaft is 3.3 KW

The maximum shear stress developed in the shaft is 157.5 KPa

Explanation:

d = 60 mm = 0.06 m ⇒ r = 0.03 m, N = 300 rev/min, F = 800 N, F = 450 N

To calculate the Power (Work done), we have:

Power (Work done per minute) = Force * distance

Power (Work done per minute) = Average torque * Angular displacement

P = T * 2πN/60

T = (F - F) * r = (800 - 450) * 0.3 = 105 N·m

Hence, P = 105 * 2π * 300 ÷ 60 = 3298.67 W

P = 3.299 KW ≈ 3.3 KW

To calculate the maximum shear stress, we have:

Maximum shear stress = twisting moment * radius ÷  Polar Moment of Inertia of Area

Mathematically,

τ = T · r / J

For this question T ⇔ T

The formula for Polar Moment of Inertia of a circular solid shaft is  J = π· / 2

J = π * ÷ 2 = 2 x

substitute J into formula for τ, we have:

τ = 105 * 0.03 ÷ 2 x = 157,500 Pa = 157.5 KPa

τ = 157.5 KPa

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