Answer :
Answer:
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[tex]_{1}[/tex] = 800 N, F[tex]_{2}[/tex] = 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[tex]_{1}[/tex] - F[tex]_{2}[/tex]) * 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[tex]_{max}[/tex] = T[tex]_{max}[/tex] · r / J
For this question T ⇔ T[tex]_{max}[/tex]
The formula for Polar Moment of Inertia of a circular solid shaft is J = π·[tex]r^{4}[/tex] / 2
J = π * [tex]0.06^{4}[/tex] ÷ 2 = 2 x [tex]10^{-5}[/tex] [tex]m^{4}[/tex]
substitute J into formula for τ[tex]_{max}[/tex], we have:
τ[tex]_{max}[/tex] = 105 * 0.03 ÷ 2 x [tex]10^{-5}[/tex] = 157,500 Pa = 157.5 KPa
τ[tex]_{max}[/tex] = 157.5 KPa