A solid shaft is to transmit 370 kW at 120 rpm. If the shearing stress of the material must not exceed 92 MPa, find the diameter required.
Given:
P = 370 kW = 370 x103 watts ((N-m)/s)= 3.7 〖x10〗^8 ((N-m)/s)
f = 120 rev/min x 1min/60s=2 rev/s
τ=92 N/mm^2
Solution:
1. Calculate Twisting Moment/ Torsion, T P=2πT f
3.7 〖x10〗^8 ((N-mm)/s)=2 π T (2 rev/s)
T = 2.944〖 x 10〗^7 N-mm
2. Solve for diameter, D of solid shaft using the following formulas: τ = Tr/J
J= π/32 D^4
r= D/2
τs= Tr/J
92 N/mm^2 = (2.944〖 x 10〗^7 N-mm (D/2))/(π/32 D)
D=117.69mm of 4.72inches
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Given:
P = 370 kW = 370 x103 watts ((N-m)/s)= 3.7 〖x10〗^8 ((N-m)/s)
f = 120 rev/min x 1min/60s=2 rev/s
τ=92 N/mm^2
Solution:
1. Calculate Twisting Moment/ Torsion, T
P=2πT f
3.7 〖x10〗^8 ((N-mm)/s)=2 π T (2 rev/s)
T = 2.944〖 x 10〗^7 N-mm
2. Solve for diameter, D of solid shaft using the following formulas:
τ = Tr/J
J= π/32 D^4
r= D/2
τs= Tr/J
92 N/mm^2 = (2.944〖 x 10〗^7 N-mm (D/2))/(π/32 D)
D=117.69mm of 4.72inches
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