Required diameter of solid shaft

Submitted by Alexander on

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.
 

Tags
Shearing Stress
Shaft

jayborre Mon, 10/03/2022 - 00:10

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