Question

The design of a machine element calls for a 40-mm-outer-diameter shaft to transmit 46 kW. If the speed of rotation is 760 rpm, determine (a) the maximum shear stress in shaft (a). (b) the maximum shear stress in shaft (b) with inner diameter 25 mm.

Answer 1

54.52 MPa

Step-by-step explanation:

power P = 46 kW = 46000 W

speed N = 760 rpm

outer diameter of shaft D = 40 mm = 0.04 m

inner diameter of shaft d = 25 mm = 0.025 m

torque T = P/Ω

where Ω = angular speed in rad/s

Ω = 2πN/60 = (2 x 3.142 x 760)/60

Ω = 79.59 rad/s

from this,

torque T = 46000/79.59 = 577.96 N-m

the relationship between torque T, maximum shear stress τmax, and shaft diameters D and d is stated as

T = (π / 16) τmax (
`D^(4)` -
`d^(4)`)/D

imputing the values, we have

577.96 = (3.142/16) x τmax x (
`0.04^(4)` -
`0.025^(4)`)/0.04

577.96 = 0.196 x τmax x (5.4 x
`10^(-5)`)

577.96 = 1.06 x
`10^(-5)` x τmax

τmax ≅ 54.52 MPa