The design of a machine element calls for a 40-mm-outer-diameter shaft to transmit 50 kW. If the speed of rotation is 730 rpm, determine (a) the maximum shear stress in shaft (a). (b) the maximum shear stress in shaft (b) with inner diameter 30 mm.

52Mpa

76Mpa

Explanation:

a. We need to perform a two step process to obtain the maximum shear stress on the shaft. For the solid shaft,

P=2×pi×N×T/60 or T=60×p/2×pi×N

Where P=power transmitted by the shaft=50×10³W

N=rotation speed of the shaft in rpm=730rpm

Pi=3.142

T is the twisting moment

By substituting the values for pi, N and P, we get

T=654Nm or 654×10³Nmm

Also, T=pi×rho×d³/16 or rho=16×T/pi×d³

Where rho=maximum shear stress

T = twisting moment=654×10³Nmm

d= diameter of shaft= 40mm

By substituting T, pi and d

Rho=52Mpa

b. For a hollow shaft, the value for rho is unknown

T=pi×rho(do⁴-di⁴/do)/16

Rho=T×16×do/pi×(do⁴-di⁴)

Where

T= twisting moment=654×10³Nmm gotten above

do=outside shaft diawter=40mm

di= inside shaft diameter =30mm

Pi=3.142

Substituting values for pi, do, di and T.

Rho=76Mpa

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explanation:

answer:

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