Estimate the endurance strength of a 1.5-in-diameter rod of AISI 1040 steel having a machined finish and heat-treated to a tensile strength of 110 kpsi, loaded in rotating bending.

36.0 kpsi2.

Explanation:

From the question

Sut=110 kpsi

Se’=0.5(110)=55 kpsi

For surface factor ka,

a=2.70, b= -0.265, ka=a(Sut)b=2.70(110)-0.265=0.777

Assuming the worst case for size factor kb, and since 0.11 less than or equal to d less than or equal to 2in,

Therefore:

kb=0.879d-0.107 = 0.879(1.5)-0.107=0.842

Hence, the endurance strength is Se= ka kb Se’ = 36.0 kpsi2.

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