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.