Question

An AISI 1040 steel rod of diameter 3 inches has a machined surface finish. It is heat-treated to a tensile strength of 150 kpsi and loaded in rotating bending. Determine the endurance strength of the rod, in kpsi.

Answer 1

≈ 41.95 Kpsi

Step-by-step explanation:

Note : We can Obtain parameters of Marin surface modification factor from Table 6-2

Given data:

Diameter of rod ( d ) = 3 inches

Tensile strength( Sut ) = 150 kpsi

Calculate the endurance strength

endurance strength ( Se ) = Ka*Kb*Kc*Kd *Ke*Kf * Se' -------- ( 1 )

Ka = aSut^b = ( 2.7 ) 150^-0.265 = 0.7157

Kb = 0.879 ( d )^0.107 = 0.879 ( 3 )^-0.107 = 0.7815

Kc, Kd , Ke , Kf = 1 , 1 , 1 , 1

Se' = 0.5 * 150 = 75

Back to equation 1

Se = 0.7157 * 0.7815 * 1 * -------- * 75

≈ 41.95 Kpsi