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20 December, 01:33

Estimate the endurance strength, Se, of a 37.5-mm - diameter rod of AISI 1040 steel having a machined finish and heat-treated to a tensile strength of 760 MPa.

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  1. 20 December, 03:50
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    endurance length is 236.64 MPa

    Explanation:

    data given:

    d = 37.5 mm

    Sut = 760MPa

    endurance limit is

    Se = 0.5 Sut

    = 0.5*760 = 380 MPa

    surface factor is

    Ka = a*Sut^b

    where

    Sut is ultimate strength

    for AISI 1040 STEEL

    a = 4.51, b = - 0.265

    Ka = 4.51*380^{-0.265}

    Ka = 0.93

    size factor is given as

    Kb = 1.29 d^{-0.17}

    Kb = 0.669

    Se = Sut * Ka*Kb

    = 380*0.669*0.93

    Se = 236.64 MPa

    therefore endurance length is 236.64 MPa
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