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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.

Respuesta :

Answer:

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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