Do all parts b) Calculate the corresponding life, in 10° cycles, (tolerance of 0.1106 cycles)N :In practice, the load is not purely alternate, andthe minimum stress is omin =-132 MPa,the maximum stress is omax = 294 MPa.c) Calculate the new stress ratio R, the stress amplitude o, in MPaand the mean stress o, in MPa.The answers are acceptable with a tolerance of 1 MPa forstresses.R:MPaMPid) Calculate the corresponding life, in 10 cycles, (tolerance of 0.1106 cycles)N : A steel component is subjected to alternate cyclical loading.The steel follows Basquin's law for high cycle fatigue, o, x N = C1(where the stress amplitude is in MPa).Ignore the geometric detail and assume that Marin's modifyingfactors are all equal to 1.You are giventhe minimum stress ain = -213 MPa,the maximum stress amax = 213 MPa.The material data areTensile strength oUTS = 539 MPa,Basquin's constant c, = 875 MPa,Basquin's exponent a = 0.085.a) Calculate the stress ratio R, the stress amplitude o, in MPa andthe mean stress am in MPa.The answers are acceptable with a tolerance of 0.01 for R and of1 MPa the stresses.R :MPaMPаb) Calculate the corresponding life, in 10° cycles, (tolerance of 0.1106 cycles)N :

Answer: (a) The stress ratio, the stress amplitude and the mean stress: R = -1 σa = 213 MPa σm = 0…

A steel component is subjected to alternate cyclical loading. The steel follows Basquin's law for high cycle fatigue, o, x Nº = C, |(where the stress amplitude is in MPa). Ignore the geometric detail and assume that Marin's modifying factors are all equal to 1. You are given the minimum stress oin= -213 MPa, the maximum stress am = 213 MPa. The material data are Tensile strength oUTS = 539 MPa, Basquin's constant c, = 875 MPa, Basquin's exponent a = 0.085. a) Calculate the stress ratio R, the stress amplitude a, in MPa and the mean stress am in MPa. The answers are acceptable with a tolerance of 0.01 for R and of 1 MPa the stresses. R: MPa MPa b) Calculate the corresponding life, in 10° cycles, (tolerance of 0.1 1o6