√ME 5463-001 Fracture Mechanics, Homework #24

— David Wagner 2009/11/30 03:38

9.24 At a location of interest in a member made of the AISI 1015 steel of Table 9.1, the material is repeatedly subjected to the uniaxial stress history shown in Fig. P9.24. Estimate the number of repetitions necessary to cause fatigue failure.

MaterialYield StrengthUltimate StrengthTrue Fracture Strengthsigma_a = {sigma prime}_f(2N_f)^b = A{N _f}^B
sigma_0sigma_noverline{sigma}_{fB}{sigma prime}_fAb = B
AISI 1015 (Normalized) 227 415 725 976 886 -0.14
sigma_asigma_m
B-C 120 25
A-D 180 30
51x 72.5 72.5

N_f1 = {1/2}(sigma_a1/{{sigma prime}_f sigma_m1})^{1/b} ={1/2}({120*10^6}/{976*10^6 -25*10^6})^{-1/0.14} =1.32*10^6 cycles.

N_f2 = {1/2}(sigma_a2/{{sigma prime}_f sigma_m2})^{1/b} ={1/2}({180*10^6}/{976*10^6-30*10^6})^{-1/0.14} =70.2 *10^3 cycles.

N_f3 = {1/2}(sigma_a3/{{sigma prime}_f sigma_m3})^{1/b} ={1/2}({72.5*10^6}/{976*10^6-72.5*10^6})^{-1/0.14} =33.5 *10^3 cycles.

{1/{1.32*10^6}} + {1/{70.2 *10^3}} + {51/{33.5 *10^3}} =0.0015374 ⇒

N_f = 650 cycles.

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