√ME 5463-001 Fracture Mechanics, Homework #23

— David Wagner 2009/11/30 03:38

A mechanical component of AISI 4142 steel treated to a harness of 450 NB is subjected to a cyclic loading. By measurements, the stress amplitudes at its weakest point are

  • sigma_ax = 300 MPa;
  • sigma_ay = 380 MPa;
  • tau_axy = 250 MPa;

and the mean stress values are

  • sigma_xm = 100 MPa;
  • sigma_ym = 50 MPa;
  • tau_xym = 150 MPa.

Estimate its fatigue life N_f. (Use the table above to find the material parameters needed.)

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 4142 (Q & T, 450 HB) 1584 1757 1998 1937 1837 -0.0762

overline{sigma}_a = {1/sqrt{2}} sqrt{ (sigma_xa - sigma_ya)^2 + (sigma_xa - sigma_za)^2 + (sigma_ya - sigma_za)^2 + 6(tau_xya + tau_xza + tau_zya)^2 } = {1/sqrt{2}} sqrt{ (80x10^6)^2 + 6(250*10^6)^2 } = 436.7*10^6

overline{sigma}_m = sigma_xm + sigma_ym + sigma_zm = 150*10^6

sigma_ar = overline{sigma}_a(1-{{overline{sigma}_m}/{{sigma prime}_f}})^{-1} = 436.7*10^6(1-{{150*10^6}/{1937*10^6}})^{-1} =473.3*10^6

N_f = {1/2} ({sigma_ar}/{{sigma prime}_f})^{1/b} = N_f = {1/2} ({473.3*10^6}/{1937*10^6})^{-1/0.0762} =53.75*10^6

53.75 Million Cycles

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