A through crack in a thick plate has the initial length of a_0 = 10 mm. The crac

Question

A through crack in a thick plate has the initial length of a_0 = 10 mm. The crack growth is approximately described by the Paris law with A = 5.5 middot 10^-12 m/cycle and m = 2.5, when Delta K is expressed in MPa middot Squareroot m. The plate is subjected to the cyclic load with a random distribution of cycles. In every 500 cycles there are approximately 80 cycles with sigma_max = 180 MPa, 220 cycles with sigma_max = 120 MPa, 140 cycles with sigma_max = 90 MP a and 60 cycles with simga_max = 50MPa. All cycles have the stress ratio R = 0.1 Estimate the number of cycles to failure. Material properties are: Young's Modulus, E = 210 GPa, Poisson's ratio, v = O.33, yield strength sigma_ys = 390 MPa, ultimate tensile strength, UTS = 485MPa, and K_Ic = 95 MPa Squareroot m. Assume that K_I = sigma middot Squareroot pi middot a/2.

Explanation / Answer

R = Smin/Smax =0.1 ==> DS =0.9 Smax

Integratio of Paris eqn gives Nf = [ af (-n/2)+1    - ai(-n/2)+1 ]/ (-n/2+1) DSn pi n/2

Find Nf for each material separately.. Since Ki given, use it directly to get af = 1/pi( 95/smax)2 =12.2 mm

ai given as 10 mm

Using the integration formula get for R=.1, N1 = 79830 for 180 Mpa stress

N2 = 220000 for 120 MPa

N3= 451700 for 90

N4= 1244600 for 60 MPa

Use Miners law to get cycles to failur from proportionate loading, ie 80/500*N1 + 220/500*N2 + 140/500*N3+ 60/500*N4

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