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(Solved): A uniaxially loaded very wide sheet of the material from Problem 18 is subjected to constant ampli ...



A uniaxially loaded very wide sheet of the material from Problem 18 is
subjected to constant amplitude loading at R=0 with SmA medium-strength steel displays the following region II Paris relationship
for long crack behavior at R = 0:
da
region I
= 2

A uniaxially loaded very wide sheet of the material from Problem 18 is subjected to constant amplitude loading at R=0 with Smax = 110 MPa. Let K = 95 MPa V/m and S, = 440 MPa. Laboratory experiments have shown that for this material, physically small crack growth occurs up to a length of 1 mm. If the wide sheet contains an initial edge crack with a = 0.3 mm, (a) Calculate the fatigue life of the sheet using only the long crack growth behavior. (b) Calculate the fatigue life, taking into consideration the small crack growth behavior. (c) Comment on your results and the use of this life prediction technique. A medium-strength steel displays the following region II Paris relationship for long crack behavior at R = 0: da region I = 2.4 x 10-¹1 (AK)2.75 dN 10%~ 105m/cycle where da/dN is in m/cycle and AK is in MPaV/m. Data on physically small cracks were generated for the same material at R = 0, and fitting a power law expression to the data yielded the following relationship: 2 physically small da = 1.8 x 10-¹0 (AK)1.75 Cracks dN ak: /MPajun ~ ok in ~ ok in region I above. (a) Plot the equation for these two relationships, the long crack equation between 10-8 m/cycle < da/dN< 10-5 m/cycle and the small crack equation between 1 MPaV/m and where it merges with the long crack Paris equation. If AKth = 5 MPaVm for the long crack data, also sketch the approximate sigmoidal portion of the long crack growth curve in region I. Complete the approximate sigmoidal long crack growth curve if K = 95 MPaV/m. (b) Based on your plot, will extrapolation of the Paris equation to region I predict conservative or nonconservative fatigue life if a physically small crack exists in a component made from this material? Linkedin fain A ? 19


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Matlab Code: %% %Long crack dadnl=[2.006220915*10^-9 10^-8:9.99*10^-8:10^-5 6.6*10^-6]; delkl=(1/(2.4*10^-11))*((dadnl).^(1/2.75)); %short crack delks=0.001:10^-4:1; dadns=1.8*10^-10.*delks.^1.75; figure; set(gca,'FontSize',20); subplot(121); plot(de
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