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— David Wagner 2009/10/02 21:08

√ME 5463-001 Fracture Mechanics, Homework #9

3.15 Using the Newman-Raju empirical equations for a semi-elliptical surface flaw [Eq.(3.91)], tabulate and plot the distribution of Y(a/W) around the flaw border (i.e. as a function of the flaw parameter, φ) for both tension and pure bending for the following flaw-geometry parameters:

(a) a/c = 1.0 and a/t = 0.3

Eq.(3.91) K = (sigma_t + H sigma_b)sqrt{{pi a} /Q} F({a/t},{a/c},{c/W},phi)

K = sigma sqrt{pi a} Y(a/W)

sigma_b = {6M}/{Wt^3}

a/c=1 ⇒ Q = Phi^2 = 1+ 1.464(a/c)^{1.65} = 2.464

Find F

a/c = 1.0 and a/t = 0.3

  • M_1 = 1.13 - 0.09(a/c) = 1.04
  • M_2 = -0.54 +{{0.89}/{0.2+{a/c}}} = {{0.89}/{1.2}} -0.54 = 0.20167
  • M_3 = 0.5 - {{1.0}/{0.65 +{a/c}}} + 14(1.0-{a/c})^24 = 0.5 - {{1.0}/{0.65 +1}} + 14 = 0.5 - {{1.0}/{1.65}} + 14 = 13.8939
  • f_phi = [(a/c)^2 cos^2phi sin^2phi]^{1/4} = root{4}{cos^2phi sin^2phi}
  • g = 1+[0.1+ 0.35(a/t)^2](1-sin phi)^2 = 1+[0.1+ 0.35(0.3)^2](1-sin phi)^2 = 1 + 0.1315(1-sin phi)^2
  • f_W = sqrt{ sec({{pi c}/{W}} sqrt{a/t}) } = sqrt{ sec({{0.5477 pi c}/{W}} ) }

F = [M_1 + M_2(a/t)^2 +M_3(a/t)^4] f_phi g f_W = [1.04 + 0.20167(0.3)^2 + 13.8939(0.3)^4] root{4}{cos^2phi sin^2phi} (1 + 0.1315(1-sin phi)^2) sec^{1/2}({{0.5477 pi c}/{W}} ) = [1.04 + 0.01815 + 0.11254] root{4}{cos^2phi sin^2phi} (1 + 0.1315(1-sin phi)^2) sec^{1/2}({{0.5477 pi c}/{W}} )

= F = 1.1707 root{4}{cos^2phi sin^2phi} (1 + 0.1315(1-sin phi)^2) sec^{1/2}({{0.5477 pi a}/{W}} )

Find H

a/c = 1.0 and a/t = 0.3

  • P = 0.2+ {a/c}+ 0.6{a/t} = 0.2+ {1.0}+ 0.18 = 1.38
  • H_1 = 1 - 0.34{a/t} -0.11{a/c}{a/t} = 1 - 0.102 -0.033 = 0.865
  • G_1 = -1.22- 0.12{a/c} = -1.34
  • G_2 = 0.55 - 1.05(a/c)^{3/4} + 0.47(a/c)^{3/2} = 0.55 - 1.05 + 0.47 = -0.03
  • H_2 = 1 + G_1{a/t} + G_2(a/t)^2 = H_2 = 1 -0.402 -0.0027 = 0.5953

H = H_1 +(H_2 - H_1)sin^P phi = 0.865 +( 0.5953 - 0.865)sin^{1.38} phi = 0.865 -0.2697sin^{1.38} phi

Find Yt(a/w) for Pure Tension

K = sigma_t sqrt{{pi a} /Q} F({a/t},{a/c},{c/W},phi) = sigma_t sqrt{pi a} Y_t(a/W) ⇒ Y_t(a/W) = 1/sqrt{Q} F = 1/sqrt{2.464} 1.1707 root{4}{cos^2phi sin^2phi} (1 + 0.1315(1-sin phi)^2) sec^{1/2}({{0.5477 pi a}/{W}} )

= Y_t(phi, a/W) = 0.7458 root{4}{cos^2phi sin^2phi} (1 + 0.1315(1-sin phi)^2) sec^{1/2}({{0.5477 pi a}/{W}} )

Find Yb(a/w) for Pure Bending

K = H sigma_b sqrt{{pi a} /Q} F({a/t},{a/c},{c/W},phi) = sigma_b sqrt{pi a} Y_b(a/W) ⇒ Y_b(a/W) = H/sqrt{Q} F = {0.865 -0.2697sin^{1.38} phi}/sqrt{2.464} 1.1707 root{4}{cos^2phi sin^2phi} (1 + 0.1315(1-sin phi)^2) sec^{1/2}({{0.5477 pi a}/{W}} )

= Y_b(phi, a/W) = {(0.6451 -0.2011) sin^{1.38} phi} root{4}{cos^2phi sin^2phi} (1 + 0.1315(1-sin phi)^2) sec^{1/2}({{0.5477 pi a}/{W}} )


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