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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

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

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}} )