— David Wagner 2009/10/07 12:42

ME 5463-001 Fracture Mechanics, Homework #10

Exercise 4.6 For 2h = W and 2a / W = 0.5, determine the geometric stress intesity factor(s) for the problem posed in Exercise 4.5 by the boundary collocation method. Let P / t = 500 lb/in.

For an infinite body, Z_inf (z) = { {2P}/{pi} } { {z sqrt{a^2-b^2}} / {(z^2 - b^2)sqrt{z^2 - a^2}} }. [This was given in Exercise 3.12, and can be found in Appendix C, Case 5.11.]

Z(z) = Z_inf(z) P(z)

P(z) = sum{j=0}{J}{A_j z^j}

Z(z) = { {2P}/{pi} } { {z sqrt{a^2-b^2}} / {(z^2 - b^2)sqrt{z^2 - a^2}} } sum{j=0}{J}{A_j z^j}

K depends on only the first term (j=J=0) of this expansion.

K = lim{delta^{+} right 0}{ Re Z |_{y=0} } · sqrt{2 pi delta}

Because of symmetry, K_A = K_B, so delta can be delta = x-a, and

K = lim{x^{+} right a}{ Re Z |_{y=0} } · sqrt{2 pi (x-a)}; at y=0, z=x, so

= K = lim{x^{+} right a}{ Re { {2P}/{pi} } { {x sqrt{a^2-b^2}} / {(x^2 - b^2)sqrt{(x+a)(x-a)}} } {A_0 x^0} } · sqrt{2 pi (x-a)}

= K = { Re { {2P}/{pi} } { {a sqrt{a^2-b^2}} / {(a^2 - b^2)sqrt{a+a}} } {A_0} } · sqrt{2 pi }

= K = { Re { {2P}/{pi} } { {sqrt{a^2-b^2}} / {(a^2 - b^2) } } {A_0} } · sqrt{pi a}.

Since all these terms are real,

K = { {2P sqrt{pi a}}/{pi} } { {sqrt{a^2-b^2}} / {(a^2 - b^2) } } {A_0}, where A_0 is found by satisfying static equilibrium, int{0}{W-b}{sigma_y |_{theta=0} dx} = P.

Assuming higher terms are complex or not significant,

int{0}{W-b}{ { { {2P}/{pi} } { {x sqrt{a^2-b^2}} / {(x^2 - b^2)sqrt{(x+a)(x-a)}} } {A_0 x^0}} dx} approx 2 P ⇒ { {A_0 sqrt{a^2-b^2}}/{pi} } int{0}{W-b}{ { { {x } / {(x^2 - b^2)sqrt{x^2-a^2}} } } dx} approx 1

= { {A_0 sqrt{a^2-b^2}}/{pi} } sqrt{a^2-b^2} (asin{{2bx+2a^2}/{2ax+2ab} } -asin{ {2bx-2a^2}/{2ax-2ab}} )/{2b^2-2a^2} {|_0}^{W-b} = { {A_0(a^2-b^2)}/{pi} } (asin{{2bx+2a^2}/{2ax+2ab} } -asin{ {2bx-2a^2}/{2ax-2ab}} )/{2b^2-2a^2} {|_0}^{W-b} = { {- A_0}/{ 2 pi} } (asin{{2bx+2a^2}/{2ax+2ab} } -asin{ {2bx-2a^2}/{2ax-2ab}} ) {|_{x=0}}^{W-b} approx 1

Substitute in the given values and solve for A_0.

…

K = { {2P sqrt{pi a}}/{pi} } { {sqrt{a^2-b^2}} / {(a^2 - b^2) } } {A_0}


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