— David Wagner 2009/09/30 16:43

√ME 5463-001 Fracture Mechanics, Homework #8

Problem 3.13 From the results of Exercise 3.12, derive an expression for K for the problem shown in Figure E.3.3, using the principle of superposition.

Figure E3.3 is a combination of a central load of 2P only downward, and a load of P at location b only upward on both sides of the crack.

From the last problem is, a load case of 2P at the center both up and down is at b=0, K = 4 P { {sqrt{a}}/{pi a } }.

A load case of P at distance b from the center both up and down is K = 2 P {sqrt{a}}/{pi sqrt{a^2 - b^2} }.

These two load cases can be combined, then decomposed as a sum of two forces P up and one 2P down, plus two forces P down and one 2P up. These decomposed forces have the same K, which must each be half the combination of the cases above.

K = 2 P { {sqrt{a}}/{pi a } } + P {sqrt{a}}/{pi sqrt{a^2 - b^2} }


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