√ME 5463-001 Fracture Mechanics, Final Exam #2

— David Wagner 2009/12/08 16:52

A single edge-notched plate (W wide, B thick, and a in crack length) is subject to a constant tensile load, P (a « W). Assuming that the material behaves elastic-plastically, determine the plastic portion of the J-integral of the specimen if the compliance of the plate has the following relationship with the crack length: C = ka.

  • Assume the length is L » B and L » a.
  • Assume k is constant.
  • The plate is under load control.

A=Ba

  • ⇒ dA = Bda
  • ⇒ a=A/B

C = Delta/P ⇒ Delta= CP=kaP= k/B A P.

Under constant load, Pi = int{0}{P}{Delta(P)} dP =int{0}{P}{k/B A P} dP ={k}/{2B} A P^2.

J = delim{|}{ {partial Pi}/{partial A} }{|} = delim{|}{ {k}/{2B} P^2 }{|}.

Assuming a Ramberg-Osgood stress-strain relationship,

J=J_el+J_pl ⇒J_pl = {k}/{2B} P^2 -{{K_I}^2}/E;

{{K_I}^2}=1.12^2 sigma^2 pi a =1.12^2 pi a {P^2}/{W^2 B^2};

⇒ J_pl = ({k}/{2B} - {1.25 pi a}/{E W^2 B^2}) P^2.

Noting the given compliance relationship implies zero compliance when the crack length is zero, the (uncracked) material's modulus of elasticity, E, can be considered infinite, and

J_pl = {{k}/{2B}} P^2.

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