√ME 5463-001 Fracture Mechanics, Final Exam #1

— David Wagner 2009/12/08 16:52

A plate 2W wide contains a center through-crack 2a long and is subject to a tensile load, P (a « W). Assuming that the material is linear elastic, then G=pi sigma^2 a/E is valid. Derive an expression for the elastic compliance (C = Delta/P) in terms of the plate dimensions and elastic modulus E, the remote stress. (Hint: G = {P^2}/{A} {partial C}/{partial a}).

  • Assume this is a thin plate, thickness B « W, ⇒ plane stress.
  • Assume this is a long plate, Length L » W?
  • Assume the load is applied uniformly across BW.

Determined

  • A = 2aB.
  • sigma = P/{BW};

G = {P^2}/{A} {partial C}/{partial a} ={P^2}/{2aB} {partial C}/{partial a} = pi sigma^2 a/E ⇒

{partial C}/{partial a} = pi sigma^2 {2Ba^2}/{EP^2} = {2 pi B P^2 a^2}/{B^2 W^2 EP^2} = {2 pi a^2}/{B W^2 E}

⇒ C = {2 pi B sigma^2 a^3}/{3 E P^2} + C_0 = {2 pi a^3}/{3B W^2 E} + C_0

E = {sigma L}/Delta_0 ⇒Delta_0 = {sigma L}/E = {PL}/{EBW}.

The compliance of the uncracked plate C_0 = Delta_0/P = {L}/{EBW}.

C = {2 pi B sigma^2 a^3}/{3 E P^2} + {L}/{EBW}=

C = {2 pi a^3}/{3 B W^2 E} + {L}/{EBW}

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