Substitute the known expression for σ xx , σ yy and σ xy in the Mode I crack problem (derived last time) and find that:        2 sin1 2 cos 2 1    r K I     K 6 (plane strain); σ 3 = 0 (plane stress). 2 cos 2 2 3    r K I         2 sin1 2 cos 2 2     r K I Substitute in to the Mises yield condition: Plane strain:     2 2 2 2 2cos121sin 2 3 2 ys I r K           7 2 2 r      Plane stress: 22 2 2cossin 2 3 1 2 ys I r K           These expressions can be used to solve for the radius of the plastic zone r p as a function of θ: Plane strain:                           cos121sin 2 3 4 1 2 2 2 I p K r 8           2 4 ys p Plane stress:                      cossin 2 3 1 4 1 2 2 ys I p K r Check: We note that the Plane Stress case reduces to our first order estimate for θ = 0. Also note that (K I / σ ys ) 2 has dimensions of length. Next we will compare the extent of the plastic zone 9 Next we will compare the extent of the plastic zone in the two situations, plane stress and plane strain, for two cases, θ = 0 and θ = 45˚. For θ = 0, , 3 1       9 1  stressplaner strainplaner p p For θ = 45˚, , 3 1   10     8.2 1 381.0  stressplaner strainplaner p p 3 Extent of the plastic zone is significantly larger for the plane stress case. Plastic Zone Shape Plane stress/plane strain 11 Plastic Zone Shape Plane stress/plane strain 12 Plastic Zone Size Engineering Formulae For Plane Stress: 2 1          ys I p K r  For Plane Strain: 13 For Plane Strain: 2 3 1          ys I p K r  Similar analyses can be done to determine the plastic zone size and shape for Mode II and Mode III loading. Specimen Thickness Effects Plane stress/plane strain 14 Thickness B Meaning of ς Recall the Strain Energy Release Rate ς . What does it physically represent? It is the rate of decrease of the total potential energy with respect to crack length (per unit thickness of crack front),i.e. 15   a PE     What is the connection between ς and K? . physically represent? It is the rate of decrease of the total potential energy with respect to crack length (per unit thickness of crack front),i.e. 15   a PE     What is the connection between. stress and plane strain, for two cases, θ = 0 and θ = 45 . For θ = 0, , 3 1       9 1  stressplaner strainplaner p p For θ = 45 , , 3 1   10     8.2 1 381.0  stressplaner strainplaner p p 3 Extent.                     cossin 2 3 1 4 1 2 2 ys I p K r Check: We note that the Plane Stress case reduces to our first order estimate for θ = 0. Also note that (K I / σ ys ) 2 has dimensions of length. Next