From Hooke’s law, the strains are linearly related to the stresses so that Since the strains are calculated from the . 2 r K I ij   31 Since the strains are calculated from the displacement gradient, . 22 r K r r K u II i     Plane Crack Problem Stress Intensity Factors The stress intensity factors for Modes I, II and III are defined as follows:   ,02lim 0    yy r I rK   , 0 2 lim 0       xy r II r K 32   , 0 2 lim 0       xy r II r K   .02lim 0    yz r III rK The stress intensity factor K depends on loading and geometry. Many different geometries have been evaluated, either analytically or numerically, and are available in the literature, e.g., Compendium of Stress Intensity Factors, D. P. Rooke. Plane Crack Problem Similitude For a crack of length 2a 1 in an infinite plate, subjected to an applied stress σ 1 the stress intensity factor is known to be . Consider two large plates, one with a center crack of length 2a 1 , the other with a center crack of length 2a 2 . A stress σ 1 is applied to the first 11 aK I   33 crack of length 2a 2 . A stress σ 1 is applied to the first plate, and a stress σ 2 is applied to the second plate. If we choose σ 1 , σ 1 , σ 2 and σ 2 so that then the fields at the crack tip are identical in both cases. This is the principle of similitude, which is very important in fracture mechanics as it allows results from laboratory scale tests to be applied to large scale fracture problems. )2()1( II KK  Plane Crack Problem Stress Intensity Factors How do we apply this analysis to the failure of actual materials? It has been found experimentally that when the stress intensity factor K (which depends on the geometry and loading) attains a critical value K C (a material property) the crack begins to grow, i.e., the 34 critical condition for the onset of fracture is K → K c . The condition can also be expressed in terms of the energy release rate, i.e., ς → ς c . What are some typical values for K C ? Si 3 N 4 Al 2 O 3 Glass K c (MPa )Material m 1  43   8 4   35 Steels Al alloys Polymers Si 3 N 4 8 4   25.0   10010   30030   Fracture Mechanics #2: Role of Crack Tip Plasticity 1 Role of Crack Tip Plasticity Plastic Zone Size Estimate Consider inelastic and permanent deformation at the crack tip (stresses are too high for the material to remain elastic). First order estimate of plastic zone size: 2 Assume: plane stress, and the material behavior is elastic-perfectly plastic. Set the stress σ yy = σ ys (along the line θ = 0). ys p yy r K     * 1 2 2 2 2 2 1 * 22 ysys p a K r     Where we have used the result that for a semi-infinite crack in a very large plate .aK I   What about the details of the plastic zone shape ? The 3 What about the details of the plastic zone shape ? The shape of the plastic zone is obtained by examining the yield condition, in conjunction with asymptotic K-field results, for all angles θ around the crack tip. Either the Mises or the Tresca criterion can be applied. Plastic Zone Shape Recall that for the Tresca yield condition yielding occurs when .2/ max ys   We will use the Mises yield condition. The Mises condition in terms of principal stresses is given as 4 condition in terms of principal stresses is given as       2 2 13 2 32 2 21 2 ys   where σ y s is the uniaxial yield stress. (For a tension test, σ 2 = σ 3 =0, σ 1 = σ ys ). On the plane θ = 0, σ xy = 0 and thus σ xx and σ yy are the principal stresses σ 1 and σ 2 . The stresses σ z ≡ σ 3 ; σ z =0 for plane stress, σ z = ν(σ xx + σ yy ) for plane strain. However, in general the shear stress σ xy is not zero and the principal stresses σ 1 and σ 2 cannot be determined so easily. 5 The principal stresses σ 1 and σ 2 are evaluated as follows (can use Mohr’s circle, for example): 2 2 21 22 , xy yyxxyyxx                 . important in fracture mechanics as it allows results from laboratory scale tests to be applied to large scale fracture problems. )2()1( II KK  Plane Crack Problem Stress Intensity Factors How. typical values for K C ? Si 3 N 4 Al 2 O 3 Glass K c (MPa )Material m 1  43   8 4   35 Steels Al alloys Polymers Si 3 N 4 8 4   25.0   10010   30030   Fracture Mechanics #2: Role of Crack. Intensity Factors, D. P. Rooke. Plane Crack Problem Similitude For a crack of length 2a 1 in an infinite plate, subjected to an applied stress σ 1 the stress intensity factor is known to be . Consider