For the isotropic self-consistent crack distribution case in Example 15.12, show that for the case v = 0.5, relation (15.3.4)₃ reduces to:

Verify the total loss of moduli at ε = 9/
16. Using these results, develop plots of the effective moduli ratios v̅/v,E̅/
E,µ̄/µ versus the crack density. Compare these results with the corresponding values from the dilute case given in Example 15.10.

Data from example 15.10

Data from example 15.12

Using the self-consistent method, effective moduli for the random distribution case can be developed. The results for this case are given by:

It is interesting to note that as ε→9/16, all effective moduli decrease to zero. This can be interpreted as a critical crack density where the material will lose its coherence. Although it would be expected that such a critical crack density would exist, the accuracy of this particular value is subject to the assumptions of the modeling and is unlikely to match universally with all materials.

Equation 15.3.4

8 = 9 16 - 27 12