(a) Debonding force: F c ∝ d x, x = m for PSA 6A, p for PSA 6B (b) Debonding displacement: δ c ∝ d y, y = n for PSA 6A, q for PSA 6B, where x and y are function of fracture energy used to derive the power law.
Patel et al. (Thu,) studied this question.