Q1 Show that the nondimensional Energy Release Rate (ERR) is given by G (μ) = – ∂ / ∂ μ_{1} (s (μ)). Find the corresponding nondimensional SIF [1].
Q2 Write a matlab script which evaluates the ERR and SIF as a function of crack halflength μ_{1} and specimen halflength μ_{2}.
Hint: Evaluate the derivative by a simple finitedifference approximation — the "virtual crack extension method" [2].
Q3 We next address a fracture application. We consider for a Zr–Ti–Ni–Cu–Be bulk metallic glass material [3] with Young's modulus = 96 GPa (and Poisson ratio ν = 0.36), a specimen of halfwidth = 5 cm and aspect ratio μ_{2} = 2, and tensile loading of = 4 kPa.

(a) Plot the dimensional ERR as a function of crack length ã.
(b) If the material Zr–Ti–Ni–Cu–Be is modeled as brittle with Fracture Toughness K_{IC} = 55 MPa/m^{1/2}, and we invoke Griffith's criterion [3], for what crack length ã will the specimen fail?

(a) Invoke Paris's law to simulate fatigue crack growth for several initial crack length values ã_{0}. How do different initial crack length values affect the rate of crack growth?
(b) Verify your numerical crack growth result by comparing it with the result obtained based on the analytical SIF [1,3] for the case ã_{0} = 1.5 cm.
References
[1] DBP Huynh and AT Patera, Reduced basis approximation and a posteriori error estimation for stress intensity factors. Int J Numer Meth Engng, 72(10):1219–1259, 2007.
[2] DM Parks, A stiffness derivative finite element technique for determination of crack tip stress intensity factors. Int J Frac, 10(4):487–502, 1974.
[3] JM Barsom and ST Rolfe, Fracture and Fatigue Control in Structures. American Society for Testing and Metals, 1999.