Crack growth life and the inspection interval: 22-Mec-B9 Advanced Engineering Structures, May 2018, Q2

Question 2: Crack growth life and the inspection interval (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. A wing skin panel is modelled as a semi-infinite plate carrying an edge crack.

Quantity	Value
Initial edge-crack length a₀	0.55 mm
Constant-amplitude stress range Δσ normal to the crack	200 N/mm²
Fracture toughness K_IC	2700 N/mm^3/2
Growth law da/dN	31 × 10⁻¹⁵(ΔK)⁴ mm/cycle
Edge-crack geometry factor β	1.12 (assumed)

Find. The number of service cycles — the maintenance interval — in which the crack grows from 0.55 mm to half the critical length that would cause fast fracture.

Approach. Fix the critical crack length from $K_{IC}$, halve it to get the inspection length, then integrate the Paris law between the two lengths; with $m = 4$ the integral is elementary.

Stress-intensity range for an edge crack. For a crack of length a at the edge of a semi-infinite plate, $$\Delta K = \beta\,\Delta\sigma\sqrt{\pi a}, \qquad \beta = 1.12$$ At the starting size, $\Delta K = 1.12(200)\sqrt{\pi(0.55)} = 294\ \text{N/mm}^{3/2}$, giving an initial growth rate of only $2.3\times10^{-4}$ mm/cycle.
Critical crack length. Fast fracture occurs when $\Delta K$ reaches the toughness, so setting $\beta\,\Delta\sigma\sqrt{\pi a_c} = K_{IC}$, $$a_c = \frac{1}{\pi}\left(\frac{K_{IC}}{\beta\,\Delta\sigma}\right)^2 = \frac{1}{\pi}\left(\frac{2700}{1.12\times200}\right)^2 = \boxed{46.2\ \text{mm}}$$
Detection length. The inspection must catch the crack before it reaches half of that: $$a_f = \tfrac{1}{2}a_c = 23.1\ \text{mm}$$
Integrate the Paris law. Substituting $\Delta K$ into $\mathrm{d}a/\mathrm{d}N = C(\Delta K)^4$ and separating variables, $$N = \int_{a_0}^{a_f}\frac{\mathrm{d}a}{C\left(\beta\Delta\sigma\right)^4\pi^2 a^2} = \frac{1}{C'}\left(\frac{1}{a_0}-\frac{1}{a_f}\right), \qquad C' = C(\beta\Delta\sigma)^4\pi^2$$ The exponent $m = 4$ is what makes the integrand $a^{-2}$ and the integral closed-form.
Evaluate the lumped constant. $$C' = (31\times10^{-15})(1.12\times200)^4\pi^2 = 7.703\times10^{-4}\ \text{mm}^{-1}\text{/cycle}$$
Maintenance interval. $$N = \frac{1}{7.703\times10^{-4}}\left(\frac{1}{0.55}-\frac{1}{23.1}\right) = 1298\,(1.8182-0.0432) = \boxed{2304\ \text{cycles}}$$

Question 2 — crack length against cycles from the integrated Paris law. Almost all of the life is spent while the crack is small; once it passes the inspection length the remaining life collapses.

The shape of that curve is the whole engineering point. Of the 2332 cycles available before fast fracture, 2304 are spent getting from 0.55 mm to 23.1 mm and only about 28 more are needed to run from 23.1 mm to the critical 46.2 mm. An inspection interval set at the full critical length would therefore be worthless: the "safety margin" would be one per cent of the life. Halving the critical length before setting the interval, as the question requires, is the standard damage-tolerance device for buying back that margin.

Quantity	Value
Initial stress-intensity range ΔK	294 N/mm^3/2
Critical crack length a_c	46.2 mm
Crack length to be detected a_f	23.1 mm
Lumped Paris constant C′	7.703 × 10⁻⁴ mm⁻¹/cycle
Maintenance interval	2304 cycles
Cycles remaining from a_f to a_c	28 cycles

Topic. Damage-tolerant life from the Paris crack-growth law.

Key relations. $\Delta K = \beta\,\Delta\sigma\sqrt{\pi a}$; $\mathrm{d}a/\mathrm{d}N = C(\Delta K)^m$; for $m=4$, $N = \left(1/a_0 - 1/a_f\right)/\left[C(\beta\Delta\sigma)^4\pi^2\right]$; fast fracture at $\beta\Delta\sigma\sqrt{\pi a_c} = K_{IC}$.

Why this works. Damage tolerance assumes a crack is already present and asks how long it takes to become dangerous. The Paris law makes the growth rate a power function of the stress-intensity range, so life is dominated by the small-crack end where $\Delta K$ is low; integrating that rate gives cycles directly. Because $\Delta K$ grows as $\sqrt{a}$ while the rate grows as the fourth power of $\Delta K$, the rate grows as $a^2$ and the curve turns almost vertical near $a_c$. The inspection interval must therefore be set well inside that knee, which is why the question halves the critical length.

Common pitfall. Using the crack length as the semi-length (the $\beta = 1.12$ edge-crack factor already assumes a is the full surface length), or forgetting $\beta$ altogether — it enters to the fourth power here and changes the life by more than 50%. Also watch the units: with stress in N/mm² and a in mm the toughness must be in N/mm^3/2, not MPa√m.

Source. Dowling, Mechanical Behavior of Materials, 4th ed., ch. 11 (fatigue crack growth); Megson, Aircraft Structures for Engineering Students, 6th ed., §15.5 (damage tolerance and inspection intervals).

Question 2 of 8: Crack growth life and the inspection interval

Question 2: Crack growth life and the inspection interval (20 marks)