Crack growth and the maintenance interval for a cracked wing skin: 22-Mec-B9 Advanced Engineering Structures, Undated paper, Q5

Question 5: Crack growth and the maintenance interval for a cracked wing skin (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. An aircraft wing skin panel treated as a semi-infinite plate with an edge crack under constant-amplitude cyclic loading.

Quantity	Symbol	Value
Initial (detectable) crack length	a₀	0.28 mm
Crack-driving stress range	Δσ	248 N/mm²
Fracture toughness	K_IC	2300 N/mm^3/2
Growth law	da/dN	35 × 10⁻¹⁵ (ΔK)⁴ mm/cycle
Edge-crack geometry factor	β	1.12 (semi-infinite plate)

Find. The maintenance (inspection) interval, in cycles, that guarantees the crack is detected before it reaches half the critical length at which the panel fractures.

Figure 5.1 — left: the edge-cracked panel under a remote stress range Δσ. Right: the integrated crack-growth history a(N). Growth is almost flat for most of the interval and then runs away as a approaches a_c, which is why the inspection interval is set on the shallow part of the curve.

Approach. Find the critical crack length from the fracture toughness, halve it to get the inspection target, then integrate the Paris law between the initial and target lengths — with $m = 4$ the integral is elementary.

Stress-intensity factor for the geometry. For an edge crack of length $a$ in a semi-infinite plate under a remote stress $\sigma$ normal to the crack, $$K = \beta\,\sigma\sqrt{\pi a},\qquad \beta = 1.12,$$ the 1.12 accounting for the free surface at the crack mouth. At the detectable length, $K_0 = 1.12 \times 248 \times \sqrt{\pi \times 0.28} = 260.5$ N/mm^3/2, comfortably below $K_{IC}$, so linear-elastic fracture mechanics applies from the outset.
Critical crack length. Fracture occurs when $K$ reaches the toughness: $$a_c = \frac{1}{\pi}\left(\frac{K_{IC}}{\beta\,\Delta\sigma}\right)^2 = \frac{1}{\pi}\left(\frac{2300}{1.12 \times 248}\right)^2 = 21.83 \text{ mm.}$$
Set the inspection target. The requirement is to detect the crack before it reaches half the critical length, so the interval is sized on $$a_f = \tfrac{1}{2}a_c = 10.913 \text{ mm.}$$ Stopping at half of $a_c$ rather than at $a_c$ itself is what turns a fracture calculation into an inspection schedule: it leaves a factor-of-two margin on crack length against scatter in the growth law and in the detection threshold.
Substitute the growth law and separate variables. With $da/dN = C(\Delta K)^4$ and $\Delta K = \beta\Delta\sigma\sqrt{\pi a}$, $$\frac{da}{dN} = C\,\beta^4 \Delta\sigma^4 \pi^2 a^2 = C' a^2, \qquad C' = C\,(\beta\Delta\sigma)^4 \pi^2.$$ Numerically, $C' = 3.5e-14 \times (1.12 \times 248)^4 \times \pi^2 = 2.0561e-03$ per mm per cycle. The exponent $m = 4$ is what makes the integral elementary: the crack length appears simply as $a^2$.
Integrate from the detectable length to the target length. $$N = \int_{a_0}^{a_f}\frac{da}{C'a^2} = \frac{1}{C'}\left[\frac{1}{a_0} - \frac{1}{a_f}\right] = \frac{1}{2.0561e-03}\left[\frac{1}{0.28} - \frac{1}{10.913}\right].$$ The bracket evaluates to $3.5714 - 0.0916 = 3.4798$ mm⁻¹, giving $$\boxed{\;N = 1692 \text{ cycles}\;}$$
Interpret the result as a maintenance rule. Because $1/a_f$ contributes only 2.6 % of the bracket, almost the whole interval is spent growing the crack through its first millimetre — the life is governed by how small a crack the inspection method can find, not by how large a crack the panel can tolerate. The panel must therefore be inspected at intervals no longer than 1692 cycles; halving the detectable crack size to 0.14 mm would roughly double the permissible interval, whereas doubling the tolerable final length would change it by only a few percent.

Quantity	Symbol	Value
Stress intensity at the initial crack	K₀	260.5 N/mm^3/2
Critical crack length	a_c	21.83 mm
Inspection target length	a_f = a_c/2	10.913 mm
Modified growth constant	C′	2.0561e-03 mm⁻¹/cycle
Maintenance interval	N	1692 cycles

Topic. Fatigue crack propagation by the Paris law, and the damage-tolerance inspection interval it defines.

Key relations. $K = \beta\sigma\sqrt{\pi a}$ with $\beta = 1.12$ for an edge crack in a semi-infinite plate; $a_c = (K_{IC}/\beta\Delta\sigma)^2/\pi$; $da/dN = C(\Delta K)^m$; for $m = 4$, $N = \dfrac{1}{C(\beta\Delta\sigma)^4\pi^2}\left[\dfrac{1}{a_0} - \dfrac{1}{a_f}\right]$.

Why this works. Damage-tolerant design accepts that cracks exist and requires only that they be found before they become dangerous. The Paris law makes that quantitative: it says the growth rate depends on the stress state only through the cyclic stress-intensity range, so a single laboratory calibration of $C$ and $m$ transfers to any geometry for which $\beta$ is known. For $m = 4$ the integral collapses to the difference of two reciprocals, which shows immediately that the initial length dominates the life — the small-crack end of the integral is where all the cycles are.

Common pitfall. Forgetting the 1.12 free-surface factor raises the computed life by about 57 % (it enters as $\beta^4$), and omitting it inside $a_c$ moves the critical length by 25 % (it enters there as $\beta^2$). The other frequent slip is integrating all the way to $a_c$ when the question asks for a fraction of it: here that would give 1715 cycles instead of 1692, an unconservative inspection interval.

Source. Dowling, Mechanical Behavior of Materials, 4th ed., §11.3–11.5; Megson, Aircraft Structures, 6th ed., §12.5 (crack propagation and inspection intervals).

Question 5 of 8: Crack growth and the maintenance interval for a cracked wing skin

Question 5: Crack growth and the maintenance interval for a cracked wing skin (20 marks)