Question 7 of 8: Inspection interval from fatigue crack growth

Question 7: Inspection interval from fatigue crack growth (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.

Find. The number of cycles for the crack to grow from $a_0$ to half the critical length — that is, the longest inspection interval that guarantees the crack is found before it reaches that size.

Approach. Find the critical crack length from the fracture toughness, halve it to get the detection size, then integrate the Paris law between the two crack lengths — which for an exponent of 4 has a closed form.

1. Write the stress-intensity factor. For an edge crack in a semi-infinite plate the standard geometry factor is $\beta = 1.12$, so$$K = 1.12\,\sigma\sqrt{\pi a}$$At the starting crack this gives $K = 1.12(210)\sqrt{\pi(0.28)} = 220.6$ N/mm$^{3/2}$, about a tenth of the toughness — comfortably in the stable-growth regime, which justifies using a growth law rather than declaring immediate failure.
2. Find the critical crack length. Fracture occurs when $K$ reaches $K_c$:$$a_c = \frac{1}{\pi}\left(\frac{K_c}{1.12\,\sigma}\right)^2 = \frac{1}{\pi}\left(\frac{2150}{235.2}\right)^2 = 26.60\ \text{mm}$$The question asks for detection at half of this, so the target crack length is$$a_f = \frac{a_c}{2} = 13.30\ \text{mm}$$
3. Reduce the growth law to a separable form. With $\Delta K = 1.12\,\sigma\sqrt{\pi a}$ and an exponent of 4, the square roots disappear and the crack length appears only as $a^2$:$$\frac{\mathrm{d}a}{\mathrm{d}N} = C\big(1.12\sigma\big)^4\pi^2 a^2$$This is the reason the examiner chose $m = 4$; for a general exponent the integral is still elementary but far messier.
4. Integrate between the two crack lengths. Separating and integrating,$$N = \frac{1}{C(1.12\sigma)^4\pi^2}\int_{a_0}^{a_f}\frac{\mathrm{d}a}{a^2} = \frac{1}{C(1.12\sigma)^4\pi^2}\left(\frac{1}{a_0}-\frac{1}{a_f}\right)$$The denominator evaluates as $(235.2)^4 = 3.0602\times10^{9}$, so $C(1.12\sigma)^4\pi^2 = 1.0568\times10^{-3}$, and the bracket is $3.5714-0.0752 = 3.4962$ mm$^{-1}$.
5. Evaluate the interval. Dividing,$$\boxed{N = \frac{3.4962}{1.0568\times10^{-3}} = 3308\ \text{cycles}}$$ The maintenance interval must therefore not exceed roughly 3300 cycles.
6. Check how much reserve that leaves. Integrating on from $a_f$ to $a_c$ with the same expression gives only 36 further cycles, so the full life from $a_0$ to fracture is 3343 cycles. The second half of the crack’s journey — from 13.3 mm to 26.6 mm — occupies just 1.1 per cent of the life, because the growth rate has risen from $8.3\times10^{-5}$ mm/cycle at $a_0$ to 0.19 mm/cycle at $a_f$.

That last figure explains the whole logic of the $a_c/2$ criterion and is the substantive engineering point of the question. Because the growth rate scales as $a^2$ here, the crack spends almost its entire life short and then becomes critical very suddenly. Setting the inspection threshold at half the critical length does not buy half the remaining life — it buys about one per cent of it. What the criterion really buys is detectability: a 13 mm crack is reliably found by routine non-destructive inspection, whereas the 0.28 mm starting flaw is at or below the detection threshold of most field methods. The interval must therefore be short enough that two successive inspections bracket the window in which the crack is both detectable and still safe, and in practice a further factor of two or three would be applied to the 3308 cycles computed here so that any crack is seen at least twice before it matters.