Paris-law inspection interval for an edge crack: 22-Mec-B9 Advanced Engineering Structures, May 2016, Q5

Question 5: Paris-law inspection interval for an edge crack (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 semi-infinite plate carrying a single edge crack under constant-amplitude cyclic tension, with a Paris exponent of exactly four.

Given data — Question 5
Quantity	Symbol	Value
Initial edge crack length	$a_{0}$	0.25 mm
Constant-amplitude stress range	$\Delta\sigma$	130 N/mm²
Fracture toughness	$K_{IC}$	2 950 N/mm^3/2
Paris coefficient	$C$	$34\times10^{-15}$ mm/cycle
Paris exponent	$m$	4
Edge-crack geometry factor	$Y$	1.12

Find. The number of cycles taken for the crack to grow from 0.25 mm to half its critical length, which is the longest permissible interval between inspections.

Approach. Get the critical crack length from the toughness, halve it to set the detection target, then integrate the Paris law between the two lengths — which for $m = 4$ has a closed form.

Write the stress intensity factor for an edge crack. For a single edge crack in a semi-infinite plate the standard result is $$K = Y\sigma\sqrt{\pi a} = 1.12\,\sigma\sqrt{\pi a}$$ The 1.12 factor is the free-surface correction; a centre crack of the same length would give $Y = 1$ and a longer life.
Find the critical crack length. Fast fracture occurs when the peak stress intensity reaches the toughness, $1.12\Delta\sigma\sqrt{\pi a_{c}} = K_{IC}$, so $$a_{c} = \frac{1}{\pi}\left(\frac{K_{IC}}{1.12\Delta\sigma}\right)^{2} = \frac{1}{\pi}\left(\frac{2950}{1.12\times130}\right)^{2} = \frac{1}{\pi}\left(20.261\right)^{2} = \boxed{130.67\ \text{mm}}$$ so the detection target is $$a_{f} = \frac{a_{c}}{2} = \boxed{65.33\ \text{mm}}$$
Set up the Paris integral. Substituting the stress intensity range into $da/dN = C\left(\Delta K\right)^{4}$ gives $$\begin{aligned} \frac{da}{dN} &= C\left(1.12\Delta\sigma\right)^{4}\pi^{2}a^{2} = C'a^{2}, \\ C' &= C\left(1.12\Delta\sigma\right)^{4}\pi^{2} \end{aligned}$$ Because the exponent is exactly four, the crack length appears only as $a^{2}$ and the integral is elementary.
Evaluate the lumped constant. $$C' = 34\times10^{-15}\times\left(145.6\right)^{4}\times\pi^{2} = 34\times10^{-15}\times4.4941\times10^{8}\times9.8696 = \boxed{1.5081\times10^{-4}}$$ in units of reciprocal millimetre-cycles.
Integrate between the two crack lengths. Separating variables, $$N = \int_{a_{0}}^{a_{f}}\frac{da}{C'a^{2}} = \frac{1}{C'}\left(\frac{1}{a_{0}}-\frac{1}{a_{f}}\right)$$ $$N = \frac{1}{1.5081\times10^{-4}}\left(\frac{1}{0.25}-\frac{1}{65.33}\right) = 6631.2\times\left(4.0000-0.0153\right) = \boxed{26\,422\ \text{cycles}}$$
Sanity-check the number against the full life. Repeating the integral to the full critical length gives 26 473 cycles, so the last half of the crack’s growth — from 65 mm to 131 mm — takes only 51 cycles, two tenths of one per cent of the life. The stress intensity range rises from 129 N/mm^3/2 at $a_{0}$ to 2086 N/mm^3/2 at $a_{f}$, and since the growth rate goes as the fourth power of that, the final stage is effectively instantaneous.
State the maintenance interval. The inspection interval must be no longer than $$\boxed{N_{\text{inspect}} = 26\,400\ \text{cycles}}$$ rounded down from 26 422, and in practice a factor of two or more would be applied to allow for scatter in $C$, for the detection threshold of the chosen inspection method, and for the fact that a crack shorter than 0.25 mm may already exist at the start of service.

Figure 5.1 — crack length against cycles from the Paris law. The curve is almost flat for the first twenty thousand cycles and then turns sharply upward, which is why the inspection interval must be set well before the target length is reached.

Question 5 — results
Quantity	Result
Critical crack length	130.67 mm
Detection target (half critical)	65.33 mm
Lumped Paris constant $C'$	$1.5081\times10^{-4}$ mm⁻¹ per cycle
$\Delta K$ at the initial crack	129.0 N/mm^3/2
$\Delta K$ at the detection target	2 086 N/mm^3/2
Cycles from 0.25 mm to 65.33 mm	26 422 cycles
Cycles to full critical length	26 473 cycles
Recommended maintenance interval	26 400 cycles (before any additional scatter factor)

Concept & Theory

Topic. Damage-tolerant life prediction: integrating the Paris law between an initial flaw and a detectable crack length to set an inspection interval.

Key relations. $K = Y\sigma\sqrt{\pi a}$ with $Y = 1.12$ for an edge crack in a semi-infinite plate; $a_{c}=\left(K_{IC}/Y\Delta\sigma\right)^{2}/\pi$; $da/dN = C\left(\Delta K\right)^{m}$; and for $m = 4$ the closed form $N = \left(1/C'\right)\left(1/a_{0}-1/a_{f}\right)$ with $C'=C\left(Y\Delta\sigma\right)^{4}\pi^{2}$.

Why this works. The Paris law is an empirical statement that crack growth per cycle depends only on the stress intensity range, which bundles geometry, crack size and load into one parameter and therefore lets a single laboratory calibration cover any component. When the exponent is exactly four the $\sqrt{\pi a}$ inside $\Delta K$ raises to $a^{2}$, the integral separates, and the result depends on the reciprocal of the initial crack length — which is why the small initial flaw, not the large final one, dominates the answer. Damage-tolerant design turns that arithmetic into an operational rule: inspect at least twice within the interval it takes a just-detectable crack to reach the critical size.

Common pitfall. Forgetting the 1.12 free-surface factor, which enters to the fourth power and would overstate the life by 57 per cent. The second trap is integrating to $a_{c}$ instead of $a_{c}/2$; here that happens to change the answer by only 0.2 per cent because the last stage is so fast, but on a material with a lower Paris exponent the difference is substantial and the question asks explicitly for half.

Source. Dowling, Mechanical Behavior of Materials, 4th ed., Ch. 8 (fracture mechanics) and Ch. 11 (fatigue crack growth); Megson, Aircraft Structures, 6th ed., Ch. 15.

Question 5 of 8: Paris-law inspection interval for an edge crack

Question 5: Paris-law inspection interval for an edge crack (20 marks)