Paris-law crack growth and the resulting inspection interval: 22-Mec-B9 Advanced Engineering Structures, December 2014, Q6

Question 6: Paris-law crack growth and the resulting 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. An edge-cracked semi-infinite plate under constant-amplitude tension, with a Paris-law growth rate of exponent four. The geometry factor for a single edge crack in a semi-infinite plate is the standard $Y = 1.12$.

Quantity	Symbol	Value
Initial (detectable) crack length	$a_0$	1 mm
Constant-amplitude stress range	$\Delta\sigma$	100 N/mm²
Fracture toughness	$K_{IC}$	3 000 N/mm^3/2
Paris coefficient and exponent	$C,\ m$	$35\times10^{-15}$ mm/cycle, 4
Edge-crack geometry factor	$Y$	1.12

Find. The number of load cycles — the inspection interval — in which the 1 mm crack grows to one half of its critical length.

Figure Q6 — crack length against cycles from the integrated Paris law. Because the growth rate scales with $a^{2}$ for $m = 4$, almost the entire interval is spent while the crack is short; the last few millimetres take a negligible fraction of the life.

Approach. Find the crack length at which the stress intensity reaches the toughness, halve it to get the target, then integrate the Paris law between the initial and target lengths — which, for $m = 4$ and a constant geometry factor, has a closed form.

Critical crack length. Fast fracture occurs when the stress intensity reaches the toughness, $K = Y\sigma\sqrt{\pi a} = K_{IC}$, so $$a_c = \frac{1}{\pi}\left(\frac{K_{IC}}{Y\Delta\sigma}\right)^{2} = \frac{1}{\pi}\left(\frac{3000}{1.12\times100}\right)^{2}$$The bracket is $26.786$, so $a_c = 717.5/\pi$ and $$\boxed{a_c = 228.4\ \text{mm}}$$The inspection target is half of this, $a_f = 114.2$ mm.
Set up the Paris integral. With $\Delta K = Y\Delta\sigma\sqrt{\pi a}$ and $m = 4$, the growth law becomes $$\frac{da}{dN} = C\,(Y\Delta\sigma)^{4}\pi^{2}a^{2} = C'a^{2}$$a separable equation in which the growth rate is proportional to the square of the crack length. Grouping the constants, $C' = 35\times10^{-15}(112)^{4}\pi^{2} = 5.436\times10^{-5}$ per mm per cycle.
Integrate between the two crack lengths. Separating and integrating $da/a^{2}$ from $a_0$ to $a_f$, $$N = \frac{1}{C'}\left(\frac{1}{a_0}-\frac{1}{a_f}\right) = \frac{1}{C'}\left(\frac{1}{a_0}-\frac{2}{a_c}\right)$$The second form is the convenient one when the target is expressed as a fraction of the critical length.
Evaluate the interval. Substituting, $1/a_0 - 1/a_f = 1 - 0.008757 = 0.99124$ per mm, so $$\boxed{N = \frac{0.99124}{5.436\times10^{-5}} = 18\,236\ \text{cycles}}$$ Note that the $1/a_f$ term removes less than one per cent of the answer — virtually the whole interval is consumed growing the crack through its first few millimetres.
Interpretation as a maintenance interval. The panel must be inspected at least once every 18 236 load cycles for the crack to be found before it reaches half the critical length. In service practice this figure is halved again, so that two inspections occur within the growth period and a single missed detection is not catastrophic; the resulting 9 100-cycle interval is what would appear in a maintenance schedule.

Check: the calculation assumes the geometry factor stays at the semi-infinite-plate value $Y = 1.12$ over the whole growth range, as the question’s "semi-infinite plate" wording directs. On a real finite panel $Y$ rises as the crack approaches the far edge, which shortens the interval; the assumption is therefore non-conservative and is why real schedules apply the additional factor of two noted above.

Quantity	Result
Critical crack length	228.4 mm
Inspection target, $a_c/2$	114.2 mm
Lumped Paris coefficient $C'$	$5.436\times10^{-5}$ mm^-1 per cycle
Maintenance interval	18 236 cycles
Recommended scheduled interval (two inspections per growth period)	about 9 100 cycles

Concept & Theory

Topic. Linear elastic fracture mechanics and Paris-law life prediction.

Key relations. $K = Y\sigma\sqrt{\pi a}$; critical length $a_c = (K_{IC}/Y\Delta\sigma)^{2}/\pi$; Paris law $da/dN = C(\Delta K)^{m}$; for $m = 4$, $N = (1/C')(1/a_0 - 1/a_f)$ with $C' = C(Y\Delta\sigma)^{4}\pi^{2}$.

Why this works. Damage-tolerant design accepts that cracks exist and asks instead how long they take to become dangerous. The stress-intensity factor is what makes this tractable: it collapses geometry, crack size and remote stress into one number that both governs the growth rate and defines failure. With an exponent of exactly four the integral is elementary, and the $1/a$ form reveals the essential feature of crack growth — life is dominated by the smallest crack size, so detection capability, not toughness, usually sets the inspection interval.

Common pitfall. Integrating with $\Delta K$ evaluated at the initial crack length instead of as a function of $a$, which turns a decaying integrand into a constant and grossly overestimates life. The other frequent error is dropping the geometry factor, or forgetting that it is raised to the fourth power here: ignoring $Y = 1.12$ inflates the predicted interval by a factor of $1.12^{4} = 1.57$.

Source. Dowling, Mechanical Behavior of Materials, 4th ed., §8.3 (stress-intensity factors) and §11.3–11.5 (fatigue crack growth and life estimates); Megson, 6th ed., §15.4.

Question 6 of 8: Paris-law crack growth and the resulting inspection interval

Question 6: Paris-law crack growth and the resulting inspection interval (20 marks)