Paris-law crack growth and the required inspection interval: 22-Mec-B9 Advanced Engineering Structures, December 2017, Q6

Question 6: Paris-law crack growth and the required 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 panel treated as a semi-infinite plate, for which the stress-intensity factor is $K = 1.12\,\sigma\sqrt{\pi a}$, under constant-amplitude cyclic loading.

Given data — panel, crack and material
Quantity	Symbol	Value
Initial (detectable) crack length	$a_0$	0.25 mm
Cyclic stress range normal to the crack	$\Delta\sigma$	105 N/mm$^{2}$
Fracture toughness	$K_{IC}$	2780 N/mm$^{3/2}$
Paris coefficient	$C$	$39\times10^{-15}$
Paris exponent	$m$	4
Edge-crack geometry factor	$Y$	1.12

Find. The number of service load cycles between inspections such that a crack is certain to be found before it reaches half its critical length.

Crack length against accumulated cycles. Growth is extremely slow at first — the rate goes as $a^{2}$ for $m = 4$ — and then accelerates sharply, which is why almost the whole interval is consumed while the crack is still short.

Approach. Find the critical crack length from the fracture toughness, halve it to get the inspection limit, then integrate the Paris law in closed form between the initial and limiting crack lengths.

Write the stress-intensity factor. For an edge crack in a semi-infinite plate the standard result is $K = 1.12\,\sigma\sqrt{\pi a}$, the factor 1.12 being the free-edge correction to the through-crack solution.
Find the critical crack length. Fast fracture occurs when $K$ reaches $K_{IC}$ at the peak of the cycle, so $$a_c = \frac{1}{\pi}\left(\frac{K_{IC}}{1.12\,\Delta\sigma}\right)^{2} = \frac{1}{\pi}\left(\frac{2780}{1.12 \times 105}\right)^{2} = 177.9\ \text{mm}.$$ Substituting back, $1.12(105)\sqrt{\pi(177.9)} = 2780$ N/mm$^{3/2}$, which confirms the inversion.
Set the inspection limit. The maintenance policy requires detection before the crack reaches half its critical length, so the limiting size is $$a_f = \tfrac{1}{2}a_c = 88.94\ \text{mm}.$$
Substitute the stress-intensity range into the Paris law. With $\Delta K = 1.12\,\Delta\sigma\sqrt{\pi a}$, $$\frac{\mathrm{d}a}{\mathrm{d}N} = C\left(1.12\,\Delta\sigma\right)^{4}\pi^{2}a^{2} \equiv C' a^{2}, \qquad C' = 39\times10^{-15}(117.6)^{4}\pi^{2} = 7.3620\times10^{-5}.$$ The exponent $m = 4$ is what makes the right-hand side a clean power of $a$ and the integration elementary.
Integrate between the two crack lengths. Separating variables, $$N = \int_{a_0}^{a_f}\frac{\mathrm{d}a}{C' a^{2}} = \frac{1}{C'}\left(\frac{1}{a_0}-\frac{1}{a_f}\right) = \frac{1}{7.3620\times10^{-5}}\left(\frac{1}{0.25}-\frac{1}{88.94}\right).$$
Evaluate the interval. The bracket is $4.000 - 0.0112 = 3.9888$ mm$^{-1}$, so $$\boxed{N = 5.42\times10^{4}\ \text{cycles} \ \ (54\,181\ \text{cycles})}$$ and this is the maximum permissible interval between inspections.
Notice how little the final crack size matters. The term $1/a_f$ contributes only $0.28$ % of the bracket. Almost the entire life is spent growing the crack through its first few millimetres, so the answer is governed overwhelmingly by the smallest crack the inspection method can reliably detect — halving $a_0$ from 0.25 to 0.125 mm would roughly double the permissible interval.
Cross-check by numerical integration. Evaluating $\int_{a_0}^{a_f}\mathrm{d}a/\left[C(1.12\Delta\sigma\sqrt{\pi a})^{4}\right]$ numerically with 400 000 strips returns 54 180.6 cycles, agreeing with the closed form to six significant figures.
Express the result as an operating rule. The panel must be inspected at least every 54 000 cycles, and in practice a scatter factor of two or three would be applied to that figure, giving a scheduled interval of roughly 18 000–27 000 cycles so that any crack is found on at least two successive inspections before it becomes critical.

Final results — Question 6
Quantity	Value
Critical crack length	$a_c = 177.9$ mm
Inspection limit (half critical)	$a_f = 88.94$ mm
Lumped Paris constant	$C' = 7.362\times10^{-5}$ mm$^{-1}$/cycle
Growth rate at $a_0$	$4.60\times10^{-6}$ mm/cycle
Required maintenance interval	54 181 cycles ($5.42\times10^{4}$)

Topic. Damage-tolerant design: linear elastic fracture mechanics and the Paris law.

Key relations. $K = Y\sigma\sqrt{\pi a}$ with $Y = 1.12$ for an edge crack in a semi-infinite plate; critical size $a_c = (K_{IC}/Y\Delta\sigma)^{2}/\pi$; Paris law $\mathrm{d}a/\mathrm{d}N = C(\Delta K)^{m}$; for $m = 4$ and constant $Y$, $N = \left(1/a_0 - 1/a_f\right)/\left[C(Y\Delta\sigma)^{4}\pi^{2}\right]$.

Why this works. Damage-tolerant design assumes a crack is already present at entry into service and asks how long it takes to become dangerous. The Paris law captures the middle, stable regime of crack growth, in which the rate depends only on the stress-intensity range. Because $\Delta K$ grows as $\sqrt{a}$, the growth rate for $m = 4$ grows as $a^{2}$, so the crack accelerates catastrophically near the end — the reason the schedule must be set on the early part of the curve and why halving the crack-size criterion barely changes the answer while halving the detectable size nearly doubles it.

Common pitfall. Forgetting the 1.12 edge correction: it enters $C'$ to the fourth power and would change the interval by 57 %. A second trap is integrating to $a_c$ rather than to $a_c/2$ as the question asks, and a third is dropping the $\pi^{2}$ that comes from $(\sqrt{\pi a})^{4}$.

Source. Dowling, Mechanical Behavior of Materials, 4th ed., §11.2 (stress intensity for common geometries) and §11.5–11.6 (fatigue crack growth, life integration).

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

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