Damage-tolerant inspection interval for an edge-cracked panel: 22-Mec-B9 Advanced Engineering Structures, May 2013, Q1

Question 1: Damage-tolerant inspection interval for an edge-cracked panel (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 normal to the crack.

Quantity	Symbol	Value
Initial (detected) crack length	$a_{0}$	0.25 mm
Stress range normal to the crack	$\Delta\sigma$	200 N/mm$^{2}$
Fracture toughness	$K_{\mathrm{IC}}$	3000 N/mm$^{3/2}$
Crack growth law	$\mathrm{d}a/\mathrm{d}N$	$35\times10^{-15}(\Delta K)^{4}$ mm/cycle
Edge-crack geometry factor	$Y$	1.12

Find. The number of load cycles available between the initial crack size and one half of the critical crack size, and hence the longest defensible interval between inspections.

Crack length against cycles from the integrated Paris law. Almost the whole life is consumed while the crack is small, which is exactly why the inspection threshold must be set well below the critical size.

Approach. Fix the critical crack length from the fracture toughness, reduce the Paris law to a function of crack length alone, then separate the variables and integrate between the initial and target crack sizes.

Write the stress-intensity factor for the geometry. For a single edge crack in a semi-infinite plate the standard result is $K = Y\,\sigma\sqrt{\pi a}$ with $Y = 1.12$, a factor that is constant with crack length for this geometry — which is what makes the integration below closed-form. The applied range is therefore $\Delta K = 1.12\,\Delta\sigma\sqrt{\pi a}$.
Locate the critical crack length. Fast fracture occurs when $K$ reaches $K_{\mathrm{IC}}$ at the peak of the cycle:$$a_{c} = \frac{1}{\pi}\left(\frac{K_{\mathrm{IC}}}{1.12\,\Delta\sigma}\right)^{2} = \frac{1}{\pi}\left(\frac{3000}{1.12\times200}\right)^{2} = 57.09\ \text{mm}$$The panel must therefore be inspected before the crack reaches $\boxed{a_{c}/2 = 28.55\ \text{mm}}$.
Reduce the growth law to a function of $a$ alone. Substituting $\Delta K$ and using $m = 4$ collapses the constants into a single coefficient:$$\frac{\mathrm{d}a}{\mathrm{d}N} = 35\times10^{-15}\left(1.12\,\Delta\sigma\right)^{4}\pi^{2}a^{2} = 8.6968 \times 10^{-4}\,a^{2}\ \text{mm/cycle}$$The exponent $m = 4$ is what makes the crack length appear squared, and it is that square that concentrates the life at small crack sizes.
Separate the variables and integrate. With $C' = 8.6968 \times 10^{-4}$,$$N = \int_{a_{0}}^{a_{c}/2}\frac{\mathrm{d}a}{C'a^{2}} = \frac{1}{C'}\left(\frac{1}{a_{0}} - \frac{2}{a_{c}}\right)$$
Evaluate. Substituting $a_{0} = 0.25$ mm and $a_{c}/2 = 28.55$ mm gives $1/a_{0} = 4.000$ and $2/a_{c} = 0.0350$ mm$^{-1}$, so$$N = \frac{4.000 - 0.0350}{8.6968 \times 10^{-4}} = \boxed{4\,559\ \text{cycles}}$$Note how little the upper limit matters: dropping the second term entirely would change the answer by under one per cent, because the crack spends its life growing through the first millimetre.
Convert the growth life into a maintenance interval. The inspection programme must guarantee that a crack of detectable size is found before it reaches $a_{c}/2$. Setting the interval at half the available life places at least two inspections inside the growth window, giving an interval of about $2\,280$ cycles — the usual damage-tolerance factor of two on inspection frequency.

Quantity	Result
Critical crack length $a_{c}$	57.1 mm
Inspection threshold $a_{c}/2$	28.5 mm
Cycles from 0.25 mm to $a_{c}/2$	4 559 cycles
Recommended maintenance interval	≈ 2 280 cycles (factor of two)

Check

The solution assumes the geometry factor stays at 1.12 over the whole growth range (exact for a semi-infinite plate), that the whole stress range is effective so that no threshold $\Delta K_{th}$ and no crack closure need be deducted, and that the quoted growth law already carries units of mm/cycle with $\Delta K$ in N/mm$^{3/2}$. Any one of these assumptions being relaxed lengthens rather than shortens the predicted life, so the interval above is the conservative one.

Concept & Theory

Topic. Damage-tolerant life prediction by integration of the Paris law.

Key relations. $K = Y\sigma\sqrt{\pi a}$; fast fracture at $K = K_{\mathrm{IC}}$; $\mathrm{d}a/\mathrm{d}N = C(\Delta K)^{m}$; for $m = 4$ and constant $Y$, $N = (1/C')\,(a_{0}^{-1} - a_{f}^{-1})$ with $C' = C(Y\Delta\sigma)^{4}\pi^{2}$.

Why this works. Damage tolerance accepts that a crack exists and asks only how long it takes to become dangerous. Because the Paris exponent for structural alloys is close to four, the growth rate scales with the square of crack length, so the integral is dominated by its lower limit: the life is set almost entirely by the smallest crack the inspection method can reliably find. This is why the flaw size assumed in certification is the non-destructive-inspection detection limit rather than anything measured on the structure, and why halving the detectable flaw size buys far more life than raising the toughness.

Common pitfall. Two errors recur. The first is dropping the 1.12 edge-crack factor and using the centre-crack form $K = \sigma\sqrt{\pi a}$, which raises the predicted life by $1.12^{4} \approx 1.57$, an unconservative 57 per cent. The second is integrating to $a_{c}$ instead of the required $a_{c}/2$; here that mistake is nearly invisible in the arithmetic, which is precisely what makes it dangerous to trust the number without checking what the question asked for.

Source. Dowling, Mechanical Behavior of Materials, 4th ed., §8.4 (stress-intensity solutions) and §11.4–11.6 (fatigue crack growth and life estimates); Megson, Aircraft Structures, 6th ed., §10.7 (fatigue and damage tolerance).

Question 1 of 8: Damage-tolerant inspection interval for an edge-cracked panel

Question 1: Damage-tolerant inspection interval for an edge-cracked panel (20 marks)