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22-Mec-B9 Advanced Engineering Structures · May 2014

Question 2 of 8: Damage-tolerant inspection interval for an edge-cracked skin panel

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

Paper format. National Exams, May 2014 — 07-Mec-B9 Advanced Engineering Structures. Three hours, open book, any non-communicating calculator permitted. Eight questions of equal total value (20 marks each); the rubric states that any five constitute a complete paper. All eight are solved here, because the complete set is the study resource.

Reference texts.

Axes and sign convention used throughout. For every thin-walled question a right-handed set is used: the span axis runs along the member, $Y$ is vertical (upward) in the cross-section, and $Z$ is horizontal. A vertical shear $S_y$ acting on a section whose product of inertia vanishes produces the open-section shear flow $q_s = -\dfrac{S_y}{I_{zz}}\displaystyle\int_0^s y\,t\,\mathrm{d}s$, measured positive in the direction of increasing $s$. Closed cells add the constant redundant flow $q_{s,0}$ fixed either by zero twist (shear-centre problems) or by moment equilibrium (a load of known line of action). All shear flows below are quoted with respect to a stated walking direction, and each question closes with the two statical checks — the resolved vertical force must return $S_y$ and the moment of the flows about a chosen origin must return the moment of the applied load.

Question 2: Damage-tolerant inspection interval for an edge-cracked skin 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. An edge crack in a semi-infinite plate under constant amplitude loading, with a Paris-law growth exponent of four.

QuantitySymbolValue
Initial (detected) crack length$a_0$0.22 mm
Stress range normal to the crack$\Delta\sigma$160 N/mm$^{2}$
Fracture toughness$K_{\mathrm{IC}}$2900 N/mm$^{3/2}$
Growth law$\mathrm{d}a/\mathrm{d}N$$32\times10^{-15}(\Delta K)^{4}$ mm/cycle
Geometry factor, edge crack in a semi-infinite plate$Y$1.12

Find. The number of load cycles available before the crack reaches half of its critical length — that number is the maintenance (inspection) interval.

Question 2 — crack growth from the detected size to half the critical length03 0006 0009 00012 00014 00001020304045cycles Ncrack length a (mm)a grows slowly for most of the life, then runs awayN = 13 883 cycles at a = 41.68 mm
Integrated Paris-law growth. The crack spends nearly the whole interval at very small sizes: 10 886 cycles to reach 1 mm and 13 649 to reach 10 mm, after which the last 32 mm of growth costs only 233 cycles.

Approach. Find the critical crack length from the fracture toughness, halve it to get the target size, then integrate the Paris law between the detected size and that target; because the exponent is four and $\Delta K$ varies as $\sqrt{a}$, the integral is elementary.

  1. Stress intensity factor for the geometry. An edge crack of length $a$ in a semi-infinite plate loaded normal to the crack has the standard free-edge correction $Y = 1.12$, so $$\Delta K = Y\,\Delta\sigma\sqrt{\pi a} = 1.12(160)\sqrt{\pi a} = 179.2\sqrt{\pi a}\ \ \text{N/mm}^{3/2}$$ with $a$ in mm. The units are consistent throughout because stress is in N/mm2 and length in mm.
  2. Critical crack length. Fast fracture occurs when $\Delta K$ reaches the toughness: $$1.12(160)\sqrt{\pi a_c} = 2900 \;\Rightarrow\; \sqrt{\pi a_c} = \frac{2900}{179.2} = 16.183 \;\Rightarrow\; a_c = \frac{(16.183)^{2}}{\pi} = \boxed{83.36\ \text{mm}}$$ The crack must therefore be found before it reaches $a_f = a_c/2 = 41.68$ mm.
  3. Reduce the Paris law to a form in the crack length alone. Raising $\Delta K$ to the fourth power collapses the square roots: $$\begin{aligned}\frac{\mathrm{d}a}{\mathrm{d}N} &= C(\Delta K)^{4} = C\bigl(Y\Delta\sigma\bigr)^{4}\pi^{2}a^{2} = C^{\prime}a^{2}, \\ C^{\prime} &= 32\times10^{-15}(179.2)^{4}\pi^{2}\end{aligned}$$$$C^{\prime} = 32\times10^{-15}\bigl(1.0312\times10^{9}\bigr)(9.8696) = 3.2569\times10^{-4}\ \text{mm}^{-1}\text{/cycle}$$ The exponent $m = 4$ is what makes this integrable in closed form; any other value would give a different power of $a$ but the same method.
  4. Integrate between the detected size and the target size. Separating variables, $$N = \int_{a_0}^{a_f}\frac{\mathrm{d}a}{C^{\prime}a^{2}} = \frac{1}{C^{\prime}}\left[\frac{1}{a_0}-\frac{1}{a_f}\right] = \frac{1}{0.00032569}\left[\frac{1}{0.22}-\frac{1}{41.681}\right]$$$$N = 3070.4\,(4.54545 - 0.02399) = \boxed{13\,883\ \text{cycles}}$$
  5. Interpret the answer as a maintenance interval. Because $1/a_f$ is only 0.5 per cent of $1/a_0$, virtually the whole life is spent growing the crack out of the small-crack regime: $\Delta K$ rises from 149 N/mm$^{3/2}$ at detection to 2,051 N/mm$^{3/2}$ at the target size. Inspecting the panel at least every 13,883 cycles therefore guarantees that a crack of the assumed initial size is found while it is still shorter than half the critical length.
QuantitySymbolResult
Stress intensity range$\Delta K$$179.2\sqrt{\pi a}$ N/mm$^{3/2}$
Critical crack length$a_c$83.36 mm
Target (half-critical) length$a_f$41.68 mm
Collapsed growth constant$C^{\prime}$3.2569 × 10−4 mm−1/cycle
Maintenance interval$N$13,883 cycles
Check: 13,883 cycles is the whole life from the detected size to half-critical, so it is the longest defensible interval. Damage tolerance practice (and CCAR/FAR continued-airworthiness reasoning) normally divides it by two or more, so that at least two inspections occur before the crack could reach the target size; that would give an interval of about 6,941 cycles. The question asks only for the interval implied by the growth calculation, which is the boxed value.