22-Mec-B9 Advanced Engineering Structures · May 2017
Question 7 of 8: Paris-law crack growth and the inspection interval
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. National Exams, May 2017 — 16-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.
T. H. G. Megson, Aircraft Structures for Engineering Students, 6th ed. — unsymmetrical bending, shear flow in open and closed thin-walled sections, shear centre, structural idealisation, single- and multi-cell torsion (Ch. 15, 16, 17, 20, 23).
A. C. Ugural and S. K. Fenster, Advanced Strength and Applied Elasticity, 5th ed. — three-dimensional stress states, the Tresca and von Mises yield criteria, torsion of non-circular prismatic bars (Ch. 1, 4, 6).
A. P. Boresi and R. J. Schmidt, Advanced Mechanics of Materials, 6th ed. — torsion of solid rectangular sections and of multiply connected cells (Ch. 4, 6).
R. M. Jones, Mechanics of Composite Materials, 2nd ed. — orthotropic lamina stiffness, the reduced stiffness matrix and its transformation (Ch. 2).
N. E. Dowling, Mechanical Behavior of Materials, 4th ed. — strain-life fatigue, the Coffin–Manson relation, the Palmgren–Miner rule, and Paris-law crack growth (Ch. 9, 11, 14).
Question 7: Paris-law crack growth and the inspection interval (20 marks)
Given. An edge-cracked panel under constant-amplitude cyclic tension, with a measured Paris-law growth rate and a known fracture toughness.
Given data
Quantity
Symbol
Value
Initial edge-crack length
$a_0$
0.25 mm
Applied stress range
$\Delta\sigma$
238 N/mm2
Fracture toughness
$K_{IC}$
2400 N/mm3/2
Paris coefficient
$C$
$37 \times 10^{-15}$
Paris exponent
$m$
4
Geometry factor, edge crack in a semi-infinite plate
$\beta$
1.12
Find. The number of cycles in which the crack grows from 0.25 mm to half its critical length — the interval within which an inspection must catch it.
Crack length against cycles for this panel. Almost all of the life is spent while the crack is small; once it passes about half the critical length the curve turns nearly vertical, which is exactly why the inspection threshold is set at $a_c/2$.
Approach. Find the critical crack length from the fracture toughness, halve it to get the detection threshold, then integrate the Paris law analytically between the initial and threshold lengths.
Write the stress-intensity factor for this geometry. For an edge crack of length $a$ in a semi-infinite plate under remote tension, $$K = \beta \sigma \sqrt{\pi a}, \qquad \beta = 1.12,$$ the 1.12 being the free-surface correction that distinguishes an edge crack from a central crack of the same half-length.
Find the critical crack length. Fracture occurs when the peak stress intensity reaches the toughness, $\beta \Delta\sigma \sqrt{\pi a_c} = K_{IC}$, so $$a_c = \frac{1}{\pi}\left(\frac{K_{IC}}{\beta \Delta\sigma}\right)^2 = \frac{1}{\pi}\left(\frac{2400}{1.12 \times 238}\right)^2 = 25.80\ \text{mm}.$$ The detection threshold set by the question is half of this, $$a_f = \tfrac{1}{2}a_c = 12.90\ \text{mm}.$$
Substitute the stress-intensity range into the Paris law. With $\Delta K = \beta \Delta\sigma \sqrt{\pi a}$ and $m = 4$, $$\frac{\mathrm{d}a}{\mathrm{d}N} = C(\Delta K)^4 = C\left(\beta \Delta\sigma\right)^4 \pi^2 a^2 = C^{\prime} a^2,$$ where the whole constant collapses into $$C^{\prime} = C(\beta\Delta\sigma)^4\pi^2 = 37 \times 10^{-15} \times (266.56)^4 \times \pi^2 = 1.8437 \times 10^{-3}.$$ The exponent $m = 4$ is what makes the integral elementary: the crack length appears only as $a^2$.
Integrate between the two crack lengths. Separating variables, $$N = \int_{a_0}^{a_f} \frac{\mathrm{d}a}{C^{\prime}a^2} = \frac{1}{C^{\prime}}\left[\frac{1}{a_0} - \frac{1}{a_f}\right] = \frac{1}{C^{\prime}}\left[\frac{1}{a_0} - \frac{2}{a_c}\right].$$ Substituting the numbers, $$N = \frac{1}{1.8437 \times 10^{-3}}\left[\frac{1}{0.25} - \frac{1}{12.90}\right] = 542.4 \times 3.9225.$$
State the maintenance interval. $$\boxed{N = 2128\ \text{cycles}}$$ is the number of load cycles available before the crack reaches half its critical length, so an inspection capable of finding a 12.9 mm crack must be scheduled at least this often. In practice a factor is applied to this figure — typically the interval is set at half the computed life, here about 1060 cycles, so that two inspections occur before the threshold is reached.
Check the shape of the answer. Because the growth rate goes as $a^2$, the term $1/a_0 = 4.00$ dominates the bracket and $1/a_f = 0.0775$ contributes under 2 %. Growing the crack the rest of the way, from 12.9 mm to the critical 25.8 mm, takes only a further 21 cycles — 1 % of the life computed above. That extreme asymmetry is the justification for the $a_c/2$ threshold: past it, there is effectively no warning left.
It is worth being explicit that the whole calculation rests on the initial flaw size. Had the assumed initial crack been 0.5 mm instead of 0.25 mm, the interval would very nearly halve, to 1043 cycles; the detectable-flaw size assumed by the inspection method is therefore as much a design decision as the material choice.
