Fatigue-crack growth in an edge-cracked plate: 22-Mec-B9 Advanced Engineering Structures, December 2019, Q5

Question 5: Fatigue-crack growth in an edge-cracked plate (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 single edge crack in a plate wide enough to be treated as semi-infinite, under constant-amplitude repeated loading, with a Paris growth law of exponent four.

Fatigue data
Quantity	Value
Initial crack length a₀	0.4 mm
Repeated stress range Δσ	150 N/mm²
Fracture toughness K_IC (as printed)	180 N/mm³
Paris coefficient C, exponent m	28.5 × 10⁻¹⁵, 4
Edge-crack geometry factor Y	1.12

Find. (a) the crack length at which fast fracture occurs, and (b) the number of constant-amplitude cycles taken to grow the crack from 0.4 mm to that length.

Figure 5.1 — Edge-cracked plate and the resulting crack-length history. With an exponent of four the crack spends most of its life almost stationary and then runs away over the last few hundred cycles.

Check: the printed unit for fracture toughness is dimensionally impossible, and at the printed magnitude the question has no answer. Stress intensity has units of stress × √length, so N/mm³ cannot be a toughness; the only self-consistent unit here is N/mm^3/2, which is also what the Paris coefficient of 28.5 × 10⁻¹⁵ implies. Taken at face value, K_IC = 180 N/mm^3/2 gives a critical crack length of 0.365 mm, which is shorter than the 0.4 mm crack the plate starts with: the stress intensity at the initial crack is already 188.3 N/mm^3/2, so the plate would fracture on the first application of load and both parts would be vacuous. This solution therefore adopts K_IC = 1800 N/mm^3/2, and tabulates the two other candidate readings in step 5. The reassuring outcome is that the cycle count barely moves between the physically admissible readings — only the critical crack length does.

Approach. Set the stress-intensity range at the crack tip equal to the toughness to obtain the critical crack length, then integrate the Paris law in closed form between the initial and critical lengths.

Write the stress-intensity range. For a single edge crack in a semi-infinite plate the geometry factor is 1.12, so $$\Delta K=Y\,\Delta\sigma\sqrt{\pi a}=1.12(150)\sqrt{\pi a}$$ At the starting crack this is $\Delta K=188.3\ \text{N/mm}^{3/2}$, a number worth computing early because it is what exposes the units problem discussed above.
Part (a) — find the critical crack length. Fast fracture occurs when the peak stress intensity reaches the toughness. The loading is "repeated", meaning it cycles between zero and the maximum, so the peak intensity equals the range and $$a_c=\frac{1}{\pi}\left(\frac{K_{IC}}{Y\,\Delta\sigma}\right)^{2}=\frac{1}{\pi}\left(\frac{1800}{1.12\times150}\right)^{2}$$ $$\boxed{a_c = 36.54\ \text{mm}}$$
Reduce the Paris law to a single power of a. Substituting the stress-intensity expression into $da/dN=C(\Delta K)^m$ with $m=4$ makes the bracket a clean square: $$\frac{da}{dN}=C\,(Y\Delta\sigma)^4\pi^2a^2=C'a^2,\qquad C'=C(Y\Delta\sigma)^4\pi^2=2.2407{\times}10^{-4}$$ The exponent of four is what makes the integral elementary; any other exponent would leave a fractional power of $a$.
Part (b) — integrate to failure. Separating variables between the initial and critical lengths, $$N=\int_{a_0}^{a_c}\frac{da}{C'a^{2}}=\frac{1}{C'}\left(\frac{1}{a_0}-\frac{1}{a_c}\right)=\frac{1}{2.2407{\times}10^{-4}}\left(\frac{1}{0.4}-\frac{1}{36.54}\right)$$ $$\boxed{N = 11\,035\ \text{cycles}}$$ The bracket is $2.500-0.027$, so the critical length contributes barely one per cent of the answer: the life is set almost entirely by how small the starting crack is.
Show how little the reading of K_IC matters. Because the second term in the bracket is so small, the life is nearly independent of the toughness. Taking the alternative admissible reading of 180 MPa·√m — that is 5692 N/mm^3/2 — moves the critical length by a factor of ten but changes the life by only one per cent, from 11 035 to 11 145 cycles. Only the literal 180 N/mm^3/2 is inadmissible, for the reason set out above.
Sense-check the growth rates. At the initial crack the rate is $3.59{\times}10^{-5}$ mm/cycle; at the critical length it is 0.299 mm/cycle, eight thousand times faster. Practically, the crack takes about ten thousand cycles to reach a few millimetres and then covers the remaining thirty millimetres in a few hundred. This is why inspection intervals for fatigue-critical structure are set from the detectable crack size rather than from the critical size.

Question 5 — results
Quantity	Adopted K_IC = 1800 N/mm^3/2	If K_IC = 180 MPa·√m	Literal 180 N/mm^3/2
(a) Crack length at failure	36.54 mm	365.4 mm	0.365 mm (below a₀)
(b) Cycles to failure	11 035	11 145	0 — fractures on loading
Lumped Paris coefficient C′	2.2407 × 10⁻⁴ mm⁻¹/cycle
ΔK at a₀ = 0.4 mm	188.3 N/mm^3/2
Growth rate at a₀ / at a_c	3.59 × 10⁻⁵ / 0.299 mm per cycle

Topic. Linear elastic fracture mechanics applied to fatigue life — the Paris regime between threshold and fast fracture.

Key relations. $\Delta K=Y\Delta\sigma\sqrt{\pi a}$ with $Y=1.12$ for a semi-infinite edge crack; critical length $a_c=(K_{IC}/Y\Delta\sigma)^2/\pi$; Paris law $da/dN=C(\Delta K)^m$; for $m=4$, $N=\left(\frac{1}{a_0}-\frac{1}{a_c}\right)\big/\left[C(Y\Delta\sigma)^4\pi^2\right]$.

Why this works. Under small-scale yielding the whole crack-tip field is controlled by the single parameter $K$, so both the growth rate and the onset of fast fracture can be written as functions of it. The Paris law then becomes an ordinary differential equation in crack length whose right-hand side is a pure power of $a$, and separation of variables completes the job. The exponent controls how the life is distributed: with $m=4$ the rate rises as $a^2$, so almost all the cycles are consumed while the crack is still small, and the integral is dominated by its lower limit.

Common pitfall. Forgetting the 1.12 edge-crack factor, which appears to the fourth power in the life and therefore changes the answer by more than fifty per cent. A second pitfall is integrating a mean stress rather than a stress range: for repeated loading between zero and maximum the two coincide, but for any other load ratio they do not.

Source. Dieter, Mechanical Metallurgy, 3rd ed., ch. 11–12 (fatigue, fracture mechanics of fatigue-crack propagation); Boresi and Schmidt, Advanced Mechanics of Materials, 6th ed., ch. 15.

Question 5 of 7: Fatigue-crack growth in an edge-cracked plate

Question 5: Fatigue-crack growth in an edge-cracked plate (20 marks)