Question 3 of 8: Definitions of Toughness; Fatigue Life of a Cracked Plate

Question 3: Definitions of Toughness; Fatigue Life of a 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.

3.1 — (a) Three meanings of “toughness”

The word “toughness” is used for three different quantities, with three different units, and the examiner is asking for all three to be separated cleanly. What they share is that each is an energy, not a stress; what distinguishes them is the volume or area over which that energy is reckoned, and how much of the material takes part.

(i) Toughness in elastic deformation — the modulus of resilience. If the deformation is entirely elastic, the only energy the material can absorb is the recoverable strain energy stored in the stretched bonds. Per unit volume this is the area under the elastic part of the stress–strain curve up to yield,

with units of joules per cubic metre (equivalently, pascals). It is the property that matters for a spring: the design goal is to store and return as much energy as possible without permanent set, so a spring material wants a high $\sigma_y$ and a low $E$, which is why hard-drawn spring steel, beryllium copper and fibre-reinforced polymers all appear in leaf and coil springs. Note that resilience says nothing about resistance to failure — it measures energy stored, all of which is given back.

(ii) Toughness in plastic deformation — tensile toughness or the work of fracture. Once the material yields, the energy going into it is largely dissipated rather than stored, and the relevant measure is the total area under the nominal stress–strain curve to fracture,

again in joules per cubic metre. Because $\varepsilon_f$ can be tens of per cent while $\varepsilon_y$ is a fraction of one per cent, plastic toughness is typically two or three orders of magnitude larger than resilience — it is dominated by ductility, not by strength. This is the quantity a Charpy or Izod impact test estimates (reported as an absorbed energy in joules rather than per unit volume, since the deforming volume is fixed by the specimen), and it is what “tough” means in the everyday engineering sense of a material that gives warning, absorbs a crash, and deforms rather than shattering. It also embodies the classic strength–ductility trade-off: every strengthening mechanism in Question 1 raises $\sigma_y$ and lowers $\varepsilon_f$, so the product passes through a maximum.

(iii) Toughness in fast fracture — fracture toughness. Neither of the first two definitions helps once a sharp crack is present, because then almost all of the material is elastic and only a small process zone at the crack tip does any work. The correct measure is the energy dissipated per unit area of new crack surface, the toughness or critical strain-energy release rate $G_c$, in joules per square metre. Its stress-based equivalent is the critical stress-intensity factor,

with $E'=E$ in plane stress and $E/(1-\nu^{2})$ in plane strain; $K_c$ has the unusual units MPa m${}^{1/2}$. Under plane-strain constraint the value falls to a geometry-independent minimum, $K_{Ic}$, which is the number quoted as a material property (ASTM E399). This is the toughness that governs the fast, unstable fracture of a cracked structure, and — as part (b) shows — it also fixes the crack length at which a fatigue crack stops growing stably and runs.

3.2 — (b) Fatigue life of the cracked plate

Given. A large flat aluminium-alloy plate carries fully reversed tensile–compressive cycles and contains a surface crack; the crack-growth law and the fracture toughness are supplied:

Find. The number of cycles $N_f$ needed to grow the 0.75 mm crack to the length at which the plate fractures in one cycle.

Approach. Establish the crack length $a_c$ at which the peak stress intensity reaches $K_c$, then separate the variables in the Paris law and integrate $\mathrm{d}N = \mathrm{d}a / A(\Delta K)^n$ from $a_0$ to $a_c$, with $\Delta K = Y\Delta\sigma\sqrt{\pi a}$ held at its effective range for reversed loading.

1. Evaluate the starting stress-intensity range and confirm the Paris regime applies. With $\Delta K = Y\Delta\sigma\sqrt{\pi a}$, $$\Delta K_0 \;=\; (1.0)(150)\sqrt{\pi \times 7.5\times10^{-4}} \;=\; 7.28\ \text{MPa}\,\text{m}^{1/2}$$ This is comfortably above the threshold $\Delta K_{th}$ of a typical aluminium alloy (2–4 MPa m${}^{1/2}$) and well below $K_c$, so the crack is growing steadily in the Paris (stage II) regime from the first cycle and the given power law may be integrated over the whole range.
2. Find the critical crack length. Fracture occurs when the stress intensity at peak load reaches the fracture toughness, $Y\sigma_{\max}\sqrt{\pi a_c} = K_c$, so $$a_c \;=\; \frac{1}{\pi}\left(\frac{K_c}{Y\sigma_{\max}}\right)^{2} \;=\; \frac{1}{\pi}\left(\frac{35}{1.0 \times 150}\right)^{2} \;=\; 1.7330\times10^{-2}\ \text{m}$$ $$\boxed{a_c \;=\; 17.33\ \text{mm}}$$ The crack therefore has to grow by a factor of 23 in length before the plate fails — a large tolerance, which is what makes the aluminium airframe a damage-tolerant structure in the first place.
3. Separate the variables in the crack-growth law. Substituting $\Delta K = Y\Delta\sigma\sqrt{\pi a}$ into $da/dN = A(\Delta K)^n$ and rearranging, $$\mathrm{d}N \;=\; \frac{\mathrm{d}a}{A\left(Y\Delta\sigma\sqrt{\pi a}\right)^{n}} \;=\; \frac{a^{-n/2}\,\mathrm{d}a}{A\,Y^{n}\,\Delta\sigma^{n}\,\pi^{n/2}}$$ Everything except $a$ is constant, because the plate is large compared with the crack and $Y$ does not change as the crack grows.
4. Integrate from the initial to the critical crack length. For $n \neq 2$, $$N_f \;=\; \frac{1}{A\,Y^{n}\,\Delta\sigma^{n}\,\pi^{n/2}}\int_{a_0}^{a_c} a^{-n/2}\,\mathrm{d}a \;=\; \frac{a_0^{-(n/2-1)} - a_c^{-(n/2-1)}}{\left(\tfrac{n}{2}-1\right)A\,Y^{n}\,\Delta\sigma^{n}\,\pi^{n/2}}$$ With $n = 2.5$ the exponent $n/2 - 1 = 0.25$, so the bracket is a difference of inverse fourth roots.
5. Substitute the numbers. The two crack-length terms are $$a_0^{-0.25} = (7.5\times10^{-4})^{-0.25} = 6.0428, \qquad a_c^{-0.25} = (1.7330\times10^{-2})^{-0.25} = 2.7561$$ and the denominator is $$0.25 \times (2\times10^{-12}) \times (1.0)^{2.5} \times (150)^{2.5} \times \pi^{1.25} \;=\; 5.7628\times10^{-7}$$ Hence $$N_f \;=\; \frac{6.0428 - 2.7561}{5.7628\times10^{-7}} \;=\; 5.703\times10^{6}$$ $$\boxed{N_f \;\approx\; 5.70\times10^{6}\ \text{cycles}}$$
6. Check the result against the growth rates at each end. At the initial crack length $da/dN = 2.86\times10^{-10}$ m/cycle — about one atomic spacing (0.286 nm in aluminium) per cycle, which is physically sensible for near-threshold growth — while at $a_c$ it has risen to $1.45\times10^{-8}$ m/cycle, a factor of 50.7. Because the rate is so much lower over most of the range, 69.4 per cent of the life is used up before the crack reaches 5 mm, and the last few millimetres pass in a small fraction of the total. This is the quantitative basis of damage-tolerant inspection intervals: the inspection must be capable of finding the crack while it is still short, because once it is long it has almost no life left.