Question 1 of 8: Fracture-Toughness Design Check and Fatigue-Limited Flaw Size

Question 1: Fracture-Toughness Design Check and Fatigue-Limited Flaw Size (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.

(a) Static Fracture Margin at the Existing Crack

Given.

Find. The critical (fracture) stress at the existing crack size, and whether the alloy should be recommended for this cyclic application.

Approach. Use the LEFM relation $\sigma_c=K_{Ic}/(Y\sqrt{\pi a})$ to find the stress that would fracture the plate at its existing crack length, then compare that critical stress to the design maximum stress and judge the margin under repeated loading (the crack does not stay fixed once cycling begins).

1. Critical (fracture) stress at $a=0.7$ mm. $$\sigma_c=\frac{K_{Ic}}{Y\sqrt{\pi a}}=\frac{20}{1\times\sqrt{\pi(0.0007)}}=\boxed{426.5\ \text{MPa}}$$
2. Static margin against the design maximum stress. The design cycle peaks at $\sigma_{max}=410$ MPa, so $$\text{margin}=\frac{426.5-410}{410}\times100\%=4.0\%$$ — the plate does not fracture on the first cycle, but the margin is very thin.
3. Engineering judgement. Because $\sigma_c\propto a^{-1/2}$, any sub-critical crack growth under the 0–410 MPa cycling (fatigue crack growth, Paris' law) continuously lowers $\sigma_c$ toward the applied 410 MPa. A 4% static margin leaves essentially no room for crack growth before $\sigma_c$ drops to the applied stress and the component fractures. Recommendation: do NOT use this alloy in this cyclic application as configured — either reduce the design stress substantially, use a higher-toughness alloy, or guarantee (by inspection) that no crack approaching 1.4 mm can be present.

(b) Largest Tolerable Initial Flaw for Infinite Fatigue Life

Given.

Find. The largest initial surface crack length $a_i$ that would still deliver $N_f\geq1\times10^{7}$ cycles.

Approach. Integrate the Paris law $da/dN=A(\Delta K)^n$ with $\Delta K=Y\Delta\sigma\sqrt{\pi a}$ from the unknown initial flaw $a_i$ to the eventual critical crack size $a_c$. Because $n=2.5\gt2$, the integral converges as $a_c\to\infty$ — almost all of the fatigue life is spent while the crack is still small, so $N_f$ becomes essentially independent of $a_c$ and is governed entirely by $a_i$. Setting this converged life equal to the $10^{7}$-cycle target and solving for $a_i$ gives the design flaw-tolerance limit directly.

1. Closed-form life for $n\gt2$, $a_c\to\infty$. $$N_f=\int_{a_i}^{\infty}\frac{da}{A\left(Y\Delta\sigma\sqrt{\pi a}\right)^n}=\frac{a_i^{\,1-n/2}}{A(Y\Delta\sigma)^n\pi^{n/2}\left(\dfrac{n}{2}-1\right)}$$
2. Solve for $a_i$ with $N_f=1\times10^{7}$. With $n=2.5$, $Y\Delta\sigma=1.75(30{,}000)=52{,}500$ psi: $$a_i=\left[\frac{N_f\,A(Y\Delta\sigma)^n\pi^{n/2}\left(\dfrac{n}{2}-1\right)}{1}\right]^{\frac{1}{1-n/2}}=\boxed{1.04\times10^{-4}\ \text{in.}\ \;(=2.64\ \mu\text{m})}$$
3. Consistency check against the reported failure. The actual implant carried an initial crack of 0.01 in. (254 μm) — about 96× larger than the $1.04\times10^{-4}$ in. tolerable limit computed above. That is fully consistent with the problem statement that the component "failed prematurely": its as-manufactured flaw was already far beyond the size an infinite-life design could tolerate.