Question 1 of 8: Fatigue-Safety Margin for a Cracked Component; Infinite-Life Flaw-Size Design for a Skeletal Implant

Question 1: Fatigue-Safety Margin for a Cracked Component; Infinite-Life Flaw-Size Design for a Skeletal Implant (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.

1.1 — (a) Is the margin against fast fracture adequate for this cyclic application?

Given.

Find. Whether the existing crack leaves an adequate safety margin against fast fracture at the maximum design stress, and hence whether the alloy is recommended for this cyclic application.

Approach. Compute the critical crack length $a_c$ at which $K_{\max}$ (governed by $\sigma_{\max}=410$ MPa, the largest stress the crack ever sees) reaches $K_{Ic}$, and compare it directly with the existing crack size $a=0.7$ mm; equivalently, compute the stress intensity the existing crack already experiences at $\sigma_{\max}$ and compare it with $K_{Ic}$.

1. Critical crack length at the maximum design stress. $$K_{\max}=Y\sigma_{\max}\sqrt{\pi a_c}=K_{Ic} \quad\Longrightarrow\quad a_c=\frac{1}{\pi}\left(\frac{K_{Ic}}{Y\sigma_{\max}}\right)^2=\frac{1}{\pi}\left(\frac{20}{1\times410}\right)^2.$$ $$\boxed{a_c \approx 0.757\ \text{mm}}$$
2. Compare the existing crack to $a_c$, and to $K_{Ic}$ directly. The existing crack, $a=0.7$ mm, is smaller than $a_c=0.757$ mm, so the component does not fracture on the very first cycle — but only by a margin of $$\frac{a_c-a}{a}=\frac{0.757-0.700}{0.700}=8.2\%.$$ Equivalently, at $\sigma_{\max}=410$ MPa the existing crack already drives $$K=Y\sigma_{\max}\sqrt{\pi a}=1\times410\sqrt{\pi(0.0007)}=19.23\ \text{MPa}\sqrt{\text{m}},$$ $$\frac{K}{K_{Ic}}=\frac{19.23}{20}=0.961,$$ i.e. every peak of the load cycle already drives the crack tip to 96% of the material's fracture toughness. The equivalent stress-based factor of safety is $\sigma_{\text{allow}}/\sigma_{\max}=426.5/410=\boxed{1.04}$.
3. Weigh the margin against what cyclic loading will do to it. A factor of safety of 1.04 on stress (equivalently, 8% on crack length) is far below what is normally required for a fatigue application — typical design practice looks for a factor of at least 1.5–2 against fast fracture, precisely because the crack is not static: every one of the design cycles drives sub-critical fatigue crack growth ($da/dN=A(\Delta K)^n$, Question 3(a) of this paper), so $a$ only grows toward $a_c$ from here, it never shrinks. A margin this thin will be consumed by ordinary crack growth well before any meaningful service life is accumulated, and material/measurement scatter in $K_{Ic}$, the 1.4 mm crack-length reading, or the peak stress itself could each independently erase the remaining 4–8%.

1.2 — (b) Largest tolerable surface-crack size for an infinite (10⁷-cycle) fatigue life

Given.

Find. The crack size $a_c$ to which the initial 0.01 in. flaw grows in $1\times10^7$ cycles — the largest surface crack the implant must be able to tolerate without fast fracture if it is to reach an infinite fatigue life.

Approach. This is the standard fatigue-crack-growth design calculation (Callister & Rethwisch, §8.9, critical surface-crack-length design problems). Integrate the Paris law from the known initial crack $a_0$ to an unknown final crack $a_c$, set the number of cycles equal to the required $10^7$, and solve for $a_c$. Because the loading is tensile–compressive, the crack is closed and does not grow during the compressive half-cycle, so the stress range that drives growth is $\Delta\sigma=\sigma_{\max}=15{,}000$ psi, not $\sigma_{\max}-\sigma_{\min}$.

1. Integrate the Paris law between $a_0$ and $a_c$. $$\frac{da}{dN}=A(\Delta K)^n=A\left(Y\Delta\sigma\sqrt{\pi a}\right)^n \;\Longrightarrow\; N_f=\int_{a_0}^{a_c}\frac{da}{A(Y\Delta\sigma)^n\pi^{n/2}a^{n/2}}=\frac{a_0^{-(n/2-1)}-a_c^{-(n/2-1)}}{\left(\tfrac{n}{2}-1\right)A(Y\Delta\sigma)^n\pi^{n/2}}\quad(n\ne2).$$ Rearranging for the final crack size: $$a_c^{-(n/2-1)}=a_0^{-(n/2-1)}-N_f\left(\tfrac{n}{2}-1\right)A(Y\Delta\sigma)^n\pi^{n/2}.$$
2. Evaluate the growth term. $\tfrac{n}{2}-1=0.25$; $Y\Delta\sigma=1.75\times15{,}000=26{,}250$ psi; $(26{,}250)^{2.5}=1.1164\times10^{11}$; $\pi^{1.25}=4.1825$. $$N_f\left(\tfrac{n}{2}-1\right)A(Y\Delta\sigma)^n\pi^{n/2}=(1\times10^7)(0.25)(1.5\times10^{-18})(1.1164\times10^{11})(4.1825)=1.7510\ \text{in.}^{-0.25}$$
3. Solve for $a_c$. $a_0^{-0.25}=(0.01)^{-0.25}=3.1623\ \text{in.}^{-0.25}$, so $$a_c^{-0.25}=3.1623-1.7510=1.4113 \;\Longrightarrow\; a_c=(1.4113)^{-4}.$$ $$\boxed{a_c \approx 0.252\ \text{in.} \approx 6.40\ \text{mm}}$$
4. What toughness this demands. For the implant to tolerate that crack at the peak stress, its fracture toughness must satisfy $$K_c\ge Y\sigma_{\max}\sqrt{\pi a_c}=1.75(15{,}000)\sqrt{\pi(0.2521)}=2.34\times10^{4}\ \text{psi}\sqrt{\text{in.}}\;(\approx25.7\ \text{MPa}\sqrt{\text{m}}).$$ A premature failure therefore means the implant's actual critical crack size, $(1/\pi)(K_c/Y\sigma_{\max})^2$, was smaller than 0.252 in. — its toughness (or the stress and geometry assumed in design) was inadequate for the flaws it contained.