21-Mat-A5 Phase Transformations and Thermal Treatment · May 2017

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

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

Paper format. National Exams, May 2017 — 10-Met-A5, Mechanical Behaviour and Fracture of Materials. Three hours, closed book, any non-communicating calculator permitted. Eight questions of 20 marks each; the rubric states that five questions constitute a complete paper and that only the first five appearing in the answer book are marked. All eight are answered here, because this set is a study resource rather than an exam script. Several questions ask explicitly for essay-format answers, and the marking scheme rewards clarity and organisation, so the discursive answers below are written as structured prose rather than as note form.

Note on the paper's subject

The printed exam header reads 10-Met-A5, Mechanical Behaviour and Fracture of Materials. The paper has no phase-transformation or heat-treatment question in the classical (TTT/CCT diagram, hardenability, tempering-curve) sense; the syllabus actually examined is fracture mechanics and fatigue-crack-growth life, strengthening and toughening of engineering materials, creep and fatigue testing, deformation processing, and elastic–plastic forming behaviour.

Reference texts. The answers below are keyed to the works normally recommended for this syllabus code:

G. E. Dieter, Mechanical Metallurgy, 3rd ed. — strengthening and plastic instability (Ch. 3, 4, 6), fracture (Ch. 7), fatigue (Ch. 12), creep and stress rupture (Ch. 13), forming processes (Ch. 15–19).
W. D. Callister and D. G. Rethwisch, Materials Science and Engineering: An Introduction, 10th ed. — strengthening (Ch. 7), failure: fracture, fatigue and creep (Ch. 8), composites (Ch. 16), corrosion and degradation (Ch. 17).
M. F. Ashby and D. R. H. Jones, Engineering Materials 1 & 2 — fast fracture, the transition flaw size, toughening mechanisms, and creep-mechanism maps.
ASM Handbook, Vol. 11 (Failure Analysis and Prevention), Vol. 14A (Metalworking: Bulk Forming) and Vol. 19 (Fatigue and Fracture) — process selection, fractography, single-crystal casting and fatigue design.
Test standards cited where relevant: ASTM E8/E8M (tension), E139 (creep and stress rupture), E466 (constant-amplitude axial fatigue), E647 (fatigue crack growth rate), E399/E1820 (fracture toughness).

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.

Quantity	Symbol	Value
Ultimate tensile strength	$\sigma_{UTS}$	800 MPa
Fracture toughness	$K_{Ic}$	20 MPa$\sqrt{\text{m}}$
Through-thickness central crack length	$2a$	1.4 mm ($a=0.7$ mm)
Geometry correction factor	$Y$	1 (given)
Design stress range	$0 \to \sigma_{\max}$	0 to 410 MPa

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.

Fig. 1.1 — Central through-thickness crack of length 2a = 1.4 mm in a large plate under remote cyclic tension, Y = 1 as stated.

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 $K_{Ic}$.

Critical crack length at the maximum design stress. $$K_{\max}=Y\sigma_{\max}\sqrt{\pi a_c}=K_{Ic} \quad\Longrightarrow\quad a_c=\frac{1}{\pi}\left(\frac{K_{Ic}}{Y\sigma_{\max}}\right)^2=\frac{1}{\pi}\left(\frac{20}{1\times410}\right)^2.$$ $$\boxed{a_c \approx 0.757\ \text{mm}}$$
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 $\sigma_{\text{allow}}/\sigma_{\max}=426.5/410=\boxed{1.04}$.
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%.

Quantity	Result
Critical crack length at $\sigma_{\max}=410$ MPa, $a_c$	0.757 mm
Existing crack length, $a$	0.700 mm
$K/K_{Ic}$ at $\sigma_{\max}$ on the existing crack	0.961
Stress-based factor of safety	1.04
Recommendation	$\boxed{\text{Not recommended}}$

Check: engineering judgement

The calculation shows the component survives the very first application of $\sigma_{\max}$, so a purely static (single-overload) fast-fracture check alone would not condemn the design. The recommendation against use rests on the combination of (i) an unusually thin 4–8% margin against $K_{Ic}$/$a_c$ at first load and (ii) the fact this is explicitly a cyclic application, so the crack is guaranteed to grow toward $a_c$ over service life rather than remain fixed. A responsible design would either specify a larger $K_{Ic}$ alloy, tighten the inspection/rejection crack-length limit well below 0.7 mm, or reduce the design stress range substantially before qualifying this material.

Concept & Theory

Topic. Linear-elastic fracture mechanics (LEFM) safety-margin assessment for a component that already contains a known crack, ahead of any explicit fatigue-crack-growth calculation.

Key relations. $K=Y\sigma\sqrt{\pi a}$; critical crack length $a_c=(1/\pi)(K_{Ic}/Y\sigma_{\max})^2$; stress-based factor of safety $=\sigma_{\text{allow}}/\sigma_{\max}$ where $\sigma_{\text{allow}}=K_{Ic}/(Y\sqrt{\pi a})$.

Why this works. $K_{\max}=K_{Ic}$ defines the boundary between "safe under the peak load" and "fractures immediately"; because the peak of a 0–$\sigma_{\max}$ cycle is the highest $K$ the crack ever sees, checking against $\sigma_{\max}$ (rather than a mean or amplitude stress) captures the worst instant of every cycle. A thin margin at that single worst instant is a red flag independent of any fatigue-crack-growth-rate law, because such a law can only move the crack closer to $a_c$, never away from it.

Common pitfall. Concluding "safe" purely because $a

Source. Ashby & Jones, Engineering Materials 1, Ch. 13 (fast fracture and safety factors); Dieter, Mechanical Metallurgy, 3rd ed., Ch. 7.

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

Given.

Quantity	Symbol	Value
Maximum tensile stress, tensile–compressive cyclic loading	$\sigma_{\max}$	15,000 psi
Maximum initial surface crack length	$a_0$	0.01 in.
Geometry correction factor	$Y$	1.75 (independent of $a$)
Paris-law exponent	$n$	2.5
Paris-law coefficient (psi, in.)	$A$	$1.5\times10^{-18}$
Required ("infinite") life	$N_f$	$1\times10^7$ cycles

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 $\sigma_{\max}-\sigma_{\min}$.

Assumption stated

"Largest tolerable surface crack size" is read as the critical crack length the component must tolerate at the end of the required life. The given initial crack length, stress, $Y$, $A$ and $n$ are exactly the data this calculation needs, and no fracture toughness is supplied because the question asks for the crack size the material must tolerate, not the one it can.

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: $$a_c^{-(n/2-1)}=a_0^{-(n/2-1)}-N_f\left(\tfrac{n}{2}-1\right)A(Y\Delta\sigma)^n\pi^{n/2}.$$
Evaluate the growth term. $\tfrac{n}{2}-1=0.25$; $Y\Delta\sigma=1.75\times15{,}000=26{,}250$ psi; $(26{,}250)^{2.5}=1.1164\times10^{11}$; $\pi^{1.25}=4.1825$. $$N_f\left(\tfrac{n}{2}-1\right)A(Y\Delta\sigma)^n\pi^{n/2}=(1\times10^7)(0.25)(1.5\times10^{-18})(1.1164\times10^{11})(4.1825)=1.7510\ \text{in.}^{-0.25}$$
Solve for $a_c$. $a_0^{-0.25}=(0.01)^{-0.25}=3.1623\ \text{in.}^{-0.25}$, so $$a_c^{-0.25}=3.1623-1.7510=1.4113 \;\Longrightarrow\; a_c=(1.4113)^{-4}.$$ $$\boxed{a_c \approx 0.252\ \text{in.} \approx 6.40\ \text{mm}}$$
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.

Quantity	Result
Crack-growth term, $N_f(n/2-1)A(Y\Delta\sigma)^n\pi^{n/2}$	1.751 in.$^{-0.25}$
Largest tolerable surface crack for $10^7$ cycles, $a_c$	$\boxed{0.252\ \text{in.}\ (6.40\ \text{mm})}$
Minimum toughness implied, $K_c$	$2.34\times10^{4}$ psi$\sqrt{\text{in.}}$ ($\approx25.7$ MPa$\sqrt{\text{m}}$)

Check: convergence and loading convention

A finite $a_c$ exists only because the growth term (1.751) is less than $a_0^{-0.25}$ (3.162). Letting $a_c\to\infty$ instead gives the largest initial flaw that could ever reach $10^7$ cycles, $a_0=(1.751)^{-4}=0.106$ in.; the actual 0.01 in. flaw is well inside that. Had the full range $\Delta\sigma=2\sigma_{\max}=30{,}000$ psi been used, the growth term would rise by $2^{2.5}=5.66$ to 9.90, exceeding 3.162 — no crack size would survive $10^7$ cycles, so the loading convention is decisive here.

Concept & Theory

Topic. Fatigue-crack-growth (Paris-law) life integration used as a design tool: the critical crack size a component must tolerate to reach a required life from a known initial flaw.

Key relations. $da/dN=A(\Delta K)^n$, $\Delta K=Y\Delta\sigma\sqrt{\pi a}$; for $n\ne2$, $N_f=[a_0^{-(n/2-1)}-a_c^{-(n/2-1)}]/[(n/2-1)A(Y\Delta\sigma)^n\pi^{n/2}]$; required toughness $K_c=Y\sigma_{\max}\sqrt{\pi a_c}$; tensile–compressive loading uses $\Delta\sigma=\sigma_{\max}$ (crack closure in compression).

Why this works. Almost all of the fatigue life is spent while the crack is small and $\Delta K$ is low, so the life is set mainly by $a_0$, and the end-of-life crack size is very sensitive to the required cycle count. Pairing $a_c$ with $K_c=Y\sigma\sqrt{\pi a_c}$ turns a crack-growth result into a material-selection requirement.

Common pitfall. Using $2\sigma_{\max}$ as the stress range for tensile–compressive loading; or ignoring the given initial crack size and taking $a_c\to\infty$ — that answers a different question (the largest permissible initial flaw), not the tolerable crack size for the stated flaw.

Source. Callister & Rethwisch, Materials Science and Engineering, 10th ed., §8.9 (crack propagation rate and its integration); Dieter, Mechanical Metallurgy, 3rd ed., Ch. 12.

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