21-Mat-A5 Phase Transformations and Thermal Treatment · December 2017
Question 5 of 8: Leak-Before-Break Wall Thickness and Required Fracture Toughness for a Target Fatigue Life
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. National Exams, December 2017 — 10-Met-A5, Mechanical Behaviour and Fracture of Materials. Three hours, closed book, any Casio- or Sharp-approved 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 sub-parts explicitly call for an essay-format answer, and the rubric rewards clarity and organisation, so those answers are written as structured prose rather than as note form.
Note on the exam title
Nothing on the paper is a phase-transformation or heat-treatment question in the TTT/CCT, hardenability or tempering sense; the syllabus actually examined is dislocation theory, slip and twinning, strengthening mechanisms, creep, fatigue, toughness and fracture mechanics, and deformation processing.
Reference texts. The answers below are keyed to the works normally recommended for this syllabus code:
G. E. Dieter, Mechanical Metallurgy, 3rd ed. — dislocation theory and slip (Ch. 4–5), strengthening mechanisms (Ch. 6), fracture (Ch. 7), fatigue (Ch. 12), creep and stress rupture (Ch. 13).
W. D. Callister and D. G. Rethwisch, Materials Science and Engineering: An Introduction, 10th ed. — dislocations and slip systems (Ch. 7), failure: fracture, fatigue and creep (Ch. 8).
M. F. Ashby and D. R. H. Jones, Engineering Materials 1 & 2 — fast fracture and toughness, fatigue crack growth, creep mechanisms and deformation-mechanism maps.
R. E. Reed-Hill and R. Abbaschian, Physical Metallurgy Principles — crystallography of slip and twinning, stereographic projections, Schmid's law.
ASM Handbook, Vol. 11 (Failure Analysis and Prevention) and Vol. 19 (Fatigue and Fracture) — fractography and environmental cracking.
Test standards cited where relevant: ASTM E8/E8M (tension), E139 (creep and stress rupture), E466/E647 (fatigue and fatigue crack growth rate), E399/E1820 (fracture toughness).
Question 5: Leak-Before-Break Wall Thickness and Required Fracture Toughness for a Target Fatigue Life (20 marks)
The Paris-law coefficient is printed as "$2\times10^{12}$", which for typical fatigue-crack-growth data in these units (m/cycle, MPa$\sqrt{\text{m}}$) is physically implausible — it would predict crack growth many orders of magnitude faster than any real alloy and gives no finite solution to part (b) at all. This is read as a sign-of-exponent printing slip for $A=2\times10^{-12}$, the standard order of magnitude quoted for steel Paris-law coefficients in these units. Part (b) is solved with $A=2\times10^{-12}$.
5.1 — (a) Maximum leak-before-break wall thickness
Given.
Given data — Question 5(a)
Quantity
Symbol
Value
Wall hoop stress
$\sigma$
460 MPa
Fracture toughness
$K_{Ic}$
98.9 MPa$\sqrt{\text{m}}$
Configuration correction factor
$Y$
1.0 (given)
Find. The maximum wall thickness $t$ for which a through-wall crack becomes a leak (a through-thickness flaw growing stably) rather than a fast fracture.
Leak-before-break design logic: for a thin-walled vessel the governing, most dangerous flaw is a through-wall crack whose length equals the wall thickness $t$. If the toughness is reached only once the crack has grown all the way through ($a_c\ge t$), the vessel leaks safely first; the design requirement is the thickness at which the critical crack length exactly equals $t$.
Approach. Set the critical (through-thickness) flaw length from the fracture-toughness criterion $K_{Ic}=Y\sigma\sqrt{\pi a_c}$, then apply the leak-before-break condition $a_c=t$.
Solve the fracture-toughness criterion for the critical crack length. $$K_{Ic}=Y\sigma\sqrt{\pi a_c}\ \Rightarrow\ a_c=\frac{1}{\pi}\left(\frac{K_{Ic}}{Y\sigma}\right)^2$$
Substitute the given values. $$a_c=\frac1\pi\left(\frac{98.9}{1.0\times460}\right)^2=\frac1\pi(0.2150)^2\text{ m}$$
Any wall thicker than 14.71 mm would reach $K_{Ic}$ before a crack has grown all the way through the wall, so the vessel would fracture catastrophically while the flaw is still buried; any wall at or below 14.71 mm leaks (the crack becomes through-wall, at which point it is stable and detectable) before it can reach the critical length for fast fracture.
5.2 — (b) Fracture toughness required for a minimum fatigue life
Given.
Given data — Question 5(b)
Quantity
Symbol
Value
Initial surface crack length
$a_0$
0.1 mm = $1\times10^{-4}$ m
Maximum cyclic stress
$\sigma_{\max}$
310 MPa
Paris-law coefficient
$A$
$2\times10^{-12}$ (see check box)
Paris-law exponent
$n$
4
Minimum required life
$N_f$
$2.7\times10^4$ cycles
Configuration correction factor
$Y$
1.0 (given)
Find. The fracture toughness $K_{Ic}$ that would just allow the crack to grow from $a_0$ to its critical (fast-fracture) length in exactly $N_f=2.7\times10^4$ cycles.
Check: assumed loading ratio
The question states only that the cyclic stress "reaches a maximum of 310 MPa per cycle," without stating a minimum. This is read as tension-tension cycling from zero ($R=0$), so the stress-intensity range is $\Delta K=Y\Delta\sigma\sqrt{\pi a}$ with $\Delta\sigma=\sigma_{\max}-\sigma_{\min}=\sigma_{\max}=310$ MPa.
Approach. Integrate the Paris law from $a_0$ to the (unknown) critical length $a_c$ to relate $N_f$ to $a_c$, then convert $a_c$ to a required $K_{Ic}$ via the fracture criterion.
Substitute $\Delta K=Y\Delta\sigma\sqrt{\pi a}$ into the Paris law for $n=4$. $$\frac{da}{dN}=A\left(Y\Delta\sigma\sqrt{\pi a}\right)^4=A\,\Delta\sigma^4\pi^2a^2$$
Integrate from $a_0$ to $a_c$ over $N_f$ cycles. $$N_f=\int_{a_0}^{a_c}\frac{da}{A\,\Delta\sigma^4\pi^2a^2}=\frac{1}{A\,\Delta\sigma^4\pi^2}\left(\frac1{a_0}-\frac1{a_c}\right)$$
Solve for the critical crack length $a_c$. $$\frac1{a_c}=\frac1{a_0}-N_f\,A\,\Delta\sigma^4\pi^2 = 10000 - (2.7\times10^4)(2\times10^{-12})(310)^4\pi^2 = 5079\ \text{m}^{-1}$$ $$a_c=\boxed{0.197\ \text{mm}}$$
Convert $a_c$ to the required toughness via the fracture criterion. $$K_{Ic}=Y\sigma_{\max}\sqrt{\pi a_c}=1.0\times310\times\sqrt{\pi\times1.97\times10^{-4}}$$ $$K_{Ic}=\boxed{7.71\ \text{MPa}\sqrt{\text{m}}}$$
A toughness of at least 7.71 MPa$\sqrt{\text{m}}$ is therefore required: any lower toughness would let the initial 0.1 mm crack reach the (lower) critical length before accumulating $2.7\times10^4$ cycles, failing early; any higher toughness gives the crack further to grow and a longer life than the 2.7×10$^4$-cycle minimum.
Final Results — Question 5
Quantity
Value
(a) Maximum leak-before-break wall thickness, $t_{\max}$
14.71 mm
(b) Critical crack length at $N_f=2.7\times10^4$ cycles, $a_c$
