Question 6 of 7: Centre-Cracked Panel — Critical Crack Length and Fatigue Life
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exam 04-BS-11, Properties of Materials — December 2016. 3 hours,
closed-book examination (approved Casio or Sharp calculator only). Candidates attempt any
five of the seven questions for a complete paper, all questions of equal value. All seven
questions are solved below for completeness.
Reference texts: Callister & Rethwisch, Materials Science and
Engineering: An Introduction, 9th ed. (crystal structure, mechanical behaviour, phase
diagrams, polymer viscoelasticity, composites, electrochemistry, fatigue/fracture, heat
treatment and hardenability, concrete).
Question 6: Centre-Cracked Panel — Critical Crack Length and Fatigue Life (20 marks)
Check: the source prints the Paris-law constant as
“$C=1.8\times10^{48}$”, a misprinted exponent (the minus sign lost). Solved with
$C=1.8\times10^{-18}$, since this is the physically sensible reading: converted to the more
commonly tabulated ksi$\sqrt{\text{in}}$ units ($C_{ksi}=C_{psi}\times10^9=1.8\times10^{-9}$), it
falls squarely in the literature range for steel Paris-law constants ($m=3$), and it returns a
realistic fatigue life (10$^5$–10$^6$ cycles) — a raw $10^{+48}$ exponent would give a
physically meaningless answer.
Fig. Q6 — centre-cracked panel under cyclic remote tension
$\sigma$.
Given. $2a_0=0.2$ in ($a_0=0.1$ in), $w=20$ in, $t=0.5$ in,
$K_{Ic}=24{,}000$ psi$\sqrt{\text{in}}$, $\sigma_{max}=13{,}000$ psi, $\sigma_{min}=0$,
$Y=1$, $C=1.8\times10^{-18}$, $m=3.0$.
Find. (i) Critical crack length at fracture. (ii) Fatigue cycles to failure.
Approach
(i) is a direct fracture-toughness calculation: failure occurs when $K$ reaches $K_{Ic}$ under
the peak cyclic stress. (ii) integrates the Paris law from the initial half-crack length $a_0$ to
the critical half-crack length $a_c$ found in part (i), using $\Delta K=K_{max}$ since
$\sigma_{min}=0$ gives $K_{min}=0$.
(i) Critical half-crack length. Failure occurs when
$K_{max}=\sigma_{max}\sqrt{\pi a_c}=K_{Ic}$:
$$a_c=\frac{1}{\pi}\left(\frac{K_{Ic}}{\sigma_{max}}\right)^2=\frac{1}{\pi}\left(\frac{24{,}000}{13{,}000}\right)^2
=\frac{1}{\pi}(1.8462)^2$$
$$\boxed{a_c\approx1.085\ \text{in}\quad(2a_c\approx2.17\ \text{in, full crack length})}$$
(well inside $w=20$ in, so the $Y=1$, $a\ll w$ assumption remains self-consistent at
failure).
(ii) Set up the Paris-law integral. With $\sigma_{min}=0$, $K_{min}=0$ and
$\Delta K=K_{max}=\sigma_{max}\sqrt{\pi a}$:
$$\frac{da}{dN}=C(\Delta K)^3=C\,\sigma_{max}^3\,\pi^{1.5}\,a^{1.5}$$
$$N=\int_{a_0}^{a_c}\frac{da}{C\,\sigma_{max}^3\,\pi^{1.5}\,a^{1.5}}
=\frac{1}{C\,\sigma_{max}^3\,\pi^{1.5}}\int_{a_0}^{a_c}a^{-1.5}\,da$$
(ii) Evaluate the integral.
$$\int_{a_0}^{a_c}a^{-1.5}\,da=\Big[-2a^{-0.5}\Big]_{a_0}^{a_c}=2\left(\frac{1}{\sqrt{a_0}}-\frac{1}{\sqrt{a_c}}\right)
=2\left(\frac{1}{\sqrt{0.1}}-\frac{1}{\sqrt{1.085}}\right)$$
$$=2(3.1623-0.9600)=4.4046$$
(ii) Fatigue life.
$$N=\frac{4.4046}{(1.8\times10^{-18})(13{,}000)^3(\pi^{1.5})}
=\frac{4.4046}{(1.8\times10^{-18})(2.197\times10^{12})(5.568)}$$
$$\boxed{N\approx2.00\times10^5\ \text{cycles}}$$
(agreement to 5 significant figures).
