Center-Cracked Panel — Fracture and Fatigue Life: 04-BS-11, December 2013, Q6

Question 6: Center-Cracked Panel — Fracture and Fatigue Life (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.

Given. Initial half-crack length $a_0=0.1$ in ($2a_0=0.2$ in); panel width $W=20$ in, thickness $=0.5$ in; $K_{Ic}=24{,}000$ psi$\sqrt{\text{in}}$; cyclic stress $0\to13{,}000$ psi ($\Delta\sigma=13{,}000$ psi, $R=0$); $Y=1$; Paris-law constants $C=1.8\times10^{-18}$, $m=3.0$.

Find. (i) Critical half-crack length $a_c$ at failure. (ii) Number of fatigue cycles $N_f$ to grow the crack from $a_0$ to $a_c$.

Fig. Q6 — centre-cracked panel: through-thickness crack of length $2a$, cyclically loaded between $\sigma=0$ and $\sigma=13{,}000$ psi.

Approach

Since $a\ll W$ the panel behaves as an infinite plate ($Y=1$); the critical crack length at failure comes directly from setting $K=K_{Ic}$ at the maximum applied stress. The fatigue life is obtained by integrating the Paris law from the initial to the critical half-crack length, using $\Delta K=\Delta\sigma\sqrt{\pi a}$ since $K_{min}=0$ (the stress ratio $R=0$).

(i) Critical crack length. Failure occurs when $K$ at the peak stress ($\sigma_{max}=13{,}000$ psi) reaches $K_{Ic}$: $$K_{Ic}=\sigma_{max}\sqrt{\pi a_c}\ \Rightarrow\ a_c=\frac{1}{\pi}\left(\frac{K_{Ic}}{\sigma_{max}}\right)^2 =\frac{1}{\pi}\left(\frac{24{,}000}{13{,}000}\right)^2=\boxed{1.085\ \text{in}}.$$ The full critical crack length is $2a_c=2.170$ in $\ll W=20$ in, confirming the $Y=1$ (infinite plate) assumption remains reasonable through to failure.
(ii) Set up the fatigue-crack-growth integral. With $K_{min}=0$, $\Delta K=\Delta\sigma\sqrt{\pi a}$, so $$\frac{da}{dN}=C\left(\Delta\sigma\sqrt{\pi a}\right)^3=C\,\Delta\sigma^3\,\pi^{1.5}\,a^{1.5}.$$ Separating variables and integrating from $a_0$ to $a_c$: $$N_f=\int_{a_0}^{a_c}\frac{da}{C\,\Delta\sigma^3\,\pi^{1.5}\,a^{1.5}} =\frac{2}{C\,\Delta\sigma^3\,\pi^{1.5}}\left(\frac1{\sqrt{a_0}}-\frac1{\sqrt{a_c}}\right).$$
Evaluate. $\Delta\sigma^3=(13{,}000)^3=2.197\times10^{12}$; $\pi^{1.5}=5.568$; $C\Delta\sigma^3\pi^{1.5}=2.202\times10^{-5}$; $1/\sqrt{a_0}=1/\sqrt{0.1}=3.162$, $1/\sqrt{a_c}=1/\sqrt{1.085}=0.960$: $$N_f=\frac{2}{2.202\times10^{-5}}\left(3.162-0.960\right)=\boxed{2.00\times10^{5}\ \text{cycles}}.$$

Quantity	Result
Critical half-crack length, $a_c$	1.085 in (2$a_c$ = 2.170 in)
Fatigue cycles to failure, $N_f$	$2.00\times10^5$ cycles

Concept & Theory

Topic. Linear elastic fracture mechanics and Paris-law fatigue crack growth.

Key relations. $K=\sigma\sqrt{\pi a}\,Y$; fracture when $K=K_{Ic}$; Paris law $da/dN=C(\Delta K)^m$; for $m=3$ the crack-growth integral has the closed form $N=\dfrac{2}{C\Delta\sigma^3\pi^{1.5}}\left(a_0^{-1/2}-a_c^{-1/2}\right)$.

Why this works. Because $R=0$ (stress cycles from zero), $K_{min}=0$ and $\Delta K$ equals the full stress-intensity range at the peak stress; the resulting power-law ODE in $a$ is separable and integrates in closed form for any $m\ne2$, with the crack's fatigue life dominated by the many cycles spent as a small, slow-growing crack (the $a_0^{-1/2}$ term), not the final rapid growth near $a_c$.

Common pitfall. Using the full crack length $2a$ rather than the half-length $a$ in $K=\sigma\sqrt{\pi a}$ — the standard center-crack formula is defined in terms of the half-crack length $a$, and doubling it inflates $K$ by $\sqrt2$ and badly under-predicts both $a_c$ and the fatigue life.

Source. Callister & Rethwisch, Materials Science and Engineering, 9th ed., Ch. 8 (fracture mechanics, fatigue crack propagation, the Paris law).

Question 6 of 8: Center-Cracked Panel — Fracture and Fatigue Life

Question 6: Center-Cracked Panel — Fracture and Fatigue Life (20 marks)

Approach