Worked examples — Fracture mechanics — stress intensity factor K, toughness K_IC

We will not use a single symbol above without the one-line meaning just given.

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The scenario matrix

Think of fracture problems as a grid. Each row is a kind of question the universe (or an exam) can ask. If you can solve one example from each row, you can solve any problem in this topic.

#	Cell (case class)	What makes it different	Example
A	Edge crack, find $K_I$	$Y=1.12$, use full length $a$	Ex 1
B	Centre crack, find $K_I$	$Y=1.0$, use half-length $a$	Ex 2
C	Failure decision ($K_I$ vs $K_{IC}$)	compare, output "safe / breaks"	Ex 3
D	Solve for critical crack size $a_{\text{max}}$	invert the formula	Ex 4
E	Solve for critical stress ${\sigma }_c$	invert the other way	Ex 5
F	Degenerate / limiting input	$a\to 0$, huge $\sigma$, very brittle material	Ex 6
G	Mixed-mode ($K_I$ and $K_{II}$)	combine with $K_{\text{eff}}$	Ex 7
H	Real-world word problem (inspection interval)	fracture + crack growth in time	Ex 8
I	Exam-style twist (units / doubling trap)	scaling, unit landmines	Ex 9

We cover every cell below. Each example makes you forecast the answer's ballpark first — guessing sharpens intuition far more than reading.

How to read the map figure

How to read this figure. Two identical plates are pulled top-and-bottom by the yellow stress arrows $\sigma$ — that vertical pull is what opens a horizontal crack. On the left, the red line is an edge crack: it starts at the plate's free left surface and its labelled length is the full $a$ you plug in, with $Y=1.12$. On the right, the blue line is a centre crack: it is buried in the middle, its total span is $2a$ (green arrows), and you must plug in the half-length $a$ (blue label) with $Y=1.0$. This left-vs-right split is the #1 thing students mix up — glance back here whenever a problem says "edge" or "centre". We use the left panel in Ex 1 and Ex 3, the right panel in Ex 2 and Ex 5.

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Cell A — Edge crack, find $K_I$

Forecast: $\sigma$ is a bit bigger than the $150\text{MPa}$ example in the parent note, and the crack a bit longer, so expect $K_I$ somewhere in the high-teens $\text{MPa}\sqrt{\text{m}}$.

1. Convert the crack length to metres. $a=3\text{mm}=0.003\text{m}$. Why this step? The $\sqrt{\text{m}}$ in the unit of $K$ only works if $a$ is in metres. Millimetres would give an answer $\sqrt{1000}\approx 31.6\times$ too small.
2. Compute $\sqrt{\pi a}$. $\pi a=3.1416\times 0.003=0.009425$, so $\sqrt{0.009425}=0.09708{\text{m}}^{1/2}$. Why this step? This is the "size factor" — how much the crack's length alone amplifies stress, before geometry.
3. Multiply everything. $K_I=1.12\times 180\times 0.09708=19.6\text{MPa}\sqrt{\text{m}}$. Why this step? $K_I=Y\sigma \sqrt{\pi a}$ — that's the whole formula, assembled.

$\boxed{K_I=19.6\text{MPa}\sqrt{\text{m}}}$

Verify: Units: $[\text{MPa}]\times [{\text{m}}^{1/2}]=\text{MPa}\sqrt{\text{m}}$ ✓. Ballpark matched the forecast (high-teens) ✓. Against aluminium's $K_{IC}=35$, this hull is safe (19.6 < 35).

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Cell B — Centre crack, find $K_I$ (the halving trap)

Forecast: They gave total length $10\text{mm}$, but centre cracks use the half-length. If you forget and use $0.010\text{m}$ you'll get an answer $\sqrt{2}\approx 1.41\times$ too big. Guess ~$27\text{MPa}\sqrt{\text{m}}$.

1. Halve the width. Total $2a=10\text{mm}\Rightarrow a=5\text{mm}=0.005\text{m}$. Why this step? The formula's $a$ is measured from the centre to one tip. The whole crack is symmetric, and $K_I$ describes one tip's field. Using the full width double-counts.
2. Size factor. $\sqrt{\pi \times 0.005}=\sqrt{0.015708}=0.12533{\text{m}}^{1/2}$. Why this step? Same size factor as before — this is the geometry-independent core.
3. Assemble. $K_I=1.0\times 220\times 0.12533=27.6\text{MPa}\sqrt{\text{m}}$. Why this step? $Y=1.0$ because a buried crack has no free surface helping it open.

$\boxed{K_I=27.6\text{MPa}\sqrt{\text{m}}}$

Verify: If you wrongly used $a=0.010$ you'd get $220\times \sqrt{\pi \cdot 0.010}=39.0$ — a full $41\%$ high. The halving matters. Against Ti's $K_{IC}=55$, still safe ✓.

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Cell C — Does it break? (the decision)

Forecast: Ceramics are very brittle (tiny $K_{IC}$). Even a small crack under moderate stress might blow past $4$. Guess: it breaks.

1. Compute the driving force $K_I$. $a=0.001\text{m}$, $\sqrt{\pi \cdot 0.001}=0.05605{\text{m}}^{1/2}$; $K_I=1.12\times 60\times 0.05605=3.77\text{MPa}\sqrt{\text{m}}$. Why this step? Before comparing, we need the number the material is feeling.
2. Compare to toughness. Is $K_I\ge K_{IC}$? $3.77$ vs $4.0$ → $3.77<4.0$. Why this step? This inequality is the failure criterion — the only test that decides survival.
3. State the verdict. $K_I<K_{IC}$, so the tile just survives — but with almost no margin.

$\boxed{K_I=3.77<K_{IC}=4.0\Rightarrow \text{no fracture (barely)}}$

Verify: A margin of only $4.0/3.77=1.06$ means a $6\%$ stress increase (or the crack growing to $\sim 1.1\text{mm}$) tips it into failure. This is why ceramics demand Non-destructive testing (NDT) and a Damage tolerance philosophy with generous safety factors ✓.

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Cell D — Solve for the critical crack size $a_{\text{max}}$

Forecast: High stress + safety factor of 2 will squeeze $a_{\text{max}}$ down to a few millimetres.

1. Write the safe condition. $K_I\le K_{IC}/n$, i.e. $Y\sigma \sqrt{\pi a}=K_{IC}/n$ at the limit. Why this step? Dividing $K_{IC}$ by $n$ builds the margin in — we design as if the material were half as tough.
2. Isolate the square root. Divide both sides by $Y\sigma$: $\sqrt{\pi a}=\frac{K_{IC}}{Y\sigma n}$ Why this step? We want $a$ alone, so we first strip away everything multiplying the $\sqrt{\pi a}$ block.
3. Square both sides to kill the square root: $\pi a={\left(\frac{K_{IC}}{Y\sigma n}\right)}^2$ Why this step? $a$ is trapped under a square root; squaring is the exact inverse operation that frees it — and it also squares the right-hand side.
4. Divide by $\pi$ to finish isolating $a$: $a_{\text{max}}=\frac{1}{\pi }{\left(\frac{K_{IC}}{Y\sigma n}\right)}^2$ Why this step? The $\pi$ was multiplying $a$ on the left, so it moves downstairs on the right — that is where the "$1/\pi$" comes from.
5. Plug in numbers. Inner fraction: $\frac{50}{1.12\times 300\times 2}=\frac{50}{672}=0.07440$. Square: $0.005535$. Divide by $\pi$: $a_{\text{max}}=0.001762\text{m}=1.76\text{mm}$. Why this step? Numbers turn the design equation into an inspection spec.

$\boxed{a_{\text{max}}=1.76\text{mm}}$

Verify: Back-substitute $a=0.001762$: $K_I=1.12\times 300\times \sqrt{\pi \cdot 0.001762}=1.12\times 300\times 0.07440=25.0\text{MPa}\sqrt{\text{m}}=K_{IC}/n=50/2$ ✓. Inspection (ultrasonic ~1 mm) must reliably find cracks below $1.76\text{mm}$ ✓.

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Cell E — Solve for the critical stress ${\sigma }_c$

Forecast: Small crack + decent toughness → high critical stress, likely a few hundred MPa.

1. Set $K_I=K_{IC}$ at fracture. $Y{\sigma }_c\sqrt{\pi a}=K_{IC}$. Why this step? Fracture happens exactly when driving force equals toughness — that's the boundary we solve on.
2. Isolate ${\sigma }_c$. ${\sigma }_c=\frac{K_{IC}}{Y\sqrt{\pi a}}$. Why this step? Here $\sigma$ is the unknown, not $a$ — same equation, different variable freed.
3. Compute. $\sqrt{\pi \cdot 0.002}=0.07927{\text{m}}^{1/2}$; ${\sigma }_c=\frac{25}{1.0\times 0.07927}=315.4\text{MPa}$. Why this step? This is the load the boom must never exceed with this crack present.

$\boxed{{\sigma }_c\approx 315\text{MPa}}$

Verify: Feed ${\sigma }_c$ back: $1.0\times 315.4\times 0.07927=25.0=K_{IC}$ ✓. Matches Griffith energy criterion intuition — smaller crack tolerates larger stress ✓.

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Cell F — Degenerate & limiting inputs

Forecast: (a) should be zero; (b) tiny toughness → tiny critical stress; (c) unbounded.

1. (a) $a=0$. $K_I=1.12\times 100\times \sqrt{\pi \cdot 0}=0$. Why this step? A body with no crack has no stress-intensity singularity — fracture mechanics simply doesn't apply, and $K_I\to 0$ confirms the formula degrades gracefully. Failure then reverts to ordinary yield-strength checks.
2. (b) Critical stress. ${\sigma }_c=\frac{K_{IC}}{Y\sqrt{\pi a}}=\frac{2}{1.12\times \sqrt{\pi \cdot 0.001}}=\frac{2}{1.12\times 0.05605}=31.9\text{MPa}$. Why this step? Shows how brutally a low-$K_{IC}$ material is limited — a 1 mm flaw caps stress at ~32 MPa, far below typical structural loads.
3. (c) $\sigma \to \infty$. Since $K_I$ is linear in $\sigma$, $K_I\to \infty$: any crack fails at sufficiently high stress. Why this step? Confirms there is no infinitely strong cracked body — every crack has a finite critical stress ${\sigma }_c=K_{IC}/(Y\sqrt{\pi a})$.

$\boxed{(a)K_I=0(b){\sigma }_c=31.9\text{MPa}(c)K_I\to \infty }$

Verify: (b) back-check: $1.12\times 31.9\times 0.05605=2.00=K_{IC}$ ✓. Limiting behaviours are monotone and sign-consistent ✓.

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Cell G — Mixed-mode fracture

Forecast: Neither mode alone exceeds $22$, but combined they might. Guess: unsafe.

1. Identify the modes present. Opening $K_I=18$, sliding $K_{II}=10$, tearing $K_{III}=0$ (defined in the callout above). Why this step? You must know which physical loading each $K$ represents before combining them.
2. Combine with $K_{\text{eff}}$. $K_{\text{eff}}=\sqrt{K_I^2+K_{II}^2+K_{III}^2}=\sqrt{18^2+10^2+0^2}=\sqrt{324+100}=\sqrt{424}=20.6\text{MPa}\sqrt{\text{m}}$. Why this step? The modes add as perpendicular components — root-sum-of-squares.
3. Compare. $K_{\text{eff}}=20.6<K_{IC}=22$ → safe, but only $1.07\times$ margin. Why this step? Mixed-mode is judged against the same toughness limit as pure Mode I.

$\boxed{K_{\text{eff}}=20.6\text{MPa}\sqrt{\text{m}}<22\Rightarrow \text{safe (thin margin)}}$

Verify: $18^2+10^2=424$, $\sqrt{424}=20.59$ ✓. Note $K_{\text{eff}}>K_I$: ignoring Mode II would have falsely declared a $22/18=1.22$ margin — mixed-mode analysis is not optional ✓.

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Cell H — Real-world word problem (inspection interval)

Forecast: Gap is ~$1.16\text{mm}$ at $0.25\text{mm/yr}$ → roughly 4–5 years to critical; inspect sooner.

1. Find the growth allowance. $\Delta a=a_{\text{max}}-a_0=1.76-0.60=1.16\text{mm}$. Why this step? The crack is already part-grown; only the remaining margin buys us time.
2. Time to critical. $t=\frac{\Delta a}{\text{rate}}=\frac{1.16}{0.25}=4.64\text{years}$. Why this step? Constant growth rate makes this a simple distance/speed calculation. (Real fatigue follows Fatigue crack growth (Paris law), which accelerates — so this linear estimate is optimistic.)
3. Set the interval. Inspect at half the time-to-critical: $4.64/2=2.32\text{years}$, rounded down to a 2-year inspection interval. Why this step? Halving guarantees at least one inspection occurs before the crack ever reaches $a_{\text{max}}$ — the standard Damage tolerance philosophy "two-lives" rule.

$\boxed{t_{\text{crit}}\approx 4.6\text{yr}\Rightarrow \text{re-inspect every 2 years}}$

Verify: After 4 years: $a=0.6+4\times 0.25=1.6\text{mm}<1.76$ still safe; after 5 years: $1.85>1.76$ failed — so critical lies between year 4 and 5, consistent with $4.64$ yr ✓.

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Cell I — Exam-style twist (scaling & unit traps)

Forecast: (a) doubling $a$ multiplies by $\sqrt{2}\approx 1.41$; (b) 531 is absurdly large — a unit error.

1. (a) Scaling. $K_I\propto \sqrt{a}$. Double $a$ → factor $\sqrt{2}$: $K_I^{\prime }=12\times \sqrt{2}=16.97\text{MPa}\sqrt{\text{m}}$. Why this step? Because $a$ lives under a square root, $K$ grows sub-linearly — a crucial and frequently-tested fact. To double $K$ you'd need $4\times$ the crack.
2. (b) Spot the error. The student used $a$ in millimetres. Correct: $a=0.004\text{m}$. Why this step? $\sqrt{\text{m}}$ vs $\sqrt{\text{mm}}$ differ by $\sqrt{1000}\approx 31.6$ — exactly the factor $531/16.8\approx 31.6$ that inflated the answer.
3. (b) Correct value. $K_I=150\times \sqrt{\pi \times 0.004}=150\times 0.11210=16.8\text{MPa}\sqrt{\text{m}}$. Why this step? A physically sane number, comparable to real toughness values.

$\boxed{(a)K_I^{\prime }=17.0(b)K_I=16.8\text{MPa}\sqrt{\text{m}}}$

Verify: (a) $12\sqrt{2}=16.97$ ✓. (b) The student's number $150\sqrt{\pi \cdot 4}=531.7$ is $\sqrt{1000}=31.6\times$ the correct $16.8$ — the tell-tale metre/millimetre signature ✓. Related to Stress concentration factors, and the $J$-integral generalises this in J-integral.

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Recall Self-test (reveal after guessing)

An edge crack $a=2$mm uses which Y? ::: $Y=1.12$ A centre crack of total width $2a=6$mm uses what $a$ in metres? ::: $a=0.003$m (half of 6 mm) Doubling crack length multiplies $K_I$ by? ::: $\sqrt{2}\approx 1.41$ Fracture happens when? ::: $K_I\ge K_{IC}$ Two modes with $K_I,K_{II}$ combine as? ::: $K_{\text{eff}}=\sqrt{K_I^2+K_{II}^2}$ Getting $K_I\approx 500\text{MPa}\sqrt{\text{m}}$ usually means? ::: crack length left in mm, not m