Fracture mechanics — stress intensity factor K, toughness K_IC: Exercises

Before we start, one reminder in plain words so no symbol is a mystery:

Figure — Fracture mechanics — stress intensity factor K, toughness K_IC

Look at the figure: the same formula, two crack pictures. On the left the crack starts at the edge and reaches depth $a$ . On the right the crack sits in the middle, total mouth-to-tip length $2 a$ , so we feed in $a$ = half of that. Getting this "which length do I use" right is the single most common slip — the figure is your anchor for every problem below.

L1 · Recognition

Exercise 1.1

State the units of $K_{I}$ , and say in one sentence what $K_{I C}$ is.

Recall Solution

$K_{I}$ has units of stress × √length, i.e. $MPa m$ . $K_{I C}$ is the material's fracture toughness: the critical value of $K_{I}$ at which a Mode-I crack begins unstable (runaway) propagation. It is measured, not calculated, and does not depend on the part's shape.

Exercise 1.2

A centre crack in a wide plate has total length $2 a = 10 mm$ . What value of $a$ goes into $K_{I} = σ π a Y$ , and what is $Y$ ?

Recall Solution

What we do: halve the total length, because the formula uses the half-length for a centre crack (see figure, right panel). $a = \frac{1}{2} (10 mm) = 5 mm = 0.005 m, Y = 1.0.$

L2 · Application

Exercise 2.1

A titanium panel ( $Y = 1.0$ , centre crack) carries $σ = 300 MPa$ over a centre crack of total length $2 a = 6 mm$ . Compute $K_{I}$ .

Recall Solution

Step 1 — half length: $a = 3 mm = 0.003 m$ . Step 2 — plug in: $K_{I} = 300 π \times 0.003 \times 1.0.$ $π a = 3.1416 \times 0.003 = 0.009425 m$ . $0.009425 = 0.09708 m^{1/2}$ . $K_{I} = 300 \times 0.09708 = 29.1 MPa m .$ (Working in MPa gives the answer directly in $MPa m$ — no need to convert to pascals and back.)

Exercise 2.2

An edge crack of depth $a = 2.5 mm$ ( $Y = 1.12$ ) sits in a hull skin under hoop stress $σ = 180 MPa$ . Find $K_{I}$ .

Recall Solution

Edge crack: use the full depth, $a = 0.0025 m$ . $π a = 3.1416 \times 0.0025 = 0.007854 m$ . $0.007854 = 0.08862 m^{1/2}$ . $K_{I} = 180 \times 0.08862 \times 1.12 = 17.9 MPa m .$

L3 · Analysis

Exercise 3.1

Aluminium 2024-T3 has $K_{I C} = 35 MPa m$ . A pressure hull region carries $σ = 150 MPa$ with an edge crack, $Y = 1.12$ . Ignoring any safety factor ( $n = 1$ ), what crack depth $a_{c}$ makes the part fracture?

Recall Solution

What we do: set the applied $K_{I}$ equal to the material limit and solve for $a$ . $σ π a_{c} Y = K_{I C} \Rightarrow a_{c} = \frac{1}{π} (\frac{K _{I C}}{Y σ})^{2} .$ Why this rearrangement: we want the crack size, so we isolate $a$ — square both sides after dividing out $Y σ$ . $\frac{K _{I C}}{Y σ} = \frac{35}{1.12 \times 150} = \frac{35}{168} = 0.20833 m^{1/2} .$ $(0.20833)^{2} = 0.043403, a_{c} = \frac{0.043403}{3.1416} = 0.013815 m \approx 13.8 mm .$ A crack deeper than ~13.8 mm fractures this hull at 150 MPa.

Exercise 3.2

For the same hull, add a safety factor $n = 2$ . What is the maximum allowable crack size $a_{m a x}$ that inspectors must be able to catch?

Recall Solution

What changes: we now require $K_{I} \leq K_{I C} / n$ , so replace $K_{I C}$ by $K_{I C} / n$ : $a_{m a x} = \frac{1}{π} (\frac{K _{I C}}{Y σ n})^{2} .$ $\frac{K _{I C}}{Y σ n} = \frac{35}{1.12 \times 150 \times 2} = \frac{35}{336} = 0.104167 m^{1/2} .$ $(0.104167)^{2} = 0.010851, a_{m a x} = \frac{0.010851}{3.1416} = 0.003454 m \approx 3.45 mm .$ Notice $a_{m a x}$ is one quarter of $a_{c}$ (13.8/3.45 ≈ 4). That is no accident — a factor $n = 2$ inside a squared term shrinks the allowable crack by $n^{2} = 4$ .

L4 · Synthesis

Exercise 4.1

A titanium truss (Ti-6Al-4V, $K_{I C} = 55 MPa m$ ) has an edge crack $Y = 1.12$ and operates at $σ = 200 MPa$ , safety factor $n = 2$ . (a) Find $a_{m a x}$ . (b) NDT (see Non-destructive testing (NDT)) reliably finds cracks down to $1.5 mm$ . Is there a safe margin?

Recall Solution

(a) $\frac{K _{I C}}{Y σ n} = \frac{55}{1.12 \times 200 \times 2} = \frac{55}{448} = 0.122768 m^{1/2} .$ $(0.122768)^{2} = 0.015072, a_{m a x} = \frac{0.015072}{3.1416} = 0.004797 m \approx 4.80 mm .$ (b) The smallest detectable crack (1.5 mm) is well below the maximum tolerable crack (4.80 mm), so yes, there is margin — inspection catches cracks roughly $4.80/1.5 = 3.2 \times$ smaller than the danger size. This is the essence of Damage tolerance philosophy: design so the critical crack is comfortably larger than what inspection can see.

Exercise 4.2

The same truss crack grows by fatigue at a steady rate of $0.4 mm/year$ (a Fatigue crack growth (Paris law) regime we treat as constant here). It starts at the just-detected size $1.5 mm$ . How many years until it reaches $a_{m a x} = 4.80 mm$ , and what inspection interval keeps us safe with one visit to spare?

Recall Solution

Step 1 — remaining crack budget: $Δ a = 4.80 - 1.5 = 3.30 mm$ . Step 2 — time to reach $a_{m a x}$ : $t = \frac{Δ a}{rate} = \frac{3.30 mm}{0.4 mm/yr} = 8.25 years .$ Step 3 — inspection interval: to guarantee at least one inspection before the crack becomes critical, inspect at half that life: $8.25/2 \approx 4.1 years$ . Rounding down for conservatism, inspect every 4 years.

L5 · Mastery

Exercise 5.1

You must pick a material for a bracket at $σ = 250 MPa$ , edge crack $Y = 1.12$ , safety factor $n = 1.5$ , and NDT that detects cracks down to $2.0 mm$ . Two candidates:

Steel 4340: $K_{I C} = 50 MPa m$
CFRP: $K_{I C} = 25 MPa m$

Compute $a_{m a x}$ for each. Which materials give a genuine inspection margin (i.e. $a_{m a x} > 2.0 mm$ )?

Recall Solution

Use $a_{m a x} = \frac{1}{π} (\frac{K _{I C}}{Y σ n})^{2}$ with $Y σ n = 1.12 \times 250 \times 1.5 = 420$ .

Steel 4340: $\frac{50}{420} = 0.119048, (0.119048)^{2} = 0.014172, a_{m a x} = \frac{0.014172}{3.1416} = 0.004511 m \approx 4.51 mm .$ CFRP: $\frac{25}{420} = 0.059524, (0.059524)^{2} = 0.003543, a_{m a x} = \frac{0.003543}{3.1416} = 0.001128 m \approx 1.13 mm .$

Decision: Steel's $a_{m a x} = 4.51 mm$ is comfortably above the 2.0 mm detection floor — safe with margin. CFRP's $a_{m a x} = 1.13 mm$ is smaller than what NDT can even see (2.0 mm) — a critical crack could exist and pass inspection undetected. CFRP fails the damage-tolerance test here and must be rejected (or redesigned to lower $σ$ ).

Exercise 5.2 (mixed-mode capstone)

A crack in an aluminium fitting ( $K_{I C} = 35 MPa m$ ) experiences combined loading: $K_{I} = 20 MPa m$ and $K_{I I} = 12 MPa m$ (Mode III negligible). Using the effective stress intensity $K_{eff} = K_{I}^{2} + K_{I I}^{2} + K_{I I I}^{2}$ from the parent note, does the fitting fracture?

Recall Solution

What we do: combine the mode contributions into one number and compare to toughness. $K_{eff} = 2 0^{2} + 1 2^{2} + 0^{2} = 400 + 144 = 544 = 23.32 MPa m .$ Since $K_{eff} = 23.3 < K_{I C} = 35$ , the fitting does not fracture — it sits at $23.3/35 = 67%$ of toughness, i.e. an effective safety factor of about $1.5$ .

Recall Self-test cloze

The formula for the stress intensity factor ::: $K_{I} = σ π a Y$ For a centre crack of total length $2 a$ , the value used in the formula is ::: the half-length $a$ The maximum allowable crack size scales with the safety factor as ::: $(1/ n)^{2}$ — doubling $n$ quarters the crack budget A safe inspection interval is at most ::: half the crack's remaining life (the two-inspection rule) Effective stress intensity for mixed mode ::: $K_{eff} = K_{I}^{2} + K_{I I}^{2} + K_{I I I}^{2}$