Interactive Lecture
Stress Intensity Factors, Crack Tip Fields, and Fatigue Crack Growth
by Dr Arnaldo Delli Carri
Classical strength of materials assumes a defect-free solid: the material fails when the applied stress reaches its strength. Engineering structures, however, always contain defects — machining scratches, weld toes, porosity, inclusions. Fracture mechanics replaces the question "Is the stress too high?" with a more powerful one: "Given a crack of known size, will it grow?"
The discipline was born from a sequence of structural disasters. The Liberty Ships of World War II fractured in cold harbour water, with cracks propagating across entire hulls in minutes. Post-war analysis by Irwin and others formalised the field: a crack concentrates stress at its tip, and it is the intensity of that local stress field — not the nominal applied stress — that governs whether the crack advances.
A crack will propagate when the stress intensity factor \(K\) at its tip reaches the material's fracture toughness \(K_\text{Ic}\). The safe operating condition is therefore \(K < K_\text{Ic}\), which links applied load, crack size, and material resistance in a single inequality.
Crack-tip loading is decomposed into three independent modes, each with its own stress intensity factor:
Mixed-mode loading occurs in many practical situations (e.g. angled cracks, shear-dominated geometries). Fracture criteria under mixed-mode loading are an active research area, but Mode I dominates in most pressure vessel and structural applications and is the focus of this lecture.
For a through crack of half-length \(a\) in an infinite plate under remote tensile stress \(\sigma\), Irwin showed that the near-tip stress field has a universal form:
\[\sigma_{ij} = \frac{K_\text{I}}{\sqrt{2\pi r}} \, f_{ij}(\theta)\]
Where:
For the central through crack in an infinite plate (the Griffith geometry), the stress intensity factor is:
\[K_\text{I} = \sigma \sqrt{\pi a}\]
Where:
For real finite geometries, a dimensionless geometry correction factor \(Y\) is introduced:
\[K_\text{I} = Y \, \sigma \sqrt{\pi a}\]
\(Y\) depends on the crack geometry, loading configuration, and specimen dimensions. For an edge crack in a semi-infinite body, \(Y \approx 1.12\). Handbook solutions exist for hundreds of standard configurations (Stress Intensity Factor Handbook, Murakami; ASTM E399).
The critical stress intensity factor \(K_\text{Ic}\) is a material property measured under plane-strain conditions (thick specimens, small plastic zone relative to thickness). It is the resistance of the material to unstable crack extension. Representative values:
High-strength alloys are often less tough than lower-strength ones — a fundamental strength–toughness trade-off that drives many materials engineering decisions.
The \(1/\sqrt{r}\) singularity predicts infinite stress at the crack tip, which is physically impossible. The material yields in a small region called the plastic zone. For plane stress (thin plate) and plane strain (thick plate), the plastic zone radii are:
\[r_p^\text{plane stress} = \frac{1}{2\pi}\left(\frac{K_\text{I}}{\sigma_y}\right)^2\]
\[r_p^\text{plane strain} = \frac{1}{6\pi}\left(\frac{K_\text{I}}{\sigma_y}\right)^2\]
Where \(\sigma_y\) is the yield strength of the material (MPa).
Linear elastic fracture mechanics (LEFM) is valid when the plastic zone is small compared to other dimensions (crack length, specimen thickness, ligament). When the plastic zone becomes large, elastic-plastic fracture mechanics (EPFM) — using the J-integral or CTOD — must be used.
Griffith (1921) approached the crack extension problem from an energy perspective. A crack advances when the elastic strain energy released per unit crack advance equals or exceeds the energy consumed in creating new fracture surfaces:
\[G = \frac{K_\text{I}^2}{E'}\]
Where:
Fracture occurs when \(G \geq G_c = K_\text{Ic}^2 / E'\).
This simulation visualises the Mode I stress field around a central through crack in an infinite plate. The colour map shows \(\sigma_{yy}\) (the crack-opening stress) normalised against the remote applied stress. Adjust the crack length and applied stress to see how the stress intensity factor and plastic zone radius change. Toggle the plastic zone boundary to observe LEFM validity.
Under cyclic loading, a crack can grow incrementally even when \(K_\text{max} < K_\text{Ic}\). Paris and Erdogan (1963) showed that the crack growth rate in the stable propagation region follows a power law in the stress intensity factor range \(\Delta K\):
\[\frac{da}{dN} = C \, (\Delta K)^m, \qquad \Delta K = \Delta\sigma \sqrt{\pi a}\]
where \(C\) and \(m\) are material constants (Paris constants), \(\Delta\sigma = \sigma_\text{max} - \sigma_\text{min}\) is the stress range, and \(N\) is the number of load cycles. The left panel shows the crack growing in the specimen; the right panel plots crack length vs. cycles on a linear scale.
Set the initial crack size and loading parameters, then press Start to watch the crack grow. The simulation stops when \(K_\text{max}\) reaches \(K_\text{Ic}\) (fast fracture). Notice how crack growth is slow at first and then accelerates dramatically — the crack is racing towards a critical size.
Fracture mechanics provides the quantitative framework for damage-tolerant design — the philosophy that structures can operate safely with cracks, provided those cracks are found and managed before they reach critical size.
| Concept | Equation | Notes |
|---|---|---|
| Stress intensity factor (general) | \(K_\text{I} = Y\,\sigma\sqrt{\pi a}\) | \(Y\) = geometry factor; \(a\) = crack size |
| Central through crack, infinite plate | \(K_\text{I} = \sigma\sqrt{\pi a}\) | \(Y = 1\); \(a\) = half crack length |
| Edge crack, semi-infinite body | \(K_\text{I} \approx 1.12\,\sigma\sqrt{\pi a}\) | \(Y \approx 1.12\) |
| Near-tip stress field | \(\sigma_{ij} = \dfrac{K_\text{I}}{\sqrt{2\pi r}}\,f_{ij}(\theta)\) | Valid for \(r \ll a\) |
| Fracture criterion | \(K_\text{I} \geq K_\text{Ic}\) | \(K_\text{Ic}\) = plane-strain fracture toughness |
| Energy release rate | \(G = K_\text{I}^2 / E'\) | \(E' = E\) (plane stress); \(E/(1{-}\nu^2)\) (plane strain) |
| Plastic zone radius (plane stress) | \(r_p = \dfrac{1}{2\pi}\!\left(\dfrac{K_\text{I}}{\sigma_y}\right)^{\!2}\) | Irwin correction; thin plates |
| Plastic zone radius (plane strain) | \(r_p = \dfrac{1}{6\pi}\!\left(\dfrac{K_\text{I}}{\sigma_y}\right)^{\!2}\) | Factor of 3 smaller; thick plates |
| Paris Law (crack growth rate) | \(da/dN = C\,(\Delta K)^m\) | Region II (stable propagation) |
| \(\Delta K\) for through crack | \(\Delta K = \Delta\sigma\sqrt{\pi a}\) | \(\Delta\sigma = \sigma_\text{max} - \sigma_\text{min}\) |
| Fatigue life (Paris integration, \(m \neq 2\)) | \(N_f = \dfrac{a_c^{1-m/2} - a_0^{1-m/2}}{C(\Delta\sigma\sqrt{\pi})^m\,(1-m/2)}\) | Integration from \(a_0\) to critical size \(a_c\) |
| Critical crack size | \(a_c = \dfrac{1}{\pi}\!\left(\dfrac{K_\text{Ic}}{\sigma_\text{max}}\right)^{\!2}\) | For infinite plate; \(Y = 1\) |
A large steel pressure vessel is fabricated from a plate with fracture toughness \(K_\text{Ic} = 55 \text{ MPa}\sqrt{\text{m}}\) and yield strength \(\sigma_y = 620 \text{ MPa}\). During hydrostatic proof testing, the hoop stress reaches \(\sigma = 310 \text{ MPa}\). Assume a central through crack in a plate of sufficient width that \(Y = 1\).
Show Solution
Part (a) — Critical crack size
At fracture, \(K_\text{I} = K_\text{Ic}\), so:
\[K_\text{Ic} = \sigma\sqrt{\pi a_c} \implies a_c = \frac{1}{\pi}\left(\frac{K_\text{Ic}}{\sigma}\right)^2\]
Substituting:
\[a_c = \frac{1}{\pi}\left(\frac{55}{310}\right)^2 = \frac{1}{\pi}(0.17742)^2 = \frac{0.031478}{\pi}\]
\[a_c = 0.01002 \text{ m} \approx \mathbf{10.0 \text{ mm}}\]
The full crack length at fracture is \(2a_c \approx 20 \text{ mm}\).
Part (b) — Plane-strain plastic zone radius
\[r_p = \frac{1}{6\pi}\left(\frac{K_\text{Ic}}{\sigma_y}\right)^2 = \frac{1}{6\pi}\left(\frac{55}{620}\right)^2\]
\[\left(\frac{55}{620}\right)^2 = (0.08871)^2 = 7.869 \times 10^{-3}\]
\[r_p = \frac{7.869 \times 10^{-3}}{6\pi} = \frac{7.869 \times 10^{-3}}{18.85} = 4.17 \times 10^{-4} \text{ m} \approx \mathbf{0.42 \text{ mm}}\]
LEFM validity check: The ratio \(r_p / a_c = 0.42 / 10.0 = 0.042\), which is well below the LEFM validity limit of approximately 0.1 (i.e. the plastic zone is less than 10% of the crack length). The small-scale yielding assumption is satisfied, and \(K_\text{Ic}\) characterises the fracture event correctly.
An aluminium alloy component contains an initial edge crack of length \(a_0 = 2 \text{ mm}\). The geometry factor is \(Y = 1.12\) (edge crack). The Paris constants for this alloy are \(C = 3.0 \times 10^{-12}\) (units: m/cycle with \(\Delta K\) in MPa√m) and \(m = 3.0\). The stress range is \(\Delta\sigma = 100 \text{ MPa}\) and the fracture toughness is \(K_\text{Ic} = 28 \text{ MPa}\sqrt{\text{m}}\). Calculate:
Show Solution
Part (a) — Critical crack length
At fast fracture: \(K_\text{Ic} = Y\,\sigma_\text{max}\sqrt{\pi a_c}\)
With \(R = 0\), \(\sigma_\text{max} = \Delta\sigma = 100 \text{ MPa}\) and \(Y = 1.12\):
\[a_c = \frac{1}{\pi}\left(\frac{K_\text{Ic}}{Y\,\sigma_\text{max}}\right)^2 = \frac{1}{\pi}\left(\frac{28}{1.12 \times 100}\right)^2\]
\[\frac{28}{112} = 0.25 \implies a_c = \frac{(0.25)^2}{\pi} = \frac{0.0625}{\pi} = 0.01989 \text{ m} \approx \mathbf{19.9 \text{ mm}}\]
Part (b) — Fatigue life
Paris Law with \(\Delta K = Y\,\Delta\sigma\sqrt{\pi a}\):
\[\frac{da}{dN} = C\,(Y\,\Delta\sigma\sqrt{\pi})^m \, a^{m/2}\]
Separating variables and integrating from \(a_0\) to \(a_c\) with \(m = 3\), \(m/2 = 1.5\), \(1 - m/2 = -0.5\):
\[N_f = \int_{a_0}^{a_c} \frac{da}{C\,(Y\,\Delta\sigma\sqrt{\pi})^m\,a^{m/2}} = \frac{1}{C\,(Y\,\Delta\sigma\sqrt{\pi})^3}\int_{a_0}^{a_c} a^{-1.5}\,da\]
\[\int_{a_0}^{a_c} a^{-1.5}\,da = \left[\frac{a^{-0.5}}{-0.5}\right]_{a_0}^{a_c} = -2\left[a^{-0.5}\right]_{a_0}^{a_c} = 2\left(a_0^{-0.5} - a_c^{-0.5}\right)\]
Computing the prefactor (converting \(a\) to metres):
\[Y\,\Delta\sigma\sqrt{\pi} = 1.12 \times 100 \times \sqrt{\pi} = 112 \times 1.7725 = 198.5 \text{ MPa}\sqrt{\text{m}}\]
\[(Y\,\Delta\sigma\sqrt{\pi})^3 = (198.5)^3 = 7.821 \times 10^6 \text{ (MPa}\sqrt{\text{m}})^3\]
The denominator: \(C \times 7.821 \times 10^6 = 3.0\times10^{-12} \times 7.821\times10^6 = 2.346\times10^{-5}\)
Computing the integral with \(a_0 = 0.002 \text{ m}\) and \(a_c = 0.01989 \text{ m}\):
\[a_0^{-0.5} = (0.002)^{-0.5} = 22.36 \text{ m}^{-0.5}\]
\[a_c^{-0.5} = (0.01989)^{-0.5} = 7.09 \text{ m}^{-0.5}\]
\[2\,(22.36 - 7.09) = 2 \times 15.27 = 30.54 \text{ m}^{-0.5}\]
\[N_f = \frac{30.54}{2.345\times10^{-5}} \approx \mathbf{1.30 \times 10^6 \text{ cycles}}\]
A compact tension (CT) specimen of high-strength steel is tested to measure \(K_\text{Ic}\). The measured value is \(K_Q = 70 \text{ MPa}\sqrt{\text{m}}\) and the yield strength is \(\sigma_y = 1400 \text{ MPa}\). The ASTM E399 standard requires the specimen thickness \(B\), crack length \(a\), and uncracked ligament \(W - a\) each to satisfy:
\[B,\; a,\; (W-a) \;\geq\; 2.5\!\left(\frac{K_Q}{\sigma_y}\right)^2\]
The specimen dimensions are: \(W = 60 \text{ mm}\), \(a = 30 \text{ mm}\), \(B = 15 \text{ mm}\). Determine whether \(K_Q\) is a valid \(K_\text{Ic}\) measurement.
Show Solution
Compute the minimum required dimension:
\[2.5\left(\frac{K_Q}{\sigma_y}\right)^2 = 2.5\left(\frac{70}{1400}\right)^2 = 2.5\,(0.05)^2 = 2.5 \times 2.5\times10^{-3} = 6.25\times10^{-3}\text{ m} = \mathbf{6.25 \text{ mm}}\]
Check each dimension:
All three dimensions exceed the minimum requirement. The plastic zone is small relative to all specimen dimensions, confirming plane-strain dominance throughout the thickness. The measured \(K_Q = 70 \text{ MPa}\sqrt{\text{m}}\) is a valid plane-strain fracture toughness value: \(K_\text{Ic} = \mathbf{70 \text{ MPa}\sqrt{\text{m}}}\).
An inspection finds a surface crack of depth \(a_0 = 0.5 \text{ mm}\) in a gas pipeline weld. The pipe operates at a cyclic hoop stress range of \(\Delta\sigma = 60 \text{ MPa}\). For this geometry, \(Y = 1.12\). Material Paris constants: \(C = 6.0\times10^{-12}\) m/cycle (with \(\Delta K\) in MPa√m), \(m = 3.0\). Fracture toughness \(K_\text{Ic} = 80 \text{ MPa}\sqrt{\text{m}}\). The pipeline pressurises once per day.
Show Solution
Part (a) — Critical crack depth
\[a_c = \frac{1}{\pi}\left(\frac{K_\text{Ic}}{Y\,\sigma_\text{max}}\right)^2\]
With \(R \approx 0\), \(\sigma_\text{max} = 60\text{ MPa}\):
\[a_c = \frac{1}{\pi}\left(\frac{80}{1.12\times60}\right)^2 = \frac{1}{\pi}\left(\frac{80}{67.2}\right)^2 = \frac{1}{\pi}(1.1905)^2 = \frac{1.4173}{\pi} = 0.4511\text{ m}\]
\[a_c \approx \mathbf{451 \text{ mm}}\]
This is very large, reflecting the high toughness relative to the moderate stress. In practice, other criteria (leak-before-break, wall thickness) would govern first.
Part (b) — Fatigue life in cycles (Paris integration, \(m = 3\))
\[N_f = \frac{2\!\left(a_0^{-0.5} - a_c^{-0.5}\right)}{C\,(Y\,\Delta\sigma\sqrt{\pi})^3}\]
Prefactor denominator:
\[Y\,\Delta\sigma\sqrt{\pi} = 1.12\times60\times1.7725 = 119.1 \text{ MPa}\sqrt{\text{m}}\]
\[(119.1)^3 = 1.690\times10^6\]
\[C\times(119.1)^3 = 6.0\times10^{-12}\times1.690\times10^6 = 1.014\times10^{-5}\]
Integral (with \(a_0 = 5\times10^{-4}\text{ m}\), \(a_c = 0.451\text{ m}\)):
\[a_0^{-0.5} = (5\times10^{-4})^{-0.5} = 44.72\text{ m}^{-0.5}\]
\[a_c^{-0.5} = (0.451)^{-0.5} = 1.49\text{ m}^{-0.5}\]
\[2\,(44.72 - 1.49) = 2\times43.23 = 86.46\text{ m}^{-0.5}\]
\[N_f = \frac{86.46}{1.014\times10^{-5}} \approx \mathbf{8.53\times10^6 \text{ cycles}}\]
Part (c) — Years and inspection recommendation
\[t = \frac{N_f}{365\text{ cycles/year}} = \frac{8.53\times10^6}{365} \approx \mathbf{23{,}400 \text{ years}}\]
The predicted life far exceeds any practical service life. However, the calculation assumes perfectly constant \(\Delta\sigma\), no corrosion, and no defect coalescence. In practice, BS 7910 would apply safety factors (typically 2–10 on life), reducing the acceptable operating period substantially. Re-inspection would be scheduled on regulatory grounds rather than fracture mechanics necessity in this case.