Fracture mechanics and crack growth

Why it matters

Damage-tolerant design assumes that cracks exist in the structure from the day it is built, from manufacturing flaws, fatigue or corrosion. The questions are then: how big can a crack get before the part breaks, and how many flights does it take to grow to that size? Linear elastic fracture mechanics answers both, and its answers set the inspection intervals written into every airliner's maintenance programme.

Key ideas

Stress intensity factor. Near the tip of a sharp crack in a linear elastic body the stresses vary as 1/√r, so they are infinite at the tip and stress alone cannot be a failure criterion. The strength of the singular field is the stress intensity factor K = Y·σ·√(π·a), where σ is the remote stress, a the crack length (half-length for a central crack, depth for an edge crack) and Y a geometry factor: Y = 1 for a central crack in a wide plate, about 1.12 for an edge crack in a wide plate. Y for finite widths and other geometries comes from handbooks. Units: MPa√m.

Modes. Mode I (opening, tension normal to the crack) is the most important; mode II is in-plane sliding, mode III tearing (anti-plane shear).

Fracture toughness. Fracture occurs when K reaches the material's fracture toughness K_c. K_c depends on thickness: thin sheets fail in plane stress with a higher K_c; thick sections approach the plane-strain minimum K_IC, valid when the thickness B ≥ 2.5·(K_IC/σ_y)². K is a load-and-geometry quantity; K_c is a material property (a data-book value).

Critical crack size and residual strength. Setting K = K_c gives the critical crack length a_c = (1/π)·(K_c/(Y·σ))² at a given stress, or the residual strength σ_c = K_c/(Y·√(π·a)) for a given crack. A damage-tolerant structure must carry limit load with the largest crack that could be missed by inspection.

Energy view (Griffith–Irwin). The energy release rate G is the energy available per unit area of crack growth; fracture occurs when G reaches the material's toughness G_c. For mode I, G = K²/E in plane stress (K²(1 − ν²)/E in plane strain), so the K and energy criteria are equivalent.

Crack-tip plasticity. LEFM is valid only when the plastic zone is small compared with the crack and the section. The plane-stress plastic zone size is about r_p = (1/2π)·(K/σ_y)². Tough, thin aluminium sheet may need elastic–plastic methods (R-curves, J-integral) beyond this topic.

Fatigue crack growth (Paris law). Under cyclic load the crack grows a little every cycle. In the mid-range of ΔK = Y·Δσ·√(π·a), growth follows da/dN = C·(ΔK)^m, with m typically 2–4 for metals and C a material constant (with stated units). Below a threshold ΔK_th cracks do not grow; near K_max = K_c growth accelerates to fracture. Integrating from the initial (detectable) crack a_i to the critical crack a_c gives the crack growth life; the inspection interval is this life divided by a factor (often 2 or more) so that the crack is found on at least one inspection before it becomes critical.

Influences on growth. Stress ratio R (higher R speeds growth), overloads (retard growth), environment (humid air and corrosive media accelerate it), temperature and material temper. 2024-T3 is preferred for lower wing and fuselage skins because of its slow crack growth and high toughness; 7075-T6 is stronger but less damage tolerant.

Formulas

K = Y·σ·√(π·a)

• K: stress intensity factor (MPa√m), Y: geometry factor, σ: remote stress (MPa), a: crack length (m).

a_c = (1/π)·(K_c / (Y·σ))²

• Critical crack length (m); K_c: fracture toughness (MPa√m), a data-book value.

σ_c = K_c / (Y·√(π·a))

• Residual strength (MPa) with a crack of length a.

G = K² / E

• Energy release rate (J/m²) in plane stress; E: Young's modulus (Pa). Plane strain: G = K²(1 − ν²)/E.

r_p = (1/(2·π))·(K/σ_y)²

• Plane-stress plastic zone size (m); σ_y: yield stress.

B ≥ 2.5·(K_IC/σ_y)²

• Thickness (m) for plane-strain conditions.

da/dN = C·(ΔK)^m with ΔK = Y·Δσ·√(π·a)

• Paris law; da/dN in m/cycle, Δσ = σ_max − σ_min (MPa), C and m material constants.

N = [a_i^(1 − m/2) − a_f^(1 − m/2)] / [C·(Y·Δσ·√π)^m·(m/2 − 1)]

• Cycles to grow from a_i to a_f (m), constant Y, m ≠ 2.

Worked examples

Example 1 (standard): critical crack size. Given: wide aluminium sheet with a central through crack, Y = 1, K_c = 70 MPa√m (data), applied stress σ = 150 MPa.

1. a_c = (1/π)·(K_c/(Y·σ))² = (1/π) × (70/150)² = (1/π) × 0.2178 = 0.0693 m.
2. Total critical crack length: 2a_c = 0.139 m = 139 mm.
3. If the crack is 2a = 60 mm (a = 0.03 m), residual strength: σ_c = 70/√(π × 0.03) = 70/0.3070 = 228 MPa. Answer: critical total length ≈ 139 mm; with a 60 mm crack the sheet still carries 228 MPa.

Example 2 (GATE level): crack growth life and inspection interval. Given: central crack, Y = 1, cycles between 0 and 100 MPa (Δσ = 100 MPa, R = 0), Paris constants C = 1 × 10⁻¹¹ (m/cycle with ΔK in MPa√m) and m = 3, initial crack a_i = 2 mm, K_c = 70 MPa√m.

1. Critical crack at σ_max = 100 MPa: a_c = (1/π) × (70/100)² = 0.156 m.
2. Constant: Y·Δσ·√π = 100 × 1.7725 = 177.2; 177.2³ = 5.568 × 10⁶.
3. a_i^(−1/2) = 0.002^(−0.5) = 22.36; a_c^(−1/2) = 0.156^(−0.5) = 2.532.
4. N = (22.36 − 2.532)/(1 × 10⁻¹¹ × 5.568 × 10⁶ × 0.5) = 19.83/2.784 × 10⁻⁵ = 7.12 × 10⁵ cycles.
5. With one cycle per flight and a factor of 2: inspection interval ≈ 3.56 × 10⁵ flights. Answer: a_c = 156 mm, crack growth life ≈ 7.1 × 10⁵ cycles, inspection interval ≈ 3.6 × 10⁵ cycles. Note that most of the life is spent while the crack is small: halving a_i matters far more than raising K_c.

Common mistakes

• Using the total crack length 2a in place of a for a central crack.
• Mixing units: K in MPa√m needs a in metres, not millimetres.
• Treating K as a material property; K_c (or K_IC) is the property, K the driving force.
• Using K_IC for thin sheet, which is conservative but may be very pessimistic; check the thickness criterion.
• Applying Paris constants with ΔK in different units from those used to fit them.
• Taking the maximum stress instead of the stress range in ΔK.

For GATE AE

Expect calculations of K, critical crack size, residual strength, plastic zone size, crack growth rate from Paris law and simple life integrations with constant Y. Conceptual questions cover modes, plane stress versus plane strain, and the link between damage tolerance and inspection intervals. Practise the Paris integration with m = 3 and m = 4.

Quick check

1. For an edge crack (Y = 1.12) of depth 5 mm under 200 MPa, what is K?
2. If the stress doubles, how does the critical crack length change?
3. With m = 4, how much faster does a crack grow when ΔK doubles?
4. Which is larger for the same alloy: K_c for thin sheet or K_IC? Answers: 1. K = 1.12 × 200 × √(π × 0.005) = 28.1 MPa√m; 2. it falls to one quarter; 3. 16 times; 4. K_c for thin sheet.