Questions: Stress Intensity Factor and Fracture Mechanics

Question 1

A crack in a structural component has a half-length of 4 mm and the applied stress intensity factor K equals 0.8 K_IC. The crack then grows so its half-length becomes 16 mm, while the applied stress remains the same. Is the component now safe?

• A. Yes — the crack is still below the critical length, so K < K_IC.
• B. No — quadrupling the crack half-length doubles K (since K ∝ √a), so K is now 1.6 K_IC and the crack propagates unstably.
• C. No — quadrupling the crack half-length quadruples K, so K is now 3.2 K_IC.
• D. Yes — K_IC is a material constant and does not change with crack size, so the same margin remains.

Answer: (B)

K = Yσ√(πa), so K scales with √a. When a increases by a factor of 4 (from 4 mm to 16 mm), K increases by √4 = 2. The initial K was 0.8 K_IC, so the new K is 1.6 K_IC — above the critical threshold. This square-root dependence is the non-obvious core result: a crack four times longer is only twice as severe in terms of K, not four times.

Question 2

What does the fracture toughness K_IC represent in practice?

• A. The maximum stress a material can sustain before any crack nucleates.
• B. The stress concentration factor at the crack tip for Mode I loading.
• C. A material property giving the critical stress intensity at which a crack propagates unstably under Mode I loading.
• D. The energy required to create a unit area of new crack surface, equivalent to surface energy.

Answer: (C)

K_IC is a material constant — like yield strength — measured experimentally using standardized specimens. It is not a stress, not a stress concentration factor, and not a surface energy directly (though it relates to Griffith energy via K_IC² = G_IC·E for plane stress). The design criterion is K < K_IC for safe operation. When K reaches K_IC, the crack grows unstably and fracture occurs.

Question 3

A structure can remain safe in service even if it contains cracks, as long as the stress intensity factor K at the largest expected crack remains below K_IC.

• T. True
• F. False

Answer: True

This is the foundation of damage-tolerant design. LEFM provides a critical crack size a_c = (K_IC / Yσ)² / π — the largest crack that can exist without catastrophic failure at the operating stress. Inspection intervals are set to ensure no crack grows beyond a_c between checks. The assumption is not zero defects, but bounded defects with known K.

Question 4

Doubling the crack area in a uniformly loaded plate doubles the stress intensity factor K.

• T. True
• F. False

Answer: False

Crack area scales as a² (for a penny-shaped crack) or as a·thickness (for a through-crack in a plate). For a through-crack of half-length a, doubling the crack area means a → √2·a (not 2a). Since K ∝ √a, this gives K → (√2)^(1/2) · K_original = 2^(1/4) · K ≈ 1.19 K_original. Even if we interpret 'doubling area' as doubling a (quadrupling the area), K only doubles — not quadruples. The nonlinear √a relationship is the key insight.

Question 5

Why does LEFM use the stress intensity factor K to characterize crack severity rather than computing the actual stress at the crack tip?

Model answer: Classical elasticity theory predicts that stress approaches infinity at the crack tip (a true mathematical singularity for a zero-radius notch). Computing a meaningful stress value there is impossible. K instead characterizes the amplitude of this singular stress field — how fast stresses rise as you approach the tip. The stress field takes the form σ ∝ K/√(2πr); K sets the scale. Two cracks with the same K have identical stress fields in the surrounding region, regardless of how they got there, so K fully determines fracture behavior. This allows a finite, measurable quantity to govern a physically infinite stress.

The singularity is not a flaw in the theory — it is the key feature. LEFM exploits the fact that the singular field is completely characterized by a single parameter K. This makes it possible to tabulate K_IC as a material constant and to use K = Yσ√(πa) as a design tool, without needing to resolve the actual stress distribution at the nanometer-scale crack tip.