Questions: Fracture Mechanics: Brittle and Ductile Failure

Question 1

An engineer selects steel for a pressure vessel that may contain small manufacturing flaws and will operate at low temperatures. Steel A has yield strength 800 MPa and KIc = 150 MPa√m. Steel B has yield strength 1400 MPa and KIc = 50 MPa√m. Which should she choose?

• A. Steel B, because higher yield strength means the vessel can withstand higher operating pressures
• B. Steel A, because higher fracture toughness means it can tolerate larger cracks before fracturing — critical when flaws are present
• C. Steel B, because higher strength always means better resistance to all failure modes
• D. Steel A, because lower strength means the material will deform plastically instead of fracturing

Answer: (B)

When cracks are present, fracture toughness KIc — not yield strength — governs failure. From K = Yσ√(πa), the critical crack size at a given stress scales as (KIc/σ)². Steel A's KIc is three times higher, meaning it tolerates dramatically larger cracks before fracture. Steel B's high yield strength is irrelevant if a crack reaches critical length first. Low temperatures exacerbate this: many high-strength steels undergo a ductile-to-brittle transition at low temperature, further reducing toughness. This is precisely the strength-toughness tradeoff at the heart of structural materials selection.

Question 2

A glass panel is rated to withstand 50 MPa. A surface scratch doubles the effective crack length (a → 2a). According to Griffith's theory (σ_c = √(2Eγ/πa)), what happens to the critical fracture stress?

• A. It doubles — longer cracks require more force to propagate
• B. It remains the same — crack length does not affect fracture stress in brittle materials
• C. It decreases by a factor of √2 — longer cracks require lower stress to propagate
• D. It increases — longer cracks have more surface area to absorb energy

Answer: (C)

From σ_c = √(2Eγ/πa), critical stress is proportional to 1/√a. Doubling a gives σ_c(new) = √(2Eγ/π·2a) = σ_c / √2 ≈ 0.71 × σ_c. The critical stress *decreases* — longer cracks are more dangerous, not less. This explains why tiny scratches on glass can cause catastrophic failure at stresses far below the theoretical crystal strength: a crack tip concentrates stress locally, and longer cracks lower the globally applied stress needed to reach that critical local concentration.

Question 3

A stronger material (higher yield strength) is typically more resistant to fracture than a weaker material.

• T. True
• F. False

Answer: False

Strength and fracture toughness are distinct properties that often trade off against each other. High-strength alloys achieved through cold work, precipitation hardening, or other microstructural refinement have high yield strength but restricted plastic zone sizes at crack tips — less energy is absorbed before fracture, giving lower KIc. Many high-strength steels fail catastrophically at lower applied stresses than lower-strength variants when cracks are present. The strength-toughness tradeoff is one of the central constraints of structural materials selection.

Question 4

According to Griffith's theory, the critical stress required to propagate a crack decreases as crack length increases — meaning longer pre-existing cracks make a material more vulnerable to fracture at lower applied stresses.

• T. True
• F. False

Answer: True

From σ_c = √(2Eγ/πa), critical stress is inversely proportional to √a. A longer crack lowers the stress needed to release enough elastic strain energy to create new crack surfaces. This is the foundation of damage-tolerant design: given an operating stress level, the maximum allowable crack size before the part must be retired can be calculated from K = Yσ√(πa) = KIc. It also explains why crack inspection and early detection matter — cracks found early can be repaired before they grow to critical length.

Question 5

Explain Griffith's energy balance: why does a crack propagate spontaneously once it reaches a critical length, rather than requiring continuously increasing applied stress?

Model answer: Griffith's insight is that crack propagation is governed by energy balance. When a crack of half-length a extends slightly, elastic strain energy stored in the surrounding stressed material is released. This released energy must either create new crack surfaces (costing energy proportional to surface energy γ) or drive further crack growth. For small cracks, each small extension releases less energy than it costs to create new surfaces, so the crack is stable. But as crack length grows, the energy released by extension scales with crack area (∝ a²) while the surface creation cost scales linearly — so there is a critical length above which each small extension releases more energy than it costs. Beyond this length, crack growth becomes self-sustaining: energy surplus from each extension drives further extension without requiring additional applied stress, and the crack runs catastrophically.

This energy perspective explains something that the simple stress-strain picture cannot: why small flaws are far more dangerous than their size suggests, and why fracture is often sudden and complete rather than gradual. The crack doesn't need the applied stress to increase; it only needs to reach the length at which the energy balance tips. The equation σ_c = √(2Eγ/πa) is the direct result of setting the energy release rate equal to the surface energy cost.