Yield strength vs fracture toughness: why can a stronger alloy tolerate only a smaller crack?

Strength describes resistance to yielding in material without flaws; toughness describes resistance to the growth of a crack that is already there. The link between them is the stress intensity factor K = Y × σ × √(π × a), where σ is the applied stress, a the crack size (the depth of an edge crack, or half the length of a through crack in the middle of a plate) and Y a geometry factor near 1. Fracture occurs when K reaches the toughness K_Ic.

Rearranged, the critical crack size is a_c = (K_Ic / (Y × σ))² / π. Both the toughness and the stress enter squared, which is why a strong, less tough alloy worked at high stress ends up with a very small tolerable crack.

With your numbers, taking Y = 1 and designing each alloy to half its yield strength. Alloy A at 250 MPa: a_c = (25 / 250)² / π = 0.0032 m, or 3.2 mm. Alloy B at 175 MPa: a_c = (40 / 175)² / π = 0.0166 m, or 16.6 mm. Alloy B tolerates a crack about five times larger.

Even if both are run at the same 175 MPa, alloy A reaches a_c = (25 / 175)² / π = 6.5 mm, still well under half of alloy B. For an edge crack Y is about 1.12, which reduces all of these sizes by about 20 percent.

The practical meaning is in inspection. A part is safe against fast fracture only if any crack is found before it reaches the critical size. If your inspection method reliably finds cracks of, say, 2 mm, alloy A at 250 MPa leaves very little growth between detectable and critical, so inspections would need to be frequent. With a 16 mm critical size there is room for a crack to grow for a long time and be found at a routine check.

This is the basis of damage-tolerant design: assume a flaw of the largest size that could be missed, calculate its growth under the service loading, and set the inspection interval at a fraction of the time it takes to become critical.

Check that the toughness value applies to your thickness. K_Ic is the plane-strain value, the lower bound reached in thick sections. The usual thickness requirement is B ≥ 2.5 × (K_Ic / σ_y)². Alloy A: 2.5 × (25 / 500)² = 6.3 mm. Alloy B: 2.5 × (40 / 350)² = 33 mm.

Thinner material than that fails with more plastic deformation at the crack tip and shows a higher apparent toughness, so thin sheet of alloy B would behave better than the K_Ic figure suggests. Toughness also varies with grain direction in plate and extrusions, so take the value for the orientation in which a crack would actually run in your part.