BS 7910 failure assessment diagram
A structure containing a crack can fail in two quite different ways. It can fail by brittle fracture, when the crack driving force reaches the material’s fracture toughness, and it can fail by plastic collapse, when the remaining ligament yields through. Classical linear elastic fracture mechanics only describes the first. Real steels at service temperature usually fail somewhere between the two, with plasticity at the crack tip raising the effective driving force well above the elastic value.
The failure assessment diagram handles both at once. Two coordinates locate the assessment point. The vertical axis is the fracture ratio K_r = K_I/K_mat + ρ, the applied stress intensity factor divided by the material toughness, with ρ a plasticity correction for secondary (residual and thermal) stresses. The horizontal axis is the load ratio L_r = σ_ref/σ_Y, the reference stress in the remaining ligament divided by yield. The FAD curve running between them is the locus of failure; a point inside it is acceptable, a point on or outside it is not.
The BS 7910 Option 1 curve is a material-generic shape that needs only yield and tensile strength: f(L_r) = (1 + 0.5L_r²)^-0.5 · [0.3 + 0.7·exp(−μL_r⁶)] for L_r ≤ 1, with μ = min(0.001E/σ_Y, 0.6), and f(L_r) = f(1)·L_r^((N−1)/2N) beyond, where N = 0.3(1 − σ_Y/σ_U). The cut-off L_r,max = (σ_Y + σ_U)/2σ_Y is the flow-stress limit — past it the ligament has collapsed regardless of toughness. The shape of the curve tells the story: near L_r = 0 the assessment is pure LEFM and K_r may reach 1; as L_r rises the permitted K_r falls away because plasticity is inflating the true driving force.
The quality of the answer is set entirely by the quality of the four inputs, and each is a piece of work in its own right. K_I comes from a stress intensity solution for the actual flaw geometry, orientation and stress distribution — BS 7910 Annex M. σ_ref comes from a reference stress solution for the same geometry and must include primary membrane and bending stress plus any misalignment. K_mat comes from measured toughness, converted from Charpy data only with the specified statistical treatment and with the correct constraint and thickness corrections. Residual stress must be included as a secondary stress, at yield magnitude for an as-welded joint unless a measured or relaxed profile can be justified. Flaws must first be characterised and recategorised per BS 7910 Clause 7 — an embedded flaw close to the surface must be recategorised as surface breaking.
Finally, this is an Option 1 assessment, not a design margin. Partial safety factors on load, flaw size and toughness must be applied per BS 7910 Annex K, or an equivalent sensitivity analysis carried out, before a result of this kind supports a decision to continue operating.
Worked example
| Yield strength σ_Y | 350 MPa |
| Tensile strength σ_U | 500 MPa |
| Young's modulus E | 210000 MPa |
| Applied stress intensity K_I | 40 MPa·m^0.5 |
| Material fracture toughness K_mat | 100 MPa·m^0.5 |
| Reference stress σ_ref | 245 MPa |
| Plasticity correction ρ | 0 |
| Load ratio L_r | 0.7 |
| Fracture ratio K_r | 0.4 |
| Curve parameter μ | 0.6 |
| Strain hardening exponent N | 0.09 |
| Cut-off L_r,max | 1.214 |
| FAD ordinate f(L_r) | 0.853 |
| Utilisation K_r / f(L_r) | 0.469 |
L_r = 245/350 = 0.700 and K_r = 40/100 + 0 = 0.400. mu = min(0.001 x 210000/350, 0.6) = min(0.600, 0.6) = 0.600. Since L_r <= 1: L_r^6 = 0.117649, exp(-0.6 x 0.117649) = exp(-0.0705894) = 0.931844, so the bracket = 0.3 + 0.7 x 0.931844 = 0.952291. (1 + 0.5 x 0.49)^-0.5 = 1.245^-0.5 = 0.896221. f(L_r) = 0.896221 x 0.952291 = 0.853464 -> 0.853. Utilisation = 0.400/0.853464 = 0.4687 -> 0.469, so the point sits well inside the diagram. N = 0.3(1 - 350/500) = 0.090 and L_r,max = 850/700 = 1.214.